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UBC Theses and Dissertations

Algebraic properties of certain rings of continuous functions Su, Li Pi 1966-12-31

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ALGEBRAIC PROPERTIES RINGS  OF CONTINUOUS  OF CERTAIN FUNCTIONS  by  LI PI SU B.Sc.  Taiwan Normal University, i960  A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY !  in the. Department of Mathematics  We accept this thesis as conforming to the required standard.  THE UNIVERSITY OF BRITISH COLUMBIA August, I966.  In presenting for  t h i s t h e s i s i n p a r t i a l f u l f i l m e n t o f the r e q u i r e m e n t s  an advanced degree a t t h e U n i v e r s i t y o f B r i t i s h Columbia, I agree  t h a t t h e L i b r a r y s h a l l make i t f r e e l y a v a i l a b l e f o r r e f e r e n c e study,  and  I f u r t h e r agree t h a t permission., f o r e x t e n s i v e c o p y i n g o f t h i s  t h e s i s f o r s c h o l a r l y purposes may be g r a n t e d by the Head o f my Department o r by h i s r e p r e s e n t a t i v e s .  I t i s understood that  copying  or p u b l i c a t i o n o f t h i s t h e s i s f o r f i n a n c i a l g a i n s h a l l not be a l l o w e d w i t h o u t my w r i t t e n  permission.  The U n i v e r s i t y o f B r i t i s h Vancouver 8, Canada  Columbia  Supervisor: L I PI SU.  Dr. J  0  V. W h i t t a k e r .  i  i  .  ALGEBRAIC PROPERTIES OF CERTAIN RINGS OF CONTINUOUS FUNCTIONS.  ABSTRACT  We s t u d y t h e r e l a t i o n s between a l g e b r a i c p r o p e r t i e s o f ' c e r t a i n r i n g s o f f u n c t i o n s and t o p o l o g i c a l p r o p e r t i e s o f t h e spaces on w h i c h t h e f u n c t i o n s a r e d e f i n e d . We b e g i n b y c o n s i d e r i n g t h e r e l a t i o n between i d e a l s o f r i n g s o f f u n c t i o n s and z»filters.  L e t C ( X ) be t h e r i n g o f m  a l l m-times d i f f e r e n t i a b l e f u n c t i o n s on a C — d i f f e r e n t i a b l e n-manifold space  X  X' ,  subset  X  L _ ( X ) t h e r i n g o f a l l L c - f u n c t i o n s on a m e t r i c  3  and OtiX)  - the r i n g o f a l l a n a l y t i c f u n c t i o n s , on a  o f t h e complex p l a n e .  I t i s p r o v e d t h a t two m~(resp. L c - ) r e a l c o m p a c t spaces X  and • Y  C^Vx)  a r e c P - d i f f e o m o r p h i c ( r e s p . Lc-homeomorphic) i f f  and C ( Y ) (resp. L „ ( X ) and L„(Y)) m  Again* i f X then  X  can be  subset i n Y homomorphic  and Y  are r i n g  isomorphic.  a r e m»(resp. L c ~ ) r e a l c o m p a c t s p a c e s ,  C »(resp.Lc~) embedded as an open [ r e s p . m  i f f (^(X) ( r e s p . L _ ( X j ) image o f C ( Y ) ( r e s p .  The s u b r i n g s  m  of C  m  i sa  closed]  aG-f>esp. 6 F - ]  L^Y)).  ( r e s p . L ) w h i c h determine t h e d®-  d l f f e o m b r p h i s m ( r e s p . Lc~homeomorphism) o f t h e spaces a r e studied. We a l s o e s t a b l i s h ' a r e p r e s e n t a t i o n f o r a t r a n s f o r m a t i o n , more g e n e r a l t h a n homomorphlsm, f r o m a r i n g o f C »differentiable m  f u n c t i o n s t o another r i n g o f  C ^ d i f f e r e n t i a b l e functions.  F i n a l l y , we show t h a t , f o r a r b i t r a r y s u b s e t s  X  and Y  iii. of  the  onto then  complex p l a n e ,  OiXY) X  and  I f there i s " a r i n g isomorphism from  which i s the Y  are  identity  conformally  on t h e  constant f u n c t i o n s ,  equivalent.  (%(X)  iv. TABLE OF CONTENTS  page INTRODUCTION  1  PART I . R i n g s o f B i f f e r e n t i a b l e  -Functions  li  R i n g s o f d i f f e r e n t i a b l e f u n c t i o n s and i d e a l s  2.  Zero-sets  3.  F i x e d , f r e e i d e a l s , and compact spaces  17  4.  R e a l i d e a l s , m-realcompact  23  5.  The l o n g l i n e  6.  Homomorphi'sms, C ~mapping, and ( f - d i f f e o m o r p h i s m  34  7.  The embedding theorems  45  8.  A r e p r e s e n t a t i o n theorem f o r t r a n s f o r m a t i o n s  and m-completely r e g u l a r spaces  spaces  12  26 m  of r i n g s o f C ~ d i f f e r e n t i a b l e f u n c t i o n s m  9.  5  53  The r i n g s o f C <»differentiable f u n c t i o n s on m  spaces w h i c h a r e n o t m-realcompact, and some algebraic properties of  C  not applicable i n  (f  PART I I . The 'Rings o f Lc{ ( o r 10.  64  L-) Functions :  R i n g s , i d e a l s and some p r o p e r t i e s o f L i p s c h i t z i a n o r Lc»functions  72  11.  L~complete r e g u l a r i t y and L°normality  80  12.  F i x e d , f r e e i d e a l s and compact spaces  84  13-  The Banach space L ( X ) and L e - r e a l c o m p a c t n e s s  86  14.  L c , L°mapping and L e , ]>homeOmorphisms  9^  V.  page  15.  Embedding theorems  99  16.  The r i n g s o f L c - f u n c t i o n s d e f i n e d on the m e t r i c spaces other than Lc-realcompact spaces, and r i n g s of L i p s c h i t z i a n f u n c t i o n on compact subsets o f  PART I I I . IT.  E  n  102  The Rings o f A n a l y t i c F u n c t i o n s Rings of a n a l y t i c f u n c t i o n on any subset of complex p l a n e  BIBLIOGRAPHY  106  117  vi.  ACKNOWLEDGEMENTS  I am g r e a t l y indebted t o P r o f e s s o r J . V. Whittaker f o r s u g g e s t i n g the t o p i c o f t h i s t h e s i s , f o r a l l o w i n g me a generous amount o f h i s time and f o r h i s many c o n s t r u c t i v e comments the p r e p a r a t i o n o f t h i s t h e s i s .  during  I a l s o wish t o thank P r o f e s s o r  D. Bures f o r h i s c r i t i c i s m o f the d r a f t form o f t h i s work, and Miss S a l l y Bate f o r t y p i n g i t . The f i n a n c i a l support o f the N a t i o n a l Research C o u n c i l of Canada and t h e U n i v e r s i t y o f B r i t i s h Columbia i s g r a t e f u l l y acknowledged.  1. INTRODUCTION  Let  (^(X) and C ( Y ) be the r i n g s o f  functions  defined  on the  C^-differentiable  and Y, r e s p e c t i v e l y ,  LipSchitzian functions respectively, defined and  L  ( 2.)  L  and (^(Yg)  n (n ) 1  , = m a  2  nifbids  and L ( Y ) the r i n g s o f  1  defined  on m e t r i c spaces  C^(X^  L(X )  X  C  m  i  ...  X,  C -differentiable  m  1  on m e t r i c spaces  c^ l^ Y  t t l e  r i n  &  s  o  f  X^ and Y^ , L -functions" c  X^ and Y^ , r e s p e c t i v e l y  (see ( 1 0 . l ) ) ,  the r i n g o f a n a l y t i c f u n c t i o n s  defined  on domains o f the complex plane (orRiemann s u r f a c e s ) Y  , r e s p e c t i v e l y , where  2  not  0  m  <, »  Xg  and  i f the s p e c i f i c case i s  mentioned. D u r i n g the l a s t twenty y e a r s , the r e l a t i o n s between the  algebraic  properties  and °c_(Y )  and^(X ) 2  and  Y ,  of 2  C ( X ) and (^(Y) ,' L ( X j J m  and L ( Y ) , X  and the t o p o l o g i c a l p r o p e r t i e s  X-^ and Y-j^ and X  g  of X  and. Yg , r e s p e c t i v e l y / have been  investigated.  Hewitt (1948) [8] showed t h a t two realcompact  (or Q-) spaces  X» and Y' (see § 7 [8]) are homeomorphic i f f  C(X')  m =0)  and C(Y') ( t h a t i s , when  are isomorphic  (Theorem 57"[8], Theorem (8.3) [7]) by means of the S t r u c t u r e space (Stone topology) t h a t two compact  (4.9) [7]).  C —differentiable m  Myers (1954) [16] proved a-manifOlds  X  and Y  (1_< m < » ) p r o v i d e d w i t h a Riemanniari m e t r i c t e n s o r o f c l a s s C™"  1  a r e C^-diffeomorphic i f f  morphic. two  P u r s e l l (1955) [20] e s t a b l i s h e d  (^-differentiable  hood-finite  (^(X) and (^(Y) a r e i s o -  n-manifolds  X  a stronger  and Y  result:  w i t h neighbor-  c o v e r i n g o f coordinate'neighborhoods are C -  diffeomorphic i f f  m  C ( X ) and C ( Y ) a r e isomorphic, where m  m  2. 1 < m < » .  N a k a i (1959) [ 1 7 ] , showed a g a i n b y u s i n g t h e  Riemannlan m e t r i c t e n s o r a s t r o n g e r r e s u l t t h a n Myers' w i t h  <^  1 _< m  oo  o  L a t e r , i n i 9 6 0 , i n " R i n g s o f Continuous F u n c t i o n s " [7] G i l l m a n and J e r i s o n gave a s y s t e m a t i c s t u d y o f t h e r i n g C ( X ) on an a r b i t r a r y t o p o l o g i c a l space  X .  They s t u d y t h e r e l a t -  i o n s between a l g e b r a i c p r o p e r t i e s o f C(X) and t o p o l o g i c a l properties of X  by examining the s p e c i a l f e a t u r e s of the  f a m i l y o f z e r o - s e t s (1.6) o f an i d e a l o f f u n c t i o n s .  The method,  used i n t h e book, w i l l p l a y t h e most i m p o r t a n t r o l e o f t h i s work. For  L ( X ) , S h e r b e r t (1963) [ 2 6 ] has shown t h a t two com-  p a c t m e t r i c spaces L(X)  X.^ and Y  1  a r e L-homeomorphic  and L ( Y ) a r e i s o m o r p h i c (Theorem 5.1  {Ik. 2) i f f  [26]).  I t a l s o h a s been known f o r some t i m e t h a t t h e c o n f o r n i a l s t r u c t u r e o f a domain i n t h e complex p l a n e i s d e t e r m i n e d t h e a l g e b r a i c s t r u c t u r e o f c e r t a i n r i n g s o f a n a l y t i c f u n c t i o n s on it.  B e r s (1948) [3] p r o v e d t h a t i f X  domains, t h e n C M ( X ) 2  Y  2  and J l ( Y ) 2  areconformally equivalent.  2  and Y  2  a r e two p l a n e  are isomorphic i f f R u d i n (1955) [25]  X  and  2  and Royden  (1956) [22] have extended t h i s theorem t o t h e case i n w h i c h and  Y  2  X  2  a r e a r b i t r a r y open Riemann s u r f a c e s .  R u d i n (1955) [25] a l s o showed t h a t i f X p l a n e domains w i t h no  2  and Y  2  a r e two  AB-removable p o i n t s , t h e n t h e y a r e  f o r m a l l y e q u i v a l e n t i f the r i n g s & t * ( X ) 2  and CK~*(Y ) 2  con-  of  bounded a n a l y t i c f u n c t i o n s on them a r e i s o m o r p h i c (*) (*) B e r s and R u d i n do n o t assume a p r i o r i t h a t t h e complex c o n s t a n t s a r e p r e s e r v e d under t h e g i v e n isomorphism. Royden, however, h a s t h i s a p r i o r i a s s u m p t i o n on t h e complex c o n s t a n t s .  3.  Later,- Ozawa and Mizumoto (1959) [18] proved t h a t when X  2  and Y  2  are two plane domains whose complementary s e t s have  p o s i t i v e c a p a c i t y , r e s p e c t i v e l y , i f there isomorphism  cp o f Log" ( X )  onto  2  f o r every complex constant  exists a direct ring  Log^ ( Y )  such t h a t  2  c j, then  X  and Y  2  cp(_c) = ;_c  a r e conformally  2  equivalent. In other a s p e c t s ,  we are a l s o i n t e r e s t e d i n the  a t i o n o f the t r a n s f o r m a t i o n s  represent-  of r i n g s of c e r t a i n kinds of  Whittaker (1961) [29.] and Kohls (1962) [12] gave a  functions.  representation  f o r the t r a n s f o r m a t i o n s  o f r i n g s o f continuous  f u n c t i o n s on d i f f e r e n t c l a s s e s o f t o p o l o g i c a l spaces. In 1965, M a g i l l [14] has obtained the a l g e b r a i c relating  C(X) and C(Y) which are both n e c e s s a r y and s u f f i c -  i e n t f o r embedding realcompaet The  conditions  Y'  i n X' , where  X'  and Y'  are two  spaces.  p r i m a r y aim o f t h i s t h e s i s i s t o u t i l i z e the method o f  G i l l m a n and J e r i s o n f o r i n v e s t i g a t i n g the a l g e b r a i c ( § f . l - 4 and § 6 ) ,  of  C (X)  1*0  , and how they are r e l a t e d w i t h the t o p o l o g i c a l p r o p e r t i e s  of  X  m  and X-^,  respectively.  r e s u l t s t o the r i n g s  (^(X)  and L ( X )  properties  C  or L ( X ) {\§ 10 -  1  X  Secondly, we g e n e r a l i z e (§7)  and L ( X ) ( § 1 5 ) . C  1  Magill's The  other o b j e c t i v e o f t h i s work i s t o e s t a b l i s h a r e p r e s e n t a t i o n f o r the t r a n s f o r m a t i o n s ( I 8).  In § 9 and § l 6  X  (or  L ( X ) and L ( Y ) )  (or  X  The  G  x  ±  )  1  are connected; then t h a t  C  and Y )  we a l s o e s t a b l i s h t h a t i f  5  (or  x  and Y  of rings of G^-differentiable  are C - d i f f e o m o r p h i c m  x  subrings o f C ( X ) m  ,  and Y."  C ( X ) and C ( Y ) m  are Isomorphic, i m p l i e s  1  X  functions  m  X  and Y  (L -homeomorphic). c  C ( Y ) ( o r L ( X ) and L ( Y , ) ) which m  1  can determine the studied.  We  applicable i n  C^-diffedmorphism for L»homeomorphism) are  a l s o give some a l g e b r a i c p r o p e r t i e s of G  F i n a l l y , we  m  C  not  i n § 9. d i s c u s s the r i n g pf a n a l y t i c f u n c t i o n s  on an a r b i t r a r y subset of the complex plane (  §17).  defined  5. PART  I  RINGS OP DIFFERENTIABLE FUNCTIONS  § 1.  Rings• of i d l f f e r e n t i a b l e f u n c t i o n s and i d e a l s . In order t o study t h e r e l a t i o n s b e t w e e n a l g e b r a i c prop-  e r t i e s of  (^(X)  and t o p o l o g i c a l p r o p e r t i e s of  X , we  shall  examine t h e s p e c i a l f e a t u r e s o f the f a m i l y o f z e r o - s e t s (see  (1.6))  o f an i d e a l , o f  n e g a t i v e .integer or m  C ( X ) , where m  » .  m  w i l l be a f i x e d non-  H e r e a f t e r we w i l l always r e f e r t o  as an a r b i t r a r y i n t e g e r such t h a t  0 < m < '« .  As we have  l e a r n e d i n Rings of continuous f u n c t i o n s [73, such a f a m i l y possesses p r o p e r t i e s analogous t o those o f a f i l t e r to  (1.15)).  (see (1.8)  T h i s f a c t w i l l p l a y an important r o l e i n the  development o f t h i s work. (1.1)  Notation:  We s h a l l w r i t e a l l equations  involving  v a r i a b l e s as i f there were a s i n g l e v a r i a b l e p r e s e n t .  For  i n s t a n c e , we w r i t e f (x)  f o r f ...  Q  yCx')  for ^  (J)  for  oX^  F o r any a  By  k  d(x,y)  - \  (x^...,  1...  ) *  x n  * \ f(xj_, 3x^ x i  x^),  ... (*«)•,  1 etc.  Q  n  n-fold subscript + ... + k  n  , and  k,  0  f  c  H  we a l s o l e t = a  k  + o^  we s h a l l mean the d i s t a n c e between  #  x  and  n  y  f ,.(x) t  Note t h a t  f (x') = £  •J|  k  (x' - x ) * + R ( x ' , x )  is  k  short f o r .  f I'"  (x»,...,x«) = E ri 1^ + ... +l x  • • • ' V ^ J * <  n  n  1  ' *' "  l i  r  n  m-(kT + . .. +k .) l n x  (x. - x ) * l  .,..(x  x  (1.2)  Definition:  A  E  of  C®)  .  n  ¥e  - ^ 5 ^  n  Let  say  +  f(x) =  f(x)  f  f Q  (x)  f  be  1  defined C  i ^ ^ l  k  ; x ,...,x )  ( x i , . . . . , ^  i s of c l a s s  i n terms of the f u n c t i o n  i f the f u n c t i o n s  V  m  i n the  ( will  k  1  subset  simply  say  *' * ' n U l * '' n  5  x  k  ( x , . . . x ) are d e f i n e d 1" * * n a l l n-fold subscripts w i t h k +..'.+k j< m , k  n  +  in  n  —  + k  A  m  ^  for  v  1  f .,(x) *™  and  n  v  (1-1)  f (x») = E °l k  f o r each  f (x)  property.  For  a 6 > 0 with  k  f(x)  m-o m)*  and  x  R (x'sx)  has  k  in  A  and  x'  are any  and  any  the  following  e > 0 , there  two  p o i n t s of  is  A  d(x',x°) < 6 , then m  k  If  k  x°  |R (x*;x)| < d ( x , x ' ) " (!)  (x' -.x)** + R ( x ' j x )  k  where  any p o i n t  such t h a t i f  Remarks:  o k  e .  m = 0 , then ( l - l ) and  (1=2)  s t a t e merely, t h a t  i s continuous. (2)  may  (o .<  k  d(x°,x) < 6  (1-2)  <  :  choose  satisfying Hence  For any  6 > 0  isolated point  so small t h a t the  d(x,x°) < 6  ( l - l ) becomes  and  x°  and  3  any  only p o i n t s  d(x»-,x°)" <' 6  is  x x°  f (x») = f (x>) + R (x';x ) !  k  k  k  £ > 0 , we and  x'  itself. so  R (x'jx) = 0 . k  Thus  ( l - l ) and  (1-2)  are  s a t i s f i e d even at the  isolated  points  7'. (3) • 0»  Moreover, i f A  i s an open s e t , then f ( x )  i n the o r d i n a r y sense, and the  f ( x ) are the p a r t i a l k  of f ( x ) (see [30] § 3 ) .  derivatives  'is  The converse i s t r u e ,  by T a y l o r ' s Theorem. (1.3)  Definition;  Let X  = ( f : f i s a function  be any subset o f E CP i n X}.  which i s  n  .  C (X) = m  C *(X) = m  = { f 6 C ( X ) : f i s bounded}. m  (1.4)  Propositioni  pointwlse addition, by  F o r any  f  subtraction  and g e C ( X ) , we denote the m  and m u l t i p l i c a t i o n  f+g , f - g , and f»g , r e s p e c t i v e l y .  commutative r i n g w i t h u n i t y zero element) and C * ( X ) C (X).  (1.5)  inverse i n  (feC *(X))  m  C (X) {(f *(X)) m  and g  is a  m  (©  i s the  m  has a m u l t i p l i c a t i v e  i s s a i d t o be a u n i t  i  U  (see Theorem 4 [31]).  If feC (x)  Definition:  C (X)  demerit, denoted by U ,  Proof i s obvious,  m  f  i s a commutative Subring w i t h  m  of  Then  of  in  C (X). m  (C *(X)). m  (1.6)  Definition;  I f f £ C ( X ) , then m  i s s a i d t o be the z e r o - s e t o f r : Remarks:  L e t Z(X) = ( Z ( f ) : f 6 (^(X)} .  i t i s clear that  Z ( f ) = <|) (see Theorem 4 [31]').  iff of  ( l ) Note that  C * ( X ) , then m  Z(f) = § .  as the m u l t i p l i c a t i v e  inverse  Z ( f ) = {x X : f ( x ) = 0}  f £ C (X) m  i s a unit  Likewise, i f f  i s a unit  But the converse need not h o l d , f  =  1  of f  i n C ( X ) may not m  2 be a bounded f u n c t i o n . f €C (E ) m  -  1  F o r example:  and Z ( f ) = <}) .  L e t f ( x ) = e'" .  But f " ( x ) = e+* 1  Then  x  2  , f"  = = 1  ec (E ) r a  1  C^E ). 1  (2)  We s h a l l show t h a t  every c l o s e d  subset o f E  1 1  is  8. a z e r o - s e t of (1.7)  C^-differentiable  Definition:  to be a z - f i l t e r  function.  A nonempty s u b f a m i l y  on  X ,  gf  Z(X)  of  i s said  i f i t s a t i s f i e d the f o l l o w i n g c o n d i t -  ions.  (1)  (1.8) the  M s ;  (ii)  If  Z ,-Z e 3 , then  (ill)  If  Z£3  Z'6  3?  1  Proof: ( i ) (li)  If  Since  Z  I  I  Z« z> Z , then  such t h a t  Z  1 3  Z £ Z [ l ] , and  That i s ,  Z  = Z(f ). 1  Z'eZ(X)  ,  Z -  i s an i d e a l ,  I  Z(f)  and  ff» e I ,.  X .  i s an i d e a l , 2  2  .  / then  such t h a t  z(f ) n Z(f )= Z(f  n Z €Z[I]  Z6Z[l] ,  such t h a t  Since  2  2  X  m  ^ $,Z[l] .  ^^fgg I  2  2  C (X)  i s a ' z - f l i t e r on  c o n t a i n s no u n i t ,  + f | e i and  Let  and  i s a proper i d e a l i n  = Z^f^) , Z  1  ff  (iii)  Z« e Z(X)  and  Z[-l] = ( Z ( f ) : f £ 1}  Let  n Zg e 3 ;  1  .  Proposition: family  Z  2  + f|)6Z[l].  f e l , and Z  8  -  Z(f«)  f'eC (X) m  .  Since  Z' = Z  However,  we have  Z» = Z» o"Z = Z(f') U Z ( f ) = Z ( f f ' ) e Z [ l ] . Remark: (^(X)  i s f a l s e , i n general.  C *(E^) m  the Z[I]  The analogue of Prop.  and l e t  f(x) =  i d e a l generated by satisfies  ( i i ) and  (1.8)  in (iii).  C *(X) m  Q.E.D.  i n p l a c e of  For example, l e t us c o n s i d e r  . 1+x^ f  with  I  Then  fe.C *(E ).  C *(E ). m  1  However,  m  1  Set  Then, i t i s c l e a r  I =(f), that  Z ( f ) = <J)fiZ[l].  Note t h a t " i d e a l " always means proper i d e a l , u n l e s s the c o n t r a r y i s mentioned.  (1.9)  Proposition:  If  3  is a z-filter  on  X , then the  family  Z [??] = ( f e C ( X ) : Z ( f ) e 3 ]  i s an i d e a l i n  Proof:  (l)  <>t  _1  m  Let  J = Z" ^]-.  no u n i t so t h a t g£J  , and  property  Since  1  J  3 ,  i s a proper subset of  h6C (X).  Then  m  ( i i ) of z - f i l t e r  Z(f.+  g) o  (^(X)  J  must  C (X).  Let  m  .  contain f,  Z(f) n Z(g)fi3 .  Z ( f + g) e 3 .  That i s ,  By  f + ge  J  - TTH*]. (2) of z - f i l t e r Z"" ^] Remarks:  f  and + g  2  again  Z(hf)e3  (l) g  For  f geC  f f i  ;  i s proper i f f  ZlZ'Hv]]  {feC (xT : Z(f)e3)  ^(X)  Z(f)  . (f,g)  Z(g)  .  Equivalently  For, l e t  I =  =  ZlZ" ^]] 1  = Z [ l ] = £Z(f)  some i d e a l  I  (3) may  i  in  C (X) m  For i n s t a n c e ,  I s any p o s i t i v e i n t e g e r . (x) = x  for a l l  a l l functions some  g€C  f  (E )  xfE  in  1  .  cf^E ) 1  i s of the form Z [ l ]  Z " [ Z [ l ] ] r> I .  matter of f a c t , s i n c e  In  Z[l] Z[l]  The  1  (a)  we  Evidently If  consider m  contains  , where  1  I = ( i ) , then  such t h a t  I  c o n s i s t s of  f ( x ) = x»g(tf) I  for  v a n i s h e s at  the p o i n t  Is a z - f i l t e r  inclusion  C (E ) , where  ieC (E )  so t h a t every f u n c t i o n i n  Hence every z e r o - s e t  {0}  :  .  It i s clear that  be proper.  m  by  = (Z(f) : Z(f)^af) = 3 . Remark (2) shows t h a t every z - f l i t e r  for  generated  .  m  Then  Hence  1  meets  C (X)  By p r o p e r t y ( i i i )  hf e J = Z " [ 5 ] .  Thus  = 3 .  .  m  Z(f)£3 .  ( X ) , the i d e a l  Is not a u n i t .of  2  (2)  f £1.}  .  i s a proper i d e a l i n  1  f  Z [ h f ] = Z(h) U Z ( f ) O  0.  As  0. a  t h a t i n c l u d e s the  , i t must be the f a m i l y of a l l z e r o - s e t s  containing  0.  set  10.  Let  M  = Z° [Z[I]] •  functions i n contains 3m+l  C (E )  I .  Ml 4 I , f o r °  However, 3m+l  € I , then  then  3m-r2 g = i~^ 4 C^E )  i ~ ^  = g»i  Note t h a t here  C^E ) 1  , then .  words,  f o r some  .  Hence  i  C ^)  (M ,f) Q  & M  £ M  5  For, i f  From Remark ( 1 )  xeE  m  Let  f^x) | 0  1  I = ( f ^ ) ..  such t h a t  I  C (E ) , and l e t  for xfiE  f o r some  vanishes at  Z[.l] c o n t a i n s the p o i n t . Z[l]  0 .  0 .  that vanish at  0 .  Moreover,  .  1  f (x) = 0 x  x  _0+  point  i&M .  Hence If  M  Q  Hence every  i £ I , then  o n l y when  x = 0), Hence  g  and  so zero-  {0} , i t  G.  x = g(x) f- (x) L  Now, (^(E ) 1  I . f o r some  , i f x + 0 (as  ^ J j . g(x) = - » ,  cannot be continuous a t the  This i s a contradiction.  in  As a matter of f a c t ,  c e r t a i n l y contains  Consequently, g(x) = x/f^x)  g(x) = + » . 0 .  f  e v i d e n t l y c o n s i s t s of a l l f u n c t i o n s i n  1  x + .0.  and  1  must be the f a m i l y of a l l z e r o - s e t s c o n t a i n i n g  geC^E )  1  g e'C^E )  i s a z - f l i t e r t h a t i n c l u d e s the s e t  = Z" [Z[I]]  1  T h i s c o n s i s t s of a l l f u n c t i o n s  f ( x ) = f - ^ x ) g(x)  t h a t every f u n c t i o n i n  Q  In other  .  f ^ e b ( E ) , and  Then  M  (i,f)=  .  1  Q  m = « , l e t us consider  for a l l  1  since  But,  - I .  * (M f) = C ^ E )  1  2  In case  f (x) = e  set i n  .  1  i s maximal.  Q  (b)  1  and i f  ggC^E )  171  C^E )  certainly  ,  °  i s a maximal i d e a l .  Q  Z ( f ) n Z ( i ) = <j> .  Since  M  M  i ^  M  3m+l  1  Q  0 . Hence 3m+l  t h a t v a n i s h at  i"^  f t M  E v i d e n t l y i t c o n s i s t s of a l l  1  Q  11. In g e n e r a l , x e M_ - I , f o r a l l p o s i t i v e i n t e g e r s n  hence f o r any a constant  f € (^(E )  n ,  which I s 'analytic at the o r i g i n without  1  terra i s i n M „ - I . o  Here, contains  Q  I  C^E ) .  properly.  For, I f  1  Z(f£)  Z:[l] = Z[M ]  and so  i n s p i t e of the f a c t t h a t  Besides,  f £ M  Q  M  , then  Q  i s a maximal i d e a l i n  Q  Z ( f ) i s d i s j o i n t from  (f-^f) = G (E ) . m  M  Hence  1  (M ,f) = ( f - ^ f ) = Q  C^E ). 1  (1.10) D e f i n i t i o n s  A z-ultrafliter  Note t h a t every subfamily i n t e r s e c t i o n property, z~ultrafliter  on  If  M  3^  and  such t h a t  m  3^ c 3  2  3  2  on  I  x  e I  2  Z" ^] 1  (1.12) P r o p o s i t i o n : _1  I ] ^ ^  & r t e  If ^  1  2  m  f c w o  a r b i t r a r y Ideals  Z(X)  Also, i f  such t h a t  Hence the r e s u l t f o l l o w s  on  X , then  C^X) .  I t f o l l o w s from P r o p o s i t i o n s  C (X)  '  i s a z-ultrafliter  i s a maximal I d e a l i n  Z  m  (1.8) and (1.9) •  Proof i s s i m i l a r t o t h a t o f Prop.  mapping  C ( X ) , then  Z [ l J c Z[lg] .  , then  c Z" [3? ] .  immediately from P r o p o s i t i o n s  Z [>£]  i n some  X .  are two a r b i t r a r y z - f l i t e r s In  , then  z-filter.  w i t h the f i n i t e  i s a maximal i d e a l i n  I t i s obvious t h a t I f  C (X)  Z(X)  by Zorn's Lemma, i s contained  Z[M] i s a z - u l t r a f l i t e r  in  of  i s a maximal  X .  (1.11) P r o p o s i t i o n :  Proof:  X  on  (l.ll). ( l . l l ) and (1.12) t h a t the  i s one-one from the s e t o f a l l maximal i d e a l s i n  onto the s e t of a l l z - u l t r a f l i t e r s  on  X .  12.  (1.13) P r o p o s i t i o n : Z(f)  Let  be a maximal i d e a l i n (^(X) . I f  M  meets every member of Z[M] , then  Proof: ideal,  Suppose t h a t  Hence  M .  = (^(X)...  (M,f)  (M,f).  f  Hence  Then, s i n c e  f o r some  Z(h + g.f) = Z(U) = <|) .  By h y p o t h e s i s Z(g)] U [Z(h)  zero-set Proof:  UeC (X) = m  h e M and g e C ( X ) . m  u z ( f ) ] = [ ' z ( h ) n z(g)} u f z ( h ) n z ( f ) ] .  0 Z(f)]<l>  •  Thus, we have  <|) o [Z(h) n  T h i s i s impossible.  Hence  Let ST be a z - u l t r a f l i t e r on X...  Z meets every member of > f , then Since  i s a maximal  But (j) = Z(h + g o f ) 3 Z(h)  Z(h) ft Z ( f ) 4 =$ •  (1.14) P r o p o s i t i o n :  M  We know t h a t the u n i t y  ( X = h + gof  n z ( g - f ) = z(h) n [ z ( g )  f£M .  Z € ^  If a  .  Z meets every member o f ?fr , & U {Z} has the  f i n i t e i n t e r s e c t i o n property.  By Zorn's Lemma, ^  generates a z - u l t r a f l i t e r  which contains ^f" .  is a z-ultraf l i t e r , y f '  = A'.  Hence  (1.15) D e f i n i t i o n : ' An i d e a l z-ideal i f  f e M.  Z(f) e Z [ l ]  I  implies  U (Z) But •4~ /  Z : ^ ^ = ./f.  i n C (X) i s s a i d t o be a m  f £.1 .  That i s ,  I = Z"  1  [Z[I]] . I t i s obvious t h a t every maximal i d e a l i s a z - i d e a l , while the i d e a l s  I = (i)  and 1^ = (f^) , given i n Remark (3) o f  Prop. (1.9) are n o t z - i d e a l s .  Zero-sets  arid m-completely Regular Spaces.  In t h i s s e c t i o n , we w i l l show t h a t every c l o s e d subset o f E  n  i s a zero-set.  (2> 1)  Lemma:  Letf (x) = x e " m  n  n x  f o r each  x £ . [ 0 > . ) , where  13m  i s a ..-•'fixed p o s i t i v e i n t e g e r . 0  uniformly t o  [0,»),  on  Then  where  { f ' : n e N]  N  always  converges  denotes the s e t  of a l l p o s i t i v e integers. f < (x) = m x - V n>  Proof:  m  "f(x) = 0 . n  Set  - nx e-  n x  m  x = 0  We have  v  n x  = x - e - ( m - nxj . m  or  1  n x  x = ^ . n  • f»(x) = ( m - i ) x - e " " ( m - nx) - n x " " e ( m - nx) m  ^m-l^nx f " g ) = e) n n' n m  v  f (Xi)  Hence  n x  m  (m  m  nx  2  2  2  - m - 2m + m ) ^ m-^ 2  2  e"  2  m  < 0 .  n  h&s i t s maximum a t  n  1  TO-2g-nx 2 _ _ mnx + n x ) .  v  e  m  ;  x  =  - (m  2  v  2  x « •£ w i t h value  Now, f o r any g i v e n e > 0 , take i n t e g e r such t h a t  N  > m/e-Je  Q  .  N  Then  ( B ) e ~ ..' m  m  t o be the p o s i t i v e  Q  (|) e" m  m  _<  ) e~ m  <_  m  •o o. \ } „ 'o m /e e  m m  whenever  e** = e ,  whenever  m  n _> N  .  Q  Hence  {g  n > N  : n € N}  n  .  |f I < e  That i s  converges u n i f o r m l y t o  0 .  Q.E.D. Let  = {x € E  n  ci  E  n  B ( a ) = {x e E  n  r  : Ux - al! < r )  : ||x - b|! _< r'} , where  x = (x^,. .. , x ) , c-tgn B ( a )  n  r  and  a=  r / r ' = k: .  B ,(b) r  Then, i t i s e v i d e n t t h a t  i s C^-diffeomorphic w i t h  ci^ 1  n  n  B ,(h)  under  r  1  ±  ±  - a ) + b ±  ±  ,  n  Now, i f f  i s a f u n c t i o n d e f i n e d on  cl^, B ( a ) , n  r  then  f ( k ^ •--•'•b ) + a ^ . . ; , k ( x - b ) -+ a ) = f . f ^ x ) i s  g(x)  1  a f u n c t i o n d e f i n e d on  <t> ()) x  n  E n  n  <t>(x) = (<^(x),... ,<j) (x)) , where' • 4> (x) = ^(x 1 i • 1  ct  ('ap..'. ,a ), b = ( b ^ . . . , b ) ,  '. 1  and  , with  n  n  c t n B , (b) , where E  r  n  <j)"^"(x) =  tjj^ * k ( x - \>^) + ' a . y I X i _< n . i  1  (<$>j(x),...,  Moreover,  14.  If M  f'./e. C* , t h e n  oi  *  then  (D^g^x)!  positive  integer  shown i n  (1.1).  (2,2)  g e C  F  is a  |D f(x)|  Let  F  k M  > f o r an a r b i t r a r y  m  D^f(x),  c r i  D g(x)  and  i  be an a r b i t r a r y  C%zero=set,  i s hounded b y  i  i s hounded b y  m , where  Theorem:  Then  , and i f  89  I.e. there  <ji ' a r e as  closed  is  subset of  f € C^E  1 1  )  E  such  1 1  .  that  Z(f j = F . PrOof:  We know t h a t  denotes  i+1  > O  s  f^x)  l  Let 0 <  Choose  =  ±+1  <  r c  t  >  0  0  t  <  0  cp e. ^ ( E ) .  tp(s-r ) Q  - g(j|x - a l l ) .  {1 0  '  Then  Then we  i  r  +  i  f  for  x € E  11  >  1  (a ) i  r i  1  Then  • define  define Finally,l e t  C^E ) 1  and  B  r  (a )  - B  r  (a )  m  r  B  center'  We now f i r s t -  1  1  elsewhere  + aj,...,t (x  2  o  n  x e cl  2  1  j /  , where  4= Q , and  e C*(E )  x  2  i  S c  ^  for  f ( x ) = t (i {Xia .)  r  (a ) i  ^  0 < f (x) < 1 Let  r i  Next  1  = p ^ - s T *  f-Jx) =  r  B  i •  r  ep(t) =|e  It i s clear that  S( >  - F  1 1  t h e open b a l l w i t h r a d i u s  - ( a * , . ..,a^)...  1  E  2  . - a ) ni a*) . 2  n  have f l  (x)  1  for  G  for x e E  x e c  ^ 1 1  B ^ ^ a  2  )  -• B - ( a ) 2  r  G < f (x) < 1  2  2  elsewhere,  and  f  2  e  C*(E ) n  15. I n g e n e r a l , we  define  - a j ) + a*" .,... , l ( x  f (x) =  ±  i = 2,3,  for  .  f (x)  =  ±  T h e n , we  1  for  x e oe^n  0  for  x e  B„ -  0 < f^(x) < 1 choice  of  ( f ) , we  ,-,.  k e N , and compact .  vanishes Thus  1  bounded on  E  1 1  I  .  Let  k  I  show t h a t  I  D^f-^  -  a  °  r  i=l  *~ 1=1  c  Thus we  the  i  3  .  C, = e  * i J /  e  *  (  1  e  (i+lUi+l  , (Jg  =  0  <  B  B  y  y  V  a  t  U  so i s  Then,  we  •' ^ i ^ k ' J-t-?  x  J  ,  for  i e N .  e  i +  o  corollary  f  &  i) L  e  / m  ([5],  m  H °i  1=1  (2' ) 1  P.108),  converges  °i  converges.  (jli ^Vcj) a  , we  Vi(x)  § -^-i—^i=l °i  c  (7.17) [23]  1=1  Q  k  r  J  e N , 2  v  i  -tj / i  *  N  1  ±  By Theorem  o n l y h a v e t o show t h a t  "Ratio t e s t "  €  1  1  n  G  k  ( a ) which i s  r  o  (a )  r  \\? \  K  However,  a  '1  1  E  f  (a )-B 1  r  4^4  t h e n h a v e t o show t h a t f o r e a c h uniformly.  B  E n  c-t n B  k  where  f € G (E ) w  on  M  i s continuous fOr a l l c-t  general,  n  f .(X) f(x) = E - T r — i=l °i  Let We w i l l  -  2  i  1  e x c e p t on  l ^ ^ l l  | D f | <. if^^  have  (a )  N  elsewhere  hounded  s  .  (a )  B_  know t h a t  n  1  n  have  r  By  - a * ) + a*" ) .  1  ±  >  iS  =  * i V  Use  16.  2  i s convergent.  1=1  converges u n i f o r m l y Moreover,  (2.35)  f o r each  OS  Therefore  s =1  ak s N , so t h a t  Z ( f ) = ff Z ( f . ) , as each f u n c t i o n i=l  Definitions  Let  X  i f e C (E ) .  f.  i s a nGn-negative  be a t o p o l o g i c a l space.  X  is  s a i d t o be m-completely r e g u l a r i f f o r every c l o s e d subset of  X  and  x e X-F  f(x) = 1  that (2.4)  , there e x i s t s a f u n c t i o n  and  Theorems  Z(X)  c l o s e d subsets of  X .  For any  c l o s e d subset  Suppose t h a t F  x f Z(f) .  the c l o s e d subsets of Sufficiency: c l o s e d subsets of x e X-F  where Then and  r  .  -1  Let  f e C (X) m  f(F) =  {0}  , t h e r e e x i s t s an  f ( F ) = {0}  [7])  Z(X)  1  and  Z(X)  Hence  i s abase for  i s a base f o r the  such t h a t Then  F  Z(g) o F  r + P . • Let  f u n c t i o n of. value ge  .  .  For each c l o s e d subset  constant f  x e X-F  and  Suppose t h a t  g(x) = r .  as  i s m~completely r e g u l a r .  Consequently  g e (^(X)  , i s the  X  X' and  X . (See  X .  , there i s  x 4 Z(g)  of  f ( x ) =1  such t h a t and  i s a base f o r the  m  Necessity:  Z(f) 3 F  .  = (Z(f) s f c C (X)]  Proof:  m  such  m  A t o p o l o g i c a l space i s m-completely r e g u l a r  i f f the f a m i l y  f e C (X)  [0}  f(F) •  f e C (X)  F  C (X) m  .  of  X  and  and f = gor"  1  ,  - i r <. .  Moreover, f ( x ) = 1 Q.E.D.  17. (2.5)  Definition:  A t o p o l o g i c a l space I s s a i d t o be m-normal  I f f o r any d i s j o i n t c l o s e d subsets  F-^ and  f (P^), = ( 0 }  f € C ( X ) such t h a t m  and  F  , t h e r e i s an  2  f ( F ) - {1}  Having proved t h a t every c l o s e d subset o f s e t , we can show t h a t every subset of  .  2  E  E  Is a zero-  11  I s m-completely  11  r e g u l a r as w e l l as m-normal. (2.6)  Proposition:  Proof: of  L e t F.^ and F  X .  = 1,2.  that  f  F£  Z(t ) = F j . ±  = <j> .  E  i s m-normal.  11  be any two d i s j o i n t c l o s e d  2  and  F  of  2  E  L e t f \X = g ±  such t h a t  11  ±  .  subsets  g  f ( Z ( g ) ) = (1) 2  .  x  g .  m  Hence  m  g  ,  such  n  € C (X) m  ±  Z(g ) n Z ( g ) ±  2  Z ( g ) n Z ( g ) = <|) ,  Since  f e G (X) .  C (E )  €  so t h a t  ±  Now, s e t f =  F | fl X =  Then,  Z ( f ) n X = F [ n X -. F  i s w e l l - d e f i n e d and  and  of  By Theorem (2.2), t h e r e a r e f ^ f g  Z ^ ) =  and  X'  By p r o p e r t y o f the r e l a t i v e t o p o l o g y , we have two  c l o s e d subsets i  Every subset  1  g  Moreover  Z(f) =  f ( F ) = { 0 ) , and x  Z ^ )  f ( F ) = [1} 2  .  Q.E.D. (2.7)  Corollary:  Every subset  X  of  E  11  I s m-completely  regular. Proof:  Since  X  i s Hausdorff and m-normal,  regular.  §3  X  i s m-completely Q.E.D.  F i x e d , f r e e i d e a l s and compact spaces. In t h i s s e c t i o n we s h a l l see the c h a r a c t e r i z a t i o n o f  f i x e d maximal I d e a l s of  C^(X)  (see ( 3 » 3 ) )  and how they a r e  18. related to a compact space (see (3.11)). (3.1)  Definition:  Then  I  and  I  be any ideal in  is said to be a fixed ideal i f  I  is said to be a free ideal i f  Remarks: that I  Let  (l)  I  that does not vanish at If  (2)  m  is empty.  I  x e X , there is a function in  x .  is a fixed ideal in  is not empty, and the set  ideal must be of this form. enlarged by making S  C^X) , then the set  I« = [f e (^(X) : f[S] •  is a fixed ideal Containing I .  (0)}  C *(X) .  is not empty;  nz[l] nZ[I]  or  In view of the definition of free ideal, we know  is free i f f for each  S = flZ[I]  (^(X)  Hence a fixed maximal  Furthermore, since  I'  can be  smaller, the only candidates for fixed  maximal ideals are the ideals  I'  for which  S  contains just  one point. The  corresponding statements hold for (3)  the z-ideal  The ideal Z°* [Z[l]] .  the set  if  is a zero-set.  S = nZ[l]  X c E , where isolated, say which  z-ideal. and  s a  a .  Z(f)  f e C (X) m  b 4 Z(f) .  •  need not be a member of  Let  0,. denote the set of a l l  i s a neighborhood of  Since  a .  nz[0 1 = {a} .  f e C (X) m  is then a  0  Suppose  b € nZ[0_]  X is m-completely regular there i s an  such that That i s ,  Z(f) fe  is a neighborhood of 0  Q  so that  Hence  n Z [ O ] «= {a} a  a but  b.| n z [ 0 j  which is a  cl  EL  contradiction.  Z [ l ] , even  To see this, let us consider a space  We claim that  b 4  mentioned above evidently contains  X contains at least one point which is not  1  for  m  In general, the two are not the  1  samej S  I'  C *(X) .  .  Now, since  {a}  is a  19. c l o s e d s e t , by Theorem (2.2), i t i s a z e r o - s e t . "a"  However,  I s n o t an I s o l a t e d p o i n t i t cannot be a neighborhood o f  itself.  {a} i s not a member o f Z[0 ]  Hence  n Z [ l ] i s a c l o s e d subset  Note t h a t s i n c e  (2.2), I t i s a z e r o - s e t . (3.2)  Notation:  X  i n X , by  (Compare [73 P.55).  If I  denote the r e s i d u e f u n c t i o n on  •  a  1(f) w i l l  i s an i d e a l i n C ( X ) , m  class o f f , f+Ij r  of value  r  for a l l  will  denote the constant  r e /R, the f i e l d  of r e a l  numbers. (3.3)  Theorem:  (1)  p r e c i s e l y the s e t s  The f i x e d maximal i d e a l s i n C ( X ) m  ^  are  = {f e C ( X ) : f ( p ) > 0} . m  (2)  The I d e a l s  (3)  F o r each  are d i s t i n c t f o r d i s t i n c t  P •  the r e a l f i e l d  fR.  I n f a c t , t h e mapping  unique isomorphism;.of Proof:  ( l ) Consider  C (X)/Mp m  i t s k e r n a l i s Hp .  maximal.  fR 1  (^(X) , t h e r e e x i s t s a p o i n t  ideal.  in  p € X .  (^(X) t o fR ,  r e fR.  9  r_(p)-= r , so  Hence i t s k e r n a l  Is  i s any f i x e d i d e a l i n  p e flZ[M] .  Clearly,  M is  which has j u s t been shown t o be a maximal  Hence i f M (2)  Is the  E v i d e n t l y i t i s a homomorphism, and  On t h e other hand, i f M  contained  Mp(f) -• f ( p )  cp from  Since f o r each  ep i s onto t h e f i e l d  i s isomorphic w i t h  onto fR , where  t h e mapping  d e f i n e d b y ep(f) = f ( p ) .  that  p , (^(Xj/Mp  i s maximal, then we must have  I f p 4 P' » s  X , there i s f e C ^ X )  M =  .  hy t h e m-completely r e g u l a r i t y o f  such t h a t  f(p') = 1  and f ( p ) = 0  20. (Prop. (2.3)).  (3)  Hence  Form; the proof of  homomorphism of with  f e Mp but  fk .  C (X) m  to  f {Mp, .  (1) ,  That i s ,  Mp is the kernal of a  , so. that  C (X)/Mp  is isomorphic  m  The uniqueness of the isomorphism follows from the  fR is the identity  fact that the only non-zero automorphism of [7, -(0.23)3 . ! (3.4)  Theorem:  (l)  precisely the sets  The fixed maximal ideals in  are  m  M*• « {f e C^X) : f(p) = 0} .  (2)  The ideals  (3)  For each  with the real f i e l d  C *(X)  fR .  M* are distinct for p, C *(X)/M* m  distinct p.  is isomorphic  In fact, the mapping M*(f) -» f(p)  is the unique isomorphism of  G *(X)/M* m  onto  , where  p "e X .  Proof is identical with that of Theorem (3.3) except for the notation. (3.5) Proposition: C *(X)  is fixed.  Proof:  We know that  m  If  X is compact, then every ideal  Z[l]  I  in  i s a family of closed subset with  f i n i t e intersection property i n the compact space  X  Hence  nz[i] + (|) . (3.6)  Propositions  pondence  p  ideals in Proof:  If  X is a compact space, then the corres-  Hp is one-one from X onto the set of a l l maximal C^X) .  In view of Prop. (3.5) , every ideal is fixed.  Prop. (3.3), each fixed maximal ideal in  C (X) m  By  is of the form  21.  Mp for some p € X , and  •« J ^  Iff  t  p = p« .  Hence the  assertion holds.  Q.E.D.  We now shift our emphasis from Ideal to z=fliter, ( 1 . 8 ) to  (3.7)  (see  (1.12)).  Definition?  Let  3  be a z - f i l t e r i n  Z(X) .  i s said to be a free or fixed z - f i l t e r according as  Then 3 03 = <j)  or + 0 • The following property Is the immediate consequence of (3.7) .  (3.8)  Proposition:  Every ideal of  G (X) i s fixed i f f every m  z - f i l t e r Is fixed. (3.9)  Lemma:  A zero-set  Z e Z(X) is compact i f f i t belongs  to no free z - f l i t e r . Proof:  Necessity:  z-filter  3.  Set  Suppose  Z Is compact and  3' = {Z 0 P : F e 3) .  Then  (1)  (|) 4 3' * since  (ii)  ( z n F) n . ( z n  F«) =  z n  (F n F«) e an  F n F' € 3 .  Thus  3'  Is a family of closed  sets i n  Z and F are i n  That i s , Sufficiency:  3 Let  03 4 $ • 8  s  as  £  But  H3 = 03' .  i s fixed. B be any family of closed subsets of  Z with the finite intersection property. X , the members of  3, Z fl P f f ;  Z with the f i n i t e intersection property.  It follows that  In  Z belongs to a  are closed i n  collection of a l l elements In  Since X.  Let  Z is closed 3  be the  Z(X) each of which contains  22,  a finite  i n t e r s e c t i o n o f members o f  z-fliter  and  flSP 4 4^ •  i s in  We know t h a t  s  vice  Z  versa.  Hence  3 .  (Since  every  closed  Suppose t h a t  so  i s a zero-set.  P  » "by P r o p .  G  (3.9)  •  Remark:  consider  Let  z  (  ^  Z(f)  {p } ejdf Q  3 , i s noncompact,  f e I  g e G (E )  .  1  i s n o t bounded.  e I .  However,  f  € C^E ) 1  (as  Then  tp }  X , and e a c h  i s a closed set  Q  .  This  that  at  i s a contradiction  3  i f e v e r y member o f  i s free.  Then  For instance,  Then  f  e (^(E ) , 1  Z ( f ) i s n o t bounded). Q  in  C (E^") m  c a n be w r i t t e n  as  Z ( f ) contains  Thus  generated by f(x) =  f  Q  0  Z ( f ) so t h a t Q  In general,  f o r any  Z ( f ) i s n o t bounded, t h e z - f i l t e r  I = ( f ) , i s fixed i n spite of the fact  that  each  member o f i t i s noncompact.  ( J . - l l ) Theorem:  In  X c E  (1)  X  (2)  Every i d e a l is  1 1  , the following are equivalent:  i s compact.  fixed.  i n  C (X) m  .  f (x)g(x)  Z ( f ) i s n o t compact f o r a n y  n z [ I ] = Z ( f ) 4 ty •  such that  , where  then  ideal  0  m  on  i s compact.  i s free.  f (x) = S i n x .  I = ( f ) > the p r i n c i p a l  some  Z  Q.E.D.  ) 4 ty i s n o t compact  Q  Therefore,  i ti s not true  1  f  hypothesis,  meets e v e r y member o f  Q  C ^ E ) , and l e t  f  Z[l]  (p }  •  isa  s e t i s a z e r o - s e t and  Then  e fts4"  3  i s free.  Hence e v e r y f u n c t i o n for  Q  Now,  Nevertheless,  a z-filter  but  p  (1.14) ,  Hence  By  L e t Mr b e a z - u l t r a f l i t e r  i t s members b e noncompact.  Proof:  Evidently,  Z € B).  nfl = A3 4 ty •  (3.10) P r o p o s i t i o n : of  e .  i s f i x e d , i . e . every  z-filter  (2*)  Every i d e a l i n  C  (X)  (3)  Every maximal i d e a l i n  ra  i s fixed, (^(X)  i s f i x e d , I.e. every  z-ultrafliter i s fixed. (3*) Proof: with  Z = X  and the f a c t  X  X  Let  n z [ l ] 4=  I s compact  •  Hence  (2*).  3  C (X) = m  i s any f a m i l y  Z ( f ) ^ F, f o r some Then, i t i s c l e a r t h a t  I = (A) , the i d e a l i n 0 Z [ l ] =f= $> • 03 + $ •  T h i s shows t h a t (2*) i m p l i e s  But  C *(X)  Z [ l ] 3 8 , Ofi 2  In other words (l).  generated by  m  X  i s compact.  Hence ( l ) I s e q u i v a l e n t  Consequently (2) and (2*) are e q u i v a l e n t .  (2) i s e q u i v a l e n t  t o (3)  every f r e e i d e a l i s contained  §4  and  e = ( Z ( f ) : f e A} .  By h y p o t h e s i s ,  Finally,  X  with f i n i t e i n t e r s e c t i o n property.  m  DB = 03 .  (3-9)  belongs t o every z - f i l t e r i n X .  A = {f e C ( X ) : 0 _< f ^ 1  F e 3} , and  with  (2) f o l l o w s from Lemma  Now, suppose t h a t (2*) i s t r u e ; and  of c l o s e d subsets of  A .  Is fixed.  m  ( l ) i m p l i e s (2*) because when  (X) .  Let  C *(X)  That ( l ) i s e q u i v a l e n t w i t h  Likewise, C  Every maximal i d e a l i n  and (2*) w i t h  because  i n a f r e e maximal i d e a l .  R e a l I d e a l s , m-realcompact Spaces. In 19^8,  (3*),  Q.E.D.  '  E. Hewitt d e f i n e d r e a l maximal I d e a l s and r e a l -  compact spaces (Q-spaces) (see [8],  §7 and [7],  Ch. 5).  He  a l s o c o n t r i b u t e d many I n t e r e s t i n g p r o p e r t i e s about r e a l maximal I d e a l s and realcompact spaces ( [ 8 ]  and [7]).  Unfortunately,  those p r o p e r t i e s can not be c a r r i e d t o the r i n g s o f  C m  d i f f e r e n t i a b l e f u n c t i o n s , s i n c e they are not l a t t i c e - o r d e r e d rings.  (See [7] (0.19) and Ch. 5).  not nonnegative or n o n p o s i t i v e , then  Indeed* i f f e (^(X) i s |f| 4 C (X) . m  24.  Recently/(1964), R. Bkouche has shown that every paracompact Hausdorff dlfferentiable n-manlfold is m-realcompact (see ( 4 . 2 ) and [4] Theorem 2 ) . closed subset of that every  E  Here, we w i l l show that every  is m-realcompact.  11  Moreover, we know  -differentiable n-manifold with countable basis  can be C^-embedded in a closed subset of  E  (cor. 1.J2 [ l 6 ] ) .  2 n + 1  We know that every residue class f i e l d of C *(X)  C (X) or m  module a maximal Ideal contains a canonical copy of  m  fR : the set of Images of the constant functions  the real f i e l d  under the canonical homomorphism. Ideal In (^(X) , and  r^ + r  For, let  M be a maximal  be arbitrary constant functions.  2  M(r^) - M(r ) , then 'M(r - r ) = M(r_ ) - M(r ) = 0 .  If  2  That is  - r  unit.  Hence  2  x  g  1  This is impossible for x  e M .  2  M(r_ ) =f M(* ) * ' 1  6 1 1 ( 1  2  that the set  r e fR}-, forms a field is clear. f i e l d with fR . write for a l l (4.1)  l  s  a  K = {M(rJ :  (3.*0*  we can  f e C (X) , and M*(f) = f(p) m  f e C *(.X) . m  Definition:  C *(X)). m  for a l l  ~  We shall Identify this sub-  Thus, in Theorems ( 3 . 3 ) and  Mp(f) - f(p)  2  Then  Let  M be any maximal Ideal in  C (X) m  (or  M is said to be a real ideal i f the canonical  copy of fR is the entire f i e l d  C (X)/M (respectively m  C *(X)/M) , land M is said to be a hyper-real ideal If the m  canonical copy of fR is not the entire f i e l d (respectively Remark: in  m  C *(X)/M). m  By Theorems (3.3) and ( 3 . 4 ) , every fixed maximal Ideal  C (X) or m  C (X)/M  C *(X) m  Is real .  25. (4.2)  Definition:  A topological space  X Is said to be  m-realcompact, i f every real maximal ideal i n It is clear that i f  (^(X) is fixed.  X i s compact, then  X is m-realcompact.  We Will show next that every closed subset of  E  Is  11  m-realcompact. (4.3)  Lemma:  An ideal i n  compact subset zero in  A of  m  X there exists an  f e I  having no  A.  Proof:  Necessity:  compact subset of then the family  Suppose X .  property. for some f  Is free and f e l  A is an arbitrary , Z(f) n A +ty>  3 = {F = Z(f) n A : for some  f e 1} is a  A having the f i n i t e intersection  Indeed,' for each ±  I  If for each  family of closed sets i n  z(f )  C (X) is free i f f for every  e I , I = 1 2. ;  P , P e 3 > we have 1  F± = Z(t^) 0 A  2  Then  P ^ P g = ( Z ^ ) n A) n  n A) - ( z ( f ) n z ( f ) ) n A = z(tf + t\) n A +ty, as  2  x  f?.+ f f e I . *  2  Thus  m \ty.  Hence  1  nZ[l] = n Z(f) 3 f€l ~*  n Z(f) n A = 03 + ty'• This i s a contradiction. fel is  f e l  with  Z(f) n A = ty..  Sufficiency: there i s an  Hence there  f e l  Suppose for every compact set such that  Z(f) A A = ty .  Then, for any  {x}  which is compact, there Is an f e l  such that  Z(f)  n {x} mty.  Hence  That Is,  x 4 Z(f) .  Ac X  nz[l] = ty . Q.E.D.  (4.4) of  Proposition:  E . 11  Then  Let  X be a closed (unbounded) subspace  X is m-realcompact.  26. Proof:  Suppose t h a t  M  fR  i s the r e a l f i e l d  i s a f r e e maximal I d e a l and  .  Let  g(x) = • - — j ; M *  .  2  g € C (X)  and  m  s m a l l number 11  i s a unit i s clear.  M(g) 4 0 .  is,  E  g  , say  Ay .  closed.  Then  Let  closed i n that  e > 0 , g < x - £  E  .  = A 'fiX  Thus, t h e r e  Z ( f ) = A' .  we know t h a t t  1  e M  B  c X .- B  6  fore,  g < r - 6  Let  h  =• |l .  on the z e r o - s e t on  Z(f) .  E* .  s r - g  (mod M).  M(r) - M(g) = r - M(g) . o p M(h ) = (M(h)) J> p o s i t i v e number,  But, s i n c e  * so we have  0  M(g)  However,  Z(f ) ] L  Z ( f ) € Z[M] .  h  ±  is  Z(f) .  There-  r - g j> e . c"  on  1  Z(f)  Extension  h .  That i s  Therefore,  M(h ) = M(r - g) = C (X)/M m  M(g) <_ r .  i s I n f i n i t e l y small.  diction. §5  It i s  By Lemma (4.35), t h e r e I s  extension  In other words  is  such  m  By Whitney's A n a l y t i c  on  X  so i s  In other words,  Then  m  X  Z e Z[M] .  Z ( f ) , and  Theorem (see [ 3 0 ] ) , we have a C o h | Z ( f ) = hj, . Hence h = r - g  as  Z ( f ) e Z[M] .  Hence  £  which i s c l o s e d i n  h  11  X .  e ct (X - B ) = Z(f) . x  X  f e (^(E ) c C ( X )  f o r some  1  = ( f - gJ2  ±  i s an  Z ( f ) (1 B  such t h a t  That  and a s u f f i c i e n t l y  which Is c l o s e d i n  i s compact i n  £  r  i s compact i n  We w i l l show t h a t  enough t o show t h a t Z ( f ) D Z  g 4 M .  f o r a l l but a compact subset of  A' = c-t'(X - B )  1 1  Then t h a t  1  Hence  For any p o s i t i v e number  (^(Xj/M  i s real Since  r  i s any  This i s a contraQ.E.D.  The Long L i n e . We now w i l l give an example t o show t h a t a non-paracompact  space may not be an m-xealcompact  space.  • 2 7 . Let  W  be the s e t of a l l o r d i n a l 'members l e s s than the  f i r s t uncountable o r d i n a l ,  {I : a e ¥} be a c o l l e c t i o n of a indexed by t h e s e t W , t h a t i s , . I I s an open  open i n t e r v a l s  and  vX  i n t e r v a l p a i r e d w i t h an element Interval  I  between  a  the immediate successor  and of  immediate predecessor of and  L  Let  y  x  and x  and  (2)  x  is in  y  x  x  x < a  in W  denotes  w i l l denote the  i s not a l i m i t  L .  Then  and  x <, y  y  is in  in W  ordinal), open  x < y In  conditions. If  W ,  an i n t e r v a l  I  , and  ?  I  , and  y  Is i n  W ,  and  I  , and  y  I s i n an i n t e r v a l  y  i s i n interval  Ip , and  a +1  by the f o l l o w i n g f i v e  W , and  i s i n interval  a < y (4)  L  We i n s e r t t h e open  and a l l the I n s e r t e d  are i n W  x = a or (3)  W  be any p o i n t s of  (1)  W .  a + 1 , where  i f a  denote the union o f Now, we order  of  (and a -1'  a  a  intervals.  a  a < 0  in  W  )  or (5)  x  and  and  are b o t h i n the same i n t e r v a l  x < y  We then t o p o l o g l z e resulting  y  L  In  I  and  X * a  by means of the order t o p o l o g y and the  space I s the " l o n g l i n e " . [9] . 1  We s h a l l show now f i v e p r o p e r t i e s o f the l o n g l i n e . (5.1)  Proposition:  L  s a t i s f i e s the f i r s t  axiom o f count-  ability (5.2)  Proposition: L The  proofs  i s a Hausdorff space.  of these two p r o p o s i t i o n s  are s t r a i g h t f o r w a r d  28. from the d e f i n i t i o n o f the l o n g l i n e . (5.3)  Definition:  order p r e s e r v i n g .  An i s o t o n e mapping i s a mapping which i s An isotone-homeomorphism  which i s a homeomorphism and b o t h isotone.  f  f  i s a mapping  and I t s i n v e r s e  And two ordered spaces are c a l l e d  are  1  isotohe-  homeomorphic i f t h e r e I s an isotone-homeomorphism the  f "  from one onto  other.  (5.4)  Proposition:  a e L , a + 1''.* [ l * a ] I s  For each  isotone-homeomorphic t o the u n i t i n t e r v a l [0,1], each p o i n t o f L , not the f i r s t element  1 , has an open  neighborhood which i s homeomorphie t o an open Proof:  consequently-  interval.  We w i l l show t h i s p r o p o s i t i o n / b y t r a n s i n f i n i t e  induc-  tion. We know t h a t [0,1]  I n TS?~  .  [1,2]  i s Isotone-homeomorphic t o  Now, assume t h a t  homeomorphic t o [0,1]  [l,p] In L  f o r each  p < a .  i s Isotone-  We have t o show t h a t  i s a l s o isotone-homeomorphic t o [0,1]  [l,a] a  in L  Case 1, i f  .  i s a n o n - l i m i t o r d i n a l , t h e n by the I n d u c t i o n h y p o t h e s i s i s Isotone homeomorphic t o [0,1]  [l,a-l] that  1  and [0,1],  [a-l,a]  [•2*1]  [0,1]  .  F i r s t of all, functions [l,a ] n  If a  i s a limit  ordinal,  [ a ] which converges t o a . n  we s h a l l show t h a t t h e r e e x i s t s a sequence of  [ f 3 such t h a t  onto  and  Thus, [ l , a ] i s i s o t o n e -  Case 2,  then t h e r e e x i s t s a sequence  However, we know  and [O,-^],, and [0,1]  are isotone-homeomorphic.  homeomorphic t o [0,1]  .  n  f  [0, -JJ-^-] , and  i s an isotone-homeomorphlsm o f  n  f n  \ l  1  »  a n  a  i ^  f n  - l  *  f  o  r a  1  1  n  €  N  o  29. By our i n d u c t i o n hypothesis.,  there e x i s t s  i s an isotone-homeomorphism o f [0,1]  map  t o [0, -~j^;]  n e N .  Evidently  e  onto  [1,0]  e (x)=  x  [l,a ] n  such t h a t  [g^] such t h a t  n  .  Let e  n  f o r each  i s an Isotone-homeomorphism  [0,1]  of  XX  onto  [0, -—^j-] .  h  n  = e og  n  on  n  Let h  [l,a ]  f o r each  R  isotone-homeomorphism o f f ^ ( x ) = h^(x)  for x €  x e [ a ^ , a ] > where  g  2  1  1  h  2  ^  i s an  [0, •=§_] .  onto  Now, l e t  and f ( x ) = g <>h (x)  [0,0^]  2  2  f o r each  2  i s d e f i n e d as f o l l o w s ;  2  i , h ( ) .  Then  a  2  1  n e N .  [l,a]  y~ 2^ i)  e^y) — g +s [ h ( a ) , 2/3]  he a mapping d e f i n e d as f o l l o w s :  n  f  o  r  y  * 2^ 2)  €  h  a i  Since  g  a  -  ]  Is a l i n e a r transformation,  2  1 an isotone-homeomorphism o f Therefore  f  [0,2/3]  t^[0 a^] = f-^ .  n , l e t f (x) = ^ „ (x)  g^h^x)  for x e [a  n  h  h  n-l^ —  a  —  u /„  n< n-l a  nTX^  ^ n^ n-l^ h  a  o  n  homeomorphism o f = n-l. f  n = > 1  1  has a l i m i t  [0,a ]  n  In general, n  ^(y)  n  = ^  n  +  -  i s an Isotone-homeomorphism o f  }  t  ^ n r * nTT^  o  onto  Thus,  [0,-^j]  f  n  i s an i s o t o n e -  such t h a t  ^nK * ^!^ 1  0  the r e q u i r e d sequence o f  Next, we s h a l l show t h a t the sequence [ f ] n  f , say, such t h a t  [ l , a ] onto  onto  a  , a ] . where  [l,a]  [f 3 .  2  f o r x e [°» -il » f ( x ) =  T h e r e f o r e , we have obtained  functions  of  \  n  2  [•£, ^ ] .  And so on.  3  f o r each  y  onto  i s an isotone-homeomorphism o f  2  such t h a t  n-lx n( •=-—=•) • n n n+I "  2/3]  [h^a-jO,  i t is  [0,1]  .  Define  x € [ l , a ) , and f ( a ) = 1 .  f  i s an isotone-homeomorphism f(x) =  f° (x)  Then the f u n c t i o n  n  f  f o r each i s well-  30. defined. that  Indeed, f o r each  n > n  implies  Q  x € [ l , a ) , there i s an  x < a  so t h a t  n  n > n  n  implies  Q  i s c o n s t a n t , and hence t h e l i m i t always e x i s t s .  such  o f n  (  x  )  Then t h e  f o l l o w i n g f i v e p r o p e r t i e s f o l l o w Immediately from t h e d e f i n i t i o n of  f  and t h e choice o f  {f 1 s  (1)  f i s one-one .  (2)  f"  (3)  f(a ) = f (a ) = n  iff n-«* (4) and  f  (By (1))'.  i s well-defined„  1  n  ( n' 0  f  -  v  and  f  f o r each  n  1 =  n , so t h a t  ,  1  are Isotone mappings,  1  are continuous a t each p o i n t of  finally  (5)  f  and  [l,a)  f"  [ 0 , 1 ) respectively.  and  To show t h a t  f  i s continuous a t a , we have t o show *1  CP }  t h a t f o r each sequence f ( a ) (= 1)  .  n £ N .  Since  f  1 .  i s hounded above by b = JJjJ ( P ) •  The*  f  n  f  which i m p l i e s  that  f "  1  b  1  1  •  f  I  ^  6  f  n  « f  = 1  (b)  Hence  f  b  n  t  ( ^ n ^ exists-  for i sa  (f(P )3 n  I»et  < 1 / t h e n by (5)> we  But  b = 1 , or  1 .  n + 1  {f(£„)} n  b , so t h a t  i s continuous a t  I s continuous a t  )•  I t i s clear that  Thus  I s continuous a t  This i s impossible. Consequently,  n  |3 < P  i s an i s o t o n e mapping  monotonic, n o n d e c r e a s i n g sequence.  know t h a t  f(P ) =  (By Prop. (5.1) and theorem (5.3*0 [19]  Without l o s s o f g e n e r a l i t y , we may assume t h a t each  J yy.  converging t o a ,  n  a .  f(p„) = b P  f° (b)=^Cand 1  £jj O ) f  n  =  1  n  = a .  = (a) • f  Now, we want t o show  We know t h a t t h e space  [0,1]  51. s a t i s f i e s the second axiom of c o u n t a b i l i t y , hence a g a i n i t i s enough t o show t h a t f o r any sequence ^'"^"^n^  =  =  *  a  W  e  m  a  [b ]  which converges t o  n  a l s o assume t h a t  y  {b ) n  1.  i s a monotonic non-decreasing sequence converging t o Since  f°°^  non-decreasing sequence. 1 .  bounded above by  0 f"  =  f" (b„) . xi  Thus  I s continuous at  1  f(0)  .  1  f ( 0 ) + 1 > and  [l,a]  onto  J^JJ b  [0,1]  Is  Let  b  =  = 1 .  R  1  x  N  n  Therefore .  Urn  l i m --I/*. „ ^ f~ (b ) = a = f" (l).  0 - a , or 1 .  )}  1  implies  n  Hence  {f" (b  0 < a , then by (5) as  If  1  i s continuous a t  homeomorphlsm of  i s a monotonic  exists .  1  0 , ^JJJ f" (h ) = 0  This i s impossible. f""  J^JJ f°° (b)^  0 —• < a .  So t h a t we have  That i s ,  n  I t i s also clear that  Then  1  iifj  n~*°»  {f~^"(b )3  i s an i s o t o n e mapping  f  i s an ISotone-  By t r a n s f i n i t e i n d u c t i o n , the  f i r s t p a r t o f the p r o p o s i t i o n i s proved. Let  a e L , a 4 I •  homeomorphic t o of  [0,1]  Then [ l , a + l ]  , so t h a t  (l,a+l)  a , i s isotone-homeomorphic t o  (5.5)  Proposition:  L  Is isotone-  (0,1)  an open neighborhood .  Q.E.D.  I s countably compact but I s not p a r a -  compact hence i t I s not a compact space. Proof:  Let  A  be any countably i n f i n i t e  w i l l be contained i n the union of  A  has at most countably many elements.  Hence  [l,a ] Q  l i m i t point i n  A .  Then  i s compact.. [l,a ] c L . Q  [l,a ] Q  How  L .  Then,  [I. : a e A c W] , where a  A  upper bound of  subset of  Let  a_  be the l e a s t  i s homeomorphic t o Ac  [l,a ] ,  That I s ,  D  A  [0*1]  .  must have a  L, i s countably compact.  32. Now, we w i l l see t h a t of P r o p . ( 5 . 4 ) , every p o i n t hood home omorphic t o L  E  L  a 4 1  a e L,  .  1  i s n o t paracompact.  has an open neighbor-  [1,2)  Moreover,  which I s a h a l f open i n t e r v a l i n E .  2-68  Theorem it  (see [9] ([9],  p. 80) p. 8l)  . L  But  L  i s an open s e t o f  Thus,  1  metrizable  I n view  L  i s locally  i s not metrizable.  i s not paracompact.  Consequently,  i s n o t compact.  (5.6)  By  Q.E.D.  Proposition:  Gf any two d i s j o i n t c l o s e d s e t s i n L ,  one i s bounded. Proof:  I f F^  and  F  are two c o f i r i a l c l o s e d s e t s , we -can  2  choose ah i n c r e a s i n g sequence  {a : n e N] where a € F, i f n n 1 i f n i s even. Then, s i n c e - F^ and 1  n F  i s odd, and  d 0  are closed  a  n  e F  f a nsN  n  2  = i * " * a e F, n F . n^» n 1 2  This i s a contra-  0  diction.  Q.E.D.  From  Prop.  (5.4), we know t h a t each p o i n t  a"+ 1  o  f  the l o n g l i n e has an open neighborhood which i s homeomorphic t o an open i n t e r v a l , while each open neighborhood o f morphic t o the h a l f open i n t e r v a l then can be considered boundary p o i n t 1 . f u n c t i o n s on (5.7)  , say.  i s homeo-  The l o n g  line  as a 1-dimensional m a n i f o l d w i t h a  Hence we can d e f i n e t h e d i f f e r e n t i a b l e  L .  Proposition:  a tail  [0,1)  1  Every f u n c t i o n  L - L ( a ) , where  a  f e C(L) i s a constant on  depends on  f , and  L(a) = [a e L :  a < a} . Proof:  I t i s c l e a r t h a t every t a i l  L - L(a)  compact, ( i n f a c t , i t i s homeomorphic t o  L  i s countably  itself).  Thus,  33. each image s e t ! f [ L - L ( a ) ] E  1  and hence compact, (as  E  Choose a number  r  i s cofinal i n  1  n e N  bounded. all  ,f  That i s ,  L .  Now, f o r  Hence, by Prop. ( 5 . 6 ) ,  there I s  Is  a  n  Q.E.D.  f [ L - L ( a ) 3 = {r} . Q  Let the f i r s t  L*  be the union space of  uncountable o r d i n a l .  f*(fi)  f  to a function  i s the f i n a l  f * e C ( L * ) , and of  f .  ion of  f*  on  L*  to  L  belongs t o  i s isomorphic w i t h Since  L*  C (L) . m  f .  cr e L* .  the Isomorphism o f  C (L*)  g e C ( L * ) , the r e s t r i c t m  I t follows that  m  C ( L * ) t every i d e a l i s m  m  M  ff  = ( f * e C (L*) :  (By Theorem ( 3 . 3 ) . )  with  C (L*) m  f o r each  m  By v i r t u e o f '  C ( L ) , the maximal i d e a l s i n m  Moreover, the f i x e d maximal i d e a l s i n in  C (L)  f -» f * .  are i n one-one correspondence w i t h those o f  M^  extension  I s compact, we a l r e a d y have a complete des-  f * ( g ) = 0} , where  Ideals  m  Evidently,  f i x e d , and the maximal I d e a l s assume the form  m  L*  by d e f i n i n g t h a t  (^(L*) , under the mapping  c r i p t i o n o f the maximal i d e a l s i n  C (L)  0 ,  f e C (L) ,  i s the unique d i f f e r e n t i a b l e  On the other hand, f o r each g  and t h e p o i n t  F o r each  constant v a l u e o f f*  m  L  Then, i t i s c l e a r t h a t  i s a compact 1-dimensional m a n i f o l d . we extend  is  e L such t h a t a_ < a f o r n n — n L e t a e L , and a„ > f ^ E a„ . We have o o — nsN n  e A . n  n  (r) .  intersection.  = {x e L : | f ( x ) - r | >. |j}  the c l o s e d s e t  d i s j o i n t from  of the nested f a m i l y i s non-  belonging to t h i s  Then the c l o s e d s e t f ° ( r ) each  has a countable base)*- so t h a t  1  0 f[L.- L(a)] aeL  the i n t e r s e c t i o n empty.  i s a c o u n t a b l y compact subset o f  C (L) m  C (L*) . m  correspond t o the  a e L , l e a v i n g j u s t one f r e e  34. maximal i d e a l i n  M  C ^ L ) , namely,  the one t h a t corresponds t o M . C (L)/M  FI  ~ C (L*)/M  m  .  m  Q  M  Though  FI  not h y p e r - r e a l , f o r  = {f £ C ( L ) : f * £ M ) , m  Q  N  Is free, i t i s  Q  Hence  L  I s not  m-realcompaet.  § 6.  Homomorphisms, C -mappings and G^-diffeombrphisms. m  In t h i s s e c t i o n we w i l l d e s c r i b e the r e a l t i o n between any C -mappings (see (6.1)) from  I c E^  m  homomorphisms  from  C ( Y ) t o G (X) . m  1  into  Y c  and  We s h a l l f i n d  m  that,  i n a sense, every homomorphism from one r i n g o f C - d i f f e r e n t i a b l e m  f u n c t i o n s i n t o another i s induced by a C -mapping (see (6.20)). m  We w i l l a l s o show t h a t any two m-realcompact  spaces  X  and Y  are C - d i f f e o m o r p h i c (see (6.1)). i f f C ( X ) and G ( Y ) m  m  m  are  isomorphic (see (6.19)). (6.1)  Definition:  T : X -* Y  L e t X c E**  1  and Y c E ^ .  I s s a i d t o be a C -mapping a t a p o i n t 1  at p .  m  If T  1  n  is  C  m  T-  T  and Y  a  And T i s C - d i f f e o m o r p h i s m m  We w i l l say then  are C - d i f f e o m o r p h i c i m p l i e s m  Definition:  An f e C ( X )  projection at a point of  n  such t h a t  X  and Y  are  Note t h a t by def. (6.1), i t i s c l e a r - t h a t  m  (6.2)  1  m  C -diffeomorphic. X  n 2  i s a C -mapping, one-one, onto and i t s i n v e r s e mapping  i s a l s o a (f-mapplng.  1  T (x ,...x ^)) i s  a t each p o i n t o f X , then T  m  i s s a i d t o be a C -mapping on X . if  p , i f each  m  component o f T(X) = ( T ( X , . . . , x ^ ) , j . . , C  A mapping  2  m  a  =n . 2  i s d a i d t o be a l o c a l i - t h  i f there e x i s t s a neighborhood U  f |U~= Jj  i = t h p r o j e c t i o n o f the space  , where E . n  M w i l l always denote the  35. (6.3)  Lemma:  and  Let X  be any subset o f  = x  i  for a l l  x e cl  x  B (a)  .  r  the bounded l o c a l p r o j e c t i o n s a t Proof:  Choose  n  .  F o r each  Then we c a l l  such t h a t  g(x) =  = x  3  n  i f x e E  ±  - B ,(a)  n  r  E  Then, i t i s c l e a r t h a t  1 1  he ±  .  Set  C**^)  (X) and s a t i s f i e s t h e r e q u i r e d c o n d i t i o n . Let  C  m  be a subset o f  a mapping from  X c E  g € C^ (6.4)  implies Theorem:  be a subset o f  T  t o . Y .c E  x  Proof:  m  for a l l  m  X  be a mapping from  into X  Y .  to  Y  and  C  m  Then  (2)  I f g.T € C ( X ) f o r each  i s a C -mapping i m p l i e s  g«r £ C ( X ) f o r a l l  m  m  g e C  m  X , then  T  ( l ) I t i s clear that  m  , and  C  i s a C «mapping on  g«T e C ( X ) f o r each m  m  g e C  X , we have, i n p a r t i c u l a r ,  m  g  .  Hence, by Def.  m  m  includes  X .  m  g«T € C ( X ) f o r each  e C (X) f o r 1 < i < n  a C -mapping. m  g«T e C ( X ) ( o r C * ( X ) )  T  Since  be  Then we w i l l see what  m  (^(Y).  a l l p r o j e c t i o n s on i  .  (1)  (2)  T (x)  c  i s a C -mapping from  Let T  a l l p r o j e c t i o n s of  T  m  ho  should be i n order t h a t  m  m  c  Q.E.D.  C ( Y ) ( o r C * ( Y ) ) , and  T  n  C  r  < 1 , elsewhere .  1 <_ 1 <_ri, the p r o j e c t i o n s o f  o g(x) .  b- (x) = ^ ( x ) C  ±  , there  f l i f x e ci^ B ( a )  V 0 < g(x) -0(x)  h ^ ( l _< I _< n ) ,  a .  G  Let  a e X  suclv t h a t  As shown i n Theorem (2.2)  r ' >.r .  e x i s t s - g e C*^E )  31  <. i < n) , h ^ e C**(X)  r > G , there are ^ , ( 1  h^x)  E  g e C™ .  which i n c l u d e s >£ T(X) = O  (6.1)  T is  56. (6.5)  Theorem:  Let  be a subset of (1) g € C  T  be a mapping from  C *(Y) .  to  Y  and  C  m  Then  m  T is  X  C -mapping i m p l i e s  g . T j C *(X)  m  m  for a l l  ,  m  (2)  g.T € C (X)  If  f o r each  m  g .€ C  , and  m  C  m  i n c l u d e s a l l l o c a l p r o j e c t i o n s d e f i n e d as i n Lemma (6.3) then  T  i s a C -mapping on  ,  X .  m  Proof:  ( 1 ) I t Is clear that g<»T e , C ( X ) f c ^ each g e.C . Tfl 9 ^ yy. Moreover, g e G (Y) and r [ X ] c Y , hence g o T e C (X) . m  (2)  m  The p r o o f i s s i m i l a r t o t h a t o f Theorem (6.4)  (2)  w i t h the p r o j e c t i o n s r e p l a c e d by the bounded l o c a l p r o j e c t i o n s . (6.6)  Definition:  a set  B.  tion  g«©  ep'  5  : D  ep  Let  be g i v e n mapping from a s e t  F o r each mapping A  carries  into  - D*" , e x p l l e i t y  g  B  from  into  Thus, ep  D .  A into  D , the composi  induces a mapping  ep'(g) = g»ep , and  ep»  I s s a i d t o be  cp .  an induced mapping of  There i s a d u a l i t y between the p r o p e r t i e s one-one and onto ( p r o v i d e d t h a t one I f f cp  D  i s onto, and  cp'  I s onto i f f cp  We a r e concerned w i t h a C a p p i n g where the r o l e of set of  E  1  D  the induced mapping  C (Y)  g s C (Y)  [resp.  into  C (X) [resp. C *(Y) i n t o m  C *(Y)]  m  T  of  X  into  The a p p r o p r i a t e C^Y)  .  m  m  .  Y ., sub-  Evidently,  t ' ( g ) = g.T e C ( X )  C *(X)3  i s one  i s one-one.  i s a homomorphism from  m  m  g,g" e C ( Y ) and any  or  m  T«, d e f i n e d by  each  m  E*~ .  i s taken by  w i l l be e i t h e r  Y  cp»  has more than one element);  for C (Y) ra  Indeed, f o r any  x e X .  (g+g')(T(X)) = g o ( x ) + g « »(T(X)) = (g»t)(x)-r(g».T)(x) = (g.T+g'o T  37. SO t h a t  T'(g+g') = (g+g' ) »T = goT+g' oT = T'(g) + T ' ( g ) . \  Similarly, (6.7)  T'(g.g*) = (g.T). (g'j »T) « (T« (g)) »(T ' (g)) .  Proposition:  The homomorphism  T'  c a r r i e s the constant  f u n c t i o n s onto the constant f u n c t i o n s of Proof:  F o r any  x € X  r(T(x)) = r . (6.8)  Hence  Proposition:  mapping Proof:  T  r £ [R  and  T'(_r) = r  (^(ftX])  identically.  T'(r)(X) =  / we have  on  1  T[X] .  The homomorphism  Q. E. D.  t ' determines the 1  uniquely.  I f CT i s a l s o determined by  » then  t'  f o r a l l g e C (Y) .  x e X , g(a(x)) - g ( T ( x ) )  Thus f o r each  By m-complete r e g u l a r i t y of  a' = T' . m  X , CT(X) = T(X) .  a = T .  Hence  Q.E.D. We now examine the d u a l i t y r e l a t i o n between (6.9)  Definition:  embedded i n  X  A subset  C^X) ]  (6.10) Theorem:  Let  m  n  C (Y) m  into  C  m  such t h a t  1  m  there i s  m  g | A = f .  be a C -mapping from m  T'  be the induced homomorphism  C ( X ) [resp. C * ( Y ) m  into  m  f  I s an isomorphism  (2)  T'  i s onto i f f T  © J  C -emhedded [ r e s p . m  ( l ) T'  X e E**  1  into  g -» g«T  from  C *(X)] . m  ( i n t o ) i f f T[X]  i s dense  i s a C - d i f f e o m o r p h i s m whose m  C *-embedded]. m  i s an isomorphism i f f  But the ' l a t t e r means  T'(g) =9  (T'(g))(x) = 0  .  [resp. C * ] -  Y .  Proof: =  is  m  (1)  image i s  g  X  T  c  Y c E ^ , and  in  T  of  and  f e C (A) [resp. C *(A)J,  i f f o r each  g e C ( X ) [resp,  A  T  implies  for a l l x e X .  38.  g = Q  implies g —-©  exists  that  T  m  for a l l  (T (g))(x ) 2  f(x )  for a l l  2  = x  o  2  X .  to  ge  =  C (Y).  for a l l  m  Since  m  T T  T(g) = f .  g s C (Y) .  1  T  i s w e l l defined  .  T[X]  as a mapping from T [ X ]  words, there  exists  The l a t t e r i s Hence  G  M  ( X ) includes  T[X] €  (^(TCX])  that T'  onto  he  f € C^X)  (^-embedded I n  Sufficiency: X .  that  a l l pro-  T"  C ( T [ X ] ) , h«T M  such t h a t  , by h y p o t h e s i s ,  g | T[X] = fOT" . 1  onto  s C ( X ) , i n other M  = f  or  g € C ( Y ) such t h a t m  h = fO  T  T~  !  i s a C -mapping from  f e C ( X ) , the f u n c t i o n M  has an e x t e n s i o n That i s  g € C (Y) M  f s g»T = T»(g) .  fOT"  1  such Hence  i s onto. m# The p r o o f f o r  by Lemma ( 6 „ 3 ) t h a t  C  i s e x a c t l y the same, s i n c e we know,  C *(X) M  =  1  .  T (g) = f.  Y .  By h y p o t h e s i s ,  Consider any  h<>T  isa  1  X  m  g | T [ X ] ' f o r some  T[X] i s  g j T[X] e  i s a C - d i f f e o m o r p h i s m from  Moreover, f o r any  X ,  the i n v e r s e  i s t h e 'function  _ 1  M  T  i s onto, _ f ( x :) =  !  Therefore,  f € C ( X ) , and  Hence  m  g(T(x^)) =  (T»(g))(x ) =  j e c t i o n s , i t f o l l o w s from Theorem (6ik)(2) C -mapping.  We s h a l l show  By m-complete r e g u l a r i t y o f  i s one-one.  f»t  f € ( ^ ( X ) ,.  f o r each  That i s ,  m  We know t h a t  C (T[X])  i s an isomorphism i f f  By m-complete  2  M  of  1  T<  ;  T ( X ^ ) = T ( X ) , then  f e C (X) .  Hence,  T  mapping  Y- .  that  s u c h  If  for a l l  !  on  By h y p o t h e s i s ,  g e G (Y)  i s one-one.  2  i s an isomorphism i f f  5  Y .  -Necessity:  there  g{T(x ))  g = Q  Y , we have t h a t  i s dense i n (2)  T  That i s ,  T [ X ] implies  on  r e g u l a r i t y of T[X]  on , Y .  c o n t a i n s bounded l o c a l p r o j e c t i o n s .  59.  (6.11) C o r o l l a r y : then  T'  Proof:  If T  m  M  m  T [ X ] = Y , by the theorem,  T  onto  C (X) .  i s an isomorphism o f C ( Y ) onto  Since  X  i s a C - d i f f e o m o r p h i s m from  i s both i s o -  morphism and onto.  (6.12) C o r o l l a r y :  If T  n, X. c E -  space  i s a C - d i f f e o m o r p h i s m o f a compact no m  t o Y c E-  1  , then the induced mapping  T' i s  onto. P r o o f : By h y p o t h e s i s , T [ X ] i s compact i n Y , so i t i s compact no no in E , hence i t i s c l o s e d i n E *• . By Whitney's A n a l y t i c E x t e n s i o n Theorem [30], f o r each f € C (E m  n 2  ) c C (Y)  such t h a t  m  g e C (T[X]) f | T[X] = g . '  i s C -embedded i n hence  T'  Y .  there  M  Moreover,  T  exists  Hence  T[X]  m  i s a C -diffeomorphism  (By Theorem (6.10)).  I s onto.  Next, we examine the I n v e r s e problem o f d e t e r m i n i n g when a given homomorphism o f C ( Y ) i n t o m  X  C -mapping from m  into  the homomorphisms from X  which  [or  C *(Y)3  all  r €  m  m  Into  , and cp m  be  Here, we s h a l l f i r s t  consider  i n t o fR , t h a t i s , the case i n  fR i s onto  fR .  cp from C ( Y )  In f a c t  m  cp(r) = r f o r  fR .  the u n i t y o f C ( Y ) That i s ,  ra  i s induced by some  Any nonzero homomorphism  cp(g) = cp( g°u) = cp(g)<>cp(u)  We know t h a t  g e C (Y)  C (Y)  M  c o n s i s t s o f j u s t one p o i n t .  (6.13) P r o p o s i t i o n :  Proof:  Y .  C (X)  i s not i d e n t i c a l l y zero.-, where .  cp(u) = 1 .  r -» cp(r)  for a l l  Hence  u is  cp(u) i s the u n i t y i n fk .  The mapping from fR t o fR  i s a nonzero homomorphism.  9  defined  By (0.22) i n [VJ.  40.  it  i s the i d e n t i t y .  In o t h e r words, cp(r) = r .  Hence  cp i s  onto.  (6.14) Propositions of  The correspondence between the homomorphisms  C '(Y) [ o r C i ( Y ) ]  onto  m  ff( , and t h e r e a l maximal i d e a l s i s  one-one. The k e r n e l  Proofs fR  o f a homomorphism,  i s a maximal i n ffl ) .  C (Y) m  the  k e r n e l of such a homomorphism.  any  ,  C (Y)  onto  m  C (Y)/ m  On t h e o t h e r hand, each r e a l maximal i d e a l i s  cp ~  IH  of  and a r e a l I d e a l (as  ker  M , l e t  imal i d e a l  cp ,  and d e f i n e  ?  Indeed, f o r each r e a l max-  be t h e isomorphism from  cp s G ( Y ) - / £ by m  f e C (Y) .  Clearly,  m  G (Y)/M m  onto  cp(f) = f(M(f).) e fl> , f o r  cp i s onto, and  k e r cp ='{f.e C ( Y ) m  cp(f) = 0} = [ f £ C ( Y ) : j>(M(f)) = 0} = {f e C ( Y ) : f € M} . m  Row, f o r any  m  f ,g e C (Y), cp(f+g) = J>(M(f+g)) = P ( M ( f ) + M(g)) = m  ? ( M ( f ) ) + J ( M ( g ) ) = cp(f) + cp(g) , and 5  P(M(f).M(g))  = j H M ( f ) ) « ? ( M ( g ) ) . = cp(f)«cp(g) .  (0.23) [7] , d i s t i n c t kernels.  (6.15) morphism  cp from  such t h a t  fR  ideal i n  1  Y is.m-realcompact i f f t o each homo-  C ( Y ) onto fR <= i . e . each nonzero homom  - t h e r e corresponds a unique p o i n t  q>(g) = g(y)  Necessity:  fR .  unique  have d i s t i n c t  (f * i s s i m i l a r .  Proposition:  morphism i n t o  onto  homomorphism onto fR  Moreover, by  Hence i t i s one-one.  The p r o o f f o r  Proof:  cp(f.g) = j>.(M(f . g ) ) ' -  Y  m  L e t cp be any homomorphism from  (6.l4),  ker. cp  (^(Y) . By m-realcdmpactness Hence  Of  f o r a l l g € C (Y) .  Then, by Prop.  y € Y .  y  of  C (Y) m  i s a r e a l maximal Y , ker. cp =  cp(g) = cp(M (g)> = g ( y )  ,  for  :  41. Sufficiency:  The h y p o t h e s i s  each homomorphism o f  says t h a t the k e r n e l o f  fR  (^(Y) onto  i s a fixed ideal. (^(Y)  view of Prop. (6.14), each r e a l maximal i d e a l o f then f i x e d .  Hence  m  Q. E.D.  L e t cp be a homomorphism from ep(u) = u .  C ( X ) such t h a t m  If Y  then there e x i s t s a unique C -mapping  T  m  N o t i c e t h a t the c o n d i t i o n Proof o f the Theorem: a : g -• (cp(g)-)(x) Since  (^(Y)  i s m-realcompact,  of  cp(u) = u  F o r each p o i n t  X  into  Y  such  (cp(u))(x) - u(x) = 1 = J =  a  ,  0  g(y)  f o r a l l g e C (Y) . T  from  cp(g) = g°T f o r each f o r each  X  into  i s n o t z e r o homomorphism.' ep(g)(x) =  Then t h a t  Y , thus defined,, s a t i s f i e s  g € C (Y) m  i s clear.  Since  g, e C ( Y ) , Theorem (6.4) shows t h a t m  In view o f Prop. (6.8),  f o r which  T  T  T  cp(g) € C ( X ) m  i sa C m  i s a unique C^-mappIng  = cp .  1  Q.E.D.  (6.17) C o r o l l a r y ; An in-real compact space  Y  c o n t a i n s a C - image m  i f f C ( X ) c o n t a i n s a homomorphic image o f m  i n c l u d e s the constant f u n c t i o n s on Proof:  m  y € Y -such t h a t  mapping.  X  C ( Y ) i n JR.  We s e t y = T(X) .  m  the mapping  i s necessary.  x e X , t h e mapping  i s a homomorphism from  By Prop. (6.15), there i s a p o i n t  Then  m  T»'•=cp .  that  of  C (Y) into  i s a g e n e r a l i z a t i o n o f Prop. (6.15).  for X  (6.16) Theorem: into  Is  Y is"m-realcompact.  Our f i r s t r e s u l t about homomorphisms from  C (X)  In  Necessity:  Let T  T' , t h e induced mapping of  m  X .  be a C -mapping from m  C ( Y ) that  X  into  Y .  T , i s a homomorphism from  42. C (Y) into  G (Y)  m  m  By Prop. (6.7)  .  C ( X ) => T ' ( C ( Y ) ) m  which  m  induces the constant f u n c t i o n s on X . T'(C (Y))  Sufficiency: on  X , there i s  T'(U)OU  g  c o n t a i n s the constant f u n c t i o n s  m  e C (Y)  such t h a t  m  = T'(u)oT'(g)  T'(uog)  =  =  .  *r'(g) = u  T'(g)  = u  Then,  Thus, the r e s u l t  .  f o l l o w s immediately from Theorem (6.16).  (6.18) C o r o l l a r y ; /Inm-realcompact space image o f X  T'(U) =  Q.E.D. Y  c o n t a i n s a dense  (f  1  i f f (^(X) c o n t a i n s an isomorphic image o f C ( Y ) m  t h a t i n c l u d e s the constant f u n c t i o n s on X .  (6.19) The Main Theorem:  Two m-realcompact  are  C ( X ) and C ^ Y ) are isomorphic.  C^-diffeomorphic i f f  .  m  Then  cp"  1  i s an isomorphism o f C ( X ) onto  Into  « p ( g ) '= g<»T  f-e  C (X) m  3  Y  and from  and c p ( f ) =  .  - 1  Then, 1  into  X , r e s p e c t i v e l y , such t h a t  , f o r each  feT-j^  g(y) =  go(ToT )(y) = g»(ToT )(y) 1  Y  cp° (go )(y) 1  T  £ (^(Y) and  g  = (g.-T  for a l l y e Y .  ).•»-.,_ ( y )  Similarly,  i d e n t i t y mapping o f X  Thus  onto i t s e l f . Hence  X  T  [17]  0  and -r^ are the  and Y  are C m  Q. E. B„  S. Bo Myers [ l 6 ] ,  have d e a l t w i t h  T ^  T-^T i s the  diffeomorphic. Remark:  =  That i s ,  i s the i d e n t i t y mapping onto i t s e l f .  i n v e r s e mappings o f each other.  and T-^  T  m  X  (^(Y) .  m  By Theorem (6.16) there e x i s t unique C -mappings from  and Y  L e t ep he an isomorphism o f (^(Y) onto  Sufficiency: m  X  The n e c e s s i t y f o l l o w s from (6.11).  Proof:  C (X)  spaces  L„ E„ P u r s e l l [20] and M. Nakai  C - d i f f e r e n t i a b l e n-manifolds. m  Theorem i s a p p l i c a b l e t o any c l o s e d subset o f E  1 1  .  This  43. In s p i t e o f the remark made i n Theorem (6.l6), every ;  homomorphism i s induced, i n essence, by a  (6.20) Theorem: C (X), Y m  L e t cp be a homomorphism from  be m-realcompact.  cp(u)(x) = 1}  T  e x i s t s a unique C -mapping m  m  Proof:  from  E  Y , such t h a t  f o r a l l x e E , and  for a l l x e X - E .  As w i t h any homomorphism, the element  idenpotent i n C ^ X ) .  e = cp(u)  i s an  Hence i t i s t h e c h a r a c t e r i s t i c f u n c t -  i o n o f the s e t E = e " ( l ) .  Since  _:i  matter o f f a c t i t i s C ) ,  E  m  Moreover,  into  Moreover, t h e r e  into  cp(g)(x) = g ( T ( x ) )  for a l l g e C (Y)  C^Y)  Then the s e t E = (x € X :  i s open-and-^closed i n X .  cp(g)(x) = 0  -mapping.  e  i s continuous (as a  i s open-and-closed i n X .  again, as w i t h any homomorphism, cp[G (Y)] .  element o f the image r i n g  e  i s the u n i t y  I t follows that  m  cp(g)[X - E ] = cp(g)cp(u)[X - E] = cp(g)[X - E]ocp( )[X - E] = LT.' u  {G}  for a l l g € C (Y) . m  If into  E I <| , c o n s i d e r the homomorphism  C ( E ) , d e f i n e d by m  a(u) = u e C ( E ) f o r m  a'(g). = cp(g) | E .  = g°T(x)  T  of E  from C ( Y ) m  I t i s clear that By Theorem (6.16),  a(u) = e | E = u .  t h e r e e x i s t s a unique C^-mapping T' = a .  a  Into  Y  such t h a t  We have then f o r any g e C ( Y ) , cp(g)(x) = ( a ( g ) ( x ) ) m  for a l l x e E  and cp(g)fx) = 0  for a l l x e X - E . Q.E.D. n„  (6.21) Theorem:  Let Y  be a homomorphism from  be a compact subspace o f E • , and cp C ^ Y ) = C™*(Y)  the s e t E - {x e X : cp(u)(x) = 1} Moreover,  into  C *(X) . m  Then  i s open-and-closed i n X .  t h e r e e x i s t s a unique cP-mapping  T  from  E  into  Y ,  4 4 .  such t h a t f o r any  The  g e C (Y) ,  cp(g)(x) = g ( r ( x ) )  for a l l  x e E , and  ep(g)(x) = 0  for a l l  x e X - E .  proof  L e t cp be a homomorphism from  (6.22) C o r o l l a r y : a r i n g of  d* f u n c t i o n s .  If Y  1  e x i s t s a unique c l o s e d subset of  Let ®  a subring of  m  for a l l  (1.15) )•  : Z(g) D F ] .  «P(S )( ) = g ( T ( x ) ) = 0 x  0  p  Z ( g ) - 5 f['E]  so t h a t  Q  g e l .  choice  (6.23) P r o p o s i t i o n : embedded C  m  image o f  g  y  e ker cp ,  Q  .  That i s ,  g (-r[E]) = {0}  or  0  C (X) ra  .  = F .  Q  The uniqueness o f  of F .  iff  (see  : cp(g) = 0) .  = a Z(g ) 5 c y [ E ]  Q.E. D.  An m-realcompact space X  i s a z-ideal  .For any  k e r cp = I .  Consequently  f o l l o w s from the  I  x £ E ,  Q  cp(g)(x) = 0 Y  Q  Z(g )  Y  Let F = C ^ T [ E ] ,  cp(g )[X] = {0}  for a l l  By into  E  x e E , and  m  so t h a t  m  0  from  T  ker cp = {g e C ( Y )  cp(g ) = © € ft c C ( X )  F  m  Then  ker cp => I .  F .  C ( Y ) onto ft which i s  m  I t i s then c l e a r t h a t  Hence  such t h a t the k e r n e l  g e C (Y).  On the other hand,  there  f u n c t i o n s t h a t v a n i s h on  m  for a l l  x is X - E , f o r any  I = (g e (^(Y)  into  m  E = {x e X : cp(u)(x) = l ] .  cp(g)(x) = g(*r(x))  such t h a t  Thus  Y  (6.20), t h e r e e x i s t s a (^-mapping  Theorem  and  of  be a homomorphism o f C ( X ) , and  C (Y)  i s m-realcompact, then  F  cp i s t h e z - i d e a l o f a l l C  Proof:  (6.20).  i s s i m i l a r t o Theorem  Y  contains  a C m  i s a homomorphic image o f  C (Y). m  Proof:  The n e c e s s i t y I s c l e a r .  homomorphism o f  C ( Y ) onto m  For s u f f i c i e n c y , l e t  C ( X ) , and. k € C ( Y ) m  m  cp be a such t h a t  45. cp(k) = u .  Then  cp(u) = cp(k)ep(u) = cp(kou) = cp(k) = u .  Theorem (6.16), we have a unique that  T ' = cp .  T[X]  i s C^-embedded.  Hence  (6.24) P r o p o s i t i o n : C  m  image o f X  C ~mapping o f m  T ' i s onto.  A compact Space  into  X  ByY  such  By Theorem (6.10)(2), then  Y  c o n t a i n s a C^-embedded  i f f C^IX) i s a homomorphic image o f  C *(Y) . m  Proof I s s i m i l a r t o t h a t o f Prop. (6.23).  §7  The embedding Theorems. K. D. M a g i l l ,  conditions r e l a t i n g and  i n 1965,  X  s u f f i c i e n t f o r embedding  (7.1)  Definition:  ring. (7.2)  Then  we say t h a t  in Y .  0  ¥6 w i l l  of C -differentiable m  algebraic  generalize  functions.  L e t Ot be a c o l l e c t i o n o f r e a l i d e a l s i n a  f*l(X i s s a i d t o be a  Definition:  the  C(Y) and C ( X ) which are b o t h n e c e s s a r y  h i s r e s u l t s t o the r i n g s  of  [14], established  Let 0  be a s u b r i n g o f a r i n g  i s 8-dense i n A  6-real i d e a l s of A  6-real i d e a l .  with  A .  i f f o r every p a i r  I - I ' + ^ j I - I '  I  Then and I '  c o n t a i n s an  element o f p . (7.3) A is  Definition:  A homomorphism  from a r i n g  A  1  into a r i n g  i s a 6-homomorphism I f i t i s n o n t r i v i a l and the image o f A' 6-dense i n A. For  (7.4)  example, a homomorphism  Definition:  onto i s a  6-homomorphism.  A s e t o f elements o f a r i n g i s s a i d t o be a  s u b r e a l , i f i t i s contained i n a r e a l i d e a l o f the r i n g .  46. (7-5) if  Definition:  A  5-homomorphism i s a  5 F-homomorphism,  the image o f every r e a l i d e a l c o n t a i n i n g the k e r n e l i s sub-  real. (7.6)  Definition:  ring A'  A  is a  A 5-homomorphism from a r i n g  A'  into a  6 G-homomorphism i f f o r every r e a l i d e a l  whose image i s s u b r e a l , t h e r e e x i s t s an element  M  of  a ^ M  such t h a t the image o f every r e a l i d e a l not c o n t a i n i n g - a  is  subreal. cp  Let  be a homomorphism of a r i n g of f u n c t i o n s  i n t o another r i n g of f u n c t i o n s property  A .  We w i l l  (7-1) t o mean t h a t f o r every  t h e r e i s an  f e A'  cp  say t h a t  g e A  has  x 4 Z(g) ,  and  x 4 Z(cp(f)) , and  such t h a t  A'  Z(g) cr Z(cp(f)).  . (7.7) and  ni -Theorem:  such t h a t  mapping. of  Proofs  $(f)  into  Y  x  4 Z(g)  M  T  m  Suppose . T  ge  C  R A  (X)  .  cp  has  from  X  into  is a C -  a l l projections Y .  m  f o r a l l f e C (Y) m  T  Then  T(X) 4  T[V] in  Y  Thus  i s open i n such t h a t  V . of  T[V] Tl T[Z(g) ] = <|) (as T[X] .  and t h a t For  Y  U  x 4 Z(g)  However, we know t h a t  e x i s t s an open neighborhood  from  ct T[Z(g)] .  Y  V n Z(g) = ({) .  m  i s a C -diffeomorphism  3>(f) = f «  ,  Then  T  and  contains  i s a C -diffeomorphlsm i n t o  such t h a t where  ^(^(Y))  T  X c E  T(X) e c,t «r[Z(g)] , then f o r each open neighborhood  t h a t there  G  G (X) .  m  T(X), U H T[ Z( g) ]=)=({) .  and  into  m  = f o T f o r a l l f e C (Y)  Sufficiency:  X  C (Y)  ( 7 - 1 ) i f f there i s a homeomorphism  In a d d i t i o n , i f  X , th£n  if  Y be an m-realcompact space and  cp a homomorphism fr'om  the p r o p e r t y Y  Let  x  of so  such t h a t T  i s one-one),  Hence, there e x i s t s an © p e n s e t  G n T[X] = T[V] .  We know then t h a t  47. t ( x ) e T[V] e G , and G O T [ Z ( g ) ] = G fl «r[Z(g)] n T[X] = (G n T[.X]) n T [ Z ( g ) ] = T[V] diction.  n r [ Z ( g ) ] = (j).'which i s a c o n t r a -  r[x) 4 . C^TI  Now, s i n c e  Z(g) ] , by m-complete  r e g u l a r i t y o f Y , t h e r e i s an f e C ^ Y ) ctyT[Z(g)3  n T[X]] 3 z(g) .  " [z(f) 1  T  4 Z(f) .  and T(X)  Y  (f  m  x  (7-1).  satisfying  onto  as f o l l o w s .  L e t x e X be given.  Moreover,  x.4 Z(u)(=(j))  such t h a t  x 4 Z(q>(f )) '(p  That I s , Y  .  <P (t )  Hence  x  for a l l  Then we have  <p(f) = f o T f  f € C (Y)  .  m  for a l l  is a  C ^ Y ) Into  i s a homomorphism. f  e C (Y) m  Q  By Prop, (6.15) y .€ Y  f e C (Y) m  T"  1  .  since  such t h a t  We then d e f i n e  -mapping,  X  - < P ( f ) ( x ) +-0 .  Q  I s m-realcompact,. t h e r e e x i s t s a unique  us t o show t h a t  sufficiency.  (7-1), there e x i s t s  i s a non-zero homomorphism.  <j5• ( f ) = f ( y )  and  'Then the mapping  <j^(f)'= q ? ( f ) ( x )  so by  Q  t  T from  We d e f i n e a mapping  - fR d e f i n e d by  : C (Y)  T h i s proves the  L e t (p be a homomorphism o f  Necessity: M  1  ;But, ^ [ z f f ) n T[X]] = (x € x :  f o r ( x ) = 0} = Z ( f o t ) = Z((P(f)).  C (X)  x 4 T" (Z'(f))  Hence  Z(f) 3  such t h a t  T(X)  = y.  I t remains f o r  e x i s t s and i s con-  tinuous. Since T  f o e C (X)  Now, i f  i s a C -mapping. m  g e C^X)  m  That i s ,  T(X) + T(X')  i s well-defined.  that  T  i s open.  so t h a t  T"  1  1  , b y Theorem (6.4)  i n X , there i s a = 1.  x ' 4 Z(<j?(f))  + 0 , while  Now, T "  In order t o show t h a t  m  and g(x>)  such t h a t  f ( r ( x ' ) ) » [cp(f))(x'>  f e C (Y)  x == f x'  g(x) = 0  such t h a t  there i s an f e C ( Y ) Hence  for a l l  M  T  T  f(r(x))  i s one-one.  i s a mapping from  By  (7-1),  and x e Z ( t p ( f ) ) . = (<p(f))(x) = 0. Therefore  T  T [ X ] onto  - 1  X.  I s continuous, i t i s enough t o show  L e t U be any open set I n X , and x e U !  48.  be  arbitrary.  Then  t h e r e i s an f e C ( Y )  is,  T(X)  i s an i n t e r i o r p o i n t o f  in  Y).  Thus  T"  f o r every goT-  =  1  T"  contains  (p(C (Y)) c o n t a i n s  g  m  f . t f o r some  s  m  Since  be a c l o s e d subset of C (X) .  X .  x e P} .  Hence i t i s a 6 - r e a l i d e a l .  Then  If  For, l e t  i d e a l of  : M_  C ( X ) , then m  However, f o r each  VL = D{M  a , there  exists  Thus,  M, = n{M  : x e E} , where  to  f o r some  a e A] .  = n{M  f e Mg have  x  : x e E} = n (M  i s clear.  mined by  Consequently, (^(X)  Thus  Therefore, A  and X  x  Hence  Z(g) r> E .  g € Mp .  T"  P  is  1  Let  E  X  x e X  i s uniquely  M  x  = M  F = ct E x  .  We w i l l  Mp  .  f e Mp ,  Indeed, each  Take any  Since  show t h a t  E  g € Mg.  We  = P .  That i s ,  i s uniquely  deter-  determined by  E ,  P  i s unique.  i s m-realcompact i f f every 8 - r e a l i d e a l of  i s of the form  Q  E = {x e X : x corresponds  r> Mp .  = Mp .  a e A).  such t h a t  x  E  :  be any 5 - r e a l  c-t Z(g) = Z(g) 3 M  x  i s m-realcompact,  i s r e a l and  : x e F ] = Mp . M  Mp = ( f e C (X) Mp = n{M  Moreover,  m  then the converse i s a l s o t r u e .  E  m  Q. E. D.  i s an i d e a l i n  M  cp(c (Y))  X , by P r o p . ( 6 . 4 ) ,  Z ( f ) 3 P}  a  so t h a t  m  T[X] .  a .C -mapping on F  X , and  f e C (Y)  .  m  a l l l o c a l p r o j e c t i o n s of  Let  T [ X ] onto  = f € C^Y) c C ( T [ X ] )  1  a l l projections  m  i s a mapping from  1  g e 9(C (Y)),  (foT)oT"  Hence i t i s  i s continuous.  1  Since  That  T [ X ] (as Y - - Z ( f ) i s open  T [ U ] i s a set of I n t e r i o r points.  Now, we assume t h a t X .  Z ( g ) c Z(q>(f)).  and  T ( X ) e T [ X ] n [Y - Z ( f ) ] c T [ X - Z ( g ) ] .  It follows that  of  By (7-1),  M  x 4 Z(cp(f))  such t h a t  m  Open so t h a t  g € C ( X ) ..  x | Z(g) f o r some  f o r some c l o s e d subset  P  of  X .  49.  (7.8) Theorem: X  Let Y  he an a r b i t r a r y t o p o l o g i c a l space and  be an m-realcompact space.  concerning  a homomorphism  Then the f o l l o w i n g statements  (jp from  C (Y)  into  m  C (X) are M  equivalent. (7-1).  (1)  cp  (2)  cp i s a 6=homomorphism.  (3)  The image o f  s a t i s f i e s property  c ( Y ) separates p o i n t s and c l o s e d m  (1) implies  Proof:  ideals of  m  x.,e F' - F  x 4 Z(g) .  Mp - Mp. .  such t h a t Z(g)  hypothesis, M  z(g)  M  T h u s  M  z  (2) i m p l i e s i d e a l i n C (X) .  cp(C (Y)) . m  an(  M  p  cp(f) €  £  f  (3).  Mp, Mp,  I n p a r t i c u l a r , f o r any z e r o - s e t  f  ^(g)  €-C (Y) M  ( p{ )) 3 (g) z  b  x ^ ^ *  such t h a t  B  y  cp(f) e  * 4 Z(q>(f)) .  t  be an a r b i t r a r y 6 - r e a l  L e t Mp  (^(Y)  u  M  Mp ~ (0)  and s i n c e  cp  can not be contained i n x 4 F .  2 hence must c o n t a i n an element o f  In e i t h e r case,  f o r any c l o s e d subset  cp(C (Y))  Z(g) 3 F  f e (^(Y)  Hence  Cn the other hand, i f F 4 X ,, choose  M^. - Mp 4 ty  and there  4 ty , there e x i s t s an f € (^(Y)  - M,  p  i s nonzero, the Image of  Then  F' e£ F such that.  m  I f F = X , then  m  Mp( = (0)) .  g s C (X)  x 4 Z(g) , we have  >  Then  F o r any two 5 - r e a l i d e a l s  there e x i s t s an  " x '  be two 6 - r e a l  c  Z(g) c Z(cp(f)) .  cp(f) e Mp - Mp, .  such t h a t  M  cp i s a 6-homomorphism.  (l).  (^(X) such t h a t  M  t o ( l ) there e x i s t s an  and  That i s ,  (2) i m p l i e s  and  and a  According  x 4 Z(cp(f))  such t h a t  in  Let  Mp - Mp, 4 ty .  C (X) with  i s an element but  (2).  C (X).  i n no 6 - r e a l i d e a l of  s e t s and i s contained  F  of  cp[C (Y)] m  X .  i s not contained  i n Mp  We w i l l now dhow t h a t  separates p o i n t s and c l o s e d subsets.  If F  i s any  50. c l o s e d subset and x 4 F -> then there e x i s t s cp(f)(x) + 0  Mp » M  f £ C ( Y ) such t h a t  cp(f) e M  m  and Z(cp(f)) => F .  (3>) i m p l i e s ( 2 ) . homomorphism.  4 =ty•  x  That i s ,  ?  By h y p o t h e s i s  - M  x  .  Hence  (cp(f))(x) 4 (cp(f))(F).  Suppose (3) and cp i s not a 6i s n o t 6-dense i n C ( X ) . Hence,  That i s , cp(C (Y)) m  m  f o r some, Mp,Mp, we have , Mp-Mp4 ty b u t (Mp-Mp, )H cp(C (Y)) = ty . s  m  1  However, Mp - M  Mp, c M f o r each  x  U{M-p - M x  e F»  Q  f o r each  x  x e F«  : x e F«} 4 =  • X  Q  4 ty •  f e C (Y) m  But  cp(f) £ Mp - M  such that  - n{M  :x € F } = 1  x  (Mp - M  ) H cp(C (Y)) m  x  That i s , there does n o t e x i s t  m  p  any  p  Mp - Mp, r>  Hence, t h e r e e x i s t s a t l e a s t one  (Mp - M , ) n cp(C (Y)) = ty .  c  Thus  and M p - M p , = M  Mp - M  such t h a t  x e F* .  there I s not any f e C ( Y ) such t h a t  X  q  .  In other words,  ep(f)(x) 4 c t e p ( f ) ( F ) .  m  x  this i s a contradiction. (7-9)  Theorem:  morphism from  Let X  and Y  if  be m-realcompact.  A homo-  C ( X ) i s a 6-homomorphism i f f  C (Y) into m  t h e r e i s a homeomorphism foT  Q.E.D.  m  T  from  f o r a l l f € C ( Y ) and T m  X  into  Y  such t h a t  i s a C -mapping.  Moreover,  m  cp(C (Y)) c o n t a i n s a l l p r o j e c t i o n s o f X , then  T"  m  C -mapping, t h a t i s T m  cp(f) =  1  is a  i s a C ^dlffeomoprhism. ra  Proof f o l l o w s from Theorems ( 7 - 7 ) and ( 7 - 8 ) . (7.10) Lemma:  Let X  r e s p e c t i v e l y , and T a homomorphism  and Y  be a C -mapplng from m  cp from  C (Y) into m  Then f o r f i x e d i d e a l s ively,  cp [ M l c M  Proof:  Sufficiency:  be any subsets o f E into  C ( X ) by  M and u of y i f f T(X) = y . Suppose  X  m  and E  1  Y .  2  ,  Define  ep(f) = f«T .  C ( Y ) and C ( X ) r e s p e c t -  T(X) = y .  m  m  Then f o r  f e M Cr  ,  51.  f(y) = 0 .  ep(f) e M  Thus  .  x  f(y) = r .  cp(M ) c M  Hence  Necessity:  y  ep(M ) c M  Suppose  Tr  T(X) = y .  m  Theorem:  (7.11)  Let  A homomorphism there  c E^  X  = foT  and  T  cp from  m  C (X) i sa M  m  T  X  from  Y  such t h a t Y ,  in  1  X , then  T~  i s a l s o a C -mapping.  Let M  be a r e a l i d e a l o f  otherwise, there would be an  contradiction.  y  m  M = M  f o r some  y e T[X] .  f € k e r cp - M f o r some  For  Z ( f ) => T [ X ]  m  T(X) = y  cp(M ) c M  (7.10),  C ( Y ) which  f e C ( Y ) such t h a t  T h i s would imply Hence  Then  T[X] i s closed,  Moreover, s i n c e  y 4 Z(f) .  contains a l l  m  the k e r n e l of. cp , k e r cp .  y e Y .  into  6F-homomorphism  Moreover, i f cp(C (Y))  m  contains  be m-realcompact.  2  m  Sufficiency:  Lemma  Y c E**  regularity).  f o r a l l f e C ( Y ) , T [ X ] i s a c l o s e d subset o f  p r o j e c t i o n s of  and  (By m-complete  C (Y) into  i s a C -mapping.  Proof:  and  1  i s a homeomorphism  cp(f)  in  Let f € C (Y) ,  .  v  Then  e C ( Y ) which i m p l i e s  iff  .  x  f - r e M and c p ( f ) - r = c p ( f - r ) e M_ . — y — — x ep(f)(x) = r . Hence cp(f)(x) = f o < r ( x ) = f ( y ) f o r a l l  Thus f  c p ( f ( x ) ) = f(«r(x)) = f ( y ) = 0 .  On the other hand,  which i s a  x e X , and by  , t h a t i s , the image  M  i s subreal  ^  C (X). M  Necessity: Theorem  cp i s a 6F-homomorphism.  Suppose  t h e r e i s a homeomorphism  (7.9)>  such t h a t  cp(f) = f o r  mapping.  We w i l l show t h a t  Choose there  M  Y  By Lemma  (7.10)  from  f o r a l l f e C ( X ) , and  y e C^ (T[X] ) = F . i s a real ideal  T  M  x  , we have  of  into  X  T  By Y  i sa C m  T [ X ] i s a c l o s e d subset o f Then,  k e r cp = Mp, er M  C ( X ) such t h a t M  y = T(X) .  Thus  .  Hence,  ep(M ) c M y  y e T[X] .  Y .  x  . That  52. is,  T[X] = c * - r [ X ] . y  The l a s t p a r t f o l l o w s from t h a t o f Theorem (7.7).  n-. o Let X c E and Y c E he m-realcompact. n  (7-12) Theorem:  1  cp from  A homomorphism  2  T{X]  M  m  X  T from  m  such t h a t  Sufficiency:  m  Let M  T~  i s a C -mapping.  be a r e a l i d e a l o f  M  From Lemma (7.10) again,  T(X) = y .  Z ( f ) r> Y - T [ X ]  C ( Y ) whose m  x € X , cp[M ] c Then  y 4 Y-  and there i s an f e C (Y)  which i s a c l o s e d subset o f Y  such t h a t  and  Moreover, i f cp(C (Y)) con-  image i s s u b r e a l i n C ( X ) . Then, f o r some  T[X]  Y  m  t a i n s a l l p r o j e c t i o n s of X / then  M^ .  into  and T i s a C ~mapping f o r a l l f e C ( Y )  i s an open subset o f Y .  Proof:  C ( X ) i s a 6G-homomorphism  C (Y) into  i f f there e x i s t s a homeomorphism cp(f) = f «T  QoE„B„  ra  and y 4 Z ( f ) .  Thus,  f 4 M and  i t f.olloi/s t h a t t h e image o f r e a l i d e a l not c o n t a i n i n g  fi s  subreal. Necessity: i s 6-homomorphism homeomorphism that  an  so that,by  T from  cp(f) = f o r .  x eX .  X  i s subreal.  Theorem (7.9)s  into  Y  cp(M ) c M y  Then i t  there e x i s t s a  which i s a C -mapping such m  Let y € T[X] .  By Lemma (7.10),  f 4 My  cp i s a 6 G-homomorphism.  Suppose  Then, x  .  y = T ( X ) f o r some  Hence, there e x i s t s  and the image o f every r e a l i d e a l n o t c o n t a i n i n g  f  y € Y.- Z ( f ) c T [ X ] .  From t h i s i t f o l l o w s t h a t  Indeed, f o r each  y' € Y - Z ( f ) , f $ M , , cp(M , ) c M ,. f o r some x'  By Lemma -(7-10),  y  y  f  = T(X') e T[X] .  y  Hence  x  5  «r[X] i s open.  T h e ' l a s t p a r t f o l l o w s frbm t h a t o f Theorem (7.7). QoEoDo  53§8  A R e p r e s e n t a t i o n Theorem f o r T r a n s f o r m a t i o n s o f Rings o f C ^ d i f ^erentiable Functions. Let  T  X  and Y  he completely r e g u l a r Hausdorff  he a t r a n s f o r m a t i o n o f  C(Y) i n t o  g e n e r a l than a homomorphism.  C(X) which i s much more  We know t h a t  T  can he r e p r e s e n t e d  by means o f a continuous mapping from a subset o f E a continuous mapping  T  from a s u b s e t o f  s a t i s f i e s certain conditions. observe t h a t i f  T  (Theorem  X  Y  if  s a t i s f i e s c e r t a i n c o n d i t i o n s (see Theorem  Lemma:  subset and  G  x. j  2  L e t f ( x ) = f (x- ,... ,x ) L  of  E  1 1  Then  A = f (x^,.  x  G .  •'-i )]• x  n  x  .  x  f  2  a f =— j i 9x^9Xj  .(.Here we may assume t h a t i < j ) .  . . iX^+h, •. . . ,Xj+k,. .. , x )  - f (x^,.  n  ±  difference,  . . ,x.j+h . . . , x ) R  (8-1)  + f(x ,...,x )  n  x  i  n  n  becomes  i  ,  ( l > * *•• > ± x  x  1  A = P(x +h) - F ( x ) .  +@]s  i  x  + k  0 < ©'< 1 .  , where  ->' • - > j > • • • > ) ~ x  n  n  By the Mean V a l u e  i  A = hF (x 4-Gh)  theorem, we have f  We  P ( x ) = f(x^,...,x^+k,...,x ) - f ( x , . . . , x ) .  Then (8-1)  h  (8.5)).  be an a b b r e v i a t i o n f o r the mixed second  1  A = [  C^X) ,  Then, i f f„ „ _ i j ~  - f(x ,...,x ,...,Xj+k,...,x ) We w r i t e  into  , and the f i r s t p a r t i a l d e r i v a t i v e s i n x.  ex'ists and i s continuous I n Let A  i f T  results.  • e x i s t s and i s continuous i n G , then  Proof:  vY  be def i h e d i n an open  n  e x i s t and be continuous.  a f —2—t— 9Xj9x^  x X and  w i l l have a r e p r e s e n t a t i o n  s t a r t w i t h some g e n e r a l i z a t i o n s o f c l a s s i c a l (8.1)  1  We w i l l  C^Y)  where T  T  into  1, [12]).  i s a transformation of  i s m-realcompact, then  spaces,  f  ( i * " *' > ± x  X  i  x  That i s , + @  h  >  54. Again, by the Mean Value Theorem, we have A = hkf /  0 < 0' < 1 .  ~(x,,. . . ,x.+©h,. . . ,x .+©'k, .. . ,x„ ) ,  Xj^x^j  JL  l  j  n  = f (x-^,. .. ,x^+h,. . . ,x^+k,. . . ,x^)  - f (x-^,. . . ,x^+h,. .. *  - f(x ...,x ,...,Xj+k,...,x ) + f(x ,...,x ) ± 3  ±  n  1  n  )  (8-1))  (by  n  x  (8-2). (8-2)  Dividing h f  x , x .'( l* * * ' J i x  x  ~ f ( 2.  .. . ) x  X  n  x n  by k , we have  )  > • • • X j + © ' k , . . . ,x )  + € ) k  3  5  n  ^Xj^h,  * * *  •  = -g [ f ( x , . . .x +h,. . . , X j + 1  • • JX^) ] -  [ f (x-^,.. . , x.^,.  - f(x ,...,x )] . 1  (x^,...,x )  as a c e r t a i n f i x e d p o i n t , and take  n  l i m i t o f (8-3)  as k -• G .  Since  f  and f  X j  have  f (x ...,x +h ...,x ) x  1>  ±  = hf ^ x  G  x  J  (x- i...,x ) L  n  (8-4).  n  by h , and take the l i m i t off i t as h - 0 .  * ' " >± > x  X  ±  so i s  ±  +h  ^  • • • > n^ x  " Xj( l*' ' ' f  x  We know t h a t s i n c e  f  x  n^  =  f  x  ±  x ^  '  i s continuous X  „ j i  ,  f  X j  X  (8.2)  e x i s t we  i s continuous, we have  = f„  X j X  in  (8-4)  3  x  f  f  x  .  I j  h^O Ti J - X j ( l Or  - f  n  n.  X j X j  ±  f x  i  Xj^-k,  the  (x^ ... ,x +©h,... ,Xy . . . , x )  x  Next, d i v i d e Since  . .  (8-3).  '  n  Consider  ±  ±  X j  .  X  Corollary:  Let G  be a d i f f e r e n t i a b l e  n-manifold.  a f 2  Then,  f' i j x  x  =— BXjax  exists  and i s continuous  on each c o o r d i n a t e  ±  a f 2  neighborhood system i m p l i e s t h a t  f X  =— <T  exists  ax ax i  J  and i s  55.  continuous on each coordinate neighborhood system. Proof:  Note that the notations  f(x, , . . . , x ), f  , etc.  refer only to the local coordinate neighborhoods.  The proof is  the same as that of Lemma (8.1). (8.3)  Lemma:  Let  f(x)  be defined in an open subset  G of  k,+. ..+k E  , and  f  .  ,  =  —-— n  in  G.  1  Then each partial derivative of  permutations of  x- ,...,x L  k  n (m = Z 1=1  f  in one of the ' n>''' x  ,  Xg,.. . j X ^  1  k\  f  exist and be continuous  k  2  , x  n  '  n  k. letters) exists, is continuous, and equal to 1  x1 l ' *' "> x *n ' k  k  Proof:  We w i l l f i r s t prove this i n the case of  f  . x  P(x) = f  x  (x) .  i  tinuous.  Then, by assumption,  By Lemma (8.1),  F x  Thus Since  f  x  „ „ i k j x  f  x  i j k x  f  x  j i k x  x  = F„ _ J k  v  x  x  f „ v  exists and is con-  so i t is continuous,  x  f  x  i k j x  x  = f  x  „ . i j k x  x  is differentiable in  x. . K  ±  f„ = f j i i j x  Thus,  x  x  X Xj  x  By Lemma (8.1)  x  exists, is continuous and is continuous,  v x  „ k j  F ._ j k  Let  i j k x  = f  x  x  x  i J k x  x  and i s differentiable in  x  is continuous.  other permutation is similar.  x, . *  The proof for the  56. Now, assume t h a t i t i s t r u e f o r a l l show t h a t i t i s t r u e f o r l e t us assume t h a t  m = p .  f k' k x, i.-.x 1 n  f  m _< p - I .  We w i l l  Without l o s s of g e n e r a l i t y , e x i s t s and i s continuous,  n  where  § k' = p . i=l  We know t h a t every permutations i s a  product of t r a n s p o s i t i o n s  ( I , j ) -* ( j , i ) .  argument a p p l i e s t o each t r a n s p o s i t i o n . d e r i v a t i v e of  f  Moreover, the above  Hence, each p a r t i a l  i s one of the permutations of  x^,...,x^, n  e x i s t s and i s continuous, and equal t o  n*  ki  f  kl  v - l  V  ±  •  manifold  n .  t h a t t h i s r e s u l t i s t r u e I n a d i f f e r e n t i a b l e n-  i n each l o c a l coordinate  Lemma:  open subset set of  t h e Lemma i s true- f o r a l l  n Notice  (8.4)  Therefore  «  E  n  Let f ( t , x ) = f ( t , x , — , x ) 1  X c E .  the open subset  , where  X = E  n  x X  be d e f i n e d i n an , X  n  i s an open sub-  n  L e t g(x) be a r e a l v a l u e d f u n c t i o n d e f i n e d on X_ , -2— , n  a l l p o i n t s of  neighborhood.  denoted by  f  at  X , and  , be non-zero f o r  G  f  v  i  •  v  i  v  t  =  —  1*1 • • • k» < 2  k. = m .  Then  —  —  •  •••••'—•  ^ . . . ^ i ^ i  e x i s t and be continuous f o r each choice o f 1 < m'« - §  -  k£,...,k^  5 f (g(x) x),  such that  m  ;  i  n  e  x  i  s  t  s  and  57. i s continuous i f f  k  k .. . Sx„  9x,  1 Proof:  e x i s t s and i s continuous.  n  The s u f f i c i e n c y f o l l o w s immediately from the Chain  r u l e f o r p a r t i a l d e r i v a t i v e s and the h y p o t h e s i s . Necessity:  and Lemma (8.3\ without  By assumption  loss  of g e n e r a l i t y , we may c o n s i d e r the f i r s t p a r t i a l d e r i v a t i v e o f f(g(x),x)  with respect t o  x  .  1  (Then we have:  = f(g(x 4-h,.../x ),  a (g( h ) , Sx-^ f  x  x  1  h  x +h ...,x )»f(g(x ...,x ),  n  x^,..-.,x ) + e h where  e•  ;  0  hand s i d e can be w r i t t e n as  1  as  5  n  h -• 0 .  1 V  n  However, the r i g h t  [ f (g(x^4li,.. . ,x ),x^+h,.. . , x ) n  n  f (g(x +h,.. . x ) , x , . . . , x ) ] + [ f (g(x +h,. .. , x ) , x , . .. , x ) 1  5  n  1  n  f(g(x ,.i,x ),x ,.:.,x )] 1  n  1  1  J  n  1  n  + eh = f ( g ( x + h , . . . , x ) , x + © h , . . . , x )  n  1  1  n  1  n  «h+f ( g ( x + © » h , . . . , x ) , x , . . . , x ) • [ g ( x + h , . , x ) - g ( x , . . . , x ) ] Q  1  + €h .  n  where  :3*(g(*?'*)  x  n  x  n  x  R  0 <.©,.©' < 1 , by the Mean Value Theorem.  Thus  f (g(x +h,...,x ),x +©h,.,..,x ) • + ^ ( g C x ^ d ' h , . . . ^ ) ,  =  1  :L  n  1  n  3x^  x  Let  k  n^°  l>'''  >x  [ g (  h - 0 .  x  + 1  n  ^ - -  We have  l  - 3  3 f  x n  )  - g(x ,...,x )] + € . 1  (g( ^ )? x  x  n  ,= f ( g ( x ) , x ) lfitil-'+ f ( g ( x ) , x ) . o  1  9x^  By h y p o t h e s i s , we have t  B g  ( l = [ (%( )> ?? - f'^gCx),*) ]• / x  SX-^ f (g(x),x) .  Ml£i2Lh*l  ax.  d%( ) x  x  x  SX-^  e x i s t s and i s continuous.  . f ( g ( x ) , x ) , and 9  9f  (8-5)  '  0  Hence  9x^  f (g(x),x)  We know t h a t  have continuous 1 s t  58. p a r t i a l ^ d e r i v a t i v e s bec'ause  2nd p a r t i a l d e r i v a t i v e s . g(x)  i s continuous.  ( 8 - 5 ) , we can show t h a t  by d i f f e r e n t i a t i n g  that,  dg^J/dx-^  g(x)  has  Hence continuous  P r o c e e d i n g by i n d u c t i o n , we  has continuous m-th  conclude  p a r t i a l d e r i v a t i v e s as we Q. E. D.  require. If  , i n the lemma, i s r e p l a c e d by a d i f f e r e n t i a b l e  n-manifold, the r e s u l t i s a l s o t r u e i n each l o c a l c o o r d i n a t e neighborhood. Now,  we  are i n the p o s i t i o n t o show the r e p r e s e n t a t i o n  theorem. (8.5)  Theorem:  Let  X  be a d i f f e r e n t i a b l e n-manifold and  an m-realcompact spaCe, C (X)  (-a) (Tg)(x)  For  then  (T(goh))(x) (b) from  E  (Tt)( ) k  Set  for a l l h e  into  m  E  is  1  G (Y) m  t , where  G  m  (Tf)(x) »  m  and  in  (T(f»h))(x) =  . Q  x e X , the mapping  F o r each  t  onto  : t -• ( T t ) ( x )  (T[ C ( Y ) ]) (x) . m  x e X , the mapping  t - (Tt) (x)  0 _< k < m , and s i n c e  Tt e C (X)  i s a k - t h p a r t i a l d e r i v a t i v e of (d)  C  C (Y)  f , g e C ( Y ) . such t h a t  ( T ( f + h ) ) ( x ) = (T(g+h))(x)  into  1  in  k  x € X , if  F o r each  (c) m  be a t r a n s f o r m a t i o n of  such t h a t  m  C "  T  Y  C'(t) + 0  for a l l t € E  ft  1  is  k  m  ,  Tt . and  x € X , where  i s d e f i n e d i n (b) . E = {x e X :, ( T [ C ( Y ) ] ) ( x )  Then t h e r e i s a C -mapping m  mapping  eontaihs more than one p o i n t ) .  m  oo  from  homeomorphism onto  E  1  x E  T into  of E  E 1  (T[ C ( Y ) ]) ( x ) m  Q  into  Y , and a continuous  such t h a t f o r each  tu(t,x ) 0  x  Q  is a  e E , i»(t,x) i s ;  59. C  in  m  t  and  (1)  x  separately,  and  satisfying  ( T f ) ( x ) = u)(f O T ( X ) , X ) =  if  (T©)(x)  if  Moreover, E i s an open subset of iff  T  T  X  m  Proof:  Let  oo  be  implies that  «)(t,x )  E  .  q  That i s ,  of a l l , we  0  .  Q  (T(r-s .  We  will  m  =  a(t,x ) Q  (Tt)(x ) Q  are  is  C  jointly  by  in  e E , where  Q  t  r 4= s ,  (T(r-s))(x ) = Q  (T(«.'  -i-))(x ) = r-s  .  t .  x  x  j E , which  Q  i s one-one, so t h a t i t s  , exists.  where  Q  T h i s shows t h a t  By h y p o t h e s i s (b) a(t,x )  By h y p o t h e s i s (b)  and  i s continuous.  Q  i s variable,  t«(r,x)  is  C  m  ut>(t,x ) = 0  By Lemma (P.675 [12])  x e E , we  v  Given any  ep  d e f i n e a mapping  ep (f) = a((Tf)(x),x) .  a homomorphism. so t h a t  m  1  (a)  both  uu  and  continuous.  For each  fR  r e E  show t h a t  f o r some  (Tu)(xJ  of Domain Theorem,  f o r each  x  defined  1  0  0  in  Y .  m  Thus' «)(t,x )  <»(r,x) = (Tr_)(x)  E  '  Q  Since  T  h e C ( Y ) , (T(h»u))(x ) = (T(h««))u ) .  Q  the Brouwer I n v a r i a n c e  x  would have  ( T h ) ( x ) = (T©)(x ) .  i n v e r s e , denoted by  into  0  -A-))'(x )  Thus, f o r each  i s a contradiction.  into  C (Y) Y  , then  m  uo(r,x ) = oo(s,x )  (Tr)(x ) = (Ts)(x ) Q  C (X)  x E  1  i s one-one f o r each  0  ), and  onto  m  r-s  a  i s dense i n  onto a C^-embedded subset of  First  i s v a r i a b l e , indeed, i f  (T«)(x  C (Y)  the mapping from  <u(t,x) = ( T t ) ( x ) .  (T©)X  f o r each f e  X , and ' T[E]  maps  i s a C - d i f f e o m o r p h i s m of  then  x € X - E  i s one-one. In a d d i t i o n , i f  by  x e E  f,g  in  ( T f ) ( x ) - u)(r,x) = ( T r ) ( x )  We  will  C (Y) m  and  of  (^(Y)  show t h a t  , let  cp  «P (g) =  is 8  x  ( T g ) ( x ) = ui(s,x) =  >  60. (Ts)(x)  .  Then, b y ( a ) , ( T ( f + g ) ) ( x )  (T(r+s))(x)  = «j(r+s,x) .  a((T(r+s))(x),x)  Hence  = (T(r+g))(x)  c? (f+g) = a ( ( T ( f + g ) ) ( x ) , x )  x  Also  x  (T(f«_s))(x) = ( T ( r . s ) ) ( x ) = a j ( r . s , x ) , h e n c e a ( ( T ( f o g ) ) ( x ) , x ) = r.s = cp (f)°cp (g) . x  x  i s fixed.  cx(t,x)  the say, can  m-realcompactness o f  define  a mapping  x € E , T(X)  i s the  f(y)(= foT(x)) a((Tf)(x),x)  T  of  for a l l  .  ( T f ) ( x ) = (T©)(x)  m  f o r each  f o r any f > g e C (Y)  g = 0 , we h a v e  the  ( l ) holds.  X - E  set  .  From the  first  Y  paragraph  so that  E  For  That i s ,  each  ep ( f ) = x  such t h a t  f(«r(x)).= .  definition  of  a  that  We now show t h a t  We know t h a t f o r  one p o i n t .  for a l l definition o f the  each  That i s , ( T f ) ( x )  f e (^(Y) of  proof,  ( x e X : ( T u ) ( x ) = (T©)(x)}  i sclosed  T h u s , we  I n p a r t i c u l a r , i f we l e t  ( T f ) ( x ) = (T©)(x)  d i s c u s s i o n i n the that  .  m  equation  of  and  y e Y ,  as follows:  x € E .  i s only  m  is a  x  (6.15)  .  m  for x e X - E .  x e X - E , (T[C (Y)])(x)  the  y  f e C (Y)  From.  f € C (Y) Y  I t f o l l o w s from the  ( T f ) ( x ) = u)(f<>T(x),x)  (Tg)(x)  into  unique p o i n t  cp  i s a unique p o i n t  for a i l E  x  i s onto.-  Y , there  x  ep (f.g) =  i s one-one m a p p i n g , where  cp__  cp (f) = f ( y )  such t h a t  (T(feg))(x) =  Moreover,  x  (6.13)  By P r o p .  =  x  = r + s = c p ( f ) + cp (g) .  n o n - z e r o homomorphism, f o r  =  .  Thus,  E , and t h e we c o n c l u d e  which coincides  i s open i n X .  Hence  E  1  with  x E i s  a d i f f e r e n t i a b l e (n+1)-manifold. Now, the (see  we a r e  definition the  of  notation  r e a d y t o show t h a t  m,  we h a v e  i n Lemma  T  is C  ( T t ) ( x ) = to k  (8.4)),  .  m  From (c) and  *n+l^  iX  k  0  i s  c " m  ic  2  i n t  for  each  61, n±l 2  x_ € 'X , where  i  k. = k .  1=2  That i s , i t s (m-k)-th d e r i v a t i v e  1  e x i s t s and i s continuous i n t . fixed  r e E  1  On the other hand, f o r every  , (Tr)^ ^(x)'» w v. v 2i ...(n+l)^ 2  By Lemma (P.675 [12])  tiable i n x . the p r o o f )  (r,x) i s differen-  k  m K k 2 . . . (n+1)  _(t,x)  1 + 1  (the f i r s t paragraph o f  i s j o i n t l y continuous i n t  2  ahd  x .  Thus, hy Lemma ( 8 . 1 ) , a l l mixed p a r t i a l d e r i v a t i v e s  e x i s t and are continuous. for  (t,x) e E  1  x E .  ( T f ) ( x ) = uo(f.T(X),X) f  From  (d) we have  (» (t,x) = } = , 0  1  By Lemma ( 8 . 3 ) , and t h e f a c t is C  , we have  m  i s a r b i t r a r y i n C ( Y ) , by Theorem m  that  f« e C ( E ) .  Since  M  T  (6.4)  T  i s a C -mapping. m  On t h e other hand, by Lemma ( 8 . 3 ) , and t h e c o n t i n u i t y o f ^ k« v (t,x) i n t 2 . . . (n+1)^i+l  and x , we have f o r each  g e C ( X ) , t»(g(x),x)  on  2  m  is C  m  E.-. t h i s f a c t w i l l be needed  later. In v i r t u e of m-complete i s n o t dense i n Y t h a t • c o i n c i d e on i s equivalent  i f o n l y i f there T [ E ] h u t not on  m  Hence (Tg)(x) iff  T  m  and  i s not one-one  g e C (Y) m  However, t h e l a t t e r  ( T f ) ( x ) = (Tg)(x)  (as we know t h a t  fora l l x € X - E ) .  Therefore  f and  fora l l  x e E .  ( T f ) ( x ) = (T©)(x) = T [ E ] i s dense i n Y  i s one-one. In a d d i t i o n , we assume now  C (X)  Y .  f  b u t t»(f ° T ( X ) , X ) = a)(g«T(x),x)  f + g  x € E , that i s ,  T  exist  t o t h e c o n d i t i o n t h a t there e x i s t  g e C ( Y ) such t h a t for a l l  r e g u l a r i t y , we know t h a t T [ E ]  which i n c l u d e s a l l constant  T  i s onto.  functions.  Then  T[G (Y)] =  Thus, f o r each  x e, X , ( T [ C ( Y ) ] ) ( x ) = C ( X ) ( x ) 3 ( r ( x ) = r : r e E } = E m  m  m  1  1  .  62.  f e C (X) , f(x) = r  On the other hand, f o r each r  € E  .  1  Hence  X = E .  Consequently T(X,)  (T[ C ^ Y ) ]) (x) = E  = T(X ) = y q  Since  T  We w i l l  show t h a t  but  f  That i s , f ( y ) = g ( y ) Q  and  so t h a t  ( T f ) ( x ) + (Tg)(x ) = tn(g(y ),X ) Q  we have  f ( y )4 g(y )  T  I.e. Now,  i s one-one.  f  € (^(Y) Q  c {h«T : h e C ( Y ) } .  all  m  h e C (Y) .  f e ( ^ ( X ) , f = h»r  each h  |  T[X]  e  .  C (T[X]) m  F i n a l l y , we w i l l g e C (*r[X])  F o r each  m  ce(goT(x),x) exists  g  f o r each € C (Y)  f o r each  an e x t e n s i o n of  g .  T  m  show t h a t  T[X]  In other words, h*T e C ( X )  g  Q  M  T"  hence is  1  C  m  f^r" -.1  on  T[X]  m  C (X) .  x e X .  Y .  f(x) =  Thus, t h e r e  M  = f , or  for  Now, f o r  i s C -embedded i n  f , d e f i n e d by  x e T [ X ] , i s In Q  i n onto,  m  (6.4),  Tg  Q  f(x) =  : h £ C (Y)} .  T  h £ C (Y)  , the f u n c t i o n  x = x ,  We then have  But, We know t h a t  By Theorem  x e X , or  Since  for a l l x e X .  0  Q  d e f i n e d by  Q  t«(g oT(x),x) = « ) ( g o T ( x ) , x ) f o r each g oT(x)  Hence  T(f )= f .  f o r some  such t h a t  m  Q  i s one-one, so  M  (^(X) = { h o  Hence  m  Q  C ( X ) (as shown above).  is in  such t h a t  Q  M  l  0  I S w e l l - d e f i n e d on T [ X ] .  1  f o r a l l x e_ X .  g(x) = a ( ( T f ) ( x ) , x ) = f o T ( x ) C (X)  oo(f(y ),x ) = 0  M  g(x) = <x(f(x),x)  t h e r e i s an  Q  eo(t,y )  g e C ( X ) , the mapping > f  for a l l x e E ='X  <»(g(x),x)  T "  Therefore  a>(f.T(x),x) =  g«T(x ) .  T*T(X  and  0  Q  i s one-one f o r  which'is a contradiction.  0  f o r each  Thus,  0  0  i»(t,x)  On the other hand,  c  Q  Then  Q  x + x ,  such t h a t  m  and as  If  Suppose  g e C (Y)  Q  f ( r ( x ) ) = g«T(x)  x ,  i s one-one.  ( T f ) ( x ) 4= ( T g ) ( x ) .  ( T f ) ( x ) = ( T g ) ( x ) = uu(goT(x),x) fixed  T Q  i s onto/ there e x i s t  ( T f ) ( x ) = (Tg)(x)  x c E .  f(r(x)) = f(T(x )) .  , then  Q  so t h a t  1  f o r some  ( T g ) ( x ) = f (x) Q  Hence  | T [ X ] = g , that i s  g°T(x) = g  Q  is  Q.E. D.  .  63. (8.6)  Theorem:  Let  X  and  Y  be any two d i f f e r e n t i a b l e n-i n 0  n  l  T E  *  N  2"  MANIF,OLDS  *  o  subspaces  r  be a C^-mapping from 1  x X  E"*"  into  E  f o r each  x  each and  € X .  Q  Y , and  «j(t,x )  i s one-one i f f T [ X ] Moreover, i f  That  Tf  m  .  i s dense i n T  Y , then is in  T  defined f o r C (X) , M  (8.5),  (b) of Theorem  T  ' of  maps onto  X  onto a  C m  (^(X) .  ( ^ ( X ) f o l l o w s immediately ftom the uo(t,x)  (8.5).  ,If  i s one-one f o r f i x e d f o r each  he  and T  T , and the f a c t  1  that  s a t i s f i e s c o n d i t i o n s (a) and  f>g e C ( Y ) m  x e X , then  uj(t,x )  Now,  T  i s into  m  f o r any  goT(x) .  m  i s a C -diffeomorphism  (Tg)(x) 0  a C -mapping from  Y .  Next, we w i l l show  (b) of Theorem  , respectively,  Then the t r a n s f o r m a t i o n  Chain r u l e , d e f i n i t i o n s of f e C (Y)  uo  ,  i s a homeomorphism i n t o  Q  s a t i s f i e s c o n d i t i o n s (a) and  Proof:  E  (Tf)(x) = t»(foT(x),x)  by  embedded subset of  and  E  into  such t h a t  1  f e (^(Y) T  X  of  are such t h a t  (Tf)(x) =  «j(f«>T(X),X) = uo(g«>T(x),x). x  Q  fo-r(x) =  e X , so we have  (^(Y),  But  ( T ( f + h ) ) ( x ) = t»( ( f + h ) T ( x ) , x ) =  t » ( f o T ( x ) + h«>T(x),x) = «j(goT(x) + h o T ( x ) , x ) = u3((g+h)T(x),x) =  (T(g+h))(x) . T  satisfies  d e f i n i t i o n of  Similarly (a) .  That  T  i s s i m i l a r to that  (8.5).  be the i n v e r s e of  T  (b) i s a consequence of  T  i s one-^one i f f T [ X ]  a((Tf)(x),x)  i s dense i n  i s one-one i f f T [ E ] i s dense i n By h y p o t h e s i s o f ; uu , f o r  «j(t,x ) .  Now,  Q  (Tf)(x) = oo(fOT(X),X) . Hence  satisfies  Hence  T .  The p r o o f t h a t  "Theorem  (T(f»h))(x) = (T(g»h))(x) .  is  Thus, C  01  on  f o r each  x  Q  e  x e X , we  f«T(x) = a((Tf)(x),x) X , f o r each  X  Y  in  , let have on  f e C (Y) . m  Y  X .  a(t,x ) 0  64.  Filially,  assume t h a t  onto a .^embedded subset of g  As shown i n the T[X]  that g  l a s t paragraph,  1  € G (Y)  such t h a t  g  _ 1  Hence is,  c  = f  or  T  maps  C  R  A  C  M  X'=f  - 1  (y)  We  ( T [ X ] ) .  g  and  know  (Tg )(x)  ( ^ ( X )  =  Q  a , we  have.  y = T(X) e T . [ X ] x ,e X .  f o r each  onto  (Y)  , define,  ( X ) .  know t h a t  and  1  Q  Tg  x € X  for a l l  = F « T " ' ( T ( X ) ) = F(x)  (Tg )(x)  We  By d e f i n i t i o n s Of  ( y ) = o)(g(y),x)  M  Y ', so there e x i s t s  | T[X] = g .  Q  C  for a l l  g e  M  t»(g«T(x), T ^ O T ^ X ) ) . FoT  F €,  1  i s a C -embedded subset of  M  Q  Given  1  .  X  .  Y  X  of  m  g(y) = g o r ( x ) - a ( F c T " ( y ) , T " ^ ) )  hy  e  i s a C -diffeomorphism  T  . That  Q.E.D. •  .  X  Note t h a t no e x t r a c o n d i t i o n i s needed f o r  Y .  or  i  §9  The  Rings of C ^ d i f f e r e n t i a b l e F u n c t i o n s  are not m-realcompact, and not We  C  M  ( X ) and  are C - d i f f e o m o r p h i c . m  C We  M  X  and  Y  are two  will first  X c E  where  n  .  ( Y ) are isomorphic i f f  X  and  s h a l l make some o b s e r v a t i o n s  d e f i n e d on  X  of a l l c o n s t r u c t a s u b r i n g of  Let  defined i n (6.2)),  •c(x ,...,x ) « x 1  S-^ = {r : r  with value  3 = 3^^1)32  and  i s the  r e /R),  for  ±  n  Y  about  ahd  S  in  the  f , g € Ot  ring,  and  , ,(as  function  = {^6-:  00 be the r i n g generated by  i s c l e a r t h a t Ot i s a commutative r i n g . quotient  2  (^(x)  1 < 1 < n  constant  it  f/g*  m-realcompact  cases i n t h i s s e c t i o n . We  Let  C  some A l g e b r a i c P r o p e r t i e s of  .  M  have shown t h a t i f  spaces, then  other  C  applicable i n  on Spaces which  1 ..'<, h}.  1 S .  Next, we  Then, embed Ot  R ( X ) , which c o n s i s t s of a l l q u o t i e n t s  Z(g) = {x e X : g(x) = 0} = <}) .  Evidently  65. X  ft(x) i s a commutative r i n g o f r a t i o n a l f u n c t i o n s on  with  u , the constant f u n c t i o n o f v a l u e 1 , z e r o element  unity  © ,-  0 .  the constant f u n c t i o n o f v a l u e  h-^ and  In the f o l l o w i n g lemmas and theorems  Ag are  r i n g s o f f u n c t i o n s w i t h the f o l l o w i n g p r o p e r t i e s : ,ft(X)c ^ C (X)  R ( Y ) c Ag e C ^ Y )  ,  M  Z ( f ) = $ , then (9.1) for  Lemma: a e X  f"" e A  and i f  x  There i s a f u n c t i o n  n  Then, i t i s c l e a r t h a t  that  f e M  Suppose t h a t  f e M .  Since  M M  g(a) + 0 .  Let h = f  i t s inverse  h " e A-^ .  That i s ,  2  1  A^ .  » {f e A  f  f  x  : f(a) = 0 ] ,  . n 2 f ( x ) = X ( x - a^) . ±  and belongs t o no other f i x e d  i s a f r e e maximal i d e a l i n A^  i s f r e e , there i s + g  2  .  We have  Since  M  g e M  Z(h)  i s an i d e a l  belongs t o M  such  such t h a t  <f) , so t h a t u = hh" € M 1  M i s t h e whole r i n g which i s impossible'.  f 4 M , or  with  belongs t o no other  _n e X c r , and  L e t a = (a.^,.. ,,a )  maximal.  f e.M  Z ( f ) = {a] , and  f r e e or f i x e d maximal i d e a l o f Proof:  ( o r Ag)  ±  (or Ag) .  1  such t h a t  f e A  c  .  Hence  -only .  Q.E. D.  n' Y  Let n .  n«  may be equal t o  Y , the r i n g of r a t i o n a l f u n c t i o n s on  Lemma:  f o r any M Proof:  where  L e t R ( Y ) be the r i n g d e f i n e d i n the same manner as a ( X )  w i t h domain (9.2)  be a subset o f E  &  I f cp i s an isomorphism from  c A  Since  i d e a l i n Ag .  1  , cp(M ) &  A^  Y .  onto  Ag , then  i s a f i x e d maximal i d e a l i n Ag .  cp i s an isomorphism onto,  cp(M ) Q  i s a maximal  By Lemma ( 9 . 1 ) there i s f ( j c ) = 2 0  ( x - a.^) i  2  66. such t h a t z  (^(  f 0  Z ( f ) = {a}  )) •  so t h a t Hence  belongs t o  M  only. ep(f )  Z ( ^ ( f ) ) = <j) , then  I f  Q  «p(M )  i s the whole r i n g  &  Z(cp(f" )) + § . .  i s a unit i n  Q  Ag  Consider  which i s  Gn the other hand, i f  0  impossible. Z ( ^ ( f ) ) contains Q  cp(f ) e  more than one p o i n t , say  b, b' e Y , then  c p ( f ) e M^,  would b e l o n g t o at l e a s t two  so t h a t  Q  f  0  ideals which again i s i m p o s s i b l e maximal i d e a l . (9.3)  Lemma:  f  and  Bg  be  subrings  of  C(X)  f u n c t i o n s , cp  isomorphism from  be  cp  into  B^  , and  i s the i d e n t i t y on the constant  X  of isomorphism, we have cp(u-u) = cp(©) = ©  Now,  -cp(u) = »u ,  so t h a t  Moreover, c p ^ )  cp(n) = n  cp(-n) - -n  k-r 4  G  f°  r  a  H  Indeed, i f  r a t i o n a l numbers  a l l r a t i o n a l numbers  r-...  1 1  <P(kJ-r  have  Then  By  property  Thus  n .  cp(-u) = n .  (by p r o p e r t y of isomorphism),  k  p  and  any  k , cp(k)  function  integer  i s also  i s an i r r a t i o n a l number,  r .  Moreover,  cp(k-r » -j^p ) = q>(u) = u , we  an  f o r a l l p o s i t i v e integer  f o r any p o s i t i v e i n t e g e r  function.  C(Y)  f o r a l l p o s i t i v e integer  We w i l l show t h a t f o r constant a constant  and be  ep(©)•=•:© .  and  = _g = cp(p_«3) = ja ^ ( j f )  That i s , ep(j^) = J[  .  connected.  cp(u) + cp(-u) = © .  or  =  functions.  I t i s c l e a r t h a t - cp(u) = u  Proof:  one  ^CW^)  Hence  r e s p e c t i v e l y , which c o n t a i n a l l constant Bg  maximal  belongs t o o n l y  G  B^  and  Q  Z ( c p ( f ) ) = b , say.  Thus Let  for  Ag  Thus  exists for  cp(k^r) »cp(  cp(-j£~)  , f o r a l l r a t i o n a l numbers' r .  =  Now,  ) =  -^rjj  if  cp(k)  i s not  67. a constant f u n c t i o n on cp(k)  i s continuous, we must have  1  set i n E say  X , then s i n c e  r  .  .  6  .  Thus  Then  x  cp(k)(X)  X  i s connected and  cp(k)(X)  i s a connected sub-  c o n t a i n s some r a t i o n a l  numbers  c p ( ^ ) -  with  cp(k)(x) = r  cp(k)  must be constant, so t h a t  Q  .  This i s a contradiction.  Consequently,  cp , r e s t r i c t e d t o t h e s u b r i n g  Y , i s a non-zero homomorphism o f fR  of constant f u n c t i o n s on  By Theorem ( 0 . 2 2 ) [ 7 ] , cp i s the i d e n t i t y .  into i t s e l f .  Q.E.D. (9-4) E  n  Theorem:  , and  Let X  and  Y  be two a r b i t r a r y subsets of  cp be an isomorphism from  constant f u n c t i o n s unchanged. T : X.-*'Y Proof:  d e f i n e d by  Define  T  Since  x  i t s i n v e r s e mapping  x  x  ^(My  again  y, y' e Y .  which i m p l i e s  in  and  y ^ y Y . ) =  M .  f  M  f x  o  s  o  be i n X  i s a C -diffeomorphism. m  X  to  . T  and  and  y  I f y - y« , then  My  r  x' x  Hence  Then  T  Y  e  X  6  6  onto  X  M T  Ag .  '  T  h  u  s  y  X  )  =  Then,  cp° (M , ) = My, f o r 1  x  cp~ (M ) = My = cp" (M 1  1  x  T h i s i s impossible f o r i s one-one.  (  x | x' .  Let y  Q  i s a f i x e d maximal i d e a l i n  m  as f o l l o w s :  A^^  Evidently,  1  x  all  cp i s an isomorphism, and i t onto,  cp" (M ) = M  = M,  x  leaving  i s a f i x e d maximal i d e a l i n Ag .  1  by Lemma ( 9 . 2 ) ,  Thus  and  i s an isomorphism o f  1  cp"° (M )  Let x  1  some  cp"  A^  cp induces a mapping  i s a s i n g l e v a l u e d mapping.  cp~ (M ) .  onto  t o be' a mapping from  1  Thus, T  Then,  cp(g) = g«T  T(X) = n Z [ c p " ( M ) ] .  By Lemma ( 9 . 2 ) ,  Ag  xt  )  x ^ x' .  be a r b i t r a r y Ag and  o " n Z t t p ^ C M ^ J ] . - T(X Q ),  68. T h i s shows t h a t  xe  T  ep(g)(x)  X , let  M  cp (cp(g) - r ) e _1  (cp(g)Xx) where  •  T"  i s onto, = r .  f o r each  ip'(g) -  Then  cp(g) = g ° T  : Y - X  .  Similarly,  That i s , (9.5)  T  cp  Let  X  Proof:  We  .  i o n s between  Hence Ag  and  (9.'4) are s a t i s f i e d .  We  In p a r t i c u l a r , i f is  C  T~  for  m  is  G  Theorem:  T»  be two A  connected subsets of  onto  2  A  1  .  Then  cp cp(g) =  and  Ag  s a t i s f y the c o n d i t i o n I n  cp  i s the I d e n t i t y on the constant -funct-  A  .  1  Then the c o n d i t i o n s i n  Theorem  The r e s u l t f o l l o w s immediately from & . E , D.  Let  X  and  d e f i n e d by  X  Y  he two subsets of  onto  T'(g) = g o T  Y .  E  n  , and  i s an isomorphism of  fat"*  e Ag  1  •T" ')T 1  then  g e Ag , g ° T e A^ , and f o r each  F o r each  are c l e a r .  = f .  Thus,  (g»T)(x) = 0  Now,  Ag  T'  f o r each  1  i s onto. x e X  f € A  f e Ag , T ^ f o r " )  f o r any  Now, or  T  Then the induced mapponto  which i s the i d e n t i t y on the constant f u n c t i o n s .  Proof:  .  Q/-E.D.  T, from X onto Y such t h a t  A^  be a C^-diffeomorphism of  •  .  (9.4).  Theorem  A  Y  »  1  g e Ag .  know t h a t b o t h  Lemma (9.3)  (9.6)  for"  1  x  be an isomorphism o f  f o r each  ing  cp" (f) s  each component of  and  r =  T  m  i n d u c e s a cP-diffeomorphism, goT  r( (x)) =  i s a C -diffeomorphism.  Corollary:  , and  11  Similarly,  each 1  T~ (y) = n z M M y ) ]  i s d e f i n e d by  1 < i < n .  and  , g - cp" ^) =  (as d e f i n e d above) , 1 < i _< n , then  each  E  'e  ,g(r(x)) = ( q f ^ )J(T(X) ) =  know t h a t f o r a l l f e Ag , f c T . e . A ^ . f = M  T  g € Ag  1  T(x)  Hence  1  How,  i f  T'(g)  ,  1  =  = S ,  g ° T [ X ] = g[Y] =  {G}  69. as  T  i s onto.  one.  That i s ,  T h i s shows t h a t  Remark:  If  only  C  x .  A-^ = ft(X) and  T  1  T»(r) = r .  That i s  T  Ag = & ( Y ) , then  Q.E.D.  and  T"  are  1  , each o f t h e i r components i s a r a t i o n a l f u n c t i o n .  m  We w i l l name t h i s mapping as rational-homeomorphlsm. know t h a t t h e r e is,  i s one-  r , T (r)(x) =  F i n a l l y , f o r each constant f u n c t i o n  r(-r(x)) = r , f o r each  not  g = 0 .  i s a non-linear  let X = Y = E  -  1 1  We a l s o  rational-homeomorphism. and  ( 0 , . . . , 0 )  T ( X ) = (T-JCX),  That  ... ,T  (X)  )  x. be  defined  T.£K) = — — f o r x ...... x  as  1 +  inverse ^ (y)  =  +  1  n  ^ ,1<  -g—d  y +. ..+y x  Then i t s  n  T " ( y ) = (<|)^(y),... , ^ ( y ) )  i s known t o be  d  1 <: i < n .  ±  with  j < n ,  2  Next, we w i l l  see some a l g e b r a i c p r o p e r t i e s o f t h e r i n g s  of continuous f u n c t i o n s which are i n a p p l i c a b l e i n t h e r i n g s of C - d i f f e r e n t i a b l e f u n c t i o n s , where  1. < m < « .  r a  (1)  The f i r s t  one i s t h a t t h e r i n g s of continuous  are l a t t i c e - o r d e r e d . are not.  neither  nor  of  i d e a l implies that  I).  , i e C (X) m  1  . but  f v 0 , In general,  1(f) _> 0  I ,  on some z e r o - s e t  (mod  let X = E  Consider |i|  (^(X) .  4 C (X> .  Thus  m  is in  Then  (^(X) .  We know that,' i n the r i n g of continuous f u n c t i o n s , ,  a z-ideal  I ( f ) _> 0  m  I(x) = x  f A 0 (2)  for  But t h e r i n g s of C - d i f f e r e n t i a b l e f u n c t i o n s  For instance,  we know t h a t  functions  i f there  I I  i f and o n l y i f f  (see [73(5.4)(a)). i s convex.  i s g e C(X)  f  Also,  i s non-negative I  i s a z-  Hence, by Theorem (5.2)[7], such t h a t  g j> 0  and  g = f  I n the r i n g o f d i f f e r e n t i a b l e f u n c t i o n such a  need n o t e x i s t .  F o r example:  Consider  X = E  1  and  g  C (X). 1  70.  Let  I = {f e C ^ X )  : Z ( f ) 3 [0,1]}.  i d e a l , i t i s convex. 1  G (X) is  so t h a t  f  +  g  agrees w i t h  f  on  does not e x i s t a t  0  lim ^  =  Ax  Thus,  g'(x)  ,  fi  L  lim  x  g(AXH-l) - , g ( l )  * -°~  lim Ax-0  =  +  g(Ax+l)  lim  =  It  g €  C (X) 1  ?  Indeed,  but  as  g(x) > 0 f o r a l l x.  Similarly, Q  b  t  *  x  1 + AX - (1+Ax)  Ax  x  ,  But, i f  1 .  x = 0 .  A  .  [ Q l ] , then the  or  = i  2  <o  l±m.  does not e x i s t at  g ( A x i ) , - gp.).-  Ax-0  Ax - A x Ax  2  Q  on a z e r o - s e t g  g(Ax).-.-s(0),  i s a z-  let f (x) = x - x I .  - g(0)  JLl*.  Now,  of  and  d e r i v a t i v e of l i m  _> 0  g _> 0  I  However i t i s not a b s o l u t e l y convex, f o r  i s not l a t t i c e - o r d e r e d .  clear that  Then s i n c e  lim, Ax-Ax - 2 A X  _  AX  A  X  ^ ° "  Ax  = - 1.  That i s ,  g  any  1  does not e x i s t a t  g e C (X)  know t h a t  with  f  (3)  Then I fl J  If  I  I n J = M  Accordingly,  f  s g  (mod I ) though we  are z - i d e a l s i n  T h i s i s not t r u e i n 1  h e J , then  J  and  in. C (E ) ,  where  and  Hence, there i s not  i s nonnegative on a z e r o - s e t  I J = I fl J . consider  g >_ 0  x = 1 .  1  Q  .  I = J = M  But  i(x) = x .  ;  a'x + e'x , where  a  Thus,  X = E  i = gh  Indeed, f o r gome  g,h  ,  1  i e M g e l  Q  = and  are C ^ - d i f f e r e n t i a b l e .  g(x) = g ( 0 ) .+•'. ax + ex , h ( x ) = h ( 0 ) + and  a'  are constant and 2  x - 0 .  Let  = [f e ( ^ ( E ) : f ( 0 ) - 0} .  IJ + I 0 J . If  C(X) , then  .  m  I .  1  Q  g ( 0 ) = 0 = h ( 0 ) , and we have  C (X)  of  g(x)h(x) = (a+e^a'+ejx  •  .  e, e' -• 0  as  1  Now,  i'(x) = 1  71. for a l l x e E  1  .  However,  g(xUh(x) - g(0)h(0)  (g.h)«(x)!  l i m ^a+€.j(a»4:€°,)x  m  =  x = 0  2 =  Q  That I s , ; i s ( I n J)»  which i s i m p o s s i b l e .  ;  H  e  n  c  e  IJ.  This  a l s o shows t h a t the f o l l o w i n g i s i n a p p l i c a b l e t o  (^(X)  C *(X). m  If  PQ = P n Q .  P  and  Q  are prime i d e a l s i n  For ^ C * ( X ) , we t a k e M  i  same argument  as above.  l  m  Q  example or  C (or C*), then  X = (°n,n)  and use the  72. PART I I THE RINGS OP La-(OR L~) FUNCTIONS  We now study the same s o r t o f p r o p e r t i e s when  (^(X)  is  r e p l a c e d b y the r i n g o f f u n c t i o n s s a t i s f y i n g a L i p s c h i t z c o n d i t i o n on each compact- subset o f a m e t r i c  £10.  space.  Rings, I d e a l s and Some P r o p e r t i e s _of L i p s c h i t z i a n o r Lc° Functions  (10.1) D e f i n i t i o n s metric  space  A r e a l <=valued f u n c t i o n  f  (X,d) i s s a i d t o be a f u n c t i o n s a t i s f y i n g a  L i p s c h i t z c o n d i t i o n on each compact subset o f each compact  d e f i n e d on a  A c X , there  fOr any two p o i n t s  (X,d),  i s a p o s i t i v e number  x x ' € A* | f ( x ) - f(x«)|  K  A  i f for such t h a t  <_ K d ( x , x » ) .  3  A  For  the sake o f b r e v i t y , we w i l l c a l l such a f u n c t i o n L e - f u n c t i o n , (see u n i f o r m L i p s c h i t z c o n d i t i o n on each compact subset P.35^[6].)  (10.2) D e f i n i t i o n s metric  space  A real-valued function  f  defined  on a  (X,d) I s s a i d t o be a L i p s c h i t z i a n f u n c t i o n ( o r  t o s a t i s f y a L i p s c h i t z c o n d i t i o n ) o f constant e x i s t s a p o s i t i v e number  K  K , i f there  such t h a t f o r any two p o i n t s  x,x« € X, |f(x)_ - f ( x ' ) | < K d(x,x») .  Note t h a t i f t h e m e t r i c  space i s  compact,  Remarks  ( l ) The L i p s c h i t z c o n d i t i o n i m p l i e s u n i f o r m c o n t i n u i t y .  Indeed, i f f on  then an L c - f u n c t i o n i s , c l e a r l y / a n L=function.  s a t i s f i e s a L i p s c h i t z c o n d i t i o n o f constant  (X,d) , then f o r any given p o s i t i v e number  K  e , choose  6 = e/K , we have f o r any x,x> e X, |f (x)-T(x».) | <_ Kd(x,x' )\<  73. K«e/K = €  whenever (2)  d(x,x«) < 5 = c/K  However, a continuous f u n c t i o n may not be a  Lipschitzian function. on -  A = {  <. x < 1) .  1  |x - x« | / | x  | */3 x  x  2 / 3  + x / 1  d(x,x') +  x  l/3 ,V3 X  +  i s hot bounded. (3)  x!  But  x' -/  5  3  Hence  + X» / | 2  3  = '  1^7-5 +  X  V3 ,l/3 x  X = E  1  and  .  1  f o r any compact subset  Let  (X,d)  be a m e t r i c  f i s a r e a l - v a l u e d L c - f u n c t i o n on  a r e a l - v a l u e d , bounded f u n c t i o n on f u n c t i o n of some constant'  ,*/3|  Then, containing  s a t i s f y i n g the  space.  X).  The f a m i l y  L(X) = [t s f i s  X, and i s "a L i p s c h i t z i a n  Lc(X)  and  L(X)  1  9  Q the constant  and  are commutative  a d d i t i o n , s u b t r a c t i o n and  I n such case, the u n i t y i s  value  Lc(X) •  K}. L - f u n c t i o n means L i p s c h i t s i a n f u n c t i o n .  r i n g s w i t h u n i t y under p o i n t w i s e  w i t h constant  x  1  (10.3) D e f i n i t i o n s  multiplication.  +  f(x) = x^  the o r i g i n , t h e r e does not e x i s t any constant  (10.4) Theorems.  u , the f u n c t i o n function with  value  0 , i s the zero'element o f them. Remarks  The c o n d i t i o n t h a t  omitted i n space.  L(X).  Then,  f  5  t o see t h a t an L e - f u n c t i o n i s a  as i t was shown i n Remark ( 2 ) ,  Cf  1  But, a continuous f u n c t i o n need not be  Consider  required condition.  - x' / !  5  i s not a L i p s c h i t z i a n f u n c t i o n .  I t i s also,easy  an L c - f u n c t i o n .  1  1  * f  5  iavcontinuous  f ( x « ) | - Ix /  |f (x)  W h e r e  '^/3j  continuous f u n c t i o n .  f(x) = x V ^  F o r example,  f  For,' l e t  p ( ) = d('p,x) x  i s bounded on (X,dO  X  can not be  be an unbounded  metric  i s a L i p s c h i t z i a n function with  74.  1 .  constant  If  For  (x) - f ( x ' ) l = |d(p,x)---- d(p,x« ) I _< &  IT  2  d(x,x» ) .  However, l e t g ( x ) = f ( x ) * f (x) = (d(p,x))  not L i p s c h i t z i a n of any constant. |d(p,x))  Indeed,  |g(x) - g ( x ' ) | =  - ( d ( p , x ) ) | = [d(p,x) + d ( p , x ' ) ] . | d ( p , x ) . -  2  d(p,x)| >  2  d(x,X')•|d(p,x) - d(p,x«)|. d(p,x')|  i s unbounded.  Here.we know t h a t  That i s ,  g  |d(p,x) -  i s not L i p s c h i t z i a n .  The reason t h a t we d i s c u s s t h i s p a r t i c u l a r r i n g t h a t i n § 1 3 , we s h a l l d i s c u s s t h a t and  L(X) i s  L(X) i s a Banach a l g e b r a ^  Lc-realcompactness.  Proof of t h e P r o p o s i t i o n : g  is  be a r b i t r a r y from  consider  L(X) o f constants  |f | _< K ' ,  i v e l y , and  First  |g| _< K"  .  L(X). K^  Let f  Kg , r e s p e c t -  and  Then, f o r any  and  x,x' € X .  | ( f + g ) ( x ) - ( f + g)(x«)| = | ( f ( x ) - f ( x « ) ) + (g(x) - g ( x ' ) ) | < | f ( x ) - f ( x ' ) | + |g(x) - g ( x « ) l  < Kjd(x,x».) + Kgd(x,x«) <  (Kj^ + Kg)d(x,x> ) .  + Kg .  K» + K" .  bounded by  |fog(x) - f o g ( x ' ) l = t  K =  Take  Hence  Moreover  f + g e L(X) .  f + g is  Now,  | f ( x ) g ( x ) - f(x»).g(x»)| < | f ( x ) | l g ( x ) ........  g ( x ' ) | + |g(x» ) | o |f (x) - f ( x ' ) | _< K ' o K g d ( x , x « ) + K " K ' o d ( x , x ' ) = ( K ' K g + K"K )«d(x,x') . 1  f«g e L(X) . For A  of  Hence, subset  X .  Evidently,  u  Q e L(X) .  arid  L ( X ) , take any  f ? g € L ( X ) and any compact  C  C  Then, by t h e f i r s t p a r t we know t h a t  ( f + g)|A, (fg)|.A € L(A) . of  X , by d e f i n i t i o n o f  I t i s obvious t h a t Remark:  | f o g | '« |f | . | g| < K ' . K " . . Hence  Also  L (X) , C  L(X) c L ( X ) .  I f f e L ( X ) , then  _< | f ( x ) - f ( x ' ) l  Since  < Kod(x,x»)  C  f|A,g|A e L ( A ) .  i s an a r b i t r a r y  f + g , f g e L (X).  Hence  | f | e L(X) . fora l l  A  subset  ;  C  u, Q e L ( X ) . C  For  Q.E.D.  ||f(x)| - |f(x«)ll  x,x» e X , and  |f|i s  75. bounded.  Hence, i f f , g e L(X.) , t h e n  f v g-- "+ *  that  f  '  +  ° g<  f  and  | f - g| e L ( X ) so  L±£^£^Ji.  f A g -  2  e L(X) .  2  That i s ,  L ( X ) i s a l a t t i c e - o r d e r e d r i n g (p.7 [ 7 ] ) .  S i m i l a r l y , we can show t h a t (10.5) P r o p o s i t i o n : w i t h constant  If f  K , then  We know t h a t  and  are L i p s c h i t z i a n .  -  i s a L i p s c h i t z i a n f u n c t i o n on (X,d)  j f | A n <£ L ( X ) w i t h t h e same  Proof: n  Lc(X) i s a lattice-ordered r i n g .  |f|A n  i s bounded by  n . f I .+  |fj A n =  Hence  -  constant.  Both | f | £  /  I If I  .-.  -  £1  2  is Lipschitzian. (10.6) Lemma:  L e t f e L ( X ) and  number  r .  Then  Proof:  We know t h a t  p o s i t i v e number =  f(x)  f(x')  Moreover  f " (=l/f) e L(X) . 1  | f ( x ) - f(x«)l < K d(x,x«)  K .  We a l s o have  |f(x) - f(xv)|  <  | f ( x ) | . | f ( x « ) l  ~  l / f _< l / r .  (10.7) Lemma:  r _< | f | f o r some p o s i t i v e ,  Hence  f o r some  |-y ( x ) - ^ ( x ) | = !  •|f(x)-.f.(x»;)|; T  K  <  ~~  D(X  ^ , X  ) ;  T  1/f e L ( X ) .  L e t f e L*(X) - { f e L ( X ) : f i s ' b o u n d e d } , and C  r < | f | f o r some p o s i t i v e number  r .  Then  f  X .  Then, as shown  =  1  (= l / f ) e  LJ(X) • Proof:  Let A  be any compact s u b s e t o f  i n Lemma ( 1 0 . 6 ) , we know (10.8) P r o p o s i t i o n : Then  f  function  l / f |A € L(A) .  F o r any p e X , d e f i n e  e L ( X ) i f f (X,d) i s bounded. t  Hence  a distant function.  l / f e L£(X).  f (X) = d ( p , x ) .  We ; s h a l l c a l l such a  76. Proof:  We have a l r e a d y Shown t h a t  function  i n the Remark o f Theorem  iff  i s hounded.  f  Hence  is  f  a Lipschitzian  (10.4).  Thus  f  € L(X)  t h e r e s u l t f o l l o w s immediately  from the boundedness of a m e t r i c space. (10.9) D e f i n i t i o n :  The s e t  -to be the z e r o - s e t o f [Z(f)  unit i n  L (X) C  Remarks:  : f e L(X)}  Z(X) = ( Z ( f )  -  L ( X ) , ( L * (X)  ( l ) I t i s clear that f e L* (X),  i f f |f | > r (2)  =  C  '  f o r some  f e L_(X) i s a u n i t i n L_(X) (L(X)) r e  is  ,  a u n i t i n L*(X)  r >,0.  Consider an unbounded m e t r i c space  •• . 1.+ d(,p,x) 1  or L(X)) i s s a i d t o be a  (L*(X) o r L ( X ) ) .  5  Z ( f ) = ty and  (L(X>)  x  Let  i s said  f e L (X),(L*(X) or L(X)) has a m u l t i -  If  p l i c a t i v e inverse i n  f( )  Z ( f ) = {x 6 X : f ( x ) = 0}  : f e L(X)} ( s e e ( l l . S ) ) .  (10.10) D e f i n i t i o n :  iff  f .  Q.E.D.  1  |f | < 1 , and  Then  ld(p>x) - d(p,x»)|  < d(x,x») .  (X,d), and  |f (x) - f (x> )| Hence  f e L(X) and  (1 + d ( p , x ) ) ( l + d(p,x«)) Z ( f ) = ty b u t f"  1  4 L(X)  |f|} r  so t h a t  f o r a l l p o s i t i v e number  f  Consequently  i s not a unit.  (10.11) D e f i n i t i o n :  A nonempty s u b f a m i l y  t o be a z - f i l t e r  (X,d),  on  r .  g  of  Z(X)  i s said  i f i t s a t i s f i e s t h e c o n d i t i o n s (1),  ( i i ) and ( i i i ) o f (1.7). (10.12) P r o p o s i t i o n : the  family  If  I  i s a proper i d e a l i n  Z [ l ] - {Z(f) : f e l }  Proof i s s i m i l a r t o (1.8).  i s a z-filter  I» (X) , then  on  C  X .  77. Remark:  If  r e s u l t need n o t be t r u e . f(x) = - — -' d(x,p) +1  Set  f e L(X) (and L*(X))  p e X .  but  f " 4 L(X)  That i s ,  (10.13) P r o p o s i t i o n :  I f 3?  f .  Z [ l ] i s not a z - f i l t e r . i s a z-filter  on  1  Q  X , then t h e  Z £ i [ 3 ] = {f e L * ( X ) :  c Z(f) € 3} L*(X)  c Z^Es]  and  and  L*(X)).  We know then t h a t  Z ^ ^ ] = {f e L ( X ) : Z ( f ) € 3 } ,  familys  Then, we know  (nor  1  I = ( f ) , the i d e a l generated by  <|) = Z ( f ) e Z [ l ] .  L*(X) , then the  ( X , d ) be unbounded, and  Let  f o r a f i x e d point  1  that  L ( X ) or  L_(X) i s r e p l a c e d by  = ffeL(X) :Z(f )e3?}  are ide,als i n  L (X),  L(X) , r e s p e c t i v e l y .  Proof i s s i m i l a r t o (1.9). Notice n o t a t i o n of  t h a t we w i l l use Z" -1 -1 „i Zj~ , Z # or Z^ . c c 1  (2)  T h i s shows t h a t every  I t i s clear that  may be proper.'  Z " ^ ! ] ] =>'I .  For. example,  f ( x ) = ( f (x)) Q  (L*(X) o r L ( X ) ) . f  l e t (X,d~) b e a (bounded) m e t r i c  2  _p .  » [d(p,x)]  I = (f ).  i n L ( X ) such t h a t  (L*(X) or L ( X ) ) . at  Let  The i n c l u s i o n  1  o  ions  z-filter  Z [ l ] f o r some i d e a l i n L_(X) , L*(X) or L ( X ) .  i s o f the form  space, and  1  L  ( l ) Ztz" ^]] - 3 .  Remarks:  t o denote the a p p r o p r i a t e  1  Q  .  Then  f o r some  As a matter o f f a c t , s i n c e  e L (X) e  g e L„(X)  I n p a r t i c u l a r , every f u n c t i o n i n in  Q  This consists of a l l funct-  f = f »g  Hence every z e r o - s e t  f  I  vanishes  Z [ l ] contains the point  Z [ l ] i s a z - f i l t e r t h a t i n c l u d e s the  set  {p} = Z ( f ) , I t must be t h e f a m i l y o f a l l z e r o - s e t s  ing  p .  Q  p .  contain-  78. The i d e a l  M  Q  = Z'-^Zfl]]  evidently consists of a l l  functions  i n L ( X ) (L*(X) or L ( X ) )  Hence  p I .  M  Q  vanishes a t p Suppose  f  However, and f  € I .  M  Q  which v a n i s h a t p .  4= I •  For i n s t a n c e ,  € L ( X ) (LJ(X) or L ( X ) ) . f  (L*(X) o r L ( X ) ) .  But, then  uous a t the p o i n t  p .  = f «g 0  f o r some  g(x) = [ d ( p , x ) ] "  Thus  g 4 Iv( ) x  p  Thus  C  Then,  (  f  )  x  f  p  e M. Q  g e L (X) C  1  i s discontin-  (L*(X) or L ( X ) ) .  This i s a contradiction. Note t h a t M  Q  Z[M ] = Z [ l ] , i n s p i t e o f the f a c t t h a t Q  + I• I t i s a l s o obvious t h a t  (10.14) D e f i n i t i o n : on  M  Q  i s a f i x e d maximal i d e a l ;  A z - u l t r a f l i t e r on  X  i s a maximal  z-filter  X .  (10.15) P r o p o s i t i o n : Z[M]  If M  i s a z - u l t r a f l i t e r on  i s a maximal i d e a l iri - L ( X ) , C  then  X .  Proof i s s i m i l a r t o (1.11).  (10.16) P r o p o s i t i o n : Z" ^ 1  I f Mr-is  a z - u i t r a f l i t e r on  X , then •  i s a maximal i d e a l i n L (X) .  Proof i s ' s i m i l a r t o (1.12). (10.17) D e f i n i t i o n : ideal i f  An i d e a l  Z ( f ) e Z [ l ] implies  Example:  i n L ( X ) i s s a i d t o be a z~  f e l .  C  That i s ,  I = Z^tZtl]].  Every f i x e d maximal i d e a l i s a z - i d e a l .  Note t h a t i f t h a t t h e r e i s an  I  I  i s an i d e a l i n L*(X) (or L ( X ) ) such  f e l with  Z ( f ) = (j) and f "  1  \ L*(X) (or L ( X ) ) ,  79then  I  i s clearly not a z-ideal,  (10.18) Proposition:  If  M Is a maximal ideal in L.(X)  (L*(X), L(X)), and Z(f)  meets every member of  Z[M] , then  f € M. Proof is similar to (1.13). (10.19) Proposition: set  If > f i s a z-ultraf l i t e r on X , and a zero-  Z meets every member of  , then  Zs  Proof is similar to (1.14). (10.20) Definition:  i n ideal  said to be a prime ideal i f  I  in  f <»g e I  (10.21) Theorem: For any z-ideal  L (X) (L*(X) or L(X)) is implies  I  in  f e I  or  gel.  L (X) (L*(X) or L(X)), Q  the following are equivalent. (1)  I  Is prime.  (2)  I  contains a prime ideal.  (3)  For a l l  then (4)  g»h e L (X) (L*(X) or L ( X ) ) , g.h = S , e  g e l  For every  or  f e L.(X) (L*(X) or L(X)), there i s a  zero-set in Proof: ideal or  h € I .  Z[l] on which  That (l) implies (2) is clear. P , and g<>h = £ , then  h e Pc I .  A  Q)  m  Qe I .  Either case Implies (4). (l).  Given  If  goh e P .  does not change sign. I contains a prime Hence  g e Pc 1  This shows that (2) implies (3).  that (3) implies (4), for every (f v Q)°(t  f  g»h e l ,  f e L  By (3)  C  To see  W , consider  f v <9 € I  or  f  A  Q  e I .  Finally, we w i l l show that (4) implies consider the function  |gj - | h | .  By  80. hypothesis, there i s a zero-set i s n o n p o s i t i v e , say. g .  o f Z [ l ] on which | g | | h  Then every zero of h  Since  I  i s a z-ideal,  g e l .  The p r o o f f o r L*(X) o r L(X) L*(X)  on Z  Z(g) => Z n Z(g) = Z 0 Z ( g o h ) e Z [ l ]  Hence  e Z[l] .  Z  or L(X)  Thus  i s zero o f  so t h a t Z [ g ] I  i s prime.  i s e x a c t l y the same w i t h  t a k i n g the p l a c e o f L ( X ) .  %11 L^Oomplete R e g u l a r i t y and L-Normality. (11.1) D e f i n i t i o n :  A m e t r i c space  (X,d) i s s a i d t o be L-  completely r e g u l a r i f , f o r every c l o s e d subset x e X - F , there i s a f u n c t i o n  f e L(X)  F  o f X , and  such t h a t  f(x) = 1  f ( F ) = {0} .  and  (11.2) D e f i n i t i o n :  A m e t r i c space  normal i f , f o r any two d i s j o i n t i s an f e L(X)  such t h a t  (X,d) i s s a i d t o be L-  c l o s e d subsets  f ( F ) = {0}  and f ( F > ) = Cl]  Note t h a t a m e t r i c space i s a Hausdorff n o r m a l l t y i m p l i e s L-complete  (11.3) Theorem:  F , F' , t h e r e  space so t h a t L-  regularity.  A m e t r i c space  (X,d) i s L-'completely r e g u l a r i  i f f the f a m i l y  Z(X) o f a l l z e r o - s e t s o f L ( X ) i s a base f o r  the c l o s e d subsets o f X . Proof i s s i m i l a r t o (2.4). (11.4) Lemma: (X,d)  L e t A be a nonempty subset o f a m e t r i c space  and f e L(A) .  Then t h e r e i s a g e L(X)  such t h a t  g|A = f . Proof: of  f .  Suppose  | f | _< n , and K  For each  x e X , define  i s the ' L i p s c h i t z i a n  constant  f _ ( x ) = Sup {f(x«) - Kd(x»,x  8l. Then f o r each  x  We know t h a t  e A , f ( x 0 = J f | * ( ' ) " Kd(x«x )} . f  Q  o  K > 0  and  Q  x  A  Q  d(x',x) j> °'  I f there e x i s t s , x^ e A  such t h a t  f ( x ) - K o d ( x , x ) > f ( x ) - K d ( x , x ) = f ( x ) , then  f(x ) - f (  x  x  1  0  1  Q  ) > K«d(x ,x ) 1  Q  Q  Q  which c o n t r a d i c t s the f a c t  G  < Kod(x,x')  for all  x,x» e A .  f ( x ) > f ( x ' ) - K>d(x«,x )  fora l l  x« e A .  | f ( x ) - f(x«)l Q  xf  0  ( f ( x ' ) - K«d(x,x')} = f ( x )  A  for x  0  f |A = f .  We w i l l show t h a t  Q  constant  K .  For arbitrary  f ( x ) - Kod(x ,x«) Q  Hence  o  { f  ( o ) " Kod(x ,x')} +  Q  i s L i p s c h i t z i a n on x  of  Kod(x ,x) > f ( x ) -  D  Q  Q  f ( x ' ) + K>(x«,x) = Q  { f  G  X  e A.,  Kod(x,x«)=nA ^ J"  X  ^ o  A  t f ( x ) - Kcd(x ,x)} = f ( x ) . o  Q  Q  f ( x « ) - K»d(x',x) = x ^ U  0  J  That i s ,  X  K e d  ( o^') x  + K o d  ( ^ x  x  O U  x  e A .  > f(x )  o  >  Thus, Hence  x,x' e X , and  + Kod(x,x»)  0  K d ( x , x » ) - K«d(x',x) . xeA  f  Q  that  A  On the other hand,  t f ( x ) - Kod(x ,x')} - K»d(xSx) = Q  o  ( f ( x ) - Kod(x ,xi) - K»d(x',x)} < ^  A  o  = f (x) . 6  g = f  Q  Hence  A n .  X  A  Cf(x ) - Kod(x,x )} Q  | f ( x ) - f j x ' ) | < K«d(x,x>) . Q  By Prop. (10.5)  g € L(X) .  Hence  0  Finally, l e t g  i s required. Q.E.D.  (11.5) Theorem:  ( l ) Every m e t r i c space  (X,d) i s L-completely  regular. (2)  Every compact m e t r i c space (X,d) i s L-  normal. Proof: Then  (1) L e t F d(F,p) 4 0 .  be a c l o s e d subset o f Let f  X  and  p € X - F .  be a f u n c t i o n d e f i n e d as f o l l o w s :  .82. f [ F ] - {1}, show t h a t If  and  f  f(p) = 0 .  |f(x) -  x' » p , then  d(x,x') d(F p)  -  F U {p} .  1  •  = 0 _<  f(x')l  f  f | F U {p} = f .  Hence  Q  F'  d i s j o i n t from  Remark:  f [ F ] = {1}  Let  2  X = E  .  ,  e L(X)  F = {(x^y) € E F  and  f [ F ] = {l) Let B  0 _< r ^ < r  2  such  f (p) = G . Q  {p} by a c l o s e d s e t a. E. D.  p  Then  (11.6) Lemma:  F'  and  : xy = 1}  and  For instance  - p  P» = {(x,y) e E  2 f e L(E )  f [ F ' ] = {0} .  Q  .  B  for x e B  ( x ) - cl^  r  W(x') « 1  f e L(X)  f o r x e cl  x  r i  i = 1,2,  ±  6  Then, t h e r e i s an  (11-1) J O < f ( x ) < 1  :  a r e two d i s j o i n t c l o s e d s e t s i n  ( x ) = {x e X : d ( x , x ) < r 3  r  f(x) = 0  Proof:  Q  However, i t i s c l e a r t h a t t h e r e i s n o t any  such t h a t  where  f  That i s ,  The compactness can not be omitted i n ( 2 ) .  xy «= -1} . E  1 .  and  Q  f o r x,x' e  F .  1 P  If x e F  K>d(x,x») =  there i s an  For ( 2 ) , we o n l y have t o r e p l a c e  K .  ?  i s bounded by  By Lemma (11.'4)  ¥e w i l l  K>d(x,x ) .  Hence ' | f ( x ) - f(x')(<_ Kod(x,x«)  I t i s clear that  .  1  w i t h constant  | f ( x ) - f ( x ' ) | = 1 , and  f e L ( F U {p}). that  F u [p]  i s L i p s c h i t z i a n on  x,x» e F , then  and  L e t K = (d(F,p))°  such t h a t :  (x ) g  Q  (x ) Q  f o r x,e X - B ^(x' ) r  L e t cp(t) = J O . t-r r  f o r t _< r  1  n  2"^l  ^1  0  for  r-, < t < r , ~ 1  for t > r  2  2  , and  f ( x ) = cp(d(x,x  Q  83. Then, I t i s obvious t h a t any  x,x' e X ,  rtFrT  s a t i s f i e s (ll°l).  d  x  0  d  x  f  Moreover, f o r  | f ( x ) - f ( x « ) | = |ep(d(x,x )) - cp(d(x' ,x) )|_<  l ( ^ o) ' ( '* o)l < r t f e "  Moreover,  x  x  d  1 .  i s bounded by  (11.7) P r o p o s i t i o n : of  f  ( ^ )» x  8  Hence  f € L(X) .  F o r every neighborhood  U  Q.E. D.  of a point  (X,d) , t h e r e i s a z e r o - s e t which i s a neighborhood o f  x  Q  x  Q  contained i n U . Proof: i s an  Since ^  > 0  U  i s a neighborhood o f  such t h a t  C J  ^^  by Lemma (11.6), t h e r e i s an Hence  Z(f) = ^ ^ r (  (11.8) P r o p o s i t i o n :  x 0  )  c  u  r  (  X Q  )  c  x , by r e g u l a r i t y , •  u  Choose  r  f e L(X) s a t i s f y i n g  •  2  > r  there 1  ,  (ll-l). Q.E.D.  Every c l o s e d subset  A  of  (X,d) i s an L-  zero-set. Proof:  L e t f ( x ) = d(A,x)  f o r each  Z(f) = A  i s clear.  Lipschitz  c o n d i t i o n w i t h constant  d(x,y)  x e X .  We;will show t h a t  f o r a l l x,y s X. .  1 .  F o r any  f  Then,, t h a t  satisfies a  That i s , x,y e X ,  | f ( x ) - f (y) | _<  d(A,x) =  i n f [d(a,x) : a e A] < i n f (d(a,y) + d(y,x) s a £ A} = i n f {d(a,y) a e A} + d(y,x) = d(A,y) + d(x,y) . d(x,y) .  Similarly,  |f(x) - f ( y ) | _< d(x,y) By Prop. ( 1 0 . 5 ) *  Thus  d(A,x) - d(A,y) _<  d(A,y) - d(A,x) < d(x,y) . f o r a l l x,y e X .  g e L(X) .  Moreover,  Hence  Let g = f A l .  Z(g) = Z ( f ) = A . Q.E.D.  84.  The in  f i x e d i d e a l s and f r e e i d e a l s are d e f i n e d the same as  (3.1).  with  Moreover, the t h r e e Remarks of (*3. l ) are v e r i f i e d  L (X), ( L * ( X ) or L ( X ) )  use  1(f)  L(X))  (^(X).  t o denote the r e s i d u e c l a s s of  (12.1) Theorem? or L ( X ) )  i n p l a c e of  (l)  The  f mod  We w i l l I .  f i x e d maximal i d e a l s i n  are p r e c i s e l y the s e t s  also  ^ ( X ) ^ (L*(X)  Mp = {f € L ( X ) ^ ( L * ( X ) C  or  s f ( p ) = 0} . (2)  The  Ideals  M  are d i s t i n c t f o r d i s t i n c t p P  (3) L(X)/Mp) mapping  i s isomorphic Mp(f)  (L*(X)/Mp  or  - f(p)  (12.2) P r o p o s i t i o n s MX)  = L(X)  L (X)/Mp c  ?  (L*(X)/Mp  fR .  In f a c t ,  I s t h e unique isomorphism of onto 1R, where  or the  L (X)/Mp^ c  p e X .  (3.3). If  (X,d)  i s compact, t h e n e v e r y i d e a l  (12.3) P r o p o s i t i o n s  (3.5). If  X  i s compact, t h e n t h e  I s one-one f r o m  X  onto the s e t o f a l l maximal i d e a l s  p  Mp  in  L (X)  I  i s fixed..  Proof i s s i m i l a r t o  C  p ,  w i t h the r e a l f i e l d  L(X)/Mp)  Proof i s .similar t o  in  F o r each  correspondence  (L(X)).  Proof i s s i m i l a r t o (3.6). (12.4) Lemmas belongs  A zero-set  Z s Z(X)  t o no f r e e z - f i l t e r . f  Proof i s s i m i l a r to (3.9).  (See  , Z + $ (3.7)).  i s compact i f f i t  .85. (12.5) P r o p o s i t i o n :  L e t Ar be a ' z - u l t r a f l i t e r on Then Ar I s f r e e .  of I t s members be noncompact. Proof:  Suppose  We know t h a t  p e n  Z ( f ) = {p} p  member o f Ar- a t p . a  .  (X,d) and eafih  f^C^) = d(p,x).  Then, c o n s i d e r  which i s compact and meets each  Hence  Z ( f ) & A r (by (10.19)).  ' This i s  contradiction.  Remark:  z-ultrafilter.  We c o n s i d e r  the Mean Value Theorem, x^ |cos  i s between  x  and x«.  We know t h a t  By  Q  f  That i s ,  where  |f(x) - f ( x ' ) l  =  'e L ( X ) , (L£(X) o r C  Z [ l ] i s a z - f i l t e r but not a  each member o f  Z [ l ] i s unbounded  But f l Z [ l ] = Z(f ) .  i t i s noncompact.  (12.6) Theorem:  and f ( x ) = s i n x .  Hence we have  Then  Q  so t h a t  1  1  L e t I = ( f ) .:.  z-ultrafilter.  X = E  s i n x - s i n x« = (cos x ) ( x - x ' )  x - J o l x - x ' ) _< d(x,x« ) .  L(X)).  (3.10), A must be a  As we have shown i n the Remark o f  F o r a m e t r i c space  (X,d), t h e f o l l o w i n g a r e  equivalent. (1)  X  (2)  Every i d e a l i n L ( X ) i s f i x e d , i . e . every z - f i l t e r C  is (2*) (3)  i s compact.  fixed.  Every I d e a l  i n L*(X)•(©r L ( X ) )  Every maximal I d e a l  i s fixed.  i n L ( X ) i s f i x e d , i . e . every C  z - u l t r a f i l t e r i s fixed. (3*)  Every maximal i d e a l I n  Proof i s s i m i l a r t o  (3.11).  L*(XV (or L ( X ) ) i s f i x e d .  86.  §13  The Banaeh Space L(X) and Lc-realcompactness. As we have shown In § 4 ,  L*(X)  or L(X)  modulo a maximal i d e a l c o n t a i n s . a fR .  of the r e a l f i e l d imal i d e a l  every r e s i d u e f i e l d of  c a n o n i c a l copy  We s h a l l show l a t e r t h a t f o r every max-  o f L*(X\(or  M  L (X),  L*(X)/M,  L(X)J,  ( o r L(X)/M)  i s iso-  fk .  morphic w i t h  (13.1) D e f i n i t i o n :  f e L(X) , Hfll^  For  x e X] , l l f L = sup { l ( ? " (*'•) 1 d(x,x») f  x  .  f  X j ( X  »  - sup {|f (x) | : X , x + x'} , and  €  a  lift! = M  +M  m  •  d  (13.2) P r o p o s i t i o n :  L(X)  i s a Banaeh a l g e b r a under the above-  d e f i n e d norm. Proof:  ||f|j  (,i) That (ii)  = G  i f f f = 0  f , g e.L.(X),  For  (iv) ||f . g L  d  d  w  d  = sup { l ( > g ( K ( ? ° g ( d(x,x») x  x  x C  |f(x)g(x) - f ( x « ) g ( x ' ) l |f(x)  - f(x>)|  iif-gi!  <  )l  +  (v)  [f ]  ||f.g|| <t|f|IJlgl1 d  d  llg!ijlft!  +  d  / hence  : be an a r b i t r a r y Cauehy sequence under the  n  Then, i t i s c l e a r t h a t  a l g e b r a , hence there i s  f € C*(X)  j| »|  d  d  11°II  under  x + x« , x,x« 6 X} , and  s  :  sequence under the norm  f  But,  Hfiijgn ,+ ii gn.it * ii < i i f M i s i i . .  iifii-iigt Let  x 8  .  d  < | f ( x ' ) h | g ( x ) - g(x>)| + !g(x*) | •  We have  above^defined norm.  to  Is clear,  llfogjj = ||fogl| + l|fogH < llfl^Hgii. + !Ifgl) f  a  ||f+gll = ||f+gl!„ + !!f+gll < It fit. +  llsll. + l!fll + k ! l = llfli + ltd • HxflJ = ,|x|o||f|l for each X e fR d  (iii)  i s clear.  .  ->.  We know t h a t such t h a t  We w i l l show t h a t  {f }  i s a Cauchy  n  G*(X) (f 3 n  f e L(X).  i s a Banaeh  converges We know t h a t  87. every Cauchy sequence i s u n i f o r m l y bounded. e > 0 , there e x i s t s a p o s i t i v e Integer  IIf  f o r a l l n,m>  - f j <.€  n  = sup Cljfjj  Let  - fjll  N| j .  Indeed, f o r each  N ^  such t h a t ||f  Thus  g  i J <  *• and  ^ - f j < e  (e.,^).  K = max  (:€ )  Then, l l , - f j < K fora l l m . Hence \ \ f j - \\f^ 1 | < K (by p r o p e r t y ( I I ) ) , o r 11 f J | < | | f II + K f o r a l l m . f  N  ; ( £ )  w  (<s)  i  d(x,x')  d(x,x')  |f (x») - f(x')l '• — — • d(x,x«) n  and a l l n . number  i s true f o r  Suppose  f f L(X).  K , there exist  l f ( x ) - tlx  a l l x,x  ?  € X , and  That I s , f o r any p o s i t i v e  x,x' e X , x ^ x'  such t h a t  1  > K .  °  | f ( x ) - f(x»)| • ' .-o , d(x ,x») Q  > KL + 1 °  0  x »x« € X , x ° °  f o r some  d ( x , x J ) > 0 , there i s a p o s i t i v e integer  1  )  |f - fJ L  <*(x ,x') < -— -•  „ \*(*p)-- n ( o ) l f  x  d(x ,x ! ) Q  jC 1 + K  Q  (  .  )  X  fora l l n > N . ,  l n( o) - n( o)' f  x  »Q  such t h a t l C  x  f  d ( *  0  x  Thus ,  l n( o> f  x  , x i )  This i s a contradiction.  >  X  Q  f  rt  4= A • °  n  0  However, on t h e other hand, f o r these two p o i n t s (  1 ,  In p a r t i c u l a r , f o r K = K +  d(x,x«)  Q  x f x'  J - f (x£) | ° — — —  X  "  f  '  d(*o* b> x  Hence  f € L(X).  F i n a l l y we have t o show t h a t flf ~ f|| -* 0  as  n -» » .  88. Since  ||f  n -» •  - f j | «. | | f  n  + ||f  w  we o n l y have t o show  [f 3  - f||  n  f|f  d  giveh any  whenever  m  f ll n  < €  d  |  °  n  d(x,x )  tjofj = || ©H^ + ||e|| N  ,  d  such t h a t  That i s  (13°1L  | < e  f o r m,n > N  and  =| < s , f o r rn^n > J  n  arid a l l  d(x,x« )  8  e X .  .  x =f= x« € JL  Since  R e w r i t i n g (13-1), we have  € ;X .  !  n -* »  as  This implies that  > I .  ••  - ffl. - 0  n  d  H .  m,n  ••  d(x,x«)  x 4= x  m,n  whenever  |———-—:~-~ x ?  ||f  € > 0 , there i s a p o s i t i v e integer  |'|.f ~ f H < € II f  and  f|| -* .0 .as  n  i s a Cauchy sequence under the norm  n  x 4  - f||  n  Now,  we h o l d f  we have  s  m  mW  ~  f i x e d , f o r any f i x e d p a i r of f  .-  J  < )  x  i s fixed.  L e t i t be  d(x,x») f'(.x) ..- f ( x « ) n  r  , and  r  =..  ' .-  ••  i  <'•<-;  d(x,x») {f^} t o f  convergence of £&2^ .Z.£L ll d(x,x»)  as  Thus, we have  |r - • • X ? -  x  r  •,  „  under the sup norm,  n °* » „  ( ) " ( ')| d(x,x») f  we have  f  x  |  < c for ~ .  x  )| <  ;  m  x =f x  1  e X .  -  '  f  for  Or { m ^ ° m f  n > N .  x  Since  x  )  f  ( x 8 )  _  d(x,x«) x,x« e X  ( ) ° v /[< d(x,x«) ~  That i s ,  The p r o o f i s then complete  .  e  '""  m > N . '  -fjx')  111  d(x,x») and a l l  n  n  5  x  {r } -  |r « r | < e  Moreover  d(x,x ) f  we know, by the u n i f o r m  A  £  ||f - f | )  d  5  for '  < e ,  are a r b i t r a r y ,  m > H ' for  m>J  .  89. L(X)  We are now back t o show t h a t every maximal I d e a l i n or  i s real.  L*(X)  We s h a l l d i s c u s s I n the case o f  L*(X).  The p r o o f f o r L ( X ) i s e x a c t l y the same except f o r the n o t a t i o n . In order t o show t h a t every the r e a l f i e l d , /R (see P. 209 [28])  9  i s isomorphic t o  L*{X)/M  "by the t h e o r y o f t o t a l l y ordered  we f i r s t c l a s s i f y the elements of L*(X)/M  as p o s i t i v e , • n e g a t i v e o r zero i n such a way t h a t fog  are p o s i t i v e and - f i s n e g a t i v e when  (13.35) D e f i n i t i o n s  -f  f  and f 4 0  M)  f  f + g and  and g.  are  (See [27]).  positive.  (mod  fields  i s positive.  i s s a i d t o be p o s i t i v e I f (mod M).  And f  f s jf|  i s s a i d t o he n e g a t i v e i f  f  i s s a i d t o he zero i f f e M .  (fa and 7PC denote the c l a s s e s o f p o s i t i v e and n e g a t i v e  Let  elements o f L * ( X ) ( o r L(X) , r e s p e c t i v e l y ) .  I n order t o j u s t i f y  t h i s c l a s s i f i c a t i o n , we s h a l l show the f o l l o w i n g lemmas.  (13°4) Lemmas of  For each  f € L * ( X ) • ( o r L ( X ) ) , one and o n l y one f £ <P ,  the three r e l a t i o n s  Proofs  We know t h a t  - f + |f| = 0  both h o l d , then Thus  f s 0  Proof:  i s a field,  and f + | f j ' H 0 (mod M) i s v a l i d .  2f = ' ( f ' +  |f|) - (-f + |f|) H 0  I f f 4 M , then  If  (mod M). Q.E.D.  F o r any two elements  and g e M,  Since  a t . l e a s t one o f the  (mod M).  (13.5) Lemmas 0 <_ f £ g  o r f € TC ''holds.  (f + | f| ) = • © s M .  (- f + | f | )  + f + | f | ' € L * ( X ) and L*(X)/M relation  f s M  then  f,g€  L*(X)  (or L ( X ) ) i f S  f € M.  f £ 0 (mod M)  so t h a t t h e r e e x i s t s an  90, element  h e L*(X)  and h s jhj (mod M) holds.  f _> 0 , 1 + f |h| . >_ 1  (by (10.7)).  diction.  Thus  so t h a t  u = J^LJLJLlMi 1 + f|h|  ~ h & |h|  I f the f i r s t should h o l d  1 + f | h | s 1 - f »h s . 0 (mod M).  we would have hypothesis,  By (13.4), and  f°'h s i (mod M).  f » h = 1 , a t l e a s t one o f the r e l a t i o n s  the f a c t t h a t (mod M),  such t h a t ,  €  M .  l/l  However, hy + f |h| e Lg(X)  This I s a contra-  I f t h e second should h o l d , we would have  1 - f | h | s i - f . h — 0 (mod M) o r g|h| + (1 - f|h|) f <_ g , so. we have  But  s g (mod M), then  g e  g's f*  - g = j g | (mod M)  H  Q.E.D.  0  (mod M)  g s |g| (mod M).  Suppose  - g s |g| (mod M)  f s | f j (mod M) t o o b t a i n '0-<_ |g| <, | f | + |g| , by  However,  which i s a c o n t r a d i c t i o n .  That i s ,  By  Thus, we have o n l y e i t h e r  or g s | g | (mod M ) .  | f | + |g| s 0 (mod-M). |.g|  and f ^ 0 (mod M).  0 (mod M).  holds, we would combine I t w i t h  (13.5)  f s 0 (mod M).  -  f € p" , f a j f j  Since  hypothesis  Hence  F o r any f , g <s L * ( X ) (or L ( X ) ) , i f f e P" , and  (13.6) Lemma:  Proof:  (mod M ) .  0  g|h| + (1 - f |h|) = i + ( g .- f ) | h | _>.l .  Again, we have a c o n t r a d i c t i o n .  f  B  Consequently  g <s 7^ , as we showed above  g +0  (mod - M ) .  Q.E.D.  (13-7) Lemma:  F o r any , f , g e L*(X) ( o r L(X>), I f f ' e P* f + g , f og £ P "  g e ^  , then  Proof:  By D e f i n i t i o n (13.3)  g  s  |g'U g k 0 (mod M).  |fog|  .  f s | f | , f + 0 (mod M) and  We then have t h a t  and f o g =}= 0 (mod M)  and  as L * ( X ) / M  f g s |f||g| = Isa f i e l d .  That i s ,  91. fog  £ f J .  Suppose  .-(f •'+ g) s j f + g|  |f + g| s f + g - ( f + g) = 0 (mod M). | f | + |g| +  |f + g| ,  | f | -m 0  imply  and  |g| = 0 (mod M). .  This i s a contradiction.  (mod M ) .  I f f + g s 0 (mod M), then  (mod M)o  Again, from We have  contradiction. (mod M).  0 < |f| <  f = 0, g = 0  Thus,  Hence  f + g B | f + gj  |.f |.+ |g| = f + g  0 < | f j < | f | + -|g| and 0._< jfj = 0  Therefore  That i s ,  We have  |f | + j g | +  0 < |gj < | f | + |gj + |f + g| which  (mod M ) .  f f j + |g|.  Then  0  and  |g| E 0 (mod M) i  f + g = | f + gj  and  |g|£ which i s a  |f + g| f 0  f + g ep"  Combining these f o u r Lemmas and the d e f i n i t i o n o f a t o t a l l y ordered f i e l d  (13.8) P r o p o s i t i o n s  (see P. 2 0 9 , [ 2 8 ] ) , we haves L*(X)/M  (.or L(X)/M)'' I ® a t o t a l l y ordered  field. (13.9) Lemmas and M  be a maximal i d e a l i n L * ( X ) , ( o r L ( X ) )  L^^X^ (or L(X))' be. normed by the sup norm  lUH^ .  i s "closed i n L*(X) ( b r L(X) r e s p e c t i v e l y ) under  Proofs of  Let M  I n view o f  ( 2 M . 1 ) . [7'3,  L*(X> o r L*.(X)"' i t s e l f .  Then  || oj|  c £ M I s e i t h e r a proper  Suppose . q*M = L*(x)'.  w  ideal  .Then  u € GiM and f o r any neighborhood o f u:, H ( u ) , H (u) •(! M + <j) . e  In p a r t i c u l a r , take  € = 1/2 .  i s , t h e r e i s an f € M |u(x) ••- f ( x ) | x e X  In  such t h a t  < 1/2 , s© t h a t  or u(x) - |f(x)|'< 1/2  1/2 < | f ( x ) |  f o r each  other, words,  Then  M  x € X .  ^ ^ ( u ) 0- M + $ •  ||u - f|| < 1/2 .  |u(x) - f (x) | < 1/2 x € X .  By Lemma (10.7)  ha®, a u n i t so t h a t  '  f c  Thus  oo  f o r each  T n a  f o r each  Hence l / f £ L*(X) .  M = L*(X) .  This i s  92. Impossible. must have  Hence  atM I s a proper I d e a l c o n t a i n i n g  M = c-tM .  L(X)),.  L*(l)/M,(or Proof:  (as f  . such t h a t  f = > 1/n € 'TL  Then we would have  i s not a constant  function).  l i m i t under the norm  |f  for a l l  IMI^ .  0 •  f + jf|  0 1 1t  n  e  other hand,  f + |f | € clM = M .  This i s a contradiction.  We have proved t h a t every maximal i d e a l i n i s real.  L*(X)  In view o f Theorem (12.6), we have t h a t  every r e a l maximal i d e a l i n (X,d)  as the  0 < f + | f | - [ ( f ° 1/n)  By (13-9) Lemma,  =f s | f | (mod M).  fora l l  f + | f | - [ ( f - 1/n) +  Indeed,  + 1/n - | f - 1/n I  + |f, - 1/n I 3 <, 2/n .  ))  such  n . Consider now the sequence  1/n I > _ | f | - 1/n , hence we have  T h i s show t h a t  n  That l s ^  = (f - 1/n) + |f - 1/n I ° n e N] which has  |f - l/n|3 =  f  Suppose t h a t there does not e x i s t such a  ( f - 1/n) + |f - 1/n I c M n  That i s ,  f u n c t i o n , there i s a p o s i t i v e integer  p o s i t i v e integer.  {g  i n L*(X)^  I t i s enough t o show t h a t f o r any f €  f ~ 1/n € P " ';'  that  M  i s an archimede&n ordered f i e l d .  L(X)/M) = ^  Is not a constant  n  F o r each maximal i d e a l  L*(X)/M  We  Q.E„D„  (13.10) P r o p o s i t i o n ; (or  M .  L ( X ) i s f i x e d i f and o n l y i f  i s compact.  (13.11) D e f i n i t i o n :  A metric  space  (X,d) i s s a i d t o he  Lc-  realcompact i f every r e a l maximal I d e a l i s f i x e d . Now,  we w i l l give an example t o show t h a t there  Lc-realcompact space which i s not compact.  i s an  However, t h e  93. existence  o f non~Lc~realeompact 'spaces remains as an open  question.  (13.12)  Lemma.  subset  A  An i d e a l i n L ,(X)  of  X  Then  Let X  Suppose t h a t  g(x) =. -. ^4  Hence  be any c l o s e d subspace o f  g |.M  L .  and  i s compact I n  Then  M(g) 4  X  as  X  which i s c l o s e d i n X f €. L ( E ) c L j E * ) n  1  Z ( f ) e.Z(M) . Z €-Z(M) .  Z(M) .  Z{f ) ±  g <, £  - g) > 0 .  Since  .  »  0  f r e e maximal i d e a l  g <s.L«(X)  and i s a unit/ i s c l e a r .  F o r any  r  ?  Is closed.  f  1  Z(f) = A  e M  8  11  .  B  .  Thus there i s an We w i l l show t h a t  Z ( f ) z> Z  such t h a t  x  Or  Z ( f ) arid  M(r) - M(g) > 0 „•  i s any p o s i t i v e number,  This Is a contradiction.  M(g)  Therefore  f o r some  I s compact I n  M  X .  By  Z ( f ^ ) fl B = |) .  e X ° B c c £ ( X - B) = Z ( f ) . ' ®n the s» e r e - s e t  B = A n X  x  I t I s . enough t o show t h a t  there i s  Then  L e t A" = e £ ( X - B)  so i s c l o s e d i n E  such t h a t  M .  r e / ^ , r > 0 , g < r  , say A .  However, we know t h a t  (13.12),  That i s ,  M(r  11  ~ /R for'a  f o r a l l b u t a compact subset o f  Lemma  E  i s Le=>reale©mpaet,  Proofs Let  h a v i n g n© zero i n A .  (4.3).  Propositions  X  f € I  there e x i s t s an  Proof i s s i m i l a r t o  (13.13)  i s f r e e i f f f o r every compact  in  Hence  r - g  That i s ,  Z(f). £  .0 .  Hence  r 2 M(g) .  i sInfinitely must be f i x e d .  small. Q E„B, 0  •9*. L c , L°Mappings and..Lc^ I^Homeomor^MsmS o  §14  (14.1) D e f i n i t i o n : s a i d t o be an of  A  A mapping  (X,d )  from  to  1  <,. K  T(X»))  2  a p o s i t i v e number for a l l  d(x,x ) 5  A  2  such t h a t  x,x« e A , f i f there i s -  for a l l  such that  K  < K d(x,x')  d (ir(x), T(X»)) 2  x,x» € X ) .  (14.2) D e f i n i t i o n :  T  both  Proof:  (X,d )  Let (X,d ) 1  T  and  be a mapping  f e L(Y) .  f o r each  X  into  Then  Y  T  L(Y) into  L(X) d e f i n e d by  cp I s continuous under t h e norm  cp i s c l e a r l y a homomorphism.  d e f i n e d i n (13.1), so t h a t How, f o r q e Y  ||fJL  q  a  Then,  2  ||»||  ||fJ1 q  f o r each  --- H f ^ + Hf ll  Jd (q^y) - d (q,y')) = sup ( s d (y,y«) 2  By Theorem  = || eH^ +||<>|!  d  cp I s bounded ( f l 3 l Theorem 7A).  l e t f ( Y ) » d(;q,y)  's q 6 Y} .  ..  such t h a t  i s an L-mapping.  (2.5,17) [21],  For  Is  2  Then  q  2  i s one-one, onto,  cp(f) = f CT .  {f  (Y,d )  ( Y , d ) be two compact m e t r i c  from  L e t cp be a mapping from  D =  to  1  c  and  f « T € L(X)  -r from  and i t s i n v e r s e a r e L - [ r e s p . I>3 mapping.  (14.3) Lemma: spaces,  A mapping  L c ( L j - homeomorphism, i f T  s a i d to be an  q  (Y,d ) i s  L - [ r e s p . L-] mapping i f , f o r each compact subset  (X,d^),' t h e r e i s a p o s i t i v e number  d (T(x),  and  T  q  2  y , y '  y e Y and d  e Y ,  = Diam (Y) + 1.  y  +  y ' }  2  d (y*y') 0  ^< gup  {r-"~;;nn  > ^ ,•  n  =1} .  Hence the s e t D  I s bounded by  d (y,y') 2  DIam(Y)+ 1 K .  9  so t h a t  In other words  cp(D) i s a bounded subset i n L(X) , say by ||f . ( T ( X ) ) - f . ( T ( X » ) ) | | < K  for a l l  q e Y ,  95. .T(X) - f  ||f  so t h a t  >L  H  •T(x»)li  d  < K'•  2  |d (q,T(x)) - d2(a,TM)| • — . • < K d (x,x»)  Hence  2  f o r a l l x,x» s I , x =^ x .  In  1  d (T(x), (k»)) 2  q = T ( X ) , then  p a r t i c u l a r , i f we take  T  < K .  d (x,x ) !  1  for  a l l x^x* €.X-, x 4 X s  .  S  If  d ( T ( x ) j T ( x ' " ) ) = d (x x»') = 0 . 2  1  K d (x,x 1  8  )  x"= x' , then we have d (?(X),T(X«) ) <  Thus  i  2  f o r a l l x>.x» e. X.  (14.4) P r o p o s i t i o n : be a mapping from  L e t (X,&-} (Y,d ) 3  (X,d^)  I f T  (1)  Q. E.D.  i s an  to  he m e t r i c spaces and  2  T  (Y,.dg) . f « T s L_(X) f o r a l l  L -mapping, then  t € L -(Y) . C  (2)  If  L -mapping of Proofs  f o r e L„(X) f o r a l l f € L_(Y) , then (X,d )  and  A  of  i s compact..  f o r x,x« e A  for  Then we have  2  -Hence, we have  !,  X .  d ( T ( x ) , T ( x ' ) ) , . <_ K ^ d ^ X j X ' )  T [ A ] is' a compact subset o f  Kgdg(T(3f:),T(x J) < KgKjd-j^xyx' ) f.T  A  such t h a t  | f ( y ) - f (y«) | _< 'Kgdg(y,y»)  T [ A ] , where and  Kg  i s an  (Yjdg)-.  ( l ) Take any compact subset  two constants and  into  x  T  Y as  |f©<r(x). any  and T  y,y> e i s continuous  f»T(x')l  S  x,x«'e A . • T h e r e f o r e ,  e L_(X) . (2)  Consider any.compact  w i l l show t h a t (3.8) [7]  that  T T  subset  I S an L-mapping on I s continuous.  A .  Hence  A 4 $  of  X .  We know from  We Theorem  T [ A ] i s compact.  Let  cp be a mapping from  L_(Y) t o  L_(X) d e f i n e d by  for-  f o r a l l f e L_(Y).  Then, I t I s obvious t h a t  cp(f) -•  cp i s a  96. homomorphism of  L (Y)  L (X) .  Into  We r e s t r i c t  L . ( Y ) | T [ A ] = {f I T [ A ] : f € L ( Y ) 3 , then £g|A  i g e L (X)3 .  A l s o f o r each L (X)  € L ( A ) , hy  f  such t h a t  C  f|A = f  L(A) = L ( X ) | A . By Lemma ( 1 4 , 3 ) , T T  f e L(X) c  L ( A ) c L (X)JA  Thus  so t h a t  C  L (Y)|T[A] q  i s an remapping on  L (X) | A =  Is i n t o  ( 1 1 . 4 ) , there, i s • a n  .  Q  to  L ( X ) | A e. L ( A ) .  i s compact,,  S i m i l a r l y , we have  C  arbitrary,  A  Since  C  cp  ep  A .  =  L(T[A]).  Since  A  is  i s then an Lc-mapping.  Q.E.D.  We w i l l i n v e s t i g a t e the r e l a t i o n between a r b i t r a r y L c mappings from  (X,d )  into  L_(Y)  L_(X).  We  of  a ring  1  to  L (X)  J  C  induced by  T s X -» Y  T  mapping  Lc-mapping  1  L ( Y ) , In C  I-  (see (6.1) and  he an  , d e f i n e d hy  and the r i n g homomorphisms  s h a l l see t h a t e v e r y homomorphism from  I n t o another  an  Let  (Y,dg)  same sense, IS  (14.13)).  Lc-mapplng.  Then, the i n d u c e d  T « ( g ) - g « T € L_,(X)  g € L „ ( Y ) i s a homomorphism from  L (Y)  f o r each  into  L„(X) .  c a r r i e s the constant f u n c t i o n s onto constant f u n c t i o n s I,.  I.  Moreover,  'proofs are s i m i l a r t o (6.7) and (14.5) Theorems (Y,d )  and  2  L  C  into  ( Y )  T  Let  T'  The  (Xjd.^) g -*  g«>T  to from  L (X). E  T'  i s an isomorphism  (2)  T»  i s onto i f f  T  C  uniquely.  be an LC~mapplng from  be the Induced homomorphism  an Lc embedded subset of 0  T  ;  (6.8) r e s p e c t i v e l y ^  (1)  an f e L  identically.  •  1  i t a l s o determines the mapping  It  Y  T  T  Proof i s s i m i l a r t o ( 6 . 1 0 ) . 1  i s dense i n  i s an Lc-homeomorphism of  «, i . e . f o r each f €.•  ( Y ) with f | [ X ] = f . 0  (into) i f f T [ X ]  ;  L  (T[X].),  '  X  Y .  onto  there i s  97.  We how examine the inverse problem of determining when a given homomorphism of  L (Y)  mapping from  Y.  X into  morphisms from  L (Y)  into  L.(X) is induced by some Lc-  We w i l l f i r s t consider the homo-  into fR , that i s , the case In which X  is a single point. (14.6) Proposition:  Every nonzero homomorphism cp from L (Y) C  into fR is onto fR . " In fact,  cp(r) = r  for a l l  r e fR .  Proof is similar to (6.13). (14.7) Propositions of  L-_(Y)  The correspondence between the homomorphisms  onto fR, , and the real maximal Ideals is one-one.  Proof is similar to (6.14). (14.8) Propositions  Y Is Lc-realcompact i f f to each homoL_(Y) onto fR - i . e .  morphism ' cp from  each nonzero homomorphism  fR - there corresponds a.unique point  into  cp(g) - g(y)  for a l l  y  of  Y such that  g e L (Y) . C  Proof Is similar to (6.15). (14.9) Theorems L (X)  Let  such that  cp be a homomorphism from  ep(u) = u .  If  there exists a unique Lc-mapping T'  L (Y)  into  Y is Lc-realcompact, then T of  X into  Y such that  = cp .  Proof is similar to (6.l6). (14.10) Corollary:  An Lc-realcompact metric space  tains an image of ah Lc-mapping of a homomorphic image of on  X.  (X^d.^)  iff  (Y,d ) 2  L (X) C  con-  contains  L (Y) that includes the constant functions Q  98. Proof I s s i m i l a r t o (6.17). (14.11) C o r o l l a r y ?  Ah Lc-realeompact  a dense image of an Lc=mapping of an isomorphic image of f u n c t i o n s on  L (Y)  space,  (X,d )  contains  C  t h a t i n c l u d e s the constant  X . ( l 4 . 5 ) and  (14.12) The Main Theorem:  LC-realeompact  and  contains  2  i f f L ('X)  1  Proof f o l l o w s Immediately from  (X,^)  (Y,d )  (Y,d ) 2  Two  (14.9). m e t r i c fepaeeS  are Lc-homeomorphic i f f L ( Y )  and  C  L (X) C  are isomorphic. Proof i s s i m i l a r t o  (6.19).  Note t h a t , i n p a r t i c u l a r , t h a t i f we (Y,d )  he  1  (X,d )  L (X) = L(X)  and  L (Y) = L(Y)  D.R.  Sherbert has a s i m i l a r r e s u l t .  and  1  compact m e t r i c spaces, then they are L  iff  In  let  c  = L-homeomorphic  are isomorphic. (See Theorem 5.1  [26].)  s p i t e o f the Theorem (14.9), every, homomorphism i s  induced, i n essence, by an Le-mapplng. (14.13) Theorem: realcompact from  L„(Y)  cp(u)(x) = 1)  For any m e t r i c space  m e t r i c space into  L„(X)  (Y,d ) , l e t 2  .  for a l l  g e L.(Y)  <¥>(g)(x) = 0  T  in  cp  from  X . E  , cp(g)(x) = g ° ( T ( x ) )  f o r a l l x. e X - E .  Proof i s s i m i l a r t o (6.20).  1  Then-the s e t  i s open-and-closed  e x i s t s a U n i q u e Lc^mapping  (X,d )  and an  Le°  be a homomorphism E = {x e X : Moreover, t h e r e  into  Y ,  such t h a t  f o r a l l x e E , and  99.  (14.14) Theorem;  Let  homomorphism from  (Y,d ) 2  he a compact space and ep he a  L*(Y) = L(Y)  E = {p e X s tp(u)(p) = 1}  into  L*(X) .  Then the set  i s open-and^clpsed i n  there exists a unique Lc-mapping from  E into  X.  Moreover,  Y , such that  for any g e L(Y) cp(g)(x) = g(t(x))  for a l l  x eE  cp(g)(x) = 0  for a l l  x e X- E .  and  Proof i s similar to (6.20). (14.15) Corollarys  Let cp be a homomorphism from  a ring of Lc-functions.  If  exists a unique closed set  L (Y)  into  Y is Lc-realcompact, then there  P in  Y such that the kernel of cp  is the z-ideal of a l l Lc-functions that vanish on P . Proof i s similar to (6.22). (14.16) Propositions  An Lc-realcompact space  an Lc-embedded image of an Lc-mapping of homomorphic image of  (Y,d ) 2  (Xid-^)  iff  contains L (X) i s a C  L (Y) .  PrOof i s similar to (6.23). (14.17) Propositions  A compact space  L*- embedded image of an Lc-mapping of a homomorphic image of  (Y,dg) (X^)  contains an iff  L*(X) Is  L*(Y) = L(Y) .  Proof is similar to (6.23).  §15  Embedding Theorems The definitions (7.1) to (7.6) are applicable In this  section.  IGO. (15.1) Theorem: (X^c^) into  Then,  homeomorphism  cp a homomorphism from L (Y)  C  and  X into  Y such that  T is an Lc-mapping. T  (Y)) = L (X) , then  Proof:  C  ep has the property (7^1) i f f there is a  T from  f e L (Y)  ep(f) = f ° T for  In addition, i f  i s an Lc-homeomorphism.  The proof of the f i r s t part is the same as Theorem ( 7 . 7 ) .  Now, let  ep° he a mapping from  defined by form  he an Lc-realcompact metric space,  1  C  cp(L  (Y,dg)  be any metric space, and L (X) .  all  Let  f => f  (f oT)oT  cp°(f) = f o r " Q  for some  oT  = f  r  Q  1  Q  .  . f  cp(L. (Y)) = L (X)  Since each Q  f e cp(L_(Y))  e L (Y) , we have  L.(Y), has the  ep°(f) .='  C  We then show that  into  is an Lc-mapping>  T  by Prop. (14.4)  (15.2) Theorem:  Q.E.D.  Let  (Y,d ) 2  be a metric space, (X,d )  Le-realcompact metric space. concerning a homomorphism  be ah  1  Then the following statements  cp from  L (Y) C  into  L (X) Q  are  equivalent.  and  (1)  ep has the property (7-1).'  (2)  cp is a 6-homomorphism.  (3)  The image of  separates points and closed sets  i t is contained i n no 6-real ideal of  Proof: (15-. 3) L (X) into  L_(X).  The proof i s similar to Theorem ( 7 . 8 ) . Theorem:  metric spaces.  X  L (Y)  Let  (Xjd.^)  and  (Y,dg)  Then a homomorphism,  be two Lc-real comp act  cp' , from  L (Y)  is a 6-homomorphism i f f there i s a homeomorphism Y such that  is an Lc-mapping.  cp(f) - f » T  In addition, i f  for a l l  T from  f c L_(.X) , and T  cp(L (Y)) c  into  is  L (X) C  itself.  101. then  T  Proof:  i s an Lc-home ©morphism. The f i r s t p a r t  i s s i m i l a r t o Theorem (7.9).  The second  p a r t i s the same as t h a t i n Theorem (15.1).  (15.4) Lemma: and  T  be an Lc-mapping from  morphism for  L e t ( X ^ d ^ and (Y,dg)  op from  any i d e a l s  cp[My] c  L ( Y ) into C  M^. and My  he any m e t r i c spaces,  X into  Y .  Define a homo-  L ( X ) by cp(f) = f o T . C  Then  o f L ( X ) and L ( Y ) , r e s p e c t i v e l y , C  C  i f f (x) = y . T  Proof i s s i m i l a r t o Lemma (7.10).  (15.5) Theorem:  L e t (X,d-j^)  m e t r i c spaces.  Then a homomorphism, cp , from  L (X) C  and (Y,dg) be Lc-realcompact  X  T[X]  i s a closed  addition,  into  Y  cp(f)  such t h a t  s  f«T f o r a l l f e L ( Y ) , C  subset o f Y^ and T  i f cp[L_(Y)] = L ( X ) , then  i s an Lc-mapping. T  In  i s an L c - homeomorphism.  Proof i s s i m i l a r t o (7.11) and the l a s t p a r t  o f Theorem (15.l).  (15.6) Theorem:  Let (X,d )  m e t r i c spaces.  Then a homomorphism, ep , from  C  c  i s a 6F-homomorphism i f f t h e r e e x i s t s a homeomorphism^  from  L (X)  L ( Y ) into  and ( Y , d ) be two Lc-realcompact g  1  L_(Y) i n t o  i s a 6G-homomorphism i f f t h e r e e x i s t s a homeomorphism^ T  from  X  T[X]  i s an open subset o f Y , and T  addition,  into  Y  i f ep  such t h a t  cp(f) = f O T  i s onto, then  f o r a l l f € L (Y) , I S an Lc-mapping.  In  T. i s a homeomorphism.  Proof i s s i m i l a r t o (7.12) and t h e l a s t p a r t o f Theorem (15.l).  ;  102. 16  The Rings o f L c - F u n e t i o n s Defined on the -Metric Spaces other than Lc-Realcompact Spaces, and R i n g s o f L i p s c h i t z i a n F u n c t i o n s on Compact Subsets o f E . n  (16.1) Lemma: such t h a t  Suppose  p e X .  Z ( f ) = (p) and such t h a t  i d e a l other than  Mp.  Proof:  f ( x ) = d(p,x)  f  Consider  belongs t o Mp  Suppose  f  for all  x.e X .  We know t h a t  g e M  M .  Then by the  g(p)'-4  such such t h a t  h = f  + g  .  We have  Z(h) = ty .  Take any compact subset  A  of  X .  |h(x) - h ( x ' ) | < K od(x,x»)  |h(x)| > m  A  and  A  and  |l/h<*) '* l/h<*'> | .-/M*fh(x» ) j |h(X)oh(X')l -4y d(x,x») . Therefore 1/h e L_(X) . m u = h»l/h € M .  For t h i s  A ,  in such t h a t  Hence  i d e a l , we would have  x,x« e A .  for all 1  <  /  m  2  Since  That i s ,  |  h  M  (  x  is,  M  cp(Mq)  Proof: ideal.  I f cp  i s an isomorphism from cp(M^) = Mp  C  cp  i s an Isomorphism onto,  By Lemma ( l 6 . l )  slder  Z(cp(f )) .  lemma  cp(f ') q  Q  f (,y) = d(q,y) Q  x  , , }  M = L (X) . 0  L.(X),  p e X , that  L (X). C  cp(M^) f  e M  i s a maximal only.  Con-  I f Z(ep(f )) = ty , then as shown i n the l a s t  i s a u n i t so t h a t ' cp(M ) q  which i s i m p o s s i b l e .  (  i s an  L„(Y) onto  f o r some  i s a f i x e d maximal i d e a l i n  Since  h  Q„E.D  c L ( Y ) , and  q  _  )  This i s impossible. (16.2) Lemma:  1/h  Thus i t s i n v e r s e  K  then f o r  °-  p  t h e r e e x i s t two p o s i t i v e numbers  X  Q  belongs t o no maximal  belongs t o a f r e e maximal i d e a l  p  exists.  f  f.e L ( X )  and n o t any other f i x e d maximal i d e a l .  p r o p e r t y o f f r e e i d e a l s , we have Let  Then t h e r e e x i s t s  Hence  i s the whole r i n g  Z(cp(f )) 4 0 •  L (X) c  On the other hand,  103. if  Z(op(f )) c o n t a i n s more than one p o i n t , say p-^ and pg ^ q  then  tp(fq)  and Mp  e  so t h a t  f  would b e l o n g t o a t  q  l e a s t two maximal i d e a l s which a g a i n i s i m p o s s i b l e f o r t o o n l y one maximal i d e a l . Therefore  Hence  Z(ep(f_)) = { p } ,  Q.E. D.  q  L e t (X d-^) 3  and (Y,dg)  spaces, and cp be an isomorphism from leaving a l l  homeomorphism. X  onto  L (Y)  T  Conversely, i f  (16.4) C o r o l l a r y :  L (X)  cp i n d u c e s a  and T  I s an Lc-  i s an Lc-homeomorphism o f cp , d e f i n e d by  Y , then the induced mapping  Proof i s s i m i l a r t o (9.4)  onto  Then  cp(g) = g«T  d e f i n e d by  go*r , i s an isomorphism from  L ( Y ) onto  cp(g) =  L (X) .  C  C  and the converse i s obvious.  L e t ( X , d ) and ( Y ^ d ) be two connected -metric 1  spaces and cp be an Isomorphism  2  o f L (Y)  cp induces a Lc-homeomorphism, T , from «p(g)  be any two m e t r i c  constant f u n c t i o n s unchanged.  T : X -* Y  mapping  belongs  say.  cp(M ) = Mp .  (16.3) Theorem:  f  ;  g ° T f o r each  onto  X  L (X) .  on Y  Then  such t h a t  g e L (Y). C  Proof i s s i m i l a r t o (9.5). Now we d i s c u s s the s i m i l a r p r o p e r t i e s as we have done i n §9  for  (^(X)  .  H e r e a f t e r , we c o n s i d e r  X  and Y  as compact  subsets o f E . n  (16.5) Lemma: 1 C i _< n Proof:  The p r o j e c t i o n f u n c t i o n s  ^(x ,...,x ) = x for 1  n  i  b e l o n g t o L(X) .  We know t h a t  1 < i < n ,  and 4>(X)  |X(x) -^o(x» ) I = Ix.^ - x £ | _< d(x,x') f o r i s bounded.  Hence  L(X)  for  104. <  1  <  1  Q . E . D.  •  n  (16.6) P r o p o s i t i o n s X  be a mapping from If T  (1)  L e t L» into  be a s u b f a m i l y ©f  L(X)  and T  3  Y . f o r € L(X)  i s an L-mapping, then  for a l l  f e L(Y) '. (2) f o r each Proofs  c o n t a i n s a l l p r o j e c t i o n s and f»T e L ( X )  i f L'  f € • L* , then  T  X  I s an L=mapping from  into  Y .  ( l ) i s obvious. (2) F o r any x,x' € X••, and any f € L« , |f<>T(x) -  fo (x')l\<  K  T  if  f  =Z  foT  K^.. = l | f ° T | l  <"d(x,x») , where  for 1 C i X h ,  IZ(T(X)) ->1(T(X'))I  .  d  |(T(X)') - (•r(x )) \  then  < K j o d ( x , x > ) , where oT  T(X»)) = (2  1(T(X))  i=l Let x'  K = ( E  1=1  as each  all  )  m  1  (T(X))  K  f e B(X) R(X)  B(X)  with  K  i s . That i s , Hence  T  B ( X ) of L ( X )  A subring  (16-1) i f  Then  2  and x* £ X .  x  2  = [f €  L ( X )  f e  ft(X) c B ( X ) c L ( X ) Q  1  ( ' )<  implies  6 B (X) N  .  C (X)} 3  d { T i x ) , T ( x ' )) < K>d(x,x« ) f o r  L(X)  Q.EjDi  3  B { X ) z> R ( X ) .  and  f " ( = l / f ) e B ( X ) , where 1  For instance, l e t  Then, i t i s obvious  and f e B ( X ) Q  X?  i s independent o f x and  exists.  .  X  T  i s d e f i n e d as i n §9 >  B (X)  f"  d  i s s a i d t o have the p r o p e r t y  We know t h a t such a B ( X ) Q  l / 2  T  i s an L-mapping.  i s a subring of Z(f) = §  .'•'.(•  4o )  i=l ^  x  denotes  ±  Hence  (T(X«))J )*< ( 2  • -  x  ^ .  1  =  ,  ±  the i - t h c o o r d i n a t e of. T ( X ) , f o r 1< I _< n . d(T(x),  In p a r t i c u l a r ,  with  that  Z ( f ) = <J) i m p l i e s  !  '  105. (16.7) Lemma; such t h a t  There i s a f u n c t i o n  Z ( f ) = (x) and f  = [f e B(X) : f ( x ) = 0]  f €  belongs t o no o t h e r f r e e or f i x e d  maximal i d e a l . Proof i s s i m i l a r t o (9.1).  (16.8) Lemm'a;  I f cp I s an isomorphism from  then f o r ML c B(Y) , cp(M ) where  B(Y) onto B ( X ) ,  I s a f i x e d maximal i d e a l i n  B(X) and B(Y) have p r o p e r t y  B(X),  (l6-l).  Proof I s s i m i l a r t o (9.2).  (16.9) Theorems, / L e t X  and Y he two compact subsets o f E  cp he an isomorphism from  and  B(Y) onto  constant f u n c t i o n s unchanged, where p r o p e r t y (l6»l). from  X  onto  Then  cp  :  9  B(X) l e a v i n g t h e  B(X) and B(Y) have the  induces an  Y , d e f i n e d by  n  L-homeomorphism,  T ,  cp(g) = g o T .  Proof, i s . S i m i l a r t o (9.^).  (16.10) C o r o l l a r y s of  E  B(X)  n  Let X  and Y  be connected compact subsets  and cp be an isomorphism from  B(Y) onto  and B(Y) have t h e p r o p e r t y (16-.I).  L-homeomorphism,  T , from  Proof I s s i m i l a r t© (9.5).  X  onto  Then  Y , d e f i n e d by  B(X) , where  ep induced an J  ep(g) = g » T .  106. PART I I I THE •RINGS OP ANALYTIC FUNCTIONS  §!7  Rings o f A n a l y t i c F u n c t i o n on any Subset o f t h e Complex Plane A l l the f u n c t i o n s c o n s i d e r e d i n t h i s s e c t i o n a r e complex  single-valued.  (£ w i l l denote the complex p l a n e .  (17.1) D e f i n i t i o n s function f o r each  foril  L e t G be an open subset o f  D = [Z 1 |Z - p j < R)  and f ( z ) = 2  a_(z-p)  and D c G , and  a  (17.2) D e f i n i t i o n s f u n c t i o n f on X  LetX  | z - p| < R , and  and |z - p| < R  s  (17.3) D e f i n i t i o n s (£.. A mapping  a n a l y t i c mapping i f in  Y .  onto.  T  f{z) - E  where  LetX T T  (See § 2  X , i f n  which  a_(z - p )  n  for a l l  n  R > 0 .  and Y  from  A  n  X  be two a r b i t r a r y subspaces  to Y  i s s a i d t o be an  i s an a n a l y t i c f u n c t i o n  on  i s s a i d t o be a conf©rmal mapping i f  X T  and v a l u e d i s one-one,  Ch. I I [2].)  (17.4) D e f i n i t i o n s Oi(x) = {f : f  (D.  E a_(z - p )  n=0  of  i s a complex  i s s a i d t o be an a n a l y t i c f u n c t i o n on  n=0  z £X  n  be an a r b i t r a r y subset o f  p € X , t h e r e i s a power s e r i e s  converges f o r  n  n  n = 0,1,2,...  number f o r each  f o r each  R > 0  which  n  n=0 z € B , where  G. , i f  i ; a (z - p)  n=0  for  A  G i s s a i d ^ t o be an a n a l y t i c f u n c t i o n on  p € G , t h e r e i s a power s e r i e s  converges on  (C.  Let X  be an a r b i t r a r y subset of (£,, and  i s an a n a l y t i c f u n c t i o n  on  X] .  107. Let  f  f f  g , and  of  f  and fog  and  g  be any two elements from  be the p o i n t w i s e sum,  g , respectively.  power s e r i e s  E a(z - p) n=0 n  |z - p| < R-^ , and  |z - p| < R  § n=0  n  n=0  n=0  |z - p| < R Thus  and  E n=0 2 n=0  (a  n=0  ( 2 a p _ )(z - p ) k=0 k  ( n± nH a  P  n  n  that  z  for  f + g  arbitrary,  n  ~ P)  R  > 0  x  a f z - p)  |z - p| < R ,  f + g  and  f g e 0[(X) .  and  a (z - p )  2  n=0  p .  For  ,  n  B ( z - p) ) n  n  ( E k=0  cuB " )• * " n  K  R = minfR^Rg} . Since  p  f e 0t{X)  is such  |z - p| < R ,  a  n  for  n  z e X  (See P. 145 [ 1 ] ) , and i s clear.  and  |z - p| < R.  n  4= 0 .  has a power s e r i e s which  < R  - p)  p e X , there e x i s t s  Then, t h a t  I  --.  n=0  |z-p|  6(z n  which converges on  n  > 0 .  2  n  f(z) =  I t i s clear that  " P)  1  E n=0  where  Suppose  Z ( f ) = [z € X : f ( z ) = 0} = 0 . z  + I n=0  |z - p| < R,,  ( f + g ) ( z ) = f ( z ) . + g(z) =  are a n a l y t i c a t  „(  R  n  fog  a  and n  f(z) =  and  f°g(z) = f(z)«g(z) =  and  and  - p) >( E n=0  n  and  § n=0  c o n v e r g i n g on  n  z e X  n  and  ^  n  z € X  a power s e r i e s R > 0  for  n  =( E ajz n=0  n  k  |z - p| < R  n  (z - p ) Hence  = 2 n=0  n  n  we have  n  n  + 0 )(z - p )  n  converges f o r  "  P,-(z - p )  , r e s p e c t i v e l y , where  2  p € X  , respectively,  2  It a (z - p ) , g ( z ) = 2 6 (z - p ) n  d i f f e r e n c e , and product  Then, f o r each and  n  (%(X) , and  —-—~ a (z - p )  Hence  = / z « (z - - p ) ) " n=0 11  l  n  1/f €  n  n n  converges a t l e a s t on  - i — f(z)  also  1  1  |z - p| < R, for  z e X  y  Moreover,  © , and  108. u , the constant are two C%{X)  function with values  1 , respectively,  0 , and  ©, u e  e n t i r e f u n c t i o n s on the complex p l a n e , so t h a t  •  The  operations  of a d d i t i o n and m u l t i p l i c a t i o n thus  d e f i n e d are a s s o c i a t i v e and laws h o l d .  Hence we  ( 1 7 . 5 ) Theorem;  The  commutative, and  the d i s t r i b u t i v e  have the f o l l o w i n g theorem? 0 t ( X ) , d e f i n e d i n (17.4), forms a  family  commutative r i n g w i t h u n i t y , (17.6) Lemma; f ( p ) = 0}  For  such t h a t  i d e a l other than ProOf;  p e l ,  Let  t h e r e i s an  Z(f) =  Mp  {p}  3  and  f ? ^ f  .  f(z) = z - p .  Then t h a t  i s a f r e e maximal i d e a l such t h a t  f r e e , there i s that S n=0  g  - P)  z  n  g(z) =  g e M eC%|X)  i s a n a l y t i c at  ot (  f e Mp  ,£ h=0  p  a_(z  p)  Now,  such t h a t  0  M  •  We  is know  z s X  and  |z - p| < R , where  n  Q  4 0 .  Let  p o s i t i v e i n t e g e r such t h a t  a  fc  + 0 .  Then  (z - p ) + . . . ,  for  g(.z),..-. <* ••!• ( z p )  Then f o r each  k  z c X  .  power s e r i e s > 0 , and  S n=0  P„(z  s  and  G  1  Since  \z - p| < R , and  a  R,  belongs  suppose t h a t  g(p) +  I t i s clear that  R > -0 .  f  so t h e r e I s a power s e r i e s  for  n  and  f € M .  which converges f o r  n  [f e.#{X) •  belongs t o no maximal  t o no other f i x e d maximal I d e a l i s c l e a r . M  =  - p«)  K  be the  a 4c^.  g(z) =  |z.- p| < R .  first Q  Let  h(z)  =  , p' € X , p + P > t h e r e i s a r  which converges f o r  |z - p'|  <  n  g(z) =  R,, x  ? n=0  B (z•- p » ) n  n  for  z e X  and  |z - p'|  <  109.  •, — •  Thus'  •• • / / ••  V N T z e X  and  •-  ••  _.  1 = 1,2,..„,  where  1  P  = P  Q  Q  ~ %  2 Y_(z  - p')  z € X  and  |zp«|  E "Ct^^ (z°p)  the power s e r i e s  Suppose  which  r / _,\n n=0 n ^ "  p' = p .  z  p  ;  Then  | z - p| < R ,  converges f o r  n  ,  n  q  .  r  h(z) = 2  , and e v i d e n t l y  <  Q  k  n  x  for  £  6^ =  and  n=0  [z = p ' | < L  •..  i(p' -pJ+(*-p»)l  w i l l he a power s e r i e s , say  converges f o r  •  I  ((P<-P)+(Z-P'))*  / z - p» | < R  x i r ' v . .-•  n=0  h(z) =  and Hence a  Q  o u ( z - p)  E  n=0  K  +  h e £%(X) .  = g(z) -  Since  for z e X  g(z) = a  Now,  i s a unit.  s  Q  + (z - p ) h ( z )  f o h ( z ) = g ( z ) - (z - p ) * "  1  4®  a  |z, - p | < R . , so t h a t  k  Q  which i s an i d e a l ,  i m p o s s i b l e , as  and  n  (z - p ) ^  f , g <s M  n  a  € M .  Q  1  f(z)oh(z) .  However, i t i s  Hence t h e a s s e r t i o n i s  proved. ( 1 7 . 7 ) Lemmas then  cp(Mp)  Proofs  I f cp i s an isomorphism from (%(X)  ep(Mp)  That  i s a maximal i d e a l i s c l e a r . f Ve Q  Z(cp(f )) = <j) , then Q  then  (%(X)  .  cp(f ) 0  Q  ZCP(f )).  Consider  cp(Mp)  Hence  0  <p(f ) 0  e  M  q  and  so t h a t  f  Q  i s the whole  Z(q?(f )) 4 (j) .  Z ( ( j ? ( f ) ) c o n t a i n s more than one p o i n t , say  and  q  2  would belong t o a t  l e a s t two maximal i d e a l s which i s again Impossible,  Q  If  0  i s a u n i t so t h a t  This i s impossible.  PrOm Lemma  Z ( f ) = [ p | , and f  such t h a t  b e l o n g s - t o n© other maximal i d e a l .  if  ,  i s a f i x e d maximal i d e a l .  (17.6), t h e r e i s ' a n  ring,  onto (%(Y)  for  f  But ,  110.  b e l o n g s t a o n l y one maximal i d e a l . q € Y . and  Now, s i n c e  <p{f )  k  i s a maximal i d e a l ,  b e l o n g s t o o n l y one maximal  0  ep(Mp)  «p(Mp)  Hence - Z(<^(f ))' = {q} ,  Let X  cp(Mp) = M  T t X -» Y , mapping o f Proofs  d e f i n e d by  Befine  T  Q(JX) such, t h a t i t i s t h e  onto  ep(g) = g ® T ,  Since  1  cp"  =  c p  Thus  T  " ^ p^ * 1  ep'^Mjj,,)  L  M  E  T  Thus,  q  q,q ..£ Y . !  o  P  one-One.  A  T  q  v  and so  L e t q^  i d e a l i n (Jl{'f)  X  cp" ^)  N  as f o l l o w s ;  o f Qi{X)  D  P T  B  E  1  p  onto  i s a f i x e d maximal i d e a l i n  1  q  I  N  X  Evidently, P 4 P' •  9 X 1 ( 1  , and  I f q = q ', t h e n  S  cp" (M ,) = M 1  p  cp°* (M ) = M  !  =  1  p  ql  q  This I s Impossible f o r p 4 P* • S  f ( p ) = q + f* = T ( P " ) .  be a r b i t r a r y  in Y *  , and ep(M ) = M ^© o a  = nZlcp' * (Mp ) ] * T ( P ) . 1  0  f o r each  i s a conformal  to Y  f o r some  M^  T  X .  Thus  I s onto.  cp(g)(p) = a »  be t h e c o r r e s p o n d i n g c o n s t a n t f u n c t i o n on  i s a maximal  "p € X . o  T h i s shows t h a t  g e 0C{Y) , and p <s X; , l e t  :T-;.;JS -  Hence  Then  p  =  a mapping  cp i s an Isomorphism, and o n t o ,  cp~ (M ) = M  = ^p? •  r  4=  and  i s a s i n g l e - v a l u e d mapping*  Then, b y Lemma ( 1 7 - 7 ) , f o r some  cp Induces  i s an isomorphism  1  O By Lemma ( 1 7 . 7 ) ,  (%(Y) .  Then  t@ be a mapping f r o m  i t s I n v e r s e mapping  Q  Hence  X • ©**t®. Y .  = nz[cp" (Mp)3 .  (%(Y)  ,  be two s u b s e t s o f C, * and ep  i d e n t i t y .on t h e - c o n s t a n t f u n c t i o n s .  q  .  q  q  Q.E.D.  and. Y  be an Isomorphism f r o m (7[|Y)  ^T(P)  0  I s a f i x e d maximal i d e a l .  (17.8) Theorem;  T(P)  cp(f ) € M  Then,  Now,  and a ep(g) -  a € M , g - c p " ^ ) = cp° (cp(g) - a ) e M ^ j , g ( ( p ) ) = c p " ( a ) ( T ( p ) ) 1  1  1  p  - S ( T ( P ) ) = a = cp(g)(p) .  f  Hence  p  T  cp(g) - g o t .  Similarly,  111. cp~ (f) « f . T " 1  ., where,  1  T"  nZ[cp(M )] .  I f we Choose  on  T ( P ) = g»T(p)  X , then  Hence  T  1  : Y -» X  i s d e f i n e d hy  g( w) = w  on  T" (q) = 1  Y , and  f(z)= z  T~ (q) = f»T~ (q)  and  1  are a n a l y t i c .  1  i s a conformal mapping.  (17-9) C o r o l l a r y ; (£y , and  Let  X  and  Q. E.D.  Y  be two connected subsets of  cp be an isomorphism from  onto (^%(X)  such  t h a t i t i s the i d e n t i t y on the r e a l constant f u n c t i o n s . cp  induces a mapping  mapping  "T  T : X >• Y .  Either  T  Then  or i t s conjugate cp(g) = g o r  i s a conformal mapping a c c o r d i n g as  or  cp(g) = g°T .  By Theorem (17.8), the mapping  Proof:  nZ[cp"' (Mp) ] 1  : f  i s one-one and onto.  f u n c t i o n of v a l u e r e a l and Hence, let  or  ep(g)(p) -.a  a.  - OQ) €  . o •  for" 1  or  1  and  Then  cp(g) - a  ~ d  cp° (f) • f•T 1  u  l  are a n a l y t i c .  n  or or  e M  X  .  a .  a  .  i s connected.  and  p € X , 1  0  Thus  To show t h a t  T  cp(a) = a  f ( z ) =» z on X , then  T  and  T  Hence  cp" (f) 1  or  or =  a , where  In p a r t i c u l a r ,  and  or  Q  cp(g) = g « T '  Similarly  ; T ( P ) = g»T(p|  is  , g - ep" ^) = » ^ vs©'  -"p  a c c o r d i n g as  t o what we have shown i n ( 1 7 . 8 ) . f o r m a l mapping.  Q  1  1  as  0  g(w) = w on Y , and 1  cp(r) = r_ i f r  T ™ ( q ) = nZ[cp(M ) ] .  T" (q) = f » T " ( q )  be the constant  g e (/t(X) o  r(p') =  g ( T ( p ) ) = cp" ( a ) ( ( p ) ) = a  , or  cp(a) = a  i s d e f i n e d by  choose  F o r each  cp(a_)  a c c o r d i n g as  a  , cp(i) = + i.  a .  Mv^j  a c c o r d i n g as  goT  t"  cp(a) = a  Let  We assume t h a t  = cp(°l) = -1  (cp(i.))  (cp(s)  «P°  a .  T d e f i n e d by  we  r(p):«. g«t(p)  " ^ ( q ) = f«T**- (q) L  are one-one i s s i m i l a r T  or  T""'  i s a conQ. E.D.  112. Remarks:  ( l ) I n Theorem ( 1 7 . 8 ) , t h e c o n d i t i o n t h a t  cp i s t h e  i d e n t i t y on t h e c o n s t a n t f u n c t i o n s can n o t he o m i t t e d . t h a t an isomorphism o f  (%(Y) onto (%(X) always l e a v e s t h e con-  s t a n t f u n c t i o n s w i t h r a t i o n a l v a l u e s unchanged. r e s t r i c t i o n on  We know  Without t h i s  cp , t h e r e s u l t o f t h e Theorem w i l l n o t be t r u e .  For  example, c o n s i d e r  X = [ p ] , Y=[<2}. Then#£x)={a :" ci '.€. (D  and  (%(Y) = {<x' : a* e (£} .  I n order t o e s t a b l i s h the required  r e s u l t , l e t us f i r s t prove t h e f o l l o w i n g Lemma: (17.10) Lemma:  There e x i s t s a non-zero automorphism  i t s e l f w h i c h i s d i f f e r e n t f r o m t h e mappings i d e n t i t y , and Proof:  2  L e t cp : (&£y2) - (&C/2)  «p(r) = r  of  cJ3 >  Z , the  1  cp : z -• "E , t h e c o n j u g a t e mapping.  i f r e (bX .  (Q,(»/2) onto i t s e l f . ep'  ep  L e t © and Ik be t h e r a t i o n a l and r e a l f i e l d s , r e s p e c t i v e l y *  and so on. and  o f (C onto  (Bl(,/2 , JZ>) or  Then  be d e f i n e d b y cp(,y2) » -J2 cp i s ah isomorphism o f  cp can be extended t o an Isomorphism, onto i t s e l f  a - b^2 - c/5 »  ( s a y a + \>J2 + c/3 --> a - h/2 +  where a, b  n o t t h e i d e n t i t y on t h e r e a l numbers.  and  c e (£L )  And so on.  which i s We have a  l i n e a r l y o r d e r e d s e t o f f i e l d s and a l i n e a r l y o r d e r e d s e t o f By Zorn's Lemma [ 1 0 ] , t h e r e e x i s t s a maximal  isomorphisms. f i e l d , that i s itself.  ^  , and a maximal isomorphism  i s t h e n n o t t h e i d e n t i t y on t h e r e a l f i e l d .  i s neither  cp^ n o r cp •  Now, d e f i n e cp(a' ) I a  1  f r o m (J^ o n t o  Q. E. D.  2  cp : Ot{X)  f o r some  a  1  Hence  - Ot{Y)  e (%(Y) .  i n t h e o b v i o u s way. On t h e o t h e r hand  ?  Then  113. a'oTsaS  where T S I - * I i s d e f i n e d by ( p ) = q.  Hence ep(a« )  T  L. Bers showed i n 1948  However,  two domains i n (D and', i f &L(X)  that i f X  and £%(Y.)  4a  1  «T .  and Y a r e  are isomorphic,  then.there e x i s t s e i t h e r a conformal o r an a n t i - c o n f o r m a l mapping which maps  X  (2) "cp  onto  (Theorem 1,[3l).  Y  In C o r o l l a r y ( 1 7 « 9 ) , n e i t h e r o f the c o n d i t i o n s  l e a v e s a l l r e a l Constant f u n c t i o n s unchanged" and "X i s  connected" can be omitted.  The Remark ( l ) above Shows t h a t the  r e s u l t i s not v a l i d i f X  I s connected and cp does riot l e a v e  a l l the r e a l constant f u n c t i o n s unchanged. X = {p-j^Pgl Then  and Y =  { q  1  5 q  2  }  cp s Ol{Y) - (%(X.)  (a,"F) .  That  p  A  .  Choose  number.  + P  2  be d e f i n e d as f o l l o w s ?  » and q  x  + q . 2  cp((a,3)) =  cp I s an isomorphism and l e a v e s a l l the r e a l con-  s t a n t f u n c t i o n s unchanged i s c l e a r . q  x  ' and 01{Y) = {(<x',p« ) : a',B« e £ } .  # ( X ) - C(oi/B) s a,p e.  Let  where  Wow, c o n s i d e r  g = (a>p) e (X(Y),  Then  cp(g) = (ex,"^) 4 g  Then where  where  T(P^) = n(Z[$~^(Mp £  )] =  i s a complex  g»t = (a,P) .  Hence  «p(g),4-g»T . L. Bers h a s shown t h a t l e t X  X  induces a conformal or an -anti-conformal mapping o f  onto  Y  (See Theorem 2  (17.11) Theorems  Let X  [3]). and Y  be a conformal mapping o f X T  j  be two domains  Then, every isomorphism o f CPt{Y)  p o s s e s s i n g boundary p o i n t s . onto  and Y  d e f i n e d by  (X(X)  t ' ( g ) = g»T  be two subsets o f  onto  Y .  CI/,  and T  Then the induced mapping  I s an Isomorphism o f 0\%Y)  which l e a v e s the constant functions, unchanged.  onto  114. Proof:  Let  *r'(g) = g»T  T(P) e Y .  p e X ,  g € Oi(Y)  Since 2  e x i s t s a power s e r i e s  g 6,^(Y)'.  f o r each  T ( P ) e Y , there  and  •T'(p) )  For each  which converges f o r  n  n=0  |w - T ( P ) |  < R-,  for  g(w) =*2  R-, > 0 , and  where  x  |w - T ( P ) | < R^  and  w € Y .  Since  2  X , t h e r e e x i s t s a power s e r i e s |z - p| < Ro , where -  | z - H p| (,R  for  R  i s a n a l y t i c on which converges  n  n  = 2  T(Z)  *  2  Therefore  i s convergent and e q u a l t o  m  such t h a t  |z - p| < R  g(T(z))  , T ( Z ) e Y , and  2  Moreover, t h e r e e x i s t s  R >  0  converges on  m  n  n=0  m  S  Y„(z  n=G  -• P )  -  n  | T ( Z ) - T ( P ) | < R-^ .  , such t h a t  | z - p | < R , and  T(p))  for  m  |z• - p |  -  z e X  I  B (  2 Y (z-p)  -  n=0  2 B  g(T(z)) =  < R  n  n  m=0 (  11  n  f o r each  m=0  T(p))  - P)**  z  B ( £ Y^Cz-p)  m=0  T(p))  Y„(  n=0  and z € X .  9  n  n  n  > G , and  0  t  y (z' - p )  ' n=0 for  B_(w - T ( p ) )  n=0  x  and  z  €  X  m  .  By  n  2 B ( 5  the W e i e r s t r a s s ' s d o u b l e - s e r i e s theorem [ l l ] ,  (z - p ) say,  n  - T(p))  E oc (z - p )  which converges f o r each  n  n  and • g ( T ( z ) ) =  2 a(z n=0  - p)  for  n  |z - p |  n  n  n=0  w i l l he a power s e r i e s i n terms o f  m  Y  m  m=0  (z - p ) ,  z, | z - p |  <R,  < R  n  and  z e X .  g»T e Ot(X)  We know t h a t .  T ^ f ' T " ^  T ' (g)  = © , then as  T  i s arbitrary i n  S i m i l a r l y , f o r each  Also  = {0}  p  ( f o  T  "  X  ) T  g<>T(z)  i s onto.  =  f  = G  .  f e Ql{X),  Hence  f o r each  That i s ,  T'  z € X  g = Q .  X . for"  Hence 1  i s on,t^or  e (X(Y)  .  Now, i f  g°T[X]  =  T h i s shows t h a t  g[Y]  T  1.15. i s one-one.  -£L  F i n a l l y , f o r any constant  (T(Z)) = a Let  fora l l  z'e X .  function  That i s ,  a ., T'(a)(z)  T« (a) * a . Q.E.D.  Z ^ ( X ) = ( Z ( f ) : f e (X(X)} , where  {z e X : f ( z ) = G}  as d e f i n e d b e f o r e .  Z(f) =  We know t h a t n o t every  c l o s e d s e t i s a z e r o - s e t of some a n a l y t i c f u n c t i o n . Z ©o(X)  i s not a base f o r the c l o s e d subset of  r e l a t i v e topology i f  X  Hence  X^ i n i t s  i s n o t a d i s c r e t e subset of (£).  How-  ever, we have the f o l l o w i n g r e s u l t .  (17.12) Theorem:  Let X  every maximal i d e a l i n (X(X) Proof:  i n 0t(X)  i s compact i f f  Is fixed.  ,  M  Suppose t h a t f o r each maximal i d e a l  i s f i x e d , and  e x i s t s a sequence o f p o i n t s * verges t o a p o i n t ' Z n  X  The n e c e s s i t y I s c l e a r . Sufficiency:  {Z  (C.  be a subset o f  : n € W]  X .  say  {Z^ : n € N} c X  Case I .  If  Z  € OLW  .  conformal mapping  Q  Then t h e r e which con-  = » , then  By the W e i e r s t r a s s  t h e r e e x i s t s an e n t i r e f u n c t i o n ,  which has zeros e x a c t l y a t f  i s n o t compact.  has no f i n i t e l i m i t p o i n t .  f a c t o r Theorem [11],  Clearly  X  {Z^ : k J> n)  Case I I .  If  Z  Q  M  f o r each 4  00  f  ,  n  n .  > then consider the  f ('Z) = — ^ — which I s a n a l y t i c except a t the ° : Z-.Z o rt  rt  point {W  n  Z  .  D  : n € N]  Then  I n the  h  = (gof )|X  Q  {Z  g Q  n  y maps  [Z  n  : n e,N]  : n e N] .  Then  h  Q  e OL{X)  By  there e x i s t s an e n t i r e  which has zeros e x a c t l y a t .  Into a sequence  w-plane which has no f i n i t e l i m i t .  f a c t o r Theorem [11],  the W e i e r s t r a s s function  f  {W  : n;:e ,H} .  n  Let  which has zero e x a c t l y a t  S i m i l a r l y , we can f i n d an  f  e OliX)  s  u  c  n  that  116. f  has zeros e x a c t l y at  case we have £ k Z  {f  :  k  2. 3 n  •  : n e N] .  {f •  {Z  : n e N}  n  L e t  M  fee  a  : k _> n]  fc  with m  a  x  i  m  f a  l  0  Then  f|Z[M] e  f o r each  n .  has zeros e x a c t l y a t i d e a l which c o n t a i n s  n Z ( f ) = <}) . n=l  Thus,  n  free.  This' i s i m p o s s i b l e .  Hence  In e i t h e r  X  must be compact.  M ° 0  is  Q.E.D.  117. BIBLIOGRAPHY  [l]  L. V. A h l f o r s , Complex A n a l y s i s . Inc., New York, 1953.  [2]  L. v . A h l f o r s and L. S a r i o , Riemann S u r f a c e s . Univ. Press," New J e r s e y , i960.  [33 [4 3  McGraw-Hill Book Company,  L. B e r s , On Rings o f A n a l y t i c F u n c t i o n s .  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