UBC Theses and Dissertations

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UBC Theses and Dissertations

Variational problems with thin obstacles Richardson, David 1978

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VARIATIONAL PROBLEMS WITH THIN OBSTACLES B.CA.., U n i v e r s i t y o f B r i t i s h C o l u m b i a , 1971 M . S c , U n i v e r s i t y o f T o r o n t o , 1972 A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF THE FACULTY OF GRADUATE STUDIES Department of M a t h e m a t i c s U n i v e r s i t y o f B r i t i s h C o l u m b i a V a n c o u v e r , B r i t i s h C o l u m b i a We a c c e p t t h i s t h e s i s as c o n f o r m i n g t o t h e r e q u i r e d s t a n d a r d THE UNIVERSITY OF BRITISH COLUMBIA A u g u s t , 1978 ( c * ) D a v i d R i c h a r d s o n , 1978 by D a v i d R i c h a r d s o n DOCTOR OF PHILOSOPHY i n In presenting this thesis in partial fulfilment of the requirements for an advanced degree at the University of British Columbia, I agree that the Library shall make it freely available for reference and study. I further agree that permission for extensive copying of this thesis for scholarly purposes may be granted by the Head of my Department or by his representatives. It is understood that copying or publication of this thesis for financial gain shall not be allowed without my written permission. Department of Mathematics The University of British Columbia 2075 Wesbrook Place Vancouver, Canada V6T 1W5 D a t e May 1, 197S A b s t r a c t I n t h i s t h e s i s the s o l u t i o n t o the v a r i a t i o n a l p r o b l e m of S i g n o r i n i i s s t u d i e d , namely: ( i ) Av = 0 i n ( i i i ) p — § o n 3fi ( i i ) v > ty on 3fi ( i v ) (v-ty) (p- - g) = 0 on 3ft where i s a domain i n R n , and v i s the u n i t i n n e r n o r m a l v e c t o r t o 8fi . I n the c a s e n = 2 a r e g u l a r i t y theorem i s proved'. I t i s shown t h a t i f tj; & C1,a(dtt) , g € L i p a(8fi) t h e n v e C1,a(dQ) i f a < 1/2 . An example i s g i v e n t o shown t h a t t h i s r e s u l t i s o p t i m a l . The method o f p r o o f r e l i e s on t e c h n i q u e s of complex a n a l y s i s and t h e r e f o r e does n o t e x t e n d t o h i g h e r d i m e n s i o n s . F o r n > 2 the c a s e where Q, i s unbounded, o r e q u i v a l e n t l y , where ty i s unbounded i n a n e i g h b o u r h o o d o f some p o i n t o f dQ. i s c o n s i d e r e d . T h i s i s a s i t u a t i o n where known e x i s t e n c e theorems do n o t a p p l y . Some s u f f i c i e n t c o n d i t i o n s f o r the p a i r (ty,g) a r e d e r i v e d t h a t w i l l e n s u r e the e x i s t e n c e o f a s o l u t i o n i n t h i s c a s e , t h e r e b y e x t e n d i n g some r e s u l t s o b t a i n e d by A. B e u r l i n g and P. M a l l i a v i n i n the two d i m e n s i o n a l c a s e . The p r o o f i n v o l v e s a v a r i a t i o n a l p r o b l e m i n a H i l b e r t space a n a l o g o u s t o t h e one c o n s i d e r e d by B e u r l i n g and M a l l i a v i n , and some p o i n t w i s e e s t i m a t e s o f R i e s z t r a n s f o r m s . ( i i i ) C o n t e n t s N o t a t i o n ( i v ) Acknowledgement (v) C h a p t e r 0 - I n t r o d u c t i o n 1 0. P r e l i m i n a r i e s 1 1. A p r o b l e m i n mechanics 3 2. The r e g u l a r i t y o f the s o l u t i o n 5 3. A S i n g u l a r Case: E x i s t e n c e 6 C h a p t e r I - The R e g u l a r i t y o f the S o l u t i o n 10 0. R e d u c t i o n s 10 1. C o n t i n u i t y o f the d e r i v a t i v e 11 2. H i g h e r r e g u l a r i t y 16 3. The l o w e r d i m e n s i o n a l ( t h i n ) i n t e r i o r o b s t a c l e 25 4. An example 27 5. A one d i m e n s i o n a l v a r i a t i o n a l p r o b l e m 30 6. Remarks, p r o b l e m s , and guesses 33 C h a p t e r I I - A S i n g u l a r Case: E x i s t e n c e 36 0. R e s t a t i n g t h e p r o b l e m 36 1. C o m p a t i b i l i t y f o r the p a i r (^,c) 38 2. C o m p a t i b i l i t y f o r t h e p a i r ( N ( r - g ) , c ) 68 3. The S m a l l e s t Superharmonic 69 REFERENCES 71 Notation Unless otherwise specified, we have used the notation that has become standard in the theory of partial differential equations and we refer to Adams [1] page 1, for definitions. (v) Acknowledgement I w i s h t o e x p r e s s my s i n c e r e g r a t i t u d e t o P r o f e s s o r L e n n a r t C a r l e s o n , f o r t h e many hours he sp e n t t e a c h i n g me M a t h e m a t i c s . I w o u l d l i k e t o thank P r o f e s s o r s D a v i d D r a s i n and John F o u r n i e r f o r t h e i r i n t e r e s t and encouragement. I am a l s o g r a t e f u l t o U.B.C, the Swedish Academy o f S c i e n c e s , and the C a n a d i a n - S c a n d i n a v i a n F o u n d a t i o n f o r f i n a n c i a l s u p p o r t . .1 Introduction §0. Preliminaries Let fi be a bounded domain i n R n with smooth boundary T . If *K Y) i - s a n integrable function on T , we may solve the c l a s s i c a l D i r i c h l e t problem for fi , and we w i l l denote by H(^) the harmonic extension of ^ to : that i s A H( I J J ) =0 i n fi , H(<JJ) = i> on T . I f g(y) i s an integrable function on T such that g(y)dy = 0 , then we can solve the Neumann problem with boundary h condition g , namely, there i s a harmonic function N(g)(x) on fi such that A N(g) =0 i n fi , = g on Y . Here -— denotes p a r t i a l d i f f e r e n t i a t i o n i n the d i r e c t i o n v , dV where v i s the unit inner normal vector to fi . The function N(g)(x) i s , of course, only determined up to a constant, so denote by N^(g)(x) the s o l u t i o n to the Neumann problem that vanishes at some fixedd point x e fi . o The solutions of the Neumann and D i r i c h l e t problems are unique so that one cannot specify both the boundary values and the normal d e r i v a t i v e of a harmonic function. However, i n t h i s thesis we s h a l l c o n s i d e r t h e f o l l o w i n g " h y b r i d " p r o b l e m , known i n the l i t e r a t u r e as S i g n o r i n i ' s p r o b l e m . P r o b l e m (1) I f if and g a r e i n t e g r a b l e f u n c t i o n s on V f i n d v such t h a t ( i ) v >_ on Y 9v 9v -( i i ) 777 <_ g on Y where -777 9v _ 3 H(v) 3v ~ 3v ( i i i ) (v-i|») (|TJ - g) = 0 on r Remark 0.1: F o r any f u n c t i o n H(x) , harmonic i n fi , we have t h a t 3H 3v = 0 , so i n o r d e r f o r ( i i ) t o be s a t i s f i e d i t i s n e c e s s a r y t h a t t h e mean v a l u e g > 0 I f 3v g = 0 we have the Neumann pr o b l e m ; — = g on r I f t he ha r l i f n t t h e e h a r m o n i c e x t e n s i o n of, ty.ast©. f^nxhast-a f i n i t e B i r . i g h l e t i n t e g r a l : V(H(ip)) I < * t h e n p r o b l e m (1) has a v a r i a t i o n a l f o r m u l a t i o n . P r o b l e m (!') F i n d v e' K such t h a t v m i n i m i z e s the f u n c t i o n a l J ( u ) = v ur + 2 gu 1 n where K = {u e W ' (fi) such t h a t u l 3 Here u sense . >_ ty i s t o be i n t e r p r e t e d i n an a p p r o p r i a t e r Remark 0.2: The a s s u m p t i o n t h a t Av = 0 i n fi can be r e p l a c e d by t h e a s s u m p t i o n t h a t Av = f f o r some f i x e d f . Indeed, i f f e L P ( f i ) p > > l e t ty e W^"' 2(fi) — n+z 0 be the s o l u t i o n o f the D i r i c h l e t p r o b l e m L\ty = f . L e t ty = ty , and g.. = g + - j ^ - t h e n i f v i s a 1 1 oV s o l u t i o n t o p r o b l e m (1) t h e n v ^ = v + ty i s a s o l u t i o n t o ( i ) A v 1 = f i n ( i i ) v ^ >^  ty on r ( i i i ) _^1 <_ g on T 9v 9v ( i v ) ( V 1 T * 1 ) - g±) = 0 on T . The c o m p a t i b i l i t y c o n d i t i o n §1. A P r o b l e m i n Mechanics g > 0 can be r e s t a t e d as r F o r a complete p h y s i c a l i n t e r p r e t a t i o n o f p r o b l e m (1) and i t s v a r i a n t s we r e f e r t o Duvaut and L i o n s [11]. However, the f o l l o w i n g s p e c i a l c a s e i s e l e m e n t a r y and o f i n t e r e s t i n i t s e l f . Suppose n = 2 and l e t c T be an open s u b s e t o f T . L e t ty(y) be a c o n t i n u o u s f u n c t i o n on Y such t h a t ty < 0 i n a n e i g h b o u r h o o d o f . L e t v be t h e s m a l l e s t c o n t i n u o u s s o l u t i o n t o the problem: ( i ) v = 0 on YQ ( i i ) v >_ ty on T ( i i i ) |J< o on r \ r Q . I n v a r i a t i o n a l f o r m , v m i n i m i z e s the D i r i c h l e t i n t e g r a l i n f |Vu| 2 12 + where = {u e: W^ ' (fi) such t h a t u ty } and where r 12 12 i W ' (fi) c o n s i s t s o f t h o s e f u n c t i o n s u e-: W ' (fi) such t h a t u = " ' r o We may i n t e r p r e t t h i s p r o b l e m m e c h a n i c a l l y by t h i n k i n g o f the s o l u t i o n v as a homogeneous membrane spanned by V , a d h e r i n g t o , and f o r c e d t o l i e above an o b s t r u c t i o n ty on rji;^ , a c h i e v i n g t h i s w i t h a minimum expense of e n e r g y , i . e . a t t a i n i n g an e q u i l i b r i u m p o s i t i o n . I n t u i t i v e l y , t h e n , ( w i t h t h e e x c e p t i o n o f a f i n i t e number o f p o i n t s o f , where the t a n g e n t i a l d e r i v a t i v e o f v w i l l have a jump d i s c o n t i n u i t y where the membrane " l i f t s o f f " ) one w o u l d e x p e c t t h a t v w i l l have a c o n t i n u o u s t a n g e n t i a l d e r i v a t i v e i f ty has a c o n t i n u o u s d e r i v a t i v e . I n the 5 n e x t s e c t i o n we show t h a t t h i s i s , i n f a c t , the c a s e . M o r e over, the modulus o f c o n t i n u i t y o f the t a n g e n t i a l d e r i v a t i v e o f v i s the same as t h a t f o r the d e r i v a t i v e o f ty up t o H o l d e r exponent 1/2 . More g e n e r a l l y , the q u e s t i o n t o be t r e a t e d i n s e c t i o n I i s t h e f o l l o w i n g : I f ty and g a r e a r e smooth i n p r o b l e m (1) does i t f o l l o w t h a t the c o r r e s p o n d i n g s o l u t i o n v i s smooth? §2. The R e g u l a r i t y o f the S o l u t i o n The r e g u l a r i t y o f s o l u t i o n s o f v a r i a t i o n a l problems w i t h i n e q u a l i t i e s on the boundary, as d e s c r i b e d above, o r more g e n e r a l l y , w i t h i n e q u a l i t i e s p r e s c r i b e d on l o w e r d i m e n s i o n a l s u r f a c e s , has been s t u d i e d by s e v e r a l a u t h o r s i n r e c e n t y e a r s . Hans Lewy t r e a t e d the c a s e i n v o l v i n g t h e D i r i c h l e t i n t e g r a l [ 1 8 ] , [19] and J . N i t s c h e d e a l t w i t h t h e m i n i m a l s u r f a c e p r o b l e m i n [ 2 4 ] . S u b s e q u e n t l y t h e p r o b l e m was s t u d i e d by K i n d e r l e h r e r [ 1 5 ] , B r e z i s [ 4 ] , [ 5 ] , and most r e c e n t l y by F r e h s e [ 1 2 ] , [13] and C a f f a r e l l i [ 8 ] . I n 1968 Shamir [25] gave an example where the i n i t i a l d a t a ty and g a r e smooth b u t where th e t a n g e n t i a l d e r i v a t i v e o f t h e s o l u t i o n v was o n l y L i p 1/2 . F r e h s e ( [ 1 3 ] , 1977) s t u d i e d a g e n e r a l c l a s s o f v a r i a t i o n a l problems and under the a s s u m p t i o n t h a t t h e i n i t i a l d a t a a r e smooth, p r o v e d t h a t t h e modulus o f c o n t i n u i t y w(6) o f t h e t a n g e n t i a l d e r i v a t i v e s o f t h e s o l u t i o n v was 6 O ( | l o g 6| q) . R e c e n t l y , C a f f a r e l l i [ 8 ] used a d i f f e r e n t t e c h n i q u e t o show t h a t under s u i t a b l e c o n d i t i o n s on ij; , the s o l u t i o n i s i n C 1 , a ( r ) f o r some a > 0 . However, f o r n = 2 , and f o r p r o b l e m ( 1 ) above, we can g i v e a complete answer t o the q u e s t i o n o f r e g u l a r i t y . I n d e ed, i f 1 a 4> t£• C ' (T) , g re L i p a ( r ) and v i s the s o l u t i o n t o p r o b l e m ( 1 th e n v <e C 1 , a ( r ) i f a < 1/2 , and v e C 1 , 1 / 2 ( r ) i f a > 1/2 §3. A S i n g u l a r Case: E x i s t e n c e The q u e s t i o n o f e x i s t e n c e o f a s o l u t i o n t o p r o b l e m ( 1 ) has been s t u d i e d by Duvaut [ 1 0 ] , B r e z i s [ 4 ] , [ 5 ] and F i c h e r a [ 1 4 ] . I n o r d e r t o e n s u r e the e x i s t e n c e o f an e x t r e m a l i t s u f f i c e s t o assume H(IJJ) k W 1 ' 2 ^ ) and g e W - 1 ^ 2 ' 2 ( r ) . Suppose now t h a t re L^ ' ( r ) , f i x y^ e T , and assume t h a t ^ i s c o n t i n u o u s l y d i f f e r e n t i a b l e on T \ {JQ} b u t unbounded i n a n e i g h b o u r h o o d o f YQ re T . C e r t a i n l y t h e r e i s a s o l u t i o n H(IJJ) t o the D i r i c h l e t p r o b l e m , and w i l l be a measurable f u n c t i o n on T \ ( Y Q } • Thus i f g i s a measurable f u n c t i o n on T \ ( Y Q ) we may l o o k f o r a s o l u t i o n t o p r o b l e m ( 1 ) f o r t h e p a i r ^ and g . However, i f the D i r i c h l e t i n t e g r a l o f H(IJJ) i s i n f i n i t e , the r e f o r m u l a t i o n as p r o b l e m ( 1 ' ) may n o t make sense s i n c e the s e t K may be empty] so i t i s c l e a r t h a t c e r t a i n e x t r a a s s u m p t i o n s on \p and g w i l l be n e c e s s a r y t o e n s u r e the e x i s t e n c e 7 o f a s o l u t i o n . The p r o b l e m t r e a t e d i n s e c t i o n I I i s t o d e r i v e some c o n d i t i o n s on t h e f u n c t i o n s ip and g t h a t w i l l g u a r a n t e e the e x i s t e n c e o f a s o l u t i o n . F o r s i m p l i c i t y we c o n s i d e r the c a s e where 0, i s t h e h a l f s p a c e , R ^ + 1 = fx ' <e R N + 1 , X. = (x,y) , x e R N , y > 0} , YQ i s the p o i n t a t i n f i n i t y , and Y i s the h y p e r p l a n e R N . I f if) i s a l o c a l l y i n t e g r a b l e f u n c t i o n d e f i n e d on R N s u c h t h a t (0.1) R .100. n 1+1 x n+1 dx < 0 0 nH~X t h e n the harmonic e x t e n s i o n o f t o R + i s g i v e n by t h e P o i s s o n i n t e g r a l i K x , y ) = P * M x ) = — n R y M t )  , i ,..2. 2(n+l)/2 n (| x-1 j +y ) d t I f g (x) i s a l o c a l l y i n t e g r a b l e f u n c t i o n on R and s a t i s f i e s (0.2) i n f —CO< Q<00 < oo R n 1+ t t h e n we can e x p l i c i t l y s o l v e t h e Neumann p r o b l e m w i t h boundary c o n d i t i o n g ( x ) : N(g)(x,y) = i 1 l o g R |1 - ^ | g ( t ) d t n = 1, z = x +iy Assume t h a t 4Kx) s a t i s f i e s (0.1) and t h a t g(x) s a t i s f i e s (0.2) t h e n p r o b l e m (1) becomes: P r o b l e m (1") F i n d v(x) s a t i s f y i n g (0.1) s u c h t h a t ( i ) v(x) >_ I(J(X) on R n ( i i ) f r (x,0) < g(x) on R n d y — ( i i i ) (v(x)-iKx))(|r (x,0)-g(x)) = 0 on R n d y ! 8v, » x , . . , . . -, . d V . . where —(x,0) xs the p o m t w i s e l x m x t l x m —— (x,y) , 9 y y=0+ 8 7 w h e r e v e r t h i s l i m i t e x i s t s , and where v(x,y) i s the P o i s s o n i n t e g r a l o f v(x) . D e f i n i t i o n . The p a i r (ifsg) i s s a i d t o be c o m p a t i b l e i f t h e above p r o b l e m has a s o l u t i o n . Smoothness a l o n e i s n o t enough t o gu a r a n t e e c o m p a t i b i l i t y . I n d e e d , l e t iKx) be an unbounded, smooth, n o n n e g a t i v e f u n c t i o n s a t i s f y i n g (0.1) t h e n the o n l y f u n c t i o n s a t i s f y i n g ( i ) and ( i i ) above f o r the p a i r (^ ,0) i s v = +°° . 9 T h i s p r o b l e m o r i g i n a t e s w i t h two i m p o r t a n t lemmas of B e u r l i n g and M a l l i a v i n ( [ 2 ] Lemmas 3 and 5) where c o m p a t i b i l i t y o f the p a i r ( f , c ) i s e s t a b l i s h e d f o r n = 1 , f e L i p 1(R) and c > 0 i s any p o s i t i v e c o n s t a n t . W i t h some a l t e r a t i o n s t h e i r r e s u l t s e x t e n d t o h i g h e r d i m e n s i o n s . Once c o m p a t i b i l i t y has been e s t a b l i s h e d f o r the p a i r (i|>,c) f o r a l l c > 0 , i t w i l l be shown t h a t (^,g) w i l l be c o m p a t i b l e f o r a s u i t a b l e c l a s s o f f u n c t i o n s g(x) . 10 Chapter I In t h i s chapter we prove a r e g u l a r i t y theorem for the so l u t i o n of problem (1) mentioned i n the introduction i n the case n = 2 . The technique used i n the proof involves methods of complex analysis and therefore does not extend to higher dimensions. In this d i r e c t i o n we r e f e r the reader to C a f f a r e l l i [8] and Frehse [13] where d i f f e r e n t techniques are used to treat the general case but weaker r e g u l a r i t y conditions are obtained from more general hypotheses. §0. Reductions 2 2 We s h a l l assume that <= R = { Z & R s z = x+iy, y > 0} , that the semi di s c (|z| < 1, y > 0} <= fi and that the i n t e r v a l 2 [-1,1] c 3fi . The general case where 3fi i s , say, C , can obviously be reduced to this case by a conformal mapping. Let 1 a ty e. C ' (T) and g e L i p a(T) . To avoid a te c h n i c a l d i f f i c u l t y we s h a l l assume that the points -1 and 1 are contained i n the set {y & r : v(y) > ty(y)) . For z e {|z| < 1, y > 0} l e t v ^ z ) = v(z) + 1 log — - — g(t)dt and extend v (z) to the unit disc by -1 Iz-tI 8 V 1 r e f l e c t i o n i . e . v (z) = v., (z) . Since - — <_ 0 on (-1,1) we 1 -L °y conclude that Vj. i s superharmonic i n the unit disc |z| < 1 . 11-Thus by the Riesz representation theorem v^z) = log S Iz-t du(t) - H(z) = u(z) - H(z) where dy is a positive measure carried on a compact set Sc.(-1,1) and H(z) is harmonic in |z| < 1 . Since v(x) > ^ (x) , we have that for x re (-1,1) u(x) = log dy(t) = v(x) + -S |x-t| log -1 Ix-tl g(t)dt + H(x) > ijj(x) + -— IT log -1 |x-t| g(t)dt + H(x) = f(x) moreover S <= { x e (-1,1) : v(x) = IJJ(X) } = {x e (-1,1) : u(x) = f(x)} Our assumptions on and g imply that the function f defined 1 a above is C ' (A,B) where (A,B) is an interval containing S in (-1,1) . Thus the problem of determining the regularity of v(x) is equivalent to the problem of determining the regularity of u(x) . We start with a proof of Hans Lewy's result that includes a new proof of continuity of the derivative. §1. Continuity of the Derivative Theorem 1.1 (Hans Lewy). Let dy(x) be a positive measure of 12' compact support S c (A,B) , and l e t u(x) be i t s l o g a r i t h m i c  p o t e n t i a l . I f there e x i s t s a. f u n c t i o n f e C^(A,B) such that f ( x ) £ u(x) on (A,B) and f ( x ) = u(x) on S , then u(x) €• C 1(A,B) . Proof. Since u(x) i s c e r t a i n l y continuous, (Evan's law), the set {x:u(x) > f ( x ) } i s open and contained i n CS which t h e r e f o r e c o n s i s t s of a countable union of i n t e r v a l s (a ,b.) . 1. Existence of d e r i v a t i v e s : Let (a,b) be such a complementary i n t e r v a l . Since u(x,y) = u(z) = l o g — - — dy(t) i s harmonic i n z-t c \ ((-°°>a] u [b,°°)) , u'(x) i s c e r t a i n l y continuous on (a,b) Note that f o r x e (a,b) u"(x) ^ ^ ^ > 0 , so that u(x) i s convex on (a,b) S (x-t) so u'(x) i s i n c r e a s i n g on (a,b) Therefore u . ( x ) < u(b)-u(x) = f ( b ) - u ( x ) < f ( b ) - f ( x ) — b-x b-x — b-x Hence l i m u' (x) = 4^_(b):-< f'(b) x->b" d 5 r ~ " ' S i m i l a r l y l i m u'(x) = x-*a+ 13. We assert now that —,(b) exists and equals f(b) . dx + If there is an interval [b,x], x > b , belonging to S , then since u(x) = f(x) there, the assertion is trivial. Suppose then that b is the limit of complementary intervals (a^,b^) e CS , b. > a. > b . By the monotonicity of u'(x) on (a.,b.) , we have for x e (a ±,b ) , f'(a±) <_ u' (x) < f (b±) . Therefore (x-a i)f'(a i) < u(x) - u(a ±) < (x^a )f'(b ) so that f(a.)+f'(a,)(x-a.)-f(b) , . (x-a.) H = i 1 < u(*>-^b> < H + (f(b.)-f'(a.)) x-b — x-b — (x-b) l l Now H = £ ^ X ) _ K £ (~ b^ + °(1) by the continuity of f'(x) . So as x, a., b. tend to b , x § CS , we have lim u ( x ) ~ " ( b ) = f'(b) . 1 1 x+b+ X" b xe S This is obvious i f x tends to b through S . By the same argument (a) = f'(a) . dx 2. Continuity In CS , u' exists and is continuous because u is harmonic in C \ S . At the endpoints of a complementary interval (a,b) we have that 4^ -(a.) = f'(a) < 4^4-(a) and dx — dx 4^_(b) < f'(b) = 4^4-(b) . At any point s on the boundary of S dx — d x T that i s not an endpoint of a complementary i n t e r v a l we may use the above argument to conclude that u'(s) = f'(s) . Thus u'(x) ex i s t s everywhere except at the endpoints of the complementary i n t e r v a l s where only ^ + and are known to e x i s t . Suppose (-T,0) e CS . We remark again that u(x,y.) i s harmonic i n C\ ((-°°,-T] U [O, 0 0)) , u(x,y) = u(x,-y) , i . e . u(x,y) i s even with respect to the x-axis. So since -jr^ -(x,y) dy du i s also harmonic i n C \((-°°,-T] U [0,°°)) , — (x,y) i s odd with °y 3 u respect to the x-axis, so — (x,0) =0 on (-T,0) . We have the dy representation |r-(x,0) = l i m — 9 y 6=0 17 t>6 u(x+t)+u(x-t)-2u(x) , 2 dt t 2 Since u(x+t) + u(x-t) - 2u(x) <_ t u M(£) , i f t i s s u f f i c i e n t l y small, £ e (x-t, x+t) c ( - x,0) , there i s no d i f f i c u l t y i n the convergence of the i n t e g r a l for x e (-T,0) . Suppose now that 4^-(0) < 4^+C0)->-f1 (0) Let dx dx A = ^ + ( 0 ) " 4^.(0) and choose e so that f'(5) > f' (0) - 4 f o r dx dx 2 0 < 5 < e Let <S < -|- min ( e , x ) . Then 15 G = u(t)+u(-t)-2u(0) dt = t><5 + = c + J ° J (u(t)-u(0))-(u(0)-u(-t)) dt t>£ 6 > C + — o (f(t)-f(0))-(u(0)-u(-t)) 2 > Co + | log(f) , by the mean value theorem and the monotonicity of u'(x) on (-t,0) Choose x such that 26 > x > 6 then u(-x+t)+u(-t-x)-2u(-x) dt t>6 (u(-x+t)-u(t))+(u(-x-t)-u(-t))-2(u(-x)-u(0)) dt + t>6 t>6 u(t)+u(-t)-2u(0) 2 t ^ " f C l + Co + \ l 0 ^ f > Since (-x-6,-x+6) c (-x,0) , and u(x) is convex on (-x,0) u(-x+t)+u(-x-t)-2u(-x) dt > -t>0 This implies |^ -(-x,0) > 0 where contradiction. 2C1 + C q + | log(|) > 0 for 6 small enough, -x e (-x,0) and we have a The continuity of u'(x) now follows easily. If xe S 16/ and x. + x , x. e S t h e n u'(x.) = f ' ( x . ) -> f ' ( x ) . I f x, e CS , x. + x , t h e n t h e r e a r e i n t e r v a l s (a.,b.) i n CS , 1 x i x s . t . x. £ (a.,b.) . However, as x. ->- x b o t h a. and b. -> x , x x x i x x and s i n c e f ' ( a ± ) < u ' ( x ± ) <. f 1 (b ) , b o t h f * ( a ) and f ' ( b ) tend t o f ' ( x ) . The c o n t i n u i t y o f u'(x) on CS i s t r i v i a l . § 2. H i g h e r r e g u l a r i t y As b e f o r e we assume dy i s a p o s i t i v e measure o f compact s u p p o r t S , and u ( x ) i s i t s l o g a r i t h m i c p o t e n t i a l . We suppose 1 a too t h a t t h e r e e x i s t s f e C ' (A,B) , 0 < a <_ 1 , such t h a t u ( x ) >^  f (x) on (A,B) and u ( x ) = f ( x ) on S c (A,B) . Theorem 1.2. i ) I f a < 1/2 t h e n u ( x ) e C 1 , a ( R ) and i i ) I f a > 1/2 t h e n u ( x ) e C 1 , 1 / 2 ( R ) . Remarks. A t t h e c r i t i c a l i n d e x a = 1/2 , Theoremlr2(i) i m p l i e s 1 8 t h a t u ( x ) e C (R) f o r a l l 6 < 1/2 . We o b s e r v e a l s o t h a t m(x) = d y ( x ) i s t h e H i l b e r t t r a n s f o r m o f u'(x) . So as a consequence o f Theorem 1.2, and t h e f a c t t h a t t h e H i l b e r t t r a n s f o r m i s bounded on L i p a , we may r e d e f i n e m(x) on a s e t o f measure z e r o t o be a c o n t i n u o u s f u n c t i o n i n L i p a . T h e r e f o r e S° c {x: m(x) > 0} i s a non-empty u n i o n o f open i n t e r v a l s and y(9 S ) = 0 . P r o o f . I t s u f f i c e s t o show t h a t u ' ( x ) i s L i p a i n a complementary i n t e r v a l (a,b) . Once t h i s has been e s t a b l i s h e d , f o r any X l < X 2 6 C S ' x l 6 ( a i ' b i ) » x 2 e ( a 2 ' b 2 ^ t h e n u'(x 1)-u ,(x 2)I < |u'(x1)-u'(b1)|+| u'(b1)-u'(a2)|+|u'(a2)-u'(x2) a a For x^ , x2 e S this relation is trivial. Let r and I be the upper and lower bounds for S in (A,B) i.e. S c [£,r] c (A,'B) . By Theorem 1, both u and its derivative u' coincide with f and f' at the points r and I . Lemma 1.1. Let x^ e u (r,°°) and x 2 e [£,r] then |u' (x )-f'(x ) | <. C |x - x w h e r e 3 = a if a < 1/2 , (3 = 1/2 X z ot j_ 2. if_ a > 1/2 . To avoid repetition of arguments we defer the proof of Lemma 1.1 until later. Without loss of generality we suppose (-p,p) is a complementary interval. Since u(x,y) = u(z) = log 1 du(t) is harmonic in z-t V = C\ ( ( — , - p ] u [ P , » ) ) then, in V , 9x" ( x ' y ) = (t-x) du(t) is also harmonic. Hence for Js (x-t)2+y2 "18 x e (-p,p) , u'(x) can be represented as the Poisson i n t e g r a l of i t s values on ( -°°,-p]u [p, 0 0) • We do t h i s e x p l i c i t l y . For z = x + i y i n V make a conformal mapping of V onto the upper h a l f of the u) = £ + in plane i n such a way that p -> 0 , 0 -> i , and -p -> i°°: e x p l i c i t l y u> = J^~7 frz-p z+p A r e f l e c t i o n i n the i n t e r v a l ( - p , p ) o f the z-p l a n e c o r r e s p o n d s t o a r e f l e c t i o n o f the c o n f o r m a l map i n the p o s i t i v e , i m a g i n a r y n a x i s o f the u - p l a n e . S e t V(u>) = | ^ ( z ) . S i n c e | ^ ( x , 0 + ) = | ^ ( x , 0 _ ) , t h e n V(£,0) = V ( - ? , 0 ) . F o r x e (-p,p) , u s i n g the P o i s s o n i n t e g r a l r e p r e s e n t a t i o n f o r V (O J ) , we have V P+X 7T V p + X J ^ 2 + p - X TT p+x V ( g ) d g ( p + x ) g + p - x = 2 / 2 2 p -x 2 (p+x) 5 +P-x 2 / 2 2 = — y p -x + 2 / 2 2 . = -7 P -x U j + I j By a change of v a r i a b l e s s = 'p;>'f — — + V l " 5> I =1 1 2 u' (s) ds (s-x) 12 • 2' V s -p- • Similarly 19 I = u'(-s) ds_ (s+x) Vs - p Therefore u*(x) * J P - 2 u' (s.) u' (-s) ds s-x s+x f l 2 v s -p Suppose now that 0 < x < p , and denote dv(s) = ds J s 2 - f note that We (a) u '(p)-u'(x) _ 1 / 2 2 V p -x u ' (p)-u'(s) ( u '(p)-u'(-s) d v s-x s+x and (b) u' (-x)-u '(-p) _ 2 2 p -x u'(s)-u ' ( -p) + u'(-s)-u ' ( -p) s+x s-x dv(s) Observe that the integrals appearing in (a) and (b) converge absolutely for 0 < x < p , and since u'(x) is monotone increasing on ( - p , p ) both terms are positive. Adding (a) to (b) and performing some sleight of hand yields the following identity: u' (p)-u'(x) + u' (-x)-u '(-p) / 2 2 / 2 2 / P -x V p -x u ' ( p ) - u ' ( - p ) dv(s) + ^ S+X TT u'(p)-u'(s) 2 2 s -x dv(s) formula continues. 20 + 2x u ' ( - s ) - u ' ( - p ) 2 2 s - x dv(s) - J l + J 2 + J 3 E s t i m a t e s . a a J : By a s s u m p t i o n 0 < u ' ( p ) - u ' ( - p ) = f ' ( p ) - f ' ( - p ) <_ 2 p so t h a t 0 l + a T ^ 2 a J i < —Z~ P 1 — TT < c £-p p (s+x)Vs - p / 2 2 J ^ : We w r i t e J as the sum 2x TT i l i s i z H l i s l + 2 x Z Z TT s -x U'y-!,(8)dv(8)4S U ' ( p ) - U ' ( 8 ) d v ( s ) Z Z TT Z Z s - x - ' s - x r = I , + I 2 + I 3 S i n c e f ' ( s ) = u ' ( s ) on S we can e x p r e s s t h e f i r s t i n t e g r a l I as a sum o v e r the complementary i n t e r v a l s c o n t a i n e d i n ( p , r ) , and i n t e g r a t e by p a r t s : 1 TT (a±,b ) c CS a ± > p b.<r l ( f ' ( s ) - u ' ( s ) ) - ds (s I 2, j l 2 -x X/s - p J 21 2 x f ( s ) - u ( s ) u (a.,b.) c 'CS ,2 2. [~2 2 'a. x i (s -x Vs -p X a^>p a. l ( f ( s ) - u ( s ) ) -A (- 0'^j_^)^s . v . 2 " 2.[2=~2-(sz-xzi/s 3p 2 s -x J s -p But f ( b . ) = u(b.) , f ( a . ) = u ( a . ) , a l s o u ( s ) > f ( s ) on (a.,b.) X X X X X X and ds (s I 2. 1 2 2 -x X/s -p < 0 , so t h a t 1 ^ i s n e g a t i v e . T h e r e f o r e J 2 < T 2 + 1 3 -1. a > 1/2 We o b s e r v e t h a t u' (s) i s i n c r e a s i n g on [r,°°) so t h a t u ' ( p ) - u ' ( s ) < u ' ( p ) - u ' ( r ) < ( r - p ) a . Thus 2x 3 TT u ' ( p ) - u ' ( s ) ds s -x . / 2 2 V s -p 2 x , . a < — ( r - p ) — TT ds , 2 2 ,/~2 2 r (s -x Vs -p < C ( r - p ) a - 1 / 2 1/2 ds ( r - " ) a - 1 / 2 c ~ 73/2 - C _ £ I 7 2 ^ ^ 7 2 s i n c e a > 1 / 2 ' (s-p) p p s i n c e by a s s u m p t i o n i ~ r <_1 . 2 2 I „ = — 2x IT u ' ( p ) - f ' (s) ds < 2x - ( S 2 - x 2 ) n ~ 2 - * p y s -p p , N a - l / 2 , ( s - p ) ds Js+p ( s 2 - x 2 ) 1/2 ds , ,3/2-a - 1/2 ( s - p ) p < C ( r - p ) a - 1 / 2 < C 1/2 So f o r a > 1/2 , J . < 2 1/2 P 2. a < 1/2, We i n v o k e Lemma 1.1, so f o r s e [r,°°) , u ' ( p ) - u ' ( s ) <C ( s - p ) '— a T h e r e f o r e J 2 < I 2 + I 3 < C x . N a - l / 2 ( s - p ) ds_ 2 2, p (s -x )Js+p ( p - x ) < C X ( p - x ) p 3/2 t a - 1 / 2 d t + c x 3/2 J t a - 3 / 2 d t ( P - x ) < C ( P - x ) -1/2 1/2 The v a r i o u s c o n s t a n t s a p p e a r i n g above depend o n l y on a and the H o l d e r c o n s t a n t a s s o c i a t e d w i t h f . The e s t i m a t e s f o r a r e s i m i l a r t o t h o s e o f J 2 . We a r e t h e r e f o r e a b l e t o o b t a i n the e s t i m a t e f o r u ' ( p ) - u ' ( x ) ( r e s p e c t i v e l y u ' ( - x ) - u ' ( - p ) ) 23 111, xd-1/2 J~2 2 a a < 1/2: u ' ( p ) - u ' ( x ) < C \iU + ~ ^ — < C (p-x) — 1/z — a P 2 a J 2_ 2 J 2 a > 1/2: u * ( p ) - u ' ( x ) < L p x + L p x p < C / p ^ — 1/2 — a P P Now f o r x e (-p,p) n o t i c e (x) = 6 4^ V> 0 So u "(x) i s convex on (-p,p) . Hence u"(x) a t t a i n s i t s minimum a t a p o i n t £ e (-P>P) o r u " ( x ) i s monotone on the whole i n t e r v a l . I n e i t h e r c a s e we have t h a t f o r x^ < x^ , x^ e (~P,p) : u ' ( x 2 ) - u ' ( x 1 ) < [ u ' ( p ) - u ' ( p - ( x 2 - x 1 ) ) ] + [ u ' ( - p + ( x 2 - x 1 ) ) - u ' ( - p ) ] ^ C a ( V V 1 / 2 • I n o r d e r t o complete the p r o o f o f Theorem 1.2, we r e t u r n t o t h e p r o o f o f Lemma 1.1. S i n c e the t e c h n i q u e i s e s s e n t i a l l y the same as t h a t employed above we o n l y s k e t c h t h e p r o o f . To r e c a l l t he s i t u a t i o n , we have S c [£,r] c (A,B) , e ( - o o 9 £ ) u (r,°°) and x 2 e [^>r] a n ( i w e w i s h t o show | u' ( x ^ ) - f 1 ( x 2 ) | <_ | x-^_x2 I 0 1 • Suppose t h a t x.^  e (-°°,£) and H = 0 . We can a l s o assume x 2 = 0 s i n c e f o r any x e [&,r] | f ' ( x ) - f ' ( 0 ) | = |f'(x)-u'(0)| < | x | a by a s s u m p t i o n . We n o t e t h a t u' (z) i s a harmonic f u n c t i o n i n C\ [0,°°) w i t h u ' ( x , y ) = u'( x , - y ) . U s i n g the c o n f o r m a l mapping co = /z of C\ [0,°°) t o the upper h a l f p l a n e co = 5+in , n > 0 , and the P o i s s o n i n t e g r a l r e p r e s e n t a t i o n we w r i t e : 24 u'(g 2)dg = 1 5 + X , 1^1 u' (t)dt (t+| X l|)/t u' (x) is increasing on (-°°,0) so that 0 < u'(0)-u ,(x 1) (u'(0)-u'(t))dt _ 0 Again since u'(x) is increasing on (r,°°) , u'(O)-u'(t) _< u'(0)-u'(r) for t e (r,°°) . Thus u'(O)-u'(t)dt ^ 1 u'(O)-u'(r)dt < 2 a-1/2 Now we write the integral J 0 as two integrals and employ the same estimates as those for 1^ and I above to obtain: u'(O)-u'(t) d t = 1 u'(0)-f'(t) d t + I f (t)-u'(t) I ( t + I ^ D / t * J ( t+lx^)/? 11 Q ( t+lx^)/? a-1/2 dt C x. a 1 1 ia-1/2 2 r (l-2a) C r a a-1/2 i f a < 1/2 i f a > 1/2 Thus 25 u ' ( 0 ) - u ' ( x 1 ) < { a 1 1 C x, a 1 1/2 i f a < 1/2 i f a > 1/2 where i s a c o n s t a n t d e p e n d i n g o n l y on f and a , n o t on r . T h i s c o m p l e t e s the p r o o f o f Theorem 1.2. Remark. As we o b s e r v e d e a r l i e r , a t t h e c r i t i c a l i n d e x a = 1/2 , 1 3 we o b t a i n u e C ' f o r a l l 3 < 1/2 . I f we impose the s t r o n g e r c o n d i t i o n t h a t f ' ( x ) s a t i s f i e s : m(f ',5)d6 ^ ,3/2 where w(f ' , 6 ) = sup sup | f 1 ( x ) - f ' (y,) | , t h e n i t f o l l o w s f r o m xeR |x-y|<6 1 1/2 t h e p r o o f o f Theorem 1.2 t h a t u e C ' I t i s u n l i k e l y , however, t h a t t h i s i s a n e c e s s a r y c o n d i t i o n b u t the a m b i g u i t y o f t h e c o i n c i d e n c e s e t makes t h i s d i f f i c u l t t o v e r i f y . §3. The Lower D i m e n s i o n a l I n t e r i o r O b s t a c l e I n t h i s p a r a g r a p h we s h a l l a g a i n c o n f i n e our remarks to the c a s e n = 2 , f o r the g e n e r a l c a s e we r e f e r t o F r e h s e [12] and C a f f a r e l l i [ 8 ] . 26 2 I f fi i s a bounded domain i n R , ^ ( z ) a c o n t i n u o u s l y d i f f e r e n t i a b l e f u n c t i o n d e f i n e d on fi , n e g a t i v e on 3fi , i t i s a c l a s s i c a l p r o b l e m t o d e t e r m i n e the r e g u l a r i t y o f t h e s m a l l e s t s u p e r h a r m o n i c v ( z ) w i t h v ( z ) >_ ip (z) i n fi , v ( z ) >_ 0 on 3fi . I t i s known t h a t i f <Kz) i s s u f f i c i e n t l y smooth, f o r example i f |A\p| i s bounded, t h e n v ( z ) has H o l d e r c o n t i n u o u s d e r i v a t i v e s [ 2 0 ] , [ 2 1 ] . Suppose now t h a t ^ ( x ) i s a c o n t i n u o u s f u n c t i o n d e f i n e d o n l y on R n fi , and n e g a t i v e on 3fi n R . L e t v ( z ) be the s m a l l e s t c o n t i n u o u s s u p e r h a r m o n i c , v ( z ) >_ 0 on 3fi and v ( x ) _> i|>(x) on R n fi t h e n we have the f o l l o w i n g consequence o f Theorem 1.2. l a Theorem 1.3. W i t h ^ ( x ) and v ( z ) as above, suppose i|)(x) e C ' (Rn ( i ) i f a < 1/2 t h e n v ( x ) e C 1 , a ( R n f i ) ( i i ) i f a > 1/2 t h e n v ( x ) e C 1 , 1 / 2 ( R n f i ) P r o o f : We have the r e p r e s e n t a t i o n f o r v ( z ) : v ( z ) = G ( z , t ) d y ( t ) = l o g — ^ - d p ( t ) - h ( z , t ) d u ( t ) Rnfi Rnfi >z fc' Rnfi - u ( z ) - H(z) where dp i s a p o s i t i v e measure o f compact s u p p o r t S c R n f i , and G(z,w) = l o g — - — - h(z,w) i s the Greens f u n c t i o n f o r fi . | Z - 0 ) | The f u n c t i o n h(z,w) i s harmonic i n z and ca f o r z, ca e fi and has bounded d e r i v a t i v e s o f a l l o r d e r s on compact s u b s e t s of fi . Thus H(z) = 2 h ( z , t ) d y ( t ) i s c e r t a i n l y a C f u n c t i o n S i n a n e i g h b o u r h o o d o f S . S i n c e v ( x ) >_.ty(x) we have t h a t f o r x e R n fi . u( x ) = l o g — - — d y ( t ) = v ( x ) + H(x) >_ i|>(x) + H(x) = f (x) I x - t I and S c { x e Rn fi : v ( x ) = Kx)} = ( x e R : u ( x ) = f(x)} . 1 rv. 1 a S i n c e ip(x) e C (Rn!!) i t f o l l o w s t h a t f e C (A,B) where (A,B) i s an i n t e r v a l c o n t a i n i n g S i n R . Theorem 1.3 now f o l l o w s f r o m Theorem 1.2. §4. An Example. We o b s e r v e d t h a t as an e l e m e n t a r y consequence of Theorem 1.2, S° i s non-empty and y(9S) = 0 . The r e g u l a r i t y o f u( x ) depends o n l y on the r e g u l a r i t y o f f ( x ) on the s e t S° . O b v i o u s l y i f f misbehaves on the s e t CS the r e g u l a r i t y o f u i s n o t a f f e c t e d . I n [18] Han Lewy p r o v e s t h a t i f f i s r e a l a n a l y t i c t h e n S c o n s i s t s o f a f i n i t e u n i o n o f i n t e r v a l s , b u t i n g e n e r a l the b i z a r r e n a t u r e o f the c o i n c i d e n c e s e t S makes i t d i f f i c u l t t o d e t e r m i n e i f the r e s u l t s o f Theorem 1.2 a r e s h a r p . However, t h e f o l l o w i n g e l e m e n t a r y example d e m o n s t r a t e s t h a t f ( x ) 1 1/2 can be a n a l y t i c b u t u ( x ) remains i n C ' ( R) . i . 28 Suppose f (x) ^ 0 i s even, smooth and concave on S . Here a g a i n we assume t h a t u ( x ) i s t h e l o g a r i t h m i c p o t e n t i a l o f a measure dy s u p p o r t e d on S c { f ( x ) = u ( x ) } , and u ( x ) >^  f ( x ) . S u r e l y u ( x ) i s even and s i n c e u ( x ) i s convex on CS , we i n f e r t h a t S c o n s i s t s o f e x a c t l y one c l o s e d i n t e r v a l [-T,T] , (see a l s o Lewy [ 1 9 ] ) . By Theorem 1.1, u ( x ) e C 1 ( R) and u ' ( x ) = f ' ( x ) f o r x e [T,T] . By a f a m i l i a r argument we have the r e p r e s e n t a t i o n , f o r x < - x : u'(x) = X -T u ' ( s ) ds ( s - x ) /S+X' T h e r e f o r e 0 < u' (-T)-U'(x) /-T-2 1 TT U ' ( - T ) - U ' ( S ) ds = -T ( S - x ) /s+T 1 + i TT -T By a s s u m p t i o n u'(-x) > 0 , and s i n c e u ' ( s ) i s d e c r e a s i n g on [-T,T] , u ' ( - x ) - u ' ( s ) > 0 f o r s e [-T,T] . A l s o u ' ( s ) i s n e g a t i v e f o r s > x . T h e r e f o r e u ' ( - x ) - u ' ( x ) > 1 u'(-x)-u'(s) _ds > 1 (s - x ) /s+x 7 7 u' (-x)ds ( s - x ) /s+x /2 u'(-x) TT l I 1/2 > 0 Thus u 1 ( x ) i s no b e t t e r t h a n L i p 1/2 i n a n e i g h b o u r h o o d o f x = -x 29 We r e c a l l the s e t t i n g o f t h e p r o b l e m i n §3 above, i . e . IJJ(X) i s a g i v e n f u n c t i o n d e f i n e d on fi n R , <Kx) £ 0 on R n 9fi and v ( z ) i s the s m a l l e s t s u p e r h a r m o n i c c o n t i n u o u s on fi , v ( z ) ^ 0 on 9fi , and v(x) > *Kx) • v(z) = l o g — - — d y ( t ) -I z—11 , h ( z , t ) d y ( t ) = u ( z ) - H ( z ) S We t h e n i n t e r p r e t e d the i n e q u a l i t y v ( x ) >_ ty(-x) as u ( x ) >_ ^ ( x ) + H ( x ) = f ( x ) . Thus f ( x ) i s n o t i n d e p e n d e n t o f t h e measure dy . I t i s t h e r e f o r e n e c e s s a r y t o p r o v i d e an example where the s i t u a t i o n d e s c r i b e d above ( f concave on S) a r i s e s . We t a k e f o r fi t h e u n i t d i s c |z| < 1 . S e t D>(x) = l - 4 x f o r |x| < 1/2 , ty(x) = 0 f o r |x| > 3/4 , and ty even and smooth L e t v ( z ) be the s o l u t i o n o f the e x t r e m a l p r o b l e m . Then 2 v(z) = l o g — — d y ( t ) -J S I z-tI l o g — - d y ( t ) = u ( z ) - H ( z ) S l 1 - z t 1 S i n c e <JJ(X) i s even, v ( x ) i s even, so t h a t S i s symmetric about th e o r i g i n . A l s o , f o r xe (-1,1) , v ( x ) > 0 so t h a t S c (-1/2,1/2) Fo r xe S , an e l e m e n t a r y c o m p u t a t i o n shows t h a t v " ( x ) > 0 , so t h a t S c o n s i s t s o f one c l o s e d i n t e r v a l [-T,T] where x < 1/2 . I f f ( x ) = <Kx)+H(x) t h e n u ( x ) = f ( x ) on [-T,T] and u ( x ) _> f ( x ) on (-1,1) . I t remains t o show t h a t f ( x ) i s concave on [-T,T] . We n o t e the f o l l o w i n g e a s i l y checked f a c t s : H(x) i s even, H(0) = 0 , H'(0) = 0 , H(x) i s an i n c r e a s i n g p o s i t i v e convex (3) f u n c t i o n (0,1) , and H (x) i s p o s i t i v e on (0,1) . F i n a l l y s i n c e v ( l ) = 0 , H ( l ) = u ( l ) . We r e s t r i c t o ur a t t e n t i o n t o (0,1/2) . I f f = tJH-H i s convex on (0,1/2) we have an immediate c o n t r a d i c t i o n s i n c e f ' ( 0 ) = 0 and f ' ( T ) i s n e g a t i v e . Now s i n c e f ^ ( x ) = H ^ 3 \ x ) > 0 on (0,1/2) , f ' ( x ) i s convex and t h e r e f o r e must be n e g a t i v e on (0,T) . We c o n c l u d e t h a t f ( x ) _< f ( 0 ) = 1 on ( 0 , x ] . A l s o s i n c e f ( x ) < u ( x ) on ( x , l ) and u ( x ) i s d e c r e a s i n g h e r e we have t h a t f (x) <_ 1 everywhere. Suppose now t h a t f " ( 0 > 0 f o r some £ < T . Then H"(5) > -ty"(0 = 8 . Now H"(x) i s i n c r e a s i n g on (0,1) so t h a t f o r x > 1/2 we have x H'(x) = H " ( t ) d t > 8dt = 8 ( x - l / 2 ) 1/2 S i m i l a r l y H(x) = H' ( t ) d t > 8 x 1/2 ( t - l / 2 ) d t = 4 ( x - l / 2 ) ' Thus H ( l ) = f ( l ) > 1 and we have a c o n t r a d i c t i o n . §5. As. One D i m e n s i o n a l V a r i a t i o n a l P r o b l e m Denote by CQ(-1,1) the c o n t i n u o u s l y d i f f e r e n t i a b l e f u n c t i o n s o f compact s u p p o r t c o n t a i n e d i n (-1,1) . F o r (j) (x) e CQ(-1,1) d e f i n e the harmonic e x t e n s i o n o f (f> t o C'\ R by <Kx,-y) = <i>(x>y) = -1 y U ( t ) 2 2 (x-t) Z+y d t . 31 D e f i n e a norm on c J c - 1 , 1 ) by ||c(,|| 2 g r a d (J>(x,y) dxdy . T h i s norm c a n be e x p r e s s e d d i r e c t l y i n 0 - o terms o f <Kx) • One such e x p r e s s i o n i s g i v e n by the Douglas f u n c t i o n a l : 1 TT ^(s)-Kt)j d s d t — CO —CO Or, e q u i v a l e n t l y , i n terms o f the F o u r i e r t r a n s f o r m : 2 = — 2TT L e t H be the c o m p l e t i o n o f CQ(-1,1) w i t h r e s p e c t t o ||•|| . H i s a H i l b e r t s p a c e . F o r <p^,<p^e tf n CQ(-1,1) , the i n n e r p r o d u c t on H may be r e a l i z e d as (4>l5<i>2) = * i ( t ) ( ' " s f ( t ' 0 ) ) d t = 34> • 2 ( 0 ( --^<t,o)) d t - l - l A v a r i a t i o n a l p r o b l e m i n a s i m i l a r H i l b e r t space p l a y e d a c r u c i a l r o l e i n a r e s u l t o f B e u r l i n g and M a l l i a v i n [ 2 ] . We have the f o l l o w i n g e l e m e n t a r y Lemma 1 „ Lemma 1.2. F o r <|> e H , ^ - d x < TT M c O l l 2 -1 1-x 32 P r o o f . L e t f ( x ) = c))(x-l) • C e r t a i n l y | | f | | = Now f ( x ) i s s u p p o r t e d on [0,2] so t h a t f o r x > 0 f 2 ( x ) = ( f ( x ) - f ( - x ) ) ' i = ( 3 f 2 ^ - ( X , 0 ) d 9 ) < TT a f 7 (fe(x,e)) de Hence 2 . r f 2 ( x ) oo TT r < TT - | ( | f ( x , 6 ) ) 2 d 6 x d x < | |f | 0 0 0 C o n s e q u e n t l y 1 <i>2(x) x+1 dx < TT S i m i l a r l y 4> 2(x) 1-x dx < TT -1 1 a Now l e t *Kx) 6 C ' (-1,1) be c o m p a c t l y s u p p o r t e d on (-1,1) , 0 <_ a < 1 . D e f i n e Q.E.D. . + . = {<f>:cf> e tf and tyU) >ty (x) a.e. f o r x e (-1,1)} i s a convex s e t i n H , nonempty s i n c e <f +(x) e , and i n v i e w o f Lemma 2, i s c l o s e d . By a c l a s s i c a l theorem K has a uni q u e element o f m i n i m a l norm, w h i c h we w i l l denote v ( x ) . 1 a Theorem 1.4. W i t h ^ ( x ) and v ( x ) as above, i f ^ ( x ) e C ' (-1,1) t h e n v ( x ) e C 1 , 0 t ( - l , l ) i f a < 1/2 , v ( x ) e C 1 , 1 / 2 ( - l , l ) I f a > 1/2 33 P r o o f . I f (j) e CI* (-1,1) , <j> _> 0 , then v + \<p e i f A _> 0 . 2 2 Thus ||v+Ac|>|| - | | v | | > 0 , w h i c h i m p l i e s (v,tj)) >_ 0 . F o r m a l l y 9v rt . 9v , . t h i s means — < 0 i . e . - — = dy i s a p o s i t i v e measure. 9y — 8y I f supp <j> c { v ( x ) > i^(x)} a s i m i l a r argument y i e l d s (v,<j>) = 0 . T h e r e f o r e i f v ( x , y ) = v ( x , - y ) = item- dt is the ,2, 2 ( x - t ) +y harmon i c e x t e n s i o n o f v ( x ) t o C \ R , we c o n c l u d e t h a t v ( z ) i s s u p e r h a r m o n i c i n C \ ((-°°,-l]u [1,°°)) and t h a t the measure dy a s s o c i a t e d to v ( z ) has compact s u p p o r t S c (-1,1) . Moreover S c {x e (-1,1): v ( x ) = *Kx) } • We now make a c o n f o r m a l mapping o f C \ ((-°°,-l] U [1,°°)) t o t h e d i s c |w| < 1 and a p p l y Theorem 1.2. T h i s c o m p l e t e s the p r o o f . §6. Remarks, P r o b l e m s , and Guesses The example i n § 5 s h o u l d be compared w i t h the f o l l o w i n g f a m i l i a r v a r i a t i o n a l p r o b l e m . 2 L e t ip e CQ(-1,1) and suppose v ( x ) m i n i m i z e s the f u n c t i o n a l J-^u) = ( u ' ( x ) ) 2 d x = ^ 2TT -1 1 2 on the c l o s e d convex s e t = {u e W^ ' (-1,1), u _> ip] . I t i s e l e m e n t a r y - t o show t h a t the graph o f v ( x ) c o n s i s t s o f p i e c e s o f the graph o f \p and i t s t a n g e n t s , i . e . v ( x ) e C^'^(-1,1) • a r x s e s , On c omparing the two examples above the o b v i o u s q u e s t i o n 2 I f ty e CQ(-1,1) , l e t v ( x ) m i n i m i z e the f u n c t i o n a l u(£) | 2 k | 2 a d ? 0 < a < 1 on the s e t K = {u e H : u > iii} where H i s the H i l b e r t space a a — a formed by t h e c o m p l e t i o n of CQ(-1,1) w i t h r e s p e c t t o the norm \HO | 2 | c | 2 a d 5 , i . e . Ha = WJJ' 2(-1,1) . Would i t 1 a be t r u e , f o r example, t h a t v e C ' (-1,1) ? Or, more g e n e r a l l y , what happens i f the w e i g h t |5| i s r e p l a c e d by a c o n t i n u o u s i n c r e a s i n g f u n c t i o n < K | 5 | ) > w i t h <$>(0) = 0 , l i m $ (| £ | ) = °° ? UI =°° A r e g u l a r i t y theorem p r o v e d by e x a m i n i n g the decay o f the F o u r i e r t r a n s f o r m o f the s o l u t i o n w o u l d be of c o n s i d e r a b l e i n t e r e s t . S i n c e the example mentioned above i s the one d i m e n s i o n a l a n a l o g u e of the " S m a l l e s t S u p e r h a r m o n i c " p r o b l e m , ( S t a m p a c c h i a and Lewy [ 2 0 ] , [ 2 1 ] ) i n R n where the r e g u l a r i t y of t h e s o l u t i o n was f o u n d to be C^~'^~(Q) , i t i s r e a s o n a b l e t o c o n j e c t u r e t h a t a l t h o u g h our t e c h n i q u e i n the p r o o f o f theorem 1.2 i s o n l y two d i m e n s i o n a l , the r e s u l t s h o u l d be t r u e on h i g h e r d i m e n s i o n s . The r e s u l t s h o u l d a l s o be t r u e i f the L a p l a c i a n i s r e p l a c e d by an a r b i t r a r y s t r o n g l y , u n i f o r m l y , e l l i p t i c o p e r a t o r w i t h , s a y , smooth c o e f f i c i e n t s , o r , f o r example, under the h y p o t h e s e s o f F r e h s e [ 1 3 ] . - 35 I n two d i m e n s i o n s , the o t h e r i m p o r t a n t case i s t h e m i n i m a l s u r f a c e p r o b l e m where the c o r r e s p o n d i n g v e c t o r f i e l d o f t h e D i r i c h l e t f o r m i s o n l y l o c a l l y c o e r c i v e . However, s i n c e i t has been shown ( S t a m p a c c h i a and Lewy [22]) t h a t i n most c a s e s t h e r e g u l a r i t y o f the s o l u t i o n o f m i n i m a l s u r f a c e p r o b l e m i s the same 1 1/2 as t h a t f o r the harmonic c a s e , i t i s l i k e l y t h a t C ' r e g u l a r i t y i s the answer h e r e as w e l l . We r e f e r t o Duvaut and L i o n s [11] f o r v a r i o u s r e l a t e d p r o b l e m s . The f o l l o w i n g theorem i s an e l e m e n t a r y consequence o f Theorem 1.2 and we omi t t h e p r o o f . 1 a Theorem 1.5. L e t ^ ( y ) < ^(Y.) > , e C " (r) . Suppose  t h a t g e L i p a (T) and t h a t f e L P(fi) , p >_ I f v i s the s o l u t i o n t o the p r o b l e m ( i ) Av = f on fi ( i i ) ip1(y) <_ v ( y ) <_ty2(yY~ on T ( i i i ) Oj^-v) (*2-v) ( | J - g) = 0 on r t h e n i f a < 1/2 , v e C1,a(T) , i f a > 1/2 , v e C 1 , 1 / 2 ( r ) . 36 C h a p t e r I I §0. I n t h i s c h a p t e r we d e r i v e some s u f f i c i e n t c o n d i t i o n s f o r t h e e x i s t e n c e o f a s o l u t i o n t o p r o b l e m (1") mentioned i n the i n t r o d u c t i o n f o r n >_ 2 . F o r the case n = 1 we r e f e r t h e r e a d e r t o B e u r l i n g and M a l l i a v i n [2]. L e t IJJ(X) be a l o c a l l y i n t e g r a b l e f u n c t i o n d e f i n e d on R n s u c h t h a t (2.1) * K x ) R n 1+ x n+1 dx < 0 0 and denote by i [ i(x,y) the harmonic e x t e n s i o n o f ip t o R n+1 + i K x , y ) = Pv" * i K x ) = - d t x e R n , y > 0 , .. .2. 2.n+l/2 n ; R n (|x-x| +y ) L e t g ( x ) be a l o c a l l y i n t e g r a b l e f u n c t i o n on R s a t i s f y i n g (2.2) e ( t ) - r 1 5 V ' L d t < °° , f o r some r > 0 , , n 1+ t and denote by N ( r - g ) ( x , y ) the s o l u t i o n o f the Neu;mann p r o b l e m w i t h boundary c o n d i t i o n r - g ( x ) , namely Assume t h a t ty(x) i s c o n t i n u o u s and s a t i s f i e s ( 2.1) and t h a t g(x) s a t i s f i e s ( 2 .2 ) t h e n p r o b l e m ( 1 " ) r e a d s as f o l l o w s . P r o b l e m ( 1 " ) F i n d v ( x ) c o n t i n u o u s , s a t i s f y i n g ( 2.1) s u c h t h a t ( i ) v ( x ) > i|)(x) ( i i ) |^(x,0) < g ( x ) — dy — and ( i i i ) (v-i|0 (|^ - g) = 0 dy where v ( x , y ) i s the P o i s s o n i n t e g r a l o f v ( x ) . I f t h e r e e x i s t s a v ( x ) s a t i s f y i n g ( i ) and ( i i ) above we say t h a t t h e p a i r (ty,g) i s c o m p a t i b l e . I n §3 below we s h a l l e s t a b l i s h t h a t i f t h e r e e x i s t s a v ( x ) s a t i s f y i n g ( i ) and ( i i ) above, t h e n t h e r e e x i s t s a s o l u t i o n t o p r o b l e m ( 1 " ) . Thus our a p p a r e n t l y weaker d e f i n i t i o n o f c o m p a t i b i l i t y i s e q u i v a l e n t t o t h e d e f i n i t i o n g i v e n i n the i n t r o d u c t i o n . I f ty e C^(R n) and i f g i s t h e n o r m a l d e r i v a t i v e of the P o i s s o n i n t e g r a l o f ty , i . e . g = -j^-(x,0) , t h e n g i s a dy m e a s u r a b l e f u n c t i o n on R n . O b v i o u s l y (ty,g) would be a c o m p a t i b l e p a i r . However, our a i m i s t o e s t a b l i s h c o n d i t i o n s on ty and g , i n d e p e n d e n t of each o t h e r , so t h a t f o r a f i x e d ty , (ty,g) w i l l be c o m p a t i b l e f o r a c l a s s o f f u n c t i o n s g and v i c e v e r s a . I n §1 and §2 i t i s shown t h a t under s u i t a b l e c o n d i t i o n s on ty and g , t h e p a i r s (ty,c) and ( N ( r - g ) , c ) w i l l be 38 c o m p a t i b l e f o r any c > 0 . Under t h i s a s s u m p t i o n , c e r t a i n l y t he p a i r (ijH-N(r-g) , r ) i s c o m p a t i b l e i . e . t h e r e e x i s t s v > \> + N ( r - g ) 9 V1 such t h a t - — < r . C l e a r l y v = v - N ( r - g ) s a t i s f i e s ( i ) and dy — 1 ( i i ) above so t h a t (ijj,g) i s a c o m p a t i b l e p a i r . Once c o m p a t i b i l i t y f o r t h e p a i r ( 4 s g ) has been e s t a b l i s h e d p a r t ( i i i ) f o l l o w s f r o m an argument o f St a m p a c c h i a and Lewy [21] t h a t w i l l be d i s c u s s e d i n §3 . § 1. C o m p a t i b i l i t y f o r the p a i r (^,c) I n o r d e r t o c l a r i f y i d e a s we b e g i n w i t h an e l e m e n t a r y example. P r o p o s i t i o n 2.1 F o r any c > 0 , 0 < Y < l > the p a i r ( j x j ,c) ' i s c o m p a t i b l e . P r o o f . By homogeneity ( n - l ) c n n R n x-T n - l i n - l ) | T |Y 1 d T = x ( n - l ) c T i J R n Y-T n - l , n - l Y - l ) T dT |x | Y K where Y = and K i s some c o n s t a n t depending on y 39 I f c > 0 , l e t R. = ( K c) c Y - l L e t v (x) = 7 — 1 ( n - l ) c n x-T n-1 T >Rc c 1 j IT Y - l m-l ' K = x Y + C(c) dT where C ( c ) i s a c o n s t a n t d e p e n d i n g on c Thus v(x) = v.(x) - C(c) > xl and |^-(x,0) < c 3y — Q.E.D. I n f a c t , any i n c r e a s i n g , r a d i a l l y s y m m e t r i c , concave f u n c t i o n ty , s a t i s f y i n g (2.1) can be d e a l t w i t h "by hand'-' i n t h i s way. However, i t w i l l be shown t h a t a h i g h degree o f smoothness f o r ty i s n e i t h e r n e c e s s a r y n o r s u f f i c i e n t . The p r o b l e m i s b o t h g l o b a l and l o c a l i n n a t u r e and we s h a l l make use o f the f o l l o w i n g m a c h i n e r y . D e f i n i t i o n ( 2 . 1 ) . L e t B denote the space o f f u n c t i o n s on R n s a t i s f y i n g f e B i f f 'D f ( c ) | 2 k| nd5 < n R where f ( ? ) denotes the F o u r i e r t r a n s f o r m o f f ( x ) P r o p o s i t i o n ( 2 . 2 ) . B i s a subspace o f B.M.O. Fo r a complete d e s c r i p t i o n o f B.M.O. and i t s p r o p e r t i e s we r e f e r t o F e f f e r m a n and S t e i n [ 2 8 ] . 40 F o r f e B i t f o l l o w s f r o m the H a r d y - L i t t l e w o o d - S o b o l o v theorem t h a t g r a d f '€,, L n ( R n ) . P r o p o s i t i o n (2.2) the n f o l l o w s f r o m t h e more g e n e r a l r e s u l t : P r o p o s i t i o n ( 2 . 3 ) : I f <j) e; L n ^ a ( R n ) , 0 < a < n , t h e n the R i e s z ' • p o t e n t i a l !„(<(>) (x) = R n x-T <|>(T)dT e- B.M.O. w i t h I (<|>) L M . n-a 1 1 a 1'B.M.O < C U . n/a P r o o f . (See : a l s d ;Neri ^ [.23.].);.= .Suppose t h a t e r a . <. 1 and t h a t Q i s a cube o f s i d e h , c e n t r e d a t the o r i g i n . L e t I denote the mean v a l u e o f I (<()) o v e r Q . Then: a I a ^ ) ' ( x ) - I Q | d x R n. I x-T .n-a - ' T) dT_ -«T)dT - -r4 ,n S-T n-a <))(T).dTdS dx ;. Cp , ' ' T " , r, , j j x ' |Q| J Q Q R n x-T i n - a S-T n - a >(T) dTdSdx dTdSdx Q Q T >2h f o r m u l a c o n t i n u e s . 41 + <KT) Q T <2h x-T n-a dxdTdS I x - S | i n-a+1 <()(T) dTdxdS Q Q T >2h + Ch a l< K T ) b i T T <2h < C Q U , . F o r a > 1 , l e t ty(x) = I. _ . (cf>) (x) . Then by the H a r d y - L i t t l e w o o d - S o b o l o v theorem, ty e L n ( R n ) . T h e r e f o r e I-, C^) = I . * ! , = I (<))) and I., (.ty) B.M.O. by the p r e v i o u s 1 1 (a-1) a 1 argument. Remarks. S i n c e t he d u a l space o f ( R n ) i s B.M.O. ( F e f f e r m a n and S t e i n ) we o b s e r v e t h a t the d u a l s t a t e m e n t o f P r o p o s i t i o n (2.3) i s t h a t I i s a c o n t i n u o u s mapping f r o m H^R 1 1) n i n t o L n a ( R n ) . P r o p o s i t i o n (2.3) t h e r e f o r e e x t e n d s t he c l a s s i c a l H a r d y - L i t t l e w o o d - S o b o l o v theorem t o the e n d p o i n t s " p = l , and q ™ . " (See S t e i n [ 2 7 ] , pp. 237). D e f i n i t i o n ( 2 . 2 ) . L e t B^ be t h a t subspace o f B c o n s i s t i n g o f f u n c t i o n s f t h a t s a t i s f y : ( i ) f e B i s s y m m e t r i c w i t h r e s p e c t t o the h y p e r p l a n e s x ^ = 0 , i = l , . . . , n i . e . f ( x ,...,x ) = f ( + x ,...,+x ) 1 n — I — n x. ( i i ) the f u n c t i o n s g^(x) = "["^"p^3^ a r e a - * - s o ^ n ^ > i = 1, . .., n . D e f i n i t i o n ( 2 . 3 ) . F o r each i n d e x i = 1,...,n , l e t be the H i l b e r t space o b t a i n e d by c o m p l e t i n g the space x. {g. = - i — r f ( x ) , f e B } w i t h r e s p e c t t o the norm l I x | \^ J_ ho | 2 k| n de P r o p o s i t i o n ( 2 . 1 ) . F o r <j> e H. , i 4 m l ^ d T < c , k M ^ i n it r -R 1 i 1 P r o o f . As o b s e r v e d e a r l i e r |v <j> | e L u(R") . L e t <j>(x,y) denote the P o i s s o n i n t e g r a l of <Kx) i . e . cj>(x,y) = P^ * <f>(x) . S i n c e the R i e s z t r a n s f o r m s a r e bounded on L n ( R n ) we have t h a t ||v c|)| I <c||v<j)|| where V denotes X ' y L n ( R n ) - X L n ( R n ) th e o r d i n a r y g r a d i e n t on R n and V denotes the o p e r a t o r xy V <(> = ( — ( x , y ) , V c()(x,y)) . Now |v <j>(x,y) I i s subharmonic xy dy x ' xy 1 „n+l xn R_j_ so t h a t |V cj)(x,y) I < P * |V (() | (x) i xy y 1 x y Y 1 I f Q i s any cube i n R o f s i d e h l e t denote the s e t = { ( x , y ) f- R^ + 1 , x Q , 0 < y < h} . Then as a consequence o f H o l d e r ' s i n e q u a l i t y V cj) (x,y) dxdy <_ xy P * V <j> ( x ) d x d y y 1 xy 1 J < C h" V ^ L n ( R n ) T h e r e f o r e the measure d p ( x , y ) = |v (f>(x,y) |dxdy xy s a t i s f i e s C a r l e s o n ' s c o n d i t i o n on cubes. T h e r e f o r e by C a r l e s o n ' s theorem (see S t e i n [27] pp. 236) we have t h a t R n + 1 ;;+ |v <|>(x,y) | n + ^ d x d y <_ (P * I ? * I(x))ndp(x,y) R n+1 + xy < C | |V 4>| | - 1 1 xy " L n ( R n ) j V cj)(x) dx xy R < c ||v <H|n+1 " X L n ( R n ) , 44 Observe t h a t <(>(x,y) v a n i s h e s on the h y p e r p l a n e x^ = 0 . We i n t r o d u c e the c y l i n d r i c a l c o o r d i n a t e s x_^  = r Cos 0 , y = r S i n 0 , X \ X_ • • • X , -5 X , ( - * • • X ) . Then f o r a l m o s t a l l x e- R 1 l - l l + l n n, <j)(x) | n + 1 < l i m |<j>(r,0,x) | n + 1 < ( 0=0 IT/2 I i / "\ J sn+1 • ^ - ( r , a , x ) da) ,TT- n ±<2> l x i l TT/2 ,1 8<j> - ,n+l , | - — ( r , a , x ) | r d a '0 T h e r e f o r e <j>(x) V l X i ' m + l dx < 2 (£) R n+1 V ^ t j ) (x,y) | n + ^ dxdy Q.E.D. We can now s t a t e the main theorem i n t h i s s e c t i o n . Theorem 2.1. L e t f be_ an element o f B^ . Suppose f bounded, | f ( T ) ' c o n t i n u o u s , and t h a t dT < 0 0 R n T I f ty (x) i_s_ a. measurable f u n c t i o n s u c h t h a t ty(x) "i '|xT|-'f •'(x5-i+ac%ns-Sant,ca.e."^. ^ £ , ^ § n the^pair* (tf»,a) i s compati"bIe±-f o r a l l 0 a > 0 . P r o o f . L e t i = 1 and l e t g = -f (x) , t h e n g e ; H 1 . 1 - l Suppose a > 0 i s g i v e n and choose b , 0 < b < a D e f i n e the s e t K = { v ( x ) , v e. f i \ ^ , v ( x ) _> g(x) a.e. f o r x^ >0} 45 K i s convex, and i f {v } i s a sequence i n K c o n v e r g i n g n to v f , th e n by the p r e c e d i n g p r o p o s i t i o n : v ( T ) - v ( T ) n m+l dT < C ||v - v | | ^ + 1 - 1 1 n M D so t h a t K i s c l o s e d . D e f i n e the f u n c t i o n a l $(v) ~ v + 2b R n T in+1 dT L e t m = i n f $(v) . S i n c e g e K , m i s f i n i t e . -veK Suppose v^, e K such t h a t $ ( v ^ ) , $ ( v 2 ) _< m + e , f o r some e > 0 . Then \^v-\+v2) ^ K s o *("2( v^ + v2^ — m " v +v T h e r e f o r e | - ( $ ( v 1 ) + $ ( v 2 ) ) - 3>( 1 2 2 ) <_ e . But H i s a H i l b e r t space so v,+v r |(*(v1)+*(v2)) - H-\h =|(||v 1H 2) - r i + V 2 n 2 M 2 < C o n s e q u e n t l y i f {v } i s a sequence i n K such t h a t $(v, ) -»-»• m , th e n {v, } i s a Cauchy sequence. T h e r e f o r e {v, } w i l l k k k converge t o an element u e K s i n c e K i s c l o s e d . For 0 < R^ < R 2 , and f o r any i n t e g e r k |u| |* + J,-. 2b R i l l x l l R 2 x u ( x ) 2 —- dx < (| |v | | +| |v - u | | ) + . 2b I in+1 k M D k M D R ; i i l x l l R 2 f o r m u l a c o n t i n u e s . ,46 +• 2b x | v k ( x ) - u ( x ) i i n+1 dx RT1|X|IR 2 R n/n+1 < »(v k) + o ( l ) + C l o g (-) Mv k-u|| D by P r o p o s i t i o n ( 2 . 4 ) . L e t t i n g k t e n d t o i n f i n i t y we o b t a i n U • + +b / 2b D x l U ( x ) i i n+1 dx < m + o ( l ) . Thus $(u) = m L e t <j)(x) be a smooth f u n c t i o n , even w i t h r e s p e c t t o the axes x^ = 0, i = 2...n, and odd w i t h r e s p e c t t o the h y p e r p l a n e = 0 . I f <j)(x) J> 0 f o r x^ _> 0 , t h e n u + £<j> e K f o r any e > 0 . T h e r e f o r e $(u+ecj>) - $(u) > 0 . That i s (2.3) 2E (u,<f>) - + e' 2eb x <()(x) i n+1 dx > 0 ,n x D i v i d e (2.3) by 2e and l e t e tend t o z e r o t o o b t a i n (2.4) ( u , 4 0 D B > - b x 1<j)(x) i i n+1 dx F o r any smooth f u n c t i o n s o f compact s u p p o r t , <j>, IJJ € , the i n n e r p r o d u c t on may be r e a l i z e d as (2.5) ( ^ > D D " a n ( - X ) R <J>(x) ( x , 0 ) d x Sy 1 1 47 where — — ( x ) denotes the boundary v a l u e s o f the n normal 3y d e r i v a t i v e o f the P o i s s o n i n t e g r a l P * ^ ( x ) d e f i n e d on R*1"1"**' . y + By p r o p o s i t i o n ( 2 . 2 ) , the s o l u t i o n o f the e x t r e m a l p r o b l e m , u(x) , i s i n B.M.O., and by t h e i n t e g r a b i l i t y c o n d i t i o n ; t , u ( T ) | dT < o o , u ( x ) has a w e l l d e f i n e d R i e s z t r a n s f o r m R , ( u ) ( x ) = l i m — e=0 n t T n + 1 T|>e 1 u ( x - T ) d T Lemma 2.1. R ( u ) ( x ) i s bounded and l i m R..(u)(x) = 0 . | x| =°° The p r o o f o f the Lemma i s somewhat t e c h n i c a l so we s h a l l d e f e r t he p r o o f u n t i l l a t e r . I n the f o l l o w i n g l e t u ( x , y ) = Py * u(x ) and f o r x e R n , 8 n u (x) w i l l denote the p o i n t w i s e l i m i t o f t h e harmonic f u n c t i o n ~ n v * 9y i n R ^ + i , l i m ^ - ^ ( x , y ) , y > 0 , whenever t h i s l i m i t e x i s t s . y=0 9y 9 n u However, (x) e x i s t s as a d i s t r i b u t i o n , and by (2.4) and ( 2 . 5 ) , 3 y n n 9 n u b x l (-1) ( x ) d x H rvdx i s a p o s i t i v e measure f o r x n > 0 and . n I I n+1 1 — 9y | x| n 9 n u we w i l l i n t e r p r e t i t i n t h i s s e n s e . A l s o [ ( - 1 ) (T)dT] w i l l d enote the n e g a t i v e ( c o n t i n u o u s ) p a r t o f the s i g n e d measure n 9 n (-1) — — ( x ) d x and we w i l l make f r e q u e n t use of the i n e q u a l i t y (2.4) 9 y n [(-D n ±-r(x)dx] < 77dx f o r x > 0 . - n — I I n+1 1 °y x '48 I n o r d e r t o f a c i l i a t e n o t a t i o n l e t d y ( x ) denote t h e ( s i g n e d ) n 9 n measure (-1) i L - y - ( x ) d x so t h a t d y ( x ) w i l l d enote t h e p o s i t i v e , 3y 9 n u a b s o l u t e l y c o n t i n u o u s measure [ ( - 1 ) ( x ) d x ] . 9 y n To c o m p l e t e the p r o o f of t h e theorem, n o t i c e t h a t f o r 3u, , _ -1 T (x) = 9y Y n 1 -1 i - dy(T) = — R n l X " T l n t l>0 )dy ( T ) x-T x-T where T = (-t.. , t„,. . ., t ) and Y ± z n n Thus |^-(x) = 3y Y n tL>o i i 2 , ~ , | I , ~,2 x-T x-T + x-T x-T dy(T) 4x.. r l i "1 Y. ? i i ~ 12 n ' t ?o' l x _ T ' l x " T l + l x ~ T M x _ T l 1.1— dy(T) C(n)b by ( 2 . 4 ) , where C(n) i s a p o s i t i v e c o n s t a n t depending o n l y on n . T h e r e f o r e 3u a i f b were chosen s m a l l enough t h e n - ^ ( x ) < — f o r x_ > 0 . 3y — 2nx^ 1 L e t F ( x , y ) = x ^ u ( x , y ) - y R ^ ( u ) ( x , y ) . F ( x , y ) i s harmonic i n R^ + 1 , F ( x , y ) = o ( ( | x | 2 + y 2 ) 1 / 2 ) i n R*+1 and l i m F ( x , y ) = x u ( x ) a l m o s t everywhere on R . Thus F ( x , y ) can y+0 49 be r e p r e s e n t e d as a s o l u t i o n t o the Neumann pr o b l e m w i t h boundary d a t a -j^-(x) = x.. - j ^ ( x ) - R.. (u) (x) . T h e r e f o r e dy x dy i ( n - l ) c < | n_l " " ~nTI ) ( ti | y i(T)-R 1(u)(T) >)dT= x^x) a.e. on R n R n T x-T As a consequence o f Lemma 2.1 we can choose R s u c h t h a t RT (u) (X) < -p- f o r x > R 1 — 2n — D e f i n e p ( T ) = { t 1 | ^ ( T ) - R 1 ( u ) ( T ) f o r |T| > R 8 u / m . 4 8 y - ( T ) f o r T < R . Then we o b t a i n f o r a l m o s t a l l x e R X 1 U ( X ) = T^I)c-R i m i n - l I m i n - l n T x-T ) p ( T ) d T + ( n - l ) c <R Ix-Tl1 1" 1 ITT"1 1 )R, ( u ) ( T ) d T - ( n - l ) c R' n I T T "1 I X - T T - 1 ) p ( T ) d T + C o n s t a n t , by Lemma 2.1 v 1 ( x ) 1 T h e r e f o r e v n ( x ) > x,u( x ) > x,g(x) = f ( x ) a.e. on R n , l — l — 1 I I 3v. X 3. and (x) <_ — . We r e p e a t t h i s p r o c e d u r e f o r i = 2,...,n x. 3v. o b t a i n i n g f u n c t i o n s v . ( x ) > f ( x ) such t h a t - — ( x ) < — . i I I 9y — n l x l n 3 v Then v ( x ) = £ v. (x) > | x | f ( x ) and — (x) <_ a , c o m p l e t i n g i = i 1 - 8 v the p r o o f o f the theorem. We r e t u r n now t o the p r o o f o f Lemma ( 2 . 1 ) . We have the r e p r e s e n t a t i o n f o r R ^ ( u ) ( x ) : (2.6) R , ( u ) ( x ) = l i m — 1 n=0 C n x 1 - t 1 R n ( | x - T |2+ n 2 / n + ] ) / 2 u(T , n)dT = l i m — n=0 a n x 1 - t 1 R n ( | x - T | 2 + n 2 ) 1 / 2 ( _ l } n 3 ^ ( X j n ) d T 3y n R n I x-T I dy(T) J u s t o b s e r v e t h a t l i m ( - l ) n 9 u ( x , y ) d x = l i m (-A ) n ^ 2 ( u ) ( x , y ) d x y=0 Sy 1 1 y=0 X ? 8 2 where A = £ — 9 > n even , and i f n i s odd t h e n 1=1 3 x. l i m ( - l ) n ^ ( x , y ) d x = l i m / ((-A fn ^ / 2 ( u ) ) ( x , y ) d y=0 dyn y=0 3 y x F o r a smooth f u n c t i o n i n , f o r m u l a (2.6) f o l l o w s f r o m r e p e a t e d d i f f e r e n t i a t i o n . We remark t h a t s i n c e u i s odd w i t h r e s p e c t t o the h y p e r p l a n e x^ = 0 , f o r x e R n an e l e m e n t a r y c a l c u l a t i o n y i e l d s \51 dy(T) = 0 and T <2 x T >2 x l o g — ^ — dp(T) I x-T I < C x T h e r e f o r e the l o g a r i t h m i c p o t e n t i a l o f t h e ( s i g n e d ) measure dy(T) 1 1 XI i s w e l l d e f i n e d . Thus u ( x ) = — l o g * dy a.e. on R . % |x| For n = 2 , l o g — — i s t h e Newtonian p o t e n t i a l so t h a t t h i s 1*1 r e p r e s e n t a t i o n i s c l a s s i c a l . F o r g e n e r a l n we r e f e r t o Landkof [16] pp. 50-51 f o r d e t a i l s . Observe t h a t R ^ ( u ) ( x ) i s symm e t r i c w i t h r e s p e c t t o the h y p e r p l a n e s x_^  = 0 , i = 1, ... ,n . So l e t n > 0 be g i v e n and suppose t h a t f o r some x = (x, ...x ) , x. > 0 f o r a l l i ,: c c 1 n x — j. x r t i . I x - T | dy(T) > n L e t x' = (x-,x„,...,x ) w i t h x- < x, < x and d e f i n e 1 L n ± — J- — D(x,x. ,u) = R x l _ t l X l _ t l (-^-^ - 1 1 ) d y ( T ) = n I x-T I ihx'-a? I F ( x , x ,T)dy(T) Observe t h a t F ( x , x ' , - ) i s n e g a t i v e s i n c e x | > x^ » X l + X l and s y m m e t r i c about t h e h y p e r p l a n e t ^ = — - — . A l s o F ( x , x ,T) x i + x i i s d e c r e a s i n g i n t ^ , f o r t ^ _< — - — . T h e r e f o r e G(x,x',T) = F ( x , x ' , T ) - F ( x , x ' , T ) i s n e g a t i v e f o r t1 > 0 , where, 52 as b e f o r e , T w i l l denote ( " t p t 2 , . . . , t n ) . We have the f o l l o w i n g e l e m e n t a r y e s t i m a t e s ; f o r x, x' f i x e d | G ( x,x',T)| <_ C ( x ^ - x 1 ) max-. 1 1_ x-T x'-T < 2 and D"G( x,x',T) < C ( x ' - x ) max-j — a 1 1 |x-T|H + 1 ' | x ' - T | l a l + 1 ct I I where D d e n o t e s a d i f f e r e n t i a l monomial i n T of o r d e r a Now F ( x , x ' , T ) d y ( T ) = G ( x , x ' , T ) d y ( T ) t±>0 x G(x,x' ,T)dy(T)' |T|> t,>0 1 < b G ( x , x ' , T ) | 1 n + 1 dT by (2.4) T >-C ( x | - X l ) x We i n t r o d u c e the o p e r a t o r s A ' A(n-])/2 r & f o r n odd V ( ^ n - 2 V 2 ) f o r n e v e n 53 and B(u) = < 8u dy f o r n odd Vu f o r n even L e t <J)(T) be a p o s i t i v e smooth f u n c t i o n , 0 <_ cf) (T) <_ 1 , s u c h t h a t 1 f o r | T | 1 1/41x| , 0 f o r | T | >_ 1 / 2 |x| *(T) = <|.(T) , <f>(T) = { and suc h t h a t A I L Then F ( x , x \ T ) d y ( T ) <f> F dy(T) + (l-<j>(T))F(x,x',T)dy(T) T <- L?l< I TI <_1XJ 4 — 1 ' — 2 = 1 1 + 1 2 By means o f a p a r t i a l i n t e g r a t i o n X l = A(<j>F) (T)B(u) (T)dT < C ( |A(*F) r ' ^ d T ) ( 1/n |grad u dT) T <J R n C ( x | - X ; L ) x (l-<j) (T).)!G:(«yx: ?T)ldy' (.T) •54, G ( x , x ' , T ) d y ( T ) ' LiU|T|«lil t l>o C ( x | - X l ) B ( x | - X ; L ) Thus D(x,x',u) < :—- where 3 i s a c o n s t a n t d e p e n d i n g o n l y on the d i m e n s i o n and ||u||^ B(x'-x.) I F L_JL< R t h e n l l — 2 x ' - t — — — dy(T) > ^ . T h e r e f o r e |x'-T| i f R ^ u X x ) >.n t h e n ^ ( ^ ( x ' ) >_ y where x { _ a n d b y t h e same argument as above i f 0 < A - l < t h e n R ^ ( u ) ( x ^ ) >_ ^ where x^ = (Ax', x o J . . . , x ) . 1 z n Remark. What we have shown i s t h a t i f R ^ ( u ) ( x ) i s l a r g e t h e n i t r e m a i n s l a r g e a t x' = ( A x ^ , x 2 , . . . > x n ) f ° r some A > 1 . We have a l s o shown t h a t i f R ^ ( u ) ( x ) i s l a r g e t h e n we c a n assume x^ ~ |x| . Suppose t h e n t h a t x^ ~ |x| and l e t = ( x ^ , . ' . . , x ^ + ( A - l ) x ^ , . . . , x ^ ) and Y = (x.. ,. .. ,x. + ^ \ ^ x 1 , . .. ,x ) where 0 < A - l < 77- , and X 1 z ± n f p suppose t h a t R ^ ( u ) ( x ) > n . 55 C o n s i d e r D(x,X^,u) = R x-T L X : - T ) d y ( T ) = F ( x , X ' T)dy(T) L e t <|>(T) = i x 1 1 f o r | Y-T| > - i 0 f o r IY-TI < -4 1 — 4 s u c h t h a t 0 < <|>(T) < l a n d I lO 0 4^ I 1 < — ? -— — 1 1 T | l c o — c L e t i|i(T) = \ 1 - <|)(T) f o r |Y-T| >_.2(X-l)x 2 0 f o r |Y-TI < ( X - l ) x , s u c h t h a t 0 <_^(T) <_ 1 and | |D I|I | \ M <_ X ( T ) = 1 - <KT) " (T) . Then ( . ( ^ - l ) x 1 ) F i n a l l y l e t D(x,X.[u) = <KT)F(x,.;X£,T)dy(T) + *(T)F(x,.X'T)dy(T) + X ( T ) F ( x , X j , T ) d y ( T ) = h + 1 2 + 1 3 Observe t h a t : |F(x,X[T) £ 1 and D F(x,';X|,T) < C x ^ X - D m a x ^ i I oi +1 ' I 1 I a +1 | x-T I1 1 i x i _ T l 1 Then ' I , A((j)F) (T)B(u) (T)dT < A(<|>F)(T)B(u)(T) dT I C | | u | | D ||A(4>F)|| l_i < C ( X - l ) A l s o I <_ |A(i|;F)B(u)|dT < | |A(*F) | | ( 56 > l / n |B(u)(T) |"dT ) Y-T <-= a^ ( | x | ) = o ( l ) as |x| tends t o i n f i n i t y , s i n c e x^ ~ |x| To e s t i m a t e the f i n a l i n t e g r a l , we d i v i d e t h e b a l l |Y—T| <_ 2 ( X - l ) x 1 i n t o 4 q u a d r a n t s where F(x,X_I,T) i s o f c o n s t a n t s i g n . I f i s a q u a d r a n t where F i s p o s i t i v e t h e n x(T)F(x,X',T)dy(l) 1 X ( T ) d y ( T ) + x(T)du(T)" I f i s a q u a d r a n t where F i s n e g a t i v e t h e n c e r t a i n l y X(T)F(x , > x[,T)dp(T) < x(T)dy(T) + 2 X ( T ) d y ( T ) ' T h e r e f o r e X 3 ± X ( T ) d y ( T ) + 2 X ( T ) d y ( T ) and u s i n g t h e same t e c h n i q u e as above we o b t a i n X 3 1 a 2 ( | x | ) + C ( X - l ) where a 2 ( | x | ) -* 0 as |x| » T h e r e f o r e D C X ^ X ^ . U ) <_ C ( X - l ) + a Q ( | x | ) where a Q ( | x | ) = o ( l ) A Summary I f R ^ ( u ) ( x ) > n and i f x i s s u f f i c i e n t l y l a r g e t h e n we can f i n d a X , i n d e p e n d e n t o f x , such t h a t X > 1 , and :57' we can f i n d a cube Q , w i t h Q = C ( n ) ( X - l ) n x such t h a t R ( u ) ( T ) > ^ j -1 „n+l. f o r a l l T e Q Suppose now t h a t R n I x-T I du(T) < -n , x ± >_ 0 , i = 1,.. . ,n We can assume x = 0 , beca u s e i f x.. > * ^ r ~ 1 X Z p , then using the same argument as above, if x' = (—K x„,...,x ) then R_(u)(x') < A I n 1 z f o r some X , 0 < X - l < — and we c o u l d p r o c e e d as b e f o r e . I f zp i x x < — 1 — ' 1 - 2B then if x' = (0,x ,...,x ) we will have that n R 1 ( u ) ( x ' ) < \ T h e r e f o r e assume t h a t R., (u) (x) < -n f o r x = (0,x o,...,x ) 1 z n x. > 0 , x.e. assume t h a t — x — c n x-T i n+1 u(T)dT > n . L e t X! = (0,x„,...,x. + ( X - l ) | x ,...,x ) f o r X c l o s e t o 1 and X • ' *2 c o n s i d e r n D(x,X,:,u) = — X c t x u ( T ) ( R x - T | n + 1 | x ! - T | n + 1 x ) dT 'n x.+^ t.< 1 ( X - l ) x t x u ( T ) ( x - T | n + 1 |x!-T| n + 1 x ) dT x— tx>o + n dT t . >-x x ± + ( X - l ) | x l tx>o 58 E s t i m a t e o f I , L e t E. denote t h e s e t {T e R n, t., > 0, t . < — 1 l — i x,.+(X-l) x Now — c n E 1H{|X-T|>- L|L } X'-T i i n+1 ) dT < C ( X - l ) x tjU(T) x-T n+2 dT x-T >-< C ( X - l ) ( f ^ ( T ) ! n + 1 d T ) 1 / n m < c u D a -D From t h e same e s t i m a t e we g e t t h a t t h e i n t e g r a l o v e r t h e s e t E.^  n {-^- _> | x-T | >_ ( X - l ) | x | ) t e n d s t o z e r o as |x| t e n d s t o i n f i n i t y . F i n a l l y we s h a l l assume t h a t f o r T e E^ n {|x-T| <_ ( X - l ) | x | } we have t h e e s t i m a t e u(T) < C (See Lemma 2.2 b e l o w ) . Then E n{ |x-T|<.(X-l) |x|} l X l |x-T|<_(X-l) | X x-T I n+1 dT < C ( X - l ) . E s t i m a t e f o r I , L e t E 2 denote the s e t {T 6 R n, t 1 > 0, t ^ > — x. + ( X - l ) x Observe t h a t f o r T e t h e k e r n e l o f t h e i n t e g r a l I ^ , ' 5 9 x _ T | n + l | X ! - T | n + 1 ) , i s n e g a t i v e . T h e r e f o r e s i n c e u(T) >_ g(T) f o r t > 0 "2 c t ^ C T ) (• x-T ) dT n J E 2 t l 8 ( T K | x - T | n + 1 " |x!-T| n + 1 ) d T 2 2 1 1 1 1 — f ( T ) ( ^ — ^ " T T - ) d T C n |T| | x - T | n + 1 | X ' - T | n + 1 <_ C ( X - l ) + o ( l ) by u s i n g t h e same e s t i m a t e s as above. Thus D(x,x:,u) < G(IA-1) +3 (Ixl) where 3 (|x|) = o ( l ) 1 — ' o 1 o Second Summary I f R^(u) (x) < - T i t h e n we have shown t h a t p r o v i d e d |x| i s l a r g e enough, X > 1 can be chosen, i n d e p e n d e n t o f |x| , so t h a t R , ( u ) ( T ) < —5T F O R A 1 1 T = (0,t„,...,t ) , x . + ( X - l ) | x | > t . > x. i 2n"1"-'- ^ n 1 — 1 — 1 We can now c o m p l e t e t h e p r o o f o f t h e Lemma. I f l i m | R 1 ( u ) ( x ) | ^ 0 t h e n f o r some n > 0 and | x | =°° a r b i t r a r i l y l a r g e v a l u e s o f |x| , we s h a l l have t h a t | R ^ ( u ) ( x ) | > n 60. Suppose t h a t R^CuMx) < -n ( t h e cas e R^CuKx) > n b e i n g s i m i l a r ) . Observe t h a t s i n c e tjuCT) R n T , n+1 dT < 0 0 , t h e harmonic e x t e n s i o n o f R ^ ( u ) ( x ) t o R^ +^ i s w e l l d e f i n e d : R ( u ) ( x , y ) = i c T f c i n J R n (|x-T| Z+y Zj 2^ 2(n+l)/2 u(T)dT . Moreover l i m R . ( u ) ( x , x ) = 0 1 1 1 x =°° Suppose x^ = 0 , and t h a t |x| i s l a r g e enough so t h a t ( |R 1(u) ( x , |x|) I < , and i f Q x = {T e R n, x. < t . < x. + ( A - l ) | x | , i = 2,...,n} t h e n R ^ u X T ) < f o r T e , t ^ = 0 , where X i s chosen a p p r o p r i a t e l y . Now Q x 1 v x i R ^ u M T , |X|) - R 1 ( u ) ( 0 , t 2 , . . . , t n ) |dT >_ -^2" However Q X 1 X R ( u ) ( T , | x | ) - R ( u ) ( T ) | d T < — Q._ |Q , Q ' X 1 X X 9y (T,s) dsdT M X ^In+15( < i'gradR (u) (T,y) | u x d T d y ) A l s o QJ J R 1 ( u ) ( T ) - R . ( u ) ( 0 , t , t ) dT 1 1 Z n x 1 "x c 0 3Rj_(u) 9x.. ( s , t 2 , . . . , t n ) dsdT X 1 X Thus < ( 1/n gr a d R 1 ( u ) ( T ) dT) x 2"+2 - |Q R 1 ( u ) ( T , x ) - R 1 ( u ) ( 0 , t 2 , . . . , t n ) dT = o ( l ) x' and we have a c o n t r a d i c t i o n . The c a s e x^  ~ |x| i s t r e a t e d s i m i l a r l y . Observe t h a t s i n c e a (Ixl) and 3 (|x|) a r e bounded and a l s o s i n c e R,(u)(0) i s o o 1 f i n i t e by a s s u m p t i o n , t h e same arguments as g i v e n above show t h a t R^(u) i s l o c a l l y bounded. A p o i n t l e f t open I t was assumed t h a t f o r |x-Y| <_ C|x| , x^ = 0 , and y± > 0 t h e n u(Y) <_ C — - . x We have the r e p r e s e n t a t i o n f o r u; y^ >_ 0 , 62 u(Y) = n \n l o g dy(T) = I Y-T I R n, t l>0 Y-T l o g - L i ^ J - d y(T) , | Y-T | Y = (- y i,y 2,...,y n) L e t F = { T e R11, |T - X | <_ 3c|x|} and l e t <|>(T) = { 1 F O R T 6 C ' F 0 f o r |X-T|<2C|X| suppose 0 < <|>(T) < 1 , <j>(T) = <(>(T) , and ||Da<()|| •<_ — Observe t h a t l o g l Y T ' = l°g ( o " = 4 l o g ( T o +1) Y-T Y-T Y-T 1 ^ 1 ^ i h = i l o g ( - i - h + 1) < — ^ , t , > 0 . Y-T Y-T Then G F . t ^ O Y-T l o g J dy(T) £ I Y-T I <KT)log dy(T) Y-T +(Tj l o g l Y T ' dy(T)" Y-T 2C|X|<JX-T|<3C|X ± y l R <j>(T) ~ dy(T) |Y-T| 2 + 4y. dy(T)" x-T >2c x Y-T 63 < C i f we e s t i m a t e as b e f o r e . x C l e a r l y l o g F IY-T I dy(T) < F n t ^ O Y-T + l o g 1 1 - L L dy(T) I Y-T I L e t B = {T e F , t 1 >_ 0 such t h a t u(T) = g ( T ) } . We c l a i m t h a t - b t , B i s a c l o s e d s e t and t h a t dy(T) F YB3~ i n+1 dT . 9u C Indeed, we o b s e r v e d i n (2.6) t h a t f o r x, > 0, — ( x ) < — . 1 3y — x 1 Thus f o r x > 0 , l i m u ( x , y ) = u ( x ) e x i s t s and i s l o w e r s e m i -y=0 c o n t i n u o u s b e i n g t h e i n c r e a s i n g l i m i t o f c o n t i n u o u s f u n c t i o n s . A l s o u ( x ) i s i n t h e same e q u i v a l e n c e c l a s s as u ( x ) , i . e . ||u|1^  = ||u|1^  . We may t h e r e f o r e assume u ( x ) i s l o w e r s e m i -c o n t i n u o u s . By a s s u m p t i o n , g i s c o n t i n u o u s so t h a t B i s c l o s e d and hence { x ^ > 0 , u ( x ) > g ( x ) } i s open. T h e r e f o r e i f cj> >;.0 i s a smooth f u n c t i o n c o m p a c t l y s u p p o r t e d on t h e s e t {x^ > 0, u ( x ) > g ( x ) } t h e n u-A <f> >_ g f o r A s u f f i c i e n t l y s m a l l , so t h a t u-Xty i s an element o f the c l o s e d convex s e t K . Hence i n t h e same way t h a t (2.4) was d e r i v e d we o b t a i n : •(u,<j)),D < -b x1<J)(x) i...i n+1 dx I n v i e w o f (2.4) i t f o l l o w s t h a t (u,ty) = -b x <|>(x) i n+1 dx , p r o v i n g x t h e c l a i m . L e t dp(T) be t h e r e s t r i c t i o n o f t h e measure du(T) ' t o B . S i n c e g i s bounded, c e r t a i n l y u i s bounded so t h a t t h e measure dp(T) a s s i g n s z e r o measure t o a p o i n t i n R n . So we may f i r s t assume B i s c l o s e d i n F . (2.8) I f Y e B t h e n l o g i L Z l . d p ( X ) < I Y-T I Rn,tl>0 Y-T l o g J J L ^ L dy(T) I Y-T I + C < c u(Y) + C x < C s i n c e f o r Y e B , u(Y) = g(Y) f ( Y ) where f was assumed t o be bounded and |Y| >_ c|x| . I f Y i B we need t h e f o l l o w i n g g e n e r a l Lemma. Observe t h a t f o r any Y e R t h e r e a r e N ( o v e r l a p p i n g ) cones , depend i n g on n , w i t h v e r t i c e s a t Y such t h a t i f E,V i s t h e p o i n t o f n B w h i c h i s c l o s e s t t o Y , t h e n any o t h e r p o i n t Z e B i s c l o s e r t o some t h a n t o Y [ C a r l e s o n [6]]. N Lemma 2.2. F o r Z e Q so t h a t |Z-£ < Z-Y t h e n l o g Jl=zL < c |Y-Z| - 1? — l o g -"-s where p = (p ... K -z| 65 and C i s an a b s o l u t e c o n s t a n t . P r o o f . We f i r s t suppose t h a t £^ >_ y^ . Then c l e a r l y we can assume 5^  _^ 2y^ , o t h e r w i s e j u s t t a k e C = 2 . I Y-Z I 1 4 y l Z l 2 y l z l (2.9) Now l o g J ^ = ± l o g (1 + —^Ar) < — . A l s o n o t e t h a t |Y-Z| 2 |Y-Z| 2 |Y-Z| 2 l o g 1 1 = ^ l o g (1+ — T) > T l o g (1 + =•) . But k V " Z | 2 | ? V - Z | 2 2 2 | C V - Z | 2 h z i > a r y i ) z i . ( ? r y i } < V y i < . . > — where < : < 1 and 2 | ? V - z | 2 21Y-Z|2 | Y-Z| |S V-Y y +|Y-Z| ( d - y ^ i i Y ^ | | \ x < _±. < — ± : < i 2|Y-Z| 2|Y-Z| 2|Y-Z ( 5 r y i ) z i t T h e r e f o r e — 0 < 1 . S i n c e l o g (1 + t ) > ^ f o r 2|Y-Z| 2 ~ ~ 2 0 < t < 1 we have t h a t 1 l € V - Z l > 1 , M + ( g l " y i ) z i , ^ ( € r y i ) z l ^ fol l o g J J z - L > l o g (1 + — ) > =- > U -Z| 2 2|Y-Z|2 ~ 8|Y-Z| 2 ~ i 6 | Y - z | 2 Combining t h i s w i t h (2.9) y i e l d s . Y-Z ^ , 0 yl . k -Z l o g J L < 32 — l o g J L . IY-Z| " il | ? V - Z | I f — Yj. ' w e c a n a s s u m e ^1 < ^ y l ' O t h e r w i s e s i n c e 66 l o g = \ l o g (1 + 4 ? 1 , Z l 0 ) 1 - l o g ( l + 2 ! l Z l 0 ) 5 -z I 2 y i z i 2 Y-Z 1 4 y i Z l >_ — l o g (1 -I j) we c o u l d t a k e C = 2 . I Y-Z I I f 5^  < \ yx , t h e n | Y-Z | > |£ V-Y| > | y x Then 4£ z E z 5 z l o g (1 + •„ 1 0 ) > l o g (1 + ^LJ^) 1 l o g (1 H ^ ^ o ) • N o w i v ,2' -? -z 35 -z 3 Y-Z Y-Z Z-L Y x + I Y-Z I ^ z < 1 and — < — < 1 . T h e r e f o r e < 1 3 Y-Z 3 Y-Z 3 Y-Z i l i v - z l i ' i z i Hence l o g J > — • .v „ i - 12 r - z Y - 7 y^ I -z 2 . Thus l o g 1 1 < 24 - i l o g —=J-. Q.E.D, Y-ZI IY-ZI H k -zl Hence l o g Y-T v = l Y-T l o g I J L - L L dp(T) I Y-T I N y — L v v = l 5 1 H v l o g ^ 1 dp(T) U - T | N C CY± I ~ f j by (2.8) s i n c e £ V e B. v = l |x| = C I f B i s n o t compact i n F t h e n by a s t a n d a r d argument we a p p r o x i m a t e dp(T) by a sequence of measures dp^(T) c o m p a c t l y s u p p o r t e d i n B . T h i s c o m p l e t e s t h e p r o o f of Lemma 2.1. I t i s n o t always easy t o d e t e r m i n e whether a g i v e n ty s a t i s f i e s t h e h y p o t h e s i s of Theorem 2.1. However i f ty b e l o n g s t o t h e f o l l o w i n g w e i g h t e d S o b o l o v space t h e n i t i s e l e m e n t a r y t o show t h a t t h e c o n d i t i o n s of Theorem 2.1 a r e f u l f i l l e d . D e f i n i t i o n ( 2 . 4 ) . L e t W d e n o t e t h e s p a c e : n 2 (V) F ( x ) • V ' I l n + 2 • ' ' T 2 1=0 x — ^ — 1 L < oo n even n+1 | | [WVx) , , n o d d i=0 | x | 5 t l - i L Theorem 2.2. L e t ty be a p o s i t i v e f u n c t i o n i n W s a t i s f y i n g Kx) i n+1 dx < x Then t h e r e e x i s t s f e B.^  such t h a t ty(x) <_ | x | f ( x ) . Hence (ty,c) i s a c o m p a t i b l e p a i r f o r a l l c > 0 . P r o o f : The p r o o f i s e l e m e n t a r y so f o r s i m p l i c i t y we s h a l l c o n s i d e r 1 ? t h e c a s e n = 3 . L e t f ( x ) = ) iK+x.. , +x„, +x„) . |x| 1 " 1 - 2 - 3 Our a s s u m p t i o n s on ty i m p l y t h a t R £(x) 3 | x | 3 dx < 0 0 and t h a t f ( x ) dx +•• R V f ( x ) dx + R~ x l v 2 f ( x ) | 2 | x | d x < R~ S i n c e the w e i g h t x s a t i s f i e s t h e c o n d i t i o n A^ , namely sup Q IQI dx Q- x x dx < 0 0 where Q i s any cube i n R , t h e R i e s z t r a n s f o r m s a r e bounded on L 2 ( R 3) . [See f o r exampl x F e f f e r m a n and Coifman [9]]. T h e r e f o r e we have t h a t i ^ c x ) ! 2 ^ 3yv < 0 0 . Hence R" x 9f 3y ( x ) A f (x) |dx <_ R R [ 9fj c __dx 3 ' 3 y ' I x l 1/2 t 1/2 ) ( |Af | |x|dx) < oo go t h a t f e B R" S i m i l a r l y g l x. — f ( x ) i = 1, 2, 3 a r e shown to be e l e m e n t s o f x B . Hence f e B^ and ^ ( x ) _< | x | f ( x ) . The case where n i s even i s t r i v i a l and t h e g e n e r a l i z a t i o n t o a r b i t r a r y odd n , i s o b v i o u s . §2. C o m p a t i b i l i t y f o r t h e p a i r ( N ( r - g ) , c) As n e c e s s a r y c o n d i t i o n s we s h a l l assume g s a t i s f i e s (2.2) and N ( r - g ) s a t i s f i e s (2.1). I f g > e f o r some p o s i t i v e and (^,c) i s c o m p a t i b l e f o r a l l c > 0 t h e n c e r t a i n l y (^,g) i s a c o m p a t i b l e p a i r . A v a r i e t y o f c o n d i t i o n s on g w i l l i m p l y t h a t N ( r - g ) s a t i s f i e s t h e h y p o t h e s e s o f Theorem (2.1) o r Theorem ( 2 . 2 ) . We s h a l l g i v e o n l y t h e s i m p l e s t . Theorem 2.3. Suppose g(x) s a t i s f i e s (2.2) f o r some r > 0 and  suppose t h a t N ( r - g ) ( x ) s a t i s f i e s (2.1) . Assume t h a t ( i ) g i s even w i t h r e s p e c t t o t h e a x i s x^ = 0 , i = l , . . . , n 1-e. g ( x 1 5 . . . , x n ) = g(+x 1,...,+x n) ( i i ) R n x 2 .. .. n r ^ I— dx < 0 0 and 7 • i n — L n-2 ( i i i ) _ i f n Is. e v e n I n - l 1=1 V (g) R | R ± ( r - g ) n x dx < 0 0 i f n i s odd £ ~ ~ i = l m/2-d v x (g) < co X i n / 2 - i Then N ( r - g ) s a t i s f i e s t h e h y p o t h e s e s o f Theorem ( 2 . 2 ) , hence ( N ( r - g ) , c ) i s c o m p a t i b l e f o r e v e r y c > 0 . The p r o o f i s o b v i o u s and we s h a l l o m i t i t . §3. The S m a l l e s t Superharmonic Suppose t h a t *Kx) s a t i s f i e s t h e c o n d i t i o n s o f Theorem 2.1 o r Theorem 2.2, so t h a t (^»c) i s c o m p a t i b l e f o r e v e r y c > 0 . Then t h e r e e x i s t s v ^ x ) s a t i s f y i n g (2.1) such t h a t >_ and 9v-^  •g^ -- <_ c . As a consequence of the proof of Theorem (2.1) and Evans Law, i f <Kx) i s continuous then v^(x) i s continuous. Theorem 2.4. There exists a continuous function v ( x ), v(x) <^  v^(x) such that v(x) > ty(x) , -j^-(x) < c and (v-iJO (|^- - c) =0 . — dy — dy Remark. The argument i s i d e n t i c a l to the one employed by Lewy and Stampacchia [21] so we s h a l l only o u t l i n e the proof. Proof. Let iKx,y) = ~ n I y I * ( T ) ; = D T ,1 ^,2, 2.(n+])/2 ' R n (| x-T I +y ) v Let C. denote the cla s s of continuous superharmonic functions n"l"l defined on R having the property that i f u(x,y) e C; , x e R n , y e R , then u(x,y) >_ty(x,y) - c|y| a n ^ such that u(x,0) s a t i s f i e s (2.1). Since 1 v 1(x,y) - c|y| = - c|y| + c n yJv^COdT e C< , we have that C i s not empty. Let V(x,y) = i n f (u)(x,y) . By u e C the arguments i n [21].it follows that v^(x,y) - c|y| >^'(x,y) >_. i()(x,y) - c | y | . Moreover V(x,y) i s continuous i n Rn+"^ and harmonic on the set { (x,y) :V(x,y) > *Jj(x,y) — c |y | } . Hence V(x,y) = v(x,y) - c|y| where v(x) = v(x,0) 9 v s a t i s f i e s (2.1), v(x) > ty(x) , — (x) < c and i f v(x) > ^(x) then — dy — 9v This completes the proof of existence of a s o l u t i o n to problem (1"). 71 REFERENCES [1] Adams, R. A. S o b o l e v Spaces, Academic P r e s s , New Y o r k , (1975), [2] B e u r l i n g , A. and M a l l i a v i n , P. On F o u r i e r t r a n s f o r m s o f measures of compact s u p p o r t , A c t a Math. 107 ( 1 9 6 2 ) . [3] Boas, R. P. E n t i r e F u n c t i o n s , Academic P r e s s . New Y o r k , N.Y. ( 1 9 5 4 ) . [4] B r e z i s , H. I n e q u a t i o n s V a r i a t i o n e l l e s , J . Math. P u r e s A p p l . (1971) . [5] . Problemes U n i l a t e r a u x , J . Math. P u r e s A p p l . (1972) . [6] C a r l e s o n , L. S e l e c t e d Problems on E x c e p t i o n a l S e t s , No. 13, D. Van N o s t r a n d Co. I n c . , P r i n c e t o n , N.J. ( 1 9 6 7 ) . [7] . Removeable s i n g u l a r i t i e s o f c o n t i n u o u s harmonic f u n c t i o n s i n R m , Math. Scandv 12 ( 1 9 6 3 ) . [8] C a f f a r e l l i , L. A. F u r t h e r r e g u l a r i t y f o r t h e S i g n o r i n i p r o b l e m , t o appear. [9] C o i f m a n , R. and F e f f e r m a n , C. Weighted norm i n e q u a l i t i e s f o r maximal f u n c t i o n s and s i n g u l a r i n t e g r a l s , S t u d i a Math. 15, ( 1 9 7 4 ) . [10] Duvaut, G. Problemes u n i l a t e r a u x en mechanique des m i l i e u x c o n t i n u s . I n t e r n a t i o n a l Congress of M a t h e m a t i c i a n s , N i c e ( 1 9 7 0 ) . [11] Duvaut, G. and L i o n s , J . L. Les i n e q u a t i o n s en mecanique e t en p h y s i q u e , P a r i s , Dunod ( 1 9 7 2 ) . [12] F r e h s e , J . Two d i m e n s i o n a l v a r i a t i o n a l problems w i t h t h i n o b s t a c l e s , Math. Z. 143 ( 1 9 7 5 ) . [13] . On S i g n o r i n i ' s p r o b l e m and v a r i a t i o n a l problems w i t h t h i n o b s t a c l e s , Ann. Sc. Nor. P i s a S e r . I V V o l . I V , 2, pp. 343 ( 1 9 7 7 ) . [14] F i c h e r a , G. 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Mech. and A n a l . 35 (1969). [25] Shamir, E. R e g u l a r i t y o f mixed second o r d e r e l l i p t i c p r o b l e m s , I s r a e l J . Math 6 ( 1 9 6 8 ) . [26] S t a m p a c c h i a , G. V a r i a t i o n a l i n e q u a l i t i e s , Theory and A p p l i c a t i o n o f Monotone O p e r a t o r s , G h i z z e t t i ( e d . ) , P r o c e e d i n g of NATO Advanced Study I n s t i t u t e , V e n i c e (1968). [27] S t e i n , E. M. S i n g u l a r I n t e g r a l s and d i f f e r e n t i a b i l i t y p r o p e r t i e s o f f u n c t i o n s , P r i n c e t o n U n i v e r s i t y P r e s s , P r i n c e t o n , N.J. ( 1 9 7 0 ) . [28] and F e f f e r m a n , C. H^-spaces of s e v e r a l v a r i a b l e s , A c t a Math 129 ( 1 9 7 2 ) . [29] Widman, K. 0. H o l d e r c o n t i n u i t y o f s o l u t i o n s o f e l l i p t i c s y s t e m s , M a n u s c r i p t a Math 5 (1971) . 

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