g(z, w; D) \u00E2\u0080\u0094 log . j[ . remains harmonic in a neighbourhood of w. xi i Introduction y y D X Figure 0.1: An example of circular symmetrization Baernstein's theorem [7] then states that for every r > 0 we have / ^(g{reie,w;D))M< / ^(g(reie,\w\;D\u00C2\u00AE)) d0, Jo Jo for every convex increasing function tp. This is of particular interest in the case of w = 0. The case of w = 0 and ip(t) = t implies the above-mentioned result on the increase of the Tj, under circular symmetrization. An interesting question is whether we may take to be any increasing function, not necessarily convex. This was asked by Hayman [59]. The answer turns out to be negative, even in the case of w \u00E2\u0080\u0094 0, as we shall prove in \u00C2\u00A71.7. A second set of interesting functionals are the wr harmonic measure functionals. Fix r > 0. Let D be a domain containing the origin. Let Dr be the intersection of D with the disc of radius r centred at 0. Then, let h be the harmonic function on Dr which equals 1 on the parts of the boundary of Dr which lie on the circle of radius r about the origin (i.e., h equals 1 on (dDr) fl {z G C : \z\ = r}) and which equals 0 on all the other parts of the boundary of Dr. (See Figure 0 . 2 . ) Put wr(D) = h(0). A probabilistic interpretation of this functional is that it is the probability that a Brownian motion starting at the origin reaches the circle of radius r centred at 0 before impacting on any other part of the boundary of D. Hence, intuitively it measures how easy it is for a Brownian particle to reach the circle of radius r while staying in D. Once again, Baernstein [7] has shown that wr(D) < wr(D\u00C2\u00AE) and hence circular symmetrization increases the wr functionals. The intuitive reason for this in terms of the probabilistic interpre-tation is that the circular symmetrization straightens out and consolidates the roads leading from 0 to the circle of radius r; see Figure 0.1. xiii Introduction k.y Figure 0.2: Boundary values for the Dirichlet problem associated with the wr functional on D In Chapter I we shall set up definitions and give some theorems on general and specific sym-metrization methods, as well as cite and summarize some material on the notions needed to define and study our functionals. We shall also give the most important results of Baernstein's famous paper [7], as well as our answers to the question of Hayman [59] mentioned above. Finally, we shall conclude Chapter I by obtaining a lower bound on the size of the set on which the function g(-,0;D) is radially decreasing; this will be useful to us in Chapter IV. In Chapter II, we shall consider symmetrization theory in discrete settings. For instance, we shall prove a generalization of a full analogue of Baernstein's above-cited results on the increase in Green's functions and tor-functionals under circular symmetrization in the setting of subsets of the discrete cylinder Z X Z T O . (Note that special cases of our results can be proved by the methods of Quine [90].) Of course the discrete cylinder is a discrete version of the continuous cylinder E X {z \u00E2\u0082\u00AC C : \z\ = 1}, which is conformally equivalent to the punctured plane C\{0} under a natural exponential conformal equivalence, and Baernstein's results can be lifted to the continuous cylinder, which is why we can say that our results on the discrete cylinder are analogous to his theorems. The method of Chapter II proceeds by proving convolution-symmetrization inequalities of the form \u00C2\u00A3 f^)K(d(x,y))g(y)< \u00C2\u00A3 f*(x)K(d(x,y))g*(y), x,y\u00C2\u00A3G x,yeG where G is a specific connected graph such as Z m , d is the shortest-distance metric on G, K is a decreasing function and / * and indicate symmetrizations of arbitrary functions / and g on xiv Introduction G. Symmetrizations inducing such inequalities do not exist for all graphs; indeed, we shall use a computer-based proof to show that they do not exist for the cube G = 7i\ or for the ternary plane G \u00E2\u0080\u0094 Z 3 . However, we shall prove the inequalities in the special cases of the circular graph G = Z m , the p-regular tree G = Tp and the octahedron edge graph G = Hs- And, of course, the case of the linear graph G = Z goes back to Hardy and Littlewood (see [58, Thm. 371]). Our method for proving these kinds of convolution-symmetrization inequalities will be a discrete version of a method of Beckner [18, 19, 20, 21], and Baernstein and Taylor [15], as presented by Baernstein [11]. This method in the discrete setting is in fact completely elementary. Given appropriate convolution-symmetrization inequalities, we then get several consequences. First, given the case of G = Tp we obtain a full Faber-Krahn inequality for subsets of the p-regular tree. Recall that the classical Faber-Krahn [47, 68] inequality stated that out of all domains D in K n of a fixed area, the first non-zero Dirichlet1 eigenvalue of the negated Laplacian \u00E2\u0080\u0094A = \u00E2\u0080\u0094 57J\u00E2\u0084\u00A2=i ~\u00C2\u00A7xl 1 S minimized precisely for a ball. Our Faber-Krahn inequality will provide a characterization of the subsets D of Tp of a given size which minimize the first non-zero eigenvalue of the negated combinatorial Laplacian with the Dirichlet boundary condition that our functions vanish outside our subset D. Not surprisingly, these subsets look like a discrete tree version of the balls which were optimal in the classical Faber-Krahn case (see Figure II.6.3 on p. 104). Our proofs, are completely elementary, although the proof of the uniqueness of the optimal subsets is quite involved. Secondly, given a convolution-symmetrization inequality on a constant degree graph G, we can, as already mentioned, obtain analogues of Baernstein's results. Indeed, we shall obtain results on symmetrization for domains which are subsets of Z x G (where G is Z, Z m , Tp or Hs); these will concern the increase in certain generalized harmonic measures and generalized Green's functions. In fact, we shall proceed by two methods. Firstly, we shall prove these results via a probabilistic approach going back to Haliste [56]. Secondly, we shall show even more general results for difference equations onZxG (even for some non-linear ones) by a modified version of a method of Baernstein [11] and Weitsman [99] who proved such results for partial differential equations. In Chapter III we shall engage in an analysis of various functionals on some collections of holomorphic functions. For instance, let 2$ be the collection of all functions / holomorphic on the unit disc {z G C : \z\ < 1} which satisfy /(0) = 0 and have \f'(x + iy)\2dxdy < 1. Note that \f'(x + iy)\2 is the Jacobian of the mapping / , and hence the displayed condition can be interpreted as saying that the area of the image of / counting multiplicities does not exceed 7r. If / is one-to-one then this condition simply says that the area of the image of / does not exceed w. Let $ be a Borel-measurable function on C. Put 1 f2ir A * ( / ) = ^y o The Dirichlet eigenvalues are obtained by having \u00E2\u0080\u0094A act on those functions on our domain D which vanish on the boundary of D. xv Introduction for / G 03. (This makes sense since functions in 03 have radial limits almost everywhere on the unit circle.) If . Given an extremal for A$, we will under some hypotheses on $ be able to obtain a variational equation that this extremal will have to satisfy. The variational equation is not a differential xvi Introduction equat ion, but involves a pseudo-differential operator, which complicates the analysis. However, the equation is good enough to yield qual i tat ive results. For example, jo int ly wi th A lec M a t h -eson [75] it has been shown that if $ is inf initely d i f ferent ia te on C then the extremals, if they exist, must extend to be infinitely differentiable functions on the unit circle (recall that they were assumed to be holomorphic inside the unit disc; this did not by itself say much about their regulari ty on the boundary of the unit disc). We wi l l see that some of the results mentioned in this paragraph continue to work in more general settings such as a-weighted Dir ichlet spaces. F ina l ly , we wrap up Chapter III by connecting the results wi th symmetr izat ion theory. We shal l prove a result on the relation of a-weighted Dir ichlet norms (0 < a < 2) and symmetr ic decreasing rearrangement, and use this result to prove that if / is extremal for A $ where $(z) is of the form t} = (f*)t = (ft)*, by Theorem 2.1. B u t by Theorem 2.1 and (2.4) we have ( ^ o / ) * = ( ( 0 o / ) A ) * = ( / t ) * so that ((4> o / ) # ) A = (4> o (f*))\ as desired. \u00E2\u0080\u00A2 D e f i n i t i o n 2 .4 . Suppose that we have a measure p on X such that al l elements of T are p-measurable and a measure v on Y such that al l elements of Q are immeasurable. We say that a rearrangement # : T \u00E2\u0080\u0094> Q is m e a s u r e - p r e s e r v i n g providing p(A) = v(A#) for al l A \u00C2\u00A3 T. Example 2.1 (Decreasing rearrangement to B.Q). Let Y be the interval [0, oo) equipped wi th Lebesgue measure. Let (X, T, p) be any measure space. G iven A \u00E2\u0082\u00AC T, let A* = [0,p(A)). Then * is easily seen to be a rearrangement, and it is also clearly measure-preserving. Since A* is always open, by Remark 2.4 it follows that / * is lower semicontinuous on [0, oo) whenever / is ^\"-measurable. Example 2.2 (Decreasing rearrangement on Z Q j . Consider the discrete space X = Y = Z+. P u t T = Q = 2 z o + . Define 5* = { i e Z + : i < |5|}, where \S\ is the cardinal i ty of a subset S of X. Then , clearly \S*\ \u00E2\u0080\u0094 \S\ and it is easy to see that * is a measure preserving rearrangement. 8 Chapter I. Definitions, background material and introductory results Given an extended real / on Z Q , we may describe the / * in a very intuit ive way. Indeed, for any n \u00C2\u00A3 Z + , the numbers / * ( 0 ) , . . . , f*(n - 1) are a list of the n largest values of / . We cal l / * the d e c r e a s i n g r e a r r a n g e m e n t of / . The reader is advised to often keep in mind the previous two examples which are very typical and are really the most basic types of rearrangement. Example 2.3 (Schwarz symmetrization in Kn). Let U be an arbi t rary Lebesgue measur-able subset of R\u00E2\u0084\u00A2. Let U\u00C2\u00AE be an open ball in R r a centred on the origin and having the same volume as U. (If U has infinite volume, then put U\u00C2\u00AE = R n . ) It is easy to see that \u00C2\u00AE is a measure preserving rearrangement on the Q be a rearrangement. Then, /* is a sequence of extended-real functions increasing pointwise to /*. Proof of Lemma. It is clear f rom the definit ion of ff and property (i) of rearrangements that the sequence increases pointwise. Let g be the pointwise l imit of the fn-F i x A \u00E2\u0082\u00AC R . Note that ( / I ) A Q (/2)A Q \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 since we have fx < fa < \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 pointwise. Moreover, note that oo / A = U (/\u00C2\u00BB)*\u00E2\u0080\u00A2 (2-9) 71=1 For , if x G / A then f(x) > A and hence for sufficiently large n we must have fn(x) > A so that x \u00C2\u00A3 (/n)A) while, conversely, if x e (fn)\ then f(x) > fn(x) > A and so x 6 f\. In the same way, oo 9X = (J [f*)x. 71=1 Bu t , by Theorem 2.1 we have ( / * ) A = ( ( / T I ) A ) * - Hence, oo 9X=\J{{fn)x)*. n=l 14 Chapter I. Definitions, background material and introductory results But ( / I ) A C ( / 2 )A C \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 so that by property (ii) of rearrangements we have ( CO U(f^x n=l Thus, gx = (fx)* by (2.9). By Theorem 2.1 it follows that gx = (f*)x and, since A was arbitrary it follows that g = f#, which completes the proof. \u00E2\u0080\u00A2 Proof of Theorem 2.3. We have already proved the result in the positive case in Theorem 2.4. Suppose now that / \u00C2\u00A3 Lp(p) and g 6 Lq(p) for 1 < p < oo, and that if p 6 {l,oo} then p, is c-finite. Let 4>(t) = t+ and ip(t) = \u00E2\u0080\u0094 (t~). Then, (p and ip are continuous monotone functions, and hence commute with rearrangements by Theorem 2.2. It follows that f* = cp o (/#) + ip o (/#) = ( o /)# + (iP o /)# , while of course f = __(Rez) for some function (j) on K. Then <& is subharmonic on {z : Im z \u00C2\u00A3 (a, b)} for (a, b) a non-empty open interval in K if and only if is convex on K. Proof. Suppose first that __

(p(\z\) is subhar-monic on C. To see this, note that in such a case t t-\u00C2\u00BB (p(et) must also be convex since t \u00C2\u00BB-> et is convex. 5. Least harmonic majorants, harmonic measures and uniformiz-for every r 6 (0,oo), and so we see that indeed $ must be subharmonic everywhere. \u00E2\u0080\u00A2 ers This section still contains no really new results, but is intended to give a precise meaning to our terminology and to collect some known facts which we will later use. 26 Chapter I. Definitions, background material and introductory results We work all the time on domains D in the complex plane C. For harmonic measures, Dirichlet problems, Brownian motions, etc., our basic reference is the book of Doob [39]. 5.1. Dirichlet problem and harmonic measure Definition 5.1. Let D be a domain and / a function on the Euclidean boundary 3D. Then we say that the Dirichlet problem on D with boundary value / is solvable if / is resolutive, i.e., if there exists a PWB solution [39, \u00C2\u00A71.VIII.2] F on D. We then say that F is the solution to the Dirichlet problem on D. The above definition is rather technical, but in practice this shall not concern us. All that is necessary for intuition is to note that F is a harmonic function on D which in some sense (which sense is made precise by the invocation of the PWB method) has the limit / at the boundary of D. Definition 5.2. A domain D is Greenian if there exists a positive nonconstant superharmonic function on D. Remark 5.1. As Doob [39, \u00C2\u00A71.11.13] notes, any domain D which is not dense in C is Greenian. For, if w G C\D, then the function f(z) = c + log \z \u00E2\u0080\u0094 w\ is harmonic (hence superharmonic) and nonconstant on D, while for large enough c it will be positive. Remark 5.2. By Doob [39, \u00C2\u00A71.V.6], the plane C is not Greenian. The domains with which this thesis will be concerned are primarily the domains of finite area. The following proposition implies that all domains of finite area are Greenian. Proposition 5.1. Any domain D which is simply connected or whose complement has positive Lebesgue area measure must necessarily be Greenian. The proof of the case of the complement having positive Lebesgue area measure will be delayed until \u00C2\u00A75.4. 27 Chapter I. Definitions, background material and introductory results Proof in simply connected case. Suppose first that D is simply connected. Let / be a Riemann map from D onto O, i.e., a univalent map sending D onto D whose existence is guaranteed by the Riemann mapping theorem [94, Thm. 14.8]. Then z 4 - log is superharmonic on D Definition 5.3. Let A be a subset of dD where D is domain. Then, the harmonic measure of A in D is defined to be the function z \\u00E2\u0080\u0094>\u00E2\u0080\u00A2 u>(z, A; D) which is the solution of the Dirichlet problem on D with boundary value 1 on A and 0 on dD\A, if this solution exists. Harmonic measure exists for any Borel set [39, \u00C2\u00A7\u00C2\u00A71.VIII.4 and 1.VIII.6]. Remark 5.3. Harmonic measure is monotone with respect to A and D. More precisely, if A C A' and D C D' with A C dD and A' C dD' then u(z, A; D) < u(z,A';D'). This follows from the fact that o>(-, A; D) is easily seen to lie in the lower PWB class for the Dirichlet problem of which u(-t A';D') is the solution. (See [39, \u00C2\u00A71.VIII.2] for definitions.) For convenience, we will sometimes write = OJ(Z, \u00E2\u0080\u00A2; D). The following result is very impor-tant, although rather technical. Theorem 5.1 (cf. Doob [39, \u00C2\u00A71 .VIII .8 ] ) . Harmonic measure exists for every Borel subset A of dD where D is a Greenian domain. The set of subsets A for which harmonic measure exists is a a-algebra %D> and for each fixed z, the function A i-\u00C2\u00BB u(z, A; D) is a measure on Tip. Technical remark 5.1. This is given by Doob [39, \u00C2\u00A71.VIII.8] in the case of dD being given by a metric compactification, and not the Euclidean boundary. However, the case of the Euclidean boundary follows from the fact that for an unbounded Greenian domain, the singleton {oo} has null harmonic measure [39, Example 1.VIII.5(a)]. Definition 5.4. Let D be a Greenian domain. We say that / \u00E2\u0082\u00AC L1 (u>D) if / is a %\u00C2\u00A3> measurable function on dD such that (see [55, p. 34]) and clearly positive and nonconstant. \u00E2\u0080\u00A2 (5.1) 28 Chapter I. Definitions, background material and introductory results for every z \u00C2\u00A3 D. Remark 5-4- In fact, it suffices to verify (5.1) for any single point z \u00C2\u00A3 D; see [39, \u00C2\u00A71.VIII.8]. The following result is quite important. Theorem 5.2 ([39, \u00C2\u00A71 .VIII .8 ] ) . Let f \u00C2\u00A3 Ll(uD) for a Greenian domain D. Then, the Dirichlet problem with boundary value f on dD has a solution F given by F(z) = f fdu,?. JdD Remark 5.5. The harmonic measure UQ coincides with normalized Lebesgue measure on T for Borel sets. The easiest way to see this is to simply note that both measures are rotation invariant finite measures, and hence must coincide on the Borel sets by the uniqueness of Haar measure (see, e.g., [93, Thm. 14.19]). 5.2. Regularity for the Dirichlet problem Definition 5.5. Let D be a Greenian domain. Call a point z of dD regular if for any / \u00C2\u00A3 Ll(uD) on dD which is continuous at z \u00C2\u00A3 dD we have F(C)-\u00E2\u0080\u00A2/(*), as C \u00E2\u0080\u0094>\u00E2\u0080\u00A2 z from within D, where F is the solution of the Dirichlet problem on D with boundary values / . A domain D is said to be regular if every point of its boundary is regular. The results given in Doob's book [39, \u00C2\u00A71.VIII] show that these definitions are equivalent to the standard definitions in the case of Greenian domains. Note that if / \u00C2\u00A3 Ll[uD) is continuous on dD for a regular domain D, then the solution to the Dirichlet problem on D extends to a continuous function on D which agrees with / on dD. We shall have occasion to use the following simple criterion for regularity and for the Greenian character of a domain. 29 Chapter I. Definitions, background material and introductory results Theorem 5.3. Let D be any domain. Let z \u00C2\u00A3 dD. Suppose that there is an curve w(r) = z + reie(r\ 0 < r < e, lying outside D, for some e > 0 and a continuous function 9. Then D is Greenian and z is regular. Only the Greenian character needs to be proved, since, given this Greenian character, the regularity is a consequence of [60, Thm. 2.11]. We shall prove the Greenian character in \u00C2\u00A75.4. Definition 5.6. We say that a domain D is a C 1 domain if for every z \u00C2\u00A3 dD there exists a function

__(t) ^ z for all t \u00C2\u00A3 (0, <5i). (Such a 8i exists since 4>'(0) ^ 0.) Define p(t) = \ (t) - z\ for t \u00C2\u00A3 [0, o\"i). Using the fact that (f)'(0) ^ 0 it is easy to see that p is continuously differentiable on [0,$i), with p'(0) > 0. Thus, there is an e > 0 such that there is a continuous function p~x: [0,s] \u00E2\u0080\u0094> [0,5) with p(p~l(r)) = r for all r \u00C2\u00A3 [0',e], and S \u00C2\u00A3 (0,<$i). We have p(t) > 0 for t \u00C2\u00A3 [0, 6]. For t \u00C2\u00A3 (0, 8} let 4>(t) ~ z m = \__

.o+ ip(t) exists by L'Hopital's rule, so that we may extend ip to a continuous function from [0, 8] to T. We may then choose a continuous logarithmic function L: [0,8] \u00E2\u0080\u0094> C such that ip(t) = eLM for all t \u00C2\u00A3 [0,8]. In fact L will map [0,8] into iR. Define 9(r) = L(p 1(r))/i. Then, for 0 < r < e we have z + re 30 Chapter I. Definitions, background material and introductory results \u00E2\u0080\u00A2 5.3. Least harmonic majorants Let $ be a subharmonic function on a domain D. Write\" LHM(z, <&; D) = inf{/i(z) : h is a harmonic function on D with h > <&}, for the least harmonic majorant of $ at z \u00C2\u00A3 D. Note that LHM(-,$;/J) is either identically +00 on D or else it is harmonic there [39, \u00C2\u00A71.111.1]. If it is harmonic, then we say that $ has a harmonic majorant on D. The following result is quite well-known. Theorem 5.4 (cf. [46, equation (1.4)] and [39, Example l.VIII.3(a)]). Assume that\u00C2\u00AE is a subharmonic function on D with an extension to D such that the extended function is continuous at every point of D while $\QD \u00E2\u0082\u00AC Ll(uD). Assume that $ (as a function on D) is continuous at every point of dD. Then, Proof. It is clear that $ is in the lower PWB class for cj>=$\dD (see [39, \u00C2\u00A71.VIII.2] for defini-tions). Let $ be the solution of the Dirichlet problem on D with boundary value 4> on dD (this exists by Theorem 5.2). Since $ is in the lower PWB class, it follows that f < $ on D. I now claim that LHM(-, <1>; D) is in the upper PWB class for . If this is true then on D. But since LHM(\u00E2\u0080\u00A2,<&; D) is the least harmonic majorant while ^ is harmonic, it then follows from the above inequalities that $ = LHM(-, $; D). The conclusion of the theorem then follows from Theorem 5.2. $ < * < LHM(-,$;D) 31 Chapter I. Definitions, background material and introductory results Thus we must prove that LHM(-, D) is in the upper PWB class for cj). To do this, let zn \u00E2\u0080\u0094> w where zn \u00C2\u00A3 D and w \u00C2\u00A3 dD. We must prove that liminf LHM(zn,$;D) > Mw). n-\u00C2\u00A5oo But LHM (2 n , D) > <&(zn) \u00E2\u0080\u0094$(w) = (w), where we have used the continuity of $ on 3D. \u00E2\u0080\u00A2 5.4. Brownian motion and harmonic measure Standard Brownian motion on R\" is a Markov process {5i}ie[o,oo) with almost surely continuous paths, values in E\" and Gaussian increments; see [39, \u00C2\u00A72.VII.2] for a rigorous definition (in the case n = 2, see also [38]). We often drop the word \"standard\" from the term \"standard Brownian motion\". We use Pz(-) and Ez[-] to indicate probabilities and expectations when the Brownian motion is conditioned to start from the point z \u00C2\u00A3 K\" at time 0. Given a domain D C K 2 , let TD = inf {t >0:Bt\u00C2\u00A3D} be the first exit time of Brownian motion from the domain D. We then have the follow-ing very useful connection between Brownian motion, harmonic measure and PWB solutions of Dirichlet problems. Theorem 5.5 (see, e.g., [39, \u00C2\u00A72 . IX .10 and \u00C2\u00A72 . IX.13] ) . LetD bea Greenian domain. Then P(rr) < oo) = 1. Let A be an rio-measurable subset of dD. Then, u(z,A;D) = Pz(BTDeA). (5.2) Moreover if f \u00C2\u00A3 LM^OJ0) and F is the solution of the Dirichlet problem on D with boundary value f on dD, then F(z) = E*[f(BTD)}. (5.3) 32 Chapter I. Definitions, background material and introductory results Note that (5.2) is equivalent to (5.3) by Theorem 5.2. The above result shows that harmonic measure of A at z in D is the probabi l i ty that when a Brownian mot ion started at z hits dD, it hits i t wi th in the set A. Thus , it measures how large A is compared to the rest of dD as seen f rom the point of view of z. A very impor tant result in two dimensions is the theorem of Levy that if / is a non-constant analyt ic funct ion and Bt a Brownian mot ion in the plane, then f(Bt) is a Brownian motion moving perhaps at a variable speed. Th is is known as the conformal invariance of Brownian mot ion. M o r e precisely, we have the fol lowing result. Theorem 5.6. If f is a non-constant analytic function on a domain D and z a fixed starting point in D then there exists a strictly increasing continuous function a (depending on z and f) such that the process {f(Ba^)}0 x_ Xt with probabi l i ty 1. I now cla im that under this extension T = mf{t > 0 : Xt i D} w i th probabi l i ty 1. (5.6) It is clear that T > in f {\u00C2\u00A3 > 0 : Xt \u00C2\u00A3 D} since Xt G D whenever t < T. Thus , to prove our c la im it wi l l suffice to show that l im XT 4: D w i th probabi l i ty 1. To do this, let 5 be the set of points of our underly ing probabi l i ty space such that on S we have: (i) T < oo and ru < oo (ii) t i\u00E2\u0080\u0094>\u00E2\u0080\u00A2 Xt is continuous on [0, T] (iii) t H-> BT is continuous on [0, TO] (iv) XT = f(B^). We have already seen that (i)-( i i i) happen wi th probabi l i ty 1. We now remark that so does (iv). To see this, note that BT G T since BT is continuous on [0,Tp], and remark that the \" / \" in \" / (BT )\" is short for the non-tangential l imit n.t. l i m / . (Of course, the non-tangential l imi t of / exists almost everywhere by Theorem 3.5.) Bu t the standard connection between non-tangential l imi ts of harmonic functions and their l imits along Brownian paths (see Brelot and Doob [27], Constant inescu and Cornea [37], as well as Burkholder and Gundy [29]) then shows that /(B^) = l i m ^ . ^ - f(BT) almost surely, f rom which the desired result follows upon replacing t w i th cx(f). We shall prove that everywhere on S we have l im t _, .x- XT \u00C2\u00A3 D. For suppose that we are work ing at a point u> G 5 and that all our random variables are sampled at precisely CJ, and 36 Chapter I. Definitions, background material and introductory results finally assume that lim^x- XT \u00E2\u0082\u00AC D. We shall obtain a contradiction. To do this, let T(t) = BT and j(t) = f(BT) for t \u00C2\u00A3 [0, T p ) . Set 7(715) = XT- Since lim XT = lim y(t), it follows that 7 is a continuous function on [0, TJJ]. Because w. D = D \u00E2\u0080\u0094>\u00E2\u0080\u00A2 D is a universal covering map, it follows that there is a continuous lifted path 7 : [0, ru] \u00E2\u0080\u0094> D such that ^ 0 7 = 7 on [0,7Tu] and 7(0) = 0 (see, e.g., [17, Chapter 7]). By uniqueness of lifts [17, Thm. 7.4.3], since 7(0) = T(O) and 7r o 7 = 7 = n o T on [0, T U ) , it follows that 7 = T on [0, TJU). But lim j(t) = 7(7D) \u00E2\u0082\u00AC D, *\u00E2\u0080\u0094*-n>\u00E2\u0080\u0094 while lim r(t) = B ^ ^ D , a contradiction. Hence, (5.6) is valid. Applying Theorem 5.5 to the standard Brownian motion X T , we see that LU(0,A;D) = P(Xt \u00C2\u00A3 A). But, given the set S defined before, which event has probability 1, we have XT = f(BTD). Thus, u(0,A;D) = P(f(BTD)eA). But B T D has uniform distribution on T when BQ = 0. (This can be seen from rotation invariance of Brownian motion and uniqueness of Haar measure on T [93, Thm. 14.19].) Hence u(0,A;D) = P(f(BTD)\u00C2\u00A3A) = \u00C2\u00B1- lA{f^d)) d6. It Jo \u00E2\u0080\u00A2 5.6. Green's functions There is more than one equivalent definition of a Green's function that could be given. The one that we shall give will be a two step definition. First suppose that D has regular boundary and 37 Chapter I. Definitions, background material and introductory results that C\D is non-empty. Then, a Green's function g(-, w; D) for D is defined to be any function such that: (i) g(z, w;D) \u00E2\u0080\u0094 0 whenever z fi D or w fi D (ii) g(-,w;D) is continuous on C\{u;} for each fixed w (iii) g(-,w;D) is harmonic on for each fixed w (iv) z \-t g(z,w;D) \u00E2\u0080\u0094 log is a harmonic function in a neighbourhood of w if to G J D is fixed. For uniqueness and existence see [60, Thms. 1.14 and 3.13]. Now suppose that D is any Greenian domain. It is easy to see that we can find a sequence Dn of regular domains with compact closures such that Di C D2 C \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 and D = U^Li Dn. (It is easy to construct such domains, using Theorem 5.3 for this purpose. As Hayman and Kennedy [60, p. 253] remark, it is easy to even make sure that each Dn is a union of finitely many discs.) We then define the Green's function of D as g(z,w;D)= lim g(z,w;Dn). n\u00E2\u0080\u0094>oo This limit exists and is independent of the choice of the Dn [60, Lem. 5.6]. It is worth noting that g(z, w; D) = g{w, z; D) for all z and w [60, Thm. 5.26]. Remark 5.8. Note that \u00C2\u00ABjf(z,0;D) = log T^T , (5.7) \z\ for z G D. This is easily verified since log ^ is locally the real part of an analytic function away from the origin, and hence harmonic there, since condition (iv) is trivial and since log ||r vanishes for z G #B. More generally, for z and w in D we have (5.8) g(z,w;ty = log 1 \u00E2\u0080\u0094 zw w 38 Chapter I. Definitions, background material and introductory results To see this, note that the function vanishes as desired for z and w in dD and that it is locally the real part of an analytic function of z on B\{u;}. The only thing left to verify is condition (iv). But, g(z, w; D) \u00E2\u0080\u0094 log j\u00E2\u0080\u0094-\u00E2\u0080\u0094j- = log |1 \u00E2\u0080\u0094 zw\ \u00E2\u0080\u0094 Relog(l - zw), and hence is locally the real part of an analytic function for z near w. The following very well known result is worth noting. Theorem 5.8. Let U ^ C be a simply connected domain. Let w G U and suppose that / : D \u00E2\u0080\u0094> U is a Riemann map from D onto U with /(0) = w. Then U is Greenian and g(z,w;U) = log I for all z G U. 5.7. Riesz' theorem and representation of least harmonic majorants Theorem 5.9 (Riesz; see [60, Thm. 3.9]). Let $ be subharmonic in a domain D, with $ ^ \u00E2\u0080\u0094oo. Then there exists a unique positive Borel measure p in D such that for any compact subset E of D the function z H-> $(z) \u00E2\u0080\u0094 / log\z \u00E2\u0080\u0094 w\dp(w) JE is harmonic on the interior of E. We shall write p = p\u00C2\u00A7 and call it the Riesz measure of Remark 5.9. The uniqueness of /x$ implies that //$ depends only locally on $. More precisely, given $ on D and given a subdomain U C D we have p$\v = p$ on U. If $ G C2{D) then = ^ A $ , where A = \u00C2\u00A3 ^ + (See [39, \u00C2\u00A71.8].) This together with the local dependence implies that if U C D is a subdomain in which $ is harmonic, then p has support in D\U since A$ = 0 on U. 39 Chapter I. Definitions, background material and introductory results The following result, although very trivial, will in fact be useful. Theorem 5 . 1 0 . Let 3>(z) = *{t) by max(<^ (i), 4>(L)), which does not change h (this can be seen from Theorem 5.4 and the fact that the change does not affect on dD) and which preserves subharmonicity as can be easily seen, we may assume that $ is constant on D(L). By Theorems 5.10 and 5.11 we have roo r-2it h(z) = (z) + / g(z,reie;D)ded^(r) Jo Jo for a positive measure u^. Since $ is constant on D(\u00C2\u00A3), the support of lies in [L, oo). It thus suffices to show that for any fixed r \u00C2\u00A3 [I, oo) we have gr monotone increasing on (\u00E2\u0080\u0094L, 0], where 9r{z) = ^J2J g{z,re*6;D). I claim that gr is circularly symmetric on O(L) if r > L. To see this note that for z G B(\u00C2\u00A3) we have i r2n / 1 \ 1 f2v 1 since g(z,-;D) \u00E2\u0080\u0094 log j-^ -q is harmonic on D(L). Now, on the right hand side of the above displayed equation, the only term which has any dependence on argz is g(z,0;D). But z i-4 g(z,0;D) is circularly symmetric (Theorem 6.1) so that indeed gr is circularly symmetric on D(L). Moreover, it is harmonic on D(L). The desired monotonicity of gr follows by scaling and an application of Corollary 6.3. \u00E2\u0080\u00A2 6.3. Steiner symmetrization Recall the definition of Steiner symmetrization as given in Example 2.4. We call a set U with U = UB Steiner symmetric about the real axis. We shall sometimes omit the words \"about the real axis\". Note that UB is always Steiner symmetric. Note also that Steiner symmetrization is a rearrangement in the sense of Definition 2.2 on the cr-algebra of all Lebesgue measurable subsets of C. 51 Chapter I. Definitions, background material and introductory results Note that a set is Steiner symmetric about the real axis if and only if every vertical line meets this set in an open interval symmetric about the real axis. As in Proposition 6.1, we easily see that if D is open then Y(-;D) is lower semicontinuous on E and DB is open. (Actually, the proofs are now easier because we do not need to distinguish any exceptional set such as the previously exceptional set of those r where 6(r; D) = oo.) Analogues of Theorems 6.1 and Theorem 6.2 can also be proved in this setting by the methods of Baernstein [7]. We shall not give the proofs, but we do state the following two results. T h e o r e m 6 .6 . Let Q(z) = (f>(Rez) be subharmonic and let D be Greenian. Then, LHM(z, D) < LHM(Re z, DB), where DB denotes Steiner symmetrization. The proof of this result follows via the Steiner analogue of Theorem 6.1 analogously to Corol-lary 6.1. We only give this result because in \u00C2\u00A7IV.10.3 we shall parenthetically mention that it could be used in a slightly modified proof of Theorem IV. 10.1. The following result is an analogue of Theorem 6.2. It could also be proved by the same methods of Baernstein [7], although it was first proved by Haliste [56] by Brownian motion methods similar to the methods we shall employ in \u00C2\u00A711.9. Haliste actually only gave the result for certain sufficiently regular domains. However, an approximation such as the one in the proof of [7, Thm. 7] easily yields the general case. T h e o r e m 6.7 ( H a l i s t e [56]). Let D be a Greenian domain. Assume that D C {z G C : Re z < M}. Then, for any z \u00C2\u00A3 D we have co(z, {Re z = M} C\D;D) < u(Re z, {Re z \u00E2\u0080\u0094 M} n LP; DB). Remark 6.8. A Steiner symmetric domain is necessarily simply connected. (This does not hold for a circularly symmetric domain. For instance, D\D(\u00E2\u0080\u0094\; \) is circularly symmetric but 52 Chapter I. Definitions, background material and introductory results evidently not simply connected.) This observation will be of great importance in \u00C2\u00A7IV.10. To see the validity of this observation, let 7 be an arbitrary curve in a Steiner symmetric domain D. Define Ft(z) = Rez + i(l \u00E2\u0080\u0094 t) Imz for t \u00C2\u00A3 [0,1] and z \u00C2\u00A3 C. Then, Ft is a homotopy with Ft o 7 \u00C2\u00A3 D for all t \u00C2\u00A3 [0,1], Fo o 7 = 7, and F\ o 7 C R. Hence, 7 is homotopic in D with a curve lying on the real axis. But clearly any such curve is homotopic with the trivial curve as can be seen by applying the homotopy Gt(z) = (1 \u00E2\u0080\u0094 i) Rez + iQz for t \u00C2\u00A3 [0,1] and z \u00C2\u00A3 C . 7. Counterexamples to a question of Hayman 7.1. Hayman's problem The material of the present section is basically taken from the author's paper [89]. Let U be a Greenian domain with w \u00E2\u0082\u00AC U. UWT\ = {z ell : g(z, w; U) > A}. Then, UW 0 if and only if U is circularly symmetric and w > 0 (one implication follows from the fact that UQ \u00E2\u0080\u0094 U; the other is due to Baernstein [7, Corollary on p. 154].) Hayman [59, Question 5.17] had asked whether we necessarily have (Uw,x)\u00C2\u00AE C (U\u00C2\u00AE)\W^\. As Baernstein [7] notes, this is the same as asking whether we always have g(ret6,\w\;U\u00C2\u00AE)>g(re'e, w;U), (7.1) where g(relB', w; U) is u\u00C2\u00AE(ret6) for u(retd) = g(retd, w; U). The equivalence of the two questions then follows from the fact that {z \u00C2\u00A3 U : g(z, w; U) > A} = (UWJ\)\u00C2\u00AE (see Theorem 2.1 as g(-, w; U) is lower semicontinuous and \u00C2\u00A9 is a rearrangement when confined to open sets). Recall that Baernstein [7] had proved the weaker inequality that g(rel9,\w\;U\u00C2\u00AE)d6> / g(rel6\w;U) d0, Jo for 0 < 61 < IT (Theorem 6.1, above). However, we will show that in general the stronger inequality (7.1) is not valid, and the answer to Hayman's question is negative, even when 53 Chapter I. Definitions, background material and introductory results Figure 7.1: The circularly symmetric domain Ua. The pole will be at a or at \u00E2\u0080\u0094a. restricted to U being simply connected and w = 0. In one of our examples, (7.1) will be false even though U is circularly symmetric (but of course w cannot lie on the non-negative real axis then). 7.2. The three counterexamples We give three counterexamples. The first is the easiest, and this is the one with U circularly symmetric. Fix any 0 < a < | . Let Ua be a disc of unit radius centred on the point a. Clearly, Ua is circularly symmetric and U\u00C2\u00AE = Ua. (See Figure 7.1.) Theorem 7.1. There exists r\ S (a, 1 \u00E2\u0080\u0094 a) such that for any r 6 (a, r j we have min g{rel6, a; Ua) = g(-r, a; Ua) < min g{rel6, -a; Ua). 0 0 The completely elementary proof will be given later. This gives a counterexample to (7.1) since Ua = U\u00C2\u00AE and since min# g(rel6, -a; Ua) = g{-r, -a; Ua) by definition of g. We now restrict the pole to lie at zero. This will make things a little more difficult. Theorem 7.2. There exists a domain U in the plane and strictly positive numbers r and e such that g(reie,0;U\u00C2\u00AE)*

0 and

* Q is a symmetrization in the technical sense of \u00C2\u00A71.2. Moreover, f* as defined above is easily seen to agree with the definition in (1.2.1) since it is easy to see that in our current definition we have (/#)A = (fx)*, which agrees with Theorem 1.2.1. Definition 1.1. Two real functions / and g on a set X are said to be similarly ordered providing that for every x and y in X we have f(x) < f(y) if and only if g(x) < g(y). A function F is -^decreasing if x -< y always implies F(x) > F(y). 70 Chapter II. Discrete symmetrization The fol lowing result is very well known [58, T h m . 368] 2 and has sl ightly less restrictive hy-potheses than Theorem 1.2.3. T h e o r e m 1.1 ( H a r d y - L i t t l e w o o d ) . For a pair of real functions f and g we have E fWsw ^ E t1-1) assuming that both sides make sense. If moreover the left hand side is finite, then equality holds if and only if f and g are similarly ordered. 2. A general framework for proving discrete master inequalities Let ( M , d) be a metr ic space such that M is at most countable. G iven two real functions / and g on M, an increasing posit ive convex funct ion $ on [0, oo) and a decreasing posit ive funct ion K on [0, oo), define Q(f,g;*,K)= E *(!/(*) \" S(y)\)K(d{x,y)). Then , we say that t h e m a s t e r i n e q u a l i t y h o l d s f o r M w i t h t h e o r d e r i n g -< providing that for al l real / and g, and al l $ and K as above, we have Q(f,g;$,K)>Q(f*,g#;$,K). (2.1) Such inequalit ies were considered in continuous cases by Beckner [18, 19, 20, 21] who generalized the work of Baernstein and Taylor [15]; see also [11]. We now outl ine a general approach to proving master inequalit ies in our discrete sett ing v ia an adaptat ion of the Baernstein-Taylor-Beckner approach; our proofs are based on the descript ion of the approach as given by Baernstein [11]. Because we cannot hope to get (2.1) for a general M and a general -<, we must make a number of assumptions. (Indeed, not every M has a well-ordering -< under which the master inequal-i ty holds\u00E2\u0080\u0094see Theorem 7.1 in \u00C2\u00A77, below, for two very simple and natural counterexamples.) 2 T o see that [58, Thm. 368] implies our result, note that ( M , -<) is order isomorphic to ([0, N] f l Z j , <) for some ./v e zJu {oo} . 71 Chapter II. Discrete symmetrization Basically, we shall make assumptions which will ensure that the method of proof given for the continuous case in [11] works. Then, to prove the master inequality in any concrete case, we will only need to prove the assumptions in that case. Recall that an isometry p of M onto itself is said to be an involution if p o p is the iden-tity function. Whenever we say \"involution\" we shall mean \"isometric involution\"; the term \"isometric\" will sometimes be explicitly given and sometimes dropped. Given an involution p of M, we set ftp = {x G M : x -< px}, and Fixp \u00E2\u0080\u0094 {x G M : px = x}. Then, fip, Fixp and pf)p are three disjoint sets whose union is M. Assumption A. There is a transitive set 3 of isometric involutions of M such that whenever p is in 3, and the points x and y are both in S)p, then the inequality d(x,y) Zo } , while for x0 > 0 we have 9yp = {x G Z : x < x0}. Finally, it is likewise easy to verify that Assumption A holds. The principal result of the present section is as follows and as mentioned before is essentially due to Baernstein, Taylor and Beckner. A proof will be given later in this section. 72 Chapter II. Discrete symmetrization Theorem 2.1 (Master Inequality). If Assumption A holds, then the master inequality for M with ordering -< must also hold. I.e., if Assumption A holds, then for any real f and g, and for any functions $ and K from [0, oo) to [0, oo) such that $ is increasing and convex and K is decreasing, we must have Q(f,g;*,K)>Q(f*,g*;*,K). (2.3) Now, for x \u00C2\u00A3 M and r \u00C2\u00A3 [0, oo), let V(x;r) = \{y : d(x,y) < r}\. In order to get an appropriate convolution inequality, we make the following assumption. Assumption B. For every fixed r \u00C2\u00A3 [0, oo) the number V(x;r) is finite and independent of x \u00C2\u00A3 M. An easy immediate consequence of Assumption B is that if K is any positive real function then the function x E K(d(x,y)) y\u00E2\u0082\u00ACM is actually constant. -Remark 2.1. Note that if V(x; r) is always finite and there exists a transitive set of isometric automorphisms of M, then Assumption B holds automatically. In particular, if V(x; r) is always finite and Assumption A holds, then Assumption B holds likewise. Our convolution-rearrangement inequality is then as follows. Theorem 2.2. Assume that the master inequality holds for M with the ordering -<, and that Assumption B also holds. Let K be a decreasing positive function on [0, oo). Then, for any positive functions f and g we have E f(*)KWx,y))g(y)< E f*(x)K(d(x,y))g*(y). (2.4) x,y\u00C2\u00A3M x,y\u00C2\u00A3M Moreover, this also holds in the case where f and g are not necessarily positive, providing that they have finite support. 73 Chapter II. Discrete symmetrization Proof. B y approximat ion (monotone convergence theorem) it suffices to consider the case where / and g have finite support . In that case we may also assume that K has bounded support ; indeed, if we set K(t) = 0 for t sufficiently large that t > d(x, y) for all x \u00E2\u0082\u00AC supp / U s u p p / * and y G supp 17 U supper* , then neither the left nor the right side of (2.4) wi l l change. Wr i te yeM B y Assumpt ion B , as noted above, ot does not depend on x and is finite (the latter fact uses the boundedness of the support of K). Let $(\u00C2\u00A3) = t2. Then we have Q(f,g;*,K) = -2 \u00C2\u00A3 f(x)K(d(x,y))g(y) + a j ^ f2(x) + a j ^ / ' f a ) , (2-5) x,y\u00C2\u00A3M x\u00C2\u00A3M y(zM and the analogous identi ty for Q(f#, g#;$>, K) also holds. Bu t , \u00C2\u00A3/ 2(*)=E(/ #) 2(*)> (2-6) x\u00C2\u00A3M xeM since / and are equimeasurable. The identi ty (2.6) also holds wi th g in place of / . Combin ing (2.6) and its analogue for g w i th the master inequali ty Q(f,g;$,K) > <3( /* , 3>, K), and wi th (2.5) and its analogue for / * and g*, we obtain (2.4) as desired. \u00E2\u0080\u00A2 Note that we have seen that if M \u00E2\u0080\u0094 Z then Assumpt ion A holds. Assumpt ion B is tr iv ial in this case, so that (2.4) must also hold. In fact, (2.4) in this case precisely coincides with [58, T h m . 371]. T h e o r e m 2 .3 (cf . [58, T h m . 375 ] ) . Suppose that Assumptions A and B are both satisfied. Given a symmetrically decreasing real valued function g on M which either has finite support or is positive, and given a decreasing function K on [0, 00), the function K * g defined by {K*g)(x) = K(d(x,y))g(y) yeM is symmetric decreasing on M. 74 Chapter II. Discrete symmetrization Proof. Our proof is adapted f rom that of [58, T h m . 375]. F i x any positive function / on M wi th finite support . Then , by Theorem 2.2 we have E* K(5) where S = d(\u00C2\u00A3, v) is the distance between the two points of M, then Q(f,g;$>,K) = Q(f#,g#;G>, K) if and only if f and g are similarly ordered. The straightforward proof by consideration of the various cases is left to the reader. Now, for p \u00C2\u00A33, define m a x ( / ( z ) , f(px)), if x \u00E2\u0082\u00AC f)p fP(x) = *

* g(W). The proof of this shall be given later. C l a i m B 2 . Assume that p, v, V, w and W are as in Claim B l . Let (\u00C2\u00A3) = t2. Define K{t) = 1 for t < 1 and K(t) = 0 forty 1. T/jen, Q{g,g;$,i<) > Q(gp,gp;$,K). Proof of Claim B 2 . L e m m a 2.2 guarantees that Q(g,g;^,K)>Q(gp,gp;^,K). Let x \u00E2\u0080\u0094 v and y = V. We shall show that condit ions (i) and (ii) of L e m m a 2.2 are satisfied. Then the desired str ict inequal i ty wi l l follow since $(t) = t2 is str ict ly convex. The facts that g(v) < g(w) = g(pv) and g(V) > g{W) = g{pV) imply that (ii) holds. Now K(d(v,V)) = K(l) = 1. If we could show that d(v,pV) > 1 then it would immediately follow that K(d(v, pV)) = 0 and (i) would necessarily hold, so that the proof of the claim would be complete. Hence, to obtain a contradict ion, suppose that d(v,pV) < 1. Now, pV = W. B y condit ion (b) we have v ^ W. Thus , the only way we can have d(v,W) < 1 is if d(v, W) = 1, i.e., if W is either the parent of v or a chi ld of v. Suppose first that W is a child of v. Then , since v ^ w and W is adjacent to w, it follows that to is a descendant of v. Bu t were w to be a descendant of v then W would have been a child of w, which would have made it impossible for W to be adjacent to v, since \u00C2\u00BB / w. Suppose now that W is the parent of v. Bu t then W = V, which is impossible since V \u00E2\u0080\u00A2< W. Thus in both cases we have a contradict ion and the claim is proved. \u00E2\u0080\u00A2 114 Chapter II. Discrete symmetrization Cont inue to assume C l a i m B l . Proof of the condition for equality in Theorem 6.2. Fol lowing the proof of Theorem 2.2, we see that our str ict inequal i ty Q{g,g; ,K) > Q(gp,gp;$,K) implies the str ict inequali ty ]T g(x)K(d(x,y))g(y) < \u00C2\u00A3 gp(x)K(d{x,y))gp{y). x,yeTp x,yeTp Bu t , on the other hand, Theorem 2.2 says that J2 gP(x)K(d(x,y))gp(y)< \u00C2\u00A3 (gp)* (x)K (d(x,y))(gp)* (y), x,y\u00E2\u0082\u00ACTp x,yeTp and the equimeasurabi l i ty of gp wi th g implies that (gp)* = g* = f'\u00E2\u0080\u00A2 Hence, Y g(x)K(d(x,y))g(y)< ]T f(x)K(d(x, y))f(y). x,y\u00C2\u00A3Tp x,yeTp Bu t by (6.6) and (6.7) it then follows that vi(G) = K(g)>K(f) = v1(G*), as desired. \u00E2\u0080\u00A2 A l l that remains to be proved is C l a i m B l . Let v be the smallest (with respect to -<) element of G * \ 5 . Let w be an element of G&\S such that f(v) \u00E2\u0080\u0094 g{w) (such a w exists because / = g& are equimeasurable while / | 5 = g\s)- We have f(v) / g{v) since if f{v) = g(v) then we could set S' = S U {v} and we would have / | s , = g\s,. The minimal i ty and choice of v would then ensure that \S'\ > \S\ and that al l ancestors of elements of S' are in 5 , thereby yielding a contradict ion. Then , we must have v -< w (we cannot have v = w since f(v) / g{v) and we cannot have w < v because of the min imal i ty of v). Claim B3. We have f{v) > f(w). 115 Chapter II. Discrete symmetrization Proof. To obtain a contradiction, suppose instead that f(v) < f(w). But v -< w so that f(v) > f(w) and so f(v) = f(w). Then, by Proposition 6.1 there exists an involution p swapping v and w, and satisfying / o p = f. Then, f(x) = f(px) = g(px) for every x G pS since g\s = f\s and p = p~l. Moreover, g(pv) = g(w) = f(v). Then, (go p)\s, \u00E2\u0080\u0094 f\s,, where S' = {v} U pS. We have \S'\ > \S\ since v^pSaspv = w^S. If we can prove that the ancestor of every element of S' lies in S' then we will have obtained a contradiction to the maximal cardinality of S. (It is clear that we must have pS C G* since / does not vanish on S, hence / = / o p does not vanish on pS, while the support of / is precisely G*.) We now prove the above statement about ancestors of elements of S'. Let x G S'. First consider the case where x G pS so that px G S. We must show that if x ^ O and X is the parent of x, then X G pS. But, if X \u00C2\u00A3 pS then pX \u00C2\u00A3 S. Since X is adjacent to x, we have pX adjacent to px. As the parent of px must lie in S, it follows that pX is not the parent of px, but must instead be a child of it. Thus, f(pX) < f(px). But / = fop so that f(pX) = f(X) and f(px) = f(x), while, since X is the parent of x, we have f(X) > f(x). Thus, f(pX) = f(px) \u00E2\u0080\u0094 f(X) = f(x). From Proposition 6.1 it follows that X = O and h(x) = 1, and likewise that px \u00E2\u0080\u0094 0 and h(pX) = 1. Since / evidently thus attains its maximum at x and at pX and also at O, while O \u00C2\u00A3 {x,pX}, it follows from Corollary 6.2 that x = pX. Thus, pO = p2x = pX \u00C2\u00A3 S. Since, f(pO) = f(0) is the maximum of / while pO \u00C2\u00A3 S, it follows from the minimality of v that v -< pO so that f(v) = f(pO) likewise. But since the maximum of / is attained on a set of cardinality at most 2 and v ^ O, it follows that in fact pO = v. Hence, O = pv = w, which contradicts the choice of w G G*\S since O G S. It only now remains to show that the parent of x lies in S' if x = v. Let V be the parent of v. IfV^pS then pV \u00C2\u00A3 S. The minimality of v shows that we have v \u00E2\u0080\u00A2< pV. Thus, f{v) > f(pV). But f(pV) = f(V) as / = fop. Thus, f(v) > f(V). Since V is the parent of v it follows that f(v) < f(V) so that f(v) = f(V), and then Proposition 6.1 implies that V = O. But we have assumed that f(v) = /(to) and there are at most two points at which / attains its maximum, 116 Chapter II. Discrete symmetrization while it evidently attains it at V = O, v and w, so that w = v or w immediately yields a contradict ion. = O, and either option \u00E2\u0080\u00A2 Completion of proof of Claim B l . Let V be the parent of v. Note that V \u00C2\u00A3 S by minimal i ty of v. If w is a descendant of v then let W be any child of w; otherwise, let W be the parent of w. Thus condit ions (a) and (b) are satisfied. I c laim that V ^ W. If w is a descendant of v then this is easy. Otherwise, suppose that V is the parent of both v and w. Then it is easy to see that there exists an involut ion p which fixes S (use here the fact that ancestors of elements of S lie in S) but swaps v and w. Then , we have f\su{v} = (d \u00C2\u00B0 P)\su{v}- B u t the parent of v lies in S by minimal i ty of v, and so we have a contradict ion to the max ima l cardinal i ty of S. Hence indeed V ^ W. It is clear that from the construct ion of V and W we obtain the fact that V -< W as v < w. Thus, V *

* sup / , s A\S then we may choose *

*: A* \u00E2\u0080\u0094>\u00E2\u0080\u00A2 A, and prove that all the above results hold. Then, we may piece these things together via Proposition 8.2 and via the fact that the {A*}AeA are disjoint as are the {A}^e^. But # restricted to subsets of A 6 A is of Schwarz type. Hence, it suffices to prove Proposition 8.6 for rearrangements of Schwarz type. Thus, assume that # is of Schwarz type. Then A = {X}. Without loss of generality, the support of / is non-empty (otherwise the result is trivial, since we can let 0 we may construct a sequence x0, x\, x2,... with the property that f(x0) = maxx / and f[xi) = max{/(a;) : x \u00E2\u0082\u00AC A^ \{a;o, Z i , . . . , a > ; - i } } for i > 0. Moreover, if supp / is finite, we may easily ensure that XQ, X \ , x2, \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 is an enumeration of X (since after having chosen those X{ which form the support of / , we then choose all the 132 Chapter II. Discrete symmetrization other X{ arbitrarily in such a way that we enumerate all of X.) Let X' = {xo,xi, x2,.. \u00E2\u0080\u00A2}\u00E2\u0080\u00A2 It is clear that because of the finiteness of fx for all A, the set X' contains at least all the points x of X such that f(x) > 0. Hence, if either / > 0 everywhere or supp/ is finite, we have X' = X. Now define for i \u00E2\u0082\u00AC Z Q . It is clear that *

)x. But (f*)x = (fx)* by Theorem 1.2.1. Suppose first that A < 0. Then, fx = X since / is positive, so that (fx)* = Y. Likewise .(/ \u00C2\u00B0 4>)x = y then, since / o d> is also positive. Suppose now that A > 0. Suppose that the cardinality of the set fx is i. Consider first the case where i = oo; necessarily we have A = 0. Then, / takes on infinitely many strictly positive values. The choice of X' implies that / is strictly positive on X', hence (/ o cj>)x = Y. But (fx)* = Y as well, since we have a Schwarz rearrangement, and the Schwarz rearrangement of an infinite subset S of X must equal Y since X# = Y and |X| = \S\. Hence, in that case we have verified the desired result. Suppose now that i < oo. Then, (/A)# = {yo,yi,---,Vi-i} by Proposition 8.1. But fx consists of points at which / takes on its i largest values. By the choice of the x{ we have fx = { a ; o , a ; i , . . . , a ; j _ i } . Thus, (fo 4>)x = rl[fx] = {yo, yi, \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 = (/A)#, as desired, where the second last equality follows from the definition of __ then gs corresponds to the usual definition of the discrete Green's function go for D. We say that an extended real function G on V2 is Steiner bi-symmetric if for every pair of integers i and j we have E G((i,a),{j,fi))< E G((i,a),(j,(l)), for all subsets K and L of X. Now, given two positive extended-real functions F and G on V2, write F [0,1] we have 9s <9s*, and, moreover, gs# is Steiner bi-symmetric. In particular, if # is of Schwarz type on X with initial element O, then for any integers i and j , any y \u00C2\u00A3 X and any convex increasing function $ we have pex pex Remark 9.4- Th is is a discrete version of a generalization of a result of Baernstein [7, T h m . 5] (see Theorem 1.6.1 of the present thesis). In the special case of the simple random walk on Z x Z m and s \u00E2\u0080\u0094 I D , this can probably be proved by using Quine's result [90] on the subharmonici ty of the discrete ^-funct ion. Our proofs, like those of Bore l l [25], are based on the probabi l ist ic method of Haliste [56, proof of T h m . 8.1]. 9.2. Reducing to the case A = 0 in Assumption 9.3 We shall show that if we can prove Theorems 9.1 and 9.2 for K satisfying Assumpt ion 9.3 with A = 0, then we can prove them for K satisfying Assumpt ion 9.3 wi th general A > 0. For , suppose that K satisfies Assumpt ion 9.3 wi th some A > 0. Define L = c(K + XS), where c > 0 is chosen so that (9.1) holds wi th L in place of K, and where S(v, w) is 1 if v = w and 0 otherwise. It is easy to see that c = (1 + A ) - 1 . It is clear that L satisfies Assumpt ions 9.1 and 9.2. Use superscripts K and L to dist inguish quantit ies denned in terms of the random walk {R^ } and the random walk {R^} defined by the transi t ion kernels K and L respectively. Professor Gregory Lawler would cal l R^ \"the walk R^ wi th geometric wait ing t imes\" . 149 Chapter II. Discrete symmetrization Assume now that Theorems 9.1 and 9.2 have been proved for the kernel L. We shall prove them for the kernel K. To do this, proceed as follows. We may describe the random walk {R%} slightly differently from before. A step of this random walk consists of first flipping a coin and with probability p = A/(l + A) staying put, while with probability 1 \u00E2\u0080\u0094 p = 1/(1 + A) taking a step with transition probabilities defined by the kernel K. Let Sn be the event that the flip of the coin was such that we took the step with transition probabilities defined by K. Now, note that the walk R^ will eventually take a step defined by the kernel K, and that the distribution of R^+1 conditioned on the event Sn is the same as the unconditioned distribution of R^+i \u00E2\u0080\u00A2 Moreover, the probability that the random walk R^ will survive until one of the events Sn happens is equal to (1 - p)s(z) + (1 - p)p(s(z))2 + (1 - p)p2(s(z)f + \u00E2\u0080\u00A2\u00E2\u0080\u00A2\u00E2\u0080\u00A2= {lZpSS{^ = where __

)) = S1(ip(v))cip(u)). Then, the left hand side of (9.15) equals J2E(vi)S\v)c(v), where E{v) = E(ip~x{v)). Applying Theorem 1.2.4, we see that YJ E ( V 1 ) S 1 ( P ) C ( V ) {z) = ^ SD(z,w)ip(w), for a positive function tp on H. More precisely, we may take Tb(W) = PW{fu = fD). Proof. For ij)(w) = Pw(fu = Tp), by the definition of the Green's function and by Fubini's theorem we have ^2QD(Z, w)ih(w) weH weH CO 71=0 Pw(fu = fD) = J2J2 pz(Rn = w a n d t d > n)p\" ;( ft/ = fD) n=0 w\u00C2\u00A3H oo = J2pZ (R\u00E2\u0084\u00A2 eH,rD>n and (Rk G U, V& G {n + 1,... , T D - 1})). n=0 But it is easy to see that the events within the Pz(-) are disjoint for distinct values of n since H C Uc. Moreover, it is easy to see that the union over n \u00C2\u00A3 Z Q of these events is the event {3n G Z Q . (Rn G H and n < Tp)}. But clearly the probability of this event if the random walk starts at z is precisely u>(z, H; D) = (f>(z). \u00E2\u0080\u00A2 Proof of Theorem 10.1. For conciseness, given a subset L of Z Q X {0}, write inf Id=inf{/ : (Z,0) G L}. Let t0 = inf H. We proceed by induction on N \u00E2\u0080\u0094 \t0\ - \H\. First, if iV = 0 then H \u00E2\u0080\u0094 H' and we are done. Suppose now that the result has been proved whenever \to\ \u00E2\u0080\u0094 \H\ < N and that 161 Chapter II. Discrete symmetrization H = \u00E2\u0080\u00A210,0) (-9,0) \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 t to (-8,0) (-7,0) (-6,0) \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 -5,0) (-4,0) (-3,0) (-2,0) (-1,0) \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 (-10,0) (-9,0) (-8,0) (-7,0) (-6,0) (-5,0) (-4,0) (-3,0) (-2,0) (-1,0) \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 ' \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 Figure 10.1: An example of the sets if and H\ (indicated with B's) in a case where to = -9, ti = \u00E2\u0080\u00947, |if | = 6 and N = 3. The symbol Aj indicates a point contained in the set Aj. N > 1. Let tx = inf{t G Z~ : t > t0,t fi if). Since N > 1, we have tx G {t0 + l,..., 1}, and moreover {t0,..., ti - 1} C H. (See Figure 10.1.) Define Hi = ( # n [ * i , l ] ) U { * o + l , . . . , i i + l}. (See Figure 10.1.) It is easy to see that \Hi \ = \H\ and that i i i is in fact just H with the hole at ti deleted. Moreover, | inf Hx\ = t0 + 1 so that | inf Hi\ \u00E2\u0080\u0094 \H\ < N as t0 < 0. Thus, if we form (iii)' from iii in the same way that H' is formed from H, by our induction hypothesis we will have w((t, 0), T; ZAffO < w((0, t), T; D\(ffi)'), whenever i < inf Hi, and in particular whenever t < inf i i . But \Hi\ = \H\ so that (iii)' = H'. Thus, the desired inequality (10.1) will follow as soon as we establish the fact that c((t,0),T;D\H)__ gz>(z,w) is a discrete harmonic function on D\{w} for any fixed w G V. Th is is easiest seen direct ly f rom the definit ion of go \u00E2\u0080\u0094 giD and from equa-t ion (10.2). Example 10.2. The funct ion z >->\u00E2\u0080\u00A2 u>(z, A; D) is a harmonic function on D for A C 3D. Th is is also easy to see f rom (10.2) and the definit ion of u(z, A; D) = u(z, A; l r j ) . 159 Chapter II. Discrete symmetrization Clearly sums and scalar multiples of harmonic functions are harmonic. A central result about discrete harmonic functions is the following very well-known maximum principle. T h e o r e m 1 0 . 2 . Let f be a harmonic function on U C V which is bounded above and let C be a real constant such that f(z) < C for all z \u00C2\u00A3 dU. Assume that U ^ V. Suppose that there exists w \u00C2\u00A3 U such that f(w) > C. Then f is constant on U. We now state our two lemmas which provide the keys to the proof of Theorem 10.1. L e m m a 1 0 . 1 . Leth{t\,t2) = 0), (t2, 0)). Then for fixed t2 \u00C2\u00A3 Z~ the function h(-,t2) is increasing on (\u00E2\u0080\u0094oo, t2] n Z~, and decreasing on [t2, \u00E2\u0080\u0094 1] fl Z _ . Similarly, for fixed t\ \u00C2\u00A3 Z _ the function h{t\, \u00E2\u0080\u00A2) is increasing on (\u00E2\u0080\u0094oo, t{\ fl Z~, and decreasing on [t2, \u00E2\u0080\u00941] fl Z~. Proof. Fix t2 \u00C2\u00A3 Z~. First suppose t\ > t2. We shall show that h{t\ - l,t2) > h(ti,t2). Let Di = {ti \u00E2\u0080\u0094 1, ti,... , \u00E2\u0080\u00941} X Z m Then, it is easy to see that Hh,t2)= J2 ^{(h,0),{(ti-l,a)};D1)gD((t1-l,a),{t2,0)). But gjj is bi-symmetric because of Theorem 9.2 so that go{{ti \u00E2\u0080\u0094 1, a)i (^2, 0)) < 5z?((ii \u00E2\u0080\u0094 l,0),(i2,0)). Thus, h(h,t2) KgDi^-1,0), (t2,0)) w((ti,0),{(ti- l,a)};Di) = w((tx, 0), {ti - 1} x Zm}; 7Ji)/i(ii - 1, t2) < /l(t! - l,t2). The inequality + l,t2) > h(ti,t2) in the case t\ < t2 is proved very similarly. The case of t\ fixed can be handled just as above (or else it can be noted that it follows from the fact that h{t\,t2) = h{t2,t\) for the simple random walk.) \u00E2\u0080\u00A2 Now, for a subset S of Z x Z m , let r s = inf {n > 0 : Rn fi S} 160 Chapter II. Discrete symmetrization and fs = inf {n >0:Rn(\u00C2\u00A3S}. The following lemma then is valid for any random walk, not just the simple random walk. It extends in an easy way to a number of situations. Lemma 10.2. Let ____(z) = u{z,T; D\H). Then, by Lemma 10.2 we have cj){z) = (f)i(z) + fo(z), 162 Chapter II. Discrete symmetrization where i(z) = Y 9D(z,w)ib(w), u>\u00E2\u0082\u00ACA, for i = 1,2. For (x, y) 6 ZQ X Z M , let 0(a;, y) = fa (x \u00E2\u0080\u0094 1, y) + <^ 2(a:, y)- I claim that faz) > l-u{z,T;D\Hx) (10.4) for all z G ZQ X Z M . Suppose for now that this claim is just. Then, for t < inf H we have co((t,o),T;D\Hx) > l - M * - i , o ) - & ( t , o ) . But 0 x(i \u00E2\u0080\u0094 1,0) < ^i(i, 0) since #\u00C2\u00A3)((\u00C2\u00A3,0), (\u00C2\u00AB, 0)) is increasing in t for i < u (Lemma 10.1) and since tp is positive while Ax C [\u00E2\u0080\u0094to, -1] X {0}. Thus, co((t, 0), T; Z A # i ) > 1 - &( i , 0) - fa(t, 0) - 1 - 0(i, 0) = u((t, 0), T; D\#), which is precisely what we were supposed to prove. Thus, we need only verify (10.4). But 4> is a bounded3 discrete harmonic function on D\HX since fa and fa are harmonic on D\AX and J D\A 2 , respectively, (since the fa are sums of Green's functions to which we can apply Example 10.1), and u(-,T; D\HX) is harmonic in D\Hi (Example 10.2), while Hx = {(1,0) + A1)UA2. Thus, the maximum principle (Theorem 10.2) implies that to show (10.4) it suffices to verify that (10.4) holds on d(D\Hx) = Hx U H2 and T. But on T, the inequality (10.4) holds trivially as its right hand side vanishes while the left is positive. Suppose now that z \u00E2\u0082\u00AC Hx = ((1, 0) + Ax) U A2. Then the right hand side of (10.4) equals 1. There are two cases to consider. First suppose that z 6 ((1,0) + Ax). Then, z \u00E2\u0080\u0094 (x + 1, 0), where (x, 0) G Ax. We have 4>{z) = fa(x +1-1,0)+ fa(x + 1, 0) = fa(x, 0) + fa(x + 1, 0). 3The function __

0 for i \u00C2\u00A3 {0,1,..., N \u00E2\u0080\u0094 1}. Of course W(0) = E^\u*(yi)-v{yi)) = Nv(F) > 0. Let Fn = n-H^^y. Then, Fn \u00C2\u00A3 M. Thus, for n = 1,..., N \u00E2\u0080\u0094 1 we have a{Fn) < a(F). Note that FN = / . But *os and \u00E2\u0080\u0094 d = cp o s#, where ct> is the monotone function defined by cp{t) = I \u00E2\u0080\u009400, if t < 0, 1 - i - 1 , if t > 0. Hence, the relation \u00E2\u0080\u0094c< \u00E2\u0080\u0094d follows trivially. 182 Chapter II. Discrete symmetrization Now, u is a positive funct ion. Let A be a fibre of V. Then , there is some TO G Z such that A C {TO} X X. (This is because each fibre of V is of the form of a cartesian product of a singleton {m} wi th a fibre of X.) If TO = k or TO G T then A does not meet Q^. Otherwise, A f l QM = A f l supp SAT, and (11.6a) (with UN in place of u and in place of f2) follows by posi t iv i ty of upi and the fact that UN vanishes wherever SN vanishes. To verify (11.6b) (with UN in place of u and fi^ in place of 0 ) , it suffices for us to verify it in the case of U being a subset of some fibre A of V. There are two cases. E i ther A and (T U {k}) x X are disjoint, or A is a subset of (T U {k}) x X. Consider first the former case. Then , U\QN lies outside the support of SN and hence UN vanishes there and (11.6b) is t r iv ia l as v is posit ive. Consider now the case when A is a subset of ( T U {k}) x X. Then , since k fi T it follows that A is either a subset of T X X or of {k} x X. In the former case, upi and v both vanish identical ly on T and we are done. Suppose then that A C {k} X X. Now, on {k} X X, we have UN equal to the indicator funct ion of SN and v equal to the indicator funct ion of S*. Now, then, on A , the funct ion is smaller than the indicator funct ion of S. Hence, (11.6b) is a consequence of the Hardy-L i t t lewood inequali ty (Theorem 1.2.3) applied wi th / = I 5 and g = u, restricted to the fibre A. Note that u has finite support (since s does), so that u G w(V). O f course Assumpt ion 11.4 follows from the choice of T and the finiteness of Q]y. Hence, the assumptions of Theorem 11.1 are satisfied. Hence, < V . Let t ing \u00E2\u0080\u0094> 00, we conclude that u < v (use L e m m a 1.2.1). This easily implies the conclusion of Theorem 9.1. For we can use condit ion (9.3) to prove that u \u00E2\u0080\u00A2< v. Then , all that remains to be proved is the Steiner symmetry of v. Bu t this follows by noting that v < v by a second appl icat ion of the above work, and then using L e m m a 11.1. The crucial fact in the above work was the formula -Au(z, 5; s) = -[(s(z))-1 - l]u(z, S;s), valid outside on {z : s(z) > 0} and it is essentially this formula which together wi th Theo-rem 11.1 implies Theorem 9.1. 183 Chapter II. Discrete symmetrization Remark 11.3. We can use a similar technique for proving Theorem 9.2. There, the necessary formula is Agsu(z) = lu-[(s(z)r1-l}gsu(z), on {z : s(z) > 0}, where gfj{z) = Y^weU dizi w'is) f\u00C2\u00B0 r U C V. The above equation is also of a type that can be handled by Theorem 11.1. Since we have already given one rigorous proof of Theorem 9.2 and have shown how to use Theorem 11.1 to prove Theorem 9.2, we leave it to the interested reader to work out the details of a proof of Theorem 9.2 via Theorem 11.1. 11.3.3. Exit times By Haliste's method, we were also able to prove a result about exit times, namely Theorem 9.4. It is natural to ask whether that result can also be proved to be a consequence of Theorem 11.1. The answer is positive. Our method here is actually very simple: since we have already shown that Theorem 11.1 implies Theorem 9.1, we need only prove that Theorem 9.1 implies The-orem 9.4. To do this, proceed as follows. Assume that the hypotheses of Theorem 9.4 are verified. Let 53 = Z x V. Let # be the Z-product symmetrization on QJ obtained from the Steiner symmetrization # on V. Given the kernel K on V, let \u00C2\u00A3((m,6), (m>') = 5m+limlK(v, v'). Then, the hypotheses of Theorem 9.1 are satisfied if we put .\u00C2\u00A3 and 03 in the place of K and V, respectively. Let \u00C2\u00A9 = {N} x V. For m \u00C2\u00A3 Z and v \u00E2\u0082\u00AC V let s(m, v) = s(v), where s was our survival function on V. Then, it is not hard to see that u*({0,z),&;s) = Pz(rs > AT), (11.18) for z \u00E2\u0082\u00AC V, where is generalized harmonic measure on QJ defined with respect to the kernel Let J = {i} X I. Then, by Theorem 9.1 we see that 5 > * ( ( 0 , j ) , \u00C2\u00A9 ; s ) < E>*((0,j),e#;s#). 184 Chapter II. Discrete symmetrization Now, @# = (5 by definition of the Z-product rearrangement on QJ, since V# = V. Moreover, s * ( m , v) = s*(v) as is easily seen, and 3# = {i} x 7*. Hence, by (11.18) (and its analogue for -^rearranged sets and functions) we see that \u00C2\u00A3 pi(Ts>N)< \u00C2\u00A3 P>(rs>N). )e{i}xT# But this is precisely the conclusion of Theorem 9.4, and so we see that Theorem 9.1 indeed implies Theorem 9.4. 185 Chapter I I I Chang-Marshall inequality, harmonic majorant functionals, and some nonlinear functionals on Dirichlet spaces Overview The purpose of the present chapter is to study the A $ and T$ functionals. We shall begin by defining the non-l inear functionals A $ act ing on functions on an arbi t rary finite measure space (\u00C2\u00A71.1). Then , in \u00C2\u00A71.2, we shall define various Dir ichlet spaces on which our functionals are to act; we shal l also review some very basic results about these Dir ichlet spaces. In \u00C2\u00A71.3, we shall define the r$ funct ionals and describe the connection between them act ing on domains of area at most VT and the A $ act ing on functions f rom the unit bal l of the holomorphic Dir ichlet space. In \u00C2\u00A72 we give the impor tant Chang-Marsha l l [32] inequality which in fact started our whole invest igat ion. We also give Essen's improvement [44] of the Chang-Marsha l l inequality, and state the Moser-Trudinger [78] inequali ty which motivated the Chang-Marsha l l inequality. Then , we prove that in a strong sense the Moser-Trudinger and Chang-Marsha l l (and a fortiori Essen) inequalit ies are unimprovable (Theorem 2.1). Section 3 wi l l be pr imar i ly devoted to the study of the A $ functionals on functions on an ar-bi t rary finite measure space. In \u00C2\u00A73.1 we shall give some results on the existence of extremal functions for A $ functionals on balls of Dir ichlet spaces by giving an important upper semi-cont inui ty result (Theorem 3.2) f rom the author 's joint paper wi th Matheson [75]. Th is result improves on a theorem of Matheson [73]. In \u00C2\u00A73.2 we examine the A $ functionals acting on 186 Chapter III, Functionals on a set of domains and on Dirichlet spaces the unit balls of Hi lber t spaces of measurable funct ions. We wi l l be able to obtain some gen-eral results which wi l l al low us to give a proof of C i m a and Matheson 's theorem [35] on the weak cont inui ty of the Chang-Marsha l l funct ional on the punctured unit bal l of the Dir ichlet space. O u r results wi l l also be useful for proving weak continuity results in the case of the Moser-Trudinger funct ional . Then , in section 3.3 we consider the general notion of a \"cr i t ical ly sharp inequal i ty\" . M u c h of the mater ia l in \u00C2\u00A73.3 comes f rom the author 's paper [88]. Par t icu lar examples of cr i t ical ly sharp inequalit ies wi l l be the Chang-Marsha l l and Moser-Trudinger inequalit ies. We shall prove that given a cr i t ical ly sharp inequal i ty for A $ we may perturb $ in such wise as to, depending on our wishes, either gain (Theorem 3.9) or destroy (Theorem 3.7) the existence of a function at which the max ima l value in the inequali ty is at ta ined. Our results are new even in the cases of the Chang-Marsha l l inequal i ty and the Moser-Trudinger inequality, al though in the former case the result was strongly suspected by C i m a and Matheson (personal communicat ion) and in the latter it was conjectured by M c L e o d and Peletier [77]. We also obtain a part ia l converse (Theorem 3.8) to the upper semicontinuity result of Theorem 3.2. In \u00C2\u00A74 we come back to the specific case of the A $ functionals on balls of Dir ichlet spaces. In \u00C2\u00A74.1 we shal l give a var iat ional equation for the extremals of our funct ionals. Th is is due to the author, refining a part ia l result of Andreev and Matheson [5], and is taken from a joint paper wi th Matheson [75]. In \u00C2\u00A74.2 we give a joint result of the author wi th Matheson [75] which shows that under appropriate condit ions on $ the extremals of the A $ functionals automatical ly satisfy some regularity condit ions. In \u00C2\u00A74.3 we give some important assumptions on the functions 4> to which our methods are appl icable. Then , in \u00C2\u00A74.4 we give some useful extensions of the results of sections 4.1 and 4.2. We shall use a variat ional equation from \u00C2\u00A74.4 later in \u00C2\u00A7IV.10. We have noted that some of the results of the present chapter come from the paper of Matheson and Pruss [75], and the proofs are sometimes taken from that paper. When this is done, the proof and the result can be assumed to have been due to the author, unless otherwise noted (as in the case of Theorem 4.2). 187 Chapter III. Functionals on a set of domains and on Dirichlet spaces In \u00C2\u00A75 we return to symmetr izat ion theory, and study the symmetr ic decreasing rearrangement act ing on boundary values of real parts of functions from the Dir ichlet spaces. We prove a str ict symmetr izat ion theorem (Theorem 5.1) and use it to prove that if $(z) = (Rez) for an appropriate funct ion *

*a be the Hi lber t space of al l functions / holomorphic on the unit disc D wi th / (0 ) = 0 and CO ii/iik=E i^/wi2<~-71 = 1 D e f i n i t i o n 1.4. Let the a - w e i g h t e d r e a l h a r m o n i c D i r i c h l e t s p a c e o a be the real Hi lbert space of al l real functions / harmonic on the unit disc D wi th / (0 ) = 0 and CO n/iiL = En a(^(/)+-\u00C2\u00BB(/))< \u00C2\u00B0\u00C2\u00B0-71=1 Remark 1.1. It is clear that \u00C2\u00A3 ) a has the inner product CO ( / , 5 ) = \u00C2\u00A3 n \u00C2\u00AB ( / ( n ) , ^ ) ) , 71=1 whereas u Q has the inner product CO (f,9) = M / ) C n ( < / ) + * \u00E2\u0080\u009E ( / ) * \u00E2\u0080\u009E ( < / ) ) \u00E2\u0080\u00A2 71 = 1 Remark 1.2. The map / H-\u00C2\u00BB Re / is an isometry sending X>a onto Ha. To see this, it suffices to note that c n ( R e / ) = R e / ( n ) and s n ( R e / ) = \u00E2\u0080\u0094 I m / ( n ) . 189 Chapter III. Functionals on a set of domains and on Dirichlet spaces D e f i n i t i o n 1.5. The spaces D i and are known as the h o l o m o r p h i c and r e a l h a r m o n i c . . . . . def def D i r i c h l e t s p a c e s , respectively. We wi l l wri te 13 = 1)1 and D = 0i for short. It is easy to verify by expanding / in a series that = - ff \f'(x + iy)\2dxdy, * J J B |2 Is and 11/112 = \" ff \Vf(x,y)\2dxdy. K JJ D Since for a holomorphic funct ion / the quant i ty \f'(x + iy)\2 is the Jacobian of the mapping / , it follows that 7r||/||ijj is the area of the image of D under / counting multiplicities, and if / is univalent (i.e., one-to-one) then 7r||/|||, is precisely equal to the area of the image of / . It is in the space 1) that our greatest interest lies. Technical remark 1.1. To place \u00C2\u00A9 in a larger picture, it should be noted that D contains un-bounded functions (any univalent map to an unbounded region of finite area wi l l be in D ) , and that not every funct ion f rom the disc algebra lies in X). Indeed, let CO / ( * ) = ^ 2 - m / V \" . m=0 Evident ly , this series converges uniformly on T and thus / lies in the disc algebra. However, CO oo \u00E2\u0080\u00A2 \u00C2\u00A3 n | / > ) | 2 = \u00C2\u00A3 2 \u00E2\u0084\u00A2 2 - m = oo, n=l m=0 and so / ^ X). O n the other hand, X> C B M O A . One of the easier ways to prove this is to note that f rom the definit ion of the norm on Dir ichlet space it easily follows that if / \u00E2\u0082\u00AC X> then oo / m \u00E2\u0080\u0094 1 \ s u p \f(mn + J) I <'oo. m > l 1 \ \u00E2\u0080\u00A2 , I - n=l \j=l j which is C . Fefferman's sufficient (and, if we have f(k) > 0 for al l k > 0, then also necessary) condit ion for a holomorphic funct ion / on D to be a member of B M O A [24, 97]. D e f i n i t i o n 1.6. Let 25, b, 2$ a and ba be the unit balls of D , o, D a and o a , respectively. 190 Chapter III. Functionals on a set of domains and on Dirichlet spaces These unit balls will be the ranges of the A $ functionals in which we will be interested. The following result will also be useful. The most useful case is where a = 0 and (3 = 1 and that case was already proved by Andreev and Matheson [5]. T h e o r e m 1.1. The inclusions XJ/3 C D a and D/3 C T)a are compact for 0 < a < (3 < 00. In particular, if fn is a weakly null sequence in T>a or oa for a \u00C2\u00A3 (0, 00), then / \u00E2\u0080\u009E \u00E2\u0080\u0094> 0 in L2(T) and hence fn tends to 0 in measure on T. ll/ll\u00C2\u00AEo \u00E2\u0080\u0094 II/IIL2(T) f\u00C2\u00B0 r a function / with mean 0 on T, The following lemma is doubtless well-known but is easier to prove than to attribute. L e m m a 1.1. Fix 1 < p < 0 0 . Consider the space \u00C2\u00A3p(w), where w is a non-negative weight with wn \u00E2\u0080\u0094> 00 as n \u00E2\u0080\u0094y 0 0 . Let ak be a norm-bounded sequence in \u00C2\u00A3p(w), converging pointwise to 0. Then ak \u00E2\u0080\u0094> 0 in P-norm. Assume this for now. Proof of Theorem 1.1. It suffices to prove that if fk -> 0 weakly in then fk \u00E2\u0080\u0094> 0 in T)a norm. The assertion for fk\u00E2\u0080\u0094>f weakly where / ^ 0 then follows by considering instead the sequence fn \u00E2\u0080\u0094 f 1 while the assertion for the harmonic Dirichlet spaces follows from Remark 1.2. Now, let p = 2, and put (ak)n = nal2fk(n). Then, I I K ) l l ^ = I I M b Q . (1.1) Let w(n) = n^la. Note that w(n) \u00E2\u0080\u0094> 00 as n \u00E2\u0080\u0094>\u00E2\u0080\u00A2 0 0 . Moreover, l l ( a * ) l l * ( u , ) = ll/fcllsv Hence, ak satisfies the conditions of Lemma 1.1, and thus ak \u00E2\u0080\u0094> 0 in I2 norm. In light of (1.1), we conclude that fk \u00E2\u0080\u0094> 0 in u a norm, as desired. The assertion that ||/||i,2(x) \u00E2\u0080\u0094>\u00E2\u0080\u00A2 0 follows since ll/lloo = II/IIL2(T) f\u00C2\u00B0 r a function / with mean 0 on T. \u00E2\u0080\u00A2 191 Chapter III. Functionals on a set of domains and on Dirichlet spaces Proof of Lemma 1.1. W i thou t loss of generality assume that ||afc||^P(w) < 1 for al l k. F i x e > 0. Choose N sufficiently large that wn > 1/e for n > N. Then , oo oo J2 Kafc)\"lP ^ \u00C2\u00A3 E wn\Mn\p < (e/2)\\ak\\ n=N n=N O n the other hand, for fixed N N-l iK)\u00C2\u00ABr^\u00C2\u00B0 as k \u00E2\u0080\u0094> oo. It follows that l i m s u p ^ o Q ||OA;||#> < \u00C2\u00A3\u00E2\u0080\u00A2 Since e > 0 was arbi trary, the proof is 1.3. The 1\"$ functionals acting on domains and the A$ acting on holomorphic and harmonic functions We now establish the convention that when A $ acts on a holomorphic or harmonic function / on D, then A $ ( / ) is defined in terms of the normalized Lebesgue measure on T , i.e., where f(et$) as usual is short for n.t. l im f{elB). D e f i n i t i o n 1.7. Let B denote the collection of al l domains in the plane which contain the origin and whose area does not exceed ir. Given a domain D \u00C2\u00A3 B and a Bore l measurable funct ion $ on C , let h(-; D) and $ \ d D \u00E2\u0082\u00AC Lx(uD) then write r*(D) = oo; if $ + | a \u00C2\u00A3 , 6 Ll{uD) and $ - | 8 \u00C2\u00A3 ) g Ll(uD) then write T*(D) = - o o ; if ^ Ll(uD) and \u00C2\u00A3 Ll{uD) then say that T$(D) is undefined. Say that the complete. \u00E2\u0080\u00A2 192 Chapter III. Functionals on a set of domains and on Dirichlet spaces funct ional T$ is b o u n d e d providing it is defined (i.e., the Dir ichlet problem in question has a solution) and sup T$(Z)) < oo. DeB Say that a domain D is e x t r e m a l for T$ providing DeB and U(D)>U(D') for all D' 6 B. B y abuse of notat ion, if cf> is defined on [0,oo) then we write for T$ where * ( * ) = * ( | z | ) . The case that wi l l interest us the most is when $ is a continuous subharmonic funct ion, in which case r * (D) = LHM(0, D) (Theorem 1.5.4). The r $ funct ionals are actual ly closely related to the A$ functionals on 53. F i rs t note that the map / H-> / [B ] sending a funct ion to its image maps 53\{0} into B, since if / \u00C2\u00A3 53 then /[O] has area at most n (since the area of the image counting multiplicities is at most TT). The fol lowing result is essentially well known (the first part is impl ic i t in , e.g., [44]). T h e o r e m 1.2. Let f be any Nevanlinna class function whose image is contained in a domain D. Let $ be subharmonic on C and continuous at every point of dD. We then have A * ( / ) < r * ( D ) . /// is univalent and D = / [D] , then equality holds. Proof. P u t h(z) = L H M ( z , D). A s in [44], note that 2n \u00C2\u00B1 j ' * * ( / ( r c w ) ) d0<\u00C2\u00B1 fj h(f(re\u00C2\u00AB)) dO = h(f(0)) = fc(0), for every 0 < r < 1, since h > $ on D. Tak ing the l imit as r f 1 we conclude that A$(/) < /i(0) (use Fatou 's lemma together wi th the existence of radial l imits of / due to Theorem 1.3.5 and the assumption that / \u00E2\u0082\u00AC N). 193 Chapter III. Functionals on a set of domains and on Dirichlet spaces If / is univalent then / is a uniformizer and we may apply Theorem 1.5.7 to obtain the desired equality. (Of course, the domain in this case is Greenian since all s imply connected domains are Greenian, while al l images of univalent functions are simply connected.) \u00E2\u0080\u00A2 C o r o l l a r y 1 .1 . Fix p \u00C2\u00A3 [ l ,oo) . Let D be a Greenian domain containing the origin and let f: D D be a uniformizer for D. Then D \u00C2\u00A3 Hp if and only if f \u00C2\u00A3 HP(B). Proof. It is clear that if D \u00C2\u00A3 Hp then / \u00C2\u00A3 Hp. Conversely, suppose that / \u00C2\u00A3 Hp. Let g be any holomorphic funct ion on D wi th image in D. Let w = g(0). Let U = \u00E2\u0080\u0094w + D. O f course, / is surjective so that there exists w' \u00C2\u00A3 D such that f(w') \u00E2\u0080\u0094 w. Let *

*(z)) \u00E2\u0080\u0094 w'. Then , F is a uniformizer for U (since is a conformal automorphism of D) wi th F(0) = 0. We have F \u00C2\u00A3 Hp. Let G(z) = g(z) - w. Then the image of G lies in U and G(0) = 0. Let $ ( z ) = \z\p. We have A*(G) < r*(t/). Bu t by Theorem 1.5.7, we have r\u00C2\u00AB(tf) = A * ( F ) . Bu t A $ ( F ) < oo, so that A$(G) < oo and hence G \u00C2\u00A3 HP(B) so that g \u00C2\u00A3 HP(D) as desired. \u00E2\u0080\u00A2 2. The Chang-Marshall, Essen and Moser-Trudinger inequalities The Chang-Marsha l l [32], Essen [44] and Moser-Trudinger [78] inequalit ies are al l closely related. We now state them, using the notat ion of the previous sections of this chapter. E s s e n I n e q u a l i t y ( [44]). Let $(\u00C2\u00A3) = e*2 for t \u00C2\u00A3 [0,oo). Then T$ is bounded on the set B of all domains in the plane containing 0 and having area at most TT. C h a n g - M a r s h a l l I n e q u a l i t y ( [32]). Let $(\u00C2\u00A3) = e*2 for t \u00C2\u00A3 [0,oo). Then A$ is bounded on the closed unit ball 23 of the Dirichlet space. 194 Chapter III. Functionals on a set of domains and on Dirichlet spaces M o s e r - T r u d i n g e r I n e q u a l i t y ( [78]). Let $(\u00C2\u00A3) = e*2. Then A $ is bounded on the set T of 1, where the integral in the definition of A $ is taken with respect to the measure dp(x) = e x dx on [0, oo). In other words, the Chang-Marsha l l inequal i ty states that Note that in l ight of Theorem 1.2 and its preceding remarks, the Chang-Marsha l l inequali ty is actual ly a consequence of the Essen inequal i ty while the Essen inequali ty restricted to U instead of on al l of B is a consequence of the Chang-Marsha l l inequality. The Chang-Marsha l l and Essen inequalit ies are also closely connected wi th the Moser-Trudinger inequal i ty since Marsha l l [72] has found a fair ly easy proof of the Chang-Marsha l l inequali ty using the Moser-Trudinger inequal i ty 1 , while Essen's proof of his inequali ty [44] also uses the Moser-Trudinger inequali ty in an essential way. For more work related to the Moser inequality, see, e.g., [1, 2, 6, 18, 19, 20, 21, 30, 31, 33, 34, 49, 52, 53, 69, 77, 80, 81, 86]. For more work related to the Chang-Marsha l l inequality, see [5, 35]. A l l three inequalit ies (Essen, Chang-Marsha l l and Moser-Trudinger) are unimprovable in that the funct ion *

/ e~xdt+l. J0 J0 Jn Jo Now, the right hand side converges to 2 as n \u00E2\u0080\u0094> oo. On the other hand, it is easy to verify that /\u00E2\u0080\u009E \u00E2\u0080\u0094> 0 in measure and A$(0) = 1 so that A$ indeed fails to be upper semicontinuous at 0 \u00E2\u0082\u00AC T. Hence = e*2 is critical for T. The following result which was conjectured by McLeod and Peletier [77] then follows immediately from Theorem 3.7. Theorem 3.10. There exists a convex, increasing and smooth function T with 0 < T(y) < y for every y 6 [0,oo) and with lim^oo = 1, such that the supremum roo sup / r(e / 2W)e-^ (3.9) is not achieved over T. Of course it should be noted that (3.9) is finite. Theorem 3.10 shows that the existence of the extremal for Moser's inequality is in some way accidental, relying on non-asymptotic properties of the function e*2. 3.3.3. Application to the Chang-Mars hall inequality Now consider the setting of the Chang-Marshall inequality, with I = T and p being normalized Lebesgue measure. Again let <\u00C2\u00A3(\u00C2\u00A3) = e*2. We have already noted in Theorem 1.1 that weak convergence in 23 implies convergence in measure on T (and we have remarked that this was already known to Andreev and Matheson [5]), so that 23 is indeed compact with respect to convergence in measure on T. Cima and Matheson [35] have shown that A$ is weakly continuous on 23\{0}. This is also a direct consequence of the Chang-Marshall inequality and 3.6 with r = 0. Thus condition (i) of criticality is satisfied. 215 Chapter III. Functionals on a set of domains and on Dirichlet spaces O n the other hand, C i m a and Matheson [35] have shown that A $ fails to be weakly upper semicontinuous at 0 G 53. We give a concise proof of this fact. For , if A $ were upper semi-continuous on al l of 23, then by Theorem 3.8 there would be a \P = T o $ which grows str ict ly faster than $ and such that A

~~ are measurable on [0,oo), then wri te Then , the main step in the construct ion of for Theorem 1 is encapsulated in the following result. 216 Chapter III. Functionals on a set of domains and on Dirichlet spaces L e m m a 3 .2 . Let Q be a subset of L1[0,oo) containing the zero function, such that for each finite number T we have s u p g \u00C2\u00A3 g \\g \u00E2\u0080\u00A2 1[O,T]||L\u00C2\u00B0\u00C2\u00B0 < \u00C2\u00B0 \u00C2\u00B0 - Assume that for every sequence gn of elements of Q, there exists a subsequence gnk which converges in measure to some g G L1[0,co) such that either g is almost everywhere null or else has \\g\\Li > l im sup^. \\gnk Hz,1 \u00E2\u0080\u00A2 Suppose further that \\ \u00E2\u0080\u00A2 \\Li fails to be upper semicontinuous at 0 G Q with respect to convergence in measure. Then, there exists a positive increasing function ~~

1. Then let gk be a sequence of elements of Q with ||fffc||\u00C2\u00A3,i \u00E2\u0080\u0094> oo. Passing to a subsequence we can assume that for some g \u00C2\u00A3 ^[0,00) we have gk g in measure. If g is almost everywhere null, then it is easy to see that the proof is complete since by the bounded convergence theorem (which is applicable because the {gk} are almost everywhere uniformly bounded on [0,1] by the hypotheses of the Lemma) we have \gk\ \u00E2\u0080\u0094> 0 so that Il5fc||\u00C2\u00A3i(0) > \dk\ = Ikfelli i - Jo1 \9k\ a n d the r i g n t h a n d s i d e tends to 00, so that 00 as desired. Choosing a subsequence if necessary, then, we may assume that gk \u00E2\u0080\u0094> 0 almost everywhere and the Lemma follows. On the other hand, if g is not almost everywhere null then H^H i^ > limsupfc HS^ HL1 by our hypotheses. But, the right hand side is infinite, and this contradicts the fact that g \u00C2\u00A3 L1. Now, assume that where the supremum is to be understood as taken over all sequences {<%} in Q tending to zero in measure. Since || \u00E2\u0080\u00A2 ||\u00C2\u00A3,i fails to be upper semicontinuous at 0 \u00C2\u00A3 Q with respect to convergence in measure, we have A > 0. Obviously, A < M. Replacing Q by {\g\ : g \u00C2\u00A3 Q} if necessary, we may assume all functions in Q are positive. Choose 0 < a < 1 such that aM < A. For g \u00C2\u00A3 Q, let M = sup || x}. Let I claim that f sup / geSxJo l imsupM^A. (3.13) x\u00E2\u0080\u0094too To show this, it suffices to prove that for any sequences xk \u00E2\u0080\u0094>\u00E2\u0080\u00A2 co and gk \u00C2\u00A3 QXk such that JQ\u00C2\u00B0\u00C2\u00B0 gk converges, we have \ i m k J0\u00C2\u00B0\u00C2\u00B0 gk < A. Fix such sequences xk and gk. Passing to subsequences, if necessary, by our hypotheses we may assume that gk either converges to 0 in measure, or else it converges in measure to some nonzero g \u00C2\u00A3 L^OjOo) with ||y||ii > lirnsup^ . ||<7fc||\u00C2\u00A3,i. If it converges to 0 in measure then lim sup^ J0\u00C2\u00B0\u00C2\u00B0 gk < A by definition of A. Otherwise note that since 9k \u00C2\u00A3 Gxk, we have / ~ gk > a\\gk\\Li. Let hk{x) = gk(x) \u00E2\u0080\u00A2 l{x>Xky. We have hk -> 0 pointwise since xk \u00E2\u0080\u0094> 00. By Fatou's lemma then, roo roo lim inf / (gk - hk) > / g. k Jo Jo But roo rxk rr9k / {9k-hk)= / gk< gk = (1 - a)\\gk\\Li, Jo Jo Jo where we have used the fact that gk \u00C2\u00A3 QXk together with (3.12). Thus, lim inf (1 - a)||flf*||Li > NL&i. k But since a < 1, this contradicts the facts that \\gW11 > limsup^ . ||<7fc||\u00C2\u00A3i a R d that g does not almost everywhere vanish. Hence, the case where gk does not converge to zero in measure is impossible, and the claim is proved. 220 Chapter III. Functionals on a set of domains and on Dirichlet spaces Now define tb(x) = (-^-] (1 A inf A ~ / T \u00C2\u00BB 9 \ . I c la im that tp(x) \u00E2\u0080\u0094>\u00E2\u0080\u00A2 1 as x \u00E2\u0080\u0094> oo. To prove this, consider the funct ion A \u00E2\u0080\u0094 at h{t) :i-a)f which is easily seen to be decreasing for t \u00C2\u00A3 [0, M] since aM < A, and which satisfies h(X) = 1. B y (3.11) and (3.12) we then have *w=(ifi)(^\u00C2\u00AB\u00C2\u00A3(,.}*Gr'))-Bu t for g \u00C2\u00A3 Qx we have J0\u00C2\u00B0\u00C2\u00B0 g < Mx so that by the monotonic i ty of h on [0, M] we have il>{x) > j^{lAh(Mx)). Now by (3.13) we have l im infx^.oo h(Mx) > h(X) = 1 and hence l i m ^ o o ip(x) = 1 as desired. Note that tp is measurable as it is increasing. It is easy to verify that ||<7||\u00C2\u00A3,i(,/,) < A for every g \u00C2\u00A3 Q. For , given g \u00C2\u00A3 Q w i th ||ff||\u00C2\u00A3,i 7^ 0, we have roo rTg roo rrg roo roo roo / W>< / 9i>+ 9<^{rg) g+ flf < A - / g+ g = X, JO Jo J-rg Jo Jrg Jrg Jrg where we have used the monotonic i ty and choice of tp. The first inequali ty came from the facts that tp < 1 everywhere and that JT\u00C2\u00B0\u00C2\u00B0 g > 0 if ||<7||\u00C2\u00A3,i / 0. If we do not need ~~ may be taken to be in C\u00C2\u00B0\u00C2\u00B0[0, co), with support bounded away from 0. Moreover if supgeg | |\u00E2\u0080\u00A2 0 almost everywhere and l imsup f c ||<7A;||LI = suPa\u00E2\u0082\u00ACff l l5l l i i (0)-Assuming the lemma for now, we may proceed to prove Theorem 3.7. Proof of Theorem 3.7. W i t hou t loss of generality assume that $(0) = 0. For / \u00E2\u0082\u00AC T and t G [0, oo), let mf(t) = p{x : $ ( | / ( x ) | ) > *}. Let Q = {mj : f G J7}. We shall apply L e m m a 3.2 to G. Let us verify its condit ions. Clear ly, every element of Q is pointwise bounded by p(I) < oo. Fur thermore, for TO/ G G we have poo I K I I L ' = / mf{t)dt = A*{f). Jo Then , using the lack of upper semicontinuity of A $ at zero, we may choose a sequence fk G T such that fk \u00E2\u0080\u0094> 0 in measure and l im supk A$(fk) > A$(0) = 0. Passing to a subsequence if necessary, we may assume that fk \u00E2\u0080\u0094> 0 almost everywhere. Then , l im supk $(|/fc|) = $(0) = 0 almost everywhere, by the cont inui ty of Hence, $(|/fc|) \u00E2\u0080\u0094> 0 almost everywhere, too, and hence also in measure. Thus , for every t > 0 we have mjk(t) \u00E2\u0080\u0094>\u00E2\u0080\u00A2 0, and in part icular nxjk \u00E2\u0080\u0094V 0 in measure while l imsup f c | | T r i / f c ( t ) | | i , i = l im sup f c A$(fk) > 0, so that || \u00E2\u0080\u00A2 H^ i fails to be upper semicontinuous at zero in Q. Now, given any sequence TOjn of elements of Q, we may choose a subsequence mjn such that 217 Chapter III. Functionals on a set of domains and on Dirichlet spaces fnk converges in measure, using the compactness with respect to convergence in measure of T. Choosing a further subsequence if necessary, we may assume fnk converges almost everywhere. If the limit is almost everywhere zero then we are done. On the other hand, if fnk \u00E2\u0080\u0094> / where / does not vanish almost everywhere then we first of all have lim^ $(|/\u00E2\u0080\u009E f c |) = $(|/|) by continuity of <&, and secondly, by the upper semicontinuity of A$ with respect to convergence in measure away from zero, we have lim supfc A$(|/\u00E2\u0080\u009EJ) < A$(|/|). Then, since 3>(|/nJ) \u00E2\u0080\u0094> $(1/1) in measure, it follows that mj \u00E2\u0080\u0094> mj almost everywhere (in fact at all points of [0,oo) other than the at most countably many discontinuities of rrty), as can be easily verified. Also, |J7Ti/Hx,1 > limsupk H^ i . Hence, the conditions for the Lemma are satisfied. Choose ~~

~~ = tp. Otherwise, since tp is a increasing funct ion [0, 00) wi th l imit 1, we may easily choose a increasing C\u00C2\u00B0\u00C2\u00B0[0,oo) funct ion cp w i th support bounded away f rom 0 and wi th the properties that 0 < 4> < ip everywhere and that ep(x) \u00E2\u0080\u0094> 1 as x \u00E2\u0080\u0094> 00. We wi l l then necessarily st i l l have llffllno) < A f o r each-5 \u00C2\u00A3 Q. We shal l now show that SUPII#IILI(0) = X. (3.14) 221 Chapter III. Functionals on a set of domains and on Dirichlet spaces To do this, fix e > 0. By definition of A, let gk be a sequence in Q with gk \u00E2\u0080\u0094> 0 in measure and II^IIL1 \u00E2\u0080\u0094* A- Choose T sufficiently large that 4>(x) > 1 \u00E2\u0080\u0094 e for x > T. Since gk \u00E2\u0080\u0094^ 0 in measure and, by our hypotheses, the gk \u00E2\u0080\u00A2 l[o,x] a r e uniformly bounded in L\u00C2\u00B0\u00C2\u00B0, the bounded convergence theorem tells us that JQT gk \u00E2\u0080\u0094>\u00E2\u0080\u00A2 0 as k \u00E2\u0080\u0094> oo. Since ||fffc||jr,i \u00E2\u0080\u0094>\u00E2\u0080\u00A2 A, we may choose K sufficiently large that gk > A \u00E2\u0080\u0094 e for k > K. Then, for k > A' we have roo roo \\9k\\m*)>J 9k~~*.) I claim that attains its maximum on T. For, A$ is upper semicontinuous on .F\{0} with respect to convergence in measure. To see this, note that A$ = A$ + A$_$. But $ \u00E2\u0080\u0094 $ is a bounded function by condition (iii) of the choice of \P, so that A$_$ is continuous with respect to convergence in measure on all of T by the bounded convergence theorem and the continuity of $ \u00E2\u0080\u0094 $. On the other hand, we had assumed the upper semicontinuity of A$ on ^\{0}. Let fn be a sequence such that lim Ay(fn) = supAy(g). (3.15) n-i-co gejr Passing to a subsequence, by compactness with respect to convergence in measure, we may assume that /\u00E2\u0080\u009E \u00E2\u0080\u0094>\u00E2\u0080\u00A2 / in measure for some / G T. If / is not almost everywhere null, then X\mn^roo A\$(fn) < A$(/) by upper semicontinuity of A*

*oo 224 Chapter III. Functionals on a set of domains and on Dirichlet spaces Now, * \u00E2\u0080\u0094 <& is a bounded continuous function and /\u00E2\u0080\u009E \u00E2\u0080\u0094>\u00E2\u0080\u00A2 0 in measure so that A$_$(/\u00E2\u0080\u009E) \u00E2\u0080\u0094>\u00E2\u0080\u00A2 \P(0) \u00E2\u0080\u0094 $(0) as n \u00E2\u0080\u0094>\u00E2\u0080\u00A2 oo by the bounded convergence theorem. Since $(0) \u00E2\u0080\u0094 (^0) = 0, we see that A \u00C2\u00A3~\l + M)\\-1p{x : A\"1 < $(|/\u00E2\u0080\u009E|) < A} > 1 + M, contradicting (3.16). Thus we see that the limit of the fn cannot be 0 and so A$ does attain its maximum on T. \u00E2\u0080\u00A2 4. Properties of extremals of the A$ on Dirichlet spaces In this section we shall discuss properties of A$ functionals acting on the unit balls of Dirichlet spaces. 4.1. A variational equation Recall that if $ was a function from [0, oo) to R then A $ was an abbreviation for A $ ( | . Q . The following result due to the author is taken from the author's joint paper with Alec Matheson [75]. Theorem 4.1. Let u>: [0,oo) -> R . Write \P(i) = $(\/t)- Assume that \I> is differentiable on (0, oo), with \9'(t)\ < ceCt for some finite constants c and C, and every t > 0. Suppose f 6 2$ is an extremal function for A$. Write s*(f)= f | / | V ( | / | V 225 Chapter III. Functionals on a set of domains and on Dirichlet spaces Then if Sq,(f) vanishes, it follows that /$'(|/|2) vanishes almost everywhere on T. Assuming that S\$(f) does not vanish, we have f \u00C2\u00A3 H2 and S*(f)zf' = V0{fV(\f\2)) onT, (4.1) where z stands for the identity function on T, and Vo is the orthogonal projection from L2(T) to H2(T)d^{f \u00C2\u00A3 H2(T) : /(O) = 0}. Moreover, if we also have < 1, then 5$(/) must vanish. Remark 4-1- The functional A$ makes sense on 23 since the hypotheses guarantee that | \ P ( \u00C2\u00A3 ) | < cC~l(eCt \u00E2\u0080\u0094 1) -(- |$(0)|, which, together with Corollary 3.5, guarantees the finiteness of A$(/) for every / \u00C2\u00A3 23 (indeed, for every / \u00C2\u00A3 D). Similarly, the right hand side of (4.1) and the integral defining S\$ (/) make sense because of Corollary 3.5 and our assumption on the size of |*'|. Also, we note that the identity function on D always satisfies (4.1) even though it may not be extremal. (For example, it will not be extremal if $(\u00C2\u00A3) = t2n and n is sufficiently large.) The interested reader may, of course, translate the conditions on $ into conditions on $, but they are perhaps more naturally stated for This will be even more true in the next theorem. If VP' (or, equivalently, $') does not vanish on (0,oo), and 5$(/) = 0, then the vanishing of /v[r'(|/|2) on T implies that / = 0, and (4.1) trivially continues to hold. The criterion (4.1) was shown by Andreev and Matheson (see [5, Cor. 3 and remarks following it]) in the special cases of the Chang-Marshall functions $a(i) = ea*2 and of the functions $(\u00C2\u00A3) = t2n, under the auxiliary assumption that / ' \u00C2\u00A3 H1. Our result above shows that this assumption will automatically be satisfied whenever / is extremal and does not vanish on (0,oo). Note that for any $ the functions / of the form f(z) = azn where n\a\2 = 1 (note that in particular this includes the identity function) provide solutions to (4.1). However, in general 226 Chapter III. Functionals on a set of domains and on Dirichlet spaces they may not be the only solutions. For instance, if = t2n where 2n > 4, then an extremal exists as noted before, and clearly the hypotheses of Theorem 4.1 are satisfied. But we have seen that the extremal in that case is not the identity function. Nor can it be of the form f(z) \u00E2\u0080\u0094 azn with n|a|2 = 1 since for any such function we have A$(/) < A$(z), where z indicates the identity function. If one could prove that for all a < 1 sufficiently close to 1 all solutions / of (4.1) are of the form f(z) = az11 with n\a\2 = 1, then it would follow that sups6*

ip(x2). By convexity, the line joining (xi, ip{x{)) with (x2, f(x2)) will be below the graph of / except over the interval [xi,^]- In particular, this line will be below (0,ip(0)). But since ip(xi) > ip(x2), it follows that this line is above the point (0,ip(x2)). Hence, ip(x2) < ip(0) = 0, which is a contradiction to the strict positivity of ip away from 0. Hence, ^ (z) = ^(Rez) has SARIP by Example 4.2. But $ \u00E2\u0080\u0094 3>(0) differs from \P only by a linear functional so that by Remark 4.3 it follows that $ - $(0) has SARIP. It follows that $ has SARIP. We may modify the preceding example as follows. Example 4-4- Let cp \u00C2\u00A3 C 1 ( R ) be convex and strictly convex at 0. Then &(z) = cp(Rez) has SSARIP. To see this, note first that by the previous example it has SARIP. Then, by adding a linear function if necessary, we may assume that '(0) = 0. Let / G 23 be such that / ^ 0. Then, Now, because (p'(0) = 0 and cp is convex, it follows that cp'(x) > 0 for x > 0 and cp'(x) < 0 for x < 0. Hence, (Re f)cp'(Re /) is everywhere non-negative on T and vanishes only where cp'(Ref) vanishes. Thus, (4.19) is non-negative. To obtain a contradiction, assume that it vanishes. Then, cp'(Ref) vanishes almost everywhere on T. Now, strict convexity of cp at 0 implies that either cp'(x) > 0 for all x > 0 or cp'(x) < 0 for all x < 0 or both. Moreover, / does not almost everywhere vanish on T and / has mean zero over T so that neither / + nor / ~ can almost everywhere vanish on T . Hence, cp'(Ref) cannot almost everywhere vanish on (4.19) 239 Chapter III. Functionals on a set of domains and on Dirichlet spaces T because 0. Let $ : C \u00E2\u0080\u0094>\u00E2\u0080\u00A2 R . Assume that A $ is defined on *B, and that $ has first partials $ j everywhere on R 2, satisfying |$j(,z)| < ceG^ for every z \u00C2\u00A3 K 2 , where 241 Chapter III. Functionals on a set of domains and on Dirichlet spaces 1 < j < 2 and c andC are any finite constants. Assume that 4> satisfies SSARIP. Suppose that all the (pure and mixed) partials of $ of order < n exist. If n > 0 then assume further that all the (pure and mixed) nth order partials are Lipschitz on compact subsets of R 2 . Suppose f G 25 is an extremal function for A $ . Then f is n times continuously differentiable on T . Furthermore, the nth derivative of f on T is absolutely continuous and lies in A * ( T ) . Remark 4-5. It is in fact possible to obtain analogues of Theorems 4.3 and 4.4 for 25 a in place of 25, for a > 1. The methods are essentially the same, except that the variat ional equation (4.20) has OO Xy/(n)*n n=l in place of zf. Indeed, this much wi l l even work for ct G (0,1), and so we can f ind an analogue of Theorem 4.3 for al l a > 0. O f course, for a < 1, the square exponential growth condit ion on $>tj(z) wi l l have to be replaced by a polynomial condi t ion, wi th the degree of the polynomial depending on a and determined by use of Theorem 3.3 and the requirements of the proof. O n the other hand, for a > 1 the square exponential growth condit ion can be dropped, and replaced by boundedness on compacta . The proof of the modif ied Theorem 4.4 for 2 5 a , a > 1, is much the same as that of the ordinary Theorems 4.2 and 4.4. The only th ing to note is that on the left hand sides of various equations instead of having Akf we wi l l have oo A f c _ 1 ^ naf(n)zn. 71=1 However, this does not affect the conclusions very much, since if this quant i ty lies in Z. 2 (T) then a fortiori so does Akf. Ac tua l ly , we can do better: we can conclude that G L2(T), where [a\ is the smallest integer greater than or equal to a. We can also conclude that / is n + [a\ \u00E2\u0080\u0094 1 t imes continuously differentiable. Since these various extensions are not of great interest to us at present, we leave the details to the reader. 242 Chapter III. Functionals on a set of domains and on Dirichlet spaces 5. Symmetric decreasing rearrangement and Dirichlet norms Let / \u00C2\u00A3 Da for a \u00C2\u00A3 [0,oo). Then / has finite nontangential boundary values almost everywhere on T since o a C o 0 C h2. Let f\u00C2\u00AE be the funct ion on T given by the symmetr ic decreasing rear-rangement of n.t . l im / (see \u00C2\u00A71.6.2). Since / \u00C2\u00A3 L2(T), we must likewise have / \u00C2\u00AE \u00C2\u00A3 L2(T). Moreover , / \u00C2\u00AE has mean zero on T since / does and since /\u00C2\u00AE and / are equimeasurable. (Given equimeasurabi l i ty, use Remark 1.2.2. Equimeasurabi l i ty follows from Remark 1.6.7.) Let / \u00C2\u00AE : D -> E be the Poisson extension of f\u00C2\u00AE : T -> E v ia the Poisson integral (Theorem 1.3.3); we wi l l then have / \u00C2\u00AE \u00C2\u00A3 h2(TS>). T h e o r e m 5 . 1 . Let f E.Da for 0 < a < 2. Let f\u00C2\u00AE be as above. Then, l l / @ l k < l l / l k - (5-1) Suppose moreover that 0 < a < 2 and that equality holds in (5.1). Then f is a rotation of f\u00C2\u00AE, i.e., there exists w \u00C2\u00A3 T such that f(z) = f\u00C2\u00AE(zw) for all z \u00C2\u00A3 D. The proof wi l l be given later. Remark 5.1. For a = 0, we always have equali ty in (5.1), since both sides of (5.1) are then equal to the L2(T) norm, and symmetr ic decreasing rearrangement preserves L2 norms. Remark 5.2. If ce = 2 then it is well known that there exists / \u00C2\u00A3 da such that / is not a rotation of / \u00C2\u00AE but | | /\u00C2\u00AE| |o Q = | | / | | n Q . To see this, note that for any function g \u00C2\u00A3 C ^ D ) D U 2 we have where g'(eie) = j\u00C2\u00A7g(el6). (This is easy to see by wr i t ing the right hand side of (5.2) in terms of the Fourier series expansion of g'.) Now, let fi and / 2 be functions in CX(T) wi th the following propert ies: (i) fi and / 2 are symmetr ic decreasing (ii) fi is constant on Aid={el6 : |0| < 7r /2}, while / 2 is constant on A 2 = f { e ^ : |0 \u00E2\u0080\u00947r| < 3VT /4 } (iii) fi + fi has mean zero. 243 Chapter III. Functionals on a set of domains and on Dirichlet spaces Now, define f(eie) = fi{eie) + f2(e^9+^^^). Then , f\u00C2\u00AE=f1 + f2. Th is is easiest seen in light of F igure 5.1 by compar ing sizes of level sets of / and of f\ + f2. It is moreover clear that / is not a rotat ion of / \u00C2\u00AE . Let g2(et6) - f2(ei(e+(>*/4V). Then g2 is constant on e~iw/4A2, while / i is constant on Ai, so that the supports of g'2 and /{ are disjoint. Likewise, f2 is constant on A2) so that the supports of f2 and f[ are disjoint. Hence, r2ir r2ir r2TV / \f'(ei8)\2d6= \fl(eie)\2d0+ \9'2(ei6)\2d9 Jo Jo Jo = r\fi(el9)\2d9+ r\f'2(ew)\2 d6 Jo Jo = r\U@)'(e^)\2d6. Jo From (5.2) we conclude that | | /\u00C2\u00AE||o 2 = | | / | |o 2 a s desired. Now, given an analyt ic funct ion F \u00C2\u00A3 S a for some finite a \u00C2\u00A3 [0, 2], let f = KeF. Then , / \u00C2\u00A3 0 a , and ||/ | |?) a = | |F | | iD a (Remark 1.2). Let / \u00C2\u00AE be the symmetr ic decreasing rearrangement of / as before; then / \u00C2\u00AE \u00C2\u00A3 Da and there exists a unique funct ion G \u00C2\u00A3 D \u00E2\u0080\u009E such that R e G = / \u00C2\u00AE ; we wil l have | |G| | io a = | | /\u00C2\u00AE| |o Q - (We put G = Vf\u00C2\u00AE in the notat ion of \u00C2\u00A71.3.3.) We then define F\u00C2\u00AE = G and can reformulate Theorem 5.1 as follows. C o r o l l a r y 5 . 1 . Let F \u00C2\u00A3 \u00C2\u00AE a for 0 < a < 2. Let F\u00C2\u00AE be as above. Then, l l ^ \u00C2\u00AE l k < l l ^ l k -Suppose moreover that 0 < a < 2 and that equality holds in (5.1). Then F is a rotation of F\u00C2\u00AE, i.e., there exists w \u00C2\u00A3 T such that F(z) = F\u00C2\u00AE(zw) for all z \u00C2\u00A3 ED. The funct ion F\u00C2\u00AE is in fact univalent on D and the image F\u00C2\u00AE[D] is Steiner symmetr ic. Th is follows f rom the fol lowing Propos i t ion . P r o p o s i t i o n 5 . 1 . Let f \u00C2\u00A3 Hr(D) be such that Re / is symmetric decreasing on T . Assume f is not constant. Then f is univalent and the image /[D] is Steiner symmetric about the real axis. 244 Chapter III. Functionals on a set of domains and on Dirichlet spaces -71 -371/4 -71/2 -7l/4 0 7T/4 7T/2 37l/4 7C -71 -371/4 -7T/2 -7l/4 0 7l/4 7C/2 37T/4 71 -71 -371/4 -7C/2 -7l/4 0 7C/4 7T/2 37T/4 71 Note: appropriate constants should be added to the functions to ensure that/ has mean zero. Figure 5.1: The functions fi, f2, f and / \u00C2\u00AE 245 Chapter III. Functionals on a set of domains and on Dirichlet spaces The proof will be given later. Corollary 5.2. Let cp be a function on R such that $(z) = (^Re z) has SARIP. Fix 0 < a < 2. Suppose that A$ attains its maximum over Q3a at f \u00C2\u00A3 2$a. Then, there exists a w \u00C2\u00A3 T suc/i i/iai z H-7- Re/(zw) symmetric decreasing on T (i.e., Re/ is the rotation of a symmetric decreasing function). Moreover, the image /[D] is Steiner symmetric and f is univalent. Proof of Corollary 5.2. Suppose that Re / is not the rotation of a symmetric decreasing function in the sense specified in the statement of the Corollary. Then, ||/\u00C2\u00AE||:oQ < ||/||5)Q < 1 in light of Corollary 5.1. But A$(/\u00C2\u00AE) = A$(/) by Proposition 1.2.1 since T has finite measure. Hence, A$ attains a maximum at /\u00C2\u00AE. But by Proposition 4.1 we then obtain a contradiction since Il/lb. < I-Hence Re / is the rotation of a symmetric decreasing function, i.e., g(z) = Re f(zw) is symmetric decreasing. Moreover, by Proposition 4.1 we have / ^ 0 so that the rest of the conclusions of the theorem follow by applying Proposition 5.1 to g. \u00E2\u0080\u00A2 Proof of Proposition 5.1. First, fix r \u00C2\u00A3 (0,1). Let fr(z) = f(zr). Put u = Re/ r and v = Im gr. By Corollary 1.6.3, the function u is symmetrically strictly decreasing on T. Now, let Dr = /r[D]. We shall prove that Dr is Steiner symmetric about the real axis and that fr is univalent on D. It will then follow that / is univalent on D(r). Taking r \u00E2\u0080\u0094> 1\u00E2\u0080\u0094 it will follow that / is univalent on D. Moreover, /[D] = Uo ) = r*(\u00C2\u00A3>) = A*(G) = f max(0, (ReG) - x). JT The right hand side is finite since G G HP(B) for some p > 1 so that G G LP(T) (all we need is G G H1 here in fact). Since $(0) is finite, it follows from (6.3) that /oo g(z, x + iy; D) dy < oo. (6.4) -oo But now G is univalent. Thus, n(r, w;G) = l { r > | G - i ( ^ ) | } , and so * c ( w ) = logJG=W Hence NQ(W) = g(w,0;D) by Theorem 1.5.8. Since g(w,0;D) = g(0,w;D), we are done by (6.4). \u00E2\u0080\u00A2 Proof of Theorem 6.2. We need only prove the \"Moreover\", since the rest follows from the special case where / is a uniformizer of D sending 0 to 0. Fix XQ G E. Let LQ = {xo+iy : y G E}. Put L \u00E2\u0080\u0094 LQ n DB and M = LQ D D B . Write Ai for one-dimensional Lebesgue measure. We must prove that L C M. Because both DB and DB are Steiner symmetric, it suffices to prove that Ai(L) < Ai(M). 255 Chapter III. Functionals on a set of domains and on Dirichlet spaces If Ai(M) is infinite, then we are done. Hence, suppose that Yi= f|Ai(M) is finite. Since Nj vanishes outside D it follows that Nf can be nonzero on a set of measure at most 2Yi, and hence NB vanishes on L\{xo + iy : \y\ > Yi} by definition of the Steiner rearrangement. But we have equality in (6.2) for Y = oo so that it follows that On the other hand by (6.2) with Y \u00E2\u0080\u0094 Y\. Since Nj\u00C2\u00AE > 0 and by Lemma 6.1 all our integrals are finite, it follows that Now, L is precisely the set of points of LQ at which Nj\u00C2\u00AE is non-zero. It follows that AI(JL) < 2YX = Ai(M) as desired. \u00E2\u0080\u00A2 256 Chapter I V Radial rearrangement Overview In this chapter, \"doma in \" shall mean \"Greenian doma in \" . Our interest in this chapter is a conjecture of Matheson and Pruss [75] to the effect that there exists a way of replacing a domain D by a simply connected and star-shaped domain D of not bigger area such that the T$ functionals are increased by the replacement, where $ is a continuous funct ion on [0, oo) such that t 1-4 <&(e*) is increasing and convex. A positive answer to this conjecture would allow for a t ighter connection between the A $ and T$ functionals. Unfortunately, we do not manage to obtain an answer to the conjecture, although we do get some part ia l results and some interesting evidence. We state the relevant conjecture and discuss it in \u00C2\u00A71.1. In \u00C2\u00A71.2 we discuss some consequences of our conjectures. Then , in \u00C2\u00A71.3, we state a conjecture (Conjecture 1.4) as to how we think D should be defined v ia circular symmetr izat ion and Marcus ' radial rearrangement [70], and we discuss a few cases in which the conjecture holds. We state a weaker conjecture concerning harmonic measures (Conjecture 1.6). We also give a few counterexamples contradict ing some possible extensions of our various conjectures. It is unfortunate that al l our main conjectures remain open. However, in \u00C2\u00A72 we do give some part ia l results. For instance, we show that the harmonic measure functionals wr are increased by our conjectured choice of D providing the original D is simply connected (Theorem 2.3). The proof uses simple connect iv i ty in an essential way. 257 Chapter IV. Radial rearrangement In \u00C2\u00A73 we discuss our problems as transferred to the cylinder. This makes the constructions and conjectures a little more intuitive. We also discuss two-sided lengthwise Steiner symmetrization, and obtain a partial result (Theorem 3.1) analogous to our Conjecture 1.6. Finally, we discuss a \"cutting\" operation which one might conjecture to increase the wr functionals for circularly symmetric domains. In \u00C2\u00A74 we discuss a formulation of our Conjecture 1.4 in terms of Green's functions, and discuss why our Conjecture 1.6 on harmonic measures is weaker than Conjecture 1.4, and how Marcus' result [70] on the increase of the inner radius under radial rearrangement is connected with our Conjecture 1.4. Then, in \u00C2\u00A75 we generalize Marcus' above-mentioned result, proving our Conjecture 1.4 in the special case of t ^ ^$(e*) being concave. In \u00C2\u00A76 we discuss a one-sided Steiner rearrangement due to Haliste [56] and state her result on the effect of this rearrangement on harmonic measures (Theorem 6.1). Then, in \u00C2\u00A77, we discuss a Brownian motion formulation of our conjectures. Having done so, we state a rather natural one-dimensional discrete version of Conjecture 1.6 which we shall prove in \u00C2\u00A79. We note that while this one-dimensional discrete version holds not only for a simple random walk, but also for a random walk with probability p of going to the right and 1 \u00E2\u0080\u0094 p of going to the left, this is not the case with Conjecture 1.6. Indeed, in \u00C2\u00A77.1 we prove that the analogue of the cylindrical Brownian motion version of Conjecture 1.6 fails when the lengthwise component of the Brownian motion is replaced by a uniform motion to the right. This seems to provide some evidence against Conjecture 1.6, although the particular counterexample domain produced here does have Conjecture 1.6 satisfied because of simple connectivity and Theorem 2.3. Finally, in \u00C2\u00A77.2 we prove that a somewhat natural conjecture about Brownian motion exit times, analogous to Conjecture 1.4, is false. However, this too does not really provide evidence against Conjecture 1.4, but is mainly a consequence of the fact that radial rearrangement does not preserve areas. Most of the material in sections 1-7 is taken from the author's paper [85]. 258 Chapter IV. Radial rearrangement In \u00C2\u00A78 we prove extensions of the Beur l ing shove theorem on the harmonic measure of slit discs [23, pp. 58-62]. In part icular , we shall prove Conjecture 1.6 in the case of a circularly symmetr ic and starshaped domain slit along the negative real axis. Our extensions of the Beur l ing shove theorem in the 2-dimensional case go further than the extensions of Essen and Hal iste [45]; however, their methods are also valid in higher dimensions, while confine ourselves to domains in R 2 . Par ts of \u00C2\u00A78 are taken from the author 's paper [84]. Then , in \u00C2\u00A79 we give the discrete one-dimensional version of Conjecture 1.6 and prove it. We also give a few other results. The discrete one-dimensional result is actual ly completely elementary, as are its proofs (broken up between sections 9.2 and 9.3). The proofs are based on an explicit formula (Theorem 9.6) for the probabi l i ty of a random walk surviv ing various dangers on its route so as to exit a bl ind alley. O u r results in \u00C2\u00A79 are simi lar to a discrete result of Essen [43], and in \u00C2\u00A79.4 we discuss the connection wi th a continuous result of Essen [42]. A l l the mater ial in \u00C2\u00A79, excepting Theorem 9.2, is taken f rom the author 's paper [87]. F ina l ly , in \u00C2\u00A710 we prove a part ia l result concerning a horizontal-convexity analogue of a weaker version of the Matheson-Pruss Conjecture 1.1. Th is result provides some evidence for that conjecture. Our proof uses the var iat ional equation for extremals of A $ given in \u00C2\u00A7111.4.4. It also uses Baernstein 's sub-Steiner rearrangement f rom \u00C2\u00A7111.5, though the proof could also be done wi th standard Steiner rearrangement and Steiner analogues of the results of Baernstein [7]. On first reading, the reader may wish to omit sections 5, 6 and 9.3, the details of the con-struct ions in sections 7.1 and 7.2, as well as some of the details in the proof of L e m m a 10.1 of \u00C2\u00A710.1. 259 Chapter IV. Radial rearrangement 1. Conjectures and counterexamples 1.1. The primary conjectures Throughout this chapter, let T be the collection of all functions # on [0, oo) such that t >->\u00E2\u0080\u00A2 __), and so T^(U) < r*(D) for all U \u00E2\u0082\u00AC B. Hence T$ attains its maximum over B. Moreover, if it attains this maximum at D, it likewise attains it at D and hence there exists an extremal simply connected domain. \u00E2\u0080\u00A2 Sakai [95] had conjectured that T$p attains a maximum over B where &p(t) = tp for 0 < p < 00. Hence, an affirmative answer to Conjecture 1.3 implies an answer to Sakai's conjecture. P r o p o s i t i o n 1.2. Let $p(i) = tp. Then T^p(D) < T$(D) for every D \u00C2\u00A3 B providing p \u00C2\u00A3 [0, 2]. If Conjecture 1.3 holds for p then this is also true for p \u00C2\u00A3 (2,4]. Proof. The case of p \u00C2\u00A3 [0,2] is the well-known Alexander-Taylor-Ullman inequality [3] (see Kobayashi [66] for another proof). (More precisely, the Alexander-Taylor-Ullman inequality is the case p \u00E2\u0080\u0094 2, and, as Sakai [95] notes, the case p < 2 follows from Holder's inequality.) The case p \u00C2\u00A3 (2, 4] follows from Conjecture 1.3 and the fact that the inequality is valid for simply connected domains D. To see the validity for simply connected domains, it suffices to use The-orem III.1.2 and the fact that A$ (/) < A$ (Id) for p \u00C2\u00A3 [0, 4], where Id is the identity function and / is any function in 23. This latter inequality has been proved by Matheson [74]. Professor 261 Chapter IV. Radial rearrangement Sakai has kindly informed the author that the desired inequality in the simply connected case was also obtained by Professors N. Suita and S. Kobayashi. \u00E2\u0080\u00A2 The inequality in the above proposition was conjectured for p 6 (2,4] by Sakai [95]. Hence, an affirmative answer to Conjecture 1.3 would imply an affirmative answer to yet another conjecture of Sakai. As can be seen, it would also simplify the proof of the Alexander-Taylor-Ullman inequality, since the inequality A$p(/) < A$(Id) for / 6 23 is quite easy to prove for p G [0,2]. Indeed, it suffices to prove the latter inequality for p = 2 since the general case then follows by Holder's inequality. But for p = 2 the inequality is essentially trivial since then A$2(/) = T,n=i \f(n)\2 while H/lli, = E^\u00C2\u00B0=i n\f(n)|2 < 1, which easily shows that A$2 attains Indeed, it follows from the Chang-Marshall inequality for univalent functions. Since the proof of the Chang-Marshall inequality given in [72] is much simpler (and it also simplifies considerably in the univalent case) than the proof of Essen's inequality [44], we see that an affirmative answer to Conjecture 1.2 would imply a simpler proof of Essen's inequality. 1.3. Radial rearrangement A tool which one would think is very natural for attacking Conjecture 1.1 is Marcus' radial rearrangement [70]. The author is grateful to Professor Albert Baernstein II for having sug-gested the use of Marcus' radial rearrangement for this purpose. Let U be a set in the plane containing a neighbourhood of the origin and choose e > 0 such that D(e) C U, where D(r) indicates an open disc of radius r about 0. Define its maximum over 23 precisely at the functions of the form c \u00E2\u0080\u00A2 Id where |c| = 1. Finally, we note that if Conjecture 1.2 holds, then the Essen inequality sup T$(D) < oo DeB for = e*2 follows from the Chang-Marshall inequality sup A$(/) < oo. 262 Chapter IV. Radial rearrangement Figure 1.1: A multiply connected domain for Example 1.1, together with its radial rearrange-ment. Then the set U* = {re* : J\~Ldp < Re{9;U)^ is called the (Marcus) radial rearrangement of U. Observe that we are using a logarithmic metric in the definitions. Note that D(e) C [/* and that if U is open then 6 H> RS(9;U) is lower semicontinuous (cf. Proposition 1.6.1) and hence \u00C2\u00A37* is open. It is easy to verify that \u00C2\u00A37* is independent of the choice of \u00C2\u00A3 and that Area(\u00C2\u00A3/*) < Area(\u00C2\u00A37) with equality if and only if \u00C2\u00A37* and U coincide almost everywhere with respect to Lebesgue area measure. Note that U* is always star-shaped and that \u00C2\u00A3/* = U if and only if U is star-shaped. Finally, one may observe that is a rearrangement in the sense of Definition 1.2.2 on the rj-pseudotopology of all measurable subsets of C containing a neighbourhood of the origin. Marcus [70] has shown that if decreases neither inner radii (see inequality (4.2), below) nor certain capacities. Example 1.1. One might naturally conjecture that [7* has the desired property that r$(<7*) > r$(C7). However, this is not the case. Let U be any domain with the following properties for some positive r and 6, where \u00C2\u00A3 is as before: 263 Chapter IV. Radial rearrangement 264 Chapter IV. Radial rearrangement (i) The harmonic measure of {\z\ = r} n d(D(r) f l U) at zero in U does not vanish. (ii) RE(9; U) < p~ldp for every 9. Such a domain can easily be exhibi ted; see F igure 1.1 (left) for a mult iply-connected example, and F igure 1.2 for a simply-connected example. G iven such a domain , (ii) implies that we wi l l have \u00C2\u00A3 / * C D( r ' ) where r' = r \u00E2\u0080\u0094 5 (see F igure 1.1, r ight). Choose x \u00C2\u00A3 (r', r ) . Define $(\u00C2\u00A3) = max(0, t \u00E2\u0080\u0094 x). Then $ is convex, hence in J7, and it is easy to verify that r$([/) > 0 by (i). O n the other hand, as J/* C D(r') we must have <&(|z|) vanishing on the closure of U* so that r$([/*) = 0, and thus radial rearrangement by itself does not give a solut ion to Conjecture 1.1. Now, given a domain U, let U\u00C2\u00AE be its circular symmetr izat ion (see \u00C2\u00A71.6.1). We have Area(f /\u00C2\u00AE) = Area ( t f ) . A n d TQ{U\u00C2\u00AE) > T9(U) for every $ \u00E2\u0082\u00AC T by Coro l la ry 1.6.1. Note that if U is c i rcular ly symmetr ic then so is U* though the converse does not in general hold. It is easy to see that no domain satisfying (i) and (ii) of Example 1.1 can be circularly symmetr ic . C o n j e c t u r e 1.4. Conf ined to the class of circular ly symmetr ic domains, radial rearrangement does not decrease any of the funct ionals T$ for $ G f . A n affirmative answer to this question would give an affirmative answer to Conjecture 1.1 since we could then let U = (U\u00C2\u00AE)*. Wh i le Conjecture 1.4 is in general open, we can prove it for a very simple class of domains. Example 1.2. Consider the closed polar rectangle H = {reie :rl__