Appendix A. The Dirichlet Problem and Harmonic Measures In this appendix we prove Wiener's theorem on regular boundary points for Green functions and Dirichlet problems. In particular, it will follow that these two con­ cepts are identical. A.I Regularity with Respect to Green Functions Let Gee be a domain such that aG is of positive capacity and, for a E G, let gG(z, a) be the Green function of G with pole at a. Recall from Sections 1.4 and 11.4 that gG(z, a) is defined as the unique function on G satisfying the following properties: (i) ga(z, a) is nonnegative and harmonic in G \ {a} and bounded as z stays away from a, 1 (ii) gG(z, a) - log -- is bounded in a neighborhood of a, Iz -al (iii) lim gG(z, a) = 0 for quasi-every x E aG z~x,ZEG with (ii) replaced by (ii)' ga(z, a) - log Izl is bounded in a neighborhood of 00 when a = 00. We call a point x E aG on the boundary of G a regular point (with respect to the Green function gG(z, a» if lim gG(z, a) = O. z~x, ZEG Soon we shall see that this notion is independent of the choice of a; therefore, we shall just speak of regular boundary points. Note also that, by definition (which, however, depends on the existence theorem for Green functions), quasi-every point on the boundary of G is a regular point. Theorem 1.1 (Wiener's Theorem). Let 0 < A < 1 and set 450 Appendix A. The Dirichlet Problem and Harmonic Measures Then x E aG, x i= 00, is a regular boundary point of G if and only if 00 n (1.1) ?; 10g(l/cap(An(x» = 00. In particular, regularity is a local property. Proof. First we consider the case when G is an unbounded domain and a = 00. Set E := C \ G and 1 V(E):= log--. cap(E) It is immediate that the condition (1.1) does not change if An (x) is defined as (note that equality is allowed at both places on the right); hence in what follows we can work with this definition of An (x), which is more convenient than the original one, for then An (x) is compact. We may assume without loss of generality that x = O. We start with the proof of the sufficiency of condition (1.1). In view of the representation (1.2) (see (1.4.8», which we use to extend gG to the whole plane, the inequality UILE (z) ::::; V (E), (1.3) and the lower semi-continuity of UILE, it suffices to show that (1.1) implies UILE(O) = V(E). (1.4) Assume to the contrary that (1.4) is not true, i.e. fJ := V(E) - UILE(O) > O. Then, in view of Theorem 1.4.1, we will have V (E*) - UILE' (0) ~ fJ (1.5) for every compact subset E* of E of positive capacity. One can easily verify from the discussion at the end of the present proof that A can be replaced by Ak with any fixed k, so we can assume A as small as we like. In particular, we can assume that A < 1/4 is so small that log(l/(l - A» < fJ/2 is satisfied. If 00 n ?; log(l/cap(A2n(0» = 00, then we set En = A2n(0); otherwise, we choose En = A2n-l(0). In any case the sets En are disjoint and satisfy A.l Regularity with Respect to Green Functions 451 00 n " -- - 00 (1.6) ~ V(En) - . Finally, we choose E* = {OJ U (Un~noEn) with an no so large that we have E* C {z I Izl :s 1/2}. Then E* is compact and E* C E. Now let z E Ei and t E Ej with j =1= i and cap(Ei) > O. If j < i then Iz - tl ::: (1 - AWl, while for j > i we have Iz - tl ::: Itl. Thus, in any case 1 1 1 log - + log -- > log --. Itl 1 - A - Iz - tl Let us integrate this inequality with respect to d/-LE*(t) on E* \Ei . Then, observing that both z and t lie in the disk DI/2(0), the left-hand side will be at most UIlE'(O) + log 1 ~ A < V(E*) - f3 + i, while the right-hand side is UIlE' (z) - [ log _1-d/-LE* (t). lEi Iz-tl Since UIlE' (z) = V (E*) for quasi-every z E Ei , we obtain [ log _1-d /-LE.(t) ::: ~, for q.e. z E E i • lEi Iz-tl 2 Let Vi be the restriction of /-LE' to Ei . Then the preceding inequality takes the form UVi(Z) > ~ for q.e. z E E . - 2' i Thus, UllEi 1 f f3 /-LE·(E i ) = vi(Ei) = f V(Ei) dVi = V(Ei ) UVid/-LEi::: 2V(Ei )· But this implies UIlE' (0) = log ~ d/-LE* (t) f It I l"i-l ::: 2,8 log - ~ -- = 00, A i V(Ei ) which is a contradiction by (1.5) and (l.6). This contradiction was the result of the assumption that (Ll) holds but (1.4) does not; hence (Ll) implies regularity. 452 Appendix A. The Dirichlet Problem and Harmonic Measures Now we turn to the necessity, and let us assume that (1.1) does not hold. Then there are arbitrary small r's such that the circle {z I Izl = r} does not intersect E := C \ G (see Lemma 1.2.1). For such an r, let K\ be the intersection of E with the disk Dr(O) := {z Ilzl ::: r}, and set K2 := E \ K\. Then both K\ and K2 are compact, and we can choose a bounded closed neighborhood K; of K2 such that also K; is disjoint from K \ . On applying Theorem 1.4.1 as before, it is enough to prove the irregularity of x = °with respect to any domain G\ ~ G containing x on its boundary; thus we can assume without loss of generality that infinitely many (or all if we like) An (0) are of positive capacity. This implies, in particular, that K\ has positive capacity. Let ko be the smallest integer with the property that for k > ko the sets Ak := Ak(O) are disjoint from K2. In what follows we shall assume that Ako is entirely contained in the disk Dr(O); if this is not the case, then in the following discussion we have to split Ako into Ako n Dr (0) and Ako \ Dr (0) and make the necessary changes. Finally, without loss of generality we may assume r < 1/2 is so small that 00 k 1 ~ V(Ak ) < 210g(1/).) is satisfied. Since for cap(Ak) > 0, k ::: ko, we have with v := ILK) IAk ( UiLAk 1 f V(K\) V ILK) (Ak) = JAk V (Ak) dILK) = V (Ak) U dILAk ::: V (Ak) , it follows that 00 ( 1 ) look 1 UiLK) (0) ::: L log k ILK) (Ak) ::: V(K\) log - L -- < - V(Kd, k=ko ). ). k=ko V (Ak) 2 and this means that gC\K/O, 00) > 0. Using the mean value inequality for the subharmonic function gC\K) (z, 00) on circles {z I Izl = r} that do not intersect E (as we have seen, there are such circles with arbitrary small r > 0) we can conclude lim sup gC\K) (z, 00) > 0. (1.7) Z--->O,ZEG Now let M be larger than the maximum of gC\K) on the boundary of K; and m the minimum of gG on the same set. Since K; is a neighborhood of K2 disjoint from K\, we have the relation aK; c G; therefore m > ° by the minimum principle. Replacing M by a larger number if necessary, we can assume that M 1m::: 1. Let us now apply the generalized minimum principle (Theorem 1.2.4) to the function (Mlm)gG - gC\K) on the set G* := C \ (K\ U KD. This is a superharmonic function in G* which is bounded from below (around infinity it behaves like «M 1m) - 1) log Izl) and which has nonnegative boundary limits A.I Regularity with Respect to Green Functions 453 quasi-everywhere on 3G* = 3K1 U3Ki (recall also that on 3Ki both gC\K[ and gG are continuous). Therefore we can conclude that (M jm)gG - gC\K[ is nonnegative on G*, and we have in view of (l. 7) limsup gc(z,oo) 2:: lim sup m gC\K[(Z, oo) > 0, Z-+O.ZEG Z-+O.ZEG M which verifies that 0 is not a regular boundary point of G. Thus, the proof of Wiener's theorem is complete in the case when a = 00. Now let us tum to the case of an arbitrary domain G with cap(3G) > O. The mapping z -+ z' := 1/(z - a) maps G into an unbounded domain G' and gG(z, a) is transformed to gG'(z', (0). Thus, x E 3G is a regular boundary point with respect to G if and only if x' = I j (x - a) is a regular boundary point with respect to G'. Therefore, we only have to show that the Wiener condition (l.I) is also preserved under this mapping. Let D be a disk around x not containing a on its boundary, and let D' be its image. Then on D the mapping z -+ z' is a constant times a nonexpansive mapping (i.e. which can only shrink distances), and on D' the same is true for its inverse.