Exposed sets in potential theory

Transcription

1 Bull. Sci. math. 130 (2006) Eposed sets in potential theory Jaroslav Lukeš, Tomáš Mocek, Ivan Netuka Charles University, Faculty of Mathematics and Physics, Sokolovská 83, Praha 8, Czech Republic Received 22 December 2005 Available online 9 February 2006 Abstract Let U be a relatively compact open subset of a harmonic space, and H(U) be the function space of all continuous functions on U which are harmonic on U. We give a complete characterization of the H(U)- eposed subsets of U. This etends the results of [J. Lukeš, T. Mocek, M. Smrčka, J. Spurný, Choquet like sets in function spaces, Bull. Sci. Math. 127 (2003) ] Elsevier SAS. All rights reserved. Résumé Etant donné un ouvert relativement compact dans un espace harmonique, H(U) est l espace des fonctions continues sur l adhérence U de U qui sont harmoniques dans U. Dans cet article, nous donnons une caractérisation complète des ensembles H(U)-eposés de U. Cela étend les résultats de [J. Lukeš, T. Mocek, M. Smrčka, J. Spurný, Choquet like sets in function spaces, Bull. Sci. Math. 127 (2003) ] Elsevier SAS. All rights reserved. MSC: primary 31D05, 46A55; secondary 31B05 Keywords: Function space; Abstract potential theory; Choquet s theory; Eposed sets; Absorbent sets The work is part of the research project MSM financed by MSMT. * Corresponding author. addresses: lukes@karlin.mff.cuni.cz (J. Lukeš), mocek@karlin.mff.cuni.cz (T. Mocek), netuka@karlin.mff.cuni.cz (I. Netuka) /$ see front matter 2006 Elsevier SAS. All rights reserved. doi: /j.bulsci

2 J. Lukeš et al. / Bull. Sci. math. 130 (2006) Introduction In classical potential theory the following assertion is now known as the Keldysh lemma: Given a regular boundary point z of a bounded open subset U of R m, there is a continuous function h on the closure U of U which is harmonic on U and satisfies h(z) = 0 and h>0 onu \{z}. In other words, at any regular point there is an eposed function. The original construction of M.V. Keldysh is rather complicated. Using Choquet s theory, there is a simple argument leading to the Keldysh lemma. Namely, if H(U) denotes the function space of all continuous functions on U which are harmonic on U, then the function space H(U) is simplicial and the set reg U of all regular points of U coincides with the Choquet boundary of H(U) (see, for eample, [7] and references therein; cf. also [11]). Under these circumstances, any point of the Choquet boundary is eposed. The paper [8] presents a characterization of the eposed sets in U, that is, those sets F U for which there eists a function h H(U) such that h = 0 onf and h>0 onu \ F. The method used there is based on deeper properties of Choquet sets in simplicial spaces. Here we present a direct proof of this characterization. Theorem 1.1. Let U be a bounded domain in R m and F a proper subset of U. Then F is H(U)- eposed if and only if F is a proper closed subset of reg U. Proof. Let F be a proper closed subset of reg U. Solving the weak Dirichlet problem we see that, for any z reg U \ F, there is a function h z H(U) such that h z (z) > 0, h z = 0 onf and 0 h z 1. The eistence of h z is guaranteed by the simpliciality of the space H(U) (cf. [9, p. 242]). The set reg U \ F is separable, so there eists a sequence {z n } of points in reg U \ F such that reg U \ F { U: h zn () > 0 }. The function h := 2 n h zn belongs to H(U), h 0onU, h = 0onF and h>0on reg U \ F. Given U, letε U stand for the balayage of the Dirac measure ε to the complement U of U (cf. Section 3 below). Then, for any U \ reg U,wehave spt ε U = reg U by a result of J. Bliedtner and W. Hansen in [4] (cf. Lemma 8.2 below). Hence, for any U \F, h() = U h dε U = U\F h dε U reg U\F h dε U > 0, and we see that F is an H(U)-eposed set. Conversely, any H(U)-eposed set F U is a proper closed subset of reg U, by basic properties of harmonic functions and regular points. Indeed, any such set is clearly closed, and also disjoint from U as a consequence of the strong minimum principle for harmonic functions. If F, then U and any H(U)-eposed function for F is a barrier for, and thus is a

3 648 J. Lukeš et al. / Bull. Sci. math. 130 (2006) regular point. Finally, if F = reg U, then any function from H(U), being zero on F, vanishes everywhere on U. Corollary 1.2 (The Keldysh lemma). Any regular boundary point of U is H(U)-eposed. The remainder of this paper is devoted to developing a more general approach to eposed sets in the framework of abstract potential theory. 2. Preliminaries Before proceeding, some notation will be established. Let K be a compact topological space and C(K) be the space of all real-valued continuous functions on K equipped with the sup-norm. We will identify the dual of C(K) with the space M(K) of all Radon measures on K. LetM + (K) denote the set of all positive Radon measures on K, and M 1 (K) the set of all probability measures in M + (K). Then M 1 (K) is a conve w -compact subset of M(K). We denote by ε the Dirac measure at K. Given a μ-integrable function f on K, we often simply write μ(f ) instead of K f dμ.ifg is a numerical function, then g + denotes the function ma(g, 0). Lower semicontinuous functions are assumed to be lower finite, and upper semicontinuous functions to be upper finite. We denote by χ A the characteristic function of a set A and by A its boundary. 3. Harmonic spaces In what follows, (X, H) is a P-harmonic space with a countable base in the sense of the aiomatics of Constantinescu and Cornea [6] or of Bauer [1]. For a set A X we shall use the notation ba for the set of all points of X where A is not thin and A for the complement of A. Further, given a positive hyperharmonic function v on X, Rv A stands for the reduced function of v on A while R v A denotes the balayage of v on A. It is known that, for any X, there eists a unique Radon measure ε A on X such that ε A (v) = R v A () for any positive hyperharmonic function v on X. For a relatively compact open set U X and a continuous function f on U, Hf U denotes the Perron Wiener Brelot solution of the Dirichlet problem on U with boundary data f.the harmonic measure μ U at a point U is defined as the unique Radon measure on U for which Hf U () = f dμ U for any f C( U). U A point z U is called regular if lim H U z f () = f(z) for any f C( U). We denote by reg U the set of all regular points of U and by irr U the set U \ reg U of all irregular points of U. If reg U = U,thesetU is called regular. A closed subset F of X is called an absorbent set if the function { 0 onf, v(z) := on X \ F,

4 J. Lukeš et al. / Bull. Sci. math. 130 (2006) is hyperharmonic on X. According to Proposition of [6], a set F is absorbent if and only if for any relatively compact resolutive set V X and any F V,wehaveμ V (X \ F)= 0. Proposition 3.1. For any relatively compact open set U and any U, we have and μ U = ε U reg U = U b ( U ) = { z U: ε U z = ε z }. Proof. See [6, Theorem 7.1.2, Corollary and Theorem 7.1.1]. Let A X. Bytheessential base βa of A we mean the smallest finely closed subset F of X such that the set A \ F is semipolar. If A is finely closed, then βa is the greatest subset F of A such that F bf (cf. [5, Proposition VI.6.1 and Corollary VI.6.6]). Lemma 3.2. The operator β : A βa is idempotent, that is, β(βa) = βa for any A X. Moreover, given A X, βa = A if and only if ba = A. In particular, b(βa) = βa for any A X. Proof. See [2, Proposition 1.1]. We will also need a slight generalization of Proposition from [6]. Lemma 3.3. Let A be a subset of X and f be a lower semicontinuous function on X with compact support. Then the function g : ε A (f ), X, is hyperharmonic on X \ A. Proof. There is an increasing sequence {ϕ n } of continuous functions on X, each with compact support, such that ϕ n f. According to Proposition of [6], the functions g n : ε A (ϕ n), X, are harmonic on X \ A. Since g n g, the assertion follows. 4. Choquet s theory of function spaces Let H be a function space on a metrizable compact space K. By this we mean a closed linear subspace of C(K) containing the constant functions and separating the points of K. Let M (H) be the set of all H-representing measures for K, namely, { } M (H) := μ M 1 (K): f()= f dμ for any f H. The set Ch H K := { K: M (H) ={ε } } K

5 650 J. Lukeš et al. / Bull. Sci. math. 130 (2006) is a G δ -set and is called the Choquet boundary for H. A function space H is called simplicial if, for any K, there eists a unique representing measure δ in M (H) carried by the Choquet boundary Ch H K. The space A c (H) of all continuous H-affine functions is defined as the family of all continuous functions f on K for which f()= f dμ for any K and any μ M (H). K The set A c (H) is a function space containing H which, in general, can differ from H. We shall denote by K c (H) the set of all continuous functions f satisfying f() f dμ for any K and any μ M (H). K The elements of K c (H) are called continuous H-conve functions and they determine a partial ordering, the so-called Choquet ordering, on the space M + (K): μ ν means that μ(f ) ν(f) for each f K c (H). Choquet sets. A subset M of K is called a Choquet set if M is closed and the following two conditions are satisfied: (a) if M and μ M (H), then spt μ M, (b) if K \ M and μ M (H), then spt μ is not contained in M. AsetF K is called H-etremal if F is closed and, for any F and any μ M (H),the measure μ is supported by F.AsetC K is called H-conve if C is closed and C whenever K, μ M (H) and spt μ C. Thus, a closed set M K is a Choquet set if and only if M is simultaneously H-etremal and H-conve. H-eposed sets.asete K is said to be H-eposed if there eists a positive function h H such that E = h 1 (0). The following theorem follows easily from results in [8] (see Proposition 12.4 and Corollary 14.5). Theorem 4.1. Let H be a simplicial function space on a metrizable com pact space K such that H = A c (H). Then a set M K is H-eposed if and only if M is a Choquet set. H-conve hull.letf be a closed subset of K. We define co H F := { K: there eists μ M (H) such that spt μ F }, the H-conve hull of F. For the reader s convenience we sketch the proof of the following lemma. Full details may be found in [8]. Lemma 4.2. If μ and ν are positive Radon measures on K and μ ν, then spt μ co H spt ν. Proof. Assume that spt μ \ co H spt ν. By a Hahn Banach type argument, there eists a function h H such that h() > ma { h(z): z co H spt ν }.

6 J. Lukeš et al. / Bull. Sci. math. 130 (2006) Adding a suitable constant, we may assume that ma h = 0onco H spt ν. Then h + is a continuous H-conve function and μ(h + )>0 ν(h + ) which contradicts the assumption that μ ν. Hereafter we assume that H is a simplicial space on a metrizable compact space K. For each K, the measure δ denotes the unique representing measure from M (H) carried by Ch H K. Observe that, given K and μ M (H), wehaveμ δ. Indeed, let λ be a maimal measure such that μ λ. Then λ M (H). AsH is simplicial, δ = λ. Admissible sets. ForaclosedsetF K, define A F := { K: δ is supported by F }. Obviously, F A F co H F. A subset F of K is said to be admissible if F is closed and F A F. Lemma 4.3. Let F be a closed subset of K. IfA F is closed, then A F = A AF. Proof. We observe that F Ch H K = A F Ch H K. Indeed, if A F Ch H K, then by the definition of A F we have spt δ F. Since δ = ε,we see that F. Conversely, assume that F Ch H K. Then again δ = ε,sosptδ ={} F. By definition, A F. The assertion now follows immediately. Remark 4.4. The set A F need not be closed, as the following eample from classical potential theory shows. Fi a bounded domain U in R m possessing eactly one irregular point z which is not an isolated point of U. Take an open set V R m such that z/ V and V U. Then, for F = U \ V,wehaveA F = F \{z} (cf. Proposition 8.3). Hence A F is not closed. Lemma 4.5. Let F be an admissible subset of K and K. Assume that there eists a representing measure μ M (H) supported by F. Then spt δ F. Proof. Define T : f δ z (f ) dμ(z), f C(F ). F (Note that the function z δ z (f ) is a bounded Borel function on F ; cf. Proposition 6.1 in [7].) Then T is a positive linear functional on C(F ). By the Riesz representation theorem there eists a Radon measure ν on F such that Tf = ν(f) for each f C(F ). If h H, then h() = h dμ = h dμ. K Hence ν(h) = Th= F F δ z (h) dμ(z) = F h(z) dμ(z) = h()

7 652 J. Lukeš et al. / Bull. Sci. math. 130 (2006) for any h H, and therefore ν M (H). Fi a compact set L K \ Ch H K. Since ν(g) = δ z (g) dμ(z) F for any bounded semicontinuous function g on F, by the Lebesgue monotone convergence theorem we get ν(l) = δ z (χ L ) dμ(z) = 0. F By the regularity of the measure ν, we see that ν is carried by the Choquet boundary Ch H K. Since the function space H is simplicial, we obtain δ = ν. We conclude that spt δ = spt ν F. Lemma 4.6. If F is an admissible subset of K, then co H F = A F. Proof. Pick co H F. There eists μ M (H) such that spt μ F. By Lemma 4.5, spt δ F. Hence A F. The reverse inclusion A F co H F being obvious, the proof is complete. Lemma 4.7. If F is a closed subset of K such that F = A F, then F is a Choquet set. Proof. Since F = A F, F is admissible. Taking into account Lemma 4.6, we have F co H F = A F = F. Thus F is H-conve. We now choose F and μ M (H). Since μ δ and spt δ F, we can use Lemma 4.2 to obtain spt μ co H spt δ co H F = F. It follows that F is H-etremal and the proof is complete. 5. Choquet s theory in harmonic spaces The harmonic function space is one of the most important eamples for Choquet s theory. In this case, U is a relatively compact open subset of a harmonic space and the corresponding function space H is H(U), that is, the family of all continuous functions on U which are harmonic on U. We tacitly assume that H(U) contains the constants and separates the points of U. In the classical harmonic space associated with the Laplace equation, the Choquet boundary of H(U) coincides with the set reg U of all regular points of U. In general harmonic spaces the situation is more complicated. Theorem 5.1. The function space H(U) is simplicial, H(U) = A c (H(U)) and Ch H(U) U = U β ( U ) = { z U: ε β( U) z = ε z }.

8 J. Lukeš et al. / Bull. Sci. math. 130 (2006) For any U, ε β( U) ε β( U) = ε Ch H(U) U. is the unique maimal H(U)-representing measure for and Proof. The proofs of these assertions are given in the paper of J. Bliedtner and W. Hansen [2]. The net assertion can be deduced, for eample, from Lemma 1.1 in [4]. We provide a direct proof here. Lemma 5.2. Assume that U, B A and B b(b). Ifε B z M z(h(u)) for any z U, then ε A M (H(U)). Proof. Let p be a continuous potential on X and U. Since B b(b), wehaverp B = R p B. Hence R A Rp B = R p B. In particular, using Proposition of [6], (ε A )B (p) = εz B (p) dεa (z) = R p B (z) dεa (z) = R B p (z) dεa (z) = R A R B p () = R B p () = εb (p), which yields ε B = (εa )B. Hence, given h H(U),wehave h() = ε B (h) = (εa )B (h) = ε B z (h) dεa (z) = h(z) dε A (z). The following important result can be stated in various degrees of generality. is an H(U)-representing mea- Corollary 5.3. Assume that U and Ch H(U) U A. Then ε A sure for. Proof. Set B := Ch H(U) U. By Theorem 5.1, ε B z M z(h(u)) for any z U and B ={z U: εz B = ε z}, so B = b(b). The statement now follows from Lemma β-absorbent sets β-absorbent sets. LetU be a relatively compact open subset of a harmonic space. Following a suggestion of W. Hansen, a subset F of U is said to be β-absorbent if F is closed and spt ε K F for any F and any compact set K U \ Ch H(U) U containing. If F is a β-absorbent set and M a closed subset of Ch H(U) U, then F M is also β-absorbent. This is an immediate consequence of the definition.

9 654 J. Lukeš et al. / Bull. Sci. math. 130 (2006) Proposition 6.1. If F is a β-absorbent subset of U, then F U is an absorbent subset of U. Proof. Pick F U and an open neighborhood V of (in U). Let W be a resolutive set and W W V. Then d := dist(, W)>0. Set K n := { z U: dist ( z, W 1/n )}, n N, n 1/d. Then {K n } is an increasing sequence of compact subsets of U, K n for each n 1/d, and { Kn : n 1/d } = W. According to Proposition of [6], ε K n n, sptε K n ε W F.As K n W,wealsohavesptε K n spt ε W F W F W F U. We see that μ W (U \ F)= ε W (U \ F)= 0, so F U is an absorbent subset of U. vaguely. Since K n U \ Ch H(U) U for any W. Hence Lemma 6.2. Let F be a β-absorbent set and F.IfasetA contains Ch H(U) U, then the measure ε A is supported by F. In particular, ε β( U) is supported by F. Proof. We remark that, for every continuous potential p and every, ε U (p) ε β( U) (p) = ε Ch H(U) U (p) ε A (p). Thus, by Proposition VI.9.9 of [5], ε A U = ε A. Hence, without loss of generality, we may suppose that A contains β( U). Theassertionisobviousfor F Ch H(U) U, since ε A = εβ( U) = ε. Suppose that F \ Ch H(U) U. According to Corollary 2.3 in [3], ε A = ( ) εa {} ε + ( 1 ε A ( )) A\{} {} ε. (1) If ε A\{} = ε, then by (1) we have ε A = ε and the assertion is again obvious. So we may assume that ε A\{} ε. Since ε (A \{}) = 0, an application of Theorem VI.10.5 of [5] yields the eistence of a decreasing sequence {V n } of open neighborhoods of A \{} such that ε V n εa\{} vaguely. Set K n := U V n. Since ε V n = ε whenever V n and ε A\{} ε, we may assume that K n for every n. AsF is a β-absorbent set, spt ε K n F. It follows that spt ε A\{} F. Consequently, using (1) again, spt ε A F, finishing the proof. Corollary 6.3. Any β-absorbent set is admissible. Proof. Let F be a β-absorbent set and F. Since ε β( U) is H(U)-maimal, the assertion follows immediately from Lemma 6.2 and the definition of admissibility. We finish this section with a lemma resembling Proposition 6.1.

10 J. Lukeš et al. / Bull. Sci. math. 130 (2006) Lemma 6.4. Let F be a closed subset of U. Then A F U is an absorbent subset of U. Proof. Putting h : ε β( U) (χ F ), U, the function h is positive and hyperharmonic on U according to Lemma 3.3. Since A F U = { U: h() = 0 }, the assertion follows. 7. Eposed sets Proposition 7.1. Any H(U)-eposed set is β-absorbent. Proof. Let F be an H(U)-eposed set. We may suppose that F U. Assume that there eist F and a compact set K U \ Ch H(U) U such that K and ε K is not supported by F.Let h H(U) epose the set F, that is, h = 0onF and h>0onu \ F. Using Corollary 5.3, we get 0 = h() = h dε K = h dε K > 0. X U\F This contradiction completes the proof. β-modification. For any subset F of X, we define the β-modification F β of F as F β := { B U: isβ-absorbent such that B ChH(U) U F Ch H(U) U }. Lemma 7.2. Let F be a closed subset of U. Then F β A F. Proof. Pick F β and choose a β-absorbent set B such that B and B Ch H(U) U F Ch H(U) U. By Corollary 6.3, B is admissible. Hence δ is supported by B, from which it follows that δ is supported by B Ch H(U) U. Thus δ is supported by F Ch H(U) U F. We now see that A F. Theorem 7.3. Let F be a closed subset of U. Then the following assertions are equivalent: (i) F is H(U)-eposed, (ii) F = A F, (iii) A F is closed and F = F β. Proof. (i) (ii): If F is an H(U)-eposed set, then F is admissible by Proposition 7.1 and Corollary 6.3. Hence F A F.AnyH(U)-eposed set is H(U)-conve, so A F co H(U) F = F. The converse implication (ii) (i) follows from Lemma 4.7 and Theorem 4.1. Assume now that F = A F. Then A F is closed and, by Proposition 7.1 and the equivalence (i) (ii), the set F is β-absorbent. Hence F F β. According to Lemma 7.2, F β A F.This establishes (ii) (iii).

11 656 J. Lukeš et al. / Bull. Sci. math. 130 (2006) To see that (iii) (ii), suppose that F = F β and A F is closed. Again, in accordance with Lemma 7.2, F = F β A F.AllthatremainstobeprovedisthatA F F. Lemma 4.3 ensures that A F = A AF. We now invoke the equivalence (i) (ii) above and Proposition 7.1 to deduce that the set A F is β-absorbent. Since A F Ch H(U) U = F Ch H(U) U (cf. the proof of Lemma 4.3), we can conclude that A F F β = F. Remark 7.4. In Remark 4.4 we have already shown that the set A F need not be closed. We do not know whether the set A F is closed under the additional assumption that F = F β. 8. The Laplace equation Consider now the harmonic space associated with classical potential theory. In this case, U is a bounded open subset of R m and the function space H(U) consists of the continuous functions on U which are harmonic on U. Any absorbent subset of a domain U is either empty or equal to U (cf. [6, Corollary and Theorem 3.2.1]). Proposition 8.1. Let U be a bounded open subset of R m. Then the Choquet boundary Ch H(U) U coincides with the set reg U of all regular points of U and δ = ε U for any U. Proof. See [5, Proposition VII.4.1]. Lemma 8.2 (Bliedtner Hansen). If U is a bounded domain in R m and U \ reg U, then spt δ = reg U. Proof. See [4, Proposition 1.2]. Proposition 8.3. Let U be a bounded domain in R m and F be a proper closed subset of U. Then (a) F is admissible if and only if either F is contained in reg U,orF contains reg U, (b) F is β-absorbent if and only if F reg U, (c) A F = U when reg U F, and A F = F reg U when reg U \ F, (d) F is H(U)-eposed if and only if F is a proper subset of reg U. Proof. (a) Assume that F reg U. Then F is admissible since δ = ε for any F.IfF reg U and F, then δ is supported by reg U,soF is admissible. Conversely, assume that F is admissible and that there eists F \ reg U. Then reg U = spt δ F according to Lemma 8.2. (b) Since reg U = Ch H(U) U, any closed subset of reg U is obviously β-absorbent. Conversely, suppose F is β-absorbent. By Corollary 6.3, F is admissible. Hence, either F is contained in reg U,orF contains reg U. In the latter case, according to Proposition 6.1, either F U = U, orf U =.IfF U = U, we would have F = U, which is impossible.

12 J. Lukeš et al. / Bull. Sci. math. 130 (2006) So we assume that F U and will show that the assumption F \ reg U leads to a contradiction. Choose a closed ball L U and put K := L {}. Notice that spt ε K is contained in L, and so is disjoint from F. This contradiction completes the proof of (b). (c) The first statement follows from Proposition 8.1 and Lemma 8.2. Suppose that reg U \ F.If F reg U, then δ = ε,so A F. For the opposite inclusion, assume first that / reg U. Then spt δ = reg U by Lemma 8.2 and therefore / A F. If / F, then either δ = ε or spt δ = reg U, which implies that / A F. (d) Any H(U)-eposed set is β-absorbent by Proposition 7.1, and therefore is contained in reg U. Obviously, reg U is not an H(U)-eposed set. To prove the converse, suppose that F is proper subset of reg U. By invoking the preceding assertion (c), F = A F and we see by Theorem 7.3 that the set F is H(U)-eposed. 9. The heat equation Consider now the harmonic space associated with the heat equation in R m+1. Harmonic functions are caloric functions here. Recall that, given an open set V R m+1, a function h : V R is said to be caloric if h C 2 (V ) and m 2 h h = 0 onv. m+1 2 j=1 j Where appropriate, points of R m+1 will be written in the form z = (, t), where R m and t R.Ifz 1,z 2 R m+1, then [z 1,z 2 ] stands for the segment with end-points z 1 and z 2. Let M R m+1 and z, w R m+1. The point w is said to be subordinated to z in M if there eist points 0,..., n R m and real numbers t 0 >t 1 > >t n such that, for the points z j = ( j,t j ), j = 0,...,n, the following holds: z 0 = z, z n = w and [z j 1,z j ] M, j = 1,...,n. Given z R m+1,lets M (z) denote the set of all points which are subordinated to z in M. In what follows, U R m+1 will be considered as a harmonic space associated with the heat equation. Proposition 9.1. Let U R m+1 be a relatively compact open set and F U be a closed set. The following statements are equivalent: (i) F is an absorbent set in U, (ii) S U (z) F, whenever z F. Proof. Assume that (i) holds, z F and there eists a point w S U (z) \ F.Letz j have the same meaning as above. There eists a greatest k {1,...,n} such that [z k 1,z k ] F.Let s 0 = sup { s [0, 1]: (1 s)z k 1 + sz k F }. Obviously, s 0 < 1. Let v = (1 s 0 )z k 1 + s 0 z k. Then v F because F is closed and ( [v,zk ]\{v} ) F =.

13 658 J. Lukeš et al. / Bull. Sci. math. 130 (2006) Since F is absorbent, there eists a hyperharmonic function u on U such that u = 0 onf and u>0 onu \ F (see [6, Proposition 6.1.1]). Let V = m+1 l=1 (a l,b l ) be an open interval in R m+1 such that V U, v V and the lower face m L = (a l,b l ) {a m+1 } l=1 of V has a non-empty intersection with [v,z k ]. It is known that the harmonic measure μ V v has a strictly positive density on L with respect to m-dimensional Lebesgue measure on L; cf. [10]. Since u>0 on a neighborhood of the singleton L [v,z k ],wehave u(v) u dμ V v > 0, V which contradicts to u(v) = 0. Thus (ii) holds. To prove the converse, suppose that z = (ξ 1,...,ξ m+1 ) F.Forη>0 define m+1 V(η)= (ξ l η,ξ l + η) and l=1 W = { (, t) R m+1 : t<ξ m+1 }. Choose η 0 > 0 such that V(η 0 ) U and fi δ (0,η 0 ).Ifw V(δ) W, then w S U (z), so w F by hypothesis. It follows that V(δ) W F because F is closed. It is known that spt μ V(δ) z V(δ) W (cf. [6, 3.3]). Hence μ V(δ) z (U \ F)= 0 whenever 0 <δ<η 0. We conclude that F is absorbent by [6, Proposition 6.1.1]. Corollary 9.2. Let U R m+1 be a relatively compact open set, F U be a β-absorbent set and z F U. Then S U (z) F U. In particular, the conclusion holds if F is an H(U)-eposed set. Proof. The assertion immediately follows from Proposition 6.1 and the preceding Proposition 9.1. Problem 9.3. It remains to characterize spt ε β( U) make the following conjecture. For U,set C := {B U: B,B is β-absorbent}. (cf. Lemma 8.2) in the heat equation case. We

14 J. Lukeš et al. / Bull. Sci. math. 130 (2006) Our conjecture is that spt ε β( U) = C Ch H(U) U. Since spt ε β( U) B whenever B is β-absorbent and B (cf. Lemma 6.2), and ε β( U) supported by Ch H(U) U,wehave References spt ε β( U) C Ch H(U) U. [1] H. Bauer, Harmonische Räume und ihre Potentialtheorie, Lecture Notes in Math., vol. 22, Springer-Verlag, Berlin, [2] J. Bliedtner, W. Hansen, Simplicial cones in potential theory, Invent. Math. 29 (1975) [3] J. Bliedtner, W. Hansen, Cones of hyperharmonic functions, Math. Z. 151 (1976) [4] J. Bliedtner, W. Hansen, Simplicial characterization of elliptic harmonic spaces, Math. Ann. 222 (1976) [5] J. Bliedtner, W. Hansen, Potential Theory an Analytic and Probabilistic Approach to Balayage, Springer-Verlag, Berlin, [6] C. Constantinescu, A. Cornea, Potential Theory on Harmonic Spaces, Springer-Verlag, Berlin, [7] J. Lukeš, J. Malý, I. Netuka, M. Smrčka, J. Spurný, On approimation of affine Baire-one functions, Israel J. Math. 134 (2003) [8] J. Lukeš, T. Mocek, M. Smrčka, J. Spurný, Choquet like sets in function spaces, Bull. Sci. Math. 127 (2003) [9] J. Lukeš, I. Netuka, J. Veselý, Choquet s theory and the Dirichlet problem, Epo. Math. 20 (2002) [10] I. Netuka, Thinness and the heat equation, Čas. Pěst. Mat. 99 (1974) [11] I. Netuka, The Dirichlet problem for harmonic functions, Amer. Math. Monthly 87 (1980) is