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ON THE MEAN VALUE PROPERTY OF SUPERHARMONIC AND SUBHARMONIC FUNCTIONS ROBERT DALMASSO Received 19 October 2005; Revised 25 January 2006; Accepted 16 February 2006 We prove a converse of the mean value property for superharmonic and subharmonic functions. The case of harmonic functions was treated by Epstein and Schiﬀer. Copyright © 2006 Hindawi Publishing Corporation. All rights reserved. Recall that a function u is harmonic (superharmonic, subharmonic) in an open set U ⊂ Rn (n ≥ 1) if u ∈ C 2 (U) and Δu = 0 (Δu ≤ 0,Δu ≥ 0) on U. Denote by H(U) the space of harmonic functions in U and SH(U) (sH(U)) the subset of C 2 (U) consisting of superharmonic (subharmonic) functions in U. If A ⊂ Rn is Lebesgue measurable, L1 (A) denotes the space of Lebesgue integrable functions on A and |A| denotes the Lebesgue measure of A when A is bounded. We also recall the mean value property of harmonic, superharmonic, and subharmonic functions in U ([2]): if x ∈ U and B(x,r) = { y ∈ Rn ; y − x < r }, r > 0, is such that B(x,r) ⊂ U, then for all u ∈ H(U) (SH(U),sH(U)), 1 u(x) = (≥, ≤) B(x,r) B(x,r) u(y)d y. (1) Using the Lebesgue-dominated convergence theorem we see that the conclusion above holds whenever B(x,r) ⊂ U if u ∈ H(U) ∩ L1 (B(x,r)) (SH(U) ∩ L1 (B(x,r)),sH(U) ∩ L1 (B(x,r))). Epstein and Schiﬀer [1] proved the following converse. Theorem 1. Let Ω ⊂ Rn (n ≥ 1) be a bounded open set. Suppose that there exists x0 ∈ Ω such that u x0 = 1 |Ω| Ω u(x)dx (2) for every u ∈ H(Ω) ∩ L1 (Ω). Then Ω is a ball with center x0 . A more general result was obtained by Kuran [3]. In this note we give a proof of the following converse. Hindawi Publishing Corporation International Journal of Mathematics and Mathematical Sciences Volume 2006, Article ID 57340, Pages 1–3 DOI 10.1155/IJMMS/2006/57340 2 Mean value property of super (sub) harmonic functions Theorem 2. Let Ω ⊂ Rn (n ≥ 1) be a bounded open set. Suppose that there exists x0 ∈ Ω such that u x0 ≥ (≤) 1 |Ω| Ω u(x)dx (3) for every u ∈ SH(Ω) ∩ L1 (Ω)\H(Ω) (sH(Ω) ∩ L1 (Ω)\H(Ω)) . Then Ω is a ball with center x0 . Proof. Clearly it is enough to consider the case of superharmonic functions. Since Ω is bounded, there exists a largest open ball B centered at x0 of radius r1 which lies in Ω. The compactness of ∂Ω implies that there is some x1 ∈ ∂Ω such that x1 − x0 = r1 . We will show that Ω = B. Define 2 −n h(x) = r1n−2 x − x0 − r12 x − x1 (4) for x ∈ Rn \{x1 }. Then h ∈ H(Rn \{x1 }) and h(x0 ) = −1. Now let R > r1 be such that Ω ⊂ B(x0 ,R). For k ∈ N∗ we set uk (x) = 1 + h(x) + 2 1 2 R − x − x 0 , 2nk x ∈ Ω. (5) Obviously uk ∈ C 2 (Ω) and Δuk = −1/k in Ω, hence uk ∈ SH(Ω)\H(Ω). Moreover uk ∈ L1 (Ω) and uk (x) ≥ 1 for x ∈ Ω\B. Since 1 + h ∈ H(Ω) ∩ L1 (Ω), we have 0 = 1 + h x0 = B 1 + h(x) dx. (6) Now using (6) we can write 1 1 1 R2 = uk x0 ≥ uk (x)dx = uk (x)dx + uk (x)dx 2nk |Ω| Ω |Ω| Ω\B |Ω| B 2 1 1 = uk (x)dx + R2 − x − x0 dx |Ω| Ω\B 2nk|Ω| B |Ω\B | ωn r1n R2 r12 ≥ + − |Ω| 2nk|Ω| n n + 2 |Ω\B | ωn r1n R2 r12 + − ≥ |Ω| 2nk|Ω| n n + 2 (7) for all k ∈ N∗ , where ωn denotes the measure of the unit sphere in Rn . This implies that |Ω\B | = 0. Then the open set Ω\B must be empty, hence Ω ⊂ B. Since Ω is open and B ⊂ Ω ⊂ B, we deduce that Ω = B. References [1] B. Epstein and M. Schiﬀer, On the mean-value property of harmonic functions, Journal d’Analyse Mathématique 14 (1965), 109–111. [2] D. Gilbarg and N. S. Trudinger, Elliptic Partial Diﬀerential Equations of Second Order, Grundlehren der Mathematischen Wissenschaften, vol. 224, Springer, Berlin, 1977. Robert Dalmasso 3 [3] Ü. Kuran, On the mean-value property of harmonic functions, The Bulletin of the London Mathematical Society 4 (1972), 311–312. Robert Dalmasso: Laboratoire LMC-IMAG, Equipe EDP, Tour IRMA, BP 53, 38041 Grenoble Cedex 9, France E-mail address: [email protected]