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Convex bodies via gravitational potentials
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ScienceDirect Expo. Math. 35 (2017) 478–482 www.elsevier.com/locate/exmath
Convex bodies via gravitational potentials S. Hou a , b , J. Xiao b , ∗ a College of Mathematics and Information Science, Hebei Normal University, Shijiazhuang, Hebei 050016, China b Department of Mathematics and Statistics, Memorial University, St. John’s, NL A1C 5S7, Canada
Received 11 May 2017
Abstract This note geometrically generalizes the main theorem in Shahgholian’s 1992 paper [2], from R3 to Rn≥2 via the Minkowski functional of a convex body with the origin in its interior. c 2017 Elsevier GmbH. All rights reserved. ⃝
MSC 2010: primary 31B15; secondary 52A38 Keywords: Convex body; Gravitational potential
Working on a convex-geometric analogue of [2, Theorem] (characterizing a uniform hollow sphere by means of the gravitational potential) from the physical space R3 to the Euclidean 2 ≤ n-dimensional space Rn , we prove Theorem. Suppose that: (i) K ⊂ Rn is a convex body (nonempty, compact and convex set) with the origin o being in its non-empty interior and { } ∥x∥ K = inf λ > 0 : x ∈ λK being its Minkowski functional where λK = {λy : y ∈ K }; ∗ Corresponding author.
E-mail addresses: [email protected] (S. Hou), [email protected] (J. Xiao). http://dx.doi.org/10.1016/j.exmath.2017.08.003 c 2017 Elsevier GmbH. All rights reserved. 0723-0869/⃝
S. Hou, J. Xiao / Expo. Math. 35 (2017) 478–482
479
(ii) Ω ⊂ Rn is a star-shaped body at the origin o (nonempty compact set satisfying [o, x] ⊂ Ω , where x ∈ Ω and [o, x] denotes the segment connecting o and x) with non-empty interior, and f is a continuous function on Rn enjoying the following two properties: (ii-a) maxx∈Ω f (x) and minx∈Ω f (x) are only attainable on ∂Ω , the boundary of Ω . Moreover, let ⎧ ∂ f (x) f (x) − f (x − t x) ⎪ ⎪ = lim ; ⎨ − ∂ K x⃗ ∥t x∥ K t→0+ ∂ f (x) f (x + t x) − f (x) ⎪ ⎪ ⎩ = lim . ∂ K x⃗+ ∥t x∥ K t→0+ ∃x1 , x2 ∈ ∂Ω , such that ⎧ max f (x) = f (x1 ); ⎪ ⎪ x∈Ω ⎪ ⎪ ⎨ min f (x) = f (x2 ); x∈Ω ⎪ ⎪ ∂ f (x1 ) ∂ f (x1 ) ∂ f (x2 ) ∂ f (x2 ) ⎪ ⎪ ⎩ − . − − + ≤ ∂ K x⃗1 ∂ K x⃗1 ∂ K x⃗2 − ∂ K x⃗2 + (ii-b) κn > 0 is a dimensional constant with { κ ∥x∥2−n as x ∈ Ω c = Rn≥3 \ Ω ; K f (x) = n κ2 ln ∥x∥−1 as x ∈ Ω c = R2 \ Ω . K Then Ω is a dilation of K , namely, Ω = λK for a constant λ > 0. Proof. Let ⎧ ⎪λ1 = sup{λ : λK ⊆ Ω }; ⎪ ⎨ x1 ∈ λ1 K ∩ ∂Ω ; λ ⎪ 2 = inf{λ : Ω ⊆ λK }; ⎪ ⎩ x2 ∈ λ2 K ∩ ∂Ω . Obviously, ∥x2 ∥ K = λ2 ≥ λ1 = ∥x1 ∥ K . If we can verify ∥x2 ∥ K ≤ ∥x1 ∥ K , then we will have λ2 = λ1 = λ
& Ω = λK ,
as desired. So, it remains to validate the last inequality. This validation consists of the three steps as seen below. Step 1. We claim { max f (x) = f (x1 ); x∈Ω
min f (x) = f (x2 ), x∈Ω
480
S. Hou, J. Xiao / Expo. Math. 35 (2017) 478–482
and then we can choose x1 , x2 as the points satisfying (ii-a). Evidently, it is enough to check the maximum case and the minimum case follows in a similar way. Since Ω is a star-shaped body at the origin and f is continuous and (ii-b), we have { as x ∈ Ω c ∪ ∂Ω ; κ ∥x∥2−n K f (x) = n −1 as x ∈ Ω c ∪ ∂Ω . κ2 ln ∥x∥ K If the first equation was not valid, then maxx∈Ω f (x) would be attained by another point x1′ ∈ ∂Ω , and hence: ⎧ ⎨κn ∥x1 ∥2−n = f (x1 ) < max f (x) = f (x1′ ) = κn ∥x1′ ∥2−n as n ≥ 3; K K x∈Ω
′ ′ −1 ⎩κ2 ln ∥x1 ∥−1 K = f (x 1 ) < max f (x) = f (x 1 ) = κ2 ln ∥x 1 ∥ K as n = 2. x∈Ω
This would imply ∥x1 ∥ K > ∥x1′ ∥ K , whence violating λ1 = sup{λ : λK ⊆ Ω } & x1 ∈ λ1 K ∩ ∂Ω ⇒ ∥x1 ∥ K ≤ ∥x1′ ∥ K . Consequently, we get { f (x1 ) = max f (x) ≥ f (x1 − t x1 ) ∀ t ∈ [0, 1]
&
x1 − t x1 ∈ λ1 K ⊆ Ω ;
f (x2 ) = min f (x) ≤ f (x2 − t x2 ) ∀ t ∈ [0, 1]
&
x2 − t x2 ∈ Ω ,
x∈Ω
x∈Ω
where x2 − t x2 ∈ Ω since Ω is a star-shaped body at the origin, thereby arriving at ⎧ ∂ f (x1 ) f (x1 ) − f (x1 − t x1 ) ⎪ ⎪ ≥ 0; ⎨ − = lim+ ∥t x1 ∥ K t→0 ∂ K x⃗1 f (x2 ) − f (x2 − t x2 ) ∂ f (x2 ) ⎪ ⎪ ⎩ = lim ≤ 0. ∥t x2 ∥ K t→0+ ∂ K x⃗2 − Step 2. Using the fact that ∥ · ∥ K is homogeneous 1, we have f (x + t x) − f (x) ∂ f (x) = lim ∂ K x⃗+ ∥t x∥ K t→0+ f (x + t x) − f (x) = lim t∥x∥ K t→0+ f (x + t x) − f (x) = lim t→0+ (1 + t)∥x∥ K − ∥x∥ K f (x + t x) − f (x) = lim + t→0 ∥(1 + t)x∥ K − ∥x∥ K ∂ f (x)+ = , ∂ ∥x∥+ K where ∂ f (x)+ and ∂ ∥x∥+ K denote respectively the directional derivatives of these functions in the direction x⃗ at the point x. Similarly, we also have ∂ f (x) ∂ f (x)− = , − ∂ K x⃗ ∂ ∥x∥− K
S. Hou, J. Xiao / Expo. Math. 35 (2017) 478–482
481
where ∂ f (x)− and ∂ ∥x∥− K denote respectively the directional derivatives of these functions in the direction −⃗ x at the point x. Step 3. With the help of the previous computations, we obtain ⎧⎧ ∂ f (x1 ) ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ − ⎪ ∂ ⎪ K x⃗1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∂ f (x2 ) ⎪ ⎪ ⎨⎩ ∂ K x⃗2 − ⎧ ∂ f (x1 ) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ∂ ⎪ K x⃗1 − ⎪ ⎪ ⎪⎪ ∂ f (x2 ) ⎪ ⎪ ⎩⎪ ⎩ ∂ K x⃗2 −
−
∂ f (x1 ) ∂ K x⃗1 + ∂ f (x2 )
=
∂ f (x1 ) ∂ K x⃗1 − ∂ f (x2 )
1−n + κn (n − 2)∥x1 ∥1−n K ≥ κn (n − 2)∥x 1 ∥ K
as n ≥ 3;
1−n = + κn (n − 2)∥x2 ∥1−n K ≤ κn (n − 2)∥x 2 ∥ K ∂ K x⃗2 + ∂ K x⃗2 − ∂ f (x1 ) ∂ f (x1 ) −1 − = + κ2 ∥x1 ∥−1 K ≥ κ2 ∥x 1 ∥ K ∂ K x⃗1 + ∂ K x⃗1 − as n = 2. ∂ f (x2 ) ∂ f (x2 ) −1 −1 − = + κ ∥x ∥ ≤ κ ∥x ∥ 2 2 2 2 K K ∂ K x⃗2 + ∂ K x⃗2 −
−
As a consequence, we achieve { κn (n − 2)∥x1 ∥1−n ≤ κn (n − 2)∥x2 ∥1−n as n ≥ 3; K K −1 κ2 ∥x1 ∥−1 K ≤ κ2 ∥x 2 ∥ K as n = 2, via (ii-a), thereby getting the required inequality ∥x2 ∥ K ≤ ∥x1 ∥ K .
□
Remark. Here, it is perhaps appropriate to give some comments on the hypotheses (i)–(ii) in Theorem. (i) If K is the closed unit ball B2n of Rn , then ∥x∥ K = |x|, the Euclidean norm of x ∈ Rn , and hence the above theorem says essentially that a uniform sphere behaves like a point mass as far as gravity is concerned. (ii) If f is a non-constant harmonic function in Ω ◦ , the interior of Ω , then the maximum principle for harmonic functions derives that maxx∈Ω f (x) and minx∈Ω f (x) are only attainable on ∂Ω . For example, if K is an ellipsoid or a two-dimensional ellipse and f is the gravitational potential (based on K ) ⎧∫ ⎪ ⎪ ∥x − y∥2−n as n ≥ 3; ⎨ K dσ (y) ∂Ω f K (x) = ∫ ⎪ ⎪ ⎩ ln ∥x − y∥−1 K dσ (y) as n = 2, ∂Ω
where σ denotes the surface area measure, then f K is harmonic in Ω ◦ , i.e. it satisfies the Laplace equation ∆ f K = 0. This can be seen from the following computation: f K (x) = f T (B2n ) (x) ∫ ⎧∫ det(T ) dσ (y) ⎪ ⎪ = dσ ( y˜ ) as n ≥ 3; ⎪ ⎨ ∂Ω ∥x − y∥n−2n ˜ |n−2 T −1 (∂Ω) | x˜ − y T (B2 ) = ∫ ∫ ⎪ det(T ) dσ (y) ⎪ ⎪ dσ ( y˜ ) as n = 2, = ⎩ n −1 ˜| T (∂Ω) ln| x˜ − y ∂Ω ln ∥x − y∥T (B2 )
482
S. Hou, J. Xiao / Expo. Math. 35 (2017) 478–482
where T is a matrix with positive determinant. However, for a general K we are led to adopt [1, Theorem 1.7.2] to compute ∆ f = 0, thereby finding a geometric condition on K such that f K is harmonic – in other words – f K is not always harmonic. Moreover, if Ω = K = B2n then one has not only for x ∈ ∂ B2n , { ∂ f B2n (x) ∂ f B2n (x) (n − 2)σn−1 as n ≥ 3; − = 2π as n = 2, ∂ B n x⃗− ∂ B n x⃗+ 2
2
by [3, Theorem 1.11], but also for x ∈ (B2n )c , ⎧∫ ⎪ 2−n 2−n ⎪ ⎪ ⎨ ∂ B n |x − y| dσ (y) = σn−1 |x| f K (x) = ∫ 2 ⎪ ⎪ ⎪ ln |x − y|−1 dσ (y) = 2π ln |x|−1 ⎩
as n ≥ 3; as n = 2,
∂ B2n
by the mean value property of the harmonic functions, where σn−1 is the surface area of B2n . This is the initial reason to consider an extension of [2, Theorem].
Acknowledgments Research of SH is supported by both CSC of China (201306040135) and NSERC of Canada (202979463102000); Research of JX is supported by NSERC of Canada (202979463102000).
References [1] R. Schneider, Convex Bodies: The Brunn–Minkowski Theory, second ed., Cambridge University Press, Cambridge, 2014. [2] H. Shahgholian, A characterization of the sphere in terms of single-layer potentials, Proc. Amer. Math. Soc. 115 (1992) 1167–1168. [3] G. Verchota, Layer potentials and regularity for the Dirichlet problem for Laplace’s equation in Lipschitzdomains, J. Funct. Anal. 59 (1984) 572–611.
ScienceDirect Expo. Math. 35 (2017) 478–482 www.elsevier.com/locate/exmath
Convex bodies via gravitational potentials S. Hou a , b , J. Xiao b , ∗ a College of Mathematics and Information Science, Hebei Normal University, Shijiazhuang, Hebei 050016, China b Department of Mathematics and Statistics, Memorial University, St. John’s, NL A1C 5S7, Canada
Received 11 May 2017
Abstract This note geometrically generalizes the main theorem in Shahgholian’s 1992 paper [2], from R3 to Rn≥2 via the Minkowski functional of a convex body with the origin in its interior. c 2017 Elsevier GmbH. All rights reserved. ⃝
MSC 2010: primary 31B15; secondary 52A38 Keywords: Convex body; Gravitational potential
Working on a convex-geometric analogue of [2, Theorem] (characterizing a uniform hollow sphere by means of the gravitational potential) from the physical space R3 to the Euclidean 2 ≤ n-dimensional space Rn , we prove Theorem. Suppose that: (i) K ⊂ Rn is a convex body (nonempty, compact and convex set) with the origin o being in its non-empty interior and { } ∥x∥ K = inf λ > 0 : x ∈ λK being its Minkowski functional where λK = {λy : y ∈ K }; ∗ Corresponding author.
E-mail addresses: [email protected] (S. Hou), [email protected] (J. Xiao). http://dx.doi.org/10.1016/j.exmath.2017.08.003 c 2017 Elsevier GmbH. All rights reserved. 0723-0869/⃝
S. Hou, J. Xiao / Expo. Math. 35 (2017) 478–482
479
(ii) Ω ⊂ Rn is a star-shaped body at the origin o (nonempty compact set satisfying [o, x] ⊂ Ω , where x ∈ Ω and [o, x] denotes the segment connecting o and x) with non-empty interior, and f is a continuous function on Rn enjoying the following two properties: (ii-a) maxx∈Ω f (x) and minx∈Ω f (x) are only attainable on ∂Ω , the boundary of Ω . Moreover, let ⎧ ∂ f (x) f (x) − f (x − t x) ⎪ ⎪ = lim ; ⎨ − ∂ K x⃗ ∥t x∥ K t→0+ ∂ f (x) f (x + t x) − f (x) ⎪ ⎪ ⎩ = lim . ∂ K x⃗+ ∥t x∥ K t→0+ ∃x1 , x2 ∈ ∂Ω , such that ⎧ max f (x) = f (x1 ); ⎪ ⎪ x∈Ω ⎪ ⎪ ⎨ min f (x) = f (x2 ); x∈Ω ⎪ ⎪ ∂ f (x1 ) ∂ f (x1 ) ∂ f (x2 ) ∂ f (x2 ) ⎪ ⎪ ⎩ − . − − + ≤ ∂ K x⃗1 ∂ K x⃗1 ∂ K x⃗2 − ∂ K x⃗2 + (ii-b) κn > 0 is a dimensional constant with { κ ∥x∥2−n as x ∈ Ω c = Rn≥3 \ Ω ; K f (x) = n κ2 ln ∥x∥−1 as x ∈ Ω c = R2 \ Ω . K Then Ω is a dilation of K , namely, Ω = λK for a constant λ > 0. Proof. Let ⎧ ⎪λ1 = sup{λ : λK ⊆ Ω }; ⎪ ⎨ x1 ∈ λ1 K ∩ ∂Ω ; λ ⎪ 2 = inf{λ : Ω ⊆ λK }; ⎪ ⎩ x2 ∈ λ2 K ∩ ∂Ω . Obviously, ∥x2 ∥ K = λ2 ≥ λ1 = ∥x1 ∥ K . If we can verify ∥x2 ∥ K ≤ ∥x1 ∥ K , then we will have λ2 = λ1 = λ
& Ω = λK ,
as desired. So, it remains to validate the last inequality. This validation consists of the three steps as seen below. Step 1. We claim { max f (x) = f (x1 ); x∈Ω
min f (x) = f (x2 ), x∈Ω
480
S. Hou, J. Xiao / Expo. Math. 35 (2017) 478–482
and then we can choose x1 , x2 as the points satisfying (ii-a). Evidently, it is enough to check the maximum case and the minimum case follows in a similar way. Since Ω is a star-shaped body at the origin and f is continuous and (ii-b), we have { as x ∈ Ω c ∪ ∂Ω ; κ ∥x∥2−n K f (x) = n −1 as x ∈ Ω c ∪ ∂Ω . κ2 ln ∥x∥ K If the first equation was not valid, then maxx∈Ω f (x) would be attained by another point x1′ ∈ ∂Ω , and hence: ⎧ ⎨κn ∥x1 ∥2−n = f (x1 ) < max f (x) = f (x1′ ) = κn ∥x1′ ∥2−n as n ≥ 3; K K x∈Ω
′ ′ −1 ⎩κ2 ln ∥x1 ∥−1 K = f (x 1 ) < max f (x) = f (x 1 ) = κ2 ln ∥x 1 ∥ K as n = 2. x∈Ω
This would imply ∥x1 ∥ K > ∥x1′ ∥ K , whence violating λ1 = sup{λ : λK ⊆ Ω } & x1 ∈ λ1 K ∩ ∂Ω ⇒ ∥x1 ∥ K ≤ ∥x1′ ∥ K . Consequently, we get { f (x1 ) = max f (x) ≥ f (x1 − t x1 ) ∀ t ∈ [0, 1]
&
x1 − t x1 ∈ λ1 K ⊆ Ω ;
f (x2 ) = min f (x) ≤ f (x2 − t x2 ) ∀ t ∈ [0, 1]
&
x2 − t x2 ∈ Ω ,
x∈Ω
x∈Ω
where x2 − t x2 ∈ Ω since Ω is a star-shaped body at the origin, thereby arriving at ⎧ ∂ f (x1 ) f (x1 ) − f (x1 − t x1 ) ⎪ ⎪ ≥ 0; ⎨ − = lim+ ∥t x1 ∥ K t→0 ∂ K x⃗1 f (x2 ) − f (x2 − t x2 ) ∂ f (x2 ) ⎪ ⎪ ⎩ = lim ≤ 0. ∥t x2 ∥ K t→0+ ∂ K x⃗2 − Step 2. Using the fact that ∥ · ∥ K is homogeneous 1, we have f (x + t x) − f (x) ∂ f (x) = lim ∂ K x⃗+ ∥t x∥ K t→0+ f (x + t x) − f (x) = lim t∥x∥ K t→0+ f (x + t x) − f (x) = lim t→0+ (1 + t)∥x∥ K − ∥x∥ K f (x + t x) − f (x) = lim + t→0 ∥(1 + t)x∥ K − ∥x∥ K ∂ f (x)+ = , ∂ ∥x∥+ K where ∂ f (x)+ and ∂ ∥x∥+ K denote respectively the directional derivatives of these functions in the direction x⃗ at the point x. Similarly, we also have ∂ f (x) ∂ f (x)− = , − ∂ K x⃗ ∂ ∥x∥− K
S. Hou, J. Xiao / Expo. Math. 35 (2017) 478–482
481
where ∂ f (x)− and ∂ ∥x∥− K denote respectively the directional derivatives of these functions in the direction −⃗ x at the point x. Step 3. With the help of the previous computations, we obtain ⎧⎧ ∂ f (x1 ) ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ − ⎪ ∂ ⎪ K x⃗1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∂ f (x2 ) ⎪ ⎪ ⎨⎩ ∂ K x⃗2 − ⎧ ∂ f (x1 ) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ∂ ⎪ K x⃗1 − ⎪ ⎪ ⎪⎪ ∂ f (x2 ) ⎪ ⎪ ⎩⎪ ⎩ ∂ K x⃗2 −
−
∂ f (x1 ) ∂ K x⃗1 + ∂ f (x2 )
=
∂ f (x1 ) ∂ K x⃗1 − ∂ f (x2 )
1−n + κn (n − 2)∥x1 ∥1−n K ≥ κn (n − 2)∥x 1 ∥ K
as n ≥ 3;
1−n = + κn (n − 2)∥x2 ∥1−n K ≤ κn (n − 2)∥x 2 ∥ K ∂ K x⃗2 + ∂ K x⃗2 − ∂ f (x1 ) ∂ f (x1 ) −1 − = + κ2 ∥x1 ∥−1 K ≥ κ2 ∥x 1 ∥ K ∂ K x⃗1 + ∂ K x⃗1 − as n = 2. ∂ f (x2 ) ∂ f (x2 ) −1 −1 − = + κ ∥x ∥ ≤ κ ∥x ∥ 2 2 2 2 K K ∂ K x⃗2 + ∂ K x⃗2 −
−
As a consequence, we achieve { κn (n − 2)∥x1 ∥1−n ≤ κn (n − 2)∥x2 ∥1−n as n ≥ 3; K K −1 κ2 ∥x1 ∥−1 K ≤ κ2 ∥x 2 ∥ K as n = 2, via (ii-a), thereby getting the required inequality ∥x2 ∥ K ≤ ∥x1 ∥ K .
□
Remark. Here, it is perhaps appropriate to give some comments on the hypotheses (i)–(ii) in Theorem. (i) If K is the closed unit ball B2n of Rn , then ∥x∥ K = |x|, the Euclidean norm of x ∈ Rn , and hence the above theorem says essentially that a uniform sphere behaves like a point mass as far as gravity is concerned. (ii) If f is a non-constant harmonic function in Ω ◦ , the interior of Ω , then the maximum principle for harmonic functions derives that maxx∈Ω f (x) and minx∈Ω f (x) are only attainable on ∂Ω . For example, if K is an ellipsoid or a two-dimensional ellipse and f is the gravitational potential (based on K ) ⎧∫ ⎪ ⎪ ∥x − y∥2−n as n ≥ 3; ⎨ K dσ (y) ∂Ω f K (x) = ∫ ⎪ ⎪ ⎩ ln ∥x − y∥−1 K dσ (y) as n = 2, ∂Ω
where σ denotes the surface area measure, then f K is harmonic in Ω ◦ , i.e. it satisfies the Laplace equation ∆ f K = 0. This can be seen from the following computation: f K (x) = f T (B2n ) (x) ∫ ⎧∫ det(T ) dσ (y) ⎪ ⎪ = dσ ( y˜ ) as n ≥ 3; ⎪ ⎨ ∂Ω ∥x − y∥n−2n ˜ |n−2 T −1 (∂Ω) | x˜ − y T (B2 ) = ∫ ∫ ⎪ det(T ) dσ (y) ⎪ ⎪ dσ ( y˜ ) as n = 2, = ⎩ n −1 ˜| T (∂Ω) ln| x˜ − y ∂Ω ln ∥x − y∥T (B2 )
482
S. Hou, J. Xiao / Expo. Math. 35 (2017) 478–482
where T is a matrix with positive determinant. However, for a general K we are led to adopt [1, Theorem 1.7.2] to compute ∆ f = 0, thereby finding a geometric condition on K such that f K is harmonic – in other words – f K is not always harmonic. Moreover, if Ω = K = B2n then one has not only for x ∈ ∂ B2n , { ∂ f B2n (x) ∂ f B2n (x) (n − 2)σn−1 as n ≥ 3; − = 2π as n = 2, ∂ B n x⃗− ∂ B n x⃗+ 2
2
by [3, Theorem 1.11], but also for x ∈ (B2n )c , ⎧∫ ⎪ 2−n 2−n ⎪ ⎪ ⎨ ∂ B n |x − y| dσ (y) = σn−1 |x| f K (x) = ∫ 2 ⎪ ⎪ ⎪ ln |x − y|−1 dσ (y) = 2π ln |x|−1 ⎩
as n ≥ 3; as n = 2,
∂ B2n
by the mean value property of the harmonic functions, where σn−1 is the surface area of B2n . This is the initial reason to consider an extension of [2, Theorem].
Acknowledgments Research of SH is supported by both CSC of China (201306040135) and NSERC of Canada (202979463102000); Research of JX is supported by NSERC of Canada (202979463102000).
References [1] R. Schneider, Convex Bodies: The Brunn–Minkowski Theory, second ed., Cambridge University Press, Cambridge, 2014. [2] H. Shahgholian, A characterization of the sphere in terms of single-layer potentials, Proc. Amer. Math. Soc. 115 (1992) 1167–1168. [3] G. Verchota, Layer potentials and regularity for the Dirichlet problem for Laplace’s equation in Lipschitzdomains, J. Funct. Anal. 59 (1984) 572–611.