THE LAPLACIAN AND THE DIRICHLET PROBLEM

Transcription

1 THE LAPLACIAN AND THE DIRICHLET PROBLEM Stefano Meda Università di Milano-Bicocca c Stefano Meda 2016

2 ii A Francesco

3 Contents 0.1 Background and preliminary results Convolution The divergence theorem Linear operators The spectrum of a linear operator Complex measures The Schwartz space and the Fourier transform The classical Dirichlet problem Introduction Harmonic functions and the mean value property Maximum principle and uniqueness Green s function The Poisson kernel for the half space The Poisson kernel for the ball The higher dimensional case The two dimensional case L p boundary conditions The Poisson integral Harmonic Hardy spaces Pointwise convergence to the boundary Weak L 1 and interpolation The Hardy Littlewood maximal function Nontangential convergence Enlarging and weakening Functionals associated to measures Natural functionals not associated with measures Distributions iii

4 iv CONTENTS 3.2 Derivatives of distributions Sobolev spaces Distributions and regularity Other operations on distributions Convolution and multiplication Support of distributions Fundamental solutions Parametrices and regularity Hilbert space methods The trace operator More on the trace operator on the unit disc Poincare s inequality The dual of H0 1 (Ω) New formulation of the Dirichlet problem More general operators The spectral theorem Compact operators The adjoint of a bounded operator The spectra of compact operators The spectral theorem Bases of eigenfunctions Sine series on the unit interval Fourier series and the Laplacian on S The Laplacian on R n

5 0.1. BACKGROUND AND PRELIMINARY RESULTS Background and preliminary results Convolution Definition Suppose that f and g are Lebesgue measurable functions in R n. The convolution f g between f and g is defined by f g(x) = f(x y) g(y) dy (0.1.1) R n for every x for which the integral above exists. Notice that f g(x) = τ y f(x) g(y) dy, R n where τ y f(x) := f(x y). The last formula exhibits f g as a superposition of translates of f. Exercise Denote by 1 the characteristic function of the interval [0, 1]. Compute 1 1 e Exercise If p is in [1, ) and f belongs to L p (R n ), then lim τ yf f p = 0. y 0 (Hint: first suppose that f is continuous with compact support, and then use an approximation argument in the L p norm). This fails if p = without any additional assumption (just take 1 [0,1] ). If we assume that f is bounded and uniformly continuous, then the conclusion holds also with p =. Remark Observe that if f is Borel measurable, then the function (x, y) f(x y) is Borel measurable on R n R n. Indeed, the latter function may be viewed as the composition of f and of the continuous function (x, y) x y. It is a well known fact in Measure Theory (that we do not prove here) that if f and g are Lebesgue measurable, then there exists two Borel measurable functions f 0 and g 0 such that f 0 = f and g 0 = g almost everywhere [Fo1, Proposition 2.12, p. 48]. Clearly the value of the convolution does not change if we substitute f 0 to f and g 0 to g in the integral (0.1.1). Therefore, whenever we consider the convolution f g of two Lebesgue measurable function f and g, we may assume that f and g are, in fact, Borel measurable. This guarantees that the function (x, y) f(x y) g(y) is measurable, for it may be viewed as the composition of a continuous function and a Borel measurable function. We shall repeatedly use this observation without any further comment. Proposition Suppose that the integrals involved in (i)-(iv) below exist. Then the following hold: (i) f g = g f; (ii) (f g) h = f (g h);

6 2 CONTENTS (iii) τ z (f g) = (τ z f) g = f (τ z g) for every z in R n ; (iv) denote by C the closure of {x+y : x supp f, y supp g}. Then supp (f g) C. Exercise Prove Proposition Some basic properties of the convolution are grouped together in the following proposition. We need the so called generalised Minkowski inequality, which is stated below. Exercise Suppose that (X, µ) and (Y, ν) are σ-finite measure spaces, that 1 p, and that f is a measurable function on X Y. Prove that [ ( ) p ] 1/p [ ] dµ(x) f(x, y) dν(y) dν(y) f(x, y) p 1/p. dµ(x) X Y Y X Hint: the cases where p = 1 or p = are trivial. For the case where 1 < p < recall the Riesz representation theorem. Exercise Denote by C 0 (R n ) the space of al continuous functions on R n that vanish at infinity, endowed with the uniform norm. Prove that C 0 (R n ) is a Banach space. Proposition The following hold: (i) if f is in L 1 (R n ) and g is in L p (R n ) for some p in [1, ], then f g is in L p (R n ) and f g Lp(Rn) f L1(Rn) g L ; p (R n ) (ii) if p is in [1, ], f is in L p (R n ) and g is in L p (R n ), where p denotes the index conjugate to p, then f g is bounded and uniformly continuous, and f g f L p (R n ) g L. p (R n ) Furthermore, if p is in (1, ), then f g is in C 0 (R n ); (iii) if f is in L 1 (R n ) and g is in C k (R n ) for some k, with bounded derivatives, then f g is in C k (R n ) and for all α such that α k. α (f g) = f ( α g) Proof. Property (i) follows directly from the generalised Minkowski inequality and the translation invariance of the Lebesgue measure. The norm inequality in (ii) follows from Hölder s inequality and the translation invariance of the Lebesgue measure. Now we show that f g is uniformly continuous. Suppose first that p is in [1, ). For each x in R n we may write τ x (f g) f g = (τ x f f) g,

7 0.1. BACKGROUND AND PRELIMINARY RESULTS 3 which implies that sup y R n τ x (f g)(y) f g(y) τ x f f p g p, which tends to 0 as x tends to 0 (see Exercise 0.1.3). If, instead, p =, then we write τ x (f g) f g = f (τ x g g), and argue as before. Finally we show that if 1 < p <, then f g is in C 0 (R n ). If f and g belong to C c (R n ), then clearly f g belongs to C c (R n ), and, a fortiori, to C 0 (R n ). If f is in L p (R n ) and g is in L p (R n ), then there exist sequences {f n } and {g n } of continuous functions with compact support such that f fn p 0 and g gn p 0 as n tends to. Therefore f g fn g n (f fn ) g n + fn (g g n ) f fn p gn p + fn p g gn p, which tends to 0 as n tends to. Since C 0 (R n ) is complete with respect to the uniform norm f g is in C 0 (R n ), as required. It remains to prove (iii). We may differentiate under the integral sign, because α g is bounded for α k, and obtain that α (f g)(x) = f(y) ( α g)(x y) dy = (f α g)(x), R n as required. Therefore f g possesses all the derivatives up to the order k. The continuity of α (f g) when α = k follows from (ii) (with p = 1 and p = ) and the fact that α g are bounded. As a consequence of (iii) above, or rather of the method of its proof, notice that if φ is a smooth compactly supported function, and f is locally integrable in R n, then φ f is smooth. Definition Given φ in L 1 (R n ) and t in R + we denote by φ t the function defined by φ t (x) = t n φ(x/t) x R n. A comparison between the graphs of φ and φ t is worth doing. A simple change of variables shows that φt L = 1 (R n ) φ L for every t. 1 (R n ) Roughly speaking, the following proposition says that if φ L1 = 1, then (R n ) the operator f f φ t is close to the identity operator when t is small. In this case the family {φ t : t > 0} is sometimes called an approximate identity for the convolution. Proposition Suppose that φ is in L 1 (R n ) and that R n φ dv = 1. The following hold: (i) if p is in [1, ) and f is in L p (R n ), then lim t 0 + f φt f Lp (R n ) = 0; (ii) if f is uniformly continuous and bounded, then lim t 0 + f φt f = 0;

8 4 CONTENTS (iii) if f is in L (R n ) and agrees with a continuous function on the open set A, then f φ t is uniformly convergent to f on compact subsets of A; (iv) suppose that φ and ψ are in Cc (R n ) and let U be a neighbourhood of supp (ψ). Then supp (φ t ψ) is contained in U for all t small enough. Furthermore lim α (φ t ψ) α ψ t 0 + = 0 for every multiindex α. Proof. For δ > 0 we have that φ t (y) dy = φ(y) dy, B(0,δ) c B(0,δ/t) c which tends to 0 as t tends to 0 + because φ is in L 1 (R n ) (use the dominated convergence theorem). Since R n φ(y) dy = 1, we may write f φ t (x) f(x) = (τ y f f)(x) φ t (y) dy R n = (τ y f f)(x) φ t (y) dy + (τ y f f)(x) φ t (y) dy. B(0,δ) B(0,δ) c First we prove (i). The formula above implies that f φt f p sup τy f f p + 2 f p φ t (y) dy. y B(0,δ) B(0,δ) c Fix ε > 0. By Exercise 0.1.3, there exists δ > 0 such that τy f f p < ε/2 for every y such that y < δ. Once we have chosen δ, we may choose t small enough so that φ B(0,δ) c t (y) dy < ε/(4 f p ). Therefore f φ t f p < ε for all t small enough, as required. The proof of (ii) is very similar to that of (i). The assumption that f is uniformly continuous ensures that τy f f is small when y is small. The proof of (iii) is much as that of (ii), the only change being that we must restrict to compact sets K contained in A. Finally, we prove (iv). The statement concerning the support of the convolution follows easily from Proposition The last statement in (iv) follows from (ii) and the fact that α (φ t ψ) = φ t α ψ The divergence theorem We briefly review some definitions and results from advanced calculus. In particular, we introduce the basic notions needed for the statement of the classical divergence theorem. A subset of R n is a domain if it is an open connected set.

9 0.1. BACKGROUND AND PRELIMINARY RESULTS 5 Definition A subset S of R n is a hypersurface of class C k if for every x 0 S there exists an open subset V of R n containing x 0 and a real valued function F C k (V ) such that F does not vanish on S V and S V = { x V : F (x) = 0 }. For the sake of definiteness, suppose that n F does not vanish on S V. By the implicit function theorem, there exists a C k function ϕ such that x n = ϕ ( x 1,..., x n 1 ) for all ( ) x 1,..., x n 1, x n in S V. For convenience, denote by x the point (x 1,..., x n 1 ) in R n 1. Note that the map (x, x n ) ( x, x n ϕ(x ) ) maps V S onto a neighbourhood of the point x 0 of the hyperplane x n = 0. The inverse of the map x ( x, ϕ(x ) ), which is defined in a suitable neighbourhood of x 0, gives a local chart of S around x 0. The surface measure of S has the following expression in terms of the local coordinates (x, ϕ(x )) dσ(x, ϕ(x )) = 1 + ϕ(x ) 2 dx. (0.1.2) It is well known that a patching argument, which uses a partition of unity on S, allows to define a measure σ, called surface measure on S; if f is a (complex valued) function on S, whose support is contained in a coordinate neighbourhood with associated local chart the inverse of the map x ( x, ϕ(x ) ), then f dσ = f ( x, ϕ(x ) ) 1 + ϕ(x ) 2 dx. (0.1.3) S U Notice that the density of the measure is of class C k 1 (U). For every x S, the vector F (x) is orthogonal to S (i.e., it is orthogonal to the hyperplane tangent to S at x 0 ). We shall assume that S is oriented, i.e. there is a choice of a unit normal vector ν(x) at each point x of S that varies continuously with x. In particular F (x) ν(x) = ± F (x). This formula shows that the normal field x ν(x) is of class C k 1 on S. It may be worth recalling that a result of H. Samelson (see Proceedings A.M.S., 22 (1969), ) shows that any compact hypersurface in R n is orientable. Definition We say that a domain Ω has C k boundary, or that Ω is of class C k, if Ω is a hypersurface of class C k. Definition Suppose that Ω is an open set, and that k is a positive integer. We denote by C k (Ω) the vector space of all functions f in C k (Ω) such that α f extends to a continuous function on Ω for all multiindices of length k.

10 6 CONTENTS We recall the classical divergence theorem. Theorem Suppose that Ω is a bounded domain with C 1 boundary, and denote by ν the unit outward normal to Ω. Suppose that w C 1 (Ω). Then div w dv = w ν dσ, where σ denotes the surface measure of Ω. Ω The following corollary will be extensively used in the sequel. Corollary Suppose that Ω is a bounded C 1 domain, and that u and v are functions in C 2 (Ω). The following hold: (i) Ω u dv = Ω νu dσ; (ii) ( first Green s identity) v u dv + Ω (iii) ( second Green s identity) (v u u v) dv = (iv) ( integration by parts) v j u dv + Ω Ω Ω Ω v u dv = Ω Ω u j v dv = Ω v ν u dσ; ( v ν u u ν v ) dσ; Ω u v ν j dσ. Proof. To prove (ii), apply the divergence theorem to w = v u. Then (i) follows from this by taking v = 1. By interchanging the role of u and v in (ii), and subtracting the resulting equalities, we get (iii). Finally, (iv) follows from the divergence theorem by taking w to be the vector field, all of whose components vanish but the j th, which is equal to uv. Denote by ω n the surface measure of the sphere B 1 (0) in R n. Various formulae will involve ω n and the Lebesgue measure of the ball B 1 (0). Some of the exercises below describe an elegant way to compute these quantities. We recall the following formula of integration in polar coordinates. Proposition Suppose that f is an integrable function on R n. Then f dv = dr r n 1 f(rω) dσ(ω) R n 0 S n 1 = dr f(ω ) dσ(ω ). 0 B r

11 0.1. BACKGROUND AND PRELIMINARY RESULTS 7 In particular, if f is radial, then R n f dv = σ(s n 1 ) 0 r n 1 f(r) dr. Exercise Prove that the measure of the unit ball in R n is ω n /n. Hint: use the divergence theorem for an appropriate vector field. Exercise By following the steps below, prove that the measure of the unit ball in R n is π n/2 /Γ(n/2 + 1) (here Γ denotes Euler s Gamma function): 2 (i) show that if a > 0, then dx = ( π/a ) n/2 ; (ii) by computing e a x 2 R n R n e a x show that ω n = 2π n/2 /Γ(n/2); dx both with (i) and using polar coordinates, (iii) compute V ( B 1 (0) ), by integrating the characteristic function of B 1 (0) in polar coordinates. Exercise Suppose that Ω is a domain in R n. Prove that the divergence operator is the adjoint of the gradient, in the sense that for every smooth vector field w and every smooth function ϕ with compact support contained in Ω ϕ div w dv = ϕ w dv. Ω Exercise Suppose that u is a smooth function on R 2 \{0}. Prove that u(r, θ) = 2 r u + 1 r ru + 1 r 2 2 θu. Then find all radial solutions to the equation u = 0 in R 2 \ {0}. Exercise Suppose that u is a smooth function on R 3 \{0}. Prove that u(r, θ, ψ) = r 2 u + 2 r ru + 1 [ 1 ] r 2 (sin ψ) 2 2 θu + ψu 2 + cot ψ ψ u. Then find all radial solutions to the equation u = 0 in R 3 \ {0}. Ω Linear operators Definition A linear operator on a complex Banach space B is a pair (X, A), where X is a (possibly not closed) linear subspace of B and A : X B is a linear operator: X is called the domain of A and will be denoted by Dom(A). Note that we do not assume that Dom(A) is dense in B, unless explicitly stated. Definition Given two linear operators A and B on B, we say that B is an extension of A, and we write B A, if Dom(B) Dom(A) and the restriction of B to Dom(A) agrees with A.

12 8 CONTENTS Consider the operators A and B on L 2 (R) defined by Clearly B A. Dom(A) = C c (R) Af = i f f Dom(A) Dom(B) = C 1 c (R) Bf = i f f Dom(B). We consider two more examples of linear operators, which are one dimensional models of the Dirichlet and Neumann Laplacian. Suppose that < a < b <. Let H D and H N be the operators on L 2 ((a, b)) defined by Dom(H D ) = {f C 2 ([a, b]) : f(a) = 0 = f(b)} H D f = f Dom(H N ) = {f C 2 ([a, b]) : f (a) = 0 = f (b)} H N f = f. It is not hard to check that the domains of H D and H N are dense in L 2 ((a, b)). Exercise Suppose that A is a densely defined linear operator in a Hilbert space H such that (Af, f) = 0 for all f in Dom(A). Prove that A = 0. Note that this fails on real vector spaces (take, e.g., a rotation of π/2 on R 2 ). An important subclass of linear operators is that of bounded operators. Definition A linear operator on B is bounded if Dom(A) = B and Af sup <. The supremum above is called the operator norm of A and f 0 f it is denoted by A. Exercise Suppose that B is a finite dimensional (Banach) space. Prove that every linear operator with domain B is bounded. (Hint: It suffices to control the operator on a basis of B.) Exercise Prove that for a linear operator A on B the following are equivalent: (i) A is a bounded operator on B; (ii) A is continuous at every point of B; (iii) A is continuous at 0. Exercise Suppose that A is a linear operator on B with dense domain. Af Assume that sup <. Prove that there exists a unique bounded f f Dom(A)\{0} linear operator à that extends A. Exercise Prove that the operators A and B defined just below Definition do not admit bounded extensions to L 2 (R). It is straightforward to check that the set of bounded operators on B is a vector space, which we denote by L(B), and that the operator norm is a norm on L(B).

13 0.1. BACKGROUND AND PRELIMINARY RESULTS 9 Proposition The space L(B) is a Banach space with respect to the operator norm. Proof. Suppose that {A n } is a Cauchy sequence of operators in L(B). Then for every ε > 0 there exists an integer ν such that Consequently, A m A n ε m, n ν. A m f A n f ε f m, n ν f B. (0.1.4) Thus, for every f in B, {A n f} is a Cauchy sequence in B. Since B is complete, there exists an element in B that we denote Af such that lim n A n f = Af. It is straightforward to check that the operator A thus defined is linear. By letting m tend to in (0.1.4), we obtain Af A n f ε f n ν f B. This implies that A A n is a bounded operator on B, and that A A n ε n ν. Thus, A is bounded (because it is the sum of the bounded operators A n and A A n ), and lim n A n A = 0, as required The spectrum of a linear operator Definition Suppose that A is a linear operator on B. We say that the complex number ζ is in the resolvent set of A if ζi A maps Dom(A) in a one-to-one fashion onto B and the resolvent operator R(ζ; A) := (ζi A) 1 is a bounded operator on B. The resolvent set of A will be denoted by ϱ(a). The spectrum σ(a) of A is defined to be C \ ϱ(a). Definition Suppose that A is a (closed) operator on B. We say that a point ζ in σ(a) is in the point spectrum of A if ζ is an eigenvalue of A. We say that a point ζ in σ(a) is in the discrete spectrum of A if it is an eigenvalue and it is an isolated point in σ(a). If B is finite dimensional, then the spectrum of a linear operator A is just the set of its eigenvalues. One of the reasons for which spectral theory on infinite dimensional Banach spaces is far more difficult than in the finite dimensional case is that the following result, whose proof we leave as an exercise, is false in the infinite dimensional case. Proposition Suppose that B is finite dimensional, and that A is a linear operator defined on B. The following hold:

14 10 CONTENTS (i) A is bounded; (ii) A is invertible if and only if A is injective if and only if A is surjective; (iii) Ran(A) is dense if and only if Ran(A) = B; (iv) if A is invertible, then A 1 is bounded. Proof. Exercise. Consider the operator A defined just below Definition There exists a (purely algebraic) extension of A to a linear operator on B [TL, p. 38]. Note that this extension cannot be bounded for we have proved in Exercise that A does not admit a bounded extension to all of B. This extension provides an example of an unbounded linear operator which is defined everywhere. Exercise Find counterexamples to statements (ii)-(iv) in the case of operators on infinite dimensional Banach spaces, by looking at the operators L and R, defined on l 2 by L(x 1, x 2, x 3,...) = (x 2, x 3,...) R(x 1, x 2, x 3,...) = (0, x 1, x 2, x 3,...), ( and C, defined on l 1 by C(x 1, x 2, x 3,...) = x 1, x 2 2, x ) 3 3, Complex measures In these notes a complex measure on R n is simply a complex valued function µ on Borel sets, with the property that there exist nonnegative finite Borel measures µ 1,..., µ 4 such that µ(e) = µ 1 (E) µ 2 (E) + iµ 1 (E) iµ 1 (E). Clearly, we may integrate bounded continuous functions ϕ with respect to each µ j, and then define, for every bounded continuous function ϕ ϕ dµ := ϕ dµ 1 ϕ dµ 2 + i ϕ dµ 3 i ϕ dµ 4. R n R n R n R n R n To every complex measure µ we associate the linear functional T µ on bounded continuous functions, defined by T µ (ϕ) = ϕ dµ. R n The total variation µ of µ is the minimal solution to the problem of finding a positive measure ν that dominates µ, in the sense that µ(e) ν(e) for every measurable E. The measure µ may be defined as follows: µ (E) := sup µ(e j ) j=1

15 0.1. BACKGROUND AND PRELIMINARY RESULTS 11 where E = j E j, the measurable sets E j are pairwise disjoint, and the supremum is taken with respect to all partitions of E as countable union of pairwise disjoint measurable subsets of E. It can be shown that µ M(R n ) := µ (R n ) is a norm on M(R n ), which makes M(R n ) a Banach space. Notice that we may view L 1 (R n ) as a closed subspace of M(R n ), by identitying a function f in L 1 (R n ) with the measure f dv. Notice also the following important inequality ϕ dµ ϕ d µ R n R n which holds for all bounded continuous functions on R n. Finally, we mention the following result, which we shall use repeatedly in the sequel, which shows that M(R n ) is a dual space. Theorem The dual of the space C 0 (R n ) (continuous function vanishing at infinity, endowed with the supremum norm) is the space M(R n ) of complex measures, endowed with the norm µ M(R n ) := µ (R n ). Given a complex measure µ on R n, the linear functional T µ on C 0 (R n ), defined by (T µ )(ϕ) = ϕ dµ R n is continuous on C 0 (R n ), and Tµ C0(R n ) = µ (R n ). For these facts, and many more, the reader is referred to [Ru, Ch. 6], or to the book of Folland cited in the References The Schwartz space and the Fourier transform For every nonnegative integer N denote by ρ N the functional on C (R n ), defined by ρ N (f) = max sup x α β f(x). (0.1.5) α, β N x R n Definition The Schwartz space S is the space of all functions f in C (R n ) such that ρ N (f) < for every nonnegative integer N. Clearly Cc (R n ) is included in S. This inclusion is proper, for the Gaussian x e x 2 belongs to S. The seminorms ρ N control both the decay and the oscillations of functions. Observe that the function f(x) = e x2 cos(exp x 4 ) decays fast at infinity, but it does not belong to S, for f is unbounded. Suppose that f is in S. Then for each multiindex β the function β f decays at infinity faster than every power. Indeed, in particular ρ N (f) < for every N. This and Exercise below imply that for each nonnegative integer N there exists a constant C, independent of f such that β f(x) C ρ N (f) (1 + x ) N x R n. (0.1.6)

16 12 CONTENTS Exercise Prove that for each nonnegative integer j there exists a constant C such that Then prove (0.1.6). x j C ( x 1 j x N j). Notice that the Schwartz space is included in L 1 (R n ). Indeed, suppose that f is in S. Then, by (0.1.6), 1 f 1 C ρ n+1 (f) R 1 + x n+1 dx = C ρ n+1(f). (0.1.7) n A similar argument shows that S is included in L p (R n ) per ogni p in (1, ). Exercise Prove that the Schwartz space is closed under the following maps: (a) translations; (b) pointwise product with a Schwartz function; (c) pointwise product with a polynomial; (d) pointwise product with an exponential of the form x e ixξ, where ξ is in R n ; (e) α for any multiindex α. Exercise Give an example of a function f on R such that f 10 6, f 10 6, but f Exercise Does the function f(x) := N=1 2 N exp( (1 + x 2 ) 1/N ) decay exponentially fast at infinity? Is it a Schwartz function? What can you say about g(x) := N=1 2 N (1 + x 2 ) N exp( (1 + x 2 ) 1/N )? Exercise Prove that the convolution of two Schwartz functions is a Schwartz function. Hint: Write x α = (x v + v) α, and use Newton s formula to expand the right hand side. The Fourier transform on S Definition Suppose that f is in S. The Fourier transform of f is the function Ff : R n C, defined by (Ff)(ξ) = f(x) e 2πixξ dx ξ R n. R n Suppose that g is in S. The inverse Fourier transform of g is the function F 1 g : R n C, defined by (F 1 g)(x) = g(ξ) e 2πixξ dξ x R n. R n The functions Ff and F 1 g will also be denoted by f and ǧ, respectively. We recall the notation τ y f(x) = f(x y) and e α (x) = e iα x α, y R n.

17 0.1. BACKGROUND AND PRELIMINARY RESULTS 13 Proposition Suppose that f and g are in S. The following hold: (i) F(τ y f) = e 2πiy Ff, and F(e 2πiy f) = τ y ( Ff ) ; (ii) F( α f) = p α Ff, and F(p α f) = ( 1) α α (Ff), where p α (ξ) = (2πiξ) α ; (iii) F(f g) = (Ff) (Fg). (iv) F(f t ) = (Ff) t and F(f t ) = (Ff) t where f t (x) = t n f(x/t) and f t (x) = f(tx). Proof. These are straightforward consequences of the definition of F and F 1. We prove (iii), and leave the other verifications to the reader. Observe that F( α f)(ξ) = α f(x) e 2πixξ dx R n (by parts) = ( 1) α R n f(x) α x ( e 2πixξ ) dx = ( 1) α ( 2πiξ) α Ff(ξ) = (2πiξ) α Ff(ξ) ξ R n, as required. Exercise Prove Proposition Exercise Suppose that A is a nonsingular linear transformation of R n. Given f in S, compute the Fourier transform of f A in terms of the Fourier transform of f. Consider the n-dimensional Gaussian ψ(x) := e π x 2. We show that ψ = ψ. (0.1.8) Since the n-dimensional Fourier transform of ψ is the product of n one-dimensional Fourier transforms of Gaussian functions on R, it suffices to prove the result for n = 1. Observe that, in this case, ψ solves the differential equation ( x + 2πx ) u = 0. (0.1.9) By taking the Fourier transform of both sides, we obtain that ( 2πiξ+i ξ ) ψ = 0, i.e., ψ is still a solution of (0.1.9). Thus, we must have ψ = c ψ for some constant c. Then, it suffices to observe that c = ψ(0) ψ(0) = e πx2 dx = 1. The following result is the reason for which the Schwartz space has been invented. Proposition The Fourier transform F is an isomorphism of S, with inverse F 1.

18 14 CONTENTS Proof. First we prove that F ( S ) S, with continuous inclusion. Suppose that f is in S and denote by α and β two multiindices. By properties (iii) and (iv) above, ξ α β (Ff)(ξ) = ( 1) β (2πi) α F( α (p β f) ) (ξ), so that ρ N (Ff) = max sup 1 ( F α (p α, β N ξ R n (2π) α β f) ) (ξ) max α, β N (0.1.6) C ρ n+1+n (f), 1 (2π) α α (p β f) 1 which shows that F is in S and that F is continuous. Next we show that F 1 F = I on S. Suppose that f is in S. We must compute the iterated integral dξ e 2πiξv f(x) e 2πixξ dx. (0.1.10) R n R n Observe that the double integral above is not absolutely convergent, so that we are not entitled to use Fubini s theorem. Set ψ(x) = e π x 2. Recall that ψ is in S and that ψ(y) dy = 1 (see (0.1.8)). By the dominated convergence R n theorem, and the definition of Fourier transform e 2πiξv f(ξ) dξ = lim e 2πiξv f(ξ) ψ(εξ) dξ R n ε 0 + R n = lim e 2πiξ(v x) f(x) ψ(εξ) dx dξ. ε 0 + R n R n By Proposition (iv), the integral with respect to ξ in the last double integral is equal to ( ψ) ε (v x), i.e. to ψ ε (v x) by (0.1.8), so that e 2πiξv f(ξ) dξ = lim f ψ ε(x), R n ε 0 + which is equal to f(x), as required. Similarly, we can prove that F F 1 = I on S. Thus, F is a bijective map, with inverse F 1. Since F is continuous, it is an homeomorphism, as required to conclude the proof of the theorem. Extensions to L 1 and to L 2 Proposition (Parseval and Plancherel) Suppose that f and g are in S. The following hold: (i) R n f g dλ = R n f ĝ dλ; (ii) f 2 = f 2. Proof. First we prove (i). Since f, g, f and ĝ are continuous and rapidly decreasing at infinity, the definition of Fourier transform and Fubini s theorem

19 0.1. BACKGROUND AND PRELIMINARY RESULTS 15 imply that we have that f g dv = R n as required. R n f(x) e 2πixξ g(ξ) dx dξ = f ĝ dv, R n R n Next we prove (ii). Observe that Ff 2 2 = Ff(ξ) Ff(ξ) dξ R n (direct consequence of the def of FT) = Ff(ξ) F 1 (f)(ξ) dξ R n (by (i), with F 1 (f) in place of g) = f(ξ) f(ξ) dξ R n = f 2 2, as required. Corollary The Fourier transform, initially defined on S, extends to: (i) an isometric isomorphism of L 2 (R n ); (ii) to a contractive map from L 1 (R n ) into C 0 (R n ). Proof. Property (i) follows directly from (ii) above and the density of the Schwartz space in L 2 (R n ). Suppose that f is in S. Observe that sup f(ξ) f dv. ξ R n R n Then (ii) follows directly from this and the density of the Schwartz space in L 1 (R n ) and in C 0 (R n ). Exercise Compute the Fourier transform of 1 (a,b). Exercise Suppose that p(x) = 1/(1+x 2 ), where x is in R. Compute p, by using the residue theorem. Exercise Consider the Gaussian ψ(x) = e π x2 on R. By using Corollary ) and the fact that for each f in L 1 (R) the function ψ t f is in L 2 (R n ) and converges to f in the L 1 (R) norm (Proposition ), prove that F : L 1 (R) C 0 (R n ) is an injective map.

20 16 CONTENTS

21 Chapter 1 The classical Dirichlet problem 1.1 Introduction Suppose that n 2, and that Ω is a bounded open set in R n. We denote by the Laplace operator, which acts on a function f in C 2 (Ω) by f(x) = n j 2 f(x) x Ω. j=1 The classical Dirichlet problem is the following: given a continuous function g on Ω, find a function u in C 2 (Ω) C(Ω) such that { u = 0 u Ω = g. in Ω This is a very challenging problem, which has been considered by many outstanding mathematicians in the past two centuries. The reader is referred to the interesting article [G] of L. Gårding for an historical account of the research on the Dirichlet problem until the first half of the twentieth century. As we shall see, it is comparatively easy to prove that if a solution to the Dirichlet problem exists, then it is unique. By contrast, it is entirely nontrivial to prove that, under suitable assumptions on the domain Ω, the Dirichlet problem is solvable. There are many problems in Mathematics and in Physics that lead to consider the Dirichlet problem. Here we recall two of them. The first is concerned with the heat diffusion in a body Ω R n (n = 2, 3 are the most important cases). Assume that a fixed temperature distribution at the boundary is maintained by a heating and refrigeration system. By suitably normalising the physical constants involved, we may assume that the temperature u(x, t) of the point x Ω at time t satisfies the following equation (known 17

22 18 CHAPTER 1. THE CLASSICAL DIRICHLET PROBLEM as the heat equation) t u(x, t) u(x, t) = 0 x Ω t > 0, (1.1.1) where the Laplacian acts on the x variable. Denote by g(x) the temperature at which the heating and refrigeration system keeps the point x Ω. Then we must have u(x, t) = g(x) x Ω t > 0. (1.1.2) Of course, the body has an initial temperature u(x, 0) at each point x Ω. Given that the boundary temperature is kept at a steady state, it is plausible that the system will evolve towards an equilibrium state u(x), that is u(x) = lim t u(x, t). Since for all T > 0 the function u T (x, t) := u(x, T + t) satisfies the boundary value problem (1.1.1) (1.1.2), it is reasonable to expect that the same will happen to u. Since u does not depend on t, u will satisfy { u = 0 u Ω = g. in Ω The second problem leading to the Dirichlet boundary value problem is internal to Mathematics. Riemann tackled the problem of constructing a conformal mapping ϕ between a proper simply connected domain Ω in the complex plane and the unit disc D := {z C : z < 1}. By this we mean that ϕ is a holomorphic bijection between Ω and D. Suppose that Ω has smooth boundary. We look for a homeomorphism ϕ : Ω D, which, restricted to Ω, is a biholomorphism between Ω and D. We may assume that 0 Ω. By possibly composing ϕ with a Möbius transformation, we may assume that ϕ(0) = 0. If such a conformal mapping ϕ exists, then z ϕ(z)/z is nonvanishing (recall that ϕ (z) 0 for all z Ω, for ϕ is conformal), whence there exists a holomorphic function f on Ω such that ϕ(z) = e f(z) z Ω. z Since the left hand side extends to a continuous map on Ω that necessarily maps Ω onto D, we have that ϕ(z) = 1 for every z Ω. Therefore log ϕ(z) = log z + Re f(z) z Ω and Re f(z) = log z z Ω. Clearly Re f is harmonic in Ω, for it is the real part of a holomorphic function, and it is a solution to the boundary value problem { u(z) = 0 u(z) Ω = log z. in Ω

23 1.2. HARMONIC FUNCTIONS AND THE MEAN VALUE PROPERTY Harmonic functions and the mean value property Definition Suppose that Ω is a domain in R n. A function u in C 2 (Ω) is harmonic in Ω if u = 0 in Ω. Note that u is harmonic if and only if the real and the imaginary part of u are harmonic, because preserves the class of real functions. Thus, when studying solutions to the Laplace equation u = 0, it often suffices to consider real-valued functions. Exercise Prove that the real and the imaginary parts of a holomorphic function in a domain Ω are harmonic in Ω. Prove that the functions x 2 y 2 and xy are harmonic in R 2, and so are the functions r j cos(jθ) and r j sin(jθ) for each nonnegative integer j. Exercise Suppose that Ω is a domain in the complex plane, that φ : Ω φ(ω) is a one-to-one holomorphic mapping, and that f is harmonic in φ(ω). Prove that the function f φ is is harmonic in φ(ω). Exercise Prove that if u is harmonic in Ω, and Ω Ω, then ν u dσ = 0. Ω Conversely, show that if u C 2 (Ω) satisfies ν u dσ = 0 for every ball B Ω, then u is harmonic in Ω. B The next result relates harmonic functions with their spherical and solid means over balls. In fact, harmonic functions may be characterised by the behaviour of their means, as we shall see later. For a ball B, we denote by σ( B) the surface measure of B and by V (B) the Lebesgue measure of B. We need the following formula, which will be used repeatedly in the sequel. Lemma Suppose that Ω is a domain, that B R (y) Ω and that u C 1 (Ω). Then for ε small enough and for every ρ (0, R + ε) B ρ(y) [ 1 ] ν u dσ = ρ n 1 ρ ρ n 1 u dσ. B ρ(y)

24 20 CHAPTER 1. THE CLASSICAL DIRICHLET PROBLEM Proof. We integrate in polar co-ordinates around y and obtain that ν u dσ = u(y + ω ω ) B ρ(y) B ρ(0) ω dσ(ω ) = ρ n 1 u(y + ρω) ω dσ(ω) as required. = ρ n 1 = ρ n 1 ρ B 1(0) B 1(0) B 1(0) = ρ n 1 ρ [ 1 ρ n 1 ρ [ u(y + ρω) ] dσ(ω) u(y + ρω) dσ(ω) B ρ(y) ] u dσ, Theorem (Mean value property) Suppose that Ω is a domain. Then for every harmonic function u and every ball B Ω with centre y u(y) = 1 u dσ and u(y) = 1 u dv. σ( B) V (B) B Proof. Suppose that B = B R (y), that 0 < ρ < R + ε for some ε > 0. By Corollary (i) and Lemma = ν u dσ = ρ n 1 d [ 1 ] B ρ(y) dρ ρ n 1 u dσ. B ρ(y) Now divide both sides by ρ n 1, integrate with respect to ρ between ε and R, and obtain 0 = 1 R n 1 u dσ 1 ε n 1 u dσ. Since u is continuous in y, lim ε 0 B R (y) B ε(y) 1 ε n 1 u dσ = ω n u(y). B ε(y) By combining the last two formulae, we obtain that 1 u(y) = ω n R n 1 u dσ, as required. B R (y) To prove the second formula, we multiply by R n 1 both sides of the equality above and integrate with respect to R between 0 and r. We obtain r n n u(y) = 1 r dr u dσ, ω n which is equivalent to the required equality. 0 B R (y) It is an important fact that a converse of Theorem holds. B

25 1.2. HARMONIC FUNCTIONS AND THE MEAN VALUE PROPERTY 21 Theorem Suppose that u is a continuous function in the domain Ω and that for every ball B with closure contained in Ω the following holds: u(c B ) = 1 u dσ, σ( B) where c B denotes the centre of B. Then u is harmonic in Ω. B Proof. Denote by φ a smooth radial function with support contained in the ball B 1 (0) such that φ dv = 1, and denote by ψ its profile, i.e., ψ( x ) = φ(x). For every ε > 0, set φ ε (x) = ε n φ(x/ε). Clearly the support of φ ε is contained in B ε (0). If x belongs to Ω ε := {x : B ε (x) Ω}, then the support of y φ ε (x y) is contained in Ω. Observe that u(y) φ ε (x y) dy = u(x εy) φ(y) dy R n y <1 1 (polar coordinates) = u(x εry ) ψ(r) r n 1 dσ(y ) dr (by assumption) (polar coordinates) 0 y =1 1 = ω n u(x) ψ(r) r n 1 dr 0 = u(x) φ dv R n = u(x). Now, the left hand side, being the convolution of u with the smooth function φ ε, can be differentiated infinitely many times, whence u is in C (Ω ε ). Since ε is arbitrary, u is in C (Ω). It remains to show that u is harmonic. Suppose that B r (x) Ω. By Corollary (i) and Lemma 1.2.5, u dv = ν u dσ B r(x) B r(x) [ = ω n r n 1 1 r σ( B r (x)) = ω n r n 1 r [ u(x) ], B r(x) for u possesses the mean value property by assumption. Thus, u dv = ω n r n 1 d dr u(x) = 0 B r(x) ] u dσ for every ball B r (x) such that B r (x) Ω. Since u is continuous, u vanishes at every point in Ω, as required. Notice that the proof of Theorem 1.2.7, combined with Theorem 1.2.6, shows that if u is harmonic in Ω, then u is smooth (it possesses derivatives of any order).

26 22 CHAPTER 1. THE CLASSICAL DIRICHLET PROBLEM Exercise Show that if u is a continuous function in Ω that possesses the solid mean value property, then u satisfies the spherical mean value property. Hint: write u(x) as the solid mean over B R (x) in polar co-ordinates, multiply both sides by R n and differentiate. Exercise Show that if u is a function in C 2 (Ω) and x 0 Ω, then 2n [ 1 ] u(x 0 ) = lim r 0 r 2 u(x 0 ) σ(s n 1 u(x 0 + r ω) dσ(ω). ) S n 1 Deduce that if u satisfies the mean value property (see Theorem 1.2.6), then u is harmonic. 1.3 Maximum principle and uniqueness for the Dirichlet problem In this section we shall prove the maximum and the minimum principles for harmonic functions. As a corollary we shall obtain that the classical Dirichlet problem on a bounded domain Ω has at most one solution. The problem of the existence of solutions to the Dirichlet problem will be addressed later. Theorem Suppose that Ω is a domain in R n, and that u is a real valued harmonic function in Ω. Then u cannot assume an interior maximum or minimum, unless it is constant. Proof. Set M := sup Ω u, and Ω M := {x Ω : u(x) = M}. Suppose that there exists y in Ω such that u(y) = M. Then Ω M is nonempty, and it is closed, for u is continuous. We shall prove that Ω M is open. It will follow that Ω M = Ω, for Ω is connected by assumption, i.e. u = M in Ω, as required. To prove that Ω M is open, choose z in Ω M. Observe that u M is harmonic ( (u M) = u), so that, by the mean value property, 0 = u(z) M = 1 V (B) B (u M) dv 0 for every ball B with radius small enough. Therefore (u M) dv = 0. B If u were strictly less than M on an open subset of B, then the integral above would be strictly negative. Therefore u = M on B, and Ω M is open, as required.

27 1.4. GREEN S FUNCTION 23 Corollary Suppose that Ω is a bounded domain in R n, and that u C 2 (Ω) C(Ω), and harmonic in Ω. Then inf u u sup u. Ω Proof. Observe that sup Ω u = sup Ω u, for u is continuous in Ω. Now, if sup Ω u > sup Ω u, then there exists a point y in Ω for which u(y) = sup Ω u. Consequently u is constant in Ω by Theorem Hence sup Ω u = sup Ω u, thereby contradicting the assumption. The proof that inf Ω u u is similar, and it is omitted. A noteworthy consequence of Corollary is the following uniqueness result for the Dirichlet problem. Theorem Suppose that Ω is a bounded domain and that v and w are functions in C 2 (Ω) C(Ω) that solve the Dirichlet problem { u = f in Ω u Ω = g. Then v = w. Ω Proof. Set u = v w. Clearly u solves { u = 0 u Ω = 0. in Ω In particular, u is a harmonic function in C 2 (Ω) C(Ω) that vanishes on Ω. By Corollary (iii), u = 0 on Ω, i.e. v = w, as required. We wish to emphasise the fact that no claims are made concerning the existence of a solution to the Dirichlet problem. 1.4 Green s function In this section we establish an important representation formula for solutions of the Dirichlet problem, under the assumption that a solution exists. Definition The Newtonian potential in R n is the function N : R n \ {0} R, defined by 1 log x if n = 2 N(x) = 2π 1 (1.4.1) x 2 n if n 3. (2 n) ω n Given a point x in R n, the Newtonian potential with pole x is the function N(x ).

28 24 CHAPTER 1. THE CLASSICAL DIRICHLET PROBLEM Note that if n 3, then N is a negative function. By contrast, if n = 2, then N has not a definite sign. The reason for the normalisation of N will be clarified below (see Definition 4.2.2). It will be proved later that N is a fundamental solution of the Laplace operator: it will play an important role in finding distributional solutions to the Poisson equation u = f, where f is a given datum. All this will be discussed in Section 4.2. Exercise Prove that the Laplacian in R n acts on smooth radial functions as follows f(r) = f (r) + n 1 f (r). r Then find [ all ] radial harmonic functions in R n. Hint: [ If f] depends only on r, then j f(r) = f (r) j (r) = f (r) x j /r. Compute j 2 f(r) similarly, and then sum over j. Exercise Prove, by direct calculation, that N is harmonic in R n \ {0}. Exercise Prove that B r(x) ν N(x y) dσ(y) = 1, We now establish Green s representation formula. Theorem (Green s representation formula) Suppose that Ω is a bounded C 1 domain, that u C 2 (Ω) C 1 (Ω), and that u L 1 (Ω). The following hold: (i) for every x in Ω [ u(x) = u ν N(x ) ν u N(x ) ] dσ + Ω (ii) if the support of u is a compact set in Ω, then u(x) = N(x ) u dv x Ω; Ω Ω N(x ) u dv ; (iii) if u is harmonic in Ω, then [ u(x) = u ν N(x ) ν u N(x ) ] dσ x Ω; Ω (iv) suppose that x Ω and that there exists a function h x C 1 (Ω) which is harmonic on Ω and satisfies (h x ) Ω = N(x ) Ω. Denote by G(x, ) the function h x + N(x ). Then u(x) = u ν G(x, ) dσ + G(x, ) u dv. Ω All the normal derivatives in the integrals above are taken with respect to the variable of integration. Ω

29 1.4. GREEN S FUNCTION 25 Proof. We prove the theorem in the case where n 3. The case n = 2 is left to the reader. First we prove (i). Consider the domain Ω \ B r (x), for r small. Apply the second Green s identity (see Corollary (iii)) with N(x ) in place of v and Ω \ B r (x) in place of Ω. We take into account the fact that N(x ) is harmonic in Ω \ B r (x), and obtain ( ) [N(x N(x ) u dv = ) ν u u ν N(x ) ] dσ. Ω\B r(x) Ω B r(x) The required formula will follow from this and the dominated convergence theorem once we prove that N(x ) ν u dσ = 0 (1.4.2) and that lim r 0 B r(x) To prove (1.4.2), note that N(x y) C r 2 n and that lim u ν N(x ) dσ = u(x). (1.4.3) r 0 B r(x) ν u max u, Ω y B r (x) which is finite, because u C 1 (Ω). Thus, N(x ) ν u dσ C max u r 2 n σ ( B r (x) ), B r(x) which tends to 0 as r tends to 0, for σ ( B r (x) ) = ω n r n 1, as required. To prove (1.4.3), write the integral in (1.4.3) as [ ] u(y) u(x) ν N(x y) dσ(y) + u(x) B r(x) Since u is smooth, u(y) u(x) r max u Ω Ω B r(x) y B r (x). ν N(x y) dσ(y). Recall that N is homogeneous of degree 2 n, hence its partial derivatives are homogeneous of degre 1 n. Thus, ν N(x y) C r 1 n y B r (x). Therefore B r(x) [ u(y) u(x) ] ν N(x y) dσ(y) C r r 1 n σ ( B r (y) ), which tends to 0 as r tends to 0. Finally, B r(x) νn(x ) dσ = 1, by Exercise 1.4.4, which completes the proof of (i).

30 26 CHAPTER 1. THE CLASSICAL DIRICHLET PROBLEM Note that (ii) and (iii) are direct consequences of the representation formula established in (i). It remains to prove (iv). Write N(x ) = G(x, ) h x in (i). We obtain [ ] u(x) = u ν G(x, ) ν u G(x, ) dσ + G(x, ) u dv Ω Ω + ν u h x dσ u ν h x dσ h x u dv. Ω Ω Note that, by the second Green s identity and the fact that h x is harmonic the last three terms in the right hand side cancel out, and the required formula follows. The last part of the theorem above suggests that the function G defined therein may play an important role in the theory. This justifies the following definition. Definition Suppose that Ω is a domain in R n. Assume that for every x Ω the Dirichlet problem { u = 0 in Ω u Ω = N(x ) is solvable, and that the solution h x is in C 1 (Ω). The function G : Ω Ω\{(z, z) : z Ω}, defined by G(x, y) := h x (y) + N(x y), is called the Green function for the domain Ω. Note that G(x, ) vanishes on the boundary of Ω. Note that we do not assert the existence of a Green s function for a generic domain Ω. In fact, there are domains which do not admit a Green s function (see VECCHI APPUNTI). Exercise Prove that the Green s function for a bounded domain Ω, if it exists, is unique. The following property of the Green function G is important, and it is related to the symmetry of (with Dirichlet boundary conditions) on L 2 (Ω). Proposition Suppose that the bounded domain Ω admits a Green s function G. Then G(x, y) = G(y, x) for every x, y in Ω, x y. Ω Proof. we shall write G z in place of G(z, ) for any z in Ω. Now, fix ε > 0 so small that the balls B ε (x) and B ε (y) are contained in Ω and their closures do not intersect, and denote by Ω ε the domain Ω \ [ B ε (x) B ε (y) ]. Observe that G x and G y are harmonic in Ω ε, and vanish on Ω. By the second Green s identity (see Corollary (iii)) applied to Ω ε [ ] [ ] Gy ν G x G x ν G y dσ = Gx ν G y G y ν G x dσ. B ε(y) B ε(x)

31 1.4. GREEN S FUNCTION 27 Since G x is harmonic on B ε (y) and Gy cn y 2 n, G y ν G x dσ = 0. lim ε 0 B ε(y) Similarly, lim ε 0 B G ε(x) x ν G y dσ = 0. We claim that lim ε 0 B ε(y) G x ν G y dσ = G(x, y) and lim G y ν G x dσ = G(y, x), ε 0 B ε(x) from which the required formula follows. We prove the first relation, as the proof of the second is almost identical. Recall that G y = h y + N y, where h x is of class C 1 (Ω), it is harmonic in Ω and satisfies h y Ω = N y. It is straightforward to check that lim ε 0 B G ε(y) x ν h y dσ = 0. The proof that lim ε 0 B G ε(y) x ν N y dσ = G x (y) is almost verbatim equal to the proof of (1.4.3). We omit the details. Exercise Suppose that G is the Green s function of the bounded domain Ω in R n, with n 3. Prove the following: (i) G(x, y) < 0 for every x, y in Ω, x y; (ii) prove that G(x, y) N(x y) for every x and y in Ω; (iii) if f is bounded in Ω, then G(x, y) f(y) dv (y) 0 as x Ω. Ω Hints: (i) note that G(x, y) tends to if y tends to x. Therefore, G < 0 near x. Apply the maximum principle to the domain Ω \ B ε (x); (ii) observe that, by the minimum principle, h x > 0 in Ω and use (i) and the definition of G; (iii) use (ii) and the Lebesgue dominated convergence theorem, away from Ω, and the extra-integrability of the Newtonian potential near Ω. We explicitly state a straightforward but important consequence of Theorem (iv): a representation formula for the solution to the Dirichlet problem { u = 0 in Ω u Ω = g, under the assumption that a solution u exists and it is of class C 1 (Ω). Corollary Suppose that Ω is a bounded C 1 domain, which admits a Green s function G. If u C 1 (Ω) is a solution to the Dirichlet problem above, then u(x) = g ν G(x ) dσ x Ω. Ω Definition Suppose that G is the Green function for the domain Ω. The function P : Ω Ω R, defined by P (x, Y ) := ν(y ) G(x, Y ) is called the Poisson kernel for the domain Ω.

32 28 CHAPTER 1. THE CLASSICAL DIRICHLET PROBLEM Corollary indicates that it is reasonable to produce efforts to determine the Poisson kernel of a given domain Ω. However, the Poisson kernel may be explicitly computed only in a few, albeit important, cases. The next sections are devoted to the cases of the unit ball and the upper half space in R n. 1.5 The Poisson kernel for the half space In this section we compute the Poisson kernel for the upper half space R n + := {(x, x n ) R n 1 R : x n > 0}. First we need to compute the Green s function for R n +. Fix a point x = (x, x n ) in the upper half space, and consider the Newtonian potential N(x ) with pole x. We must find a function h x in C 1 (R n +) that is harmonic on R n + and such that N(x ) + h x vanishes identically when x n = 0. Observe that the Newtonian potential is, up to a constant, the potential generated by a unit negative charge placed at the point x =: (x, x n ). By symmetry, its values on the hyperplane x n = 0 are the same of those of a Newtonian potential generated by a unit negative charge placed at the point x := (x, x n ). Thus, the Green s function G of R n + is given by 1 x y log G(x, y) = 2π x if n = 2 y 1 [ x y 2 n x y 2 n] if n 3. (2 n) ω n (1.5.1) Exercise Check that the function G defined above is the Green s function for the upper half space. First we consider the case where n 3. By definition, for every x R n + and y R n 1 the Poisson kernel P (x, y ) is then given by P (x, y ) = ν G(x, y ) = yn G(x, y ) = 1 [ yn x n ω n x y n y n + x n x y ] y n. n=0 Observe that x y = x y when y n = 0. Hence P (x, y ) = 2 ω n = 2 ω n x n x y n x n ( x y 2 + x 2 n) n/2 x n > 0, x, y R n 1. (1.5.2) A similar computation shows that the formula above holds also in the case where n = 2. Set p(x ) := 2 1 ω ( n x ) x R n 1. n/2