WHICH SOLUTIONS OF THE THIRD PROBLEM FOR THE POISSON EQUATION ARE BOUNDED?

This paper deals with the problem ∆u= g onG and ∂u/∂n+u f = L on ∂G. Here,G⊂Rm, m> 2, is a bounded domain with Lyapunov boundary, f is a bounded nonnegative function on the boundary of G, L is a bounded linear functional on W1,2(G) representable by a real measure μ on the boundary of G, and g ∈ L2(G)∩Lp(G), p > m/2. It is shown that a weak solution of this problem is bounded in G if and only if the Newtonian potential corresponding to the boundary condition μ is bounded in G.


Suppose that G
for each v ∈ Ᏸ, the space of all compactly supported infinitely differentiable functions in R m .Here, ∂G denotes the boundary of G and cl G is the closure of G; Ᏼ k is the kdimensional Hausdorff measure normalized so that Ᏼ k is the Lebesgue measure in R k .Denote by Ᏸ(G) the set of all functions from Ᏸ with the support in G.
For an open set V ⊂ R m , denote by W 1,2 (V ) the collection of all functions f ∈ L 2 (V ), the distributional gradient of which belongs to [L 2 (V )] m .Definition 1.Let f ∈ L ∞ (Ᏼ), g ∈ L 2 (G) and let L be a bounded linear functional on W 1,2 (G) such that L(ϕ) = 0 for each ϕ ∈ Ᏸ(G).We say that u ∈ W 1,2 (G) is a weak solution in W 1,2 (G) of the third problem for the Poisson equation if for each v ∈ W 1,2 (G).
Denote by Ꮿ (∂G) the Banach space of all finite signed Borel measures with support in ∂G with the total variation as a norm.We say that the bounded linear functional L on where A is the area of the unit sphere in R m .For the finite real Borel measure ν, denote the Newtonian potential corresponding to ν, for each x for which this integral has sense.We denote by Ꮿ b (∂G) the set of all µ ∈ Ꮿ (∂G) for which ᐁµ is bounded on R m \ ∂G.
Remark that Ꮿ b (∂G) is the set of all µ ∈ Ꮿ (∂G) for which there is a polar set M such that ᐁµ(x) is meaningful and bounded on R m \ M, because R m \ ∂G is finely dense in R m (see [1, Chapter VII, Sections 2, 6], [7,Theorems 5.10 and 5.11]) and ᐁµ = ᐁµ + − ᐁµ − is finite and fine-continuous outside of a polar set.Remark that Ᏼ m−1 (M) = 0 for each polar set M (see [7,Theorem 3.13]).(For the definition of polar sets, see [4, Chapter 7, Section 1]; for the definition of the fine topology, see [4,Chapter 10].) Denote by Ᏼ the restriction of Ᏼ m−1 to ∂G.
Notation 5. Let X be a complex Banach space and T a bounded linear operator on X.We denote by KerT the kernel of T, by σ(T) the spectrum of T, by r(T) the spectral radius of T, by X the dual space of X, and by T the adjoint operator of T. Denote by I the identity operator.
Theorem 6.Let X be a complex Banach space and K a compact linear operator on X.Let Y be a subspace of X and T a closed linear operator from Y to X such that y(Tx) = x(T y) for each x, y ∈ Y .Suppose that K (Y ) ⊂ Y and KT y = TK y for each the ascent of (K − αI) is equal to 1. Since α is a pol of the resolvent and the ascent of (K − αI) is equal to 1, [5, Satz 50.2] yields that the space X is the direct sum of Ker(K − αI) and (K − αI)(X ) and the descent of (K − αI) is equal to 1. Since the descent of (K − αI) is equal to 1, we have Since the space X is the direct sum of Ker(K − αI) and (K − αI)(X ) = (K − αI) 2 (X ), the operator (K − αI) is invertible on (K − αI)(X ).If α ∈ σ(K ), then the space X is the direct sum of Ker(K − αI) and (K − αI)(X ), and the operator (K − αI) is invertible on (K − αI)(X ).Therefore, there are If α ∈ σ(K Z ), then the space Z is the direct sum of Ker(K Z − αI) and (K Z − αI)(Z), and the operator (K Z − αI) is invertible on Z. Suppose that α ∈ σ(K Z ).Since K Z is compact, then α is a pol of the resolvent by [5,Satz 50.4].Since the ascent of (K Z − αI) is equal to 1. Since α is a pol of the resolvent and the ascent of (K Z − αI) is equal to 1, [5, Satz 50.2] yields that the space Z is the direct sum of Ker(K Z − αI) and (K Z − αI)(Z) and the descent of (K Z − αI) is equal to 1. Since the descent of (K Z − αI) is equal to 1, we have Since the space Z is the direct sum of Ker(K Z − αI) and ( we obtain that y 1 = 0 and y 2 = y.Thus, y Since w(T y) = 0 for each w ∈ Ker(K − αI), [15, Chapter 10, Theorem 3] yields T y ∈ (K − αI)(X).
Denote by K the restriction of K to (K − αI)(X).If we denote by P the spectral projection corresponding to the spectral set {α} and the operator K , then P(X [14, Chapter VI, Theorem 1.3, and Lemma 1.5], there is converges.Easy calculation yields that V is the inverse operator of the operator and thus Ker(K − αI) 2 = Ker(K − αI).If α ∈ σ(K), then the space X is the direct sum of Ker(K − αI) and (K − αI)(X), and the operator (K − αI) is invertible on X. Suppose that α ∈ σ(K).Since K is compact, then α is a pol of the resolvent by [5,Satz 50.4].Since the ascent of (K − αI) is equal to 1. Since α is a pol of the resolvent and the ascent of (K − αI) is equal to 1, [5, Satz 50.2] yields that the space X is the direct sum of Ker(K − αI) and (K − αI)(X) and the descent of (K − αI) is equal to 1. Since the descent of (K − αI) is equal to 1, we have (K − αI) 2 (X) = (K − αI)(X).Since the space X is the direct sum of Ker(K − αI) and (K − αI)(X) = (K − αI) 2 (X), the operator (K − αI) is invertible on (K − αI)(X).Denote by K the restriction of K to (K − αI)(X).If we denote by Q the spectral projection corresponding to the spectral set {α} and the operator K, then Q(X) = (K − αI)(X) by [5,  Since T y ∈ (K − αX) and r(t converges.Since T is closed, x 2 = z k , and Tz k converges, then the vector x 2 lies in Y , the domain of T. Define for ϕ ∈ L ∞ (Ᏼ) and x ∈ ∂G, for each v ∈ Ᏸ.
[5,x ∈ Y , then y ∈ Y .Suppose that y ∈ Y .Since K is a compact operator, the operator K is a compact operator by [14, Chapter IV, Theorem 4.1].Suppose first that α ∈ σ(K ).Since K is compact, then α is a pol of the resolvent by[5, Satz 50.4].Since Ker and only if y ∈ Y .Proof.