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The solution operator $(-\Delta +m^2{)}^{-1}:L^2(U)\to L^2(U)$ is compact on every bounded open set because it maps boundedly into $H_0^1(U)$ and the Rellich-Kondrashov compactness theorem for H01 embeds that space compactly into $L^2(U)$.

For $s\in \mathbb{R}$, the Sobolev space $H^s({\mathbb{R}}^n)$ consists of tempered distributions $u$ such thatFor a multi-index $\alpha$,Sincewe obtain the Sobolev derivative estimateThus differentiation is bounded and linear between the stated spaces.

Taking the Fourier transform ofgivesThe only possible solution is thereforeMoreover,This proves existence, uniqueness, and bounded dependence. Conversely, $u\mapsto (-\Delta +1)u$ maps $H^{s+2}$ boundedly to $H^s$, sois a linear isomorphism with inverse multiplier $(1+|\xi {|}^2{)}^{-1}$. This is the massive Laplacian isomorphism on Sobolev spaces.

Finally, the assumed estimate and Plancherel theorem implyThus $(u_j)$ is bounded in $H^2({\mathbb{R}}^n)$. The functions and all their first derivatives are supported in the fixed bounded set $\overline{\Omega }$. Their $H^1$ bounds give uniform translation estimatesfor $v=u_j$ and $v=D_{\ell }{u}_j$. Tight support and the Rellich-Kondrashov compactness theorem for H01, equivalently the Fourier compactness criterion, therefore provide a common subsequence for which $u_j$ and every first derivative converge strongly in $L^2$. HenceThis is compactness from bounded support and an H2 bound.

A weak solution is a function $u\in H_0^1(U)$ satisfyingThe zero boundary condition is encoded by membership in $H_0^1(U)$.

On the Hilbert space $H_0^1(U)$ defineThe form is bounded and coercive:The functional ${\ell }_f$ is bounded by the Cauchy-Schwarz inequality. The Lax-Milgram theorem therefore gives a unique weak solution and a boundThus $T:L^2(U)\to H_0^1(U)$ is bounded. Composing it with the compact Rellich-Kondrashov compactness theorem for H01 embeddingshows that $T:L^2(U)\to L^2(U)$ is the compact massive-Laplacian resolvent.

The Rellich-Kondrashov compactness theorem for H01 states that if $\Omega \subset {\mathbb{R}}^d$ is bounded and open, then the inclusionis compact: every bounded sequence in $H_0^1(\Omega )$ has a subsequence converging strongly in $L^2(\Omega )$.

Let $(u_n)$ be bounded in $H_0^1(\Omega )$. Since this is a Hilbert space, Banach--Alaoglu and reflexivity give a subsequence, still denoted $(u_n)$, and $u\in H_0^1(\Omega )$ such that $u_n\rightharpoonup u$ weakly in $H_0^1(\Omega )$ and hence in $L^2(\Omega )$. Extend all these functions by zero to ${\mathbb{R}}^d$. The zero extensions belong to $H^1({\mathbb{R}}^d)$ and remain uniformly bounded there.

For each $\xi \in {\mathbb{R}}^d$,because $e^{ix\cdot \xi }{1}_{\Omega }\in L^2(\Omega )$. On each ball $|\xi |\le R$, Cauchy--Schwarz gives a uniform bound on $|{\hat{u}}_n-\hat{u}|$, so dominated convergence yields

The gradient bound controls high frequencies. By Plancherel,uniformly in $n$. First choose $R$ large and then $n$ large. The low- and high-frequency estimates show ${\hat{u}}_n\to \hat{u}$ in $L^2({\mathbb{R}}^d)$, and a final application of Plancherel gives $u_n\to u$ strongly in $L^2(\Omega )$.

Weak convergence $u_n\rightharpoonup u$ in $H_0^1(\Omega )$ impliesin $L^2$. The $L^2$ norm is weakly lower semicontinuous, so

The Rellich-Kondrashov compactness theorem for H01 upgrades the weak convergence tostrongly in $L^2(\Omega )$. Indeed, compactness gives this along every subsequence after passage to a further subsequence, and the weak limit uniquely identifies every such strong limit as $u$. Consequently,Combining the two terms proves the weak lower semicontinuity of a bounded-domain Schrodinger energy:

Suppose the claimed Poincare inequality with a partial Dirichlet boundary fails. There are $v_k\in V$ withAfter the normalization $w_k=v_k/\Vert v_k{\Vert }_{L^2(U)}$,Thus $(w_k)$ is bounded in the Sobolev space $H^1(U)$. The Rellich-Kondrashov compactness theorem and the corresponding compact embedding for a bounded $C^1$ domain give a subsequence that converges strongly in $L^2(U)$ and weakly in $H^1(U)$ to some $w$. The Sobolev space with a partial Dirichlet condition $V$ is a closed vector subspace, hence weakly closed, so $w\in V$. Moreover $\nabla w=0$, and connectedness of $U$ makes $w$ a constant function. Its trace vanishes on the positive-measure set ${\Gamma }_1$, so that constant is zero. This contradictsTherefore some $C_P$ satisfiesSince the reverse bound $\Vert \nabla v{\Vert }_2\le \Vert v{\Vert }_{H^1}$ is immediate,The gradient seminorm is a norm on $V$ because equality to zero would make $v$ a constant whose trace on ${\Gamma }_1$ is zero.

If $U$ is bounded and connected and a boundary portion ${\Gamma }_D$ has positive surface measure, thenfor every $v\in H^1(U)$ whose trace vanishes on ${\Gamma }_D$. Otherwise a normalized counterexample sequence, the Rellich-Kondrashov compactness theorem, and continuity of the trace would produce a nonzero constant with zero trace on ${\Gamma }_D$.