 Solution

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 $L^2(\Omega )$.