Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 23H (Analysis of Functions)

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 $L^2(\Omega )$.

No. Choose a nonzero $\phi \in C_c^{\infty }(\mathbb{R})$ and a number $L$ larger than the diameter of its support, and setTranslation preserves both the $L^2$ norm and the derivative norm, so $(u_n)$ is bounded in $H^1(\mathbb{R})$. Distinct $u_n$ have disjoint supports, and henceThus the sequence has no $L^2$-Cauchy subsequence and therefore no strongly convergent subsequence. The embedding $H^1(\mathbb{R})\hookrightarrow L^2(\mathbb{R})$ is not compact.

Fix $R>0$ and choose ${\chi }_R\in C_c^{\infty }((-R-1,R+1))$ with ${\chi }_R=1$ on $[-R,R]$. The sequence $({\chi }_R{f}_n)$ is bounded in $H_0^1((-R-1,R+1))$. By part (a), each subsequence has a further subsequence converging strongly in $L^2$ on this bounded interval. The original weak convergence in $L^2(\mathbb{R})$ forces every such strong limit to be zero. It follows that the whole sequence satisfiesotherwise a subsequence bounded away from zero would contradict the preceding compactness argument.

The pointwise hypothesis gives a uniform tail estimate:whose right-hand side tends to zero as $R\to \infty$, independently of $n$. Given $\epsilon >0$, first choose $R$ so that this tail is below $\epsilon /2$, and then choose $n$ so that the integral on $[-R,R]$ is below $\epsilon /2$. Therefore