Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 105 / 1 / b

With the standard Sobolev space inner product, $u_i\rightharpoonup u$ weakly in $H^1(U)$ meansfor every $v\in H^1(U)$. Equivalently, every bounded linear functional on $H^1(U)$ takes convergent values on the sequence.

The Sobolev space $H^1(U)$ is a separable Hilbert space. A bounded sequence therefore has a weakly convergent subsequence by the weak subsequence of a bounded Hilbert-space sequence; write $u_{i_j}\rightharpoonup u$ in $H^1(U)$. Since $U$ is bounded with smooth boundary, the Rellich-Kondrachov compactness theorem says that $H^1(U)\hookrightarrow L^2(U)$ is compact. Passing to a further subsequence gives