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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

Let $\Vert x_n\Vert \le C$. For each $i$, Bessel inequality givesso the $i$th coordinate sequence is bounded. By the Bolzano-Weierstrass theorem, choose a subsequence on which the first coordinate converges. From it choose a further subsequence on which the second coordinate converges, and continue.

Take the diagonal subsequence $(x_{\varphi (n)})$. For every fixed $i$, its $i$th coordinate converges, and it remains bounded by $C$. The coordinate criterion for weak convergence in a separable Hilbert space supplies an $x_{\infty }\in H$ such thatThis proves the theorem on a weak subsequence of a bounded Hilbert-space sequence.

Choose $h_j\to 0$. The assumed bound and the weak subsequence of a bounded Hilbert-space sequence result give a subsequence for whichFor every test function $\phi \in C_c^{\infty }(V)$, a change of variables yields the difference-quotient integration-by-parts identityThe right side converges to $-{\int }_{V}u{D}_i\phi$, while the left side converges to ${\int }_V{g}_i\phi$. Thus $g_i$ is the $i$th weak derivative of $u$. This holds for every $i$, so