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For each real $a$, define the sublevel set $K_a=\{u\in K:F(u)\le a\}$. Continuity of $F$ and closure of $K$ make $K_a$ a closed set in $X$; the convex function inequality makes it a convex set. The preceding result therefore makes every $K_a$ a sequentially weakly closed set. Equivalently, $F$ has weak lower semicontinuity along sequences in $K$.

Every $K_a$ is bounded. This is immediate if $K$ is bounded. Otherwise, an unbounded $K_a$ would contain $u_n$ with $\Vert u_n\Vert \ge n$, contradicting the assumed coercive function condition $F(u_n)\to \infty$.

Choose $z\in K$ and put $\ell ={\inf }_{K}F\in [-\infty ,F(z)]$. Choose a minimizing sequence $(u_n)$ entirely in $K_{F(z)}$, with $F(u_n)\to \ell$; when $\ell =-\infty$, require $F(u_n)\le -n$ for sufficiently large $n$. It is bounded by the previous paragraph.

The weak subsequence of a bounded Hilbert-space sequence theorem gives $u_{n_j}\rightharpoonup u_0$. This applies even if $X$ is not separable: the closed linear span of the sequence is a separable Hilbert space; successive extraction makes every coefficient in a countable Hilbertian basis converge, the common norm bound gives square-summable limiting coefficients by the Bessel inequality, and approximation by finite linear combinations proves weak convergence. Orthogonal decomposition extends that conclusion to all test vectors in $X$. Since $K$ is a sequentially weakly closed set, $u_0\in K$.

For any real $a>\ell$, a tail of $(u_{n_j})$ lies in $K_a$, so $u_0\in K_a$. If $\ell =-\infty$, this would give $F(u_0)\le a$ for every real $a$, impossible because $F$ is real-valued. Thus $\ell$ is finite; letting $a\downarrow \ell$ gives $F(u_0)\le \ell$, while the definition of infimum gives the reverse inequality. ThereforeThis proves minimum of a coercive convex functional on a Hilbert space by the direct method in the calculus of variations, with bounded sublevel sets supplying the required weak sequential compactness in a Hilbert space.

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