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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.