Bounded sequences in a separable Hilbert space have weakly convergent subsequences

Apply Gram–Schmidt to a countable dense sequence in $X$, omitting zero remainders, to obtain a finite or countable orthonormal family $(e_j)$ with dense span. Use the following deterministic extraction for each bounded real sequence $(b_n)$ with $\mathrm{\mid }{b}_n\mathrm{\mid }\le C$. Start with $[-C,C]$; at each bisection retain the left closed half if it contains infinitely many terms, and otherwise the right closed half, which then contains infinitely many terms. From the $k$th retained interval select the least index larger than the preceding selected index whose term lies in that interval. The selected terms lie in nested intervals of diameters tending to zero, hence are Cauchy; [F9] gives a convergent subsequence, so the entire selected sequence converges. This rule determines the extracted indices uniquely, including when $C=0$, and can therefore be recursively applied to successive coordinates $(x_n,e_j{)}_X$ without dependent choice. For finitely many coordinates use the last extracted subsequence; for infinitely many take the diagonal of the nested subsequences, whose indices are strictly increasing and whose tail lies in every earlier subsequence. If the family is empty retain the original sequence. The resulting $(x_{n_k})$ has a limit $a_j={\lim }_k(x_{n_k},e_j{)}_X$ for each coordinate present.

Let $E$ be any separable real Hilbert space. The formula $(f,g{)}_{L^2(E)}={\int }_0^T(f(t),g(t){)}_{E}dt$ is well-defined by pointwise Cauchy–Schwarz and Holder [F4], and its induced norm is the square-integral norm. To prove completeness, let $(f_n)$ be Cauchy. By [A1], choose a subsequence $(f_{n_k})$ with summable successive $L^2(E)$ distances and representatives for that countable subsequence. Put $h_k(t)=\mathrm{\parallel }{f}_{n_{k+1}}(t)-f_{n_k}(t){\mathrm{\parallel }}_E$. Minkowski bounds partial-sum $L^2$ norms by the sums of the norms; monotone convergence [F4] gives $h={\sum }_k{h}_k\in L^2(0,T)$. Its infinite-value set is null, since positive measure would force $\int h^2\ge N^2\mathrm{\mid }\{h=\mathrm{\infty }\}\mathrm{\mid }$ for every $N$. Thus $f_{n_k}(t)$ converges in complete $E$ for a.e. $t$ to $f(t)$. The limit is strongly measurable: because $E$ is separable, distances to a countable dense set are measurable; first-dense-point approximants are countably valued, and finite truncations with summable exceptional-set measures converge a.e. by [F5]–[F6]. The pointwise tail estimate and Minkowski give $\mathrm{\parallel }f-f_{n_m}{\mathrm{\parallel }}_{L^2(E)}\le {\sum }_{k\ge m}\mathrm{\parallel }{f}_{n_{k+1}}-f_{n_k}{\mathrm{\parallel }}_{L^2(E)}\to 0.$ The original Cauchy sequence then converges to $f$. This proves that $L^2(0,T;E)$ is Hilbert.

Finite-valued simple functions are dense in $L^2(0,T;E)$: take the pointwise simple approximants from [F1], replace values by zero where their norm exceeds twice the target norm, and use dominated convergence [F4]; the squared error is at most nine times the target's squared norm. Approximate the finitely many values by a countable dense subset of $E$, and each scalar indicator by the countable dense family [F3]. Since $(0,T)$ has finite measure, these approximations produce a countable dense family of finite sums in $L^2(0,T;E)$. With [F2], this proves that both $L^2(0,T;V)$ and $L^2(0,T;V^{\ast })$, and therefore their product, are separable Hilbert spaces.

For each $m$, orthogonality gives ${\sum }_{j=1}^m{a}_j^2={\lim }_k{\sum }_{j=1}^m(x_{n_k},e_j{)}_X^2\le C^2$. Thus $(a_j)$ is square-summable, so its partial sums form a Cauchy sequence; completeness [F8] gives the Hilbert-series limit $x={\sum }_j{a}_j{e}_j$ in $X$.

Inner products $(x_{n_k},y{)}_X$ converge to $(x,y{)}_X$ when $y$ lies in the finite span of the $e_j$, by coordinate convergence. For arbitrary $y\in X$, approximate it by such a finite-span vector $y_m$ using the dense span from step 1.1; the uniform bound gives $\mathrm{\mid }(x_{n_k}-x,y-y_m{)}_X\mathrm{\mid }\le (C+\mathrm{\parallel }x{\mathrm{\parallel }}_X)\mathrm{\parallel }y-y_m{\mathrm{\parallel }}_X$. First choose $m$ large, then $k$ large. Therefore $x_{n_k}\rightharpoonup x$.

The derivative graph is closed in the product: if $(u_n,g_n)\to (u,g)$ strongly in both factors, continuous embedding $V\to H$ and Cauchy–Schwarz let every scalar test identity [F1] pass to the limit. Thus the graph is a closed Hilbert subspace. It is weakly closed: a weak limit is orthogonal to the graph's orthogonal complement, so [F7] gives $(S^{\perp }{)}^{\perp }=S$ for that closed subspace. Apply steps 1.1–3.1 to the separable product Hilbert space. Every bounded graph sequence has a weakly convergent subsequence by that argument, and its limit remains in the graph. This proves the asserted weak subsequential compactness in $W(0,T;V)$.