Weak compactness of bounded space-time Sobolev gradients

If $\mathrm{\Omega }=\mathrm{\varnothing }$, the closure definition in [F1] gives $V=\{0\}$, so every $u_j$ and its $H$ image is zero. The assumed strong $L^p(0,T;H)$ limit is therefore $u=0$ even when $H$ is nonzero; taking $U=0$ proves all conclusions.

Assume $\mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }$, put $Q=(0,T)\times \mathrm{\Omega }$, and give $Y_0=(L^p(Q;\mathbb{R}){)}^d$ its maximum product norm. Each scalar coordinate space is reflexive Banach by [F2], so $Y_0$ is Banach: a Cauchy sequence converges in each coordinate and therefore in the maximum norm. A bounded sequence in $Y_0$ has bounded coordinates; apply [F3] successively to the $d$ coordinates to obtain one subsequence weakly converging in every coordinate. A bounded functional on $Y_0$ is the sum of its restrictions to the coordinate embeddings, so this is weak convergence in $Y_0$. The reverse implication in [F3] makes $Y_0$ reflexive. The Euclidean-vector norm of $Y=L^p(Q;{\mathbb{R}}^d)$ and the $Y_0$ product norm are equivalent because the Euclidean norm and the finite-dimensional ${\mathrm{\ell }}^p$ norm on ${\mathbb{R}}^d$ are equivalent. Equivalent norms have the same continuous dual and weak convergence, so $Y$ is reflexive as well.

The gradient map $\mathrm{\Gamma }:V\to L^p(\mathrm{\Omega };{\mathbb{R}}^d)$, $\mathrm{\Gamma }v=\mathrm{\nabla }v$, is an isometry for the chosen norm and has closed range: if $\mathrm{\nabla }{v}_n$ converges in $L^p$, then $(v_n)$ is Cauchy in $V$; [F1] gives a limit in $V$ whose gradient is that limit.

For a finite-valued simple $U(t)={\sum }_{\mathrm{\ell }=1}^q{\mathbf{1}}_{E_{\mathrm{\ell }}}(t)v_{\mathrm{\ell }}$ with $v_{\mathrm{\ell }}\in V$ and disjoint measurable $E_{\mathrm{\ell }}$, apply [F6] to $\mathrm{\Gamma }U(t)={\sum }_{\mathrm{\ell }}{\mathbf{1}}_{E_{\mathrm{\ell }}}(t)\mathrm{\nabla }{v}_{\mathrm{\ell }}$. It supplies the space-time representative and equality of the Bochner and product $L^p$ norms. Simple-function density from [F6] and completeness of $Y$ from [F2] extend this map to an isometry ${\mathrm{\Gamma }}_p:L^p(0,T;V)\to Y$. Its range $S$ is closed: if ${\mathrm{\Gamma }}_p{U}_n$ converges in $Y$, the isometry makes $U_n$ Cauchy in $L^p(0,T;V)$; completeness [F5] gives $U_n\to U$, and then ${\mathrm{\Gamma }}_p{U}_n\to {\mathrm{\Gamma }}_{p}U$. Thus [F4] makes $S$ a reflexive Banach space.

The sequence ${\mathrm{\Gamma }}_p{u}_j$ is bounded in $S$. Reflexivity of $S$ and [F3] give a weakly convergent subsequence with limit $G\in S$. Since $G={\mathrm{\Gamma }}_{p}U$ for some $U\in L^p(0,T;V)$ and ${\mathrm{\Gamma }}_p$ is an isometry, the same subsequence $u_j$ converges weakly to $U$ in $L^p(0,T;V)$; applying any bounded functional after ${\mathrm{\Gamma }}_p^{-1}$ verifies this assertion.

The bounded inclusion in [F5] sends this weak convergence to weak convergence of the $H$-valued images to the image of $U$. By hypothesis those same images converge strongly, hence weakly, to $u$; Hahn–Banach separates distinct points, so $u$ is the image of $U$. Weak convergence in $S$ also implies weak convergence of gradients in $Y$, by restriction of $Y^{\ast }$ functionals to $S$.

The subsequence therefore has the asserted weak $L^p(V)$ limit, that limit has $H$ image $u$, and its space-time gradient is the weak limit $G=\mathrm{\nabla }U$. This proves all conclusions.