Scalar weak equations give a weakly continuous Hilbert-space representative

Choose a countable dense set $D\subseteq H$ by [F4]. For every $d\in D$ and positive rational $r$ for which $B_H(d,r)\cap X\mathrm{\ne }\mathrm{\varnothing }$, choose one point of that intersection using [A1]; the resulting countable set $W\subseteq X$ is dense in $H$. Indeed, given $y\in H$ and $ϵ>0$, choose $d\in D$ with $\mathrm{\parallel }y-d{\mathrm{\parallel }}_H<ϵ\mathrm{/}4$, then a rational $r$ with $\mathrm{\parallel }y-d{\mathrm{\parallel }}_H<r<ϵ-\mathrm{\parallel }y-d{\mathrm{\parallel }}_H$; density of $X$ makes the selected point available and it lies within $ϵ$ of $y$. The rational linear span $W_0$ of $W$ is still countable and dense in $H$.

For $w\in W_0$, set $g_w(t)=(u(t),w{)}_H$ almost everywhere. Since $T<\mathrm{\infty }$ and $u\in L^2(0,T;H)$, $g_w\in L^1(0,T)$; the scalar weak equation implies its distributional derivative is $h_w$. By [F1] it has a unique absolutely continuous representative $G_w$ on $[0,T]$. Integration by parts for $G_w$ and comparison with the given identity give $G_w(0)=(u_0,w{)}_H$. Moreover $\mathrm{\mid }{g}_w(t)\mathrm{\mid }\le M\mathrm{\parallel }w{\mathrm{\parallel }}_H$ almost everywhere, so continuity gives $\mathrm{\mid }{G}_w(t)\mathrm{\mid }\le M\mathrm{\parallel }w{\mathrm{\parallel }}_H$ for every $t\in [0,T]$.

For $w,z\in W_0$ and rational $q$, the identities $g_{w+z}=g_w+g_z$ and $g_{qw}=q{g}_w$ hold almost everywhere. The corresponding identities for the continuous representatives hold everywhere, since two continuous functions equal almost everywhere on $[0,T]$ are equal everywhere. Thus for each fixed $t$, $w\mapsto G_w(t)$ is rational-linear on $W_0$ and bounded by $M\mathrm{\parallel }w{\mathrm{\parallel }}_H$. It extends uniquely by continuity from the dense set $W_0$ to a bounded real-linear functional $L_t$ on $H$; real linearity follows by approximating any real scalar by rationals.

By [F3], there is a unique $\tilde{u}(t)\in H$ with $(\tilde{u}(t),w{)}_H=L_t(w)$ for every $w\in H$. For $w\in W_0$, $t\mapsto (\tilde{u}(t),w{)}_H=G_w(t)$ is continuous, and $\mathrm{\parallel }\tilde{u}(t){\mathrm{\parallel }}_H\le M$. For any $z\in H$, approximate it in norm by $w_k\in W_0$; then ${\sup }_t\mathrm{\mid }(\tilde{u}(t),z-w_k{)}_H\mathrm{\mid }\le M\mathrm{\parallel }z-w_k{\mathrm{\parallel }}_H$, so $t\mapsto (\tilde{u}(t),z{)}_H$ is a uniform limit of continuous functions and is continuous. Since $G_w(0)=(u_0,w{)}_H$ on the dense set $W_0$, Riesz uniqueness gives $\tilde{u}(0)=u_0$. Choose a strongly measurable representative of $u$. For each $w\in W_0$, $(\tilde{u}(t),w{)}_H=(u(t),w{)}_H$ outside a null set; [F5] makes their countable union null. Outside that union the two vectors have the same inner products with the dense set $W_0$, hence are equal. Thus $\tilde{u}=u$ almost everywhere as $H$-valued functions.

If ${\tilde{u}}_1$ and ${\tilde{u}}_2$ are two weakly continuous representatives of $u$, then for every $z\in H$ their scalar pairings are continuous and agree almost everywhere, hence agree for every $t$. Since this holds for every $z$, the Hilbert inner product separates vectors and ${\tilde{u}}_1(t)={\tilde{u}}_2(t)$ for every $t$.