The Banach evolution-space energy identity and H trace

Fix a compact interval $K\subset (0,T)$ and a smooth unit-mass kernel ${\rho }_{ϵ}$ supported in $(-ϵ,ϵ)$, with $ϵ<\mathrm{dist}(K,\{0,T\})$. Set $g=u^{\mathrm{\prime }}$ and $u_{ϵ}(t)={\int }_0^T{\rho }_{ϵ}(t-r)u(r)dr$. This is smooth in $V$: difference quotients of the kernel and each of its derivatives converge uniformly on the compact integration region, and multiplying their uniform error by $\mathrm{\parallel }u{\mathrm{\parallel }}_{L^1(V)}<\mathrm{\infty }$ bounds the error in the differentiated Bochner integral. For $v\in V$, use the scalar test $r\mapsto {\rho }_{ϵ}(t-r)$ in [F1]; since its derivative is $-{\rho }_{ϵ}^{\mathrm{\prime }}(t-r)$, this gives $⟨Ji{\dot{u}}_{ϵ}(t),v⟩={\int }_0^T{\rho }_{ϵ}^{\mathrm{\prime }}(t-r)(iu(r),iv{)}_{H}dr={\int }_0^T{\rho }_{ϵ}(t-r)⟨g(r),v⟩dr\mathrm{.}$ Thus the pivot derivative $u_{ϵ}^{\mathrm{\prime }}:=Ji{\dot{u}}_{ϵ}$ equals ${\rho }_{ϵ}\ast g$ in $V^{\ast }$. The ordinary $H$ product rule now yields $\frac{d}{dt}\frac{1}{2}\mathrm{\parallel }i{u}_{ϵ}(t){\mathrm{\parallel }}_H^2=⟨u_{ϵ}^{\mathrm{\prime }}(t),u_{ϵ}(t)⟩\mathrm{.}$ Applying [F2] separately with $(X,r)=(V,p)$ and $(V^{\ast },p^{\mathrm{\prime }})$ proves $u_{ϵ}\to u$ in $L^p(K;V)$ and $u_{ϵ}^{\mathrm{\prime }}\to u^{\mathrm{\prime }}$ in $L^{p^{\mathrm{\prime }}}(K;V^{\ast })$. Below, $V$-valued functions in $H$ expressions are identified with their images under $i$.

For two small radii $ϵ,\delta$, put $w=u_{ϵ}-u_{\delta }$ on a fixed nondegenerate compact interval $K\subset (0,T)$. Continuity gives a time $s\in K$ with $\mathrm{\parallel }w(s){\mathrm{\parallel }}_H^p\le \mathrm{\mid }K{\mathrm{\mid }}^{-1}\mathrm{\parallel }w{\mathrm{\parallel }}_{L^p(K;H)}^p$: otherwise integration contradicts the definition of that norm. Apply the smooth energy identity of step 1.1 to the difference $w$ and integrate between $s$ and any $t\in K$. Hölder [F4] yields ${\sup }_{t\in K}\mathrm{\parallel }w(t){\mathrm{\parallel }}_H^2\le \mathrm{\mid }K{\mathrm{\mid }}^{-2\mathrm{/}p}\mathrm{\parallel }w{\mathrm{\parallel }}_{L^p(K;H)}^2+2\mathrm{\parallel }{w}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^{p^{\mathrm{\prime }}}(K;V^{\ast })}\mathrm{\parallel }w{\mathrm{\parallel }}_{L^p(K;V)}⟶0.$ Indeed [F2] makes both $w$ and $w^{\mathrm{\prime }}$ tend to zero in their indicated norms, and the continuous inclusion [F3] controls the $H$-valued $L^p$ norm. This holds for every $p>1$ without a prior $L^{\mathrm{\infty }}(H)$ bound. Thus the mollifications are uniformly Cauchy in $H$ on $K$.

Completeness of $H$ gives a uniform limit in $C(K;H)$, which agrees almost everywhere with $u$ because $u_{ϵ}\to u$ in $L^p(K;H)$. The limits agree on overlapping compact intervals, so they define $u\in C((0,T);H)$. Integrate step 1.1 between arbitrary $s,t\in K$. Uniform convergence passes the endpoint norms to those of $u$, while $\mathrm{\parallel }⟨u_{ϵ}^{\mathrm{\prime }},u_{ϵ}⟩-⟨u^{\mathrm{\prime }},u⟩{\mathrm{\parallel }}_{L^1(K)}\le \mathrm{\parallel }{u}_{ϵ}^{\mathrm{\prime }}-u^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^{p^{\mathrm{\prime }}}(V^{\ast })}\mathrm{\parallel }{u}_{ϵ}{\mathrm{\parallel }}_{L^p(V)}+\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^{p^{\mathrm{\prime }}}(V^{\ast })}\mathrm{\parallel }{u}_{ϵ}-u{\mathrm{\parallel }}_{L^p(V)}\to 0.$ Hence for all $0<s\le t<T$, $\mathrm{\parallel }u(t){\mathrm{\parallel }}_H^2-\mathrm{\parallel }u(s){\mathrm{\parallel }}_H^2=2{\int }_s^t⟨u^{\mathrm{\prime }}(r),u(r)⟩dr\mathrm{.}$

By [F4], the energy pairing is in $L^1(0,T)$. The identity in step 3.1 therefore shows that $\mathrm{\parallel }u(t){\mathrm{\parallel }}_H^2$ has finite limits as $t\downarrow 0$ and $t\uparrow T$. For each fixed $v\in V$, the distributional derivative of $(u(t),iv{)}_H$ is $⟨u^{\mathrm{\prime }}(t),v⟩\in L^1(0,T)$, so [F5] gives an absolutely continuous representative on $[0,T]$ agreeing with the interior pairing.

For $0<\delta <T$, choose a good time $s_{\delta }\in (0,\delta )$ with $u(s_{\delta })\in V$ and $\mathrm{\parallel }u(s_{\delta }){\mathrm{\parallel }}_V^p\le \frac{2}{\delta }{\int }_0^{\delta }\mathrm{\parallel }u(r){\mathrm{\parallel }}_V^{p}dr\mathrm{.}$ Such a time exists by averaging over $(0,\delta )$. For every interior $t\in (0,\delta )$, use step 3.1 for the two norm squares and [F5] for the scalar pairing against the fixed vector $u(s_{\delta })\in V$. Subtracting twice that pairing identity gives $\mathrm{\parallel }u(t)-u(s_{\delta }){\mathrm{\parallel }}_H^2=2{\int }_{s_{\delta }}^t⟨u^{\mathrm{\prime }}(r),u(r)-u(s_{\delta })⟩dr\mathrm{.}$ Hölder and the triangle inequality bound its absolute value by $2\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^{p^{\mathrm{\prime }}}(0,\delta ;V^{\ast })}(\mathrm{\parallel }u{\mathrm{\parallel }}_{L^p(0,\delta ;V)}+{\delta }^{1\mathrm{/}p}\mathrm{\parallel }u(s_{\delta }){\mathrm{\parallel }}_V),$ which tends to zero as $\delta \downarrow 0$. The bound is uniform in $t\in (0,\delta )$, so $u(t)$ is Cauchy in $H$ as $t\downarrow 0$ and has a strong limit there. The same argument on $(T-\delta ,T)$ gives a strong limit at $T$. Thus $u$ extends to $C([0,T];H)$.

Let $s\downarrow 0$ and $t\uparrow T$ in step 3.1. Strong endpoint continuity and integrability of $⟨u^{\mathrm{\prime }},u⟩$ pass the identity to all $0\le s\le t\le T$. If two continuous representatives agree almost everywhere, their difference is continuous and vanishes on a dense subset, so it vanishes everywhere. This proves uniqueness and the claimed chain rule.