Time derivatives of fixed and differentiable comparison projections

Let $g_1,g_2\in L^2(0,T;X)$ both satisfy the distributional derivative identity for $u$. Put $h=g_1-g_2$ and extend it by zero to $\mathbb{R}$. For every fixed $x^{\ast }\in X^{\ast }$ and every interior $t$ farther than $ϵ$ from the endpoints, the scalar pairing of ${\eta }_{ϵ}\ast h(t)$ with $x^{\ast }$ is zero: it is the zero distribution paired with $s\mapsto {\eta }_{ϵ}(t-s)$. Since continuous linear functionals separate points, ${\eta }_{ϵ}\ast h(t)=0$. By [F3] these convolutions converge to $h$ in $L^2$ on compact subintervals; exhausting $(0,T)$ proves $h=0$ a.e., so the weak derivative is unique.

The Bochner integral $G(t)={\int }_0^t{u}^{\mathrm{\prime }}(s)ds$ exists by [F2] and [F5]. For each $x^{\ast }\in X^{\ast }$, scalar Fubini [F4] and integration by parts show that $G$ has weak derivative $u^{\mathrm{\prime }}$. Hence $w=u-G$ has zero distribution derivative. If $\psi \in C_c^{\mathrm{\infty }}(0,T)$ has integral zero, its primitive is compactly supported, so $\int w\psi =0$ in $X$. Choose $\eta \in C_c^{\mathrm{\infty }}(0,T)$ with integral one and put $c=\int w\eta$. Subtracting $(\int \psi )\eta$ shows that $w-c$ has zero distribution. As in step 1.1 its interior convolutions vanish, and [F3] makes them converge to $w-c$ locally in $L^2$. Thus $w=c$ a.e. and $u(t)=c+G(t)$ is a continuous representative. Any two continuous representatives equal a.e. agree everywhere, because their continuous difference vanishes on the dense set where they agree; hence this representative is unique.

For fixed $P$, test the weak derivative identity for $u$ against $y^{\ast }\circ P\in X^{\ast }$ for arbitrary $y^{\ast }\in Y^{\ast }$. Boundedness of $P$ gives ${\int }_0^T{y}^{\ast }(Pu(t)){\varphi }^{\mathrm{\prime }}(t)dt=-{\int }_0^T{y}^{\ast }(P{u}^{\mathrm{\prime }}(t))\varphi (t)dt\mathrm{.}$ The right side is represented by the $L^2(Y)$ function $P{u}^{\mathrm{\prime }}$, and uniqueness from step 1.1 gives $(Pu{)}^{\mathrm{\prime }}=P{u}^{\mathrm{\prime }}$.

For the operator-norm $C^1$ family, fix $y^{\ast }\in Y^{\ast }$ and put $\mathrm{\ell }(t)=y^{\ast }\circ P(t)\in X^{\ast }$, so ${\mathrm{\ell }}^{\mathrm{\prime }}(t)=y^{\ast }\circ P^{\mathrm{\prime }}(t)$. Use the representative $u(t)=c+{\int }_0^t{u}^{\mathrm{\prime }}(s)ds$ from step 2.1. For $\varphi \in C_c^{\mathrm{\infty }}(0,T)$, scalar Fubini and integration by parts give ${\int }_0^T\mathrm{\ell }(t)u(t){\varphi }^{\mathrm{\prime }}(t)dt=-{\int }_0^T({\mathrm{\ell }}^{\mathrm{\prime }}(t)u(t)+\mathrm{\ell }(t)u^{\mathrm{\prime }}(t))\varphi (t)dt\mathrm{.}$ Indeed, substitute $c+{\int }_0^t{u}^{\mathrm{\prime }}(s)ds$ on the left, interchange the integrals over $0\le s\le t\le T$, and use ${\int }_s^T\mathrm{\ell }(t){\varphi }^{\mathrm{\prime }}(t)dt=-\mathrm{\ell }(s)\varphi (s)-{\int }_s^T{\mathrm{\ell }}^{\mathrm{\prime }}(t)\varphi (t)dt$; a second Fubini interchange combines the ${\mathrm{\ell }}^{\mathrm{\prime }}$ terms into ${\mathrm{\ell }}^{\mathrm{\prime }}(t)u(t)$. Absolute integrability follows from boundedness of $\mathrm{\ell },{\mathrm{\ell }}^{\mathrm{\prime }},\varphi ,{\varphi }^{\mathrm{\prime }}$ and [F5]. Since this identity holds for all $y^{\ast }$ and $P^{\mathrm{\prime }}u,P{u}^{\mathrm{\prime }}\in L^2(0,T;Y)$, the vector-valued weak derivative is $P^{\mathrm{\prime }}u+P{u}^{\mathrm{\prime }}$.

The displayed derivative belongs to $L^2(0,T;Y)$ because $P$ and $P^{\mathrm{\prime }}$ are bounded on $[0,T]$ and $u,u^{\mathrm{\prime }}\in L^2(0,T;X)$. Hence $P(t)u(t)\in H^1(0,T;Y)$ with the asserted formula.