Let $\Omega$ be a bounded Lipschitz domain in $\mathbb{R}^n$. $H^{-1}(\Omega)$ is the dual of $H_0^1(\Omega)$. For shorthand I write $\mathcal{H} = H^1(0,T,H_0^1(\Omega),H^{-1}(\Omega))$. I want to prove the existence of $C>0$ such that for all $u\in\mathcal{H}$
$$||u||_{L^\infty(0,T,L^2(\Omega))} \le C||u||_\mathcal{H}$$
I've proven the lemma (which I was hinted) that the derivative of $||u(t)||^2_{L^2(\Omega)}$
$$\frac{d}{dt} ||u(t)||^2_{L^2(\Omega)} =\frac{d}{dt} (u(t),u(t))_{L^2(\Omega)} = 2\langle\partial_tu(t), u(t)\rangle_{H^{-1}\times H_0^1}$$
From there I tried the following
\begin{align} ||u||^2_{L^\infty(0,T,L^2(\Omega))} &= \text{esssup}_{t\in[0,T]} ||u(t)||^2_{L^2(\Omega)} \\ &\le 2\int_0^T |\langle\partial_tu(s), u(s)\rangle_{H^{-1}\times H_0^1}| ds + ||u(0)||^2_{L^2(\Omega)} \\ &\le 2 \left(\int_0^T ||\partial_tu(s)||^2_{H^{-1}(\Omega)} ds\right)^{1/2} \left(\int_0^T ||u(s)||^2_{H^1_0(\Omega)} ds\right)^{1/2} + ||u(0)||^2_{L^2(\Omega)} \\ &\le ||\partial_tu||^2_{L^2(0,T,H^{-1}(\Omega))} + ||u||^2_{L^2(0,T,H^1_0(\Omega))} + ||u(0)||^2_{L^2(\Omega)} \\ &= ||u(0)||^2_\mathcal{H} + ||u(0)||^2_{L^2(\Omega)} \end{align}
Which is close to what I want to prove but the last term is not possible to bound for arbitrary $u\in\mathcal{H}$. I've tried to follow the proof in Evans 5.9.2 that $C([0,T])\subset L^\infty([0,T])$ in hopes that it will be similar with no luck. Any help would be appreciated.
From the formula you proved, we get $$ \|u(t)\|_{L^2}^2 - \|u(s)\|_{L^2}^2 = \int_s^t \langle u_t,u\rangle d\tau \le \|u_t\|_{L^2(H^{-1})} \|u\|_{L^2(H^1)} $$ for all $s,t$. Now integrate wrt $s\in (0,T)$ to obtain $$ T\|u(t)\|_{L^2}^2 - \|u\|_{L^2(L^2)}^2 \le T\|u_t\|_{L^2(H^{-1})} \|u\|_{L^2(H^1)}, $$ which is the claim: taking the supremum over $t\in (0,T)$ we get $$ T \|u(t)\|_{L^\infty(L^2)}^2\le T\|u_t\|_{L^2(H^{-1})} \|u\|_{L^2(H^1)} + \|u\|_{L^2(L^2)}^2 \le T\|u_t\|_{L^2(H^{-1})} \|u\|_{L^2(H^1)} + \|u\|_{L^2(H^1)}^2 . $$