Evans' PDE: Details in proving the improved regularity of weak solution to parabolic equation

Below comes from Theorem 5 (Improved regularity of parabolic equation) of Chapter 7.1 (page 383-384) in Evans' PDE, and I will paraphrase my question:

Suppose we have a sequence of solutions to projection problem $$u_m$$ such that $$u_m \to u$$ weakly in $$L^2(0,T; H^1_0(U))$$ and $$u'_m \to u'$$ weakly in $$L^2(0,T; H^{-1}(U))$$ (those are weak convergence, but I don't know how to type the weak convergence arrow). Suppose further we obtain a uniform bound: $$\sup_{0\leq t \leq T} \|u_m\|_{H^1_0}^2 \leq C(\|g\|_{H^1_0}^2 + \|f\|_{L^2(0,T;L^2(U)}^2),$$ then we have $$\|u\|_{L^\infty(0,T;H^1_0)}^2 \leq C(\|g\|_{H^1_0}^2 + \|f\|_{L^2(0,T;L^2(U)}^2)$$.

Since we have only weak convergence, why do we have such pointwise bound? Could anyone give me some hint on this?

Edit: It should be essential supremum instead of supremum, for the argument given in the below answer.

• That is Problem 6 in Evan's textbook with a hint. Have you tried to solve the problem? Jul 21, 2020 at 14:42
• The problem is easier than the answer below. You have stronger than weak convergence, since he asks you to assume that $u_m$ is uniformly bounded in $H^1_0$. A general hint is to use this to get that $u_m$ converges weakly in $H^1$ and use lower-semicontinuity of the $H^1_0$ norm.
– Jeff
Jul 21, 2020 at 15:03

First I feel like we are missing some information here. With the regularity of $$u_m$$ and $$u$$ you stated, namely $$u, u_m \in \mathcal{W}(0,T, H^1_0, H^{-1})$$ the term $$\sup\limits_{0 \leq t \leq T} \| u_m \|_{H^1_0}^2$$ (for which I assume you mean $$\sup\limits_{0 \leq t \leq T} \| u_m(t) \|_{H^1_0}^2$$) doesn't make sense since you only have the embedding \begin{align} \mathcal{W}(0,T, H^1_0, H^{-1}) \hookrightarrow \mathcal{C}([0,T], L^2). \end{align} and not an embedding \begin{align} \mathcal{W}(0,T, H^1_0, H^{-1}) \hookrightarrow \mathcal{C}([0,T], H^1_0). \end{align} So for that reason let us switch the supremum for the essential supremum. Now let us fix a measurable set $$\Xi \subseteq (0,T)$$. From the convergence of $$u_m$$ we then especially have \begin{align} u_m \rightharpoonup u \,\,\, \text{ in } \,\,\, L^2(\Xi, H^1_0). \end{align} From the weak sequential lower semicontinuity of the norm in $$L^2(\Xi, H^1_0)$$ we find \begin{align}\tag{1} \| u \|_{L^2(\Xi, H^1_0)}^2 \leq \liminf_{m} \| u_m \|_{L^2(\Xi, H^1_0)}^2 \leq |\Xi| C \left( \| g \|_{H^1_0}^2 + \| f \|_{L^2(0,T;L^2)}^2 \right). \end{align} Assuming there exists some measurable set $$M \subseteq (0,T)$$ with $$|M| >0$$ and such that \begin{align} \| u(t) \|_{H^1_0}^2 > C \left( \| g \|_{H^1_0}^2 + \| f \|_{L^2(0,T;L^2)}^2 \right) \quad \forall t \in M \end{align} we can take $$\Xi = M$$ in (1) and get an immediate contradiction. (Actually we could have straight away started with the set $$M$$ but whatever)
We can now at least conclude \begin{align} \underset{t \in (0,T)}{\mathrm{esssup}} \| u(t) \|_{H^1_0}^2 \leq C \left( \| g \|_{H^1_0}^2 + \| f \|_{L^2(0,T;L^2)}^2 \right). \end{align}
What do you think? Another approach would be to look for more information on $$u_m, u$$ which would grant $$u_m(t) \to u(t)$$ in $$H^1_0$$ for almost any $$t \in (0,T)$$. One could then use the weak sequential lower semicontinuity of the norm in $$H^1_0$$ instead of $$L^2(0,T;H^1_0)$$. With the given information I only see how one could maybe prove $$u_m(t) \to u(t)$$ in $$L^2$$ for almost any $$t \in (0,T)$$. I might be missing something though.