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.

  • $\begingroup$ That is Problem 6 in Evan's textbook with a hint. Have you tried to solve the problem? $\endgroup$
    – Student
    Jul 21, 2020 at 14:42
  • 2
    $\begingroup$ 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. $\endgroup$
    – Jeff
    Jul 21, 2020 at 15:03

1 Answer 1


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.

  • 1
    $\begingroup$ Thank you! You're correct! My bad, Evans is indeed discussing essential supremum. $\endgroup$
    – mathdoge
    Jul 22, 2020 at 14:51

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