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.

  • $\begingroup$ There seems to be a type in the definition of $\mathcal{H}$. $\endgroup$ May 19, 2022 at 22:03
  • $\begingroup$ What do you mean? $\endgroup$
    – Marcy
    May 19, 2022 at 22:09
  • $\begingroup$ $C(O\wedge T)\vee L(O\wedge T) = C(O\vee T)\wedge L(O\vee T)$ $\endgroup$
    – Barb
    May 19, 2022 at 22:16
  • $\begingroup$ You write $\mathcal{H} = H^1(0,T,X)$. What is $X$? $X = H^1_0(\Omega)$ or $X = H^{-1}(\Omega)$? $\endgroup$ May 19, 2022 at 22:17
  • 1
    $\begingroup$ I think there is a difference in notations we are using. I'm not sure what the equivalent is in your notation. The definition I've been taught: for $u\in\mathcal{H}$ we have that $u(t)\in H_0^1(\Omega), \partial_tu(t)\in H^{-1}(\Omega), $ for all $t\in[0,T]$ $\endgroup$
    – Marcy
    May 19, 2022 at 23:22

1 Answer 1


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 $$ T \|u(t)\|_{L^\infty(L^2)}^2\le T\|u_t\|_{L^2(H^{-1})} \|u\|_{L^2(H^1)} + \|u\|_{L^2(L^2)}^2 \le T\|u_t\|_{L^2(H^{-1})} \|u\|_{L^2(H^1)} + \|u\|_{L^2(H^1)}^2 . $$

  • $\begingroup$ Can you clarify how does the thing on the left relates to the $\infty$-norm? If we take the supremum wrt $t$ we obtain $T||u||^2_{L^\infty(L^2)} - ||u||^2_{L^2(L^2)}$. If I remember correctly $L^\infty$ is embedded in $L^2$ because we have a bounded domain, but I cannot see any way to control the injection constant in a useful way here since this must hold for arbitrary $u$. $\endgroup$
    – Marcy
    May 20, 2022 at 7:51
  • $\begingroup$ see edit................ $\endgroup$
    – daw
    May 20, 2022 at 8:57

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