# Proof of the embedding of time dependent Sobolev spaces

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.

• There seems to be a type in the definition of $\mathcal{H}$. May 19, 2022 at 22:03
• What do you mean? May 19, 2022 at 22:09
• $C(O\wedge T)\vee L(O\wedge T) = C(O\vee T)\wedge L(O\vee T)$
– Barb
May 19, 2022 at 22:16
• You write $\mathcal{H} = H^1(0,T,X)$. What is $X$? $X = H^1_0(\Omega)$ or $X = H^{-1}(\Omega)$? May 19, 2022 at 22:17
• 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 $t\in[0,T]$ May 19, 2022 at 23:22

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 $$T \|u(t)\|_{L^\infty(L^2)}^2\le T\|u_t\|_{L^2(H^{-1})} \|u\|_{L^2(H^1)} + \|u\|_{L^2(L^2)}^2 \le T\|u_t\|_{L^2(H^{-1})} \|u\|_{L^2(H^1)} + \|u\|_{L^2(H^1)}^2 .$$
• 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 $u$. May 20, 2022 at 7:51