1
$\begingroup$

Given a sequence $(u_n)_{n\in \mathbb{N}} \subseteq L^\infty (0,T; L^2(\Omega)) \cap H^1(0,T;L^2(\Omega))$ with

\begin{align*} u_n \overset{\ast}{\rightharpoonup} u \,\, \text{ in } \,\, L^\infty (0,T; L^2(\Omega)) \end{align*}

where $T>0$ and $\Omega \subseteq \mathbb{R}^{42}$ is an open set, does the inequality

\begin{align*} \liminf_{n \to \infty} \|u_n(T)\|_{L^2(\Omega)} \geq \|u(T)\|_{L^2(\Omega)} \end{align*}

hold? I am thinking about the weak lower semicontinuity of $\|\cdot \|_{L^2(\Omega)}$, but for this I would need weak convergence of $(u_n(T))_{n\in \mathbb{N}}$ in $L^2(\Omega)$ which feels awkward.

Note that evaluating $u_n$ at the point $T$ makes sense because one has the embedding

\begin{align*} H^1(0,T;L^2(\Omega)) \hookrightarrow \mathcal{C}([0,T],L^2(\Omega)). \end{align*}

I am happy about any kind of help. Thanks in advance.

$\endgroup$
  • 1
    $\begingroup$ The evaluation of $u(T)$ does not make sense, since $u \in L^\infty(0,T;L^2(\Omega))$ only. And weak-* in $L^\infty$ is not enough, even if you replace $L^2(\Omega)$ by $\mathbb R$. $\endgroup$ – gerw Apr 8 '19 at 19:55
  • $\begingroup$ Thank you very much for your comment. It made me realize that I forgot an assumption. Below I tried to write down a proof given the new assumption. I would be very thankful if you could give it a short look. $\endgroup$ – neca Apr 9 '19 at 8:58
2
$\begingroup$

I will use the additional assumption (mentioned in your answer):

The sequence $(u_n')_{n \in \mathbb{N}} $ is bounded in $L^\infty(0,T;L^2(\Omega))$

Under this assumption, we can check that $u_n' \stackrel*\rightharpoonup u'$ in $L^\infty(0,T;L^2(\Omega))$.

Now, we have the identity $$ u_n(T) = u_n(t) + \int_t^T u_n'(s) \, \mathrm{d}s. $$ Integration over $t$ implies $$ T \, u_n(T) = \int_0^T u_n(t) + \int_t^T u_n'(s) \, \mathrm{d}s \, \mathrm{d}t = \int_0^T u_n(t) \, \mathrm{d}t + \int_0^T s \, u_n'(s) \, \mathrm{d} s. $$ From the weak-* convergence of $u_n$ and $u_n$, we can infer $$ u_n(T) \rightharpoonup u(T) $$ in $L^2(\Omega)$. This implies the desired inequality.

$\endgroup$
  • $\begingroup$ Thank you very much! $\endgroup$ – neca Apr 11 '19 at 6:35
0
$\begingroup$

Apparently I have an additional assumption:

The sequence $(u_n')_{n \in \mathbb{N}} $ is bounded in $L^\infty(0,T;L^2(\Omega))$

Having this assumption I will make a try of proving my desired inequality.

Given that the above sequence is bounded we have $u_n' \overset{\ast}{\rightharpoonup} v$ in $L^\infty(0,T;L^2(\Omega))$ for some $v \in L^\infty(0,T;L^2(\Omega))$. It then follows that $u \in H^1(0,T;L^2(\Omega))$ with $u'=v$. So now the evaluation of $u$ in the point $T$ makes sense because we can again use the embedding into the space of continuous functions.

The weak star convergence of $(u_n)_{n \in \mathbb{N}}$ and $(u_n')_{n \in \mathbb{N}}$ in $L^\infty(0,T;L^2(\Omega))$ implies weak convergence of $(u_n)_{n \in \mathbb{N}}$ and $(u_n')_{n \in \mathbb{N}}$ in $L^2(0,T;L^2(\Omega))$. I would now assume that \begin{align*} u_n \rightharpoonup u \,\,\text{ in }\,\, H^1(0,T;L^2(\Omega)) \end{align*} holds. I know that it holds in the real valued Sobolev space case and I guess that the proof can easily be adapted to the Bochner setting.

Now the compact embedding $H^1(0,T;L^2(\Omega)) \overset{c}{\hookrightarrow} L^2(0,T;L^2(\Omega))$ (here we should maybe upgrade $\Omega$ to be a bounded Lipschitz domain) implies \begin{align*} u_n \rightarrow u \,\,\text{ in }\,\, L^2(0,T;L^2(\Omega)) \end{align*} up to a subsequence. Finally we get \begin{align*} u_n(t) \rightarrow u(t) \,\,\text{ in }\,\, L^2(\Omega) \,\,\text{ for almost any t $\in$ [$0$,$T$]} \end{align*} up to another subsequence and because of the continuity of $u$ and $u_n$ for every $n \in \mathbb{N}$ we get the convergence for all $t \in [0,T]$. Hence especially for $T$ so that the desired inequality even holds with equality.

$\endgroup$
  • 1
    $\begingroup$ The embedding from $H^1(0,T;L^2(\Omega))$ to $L^2(0,T;L^2(\Omega))$ is not compact. Consider $v_n \rightharpoonup v$ in $L^2(\Omega)$ (without strong convergence). Then, we define $u_n(t,x) := v_n(x)$. This is bounded in the first space but fails to converge in the second space. $\endgroup$ – gerw Apr 9 '19 at 10:59
  • $\begingroup$ That is unfortunate. Do you have an idea where to go from the weak convergence in $H^1(0,T;L^2(\Omega))$? $\endgroup$ – neca Apr 9 '19 at 13:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.