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.

  • 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

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.

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

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.

  • 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

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