Weak star convergent sequence in $L^\infty(0,T; L^2(\Omega))$

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*}

• 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 $\mathbb R$. – gerw Apr 8 '19 at 19:55
• 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. – neca Apr 9 '19 at 8:58

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.

• Thank you very much! – 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.

• 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. – gerw Apr 9 '19 at 10:59
• That is unfortunate. Do you have an idea where to go from the weak convergence in $H^1(0,T;L^2(\Omega))$? – neca Apr 9 '19 at 13:01