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I am trying to do an exercise and if the affirmation below is true, my exercise is done .

This is the affirmation :

Affirmation : Let $H$ a Hilbert space and suppose $u_k$ converges weakly to $u$ in $L^{2}(0,T;H)$.

Suppose that $\operatorname{ess sup}_{ 0 \leq t \leq T} \ || u_k (t)|| \leq C$. Then $u_k(t) $ converges weakly to $u(t)$ for every $t$.

I am trying to do, but nothing... . Someone can give me a hint ? (please dont give me a complete answer , just a hint ^^ )

thanks in advance ^^

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  • $\begingroup$ Do you need the convergence in the whole sequence, or only a subsequence is fine? $\endgroup$ – Tomás Jun 22 '13 at 11:22
  • $\begingroup$ @Tomás , a subsequence is fine. Do you know how to prove ? $\endgroup$ – math student Jun 22 '13 at 13:16
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Let $H=L^2(\Omega)$ where $\Omega\subset\mathbb{R^N}$ is a bounded domain and take $f\in H$ with $f\neq 0$. Let $e_k$ be a orthonormal basis of $H$. Define $u:(0,T)\to H$ by

$$ u_k(t) = \left\{ \begin{array}{ccc} tf &\mbox{ if $t\in (0,T/2)\cup(T/2,1)$} \\ e_k &\mbox{ if $t=T/2$ } \end{array} \right. $$

Note that $\int_0^T \|u_k(t)\|_2^2dt=\int_0^Tt^2\|f\|_2^2dt<\infty$, then $u_k\in L^2(0,T,H)$. Moreover, $$\int_0^T(u_k(t),g(t))dt=\int_0^T(tf,g(t))dt,\ \forall\ g\in L^2(0,T,H)$$

Hence, $u_k\to ft$ weakly in $L^2(0,T,H)$. Also, we have that (unless for $t=T/2$) $\|u_k(t)\|_2=\|tf\|_2\leq C$, where $C$ does not depends on $t$ or $k$, which implies that $$\operatorname{ess}\sup_{0\leq t\leq R}\|u_k(t)\|_2\leq C$$

To conclude, we note that $u_k(T/2)$ does not converge weakly to $(T/2)f$.

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The Hilbert space $H$ here is a red herring; this is false even when $H = \mathbb{R}$ (so we are just talking about $L^2([0,T])$.)

Indeed, as it stands it is false even if you have strong convergence in $L^2([0,T])$. There is a standard example of a sequence of functions $u_k : [0,1] \to [0,1]$ which converge in measure to 0, but for which $u_k(t)$ diverges for every $t$. See, for instance, Example 4 of these notes of Terry Tao, where he calls it the "typewriter sequence". By dominated convergence, we also have $u_k \to 0$ in $L^2$.

Strong $L^2$ convergence does imply almost everywhere convergence of a subsequence, but weak $L^2$ convergence does not. Consider $L^2([-\pi,\pi], \mathbb{C})$ and let $u_k(x) = \frac{1}{\sqrt{2\pi}}e^{ikx}$. The $u_k$ are an orthonormal sequence in $L^2$, so $u_k \to 0$ weakly in $L^2$. If there were a subsequence $u_{k_j}$ with $u_{k_j} \to 0$ almost everywhere, then by dominated convergence we would also have $u_{k_j} \to 0$ in $L^2$. But $\|u_k\|_{L^2} = 1$ for every $k$, so this is absurd.

Generally speaking, $L^2$ convergence cannot tell you much about pointwise convergence, and weak $L^2$ convergence tells you even less.

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So if we have $u_n\in L^2(0,T;L^2(\Omega))\cap L^\infty(0,T;H^{-1}(\Omega))$ and $u_n\to u$ in $L^2(0,T;H^{-1}(\Omega))$, can we deduce the weak convergence of subsequence $u_{n_k}(t)$ to $u(t)$ in $L^2(\Omega)$ for a.e $t$? $\Omega$ is bounded domain in $\mathbb R^3$.

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