Implication of weak-$^*$ convergence in $L^\infty(0,T;V)$

Question

I've got a (hopefully) small problem with the following setting: Let $V,H,V'$ be a Gelfand-Triple ($V$ be seperable and reflexive), $\lbrace u_j \rbrace_j \in L^\infty(0,T;V)$ and $u_j(0) = u_{0j}$ be bounded with

\begin{align} u_j \to u \text{ weakly star in } L^\infty(0,T;V). \end{align}

Then $u_j(0) \to u_{0} =u(0)$ weakly in $V$.

Because of the boundness of $u_{0j}$ we can extract a weakly converging subsequence. But why is the limit $u_0$?

Thanks in advance, FFoDWindow

Answer 1

This does not hold. Take $V = H = \mathbb{R}$ and $$u_j(t) = \begin{cases} j \, t & \text{if } t < 1/j, \\ 1 &\text{else}.\end{cases}$$ Then, $u_j \to 1$ weakly in $L^\infty(0,1)$ but $u_j(0) = 0 \not\to 1 = u(0)$.

By the way, for arbitrary $u \in L^\infty(0,T;V)$, the value $u(0)$ is not defined.