about weak convergence in $L^{2}(0,T;H)$

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 $(T/2)f$.