Convergent Operator, weakly convergent sequence => weakly convergent? Suppose we have a Hilbert space $X$, a weakly convergent sequence $u_k\rightharpoonup u$ and a convergent operator $T_k \rightarrow T$ in the norm of $\mathcal{L}(X)$ (bounded, linear operators).
Is the assertion $T_k u_k \rightharpoonup Tu$ correct?
Thanks in advance!

As noted by @Razieh Noori, if $f\in X^*$, then $f\circ T\in X^*$. Since $u_k\rightharpoonup u \Rightarrow  \exists M:\|u_k\|\leq M\quad \forall k\quad$. Also $T_k\rightarrow T \Rightarrow$ for $\epsilon>0\quad\|T_k-T\|_{op}<\frac{\epsilon}{2 M \|f\|_*}$ for $k>K_1$.
Then
$|f\circ T_ku_k-f\circ T u|\leq |f\circ T_k u_k-f\circ T u_k|+\underbrace{|f\circ Tu_k-f\circ Tu|}_{<\frac{\epsilon}{2}\quad \forall k> K_2}  \leq \|f\|_*\|T_k-T\|_{op}\|u_k\|$ $+\frac{\epsilon}{2}<\|f\|_*\|T_k-T\|_{op}M+\frac{\epsilon}{2}\leq \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\quad$ for  $k>\max\{K_1,K_2\}$ 
