# 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?

• Every Hilbert space is reflexive. – uniquesolution Sep 4 '15 at 16:30
• how convergent weak or..? – R.N Sep 4 '15 at 16:34
• Yes, weakly convergent. – fmeyer Sep 4 '15 at 16:36
• Could you give me a short sketch of the proof? – fmeyer Sep 4 '15 at 16:37
• oh in fist i thought it is obvious. but i start to wright it needs challenge – R.N Sep 4 '15 at 16:40

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 $k>K_1$.
$|f\circ T_ku_k-f\circ T u|\leq |f\circ T_k u_k-f\circ T u_k|+\underbrace{|f\circ Tu_k-f\circ Tu|}_{<\frac{\epsilon}{2}\quad \forall k> K_2} \leq \|f\|_*\|T_k-T\|_{op}\|u_k\|$ $+\frac{\epsilon}{2}<\|f\|_*\|T_k-T\|_{op}M+\frac{\epsilon}{2}\leq \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\quad$ for $k>\max\{K_1,K_2\}$
ok suppose $f\in{X^*}$ then $foT\in{X^*}$ since $u_k$ is weakly convergent we have $foT_k u_k \rightharpoonup foTu$