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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!

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    $\begingroup$ Every Hilbert space is reflexive. $\endgroup$ – uniquesolution Sep 4 '15 at 16:30
  • $\begingroup$ how convergent weak or..? $\endgroup$ – R.N Sep 4 '15 at 16:34
  • $\begingroup$ Yes, weakly convergent. $\endgroup$ – fmeyer Sep 4 '15 at 16:36
  • $\begingroup$ Could you give me a short sketch of the proof? $\endgroup$ – fmeyer Sep 4 '15 at 16:37
  • $\begingroup$ oh in fist i thought it is obvious. but i start to wright it needs challenge $\endgroup$ – R.N Sep 4 '15 at 16:40
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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 $k>\max\{K_1,K_2\}$

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ok suppose $f\in{X^*}$ then $foT\in{X^*}$ since $u_k$ is weakly convergent we have $foT_k u_k \rightharpoonup foTu$

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  • $\begingroup$ interesting, who has given me -1?! :O $\endgroup$ – R.N Sep 4 '15 at 18:22

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