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Suppose that $\mathcal{H}$ is a separable Hilbert space and $(e_n)_{n\in\Bbb N}$ is an orthonormal basis for $\mathcal{H}$. Let $(T_m)_{m\in\Bbb N}\subset B(\mathcal{H})$ such that the sequence $(T_me_n)_{m\in\Bbb N}$ is convergent. Could we prove that $(T_m)$ is weakly convergent?

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    $\begingroup$ It would be better if you could show where you're stuck when trying to prove this. $\endgroup$ – Saad Apr 25 '18 at 7:12
  • $\begingroup$ @AlexFrancisco I can prove it for elements in span$(e_n)$. But I have difficulty for elements in the clouser. $\endgroup$ – saeed Apr 25 '18 at 8:32
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No, this statement is false: As a counterexample take $H=l^2$ and $T_m((x_i)_{i \in \mathbb{N}}):= m x_m e_1$, where $(e_i)_{i \in \mathbb{N}}$ is the 'standard' orthonormal basis for $l^2$. Then we always have $T_m(e_n) =0$ for all $m >n$. Thus $T_m(e_n) \rightarrow 0$, but $(T_m)_{m \in \mathbb{N}}$ is unbounded; in fact, $\|T_m\|=m$. Note that $\phi_m(x) := \langle T_m(x),e_1 \rangle = m x_m$ is unbounded. However, weakly convergence would imply that this sequence is bounded. One may also argue that $x= (1/n)_{n \in \mathbb{N}} \in l^2$, but $\phi_m(x) = 1$, i.e. $\phi_m(x) \not\rightarrow 0$.

On the other hand, if $(T_m)_{m \in \mathbb{N}}$ is bounded, the statement is true. Let $y \in H$ be fixed. Just take $x= \sum_{k=1}^\infty x_k e_k \in H$ and $\|\sum_{k=n}^\infty x_k e_k \|_H < \varepsilon/C$, where $C= \sup_{m \in \mathbb{N}} \|T_m\| \|y\|$, for all $n \geq N$, then $$\tag{1}|\langle T_m x , y \rangle| \leq \sum_{k=1}^N |x_k \langle T_m e_k ,y \rangle| + \sup_{m \in \mathbb{N}} \|T_m\| \|y\| \|\sum_{k=N+1}^\infty x_k e_k \|_H \leq \sum_{k=1}^N |x_k \langle T_m e_k ,y \rangle| + \varepsilon.$$ Since $N$ is fixed, we can take $m \geq M$ such that (1) is less than $2 \varepsilon$.

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