Let $H$ be a Hilbert space with $\text{dim}=\infty$ , and $\{e_n\}$ be an orthogonal sequence of projections in $B(H)$. Show that $\{\sqrt{n}e_n ; n\geq 1\}$ does not admit a subsequence converging to zero weakly.
Suppose $\{\sqrt{n_k}e_{n_k}\}$ be a weakly convergent subsequence to zero , so $$\sqrt{n_{k}}e_{n_k}\to 0 (\text{wot}) \Longrightarrow \sqrt{n_k} \langle e_{n_k}\xi,\eta\rangle \to 0~~~(\xi,\eta\in H)$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Longrightarrow \{\sqrt{n_k}\langle e_{n_k} \xi , \eta\rangle\} \text{is norm bounded}$$ Now I need to show the above subsequence is not bounded for a contradiction.
While I'm sure that the subsequence is not bounded, but I can not show it. Please help me. Thanks.
Let $\{\sqrt{n_k}e_{n_k}\}_{k=1}^{\infty}$ be an arbitrary subsequence of $\{\sqrt{n}e_{n}\}_{n=1}^{\infty}$. It is clear that $k\leq n_k$. For each positive integer $k$, there exists $\xi_k\in H$ such that $\| \xi_k\|=1$ and $e_{n_k}\xi_k=\xi_k$. Define the sequence of numbers $\{ \alpha_k\}_{k=1}^{\infty}$ as follows: $\alpha_k=\frac{1}{\sqrt[4]{n_k}}$ if $k$ is a power of $2$, i.e., if $k=2^m$ for some positive integer $m$; for $k\ne 2^m$, let $\alpha_k=0$. It follows that $$ \sum_{k=1}^{\infty}|\alpha_k|^2=\sum_{m=1}^{\infty}|\alpha_{2^m}|^2\leq \sum_{m=1}^{\infty}\frac{1}{\sqrt[4]{2^m}}<\infty. \tag1 $$ Thus $\xi=\sum_{k=1}^{\infty}\alpha_k\xi_k$ is well defined vector in $H$. It follows from $$ \langle \sqrt{n_k}e_{n_k}\xi,\xi\rangle=\sqrt{n_k}\alpha_{k}^{2}=\left\{ \begin{array}{cc} 1; & k=2^m\\ 0; & k\ne 2^m\end{array} \right. $$ that the sequence $\{ \langle \sqrt{n_k}e_{n_k}\xi,\xi\rangle\}_{k=1}^{\infty}$ does not converge to $0$ and therefore $\{ \sqrt{n_k}e_{n_k}\}_{k=1}^{\infty}$ does not converge to $0$ in the weak operator topology.