Convergence problem with a sequence of operators Is it true that the sequence $\{T_n:H\rightarrow H\}_n$ with $H=l^2(\mathbb{N};\mathbb{R})$ ($\mathbb{N}$ is just the index set of the series) defined by $T_n(x_0,x_1,x_2,...)=(x_n,.x_{n-1},...,x_0,0,0,...)$ converges weakly to zero?
Thanks
 A: Yes, if by "weakly" you mean "in the weak operator topology". Because, for each element $e_k$ of the canonical basis,
$$\tag1
\langle T_nx,e_k\rangle=x_{n+1-k}\to0.
$$
From here you move to $\langle T_nx,y\rangle\to0$ for all $y$ in the span of $\{e_k\}_k$, and then using that $\|T_nx\|\leq\|x\|$ you can go to limits of those $y$, so for all $y$. 

Edit: details. Suppose first that $y=\sum_{k=1}^m \beta_ke_k$. Then
$$
\lim_n\langle T_nx,y\rangle=\lim_n\sum_{k=1}^m\bar{\beta_k}\,\langle T_nx,e_k\rangle
=\sum_{k=1}^m\bar{\beta_k}\,\lim_n\langle T_nx,e_k\rangle=0.
$$
For arbitrary $y$: fix $\varepsilon>0$. Then there exists $y_0\in\operatorname{span}\{e_k:\ k\}$ with $\|y-y_0\|<\varepsilon$. Then
\begin{align}
|\langle T_nx,y\rangle|&\leq|\langle T_nx,y_0\rangle|+|\langle T_nx,y-y_0\rangle|\\[0.3cm]
&\leq |\langle T_nx,y_0\rangle|+\|T_nx\|\,\|y-y_0\|\\[0.3cm]
&\leq |\langle T_nx,y_0\rangle|+\|y-y_0\|\\[0.3cm]
&\leq |\langle T_nx,y_0\rangle|+\varepsilon. 
\end{align}
It follows that 
$$
\limsup_n|\langle T_nx,y\rangle|\leq\varepsilon. 
$$
As $\varepsilon$ was arbitrary we get that $\limsup_n|\langle T_nx,y\rangle|=0$, which shows that the limit exists and is zero. 
