# Show that if $T \in B(X,Y)$ and $x_n \rightarrow x$ weakly, then $Tx_n \rightarrow Tx$ weakly

Suppose $T \in B(X,Y)$ and $x_n \rightarrow x$ weakly. Show that $Tx_n \rightarrow Tx$ weakly.

My attempt:

Let $y^* \in Y^*$. We want to show that $|y^*(Tx_n) - y^*(Tx)| < \epsilon$. Since $y^*$ is a bounded functional, we can assume that $\| y^* \|=1$. Note that $$|y^*(Tx_n) - y^*(Tx)| = |y^*(Tx_n - Tx)| \leq \| T \| \| x_n - x \|_X$$

Note that $\| x_n - x \|_X = \sup \{ |x^*(x_n) - x^*(x)| : \| x^* \| \leq 1 \}$. Since $x_n \rightarrow x$ weakly, for all $x^* \in X^*$ and $\epsilon >0$, we have $$|x^*(x_n) - x^*(x)| < \dfrac{\epsilon}{\| T \| +1}.$$

Hence, $$|y^*(Tx_n) - y^*(Tx)| \leq \| T \| \| x_n - x \|_X < \epsilon.$$

I think something is wrong in my proof. It seems I am using the fact that a weakly convergent sequence is also strong convergent.

Can anyone help me to check?

You're not using the weak convergence of $x_n$ here.
The correct (and significantly easier) approach is to note that if $y^*$ is a bounded functional, then so is $y^*T$.
• I am not using weak convergence of $x_n$? I have the inequality $|x^*(x_n) - x^*(x)| < \dfrac{\epsilon}{\| T \| + 1}$. Commented Oct 24, 2015 at 13:18
• You have to clarify that statement. For which $x^*$? For which $n$? Also, you should consider the second approach. Commented Oct 24, 2015 at 13:39