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?


1 Answer 1


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 $y^*T$.

  • $\begingroup$ I am not using weak convergence of $x_n$? I have the inequality $|x^*(x_n) - x^*(x)| < \dfrac{\epsilon}{\| T \| + 1}$. $\endgroup$
    – Idonknow
    Commented Oct 24, 2015 at 13:18
  • $\begingroup$ You have to clarify that statement. For which $x^*$? For which $n$? Also, you should consider the second approach. $\endgroup$ Commented Oct 24, 2015 at 13:39

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