# Convergence in the weak operator topology implies uniform boundedness in the norm topology?

If $\{T_n\}$ is a sequence of bounded operators on the Banach space $X$ which converge in the weak operator topology, could someone help me see why it is uniformly bounded in the norm topology?

I know how to apply the uniform boundedness principle to reach the conclusion above under the stronger assumption that the $T_n$ converge in the strong operator topology.

$\textbf{Result}:$ If for a sequence $(x_n)$ in a Banach space $X$, the sequence $(f(x_n))$ is bounded for all $f\in X^*$, then the sequence $(\|x_n\|)$ is bounded.
Now suppose that you have proved the result. Suppose $(T_n)$ converges to $T$ in weak operator topology, i.e., \begin{align*} |f(T_nx)-f(Tx)| \longrightarrow 0 & & \text{for all } x\in X, f\in X^*. \end{align*} In particular, $(f(T_nx))$ is bounded for all $x\in X, f\in X^*$. Now by the result it follows that $(\|T_nx\|)$ is bounded for all $x \in X$. And now Uniform Boundedness Principle gives you that $(\|T_n\|)$ is uniformly bounded.
Proof of Hint: We have $|f(x_n)| < \infty$ for all $f\in X^*$. Consider the sequence $(T_n)$ of linear functionls on $X^*$ defined by $T_n : X^* \to \mathbb{K}$ by $T_n(f) = f(x_n)$. Then $\|T_nf\| = |f(x_n)| < \infty$ for all $f \in X^*$. So, UBP implies that $(\|T_n\|)$ is bounded. But $\|T_n\| = \|x_n\|$ (by cannonical embedding of $X$ into $X^{**}$) and this proves our result.
• Thank you, but unfortunately I need some help proving your result above. (I was fine with using it afterwards, as you just did, but I appreciate your editing anyway.) We could take $f$ with $||f||=1$ , but this doesn't seem to help... – pitaya Apr 18 '15 at 18:17
• Oh, OK, I see, your result is a non-trivial consequence of the Hahn-Banach theorem, thank you! (I don't think that the UBP is needed for this result, for we can take $||f||=1.$) – pitaya Apr 18 '15 at 18:22
• but you can see that even without the result you can directly solve your question (just imitate the proof of the hint with $T_nx$ at place of $x_n)$. The reason why I wrote the result separately was to keep it notationally simple. – Urban PENDU Apr 18 '15 at 18:27