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.
I'd appreciate any helpful hints.

Hint: Try to prove the following result
$\textbf{Result}:$ 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.
