Showing that a map defined using the dual is a bounded linear operator from X' into X' I have trouble answering the second part of the following exercise. Any help would be appreciated!
Let $(X, \| \cdot \|)$ be a reflexive Banach space. Let $\{ T_n \}_{n = 1}^\infty$ be a sequence of bounded linear operators from $X$ into $X$ such that $\lim_{n \to \infty} f(T_n x)$ exists for all $f \in X'$ and all $x \in X$.
(a) Show that $\sup_{n} \|T_n' \| < + \infty$
(b) Show that the map $S$ defined by $(Sf)(x) := \lim_{n \to \infty} (T_n'f)(x)$ is a bounded linear operator from $X'$ into $X'$.
What I've done so far:
(a): I have done this using the Uniform Boundedness Principle twice after first showing that $\sup_n \| T_n' (f)(x) \| < + \infty$
(b): I think that $\displaystyle \sup_n\|T_n' \|$ would make a good bound since we have just seen that it is finite, but I have not succeeded in proving that so far...
Thank you very much for any hints you can offer me!

After you have shown that $A := \sup\limits_n \lVert T_n'\rVert < \infty$, you have a known bound on $S(f)(x)$ for every $x\in X$ and $f\in X'$, namely
$$\lvert S(f)(x)\rvert = \lim_{n\to\infty} \lvert T_n'(f)(x)\rvert \leqslant \limsup_{n\to\infty} \lVert T_n'(f)\rVert\cdot \lVert x\rVert \leqslant \limsup_{n\to\infty} \lVert T_n'\rVert\cdot \lVert f\rVert\cdot \lVert x\rVert.$$
From $\lvert S(f)(x)\rvert \leqslant N\lVert f\rVert\cdot \lVert x\rVert$, we deduce by the definition of the norm on $X'$ that $\lVert S(f)\rVert \leqslant N\cdot \lVert f\rVert$, and this in turn shows that $S$ is a continuous (bounded) operator on $X'$ with $\lVert S\rVert \leqslant N$.
