Let $X$ be a Banach space, and let $\{T_n\}_{n = 1}^{\infty}$ be a sequence of bounded linear operators from $X$ into $X$ such that for all $f \in X'$ and $x \in X$, the limit $\lim\limits_{n \to \infty} f(T_n(x))$ exists.
What I have to show is that $\sup\limits_{n \in \mathbb{N}} ||T_n|| < \infty$. Using the Uniform Boundedness Principle, I can show that $$ \sup_{n \in \mathbb{N}} \{ ||f \cdot T_n|| : n \in \mathbb{N} \} < \infty $$ for all $f \in X'$. But that's not enough.
What I thought is that maybe I could define some sort of new operator $K_n : X' \to X'$ by $f \mapsto f \circ T_n$, then the same Principle gives us that $$ \sup_{n \in \mathbb{N}} \{||K_n|| : n \in \mathbb{N} \} < \infty $$ hence $$ \sup_{n \in \mathbb{N}} \sup_{||f|| = 1||} ||f T_n|| < \infty $$ but I have no idea how to continue from here.
Consider $A=\{f\circ T_n, f \in X', \|f\|=1\}$. You can apply the uniform boundedness principle to $A$. This implies there exists $B>0$ such that for every $x$ with $\|x\|=1$, $\| f(T_n(x))\|<B, \|f\|\leq 1$.
Let $x\in X, \|x\|=1$. There exists $f^n_x$ such that $\|f^n_x\|=1$ and $f_x^n(T_n(x))=\|T_n(x)\|$. If $T_n(x)=0$, just take $f_x^n=0$. If not, consider $Vect(T_n(x))$ there exists $g^n_x$ define on $Vect(T_n(x))$ such that $\|g^n_x\|=1$ and $g_x^n((T_n(x))=\|T_n(x)\|. $ (Take $g_x^n({{T_n(x)}\over {\|T_n(x)\|}})=1$). Hahn Banach shows that you can extend $g_x^n$ to $f_x^n$ such that $\|f_x^n\|=1$. This implies that $f_x^n(T_n(x))=\|T_n(x)\|<B$. We deduce that $Sup_n\|T_n\|\leq B$.