Given $X$ a Banach space, to say $\{A_i\}_{i\in I}$, a family of bounded linear operators from $X$ to another normed space $Y$, is weakly bounded is to say that, fixing any $x\in X$ and $y^*\in Y^*$, the numerical set $\{y^*(A_i x)\}_{i\in I}$ is bounded. Now the result to prove is: a weakly bounded family must also be uniformly bounded, i.e., $\{\|A_i\|\}$ is bounded.
This is apparently a corollary of Banach-Steinhaus Theorem (aka uniform boundedness principle, UBP). I think proof by contradiction is the right track, but I've been stuck on the proper choice of $y^*$. My hunch is that we need to vary $y^*$ rather than just comstructing a specific one to violate the weak boundedness.
Any help is appreciated.
As I hinted at in the comments, one way to show this is as follows:
One uses the isometric embedding of $Y$ into its bidual to show that if $A \subset Y$ is such that $y^\ast (A) \subset \Bbb{K}$ is bounded for each $y^\ast \in Y^\ast$, then $A$ is a bounded set. Note: This does not even use completeness of $Y$, just that of $Y^\ast$.
Using the predecing point and the given assumptions, we conclude that for every $x \in X$, the set $\{A_i x \,:\, i \in I\}$ is bounded in $Y$. Now, another application of the uniform boundedness principle yelds the claim. This uses completeness of $X$.