Let $X$ be a Banach space, $Y$ a normed vector space and $\tau\subset L(X,Y)$.
Show that if $\sup_{T\in \tau}|y^*(Tx)|<\infty$ for all $x\in X,y^*\in Y^*$
then $\tau$ is bounded in $L(X,Y)$.
Seems that we have to use the Banach-Steinhaus theorem which would yield that $\tau\subset L(X,Y)$ is bounded.
But I don't know how to show that $\tau$ is points wise bounded. I don't see any connection to the assumption.
First let us fix $x\in X$. Now we consider the family of functionals $Y^*\to\mathbb{C}$ given by $$\tau' = \{ y^*\mapsto y^*(Tx)\,:\,T\in\tau\}.$$ By assumption we have $$\sup_{T'\in\tau'} |T' y^*|=\sup_{T\in\tau} |y^*(Tx)|<\infty$$ Also note that $Y^*$ is complete. Thus we can apply Banach-Steinhaus to $\tau'$ and get $$\sup_{T'\in\tau'} \|T'\| = \sup_{T\in\tau} \sup_{\|y^*\|_{Y^*}=1} |y^*(Tx)| < \infty$$
Claim: Let $B$ be a Banach space. Then for all $b\in B$ we have $$\|b\|_B = \sup_{\|b^*\|_{B^*}=1} |b^*(b)|$$
Proof: The canonical embedding map $\mathrm{ev}:B\to B^{**}$, $b\mapsto (b^* \mapsto b^*(b))$ is injective and (most importantly for us) norm preserving (follows from Hahn-Banach, you can look it up in any introductory text book). Thus $$\|b\|_B=\|\mathrm{ev}(b)\|_{B^{**}}=\sup_{\|b^*\|_{B^*}=1} |b^*(b)|.$$ $\square$
By the claim it follows that
$$\sup_{T\in\tau} \|Tx\|_Y < \infty$$
for all $x\in X$. Thus we can apply Banach-Steinhaus to see that $\tau$ is bounded.