I just read an alternative proof of the Uniform Boundness Principle when $Y$ is banach space that goes like this; Suppose we have $X$ and $Y$ Banach spaces and $F \subset L(X,Y)$ such that for every $x\in X$, $\sup_{T\in F}\|Tx\|<\infty $. Then it follows that $\sup_{T\in F}\|T\|< \infty$.
Let $B(F,Y)$ be the vector space formed by the applications $f$ defined in $F$ and that take value in $Y$, since $Y$ is a complete space this will be a complete hence a Banach space. Now define $S : X\rightarrow B(F,Y)$ as $(Sx)T=T(x), T\in F$. This will be a linear operator and using the closed graph theorem its easy to show that this will be a limited operator, because it will be a closed operator.
Now we have that for $T \in F $ and $x\in X$ $\|Tx\|=\|(Sx)T\|\leq \|Sx\|\leq \|S\| \cdot \|x\|$ and so we get our result.
My doubt is when we make the step $\|(Sx)T\|\leq \|Sx\|$ I dont quite see why this is true , shouldnt we have that $\|(Sx)T\|\leq \|(Sx)\|\cdot \|T\|$?
Thanks in advance !
Well I am not quite familiar with this concept but it looks like $||Sx||$ is really more of an operator norm, since $S$ maps to a space of operators. Hence $||Sx||=\sup_{||T||_{\text{op}}\le1}||(Sx)T||\ge ||(Sx)T||$ for all $||T||_{\text{op}}\le 1$. However, I don't see, why this should be true for all $T$ regardless of their norm.