I am trying to better and more rigorously understand the Principle of Uniform Boundedness (PUB). Recall that the statement of PUB reads:
Let $X,Y$ be Banach Spaces and suppose that $(T_n)_{n\in\mathbb N}\subset\mathcal B(X,Y)$. If for every $x\in X$ it follows that $\sup_{n\in\mathbb N}\|T_nx\|<\infty$, then $\sup_{n\in\mathbb N}\|T_n\|<\infty$.
Consider, in particular, the part which reads "... for every $x\in X$ it follows that $\sup_{n\in\mathbb N}\|T_nx\|<\infty$ ...". Now this precisely means that, $$\forall x\in X,\,\sup\{\|T_nx\|:n\in\mathbb N\}<\infty,$$ which is equivalent to the statement that, $$\forall x\in X,\,\exists M\in\mathbb R_{\ge0}\,\forall\, n\in\mathbb N:\|T_nx\|\le M.$$ My question: how, exactly, is this equivalent to the statement that, $$\forall x\in X:\|x\|\le1,\,\exists M\in\mathbb R_{\ge0}\,\forall\, n\in\mathbb N:\|T_nx\|\le M?$$ Note that the set of those $x\in X:\|x\|\le1$ is precisely the ball in $X$ centered at $0_X$ of radius $1$. I think I am close in making the connection, and I am convinced that it is to do with the fact that the operator norm can be defined as, $$\|T\|=\sup\{\|Tx\|:\|x\|\le1\},$$ but I am not sure how I should deal with the quantifiers that slipping this definition into the above brings with it. Can anybody indicate or highlight how the equivalence should be drawn out?
One implication is trivial since if we can find an $M$ for every $x$ then we can certainly find an $M$ for every $x$ satisfying $\|x\| \leq 1$.
For the non-trivial implication, pick an arbitrary $x \in X$. Then $y = \frac{x}{\|x\|}$ satisfies $\|y\| = 1$ and so there is an $N$ such that $$\sup_{n \in \mathbb{N}} \|T_n y \| \leq N.$$ However, $\|T_ny\| = \frac{1}{\|x\|} \|T_nx\|$ and so, setting $M = N \|x\|$, we get $$\sup_{n \in \mathbb{N}} \|T_n x \| \leq M$$ as desired.