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Consider the uniform boundedness principle:

UBP. Let $E$ and $F$ be two Banach spaces and let $(T_i)_{i \in I}$ be a family (not necessarily countable) of continuous linear operators from $E$ into $F$. Assume that $\sup_{i \in I} \|T_ix \| < \infty$ for all $x \in E$. Then $\sup_{i \in I} \|T_i\|_{\mathcal{L}(E,F)} < \infty$.

I don't understand the statement of the UBP. The assumption tells us that, fixed an element $u$, we surely find a $\|T_ku\|< \infty$ (in particular, for that fixed $u$ each other $T$ is limited in $u$ too). The conclusion tells us that the sup over the $i$'s of the set $$ \biggl\{ \sup_{\|x\|\leq 1} \|Tx\| \biggr\} $$ is limited. But isn't that clear from the assumption? I mean, if each $T$ is bounded, a fortiori the conclusion must hold... please explain me where I am wrong.

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up vote 7 down vote accepted

To illustrate the point which Davide explained in his answer, let us have a look at a concrete example.

Let us take $E=$ the set of all real sequences with finite support (i.e., only finitely many terms are non-zero). Let us use the norm $\|x\|=\sup_n |x_n|$. The space $E$ is a linear normed space, but it is not a Banach space. (So the assumptions of Banach-Steinhaus theorem are not fulfilled.)

Let us take $F=\mathbb R$ and $T_n(x)=\sum_{k=1}^n x_k$.

For every $x\in E$ we have $|T_n(x)| \le \sum_{k=1}^n |x_k|$, which is a finite number. So $\sup_n |T_n(x)|<+\infty$ for any fixed $x\in E$.

But if we take $x_n=(\underset{\text{$n$-times}}{\underbrace{1,\dots,1,}}0,0,\dots)$, then $T_n(x_n)=n$ and $\|x_n\|\le 1$. So we see that $T_n(x)$ is not bounded on the unit ball, i.e. $$\sup_{\|x\|\le 1, n\in\mathbb{N}} T_n(x)=+\infty.$$

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The problem is that $\sup_{i\in I}\lVert T_ix\rVert=:c(x)$ depend on $x$. The fact that it's bounded of each $x$ of the unit ball doesn't imply that it can be bounded uniformly on the unit ball (otherwise we would deduce that each linear map is continuous).

So the aim of the UBP, is, its name suggests, to show $\sup_{\lVert x\rVert=1}c(x)$ is finite.

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