Uniform boundedness principle statement 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.

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. $$\sup_{\|x\|\le 1, n\in\mathbb{N}} T_n(x)=+\infty.$$