Complete domain is necessary for Banach-Steinhaus Theorem

Question

Theorem: Let $E$ Banach space, $F$ normed space, $\{T_{\alpha}\}_{\alpha \in I} \subseteq L(E;F) = \{ T:E \to F ; T $ is linear and continuous $ \}$. If $\forall x \in E \; \exists M_x \geq 0$ st. $\| T_{\alpha} (x) \| \leq M_x \; \forall \alpha \in I$, then $\exists M \geq 0$ st. $ \| T_{\alpha} \| \leq M \; \forall \alpha \in I$.

I want to see an example where $E$ is just normed and then the Theorem won't work.

Thanks.

Answer 1

Quote from Counterexamples around Banach-Steinhaus theorem:

Let $c_{00}$ be the vector space of real sequences $x = (x_n)$ eventually vanishing, equipped with the norm $$\Vert x\Vert = \sup_{n \in\mathbb N}\vert x_n\vert$$ For $n\in\mathbb N$, $T_n : E \to E$ denotes the linear map defined by $$T_n x = (x_1,2 x_2,\dots, n x_n,0,0,\dots).$$ $T_n$ is continuous as for $\Vert x\Vert\le 1$, we have $$ \begin{align*} \Vert T_n x \Vert &= \Vert (x_1,2 x_2,\dots, n x_n,0,0,\dots) \Vert\\ & = \sup_{1 \le k \le n} \vert k x_k \vert \le n \Vert x \Vert \le n \end{align*} $$ For any $x\in E$ there exists $k\in\mathbb N$ such that $x = (x_1,\dots ,x_k,0,0,\dots)$. Hence for $n\ge k$, $T_n x$ is constant and equal to $(x_1,2 x_2,\dots,k x_k,0,0,\dots)$, which proves that $(T_n x)$converges. However $(T_n)$ is not uniformly bounded as we can see looking at the elements $\mathbf{1}_n\in E$ which have all terms equal to $0$ except the first $n$ ones which are equal to $1$: $$ \Vert\mathbf{1}_n\Vert = 1\text{ and }\Vert T_n\mathbf{1}_n\Vert = \Vert(1,2,\dots,n,0,0,\dots)\Vert = n. $$