A counterexample of Banach Steinhaus Theorem I was reading about a consequence of Banach-Steinhaus theorem which states that:

Let $E$ be a Banach space and $F$ be a normed space, and let $\{T_n\}_{n\in \mathbb{N}}$ be a sequence of bounded linear operators from $E$ to $F$, if the sequence $\{T_n x\}_{n\in \mathbb{N}}$ converges for each $x\in E$, then if we define:
$$
T: E\longrightarrow F
$$
$$
x \mapsto Tx = \lim_{n\to \infty} T_n x
$$
then

*

*$\displaystyle \sup_{n\in \mathbb{N}} || T_n || <\infty$

*$T$ is a bounded linear operator

*$\displaystyle || T || \leq \liminf_{n\to \infty} ||T_n ||$



So, I was wondering when this doesn't hold.


I tried the following example: Let $E=F=c_{00}$ the space of bounded sequences with a finite number of non-zero terms. Obviously $c_{00}$ is not a Banach space, so there is the reason the statement above is not verified, but in order to see that, I defined a sequence of bounded linear operators as follows:
For each $n\in \mathbb{N}$, let $ T_n: E\longrightarrow F
$ such that
$$x=(x_1, x_2, ..., x_n, 0,0,...) \mapsto T_n x = (x_1,2 x_2,..., n x_n, 0, 0,..)
$$
then $T_n$ is a bounded linear operator for every $n\in \mathbb{N}$, but if we define $T$ as above, $T$ is a linear unbounded operator.

I tried to see why is an unbounded operator, this was my attempt:
Suppose by contradiction that $T$ is a bounded operator, then exist  $C>0$ such that
$$
||Tx ||\leq C ||x ||
$$
for every $x\in E$
if we consider $x=e_k=(0,0,...,0,1,0,0,..)$, $1$ on the $k$-th position. We have that
$$
T e_k = \lim_{n\to \infty} T_n e_k = k e_k
$$
then
$$
|| T e_k || =k \leq C 
$$
but this says that $T$ is bounded.

Did I miss something in this proof?.


The norm of $T$ is the supremum of $\lvert\lvert Tx \rvert\rvert$ over all unit vectors $x \in E$ and this is at least the supremum of $\lvert\lvert Te_k \rvert\rvert$ over all elementary basis vectors $(e_k)_{k=1}^\infty$.
