# 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

1. $$\displaystyle \sup_{n\in \mathbb{N}} || T_n || <\infty$$
2. $$T$$ is a bounded linear operator
3. $$\displaystyle || T || \leq \liminf_{n\to \infty} ||T_n ||$$

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?.

• Please edit the question to include what you were reading this from. – Shaun Dec 9 '18 at 3:11
• It seems you are trying to show $T$ is unbounded by contradiction. You begin "Suppose $T$ is bounded" and you end with "but this says $T$ is bounded". Where is the contradiction? – DanielWainfleet Dec 9 '18 at 3:44
• You right, I tried to prove that by contradiction. I forgot to include that part. Thanks. The question is that whether my attempt of proof was right or this example is not valid either. I put that T is bounded since I couldn’t het anything more from that. A partner says that contradiction occurs because T is bounded for every k and that’s all. Buy I don’t see that clearly. – Jeff Dec 9 '18 at 3:59
• Isn't $k \leq C$ for all $k$ a contradiction? – Kavi Rama Murthy Dec 9 '18 at 4:54
• it is, but I already solve this problem with your help, thank you for helping me. – Jeff Dec 16 '18 at 1:36

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 $$(e_k)_{k=1}^\infty$$.
• Your comment give me the answer, because if it suffices for all $x\in E$ then it suffices for a basis vector $e_k$, for every $k\in \mathbb{N}$, then it follows the contradiction I was looking for. – Jeff Dec 16 '18 at 1:37