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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?.

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  • $\begingroup$ Please edit the question to include what you were reading this from. $\endgroup$ – Shaun Dec 9 '18 at 3:11
  • $\begingroup$ 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? $\endgroup$ – DanielWainfleet Dec 9 '18 at 3:44
  • $\begingroup$ 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. $\endgroup$ – Jeff Dec 9 '18 at 3:59
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    $\begingroup$ Isn't $k \leq C$ for all $k$ a contradiction? $\endgroup$ – Kavi Rama Murthy Dec 9 '18 at 4:54
  • $\begingroup$ it is, but I already solve this problem with your help, thank you for helping me. $\endgroup$ – Jeff Dec 16 '18 at 1:36
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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 $(e_k)_{k=1}^\infty$.

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  • $\begingroup$ 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. $\endgroup$ – Jeff Dec 16 '18 at 1:37

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