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I'm studying funcional analysis and our professor left the following problem:

Let $E$, $F$ be normed spaces, and $\left( A_n\right)$ be a sequence of limited linear operators from E to F, such that, for every $x \in E$ the limit $$\lim_{n\to\infty}A_nx$$ exists.

Prove that $A:E\rightarrow F$ given by $Ax=\lim_{n\to\infty}A_nx$ is limited if and only if $\sup_n \lVert A_n \rVert\lt\infty$

It reminded me of the Uniform Boundedness Principle, but I don't have the Banach hypothesis.

I was able to prove the $\Leftarrow$ direction of the equivalence, but I could not prove the other way around.

My friend and I tried the following:

$\forall x\in E, \exists n_0$ such that $\lVert A_n x\rVert\le \lVert A_nx-Ax\rVert+\lVert Ax\rVert\lt1+\lVert A\rVert\lVert x\rVert$, $\forall n\gt n_0$

And now we just wanted to pass the supremum over $x\in E, \lVert x\rVert=1$ so that we get $\lVert A_n\rVert\le 1+\lVert A\rVert$, for $n\gt n_0$. But I strongly think we can't use the supremum because $n_0$ depends on each $x$, since we have the pointwise convergence.

Does anyone has any suggestions or thoughts on how to improve this?

(I've searched for this problem here, and couldn't find, sorry if it is a duplicate)

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  • $\begingroup$ what does limited mean $\endgroup$ – mathworker21 Sep 18 '18 at 22:35
  • $\begingroup$ @mathworker21 Presumably a synonym "bounded." $\endgroup$ – Mike Earnest Sep 18 '18 at 22:43
  • $\begingroup$ @mathworker21 $\lVert A_n x\rVert \le \lVert A \rVert \lVert x \rVert, \forall x \in E$. $\endgroup$ – Andre Antunes Sep 18 '18 at 22:46
  • $\begingroup$ I wasn't sure of what is the correct term in English for this. I just translated in my mind and forgot to look in the book. $\endgroup$ – Andre Antunes Sep 18 '18 at 22:48
  • $\begingroup$ bounded or continuous $\endgroup$ – mathworker21 Sep 18 '18 at 22:51
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I guess that this claim is equivalent to the Uniform Boundedness Principle. What you have written is already a complete proof if we are allowed to use this result. But as you said, we need $E$ and $F$ to be both Banach spaces.

Here is a counterexample in the non-complete case. Take $E=c_{00}$ i.e. the space of eventually vanishing sequences $(x_1,\dots, x_N,0,0,\dots)$ with the supremum norm, $F=\mathbb{R}$, and $$A_nx:=nx_n $$ So for each $x\in c_{00}$ we have $A_nx=0$ for large enough $n$, and hence $$\lim_{n\to +\infty}A_nx=0=A(0) $$ where $A=0$, which is a bounded operator. However, $$\|A_n\| =n\sup_{\|x\|_{\infty}\leq 1}|x_n|=n$$ and hence $\sup_{n\in \mathbb{N}}\|A_n\|=\infty$, contradicting the statement.

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  • $\begingroup$ Very nice. I don't think the claim for fixed $E,F,(A_n)_n$ is equivalent to UBP, but correct me if I'm wrong. $\endgroup$ – mathworker21 Sep 18 '18 at 22:49
  • $\begingroup$ Actually I don't know, that's why I wrote "I guess" :) $\endgroup$ – Lorenzo Quarisa Sep 18 '18 at 22:50
  • $\begingroup$ @LorenzoQuarisa Thank you very much! Such a elegant counterexample! Just to be sure: I can show that the norm of $A_n$ is $n$, by taking the sequence with entries equal to 1 up to the n-th entry, right? So that, the norm of this sequence is 1 and $A_n x=n$? $\endgroup$ – Andre Antunes Sep 18 '18 at 23:02
  • $\begingroup$ @AndreAntunes Yes (you just need $x_n=1$ and $|x_k|\leq 1$ for all $k$) $\endgroup$ – Lorenzo Quarisa Sep 19 '18 at 12:04

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