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)
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.