If a sequence of bounded operator converges pointwise, then it is bounded in norm
The last update contains a correct solution to the exercise.
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Updated on December 01, 2020Comments

sequence almost 2 years
$\newcommand{\vertiii}[1]{{\left\vert\kern0.25ex\left\vert\kern0.25ex\left\vert #1 \right\vert\kern0.25ex\right\vert\kern0.25ex\right\vert}}$
Let $\left(V, \\cdot\_V \right)$ be a Banach space and $\left(W, \\cdot\_W \right)$ a normed vector space. Show that if a sequence $(T_n)_n\subset \mathcal{B}(V,W)$ converges pointwise then $\sup \vertiii{T_n}<\infty$.
So we know that $\sup\limits_{x\in V} \frac{\T_n x\}{\x\}=\vertiii{T_n}$, thus
$$\lim_{n\to\infty}\sup\limits_{x\in V} \frac{\T_n x\}{\x\}=\sup\limits_{x\in V} \frac{\T x\}{\x\}=\vertiii{T}<\infty$$
I'm wondering if I didn't miss something, since this seems a bit too simple. Please let me know.
Update (version 3):
Since $V$ is a complete metric space, it is Banach, and $T$ is defined on $V$. Since $\{(T_n(x))\}_n$ converges pointwise, it is bounded, so that there exists a function $C:V\to \mathbb{R}$ such that
$$\sup_{n\in \mathbb{N}} \T_n(x)\<C(x)$$
for each $x\in V$. By the Uniform Boundedness Principle, we deduce that there exists $M>0$ such that
$$\sup_{n\in\mathbb{N}} \vertiii{T_n} \le M <\infty.$$

uniquesolution almost 5 yearsYou are missing the fact that in general pointwise convergence does not imply uniform convergence, and so you need an argument, or a theorem that already made such an argument, to justify why you can take the limit of the sup when you are only given pointwise convergence. Consult the uniform boundeness theorem.

Giuseppe Negro almost 5 yearsWell, yes, you cannot interchange lim and sup like that. That's a general fact of analysis, it is not limited to linear operators. Take for example the sequence $$a_n=(\underbrace{0\ldots 0}_{n\, \text{places}}, 1, 1\ldots).$$ Then $\lim_{n\to \infty} a_n = (0,0,0\ldots)$ but $\lim_{n\to \infty} \sup a_n =\lim_{n\to \infty} 1 = 1.$

sequence almost 5 years@GiuseppeNegro I don't seem to be interchanging lim and sup. Can you please clarify?

Giuseppe Negro almost 5 yearsHow can you say that $\lim( \sup \T_n x\)=\sup \Tx\$? (forgive my sloppiness, please). Since $\Tx\=\lim \T_n x\$, you are interchanging lim and sup.

sequence almost 5 years@GiuseppeNegro I see it now, thanks.

sequence almost 5 years@GiuseppeNegro I've updated my post. Can you please check if it's okay now?

Giuseppe Negro almost 5 yearsIt is not ok. The uniform boundedness principle tells you that if you can bound T_n pointwise then you have a uniform bound on $\T_n\$.

sequence almost 5 years@GiuseppeNegro Can you please clarify which part of my proof is invalid?

Giuseppe Negro almost 5 yearsAll of the Update is nonsense. What is $B$? How do you know that $T$ is a bounded operator? Actually, you did not even explicitly define $T$. I guessed that $Tx:=\lim_{n\to \infty} T_n x$. I would erase everything of the "Update" and start again. To apply the UBP you need to show that $\ T_n x\\le C(x)<\infty$ for all $x\in V$, where $C(x)$ is a number that is independent of $n$ (but may be dependent on $x$).

sequence almost 5 years@GiuseppeNegro I have redone the proof. Can you please take a look?

Giuseppe Negro almost 5 yearsGetting closer, but you don't know if $C(x)$ is bounded. That's the whole point of the UBP. A pointwise bound $\T_n x\\le C(x)$, with the right hand side depending on $x$, suffices to give a uniform bound $\T_n\\le M$ for some $M>0$. That's a miracle.

sequence almost 5 years@GiuseppeNegro Fixed again.

sequence almost 5 yearsBut if $T_n(x)$ is pointwise convergent then it is boudned, so for every $x\in V$, shouldn't it be true that $C(x)<\infty$? I'm realizing that the supremum of $C(x)$ does not necessarily exist, like for $x^2$ on all of $\mathbb{R}$.
