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$\newcommand{\vertiii}[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1 \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}}$

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

$$\sup_{n\in\mathbb{N}} \vertiii{T_n} \le M <\infty.$$

zhw.
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sequence
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    You 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. – uniquesolution Nov 29 '17 at 14:41
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    Well, 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 $\lim_{n\to \infty} \sup a_n =\lim_{n\to \infty} 1 = 1.$ – Giuseppe Negro Nov 29 '17 at 14:42
  • @GiuseppeNegro I don't seem to be interchanging lim and sup. Can you please clarify? – sequence Nov 29 '17 at 20:26
  • How 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. – Giuseppe Negro Nov 29 '17 at 21:24
  • @GiuseppeNegro I see it now, thanks. – sequence Nov 29 '17 at 23:44
  • @GiuseppeNegro I've updated my post. Can you please check if it's okay now? – sequence Nov 30 '17 at 00:07
  • It is not ok. The uniform boundedness principle tells you that if you can bound T_n pointwise then you have a uniform bound on $|T_n|$. – Giuseppe Negro Nov 30 '17 at 14:03
  • @GiuseppeNegro Can you please clarify which part of my proof is invalid? – sequence Nov 30 '17 at 19:12
  • All 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 $x$). – Giuseppe Negro Nov 30 '17 at 20:19
  • @GiuseppeNegro I have redone the proof. Can you please take a look? – sequence Dec 01 '17 at 00:03
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    Getting 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. – Giuseppe Negro Dec 01 '17 at 00:15
  • @GiuseppeNegro Fixed again. – sequence Dec 01 '17 at 00:30
  • But 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 $\mathbb{R}$. – sequence Dec 01 '17 at 00:35

1 Answers1

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The last update contains a correct solution to the exercise.

Giuseppe Negro
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