If a sequence of bounded operator converges pointwise, then it is bounded in norm

real-analysis functional-analysis proof-verification operator-theory

1,181

The last update contains a correct solution to the exercise.

1,181

sequence

Updated on December 01, 2020

Comments

sequence over 2 years

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

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

$$\sup_{n\in\mathbb{N}} \vertiii{T_n} \le M <\infty.$$
- uniquesolution over 5 years
  
  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.
- Giuseppe Negro over 5 years
  
  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 $\lim_{n\to \infty} \sup a_n =\lim_{n\to \infty} 1 = 1.$
- sequence over 5 years
  
  @GiuseppeNegro I don't seem to be interchanging lim and sup. Can you please clarify?
- Giuseppe Negro over 5 years
  
  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.
- sequence over 5 years
  
  @GiuseppeNegro I see it now, thanks.
- sequence over 5 years
  
  @GiuseppeNegro I've updated my post. Can you please check if it's okay now?
- Giuseppe Negro over 5 years
  
  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 $\|T_n\|$.
- sequence over 5 years
  
  @GiuseppeNegro Can you please clarify which part of my proof is invalid?
- Giuseppe Negro over 5 years
  
  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 $x$).
- sequence over 5 years
  
  @GiuseppeNegro I have redone the proof. Can you please take a look?
- Giuseppe Negro over 5 years
  
  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.
- sequence over 5 years
  
  @GiuseppeNegro Fixed again.
- sequence over 5 years
  
  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 $\mathbb{R}$.