Exercise on dual / second dual spaces (functional analysis)

[TaPaKaH]

Homework Statement

Let (X,\|\cdot\|) be a reflexive Banach space. Let \{T_n\}_{n\in\mathbb{N}} be a sequence of bounded linear operators from X into X such that \lim_{n\to\infty}f(T_nx) exists for all f\in X' and x\in X.
Use the Uniform Boundedness Principle (twice) to show that \sup_{n\in\mathbb{N}}\|T_n'\|<\infty.

Homework Equations

For operators between normed spaces we have \|T'\|=\|T\|, but I'm not sure if this can help in this case.

The Attempt at a Solution

I am currently at loss how to deal with information on functions f to apply UBP.
Any hints welcome.

 

[micromass]

What is T_n^\prime??

It might be a good idea to write f(T_n x) = \widehat{T_n x}(f), where \widehat{T_n x} \in X^{\prime\prime}. Apply UBT on this.

 

[TaPaKaH]

T_n' is the dual operator of operator T, i.e. T'\in B(X',X') s.t. T'(f)(x)=f(Tx) for x\in X and f\in X'.

Do I get this right that you denote the dual operator with "hat" symbol?

 

[micromass]

Yes, I mean the dual operator.

Ok, so you need to prove \sup_n \|T_n^\prime\|<+\infty. What does the UBP tell you?? Is it sufficient to prove something easier?

 

[TaPaKaH]

I came up with (what I think is) a proof in 3 steps using UBP only once.

1) Knowing that \lim_{n\to\infty}f(T_nx)=\lim_{n\to\infty}T_n'(f)(x) exists for all f\in X' (is given), one can say that for any n\in\mathbb{N}: \|T_n'(f)\|<\infty hence \sup_{n\in\mathbb{N}}\|T_n'(f)\|<\infty for all f\in X'.

2) Since T_n\in B(X,X) then T_n'\in B(X',X') for all n\in\mathbb{N} (there is a Theorem I can refer to).

3) "2" allows to apply UBP to "1" and obtain \sup_{n\in\mathbb{N}}\|T_n'\|<\infty.

I'm not sure whether my argument in "1" is correct, can you please comment?

 

[micromass]

1) Knowing that \lim_{n\to\infty}f(T_nx)=\lim_{n\to\infty}T_n'(f)(x) exists for all f\in X' (is given), one can say that for any n\in\mathbb{N}: \|T_n'(f)\|<\infty

OK, I see how you can conclude that for any n seperately (you don't even need the limit for that, you are now just saying that T_n^\prime(f) is bounded), but

hence \sup_{n\in\mathbb{N}}\|T_n'(f)\|<\infty for all f\in X'.

I don't see how you can conclude this. Right now you are making a statement for all n. Before, you have made the statement for particular n.

For example, why can't you have

\|T_n^\prime(f)\|=2^n<+\infty

but then

\sup_n \|T_n^\prime(f)\|=+\infty

So I don't see how you can infer your statement about the supremum from the previous. You should clarify. (you will need UBP here)

 

[TaPaKaH]

In 1) one can apply Banach-Steinhaus Theorem and get that T'(f):=\lim_{n\to\infty}T_n'(f) is in B(X',X') so \|T'(f)\|<\infty for any f\in X' since \|T'\|<\infty and \|f\|<\infty.

So if for all f\in X' we have \lim_{n\to\infty}\|T_n'(f)\|=\|T'(f)\|<\infty then it implies that \sup_{n\in\mathbb{N}}\|T_n'(f)\|<\infty for all f\in X', no?

 

[micromass]

In 1) one can apply Banach-Steinhaus Theorem

What is the Banach-Steinhaus theorem to you?? Usually it is the same as the UBP, but you seem to imply that there is a difference.

and get that T'(f):=\lim_{n\to\infty}T_n'(f)

It is not obvious to me that you can actually make that definition. All you know is that

\lim_{n\rightarrow +\infty} T_n^\prime(f)(x)

converges for all f and all x. Why does there exist an operator T such that

T^\prime (f)(x)=\lim_{n\rightarrow +\infty} T_n^\prime (f)(x)

 

[TaPaKaH]

in the literature I'm using, first Theorem from here is called UBP and Corollary is called BST, excuse me for possible confusion

Now I am slightly confused. You suggested me to apply UBP to T_n'(f)(x) but I can't see a way to do this to get \sup_n\|T_n'(f)\|<\infty.

 

[micromass]

All you have to show is that if f is fixed and x is fixed but arbitrary, then

\sup_n \|T_n^\prime (f)(x)\|<+\infty

then UBP would imply

\sup_n \|T_n^\prime(f)\|<+\infty

 

[TaPaKaH]

Do I now get the idea correctly?

1. For any f\in X' and x\in X: existence of \lim_{n\to\infty}T_n'(f)(x) implies existence of \lim_{n\to\infty}\|T_n'(f)(x)\|. Every element of sequence is bounded, so \sup_n\|T_n'(f)(x)\|<\infty.

2. Applying UBP on x and then on f to \sup_n\|T_n'(f)(x)\|<\infty we get \sup_n\|T_n'\|<\infty.

 

[micromass]

That seems right!