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 separately (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!