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Homework Help: Exercise on dual / second dual spaces (functional analysis)

  1. Dec 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex](X,\|\cdot\|)[/itex] be a reflexive Banach space. Let [itex]\{T_n\}_{n\in\mathbb{N}}[/itex] be a sequence of bounded linear operators from [itex]X[/itex] into [itex]X[/itex] such that [itex]\lim_{n\to\infty}f(T_nx)[/itex] exists for all [itex]f\in X'[/itex] and [itex]x\in X[/itex].
    Use the Uniform Boundedness Principle (twice) to show that [itex]\sup_{n\in\mathbb{N}}\|T_n'\|<\infty[/itex].

    2. Relevant equations
    For operators between normed spaces we have [itex]\|T'\|=\|T\|[/itex], but I'm not sure if this can help in this case.

    3. The attempt at a solution
    I am currently at loss how to deal with information on functions [itex]f[/itex] to apply UBP.
    Any hints welcome.
  2. jcsd
  3. Dec 22, 2012 #2
    What is [itex]T_n^\prime[/itex]??

    It might be a good idea to write [itex]f(T_n x) = \widehat{T_n x}(f)[/itex], where [itex]\widehat{T_n x} \in X^{\prime\prime}[/itex]. Apply UBT on this.
  4. Dec 22, 2012 #3
    [itex]T_n'[/itex] is the dual operator of operator [itex]T[/itex], i.e. [itex]T'\in B(X',X')[/itex] s.t. [itex]T'(f)(x)=f(Tx)[/itex] for [itex]x\in X[/itex] and [itex]f\in X'[/itex].

    Do I get this right that you denote the dual operator with "hat" symbol?
  5. Dec 22, 2012 #4
    Yes, I mean the dual operator.

    Ok, so you need to prove [itex]\sup_n \|T_n^\prime\|<+\infty[/itex]. What does the UBP tell you?? Is it sufficient to prove something easier?
  6. Dec 22, 2012 #5
    I came up with (what I think is) a proof in 3 steps using UBP only once.

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

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

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

    I'm not sure whether my argument in "1" is correct, can you please comment?
  7. Dec 22, 2012 #6
    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 [itex]T_n^\prime(f)[/itex] is bounded), but

    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


    but then

    [tex]\sup_n \|T_n^\prime(f)\|=+\infty[/tex]

    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)
  8. Dec 22, 2012 #7
    In 1) one can apply Banach-Steinhaus Theorem and get that [itex]T'(f):=\lim_{n\to\infty}T_n'(f)[/itex] is in [itex]B(X',X')[/itex] so [itex]\|T'(f)\|<\infty[/itex] for any [itex]f\in X'[/itex] since [itex]\|T'\|<\infty[/itex] and [itex]\|f\|<\infty[/itex].

    So if for all [itex]f\in X'[/itex] we have [itex]\lim_{n\to\infty}\|T_n'(f)\|=\|T'(f)\|<\infty[/itex] then it implies that [itex]\sup_{n\in\mathbb{N}}\|T_n'(f)\|<\infty[/itex] for all [itex]f\in X'[/itex], no?
    Last edited: Dec 22, 2012
  9. Dec 22, 2012 #8
    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.

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

    [tex]\lim_{n\rightarrow +\infty} T_n^\prime(f)(x)[/tex]

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

    [tex]T^\prime (f)(x)=\lim_{n\rightarrow +\infty} T_n^\prime (f)(x)[/tex]
  10. Dec 22, 2012 #9
    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 [itex]T_n'(f)(x)[/itex] but I can't see a way to do this to get [itex]\sup_n\|T_n'(f)\|<\infty[/itex].
  11. Dec 22, 2012 #10
    All you have to show is that if f is fixed and x is fixed but arbitrary, then

    [tex]\sup_n \|T_n^\prime (f)(x)\|<+\infty[/tex]

    then UBP would imply

    [tex]\sup_n \|T_n^\prime(f)\|<+\infty[/tex]
  12. Dec 22, 2012 #11
    Do I now get the idea correctly?

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

    2. Applying UBP on x and then on f to [itex]\sup_n\|T_n'(f)(x)\|<\infty[/itex] we get [itex]\sup_n\|T_n'\|<\infty[/itex].
  13. Dec 22, 2012 #12
    That seems right!
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