1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    [tex]\|T_n^\prime(f)\|=2^n<+\infty[/tex]

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    That seems right!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Exercise on dual / second dual spaces (functional analysis)
  1. Dual space (Replies: 4)

Loading...