Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question Regarding L^p Spaces

  1. Jul 28, 2011 #1
    This is a functional analysis qualifying exam problem that I can't figure out. Any assistance would be appreciated since I have to take a similar qual soon. I was able to make some limited progress in the p=2 case using Holders inequality.

    Suppose [itex]f_n, f\in L^p[/itex] where [itex]1\le p <\infty[/itex] and that [itex] f_n \rightarrow f[/itex] a.e. Show that [itex]\|f_n-f\|_p \rightarrow 0[/itex] iff [itex] \|f_n\|_p \rightarrow \|f\|_p [/itex].
     
    Last edited: Jul 28, 2011
  2. jcsd
  3. Jul 28, 2011 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Hi tornado28! :smile:

    First reduce everything to the case p=1. That is, assume that you know the result for p=1, show that it also holds for p>1.

    In reducing this case, you will need the inequality:

    For [itex]x,y\geq 0[/itex], then [itex]|x-y|^p\leq |x^p-y^p|[/itex]. Try to prove this.
     
  4. Jul 29, 2011 #3
    This is in fact a difficult question. I had the same when I took my analysis qual a few years ago. Here is how you prove it. First, notice that [itex] \|f_n\|_p =\| f_n-f+f \|_p \leq \| f_n-f \|_p + \| f\|_p[/itex] This implies that [itex] \| f_n\|_p-\| f\|_p \leq \| f_n-f \|_p [/itex]. Thus if lim[itex] (\| f_n-f \|_p)=0 [/itex] then lim[itex] (\| f_n\|_p - \| f \|_p)=0 [/itex]
    Next, to show the second part, use the fact that [itex]2^p( |f_n|^p+|f|^p-|f_n - f|^p) \geq 0[/itex] and lim[itex]( 2^p( |f_n|^p+|f|^p-|f_n - f|^p))=2^{p+1}|f|^p [/itex] and apply Fatou's lemma. I will let you finish the rest.
    Vignon Oussa
     
    Last edited: Jul 29, 2011
  5. Aug 4, 2011 #4
  6. Aug 5, 2011 #5
    Ok, I understand why [itex]\|f_n-f\|_p \rightarrow 0 \Rightarrow \|f_n\|_p \rightarrow \|f\|_p [/itex], but I still don't have a solution for the other direction. In your solution, Vig, it is not necessarily true that [itex]2^p( |f_n|^p+|f|^p-|f_n - f|^p) \geq 0[/itex] since we could, for instance, have [itex]f_n=1[/itex], [itex]f = -1[/itex], and [itex]p=2[/itex].

    I think I could do it using Egorov's Theorem in the case that the underlying space is sigma compact, but since both of you seem to think that it can be done in general I wonder if I could have another hint.

    I thought I had a proof using the inequality you suggested Micro, but I'm running into the same problem. It's not necessarily the case that [itex]|f_n - f|^p \le \left| |f_n|^p - |f|^p \right|[/itex] when [itex]f_n[/itex] and [itex]f[/itex] have opposite signs.
     
  7. Aug 5, 2011 #6

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Ok, let's do this in the way I know works (I was trying to find an easy way out). Firstly, establish that

    [tex]\lim_{n\rightarrow +\infty}{2^p(|f_n|^p+|f|^p)-|f_n-f|^p}=2^{p+1}|f|^p[/tex]

    Now, apply Fatou's lemma to calculate

    [tex]\int{2^{p+1}|f|^pd\mu}[/tex]

    (note: to be able to apply Fatou's lemma, you'll need to know that [itex]2^p(|f_n|^p+|f|^p)-|f_n-f|^p\geq 0[/itex]. To show that this is the case, apply that the function [itex]\Phi(x)=x^p[/itex] is convex)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Question Regarding L^p Spaces
  1. L^p Spaces (Replies: 17)

  2. For 1 < p < oo l^p (Replies: 1)

Loading...