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Lp space question

  1. Apr 13, 2009 #1
    Suppose [tex]0 < p < q < \infty[/tex]. Then [tex]L^p \nsubseteq L^q[/tex] iff [tex]X[/tex] contains sets of arbitrarily small positive measure.

    I have proved one part, namely, if [tex]X[/tex] contains sets of arbitrarily small positive measure then [tex]L^p \nsubseteq L^q[/tex]

    Can anyone give some hints to solve the other part?

    Thanks
     
  2. jcsd
  3. Apr 13, 2009 #2
    Try this? If [tex]\int|g|<\infty[/tex] consider [tex]E_n=\{x:|g(x)|>n\}[/tex].
     
  4. Apr 14, 2009 #3
    I don't know how is the above trure.

    We know only that there is a function [tex]f[/tex] in [tex]L^p[/tex] but not in [tex]L^q[/tex]. From this we have to show that [tex]X[/tex] contains sets of arbitrarily small measure.
     
  5. Apr 14, 2009 #4
    Let f be in L^p and suppose X does not contain sets of arbitrarily small measure. Prove f is in L^q.

    Intuitively, to show integral of |f|^q is finite, you have to show (1) |f| can't be too large, and (2) in the places where |f| is small, the integral is still finite.

    Use my earlier hint to deal with (1). Either use g=f^p or g=f^q.

    For (2), you'll simply use p<q.
     
  6. Apr 15, 2009 #5
    Yes using the fact that [tex]p<q[/tex], I proved that [tex]\int |f|^q<\infty[/tex] on [tex]{|f|\le 1[/tex]
    But I don't know how to make use of the fact the [tex]X[/tex] doesn't contain sets of arbitrarily small positive measure in proving (2).
     
  7. Apr 15, 2009 #6
    Let f be in L^p and suppose X does not contain sets of arbitrarily small measure. Use the hint with g=f^p to prove |f| must be bounded.


    You must mean (1), since you proved (2) already. The condition on X is only needed to prove (1).
     
  8. Apr 15, 2009 #7
    Yes absolutely.
    If [tex]E_n=\{x:|f(x)|^p>n\}[/tex] then [tex]E_n=\{x:|f(x)|>n^{1/p}\}[/tex] and [tex]E_1\subset E_2\subset...}[/tex]
    Moreover, since [tex]X[/tex] does not contain sets of arbitrarily small measure, [tex]\exists \epsilon >0[/tex] s.t. [tex]\mu(E)\ge \epsilon[/tex] for all [tex]E\subset X[/tex]
    From these facts I m unable to figure it out.
    Thanks for your replies Billy Bob.
     
  9. Apr 15, 2009 #8
    [tex]E_1\supset E_2\supset\dots[/tex]
     
  10. Apr 15, 2009 #9
    yes absolutely..sorry. Then how..?
     
  11. Apr 15, 2009 #10
    Suppose |f|^p was not bounded.

    Consider [tex]\int_{E_n}|f|^p[/tex]

    How small, in measure, can E_n get, anyway?
     
  12. Apr 15, 2009 #11
    Thank you very much Billy Bob! I completely got it now.
    [tex]\int_{E_n}|f|^p\ge n\mu(E_n)[/tex]
    Therefore, [tex]\mu(E_n)\le 1/n\int_{E_n}|f|^p\le 1/n\int |f|^p[/tex]
    So [tex]\mu(E_n)=0[/tex] for some n or [tex]\mu(E_n)\to 0[/tex]
    The first one implies that [tex]f[/tex] is bounded and the second one implies X contains sets of arbitrarily small positive measures.
    Am I right?
     
  13. Apr 16, 2009 #12
    Very nice
     
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