Lp space question

  • Thread starter Edwinkumar
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  • #1
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Main Question or Discussion Point

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
 

Answers and Replies

  • #2
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Try this? If [tex]\int|g|<\infty[/tex] consider [tex]E_n=\{x:|g(x)|>n\}[/tex].
 
  • #3
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Try this? If [tex]\int|g|<\infty[/tex] consider [tex]E_n=\{x:|g(x)|>n\}[/tex].
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.
 
  • #4
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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.
 
  • #5
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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).
 
  • #6
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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.


But I don't know how to make use of the fact the X doesn't contain sets of arbitrarily small positive measure in proving (2).
You must mean (1), since you proved (2) already. The condition on X is only needed to prove (1).
 
  • #7
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You must mean (1), since you proved (2) already. The condition on X is only needed to prove (1).
Yes absolutely.
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.
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.
 
  • #8
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[tex]E_1\supset E_2\supset\dots[/tex]
 
  • #9
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[tex]E_1\supset E_2\supset\dots[/tex]
yes absolutely..sorry. Then how..?
 
  • #10
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Suppose |f|^p was not bounded.

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

How small, in measure, can E_n get, anyway?
 
  • #11
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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?
 
  • #12
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