- #1

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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

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- Thread starter Edwinkumar
- Start date

- #1

- 23

- 0

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

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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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I don't know how is the above trure.Try this? If [tex]\int|g|<\infty[/tex] consider [tex]E_n=\{x:|g(x)|>n\}[/tex].

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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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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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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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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Yes absolutely.You must mean (1), since you proved (2) already. The condition on X is only needed to prove (1).

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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yes absolutely..sorry. Then how..?[tex]E_1\supset E_2\supset\dots[/tex]

- #10

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

How small, in measure, can E_n get, anyway?

- #11

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[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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Am I right?

Very nice

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