Solving L^p, L^q Subset Inequalities in X Sets of Arbitrary Size

[Edwinkumar]

Suppose 0 < p < q < \infty. Then L^p \nsubseteq L^q iff X contains sets of arbitrarily small positive measure.

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

Can anyone give some hints to solve the other part?

Thanks

 

[Billy Bob]

Try this? If \int|g|<\infty consider E_n=\{x:|g(x)|>n\}.

 

[Edwinkumar]

Try this? If \int|g|<\infty consider E_n=\{x:|g(x)|>n\}.

I don't know how is the above trure.

We know only that there is a function f in L^p but not in L^q. From this we have to show that X contains sets of arbitrarily small measure.

 

[Billy Bob]

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.

 

[Edwinkumar]

Yes using the fact that p&lt;q, I proved that \int |f|^q&lt;\infty on {|f|\le 1
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).

 

[Billy Bob]

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).

 

[Edwinkumar]

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 E_n=\{x:|f(x)|^p&gt;n\} then E_n=\{x:|f(x)|&gt;n^{1/p}\} and E_1\subset E_2\subset...}
Moreover, since X does not contain sets of arbitrarily small measure, \exists \epsilon &gt;0 s.t. \mu(E)\ge \epsilon for all E\subset X
From these facts I m unable to figure it out.
Thanks for your replies Billy Bob.

 

[Billy Bob]

E_1\supset E_2\supset\dots

 

[Edwinkumar]

E_1\supset E_2\supset\dots

yes absolutely..sorry. Then how..?

 

[Billy Bob]

Suppose |f|^p was not bounded.

Consider \int_{E_n}|f|^p

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

 

[Edwinkumar]

Thank you very much Billy Bob! I completely got it now.
\int_{E_n}|f|^p\ge n\mu(E_n)
Therefore, \mu(E_n)\le 1/n\int_{E_n}|f|^p\le 1/n\int |f|^p
So \mu(E_n)=0 for some n or \mu(E_n)\to 0
The first one implies that f is bounded and the second one implies X contains sets of arbitrarily small positive measures.
Am I right?

 

[Billy Bob]

Am I right?

Very nice