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 $$p<q$$, I proved that $$\int |f|^q<\infty$$ 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 $$E_n=\{x:|f(x)|^p>n\}$$ then $$E_n=\{x:|f(x)|>n^{1/p}\}$$ and $$E_1\subset E_2\subset...}$$
Moreover, since $$X$$ does not contain sets of arbitrarily small measure, $$\exists \epsilon >0$$ 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