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Let $1\leq p<q<\infty$. Which of the following inclusions are true?

  1. $L^p(0,1)\subset L^q(0,1)$
  2. $L^q(0,1)\subset L^p(0,1)$
  3. $L^p(0,\infty)\subset L^q(0,\infty)$
  4. $L^q(0,\infty)\subset L^q(0,\infty)$

I already know that 1. is false (consider $f(x)=1/\sqrt{x}$ with $p=1$, $q=2$) and 2. holds, which can be shown using the Hölder-inequality.

Now I'm not sure about 3. and 4. I think 3. doesn't hold either, but cannot think of an example to show this. Finally 4. I think is wrong either, since as far as I know the inclusion $L^q(\Omega)\subset L^p(\Omega)$ only holds if $\lambda(\Omega)<\infty$. But again, I cannot think of a counter-example for this either.

Can anyone think of good examples for this? (Or correct my answer if I'm wrong) Thanks!

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    $\begingroup$ A counterexample for 3. is $f(x) = x^{-1/q} \chi_{(0,1)}$. A counterexample for 4. is $f(x) = x^{-1/p} \chi_{(1,\infty)}$. $\endgroup$
    – Umberto P.
    Apr 24, 2014 at 15:04

1 Answer 1

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Suppose that $(X,\mathcal M,\mu)$ is a measure space and that $1 \le p < q < \infty$. Then

  1. $L^q(\mu) \subset L^p(\mu)$ if and only if $X$ does not contain sets of arbitrarily large finite measure, and
  2. $L^p(\mu) \subset L^q(\mu)$ if and only if $X$ does not contain sets of arbitrarily small positive measure.

In your case, 2 is true but the others are all false.

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  • $\begingroup$ Oh, this is very useful! Thank you! $\endgroup$
    – dinosaur
    Apr 24, 2014 at 16:42

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