Show that if $\mu(f>0)<1$ then $\lim\limits_{p\to 0^+}||f||_p=0$

Hint: Use Hölder's inequality.

But I can't see where I should use it.

I'm trying to use it in $\displaystyle\int |f|^p\,d\mu = \int |f|^{p-q}|f|^{q}$, but it didn't help.

  • $\begingroup$ What is $\mu (f>0)$? $\endgroup$ – Owen Aug 21 '15 at 5:21
  • 1
    $\begingroup$ Are you sure that it's $\lim\limits_{ p \to 0^+}$? $\| \cdot \|_p$ is only a norm for $p \geq 1$. I know that often the same definition is used even for when $p < 1$, but since Holder's inequality holds only for $p > 1$, it seems a bit odd. $\endgroup$ – Marcus M Aug 21 '15 at 5:22
  • 1
    $\begingroup$ @Owen $\mu(f > 0) = \mu\left(\left\{ \omega \in \Omega : f(\omega) > 0 \right\} \right)$. $\endgroup$ – Marcus M Aug 21 '15 at 5:33
  • $\begingroup$ @MarcusM It's just like I said. It seems odd to me to. $\endgroup$ – Andre Gomes Aug 21 '15 at 5:37
  • 1
    $\begingroup$ $\lim\limits_{p\to 0^+}||f||_p=0$ or $\lim\limits_{p\to 0^+}||f||_p=1$ ? $\endgroup$ – hermes Aug 21 '15 at 6:01

You're leaving out a hypothesis - as stated the problem is false, it can happen that $||f||_p=\infty$ for every $p>0$.

If you assume that there exists $p>0$ with $||f||_p<\infty$ then yes, the limit is $0$ (not $1$ as has been claimed), and yes this follows from Holder's inequality, even though we're talking about $p<1$ and Holder only applies to $p\ge1$. The $p$ in $||f||_p$ is not the $p$ we're going to use in Holder.

Also any proof that doesn't use the fact that $\mu(f>0)<1$ is wrong; without that assumption the limit could be more or less anything (in general the limit is $\exp(\int\log f)$).

I'm not going to do the whole problem, just an illustration of how the proof is going to work. Say $E$ is the set where $f>0$. Assume that $||f||_1<\infty$.

Note that $f^{1/2} = \Bbb 1_E f^{1/2}$. The Cauchy-Schwarz inequality shows that $$\int f^{1/2}=\int\Bbb 1_Ef^{1/2}\le||\Bbb1_E||_2||f^{1/2}||_2 =\mu(E)^{1/2}\left(\int f\right)^{1/2}.$$Square both sides and you get $$||f||_{1/2}\le\mu(E)||f||_1.$$

Note how we proved something about $||f||_{1/2}$ using Cauchy-Schwarz, which only talks about $p=2$. You can prove similar but more general inequalities using Holder, which will show that $||f||_p\to0$. (The fact that $\mu(E)<1$ is going to be important...)

| cite | improve this answer | |
  • $\begingroup$ Assuming that $||f||_1<\infty$ isn't equivalent to assuming that $f$ is integrable? $\endgroup$ – Andre Gomes Aug 21 '15 at 17:05
  • $\begingroup$ Yes of course those two are equivalent. What did I say that you think says this is not so? $\endgroup$ – David C. Ullrich Aug 21 '15 at 17:09
  • $\begingroup$ Then I did't get this part "If you assume that there exists $p>0$ with $||f||_p<\infty$ then yes, the limit is 0 (not 1 as has been claimed)" $\endgroup$ – Andre Gomes Aug 21 '15 at 17:24
  • $\begingroup$ I don't understand what you don't get about that... $\endgroup$ – David C. Ullrich Aug 21 '15 at 17:28
  • 1
    $\begingroup$ Yes, it suffices to assume that $f$ is integrable! II didn't say otherwise. As you pointed out, assuming that $f$ is integrable is the same as assuming that $||f||_1<\infty$. And if $||f||_1<\infty$ then there exists a $p>0$ such that $||f||_p<\infty$. Assuming that $f$ is integrable is a stronger assumption. $\endgroup$ – David C. Ullrich Aug 21 '15 at 17:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.