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

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.

• What is $\mu (f>0)$? – Owen Aug 21 '15 at 5:21
• 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. – Marcus M Aug 21 '15 at 5:22
• @Owen $\mu(f > 0) = \mu\left(\left\{ \omega \in \Omega : f(\omega) > 0 \right\} \right)$. – Marcus M Aug 21 '15 at 5:33
• @MarcusM It's just like I said. It seems odd to me to. – Andre Gomes Aug 21 '15 at 5:37
• $\lim\limits_{p\to 0^+}||f||_p=0$ or $\lim\limits_{p\to 0^+}||f||_p=1$ ? – 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...)

• Assuming that $||f||_1<\infty$ isn't equivalent to assuming that $f$ is integrable? – Andre Gomes Aug 21 '15 at 17:05
• Yes of course those two are equivalent. What did I say that you think says this is not so? – David C. Ullrich Aug 21 '15 at 17:09
• 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)" – Andre Gomes Aug 21 '15 at 17:24
• I don't understand what you don't get about that... – David C. Ullrich Aug 21 '15 at 17:28
• 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. – David C. Ullrich Aug 21 '15 at 17:38