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.
 A: 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...)
