# How compute $\lim_{p\rightarrow 0} \|f\|_p$ in a probability space?

I not solve the follow limit $$\lim_{p\rightarrow 0} \bigg[\int_{\Omega} |f|^p d\mu \bigg]^{1/p} = \exp\bigg[ \int_{\Omega} \log|f|d\mu \bigg],$$

where $(\Omega, \mathcal{F}, \mu)$ is a probability space and $f,\log |f| \in L^1(\Omega).$

Can someone help me?

Thank you!

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$|f|^p =e^{\ln|f| p}\approx 1+\ln|f| p$, the integral of this is $1+p\int \ln|f|$. Now use $\lim (1+a\epsilon)^{1/\epsilon} = e^a$. –  Maesumi Jan 29 '13 at 4:22
Defenitely this is a duplicate, but I can't find the original question –  Norbert Jan 29 '13 at 5:08
@Maesumi: When you say $|f|^p \approx 1+p\ln|f|$, you mean $\lim_{p\to 0} \frac{|f|^p}{1+p\ln|f|}=1$, right? How do you then apply this to the integral? I don't know whether what you say can be made rigorous. Can you make it a bit precise? –  Kumara Jan 29 '13 at 7:01
For $p<1$ we do not have a norm to call it $||\cdot||_p$. But the limit question is legitimate and a similar limit is done for ordinary vectors in calculus. Unfortunately I do not have the details for a probability level course. –  Maesumi Jan 29 '13 at 13:29
–  Byron Schmuland Jan 29 '13 at 14:39

By Jensen's inequality $$\int \log |f|\le \log \|f\|_q.$$ Hence $$\int \log |f| \le \log \|f\|_q = \frac{1}{q} \log \int |f|^q \le \frac{\int (|f|^q-1)}{q}.$$ Now apply DCT to conclude that the right hand side goes to $\int \log |f|$ as $q$ tends to $0$ from the right.
Thus $$\|f\|_q \to \exp(\int \log |f|) \text{ as } q\to 0^+.$$