# $L^p$ norm and integral equality prove [duplicate]

Possible Duplicate:
Integrate and measure problem.

Assume $\mu(X)=1$, $f \in L^{p} (X,M,\mu)$ for some $0<p \le \infty$

I want to prove that: $$\lim_{p\to 0}||f||_p = e^{\int_X \log|f|d \mu}$$

I'm going to prove $\ge$ part using Jensen inequality, but I cannot go opposite side. How can I make it?

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## marked as duplicate by Willie WongJun 14 '12 at 9:20

The discrete (finite) version of this statement can be shown directly by continuity of exponentiation and l'Hospital rule, but I'm not sure how to generalize it to arbitrary spaces. –  tomasz Jun 14 '12 at 1:55
The same question here –  leo Jun 14 '12 at 4:15

(I didn't have time to consider the case $\int_X\log|f|=-\infty$, so I post what I have)
Assume first that $\int_X\log|f|\,d\mu$ is finite.
Using that $\mu(X)=1$, that $p$ can be assumed small, and Taylor approximations around $0$ for $\log(1+t)$, $e^t$ (i.e. $\log(1+t)\simeq t$, $e^t\simeq1+t$), $$e^{\int_X\log|f|\,d\mu}=\lim_{p\to0}e^{\frac1p\int_Xp\log|f|}=\lim_{p\to0}e^{\frac1p\log\left(1+\int_Xp\log|f|\right)}=\lim_{p\to0}\left(1+\int_Xp\log|f|\right)^{1/p} =\lim_{p\to0}\left(\int_X1+p\log|f|\right)^{1/p} =\lim_{p\to0}\left(\int_Xe^{p\log|f|}\right)^{1/p}\\ =\lim_{p\to0}\left(\int_X{|f|^p}\,\right)^{1/p} =\lim_{p\to0}\|f\|_p.$$
In the case where $\int_X\log|f|\,d\mu=\infty$, then $\|f\|_p=\infty$ for all $p$ and so the equality holds. Indeed, if $\|f\|_p<\infty$ for some $p$, using that there exists $k>0$ such that $\log t\leq t^p$ if $t>k$, we get $$\int_X\log|f|=\int_{|f|\leq k}\log|f|+\int_{|f|>k}\log|f|\leq\log k + \int_{|f|>k}|f|^p\leq\log k +\|f\|_p<\infty.$$
I am wondering how to push this to the case $\int_{X}Log|f|=-\infty$. But the other post solved this issue. –  Bombyx mori Dec 25 '12 at 7:55