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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 Wong Jun 14 '12 at 9:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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 $$ \int_X\log|f|=\int_{|f|\leq k}\log|f|+\int_{|f|>k}\log|f|\leq\log k + \int_{|f|>k}|f|^p\leq\log k +\|f\|_p<\infty. $$

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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

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