# Limit of $L^p$ norm when $p\to0$

Let ($\Omega$, $\cal{F}$, $\mu$) be a probability space and $f\in L^1(\Omega)$. Prove that

$$\displaystyle\lim_{p\to 0} \left[ \int_{\Omega}|f|^pd\mu \right]^{\frac{1}{p}}=\exp \left[ \int_{\Omega}\log|f| d\mu \right],$$

where $\exp[-\infty]=0$. To simplify the problem, we may assume $\log|f|\in L^1(\Omega).$

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– Byron Schmuland Jan 30 '13 at 23:31

Assume that $\int_{\Omega}-\log|f|d\mu<\infty$. Let $g(p):=\frac 1p\log\int_{\Omega}|f|^pd\mu-\int_{\Omega}\log|f|d\mu$.
Since $t\mapsto \log t$ is concave, by Jensen inequality we get $g(p)\geqslant 0$. Using the inequality $\ln(1+t)\leqslant t$ we have $$0\leqslant g(p)\leqslant \frac 1p\left(\int_{\Omega}|f|^pd\mu-1\right)-\int_{\Omega}\log|f|d\mu.$$ Now the problem reduces to show that $\lim_{p\to 0}\frac 1p\left(\int_{\Omega}|f|^pd\mu-1\right)-\int_{\Omega}\log|f|d\mu=0$. To see that, take a sequence $\{p_n\}$ which converges to $0$ and put $f_n(x):=\frac{|f(x)|^{p_n}-1}{p_n}-\log |f(x)|$. The sequence $\{f_n\}$ converges almost everywhere to $0$ and we have, if $t\geq 1$, $0<p<1$ $$\left|\frac{t^p-1}p\right|=\int_1^t s^{p-1}ds\leqslant t-1$$ since the map $s\mapsto s^{p-1}$ is decreasing, and if $0<t<1$ $$\left|\frac{t^p-1}p\right|=\int_t^1s^{p-1}ds\leqslant \int_t^1s^{-1}ds=-\log t$$ so denoting $A=\{x, |f(x)|\geqslant 1\}$, $$\left|f_n(x)\right|\leqslant (|f(x)|-1)\mathbf 1_A(x)-\log|f(x)|\mathbf 1_{A^c}(x),$$ which is integrable. We can conclude by the dominated convergence theorem.
Now assume that $\int_{\Omega}\log|f|d\mu=-\infty$. Consider $f_R:=|f|\mathbf 1_{\{|f|\gt 1/R\}}$. Then $-\log |f_R|\leqslant \log R$, hence by the previous case, $$\tag{*} \lim_{p \to 0}\left[ \int_{\Omega}\left|f_R\right|^pd\mu \right]^{\frac{1}{p}}=\exp\left(\int_\Omega\log|f_R|\mathrm \mu\right).$$ Fix a positive $\varepsilon$ and by monotone convergence, we may choose $R_0$ such that $\exp\left(\int_\Omega\log|f_{R_0} |\mathrm \mu\right)\lt \varepsilon$ and $1/R_0\lt \varepsilon$. Then $$\left(\int_{\Omega}|f|^p\mathrm d\mu\right)^{1/p} \leqslant \frac 1{R_0}+ \left[ \int_{\Omega}\left|f_{R_0} \right|^pd\mu \right]^{\frac{1}{p}},$$ so that $$\limsup_{p\to 0}\left(\int_{\Omega}|f|^p\mathrm d\mu\right)^{1/p} \leqslant \frac 1{R_0}+\exp\left(\int_\Omega\log|f_{R_0} |\mathrm \mu\right)\leqslant 2\varepsilon.$$
Thank you for the nice proof! Do you know how to extend the argument in case of $\int \log |f| d\mu =-\infty$? – PhoemueX Jan 17 at 18:57