# 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 $\log|f|\in L^1(\Omega).$

• – user940 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 '16 at 18:57
• @PhoemueX I think that a truncation argument can do the trick (I have edited). – Davide Giraudo Jan 17 '16 at 22:25
• @DavideGiraudo Why is the sequence $\{f_n\}$ converging to $0$ a.e.? – nekodesu Feb 25 '18 at 19:23
• How can we apply the first case to $f_R$? I think $\int_\Omega \log |f_R|d\mu =-\infty$ – probably123 May 3 at 7:30
• How about $f_R=|f|\textbf{1}_{\{|f|>1/R\}}+\textbf{1}_{\{|f|\leq1/R\}} /R$? – probably123 May 3 at 9:04

When $|f|,\log|f| \in L^1$, we may prove this by recognizing the definition of the derivative:

Indeed, we can take logs to see that $$\lim_{p \to 0} \frac{\log\int |f|^p d\mu}{p}=\frac{d}{dp} \int |f|^pd\mu \bigg|_{p=0} = \int \frac{d}{dp}|f|^p\bigg|_{p=0}d\mu = \int \log|f|d\mu.$$

The reason we can put the derivative inside the integral sign is because we know that $\frac{d}{dp} |f|^p = |f|^p\log|f|$ which is bounded (uniformly in $p \leq 1/2$) by $2|f| + |\log|f||\in L^1$. Indeed, $|f|^p|\log|f|| \leq |\log|f||$ when $|f|\leq 1$, and $|f|^p |\log |f||\leq |f|^{1/2}|\log|f||\leq2|f|$ when $|f|\geq 1$ (since $|\log u| \leq 2u^{1/2}$ for $u \geq 1$). Thus applying Theorem 3.5.1 in these notes gives the second equality above.

Note that this is essentially the same proof given in the other answer above, the main point is to use dominated convergence. I just wanted to exposit it in a slightly different way.

Alternative solution

Suppose f>0. Indeed $$\displaystyle \lim\limits_{p\to 0^+} \dfrac{f^p-1}{p} = \ln f \Rightarrow \int_{\Omega}\dfrac{f^p-1}{p}\to \int_{\Omega} \ln f,$$cause for $$p\ll 1$$, we have $$\left|\dfrac{f^p-1}{p}\right|\le 1+\left|\ln f\right|.$$

Now observe that $$\displaystyle \left(\dfrac{1}{|\Omega|}\int_{\Omega}f^p\right)^{1/p}=\displaystyle \left(1+\dfrac{p}{|\Omega|}\int_{\Omega}\dfrac{f^p-1}{p}\right)^{1/p}.$$

Write $$\displaystyle \dfrac{1}{|\Omega|}\int_{\Omega}\dfrac{f^p-1}{p}=g(p)$$, and see that $$\lim\limits_{p\to 0^+}\left(1+g(p)p\right)^{1/p}=\lim\limits_{p\to 0^+}\exp\left(g(p)\ln(1+g(p)p)^{1/g(p)p}\right)\\ =\exp\left(\lim_{p\to 0^+}g(p)\right)=\exp\left(\frac{1}{|\Omega|}\int_{\Omega}\ln f\right)$$