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


4 Answers 4


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 $$\limsup_{p\to 0}\left(\int_{\Omega}|f|^p\mathrm d\mu\right)^{1/p} \leqslant \frac 1{R_0}+\exp\left(\int_\Omega\log|f_{R_0} |\mathrm \mu\right)\leqslant 2\varepsilon.$$

  • $\begingroup$ Thank you for the nice proof! Do you know how to extend the argument in case of $\int \log |f| d\mu =-\infty$? $\endgroup$
    – PhoemueX
    Jan 17, 2016 at 18:57
  • $\begingroup$ @PhoemueX I think that a truncation argument can do the trick (I have edited). $\endgroup$ Jan 17, 2016 at 22:25
  • $\begingroup$ @DavideGiraudo Why is the sequence $\{f_n\}$ converging to $0$ a.e.? $\endgroup$
    – nekodesu
    Feb 25, 2018 at 19:23
  • 2
    $\begingroup$ How about $f_R=|f|\textbf{1}_{\{|f|>1/R\}}+\textbf{1}_{\{|f|\leq1/R\}} /R$? $\endgroup$
    – blancket
    May 3, 2020 at 9:04
  • 2
    $\begingroup$ The proof for $\int\log|f|\,d\mu=-\infty$ is incorrect. to begin with $-\log|f_R|\leq\log R$ is false since $-\log|f_R|=\infty$ on $\{|f|<1/R\}$. $\endgroup$
    – Mittens
    Jan 24, 2021 at 2:41

@DavideGiraudo: This is a long comment an so I laid it as an answer.

  1. I did not find a direct way to fix the problem for the case where $\log|f|\notin L_1(\mu)$ in the spirit of our solution.
  2. I did however obtained a slightly different solution that applies to whether $\log|f|$ is integrable or $\int(\log|f|)_-=\infty$. Here is a sketch:

Basically one notices that for any $a>0$, the map $\phi_a(p)=\frac{a^p-1}{p}$ is monotone nondecreasing on $(0,\infty)$ (due to convexity of $p\mapsto a^p$) and so, $g_p:=(|f|-1)-\frac{|f|^p-1}{p}$ is positive and nondecreasing as $p\searrow0$. Monotone convergence implies that $$\lim_{p\rightarrow0+}\int g_p=\int \lim_{p\rightarrow0+}g_p=\int(|f|-1-\log|f|)$$ Thus, $\lim_{p\rightarrow0+}\int\frac{|f|^p-1}{p}=\int\log|f|$ regardless of integrability of $\log|f|$. The conclusion follows now from the inequality $\log(a)\leq a-1$ for all $a>0$ and Jensen's inequality: \begin{align} \int_\Omega\log|f|\,d\mu&= \frac{1}{p}\int_\Omega\log(|f|^p)\,d\mu\leq \frac{1}{p}\log\Big(\int_\Omega|f|^p\,d\,\mu\Big)=\log\|f\|_p\\ &\leq \frac{\|f\|^p_p-1}{p}=\int_\Omega\frac{|f|^p -1}{p}\,d\mu\xrightarrow{p\rightarrow0+}\int_\Omega\log|f|\,d\mu \end{align}


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.

  • $\begingroup$ I would just like to point out that, with some technical pain, this method also works even if we do not assume that $\log|f|\in L^1$. $\endgroup$
    – Lentes
    Oct 23, 2020 at 1:07
  • $\begingroup$ -1. The first equation does not hold --- at least not proved to hold. You need Jensen's inequality then apply the last line of @OliverDiaz. Or you need to find an alternative proof. $\endgroup$
    – Hans
    Mar 1, 2021 at 22:17
  • $\begingroup$ @Hans What exactly does not hold? The first equality is nothing more than the definition of the derivative. $\endgroup$
    – shalop
    Mar 1, 2021 at 23:30
  • $\begingroup$ No. The definition of the derivative gives only $\displaystyle\lim_{p \to 0} \frac{\log\int |f|^p d\mu}{p}=\frac{d}{dp}\log \int |f|^pd\mu \bigg|_{p=0}.$ You have to prove $\displaystyle \frac{d}{dp}\log \int |f|^pd\mu =\frac{\frac{d}{dp}\int |f|^pd\mu}{\int |f|^pd\mu}=\frac1{\int |f|^pd\mu}\int\frac{d}{dp} |f|^pd\mu$ and is continuous for $p\in[0,1]$ first. $\endgroup$
    – Hans
    Mar 2, 2021 at 8:32
  • $\begingroup$ Everything else looks fine. I guess downvoting it was too harsh. I will upvote once you fix the little flaw. $\endgroup$
    – Hans
    Mar 2, 2021 at 14:49

Alternative solution

Suppose $f>0$. Indeed by the monotone convergence theorem $$\displaystyle \dfrac{f^p-1}{p}\searrow \ln f\in L^1(\Omega)\,\,\,\,\mbox{when}\,\,\,\, p\to 0^+\Rightarrow \int_{\Omega}\dfrac{f^p-1}{p}\searrow\int_{\Omega} \ln f \,\,\,\,\mbox{when}\,\,\,\, p\to 0^+. $$

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 $$\lim\limits_{p\to 0^+}\left(1+g(p)p\right)^{1/p}=\lim\limits_{p\to 0^+}\exp\left(g(p)\ln(1+g(p)p)^{1/g(p)p}\right)\\ =\exp\left(\lim_{p\to 0^+}g(p)\right)=\exp\left(\frac{1}{|\Omega|}\int_{\Omega}\ln f\right)$$

Edit: On my first version of this answer, i use a wrong step noted by @Hans to justify the interchange of the limit with the integral, it can be corrected using that $h(p)=(a^p-1)/p$ is a nondecreascing function of $p>0$ for each fixed $a>0$, indeed with these restrictions $h’(p)\geq 0$.

  • 1
    $\begingroup$ -1. "for $p\ll 1$, we have $\left|\dfrac{f^p-1}{p}\right|\le 1+\left|\ln f\right|.$" This is wrong. $\endgroup$
    – Hans
    Mar 1, 2021 at 19:06
  • $\begingroup$ I really thank u to point my mistake, I'll correct this right now!. $\endgroup$ Jul 21, 2021 at 15:05
  • $\begingroup$ How are we using MCT on $\{(f^{p_n}-1)/p_n\}_n$ if it is decreasing and is also not non-negative (e.g. for |f|<1)? Also, if we are using MCT, do we need to assume $\ln f$ is integrable? $\endgroup$
    – ciru_4011
    Nov 20, 2021 at 11:27
  • $\begingroup$ For you first question i cite the Theorem 5.2 of Zygmund : Measure and Integral - An introduction to real analysis by using that f is in $L^1$, i take $f>0$ because $f$ is in $L^1$ if and only $|f|$ is in $L^1$. For the second quest in my first look we need take care, but we try justify the passage of the limit on the second equality by using the characterization of “e^x” in terms of limits going to infinity.. $\endgroup$ Nov 24, 2021 at 1:35
  • $\begingroup$ PS 1: in my question i assume ln f in $L^1$. $\endgroup$ Nov 24, 2021 at 1:43

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