Limit of norm $L^p$ when $p\to 0$ [duplicate]

Solution 1:

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

Solution 2:

@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}

Solution 3:

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.