For $0

Question

In this problem Limit of $L^p$ norm when $p\to0$, the writer states that for $0<p<1$ we have that $$\Big(\int_\Omega |f|^pd\mu\Big)^{1/p}\leq \frac{1}{R_0} + \Big(\int_\Omega |f_{R_0}|^p\Big)^{1/p} $$ where $\mu$ is a positive measure such that $\mu(\Omega)=1$ and $f_{R_0}=|f|1_A$ where $A={\{x: |f(x)|\geq \frac{1}{R_0}\}}$. I see that $$\int_\Omega |f|^pd\mu = \int_A |f|^pd\mu + \int_{A^c}|f|^pd\mu \leq \int_\Omega |f_{R_0}|^pd\mu + \frac{1}{R_0^p}$$ But I know that $a^{1/p} + b^{1/p}\leq (a+b)^{1/p}$ where $a,b\geq0$. So I am not sure how to distribute the the $1/p$ power.

Answer 1

It seems to me that the solution for the case $\int_\Omega\log|f|\,d\mu=-\infty$ quoted by the OP is incorrect. For example, the responder defines $f_R=|f|\mathbb{1}(|f|\geq 1/R)$ and claims that $$-\log f_R \leq \log(R)$$ This is valid on $\{|f|\geq1/R\}$ but false on $\{|f|<1/R\}$ for there $$-\log f_R =\infty$$ Also the use of Minkowski's inequality for $0<p<1$ is not valid.

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Anyway, if the OP is primarily interested in original problem, i.e. show that $$\lim_{p\rightarrow0}\Big(\int_\Omega|f|^p\,d\mu\Big)^{1/p}=\exp\Big(\int_\Omega\log|f|\,d\mu\Big)$$ then one may proceed as follows:

For any $a>0$, the function $p\mapsto a^p$ is convex and so the map $\phi_a(p)=\frac{a^p-1}{p}$ is monotone nondecreasing on $(0,\infty)$, and $\lim_{p\rightarrow0+}\phi(p)=\log(a)$. It follows that if $$g_p(\omega):=\Big(|f(\omega)|-1\Big) - \frac{|f(\omega)|^p -1}{p},$$ then $0\leq g_{p'}\leq g_p$ for all $0<p<p'\leq1$, and $g_p\xrightarrow{p\rightarrow0+}|f|-1-\log(|f|)$. An application of monotone convergence gives \begin{align} \lim_{p\rightarrow0+}\Big(\int_\Omega(|f|-1)\,d\mu-\int_\Omega\frac{|f|^p-1}{p}\,d\mu\Big)&=\lim_{p\rightarrow0+}\int_\Omega g_p\,d\mu\\ &=\int_\Omega\lim_{p\rightarrow0+}g_p\,d\mu=\int_\Omega(|f|-1)-\log|f|\,d\mu \end{align} whence we conclude that $$\lim_{p\rightarrow0+}\int_\Omega\frac{|f|^p-1}{p}\,d\mu=\int_\Omega\log|f|\,d\mu$$ regardless of whether $\int_\Omega\log|f|\,d\mu=-\infty$ or $\log|f|\in\mathcal{L}_1(\mu)$.

The conclusion follows from this along with the inequalities \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}=\frac{\int_\Omega|f|^p\,d\mu -1}{p}\xrightarrow{p\rightarrow0+}\int_\Omega\log|f|\,d\mu \end{align}

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A short proof that $\phi_a(p)=\frac{a^p-1}{p}$ is monotone nondecreasing in $(0,\infty)$

Claim: If $\phi$ is convex in $(a,b)$, then for any $a<x<u<y<b$ \begin{align} \frac{\phi(u)-\phi(x)}{u-x}\leq \frac{\phi(y)-\phi(x)}{y-x}\leq \frac{\phi(y)-\phi(u)}{y-u}\tag{1}\label{one} \end{align}

Indeed, if $u=(1-\lambda)x+\lambda y$, then $\lambda=\frac{u-x}{y-x}$ and from $$\phi((1-\lambda)x+\lambda y)\leq (1-\lambda)\phi(x)+\lambda\phi(x)$$ we get $$ \frac{\phi(u)-\phi(x)}{u-x}\leq \frac{\phi(y)-\phi(x)}{y-x} $$ A similar argument gives the second inequality in \eqref{one}. The converse holds to, that is if \eqref{one} holds for any $a<x<u<y<b$, then $\phi$ is convex in $(a,b)$.

In our case, let $\phi(p)=a^p$ ($a>0$). This is a convex function and with $x=0$ and $0<u<y$ $$ \frac{\phi(u)-\phi(0)}{u-0}=\frac{a^u-1}{u}\leq \frac{\phi(y)-\phi(0)}{y-0}=\frac{a^y-1}{y} $$