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If $f \in L^{p_0}(X,M,\,u)$ for some $0, then $1. \lim_{p\to0}\int_X |f|^p \, d\mu=\mu(\{x \in X \mid f(x) \ne0\}).$

And if additional assume $\mu(X)=1$,

then I wanna prove that $f \in L^p(X,M,\,u)$ for some $0

, and the equation below. $2. \lim_{p\to0}\|f\|_p=e^{\int_X\log|f|\,d\mu}$

I want to know how can I conclude those results. First, I'm trying to use integrate over the set on which $0<|f(x)|\le1$ and use MCT. and the set on which $|f(x)|>1$ use LDCT to prove that. But I can't conclude to measure $\mu$ and how can I approach second fact?

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    Oh thanks again for your kind explain @uncookedfalcon.2012-06-11

2 Answers 2

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For the second equation: first assume that $f$ does not vanish on a set of positive measure. We have

$\log\|f\|_p = \frac{1}{p} \log(\int_X |f|^p d\mu) .$

We apply L'hopital's rule to take the limit as $p\rightarrow 0$. Since $|f|^p \log |f|$ is bounded by either a constant or $|f|^{p_0}$ for small $p$, we can differentiate under the integral sign to get

$\frac{d}{dp} \log(\int_X |f|^p d\mu) = \frac{\int_X \log |f| * |f|^p d\mu}{\int_X |f|^p d\mu}.$

Of course the derivative of the denominator $p$ is just 1. Therefore

$\lim_{p\rightarrow 0} \log\|f\|_p = \lim_{p\rightarrow 0}\frac{\int_X \log |f| * |f|^p d\mu}{\int_X |f|^p d\mu} = \frac{\int_X \log |f| d\mu}{\int_X 1 d\mu} = \int_X \log |f| d\mu ,$

by dominated convergence. It follows that

$\lim_{p\rightarrow 0} \|f\|_p = e^{\int_X \log |f| d\mu } .$

If $f = 0$ on a set $E$ of positive measure, then by H\'older's inequality (with $p_0^* = p_0/(1-p_0)$),

$\int_X |f|^p d\mu = \int_X \chi_{E^c} |f|^p d\mu \leq \||f|^p\|_{p_0} \|\chi_{E^c}\|_{p_0^*} = (\int_X |f|^{p p_0})^{1/p_0} \mu(E^c)^{1/p_0^*}.$

Thus,

$\|f\|_p \leq \|f\|_{p p_0} \mu(E^c)^{1/pp_0^*}.$

The first term here is bounded for small $p$ and the second tends to 0 as $p\rightarrow 0$ since $\mu(E^c) < 1$. Thus $\|f\|_p \rightarrow 0$, which equals $e^{\int_X \log |f| d\mu}$ if we interpret $e^{-\infty}$ as 0.

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    @wowhapjs Sorry, Hölder Inequality is NOT valid for p_0<1. So, for the proof above to work in the case "$f=0$ on a ser $E$ of positive measure", we need to include the assumption that $p_0 \geqslant 1$.2015-09-13
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$1$. Define $ \begin{align} E^>&=\{x\in X:|f(x)|\ge1\}\\ E^<&=\{x\in X:0<|f(x)|<1\}\\ E^=&=\{x\in X:|f(x)|=0\} \end{align}\tag{1a} $

On $E^>$, $|f(x)|^p$ decreases to $1$ as $p$ decreases to $0$; on $E^<$, $|f(x)|^p$ increases to $1$ as $p$ decreases to $0$; and on $E^=$, $|f(x)|=0$ as $p$ decreases to $0$.

Therefore, by monotone convergence on $E^<$ and dominated convergence on $E^>$, $ \begin{align} \lim_{p\to0^+}\int_{E^>}|f(x)|^p\,\mathrm{d}x&=\mu(E^>)\\ \lim_{p\to0^+}\int_{E^<}|f(x)|^p\,\mathrm{d}x&=\mu(E^<)\\ \lim_{p\to0^+}\int_{E^=}|f(x)|^p\,\mathrm{d}x&=0 \end{align}\tag{1b} $ Summing these yields $ \lim_{p\to0^+}\int_X|f(x)|^p\,\mathrm{d}x=\mu(\{x\in X:|f(x)|\not=0\})\tag{1c} $


$2$. Preliminaries

For $p\gt0$ and $t\ge0$, define $ g_p(t)=\frac{t^p-1}{p}\tag{2a} $ Claim: $g_p(t)$ is non-decreasing in both $p$ and $t$.

$g_p(t)$ is non-decreasing in $t$: This follows from $ g_p^\prime(t)=t^{p-1}\ge0\tag{2b} $

$g_p(t)$ is non-decreasing in $p$: As Didier commented, this follows from $ g_p(t)=\int_1^tu^{p-1}\,\mathrm{d}u\tag{2c} $ and because $u^{p-1}$ is non-decreasing in $p$ when $u\ge1$ and non-increasing in $p$ when $0\le u\le1$.

Furthermore, L'Hopital says $ \lim_{p\to0^+}g_p(t)=\log(t)\tag{2d} $

$\hspace{1pt}$

Jensen's Inequality says that $h(p)=\|f\|_p$ is non-decreasing in $p$.

$\hspace{1pt}$

Consider an $\epsilon$ neighborhood of $-\infty$ to be $(-\infty,-\frac1\epsilon)$ and let $L=\lim\limits_{p\to0^+}\log(h(p))$.

For any $\epsilon>0$, choose $q>0$ so that $\log(h(q))$ is within an $\frac{\epsilon}{2}$ neighborhood of $L$.

Choose $r>0$ so that $g_r(h(q))$ is within an $\epsilon$ neighborhood of $L$.

If $p<\min(q,r)$, then both $\log(h(p))$ and $g_p(h(p))$ will be within an $\epsilon$ neighborhood of $L$. Therefore, $ \lim_{p\to0^+}\log(h(p))=\lim_{p\to0^+}g_p(h(p))\tag{2e} $

Main Result

Define $E=\{x:|f(x)|>1\}$, then the results above yield $ \begin{align} \lim_{p\to0^+}\log\left(\|f\|_p\right) &=\lim_{p\to0^+}\frac{\|f\|_p^p-1}{p}\\ &=\lim_{p\to0^+}\int_X\frac{|f(x)|^p-1}{p}\,\mathrm{d}x\\ &=\color{#C00000}{\lim_{p\to0^+}\int_{E}\frac{|f(x)|^p-1}{p}\,\mathrm{d}x} +\color{#00A000}{\lim_{p\to0^+}\int_{X\setminus E}\frac{|f(x)|^p-1}{p}\,\mathrm{d}x}\\ &=\color{#C00000}{\int_{E}\log|f(x)|\,\mathrm{d}x} +\color{#00A000}{\int_{X\setminus E}\log|f(x)|\,\mathrm{d}x}\\ &=\int_{X}\log|f(x)|\,\mathrm{d}x\tag{2f} \end{align} $ The left limit, in red, is by Dominated Convergence, while the right limit, in green, is by Monotone Convergence. Exponentiate to get $ \lim_{p\to0^+}\|f\|_p=e^{\int_{X}\log|f(x)|\,\mathrm{d}x}\tag{2g} $

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    Yes. When you first brought my attention to this, I forgot that the integral requiring Monotone Convergence was negative and thought that since it was a decreasing function, Dominated Convergence was what to use. However, as we both noticed, when a negative function is decreasing, we need to use Monotone Convergence. I have reverted the answer to where it was before.2015-09-14