# Integrate and measure problem.

If $f \in L^{p_0}(X,M,\,u)$ for some $0<p_0 \le\infty$, then $$1. \lim_{p\to0}\int_{X} |f|^pd\mu=\mu(\{x \in X | f(x) \ne0\}).$$

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

then I wanna prove that $f \in L^{p}(X,M,\,u)$ for some $0<p \le p_0$, and the equation below. $$2. \lim_{p\to0}||f||_p=e^{\int_Xlog|f|d\mu}$$

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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for the first problem, using $MCT$ and $LDCT$ as you say lets you interchange the limit and integral, so you get $\int_X \lim_{p \rightarrow 0} |f|^p = \int_X 1(f(x) \neq 0)$ –  uncookedfalcon Jun 11 '12 at 9:20
and I think that if $\mu(X) = 1$, you mean it follows that $f \in L^p$ for every $p \leqslant p_0$, because we have $$\int |f|^p \leqslant 1 + \int_{|f| \geqslant 1} |f|^{p_0} < \infty$$ –  uncookedfalcon Jun 11 '12 at 9:25
I followed your nice approach so i concluded first equation. But I still cannot understand your second answer so I cannot make it for second equation. Can you explain second inequality more precisely? Thanks @uncookedfalcon –  wowhapjs Jun 11 '12 at 16:11
oh not really an answer, all I was saying was that to talk about $\lim_{p \rightarrow 0} \| f \|_p$, you need that $f$'s in every $p$ smaller than $p_0$, and that's true because $$\int |f|^p = \int_{|f| < 1} |f|^p + \int_{|f| \geqslant 1} |f|^p$$$$\leqslant \int_{|f| < 1} 1 + \int_{|f| \geqslant 1} |f|^p \leqslant 1 + \int_{|f| \geqslant 1} |f|^{p_0} < \infty$$ –  uncookedfalcon Jun 11 '12 at 19:13
Oh thanks again for your kind explain @uncookedfalcon. –  wowhapjs Jun 11 '12 at 22:47

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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It's no doubtfully perfect nice answer! –  wowhapjs Jun 11 '12 at 22:49

$1$. 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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We can lift the restriction... Hmmm... :-) If you can then do it, don't you think? –  Did Jun 15 '12 at 7:31
@did: I completely reworked answer 2 using properties of $g_p(t)=\frac{t^p-1}{p}$ and $\|f\|_p$. –  robjohn Jul 19 '12 at 16:33
To show that $h_p(t)\geqslant0$, note that $h_p(t)=(t^p/p^2)\cdot(s-\log(1+s))$ with $s=t^{-p}-1$ and that $\log(1+s)\leqslant s$ for every $s\gt-1$, by concavity of log (or by some other argument). –  Did Jul 19 '12 at 16:46
@did: thanks! That looks a bit simpler. –  robjohn Jul 19 '12 at 16:48
Still simpler, $g_p(t)=\int\limits_1^ts^{p-1}\mathrm ds$ and $p\mapsto s^{p-1}$ is nonincreasing on $s\leqslant1$ and nondecreasing on $s\geqslant1$. –  Did Jul 19 '12 at 16:57