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Again I am stuck with this problem in Rudin:

Assume that $|f|_{r}<\infty$ for some $r<0$. Prove that $$\lim_{p\rightarrow 0}|f|_{p}=e^{\int_{X}\log |f|d\mu}$$

Let $r<p<0$ and we have $K=\frac{r}{p}>1$. Denote its conjugate by $K'$. Then we have $$\int f^{p}d\mu=\int (f^{p})*1d\mu\le (\int [f^{p}]^{\frac{r}{p}})^{\frac{p}{r}}$$ since by assumption $\mu(X)=1$. So in particular we have $$|f|_{p}\le |f|_{r}<\infty$$ Since $|f|_{p}$ is monotonely decreasing with $p\rightarrow 0$, it must have a limit.

We now apply Jensen's inequality, which gives us $$\log^{\int_{X}Fd\mu}\ge \int_{X}\log[F]d\mu$$ Here $F=f^{p}$. So we have $$ \int_{X}f^{p}d\mu\ge (e^{\int_{X}\log[f]d\mu})^{p} $$ taking the $p$-th root on both sides we conclude that $$|f|_{p}\ge e^{\int \log|f|d\mu}$$

But then I got totally stuck. It is worth pointing that Jensen's inequality is only an equality when $f^{p}=c$ is a constant. Therefore $f$ has to be a constant as well.

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This has already been asked on the site. See, for example, math.stackexchange.com/questions/158049/… –  Martin Argerami Dec 25 '12 at 6:34
    
.......:( sorry! I am not good at searching... –  Bombyx mori Dec 25 '12 at 6:35
    
But there is a difference; here the $p$ is approaching from negative instead of positive. –  Bombyx mori Dec 25 '12 at 6:35
    
No worries. It actually took me several minutes to find the duplicate, and that was because I vaguely remembered having answered a similar question. –  Martin Argerami Dec 25 '12 at 6:36
    
I see. In your arguments you seem to use Hölder's inequality for negative exponents: what is your source for that? –  Martin Argerami Dec 25 '12 at 6:39

1 Answer 1

up vote 2 down vote accepted

Note that for $u>0$, $$ k(u)=u\log(u)-u+1\ge0\tag{1} $$ This is because $k'(u)=\log(u)$ and $k''(u)=1/u$ show that $k(u)$ has a minimum at $u=1$.

Now, by L'Hospital, we have $$ \lim_{p\to0}\frac{x^p-1}p=\log(x)\tag{2} $$ Furthermore, applying $(1)$ yields $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}p}\frac{x^p-1}p &=\frac{p\log(x)x^p-x^p+1}{p^2}\\ &=\frac{k(x^p)}{p^2}\\ &\ge0\tag{3} \end{align} $$ Thus, by Dominated Convergence, we have $$ \begin{align} \log\left(\lim_{p\to0}\left(\int_X|f(x)|^p\,\mathrm{d}\mu\right)^{1/p}\right) &=\lim_{p\to0}\frac1p\log\left(\int_X|f(x)|^p\,\mathrm{d}\mu\right)\\ &=\lim_{p\to0}\frac1p\log\left(1+p\int_X\frac{|f(x)|^p-1}{p}\,\mathrm{d}\mu\right)\\ &=\lim_{p\to0}\frac1p\log\left(1+p\int_X\log|f(x)|\,\mathrm{d}\mu\right)\\ &=\int_X\log|f(x)|\,\mathrm{d}\mu\tag{4} \end{align} $$ Therefore, $$ \lim_{p\to0}\left(\int_X|f(x)|^p\,\mathrm{d}\mu\right)^{1/p} =\exp\left(\int_X\log|f(x)|\,\mathrm{d}\mu\right)\tag{5} $$

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Thank you so much for the help! –  Bombyx mori Dec 25 '12 at 12:35
    
You're welcome. Each time I prove this theorem, I see some new underlying idea that makes it simpler. –  robjohn Dec 25 '12 at 12:41

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