# Let $f:\Omega\to(0,+\infty)$ and $\ln(x)$ be $\mu$-integrable

Show that $\displaystyle \lim\limits_{p\to 0^+} ||f||_p = \exp(\int\ln(f)\,d\mu)$. In case it comes to be helpful.

So far I've shown that $\displaystyle\lim\limits_{p\to 0^+}\int\frac{f^p-1}{p}\,d\mu=\int\ln(f)\,d\mu$.

My approach is trying this:

$$||f||_p=(\int f^p\,d\mu)^\frac{1}{p}=\exp(\frac{1}{p}\ln(\int f^p\,d\mu)).$$

And then I'm trying to show that $$\frac{1}{p}\ln(\int f^p\,d\mu)\xrightarrow{p\to 0^+}\int\ln(f)\,d\mu$$

(if needed, we can assume that $\mu$ is a probability measure)

• First hint: can you show that $\limsup_{p \to 0^+} \|f\|_p \leq e^{\int \ln (f) \ d \mu}$? This is the "easy" inequality, and can be obtained quickly from what you have. – D. Thomine Aug 23 '15 at 15:55
• @D.Thomine I can't see much difference. I mean, obviously I see that it's different, but I can't prove it either. – Andre Gomes Aug 23 '15 at 15:57

Hint. Assume that $\mu$ is a probability measure, and write

$$\left( \int f^p \, d\mu \right)^{1/p} = \left( 1 + p \int \frac{f^p -1}{p} \, d\mu \right)^{1/p}.$$

Now can you combine your first result with the following lemma?

Lemma. Assume that $g(p) \to \ell$ as $p \to 0^+$. Then $$\lim_{p \to 0^+} (1 + pg(p))^{1/p} = e^{\ell}.$$

• Excellent. But it seems to me there's a slight gap. Taking $x^{1/p}$ is problematic if $x<0$. If $\ell\in\Bbb R$ this is no problem, since then $1+pg(p)>0$ for small enough $p>0$. But what about the case $\ell=-\infty$? – David C. Ullrich Aug 23 '15 at 18:15
• @DavidC.Ullrich, You are right. I was a little bit careless when stating the lemma. Luckly for our purpose, simply imposing the condition $1+pg(p) > 0$ is sufficient. – Sangchul Lee Aug 23 '15 at 18:34

When one wants to prove that an equality holds, it is often fruitful to try to prove two inequalities. Here, we would like to show that:

$$\limsup_{p \to 0^+} \ \frac{1}{p} \ln \left( \int f^p \ d \mu\right) \leq \int \ln (f) \ d \mu,$$

and:

$$\liminf_{p \to 0^+} \ \frac{1}{p} \ln \left( \int f^p \ d \mu\right) \geq \int \ln (f) \ d \mu.$$

Here, it is interesting, because the arguments behind both inequalities are quite different. Note that $\ln$ is concave. It is quite annoying, but we know one thing about probability measures and concave functions: jensen's inequality. Let's try this! For any $p > 0$,

$$\frac{1}{p} \ln \left( \int f^p \ d \mu\right) \geq \frac{1}{p} \int \ln \left(f^p\right) \ d \mu = \int f \ d \mu.$$

Taking the liminf, we get:

$$\liminf_{p \to 0^+} \ \frac{1}{p} \ln \left( \int f^p \ d \mu\right) \geq \int \ln (f) \ d \mu.$$

That's one side. For the other, we need a upper bound. There is a weel-known upper bound for the log, that is, $\ln (x) \leq x-1$. And this looks like something you already used. Let's try this! For any p > 0\$,

$$\frac{1}{p} \ln \left( \int f^p \ d \mu\right) \leq \frac{1}{p} \left( \int f^p \ d \mu -1\right) = \int \frac{f^p-1}{p} \ d \mu.$$

The RHS is something whose limit we know. Taking the limsup, we get:

$$\limsup_{p \to 0^+} \ \frac{1}{p} \ln \left( \int f^p \ d \mu\right) \leq \int \ln (f) \ d \mu,$$

which is what was needed.