# Show the equality

Let $$(X, \mathcal{A}, \mu)$$ space with measure, $$\mu(X) = 1$$, $$\epsilon > 0$$ and $$f: X \rightarrow [\epsilon,\infty)$$ a $$\mathcal{A}$$-measurable and bounded function, I've tried show $$\lim_{p \rightarrow 0^{+}} \left(\int_{X} f^p(x)d\mu(x) \right)^{\frac{1}{p}} = \exp\left(\int_X \ln(f(x))d\mu(x)\right)$$

My attempt: I know is true, in this case, $$\displaystyle \int_X \ln(f(x))d\mu(x) \leq \ln\left(\int_Xf(x)d\mu(x)\right)$$, because $$\ln(t) \leq t - 1$$, $$\forall t > 0$$, take $$t = \frac{f(x)}{\int_Xf(x)d\mu(x)}$$ and integrate over $$X$$. So $$\left(\int_{X} f^p(x)d\mu(x) \right)^{\frac{1}{p}} = \exp\left(\frac{1}{p}\ln\left(\int_X f^p(x)d\mu(x)\right)\right) \geq \exp\left(\frac{1}{p}\int_X \ln(f^p(x))d\mu(x) \right) = \exp\left(\int_X \ln(f(x))d\mu(x) \right)$$, then

$$\lim_{p \rightarrow 0^{+}} \left(\int_{X} f^p(x)d\mu(x) \right)^{\frac{1}{p}} \geq \exp\left(\int_X \ln(f(x))d\mu(x)\right)$$

If $$\left(\int_{X} f^p(x)d\mu(x) \right)^{\frac{1}{p}}$$ is decreasing, the equality it's ok, but I don't know how show this.

There is a hint in exercise, $$\displaystyle \lim_{t \rightarrow 0^+} \frac{a^t - 1}{t} = \ln(a)$$. I've tried use this limit for show decreasing, but it wasn't successful.

Thank you.

## 1 Answer

Define $G\colon \Bbb R_{\geq 0}\to \Bbb R$ by $G(p) = \int_Xf^p\,{\rm d}\mu$. We have $G(0) = \mu(X) = 1$ and so $\ln G(0) = 0$. So:

\begin{align} \lim_{p \to 0^+}\left( \int_X f^p\,{\rm d}\mu\right)^{1/p} &= \lim_{p \to 0^+} G(p)^{1/p} = \lim_{p \to 0^+} \exp \ln G(p)^{1/p} \\ &= \exp \lim_{p \to 0^+} \frac{\ln G(p)}{p} = \exp \lim_{p \to 0^+} \frac{\ln G(p)-\ln G(0)}{p-0} \\ &= \exp \frac{{\rm d}}{{\rm d}p}\bigg|_{p=0} \ln G(p) = \exp \frac{G'(0)}{G(0)} = \exp G'(0). \end{align}

Now we only have to check that $G'(0) = \int_X \ln f\,{\rm d}\mu$. We have:

\begin{align} G'(0) &= \lim_{p \to 0^+}\frac{G(p)-G(0)}{p-0} = \lim_{p\to 0^+}\frac{\int_X f^p\,{\rm d}\mu - 1}{p} \\ &\stackrel{(\ast)}{=} \lim_{p \to 0^+} \int_X \frac{f^p - 1}{p}\,{\rm d}\mu \stackrel{(\ast\ast)}{=} \int_X \lim_{p \to 0^+}\frac{f^p-1}{p}\,{\rm d}\mu \\ &= \int_X \ln f\,{\rm d} \mu,\end{align}

where in $(\ast)$ we use that $\mu(X) = 1$ and in $(\ast\ast)$ we use the bounded convergence theorem.