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What is $$\lim_{p\to0}\left(\int_0^1\left(f(x)^p\right)dx\right)^{1/p},$$

where $f$ is continuous?

It was an exam question and I don't know how to even get started with it, could you help? (It is not $f(x^p) $ but $\left(f(x)^p\right)$ so the $p$-th power of $f(x)$ )

neander
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Nesa
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1 Answers1

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This is what I tried. I'm not sure though. $$A:=\lim_{p\to0}\left(\int_0^1\left(f(x)^p\right)dx\right)^{1/p}$$ $$\ln A = \lim_{p\to0} \dfrac{\ln \left(\int_0^1f(x)^pdx\right)}{p}$$ $$\ln A = \lim_{p\to0} \dfrac{\int_0^1f(x)^p \ln (f(x))dx}{\int_0^1f(x)^pdx}=\frac{\int _0^1\ln f(x) dx}{1}$$ $$A = e^{\int _0^1\ln f(x) dx}$$

I'm seeing that if this were a (riemann) summation instead of an integration, we would have (since exponential of a sum is a product)

\begin{align*} A &= e^{\sum_{r=0}^n \ln f(\frac{r}{n}) \frac1{n}}\\ &= \prod_{r=0}^n e^{\ln f(\frac{r}{n}) \frac1{n}}\\ &= \left( \prod_{r=0}^n f\left(\frac{r}{n}\right)\right)^{\frac{1}{n}}\\ \end{align*}

This seems to be some sort of a geometric mean. Am I correct? Does this have some implications?

Aritra Das
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  • Perhaps expand on your answer and complete the question. – Arbuja May 13 '16 at 19:34
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    Your argument works only when $\ln f(x)$ is integrable. The result is correct (assuming $f\geqslant 0$), with the natural $\ln 0 = -\infty$ and $e^{-\infty} = 0$, but when $\ln f(x)$ isn't integrable, you can't differentiate under the integral. – Daniel Fischer May 13 '16 at 19:36
  • @DanielFischer It was mentioned that $f(x)$ is continuous. Am I missing something? – Aritra Das May 13 '16 at 19:39
  • What if $f(x) = 0$ on the subinterval $[0,c]$ for some $c > 0$? Even if $f$ has only isolated zeros, it can happen that the logarithm is not integrable. Consider $\exp(-1/x^2)$ for an example. If such bad things don't happen, you have a nice argument. But it doesn't cover all (bad) cases. – Daniel Fischer May 13 '16 at 19:44
  • @DanielFischer Ohh, never thought about that. Thanks for all the perspective you provided. (I'm a high school student, so half the things that are mentioned here are unknown to me, like $L^p$ norms) – Aritra Das May 13 '16 at 19:56
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    You're doing fine. You already know a lot. – Daniel Fischer May 13 '16 at 20:01
  • @DanielFischer co could you tell me how to handle the bad cases? – Nesa May 13 '16 at 20:48
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    @Nesa There's a proof in the answer at the duplicate. Dealing with the bad cases is non-trivial. – Daniel Fischer May 13 '16 at 20:51