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This is something I was thinking about, so I'm going to post it as a question and post my own answer. I hope that anyone who wants will comment, correct me if I'm wrong, and add their own knowledge and thoughts.

Suppose $f:[a,b]\to \mathbb{R}$ is a positive, continuous function. The $L^p$ norms of $f$ are the analogs to the weighted power means of a group of positive numbers $a_1, \dotsc, a_n$. (Perhaps it would be more precise to say that the expressions $\left(\int_a^b\frac{f^p}{b-a}dx\right)^{1/p}$ are the analogs.) The power means are given by $$M_p(a_1, \dotsc, a_n)=(w_1a_1^p + \dotsb + w_na_n^p)^{1/p},$$

where $w_i$ such that $\sum w_i =1$ are any weights we like (the simplest case is when $w_i=1/n$ for all $i$). It is well-known (see here) that the power means $M_p$ tend to the geometric mean as $p\to 0$. What does $$\left( \frac{\int_a^b f(x)^p}{(b-a)}\right)^{1/p}$$

tend to as $p\to 0$?

Eric Auld
  • 28,127

2 Answers2

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Assume the limit exists and call it $L$. Then $$\log L = \lim_{p\to 0}\frac{\log \left( \int_a^b f(x)^p dx \right)-\log (b-a)}{p},$$

which is just $$\left.\frac{d}{dp}\right|_{p=0}\int_a^b f(x)^p dx=\int_a^b \log f(x) dx, \quad \text{so}\\L = \exp\left[ \int_a^b \log f(x) dx\right].$$

This jibes with what we know from the power means, because $\exp\left(w_1 \log a_1 + \dotsb + w_n \log a_n\right)$ is just the geometric mean.

More questions: are the $p$ norms in ascending order of $p$ like the power means are?

Eric Auld
  • 28,127
7

Since you normalized the measure space to have total measure $1$, Jensen's inequality immediately implies the monotonicity of $L^p$ norm (quasinorm for $p<1$) with respect to $p$. Namely, for all $p>0$ and all $r>1$ we have (using the convexity of $t\to t^r$ on $[0,\infty)$)
$$ \left(\int_X |f|^{p}\right)^{r} \le \int_X (|f|^p)^r $$ hence $\|f\|_p\le \|f\|_{pr}$.

You can also use Hölder's inequality to the same purpose.

Thus, the limit $\lim_{p\to 0}\|f\|_p$ exists under the rather weak assumption that some $L^p$ norm of $f$ is finite. As you noted, it is equal to $\exp\left(\int_X \log|f| \right)$. Continuity and the particular form of measure space are irrelevant.

you can call me Al
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