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I need to measure a mean value of a function $f(\mathbf{x})\geq0$ defined over a domain $\Omega$.

It would be very useful to me to be able to parametrize this mean value so that:

$$M_{-\infty}(f)=\min\limits_{\Omega}(f)$$ $$M_{\infty}(f)=\max\limits_{\Omega}(f)$$ $$p < q \Rightarrow M_p(f) \leq M_q(f)$$

Well, power means have these properties, but they are defined for discrete data:

$$M_p(k_1,...,k_n) = \left[\frac{1}{n}\sum\limits_{i=1}^{n} k_i^p\right]^{\frac{1}{p}}$$

$$M_0(k_1,...,k_n) = \lim_{p\rightarrow0}M_p(k_1,...,k_n) = \left[\prod\limits_{i=1}^{n}k_i\right]^{\frac{1}{n}}$$

By generalizing this to the continuous case, I believe we would have:

$$M_p(f) = \left[\frac{1}{V(\Omega)}\int_{\Omega}f(\mathbf{x})^p\,dV\right]^{\frac{1}{p}}$$

Does this have the desired properties? What would be the expression for $M_0(f)$? Does this integral power mean have a name? I am having trouble in searching for references about it, everything I find is about the discrete case.

EDIT: Well, although I am still not sure about the other properties, I tried to obtain the expression of $M_0(f)$.

$$M_p(f) = \exp\left(\log\left(M_p(f)\right)\right) = \exp\left(\frac{1}{p}\left[\log\left(\int_\Omega f(\mathbf{x})^p\,dV\right) - \log\left(V(\Omega)\right)\right]\right)$$

The following limit can be simplified using the L'Hôpital's rule: $$\lim_{p\rightarrow 0} \frac{1}{p}\left[\log\left(\int_\Omega f(\mathbf{x})^p\,dV\right) - \log\left(V(\Omega)\right)\right] = \lim_{p\rightarrow 0} \frac{d}{dp}\left[\log\left(\int_\Omega f(\mathbf{x})^p\,dV\right)\right] =\\= \lim_{p\rightarrow 0} \left[\frac{1}{\displaystyle \int_\Omega f(\mathbf{x})^p\,dV}\,\int_\Omega \log(f(\mathbf{x}))\,f(\mathbf{x})^p\,dV\right] = \frac{1}{V(\Omega)}\,\int_\Omega \log(f(\mathbf{x}))\,dV =\\= M_1(\log(f))$$

Therefore, $\boxed{M_0(f) = \exp(M_1(\log(f)))}$.

I had stated that: "If the expression is valid, it seems to me that, if exists $\mathbf{x_0} \in \Omega$ such that $f(\mathbf{x_0}) = 0$, then all means with $p\leq 0$ will be the same: $M_{-\infty} = ... = M_{-2}(f) = M_{-1}(f) = M_{0}(f) = \min\limits_{\Omega}(f) = 0$.". However, I see now that this is not true.

For example, it seems that $f(x) = |x|^n$, with $n>0$, will have $M_{p} = 0$ in any domain containing the origin only for $p \leq -1/n$. It seems that for a single point $\mathbf{x_0} \in \Omega$ such that $f(\mathbf{x_0}) = 0$ to be enough to all means with $p\leq 0$ be $0$, $f$ would have to be flat in $\mathbf{x_0}$.

Still, I am not sure if the inequalities are valid for a function which can be $0$ at some point. So I add to the original question: are these properties valid for a function $f(\mathbf{x})\geq0$ defined over a domain $\Omega$?

Daniel Cunha
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    Strongly related: https://en.wikipedia.org/wiki/Lp_space – Paul Sinclair Jan 29 '22 at 15:28
  • @PaulSinclair Thank you for your comment. I understand there is a relation with measures. However it is not clear to me if the inequalities are still valid when considering the average value (dividing by $V^{\frac{1}{p}}$). I mean, $M_p \leq M_q$ when $p<q$. Moreover, it is not clear to me if this is well defined for any value of $p$ ranging from $-\infty$ to $\infty$, usually I find restrictions like $p\geq 1$, so the ingregral can be seen as a norm. – Daniel Cunha Jan 29 '22 at 15:36
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    Both true, which is why I only said it was strongly related, rather directly applying to your question. But this relation isn't so much about "measure" as just summing $p$-th powers. Measure is just a means of defining that summation. – Paul Sinclair Jan 29 '22 at 15:43
  • So the properties are valid? Could you write an answer with some references or demonstrations? – Daniel Cunha Jan 29 '22 at 15:50
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    My apologies. I meant that both of your concerns are real issues. I don't have immediate answers for them. Though the reason we don't talk about $L^p$ spaces for $p < 1$ is that $|\cdot|_p$ ceases to be a norm, which is not an issue for your question. – Paul Sinclair Jan 29 '22 at 16:06

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Yes, the power mean inequality $M_p(f) \le M_q(f)$ is valid for all $p, q \in [-\infty, \infty]$ and all measurable functions $f \ge 0$, provided we adopt the conventions $0^{-1} = \infty, \infty^{-1} = 0, \log(0) = -\infty$, etc. For real $p, q \ne 0$ this can be shown using Jensen's inequality applied to the function $x \mapsto x^{q/p}$, which is convex when $q > 0$ and concave when $q < 0 $. In the case where $p$ or $q$ is zero, assuming $M_0(f) = \exp(M_1(\log f))$ exists, this follows from Jensen's inequality applied to the $\log$ function.

However, the limit of $M_p$ as $p \to 0$ is not necessarily equal to $M_0(f)$, which also may not exist since $\log f$ can take on both positive and negative values. A sufficient condition for $\lim_{p \to 0^+} M_p(f) = M_0(f)$ is that $M_q(f) < \infty$ for some $q \in (0, \infty)$. Similarly, a sufficient condition for $\lim_{p \to 0^-} M_p(f) = M_0(f)$ is that $M_q(f) > 0$ for some $q \in (-\infty, 0)$. A proof of this fact can be found at this M.SE answer (the comments contain some corrections on the application of convergence theorems).

lily
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