The $f$-mean, where $f$ is a continuous monotonically-increasing function, is defined as:

$$ M_f(x_1, \dots, x_n) = f^{-1}\left( \frac{f(x_1)+ \cdots + f(x_n)}n \right). $$

For any $f$, $M_f$ has the following nice properties:

  1. Continuous;
  2. Monotonically-increasing in each argument;
  3. Symmetric -- attains the same value for any permutation of the arguments;
  4. Fixed point: for each $x\in \mathbb{R}$, $M_f(x,\ldots,x) = x$.

Is it true that any function satisfying these four properties is an $f$-mean for some function $f$?

If not - what properties should be added in order to characterize $f$-mean?

  • $\begingroup$ I advise you to read the works of the Belgrade team lead by Mitrinovic. One book (at least) of Mitrinovic is on line : isinj.com/mt-usamo/…. See as well "Handbook of Means and Their Inequalities" by P.S. Bullen. $\endgroup$ – Jean Marie Jun 13 '19 at 21:14

There is a result obtained by Kolmogorov in [1]. It requires one more condition: you may replace some subgroup of arguments by their mean values and this must not change value of the whole mean. Formally, we say that sequence of functions $\{M_n: \mathbb{R}^n\to \mathbb{R}\}_{n=1}^\infty$ defines a regular type of mean if

  1. $M_n$ is continuous and monotonically increasing by each argument for all $n$;
  2. $M_n$ is symmetric for all $n$;
  3. $M_n(x,x,\dots,x) = x$ for all $n$;
  4. $M_{n+m}(x_1,\dots,x_n,y_1,\dots,y_m) = M_{n+m}(x,\dots,x,y_1,\dots,y_m)$ where $x = M_n(x_1, \dots, x_n)$ for all $m,n$.

The state is following:

Means of regular type are $f$-means.

Here we prove the state for means of regular type with bounded domain, i.e. consider only arguments from some finite interval $[a,b]$. Then the proof may be easily generalized to the case of infinite domain.

Proof: Let's use notation $M(m[x], n[y]) = M_{m+n}(x_1,\dots,x_m,y_1,\dots,y_n)$ where $x_1 = \dots = x_m = x$ and $y_1 = \dots = y_n = y$ (i.e. mean of $m$ values $x$ and $n$ values $y$). Using properties 3 and 4 we get $$ M(pm[x], pn[y]) = M\big(p\big[M(m[x], n[y])\big]\big) = M(m[x], n[y]). $$ Thus for $mn' = nm'$ we have $$ M(m[x], n[y]) = M(mn'[x], nn'[y]) = M(nm'[x], nn'[y]) = M(m'[x], n'[y]).\tag{1} $$ Now for every rational number $0 \leq z = \frac{p}{q} \leq 1$ one may define function $\psi(z)$ in the following way: $$ \psi(z) = M(p[b], (q-p)[a]) $$ ($a$ and $b$ are bounds of the arguments domain). This definition is correct: indeed, for all pairs $p/q = p'/q'$ we have $p(q' - p') = (q - p)p'$, so by $(1)$ value of $\psi$ is unique for any $z$.

Let's now show that monotony of $M$ induce monotony of $\psi$. Consider two rational numbers $1 \ge z' > z \ge 0$. One may represent them in form $z = p/q$, $z' = p'/q$ where $p' > p$. Then using property 1 we get $$ \psi(z') = M(p'[b], (q - p')[a]) > M(p[b], (q-p)[a]) = \psi(z). $$ Now, since $\psi$ is monotonically increase there is inverse function $\psi^{-1}$ which is monotonically increasing too.

Consider now set of numbers $x_i = \psi(z_i)$, $i = 1,\dots,n$, where $z_i$ are rational. Again, we use representation $z_i = p_i/q$ with common denominator. Then $x_i = M(p_i[b], (q-p_i)[a])$ and (by property 4) $$ \begin{align} M(x_1,\dots, x_n) = M\big((p_1 + \dots + p_m)[b],(nq - p_1 - \dots - p_n)[a]\big) = \\ =\psi\left(\frac{p_1 + \dots + p_n}{nq}\right) = \psi\left(\frac{z_1 + \dots + z_n}{n}\right) =\\ = \psi\left(\frac{\psi^{-1}(x_1) + \dots + \psi^{-1}(x_n)}{n}\right) \tag{2} \end{align} $$ which is form of $\psi^{-1}$-mean.

Finally, let's prove that $\psi$ is continuous for all $0 < z < 1$. Suppose it is not true in some point $z'$, so that $u = \psi(z' - 0) \neq \psi(z' + 0) = v$. From $(2)$ for two rational numbers $z_1, z_2$ we have $$ M(\psi(z_1), \psi(z_2)) = \psi\left(\frac{z_1 + z_2}{2}\right). $$ Now we may make a passage $z_1 \to z'- 0$, $z_2 \to z' + 0$ which implies $$ \psi(z') = \lim_{\array{z_1\to z' - 0 \\ z_2\to z' + 0}} \psi\left(\frac{z_1 + z_2}{2}\right) = M(u,v) > u. $$ But it is always possible to make $\frac{z_1 + z_2}{2} < z'$ so that $\frac{z_1 + z_2}{2} \to z' - 0$ and then $\lim\psi\left(\frac{z_1 + z_2}{2}\right) = u$. This contradiction shows that $\psi$ is continuous in point $z'$. The same may be concluded for points $0$ and $1$.

So we see that values of $\psi(z)$ for rational $z$ form a dense set in the interval between $a = \psi(0)$ and $b = \psi(1)$, and so we may extend domain of $\psi$ to the whole interval $[0,1]$ remaining result $(2)$ holds by continuity.


[1] Колмогоров А.Н. Избранные труды. Математика и механика. - М.: Наука, 1985 (in Russian)

  • $\begingroup$ This also seems to answer this question: math.stackexchange.com/questions/2890629 $\endgroup$ – colt_browning Jun 13 '19 at 21:04
  • $\begingroup$ @colt_browning I don't see right now how exactly but maybe it is $\endgroup$ – Anton Grudkin Jun 13 '19 at 21:10
  • $\begingroup$ Very interesting, thanks! is there a simple example showing that property 4 is necessary for the characterization? $\endgroup$ – Erel Segal-Halevi Jun 14 '19 at 5:46
  • $\begingroup$ @ErelSegal-Halevi actually no example needed to see the necessity of property 4, you may just check that each $f$-mean is one of regular type which is pretty easy to see $\endgroup$ – Anton Grudkin Jun 14 '19 at 15:05
  • $\begingroup$ @AntonGrudkin Sure, I see this, but one could think that maybe property 4 is in some way implied by the previous properties. So it could be interesting to see an example in which properties 1,2,3 hold but property 4 does not hold. $\endgroup$ – Erel Segal-Halevi Jun 15 '19 at 19:03

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