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, 2019 at 21:14

1 Answer 1


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 $\endgroup$ Jun 13, 2019 at 21:04
  • $\begingroup$ @colt_browning I don't see right now how exactly but maybe it is $\endgroup$ Jun 13, 2019 at 21:10
  • $\begingroup$ Very interesting, thanks! is there a simple example showing that property 4 is necessary for the characterization? $\endgroup$ Jun 14, 2019 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 $\endgroup$ Jun 14, 2019 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. $\endgroup$ Jun 15, 2019 at 19:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.