# Characterization of the quasi-arithmetic mean

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?

• 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. Jun 13, 2019 at 21:14

There is a result obtained by Kolmogorov in . 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.

References:

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

• This also seems to answer this question: math.stackexchange.com/questions/2890629 Jun 13, 2019 at 21:04
• @colt_browning I don't see right now how exactly but maybe it is Jun 13, 2019 at 21:10
• Very interesting, thanks! is there a simple example showing that property 4 is necessary for the characterization? Jun 14, 2019 at 5:46
• @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 Jun 14, 2019 at 15:05
• @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. Jun 15, 2019 at 19:03