Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I would like to know which properties a function $f:\mathbb{R}\rightarrow\mathbb{R}$ must have so that I can say: $$f(\overline{x}) = \overline{f(x)}$$

with $\overline{x}$ being the mean of the $x$'s. Or, more explicitly written:

$$f \left(\frac{x_1 + x_2 + \cdots + x_n}{n}\right) = \frac{f(x_1) + f(x_2) + \cdots + f(x_n)}{n}$$

Thanks in advance.

share|cite|improve this question
up vote 2 down vote accepted

Given a function $f$ with the property $$\tag{P} f\left(\frac{1}{n}\sum_{i=1}^nx_i\right)=\frac{1}{n}\sum_{i=1}^nf(x_i) \quad \forall\ n \in \mathbb{N},\ x_1,\ldots, x_n \in \mathbb{R}, $$ we set $$ F(x)=f(x)-f(0). $$ Then $$\tag{1} F(0)=0, \quad F\left(\frac{x}{n}\right)=\frac{1}{n}F(x) \quad \forall n \in \mathbb{N},\ x \in \mathbb{R}. $$ Since $$ 0=F\left(\frac{x-x}{2}\right)=f\left(\frac{x-x}{2}\right)-f(0)=\frac{f(x)+f(-x)-2f(0)}{2}=\frac{F(x)+F(-x)}{2} \quad \forall\ x \in \mathbb{R} $$ we have $$\tag{2} F(-x)=-F(x) \quad \forall\ x \in \mathbb{R} $$ Thanks to (1) and (2) we get $$ F\left(\frac{x}{n}\right)=F\left(\frac{-x}{-n}\right)=\frac{1}{-n}F(-x)=\frac{1}{n}F(x) \quad \forall n \in -\mathbb{N},\ x \in \mathbb{R}. $$ Hence $$\tag{3} F\left(\frac{x}{m}\right)=\frac{1}{m}F(x),\ F(nx)=nF(x) \quad \forall m \in \mathbb{Z}\setminus\{0\},\ n \in \mathbb{Z},\ x \in \mathbb{R}. $$ It follows from (3) that $$\tag{4} F(rx)=rF(x) \quad \forall r \in \mathbb{Q},\ x \in \mathbb{R}. $$ Again, thanks to (3) we have $$ F(x+y)=F\left(\frac{2x+2y}{2}\right)=\frac{F(2x)+F(2y)}{2}=F(x)+F(y) \quad \forall\ x \in \mathbb{R}. $$ Thus, $F: \mathbb{R} \to \mathbb{R}$ is a $\mathbb{Q}$-linear map, and any function of the form $f=F+\alpha$ satisfies (P), where $\alpha$ is a real constant, and $F$ a $\mathbb{Q}$-linear on $\mathbb{R}$.

share|cite|improve this answer
You do not neccesarily get $f(0)=0$. For example every constant function fulfills the functional equation, because then $f\left(\frac{1}{n}\sum_{i=1}^nx_i\right)=c=\frac{nc}{n}=\frac{1}{n}\sum_{i=1}^n‌​f(x_i)$ holds. – Dominik Dec 5 '12 at 19:11
My bad, I see the point! – Mercy King Dec 5 '12 at 20:27

Define $g(x)=f(x)-f(0)$. Then, setting $x_2=...=x_n=0$ you see $f(x/n)=f(x)/n+f(0)-f(0)/n$ or equivalently $g(x/n)=g(x)/n$. This implies $g(x+y)=2(g(x+y)/2)=2g((x+y)/2)=2(g(x)+g(y))/2=g(x)+g(y)$.

Furthermore, we obtain $$g(m/n x)=g(mx)/n=m/n (g(mx)/m)=m/n g(x) $$ for all $m,n \in \Bbb N$. So $g(qx)=qg(x)$ for every positive rational $q$. Indeed, this holds for any rational $q$ since $g(-x)+g(x)=g(0)=f(0)-f(0)=0$, which implies $g(-qx)=-g(qx)=-qg(x)$.

In general, $g$ can be any linear map of $\Bbb R$ to $\Bbb R$, where $\Bbb R$ is viewed as a $\Bbb Q$ vector space, but if you demand that $f$ is continuous, $g(q)=qg(1)=cq$ implies $g(r)=cr$ for every real $r$. So $f(x)=cx+f(0)=cx+b$ are the only continuous solutions.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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