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

Let f be a function from $\mathbb R$ to $\mathbb R$ satisfying $f(\frac{x_1+x_2}{2})=\frac{f(x_1)+f(x_2)}{2}$

Prove that for any positive integer $n$ we have $f(\frac{x_1+x_2\dots+x_n}{n})=\frac{f(x_1)+f(x_2)+\dots+f(x_n)}{n}$


I managed to prove it for all powers of 2. Mabye it is a case of cauchy induction? Thanks in advance.


share|cite|improve this question
I can't seem to find the induction step. I tried direct normal induction but I can't do it. – Carry on Smiling Feb 11 '14 at 3:27
up vote 2 down vote accepted

Put $a=\frac{x_1+...+x_n}{n}$ and $m=2^k>n$.

Then $$\begin{align}f(a)&=f\left(\frac{\frac{m}{n}(x_1+...+x_n)}{m}\right)\\&=f\left(\frac{x_1+...+x_n+(m-n)a}{m}\right)\\&=f\left(\frac{x_1+...+x_n+a+a+...+a}{m}\right)\\&=\frac{f(x_1)+...+f(x_n)+f(a)+f(a)+...+f(a)}{m}\\&=\frac{f(x_1)+...+f(x_n)+(m-n)f(a)}{m}\\&=\frac{f(x_1)+...+f(x_n)}{m}+\frac{(m-n)}{m}f(a)\end{align}$$

The third equality is using that you already know that it is true for powers of $2$, and $m=2^k$.

Solve for $f(a)$ and you get it.

share|cite|improve this answer
I think 6our answer has a tiny need to change m for n+1. It is a regular cauchy induction. – Carry on Smiling Feb 11 '14 at 15:09
@user4140 It is indeed just Cauchy's induction, that is why $m$ cannot be $n+1$. We need $m$ to be a power of $2$ larger than $n$ for which you have already proved the formula by induction. – user127249 Feb 11 '14 at 16:11
I think it needs to be n+1. For the argument to work. Mainly because: How do you know f((m-n)a)=(m-n)(f(a)) And there aren't necessary m terms in the second to last sum. – Carry on Smiling Feb 11 '14 at 22:34
@user4140 We are never using that. We don't even know if that is true. What we are using is that $f(\frac{x_1+...+x_n+(m-n)a}{m})=f(\frac{x_1+...+x_n+a+a+...+a}{m})=\frac{f(x_1)‌​+...+f(x_n)+f(a)+f(a)+...+f(a)}{m}$, which is the know property for the average of $m=2^k$ numbers. I though you knew how Cauchy's induction goes. – user127249 Feb 11 '14 at 22:58
I added those intermediate steps above in the proof so you understand what is going on. – user127249 Feb 11 '14 at 23:01

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.