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.


  • $\begingroup$ I can't seem to find the induction step. I tried direct normal induction but I can't do it. $\endgroup$ – Jorge Fernández Hidalgo Feb 11 '14 at 3:27

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.

  • $\begingroup$ I think 6our answer has a tiny mistake.you need to change m for n+1. It is a regular cauchy induction. $\endgroup$ – Jorge Fernández Hidalgo Feb 11 '14 at 15:09
  • $\begingroup$ @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. $\endgroup$ – user127249 Feb 11 '14 at 16:11
  • $\begingroup$ 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. $\endgroup$ – Jorge Fernández Hidalgo Feb 11 '14 at 22:34
  • $\begingroup$ @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. $\endgroup$ – user127249 Feb 11 '14 at 22:58
  • $\begingroup$ I added those intermediate steps above in the proof so you understand what is going on. $\endgroup$ – user127249 Feb 11 '14 at 23:01

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