Functions $f:\mathbb Z\to\mathbb Q$ satisfying $f\left(\frac{x+y}3\right)=\frac{f(x)+f(y)}2$ knowing that $\frac{x+y}3\in\mathbb Z$

Question

What is the solution of the following functional equation? (I must confess it is a headache for me.)

Find all the functions $ f : \mathbb Z \to \mathbb Q $ such that $$ f \left( \frac { x + y } 3 \right) = \frac { f ( x ) + f ( y ) } 2 $$ fora ll $x , y \in \mathbb Z $ such that $ \frac { x + y } 3 \in \mathbb Z $.

I'd appreciate your help and comments. I've tried it a lot but until today I'm not able to solve it... :/

Martin Argerami · Accepted Answer · 2014-01-23T15:41:45.227

9

The function $f$ is necessarily constant.

We proceed by induction. Note that $$ f(3k)=\frac{f(3k)+f(3k)}2=f\left(\frac{3k+3k}3\right)=f(2k), $$ and $$ f(k)=f\left(\frac{2k+k}3\right)=\frac{f(2k)+f(k)}2, $$ from where $f(2k)=f(k)$. In particular, $f(3)=f(2)=f(1)$.

We have $f(1)=f(\frac{0+3}3)=\frac{f(0)+f(3)}2=\frac{f(0)+f(1)}2$, so $f(0)=f(1)$. Now assume that $f(0)=f(1)=\cdots=f(k)$. Then we do the inductive step: $$ f(k+1)=f\left(\frac{3k+3}3\right)=\frac{f(3k)+f(3)}2=\frac{f(k)+f(1)}2=f(k). $$

As mentioned by Hagen, from $f(0)=\frac{f(k)+f(-k)}2$ we get $f(-k)=2f(0)-f(k)$; as $f(k)=f(0)$ for positive $k$, we get that $f(k)=f(0)$ for all $k\in \mathbb Z$.

edited Jan 23 '14 at 15:41

answered Jan 17 '14 at 19:21

Martin Argerami

205,756

3

Or use $f(0)=\frac{f(x)+f(-x)}{2}$ for the negatives. – Hagen von Eitzen Jan 17 '14 at 19:24
I think you have only proved that a constant $f$ is a valid solution, not that it is the only solution. – Jeff Snider Jan 17 '14 at 19:28
1

@JeffSnider The proof shows $f(k+1)=f(k)$ for all $k\in\mathbb N$ and then also for negative $k$. Then $f$ must be constant. – Hagen von Eitzen Jan 17 '14 at 19:30
1

This one is okay, it proves by induction that $f$ is constant. – Sungjin Kim Jan 17 '14 at 19:30
Your statement "Assume that $f(0)=f(1)=\cdots=f(k)$" is unnecessary and what led me to misread the proof. – Jeff Snider Jan 17 '14 at 19:34
@Jeff: it depends on what you consider "unnecessary". How do you show that if $f(k+1)=f(k)$ then $f$ is constant? – Martin Argerami Jan 17 '14 at 19:51

Functions $f:\mathbb Z\to\mathbb Q$ satisfying $f\left(\frac{x+y}3\right)=\frac{f(x)+f(y)}2$ knowing that $\frac{x+y}3\in\mathbb Z$

1 Answers1