Find all functions $f:\Bbb{Q}\rightarrow \Bbb{Q}$ such that $f(x)+f(t)=f(y)+f(z)$ for all rational numbers $x

Question

Find all functions $f:\Bbb{Q}\rightarrow \Bbb{Q}$ such that $f(x)+f(t)=f(y)+f(z)$ for all rational numbers $x<y<z<t$ that form an arithmetic progression.

The handout asks me to use to Cauchy equations.

Note: I couldn't make any nice progress, so I tried to add everything I did.

I got $$ f(a)+f(a+3d) = f(a+d) + f(a+2d) \\ f(a+d) + f(a+4d) = f(a+2d) + f(a+3d).$$

and then I got $f(a)+f(a+4d) = 2f(a+2d)$

then I noticed that $a+2d$ is the mean of $a$ and $a+4d$ , so we can say that for any $m,n \in \Bbb Q$ , we get $f(m)+f(n) = 2f(\frac {m+n} {2})$ .

Then , I tried substituting $m=n=0 $, but we don't get anything nice.

Moreover, I still can't figure out how to use Cauchy since I haven't got anything in the form $f(x+y)=f(x)+f(y)$, though $f(m)+f(n) = 2f(\frac {m+n} {2})$ looks quite similar.

Can anyone give me some hints ? Thanks in advance !

Answer 1

Let $$S=\{\,s\in\Bbb Q\mid \forall a\in\Bbb Q\colon f(sa)-f(0)=s\cdot(f(a)-f(0))\,\}. $$ Trivially, $0\in S$ and $1\in S$. From the result (with $m=-2$ and $n=a$), we find $-1\in S$.

Next we verify: If $s_1,s_2,s_3,s_4$ are in AP and three of them are $\in S$, then so is the fourth. Indeed, assume three of the $s_i$ are $\in S$. Then in the functional eequation evaluated at $(s_1a,s_2a,s_3a,s_4a)$, the contributions of $f(0)$ cancel, as do those of the $s_if(a)$, and the cancellation of the $f(s_ia)$ is equivalent to the fourth $s_i$ being $\in S$.

In particular, starting from $-1,0,1\in S$, we find by induction that $\pm n\in S$ for all $n\in\Bbb N$ (in other words, $\Bbb Z\subseteq S$). Then if $\frac mn\in\Bbb Q$, we have $$f(\tfrac mn)-f(0) =m(f(\tfrac 1n)-f(0))$$ and $$f(1)-f(0) =n(f(\tfrac 1n)-f(0))$$ so that $$f(\tfrac mn)-f(0)=\frac mn(f(1)-f(0) $$ and ultimately $$ f(x)=ux+v$$ for all $x\in\Bbb Q$, where $v=f(0)$ and $u=f(1)-f(0)$.

Answer 2

@Hagen von Eitzen 's answer looks good, but given the OP's work there is a shorter path to a solution.

Indeed, they saw that for all rationals $m,n$, $f(m)+f(n)=2f\left(\frac{m+n}{2}\right)$.

Now, let $g(x)=f(x)-f(0)$, then $g$ satisfies the equation $\frac{g(m)+g(n)}{2}=g\left(\frac{m+n}{2}\right)$. Set $m=2y$, $n=0$, if follows that $g(2y)=2g(y)$ for all rationals $y$, and thus $g(m)+g(n)=g(m+n)$, hence a Cauchy equation.

Answer 3

So we have for all $d\neq 0$ and $x$: $$f(x-2d)+f(x+2d)= f(x-d)+f(x+d)$$

Let $g(x) = f(x+2d) -f(x+d)$ then $$g(x)=g(x-3d)$$ so $g(x)$ is constant function. So $$f(x+2d) = f(x+d)+c \;\;\;(*)$$ So for $x=-d$ we get $$f(d)=f(0)+c$$ so $f$ is constant function for all $d\ne 0$. So from $(*)$ we get $c=0$. But now also for $x=-d$ we get $f(d)= f(0)$ for all $d\ne 0 $, so $\boxed{f(x)=f(0)}$ for all $x$