Let $Q^+$ denote the set of all positive rational number and let $\alpha\in Q^+.$ Determine all functions $f:Q^+ \to (\alpha,+\infty )$ satisfying $$f(\frac{ x+y}{\alpha}) =\frac{ f(x)+f(y)}{\alpha}$$for all $x,y\in Q^+ .$
Problem
Source: MEMO 2018 I1
Tags: algebra, function, functional equation
MNJ2357
02.09.2018 14:15
It's easy to see that if $\alpha \neq 2$, there are no solutions. For $\alpha =2$, $f(x)=ax+b$, where $a\ge 0, b>2$, which is clearly a solution. Full solution coming soon
Tintarn
05.09.2018 11:06
Put $x=y$ to see that $f\left(\frac{2x}{\alpha}\right)=\frac{2}{\alpha} f(x)$. So if $z$ is in the image of $f$, also $\frac{2}{\alpha} \cdot z$ and $\frac{\alpha}{2} \cdot z$ are. So if $\alpha \ne 2$, the image of $f$ contains arbitrarily small positive numbers contradicting the assumption that $f(x)>\alpha$ for all $x$. So we must have $\alpha=2$.
But then $f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}$ which is a classical equation. For the sake of completeness, here is the solution.
So $f(x+y)=\frac{f(2x)+f(2y)}{2}$ for all $x,y>0$. So $f(x+2)-f(x+1)=\frac{f(4)-f(2)}{2}=:a$ i.e. $f(x+1)=f(x)+a$ whenever $x>1$.
But $f(x+1)+f(x-1)=2f(x)$ for all $x>1$ whence $f(x-1)=f(x)-a$ for all $x>1$ i.e. $f(x+1)=f(x)+a$ whenever $x>0$.
Finally, putting $y=1$ above we find $f(x)+a=\frac{f(2x)+f(2)}{2}$ so that $f(2x)=2f(x)-b$ for some constant $b$ and all $x>0$. But then the equation is $f(x+y)=f(x)+f(y)-b$ for all rationals $x,y>0$. From here $f(nx)=nf(x)-(n-1)b$ for all $n \in \mathbb{N}$ and hence $f(qx)=qf(x)-(q-1)b$ for all rationals $q,x>0$. Putting $x=1$ we find that $f(q)=qf(1)-(q-1)b$ for all rationals $q>0$. But then $f(1)=a+b$ and hence $f(x)=ax+b$ for all rationals $x>0$. Finally, the boundedness condition implies that $a \ge 0$ and $b>2$. All of these are of course indeed solutions.
Tintarn
05.09.2018 12:58
Actually in the same way as above we can solve the problem for general $\alpha>0$ without the restriction of the range i.e. for $f:\mathbb{Q}^+ \to \mathbb{R}$. As above we have $f(x+1)+f(2)=f(x+2)+f(1)$ and hence $f(x+2)-f(x+1)=a$ is constant i.e. $f(x+1)=f(x)+a$ for all $x>1$. But then again $f(x+1)+f(x-1)=2f(x)$ implis $f(x+1)=f(x)+a$ for all $x>0$. But then $f(x+y)=\frac{f(\alpha x)+f(\alpha y)}{\alpha}$ with $y=1$ implies that $f(\alpha x)=\alpha f(x)+b$ for some constant $b$. But then again $f(x+y)=f(x)+f(y)-c$ for some $c$ and from here as above we find that $f(x)=ax+c$ for all $x$.
lazizbek42
10.01.2022 21:52
from the simple inequality $a = 2$.then Jenson's equation
Tintarn
02.02.2022 18:09
Actually, there is a little catch in my solution above, for when $a>0$, we can also allow $b=2$. So the result really is $f(x)=ax+b$ where either $a>0, b \ge 2$ or $a=0, b>2$.