Solving the functional equation $ xf(x) - yf(y) = (x - y)f(x + y) $

Question

I found the following functional equation:

Find all functions $f : \Bbb R \rightarrow \Bbb R $ such that: $$ xf(x) - yf(y) = (x - y)f(x + y) \text{ for all }x, y \in \mathbb R $$

Could you please help me? I think I proved that if $f(0) = 0$ then for each $x \in \Bbb Q$ and every integer $k$ we have $f(kx) = kf(x)$, but I don't know how to continue. Thanks in advance.

Answer 1

It's obvious that every function of the form $f(x)=ax+b$ satisfies the equation: $$xf(x)-yf(y)=(x-y)f(x+y)\tag0\label0$$ It can be shown that those are the only solutions indeed. To show that, let $a=f(1)-f(0)$ and $b=f(0)$, and define $g(x)=f(x)-ax-b$. It's easy to see that by \eqref{0}, $g$ satisfies $$xg(x)-yg(y)=(x-y)g(x+y)\tag1\label1$$ and we have $g(0)=0$ and $g(1)=0$. Letting $x=1$ and $y=-1$ in \eqref{1} we get $g(-1)=0$. Now, letting $y=1$ and $y=-1$ in \eqref{1}, we respectively get: $$xg(x)=(x-1)g(x+1)\tag2\label2$$ $$xg(x)=(x+1)g(x-1)\tag3\label3$$ Substituting $x+1$ for $x$ in \eqref{3} we have: $$(x+2)g(x)=(x+1)g(x+1)\tag4\label4$$ Subtracting \eqref{4} and \eqref{2} we get $2g(x)=2g(x+1)$ and thus $g(x)=g(x+1)$. Hence by \eqref{4} we have $(x+2)g(x)=(x+1)g(x)$ and therefore $g$ is the constant zero function. So $f(x)=ax+b$.

Answer 2

For $x=0$ we find that $yf(y)=-yf(y)$ for all $y\in\Bbb{R}$, so $yf(y)=0$ for all $y\in\Bbb{R}$. It follows that $f(y)=0$ whenever $y\neq0$. Then taking $x=1$ and $y=-1$ shows that $f(1)-f(-1)=2f(0)$, where $f(1)=f(-1)=0$, hence also $f(0)=0$. So $f=0$.