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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) $ for all $x, y \in \mathbb R $

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

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  • $\begingroup$ Hint: Take $y=-x$ . $\endgroup$ – z100 Dec 8 '15 at 17:38
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    $\begingroup$ Can you explain how that would help? I tried it multiple times before, but nothing came out. $\endgroup$ – thefunkyjunky Dec 8 '15 at 17:41
  • $\begingroup$ For a start, the functions $f(x)=x$ and $f(x)=c$ for any $c\in\Bbb{R}$ work. $\endgroup$ – Servaes Dec 8 '15 at 17:43
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    $\begingroup$ Yeah, also every $f(x) = a * x + b$ I think $\endgroup$ – thefunkyjunky Dec 8 '15 at 17:44
  • $\begingroup$ You get $\frac{f(x)+f(-x)}{2}=f(0)$ . Set $f(0)$ whatever you want and use the geometricic meaning of the left side and see if your solution is complete. $\endgroup$ – z100 Dec 8 '15 at 17:54
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It's obvious that every function of the form $f(x)=ax+b$ satisfies the equation: $$xf(x)-yf(y)=(x-y)f(x+y)\qquad(1)$$ 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 (1), $g$ satisfies $$xg(x)-yg(y)=(x-y)g(x+y)\qquad(2)$$ and we have $g(0)=0$ and $g(1)=0$. Letting $x=1$ and $y=-1$ in (2) we get $g(-1)=0$. Now, letting $y=1$ and $y=-1$ in (2), we respectively get: $$xg(x)=(x-1)g(x+1)\qquad(3)$$ $$xg(x)=(x+1)g(x-1)\qquad(4)$$ Substituting $x+1$ for $x$ in (4) we have: $$(x+2)g(x)=(x+1)g(x+1)\qquad(5)$$ Subtracting (5) and (3) we get $2g(x)=2g(x+1)$ and thus $g(x)=g(x+1)$. Hence by (5) we have $(x+2)g(x)=(x+1)g(x)$ and therefore $g$ is the constant zero function. So $f(x)=ax+b$.

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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 $f=0$.

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  • $\begingroup$ Sorry, I saw I have made a mistake in the statement. I'll correct it shortly. Anyways, thank you $\endgroup$ – thefunkyjunky Dec 8 '15 at 17:34

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