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Find all real functions $f\colon \mathbb{R}\rightarrow \mathbb{R}$ so that $(x+y)(f(x)-f(y))=f(x^2)-f(y^2)$.

Can someone at least find the value of $f(1)$ if it is possible, it would help me.

Peter Woolfitt
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CryoDrakon
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    It's easy to show that all constant functions along with the identity function meet the definition. Though I am not yet sure whether there are more. What do you mean by $f(1)$? There are many functions that fit the description.. – benji Dec 23 '14 at 21:24
  • @benji that's not the purpose of solving a functional equation. By $f(1)$ he means we should find the value of the function at the point $1$ deduced by this equation. – Lukas Juhrich Dec 23 '14 at 21:28
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    If it's of help to anyone, plugging in $0$ and $-1$ to the equation gives us $2f(0)=f(1)+f(-1)$. – Peter Woolfitt Dec 23 '14 at 21:29
  • One of possible solutions: $f(x)=kx$, where $k$ is the constant. – Oleg567 Dec 23 '14 at 21:32
  • @Oleg567 to be more precise, $f(x)=kx$, where k is some constant value. – CryoDrakon Dec 23 '14 at 21:34
  • @Lukas what benji is saying, I think, is that for any real number $r$ there is a solution $f$ satisfying $f(1)=r$ (the constant solution). So the equation doesn't unambiguously determine the value of a solution at 1. – Teri Dec 23 '14 at 21:34
  • @SoulEater, yes, i edited the post. – Oleg567 Dec 23 '14 at 21:34

2 Answers2

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Without loss of generality, lets look for solutions to the equation satisfying $f(0)=0$. The equation then becomes $$xf(x)=f(x^2),\ x\in \mathbb R. $$ Therefore $$(x+y)(f(x)-f(y))=f(x^2)-f(y^2)=xf(x)-yf(y),$$ from which we get (by expanding the LHS) $$yf(x)=xf(y) $$ for all $x,y\in \mathbb R$. Setting $y=1$ we see that all solutions are of the form $$f(x)=kx.$$

Teri
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    How can we assume $f(0)=0$ w/o loss of generality? – Lukas Juhrich Dec 23 '14 at 21:56
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    If $f$ is any solution, then $\tilde f$, given by $\tilde f(x)=f(x)-f(0)$ also is (you can easily check this) and satisfies $\tilde f(0)=0$. Therefore a general solution $f$ is of the form $\tilde f+f(0)$ where $\tilde f$ is a solution satisfying $\tilde f(0)=0$. – Teri Dec 23 '14 at 21:58
  • What about $f(x)=a$ for $a\in\mathbb{R}-{0}$ which fits criteria,than $f(0)=a$, – kingW3 Dec 23 '14 at 22:00
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    @KingW3 for that particular $f$, the function $\tilde f$ is the constant function zero (using notation from my comment above). – Teri Dec 23 '14 at 22:03
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    You could also note that functional equation is linear (in that any linear combination of solution is a solution) and that constant functions satisfy it, so we can safely "quotient out" (in the vector space sense) constant terms by requiring $f(0)=0$. – Milo Brandt Dec 24 '14 at 04:23
  • A function $f(x)=ax+b$ is also a soultion, but $f(0)=b$. – CryoDrakon Dec 24 '14 at 14:08
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    @SoulEater yes, in this case $\tilde f(x)=f(x)-f(0)=ax$.. – Teri Dec 24 '14 at 16:20
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I made a mistake last time. Now everything should be fine.

Putting $y=0$ we get $f(x^2)=x(f(x)-f(0))+f(0)$. Putting that in the first equation leaves us with $(x+y)(f(x)-f(y))=x(f(x)-f(0))+f(0)-y(f(y)-f(0))-f(0)$

$y f(x)-x f(y)=-f(0)(x-y)$

With $y=1$ we get: $f(x)-x f(1)=-f(0)(x-1) \Rightarrow f(x)=x(f(1)-f(0))+f(0)$ If we put $f(1)-f(0)=a$ and $f(0)=b$ and then we can write: $f(x)=ax+b$.

Martin Sleziak
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CryoDrakon
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