Can anything be said about the solutions of the following functional equation? $ f(x, y + z) = f(x, y) + f(x + y, z) $
I don't seem to be able to find much in what I think are the standard references in these cases.
Thanks for any tip.
Can anything be said about the solutions of the following functional equation? $ f(x, y + z) = f(x, y) + f(x + y, z) $
I don't seem to be able to find much in what I think are the standard references in these cases.
Thanks for any tip.
Following the beautiful idea of Robert Israel, we will show that the solutions of the equation are precisely functions of the form $f(x,y)=g(x+y)-g(x)$, where $g$ is an arbitrary function.
First, we plug $z = -y$ into the equation. This yields $f(x, 0) = f(x, y) + f(x + y, -y).\tag{1}$
In the case $y=0$, this tells us that $f(x,0)=0$ for all $x$, as Robert Israel already noticed. Using this fact in $(1)$, we have that $f(x,y)=-f(x+y,-y)\tag{2}$ must hold for all $x,y$ in order for $f$ to be a solution. We will now use this fact in the original equation. The original equation says that $f(x, y) = f(x, y + z) - f(x + y, z)$ holds for all $x,y,z$. Using $(2)$ twice we may rewrite this as $f(x, y) = - f(x+y+z, - y - z) + f(x+y+z, -z)$ Now, plug in $z=-x-y$ and get $f(x, y) = - f(0, x) + f(0, x+y).$ This means that if $f$ solves the original equation, we may define the function $g$ by $g(w)=f(0,w)$ and then $f(x,y)=g(x+y)-g(x)$ will hold. This shows that indeed the solutions of the equation are precisely of the form suggested by Robert Israel.
Take $y=0$ to get $f(x,z) = f(x,0) + f(x,z)$, so $f(x,0) = 0$.
$f(x,y) = g(x+y) - g(x)$ is a solution for arbitrary functions $g$. I don't know if those are all the solutions: they seem to be all the ones that are low-degree polynomials.
EDIT: Note that the transformation $f(x,y) = F(x,x+y)$ and change of variables $x = X, y = Y-X, z = Z - Y$ makes the equation $ F(X,Z) = F(X,Y)+F(Y,Z)$ Thus (taking, say, $Y=0$, $G(X) = F(X,0)$ and $H(Z) = F(0,Z)$) we have $F(X,Z) = G(X) + H(Z)$. But then $G(X) + H(Z) = G(X) + H(Y) + G(Y) + H(Z)$ which says $G(Y) = -H(Y)$. So we have $F(X,Z) = H(Z) - H(X)$, which transforms back to $f(x,y) = H(x+y) - H(x)$.