Find all functions $ f,g : \mathbb R \to \mathbb R $ that satisfy the functional equation $$g(x)f(y) = f \left(\frac{x + y}{2}\right)^2-f\left(\frac{x-y}{2}\right)^2$$ for all $x,y \in \mathbb R$.
I need hints for this problem.
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Mohsen Shahriari
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Doesn't this force $$\frac{f \left(\frac{x + y}{2}\right)^2-f\left(\frac{x-y}{2}\right)^2}{f(x)}$$ to be a constant function of $y$? – Gregory Grant May 15 '15 at 17:53
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Some hints/remarks about necessity: Play around with the substitutions $y = x$, $y = -x$ to show that necessarily either
Case 1: $f$ must be an odd function, or
Case 2: $g \equiv 0$.
If in Case 1, go on to show either
Case 1(a): $g \equiv f$, or
Case 1(b): $f \equiv 0$. In this case $g$ can be any arbitrary function.
If in Case 2, show that necessarily $f = c$ for some constant $c$.
The remaining bit which I haven't any suggestions is Case 1(a). In this case, it is sufficient if $f$ is a linear function: $f(ax + y) = af(x) + f(y)$ for $a,x,y \in \mathbb{R}$, but I don't know if it is necessary.
Lost in a Maze
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For more about Case 1(a), see here. Note that substituting $x+y$ for $x$ and $x-y$ for $y$ in $f(x)f(y)=f\left(\frac{x+y}2\right)^2-f\left(\frac{x-y}2\right)^2$, one gets $f(x+y)f(x-y)=f(x)^2-f(y)^2$. – Mohsen Shahriari Nov 06 '20 at 00:35