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Consider the functional equation $f(y,z)g(y+z)=h(y)$, where the functions $f,g,h$ are all continuously differentiable and where $h$ is not constant.

I have just about convinced myself that any solution to this must take the form $g(x)=e^{cx}$ for some constant $c$ and $f(y,z)=\tilde f(y)e^{-cz}$, for some function $\tilde f$. But I don't have a proper proof.

I suspect the proof is so obvious (except to me) that it's unworthy of being documented in, e.g., Aczel-Dhombres, or else it's a well-known result (again, except to me). Or maybe there's another solution I'm missing?

A pointer to an existing result or proof or counterexample would all be welcome!

Ethan Ligon
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  • Observe that if $f(y,z)=f_1(y)f_2(z)$ and $g(y+z)=g_1(y)g_2(z)$, then $f_2(z)=\frac1{g_2(z)}$ and $f_1(y)g_1(y)=h(y)$ – Simply Beautiful Art Oct 09 '16 at 00:48
  • Certainly, @SimpleArt. And this plus the fact that $\log g(y+z)=\log g_1(y)+\log g_2(z)$ takes the form of Pexider's equation ($a(x+y)=b(x)+c(y)$) is what leads to my conclusion that $g(x)=e^{cx}$. But I don't see how to rule out the possibility that some other solution exists, with $f(y,z)\neq f_1(y)f_2(z)$. – Ethan Ligon Oct 11 '16 at 00:52
  • Well, we could have $g(x)=1$ and $f(y,z)=h(y)$. – Simply Beautiful Art Oct 11 '16 at 00:59
  • That would be a special case of $g(x)=e^{cx}$, of course (with $c=0$). – Ethan Ligon Oct 11 '16 at 16:33
  • Then have $g(x)=1/2$. You can't make it work out in that case. – Simply Beautiful Art Oct 11 '16 at 20:35
  • Taking $g(x)$ equal to any constant $k$ will just mean that the solution for $f(y,z)$ will be scaled by $1/k$; that doesn't change the form of the solution I propose. You're certainly correct that /if/ $g(x)$ is a constant then $f$ can't be a function of $z$. The remaining question is whether it's possible to have a non-trivial function $g$ and a non-separable function $f$ that satisfies the original equation. – Ethan Ligon Oct 12 '16 at 18:30

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