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Suppose we have an identity of the form

$$x e^{f(x,y)}+y e^{g(x,y)} \equiv (x+y)e^{h(x,y)},$$ for all $x,y\in D$ where $D$ is some domain. Does this imply that $f(x,y)\equiv g(x,y)\equiv h(x,y)$ in general?

pshmath0
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    Note that multiplying through by $e^{-h}$ shows that, without loss of generality, one can assume $h\equiv 0$. The problem is equivalent to finding $f(x,y)$ and $g(x,y)$ such that $x e^f + y e^g = x + y$. Rearrange: $x(e^{f} - 1) + y (e^{g} - 1) = 0$. You can then set $e^f-1\equiv y$ and $e^g-1\equiv -x$ and solve for $f,g$ to obtain $f=\log(1+y)$ and $g=\log(1-x)$. That's only valid for $x<1$ and $y>-1$, of course. – Will Nelson Aug 28 '14 at 20:02

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I'll expand on my comment to construct distinct functions $f,g,h$ on all $\mathbb{R}^2$ that satisfy the given constraint.

First, define $\tilde{f}\equiv f-h$ and $\tilde{g}\equiv g -h$. The constraint can be rearranged to yield $$ (e^{\tilde{f}(x,y)} - 1)x + (e^{\tilde{g}(x,y)}-1)y = 0. $$ Ignoring division by $0$ issues, we can write $$ \frac{e^{\tilde{g}(x,y)} - 1}{e^{\tilde{f}(x,y)} - 1} = - \frac{x}{y}. $$

Staring at this, it's not too difficult to construct appropriate $\tilde{f}$ and $\tilde{g}$. Take, for example, \begin{eqnarray} \tilde{f}(x,y) &=&\left\{\begin{array}{ll} \log(1 + \frac{1}{2}\left|\frac{y}{x}\right|) & \text{ if $xy\ne 0$} \\ 0 & \text{ otherwise} \end{array}\right. \\ \tilde{g}(x,y) &=&\left\{\begin{array}{ll} \log(1 - \frac{1}{2}\text{sgn}(xy)) & \text{ if $xy\ne 0$} \\ 0 & \text{ otherwise}. \end{array}\right. \end{eqnarray}

Here, $\text{sgn}(\cdot)$ is the sign function. One can then take $f\equiv \tilde{f}+h$ and $g=\tilde{g}+h$ for arbitrary $h$.

Will Nelson
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