# Uniqueness of Pexider's functional equation

Let $f:\mathbb{R}\rightarrow\mathbb{R}$, $g:\mathbb{R}\rightarrow\mathbb{R}$, and $h:\mathbb{R}\rightarrow\mathbb{R}$ and consider Pexider's equation, $$f(x) + g(y) = h(x + y) \qquad \qquad (1)$$ where $f$, $g$ and $h$ are unknown. I assume (for simplicity) that $f$, $g$, and $h$ are twice continuously differentiable. Then, we can find a solution by differentiating with respect to $x$ and $y$, $$h''(x+y)=0$$ Integrating out, substituting back into $(1)$ and equating coefficients we get, $$h(z)=c_1z +c_2 + c_3 \qquad f(x)=c_1x +c_2 \qquad g(y)=c_1y +c_3$$ for arbitrary constants $c_1$, $c_2$ and $c_3$.

I have now found a solution to $(1)$. How do I prove that this solution is unique? It seems immediate, but can't quite convince myself.

• Well, I am no expert on partial derivatives, but what you have got is $\dfrac{\partial^2(h)}{\partial x \partial y}=0$. Is it enough to conclude what you have concluded? – shadow10 Jul 22 '14 at 15:05
• I'm not sure. There are several theorems on uniqueness of solutions of PDEs and ODEs but my suspicion is that uniqueness doesn't even need to use any of this fancy machinery. – user103828 Jul 22 '14 at 15:08

Suppose $h$ is twice continuously differentiable. Therefore,
$h''(x+y)=0$ if and only if $h(x+y)=(x+y)k+l$, where $k,l$ are constants （1）
Since $f(x)+g(y)=h(x+y)$, we have $0=\frac{\partial^2}{\partial x \partial y}(L.H.S.) = h''(x+y)$ （2）
Therefore by (1) and (2) we have $f(x)+g(y)=(x+y) k+l$, and since $x,y$ are arbitrary, $f,g$ have to be linear functions.
• Yes, uniqueness seems immediate because $h''(x+y)=0 \Leftrightarrow h(x+y)=c_1 (x+y) +c_2$. – user103828 Jul 22 '14 at 20:33