I have the following problem. I have a function $v(x, \theta)$ that can be expressed in two ways, for all $x, \theta \in \Re$:


1. $v(x, \theta) = u(x - \theta)$, where $u$ is strictly concave and symmetric about $0$, and
2. $v(x, \theta) = g\_1 (x) f\_1 (\theta) + g\_2(x) f\_2 (\theta) + f\_3 (\theta)$


It is clear that $v(x, \theta) = - (x - \theta)^2$ is a solution, as I can set $g\_1 = - x^2$, $g\_2 = 2x$, $f\_1 = 1$, $f\_2 = \theta$, and $f\_3 = - \theta^2$. Similarly, it is easy to check that $v(x, \theta) = - k\_1 (x - \theta)^2 + k\_2$ is a solution for all $k\_1 > 0$ and $k\_2 \in \Re$.


I have a heuristic argument that these are the *only* solutions, but I have been unable to prove it, or find a reference that would provide an answer.


So the question is to find all functions $v(x, \theta)$ that satisfy (1) and (2). 


(I am happy to impose sufficient smoothness on $v(x, \theta)$ if that helps find answers, and I am also willing to assume $f\_1$ and $f\_2$ are monotonic).


Any guidance would be appreciated!



A functional equation with a quadratic solution
Let us slightly generalize your functional equation as $u(x,y)=v(x-y)$,
$$u(x,y)=\sum\_{j=1}^3f\_j(x)g\_j(y).$$
In your case, $f\_3=1$. According to a theorem of Rubel and Gauchman, for a sufficiently smooth function $u$, the necessary and sufficient condition
for such representation of sums of products of smooth functions is that
the determinant
$$\det\left(\frac{\partial^{i+j}u}{\partial x^i\partial y^j}\right)\_{i,j=0,\ldots,3}\equiv0.$$
Using $u(x,y)=v(x-y)$ we obtain $u\_x=v'$ and $u\_y=-v'$, and in general,
$$\frac{\partial^{i+j}}{\partial x^i\partial y^j}=(-1)^jv^{(i+j)}.$$
Substituting this to our determinant we obtain the Wrosnkian
$$W(v,-v',v'',-v''')\equiv 0.$$ 
This means that $v,v',v'',v'''$ are linearly dependent, therefore
$v$ satisfies a linear differential equation of at most $3$-d order with constant coefficients. So is a generalized exponential sum:
$$v(x)=c\_1e^{\lambda\_1 x}+c\_2e^{\lambda\_2 x}+c\_3e^{\lambda\_3 x},$$
if the roots of the characteristic equation are distinct, or
$$v(x)=c\_1e^{\lambda\_1 x}+(c\_2+c\_3)e^{\lambda\_2 x},$$
if there are two roots, or
$$v(x)=(c\_1+c\_2x+c\_3x^2)e^{\lambda x},$$
if there is only one root.
These are all possible forms of $v$. I leave it to you to determine exactly which of these functions
satisfy your specific equation (with $f\_3=1$) and which of them are concave or
whatever other conditions you want to impose. Solutions that you found correspond to
case 3, with $\lambda=0$. Solution that I gave in the comment belongs to case 1 with $c\_3=0$ or to case 2 with $c\_3=0$.


Ref. MR1024482
Gauchman, Hillel; Rubel, Lee A.
Sums of products of functions of x times functions of y. 
Linear Algebra Appl. 125 (1989), 19–63. 


