Which functions $f$ satisfy $f(x-a-c)f(x-b)=\alpha f(x-c-\tfrac{a+b}{2})f(x-\tfrac{a+b}{2})$ for every $x$ where $a,b,c,\alpha$ are constants? Let $a,b,c \in \mathbb R$ be real constants. I'm searching for functions $f : \mathbb R \to \mathbb C$ with the following property: For every such $a,b,c$ there exists a constant $\alpha = \alpha(a,b,c)$ depending on $a,b,c$ such that the following identity holds true:
$$f(x-a-c)f(x-b)=\alpha f(x-c-\tfrac{a+b}{2})f(x-\tfrac{a+b}{2})\ \ \ \ \forall x \in \mathbb R.$$
The only function I found so far satisfying the previous identity is a Gaussian. This follows from the fact that the product of two Gaussians is again (up to a constant) a Gaussian. Does someone have an idea if there are other maps having the above property?
 A: Define $g: \Bbb R \to \Bbb C$ such that $g(x)=\ln{f(x)}$, then
$$g(x-a-c)+g(x-b)=\ln(\alpha)+ g(x-c-\tfrac{a+b}{2})+g(x-\tfrac{a+b}{2})$$
Denote $a = u+v$ and $b = u-v$ then
$$g((x-u)-c-v)+g((x-u)+v)=\ln(\alpha(u+v,u-v,c))+ g((x-u)-c)+g(x-u)  \quad \forall x \in \Bbb R$$
$$\iff g(x-c-v)+g(x+v)=\ln(\alpha(u+v,u-v,c))+ g(x-c)+g(x) \quad \forall x \in \Bbb R \tag{1}$$
Define $h: \Bbb R \to \Bbb C$ such that $h_v(x)=g(x+v)-g(x)$ then
$$ h(x) - h(x-c-v)=\ln(\alpha(u+v,u-v,c)) \quad \forall x \in \Bbb R  \tag{2}$$
The functional equation $(2)$ has the solution  (I don't remember the proof but this functional equation is well known)
$$h(x) = d + \frac{\ln(\alpha(u+v,u-v,c))}{c+v}x $$
with a constant $d$.
From $(1)$ we have
$$g(x+v)-g(x) = d + \frac{\ln(\alpha(u+v,u-v,c))}{c+v}x$$
we can deduce that
$$g(x) = \frac{\ln(\alpha(u+v,u-v,c))}{v(c+v)}x^2+\left(\frac{d}{v}- \frac{\ln(\alpha(u+v,u-v,c))}{2(c+v)}\right)x + e$$
with a constant $e$
Return back to the function $f(x) = e^{g(x)}$, the only possible function of $f$ is
$$f(x) = \exp{(Ax^2+Bx+C)}$$
We can conclude that, $f$ can only be the gaussian function.
