Restrictions on $f(x) f(y)$ as a function of $x-y$. Consider a function of the form $g(x,y) = f(x) f(y)$. This function can just as well be written as $h(x-y) k(x+y)$ for some functions $h$ and $k$. Does the form of this function put any restriction on $h(x-y)$, or can $h(x-y)$ be an arbitrary function given a suitable choice of $f$?
 A: You cannot always decompose $g(x,y)$ using your suggestion of it being a product of $2$ alternate functions. Consider the quite simple case of $f(x) = x$. You get
$$g(x,y) = f(x)f(y) = xy \tag{1}\label{eq1A}$$
Consider functions $h(x-y)$ and $k(x+y)$ existing such that
$$g(x,y) = h(x-y)k(x+y) = xy \tag{2}\label{eq2A}$$
Set $y = x$ to get
$$h(0)k(2x) = x^2 \implies h(0)k(x) = \left(\frac{x}{2}\right)^2 \implies k(x) = \frac{x^2}{4h(0)} \tag{3}\label{eq3A}$$
Substituting this into \eqref{eq2A} gives
$$h(x-y) = \frac{xy}{k(x+y)} = \frac{4h(0)xy}{(x+y)^2} \tag{4}\label{eq4A}$$
Using $x = 2y$ gives
$$h(y) = \frac{4h(0)(2y^2)}{9y^2} = \left(\frac{8}{9}\right)h(0) \implies h(0) = \left(\frac{8}{9}\right)h(0) \implies h(0) = 0 \tag{5}\label{eq5A}$$
However, from \eqref{eq3A}, this gives that $x^2 = 0$, which is incorrect.
This provides one simple example where you can't decompose a function differently as you request. I believe there are many other such cases (in fact, most of the time), but I'm unsure how to classify, in general, when it will, and will not, work.
In fact, I have trouble coming up with any non-trivial examples which do work. Nonetheless, one relatively simple one is $f(x) = e^x$, so $f(x)f(y) = \left(e^x\right)\left(e^y\right) = e^{x+y}$. You can then have $h(x-y) = 1$ and $k(x+y) = e^{x+y}$.
