Rotationally invariant function after tensorizing Suppose that $f$ is a real valued function such that $$f(x)f(y)=f((x+y)/\sqrt 2) f((x-y)/\sqrt 2)$$ for all real numbers $x,y$. Then is it necessarily true that $f(x)=Ae^{bx^2}$ for some real numbers $A,b$?
I think I can prove it to be true if $f$ is nonnegative and integrable, though I’m not sure of a strategy to tackle less well behaved functions, perhaps there’s a weird $f$ that changes sign or has local blowups or is even nonmeasurable but satisfies that identity.
 A: Given a function $\,f:\mathbb{R}\to\mathbb{R},\,$ define the
two variable function
$$ F(x,y) := f(x)f(y)-f\Big(\frac{x+y}{\sqrt 2}\Big)
 f\Big(\frac{x-y}{\sqrt 2}\Big). \tag{1}$$
Suppose given that $\,F(x,y) = 0 \,$ for all real $\,x\,$ and $\,y.\,$
What does this imply? To avoid triviality, assume that the
function $\,f(x)\,$ is not identically zero.
First, use equation $(1)$ to get
$$ 0 = F(x,x) = f(x)^2 - f(\sqrt{2}x)f(0). \tag{2} $$
If $\,f(0)=0\,$ then this implies that $\,f(x)=0\,$ for all
$\,x\in\mathbb{R}\,$ which is excluded by assumption. Thus,
let $\,A:=f(0)\ne 0.$
Second, use equation $(1)$ to get
$$ 0 = F(\sqrt{2}x,0) = f(0)f(\sqrt{2}x) - f(x)^2. \tag{3} $$
This implies that
$$ f(\sqrt{2}x)/A = (f(x)/A)^2. \tag{4} $$
Since $\,x\,$ was arbitrary, this implies that
$\,f(x)/A \ge 0\,$ for all  $\,x\in\mathbb{R}.\,$
Third, use equation $(1)$ to get
$$ 0 = F(x,-x) - F(-x,x) = f(0)(f(-\sqrt{2}x)-f(\sqrt{2}x)). \tag{5}$$
Since $\,x\,$ was arbitrary, then this implies that
$\,f(x) = f(-x)\,$ for all $\,x\in\mathbb{R}.\,$
To summarize, assuming $\,f(x)\,$ is not identically zero,
then $\,f(x) = f(-x),\,$ and
$\,f(\sqrt{2}x)/A = (f(x)/A)^2 \ge 0\,$
for all $\,x\in\mathbb{R}.\,$
Use equation $(1)$ and previous results to get
$$ f(x)^2 f(y)^2 = A^2f(x+y)f(x-y),\quad \forall x,y\in\mathbb{R}. \tag{6} $$
With $\,A=1\,$ this is the multiplicative form of the parallelogram law.
At this point, from the example of
Cauchy's functional equation, any reasonable
solution of equation $(6)$ must be of the form
$\,f(x) = Ae^{bx^2}\,$ although there are badly behaved
solutions also.
