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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.

shalop
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    Ok, here is a partial progress which shows that $f$ cannot change sign. If $f(0)=0$ then $0=f(x)f(0) = f(x/\sqrt{2})^2$ so $f$ is identically 0. Thus we may assume wlog that $f(0)=1$ (else replace $f(x)$ by $f(x)/f(0)$ which satisfies the same relation). Then $f(x) = f(x)f(0) = f(x/\sqrt{2})^2$ and thus $f$ must be non-negative. – shalop Apr 24 '22 at 16:09

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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.

Somos
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