I am looking for all real-valued continuous functions f, on R, which satisfy

$$ f(x)*f(y) = f(x_1)f(y_1) $$ for all x,y, $x_1$, $y_1$, such that $$x^2 + y^2 = (x_1)^2 +(y_1)^2.$$

I don't have much idea on how to solve this problem. The only thing that comes to mind, which doesn't help very much, is the fact that if I let g(x,y) = f(x)*f(y), then since the function g factorizes into two functions of a single variable, we have that the integral of g is the product of the single-variable integrals of f(x)dx and f(y)dy.


Edit: This is not a homework problem. It is a problem that dates back to 2007, as far as I know, and there is a not-so-good solution to it that basically says "guess that the function is Guassian and let's force it to be Guassian." I am looking for another solution to this problem. Thanks.


If $f(x)=0$ then $$f(x)\cdot f(x)=f(0)f\bigl(\sqrt{2} x\bigr)=0$$ for all $x\in{\mathbb R}$. Therefore assume $f(0)=C\ne0$ and define $$g(u):={1\over C}f\bigl(\sqrt{u}\bigr)\qquad(u\geq0)\ .$$ Then $$g(u)\cdot g(v)={1\over C^2}f\bigl(\sqrt{u}\bigr)\cdot f\bigl(\sqrt{v}\bigr)={1\over C^2}f\bigl(\sqrt{u+v}\bigr)f(0)=g(u+v)$$ for arbitrary nonnegative $u$, $v$. Since $g$ is continuous on ${\mathbb R}_{\geq0}$ it follows that $$g(u)=e^{\lambda u}\qquad(u\geq0)$$ for a certain constant $\lambda\in{\mathbb R}$. This then implies $$f(x)=C\>e^{\lambda x^2}\qquad(x\geq0)\ ,$$ and from $$f(x)\cdot f(-x)=f(0)\cdot f\bigl(\sqrt{2}x\bigr)$$ it is then easy to conclude that $$f(-x)=C\>e^{\lambda x^2}\qquad(x\geq0)$$ holds as well.


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