2
$\begingroup$

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.

Thanks,

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.