question:
Find all continuous functions $f:[0,\infty)\to(0,\infty) $such that
$f\left(\sqrt{\dfrac{x^2+y^2}{2}}\right)=\sqrt{f(x)\cdot f(y)},\;\forall x,y\geq0$
my attempt:
Since it doesn't satisfy homgeneity so, it must be some non linear function.I Tried substitutions like $(0,0), (x,x)$ but everything leads to equality
i don't know how to even start solving it.Any help will be appreciated. Try giving simple solutions which are less rigorous & intuitive.
note: the range excludes $0$
Arguably, there is no such thing as basic functional equation. Each one is unique; you may or may not be able to solve it, but even a success does not help you much with other beasts.
This, for one, has solutions of the form $$f(x)=C\cdot e^{a\cdot x^2}\tag1$$ where $a\in\mathbb R$ and $C\in\mathbb R_+$ are arbitrary constants.
Now that we know it, let's work out some post factum explanation.
Say, $f(x)=e^{g(x)}$. Then the equation becomes $$g\left(\sqrt{\dfrac{x^2+y^2}{2}}\right)=\dfrac{g(x)+g(y)}2\tag2$$
Now the obvious next step is to assume that $g(x) = u(x^2)$. Then
$$u\left(\dfrac{x^2+y^2}2\right)=\dfrac{u(x^2)+u(y^2)}2\tag3$$
With $x^2$ and $y^2$ being basically any numbers from the function's domain, this can be rewritten in a simpler form: $$u\left(\dfrac{x+y}2\right)=\dfrac{u(x)+u(y)}2\tag4$$ which (with the assumption of continuity) can easily be shown to only have linear solutions.
So it goes.