4

Find all function $f:\mathbb{R^+}→\mathbb{R^+}$ that satisfies the given condition: $$f(xy)^{xy} =f(x)^x f(y)^y$$

If $f:\mathbb{R}→\mathbb{R^+}$ the question would be rather simple, as putting in $y=0 $ yields that $1=f(x)^x$ thus implying $f(x)=1$ for all $x$.

However, since $f:\mathbb{R^+}→\mathbb{R^+}$, I first tried $y=1$, which yields that $f(1)=1 $.

Also note that $f(2x)^{4x}=f(4x^2)^{4x^2}=f(x)^{x}f(4x)^{4x}$.

Putting $x=1$ yields that $f(2)=f(4)$

I also tried differentiating $f(xy)^{xy} =f(x)^x f(y)^y$, but that did not prove very useful.

Any help would be appreciated.

Chad Shin
  • 2,132

1 Answers1

5

Let $g := \ln \circ f$. Then $$ xy g(xy) = xg(x) + yg(y). $$

This implies that $h(x) := x g( x)$ satisfies $$ h(xy) = h(x) + h(y), $$ so that finally, the function $\alpha : \Bbb{R}\to \Bbb{R}, x \mapsto h(e^x)$ satisfies $$ \alpha (x+y) = h(e^x e^y) = h(e^x)+ h(e^y)= \alpha (x) + \alpha (y). $$

Now, additive functions on $\Bbb{R}$ are relatively well understood. If we assume $f$ (and thus $\alpha$) to be continuous, there is some $a\in\Bbb{R}$ with $\alpha(x)= ax$ for all $x$.

I leave it to you to reverse my substitutions to find out what this yields for $f$.

PhoemueX
  • 35,087