Solution to the functional equation $f(x^y)=f(x)^{f(y)}$ Consider the functional equation problem
$$ f : \mathbb R \to \mathbb R \text , $$
$$ f \left( a ^ b \right) = f ( a) ^ { f ( b ) } \text , $$
when $ a , b \in \mathbb R $, $ a , b \ge 0 $.
So far the only solution I have is the trivial
$$ f ( x ) = x \text . $$
Does there exist any other possible solution?
Even if the $ f $ doesn't have a closed form in terms of elementary functions, is there some way I could derive a series or an alternative expression for it?
I tried to get a little more creative:
$$ f ( w ) = f \left( w ^ { \frac 1 w } \right) ^ { f ( w ) } \text , $$
which nests deep into
$$ f ( w ) = f \left( w ^ { \frac 1 w } \right) ^ { f \left( w ^ { \frac 1 w } \right) ^ { f \left( w ^ { \frac 1 w } \right) ^ \vdots } } \text , $$
which is equivalent to
$$ \ln \big( f ( w ) \big) = \ln \bigg( f \left( w ^ { \frac 1 w } \right) \bigg) f ( w ) \text . $$
I was hoping to generate some sort of series using this.
 A: We can show that the only nice solutions to
$$f\left(x^y\right)=f(x)^{f(y)}\tag0\label0$$
are the identity function, and the constant functions $1$ and $-1$. There are also a few ugly solutions.
To avoid problems with $0^0$, let's first find all functions $f$ such that $f:(0,+\infty)\to(-\infty,0)\cup(0,+\infty)$. (Note that because the question is about nonnegative $x$ and $y$ satisfying \eqref{0}, therefore if the domain of $f$ included a negative $x$, then $f(x)$ would be arbitary.)
In the case that $|f(x)|=1$ for every $x>0$, we have $f(x)^{+1}=f(x)^{-1}=f(x)$; so for every $x,y>0$ we get:
$$f\left(x^y\right)=f(x)^{f(y)}=f(x)^{\pm1}=f(x)$$
Thus by fixing $x$ to have a value greater than $1$ and letting $y$ to range over all positive numbers, we conclude that $f$ is constant on $(1,+\infty)$. By the same argument, $f$ is constant on $(0,1)$. So, $f$ must be of the form:
$$\cases{f(x)=a\qquad0<x<1\\f(x)=b\qquad x=1\\f(x)=c\qquad x>1}$$
where $|a|=|b|=|c|=1$. In fact, we can show that every choice of $(a,b,c)$ gives a solution; and the only nice ones are the constant functions $1$ and $-1$.
Now, if there is a $z$ such that $|f(z)|\neq1$, then by \eqref{0}:
$$f(z)^{f(xy)}=f\left(z^{xy}\right)=f\big(\left(z^x\right)^y\big)=f\left(z^x\right)^{f(y)}=\left(f(z)^{f(x)}\right)^{f(y)}=f(z)^{f(x)f(y)}$$
$$\therefore f(xy)=f(x)f(y)\tag1\label1$$
$$\therefore f(z)^{f(x+y)}=f\left(z^{x+y}\right)=f\left(z^xz^y\right)=f\left(z^x\right)f\left(z^y\right)=f(z)^{f(x)}f(z)^{f(y)}=f(z)^{f(x)+f(y)}$$
$$\therefore f(x+y)=f(x)+f(y)\tag2\label2$$
Thus by \eqref{1}, for every $x>0$ we get $f(x)=f\left(\sqrt x\right)^2>0$. So by \eqref{2}, we conclude that $f$ is strictly increasing.
Using \eqref{0}, we get $f(1)=f(1)^{f(1)}$ so $f(1)=1$ which inductively yields $f(r)=r$ for every positive rational number $r$ (by using \eqref{2}). Because $f$ is increasing, it must be the identity function.
Now if we want to find all the functions $f:[0,+\infty)\to\mathbb{R}$ satisfying \eqref{0}, then we should deal with $0^0$ and this expression must be defined. If we define $0^0$ to be something other than $0$ or $1$, then we'll lose the usual algebraic properties of exponentiation. Here, I omit the proof, and just claim that if $0^0=0$ then we'll have these ugly solutions:

*

*$f(x)=0\qquad x\geq0$

*$\cases{f(x)=-1\qquad x=0\\f(x)=1\qquad x>0}$

*$\cases{f(x)=1\qquad x=0\\f(x)=a\qquad x>0}$ where $a=0$ or $a=-1$
and if $0^0=1$ then the only solutions are the nice ones.
If we want to include negative numbers in the domain of \eqref{0} (and avoid complex numbers) then $y$ must only take rational values with odd denominator (to avoid problems with defining exponentiation). In this case, we have the nice solutions and there may be other solutions.
There are problems with extending the domain and codomain to complex numbers, because exponentiation is a many-valued relation in this case, and also the algebraic properties become complicated. Even it's not clear how to find all the constant solutions! I've asked a related question.
