Consider the functional equation problem

$$ f: \Bbb{R} \rightarrow \Bbb{R}$$

$$ f(a^b) = f(a)^{f(b)},$$

when $a,b \in \Bbb{R}, a,b \ge 0.$

So far the only solution I have is the trivial

$$ f(x) = x$$

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 alternative expression for it?

I tried to get a little more creative:

$$ f(w) = f(w^{\frac{1}{w}})^{f(w)}$$

Which nests deep into

$$ f(w) = f(w^{\frac{1}{w}})^{f(w^{\frac{1}{w}})^{f(w^{\frac{1}{w}})^{\vdots}}} $$

Which is equivalent to

$$ \ln(f(w)) = \ln(f(w^{\frac{1}{w}}))f(w) $$

I was hoping to generate some sort of series using this.

  • 2
    $\begingroup$ Don't you want the functional equations form $\mathbb R_{\geq 0}$ to $\mathbb R$? Otherwise the negative values can have any value $\endgroup$ – wythagoras May 26 '15 at 19:28
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    $\begingroup$ $f(x)=1$ for all $x$ is another trivial answer. $\endgroup$ – Thomas Andrews May 26 '15 at 19:28
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    $\begingroup$ $f(2)=f(4)=1$ or $f(2)=2$, $f(4)=4$. By looking to $(a,b)=(4,2)$ and $(a,b)=(2,4)$ respectively. $\endgroup$ – wythagoras May 26 '15 at 19:36
  • $\begingroup$ For any $a>0$ we observe that $f(a) \neq 0$, for then we would have $f(a^a)=f(a)^{f(a)}=0^0$ which is not defined. $\endgroup$ – Indrayudh Roy May 26 '15 at 19:44
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    $\begingroup$ @IndrayudhRoy Many mathematicisn define $0^0=1$. There are strong reasons to do this. $\endgroup$ – Thomas Andrews May 26 '15 at 19:48

We can show that the only $nice$ solutions to $$f(x^y)=f(x)^{f(y)}\qquad(1)$$ 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 $(1)$, 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: $$f(x^y)=f(x)^{f(y)}=f(x)^{\pm1}=f(x)$$ 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 $(1)$: $$f(z)^{f(xy)}=f(z^{xy})=f((z^x)^y)=f(z^x)^{f(y)}=\left(f(z)^{f(x)}\right)^{f(y)}=f(z)^{f(x)f(y)}$$ $$\therefore f(xy)=f(x)f(y)\qquad(2)$$ $$\therefore f(z)^{f(x+y)}=f(z^{x+y})=f(z^xz^y)=f(z^x)f(z^y)=f(z)^{f(x)}f(z)^{f(y)}=f(z)^{f(x)+f(y)}$$ $$\therefore f(x+y)=f(x)+f(y)\qquad(3)$$ Thus by $(2)$, for every $x>0$ we get $f(x)=f(\sqrt{x})^2>0$. So by $(3)$, we conclude that $f$ is strictly increasing. Using $(1)$, 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 $(3)$). 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 $(1)$, 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 $(1)$ (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.

  • 1
    $\begingroup$ It's interesting that equations (2) and (3) look like the equations for the automorphisms of a field. But the positive real numbers aren't a field, so I'm not sure if this observation gains us anything. $\endgroup$ – Michael Seifert Jul 23 '15 at 14:20
  • $\begingroup$ @MichaelSeifert, its interesting how over the course of months I've come to really appreciate your comment, (mainly because I know now what automorphisms are!). I suppose thats an indicator of slow but steady progress $\endgroup$ – frogeyedpeas Oct 24 '15 at 3:01

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