# Which functions commute with exponentials?

If $f : \mathbb{R} \to \mathbb{R}$ satisfies $$f(e^x) = e^{f(x)},$$ must it also be an exponential function?

• It could at least be the identity, or an iterated exponential function. – Henning Makholm Apr 6 '16 at 7:44
• Is $f$ supposed to be continuous or something? – Eric Wofsey Apr 6 '16 at 8:03

If we substitute $y=e^x$ we get $$f(y)=e^{f(\log y)}$$ which means that you can choose arbitrary values for $f$ on $(-\infty,0]$, and the values for every positive argument will then be given by applying the above recursion a finite number of times.
If you want the solution to be continuous (or smooth) at $x=0$, the initial values need to fit together properly, but there's still lots of freedom to choose them. For example you can choose a smooth shape of the function for $f$ on $[-\frac12,\frac12]$ with the constraint that $f$ is positive on $(0,\frac12]$. Then the recurrence can be run backwards on to give values on $(-\infty,-\log2]$, and you can then choose a smooth bridging segment on $(-\log2,-\frac12)$.
If you want $f$ to be real analytic, considerably more finesse might be needed -- it is not clear to me offhand whether there are then any solutions other than the identity and $\exp^n$.
Note that for any $x\in\mathbb{R}$, there is a unique $n\in\mathbb{N}$ and $y\in (-\infty,0]$ such that $x=\exp^n(y)$ (where $\exp^n$ means you iterate the exponential function $n$ times): to find $n$ and $y$, take iterated logarithms of $x$ until you reach a nonpositive number; that nonpositive number is $y$, and $n$ is the number of logarithms you had to take. So now let $g:(-\infty,0]\to\mathbb{R}$ be any function at all and define $f(x)=\exp^n(g(y))$, where $n$ and $y$ satisfy $x=\exp^n(y)$. Then $f(\exp(x))=\exp(f(x))$ for all $x$, since $\exp(x)$ will have the same $y$ and its $n$ will be incremented by $1$.
So there are many many such functions $f$, one for every function $g:(-\infty,0]\to\mathbb{R}$. It is also not difficult to see that every such $f$ must be obtained in this way (since you can recover $g$ from $f$ as the restriction of $f$ to $(-\infty,0]$, and this determines all other values of $f$ since $f$ must commute with $\exp$).