Is there a function $f(x)$ that satisfies $f(e^x) = e^x f(x)$?

Does there exist any function $f$ such that

$$f(e^x) = e^x f(x)?$$

If so, I believe it could be used to create an analytics versions of the function $\exp_{n+1}(x) = \exp(\exp_{n}(x))$ (where $n$ is an integer and $\exp_0(x)=x$ as an initial condition; I don't know what combinations of $x$ and $n$ that would work) by letting $\exp_n(x)$ be a solution to the differential equation

$$\frac{\partial\, \exp_n(x)}{\partial\, n} = f(\exp_n(x)).$$

Solution 1:

You asked about a function such that $$ f(e^x) = e^x f(x). $$ Define the function arbitrarily for $\,x\le 0.\,$ Then define $$ f(x) = x f(\ln x)\quad \text{ for }\quad 0 < x \le 1 $$ since we already know $\,f(x)\,$ for $\,x\le 0.\,$ Similarly, define $$ f(x) = x f(\ln x)\quad \text{ for }\quad 1 < x \le e $$ since we already know $\,f(x)\,$ for $\,0 < x\le 1.\,$ Continuing this process, we can find the value of $\,f(x)\,$ for all real numbers such that the functional equation is satisfied.

The bounty question is

Is there an elementary function that satisfies the functional equation?

The answer to that is that if there was, then we would already know about it. For example, the Elementary Functions chapter of DLMF has details about the standard elementary functions. The last is the Lambert W function which satisfies $\,W(x)e^{W(x)}=x\,$ and $\,W(x)=x\,e^{-W(x)}.\,$ If the requirement is that the function must be differentiable, then that can be arranged similarly by placing constraints on the values of the function and its derivative for negative reals.