# 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

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

• One can start with any function whatsoever defined on $(-\infty,0]$ and then use the desired functional equation to define it on $(0,1]$, then $(1,e]$, then $(e,e^e]$, and so on. So there are countless functions of this type. With only slightly more work one can arrange for them to be continuous, differentiable, and so on. Real analytic is another story.... Commented May 4, 2021 at 17:11
• To get the trivial answer out of the way: $f \equiv 0$ works. Commented May 4, 2021 at 17:12
• Let $g(x) = e^x$ and let $h(x) = [f\circ g](x) = f[g(x)].$ Then, $h'(x) = f'[g(x)] \times g'(x) = f'(e^x) \times e^x.$ By the problem's premise, $h(x) = g(x) \times f(x).$ This implies that $h'(x) = [g'(x) \times f(x)] + [f'(x) \times g(x)] = e^x[f(x) + f'(x)].$ So, one solution that does not work is $f(x) = 1 \implies f'(x) = 0.$ Commented May 4, 2021 at 17:39
• Seems tough to find non-trivial solutions to $f'(e^x) = f(x) + f'(x).$ Commented May 4, 2021 at 17:45
• The set of solutions form a vector space: the sum of two solutions is a solution and if $f$ is a solution, then so is $\lambda f$. Commented May 9, 2021 at 9:28

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 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.