# Does $f'(x) = f(\ln(x))$ have an easily expressed solution?

This is motivated by this question, where we can consider a bound of the form $f((n+1)!) - f(n!) \le 2f(n)$. To find a function with a similar growth rate, I wondered if there's a technique for solving differential equations of the form $$f'(x) = f(\ln(x))$$ or equivalently $$f'(e^y) = f(y).$$

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I'm working over the reals. I think your function would be very complicated if it exists. It certainly wouldn't be smooth. To see that, consider the second derivative of $f$, for example. $$f^{\prime\prime}(x) = f^\prime(\ln x)/x = f(\ln\ln x)/x\,.$$ Further differentiation will produce a term in the expression for $f^{(n)}(x)$ which involve $n$ iterated logarithms of $x$. For any arbitrary $x$, some number of logarithms will be "too many" and result in taking the logarithm of a negative number. So, for any $x$, we can find $n$ such that the $n$th derivative of $f$ does not exist, and hence $f$ couldn't be smooth, or even $C^n$.
Define $f$ arbitrarily in $[1,e]$, then for $x>e$ define $$f(x) = f(e) + \int_e^x f'(t)\,dt = f(e)+\int_e^x f(\log t)\,dt ,$$ and for $y<1$ define $f(y) = f'(e^y)$. The problem then is to make it match at the endpoints.