I recently read an article in which the author describes how to find some functions that obey to certain recursion relationships. If we want to find a function that satisfies, for example, $f(x^a) = x \cdot f(x)$, then the author explains we can proceed as follows:
Consider $ x > a > 1$ . Then $f(x) = f((x^{(1/a)})^a) = x^{1/a}f(x^{(1/a)})$
$= x^{1/a}f((x^{(1/a^2)})^a) = x^{1/a + 1/a^2}f(x^{(1/a^2)}) = x^{1/a + 1/a^2 + ... + 1/a^n}f(x^{1/a^n})$ .
We know that the limit of $1/a^n$ is zero, when $n$ tends to infinity. The equality $ r + r^2 + r^3 + ... = r/(1-r)$ is also useful. When we finally set $f(1) = 1$, we may write:
$f(x) = x^{\frac{1/a}{1-1/a}} f(x^0) = x^{\frac1{a-1}}$.
Now, the author and I wonder if a function can be found that satisifies the recurrence relationship $f(\log(x)) = x \cdot f(x)$ . For me, the main motivator for asking this question is plain curiosity. As always, pointers to relevant literature are very much appreciated.
Thanks,
Max
NB: log(x) is the natural logarithm of x.