Find a function such that $f(\log(x)) = x \cdot f(x) $ 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. 
 A: Claim given any function $g:(0,1]\to \mathbb{R}$, there exists a unique extension $f:\mathbb{R}\to\mathbb{R}$ satisfying your relation. 
Proof. For $x > 1$, there exists a unique $n\in\mathbb{N}$ such that taken the $n$-fold logarithm of $x$ gives you a number in $(0,1]$. And for $x \leq 0$, $e^x \in (0,1]$. So by iteration the function is well-defined. 

Edit Okay, let me do this explicitly. Fix your favourite function $g(x)$ on $(0,1]$. Mine happens to be the Cantor function. It doesn't matter at all for the construction what this function is. 
Let your $f(x)$ be defined piecewise. For $x\in (0,1]$ define $f(x) = g(x)$. 
For $x\leq 0$, define $f(x) = e^{x} g(e^{x})$. Notice that $e^x$ for $x\leq 0$ is a number in $(0,1]$. 
For $1 < x \leq e$, define $f(x) = \frac1x g(\log x)$. For $e < x \leq e^e$, let $f(x) = \frac1x f(\log x) = \frac1{x\log x} g(\log \log x)$. For $e^e < x \leq e^{e^e}$, let $f(x) = \frac1x f(\log x) = \frac1{x \cdot \log x \cdot \log\log x} g(\log \log \log x)$. And so on. 
If your favourite function is $g(x) = 0$, then when you run this procedure you get Chandru's example where $f(x) = 0$ everywhere. If your favourite function is $g(x) = 1$, you have that $f(x) = e^{x}$ for $x \leq 0$, $f(x) = 1$ for $0 < x \leq 1$, $f(x) = \frac1x$ for $1 < x \leq e$, $f(x) = \frac{1}{x\log x}$ for $e < x \leq e^e$ and so on and so forth. 
For whatever function you choose to start with as $g(x)$ defined on $(0,1]$, you get one corresponding function $f(x)$ that solves your recurrence relation. Since there are uncountably many functions on $(0,1]$, there are also uncountably many possible solutions to your recurrence relation. 
