TITLE: A function satisfying $f(\frac1{x+1})\cdot x=f(x)-1$ and $f(1)=1$?
QUESTION [10 upvotes]: $f:[0,\infty)\to\mathbb{R}$ is a continuous function which satisfies $f(1)=1$ and:
$$f(\frac1{x+1})\cdot x=f(x)-1$$
Does there exist such a function, if they do, are there infinitely many? And is there any explicit example? I have been able to derive these:

$f(1)\cdot0=f(0)-1$, which means $f(0)=1$.
$f(\frac1{2})=f(1)-1=0$
Similarly, $f(\frac23)=-2$ and then $f(\frac35)=\frac{-9}2$ and so on. 

This is going on without any pattern I can see. Is there anything special in the sequence $(1,\frac12,\frac23,...,a_n,...)$ where $a_n=\frac1{a_{n-1}+1}$? I have not been able to use continuity as well. Please help!

REPLY [4 votes]: There is no such function.  Let $x_0=1$, $x_{n+1}=\frac1{x_n+1}$.  Note that all $x_n\in(0,1]$.  Now, consider
$$ f(x_{n+1}) = \frac{f(x_n)-1}{x_n} $$
as a recurrence for the sequence $f(x_n)$.  We compute that $f(x_2) = -2 < 0$.  Since all $x_n > 0$, this implies (by induction, say) that $f(x_n)<0$ for $n\ge 2$.  But this means the numerator of the RHS is negative; thus the fact that $x_n\le 1$ implies
$$ f(x_{n+1}) \le f(x_n)-1 $$
and so $f(x_n)\to -\infty$, violating continuity.  (I'm taking for granted here that $x_n$ converges; indeed, $x_n$ is the sequence of convergents of $\frac{\sqrt5-1}{2}$.)