A function satisfying $f(\frac1{x+1})\cdot x=f(x)-1$ and $f(1)=1$? $f:[0,\infty)\to\mathbb{R}$ 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!
 A: Such a function doesn't exist. 
Forget $f(1) = 1$. Just assume $f(x)$ satisfy the functional equation on $[0,\infty)$. We have
$$
\begin{align}
f(x) &= 1 + x f\left(\frac{1}{x+1}\right)\\
&= 1 + x \left[ 1 + \frac{1}{x+1}f\left(\frac{x+1}{x+2}\right)\right]\\
&= 1 + x \left[ 1 + \frac{1}{x+1} + \frac{1}{x+2} f\left(\frac{x+2}{2x+3}\right) \right]\\
&\;\vdots\\
&= 1 + x \left[ 1 + \left( \sum_{k=2}^{p-1} \frac{1}{F_{k-1} x + F_k} \right) + \frac{1}{F_{p-1} x + F_p}f\left(\frac{F_{p-1} x + F_p}{F_p x + F_{p+1}}\right)\right]\\
&\;\vdots
\end{align}
$$
where $F_k$ is the $k^{th}$ Fibonacci number. Notice 
$$\lim_{p\to\infty} \frac{F_{p+1}}{F_p} = \phi = \frac{1+\sqrt{5}}{2}
\quad\implies\quad
\lim_{p\to\infty} \frac{F_{p-1} x + F_p}{F_p x + F_{p+1}} = \frac{1}{\phi}
$$
If we further assume $f(x)$ is continuous at a single point $\displaystyle\;\frac{1}{\phi}$,
the last expression converges as $p \to \infty$ and we get:
$$f(x) = 1 + x + \sum_{k=2}^\infty \frac{x}{F_{k-1} x + F_k}$$
Substitute $x$ by $1$, we get
$$f(1) = 1 + 1 + \sum_{k=2}^\infty\frac{1}{F_{k-1}+F_k} = \frac{1}{F_1} + \frac{1}{F_2} + \sum_{k=3}^\infty \frac{1}{F_k}
= \sum_{k=1}^\infty \frac{1}{F_k}$$
The number at the RHS is known as the 
Reciprocal Fibonacci constant
$\psi$ with a value $$\approx 3.359885666243177553172011302918927179688905133731$$
differs from $1$. 
A: 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 $\frac{\sqrt5-1}{2}$.)
A: This is not a full solution, but instead I wanted to give two thoughts that are too long for the comments section. 
1) Note that since $f(0) = 1$ and $f(1) = 1$ you can rearrange your expression to 
$$
\frac{f(x) - f(0)}{x-0} = f \Big( \frac{1}{x+1} \Big)
$$
for any $x \neq 0$. Taking a limit on both sides and using the continuity of $f$, you find that $f'(0) = f(1) = 1$. So, at very least, any function satisfying your criteria will be differentiable at $0$. In particular, if a function exists which is smooth enough that you can perform a Taylor expansion near $0$, you will have $f(x) = 1 + x + O(x^2)$. 
2) If you can find $f(x)$ for all $x \in (0,1)$ you'll be done since for any $y \in (1, \infty)$ you have $\frac{1}{1+y} \in (0,\frac{1}{2})$ (here you can see $(0,1)$ is even more than you actually need, you really only need $(0,\frac{\sqrt{5}-1}{2}]$, but the idea is the same). So, if you are able to use your reasoning above to find $f(q)$ for every $q \in \Bbb{Q} \cap (0,1)$, then using continuity and density arguments, you can find $f(x)$ for every $x \in (0,1)$ and then be done.
Hopefully this will help.
A: Let $T(x)=1/(x+1)$.  Split up $\mathbb{R}$ into subsets $S_a=\{T^n(a)(n=-\infty..\infty)\}$.  The values of $f(S_a)$ are independent from the values of $f(S_b)$ if $b\not\in S_a$.
$T^n(a)$ approaches $1/\phi$ as $n\rightarrow\infty$ and $-\phi$ as $n\to-\infty$, so $f(T^n(a))$ need to approach a limit for all $a$, by continuity.
But as $T^n(a)$ approaches $1/\phi$, and $f(T(a))=f(a)/a-1/a\approx \phi f(a)$, so $f(a)$ grows by powers of $\phi$ as $a\to 1/\phi$.  I don't think it converges
