real analysis - A function satisfying a given condition

Is there a continuous real valued function $f$ satisfying $f(x+1)(f(x)+1)=1$ for all $x$ in the domain of $f$(possibly $\mathbb{R})$? Clearly the image of $f$ does not include $0$ and $-1$. By the continuity of $f$ and the intermediate value theorem, $f$ has to be always positive or always negative since otherwise there would be a zero of $f$. Does such a continuous $f$ exist? How about discontinuous functions?

Please give a hint to proceed. Thank you.

Answer

Suppose $f:\mathbb{R}\to\mathbb{R}$ is a continuous solution. Let us consider the sequence $f(n)$, $n\in\mathbb{Z}$. If we write $$f(n)=\frac{a_n}{a_{n+1}},\tag{*}$$ by the given recursion formula, we have
$$f(n+1) = \frac{a_{n+1}}{a_n+a_{n+1}}.$$ This implies the sequence $a_n$ is a Fibonacci sequence, i.e. $a_{n+2} = a_{n+1}+a_n$. It is well-known that the general solution of $a_n$ is given by
$$a_n = A\alpha^n + B\beta^n$$ for some $A,B\in\mathbb{R}$ where $\alpha = \frac{1+\sqrt{5}}{2}$ and $\beta=\frac{1-\sqrt{5}}{2}$. Plugging this into the expression in $(*)$, we get
$$f(n) = \frac{A\alpha^n + B\beta^n}{A\alpha^{n+1} + B\beta^{n+1}}.$$ Assume that $A=0$. Since not both $A,B$ are $0$, we have $f(n) = \beta^{-1}$. In the same way, $f(n)=\alpha^{-1}$ holds if $B=0$. Now, assume that $A$ and $B$ are not zero. By taking limit as $n\to\pm\infty$, we have
$$\lim_{n\to\infty}f(n) = \alpha^{-1} = -\beta>\frac{1}{2},$$ and
$$\lim_{n\to-\infty}f(n) =\beta^{-1} = -\alpha<-\frac{3}{2}.$$ But this implies by IVP that there exists $c\in\mathbb{R}$ such that $f(c)=0$. Since this is impossible as $f(x)\ne 0$, there are only $2$ possibilities: Either $f(n) \equiv \alpha^{-1}$ or $f(n)\equiv \beta^{-1}$. Finally, notice that the above argument applies in the same way to $f(x+n)$ for arbitrary $x\in \mathbb{R}$. This implies $f(\mathbb{R})\subset \{\alpha^{-1}, \beta^{-1}\}$. By the continuity of $f$, this says that $f$ must be a constant function. Thus the only continuous solutions to the equation are
$$f(x) \equiv \alpha^{-1} = \frac{-1+\sqrt{5}}{2}$$ or
$$f(x) \equiv \beta^{-1} = \frac{-1-\sqrt{5}}{2}.$$

Note: Note that for all $n\in\mathbb{Z}$, $$f(n)=\frac{a_n}{a_{n+1}}\notin\mathbb{Q} \Leftrightarrow f(n+1)=\frac{a_{n+1}}{a_n+a_{n+1}}\notin\mathbb{Q}.$$ Suppose that $f(0)= \frac{a_0}{a_{1}}$ is given by an irrational number. If $a_{n+2}=0$ for some $n\ge 0$, then $f(n)=-1$ holds and this contradicts $f(0)$ is irrational. In the same way, if $a_n=0$ for some $n<0$, then $f(n)=0$ holds and this contradicts $f(0)$ is irrational. So, the irrational initial data $f(0)$ guarantees that $f(n)$ is well-defined for all $n\in\mathbb{Z}$.

Let $a_0 = 1$ and $a_1=\pi$. If we solve it for $A,B$, we get
$$A=\frac{\beta-\pi}{\beta-\alpha},\quad B=\frac{\pi-\alpha}{\beta-\alpha}.$$ If we let $f(x) = \frac{1}{\pi}$ for $x\in [0,1)$, this implies
$$f(x+n) = \frac{(\beta-\pi)\alpha^n + (\pi-\alpha)\beta^n}{(\beta-\pi)\alpha^{n+1} + (\pi-\alpha)\beta^{n+1}}.$$ is well-defined for $n\in\mathbb{Z}$ and satisfies the given functional equation. This shows that

$$f(x)=\frac{(\beta-\pi)\alpha^{\lfloor x\rfloor} + (\pi-\alpha)\beta^{\lfloor x\rfloor}}{(\beta-\pi)\alpha^{{\lfloor x\rfloor}+1} + (\pi-\alpha)\beta^{{\lfloor x\rfloor}+1}}$$ is an example of a discontinuous solution $f:\mathbb{R}\to\mathbb{R}$.