This question is inspired by the big MO question [here](https://mathoverflow.net/questions/45608/does-the-formal-power-series-solution-to-ffx-sin-x-converge); and also inspired by the big MO question [here](https://mathoverflow.net/questions/17605/how-to-solve-ffx-cosx?noredirect=1&lq=1). The premise of this question requires a backdrop on fractional iteration; so I'll start slow.


Suppose we have a holomorphic function $\phi : G \to G$ where $G$ is an open and connected set. Assume that $\phi^{\circ n}(\xi) \to \xi\_0$ as $n \to \infty$ where $\xi \in G$ and $\xi\_0 \in G$. There then exists the Schroder function $\Psi: G \to \mathbb{C}$ where $\Psi(\phi(\xi)) = \phi'(\xi\_0)\Psi(\xi)$; and $\Psi$ is non-trivial.


When searching for root functions of $\phi$; i.e. searching for functions $f = \sqrt[n]{\phi}:G \to G$ such that $\phi = f(f(\cdots(n\text{ times})\cdots f(\xi)$; it is a simple exercise that there are $n$ solutions. And that each solution is an analytic continuation of $\Psi^{-1}(\sqrt[n]{\phi'(\xi\_0)}\Psi(\xi))$, from a tiny neighborhood about $\xi\_0$ to all of $G$, for one of the $n$ values of $\sqrt[n]{\phi'(\xi\_0)}$.


So since $\cos$, for example, is entire and about the immediate basin $I$ of its only real fixed point $\xi\_0$, $\cos : I \to I$ and $\cos^{\circ n} \to \xi\_0$; there equals two different functions $f\_0,f\_1 : I \to I$ such that $f\_{01}(f\_{01}(\xi)) = \cos(\xi)$. Interestingly enough, it can be showed neither of these $f\_0$ and $f\_1$ are real valued on the reals, proving there exists no analytic $f:\mathbb{R} \to \mathbb{R}$ such that $f(f(x)) = \cos(x)$.


Moving on, we come to the $\sin$ function that has a fixed point at zero. This is a vastly different scenario though. With this function there is no open set satisfying the above criterion; $\sin$'s fixed point at zero is neutral (its multiplier $\sin'(0) = 1$) and thus no neighborhood about zero where $\sin(\sin(...\sin(\xi) \to 0$. But nonetheless we have that $\sin(\sin(...\sin(\mathbb{R}) \to 0$.


Now, just like with $\cos$, there should be two solutions to the equation $f(f(x)) = \sin(x)$ for $ x \in \mathbb{R}$. In the first question I'm referencing, we're only talking about one of these solutions. In fact, they are solving for the solution where, $f'(0) = 1$. But technically there should also be a solution $f'(0) =-1$. Namely, by all lawful right, there should be a second solution that is actually decreasing in a neighborhood of $0$. By all rights, it'll probably look incredibly chaotic on $\mathbb{R}$!


To better explain this, assume that $0<\phi'(0) < 1$ and $\phi:\mathbb{R} \to \mathbb{R}$ where $\phi(\phi(...\phi(x) \to 0$ for all $x$ ($\phi$ analytic). Then there exists $f\_+$ and $f\_-$ where $f\_{\pm} = \Psi^{-1}(\pm\sqrt{|\phi'(0)|}\Psi(x))$; and since both $\Psi$ and $\Psi^{-1}$ are $\mathbb{R} \to \mathbb{R}$ this means that $f\_{\pm}:\mathbb{R} \to \mathbb{R}$. Obviously $-1 < f\_{-}'(0) < 0$ which implies $f\_{-}(x)$ is decreasing in a neighborhood of $0$.


Adding to this; I'll be rough here; taking $\phi\_n = (1-\frac{1}{n})\sin(x)$; and solving for $f\_{\pm}^{(n)} = \pm\sqrt{\phi\_n}$ it can be shown that $f^{(n)}\_{+} \to f\_{+}$ where $f\_{+}$ is presented in [the original question](https://mathoverflow.net/questions/45608/does-the-formal-power-series-solution-to-ffx-sin-x-converge) (It is not the formal power series; but instead in the answers mentioned). However all the methods there fail at producing $f\_{-}$; in fact they make no mention of it. Whereas, to me, it appears to be intuitive to assume $f\_{-}^{(n)} \to f\_{-}$; and that in fact there is a functional square root of $\sin$ that is decreasing in a neighborhood of zero.



> 
> Does there exist an analytic function $f\_{-}: \mathbb{R}\_{\neq 0} \to
> \mathbb{R}$ where $f\_{-}(f\_{-}(x)) = \sin(x)$ and $f\_{-}'(0) =-1$?
> 
> 
> 


Edit: I forgot to add that $f\_{-}(x)$ need not be analytic at $0$. This is because $f\_{-}$ isn't necessarily defined in a neighborhood of zero in the complex plane; where as it is more obvious that there is a neighborhood of $\mathbb{R}\_{\neq 0}$ in the complex plane that $f\_{-}$ is defined. This involves talking about Julia sets and such not, I'd like to avoid that as much as I can.



What about the other $f$ such that $f(f(x)) = \sin(x)$?
Clearly, if $f\_+(z)$ is a functional square-root of $\sin(z)$, then so is $g(z) := -f\_+(-z)$; moreover, they have the same derivative at the origin. Hence, your post implies that $g(z) = f\_+(z)$, so $f\_+(z)$ is an odd function.


It follows that $-f\_+(z)$ is a functional square-root of $\sin(z)$, so $-f\_+(z) = f\_-(z)$.


