Proof that sequence converges to staircase function Let $ \ f: \mathbb{R} \to \mathbb{R} \ $ be the function $ \ f(x) = x + \sin(x) \ $, for all $ \ x \in \mathbb{R} \ $. Define a sequence of functions $ \ (f_n) \ $ such that $ \ f_1 = f \ $ and, $\forall n \in \mathbb{N} \ $, $ \ f_{n+1} = f \circ f_n \ $. We get $ \ f_1 = f \ $, $f_2 = f \circ f = f^2 \ $, $f_3 = f \circ f \circ f = f^3 \ $, ...
Prove that this sequence converges to a (shifted)
staircase function.
Thanks.
 A: In the following I assume you are interested in point-wise convergence.
Notice that points of the form $k\pi$ are fixed points of $f$, and so are fixed by every function in your sequence $\{ f_n \}_n$. Moreover, notice that $f(x)=-f(-x)$, and so this also holds for every function in the sequence. As a consequence, it is enough to focus on points $x\geq 0$, and the rest follows by symmetry. Moreover, by an argument similar to the one above, it can easily be seen that it is sufficient to study the sequence in the interval $[0,\pi]$.
It is enough to prove that, if $x=(2k+1)\pi+y$, with $-\pi<y<\pi$, then $f^n(x)\to (2k+1)\pi$, which can be easily shown by working in $[0,\pi]$.
A: Note that for $k \in \mathbb{Z}$;
$\lim_{n \rightarrow \infty}f_n(k\pi) = k\pi$
Now there are two remaining cases where the domain can belong
Case 1) $x \in ((2k-1)\pi,2k\pi)$
Then $(2k-1)\pi < f(x) < x$; Hence $f_n(x)$ is a decreasing sequence bounded below by $(2k-1)\pi$; Thus $f_n(x)$ must converge to some $p_x$ in $((2k-1)\pi,2k\pi)$ that must be a fixed point; thus we must have $p_x = (2k-1)\pi$
Case 2) $y \in (2k\pi,(2k+1)\pi)$
Note that in this case $ y < f(y) < (2k+1)\pi$; Hence $f_n(y)$ is a increasing sequence bounded above by $(2k+1)\pi$; by similar reasoning to Case 1) we have that $f_n(y)$ converges to $(2k+1)\pi$
A: It suffices to show the following:
Let $\{a_n\}$ be recursively defined as
$$
a_1=x, \quad a_{n+1}=a_{n}+\sin a_{n}=f(a_n).
$$
Then:
(i) If $x\in k\pi)$, then $\{a_n\}$ is constant.
(ii) If $x\in \big(2k\pi,(2k+1)\pi\big)$, then $a_n\to (2k+1)\pi$.
(iii) If $x\in \big((2k-1)\pi,2k\pi\big)$ then $a_n\to (2k-1)\pi$.
Proof.
(i) is obvious. For (ii), if $x\in \big(2k\pi,(2k+1)\pi\big)$, then $\sin x>0$ and
$$
2k\pi<x<x+\sin x=x-\int^{(2k+1)\pi}_x\cos t<x+\big((2k+1)\pi -x\big)=(2k+1)\pi.
$$
Hence, $\{a_n\}$ is strictly increasing and upper bounded by $(2k+1)\pi$ and hence convergent to a $y\in (2k\pi,(2k+1)\pi]$. Now if $a_n\to y$, then
$$
a_{n+1}=a_n+\sin a_n\to y+\sin y.
$$
But $a_{n+1}\to y$, and hence $y$ is of the form $(2\ell+1)\pi$. The only such number in the interval $(2k\pi,(2k+1)\pi]$ is $(2k+1)\pi$. Hence $a_n\to (2k+1)\pi$.
(iii) is dealt to in a similar fashion.
