Given the sequence of functions, $f_1(x):=\sin(x)$ and $f_{n+1}(x):=\sin(f_n(x))$, why $|f_n(x)|\leq f_n(1)$? I'm analyzing this sequence of functions (for $x\in \Bbb R$):
$$\begin{align}f_1(x)&:=\sin(x)\\f_{n+1}(x)&:=\sin(f_n(x))\end{align}$$
to show if it converges uniform or pointwise. My book suggests that since $|f_1(x)|\leq 1$ (which I do understand) then $|f_n(x)|\leq f_n(1)$ for every $n>1$. I'm not sure if this is correct for example $|f_2(x)|=|\sin(\sin(x))|$ is not lower or equal than $f_2(1)=\sin(\sin(1))$ for every $x>1$. 
 A: For simplicity we introduce $f_0(x)=x$. Note that $f_1(x)=\sin(f_0(x))$, i.e., the recursion formula holds
.
Also, we could replace the recursion with $f_{n+1}(x)=f_n(\sin x)$. (Boith these observations might be shown by induction).
Claim. Let $n\in\Bbb N_0$. Then $f_n$ is an odd function.
Proof. This is clear for $n=0$. As the composition of odd functions is odd, the recursion formula allows us to conclude the claim by induction. $\square$
Claim. Let $n\in\Bbb N_0$. Then $f_{n+1}(x)=f_n(\sin(x))$. (In particular, $f_n(0)=0$)
Proof. This is clear for $n=0$.
Assume the claim holds for some $n\in\Bbb N_0$.
Then $$f_{n+2}(x)=\sin(f_{n+1}(x))=\sin(f_n(\sin x))=f_{n+1}(\sin x)$$
so that the claim for all $n$ follows by induction. $\square$
Claim. Let $n\in\Bbb N_0$. Then $f_n$ is strictly increasing on $[0,1]$.
Proof. Once again clear for $n=0$.
Using the preceding claim and that the sine function is a strictly increasing map $[0,1]\to[0,1]$, the claim follows once again by induction. $\square$
Assume $f_n$ is strictly increasing on $[0,1]$. As the sine is strictly increasing on $[0,1]$ and maps $[0,1]$ into $[0,1]$, we conclude that $f_{n+1}=f_n\circ \sin$ is strictly increasing. $\square$
Corollary.
For any $n\in\Bbb N_0$ and $x\in[-1,1]$ we have $|f_n(x)|\le f_n(1)$.
Proof. As $f_n$ is odd, we need only consider $0\le x\le 1$. By monotonicity, we have $$0=f_n(0)\le f_n(x)\le f_n(1).$$ $\square$
