$f_1(x)=sin(x),\; f_{n+1}(x)=sin(f_n(x))$, $(f_n)$ converges to zero uniformly

Question

Consider for $n\in\mathbb{N}$ the function $f_n:\mathbb{R}\to\mathbb{R}$ given by $$f_1(x)=sin(x),\; f_{n+1}(x)=sin(f_n(x)).$$

I'm stuck to prove that $(f_n)_n$ converges to $f=0$ uniformly.

First of all, $(f_n)_n$ converges to $f=0$ uniformly means $\sup_{x\in\mathbb{R}}|f_n(x)-0|\to 0$ für $n\to\infty$. I know that $|sin(f_n(x))|< |x|$ for all nonzero $x\in\mathbb{R}$, does this estimation help? Furthermore I found this answer Given the sequence of functions, $f_1(x):=\sin(x)$ and $f_{n+1}(x):=\sin(f_n(x))$, why $|f_n(x)|\leq f_n(1)$? but I don't know, how it helps.. I don't know how to continue.

Best

Answer 1

Hint

You have $-1 \leq f_1(x) \leq 1$.

Prove by induction that $$ | f_n(x) | \leq \sin(\sin (\sin(...(\sin(1)))))$$

Hint 2

Prove that the sequence $f_n(1)=\sin(\sin (\sin(...(\sin(1)))))$ is decreasing and bounded. Deduce that it is convergent to a limit $l$ which satisfies $$\sin(l)=l$$

Which is the only real number which satisfies this relation?