Iteration Sets, uniformly convergence of functionsets I am working on this problem:
let $f:R \to R$  with $f_0(x) = x$ and $f_{n+1}(x)= cos(f_n(x))$ for all $x \in R$ and $n \in N_0$.
(a) Show for all $x \in R$ and $n \in N$ using the fact that $\pi \gt 3 $ :
$$|f_{n+1}(x)-f_n(x)| \le 2(\frac{\sqrt3}{2})^{n-1}$$
(b) Show that the set $(f_n)_{n\in N}$ converges uniformly against a function $f$.
I tried to show $(a)$ with Induction but it did not work.
Do you know how to approach $(a)$ ?
For $(b)$ i'm gonna use Cauchy criteria but i'm not sure how to do it properly?
If you know how to approach this problem, let me know please.
Thank you for your Answers. :)
 A: For $n\in \Bbb N$ we have $f_n(x) \in [-1,1]$.
With $\pi > 3$ it follows $f_n(x) \in\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]$
So we get with by mean value theorem $$|f_{n+1}(x)-f_n(x)| = |\cos(f_n(x))-\cos(f_{n-1}(x))| = |\sin(\xi)|\cdot|f_{n}(x)-f_{n-1}(x)|$$ for a $\xi \in\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]$
So in total we get:
\begin{align*}|f_{n+1}(x)-f_n(x)| &= \sin(\xi)|\cdot|f_{n}(x)-f_{n-1}(x)| \\  &\le \sin\left(\frac{2\pi}{3}\right)\cdot|f_{n}(x)-f_{n-1}(x)| \\ &=  \frac{\sqrt{3}}{2} |f_{n}(x)-f_{n-1}(x)| \end{align*}
By induction we get:
\begin{align*}|f_{n+1}(x)-f_n(x)| &\le \left(\frac{\sqrt{3}}{2}\right)^{n-1} |f_{2}(x)-f_{1}(x)| \\ &\le \left(\frac{\sqrt{3}}{2}\right)^{n-1}\left(|f_{2}(x)| + |f_{1}(x)|\right) \\&\le 2\left(\frac{\sqrt{3}}{2}\right)^{n-1}\end{align*}
For b) first consider that $1 \ge f_{n+1}(x) \ge f_n(x)$ for $n>1$. So $f_n(x)$ converges at all. Then we use again the fact that have $f(x) \in [-1,1]$ from which it follows $$1 \ge f_n(x) \ge f_n(0)$$
for $n>1$ and $f_n(0) \to 1$ independent of $x$ so the convergence is uniformly.
