Let f : R → R be a function such that | f(x) - f(y)| ≤ | sin x - sin y|, for each x,y Є R ..... Let f : $\Bbb R$  → $\Bbb R$ be a function with the property | f(x) - f(y)| ≤ | sin x - sin y| , for every real numbers x,y.


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*Prove that that there exists an unique real number c such that f(c) = c.

*Consider the sequence ($x_{n}$) with $x_{0}$=0 and $x_{n+1}$= f($x_{n}$) for every positive integer n. Prove that sequence converges and its limit is c.


So far I have come up with this: since sin x = 0 implies x= n$\pi$, where n is an integer, we get that f(n$\pi$)=f(0). Since the function sin is periodic with period 2$\pi$, we obtain | f(x+2$\pi$) - f(x)|  ≤ | sin(x+2$\pi$) - sin x|, which is equivalent to f(x+2$\pi$) = f(x), for every real x, so the function f has period 2$\pi$.I've also thought about restricting to the interval [0,2$\pi$], since f is periodic, and apply Rolle's theorem on the intervals [0,$\pi$] and [$\pi$,2$\pi$], but I got nothing.
 A: Since $|f(x)-f(y)| \le |\sin x - \sin y| \le |x-y|$, we see that $f$ is continuous.
Since $|f(x)-f(0)| \le |\sin x| \le 1$, we see that $f$ is bounded.
It follows from the intermediate value theorem that the function $\phi(x) = f(x)-x$ must have at least one zero, that is $f$ has at least one fixed point.
Since $\sin x -\sin y = \int_x^y \cos(t) dt$, and $|\cos t| <1$ for ae. $t$, we see that if $x \neq y$, then
$|\sin x - \sin y| < |x-y|$. Consequently $f$ can only have one fixed point, since if $f(x)=x$, $f(y) = y$, then if $x\neq y$, we have
$|x-y|=|f(x)-f(y)| < |x-y|$, a contradiction.
Let $c$ denote the unique fixed point.
Now suppose $x_{n+1} = f(x_n)$. We see from above that $x_n \in [f(0)-1,f(0)+1]$, which is compact. Let $\delta_n = |x_{n+1}-x_n|$, and if $\delta_n >0$, we see that $0 \le \delta_{n+1} < \delta_n \le 2$. 
I need a technical estimate: Suppose $0 < \delta \le |x-y| \le \pi$, then 
$|\sin x - \sin y| \le (\operatorname{sinc} {\delta \over 2})|x-y|$.
Since $\sin x - \sin y = 2 \sin { x-y\over 2} \cos { x+y \over 2}$, then
$|\sin x-\sin y| \le 2 |\sin { x-y\over 2}| = 2 |\operatorname{sinc} {x - y \over 2} ||{x - y \over 2}| $. Since $\max_{t \in [\delta, \pi]} \operatorname{sinc} {t \over 2}  = \operatorname{sinc} {\delta \over 2}$, we obtain the desired result.
Returning to the proof, we have $\delta = \lim_n \delta_n = 0$.  To see this, suppose $\delta  >0$. Then the above shows that $\delta_{n+1} \le \operatorname{sinc} {\delta \over 2} \delta_n$, which gives a contradiction, hence $\delta = 0$, and so $\lim_n (f(x_n) - x_n) = 0$.
Since $x_n \in [f(0)-1,f(0)+1]$  is compact, we see that $x_{n}$ must have an accumulation point $x$, and we must have $f(x)-x = 0$, and so $x=c$ (since the fixed point is unique). It follows that $x_n \to c$.
