Calculus - Sequence and Function - hard exercise EDIT: With your help I have succeed to prove it! Thanks! I also have another question related to this one - written down here.
Let $a,b \in \mathbb{R}$ such that $a<b$ and let $f:[a,b] \to [a,b]$ be a differentiable function, let $t \in [a,b]$, and consider the sequence $(x_n)_{n=1}^{\infty}$ defined by: $$\left\{\begin{matrix}
x_1=t & \\ 
x_{n+1}=f(x_n) & \forall n\geq 1 
\end{matrix}\right.$$
Suppose that there exists a point $\alpha \in [a,b]$ such that $f(\alpha)=\alpha$.
Prove, while using MVT, that if there exists a $0 \le q <1$ such that $|f'(x)|\le q$ for every $x \in [a,b]$, then $\lim_{n \to \infty}x_n=\alpha$
-
I don't know how to use MVT here, since I don't know nothing about $\alpha$ and $x_n$, I can't take the interval $[\alpha,x_n]$ or $[x_n,\alpha]$, ($\alpha$ and $x_n$ can be equal).
Any help will be awesome!
Thanks a lot!
EDIT:
Now I have a $y \in \mathbb{R}$, and the sequence $(x_n)_{n=0}^{\infty}$:
$$x_0=y ~ ~ , ~ ~ x_{n+1}=cos(x_n), \forall n \ge 0$$
Prove thatt the sequence converge to a limit $0<\alpha<1$.
Any tips?
Thanks again!
 A: Take any two points $x_1, x_2 \in [a,b]$ such that $x_1<x_2$. Using MVT on $[x_1,x_2]$ we get
\begin{equation}\label{eq1}
|f(x_1)-f(x_2)| \leq q|x_1-x_2|
\end{equation}
. If there exists another $\beta \in [a,b]$ such that $f(\beta)=\beta$ then $|\alpha-\beta|\leq q |\alpha-\beta|$ which implies such $\alpha$ is unique.
\begin{equation}
|x_{n+1}-x_n|=|f(x_n)-f(x_{n-1})|\leq q|x_n-x_{n-1}| \leq \cdots\leq q^{n-1}|x_2-x_1|=q^{n-1}|f(t)-t|
\end{equation}
If $n>m$
\begin{align}
|x_{m}-x_n| &\leq |x_{m+1} - x_{m+2}|  + \cdots + |x_{n+1} - x_n |\\
              &\leq  (q^{m} +\cdots +q^ {n-1} )|f (t )- t |\\
              &\leq \frac{q^{m}}{1-q} |f (t )- t |
\end{align}
Since $q<1$, $\{x_n\}_n$ is a Cauchy sequence. As $[a,b]$ is complete with respect to $|\cdot|$, there exists $\gamma \in [a,b]$ such that $x_n \rightarrow \gamma$ as $n \rightarrow \infty$. By continuity of $f$, $\lim_{\rightarrow \infty}f(x_n) =f(\gamma)$ also $\lim_{\rightarrow \infty}f(x_n)=\lim_{\rightarrow \infty}x_{n+1}= \gamma$. Thus $\gamma=f(\gamma)$. But since $\alpha$ is unique $\alpha=\gamma$ and $\lim_{n \rightarrow \infty}x_{n}= \alpha$.
A: By using the MVT you can show:
$$
|f(b) - f(a)| < b-a.
$$
Now, have a look at the Banach fixed point theorem for some inspiration.
https://en.wikipedia.org/wiki/Banach_fixed-point_theorem
A: Since $f$ is continuous and applies the interval $[a,b]$ onto itself, the mean value theorem already tells you that there is at least one point in those conditions. The condition on the derivative gives you uniqueness. If there were two distinct $\alpha_1, \alpha_2$ in those conditions, then
$$
|\alpha_1-\alpha_2| = |f(\alpha_1)-f(\alpha_2)| \leq q |\alpha_1-\alpha_2|
$$
and so,
$$
(1-q)|\alpha_1-\alpha_2|\leq 0.
$$
since $0< q < 1$ and $|\alpha_1-\alpha_2|\ge 0$, the only possibility is that $\alpha_1=\alpha_2$.
The convergence of the sequence can now be derived from the inequality
$$
|x_n - \alpha|\leq q|x_{n-1}-\alpha| \leq \cdots \leq q^{n-1}|x_1-\alpha| \leq (b-a) q^{n-1} \to 0.
$$
