convergence sequence 
Suppose that $g$ is continuous on an interval $[a,b]$ and that  $g(x) ∈ [a,b]$ for all $x ∈ [a,b]$.
(a) Use the intermediate value theorem to prove that is at least one number $c ∈ [a,b]$ with $g(c) = c$.

Here is my attempt:
Define $G: [a,b] \to $$\mathbb R$ by $G(x) = x - g(x).$  Then $G$ is continuous on $[a,b].$ Since $a  \leq g(a)\leq b$ and  $a  \leq g(b)\leq b$ we find $G(a) = a -g(a) \leq 0 $ and $G(b) = b -g(b) \geq 0 $. By the intermediate value theorem, there is a $c ∈ [a,b]$ such that $G(c) = 0$ or $c - g(c) = 0.$ Thus, $g(c) = c.$

(b) Suppose further that $g$ is differentiable and that there exists a $\lambda <1$ with  $|g'(x)|\leq \lambda$ for all $x ∈ [a,b]$. Prove that there is exactly one number $c ∈ [a,b]$ with $g(c) = c$.

Here is my attempt: 
Since there exists two different fixed points $\xi<\xi'$. Then as stated we can use Lagrange's theorem on $[\xi,\xi']$:
We get $$1=\frac{g(\xi)-g(\xi')}{\xi-\xi'}=f'(\nu)$$ for some  $\nu$, which contradicts $|f'(x)|<1$.

(c) For any initial value $x_{0} ∈ [a,b]$, define a sequence ${x_{n}} = x_{0}, x_{2}, x_{3} ...$ by $x_n = g(x_{n-1})$ for $n \geq 0$, define $E_{n} = |x_{n}-c|$ and $D_{n} = |x_{n+1} - x_{n}|$.
(i) Prove that $E_{n} < \lambda^{n} E_{0},$ which would make $(x_{n})$ converge to $c$? 
(ii) Prove that $D_{n} < \lambda^{n} D_{0}$?
(iii) Prove that for $n < k$, $|x_k - x_n|\leq\dfrac{\lambda^{n} - \lambda^{k}}{1-\lambda}(D_{0}).$
(iv) Prove that $E_{n} \leq \dfrac{\lambda^{n}}{1-\lambda}(D_{0}).$

 A: The key result here is the mean value theorem which states that for $x, \in [a,b]$, with $x \neq y$, then there exists some $\xi \in (a,b)$ such that $g(x) -g(y) = g'(\xi)(x-y)$.
It follows from this that $|g(x)-g(y)| \le \lambda |x-y|$ for all $x,y \in [a,b]$.
(i) You have some $c$ such that $g(c) = c$. Then the above gives: $|g(x)-c| = |g(x)-g(c)| \le \lambda |x-c|$. It follows that $E_{n+1} \le \lambda E_n$, and so $E_n \le \lambda^n E_0$.
(ii) The same sort of idea gives $|g(g(x))-g(x)| \le \lambda |g(x)-x|$, letting $x=x_n$ gives $|x_{n+2}-x_{n+1}| \le \lambda | x_{n+1}-x_n|$, from which $D_n \le \lambda^n D_0$ follows.
For (iii), you have $x_k-x_n = x_k -x_{k-1} + x_{k-1} -x_{k-2} + \cdots + x_{n+1}-x_n$, from which we get $|x_k-x_n| \le |x_k -x_{k-1}| + |x_{k-1} -x_{k-2}| + \cdots + |x_{n+1}-x_n|$, hence $|x_k-x_n| \le D_{k-1}+\cdots + D_n \le (\lambda^{k-1}+\cdots \lambda^n) D_0 = \frac{\lambda^n-\lambda^k}{1-\lambda} D_0$.
For (iv), notice that $\lim_n E_n = 0$, hence $x_n \to c$. So, fix $n$ in (iii) and let $k \to \infty$ to get $E_n = |x_n-c| \le \frac{\lambda^n}{1-\lambda} D_0$ (remember $\lambda <1$, so $\lambda^k \to 0$).
