Problems to demonstrate that the succession is convergent

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The exercise seems easy but I have problems to resolve it: We define a function

$$f : \mathbb{R} \to\mathbb{R}$$

and a $c \in(0, 1)$ such that $\forall x, y \in\mathbb{R} : |f(x) − f(y)| \leq c|x − y|$. We are asked to demonstrate that the sequence defined by $X_1 = a$ and $X_{n+1} = f(X_n)$ is convergent.

I tried to demonstrate that the sequence is contractive so the sequence will also be of cauchy so the sequence will be convergent but im having problems to prove it by induction. Thanks.

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You can prove that $x_n$ is Cauchy by letting $m>n$ and

$$\begin{align}|x_m-x_n|&=|f^m(x_1)-f^n(x_1)|\\&\leq |f^m(x_1)-f^{m-1}(x_1)|+...+|f^{n+1}(x_1)-f^n(x_1)|\\&=: A.\end{align}$$

Now $|f^{k+1}(x_1)-fk(x_1)|\leq c|f^k(x_1)-f^{k-1}(x_1)|\leq...\leq c^k|f(x_1)-x_1|$.

Therefore $$\begin{align}A&\leq (c^{m-1}+c^{m-2}+...+c^n)|f(x_1)-x_1|\\&\leq c^n(1+c+c^2+...)|f(x_1)-x_1|\\&\leq \frac{c^n}{1-c}|f(x_1)-x_1|.\end{align}$$

Since $c\in(0,1)$, then $\frac{c^n}{1-c}|f(x_1)-x_1|$ can be made as small as you want, by taking $n$ large enough.