Prove the contracting mapping theorem (the version on $\mathbb R$): Suppose $f : [a,b] \to [a,b]$ is continuous and satisfies $$|f(x) - f(y)| < c|x - y|$$ for some $0 < c < 1$, then $f$ has a fixed point.
I know that if we let $f : X \to X$ be a mapping from a set $X$ to itself, we call a point $x \in X$ a fixed point of f if $f(x) = x$. For example, if $[a, b]$ is a closed interval then any continuous function $f : [a, b] \to [a, b]$ has at least one fixed point. I know that it must follow somehow from the intermediate value theorem but do not know how to construct a general proof. Also, I believe there is a way to do it without using the intermediate value theorem but have no idea how.
ii) Show that if $f$ has a fixed point, then it is unique. This I know must be proved through contradiction and probably follows from the previous proof
iii) Suppose $\lim x_n = x$. Show that $x \in [a,b]$.
Thanks for the help
The uniqueness of the fixed point is quite easy: if $x$ and $y$ are two such points in $[a,b]$, and if we assume that they are different, then $$|x - y| = |f(x) - f(y)| < c |x-y|, $$ hence the contradiction.
For the existence of the fixed point, the classical proof is the following: start from any point $x \in [a,b]$, then consider the sequence given by $x_0 = x, x_1 = f(x), x_2 = f(x_1), \dots, x_{n+1} = f(x_n). $
You can find out what property the limit of that sequence will have, and why this limit exists (i.e. why does the sequence converge in $[a,b]$.) To be a bit more specific I would need the order of the questions: the proof of the contraction mapping is i) ? If so, what is the definition of the sequence $(x_n)_n$ in iii) ?