# Contraction Principle: Existence of a fixed point proof.

Definition of contraction:

A function $$f : M \mapsto M$$ is a contraction $$\iff$$ $$\exists r \in (0, 1)$$ such that $$d\left( f(x), f(y) \right) \leq r d(x, y)$$ $$\forall$$ $$x, y \in M$$.

Statement of the theorem:

Let $$M$$ be a complete metric space and $$f : M \mapsto M$$ be a contraction. Then, there exists a unique fixed point of $$f$$ (i.e. $$\exists$$ $$x_0 \in M$$ such that $$f(x_0) = x_0$$). Moreover, $$x_0 = \lim \limits_{m \to \infty} f^m(x)$$ $$\forall x \in M$$.

Notation: $$f^m = \underbrace{f \circ f \circ \cdots \circ f}_{\text{m times }}$$

I just need help with the existence part. I already showed for any $$x \in M$$, $$\left( f^m(x) \in M \right)_{m \in \mathbb{N}}$$ is a Cauchy sequence and hence convergent by completeness of $$M$$.

From this, how do I get to $$f(x_0) = x_0$$?

• Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. Commented Feb 16, 2020 at 1:51
• $f$ is a contraction so it's continuous. Then, the result is immediate because $x_n=f(x_{n-1})$ Commented Feb 16, 2020 at 1:53
• What the sign $!$ in the title means?
– Nick
Commented Feb 16, 2020 at 2:09
• @Nick: it means “there exists a unique...” Commented Feb 16, 2020 at 2:24
• @ironX The definition of contraction is $|f(x)-f(y)|<\alpha |x-y|$ for some $0<\alpha<1$ for all $x,y$ so it is immediate that $f$ is not only continuous but uniformly continuous: for $\epsilon>0$ take $\delta= \epsilon.$ Or what is your definition of continuity? Commented Feb 16, 2020 at 2:59

Since $$f$$ is a contraction, $$f$$ is uniformly continuous, for if $$\varepsilon>0$$ is arbitrary, and if $$x$$ and $$y$$ are two points of $$M$$ such that $$d(x,y) < \delta := \varepsilon/r$$ then $$d(f(x),f(y)) \leq r \cdot d(x,y) < \varepsilon.$$ With this, follows the desired result: \begin{align} f(x_0) = f\Big( \lim_{m\to\infty} f^m(x) \Big) &\stackrel{(1)}{=} \lim_{m\to\infty} f(f^m(x)) \\ &= \lim_{m\to\infty} f^{m+1}(x) = x_0 \end{align} where, in $$(1)$$, we use the fact that $$f$$ is continuous.

To prove the existence, do similar work as you did.

• Take any $$x\in M$$, and $$x_1 = x$$.
• Inductively, $$x_{n+1} = f(x_n)$$ for $$n\geq 1$$.

Then by the contraction property, the sequence $$(x_n)$$ is Cauchy. By the completeness of $$M$$, there is $$x_0 = \lim x_n$$.

To prove the uniqueness, assume there are two distinct fixed points $$x_0$$ and $$y_0$$. Then $$d(f(x_0),f(y_0)) = d(x_0,y_0)$$, which contradicts to the contraction property.

• This does not answer the question. Commented Feb 16, 2020 at 3:11

It's easy to see that $$f$$ is continuos funtion

Consider $$\{x_n\}_{n\ge 0}$$: $$x_0=x$$ and $$x_{n+1}=f(x_n)$$

We found that $$d(x_m,x_n)\le rd(x_{m-1},x_{n-1})$$ (1)

So with $$\epsilon >0$$, we choose $$n>log_r(\frac{(1-r)\epsilon}{d(x_1,x_0})$$ then $$\forall p\in\mathbb{N}$$:

$$d(x_{n+p},x_n)$$

$$\le d(x_{n+p},x_{n+p-1})+d(x_{n+p-1},x_{n+p-2})+...+d(x_{n+1},x_n)$$

$$\le (r^{n+p-1}+r^{n+p-2}+...+r^n)d(x_1,x_0)$$

=$$r^n\frac{1-r^p}{1-r}d(x_1,x_0)$$

$$\le r^n\frac{d(x_1,x_0)}{1-r}$$

$$\le \epsilon$$

So $$\{x_n\}_{n\ge 0}$$ is a Cauchy sequence, in complete metric space $$M$$ it converges at $$N$$

Then $$N=lim_{n\rightarrow \inf}x_n=\lim_{n\rightarrow \inf}x_{n+1}=\lim _{n\rightarrow}f(x_n)=f(N)$$, by the continuty of $$f$$