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 $f(x_0) = x_0$?

  • $\begingroup$ 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. $\endgroup$
    – Shaun
    Commented Feb 16, 2020 at 1:51
  • $\begingroup$ $f$ is a contraction so it's continuous. Then, the result is immediate because $x_n=f(x_{n-1})$ $\endgroup$ Commented Feb 16, 2020 at 1:53
  • $\begingroup$ What the sign $!$ in the title means? $\endgroup$
    – Nick
    Commented Feb 16, 2020 at 2:09
  • $\begingroup$ @Nick: it means “there exists a unique...” $\endgroup$
    – Clayton
    Commented Feb 16, 2020 at 2:24
  • 2
    $\begingroup$ @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? $\endgroup$ Commented Feb 16, 2020 at 2:59

3 Answers 3


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.

  • 1
    $\begingroup$ This does not answer the question. $\endgroup$
    – azif00
    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 $\forall p\in\mathbb{N}$:


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

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


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

$\le \epsilon$

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 $f$


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