Metric spaces and fix point

Question

I saw this problem in my course of Intr. to the topology:

Let $(X,d)$ be a compact metric space and $$f :(X,d) \rightarrow (X,d)$$ a continuous function such that: $\quad d(f(x);f(y)) < d(x;y)$, if $x\neq y$.

Prove that $f$ has an unique fix point in $X$.

Well, after proving these statements I could resolve the problem, I think,

$\forall k\in \mathbb{N} : d(x_{k+1};x_k)\leq \alpha^k d(x_1;x_0)$
$\forall k\in \mathbb{N} : d(x_k,x_0)\leq (1+ \dots + \alpha^{k-1}) d(x_1;x_0)$

where: $x_1 = f(x_0)$ and $$\forall n\in \mathbb{N} : x_{n+1}=f(x_n)$$

but it's the right way?

-Thank you for all.

for uniqueness use two fixed points into your inequality – Arpit Kansal May 12 '15 at 06:07 — Arpit Kansal, May 12 '15 at 06:07

score 21 · Accepted Answer · answered May 12 '15 at 06:12

21

Consider the real valued continuous function $g:X \to \mathbb R$ defined as $$ g(x) = d(x, f(x)) $$ As the space $X$ is compact, $g$ achieves its minimum, say at $x_0.$

If we assume $g(x_0) > 0,$ then calculate $$ g(f(x_0)) = d(f(x_0), f(f(x_0)) ) < d(x_0, f(x_0)) = g(x_0), $$ which is a contradiction of the minimality of $g(x_0).$ So, in fact, $g(x_0) = 0$ and so $f(x_0) = x_0$ shows the fixed point. For uniqueness see my comment.

answered May 12 '15 at 06:12

Arpit Kansal

10,268

3

Thank you very much! This proof is simple and elegant. – Ysaac May 12 '15 at 06:17
wow, interesting. – R.N Sep 16 '15 at 17:05
@RaziehNoori glad that you enjoyed it! – Arpit Kansal Sep 17 '15 at 16:02

Metric spaces and fix point

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