3
$\begingroup$

LEt $X\subseteq \mathbb{R}^d$ be a closed set, and $f: X \to X$ be a function such that

$$ ||f(x_1) - f(x_2)|| < ||x_1 - x_2|| \; \; \forall x_1, x_2 \in X $$

If $X$ is compact, then there exists a unique $p \in X$ such that $f(p) = p $.

How Can I approach this problem? My thought was to assume by contradiction that $\forall x $, $f(x) \neq x$. But, then how can I obtain a contradiction?

$\endgroup$
2
  • $\begingroup$ Isn't your function a contraction on a complete space? $\endgroup$
    – Mhenni Benghorbal
    Feb 14, 2014 at 5:24
  • 1
    $\begingroup$ @MhenniBenghorbal No, it's what is sometimes known as a weak contraction. For Banach's fixed point theorem, you need a $q < 1$ with $\lVert f(x)-f(y)\rVert \leqslant q\cdot \lVert x-y\rVert$, here $$\sup_{x\neq y} \frac{\lVert f(x)-f(y)\rVert}{\lVert x-y\rVert} = 1$$ is possible. If a fixed point exists, it is of course unique, but a weak contraction on a complete metric space need not have a fixed point. You need compactness (or a stronger contraction property) to guarantee the existence of a fixed point. $\endgroup$
    – Daniel Fischer
    Feb 14, 2014 at 13:18

1 Answer 1

Reset to default
3
$\begingroup$

The function $x\mapsto \lVert x-f(x)\rVert$ is continuous. Since $X$ is compact, it attains its minimum, say in $x_0\in X$. What follows for $x_0$ and $f(x_0)$?

$\endgroup$
13
  • $\begingroup$ Then, uniqueness is pretty much a straightforward application of the assumptions. $\endgroup$
    – Jonathan Y.
    Feb 13, 2014 at 23:01
  • 1
    $\begingroup$ Right. The uniqueness doesn't even need compactness. $\endgroup$
    – Daniel Fischer
    Feb 13, 2014 at 23:02
  • $\begingroup$ That's a really nice way to do it. My first thought was to use Banach's fixed-point theorem (as I stated in my answer). $\endgroup$
    – Daniel Robert-Nicoud
    Feb 13, 2014 at 23:05
  • $\begingroup$ So, by compactness, then we have $\inf_{x \in X} g(x) = g(x_0) = ||x_0 - f(x_0)|| $ for some $x_0 \in X$. is this correct? @Daniel Fischer $\endgroup$
    – user124140
    Feb 14, 2014 at 0:01
  • $\begingroup$ @Learner Yes. And what follows for $g(f(x_0))$? $\endgroup$
    – Daniel Fischer
    Feb 14, 2014 at 0:08

You must log in to answer this question.