Contraction Mapping to Compact Metric.....

Question

Let: $f:X \rightarrow X$ be a contraction on a compact metric space $X$. Show that $\bigcap_{i=1}^\infty f^n(X)$ is a single point, where $f^n=f \circ f \circ \cdot \cdot \cdot \circ f (n$ times). Is this true if $X=\mathbb R$?

Answer 1

Assume $d(f(x),f(y))\le kd(x,y)$ with $k<1$. You should know that $f$ has a unique fixpoint, $x_0$ say. As the continous map $x\mapsto d(x_0,x)$ attains its maximum, $R$ say, we find that $d(x_0,f^n(x))=d(f^n(x_0),f^n(x))\le k^nR$, hence $f^n(X)\subseteq \overline B(x_0,k^nR)$. For any $x\ne x_0$, we find $n$ with $k^nR<d(x_0,x)$, thus showing $x\notin\bigcap_n f^n(X)$ (whwreas clearly $x_0\in \bigcap_nf^n(X)$).

For $X=\mathbb R$ consider the contraction $f(x)=\frac12 x$. Then $f^n(X)=X$ for all $n$, hence the intersection is not a singleton set.

Answer 2

@YoungCHOI,

If $f$ is a contraction on $X$, then we know, by definition, $d(f(x),f(y)) \leq k \cdot d(x,y)$, for $x,y \in X$ and $0 \leq k < 1$. We can restate this, without loss of generality, that $d(f(x), f(y)) < d(x,y)$.

Then we assert that [Lemma] such an $f$ must have a unique fixed point $x \in X$ such that $f(x) = x$.

Firstly, uniqueness of $x$ is fairly straightforward, for let there be some other fixed point $y \in X$ such that $f(y) = y$. Then we would have $d(f(x), f(y)) = d(x,y)$, which contradicts the contractive property of $f$ on $X$.

We proceed by contradiction: assume no such unique, fixed point exists. Since continuity of $f$ follows from the fact that its a contraction mapping, we define $h(x) = d(f(x),x)$. $h$ is readily seen to be positive, i.e. $h(x) > 0$ (X is a metric space) and continuous (via continuity of $f$ and the properties of the metric $d$). Compactness of $X$ also tells us, via the Maximum Value Theorem, that there exists some $p \in X$ such that $h(x) \leq h(p), \space \forall x \in X$. Now we are prepared to derive our contradiction, for notice:

$h(f(p)) = d(f(f(p)),f(p)) < d(f(p),p) = h(p)$.

[We derive the central inequality from the contractive property $f$ induces on $X$.] But we assumed there was no such fixed point!

Now we extend what we have just proven to demonstrate your hypothesis. Given a contraction mapping $f: X \to X$, define $f^n = f \circ f \circ ... [n \space times]$. Then we know that $f$ has a unique fixed point, say $x \in X$. Then, inductively, we have that $f^{n}(x) = x$ as well, as $f(f(f(f(\dots f(x)))))$ reduces to $f(x) = x$ for all such $n$. Therefore we know that $\bigcap_{n=1}^{\infty}f^{n}(X)$ is non-empty, consisting of at least the point $f(x) = x$. To show uniqueness of this point, i.e. that there is at most one such fixed point, would contradict the contractive properties of 'each' concatenated $f$ on $X$, as demonstrated in the proof above, by inducing an equality $d(f(x),f(y)) = d(x,y)$ for another fixed $f(y)=y \in X$.

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This would not work for $X = \mathbb{R}$, as $X$ would not be compact, so the functions $f$ may not attain a maximum, which, as you saw in our proof above, 'forces' $f$ to retain a fixed point.