Unique fixed point

• Mar 12th 2011, 11:29 AM
BenH
Unique fixed point
I'm having a problem understanding this:

Show that if $U$ is a continuous map such that $U^n:M \rightarrow M$ is a strict contraction for some fixed integer $n$, then $U$ has a unique fixed point.

I'm not sure I understand what they mean by $U^n$.
• Mar 12th 2011, 11:44 AM
Drexel28
Quote:

Originally Posted by BenH
I'm having a problem understanding this:

Show that if $U$ is a continuous map such that $U^n:M \rightarrow M$ is a strict contraction for some fixed integer $n$, then $U$ has a unique fixed point.

I'm not sure I understand what they mean by $U^n$.

They mean the $n$-fold composition with itself. What is $M$? Hopefully (for my solution anyways) it's a compact metric space.
• Mar 12th 2011, 12:01 PM
BenH
Quote:

Originally Posted by Drexel28
They mean the $n$-fold composition with itself. What is $M$? Hopefully (for my solution anyways) it's a compact metric space.

Does the $n$-fold composition with itself mean $U^n=U\circ U\circ U \circ \cdots \circ U$? It's a metric space, but I don't think you can assume that it is compact.
• Mar 12th 2011, 03:24 PM
Tinyboss
That's what n-fold composition is, yes. And I believe it's completeness, not compactness, that you need for this result. To get a fixed point, make a sequence $\{x,U(x),U^{(2)}(x),\dots\}$ and show that it's Cauchy (this is where we need the space to be complete). For uniqueness, suppose there are two distinct fixed points and look at their images under U--do they get closer together?.
• Mar 12th 2011, 07:16 PM
Drexel28
Quote:

Originally Posted by Tinyboss
That's what n-fold composition is, yes. And I believe it's completeness, not compactness, that you need for this result. To get a fixed point, make a sequence $\{x,U(x),U^{(2)}(x),\dots\}$ and show that it's Cauchy (this is where we need the space to be complete). For uniqueness, suppose there are two distinct fixed points and look at their images under U--do they get closer together?.

That is the classic method--but if it's compact there is an eaiser way by considering the map $\text{dist}_U:M\to\mathbb{R}:d(x,U(x))$. It's easy to show that this mapping is continuous since $d:M\times M\to\mathbb{R}$ and $j:M\to M\times M:x\mapsto (x,U(x))$ are both continuous (the first obviously and the second since each coordinate map is continuous) and $\text{dist}_U=d\circ j$. Thus, since $M$ is compact one has that $\text{dist}_U$ has a minimum point. Use the $n$-fold composition contraction fact then to show (by contradiction) that this point must be a fixed point--for uniqueness see Tinyboss's response.
• Mar 13th 2011, 06:22 AM
BenH
Thanks for your replies! Compactness can not be assumed. I'm not sure, but I think completeness can be assumed. Anyway, lets assume $M$ is complete. Then since $U^n$ is a strict contraction, by Banach fix point theorem there exists a unique fixed point $a\in M$ s.t $U^n(a)=a$. Performing $U$ on both sides we get
$U(U^n(a))=U(a)\Leftrightarrow U^n(U(a))=U(a)$.
This means that $U(a)$ is a fixed point to $U^n$ as well.
Uniqueness implies that $U(a)=a$. Thus $a$ is a fixed point to $U$ as well. As for uniqueness, assume that $a$ and $b$ both are fixed points to $U$
$d_M (a,b)=d_M(U(a),U(b))=d_M(U(U(a)),U(U(b)))=\cdots =d_M(U^n(a),U^n(b))\leq Ld_M(a,b)\Rightarrow d_M(a,b)=0 \Leftrightarrow a=b$.
$0 because of that $U^n$ is a strict contraction.
Is this correct?