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Let X be a complete metric space and let $S: X \to X$ be a map. Assume that there exists $m\geq 1$, so: $S^m=\underbrace{S∘S∘··∘S}_{m\text{}}$ is a contraction.

1) Show that S has a unique fixed point.

2) Show that for $m=2$ we can take $S=\cos:[0,\pi/2]→[\pi/2]$

Definition:

Let $X = (X,d)$ be a metric space.

A map $S:T\to T$ is a contraction if there exists a number $0≤\beta<1$, so:

$d(Sx,Sy)\leq \beta d(x,y)$ for all $x,y\in X$

My solutions:

1) First we have to show that $S$ has a fixed point by showing that $S^m$ has a fixed point. But since $S^m$ is a contraction on a complete metric space then it follows from Banach's theorem that $S^m$ has a fixed point (and it's unique).

If $x$ is a fixed point for $S^m$ then $S^mx=x$ and $S^{m+1}x=Sx$.

Now we have to show that $Sx=x$.

$Sx = S(S^mx)= S^m(Sx) \implies Sx = x$

2) This part I'm not sure about how to understand the question. Is there anyone who can help me with this part? I'll appreciate any hint.

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    $\begingroup$ @Clayton The OP proved that $Sx$ is a fixed point of $S^m$, but $S^m$ has a unique fixed point which is $x$, so $Sx=x$. I may be missing something, but seems good to me. However, it remains to prove that this is the unique fixed point of $S$. $\endgroup$ – C. Falcon Mar 7 '18 at 16:01
  • $\begingroup$ @C.Falcon: Thanks. I was being dense for a moment. $\endgroup$ – Clayton Mar 7 '18 at 16:03
  • $\begingroup$ We must have $\beta<1,$ not $\beta\leq 1.$ Maybe this is just a typo. But if $X=\Bbb R$ with the usual metric and $S(x)=x+1,$ then $\beta =1$ and we are stuck. $\endgroup$ – DanielWainfleet Mar 7 '18 at 17:00
  • $\begingroup$ @DanielWainfleet. Ups my bad! It's right that $ 0≤β<1 $ $\endgroup$ – Gabarta123 Mar 7 '18 at 17:05
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$1.$ As it is right now, you have (correctly) proved that $S$ has a fixed point, but it remains to show that it is unique. For that matter, let $y$ be another fixed point of $S$, then recursively, one has $S^my=y$, but $x$ is the unique fixed point of $S^m$, so that $x=y$.

$2.$ You have to prove that $\cos\circ\cos$ is a contraction of $[0,\pi/2]$, for example see the answer here.

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