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Complete metric space and unique fixed point

MathReer1

New member
Feb 28, 2018
2
Let X be a complete metric space and let S: X → X be a map.
Assume that there exists m≥1, so:
${S}^{m}=S∘S∘··∘S $,where the length is m
is a contraction.

1) Show that S has a unique fixed point
2) Show that for $m=2$ we can say that $S=cos:[0,\frac{π}{2}]→[\frac{π}{2}]$

Definition:
Let X = (X,d) be a metric space.
A map $S: X → X$ is contraction if there exists a number $0≤β≤1$ so:
$d(S_x,S_y )≤βd(x,y)$ for all $x,y∈X$ [1]

A fixed point for S is a $x∈X$ with $ T_x=x$

1) Assume that x and y are fixed points for S.
[1] implies that $d(x,y)=d(S_x,S_y )≤βd(x,y)$ [2]

Since$ β≤1$ and $d(x,y)≥0$ is [2] only true when d(x,y)=0, which implies that x = y.

Therefore S has a unique fixed point.

2) I'm not sure about this part. Can someone help me with this part, please.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,657
Leeds, UK
Let X be a complete metric space and let S: X → X be a map.
Assume that there exists m≥1, so:
${S}^{m}=S∘S∘··∘S $,where the length is m
is a contraction.

1) Show that S has a unique fixed point
2) Show that for $m=2$ we can say that $S=cos:[0,\frac{π}{2}]→[\frac{π}{2}]$

Definition:
Let X = (X,d) be a metric space.
A map $S: X → X$ is contraction if there exists a number $0≤β≤1$ so:
$d(S_x,S_y )≤βd(x,y)$ for all $x,y∈X$ [1]

A fixed point for S is a $x∈X$ with $ T_x=x$

1) Assume that x and y are fixed points for S.
[1] implies that $d(x,y)=d(S_x,S_y )≤βd(x,y)$ [2]

Since$ β≤1$ and $d(x,y)≥0$ is [2] only true when d(x,y)=0, which implies that x = y.

Therefore S has a unique fixed point.

2) I'm not sure about this part. Can someone help me with this part, please.
1) Your proof shows that if a contraction has a fixed point then that point is unique. This falls short of what the question is asking for, in two ways. First, you are not told that $S$ is a contraction, but only that some power $S^m$ is a contraction. Second, you have not shown that $S$ need have any fixed points at all.

To show that $S$ has a fixed point, start by showing that $S^m$ has a fixed point. (You may know that already, because it is a standard theorem that a contraction on a complete metric space has a (unique) fixed point.) If $x$ is a fixed point for $S^m$ then $S^mx = x$ and $S^{m+1}x = Sx$. Use the contraction property of $S^m$ to deduce that $Sx=x.$

2) I think that what this question is asking you to show is that the map given by $Sx = \cos x:[0,\frac\pi2] \to[0,\frac\pi2]$ has the property that $S$ is not a contraction but that $S^2$ is a contraction.