[SOLVED] Fixed Point

**redsoxfan325** · October 1st 2009, 08:29 PM

Let $M$ be a compact metric space and $\Phi:M\longrightarrow M$ be such that $d(\Phi(x),\Phi(y))<d(x,y)$ for all $x,y\in M$ , $x\neq y$ .

Show that $\Phi$ has a unique fixed point. [Hint: Minimize $d(\Phi(x),x)$ .]

I'm not really sure how to use the hint, or even start the problem. This is in the chapter on the contraction mapping principle, so it seems like I have to get it so that I can apply the CMP. Any suggestions would be most welcome. I don't really want the whole problem solved. If you solve the whole problem, at least put some of it in a spoiler, because I'd like to solve as much of this on my own as I can.

**Jose27** · October 1st 2009, 08:49 PM

Suppose $f:M \rightarrow [0, \infty )$ is such that $f(x)=d( \Phi(x),x)$ then $f$ is is continous. $M$ is compact. If in $y$ $f$ attains a minimum, what can you say about $f(y)$ ?

**redsoxfan325** · October 1st 2009, 09:07 PM

Originally Posted by Jose27

Suppose $f:M \rightarrow [0, \infty )$ is such that $f(x)=d( \Phi(x),x)$ then $f$ is is continous. $M$ is compact. If in $y$ $f$ attains a minimum, what can you say about $f(y)$ ?

So $f$ is a (Lipschitz) continuous function on a compact set. It has a min, max, and it's uniformly continuous.

What do you mean by "If in $y$ $f$ attains a minimum"?

$y$ is a point. I'm not sure what you mean. Did you mean that if $f(y)$ is the minimum value of $f$ ?

**Jose27** · October 2nd 2009, 06:03 AM

Originally Posted by redsoxfan325

$y$ is a point. I'm not sure what you mean. Did you mean that if $f(y)$ is the minimum value of $f$ ?

Yes

**redsoxfan325** · October 2nd 2009, 09:27 AM

Thank you for your help. I have solved the problem.

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