Thread: [SOLVED] Fixed Point

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

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

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.

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

Originally Posted by Jose27

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

So $\displaystyle f$ 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 $\displaystyle y$ $\displaystyle f$ attains a minimum"?

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

Originally Posted by redsoxfan325

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

Yes

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