2

Suppose $(X,d)$ is compact and we have a mapping $T:X \rightarrow X$ such that $d(Tx,Ty)<d(x,y)$ for every $x,y\in X$ with $x\neq y$. The question is that to show $T$ has a unique fixed point.

Our professor gave us the following hint: Consider $inf d(x,Tx)$.

What I tried:

I defined $f: X\rightarrow \mathbb{R}$ such that $f(x)=d(x,Tx)$. Since $X$ is compact, it attains its infimum. But, I could not show that the infimum of this function must be zero. Can you help me to verify this?

Thanks for any help.

Mahmut Esat Akın
  • 456
  • 2
  • 12
  • You might want to investigate that "it attains its infimum" sentence of yours. Perhaps through an interpretation as of content. –  Jan 03 '20 at 16:54
  • Does this answer your question? Prove the map has a fixed point – Jan Jan 03 '20 at 17:04
  • If $d(x_1, T(x_1)) = \inf d(x, Tx)$ then what what is $d(T(x_1), T(T(x_1))$? Is it possible that $d(T(x_1), T(T(x_1)) < d(x_1, T(x_1)) = \inf d(x,Tx)$? .... and if $x\ne y$ then $d(x,y)> d(Tx, Ty) \ne d(x,y)$. Can both $x,y$ be fixed points. – fleablood Jan 03 '20 at 17:31

1 Answers1

4

There are two parts to this question: (1) proving there is a fixed point, and (2) proving it is unique.

(2) is easy: suppose $x\ne y$ are both fixed points then $d(x,y)=d(Tx,Ty)<d(x,y)$. Contradiction.

(1). You made a good start. Suppose the infimum is $a\ne 0$. Then we have $x_0$ with $d(x_0,Tx_0)=a$. But now take $x_1=Tx_0$. We have $d(x_1,Tx_1)<d(x_0,Tx_0)=a$. Contradiction.

almagest
  • 18,650