# Prove there is no contraction mapping from compact metric space onto itself

This question is from Foundations of mathematical analysis by Richard Johnsonbaugh

The thing with this question is that there is a question that seems to prove the opposite claim Prove the map has a fixed point - someone look into this

How should one go about dealing with this question?

• The magic word is "onto": the contraction mapping in (c) can't be surjective. Feb 29, 2016 at 22:10
• @RobArthan Are you saying that no contraction mapping can be surjective? Feb 29, 2016 at 22:10
• Not if the space is compact. Feb 29, 2016 at 22:11
• And actually any 1-Lipschitz surjective function from a compact to itself is an isometry. Feb 29, 2016 at 22:15
• @CaptainLama What would be an example of such function and what is 1 lipschitz Feb 29, 2016 at 23:27

Suppose that $f: M \to M$ was onto. Then for every $x,y \in M$ $x \not=y$, there exists an $x',y' \in M$ s.t. $f(x')=x$ and $f(y')=y$. Then $$d(x,y)=d(f(x'),f(y'))\leq c d(x',y')<d(x',y').$$

Let $B=\max_{x,y \in M^2} d(x,y)$, this exists since $M$ is compact and $d:M^2 \to \mathbb{R}$ is continuous. But, by the above fact, for any $x,y \in M$ there exist an $x',y'$ s.t. $$d(x,y)<d(x',y'),$$ which contradicts the existence of a maximizer $B$.

• What is the purpose of making the function "on-to" specifically. How does it fail if it were not onto? Feb 29, 2016 at 22:22
• The function needs to be onto for the existence of $x'$ and $y'$. Without that step, we wouldn't be able to guarantee that $d(x,y)$ wasn't the maximizer for any $x$ and $y$.
– Nick
Feb 29, 2016 at 22:53

This probably works. Define a distance function $r:M\rightarrow M$ such that $$r(x,y) = d(x,y).$$ Note that $r(\cdot)$ is a continuous function, whose proof can be seen here: Is the distance function in a metric space (uniformly) continuous?

Thus, since $r$ is continuous on compact $M$, it attains its supremum, say at $(x^\ast, y^\ast)$. Note that $f(x^\ast),f(y^\ast)\in M$, which means that $$r(f(x^\ast),f(y^\ast)) = d(f(x^\ast),f(y^\ast)) \leq d(x^\ast, y^\ast)$$ by definition of $(x^\ast, y^\ast)$ resulting in a contradiction.

HINT: Let $p$ be the fixed point guaranteed by the earlier question. Show that there is an $x\in X$ that maximizes $d(p,x)$. Then show that $x\notin f[X]$.