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This question is from Foundations of mathematical analysis by Richard Johnsonbaugh enter image description here

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?

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Fraïssé
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    The magic word is "onto": the contraction mapping in (c) can't be surjective. – Rob Arthan Feb 29 '16 at 22:10
  • @RobArthan Are you saying that no contraction mapping can be surjective? – Fraïssé Feb 29 '16 at 22:10
  • Not if the space is compact. – Rob Arthan Feb 29 '16 at 22:11
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    And actually any 1-Lipschitz surjective function from a compact to itself is an isometry. – Captain Lama Feb 29 '16 at 22:15
  • @CaptainLama What would be an example of such function and what is 1 lipschitz – Fraïssé Feb 29 '16 at 23:27
  • In general, $k$-Lipschitz means that $d(f(x),f(y)) \leqslant kd(x,y)$ for all $x,y$. So a contraction is just a $k$-Lipschitz function for some $k<1$, and $1$-Lipschitz means that $d(f(x),f(y)) \leqslant d(x,y)$ ; in particular any contraction is $1$-Lipschitz, as is any isometry. My statement is a little stronger than what you asked because a function that is $1$-Lipschitz but not an isometry needs not be a contraction either. In fact, even if $d(f(x),f(y))<d(x,y)$ for all $x,y$, $f$ still needs not be a contraction. – Captain Lama Feb 29 '16 at 23:39

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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 $B$.

Nick
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  • What is the purpose of making the function "on-to" specifically. How does it fail if it were not onto? – Fraïssé Feb 29 '16 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 '16 at 22:53
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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.

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Sarah
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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 $x\notin f[X]$.

Brian M. Scott
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