Function on compact set does not expands distance

Question

I have the following question.

Let $K$ be compact subset of $\mathbb{R^n}$. Prove that there does not exists any continuous $f: K \rightarrow K$ which expands distance i.e. $\forall x,y \in K$ $|f(x)-f(y)| > |x-y|.$

Any help will be appreciated.

Thank you.

Answer 1

Define $g:K\times K$ by $g(x,y)=\left \| x-y \right \|$.

As $g$ is continuous, and $K\times K$ is compact, there is an $(x_0,y_0)\in K\times K$ such that $g(x_0,y_0)=\left \| x_0-y_0 \right \|$ is a maximum.

But then, since by hypothesis, $(f(x_0),f(y_0))\in K\times K$ we have

$\left \| \frac{f(x_0)-f(y_0)}{x_0-y_0} \right \|\leq 1\Rightarrow \left \| f(x_0)-f(y_0) \right \|\leq \left \| x_0-y_0 \right \|$