Let $f:(E,d)\to (E,d)$ such that $$d(f(x),f(y))<d(x,y), \forall x,y\in E ,x\neq y$$ where $E$ is compact
I want to prove that $f$ has a unique fixed point
I consider the recurrent sequence $$\begin{cases} x_{n+1}=f^{n+1}(x_0)\\ x_0\in E\end{cases}$$
But how to prove that it is a Cauchy sequence?
if $d(f(x),f(y))<d(x,y)$ and $x_{n+1}=f^{n+1}(x_0)$ if for example i want to to see $\lim_{n\to+\infty} d(x_n,x_{n+1})$
$d(x_n,x_{n+1})= d(f(x_{n-1}),f(x_n))<d(x_{n-1},x_n)<...<d(x_0,x_1)$ then the limit is not 0 how to prove that it is a Cauchy sequence ?
The continuous map $g\colon E\to\Bbb R$, $x\mapsto d(x,f(x))$ assumes its minimum at some point $\hat x$. Assume $\hat x\ne f(\hat x)$. Then $g(f(\hat x))=d(f(\hat x),f(f(\hat x))<d(\hat x,f(\hat x))=g(\hat x)$, contradiction. We conclude that $\hat x$ is a fixpoint.
Now that we have a fixpoint, it also follows that the recurrence $x_{n+1}=f(x_n)$ converges to $\hat x$ (nd of course uniqueness of the fixpoint): On the compact set $\overline{\{\,x_n\mid n\in\Bbb N\,\}}$, the continuos function $d(\cdot,\hat x)$ assumes its minimum at some point $y$. As also $f(y)$ is in that closure, it follows $d(f(y),\hat x)\ge d(y,\hat x)$, which contradicts the given inequality unless $y=\hat x$. Hence there is a subsequence $x_{n_k}\to y=\hat x$, which implies $x_n\to x$ because $d(x_n,\hat x)$ is decreasing.