Let $f: [a,b]\to [a,b]$ be continuous, and satisfy $\forall\ x\neq y\in [a,b], |f(x)-f(y)|<|x-y|$.
For any $x_0\in [a,b], x_{n+1}=f(x_n)$, show that $\lim_{n\to\infty} x_n=\xi$, where $\xi$ is the unique fixed point of $f$: $f(\xi)=\xi$.
My attempt: Consider $y_n=|x_n-\xi|$, then $y_n\searrow, \geq 0$. So $\lim y_n=l\geq 0$ exists. If $l=0$, then we are done.
But I could not derive a contradiction when $l>0$. Any ideas? Thank you.
