Prove that $f(x)=x^2$ is a contraction on each interval in $[0,0.5]$ I need to formally prove that $f(x)=x^2$ is a contraction on each interval on $[0,a],0<a<0.5$. From intuition, we know that its derivative is in the range $(-1,1)$ implies that the distance between $f(x)$ and $f(y)$ is less then the distance between $x$ and $y$. 
But now I need an explicit $\lambda$ such that $0\le \lambda<1$ and $d(f(x),f(y))\le \lambda d(x,y)$, where $d$ is the standard metric on $\Bbb R$.
Thanks a lot!

Remark that 
$$
|f(x)-f(y)| = |(x+y)(x-y)| \leq (|x|+|y|)|x-y| < 2a |x-y|
$$
if $x$, $y \in [0,a]$. Now $2a<1$ by assumption.
