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!

If $x\in[0,a]$ we get 
$$
\sup_{t\in[0,a]}|f'(t)|=\sup_{t\in[0,a]}|2t|=2a
$$ 
So 
$$
d(f(x),f(y))\leq 2a d(x,y) \qquad \forall x,y\in [0,a]
$$
and we can't find a constant less than $2a$, in fact for $x=a$ and $y=a-\epsilon$ for $a>\epsilon>0$ we get :
$$
\frac{d(f(x),f(y))}{d(x,y)}=\frac{d(a^2,(a-\epsilon)^2)}{\epsilon}=\frac{2a\epsilon -\epsilon^2}{\epsilon}=2a-\epsilon
$$
