Prove that $f(x)=x^2$ is a contraction on each interval in $[0,0.5]$

Question

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!

Answer 1

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.

Answer 2

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 : $$ \frac{d(f(x),f(y))}{d(x,y)}=\frac{d(a^2,(a-\epsilon)^2)}{\epsilon}=\frac{2a\epsilon -\epsilon^2}{\epsilon}=2a-\epsilon $$