Where is $\,\,f(x)=x^2\,\,$ a contraction mapping? I can't understand this:
My notes define a function $f$ to be a contraction if 
$$\lvert\, f(x)-f(y)\rvert\leq c\lvert x-y\rvert,$$ 
for some 0$\lt c\lt 1$.
But then I have a question in my assignment which states:
For what value of $r$,the mapping $f(x)=x^2$ is contraction on $[0,r]$.
I can't understand what the question wants.
Please explain....
 A: Let $r\in (0,1/2)$, then $2r<1$, and let $x,y\in [0,r]$, then
$$
\lvert\,x^2-y^2 \rvert=\lvert\,x+y \rvert\cdot\lvert\,x-y \rvert \le 2r\,
\lvert\,x-y \rvert
$$
Hence $f(x)=x^2$ is a contraction in $[0,r]$.
Note. How do we come up with this $r$?  We first observe that, due to the Mean Value Theorem
$$
f(x)-f(y)=f'(\xi)(x-y), \quad\text{for some $\xi\in(x,y)$.}
$$
Hence, if $f$ is defined on $[a,b]$, then
$$
\lvert\,f(x)-f(y)\rvert=\lvert f'(\xi)\rvert\,\lvert x-y\rvert\le
\sup_{\xi\in[a,b]}\lvert\, f'(\xi)\rvert\,\lvert x-y\rvert.
$$
Thus, if we want $f$ to be a contraction mapping, then $[a,b]$ should be picked so that
$\sup_{\xi\in[a,b]}\lvert\, f'(\xi)\rvert<1$ and if $f(x)=x^2$, then 
$$
\sup_{\xi\in[0,r]}\lvert\, f'(\xi)\rvert=\sup_{\xi\in[0,r]}2\xi=2r.
$$
A: Well, just use your definition. The question asks for what $r$ does there exist a $c\in (0,1)$ such that
$$\|x^2-y^2 \| \leq c\|x-y \| $$
for all $x,y\in[0,r]$.
By the way, I answered here what the meaning of the problem is, and not how to solve it, since this seems to be what you are asking.
A: $$\ ||f(x)-f(y)||=||x^2-y^2||=||(x-y)(x+y)||≤c\cdot||x-y||$$
For $\ x=y $ it is trivial.
For $\ x≠y$ you have:
$$\ ||x+y||≤c<1$$
So the interval must be $\ [0,\frac{1}{2}]$. In this way $\ x $ and $\ y $ cannot be larger than $\ \frac{1}{2}$ so their sum must be less than 1 and the previous inequality is respected.
A: The Lipschitz condition is equivalent $|f'(x)|\le c$, that is, $|2x|\le c < 1$ on $[0,r]$, which translates to $0\le r<1/2$.
