Why is the function $f(x)=x^2$ is not a contraction on $[0,0.5]$? Let $(X,d)$ be a metric space and let $F:A(\subset X)\to X$. We say $F$ is a contraction if there exists $\lambda$ where $0\leq\lambda<1$ such that
$$d(F(x),F(y))\leq\lambda d(x,y)$$
for all $x,y\in X$.
My question is:
I understand that the function $f(x)=x^2$ is a contraction on each interval on $[0,a], 0<a<0.5$.
But my doubt is why is it NOT a contraction on $[0,0.5]$?

As we can see any distance on any interval on the horizontal axis is less than those on the vertical axis, so there should be some $\lambda$ that satisfy the inequality.
I don't really know the reason. Maybe is there any counterexample that makes it not a contraction?
Many thanks in advance for the help.
 A: Suppose by contradiction that $x^2$ is a contraction of $[0, 0.5]$ with some constant $0<c<1$ satisfying for all $x \neq y \in [0,0.5]$
$$|x^2-y^2|< c|x-y|$$
As a simple consequence of the MVT, $c > 0.5$.
But now,
$$c|x-y|>|x^2-y^2|=|x+y| \cdot |x-y|$$
which implies $c > |x+y|=x+y$. Now, take simply $x=0.5$ and $y= c - 0.5$ to get a contradiction $c>c$.
The fact is that $x^2$ is a contraction of $[0,a]$ for all $a \in (0,0.5)$ because of the MVT. But as $a \to 0.5$, the best Lipschitz constant $c$ approaches to $1$, so that there is no constant $c<1$ working for the whole interval $[0,0.5]$. For this limit case, necessarily $c\ge 1$ so that $x^2$ is Lipschitz but not a contraction.
A: A quick argument using derivatives: note that $f'(0.5) = 1$.  So, for any $c < 1$, we have
$$
\lim_{a \to 0.5^-} \frac{f(0.5) - f(a)}{0.5 - a}> c
$$
So, for any such $c$, there exists an $a \in (0,0.5)$ such that
$$
\frac{f(0.5) - f(a)}{0.5 - a} > c \implies\\
|f(0.5) - f(a)| > c|0.5 - a|
$$
The conclusion follows.
A: Let me add a point that might help for the understanding. In the other answers, we already have seen that
$$|x^2 - y^2| \le C \, |x - y| \quad\forall x,y\in [0,0.5]$$
implies $C \ge 1$.
Now, the OP correctly observed
"As we can see any distance on any interval on the horizontal axis is less than those on the vertical axis"
Indeed, we have
$$|x^2 - y^2| < |x - y| \quad \forall x,y\in[0,0.5], x\ne y.$$
But now, he OP concludes "so there should be some $\lambda$ that satisfy the inequality." And this conclusion is not valid.
In fact, for every $x,y \in [0,0.5]$, $x\ne y$, we find $\lambda(x,y) < 1$, such that
$$|x^2 - y^2| \le \lambda(x,y) \, |x - y|$$
However, you cannot choose $\lambda(x,y)$ to be uniformly smaller than $1$.
