Confusion about proving $f$ is not a contraction

Question

Let $f:[0,2]\to\mathbb R, f(x)=\frac{1}{3}(4-x^{2})$

I know that

$f$ is a contraction $\iff$ $\exists L \in ]0,1[$ such that for all $x, y \in [0,2]: |f(x)-f(y)|\leq L|x-y|$

I am having trouble disproving that $f$ is a contraction (i.e. $f$ is not a contraction), because using the definition would lead to:

$f$ is not a contraction $\iff\forall L \in ]0,1[, \exists x, y \in [0,2]: |f(x)-f(y)|> L|x-y|$ (*)

This condition seems pretty steep.

Idea:(Although I do not believe I am proving (*).)

Let $x,y \in [0,2]$ then $|f(x)-f(y)|=\frac{1}{3}|y^2-x^2|$ and then using the Fundamental Theorem of Calculus, we get $\frac{1}{3}|y^2-x^2|=\frac{1}{3}|y-x||2t|$ where $t \in [y,x]\subset[0,2]$, thus $|f(x)-f(y)|\leq\frac{4}{3}|x-y|$ which seems to show that it is not a contraction, however using the definition (*), I still have not proven it right?

What am I missing to prove that it is not a contraction?

Answer 1

Hint: $|f(x)-f(y)|\le L|x-y|$ implies $|f'(x)| \le L < 1$ for all $x \in (0,2)$.

Answer 2

Idea: $f'(x) = -\frac23x$, so $f'(2) = -\frac43 < -1$, so we need to pick a neighbourhood of $2$.

It turns out that $1.5$ and $2$ works: $|f(1.5)-f(2)| = \frac7{12} > \frac12 = |1.5-2|$.