Determine whether or not the function $f:(0,1) \to\mathbb R$ defined by $f(x)=x^2-2x$ is uniformly continuous

Question

My Attempt: $|f(x)-f(y)| = |x^2$-$y^2| = |x-y| |x+y|<\delta = \epsilon$

Answer 1

For any $\epsilon>0$, take $\delta=\epsilon/2$.
Now, whenever $x,y\in(0,1)$ such that $|x-y|<\delta$, we have

$$|f(x)-f(y)| =|(x^2-2x)-(y^2-2y)|= |x^2-y^2-2(x-y)|\\ = |x-y| (2-x-y)\leq 2|x-y|<2\delta = \epsilon.$$

Answer 2

For $x\in (0,1)$ we have $|f'(x)|=|2x-2| \le 2$, thus, by the mean value theorem:

$$|f(x)-f(y)| \le 2|x-y|$$

for all $x,y \in (0,1).$

This shows that $f$ is Lipschitz- continuous on $(0,1)$, hence uniformly continuous on $(0,1).$

Answer 3

Or:

$x,y \in (0,1)$:

$|f(x)-f(y)|=$

$|(x-1)^2-1-(y-1)^2+1|=$

$|x-y||x+y-2|\lt 2|x-y|.$