Mapping of an open interval under Lipschitz function

Question

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a Lipschitz function with Lipschitz constant $A$. We want to show that for any finite open interval $I=(a,b)$, $f(I)$ is contained in an open interval $J=(c,d)$ such that $$ d-c \leq A(b-a).$$

The only difficulty of this problem is the fact that the interval $J$ must be open. Changing the endpoints of any half-open or closed interval would definitely change its length. How can we handle this problem?

Answer 1

Define $x_m := (a+b)/2$ and $y_m := f(x_m)$. Then $|x -x_m | < (b-a)/2 $ for all $x \in I$.

What can you conclude about $|f(x) - y_m|$, using the Lipschitz condition?

This gives you a suitable open interval $J = (c, d)$.