# Showing that a property $\implies$ Lipschitz

Suppose $f$ has the following property on $[a,b]$: For any $\delta>0$, whenever you choose a countable collection of pairwise disjoint intervals $\lbrace (a_k,b_k) \rbrace_{i=1}^{\infty}$ such that $\sum_{i=1}^\infty b_k-a_k < \delta$, you get that

$$\sum_{i=1}^\infty |f(b_k)-f(a_k)| < \sum_{i=1}^\infty b_k-a_k < \delta$$

I am motivated to show that $f$ is Lipschitz because of the form of this last line. It is remarkably similar to $|f(b)-f(a)|\leq C(b-a)$.

## 1 Answer

First observe that the condition you give implies that $|f(x) - f(y)| \le |x-y|$ whenever $x,y \in [a,b]$ and $|x-y| < \delta$. Also observe this implies $f$ is bounded.

Let $M$ be an upper bound of $|f|$.

Let $x,y \in [a,b]$. If $|x-y| < \delta$ then $$|f(y) - f(x)| \le |y - x|.$$On the other hand, if $|x-y| \ge \delta$ then $$|f(y) - f(x)| \le 2M = \dfrac{2M}{\delta} \delta \le \dfrac{2M}{\delta}|x-y|.$$

It follows that $f$ is Lipschitz with constant $\max\{1, 2M/\delta\}$.

• Fantastic. Well done! – Prototank Apr 26 '14 at 13:44