2
$\begingroup$

A function $f: [a,b] \rightarrow \mathbb{R}$ is Lipschitz continuous if and only if $\forall \epsilon >0, \exists \delta>0$ such that for every finite collection of closed intervals in $[a,b]$, $\{[a_k,b_k]\}_{k=1}^N$ (not necessarily disjoint) with

$\sum_{k=1}^N b_k-a_k < \delta$ then $\sum_{k=1}^N |f(b_k)-f(a_k)|<\epsilon$

The easy part is to prove that a Lipschitz function satisfy the property. But I'm having trouble with the other part.

I tried to mimic the proof that $\sqrt{x}$ does not satisfy the property with not necessarily disjoint intervals but unfortunately I got stuck. I saw this exercise on chapter 5 of Bruckner, real analysis.

I would really appreciate any hints or suggestions.

$\endgroup$
  • $\begingroup$ No, the definition of absolute continuity needs disjoint intervals, but if we change it to just intervals (not necessarily disjoint) then it is the same as the Lispchitz condition. The book measure theory of Frank Jones also has this as an exercise $\endgroup$ – user128422 Sep 1 '17 at 1:54
1
$\begingroup$

Suppose $f$ is not Lipschitz continuous. Let $\epsilon>0$ be given. I will construct intervals with $\sum |a_k-b_k|<\epsilon$ but $$\sum|f(a_k)-f(b_k)|\ge 1/2\tag1$$

Since $f$ is not Lipschitz, there exist points $s<t$ such that $$|f(s)-f(t)|>\epsilon^{-1}|s-t|\tag2$$

Case 1: $|s-t|<\epsilon$.

Use the interval $[s,t]$ as $[a_k,b_k]$, repeating it $N$ times where $N$ is chosen so that $$\frac{\epsilon}{2} \le N|s-t| < \epsilon\tag3$$ (Such $N$ always exists, why?) Observe that (2) and (3) imply (1).

Case 2: $|s-t| \ge \epsilon$.

Divide the interval $[s,t]$ into $N$ equal subintervals $[s_k,t_k]$; choose the number of pieces $N$ so that $$\frac{\epsilon}{2}\le \frac{|s-t|}{N} < \epsilon\tag4$$ (This is always possible, why?) The triangle inequality implies that at least one piece $[s_k,t_k]$ satisfies $$|f(s_k)-f(t_k)|>\epsilon^{-1}|s_k-t_k|\tag5$$ Use the single interval $[s_k,t_k]$ to achieve (1), using (4) and (5).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.