A function $f(x)$ on $[0,1]$ is said to satisfy a Lipschitz condition if there exists a constant $M$, such that $$|f(x)-f(y)|\leqslant M|x-y| ~\forall~x,y\in[0,1]. $$

I want to show the following: If $f$ is Lipschitz, then
(i) $f$ is absolutely continuous
(ii) $f$ is of bounded variation.
(iii) $f'(x)$ exists almost everywhere on $[0,1]$ and $f(x)=\int_0^x f'(t)\,\mathrm{d}t$, provided that $f(0)=0$.

(i) Let $\varepsilon \gt 0$. Choose $\delta=\varepsilon /M$. Then for any collection $\{[x_i,y_i]\}$ of non-overlapping intervals with total length $\sum |x_i-y_i|\lt \delta$, we have $$\sum|f(x_i)-f(y_i)|\lt M\sum|x_i-y_i|\lt \varepsilon.$$ Hence $f$ is absolutely continuous.

(ii) For this, can I simply use the fact that an absolutely continuous functions is of bounded variation or I would have to prove it formally?

(iii) I also know that if $f$ absolutely continuous $\implies$ $f$ is of bounded variation $\implies f'(x)$ exists a.e. and $$ f(x)=f(0)+\int_0^x f'(t)\,\mathrm{d}t. $$ Thus, $f(x)=\int_0^x f'(t)\,\mathrm{d}t \text{ if } f(0)=0.$

Can I use these known facts to prove (ii) and (iii)?

I am also asked to find a function $h(x)$ on $[0,1]$ such that $$\tag{​*​}\int_0^1 h'(x)\text{d}x\lt h(1)-h(0).$$
This is what I did: Let $h$ be the Cantor function is monotone increasing of $[0,1]$. $h(0)=1,~h(1)=1$. $h'=0,~\forall~x$ not in the Cantor set. Then the inequality holds.
Are there other examples where (*) holds?

  • $\begingroup$ Yep - your (i) is good. Apparently, your (ii) and (iii) follow from that. $\endgroup$ – davidlowryduda Nov 23 '11 at 18:56
  • $\begingroup$ (i) is indeed fine. I would think that whether (ii) and (iii) are allowed to follow automatically would be up to your professor. If you proved that every absolutely continuous function is of bounded variation, then it would certainly seem fair game. I would still ask your professor to be sure, however. $\endgroup$ – JavaMan Nov 23 '11 at 18:57

As was pointed out in the comments, your argument for (i) is perfectly fine.

Concerning (ii), yes, that's true, you could apply that result. On the other hand, the direct proof is straightforward:

We want to prove that $$\tag{#} \sup{ \left\{ \sum_{k=1}^n |f(a_k) - f(a_{k-1})| \,:\, 0\leq a_0 \leq a_1 \leq \cdots \leq a_{n-1} \leq a_n \leq 1 \right\} } \lt \infty $$ where the supremum is taken over all finite increasing sequences in $[0,1]$. Finiteness follows from the Lipschitz condition $|f(x) - f(y)| \leq M\cdot|x-y|$ because for a finite increasing sequence $0\leq a_0 \leq a_1 \leq \cdots \leq a_{n-1} \leq a_n \leq 1$ we have $$ \sum_{k=1}^n |f(a_k) - f(a_{k-1})| \leq \sum_{k=1}^n\phantom{|} M \cdot |a_k - a_{k-1}| \leq M $$ because $|a_k-a_{k-1}| = a_{k}-a_{k-1}$ and $\sum_{k=1}^n (a_k-a_{k-1}) = a_n -a_0 \leq 1 - 0 = 1$. This proves that the supremum in $(\#)$ is in fact bounded by $M \lt \infty$.

This seems a bit easier than making the detour via absolute continuity (you'll notice that I hardly did anything else than what you did in your proof for (i), and that's why I gave the fully detailed argument).

Compare this with the proof of the fact that absolute continuity of a function $f$ implies that it is of bounded variation (which isn't much more difficult, but somewhat more fiddly, I think).

Now for (iii) I don't think that it is intended that you do this by hand, as a direct and detailed proof of this involves a considerable amount of work. So what you do is almost certainly the intention of this exercise.

Be a little careful, though: you say

I also know that if $f$ absolutely continuous $\implies$ $f$ is f bounded variation $\implies f'(x)$ exists a.e. and $$ f(x)=f(0)+\int_0^1 f'(x)\text{d}x. $$ Thus, $f(x)=\int_0^1 f'(x)\text{d}x~~~\text{if}~~f(0)=0.$

The second implication you write here is not true: bounded variation alone doesn't give that $f(x) = f(0) + \int_{0}^x f'(t)\,dt$ — note that there are two typos here as well:

  1. the upper bound of the integral should be $x$ instead of $1$ and
  2. your equality $f(x) = f(0) + \int_{0}^1 f'(x)\,dx$ involves $x$ with two different meanings: on the left hand side you have some point $x \in [0,1]$ and on the right hand side you have $x$ as a “dummy variable” for integration.

To see that the implication that bounded variation of $f$ doesn't imply that $f(x) = f(0) + \int_{0}^x f'(t)\,dt$, just notice that the Cantor function $h$ is monotone, hence of bounded variation, but $\int_{0}^x h(t)\,dt = 0$ for all $x \in [0,1]$.

However, I think that you meant to say that $f$ absolutely continuous implies $f(x) = f(0) + \int_{0}^x f(t)\,dt$ without interrupting by "$f$ has bounded variation", and then the rest of the argument here is fine too.

To get other (trivial examples) for your last question you can of course take any continuously differentiable (or even only absolutely continuous) function $g$ and add $h$ to it. Genuine examples not involving the Cantor function or simple variations of it are a bit cumbersome to come up with, so I think you solved that exercise satisfactorily, too.

When looking up something else, I recently found this nice and careful write-up by Noella Grady, which is about the present and many related matters.

  • $\begingroup$ Thanks very much for your response. For (iii), what I wanted to say was that If $f$ is absolutely continuous, the $f'$ exists a.e. and $f(x)=f(a)+\int_a^x f'(t)~\text{d}t $. $\endgroup$ – Colin Nov 23 '11 at 23:25
  • $\begingroup$ I was pretty sure that this was what you meant, that's why I didn't expand on what is correct. Just be careful with such formulations, because it is hard to tell such slips from genuine misconceptions (when correcting homework or an exam). $\endgroup$ – t.b. Nov 23 '11 at 23:29
  • $\begingroup$ Thanks also for pointing out the typos. I'm most grateful. $\endgroup$ – Colin Nov 23 '11 at 23:32

For the counterexample, since you are not requiring differentibility everywhere in $[0,1]$, you can always use a step function like: $$h(x):=\begin{cases} 0 &\text{, if } 0\leq x<1/2 \\ 1&\text{, if } 1/2\leq x\leq 1\; .\end{cases}$$ In fact function $h$ is differentiable a.e. with $h^\prime=0$ a.e., hence: $$0=\int_0^1 h^\prime (x)\ \text{d} x< h(1)-h(0)=1\; .$$


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