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Let $f: [0,1] \to \mathbb{R}$ be a continuous monotone function such that $f$ is differentiable everywhere on $(0,1)$ and $f'(x)$ is continuous on $(0,1)$.

(a) Prove that $f$ is absolutely continuous on [0,1].

(b) Construct a counter example where this does not hold if we drop the monotonic assumption.

My attempt:

Let $\epsilon > 0$, and $\{(a_n,b_n)\}_{n=1}^{k}$ be a disjoint sequence of open sets in $[0,1]$, then by the continuity of $f$ in $[0,1]$ we might pick $\delta_n > 0$ such that $|b_n - a_n| < \delta_n \Rightarrow |f(b_n) - f(a_n)| < \frac{\epsilon}{k}$ $\forall$ $n = 1,...,k$. Then we can choose $\delta = \sum_{n=1}^{k}\delta_n$, then

$\sum_{n=1}^{k}|b_n - a_n| < \delta \rightarrow \sum_{n=1}^{k}|f(b_n) - f(a_n)| < \sum_{n=1}^{k}\frac{\epsilon}{k} = \epsilon$,

Hence, $f$ is absolutely continuous. I know this is not correct because there are continuous functions which are not absolutely continuous if we drop the monotone condition as part $2$ suggest but I never use any of the other conditions. What am I doing wrong? What am I missing?

Part $(2)$ The function is $x\sin(\frac{1}{x})$ if $x \neq 0$, $0$ if $x = 0$.

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2 Answers 2

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Hint for (a): WLOG, $f$ is increasing. Then $f'\ge 0.$ For $b\in (1/2,1),$

$$f(b)-f(1/2) = \int_{1/2}^b f'.$$

As $b\to 1^-,$ the limit on the left is $f(1)-f(1/2)$ by continuity. The limit on the right is $\int_{1/2}^1 f'$ by the MCT.

For (b), think about $f(x)=x\sin(1/x).$

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  • $\begingroup$ This approach makes the use of a characterization of absolutely continuous functions. I would like a more direct proof but this should also work. Thanks!!! $\endgroup$ Jul 3, 2018 at 21:43
  • $\begingroup$ Anyway, why did you choose (1/2,1) and how this generalizes the absolute continuity to [0,1] ? $\endgroup$ Jul 3, 2018 at 21:46
  • $\begingroup$ We need an anchor. Why not $1/2$? This and the same argument near $0$ give $\int_0^1 f' < \infty.$ I am not using any characterization of AC other than the direct definition. $\int_0^1 f' < \infty $ implies all you need. Note I've added a hint for (b). $\endgroup$
    – zhw.
    Jul 3, 2018 at 21:52
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The point which is wrong is the implication

$$\sum_{n=1}^k |b_n-a_n| <\delta \rightarrow \sum_{n=1}^k |f(b_n)-f(a_n)|$$

here you "implicitely" assume the wrong claim that if $$\sum_{n=1}^k |b_n-a_n| <\sum_{n=1}^k \delta_n$$ then each $|b_n-a_n|<\delta_n$.

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  • $\begingroup$ It was the other way. I choose all $\delta_n$ to satisfy that $|b_n - a_n| < \delta_n$ but each delta is positive, so if we sum both sides, the equality holds. I don't see a problem with that. $\endgroup$ Jul 3, 2018 at 21:33
  • $\begingroup$ @RichardClare I understand, but I still certain that this is the wrong step. You should note that if you choose $\delta_n$ to be dependent on $b_n,a_n$ then you don't follow the definition of continuity. The $\delta$ has to be universal for all such intervals (i.e. for all $\varepsilon>0$ there exists $\delta>0$ such that for all intervals...) $\endgroup$
    – Yanko
    Jul 3, 2018 at 21:36
  • $\begingroup$ The $\delta$ is universal for all such intervals if $f$ is absolutely continuous but this is not the case. $\endgroup$ Jul 3, 2018 at 21:37
  • $\begingroup$ @RichardClare You want to prove that $f$ is absolutely continuous. Then you want to find a universal $\delta$. In your argument (which is not completely clear) you either make a wrong implication as stated in my answer, or you only prove the existence of such $\delta$ for a given set of intervals. $\endgroup$
    – Yanko
    Jul 3, 2018 at 21:39
  • $\begingroup$ Yes, but my collection was chosen arbitrarily as my $\epsilon$ $\endgroup$ Jul 3, 2018 at 21:40

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