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If a function is monotonic on set E. Is f differentiable almost everywhere? I have proved for case E closed bounded or open intervals, hence all open sets. But in general I am not able to figure it out.

And this I know that derivative of f at isolated points is not defined(or is infinity). But there can be atmost countable no. of isolated points. So what we can say about general E whether it contains isolated points or not.

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  • $\begingroup$ I am not sure I understand. Are you assuming $\mathrm{dom}(f)=E$? $\endgroup$ – Andrés E. Caicedo Dec 3 '14 at 3:48
  • $\begingroup$ Anyway, assuming that's what you meant, can you extend $f$ to a function monotone on $\bar E$ and then further to a function monotone everywhere? $\endgroup$ – Andrés E. Caicedo Dec 3 '14 at 3:55
  • $\begingroup$ @AndresCaicedo If it is possible then how my problem will get solve? As someone has pointed the way of extension. Yes I am assuming dom(f) = E $\endgroup$ – Sushil Dec 3 '14 at 13:58
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First, you should note that the set of isolated points of $E$ is countable. This is in fact a general property of $\mathbb{R}$

Theorem: Let $E$ be a subset of $\mathbb{R}$ and let $F$ be the set of isolated points of $\mathbb{R}$. Then $F$ is at most countable.

Proof: Suppose otherwise, that is, that $F$ is uncountable. Then there exists some interval $[k,k+1]$ such that $F\cap[k,k+1]$ is uncountable. For each $x\in F\cap [k,k+1]$, choose a rational number $q_x$, $0<q_x<1$ such that $(x-2q_x,x+2q_x)\cap F=\varnothing$. Since the set $\left\{q_x:x\in F\cap[k,k+1]\right\}$, then there exists some $q$ such that $X=\left\{x:q_x=q\right\}$ is uncountable, in particular infinite. The choice of $q_x$ implies that the sets $(x-q,x+q)$ are all disjoint for $x\in X$, and they are all contained in $[k-1,k+2]$. Therefore, we constructed an infinite family of disjoint intervals of length $2q$, all of which are contained in the bounded interval $[k-1,k+2]$, a contradiction. QED

(Probably, there is a nicer proof of this theorem somewhere in this site.)

Therefore, we should not worry about the isolated points of $E$ when analysing derivatives: the set of isolated points has null measure.

A trick that works here is to extend your function $f$ to an interval containing $E$. We can do this in the following manner:

Let $E\subseteq\mathbb{R}$ and $f:E\to\mathbb{R}$ be monotonic. The function $\hat{f}:(\inf E,\sup E)\to\mathbb{R}$ given by $\hat{f}(x)=\sup_{y\in E,y\leq x}f(y)$ is an extension of $f$ (if $\sup E$ or $\inf E\in E$, define $\hat{f}(\sup E)=f(\sup E)$ or $\hat{f}(\inf E)=f(\inf E)$).

Another extension is given by $\overline{f}(x)=\inf_{y\in E,y\geq x}f(x)$. In fact, you can check that if $g$ is any other extension of $f$ defined on $[\inf E,\sup E]\cap E$, then $\hat{f}(x)\leq g(x)\leq\overline{f}(x)$ for all $x$.

Alternatively, you can prove this with Zorn's Lemma, but the argument is basically the same: Zorn's lemma gives you a maximal extension of $f$ to a monotonic function $\widetilde{f}:F\to \mathbb{R}$ defined on some subset $F\supseteq E$. To show that $F$ is an interval you apply the argument above and extend $\widetilde{f}$ to some interval containing $F$. Maximality implies that $F$ is that interval.

Now, about your question of differentiability of $f$: For almost every point $x$ of $(\inf E,\sup E)$, the function $\hat{f}$ is differentiable at $x$. But we also know that almost every point of $E$ is not isolated. Using these two fact, we conclude that almost every point $x$ of $E\cap(\inf E,\sup E)$ is not an isolated point of $E$, and $\hat{f}$ is differentiable at $x$. You can then check that for such $x$, $f$ is differentiable at $x$, and $f'(x)=\hat{f}'(x)$.

Therefore, $f$ is differentiable at almost every point of $E$.

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  • $\begingroup$ Are points of differentiability of f and f^ are same? And I think f^ you defined for case f is increasing $\endgroup$ – Sushil Dec 3 '14 at 13:46
  • $\begingroup$ Using Zorn's Lemma can you please give some reference $\endgroup$ – Sushil Dec 3 '14 at 13:46
  • $\begingroup$ @Sushil I put lots of details. I hope it gets clearer now. $\endgroup$ – Luiz Cordeiro Dec 3 '14 at 21:16

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