1
$\begingroup$

As a continuation to Monotone Function, Derivative Limit Bounded - Differentiable?,

is there an example of a function $f \colon [a,b] \to \mathbb{R} $ that satisfies the following conditions:

  1. $f$ is strictly monotone.
  2. $\exists r>0 \colon \forall x \in [a,b], \forall h \in \mathbb{R} \colon |\frac{f(x+h) - f(x)}{h}| < r $

which does not have right (or left) derivative at some point in $[a,b]$? In $(a,b)$?

In addition, can such a function's (regular two-sided) derivative be undefined over a dense set of points in $[a,b]$? Or must they be isolated, as in the answer to the linked question?

$\endgroup$

1 Answer 1

3
$\begingroup$

Below are some results pertaining to your second question. I would have to dig through some research papers I have copies of to give a similarly detailed answer to your first question. I don’t have time to do this now, but I might return to this question at a later time if no one posts such an answer.

Your second question has a very precise answer in the following paper.

Zygmunt Zahorski, Sur l'ensemble des points de non-dérivabilité d'une fonction continue [On the set of points of non-differentiability of a continuous function], Bulletin de la Société Mathématique de France 74 (1946), 147-178.

Let $E \subseteq {\mathbb R}$ be such that $E$ has Lebesgue measure zero and $E$ is a $G_{\delta \sigma}$ set. On the middle of p. 176 of Zahorski's paper, in his Remarque, Zahorski gives a function $S:{\mathbb R} \rightarrow {\mathbb R}$ such that $S(x)$ is strictly increasing AND $S(x)$ is Lipschitz continuous AND $E$ is equal to the set of points at which $S(x)$ does not have a finite two-sided derivative.

Among the possibilities for the set $E$ will be any $F_{\sigma}$ measure zero set, since any $F_{\sigma}$ set is a $G_{\delta \sigma}$ set. Thus, $E$ can be any finite or countable union of singleton sets (i.e. $E = {\mathbb Q}$ is possible) and $E$ can be any finite or countable union of measure zero Cantor-like sets. More generally, the $F_{\sigma}$ measure zero possibilities for $E$ include any finite or countable union of sets each of which is either a singleton set or a measure zero Cantor-like set.

EXAMPLE 1: $E$ can be such that the intersection of $E$ with any nonempty open interval has cardinality continuum (i.e. $E$ can be $c$-dense in ${\mathbb R}).$

Note that this is "bigger" than saying both $E$ is dense and $E$ has cardinality continuum, since the rationals union the Cantor middle thirds set is both dense and has cardinality continuum, but is not $c$-dense in the reals. To obtain such a set $E,$ place a scaled and translated copy of the Cantor middle thirds set in each interval of the form $(r,s)$ where $r$ and $s$ are rational numbers, and then let $E$ be the union of all these Cantor sets. Note that there are only countably many such open intervals with rational endpoints, so the resulting union of Cantor sets will be a countable union of closed sets (hence, the union is an $F_{\sigma}$ set) each of which has measure zero (hence, the union has measure zero).

EXAMPLE 2: $E$ can be such that the intersection of $E$ with any nonempty open interval has Hausdorff dimension $1.$

Note that this is stronger than saying that every such intersection has cardinality continuum. To obtain such a set $E,$ place in each open interval with rational endpoints a Cantor-like set that has measure zero and Hausdorff dimension $1$ (see Measure 0 sets on the line with Hausdorff dimension 1, for example), and then let $E$ be the union of all these Cantor sets.

(ADDED NEXT DAY)

Also among the possibilities for the set $E$ will be any $G_{\delta}$ measure zero set (whether dense or not), since any $G_{\delta}$ set is a $G_{\delta \sigma}$ set. Thus, $E$ can be a co-meager set, since there exist dense $G_{\delta}$ measure zero sets (see here and here also). Incidentally, in ${\mathbb R}$ each co-meager set (whether $G_{\delta}$ or not) is $c$-dense in ${\mathbb R}.$ Thus, any dense $G_{\delta}$ measure zero set could also be used for Example 1. However, dense $G_{\delta}$ measure zero sets can be small relative to Hausdorff dimension, so dense $G_{\delta}$ measure zero sets in general could not be used for Example 2. For example, the set of Liouville numbers is a dense $G_{\delta}$ measure zero set that has Hausdorff dimension $0$ --- for a proof see p. 13 in these notes.

EXAMPLE 3: $E$ can be such that $E$ is co-meager and the intersection of $E$ with any nonempty open interval has Hausdorff dimension $1.$

Note that none of the types of sets considered in either Example 1 or Example 2 will work here, since any $F_{\sigma}$ measure zero set is automatically meager, and hence none of those previous sets are even remotely big enough to be co-meager (see my discussion here). This is because each $F_{\sigma}$ measure zero set is a countable union of closed measure zero sets, and each closed measure zero set is nowhere dense. To obtain such a set $E$ for Example 3, let $E$ be the union of a set from Example 2 and a dense $G_{\delta}$ measure zero set, and recall that the union of an $F_{\sigma}$ set and a $G_{\delta}$ set is a $G_{\delta \sigma}$ set.

$\endgroup$
1
  • $\begingroup$ Thank you Dave! Problem wasn't trivial after all! And thank you for the good explanation about constructing an example set. $\endgroup$
    – co.sine
    Commented Apr 21, 2018 at 14:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .