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A function $f:[0,1]\rightarrow\mathbb{R}$ is called singular continuous, if it is nonconstant, nondecreasing, continuous and $f^\prime(t)=0$ whereever the derivative exists.

Let $f$ be a singular continuous function and $T$ the set where $f$ is not differentiable.

Question: Is $T$ nowhere dense?

Examples:

  • A classical example of such a function is the so-called devil's staircase, obtained as a limit of increasing step functions constructed in a similar way as the classical Cantor set. In that case, the set in question is the Cantor set, which is closed and of measure zero, therefore nowhere dense.

  • Another example, as suggested by David Mitra below, is the Minkowski question mark function. In that case I'm not aware of any characterization of $T$ which allows to decide whether its nowhere dense or not.

  • If you have another (interesting) example for a singular continuous function, please give a comment.

Edits/Notes/Progress:

  • It is clear that $T$ has Lebesgue measure $0$.

  • In view of the correspondence of nondecreasing functions with positive measures, singular continuous functions correspond to singular continuous measures, i.e. an atomless positive Borel measures concentrated on a set of Lebesgue measure zero. The set $T$ corresponds very roughly to that "set of concentration".

Despite of that I would like an answer avoiding measure theory or probability language.

  • My entire knowledge on singular continuous functions is spanned by the information contained in this question. If you possess any further insights into the concept or references containing such, please don't hesitate too leave a comment. It is highly appreciated!
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@Lily: that function is discontinuous at each $x\in\mathbb{Q}$. –  robjohn Feb 8 at 0:51
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Have you looked at the Minkowski Question Mark function? Here is a paper giving results concerning the set $T$ for this function. –  David Mitra Feb 10 at 14:52
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I'll have to think for a bit to decide if the results there preclude $T$ from being nowhere dense. –  David Mitra Feb 10 at 14:55
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I'll look over some stuff I have at home tomorrow and try to get something written up in a day or two that hopefully will be of use. If nothing else, I can certainly provide a lengthy list of references. For now, you might find my November 2000 sci.math essay ESSAY ON NON-DIFFERENTIABILITY POINTS OF MONOTONE FUNCTIONS of some use. –  Dave L. Renfro Feb 10 at 22:02
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See also this math StackExchange post: How prove or disprove $f$ Non-differentiable points countably infinite. In your case, you don't just want the non-differentiability set for a strictly increasing continuous function, but for such a function with the additional requirement that the function has a zero derivative almost everywhere. For this, see the Zbl review (in English) of K. M. Garg's 1969 paper On singular functions. –  Dave L. Renfro Feb 10 at 22:14

3 Answers 3

up vote 4 down vote accepted
+50

I will use the following terms to avoid confusion with standard terminology:

$T$-singular: A non-constant function $f:[0,1] \rightarrow \mathbb R$ that is non-decreasing and has no nonzero finite derivative.

singular: A function with a zero derivative almost everywhere.

A natural question is whether these are equivalent in the realm of strictly increasing continuous functions. Until recently, all known examples of strictly increasing continuous singular functions were also $T$-singular. In Martinez-Blanco's 1999 Ph.D. Dissertation [51] (see my comments in Paradís/Viader/Bibiloni (2001) [57] below) and in Sáncheza/Viaderb/Paradísb/Díaz Carrillo (2012) [69], one can find examples of strictly increasing continuous singular functions that are not $T$-singular.

Since these notions are not equivalent for strictly increasing continuous functions, I will raise the bar and seek functions that are both $T$-singular and singular.

It turns out that MOST (in the Baire category sense) non-decreasing continuous functions $f:[0,1] \rightarrow \mathbb R$ are strictly increasing and $T$-singular and singular and have a non-differentiability set that is co-meager in $[0,1]$ (which is way, way bigger than just saying "not nowhere dense in $[0,1]$). In fact, most such functions also satisfy a stronger notion of $T$-singular, namely they have no nonzero finite one-sided derivative.

Let $D^{+}f$ and $D_{+}f$ be the right upper and right lower Dini derivates of $f,$ and let $D^{-}f$ and $D_{-}f$ be the left upper and left lower Dini derivates of $f.$

Nothing advanced is needed to understand these derivates in what follows. You know what it means for the right derivative to exist at a certain point. O-K, so suppose the right derivative doesn't exist at $x=a.$ Then there is not a unique limit (of difference quotients based at $x=a$) on the right at $x=a,$ which means you can get different limits as you approach $x=a$ on the right using different sequences. $D^{+}f(a)$ is simply the largest possible limit (of the difference quotients) you can get by approaching $x=a$ on the right, and $D_{+}f(a)$ is simply the smallest (in the real number ordering, so $-15$ is smaller than $3$) possible limit (of the difference quotients) you can get by approaching $x=a$ on the right. Either or both of these could be $+\infty$ or $-\infty.$

Theorems 3 & 4 (p. 48 & p. 49) in Zamfirescu (1981) [81] tell us that all but a first Baire category of functions in the space of non-decreasing continuous functions $f:[0,1] \rightarrow \mathbb R$ (sup metric) have all of the following properties:

$$\text{(a)} \;\;\;\; D_{-}f(x) = 0 \;\; \text {or} \;\; D^{-}f(x) = +\infty \;\;\; \text {for all} \; x \in (0,1]$$ $$\text{(b)} \;\;\;\; D_{+}f(x) = 0 \;\; \text {or} \;\; D^{+}f(x) = +\infty \;\;\; \text {for all} \; x \in [0,1)$$ $$\text{(c)} \;\;\;\; f'(x) = 0 \;\;\; \text {for almost all} \; x \in [0,1]$$

Statement (a) implies that for any such function the left derivative (if it exists, finitely or infinitely) can only be $0$ or $+\infty.$ Statement (b) implies that for any such function the right derivative (if it exists, finitely or infinitely) can only be $0$ or $+\infty.$ Note that either (a) or (b) alone is enough to imply that any such function is $T$-singular.

Statement (c) implies that any such function is singular. Incidentally, using the fact that non-decreasing functions have a finite two-sided derivative almost everywhere, we get Statement (c) immediately from Statement (a) [and we also get Statement (c) immediately from Statement (b)].

At this point we know that most non-decreasing continuous functions have no nonzero finite one-sided derivative and they have a zero two-sided derivative almost everywhere.

The next result also holds for all but a first Baire category set of functions, and since the intersection of two such sets has the same property (intersection of two co-meager sets is co-meager, this being an application of De Morgan's laws to the fact that the union of two meager sets is meager), we get all but a first Baire category of functions satisfying all of (a), (b), (c) above along with both of (d) and (e) below.

Theorem 3 (p. 153) in Zamfirescu (1984) [82] tells us that all but a first Baire category set of functions in the space of non-decreasing continuous functions $f:[0,1] \rightarrow \mathbb R$ (sup metric) have both of the following properties:

$$\text{(d)} \;\;\;\; D_{-}f(x) = 0 \;\; \text {and} \;\; D^{-}f(x) = +\infty \;\;\; \text {for most} \; x \in [0,1]$$ $$\text{(e)} \;\;\;\; D_{+}f(x) = 0 \;\; \text {and} \;\; D^{+}f(x) = +\infty \;\;\; \text {for most} \; x \in [0,1]$$

Here,$\;$ for most $x \in [0,1]$ means all but a first Baire category subset of $[0,1]$ (i.e. for co-meagerly many $x \in [0,1]$). In particular, (d) and (e) hold for a dense set of $x \in [0,1]$ (in fact, they both hold for a $c$-dense set of $x \in [0,1],$ but we don't need to be that greedy at this moment), which tells us two things of relevance. First, each of (d) and (e) implies that any such function has to be strictly increasing. [Note that the zero function satisfies all of (a), (b), (c), so the earlier result doesn't force strictly increasing.] For example, (d) implies that for densely many $x \in [0,1]$ we have $D^{-}f(x) = +\infty,$ which prevents the possibility of there being any intervals over which $f$ is constant. (If $f$ were constant on any interval, then all four Dini derivates of $f$ would be zero at each point in the interior of any such interval.) Second, (d) implies that $f$ has no left derivative (finite or infinite) at densely many points in $[0,1]$ and (e) implies that $f$ has no right derivative (finite or infinite) at densely many points. Compare this with what you were looking for: All you wanted was a non-nowhere dense set of points where the function fails to have a finite two-sided derivative.

In fact, since the intersection of two co-meager subsets of $[0,1]$ is a co-meager subset of $[0,1],$ we actually have densely many points (in fact, a $c$-dense set of such points; in fact, co-meagerly many such points) in $[0,1]$ where there is no finite or infinite one-sided derivative (on either side). [This is how Zamfirescu actually stated Theorem 3, namely that all equalities in (d) and (e) hold for co-meagerly many points.] In fact, on each side of each such point, the limits of the secant slopes fully fill up the interval $[0, \, +\infty].$ (To have a left derivative, for example, you can't even have two different limits of secant slopes as you approach on the left side.)

To Summarize: You wanted $T$-singular, and I gave you this along with strictly increasing and singular. You wanted a non-nowhere dense set, and I gave you a dense set (in fact, a set with cardinality $c$ in every interval, and even more than this). You wanted non-differentiable in the sense of not having a finite two-sided derivative, and I gave you non-differentiable in the sense of not having a finite or infinite derivative on either side (and even more than this). You wanted one function, and I gave you continuum many such functions in every $\epsilon$-neighborhood of each non-decreasing continuous function (and even more than this).

See my "question" Bibliography for Singular Functions regarding your comment: If you have another (interesting) example for a singular continuous function, please give a comment.

[51] José M. Martinez-Blanco, Representaciones de Cantor y Funciones Singulares Asociadas [Representations of Associated Cantor Singular Functions], Ph.D. Dissertation (under Eusebio Corbacho Rosas), Universiad de Vigo (Spain), 1999.

[57] Jaume Paradís, Pelegrí Viader, and Lluís Bibiloni, The derivative of Minkowski's $?(x)$ function, Journal of Mathematical Analysis and Applications 253 #1 (1 January 2001), 107-125.

The main result proved is that for each $0 < x < 1$ the Minkowski $?(x)$ function satisfies one of the following: (1) The two-sided derivative equals $0.$ (2) The two-sided derivative equals $\infty .$ (3) The two-sided derivative does not exist, finitely or infinitely. As far as I can tell, the paper does not consider one-sided derivative behavior or Dini derivate behavior. 2nd to last paragraph on p. 114: "These results could lead us to conjecture that this behavior is common to all singular functions but there exist families of singular functions for which there are points in which the derivative is finite and different from zero; see [12]. Note: Their reference [12] is Martinez-Blanco (1999).

[69] Juan Fernández Sáncheza, Pelegrí Viaderb, Jaume Paradísb, and Manuel Díaz Carrillo, A singular function with a non-zero finite derivative, Nonlinear Analysis: Theory, Methods & Applications 75 #13 (September 2012), 5010–5014.

Authors' Abstract: This paper exhibits, for the first time in the literature, a continuous strictly increasing singular function with a derivative that takes non-zero finite values at some points. For all the known "classic" singular functions--Cantor's, Hellinger's, Minkowski's, and the Riesz-Nágy one, including its generalizations and variants--the derivative, when it existed and was finite, had to be zero. As a result, there arose a strong suspicion (almost a conjecture) that this had to be the case for any singular function. We present here a singular function, constructed as a patchwork of known classic singular functions, with derivative $1$ on a subset of the Cantor set.

[81] Tudor Zamfirescu, Most monotone functions are singular, American Mathematical Monthly 88 #1 (January 1981), 47-49.

[82] Tudor Zamfirescu, Typical monotone continuous functions, Archiv der Mathematik 42 #2 (1984), 151-156.

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Awesome! This is even more than I needed. Thank you so much for sharing your expertise! –  Your Ad Here Feb 16 at 17:08

Consider a sequence of independent identically distributed random variables $\left(X_n\right)_{n \ge 1}$, all following a Bernoulli law of parameter $p \in (0,1)$.

Set $X = \sum_{n \ge 1} X_n 2^{-n}$, and denote by $\mu_p$ the law of $X$. All those measures are atomless and have full support, but they are all mutually singular. On the other hand, $\mu_{\frac{1}{2}}$ is simply the Lebesgue measure, so for any $p \neq \frac{1}{2}$, $\mu_p$ is an atomless, singular measure with respect to Lebesgue, and has full support.

If you don't like the presentation, you can translate everything in a more measure-theoritic language.

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Thanks for your answer, but I don't see how that translates into a singular continuous function. Note that $T$ has to be of Lebesgue measure $0$, it can impossibly be the entire interval (or what is meant by "full support"?). –  Your Ad Here Feb 10 at 12:07
    
By support, I meant the essential support of the measure, namely the complement of the biggest open set of zero measure. This is the usual definition of the support of a measure. Maybe by support, you meant a set of full measure, but in this case, you cannot speak about THE support, since such a set is by no mean unique. –  Ahriman Feb 10 at 20:30
    
Yes, I'm aware of that problem. That is why I also used the word "concentrated" above. But maybe I should reword it above. I still don't see the singular continuous function though. –  Your Ad Here Feb 10 at 20:54
    
You mentionned you know the connection between singular functions and singular measures. So you can pass from the singular measures I've constructed to singular functions. –  Ahriman Feb 10 at 22:01
    
Yea, I "know" the connection, but I don't have a feeling for how the set $T$ interacts with it. That is, I have no clue how to actually really prove then that the function I get in the end is actually really a counterexample to my question above. –  Your Ad Here Feb 10 at 22:12

This may not be completely rigorous but I think it will provide an outline of an example where the answer is "no". In a nutshell, start adding together lots of Cantor-Lebesgue functions.

Start by letting $f_1$ be the Cantor-Lebesgue function. Its derivative is zero almost everywhere and its set of points of nondifferentiability is the Cantor set.

Now let $f_2$ be a modified Cantor-Lebesgue function that increases from $0$ to $1/2$ on the interval $(1/3,2/3)$ and is otherwise constant. Its set of points of nondifferentiability is a Cantor-type set inside $(1/3,2/3)$.

There is no overlap between the points of non-differentiability of $f_1$ and $f_2$, so their sum is nondifferentiable on the union of the two sets of nondifferentiability. (At least away from the points $1/3$ and $2/3$). Moreover, $(f_1 + f_2)' = 0$ almost everywhere.

$f_1 + f_2$ is constant on lots of open intervals, three of which have length $1/9$. Let $f_3$, $f_4$, $f_5$ be modified Cantor-Lebesgue functions that increase from $0$ to $1/(3 \cdot 2^2)$ on these intervals separately. Now consider $f_1 + f_2 + f_3 + f_4 + f_5$.

Proceed in this manner, filling the largest intervals of constancy with more Cantor-Lebesgue functions.

The derivative of the infinite sum at the end can be treated using e.g. Fubini's lemma.

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So what will be $T$ in the end and why is it not nowhere dense? –  Your Ad Here Feb 15 at 12:38

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