Infinite derivatives of the Takagi-Van der Waerden functions

TL;DR

This paper investigates the Takagi-Van der Waerden functions, characterizing points with infinite lateral derivatives and establishing that this set has Hausdorff dimension one but zero Lebesgue measure.

Contribution

It provides a detailed characterization of the points with infinite derivatives and proves the measure-theoretic properties of this set.

Findings

Set of points with infinite derivatives has Hausdorff dimension one.

This set has Lebesgue measure zero.

Lateral derivatives are infinite at these points.

Abstract

In this paper we characterize the set of points where the lateral derivatives of the Takagi-Van der Waerden functions are infinite. We also prove that the set of points with infinite derivative has Hausdorff dimension one and Lebesgue measure zero.

Equations

$$f_r(x)=\sum _{n=0}^{\infty }\frac{1}{r^n}\varphi (r^{n}x)$$

$$\mathcal{L}(I_n)\le {\alpha }_n,$$

$$T_{\mathcal{D}}(x)=\sum _{n=1}^{\infty }{g}_n(x)$$

$$\partial f(x)=\left[D^{-}f(x),d_{+}f(x)\right]\cap \mathbb{R}.$$

$$D^{-}f(x)=\underset{t\uparrow 0}{\mathrm{limsup}}\frac{f(x+t)-f(x)}{t}\text{and}{d}_{+}f(x)=\underset{t\downarrow 0}{\mathrm{liminf}}\frac{f(x+t)-f(x)}{t}.$$

$$D_n=\left\{\frac{k}{r^{n-1}}\in [0,1]:k\in \mathbb{Z}\right\},$$

$$f_r(x)=\sum _{n=1}^{\infty }{g}_n(x).$$

$$\Big{\{}x\in[0,1]:\sum_{n=1}^{\infty}g^{\prime}_{n}(x)=+\infty\Big{\}}\quad\text{and}\quad\Big{\{}x\in[0,1]:\sum_{n=1}^{\infty}g^{\prime}_{n}(x)=-\infty\Big{\}}$$

$$\{x\in [0,1]:f_r^{\prime }(x)=+\infty \}\text{and}\{x\in [0,1]:f_r^{\prime }(x)=-\infty \}$$

$$x=\sum _{n=1}^{\infty }\frac{{\epsilon }_n}{r^n},{\epsilon }_n\in \{0,1,\dots ,r-1\}.$$

$${\hat{x}}_n=\sum _{k=1}^{n-1}\frac{{\epsilon }_k}{r^k},$$

$$O_n=O_n(x):=\#\left\{k\le n:{\epsilon }_k<\frac{r-1}{2}\right\}$$

$$I_n=I_n(x):=\#\left\{k\le n:{\epsilon }_k>\frac{r-1}{2}\right\}.$$

$${\phi }_r(x)=\sum _{n=1}^{\infty }\frac{{\tilde{\epsilon }}_n}{r^n}$$

$$A^{+}=\Big{\{}x:\sum_{n=1}^{\infty}g^{\prime}_{n}(x)=+\infty\Big{\}}\quad\textrm{and}\quad A^{-}=\Big{\{}x:\sum_{n=1}^{\infty}g^{\prime}_{n}(x)=-\infty\Big{\}}$$

$$J_k^n\cap A^{+}=\frac{k}{r^n}+\left(J_0^n\cap A^{+}\right).$$

$$\mathcal{L}(J_k^n\cap A^{+})=\frac{1}{r^n}\mathcal{L}(A^{+})=\mathcal{L}(J_k^n)\mathcal{L}(A^{+}).$$

$$\mathcal{L}(A^{-})=\mathcal{L}(A^{+})=0.$$

$$T_{\mathcal{D}}(x)=\sum _{n=1}^{\infty }{\tilde{g}}_k(x)$$

$${\tilde{g}}_n(x)=\frac{1}{2}\frac{1}{r^{n-1}}-g_n(x)$$

$$T_{\mathcal{D}}(x)=\frac{r}{2(r-1)}-f_r(x).$$

$$\frac{f_r(b_n)-f_r(x)}{b_n-x}\le \sum _{k=1}^n{g}_k^{\prime }(x)\le \frac{f_r(a_n)-f_r(x)}{a_n-x}.$$

$$\begin{gathered} \frac{f_r(b_n)-f_r(x)}{b_n-x}=\sum _{k=1}^n\frac{g_k(b_n)-g_k(x)}{b_n-x}-\sum _{k=n+1}^{\infty }\frac{g_k(x)}{b_n-x} \\ \le \sum _{k=1}^n\frac{g_k(b_n)-g_k(x)}{b_n-x}\le \sum _{k=1}^n{g}_k^{\prime }(x) \end{gathered}$$

$$d_n=\min \{y-x:y\in D_n\cup {\tilde{D}}_n,y>x\}.$$

$$x_k\in D_n\cup {\tilde{D}}_n,\text{and}x-x_k=d_n,\text{for mk−1<n≤mk.}$$

$$\frac{f_r(x+h_n)-f_r(x)}{h_n}\ge$$

$$\frac{f_r(x-h_n^{\prime })-f_r(x)}{-h_n^{\prime }}\le$$

$$\begin{gathered} \frac{f_r(x+h)-f_r(x)}{h}=\sum _{m=1}^{m_k}\frac{g_m(x+h)-g_m(x)}{h}+\sum _{m=m_k+1}^{\infty }\frac{g_m(x+h)-g_m(x)}{h} \\ \ge \sum _{m=1}^{m_k}{g}_m^{\prime }(x)-\frac{1}{2h}\sum _{m=m_k+1}^{\infty }\frac{1}{r^{m-1}}\ge \sum _{m=1}^{m_k}{g}_m^{\prime }(x)-r^{j-m_k}\frac{r}{r-1}. \end{gathered}$$

$$\frac{f_r(x+h_n)-f_r(x)}{h_n}\ge \sum _{m=1}^{m_k}{g}_m^{\prime }(x)-\frac{r}{r-1}\ge \sum _{m=1}^n{g}_m^{\prime }(x)-1-\frac{r}{r-1}.$$

$$\frac{f_r(x+h_n)-f_r(x)}{h_n}\ge \sum _{m=1}^{m_{k+1}}{g}_m^{\prime }(x)-\frac{r}{r-1}\ge \sum _{m=1}^n{g}_m^{\prime }(x)-\frac{r}{r-1}$$

$$\begin{gathered} \frac{f_r(x+h)-f_r(x)}{h} \\ \ge \sum _{m=1}^{m_k}{g}_m^{\prime }(x)+\sum _{m=m_k+1}^{m_{k+1}}\frac{g_m(x+h)-g_m(x)}{h}-\frac{1}{2h}\sum _{m=m_{k+1}+1}^{\infty }\frac{1}{r^{m-1}} \\ =\sum _{m=1}^{m_k}{g}_m^{\prime }(x)+\sum _{m=m_k+1}^{m_{k+1}}\frac{g_m(x+h)-g_m(x)}{h}-\frac{1}{2h{r}^{m_{k+1}}}\frac{r}{r-1} \\ \ge \sum _{m=1}^{m_k}{g}_m^{\prime }(x)+\sum _{m=m_k+1}^{m_{k+1}}\frac{g_m(x+h)-g_m(x)}{h}-\frac{r}{r-1}. \end{gathered}$$

Full text

The authors were partially supported by Grant MTM2015-65825-P

1. Introduction

Takagi-Van der Waerden functions are an immediate generalization of the Takagi function (see [15]) and they constitute a family of continuous nowhere differentiable functions. The first proof that we know of this fact can be found in [5], other proofs can be found in [1] or in a more general setting in [10]. The surveys [2] and [13] are very good references on the Takagi function, its properties and generalizations.

For every integer $r\geq 2$, the Takagi-Van der Waerden function $f_{r}:[0,1]\to\mathbb{R}$ is defined as

[TABLE]

where $\phi(x)$ denotes the distance from the point $x$ to the nearest integer. Let us observe that $f_{2}$ is the Takagi function, and $f_{10}$ is the Van der Waerden function (see [16]). This family of functions has been studied by many authors such as H. Whitney [17], J. B. Brown and G. Kozlowski [6], A. Shidfar and K. Sabetfakhri [14], Y. Baba [4] or the authors we will mention hereunder.

The first two authors in [10] introduce a generalization of the Takagi-Van der Waerden functions, named Generalized Takagi-Van der Waerden functions, which is defined on a separable real Hilbert space, and they also study the differentiation of the functions belonging to this generalization.

It is natural to ask about the set of points at which the Takagi-Van der Waerden functions possess a left-sided, right-sided or two-sided infinite derivative (see [2]). That is the issue we will try to address in this paper. Regarding this topic, M. Krüppel [12] and P. C. Allaart and K. Kawamura [3] provide a complete characterization of the sets of points where the Takagi function possesses an infinite derivative. Also, P. C. Allaart and K. Kawamura [3] prove that those sets of points have Hausdorff dimension one.

The aim of this paper is to fully characterize the set of points at which the Takagi-Van der Waerden function has an infinite derivative. In this sense, we will generalize the results obtained by M. Krüppel [12] and P. C. Allaart and K. Kawamura [3]. Besides, we will prove that the set of points with infinite derivative has Hausdorff dimension one and Lebesgue measure zero.

Throughout this paper, we will write $f_{r}$ in terms of the corresponding Generalized Takagi-Van der Waerden function. For this reason, we briefly present this generalization defined on $[0,1]$.

Let $D$ be a countable and dense subset of $[0,1]$. Let us consider $\mathcal{D}=\{D_{n}\}_{n}$ an increasing sequence of finite subsets of $D$ satisfying that $D=\cup_{n}D_{n}$. We will say that $\mathcal{D}$ is a decomposition of $D$. Furthermore, we will denote the family of all connected components of $[0,1]\setminus D_{n}$ by $\mathcal{F}_{n}$ and by $\mathcal{F}$ the union of all the families $\mathcal{F}_{n}$.

If $\mathcal{L}$ denotes the Lebesgue measure on $\mathbb{R}$, we will also require the following restriction on the decomposition $\mathcal{D}$:

[TABLE]

for every $I_{n}\in\mathcal{F}_{n}$, where $\{\alpha_{n}\}_{n}\in\ell^{1}$ is a sequence of positive numbers satisfying the inequalities $2\alpha_{n+1}\leq\alpha_{n}$ for every $n$.

For every decomposition $\mathcal{D}=\{D_{n}\}$ of $D$, the Generalized Takagi-Van der Waerden function $T_{\mathcal{D}}:[0,1]\to\mathbb{R}$ is defined as

[TABLE]

where $g_{n}(x)=\text{dist}(x,D_{n})$ denotes the distance from $x$ to the set $D_{n}$. In [10], the following result was formulated for a separable real Hilbert space when some restrictions on the decomposition $\mathcal{D}$ are required. However, in the case of $\mathbb{R}$ these restrictions are not necessary.

Theorem 1.1**.**

If $x\in D$, then $\partial T_{\mathcal{D}}(x)=\mathbb{R}$, otherwise $\partial T_{\mathcal{D}}(x)=\emptyset$.

Here $\partial T_{\mathcal{D}}(x)$ denotes the Fréchet subdifferential of $T_{\mathcal{D}}$ at $x$ (see [9]). In the one dimensional case, the subdifferential of a function $f:\mathbb{R}\to\mathbb{R}$ at a point $x$ is characterized in terms of the Dini derivatives as follows

[TABLE]

Let us remember that the Dini derivatives are defined by

[TABLE]

The other two Dini derivatives $D^{+}f(x)$ and $d_{-}f(x)$ are defined analogously.

If we consider the set $D=\{kr^{-n}\in[0,1]:k,n\in\mathbb{Z}^{+}\}$ and its decomposition

[TABLE]

we may rewrite the function $f_{r}$ as the corresponding Generalized Takagi-Van der Waerden function given by

[TABLE]

Moreover, we will denote by $\widetilde{D}_{n}$ the set of middle points of consecutive points of $D_{n}$ and by $\widetilde{D}$ their union, $\cup_{n}\widetilde{D}_{n}$.

Let us describe the body of this paper. In Section 2 we will examine the behavior of the functions $g_{n}$, as well as the derivatives series $\sum_{n=1}^{\infty}g^{\prime}_{n}(x)$ when $x\not\in D\cup\widetilde{D}$. Let us observe that $g^{\prime}_{n}(x)$ does not exist for $n$ big enough when $x\in D\cup\widetilde{D}$. Additionally, we will prove that the sets

[TABLE]

have Lebesgue measure zero.

At the beginning of the Section 3 we will characterize the lateral derivatives when $x\in D$ and when $x\in\widetilde{D}$. Secondly, we will investigate the relation of the derivatives series with lateral derivatives and Dini derivatives when $x\not\in D\cup\widetilde{D}$. It is important to highlight the role of $r$ parity when considering this relation. We will illustrate it with an example. Subsequently, we dedicate the rest of this section to present the results that completely characterize the set of points where the Takagi-Van der Waerden functions possess a left-sided or right-sided infinite derivative.

Finally, in Section 4 we will prove that the sets

[TABLE]

have Hausdorff dimension one and Lebesgue measure zero.

2. Behavior of the functions $g_{n}$ and their derivatives series

We will begin by introducing some notation and basic results. For a real number $x\in[0,1]$ we consider its basis $r$ expansion

[TABLE]

It is immediate that $x\in D$ if and only if there exists $n_{0}$ such that either $\varepsilon_{n}=0$ for every $n\geq n_{0}$ or $\varepsilon_{n}=r-1$ for every $n\geq n_{0}$. In this case, unless expressly stated otherwise, we will choose the representation ending in all zeros. On the other hand, regarding $\widetilde{D}$, we have to distinguish two cases: if $r$ is even then $\widetilde{D}\subset D$, but for $r$ odd, $x\in\widetilde{D}$ if and only if $\varepsilon_{n}=\frac{r-1}{2}$ for every $n\geq n_{0}$ for some $n_{0}$.

It is clear that when $x\in D\cup\widetilde{D}$, $g^{\prime}_{n}(x)$ does not exists for $n$ big enough, meanwhile if $x\not\in D\cup\widetilde{D}$, then the derivatives $g^{\prime}_{n}(x)\in\{-1,1\}$, are determined as follows:

1. (1)

If $\varepsilon_{n}<\frac{r-1}{2}$, then $g^{\prime}_{n}(x)=1$. 2. (2)

If $\varepsilon_{n}>\frac{r-1}{2}$, then $g^{\prime}_{n}(x)=-1$. 3. (3)

If $\varepsilon_{n}=\frac{r-1}{2}$, then $g^{\prime}_{n}(x)=g^{\prime}_{k_{n}}(x)$ where $k_{n}=\min\left\{j>n:\varepsilon_{j}\neq\frac{r-1}{2}\right\}$.

Let us observe that the third case can only occur when r is odd. If $n\geq 2$ we denote by

[TABLE]

the biggest element of $D_{n}$ smaller than $x$, $\hat{x}_{1}=0$. It is immediate that $g_{n}(x)=\min\left\{x-\hat{x}_{n},\hat{x}_{n}+\frac{1}{r^{n-1}}-x\right\}.$ More precisely, if $x\not\in D\cup\widetilde{D}$ we have

1. (1)

If $\varepsilon_{n}<\frac{r-1}{2}$, then $g_{n}(x)=x-\hat{x}_{n}$. 2. (2)

If $\varepsilon_{n}>\frac{r-1}{2}$, then $g_{n}(x)=\hat{x}_{n}+\frac{1}{r^{n-1}}-x$. 3. (3)

If $\varepsilon_{n}=\frac{r-1}{2}$, then $g_{n}(x)=x-\hat{x}_{n}$ if and only if $\varepsilon_{k_{n}}<\frac{r-1}{2}$.

The functions $g_{n}$, and consequently $f_{r}$ are symmetric with respect to $x=1/2$, hence if we consider the function $S:[0,1]\to[0,1]$ defined by $S(x)=1-x$, it is immediate that $g_{n}(x)=g_{n}(S(x))$ and $g^{\prime}_{n}(x)=-g^{\prime}_{n}(S(x))$ for every $x\notin D\cup\widetilde{D}$. The following fact will be useful later on.

Fact 2.1**.**

For every $x\notin D\cup\widetilde{D}$ we have that $\sum_{n=1}^{\infty}g^{\prime}_{n}(x)=+\infty$ if and only if $\sum_{n=1}^{\infty}g^{\prime}_{n}(S(x))=-\infty$.

We denote

[TABLE]

and

[TABLE]

Let us observe that $O_{n}+I_{n}\leq n$ with equality whenever $r$ is even. In this case $\frac{r-1}{2}\notin\mathbb{Z}$ and consequently $\varepsilon_{n}\neq\frac{r-1}{2}$ always. However, if $r$ is odd, we may have $\varepsilon_{n}=\frac{r-1}{2}$ for some $n$. For this reason, we introduce the function $\varphi_{r}:[0,1]\smallsetminus(D\cup\widetilde{D})\to[0,1]$ defined by

[TABLE]

where, if we denote, as above, by $k_{n}$ the smallest index greater than $n$ such that $\varepsilon_{k_{n}}\neq\frac{r-1}{2}$, then $\widetilde{\varepsilon}_{n}=\varepsilon_{k_{n}}$ whenever $\varepsilon_{n}=\frac{r-1}{2}$ and $\widetilde{\varepsilon}_{n}=\varepsilon_{n}$ otherwise. The following result is immediate:

Proposition 2.2**.**

Assume that $x\notin D\cup\widetilde{D}$. If $r$ odd, then $g^{\prime}_{n}(x)=g^{\prime}_{n}(\varphi_{r}(x))$ for every $n$. Consequently

1. (1)

$\sum_{n=1}^{\infty}g^{\prime}_{n}(x)=+\infty$* if and only if $\lim_{n}(O_{n}(\varphi_{r}(x))-I_{n}(\varphi_{r}(x)))=+\infty$.* 2. (2)

$\sum_{n=1}^{\infty}g^{\prime}_{n}(x)=-\infty$* if and only if $\lim_{n}(O_{n}(\varphi_{r}(x))-I_{n}(\varphi_{r}(x)))=-\infty$.*

Theorem 2.3**.**

The sets

[TABLE]

have Lebesgue measure zero.

Proof.

These sets are obviously disjoint, and they have the same measure since, by Corollary 2.1, A^{+}=S\bigl{(}A^{-}\bigr{)}.

On the other hand, if we denote $J^{n}_{0}=(0,\frac{1}{r^{n}})$ and $J^{n}_{k}=\frac{k}{r^{n}}+J_{0}$, we have that

[TABLE]

Hence $\mathcal{L}(J^{n}_{k}\cap A^{+})=\mathcal{L}(J^{n}_{0}\cap A^{+})$ for every $k=0,\dots,r^{n}-1$, which implies

[TABLE]

Now, let $I\subset[0,1]$ be an arbitrary interval. We have that $I$ is union of a countable amount of disjoint intervals $J^{n}_{k}$ with $n\in\mathbb{N}$ and $k\in\{0,\dots,r^{n}-1\}$ plus a null (countable) set. This implies that $\mathcal{L}(I\cap A^{+})=\mathcal{L}(I)\mathcal{L}(A^{+})$ for every interval $I\subset[0,1]$. It is well known that a set enjoying that property measures either [math] or $1$ necessarily, but it cannot measure $1$ since $A^{+}$ and $A^{-}$ have the same measure. Therefore

[TABLE]

∎

3. Characterization of infinite derivatives

This section includes the main results of the paper. First we characterize the lateral derivatives of $f_{r}$ when $x\in D$ and when $x\in\widetilde{D}$. The first result that we present in this section is an immediate consequence of Theorem 1.1.

Proposition 3.1**.**

If $x\in D$, then $f^{\prime+}_{r}(x)=+\infty$ and $f^{\prime-}_{r}(x)=-\infty$.

Let us observe that if $r$ is even, then $\widetilde{D}_{n}\subset D_{n+1}$ and therefore $\widetilde{D}\subset D$. However, if $r$ is odd then $\widetilde{D}\cap D=\emptyset$. Nevertheless we have the following result:

Proposition 3.2**.**

If $r$ is odd and $x\in\widetilde{D}$, then $f^{\prime+}_{r}(x)=-\infty$ and $f^{\prime-}_{r}(x)=+\infty$.

Proof.

Let us consider the Generalized Takagi-Van der Waerden function associated to the decomposition $\mathcal{D=}\{\widetilde{D}_{n}\}$ of $\widetilde{D}$ defined as

[TABLE]

where $\widetilde{g}_{k}(x)=d(x,\widetilde{D}_{k})$ denotes the distance of $x$ to the set $\widetilde{D}_{k}$. Let us observe that

[TABLE]

and consequently

[TABLE]

From Theorem 1.1, if we observe that $D^{-}T_{\mathcal{D}}(x)=-d_{-}f_{r}(x)$ and $d_{+}T_{\mathcal{D}}(x)=-D^{+}f_{r}(x)$ we obtain the result. ∎

In the sequel we will consider the case $x\not\in D\cup\widetilde{D}$.

Lemma 3.3**.**

For all n if $x\in(a_{n},b_{n})\in\mathcal{F}_{n}$, then

[TABLE]

Proof.

It is enough to observe that

[TABLE]

since $g_{k}$ is linear on $[x,b_{n}]$ for $k\leq n$ provided that $g^{\prime}_{k}(x)=-1$. The other inequality is similar. ∎

Let $x\notin D\cup\widetilde{D}$. For every $n$, we denote

[TABLE]

Remark 3.4**.**

Let us observe that $0<d_{n}<\frac{1}{2r^{n-1}}$. Furthermore, we have that if $d_{n+1}<d_{n}$ then $d_{n}\geq\frac{1}{2r^{n}}$.

We also observe that $d_{n}=d_{n+1}$, if and only if, one of the following situations occurs:

1. (1)

$\varepsilon_{n}=r-1$* and $\varepsilon_{k_{n+1}}>\frac{r-1}{2}$.* 2. (2)

$\varepsilon_{n}=\frac{r-2}{2}$* and $\varepsilon_{n+1}\geq\frac{r}{2}$.* 3. (3)

$\varepsilon_{n}=\frac{r-1}{2}$* and $\varepsilon_{k_{n+1}}<\frac{r-1}{2}$.*

where $k_{n}$ is defined as above.

Moreover, $g_{n}$ is linear in $[x,x+d_{n}]$ for every $n$.

There exist an strictly increasing sequence of integers $\{m_{k}\}_{k\geq 0}$, with $m_{0}=0$, and a strictly decreasing sequence $\{x_{k}\}_{k\geq 1}\downarrow x$ such that:

[TABLE]

Let us observe that $\frac{1}{2r^{m_{k}}}<d_{m_{k}}<\frac{1}{2r^{m_{k}-1}}$.

Lemma 3.5**.**

Let $x\not\in D\cup\widetilde{D}$. There exist sequences $\{h_{n}\}\downarrow 0$ and $\{h^{\prime}_{n}\}\downarrow 0$ such that:

[TABLE]

Proof.

We observe first that if $j\geq m_{k}$ and $\frac{1}{2r^{j}}<h\leq d_{m_{k}}$ then

[TABLE]

Let $n\geq 1$. If $n=m_{k}$ for some $k$ the result is immediate by (3.1) if we take $h_{n}=d_{m_{k}}$. If $m_{k}<n<m_{k+1}$ and $x_{k+1}\in D_{m_{k}+1}$, let $h_{n}=d_{m_{k}}$. In this case $g^{\prime}_{m}(x)=-1$ for $m_{k}+2\leq m\leq m_{k+1}$ and therefore, by (3.1),

[TABLE]

If $m_{k}<n<m_{k+1}$ and $x_{k+1}\in\widetilde{D}_{m_{k+1}}$, then we take $h_{n}=d_{m_{k+1}}$. Again by (3.1), we have that

[TABLE]

since $g^{\prime}_{m}(x)=1$ for all $m_{k}<m\leq m_{k+1}$.

The second inequality is obtained by applying the previous one at $1-x$ and using the symmetry of $f_{r}$.

∎

The following proposition is an immediate consequence of lemmas 3.3 and 3.5.

Proposition 3.6**.**

Let $x\not\in D\cup\widetilde{D}$.

1. (1)

If either $f^{\prime+}_{r}(x)=+\infty$ or $f^{\prime-}_{r}(x)=+\infty$, then $\sum_{k=1}^{\infty}g^{\prime}_{k}(x)=+\infty$. 2. (2)

If either $f^{\prime+}_{r}(x)=-\infty$ or $f^{\prime-}_{r}(x)=-\infty$, then $\sum_{k=1}^{\infty}g^{\prime}_{k}(x)=-\infty.$ 3. (3)

If $\sum_{k=1}^{\infty}g^{\prime}_{k}(x)=+\infty$ then $D^{+}f_{r}(x)=D^{-}f_{r}(x)=+\infty$. 4. (4)

If $\sum_{k=1}^{\infty}g^{\prime}_{k}(x)=-\infty$ then $d_{+}f_{r}(x)=d_{-}f_{r}(x)=-\infty$.

From Theorem 2.3, Proposition 3.6 (1) and Proposition 3.6 (2), we deduce the next result.

Corollary 3.7**.**

The sets $\{x:f^{\prime+}_{r}(x)=+\infty\}$, $\{x:f^{\prime-}_{r}(x)=+\infty\}$, $\{x:f^{\prime+}_{r}(x)=-\infty\}$ and $\{x:f^{\prime-}_{r}(x)=-\infty\}$ are nulls for the Lebesgue measure.

As we will see in the example below, we cannot improve Proposition 3.6 (3) getting $f^{\prime+}_{r}(x)=+\infty$. However, if $r$ is even we have the following result that extends Theorem 3.1 in [3].

Proposition 3.8**.**

If $r$ is even, then

1. (1)

$f^{\prime+}_{r}(x)=+\infty$* provided that $\sum_{k=1}^{\infty}g^{\prime}_{k}(x)=+\infty$.* 2. (2)

$f^{\prime-}_{r}(x)=-\infty$* provided that $\sum_{k=1}^{\infty}g^{\prime}_{k}(x)=-\infty$.*

Proof.

It suffices to prove the first statement since $f^{\prime-}_{r}(x)=-\infty$ if and only if $f^{\prime+}_{r}(1-x)=+\infty$. If $d_{m_{k+1}}<h\leq d_{m_{k}}$ then

[TABLE]

since $\frac{1}{2r^{m_{k+1}}}<d_{m_{k+1}}$. If $m_{k+1}>m_{k}+1$ then

[TABLE]

If $m_{k+1}=m_{k}+1$ (3.3) is immediate from (3.2). ∎

Observe that it is not possible to have a full converse of Proposition 3.6 since, for $r=2$, in [11] it is provided an example of a point such that the series of the derivatives converges to $+\infty$ but the function $f_{2}$ has not $+\infty$ derivative at that point.

Next example for $f_{3}$ was the first clue that we had of the importance of $r$ parity while dealing with these properties. We omit the proof because the result will be an immediate consequence of a subsequent theorem.

Example 3.9**.**

Let

[TABLE]

where $\varepsilon_{n}=0$ if $n=10^{k}$ for some $k$, and $\varepsilon_{n}=1$ otherwise. We have that $g^{\prime}_{n}(x)=1$ for every $n$, but $f^{\prime}_{3}(x)\neq+\infty$.

Furthermore, in the previous example it is not difficult to see directly that $d_{+}f_{3}(x)=-\infty$ although $\sum_{k}g^{\prime}_{k}(x)=+\infty$. Consequently, Proposition 3.8, and Theorem 3.1 in [3] do not hold for $f_{3}$ since $\lim_{n}(O_{n}-I_{n})=+\infty$ and $f^{\prime+}_{3}(x)\neq+\infty$, even if we define $O_{n}$ not as $O_{n}(x)$ but as $O_{n}(\varphi_{3}(x))$. Finally, observe that similar examples exist for every $r$.

Now, we present Theorem 3.11 that characterizes the set of points where $f^{\prime+}_{r}(x)=+\infty$, and therefore it extends Proposition 3.8 (1) and Proposition 3.6 (1) for all $r\geq 2$. On the other hand, with respect to conditions that guarantee that $f^{\prime-}_{r}(x)=+\infty$, Theorem 3.13 generalizes the results that appear in [3] and [11].

Let $r\geq 2$ and $x\not\in D\cup\widetilde{D}$, we arrange the infinite set $\{i:\varepsilon_{i}\neq\frac{r-1}{2}\}$ as an increasing sequence $\{i_{n}\}_{n}$. Observe that if $r$ is even, that set is $\mathbb{N}$.

Lemma 3.10**.**

Let $r\geq 2$ be an integer and $\ell\geq 2$. Then, there exists $x_{0}\in(0,\ell)$ satisfying that $r^{-x_{0}}(\ell-x_{0})+x_{0}<\log_{r}\ell+3$.

Proof.

We consider the function $\varphi(x)=r^{-x}(\ell-x)+x-\frac{\log\ell}{\log r}$ defined on $[0,\ell]$ and we observe that $\varphi(0)=\varphi(l)=l-\frac{\log\ell}{\log r}$. There exists $x_{0}\in(0,\ell)$ such that $(\ell-x_{0})\log r+1=r^{x_{0}}$. Hence,

[TABLE]

∎

Theorem 3.11**.**

$f^{\prime+}_{r}(x)=+\infty$* if and only if*

[TABLE]

Proof.

We only have to prove the $r$ odd case, since for $r$ even, the condition reduces to the convergence to $+\infty$ of the series, and then the result follows from Proposition 3.8 and Proposition 3.6 (1).

“If” part: Let $\frac{1}{r^{p+1}}\leq 2h<\frac{1}{r^{p}}$ and let $n$ be such that $i_{n}<p\leq i_{n+1}$. We denote

[TABLE]

For $k=1,2,\ldots$ we have that $\Delta_{k}(h)\in[-1,1]$. Now, we observe that if $m\geq p$ then

[TABLE]

Let $1\leq k\leq i_{n}$. If $\varepsilon_{i_{n}}\neq r-1$ then, by Remark 3.4, we have that $d_{i_{n}+1}<d_{i_{n}}$ and therefore $d_{k}\geq d_{i_{n}}\geq\frac{1}{2r^{i_{n}}}\geq\frac{1}{2r^{p-1}}>h$. Hence, $g_{k}$ restricted to $[x,x+h]$ is linear for all $1\leq k\leq i_{n}$ and consequently $\Delta_{k}(h)=g^{\prime}_{k}(x)$.

If $\varepsilon_{i_{n}}=r-1$ then let $k_{0}=\max\{k<i_{n}:\;\varepsilon_{k}<r-1\}$. In this case, again by Remark 3.4, we have that $d_{k_{0}}>d_{k_{0}+1}$ and $d_{k_{0}}\geq\frac{1}{2r^{k_{0}}}\geq\frac{1}{2r^{i_{n}+1}}>h$. Therefore, $g_{k}$ restricted to $[x,x+h]$ is linear for all $1\leq k\leq k_{0}$ and then $\Delta_{k}(h)=g_{k}^{\prime}(x)$. As $g_{k}^{\prime}(x)=-1$ if $k_{0}<k\leq i_{n}$ then $\Delta_{k}(h)\geq g_{k}^{\prime}(x)$.

Thus, in both cases we obtain that

[TABLE]

In particular, if $p=i_{n+1}=i_{n}+1$ we have, by (3.4),

[TABLE]

In what follows, we assume that $i_{n+1}-i_{n}>1$.

If $\varepsilon_{i_{n+1}}>\frac{r-1}{2}$ then $g^{\prime}_{k}(x)=-1$ for all $i_{n}<k\leq i_{n+1}$ and hence, by (3.4), we obtain

[TABLE]

If $\varepsilon_{i_{n+1}}<\frac{r-1}{2}$ then $x+d_{k}\in\widetilde{D}_{k}$ for all $i_{n}<k\leq i_{n+1}$ and, by Remark 3.4, we obtain that $d_{i_{n}+1}=\dotsb=d_{i_{n+1}}<\frac{1}{2r^{i_{n+1}-1}}$. If $h\leq d_{i_{n+1}}$ then

[TABLE]

If $h>d_{i_{n+1}}$ we have that $x+d_{k}<x+h\leq x+\frac{1}{2r^{p}}$ and then

[TABLE]

for every $k$ such that $i_{n}<k\leq p$. By Remark 3.4, we have that $d_{i_{n+1}}>d_{i_{n+1}+1}$ and then $d_{i_{n+1}}\geq\frac{1}{2r^{i_{n+1}}}$. Therefore,

[TABLE]

Finally, we conclude that

[TABLE]

which gives us the result.

For the converse we may assume that $i_{n+1}-i_{n}\geq 7$, since otherwise

[TABLE]

and the result follows from Proposition 3.6.

First, assume that $\varepsilon_{i_{n+1}}>\frac{r-1}{2}$. By Remark 3.4, we have that $d_{i_{n+1}-1}>d_{i_{n+1}}$ and then $d_{i_{n+1}-1}\geq\frac{1}{2r^{i_{n+1}-1}}$. Therefore, $\Delta_{k}\left(\frac{1}{2r^{i_{n+1}-1}}\right)=g^{\prime}_{k}(x)$ for all $1\leq k\leq i_{n+1}-1$ and, by (3.4), we obtain

[TABLE]

On the other hand, if $\varepsilon_{i_{n+1}}<\frac{r-1}{2}$ and $i_{n}<p\leq i_{n+1}-5$ then $d_{i_{n+1}}<d_{i_{n}}$ and, as we have seen in the first part of the proof that, for $\frac{1}{r^{p+1}}\leq 2h=\frac{1}{r^{p+\alpha}}<\frac{1}{r^{p}}$ we have

[TABLE]

Let $x_{0}$ be as in Lemma 3.10 for $\ell=i_{n+1}-i_{n}-5$ and let $p=i_{n+1}-5-[x_{0}]$. It is clear that $i_{n}<p\leq i_{n+1}-5$. Let $\alpha=[x_{0}]+1-x_{0}$ and $2h=r^{-(p+\alpha)}$. Then

[TABLE]

and therefore

[TABLE]

Hence, by (3.4), we have

[TABLE]

Letting $n$ to infinite, and therefore $h$ to $0^{+}$, we obtain

[TABLE]

∎

Remark 3.12**.**

In Example 3.9, $i_{n}=10^{n}$. Hence

[TABLE]

Therefore $f^{\prime+}_{3}(x)\neq+\infty$. As a matter of fact it is not difficult to prove directly that $d^{+}f_{3}(x)=-\infty$.

Let $r\geq 2$ and $x\not\in D\cup\widetilde{D}$, we arrange the infinite set $\{n:\varepsilon_{n}\neq 0\}$ as an increasing sequence $\{n_{k}\}_{k}$.

Theorem 3.13**.**

$f^{\prime-}_{r}(x)=+\infty$* if and only if*

[TABLE]

Proof.

“If” part: Let $\frac{1}{r^{p+1}}\leq 2h<\frac{1}{r^{p}}$, let $k$ such that $n_{k}\leq p<n_{k+1}$. We denote

[TABLE]

Now, as in Theorem 3.11, we have that if $m\geq p$ then

[TABLE]

Let $1\leq n\leq n_{k}$. If $g^{\prime}_{n}(x)=1$ then $g_{n}(x)=x-\hat{x}_{n}$, and $g_{n}(x-h)=x-h-\hat{x}_{n}$ since $h<\frac{1}{r^{p}}\leq\frac{1}{r^{n_{k}}}\leq x-\hat{x}_{n_{k}}\leq x-\hat{x}_{n}$. Consequently, $\Delta_{n}(h)=g^{\prime}_{n}(x)=1$. Thus, we obtain that

[TABLE]

In particular, if $n_{k+1}=n_{k}+1$ we have, by (3.6)

[TABLE]

In what follows, we assume that $n_{k+1}-n_{k}>1$.

On the other hand, let $n_{k}<n\leq p$. We have that $g^{\prime}_{n}(x)=1$ and $\hat{x}_{n}=\hat{x}_{p}$. If $x-\hat{x}_{p}\leq h$ then $g_{n}(x-h)=\hat{x}_{p}-(x-h)$ and

[TABLE]

meanwhile if $x-\hat{x}_{p}>h$ then $g_{n}(x-h)=x-h-\hat{x}_{p}$ and $\Delta_{n}(h)=1$. In both cases, proceeding in a similar way as in (3.5), we deduce

[TABLE]

Finally, we conclude that

[TABLE]

which gives us the result.

For the converse we may assume that $n_{k+1}-n_{k}\geq 7$ since for the indices that not satisfy this inequality, the necessary estimation follows from Proposition 3.6.

Let $n_{k}<p\leq n_{k+1}-5$ and $\frac{1}{r^{p+1}}\leq 2h=\frac{1}{r^{p+\alpha}}<\frac{1}{r^{p}}$. First, we observe that if $\Delta_{n}(h)=g^{\prime}_{n}(x)$ for some $n$, then $\Delta_{k}(h)=g^{\prime}_{k}(x)$ for all $1\leq k\leq n$. As we have seen in the first part of the proof, if $g^{\prime}_{n_{k}}(x)=1$ then $\Delta_{n_{k}}(h)=g^{\prime}_{n_{k}}(x)=1$. However, we can say more:

If $\varepsilon_{n_{k}}>\frac{r}{2}$ then $g^{\prime}_{n_{k}}(x)=-1$ and $g_{n_{k}}(x-h)=\hat{x}_{n_{k}}+\frac{1}{r^{n_{k}-1}}-(x-h)$ since

[TABLE]

Therefore, $\Delta_{n_{k}}(h)=g^{\prime}_{n_{k}}(x)=-1$.

If $\varepsilon_{n_{k}}=\frac{r}{2}$, what happens only when $r$ is even, we distinguish the following cases: If $\varepsilon_{n_{k}-1}=\frac{r}{2}$ then $\Delta_{n_{k}-1}(h)=g^{\prime}_{n_{k}-1}(x)=-1$ since

[TABLE]

meanwhile if $\varepsilon_{n_{k}-1}\neq\frac{r}{2}$ we are in one of the previous cases and then $\Delta_{n_{k}-1}(h)=g^{\prime}_{n_{k}-1}(x)$.

Lastly, let us observe that if $\varepsilon_{n_{k}}=\frac{r-1}{2}$ then $g^{\prime}_{n_{k}}(x)=1$ since $\varepsilon_{n_{k}+1}=0$. On the other hand, if $n_{k}<n\leq p$ then $x-\hat{x}_{n}=x-\hat{x}_{p}\leq h$ and, as we have seen in the first part of the proof, we have that

[TABLE]

Finally, we conclude that

[TABLE]

and the result follows as in Theorem 3.11. ∎

Remark 3.14**.**

Let $x$ be the point defined in Example 3.9. Then $f^{\prime-}_{3}(x)=+\infty$.

Theorems 3.13 and 3.11 allow us to deduce the following result for the derivatives $-\infty$.

Corollary 3.15**.**

If we define $\{p_{n}\}_{n}$ as the sequence on indices such that $\varepsilon_{p_{n}}\neq r-1$, then

1. (1)

$f^{\prime+}_{r}(x)=-\infty$* if and only if*

[TABLE] 2. (2)

$f^{\prime-}_{r}(x)=-\infty$* if and only if*

[TABLE]

Proof.

It is enough to observe that $1-x=\sum_{k=1}^{\infty}\frac{(r-1)-\varepsilon_{k}}{r^{k}}$ and the result follows immediately from the previous theorems. ∎

4. Hausdorff dimension

The aim of the results that we present in this section is to prove that the set of points that have infinite derivative has Hausdorff dimension one. Let $n\geq 3$ be an odd integer. We define the set

[TABLE]

and we observe that $B^{+}_{n}\subset D_{n+1}\setminus\{0,1\}$. We will consider

[TABLE]

and $\overline{x}_{m}=\hat{x}_{m}+\varepsilon_{m}r^{-m}$. On the other hand, we define the set

[TABLE]

Observe that $A^{+}_{n,k}$ is a finite union of closed intervals whose endpoints belong to $D_{n+1}$. In this case, we choose the representation ending in all $r-1$ for the right endpoint of each interval.

Concerning the Hausdorff dimension we have the following results.

Lemma 4.1**.**

If $r$ is odd, then the Hausdorff dimension of the set

[TABLE]

is greater than or equal to $\frac{(n-1)\log r-1}{n\log r}$.

Proof.

Let us consider the $r^{n}$ intervals $J_{d}=[d,d+\frac{1}{r^{n}}]$ for $d\in D_{n+1}$ and let $\varphi_{d}:[0,1]\to J_{d}$ be the contractive mapping defined by $\varphi_{d}(x)=d+\frac{x}{r^{n}}$. We observe that the family $\{\varphi_{d}\}_{d\in B^{+}_{n}}$ satisfies the Moran’s open set condition (see [7] or [8] for instance). Let us remember that the symmetry function $S$ is defined by $S(x)=1-x$. We have that $S(J_{d})\cap J_{d}=\emptyset$ for every $d\in B^{+}_{n}$ since $\varepsilon_{n}\neq\frac{r-1}{2}$. It is easy to see that for $d_{1},d_{2}$ satisfying $S(J_{d_{1}})=J_{d_{2}}$, we have that if $d_{1}\in B_{n}^{+}$ then $d_{2}\notin B^{+}_{n}$, meanwhile if $d_{1}\notin B_{n}^{+}$ then either $d_{2}\in B^{+}_{n}$ or $d_{2}=0$. From these facts we deduce that

[TABLE]

Finally, let us observe that $A_{n}^{+}$ is the unique non empty compact set that satisfies

[TABLE]

Indeed, it is enough to realize that if $y\in A^{+}_{n,k}$ then $\varphi_{d}(y)\in A^{+}_{n,k+1}$, meanwhile if $x\in A_{n}^{+}$ then $y=r^{n}\left(x-\overline{x}_{n}\right)\in A_{n}^{+}$ and $\varphi_{\overline{x}_{n}}(y)=x.$ Consequently, the Hausdorff dimension of $A_{n}^{+}$ is

[TABLE]

∎

Lemma 4.2**.**

If $r$ is even, then the Hausdorff dimension of the set

[TABLE]

is greater than or equal to $1-\frac{2+\log 2}{n\log r}$.

Proof.

Using the same argument as in the previous lemma, it is enough to observe that when $r$ is even we have

[TABLE]

∎

Theorem 4.3**.**

The set $\{x:f^{\prime}_{r}(x)=+\infty\}$ has Lebesgue measure zero and Hausdorff dimension one.

Proof.

It is immediate to see that the Lebesgue measure is zero from Corollary 3.7. Concerning the Hausdorff dimension, this is a consequence of the fact that

[TABLE]

for every $n$. Indeed, if $x\notin D\cup\widetilde{D}$ belongs to $A^{+}_{n}$ then

[TABLE]

On the other hand, with the notation of Theorems 3.11 and 3.13, if $r$ is odd we have that $i_{m+1}-i_{m}\leq n$ and $n_{m+1}-n_{m}<2n$ for every $m$, meanwhile $i_{m+1}-i_{m}=1$ and $n_{m+1}-n_{m}<2n$ when $r$ is even. Finally, the result follows from Lemmas 4.1 and 4.2. ∎

By defining the sets $B_{n}^{-}$, $A_{n,k}^{-}$ and $A_{n}^{-}$ in a similar way, the same arguments allow us to obtain the following result.

Theorem 4.4**.**

The set $\{x:f^{\prime}_{r}(x)=-\infty\}$ has Lebesgue measure zero and Hausdorff dimension one.