Hamilton–Jacobi flows with nowhere differentiable initial data

Abstract

We deal with a sort of inverse problem. Namely, we aim at deriving features of initial data of an Hamilton-Jacobi flow starting from specific assumptions on the flow. To this scope, we introduce a class of nowhere differentiable functions, and characterize a function in this class through a property of the Hamilton–Jacobi flow issued from this function. This analysis shows that, despite the regularizing effect of the flow, nowhere differentiability of initial data can be nevertheless detected looking at the flow for positive times.

1 Introduction

We call partition of [0, 1] any finite strictly increasing sequence $a_k$, $k=0,1, \cdots , N$ of points in [0, 1] with $a_0=0$, $a_N=1$. Given a partition $\Gamma =\{a_k\}_{k=0}^N$, we set

$$\begin{aligned} |\Gamma |=\displaystyle {\max _{1\le k\le N}(a_{k}-a_{k-1})}. \end{aligned}$$

Set $\mathbb N_0=\{0,1,2,\cdots \}$.

Definition 1.1

A sequence $\{\Gamma _n\}_{n\in \mathbb N_0}$ of partitions of [0, 1] is called admissible if the following hold:

$(\mathrm{D}1)$:

$\Gamma _0=\{0,1\}$.

$(\mathrm{D}2)$:

For each $n\in \mathbb N_0$, $\Gamma _n\subset \Gamma _{n+1}$.

$(\mathrm{D}3)$:

$|\Gamma _n|\rightarrow 0$ as $n\rightarrow \infty$.

A typical example of admissible sequence of partitions of [0, 1] is given by

$$\begin{aligned} \Gamma _n=\biggl \{\frac{k}{r^n}\, \bigg | \, k\in \mathbb N_0,\ 0\le k \le r^n\biggr \},\quad n\in \mathbb N_0, \end{aligned}$$

(1.1)

where r is an integer such that $r\ge 2$.

We denote by $C_p(\mathbb R)$ the set of all continuous and 1–periodic functions f on $\mathbb R$ satisfying $f(0)=0$. For a given function $f\in C_p(\mathbb R)$, we define

$$\begin{aligned} H_tf(x):=\inf _{y\in \mathbb R}q_f(t,x;y),\quad (t,x)\in (0,\infty )\times \mathbb R, \end{aligned}$$

(1.2)

where $\{q_f(t,x;y)\}_{y\in {\mathbb R}}$ is the family of parabolas defined by

$$\begin{aligned} q_f(t,x;y)=f(y)+\frac{1}{2t}(x-y)^2,\quad (t,x,y)\in (0,\infty )\times \mathbb {R} \times \mathbb {R}. \end{aligned}$$

The function $H_t f$ is called the Moreau envelope of f, which is a special case of inf-convolution. It is known that $H_{t} f$ is a unique viscosity solution to the following Cauchy problem of the Hamilton–Jacobi equation in the class of bounded and uniformly continuous functions on $[0,T) \times \mathbb {R}$ for each $T \in (0,\infty )$:

$$\begin{aligned}&\frac{\partial u}{\partial t}(t,x) +\frac{1}{2}\Bigl (\frac{\partial u}{\partial x}(t,x)\Bigr )^2 =0,&\quad&(t,x) \in (0,T) \times \mathbb {R}, \nonumber \\&u(0,x)=f(x),&x \in \mathbb {R} \end{aligned}$$

(cf. [11, Propositions A.2 and A.4]).

We introduce the following condition for the Hamilton–Jacobi flow $\{H_tf\}$:

$\hbox {{(HJ)}}$:

There exist a constant $\theta \in (0,\infty )$ and an admissible sequence $\{\Gamma _n\}$ of partitions of the interval [0, 1] satisfying

$$\begin{aligned} H_tf(x)=\min _{y\in \Gamma _n} q_f(t,x;y),\quad (t,x)\in \left[ \frac{|\Gamma _n|}{2\theta },\infty \right) \times [0,1] \end{aligned}$$

(1.3)

for every $n\in \mathbb N_0$.

We immediately derive from (1.3) that any $H_tf$ satisfying $\text{(HJ) }$ is piecewise quadratic on the interval [0, 1] for all $t>0$, since $|\Gamma _n|\rightarrow 0$ as $n\rightarrow \infty$. Our aim is to determine all initial data $f\in C_p(\mathbb R)$ such that $\{H_tf\}$ satisfies $\text{(HJ) }$. For this purpose, we introduce the following class:

Definition 1.2

(The class ${\mathcal R}$) Let $\theta \in (0,\infty )$. A function $f \in C_{p}({\mathbb R})$ belongs to ${\mathcal R}_\theta$ if there exists an admissible sequence $\{\Gamma _n\}$ of partitions $\Gamma _n= \{a_k^{(n)}\}_{k=0}^{N_n}$ of [0, 1] satisfying

• $$\begin{aligned} \frac{f(a_{k}^{(n)})-f(x)}{a_{k}^{(n)}-x}-\frac{f(x)-f(a_{k-1}^{(n)})}{x-a_{k-1}^{(n)}} \le -\theta ,\quad x\in (a_{k-1}^{(n)},a_{k}^{(n)}),\ k\in \{1,2,\cdots ,N_n\} \end{aligned}$$

for each $n\in \mathbb N_0$. We set ${\mathcal R}= \bigcup _{\theta >0}{\mathcal R}_\theta$.

The above class $\mathcal R$ reduces to $\mathcal P=\bigcup _{\theta >0}{\mathcal P}_\theta$, already defined in [7], if $\{\Gamma _n\}$ is taken as in (1.1). We recall that each function of $\mathcal P_\theta$ is nowhere differentiable, and that it has been proven in [7] that if $f\in {\mathcal P}_\theta$, for some $\theta \in (0,\infty )$, then $\{H_tf\}$ satisfies $\text{(HJ) }$ with this $\theta$ and $\{\Gamma _n\}$ taken as in (1.1). Conversely, if $\{H_tf\}$ satisfies (1.3), for $f\in C_p(\mathbb R)$, with some $\theta >0$ and $\{\Gamma _n\}$ of the form (1.1), we have already in [7] a partial result showing that f is non-differentiable at each point of a dense subset of [0, 1]. We make here a step forward, as illustrated by the forthcoming Proposition 1.4 and Theorem 1.5.

Functions in $\mathcal R$, for general admissible sequences of partitions of the interval [0, 1], satisfy the following relevant properties:

Proposition 1.3

Each function in ${\mathcal R}$ is nowhere differentiable on $\mathbb R$.

Proposition 1.4

If $f\in {\mathcal R}$ then $\{H_tf\}$ satisfies (HJ).

Proofs of the above results will be given in Sect. 3.

For a given admissible sequence $\{\Gamma _n\}$ of partitions of the interval [0, 1], we consider the following condition:

$\mathrm{(A1)}$:

There exists a constant $\delta \in (0,1]$ such that

$$\begin{aligned} |\Gamma _n|\le \frac{1}{\delta }\min \{|x-y|\,|\, x, y\in \Gamma _n,\ x\ne y\},\quad n\in \mathbb N_0. \end{aligned}$$

Note that when $\Gamma _n$ is given by (1.1), (A1) is fulfilled with $\delta =1$. As for some result for (A1) and other examples of admissible sequences of partitions of [0, 1] satisfying (A1) or not satisfying (A1), see Sect. 4.

Our main result is:

Theorem 1.5

Let $f\in C_p(\mathbb R)$. Assume that $\{H_tf\}$ satisfies (HJ) and that the admissible sequence $\{\Gamma _n\}$ of partitions of [0, 1] appearing in (HJ) satisfies (A1). Then $f\in \mathcal R$. In particular, f is nowhere differentiable on $\mathbb R$.

Condition (A1) seems unavoidable for proving Theorem 1.5. Proposition and Theorem 1.5 together say that the Hamilton–Jacobi property $\text{(HJ) }$ almost characterizes functions in ${\mathcal R}$, although we need adding (A1) in Theorem 1.5. In Theorem 1.5, we perform a sort of inverse analysis for the Hamilton–Jacobi flow, in the sense that we determine some features of initial data starting from properties of the corresponding flows. As for another type of inverse problems for Hamilton–Jacobi equations, see [4, 13] etc.

In Sect. 3, we give a more general version of condition (1.3) in $\text{(HJ) }$, showing that Theorem 1.5 still holds even if (1.3) in (HJ) is replaced by

$$\begin{aligned} H_{t_n}f(x)=\min _{y\in \Gamma _n} q_f(t_n,x;y),\quad x\in [0,1] \end{aligned}$$

for some positive sequence $\{t_n\}$ converging to 0 as $n\rightarrow \infty$ (see Theorem 3.16).

Next, we provide examples of functions belonging to the class $\mathcal R$ for a better understanding of the theory. In [7], we already gave several examples of functions in the class $\mathcal P$, such as the generalized Takagi function $\tau _r$, with an integer r such that $r\ge 2$. It is defined by

$$\begin{aligned} \tau _r(x) =\sum _{n=0}^{\infty }d( x,\Gamma _n),\quad x\in [0,1] \end{aligned}$$

with $\Gamma _n$ of the form (1.1). Here $d(\cdot ,E)$ is the Euclidean distance function from a subset E of $\mathbb R$, that is,

$$\begin{aligned} d(x,E)= \min \{|x -z| \,|\, z \in E \},\quad x\in {\mathbb R}. \end{aligned}$$

(1.4)

Note that $\tau _r$ can be equivalently written as

$$\begin{aligned} \tau _r(x) =\sum _{n=0}^{\infty }\frac{1}{r^n}d(r^nx,\mathbb Z),\quad x\in [0,1]. \end{aligned}$$

When $r=2$, it is just the classical Takagi function (see [1, 12, 14]).

In this paper, adapting the procedure given in [7], we systematically construct, for a given admissible sequence of partitions $\{\Gamma _n\}$ of [0, 1], examples of functions in the class $\mathcal R$. As a special case, this class includes the functions

$$\begin{aligned} f(x) =\sum _{n=0}^{\infty }d(x,\Gamma _n),\quad x\in [0,1], \end{aligned}$$

(1.5)

where $\{\Gamma _n\}$ is any admissible sequence of partitions of [0, 1] with $\sum _{n=0}^{\infty } |\Gamma _n|<+\infty$. The function of (1.5) was studied by Ferrera and Gómez Gil [6] (see Remark 3.4).

The paper is organized as follow: In Sect. 2, we give some preliminaries. In Sect. 3, we state and prove our main results. Theorem 1.5 is proved here. In Sect. 4, we give examples of admissible sequences of partitions of [0, 1] satisfying (A1) and functions belonging to $\mathcal R$.

2 Preliminaries

We first recall some definitions we need in what follows. Given a lower semicontinuous function f defined on $\mathbb R$, we say that a $C^1$ function $\psi$ is subtangent to f at a point $x_0$ if $\psi (x_0)=f(x_0)$ and $\psi (x) \le f(x)$ in a neighborhood of $x_0$. The subdifferential $D^-f(x_0)$ of f at $x_0$ is defined as follows:

$$\begin{aligned} D^-f(x_0)&=\biggl \{p\in \mathbb R\, \bigg | \, \liminf _{y\rightarrow x_0}\frac{f(y)-f(x_0)-p(y-x_0)}{|y-x_0|}\ge 0\biggr \} \\&= \biggl \{ p\in \mathbb R\, \bigg | \, p=\psi '(x_0) \;\mathrm{with}\, C^1 \mathrm{subtangent}\, \psi \,\mathrm{to}\, f\, \mathrm{at}\, x_0 \biggr \}. \end{aligned}$$

This set is closed, convex, possibly empty. It is clear that if $D^-f(x_0)$ is empty or multivalued, then f is not differentiable at $x_0$ (see [2, p.29] and [3, p.50]).

It is now convenient to introduce the Dini derivatives of a function $f:\mathbb R\rightarrow \mathbb R$ at a point $x_0$, and to put them in relation with the subdifferential $D^-f(x_0)$. Dini derivatives at $x_0$ are four, possibly infinite:

$$\begin{aligned} \mathfrak D_-f(x_0)&= \liminf _{h \rightarrow 0^-} \frac{f(x_0+ h)-f(x_0)}{h}, \\ \mathfrak D^-f(x_0)&= \limsup _{h \rightarrow 0^-} \frac{f(x_0+ h)-f(x_0)}{h}, \\ \mathfrak D_+ f(x_0)&= \liminf _{h \rightarrow 0^+} \frac{f(x_0+ h)-f(x_0)}{h}, \\ \mathfrak D^+ f(x_0)&= \limsup _{h \rightarrow 0^+} \frac{f(x_0+ h)-f(x_0)}{h}. \end{aligned}$$

It is clear that f is differentiable at $x_0$ if and only if all the Dini derivatives are finite and coincide. For $a\in \mathbb R\cup \{\pm \infty \}$ and $b\in \mathbb R\cup \{\pm \infty \}$, we define the set $\llbracket a,b \rrbracket$ by

$$\begin{aligned} \llbracket a,b\rrbracket =\left\{ \begin{array}{ll} {[a,b]} &{}\text{ if }\ a,b\in \mathbb {R} \text { and }a\le b,\\ {[a,\infty )} &{}\text{ if }\ a\in \mathbb {R}\text { and } b=+\infty ,\\ {(-\infty ,b]} &{}\text{ if }\ a=-\infty , \text { and } b\in \mathbb {R},\\ {\mathbb R} &{}\text{ if }\ a=-\infty , b=+\infty ,\\ \emptyset &{}\text{ otherwise }. \\ \end{array} \right. \end{aligned}$$

(2.1)

Proposition 2.1

Given a lower semicontinuous function $f:\mathbb R\rightarrow \mathbb R$ and $x_0 \in \mathbb R$, we have

$$\begin{aligned} D^-f(x_0)= \llbracket {\mathfrak {D} }^-f(x_0), {\mathfrak {D} }_+f(x_0)\rrbracket . \end{aligned}$$

(2.2)

Proof

Given $p\in \mathbb R$, we have

$$\begin{aligned}&\liminf _{y\rightarrow x_0-}\frac{f(y)-f(x_0)-p(y-x_0)}{|y-x_0|}=\liminf _{y\rightarrow x_0-}\biggl (-\frac{f(y)-f(x_0)}{y-x_0}+p\biggr )\\&~~~~~~~~~~~~=-\limsup _{y\rightarrow x_0-}\frac{f(y)-f(x_0)}{y-x_0}+p=-\mathfrak D^-f(x_0)+p \end{aligned}$$

and

$$\begin{aligned} \liminf _{y\rightarrow x_0+\!}\frac{f(y)-f(x_0)-p(y-x_0)}{|y-x_0|}\!=\!\liminf _{y\rightarrow x_0+}\biggl (\frac{f(y)-f(x_0)}{y-x_0}-p\biggr )\!=\!\mathfrak D_+f(x_0)-p. \end{aligned}$$

From these equalities, it is not difficult to show (2.2). The proof is complete. $\square$

The following result is obvious.

Lemma 2.2

Let $a,b,\theta \in \mathbb R$ with $a<b$ and $\theta >0$. Then, for a function $f:[a,b]\rightarrow \mathbb R$, the inequality

$$\begin{aligned} \frac{f(b)-f(x)}{b-x}-\frac{f(x)-f(a)}{x-a} \le -\theta ,\quad x\in (a,b) \end{aligned}$$

is equivalent to the inequality

$$\begin{aligned} f(x)\ge \frac{\theta }{b-a}{(x-a)(b-x)}+ \frac{1}{b-a}\,[(b-x) f(a)+ (x-a) f(b)],\ x\in (a,b). \end{aligned}$$

The next lemma can be proved similarly to [7, Proposition 2.2]. For given $a, b, t \in \mathbb R$ with $a <b$ and $t>0$, we denote by $x_{a,b}(t)$ the unique point of intersection between $q_f(t,\cdot ;a)$ and $q_f(t,\cdot ;b)$, that is,

$$\begin{aligned} x_{a,b}(t) = \frac{a+b}{2}+ t\, \frac{f(b)-f(a)}{b-a}. \end{aligned}$$

(2.3)

Lemma 2.3

Let $a,b, t \in \mathbb R$ with $a <b$ and $t>0$. Then the inequality

$$\begin{aligned} \min \{q_f(t,x;a), q_f(t,x;b)\}\le q_f(t,x;y),\quad x\in \mathbb R, \ y\in (a,b) \end{aligned}$$

(2.4)

is equivalent to the inequality

$$\begin{aligned} \frac{f(b)-f(y)}{b-y}-\frac{f(y)-f(a)}{y-a} \le -\frac{b-a}{2t},\quad y\in (a,b). \end{aligned}$$

(2.5)

Proof

A geometrical investigation shows that inequality (2.4) holds if and only if the inequality $x_{y,b}(t)\le x_{a,y}(t)$ holds for all $y\in (a,b)$. It is easy to see that this inequality is equivalent to that in (2.5).$\square$

We record for later use:

Lemma 2.4

Let $f\in C_p(\mathbb {R})$ and $a, b, p, t\in \mathbb R$ with $t>0$. Then

$$\begin{aligned} q_f(t,x;b)-q_f(t,x;a)= q_f(t,p;b)-q_f(t,p;a) -\frac{b-a}{t}(x-p),\quad x\in \mathbb R. \end{aligned}$$

The proof is trivial.

As a consequence, we have

Lemma 2.5

Let $f\in C_p(\mathbb {R})$ and $a, b, p, t\in \mathbb R$ with $a<b$ and $t>0$. If $H_tf(p)=q_f(t,p;a)= q_f(t,p;b)$, then

$$\begin{aligned} \frac{f(b)-f(y)}{b-y}-\frac{f(y)-f(a)}{y-a} \le -\frac{b-a}{2t},\quad y\in (a,b). \end{aligned}$$

Proof

Let $y\in (a,b)$. Since $H_tf(p)=q_f(t,p;a)$, we have $q_f(t,p;a) \le q_f(t,p;y)$. By Lemma 2.4, we have $q_f(t,\cdot ;a)\le q_f(t,\cdot ;y)$ on $(-\infty ,p]$. On the other hand, since $H_tf(p)=q_f(t,p;b)$, we have $q_f(t,p;b) \le q_f(t,p;y)$. By Lemma 2.4, we have $q_f(t,\cdot ;b)\le q_f(t,\cdot ;y)$ on $[p,\infty )$. Thus, (2.4) holds. We conclude the proof exploiting Lemma 2.3. $\square$

The following lemma will be repeatedly used in the paper. For the proof, see [7, Lemma 4.1, p. 356].

Lemma 2.6

Let $f \in C_p(\mathbb {R})$. If $f(z)\ge 0$ for all $z\in \mathbb R$, then

$$\begin{aligned} H_t f(x)=\min _{z \in [0,1]}q_f(t,x;z), \quad (t,x) \in (0,\infty ) \times [0,1]. \end{aligned}$$

(2.6)

Next lemma will be crucial for proving Theorem 1.5.

Lemma 2.7

Let $\{\Gamma _n\}_{n\in \mathbb N_0}$ be a sequence of partitions of [0, 1] satisfying (D1), (D2), and with $|\Gamma _n|\not \rightarrow 0$ as $n\rightarrow \infty$. Then there exist constants $\alpha ,\beta \in [0,1]$ with $\alpha <\beta$ such that

$$\begin{aligned} (\alpha ,\beta )\,\cap \,{\Gamma }=\emptyset , \end{aligned}$$

where $\Gamma = \bigcup _{n=0}^{\infty }\Gamma _n.$

Proof

By (D2), we have $|\Gamma _n|\ge |\Gamma _{n+1}|$ for each $n\in \mathbb N_0$. Thus, $\delta :=\lim _{n\rightarrow \infty }|\Gamma _n|>0$ does exist. We can in addition find sequences $a_n \in \Gamma _n$, $b_n \in \Gamma _n$ converging respectively, up to subsequences, to $\alpha$, $\beta$ in [0, 1] with $|\alpha - \beta |=\delta$.

Now, suppose for purposes of contradiction that $\gamma \in (\alpha ,\beta )\,\cap \,{\Gamma }\ne \emptyset$. Given any $\varepsilon >0$, we have $|a_n- \alpha | < \varepsilon$, $|b_n - \beta | < \varepsilon$ for n large, and consequently

$$\begin{aligned} |a_n-\gamma | + |b_n-\gamma |&\le |a_n-\alpha | + |\alpha - \gamma | + |b_n-\beta | + |\beta - \gamma | \\&\le \delta + 2 \, \varepsilon . \end{aligned}$$

Taking $\varepsilon < \frac{\delta }{2}$, we see that

$$\begin{aligned} \hbox {either}\;\; |a_n - \gamma |< \delta \quad \hbox {or} \; \; |b_n - \gamma | < \delta \end{aligned}$$

which is in contradiction with the very definition of $\delta$. $\square$

For given $a\in [0,1]$ and $s,t\in \mathbb R$ with $0<s<t$, let

$$\begin{aligned} \rho (x)=\frac{(t-s)a+sx}{t},\quad x\in [0,1]. \end{aligned}$$

Note that $0\le \rho (x)\le 1$ and $\rho$ is strictly increasing on [0, 1]. Note further that $y= \rho (x)$ is the unique minimizer over [0, 1] of the function $F:[0,1]\rightarrow \mathbb R$ defined by

$$\begin{aligned} F(y)= q_f(s,y;a)+ \frac{(x-y)^2}{2(t-s)},\quad y\in [0,1] \end{aligned}$$

(2.7)

and the value of the minimum is $q_f(t,x;a)$.

A relevant fact is:

Lemma 2.8

Let $a\in [0,1]$ and $s,t\in \mathbb R$ with $0<s<t$. Let $f\in C_p(\mathbb R)$ be a nonnegative function on $\mathbb R$. If there exist constants $\alpha ,\beta$ with $0\le \alpha <\beta \le 1$ such that

$$\begin{aligned} q_f(t,x;a)=\min _{y\in [0,1]} \left[ H_sf(y)+ \frac{(x-y)^2}{2(t-s)}\right] ,\quad x\in [\alpha ,\beta ], \end{aligned}$$

(2.8)

then

$$\begin{aligned} H_sf(y)=q_f(s,y;a),\quad y\in [\rho (\alpha ),\rho (\beta )]. \end{aligned}$$

(2.9)

Proof

Suppose that (2.9) is false. Then, by the definition of $H_sf$ in (1.2), there exist constants $\alpha ',\beta '$ with $0\le \alpha<\alpha '<\beta '< \beta \le 1$ such that

$$\begin{aligned} H_sf(y)<q_f(s,y;a),\quad y\in [\rho (\alpha '),\rho (\beta ')]. \end{aligned}$$

Take $\varepsilon >0$ so that

$$\begin{aligned} H_sf(y)+\varepsilon <q_f(s,y;a),\quad y\in [\rho (\alpha '),\rho (\beta ')]. \end{aligned}$$

(2.10)

Set $J=[\rho (\alpha '),\rho (\beta ')]$. Then, by (2.8), we have

$$\begin{aligned} \begin{aligned} q_f(t,x;a) \le&\min _{y\in J}\left[ H_sf(y)+ \frac{(x-y)^2}{2(t-s)}\right] ,\quad x\in [\alpha ,\beta ]. \end{aligned} \end{aligned}$$

(2.11)

We choose $\delta >0$ so that $0<\delta <\frac{\beta '-\alpha '}{2}$. Then $\alpha '< \frac{\alpha '+\beta '}{2}-\delta< \frac{\alpha '+\beta '}{2}+\delta <\beta '$. Let us choose x so that

$$\begin{aligned} \frac{\alpha '+\beta '}{2}-\delta<x< \frac{\alpha '+\beta '}{2}+\delta . \end{aligned}$$

We note that $\rho (x)\in J$, since $\alpha '<x<\beta '$. Since F of (2.7) takes its minimum $q_f(t,x;a)$ over [0, 1] only at $y= \rho (x)$, we have, by (2.10),

$$\begin{aligned}&\min _{y\in J} \left[ H_sf(y)+ \frac{(x-y)^2}{2(t-s)}\right] \\&\quad \le \ \min _{y\in J}\left[ q_f(s,y;a)-\varepsilon + \frac{(x-y)^2}{2(t-s)}\right] =\min _{y\in J}F(y)-\varepsilon =q_f(t,x;a)-\varepsilon . \end{aligned}$$

This is a contradiction by (2.11). Therefore, (2.9) holds true. $\square$

3 The main results

In this section, we state and prove our main results. We start by Propositions 1.3 and 1.4. A preliminary fact is:

Lemma 3.1

Let $f\in {\mathcal R}$. Then $f(x)>0$ for $x\in (0,1)$.

Proof

Let $f\in {\mathcal R}_\theta$ for some $\theta \in (0,\infty )$, and $\{\Gamma _n\}$ the admissible sequence of partitions of the interval [0, 1] appearing in Definition 1.2. Thus $\Gamma _0 =\{0,1\}$. By Lemma 2.2 and (R1) with $n=0$, we have $f(x)\ge \theta x(1-x)>0$ for $x\in (0,1)$, since $f(0)=f(1)=0$.$\square$

In the following two lemmas, we let $\{\Gamma _n\}$ be the admissible sequence of partitions of the interval [0, 1] appearing in Definition 1.2 and we set, for each $n\in \mathbb N_0$,

$$\begin{aligned}&\alpha _n(x)&=\max \{a_k^{(n)}\in \Gamma _n\,| a_k^{(n)}<x\},\quad x\in (0,1], \end{aligned}$$

(3.1)

$$\begin{aligned}&\beta _n(x)&=\min \{a_k^{(n)}\in \Gamma _n\,| x< a_k^{(n)}\},\quad x\in [0,1). \end{aligned}$$

(3.2)

We set $\alpha _n(0)=\alpha _n(1)-1$ and $\beta _n(1)=\beta _n(0)+1$. By (D3), we see that

$$\begin{aligned} \lim _{n\rightarrow \infty } \alpha _{n}(x)= \lim _{n\rightarrow \infty } \beta _{n}(x)=x \qquad { \text {for\, each } x\in [0,1].} \end{aligned}$$

(3.3)

We set

$$\begin{aligned} \Gamma =\bigcup _{n=0}^{+\infty }\Gamma _n. \end{aligned}$$

It is apparent that $\Gamma$ is a countable and dense set of [0, 1].

We break through the proof of Proposition 1.3 by the following two lemmas.

Lemma 3.2

Let $f\in {\mathcal R}$. If $x \in [0,1]\setminus \Gamma$, then $D^-f(x)= \emptyset$. Thus f is non-differentiable at each point of $[0,1]\setminus \Gamma$.

Proof

For $x \in [0,1]\setminus \Gamma$, let $\{\alpha _{n}(x)\}$ and $\{\beta _{n}(x)\}$ be the sequences defined in (3.1) and (3.2), respectively.

Let $n\in \mathbb N_0$. Since $x \in [0,1]\setminus \Gamma$, there exists a unique integer $j\in \{0,1,2,\ldots ,N_n-1\}$ such that $\alpha _{n}(x)=a_j^{(n)}$ and $\beta _{n}(x)=a_{j+1}^{(n)}$. Thus, by Definition 1.2, we have

$$\begin{aligned} \frac{f(\beta _{n}(x))-f(x)}{\beta _{n}(x)-x}-\frac{f(x)-f(\alpha _{n}(x))}{x-\alpha _{n}(x)} \le -\theta \end{aligned}$$

for a suitable $\theta >0$. By (3.3), this implies

$$\begin{aligned} {\mathfrak {D} }_+ f(x)&\le \liminf _{n\rightarrow \infty } \frac{f(\beta _{n}(x))-f(x)}{\beta _{n}(x)-x} \le \limsup _{n\rightarrow \infty } \frac{f(\alpha _{n}(x))-f(x)}{\alpha _{n}(x)-x} -\theta \\&\le {\mathfrak {D} }^-f(x) - \theta . \end{aligned}$$

We then invoke Proposition 2.1 to get the assertion. $\square$

Lemma 3.3

Let $f\in {\mathcal R}$. If $x_0 \in \Gamma$, then $D^-f(x_0)= \mathbb R$. Thus f is non-differentiable at each point of $\Gamma$.

Proof

Let $n\in \mathbb N_0$ and $x_0\in \Gamma \cap (0,1)$. By (R1), we have

$$\begin{aligned} \frac{f(x) -f(x_0)}{x-x_0} \ge \frac{f(x) -f(\beta _{n}(x_0))}{x-\beta _{n}(x_0)} + \theta ,\quad x\in (x_0,\beta _n(x_0)) \end{aligned}$$

for a suitable $\theta >0$. By freezing n and letting x go to $x_0$, we get

$$\begin{aligned} {\mathfrak {D} }_+f(x_0) \ge \frac{f(x_0) -f(\beta _{n}(x_0))}{x_0-\beta _{n}(x_0)} + \theta \mathrm{~for \,any}\, n. \end{aligned}$$

(3.4)

By letting n go to $\infty$, we get, by (3.3),

$$\begin{aligned} \liminf _{n\rightarrow \infty } \frac{f(x_0) -f(\beta _{n}(x_0))}{x_0-\beta _{n}(x_0)} \ge {\mathfrak {D} }_+f(x_0). \end{aligned}$$

(3.5)

Relations (3.4), (3.5) are compatible if and only if ${\mathfrak {D} }_+f(x_0)= + \infty$. Arguing as above with obvious adaptations we also get ${\mathfrak {D} }^-f(x_0)= - \infty$. Finally, by Proposition 2.1,

$$\begin{aligned} D^-f(x_0)=\llbracket {\mathfrak {D} }^-f(x_0), {\mathfrak {D} }_+f(x_0)\rrbracket = \mathbb R, \end{aligned}$$

as was asserted. When $x_0=0$ or $x_0=1$, we can show that $D^-f(x_0)= \mathbb R$ similarly, since $f\in C_p(\mathbb R)$. $\square$

Remark 3.4

When $f\in C_p(\mathbb R)$ is given by (1.5), Ferrera and Gómez Gil [6] showed that f satisfies

$$\begin{aligned} D^-f(x_0)= \emptyset \quad (x_0 \in [0,1]\setminus \Gamma ),\quad D^-f(x_0)= \mathbb R\quad (x_0 \in \Gamma ). \end{aligned}$$

Next, we prove Proposition 1.4.

Proof of Proposition 1.4

Let $f\in \mathcal R$. By Lemmas 3.1 and 2.6, we obtain (2.6).

Choose a constant $\theta \in (0,\infty )$ and an admissible sequence $\{\Gamma _n\}$ of partitions $\Gamma _n= \{a_k^{(n)}\}_{k=0}^{N_n}$ of the interval [0, 1] appearing in Definition 1.2. Now, fix $(n,t)\in \mathbb N_0\times [\frac{|\Gamma _n|}{2\theta },\infty )$ arbitrarily. For any $y\in (a_{k}^{(n)},a_{k+1}^{(n)})$ and $k\in \{0,1,2,\cdots ,N_n-1\}$, we have

$$\begin{aligned} \frac{f(a_{k+1}^{(n)})-f(y)}{a_{k+1}^{(n)}-y}-\frac{f(y)-f(a_{k}^{(n)})}{y-a_{k}^{(n)}} \le -\theta \le -\frac{a_{k+1}^{(n)}-a_k^{(n)}}{2t}, \end{aligned}$$

since $\displaystyle {t\ge \frac{|\Gamma _n|}{2\theta } \ge \frac{a_{k+1}^{(n)}-a_{k}^{(n)}}{2\theta }}$. By Lemma 2.3 and (2.6), it is easy to see that $\{H_tf\}$ satisfies $\text{(HJ) }$. $\square$

The proof of Theorem 1.5 passes through the definition of a more general framework. For this purpose, we introduce a new class of functions contained in $C_p(\mathbb R)$.

Definition 3.5

(The class ${\mathcal H}$) A function $f \in C_{p}({\mathbb R})$ is said to belong to ${\mathcal H}$ if there exist an admissible sequence of partitions $\{\Gamma _n\}$ of the interval [0, 1] and a positive and bounded sequence $\{K_n\}_{n\in \mathbb N_0}$ such that $t_n:=K_n|\Gamma _n|$ satisfies

$(\mathrm{H1})$:

$t_0>t_1>t_2>\cdots>t_n>t_{n+1}\rightarrow 0$ as $n\rightarrow \infty$.

$(\mathrm{H2})$:

For each $n\in \mathbb N_0$,

$$\begin{aligned} H_{t_n}f(x)=\min _{y\in \Gamma _n} q_f(t_n,x;y),\quad x\in [0,1]. \end{aligned}$$

Proposition 3.6

Let $f\in C_p({\mathbb R})$. If $\{H_tf\}$ satisfies (HJ), then $f\in \mathcal H$. In particular, ${\mathcal R}\subset \mathcal H$.

Proof

Let $f\in C_p({\mathbb R})$ such that $\{H_tf\}$ satisfies $\text{(HJ) }$. Choose a constant $\theta \in (0,\infty )$ and the admissible sequence of partitions $\{\Gamma _n\}$ of the interval [0, 1] appearing in $\text{(HJ) }$.

Set

$$\begin{aligned} K_n =\frac{1}{2\theta }+\frac{1}{n+1},\quad n\in \mathbb N_0. \end{aligned}$$

Then $\{K_n\}$ is a positive and bounded sequence. Since $\Gamma _n\subset \Gamma _{n+1}$ for $n\in \mathbb N_0$, we have $|\Gamma _n|\ge |\Gamma _{n+1}|$ for $n\in \mathbb N_0$. Let $\{t_n\}$ be the sequence in Definition 3.5. For $n\in \mathbb N_0$, we have

$$\begin{aligned}&t_n-t_{n+1} =\bigl (\frac{1}{2\theta }+\frac{1}{n+1}\bigr )|\Gamma _n| -\bigl (\frac{1}{2\theta }+\frac{1}{n+2}\bigr )|\Gamma _{n+1}|\\ \ge \&\bigl (\frac{1}{2\theta }+\frac{1}{n+1}\bigr )|\Gamma _{n+1}| -\bigl (\frac{1}{2\theta }+\frac{1}{n+2}\bigr )|\Gamma _{n+1}|=\frac{|\Gamma _{n+1}|}{(n+1)(n+2)}>0. \end{aligned}$$

Since $|\Gamma _n|\rightarrow 0$ as $n\rightarrow \infty$, we have $t_n\rightarrow 0$ as $n\rightarrow \infty$. Hence $\{t_n\}$ satisfies (H1) of Definition 3.5. Since $t_n >\frac{|\Gamma _n|}{2\theta }$ and $\{H_tf\}$ satisfies $\text{(HJ) }$, f satisfies (H2) of Definition 3.5. Therefore, $f\in \mathcal H$.

When $f\in \mathcal R$, we see that $f \in \mathcal H$ by Proposition 1.4 and the arguments above.$\square$

Lemma 3.7

Let $f\in \mathcal H$. Then $f(x)>0$ for $x\in (0,1)$.

Proof

Let $f\in \mathcal H$. By (H2) with $n=0$ in Definition 3.5, we have

$$\begin{aligned} H_{t_0}f(x)= \min \left\{ \frac{x^2}{2t_0},\frac{(1-x)^2}{2t_0}\right\} >0,\quad x\in (0,1). \end{aligned}$$

Since $f(x)\ge H_{t_0}f(x)$ for $x\in \mathbb R$ by (1.2), we conclude the lemma. $\square$

In the following, we always consider a function $f\in \mathcal H$ and the admissible sequence $\{\Gamma _n\}$ of partitions of [0, 1] appearing in Definition 3.5. We set

$$\begin{aligned} E_n(y)=\bigl \{x\in [0,1]\,\bigl |\bigr .\, H_{t_n}f(x)=q_f(t_n,x;y)\bigr \},\quad n\in \mathbb N_0,\ y\in \Gamma _n, \end{aligned}$$

where $\{t_n\}$ is the sequence appearing in Definition 3.5.

Lemma 3.8

For every $n\in \mathbb N_0$ and $y\in \Gamma _n$, $E_n(y)$ is a closed interval in [0, 1], possibly reduced to a singleton or empty.

Let $x, y,z, t\in \mathbb R$ with $y\ne z$ and $t>0$. Note that when $z>y$, the inequality $q_f(t,x;z)\ge q_f(t,x;y)$ holds if and only if the inequality

$$\begin{aligned} x\le \frac{y+z}{2}+ t\, \frac{f(z)-f(y)}{z-y} \end{aligned}$$

holds. Note also that when $z<y$, the inequality $q_f(t,x;z)\ge q_f(t,x;y)$ holds if and only if the inequality

$$\begin{aligned} x\ge \frac{y+z}{2}+ t\, \frac{f(z)-f(y)}{z-y} \end{aligned}$$

holds.

Proof

Let $x,\hat{x}\in E_n(y)$ and $\lambda \in (0,1)$. Then, for any $z\in \Gamma _n\setminus \{y\}$, we have

$$\begin{aligned} q_f(t_n,x;z)\ge q_f(t_n,x;y),\quad q_f(t_n,\hat{x};z)\ge q_f(t_n,\hat{x};y). \end{aligned}$$

(3.6)

When $z>y$, we have, from these inequalities,

$$\begin{aligned} x, \hat{x}\le \frac{y+z}{2}+ t_n\frac{f(z)-f(y)}{z-y}, \end{aligned}$$

so that

$$\begin{aligned} \lambda x+(1-\lambda ) \hat{x}\le \frac{y+z}{2}+ t_n\frac{f(z)-f(y)}{z-y}. \end{aligned}$$

Thus the last inequality implies that

$$\begin{aligned} q_f(t_n,\lambda x+(1-\lambda ) \hat{x};z)\ge q_f(t_n,\lambda x+(1-\lambda ) \hat{x};y). \end{aligned}$$

(3.7)

Similarly, when $z<y$, we have also the inequality (3.7) from (3.6). This implies that $\lambda x+(1-\lambda )\hat{x}\in E_n(y)$. Hence $E_n(y)$ is a convex set in $\mathbb R$. Since $H_{t_n}f$ is continuous on $\mathbb R$, $E_n(y)$ is a closed set in $\mathbb R$, so that it is a closed interval. The proof is complete. $\square$

Next, we set

$$\begin{aligned} \hat{\Gamma }_n=\bigl \{y\in \Gamma _n\,\bigl |\bigr .\, \text{ the } \text{ interior } \text{ of } E_n(y) \text{ is } \text{ not } \text{ empty }\bigr \}. \end{aligned}$$

By (H2) of Definition 3.5, we see that

$$\begin{aligned} H_{t_n}f(x)=\min _{y\in \hat{\Gamma }_n} q_f(t_n,x;y),\quad n\in \mathbb N_0,\ x\in [0,1], \end{aligned}$$

(3.8)

since a point $y\in \Gamma _n$ such that $E_n(y)$ is the empty set or a singleton never contributes to the minimization of $q_f(t_n,\cdot ,y)$ over $\Gamma _n$. We proceed showing that $\{\hat{\Gamma }_n\}$ is an admissible sequence of partitions of [0, 1].

Lemma 3.9

For every $n\in \mathbb N_0$, $\min \hat{\Gamma }_n=0$ and $\max \hat{\Gamma }_n =1$. In particular, $\hat{\Gamma }_0=\{0,1\}$.

Proof

Let $y_0:=\min \{y\in \Gamma _n| y>0\}$. Then $0 <y_0\le 1$. We show that $[0,\frac{y_0}{2})\subset E_n(0)$, which shows that $0\in \hat{\Gamma }_n$ and $\min \hat{\Gamma }_n=0$. Indeed, let $y\in \Gamma _n$ with $0<y\le 1$. Since $f\in \mathcal H$, we have $f\ge 0$ on [0, 1] by Lemma 3.7. Then, for any $x\in [0,\frac{y_0}{2})$, we have

$$\begin{aligned} \frac{y}{2t_n}(2x-y)<0\le f(y), \end{aligned}$$

since $0\le 2x <y_0\le y$. Thus it is easy to see that

$$\begin{aligned} \frac{x^2}{2t_n}<f(y)+\frac{1}{2t_n}(x-y)^2. \end{aligned}$$

Therefore, $q_f(t_n,x;0)< q_f(t_n,x;y)$ and $H_{t_n}f(x)=q_f(t_n,x;0)$. This implies that $x\in E_n(0)$, so that $[0,\frac{y_0}{2})\subset E_n(0)$ as was desired.

Next, let $y_1:=\max \{y\in \Gamma _n| y<1\}$. Then $0 \le y_1< 1$. Similarly to the proof above, we can show that $(\frac{y_1+1}{2},1]\subset E_n(1)$, which shows that $1\in \hat{\Gamma }_n$ and $\max \hat{\Gamma }_n=1$.

When $n=0$, we have $\hat{\Gamma }_0\subset \Gamma _0=\{0,1\}$. Since $\min \hat{\Gamma }_0=0$ and $\max \hat{\Gamma }_0 =1$, we have $\hat{\Gamma }_0=\{0,1\}$. The proof is complete. $\square$

Lemma 3.10

For every $n\in \mathbb N_0$, $\hat{\Gamma }_n\subset \hat{\Gamma }_{n+1}$.

Proof

Since $f\in \mathcal H$, we have $f\ge 0$ on [0, 1] by Lemma 3.7. For $n\in \mathbb N_0$, let $s=t_{n+1}$ and $t=t_n$, where $\{t_n\}$ is the sequence in Definition 3.5. By (H1) of Definition 3.5, we have $0<s<t$. Then using the semigroup property $H_tf =H_{t-s}\circ H_sf$ and Lemma 2.8, we see easily that if $a\in \hat{\Gamma }_n$ then $a\in \hat{\Gamma }_{n+1}$. The proof is complete. $\square$

Lemma 3.11

$|\hat{\Gamma }_n|\rightarrow 0$ as $n\rightarrow \infty$.

Proof

Since $\hat{\Gamma }_n\subset \Gamma _n$ for each $n\in \mathbb N_0$, $\hat{\Gamma }_n$ is a finite set. By Lemmas 3.9 and 3.10, $\{\hat{\Gamma }_n\}_{n\in \mathbb N_0}$ satisfies (D1) and (D2) of Definition 1.1 of Sect. 1 where $\Gamma _n$ is replaced by $\hat{\Gamma }_n$. Suppose that the conclusion of the theorem is false. Then, by Lemma 2.7, there exist constants $\alpha ,\beta \in [0,1]$ with $\alpha <\beta$ such that

$$\begin{aligned} (\alpha ,\beta )\,\cap \,{\hat{\Gamma }}=\emptyset , \end{aligned}$$

where $\hat{\Gamma }= \bigcup _{n=0}^{\infty }\hat{\Gamma }_n.$

Choose $\delta \in (0,\frac{\beta -\alpha }{2})$, and fix $x\in (\alpha +\delta ,\beta -\delta )$ arbitrarily. Let $n\in \mathbb N_0$ and $y\in \hat{\Gamma }_n$. Then $y\le \alpha$ or $\beta \le y$. When $y\le \alpha$, we have $x-y\ge \alpha +\delta -\alpha =\delta$, so that $(x-y)^2\ge \delta ^2$. When $\beta \le y$, we have $y-x\ge \beta -(\beta -\delta )=\delta$, so that $(x-y)^2\ge \delta ^2$. Thus, for any $y\in \hat{\Gamma }_n$, we have $(x-y)^2\ge \delta ^2$.

Let $\{t_n\}$ be the sequence in Definition 3.5. Then $t_n\rightarrow 0$ as $n\rightarrow \infty$. Note that $f\ge 0$ on [0, 1] by Lemma 3.7, since $f\in \mathcal H$. Then

$$\begin{aligned} q_f(t_n,x;y)=f(y)+\frac{( x-y)^2}{2t_n}\ge \frac{\delta ^2}{2t_n},\quad x\in (\alpha +\delta ,\beta -\delta ), \ n\in \mathbb N_0,\ y\in \hat{\Gamma }_n. \end{aligned}$$

Thus

$$\begin{aligned} \min _{y\in \hat{\Gamma }_n} q_f(t_n,x;y)\ge \frac{\delta ^2}{2t_n},\quad x\in (\alpha +\delta ,\beta -\delta ), \ n\in \mathbb N_0. \end{aligned}$$

By (3.8), we have $H_{t_n}f(x)\rightarrow \infty$ as $n\rightarrow \infty$ for $x\in (\alpha +\delta ,\beta -\delta )$. However, $\{H_{t_n}f\}$ must converges to f as $n\rightarrow \infty$ uniformly on $\mathbb R$. This is a contradiction, and the conclusion holds true. The proof is complete. $\square$

From Lemmas 3.9, 3.10 and 3.11, we have

Proposition 3.12

$\{\hat{\Gamma }_n\}$ is an admissible sequence of partitions of the interval [0, 1] with $\hat{\Gamma }_n\subset \Gamma _n$ for all $n\in \mathbb N_0$.

Remark 3.13

We do not know whether $\hat{\Gamma }_n=\Gamma _n$ for each $n\in \mathbb N_0$.

Now, we set

$$\begin{aligned}&\quad N_n= (\text{ the } \text{ number } \text{ of } \text{ elements } \text{ of } \hat{\Gamma }_n)-1\text{, }\nonumber \\&\quad \hat{\Gamma }_n=\{a_0^{(n)}, a_1^{(n)}, a_2^{(n)},\cdots , a_{N_n}^{(n)}\}\ \ \text{ with }\ \ 0=a_0^{(n)}< a_1^{(n)}< a_2^{(n)}\cdots <a_{N_n}^{(n)}=1. \end{aligned}$$

(3.9)

In addition, we set

$$\begin{aligned} p_0^{(n)}=0,\quad p_j^{(n)}=x_{a_{j-1}^{(n)},\ a_{j}^{(n)}}(t_n), \quad j\in \{1,2,\ldots ,N_n\}, \end{aligned}$$

where $x_{a,b}(t)$ is defined by (2.3) for given $a, b, t \in \mathbb R$ with $a <b$ and $t>0$.

Lemma 3.14

For every $n\in \mathbb N_0$, we have

$$\begin{aligned} H_{t_n}f(p_j^{(n)})= q_f(t_n,p_j^{(n)};a_j^{(n)})=q_f(t_n,p_{j}^{(n)};a_{j-1}^{(n)}),\quad j\in \{1,2,\cdots , N_n\}. \end{aligned}$$

(3.10)

Proof

Fix $n\in \mathbb N_0$ and let $j\in \{1,2,\cdots , N_n\}$. By the definition of $p_j^{(n)}$, we have $q_f(t_n,p_j^{(n)};a_j^{(n)})=q_f(t_n,p_{j}^{(n)};a_{j-1}^{(n)})$. Supposing that

$$\begin{aligned} H_{t_n}f(p_j^{(n)})< q_f(t_n,p_j^{(n)};a_j^{(n)})=q_f(t_n,p_{j}^{(n)};a_{j-1}^{(n)}), \end{aligned}$$

(3.11)

we derive a contradiction. In this case, we find a point $c\in \hat{\Gamma }_n\setminus \{a_{j-1}^{(n)}, a_{j}^{(n)}\}$ such that $H_{t_n}f(p_j^{(n)})= q_f(t_n,p_j^{(n)};c)$ by (3.8). Note that $c< a_{j-1}^{(n)}$ or $a_{j}^{(n)}<c$.

First we consider the case $c< a_{j-1}^{(n)}$. By Lemma 2.4, we have

$$\begin{aligned}&q_f(t_n,x;a_{j-1}^{(n)})-q_f(t_n,x;c)\\&\quad =\ q_f(t_n,p_j^{(n)};a_{j-1}^{(n)})-q_f(t_n,p_j^{(n)};c) -\frac{a_{j-1}^{(n)}-c}{t_n}(x-p_j^{(n)})\\&\quad >\ -\frac{a_{j-1}^{(n)}-c}{t_n}(x-p_j^{(n)})\ge 0,\quad x \le p_j^{(n)}. \end{aligned}$$

On the other hand, by Lemma 2.4 again, we have

$$\begin{aligned}&q_f(t_n,x;a_{j}^{(n)})-q_f(t_n,x;a_{j-1}^{(n)})\\&\quad = q_f(t_n,p_j^{(n)};a_{j}^{(n)})-q_f(t_n,p_j^{(n)};a_{j-1}^{(n)}) -\frac{a_{j}^{(n)}-a_{j-1}^{(n)}}{t_n}(x-p_j^{(n)})\\&\quad = -\frac{a_{j}^{(n)}-a_{j-1}^{(n)}}{t_n}(x-p_j^{(n)})< 0,\quad x > p_j^{(n)}. \end{aligned}$$

Thus,

$$\begin{aligned} q_f(t_n,x;a_{j-1}^{(n)})>\min \{q_f(t_n,x;c), q_f(t_n,x;a_{j}^{(n)})\},\quad x\in \mathbb R. \end{aligned}$$

This implies that $E_n(a_{j-1}^{(n)})=\emptyset$ and $a_{j-1}^{(n)}\not \in \hat{\Gamma }_n$. This is a contradiction.

Next we consider the case $a_{j}^{(n)}<c$. Similarly to the arguments above, we can show that $E_n(a_{j}^{(n)})=\emptyset$ and $a_{j}^{(n)}\not \in \hat{\Gamma }_n$. This is also a contradiction.

Hence we conclude that (3.11) leads to a contradiction. Therefore (3.10) holds true. The proof is complete. $\square$

Fig. 1

The graph of $H_{t_n}f$ with $N_n=4$

Remark 3.15

Let $n\in \mathbb N_0$. By Lemma 3.14, we can show that, for each $j\in \{0,1,2,\ldots ,N_n\}$,

$$\begin{aligned} H_{t_n}f= q_f(t_n, \cdot \,; a_j^{(n)})\quad \text{ on }\quad [p_j^{(n)},p_{j+1}^{(n)}], \end{aligned}$$

where we put $p_{N_n+1}^{(n)}=1$ (see Fig. 1). However we do not use this expression in the following. So we omit its proof.

The next result is a partial converse of Proposition 3.6, where condition (A1) of Sect. 1 plays a key role. In Sect. 4, we study examples of admissible sequences of partitions of [0, 1] satisfying (A1) or not satisfying (A1).

Theorem 3.16

Let $f\in \mathcal H$. Assume that the admissible sequence of partitions $\{\Gamma _n\}$ of the interval [0, 1] appearing in the definition of $f\in \mathcal H$ satisfies (A1). Then $f\in \mathcal R$.

Proof

Fix $n\in \mathbb N_0$. Let $t_n=K_n |\Gamma _n|$ for $n\in \mathbb N_0$ and set $M=\sup \{K_n\,|\, n\in \mathbb N_0\}$ which appear in Definition 3.5 by the fact that $f\in \mathcal H$.

Let $\{\hat{\Gamma }_n\}$ be the admissible sequence of partitions of the interval [0, 1] in (3.9). Let $j\in \{1,2,\cdots , N_n\}$. By Lemmas 2.5 and 3.14, we have

$$\begin{aligned} \frac{f(a_{j}^{(n)})-f(y)}{a_{j}^{(n)}-y}-\frac{f(y)-f(a_{j-1}^{(n)})}{y-a_{j-1}^{(n)}} \le -\frac{a_{j}^{(n)}-a_{j-1}^{(n)}}{2t_n},\quad y\in (a_{j-1}^{(n)},a_{j}^{(n)}). \end{aligned}$$

Since $\hat{\Gamma }_n\subset \Gamma _n$ and $\{\Gamma _n\}$ satisfies (A1), we have

$$\begin{aligned} a_{j}^{(n)}-a_{j-1}^{(n)}\ge \min \{|x-y|\,|\, x, y\in \hat{\Gamma }_n,\ x\ne y\}\ge \min \{|x-y|\,|\, x, y\in \Gamma _n,\ x\ne y\} \ge \delta |\Gamma _n|. \end{aligned}$$

Then we have

$$\begin{aligned} \frac{f(a_{j}^{(n)})-f(y)}{a_{j}^{(n)}-y}-\frac{f(y)-f(a_{j-1}^{(n)})}{y-a_{j-1}^{(n)}} \le -\frac{\delta |\Gamma _n|}{2K_n|\Gamma _n|}\le -\frac{\delta }{2M},\quad y\in (a_{j-1}^{(n)},a_{j}^{(n)}). \end{aligned}$$

Thus $f\in \mathcal R$ by Definition 1.2 where $\{\Gamma _n\}$ and $\theta$ are, respectively, replaced by $\{\hat{\Gamma }_n\}$ and $\frac{\delta }{2M}$. Hence, we conclude the proof. $\square$

Remark 3.17

In the proof of Theorem 3.16, the condition (A1) is unavoidable to show that $\mathcal H\subset \mathcal R$. We do not know whether it is actually necessary to prove that $\mathcal H\subset \mathcal R$.

Proof of Theorem 1.5

Let $f\in C_p(\mathbb R)$. Since $\{H_tf\}$ satisfies $\text{(HJ) }$, we have $f\in \mathcal H$ by Proposition 3.6. Here, the admissible sequence $\{\Gamma _n\}$ of partitions of [0, 1] appearing in $\text{(HJ) }$ gives an admissible sequence of partitions of [0, 1] satisfying Definition 3.5. Since $\{\Gamma _n\}$ satisfies (A1), we conclude that $f\in \mathcal R$ by Theorem 3.16. By Proposition 1.3, f is nowhere differentiable on $\mathbb R$. $\square$

4 Examples

In this section, we give examples of admissible sequences of partitions of the interval [0, 1] satisfying (A1) or not satisfying (A1). We also systematically construct examples of functions belonging to the class $\mathcal R$.

First we consider (A1). As explained in Sect. 1, $\{\Gamma _n\}$ of (1.1) satisfies (A1) with $\delta =1$. We give one more example of admissible sequences of partitions of the interval [0, 1] satisfying (A1) with $\delta =1$.

Example 4.1

Let

$$\begin{aligned} \Gamma _n=\biggl \{\frac{k}{((n+1)!)^r}\, \bigg | \, k\in \mathbb N_0,\ 0 \le k \le ((n+1)!)^r\biggr \},\quad n\in \mathbb N_0, \end{aligned}$$

where r is a positive integer. It is easy to see that $\{\Gamma _n\}$ is an admissible sequence of partitions of the interval [0, 1] satisfying (A1) with $\delta =1$.

Next, for a given admissible sequence of partitions of the interval [0, 1] satisfying (A1), we give a method to construct another admissible sequences of partitions of the interval [0, 1] satisfying (A1).

Theorem 4.2

Let $\{\Gamma _n\}$ be an admissible sequence of partitions of the interval [0, 1] satisfying (A1) with a constant $\delta \in (0,1]$. Assume that $\varphi$ is a strictly increasing function on [0, 1] with $\varphi (0)=0$ and $\varphi (1)=1$ such that there exist constant $\alpha , \beta$ $(0<\alpha \le 1\le \beta )$ satisfying

$$\begin{aligned} \alpha (x-y)\le \varphi (x)-\varphi (y)\le \beta (x-y),\quad 0\le y < x\le 1. \end{aligned}$$

Then $\{\varphi (\Gamma _n)\}$ is an admissible sequence of partitions of the interval [0, 1] satisfying (A1) with a constant $\delta _\varphi \in (0,\frac{\alpha \delta }{\beta }]$.

Since the proof is clear, we omit it. We give an example of Theorem 4.2.

Example 4.3

For a given $\hat{\delta }\in (0,1]$, let

$$\begin{aligned} a=\frac{2\hat{\delta }}{1+\hat{\delta }}\in (0,1]. \end{aligned}$$

Define the function $\varphi :[0,1]\rightarrow [0,1]$ by

$$\begin{aligned} \varphi (t)=\left\{ \begin{array}{ll} at, &{}t\in [0,\frac{1}{2}], \\ &{}\\ (2-a)t+ a-1, &{}t\in (\frac{1}{2},1]. \end{array} \right. \end{aligned}$$

Then $\varphi$ satisfies the assumptions of Theorem 4.2 with $\alpha =a$ and $\beta =2-a$, since $a\le 2-a$. Let

$$\begin{aligned} \Gamma _n=\biggl \{\frac{k}{2^n}\,\bigg | \, k\in \mathbb N_0,\ 0\le k \le 2^n\biggr \},\quad n\in \mathbb N_0. \end{aligned}$$

Recall that $\{\Gamma _n\}$ is an admissible sequence of partitions of the interval [0, 1] satisfying (A1) with $\delta =1$. Thus, by Theorem 4.2, we see that $\{\varphi (\Gamma _n)\}$ is an admissible sequence of partitions of the interval [0, 1] satisfying (A1) with a constant $\delta _\varphi \in (0,\hat{\delta }]$, since

$$\begin{aligned} \frac{\alpha \delta }{\beta }=\frac{a\cdot 1}{2-a}=\hat{\delta }. \end{aligned}$$

Finally, we give an example of admissible sequences of partitions of the interval [0, 1] not satisfying (A1).

Example 4.4

Let $\Gamma _0=\{0,1\}$, and

$$\begin{aligned}&\Gamma _n=\biggl \{\frac{k}{2^n}\, \bigg | \, k\in \mathbb N_0,\ 0\le k \le 2^{n-1}\biggr \}\,\\&\qquad \qquad \qquad \cup \ \biggl \{\frac{\ell }{2^{2n}}\, \bigg |\, \ell \in \mathbb N_0,\ 2^{n-1}+1\le \ell \le 2^{2n}\biggr \} \end{aligned}$$

for a positive integer n. It is easy to see that $\{\Gamma _n\}$ is an admissible sequence of partitions of the interval [0, 1] such that

$$\begin{aligned} |\Gamma _n|=\frac{1}{2^n},\quad n\in \mathbb N_0. \end{aligned}$$

Since

$$\begin{aligned} \min \{|x-y|\,|\, x, y\in \Gamma _n,\ x\ne y\}=\frac{1}{2^{2n}},\quad n\in \mathbb N_0, \end{aligned}$$

there exists no $\delta \in (0,1]$ such that $\{\Gamma _n\}$ satisfies (A1).

Next, we systematically construct examples of functions belonging to the class $\mathcal R$. Let $\Gamma =\{\Gamma _n\}$ be a given admissible sequence of partitions of the interval [0, 1]. Here we do not necessarily assume that $\{\Gamma _n\}$ satisfies (A1). We show that all the functions of the form

$$\begin{aligned} T_{\Gamma ,\Psi }(x) =\sum _{j=0}^{\infty }|\Gamma _j|\, \psi _j(|\Gamma _j|^{-1}d(x,\Gamma _j)),\quad x\in [0,1], \end{aligned}$$

(4.1)

belong to $\mathcal R$, provided that $\sum _{j=0}^{\infty } |\Gamma _j|<+\infty$, where $\Psi =\{\psi _j\}$ is a sequence of functions satisfying:

$\mathrm{(A2)}$:

For each $j\in \mathbb N_0$, $\psi _j$ is concave and strictly increasing on $[0,\frac{1}{2}]$ with $\psi _j(0)=0$ such that

$$\begin{aligned} K:=\inf _{j\in \mathbb N_0}\psi _j\bigl (\frac{1}{2}\bigr )>0,\quad L:=\sup _{j\in \mathbb N_0} \psi _j\bigl (\frac{1}{2}\bigr ) <\infty . \end{aligned}$$

(4.2)

Note that the choice of $\psi _j(x)=x$ for all j is admissible, and in this case $T_{\Gamma ,\Psi }$ coincides with the function in (1.5). In this sense, our results includes those of [6] for a function on $\mathbb R$ (see [6] and compare to Lemmas 3.2, 3.3 and Remark 3.4).

We provide examples of $\{\psi _j\}$ satisfying (A2).

Example 4.5

For each $j\in \mathbb N_0$, let $\psi _j(x)=A_jx^{\alpha _j}$ for $x\in [0,\frac{1}{2}]$, where $\{A_j\}_{j\in \mathbb N_0}$ and $\{\alpha _j\}_{j\in \mathbb N_0}$ are sequences satisfying

$$\begin{aligned} 0< \inf _{j\in \mathbb N_0}A_j\le \sup _{j\in \mathbb N_0} A_j<+\infty ,\quad 0<\alpha _j\le 1\ (j\in \mathbb N_0). \end{aligned}$$

Since $\frac{1}{2} \le (\frac{1}{2})^{\alpha _j}<1$ $(j\in \mathbb N_0)$, $\{\psi _j\}$ satisfies (A2). In particular, the choice of $\psi _j(x)=x$ for all j is admissible.

Under (A2), since $|\Gamma _j|^{-1}d(x, \Gamma _j) \le \frac{1}{2}$, we have $0\le T_{\Gamma ,\Psi }(x)\le L \sum _{j=0}^{\infty }|\Gamma _j|$, so that if $\sum _{j=0}^{\infty }|\Gamma _j| <+\infty$, then $T_{\Gamma ,\Psi }$ can be extended as a function of $C_p(\mathbb R)$.

Theorem 4.6

Let $\{\Gamma _n\}$ be an admissible sequence of partitions of the interval [0, 1] with $\sum _{n=0}^{\infty }|\Gamma _n|<+\infty$. Assume that (A2) holds. Then the function $T_{\Gamma ,\Psi }$ defined by (4.1) belongs to $\mathcal R_K$, where K is the constant of (4.2).

In order to prove Theorem 4.6, we preliminarily give three lemmas.

Lemma 4.7

Assume (A2). Then we have the inequality

$$\begin{aligned} Ky\le \psi _j(y),\quad j\in \mathbb N_0,\ y\in [0,\frac{1}{2}]. \end{aligned}$$

Proof

Let $j\in \mathbb N_0$ and $y\in [0,\frac{1}{2}]$. Since $\psi _j$ is concave and strictly increasing on $[0,\frac{1}{2}]$, we derive

$$\begin{aligned} \psi _j(y)\ge \psi _j(\frac{y}{2})=\psi _j((1-y)0+y\cdot \frac{1}{2})\ge (1-y)\psi _j(0)+y\psi _j(\frac{1}{2})=y\psi _j(\frac{1}{2})\ge Ky. \end{aligned}$$

The proof is complete.$\square$

Recall that d is the function defined in (1.4).

Lemma 4.8

Let n be a positive integer and $a,b \in \Gamma _n$ with $a<b$ and $(a,b)\,\cap \, \Gamma _n=\emptyset$. Then, for any $j\in \{0,1,2,\ldots ,n-1\}$, we have

$$\begin{aligned} \frac{b-x}{b-a}\, d(a,\Gamma _j)+ \frac{x-a}{b-a}\, d(b,\Gamma _j)\le d(x,\Gamma _j),\quad x\in (a,b). \end{aligned}$$

Proof

Fix $j\in \{0,1,2,\ldots ,n-1\}$ arbitrarily. Then the lemma is an immediate consequence of the fact that $d(\cdot ,\Gamma _j)$ is concave on [a, b] since $(a,b)\cap \Gamma _n=\emptyset$. $\square$

Lemma 4.9

Let $n\in \mathbb N_0$ and $a,b \in \Gamma _n$ with $a< b$ and $(a,b)\cap \Gamma _n=\emptyset$. Then, we have

$$\begin{aligned} d(x,\Gamma _n) \ge \frac{(x-a)(b-x)}{b-a},\quad x\in (a,b). \end{aligned}$$

Proof

The proof is immediate. $\square$

Now, we prove Theorem 4.6.

Proof of Theorem 4.6

To ease notations we will write T in place of $T_{\Gamma ,\Psi }$.

In the following, let $n\in \mathbb N_0$ and $a,b \in \Gamma _n$ with $a< b$ and $(a,b)\cap \Gamma _n=\emptyset$. By Lemma 2.2, it is sufficient to show the inequality

$$\begin{aligned} T(x)\ge \frac{K}{b-a}{(x-a)(b-x)}+ \frac{1}{b-a}\,[(b-x) T(a)+ (x-a) T(b)],\ x\in (a,b). \end{aligned}$$

(4.3)

Let first $n=0$. In this case, $\Gamma _0=\{0,1\}$. We see that the constants $a,b \in \Gamma _0$ with $a< b$ and $(a,b)\cap \Gamma _0=\emptyset$ must be $a=0$ and $b=1$. By Lemma 4.9, we have $d(x,\Gamma _0) \ge x(1-x)$ for $x\in (0,1)$. Note that $\psi _j\ge 0$ on $[0,\frac{1}{2}]$ for each j by Lemma 4.7. Since $T(0)=T(1)=0$, we have, by Lemma 4.7,

$$\begin{aligned} T(x)&=\sum _{j=0}^{\infty }|\Gamma _j|\, \psi _j(|\Gamma _j|^{-1}d(x,\Gamma _j)) \ge |\Gamma _0|\,\psi _0(|\Gamma _0|^{-1}d(x,\Gamma _0)) \nonumber \\&\quad \ge \ Kd(x,\Gamma _0) \ge Kx(1-x)\nonumber \\&\quad =\ Kx(1-x)+ \frac{1}{1-0}\,[(1-x) T(0)+ (x-0) T(1)],\quad x\in (0,1). \end{aligned}$$

This shows (4.3) for $n=0$.

Next, fix a positive integer n arbitrarily. Since $\psi _j\ge 0$ on $[0,\frac{1}{2}]$ for each j by Lemma 4.7, we have

$$\begin{aligned} T(x)&=\sum _{j=0}^{\infty }|\Gamma _j|\, \psi _j(|\Gamma _j|^{-1}d(x,\Gamma _j))\\&\ge \ \sum _{j=0}^{n-1}|\Gamma _j|\, \psi _j(|\Gamma _j|^{-1}d(x,\Gamma _j))+ |\Gamma _n|\, \psi _n(|\Gamma _n|^{-1}d(x,\Gamma _n))\\&=:I_n+J_n,\ x\in (a,b). \end{aligned}$$

Since $\psi _j$ is strictly increasing on $[0,\frac{1}{2}]$, we have, by Lemmas 4.7 and 4.9,

$$\begin{aligned} J_n=|\Gamma _n|\, \psi _n(|\Gamma _n|^{-1}d(x,\Gamma _n)) \ge Kd(x,\Gamma _n)\ge K\frac{(x-a)(b-x)}{b-a},\quad x\in (a,b). \end{aligned}$$

On the other hand, let $j\in \{0,1,2,\cdots ,n-1\}$. We have, by Lemma 4.8 and the concavity of $\psi _j$ on $[0,\frac{1}{2}]$,

$$\begin{aligned} \frac{b-x}{b-a}\, \psi _j(|\Gamma _j|^{-1}d(a,\Gamma _j))+ \frac{x-a}{b-a}\, \psi _j(|\Gamma _j|^{-1}d(b,\Gamma _j)) \end{aligned}$$

(4.4)

$$\begin{aligned} \le \&\psi _j\biggl (\frac{b-x}{b-a}\, |\Gamma _j|^{-1}d(a,\Gamma _j)+ \frac{x-a}{b-a}\,|\Gamma _j|^{-1} d(b,\Gamma _j)\biggr )\nonumber \\ \le \&\psi _j(|\Gamma _j|^{-1}d(x,\Gamma _j)),\quad x\in (a,b), \end{aligned}$$

Fig. 2

The grapf of $T_{\Gamma ,\Psi }$ when $\psi _j(x)=x$

Fig. 3

The grapf of $T_{\Gamma ,\Psi }$ when $\psi _j(x)=\sqrt{x}$

since $\psi _j$ is strictly increasing on $[0,\frac{1}{2}]$. Since $a,b \in \Gamma _n$, we see that $d(a,\Gamma _j)=d(b,\Gamma _j)=0$ for each $j\in \{n, n+1,\cdots \}$. Hence, by (4.4), we have

$$\begin{aligned}&\frac{1}{b-a}\,[(b-x) T(a)+ (x-a) T(b)]\\ =\&\frac{b-x}{b-a}\, \sum _{j=0}^{n-1}|\Gamma _j|\,\psi _j(|\Gamma _j|^{-1}d(a,\Gamma _j))+ \frac{x-a}{b-a}\, \sum _{j=0}^{n-1}|\Gamma _j|\,\psi _j(|\Gamma _j|^{-1}d(b,\Gamma _j))\\ =\&\sum _{j=0}^{n-1}|\Gamma _j|\, \biggl (\frac{b-x}{b-a}\, \psi _j(|\Gamma _j|^{-1}d(a,\Gamma _j))+ \frac{x-a}{b-a}\, \psi _j(|\Gamma _j|^{-1}d(b,\Gamma _j))\biggr )\\ \le \&\sum _{j=0}^{n-1}|\Gamma _j|\,\psi _j(|\Gamma _j|^{-1}d(x,\Gamma _j))=I_n,\quad x\in (a,b). \end{aligned}$$

Therefore, we conclude (4.3). The proof is complete. $\square$

Finally we give two figures of functions in $\mathcal R$ constructed by Theorem 4.6. Let $\{\Gamma _n\}$ be the admissible sequence of partitions of the interval [0, 1] satisfying (A1) given in Example 4.1 with $r=1$. Figure 2 is the grapf of $T_{\Gamma ,\Psi }$ of (4.1) when $\psi _j(x)=x$ for all j. Figure 3 is the grapf of $T_{\Gamma ,\Psi }$ of (4.1) when $\psi _j(x)=\sqrt{x}$ for all j.