A Note on the Continuous and Nowhere Differentiable Function

Abstract

We continue the studies of derivatives of the non-harmonic series $\sum c_{n}exp\left(i{\lambda }_{n}x\right)$. We consider replacing $sin{\lambda }_{n}x$ with some step function $f_n:[0,1)\to \{0,1,2\}$. The purpose of this paper is to show that $f\left(x\right)=\sum {\displaystyle \frac{f_n\left(x\right)}{2^n}}$ is continuous and nowhere differentiable in $(0,1)$. All functions $f_n$, used to construct f, are created from symmetrical blocks of type M: [0,1,0] and blocks of type L: [0,1,2] and R: [2,1,0], located symmetrically with respect to M. The function $f_1$ takes the value zero in the intervals [0,1/3), [2/3,1) and the value one in the interval [1/3,2/3), i.e., it consists of the block of type M: [0,1,0]. Function $f_2$ takes the value two in the intervals [2/9,3/9) and [6/9,7/9); the value one in the intervals [1/9,2/9), [4/9,5/9), and [7/9,8/9); and the value zero in the remaining intervals. This means a composition of blocks of [0,1,2][0,1,0][2,1,0], i.e., LMR blocks. The function $f_3$ is a symmetric composition of blocks, LLMLMRMRR. These blocks are discontinuous analogs of the functions that produces Schoenberg curves.

1. Introduction

Chaundy and Jolliffe [1] proved the following:

Theorem 1.

If ${\left\{c_k\right\}}_{k=1}^{\infty }\subset R_{+}$ is decreasing to zero, then ${\sum }_{k=1}^{\infty }{c}_{k}sinkx$ converges uniformly in x if and only if $k{c}_k\to 0$ as $k\to \infty$.

Theorem 1 has had numerous generalizations. A result due to Žak and Šneider [2] holds for double sine series:

Theorem 2.

If ${\left\{c_{jk}\right\}}_{j,k=1}^{\infty }\subset R_{+}$ is a monotonically decreasing double sequence, i.e., a sequence of real numbers such that for $j,k=1,2,\dots$:

$c_{jk}-c_{j+1,k}\ge 0$, $c_{jk}-c_{j,k+1}\ge 0$ and $c_{jk}-c_{j,k+1}-c_{j+1,k}+c_{j+1,k+1}\ge 0$,

then ${\sum }_{j=1}^{\infty }{\sum }_{k=1}^{\infty }{c}_{jk}sinjxsinky$ is uniformly regularly convergent in $(x,y)$ if and only if $jk{c}_{jk}\to 0$ as $j+k\to \infty$.

Theorem 2 was generalized by Kórus [3]. He has defined new classes of double sequences ($SBVD{S}_1$) to obtain those generalizations.

Dyachenko, Mukanov, and Tikhonov [4] proved the Chaundy–Jolliffe theorem for $GM\left(\beta \right)$ sequences with majorants $\beta$ having the form ${\beta }_n={\displaystyle \frac{1}{n}}{F}_n\left(a\right)$, where $F_n$ is admissible.

A series

$$\sum c_k{e}^{i{\lambda }_k\theta }$$

was motivation for the generalization of Theorem 1. Such series were studied by Paley and Wiener, who called them non-harmonic Fourier series. They proved that [5]:

Theorem 3.

If $|{\lambda }_k-k|\le D<1/{\pi }^2$ for $-\infty <k<\infty$, then the sequence $\left\{e^{i{\lambda }_k\theta }\right\}$ is closed in $L^2(-\pi ,\pi )$ and possesses a unique biorthogonal set $\left\{h_k\left(\theta \right)\right\}$, such that the series

${\sum }_{k=-\infty }^{\infty }\{{\displaystyle \frac{e^{i{\lambda }_k\theta }}{2\pi }}{\int }_{-\pi }^{\pi }f\left(t\right)e^{-ikt}dt-e^{i{\lambda }_k\theta }{\int }_{-\pi }^{\pi }f\left(t\right)h_k\left(t\right)dt\}$ converges uniformly to zero over interval $(-\pi +\delta ,\pi -\delta )$ for any positive δ, and over any such interval, the summability properties of

${\sum }_{k=-\infty }^{\infty }{e}^{i{\lambda }_k\theta }{\int }_{-\pi }^{\pi }f\left(t\right)h_k\left(t\right)dt$ are uniformly the same as those of the Fourier series of $f\left(\theta \right)$.

One of the results of paper [6] is that for any $\alpha \in (0,2)$$\sum a_{n}sin{n}^{\alpha }x$ converges uniformly if and only if $n{a}_n\to 0$.

In paper [7], it was proved that $\sum a_{n}sin\sqrt{n}x$ converges uniformly on $[0,\pi ]$ if and only if $n{a}_n\to 0$. In paper [8], the following theorem was proved:

Theorem 4.

Let $c,d\in R$ such that $0<c<d$. If ${\left\{n{a}_n\right\}}_{n=1}^{\infty }$ is nonincreasing and ${lim}_{n\to \infty }n{a}_n=0$, then the series:

$$f\left(x\right)=\sum _{n=1}^{\infty }{a}_{n}sin\left(x\sqrt{n}\right)$$

can be differentiated term-by-term on $[c,d]\left(or\right[-d,-c\left]\right)$. The series

$$f^{\prime }\left(x\right)=\sum _{n=1}^{\infty }{a}_n\sqrt{n}cos\left(x\sqrt{n}\right)$$

converges uniformly on $[c,d]\left(or\right[-d,-c\left]\right)$.

In the recent paper [9], it was proved, among others, that if $\left\{c_n\right\}$ is a nonincreasing sequence and ${\sum }_{n=1}{c}_{n}cos\left(\sqrt{n}x\right)$ is convergent at any point $x\ne 0$, then $li{m}_{n\to \infty }\sqrt{n}{c}_n=0$.

Constructions of continuous but nowhere differentiable functions have been widely explored in the literature, particularly through the concept of fractal functions and iterated function systems. In the paper [10], some continuous interpolation functions $f:I\to R$ were introduced, where I is a real closed interval, which appear ideally suited for the approximation of naturally occurring functions that display some kind of geometrical self-similarity under magnification. The new interpolation functions were referred to as fractal because they can occur such that they are not differentiable.

The paper [11] is devoted to an example of space-filling curves: Schoenberg’s curve. Sagan showed that the proof that Schoenberg’s curve is nowhere differentiable is straightforward and does not utilize any advanced notions and techniques. To produce Schoenberg’s curve, let $I=[0,1]$ and let $f,g$ be defined by: ${\displaystyle f\left(t\right)=\frac{1}{2}\sum _{k=0}^{\infty }p\left(3^{2k}t\right)/2^k}$, ${\displaystyle g\left(t\right)=\frac{1}{2}\sum _{k=0}^{\infty }p\left(3^{2k+1}t\right)/2^k}$, where $p\left(t\right)$ is the even continuous function of period 2, which is defined in the interval (0, 1) as follows: $p\left(t\right)=0$ for $t\in (0,1/3)$, $p\left(t\right)=1$ for $t\in (2/3,1)$, and $p\left(t\right)$ is linear in $(1/3,2/3)$. It was proved that the functions $f,g$ are nowhere differentiable by the following lemma: If $f:[0,1]\to R$ is differentiable at $t\in (0,1)$, then, for any two sequences $\left\{a_n\right\}\to t$, $\left\{b_n\right\}\to t$ with $0<a_n<t<b_n<1$, by necessity, ${lim}_{n\to \infty }{\displaystyle \frac{f\left(b_n\right)-f\left(a_n\right)}{b_n-a_n}}=f^{\prime }\left(t\right)$ exists.

We will consider a function analogous to p, which is the step function $f_n:[0,1)\to \{0,1,2\}$, and we will replace $sin{\lambda }_{n}x$ in a sine series considered by Chaundy and Jolliffe. $f_n$ will be defined by assigning values 0, 1, 2 to intervals $[{\displaystyle \frac{k}{3^n}},{\displaystyle \frac{k+1}{3^n}})$ for $k=0,1,2,\dots ,3^n-1.$ We will show that $f\left(x\right)=\sum {\displaystyle \frac{f_n\left(x\right)}{2^n}}$ is continuous and nowhere differentiable on $(0,1)$. We will be choosing a small $\Delta$ directly so that ${\displaystyle \frac{f(x_0+\Delta )-f\left(x_0\right)}{\Delta }}$ is sufficiently large.

The book [12] is a comprehensive, self-contained compendium of results on continuous nowhere differentiable functions.

2. Main Results

Let ${\left(f_n\right)}_{n=1}^{\infty }$ be the sequence of functions $f_n:[0,1)\to \{0,1,2\}$, defined as follows:

$$f_1\left(x\right)={\mathbf{1}}_{[{\displaystyle \frac{1}{3}},{\displaystyle \frac{2}{3}})}\left(x\right),$$

$$f_2\left(x\right)={\mathbf{1}}_{[{\displaystyle \frac{1}{9}},{\displaystyle \frac{2}{9}})\cup [{\displaystyle \frac{4}{9}},{\displaystyle \frac{5}{9}})\cup [{\displaystyle \frac{7}{9}},{\displaystyle \frac{8}{9}})}\left(x\right)+2\ast {\mathbf{1}}_{[{\displaystyle \frac{2}{9}},{\displaystyle \frac{3}{9}})\cup [{\displaystyle \frac{6}{9}},{\displaystyle \frac{7}{9}})}\left(x\right),$$

$$\begin{array}{cc}\hfill f_3\left(x\right)& =2\ast {\mathbf{1}}_{[{\displaystyle \frac{2}{27}},{\displaystyle \frac{3}{27}})\cup [{\displaystyle \frac{5}{27}},{\displaystyle \frac{6}{27}})\cup [{\displaystyle \frac{11}{27}},{\displaystyle \frac{12}{27}})\cup [{\displaystyle \frac{15}{27}},{\displaystyle \frac{16}{27}})\cup [{\displaystyle \frac{21}{27}},{\displaystyle \frac{22}{27}})\cup [{\displaystyle \frac{24}{27}},{\displaystyle \frac{25}{27}})}\left(x\right)+\hfill \\ & +{\mathbf{1}}_{[{\displaystyle \frac{1}{27}},{\displaystyle \frac{2}{27}})\cup [{\displaystyle \frac{4}{27}},{\displaystyle \frac{5}{27}})\cup [{\displaystyle \frac{7}{27}},{\displaystyle \frac{8}{27}})\cup [{\displaystyle \frac{10}{27}},{\displaystyle \frac{11}{27}})\cup [{\displaystyle \frac{13}{27}},{\displaystyle \frac{14}{27}})\cup [{\displaystyle \frac{16}{27}},{\displaystyle \frac{17}{27}})\cup [{\displaystyle \frac{19}{27}},{\displaystyle \frac{20}{27}})\cup [{\displaystyle \frac{22}{27}},{\displaystyle \frac{23}{27}})\cup [{\displaystyle \frac{25}{27}},{\displaystyle \frac{26}{27}})}\left(x\right),\hfill \end{array}$$

$$\begin{array}{c}\hfill ⋮\hfill \\ \hfill f_n\left(x\right)=\left\{\begin{array}{ccc}2& \mathrm{if}& (x\in [{\displaystyle \frac{3k}{3^n}},{\displaystyle \frac{3k+1}{3^n}}),f_{n-1}\left({\displaystyle \frac{k}{3^{n-1}}}\right)-f_{n-1}\left({\displaystyle \frac{k-1}{3^{n-1}}}\right)=-1\hfill \\ & & \mathrm{and}k=1,2,\dots ,3^{n-1}-1)\hfill \\ & \mathrm{or}& (x\in [{\displaystyle \frac{3k+2}{3^n}},{\displaystyle \frac{3k+3}{3^n}}),f_{n-1}\left({\displaystyle \frac{k}{3^{n-1}}}\right)-f_{n-1}\left({\displaystyle \frac{k+1}{3^{n-1}}}\right)=-1\hfill \\ & & \mathrm{and}k=0,1,\dots ,3^{n-1}-2),\hfill \\ 1& \mathrm{if}& x\in [{\displaystyle \frac{3k+1}{3^n}},{\displaystyle \frac{3k+2}{3^n}})\mathrm{and}k=0,1,2,\dots ,3^{n-1}-1,\hfill \\ 0& \mathrm{for}& \mathrm{other}x\in [0,1).\hfill \end{array}\right.\hfill \end{array}$$

(1)

Observe that $\forall x\in [0,1)\exists k\in \{0,1,2,\dots ,3^{n-1}-1\}x\in [{\displaystyle \frac{3k}{3^n}},{\displaystyle \frac{3k+3}{3^n}}).$ In view of (1), depending on the value of $n,k$, the function $f_n$ is determined on $[{\displaystyle \frac{3k}{3^n}},{\displaystyle \frac{3k+3}{3^n}})$ in three different ways:

First:

$$f_n\left(x\right)=\left\{\begin{array}{ccc}1& \mathrm{for}& x\in [{\displaystyle \frac{3k+1}{3^n}},{\displaystyle \frac{3k+2}{3^n}})\hfill \\ 0& \mathrm{for}& x\in [{\displaystyle \frac{3k}{3^n}},{\displaystyle \frac{3k+1}{3^n}})\cup [{\displaystyle \frac{3k+2}{3^n}},{\displaystyle \frac{3k+3}{3^n}}).\hfill \end{array}\right.$$

(2)

Therefore,

$$f_{n+1}\left(x\right)=\left\{\begin{array}{ccc}2& \mathrm{for}& x\in [{\displaystyle \frac{9k+2}{3^{n+1}}},{\displaystyle \frac{9k+3}{3^{n+1}}})\cup [{\displaystyle \frac{9k+6}{3^{n+1}}},{\displaystyle \frac{9k+7}{3^{n+1}}})\hfill \\ 1& \mathrm{for}& x\in [{\displaystyle \frac{9k+1}{3^{n+1}}},{\displaystyle \frac{9k+2}{3^{n+1}}})\cup [{\displaystyle \frac{9k+4}{3^{n+1}}},{\displaystyle \frac{9k+5}{3^{n+1}}})\cup [{\displaystyle \frac{9k+7}{3^{n+1}}},{\displaystyle \frac{9k+8}{3^{n+1}}})\hfill \\ 0& \mathrm{for}& \mathrm{others}x\in [{\displaystyle \frac{9k}{3^{n+1}}},{\displaystyle \frac{9k+9}{3^{n+1}}}).\hfill \end{array}\right.$$

(3)

Second:

$$f_n\left(x\right)=\left\{\begin{array}{ccc}0& \mathrm{for}& x\in [{\displaystyle \frac{3k}{3^n}},{\displaystyle \frac{3k+1}{3^n}})\hfill \\ 1& \mathrm{for}& x\in [{\displaystyle \frac{3k+1}{3^n}},{\displaystyle \frac{3k+2}{3^n}})\hfill \\ 2& \mathrm{for}& x\in [{\displaystyle \frac{3k+2}{3^n}},{\displaystyle \frac{3k+3}{3^n}}).\hfill \end{array}\right.$$

(4)

Therefore,

$$f_{n+1}\left(x\right)=\left\{\begin{array}{ccc}2& \mathrm{for}& x\in [{\displaystyle \frac{9k+2}{3^{n+1}}},{\displaystyle \frac{9k+3}{3^{n+1}}})\cup [{\displaystyle \frac{9k+5}{3^{n+1}}},{\displaystyle \frac{9k+6}{3^{n+1}}})\hfill \\ 1& \mathrm{for}& x\in [{\displaystyle \frac{9k+1}{3^{n+1}}},{\displaystyle \frac{9k+2}{3^{n+1}}})\cup [{\displaystyle \frac{9k+4}{3^{n+1}}},{\displaystyle \frac{9k+5}{3^{n+1}}})\cup [{\displaystyle \frac{9k+7}{3^{n+1}}},{\displaystyle \frac{9k+8}{3^{n+1}}})\hfill \\ 0& \mathrm{for}& \mathrm{others}x\in [{\displaystyle \frac{9k}{3^{n+1}}},{\displaystyle \frac{9k+9}{3^{n+1}}}).\hfill \end{array}\right.$$

(5)

Third:

$$f_n\left(x\right)=\left\{\begin{array}{ccc}2& \mathrm{for}& x\in [{\displaystyle \frac{3k}{3^n}},{\displaystyle \frac{3k+1}{3^n}})\hfill \\ 1& \mathrm{for}& x\in [{\displaystyle \frac{3k+1}{3^n}},{\displaystyle \frac{3k+2}{3^n}})\hfill \\ 0& \mathrm{for}& x\in [{\displaystyle \frac{3k+2}{3^n}},{\displaystyle \frac{3k+3}{3^n}}).\hfill \end{array}\right.$$

(6)

Therefore,

$$f_{n+1}\left(x\right)=\left\{\begin{array}{ccc}2& \mathrm{for}& x\in [{\displaystyle \frac{9k+3}{3^{n+1}}},{\displaystyle \frac{9k+4}{3^{n+1}}})\cup [{\displaystyle \frac{9k+6}{3^{n+1}}},{\displaystyle \frac{9k+7}{3^{n+1}}})\hfill \\ 1& \mathrm{for}& x\in [{\displaystyle \frac{9k+1}{3^{n+1}}},{\displaystyle \frac{9k+2}{3^{n+1}}})\cup [{\displaystyle \frac{9k+4}{3^{n+1}}},{\displaystyle \frac{9k+5}{3^{n+1}}})\cup [{\displaystyle \frac{9k+7}{3^{n+1}}},{\displaystyle \frac{9k+8}{3^{n+1}}})\hfill \\ 0& \mathrm{for}& \mathrm{others}x\in [{\displaystyle \frac{9k}{3^{n+1}}},{\displaystyle \frac{9k+9}{3^{n+1}}}).\hfill \end{array}\right.$$

(7)

Let $x\in [0,1),n\ge 1,k\in \{0,1,2,\dots ,3^{n-1}-1\}$. Let us define:

$$M_n^k\left(x\right)=\left\{\begin{array}{ccc}1& \mathrm{if}& x\in [{\displaystyle \frac{3k+1}{3^n}},{\displaystyle \frac{3k+2}{3^n}})\hfill \\ 0& \mathrm{for}& \mathrm{others}x\in [0,1),\hfill \end{array}\right.$$

(8)

$$L_n^k\left(x\right)=\left\{\begin{array}{ccc}2& \mathrm{if}& x\in [{\displaystyle \frac{3k+2}{3^n}},{\displaystyle \frac{3k+3}{3^n}})\hfill \\ 1& \mathrm{if}& x\in [{\displaystyle \frac{3k+1}{3^n}},{\displaystyle \frac{3k+2}{3^n}})\hfill \\ 0& \mathrm{for}& \mathrm{others}x\in [0,1),\hfill \end{array}\right.$$

(9)

$$R_n^k\left(x\right)=\left\{\begin{array}{ccc}2& \mathrm{if}& x\in [{\displaystyle \frac{3k}{3^n}},{\displaystyle \frac{3k+1}{3^n}})\hfill \\ 1& \mathrm{if}& x\in [{\displaystyle \frac{3k+1}{3^n}},{\displaystyle \frac{3k+2}{3^n}})\hfill \\ 0& \mathrm{for}& \mathrm{others}x\in [0,1).\hfill \end{array}\right.$$

(10)

In view of (2)–(7), we obtain:

$$\begin{array}{cc}\mathrm{If} f_n\left(x\right)=M_n^k\left(x\right) & \mathrm{for} \mathrm{all}x\in [{\displaystyle \frac{3k}{3^n}},{\displaystyle \frac{3k+3}{3^n}}),\mathrm{then}\\ & f_{n+1}\left(x\right)=L_{n+1}^{3k}\left(x\right)+M_{n+1}^{3k+1}\left(x\right)+R_{n+1}^{3k+2}\left(x\right)and{f}_{n-1}\left(x\right)=const.\end{array}$$

(11)

$$\begin{array}{cc}\mathrm{If} f_n\left(x\right)=L_n^k\left(x\right)& \mathrm{for} \mathrm{all}x\in [{\displaystyle \frac{3k}{3^n}},{\displaystyle \frac{3k+3}{3^n}}),\mathrm{then}\\ & f_{n+1}\left(x\right)=L_{n+1}^{3k}\left(x\right)+L_{n+1}^{3k+1}\left(x\right)+M_{n+1}^{3k+2}\left(x\right)and{f}_{n-1}\left(x\right)=const.\end{array}$$

(12)

$$\begin{array}{cc}\mathrm{If} f_n\left(x\right)=R_n^k\left(x\right)& \mathrm{for} \mathrm{all}x\in [{\displaystyle \frac{3k}{3^n}},{\displaystyle \frac{3k+3}{3^n}}),\mathrm{then}\\ & f_{n+1}\left(x\right)=M_{n+1}^{3k}\left(x\right)+R_{n+1}^{3k+1}\left(x\right)+R_{n+1}^{3k+2}\left(x\right)and{f}_{n-1}\left(x\right)=const.\end{array}$$

(13)

Table of dependencies (2)–(13):

Figure 1, Figure 2 and Figure 3 show graphs of the functions $f_1, f_2, f_3.$ Colors indicate green for L-type functions, blue for M-type functions, and red for R-type functions.

Definition 5.

We define the function $f:[0,1)\to \mathbf{R}$ as:

$$f\left(x\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{f_n\left(x\right)}{2^n}} for x\in [0,1).$$

(14)

Figure 4, Figure 5, Figure 6 and Figure 7 show graphs of the functions:

$$S_w\left(x\right)=\sum _{n=1}^w{\displaystyle \frac{f_n\left(x\right)}{2^n}}\mathrm{for} x\in [0,1) \mathrm{and} w=1,2,3,4,\mathrm{respectively}.$$

(15)

Theorem 6.

The function (14) is continuous and nowhere differentiable on $(0,1)$.

3. Proof

3.1. Continuity of the Function (14)

The series (14) is uniformly convergent. This follows from the Weierstrass M-test. Let us define $A=\{i/3^m:m=1,2,3,\dots \mathrm{and} i=1,2,\dots ,3^m-1\}$, $D=(0,1)-A.$ Observe that $\forall x_0\in D\forall n\ge 1f_n$ is continuous at $x_0$. After considering that the series (14) is uniformly convergent, the function f is continuous at $x_0\in D$.

We consider $x_0\in A$. Observe that ${lim}_{x\to {\displaystyle \frac{i}{3^m}}+}{f}_m\left(x\right)=f_m\left({\displaystyle \frac{i}{3^m}}\right)$ and ${lim}_{x\to {\displaystyle \frac{i}{3^m}}-}{f}_m\left(x\right)=f_m\left({\displaystyle \frac{i-1}{3^m}}\right)$.

We consider $3\nmid i$. In view of (1), the functions $f_1, f_2, \dots ,f_{m-1}$ are continuous at $x_0={\displaystyle \frac{i}{3^m}}$; therefore:

$$\sum _{n=1}^{m-1}{\displaystyle \frac{f_n\left(x\right)}{2^n}}\mathrm{is} \mathrm{continuous} \mathrm{at} x_0.$$

(16)

Moreover, $\forall k>1f_{m+k}\left({\displaystyle \frac{3^{k}i-1}{3^{m+k}}}\right)=f_{m+k}\left({\displaystyle \frac{3^{k}i}{3^{m+k}}}\right)=0$. After considering that the series (14) is uniformly convergent, we obtain:

$$\sum _{n=m+2}^{\infty }{\displaystyle \frac{f_n\left(x\right)}{2^n}}\mathrm{is} \mathrm{continuous} \mathrm{at} x_0.$$

(17)

In view of (1): $|f_m\left({\displaystyle \frac{i-1}{3^m}}\right)-f_m\left({\displaystyle \frac{i}{3^m}}\right)|=1$.

If $f_m\left({\displaystyle \frac{i-1}{3^m}}\right)-f_m\left({\displaystyle \frac{i}{3^m}}\right)=1$ then $f_{m+1}\left({\displaystyle \frac{3i-1}{3^{m+1}}}\right)-f_{m+1}\left({\displaystyle \frac{3i}{3^{m+1}}}\right)=-2$.

If $f_m\left({\displaystyle \frac{i-1}{3^m}}\right)-f_m\left({\displaystyle \frac{i}{3^m}}\right)=-1$ then $f_{m+1}\left({\displaystyle \frac{3i-1}{3^{m+1}}}\right)-f_{m+1}\left({\displaystyle \frac{3i}{3^{m+1}}}\right)=2$. Therefore:

$${\displaystyle \frac{1}{2^m}}[f_m\left({\displaystyle \frac{i-1}{3^m}}\right)-f_m\left({\displaystyle \frac{i}{3^m}}\right)]+{\displaystyle \frac{1}{2^{m+1}}}[f_{m+1}\left({\displaystyle \frac{3i-1}{3^{m+1}}}\right)-f_{m+1}\left({\displaystyle \frac{3i}{3^{m+1}}}\right)]=0.$$

(18)

In view of (16)–(18), we obtain:

$$\sum _{n=1}^{\infty }{\displaystyle \frac{f_n\left(x\right)}{2^n}} \mathrm{is} \mathrm{continuous} \mathrm{at} x_0 \mathrm{for} 3\nmid i.$$

For cases $3\mid i,9\nmid i$ and $9\mid i$, the proof of continuity is similar.

3.2. Non-Differentiability of the Function (14)—Scheme of Proof

Let $x_0,m$ be fixed. We will look for

${\Delta }_{m+2}\in \{\pm 1/3^{m+2},\pm 2/3^{m+2},\pm 3/3^{m+2},\pm 6/3^{m+2},\pm 9/3^{m+2}\}$ so that

$\left|{\displaystyle \frac{f(x_0+{\Delta }_{m+2})-f\left(x_0\right)}{{\Delta }_{m+2}}}\right|\ge {\displaystyle \frac{1}{3}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}$ in the following way.

1. We are looking for $l\in \{0,1,2,\dots ,3^{m+2}-1\}$ so that $x_0\in [{\displaystyle \frac{l}{3^{m+2}}},{\displaystyle \frac{l+1}{3^{m+2}}})$. Taking into account only (8)–(10), we can write:

$$M_{m+3}^l{|}_{[{\displaystyle \frac{3l}{3^{m+3}}},{\displaystyle \frac{3l+3}{3^{m+3}}})}\left(x\right)=\left\{\begin{array}{ccc}0& \mathrm{if}& x\in [{\displaystyle \frac{3l}{3^{m+3}}},{\displaystyle \frac{3l+1}{3^{m+3}}})\hfill \\ 1& \mathrm{if}& x\in [{\displaystyle \frac{3l+1}{3^{m+3}}},{\displaystyle \frac{3l+2}{3^{m+3}}})\hfill \\ 0& \mathrm{for}& x\in [{\displaystyle \frac{3l+2}{3^{m+3}}},{\displaystyle \frac{3l+3}{3^{m+3}}}),\hfill \end{array}\right.$$

$$L_{m+3}^l{|}_{[{\displaystyle \frac{3l}{3^{m+3}}},{\displaystyle \frac{3l+3}{3^{m+3}}})}\left(x\right)=\left\{\begin{array}{ccc}0& \mathrm{if}& x\in [{\displaystyle \frac{3l}{3^{m+3}}},{\displaystyle \frac{3l+1}{3^{m+3}}})\hfill \\ 1& \mathrm{if}& x\in [{\displaystyle \frac{3l+1}{3^{m+3}}},{\displaystyle \frac{3l+2}{3^{m+3}}})\hfill \\ 2& \mathrm{for}& x\in [{\displaystyle \frac{3l+2}{3^{m+3}}},{\displaystyle \frac{3l+3}{3^{m+3}}}),\hfill \end{array}\right.$$

$$R_{m+3}^l{|}_{[{\displaystyle \frac{3l}{3^{m+3}}},{\displaystyle \frac{3l+3}{3^{m+3}}})}\left(x\right)=\left\{\begin{array}{ccc}2& \mathrm{if}& x\in [{\displaystyle \frac{3l}{3^{m+3}}},{\displaystyle \frac{3l+1}{3^{m+3}}})\hfill \\ 1& \mathrm{if}& x\in [{\displaystyle \frac{3l+1}{3^{m+3}}},{\displaystyle \frac{3l+2}{3^{m+3}}})\hfill \\ 0& \mathrm{for}& x\in [{\displaystyle \frac{3l+2}{3^{m+3}}},{\displaystyle \frac{3l+3}{3^{m+3}}}).\hfill \end{array}\right.$$

In view of (1), the function $f_{m+3}{|}_{[{\displaystyle \frac{3l}{3^{m+3}}},{\displaystyle \frac{3l+3}{3^{m+3}}})}$ is defined by either: $M_{m+3}^l{|}_{[{\displaystyle \frac{3l}{3^{m+3}}},{\displaystyle \frac{3l+3}{3^{m+3}}})}$ or $L_{m+3}^l{|}_{[{\displaystyle \frac{3l}{3^{m+3}}},{\displaystyle \frac{3l+3}{3^{m+3}}})}$ or $R_{m+3}^l{|}_{[{\displaystyle \frac{3l}{3^{m+3}}},{\displaystyle \frac{3l+3}{3^{m+3}}})}$. For the sake of focus, let us assume that $f_{m+3}{|}_{[{\displaystyle \frac{3l}{3^{m+3}}},{\displaystyle \frac{3l+3}{3^{m+3}}})}=R_{m+3}^l{|}_{[{\displaystyle \frac{3l}{3^{m+3}}},{\displaystyle \frac{3l+3}{3^{m+3}}})}$.

2. We choose ${\Delta }_{m+2}\in \{\pm 1/3^{m+2},\pm 2/3^{m+2},\pm 3/3^{m+2},\pm 6/3^{m+2},\pm 9/3^{m+2}\}$ so that: $f_{m+3}{|}_{[{\displaystyle \frac{3l^{\prime }}{3^{m+3}}},{\displaystyle \frac{3l+3}{3^{m+3}}})}=R_{m+3}^{l^{\prime }}{|}_{[{\displaystyle \frac{3l^{\prime }}{3^{m+3}}},{\displaystyle \frac{3l^{\prime }+3}{3^{m+3}}})}$ for $l^{\prime }=3^{m+2}{\Delta }_{m+2}+l$. The way of selection ${\Delta }_{m+2}$ is shown in the Figure 8.

After considering $x_0-{\displaystyle \frac{3l}{3^{m+3}}}=x_0-{\Delta }_{m+2}-{\displaystyle \frac{3l^{\prime }}{3^{m+3}}}$, we obtain:

$R_{m+3}^l\left(x_0\right)=R_{m+3}^{l^{\prime }}(x_0+{\Delta }_{m+2})$ and $f_{m+3}(x_0+{\Delta }_{m+2})-f_{m+3}\left(x_0\right)=0$, $f_{m+4}(x_0+{\Delta }_{m+2})-f_{m+4}\left(x_0\right)=0$, $f_{m+5}(x_0+{\Delta }_{m+2})-f_{m+5}\left(x_0\right)=0$,…. In the next section, we will determine precisely ${\Delta }_{m+2}$ so that:

${\displaystyle \left|\sum _{n=m}^{m+2}\frac{f_n(x_0+{\Delta }_{m+2})-f_n\left(x_0\right)}{2^n}\right|\ge \frac{1}{3}{\left(\frac{3}{2}\right)}^{m+2}}$. For example, in Figure 8, ${\Delta }_{m+2}={\displaystyle \frac{6}{3^{m+2}}}$ and ${\displaystyle \frac{f(x_0+{\Delta }_{m+2})-f\left(x_0\right)}{{\Delta }_{m+2}}}={\displaystyle \frac{1/2^m-1/2^{m+1}}{6/3^{m+2}}}={\displaystyle \frac{1}{3}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}$.

3.3. Non-Differentiability of the Function (14)—Detailed Proof

Let $x_0\in (0,1)$ be a fixed real number. Then:

$\forall m>1\exists k\left(m\right)\in \{0,1,2,\dots ,3^{m-1}-1\}{x}_0\in [{\displaystyle \frac{3k\left(m\right)}{3^m}},{\displaystyle \frac{3k\left(m\right)+3}{3^m}})$. Let us define ${\Delta }_{m+2}^j=j/3^{m+2}$.

Let us assume that the function $f_m$ is determined on $[{\displaystyle \frac{3k\left(m\right)}{3^m}},{\displaystyle \frac{3k\left(m\right)+3}{3^m}})$ according to (2), so that:

$$f_m\left(x\right)=M_m^{k\left(m\right)}\left(x\right).$$

(19)

In view of (11)–(13), we obtain:

$$f_{m+r}\left(x\right)=const \mathrm{for} r\in \{-m+1,-m+2,\dots ,-1\},$$

(20)

$$f_{m+1}\left(x\right)=L_{m+1}^{3k\left(m\right)}\left(x\right)+M_{m+1}^{3k\left(m\right)+1}\left(x\right)+R_{m+1}^{3k\left(m\right)+2}\left(x\right),$$

(21)

$$\begin{array}{c}{f}_{m+2}\left(x\right)=L_{m+2}^{9k\left(m\right)}\left(x\right)+L_{m+2}^{9k\left(m\right)+1}\left(x\right)+M_{m+2}^{9k\left(m\right)+2}\left(x\right)+L_{m+2}^{9k\left(m\right)+3}\left(x\right)+M_{m+2}^{9k\left(m\right)+4}\left(x\right)+\hfill \\ \hfill +R_{m+2}^{9k\left(m\right)+5}\left(x\right)+M_{m+2}^{9k\left(m\right)+6}\left(x\right)+R_{m+2}^{9k\left(m\right)+7}\left(x\right)+R_{m+2}^{9k\left(m\right)+8}\left(x\right),\end{array}$$

(22)

$$\begin{array}{c}{f}_{m+3}\left(x\right)=L_{m+3}^{27k\left(m\right)}\left(x\right)+L_{m+3}^{27k\left(m\right)+1}\left(x\right)+M_{m+3}^{27k\left(m\right)+2}\left(x\right)+L_{m+3}^{9k\left(m\right)+3}\left(x\right)+\hfill \\ +L_{m+3}^{9k\left(m\right)+4}\left(x\right)+M_{m+3}^{9k\left(m\right)+5}\left(x\right)+L_{m+3}^{9k\left(m\right)+6}\left(x\right)+M_{m+3}^{9k\left(m\right)+7}\left(x\right)+R_{m+3}^{9k\left(m\right)+8}\left(x\right)+\\ +L_{m+3}^{27k\left(m\right)+9}\left(x\right)+L_{m+3}^{27k\left(m\right)+10}\left(x\right)+M_{m+3}^{27k\left(m\right)+11}\left(x\right)+L_{m+3}^{9k\left(m\right)+12}\left(x\right)+M_{m+3}^{9k\left(m\right)+13}\left(x\right)+\\ +R_{m+3}^{9k\left(m\right)+14}\left(x\right)+M_{m+3}^{9k\left(m\right)+15}\left(x\right)+R_{m+3}^{9k\left(m\right)+16}\left(x\right)+R_{m+3}^{9k\left(m\right)+17}\left(x\right)+L_{m+3}^{27k\left(m\right)+18}\left(x\right)+\\ +M_{m+3}^{27k\left(m\right)+19}\left(x\right)+R_{m+3}^{27k\left(m\right)+20}\left(x\right)+M_{m+3}^{9k\left(m\right)+21}\left(x\right)+R_{m+3}^{9k\left(m\right)+22}\left(x\right)+\\ \hfill +R_{m+3}^{9k\left(m\right)+23}\left(x\right)+M_{m+3}^{9k\left(m\right)+24}\left(x\right)+R_{m+3}^{9k\left(m\right)+25}\left(x\right)+R_{m+3}^{9k\left(m\right)+26}\left(x\right).\end{array}$$

(23)

Case 1: $x_0\in [{\displaystyle \frac{27k\left(m\right)}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+1}{3^{m+2}}})$.

Then, $x_0+{\Delta }_{m+2}^1\in [{\displaystyle \frac{27k\left(m\right)+1}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+2}{3^{m+2}}}).$ In view of (22):

$f_{m+2}\left(x_0\right)=L_{m+2}^{9k\left(m\right)}\left(x_0\right)=0,f_{m+2}(x_0+{\Delta }_{m+2}^1)=L_{m+2}^{9k\left(m\right)}(x_0+{\Delta }_{m+2}^1)=1.$

$$\mathrm{Therefore} f_{m+2}(x_0+{\Delta }_{m+2}^1)-f_{m+2}\left(x_0\right)=1.$$

(24)

Note that $x_0\in [{\displaystyle \frac{9k\left(m\right)}{3^{m+1}}},{\displaystyle \frac{9k\left(m\right)+\frac{1}{3}}{3^{m+1}}})$ and $x_0+{\Delta }_{m+2}^1\in [{\displaystyle \frac{9k\left(m\right)+\frac{1}{3}}{3^{m+1}}},{\displaystyle \frac{9k\left(m\right)+\frac{2}{3}}{3^{m+1}}}).$

In view of (21):

$f_{m+1}\left(x_0\right)=L_{m+1}^{3k\left(m\right)}\left(x_0\right)=0, f_{m+1}(x_0+{\Delta }_{m+2}^1)=L_{m+1}^{3k\left(m\right)}(x_0+{\Delta }_{m+2}^1)=0.$

$$\mathrm{Therefore} f_{m+1}(x_0+{\Delta }_{m+2}^1)-f_{m+1}\left(x_0\right)=0.$$

(25)

Note that $x_0\in [{\displaystyle \frac{3k\left(m\right)}{3^m}},{\displaystyle \frac{3k\left(m\right)+\frac{1}{9}}{3^m}})$ and $x_0+{\Delta }_{m+2}^1\in [{\displaystyle \frac{3k\left(m\right)+\frac{1}{9}}{3^m}},{\displaystyle \frac{9k\left(m\right)+\frac{2}{9}}{3^m}}).$

In view of (19):

$f_m\left(x_0\right)=M_m^{k\left(m\right)}\left(x_0\right)=0, f_m(x_0+{\Delta }_{m+2}^1)=M_m^{k\left(m\right)}(x_0+{\Delta }_{m+2}^1)=0.$

$$\mathrm{Therefore} f_m(x_0+{\Delta }_{m+2}^1)-f_m\left(x_0\right)=0.$$

(26)

In view of (20):

$$\forall r<0f_{m+r}(x_0+{\Delta }_{m+2}^1)-f_{m+r}\left(x_0\right)=0.$$

(27)

In view of (23):

$$f_{m+3}\left(x_0\right)=L_{m+3}^{27k\left(m\right)}\left(x_0\right)=L_{m+3}^{27k\left(m\right)+1}(x_0+{\Delta }_{m+2}^1)=f_{m+3}(x_0+{\Delta }_{m+2}^1).$$

(28)

$$\mathrm{Therefore} f_{m+3}(x_0+{\Delta }_{m+2}^1)-f_{m+3}\left(x_0\right)=0.$$

(29)

In view of (11)–(13), (28), and (29), we obtain:

$$\forall r\ge 3f_{m+r}(x_0+{\Delta }_{m+2}^1)-f_{m+r}\left(x_0\right)=0.$$

(30)

In view of (14), (24)–(27), and (30), we obtain:

$${\displaystyle \frac{f(x_0+{\Delta }_{m+2}^1)-f\left(x_0\right)}{{\Delta }_{m+2}^1}}={\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(31)

Case 2: $x_0\in [{\displaystyle \frac{27k\left(m\right)+1}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+2}{3^{m+2}}})$.

Then, $x_0-{\Delta }_{m+2}^1\in [{\displaystyle \frac{27k\left(m\right)}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+1}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0-{\Delta }_{m+2}^1)-f\left(x_0\right)}{-{\Delta }_{m+2}^1}}={\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(32)

Case 3: $x_0\in [{\displaystyle \frac{27k\left(m\right)+2}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+3}{3^{m+2}}})$.

Then, $x_0+{\Delta }_{m+2}^3\in [{\displaystyle \frac{27k\left(m\right)+5}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+6}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0+{\Delta }_{m+2}^3)-f\left(x_0\right)}{{\Delta }_{m+2}^3}}={\displaystyle \frac{2}{3}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(33)

Case 4: $x_0\in [{\displaystyle \frac{27k\left(m\right)+3}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+4}{3^{m+2}}})$.

Then, $x_0+{\Delta }_{m+2}^1\in [{\displaystyle \frac{27k\left(m\right)+4}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+5}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0+{\Delta }_{m+2}^1)-f\left(x_0\right)}{{\Delta }_{m+2}^1}}={\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(34)

Case 5: $x_0\in [{\displaystyle \frac{27k\left(m\right)+4}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+5}{3^{m+2}}})$.

Then, $x_0-{\Delta }_{m+2}^1\in [{\displaystyle \frac{27k\left(m\right)+3}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+4}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0-{\Delta }_{m+2}^1)-f\left(x_0\right)}{-{\Delta }_{m+2}^1}}={\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(35)

Case 6: $x_0\in [{\displaystyle \frac{27k\left(m\right)+5}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+6}{3^{m+2}}})$.

Then, $x_0-{\Delta }_{m+2}^3\in [{\displaystyle \frac{27k\left(m\right)+2}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+3}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0-{\Delta }_{m+2}^3)-f\left(x_0\right)}{-{\Delta }_{m+2}^3}}={\displaystyle \frac{2}{3}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(36)

Case 7: $x_0\in [{\displaystyle \frac{27k\left(m\right)+6}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+7}{3^{m+2}}})$.

Then, $x_0-{\Delta }_{m+2}^2\in [{\displaystyle \frac{27k\left(m\right)+4}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+5}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0-{\Delta }_{m+2}^2)-f\left(x_0\right)}{-{\Delta }_{m+2}^2}}={\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(37)

Case 8: $x_0\in [{\displaystyle \frac{27k\left(m\right)+7}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+8}{3^{m+2}}})$.

Then, $x_0-{\Delta }_{m+2}^2\in [{\displaystyle \frac{27k\left(m\right)+5}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+6}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0-{\Delta }_{m+2}^2)-f\left(x_0\right)}{-{\Delta }_{m+2}^2}}={\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(38)

Case 9: $x_0\in [{\displaystyle \frac{27k\left(m\right)+8}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+9}{3^{m+2}}})$.

Then, $x_0+{\Delta }_{m+2}^6\in [{\displaystyle \frac{27k\left(m\right)+14}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+15}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0+{\Delta }_{m+2}^6)-f\left(x_0\right)}{{\Delta }_{m+2}^6}}={\displaystyle \frac{1}{3}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(39)

Case 10: $x_0\in [{\displaystyle \frac{27k\left(m\right)+9}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+10}{3^{m+2}}})$.

Then, $x_0+{\Delta }_{m+2}^1\in [{\displaystyle \frac{27k\left(m\right)+10}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+11}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0+{\Delta }_{m+2}^1)-f\left(x_0\right)}{{\Delta }_{m+2}^1}}={\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(40)

Case 11: $x_0\in [{\displaystyle \frac{27k\left(m\right)+10}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+11}{3^{m+2}}})$.

Then, $x_0-{\Delta }_{m+2}^1\in [{\displaystyle \frac{27k\left(m\right)+9}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+10}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0-{\Delta }_{m+2}^1)-f\left(x_0\right)}{-{\Delta }_{m+2}^1}}={\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(41)

Case 12: $x_0\in [{\displaystyle \frac{27k\left(m\right)+11}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+12}{3^{m+2}}})$.

Then, $x_0+{\Delta }_{m+2}^2\in [{\displaystyle \frac{27k\left(m\right)+13}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+14}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0+{\Delta }_{m+2}^2)-f\left(x_0\right)}{{\Delta }_{m+2}^2}}={\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(42)

Case 13: $x_0\in [{\displaystyle \frac{27k\left(m\right)+12}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+13}{3^{m+2}}})$.

Then, $x_0-{\Delta }_{m+2}^2\in [{\displaystyle \frac{27k\left(m\right)+10}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+11}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0-{\Delta }_{m+2}^2)-f\left(x_0\right)}{-{\Delta }_{m+2}^2}}={\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(43)

Case 14: $x_0\in [{\displaystyle \frac{27k\left(m\right)+13}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+14}{3^{m+2}}})$.

Then, $x_0-{\Delta }_{m+2}^2\in [{\displaystyle \frac{27k\left(m\right)+11}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+12}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0-{\Delta }_{m+2}^2)-f\left(x_0\right)}{-{\Delta }_{m+2}^2}}={\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(44)

Case 15: $x_0\in [{\displaystyle \frac{27k\left(m\right)+14}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+15}{3^{m+2}}})$.

Then, $x_0-{\Delta }_{m+2}^6\in [{\displaystyle \frac{27k\left(m\right)+8}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+9}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0-{\Delta }_{m+2}^6)-f\left(x_0\right)}{-{\Delta }_{m+2}^6}}={\displaystyle \frac{1}{3}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(45)

Case 16: $x_0\in [{\displaystyle \frac{27k\left(m\right)+15}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+16}{3^{m+2}}})$.

Then, $x_0-{\Delta }_{m+2}^2\in [{\displaystyle \frac{27k\left(m\right)+13}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+14}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0-{\Delta }_{m+2}^2)-f\left(x_0\right)}{-{\Delta }_{m+2}^2}}=-{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(46)

Case 17: $x_0\in [{\displaystyle \frac{27k\left(m\right)+16}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+17}{3^{m+2}}})$.

Then, $x_0+{\Delta }_{m+2}^1\in [{\displaystyle \frac{27k\left(m\right)+17}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+18}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0+{\Delta }_{m+2}^1)-f\left(x_0\right)}{{\Delta }_{m+2}^1}}=-{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(47)

Case 18: $x_0\in [{\displaystyle \frac{27k\left(m\right)+17}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+18}{3^{m+2}}})$.

Then, $x_0-{\Delta }_{m+2}^1\in [{\displaystyle \frac{27k\left(m\right)+16}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+17}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0-{\Delta }_{m+2}^1)-f\left(x_0\right)}{-{\Delta }_{m+2}^1}}=-{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(48)

Case 19: $x_0\in [{\displaystyle \frac{27k\left(m\right)+18}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+19}{3^{m+2}}})$.

Then, $x_0-{\Delta }_{m+2}^6\in [{\displaystyle \frac{27k\left(m\right)+12}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+13}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0-{\Delta }_{m+2}^6)-f\left(x_0\right)}{-{\Delta }_{m+2}^6}}=-{\displaystyle \frac{1}{3}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(49)

Case 20: $x_0\in [{\displaystyle \frac{27k\left(m\right)+19}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+20}{3^{m+2}}})$.

Then, $x_0+{\Delta }_{m+2}^2\in [{\displaystyle \frac{27k\left(m\right)+21}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+22}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0+{\Delta }_{m+2}^2)-f\left(x_0\right)}{{\Delta }_{m+2}^2}}=-{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(50)

Case 21: $x_0\in [{\displaystyle \frac{27k\left(m\right)+20}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+21}{3^{m+2}}})$.

Then, $x_0+{\Delta }_{m+2}^2\in [{\displaystyle \frac{27k\left(m\right)+22}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+23}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0+{\Delta }_{m+2}^2)-f\left(x_0\right)}{{\Delta }_{m+2}^2}}=-{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(51)

Case 22: $x_0\in [{\displaystyle \frac{27k\left(m\right)+21}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+22}{3^{m+2}}})$.

Then, $x_0-{\Delta }_{m+2}^2\in [{\displaystyle \frac{27k\left(m\right)+19}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+20}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0-{\Delta }_{m+2}^2)-f\left(x_0\right)}{-{\Delta }_{m+2}^2}}=-{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(52)

Case 23: $x_0\in [{\displaystyle \frac{27k\left(m\right)+22}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+23}{3^{m+2}}})$.

Then, $x_0+{\Delta }_{m+2}^1\in [{\displaystyle \frac{27k\left(m\right)+23}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+24}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0+{\Delta }_{m+2}^1)-f\left(x_0\right)}{{\Delta }_{m+2}^1}}=-{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(53)

Case 24: $x_0\in [{\displaystyle \frac{27k\left(m\right)+23}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+24}{3^{m+2}}})$.

Then, $x_0-{\Delta }_{m+2}^1\in [{\displaystyle \frac{27k\left(m\right)+22}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+23}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0-{\Delta }_{m+2}^1)-f\left(x_0\right)}{-{\Delta }_{m+2}^1}}=-{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(54)

Case 25: $x_0\in [{\displaystyle \frac{27k\left(m\right)+24}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+25}{3^{m+2}}})$.

Then, $x_0-{\Delta }_{m+2}^3\in [{\displaystyle \frac{27k\left(m\right)+21}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+22}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0-{\Delta }_{m+2}^3)-f\left(x_0\right)}{-{\Delta }_{m+2}^3}}=-{\displaystyle \frac{2}{3}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(55)

Case 26: $x_0\in [{\displaystyle \frac{27k\left(m\right)+25}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+26}{3^{m+2}}})$.

Then, $x_0+{\Delta }_{m+2}^1\in [{\displaystyle \frac{27k\left(m\right)+26}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+27}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0+{\Delta }_{m+2}^1)-f\left(x_0\right)}{{\Delta }_{m+2}^1}}=-{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(56)

Case 27: $x_0\in [{\displaystyle \frac{27k\left(m\right)+26}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+27}{3^{m+2}}})$.

Then, $x_0-{\Delta }_{m+2}^1\in [{\displaystyle \frac{27k\left(m\right)+25}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+26}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0-{\Delta }_{m+2}^1)-f\left(x_0\right)}{-{\Delta }_{m+2}^1}}=-{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(57)

Let us assume that the function $f_m$ is determined on $[{\displaystyle \frac{3k\left(m\right)}{3^m}},{\displaystyle \frac{3k\left(m\right)+3}{3^m}})$ according to (4), so that:

$$f_m\left(x\right)=L_m^{k\left(m\right)}\left(x\right).$$

(58)

In view of (11)–(13), we obtain:

$$f_{m+r}\left(x\right)=const \mathrm{for} r\in \{-m+1,-m+2,\dots ,-1\},$$

(59)

$$f_{m+1}\left(x\right)=L_{m+1}^{3k\left(m\right)}\left(x\right)+L_{m+1}^{3k\left(m\right)+1}\left(x\right)+M_{m+1}^{3k\left(m\right)+2}\left(x\right),$$

(60)

$$\begin{array}{c}{f}_{m+2}\left(x\right)=L_{m+2}^{9k\left(m\right)}\left(x\right)+L_{m+2}^{9k\left(m\right)+1}\left(x\right)+M_{m+2}^{9k\left(m\right)+2}\left(x\right)+L_{m+2}^{9k\left(m\right)+3}\left(x\right)+L_{m+2}^{9k\left(m\right)+4}\left(x\right)+\hfill \\ \hfill +M_{m+2}^{9k\left(m\right)+5}\left(x\right)+L_{m+2}^{9k\left(m\right)+6}\left(x\right)+M_{m+2}^{9k\left(m\right)+7}\left(x\right)+R_{m+2}^{9k\left(m\right)+8}\left(x\right),\end{array}$$

(61)

$$\begin{array}{c}{f}_{m+3}\left(x\right)=L_{m+3}^{27k\left(m\right)}\left(x\right)+L_{m+3}^{27k\left(m\right)+1}\left(x\right)+M_{m+3}^{27k\left(m\right)+2}\left(x\right)+L_{m+3}^{9k\left(m\right)+3}\left(x\right)+\hfill \\ +L_{m+3}^{9k\left(m\right)+4}\left(x\right)+M_{m+3}^{9k\left(m\right)+5}\left(x\right)+L_{m+3}^{9k\left(m\right)+6}\left(x\right)+M_{m+3}^{9k\left(m\right)+7}\left(x\right)+R_{m+3}^{9k\left(m\right)+8}\left(x\right)+\\ +L_{m+3}^{27k\left(m\right)+9}\left(x\right)+L_{m+3}^{27k\left(m\right)+10}\left(x\right)+M_{m+3}^{27k\left(m\right)+11}\left(x\right)+L_{m+3}^{9k\left(m\right)+12}\left(x\right)+L_{m+3}^{9k\left(m\right)+13}\left(x\right)+\\ +M_{m+3}^{9k\left(m\right)+14}\left(x\right)+L_{m+3}^{9k\left(m\right)+15}\left(x\right)+M_{m+3}^{9k\left(m\right)+16}\left(x\right)+R_{m+3}^{9k\left(m\right)+17}\left(x\right)+L_{m+3}^{27k\left(m\right)+18}\left(x\right)+\\ +L_{m+3}^{27k\left(m\right)+19}\left(x\right)+M_{m+3}^{27k\left(m\right)+20}\left(x\right)+L_{m+3}^{9k\left(m\right)+21}\left(x\right)+M_{m+3}^{9k\left(m\right)+22}\left(x\right)+\\ \hfill +R_{m+3}^{9k\left(m\right)+23}\left(x\right)+M_{m+3}^{9k\left(m\right)+24}\left(x\right)+R_{m+3}^{9k\left(m\right)+25}\left(x\right)+R_{m+3}^{9k\left(m\right)+26}\left(x\right).\end{array}$$

(62)

Case $w=1,4,10,13,19.$$x_0\in [{\displaystyle \frac{27k\left(m\right)+w-1}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+w}{3^{m+2}}})$.

Then, $x_0+{\Delta }_{m+2}^1\in [{\displaystyle \frac{27k\left(m\right)+w}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+w+1}{3^{m+2}}}).$ In view of (11)–(13) and (58)–(62):

$$f_{m+2}(x_0+{\Delta }_{m+2}^1)-f_{m+2}\left(x_0\right)=1.$$

(63)

$$f_{m+1}(x_0+{\Delta }_{m+2}^1)-f_{m+1}\left(x_0\right)=0.$$

(64)

$$f_m(x_0+{\Delta }_{m+2}^1)-f_m\left(x_0\right)=0.$$

(65)

$$\forall r<0f_{m+r}(x_0+{\Delta }_{m+2}^1)-f_{m+r}\left(x_0\right)=0.$$

(66)

$$\forall r\ge 3f_{m+r}(x_0+{\Delta }_{m+2}^1)-f_{m+r}\left(x_0\right)=0.$$

(67)

In view of (14) and (63)–(67), we obtain:

$${\displaystyle \frac{f(x_0+{\Delta }_{m+2}^1)-f\left(x_0\right)}{{\Delta }_{m+2}^1}}={\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(68)

Case $w=2,5,11,14,20.$$x_0\in [{\displaystyle \frac{27k\left(m\right)+w-1}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+w}{3^{m+2}}})$.

Then, $x_0-{\Delta }_{m+2}^1\in [{\displaystyle \frac{27k\left(m\right)+w-2}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+w-1}{3^{m+2}}})$ and we obtain:

$${\displaystyle \frac{f(x_0-{\Delta }_{m+2}^1)-f\left(x_0\right)}{-{\Delta }_{m+2}^1}}={\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(69)

Case $w=3,12.$$x_0\in [{\displaystyle \frac{27k\left(m\right)+w-1}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+w}{3^{m+2}}})$.

Then $x_0+{\Delta }_{m+2}^3\in [{\displaystyle \frac{27k\left(m\right)+w+2}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+w+3}{3^{m+2}}})$, and we obtain:

$${\displaystyle \frac{f(x_0+{\Delta }_{m+2}^3)-f\left(x_0\right)}{{\Delta }_{m+2}^3}}={\displaystyle \frac{2}{3}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(70)

Case $w=6,15,21.$$x_0\in [{\displaystyle \frac{27k\left(m\right)+w-1}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+w}{3^{m+2}}})$.

Then, $x_0+{\Delta }_{m+2}^2\in [{\displaystyle \frac{27k\left(m\right)+w+1}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+w+2}{3^{m+2}}})$, and we obtain:

$${\displaystyle \frac{f(x_0+{\Delta }_{m+2}^2)-f\left(x_0\right)}{{\Delta }_{m+2}^2}}={\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(71)

Case $w=7,8,16,17,22.$$x_0\in [{\displaystyle \frac{27k\left(m\right)+w-1}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+w}{3^{m+2}}})$.

Then, $x_0-{\Delta }_{m+2}^2\in [{\displaystyle \frac{27k\left(m\right)+w-3}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+w-2}{3^{m+2}}})$, and we obtain:

$${\displaystyle \frac{f(x_0-{\Delta }_{m+2}^2)-f\left(x_0\right)}{-{\Delta }_{m+2}^2}}={\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(72)

Case $w=9.$$x_0\in [{\displaystyle \frac{27k\left(m\right)+8}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+9}{3^{m+2}}})$.

Then, $x_0+{\Delta }_{m+2}^9\in [{\displaystyle \frac{27k\left(m\right)+17}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+18}{3^{m+2}}})$, and we obtain:

$${\displaystyle \frac{f(x_0+{\Delta }_{m+2}^9)-f\left(x_0\right)}{{\Delta }_{m+2}^9}}={\displaystyle \frac{4}{9}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(73)

Case $w=18.$$x_0\in [{\displaystyle \frac{27k\left(m\right)+17}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+18}{3^{m+2}}})$.

Then, $x_0+{\Delta }_{m+2}^6\in [{\displaystyle \frac{27k\left(m\right)+23}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+24}{3^{m+2}}})$, and we obtain:

$${\displaystyle \frac{f(x_0+{\Delta }_{m+2}^6)-f\left(x_0\right)}{{\Delta }_{m+2}^6}}={\displaystyle \frac{1}{3}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(74)

Case $w=23,24.$$x_0\in [{\displaystyle \frac{27k\left(m\right)+w-1}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+w}{3^{m+2}}})$.

Then, $x_0+{\Delta }_{m+2}^2\in [{\displaystyle \frac{27k\left(m\right)+w+1}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+w+2}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0+{\Delta }_{m+2}^2)-f\left(x_0\right)}{{\Delta }_{m+2}^2}}=-{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(75)

Case $w=25.$$x_0\in [{\displaystyle \frac{27k\left(m\right)+24}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+25}{3^{m+2}}})$.

Then, $x_0-{\Delta }_{m+2}^2\in [{\displaystyle \frac{27k\left(m\right)+22}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+23}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0-{\Delta }_{m+2}^2)-f\left(x_0\right)}{-{\Delta }_{m+2}^2}}=-{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(76)

Case $w=26.$$x_0\in [{\displaystyle \frac{27k\left(m\right)+25}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+26}{3^{m+2}}})$.

Then, $x_0+{\Delta }_{m+2}^1\in [{\displaystyle \frac{27k\left(m\right)+26}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+27}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0+{\Delta }_{m+2}^1)-f\left(x_0\right)}{{\Delta }_{m+2}^1}}=-{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(77)

Case $w=27.$$x_0\in [{\displaystyle \frac{27k\left(m\right)+26}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+27}{3^{m+2}}})$.

Then, $x_0-{\Delta }_{m+2}^1\in [{\displaystyle \frac{27k\left(m\right)+25}{3^{m+2}}},{\displaystyle \frac{27k\left(m\right)+26}{3^{m+2}}}).$ Therefore:

$${\displaystyle \frac{f(x_0-{\Delta }_{m+2}^1)-f\left(x_0\right)}{-{\Delta }_{m+2}^1}}=-{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.$$

(78)

Let us assume that the function $f_m$ is determined on $[{\displaystyle \frac{3k\left(m\right)}{3^m}},{\displaystyle \frac{3k\left(m\right)+3}{3^m}})$ according to (6), so that:

$$f_m\left(x\right)=R_m^{k\left(m\right)}\left(x\right).$$

(79)

In view of (11)–(13), we obtain:

$$f_{m+r}\left(x\right)=const \mathrm{for} r\in \{-m+1,-m+2,\dots ,-1\},$$

(80)

$$f_{m+1}\left(x\right)=M_{m+1}^{3k\left(m\right)}\left(x\right)+R_{m+1}^{3k\left(m\right)+1}\left(x\right)+R_{m+1}^{3k\left(m\right)+2}\left(x\right),$$

(81)

$$\begin{array}{c}{f}_{m+2}\left(x\right)=L_{m+2}^{9k\left(m\right)}\left(x\right)+M_{m+2}^{9k\left(m\right)+1}\left(x\right)+R_{m+2}^{9k\left(m\right)+2}\left(x\right)+M_{m+2}^{9k\left(m\right)+3}\left(x\right)+R_{m+2}^{9k\left(m\right)+4}\left(x\right)+\hfill \\ \hfill +R_{m+2}^{9k\left(m\right)+5}\left(x\right)+M_{m+2}^{9k\left(m\right)+6}\left(x\right)+R_{m+2}^{9k\left(m\right)+7}\left(x\right)+R_{m+2}^{9k\left(m\right)+8}\left(x\right),\end{array}$$

(82)

$$\begin{array}{c}{f}_{m+3}\left(x\right)=L_{m+3}^{27k\left(m\right)}\left(x\right)+L_{m+3}^{27k\left(m\right)+1}\left(x\right)+M_{m+3}^{27k\left(m\right)+2}\left(x\right)+L_{m+3}^{9k\left(m\right)+3}\left(x\right)+\hfill \\ +M_{m+3}^{9k\left(m\right)+4}\left(x\right)+R_{m+3}^{9k\left(m\right)+5}\left(x\right)+M_{m+3}^{9k\left(m\right)+6}\left(x\right)+R_{m+3}^{9k\left(m\right)+7}\left(x\right)+R_{m+3}^{9k\left(m\right)+8}\left(x\right)+\\ +L_{m+3}^{27k\left(m\right)+9}\left(x\right)+M_{m+3}^{27k\left(m\right)+10}\left(x\right)+R_{m+3}^{27k\left(m\right)+11}\left(x\right)+M_{m+3}^{9k\left(m\right)+12}\left(x\right)+R_{m+3}^{9k\left(m\right)+13}\left(x\right)+\\ +R_{m+3}^{9k\left(m\right)+14}\left(x\right)+M_{m+3}^{9k\left(m\right)+15}\left(x\right)+R_{m+3}^{9k\left(m\right)+16}\left(x\right)+R_{m+3}^{9k\left(m\right)+17}\left(x\right)+L_{m+3}^{27k\left(m\right)+18}\left(x\right)+\\ +M_{m+3}^{27k\left(m\right)+19}\left(x\right)+R_{m+3}^{27k\left(m\right)+20}\left(x\right)+M_{m+3}^{9k\left(m\right)+21}\left(x\right)+R_{m+3}^{9k\left(m\right)+22}\left(x\right)+\\ \hfill +R_{m+3}^{9k\left(m\right)+23}\left(x\right)+M_{m+3}^{9k\left(m\right)+24}\left(x\right)+R_{m+3}^{9k\left(m\right)+25}\left(x\right)+R_{m+3}^{9k\left(m\right)+26}\left(x\right).\end{array}$$

(83)

In analogy with (63)–(78), we obtain:

$$\begin{array}{cc}\forall x_0\in [{\displaystyle \frac{3k\left(m\right)}{3^m}},{\displaystyle \frac{3k\left(m\right)+3}{3^m}})& \hspace{1em}\exists {\Delta }_{m+2}\in \{\pm {\Delta }_{m+2}^1,\pm {\Delta }_{m+2}^2,\pm {\Delta }_{m+2}^3,\\ & \pm {\Delta }_{m+2}^6,\pm {\Delta }_{m+2}^9\}\left|{\displaystyle \frac{f(x_0+{\Delta }_{m+2})-f\left(x_0\right)}{{\Delta }_{m+2}}}\right|\ge {\displaystyle \frac{1}{3}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.\end{array}$$

(84)

This follows from (11)–(14) and (79)–(83).

In view of (31)–(57), (68)–(78), and (84), we obtain:

$$\begin{array}{cc}\forall x_0\in (0,1)\forall m\exists {\Delta }_{m+2}\in & \{\pm {\Delta }_{m+2}^1,\pm {\Delta }_{m+2}^2,\pm {\Delta }_{m+2}^3,\\ & \hspace{1em}\pm {\Delta }_{m+2}^6,\pm {\Delta }_{m+2}^9\}\left|{\displaystyle \frac{f(x_0+{\Delta }_{m+2})-f\left(x_0\right)}{{\Delta }_{m+2}}}\right|\ge {\displaystyle \frac{1}{3}}{\left({\displaystyle \frac{3}{2}}\right)}^{m+2}.\end{array}$$

Then, $\forall x_0\in (0,1){lim}_{m\to \infty }{\Delta }_{m+2}=0$, but ${lim}_{m\to \infty }\left|{\displaystyle \frac{f(x_0+{\Delta }_{m+2})-f\left(x_0\right)}{{\Delta }_{m+2}}}\right|=+\infty$. The proof of the theorem is finished.

4. Conclusions

This paper introduces the construction of continuous and nowhere differentiable functions using step functions. These functions are simple combinations of the three blocks [0,1,0], [0,1,2], and [2,1,0]. This allows us to directly find sufficiently small $\Delta$ to obtain sufficiently large values of ${\displaystyle \frac{f(x_0+\Delta )-f\left(x_0\right)}{\Delta }}$. This effect can be used in a future paper to construct continuous and nowhere differentiable functions as an infinite series of continuous and nowhere differentiable functions.