Estimation of the Fractal Dimensions of the Linear Combination of Continuous Functions

Abstract

In the present paper, we try to estimate the fractal dimensions of the linear combination of continuous functions with different fractal dimensions. Initially, a general method to calculate the lower and the upper Box dimension of the sum of two continuous functions by classifying all the subsequences into different sets has been proposed. Further, we discuss the majority of possible cases of the sum of two continuous functions with different fractal dimensions and obtain their corresponding fractal dimensions estimation by using that general method. We prove that the linear combination of continuous functions having no Box dimension cannot keep the fractal dimensions closed. In this way, we have figured out how the fractal dimensions of the linear combination of continuous functions change with certain fractal dimensions.

1. Introduction

Let $C_I$ be the set of all continuous functions defined on the unit interval $I=[0,1]$. It is well known that $C_I$ is a complete metric space equipped with the following distance

$$\rho \left(f,g\right)=\underset{x\in I}{max}\left|f\left(x\right)-g\left(x\right)\right|.$$

Here $f\left(x\right),g\left(x\right)\in C_I$. From Baire Category Theorem, we know that a complete distance space must be the second-class set, so $C_I$ is not a sparse set. In fact, $C_I$ is composed of differentiable functions and continuous functions which are not differentiable at certain points in I. By [1,2], there even exist continuous functions differentiable nowhere on I such as the following Weierstrass function.

Example 1

([1,3]). The Weierstrass function

Let$0<\alpha <1,\lambda >4$. The Weierstrass function is defined as

$$W\left(x\right)=\sum _{j=1}^{\infty }{\lambda }^{-\alpha j}sin\left({\lambda }^{j}x\right).$$

From [1], the Weierstrass function is a typical example of a continuous function differentiable nowhere on I. By Theorem 2.3.7 in [2], the set of continuous functions in $C_I$, such as the Weierstrass function given above, which is differentiable nowhere, is non-empty. Furthermore, the coset of this set is the first-class set.

There are many examples such as the Weierstrass function that are continuous functions and differentiable nowhere on I. In [4,5], Self-affine curves and the corresponding fractal interpolation functions have been given. The Besicovitch function has been shown in [6,7]. Barnsley made research on the linear fractal interpolation function in [8]. The examples of continuous functions with the Box dimension two, which are differentiable nowhere on I, can be found in [9,10].

The continuous functions discussed above are usually called as fractal functions. From [11,12], we know they are unbounded variation functions and length of their graphs is infinite. Moreover, their most obvious feature is that they have the fractal dimensions larger than the topological dimension. For example, the Weierstrass function has the Box dimension $2-\alpha >1$ on I. The definition of the Box dimension of a continuous function has been given as follows.

Definition 1

([1]). Let a continuous function $f\left(x\right)$ be defined on a closed interval $[a,b](a<b)$ and

$$\mathrm{\Gamma }(f,[a,b\left]\right)=\left\{\right(x,f\left(x\right)):x\in [a,b\left]\right\}$$

be the graph of $f\left(x\right)$ on $[a,b]$. Let $N_{\delta }\mathrm{\Gamma }(f,[a,b])$ be the smallest number of sets of diameter at most δ which can cover $\mathrm{\Gamma }(f,[a,b\left]\right)$. The lower and the upper Box dimension of $\mathrm{\Gamma }(f,[a,b\left]\right)$ respectively are defined as

$${\underline{dim}}_B\mathrm{\Gamma }(f,[a,b])=\underset{\delta \to 0}{\underline{lim}}\frac{log{N}_{\delta }\mathrm{\Gamma }(f,[a,b])}{-log\delta }$$

(1)

and

$${\overline{dim}}_B\mathrm{\Gamma }(f,[a,b])=\underset{\delta \to 0}{\overline{lim}}\frac{log{N}_{\delta }\mathrm{\Gamma }(f,[a,b])}{-log\delta }.$$

(2)

If (1) and (2) are equal, we refer to the common value as the Box dimension of $\mathrm{\Gamma }(f,[a,b\left]\right)$

$${dim}_B\mathrm{\Gamma }(f,[a,b])=\underset{\delta \to 0}{lim}\frac{log{N}_{\delta }\mathrm{\Gamma }(f,[a,b])}{-log\delta }.$$

(3)

From Example 1 and Definition 1,

$${dim}_B\mathrm{\Gamma }(W,I)=2-\alpha .$$

Meanwhile, self–affine curves and the linear fractal interpolation function always have the Box dimension which are larger than one. In addition, more work about the Box dimension of fractal functions can be found in [13,14,15,16,17].

As is known to all, the linear combination of continuous functions can keep some properties closed, such as the differentiability and variation of functions. Therefore, we naturally think about how the fractal dimensions of the linear combination of continuous functions with certain fractal dimensions change. In [1], if two continuous functions have the different Box dimension, the Box dimension of the sum of these two continuous functions must be the larger one of them. That is, if

$${dim}_B\mathrm{\Gamma }(f,I)\ne {dim}_B\mathrm{\Gamma }(g,I),$$

then

$${dim}_B\mathrm{\Gamma }(f+g,I)=max\{{dim}_B\mathrm{\Gamma }(f,I),{dim}_B\mathrm{\Gamma }(g,I)\}$$

when both $f\left(x\right)$ and $g\left(x\right)$ are continuous on I. If two continuous functions have the same Box dimension, the corresponding conclusion can be found in [18]. Moreover, [6] tells us

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)={\overline{dim}}_B\mathrm{\Gamma }(f,I)$$

when

$${\overline{dim}}_B\mathrm{\Gamma }(g,I)<{\underline{dim}}_B\mathrm{\Gamma }(f,I).$$

(4)

However, ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ is unknown under the condition of (4). Furthermore, if

$${\overline{dim}}_B\mathrm{\Gamma }(g,I)\ge {\underline{dim}}_B\mathrm{\Gamma }(f,I),$$

the problems will become more complicated.

In this paper, we make further research on the fractal dimensions of the sum of two continuous functions with different fractal dimensions. The majority of possible situations have been summarized in Section 3. We divide the subjects into three broad categories to consider as follows:

(1)

Both of two continuous functions have the Box dimension;

(2)

One continuous function has the Box dimension but the other one does not have the Box dimension;

(3)

Neither of two continuous functions has the Box dimension.

In the first broad category, there exist two situations by whether the Box dimensions of these two functions are equal or not. Both of their conclusions have been shown above. Moreover, we discuss five and seven situations respectively in the second and third broad category by the relationship among the fractal dimensions of these two functions. Using the main method proposed in Section 2, we obtain their corresponding fractal dimensions estimation elementarily. We find that the fractal dimensions of the linear combination of continuous functions with certain fractal dimensions may be equal to arbitrary numbers belonging to a certain interval.

2. Preliminaries

In this section, we give certain basic notations and results. First of all, we put forward a general method to calculate ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ and ${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)$ in Section 2.1.

2.1. Main Method

Let

$${\Phi }_f\left(\delta \right)={\displaystyle \frac{log{N}_{\delta }\mathrm{\Gamma }(f,I)}{-log\delta }},f\left(x\right)\in C_I.$$

Here $0<\delta <\frac{1}{2}$. Then the lower Box dimension, the upper Box dimension and the Box dimension of $\mathrm{\Gamma }(f,I)$ can be respectively written as

${\underline{dim}}_B\mathrm{\Gamma }(f,I)=\underset{\delta \to 0}{\underline{lim}}{\Phi }_f\left(\delta \right)$, ${\overline{dim}}_B\mathrm{\Gamma }(f,I)=\underset{\delta \to 0}{\overline{lim}}{\Phi }_f\left(\delta \right)$ and ${dim}_B\mathrm{\Gamma }(f,I)=\underset{\delta \to 0}{lim}{\Phi }_f\left(\delta \right)$.

It is well known that $\underset{\delta \to 0}{lim}{\Phi }_f\left(\delta \right)$ may or may not exist. In fact, the number of the accumulation points of ${\Phi }_f\left(\delta \right)$ when $\delta \to 0$ is uncertain. For $f\left(x\right),g\left(x\right)\in C_I$, now we denote ${\Omega }_f={\left\{{\mu }_j\right\}}_{j\in J_1}$ as the set of all the accumulation points of ${\Phi }_f\left(\delta \right)$ when $\delta \to 0$ and ${\Omega }_g={\left\{{\nu }_j\right\}}_{j\in J_2}$ as the set of all the accumulation points of ${\Phi }_g\left(\delta \right)$ when $\delta \to 0$. Here, $J_1$ and $J_2$ are two index sets reflecting the number of elements in ${\Omega }_f$ and ${\Omega }_g$ respectively. At this time, we know

$${\underline{dim}}_B\mathrm{\Gamma }(f,I)=\underset{j\in J_1}{inf}\left\{{\mu }_j\right\},\hspace{1em}{\overline{dim}}_B\Gamma (f,I)=\underset{j\in J_1}{sup}\left\{{\mu }_j\right\}$$

and

$${\underline{dim}}_B\Gamma (g,I)=\underset{j\in J_2}{inf}\left\{{\nu }_j\right\},\hspace{1em}{\overline{dim}}_B\Gamma (g,I)=\underset{j\in J_2}{sup}\left\{{\nu }_j\right\}$$

For $\forall j\in J_1$, we denote ${\Delta }_j$ as the set of a subsequence ${\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }$ corresponding to ${\mu }_j$ which satisfies

$$\underset{k\to \infty }{lim}{\Phi }_f\left({\delta }_{l_k}^j\right)={\mu }_j,\forall {\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }\in {\Delta }_j.$$

Here $\underset{k\to \infty }{lim}{\delta }_{l_k}^j=0$.

Now we denote ${\alpha }_j$ and ${\beta }_j$ as the minimum and the maximum value in the following set

$$A_j=\left\{\underset{k\to \infty }{lim}{\Phi }_{f+g}\left({\delta }_{l_k}^j\right):{\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }\in {\Delta }_j\right\}$$

respectively. It is obvious that ${\displaystyle \bigcup _{j\in J_1}}{\Delta }_j$ contains all possible subsequences. In other words, ${\displaystyle \bigcup _{j\in J_1}}{A}_j$ contains all the accumulation points of ${\Phi }_{f+g}\left(\delta \right)$ when $\delta \to 0$. This means ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ and ${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)$ are the minimum and the maximum value in the set ${\displaystyle \bigcup _{j\in J_1}}{A}_j$, respectively. Hence, we can obtain a calculation of ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ and ${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)$, that is

$${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{j\in J_1}{inf}\left\{{\alpha }_j\right\}$$

(5)

and

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{j\in J_1}{sup}\left\{{\beta }_j\right\}.$$

(6)

From (5) and (6), we observe that the key work to calculate ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ and ${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)$ is to figure out the values of ${\alpha }_j$ and ${\beta }_j$. In preparation for the subsequent research, we first give several basic lemmas in the next subsection.

2.2. Basic Lemmas

Given a function $f\left(x\right)$ and an interval $[a,b]$, we write $R_f[a,b]$ for the maximum range of $f\left(x\right)$ over $[a,b]$ as

$$R_f[a,b]=\underset{a\le x,y\le b}{sup}|f\left(x\right)-f\left(y\right)|.$$

(7)

With (7), we can obtain the following lemma.

Lemma 1

([1]). Let $f\left(x\right)\in {\mathit{C}}_{\mathit{I}}$. Suppose that $0<\delta <\frac{1}{2}$, and n is the least integer greater than or equal to ${\delta }^{-1}$. The range of $N_{\delta }\mathrm{\Gamma }(f,I)$ can be estimated as

$$\sum _{i=0}^{n-1}max\left\{\frac{R_f\left[i\delta ,(i+1)\delta \right]}{\delta },1\right\}\le N_{\delta }\mathrm{\Gamma }(f,I)\le \sum _{i=0}^{n-1}\left\{2+\frac{R_f\left[i\delta ,(i+1)\delta \right]}{\delta }\right\}.$$

For $f\left(x\right),g\left(x\right)\in {\mathit{C}}_{\mathit{I}}$, by simple discussion, we can obtain the relationship among $R_f[a,b]$, $R_g[a,b]$ and $R_{f+g}[a,b]$, that is

$$\left|R_f\left[a,b\right]-R_g\left[a,b\right]\right|\le R_{f+g}\left[a,b\right]\le R_f\left[a,b\right]+R_g\left[a,b\right].$$

(8)

Then an estimation of $N_{\delta }\mathrm{\Gamma }(f+g,I)$ has been presented as Lemma 2 which reveals the relationship among $N_{\delta }\mathrm{\Gamma }(f,I)$, $N_{\delta }\mathrm{\Gamma }(g,I)$ and $N_{\delta }\mathrm{\Gamma }(f+g,I)$.

Lemma 2.

Let $f\left(x\right),g\left(x\right)\in {\mathit{C}}_{\mathit{I}}$. The range of $N_{\delta }\mathrm{\Gamma }(f+g,I)$ can be estimated as

$$\frac{1}{3}\left|N_{\delta }\mathrm{\Gamma }(f,I)-N_{\delta }\mathrm{\Gamma }(g,I)\right|\le N_{\delta }\mathrm{\Gamma }(f+g,I)\le 3N_{\delta }\mathrm{\Gamma }(f,I)+N_{\delta }\mathrm{\Gamma }(g,I).$$

Proof. 

On one hand, it follows from Lemma 1 that

$$\sum _{i=0}^{n-1}\frac{R_g\left[i\delta ,(i+1)\delta \right]}{\delta }\le N_{\delta }\mathrm{\Gamma }(g,I)$$

and

$$\sum _{i=0}^{n-1}\left\{2+\frac{R_f\left[i\delta ,(i+1)\delta \right]}{\delta }\right\}\le \sum _{i=0}^{n-1}3max\left\{\frac{R_f\left[i\delta ,(i+1)\delta \right]}{\delta },1\right\}\le 3N_{\delta }\mathrm{\Gamma }(f,I).$$

(9)

Combining (8), (9) and Lemma 1,

$$\begin{array}{cc}\hfill N_{\delta }\mathrm{\Gamma }(f+g,I)& \le \sum _{i=0}^{n-1}\left\{2+\frac{R_{f+g}\left[i\delta ,(i+1)\delta \right]}{\delta }\right\}\hfill \\ & \le \sum _{i=0}^{n-1}\left\{2+\frac{R_f\left[i\delta ,(i+1)\delta \right]}{\delta }\right\}+\sum _{i=0}^{n-1}\frac{R_g\left[i\delta ,(i+1)\delta \right]}{\delta }\hfill \\ & \le 3N_{\delta }\mathrm{\Gamma }(f,I)+N_{\delta }\mathrm{\Gamma }(g,I).\hfill \end{array}$$

On the other hand, it easily follows from (8) that

$$\sum _{i=0}^{n-1}{R}_{f+g}\left[i\delta ,(i+1)\delta \right]\ge \left|\sum _{i=0}^{n-1}{R}_f\left[i\delta ,(i+1)\delta \right]-\sum _{i=0}^{n-1}{R}_g\left[i\delta ,(i+1)\delta \right]\right|.$$

(10)

Combining (9), (10) and Lemma 1,

$$\begin{array}{cc}\hfill 3N_{\delta }\mathrm{\Gamma }(f+g,I)& \ge \sum _{i=0}^{n-1}\left\{2+\frac{R_{f+g}\left[i\delta ,(i+1)\delta \right]}{\delta }\right\}\hfill \\ & \ge 2n+\left|\sum _{i=0}^{n-1}\frac{R_f\left[i\delta ,(i+1)\delta \right]}{\delta }-\sum _{i=0}^{n-1}\frac{R_g\left[i\delta ,(i+1)\delta \right]}{\delta }\right|\hfill \\ & \ge \left|N_{\delta }\mathrm{\Gamma }(f,I)-N_{\delta }\mathrm{\Gamma }(g,I)\right|.\hfill \end{array}$$

That is

$$N_{\delta }\mathrm{\Gamma }(f+g,I)\ge \frac{1}{3}\left|N_{\delta }\mathrm{\Gamma }(f,I)-N_{\delta }\mathrm{\Gamma }(g,I)\right|.$$

This completes the proof of Lemma 2. □

Lemma 2 indicates that the value of $N_{\delta }\mathrm{\Gamma }(f+g,I)$ can be controlled by certain linear combinations of $N_{\delta }\mathrm{\Gamma }(f,I)$ and $N_{\delta }\mathrm{\Gamma }(g,I)$, which implies that ${\Phi }_{f+g}\left(\delta \right)$ seems to have some kind of connection with ${\Phi }_g\left(\delta \right)$ and ${\Phi }_g\left(\delta \right)$. To figure out the values of ${\alpha }_j$ and ${\beta }_j$, we first introduce the following essential lemma which shows a conclusion about sequences.

Lemma 3.

Let $f\left(x\right),g\left(x\right)\in C_I$. For any nonnegative sequence ${\left\{{\delta }_{l_k}\right\}}_{k=1}^{\infty }$ satisfying $\underset{k\to \infty }{lim}{\delta }_{l_k}=0$, it holds

$$\underset{k\to \infty }{\underline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}\right)=\underset{k\to \infty }{\underline{lim}}{\Phi }_g\left({\delta }_{l_k}\right) and \underset{k\to \infty }{\overline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}\right)=\underset{k\to \infty }{\overline{lim}}{\Phi }_g\left({\delta }_{l_k}\right)$$

when

$$\underset{k\to \infty }{\overline{lim}}{\Phi }_f\left({\delta }_{l_k}\right)<\underset{k\to \infty }{\underline{lim}}{\Phi }_g\left({\delta }_{l_k}\right).$$

Proof. 

This follows easily from Definition 1 and Lemma 2. So the proof is omitted. □

2.3. Two Elementary Results

With preparatory work, now we show two elementary results for the fractal dimensions of the sum of two continuous functions in this subsection. If the upper Box dimensions of two continuous functions are not equal, we have the first elementary result giving a calculation of the upper Box dimension of the sum of these two functions as below.

Theorem 1.

Let $f\left(x\right),g\left(x\right)\in {\mathit{C}}_{\mathit{I}}$. It holds

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)=max\left\{{\overline{dim}}_B\mathrm{\Gamma }(f,I),{\overline{dim}}_B\mathrm{\Gamma }(g,I)\right\}$$

when

$${\overline{dim}}_B\mathrm{\Gamma }(f,I)\ne {\overline{dim}}_B\mathrm{\Gamma }(g,I).$$

Proof. 

On one hand, it follows from Definition 1 and Lemma 2 that

$$\begin{array}{cc}\hfill {\overline{dim}}_B\mathrm{\Gamma }(f+g,I)& =\underset{\delta \to 0}{\overline{lim}}\frac{log{N}_{\delta }\mathrm{\Gamma }(f+g,I)}{-log\delta }\hfill \\ & \le \underset{\delta \to 0}{\overline{lim}}\frac{log\left(3N_{\delta }\mathrm{\Gamma }(f,I)+N_{\delta }\mathrm{\Gamma }(g,I)\right)}{-log\delta }\hfill \\ & \le \underset{\delta \to 0}{\overline{lim}}\frac{log4max\left\{N_{\delta }\mathrm{\Gamma }(f,I),N_{\delta }\mathrm{\Gamma }(g,I)\right\}}{-log\delta }\hfill \\ & =max\left\{\underset{\delta \to 0}{\overline{lim}}\frac{log{N}_{\delta }\mathrm{\Gamma }(f,I)}{-log\delta },\underset{\delta \to 0}{\overline{lim}}\frac{log{N}_{\delta }\mathrm{\Gamma }(g,I)}{-log\delta }\right\}.\hfill \end{array}$$

Thus

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)\le max\left\{{\overline{dim}}_B\mathrm{\Gamma }(f,I),{\overline{dim}}_B\mathrm{\Gamma }(g,I)\right\}.$$

(11)

On the other hand, if

$${\overline{dim}}_B\mathrm{\Gamma }(f,I)>{\overline{dim}}_B\mathrm{\Gamma }(g,I),$$

there must exist an index set $J_1^{\prime }\subset J_1$ such that

$$\underset{i\in J_2}{sup}\left\{{\nu }_i\right\}<\underset{j\in J_1^{\prime }}{inf}\left\{{\mu }_j\right\}\le \underset{j\in J_1^{\prime }}{sup}\left\{{\mu }_j\right\}={\overline{dim}}_B\mathrm{\Gamma }(f,I).$$

Thus for $\forall j\in J_1^{\prime }$,

$$\underset{k\to \infty }{\overline{lim}}{\Phi }_g\left({\delta }_{l_k}^j\right)\le \underset{i\in J_2}{sup}\left\{{\nu }_i\right\}<{\mu }_j=\underset{k\to \infty }{\underline{lim}}{\Phi }_f\left({\delta }_{l_k}^j\right),\forall {\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }\in {\Delta }_j.$$

Then it follows from Lemma 3 that for $\forall j\in J_1^{\prime }$,

$$\underset{k\to \infty }{\overline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}^j\right)=\underset{k\to \infty }{\overline{lim}}{\Phi }_f\left({\delta }_{l_k}^j\right)={\mu }_j,\forall {\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }\in {\Delta }_j.$$

This means ${\beta }_j={\mu }_j$ for $\forall j\in J_1^{\prime }$. From (6), we obtain

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{j\in J_1}{sup}\left\{{\beta }_j\right\}\ge \underset{j\in J_1^{\prime }}{sup}\left\{{\beta }_j\right\}=\underset{j\in J_1^{\prime }}{sup}\left\{{\mu }_j\right\}={\overline{dim}}_B\mathrm{\Gamma }(f,I).$$

If

$${\overline{dim}}_B\mathrm{\Gamma }(g,I)>{\overline{dim}}_B\mathrm{\Gamma }(f,I),$$

similarly, we can obtain

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)\ge {\overline{dim}}_B\mathrm{\Gamma }(g,I).$$

Thus

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)\ge max\left\{{\overline{dim}}_B\mathrm{\Gamma }(f,I),{\overline{dim}}_B\mathrm{\Gamma }(g,I)\right\}$$

(12)

when

$${\overline{dim}}_B\mathrm{\Gamma }(f,I)\ne {\overline{dim}}_B\mathrm{\Gamma }(g,I).$$

Therefore, we can obtain the conclusion of Theorem 1 by (11) and (12). □

Next, we prove the second elementary result for the lower Box dimension estimation of the sum of two continuous functions. If the upper Box dimension of one function is less than the lower Box dimension of another function, the following theorem tells us the conclusion.

Theorem 2.

Let $f\left(x\right),g\left(x\right)\in {\mathit{C}}_{\mathit{I}}$. It holds

$${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)={\underline{dim}}_B\mathrm{\Gamma }(f,I)$$

when

$${\overline{dim}}_B\mathrm{\Gamma }(g,I)<{\underline{dim}}_B\mathrm{\Gamma }(f,I).$$

Proof. 

In this situation, for $\forall j\in J_1$,

$$\underset{k\to \infty }{\overline{lim}}{\Phi }_g\left({\delta }_{l_k}^j\right)\le \underset{i\in J_2}{sup}\left\{{\nu }_i\right\}<\underset{i\in J_1}{inf}\left\{{\mu }_i\right\}\le \underset{k\to \infty }{\underline{lim}}{\Phi }_f\left({\delta }_{l_k}^j\right),\forall {\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }\in {\Delta }_j.$$

Then it follows from Lemma 3 that for $\forall j\in J_1$,

$$\underset{k\to \infty }{\underline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}^j\right)=\underset{k\to \infty }{\underline{lim}}{\Phi }_f\left({\delta }_{l_k}^j\right)={\mu }_j,\forall {\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }\in {\Delta }_j.$$

This means ${\alpha }_j={\mu }_j$ for $\forall j\in J_1$. From (5), we obtain

$${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{j\in J_1}{inf}\left\{{\alpha }_j\right\}=\underset{j\in J_1}{inf}\left\{{\mu }_j\right\}={\underline{dim}}_B\mathrm{\Gamma }(f,I).$$

This completes the proof of Theorem 2. □

From Theorems 1 and 2, we find the upper Box dimension estimation of the sum of two continuous functions has a ‘good’ conclusion. It implies that a continuous function with smaller upper Box dimension can be absorbed by another continuous function with larger upper Box dimension. However, results for the lower Box dimension estimation of the sum of two continuous functions are not so ‘good’ relatively.

Up to now, we can only figure out the case when the upper Box dimension of one function is less than the lower Box dimension of another function. Theorem 2 shows that the lower Box dimension of the sum of two continuous functions is equal to the maximum one of the lower Box dimensions of these two functions under the condition of (4). If the upper Box dimension of one function is no less than the lower Box dimension of another function, we will further discuss this situation in the next section.

3. The Linear Combination of Two Functions with Different Fractal Dimensions

Let $f\left(x\right),g\left(x\right)\in C_I$. For $a,b\in \mathbb{R}$, the linear combination of $f\left(x\right)$ and $g\left(x\right)$ can be written as

$$a·f\left(x\right)+b·g\left(x\right).$$

(13)

Thus the sum of $f\left(x\right)$ and $g\left(x\right)$ means $a=b=1$ in (13) as $f\left(x\right)+g\left(x\right)$. From Definition 1, we know if the Box dimension of $\mathrm{\Gamma }(f,I)$ does not exist, the upper and the lower Box dimension of $\mathrm{\Gamma }(f,I)$ must exist. For the convenience of discussion, we first give the definition of fractal functions set.

Let $F_s$ be the set of all continuous functions whose Box dimension exists and is equal to s on I when $1\le s\le 2$. That is, $F_s$ is the set of $s—$dimensional continuous functions on I. For example, the Weierstrass function $W\left(x\right)\in F_{2-\alpha }$. The functions constructed in [9,10] belong to $F_2$. In [19], we can find a special fractal function belonging to $F_1$. In fact, for $\forall s\in [1,2]$, $F_s$ is a non-empty set.

Let $F_{s_1}^{s_2}$ be the set of all continuous functions whose Box dimension does not exist on I. Here, $s_1,s_2$ are, respectively, the lower and the upper Box dimension of the function on I as $1\le s_1<s_2\le 2$. For example, the Besicovitch function $B\left(x\right)\in F_{s_1}^{s_2}$ for suitably chosen ${\left\{{\lambda }_j\right\}}_{j=1}^{\infty }$ [6]. In addition, for $\forall s_1,s_2$ satisfying $1\le s_1<s_2\le 2$, $F_{s_1}^{s_2}$ is a non-empty set, either.

3.1. $f\left(x\right)\in F_{s_1}$ and $g\left(x\right)\in F_{s_2}$

Since $f\left(x\right)\in F_{s_1}$ and $g\left(x\right)\in F_{s_2}$, both $f\left(x\right)$ and $g\left(x\right)$ have the Box dimension on I. At this time, we mainly discuss two situations. One is that $f\left(x\right)$ and $g\left(x\right)$ have the same Box dimension. The other condition is that $f\left(x\right)$ and $g\left(x\right)$ have the different Box dimension.

3.1.1. $f\left(x\right)\in F_{s_1},g\left(x\right)\in F_{s_2}$ for $s_1\ne s_2$

If $f\left(x\right)$ and $g\left(x\right)$ have the different Box dimension, we have the following assertion adopted from [1,18].

Theorem 3

([1,18]). Let $f\left(x\right)\in F_{s_1},g\left(x\right)\in F_{s_2}$ and $s_1\ne s_2$. Then

$${dim}_B\mathrm{\Gamma }(f+g,I)=max\{{dim}_B\mathrm{\Gamma }(f,I),{dim}_B\mathrm{\Gamma }(g,I)\}.$$

Theorem 3 tells us that the Box dimension of the sum of two continuous functions with the different Box dimension exists and is equal to the larger one of the Box dimensions of these two functions. This conclusion indicates that a continuous function with smaller Box dimension can be absorbed by another continuous function with bigger Box dimension.

3.1.2. $f\left(x\right),g\left(x\right)\in F_s$

If $f\left(x\right)$ and $g\left(x\right)$ have the same Box dimension s, we should discuss two situations by whether s is equal to one or not. If $s\ne 1$, we have the following assertion adopted from Ref. [18].

Theorem 4

([18]). Let $f\left(x\right),g\left(x\right)\in F_s$ and $1<s\le 2$.

(1) 

If the Box dimension of $\mathrm{\Gamma }(f+g,I)$ exists, it may be any number between one and s.

(2) 

If the Box dimension of $\mathrm{\Gamma }(f+g,I)$ does not exist,

$$1\le {\underline{dim}}_B\mathrm{\Gamma }(f+g,I)<{\overline{dim}}_B\mathrm{\Gamma }(f+g,I)<s.$$

(14)

Here, ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ and ${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)$ could be any numbers satisfying (14).

Theorem 4 shows that the Box dimension of the sum of two continuous functions with the same Box dimension s which is not equal to one may exist or not. Even if the Box dimension of the sum of these two functions exists, its value could be any number between one and s. If the Box dimension of the sum of these two functions does not exist, its lower and upper Box dimension could be any numbers between one and s.

If $s=1$, the conclusion given below holds trivially, which implies that $F_1$ is a linear space.

Theorem 5.

Let $f\left(x\right),g\left(x\right)\in F_1$. It holds

$${dim}_B\mathrm{\Gamma }(f+g,I)=1.$$

Proof. 

On one hand, by (11),

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)\le 1.$$

On the other hand, we know the lower Box dimension of any continuous functions is no less than one. That is

$${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)\ge 1.$$

Thus

$${dim}_B\mathrm{\Gamma }(f+g,I)=1.$$

□

3.2. $f\left(x\right)\in F_{s_1}^{s_2}$ and $g\left(x\right)\in F_s$

Since $f\left(x\right)\in F_{s_1}^{s_2}$ and $g\left(x\right)\in F_s$, the Box dimension of $\mathrm{\Gamma }(f,I)$ does not exist but the Box dimension of $\mathrm{\Gamma }(g,I)$ exists. At this time, we mainly discuss five situations according to the relationship among s, $s_1$ and $s_2$ as the following subsections.

3.2.1. $s_1<s_2<s$

It is obvious that

$${\overline{dim}}_B\mathrm{\Gamma }(f,I)\ne {\overline{dim}}_B\mathrm{\Gamma }(g,I)$$

and

$${\overline{dim}}_B\mathrm{\Gamma }(f,I)<{\underline{dim}}_B\mathrm{\Gamma }(g,I).$$

So we can directly acquire the conclusion that

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)=max\left\{{\overline{dim}}_B\mathrm{\Gamma }(f,I),{\overline{dim}}_B\mathrm{\Gamma }(g,I)\right\}=s$$

by Theorem 1 and

$${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)={\underline{dim}}_B\mathrm{\Gamma }(g,I)=s$$

by Theorem 2. It means

$${dim}_B\mathrm{\Gamma }(f+g,I)=s.$$

That is

$$f\left(x\right)+g\left(x\right)\in F_s.$$

3.2.2. $s<s_1<s_2$

Similar argument with that of Section 3.2.1,

$$f\left(x\right)+g\left(x\right)\in F_{s_1}^{s_2}.$$

3.2.3. $s_1<s<s_2$

From Theorem 1, we have

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)=max\left\{{\overline{dim}}_B\mathrm{\Gamma }(f,I),{\overline{dim}}_B\mathrm{\Gamma }(g,I)\right\}=s_2.$$

Now we make research on ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$. In this situation, there must exist two index sets denoted as $J_1^{\left(1\right)}$ and $J_1^{\left(2\right)}$ such that

$$\underset{j\in J_1^{\left(1\right)}}{inf}\left\{{\mu }_j\right\}\le \underset{j\in J_1^{\left(1\right)}}{sup}\left\{{\mu }_j\right\}<s\le \underset{j\in J_1^{\left(2\right)}}{inf}\left\{{\mu }_j\right\}\le \underset{j\in J_1^{\left(2\right)}}{sup}\left\{{\mu }_j\right\}$$

and

$$J_1^{\left(1\right)}\bigcup J_1^{\left(2\right)}=J_1.$$

For the convenience of discussion, we write $\underset{j\in J_1^{\left(2\right)}}{inf}\left\{{\mu }_j\right\}={\mu }_{j_{*}}$. Here $j_{*}\in J_1^{\left(2\right)}$. Since $s_1<s<s_2$, the element s may belong to ${\Omega }_f$ or not. In other words, ${\mu }_{j_{*}}$ may be equal to s or not. So we should discuss two cases as follows.

Case 1: ${\mu }_{j_{*}}\ne s$

From Lemma 3, we check every element ${\mu }_j$ in the set ${\Omega }_f$ and then obtain the following results.

(I) For $j\in J_1^{\left(1\right)}$, we know ${\mu }_j<s$. That is

$$\underset{k\to \infty }{\overline{lim}}{\Phi }_f\left({\delta }_{l_k}^j\right)={\mu }_j<s=\underset{k\to \infty }{\underline{lim}}{\Phi }_g\left({\delta }_{l_k}^j\right),\forall {\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }\in {\Delta }_j.$$

Thus

$$\underset{k\to \infty }{\underline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}^j\right)=\underset{k\to \infty }{\underline{lim}}{\Phi }_g\left({\delta }_{l_k}^j\right)=s,\forall {\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }\in {\Delta }_j.$$

This means ${\alpha }_j=s.$

(II) For $j\in J_1^{\left(2\right)}$, we know ${\mu }_j>s$. That is

$$\underset{k\to \infty }{\underline{lim}}{\Phi }_f\left({\delta }_{l_k}^j\right)={\mu }_j>s=\underset{k\to \infty }{\overline{lim}}{\Phi }_g\left({\delta }_{l_k}^j\right),\forall {\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }\in {\Delta }_j.$$

Thus

$$\underset{k\to \infty }{\underline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}^j\right)=\underset{k\to \infty }{\underline{lim}}{\Phi }_f\left({\delta }_{l_k}^j\right)={\mu }_j,\forall {\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }\in {\Delta }_j.$$

This means ${\alpha }_j={\mu }_j>s.$

So in this case, we can assert that

$${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{j\in J_1}{inf}\left\{{\alpha }_j\right\}=min\left\{\underset{j\in J_1^{\left(1\right)}}{inf}\left\{{\alpha }_j\right\},\underset{j\in J_1^{\left(2\right)}}{inf}\left\{{\alpha }_j\right\}\right\}=min\left\{s,\underset{j\in J_1^{\left(2\right)}}{inf}\left\{{\mu }_j\right\}\right\}=s.$$

Case 2: ${\mu }_{j_{*}}=s$

Similarly, we check every element ${\mu }_j$ in the set ${\Omega }_f$.

(I) For $j\in J_1\backslash \left\{j_{*}\right\}$, the result is the same with Case 1, that is

$${\alpha }_j=s,j\in J_1^{\left(1\right)},$$

and

$${\alpha }_j={\mu }_j>s,j\in J_1^{\left(2\right)}\backslash \left\{j_{*}\right\}.$$

(II) For $j=j_{*}$,

$$\underset{k\to \infty }{lim}{\Phi }_f\left({\delta }_{l_k}^{j_{*}}\right)=\underset{k\to \infty }{lim}{\Phi }_g\left({\delta }_{l_k}^{j_{*}}\right)=s,\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\Delta }_{j_{*}}.$$

Here $s\in \left(1,2\right)$. Now we choose any two possible functions $F\left(x\right)$ and $G\left(x\right)$ with the same Box dimension s, which means

$$\underset{k\to \infty }{lim}{\Phi }_F\left({\delta }_{l_k}^{j_{*}}\right)=\underset{k\to \infty }{lim}{\Phi }_G\left({\delta }_{l_k}^{j_{*}}\right)=s,\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\Delta }_{j_{*}}.$$

From Theorem 4, if Box dimension of $\mathrm{\Gamma }(F+G,I)$ exists, its value could be any number between one and s. In other words, $\underset{k\to \infty }{\underline{lim}}{\Phi }_{F+G}\left({\delta }_{l_k}^{j_{*}}\right)$ could be any number between one and s. If Box dimension of $\mathrm{\Gamma }(F+G,I)$ does not exist,

$$1\le {\underline{dim}}_B\mathrm{\Gamma }(F+G,I)<{\overline{dim}}_B\mathrm{\Gamma }(F+G,I)<s.$$

(15)

Here, ${\underline{dim}}_B\mathrm{\Gamma }(F+G,I)$ and ${\overline{dim}}_B\mathrm{\Gamma }(F+G,I)$ could be any numbers satisfying (15). Furthermore, we know

$$1\le {\underline{dim}}_B\mathrm{\Gamma }(F+G,I)\le \underset{k\to \infty }{\underline{lim}}{\Phi }_{F+G}\left({\delta }_{l_k}^{j_{*}}\right)\le {\overline{dim}}_B\mathrm{\Gamma }(F+G,I)<s.$$

From the arbitrariness of ${\underline{dim}}_B\mathrm{\Gamma }(F+G,I)$ and ${\overline{dim}}_B\mathrm{\Gamma }(F+G,I)$ satisfying (15),

$$1\le \underset{k\to \infty }{\underline{lim}}{\Phi }_{F+G}\left({\delta }_{l_k}^{j_{*}}\right)<s,\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\Delta }_{j_{*}}.$$

(16)

Here, $\underset{k\to \infty }{\underline{lim}}{\Phi }_{F+G}\left({\delta }_{l_k}^{j_{*}}\right)$ could be any number satisfying (16). Let $g\left(x\right)=G\left(x\right)$. Then we investigate the connection between $f\left(x\right)$ and $F\left(x\right)$.

It is obvious that for $\forall i\in J_1$,

$$\underset{k\to \infty }{lim}{\Phi }_F\left({\delta }_{l_k}^i\right)=s,\forall {\left\{{\delta }_{l_k}^i\right\}}_{k=1}^{\infty }\in {\Delta }_i.$$

For $j\in J_1$, now we define ${\Psi }_j$ as the set of $F_j$ satisfying

$$\underset{k\to \infty }{lim}{\Phi }_{F_j}\left({\delta }_{l_k}^i\right)=\left\{\begin{array}{cc}{\mu }_j\prime \hfill & i=j\hfill \\ s,\hfill & i\in J_1\backslash \left\{j\right\}\hfill \end{array}\right.\prime \forall F_j\in {\Psi }_j,\forall {\left\{{\delta }_{l_k}^i\right\}}_{k=1}^{\infty }\in {\Delta }_i.$$

For $\forall F_j\in {\Psi }_j$, we note that we only change the limitation of ${\Phi }_F\left({\delta }_{l_k}^j\right)$ from s to ${\mu }_j$ when ${\delta }_{l_k}^j\to 0$ for $\forall {\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }\in {\Delta }_j$. For the convenience of notation, we denote this transformation as

$$T_j:F\underset{s\to {\mu }_j}{\overset{{\Delta }_j}{\to }}{F}_j.$$

Write $T_j\odot F=F_j$. Then we can acquire a series of transformations ${\left\{T_j\right\}}_{j\in J_1}$. We find that ${\left\{T_j\right\}}_{j\in J_1}$ can be divided into three different categories in terms of different effects on F, which have been discussed as follows.

Transformation 1. For $j\in J_1^{\left(1\right)}$, since ${\mu }_j<s$, we observe that the only different result for $F_j$ from F is that

$$\underset{k\to \infty }{lim}{\Phi }_{F_j+g}\left({\delta }_{l_k}^j\right)=\underset{k\to \infty }{lim}{\Phi }_g\left({\delta }_{l_k}^j\right)=s,\forall F_j\in {\Psi }_j,\forall {\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }\in {\Delta }_j$$

by Lemma 3. But for other sets ${\Delta }_i(i\in J_1\backslash \left\{j\right\})$, the results for $F_j$ are the same as F. Specially for ${\Delta }_{j_{*}}$,

$$1\le \underset{k\to \infty }{\underline{lim}}{\Phi }_{F_j+g}\left({\delta }_{l_k}^{j_{*}}\right)<s,\forall F_j\in {\Psi }_j,\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\Delta }_{j_{*}}.$$

(17)

Here, $\underset{k\to \infty }{\underline{lim}}{\Phi }_{F_j+g}\left({\delta }_{l_k}^{j_{*}}\right)$ could be any number satisfying (17).

Transformation 2. For $j=j_{*}$, since ${\mu }_{j_{*}}=s$, the results for $F_{j_{*}}$ are the same as F. Specially for ${\Delta }_{j_{*}}$,

$$1\le \underset{k\to \infty }{\underline{lim}}{\Phi }_{F_{j_{*}}+g}\left({\delta }_{l_k}^{j_{*}}\right)<s,\forall F_{j_{*}}\in {\Psi }_{j_{*}},\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\Delta }_{j_{*}}.$$

(18)

Here, $\underset{k\to \infty }{\underline{lim}}{\Phi }_{F_{j_{*}}+g}\left({\delta }_{l_k}^{j_{*}}\right)$ could be any number satisfying (18).

Transformation 3. For $j\in J_1^{\left(2\right)}\backslash \left\{j_{*}\right\}$, since ${\mu }_j>s$, we observe that the only different result for $F_j$ from F is that

$$\underset{k\to \infty }{lim}{\Phi }_{F_j+g}\left({\delta }_{l_k}^j\right)=\underset{k\to \infty }{lim}{\Phi }_{F_j}\left({\delta }_{l_k}^j\right)={\mu }_j,\forall F_j\in {\Psi }_j,\forall {\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }\in {\Delta }_j$$

by Lemma 3. But for other sets ${\Delta }_i(i\in J_1\backslash \left\{j\right\})$, the results for $F_j$ are the same as F. Specially for ${\Delta }_{j_{*}}$,

$$1\le \underset{k\to \infty }{\underline{lim}}{\Phi }_{F_j+g}\left({\delta }_{l_k}^{j_{*}}\right)<s,\forall F_j\in {\Psi }_j,\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\Delta }_{j_{*}}.$$

(19)

Here, $\underset{k\to \infty }{\underline{lim}}{\Phi }_{F_j+g}\left({\delta }_{l_k}^{j_{*}}\right)$ could be any number satisfying (19).

Now we do all the transformations ${\left\{T_j\right\}}_{j\in J_1}$ on F denoted as $T_j{\displaystyle \underset{j\in J_1}{⨀}}F=F_{J_1}$. Define ${\Psi }_{J_1}$ as the set of $F_{J_1}$. From discussion above, we know for $\forall i\in J_1$,

$$\underset{k\to \infty }{lim}{\Phi }_{F_{J_1}}\left({\delta }_{l_k}^i\right)={\mu }_i,\forall F_{J_1}\in {\Psi }_{J_1},\forall {\left\{{\delta }_{l_k}^i\right\}}_{k=1}^{\infty }\in {\Delta }_i$$

(20)

and for ${\Delta }_{j_{*}}$,

$$1\le \underset{k\to \infty }{\underline{lim}}{\Phi }_{F_{J_1}+g}\left({\delta }_{l_k}^{j_{*}}\right)<s,\forall F_{J_1}\in {\Psi }_{J_1},\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\Delta }_{j_{*}}.$$

(21)

Here, $\underset{k\to \infty }{\underline{lim}}{\Phi }_{F_{J_1}+g}\left({\delta }_{l_k}^{j_{*}}\right)$ could be any number satisfying (21). From (20), we note that $f\in {\Psi }_{J_1}$. Let $F_{J_1}=f$. Thus for ${\Delta }_{j_{*}}$,

$$1\le \underset{k\to \infty }{\underline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}^{j_{*}}\right)<s,\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\Delta }_{j_{*}}.$$

(22)

Here, $\underset{k\to \infty }{\underline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}^{j_{*}}\right)$ could be any number satisfying (22). This means ${\alpha }_{j_{*}}$ could be any number satisfying $1\le {\alpha }_{j_{*}}<s$. So in this case, we can assert that

$${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{j\in J_1}{inf}\left\{{\alpha }_j\right\}=min\left\{s,{\alpha }_{j_{*}},\underset{j\in J_1^{\left(2\right)}\backslash \left\{j_{*}\right\}}{inf}\left\{{\mu }_j\right\}\right\}={\alpha }_{j_{*}},$$

which implies that ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ could be any number satisfying

$$1\le {\underline{dim}}_B\mathrm{\Gamma }(f+g,I)<s.$$

From discussion of Cases 1 and 2, we can come to the conclusion that ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ could be any number satisfying

$$1\le {\underline{dim}}_B\mathrm{\Gamma }(f+g,I)\le s.$$

So when $s_1<s<s_2$,

$$f\left(x\right)+g\left(x\right)\in F_v^{s_2}.$$

Here, v could be any number belonging to $[1,s]$.

3.2.4. $s_1<s=s_2$

It follows from (11) that

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)\le max\left\{{\overline{dim}}_B\mathrm{\Gamma }(f,I),{\overline{dim}}_B\mathrm{\Gamma }(g,I)\right\}=s.$$

Let $g\left(x\right)=-f\left(x\right)+H\left(x\right)$. If we suppose

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)<s,$$

we can obtain

$${\underline{dim}}_B\mathrm{\Gamma }(g,I)={\underline{dim}}_B\mathrm{\Gamma }(-f+H,I)={\underline{dim}}_B\mathrm{\Gamma }(f,I)=s$$

by Theorem 2. This is in contradiction with $s_1<s$. Thus

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)=s.$$

Then we investigate ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$. In this situation, we know

$$\underset{j\in J_1}{inf}\left\{{\mu }_j\right\}<\underset{j\in J_1}{sup}\left\{{\mu }_j\right\}=s.$$

For the convenience of discussion, we write $\underset{j\in J_1}{sup}\left\{{\mu }_j\right\}={\mu }_{j_{*}}$. Similar argument with that in Section 3.2.3, we can obtain the following results.

(I) For $j\in J_1\backslash \left\{j_{*}\right\}$, we have ${\alpha }_j=s$.

(II) For $j=j_{*}$,

$$\underset{k\to \infty }{lim}{\Phi }_f\left({\delta }_{l_k}^{j_{*}}\right)=\underset{k\to \infty }{lim}{\Phi }_g\left({\delta }_{l_k}^{j_{*}}\right)=s,\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\Delta }_{j_{*}}.$$

Here, $s\in (1,2]$. We note that Theorem 4 also holds for $s=2$. In the same way with Case 2 in Section 3.2.3, we can conclude that ${\alpha }_{j_{*}}$ could be any number satisfying $1\le {\alpha }_{j_{*}}<s$.

So we can assert that

$${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{j\in J_1}{inf}\left\{{\alpha }_j\right\}=min\left\{s,{\alpha }_{j_{*}}\right\}={\alpha }_{j_{*}},$$

which implies that ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ could be any number satisfying

$$1\le {\underline{dim}}_B\mathrm{\Gamma }(f+g,I)<s.$$

Hence,

$$f\left(x\right)+g\left(x\right)\in F_v^s.$$

Here, v could be any number belonging to $[1,s)$.

3.2.5. $s_1=s<s_2$

It follows from Theorem 1 that

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)=s_2.$$

For ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$, in this situation, we know

$$s=\underset{j\in J_1}{inf}\left\{{\mu }_j\right\}<\underset{j\in J_1}{sup}\left\{{\mu }_j\right\}.$$

Write $\underset{j\in J_1}{inf}\left\{{\mu }_j\right\}={\mu }_{j_{*}}$. Using a similar argument as that in SubSection 3.2.3, we can obtain the following results.

(I) For $j\in J_1\backslash \left\{j_{*}\right\}$, ${\alpha }_j={\mu }_j>s$.

(II) For $j=j_{*}$, we should discuss two cases according to whether s is equal to one or not.

Case 1: $s\ne 1$

In this case,

$$\underset{k\to \infty }{lim}{\Phi }_f\left({\delta }_{l_k}^{j_{*}}\right)=\underset{k\to \infty }{lim}{\Phi }_g\left({\delta }_{l_k}^{j_{*}}\right)=s,\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\Delta }_{j_{*}}.$$

Here, $s\in (1,2)$. In the same way with Case 2 in Section 3.2.3, we know ${\alpha }_{j_{*}}$ could be any number satisfying $1\le {\alpha }_{j_{*}}<s$.

So we can assert that

$${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{j\in J_1}{inf}\left\{{\alpha }_j\right\}=min\left\{{\alpha }_{j_{*}},\underset{j\in J_1\backslash \left\{j_{*}\right\}}{inf}\left\{{\mu }_j\right\}\right\}={\alpha }_{j_{*}},$$

which means ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ could be any number satisfying

$$1\le {\underline{dim}}_B\mathrm{\Gamma }(f+g,I)<s.$$

Thus

$$f\left(x\right)+g\left(x\right)\in F_v^{s_2}.$$

Here, v could be any number belonging to $[1,s)$.

Case 2: $s=1$

In this case,

$$\underset{k\to \infty }{lim}{\Phi }_f\left({\delta }_{l_k}^{j_{*}}\right)=\underset{k\to \infty }{lim}{\Phi }_g\left({\delta }_{l_k}^{j_{*}}\right)=1,\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\Delta }_{j_{*}}.$$

From Theorem 5 and in the same way with Case 2 in Section 3.2.3, we can obtain ${\alpha }_{j_{*}}=1$.

So we can assert that

$${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{j\in J_1}{inf}\left\{{\alpha }_j\right\}=min\left\{1,\underset{j\in J_1\backslash \left\{j_{*}\right\}}{inf}\left\{{\mu }_j\right\}\right\}=1.$$

Thus

$$f\left(x\right)+g\left(x\right)\in F_1^{s_2}.$$

3.3. $f\left(x\right)\in F_{s_1}^{s_2}$ and $g\left(x\right)\in F_{s_3}^{s_4}$

Since $f\left(x\right)\in F_{s_1}^{s_2}$ and $g\left(x\right)\in F_{s_3}^{s_4}$, neither $f\left(x\right)$ nor $g\left(x\right)$ has the Box dimension. Here, we denote $m_j$ and $M_j$ as the minimum and the maximum value, respectively, in the following set

$${\Sigma }_j=\left\{\underset{k\to \infty }{lim}{\Phi }_g\left({\delta }_{l_k}^j\right):{\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }\in {\Delta }_j\right\}.$$

At this time, we mainly discuss seven situations according to the relationship among $s_1$, $s_2$, $s_3$ and $s_4$ as the following subsections.

3.3.1. $s_1<s_2<s_3<s_4$

Similar discussion with that of Section 3.2.1,

$$f\left(x\right)+g\left(x\right)\in F_{s_3}^{s_4}.$$

3.3.2. $s_1<s_2=s_3<s_4$

It follows from Theorem 1 that

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)=s_4.$$

Then we investigate ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$. In this situation,

$$\underset{j\in J_1}{inf}\left\{{\mu }_j\right\}<\underset{j\in J_1}{sup}\left\{{\mu }_j\right\}=s_3.$$

Write $\underset{j\in J_1}{sup}\left\{{\mu }_j\right\}={\mu }_{j_{*}}$. Now we check every element ${\mu }_j$ in the set ${\Omega }_f$.

(I) For $j\in J_1\backslash \left\{j_{*}\right\}$, we know ${\mu }_j<{\underline{dim}}_B\mathrm{\Gamma }(g,I)$. That is

$$\underset{k\to \infty }{\overline{lim}}{\Phi }_f\left({\delta }_{l_k}^j\right)={\mu }_j<s_3\le \underset{k\to \infty }{\underline{lim}}{\Phi }_g\left({\delta }_{l_k}^j\right),\forall {\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }\in {\Delta }_j.$$

From Lemma 3,

$$\underset{k\to \infty }{\underline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}^j\right)=\underset{k\to \infty }{\underline{lim}}{\Phi }_g\left({\delta }_{l_k}^j\right),\forall {\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }\in {\Delta }_j.$$

(23)

This means ${\alpha }_j=m_j\ge s_3$.

(II) For $j=j_{*}$, we discuss two cases according to whether $s_3$ belongs to ${\Sigma }_{j_{*}}$ or not as follows.

Case 1: $s_3\notin {\Sigma }_{j_{*}}$

In this case,

$$\underset{k\to \infty }{\underline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}^{j_{*}}\right)=\underset{k\to \infty }{\underline{lim}}{\Phi }_g\left({\delta }_{l_k}^{j_{*}}\right),\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\Delta }_{j_{*}}.$$

Combining (23), we obtain

$${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{\delta \to 0}{\underline{lim}}{\Phi }_{f+g}\left(\delta \right)=\underset{\delta \to 0}{\underline{lim}}{\Phi }_g\left(\delta \right)={\underline{dim}}_B\mathrm{\Gamma }(g,I)=s_3.$$

Case 2: $s_3\in {\Sigma }_{j_{*}}$

In this case, there must exist a subset ${\tilde{\Delta }}_{j_{*}}\subseteq {\Delta }_{j_{*}}$ such that

$$\underset{k\to \infty }{\underline{lim}}{\Phi }_f\left({\delta }_{l_k}^{j_{*}}\right)=\underset{k\to \infty }{\underline{lim}}{\Phi }_g\left({\delta }_{l_k}^{j_{*}}\right)=s_3,\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\tilde{\Delta }}_{j_{*}}.$$

Similar discussion with that of Case 2 in Section 3.2.3, we know $\underset{k\to \infty }{\underline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}^{j_{*}}\right)$ could be any number satisfying

$$1\le \underset{k\to \infty }{\underline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}^{j_{*}}\right)<s_3,\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\tilde{\Delta }}_{j_{*}}.$$

(24)

Now we consider two situations as follows:

(i) When ${\tilde{\Delta }}_{j_{*}}={\Delta }_{j_{*}}$, we know ${\alpha }_{j_{*}}$ could be any number satisfying $1\le {\alpha }_{j_{*}}<s_3$. So in this situation, we can assert that

$${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{j\in J_1}{inf}\left\{{\alpha }_j\right\}=min\left\{\underset{j\in J_1\backslash \left\{j_{*}\right\}}{inf}\left\{m_j\right\},{\alpha }_{j_{*}}\right\}={\alpha }_{j_{*}},$$

which means ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ could be any number satisfying

$$1\le {\underline{dim}}_B\mathrm{\Gamma }(f+g,I)<s_3.$$

(ii) When ${\tilde{\Delta }}_{j_{*}}\subset {\Delta }_{j_{*}}$,

$$\underset{k\to \infty }{\overline{lim}}{\Phi }_f\left({\delta }_{l_k}^{j_{*}}\right)=s_3<\underset{k\to \infty }{\underline{lim}}{\Phi }_g\left({\delta }_{l_k}^{j_{*}}\right),\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\Delta }_{j_{*}}\backslash {\tilde{\Delta }}_{j_{*}}.$$

From Lemma 3,

$$\underset{k\to \infty }{\underline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}^{j_{*}}\right)=\underset{k\to \infty }{\underline{lim}}{\Phi }_g\left({\delta }_{l_k}^{j_{*}}\right)>s_3,\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\Delta }_{j_{*}}\backslash {\tilde{\Delta }}_{j_{*}}.$$

Combining (24), we obtain

$${\alpha }_{j_{*}}=\underset{k\to \infty }{\underline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}^{j_{*}}\right),\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\tilde{\Delta }}_{j_{*}},$$

which means ${\alpha }_{j_{*}}$ could be any number satisfying $1\le {\alpha }_{j_{*}}<s_3$. Hence, the result in this situation is the same with (i).

From discussion of Case 1 and 2, we can come to the conclusion that ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ could be any number satisfying

$$1\le {\underline{dim}}_B\mathrm{\Gamma }(f+g,I)\le s_3.$$

Thus

$$f\left(x\right)+g\left(x\right)\in F_v^{s_4}.$$

Here, v could be any number belonging to $[1,s_3]$.

3.3.3. $s_1<s_3<s_2<s_4$

It follows from Theorem 1 that

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)=s_4.$$

Then we investigate ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$. In this situation, there must exist two index sets denoted as $J_1^{\left(1\right)}$ and $J_1^{\left(2\right)}$ such that

$$\underset{j\in J_1^{\left(1\right)}}{inf}\left\{{\mu }_j\right\}\le \underset{j\in J_1^{\left(1\right)}}{sup}\left\{{\mu }_j\right\}<s_3\le \underset{j\in J_1^{\left(2\right)}}{inf}\left\{{\mu }_j\right\}\le \underset{j\in J_1^{\left(2\right)}}{sup}\left\{{\mu }_j\right\}<s_4$$

and

$$J_1^{\left(1\right)}\bigcup J_1^{\left(2\right)}=J_1.$$

Write $\underset{j\in J_1^{\left(2\right)}}{inf}\left\{{\mu }_j\right\}={\mu }_{j_{*}}$. Here, $j_{*}\in J_1^{\left(2\right)}$. Now we check every element ${\mu }_j$ in the set ${\Omega }_f$.

(I) For $j\in J_1^{\left(1\right)}$, we know ${\mu }_j<{\underline{dim}}_B\mathrm{\Gamma }(g,I)$. That is

$$\underset{k\to \infty }{\overline{lim}}{\Phi }_f\left({\delta }_{l_k}^j\right)={\mu }_j<s_3\le \underset{k\to \infty }{\underline{lim}}{\Phi }_g\left({\delta }_{l_k}^j\right),\forall {\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }\in {\Delta }_j.$$

From Lemma 3,

$$\underset{k\to \infty }{\underline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}^j\right)=\underset{k\to \infty }{\underline{lim}}{\Phi }_g\left({\delta }_{l_k}^j\right),\forall {\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }\in {\Delta }_j.$$

This means ${\alpha }_j=m_j$.

(II) For $j\in J_1^{\left(2\right)}$, we discuss two cases as follows.

Case 1: ${\mu }_j\notin {\Sigma }_j$ for $\forall j\in J_1^{\left(2\right)}$

In this case, we check every element ${\mu }_j(j\in J_1^{\left(2\right)})$ by similar discussion with that of s in Section 3.2.3. It is clear that ${\mu }_{j_{*}}$ may be equal to $s_3$ or not. So we should discuss two situations.

(i) When ${\mu }_{j_{*}}\ne s_3$, for $\forall j\in J_1^{\left(2\right)}$,

$$\underset{k\to \infty }{\underline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}^j\right)={\mu }_j,\forall {\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }\in {\Delta }_j.$$

This means ${\alpha }_j={\mu }_j$ for $\forall j\in J_1^{\left(2\right)}$. Combining discussion in (I), we can assert that

$${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{j\in J_1}{inf}\left\{{\alpha }_j\right\}=min\left\{\underset{j\in J_1^{\left(1\right)}}{inf}\left\{m_j\right\},\underset{j\in J_1^{\left(2\right)}}{inf}\left\{{\mu }_j\right\}\right\}=min\left\{\underset{j\in J_1^{\left(1\right)}}{inf}\left\{m_j\right\},{\mu }_{j_{*}}\right\}.$$

Therefore, we find that the value of ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ depends on different situations of accumulation points distribution of ${\Phi }_f\left(\delta \right)$ and ${\Phi }_g\left(\delta \right)$ when $\delta \to 0$. Since choices of $f\left(x\right)$ and $g\left(x\right)$ with different situations of which are arbitrary, it means $m_j$ could be any number satisfying

$$s_3\le m_j\le s_4,\forall j\in J_1^{\left(1\right)}$$

and ${\mu }_{j_{*}}$ could be any number satisfying

$$s_3<{\mu }_{j_{*}}\le s_2<s_4.$$

Hence, we can deduce that ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ could be any number satisfying

$$s_3\le {\underline{dim}}_B\mathrm{\Gamma }(f+g,I)\le s_2.$$

(ii) When ${\mu }_{j_{*}}=s_3$, the only result different from (i) is

$$\underset{k\to \infty }{\underline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}^{j_{*}}\right)=\underset{k\to \infty }{\underline{lim}}{\Phi }_g\left({\delta }_{l_k}^{j_{*}}\right),\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\Delta }_{j_{*}}.$$

This means ${\alpha }_{j_{*}}=m_{j_{*}}$. At this time,

$${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{j\in J_1}{inf}\left\{{\alpha }_j\right\}=min\left\{\underset{j\in J_1^{\left(1\right)}\bigcup \left\{j_{*}\right\}}{inf}\left\{m_j\right\},\underset{j\in J_1^{\left(2\right)}\backslash \left\{j_{*}\right\}}{inf}\left\{{\mu }_j\right\}\right\}.$$

Here, $m_j$ could be any number satisfying

$$m_{j_{*}}>s_3 \mathrm{and} s_3\le m_j\le s_4,\forall j\in J_1^{\left(1\right)}$$

.

Moreover, ${\mu }_j$ could be any number satisfying

$$s_3<{\mu }_j\le s_2<s_4.$$

Hence, we can deduce that ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ could be any number satisfying

$$s_3\le {\underline{dim}}_B\mathrm{\Gamma }(f+g,I)\le s_2.$$

Case 2: $\exists j\in J_1^{\left(2\right)}$ such that ${\mu }_j\in {\Sigma }_j$

In this case, we denote $J^{\prime }$ as the set of all the numbers $j\in J_1^{\left(2\right)}$ satisfying ${\mu }_j\in {\Sigma }_j$. It is obvious that $J^{\prime }$ is non empty. For $\forall j\in J^{\prime }$, there must exist a subset ${\tilde{\Delta }}_j\subseteq {\Delta }_j$ such that

$$\underset{k\to \infty }{\underline{lim}}{\Phi }_f\left({\delta }_{l_k}^j\right)=\underset{k\to \infty }{\underline{lim}}{\Phi }_g\left({\delta }_{l_k}^j\right)={\mu }_j,\forall {\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }\in {\tilde{\Delta }}_j.$$

Similar discussion with that of Case 2 in Section 3.3.2, we know ${\alpha }_j$ could be any number satisfying $1\le {\alpha }_j<{\mu }_j$. Combining the results obtained in (I) and Case 1, we can assert that

$${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{j\in J_1}{inf}\left\{{\alpha }_j\right\}=min\left\{\underset{j\in J_1^{\left(1\right)}}{inf}\left\{m_j\right\},min\left\{\underset{j\in J_1^{\left(2\right)}\backslash J^{\prime }}{inf}\left\{{\mu }_j\right\},\underset{j\in J^{\prime }}{inf}\left\{{\alpha }_j\right\}\right\}\right\}.$$

(25)

From (25), we find that the value of ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ also depends on different situations of the number of the same accumulation points of ${\Phi }_f\left(\delta \right)$ and ${\Phi }_g\left(\delta \right)$ by the same subsequence. Since choices of $f\left(x\right)$ and $g\left(x\right)$ with different situations of which are arbitrary, the numbers in $J^{\prime }$ are arbitrary belonging to $J_1^{\left(2\right)}$. Now we investigate all the possible values of ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$. There exist two situations we should discuss according to whether $j_{*}$ belongs to $J^{\prime }$ or not.

(i) When $j_{*}\in J^{\prime }$, we know ${\alpha }_{j_{*}}$ could be any number satisfying $1\le {\alpha }_{j_{*}}<{\mu }_{j_{*}}$. For $j\in J^{\prime }\backslash \left\{j_{*}\right\}$, ${\alpha }_j$ could be any number satisfying $1\le {\alpha }_j<{\mu }_j$. Since ${\mu }_j>{\mu }_{j_{*}}$, we can deduce that $\underset{j\in J^{\prime }}{inf}\left\{{\alpha }_j\right\}$ could be any number satisfying

$$1\le \underset{j\in J^{\prime }}{inf}\left\{{\alpha }_j\right\}<{\mu }_{j_{*}}.$$

Furthermore, we note that

$$\underset{j\in J_1^{\left(2\right)}\backslash J^{\prime }}{inf}\left\{{\mu }_j\right\}>{\mu }_{j_{*}},$$

which implies that $min\left\{\underset{j\in J_1^{\left(2\right)}\backslash J^{\prime }}{inf}\left\{{\mu }_j\right\},\underset{j\in J^{\prime }}{inf}\left\{{\alpha }_j\right\}\right\}$ could be any number satisfying

$$1\le min\left\{\underset{j\in J_1^{\left(2\right)}\backslash J^{\prime }}{inf}\left\{{\mu }_j\right\},\underset{j\in J^{\prime }}{inf}\left\{{\alpha }_j\right\}\right\}<{\mu }_{j_{*}}.$$

Moreover, observe that $m_j$ could be any number satisfying

$$s_3\le m_j\le s_4,\forall j\in J_1^{\left(1\right)}$$

and ${\mu }_{j_{*}}$ could be any number satisfying

$$s_3<{\mu }_{j_{*}}\le s_2<s_4.$$

Combining (25) and the above discussion, we can conclude that ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ could be any number satisfying

$$1\le {\underline{dim}}_B\mathrm{\Gamma }(f+g,I)<s_2.$$

(ii) When $j_{*}\notin J^{\prime }$, we know ${\alpha }_{j_{*}}={\mu }_{j_{*}}$. This means

$$\underset{j\in J_1^{\left(2\right)}\backslash J^{\prime }}{inf}\left\{{\mu }_j\right\}={\mu }_{j_{*}}.$$

For $j\in J^{\prime }$, ${\alpha }_j$ could be any number satisfying $1\le {\alpha }_j<{\mu }_j$. Since ${\mu }_j>{\mu }_{j_{*}}$, we can deduce that $min\left\{\underset{j\in J_1^{\left(2\right)}\backslash J^{\prime }}{inf}\left\{{\mu }_j\right\},\underset{j\in J^{\prime }}{inf}\left\{{\alpha }_j\right\}\right\}$ could be any number satisfying

$$1\le min\left\{\underset{j\in J_1^{\left(2\right)}\backslash J^{\prime }}{inf}\left\{{\mu }_j\right\},\underset{j\in J^{\prime }}{inf}\left\{{\alpha }_j\right\}\right\}\le {\mu }_{j_{*}}.$$

Then similar with (i), we can obtain the result that ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ could be any number satisfying

$$1\le {\underline{dim}}_B\mathrm{\Gamma }(f+g,I)\le s_2.$$

From discussion of Cases 1 and 2, we can come to the conclusion that ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ could be any number satisfying

$$1\le {\underline{dim}}_B\mathrm{\Gamma }(f+g,I)\le s_2.$$

Thus

$$f\left(x\right)+g\left(x\right)\in F_v^{s_4}.$$

Here, v could be any number belonging to $[1,s_2]$.

3.3.4. $s_3<s_1<s_2<s_4$

It follows from Theorem 1 that

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)=s_4.$$

Then we investigate ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$. In this situation,

$$s_3<\underset{j\in J_1}{inf}\left\{{\mu }_j\right\}<\underset{j\in J_1}{sup}\left\{{\mu }_j\right\}<s_4.$$

Write $\underset{j\in J_1}{inf}\left\{{\mu }_j\right\}={\mu }_{j_{*}}$. Similarly, we discuss two cases as follows:

Case 1: ${\mu }_j\notin {\Sigma }_j$ for $\forall j\in J_1$

Similar argument with Case 1 in Section 3.3.3, we know for $\forall j\in J_1$,

$$\underset{k\to \infty }{\underline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}^j\right)={\mu }_j,\forall {\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }\in {\Delta }_j.$$

This means ${\alpha }_j={\mu }_j$ for $\forall j\in J_1$. Thus

$${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{j\in J_1}{inf}\left\{{\alpha }_j\right\}=\underset{j\in J_1}{inf}\left\{{\mu }_j\right\}={\underline{dim}}_B\mathrm{\Gamma }(f,I)=s_1.$$

Case 2: $\exists j\in J_1$ such that ${\mu }_j\in {\Sigma }_j$

Similar argument with Case 2 in Section 3.3.3, we know $\underset{j\in J_1}{inf}\left\{{\alpha }_j\right\}$ could be any number satisfying

$$1\le \underset{j\in J_1}{inf}\left\{{\alpha }_j\right\}\le {\mu }_{j_{*}}=s_1,$$

which means ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ could be any number satisfying

$$1\le {\underline{dim}}_B\mathrm{\Gamma }(f+g,I)\le s_1.$$

From discussion of Cases 1 and 2, we can come to the conclusion that

$$f\left(x\right)+g\left(x\right)\in F_v^{s_4}.$$

Here, v could be any number belonging to $[1,s_1]$.

3.3.5. $s_3=s_1<s_2<s_4$

It follows from Theorem 1 that

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)=s_4.$$

Then we investigate ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$. In this situation,

$$s_3=\underset{j\in J_1}{inf}\left\{{\mu }_j\right\}<\underset{j\in J_1}{sup}\left\{{\mu }_j\right\}<s_4.$$

Write $\underset{j\in J_1}{inf}\left\{{\mu }_j\right\}={\mu }_{j_{*}}$. Similarly, we discuss two cases as follows:

Case 1: ${\mu }_j\notin {\Sigma }_j$ for $\forall j\in J_1$

Similar argument with Case 1 in Section 3.3.3, we know

$$\underset{k\to \infty }{\underline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}^{j_{*}}\right)=\underset{k\to \infty }{\underline{lim}}{\Phi }_g\left({\delta }_{l_k}^{j_{*}}\right),\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\Delta }_{j_{*}}$$

and for $\forall j\in J_1\backslash \left\{j_{*}\right\}$,

$$\underset{k\to \infty }{\underline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}^j\right)={\mu }_j,\forall {\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }\in {\Delta }_j.$$

This means ${\alpha }_{j_{*}}=m_{j_{*}}$ and ${\alpha }_j={\mu }_j$ for $\forall j\in J_1\backslash \left\{j_{*}\right\}$. At this time,

$${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{j\in J_1}{inf}\left\{{\alpha }_j\right\}=min\left\{m_{j_{*}},\underset{j\in J_1\backslash \left\{j_{*}\right\}}{inf}\left\{{\mu }_j\right\}\right\}.$$

Here, we note that $m_{j_{*}}$ could be any number satisfying

$$s_3<m_{j_{*}}\le s_4$$

and $\underset{j\in J_1\backslash \left\{j_{*}\right\}}{inf}\left\{{\mu }_j\right\}$ could be any number satisfying

$$s_3<\underset{j\in J_1\backslash \left\{j_{*}\right\}}{inf}\left\{{\mu }_j\right\}\le s_2<s_4.$$

Hence, we can deduce that ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ could be any number satisfying

$$s_3<{\underline{dim}}_B\mathrm{\Gamma }(f+g,I)\le s_2.$$

Case 2: $\exists j\in J_1$ such that ${\mu }_j\in {\Sigma }_j$

Similar argument with Case 2 in Section 3.3.3, we can obtain the result that $\underset{j\in J_1}{inf}\left\{{\alpha }_j\right\}$ could be any number satisfying

$$1\le \underset{j\in J_1}{inf}\left\{{\alpha }_j\right\}<\underset{j\in J_1\backslash \left\{j_{*}\right\}}{inf}\left\{{\mu }_j\right\}\le s_2,$$

which means ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ could be any number satisfying

$$1\le {\underline{dim}}_B\mathrm{\Gamma }(f+g,I)<s_2.$$

From discussion of Cases 1 and 2, we can come to the conclusion that

$$f\left(x\right)+g\left(x\right)\in F_v^{s_4}.$$

Here, v could be any number belonging to $[1,s_2]$.

3.3.6. $s_1<s_3<s_2=s_4$

In this situation, there must exist two index sets denoted as $J_1^{\left(1\right)}$ and $J_1^{\left(2\right)}$ such that

$$\underset{j\in J_1^{\left(1\right)}}{inf}\left\{{\mu }_j\right\}\le \underset{j\in J_1^{\left(1\right)}}{sup}\left\{{\mu }_j\right\}<s_3\le \underset{j\in J_1^{\left(2\right)}}{inf}\left\{{\mu }_j\right\}\le \underset{j\in J_1^{\left(2\right)}}{sup}\left\{{\mu }_j\right\}=s_4$$

and

$$J_1^{\left(1\right)}\bigcup J_1^{\left(2\right)}=J_1.$$

We note that the only difference between this situation and that in Section 3.3.3 is that $\underset{j\in J_1^{\left(2\right)}}{sup}\left\{{\mu }_j\right\}=s_4$. Similar argument with Section 3.3.3, we can obtain the same conclusion as Section 3.3.3 that ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ could be any number satisfying

$$1\le {\underline{dim}}_B\mathrm{\Gamma }(f+g,I)\le s_2.$$

Then we investigate ${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)$. From (11),

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)\le s_2.$$

(26)

Write $\underset{j\in J_1^{\left(2\right)}}{sup}\left\{{\mu }_j\right\}={\mu }_{j_{*}}$. Here, $j_{*}\in J_1^{\left(2\right)}$. Now we discuss two cases according to whether ${\mu }_{j_{*}}$ belongs to ${\Sigma }_{j_{*}}$ or not as follows.

Case 1: ${\mu }_{j_{*}}\notin {\Sigma }_{j_{*}}$

In this case, it follows from Lemma 3 that

$$\underset{k\to \infty }{\overline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}^{j_{*}}\right)={\mu }_{j_{*}}=s_2,\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\Delta }_{j_{*}}.$$

This means ${\beta }_{j_{*}}=s_2$. Thus

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{j\in J_1}{sup}\left\{{\beta }_j\right\}\ge {\beta }_{j_{*}}=s_2.$$

Combining (26), we obtain

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)=s_2.$$

Case 2: ${\mu }_{j_{*}}\in {\Sigma }_{j_{*}}$

In this case, we check every element ${\mu }_j$ in the set ${\Omega }_f$ as follows.

(I) For $j\in J_1^{\left(1\right)}$, we know ${\mu }_j<{\underline{dim}}_B\mathrm{\Gamma }(g,I)$. From Lemma 3,

$$\underset{k\to \infty }{\overline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}^j\right)=\underset{k\to \infty }{\overline{lim}}{\Phi }_g\left({\delta }_{l_k}^j\right),\forall {\left\{{\delta }_{l_k}^j\right\}}_{k=1}^{\infty }\in {\Delta }_j.$$

This means ${\beta }_j=M_j$. Here, $M_j$ could be any number satisfying $s_3\le M_j\le s_2$.

(II) For $j=j_{*}$, there must exist a subset ${\tilde{\Delta }}_{j_{*}}\subseteq {\Delta }_{j_{*}}$ such that

$$\underset{k\to \infty }{lim}{\Phi }_f\left({\delta }_{l_k}^{j_{*}}\right)=\underset{k\to \infty }{lim}{\Phi }_g\left({\delta }_{l_k}^{j_{*}}\right)=s_2,\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\tilde{\Delta }}_{j_{*}}.$$

Here, $s_2\in (1,2]$. Now we choose any two possible functions $F\left(x\right)$ and $G\left(x\right)$ with the same Box dimension $s_2$, which means

$$\underset{k\to \infty }{lim}{\Phi }_F\left({\delta }_{l_k}^{j_{*}}\right)=\underset{k\to \infty }{lim}{\Phi }_G\left({\delta }_{l_k}^{j_{*}}\right)=s_2,\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\tilde{\Delta }}_{j_{*}}.$$

From Theorem 4, if the Box dimension of $\mathrm{\Gamma }(F+G,I)$ exists, its value could be any number between one and $s_2$. In other words, $\underset{k\to \infty }{lim}{\Phi }_{F+G}\left({\delta }_{l_k}^{j_{*}}\right)$ exists and its value could be any number between one and $s_2$. If the Box dimension of $\mathrm{\Gamma }(F+G,I)$ does not exist,

$$1\le {\underline{dim}}_B\mathrm{\Gamma }(F+G,I)<{\overline{dim}}_B\mathrm{\Gamma }(F+G,I)<s_2.$$

(27)

Here, ${\underline{dim}}_B\mathrm{\Gamma }(F+G,I)$ and ${\overline{dim}}_B\mathrm{\Gamma }(F+G,I)$ could be any numbers satisfying (27). Also, we know

$$1\le {\underline{dim}}_B\mathrm{\Gamma }(F+G,I)\le \underset{k\to \infty }{\overline{lim}}{\Phi }_{F+G}\left({\delta }_{l_k}^{j_{*}}\right)\le {\overline{dim}}_B\mathrm{\Gamma }(F+G,I)<s_2.$$

From arbitrariness of ${\underline{dim}}_B\mathrm{\Gamma }(F+G,I)$ and ${\overline{dim}}_B\mathrm{\Gamma }(F+G,I)$ satisfying (27),

$$1\le \underset{k\to \infty }{\overline{lim}}{\Phi }_{F+G}\left({\delta }_{l_k}^{j_{*}}\right)<s_2,\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\tilde{\Delta }}_{j_{*}}.$$

(28)

Here, $\underset{k\to \infty }{\overline{lim}}{\Phi }_{F+G}\left({\delta }_{l_k}^{j_{*}}\right)$ could be any number satisfying (28). Similar argument with Case 2 in Section 3.2.3, we can also obtain the similar result with (22) that

$$1\le \underset{k\to \infty }{\overline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}^{j_{*}}\right)<s_2,\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\tilde{\Delta }}_{j_{*}}.$$

(29)

Here, $\underset{k\to \infty }{\overline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}^{j_{*}}\right)$ could be any number satisfying (29). Similar discussion with Case 2 in Section 3.3.2, now we consider two situations as follows:

(i) When ${\tilde{\Delta }}_{j_{*}}={\Delta }_{j_{*}}$, ${\beta }_{j_{*}}$ could be any number satisfying $1\le {\beta }_{j_{*}}<s_2$.

(ii) When ${\tilde{\Delta }}_{j_{*}}\subset {\Delta }_{j_{*}}$,

$$\underset{k\to \infty }{\overline{lim}}{\Phi }_g\left({\delta }_{l_k}^{j_{*}}\right)<s_2=\underset{k\to \infty }{\underline{lim}}{\Phi }_f\left({\delta }_{l_k}^{j_{*}}\right),\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\Delta }_{j_{*}}\backslash {\tilde{\Delta }}_{j_{*}}.$$

From Lemma 3,

$$\underset{k\to \infty }{\overline{lim}}{\Phi }_{f+g}\left({\delta }_{l_k}^{j_{*}}\right)=\underset{k\to \infty }{\overline{lim}}{\Phi }_f\left({\delta }_{l_k}^{j_{*}}\right)=s_2,\forall {\left\{{\delta }_{l_k}^{j_{*}}\right\}}_{k=1}^{\infty }\in {\Delta }_{j_{*}}\backslash {\tilde{\Delta }}_{j_{*}}.$$

Combining (29), we obtain ${\beta }_{j_{*}}=s_2$.

From (i) and (ii), we know ${\beta }_{j_{*}}$ could be any number satisfying $1\le {\beta }_{j_{*}}\le s_2$.

(III) For $j\in J_1^{\left(2\right)}\backslash \left\{j_{*}\right\}$, ${\mu }_j$ may belong to ${\Sigma }_j$ or not. Similar with the previous discussion, we can obtain the result that ${\beta }_j$ could be any number satisfying $1\le {\beta }_j\le {\mu }_j$ or ${\beta }_j=M_j$. Here, $M_j$ could be any number satisfying ${\mu }_j<M_j\le s_2$. This means ${\beta }_j$ could be any number satisfying $1\le {\beta }_j\le s_2$.

Hence, combining (I), (II) and (III), we can assert that

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{j\in J_1}{sup}\left\{{\beta }_j\right\}=max\left\{\underset{j\in J_1^{\left(1\right)}}{sup}\left\{M_j\right\},\underset{j\in J_1^{\left(2\right)}}{sup}\left\{{\beta }_j\right\}\right\}.$$

From all the possible values of $M_j(j\in J_1^{\left(1\right)})$ and ${\beta }_j(j\in J_1^{\left(2\right)})$ given above, we can conclude that ${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)$ could be any number satisfying

$$s_3\le {\overline{dim}}_B\mathrm{\Gamma }(f+g,I)\le s_2.$$

From discussion above, we observe that the Box dimension of $\mathrm{\Gamma }(f+g,I)$ may exist or not. Here, we explore the possibility of the situation when the Box dimension of $\mathrm{\Gamma }(f+g,I)$ exists. We give several particular situations below to illustrate that ${dim}_B\mathrm{\Gamma }(f+g,I)$ could be any number belonging to $[s_3,s_2]$.

Situation I We choose two functions $f\left(x\right)\in F_{s_1}^{s_2}$ and $g\left(x\right)\in F_{s_3}^{s_4}$$(s_1<s_3<s_2=s_4)$ satisfying the following conditions:

(1)

${\mu }_1=s_1$ and ${\mu }_2=s_2$ are the only two elements in the set ${\Omega }_f$;

(2)

$s_4$ is the only one element in the set ${\Sigma }_1$;

(3)

$s_3$ is the only one element in the set ${\Sigma }_2$.

In this situation, we note that

$${\alpha }_1={\beta }_1={\alpha }_2={\beta }_2=s_2.$$

Thus

$${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{j\in J_1}{inf}\left\{{\alpha }_j\right\}=min\left\{{\alpha }_1,{\alpha }_2\right\}=s_2$$

and

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{j\in J_1}{sup}\left\{{\beta }_j\right\}=max\left\{{\beta }_1,{\beta }_2\right\}=s_2.$$

That is

$${dim}_B\mathrm{\Gamma }(f+g,I)=s_2.$$

Situation II We choose two functions $f\left(x\right)\in F_{s_1}^{s_2}$ and $g\left(x\right)\in F_{s_3}^{s_4}$$(s_1<s_3<s_2=s_4)$ satisfying the following conditions:

(1)

Three are only three elements ${\mu }_1$, ${\mu }_2$, ${\mu }_3$ in the set ${\Omega }_f$. Here, ${\mu }_1=s_1$, ${\mu }_3=s_2$ and ${\mu }_2$ could be any number belonging to $(s_3,s_2)$;

(2)

${\mu }_2$ is the only one element in the set ${\Sigma }_1$;

(3)

$s_3$ is the only one element in the set ${\Sigma }_2$;

(4)

$s_4$ is the only one element in the set ${\Sigma }_3$.

In this situation, it is obvious

$${\alpha }_1={\beta }_1={\alpha }_2={\beta }_2={\mu }_2.$$

From the previous discussion, we know $\underset{k\to \infty }{lim}{\Phi }_{f+g}\left({\delta }_{l_k}^3\right)$ may exist and its value could be any number between one and $s_2$, which means its value could be equal to ${\mu }_2$. Let

$$\underset{k\to \infty }{lim}{\Phi }_{f+g}\left({\delta }_{l_k}^3\right)={\mu }_2,\forall {\left\{{\delta }_{l_k}^3\right\}}_{k=1}^{\infty }\in {\Delta }_3.$$

That is

$${\alpha }_3={\beta }_3={\mu }_2.$$

Thus

$${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{j\in J_1}{inf}\left\{{\alpha }_j\right\}=min\left\{{\alpha }_1,{\alpha }_2,{\alpha }_3\right\}={\mu }_2$$

and

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{j\in J_1}{sup}\left\{{\beta }_j\right\}=max\left\{{\beta }_1,{\beta }_2,{\beta }_3\right\}={\mu }_2.$$

So

$${dim}_B\mathrm{\Gamma }(f+g,I)={\mu }_2,$$

which means ${dim}_B\mathrm{\Gamma }(f+g,I)$ could be any number belonging to $(s_3,s_2)$.

Situation III We choose two functions $f\left(x\right)\in F_{s_1}^{s_2}$ and $g\left(x\right)\in F_{s_3}^{s_4}$$(s_1<s_3<s_2=s_4)$ satisfying the following conditions:

(1)

${\mu }_1=s_1$ and ${\mu }_2=s_2$ are the only two elements in the set ${\Omega }_f$;

(2)

$s_3$ is the only one element in the set ${\Sigma }_1$;

(3)

$s_4$ is the only one element in the set ${\Sigma }_2$.

In this situation, it is obvious

$${\alpha }_1={\beta }_1=s_3.$$

From the previous discussion, we know $\underset{k\to \infty }{lim}{\Phi }_{f+g}\left({\delta }_{l_k}^2\right)$ may exist and its value could be any number between one and $s_2$, which means its value could be equal to $s_3$. Let

$$\underset{k\to \infty }{lim}{\Phi }_{f+g}\left({\delta }_{l_k}^2\right)=s_3,\forall {\left\{{\delta }_{l_k}^2\right\}}_{k=1}^{\infty }\in {\Delta }_2.$$

That is

$${\alpha }_2={\beta }_2=s_3.$$

Thus

$${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{j\in J_1}{inf}\left\{{\alpha }_j\right\}=min\left\{{\alpha }_1,{\alpha }_2\right\}=s_3$$

and

$${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)=\underset{j\in J_1}{sup}\left\{{\beta }_j\right\}=max\left\{{\beta }_1,{\beta }_2\right\}=s_3.$$

That is

$${dim}_B\mathrm{\Gamma }(f+g,I)=s_3.$$

From these three situations above, we find that ${dim}_B\mathrm{\Gamma }(f+g,I)$ could be any number belonging to $[s_3,s_2]$.

Therefore, when $s_1<s_3<s_2=s_4$, we can come to the conclusions as follows:

(1)

If the Box dimension of $\mathrm{\Gamma }(f+g,I)$ exists,

$$f\left(x\right)+g\left(x\right)\in F_u.$$

Here, u could be any number belonging to $[s_3,s_2]$.

(2)

If the Box dimension of $\mathrm{\Gamma }(f+g,I)$ does not exist,

$$f\left(x\right)+g\left(x\right)\in F_v^u.$$

Here, $u,v$ could be any numbers satisfying $s_3\le v<u\le s_2$ or $1\le v<s_3\le u\le s_2$.

3.3.7. $s_3=s_1<s_2=s_4$

In this situation,

$$s_3=\underset{j\in J_1}{inf}\left\{{\mu }_j\right\}<\underset{j\in J_1}{sup}\left\{{\mu }_j\right\}=s_4.$$

Now we discuss two cases according to whether $s_1$ is equal to one or not as follows.

Case 1: $s_1\ne 1$

In this case, we can similarly obtain the same conclusion as Section 3.3.5 that ${\underline{dim}}_B\mathrm{\Gamma }(f+g,I)$ could be any number satisfying

$$1\le {\underline{dim}}_B\mathrm{\Gamma }(f+g,I)\le s_2.$$

Moreover, for ${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)$, the result is similar with that for $j\in J_1^{\left(2\right)}$ in Section 3.3.6, that is ${\beta }_j$ could be any number satisfying $1\le {\beta }_j\le s_2$. This means ${\overline{dim}}_B\mathrm{\Gamma }(f+g,I)$ could be any number satisfying

$$1\le {\overline{dim}}_B\mathrm{\Gamma }(f+g,I)\le s_2.$$

Then similar with Section 3.3.6, we explore the possibility of the situation when the Box dimension of $\mathrm{\Gamma }(f+g,I)$ exists. We give several particular situations below to illustrate that ${dim}_B\mathrm{\Gamma }(f+g,I)$ could be any number belonging to $[1,s_1)\bigcup (s_1,s_2]$.

Situation I We choose two functions $f\left(x\right),g\left(x\right)\in F_{s_1}^{s_2}$ satisfying the following conditions:

(1)

${\mu }_1=s_1$ and ${\mu }_2=s_2$ are the only two elements in the set ${\Omega }_f$;

(2)

$s_2$ is the only one element in the set ${\Sigma }_1$;

(3)

$s_1$ is the only one element in the set ${\Sigma }_2$.

Similar with Situation I in Section 3.3.6, we have

$${dim}_B\mathrm{\Gamma }(f+g,I)=s_2.$$

Situation II We choose two functions $f\left(x\right),g\left(x\right)\in F_{s_1}^{s_2}$ satisfying the following conditions:

(1)

Three are only three elements ${\mu }_1$, ${\mu }_2$, ${\mu }_3$ in the set ${\Omega }_f$. Here, ${\mu }_1=s_1$, ${\mu }_3=s_2$ and ${\mu }_2$ could be any number belonging to $(s_1,s_2)$;

(2)

${\mu }_2$ is the only one element in the set ${\Sigma }_1$;

(3)

$s_1$ is the only one element in the set ${\Sigma }_2$;

(4)

$s_2$ is the only one element in the set ${\Sigma }_3$.

Similar with Situation II in Section 3.3.6, we know ${dim}_B\mathrm{\Gamma }(f+g,I)$ could be any number belonging to $(s_1,s_2)$.

Situation III Let

$$g\left(x\right)=-f\left(x\right)+H\left(x\right).$$

We choose $f\left(x\right)\in F_{s_1}^{s_2}$ and $H\left(x\right)\in F_s$. Here, s could be any number belonging to $[1,s_1)$. Then

$${\overline{dim}}_B\mathrm{\Gamma }(g,I)={\overline{dim}}_B\mathrm{\Gamma }(-f+H,I)=max\left\{{\overline{dim}}_B\mathrm{\Gamma }(f,I),{\overline{dim}}_B\mathrm{\Gamma }(H,I)\right\}=s_2$$

and

$${\underline{dim}}_B\mathrm{\Gamma }(g,I)={\underline{dim}}_B\mathrm{\Gamma }(-f+H,I)={\underline{dim}}_B\mathrm{\Gamma }(f,I)=s_1$$

by Theorem 1 and 2. Thus the Box dimension of $\mathrm{\Gamma }(f+g,I)$ exists and could be any number belonging to $[1,s_1)$.

From these three situations above, we find that ${dim}_B\mathrm{\Gamma }(f+g,I)$ could be any number belonging to $[1,s_1)\bigcup (s_1,s_2]$. In addition, we indicate that it is impossible for ${dim}_B\mathrm{\Gamma }(f+g,I)$ to equal $s_1$. Let

$$g\left(x\right)=-f\left(x\right)+H\left(x\right).$$

Here, we choose $f\left(x\right)\in F_{s_1}^{s_2}$ and $H\left(x\right)\in F_{s_1}$. From Case 1 in Section 3.2.5, we know ${\underline{dim}}_B\mathrm{\Gamma }(g,I)$ could be any number belonging to $[1,s_1)$. This is in contradiction with ${\underline{dim}}_B\mathrm{\Gamma }(g,I)=s_1$.

So when $s_1\ne 1$, we can come to the conclusions as follows:

(1)

If the Box dimension of $\mathrm{\Gamma }(f+g,I)$ exists,

$$f\left(x\right)+g\left(x\right)\in F_u.$$

Here, u could be any number belonging to $[1,s_1)\bigcup (s_1,s_2]$.

(2)

If the Box dimension of $\mathrm{\Gamma }(f+g,I)$ does not exist,

$$f\left(x\right)+g\left(x\right)\in F_v^u.$$

Here, $u,v$ could be any numbers satisfying $1\le v<u\le s_2$.

Case 2: $s_1=1$

In this case, combining Case 2 in Section 3.2.5 and Case 1 in the present subsection, we can directly obtain the following conclusions.

(1)

If the Box dimension of $\mathrm{\Gamma }(f+g,I)$ exists,

$$f\left(x\right)+g\left(x\right)\in F_u.$$

Here, u could be any number belonging to $[1,s_2]$.

(2)

If the Box dimension of $\mathrm{\Gamma }(f+g,I)$ does not exist,

$$f\left(x\right)+g\left(x\right)\in F_v^u.$$

Here, $u,v$ could be any numbers satisfying $1\le v<u\le s_2$.

4. Main Results

In this section, we sum up the conclusions of all the cases we have discussed in the last section. For $f\left(x\right)\in C_I$ and $c\in \mathbb{R}$, we know

$${\underline{dim}}_B\mathrm{\Gamma }(c·f,I)={\underline{dim}}_B\mathrm{\Gamma }(f,I) \mathrm{and} {\overline{dim}}_B\mathrm{\Gamma }(c·f,I)={\overline{dim}}_B\mathrm{\Gamma }(f,I)$$

when $c\ne 0$. Hence, we can obtain the results for the fractal dimensions of the linear combination of continuous functions denoted by (13). Theorem 6 shows the conclusions of the cases when one continuous function has the Box dimension but the other one does not have the Box dimension. Moreover, Theorem 7 shows the conclusions of the cases when neither of two continuous functions has the Box dimension.

Theorem 6.

Let $f\left(x\right)\in F_{s_1}^{s_2},g\left(x\right)\in F_s$ and $a,b$ be any nonzero real numbers.

(1)

If $s_1<s_2<s$,

$$a·f\left(x\right)+b·g\left(x\right)\in F_s.$$

(2)

If $s<s_1<s_2$,

$$a·f\left(x\right)+b·g\left(x\right)\in F_{s_1}^{s_2}.$$

(3)

If $s_1<s<s_2$,

$$a·f\left(x\right)+b·g\left(x\right)\in \bigcup _{v\in [1,s]}{F}_v^{s_2}.$$

(4)

If $s_1<s=s_2$,

$$a·f\left(x\right)+b·g\left(x\right)\in \bigcup _{v\in [1,s)}{F}_v^s.$$

(5)

If $1<s_1=s<s_2$,

$$a·f\left(x\right)+b·g\left(x\right)\in \bigcup _{v\in [1,s)}{F}_v^{s_2}.$$

(6)

If $1=s_1=s<s_2$,

$$a·f\left(x\right)+b·g\left(x\right)\in F_1^{s_2}.$$

Remark 1

(Remarks to Theorem 6). In Theorem 6 if $a=0$ and $b\ne 0$,

$$a·f\left(x\right)+b·g\left(x\right)\in F_s.$$

If $a\ne 0$ and $b=0$,

$$a·f\left(x\right)+b·g\left(x\right)\in F_{s_1}^{s_2}.$$

And if $a=b=0$, $a·f\left(x\right)+b·g\left(x\right)$ is of bounded variation on I as

$$a·f\left(x\right)+b·g\left(x\right)\in F_1.$$

Theorem 7.

Let $f\left(x\right)\in F_{s_1}^{s_2},g\left(x\right)\in F_{s_3}^{s_4}$ and $a,b$ be any nonzero real numbers.

(1)

If $s_1<s_2<s_3<s_4$,

$$a·f\left(x\right)+b·g\left(x\right)\in F_{s_3}^{s_4}.$$

(2)

If $s_1<s_2=s_3<s_4$,

$$a·f\left(x\right)+b·g\left(x\right)\in \bigcup _{v\in [1,s_3]}{F}_v^{s_4}.$$

(3)

If $s_1<s_3<s_2<s_4$,

$$a·f\left(x\right)+b·g\left(x\right)\in \bigcup _{v\in [1,s_2]}{F}_v^{s_4}.$$

(4)

If $s_3<s_1<s_2<s_4$,

$$a·f\left(x\right)+b·g\left(x\right)\in \bigcup _{v\in [1,s_1]}{F}_v^{s_4}.$$

(5)

If $s_3=s_1<s_2<s_4$,

$$a·f\left(x\right)+b·g\left(x\right)\in \bigcup _{v\in [1,s_2]}{F}_v^{s_4}.$$

(6)

If $s_1<s_3<s_2=s_4$,

$$a·f\left(x\right)+b·g\left(x\right)\in \left(\bigcup _{u\in [s_3,s_2]}{F}_u\right)\cup \left(\bigcup _{\begin{array}{c}v,u\in [s_3,s_2]\\ v<u\end{array}}{F}_v^u\right)\cup \left(\bigcup _{\begin{array}{c}u\in [s_3,s_2]\\ v\in [1,s_3)\end{array}}{F}_v^u\right).$$

(7)

If $1<s_3=s_1<s_2=s_4$,

$$a·f\left(x\right)+b·g\left(x\right)\in \left(\bigcup _{u\in [1,s_1)\cup (s_1,s_2]}{F}_u\right)\cup \left(\bigcup _{\begin{array}{c}v,u\in [1,s_2]\\ v<u\end{array}}{F}_v^u\right).$$

(8)

If $1=s_3=s_1<s_2=s_4$,

$$a·f\left(x\right)+b·g\left(x\right)\in \left(\bigcup _{u\in [1,s_2]}{F}_u\right)\cup \left(\bigcup _{\begin{array}{c}v,u\in [1,s_2]\\ v<u\end{array}}{F}_v^u\right).$$

Remark 2

(Remarks to Theorem 7). In Theorem 7 if $a=0$ and $b\ne 0$,

$$a·f\left(x\right)+b·g\left(x\right)\in F_{s_3}^{s_4}.$$

If $a\ne 0$ and $b=0$,

$$a·f\left(x\right)+b·g\left(x\right)\in F_{s_1}^{s_2}.$$

And if $a=b=0$, $a·f\left(x\right)+b·g\left(x\right)$ is of bounded variation on I as

$$a·f\left(x\right)+b·g\left(x\right)\in F_1.$$

5. Conclusions

Throughout the present paper, we mainly investigate the fractal dimensions estimation of the linear combination of continuous functions. The work in this paper can be summarized as the following four aspects:

(1)

We put forward a general method to calculate the lower and the upper Box dimension of the sum of two continuous functions by classifying all the subsequences into different sets, which is the key work in the present paper.

(2)

We acquire several basic results for the lower and the upper Box dimension of the sum of two continuous functions in certain situations. If the upper Box dimensions of two continuous functions are not equal, the upper Box dimension of the sum of these two functions is equal to the maximum one of the upper Box dimensions of these two functions. If the upper Box dimension of one function is less than the lower Box dimension of another function, the lower Box dimension of the sum of these two functions is equal to the maximum one of the lower Box dimensions of these two functions.

(3)

We discuss the majority of possible cases of the sum of two continuous functions with different fractal dimensions. We divide the subjects into three broad categories to consider as follows:

(i)

Both of two continuous functions have the Box dimension;

(ii)

One continuous function has the Box dimension but the other one does not have the Box dimension;

(iii)

Neither of two continuous functions has the Box dimension.

Moreover, we obtain their corresponding fractal dimensions estimation by using the main method proposed in Section 2 and sum up the results for all the cases discussed above in the end.

(4)

We find that the space which is consist of all continuous functions without Box dimension is not linear. We also find the fractal dimensions of the linear combination of continuous functions with certain fractal dimensions may be equal to arbitrary numbers belonging to a certain interval.

Meanwhile, there still exist several points worthy of improvement and further research in the present paper. We should point out that our results for fractal dimensions estimation are only based on theoretical analysis. However, specific examples of fractal functions should be given to support our theoretical results. People could make further research on this problem by carrying out the numerical simulation choosing specific examples of fractal functions with different fractal dimensions, which will make our argument more convincing.

Up to now, how the fractal dimensions of the linear combination of continuous functions with certain fractal dimensions change has been investigated only elementarily. We find that the linear combination of continuous functions cannot keep the fractal dimensions closed unless these two functions have the same Box dimension one. To further study fractal functions spaces, we finally propose the following problem that should be considered in the future:

Question. What can the fractal dimensions of the product of two continuous functions be? Under what circumstances can the product of two continuous functions keep the fractal dimensions closed?

People could discuss this question by a similar classification of the fractal dimensions of two continuous functions with the present paper. Then, three fundamental operations of addition, scalar multiplication and multiplication imposed on fractal functions spaces will be studied totally.