© Wuhan University 2021

0 Introduction

Boolean functions are important objects in discrete mathematics. They play a role in symmetric cryptography and error-correcting coding theory, and they also have a significant influence on the design and analysis of cryptographic algorithms. The Walsh transform is a vital tool to investigate cryptographic properties of Boolean functions. Some important properties of cryptographic functions, such as resiliency and nonlinearity can be characterized by their Walsh transform^[1-3]. An interesting problem is to find Boolean functions with few Walsh transform values and determine their distributions.

Bent functions introduced by Rothaus^[4] in 1976 are interesting combinatorial objects with maximum Hamming distance to the set of all affine functions, but they cannot be used in cryptography directly since they exist only in an even number of variables and are not be balanced. Such functions have been extensively studied because of their important applications in coding theory^[5,6], cryptography^[7], and sequence designs^[8]. To get balanced functions with high nonlinearity in odd or even number of variable, Carlet^[9] generalized the bent functions to plateaued functions and they take Walsh transform values 0, $\pm 2^k$ for a fixed positive integer k. Semi-bent, as a particular case, is an important kind of Boolean functions with three Walsh transform values. In Ref. [10], some classes of Boolean functions with four-valued Walsh spectra are presented by complementing the values of bent functions at two points, one of which is zero and the other is nonzero, and their Walsh spectrum distributions are determined finally. Inspired by this work, recently Jin et al^[11] presented three classes of Boolean functions with six-valued Walsh spectra, which were derived from bent functions by complementing their values at the zero and another two nonzero points, and determined their Walsh spectrum distributions with a similar method. In Ref. [12], some classes of Boolean functions with five Walsh transform values were presented by adding the product of two or three linear functions into some known bent functions, and their Walsh spectrum distributions were determined finally. In Ref. [13], Tang et al gave a generic construction of bent functions defined as $f(x)=g(x)+F({\text{Tr}}_1^n(u_{1}x),{\text{Tr}}_1^n(u_{2}x),\cdots ,{\text{Tr}}_1^n(u_{\tau }x)),$ where n=2m, g(x) is any known bent function over $F_{2^n}$ satisfying some conditions, $F(X_1,X_2,\cdots ,X_{\tau })$ is an arbitrary polynomial in $F_2[X_1,X_2,\cdots ,X_{\tau }]$. In particular, the cases of $F(X_1,X_2,X_3)=X_1{X}_2{X}_3$ and $F(X_1,X_2)=X_1{X}_2$ have been studied by Xu et al^[12] and Mesnager^[14], respectively. The purpose of this paper is to present the Walsh transform of the Boolean function defined as $g(x)=f(x){\text{Tr}}_1^n(x)+h(x){\text{Tr}}_1^n(\delta x)$(1) where f(x) and h(x) are Boolean functions over $F_{2^n}$ and $\delta \in F_{2^n}$. In particular, the case of $g(x)=f(x){\text{Tr}}_1^n(x)+(f(x)+1){\text{Tr}}_1^n(\delta x)$(2) has been studied by Pang et al^[15].

This paper is organized as follows. In Section 1, we give some basic concepts and results. In Section 2, we present the Walsh transform of the Eq. (1). In Section 3, we conclude this paper.

1 Preliminaries

Let $F_2^n$ denote the n-dimensional vector space over F₂, and $F_{2^n}$ denote the finite field with 2ⁿ elements. For any set E, $E^{\ast }=E\backslash \left\{0\right\}$. By viewing each $x=x_1{\xi }_1+x_2{\xi }_2+\cdots +x_n{\xi }_n\in F_{2^n}$ as a vector $(x_1,x_2,\cdots ,$$x_n)\in F_2^n$ where $\{{\xi }_1,{\xi }_2,\cdots ,{\xi }_n\}$ is a basis of $F_{2^n}$ over F₂, we identify $F_2^n$ with $F_{2^n}$ and then every function $f:F_{2^n}\to F_2$ is equivalent to a Boolean function. For $x,y\in F_{2^n}$, the inner product is defined as $x\cdot y={\text{Tr}}_1^n(xy)$.

For any positive integer k|n, the trace function from $F_{2^n}$ to $F_{2^k}$ is the mapping defined as ${\text{Tr}}_k^n(x)={\displaystyle \sum _{i=0}^{\frac{n}{k}-1}{x}^{2^{ik}}}=x+x^{2^k}+x^{2^{2k}}+\cdots +x^{2^{n-1}},x\in F_{2^n}$ When k=1, ${\text{Tr}}_1^n(x)={\displaystyle \sum _{i=0}^{n-1}{x}^{2^i}}=x+x^2+x^{2^2}+\cdots +x^{2^{n-1}}$ is called the absolute trace function.

Some important and useful properties of the trace function are provided in the following:

1) ${\text{Tr}}_1^n(ax+by)=a{\text{Tr}}_1^n(x)+b{\text{Tr}}_1^n(y)$, $\forall x,y\in F_{2^n}$ and $a,b\in F_2$.

2) ${\text{Tr}}_1^n(x^2)={\text{Tr}}_1^n(x)$ for any $x\in F_{2^n}$.

3) For any $\alpha \in F_{2^n}$, ${\displaystyle \sum _{x\in F_{2^n}}{(-1)}^{{\text{Tr}}_1^n(\alpha x)}}=0$ if $\alpha \ne 0$.

4) When $F_2\subset F_{2^m}\subset F_{2^n}$, the trace function ${\text{Tr}}_1^n(\alpha )$ satisfies the transitivity property, that is, ${\text{Tr}}_1^n(\alpha )={\text{Tr}}_1^m({\text{Tr}}_m^n(\alpha ))$.

5) For any $\alpha \in F_{2^n}$, ${({\text{Tr}}_1^n(\alpha ))}^{2^j}={\text{Tr}}_1^n({\alpha }^{2^j})$, $j=0,1,\cdots$.

Let f be a Boolean function from $F_{2^n}$ to F₂, and the set of which is denoted by B_n. The Walsh transform of $f\in B_n$ at $F_{2^n}$ is defined as $W_f(a)={\displaystyle \sum _{x\in F_{2^n}}{(-1)}^{f(x)+{\text{Tr}}_1^n(ax)}},a\in F_{2^n}.$

The values $W_f(a)$, $a\in F_{2^n}$ are called the Walsh coefficients of f. The Walsh spectrum of a Boolean function f is the multiset $\{W_f(a),a\in F_{2^n}\}$. A Boolean function f is said to be balanced if $W_f(0)=0$.

The nega-Hadamard transform of f(x) at $a\in F_{2^n}$ is the complex valued function $N_f(a)={\displaystyle \sum _{x\in F_{2^n}}{(-1)}^{f(x)+{\text{Tr}}_1^n(ax)+\sigma (x)}{i}^{{\text{Tr}}_1^n(x)}}$ where $\sigma (x)$ is the function defined by $\sigma (x)={\displaystyle \sum _{0\le i<j\le n-1}{(x)}^{2^i}{(x)}^{2^j}}$. A function $f\in B_n$ is negabent if $|N_f(a)|=1$ for all $a\in F_{2^n}$.

2 Main Results

Let n be a positive integer and f be a Boolean function from $F_{2^n}$ to F₂. For any $\delta \in F_{2^n}$, the Boolean function $g(x)=f(x){\text{Tr}}_1^n(x)+h(x){\text{Tr}}_1^n(\delta x)$ can be written as$\begin{array}{l}g(x)=f(x){\text{Tr}}_1^n(x)+h(x){\text{Tr}}_1^n(\delta x)\\ \begin{array}{cc}& \end{array}=\{\begin{array}{cc}0,& x\in T_{0,0}\\ h(x),& x\in T_{0,1}\\ f(x),& x\in T_{1,0}\\ f(x)+h(x),& x\in T_{1,1}\end{array}\end{array}$where $T_{i,j}=\{x\in F_{2^n}|{\text{Tr}}_1^n(x)=i\text{\hspace{0.05em}\hspace{0.05em}\hspace{0.05em}\hspace{0.05em}and\hspace{0.05em}\hspace{0.05em}\hspace{0.05em}\hspace{0.05em}}{\text{Tr}}_1^n(\delta x)=j\}$ for $i,j=0,1$.

The relationship between $W_g(b)$ and $W_f(b)$, $W_h(b)$, $W_{f+h}(b)$ is given in Theorem 1.

Theorem 1   Let $\delta \in F_{2^n}$,$g(x)=f(x){\text{Tr}}_1^n(x)+h(x){\text{Tr}}_1^n(\delta x)$Then, the Walsh transform of g(x) at $b\in F_{2^n}$ is given by$$W_g(b)\{\begin{array}{lll}\begin{array}{l}{2}^{n-2}+\frac{1}{4}[W_f(0)-W_f(1)+W_f(\delta )-W_f(\delta +1)+W_h(0)+W_h(1)-W_h(\delta )-W_h(\delta +1)\\ +W_{f+h}(0)-W_{f+h}(1)-W_{f+h}(\delta )+W_{f+h}(\delta +1)]\end{array}\hfill & \text{if}\hfill & b=0\hfill \\ \begin{array}{l}{2}^{n-2}+\frac{1}{4}[-W_f(0)+W_f(1)-W_f(\delta )+W_f(\delta +1)+W_h(0)+W_h(1)-W_h(\delta )-W_h(\delta +1)\\ -W_{f+h}(0)+W_{f+h}(1)+W_{f+h}(\delta )-W_{f+h}(\delta +1)],\end{array}\hfill & \text{if}\hfill & b=1\hfill \\ \begin{array}{l}{2}^{n-2}+\frac{1}{4}[W_f(0)-W_f(1)+W_f(\delta )-W_f(\delta +1)-W_h(0)-W_h(1)+W_h(\delta )+W_h(\delta +1)\\ -W_{f+h}(0)+W_{f+h}(1)+W_{f+h}(\delta )-W_{f+h}(\delta +1)],\end{array}\hfill & \text{if}\hfill & b=\delta \hfill \\ \begin{array}{l}{2}^{n-2}+\frac{1}{4}[-W_f(0)+W_f(1)-W_f(\delta )+W_f(\delta +1)-W_h(0)-W_h(1)+W_h(\delta )+W_h(\delta +1)\\ +W_{f+h}(0)-W_{f+h}(1)-W_{f+h}(\delta )+W_{f+h}(\delta +1)],\end{array}\hfill & \text{if}\hfill & b=\delta +1\hfill \\ \begin{array}{l}\frac{1}{4}[W_f(b)-W_f(b+1)+W_f(b+\delta )-W_f(b+\delta +1)+W_h(b)+W_h(b+1)\\ -W_h(b+\delta )-W_h(b+\delta +1)+W_{f+h}(b)-W_{f+h}(b+1)-W_{f+h}(b+\delta )+W_{f+h}(b+\delta +1)],\end{array}\hfill & \text{if}\hfill & b\in F_{2^n}^{\ast }\backslash \{1,\delta ,\delta +1\}\hfill \end{array}$$

1) When $\delta =0$,$W_g(b)=\{\begin{array}{cc}{2}^{n-1}+\frac{1}{2}[W_f(0)-W_f(1)],& \text{if\hspace{0.17em}}b=0\\ 2^{n-1}-\frac{1}{2}[W_f(0)-W_f(1)],& \text{if\hspace{0.17em}}b=1\\ \frac{1}{2}[W_f(b)-W_f(b+1)],& \text{if\hspace{0.17em}}b\in F_{2^n}^{\ast }\backslash \left\{1\right\}\end{array}$

2) When $\delta =1$, $W_g(b)=\{\begin{array}{cc}{2}^{n-1}+\frac{1}{2}[W_{f+h}(0)-W_{f+h}(1)],& \text{if\hspace{0.17em}}b=0\\ 2^{n-1}-\frac{1}{2}[W_{f+h}(0)-W_{f+h}(1)],& \text{if\hspace{0.17em}}b=1\\ \frac{1}{2}[W_{f+h}(b)-W_{f+h}(b+1)],& \text{if\hspace{0.17em}}b\in F_{2^n}^{\ast }\backslash \left\{1\right\}\end{array}$

3) When $\delta \in F_{2^n}^{\ast }\backslash \left\{1\right\}$,

Proof   For simplicity, denote$\begin{array}{l}{\theta }_t={\displaystyle \sum _{x\in T_{i,j}}{(-1)}^{f(x)+{\text{Tr}}_1^n(bx)}},{\theta }_t{}^{\prime }={\displaystyle \sum _{x\in T_{i,j}}{(-1)}^{h(x)+{\text{Tr}}_1^n(bx)}},\\ {{\theta }^{″}}_t={\displaystyle \sum _{x\in T_{i,j}}{(-1)}^{f(x)+h(x)+{\text{Tr}}_1^n(bx)}}\end{array}$where $i,j\in \{0,1\}$, $t=2i+j$, $0\le t\le 3$.

The proof proceeds in terms of three cases: $\delta =0$, $\delta =1$ and $\delta \in F_{2^n}^{\ast }\backslash \left\{1\right\}$.

1) If $\delta =0$, then one obtains$\begin{array}{c}{W}_g(b)={\displaystyle \sum _{x\in F_{2^n}}{(-1)}^{f(x){\text{Tr}}_1^n(x)+{\text{Tr}}_1^n(bx)}}\\ ={\displaystyle \sum _{x\in F_{2^n},{\text{Tr}}_1^n(x)=0}{(-1)}^{{\text{Tr}}_1^n(bx)}}\\ +{\displaystyle \sum _{x\in F_{2^n},{\text{Tr}}_1^n(x)=1}{(-1)}^{f(x)+{\text{Tr}}_1^n(bx)}}\\ =A_1+A_2,\end{array}$where $A_1={\displaystyle \sum _{{\text{Tr}}_1^n(x)=0}{(-1)}^{{\text{Tr}}_1^n(bx)}}$, $A_2={\displaystyle \sum _{{\text{Tr}}_1^n(x)=1}{(-1)}^{f(x)+{\text{Tr}}_1^n(bx)}}$.

Note that$A_1=\{\begin{array}{cc}{2}^{n-1},& \text{if\hspace{0.17em}}b=0,1\\ 0,& \text{otherwise}\end{array}$and $A_2={\theta }_2+{\theta }_3$.

Then by$\begin{array}{l}{W}_f(b)={\displaystyle \sum _{x\in F_{2^n}}{(-1)}^{f(x)+{\text{Tr}}_1^n(bx)}}\\ ={\theta }_0+{\theta }_1+{\theta }_2+{\theta }_3\end{array}$and$\begin{array}{c}{W}_f(b+1)={\displaystyle \sum _{x\in F_{2^n}}{(-1)}^{f(x)+{\text{Tr}}_1^n((b+1)x)}}\\ ={\theta }_0+{\theta }_1-{\theta }_2-{\theta }_3,\end{array}$we have$A_2=\frac{1}{2}[W_f(b)-W_f(b+1)]$

2) The proof of 2) is obvious from 1).

3) If $\delta \in F_{2^n}^{\ast }\backslash \left\{1\right\}$, then one obtains$\begin{array}{l}{W}_g(b)={\displaystyle \sum _{x\in F_{2^n}}{(-1)}^{f(x){\text{Tr}}_1^n(x)+h(x){\text{Tr}}_1^n(\delta x)+{\text{Tr}}_1^n(bx)}}\\ ={\displaystyle \sum _{x\in T_{0,0}}{(-1)}^{{\text{Tr}}_1^n(bx)}}+{\displaystyle \sum _{x\in T_{0,1}}{(-1)}^{h(x)+{\text{Tr}}_1^n(bx)}}\\ +{\displaystyle \sum _{x\in T_{1,0}}{(-1)}^{f(x)+{\text{Tr}}_1^n(bx)}}+{\displaystyle \sum _{x\in T_{1,1}}{(-1)}^{f(x)+h(x)+{\text{Tr}}_1^n(bx)}}\\ =C_1+{{\theta }^{\prime }}_1+{\theta }_2+{{\theta }^{″}}_3\end{array}$where $C_1={\displaystyle \sum _{x\in T_{0,0}}{(-1)}^{{\text{Tr}}_1^n(bx)}}$.

Note that$C_1=\{\begin{array}{cc}{2}^{n-2},& \text{if\hspace{0.17em}}b=0,1,\delta ,\delta +1,\\ 0,& \text{otherwise.}\end{array}$Together with the fact$\begin{array}{l}{W}_f(b+\delta )={\displaystyle \sum _{x\in F_{2^n}}{(-1)}^{f(x)+{\text{Tr}}_1^n((b+\delta )x)}}\\ ={\displaystyle \sum _{x\in T_{0,0}}{(-1)}^{f(x)+{\text{Tr}}_1^n(bx)}}-{\displaystyle \sum _{x\in T_{0,1}}{(-1)}^{f(x)+{\text{Tr}}_1^n(bx)}}\\ +{\displaystyle \sum _{x\in T_{1,0}}{(-1)}^{f(x)+{\text{Tr}}_1^n(bx)}}-{\displaystyle \sum _{x\in T_{1,1}}{(-1)}^{f(x)+{\text{Tr}}_1^n(bx)}}\\ ={\theta }_0-{\theta }_1+{\theta }_2-{\theta }_3\end{array}$and$\begin{array}{l}{W}_f(b+\delta +1)={\displaystyle \sum _{x\in F_{2^n}}{(-1)}^{f(x)+{\text{Tr}}_1^n((b+\delta +1)x)}}\\ ={\displaystyle \sum _{x\in T_{0,0}}{(-1)}^{f(x)+{\text{Tr}}_1^n(bx)}}-{\displaystyle \sum _{x\in T_{0,1}}{(-1)}^{f(x)+{\text{Tr}}_1^n(bx)}}\\ -{\displaystyle \sum _{x\in T_{1,0}}{(-1)}^{f(x)+{\text{Tr}}_1^n(bx)}}+{\displaystyle \sum _{x\in T_{1,1}}{(-1)}^{f(x)+{\text{Tr}}_1^n(bx)}}\\ ={\theta }_0-{\theta }_1-{\theta }_2+{\theta }_3\end{array}$Similarly,$\begin{array}{l}{W}_h(b)=({{\theta }^{\prime }}_0+{{\theta }^{\prime }}_1+{{\theta }^{\prime }}_2+{{\theta }^{\prime }}_3)\\ W_h(b+1)=({{\theta }^{\prime }}_0+{{\theta }^{\prime }}_1-{{\theta }^{\prime }}_2-{{\theta }^{\prime }}_3)\\ W_h(b+\delta )=({{\theta }^{\prime }}_0-{{\theta }^{\prime }}_1+{{\theta }^{\prime }}_2-{{\theta }^{\prime }}_3)\\ W_h(b+\delta +1)=({{\theta }^{\prime }}_0-{{\theta }^{\prime }}_1-{{\theta }^{\prime }}_2+{{\theta }^{\prime }}_3)\\ W_{f+h}(b)=({{\theta }^{″}}_0+{{\theta }^{″}}_1+{{\theta }^{″}}_2+{{\theta }^{″}}_3)\\ W_{f+h}(b+1)=({{\theta }^{″}}_0+{{\theta }^{″}}_1-{{\theta }^{″}}_2-{{\theta }^{″}}_3)\\ W_{f+h}(b+\delta )=({{\theta }^{″}}_0-{{\theta }^{″}}_1+{{\theta }^{″}}_2-{{\theta }^{″}}_3)\\ W_{f+h}(b+\delta +1)=({{\theta }^{″}}_0-{{\theta }^{″}}_1-{{\theta }^{″}}_2+{{\theta }^{″}}_3)\end{array}$Then by$\begin{array}{l}{\theta }_2-{\theta }_1=\frac{1}{2}[W_f(b+\delta )-W_f(b+1)]\\ {\theta }_1+{\theta }_3=\frac{1}{2}[W_f(b)-W_f(b+\delta )]\\ {\theta }_2-{\theta }_3=\frac{1}{2}[W_f(b+\delta )-W_f(b+\delta +1)]\end{array}$we have${\theta }_2=\frac{1}{4}[W_f(b)-W_f(b+1)+W_f(b+\delta )-W_f(b+\delta +1)]$by$\begin{array}{l}{{\theta }^{\prime }}_1+{{\theta }^{\prime }}_3=\frac{1}{2}[W_h(b)-W_h(b+\delta )]\\ {{\theta }^{\prime }}_1-{{\theta }^{\prime }}_3=\frac{1}{2}[W_h(b+1)-W_h(b+\delta +1)]\end{array}$we have${{\theta }^{\prime }}_1=\frac{1}{4}[W_h(b)+W_h(b+1)-W_h(b+\delta )-W_h(b+\delta +1)]$Similarly, we have$\begin{array}{l}{{\theta }^{″}}_3=\\ \frac{1}{4}[W_{f+h}(b)-W_{f+h}(b+1)-W_{f+h}(b+\delta )+W_{f+h}(b+\delta +1)]\end{array}$One immediately gets that$\begin{array}{l}{{\theta }^{\prime }}_1+{\theta }_2+{{\theta }^{″}}_3\\ =\frac{1}{4}[W_f(b)-W_f(b+1)+W_f(b+\delta )\\ -W_f(b+\delta +1)+W_h(b)+W_h(b+1)\\ -W_h(b+\delta )-W_h(b+\delta +1)+W_{f+h}(b)\\ -W_{f+h}(b+1)-W_{f+h}(b+\delta )+W_{f+h}(b+\delta +1)]\end{array}$when b=0,$\begin{array}{l}{W}_g(0)=2^{n-2}+\frac{1}{4}[W_f(0)-W_f(1)+W_f(\delta )-W_f(\delta +1)\\ +W_h(0)+W_h(1)-W_h(\delta )-W_h(\delta +1)\\ +W_{f+h}(0)-W_{f+h}(1)-W_{f+h}(\delta )+W_{f+h}(\delta +1)]\end{array}$when b=1,$\begin{array}{l}{W}_g(1)=2^{n-2}+\frac{1}{4}[-W_f(0)+W_f(1)-W_f(\delta )+W_f(\delta +1)\\ +W_h(0)+W_h(1)-W_h(\delta )-W_h(\delta +1)\\ -W_{f+h}(0)+W_{f+h}(1)+W_{f+h}(\delta )-W_{f+h}(\delta +1)]\end{array}$when $b=\delta$,$\begin{array}{l}{W}_g(\delta )=2^{n-2}+\frac{1}{4}[W_f(0)-W_f(1)+W_f(\delta )-W_f(\delta +1)\\ -W_h(0)-W_h(1)+W_h(\delta )+W_h(\delta +1)\\ -W_{f+h}(0)+W_{f+h}(1)+W_{f+h}(\delta )-W_{f+h}(\delta +1)]\end{array}$when $b=\delta +1$,$\begin{array}{c}{W}_g(\delta +1)=\\ 2^{n-2}+\frac{1}{4}[-W_f(0)+W_f(1)-W_f(\delta )+W_f(\delta +1)\\ -W_h(0)-W_h(1)+W_h(\delta )+W_h(\delta +1)\\ +W_{f+h}(0)-W_{f+h}(1)-W_{f+h}(\delta )+W_{f+h}(\delta +1)]\end{array}$The proof is completed.

In particular, when $h(x)=f(x)+1$, the function $g(x)=f(x){\text{Tr}}_1^n(x)+(f(x)+1){\text{Tr}}_1^n(\delta x)$ is exactly the ones studied by Pang et al^[15]. Note that our generic construction works for any f(x) and h(x). Therefore, our construction contains the previous ones in Ref. [15] as special cases.

Corollary 1   Let $\delta \in F_{2^n}\backslash \left\{1\right\}$, and$g(x)=f(x){\text{Tr}}_1^n(x)+(f(x)+1){\text{Tr}}_1^n(\delta x)$The Walsh transform of g(x) at $b\in F_{2^n}$ is given by$W_g(b)=\{\begin{array}{cc}{2}^{n-1}+\frac{1}{2}[W_f(\delta +1)-W_f(0)],& \text{if\hspace{0.17em}}b=1\\ 2^{n-1}-\frac{1}{2}[W_f(\delta +1)-W_f(0)],& \text{if\hspace{0.17em}}b=\delta \\ \frac{1}{2}[W_f(b+\delta )-W_f(b+1)],& \text{if\hspace{0.17em}}b\in F_{2^n}\backslash \{1,\delta \}\end{array}$In the following, pang proposed three new classes of Boolean functions having the form as Eq. (2) by suitable choices of f(x). The first class is obtained from bent functions, including Dillon bent, kasami bent and Gold-like bent functions, and from the definition of the dual of bent functions, the Walsh transform value distribution of such class is presented. Ref. [15] indicates that the Walsh spectrum distribution of g(x) derived from bent functions is obtained from calculating the dual function of f and the values of $\#H_1$, $\#H_2$ and $\#H_3$. Therefore, the Walsh spectrum distribution of$g(x)={\text{Tr}}_1^n(a{x}^{2^m-1}){\text{Tr}}_1^n(x)+({\text{Tr}}_1^n(a{x}^{2^m-1})+1){\text{Tr}}_1^n(\delta x)$is presented.

The second class is derived from Gold functions $f(x)={\text{Tr}}_1^n(a{x}^{2^i+1})$ and their Walsh spectrum distribution is obtained by making use of the Walsh transform property of Gold functions and the known conclusions of Weil sums in characteristic 2. The Walsh spectrum distribution of g(x) which is obtained by Gold functions is discussed in the following three cases.

Case 1  n/d is even and $a\ne {\alpha }^{t(2^d+1)}$ for any integer t. It is known that in this case $f(x)={\text{Tr}}_1^n(a{x}^{2^i+1})$ is bent. Then the Walsh spectrum distribution of$g(x)={\text{Tr}}_1^n(a{x}^{2^i+1}){\text{Tr}}_1^n(x)+({\text{Tr}}_1^n(a{x}^{2^i+1})+1){\text{Tr}}_1^n(\delta x)$is given in Ref. [15].

Case 2  n/d is even and $a={\alpha }^{t(2^d+1)}$ for some integer t. In this case the Walsh spectrum distribution of$g(x)={\text{Tr}}_1^n(a{x}^{2^i+1}){\text{Tr}}_1^n(x)+({\text{Tr}}_1^n(a{x}^{2^i+1})+1){\text{Tr}}_1^n(\delta x)$depends on whether $h(x)={(\delta +1)}^{2^i}$ is solvable.

Case 3  n/d is odd. In this case we only need to consider $f(x)={\text{Tr}}_1^n(x^{2^i+1})$, then the Walsh spectrum distribution of$g(x)={\text{Tr}}_1^n(a{x}^{2^i+1}){\text{Tr}}_1^n(x)+({\text{Tr}}_1^n(a{x}^{2^i+1})+1){\text{Tr}}_1^n(\delta x)$is presented in Ref. [15].

The last class comes from the product of linearized polynomials which have three or four Walsh transform values. With the help of k-tuple balance property, the Walsh spectrum distribution of such functions are determined. Ref. [15] present the Walsh transform of $f(x)={\displaystyle \prod _{i=1}^k{\text{Tr}}_1^n(a_{i}x)}$ together with the Walsh transform of Eq. (2), the Walsh spectrum distribution of$g(x)={\displaystyle \prod _{i=1}^k{\text{Tr}}_1^n(a_{i}x)}{\text{Tr}}_1^n(x)+({\displaystyle \prod _{i=1}^k{\text{Tr}}_1^n(a_{i}x)}+1){\text{Tr}}_1^n(\delta x)$is given.

In another particular case, when f(x)=0 and $h(x)={\text{Tr}}_1^n(ux)$, the function $g(x)={\text{Tr}}_1^n(ux){\text{Tr}}_1^n(vx)$ is studied by Wu et al^[16]. They give the necessary and sufficient conditions for g(x) to be negabent.

Corollary 2   Let $g(x)={\text{Tr}}_1^n(ux){\text{Tr}}_1^n(vx)$, where $(u,v)\in F_{2^n}^{\ast }\times F_{2^n}^{\ast }$. Then g(x) is negabent on $F_{2^n}$ if and only if one of the following conditions is satisfied:

1) ${\text{Tr}}_1^n(u)=0$ and ${\text{Tr}}_1^n(uv)=0$.

2) ${\text{Tr}}_1^n(u)=1$ and ${\text{Tr}}_1^n((u+1)v)=0$.

In Ref. [16], first they presented the necessary and sufficient conditions for the functions$f(x)={\text{Tr}}_1^k(\lambda x^{2^k+1})+{\text{Tr}}_1^n(ux){\text{Tr}}_1^n(vx)$to be negabent, where n=2k, $(u,v)\in F_{2^n}^{\ast }\times F_{2^n}^{\ast }$, $\lambda \in F_{2^k}$, when $\lambda =0$ it is the one discussed in Ref. [16]. Further, by using some permutation trinomials over $F_{2^n}$, they presented some classes of negabent functions of the form$f(x)={\text{Tr}}_1^k(\lambda x^{2^k+1})+{\text{Tr}}_1^n(ux){\text{Tr}}_1^n(vx),$where 0<k<n.