in the form $a+b\gamma$ with $a, b \in \mathbb{F}_{2^{k}}.$ We have

$$\begin{aligned}
W_{F}(\lambda,a+b\gamma)
             &\!=\!&\sum\limits_{x\in \mathbb{F}_{2^{n}}}(-1)^{{\rm Tr}^{n}_{1}((a+b\gamma)x^s)+{\rm Tr}^{n}_{1}(\lambda x)}\\
             &\!=\!&\sum\limits_{x\in \mathbb{F}_{2^{n}}}(-1)^{{\rm Tr}^{n}_{1}((ax^s+b\gamma x^s)+{\rm Tr}^{n}_{1}(\lambda x)}\\
             &\!=\!&\sum\limits_{x\in \mathbb{F}_{2^{n}}}(-1)^{{\rm Tr}^{n}_{1}(b\gamma x^s)+{\rm Tr}^{n}_{1}(\lambda x)}~(\text{recall~that}~ax^s\in \mathbb{F}_{2^{k}}).\\
             \end{aligned}$$

Let $U=\{x\in \mathbb{F}_{2^{n}} ~|~x\overline{x}=1\}$, then for any $x\in \mathbb{F}^{\ast}_{2^{n}}$, it can be uniquely written as $x=yu$, where $y\in \mathbb{F}^{\ast}_{2^{k}}$, and $u\in U$. We have

$$\begin{aligned}
W_{F}(0,a+b\gamma)
             &\!=\!&\sum\limits_{y\in \mathbb{F}^{\ast}_{2^{k}},u\in U}(-1)^{{\rm Tr}^{n}_{1}(b\gamma (yu)^s)}+1\\
             &\!=\!&\sum\limits_{u\in U}\sum\limits_{y\in \mathbb{F}^{\ast}_{2^{k}}}(-1)^{{\rm Tr}^{n}_{1}(b\gamma y^{2d})}+1~(\text{recall~ that~} u^s=1,y^s=(y\overline{y})^d=y^{2d})\\
             &\!=\!&\sum\limits_{u\in U}\sum\limits_{y\in \mathbb{F}_{2^{k}}}(-1)^{{\rm Tr}^{n}_{1}(b\gamma y^{2d})}-2^{k}\\
             &\!=\!&\sum\limits_{u\in U}\sum\limits_{y\in \mathbb{F}_{2^{k}}}(-1)^{{\rm Tr}^{k}_{1}(b(\gamma+\overline{\gamma}))y^{2d})}-2^{k}.\\
             \end{aligned}$$

Denote $\sum\limits_{y\in \mathbb{F}_{2^{k}}}(-1)^{{\rm Tr}^{k}_{1}(b(\gamma+\overline{\gamma}))y^{2d}}$ by $A$, for $b\neq 0$, we have $W_{F}(0,a+b\gamma)=(2^k+1)A-2^k=2^{k}(A-1)+A=\pm 2^{k}$. It imply that $(2-A)2^{k}=A$ or $A(2^{k}+1)=0$. If the former holds, then $A\neq 2$, and by the fact $\frac{A}{2-A}=2^k$ is an integer we have $A=1$. However, it would imply that $2^{k}(2-A)=2^{k}=A=1$, a contradiction with the assumption that $k$ is a positive integer. Then $A=0$, and it means that $\sum\limits_{y\in \mathbb{F}_{2^{k}}}(-1)^{{\rm Tr}^{k}_{1}(b(\gamma+\overline{\gamma}))y^{2d}}=0$ for any $b\in \mathbb{F}^{\ast}_{2^{k}}$. Noting that $(\gamma+\overline{\gamma})x$ is a bijective map from $\mathbb{F}^{\ast}_{2^{k}}$ to $\mathbb{F}^{\ast}_{2^{k}}$, we have for any $b\in \mathbb{F}^{\ast}_{2^{k}}$, $\sum\limits_{y\in \mathbb{F}_{2^{k}}}(-1)^{{\rm Tr}^{k}_{1}(by^{d})}=0$. Therefore, the function $y^d$ is a permutation on $\mathbb{F}_{2^{k}}$, that is, ${\rm gcd}(d,2^{k}-1)=1$. The last assertion can be seen from the fact that $F(x)\in \mathbb{F}_{2^{k}}$ for any $x\in \mathbb{F}_{2^{n}}$ and [@Pott Proposition 3].

