different differential spectrum. In fact, it is easy to see that the differential spectrum of a trivial function takes two values $0$ and $2^k$ only, and their Walsh spectrum takes the values $0, \pm2^k$ and $2^{2k}$ only.

\[walsh-1\] Let $G(x,y)=(x^{2^i}\pi(y)+g(y),h(y))$, where $\pi(y)$ is a permutation over $\mathbb{F}_{2^{k}}$, and $g(y)$ and $h(y)$ are arbitrary $(k,k)$-function and $(k, m-k)$-function respectively. Then for any $(a,b)\in \mathbb{F}_{2^{k}}^2, (0,0)\neq(u,v)\in \mathbb{F}_{2^{k}}\times\mathbb{F}_{2^{m-k}}$, the Walsh spectrum of $F$ satisfies $$\begin{aligned}
W_{G}((a,b),(u,v))\!=\!
\begin{cases}\!\pm\!~2^k, &{\rm~if~}u\!\neq\! 0,\\0, &{\rm~if~}u\!=\!0, a\neq 0,\\2^{k}W_{h}(b,v), &{\rm~otherwise.}
\end{cases}\end{aligned}$$ Moreover, $G$ is plateaued if and only if $h$ is plateaued.

Let $(a,b) \in \mathbb{F}_{2^{k}}\times \mathbb{F}_{2^{k}}$. For any $(0, 0)\neq (u,v) \in  \mathbb{F}_{2^{k}}\times \mathbb{F}_{2^{m-k}}$, we have $$\begin{aligned}
\langle (u,v), G(x,y)\rangle &\!=\!& \langle u,
x^{2^{i}}\pi(y)\!+\!g(y)\rangle\!+\!\langle v, h(y)\rangle\\
& \!=\!& {\rm Tr}^{k}_{1}(u x^{2^{i}}\pi(y))\!+\!{\rm Tr}^{k}_{1}(u g(y))\!+\!{\rm Tr}^{m-k}_{1}(vh(y)):=f(x,y).\end{aligned}$$ One has $${\begin{array}{lll}W_{f}(a,b)&\!= \! \sum\limits_{(x,y)\in \mathbb{F}_{2^{k}}\times \mathbb{F}_{2^{k}}}(\!-\!1)^{f(x,y)\!+\!{\rm Tr}^{k}_{1}(ax\!+\!by)}  \\
             &\!= \!\sum\limits_{y\in  \mathbb{F}_{2^{k}} }(\!-\!1)^{{\rm Tr}^{m-k}_{1}(vh(y))+{\rm Tr}^{k}_{1}(ug(y))\!+\!{\rm Tr}^{k}_{1}(by)}\sum\limits_{x\in  \mathbb{F}_{2^{k}}}(\!-\!1)^{{\rm Tr}^{k}_{1}(((u\pi(y))^{2^{k-i}}\!+\!a)x)}\\
             &\!=\! 2^{k}\sum\limits_{u\pi(y)=a^{2^i}}(\!-\!1)^{{\rm Tr}^{m-k}_{1}(vh(y))\!+\!{\rm Tr}^{k}_{1}(by)+{\rm Tr}^{k}_{1}(ug(y))}. \\
             \end{array}}$$ By Theorem \[keythm-1\], $f(x,y)$ is bent if and only if $u\neq 0$. Hence when $u\neq 0$ it holds $W_{G}((a,b),(u,v))=\pm 2^k$.

If $u=0$ and $v\neq 0$, then for $a=0$, the solutions to $u\pi(y)=a^{2^i}$ are exactly all $y$ in $\mathbb{F}_{2^{k}}$. Therefore in this case $W_{f}(a,b)=2^{k}\sum\limits_{y\in\mathbb{F}_{2^{k}}}(-1)^{{\rm Tr}^{m-k}_{1}(vh(y))+{\rm Tr}^{k}_{1}(by)}=2^{k}W_{h}(b,v).$ While for $a\neq 0$, there 