x\in D_1\\
    2, & x\in D_2\\
    \end{cases}$$ is an $s$-plateaued function.

The Walsh transform $$\widehat{f}(0)=\sum_{x \in \mathbb{F}_{p^{n}}} \zeta_3^{f(x)}=|D_0|+\zeta_3|D_1|+\zeta_3^2|D_2|.$$ With the conditions given in the theorem, we obtain $$\widehat{f}(0)=\begin{cases}
    0 & \text{if}\; |D_0|=|D_1|=|D_2|,\\
    3^{\frac{n+s}{2}} & \text{if}\;|D_i|=|D_j|=3^{n-1}-3^{\frac{n+s-2}{2}}, |D_k|=3^{n-1}+2.3^{\frac{n+s-2}{2}},\\
    -3^{\frac{n+s}{2}}  & \text{if}\; |D_i|=|D_j|=3^{n-1}+3^{\frac{n+s-2}{2}}, |D_k|=3^{n-1}-2.3^{\frac{n+s-2}{2}}.
    \end{cases}$$ For each $a\in \mathbb{F}_{3^n}^*$, the Walsh transform of $f(x)$ is $$\begin{aligned}
    \widehat{f}(a)&=\sum_{x \in \mathbb{F}_{p^{n}}} \zeta_3^{f(x)-Tr_n(ax)}\\
    &=\sum_{x \in D_0} \zeta_3^{Tr_n(-ax)} + \zeta_3\sum_{x \in D_1} \zeta_3^{Tr_n(-ax)} + \zeta_3^2\sum_{x \in D_2} \zeta_3^{Tr_n(-ax)}\\
    &=\chi_{-a}(D_0)+\zeta_3 \chi_{-a}(D_1) + \zeta_3^2\chi_{-a}(D_2).
    \end{aligned}$$ Then with the assumptions of the theorem and after some easy calculations one can show that $|\widehat{f}(a)|^2 \in \{0,3\lambda^2\}$.

Results on p-ary Functions
==========================

In this section we will develop some tools to characterize $p-$ary $s-$plateaued functions. Our results are mimicking the results of [@olmez2].

\[Matrix-Eq\] Let $f$ be a function from $\mathbb{F}_{p^{n}}$ to $\mathbb{F}_{p}$ and $M=(m_{x,y})$ be a $p^n \times p^n$ matrix where $m_{x,y}=\zeta_p^{f(x+y)}$. Then, $f$ is an $s$-plateaued function if and only if $$MM^*M=p^{n+s}M \label{Mfequation}$$ where $M^*$ is the adjoint of the matrix $M$.

Suppose $f$ is an plateaued function with $|\widehat{f}(x)| \in \{0, p^{(n+s)/2}\}$ for all $x \in  \mathbb{F}_{p^{n}}$. Then, $$\begin{split}
    (MM^*M)_{x,y}&=\sum_{z\in\mathbb{F}_{p^{n}}} \left(\sum_{c\in \mathbb{F}_{p^{n}}} m_{x,c}\overline{m_{z,c}}\right)m_{z,y} \\
    &=\sum_{z\in \mathbb{F}_{p^{n}}} \left(\sum_{c\in \mathbb{F}_{p^{n}}} F(x+c)\overline{F(z+c)}\right)F(z+y) \\
    &=\sum_{c\in \mathbb{F}_{p^{n}}} F(x+c)\left(\sum_{w\in \mathbb{F}_{p^{n}} } \overline{F(w)}F(w+y-c)\right)\\
    &=\sum_{c\in \mathbb{F}_{p^{n}}} (\overline{F}*F)(c-y)F(x+c) 