\\
    &=\sum_{u\in \mathbb{F}_{p^{n}}} (F*\overline{F})(u-x-y)F(u) \\
    &=((F*\overline{F})*F)(x+y).
    \end{split}$$ Let $A=(F*\overline{F})*F$. Then, the Fourier transform of $A$ is $\widehat{A}=\widehat{F}\cdot\widehat{\overline{F}}\cdot\widehat{F}$. Now by Fourier inversion $$\begin{split}
    A(x+y)&=\frac{1}{p^{n}}\sum_{\beta \in \mathbb{F}_{p^{n}}} \widehat{F}(\beta)\widehat{\overline{F}}(\beta)\widehat{F}(\beta)\zeta_p^{Tr((x+y)\beta)}\\
    &=p^{n+s}\frac{1}{p^{n}}\sum_{\beta \in \mathbb{F}_{p^{n}}} \widehat{F}(\beta)\zeta_p^{Tr((x+y) \beta)}\\
    &=p^{n+s}F(x+y).\\
    \end{split}$$ Hence the equation holds.

Suppose $MM^*M=p^{n+s}M$. This implies $((F*\overline{F})*F)(x)=p^{n+s}F(x)$ for all $x \in \mathbb{F}_{p^n}$. Apply the Fourier transform on both of the sides. Then, $$\widehat{F}(x)(\widehat{F}(x)\cdot \overline{\widehat{F}(x)}-p^{n+s})=0$$ for all $x \in \mathbb{F}_{p^n}$. Hence, $|\widehat{F}(x)| \in \{0, p^{(n+s)/2}\}$ for all $x \in  \mathbb{F}_{p^{n}}.$

An $n \times n$ complex matrix $M$ is called a Butson-Hadamard matrix if $$MM^* =nI_n.$$ It is easy to see that a $q^n \times q^n $ Butson-Hadamard matrix $M$ also satisfies $$MM^*M=q^nM.$$ Our result implies that $M$ can be associated with $0-$plateaued function. This indicates the well-known connection between Butson-Hadamard matrices and bent functions.

As a corollary of Lemma \[Matrix-Eq\], we can characterize $s$-plateaued functions with their first and second derivatives. First and second derivative of a $p-$ary function is defined by $$D_af(x)=f(x+a)-f(x)$$ and $$D_aD_bf(x)=f(x+a+b)+f(x)-f(x+a)-f(x+b)$$ respectively.

$f$ is an $s$-plateaued function from $\mathbb{F}_{p^{n}}$ to $\mathbb{F}_{p}$ if and only if the expression $\sum_{a,b\in \mathbb{F}_{p^{n}}} \zeta_p ^{D_aD_b f(u)}$ does not depend on $u \in \mathbb{F}_{p^{n}} $. This constant expression equals to $p^{n+s}$.

Since the equation $$MM^*M=p^{n+s}M$$ holds, $$M^*MM^*=p^{n+s}M^*$$ 