Definition of Bent functions over Galois rings using Fourier transform (walsh transform) First, note that the following definition is true:
Definition (Carlet): Let $R=GR(p^k,m)$. A function $f$  from $R^n$ to $R$ is bent if
$$|\sum_{x \in R^n}  w^{Tr(f(x)-ax)}|=|R|^{n/2}$$
where $a \in R^n$, $w=e^{2\pi i/p^k}$,Tr is the trace function from  $GR(p^k,m)$ to $GR(p^k,1)=Z_{p^k}$ and $ax$ is the dot product of $a$ with $x$.
According to this definition can we answer the following question:
Let f be a function from $GR(p^2,m)$ to $GR(p^2,1)=Z_{p^2}$ where $p$ is an odd prime and $m>1$ be a positive integer.
For $f$ to be a bent function, what property should its walsh transform satisfy? 
That is, is the following true?
$f$ is bent if
$$|\sum_{x \in GR(p^2,m)}  w^{Tr(f(x)-ax)}|=|GR(p^2,1)|^{m/2}$$
where $w=e^{2\pi i/p^2}$, $a \in GR(p^2,m)$,  Tr is the trace function from  $GR(p^2,m)$ to $GR(p^2,1)$ and $ax$ is the dot product of $a$ with $x$.
Many thanks in advance.
 A: It seems to me that in your special case we have $R=GR(p^2,m)$, so $|R|=p^{2m}$. Furthermore, it appears that $n=1$ (a single variable ranging over $R$. The Walsh transform on functions with domain $R$ seeks to express them in terms of the characters of the additive group of $R$. In this case all the characters $\psi:R\to\mathbb{C}^*$ of $R$ are of the form
$$
\psi_a(x)=w^{Tr(ax)},
$$
where $w=e^{2\pi i/p^2}$. So if $g:R\to\mathbb{C}^*$  is any complex-valued function on $R$, the components of its Walsh transform $\hat g$ are
$$
\hat{g}(a)=\sum_{x\in R}g(x)\overline{\psi_a(x)}.
$$
(Some authors would use a scaling factor $1/|R|$ here.)
In your case of $f:GR(p^2,m)\to GR(p^2,1)$, so we most likely want to study the Walsh transform of the `exponential' version $x\mapsto w^{f(x)}$.
The bent-property in all the contexts is to require that all the components of the Walsh transform have the same magnitude (bent = as far aways from linear, i.e. a character, as possible). Here that would translate to the requirement that for all $a\in R$ the sum
$$
\hat{f}(a)=\sum_{x\in R}w^{f(x)-Tr(ax)}
$$
should have absolute value $\sqrt{|R|}=p^m$.
