Fourier decomposition in terms of another basis Given a Boolean function $f:\{-1,1\}^n\rightarrow \{-1,1\}$, it is well know that the Fourier decomposition of $f$ can be written as $f(x)=\sum_{S\subseteq \{1,\ldots,n\}} \widehat{f}(S) \prod_{i\in S} x_i$ where $\widehat{f}(S)$ are the Fourier coefficients. So the quantity $\prod_{i\in S} x_i$ can be viewed as the parity of $x\in \{-1,1\}^n$ when restricted to $S\subseteq [n]$. 

Is there a way to write the Fourier decomposition in terms of the AND of its variables, i.e., is it possible to express every Boolean function $f$ in terms of 
$f(x)=\sum_{S\subseteq \{1,\ldots,n\}} \widehat{f}(S) AND(x_{S})$?  Can this be generalized even further? In general, can any polynomial $p:\{-1,1\}^n\rightarrow \mathbb{R}$ be written in terms of an "AND" decomposition instead of a parity decomposition (in the Fourier sense).
As pointed out in the comments if $u\in \{\pm 1\}$ then $x=x(u) \in \{0,1\}$ where $$x(u)=\frac{1-u}{2},$$ with $x(-1)=1,$ and $x(1)=0.$ This will then yield
$$f(x)=2^{n-1}f(0)-\frac{1}{2} \sum_{S \in \{1,\ldots,n\}} \hat{f}(S) \prod_{i \in S} x_i,$$ and if we denote the $\{0,1\}$ valued version of $f$ as $\tilde{f},$
$$\frac{1-\tilde{f}(x)}{2}=2^{n-1}f(0)-\frac{1}{2} \sum_{S \in \{1,\ldots,n\}} \hat{f}(S) \prod_{i \in S} x_i,$$ leading to
$$\frac{\tilde{f}(x)}{2}=\frac{1}{2}\left[1+ \sum_{S \in \{1,\ldots,n\}} \hat{f}(S) \prod_{i \in S} x_i\right]-2^{n-1}f(0),$$
or
$$\tilde{f}(x)=1-2^{n}f(0)+\sum_{S \in \{1,\ldots,n\}} \hat{f}(S) \prod_{i \in S} x_i,$$
if I haven't made any errors along the way.