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Consider functions on the discrete cube $\{-1,1\}^n$.

We consider the Discrete Fourier Transform of such functions. Suppose we denote the parity function on a subset $S \subseteq [n]$ of co-ordinates by $\Pi_S(x)=\pi_{i \in S}(x_i)$, then the Fourier coefficient $\hat{f}_S$ is simply the expectation: $\hat{f}_S=\mathbb{E}_x[f(x)\Pi_S(x)]$; and by orthonormality of the parity functions, any $f$ may be represented as $f(x)=\sum_{S \subseteq [n]}\hat{f}_S \Pi_S(x)$ (the summation being in $\mathbb{R}$.)

I am interested in knowing the difference between boolean-valued ($\{-1,1\}^n \rightarrow \{-1,1\}$) and real-valued functions ($\{-1,1\}^n \rightarrow \mathbb{R}$). More specifically, I would like to know the difference between the Fourier spectra of either class of functions.

What properties of the Fourier spectra hold for one class but not for the other?

(As an example: It can be proved that for Boolean valued functions, if all the weight is concentrated on Fourier coefficients of size at most 1 then the function is either a constant or a dictator. This is not true for real-valued functions.)

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I do not understand what caused the downvote... It would not hurt if the author would make explicit what exactly he means by Fourier transform, though (I imagine he looks at the domain as an abelian group and considers the corresponding Fourier analysis...) – Mariano Suárez-Alvarez Jan 12 at 23:21
Thanks Mariano. Yes, the domain is interpreted as an abelian group; the fourier coefficients are just the correlations with the parity functions. I am not sure what the name for this fourier transform is. – Klingonesque Jan 12 at 23:59
It's called the discrete Fourier transform, but you're going to have to specify whether you want the convolution for your first class of functions to be mod 2 or over the integers. – Qiaochu Yuan Jan 13 at 2:34
Qiaochu: Thanks! I hope my edit also answers your question. – Klingonesque Jan 13 at 15:41

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