For a distribution X over $\{0,1\}^n$, we can define the Fourier coefficient of the distribution as $\hat{Y}(s)= \textbf{E}_{y\in Y}({\chi_s(y)})$.The question I have is, do there exist distributions that can be efficiently computed such that at most polynomially many Fourier coefficients are non-zero?
Are there any distributions with only polynomially many non-zero Fourier coefficients and a small support?
reference request;pseudorandomness
If you want $Y$ to have entropy less than $0.99 n$ bits, the answer is no, by the uncertainty principle: Either $Y$ has high entropy or its Fourier transform has large support.Theorem. Let $H(Y)$ be the Shannon entropy of $Y$ and let $F  \subset \{0,1\}^n$ be the support of $\hat{Y}$. Then $H(Y) \geq n - \log |F|$.Proof. Consider the collision probability of $Y$ (the probability that two independent samples of $Y$ are the same). By Parseval's identity,$$CP(Y) = \sum_y \text{Pr}[Y=y]^2 = 2^{-n} \sum_s \hat{Y}(s)^2 \leq 2^{-n} |F|,$$ as $|\hat{Y}(s)| \leq 1$. On the other hand, $CP(Y) = 2^{-H_2(Y)}$, where $H_2(Y)$ is the Renyi entropy of $Y$. Noting that $H_2(Y) \leq H(Y)$ gives the result. Q.E.D.If $|F| = \mathrm{poly}(n)$, then $H(Y) = n-O(\log n)$. So, if $Y$ is the output of a pseudorandom generator, the seed length is at least $n-O(\log n)$.