Every bounded Boolean function admits a Fourier expansion in the basis of monomials (Ref: Boolean function analysis, Ryan O'Donnel Course) I was auditing Ryan's course and he said (05:00) that the following function :
$$ g(x_1,x_2,x_3,x_4) = \begin{cases} 1 & (x_3 =x_4=1) \vee (x_1 \leq x_2 \leq x_3 \leq x_4) \vee (x_1 \geq x_2 \geq x_3 \geq x_4) \\ -1 & \textit{otherwise} \end{cases} $$
admits the following Fourier expansion:
$$g(x_1,x_2,x_3,x_4) = \frac{1}{8} -\frac{1}{8} x_1 + \frac{1}{8} x_2 - \frac{1}{8} x_3 - \frac{1}{8} x_4 + \frac{3}{8} x_1 x _2 + \frac{1}{8} x_1x_3 - \frac{3}{8} x_1 x _4 + \frac{3}{8} x_2 x _3 - \frac{1}{8} x_2 x _4 + \frac{5}{8} x_3 x _4 + \frac{1}{8} x_1 x_2 x_3 + \frac{1}{8} x_1 x_2 x_4 - \frac{1}{8} x_1 x_3 x_4 + \frac{1}{8} x_2 x_3 x_4 - \frac{1}{8} x_1 x_2 x_3 x_4$$
But just replacing $x_3 = x_4 =1$ doesn't satisfy the definition of $g$.Was it some typo mistake or there's something I'm missing?
I'm interested particularly at this function, is there an easy way to find Fourier coefficients without calculating the expectations?