Let $f$ be a boolean function with minimum degree real polynomial representing it be of degree $d$.

Is there a relation between number of zeros $f$ or $1-f$ and degree $d$?

  • $\begingroup$ Yes. More generally, the Fourier coefficients are all multiples of $2^{-d}$. $\endgroup$ Commented May 10, 2016 at 4:36
  • $\begingroup$ ah serious???? how so? $\endgroup$
    – Turbo
    Commented May 10, 2016 at 4:40
  • $\begingroup$ You can prove it by induction. Should be in O'Donnell's book. $\endgroup$ Commented May 10, 2016 at 4:57

1 Answer 1


Exercise 12 (online version) in §1 of Ryan O'Donnell's Analysis of Boolean functions shows that all Fourier coefficients of a degree $d$ function are integer multiples of $2^{-d}$. In particular, the number of zeroes of $f$ is an integer multiple of $2^{n-d}$.

Moreover, for every integer $k \in \{0,\ldots,2^d\}$ there is a degree $d$ Boolean function having exactly $k2^{n-d}$ zeroes. This easily follows from the fact that a function on $d$ variables has degree at most $d$.

  • $\begingroup$ Sorry to nitpick, but when you said ‘degree d function’ above did you mean ‘degree d Boolean function’? If so, do you know what happens if we drop the Boolean requirement? I.e. take f to be any real valued function (of Fourier degree d) on the hypercube. $\endgroup$
    – gen
    Commented Sep 27, 2021 at 21:27
  • $\begingroup$ Right, I meant Boolean degree $d$ function. I'm not sure what happens in general – if you're interested, you can ask a question about it. $\endgroup$ Commented Sep 27, 2021 at 21:30
  • $\begingroup$ Oh yes, of course. I wonder if there still might be a way to relate the 1-norm of f to it’s Fourier degree in this case as well… What if one assumes that the 2-norm of f is 1? $\endgroup$
    – gen
    Commented Sep 27, 2021 at 21:30
  • $\begingroup$ The question is about number of zeroes. That's not a $p$-norm for any $p$, though it's related to the so-called $L_0$ "norm". $\endgroup$ Commented Sep 27, 2021 at 21:33

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