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Let $\mathbb{F}_2=\{0,1\}$ be the field with two elements, and let $u:\mathbb{F}_2^n\rightarrow \mathbb{F}_2$. Suppose that $n$ is odd.

Is it possible that $$ \sum_{x \in \mathbb{F}_2^n}(-1)^{u(x)+u(x+a)}= 0, $$ for every $a \neq 0$ in $\mathbb{F}_2^n$?

I treat the sum here as natural number, not as an element of $\mathbb{F}_2$.


This question arose in the context of this question. (Trying to derive a lower bound on the approximate multiplicativity of a Boolean function). When $n$ is even there are such functions; this is related to Bent functions.

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  • $\begingroup$ @PeterMueller The OP reask their question with the additional condition that $n$ is odd. $\endgroup$
    – Zerox
    Mar 5, 2022 at 18:53

1 Answer 1

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I think the answer is no. To see this, observe that \begin{equation*} \left(\mathbb{E}_{\mathbf{x}}[(-1)^{u(\mathbf{x})}]\right)^2=\mathbb{E}_{\mathbf{x}}[(-1)^{u(\mathbf{x})}]\mathbb{E}_{\mathbf{y}}[(-1)^{u(\mathbf{y})}]=\mathbb{E}_{\mathbf{x},\mathbf{y}}[(-1)^{u(\mathbf{x})+u(\mathbf{y})}]=\mathbb{E}_{\mathbf{x},\mathbf{a}}[(-1)^{u(\mathbf{x})+u(\mathbf{x}+\mathbf{a})}], \end{equation*} where we use independence and then rewriting $\mathbf{y}=\mathbf{x}+\mathbf{a}$, and where all expectations are uniform over $\mathbb{F}_2^n$. The assumption then is that for $\mathbf{a}=\mathbf{0}$, the expectation over $\mathbf{x}$ is clearly $1$, while for all other values of $\mathbf{a}$, the expectation over $\mathbf{x}$ is $0$. It follows that \begin{equation*} \left(\mathbb{E}_{\mathbf{x}}[(-1)^{u(\mathbf{x})}]\right)^2 = \frac{1}{2^n}\iff \mathbb{E}_{\mathbf{x}}[(-1)^{u(\mathbf{x})}] = \frac{\pm 1}{2^{n/2}} \end{equation*} However, it's not hard to see that this expectation must also be an integer multiple of $1/2^n$. But then this integer must be $\pm 2^{n/2}$, but this is integral if and only if $n$ is even.

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  • $\begingroup$ Wow, that is a very beautiful argument! Thanks. $\endgroup$ Mar 5, 2022 at 19:00
  • $\begingroup$ You might be interested in the following follow-up question. I think that I was able to use your argument to squeeze a bit more information, namely to show that the relevant quantity needs to be "far away" from zero. Here: mathoverflow.net/questions/417592/… $\endgroup$ Mar 7, 2022 at 13:34
  • $\begingroup$ BTW, I will happily give you credit for this argument in the follow-up question, if you want. (I am just studying this material and enjoy playing with it. There is no heavy goal here). $\endgroup$ Mar 7, 2022 at 13:36

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