Boolean functions violate the square-root degree bound by arbitrary factors

What OpenAI claims

Disproves the proposed square-root bound relating a Boolean function's linear Fourier coefficients to its polynomial degree. For every C > 0, there is a sign-valued Boolean function f with ∑_if̂({i}) > C√(deg(f)). Thus its total signed correlation with individual input bits can exceed the proposed bound by an arbitrary factor. (Lean)

Paper

1. Unbounded Violations of the Square-Root Degree Bound

   26 September 2026 · Preprint, not peer reviewed · PDF

   We disprove the Gopalan–Servedio square-root conjecture, even up to an arbitrary constant factor. For every real C > 0, there is a nonconstant Boolean function f:{-1,1}ⁿ → {-1,1} on a finite sign cube such that

Status

Preprint. The paper has not been independently reviewed. Treat the result as OpenAI's claim until other mathematicians have checked it. If a video exists, it illustrates the statement; it is not the proof.