An autocorrelation problem with functions that can take negative values

AlphaEvolve found a construction that improved on the best bound known at the time; the problem itself remains open.

Problem
Problem 4 in the AlphaEvolve repository: An autocorrelation problem with functions that can take negative values
Credit
AlphaEvolve (Google DeepMind) · LLM-guided evolutionary code search, with Georgiev, Gómez-Serrano, Tao and Wagner
Checked
Construction can be checked directlyPreprint (arXiv, 2025)
Timeline
Paper submitted 2025-11-03
Sources
Problem page (AlphaEvolve repository)Paper (arXiv:2511.02864)
PARTIALImproved bound · new construction

The problem

Let CC be the best constant for which one has

max⁡−1/2≤t≤1/2∣∫Rf(t−x)f(x)dx∣≥C(∫−1/41/4f(x)dx)2\max_{-1/2 \leq t \leq 1/2}\left |\int_\mathbb{R} f(t-x) f(x) dx\right| \geq C \left(\int_{-1/4}^{1/4} f(x) dx\right)^2

for all f ⁣:[−1/4,1/4]→Rf \colon [-1/4,1/4] \to \mathbb{R} (note ff can take negative values). What is CC?

Statement quoted from problem 4 of the AlphaEvolve repository of problems (Google DeepMind, CC BY 4.0).

What AI did

AlphaEvolve found a construction that improved on the best bound known at the time; the problem itself remains open. Status in the repository: New result. See Section 6.2 in "Mathematical Exploration and Discovery at Scale" paper. Records may have moved since; the repository's notes below list the latest bounds it knows of.

Notes from the repository

This is one of two very similar optimization problems, both of which have been studied in the literature, with some results for one problem incorrectly attributed to another. We will update the references on this problem (and add its sibling) soon.

How it was checked

The results are explicit constructions or programs; the repository's notebooks include the verification code used to score them.

Limitations

An improved bound is progress, not a solution: the exact answer to the problem is still unknown. The authors report that results were better when experts set up and guided the search.

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