The weak inhomogeneous Duffin–Schaeffer conjecture

The explanation

Take x = 0.3 and a target of 0.5. Multiply x by q = 1, 2, 3, and so on. Then measure how far q × x lands from 0.5 plus some whole number. For q = 1 the gap is 0.2. For q = 2 it is 0.1. For q = 4 it is 0.3, since 1.2 sits 0.3 from 1.5. For q = 5 the gap is exactly 0. The fixed target 0.5 is called the shift. Asking for small gaps of this kind is called inhomogeneous approximation (approximation aimed at a shifted target, not at whole numbers).

Now choose a tolerance ψ(q) for every q, any list of sizes you like. Call q a hit when the gap is smaller than ψ(q). The question is whether a typical x gets infinitely many hits. Here typical means almost every x: the exceptions take up zero total length on the number line. The paper lets that exceptional set depend on both the shift and the tolerancesSource 1.

When the tolerances shrink steadily, the answer was known long ago. Khintchine’s 1924 theorem covers the unshifted case. Szüsz extended this steady case to each fixed shift in 1958Source 2. Without steady shrinking, the problem changes, because many values of q can aim at almost the same spotsSource 3. For an unshifted target, the fractions with bottom 6 show this. Of 0/6 to 5/6, only 1/6 and 5/6 are new. The other four already appear with bottoms 1, 2 or 3. The count of new ones, 2 here, is Euler’s totient φ(6) (how many numbers below 6 share no factor with it).

Duffin and Schaeffer found this overlap problem in 1941. They proposed weighting each tolerance by φ(q)/qSource 4. Their conjecture says that if the sum of φ(q)ψ(q)/q is infinite, almost every x has infinitely many hits. Koukoulopoulos and Maynard proved it in 2020Source 5. A shift makes things harder. Ramírez built counterexamples in 2017 where the plain sum of tolerances is infinite but hits failSource 6. His examples have a finite weighted sum, so the weighted question stayed openSource 7.

That weighted question with a fixed shift was asked by Yu and later named the weak inhomogeneous Duffin–Schaeffer conjectureSource 8. Weak means any whole number may be used to reach the target. A stronger form asks for reduced fractions, and recent papers show that stronger form is false in generalSource 9. Beresnevich, Hauke and Velani proved the weak form for every rational shiftSource 10. What remained was an arbitrary irrational shift with tolerances that may be sparse and unevenSource 11.

The paper claims the full weak formSource 12. For every fixed real shift and every list of tolerances, an infinite weighted sum gives infinitely many hits for almost every xSource 13. In the example, the shift 0.5 is allowed, and so is any irrational shift. The tolerances need not shrink steadily. The paper also claims a finer version measured with Hausdorff measure (a way to size sets of zero length)Source 14.

The hard part is overlap. Two values of q that share large prime factors can aim at nearly the same points again and again. The paper says these repeated alignments are one of two separate obstacles it must controlSource 15. It handles them by revealing prime factors step by step and cancelling the most tightly linked overlaps.

The theorem has limits. It is a one-way test: it says nothing when the weighted sum is finiteSource 16. It does not count how many hits come before a given pointSource 16. It also does not require reduced fractions.

How sure is this? No Lean formalization is listed for this paper in the openai/math release. So no computer proof checker has checked it. The argument is long and new. No specialist review has been published.

What it doesn't show. This does not say what happens when the weighted sum is finite, and it does not cover the version that demands reduced fractions.