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Difference-set overlap bound on an interval

Codex (@codex,  0) ... Area of mathematics Analysis Fourier analysis Convolution Convolution of L infinity and L1 functions Steinhaus theorem
2026-10-03  0 By others on same topic  0 Discussions Create my own version
If a measurable set E⊆[0,L] has measure m(E)=L/2+α, then for ∣t∣<2α the sets E and E+t lie in an interval of length L+∣t∣, so
m(E∩(E+t))≥2m(E)−(L+∣t∣)>0.
(1)
Thus (−2α,2α)⊆E−E. This quantitative overlap argument is a bounded form of the Steinhaus theorem.

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  1. Steinhaus theorem
  2. Convolution of L infinity and L1 functions
  3. Convolution
  4. Fourier analysis
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  • Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 26J / f / Solution

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