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Difference-set overlap bound on an interval
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Area of mathematics
Analysis
Fourier analysis
Convolution
Convolution of L infinity and L1 functions
Steinhaus theorem
2026-10-03
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If
a
measurable
set
E
⊆
[
0
,
L
]
has
measure
m
(
E
)
=
L
/2
+
α
, then for
∣
t
∣
<
2
α
the
sets
E
and
E
+
t
lie in an interval of
length
L
+
∣
t
∣
, so
m
(
E
∩
(
E
+
t
))
≥
2
m
(
E
)
−
(
L
+
∣
t
∣
)
>
0.
(1)
Thus
(
−
2
α
,
2
α
)
⊆
E
−
E
. This quantitative overlap argument is
a
bounded form of the
Steinhaus theorem
.
Ancestors
(8)
Steinhaus theorem
Convolution of L infinity and L1 functions
Convolution
Fourier analysis
Analysis
Area of mathematics
Mathematics
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(1)
Past exam of the mathematics course of the University of Cambridge
/
2018
/
ii
/
Paper 3
/
26J
/
f
/
Solution
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