NEAR OPTIMAL BOUNDS IN FREIMAN'S THEOREM

NEAR OPTIMAL BOUNDS IN FREIMAN’S THEOREM 7
LEMMA 5
The set X + Y − X − Y contains a Bohr set B(Ɣ, γ ) such that |Ɣ| ≪K 3ε log K and
γ ≫ K −3ε log −1 K.
Proof
Define
= r ∈ Zn : | ˆX(r)| ≥K −ε |X|/2 .
We will show that B = B(, 1/6) ⊆ X + Y − X − Y .Letr(x) be the number of
representations of x in X + Y − X − Y .Ifx ∈ B, then, by (1) and Parseval’s formula,
we have
r(x) = 1 n−1
| ˆX(s)|
n
2 | ˆY (s)| 2 e 2πixs/n
≥ 1
n
≥ 1
2n
s=0
s∈
s∈
s=0
| ˆX(s)| 2 | ˆY (s)| 2 cos(2πxs/n) − 1
n
| ˆX(s)| 2 | ˆY (s)| 2 − 1
n
| ˆX(s)| 2 | ˆY (s)| 2
s∈
| ˆX(s)| 2 | ˆY (s)| 2
s∈
≥ 1 n−1
| ˆX(s)|
2n
2 | ˆY (s)| 2 − 3
| ˆX(s)|
2n
2 | ˆY (s)| 2
s∈
> E(X, Y )/2 − 3
8n K−2ε |X| 2
n−1
| ˆY (s)| 2
s=0
≥ E(X, Y )/2 − (3/8)K −2ε |X| 2 |Y | > 0.
By Chang’s spectral lemma (Lemma 4), there is a set Ɣ such that |Ɣ| ≪
K 2ε log(n/|X|) ≪ K 3ε log K and ⊆ Span(Ɣ), so that
B Ɣ, 1/(6|Ɣ|) ⊆ B(, 1/6) ⊆ X + Y − X − Y. ✷
Remark. We proved Bogolyubov-Ruzsa’s lemma in a standard way, but one can easily
strengthen the assertion by removing one summand. Indeed, by (3), we have
1 n−1
ˆX(s)| ˆY (s)|
n
s=0
2 e −2πib0s/n −ε
≥ K |X||Y |.
Therefore, using a similar argument as in the proof of Lemma 5, we infer that a shift
of B(Ɣ, γ ) is contained in X + Y − Y.