NEAR OPTIMAL BOUNDS IN FREIMAN'S THEOREM

148 MARGULIS and MOHAMMADI
all j>s,we have
1
Ct
2j
+
Bt
1
2j
+
At
1
2 j
\
L∈Qt ( 1
2j ,ε)
AL
1
t ,ε <
2j 1
. (68)
2 (1+ε/4)j
Let η1 < 1/4 be as in Proposition 5.3, andletη = η1/2. The Conclusions of
Corollaries 5.5 and 5.11 hold with this η and the corresponding τ.Now let Q < t (1/2j ,ε)
(resp., Q ≥ t (1/2 j ,ε)) be the set of quasi-null subspaces in Qt(1/2 j ,ε) with norm less
than M (resp., greater than or equal to M.)Wehave
|
L∈Q < t (1/2j ,ε) AL 1
t (
|
L∈Q ≥ t (1/2 j ,ε) AL t
( 1
2 j ,ε)|≤
i
t i+j e /(2 )
i et /(2i+j+1 )
t i+j e /(2 )
et /(2i+j+1 )
2 j ,ε)| ≤
L |It (δ)| ≤C1 i
L |It (δ)|≤C2
i
1
2 (1+η)j+i/2 +
1
2 (1+η)j+i/2 ,
e −τt
2 (1/2−τ)i+(1−τ)j
where C1 and C2 are absolute constants independent of µ1. The inequality in the
first line above follows from Corollary 5.5 and from the fact that the definition of
Qt(1/2 j ,ε) excludes exceptional subspaces. The inequalities in the second line follow
from Corollary 5.11. Wenowhave
2 j
j>s
L∈Qt
1
2j ,ε
,
(69)
AL
1
1
t ,ε ≤ C
2j 2s 1
+ . (70)
η M
Here C ′ is an absolute constant which depends on C in (70). This inequality together
with (68) gives
K
( fgr) ˜ >
s (atKqξ) dk ≤ C′
+ C
2εs/4
1 1
+
2ηs M
. (71)
Recall that M was chosen such that C/M < ɛ/6. Now we choose s large enough
such that the right-hand side of (71)islessthanɛ/2. The above estimate holds for all
r.
As we mentioned in the proof of Theorem 3.2, There exists r0 = r0(s, ɛ) such
that if r>r0, then µ(A(r)) r0(s, ɛ). We have
lim sup
t→∞
K
( fgr) ˜ ≤
s (atkqξ)ν(k) dk ≤ 2 s lim sup
t→∞
K
gr(atkqξ)ν(k) dk ≤ ɛ/2.
(72)
Thus (72) and(71) give:ifr>r0(s, ε), then
lim sup
t→∞
fgr(atkqξ) ˜ dk < ɛ. (73)
K