NEAR OPTIMAL BOUNDS IN FREIMAN'S THEOREM

CALDERÓN INVERSE PROBLEM ON RIEMANN SURFACES 117
Proof of Lemma A.3
We need to find Rh satisfying RhL 2 ≤ Ch5/4 and solving
(g + V )Rh =−(g + V )vh =: Mh.
Thanks to Corollary A.2, it suffices to show that MhB ∗ ≤ Ch5/4 . Thus, let w ∈ B;
then we have, by (A.3),
wMh dvg = I1 + I2 + I3,
where
I1 : =
I2 : = 1
h
M0
|Z|≤ √ h
I3 : = 2
h 3/2
wVivhe 2ρ dZ,
|Z|≤ √ we
h
(1/h)α·Z χ1
|Z|≤ √ wχ2
h
Z√h
e 2ρ dZ,
Z√h
e (1/h)α·Z dZ,
and χ1 = gη, χ2 = i∂xη − ∂yη. In the above equation, the term I3 has the worst
growth when h → 0. We analyze its behavior, and the preceding terms can be treated
in similar fashion. One has
I3 =−h 1/2
=−h 1/2
|Z|≤ √ wχ2
h
|Z|≤ √ (∂x − i∂y)
h
2
Z√h
(∂x − i∂y) 2 e (1/h)α·Z dZ
Z√h
wχ2 e (1/h)α·x dZ.
Notice that the boundary term in the integration by parts vanishes because w ∈ H 1 0
and ∂νw|∂M0 vanishes on the support of η. The term (∂x − i∂y) 2 (wχ2(Z/ √ h)) has
derivatives hitting both χ2(Z/ √ h) and w. The worst growth in h would occur when
both derivatives hit χ2(Z/ √ h), in which case a h −1 factor would come out. Combined
with the h 1/2 term in front of the integral this gives a total of a h −1/2 in front. By this
observation we have improved the growth from h −3/2 to h −1/2 . Repeating this line of argument
and using the Cauchy-Schwarz inequality, we can see that |I3| ≤Ch 5/4 wH 2
(an elementary computation shows that functions of the form χ(Z/ √ h)e (1/h)α·Z have
L 2 norm bounded by Ch 3/4 ). Therefore, (1/h 3/2 )χ2(Z/ √ h)e (1/h)α·Z B ′ ≤ Ch5/4 ,
and we are done.