break up the surface integral over ∂ E into the pieces ∂ B
ǫ
x, {x
1
M } ∩ ∂ D and D ∩ π
M
to end up with
Z
{x
1
M }∩∂ D
=
− Z
D ∩π
M
+ Z
∂ B
ǫ
x
¨
u y
∂ ∂ n
y
G
D
x, y
− G
D
x, y ∂ u
∂ n
y
y «
σd y 18
where now the
∂ ∂ n
y
in the ∂ B
ǫ
x integral is inward normal differentiation for B
ǫ
x. Now u = f on ∂ D, supp f ⊆ B
M
0 ∩ B
ǫ
x
c
and G
D
x, · = 0 on ∂ D, so we have Z
∂ D
f y
∂ ∂ n
y
G
D
x, y
σd y = Z
{x
1
M }∩∂ D
¨ u y
∂
∂ n
y
G
D
x, y
− G
D
x, y ∂ u
∂ n
y
y «
σd y
=
− Z
D ∩π
M
+ Z
∂ B
ǫ
x
{}, by 18.
19
Once we prove lim
M →∞
Z
D ∩π
M
¨ u y
∂
∂ n
y
G
D
x, y
− G
D
x, y ∂ u
∂ n
y
y «
σd y = 0 20
and lim
ǫ→0
Z
∂ B
ǫ
x
¨ u y
∂
∂ n
y
G
D
x, y
− G
D
x, y ∂ u
∂ n
y
y «
σd y = 2ux, 21
we can let M → ∞ and ǫ → 0 in 19 to get
ux = 1
2 Z
∂ D
f y
∂ ∂ n
y
G
D
x, y
σd y, which is exactly 13, as desired.
Formulas 20–21 will be proved in the next two subsections.
2.3 Proof of 20
The following result is a consequence of the proof of Lemma 6.5 in Gilbarg and Trudinger 1983 combined with the comments subsequent to the proof.
Lemma 2.5. Let Ω be a domain in R
d
with a C
2, α
boundary portion T . Suppose u ∈ C
2, α
Ω ∪ T is a solution of
∆
R
d
+ Hu = 0 on
Ω u = 0
on T,
2670
where Λ := |H|
0, α;Ω
= sup
Ω
|H| + sup
x, y ∈Ω
x 6= y
|Hx − H y| |x − y|
α
∞. Then for any z
∈ T there is δ 0 such that for Bz = B
δ
z ∩ Ω and T z = B
δ
z ∩ T we have
sup {dx, ∂ Bz − T z|∇ux|: x ∈ Bz ∪ T z} ≤ C sup
Bz
|u|. Here
δ depends only on the diameter of the domain of the C
2, α
diffeomorphism ψ that straightens the
boundary near z and C depends only on d, α, Λ and the C
2, α
bounds on ψ.
We now apply Lemma 2.5.
Lemma 2.6. Suppose b : D → R is bounded with
sup
M ≥1
aM
α+2
sup |bz − b y|
|z − y|
α
: z, y ∈ D; z 6= y; z
1
, y
1
∈ M − aM, M + aM ∞,
sup
M ≥1
aM
2
sup {|bz|: z ∈ D, z
1
∈ M − aM, M + aM} ∞. If v
∈ C
2, α
D \{x} is a solution of
∆ + bv = 0 on
D \{x}
with v = 0 on ∂ D outside a compact set, then for large M ,
|∇vz| ≤ C aM
−1
sup {|v y|: y ∈ D, y
1
∈ M − aM, M + aM}, z
∈ D ∩ π
M
. Proof. Define
γ
M
t = at aM + M
aM ,
t ≥
1 2
− M aM
. Then for
H
M
= z
1
, ˜ z: z
1 1
2
− M aM
, |˜z| γ
M
z
1
we have
z ∈ H
M
⇔ aMz + M x ∈ D
z ∈ H
M
∩ {−1 z
1
1} ⇔ aMz + M x ∈ D ∩ { y
1
, ˜ y: M
− aM y
1
M + aM } z
∈ ∂ H
M
∩ {−1 z
1
1} ⇔ aMz + M x ∈ ∂ D ∩ { y
1
, ˜ y: M
− aM y
1
M + aM }. 22
2671
Notice H
M
is obtained from D via translating by −M x
and then scaling by 1 aM . For any function
g on D, we define g
M
z = gaM z + M x ,
z ∈ H
M
. Then for
L
M
= ∆ + aM
2
b
M
we have L
M
v
M
= 0 on
H
M
\ x
− M x aM
. Since
aM M
→ 0 as M → ∞, for large M we have x
1
M − aM. Thus for large M ,
L
M
v
M
= 0 on
H
M
∩ {−1 z
1
1}. Since v
∈ C
2, α
D\{x} and v = 0 on ∂ D outside a compact set, by making M larger if necessary, we have that
v
M
∈ C
2, α
H
M
∩ {−1 ≤ z
1
≤ 1} v
M
= 0 on
∂ H
M
∩ {−1 ≤ z
1
≤ 1}. We are going to apply Lemma 2.5 to L
M
and v
M
on Ω = H
M
∩ {−1 z
1
1} T =
∂ H
M
∩ {−1 z
1
1}. This is legitimate because by our hypotheses on b, there exists Λ
0 such that for large M , |aM
2
b
M
|
0, α;Ω
≤ Λ. 23
By symmetry, compactness and our Blanket Assumptions on a ·, for each z ∈ ∂ H
M
∩{−
1 2
≤ z
1
≤
1 2
}, the C
2, α
bounds on the diffeomorphism straightening the boundary near z are independent of z and large M . Roughly speaking, for large M , the set H
M
∩ {−1 z
1
1} looks like the set {z
1
, ˜ z:
− 1 z
1
1, |˜z| 1}. Combined with 23, it follows that the constant C appearing in Lemma 2.5 is independent of such z and large M . Likewise, the
δ in the lemma is also independent of z and large M . The net effect is that for some
δ 0 and C 0, for Bz = B
δ
z ∩ Ω T z = B
δ
z ∩ ∂ Ω, we have
sup
y ∈Bz∪T z
d y, ∂ Bz − T z|∇v
M
y| ≤ C sup
Bz
|v
M
| ≤ C sup
Ω
|v
M
| 24
whenever z ∈ ∂ H
M
∩ {−
1 2
≤ z
1
≤
1 2
} and M is large. 2672
In particular, given y ∈ π
∩ H
M
∩ {| ˜y| 1 −
δ 2
}, choose z ∈ π ∩ ∂ H
M
such that d y, π
∩ ∂ H
M
= d y, z. Then y
∈ B
δ
z and by 24, for M large, |∇v
M
y| ≤ C d y, ∂ Bz − T z
−1
sup
Ω
|v
M
|. Since y
∈ π , d y,
∂ Bz − T z ≥
δ 2
and we get that for some C
1
0, for large M , |∇v
M
y| ≤ C
1
δ
−1
sup
Ω
|v
M
|, y
∈ π ∩ H
M
∩ | ˜y| ≥ 1 −
δ 2
. 25
On the other hand, by the Schauder interior estimates Gilbarg and Trudinger 1983, Theorem 6.2,
sup
Ω
d y, ∂ Ω|∇v
M
y| ≤ C
2
sup
Ω
|v
M
|, where C
2
0 is independent of large M . Combined with 25, we get that for some C 0, for all large M ,
|∇v
M
y| ≤ Cδ
−1
sup
Ω
|v
M
|, y
∈ π ∩ H
M
. Converting back to v and using 22, for all large M we have
|aM∇vaM y + M x | ≤ Cδ
−1
sup
z ∈Ω
|vaMz + M x |,
aM y + M x ∈ D ∩ π
M
, which is to say
|∇vz| ≤ Cδ
−1
aM
−1
sup {|vw|: w ∈ D, w
1
∈ M − aM, M + aM}, z
∈ D ∩ π
M
, as desired.
Now we can prove 20. Taking b = 0 in Lemma 2.6, v = u or G
D
x, · satisfies the required hypotheses, so for some C
0, for all large M , ∂ u
∂ n
y
y ≤ C aM
−1
sup |u|,
y ∈ π
M
∩ D ∂
∂ n
y
G
D
x, y ≤ C aM
−1
sup {G
D
x, z: z ∈ D, M − aM z
1
M + aM }, y
∈ π
M
∩ D, where
∂ ∂ n
y
is inward normal differentiation at the boundary part π
M
∩ D of D ∩ {z
1
M }. But it is well-known that
|G
D
x, y| ≤ C|x − y|
2 −d
, and since
aM M
→ 0 as M → ∞, it follows that for large M ,
∂ ∂ n
y
G
D
x, y ≤ C aM
−1
M
2 −d
, y
∈ π
M
∩ D. Since u is bounded, we end up with
u y ∂
∂ n
y
G
D
x, y − G
D
x, y ∂ u
∂ n
y
y ≤ C aM
−1
M
2 −d
, y
∈ π
M
∩ D, 2673
for all large M . Hence as M → ∞
Z
D ∩π
M
¨ u y
∂ ∂ n
y
G
D
x, y − G
D
x, y ∂ u
∂ n
y
y «
σd y ≤ C aM
−1
M
2 −d
aM
d −1
= C aM
M
d −2
→ 0, as desired.
2.4 Proof of 21
