38 2. FOUR IMPORTANT LINEAR PDE
Proof. 1. For each fixed x, the mapping y G(x, y) is harmonic, except
for y = x. As G(x, y) = G(y, x), x G(x, y) is harmonic, except for x = y.
Thus x
∂G
∂yn
(x, y) = K(x, y) is harmonic for x R+,
n
y
∂R+.n
2. A direct calculation, the details of which we omit, verifies
(34) 1 =
∂R+n
K(x, y) dy
for each x R+.
n
As g is bounded, u defined by (33) is likewise bounded.
Since x K(x, y) is smooth for x = y, we easily verify as well u
C∞(R+),n
with
Δu(x) =
∂R+n
ΔxK(x, y)g(y) dy = 0 (x
R+).n
3. Now fix
x0
∂R+,
n
ε 0. Choose δ 0 so small that
(35) |g(y)
g(x0)|
ε if |y
x0|
δ, y
∂R+.n
Then if |x
x0|
δ
2
, x
R+,n
(36)
|u(x)
g(x0)|
=
∂Rn
+
K(x, y)[g(y)
g(x0)]
dy

∂R+∩B(x0,δ)n
K(x, y)|g(y)
g(x0)|
dy
+
∂R+−B(x0,δ)n
K(x, y)|g(y)
g(x0)|
dy
=: I + J.
Now (34), (35) imply
I ε
∂R+n
K(x, y) dy = ε.
Furthermore if |x
x0|

δ
2
and |y
x0|
δ, we have
|y
x0|
|y x| +
δ
2
|y x| +
1
2
|y
x0|;
and so |y x|
1
2
|y
x0|.
Thus
J 2 g
L∞
∂R+−B(x0,δ)n
K(x, y) dy

2n+2
g
L∞
xn
nα(n)
∂R+−B(x0,δ)n
|y
x0|−n
dy
0 as xn
0+.
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