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+.
