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+.