24 2. FOUR IMPORTANT LINEAR PDE

As the expression on the right-hand side of (10) is continuous in the variable

x, we see u ∈

C2(Rn).

2. Since Φ blows up at 0, we will need for subsequent calculations to

isolate this singularity inside a small ball. So ﬁx ε 0. Then

(11)

Δu(x) =

B(0,ε)

Φ(y)Δxf(x − y) dy +

Rn−B(0,ε)

Φ(y)Δxf(x − y) dy

=: Iε + Jε.

Now

(12) |Iε| ≤ C

D2f

L∞(Rn)

B(0,ε)

|Φ(y)| dy ≤

Cε2|

log ε| (n = 2)

Cε2

(n ≥ 3).

An integration by parts (see §C.2) yields

(13)

Jε =

Rn−B(0,ε)

Φ(y)Δyf(x − y) dy

= −

Rn−B(0,ε)

DΦ(y) · Dyf(x − y) dy

+

∂B(0,ε)

Φ(y)

∂f

∂ν

(x − y) dS(y)

=: Kε + Lε,

ν

denoting the inward pointing unit normal along ∂B(0,ε). We readily check

(14) |Lε| ≤ Df

L∞(Rn)

∂B(0,ε)

|Φ(y)| dS(y) ≤

Cε| log ε| (n = 2)

Cε (n ≥ 3).

3. We continue by integrating by parts once again in the term Kε, to

discover

Kε =

Rn−B(0,ε)

ΔΦ(y)f(x − y) dy −

∂B(0,ε)

∂Φ

∂ν

(y)f(x − y) dS(y)

= −

∂B(0,ε)

∂Φ

∂ν

(y)f(x − y) dS(y),

since Φ is harmonic away from the origin. Now DΦ(y) =

−1

nα(n)

y

|y|n

(y = 0)

and

ν

=

−y

|y|

= −

y

ε

on ∂B(0,ε). Consequently

∂Φ

∂ν

(y) =

ν

·DΦ(y) =

1

nα(n)εn−1

on ∂B(0,ε). Since

nα(n)εn−1

is the surface area of the sphere ∂B(0,ε), we

have

(15)

Kε = −

1

nα(n)εn−1

∂B(0,ε)

f(x − y) dS(y)

= −−

∂B(x,ε)

f(y) dS(y) → −f(x) as ε → 0.