33.5. HELMHOLTZ DECOMPOSITIONS 1147

Lemma 33.5.2 For n≥ 2

Φ,i j (y) =Ωi j (y)|y|n

whereΩi j is Lipschitz continuous on Sn−1, (33.5.66)

Ωi j (λy) = Ωi j (y), (33.5.67)

for all λ > 0, and ∫Sn−1

Ωi j (y)dσ = 0. (33.5.68)

Proof:Proof: The case n = 2 is left to the reader. 33.5.66 and 33.5.67 are obvious from the

above descriptions. It remains to verify 33.5.68. If n ≥ 3 and i ̸= j, then this formula isalso clear from 33.5.64. Thus consider the case when n≥ 3 and i = j. By symmetry,

I ≡∫

Sn−11−ny2

i dσ =∫

Sn−11−ny2

jdσ .

Hence

nI =n

∑i=1

∫Sn−1

1−ny2i dσ =

∫Sn−1

(n−n∑

iy2

i

)dσ

=∫

Sn−1(n−n)dσ = 0.

This proves the lemma.Let U be a bounded open set locally on one side of its boundary having Lipschitz

boundary so the divergence theorem holds and let B = B(0,R) where

B⊇U−U ≡ {x−y : x ∈U,y ∈U}

Let f ∈C∞c (U) and define for x ∈U ,

u(x)≡∫

BΦ(y) f (x−y)dy =

∫U

Φ(x−y) f (y)dy.

Let h(y) = f (x−y) . Then since Φ is in L1 (B),

∆u(x) =∫

BΦ(y)∆ f (x−y)dy =

∫B

Φ(y)∆h(y)dy

=∫

B\B(0,ε)∇ · (∇h(y)Φ(y))−∇Φ(y) ·∇h(y)dy

+∫

B(0,ε)Φ(y)∆h(y)dy.