1148 CHAPTER 33. FOURIER ANALYSIS IN Rn

The last term converges to 0 as ε → 0 because Φ is in L1 and ∆h is bounded. Sincespt(h)⊆ B, the divergence theorem implies

∆u(x) =−∫

∂B(0,ε)Φ(y)∇h(y) ·ndσ −

∫B\B(0,ε)

∇Φ(y) ·∇h(y)dy+ e(ε) (33.5.69)

where here and below, e(ε)→ 0 as ε→ 0. The first term in 33.5.69 converges to 0 as ε→ 0because ∣∣∣∣∫

∂B(0,ε)Φ(y)∇h(y) ·ndσ

∣∣∣∣≤{ Cnh1

εn−2 εn−1 =Cnhε if n > 2Ch (lnε)ε if n = 2

and since ∆Φ(y) = 0,∇Φ(y) ·∇h(y) = ∇ · (∇Φ(y)h(y)).

Consequently

∆u(x) =−∫

B\B(0,ε)∇ · (∇Φ(y)h(y))dy+ e(ε).

Thus, by the divergence theorem, 33.5.65, and the definition of h above,

∆u(x) =∫

∂B(0,ε)f (x−y)∇Φ(y) ·ndσ + e(ε)

=∫

∂B(0,ε)f (x−y)

(− y

an−1 |y|n)·(− y|y|

)dσ + e(ε)

= −(∫

∂B(0,ε)f (x−y)dσ (y)

)1

an−1εn−1 + e(ε).

Letting ε → 0,−∆u(x) = f (x).

This proves the following lemma.

Lemma 33.5.3 Let U be a bounded open set in Rn with Lipschitz boundary and let B ⊇U−U where B = B(0,R). Let f ∈C∞

c (U). Then for x ∈U,∫B

Φ(y) f (x−y)dy =∫

UΦ(x−y) f (y)dy,

and it follows that if u is given by one of the above formulas, then for all x ∈U,

−∆u(x) = f (x).

Theorem 33.5.4 Let f ∈ Lp (U). Then there exists u ∈ Lp (U) whose weak derivatives arealso in Lp (U) such that in the sense of weak derivatives,

−∆u = f .

It is given by

u(x) =∫

BΦ(y) f̃ (x−y)dy =

∫U

Φ(x−y) f (y)dy (33.5.70)

where f̃ denotes the zero extension of f off of U.