33.5. HELMHOLTZ DECOMPOSITIONS 1149

Proof: Let f ∈ Lp (U) and let fk ∈C∞c (U) , || fk− f ||Lp(U)→ 0, and let uk be given by

33.5.70 with fk in place of f . Then by Minkowski’s inequality,

||u−uk||Lp(U) =

(∫U

(∫B

Φ(y)∣∣∣ f̃ (x−y)− fk (x−y)

∣∣∣dy)p

dx)1/p

≤

(∫B|Φ(y)|

(∫U

∣∣∣ f̃ (x−y)− fk (x−y)∣∣∣p dx

)1/p

dy

)≤

∫B|Φ(y)|dy || f − fk||Lp(U) =C (B) || f − fk||Lp(U)

and so uk→ u in Lp (U). Also

uk,i (x) =∫

UΦ,i (x−y) fk (y)dy =

∫B

fk (x−y)Φ,i (y)dy.

Now letwi ≡

∫B

f̃ (x−y)Φ,i (y)dy. (33.5.71)

and since Φ,i ∈ L1 (B), it follows from Minkowski’s inequality that∣∣∣∣uk,i−wi∣∣∣∣

Lp(U)

≤(∫

U

(∫B

∣∣∣ fk (x−y)− f̃ (x−y)∣∣∣ |Φ,i (y)|dy

)p

dx)1/p

≤∫

B|Φ,i (y)|

(∫U

∣∣∣ fk (x−y)− f̃ (x−y)∣∣∣p dx

)1/p

dy

≤ C (B) || fk− f ||Lp(U)

and so uk,i→ wi in Lp (U).Now let φ ∈C∞

c (U). Then∫U

wiφdx =− limk→∞

∫U

ukφ ,idx =−∫

Uuφ ,idx.

Thus u,i = wi ∈ Lp (Rn) and so if φ ∈C∞c (U),∫

Uf φdx = lim

k→∞

∫U

fkφdx = limk→∞

∫U

∇uk ·∇φdx =∫

U∇u ·∇φdx

and so −∆u = f as claimed. This proves the theorem.One could also ask whether the second weak partial derivatives of u are in Lp (U). This

is where the theory singular integrals is used. Recall from 33.5.70 and 33.5.71 along withthe argument of the above lemma, that if u is given by 33.5.70, then u,i is given by 33.5.71which equals ∫

UΦ,i (x−y) f (y)dy.