1146 CHAPTER 33. FOURIER ANALYSIS IN Rn

≤ 2n |y|C (n) |x|n−1

|x|2n =C (n)2n |y||x|n+1 .

Thus ∫|x|>2|y|

∣∣∣∣Ω(x)(

1|x−y|n

− 1|x|n

)∣∣∣∣dx

≤ C (n)∫|x|>2|y|

|y||x|n+1 dx

≤C (n)∫|u|>2

1

|u|n+1 du. (33.4.63)

From 33.4.62 and 33.4.63,∫|x|>2|y|

|K (x−y)−K (x)|dx≤C (n,LipΩ).

This proves the theorem.

33.5 Helmholtz DecompositionsIt turns out that every vector field which has its components in Lp can be written as a sumof a gradient and a vector field which has zero divergence. This is a very remarkable result,especially when applied to vector fields which are only in Lp. Recall that for u a functionof n variables, ∆u = ∑

ni=1

∂ 2u∂x2

i.

Definition 33.5.1 Define

Φ(y)≡

{− 1

a1ln |y| , if n = 2,

1(n−2)an−1

|y|2−n , if n > 2.

where ak denotes the area of the unit sphere, Sk.

Then it is routine to verify ∆Φ = 0 away from 0. In fact, if n > 2,

Φ,ii (y) =Cn

[1|y|n−n

y2i

|y|n+2

], Φ,i j (y) =Cn

yiy j

|y|n+2 , (33.5.64)

while if n = 2,

Φ,22 (y) =C2y2

1− y22(

y21 + y2

2

)2 , Φ,11 (y) =C2y2

2− y21(

y21 + y2

2

)2 ,

Φ,i j (y) =C2y1y2(

y21 + y2

2

)2 .

Also,∇Φ(y) =

−yan−1 |y|n

. (33.5.65)

In the above the subscripts following a comma denote partial derivatives.

1146 CHAPTER 33. FOURIER ANALYSIS IN R"2 lylCln) ix"! _ Cn) 2"IyIx|2” |x|"!Thus1 1Q(x) ( — ) dx[ooo [x—y|" |x|"ly|C(n)Ix|>2ly| |x|"1<C(n) | du. (33.4.63)ju|>2 Jul”*From 33.4.62 and 33.4.63,| |K (x—y) —K(x)|dx < C(n,LipQ).IxI>2IyThis proves the theorem.33.5 Helmholtz DecompositionsIt turns out that every vector field which has its components in L? can be written as a sumof a gradient and a vector field which has zero divergence. This is a very remarkable result,especially when applied to vector fields which are only in L?. Recall that for u a functionn Ouof n variables, Au = Yj_, 5.UDefinition 33.5.1 Define_Lnly|. ifn=yal “anil tn=2a i 22a WT fn > 2.where a, denotes the area of the unit sphere, S*.Then it is routine to verify A® = 0 away from 0. In fact, ifn > 2,I yz Vij® jj (y) = Ch —_ —yj—1— 5 D 5 (y) = C,—3,; (33.5.64)ly|" ly|"*? ly|"*?while if n = 2, > >YrI—Y yay®»(y) = OTS: ®ii(y) =C ETL(vi +95) (vt +93)yiy2@ jj (y)= Oo(vi +y3)Also,—yV®(y) = ———_.. (33.5.65)(y) analy!In the above the subscripts following a comma denote partial derivatives.