Kenneth Kuttler

1126 CHAPTER 33. FOURIER ANALYSIS IN Rn

LetΩ∗ ≡ ∪∞

i=1Q∗iand let

E∗ ≡ Rn \Ω∗.

Thus E∗ ⊆ E. Let x ∈ E∗. Then because of 33.3.11,∫Qi

F−1ρ (x−y)b(y)dy

=∫

[F−1

ρ (x−y)−F−1ρ (x−yi)

]b(y)dy, (33.3.15)

where yi is the center of Qi. Consequently if the sides of Qi have length 2t/√

n, 33.3.15implies ∫

E∗

∣∣∣∣∫Qi

F−1ρ (x−y)b(y)dy

∣∣∣∣dx≤ (33.3.16)∫E∗

∫Qi

∣∣F−1ρ (x−y)−F−1

ρ (x−yi)∣∣ |b(y)|dydx

=∫

∫E∗

∣∣F−1ρ (x−y)−F−1

ρ (x−yi)∣∣dx |b(y)|dy (33.3.17)

≤∫

∫|x−yi|≥2t

∣∣F−1ρ (x−y)−F−1

ρ (x−yi)∣∣dx |b(y)|dy (33.3.18)

since if x ∈ E∗, then |x−yi| ≥ 2t. Now for y ∈ Qi,

|y−yi| ≤

∑j=1

(t√n

)2)1/2

= t.

From 33.3.8 and the change of variables u = x−yi 33.3.16 - 33.3.18 imply∫E∗

∣∣∣∣∫Qi

F−1ρ (x−y)b(y)dy

∣∣∣∣dx≤C1

∫Qi

|b(y)|dy. (33.3.19)

Now from 33.3.19, and the fact that b = 0 off Ω,∫E∗

∣∣F−1ρ ∗b(x)

∣∣dx =∫

E∗

∣∣∣∣∫RnF−1

ρ (x−y)b(y)dy∣∣∣∣dx

=∫

E∗

∣∣∣∣∣ ∞

∑i=1

∫Qi

F−1ρ (x−y)b(y)dy

∣∣∣∣∣dx

≤∫

E∗

∞

∑i=1

∣∣∣∣∫Qi

F−1ρ (x−y)b(y)dy

∣∣∣∣dx

=∞

∑i=1

∫E∗

∣∣∣∣∫Qi

F−1ρ (x−y)b(y)dy

∣∣∣∣dx

≤∞

∑i=1

∫Qi

|b(y)|dy =C1 ||b||1.

1126LetQand letCHAPTER 33. FOURIER ANALYSIS IN R""= U1Q;E* =R"\QO*.Thus E* C E. Let x € E*. Then because of 33.3.11,F-Qi'p(x—y)b(y) dy= [ [Ftp xy) F'p (xy) Div)ay.where y; is the center of Q;. Consequently if the sides of Q; have length 2r/,/n, 33.3.15impliesIQiF~'p(x—y)b(y)dy|dx <[fle ey) -F "9 xy] bv) ava~ adeIAF-'p(x—y) —F-'p(x—y;,)|dx|b(y)|dysince if x € E*, then |x —y,| > 2r. Now for y € Q;,ly—yil< (£(4)) =[dea lh 2D FP yi] alo ladFrom 33.3.8 and the change of variables u = x — y; 33.3.16 - 33.3.18 implyI.QiNow from 33.3.19, and the fact that b =| |F-'p *b(x)|dxE*IAlAF~'p(x—y)b(y)dyav<ci [| \b(y)ldy0 off Q,[LF le -wb)ayRR"[. iY [Fi setv)av[.%X Iraf \b(y)|dy =) |[b||).i=l idxF'p(x—y)b(y)dyQ;of Py) bly) dydxdxdx(33.3.15)(33.3.16)(33.3.17)(33.3.18)(33.3.19)