Topics In Analysis

31.2 The Calderon Zygmund Decomposition

For a given nonnegative integrable function, ℝⁿ can be decomposed into a set where the function is small and a set which is the union of disjoint cubes on which the average of the function is under some control. The measure in this section will always be Lebesgue measure on ℝⁿ. This theorem depends on the Lebesgue theory of differentiation.

Theorem 31.2.1 Let f ≥ 0, ∫ fdx < ∞, and let α be a positive constant. Then there exist sets F and Ω such that

n ℝ = F \cup Ω,F \cap Ω = \emptyset (31.2.3)

(31.2.3)

f (x) \leq α a.e. on F (31.2.4)

(31.2.4)

Ω = ∪_k=1^∞Q_k where the interiors of the cubes are disjoint and for each cube, Q_k,

\int --1--- n α < m (Qk) Qk f (x)dx \leq 2 α. (31.2.5)

(31.2.5)

Proof: Let S₀ be a tiling of ℝⁿ into cubes having sides of length M where M is chosen large enough that if Q is one of these cubes, then

\int --1-- fdm \leq α. (31.2.6) m (Q) Q

(31.2.6)

Suppose S₀,

,S_m have been chosen. To get S_m+1, replace each cube of S_m by the 2ⁿ cubes obtained by bisecting the sides. Then S_m+1 consists of exactly those cubes of S_m for which 31.2.6 holds and let T_m+1 consist of the bisected cubes from S_m for which 31.2.6 does not hold. Now define

F \equiv {x : x is contained in some cube from Sm for all m} ,

Ω \equiv ℝn ∖F = \cup\infty \cup {Q : Q \in T } m=1 m

Note that the cubes from T_m have pair wise disjoint interiors and also the interiors of cubes from T_m have empty intersections with the interiors of cubes of T_k if k≠m.

Let x be a point of Ω and let x be in a cube of T_m such that m is the first index for which this happens. Let Q be the cube in S_m−1 containing x and let Q^∗ be the cube in the bisection of Q which contains x. Therefore 31.2.6 does not hold for Q^∗. Thus

\leqα \int ◜----◞◟\int----◝ α < --1--- f dx \leq-m-(Q) --1-- fdx \leq 2nα m (Q *) Q* m (Q *) m (Q ) Q

which shows Ω is the union of cubes having disjoint interiors for which 31.2.5 holds.

Now a.e. point of F is a Lebesgue point of f. Let x be such a point of F and suppose x ∈ Q_k for Q_k ∈ S_k. Let d_k ≡ diameter of Q_k. Thus d_k → 0.

\int \int ---1-- |f (y)- f (x)|dy \leq--1--- |f (y)- f (x)|dy m (Qk) Qk m (Qk) B (x,dk)

\int = m-(B(x,dk))----1------ |f (x) - f (y)|dy m (Qk) m (B (x,dk)) B(x,dk)

1 \int \leq Kn m-(B-(x,d-))- |f (x) - f (y)|dy k B(x,dk)

where K_n is a constant which depends on n and measures the ratio of the volume of a ball with diamiter 2d and a cube with diameter d. The last expression converges to 0 because x is a Lebesgue point. Hence

--1---\int f (x) = kli\tom\infty m (Qk ) Q f (y)dy \leq α k

and this shows f

≤ α a.e. on F. This proves the theorem.