2018年5月16日星期三

Analysis and PDE : Covering and decomposition lemmas - Vitali, Calderon-Zygmund, Whitney

Some estimations a priori are important in PDE, while they come from some functional inequality, and some of them come from harmonic analysis and require some combinatoric covering lemma to prove it. This is a subject that I have learned several years ago from Prof. Hongquan LI when I was in Fudan. Today I spent a whole day to review them.

Vitali covering

When we study the Hardy-Littlewood maximal function 
$$\mathcal{M} f(x) = \sup_{r > 0} \frac{1}{|B_r(x)|} \int_{B_r(x)} f(y) dy$$
We would like to prove that this operator is of type strong $(p,p)$. The strategy is to prove that it is weak $(1,1)$ and strong $(\infty, \infty)$ and then uses the interpolation inequality. However, the weak $(1,1)$ isn't very clear. One tool used in the proof is so called Vitali covering. It says that for a covering $\{B_i\}_{i \geq 0}$ we could abstract a disjoint sub-covering $\{B_{kj}\}_{j \geq 0}$such that 
$$\bigcup_j^{\infty} B_{kj} \subset \bigcup_i^{\infty} B_i \subset \bigcup_j^{\infty} 3B_{kj}$$
This nice structure makes some sub-additive into additive and is very  useful in the proof.

Calderon-Zygmund decomposition

Calderon-Zygmund decomposition could be seen as a dyadic version of maximal inequality. We start by the system of dyadic cubes 
$$\mathcal{D} = \bigcup_{k \in \mathbb{Z}} \mathcal{D_k}$$ 
where $\mathcal{D_k}$ is the collection of cubes of  length $2^{-k}$. Then for any point $x$, it belongs to only one cube in the system $\mathcal{D_k}$. And any two cubes could be disjoint or one included in another - they could not have non-trivial intersection and difference at the same time. This decomposition is used to study singular integral, but at first we could use it to build a generalized cubic version of maximal function defined as 
$$\mathcal{M}_Q f(x) = \sup_{x \in Q \in \mathcal{D}} \frac{1}{|Q|} \int_{Q} f(y) dy$$

The first fact is that any open set in $\mathbb{R}^d$ could be built by the disjoint union of cubes. Idea is simple, we give one cube for each point $x$ such that $x \in Q \subset \Omega$. Then we mesh them if one belongs to another.

We apply this idea to the open set 
$$A_{\lambda} = \{x | \mathcal{M}_Q f(x)  \geq \lambda \}$$
and for each point, we choose the largest $Q$ admit. This works once we suppose $f \in L^1(\mathbb{R}^d)$ where an infinite cubes make it null. Then we know that this gives a perfect partition of the open set.
$$A_{\lambda} = \bigsqcup_{i=1}^{\infty}Q_i$$
Moreover, we know that the average at each cube is less than $2^d \lambda$ if $f$ is positive by the stopping time property, and $ \mathcal{M}_Q f(x) < \lambda \text{ a.e }$ for the point outside $A_{\lambda}$ by the Lebesgue differential theorem.  

Whitney decomposition

Finally, we give a stronger version of Whitney decomposition. It is also a dyadic decomposition but has two more properties
1, The distance of one cube from $\Omega^c$ is comparable to its length.
2, If two cubes are neighbors, their lengths are comparable.
So this decomposition looks more balanced.

Its construction is a little tricky : it applies a "level by level peeling" to cover the level by the distance to the boundary. Then we do mesh. This make the decomposition always comparable by some distance. 

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