On $L^{p}$-theory for parabolic and elliptic integro-differential equations with scalable operators in the whole space

Abstract

Elliptic and parabolic integro-differential model problems are considered in the whole space. By verifying Hörmander condition, the existence and uniqueness is proved in $L_{p}$-spaces of functions whose regularity is defined by a scalable, possibly nonsymmetric, Levy measure. Some rough probability density function estimates of the associated Levy process are used as well.

Appendix

Given a function $\kappa :\left( 0,\infty \right) \rightarrow \left( 0,\infty \right)$, consider the collection ${\mathbb {Q}}$ of sets $Q_{\delta }=Q_{\delta }\left( t,x\right) =\left( t-\kappa \left( \delta \right) ,t+\kappa \left( \delta \right) \right) \times B_{\delta }\left( x\right) ,\left( t,x\right) \in {\mathbf {R}}\times {\mathbf {R}}^{d}={\mathbf {R}} ^{d+1},\delta >0$. The volume $\left| Q_{\delta }\left( t,x\right) \right| =c_{0}\kappa \left( \delta \right) \delta ^{d}.$ We will need the following assumptions.

A1 $\kappa$ is continuous, $\lim _{\delta \rightarrow 0}\kappa \left( \delta \right) =0$ and $\lim _{\delta \rightarrow \infty }\kappa \left( \delta \right) =\infty .$

A2 There is a nondecreasing continuous function $l\left( \varepsilon \right) ,\varepsilon >0,$ such that $\lim _{\varepsilon \rightarrow 0}l\left( \varepsilon \right) =0$ and

$$\begin{aligned} \kappa \left( \varepsilon r\right) \le l\left( \varepsilon \right) \kappa (r),r>0,\varepsilon >0. \end{aligned}$$

Since $Q_{\delta }\left( t,x\right)$ not “exactly” increases in $\delta$, we present the basic estimates involving maximal functions based on the system $\mathbb {Q=}\left\{ Q_{\delta }\right\}$.

1.1 Vitali lemma, maximal functions

We start with engulfing property.

Lemma 13

Let A2 hold. If $Q_{\delta }\left( t,x\right) \cap Q_{\delta ^{\prime }}\left( r,z\right) \ne \emptyset$ with $\delta ^{\prime }\le \delta$, then there is $K_{0}\ge 3$ such that $Q_{K_{0}\delta }\left( t,x\right)$ contains both, $Q_{\delta }\left( t,x\right)$ and $Q_{\delta ^{\prime }}\left( r,z\right) ,$ and

$$\begin{aligned} \left| Q_{\delta }\left( t,x\right) \right| \le \left| Q_{K_{0}\delta }\left( t,x\right) \right| \le K_{0}^{d}l\left( K_{0}\right) \left| Q_{\delta }\left( t,x\right) \right| . \end{aligned}$$

Proof

Let $\left( s,y\right) \in Q_{\delta }\left( t,x\right) \cap Q_{\delta ^{\prime }}\left( r,z\right)$ with $\delta ^{\prime }\le \delta .$ If $\left( r^{\prime },z^{\prime }\right) \in Q_{\delta ^{\prime }}\left( r,z\right)$, then $\left| z^{\prime }-x\right| \le 3\delta$, and using A2,

$$\begin{aligned} \left| r^{\prime }-t\right| \le \left| r^{\prime }-r\right| +\left| r-s\right| +\left| s-t\right| \le 2\kappa \left( \delta ^{\prime }\right) +\kappa \left( \delta \right) \le [2l\left( 1\right) +1]\kappa \left( \delta \right) . \end{aligned}$$

We choose $K_{0}\ge 3$ so that $[2l\left( 1\right) +1]l(1/K_{0})\le 1$. By A2,

$$\begin{aligned}{}[2l\left( 1\right) +1]\kappa \left( \delta \right) \le [2l\left( 1\right) +1]l(1/K_{0})\kappa \left( K_{0}\delta \right) \le \kappa \left( K_{0}\delta \right) . \end{aligned}$$

Hence $Q_{\delta ^{\prime }}\left( r,z\right) \subseteq Q_{K_{0}\delta }\left( t,x\right)$ and, obviously, $Q_{\delta }\left( t,x\right) \subseteq Q_{K_{0}\delta }\left( t,x\right)$. Also,

$$\begin{aligned} \left| Q_{K_{0}\delta }\left( t,x\right) \right| =c_{0}K_{0}^{d}\delta ^{d}\kappa (K_{0}\delta )\le c_{0}K_{0}^{d}\delta ^{d}l(K_{0})\kappa (\delta )=K_{0}^{d}l(K_{0})\left| Q_{\delta }\left( t,x\right) \right| . \end{aligned}$$

$\square$

Now, following 3.1 in [9], we prove Vitali covering lemma.

Lemma 14

Let $E\subseteq {\mathbf {R}}\times {\mathbf {R}}^{d}$ be a union of a finite collection $\left\{ Q^{\prime }\right\}$ of sets from the system $\{Q_{\delta }\left( t,x):(t,x\right) \in {\mathbf {R}}^{d+1},\delta >0\}$ and A2 hold.

There is a positive $c=\frac{1}{K_{0}^{d}l\left( K_{0}\right) }$ and a disjoint subcollection $\big \{ Q^{k}=Q_{\delta _{k}}\left( t_{k},x_{k}\right) ,1\le k\le m\big \}$ such that

$$\begin{aligned} \sum _{k=1}^{m}\left| Q^{k}\right| \ge c\left| E\right| . \end{aligned}$$

Proof

Let $Q^{1}=Q_{\delta _{1}}\left( t_{1},x_{1}\right)$ be the set of the collection $\left\{ Q^{\prime }\right\}$ with maximal $\delta$. Let $Q^{2}=Q_{\delta _{2}}\left( t_{2},x_{2}\right)$ be the set with maximal $\delta$ among remaining sets in $\left\{ Q^{\prime }\right\}$ that do not intersect $Q^{1}$. According to Lemma 13, $Q_{K_{0}\delta _{1}}\left( t_{1},x_{1}\right)$ contains $Q^{1}$ and all $Q_{\delta }$ in $\left\{ Q^{\prime }\right\}$ that intersect $Q^{1}$ and such that $\delta \le \delta _{1}.$ Continuing we get $Q_{K_{0}\delta _{k}}\left( t_{k},x_{k}\right)$ containing $Q^{k}=Q_{\delta _{k}}\left( t_{k},x_{k}\right)$ and all $Q_{\delta }$ in $\left\{ Q^{\prime }\right\}$ that intersect $Q^{k}$ and such that $\delta \le \delta _{k}.$ So we obtain a finite disjoint subcollection $\left\{ Q^{k}=Q_{\delta _{k}}\left( t_{k},x_{k}\right) ,1\le k\le m\right\}$ such that $\cup _{k=1}^{m}Q_{K_{0}\delta _{k}}\left( t_{k},x_{k}\right) \supseteq Q_{\delta }$ for any $Q^{\delta }$ in $\left\{ Q^{\prime }\right\} .$ Hence $\cup _{k=1}^{m}Q_{K_{0}\delta _{k}}\left( t_{k},x_{k}\right) \supseteq E$, and by Lemma 13,

$$\begin{aligned} \left| E\right| \le \sum _{k=1}^{m}\left| Q_{K_{0}\delta _{k}}\right| \le K_{0}^{d}l\left( K_{0}\right) \sum _{k=1}^{m}\left| Q^{k}\right| . \end{aligned}$$

$\square$

Remark 5

The statement of the Lemma 14 still holds if instead of A2 we assume that there is a constant C so that $C\kappa \left( \delta \right) \ge \kappa \left( \delta ^{\prime }\right)$ whenever $\delta \ge \delta ^{\prime }.$

Following [9], for a locally integrable function $f\left( t,x\right)$ on ${\mathbf {R}}^{d+1}$ we define

$$\begin{aligned} \left( A_{\delta }f\right) (t,x)=\frac{1}{\left| Q_{\delta }\left( t,x\right) \right| }\int _{Q_{\delta }\left( t,x\right) }f\left( s,y\right) dsdy,\left( t,x\right) \in \mathbf {R\times R}^{d},\delta >0 \end{aligned}$$

and the maximal function of f by

$$\begin{aligned} {\mathcal {M}}f\left( t,x\right) =\sup _{\delta >0}\left( A_{\delta }\left| f\right| \right) (t,x),\left( t,x\right) \in {\mathbf {R}}^{d+1}. \end{aligned}$$

We use collection ${\mathbb {Q}}$ to define a larger, noncentered maximal function of f, as

$$\begin{aligned} \widetilde{{\mathcal {M}}}f\left( t,x\right) =\sup _{\left( t,x\right) \in Q} \frac{1}{\left| Q\right| }\int _{Q}|f\left( s,y\right) |dsdy,\left( t,x\right) \in {\mathbf {R}}^{d+1}, \end{aligned}$$

where $\sup$ is taken over all $Q\in {\mathbb {Q}}$ that contain (t, x).

Remark 6

Let A2 hold and $K_{0}$ be a constant in Lemma 13 . For a locally integrable f on ${\mathbf {R}}^{d+1},$

$$\begin{aligned} {\mathcal {M}}f\le \widetilde{{\mathcal {M}}}f\le \frac{1}{K_{0}^{d}l\left( K_{0}\right) }{\mathcal {M}}f. \end{aligned}$$

Indeed, if $(t,x)\in Q^{\prime }=Q_{\delta }\left( t^{\prime },x^{\prime }\right)$, then by Lemma 13

$$\begin{aligned} \frac{1}{\left| Q^{\prime }\right| }\int _{Q^{\prime }}\left| f\right| \le \frac{K_{0}^{d}l\left( K_{0}\right) }{\left| Q_{K_{0}\delta }\left( t,x\right) \right| }\int _{Q_{K_{0}\delta }\left( t,x\right) }\left| f\right| . \end{aligned}$$

Note $\widetilde{{\mathcal {M}}}f$ is lower semicontinuous.

Theorem 1.3.1 in [9] holds for ${\mathbb {Q}}$ (we sketch its proof).

Theorem 3

Let A2 hold and f be measurable function on ${\mathbf {R}} ^{d+1}=\mathbf {R\times R}^{d}.$

(a) :

If $f\in L_{p},1\le p\le \infty$, then ${\mathcal {M}}f$ is finite a.e.

(b):

If $f\in L_{1}$, then for every $\alpha >0$,

$$\begin{aligned} \left| \left\{ {\mathcal {M}}f\left( t,x\right) >\alpha \right\} \right| \le \frac{c}{\alpha }\int |f|dtdx. \end{aligned}$$

(c):

If $f\in L_{p},1<p\le \infty$, then ${\mathcal {M}}f\in L_{p}$ and

$$\begin{aligned} \left| {\mathcal {M}}f\right| _{L_{p}}\le N_{p}\left| f\right| _{L_{p}}, \end{aligned}$$

where $N_{p}$ depends only on p, l and $K_{0}.$

Proof

(b) Let $E_{\alpha }=\left\{ \widetilde{{\mathcal {M}}}f\left( t,x\right) >\alpha \right\}$ and $E\subseteq E_{\alpha }$ be any compact subset. Since $\widetilde{{\mathcal {M}}}f$ is lower semicontinuous, $E_{\alpha }$ is open. By definition of $\widetilde{{\mathcal {M}}}f$ for each $\left( t,x\right) \in E$, there is $Q\in {\mathbb {Q}}$ so that $\left( t,x\right) \in Q$ and

$$\begin{aligned} \left| Q\right| \le \frac{1}{\alpha }\int _{Q}\left| f\right| . \end{aligned}$$

Since E is compact there exist a finite number $Q_{\delta _{1}}\left( t_{1},x_{1}\right) ,\ldots ,Q_{\delta _{n}}\left( t_{n},x_{n}\right) \in {\mathbb {Q}}$ so that $E\subseteq \cup _{j=1}^{n}Q_{\delta _{j}}\left( t_{j},x_{j}\right)$. By Lemma 14, there is a subcovering of disjoint sets $Q^{1},\ldots ,Q^{m}$ so that

$$\begin{aligned} \left| E\right| \le c\sum _{k=1}^{m}\left| Q^{k}\right| \le \frac{c}{\alpha }\sum _{k=1}^{m}\int _{Q^{k}}\left| f\right| \le \frac{c}{\alpha }\int \left| f\right| . \end{aligned}$$

with $c=K_{0}^{d}l\left( K_{0}\right)$. Taking $\sup$ over all such compacts E we get (b).

(c) Let $f_{1}=f\chi _{\left\{ \left| f\right| >\alpha /2\right\} }$. Note that $\widetilde{{\mathcal {M}}}f\le \widetilde{{\mathcal {M}} }f_{1}+\frac{\alpha }{2}.$ Hence by part (b)

$$\begin{aligned} \left| \left\{ \widetilde{{\mathcal {M}}}f>\alpha \right\} \right| \le \left| \left\{ \widetilde{{\mathcal {M}}}f_{1}>\alpha /2\right\} \right| \le \frac{2c}{\alpha }\int _{\left| f\right| >\alpha /2}\left| f\right| dm. \end{aligned}$$

On the other hand,

$$\begin{aligned} \int \left( \widetilde{{\mathcal {M}}}f\right) ^{p}= & {} p\int _{0}^{\infty }\left| \left\{ \widetilde{{\mathcal {M}}}f>\alpha \right\} \right| \alpha ^{p-1}d\alpha \\\le & {} 2cp\int \int _{0}^{2\left| f\right| }\alpha ^{p-2}d\alpha \left| f\right| =c\frac{p}{p-1}2^{p}\int \left| f\right| ^{p}. \end{aligned}$$

$\square$

Corollary 7

Let $f\in L_{1}$. Then

$$\begin{aligned} \lim _{\delta \rightarrow 0}A_{\delta }f\left( t,x\right) =f\left( t,x\right) \text { a.e.} \end{aligned}$$

and $\left| f\left( t,x\right) \right| \le {\mathcal {M}}f\left( t,x\right)$ a.e. Moreover, for every $\alpha >0$,

$$\begin{aligned} \left| \left\{ \mathcal {{\tilde{M}}}f\left( t,x\right)>\alpha \right\} \right| \le \frac{2c}{\alpha }\int _{\left\{ \mathcal {{\tilde{M}}}f\left( t,x\right) >\alpha /2\right\} }|f|dm, \end{aligned}$$

where c is a constant in Theorem 3.

Proof

Let $f\in L_{1},\varepsilon >0$. There is $g\in C_{c}\left( {\mathbf {R}} ^{d+1}\right)$ so that $\left| f-g\right| _{L_{1}}\le \varepsilon$ . Let $\eta >0$. Since g is uniformly continuous, for all $\left( t,x\right)$

$$\begin{aligned}&\left| A_{\delta }g\left( t,x\right) -g\left( t,x\right) \right| \\&\quad \le \frac{1}{\left| Q_{\delta }\left( t,x\right) \right| } \int _{Q_{\delta }\left( t,x\right) }\left| g\left( s,y\right) -g\left( t,x\right) \right| dsdy\le \eta \end{aligned}$$

if $\delta \le \delta _{0}$ for some $\delta _{0}>0$. Hence $\sup _{t,x}\left| A_{\delta }g\left( t,x\right) -g\left( t,x\right) \right| \rightarrow 0$ as $\delta \rightarrow 0.$ Now for $\left( t,x\right) \in {\mathbf {R}}^{d+1},$

$$\begin{aligned}&\lim \sup _{\delta \rightarrow 0}\left| A_{\delta }f\left( t,x\right) -f\left( t,x\right) \right| \\&\quad \le \lim \sup _{\delta \rightarrow 0}\left| A_{\delta }f\left( t,x\right) -A_{\delta }g\left( t,x\right) \right| +\left| g\left( t,x\right) -f\left( t,x\right) \right| \\&\quad \le {\mathcal {M}}\left( f-g\right) \left( t,x\right) +\left| g\left( t,x\right) -f\left( t,x\right) \right| . \end{aligned}$$

Hence for any $\alpha >0,$ by Theorem 3,

$$\begin{aligned}&\left| \left\{ \lim \sup _{\delta \rightarrow 0}\left| A_{\delta }f\left( t,x\right) -f\left( t,x\right) \right|>\alpha \right\} \right| \\&\quad \le \left| \left\{ {\mathcal {M}}\left( f-g\right)>\alpha /2\right\} \right| +\left| \left\{ \left| g-f\right| >\alpha /2\right\} \right| \\&\quad \le \frac{2c\varepsilon }{\alpha }+\frac{2\varepsilon }{\alpha }. \end{aligned}$$

Since $\varepsilon$ and $\alpha$ are arbitrary, it follows that $\lim \sup _{\delta \rightarrow 0}\left| A_{\delta }f\left( t,x\right) -f\left( t,x\right) \right| =0$ a.e. Hence for almost all $\left( t,x\right) ,$

$$\begin{aligned} \left| f\left( t,x\right) \right|= & {} \left| \lim _{\delta \rightarrow 0}A_{\delta }f\left( t,x\right) \right| \le \lim _{\delta \rightarrow 0}\frac{1}{\left| Q_{\delta }\left( t,x\right) \right| } \int _{Q_{\delta }\left( t,x\right) }\left| f\left( t,y\right) \right| dtdy \\\le & {} \sup _{\delta >0}\frac{1}{\left| Q_{\delta }\left( t,x\right) \right| }\int _{Q_{\delta }\left( t,x\right) }\left| f\left( t,y\right) \right| dtdy={\mathcal {M}}f\left( t,x\right) . \end{aligned}$$

Finally, for $f_{1}=f\chi _{\left\{ \left| f\right| >\alpha /2\right\} }$ we have $\mathcal {{\tilde{M}}}f\le \mathcal {{\tilde{M}}}f_{1}+ \frac{\alpha }{2}$, and by Theorem 3(b),

$$\begin{aligned} \left| \left\{ \mathcal {{\tilde{M}}}f>\alpha \right\} \right| \le \left| \left\{ \mathcal {{\tilde{M}}}f_{1}>\alpha /2\right\} \right| \le \frac{2c}{\alpha }\int _{\left| f\right|>\alpha /2}\left| f\right| \le \frac{2c}{\alpha }\int _{\mathcal {{\tilde{M}}}f>\alpha /2}\left| f\right| . \end{aligned}$$

$\square$

1.2 Calderon–Zygmund decomposition

Assume A1, A2 hold. Let $F\subseteq \mathbf {R\times R}^{d}$ be closed and $O=F^{c}={\mathbf {R}}^{d+1}\backslash F.$ For $\left( t,x\right) \in O$, let

$$\begin{aligned} D\left( t,x\right) =\inf \left\{ \delta >0:Q_{\delta }\left( t,x\right) \cap F\ne \emptyset \right\} . \end{aligned}$$

For each $\left( t,x\right) \in O$, $D\left( t,x\right) \in \left( 0,\infty \right)$. Let $K_{0}$ be a constant in Lemma 13. We fix $A>1$ so that $l\left( 1/A\right) <1$ and $\varepsilon >0$ so that $l\left( 2K_{0}\varepsilon \right)<1,\varepsilon \le \frac{1}{4AK_{0}^{3}}<1.$ Then, denoting $D=D(t,x),$ we have

$$\begin{aligned}&\kappa \left( \varepsilon D\right) \le l\left( 2\varepsilon \right) \kappa \left( D/2\right) \le \kappa \left( \frac{D}{2}\right) ,\kappa \left( \varepsilon D\right) \le l\left( \varepsilon \right) \kappa \left( D\right) \le \kappa \left( D\right) ,\\&\kappa (D)\le l\left( 1/A\right) \kappa \left( AD\right) <\kappa \left( AD\right) , \end{aligned}$$

and

$$\begin{aligned} \kappa \left( \varepsilon D\right) \le l\left( 2K_{0}\varepsilon \right) \kappa \left( D/2K_{0}\right) \le \kappa \left( D/2K_{0}\right) . \end{aligned}$$

Consider the covering $Q_{\varepsilon D(t,x)}\left( t,x\right) ,\left( t,x\right) \in O$, of O. Let

$$\begin{aligned} Q^{k}=Q_{\varepsilon D(t_{k},x_{k})}\left( t_{k},x_{k}\right) ,k\ge 1, \end{aligned}$$

be its maximal disjoint subcollection: for any $Q_{\varepsilon D\left( t,x\right) }\left( t,x\right)$ there is k so that $Q_{\varepsilon D\left( t,x\right) }(t,x)\cap Q^{k}\ne \emptyset .$ Let

$$\begin{aligned} Q^{*k}=Q_{D(t_{k},x_{k})/2}\left( t_{k},x_{k}\right) ,Q^{**k}=Q_{AD\left( t_{k},x_{k}\right) }\left( t_{k},x_{k}\right) . \end{aligned}$$

Note that $Q^{k}\subseteq Q^{*k}\subseteq Q_{D(t_{k},x_{k})}\left( t_{k},x_{k}\right) \subseteq O,Q^{**k}\cap F\ne \emptyset$. We will show that $\cup _{k}Q^{*k}=O$. Let $\left( t,x\right) \in O$ and $Q_{\varepsilon D\left( t_{k}x_{k}\right) }\left( t_{k},x_{k}\right) \cap Q_{\varepsilon D(t,x)}\left( t,x\right) \ne \emptyset$ for some k. Since

$$\begin{aligned} Q_{\varepsilon D\left( t_{k},x_{k}\right) }\left( t_{k},x_{k}\right)\subseteq & {} Q_{D\left( t_{k},x_{k}\right) }\left( t_{k},x_{k}\right) \subseteq Q_{AD\left( t_{k},x_{k}\right) }\left( t_{k},x_{k}\right) , \\ Q_{\varepsilon D\left( t,x\right) }\left( t,x\right)\subseteq & {} Q_{D(t,x)/(2K_{0})}, \end{aligned}$$

it follows that

$$\begin{aligned} Q_{D(t,x)/(2K_{0})}\left( t,x\right) \cap Q_{AD\left( t_{k},x_{k}\right) }\left( t_{k},x_{k}\right) \ne \emptyset . \end{aligned}$$

We show by contradiction that $AD\left( t_{k},x_{k}\right) \ge D\left( t,x\right) /2K_{0}.$ If not so, then $AD\left( t_{k},x_{k}\right) <D\left( t,x\right) /2K_{0}$, and, by Lemma 13, $Q_{D(t,x)/2K_{0}} \left( t,x\right)$ and $Q_{AD\left( t_{k},x_{k}\right) }\left( t_{k},x_{k}\right)$ are contained in $Q_{_{D\left( t,x\right) /2}}(t,x)\subseteq O$: a contradiction to $Q_{AD\left( t_{k},x_{k}\right) }\left( t_{k},x_{k}\right) \cap F\ne \emptyset .$ Therefore $AD\left( t_{k},x_{k}\right) \ge D\left( t,x\right) /2K_{0}$ and $2AK_{0}\varepsilon D\left( t_{k},x_{k}\right) \ge \varepsilon D\left( t,x\right)$. Now, $Q_{\varepsilon D\left( t_{k}x_{k}\right) }\left( t_{k},x_{k}\right) \subseteq Q_{2AK_{0}\varepsilon D\left( t_{k}x_{k}\right) }\left( t_{k},x_{k}\right)$ and $Q_{\varepsilon D\left( t_{k}x_{k}\right) }\left( t_{k},x_{k}\right) \cap Q_{\varepsilon D(t,x)}\left( t,x\right) \ne \emptyset \,$. Hence by Lemma 13, $Q_{\varepsilon D\left( t,x\right) }$ is contained in $Q_{2AK_{0}^{2}\varepsilon D\left( t_{k},x_{k}\right) }\left( t_{k},x_{k}\right)$. Since $2AK_{0}^{2}\varepsilon \le \frac{1}{ 2K_{0}}$, it follows by Lemma 13 that

$$\begin{aligned} Q_{\varepsilon D\left( t,x\right) }\left( t,x\right) \subseteq Q_{2AK_{0}^{2}\varepsilon D\left( t_{k},x_{k}\right) }\left( t_{k},x_{k}\right) \subseteq Q_{D\left( t_{k},x_{k}\right) /2}\left( t_{k},x_{k}\right) =Q^{*k}. \end{aligned}$$

So we proved the following statement.

Lemma 15

(cf. Lemma 2 in Chapter I, 3.2 of [9]) Assume A1, A2 hold. Given a closed nonempty F, there are sequences $Q^{k}$, $Q^{*k}$ and $Q^{**k}$ in ${\mathbb {Q}}$ having the same center but with radius expanded by the same factor $c_{1}^{**}>c_{1}^{*}>c_{1}\,\$ so that $Q^{k}\subseteq Q^{*k}\subseteq Q^{**k}$ (all of them are of the form $Q_{bD\left( t_{k},x_{k}\right) }\left( t_{k},x_{k}\right)$ with $b=c_{1},c_{1}^{*},c_{1}^{**}$ correspondingly) and

(a) :

the sets $Q^{k}$ are disjoint.

(b) :

$\cup _{k}Q^{*k}=O=F^{c}.$

(c) :

$Q^{**k}\cap F\ne \emptyset$ for each k.

Remark 7

Assume A1, A2 hold and $Q^{k}\subseteq Q^{*k}\subseteq Q^{**k}$ be the sequences in ${\mathbb {Q}}$ from Lemma . It is easy to find a sequence of disjoint measurable sets $C^{k}$ so that $Q^{k}\subseteq C^{k}\subseteq Q^{*k}$ and $\cup _{k}C^{k}=O$. For example (see Remark, p. 15, in [9]),

$$\begin{aligned} C^{k}=Q^{*k}\cap \left( \cup _{j<k}C^{j}\right) ^{c}\cap \left( \cup _{j>k}Q^{j}\right) ^{c}. \end{aligned}$$

Now we derive Calderon–Zygmund decomposition for ${\mathbb {Q}}.\,\$

Theorem 4

(cf. Theorem 2 in Chapter I, 4.1 of [9]) Assume A1, A2 hold. Let $f\in L_{1}\left( \mathbf {R\times R}^{d}\right)$, $\alpha >0$ and $O_{\alpha }=\left\{ \widetilde{{\mathcal {M}}}f>\alpha \right\} .$ Consider the sets $Q^{k}\subseteq C^{k}\subseteq Q^{*k}\subseteq O_{\alpha }$ of Lemma 15 and Remark 7 associated to $O_{\alpha }.$

There is a decomposition $f=g+b$ with

$$\begin{aligned} g\left( t,x\right) =\left\{ \begin{array}{ll} f(t,x) &{} \text {if }(t,x)\notin O_{\alpha }, \\ \frac{1}{\left| C^{k}\right| }\int _{C^{k}}f &{} \text { if }(t,x)\in C^{k},k\ge 1, \end{array} \right. \end{aligned}$$

(4.34)

and with $b=\sum _{k}b_{k}$, where

$$\begin{aligned} b_{k}=\chi _{C^{k}}\left[ f\left( x\right) -\frac{1}{\left| C^{k}\right| }\int _{C^{k}}f\text { }\right] ,k\ge 1, \end{aligned}$$

(4.35)

(note $C^{k}$ are disjoint, $\cup _{k}C^{k}=O_{\alpha }$). Also,

(i) :

$\left| g\left( t,x\right) \right| \le c\alpha$ for a.e. x.

(ii) :

support($b_{k})\subseteq Q^{*k},$

$$\begin{aligned} \int b_{k}=0\text { and }\int \left| b_{k}\right| \le c\alpha \left| Q^{*k}\right| . \end{aligned}$$

(iii) :

$\sum _{k}\left| Q^{*k}\right| \le \frac{c}{\alpha }\int \left| f\right| .$

Proof

The set $O_{\alpha }=\left\{ \widetilde{{\mathcal {M}}}f>\alpha \right\}$ is open. We can apply Lemma 15 and Remark 7 to it and consider the sets $Q^{k}\subseteq C^{k}\subseteq Q^{*k}\subseteq O_{\alpha }$ with $C^{k}$ disjoint and $\cup _{k}C^{k}=O_{\alpha }.$

Define g by (4.34). Hence $f=g+\sum _{k}b_{k}$ with $b_{k}$ given by ( 4.35). Obviously

$$\begin{aligned} \sum _{k}\left| Q^{k}\right| \le \left| O_{\alpha }\right| . \end{aligned}$$

(i) By Corollary 7, $\left| f\left( t,x\right) \right| \le \alpha$ a.e. on $O_{\alpha }^{c}=\left\{ \widetilde{{\mathcal {M}}}f\left( t,x\right) \le \alpha \right\}$. Hence :  so $\left| g\left( x\right) \right| \le \alpha$ a.e. on $O_{\alpha }^{c}$. On the other hand, if $Q^{**k}\in {\mathbb {Q}}$ is the sequence of Lemma 15, then

$$\begin{aligned} \frac{1}{\left| Q^{**k}\right| }\int _{Q^{**k}}\left| f\right| \le \alpha \end{aligned}$$

because $Q^{**k}\cap O_{\alpha }^{c}\ne \emptyset$ and $\widetilde{ {\mathcal {M}}}f\left( t,x\right) \le \alpha$ on $O_{\alpha }^{c}$ (the definition of $\widetilde{{\mathcal {M}}}$ implies it). Since $\left| Q^{k}\right| \le \left| C^{k}\right| \le \left| Q^{*k}\right| \le \left| Q^{**k}\right| \le l\left( \frac{ c_{1}^{**}}{c_{1}}\right) \left| Q^{k}\right|$ and $C^{k}\subseteq Q^{**k}$, it follows that

$$\begin{aligned} \left| g\right| \le c\alpha . \end{aligned}$$

(ii) Only inequality is not trivial:

$$\begin{aligned} \int \left| b_{k}\right| \le 2\int _{C^{k}}\left| f\right| \le 2\left| Q^{**k}\right| \frac{1}{\left| Q^{**k}\right| }\int _{Q^{**k}}\left| f\right| \le c\alpha \left| Q^{*k}\right| . \end{aligned}$$

(iii) We have

$$\begin{aligned} \left| O_{\alpha }\right| =\left| \left\{ \widetilde{{\mathcal {M}}} f\left( t,x\right) >\alpha \right\} \right| \ge \sum _{k}\left| Q^{k}\right| \ge {\tilde{c}}\sum _{k}\left| Q^{*k}\right| \end{aligned}$$

and the inequality follows by Theorem 3. $\square$

1.3 $L_{p}$-estimates

Let

$$\begin{aligned} \left( Tf\right) \left( t,x\right) =\int _{{\mathbf {R}}^{d+1}}K\left( t,x,s,y\right) f\left( s,y\right) dsdy,(t,x)\in {\mathbf {R}}^{d+1}, \end{aligned}$$

where K is measurable and for almost all $\left( t,x\right) \in {\mathbf {R}} ^{d+1}$ the function $K\left( t,x,\cdot \right) f$ is integrable for all $f\in C_{0}^{\infty }\left( {\mathbf {R}}^{d+1}\right) .$ We assume that T is bounded on $L_{q}$:

$$\begin{aligned} \left| Tf\right| _{L_{q}}\le C\left| f\right| _{L_{q}},f\in L_{q}. \end{aligned}$$

(4.36)

In addition, we assume that Hörmander condition holds: there are constants $c>1,A>0$ so that for any $Q_{\delta }\in {\mathbb {Q}}$,

$$\begin{aligned} \int _{{\mathbf {R}}^{d+1}\backslash Q_{c\delta }\left( s,y\right) }\left| K\left( t,x,{\bar{s}},{\bar{y}}\right) -K\left( t,x,s,y\right) \right| dxdt\le A,({\bar{s}},{\bar{y}})\in Q_{\delta }\left( s,y\right) . \end{aligned}$$

(4.37)

Theorem 5

(cf. Theorem 3 of Chapter I, 5.1 in [9] ) Let A1, A2, (4.36) and (4.37) hold. Then T is bounded in $L_{p}$-norm on $L_{p}\cap L_{q}$ if $1<p<q$. More precisely,

$$\begin{aligned} \left| T\left( f\right) \right| _{L_{p}}\le A_{p}\left| f\right| _{L_{p}},f\in L_{p}\cap L_{q},1<p<q, \end{aligned}$$

where $A_{p}$ depends only on the constant A and p.

Proof

By Marcinkiewicz interpolation theorem (see [9]), it is enough to prove that

$$\begin{aligned} m\left( \left| Tf\right|>\alpha \right) \le \frac{A^{\prime }}{ \alpha }\int \left| f\right| dxdt,f\in L_{1}\cap L_{q},\alpha >0, \end{aligned}$$

where $A^{\prime }$ depends on A.

For a large constant $c^{\prime }$ (to be determined) we estimate $m\left( \left| Tf\right| >c^{\prime }\alpha \right)$. For a fixed $\alpha >0$ we consider the decomposition $f=g+b$ in Theorem 4. First note that

$$\begin{aligned} m\left( \cup _{n}Q_{n}^{**}\right) \le \sum _{n}m\left( Q_{n}^{**}\right) \le c\sum _{n}m\left( Q_{n}^{*}\right) \le \frac{c}{ \alpha }\int \left| f\right| . \end{aligned}$$

It is enough to show that

$$\begin{aligned} \left| \left\{ \left| Tg\right|>\left( c^{\prime }/2\right) \alpha \right\} \right| +\left| \left\{ \left| Tb\right| >\left( c^{\prime }/2\right) \alpha \right\} \right| \le A^{\prime }/\alpha \int \left| f\right| dx. \end{aligned}$$

First notice that $g\in L_{q}$. Indeed (recall $\cup _{k}Q_{k}^{*}=\cup _{k}C_{k}$),

$$\begin{aligned} \int \left| g\right| ^{q}=\int _{\left( \cup _{k}Q_{k}^{*}\right) ^{c}}|g|^{q}+\int _{\cup _{k}C_{k}}\left| g\right| ^{q}\le c\alpha ^{q-1}\int |f| \end{aligned}$$

because

$$\begin{aligned} \int _{\left( \cup _{k}Q_{k}^{*}\right) ^{c}}\left| g\right| ^{q}\le & {} \alpha ^{q-1}\int _{\left( \cup _{k}Q_{k}^{*}\right) ^{c}}|f|, \\ \int _{\cup _{k}Q_{k}^{*}}\left| g\right| ^{q}\le & {} c\alpha ^{q}\sum _{k}\left| Q_{k}^{*}\right| \le c\alpha ^{q-1}\int \left| f\right| . \end{aligned}$$

By Chebyshev inequality,

$$\begin{aligned}&\left| \left\{ \left| Tg\right| >\left( c^{\prime }/2\right) \alpha \right\} \right| \le \left( \frac{c^{\prime }\alpha }{2}\right) ^{-q}\left| Tg\right| _{L_{q}}^{q}\le \left( \frac{c^{\prime }\alpha }{2}\right) ^{-q}A^{q}\left| g\right| _{L_{q}}^{q} \\&\quad \le c\left( \frac{c^{\prime }\alpha }{2}\right) ^{-q}A^{q}\alpha ^{q-1}\int \left| f\right| \le \frac{A^{\prime }}{\alpha } \left| f\right| _{L_{1}}, \end{aligned}$$

and

$$\begin{aligned} \int _{\left( \cup _{k}Q_{k}^{**}\right) ^{c}}|Tb|\le \sum _{k}\int _{\left( Q_{k}^{**}\right) ^{c}}\left| Tb_{k}\right| . \end{aligned}$$

Let $(s_{k},y_{k})$ be the center of $Q_{k}^{*}$ (and $Q_{k}^{**}$). Since for $x\notin Q_{k}^{*},$ we have, denoting $f_{k}=1/\left| C^{k}\right| \int _{C_{k}}f$,

$$\begin{aligned} Tb_{k}= & {} \int _{C_{k}}K\left( t,x,s,y\right) [f\left( s,y\right) -f_{k}]dsdy\\= & {} \int _{C_{k}}[K\left( t,x,s,y\right) -K\left( t,x,s_{k},y_{k}\right) ][f\left( s,y\right) -f_{k}]dsdy, \end{aligned}$$

and

$$\begin{aligned}&\int _{(Q_{k}^{**})^{c}}\left| Tb_{k}\right| dtdx \\&\quad \le \int ~\left| b_{k}\right| dsdy\sup _{(s,y)\in Q_{k}^{*}}\int _{(Q_{k}^{**})^{c}}|K\left( t,x,s,y\right) -K\left( t,x,s_{k},y_{k}\right) |dtdx\le cA\alpha \left| Q_{k}^{*}\right| . \end{aligned}$$

Hence

$$\begin{aligned} \int _{\left( \cup _{k}Q_{k}^{**}\right) ^{c}}|Tb|\le cA\alpha \sum _{k}\left| Q_{k}^{*}\right| \le cA\int \left| f\right| . \end{aligned}$$

$\square$