A pointwise estimate for positive dyadic shifts and some applications

Abstract

We prove a (sharp) pointwise estimate for positive dyadic shifts of complexity m which is linear in the complexity. This can be used to give a pointwise estimate for Calderón-Zygmund operators and to answer a question originally posed by Lerner. Several applications to weighted estimates for both multilinear Calderón-Zygmund operators and square functions are discussed.

Appendix: The weak-type estimate for multilinear m-shifts

Here we prove the weak-type estimate for k-linear m-shifts needed in Sect. 2. Notice that the only important point of the estimates below is the independenceof the constants from the parameter m. The proof could be more or less standard by now, but the authors have not been able to find it elsewhere. Therefore we include it for completeness.

Theorem 4.1

$$\begin{aligned} \sup _{\lambda > 0} \lambda |\left\{ x\in P_0: \mathcal {A}_{P_0,\alpha }^m \vec {f}(x) > \lambda \right\} |^{k} \le C_{W} \Vert \alpha \Vert _{{\text {Car}}(P_0)} \prod _{i=1}^k \Vert f_i\Vert _{L^1(P_0)}, \end{aligned}$$

(4.1)

where $C_W > 0$ only depends on k and d, and in particular is independent of m.

We will essentially follow Grafakos-Torres [6, 9]. We first prove an $L^{2}$ bound and then apply a Calderón-Zygmund decomposition. For the $L^{2}$ bound we will use a multilinear Carleson embedding theorem by Chen and Damián [2], from which we only need the unweighted result:

$$\begin{aligned} \left( \sum _{Q \in \mathscr {D}(P_0)} \alpha _Q \left( \prod _{i=1}^k \langle f_i \rangle _Q \right) ^p \right) ^{\frac{1}{p}} \le \Vert \alpha \Vert _{\text {Car}(P_0)} \prod _{i=1}^k p_i' \Vert f_i\Vert _{L^{p_i}(P_0)} \end{aligned}$$

(4.2)

whenever

$$\begin{aligned} \frac{1}{p} = \frac{1}{p_1} + \dots + \frac{1}{p_k}. \end{aligned}$$

Now we can prove

Proposition 4.2

$$\begin{aligned} \Vert \mathcal {A}_{P_0,\alpha }^m \vec {f}\Vert _{L^2(P_0)} \le 4\Vert \alpha \Vert _{{\text {Car}}(P_0)}\prod _{i=1}^k \Vert f_i\Vert _{L^{2k}(P_0)} \end{aligned}$$

Proof

We begin by using duality and homogeneity to reduce to showing

$$\begin{aligned} \int _{P_0} g(x) \mathcal {A}_{P_0,\alpha }^m \vec {f}(x) \, dx \le 4 \end{aligned}$$

assuming that $\Vert f_i\Vert _{L^{2k}(P_0)} = \Vert g\Vert _{L^2(P_0)} = \Vert \alpha \Vert _{{\text {Car}}(P_0)} = 1$ and $g\ge 0$. By definition and Cauchy-Schwarz, this is equivalent to

$$\begin{aligned} \left( \sum _{Q \in \mathscr {D}_{\ge m}(P_0)} \alpha _Q \left( \prod _{i=1}^k \langle f_i \rangle _{Q^{(m)}} \right) ^2 |Q| \right) ^{1/2} \left( \sum _{Q \in \mathscr {D}_{\ge m}(P_0)} \alpha _Q \langle g \rangle _Q^2 |Q| \right) ^{1/2}. \end{aligned}$$

The second term can be estimated, using (4.2) in the linear case, by

$$\begin{aligned} \left( \sum _{Q \in \mathscr {D}_{\ge m}(P_0)} \alpha _Q \langle g \rangle _Q^2 |Q| \right) ^{1/2} \le 2. \end{aligned}$$

For the first term observe that the sequence $\beta _Q$ defined by

$$\begin{aligned} \beta _Q = \frac{1}{2^{dm}} \sum _{R \in \mathscr {D}_m(Q)} \alpha _R \end{aligned}$$

is a Carleson sequence adapted to $P_0$ of the same constant. Indeed:

$$\begin{aligned} \frac{1}{|Q|} \sum _{R \in \mathscr {D}(Q)} \beta _R|R|&= \frac{1}{|Q|}\sum _{R \in \mathscr {D}(Q)} |R| \frac{1}{2^{dm}} \sum _{T \in \mathscr {D}_m(R)} \alpha _T \\&= \frac{1}{|Q|}\sum _{R \in \mathscr {D}(Q)}\sum _{T \in \mathscr {D}_m(R)} \alpha _T |T| \\&= \frac{1}{|Q|} \sum _{R \in \mathscr {D}_{\ge m}(Q)} \alpha _R|R| \\&\le \Vert \alpha \Vert _{\text {Car}(I)} \\&=1. \end{aligned}$$

Therefore, we can write the first term as

$$\begin{aligned} \left( \sum _{Q \in \mathscr {D}(P_0)} \beta _Q \left( \sum _{i=1}^k \langle f_i \rangle _Q \right) ^2 |Q| \right) ^{1/2}, \end{aligned}$$

which can also be estimated by (4.2) as follows:

$$\begin{aligned} \left( \sum _{Q \in \mathscr {D}(P_0)} \beta _Q \left( \sum _{i=1}^k \langle f_i \rangle _Q \right) ^2 |Q| \right) ^{1/2} \le \left( \frac{2k}{2k-1} \right) ^k \le 2. \end{aligned}$$

Combining both terms we arrive at

$$\begin{aligned} \int _{P_0} g(x) \mathcal {A}_{P_0,\alpha }^m \vec {f}(x) \, dx \le 4 \end{aligned}$$

which is what we wanted. $\square$

Now we can prove Theorem 4.1.

Proof

By homogeneity we can assume $\Vert \alpha \Vert _{{\text {Car}}(P_0)} = \Vert f_i\Vert _{L^1(P_0)} = 1$. We now follow the classical scheme which uses the $L^2$ bound and a standard Calderón-Zygmund decomposition, see for example Grafakos-Torres [6]. However, we need to be careful with the dependence on m, so we will adapt the proof in [9] to our operators.

Assume without loss of generality that $f_i \ge 0$. Define

$$\begin{aligned} \Omega _i = \left\{ x\in P_0: \mathcal {M}^d f_i(x) > \lambda ^{1/k}\right\} . \end{aligned}$$

If $\langle f_i \rangle _{P_0} > \lambda ^{1/k}$ then by the homogeneity assumption

$$\begin{aligned} |P_0| < \lambda ^{-1/k} \end{aligned}$$

and the estimate follows. Therefore, we can assume $\langle f_i \rangle _{P_0} \le \lambda ^{1/k}$ for all $1 \le i \le k$ and hence we can write $\Omega _i$ as a union the cubes in a collection $\mathcal {R}_i$ consisting of pairwise disjoint dyadic (strict) subcubes of $P_0$ with the property

$$\begin{aligned} \langle f_i \rangle _R > \lambda ^{1/k} \quad \text {and} \quad \langle f_i \rangle _{R^{(1)}} \le \lambda ^{1/k}. \end{aligned}$$

For each $1 \le i \le k$ let $b_i = \sum _{R \in \mathcal {R}_i} b_i^R$, where

$$\begin{aligned} b_i^R(x) := \left( f_i(x) - \langle f_i \rangle _R \right) \mathbbm {1}_R(x). \end{aligned}$$

We now let $g_i = f_i-b_i$.

Observe that we have

$$\begin{aligned} |g_i(x)| \le 2^d \lambda ^{1/k}, \end{aligned}$$

as well as

$$\begin{aligned} |\Omega _i| = \sum _{R \in \mathcal {R}_i} |R| \le \lambda ^{-1/k}. \end{aligned}$$

Define $\Omega = \cup _{i=1}^k \Omega _i$, then we have

$$\begin{aligned} |\{x \in P_0: \, \mathcal {A}_{P_0,\alpha }^m \vec {f}(x) > \lambda \}|&\le |\Omega | + |\{x\in P_0{\setminus } \Omega : \, \mathcal {A}_{P_0,\alpha }^m \vec {f}(x) > \lambda \}| \nonumber \\&\le k \lambda ^{-1/k} + |\{x\in P_0{\setminus } \Omega : \, \mathcal {A}_{P_0,\alpha }^m \vec {f}(x) > \lambda \}|. \end{aligned}$$

(4.3)

To estimate the second term observe that

$$\begin{aligned} \mathcal {A}_{P_0,\alpha }^m \vec {f}(x)&= \mathcal {A}_{P_0,\alpha }^m (\vec {g} + \vec {b})(x) \\&= \mathcal {A}_{P_0,\alpha }^m \vec {g}(x) + \sum _{j=1}^{2^k-1} \mathcal {A}_{P_0,\alpha }^m(h_1^j, \ldots , h_k^j)(x), \end{aligned}$$

where the functions $h_i^j$ are either $g_i$ or $b_i$ and, furthermore, for each $1 \le j \le 2^k-1$ there is at least one $1 \le i \le k$ such that $h_i^j = b_i$. Fix j and let $i_j$ be such that $h_{i_j}^j = b_{i_j}$, then

$$\begin{aligned}&\mathcal {A}_{P_0}^m (h_1^j, h_2^j, \ldots , h_{i_j}^j, \ldots , h_k^j)(x)\\&\quad = \sum _{Q \in \mathscr {D}_{\ge m}(P_0)} \alpha _Q \left( \prod _{i=1}^k \langle h_i^j \rangle _{Q^{(m)}} \right) \mathbbm {1}_Q(x) \\&\quad = \sum _{Q \in \mathscr {D}_{\ge m}(P_0)} \alpha _Q \langle b_{i_j} \rangle _{Q^{(m)}}\left( \prod _{1 \le i \le k, \, i \ne i_j} \langle h_i^j \rangle _{Q^{(m)}} \right) \mathbbm {1}_Q(x) \\&\quad = \sum _{R \in \mathcal {R}_{i_j}} \sum _{Q \in \mathscr {D}_{\ge m}(P_0)} \alpha _Q \langle b_{i_j}^R \rangle _{Q^{(m)}}\left( \prod _{1 \le i \le k, \, i \ne i_j} \langle h_i^j \rangle _{Q^{(m)}} \right) \mathbbm {1}_Q(x) \\&\quad = \sum _{R \in \mathcal {R}_{i_j}} \sum _{Q \in \mathscr {D}_{> m}(R)} \alpha _Q \langle b_{i_j}^R \rangle _{Q^{(m)}}\left( \prod _{1 \le i \le k, \, i \ne i_j} \langle h_i^j \rangle _{Q^{(m)}} \right) \mathbbm {1}_Q(x). \end{aligned}$$

So we deduce that $\mathcal {A}_{P_0,\alpha }^m (h_1^j, \ldots , h_k^j)(x) = 0$ for all $x \notin \Omega _{i_j}$. With this fact we can see that the second term in (4.3) is actually identical to

$$\begin{aligned} |\{x\in P_0{\setminus } \Omega : \, \mathcal {A}_{P_0,\alpha }^m \vec {g}(x) > \lambda \}|. \end{aligned}$$

Now we can use the $L^2$ bound as follows:

$$\begin{aligned} |\{x\in P_0{\setminus } \Omega : \, \mathcal {A}_{P_0,\alpha }^m \vec {g}(x) > \lambda \}|&\le \frac{1}{\lambda ^2} \Vert \mathcal {A}_{P_0,\alpha }^m \vec {g}\Vert _{L^2(P_0)}^2 \\&\le \frac{16}{\lambda ^2} \prod _{i=1}^k \Vert g_i\Vert _{L^{2k}(P_0)}^{2} \\&\le \frac{16}{\lambda ^2} \prod _{i=1}^k \left( 2^d \lambda ^{1/k} \right) ^{\frac{2k-1}{k}} \Vert g_i\Vert _{L^1(P_0)}^{1/k} \\&= \frac{16}{\lambda ^2} 2^{d(2k-1)} \lambda ^{2-1/k} \\&= 2^{4+ d(2k-1)}\lambda ^{-1/k}. \end{aligned}$$

Putting both estimates together we arrive at

$$\begin{aligned} |\{x \in P_0: \, \mathcal {A}_{P_0,\alpha }^m \vec {f}(x) > \lambda \}| \le 2^{5+ d(2k-1)}\lambda ^{-1/k} \end{aligned}$$

which yields the result with $C_{W} = 2^{k(5+ d(2k-1))}$. $\square$