The Multilinear Littlewood–Paley Operators with Minimal Regularity Conditions

Abstract

We investigate the quantitative weighted estimates for a large class of the multilinear Littlewood–Paley square operators. Our kernels satisfy the minimal regularity assumption, called $L^r$-Hörmander condition. We respectively establish the pointwise sparse domination for the multilinear square functions and their iterated commutators. Based on them, we obtain the strong type quantitative bounds and endpoint estimates. We recover lots of known weighted inequalities for Littlewood–Paley operators. Significantly, the approach is dyadic, quite elementary and simpler than that presented previously.

Appendix

In this section, we will present some observations about $L^p (\log L)^a$ functions, which will help us to comprehend the inverse of $L^p (\log L)^a$ functions. Those can not be found in any other references.

Henceforth, given a Young function A, we mean that $A^{-1}(x) = \inf \{y \in \mathbb {R}: A(y) > x\}$ for $x \in [0,\infty ]$, where $\inf \emptyset = \infty$.

Proposition 7.1

Let $-\infty<a<\infty$. Let $\Phi (x) = x(1 + \log ^+ x)^a$. Then $\Phi ^{-1}(x)\approx x(1 + \log ^+ x)^{-a}$ on $[0,\infty ]$ if $a\ge -1$, and on $[0,1]\cup [e^{-a-1},\infty ]$ if $a<-1$.

Proof

Let $\Psi (x)= x(1 + \log ^+ x)^{-a}$.

• (i) The case $a=0$. In this case tivially $\Phi ^{-1}(t)=\Psi (t)$.

• (ii) The case $a>0$. For $0\le t\le 1$ we have

  $$\begin{aligned} \Phi (\Psi (t))=t, \end{aligned}$$

  which means that

  $$\begin{aligned} \Phi ^{-1}(t)=\Psi (t)\ \text { for }0\le t\le 1. \end{aligned}$$

  For $t>1$, there holds

  $$\begin{aligned} \Phi (\Psi (t))&=t(1+\log ^+ t)^{-a}\bigl (1+\log ^+(t(1+\log ^+ t)^{-a})\bigr )^a \\&<t(1+\log ^+ t)^{-a}\bigl (1+\log ^+t\bigr )^a=t, \end{aligned}$$

  which combined with the increasingness of $\Phi (t)$ yields that

  $$\begin{aligned} \Phi ^{-1}(t)\ge \Psi (t)\ \text { for } 1<t<\infty . \end{aligned}$$

  Next, let $A\ge \max \{2^a,(2a)^a\}$. Then

  $$\begin{aligned} \Phi (A\Psi (t)) =At(1+\log ^+ t)^{-a}\bigl (1+\log ^+(At(1+\log ^+ t)^{-a})\bigr )^a. \end{aligned}$$

  (7.1)

  In the case $0<a\le \frac{1}{2}$, we see that

  $$\begin{aligned} (1+\log ^+ t)^{a}\le (1+\log ^+ t)^{1/2}\le \sqrt{t}. \end{aligned}$$

  This and (7.1) yield that

  $$\begin{aligned} \Phi (A\Psi (t))&>At(1+\log ^+ t)^{-a}\bigl (1+\log ^+\sqrt{t})\bigr )^a \\&\ge \frac{A}{2^a}t(1+\log ^+ t)^{-a}\bigl (1+\log ^+ t)\bigr )^a =\frac{A}{2^a} t>t. \end{aligned}$$

  Hence, we deduce that

  $$\begin{aligned} \Phi ^{-1}(t)\le A\Psi (t)\ \text { for }1< t<\infty . \end{aligned}$$

  (7.2)

  In the case $a>\frac{1}{2}$, let

  $$\begin{aligned} f(t)=A^{\frac{1}{a}}t^{\frac{1}{2a}}-1-\log t. \end{aligned}$$

  Then $f(1)=A^{\frac{1}{a}}-1>0$, and noting $A\ge (2a)^a$, we get

  $$\begin{aligned} f'(t)=\frac{A^{\frac{1}{a}}}{2a}t^{\frac{1}{2a}-1}-\frac{1}{t} =\frac{A^{\frac{1}{a}}}{2at}\Bigl (t^{\frac{1}{2a}}-\frac{2a}{A^{\frac{1}{a}}} \Bigr )>0\ \text { for }t>1. \end{aligned}$$

  Hence we have

  $$\begin{aligned} A\sqrt{t}> (1+\log ^+t)^a\ \text { for }t>1, \end{aligned}$$

  and so combinig (7.1) we get

  $$\begin{aligned} \Phi (A\Psi (t))>At(1+\log ^+ t)^{-a}\bigl (1+\log ^+\sqrt{t})\bigr )^a. \end{aligned}$$

  So, as in the case $0<a\le \frac{1}{2}$, we have (7.2). For $t=\infty$ we have trivially $\Phi ^{-1}(\infty )=\Psi (\infty )=\infty$. All together we obtain

  $$\begin{aligned} \Psi (t)\le \Phi ^{-1}(t)\le A\Psi (t) \ \text { for }0\le t\le \infty . \end{aligned}$$

• (iii) The case $a<0$.

  On [0, 1], $\Phi (t)=1$, and so the conclusion is clear. On $(1,\infty )$ we have

  $$\begin{aligned} \Phi (\Psi (t))&=t\frac{(1+\log ^+ t)^{-a}}{\bigl (1+\log ^+(t(1+\log ^+ t)^{-a})\bigr )^{-a}} \\&<t(1+\log ^+ t)^{-a}\bigl (1+\log ^+t\bigr )^a=t. \end{aligned}$$

  Since $\Phi (t)$ is increasing on $[e^{-a-1},\infty ]$, we see that

  $$\begin{aligned} \Phi ^{-1}(t)\ge \Psi (t) \ \text { on } [e^{-a-1},\infty ). \end{aligned}$$

  (7.3)

  For $A\ge (-2a)^{-a}$

  $$\begin{aligned} \Phi (A\Psi (t))=At\frac{(1+\log ^+ t)^{-a}}{\bigl (1+\log ^+(At(1+\log ^+ t)^{-a})\bigr )^{-a}}. \end{aligned}$$

  (7.4)

  In (ii), we showed that if $A\ge (-2a)^{-a}$

  $$\begin{aligned} A\sqrt{t}> (1+\log ^+t)^{-a}\ \text { for }t>1. \end{aligned}$$

  This combined with (7.4) gives

  $$\begin{aligned} \Phi (A\Psi (t))=At\frac{(1+\log ^+ t)^{-a}}{\bigl (1+\log ^+(A^2t^{3/2})\bigr )^{-a}} \le \frac{A}{\bigl (\frac{3}{2}(1+2\log A)\bigr )^{-a}}\,t. \end{aligned}$$

  So, if $A\ge (-3a)^{-a}$, we have $\frac{A}{\bigl (\frac{3}{2}(1+2\log A)\bigr )^{-a}}\ge 1$, and

  $$\begin{aligned} \Phi (A\Psi (t))\ge t\ \text { for }t>1. \end{aligned}$$

  Thus, since $\Phi (t)$ is increasing on $[e^{-a-1},\infty ]$, we get

  $$\begin{aligned} \Phi ^{-1}(t)\le A\Psi (t) \ \text { for }t\ge e^{-a-1}, \end{aligned}$$

  (7.5)

  if $A\ge \max \{(-2a)^{-a},\,(-3a)^{-a}\}$. By (7.3) and (7.5), we obtain

  $$\begin{aligned} \Psi (t)\le \Phi ^{-1}(t)\le A\Psi (t) \ \text { for }t\ge e^{-a-1}. \end{aligned}$$

  This completes the proof of Proposition 7.1. $\square$

Remark 7.2

Let $\Phi (t)=t(1+\log ^+ t)^a$ and $a\in \mathbb R$. If $a\ge 0$, $\Phi (t)$ is increasing and convex on $[0,\infty ]$.

If $-1\le a<0$, $\Phi (t)$ is increasing on $[0,\infty ]$. If $a<-1$, then $\Phi (t)$ is increasing on $[0,1]\cup [e^{-a-1},\infty ]$, and is decreasing on $[1,e^{-a-1}]$. Moreover, for $a<0$, $\Phi ''(t)=0$ on (0, 1), $>0$ on $(1,e^{-a})$ and $<0$ on $(e^{-a},\infty )$.

Proposition 7.3

Let $-\infty<a<\infty$ and $p>0$. Let $\Phi _p(x) = x^p(1 + \log ^+ x)^a$ and $\tilde{\Phi }_p(x) = x^p(1 + \log ^+ x^p)^a$. Then there exist $A_1, A_2\ge 1$ such that

$$\begin{aligned} {\Phi }_p(x) \le \tilde{\Phi }_p(A_1x), \ 0\le x<\infty , \end{aligned}$$

and

$$\begin{aligned} \tilde{\Phi }_p(x) \le {\Phi }_p(A_2x), \ 0\le x<\infty . \end{aligned}$$

Proof

If $x\in [0,1]$ or $a=0$, $\Phi _p(x)=\tilde{\Phi }_p(x)=x$, and so we have only to treat the case $x>1$ and $a\ne 0$. We consider the following four cases: $p\ge 1$ and $a>0$, $p\ge 1$ and $a<0$, $0<p<1$ and $a>0$, and $0<p<1$ and $a<0$.

Case 1$p\ge 1$and$a>0$. Clearly $\Phi _p(x)\le \tilde{\Phi }_p(x)$. And

$$\begin{aligned} \tilde{\Phi }_p(x)&=x^p(1 + p\log x)^a \\&\le p^ax^a(1 + \log x)^a\le (p^{a/p}x)^a(1 + \log (p^{a/p}x))^a ={\Phi }_p(p^{a/p}x). \end{aligned}$$

Case 2$p\ge 1$and$a<0$. Clearly $\Phi _p(x)\ge \tilde{\Phi }_p(x)$. And

$$\begin{aligned} {\Phi }_p(x)&= \frac{(pA)^{-a}x^p}{(pA + pA\log x)^{-a}} \le \frac{(pA)^{-a}x^p}{(pA + p\log x)^{-a}} \\&\le \frac{(pA)^{-a}x^p}{(pA -p\log (pA)^{-a/p}+ p\log (pA)^{-a/p}x)^{-a}} \\&=\frac{(pA)^{-a}x^p}{(pA -p\log (pA)^{-a/p}+ \log ((pA)^{-a/p}x)^p)^{-a}}. \end{aligned}$$

There exists $A_0>0$ such that

$$\begin{aligned} pA_0 -p\log (pA_0)^{-a/p}\ge 1. \end{aligned}$$

In this case $pA_0\ge 1$. Hence, letting $A_1=(pA_0)^{-a/p}>1$ we have

$$\begin{aligned} {\Phi }_p(x)\le \tilde{\Phi }_p(A_1x). \end{aligned}$$

Case 3$0<p<1$and$a>0$. It is easy to see that $\Phi _p(x)\ge \tilde{\Phi }_p(x)$. And

$$\begin{aligned} {\Phi }_p(x)&=x^p(1 + \log x)^a=\Big (\frac{1}{p}\Big )^{a}x^p(p + p\log x)^a\\&\le \Big (\frac{1}{p}\Big )^{a}x^p\Bigg (p +p \log \bigg (\Big (\frac{1}{p}\Big )^{a/p} x\bigg )\Bigg )^a \\&\le ((\frac{1}{p})^{a/p}x)^p\bigg (1 + \log \Big (\Big (\frac{1}{p}\Big )^{a/p} x\Big )^p\bigg )^a ={\tilde{\Phi }}_p\Bigg (\Big (\frac{1}{p}\Big )^{a/p}x\Bigg ). \end{aligned}$$

Case 4$0<p<1$and$a<0$. There holds

$$\begin{aligned} \tilde{\Phi }_p(x)=\frac{x^p}{(1 + p\log ^+ x)^{-a}} \ge \frac{x^p}{(1 + \log ^+ x)^{-a}}= \Phi _p(x). \end{aligned}$$

And for $A>1$

$$\begin{aligned} \tilde{\Phi }_p(x)&= \frac{x^p}{(1 + p\log ^+ x)^{-a}} \\&\le \frac{(\frac{1}{p})^{-a}x^p}{(1 + \log x)^{-a}} =\frac{(\frac{A}{p})^{-a}x^p}{(A + A\log x)^{-a}} \\&\le \frac{(\frac{A}{p})^{-a}x^p}{(A + \log x)^{-a}} =\frac{((\frac{A}{p})^{-a/p}x)^p}{(A+\frac{a}{p}\log \frac{A}{p} + \log ((\frac{A}{p})^{-a/p}x))^{-a}}. \end{aligned}$$

There exists $A_4>1$ such that

$$\begin{aligned} A_4+\frac{a}{p}\log \frac{A_4}{p}\ge 1. \end{aligned}$$

In this case $A_4/p\ge 1$. Hence, letting $A_2=(A_4/p)^{-a/p}>1$ we have

$$\begin{aligned} \tilde{\Phi }_p(x)\le {\Phi }_p(A_2x). \end{aligned}$$

Lemma 7.4

Let $a_1\ge 0$ and let $\Phi _1$, $\Phi _2$ be nonnegative, increasing and continuous functions on $[a_1,\infty )$. If for some $A_1\ge 1$

$$\begin{aligned} \Phi _1(t)\le \Phi _2(A_1 t),\ a_1\le t<\infty , \end{aligned}$$

(7.6)

then it holds

$$\begin{aligned} \Phi _2^{-1}(t)\le A_1\Phi _1^{-1}(t)\ \text { for }\max \{\Phi _1(a_1), \Phi _2(a_1)\}\le t<\infty . \end{aligned}$$

(7.7)

Proof

From (7.6) we get for $t\ge \Phi _1(a_1)$

$$\begin{aligned} \Phi _2(A_1\Phi _1^{-1}(t))\ge \Phi _1(\Phi _1^{-1}(t))=t, \end{aligned}$$

which implies that

$$\begin{aligned} A_1\Phi _1^{-1}(t)\ge \Phi _2^{-1}(t)\ \text { for }t\ge \max \{\Phi _1(a_1), \Phi _2(a_1)\}. \end{aligned}$$

Proposition 7.5

Let $-\infty<a<\infty$ and $p>1$. Let $\Phi _p(x) = x^p(1 + \log ^+ x)^a$. Then $\Phi _p^{-1}(x)\approx x^{1/p}(1 + \log ^+ x)^{-a/p}$ on $[0,\infty ]$ if $a\ge -1$, and on $[0,1]\cup [-ae^{-a-1},\infty ]$ if $a<-1$.

Proof

One can easily check that $\Phi _p(t)$ is increasing on $[0,\infty )$ if $a\ge -p$, and is increasing on $[0,1]\cup [e^{-a/p -1},\,\infty )$. Let $\tilde{\Phi }_p(t)=t^p(1 + \log ^+ t^p)^a$. We know that $\tilde{\Phi }_p(t)$ is increasing on $[0,\infty )$ if $a\ge -1$, and is increasing on $[0,1]\cup [e^{-(a+1)/p},\,\infty )$. Hence, if $a\ge 1$, by Proposition 7.3 and Lemma 7.4 we see that $\Phi _p^{-1}(t)\approx \tilde{\Phi }_p^{-1}(t)$ on $[0, \infty )$.

Here we see easily that $\tilde{\Phi }_p^{-1}=(\Phi ^{-1}(t))^{1/p}$, where $\Phi (t)=t(1 + \log ^+ t)^a$. Hence by Proposition 7.1 we obtain

$$\begin{aligned} \Phi _p^{-1}(t)\approx t^{1/p}(1+\log ^+ t)^{-a/p}\ \text { on }[0,\infty ]. \end{aligned}$$

In the case $a<-1$, on [0, 1] it is trivial. Next, we observe that $e^{-a/p -1}<e^{-(a+1)/p}$ and $\Phi _p(e^{-a/p -1})=-\frac{a}{p} e^{-\frac{a}{p} -1}<-ae^{-(a+1)} =\tilde{\Phi }_p(e^{-(a+1)/p})$. By the proof of Proposition 7.3, we know that $\tilde{\Phi }_p(t)\le \Phi _p(t)\le \tilde{\Phi }_p(At)$$(0\le t<\infty )$, for some $A>1$. So, by Lemma 7.4 we see that $\Phi _p^{-1}(t)\approx \tilde{\Phi }_p^{-1}(t)$ on $[-ae^{-(a+1)},\,\infty )$.

Remark 7.6

In genera, the following does not hold: Let $a_1\ge 0$ and let $\Phi _1$, $\Phi _2$ be nonnegative, increasing and continuous functions on $[a_1,\infty )$.

Then, from $\Phi _1(t)\approx \Phi _2(t)$ it follows $\Phi _1^{-1}(t)\approx \Phi _2^{-1}(t)$. In fact, let $\Phi _1(t)=\log (1+t)$ and $\Phi _2(t)=\log (1+t)^2$. Then $\Phi _1(t)\approx \Phi _2(t)$, but $\Phi _1^{-1}(t)=e^t -1$ and $\Phi _2^{-1}(t)=e^{t/2} -1$. And there exists no $C>0$ such that $\Phi _1^{-1}(t)\le C\Phi _2^{-1}(t)$.

However, if we suppose the convexty in addition, then we get a good result.

Lemma 7.7

Let $\Phi _1$, $\Phi _2$ be nonnegative, increasing and convex functions on [0, B]. If for some $A_1, A_2\ge 1$

$$\begin{aligned} \frac{\Phi _2(t)}{A_1}\le \Phi _1(t)\le A_2 \Phi _2(t),\ 0\le t<\infty , \end{aligned}$$

then it holds

$$\begin{aligned} \Phi _1(t)\le \Phi _2(A_2t)\ \text { and } \Phi _2(t)\le \Phi _1(A_1t) \ \ 0\le t<\infty . \end{aligned}$$

Proof

Since $\Phi _2(0)=0$ and $\Phi _2$ is convex on [0, B], we have

$$\begin{aligned} \frac{\Phi _2(t)}{t}\le \frac{\Phi _2(A_2t)}{A_2t},\ 0\le t<\infty , \end{aligned}$$

and hence $A_2\Phi _2(t)\le \Phi _2(A_2t)$, $0\le t<\infty$. Thus we get

$$\begin{aligned} \Phi _1(t)\le \Phi _2(A_2t)\ \ 0\le t<\infty . \end{aligned}$$

Similarly we have

$$\begin{aligned} \Phi _2(t)\le \Phi _1(A_1t)\ \ 0\le t<\infty . \end{aligned}$$