Estimates for trilinear flag paraproducts on $L^{\infty }$ and Hardy spaces

Abstract

The boundedness of trilinear flag paraproducts from products of Hardy spaces $H^{p} \times H^q \times H^r$ to $L^s$, $1/p+1/q+1/r=1/s$, is discussed in the full range $0<p,q,r \le \infty$. The range of exponents to allow the boundedness is determined.

Appendices

Appendix 1

Theorem A shows that bilinear Fourier multipliers in $\mathcal {M}(\mathbb {R}^{2n})$ induce bounded operators from $H^p \times H^q$ to $L^r$ if $0<p, q\le \infty$, $0<r<\infty$ and $1/p+1/q=1/r$. In this theorem, we cannot replace $H^p$ by $L^p$ if $p\le 1$. (By symmetry, we cannot replace $H^q$ by $L^q$ if $q\le 1$ either.) Also, for $m\in \mathcal {M}(\mathbb {R}^{2n})$, $T_{m}$ is not always bounded from $L^{\infty } \times L^{\infty }$ to $L^{\infty }$. Although these facts may be known to many people, here we give proofs.

We first give an example of $m\in \mathcal {M}(\mathbb {R}^{2n})$ for which $T_{m}$ is not bounded from $L^p \times H^q$ to $L^r$ for $0<p\le 1$, $0<q<\infty$ and $1/p+1/q=1/r$. We take a sufficiently large integer $j_0$ satisfying

$$\begin{aligned} \{\xi \in \mathbb {R}^n \mid |\xi -2^je_1| \le 2^{j-j_0}\} \subset \{\xi \in \mathbb {R}^n \mid 2^{j-1/4} \le |\xi | \le 2^{j+1/4}\}, \quad j \in \mathbb {Z}, \end{aligned}$$

(5.1)

where $e_1=(1,0,\dots ,0) \in \mathbb {R}^n$, and take functions $\widetilde{\varphi }, \varphi , \psi \in \mathcal {S}(\mathbb {R}^n)$ such that

$$\begin{aligned}&\widetilde{\varphi }(\xi )=1 \ \text {on} \ \{|\xi | \le 2^{-j_0}\},\\&\quad \mathrm {supp}\, \varphi \subset \{|\xi | \le 2^{-j_0}\}, \quad |\mathcal {F}^{-1}\varphi (x)| \ge 1 \ \text {on} \ \{|x| \le 1\},\\&\quad \mathrm {supp}\, \psi \subset \{2^{-1/2} \le |\xi | \le 2^{1/2}\}, \quad \psi (\xi )=1 \ \text {on} \ \{2^{-1/4} \le |\xi | \le 2^{1/4}\}. \end{aligned}$$

We define

$$\begin{aligned} m(\xi ,\eta )=\sum _{j=1}^{\infty } \varphi (\xi /2^j)\psi (\eta /2^j). \end{aligned}$$

This $m(\xi ,\eta )$ is a typical example of bilinear Fourier multipliers in $\mathcal {M}(\mathbb {R}^{2n})$. We shall prove that the corresponding operator $T_m$ is not bounded from $L^p \times H^q$ to $L^r$ for $0<p\le 1$, $0<q<\infty$ and $1/p+1/q=1/r$.

To show the unboundedness of $T_m$, we set

$$\begin{aligned} f_N(x)=2^{Nn}\widetilde{\Phi }(2^Nx), \quad g_{N,t}(x)=\sum _{j=1}^N r_j(t)2^{jn/q} e^{ie_1\cdot (2^jx)}\Phi (2^jx), \end{aligned}$$

where $\widetilde{\Phi }=\mathcal {F}^{-1}\widetilde{\varphi }$, $\Phi =\mathcal {F}^{-1}\varphi$ and $r_j(t)$, $j \ge 1$, $0 \le t \le 1$, are the Rademacher functions [5, Appendix C]. We show that the estimate

$$\begin{aligned} \Vert T_m(f_N,g_{N,t})\Vert _{L^r} \lesssim \Vert f_N\Vert _{L^p}\Vert g_{N,t}\Vert _{H^q}, \quad N \ge 1, \ 0 \le t \le 1 \end{aligned}$$

(5.2)

does not hold.

First, observe that $\Vert f_N\Vert _{L^p} \approx 2^{Nn(1-1/p)}$. Second, since $\mathcal {F}_{x \rightarrow \xi }\left[ e^{ie_1\cdot x}\Phi (x)\right] =\varphi (\xi -e_1)$, the condition (5.1) and the support condition of $\varphi$ imply that the support of the Fourier transform of $e^{ie_1\cdot x}\Phi (x)$ is away from the origin. This means that $e^{ie_1\cdot x}\Phi (x)$ belongs to $H^q$ even if $q \le 1$. Thus, by change of variables,

$$\begin{aligned} \Vert g_{N,t}\Vert _{H^q}&\le {\left\{ \begin{array}{ll} \sum _{j=1}^N \left\| r_j(t)2^{jn/q}e^{ie_1\cdot (2^jx)}\Phi (2^jx) \right\| _{H^q},&{}q>1\\ \left( \sum _{j=1}^N \left\| r_j(t)2^{jn/q}e^{ie_1\cdot (2^jx)}\Phi (2^jx) \right\| _{H^q}^q \right) ^{1/q},&{}q \le 1 \end{array}\right. }\nonumber \\&\approx {\left\{ \begin{array}{ll} \sum _{j=1}^N \left\| \Phi (x) \right\| _{L^q}, &{}q>1 \\ \left( \sum _{j=1}^N \left\| e^{ie_1\cdot x}\Phi (x) \right\| _{H^q}^q \right) ^{1/q}, &{}q \le 1 \end{array}\right. }\nonumber \\&\approx N^{\max \{1,1/q\}}. \end{aligned}$$

Here we emphasize that the implicit constant is independent of t. Third, since $\widehat{f_N}(\xi )=\widetilde{\varphi }(2^{-N}\xi )$ and

$$\begin{aligned} \widehat{g_{N,t}}(\eta ) =\sum _{j=1}^N r_j(t)2^{jn(1/q-1)} \varphi (2^{-j}(\eta -2^je_1)), \end{aligned}$$

we have, by (5.1) and the properties of $\widetilde{\varphi }, \varphi , \psi$,

$$\begin{aligned} T_{m}(f_N,g_{N,t})(x) =\sum _{j=1}^N r_j(t)2^{jn(1+1/q)}e^{ie_1\cdot (2^jx)}\Phi (2^jx)^2. \end{aligned}$$

Hence, if (5.2) holds, then we have

$$\begin{aligned} \left( \int _{\mathbb {R}^n} \left| \sum _{j=1}^N r_j(t)2^{jn(1+1/q)}e^{ie_1\cdot (2^j)x}\Phi (2^jx)^2\right| ^r dx \right) ^{1/r} \lesssim 2^{Nn(1-1/p)}N^{\max \{1,1/q\}} \end{aligned}$$

for $N \ge 1$ and $0 \le t \le 1$. Since the implicit constant is independent of t, taking $L^r$-norm with respect to $0 \le t \le 1$, we have

$$\begin{aligned} \left( \int _{[0,1]}\int _{\mathbb {R}^n} \left| \sum _{j=1}^N r_j(t)2^{jn(1+1/q)}e^{ie_1\cdot (2^j)x}\Phi (2^jx)^2\right| ^r dxdt \right) ^{1/r} \lesssim 2^{Nn(1-1/p)}N^{\max \{1,1/q\}}. \end{aligned}$$

Furthermore, Khintchine’s inequality [5, Appendix C] gives

$$\begin{aligned} \left\{ \int _{\mathbb {R}^n} \left( \sum _{j=1}^N \left| 2^{jn(1+1/q)}e^{ie_1\cdot (2^j)x}\Phi (2^jx)^2\right| ^2 \right) ^{r/2} dx \right\} ^{1/r} \lesssim 2^{Nn(1-1/p)}N^{\max \{1,1/q\}} \end{aligned}$$

for $N \ge 1$. Using the fact $|\Phi (x)| \ge 1$ for $|x| \le 1$, we see that the left hand side is bounded below by

$$\begin{aligned}&\left\{ \sum _{k=1}^N \int _{\frac{1}{2^{k+1}}<|x|<\frac{1}{2^k}} \left( \sum _{j=1}^N \left| 2^{jn(1+1/q)}\Phi (2^jx)^2\right| ^2 \right) ^{r/2} dx \right\} ^{1/r}\nonumber \\&\quad \ge \left\{ \sum _{k=1}^N\int _{\frac{1}{2^{k+1}}<|x|<\frac{1}{2^k}} \left| 2^{kn(1+1/q)}\Phi (2^kx)^2\right| ^r dx \right\} ^{1/r}\nonumber \\&\quad \ge \left( \sum _{k=1}^N\int _{\frac{1}{2^{k+1}}<|x|<\frac{1}{2^k}} 2^{knr(1+1/q)} dx \right) ^{1/r}\nonumber \\&\quad \approx \left( \sum _{k=1}^N 2^{knr(1-1/p)} \right) ^{1/r} \approx {\left\{ \begin{array}{ll} N^{1/r},&{}p=1\\ 1, &{}p<1. \end{array}\right. } \end{aligned}$$

Thus, if (5.2) holds, then we must have $N^{1/r} \lesssim N^{\max \{1,1/q\}}$ if $p=1$ and $1 \lesssim 2^{Nn(1-1/p)}N^{\max \{1,1/q\}}$ if $p<1$. But the last inequalities are impossible.

We next give an example of $m\in \mathcal {M}(\mathbb {R}^{2n})$ for which $T_{m}$ is not bounded from $L^p \times L^{\infty }$ to $L^p$ for $0<p \le 1$. Let

$$\begin{aligned} m(\xi ,\eta )=\frac{\xi _1}{(|\xi |^2+|\eta |^2)^{1/2}}, \quad \xi ,\eta \in \mathbb {R}^n, \end{aligned}$$

where $\xi _1 \in \mathbb {R}$ is the first component of $\xi$. This $m(\xi ,\eta )$ belongs to $\mathcal {M}(\mathbb {R}^{2n})$. To see that the corresponding operator $T_{m}$ is not bounded from $L^p \times L^{\infty }$ to $L^p$ if $p \le 1$, we take a function $\varphi \in \mathcal {S}(\mathbb {R}^n)$ satisfying $\int _{\mathbb {R}^n}\varphi (\eta )\, d\eta =(2\pi )^n$, and set $g_{\epsilon }(x)=[\mathcal {F}^{-1}\varphi ](\epsilon x)$, $\epsilon >0$. Assume that the estimate

$$\begin{aligned} \Vert T_m(f,g_{\epsilon })\Vert _{L^p} \lesssim \Vert f\Vert _{L^p}\Vert g_{\epsilon }\Vert _{L^{\infty }} \end{aligned}$$

(5.3)

holds for $f \in \mathcal {S}(\mathbb {R}^n)$ and $\epsilon >0$. Then by the Lebesgue dominated convergence theorem,

$$\begin{aligned} \lim _{\epsilon \rightarrow 0} T_m(f,g_{\epsilon })(x)&=\lim _{\epsilon \rightarrow 0} \frac{1}{(2\pi )^{2n}} \int _{\mathbb {R}^{2n}} e^{ix\cdot (\xi +\eta )} \frac{\xi _1}{(|\xi |^2+|\eta |^2)^{1/2}} \widehat{f}(\xi )\epsilon ^{-n}\varphi (\epsilon ^{-1}\eta )\, d\xi d\eta \nonumber \\&=\frac{1}{(2\pi )^n} \int _{\mathbb {R}^n} e^{ix\cdot \xi } \frac{\xi _1}{|\xi |} \widehat{f}(\xi )\, d\xi =R_1(f)(x). \end{aligned}$$

Then, by Fatou’s lemma, (5.3) implies

$$\begin{aligned} \Vert R_1(f)\Vert _{L^p} \lesssim \Vert f\Vert _{L^p}. \end{aligned}$$

Hence, we reach the contradiction, since the Riesz transform $R_1$ is not bounded on $L^p$ if $p \le 1$.

The fact that bilinear Fourier multipliers in $\mathcal {M}(\mathbb {R}^{2n})$ do not always induce bounded operators from $L^{\infty } \times L^{\infty }$ to $L^{\infty }$ follows from the second counter example by duality.

Appendix 2

In this appendix, we shall give a proof of

$$\begin{aligned} \Vert f g\Vert _{L^p} \le A\Vert f\Vert _{H^p}, \ f \in H^p \quad \Longrightarrow \quad \Vert g\Vert _{L^{\infty }} \lesssim A, \end{aligned}$$

(6.1)

where the implicit constant is independent of g and A. This fact combined with duality also shows that the multiplication operator $f\mapsto fg$ is bounded from $L^{\infty }$ to $\textit{BMO}$ only if $g\in L^{\infty }$. Since the case $p>1$ is well known, we consider only the case $p \le 1$.

Let $Q_0=[-1/2,1/2]^n$ and $P_0$ be a polynomial satisfying

$$\begin{aligned} \Vert P_0\Vert _{L^{\infty }(Q_0)}=1, \quad \int _{Q_0} P_0 (y) y^{\alpha }\, dy=0 \;\;\text {for}\;\; |\alpha |\le [n/p-n]. \end{aligned}$$

Since $P_0$ is not identically zero, we see that $|\{y\in Q_0\mid P_0(y)=0\}|=0$. Thus, we can take $\delta >0$ such that

$$\begin{aligned} |\{y\in Q_0\mid |P_0(y)|> \delta \}|\ge \frac{9}{10}. \end{aligned}$$

(6.2)

For cubes Q in $\mathbb {R}^n$, we define

$$\begin{aligned} a_{Q}(x) =|Q|^{-1/p} \mathbf {1}_{Q}(x) P_0 \left( \frac{x-c(Q)}{\ell (Q)} \right) , \end{aligned}$$

(6.3)

where we used the same notations $c_Q$ and $\ell (Q)$ as in the proof of Lemma 2.5. Then

$$\begin{aligned} \mathrm {supp}\, a_Q \subset Q, \quad \Vert a_Q\Vert _{L^{\infty }}=|Q|^{-1/p}, \quad \int a_{Q}(x)x^{\alpha }\, dx =0 \;\;\text {for}\;\; |\alpha |\le [n/p-n], \end{aligned}$$

and this means that $a_Q$ is an $H^p$-atom. By (6.2),

$$\begin{aligned} |\{x\in Q \mid |a_Q(x)|> \delta |Q|^{-1/p}\}| \ge \frac{9}{10}|Q|. \end{aligned}$$

(6.4)

We assume that a measurable function g satisfies the condition in the left hand side of (6.1). Let $\lambda >0$ be such that $|\{x\in \mathbb {R}^n\mid |g(x)|> \lambda \}|>0$. By the Lebesgue differentiation theorem, there exists a cube Q such that

$$\begin{aligned} \left| \{x\in Q \mid |g(x)|> \lambda \}\right| \ge \frac{9}{10}|Q| \end{aligned}$$

(6.5)

(see [16, Chapter 1, Proposition 1]). We consider $a_Q$ defined by (6.3) with this Q. Then, since $\Vert a_{Q}\Vert _{H^p}\lesssim 1$, our assumption (namely, the condition in the left hand side of (6.1)) implies

$$\begin{aligned} \int |a_{Q}(x) g(x)|^p\, dx \lesssim A^p. \end{aligned}$$

On the other hand, it follows from (6.4) and (6.5) that

$$\begin{aligned} \left| \{x\in Q \mid |a_Q(x)|> \delta |Q|^{-1/p}, \; |g(x)|> \lambda \}\right| \ge \frac{8}{10}|Q|, \end{aligned}$$

which gives

$$\begin{aligned} \int |a_{Q}(x) g(x)|^p\, dx \ge \int _{Q} \mathbf {1}_{\{|a_Q(x)|> \delta |Q|^{-1/p},\; |g(x)|> \lambda \}} |a_{Q}(x) g(x)|^p\, dx \ge \frac{8}{10}\delta ^p \lambda ^{p}. \end{aligned}$$

Hence, $\frac{8}{10}\delta ^p \lambda ^{p}\lesssim A^{p}$, that is, $\lambda \lesssim A$. Thus we proved that $|\{x\in \mathbb {R}^n\mid |g(x)|> \lambda \}|>0$ holds only for $\lambda \lesssim A$. This implies $\Vert g\Vert _{L^{\infty }}\lesssim A$.

Appendix 3

We shall prove the factorization (4.18). For this it is sufficient to show that for every $u\in L^1$ there exists a $g\in H^1$ such that $|g(x)|\le |u(x)|\le 2|g(x)|$ for almost all $x\in \mathbb {R}^n$ and $\Vert g\Vert _{H^1}\approx \Vert u\Vert _{L^1}$. Take a $u\in L^1$. For $j\in \mathbb {Z}$, set $E_{j}=\{x \mid 2^j< |u(x)|\le 2^{j+1}\}$ and take a collection $\{Q_{j,k}\}_{k}$ of disjoint dyadic cubes such that $|Q_{j,k}\cap E_{j}|\ge 2^{-1}|Q_{j,k}|$ and $\{Q_{j,k}\}_{k}$ covers $E_{j}$ except for a null set. We set $E_{j,k}=Q_{j,k}\cap E_{j}$. We take $E^{\prime }_{j,k}\subset E_{j,k}$ such that $|E^{\prime }_{j,k}|=2^{-1}|E_{j,k}|$ and set

$$\begin{aligned} g(x) = \sum _{j}\sum _{k} 2^{j} \left\{ \mathbf {1}_{E^{\prime }_{j,k}}(x) - \mathbf {1}_{E_{j,k}\!\setminus \! E^{\prime }_{j,k}}(x) \right\} . \end{aligned}$$

For almost every $x\in \mathbb {R}^n$, we have

$$\begin{aligned} |g(x)| = \sum _{j}\sum _{k} 2^{j} \mathbf {1}_{E_{j,k}}(x) =\sum _{j} 2^{j} \mathbf {1}_{E_j}(x) \end{aligned}$$

and thus $|g(x)|\le |u(x)|\le 2|g(x)|$. Since $|Q_{j,k}|^{-1} \left\{ \mathbf {1}_{E^{\prime }_{j,k}}(x) - \mathbf {1}_{E_{j,k}\!\setminus \! E^{\prime }_{j,k}}(x)\} \right\}$ is an $H^1$-atom, we have

$$\begin{aligned} \Vert g\Vert _{H^1} \lesssim \sum _{j}\sum _{k} 2^{j}|Q_{j,k}| \approx \sum _{j}\sum _{k} 2^{j}|E_{j,k}| \approx \Vert u\Vert _{L^1}. \end{aligned}$$

The converse inequality $\Vert g\Vert _{H^1}\gtrsim \Vert u\Vert _{L^1}$ is obvious since $|g(x)|\approx |u(x)|$ almost everywhere.