Anisotropic Mixed-Norm Hardy Spaces

Abstract

We introduce and explore Hardy spaces defined by mixed Lebesgue norms and anisotropic dilations. We prove that the definitions of these spaces in terms of smooth, non-tangential, auxiliary, grand, and Poisson maximal operators coincide. We also study the relation between anisotropic mixed-norm Hardy spaces and mixed-norm Lebesgue spaces.

Appendix

1.1 Proof of Lemma 4.2

Proof

Let $\phi \in {\mathcal {S}}$ such that

$$\begin{aligned} {\hat{\phi }}(\xi )= {\left\{ \begin{array}{ll} 1,\;\; |\xi |\le 1 \\ 0,\;\; |\xi |\ge 2 \end{array}\right. }. \end{aligned}$$

We set $\varphi (x):=\phi (x)-2^{-\nu }\phi (2^{-\vec {a}}x)$ and $\psi _k(x)=\varphi _{(k)}(x)$ for every $k\in {\mathbb {N}}$. Then $\widehat{\psi _k}(\xi )={\hat{\varphi }}(2^{-k\vec {a}}\xi )={\hat{\phi }}(2^{-k\vec {a}}\xi )-{\hat{\phi }}(2^{(1-k)\vec {a}}\xi ),\;k\in {\mathbb {N}}$. We denote by $\psi _0=\phi$ and we have $\mathrm{{supp}\, }\widehat{\psi _0}\subset \{|\xi |\le 2\}=:T_0$ and

$$\begin{aligned} \mathrm{{supp}\, }\widehat{\psi _k}\subset 2^{k\vec {a}}\{2^{-a_M}\le |\xi |\le 2\}=:T_k,\;\;k\in {\mathbb {N}}. \end{aligned}$$

(6.7)

Also for every $\xi \in {\mathbb {R}}^n$, $\lim \limits _{k\rightarrow \infty } {\hat{\phi }}(2^{-k\vec {a}}\xi )=1$, so

$$\begin{aligned} \sum \limits _{k=0}^\infty \widehat{\psi _k}(\xi ) =1. \end{aligned}$$

(6.8)

We have $|{\hat{\Phi }}(\xi )|\ge 1/2$ when $|\xi |\le 2$ (or we modify properly $\Phi$ and keep the same notation), then by (6.8)

$$\begin{aligned} {\hat{\Psi }}(\xi )=\sum \limits _{k=0}^\infty \widehat{\psi _k}(\xi ){\hat{\Psi }}(\xi )=:\sum \limits _{k=0}^\infty {\hat{\eta }}^{(k)}(\xi ){\hat{\Phi }}(2^{-k\vec {a}}\xi ), \end{aligned}$$

where ${\hat{\eta }}^{(k)}(\xi )=\widehat{\psi _k}(\xi )\Big ({\hat{\Phi }}(2^{-k\vec {a}}\xi )\Big )^{-1}{\hat{\Psi }}(\xi ),$ which gives (4.3).

By (6.7) it is $\mathrm{{supp}\, }{\hat{\eta }}^{(k)}\subset T_k,\;k\ge 0$. Let $N>\nu$ and $|\beta |< \frac{N-\nu }{a_M}$, then

$$\begin{aligned} \bigg |\partial ^\beta {\hat{\Psi }}(\xi )\chi _{T_k}(\xi ) \bigg |\le c_N (1+|\xi |)^{-N} \sum \limits _{|\gamma |\le N} \bigg |\xi ^\gamma \partial ^\beta {\hat{\Psi }}(\xi ) \bigg |\chi _{T_k}(\xi ). \end{aligned}$$

(6.9)

Let $k\in {\mathbb {N}}$. For every $\xi \in T_k$ it is $\xi =2^{k\vec {a}}\zeta$, for some $\zeta$ with $2^{-a_M}\le |\zeta |\le 2.$ By (2.8) and (2.3)

$$\begin{aligned} \langle \xi \rangle _{{\vec {a}}}\ge c \bigg (1+|2^{k\vec {a}}\zeta |_{\vec {a}} \bigg )=c \bigg (1+2^k|\zeta |_{\vec {a}} \bigg )\ge c 2^k|\zeta |_{\vec {a}}\ge c 2^k, \end{aligned}$$

which is also true for $\xi \in T_0$, so

$$\begin{aligned} (1+|\xi |)^{-N}\le c \langle \xi \rangle _{{\vec {a}}}^{-Na_m}\le c 2^{-k Na_m},\;k\ge 0. \end{aligned}$$

(6.10)

On the other hand

$$\begin{aligned} |\xi ^\gamma \partial ^\beta {\hat{\Psi }}(\xi )|= & {} |\widehat{\partial ^\gamma (x^\beta \Psi )}(\xi )|\\\le & {} \sum \limits _{\delta \le \gamma } {\textstyle \left( {\begin{array}{c}\gamma \\ \delta \end{array}}\right) }\int _{{\mathbb {R}}^n} |x^{(\beta -\delta )_+}||\partial ^{\gamma -\delta } \Psi (x)|\text {d}x, \end{aligned}$$

where for every $\gamma \in ({{\mathbb {N}}}\cup \{0\})^n,\;x^{\gamma _+}=x_1^{{\gamma _1}_+}\cdots x_n^{{\gamma _n}_+}$, and ${\gamma _j}_+=\max (\gamma _j,0).$

But using (2.6) and (2.8)

$$\begin{aligned} |x^{(\beta -\delta )_+}|\le |x|^{|(\beta -\delta )_+|}\le (1+|x|)^{|(\beta -\delta )_+|}\le (1+|x|)^{|\beta |}\le c \langle x\rangle _{{\vec {a}}}^{|\beta |a_M}. \end{aligned}$$

Then by (3.5), (2.11), and since $N>|\beta |a_M+\nu$

$$\begin{aligned} |\xi ^\gamma \partial ^\beta {\hat{\Psi }}(\xi )|\le c_N {\mathscr {P}}_N(\Psi )\sum \limits _{\delta \le \gamma } \int _{{\mathbb {R}}^n} \langle x\rangle _{{\vec {a}}}^{-(N-|\beta |a_M)}\text {d}x\le c_N {\mathscr {P}}_N(\Psi ). \end{aligned}$$

(6.11)

So combining (6.9)–(6.11)

$$\begin{aligned} \bigg |\partial ^\beta {\hat{\Psi }}(\xi )\chi _{T_k}(\xi ) \bigg |\le c_N{\mathscr {P}}_N(\Psi )2^{-kNa_m}. \end{aligned}$$

(6.12)

Furthermore $|\partial ^\alpha {\hat{\psi }}_k(\xi )|,|\partial ^\alpha ({\hat{\Phi }}(2^{-k\vec {a}}\xi ))^{-1}(\xi )|\le c_\alpha 2^{-k\vec {a}\cdot \alpha }.$ The last estimate and (6.12) give

$$\begin{aligned} \bigg |\partial ^\alpha {\hat{\eta }}^{(k)}(\xi ) \bigg |\le & {} \sum \limits _{\beta \le \alpha } {\textstyle \left( {\begin{array}{c}\alpha \\ \beta \end{array}}\right) } c_\alpha 2^{-k\vec {a}\cdot \beta } c_N {\mathscr {P}}_N(\Psi ) 2^{-kNa_m}\\\le & {} c_N {\mathscr {P}}_N(\Psi ) 2^{-kNa_m}, \end{aligned}$$

for $|\alpha |\le N_1$.

Finally (2.6) leads to

$$\begin{aligned} |\eta ^{(k)}(x)|= & {} \bigg |{\mathcal {F}}^{-1}{\hat{\eta }}^{(k)}(x) \bigg |\le c (1+|x|)^{-N_1}\sum \limits _{|\alpha |\le N_1} \big \Vert \partial ^\alpha {\hat{\eta }}^{(k)}\big \Vert _1\\\le & {} c \langle x\rangle _{{\vec {a}}}^{-N_1a_m} \sum \limits _{|\alpha |\le N_1} \big \Vert \partial ^\alpha {\hat{\eta }}^{(k)}\big \Vert _\infty |\mathrm{{supp}\, }{\hat{\eta }}^{(k)}|\\\le & {} c_N {\mathscr {P}}_N(\Psi ) 2^{k\nu }2^{-kNa_m}\langle x\rangle _{{\vec {a}}}^{-N_1a_m}, \end{aligned}$$

since $|T_k|=\int _{T_k} \text {d}y\le 2^{k\nu } \int _{\{|\xi |\le 2\}} \text {d}z =c2^{k\nu }$, where we set $y=2^{k\vec {a}}z$. $\square$

1.2 Proof of Lemma 4.4.

Proof

Let $\varphi \in {\mathcal {S}}_*$. We assume without loss of generality that ${\hat{\varphi }}(0)=1.$ Then $\lim \limits _{k\rightarrow \infty } {\hat{\varphi }}(2^{-k\vec {a}}\xi )=1,\;\xi \in {\mathbb {R}}^n$. We set $N:=M+2$ and we observe that

$$\begin{aligned} \sum \limits _{m=1}^N{\textstyle \left( {\begin{array}{c}N\\ m\end{array}}\right) }{\hat{\varphi }}(\xi )^{2m}(1-{\hat{\varphi }}(\xi )^2)^{N-m}= & {} ({\hat{\varphi }}(\xi )^2+1-{\hat{\varphi }}(\xi )^2)^N-(1-{\hat{\varphi }}(\xi )^2)^N\nonumber \\= & {} 1-(1-{\hat{\varphi }}(\xi )^2)^N. \end{aligned}$$

(6.13)

In addition

$$\begin{aligned}&\sum \limits _{m=1}^N{\textstyle \left( {\begin{array}{c}N\\ m\end{array}}\right) }\big [{\hat{\varphi }}(2^{-k\vec {a}}\xi )^2-{\hat{\varphi }}(2^{-(k-1)\vec {a}}\xi )^2\big ]^m \big (1-{\hat{\varphi }}(2^{-k\vec {a}}\xi )^2\big )^{N-m}\\&\quad = \bigg (1-{\hat{\varphi }}(2^{-(k-1)\vec {a}}\xi )^2 \bigg )^N- \bigg (1-{\hat{\varphi }}(2^{-k\vec {a}}\xi )^2 \bigg )^N, \end{aligned}$$

which implies that

$$\begin{aligned} \sum \limits _{k=1}^\infty \sum \limits _{m=1}^N {\textstyle \left( {\begin{array}{c}N\\ m\end{array}}\right) } \bigg [{\hat{\varphi }}(2^{-k\vec {a}}\xi )^2-{\hat{\varphi }}(2^{-(k-1)\vec {a}}\xi )^2 \bigg ]^m \big (1-{\hat{\varphi }}(2^{-k\vec {a}}\xi )^2 \big )^{N-m}=(1-{\hat{\varphi }}(\xi )^2)^N,\nonumber \\ \end{aligned}$$

(6.14)

since $\lim \limits _{k\rightarrow \infty }{\hat{\varphi }}(2^{-k\vec {a}}\xi )=1$. By (6.13) and (6.14) we obtain

$$\begin{aligned} 1=&\sum \limits _{m=1}^N{\textstyle \left( {\begin{array}{c}N\\ m\end{array}}\right) }{\hat{\varphi }}(\xi )^{2m}(1-{\hat{\varphi }}(\xi )^2)^{N-m}\\&+\;\sum \limits _{k=1}^\infty \sum \limits _{m=1}^N {\textstyle \left( {\begin{array}{c}N\\ m\end{array}}\right) } \bigg [{\hat{\varphi }}(2^{-k\vec {a}}\xi )^2-{\hat{\varphi }}(2^{-(k-1)\vec {a}}\xi )^2 \bigg ]^m \big (1-{\hat{\varphi }}(2^{-k\vec {a}}\xi )^2 \big )^{N-m}. \end{aligned}$$

It follows that

$$\begin{aligned} 1={\hat{\psi }}(\xi ){\hat{\varphi }}(\xi )+\sum \limits _{k=1}^\infty {\widehat{\Psi }}(2^{-k\vec {a}}\xi )\Big ({\hat{\varphi }}(2^{-k\vec {a}}\xi )- {\hat{\varphi }}(2^{-(k-1)\vec {a}}\xi )\Big ), \end{aligned}$$

(6.15)

where

$$\begin{aligned} {\hat{\psi }}(\xi ):=\sum \limits _{m=1}^N{\textstyle \left( {\begin{array}{c}N\\ m\end{array}}\right) }{\hat{\varphi }}(\xi )^{2m-1}(1-{\hat{\varphi }}(\xi )^2)^{N-m} \end{aligned}$$

and

$$\begin{aligned} {\widehat{\Psi }}(\xi ):=\Big ({\hat{\varphi }}(\xi )+{\hat{\varphi }}(2^{\vec {a}}\xi )\Big )\sum \limits _{m=1}^N{\textstyle \left( {\begin{array}{c}N\\ m\end{array}}\right) } \bigg [{\hat{\varphi }}(\xi )^2-{\hat{\varphi }}(2^{\vec {a}}\xi )^2\bigg ]^{m-1}(1-{\hat{\varphi }}(\xi )^2)^{N-m}. \end{aligned}$$

It is easy to see that

$$\begin{aligned} \partial ^\alpha {\widehat{\Psi }}(0)=0,\;\;\text {for every}\;\;|\alpha |\le N-2=M, \end{aligned}$$

so $\Psi \in {\mathcal {S}}_M$. We write (6.15) with $\xi$ be replaced by $2^{-\vec {a}}\xi$

$$\begin{aligned} 1={\hat{\psi }}(2^{-\vec {a}}\xi ){\hat{\varphi }}(2^{-\vec {a}}\xi )+\sum \limits _{k=2}^\infty {\widehat{\Psi }}(2^{-k\vec {a}}\xi )\Big ({\hat{\varphi }}(2^{-k\vec {a}}\xi )- {\hat{\varphi }}(2^{-(k-1)\vec {a}}\xi )\Big ) \end{aligned}$$

(6.16)

and subtract (6.16) from (6.15) to get

$$\begin{aligned} {\widehat{\Psi }}(2^{-\vec {a}}\xi )\Big ({\hat{\varphi }}(2^{-\vec {a}}\xi )-{\hat{\varphi }}(\xi )\Big )= {\hat{\psi }}(2^{-\vec {a}}\xi ){\hat{\varphi }}(2^{-\vec {a}}\xi )-{\hat{\psi }}(\xi ){\hat{\varphi }}(\xi ). \end{aligned}$$

(6.17)

Let $m>j$ and $f\in {\mathcal {S}}'$. Then by (6.17)

$$\begin{aligned} {\hat{\psi }}(2^{-j\vec {a}}\xi ){\hat{\varphi }}(2^{-j\vec {a}}\xi ){\hat{f}}(\xi )&+ \sum \limits _{k=j+1}^m {\widehat{\Psi }}(2^{-k\vec {a}}\xi )\Big ({\hat{\varphi }}(2^{-k\vec {a}}\xi )- {\hat{\varphi }}(2^{-(k-1)\vec {a}}\xi )\Big ){\hat{f}}(\xi )\\&={\hat{\psi }}(2^{-m\vec {a}}\xi ){\hat{\varphi }}(2^{-m\vec {a}}\xi ){\hat{f}}(\xi )\rightarrow {\hat{f}}(\xi )\;\;\text {as}\;\;m\rightarrow \infty . \end{aligned}$$

Now (4.13) follows simply by inverting the Fourier transform. $\square$