Discrete Hardy’s inequalities with $0<p⩽1$

Let $0<p⩽1$ . The discrete Hardy spaces $H^p(\mathbb{Z})$ consists of those sequence $a={\left\{a(n)\right\}}_{n\in \mathbb{Z}}$ satisfying $$\Vert a{\Vert }_{H^p(\mathbb{Z})}=\Vert a{\Vert }_{l^p(\mathbb{Z})}+\Vert H^{d}a{\Vert }_{l^p(\mathbb{Z})}<\infty \text{.}$$

Let $0<p⩽1$ . A sequence $a={\left\{a(k)\right\}}_{k\in \mathbb{Z}}$ is an $H^p(\mathbb{Z})$ -atom if it satisfies

• supp a is contained in a ball in $\mathbb{Z}$ of cardinality $2n+1\text{,}n⩾1$ .

• $\Vert a{\Vert }_{l^{\infty }(\mathbb{Z})}⩽{\left(2n+1\right)}^{-1/p}$ .

• ${\sum }_{n\in \mathbb{Z}}{n}^{\alpha }a(n)=0$ for every $\alpha \in \mathbb{N}$ with $\alpha ⩽\frac{1}{p}-1$ .

Let $0<p⩽1$ . The atomic discrete Hardy space $H_{\mathrm{at}}^p(\mathbb{Z})$ consists of those sequence $a={\left\{a(n)\right\}}_{n\in \mathbb{Z}}$ such that $$a={\displaystyle \sum _{j\in \mathbb{N}}}{\lambda }_j{a}_j$$ where $a_j$ are $H^p(\mathbb{Z})$ -atoms and $$\Vert a{\Vert }_{H_{\mathrm{at}}^p(\mathbb{Z})}=\inf \left\{{\left({\displaystyle \sum _{j\in \mathbb{N}}}|{\lambda }_j{|}^p\right)}^{\frac{1}{p}}\right\}$$ where the infimum is taken over all possible representations of a in terms of $H^p(\mathbb{Z})$ -atoms.

Let $0<p⩽1$ . Then, there exist constants $B\text{,}C>0$ such that for any sequence $a={\left\{a(n)\right\}}_{n\in \mathbb{Z}}$ , we have $$B\Vert a{\Vert }_{H^p(\mathbb{Z})}⩽\Vert a{\Vert }_{H_{\mathrm{at}}^p(\mathbb{Z})}⩽C\Vert a{\Vert }_{H^p(\mathbb{Z})}\text{.}$$

Let $0<p⩽1$ and $0⩽\mu <1$ . Suppose that $\alpha \in \mathbb{N}\cup \{0\}$ satisfies $\alpha ⩽\frac{1}{p}-1$ . If

Let $0<p⩽1$ , $0⩽\mu <1$ and $\alpha \in \mathbb{N}\cup \{0\}$ with $\alpha ⩽\frac{1}{p}-1$ . Suppose that $$\frac{1}{q}-\frac{1}{r}<\mu ⩽\frac{1}{q}$$ for some $q>1$ . If $a={\left\{a(n)\right\}}_{n\in \mathbb{Z}}$ satisfies

• $\mathrm{supp}a$ is contained in a ball B in $\mathbb{N}⧹\{0\}$ of cardinality $2n+1\text{,}n⩾1$ ,

• $\Vert a{\Vert }_{l^{\infty }(\mathbb{N})}⩽{\left(2n+1\right)}^{-1/p}$ ,

• ${\sum }_{n=0}^{\infty }{n}^{\alpha }a(n)=0$ ,

Let $B=\{i\in \mathbb{N}:M⩽i⩽N\}\text{,}M\text{,}N\in \mathbb{N}$ . We have $|B|=N-M+1$ . For any $i<M$ , we have $$(T_{\alpha \text{,}\mu }a)(i)=\frac{1}{i^{\alpha -\mu +1}}{\displaystyle \sum _{j=1}^i}{j}^{\alpha }a(j)=0\text{.}$$

Similarly, for any $i>N$ , Items (1) and (3) assure that $$(T_{\alpha \text{,}\mu }a)(i)=\frac{1}{i^{\alpha -\mu +1}}{\displaystyle \sum _{j=1}^i}{j}^{\alpha }a(j)=\frac{1}{i^{\alpha -\mu +1}}{\displaystyle \sum _{j\in \mathbb{N}}}{j}^{\alpha }a(j)=0\text{.}$$

Therefore, we find that $\mathrm{supp}(T_{\alpha \text{,}\mu }a)\subseteq B$ .

The Hölder inequality yields $$|{\displaystyle \sum _{j=1}^m}{j}^{\alpha }a(j)|⩽{\left({\displaystyle \sum _{j=1}^m}|a(j){|}^q\right)}^{1/q}{\left({\displaystyle \sum _{j=1}^m}{j}^{\alpha q\prime }\right)}^{1/q\prime }⩽C\Vert a{\Vert }_{l^q}{m}^{\alpha +\frac{1}{q\prime }}$$ for some $C>0$ . Therefore, $$|(T_{\alpha \text{,}\mu }a)(m)|⩽C\frac{1}{m^{\alpha -\mu +1}}\Vert a{\Vert }_{l^q}{m}^{\alpha +\frac{1}{q\prime }}={\mathit{Cm}}^{\mu -\frac{1}{q}}\Vert a{\Vert }_{l^q}\text{.}$$

Furthermore, as $\mathrm{supp}(T_{\alpha \text{,}\mu }a)\subseteq B$ and $\mu ⩽\frac{1}{q}$ , we obtain $$\begin{array}{cc}\hfill \Vert T_{\alpha \text{,}\mu }a{\Vert }_{l^r}^r⩽& C\Vert a{\Vert }_{l^q}^r{\displaystyle \sum _{m=M}^N}{m}^{r\mu -\frac{r}{q}}⩽C{\int }_M^{N+1}{x}^{r\mu -\frac{r}{q}}\mathit{dx}\hfill \\ \hfill ⩽& C\Vert a{\Vert }_{l^q}^r(N^{r\mu -\frac{r}{q}+1}-M^{r\mu -\frac{r}{q}+1})\text{.}\hfill \end{array}$$

Since $\frac{1}{q}-\frac{1}{r}<\mu ⩽\frac{1}{q}$ , we find that $0<r\mu -\frac{r}{q}+1⩽1$ . Consequently, $$(N^{r\mu -\frac{r}{q}+1}-M^{r\mu -\frac{r}{q}+1})⩽{\left(N-M\right)}^{r\mu -\frac{r}{q}+1}⩽|B{|}^{r\mu -\frac{r}{q}+1}\text{.}$$

Hence, $$\Vert T_{\alpha \text{,}\mu }a{\Vert }_{l^r}⩽C\Vert a{\Vert }_{l^q}|B{|}^{\mu -\frac{r}{q}+\frac{1}{r}}⩽C\Vert a{\Vert }_{l^{\infty }}|B{|}^{\frac{1}{q}}|B{|}^{\mu -\frac{1}{q}+\frac{1}{r}}⩽C|B{|}^{\mu +\frac{1}{r}-\frac{1}{p}}$$ for some $C>0$ .  □

Let $0<p⩽1$ and $\alpha \in \mathbb{N}\cup \{0\}$ with $\alpha ⩽\frac{1}{p}-1$ . If $a\in H^p(\mathbb{Z})$ with $\mathrm{supp}a\subset \mathbb{N}⧹\{0\}$ , then there exist a sequence of scalars ${\left\{{\lambda }_j\right\}}_{j\in \mathbb{N}}$ and sequences ${\left\{a_j\right\}}_{j\in \mathbb{N}}$ satisfying

• supp $a_j$ is contained in a ball in $\mathbb{N}⧹\{0\}$ of cardinality $2n+1\text{,}n⩾1$ ,

• $\Vert a_j{\Vert }_{l^{\infty }(\mathbb{Z})}⩽2{\left(2n+1\right)}^{-1/p}$ ,

• ${\sum }_{n=0}^{\infty }{n}^{\alpha }{a}_j(n)=0$ ,

In view of Theorem 1, we have a sequence of scalars ${\left\{{\lambda }_j\right\}}_{j\in \mathbb{N}}$ and $H^p(\mathbb{Z})$ -atoms ${\left\{a_j\right\}}_{j\in \mathbb{N}}$ such that $a={\sum }_{j\in \mathbb{N}}{\lambda }_j{a}_j$ and ${\left.{\sum }_{j\in \mathbb{N}}|{\lambda }_j{|}^p\right)}^{1/p}⩽C\Vert a{\Vert }_{H^p(\mathbb{Z})}$ .

We split the proof into three cases, $\alpha$ is a positive even integer, $\alpha$ is a positive odd integer and $\alpha$ equals zero.

1. $\alpha$ is a positive even integer. We consider the even part of a, $$a_e(k)={\displaystyle \sum _{j\in \mathbb{N}}}{\lambda }_j\left(\frac{a_j(k)+a_j(-k)}{2}\right)\text{.}$$

   Since $\mathrm{supp}a\subset \mathbb{N}⧹\{0\}$ , we have $a=2a_e{\chi }_{\mathbb{N}⧹\{0\}}$ . That is, $$a(k)={\displaystyle \sum _{j\in \mathbb{N}}}{\lambda }_j(a_j(k)+a_j(-k)){\chi }_{\mathbb{N}⧹\{0\}}(k)\text{.}$$

   Write $A_j={\left\{A_j(k)\right\}}_{k\in \mathbb{Z}}$ where $A_j(k)=(a_j(k)+a_j(-k)){\chi }_{\mathbb{N}⧹\{0\}}(k)$ . We are going to show that $A_j$ fulfills Item (1)–(3).

   As $\alpha$ is even, we find that $${\displaystyle \sum _{k\in \mathbb{Z}}}{k}^{\alpha }{a}_j(k)={\displaystyle \sum _{k\in \mathbb{Z}}}{k}^{\alpha }{a}_j(-k)=0\text{.}$$

   If $\mathrm{supp}{a}_j\subset \mathbb{N}⧹\{0\}\text{,}{A}_j=a_j$ satisfies Item (1)–(3). If $\mathrm{supp}{a}_j\subset \mathbb{Z}⧹(\mathbb{N}\cup \{0\})$ , then $A_j(·)=a_j(-·)$ also satisfies Item (1)–(3).

   If $0\in \mathrm{supp}{a}_j$ , we find that $${\displaystyle \sum _{n=0}^{\infty }}{n}^{\alpha }{A}_j(n)={\displaystyle \sum _{n=0}^{\infty }}{n}^{\alpha }(a_j(n)+a_j(-n))={\displaystyle \sum _{n\in \mathbb{Z}}}{n}^{\alpha }{a}_j(n)=0$$ because $\alpha$ is a positive even integer.

   We have $\Vert A_j{\Vert }_{l^{\infty }(\mathbb{N})}⩽2\Vert a_j{\Vert }_{l^{\infty }(\mathbb{N})}$ . Moreover, $\mathrm{supp}{A}_j\subseteq (\mathrm{supp}{a}_j\cup \mathrm{supp}{a}_j(-·))\cap (\mathbb{N}⧹\{0\})$ . Hence, $|\mathrm{supp}{A}_j|⩽|\mathrm{supp}{a}_j|$ .

   Consequently, $A_j$ fulfills Item (1)–(3).

2. $\alpha$ is a positive odd integer. We consider the odd part of a, $$a_o(k)={\displaystyle \sum _{j\in \mathbb{N}}}{\lambda }_j\left(\frac{a_j(k)-a_j(-k)}{2}\right)\text{.}$$

   Since $\mathrm{supp}a\subset \mathbb{N}⧹\{0\}$ , we have $a=2a_o{\chi }_{\mathbb{N}⧹\{0\}}$ . That is, $$a(k)={\displaystyle \sum _{j\in \mathbb{N}}}{\lambda }_j(a_j(k)-a_j(-k)){\chi }_{\mathbb{N}⧹\{0\}}(k)\text{.}$$

   Write $A_j={\left\{A_j(k)\right\}}_{k\in \mathbb{Z}}$ where $A_j(k)=(a_j(k)-a_j(-k)){\chi }_{\mathbb{N}⧹\{0\}}(k)$ .

   As $\alpha$ is odd, we find that $${\displaystyle \sum _{k\in \mathbb{Z}}}{k}^{\alpha }{a}_j(k)=-{\displaystyle \sum _{k\in \mathbb{Z}}}{k}^{\alpha }{a}_j(-k)=0\text{.}$$

   If $\mathrm{supp}{a}_j\subset \mathbb{N}⧹\{0\}\text{,}{A}_j=a_j$ satisfies Item (1)–(3). If $\mathrm{supp}{a}_j\subset \mathbb{Z}⧹(\mathbb{N}\cup \{0\})$ , then $A_j(·)=a_j(-·)$ also satisfies Item (1)–(3).

   If $0\in \mathrm{supp}{a}_j$ , we find that $${\displaystyle \sum _{n=0}^{\infty }}{n}^{\alpha }{A}_j(n)={\displaystyle \sum _{n=0}^{\infty }}{n}^{\alpha }(a_j(n)-a_j(-n))={\displaystyle \sum _{n\in \mathbb{Z}}}{n}^{\alpha }{a}_j(n)=0$$ because $\alpha$ is a positive odd integer.

   Therefore, $A_j$ fulfills Item (1)–(3).

3. $\alpha$ equals to zero. In this case, we also consider the even part of a, $$a_e(k)={\displaystyle \sum _{j\in \mathbb{N}}}{\lambda }_j\left(\frac{a_j(k)+a_j(-k)}{2}\right)\text{.}$$ Write $$b_j(k)=\left\{\begin{array}{cc}{a}_j(k)\text{,}\hfill & k\ne 0\text{,}\hfill \\ \frac{a_j(0)}{2}\text{,}\hfill & k=0\text{.}\hfill \end{array}\right.$$ When $k\ne 0$ , we obviously have $$a_e(k)={\displaystyle \sum _{j\in \mathbb{N}}}{\lambda }_j\left(\frac{a_j(k)+a_j(-k)}{2}\right)={\displaystyle \sum _{j\in \mathbb{N}}}{\lambda }_j\left(\frac{b_j(k)+b_j(-k)}{2}\right)\text{.}$$ Since $\mathrm{supp}a\subset \mathbb{N}⧹\{0\}$ , we have $$a(0)={\displaystyle \sum _{j\in \mathbb{N}}}{\lambda }_j{a}_j(0)=0\text{.}$$ Therefore, for $k=0$ , we have $$a_e(0)=a(0)=0={\displaystyle \sum _{j\in \mathbb{N}}}{\lambda }_j{a}_j(0)={\displaystyle \sum _{j\in \mathbb{N}}}{\lambda }_j\left(\frac{b_j(0)+b_j(0)}{2}\right)\text{.}$$ As $\mathrm{supp}a\subset \mathbb{N}⧹\{0\}$ , we have $a=2a_e{\chi }_{\mathbb{N}⧹\{0\}}$ . That is, $$a(k)={\displaystyle \sum _{j\in \mathbb{N}}}{\lambda }_j(b_j(k)+b_j(-k)){\chi }_{\mathbb{N}⧹\{0\}}(k)\text{.}$$

   Write $B_j={\left\{B_j(k)\right\}}_{k\in \mathbb{Z}}$ where $B_j(k)=(b_j(k)+b_j(-k)){\chi }_{\mathbb{N}⧹\{0\}}(k)$ . It remains to show that $B_j$ satisfies Item (1)–(3).

   When $\mathrm{supp}{a}_j\subset \mathbb{N}⧹\{0\}\text{,}{B}_j=a_j$ satisfies Item (1)–(3). When $\mathrm{supp}{a}_j\subset \mathbb{Z}⧹(\mathbb{N}\cup \{0\})$ , then $B_j(·)=a_j(-·)$ also satisfies Item (1)–(3).

   If $0\in \mathrm{supp}{a}_j$ , we find that $$\begin{array}{cc}\hfill {\sum }_{n=0}^{\infty }{B}_j(n)=& {\sum }_{n=0}^{\infty }(b_j(n)+b_j(-n))=\left({\displaystyle \sum _{n=1}^{\infty }}{a}_j(n)+{\displaystyle \sum _{n=-\infty }^{-1}}{a}_j(n)\right)+2b_j(0)\hfill \\ \hfill =& {\sum }_{n\in \mathbb{Z}}{a}_j(n)=0\text{.}\hfill \end{array}$$