$L^{p}$ Hardy type inequality in the half-space on the H-type group

Abstract

In the current work we studied Hardy type and $L^{p}$ Hardy type inequalities in the half-space on the H-type group, where the Hardy inequality in the upper half-space $\mathbf{R}_{+}^{n}$ was proved by Tidblom in (J. Funct. Anal. 221:482-495, 2005).

1 Introduction

In recent years a lot of authors studied the Hardy inequalities (see [1–5]). They are the extensions of the original inequality by Hardy [6]. The Heisenberg group, denoted by $\mathbf{H}_{n}$, is also very popular in mathematics (see [7–11]). By $\mathbf{H}_{n,+}=\{(z,t)\in\mathbf {H}_{n}| z\in C^{n}, t>0\}$ is denoted the half-space on the Heisenberg group. A Hardy type inequality on $\mathbf{H}_{n,+}$ in [4] is stated as follows. For $u\in C_{0}^{\infty}(\mathbf{H}_{n,+})$, we have

$$\begin{aligned} \int_{\mathbf{H}_{n,+}}\vert \nabla_{\mathbf{H}_{n}}u\vert ^{2} \,dz\,dt \geq \int_{\mathbf{H}_{n,+}}\frac{\vert z\vert ^{2}}{t^{2}}\vert u\vert ^{2}\,dz \,dt +\frac{(Q+2)(Q-2)}{4} \int_{\mathbf{H}_{n,+}}\rho ^{-4}\vert z\vert ^{2} \vert u\vert ^{2}\,dz\,dt, \end{aligned}$$

where $\rho=(\vert z\vert ^{4}+t^{2})^{\frac{1}{4}}$ and $Q=2n+2$ is the homogeneous dimension of the Heisenberg group. We know that the H-type group, denoted by $\mathbf{H}=\{(z,t)\in\mathbf {H}| z\in C^{n}, t\in\mathbf{R}^{m}\}$, is the nilpotent Lie group introduced by Kaplan (see [12]). We also know that $\mathbf {H}_{n}$ is a nilpotent Lie group with homogeneous dimension $2n+2$. The homogeneous dimension of H is $2n+2m$. Kaplan introduced the H-type group as a direct generalization of the Heisenberg group, which motivates us to study the H-type group.

In this paper we prove the Hardy type inequality in the half-space on the H-type group (see Theorem 2.1). The half-space on the H-type group is given by $\mathbf{H}^{+}=\{(z,t)\in\mathbf{H}| t_{m}>0\}$. For $u\in C_{0}^{\infty}(\mathbf{H}^{+})$, we have

$$\begin{aligned} \int_{\mathbf{H}^{+}}\vert \nabla_{\mathbf{H}}u\vert ^{2} \,dz\,dt \geq& \frac{1}{16} \int_{\mathbf{H}^{+}}\frac {\vert z\vert ^{2}}{t_{m}^{2}}\vert u\vert ^{2}\,dz \,dt \\ &{}+\frac{(Q-2)(Q+2)}{4} \int_{\mathbf {H}^{+}}d(z,t)^{-4}\vert z\vert ^{2} \vert u\vert ^{2}\,dz\,dt \\ &{}-(Q+2) \int_{\mathbf{H}^{+}}d(z,t)^{-4}\sum_{k=1}^{m-1} \bigl\langle U^{(k)}z, U^{(m)}z \bigr\rangle \frac{t_{k}\vert u\vert ^{2}}{t_{m}} \,dz\,dt, \end{aligned}$$

where $d(z,t)=(\vert z\vert ^{4}+16\vert t\vert ^{2})^{\frac{1}{4}}$ and $Q=2n+2m$ is the homogeneous dimension of the H-type group.

In [1], the $L^{p}$ Hardy inequalities in the upper half-space $\mathbf{R}_{+}^{n}$ were studied. So we are also interested in the $L^{p}$ Hardy type inequalities in the half-space on the H-type group.

In the remainder of this section we give a basic concept of H-type group and a useful theorem.

Let $(z,t), (z',t')\in\mathbf{H}$, $U^{(j)}$ is a $2n\times2n$ skew-symmetric orthogonal matrix and $U^{(j)}$ satisfy $U^{(i)}U^{(j)}+U^{(j)}U^{(i)}=0$, $i, j=1,2,\ldots,m$ with $i\neq j$. The group law is given by

$$\begin{aligned} (z,t) \bigl(z',t'\bigr)=\biggl(z+z',t+t'+ \frac{1}{2}\Im\bigl(zz'\bigr)\biggr), \end{aligned}$$

where $(\Im(zz'))_{j}=\langle z,U^{(j)}z'\rangle$, $\langle z,U^{(j)}z'\rangle$ is the inner product of z and $U^{(j)}z'$ on $\mathbf{R}^{2n}$.

The left invariant vector fields are given by

$$\begin{aligned}& X_{j}=\frac{\partial}{\partial x_{j}}+\frac{1}{2}\sum _{k=1}^{m}\Biggl(\sum_{i=1}^{2n}s_{i}U_{i,j}^{(k)} \Biggr)\frac{\partial}{\partial t_{k}}, \quad j=1,2,\ldots,n, \\& Y_{j}=\frac{\partial}{\partial y_{j}}+\frac{1}{2}\sum _{k=1}^{m}\Biggl(\sum_{i=1}^{2n}s_{i}U_{i,j+n}^{(k)} \Biggr)\frac{\partial}{\partial t_{k}}, \quad j=1,2,\ldots,n, \\& T_{k}=\frac{\partial}{\partial t_{k}}, \quad k=1,2,\ldots,m, \end{aligned}$$

where $s_{i}=x_{i}$ for $i=1,2,\ldots,n$ and $s_{i}=y_{i-n}$ for $i=n+1,n+2,\ldots,2n$. The sub Laplacian $\mathcal{L}$ is defined by

$$\begin{aligned} \mathcal{L}=-\sum_{j=1}^{n} \bigl(X_{j}^{2}+Y_{j}^{2}\bigr). \end{aligned}$$

We write $\nabla_{\mathbf{H}}=(X_{1},\ldots,X_{n},Y_{1},\ldots,Y_{n})$ and

$$\begin{aligned} \operatorname {div}_{\mathbf{H}}(f_{1},f_{2},\ldots,f_{2n})= \sum_{j=1}^{n}(X_{j}f_{j}+Y_{j}f_{j+n}). \end{aligned}$$

We define the Kohn Laplacian $\Delta_{\mathbf{H}}$ by

$$\begin{aligned} \Delta_{\mathbf{H}}=\sum_{j=1}^{n} \bigl(X_{j}^{2}+Y_{j}^{2} \bigr). \end{aligned}$$

(1)

On the Heisenberg group, a fundamental solution for the sub Laplacian was studied in [13]. Similarly, we give a fundamental solution for $\Delta_{\mathbf{H}}$ below. For $0< r<\infty$ and $(z,t)\in\mathbf {H}$, we define $\delta_{r}(z,t)=(rz,r^{2}t)$.

Theorem 1.1

A fundamental solution for $\Delta_{\mathbf{H}}$ with source at 0 is given by $c_{n,m}d(z,t)^{-Q+2}$, where

$$\begin{aligned} c_{n,m}^{-1}=4(n+m+1) (1-n-m) \int_{\mathbf{H}}\vert z\vert ^{2} \bigl(d(z,t)^{4}+1 \bigr)^{\frac{-n-m-3}{2}}\,dz\,dt. \end{aligned}$$

For $u(z,t)\in C_{0}^{\infty}(\mathbf{H})$, we have

$$\bigl\langle \Delta_{\mathbf{H}}u(z,t), c_{n,m}d(z,t)^{-Q+2} \bigr\rangle _{L^{2}(\mathbf{H})}=u(0,0). $$

Proof

For $\varepsilon>0$, let $d_{\varepsilon}(z,t)=(d(z,t)^{4}+\varepsilon ^{4})^{\frac{1}{4}}$, similar to [13], by equation (1) and a direct calculation, we have

$$\begin{aligned} \Delta_{\mathbf{H}}d_{\varepsilon}(z,t)^{-Q+2}=\varepsilon^{-Q} \phi \bigl(\delta_{\frac{1}{\varepsilon}}(z,t)\bigr), \end{aligned}$$

(2)

where

$$\begin{aligned} \phi(z,t)=4(n+m+1) (1-n-m)\vert z\vert ^{2} \bigl(d(z,t)^{4}+1 \bigr)^{\frac {-n-m-3}{2}}. \end{aligned}$$

From this, it follows that, for all $u(z,t)\in C_{0}^{\infty}(\mathbf{H})$,

$$\begin{aligned} \bigl\langle \Delta_{\mathbf{H}}u(z,t), c_{n,m}d(z,t)^{-Q+2} \bigr\rangle _{L^{2}(\mathbf{H})} =&\lim_{\varepsilon\rightarrow0}\bigl\langle \Delta_{\mathbf{H}}u(z,t), c_{n,m}d_{\varepsilon}(z,t)^{-Q+2} \bigr\rangle _{L^{2}(\mathbf{H})} \\ =&\lim_{\varepsilon\rightarrow0}\bigl\langle u(z,t), c_{n,m}\Delta _{\mathbf{H}}d_{\varepsilon}(z,t)^{-Q+2}\bigr\rangle _{L^{2}(\mathbf {H})} \\ =&u(0,0). \end{aligned}$$

 □

For $\varepsilon>0$, the Green’s function on the half-space on the H-type group is given by

$$\begin{aligned} G(z,t,\varepsilon) =&\frac{1}{ (\vert z\vert ^{4}+16\sum_{j=1}^{m-1}t_{j}^{2}+16(t_{m}-\varepsilon)^{2} )^{\frac {Q-2}{4}}} -\frac{1}{ (\vert z\vert ^{4}+16\sum_{j=1}^{m-1}t_{j}^{2}+16(t_{m}+\varepsilon)^{2} )^{\frac {Q-2}{4}}}. \end{aligned}$$

2 Result

We give the main results of this paper in this section.

Theorem 2.1

For $u\in C_{0}^{\infty}(\mathbf{H}^{+})$, we have

$$\begin{aligned} \int_{\mathbf{H}^{+}}\vert \nabla_{\mathbf{H}}u\vert ^{2} \,dz\,dt \geq& \frac{1}{16} \int_{\mathbf{H}^{+}}\frac {\vert z\vert ^{2}}{t_{m}^{2}}\vert u\vert ^{2}\,dz \,dt \\ &{}+\frac{(Q-2)(Q+2)}{4} \int_{\mathbf {H}^{+}}d(z,t)^{-4}\vert z\vert ^{2} \vert u\vert ^{2}\,dz\,dt \\ &{}-(Q+2) \int_{\mathbf{H}^{+}}d(z,t)^{-4}\sum_{k=1}^{m-1} \bigl\langle U^{(k)}z, U^{(m)}z \bigr\rangle \frac{t_{k}\vert u\vert ^{2}}{t_{m}} \,dz\,dt. \end{aligned}$$

The theorems below show us the $L^{p}$ Hardy type inequalities in the half-space on the H-type group.

Theorem 2.2

Let $u\in C_{0}^{\infty}(\mathbf{H}^{+})$ and $1< p<\infty$, then

$$\begin{aligned} \int_{\mathbf{H}^{+}}\vert \nabla_{\mathbf{H}}u\vert ^{p} \,dz\,dt \geq& \biggl(\frac {p-1}{p}\biggr)^{p}\frac{p}{4} \int_{\mathbf{H}^{+}}\frac {\vert u\vert ^{p}\vert z\vert ^{2}}{t_{m}^{p}}\,dz\,dt \\ &{}-\biggl(\frac{p-1}{p}\biggr)^{p}\frac{p-1}{2^{\frac{p}{p-1}}} \int_{\mathbf {H}^{+}}\frac{1}{t_{m}^{p}}\vert z\vert ^{\frac{p}{p-1}} \vert u\vert ^{p}\,dz\,dt. \end{aligned}$$

(3)

Theorem 2.3

Let $u\in C_{0}^{\infty}(\mathbf{H}^{+})$ and $1< p<\infty$, then

$$\begin{aligned} \int_{\mathbf{H}^{+}}\vert \nabla_{\mathbf{H}}u\vert ^{p} \,dz\,dt \geq& \biggl(\frac {p-1}{p}\biggr)^{p}\frac{p}{4} \int_{\mathbf{H}^{+}}\frac {\vert u\vert ^{p}\vert z\vert ^{2}}{t_{m}^{p}}\,dz\,dt \\ &{}-\biggl(\frac{p-1}{p}\biggr)^{p}\frac{p-1}{2^{\frac{p}{p-1}}} \int_{\mathbf {H}^{+}}\frac{1}{t_{m}^{p}}\bigl\vert 1-t_{m}^{p-1} \bigr\vert ^{\frac{p}{p-1}}\vert z\vert ^{\frac {p}{p-1}}\vert u\vert ^{p}\,dz\,dt. \end{aligned}$$

(4)

We also study the $L^{p}$ Hardy type inequalities in the H-type group.

Theorem 2.4

Let $u\in C_{0}^{\infty}(\mathbf{H})$ and $1< p<\infty$, then

$$\begin{aligned} \int_{\mathbf{H}}\vert \nabla_{\mathbf{H}}u\vert ^{p} \,dz\,dt \geq& \biggl(\frac{p-1}{p}\biggr)^{p}p(Q-2)^{2} \int_{\mathbf{H}}\frac {\vert u\vert ^{p}\vert z\vert ^{2}}{d^{(-Q+2)p+2Q}}\,dz\,dt \\ &{}-\biggl(\frac{p-1}{p}\biggr)^{p}(p-1) (Q-2)^{\frac{p}{p-1}} \int_{\mathbf{H}}\frac {\vert u\vert ^{p}\vert z\vert ^{\frac{p}{p-1}}}{d^{(-Q+2)p+\frac{p}{p-1}Q}}\,dz\,dt. \end{aligned}$$

(5)

Theorem 2.5

Let $u\in C_{0}^{\infty}(\mathbf{H})$ and $1< p<\infty$. Then

$$\begin{aligned} & \int_{\mathbf{H}}\vert \nabla_{\mathbf{H}}u\vert ^{p} \,dz\,dt \\ &\quad \geq\biggl(\frac{p-1}{p}\biggr)^{p-1}c_{n,m}^{-1} \bigl\vert u(0)\bigr\vert ^{p}+ \biggl(\frac {p-1}{p} \biggr)^{p}p(Q-2)^{2} \int_{\mathbf{H}}\frac {\vert u\vert ^{p}\vert z\vert ^{2}}{d^{(-Q+2)p+2Q}}\,dz\,dt \\ &\qquad {}-\biggl(\frac{p-1}{p}\biggr)^{p}(p-1) (Q-2)^{\frac{p}{p-1}} \int_{\mathbf{H}}\frac {\vert u\vert ^{p}\vert 1-d^{(-Q+2)(p-1)}\vert ^{\frac{p}{p-1}}\vert z\vert ^{\frac {p}{p-1}}}{d^{(-Q+2)p+\frac{p}{p-1}Q}}\,dz\,dt. \end{aligned}$$

(6)

3 Hardy type inequality

This section is to show the Hardy type inequality in $\mathbf{H}^{+}$.

Proof of Theorem 2.1

Let $v(z,t)=G(z,t,\varepsilon)^{-\frac{1}{2}}u(z,t)$. Write $t^{\varepsilon }=(0,\ldots,0,\varepsilon)$. We know that $G(0,t^{\varepsilon },\varepsilon)=\infty$, so we have $v(0,t^{\varepsilon})=0$ and $u(z,t)=G(z,t,\varepsilon)^{\frac{1}{2}}v(z,t)$. Then we obtain

$$\begin{aligned} \nabla_{\mathbf{H}}u=\biggl(\frac{1}{2}\frac{\nabla_{\mathbf{H}}G}{G}+ \frac {\nabla_{\mathbf{H}}v}{v}\biggr)u \end{aligned}$$

and

$$\begin{aligned} & \int_{\mathbf{H}^{+}}\vert \nabla_{\mathbf{H}}u\vert ^{2} \,dz\,dt \\ &\quad =\frac{1}{4} \int_{\mathbf{H}^{+}}\frac{\vert \nabla_{\mathbf {H}}G\vert ^{2}}{G^{2}}\vert u\vert ^{2}\,dz \,dt+ \int_{\mathbf{H}^{+}}\frac{\langle\nabla _{\mathbf{H}}G, \nabla_{\mathbf{H}}v\rangle}{Gv} \vert u\vert ^{2}\,dz \,dt+ \int _{\mathbf{H}^{+}}\frac{\vert \nabla_{\mathbf {H}}v\vert ^{2}}{v^{2}}\vert u\vert ^{2}\,dz \,dt \\ &\quad =\frac{1}{4} \int_{\mathbf{H}^{+}}\frac{\vert \nabla_{\mathbf {H}}G\vert ^{2}}{G^{2}}\vert u\vert ^{2}\,dz \,dt+ \int_{\mathbf{H}^{+}}v\langle\nabla _{\mathbf{H}}G, \nabla_{\mathbf{H}}v \rangle \,dz\,dt+ \int_{\mathbf {H}^{+}}\vert \nabla_{\mathbf{H}}v\vert ^{2}G \,dz\,dt \\ &\quad =\frac{1}{4} \int_{\mathbf{H}^{+}}\frac{\vert \nabla_{\mathbf {H}}G\vert ^{2}}{G^{2}}\vert u\vert ^{2}\,dz \,dt+\frac{1}{2} \int_{\mathbf{H}^{+}}\bigl\langle \nabla_{\mathbf{H}}G, \nabla_{\mathbf{H}}v^{2}\bigr\rangle \,dz\,dt+ \int_{\mathbf {H}^{+}}\vert \nabla_{\mathbf{H}}v\vert ^{2}G \,dz\,dt \\ &\quad =\frac{1}{4} \int_{\mathbf{H}^{+}}\frac{\vert \nabla_{\mathbf {H}}G\vert ^{2}}{G^{2}}\vert u\vert ^{2}\,dz \,dt+\frac {1}{2}c_{n,m}^{-1}v^{2} \bigl(0,t^{\varepsilon}\bigr)+ \int_{\mathbf{H}^{+}}\vert \nabla _{\mathbf{H}}v\vert ^{2}G \,dz\,dt \\ &\quad =\frac{1}{4} \int_{\mathbf{H}^{+}}\frac{\vert \nabla_{\mathbf {H}}G\vert ^{2}}{G^{2}}\vert u\vert ^{2}\,dz \,dt+ \int_{\mathbf{H}^{+}}\vert \nabla_{\mathbf {H}}v\vert ^{2}G \,dz\,dt \\ &\quad \geq\frac{1}{4} \int_{\mathbf{H}^{+}}\frac{\vert \nabla_{\mathbf {H}}G\vert ^{2}}{G^{2}}\vert u\vert ^{2}\,dz \,dt. \end{aligned}$$

Using L’Hospital’s rule, we also have

$$\begin{aligned} \lim_{\varepsilon\rightarrow0^{+}}\frac{G(z,t,\varepsilon)}{\varepsilon }=16(Q-2)t_{m}d(z,t)^{-Q-2} \end{aligned}$$

and

$$\begin{aligned} \lim_{\varepsilon\rightarrow0^{+}}\biggl\vert \frac{\nabla_{\mathbf {H}}G(z,t,\varepsilon)}{\varepsilon}\biggr\vert ^{2} =&\bigl(16(Q-2)\bigr)^{2} \bigl(t_{m}^{2} \bigl\vert \nabla_{\mathbf{H}}d(z,t)^{-Q-2}\bigr\vert ^{2} \\ &{}+2d(z,t)^{-Q-2}t_{m}\bigl\langle \nabla_{\mathbf{H}}d(z,t)^{-Q-2}, \nabla _{\mathbf{H}}t_{m} \bigr\rangle \\ &{}+\bigl(d(z,t)^{-Q-2}\bigr)^{2}\vert \nabla_{\mathbf{H}}t_{m} \vert ^{2} \bigr). \end{aligned}$$

Because

$$\begin{aligned} \nabla_{\mathbf{H}}t_{m}=\Biggl(\frac{1}{2}\sum _{i=1}^{2n}s_{i}U_{i,1}^{(m)}, \ldots,\frac{1}{2}\sum_{i=1}^{2n}s_{i}U_{i,n}^{(m)}, \frac{1}{2}\sum_{i=1}^{2n}s_{i}U_{i,1+n}^{(m)}, \ldots,\frac{1}{2}\sum_{i=1}^{2n}s_{i}U_{i,2n}^{(m)} \Biggr), \end{aligned}$$

from this we can see that

$$\begin{aligned} \vert \nabla_{\mathbf{H}}t_{m}\vert =& \Biggl(\Biggl( \frac{1}{2}\sum_{i=1}^{2n}s_{i}U_{i,1}^{(m)} \Biggr)^{2}+\cdots +\Biggl(\frac{1}{2}\sum _{i=1}^{2n}s_{i}U_{i,n}^{(m)} \Biggr)^{2} \\ &{}+\Biggl(\frac{1}{2}\sum_{i=1}^{2n}s_{i}U_{i,1+n}^{(m)} \Biggr)^{2}+\cdots+\Biggl(\frac {1}{2}\sum _{i=1}^{2n}s_{i}U_{i,2n}^{(m)} \Biggr)^{2} \Biggr)^{\frac {1}{2}} \\ =&\frac{1}{2}\vert z\vert . \end{aligned}$$

By a direct calculation, we get

$$\begin{aligned} \bigl\vert \nabla_{\mathbf{H}}d(z,t)\bigr\vert ^{2}= \frac{\vert z\vert ^{2}}{d(z,t)^{2}}. \end{aligned}$$

Thus we have

$$\begin{aligned}& \bigl\vert \nabla_{\mathbf {H}}d(z,t)^{-Q-2}\bigr\vert ^{2}=(Q+2)^{2}d(z,t)^{-2Q-8}\vert z\vert ^{2}, \\& \vert \nabla_{\mathbf{H}}t_{m}\vert ^{2}= \frac{1}{4}\vert z\vert ^{2} , \end{aligned}$$

(7)

and

$$\begin{aligned} \bigl\langle \nabla_{\mathbf{H}}d(z,t)^{-Q-2},\nabla_{\mathbf{H}}t_{m} \bigr\rangle =2(-Q-2)d(z,t)^{-Q-6}\Biggl(\sum _{k=1}^{m}\bigl\langle U^{(k)}z, U^{(m)}z \bigr\rangle t_{k}\Biggr). \end{aligned}$$

Consequently, we have

$$\begin{aligned} \lim_{\varepsilon\rightarrow0^{+}}\biggl\vert \frac{\nabla_{\mathbf {H}}G(z,t,\varepsilon)}{\varepsilon }\biggr\vert ^{2} =&\bigl(16(Q-2)\bigr)^{2}\Biggl((Q+2)^{2}d(z,t)^{-2Q-8} \vert z\vert ^{2}t_{m}^{2} \\ &{}+4(-Q-2) d(z,t)^{-2Q-8}\bigl\langle U^{(m)}z, U^{(m)}z \bigr\rangle t_{m}^{2} +\frac{1}{4}d(z,t)^{-2Q-4}\vert z\vert ^{2} \\ &{}+4(-Q-2)d(z,t)^{-2Q-8}\sum_{k=1}^{m-1} \bigl\langle U^{(k)}z, U^{(m)}z \bigr\rangle t_{m}t_{k}\Biggr). \end{aligned}$$

This finishes the proof of the theorem. □

4 $L^{p}$ Hardy type inequality

In this section, we are going to consider the $L^{p}$ Hardy type inequalities in $\mathbf{H}^{+}$ and H, respectively. Let Ω be a domain in H. We write $\varrho(z,t)=\operatorname {dist}((z,t),\partial\Omega)$. Similar to [1], we have the lemma below.

Lemma 4.1

Let $u\in C_{0}^{\infty}(\Omega)$, $l\in\{1,2,3,\ldots\}$, $1< p<\infty$, $s\in(-\infty, lp-1)$, $F_{j}\in C^{1}(\Omega)$, $j=1,2,\ldots,2n$, $F=(F_{1},F_{2},\ldots,F_{2n})$ and $w\in C^{1}(\Omega)$ be a nonnegative weight function. We write $C(p,l,s)=(\frac {lp-s-1}{p})^{p}$, then we have

$$\begin{aligned} \int_{\Omega}\frac{\vert \nabla_{\mathbf{H}}u\vert ^{p}w}{\varrho ^{(l-1)p-s}}\,dz\,dt \geq& C(p,l,s) \int_{\Omega}\frac{p\vert u\vert ^{p}\vert \nabla _{\mathbf{H}}\varrho \vert ^{2}w}{\varrho^{lp-s}}\,dz\,dt \\ &{}-C(p,l,s) \int_{\Omega}\frac{p\vert u\vert ^{p}\Delta_{\mathbf{H}}\varrho w}{(lp-s-1)\varrho^{lp-s-1}}\,dz\,dt \\ &{}+C(p,l,s) \int_{\Omega}\frac{p\operatorname {div}_{\mathbf{H}}F\vert u\vert ^{p} w}{lp-s-1}\,dz\,dt \\ &{}-C(p,l,s) \int_{\Omega}\frac{p-1}{\varrho^{lp-s}}\bigl\vert \nabla_{\mathbf {H}} \varrho-\varrho^{lp-s-1}F\bigr\vert ^{\frac{p}{p-1}}\vert u\vert ^{p} w\,dz\,dt \\ &{}+\biggl(\frac{lp-s-1}{p}\biggr)^{p-1} \int_{\Omega}\nabla_{\mathbf{H}}w\biggl(F-\frac {\nabla_{\mathbf{H}}\varrho}{\varrho^{lp-s-1}} \biggr)\vert u\vert ^{p}\,dz\,dt. \end{aligned}$$

(8)

Proof

Applying Hölder’s inequality, we can deduce that

$$\begin{aligned} &p^{p} \int_{\Omega}\frac{\vert \nabla_{\mathbf{H}}u\vert ^{p}w}{\varrho ^{(l-1)p-s}}\,dz\,dt \biggl( \int_{\Omega}\biggl\vert \frac{\nabla_{\mathbf{H}}\varrho }{\varrho^{l(p-1)+\frac{s}{p}-s}}-\varrho^{l-1-\frac{s}{p}}F \biggr\vert ^{\frac {p}{p-1}}\vert u\vert ^{p} w\,dz\,dt \biggr)^{p-1} \\ &\quad \geq p^{p}\biggl\vert \int_{\Omega}\biggl(\frac{\nabla_{\mathbf{H}}\varrho w}{\varrho ^{lp-s-1}}-Fw\biggr) \bigl(\operatorname {sign}(u) \vert u\vert ^{p-1}\bigr)\nabla_{\mathbf {H}}u\,dz\,dt \biggr\vert ^{p}. \end{aligned}$$

On the other hand, by partial integration we get

$$\begin{aligned} &p^{p}\biggl\vert \int_{\Omega}\biggl(\frac{\nabla_{\mathbf{H}}\varrho w}{\varrho ^{lp-s-1}}-Fw\biggr) \bigl(\operatorname {sign}(u) \vert u\vert ^{p-1}\bigr)\nabla_{\mathbf {H}}u\,dz\,dt \biggr\vert ^{p} \\ &\quad =\biggl\vert \int_{\Omega} \biggl(\biggl(\frac{(lp-s-1)\vert \nabla_{\mathbf{H}}\varrho \vert ^{2}}{\varrho^{lp-s}}-\frac{\Delta_{\mathbf{H}}\varrho}{\varrho ^{lp-s-1}}+ \operatorname {div}_{\mathbf{H}}F\biggr)w\\ &\qquad {}+\nabla_{\mathbf{H}}w\biggl(F-\frac {\nabla_{\mathbf{H}}\varrho}{\varrho^{lp-s-1}} \biggr) \biggr)\vert u\vert ^{p}\,dz\,dt\biggr\vert ^{p}. \end{aligned}$$

Thus we obtain

$$\begin{aligned} &p^{p} \int_{\Omega}\frac{\vert \nabla_{\mathbf{H}}u\vert ^{p}w}{\varrho ^{(l-1)p-s}}\,dz\,dt \\ &\quad \geq\biggl\vert \int_{\Omega} \biggl(\biggl(\frac{(lp-s-1)\vert \nabla_{\mathbf{H}}\varrho \vert ^{2}}{\varrho^{lp-s}}-\frac{\Delta_{\mathbf{H}}\varrho}{\varrho ^{lp-s-1}}+ \operatorname {div}_{\mathbf{H}}F\biggr)w\\ &\qquad {}+\nabla_{\mathbf{H}}w\biggl(F-\frac {\nabla_{\mathbf{H}}\varrho}{\varrho^{lp-s-1}} \biggr) \biggr)\vert u\vert ^{p}\,dz\,dt\biggr\vert ^{p} \\ &\qquad {}\times\biggl( \int_{\Omega}\biggl\vert \frac{\nabla_{\mathbf{H}}\varrho}{\varrho ^{l(p-1)+\frac{s}{p}-s}}-\varrho^{l-1-\frac{s}{p}}F \biggr\vert ^{\frac {p}{p-1}}\vert u\vert ^{p} w\,dz\,dt \biggr)^{-p+1}. \end{aligned}$$

It is clear that $\frac{\vert a\vert ^{p}}{b^{p-1}}\geq pa-(p-1)b$ for $b>0$. Then we have equation (8). □

For $F=0$, we have

$$\begin{aligned} \int_{\Omega}\frac{\vert \nabla_{\mathbf{H}}u\vert ^{p}w}{\varrho ^{(l-1)p-s}}\,dz\,dt \geq& C(p,l,s) \int_{\Omega}\frac{p\vert u\vert ^{p}\vert \nabla _{\mathbf{H}}\varrho \vert ^{2}w}{\varrho^{lp-s}}\,dz\,dt \\ &{}-C(p,l,s) \int_{\Omega}\frac{p\vert u\vert ^{p}\Delta_{\mathbf{H}}\varrho w}{(lp-s-1)\varrho^{lp-s-1}}\,dz\,dt \\ &{}-C(p,l,s) \int_{\Omega}\frac{p-1}{\varrho^{lp-s}}\vert \nabla_{\mathbf {H}}\varrho \vert ^{\frac{p}{p-1}}\vert u\vert ^{p} w\,dz\,dt \\ &{}-\biggl(\frac{lp-s-1}{p}\biggr)^{p-1} \int_{\Omega}\frac{\nabla_{\mathbf {H}}w\nabla_{\mathbf{H}}\varrho}{\varrho^{lp-s-1}}\vert u\vert ^{p}\,dz \,dt. \end{aligned}$$

Now, let us discuss the $L^{p}$ Hardy type inequalities in $\mathbf {H}^{+}$. Let $l=1$, $s=0$, and $w=1$, we have by equation (8)

$$\begin{aligned} \int_{\Omega} \vert \nabla_{\mathbf{H}}u\vert ^{p} \,dz\,dt \geq& \biggl(\frac {p-1}{p}\biggr)^{p} \int_{\Omega}\frac{p\vert u\vert ^{p}\vert \nabla_{\mathbf{H}}\varrho \vert ^{2}}{\varrho^{p}}\,dz\,dt \\ &{}-\biggl(\frac{p-1}{p}\biggr)^{p} \int_{\Omega}\frac{p\vert u\vert ^{p}\Delta_{\mathbf {H}}\varrho}{(p-1)\varrho^{p-1}}\,dz\,dt \\ &{}+\biggl(\frac{p-1}{p}\biggr)^{p} \int_{\Omega}\frac{p\operatorname {div}_{\mathbf {H}}F\vert u\vert ^{p} }{p-1}\,dz\,dt \\ &{}-\biggl(\frac{p-1}{p}\biggr)^{p} \int_{\Omega}\frac{p-1}{\varrho^{p}}\bigl\vert \nabla _{\mathbf{H}} \varrho-\varrho^{p-1}F\bigr\vert ^{\frac{p}{p-1}}\vert u\vert ^{p}\,dz\,dt. \end{aligned}$$

(9)

For $\Omega=\mathbf{H}^{+}$, we have $\varrho=t_{m}$. So we get

$$\begin{aligned} \int_{\mathbf{H}^{+}}\vert \nabla_{\mathbf{H}}u\vert ^{p} \,dz\,dt \geq& \biggl(\frac {p-1}{p}\biggr)^{p} \int_{\mathbf{H}^{+}}\frac{p\vert u\vert ^{p}\vert \nabla_{\mathbf {H}}t_{m}\vert ^{2}}{t_{m}^{p}}\,dz\,dt \\ &{}-\biggl(\frac{p-1}{p}\biggr)^{p} \int_{\mathbf{H}^{+}}\frac{p\vert u\vert ^{p}\Delta _{\mathbf{H}}t_{m} }{(p-1)t_{m}^{p-1}}\,dz\,dt \\ &{}+\biggl(\frac{p-1}{p}\biggr)^{p} \int_{\mathbf{H}^{+}}\frac{p\operatorname {div}_{\mathbf {H}}F\vert u\vert ^{p} }{p-1}\,dz\,dt \\ &{}-\biggl(\frac{p-1}{p}\biggr)^{p} \int_{\mathbf{H}^{+}}\frac{p-1}{t_{m}^{p}}\bigl\vert \nabla _{\mathbf{H}}t_{m}-t_{m}^{p-1}F\bigr\vert ^{\frac{p}{p-1}}\vert u\vert ^{p}\,dz\,dt. \end{aligned}$$

(10)

Proof of Theorem 2.2

We know that

$$\begin{aligned} \vert \nabla_{\mathbf{H}}t_{m}\vert =\frac{1}{2}\vert z \vert \end{aligned}$$

and

$$\begin{aligned} \Delta_{\mathbf{H}}t_{m}=0. \end{aligned}$$

Set $F=0$, using equation (10), then we obtain equation (3). □

Proof of Theorem 2.3

Set $F=\nabla_{\mathbf{H}}t_{m}$. Since $U^{(m)}$ is a $2n\times2n$ skew-symmetric orthogonal matrix, we have

$$\begin{aligned} \operatorname {div}_{\mathbf{H}}F =&\sum_{j=1}^{n}X_{j} \frac{1}{2}\sum_{i=1}^{2n}s_{i}U_{i,j}^{(m)}+ \sum_{j=1}^{n}Y_{j} \frac{1}{2}\sum_{i=1}^{2n}s_{i}U_{i,j+n}^{(m)} \\ =&\sum_{j=1}^{n}\frac{1}{2}U_{j,j}^{(m)}+ \sum_{j=1}^{n}\frac {1}{2}U_{j+n,j+n}^{(m)} \\ =&0. \end{aligned}$$

Using equation (10), we have equation (4). □

Now we are going to deal with the $L^{p}$ Hardy type inequalities in H.

Lemma 4.2

Let $u\in C_{0}^{\infty}(\mathbf{H})$, $l\in\{1,2,3,\ldots\}$, $1< p<\infty$, $s\in(-\infty, lp-1)$, $F_{j}\in C^{1}(\mathbf{H})$, $j=1,2,\ldots,2n$, $F=(F_{1},F_{2},\ldots,F_{2n})$ and $w\in C^{1}(\mathbf{H})$ be a nonnegative weight function. Then we have

$$\begin{aligned} & \int_{\mathbf{H}}\frac{\vert \nabla_{\mathbf {H}}u\vert ^{p}w}{(d^{-Q+2})^{(l-1)p-s}}\,dz\,dt \\ &\quad \geq C(p,l,s) \int_{\mathbf{H}}\frac{p\vert u\vert ^{p}\vert \nabla_{\mathbf {H}}d^{-Q+2}\vert ^{2}w}{(d^{-Q+2})^{lp-s}}\,dz\,dt +C(p,l,s) \int_{\mathbf{H}}\frac{p\operatorname {div}_{\mathbf{H}}F\vert u\vert ^{p} w}{lp-s-1}\,dz\,dt \\ & \qquad {}-C(p,l,s) \int_{\mathbf{H}}\frac{p-1}{(d^{-Q+2})^{lp-s}}\bigl\vert \nabla _{\mathbf{H}}d^{-Q+2}-\bigl(d^{-Q+2}\bigr)^{lp-s-1}F\bigr\vert ^{\frac{p}{p-1}}\vert u\vert ^{p} w\,dz\,dt \\ &\qquad {}+\biggl(\frac{lp-s-1}{p}\biggr)^{p-1} \int_{\mathbf{H}}\nabla_{\mathbf {H}}w\biggl(F-\frac{\nabla_{\mathbf{H}}d^{-Q+2}}{(d^{-Q+2})^{lp-s-1}} \biggr)\vert u\vert ^{p}\,dz\,dt, \end{aligned}$$

(11)

where $C(p,l,s)=(\frac{lp-s-1}{p})^{p}$.

Proof

Similar to Lemma 4.1, we have

$$\begin{aligned} & \int_{\mathbf{H}}\frac{\vert \nabla_{\mathbf {H}}u\vert ^{p}w}{(d^{-Q+2})^{(l-1)p-s}}\,dz\,dt \\ &\quad \geq C(p,l,s) \int_{\mathbf{H}}\frac{p\vert u\vert ^{p}\vert \nabla_{\mathbf {H}}d^{-Q+2}\vert ^{2}w}{(d^{-Q+2})^{lp-s}}\,dz\,dt \\ & \qquad {}-C(p,l,s) \int_{\mathbf{H}}\frac{p\vert u\vert ^{p}\Delta_{\mathbf{H}}d^{-Q+2} w}{(lp-s-1)(d^{-Q+2})^{lp-s-1}}\,dz\,dt \\ & \qquad {}+C(p,l,s) \int_{\mathbf{H}}\frac{p\operatorname {div}_{\mathbf{H}}F\vert u\vert ^{p} w}{lp-s-1}\,dz\,dt \\ & \qquad {}-C(p,l,s) \int_{\mathbf{H}}\frac{p-1}{(d^{-Q+2})^{lp-s}}\bigl\vert \nabla _{\mathbf{H}}d^{-Q+2}-\bigl(d^{-Q+2}\bigr)^{lp-s-1}F\bigr\vert ^{\frac{p}{p-1}}\vert u\vert ^{p} w\,dz\,dt \\ & \qquad {}+\biggl(\frac{lp-s-1}{p}\biggr)^{p-1} \int_{\mathbf{H}}\nabla_{\mathbf {H}}w\biggl(F-\frac{\nabla_{\mathbf{H}}d^{-Q+2}}{(d^{-Q+2})^{lp-s-1}} \biggr)\vert u\vert ^{p}\,dz\,dt. \end{aligned}$$

(12)

We know that $c_{n,m}d(z,t)^{-Q+2}$ is a fundamental solution for $\Delta_{\mathbf{H}}$. So we have

$$\begin{aligned} \int_{\mathbf{H}}\frac{\vert u\vert ^{p}\Delta_{\mathbf{H}}d^{-Q+2} w}{(d^{-Q+2})^{lp-s-1}}\,dz\,dt=c_{n,m}^{-1} \bigl\vert u(0)\bigr\vert ^{p}w(0)d(0)^{(Q-2)(lp-s-1)}=0. \end{aligned}$$

 □

For $l=1$, $s=0$, and $w=1$, we have

$$\begin{aligned} & \int_{\mathbf{H}}\vert \nabla_{\mathbf{H}}u\vert ^{p} \,dz\,dt \\ &\quad \geq \biggl(\frac{p-1}{p}\biggr)^{p} \int_{\mathbf{H}}\frac{p\vert u\vert ^{p}\vert \nabla _{\mathbf{H}}d^{-Q+2}\vert ^{2}}{(d^{-Q+2})^{p}}\,dz\,dt \\ & \qquad {}+\biggl(\frac{p-1}{p}\biggr)^{p} \int_{\mathbf{H}}\frac{p\operatorname {div}_{\mathbf {H}}F\vert u\vert ^{p} }{p-1}\,dz\,dt \\ & \qquad {}-\biggl(\frac{p-1}{p}\biggr)^{p} \int_{\mathbf{H}}\frac {p-1}{(d^{-Q+2})^{p}}\bigl\vert \nabla_{\mathbf {H}}d^{-Q+2}- \bigl(d^{-Q+2}\bigr)^{p-1}F\bigr\vert ^{\frac{p}{p-1}}\vert u \vert ^{p}\,dz\,dt. \end{aligned}$$

(13)

Set $F=0$, then we get

$$\begin{aligned} \int_{\mathbf{H}}\vert \nabla_{\mathbf{H}}u\vert ^{p} \,dz\,dt \geq& \biggl(\frac{p-1}{p}\biggr)^{p} \int_{\mathbf{H}}\frac{p\vert u\vert ^{p}\vert \nabla _{\mathbf{H}}d^{-Q+2}\vert ^{2}}{(d^{-Q+2})^{p}}\,dz\,dt \\ &{}-\biggl(\frac{p-1}{p}\biggr)^{p} \int_{\mathbf{H}}\frac {p-1}{(d^{-Q+2})^{p}}\bigl\vert \nabla_{\mathbf{H}}d^{-Q+2} \bigr\vert ^{\frac {p}{p-1}}\vert u\vert ^{p}\,dz\,dt. \end{aligned}$$

(14)

Proof of Theorem 2.4

It is obvious that

$$\begin{aligned} \vert \nabla_{\mathbf{H}}d\vert ^{2}=\frac{\vert z\vert ^{2}}{d^{2}}. \end{aligned}$$

So we have

$$\begin{aligned} \bigl\vert \nabla_{\mathbf{H}}d^{-Q+2}\bigr\vert ^{2}=(Q-2)^{2}d^{2(-Q+1)}\frac {\vert z\vert ^{2}}{d^{2}}. \end{aligned}$$

(15)

From this together with (14), we get equation (5). □

Proof of Theorem 2.5

Let $F=\nabla_{\mathbf{H}}d^{-Q+2}$. Then we have $\operatorname {div}_{\mathbf{H}}F=\operatorname {div}_{\mathbf{H}}\nabla _{\mathbf{H}}d^{-Q+2}=\Delta_{\mathbf{H}}d^{-Q+2}$. From equations (13) and (15), it follows that

$$\begin{aligned} & \int_{\mathbf{H}}\vert \nabla_{\mathbf{H}}u\vert ^{p} \,dz\,dt \\ &\quad \geq \biggl(\frac{p-1}{p}\biggr)^{p}p(Q-2)^{2} \int_{\mathbf{H}}\frac {\vert u\vert ^{p}\vert z\vert ^{2}}{d^{(-Q+2)p+2Q}}\,dz\,dt \\ & \qquad {}+\biggl(\frac{p-1}{p}\biggr)^{p} \int_{\mathbf{H}}\frac{p\Delta_{\mathbf {H}}d^{-Q+2}\vert u\vert ^{p} }{p-1}\,dz\,dt \\ &\qquad {}-\biggl(\frac{p-1}{p}\biggr)^{p}(p-1) (Q-2)^{\frac{p}{p-1}} \int_{\mathbf{H}}\frac {\vert u\vert ^{p}\vert 1-d^{(-Q+2)(p-1)}\vert ^{\frac{p}{p-1}}\vert z\vert ^{\frac {p}{p-1}}}{d^{(-Q+2)p+\frac{p}{p-1}Q}}\,dz\,dt, \end{aligned}$$

which implies that

$$\begin{aligned} & \int_{\mathbf{H}}\vert \nabla_{\mathbf{H}}u\vert ^{p} \,dz\,dt \\ &\quad \geq\biggl(\frac{p-1}{p}\biggr)^{p-1}c_{n,m}^{-1} \bigl\vert u(0)\bigr\vert ^{p}+ \biggl(\frac {p-1}{p} \biggr)^{p}p(Q-2)^{2} \int_{\mathbf{H}}\frac {\vert u\vert ^{p}\vert z\vert ^{2}}{d^{(-Q+2)p+2Q}}\,dz\,dt \\ & \qquad {}-\biggl(\frac{p-1}{p}\biggr)^{p}(p-1) (Q-2)^{\frac{p}{p-1}} \int_{\mathbf{H}}\frac {\vert u\vert ^{p}\vert 1-d^{(-Q+2)(p-1)}\vert ^{\frac{p}{p-1}}\vert z\vert ^{\frac {p}{p-1}}}{d^{(-Q+2)p+\frac{p}{p-1}Q}}\,dz\,dt. \end{aligned}$$

So we have equation (6). □