Hardy-type inequalities on a half-space in the Heisenberg group

Abstract

We prove some Hardy-type inequalities on half-spaces for Kohn’s sub-Laplacian in the Heisenberg group. Furthermore, the constants we obtained are sharp.

MSC:26D10, 35H20.

1 Introduction

The Hardy inequality in ${\mathbb{R}}^N$ reads that for all $u\in C_0^{\mathrm{\infty }}({\mathbb{R}}^N)$ and $N\ge 3$,

$${\int }_{{\mathbb{R}}^N}{|\mathrm{\nabla }u|}^{2}dx\ge \frac{{(N-2)}^2}{4}{\int }_{{\mathbb{R}}^N}\frac{u^2}{{|x|}^2}dx$$

(1.1)

and the constant $\frac{{(N-2)}^2}{4}$ in (1.1) is sharp. Recently, it has been proved by Nazarov ([1], Proposition 4.1, see also [2]) that the following Hardy inequality is valid for $f\in C_0^{\mathrm{\infty }}({\mathbb{R}}_{+}^N)$:

$${\int }_{{\mathbb{R}}_{+}^N}{|\mathrm{\nabla }u(x)|}^{2}dx\ge \frac{N^2}{4}{\int }_{{\mathbb{R}}_{+}^N}\frac{u{(x)}^2}{{|x|}^2}dx,$$

(1.2)

where ${\mathbb{R}}_{+}^N=\{(x_1,\dots ,x_n)|x_1>0\}$, and the constant $\frac{N^2}{4}$ is sharp. This shows that the Hardy constant jumps from $\frac{{(N-2)}^2}{4}$ to $\frac{N^2}{4}$ when the singularity of the potential reaches the boundary. For more information about this inequality and its applications, we refer to [3–10] and the references therein.

The aim of this note is to prove an analogous Hardy-type inequality on a half-space for Kohn’s sub-Laplacian in Heisenberg groups ${\mathbb{H}}^n$. It has been proved by D’Ambrosio ([11], Theorem 3.3) that for $u\in C_0^{\mathrm{\infty }}({\mathbb{H}}^n)$, the following holds:

$${\int }_{{\mathbb{H}}^n}{|{\mathrm{\nabla }}_{H}u|}^{2}dxdt\ge {(n-1)}^2{\int }_{{\mathbb{H}}^n}\frac{u^2}{{|x|}^2}dxdt,$$

(1.3)

where ${\mathrm{\nabla }}_H$ is the horizontal gradient associated with Kohn’s sub-Laplacian on ${\mathbb{H}}^n$ (for details, see Section 2). Furthermore, the constant ${(n-1)}^2$ in (1.3) is sharp (see [12], Theorem 3.13). In this note we shall show that when the singularity is on the boundary, the Hardy constant also jumps. In fact, we have the following.

Theorem 1.1 For all $u\in C_0^{\mathrm{\infty }}({\mathbb{H}}_{+}^n)$, the following holds:

$${\int }_{{\mathbb{H}}_{+}^n}{|{\mathrm{\nabla }}_{H}u|}^{2}dxdt\ge n^2{\int }_{{\mathbb{H}}_{+}^n}\frac{u^2}{{|x|}^2}dxdt,$$

(1.4)

where ${\mathbb{H}}_{+}^n=\{(x,t)\in {\mathbb{H}}^n:x_1>0\}$, and the constant $n^2$ in (1.4) is sharp.

In order to prove Theorem 1.1, we use a new technique which is different from that in [1, 2]. In fact, it seems that the method used in [1, 2] cannot be applied to Kohn’s sub-Laplacian.

With the same technique, we obtain the following sharp Hardy inequality on ${\mathbb{H}}_{k_{+}}^n=\{(x,t)\in {\mathbb{H}}^n:x_1>0,\dots ,x_k>0\}$.

Theorem 1.2 Let $1\le k\le 2n$. For all $u\in C_0^{\mathrm{\infty }}({\mathbb{H}}_{k_{+}}^n)$, the following holds:

$${\int }_{{\mathbb{H}}_{k_{+}}^n}{|{\mathrm{\nabla }}_{H}u|}^{2}dxdt\ge {(n+k-1)}^2{\int }_{{\mathbb{H}}_{k_{+}}^n}\frac{u^2}{{|x|}^2}dxdt.$$

(1.5)

Furthermore, the constant ${(n+k-1)}^2$ in (1.5) is sharp.

2 Proofs

Let ${\mathbb{H}}^n=({\mathbb{R}}^{2n}\times \mathbb{R},\circ )$ be the $(2n+1)$-dimensional Heisenberg group whose group structure is given by

$$(x,t)\circ (x^{\prime },t^{\prime })=(x+x^{\prime },t+t^{\prime }+2\sum _{j=1}^n(x_{2j}^{\prime }{x}_{2j-1}-x_{2j-1}^{\prime }{x}_{2j})).$$

The vector fields

$$X_{2j-1}=\frac{\partial }{\partial x_{2j-1}}+2x_{2j}\frac{\partial }{\partial t},X_{2j}=\frac{\partial }{\partial x_{2j}}-2x_{2j-1}\frac{\partial }{\partial t}$$

($j=1,\dots ,n$) are left invariant and generate the Lie algebra of ${\mathbb{H}}^n$. Kohn’s sub-Laplace on ${\mathbb{H}}^n$ is

$${\mathrm{\Delta }}_H=\sum _{j=1}^{2n}{X}_j^2=\sum _{j=1}^{2n}\frac{{\partial }^2}{\partial x_j^2}+4{|x|}^2\frac{{\partial }^2}{\partial t^2}+4\sum _{j=1}^n(x_{2j}\frac{\partial }{\partial x_{2j-1}}-x_{2j-1}\frac{\partial }{\partial x_{2j}})\frac{\partial }{\partial t}$$

and the horizontal gradient is the $(2n)$-dimensional vector given by

$${\mathrm{\nabla }}_H=(X_1,\dots ,X_{2n})={\mathrm{\nabla }}_x+2\mathrm{\Lambda }x\frac{\partial }{\partial t},$$

where ${\mathrm{\nabla }}_x=(\frac{\partial }{\partial x_1},\dots ,\frac{\partial }{\partial x_{2n}})$, Λ is a skew symmetric and orthogonal matrix given by

$$\mathrm{\Lambda }=diag(J_1,\dots ,J_n),J_1=\cdots =J_n=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right).$$

By the definition of ${\mathrm{\nabla }}_H$, we have, for $\alpha \in \mathbb{R}$ and $|x|\ne 0$,

$$\begin{array}{rcl}{\mathrm{\Delta }}_H\left(x_1{|x|}^{\alpha }\right)& =& \sum _{j=1}^{2n}\frac{{\partial }^2}{\partial x_j^2}\left(x_1{|x|}^{\alpha }\right)=\sum _{j=2}^{2n}\frac{{\partial }^2}{\partial x_j^2}\left(x_1{|x|}^{\alpha }\right)+\frac{{\partial }^2}{\partial x_1^2}\left(x_1{|x|}^{\alpha }\right)\\ =& x_1\sum _{j=2}^{2n}\frac{{\partial }^2}{\partial x_j^2}{|x|}^{\alpha }+x_1\frac{{\partial }^2}{\partial x_1^2}{|x|}^{\alpha }+2\frac{\partial {|x|}^{\alpha }}{\partial x_1}\\ =& x_1\sum _{j=1}^{2n}\frac{{\partial }^2}{\partial x_j^2}{|x|}^{\alpha }+2\frac{\partial {|x|}^{\alpha }}{\partial x_1}\\ =& \alpha (2n+\alpha )x_1{|x|}^{\alpha -2}.\end{array}$$

(2.1)

Similarly,

$$\begin{array}{rcl}{\mathrm{\Delta }}_H\left({|x|}^{\alpha }\prod _{i=1}^k{x}_i\right)& =& \sum _{j=1}^{2n}\frac{{\partial }^2}{\partial x_j^2}\left({|x|}^{\alpha }\prod _{i=1}^k{x}_i\right)\\ =& \sum _{j=k+1}^{2n}\frac{{\partial }^2}{\partial x_j^2}\left({|x|}^{\alpha }\prod _{i=1}^k{x}_i\right)+\sum _{l=1}^k\frac{{\partial }^2}{\partial x_l^2}\left({|x|}^{\alpha }\prod _{i=1}^k{x}_i\right)\\ =& \prod _{i=1}^k{x}_i\sum _{j=1}^{2n}\frac{{\partial }^2}{\partial x_j^2}{|x|}^{\alpha }+2k{|x|}^{\alpha -2}\prod _{i=1}^k{x}_i\\ =& \alpha (2n+2k+\alpha -2){|x|}^{\alpha -2}\prod _{i=1}^k{x}_i.\end{array}$$

(2.2)

Proof of Theorem 1.1 Using the substitution $u=x_1{|x|}^{-n}f$, we get

$$\begin{array}{rcl}{\int }_{{\mathbb{H}}_{+}^n}{|{\mathrm{\nabla }}_{H}u|}^2& =& {\int }_{{\mathbb{H}}_{+}^n}[{\left|{\mathrm{\nabla }}_H\left(x_1{|x|}^{-n}\right)\right|}^2{f}^2+{|{\mathrm{\nabla }}_{H}f|}^2\frac{x_1^2}{{|x|}^{2n}}+\frac{〈{\mathrm{\nabla }}_H(x_1^2{|x|}^{-2n}),{\mathrm{\nabla }}_H{f}^2〉}{2}]\\ \ge & {\int }_{{\mathbb{H}}_{+}^n}({\left|{\mathrm{\nabla }}_H\left(x_1{|x|}^{-n}\right)\right|}^2{f}^2+\frac{〈{\mathrm{\nabla }}_H(x_1^2{|x|}^{-2n}),{\mathrm{\nabla }}_H{f}^2〉}{2})\\ =& {\int }_{{\mathbb{H}}_{+}^n}{f}^2({\left|{\mathrm{\nabla }}_H\left(x_1{|x|}^{-n}\right)\right|}^2-\frac{1}{2}{\mathrm{\Delta }}_H\left(x_1^2{|x|}^{-2n}\right)).\end{array}$$

Using the following identity, for $g\in C^2({\mathbb{H}}^n)$,

$$\frac{1}{2}{\mathrm{\Delta }}_H{g}^2=\frac{1}{2}\sum _{j=1}^{2n}{X}_j^2{g}^2=g\sum _{j=1}^{2n}{X}_j^{2}g+\sum _{j=1}^m{|X_{j}g|}^2=g{\mathrm{\Delta }}_{H}g+{|{\mathrm{\nabla }}_{H}g|}^2,$$

(2.3)

we have, by (2.1),

$$\begin{array}{rcl}{\left|{\mathrm{\nabla }}_H\left(x_1{|x|}^{-n}\right)\right|}^2-\frac{1}{2}{\mathrm{\Delta }}_H\left(x_1^2{|x|}^{-2n}\right)& =& -x_1{|x|}^{-n}{\mathrm{\Delta }}_H\left(x_1{|x|}^{-n}\right)\\ =& -x_1{|x|}^{-n}\cdot (-n)\cdot n{x}_1{|x|}^{-n-2}\\ =& n^2{x}_1^2{|x|}^{-2n-2}.\end{array}$$

Therefore,

$$\begin{array}{rcl}{\int }_{{\mathbb{H}}_{+}^n}{|{\mathrm{\nabla }}_{H}u|}^{2}dxdt& \ge & {\int }_{{\mathbb{H}}_{+}^n}{f}^2({\left|{\mathrm{\nabla }}_H\left(x_1{|x|}^{-n}\right)\right|}^2-\frac{1}{2}{\mathrm{\Delta }}_H\left(x_1^2{|x|}^{-2n}\right))dxdt\\ =& n^2{\int }_{{\mathbb{H}}_{+}^n}{f}^2{x}_1^2{|x|}^{-2n-2}dxdt\\ =& n^2{\int }_{{\mathbb{H}}_{+}^n}\frac{u^2}{{|x|}^2}dxdt.\end{array}$$

(2.4)

Now we show the constant $n^2$ in (1.4) is sharp. Choosing

$$g(x,t)=\varphi (x)\omega (t),$$

where $\varphi \in C_0^{\mathrm{\infty }}({\mathbb{R}}_{+}^{2n})$ and $\omega \in C_0^{\mathrm{\infty }}(\mathbb{R})$, we have

$${\mathrm{\nabla }}_{H}g(x,t)={\mathrm{\nabla }}_{x}g(x,t)+2\mathrm{\Lambda }x\frac{\partial }{\partial t}g(x,t)=\omega (t){\mathrm{\nabla }}_x\varphi (x)+2\varphi (x){\omega }^{\prime }(t)\mathrm{\Lambda }x.$$

Therefore,

$$\begin{array}{rcl}{|{\mathrm{\nabla }}_{H}g(x,t)|}^2& =& 〈\omega (t){\mathrm{\nabla }}_x\varphi (x)+2\varphi (x){\omega }^{\prime }(t)\mathrm{\Lambda }x,\omega (t){\mathrm{\nabla }}_x\varphi (x)+2\varphi (x){\omega }^{\prime }(t)\mathrm{\Lambda }x〉\\ =& {\omega }^2(t){|{\mathrm{\nabla }}_x\varphi |}^2+4{\varphi }^2{|x|}^2{|{\omega }^{\prime }(t)|}^2+4\omega (t){\omega }^{\prime }(t)\varphi 〈\mathrm{\Lambda }x,{\mathrm{\nabla }}_x\varphi 〉.\end{array}$$

(2.5)

To get the last equation, we use the fact ${|\mathrm{\Lambda }x|}^2={|x|}^2$.

Since

$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\omega (t){\omega }^{\prime }(t)dt=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{d{\omega }^2(t)}{dt}dt=0,$$

we have, by (2.5),

$$\begin{array}{rcl}\frac{{\int }_{{\mathbb{H}}_{+}^n}{|{\mathrm{\nabla }}_{H}g(x,t)|}^{2}dxdt}{{\int }_{{\mathbb{H}}_{+}^n}\frac{g^2}{{|x|}^2}dxdt}& =& \frac{{\int }_{{\mathbb{H}}_{+}^n}{\omega }^2(t){|{\mathrm{\nabla }}_x\varphi |}^{2}dxdt+4{\int }_{{\mathbb{H}}_{+}^n}{\varphi }^2{|x|}^2{|{\omega }^{\prime }(t)|}^{2}dxdt}{{\int }_{{\mathbb{R}}_{+}^{2n}}\frac{{\varphi }^2}{{|x|}^2}dx\cdot {\int }_{\mathbb{R}}{\omega }^{2}dt}\\ =& \frac{{\int }_{{\mathbb{R}}_{+}^{2n}}{|{\mathrm{\nabla }}_x\varphi |}^{2}dx}{{\int }_{{\mathbb{R}}_{+}^{2n}}\frac{{\varphi }^2}{{|x|}^2}dx}+4\frac{{\int }_{\mathbb{R}}{|{\omega }^{\prime }(t)|}^{2}dt}{{\int }_{\mathbb{R}}{\omega }^{2}dt}\cdot \frac{{\int }_{{\mathbb{R}}_{+}^{2n}}{\varphi }^2{|x|}^{2}dx}{{\int }_{{\mathbb{R}}_{+}^{2n}}\frac{{\varphi }^2}{{|x|}^2}dx}.\end{array}$$

Notice that

$$\underset{\omega \in C_0^{\mathrm{\infty }}(\mathbb{R})\setminus \{0\}}{inf}\frac{{\int }_{\mathbb{R}}{|{\omega }^{\prime }(t)|}^{2}dt}{{\int }_{\mathbb{R}}{\omega }^{2}dt}=0,$$

(2.6)

we have

$$\underset{u\in C_0^{\mathrm{\infty }}({\mathbb{H}}_{+}^n)\setminus \{0\}}{inf}\frac{{\int }_{{\mathbb{H}}_{+}^n}{|{\mathrm{\nabla }}_{H}u|}^{2}dxdt}{{\int }_{{\mathbb{H}}_{+}^n}\frac{u^2}{{|x|}^2}dxdt}\le \underset{\varphi \in C_0^{\mathrm{\infty }}({\mathbb{R}}^{2n})\setminus \{0\}}{inf}\frac{{\int }_{{\mathbb{R}}_{+}^{2n}}{|{\mathrm{\nabla }}_x\varphi |}^{2}dx}{{\int }_{{\mathbb{R}}_{+}^{2n}}\frac{{\varphi }^2}{{|x|}^2}dx}=n^2.$$

Here we use the sharp Hardy inequality (1.2). This completes the proof of Theorem 1.1. □

Proof of Theorem 1.2 The proof is similar to that of Theorem 1.1. Using the substitution $u=f{|x|}^{-n-k+1}{\prod }_{i=1}^k{x}_i$, we get

$$\begin{array}{rcl}{\int }_{{\mathbb{H}}_{k_{+}}^n}{|{\mathrm{\nabla }}_{H}u|}^2& =& {\int }_{{\mathbb{H}}_{k_{+}}^n}\left[\right|{\mathrm{\nabla }}_H\left({|x|}^{-n-k+1}\prod _{i=1}^k{x}_i\right){|}^2{f}^2+{|{\mathrm{\nabla }}_{H}f|}^2\frac{{\prod }_{i=1}^k{x}_i^2}{{|x|}^{2n}}\\ +\frac{1}{2}〈{\mathrm{\nabla }}_H\left({|x|}^{-2n-2k+2}\prod _{i=1}^k{x}_i^2\right),{\mathrm{\nabla }}_H{f}^2〉]\\ \ge & {\int }_{{\mathbb{H}}_{k_{+}}^n}\left(\right|{\mathrm{\nabla }}_H\left({|x|}^{-n-k+1}\prod _{i=1}^k{x}_i\right){|}^2{f}^2+\frac{1}{2}〈{\mathrm{\nabla }}_H\left({|x|}^{-2n-2k+2}\prod _{i=1}^k{x}_i^2\right),{\mathrm{\nabla }}_H{f}^2〉)\\ =& {\int }_{{\mathbb{H}}_{k_{+}}^n}{f}^2\left(\right|{\mathrm{\nabla }}_H\left({|x|}^{-n-k+1}\prod _{i=1}^k{x}_i\right){|}^2-\frac{1}{2}{\mathrm{\Delta }}_H\left({|x|}^{-2n-2k+2}\prod _{i=1}^k{x}_i^2\right)).\end{array}$$

Using the identities (2.3) and (2.2), we have

$$\begin{array}{c}|{\mathrm{\nabla }}_H\left({|x|}^{-n-k+1}\prod _{i=1}^k{x}_i\right){|}^2-\frac{1}{2}{\mathrm{\Delta }}_H\left({|x|}^{-2n-2k+2}\prod _{i=1}^k{x}_i^2\right)\hfill \\ =-{|x|}^{-n-k+1}\prod _{i=1}^k{x}_i\cdot {\mathrm{\Delta }}_H\left({|x|}^{-n-k+1}\prod _{i=1}^k{x}_i\right)\hfill \\ ={(n+k-1)}^2{|x|}^{-2n-2k}\prod _{i=1}^k{x}_i^2.\hfill \end{array}$$

Therefore,

$$\begin{array}{rcl}{\int }_{{\mathbb{H}}_{k_{+}}^n}{|{\mathrm{\nabla }}_{H}u|}^2& \ge & {\int }_{{\mathbb{H}}_{k_{+}}^n}{f}^2\left(\right|{\mathrm{\nabla }}_H\left({|x|}^{-n-k+1}\prod _{i=1}^k{x}_i\right){|}^2-\frac{1}{2}{\mathrm{\Delta }}_H\left({|x|}^{-2n-2k+2}\prod _{i=1}^k{x}_i^2\right))\\ =& {(n+k-1)}^2{\int }_{{\mathbb{H}}_{k_{+}}^n}{f}^2{|x|}^{-2n-2k}\prod _{i=1}^k{x}_i^2\\ =& {(n+k-1)}^2{\int }_{{\mathbb{H}}_{k_{+}}^n}\frac{u^2}{{|x|}^2}.\end{array}$$

To see the constant ${(n+k-1)}^2$ in (1.5) is sharp, we consider the function

$$h(x,t)=\psi (x)\omega (t),$$

where $\psi \in C_0^{\mathrm{\infty }}({\mathbb{R}}_{k_{+}}^{2n})$ and $\omega \in C_0^{\mathrm{\infty }}(\mathbb{R})$. Here we denote by ${\mathbb{R}}_{k_{+}}^{2n}=\{x\in {\mathbb{R}}^{2n}:x_1>0,\dots ,x_k>0\}$. Then

$$\begin{array}{rcl}{|{\mathrm{\nabla }}_{H}h(x,t)|}^2& =& 〈\omega (t){\mathrm{\nabla }}_x\psi (x)+2\psi (x){\omega }^{\prime }(t)\mathrm{\Lambda }x,\omega (t){\mathrm{\nabla }}_x\psi (x)+2\psi (x){\omega }^{\prime }(t)\mathrm{\Lambda }x〉\\ =& {\omega }^2(t){|{\mathrm{\nabla }}_x\psi |}^2+4{\psi }^2{|x|}^2{|{\omega }^{\prime }(t)|}^2+4\omega (t){\omega }^{\prime }(t)\psi 〈\mathrm{\Lambda }x,{\mathrm{\nabla }}_x\psi 〉\end{array}$$

and

$$\begin{array}{rcl}{\int }_{{\mathbb{H}}_{k_{+}}^n}{|{\mathrm{\nabla }}_{H}h(x,t)|}^{2}dxdt& =& {\int }_{{\mathbb{H}}_{k_{+}}^n}({\omega }^2(t){|{\mathrm{\nabla }}_x\psi |}^2+4{\psi }^2{|x|}^2{|{\omega }^{\prime }(t)|}^2)dxdt\\ +4{\int }_{{\mathbb{R}}_{k_{+}}^{2n}}\psi 〈\mathrm{\Lambda }x,{\mathrm{\nabla }}_x\psi 〉dx\cdot {\int }_{\mathbb{R}}\omega (t){\omega }^{\prime }(t)dt\\ =& {\int }_{{\mathbb{H}}_{k_{+}}^n}({\omega }^2(t){|{\mathrm{\nabla }}_x\psi |}^2+4{\psi }^2{|x|}^2{|{\omega }^{\prime }(t)|}^2)dxdt\\ +4{\int }_{{\mathbb{R}}_{k_{+}}^{2n}}\psi 〈\mathrm{\Lambda }x,{\mathrm{\nabla }}_x\psi 〉dx\cdot \frac{1}{2}{\int }_{\mathbb{R}}d{\omega }^2(t)\\ =& {\int }_{{\mathbb{H}}_{k_{+}}^n}({\omega }^2(t){|{\mathrm{\nabla }}_x\psi |}^2+4{\psi }^2{|x|}^2{|{\omega }^{\prime }(t)|}^2)dxdt.\end{array}$$

Therefore,

$$\begin{array}{rcl}\frac{{\int }_{{\mathbb{H}}_{k_{+}}^n}{|{\mathrm{\nabla }}_{H}h(x,t)|}^{2}dxdt}{{\int }_{{\mathbb{H}}_{k_{+}}^n}\frac{h^2}{{|x|}^2}dxdt}& =& \frac{{\int }_{{\mathbb{H}}_{k_{+}}^n}{\omega }^2(t){|{\mathrm{\nabla }}_x\psi |}^{2}dxdt+4{\int }_{{\mathbb{H}}_{k_{+}}^n}{\psi }^2{|x|}^2{|{\omega }^{\prime }(t)|}^{2}dxdt}{{\int }_{{\mathbb{R}}_{k_{+}}^{2n}}\frac{{\psi }^2}{{|x|}^2}dx\cdot {\int }_{\mathbb{R}}{\omega }^{2}dt}\\ =& \frac{{\int }_{{\mathbb{R}}_{+}^{2n}}{|{\mathrm{\nabla }}_x\psi |}^{2}dx}{{\int }_{{\mathbb{R}}_{k_{+}}^{2n}}\frac{{\psi }^2}{{|x|}^2}dx}+4\frac{{\int }_{\mathbb{R}}{|{\omega }^{\prime }(t)|}^{2}dt}{{\int }_{\mathbb{R}}{\omega }^{2}dt}\cdot \frac{{\int }_{{\mathbb{R}}_{k_{+}}^{2n}}{\psi }^2{|x|}^{2}dx}{{\int }_{{\mathbb{R}}_{k_{+}}^{2n}}\frac{{\psi }^2}{{|x|}^2}dx}.\end{array}$$

Thus, by (2.6),

$$\begin{array}{rl}\underset{u\in C_0^{\mathrm{\infty }}({\mathbb{H}}_{k_{+}}^n)\setminus \{0\}}{inf}\frac{{\int }_{{\mathbb{H}}_{k_{+}}^n}{|{\mathrm{\nabla }}_{H}u|}^{2}dxdt}{{\int }_{{\mathbb{H}}_{k_{+}}^n}\frac{u^2}{{|x|}^2}dxdt}& \le \underset{\psi \in C_0^{\mathrm{\infty }}({\mathbb{R}}^{2n})\setminus \{0\}}{inf}\frac{{\int }_{{\mathbb{R}}_{+}^{2n}}{|{\mathrm{\nabla }}_x\psi |}^{2}dx}{{\int }_{{\mathbb{R}}_{k_{+}}^{2n}}\frac{{\psi }^2}{{|x|}^2}dx}\\ ={(n+k-1)}^2.\end{array}$$

Here we use the sharp Hardy inequality ([9], Theorem 1.1)

$${\int }_{{\mathbb{R}}_{+}^{2n}}{|\mathrm{\nabla }f|}^{2}dx\ge {(n+k-1)}^2{\int }_{{\mathbb{R}}_{+}^{2n}}\frac{f^2}{{|x|}^2}dx.$$

The proof of Theorem 1.2 is therefore completed. □