An improved Hardy type inequality on Heisenberg group

Abstract

Motivated by the work of Ghoussoub and Moradifam, we prove some improved Hardy inequalities on the Heisenberg group ${ℍ}^n$ via Bessel function.

Mathematics Subject Classification (2000):

Primary 26D10

1 Introduction

Hardy inequality in ℝ ^N reads, for all $u\in C_0^{\infty }\left({ℝ}^N\right)$ and N ≥ 3,

$$\underset{{ℝ}^N}{\int }|\nabla u{|}^2\mathsf{\text{d}}x\ge \frac{{\left(N-2\right)}^2}{4}\underset{{ℝ}^N}{\int }\frac{u^2}{|x{|}^2}\mathsf{\text{d}}x$$

(1.1)

and $\frac{{\left(N-2\right)}^2}{4}$ is the best constant in (1.1) and is never achieved. A similar inequality with the same best constant holds in ℝ ^N is replaced by an arbitrary domain Ω ⊂ ℝ ^N and Ω contains the origin. Moreover, in case Ω ⊂ ℝ ^N is a bounded domain, Brezis and Vázquez [1] have improved it by establishing that for $u\in C_0^{\infty }\left(\Omega \right)$,

$$\underset{\Omega }{\int }|\nabla u{|}^2\mathsf{\text{d}}x\ge \frac{{\left(N-2\right)}^2}{4}\underset{\Omega }{\int }\frac{u^2}{|x{|}^2}\mathsf{\text{d}}x+z_0^2{\left(\frac{{\omega }_N}{\left|\Omega \right|}\right)}^{\frac{2}{N}}\underset{\Omega }{\int }{u}^2\mathsf{\text{d}}x,$$

(1.2)

where ω _N and |Ω| denote the volume of the unit ball and Ω, respectively, and z₀ = 2.4048... denotes the first zero of the Bessel function J₀(z). Inequality (1.2) is optimal in case Ω is a ball centered at zero. Triggered by the work of Brezis and Vázquez (1.2), several Hardy inequalities have been established in recent years. In particular, Adimurthi et al.([2]) proved that, for $u\in C_0^{\infty }\left(\Omega \right)$, there exists a constant C _n,k such that

$$\underset{\Omega }{\int }|\nabla u{|}^2\mathsf{\text{d}}x\ge \frac{{\left(N-2\right)}^2}{4}\underset{\Omega }{\int }\frac{u^2}{|x{|}^2}\mathsf{\text{d}}x+C_{n,k}\sum _{j=1}^k\underset{\Omega }{\int }\frac{u^2}{|x{|}^2}{\left(\prod _{i=1}^j{log}^{\left(i\right)}\frac{\rho }{\left|x\right|}\right)}^{-2}\mathsf{\text{d}}x,$$

(1.3)

where

$$\rho =\left(\underset{x\in \Omega }{sup}\left|x\right|\right)\left(e^{e^{{.}^{{.}^{e\left(k-times\right)}}}}\right),$$

log⁽¹⁾(.) = log(.) and log^(k)(.) = log(log^(k-1)(.)) for k ≥ 2. Filippas and Tertikas ([3]) proved that, for $u\in C_0^{\infty }\left(\Omega \right)$, there holds

$$\underset{\Omega }{\int }|\nabla u{|}^2\ge \frac{{\left(N-2\right)}^2}{4}\underset{\Omega }{\int }\frac{u^2}{|x{|}^2}+\frac{1}{4}\sum _{k=1}^{\infty }\underset{\Omega }{\int }\frac{u^2}{|x{|}^2}{X}_1^2\left(\frac{\left|x\right|}{D}\right)\cdot \cdots \cdot X_k^2\left(\frac{\left|x\right|}{D}\right),$$

(1.4)

where D ≥ sup _x∈Ω|x|,

$$X_1(s)={(1-\ln s)}^{-1},X_k(s)=X_1(X_{k-1}(t))$$

for k ≥ 2 and $\frac{1}{4}$ is the best constant in (1.4) and is never achieved. More recently, Ghoussoub and Moradifam ([4]) give a necessary and sufficient condition on a radially symmetric potential V(|x|) on Ω that makes it an admissible candidate for an improved Hardy inequality. It states that the following improved Hardy inequality holds for $u\in C_0^{\infty }\left(B_{\rho }\right)$, where B _ρ = {x ∈ ℝ ⁿ :|x| < ρ},

$$\underset{\Omega }{\int }|\nabla u{|}^2\mathsf{\text{d}}x\ge \frac{{\left(N-2\right)}^2}{4}\underset{\Omega }{\int }\frac{u^2}{|x{|}^2}\mathsf{\text{d}}x+\underset{\Omega }{\int }\frac{u^2}{|x{|}^2}V\left(\left|x\right|\right)\mathsf{\text{d}}x$$

(1.5)

if and only if the ordinary differential equation

$$y^{″}\left(r\right)+\frac{y^{\prime }\left(r\right)}{r}+V\left(r\right)y\left(r\right)=0$$

has a positive solution on (0, ρ]. These include inequalities (1.2)-(1.4).

Motivated by the work of Ghoussoub and Moradifam ([4]), in this note, we shall prove similar improved Hardy inequality on the Heisenberg group ${ℍ}^n$. Recall that the Heisenberg group ${ℍ}^n$ is the Carnot group of step two whose group structure is given by

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

The vector fields

$$\begin{array}{c}{X}_{2j-1}=\frac{\partial }{\partial x_{2j-1}}+2x_{2j}\frac{\partial }{\partial t},\\ X_{2j}=\frac{\partial }{\partial x_{2j}}-2x_{2j-1}\frac{\partial }{\partial t},\end{array}$$

(j = 1,..., n) are left invariant and generate the Lie algebra of ${ℍ}^n$. The horizontal gradient on ${ℍ}^n$ is the (2n) -dimensional vector given by

$${\nabla }_{ℍ}=\left(X_1,\dots ,X_{2n}\right)={\nabla }_x+2\Lambda x\frac{\partial }{\partial t},$$

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

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

For more information about ${ℍ}^n$, we refer to [5–8]. To this end we have:

Theorem 1.1

Let B _R = {x ∈ ℝ²ⁿ: |x| < R} and Ω _H = B _R × ℝ ∈ ${ℍ}^n$. Let V(|x|) be a radially symmetric decreasing nonnegative function on B _R . If the ordinary differential equation

$$y^{″}\left(r\right)+\frac{y^{\prime }\left(r\right)}{r}+V\left(r\right)y\left(r\right)=0$$

has a positive solution on (0, R], then the following improved Hardy inequality holds for$u\in C_0^{\infty }\left({\Omega }_H\right)$

$$\underset{{\Omega }_H}{\int }|{\nabla }_{ℍ}u{|}^2\mathsf{\text{d}}x\mathsf{\text{d}}t\ge {\left(n-1\right)}^2\underset{{\Omega }_H}{\int }\frac{u^2}{|x{|}^2}\mathsf{\text{d}}x\mathsf{\text{d}}t+\underset{{\Omega }_H}{\int }\frac{u^2}{|x{|}^2}V\left(\left|x\right|\right)\mathsf{\text{d}}x\mathsf{\text{d}}t$$

(1.6)

and the constant (n - 1)²in (1.6) is sharp in the sense of

$${\left(n-1\right)}^2=\underset{u\in C_0^{\infty }\left({\Omega }_H\right)\backslash \left\{0\right\}}{inf}\frac{{\int }_{{\Omega }_H}|{\nabla }_{ℍ}u{|}^2\mathsf{\text{d}}x\mathsf{\text{d}}t}{{\int }_{{\Omega }_H}\frac{u^2}{|x{|}^2}\mathsf{\text{d}}x\mathsf{\text{d}}t}.$$

Corollary 1.2

There holds, for$u\in C_0^{\infty }\left({\Omega }_H\right)$,

$$\underset{{\Omega }_H}{\int }|{\nabla }_{ℍ}u{|}^2\ge {\left(n-1\right)}^2\underset{{\Omega }_H}{\int }\frac{u^2}{|x{|}^2}+\frac{1}{4}\sum _{j=1}^k\underset{{\Omega }_H}{\int }\frac{u^2}{|x{|}^2}{\left(\prod _{i=1}^j{log}^{\left(i\right)}\frac{R}{\left|x\right|}\right)}^{-2}$$

(1.7)

and the constant 1/4 is sharp in the sense of

$$\frac{1}{4}=\underset{u\in C_0^{\infty }\left({\Omega }_H\right)\backslash \left\{0\right\}}{inf}\frac{{\int }_{{\Omega }_H}|{\nabla }_{ℍ}u{|}^2-{\left(n-1\right)}^2{\int }_{{\Omega }_H}\frac{u^2}{|x{|}^2}-\frac{1}{4}{\sum }_{j=1}^{k-1}{\int }_{{\Omega }_H}\frac{u^2}{|x{|}^2}{\left({\prod }_{i=1}^j{log}^{\left(i\right)}\frac{R}{\left|x\right|}\right)}^{-2}}{{\int }_{{\Omega }_H}\frac{u^2}{|x{|}^2}{\left({\prod }_{i=1}^k{log}^{\left(i\right)}\frac{R}{\left|x\right|}\right)}^{-2}}.$$

Corollary 1.3

There holds, for $u\in C_0^{\infty }\left({\Omega }_H\right)$ and D ≥ R,

$$\underset{{\Omega }_H}{\int }|{\nabla }_{ℍ}u{|}^2\ge {\left(n-1\right)}^2\underset{{\Omega }_H}{\int }\frac{u^2}{|x{|}^2}+\frac{1}{4}\sum _{k=1}^{\infty }\underset{{\Omega }_H}{\int }\frac{u^2}{|x{|}^2}{X}_1^2\left(\frac{\left|x\right|}{D}\right)\cdot \cdots \cdot X_k^2\left(\frac{\left|x\right|}{D}\right),$$

(1.8)

and the constant 1/4 is sharp in the sense of

$$\frac{1}{4}=\underset{u\in C_0^{\infty }\left({\Omega }_H\right)\backslash \left\{0\right\}}{inf}\frac{{\int }_{{\Omega }_H}|{\nabla }_{ℍ}u{|}^2-{\left(n-1\right)}^2{\int }_{{\Omega }_H}\frac{u^2}{|x{|}^2}-\frac{1}{4}{\sum }_{j=1}^{k-1}{\int }_{{\Omega }_H}\frac{u^2}{|x{|}^2}{X}_1^2\left(\frac{\left|x\right|}{D}\right)\cdots X_j^2\left(\frac{\left|x\right|}{D}\right)}{{\int }_{{\Omega }_H}\frac{u^2}{|x{|}^2}{X}_1^2\left(\frac{\left|x\right|}{D}\right)\cdot \cdots \cdot X_k^2\left(\frac{\left|x\right|}{D}\right)}.$$

2 Proof

To prove the main result, we first need the following preliminary result.

Lemma 2.1

Let B _R = {x ∈ ℝ²ⁿ: |x| < R} and V(|x|) be a radially symmetric decreasing nonnegative function on B _R . If the ordinary differential equation

$$y^{″}\left(r\right)+\frac{y^{\prime }\left(r\right)}{r}+V\left(r\right)y\left(r\right)=0$$

has a positive solution on (0, R], then the following improved Hardy inequality holds for$f\in C_0^{\infty }\left(B_R\right)$,

$$\underset{B_R}{\int }|{\partial }_{r}f{|}^2\mathsf{\text{d}}x\ge {\left(n-1\right)}^2\underset{B_R}{\int }\frac{f^2}{|x{|}^2}\mathsf{\text{d}}x+\underset{B_R}{\int }\frac{f^2}{|x{|}^2}V\left(\left|x\right|\right)\mathsf{\text{d}}x,$$

(2.1)

where r = |x| and${\partial }_r=\frac{⟨x,\nabla ⟩}{\left|x\right|}$is the radial derivation.

Proof

Observe that if f is radial, i.e., f(x) = f(r), then |∇ f| = |∂ _r f|. By inequality (1.5), inequality (2.1) holds.

Now let $f\in C_0^{\infty }\left(B_R\right)$. If we extend f as zero outside B _R , we may consider $f\in C_0^{\infty }\left({ℝ}^{2n}\right)$. Decomposing f into spherical harmonics we get (see e.g., [9])

$$f=\sum _{k=0}^{\infty }{f}_k\left(r\right){\varphi }_k\left(\sigma \right),$$

where ϕ _k (σ) are the orthonormal eigenfunctions of the Laplace-Beltrami operator with responding eigenvalues

$$c_k=k\left(N+k-2\right),k\ge 0.$$

The functions f _k (r) belong to $C_0^{\infty }\left(B_R\right)$, satisfying f _k (r) = O(r^k ) and $f_k^{\prime }\left(r\right)=O\left(r^{k-1}\right)$ as r → 0. So

$$\underset{B_R}{\int }|{\partial }_{r}f{|}^2\mathsf{\text{d}}x=\sum _{k=0}^{\infty }\underset{B_R}{\int }|f_k^{\prime }{|}^2\mathsf{\text{d}}x$$

(2.2)

and

$${\left(n-1\right)}^2\underset{B_R}{\int }\frac{f^2}{|x{|}^2}\mathsf{\text{d}}x+\underset{B_R}{\int }\frac{f^2}{|x{|}^2}V\left(\left|x\right|\right)\mathsf{\text{d}}x=\sum _{k=0}^{\infty }\left({\left(n-1\right)}^2\underset{B_R}{\int }\frac{f_k^2}{|x{|}^2}\mathsf{\text{d}}x+\underset{B_R}{\int }\frac{f_k^2}{|x{|}^2}V\left(\left|x\right|\right)\mathsf{\text{d}}x\right).$$

(2.3)

Note that if f is radial, then inequality (2.1) holds. We have, since $f_k\left(r\right)\in C_0^{\infty }\left(B_R\right)$,

$$\underset{B_R}{\int }|f_k^{\prime }{|}^2\mathsf{\text{d}}x\ge {\left(n-1\right)}^2\underset{B_R}{\int }\frac{f_k^2}{|x{|}^2}\mathsf{\text{d}}x+\underset{B_R}{\int }\frac{f_k^2}{|x{|}^2}V\left(\left|x\right|\right)\mathsf{\text{d}}x.$$

Therefore, by (2.2) and (2.3),

$$\begin{array}{l}{\displaystyle \underset{B_R}{\int }{\left|{\partial }_{r}f\right|}^2\text{d}x={\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \underset{B_R}{\int }}{\left|f_k^{\text{'}}\right|}^2\text{d}x}\\ \ge {\displaystyle \sum _{k=0}^{\infty }}\left({(n-1)}^2{\displaystyle \underset{B_R}{\int }}\frac{f_k^2}{{\left|x\right|}^2}\text{d}x+{\displaystyle \underset{B_R}{\int }}\frac{f_k^2}{{\left|x\right|}^2}V(\left|x\right|)\text{d}x\right)\\ {=(n-1)}^2{\displaystyle \underset{B_R}{\int }}\frac{f^2}{{\left|x\right|}^2}\text{d}x+{\displaystyle \underset{B_R}{\int }}\frac{f^2}{{\left|x\right|}^2}V(\left|x\right|)\text{d}x\mathrm{.}\end{array}$$

This completes the proof of lemma 2.1.

Proof of Theorem 1.1

Recall that the horizontal gradient on ${ℍ}^n$ is the (2n)-dimensional vector given by

$${\nabla }_{ℍ}=\left(X_1,\dots ,X_{2n}\right)={\nabla }_x+2\Lambda x\frac{\partial }{\partial t},$$

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

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

Therefore, for any $\varphi \in C_0^{\infty }\left({ℍ}^n\right)$,

$$\begin{array}{lll}\hfill ⟨x,{\nabla }_{ℍ}\varphi ⟩& =⟨x,{\nabla }_x\varphi ⟩+2⟨x,\Lambda x⟩\frac{\partial \varphi }{\partial t}& \hfill \text{(1)}\\ =⟨x,{\nabla }_x\varphi ⟩.& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$$

(2.4)

Here we use the fact ⟨x, Λx⟩ = 0 since Λ is a skew symmetric matrix.

Since $u\in C_0^{\infty }\left({\Omega }_H\right)$, for every t ∈ ℝ,$u\left(\cdot ,t\right)\in C_0^{\infty }\left(B_R\right)$. By Lemma 2.1,

$$\underset{B_R}{\int }|{\partial }_{r}u{|}^2\mathsf{\text{d}}x\ge {\left(n-1\right)}^2\underset{B_R}{\int }\frac{u^2}{|x{|}^2}\mathsf{\text{d}}x+\underset{B_R}{\int }\frac{u^2}{|x{|}^2}V\left(\left|x\right|\right)\mathsf{\text{d}}x$$

(2.5)

Integrating both sides of the inequality (2.5) with respect to t, we have,

$$\underset{{\Omega }_H}{\int }|{\partial }_{r}u{|}^2\mathsf{\text{d}}x\mathsf{\text{d}}t\ge {\left(n-1\right)}^2\underset{{\Omega }_H}{\int }\frac{u^2}{|x{|}^2}\mathsf{\text{d}}x\mathsf{\text{d}}t+\underset{{\Omega }_H}{\int }\frac{u^2}{|x{|}^2}V\left(\left|x\right|\right)\mathsf{\text{d}}x\mathsf{\text{d}}t$$

(2.6)

By (2.4) and the pointwise Schwartz inequality, we have

$$\left|{\partial }_{r}u\right|=\frac{\left|⟨x,{\nabla }_{x}u⟩\right|}{\left|x\right|}=\frac{\left|⟨x,{\nabla }_{ℍ}u⟩\right|}{\left|x\right|}\le \left|{\nabla }_{ℍ}u\right|.$$

Therefore, we obtain, by (2.6)

$${\left(n-1\right)}^2\underset{{\Omega }_H}{\int }\frac{u^2}{|x{|}^2}\mathsf{\text{d}}x\mathsf{\text{d}}t+\underset{{\Omega }_H}{\int }\frac{u^2}{|x{|}^2}V\left(\left|x\right|\right)\mathsf{\text{d}}x\mathsf{\text{d}}t\le \underset{{\Omega }_H}{\int }|{\nabla }_{ℍ}u{|}^2\mathsf{\text{d}}x\mathsf{\text{d}}t.$$

(2.7)

To see the constant (n - 1)² is sharp, we choose u(x, t) = ϕ(|x|)w(t) with $\varphi \left(\left|x\right|\right)\in C_0^{\infty }\left(B_R\right)$ and $w\left(t\right)\in C_0^{\infty }\left(ℝ\right)$. Since ϕ is radial, we have

$$\begin{array}{l}{\left|{\nabla }_{ℍ}u(x\mathrm{,}t)\right|}^2=〈w(t){\nabla }_x\varphi (\left|x\right|)+2\varphi (\left|x\right|)\Lambda x{w}^{\prime }(t)\mathrm{,}w(t){\nabla }_x\varphi (\left|x\right|)+2\varphi (\left|x\right|)\Lambda x{w}^{\prime }(t)〉\\ ={\left|{\nabla }_x\varphi (\left|x\right|)\right|}^2{w}^2(t)+4{\left|\Lambda x\right|}^2{\varphi }^2{(w^{\prime }(t))}^2+4〈{\nabla }_x\varphi (\left|x\right|)\mathrm{,}\Lambda x〉\varphi (\left|x\right|)w^{\prime }(t)\\ ={\left|{\nabla }_x\varphi (\left|x\right|)\right|}^2{w}^2(t)+4|\Lambda x{|}^2{\varphi }^2{(w^{\prime }(t))}^2+4{\varphi }^{\prime }(\left|x\right|)〈\frac{x}{\left|x\right|}\mathrm{,}\Lambda x〉\varphi (\left|x\right|)w^{\prime }(t)\\ ={\left|{\nabla }_x\varphi (\left|x\right|)\right|}^2{w}^2(t)+4{\left|x\right|}^2{\varphi }^2(\left|x\right|)(w^{\prime }(t){)}^2\mathrm{.}\end{array}$$

Here we use the fact |Λx| = |x| since Λ is a orthogonal matrix. Therefore,

$$\begin{array}{c}\frac{{\displaystyle {\int }_{{\Omega }_H}}|{\nabla }_{ℍ}u{|}^2\text{d}x\text{d}t}{{\displaystyle {\int }_{{\Omega }_H}}\frac{u^2}{|x{|}^2}\text{d}x\text{d}t}=\frac{{\displaystyle {\int }_{{\Omega }_H}}|{\nabla }_x\varphi (|x{|)|}^2{w}^2(t)}{{\displaystyle {\int }_{{\Omega }_H}}\frac{\varphi {(|x|)}^{2}w{(t)}^2}{|x{|}^2}}+4\frac{{\displaystyle {\int }_{{\Omega }_H}}|x{|}^2{\varphi }^2(\left|x\right|)(w^{\prime }(t{))}^2}{{\displaystyle {\int }_{{\Omega }_H}}\frac{\varphi {(|x|)}^{2}w{(t)}^2}{|x{|}^2}}\\ =\frac{{\displaystyle {\int }_{B_R}}|{\nabla }_x\varphi (|x{|)|}^2\text{d}x\cdot {\displaystyle {\int }_{-\infty }^{+\infty }}{w}^2(t)\text{d}t}{{\displaystyle {\int }_{B_R}}\frac{\varphi {(|x|)}^2}{|x{|}^2}\text{d}x\cdot {\displaystyle {\int }_{-\infty }^{+\infty }}{w}^2(t)\text{d}t}+4\frac{{\displaystyle {\int }_{B_R}}|x{|}^2{\varphi }^2(\left|x\right|)\text{d}x\cdot {\displaystyle {\int }_{-\infty }^{+\infty }}{(w^{\prime }(t))}^2}{{\displaystyle {\int }_{B_R}}\frac{\varphi {(|x|)}^2}{|x{|}^2}\text{d}x\cdot {\displaystyle {\int }_{-\infty }^{+\infty }}{w}^2(t)\text{d}t}\\ =\frac{{\displaystyle {\int }_{B_R}}|{\nabla }_x\varphi (|x{|)|}^2\text{d}x}{{\displaystyle {\int }_{B_R}}\frac{\varphi {(|x|)}^2}{|x{|}^2}\text{d}x}+4\frac{{\displaystyle {\int }_{B_R}}|x{|}^2{\varphi }^2(\left|x\right|)\text{d}x}{{\displaystyle {\int }_{B_R}}\frac{\varphi {(|x|)}^2}{|x{|}^2}\text{d}x}\cdot \frac{{\displaystyle {\int }_{-\infty }^{+\infty }}{(w^{\prime }(t))}^2}{{\displaystyle {\int }_{-\infty }^{+\infty }}{w}^2(t)\text{d}t}\end{array}$$

Since

$$\underset{w(t)\in C_0^{\infty }(ℝ)\backslash \left\{0\right\}}{{\displaystyle \inf }}\frac{{\displaystyle {\int }_{ℝ}}|w^{\prime }(t{)|}^2\text{d}t}{{\displaystyle {\int }_{ℝ}}|w(t{)|}^2\text{d}t}=\mathrm{0,}$$

we obtain

$$\underset{u\in C_0^{\infty }({\Omega }_H)\backslash \left\{0\right\}}{{\displaystyle \inf }}\frac{{\displaystyle {\int }_{{\Omega }_H}}|{\nabla }_{ℍ}u{|}^2\text{d}x\text{d}t}{{\displaystyle {\int }_{{\Omega }_H}}\frac{u^2}{|x{|}^2}\text{d}x\text{d}t}\le \underset{\varphi \in C_0^{\infty }(B_R)\backslash \left\{0\right\}}{{\displaystyle \inf }}\frac{{\displaystyle {\int }_{B_R}}|{\nabla }_x\varphi (|x{|)|}^2\text{d}x}{{\displaystyle {\int }_{B_R}}\frac{\varphi {(|x|)}^2}{|x{|}^2}\text{d}x}=(n-{1)}^2\mathrm{.}$$

The proof of Theorem 1.1 is completed.

Proof of Corollary 1.2

By Theorem 1.1 and following [4], it is enough to show the constant 1/4 is sharp. Choose u(x, t) = ϕ(|x|)w(t) with $\varphi \left(\left|x\right|\right)\in C_0^{\infty }\left(B_R\right)$ and $w\left(t\right)\in C_0^{\infty }\left(ℝ\right)$. By the proof of Theorem 1.1,

$$|{\nabla }_{ℍ}u(x\mathrm{,}t{)|}^2=|{\nabla }_x\varphi (|x{|)|}^2{w}^2(t)+4|x{|}^2{\varphi }^2(|x|)(w^{\prime }(t{))}^2\mathrm{.}$$

Therefore,

$$\begin{array}{l}\frac{{\displaystyle {\int }_{{\Omega }_H}}|{\nabla }_{ℍ}u{|}^2-{(n-1)}^2{\displaystyle {\int }_{{\Omega }_H}}\frac{u^2}{|x{|}^2}-\frac{1}{4}{\displaystyle {\sum }_{j=1}^{k-1}{\displaystyle {\int }_{{\Omega }_H}}\frac{u^2}{|x{|}^2}{\left({\displaystyle {\prod }_{i=1}^j{{\displaystyle \log }}^{(i)}\frac{R}{\left|x\right|}}\right)}^{-2}}}{{\displaystyle {\int }_{{\Omega }_H}}\frac{u^2}{|x{|}^2}{\left({\displaystyle {\prod }_{i=1}^k{{\displaystyle \log }}^{(i)}\frac{R}{\left|x\right|}}\right)}^{-2}}\\ =\frac{{\displaystyle {\int }_{{\Omega }_H}}|{\nabla }_x\varphi (|x{|)|}^2{w}^2(t)-{(n-1)}^2{\displaystyle {\int }_{{\Omega }_H}}\frac{{\varphi }^2{w}^2(t)}{|x{|}^2}-\frac{1}{4}{\displaystyle {\sum }_{j=1}^{k-1}{\displaystyle {\int }_{{\Omega }_H}}\frac{{\varphi }^2{w}^2(t)}{|x{|}^2}{\left({\displaystyle {\prod }_{i=1}^j{{\displaystyle \log }}^{(i)}\frac{R}{\left|x\right|}}\right)}^{-2}}}{{\displaystyle {\int }_{{\Omega }_H}}\frac{{\varphi }^2{w}^2(t)}{|x{|}^2}{\left({\displaystyle {\prod }_{i=1}^k{{\displaystyle \log }}^{(i)}\frac{R}{\left|x\right|}}\right)}^{-2}}\\ +4\frac{{\displaystyle {\int }_{{\Omega }_H}}|x{|}^2{\varphi }^2(\left|x\right|)(w^{\prime }(t{))}^2}{{\displaystyle {\int }_{{\Omega }_H}}\frac{{\varphi }^2{w}^2(t)}{|x{|}^2}{\left({\displaystyle {\prod }_{i=1}^k{{\displaystyle \log }}^{(i)}\frac{R}{\left|x\right|}}\right)}^{-2}}\\ =\frac{{\displaystyle {\int }_{B_R}}|{\nabla }_x\varphi (|x{|)|}^2-{(n-1)}^2{\displaystyle {\int }_{B_R}}\frac{{\varphi }^2}{|x{|}^2}-\frac{1}{4}{\displaystyle {\sum }_{j=1}^{k-1}{\displaystyle {\int }_{B_R}}\frac{{\varphi }^2}{|x{|}^2}{\left({\displaystyle {\prod }_{i=1}^j{{\displaystyle \log }}^{(i)}\frac{R}{\left|x\right|}}\right)}^{-2}}}{{\displaystyle {\int }_{B_R}}\frac{{\varphi }^2}{|x{|}^2}{\left({\displaystyle {\prod }_{i=1}^k{{\displaystyle \log }}^{(i)}\frac{R}{\left|x\right|}}\right)}^{-2}}\\ +4\frac{{\displaystyle {\int }_{B_R}}|x{|}^2{\varphi }^2(\left|x\right|)}{{\displaystyle {\int }_{B_R}}\frac{{\varphi }^2}{|x{|}^2}{\left({\displaystyle {\prod }_{i=1}^k{{\displaystyle \log }}^{(i)}\frac{R}{\left|x\right|}}\right)}^{-2}}\cdot \frac{{\displaystyle {\int }_{-\infty }^{+\infty }}{(w^{\prime }(t))}^2}{{\displaystyle {\int }_{-\infty }^{+\infty }}{w}^2(t)\text{d}t}\mathrm{.}\end{array}$$

Since

$$\begin{array}{l}\underset{u\in C_0^{\infty }({\Omega }_H)\backslash \left\{0\right\}}{{\displaystyle \inf }}\frac{{\displaystyle {\int }_{{\Omega }_H}}|{\nabla }_{ℍ}u{|}^2-{(n-1)}^2{\displaystyle {\int }_{{\Omega }_H}}\frac{u^2}{|x{|}^2}-\frac{1}{4}{\displaystyle {\sum }_{j=1}^{k-1}{\displaystyle {\int }_{{\Omega }_H}}\frac{u^2}{|x{|}^2}{\left({\displaystyle {\prod }_{i=1}^j{{\displaystyle \log }}^{(i)}\frac{R}{\left|x\right|}}\right)}^{-2}}}{{\displaystyle {\int }_{{\Omega }_H}}\frac{u^2}{|x{|}^2}{\left({\displaystyle {\prod }_{i=1}^k{{\displaystyle \log }}^{(i)}\frac{R}{\left|x\right|}}\right)}^{-2}}\\ \le \underset{\varphi (|x|)\in C_0^{\infty }(B_R)\backslash \left\{0\right\}}{{\displaystyle \inf }}\frac{{\displaystyle {\int }_{B_R}}|{\nabla }_x\varphi (|x{|)|}^2-{(n-1)}^2{\displaystyle {\int }_{B_R}}\frac{{\varphi }^2}{|x{|}^2}-\frac{1}{4}{\displaystyle {\sum }_{j=1}^{k-1}{\displaystyle {\int }_{B_R}}\frac{{\varphi }^2}{|x{|}^2}{\left({\displaystyle {\prod }_{i=1}^j{{\displaystyle \log }}^{(i)}\frac{R}{\left|x\right|}}\right)}^{-2}}}{{\displaystyle {\int }_{B_R}}\frac{{\varphi }^2}{|x{|}^2}{\left({\displaystyle {\prod }_{i=1}^k{{\displaystyle \log }}^{(i)}\frac{R}{\left|x\right|}}\right)}^{-2}}\\ =\frac{1}{4}\mathrm{.}\end{array}$$

we have

$$\begin{array}{l}\underset{u\in C_0^{\infty }({\Omega }_H)\backslash \left\{0\right\}}{{\displaystyle \inf }}\frac{{\displaystyle {\int }_{{\Omega }_H}}|{\nabla }_{ℍ}u{|}^2-{(n-1)}^2{\displaystyle {\int }_{{\Omega }_H}}\frac{u^2}{|x{|}^2}-\frac{1}{4}{\displaystyle {\sum }_{j=1}^{k-1}{\displaystyle {\int }_{{\Omega }_H}}\frac{u^2}{|x{|}^2}{\left({\displaystyle {\prod }_{i=1}^j{{\displaystyle \log }}^{(i)}\frac{R}{\left|x\right|}}\right)}^{-2}}}{{\displaystyle {\int }_{{\Omega }_H}}\frac{u^2}{|x{|}^2}{\left({\displaystyle {\prod }_{i=1}^k{{\displaystyle \log }}^{(i)}\frac{R}{\left|x\right|}}\right)}^{-2}}\\ \le \underset{\varphi (|x|)\in C_0^{\infty }(B_R)\backslash \left\{0\right\}}{{\displaystyle \inf }}\frac{{\displaystyle {\int }_{B_R}}|{\nabla }_x\varphi (|x{|)|}^2-{(n-1)}^2{\displaystyle {\int }_{B_R}}\frac{{\varphi }^2}{|x{|}^2}-\frac{1}{4}{\displaystyle {\sum }_{j=1}^{k-1}{\displaystyle {\int }_{B_R}}\frac{{\varphi }^2}{|x{|}^2}{\left({\displaystyle {\prod }_{i=1}^j{{\displaystyle \log }}^{(i)}\frac{R}{\left|x\right|}}\right)}^{-2}}}{{\displaystyle {\int }_{B_R}}\frac{{\varphi }^2}{|x{|}^2}{\left({\displaystyle {\prod }_{i=1}^k{{\displaystyle \log }}^{(i)}\frac{R}{\left|x\right|}}\right)}^{-2}}\\ =\frac{1}{4}\mathrm{.}\end{array}$$

Here we use the fact that the sharp constant in inequality (1.3) is 1/4 (see [4]). This completes the proof of Corollary 1.2.

Proof of Corollary 1.3

The proof is similar to that of Corollary 1.2 and it is enough to show the constant 1/4 is sharp. Choose u(x, t) = ϕ(|x|)w(t) with $\varphi \left(\left|x\right|\right)\in C_0^{\infty }\left(B_R\right)$ and $w\left(t\right)\in C_0^{\infty }\left(ℝ\right)$. Then

$$|{\nabla }_{ℍ}u(x\mathrm{,}t{)|}^2=|{\nabla }_x\varphi (|x{|)|}^2{w}^2(t)+4|x{|}^2{\varphi }^2(|x|)(w^{\prime }(t{))}^2\mathrm{.}$$

Therefore,

$$\begin{array}{l}\frac{{\displaystyle {\int }_{{\Omega }_H}}|{\nabla }_{ℍ}u{|}^2-{(n-1)}^2{\displaystyle {\int }_{{\Omega }_H}}\frac{u^2}{|x{|}^2}-\frac{1}{4}{\displaystyle {\sum }_{j=1}^{k-1}{\displaystyle {\int }_{{\Omega }_H}}\frac{u^2}{|x{|}^2}{X}_1^2\left(\frac{\left|x\right|}{D}\right)\cdots X_j^2\left(\frac{\left|x\right|}{D}\right)}}{{\displaystyle {\int }_{{\Omega }_H}}\frac{u^2}{|x{|}^2}{X}_1^2\left(\frac{\left|x\right|}{D}\right)\cdot \cdots \cdot X_k^2\left(\frac{\left|x\right|}{D}\right)}\\ =\frac{{\displaystyle {\int }_{{\Omega }_H}}|{\nabla }_x\varphi (|x{|)|}^2{w}^2(t)-{(n-1)}^2{\displaystyle {\int }_{{\Omega }_H}}\frac{{\varphi }^2{w}^2(t)}{|x{|}^2}-\frac{1}{4}{\displaystyle {\sum }_{j=1}^{k-1}{\displaystyle {\int }_{{\Omega }_H}}\frac{{\varphi }^2{w}^2(t)}{|x{|}^2}{X}_1^2\left(\frac{\left|x\right|}{D}\right)\cdots X_j^2\left(\frac{\left|x\right|}{D}\right)}}{{\displaystyle {\int }_{{\Omega }_H}}\frac{{\varphi }^2{w}^2(t)}{|x{|}^2}{X}_1^2\left(\frac{\left|x\right|}{D}\right)\cdots X_j^2\left(\frac{\left|x\right|}{D}\right)}\\ +4\frac{{\displaystyle {\int }_{{\Omega }_H}}|x{|}^2{\varphi }^2(\left|x\right|)(w^{\prime }(t{))}^2}{{\displaystyle {\int }_{{\Omega }_H}}\frac{{\varphi }^2{w}^2(t)}{|x{|}^2}{X}_1^2\left(\frac{\left|x\right|}{D}\right)\cdots X_j^2\left(\frac{\left|x\right|}{D}\right)}\\ =\frac{{\displaystyle {\int }_{B_R}}|{\nabla }_x\varphi (|x{|)|}^2-{(n-1)}^2{\displaystyle {\int }_{B_R}}\frac{{\varphi }^2}{|x{|}^2}-\frac{1}{4}{\displaystyle {\sum }_{j=1}^{k-1}{\displaystyle {\int }_{B_R}}\frac{{\varphi }^2}{|x{|}^2}{X}_1^2\left(\frac{\left|x\right|}{D}\right)\cdots X_j^2\left(\frac{\left|x\right|}{D}\right)}}{{\displaystyle {\int }_{B_R}}\frac{{\varphi }^2}{|x{|}^2}{X}_1^2\left(\frac{\left|x\right|}{D}\right)\cdots X_j^2\left(\frac{\left|x\right|}{D}\right)}\\ +4\frac{{\displaystyle {\int }_{B_R}}|x{|}^2{\varphi }^2(\left|x\right|)}{{\displaystyle {\int }_{B_R}}\frac{{\varphi }^2}{|x{|}^2}{X}_1^2\left(\frac{\left|x\right|}{D}\right)\cdots X_j^2\left(\frac{\left|x\right|}{D}\right)}\cdot \frac{{\displaystyle {\int }_{-\infty }^{+\infty }}{(w^{\prime }(t))}^2}{{\displaystyle {\int }_{-\infty }^{+\infty }}{w}^2(t)\text{d}t}\mathrm{.}\end{array}$$

Thus

$$\begin{array}{l}\underset{u\in C_0^{\infty }({\Omega }_H)\backslash \left\{0\right\}}{{\displaystyle \inf }}\frac{{\displaystyle {\int }_{{\Omega }_H}}|{\nabla }_{ℍ}u{|}^2-{(n-1)}^2{\displaystyle {\int }_{{\Omega }_H}}\frac{u^2}{|x{|}^2}-\frac{1}{4}{\displaystyle {\sum }_{j=1}^{k-1}{\displaystyle {\int }_{{\Omega }_H}}\frac{u^2}{|x{|}^2}{X}_1^2\left(\frac{\left|x\right|}{D}\right)\cdots X_j^2\left(\frac{\left|x\right|}{D}\right)}}{{\displaystyle {\int }_{{\Omega }_H}}\frac{u^2}{|x{|}^2}{X}_1^2\left(\frac{\left|x\right|}{D}\right)\cdot \cdots \cdot X_k^2\left(\frac{\left|x\right|}{D}\right)}\\ \le \underset{\varphi (|x|)\in C_0^{\infty }(B_R)\backslash \left\{0\right\}}{{\displaystyle \inf }}\frac{{\displaystyle {\int }_{B_R}}|{\nabla }_x\varphi (|x{|)|}^2-{(n-1)}^2{\displaystyle {\int }_{B_R}}\frac{{\varphi }^2}{|x{|}^2}-\frac{1}{4}{\displaystyle {\sum }_{j=1}^{k-1}{\displaystyle {\int }_{B_R}}\frac{{\varphi }^2}{|x{|}^2}\left(\frac{\left|x\right|}{D}\right)\cdots X_j^2\left(\frac{\left|x\right|}{D}\right)}}{{\displaystyle {\int }_{B_R}}\frac{{\varphi }^2}{|x{|}^2}\left(\frac{\left|x\right|}{D}\right)\cdots X_j^2\left(\frac{\left|x\right|}{D}\right)}\\ =\frac{1}{4}\mathrm{.}\end{array}$$

since

$$\underset{w(t)\in C_0^{\infty }(ℝ)\backslash \left\{0\right\}}{{\displaystyle \inf }}\frac{{\displaystyle {\int }_{ℝ}}|w^{\prime }(t{)|}^2\text{d}t}{{\displaystyle {\int }_{ℝ}}|w(t{)|}^2\text{d}t}=0.$$

This completes the proof of Corollary 1.3.