The Hardy inequality in $ \mathbb{R}^{N} $ reads that, for all $ u\in C^{\infty}_{0}(\mathbb{R}^{N}) $ and $ N\geq 3 $,

$\int_{\mathbb{R}^{N}}|\nabla u|^{2}dx\geq\frac{(N-2) ^{2}}{4}\int_{\mathbb{R}^{N}}\frac{u^{2}}{|x|^{2}}dx$

(1.1)

and the constant $ \frac{(N-2) ^{2}}{4} $ in (1.1) is sharp.Recently, it was proved by Nazarov (see [12], Proposition 4.1 and [6]) that the following Hardy inequality is valid for $ f\in C^{\infty}_{0}(\mathbb{R}^{N}_{+}) $,

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

(1.2)

where $ \mathbb{R}^{N}_{+}=\{(x_{1},\cdots,x_{n})|x_{N}>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. Inequality (1.2) was generalized by Su and Yang [14] to the cone $ \mathbb{R}_{k_+}^{N}:=\mathbb{R}^{N-k}×(\mathbb{R}_{+})^{k}=\{(x_{1},\cdots,x_{N})|x_{N-k+1}>0,\cdots,x_{N}>0\} $.For more information about this inequality and its applications, we refer to [2, 3] and the references therein.

The aim of this note is to prove similar Hardy type inequality on half spaces for Kohn's sublaplacian in H-type groups $ G $, a remarkable class of stratified groups of step two introduced by Kaplan [11]. Let $ G=(\mathbb{R}^{m}×\mathbb{R}^{n},\circ) $ with group law defined in Section 2. It was proved by Han et al. (see [10] and [5, 8, 9, 13] for analogous inequalities on Heisenberg group) that for $ u\in C^{\infty}_{0}(G) $, there holds

$\int_{\mathbb{R}^{m}×\mathbb{R}^{n}}|\nabla_{G} u|^{2}dxdt\geq\frac{(Q-2) ^{2}}{4}\int_{\mathbb{R}^{m}×\mathbb{R}^{n}}\frac{u^{2}}{\rho^{2}}|\nabla_{G}\rho|^{2}dxdt,$

(1.3)

and the constant $ \frac{(Q-2) ^{2}}{4} $ is sharp, where $ Q=m+2n $, $ \rho(x,y)=(|x|^{4}+16|t|^{2})^{\frac{1}{4}} $ and $ \nabla_{G} $ is the the horizontal gradient associated with the Kohn's sublaplacian on $ G $ (for details, see Section 2). In this note we shall show when the singularity is on the boundary, the Hardy constant also jumps. In fact, we have the following:

Theorem 1.1  Let $ \alpha<Q-2 $. There holds, for all $ u\in C^{\infty}_{0}(\mathbb{R}^{m}× \mathbb{R}_{+}^{n}) $,

$\int_{\mathbb{R}^{m}× \mathbb{R}_{+}^{n}}\frac{|\nabla_{G}u|^{2}}{\rho^{\alpha}}dxdt\geq\left(\frac{(Q+2-\alpha)^{2}}{4}+2\alpha)\right)\int_{\mathbb{R}^{m}×\mathbb{R}_{+}^{n}}\frac{u^{2}}{\rho^{2+\alpha}}|\nabla_{G}\rho|^{2}dxdt,$

(1.4)

and the constant $ \frac{(Q+2-\alpha)^{2}}{4}+2\alpha $ in (1.4) is sharp.

We begin by describing the Lie groups and Lie algebras under consideration. For more information about H-type groups, we refer to [1, 11] and references therein. A H-type group $ G $ is a Carnot group of step two with the following properties: the Lie algebra $ \mathfrak g $ of $ G $ is endowed with an inner product $ \langle,\rangle $ such that, if $ \mathfrak z $ is the center of $ \mathfrak g $, then $ [\mathfrak z^{\perp},\mathfrak z^{\perp}]=\mathfrak z $ and moreover, for every fixed $ z\in\mathfrak z $, the map $ J_{z}: \mathfrak z^{\perp}\rightarrow\mathfrak z^{\perp} $ defined by

$\langle J_{z}(v),\omega\rangle=\langle z,[v,\omega]\rangle,\;\;\forall \omega\in\mathfrak z^{\perp}\label{eq1}$

is an orthogonal map whenever $ \langle z,z\rangle=1 $. Set $ m=\dim\mathfrak z^{\perp} $ and $ n=\dim \mathfrak z $. In the sequel we shall fix on $ G $ a system of coordinates $ (x,t) $ and that the group law has the form

$\begin{split}(x,t)\circ (x',t')=\begin{pmatrix} x_{i}+x'_{i},\;\; i=1,2,\cdots,m \\ t_{j}+t'_{j}+\frac{1}{2} <x,U^{(j)}x'>,\;\;j=1,2,\cdots,n\\\end{pmatrix},\end{split}$

(2.1)

where the matrices $ \{U^{(j)}\}^{n}_{j=1} $ have the following two properties (see [1])

(1) $ U^{(j)} $ is a $ m× m $ Skew symmetric and orthogonal matrix, for every $ j=1,2,\cdots,n $;

(2) $ U^{(i)} U^{(j)}+U^{(j)}U^{(i)}=0 $ for every $ i,j\in\{1,2,\cdots,n\} $ with $ i\neq j $.

A easy computation shows that the vector field in the algebra $ \mathfrak{g} $ of $ N=(\mathbb{R}^{m+n},\circ) $ that agrees at the origin with $ \frac{\partial}{\partial x_{j}}(j=1,\cdots,m) $ is given by

$\begin{split}X_{j} = \frac{\partial}{\partial x_{j}}+\frac{1}{2}\sum^{n}_{k=1}\left(\sum_{i=1}^{m}U^{(k)}_{i,j}x_{i}\right)\frac{\partial}{\partial t_{k}},\end{split}$

and that $ \mathfrak{g} $ is spanned by the left-invariant vector fields $ X_{1},\cdots,X_{m},\frac{\partial}{\partial t_{1}},\cdots,\frac{\partial}{\partial t_{n}} $. We use the notation $ \nabla_{G}=(X_{1},\cdots,X_{m}) $ and call it the horizontal gradient. The horizontal gradient can be written in the form

$\nabla_{G}=\nabla_{x}-\frac{1}{2}U^{(1) }x\frac{\partial}{\partial t_{1}}-\cdots-\frac{1}{2}U^{(n)}x\frac{\partial}{\partial t_{n}}$

(2.2)

with $ x=(x_{1},\cdots,x_{m})\;\;\textrm{and}\;\;\nabla_{x}=(\frac{\partial}{\partial x_{1}},\cdots,\frac{\partial}{\partial x_{m}}). $ The Kohn's sublaplacian on the H-type group $ G $ is given by

$\begin{split}\Delta_{G}&=\sum^{m}_{j=1}X^{2}_{j}=\sum^{m}_{j=1}\left(\frac{\partial}{\partial x_{j}}+\frac{1}{2}\sum^{n}_{k=1}\left(\sum_{i=1}^{m}U^{(k)}_{i,j}x_{i}\right)\frac{\partial}{\partial t_{k}}\right)^{2}\\&=\Delta_{x}+\frac{1}{4}|x|^{2}\Delta_{t}+\sum^{n}_{k=1}\langle x,U^{(k)}\nabla_{x}\rangle\frac{\partial}{\partial t_{k}},\end{split}$

where $ \Delta_{x}=\sum\limits^{m}_{j=1}\left(\frac{\partial}{\partial x_{j}}\right)^{2} $ and $ \Delta_{t}=\sum\limits^{n}_{k=1}\left(\frac{\partial}{\partial t_{k}}\right)^{2}. $ Moreover, on functions $ f(x,t)=\widetilde{f}(|x|,t) $, we have

$\langle x,U^{(k)}\nabla_{x}\rangle\widetilde{f}(|x|,t)=0,\;\;k=1,2,\cdots,n.$

Hence

$\Delta_{G}\widetilde{f}(|x|,t)=\Delta_{x}\widetilde{f}(|x|,t)+\frac{1}{4}|x|^{2}\Delta_{t}\widetilde{f}(|x|,t).$

(2.3)

We also have

$\begin{split}|\nabla_{G}\widetilde{f}(|x|,t)|^{2}=&\left|\nabla_{x}\widetilde{f}-\frac{1}{2}\sum^{n}_{j=1}U^{(j)}x\frac{\partial\widetilde{f}}{\partial t_{j}}\right|^{2}\\=&|\nabla_{x}\widetilde{f}|^{2}+\frac{1}{4}\left|\sum^{n}_{j=1}U^{(j)}x\frac{\partial\widetilde{f}}{\partial t_{j}}\right|^{2}-\sum^{n}_{j=1}\langle U^{(j)}x,\nabla_{x}\widetilde{f}(|x|,t)\rangle\frac{\partial\widetilde{f}}{\partial t_{j}}\\=&|\nabla_{x}\widetilde{f}|^{2}+\frac{1}{4}|x|^{2}|\nabla_{t}\widetilde{f}(|x|,t)|^{2}.\end{split}$

(2.4)

To get the last inequality, we use the fact ($ r=|x| $ in the equality below)

$\langle U^{(j)}x,\nabla_{x}\widetilde{f}(|x|,t)\rangle=\left\langle U^{(j)} x,\frac{x}{|x|}\right\rangle\frac{\partial \widetilde{f}}{\partial r}=0$

since $ U^{(j)} $ $ (1\leq j\leq n) $ is a skew-symmetric matric and

$\begin{split}\left|\sum^{n}_{j=1}U^{(j)}x\frac{\partial \widetilde{f}}{\partial t_{j}}\right|^{2}=&\sum^{n}_{j=1}\left|U^{(j)}x\frac{\partial\widetilde{f}}{\partial t_{j}}\right|^{2}+2\sum_{i< j}\langle U^{(i)}x,U^{(j)}x\rangle \frac{\partial \widetilde{f}}{\partial t_{i}}\frac{\partial \widetilde{f}}{\partial t_{j}}\\=&|x|^{2}|\nabla_{t}\widetilde{f}(|x|,t)|^{2}+2\sum_{i< j}\langle(U^{(j)})^{T}U^{(i)}x,x\rangle \frac{\partial\widetilde{f}}{\partial t_{i}}\frac{\partial \widetilde{f}}{\partial t_{j}}\\=&|x|^{2}|\nabla_{t}\widetilde{f}(|x|,t)|^{2}-2\sum_{i< j}\langle U^{(j)}U^{(i)}x,x\rangle \frac{\partial \widetilde{f}}{\partial t_{i}}\frac{\partial \widetilde{f}}{\partial t_{j}}\\=&|x|^{2}|\nabla_{t}\widetilde{f}(|x|,t)|^{2}\end{split}$

since $ U^{(j)}U^{(i)} $ is also skew-symmetric for every $ i\neq j $, for

$(U^{(j)}U^{(i)})^{T}=(U^{(i)})^{T}(U^{(j)})^{T}=(-U^{(i)})(-U^{(j)})=U^{(i)}U^{(j)}=-U^{(j)}U^{(i)}.$

For each real number $ λ>0 $, there is a dilation naturally associated with the group structure which is usually denoted as $ \delta_{λ}(\xi)=\delta_{λ}(x,t)=(λ x,λ^{2}t) $, $ \xi=(x,t)\in G $. However, for simplicity we will write $ λ \xi $ to denote $ \delta_{λ}(\xi) $. The Jacobian determinant of $ \delta_{λ} $ is $ λ^{Q} $, where $ Q=m+2n $ is the homogeneous dimension of $ G $. The anisotropic dilation structure on $ G $ introduces homogeneous norm $ \rho(\xi)=\rho(x,t)=(|x|^{4}+16|t|^{2})^{\frac{1}{4}}. $ With this norm, we can define the Heisenberg ball centered at $ \xi=(x,t) $ with radius $ R $,

$B(\xi,R)=\{v\in G: \rho(\xi^{-1}\circ v)<R\}.$

For simplicity, we set

$B_{R}=B(0,R)=\{v\in G: \rho(v)<R\}.$

(2.5)

Given any $ \xi=(x,t)\neq \mathbf{0} $, set $ x^*=\frac{x}{\rho(\xi)} $, $ t^*=\frac{t}{\rho(\xi)^2} $ and $ \xi^*=(x^*,t^*) $. Then $ \xi^* \in\Sigma=\{v\in G,\rho(v)=1\} $, the Heisenberg unit sphere. Furthermore, we have the following polar coordinates on $ G $ (see [7]):

$\int_{G}f(\xi)dxdt=\int^{\infty}_{0}\int_{\Sigma}f(λ\xi^{\ast})λ^{Q-1}d\sigma dr$

for all $ f\in L^{1}(G) $ and for $ \beta>-m $ (see [4]),

$C_\beta:=\int_{\Sigma}|x^{\ast}|^{\beta}d\sigma=\frac{1}{4^{n-\frac{1}{2}}}\frac{\pi^{\frac{n+m}{2}}\Gamma(\frac{m+\beta}{4})}{\Gamma(\frac{m}{2})\Gamma(\frac{Q+\beta}{4})}>0.$

(2.6)

A function $ f $ on $ G $ is said to be radial if $ f(x,t)=\widetilde{f}(\rho) $. If $ f $ is radial, it is easy to check

$|\nabla_{G}f|=|f'(\rho)|\cdot|\nabla_{G}\rho|=|f'(\rho)|\frac{|x|}{\rho}$

(2.7)

and

$\Delta_{G}f=|\nabla_{G}\rho|^{2}\left(f"+\frac{Q-1}{\rho}f'\right)=\frac{|x|^{2}}{\rho^{2}}\left(f"+\frac{Q-1}{\rho}f'\right).$

(2.8)

Before the proof of main results, we need the following lemma.

Lemma 3.1  Let $ f\in C^{\infty}(G) $ be a radial function.There holds

(1) $ |\nabla_{G}(t_{n}f)|^{2}=t^{2}_{n}|f'|^{2}|\nabla_{G}\rho|^{2}+\frac{|x|^{2}}{4}f^{2}+\frac{t_{n}|x|^{2}}{4}\cdot\frac{\partial f^{2}}{\partial t_{n}} $;

(2) $ \Delta_{G}(t_{n}f)=t_{n}(f''+\frac{Q+3}{\rho}f')|\nabla_{G}\rho|^{2}. $

Proof  (1) Since $ f $ is radial, we get, by (2.5),

$\begin{split}|\nabla_{G}(t_{n}f)|^{2}=&|\nabla_{x}(t_{n}f)|^{2}+\frac{1}{4}|x|^{2}|\nabla_{t}(t_{n}f)|^{2}\\=&t^{2}_{n}|\nabla_{x}f|^{2}+\frac{|x|^{2}}{4}\left(t^{2}_{n}|\nabla_{t}f|^{2}+f^{2}|\nabla_{t}t_{n}|^{2}+2t_{n}f\langle\nabla_{t}t_{n},\nabla_{t}f\rangle\right)\\=&t^{2}_{n}\left(|\nabla_{x}f|^{2}+\frac{|x|^{2}}{4}|\nabla_{t}f|^{2}\right)+\frac{|x|^{2}}{4}f^{2}+\frac{t_{n}|x|^{2}}{2}\cdot f\frac{\partial f}{\partial t_{n}}\\=&t^{2}_{n}|\nabla_{G}f|^{2}+\frac{|x|^{2}}{4}f^{2}+\frac{t_{n}|x|^{2}}{4}\cdot\frac{\partial f^{2}}{\partial t_{n}}.\end{split}$

It remains to use (2.7) and the desired result follows.

(2) By (2.3) and (2.8), we have

$\begin{split}\Delta_{G}(t_{n}f)=&\Delta_{x}(t_{n}f)+\frac{|x|^{2}}{4}\Delta_{t}(t_{n}f)=t_{n}\Delta_{x}f+\frac{|x|^{2}}{4}\left(t_{n}\Delta_{t}f+2\frac{\partial f}{\partial t_{n}}\right)\\=&t_{n}\left(\Delta_{x}f+\frac{|x|^{2}}{4}\Delta_{t}f\right)+\frac{|x|^{2}}{4}f'\cdot\frac{\partial\rho}{\partial t_{n}}=t_{n}\Delta_{G}f+4t_{n}\frac{|x|^{2}}{\rho^{2}}\cdot\frac{f'}{\rho}\\=&t_{n}\left(f"+\frac{Q-1}{\rho}f'\right)|\nabla_{G}\rho|^{2}+t_{n}|\nabla_{G}\rho|^{2}\cdot\frac{4f'}{\rho}\\=&t_{n}\left(f"+\frac{Q+3}{\rho}f'\right)|\nabla_{G}\rho|^{2}.\end{split}$

Now we can prove Theorem 1.1.

Proof of Theorem 1.1  First, we consider the case $ \alpha=0 $. Using the substitution $ u=t_{n}\rho^{-\frac{Q+2}{2}}f $, we get

$\begin{split} \int_{\mathbb{R}^{m}× \mathbb{R}_{+}^{n}}|\nabla_{G} u|^{2}=& \int_{\mathbb{R}^{m}× \mathbb{R}_{+}^{n}} \left[|\nabla_{G}(t_{n}\rho^{-\frac{Q+2}{2}})|^{2}f^{2}+ |\nabla_{G}f|^{2}\frac{t^{2}_{n}}{\rho^{Q+2}}+ \frac{\langle\nabla_{G}(t^{^{2}}_{n}\rho^{-(Q+2) },\nabla_{G}f^{2}\rangle}{2} \right]\\ \geq&\int_{\mathbb{R}^{m}× \mathbb{R}_{+}^{n}} \left(|\nabla_{G}(t_{n}\rho^{-\frac{Q+2}{2}})|^{2}f^{2}+ \frac{1}{2}\langle\nabla_{G}(t^{^{2}}_{n}\rho^{-(Q+2) },\nabla_{G}f^{2}\rangle\right)\\ =&\int_{\mathbb{R}^{m}× \mathbb{R}_{+}^{n}}f^{2} \left(|\nabla_{G}(t_{n}\rho^{-\frac{Q+2}{2}})|^{2}- \frac{1}{2}\Delta_{G}(t^{^{2}}_{n}\rho^{-(Q+2) })\right).\end{split}$

Notice that, for $ g\in C^{\infty}(G) $,

$\begin{split} \Delta_{G}g^{2}=\sum^{m}_{j=1}X^{2}_{j}g^{2}=2g\sum^{m}_{j=1}X^{2}_{j}g+2\sum^{m}_{j=1}|X_{j}g|^{2}=2g \Delta_{G}g+2|\nabla_{G}g|^{2}.\end{split}$

We have, by Lemma 3.1 (2),

$\begin{split}&|\nabla_{G}(t_{n}\rho^{-\frac{Q+2}{2}})|^{2}- \frac{1}{2}\Delta_{G}(t^{^{2}}_{n}\rho^{-(Q+2) }=-t_{n}\rho^{-\frac{Q+2}{2}}\Delta_{G}(t_{n}\rho^{-\frac{Q+2}{2}})\\=&-t_{n}\rho^{-\frac{Q+2}{2}}\cdot t_{n}\left(\frac{(Q+2) (Q+4) }{4}-\frac{(Q+2) (Q+3) }{2}\right)\rho^{-\frac{Q+6}{2}}|\nabla_{G}\rho|^{2}\\=&\frac{(Q+2) ^{2}}{4}t^{2}_{n}\rho^{-(Q+4) }|\nabla_{G}\rho|^{2}.\end{split}$

Therefore

$\begin{split} \int_{\mathbb{R}^{m}× \mathbb{R}_{+}^{n}}|\nabla_{G} u|^{2}\geq&\int_{\mathbb{R}^{m}× \mathbb{R}_{+}^{n}}f^{2} \left(|\nabla_{G}(t_{n}\rho^{-\frac{Q+2}{2}})|^{2}- \frac{1}{2}\Delta_{G}(t^{^{2}}_{n}\rho^{-(Q+2) })\right)\\ =&\frac{(Q+2) ^{2}}{4}\int_{\mathbb{R}^{m}× \mathbb{R}_{+}^{n}}f^{2}t^{2}_{n}\rho^{-(Q+4) }|\nabla_{G}\rho|^{2}\\ =&\frac{(Q+2) ^{2}}{4}\int_{\mathbb{R}^{m}× \mathbb{R}_{+}^{n}}\frac{u^{2}}{\rho^{2}}|\nabla_{G}\rho|^{2}.\end{split}$

(3.1)

Now we show the constant $ \frac{(Q+2) ^{2}}{4} $ in (3.1) is sharp. Consider the family of function $ g_{\varepsilon}=t_{n}f_{\varepsilon}(\rho) $, where

$f_{\varepsilon}(\rho)=\left\{ \begin{array}{ll} \varepsilon^{-(Q+2) /2},& \hbox{ $ \rho\leq \varepsilon $ ;} \\ \rho^{-(Q+2) /2},& \hbox{ $ \rho> \varepsilon $ .} \end{array}\right.$

We take $ g_{\varepsilon} $ as the test function. By (2.4) and symmetry,

$\begin{split}2\int_{\mathbb{R}^{m}× \mathbb{R}_{+}^{n}}|\nabla_{G}g_{\varepsilon}|^{2}=&2\int_{\mathbb{R}^{m}×\mathbb{R}_{+}^{n}}\left(|\nabla_{x}g_{\varepsilon}|^{2}+\frac{|x|^{2}}{4}|\nabla_{t}g_{\varepsilon}|^{2}\right)\\=&\int_{\mathbb{R}^{m}× \mathbb{R}^{n}}\left(|\nabla_{x}g_{\varepsilon}|^{2}+\frac{|x|^{2}}{4}|\nabla_{t}g_{\varepsilon}|^{2}\right)\\=&\int_{G\setminus B_{\varepsilon}}\left(|\nabla_{x}g_{\varepsilon}|^{2}+\frac{|x|^{2}}{4}|\nabla_{t}g_{\varepsilon}|^{2}\right)dxdt+\varepsilon^{-Q-2}\int_{B_{\varepsilon}}\frac{|x|^{2}}{4}dxdt\end{split}$

and

$2\int_{\mathbb{R}^{m}× \mathbb{R}_{+}^{n}}\frac{g_{\varepsilon}^{2}}{\rho^{2}}|\nabla_{G}\rho|^{2}=\int_{\mathbb{R}^{m}× \mathbb{R}^{n}}\frac{g_{\varepsilon}^{2}}{\rho^{2}}|\nabla_{G}\rho|^{2}\geq\int_{G\setminus B_{\varepsilon}}\frac{g_{\varepsilon}^{2}}{\rho^{2}}|\nabla_{G}\rho|^{2}.$

By Lemma 3.1 (1), we have, for $ \rho> \varepsilon $,

$\begin{split}|\nabla_{G}g_{\varepsilon}|^{2}=&\frac{(Q+2) ^{2}}{4}t^{2}_{n}\rho^{-(Q+4) }|\nabla_{G}\rho|^{2}+\frac{|x|^{2}}{4}\rho^{-(Q+2) }+\frac{|x|^{2}}{4}\cdot t_{n}\frac{\partial \rho^{-(Q+2) }}{\partial t_{n}}\\=&\frac{(Q+2) ^{2}}{4}t^{2}_{n}\rho^{-(Q+4) }|\nabla_{G}\rho|^{2}+\frac{|x|^{2}}{4}\rho^{-(Q+2) }-2(Q+2) \frac{|x|^{2}t^{2}_{n}}{\rho^{Q+6}}.\end{split}$

Since $ \displaystyle\int_{G\setminus B_{\varepsilon}}\left(\frac{|x|^{2}}{4}\rho^{-(Q+2) }-2(Q+2) \frac{|x|^{2}t^{2}_{n}}{\rho^{Q+6}}\right)dxdt=0 $, we have

$\begin{split}2\int_{\mathbb{R}^{m}× \mathbb{R}_{+}^{n}}|\nabla_{G}g_{\varepsilon}|^{2}=&\frac{(Q+2) ^{2}}{4}\int_{G\setminus B_{\varepsilon}}t^{2}_{n}\rho^{-(Q+4) }|\nabla_{G}\rho|^{2}+\varepsilon^{-Q-2}\int_{B_{\varepsilon}}\frac{|x|^{2}}{4}dxdt.\end{split}$

Therefore

$\begin{split}&\inf_{u\in C^{\infty}_{0}(\mathbb{R}^{m}×\mathbb{R}_{+}^{n})\setminus\{0\}}\frac{\displaystyle\int_{\mathbb{R}^{m}×\mathbb{R}_{+}^{n}}|\nabla_{G}u|^{2}dxdt}{\displaystyle\int_{\mathbb{R}^{m}×\mathbb{R}_{+}^{n}}\frac{u^{2}}{\rho^{2}}|\nabla_{G}\rho|^{2}dxdt}\leq\frac{\displaystyle\int_{\mathbb{R}^{m}×\mathbb{R}_{+}^{n}}|\nabla_{G}g_{\varepsilon}|^{2}dxdt}{\displaystyle\int_{\mathbb{R}^{m}×\mathbb{R}_{+}^{n}}\frac{g_{\varepsilon}^{2}}{\rho^{2}}|\nabla_{G}\rho|^{2}dxdt}\\=&\frac{\displaystyle\int_{\mathbb{R}^{m}×\mathbb{R}^{n}}|\nabla_{G}g_{\varepsilon}|^{2}dxdt}{\displaystyle\int_{\mathbb{R}^{m}×\mathbb{R}^{n}}\frac{g_{\varepsilon}^{2}}{\rho^{2}}|\nabla_{G}\rho|^{2}dxdt}\leq\frac{\displaystyle\int_{G}|\nabla_{G}g_{\varepsilon}|^{2}dxdt}{\displaystyle\int_{G\setminus B_{\varepsilon}}\frac{g_{\varepsilon}^{2}}{\rho^{2}}|\nabla_{G}\rho|^{2}dxdt}\\=&\frac{(Q+2) ^{2}}{4}+\varepsilon^{-Q-2}\frac{\displaystyle\int_{B_{\varepsilon}}\frac{|x|^{2}}{4}dxdt}{\displaystyle\int_{G\setminus B_{\varepsilon}}t^{2}_{n}\rho^{-(Q+4) }|\nabla_{G}\rho|^{2}}\rightarrow\frac{(Q+2) ^{2}}{4},\;\;\varepsilon\rightarrow 0.\end{split}$

This completes the proof.