# Fractional Caffarelli-Kohn-Nirenberg type inequalities on the Heisenberg
group

Rama Rawat<EMAIL_ADDRESS>,  Haripada Roy<EMAIL_ADDRESS>and
Prosenjit Roy<EMAIL_ADDRESS>Indian Institute of Technology Kanpur,
India

###### Abstract.

The aim of this work is to establish some cases of the Caffarelli-Kohn-
Nirenberg inequalities on the Heisenberg group for the fractional Sobolev
spaces. Here we work with the fractional Sobolev spaces as given by Adimurthi
and Mallick in [1]. Our inequalities also give an improvement on the range of
indices for the Hardy type inequality established in [1].

###### Key words and phrases:

Heisenberg group; Fractional order Sovolev spaces; Hardy and Sobolev
inequalities; Caffarelli-Kohn-Nirenberg inequality

###### 2023 Mathematics Subject Classification:

46E35; 26D15; 43A80; 49J52

##  1\. Introduction

In [3], L. Caffarelli, R. Kohn and L. Nirenberg established a family of
interpolation inequalities on the Euclidean space, now known as, Caffarelli-
Kohn-Nirenberg (CKN) inequalities. The Hardy and Sobolev inequalities are
included as particular cases of these CKN inequalities. Later analogous
versions of Hardy-Sobolev inequalities were obtained in [5, 6, 7, 12] in the
setting of fractional Sobolev spaces on $\mathbb{R}^{n}$, in [1] on the
Heisenberg group and in [10] for homogeneous spaces.

In this paper we establish some cases of the fractional Caffarelli-Kohn-
Nirenberg (CKN) inequalities on the Heisenberg group. We will be working with
the fractional Sobolev spaces as defined by Adimurthi and Mallick in [1] which
are more natural when the index $p$ is not necessarily $2.$ This differs from
the fractional Sobolev spaces considered in [10, 17] and the results therein
for more general homogeneous spaces including the Heisenberg group. As in the
Euclidean case, our version of CKN fractional inequalities contains in
particular fractional Hardy’s inequality and Sobolev inequality as proved in
[1]. Further, we work with decomposition of the Heisenberg group into sets
which allow us to estimate required integrals, circumventing the use of
extension domains. We are not aware if half spaces or the domains which appear
in our proofs are extension domains for the fractional Sobolev spaces we are
considering in this paper. Our results also improve on range of indices for
which the fractional Hardy’s inequality holds.

We begin by setting notations and recalling results which we need in the
sequel. Let $n\geq 1$, $p>1$, $s\in(0,1)$ and
$\eta,\,\eta_{1},\,\eta_{2}\in\mathbb{R}$ with $\eta_{1}+\eta_{2}=\eta$. For
any measurable subset $\Omega\subset\mathbb{R}^{n}$, we define the fractional
Sobolev space $W^{s,p,\eta}(\Omega)$ as

$W^{s,p,\eta}(\Omega):=\left\\{u\in
L^{p}(\Omega):[u]_{W^{s,p,\eta}(\Omega)}:=\left(\int_{\Omega}\int_{\Omega}\frac{|x|^{\eta_{1}p}|y|^{\eta_{2}p}|u(x)-u(y)|^{p}}{|x-y|^{n+sp}}dxdy\right)^{\frac{1}{p}}<\infty\right\\},$

endowed with the norm

$\lVert u\rVert_{W^{s,p,\eta}(\Omega)}=\lVert
u\rVert_{L^{p}(\Omega)}+[u]_{W^{s,p,\eta}(\Omega)}.$

When $\eta_{1}=\eta_{2}=\eta=0$, we denote the quantities $\lVert
u\rVert_{W^{s,p,0}(\Omega)}$ and $[u]_{W^{s,p,0}(\Omega)}$ by $\lVert
u\rVert_{W^{s,p}(\Omega)}$ and $[u]_{W^{s,p}(\Omega)}$ respectively.

Let $p>1$, $0<s<1$, $q\geq 1$, $\tau>0$, $0\leq a\leq 1$ and $\beta,$
$\gamma,$ $\eta,$ $\eta_{1},$ $\eta_{2}$ $\in\mathbb{R}$ be such that
$\eta_{1}+\eta_{2}=\eta$ and

$\frac{1}{\tau}+\frac{\gamma}{n}=a\left(\frac{1}{p}+\frac{\eta-s}{n}\right)+(1-a)\left(\frac{1}{q}+\frac{\beta}{n}\right).$
(1.1)

When $a>0$, we assume the following additional conditions

$\gamma=a\sigma+(1-a)\beta,\ \ 0\leq\eta-\sigma$ (1.2)

and

$\eta-\sigma\leq s,\ \mathrm{when}\ \
\frac{1}{\tau}+\frac{\gamma}{n}=\frac{1}{p}+\frac{\eta-s}{n}.$ (1.3)

The following fractional Caffarelli-Kohn-Nirenberg inequalities were proved by
Hoai-Minh Nguyen and Marco Squassina in [13].

######  Theorem 1.

_[13]_ Let $n,p,s,q,\tau,a,\beta,\gamma,\eta_{1},\eta_{2}$ and $\eta$ be as
above satisfying the conditions (1.1), (1.2) and (1.3).
_(i)_ If $\frac{1}{\tau}+\frac{\gamma}{n}>0$, then we have

$\lVert|x|^{\gamma}u\rVert_{L^{\tau}(\mathbb{R}^{n})}\leq
C[u]^{a}_{W^{s,p,\eta}(\mathbb{R}^{n})}\lVert|x|^{\beta}u\rVert^{1-a}_{L^{q}(\mathbb{R}^{n})},\
\forall u\in C_{c}^{1}(\mathbb{R}^{n}).$ (1.4)

_(ii)_ If $\frac{1}{\tau}+\frac{\gamma}{n}<0$, then we have

$\lVert|x|^{\gamma}u\rVert_{L^{\tau}(\mathbb{R}^{n})}\leq
C[u]^{a}_{W^{s,p,\eta}(\mathbb{R}^{n})}\lVert|x|^{\beta}u\rVert^{1-a}_{L^{q}(\mathbb{R}^{n})},\
\forall u\in C_{c}^{1}(\mathbb{R}^{n}\setminus\\{0\\}).$ (1.5)

If we consider $a=1$ and $\eta_{1}=\eta_{2}=\eta=0$ in (1.1), $\gamma=0$
implies $\tau=p^{*}=\frac{np}{n-p}$, and inequality (1.4) gives the well known
fractional Sobolev inequality proved earlier in [5, 12]. On the other hand,
the corresponding inequality for $\gamma=-s$ and $\tau=p$ is the fractional
Hardy inequality obtained in [7, 12].

Inequalities (1.4) and (1.5) are not true in general for the case
$\frac{1}{\tau}+\frac{\gamma}{n}=0$. Instead the following weighted version
holds in this case as was proved in [13].

######  Theorem 2.

_[13]_ Let $n,p,s,q,\tau,a,\beta,\gamma,\eta_{1},\eta_{2}$ and $\eta$ be as
above satisfying the condition (1.1) and $0\leq\eta-\sigma\leq s$.
_(i)_ If $\frac{1}{\tau}+\frac{\gamma}{n}=0$ and $u\in
C_{c}^{1}(\mathbb{R}^{n})$ with supp$(u)\subset B(0,R)$, then we have

$\left(\int_{\mathbb{R}^{n}}\frac{|x|^{\tau\gamma}}{\ln^{\tau}\left(4R/|x|\right)}|u|^{\tau}dx\right)^{\frac{1}{\tau}}\leq
C[u]^{a}_{W^{s,p,\eta}(\mathbb{R}^{n})}\lVert|x|^{\beta}u\rVert^{1-a}_{L^{q}(\mathbb{R}^{n})}.$
(1.6)

_(ii)_ If $\frac{1}{\tau}+\frac{\gamma}{n}=0$ and $u\in
C_{c}^{1}(\mathbb{R}^{n})$ with supp$(u)\cap B(0,r)=\phi$, then we have

$\left(\int_{\mathbb{R}^{n}}\frac{|x|^{\tau\gamma}}{\ln^{\tau}\left(4|x|/r\right)}|u|^{\tau}dx\right)^{\frac{1}{\tau}}\leq
C[u]^{a}_{W^{s,p,\eta}(\mathbb{R}^{n})}\lVert|x|^{\beta}u\rVert^{1-a}_{L^{q}(\mathbb{R}^{n})}.$
(1.7)

Before going further, we recall the preliminaries on the Heisenberg group. The
Heisenberg group $\mathbb{H}^{n}$ is defined as

$\mathbb{H}^{n}:=\\{\xi=(x,y,t)=(z,t):(x,y)=z\in\mathbb{R}^{n}\times\mathbb{R}^{n}\
\mathrm{and}\ t\in\mathbb{R}\\}$

with the group law

$\xi\circ\xi^{\prime}=(x+x^{\prime},y+y^{\prime},t+t^{\prime}+2\langle
y,x^{\prime}\rangle-2\langle x,y^{\prime}\rangle),$

where $\xi=(x,y,t)$,
$\xi^{\prime}=(x^{\prime},y^{\prime},t^{\prime})\in\mathbb{H}^{n}$ and
$\langle\cdot,\cdot\rangle$ denotes the usual Euclidean inner product in
$\mathbb{R}^{n}$. It is easy to see that $0\in\mathbb{H}^{n}$ is the identity
element and $-\xi$ is the inverse of $\xi\in\mathbb{H}^{n}$. A basis for the
left invariant vector fields is given by

$X_{i}=\frac{\partial}{\partial x_{i}}+2y_{i}\frac{\partial}{\partial t},\ \
1\leq i\leq n,$ $Y_{i}=\frac{\partial}{\partial
y_{i}}-2x_{i}\frac{\partial}{\partial t},\ \ 1\leq i\leq n,$
$T=\frac{\partial}{\partial t}.$

For any $C^{1}$ function $u$, $\nabla_{\mathbb{H}^{n}}(u)$ is called the sub-
gradient, defined by

$\nabla_{\mathbb{H}^{n}}(u)=\left(X_{1}(u),\cdots,X_{n}(u),Y_{1}(u),\cdots,Y_{n}(u)\right)$

and

$|\nabla_{\mathbb{H}^{n}}(u)|^{2}=\sum_{i=1}^{n}\left(|X_{i}(u)|^{2}+|Y_{i}(u)|^{2}\right).$

For any $\xi=(x,y,t)=(z,t)\in\mathbb{H}^{n}$,
$d(\xi)=\left((|x|^{2}+|y|^{2})^{2}+t^{2}\right)^{\frac{1}{4}}=\left(|z|^{4}+t^{2}\right)^{\frac{1}{4}}$
is the Koranyi-Folland non isotropic gauge and $Q=2n+2$ is the homogeneous
dimension of the Heisenberg group $\mathbb{H}^{n}$. The left-invariant Haar
measure on $\mathbb{H}^{n}$ is the Lebesgue measure on $\mathbb{R}^{2n+1}.$ We
define the ball centered at $\xi$ with radius $r$ in $\mathbb{H}^{n}$ as

$B_{r}(\xi)=\\{\xi^{\prime}=(z^{\prime},t^{\prime}):d({\xi}^{-1}\circ\xi^{\prime})<r\\},$
(1.8)

and the annular domain as

$A_{r,R}(\xi)=\\{\xi^{\prime}=(z^{\prime},t^{\prime}):r<d({\xi}^{-1}\circ\xi^{\prime})<R\\}.$
(1.9)

We also define the following set

$D_{r}(\xi)=\\{\xi^{\prime}=(z^{\prime},t^{\prime}):|z^{\prime}-z|<r\
\mathrm{and}\ \ |t^{\prime}-t-2\langle y,x^{\prime}\rangle+2\langle
x,y^{\prime}\rangle|<r^{2}\\}.$ (1.10)

Various versions of Hardy’s inequality on $\mathbb{H}^{n}$ were proved by
several authors in last two decades. The following inequality is one of the
general version of the same, proved by L. D’Ambrozio in [4]:

$\int_{\Omega}\frac{|z|^{\beta}}{d(\xi)^{\alpha}}|u(\xi)|^{p}d\xi\leq
C_{n,p,\alpha,\beta}^{p}\int_{\Omega}|\nabla_{\mathbb{H}^{n}}(u)|^{p}|z|^{\beta-p}d(\xi)^{2p-\alpha},\
\forall u\in C_{c}^{\infty}(\Omega),$ (1.11)

where $\Omega$ is an open subset of $\mathbb{H}^{n}$, $n\geq 1$, $p>1$,
$\alpha,\beta\in\mathbb{R}$ satisfy $Q>\alpha-\beta$ and $2n>p-\beta$ and
$C_{n,p,\alpha\beta}=\frac{p}{Q+\beta-\alpha}$ is the optimal constant.

Inequality (1.11) with $\alpha=4$ and $\beta=2$ was proved in [8] for the
domain $\mathbb{H}^{n}\setminus\\{0\\}$. Whereas (1.11) was proved for the
domain $\mathbb{H}^{n}$ with $\alpha=2p$ and $\beta=p$ in [14]. Adimurthi and
Sekar proved (1.11) with $\alpha=2p$ and $\beta=2$, for $u\in
FS_{0}^{1,p}(\mathbb{H}^{n})$ in [2], where the space
$FS_{0}^{1,p}(\mathbb{H}^{n})$ is the completion of
$C_{c}^{\infty}(\mathbb{H}^{n})$ under the norm

$|u|^{p}_{1,p}=\int_{\mathbb{H}^{n}}\frac{|\nabla_{\mathbb{H}^{n}}(u)|^{p}}{|z|^{p-2}}dzdt.$

In [9], Yazhou Han established Caffarelli-Kohn-Nirenberg type inequalities for
$\mathbb{H}^{n}$. Further, D. Suragan and his coauthors had extended some
cases of CKN type inequalities on stratified Lie groups (see [16, 18]) and
homogeneous groups (see [15]) with optimal constants.

For $1\leq p<\infty$, $s\in(0,1)$ and $\alpha\in\mathbb{R},$ consider the
fractional order Sobolev space on $\mathbb{H}^{n}$ as given in [1]:

$W_{0}^{s,p,\alpha}(\mathbb{H}^{n})=\mathrm{Cl}\left\\{f\in
C_{c}^{\infty}(\mathbb{H}^{n}):\int_{\mathbb{H}^{n}}\int_{\mathbb{H}^{n}}\frac{|f(\xi)-f(\xi^{\prime})|^{p}d\xi^{\prime}d\xi}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps}|z-z^{\prime}|^{(p-2)\alpha}}<\infty\right\\},$

where $\xi=(z,t)$, $\xi^{\prime}=(z^{\prime},t^{\prime}).$ The closure is
considered under the norm $\lVert f\rVert_{s,p,\alpha}^{p}=\lVert
f\rVert_{L^{p}(\mathbb{H}^{n})}^{p}+[f]_{s,p,\alpha}^{p}$, where

$[f]_{s,p,\alpha}=\left(\int_{\mathbb{H}^{n}}\int_{\mathbb{H}^{n}}\frac{|f(\xi)-f(\xi^{\prime})|^{p}d\xi^{\prime}d\xi}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps}|z-z^{\prime}|^{(p-2)\alpha}}\right)^{\frac{1}{p}}.$

Let $\Omega$ be any open set in $\mathbb{H}^{n}$. We define the space

$W^{s,p,\alpha}(\Omega)=\left\\{f\in
L^{p}(\Omega):[f]_{s,p,\alpha,\Omega}:=\left(\int_{\Omega}\int_{\Omega}\frac{|f(\xi)-f(\xi^{\prime})|^{p}d\xi^{\prime}d\xi}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps}|z-z^{\prime}|^{(p-2)\alpha}}\right)^{\frac{1}{p}}<\infty\right\\},$

with the norm $\lVert f\rVert_{s,p,\alpha,\Omega}^{p}=\lVert
f\rVert_{L^{p}(\Omega)}^{p}+[f]_{s,p,\alpha,\Omega}^{p}$.

Throughout this work, we assume $(p-2)\alpha\geq 0$. We also assume
$(p-2)\alpha<Q-2$, which follows from the non triviality of the space
$W_{0}^{s,p,\alpha}(\mathbb{H}^{n})$ (see [1], Section 3). The following
Sobolev embedding theorem for $W_{0}^{s,p,\alpha}(\mathbb{H}^{n})$ is proved
in [1].

######  Theorem 3.

_[1]_ Let $1\leq p<\infty$, $s\in(0,1)$ and $\alpha\in\mathbb{R}$ be such that
$ps+(p-2)\alpha<Q$. Then there exists a positive constant $C$, depends only on
$n$, $p$, $s$ and $\alpha$, such that

$\lVert f\rVert_{L^{p^{*}}(\mathbb{H}^{n})}\leq C[f]_{s,p,\alpha},\ \forall
f\in W_{0}^{s,p,\alpha}(\mathbb{H}^{n}),$ (1.12)

where $p^{*}=\frac{Qp}{Q-ps-(p-2)\alpha}$ is the fractional critical Sobolev
exponent.

A fractional Hardy type inequality for the functions of
$W_{0}^{s,p,\alpha}(\mathbb{H}^{n})$ has been proved in [1] which is as
follows:

######  Theorem 4.

_[1]_ Let $1\leq p<\infty$, $s\in(0,1)$ and $\alpha\in\mathbb{R}$ satisfy the
following conditions:
_(a)_ $ps>2$;
_(b)_ $ps+(p-2)\alpha<Q$.
Then there exists a positive constant $C$, depends only on $n$, $p$, $s$ and
$\alpha$, such that

$\int_{\mathbb{H}^{n}}\frac{|f(\xi)|^{p}d\xi}{d(\xi)^{ps}|z|^{(p-2)\alpha}}\leq
C[f]_{s,p,\alpha}^{p},\ \forall f\in W_{0}^{s,p,\alpha}(\mathbb{H}^{n}).$
(1.13)

Theorem 4 holds without the condition (a) only for the case $p=2$ or
$\alpha=0$ i.e., when $(p-2)\alpha=0$ (see [1], Remark 1.3.). In [10] A.
Kassymov and D. Suragan established a fractional Hardy type inequality on the
homogeneous Lie groups, which in particular on $\mathbb{H}^{n}$ gives (1.13)
with $\alpha=0$. If $(p-2)\alpha>0$, condition (a) forces $p$ to be large
enough when $s$ is very small. On the other hand, non triviality of the space
restricts the range of $p$. To be precise, Theorem 4 remains unproven for
$0<s\leq 2/\left(\frac{Q-2}{\alpha}+2\right)$.

In this paper we establish analogous version of the inequalities (1.4), (1.5),
(1.6) and (1.7) (considering $\eta_{1}=\eta_{2}=\eta=\beta=0$) for
$W_{0}^{s,p,\alpha}(\mathbb{H}^{n})$, which includes Theorem 3 and Theorem 4
as particular cases. For the case $(p-2)\alpha=0$, an analogous version of the
inequalities (1.4)-(1.7) in the Heisenberg group were established in [17].
However, $(p-2)\alpha\neq 0$ case is totally different from the case
$(p-2)\alpha=0$. In our case the corresponding dimensional balance relation is

$\frac{1}{\tau}+\frac{\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)}{Q}=a\left(\frac{1}{p}-\frac{s\left(1+(p-2)\frac{\alpha}{ps}\right)}{Q}\right)+\frac{1-a}{q},$
(1.14)

where $p>1$, $s\in(0,1)$, $\alpha\in\mathbb{R}$, $\tau>1$, $0<a\leq 1$,
$\gamma\in\mathbb{R}$ and $q\geq 1$. We also assume

$0\leq-\gamma\leq as.$ (1.15)

We prove the following results.

######  Theorem 1.1.

Let $p>1$, $s\in(0,1)$, $\alpha\in\mathbb{R}$, $\tau>1$, $0<a\leq 1$,
$\gamma\in\mathbb{R}$ and $q>(1-a)\tau$ satisfy (1.14) and (1.15).
_(i)_ If
$\frac{1}{\tau}+\frac{\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)}{Q}>\frac{2}{\tau
Q}$, then

$\left(\int_{\mathbb{H}^{n}}d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f(\xi)|^{\tau}d\xi\right)^{\frac{1}{\tau}}\leq
C[f]_{s,p,\alpha}^{a}\lVert f\rVert_{L^{q}(\mathbb{H}^{n})}^{1-a},\
\forall\,f\in C_{c}^{1}(\mathbb{H}^{n}).$ (1.16)

_(ii)_ If
$\frac{1}{\tau}+\frac{\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)}{Q}<\frac{2}{\tau
Q}$, then

$\left(\int_{\mathbb{H}^{n}}d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f(\xi)|^{\tau}d\xi\right)^{\frac{1}{\tau}}\leq
C[f]_{s,p,\alpha}^{a}\lVert f\rVert_{L^{q}(\mathbb{H}^{n})}^{1-a},\
\forall\,f\in C_{c}^{1}(\mathbb{H}^{n}\setminus\\{\xi=(z,t):z=0\\}).$ (1.17)

Comparing with the corresponding result in the Euclidean case, it is natural
to ask if (1.16) holds for
$0<1/\tau+\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)/{Q}\leq\frac{2}{\tau Q}$
as well. In general, the answer is not known to us, but as our next result
shows, under an additional conditions on indices $\tau,\gamma$ and $q$, (1.16)
continues to hold for
$1/\tau+\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)/{Q}>0$.

######  Theorem 1.2.

Let $p$, $s$, $\alpha$, $\tau$, $a$, $\gamma$ and $q$ are as considered in
Theorem _1.1_ and satisfy (1.14) and (1.15). If
$\frac{1}{\tau}+\frac{\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)}{Q}>0$ and
$\tau\gamma\leq-2$, then for $q>\frac{(1-a)\tau
Q}{Q-2+\tau\gamma(p-2)\frac{\alpha}{ps}}$ we have

$\left(\int_{\mathbb{H}^{n}}d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f(\xi)|^{\tau}d\xi\right)^{\frac{1}{\tau}}\leq
C[f]_{s,p,\alpha}^{a}\lVert f\rVert_{L^{q}(\mathbb{H}^{n})}^{1-a},\
\forall\,f\in C_{c}^{1}(\mathbb{H}^{n}).$ (1.18)

When $1/\tau+\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)/{Q}=0$, L.H.S of
(1.18) need not be finite in general. As in the Euclidean case, correction by
a weight function leads to the following analogous inequalities in this case.

######  Theorem 1.3.

Let $p$, $s$, $\alpha$, $\tau$, $a$, and $\gamma$ be as in Theorem _1.1_ and
$q>\frac{(1-a)\tau Q}{Q-2+\tau\gamma(p-2)\frac{\alpha}{ps}}$ satisfy (1.14)
and (1.15). If

$\frac{1}{\tau}+\frac{\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)}{Q}=0,$
(1.19)

then
_(i)_ For $f\in C_{c}^{1}(\mathbb{H}^{n})$ with $\mathrm{supp}\,f\subset
B_{R}(0)$, we have

$\left(\int_{\mathbb{H}^{n}}\frac{d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}}{\ln^{\tau}(4R/d(\xi))}|f(\xi)|^{\tau}d\xi\right)^{\frac{1}{\tau}}\leq
C[f]_{s,p,\alpha}^{a}\lVert f\rVert_{L^{q}(\mathbb{H}^{n})}^{1-a}.$ (1.20)

_(ii)_ For $f\in C_{c}^{1}(\mathbb{H}^{n})$ with $\mathrm{supp}\,f\cap
B_{r}(0)=\phi$, we have

$\left(\int_{\mathbb{H}^{n}}\frac{d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}}{\ln^{\tau}(4d(\xi)/r)}|f(\xi)|^{\tau}d\xi\right)^{\frac{1}{\tau}}\leq
C[f]_{s,p,\alpha}^{a}\lVert f\rVert_{L^{q}(\mathbb{H}^{n})}^{1-a}.$ (1.21)

For $a=1$ and $\gamma=-s,$ (1.16) and (1.18) are the fractional Hardy
inequality (1.13) for the corresponding indices, and they hold for $ps=2$ when
$ps+(p-2)\alpha<Q$, and for $ps\leq 2$ as well when $ps+(p-2)\alpha<Q-2$. This
an improvement on the earlier results proved in [1, 10].

The proof of our first result follows the technique used in [3] for the
Euclidean spaces. We decompose $\mathbb{H}^{n}$ into the sets

$\mathscr{A}_{k}:=\\{\xi=(z,t)\in\mathbb{H}^{n}:2^{k}<|z|<2^{k+1}\\}$ (1.22)

to deal with the term $|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}$ and then we
decompose each $\mathscr{A}_{k}$ into

$\mathscr{A}_{k,j}:=\\{\xi=(z,t)\in\mathscr{A}_{k}:j2^{2k}<t<(j+1)2^{2k}\\}$
(1.23)

so that $|\mathscr{A}_{k,j}|=C2^{kQ}$ (see Figure 1.1).

tz$\mathscr{A}_{k+1}$$\mathscr{A}_{k}$$\mathscr{A}_{k-1}$$\mathscr{A}_{k+1,0}$$\mathscr{A}_{k,0}$$\mathscr{A}_{k,1}$$\mathscr{A}_{k,2}$$\mathscr{A}_{k,3}$tz$\mathscr{A}_{k,j}$$\mathscr{B}_{k,j}$
$\mathrm{Figure}\
1.1\quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\mathrm{Figure}\ 1.2$

For any measurable set $\Omega\subset\mathbb{H}^{n}$ we denote by

$(f)_{\Omega}:=\fint_{\Omega}f(\xi)d\xi=\frac{1}{|\Omega|}\int_{\Omega}f(\xi)d\xi,$

the average of the function $f$ over $\Omega$. If $\mathscr{A}$ is a bounded
extension domain for $W^{s,p,\alpha}$ (discussed in next section), then the
following inequality holds:

$\fint_{\mathscr{A}}|f-(f)_{\mathscr{A}}|^{\tau}d\xi\leq
C[f]_{s,p,\alpha,\mathscr{A}}^{a\tau}\lVert
f\rVert_{L^{q}(\mathscr{A})}^{(1-a)\tau},\ \ \forall f\in
C(\bar{\mathscr{A}}).$ (1.24)

Such an inequality plays a crucial role in the proof of the Euclidean results.
It is not known to us whether the sets $\\{\mathscr{A}_{k,j}\\}$ are extension
domains for $W^{s,p,\alpha}$ or not, but if $K$ is a compact subset of a
bounded open set $\Omega$, the following inequality is always true:

$\fint_{K}|f-(f)_{\Omega}|^{\tau}d\xi\leq
C(\Omega,K)[f]_{s,p,\alpha,\Omega}^{a\tau}\lVert
f\rVert_{L^{q}(\Omega)}^{(1-a)\tau},\ \ \forall f\in C(\bar{\Omega}).$ (1.25)

Here the constant $C(\Omega,K)$ depends on the sets $K$ and $\Omega$. In
particular, we establish the following inequality (in Lemma 3.1):

$\fint_{\mathscr{A}_{k,j}}|f-(f)_{\mathscr{B}_{k,j}}|^{\tau}d\xi\leq
C\,2^{-\frac{k}{\tau}\left(Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)\right)}[f]_{s,p,\alpha,\mathscr{B}_{k,j}}^{a\tau}\lVert
f\rVert_{L^{q}(\mathscr{B}_{k,j})}^{(1-a)\tau},$ (1.26)

where

$\mathscr{B}_{k,j}:=\\{\xi=(z,t)\in\mathbb{H}^{n}:2^{k-1}<|z|\leq 2^{k+2},\
(j-1)2^{2k}<t<(j+2)2^{2k}\\},$ (1.27)

(see Figure 1.2). Note that $\mathscr{A}_{k,j}\Subset\mathscr{B}_{k,j}$. By
the notation $A\Subset B$ we mean that the closure of the set $A$ is a compact
subset of $B$. The constant $C$ in (1.26) is independent of $k$ and $j$. The
dependency of the constant $C(\Omega,K)$ in (1.25) on the sets is being taken
care of by the factor
$2^{-\frac{k}{\tau}\left(Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)\right)}$.

######  Remark 1.

_Note that the sets $\mathscr{A}_{k,j}$’s are mutually disjoint and_

$\sum_{k}\sum_{j}\int_{\mathscr{A}_{k,j}}f(\xi)d\xi=\int_{\mathbb{H}^{n}}f(\xi)d\xi$

_holds for any measurable function $f.$ This is not the case for
$\mathscr{B}_{k,j}$ since these sets are not mutually disjoint. However there
is a positive constant $C$, independent of $k$ and $j$ such that_

$\sum_{k}\sum_{j}\int_{\mathscr{B}_{k,j}}f(\xi)d\xi\leq
C\int_{\mathbb{H}^{n}}f(\xi)d\xi$ (1.28)

_holds for any non-negative function, which will take care the overlapping
issue._

For the case $(p-2)\alpha=0,$ like the proof in the Euclidean case, it is
sufficient to work with the annular decomposition

$A_{k}:=A_{2^{k},2^{k+1}}(0)=\\{\xi\in\mathbb{H}^{n}:2^{k}<d(\xi)\leq
2^{k+1}\\},\ \ k\in\mathbb{Z},$ (1.29)

of $\mathbb{H}^{n}$ for $Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)>0$.
Whereas, for the case $(p-2)\alpha\neq 0,$ we work with the decomposition
(1.23) under the additional condition that
$Q+\tau\gamma(1+(p-2)\frac{\alpha}{ps})>2$. In the proof of Theorem 1.2 and
Theorem 1.3 we use the decomposition (1.29) and we deal with the term
$|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}$ by including this in the integral
$\int_{\mathscr{B}_{k}}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f(\xi)|^{\tau}d\xi$
(see Lemma 4.4), where the condition $\tau\gamma\leq-2$ is crucial.

The paper is organized as follows: In Section 2 we discuss some basic
properties and embedding theorems of the space $W^{s,p,\alpha}(\Omega)$.
Section 3 contains the proof of Theorem 1.1. In Section 4 we first establish
some relevant results (Proposition 4.1 and Lemma 4.4), and then with the help
of those results we prove Theorem 1.2. We conclude the paper with the proof of
Theorem 1.3 in Section 5.

##  2\. Some basic properties of fractional Sobolev spaces

The following result gives the relation between the spaces
$W^{s,p,\alpha}(\Omega)$ and $W^{s^{\prime},p,\alpha}(\Omega)$ for
$s^{\prime}<s$. This result for the case $(p-2)\alpha=0$ is already
established in [11].

######  Proposition 2.1.

Let $1\leq p<\infty$ and $\alpha\in\mathbb{R}$, $0<s^{\prime}\leq s<1$ and
$\Omega$ be an open set in $\mathbb{H}^{n}$. Then for any measurable function
$f:\Omega\rightarrow\mathbb{R}$,

$\lVert f\rVert_{s^{\prime},p,\alpha,\Omega}\leq C\lVert
f\rVert_{s,p,\alpha,\Omega}$ (2.1)

for some positive constant $C=C(n,p,s,\alpha)$. In other word,
$W^{s,p,\alpha}(\Omega)\subseteq W^{s^{\prime},p,\alpha}(\Omega)$.

###### Proof.

We have

$[f]_{s^{\prime},p,\alpha,\Omega}^{p}=\int_{\Omega}\int_{\Omega\cap
D_{1}(\xi)}\frac{|f(\xi)-f(\xi^{\prime})|^{p}d\xi^{\prime}d\xi}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps^{\prime}}|z-z^{\prime}|^{(p-2)\alpha}}\\\
+\int_{\Omega}\int_{\Omega\cap
D_{1}(\xi)^{C}}\frac{|f(\xi)-f(\xi^{\prime})|^{p}d\xi^{\prime}d\xi}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps^{\prime}}|z-z^{\prime}|^{(p-2)\alpha}}=I_{1}+I_{2},$
(2.2)

where $D_{1}(\xi)$ is the set defined in (1.10). Since
$d({\xi}^{-1}\circ\xi^{\prime})<1$ in $D_{1}(\xi)$,

$I_{1}\leq\int_{\Omega}\int_{\Omega\cap
D_{1}(\xi)}\frac{|f(\xi)-f(\xi^{\prime})|^{p}d\xi^{\prime}d\xi}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps}|z-z^{\prime}|^{(p-2)\alpha}}\leq[f]_{s,p,\alpha,\Omega}^{p}.$
(2.3)

We now estimate the integral $I_{2}$.

$I_{2}\leq 2^{p-1}\int_{\Omega}\int_{\Omega\cap
D_{1}(\xi)^{C}}\frac{(|f(\xi)|^{p}+|f(\xi^{\prime})|^{p})d\xi^{\prime}d\xi}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps^{\prime}}|z-z^{\prime}|^{(p-2)\alpha}}.$

Using symmetry, we obtain

$I_{2}\leq 2^{p}\int_{\Omega}\left(\int_{\Omega\cap
D_{1}(\xi)^{C}}\frac{d\xi^{\prime}}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps^{\prime}}|z-z^{\prime}|^{(p-2)\alpha}}\right)|f(\xi)|^{p}d\xi.$
(2.4)

Let

$J=\int_{\Omega\cap
D_{1}(\xi)^{C}}\frac{d\xi^{\prime}}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps^{\prime}}|z-z^{\prime}|^{(p-2)\alpha}}\\\
\leq\int_{D_{1}(\xi)^{C}}\frac{dz^{\prime}dt^{\prime}}{(|z-z^{\prime}|^{4}+(t^{\prime}-t-2\langle
y,x^{\prime}\rangle+2\langle
x,y^{\prime}\rangle)^{2})^{\frac{Q+ps^{\prime}}{4}}|z-z^{\prime}|^{(p-2)\alpha}}.$

We use the change of variable $\zeta=z^{\prime}-z$ and
$\mu=t^{\prime}-t-2\langle y,x^{\prime}\rangle+2\langle x,y^{\prime}\rangle$,
and obtain

$J\leq\int_{D_{1}(0)^{C}}\frac{d\zeta
d\mu}{(|\zeta|^{4}+\mu^{2})^{\frac{Q+ps^{\prime}}{4}}|\zeta|^{(p-2)\alpha}}.$

Let us define the following sets:

$B_{1}=\\{(z,t)\in\mathbb{R}^{2n}\times\mathbb{R}:|z|>1,\,|t|>1\\},$
$B_{2}=\\{(z,t)\in\mathbb{R}^{2n}\times\mathbb{R}:|z|>1,\,|t|<1\\},$
$B_{3}=\\{(z,t)\in\mathbb{R}^{2n}\times\mathbb{R}:|z|<1,\,|t|>1\\},$

and the integrals

$J_{1}=\int_{B_{1}}\frac{d\zeta
d\mu}{(|\zeta|^{4}+\mu^{2})^{\frac{Q+ps^{\prime}}{4}}|\zeta|^{(p-2)\alpha}},$
$J_{2}=\int_{B_{2}}\frac{d\zeta
d\mu}{(|\zeta|^{4}+\mu^{2})^{\frac{Q+ps^{\prime}}{4}}|\zeta|^{(p-2)\alpha}},$
$J_{3}=\int_{B_{3}}\frac{d\zeta
d\mu}{(|\zeta|^{4}+\mu^{2})^{\frac{Q+ps^{\prime}}{4}}|\zeta|^{(p-2)\alpha}}.$

Clearly $D_{1}(0)^{C}=B_{1}\cup B_{2}\cup B_{3}$ up to a measure zero set, and
$J\leq J_{1}+J_{2}+J_{3}$.
Estimate for $J_{1}$: Let $\omega_{2n}$ be the measure of the unit sphere in
$\mathbb{R}^{2n}$.

$J_{1}=2\omega_{2n}\int_{1}^{\infty}\int_{1}^{\infty}\frac{r^{Q-3}drd\mu}{(r^{4}+\mu^{2})^{\frac{Q+ps^{\prime}}{4}}r^{(p-2)\alpha}}\\\
(\mathrm{using}\ \mu=r^{2}t)\
=2\omega_{2n}\int_{1}^{\infty}\left(\int_{\frac{1}{r^{2}}}^{\infty}\frac{dt}{(1+t^{2})^{\frac{Q+ps^{\prime}}{4}}}\right)\frac{dr}{r^{1+ps^{\prime}+(p-2)\alpha}}\\\
\leq
2\omega_{2n}\int_{1}^{\infty}\left(\int_{0}^{\infty}\frac{dt}{(1+t^{2})^{\frac{Q+ps^{\prime}}{4}}}\right)\frac{dr}{r^{1+ps^{\prime}+(p-2)\alpha}}.$

Since $Q+ps^{\prime}>2$ and $ps^{\prime}+(p-2)\alpha>0$, both the integrals of
R.H.S is finite. Hence $J_{1}\leq C_{1}$.
Estimate for $J_{2}$:

$J_{2}=2\omega_{2n}\int_{1}^{\infty}\int_{0}^{1}\frac{r^{Q-3}drd\mu}{(r^{4}+\mu^{2})^{\frac{Q+ps^{\prime}}{4}}r^{(p-2)\alpha}}\\\
(\mathrm{using}\ \mu=r^{2}t)\
=2\omega_{2n}\int_{1}^{\infty}\left(\int_{0}^{\frac{1}{r^{2}}}\frac{dt}{(1+t^{2})^{\frac{Q+ps^{\prime}}{4}}}\right)\frac{dr}{r^{1+ps^{\prime}+(p-2)\alpha}}\\\
\leq
2\omega_{2n}\int_{1}^{\infty}\left(\int_{0}^{1}\frac{dt}{(1+t^{2})^{\frac{Q+ps^{\prime}}{4}}}\right)\frac{dr}{r^{1+ps^{\prime}+(p-2)\alpha}}\leq
C_{2}.$

Estimate for $J_{3}$:

$J_{3}\leq
2\omega_{2n}\int_{0}^{1}\left(\int_{1}^{\infty}\frac{d\mu}{\mu^{\frac{Q+ps^{\prime}}{2}}}\right)r^{Q-3-(p-2)\alpha}dr.$

Since $\frac{Q+ps^{\prime}}{2}>1$ and $(p-2)\alpha<Q-2$, we have $J_{3}\leq
C_{3}$. Hence $J$ is integrable.
Hence from (2.4), we get

$I_{2}\leq 2^{p}C\lVert f\rVert_{L^{p}(\Omega)}^{p}.$ (2.5)

Combining (2.2), (2.3) and (2.5), we obtain

$\lVert f\rVert_{s^{\prime},p,\alpha,\Omega}^{p}\leq(2^{p}C+1)\lVert
f\rVert_{L^{p}(\Omega)}^{p}+[f]_{s,p,\alpha,\Omega}^{p}\leq
C(n,p,s,\alpha)\lVert f\rVert_{s,p,\alpha,\Omega}^{p}.$ (2.6)

∎

######  Definition 1.

Let $1\leq p<\infty$, $s\in(0,1)$ and $\alpha\in\mathbb{R}$, and let
$\Omega\subset\mathbb{H}^{n}$ be an open set. We say that $\Omega$ is an
extension domain, if there exists a positive constant
$C=C(n,p,s,\alpha,\Omega)$ such that, for any function $f\in
W^{s,p,\alpha}(\Omega)$, there exists $\bar{f}\in
W_{0}^{s,p,\alpha}(\mathbb{H}^{n})$ with $\bar{f}|_{\Omega}=f$, and

$\lVert\bar{f}\rVert_{s,p,\alpha}\leq C\lVert f\rVert_{s,p,\alpha,\Omega}.$
(2.7)

In the following result we extend any function of $W^{s,p,\alpha}(\Omega)$,
supported in $K\Subset\Omega$ to a function in
$W^{s,p,\alpha}(\mathbb{H}^{n})$.

######  Proposition 2.2.

Let $1\leq p<\infty$, $s\in(0,1)$, $\alpha\in\mathbb{R}$ and $\Omega$ be any
open set in $\mathbb{H}^{n}$. If $f\in W^{s,p,\alpha}(\Omega)$ and $K$ is any
compact subset of $\Omega$ such that $f\equiv 0$ in $\Omega\setminus K$, then
the extension function

$\bar{f}(\xi)=\begin{cases}f(\xi)&\text{if $x\in\Omega$}\\\ 0&\text{if
$x\in\mathbb{H}^{n}\setminus\Omega$}\end{cases}$

belongs to $W^{s,p,\alpha}(\mathbb{H}^{n})$, and there exists a positive
constant $C$ depending on $n$, $p$, $s$, $\alpha$, $K$, $\Omega$ and is such
that

$\lVert\bar{f}\rVert_{s,p,\alpha}\leq C\lVert f\rVert_{s,p,\alpha,\Omega}.$
(2.8)

###### Proof.

We have

$[\bar{f}]_{s,p,\alpha}^{p}=\int_{\mathbb{H}^{n}}\int_{\mathbb{H}^{n}}\frac{|\bar{f}(\xi)-\bar{f}(\xi^{\prime})|^{p}d\xi^{\prime}d\xi}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps}|z-z^{\prime}|^{(p-2)\alpha}}\\\
=\int_{\Omega}\int_{\Omega}\frac{|f(\xi)-f(\xi^{\prime})|^{p}d\xi^{\prime}d\xi}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps}|z-z^{\prime}|^{(p-2)\alpha}}+2\int_{K}\int_{\Omega^{C}}\frac{|f(\xi)|^{p}d\xi^{\prime}d\xi}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps}|z-z^{\prime}|^{(p-2)\alpha}}\\\
=[f]^{p}_{s,p,\alpha,\Omega}+2\int_{K}|f(\xi)|^{p}\int_{\Omega^{C}}\frac{d\xi^{\prime}}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps}|z-z^{\prime}|^{(p-2)\alpha}}d\xi.$
(2.9)

We claim that $\exists\,\beta>0$ such that
$d({\xi}^{-1}\circ\xi^{\prime})\geq\beta$ for all $\xi\in K$ and
$\xi^{\prime}\in\Omega^{C}$. If not, then $\exists\,\xi\in K$ and
$\\{\xi_{m}^{\prime}\\}\subset\Omega^{C}$ such that
$d({\xi}^{-1}\circ\xi_{m}^{\prime})<\frac{1}{m}$ $\forall m\in\mathbb{N}$. In
particular, $|z-z_{m}^{\prime}|<\frac{1}{m}$ and $|t_{m}^{\prime}-t-2\langle
y,x_{m}^{\prime}\rangle+2\langle x,y_{m}^{\prime}\rangle|<\frac{1}{m^{2}}$,
where
$\xi_{m}^{\prime}=(x_{m}^{\prime},y_{m}^{\prime},t_{m}^{\prime})=(z_{m}^{\prime},t_{m}^{\prime})$.
Now

$|t_{m}^{\prime}-t|\leq|t_{m}^{\prime}-t-2\langle
y,x_{m}^{\prime}\rangle+2\langle x,y_{m}^{\prime}\rangle|+2|\langle
y,x_{m}^{\prime}\rangle-\langle x,y_{m}^{\prime}\rangle|\\\
=|t_{m}^{\prime}-t-2\langle y,x_{m}^{\prime}\rangle+2\langle
x,y_{m}^{\prime}\rangle|+2|\langle(-y,x),z-z_{m}^{\prime}\rangle|\\\
\leq|t_{m}^{\prime}-t-2\langle y,x_{m}^{\prime}\rangle+2\langle
x,y_{m}^{\prime}\rangle|+2|z||z-z_{m}^{\prime}|\leq\frac{1}{m^{2}}+\frac{2R}{m},$

where $R>0$ is such that $K\subset B(0,R)$, which implies that the sequence
$\\{\xi_{m}^{\prime}\\}$ converges to $\xi$ and contradicts the fact $K$ is a
compact subset of $\Omega$.
Hence we have for any $\xi\in K$,

$\int_{\Omega^{C}}\frac{d\xi^{\prime}}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps}|z-z^{\prime}|^{(p-2)\alpha}}\\\
\leq\int_{B_{\beta}(\xi)^{C}}\frac{d\xi^{\prime}}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps}|z-z^{\prime}|^{(p-2)\alpha}}\leq\int_{B_{\beta}(0)^{C}}\frac{d\xi^{\prime}}{d(\xi^{\prime})^{Q+ps}|z^{\prime}|^{(p-2)\alpha}}.$

One can show the last integral is finite by an argument similar to one as in
Proposition 2.1. This gives

$[\bar{f}]_{s,p,\alpha}^{p}\leq[f]^{p}_{s,p,\alpha,\Omega}+C\lVert
f\rVert_{L^{p}(\Omega)}^{p}.$ (2.10)

This completes the proof. ∎

######  Corollary 1.

Let $1\leq p<\infty$, $s\in(0,1)$ and $\alpha\in\mathbb{R}$ be such that
$ps+(p-2)\alpha<Q$. Let $\Omega\subset\mathbb{H}^{n}$ be an open set and $K$
be any compact subset of $\Omega$. Then there exists a positive constant $C$,
which depends on $n$, $p$, $s$, $\alpha$ ,$K,$ $\Omega$ and is such that

$\lVert f\rVert_{L^{p^{*}}(K)}\leq C\lVert f\rVert_{s,p,\alpha,\Omega},\
\forall f\in C(\bar{\Omega}),$ (2.11)

where $p^{*}=\frac{Qp}{Q-ps-(p-2)\alpha}$.

###### Proof.

Let $K^{\prime}$ be an open set such that $K\Subset K^{\prime}\Subset\Omega$.
Choose a function $\phi\in C_{c}^{\infty}(\mathbb{H}^{n})$ such that
$0\leq\phi\leq 1$, $\phi\equiv 1$ in $K$ and $\phi\equiv 0$ in
$\mathbb{H}^{n}\setminus K^{\prime}$.
Clearly $f\phi\in C_{c}(\mathbb{H}^{n})$. Sobolev inequality (1.12) gives,

$\left(\int_{K}|f|^{p^{*}}\right)^{\frac{1}{p^{*}}}\leq\left(\int_{\mathbb{H}^{n}}|f\phi|^{p^{*}}\right)^{\frac{1}{p^{*}}}\leq
C[f\phi]_{s,p,\alpha}.$ (2.12)

It follows from (2.8),

$\lVert f\rVert_{L^{p^{*}}(K)}\leq C\lVert f\phi\rVert_{s,p,\alpha,\Omega}\leq
C(\lVert f\rVert_{L^{p}(\Omega)}+[f\phi]_{s,p,\alpha,\Omega}).$ (2.13)

We now estimate the quantity $[f\phi]_{s,p,\alpha,\Omega}$ in terms of $\lVert
f\rVert_{s,p,\alpha,\Omega}$.

$[f\phi]^{p}_{s,p,\alpha,\Omega}\leq
2^{p-1}\int_{\Omega}|f(\xi)|^{p}\int_{\Omega}\frac{|\phi(\xi)-\phi(\xi^{\prime})|^{p}d\xi^{\prime}d\xi}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps}|z-z^{\prime}|^{(p-2)\alpha}}\\\
+2^{p-1}\int_{\Omega}\int_{\Omega}\frac{|f(\xi)-f(\xi^{\prime})|^{p}|\phi(\xi^{\prime})|^{p}d\xi^{\prime}d\xi}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps}|z-z^{\prime}|^{(p-2)\alpha}}\\\
\leq
2^{2p-1}\int_{\Omega}|f(\xi)|^{p}\int_{B(\xi,1)}\frac{|\phi(\xi)-\phi(\xi^{\prime})|^{p}d\xi^{\prime}d\xi}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps}|z-z^{\prime}|^{(p-2)\alpha}}\\\
+2^{2p-1}\int_{\Omega}|f(\xi)|^{p}\int_{B(\xi,1)^{C}}\frac{d\xi^{\prime}}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps}|z-z^{\prime}|^{(p-2)\alpha}}d\xi+2^{p-1}[f]^{p}_{s,p,\alpha,\Omega}.$
(2.14)

In the last integral we use $\phi\leq 1$. Since $\phi\in
C_{c}^{\infty}(\mathbb{H}^{n})$, $\exists M>0$ such that
$\frac{|\phi(\xi)-\phi(\xi^{\prime})|}{d({\xi}^{-1}\circ\xi^{\prime})}\leq M$
for all $\xi,\xi^{\prime}\in\Omega$, (see [1], Proposition 3.1). Hence for any
$\xi\in\Omega$,

$\int_{B(\xi,1)}\frac{|\phi(\xi)-\phi(\xi^{\prime})|^{p}d\xi^{\prime}}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps}|z-z^{\prime}|^{(p-2)\alpha}}\leq
M^{p}\int_{B(\xi,1)}\frac{d\xi^{\prime}}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps-p}|z-z^{\prime}|^{(p-2)\alpha}}.$

Both the integrals

$\int_{B(\xi,1)}\frac{d\xi^{\prime}}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps-p}|z-z^{\prime}|^{(p-2)\alpha}}d\xi$

and

$\int_{B(\xi,1)^{C}}\frac{d\xi^{\prime}}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps}|z-z^{\prime}|^{(p-2)\alpha}}$

are shown to be finite in [1], (Proposition 3.1). Hence

$[f\phi]^{p}_{s,p,\alpha,\Omega}\leq C\left(\lVert
f\rVert_{L^{p}(\Omega)}^{p}+[f]^{p}_{s,p,\alpha,\Omega}\right)\leq C\lVert
f\rVert_{s,p,\alpha,\Omega}^{p}.$ (2.15)

Inequality (2.11) follows from (2.13) and (2.15). ∎

######  Theorem 2.3.

Let $1\leq p<\infty$, $s\in(0,1)$ and $\alpha\in\mathbb{R}$ be such
$ps+(p-2)\alpha<Q$. Let $\Omega\subset\mathbb{H}^{n}$ be an extension domain.
Then there exists a positive constant $C=C(n,p,s,\alpha,\Omega)$ such that for
any $q\in[p,p^{*}]$ and for any $f\in W^{s,p,\alpha}(\Omega)$, we have

$\lVert f\rVert_{L^{q}(\Omega)}\leq C\lVert f\rVert_{s,p,\alpha,\Omega}.$
(2.16)

In other word, $W^{s,p,\alpha}(\Omega)$ is continuously embedded in
$L^{q}(\Omega)$ for any $q\in[p,p^{*}]$.

###### Proof.

The case $q=p$ and $q=p^{*}$ is trivial. For the case $q\in(p,p^{*})$, choose
$0<\theta<1$ such that $\frac{1}{q}=\frac{\theta}{p}+\frac{1-\theta}{p^{*}}$.
Inequality (2.16) follows from Hölder’s inequality with the conjugets
$\frac{p}{\theta q}$ and $\frac{p^{*}}{(1-\theta)q}$, (1.12) and (2.7). ∎

######  Remark 2.

When $ps+(p-2)\alpha\rightarrow Q$, clearly the fractional critical Sobolev
exponent $p^{*}$ goes to $\infty$. Hence if $ps+(p-2)\alpha=Q$, one may expect
that $W^{s,p,\alpha}(\Omega)$ is continuously embedded in $L^{q}(\Omega)$ for
any $q\geq p$. Our next result assures that this embedding is true even in the
case $ps+(p-2)\alpha\geq Q$.

######  Theorem 2.4.

Let $1\leq p<\infty$, $s\in(0,1)$ and $\alpha\in\mathbb{R}$ be such
$ps+(p-2)\alpha\geq Q$. Let $\Omega\subset\mathbb{H}^{n}$ be an extension
domain. Then there exists a positive constant $C=C(n,p,s,\alpha,\Omega)$ such
that for any $q\geq p$ and for any $f\in W^{s,p,\alpha}(\Omega)$, we have

$\lVert f\rVert_{L^{q}(\Omega)}\leq C\lVert f\rVert_{s,p,\alpha,\Omega}.$
(2.17)

###### Proof.

Since $(p-2)\alpha<Q-2$, we can choose $0<s^{\prime}<s$ such that
$ps^{\prime}+(p-2)\alpha<Q$ and $q\leq\frac{pQ}{Q-ps^{\prime}-(p-2)\alpha}$.
By Theorem 2.3, we have

$\lVert f\rVert_{L^{q}(\Omega)}\leq C\lVert
f\rVert_{s^{\prime},p,\alpha,\Omega}.$ (2.18)

Inequality (2.17) then follows from Proposition 2.1. ∎

##  3\. Proof of Theorem 1.1

The following lemma, which uses Corollary 1 is key to the proof of Theorem
1.1.

######  Lemma 3.1.

Let $p>1$, $s\in(0,1)$, $\alpha\in\mathbb{R}$,
$\tau>1$,$\,\gamma\in\mathbb{R}$, $0<a\leq 1$ and $q>(1-a)\tau$ satisfy (1.14)
and (1.15). Let $\lambda>0$, $0<r^{\prime}<r<R<R^{\prime}$,
$\mu,\mu^{\prime}\in\mathbb{R}$ and $\ell,\ell^{\prime}>0$ be such that
$\mu^{\prime}<\mu$ and $\mu+\ell<\mu^{\prime}+\ell^{\prime}$ and consider the
following sets:

$\mathscr{D}_{\lambda}=\mathscr{D}_{r,R,\mu,\ell}(\lambda):=\\{\xi=(z,t)\in\mathbb{H}^{n}:\lambda
r<|z|<\lambda R\ \mathrm{and}\ \ \mu\lambda^{2}<t<(\mu+\ell)\lambda^{2}\\}$

and
$\mathscr{D}_{\lambda}^{\prime}=\mathscr{D}_{r^{\prime},R^{\prime},\mu^{\prime},\ell^{\prime}}(\lambda)$.
Then for $f\in C(\bar{\mathscr{D}}_{\lambda}^{\prime})$, we have

$\left(\fint_{\mathscr{D}_{\lambda}}|f-(f)_{\mathscr{D}_{\lambda}^{\prime}}|^{\tau}\right)^{\frac{1}{\tau}}\leq
C\left(\lambda^{ps+(p-2)\alpha-Q}[f]_{s,p,\alpha,{\mathscr{D}_{\lambda}^{\prime}}}^{p}\right)^{\frac{a}{p}}\left(\fint_{\mathscr{D}_{\lambda}^{\prime}}|f|^{q}d\xi\right)^{\frac{1-a}{q}},$
(3.1)

where the constant $C>0$ is independent of $f$ and $\lambda$, $\mu$ and
$\mu^{\prime}$.

###### Proof.

From (1.14), we have

$\frac{1}{\tau}=a\left(\frac{1}{p}-\frac{s^{\prime}\left(1+(p-2)\frac{\alpha}{ps}\right)}{Q}\right)+\frac{1-a}{q},$

where $s^{\prime}=s+\gamma/a<s$. Let $\alpha^{\prime}=\alpha s^{\prime}/s$ and
${p^{\prime}}^{*}=\frac{pQ}{Q-ps^{\prime}-(p-2)\alpha^{\prime}}$. We have
$1/\tau=a/{p^{\prime}}^{*}+(1-a)/q$ and ${p^{\prime}}^{*}>0$ follows from the
range of $q$.
For any function $f$, we define the function
$\tilde{f}(\xi)=f(D_{\lambda}\xi)$, where $D_{\lambda}\xi=(\lambda
z,\lambda^{2}t)$, is the dilation in $\mathbb{H}^{n}$. Let us define the sets

$\mathscr{D}_{1}=\\{\xi=(z,t)\in\mathbb{H}^{n}:r<|z|\leq R\ \mathrm{and}\ \
\mu<t\leq\mu+\ell\\},$
$\mathscr{D}_{1}^{\prime}=\\{\xi=(z,t)\in\mathbb{H}^{n}:r^{\prime}<|z|\leq
R^{\prime}\ \mathrm{and}\ \ \mu^{\prime}<t\leq\mu^{\prime}+\ell^{\prime}\\}.$

Clearly, $f\in C^{1}(\bar{\mathscr{D}}_{\lambda}^{\prime})$ if and only if
$\tilde{f}\in C^{1}(\bar{\mathscr{D}}_{1}^{\prime})$. Hölder’s inequality,
inequality (2.11) with $K=\bar{\mathscr{D}}_{1}$ and
$\Omega=\mathscr{D}_{1}^{\prime}$, and the fact
$[\tilde{f}]_{s^{\prime},p,\alpha^{\prime},\mathscr{D}_{1}^{\prime}}\leq
C[\tilde{f}]_{s,p,\alpha,\mathscr{D}_{1}^{\prime}}$ respectively gives

$\left(\int_{\mathscr{D}_{1}}|\tilde{f}-(f)_{\mathscr{D}_{\lambda}^{\prime}}|^{\tau}\right)^{\frac{1}{\tau}}\leq\left(\int_{\mathscr{D}_{1}}|\tilde{f}-(f)_{\mathscr{D}_{\lambda}^{\prime}}|^{{p^{\prime}}^{*}}\right)^{\frac{a}{{p^{\prime}}^{*}}}\left(\int_{\mathscr{D}_{1}}|\tilde{f}-(f)_{\mathscr{D}_{\lambda}^{\prime}}|^{q}\right)^{\frac{1-a}{q}}\\\
\leq
C\left(\lVert\tilde{f}-(f)_{\mathscr{D}_{\lambda}^{\prime}}\rVert_{L^{p}(\mathscr{D}_{1}^{\prime})}^{p}+[\tilde{f}]_{s^{\prime},p,\alpha^{\prime},\mathscr{D}_{1}^{\prime}}^{p}\right)^{\frac{a}{p}}\left(\int_{\mathscr{D}_{1}}|\tilde{f}-(f)_{\mathscr{D}_{\lambda}^{\prime}}|^{q}\right)^{\frac{1-a}{q}}\\\
\leq
C\left(\lVert\tilde{f}-(f)_{\mathscr{D}_{\lambda}^{\prime}}\rVert_{L^{p}(\mathscr{D}_{1}^{\prime})}^{p}+[\tilde{f}]_{s,p,\alpha,\mathscr{D}_{1}^{\prime}}^{p}\right)^{\frac{a}{p}}\left(\int_{\mathscr{D}_{1}}|\tilde{f}-(f)_{\mathscr{D}_{\lambda}^{\prime}}|^{q}\right)^{\frac{1-a}{q}}.$
(3.2)

Using the change of variable $\bar{\xi}=D_{\lambda}\xi$ in (3.2) and obtain

$\left(\fint_{\mathscr{D}_{\lambda}}|f-(f)_{\mathscr{D}_{\lambda}^{\prime}}|^{\tau}\right)^{\frac{1}{\tau}}\\\
\leq
C\left(\fint_{\mathscr{D}_{\lambda}^{\prime}}|f-(f)_{\mathscr{D}_{\lambda}^{\prime}}|^{p}+\lambda^{ps+(p-2)\alpha-Q}[f]^{p}_{s,p,\alpha,\mathscr{D}_{\lambda}^{\prime}}\right)^{\frac{a}{p}}\left(\fint_{\mathscr{D}_{\lambda}}|f-(f)_{\mathscr{D}_{\lambda}^{\prime}}|^{q}\right)^{\frac{1-a}{q}}\\\
\leq
C\left(\fint_{\mathscr{D}_{\lambda}^{\prime}}|f-(f)_{\mathscr{D}_{\lambda}^{\prime}}|^{p}+\lambda^{ps+(p-2)\alpha-Q}[f]^{p}_{s,p,\alpha,\mathscr{D}_{\lambda}^{\prime}}\right)^{\frac{a}{p}}\left(\fint_{\mathscr{D}_{\lambda}^{\prime}}|f|^{q}\right)^{\frac{1-a}{q}}.$
(3.3)

Now

$\int_{\mathscr{D}_{\lambda}^{\prime}}|f(\xi)-(f)_{\mathscr{D}_{\lambda}^{\prime}}|^{p}d\xi=\int_{\mathscr{D}_{\lambda}^{\prime}}\left|f(\xi)-\frac{1}{|{\mathscr{D}_{\lambda}^{\prime}}|}\int_{\mathscr{D}_{\lambda}^{\prime}}f(\xi^{\prime})d\xi^{\prime}\right|^{p}d\xi\\\
=\int_{\mathscr{D}_{\lambda}^{\prime}}\left|\frac{1}{|{\mathscr{D}_{\lambda}^{\prime}}|}\int_{\mathscr{D}_{\lambda}^{\prime}}\left(f(\xi)-f(\xi^{\prime})\right)d\xi^{\prime}\right|^{p}d\xi\leq\int_{\mathscr{D}_{\lambda}^{\prime}}\left(\fint_{\mathscr{D}_{\lambda}^{\prime}}\left|f(\xi)-f(\xi^{\prime})\right|d\xi^{\prime}\right)^{p}d\xi.$
(3.4)

Using Jensen’s inequality, we get

$\left(\fint_{\mathscr{D}_{\lambda}^{\prime}}\left|f(\xi)-f(\xi^{\prime})\right|d\xi^{\prime}\right)^{p}\leq\frac{1}{|{\mathscr{D}_{\lambda}^{\prime}}|}\int_{\mathscr{D}_{\lambda}^{\prime}}\left|f(\xi)-f(\xi^{\prime})\right|^{p}d\xi^{\prime}.$
(3.5)

Let $\xi,\xi^{\prime}\in\mathscr{D}_{\lambda}^{\prime}$, where
$\xi=(z,t)=(x,y,t)$ and
$\xi^{\prime}=(z^{\prime},t^{\prime})=(x^{\prime},y^{\prime},t^{\prime})$. We
have

$d({\xi}^{-1}\circ\xi^{\prime})=\left(|z^{\prime}-z|^{4}+(t^{\prime}-t+2(\langle
x,y^{\prime}\rangle-\langle y,x^{\prime}\rangle))^{2}\right)^{\frac{1}{4}}\\\
\leq\left(|z^{\prime}-z|^{4}+2(t^{\prime}-t)^{2}+8(\langle
x,y^{\prime}\rangle-\langle y,x^{\prime}\rangle)^{2}\right)^{\frac{1}{4}}\\\
=\left(|z^{\prime}-z|^{4}+2(t^{\prime}-t)^{2}+8(\langle(y^{\prime},-x^{\prime}),z-z^{\prime}\rangle)^{2}\right)^{\frac{1}{4}}.$

Cauchy-Schwarz inequality in $\mathbb{R}^{2n}$ gives

$d({\xi}^{-1}\circ\xi^{\prime})\leq\left(|z^{\prime}-z|^{4}+2(t^{\prime}-t)^{2}+8|z^{\prime}|^{2}|z-z^{\prime}|^{2}\right)^{\frac{1}{4}}\\\
\leq((2{R^{\prime}})^{4}\lambda^{4}+2\cdot{\ell^{\prime}}^{2}\lambda^{4}+8{R^{\prime}}^{2}(2{R^{\prime}})^{2}\lambda^{4})^{\frac{1}{4}}=C\lambda.$

Note that $|{\mathscr{D}_{\lambda}^{\prime}}|=C\lambda^{Q}$. Hence, from (3.4)
and (3.5), we get

$\int_{\mathscr{D}_{\lambda}^{\prime}}|f(\xi)-(f)_{\mathscr{D}_{\lambda}^{\prime}}|^{p}d\xi\\\
\leq
C\lambda^{ps+(p-2)\alpha}\int_{\mathscr{D}_{\lambda}^{\prime}}\int_{\mathscr{D}_{\lambda}^{\prime}}\frac{|f(\xi)-f(\xi^{\prime})|^{p}}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps}|z-z^{\prime}|^{(p-2)\alpha}}d\xi=C\lambda^{ps+(p-2)\alpha}[f]_{s,p,\alpha,{\mathscr{D}_{\lambda}^{\prime}}}^{p},$
(3.6)

which gives

$\fint_{\mathscr{D}_{\lambda}^{\prime}}|f(\xi)-(f)_{\mathscr{D}_{\lambda}^{\prime}}|^{p}d\xi\leq
C\lambda^{ps+(p-2)\alpha-Q}[f]_{s,p,\alpha,{\mathscr{D}_{\lambda}^{\prime}}}^{p}.$
(3.7)

Inequality (3.1) follows from (3.3) and (3.7). ∎

######  Remark 3.

_For $i=0,1,2$ and $3$ we consider the connected domain
$\mathscr{B}_{k+1,j}\cup\mathscr{B}_{k,4j+i}$ as defined in (1.27), and we
define the set_

$\mathscr{C}_{k,j}:=\\{\xi=(z,t)\in\mathbb{H}^{n}:2^{k-2}<|z|\leq 2^{k+3},\
(j-2)2^{2k}<t<(j+3)2^{2k}\\}.$

_Observe that
$\mathscr{B}_{k+1,j}\cup\mathscr{B}_{k,4j+i}\Subset\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i}$
and the following inequality is true:_

$\left(\fint_{\mathscr{B}_{k+1,j}\cup\mathscr{B}_{k,4j+i}}|f-(f)_{\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i}}|^{\tau}\right)^{\frac{1}{\tau}}\\\
\leq
C\left(2^{k(ps+(p-2)\alpha-Q)}[f]_{s,p,\alpha,\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i}}^{p}\right)^{\frac{a}{p}}\left(\fint_{\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i}}|f|^{q}d\xi\right)^{\frac{1-a}{q}},$
(3.8)

_where the constant $C$ is independent of $k$ and $j$._

The following elementary lemma is established in [19].

######  Lemma 3.2.

Let $\tau>1$ and $c>1$. Then for all $a,b\in\mathbb{R}$, we have

$(|a|+|b|)^{\tau}\leq
c|a|^{\tau}+\left(1-c^{-\frac{1}{\tau-1}}\right)^{1-\tau}|b|^{\tau}.$ (3.9)

###### Proof of Theorem 1.1.

Step 1: Proof of (i). Let $\mathrm{supp}f\subset D_{2^{n_{0}+1}}(0)$ for some
$n_{0}\in\mathbb{Z}$. For $k,j\in\mathbb{Z}$, we consider the sets
$\\{\mathscr{A}_{k}\\}$, $\\{\mathscr{A}_{k,j}\\}$ and $\mathscr{B}_{k,j}$
defined in (1.22), (1.23) and (1.27) respectively.
By Lemma 3.1 and by the relation (1.14) we have

$\left(\fint_{\mathscr{A}_{k,j}}|f-(f)_{\mathscr{B}_{k,j}}|^{\tau}d\xi\right)^{\frac{1}{\tau}}\leq
C\left(2^{k(ps+(p-2)\alpha-Q)}[f]_{s,p,\alpha,\mathscr{B}_{k,j}}^{p}\right)^{\frac{a}{p}}\left(\frac{1}{2^{kQ}}\int_{\mathscr{B}_{k,j}}|f|^{q}d\xi\right)^{\frac{1-a}{q}}\\\
=C\,2^{-\frac{k}{\tau}\left(Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)\right)}[f]_{s,p,\alpha,\mathscr{B}_{k,j}}^{a}\lVert
f\rVert_{L^{q}(\mathscr{B}_{k,j})}^{1-a}.$ (3.10)

Observe that
$\mathrm{supp}f\cap\mathscr{A}_{k}\subset\cup_{j=-n_{k}}^{n_{k}-1}\mathscr{A}_{k,j}$,
where $n_{k}=2^{2(n_{0}+1-k)}$ and $|\mathscr{A}_{k,j}|=C2^{kQ}$ for all $k$
and $j$. Using the fact that, $d(\xi)\geq|z|>2^{k}$ in $\mathscr{A}_{k,j}$ for
all $j$, we obtain

$\int_{\mathscr{A}_{k,j}}d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f|^{\tau}d\xi\leq
C2^{k\left(Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)\right)}\left(\fint_{\mathscr{A}_{k,j}}|f-(f)_{\mathscr{B}_{k,j}}|^{\tau}+|(f)_{\mathscr{B}_{k,j}}|^{\tau}\right)\\\
\leq C[f]_{s,p,\alpha,\mathscr{B}_{k,j}}^{a\tau}\lVert
f\rVert_{L^{q}(\mathscr{B}_{k,j})}^{(1-a)\tau}+C\,2^{k\left(Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)\right)}|(f)_{\mathscr{B}_{k,j}}|^{\tau}.$
(3.11)

For $t_{1},t_{2}\geq 0$ with $t_{1}+t_{2}\geq 1$, and for $x_{k}\geq 0$ and
$y_{k}\geq 0$, one has

$\sum_{k\in\mathbb{Z}}x_{k}^{t_{1}}y_{k}^{t_{2}}\leq\left(\sum_{k\in\mathbb{Z}}x_{k}\right)^{t_{1}}\left(\sum_{k\in\mathbb{Z}}y_{k}\right)^{t_{2}}.$
(3.12)

We have $\frac{1}{\tau}\leq\frac{a}{p}+\frac{1-a}{q}$, which follows from
(1.14) and (1.15). Summation with respect to $j$ from $-n_{k}$ to $n_{k}-1$ on
(3.11) and then using the inequality (3.12) with $t_{1}=\frac{\tau a}{p}$ and
$t_{2}=\frac{\tau(1-a)}{q}$ we get

$\int_{\mathscr{A}_{k}}d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f|^{\tau}d\xi\leq
C\left(\sum_{j=-n_{k}}^{n_{k}-1}[f]_{s,p,\alpha,\mathscr{B}_{k,j}}^{p}\right)^{\frac{a\tau}{p}}\left(\sum_{j=-n_{k}}^{n_{k}-1}\lVert
f\rVert_{L^{q}(\mathscr{B}_{k,j})}^{q}\right)^{\frac{(1-a)\tau}{q}}\\\
+C2^{k\left(Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)\right)}\sum_{j=-n_{k}}^{n_{k}-1}|(f)_{\mathscr{B}_{k,j}}|^{\tau}.$
(3.13)

Let $m\in\mathbb{Z}$ be such that $n_{0}-m\geq 2$. Again summation with
respect to $k$ from $m$ to $n_{0}$ and then (3.12) gives

$\int_{\\{\xi\in\mathbb{H}^{n}:|z|>2^{m}\\}}d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f|^{\tau}d\xi\\\
\leq
C\left(\sum_{k=m}^{n_{0}}\sum_{j=-n_{k}}^{n_{k}-1}[f]_{s,p,\alpha,\mathscr{B}_{k,j}}^{p}\right)^{\frac{a\tau}{p}}\left(\sum_{k=m}^{n_{0}}\sum_{j=-n_{k}}^{n_{k}-1}\lVert
f\rVert_{L^{q}(\mathscr{B}_{k,j})}^{q}\right)^{\frac{(1-a)\tau}{q}}\\\
+C\sum_{k=m}^{n_{0}}2^{k\left(Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)\right)}\sum_{j=-n_{k}}^{n_{k}-1}|(f)_{\mathscr{B}_{k,j}}|^{\tau}.$

It follows from Remark 1,

$\int_{\\{\xi\in\mathbb{H}^{n}:|z|>2^{m}\\}}d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f|^{\tau}d\xi\\\
\leq C[f]_{s,p,\alpha}^{a\tau}\lVert
f\rVert_{L^{q}(\mathbb{H}^{n})}^{(1-a)\tau}+C\sum_{k=m}^{n_{0}}2^{k\left(Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)\right)}\sum_{j=-n_{k}}^{n_{k}-1}|(f)_{\mathscr{B}_{k,j}}|^{\tau}.$
(3.14)

Observe that, $n_{k}=4n_{k+1}$. For $-n_{k+1}\leq j\leq n_{k+1}-1$ and for
$i=0,1,2$ and $3$, we have

$(f)_{\mathscr{B}_{k,4j+i}}-(f)_{\mathscr{B}_{k+1,j}}=\frac{1}{|\mathscr{B}_{k,4j+i}|}\int_{\mathscr{B}_{k,4j+i}}f-\frac{1}{|\mathscr{B}_{k+1,j}|}\int_{\mathscr{B}_{k+1,j}}f\\\
=\frac{1}{|\mathscr{B}_{k,4j+i}|}\int_{\mathscr{B}_{k,4j+i}}\left(f-(f)_{\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i}}\right)-\frac{1}{|\mathscr{B}_{k+1,j}|}\int_{\mathscr{B}_{k+1,j}}\left(f-(f)_{\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i}}\right).$

The set $\mathscr{C}_{k,j}$ is defined in Remark 3. Since
$|\mathscr{B}_{k+1,j}|\sim|\mathscr{B}_{k,4j+i}|$, we have

$|(f)_{\mathscr{B}_{k,4j+i}}-(f)_{\mathscr{B}_{k+1,j}}|\leq\frac{C}{|\mathscr{B}_{k+1,j}\cup\mathscr{B}_{k,4j+i}|}\int_{\mathscr{B}_{k+1,j}\cup\mathscr{B}_{k,4j+i}}\left|f-(f)_{\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i}}\right|\\\
\leq
C\left(\fint_{\mathscr{B}_{k+1,j}\cup\mathscr{B}_{k,4j+i}}\left|f-(f)_{\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i}}\right|^{\tau}\right)^{\frac{1}{\tau}},$
(3.15)

using Jensen’s inequality. Inequality (3.8) gives

$|(f)_{\mathscr{B}_{k,4j+i}}-(f)_{\mathscr{B}_{k+1,j}}|\leq
C\,2^{-\frac{k}{\tau}\left(Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)\right)}[f]_{s,p,\alpha,\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i}}^{a}\lVert
f\rVert_{L^{q}(\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i})}^{(1-a)}.$ (3.16)

For simplicity of the notation we denote
$Q^{\prime}=Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)$. Note that
$Q^{\prime}>2$, which follows from the assumption
$1/\tau+\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)/{Q}>\frac{2}{\tau Q}$. Let
$\delta=2/\big{(}1+2^{Q^{\prime}-2}\big{)}<1$. Applying Lemma 3.2 with
$c=\delta 2^{Q^{\prime}-2}>1$ we obtain

$|(f)_{\mathscr{B}_{k,4j+i}}|^{\tau}\leq\delta
2^{Q^{\prime}-2}|(f)_{\mathscr{B}_{k+1,j}}|^{\tau}\\\
+C\,2^{-kQ^{\prime}}[f]_{s,p,\alpha,\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i}}^{a\tau}\lVert
f\rVert_{L^{q}(\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i})}^{(1-a)\tau}.$

Therefore

$2^{kQ^{\prime}}\sum_{i=0}^{3}|(f)_{\mathscr{B}_{k,4j+i}}|^{\tau}\leq\delta
2^{(k+1)Q^{\prime}}|(f)_{\mathscr{B}_{k+1,j}}|^{\tau}+C\sum_{i=0}^{3}[f]_{s,p,\alpha,\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i}}^{a\tau}\lVert
f\rVert_{L^{q}(\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i})}^{(1-a)\tau}.$
(3.17)

Note that

$\sum_{j=-n_{k+1}}^{n_{k+1}-1}\sum_{i=0}^{3}a_{k,4j+i}=\sum_{j=-n_{k}}^{n_{k}-1}a_{k,j}.$
(3.18)

Summation with respect to $j$ from $-n_{k+1}$ to $n_{k+1}-1$ gives

$2^{kQ^{\prime}}\sum_{j=-n_{k}}^{n_{k}-1}|(f)_{\mathscr{B}_{k,j}}|^{\tau}\leq\delta
2^{(k+1)Q^{\prime}}\sum_{j=-n_{k+1}}^{n_{k+1}-1}|(f)_{\mathscr{B}_{k+1,j}}|^{\tau}\\\
+C\sum_{j=-n_{k+1}}^{n_{k+1}-1}\sum_{i=0}^{3}[f]_{s,p,\alpha,\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i}}^{a\tau}\lVert
f\rVert_{L^{q}(\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i})}^{(1-a)\tau}.$
(3.19)

Again summing (3.19) with respect to $k$, we obtain

$\sum_{k=m}^{n_{0}}2^{kQ^{\prime}}\sum_{j=-n_{k}}^{n_{k}-1}|(f)_{\mathscr{B}_{k,j}}|^{\tau}-\delta\sum_{k=m}^{n_{0}}2^{(k+1)Q^{\prime}}\sum_{j=-n_{k+1}}^{n_{k+1}-1}|(f)_{\mathscr{B}_{k+1,j}}|^{\tau}\\\
\leq
C\sum_{k=m}^{n_{0}}\sum_{j=-n_{k+1}}^{n_{k+1}-1}\sum_{i=0}^{3}[f]_{s,p,\alpha,\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i}}^{a\tau}\lVert
f\rVert_{L^{q}(\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i})}^{(1-a)\tau},$

which gives

$2^{mQ^{\prime}}\sum_{j=-n_{m}}^{n_{m}-1}|(f)_{\mathscr{B}_{m,j}}|^{\tau}+\sum_{k=m+1}^{n_{0}}(1-{\delta})2^{kQ^{\prime}}\sum_{j=-n_{k}}^{n_{k}-1}|(f)_{\mathscr{B}_{k,j}}|^{\tau}\\\
\leq
C\sum_{k=m}^{n_{0}}\sum_{j=-n_{k+1}}^{n_{k+1}-1}\sum_{i=0}^{3}[f]_{s,p,\alpha,\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i}}^{a\tau}\lVert
f\rVert_{L^{q}(\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i})}^{(1-a)\tau},$

and it follows that

$\sum_{k=m}^{n_{0}}(1-{\delta})2^{kQ^{\prime}}\sum_{j=-n_{k}}^{n_{k}-1}|(f)_{\mathscr{B}_{k,j}}|^{\tau}\leq
C\sum_{k=m}^{n_{0}}\sum_{j=-n_{k+1}}^{n_{k+1}-1}\sum_{i=0}^{3}[f]_{s,p,\alpha,\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i}}^{a\tau}\lVert
f\rVert_{L^{q}(\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i})}^{(1-a)\tau}.$
(3.20)

We apply (3.12) thrice on R.H.S and by a similar argument as discussed in
Remark 1, there is a constant $C$, independent of $k$ and $j$ such that

$\sum_{k=m}^{n_{0}}\sum_{j=-n_{k+1}}^{n_{k+1}-1}\sum_{i=0}^{3}[f]_{s,p,\alpha,\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i}}^{a\tau}\lVert
f\rVert_{L^{q}(\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i})}^{(1-a)\tau}\\\
\leq\left(\sum_{k=m}^{n_{0}}\sum_{j=-n_{k+1}}^{n_{k+1}-1}\sum_{i=0}^{3}[f]_{s,p,\alpha,\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i}}^{p}\right)^{\frac{a\tau}{p}}\left(\sum_{k=m}^{n_{0}}\sum_{j=-n_{k+1}}^{n_{k+1}-1}\sum_{i=0}^{3}\lVert
f\rVert_{L^{q}(\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i})}^{q}\right)^{\frac{(1-a)\tau}{q}}\\\
\leq C[f]_{s,p,\alpha}^{a\tau}\lVert
f\rVert_{L^{q}(\mathbb{H}^{n})}^{(1-a)\tau}.$ (3.21)

Combining (3.14), (3.20) and (3.21), we get

$\int_{\\{\xi\in\mathbb{H}^{n}:|z|>2^{m}\\}}d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f|^{\tau}d\xi\leq
C[f]_{s,p,\alpha}^{a\tau}\lVert f\rVert_{L^{q}(\mathbb{H}^{n})}^{(1-a)\tau}.$
(3.22)

Inequality (1.16) follows by passing the limit $m\rightarrow-\infty$.

Step 2: Proof of (ii). Choose $m\in\mathbb{Z}$ such that

$\mathrm{supp}(f)\cap\\{\xi=(z,t):|z|<2^{m}\\}=\phi.$

Let $\delta=\big{(}1+2^{Q^{\prime}-2}\big{)}/2$. Then $\delta<1$. Applying
Lemma 3.2 in (3.16) with $c=\delta/2^{Q^{\prime}-2}>1$, we get

$|(f)_{\mathscr{B}_{k+1,j}}|^{\tau}\leq\delta
2^{-\left(Q^{\prime}-2\right)}|(f)_{\mathscr{B}_{k,4j+i}}|^{\tau}+C\,2^{-kQ^{\prime}}[f]_{s,p,\alpha,\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i}}^{a\tau}\lVert
f\rVert_{L^{q}(\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i})}^{(1-a)\tau}.$

Summation with respect to $i$ gives

$2^{(k+1)Q^{\prime}}|(f)_{\mathscr{B}_{k+1,j}}|^{\tau}\leq\delta
2^{kQ^{\prime}}\sum_{i=0}^{3}|(f)_{\mathscr{B}_{k,4j+i}}|^{\tau}+C\sum_{i=0}^{3}[f]_{s,p,\alpha,\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i}}^{a\tau}\lVert
f\rVert_{L^{q}(\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i})}^{(1-a)\tau}.$

Again summing with respect to $j$ from $-n_{k+1}$ to $n_{k+1}-1$ and using
(3.18) we obtain

$2^{(k+1)Q^{\prime}}\sum_{j=-n_{k+1}}^{n_{k+1}-1}|(f)_{\mathscr{B}_{k+1,j}}|^{\tau}\leq\delta
2^{kQ^{\prime}}\sum_{j=-n_{k}}^{n_{k}-1}|(f)_{\mathscr{B}_{k,j}}|^{\tau}\\\
+C\sum_{j=-n_{k+1}}^{n_{k+1}-1}\sum_{i=0}^{3}[f]_{s,p,\alpha,\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i}}^{a\tau}\lVert
f\rVert_{L^{q}(\mathscr{C}_{k+1,j}\cup\mathscr{C}_{k,4j+i})}^{(1-a)\tau}.$
(3.23)

Finally summation over $k$ from $m-1$ to $n_{0}-1$ and (3.21) gives

$\sum_{k=m}^{n_{0}}(1-\delta)2^{kQ^{\prime}}\sum_{j=-n_{k}}^{n_{k}-1}|(f)_{\mathscr{B}_{k,j}}|^{\tau}\leq
C[f]_{s,p,\alpha}^{a\tau}\lVert f\rVert_{L^{q}(\mathbb{H}^{n})}^{(1-a)\tau}.$
(3.24)

Inequality (1.17) can be obtained by applying (3.24) in (3.14). This completes
the proof. ∎

##  4\. Proof of Theorem 1.2

For the proof of Theorem 1.2 we need the following two lemmas established in
[1].

######  Lemma 4.1.

Let $1\leq p<\infty$, $s\in(0,1)$ and $\alpha\in\mathbb{R}$ be such that
$(p-2)\alpha\geq 0$ and let $E\subset\mathbb{H}^{n}$ be any measurable set
with finite Lebesgue measure, i.e. $|E|<\infty$. Fix
$\xi=(z,t)\in\mathbb{H}^{n}$. Then there exists a positive constant $C$,
depends only on $n$, $p$, $s$ and $\alpha$ such that

$\int_{E^{C}}\frac{d\xi^{\prime}}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps}|z-z^{\prime}|^{(p-2)\alpha}}\geq
C|E|^{-\frac{ps+(p-2)\alpha}{Q}},$ (4.1)

where $\xi^{\prime}=(z^{\prime},t^{\prime})\in E^{C}$.

######  Lemma 4.2.

Let $1\leq p<\infty$, $s\in(0,1)$ and $\alpha\in\mathbb{R}$ be such that
$(p-2)\alpha\geq 0$ and $ps+(p-2)\alpha<Q$. Let $\\{e_{i}\\}$ and
$\\{d_{i}\\}$ be two non-negative sequences of real numbers satisfying the
following properties:
_(a)_ $e_{i}$ is decreasing;
_(b)_ $e_{i}=\sum_{j=i}^{\infty}d_{j}$.
Then the followings hold.
_(i)_ For any fixed $T>0$ we have

$\sum_{i\in\mathbb{Z}}e_{i}^{\frac{Q-ps-(p-2)\alpha}{Q}}T^{i}\leq
T^{\frac{Q}{Q-ps-(p-2)\alpha}}\sum_{i\in\mathbb{Z},e_{i}\neq
0}e_{i+1}e_{i}^{-\frac{ps+(p-2)\alpha}{Q}}T^{i}.$ (4.2)

_(ii)_ Moreover, if $T>1$ then

$\sum_{i\in\mathbb{Z},e_{i-1}\neq 0}\,\sum_{j\geq
i+1}T^{pi}e_{i-1}^{-\frac{ps+(p-2)\alpha}{Q}}d_{j}\leq\frac{1}{T^{p}-1}\sum_{i\in\mathbb{Z},e_{i-1}\neq
0}T^{pi}e_{i-1}^{-\frac{ps+(p-2)\alpha}{Q}}d_{i}.$ (4.3)

We next need the following lemma to prove our theorem.

######  Lemma 4.3.

Let $\beta\leq 0$ be such that $Q-2+\beta>0$. Then for any measurable set
$D\subset B_{\lambda}(0)$, we have

$\int_{D}|z|^{\beta}dzdt\leq C\lambda^{2}|D|^{\frac{Q-2+\beta}{Q}},$ (4.4)

where the constant $C$ is independent of $\lambda$.

###### Proof.

Let $R>0$ be such that $|D|=2\omega_{2n}R^{Q}=|D_{R}(0)|$, where the set
$D_{R}(0)$ is defined in (1.10). Note that $R\leq C\lambda$. Now

$I:=\int_{D}|z|^{\beta}dzdt=\int_{D\cap D_{R}(0)}|z|^{\beta}dzdt+\int_{D\cap
D_{R}(0)^{C}}|z|^{\beta}dzdt.$

Let us define the sets

$D_{R,1}:=\\{z\in\mathbb{R}^{2n}:|z|>R\\}\times\\{t\in\mathbb{R}:|t|>R^{2}\\},$
$D_{R,2}:=\\{z\in\mathbb{R}^{2n}:|z|>R\\}\times\\{t\in\mathbb{R}:|t|<R^{2}\\},$
$D_{R,3}:=\\{z\in\mathbb{R}^{2n}:|z|<R\\}\times\\{t\in\mathbb{R}:|t|>R^{2}\\},$

and the quantities

$I_{0}:=\int_{D\cap D_{R}(0)}|z|^{\beta}dzdt,$ $I_{i}:=\int_{D\cap
D_{R,i}}|z|^{\beta}dzdt,\ \ i=1,2,3.$

Clearly, $D_{R}(0)^{C}=D_{R,1}\cup D_{R,2}\cup D_{R,3}$ and
$I=I_{0}+I_{1}+I_{2}+I_{3}$. We will estimate the integrals separately.
Estimate for $I_{0}$:

$I_{0}\leq\int_{D_{R}(0)}|z|^{\beta}dzdt=C\int_{0}^{R}\int_{0}^{R^{2}}r^{Q-3+\beta}drdt\\\
=CR^{2}\frac{R^{Q-2+\beta}}{Q-2+\beta}\leq
C\lambda^{2}R^{Q-2+\beta}=C\lambda^{2}|D|^{\frac{Q-2+\beta}{Q}}.$ (4.5)

Estimate for $I_{1}$:

$I_{1}=\int_{D\cap D_{R,1}}|z|^{\beta}dzdt\leq R^{\beta}|D|=CR^{Q+\beta}=\leq
C\lambda^{2}|D|^{\frac{Q-2+\beta}{Q}}.$ (4.6)

Estimate for $I_{2}$:

$I_{2}=\int_{D\cap D_{R,2}}|z|^{\beta}dzdt\leq R^{\beta}|D|=CR^{Q+\beta}=\leq
C\lambda^{2}|D|^{\frac{Q-2+\beta}{Q}}.$ (4.7)

Estimate for $I_{3}$:

$I_{3}\leq C\int_{0}^{R}\int_{0}^{\lambda^{2}}r^{Q-3+\beta}drdt\leq
C\lambda^{2}\frac{R^{Q-2+\beta}}{Q-2+\beta}=C\lambda^{2}|D|^{\frac{Q-2+\beta}{Q}}.$
(4.8)

This completes the proof. ∎

Our next proposition is a fractional CKN type inequality in a bounded domain
which doesn’t contain a ball around origin.

######  Proposition 4.1.

Let $p$, $s$, $\alpha$, $\tau$, $a$, $\gamma$ and $q$ be as in Theorem _1.1_
satisfying the relation (1.14), (1.15) and
$Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)\geq 0$. In addition we
assume $\tau\gamma\leq-2$ when
$Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)\\\ >0$ and $\tau\gamma<-2$
when $Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)=0$. Let $\Omega$ be a
bounded open set in $\mathbb{H}^{n}$ such that $\Omega\cap B_{r}(0)=\phi$ for
some $r>0$. Then there exists a positive constant $C$ which depends on $n$,
$p$, $s$, $\alpha$, $r$ and $\Omega$ such that for $q>\frac{(1-a)\tau
Q}{Q-2+\tau\gamma(p-2)\frac{\alpha}{ps}}$, we have

$\left(\int_{\Omega}d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f(\xi)|^{\tau}d\xi\right)^{\frac{1}{\tau}}\leq
C[f]_{s,p,\alpha,\Omega}^{a}\lVert f\rVert_{L^{q}(\Omega)}^{1-a}\ \forall f\in
C_{c}(\Omega).$ (4.9)

###### Proof.

It is enough to prove (4.9) for $0\leq f\in C_{c}(\Omega)$. For
$i\in\mathbb{Z},$ define the following sets:

$E_{i}:=\\{\xi\in\Omega:f(\xi)\geq 2^{i}\\},$ $D_{i}:=E_{i}\setminus
E_{i+1}=\\{\xi\in\Omega:2^{i}\leq f(\xi)<2^{i+1}\\}.$

Let $e_{i}=|E_{i}|$ and $d_{i}=|D_{i}|$. Clearly $d_{i}\leq e_{i}$ and they
satisfy the conditions of Lemma 4.2.

Relation (1.14) gives

$\frac{Q-2+\tau\gamma(p-2)\frac{\alpha}{ps}}{\tau
Q}=a\left(\frac{Q-ps^{\prime}-(p-2)\alpha}{pQ}\right)+\frac{1-a}{q},$ (4.10)

where $0<s^{\prime}=s+\frac{\gamma}{a}+\frac{2}{a\tau}\leq s$ and
$Q-ps^{\prime}-(p-2)\alpha>0$ follows from the range of $q$. We first estimate
the quantity $[f]_{s^{\prime},p,\alpha,\Omega}$. Since $f\in
C_{c}(\mathbb{H}^{n})$, for any $i,j\in\mathbb{Z}$ with $j\leq i-2$, if
$\xi=(z,t)\in D_{i}$ and $\xi^{\prime}=(z^{\prime},t^{\prime})\in D_{j}$, then
we have $2^{j}\leq f(\xi^{\prime})\leq 2^{j+1}\leq 2^{i-1}<2^{i}\leq
f(\xi)\leq 2^{i+1}$, which gives $f(\xi)-f(\xi^{\prime})\geq 2^{i-1}.$ Hence

$[f]_{s^{\prime},p,\alpha,\Omega}^{p}\geq\sum_{i\in\mathbb{Z}}\,\sum_{j\leq
i-2}\int_{D_{i}}\int_{D_{j}}\frac{|f(\xi)-f(\xi^{\prime})|^{p}d\xi^{\prime}d\xi}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps^{\prime}}|z-z^{\prime}|^{(p-2)\alpha}}\\\
\geq\sum_{i\in\mathbb{Z}}\,\sum_{j\leq
i-2}2^{p(i-1)}\int_{D_{i}}\int_{D_{j}}\frac{d\xi^{\prime}d\xi}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps^{\prime}}|z-z^{\prime}|^{(p-2)\alpha}}\\\
=\sum_{i\in\mathbb{Z}}2^{p(i-1)}\int_{D_{i}}\int_{\cup_{j\leq
i-2}\,D_{j}}\frac{d\xi^{\prime}d\xi}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps^{\prime}}|z-z^{\prime}|^{(p-2)\alpha}}\\\
=\sum_{i\in\mathbb{Z}}2^{p(i-1)}\int_{D_{i}}\int_{E_{i-1}^{C}}\frac{d\xi^{\prime}d\xi}{d({\xi}^{-1}\circ\xi^{\prime})^{Q+ps^{\prime}}|z-z^{\prime}|^{(p-2)\alpha}}.$

Using Lemma 4.1 we obtain

$[f]_{s^{\prime},p,\alpha,\Omega}^{p}\geq C\sum_{i\in\mathbb{Z},e_{i-1}\neq
0}2^{p(i-1)}e_{i-1}^{-\frac{ps^{\prime}+(p-2)\alpha}{Q}}d_{i}.$ (4.11)

Putting $d_{i}=e_{i}-\sum_{j\geq i+1}d_{j}$ and then using Lemma 4.2 (ii) with
$T=2$, we get

$[f]_{s^{\prime},p,\alpha,\Omega}^{p}\geq\frac{C}{2^{p}}\left(\sum_{i\in\mathbb{Z},e_{i-1}\neq
0}2^{pi}e_{i}e_{i-1}^{-\frac{ps^{\prime}+(p-2)\alpha}{Q}}-\sum_{i\in\mathbb{Z},e_{i-1}\neq
0}\,\sum_{j\geq
i+1}2^{pi}e_{i-1}^{-\frac{ps^{\prime}+(p-2)\alpha}{Q}}d_{j}\right)\\\
\geq\frac{C}{2^{p}}\left(\sum_{i\in\mathbb{Z},e_{i-1}\neq
0}2^{pi}e_{i}e_{i-1}^{-\frac{ps^{\prime}+(p-2)\alpha}{Q}}-\frac{1}{2^{p}-1}\sum_{i\in\mathbb{Z},e_{i-1}\neq
0}2^{pi}e_{i-1}^{-\frac{ps^{\prime}+(p-2)\alpha}{Q}}d_{i}\right)\\\
\geq\frac{C}{2^{p}}\sum_{i\in\mathbb{Z},e_{i-1}\neq
0}2^{pi}e_{i}e_{i-1}^{-\frac{ps^{\prime}+(p-2)\alpha}{Q}}-\frac{1}{2^{p}-1}[f]_{s^{\prime},p,\alpha,\Omega}^{p},\
\left[\mathrm{using}\ \mathrm{inequality}\ \eqref{3.19}\right]$

which gives

$[f]_{s^{\prime},p,\alpha,\Omega}^{p}\geq\frac{2^{p}-1}{2^{2p}}C\sum_{i\in\mathbb{Z},e_{i-1}\neq
0}2^{pi}e_{i}e_{i-1}^{-\frac{ps^{\prime}+(p-2)\alpha}{Q}}.$

By Lemma 4.2 (i) with $T=2^{p}$ we get

$[f]_{s^{\prime},p,\alpha,\Omega}^{p}\geq
C\sum_{i\in\mathbb{Z}}2^{pi}e_{i}^{\frac{Q-ps^{\prime}-(p-2)\alpha}{Q}}.$
(4.12)

On the other hand,

$\int_{\Omega}|f(\xi)|^{q}d\xi=\sum_{i\in\mathbb{Z}}\int_{D_{i}}|f(\xi)|^{q}d\xi\geq\sum_{i\in\mathbb{Z}}2^{qi}d_{i}.$
(4.13)

Now let

$J:=\int_{\Omega}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f(\xi)|^{\tau}d\xi=\sum_{i\in\mathbb{Z}}\int_{D_{i}}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f(\xi)|^{\tau}d\xi\leq\sum_{i\in\mathbb{Z}}2^{\tau(i+1)}\int_{D_{i}}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}d\xi.$

It follows from the assumptions that $Q-2+\tau\gamma(p-2)\frac{\alpha}{ps}>0$.
Therefore, from Lemma 4.3 with $\beta=\tau\gamma(p-2)\frac{\alpha}{ps},$ we
get

$J\leq C\sum_{i\in\mathbb{Z}}2^{\tau
i}d_{i}^{\frac{Q-2+\tau\gamma(p-2)\frac{\alpha}{ps}}{Q}}.$ (4.14)

Relation (4.10) gives

$J\leq
C\sum_{i\in\mathbb{Z}}\left(2^{pi}d_{i}^{\frac{Q-ps^{\prime}-(p-2)\alpha}{Q}}\right)^{\frac{a\tau}{p}}\left(2^{qi}d_{i}\right)^{\frac{(1-a)\tau}{q}}.$
(4.15)

By inequality (3.12) and by the fact $d_{i}\leq e_{i}$ we obtain

$J\leq
C\left(\sum_{i\in\mathbb{Z}}2^{pi}e_{i}^{\frac{Q-ps^{\prime}-(p-2)\alpha}{Q}}\right)^{\frac{a\tau}{p}}\left(\sum_{i\in\mathbb{Z}}2^{qi}d_{i}\right)^{\frac{(1-a)\tau}{q}}.$
(4.16)

Inequality (4.16), (4.12) and (4.13) together imply

$\left(\int_{\Omega}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f(\xi)|^{\tau}d\xi\right)^{\frac{1}{\tau}}\leq
C[f]_{s^{\prime},p,\alpha,\Omega}^{a}\lVert f\rVert_{L^{q}(\Omega)}^{1-a}.$
(4.17)

Inequality (4.9) follows from (4.17) and the facts that $d(\xi)\geq r$ in
$\Omega$ and $[f]_{s^{\prime},p,\alpha,\Omega}\leq C[f]_{s,p,\alpha,\Omega}$.
∎

######  Corollary 2.

Let $p$, $s$, $\alpha$, $\tau$, $a$, $\gamma$ and $q$ be as in Theorem _1.1_
satisfying the relation (1.14), (1.15) and
$Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)\geq 0$. In addition we
assume $\tau\gamma\leq-2$ when
$Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)>0$ and $\tau\gamma<-2$ when
$Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)=0$. For $0<r<R$ consider
the set $A_{r,R}=A_{r,R}(0)$ as defined in (1.9). Then there exists a constant
$C>0$ depends on $n$, $p$, $s$, $\alpha$, $r$ and $R$ such that for
$q>\frac{(1-a)\tau Q}{Q-2+\tau\gamma(p-2)\frac{\alpha}{ps}}$, we have

$\left(\int_{A_{r,R}}d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f(\xi)|^{\tau}d\xi\right)^{\frac{1}{\tau}}\leq
C\lVert f\rVert_{s,p,\alpha,A_{\frac{r}{2},2R}}^{a}\lVert
f\rVert_{L^{q}(A_{\frac{r}{2},2R})}^{1-a},\ \forall f\in
C(\overline{{A}_{\frac{r}{2},2R}}).$ (4.18)

###### Proof.

Let $\phi\in C_{c}^{\infty}(\mathbb{H}^{n})$ be such that $0\leq\phi\leq 1$,
$\phi\equiv 1$ in $A_{r,R}$ and $\phi\equiv 0$ in
$A_{\frac{3r}{4},\frac{3R}{2}}^{C}$. Clearly $f\phi\in
C_{c}(A_{\frac{r}{2},2R})$. By Proposition 4.1,

$\left(\int_{A_{r,R}}d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f(\xi)|^{\tau}d\xi\right)^{\frac{1}{\tau}}\\\
\leq\left(\int_{A_{\frac{r}{2},2R}}d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f(\xi)|^{\tau}|\phi(\xi)|^{\tau}d\xi\right)^{\frac{1}{\tau}}\leq
C[f\phi]_{s,p,\alpha,A_{\frac{r}{2},2R}}^{a}\lVert
f\phi\rVert_{L^{q}(A_{\frac{r}{2},2R})}^{1-a}.$ (4.19)

Inequality (4.18) follows from (4.19), (2.15) and the fact $0\leq\phi\leq 1$.
∎

######  Lemma 4.4.

Let $p$, $s$, $\alpha$, $\tau$, $a$, $\gamma$ and $q$ be as in Theorem _1.1_
satisfying the relation (1.14), (1.15) and
$Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)\geq 0$. In addition we
assume $\tau\gamma\leq-2$ when
$Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)>0$ and $\tau\gamma<-2$ when
$Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)=0$. Let $\lambda>0$,
$0<r<R$ and denote the sets $A_{\lambda r,\lambda R}(0)$ and $A_{\frac{\lambda
r}{2},2\lambda R}(0)$ by $A_{\lambda}$ and $F_{\lambda}$ respectively. Then
for $q>\frac{(1-a)\tau Q}{Q-2+\tau\gamma(p-2)\frac{\alpha}{ps}}$ and for $f\in
C(\bar{F}_{\lambda})$, we have

$\left(\fint_{A_{\lambda}}d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f-(f)_{F_{\lambda}}|^{\tau}\right)^{\frac{1}{\tau}}\\\
\leq
C\lambda^{a\frac{ps+(p-2)\alpha-Q}{p}+\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)}[f]_{s,p,\alpha,F_{\lambda}}^{a}\left(\fint_{F_{\lambda}}|f|^{q}d\xi\right)^{\frac{1-a}{q}},$
(4.20)

where the constant $C$ is independent of $\lambda$ and $f$.

###### Proof.

For any function $f$ we define the function
$\tilde{f}(\xi)=f(D_{\lambda}\xi)$. Note that $\tilde{f}\in
C(\overline{A_{\frac{r}{2},2R}})$ if and only if $f\in C(\bar{F}_{\lambda})$.
We have by Corollary 2,

$\left(\int_{A_{r,R}}d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|\tilde{f}-(f)_{F_{\lambda}}|^{\tau}\right)^{\frac{1}{\tau}}\\\
\leq
C\left(\lVert\tilde{f}-(f)_{F_{\lambda}}\rVert_{L^{p}(A_{\frac{r}{2},2R})}^{p}+[\tilde{f}]^{p}_{s,p,\alpha,A_{\frac{r}{2},2R}}\right)^{\frac{a}{p}}\left(\lVert\tilde{f}-(f)_{F_{\lambda}}\rVert_{L^{q}(A_{\frac{r}{2},2R})}\right)^{1-a}.$

A change of variable gives

$\left(\lambda^{-\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)}\fint_{A_{\lambda}}d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f-(f)_{F_{\lambda}}|^{\tau}\right)^{\frac{1}{\tau}}\\\
\leq
C\left(\fint_{F_{\lambda}}|f-(f)_{F_{\lambda}}|^{p}+\lambda^{ps+(p-2)\alpha-Q}[f]_{s,p,\alpha,F_{\lambda}}^{p}\right)^{\frac{a}{p}}\left(\fint_{F_{\lambda}}|f|^{q}\right)^{\frac{1-a}{q}}.$
(4.21)

One can observe that $d({\xi}^{-1}\circ\xi^{\prime})\leq C\lambda$ for
$\xi,\xi^{\prime}\in F_{\lambda}$. A similar argument as for (3.7) gives

$\fint_{F_{\lambda}}|f-(f)_{F_{\lambda}}|^{p}\leq
C\lambda^{ps+(p-2)\alpha-Q}[f]_{s,p,\alpha,F_{\lambda}}^{p}.$ (4.22)

Inequality (4.20) follows from (4.21) and (4.22). ∎

The above lemma is the main ingredient of the proof of Theorem 1.2. The
technique of proof is similar to that of Theorem 1.1.

###### Proof of Theorem 1.2.

Let $f\in C_{c}^{1}(\mathbb{H}^{n})$. We choose $n_{0}\in\mathbb{Z}$ such that
$\mathrm{supp}\,f\subset B_{2^{n_{0}+1}}(0)$. Let us consider the
decomposition $\\{A_{k}\\}$ of $\mathbb{H}^{n}$ defined in (1.29) and we
define the sets $F_{k}:=A_{2^{k-1},2^{k+2}}(0)$. By Lemma 4.4 and by the
relation (1.14) we have

$\left(\fint_{A_{k}}d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f-(f)_{F_{k}}|^{\tau}d\xi\right)^{\frac{1}{\tau}}\\\
\leq
C2^{k\left(a\frac{ps+(p-2)\alpha-Q}{p}+\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)\right)}[f]_{s,p,\alpha,F_{k}}^{a}\left(\frac{1}{2^{kQ}}\int_{F_{k}}|f|^{q}d\xi\right)^{\frac{1-a}{q}}\\\
=C\,2^{-\frac{kQ}{\tau}}[f]_{s,p,\alpha,F_{k}}^{a}\lVert
f\rVert_{L^{q}(F_{k})}^{1-a}.$ (4.23)

Applying (4.23), Lemma 4.3 and the fact $|A_{k}|\sim|F_{k}|\sim 2^{kQ}$, we
obtain

$\int_{A_{k}}d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f|^{\tau}d\xi\\\
\leq
C2^{kQ}\fint_{A_{k}}d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f-(f)_{F_{k}}|^{\tau}+C2^{k\tau\gamma}|(f)_{F_{k}}|^{\tau}\int_{A_{k}}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}d\xi\\\
\leq C[f]_{s,p,\alpha,F_{k}}^{a\tau}\lVert
f\rVert_{L^{q}(F_{k})}^{(1-a)\tau}+C\,2^{k\left(Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)\right)}|(f)_{F_{k}}|^{\tau}.$
(4.24)

Summation with respect to $k$ from $m$ to $n_{0}$, and inequality (3.12) gives

$\int_{d(\xi)>2^{m}}d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f|^{\tau}d\xi\\\
\leq
C\left(\sum_{k=m}^{n_{0}}[f]_{s,p,\alpha,F_{k}}^{p}\right)^{\frac{a\tau}{p}}\left(\sum_{k=m}^{n_{0}}\lVert
f\rVert_{L^{q}(F_{k})}^{q}\right)^{\frac{(1-a)\tau}{q}}+C\sum_{k=m}^{n_{0}}2^{k\left(Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)\right)}|(f)_{F_{k}}|^{\tau}.$
(4.25)

Observe that $F_{k}\cup F_{k+1}$ is the set $A_{2^{k-1},2^{k+3}}(0)$. Let us
define the set $G_{k}:=A_{2^{k-2},2^{k+4}}(0)$. Now

$(f)_{F_{k}}-(f)_{F_{k+1}}=\frac{1}{|F_{k}|}\int_{F_{k}}\left(f-(f)_{G_{k}}\right)-\frac{1}{|F_{k+1}|}\int_{F_{k+1}}\left(f-(f)_{G_{k}}\right).$

Since $|{F_{k}}|\sim|F_{k+1}|$ and $|z|\leq d(\xi)\leq 2^{k+3}$ in $F_{k}\cup
F_{k+1}$, we have

$|(f)_{F_{k}}-(f)_{F_{k+1}}|\leq\frac{C}{|{F_{k}}\cup F_{k+1}|}\int_{F_{k}\cup
F_{k+1}}\left|f-(f)_{G_{k}}\right|\\\ \leq C\left(\fint_{F_{k}\cup
F_{k+1}}\left|f-(f)_{G_{k}}\right|^{\tau}\right)^{\frac{1}{\tau}}\\\ \leq
C2^{-k\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)}\left(\fint_{F_{k}\cup
F_{k+1}}d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}\left|f-(f)_{G_{k}}\right|^{\tau}\right)^{\frac{1}{\tau}}.$

Since $F_{k}\cup F_{k+1}\Subset G_{k}$, a similar argument as (4.23) implies

$|(f)_{F_{k}}-(f)_{F_{k+1}}|\leq
C2^{-\frac{k}{\tau}\left(Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)\right)}[f]_{s,p,\alpha,G_{k}}^{a}\lVert
f\rVert_{L^{q}(G_{k})}^{(1-a)}.$ (4.26)

Let $\delta=2/(1+2^{Q^{\prime}})<1$, where
$Q^{\prime}=Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)$. Applying Lemma
3.2 with $c=\delta 2^{Q^{\prime}}$ we obtain

$|(f)_{F_{k}}|^{\tau}\leq\delta
2^{Q^{\prime}}|(f)_{F_{k+1}}|^{\tau}+C2^{-kQ^{\prime}}[f]_{s,p,\alpha,G_{k}}^{a\tau}\lVert
f\rVert_{L^{q}(G_{k})}^{(1-a)\tau},$

which implies

$2^{kQ^{\prime}}|(f)_{F_{k}}|^{\tau}\leq\delta
2^{(k+1)Q^{\prime}}|(f)_{F_{k+1}}|^{\tau}+C[f]_{s,p,\alpha,G_{k}}^{a\tau}\lVert
f\rVert_{L^{q}(G_{k})}^{(1-a)\tau}.$

Summation with respect to $k$ from $m$ to $n_{0}$ gives

$2^{mQ^{\prime}}|(f)_{F_{m}}|^{\tau}+(1-\delta)\sum_{k=m+1}^{n_{0}}2^{kQ^{\prime}}|(f)_{F_{k}}|^{\tau}\leq
C\sum_{k=m}^{n_{0}}[f]_{s,p,\alpha,G_{k}}^{a\tau}\lVert
f\rVert_{L^{q}(G_{k})}^{(1-a)\tau},$

and it follows from (3.12),

$(1-\delta)\sum_{k=m}^{n_{0}}2^{kQ^{\prime}}|(f)_{F_{k}}|^{\tau}\leq
C\left(\sum_{k=m}^{n_{0}}[f]_{s,p,\alpha,G_{k}}^{p}\right)^{\frac{a\tau}{p}}\left(\sum_{k=m}^{n_{0}}\lVert
f\rVert_{L^{q}(G_{k})}^{q}\right)^{\frac{(1-a)\tau}{q}}.$ (4.27)

Inequality (1.16) follows from (4.25) and (4.27) and from the facts

$\sum_{k=m}^{n_{0}}\int_{F_{k}}f(\xi)d\xi\leq
3\int_{\mathbb{H}^{n}}f(\xi)d\xi\ \ \mathrm{and}\
\sum_{k=m}^{n_{0}}\int_{G_{k}}f(\xi)d\xi\leq 5\int_{\mathbb{H}^{n}}f(\xi)d\xi$
(4.28)

for any non-negative function $f$. ∎

######  Remark 4.

_If $Q-ps-(p-2)\alpha>0$ and $\tau\gamma<-2$, Theorem 1.2 can be proved
directly by the technique used in [1] to prove Theorem 4 as follows:
For $0\leq f\in C_{c}(\Omega)$ and for $i\in\mathbb{Z},$ let us define the
sets:_

$E_{i}:=\\{\xi\in\mathbb{H}^{n}:f(\xi)\geq 2^{i}\\},$ $D_{i}:=E_{i}\setminus
E_{i+1}=\\{\xi\in\mathbb{H}^{n}:2^{i}\leq f(\xi)<2^{i+1}\\},$

_and let $e_{i}=|E_{i}|$ and $d_{i}=|D_{i}|$. One can show that (see [1],
Section 5)_

$J:=\int_{\mathbb{H}^{n}}d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f(\xi)|^{\tau}d\xi\leq
C\sum_{i\in\mathbb{Z}}2^{\tau
i}d_{i}^{\frac{Q+\tau\left(1+\gamma(p-2)\frac{\alpha}{ps}\right)}{Q}}.$

_By the relation ( 1.14) and inequality (3.12) it follows that_

$J\leq
C\left(\sum_{i\in\mathbb{Z}}2^{pi}d_{i}^{\frac{Q-ps-(p-2)\alpha}{Q}}\right)^{\frac{a\tau}{p}}\left(\sum_{i\in\mathbb{Z}}2^{qi}d_{i}\right)^{\frac{(1-a)\tau}{q}}.$
(4.29)

_On the other hand, we have_

$[f]_{s,p,\alpha}^{p}\geq
C\sum_{i\in\mathbb{Z}}2^{pi}e_{i}^{\frac{Q-ps-(p-2)\alpha}{Q}}\geq
C\sum_{i\in\mathbb{Z}}2^{pi}d_{i}^{\frac{Q-ps-(p-2)\alpha}{Q}}.$ (4.30)

_and_

$\int_{\Omega}|f(\xi)|^{q}d\xi=\sum_{i\in\mathbb{Z}}\int_{D_{i}}|f(\xi)|^{q}d\xi\geq\sum_{i\in\mathbb{Z}}2^{qi}d_{i}.$
(4.31)

_Inequality ( 4.29), (4.30) and (4.31) together establish (1.18) for the case
$Q-ps-(p-2)\alpha>0$ along with $\tau\gamma<-2$._

##  5\. Proof of the limiting case

Before proving Theorem 1.3 we discuss the following remark:

######  Remark 5.

_In the proof of the limiting case
$1/\tau+\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)/{Q}=0$ we will be using
Lemma 4.4. Though the condition $\tau\gamma\leq-2$ is crucial for this, it
automatically follows from the assumption (1.19) and $(p-2)\alpha<Q-2$ as
follows:
Relation (1.14) gives_

$\frac{1}{\tau}=a\left(\frac{Q-(1+\gamma/as)(ps+(p-2)\alpha)}{pQ}\right)+\frac{1-a}{q}.$

_We have_

$\frac{aps}{-\tau\gamma}=a\left(\frac{Q-(1+\gamma/as)(ps+(p-2)\alpha)}{(-\gamma/as)Q}\right)+\frac{1-a}{(-\gamma/as)}\frac{p}{q}\\\
=a+(1-a)\frac{p}{q}+\frac{p(1+\gamma/as)}{(-\gamma/as)}\left[a\left(\frac{Q-ps-(p-2)\alpha}{pQ}\right)+\frac{1-a}{q}\right]\\\
=a+(1-a)\frac{p}{q}+\frac{p(1+\gamma/as)}{(-\gamma/as)}\left[\frac{Q+\tau\gamma\left(1+(p-2)\frac{\alpha}{ps}\right)}{\tau
Q}\right].$

_Assumption ( 1.19) implies $\frac{aps}{-\tau\gamma}\geq a$, i.e.,
$-\tau\gamma\leq ps$. Therefore we have
$Q-2+\tau\gamma(p-2)\frac{\alpha}{ps}>Q-2-(p-2)\alpha>0$. This together with
(1.19) implies $\tau\gamma<-2$._

###### Proof of Theorem 1.3.

Let $A_{k}$, $F_{k}$ and $G_{k}$ be the sets as defined earlier. Choose
$n_{0}\in\mathbb{Z}$ such that $2^{n_{0}}<R\leq 2^{n_{0}+1}$. From equation
(4.24) we get

$\int_{A_{k}}d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}|f|^{\tau}d\xi\leq
C[f]_{s,p,\alpha,F_{k}}^{a\tau}\lVert
f\rVert_{L^{q}(F_{k})}^{(1-a)\tau}+C|(f)_{F_{k}}|^{\tau}.$ (5.1)

Observe that, $\ln(4R/d(\xi))\geq(n_{0}-k+1)\ln 2,\,$ in $A_{k}$. Summing with
respect to $k$ and using (3.12) and (4.28) we obtain

$\int_{d(\xi)>2^{m}}\frac{d(\xi)^{\tau\gamma}|z|^{\tau\gamma(p-2)\frac{\alpha}{ps}}}{\ln^{\tau}(4R/d(\xi))}|f|^{\tau}d\xi\leq
C[f]_{s,p,\alpha}^{a\tau}\lVert
f\rVert_{L^{q}(\mathbb{H}^{n})}^{(1-a)\tau}+C\sum_{k=m}^{n_{0}}\frac{|(f)_{F_{k}}|^{\tau}}{(n_{0}-k+1)^{\tau}}.$
(5.2)

Here we use $\frac{1}{(n_{0}-k+1)^{\tau}}\leq 1$ in the first term of right
hand side. We now apply Lemma 3.2 in (4.26) with
$c=\left(\frac{n_{0}-k+1}{n_{0}-k+\frac{1}{2}}\right)^{\tau-1}$ and obtain

$|(f)_{F_{k}}|^{\tau}\leq\frac{(n_{0}-k+1)^{\tau-1}}{(n_{0}-k+\frac{1}{2})^{\tau-1}}|(f)_{F_{k+1}}|^{\tau}+2^{\tau-1}(n_{0}-k+1)^{\tau-1}[f]_{s,p,\alpha,G_{k}}^{a\tau}\lVert
f\rVert_{L^{q}(G_{k})}^{(1-a)\tau},$

which implies

$\frac{|(f)_{F_{k}}|^{\tau}}{(n_{0}-k+1)^{\tau-1}}-\frac{|(f)_{F_{k+1}}|^{\tau}}{(n_{0}-k+\frac{1}{2})^{\tau-1}}\leq
C[f]_{s,p,\alpha,G_{k}}^{a\tau}\lVert f\rVert_{L^{q}(G_{k})}^{(1-a)\tau}.$

Summation with respect to $k$ gives

$\sum_{k=m}^{n_{0}}\frac{|(f)_{F_{k}}|^{\tau}}{(n_{0}-k+1)^{\tau-1}}-\sum_{k=m}^{n_{0}}\frac{|(f)_{F_{k+1}}|^{\tau}}{(n_{0}-k+\frac{1}{2})^{\tau-1}}\leq
C[f]_{s,p,\alpha}^{a\tau}\lVert f\rVert_{L^{q}(\mathbb{H}^{n})}^{(1-a)\tau}.$
(5.3)

On the R.H.S we use (3.12) and (4.28). Using a change of variable in the
second sum of L.H.S, we obtain

$\frac{|(f)_{F_{m}}|^{\tau}}{(n_{0}-m+1)^{\tau-1}}+\sum_{k=m+1}^{n_{0}}\left[\frac{1}{(n_{0}-k+1)^{\tau-1}}-\frac{1}{(n_{0}-k+\frac{3}{2})^{\tau-1}}\right]|(f)_{F_{k}}|^{\tau}\\\
\leq C[f]_{s,p,\alpha}^{a\tau}\lVert
f\rVert_{L^{q}(\mathbb{H}^{n})}^{(1-a)\tau}.$ (5.4)

One can observe that
$\frac{1}{(n_{0}-k+1)^{\tau-1}}-\frac{1}{(n_{0}-k+\frac{3}{2})^{\tau-1}}\sim\frac{1}{(n_{0}-k+1)^{\tau}}$.
Hence

$\sum_{k=m}^{n_{0}}\frac{|(f)_{F_{k}}|^{\tau}}{(n_{0}-k+1)^{\tau-1}}\leq
C[f]_{s,p,\alpha}^{a\tau}\lVert f\rVert_{L^{q}(\mathbb{H}^{n})}^{(1-a)\tau}.$
(5.5)

Inequality (1.20) follows from (5.2) and (5.5) and by passing to the limit
$m\rightarrow-\infty$.
One can prove (1.21) by a similar technique used in the proof of (1.17) and
(1.20). ∎

Acknowledgements: This work is part of doctoral thesis of the second author.
He is grateful for the support provided by IIT Kanpur, India and MHRD,
Government of India (GATE fellowship). The third author acknowledges the
support provided by SERB through the Core Research Grant (CRG/2022/007867).

## References

  * [1] Adimurthi and Arka Mallick. Hardy type inequality on fractional order Sobolev spaces on the Heisenberg group., Ann. Sc. Norm. Super. Pisa Cl. Sci. (5), 18(3), 917–949, 2018.
  * [2] Adimurthi and Anusha Sekar. Role of the fundamental solution in Hardy-Sobolev-type inequalities. Proc. Roy. Soc. Edinburgh Sect. A, 136(6), 1111–1130, 2006.
  * [3] L. Caffarelli, R. Kohn, and L. Nirenberg. First order interpolation inequalities with weights. Compositio Math., 53(3), 259–275, 1984.
  * [4] L. D’Ambrozio. Some Hardy inequalities on the Heisenberg group. Differ. Uravn., 40(4), 509–521, 575, 2004.
  * [5] Eleonora Di Nezza, Giampiero Palatucci, and Enrico Valdinoci. Hitchhiker’s guide to the fractional Sobolev spaces. Bull. Sci. Math., 136(5), 521–573, 2012.
  * [6] Bartl omiej Dyda. A fractional order Hardy inequality. Illinois J. Math., 48(2), 575–588, 2004.
  * [7] Rupert L. Frank and Robert Seiringer. Non-linear ground state representations and sharp Hardy inequalities. J. Funct. Anal., 255(12), 3407–3430, 2008.
  * [8] N. Garofalo and E. Lanconelli. Frequency functions on the Heisenberg group, the uncertainty principle and unique continuation. Ann. Inst. Fourier (Grenoble), 40(2), 313–356, 1990.
  * [9] Yazhou Han. Weighted Caffarelli-Kohn-Nirenberg type inequality on the Heisenberg group. Indian J. Pure Appl. Math., 46(2), 147–161, 2015.
  * [10] A. Kassymov and D. Suragan. Fractional Hardy-Sobolev inequalities and existence results for fractional sub-Laplacians. J. Math. Sci. (N.Y.), 250(2), 337–350, 2020.
  * [11] Maria Manfredini, Giampiero Palatucci, Mirco Piccinini, and Sergio Polidoro. Hölder continuity and boundedness estimates for nonlinear fractional equations in the Heisenberg group. J. Geom. Anal., 33(3), Paper No. 77, 41, 2023.
  * [12] V. Maz’ya and T. Shaposhnikova. On the Bourgain, Brezis, and Mironescu theorem concerning limiting embeddings of fractional Sobolev spaces. J. Funct. Anal., 195(2), 230–238, 2002.
  * [13] Hoai-Minh Nguyen and Marco Squassina. Fractional Caffarelli-Kohn-Nirenberg inequalities. J. Funct. Anal., 274(9), 2661–2672, 2018.
  * [14] Pengcheng Niu, Huiqing Zhang, and Yong Wang. Hardy type and Rellich type inequalities on the Heisenberg group. Proc. Amer. Math. Soc., 129(12), 3623–3630, 2001.
  * [15] Tohru Ozawa, Michael Ruzhansky, and Durvudkhan Suragan. $L^{p}$-Caffarelli-Kohn-Nirenberg type inequalities on homogeneous groups. Q. J. Math., 70(1), 305–318, 2019.
  * [16] Michael Ruzhansky and Durvudkhan Suragan. On horizontal Hardy, Rellich, Caffarelli-Kohn- Nirenberg and $p$-sub-Laplacian inequalities on stratified groups. J. Differential Equations, 262(3), 1799–1821, 2017.
  * [17] Michael Ruzhansky and Durvudkhan Suragan. Hardy inequalities on homogeneous groups. volume 327 of Progress in Mathematics. Birkhäuser/Springer, Cham, 2019. 100 years of Hardy inequalities.
  * [18] Michael Ruzhansky, Durvudkhan Suragan, and Nurgissa Yessirkegenov. Caffarelli-Kohn-Nirenberg and Sobolev type inequalities on stratified Lie groups. NoDEA Nonlinear Differential Equations Appl., 24(5), Paper No. 56, 12, 2017.
  * [19] Adimurthi, Prosenjit Roy, and Vivek Sahu. Fractional boundary Hardy inequality for the critical cases. Preprint, https://arxiv.org/abs/2308.11956.