lemmatheo lemma corollarytheo corollary propositiontheo proposition

# Nondegeneracy of the positive solutions for critical nonlinear Hartree
equation in $\operatorname{\mathbb{R}}^{6}$

Xuemei Li  Xuemei Li
School of Mathematical Sciences,
Beijing Normal University,
Beijing, 100875, People’s Republic of China<EMAIL_ADDRESS>,
Xingdong Tang  Xingdong Tang
School of Mathematics and Statistics,
Nanjing Univeristy of Information Science and Technology,
Nanjing, 210044, People’s Republic of China<EMAIL_ADDRESS>and  Guixiang
Xu  Guixiang Xu
Laboratory of Mathematics and Complex Systems,
Ministry of Education,
School of Mathematical Sciences,
Beijing Normal University,
Beijing, 100875, People’s Republic of China<EMAIL_ADDRESS>
###### Abstract.

We prove that any positive solution for the critical nonlinear Hartree
equation

$-\operatorname{\mathop{}\\!\mathbin{\bigtriangleup}}{u}\left(x\right)-\int_{\operatorname{\mathbb{R}}^{6}}\frac{\left|{u}\left(y\right)\right|^{2}}{\left|x-y\right|^{4}}\mathrm{d}y\,{u}\left(x\right)=0,\quad\text{}\quad
x\in\operatorname{\mathbb{R}}^{6}.$

is nondegenerate. Firstly, in terms of spherical harmonics, we show that the
corresponding linear operator can be decomposed into a series of one
dimensional linear operators. Secondly, by making use of the Perron-Frobenius
property, we show that the kernel of each one dimensional linear operator is
finite. Finally, we show that the kernel of the corresponding linear operator
is the direct sum of the kernel of all one dimensional linear operators.

###### Key words and phrases:

nondegeneracy, spherical harmonics, addition formula, energy critical,
Newtonian potential

###### 2010 Mathematics Subject Classification:

35J91 ; 35B38

##  1\. Introduction

The purpose of this paper is to derive the nondegeneracy property of the
postive solutions to the following $\dot{H}^{1}$-critical nonlinear Hartree
(NLH) equation,

$\displaystyle\begin{cases}&-\operatorname{\mathop{}\\!\mathbin{\bigtriangleup}}{u}\left(x\right)-\int_{\operatorname{\mathbb{R}}^{6}}\frac{\left|{u}\left(y\right)\right|^{2}}{\left|x-y\right|^{4}}\mathrm{d}y\,{u}\left(x\right)=0,\quad
x\in\operatorname{\mathbb{R}}^{6},\\\
&\lim\limits_{{\left|x\right|\to+\infty}}\left|{u}\left(x\right)\right|=0.\end{cases}$
(1.1)

It is well-known that (see, for example [16]) any postive solution of the
equation (1.1) belongs precisely to

$\\{\,{\omega_{\lambda,z}}\left(x\right)=\lambda^{2}{\omega}\left(\lambda
x+z\right)\;|\;\lambda>0\text{~{}and~{}}z\in\operatorname{\mathbb{R}}^{6}\,\\},$
(1.2)

where $\omega$ is the radial, postive ground state, and has the explicit form

${\omega}\left(x\right)=\frac{12}{\pi^{\nicefrac{{3}}{{2}}}}\frac{1}{\left(1+\left|x\right|^{2}\right)^{2}}.$
(1.3)

On the one hand, since the function ${\omega_{\lambda,z}}\left(x\right)$,
defined by (1.2), satisfies

$\displaystyle-\operatorname{\mathop{}\\!\mathbin{\bigtriangleup}}{\omega_{\lambda,z}}\left(x\right)-\int_{\operatorname{\mathbb{R}}^{6}}\frac{\left|{\omega_{\lambda,z}}\left(y\right)\right|^{2}}{\left|x-y\right|^{4}}\mathrm{d}y\,{\omega_{\lambda,z}}\left(x\right)=0,\quad
x\in\operatorname{\mathbb{R}}^{6},~{}\lambda>0,~{}\text{and}~{}z\in\operatorname{\mathbb{R}}^{6},$
(1.4)

by differentiating (1.4) with respect to the parameters $\lambda$ and $z$ at
$\lambda=1$ and $z=0$ formlly, we obtain that,

${L\Lambda\omega}\left(x\right)=0,\quad\text{and}\quad{L\frac{\partial\omega}{\partial
x_{j}}}\left(x\right)=0,\quad 1\leq j\leq 6,$

where

${\Lambda\omega}\left(x\right):=2{\omega}\left(x\right)+2x\cdot{\nabla\omega}\left(x\right),$
(1.5)

and the operator $L$ is defined by

$\displaystyle{L\varphi}\left(x\right):=-{\operatorname{\mathop{}\\!\mathbin{\bigtriangleup}}\varphi}\left(x\right)-{\Phi}\left(\omega^{2}\right)\left(x\right){\varphi}\left(x\right)-2{\Phi}\left(\omega\varphi\right)\left(x\right){\omega}\left(x\right),$
(1.6)

with

${\Phi
f}\left(x\right):=\int_{\operatorname{\mathbb{R}}^{6}}\frac{{f}\left(y\right)}{\left|x-y\right|^{4}}\mathrm{d}y.$
(1.7)

Moreover, by the linearity of the operator $L$, if $\varphi$ belongs to the
set

$\operatorname{\mathcal{N}}=\operatorname{span}\left\\{{\Lambda\omega,~{}\frac{\partial\omega}{\partial
x_{1}},~{}\frac{\partial\omega}{\partial
x_{2}},~{}\frac{\partial\omega}{\partial
x_{3}},~{}\frac{\partial\omega}{\partial
x_{4}},~{}\frac{\partial\omega}{\partial
x_{5}},~{}\frac{\partial\omega}{\partial x_{6}}}\right\\},$ (1.8)

we get, ${L\varphi}\left(x\right)=0$.

On the other hand, an important question arising in the analysis of solutions
for (1.1) is to study the kernel of the linearized operator $L$ close to
$\omega$. More precisely, one can address the following question:

Is there any other function $\varphi$ vanishing at infinity satisfies
${L\varphi}\left(x\right)=0$, except that belongs to the set
$\operatorname{\mathcal{N}}$ defined by (1.8)?

Our main result in this paper is devoted to a negative answer to the question
aboved. More precisely, we will prove the following theorem which states that
the solution $\omega$ (see (1.3)) of (1.1) is nondegenerate.

######  Theorem 1.1.

The solution $\omega$, defined by (1.3), for the problem (1.1) is
nondegenerate. More precisely, let $L$ be defined by (1.6), if $f\in
L^{2}\left(\operatorname{\mathbb{R}}^{6}\right)$ satisfies
${{Lf}}\left(x\right)=0$, then

$f\in\operatorname{span}\left\\{\Lambda\omega,~{}\frac{\partial\omega}{\partial{x_{1}}},~{}\frac{\partial\omega}{\partial{x_{2}}},~{}\frac{\partial\omega}{\partial{x_{3}}},~{}\frac{\partial\omega}{\partial{x_{4}}},~{}\frac{\partial\omega}{\partial{x_{5}}},~{}\frac{\partial\omega}{\partial{x_{6}}}\right\\}.$

The nondegeneracy of the ground state for the nonlinear elliptic equations
plays a key role in the analysis of long time dynamics of the solution to the
corresponding evaluation equations. For example, in the context of
$\dot{H}^{1}$-critical nonlinear Schrödinger (NLS) equation,

$i{\frac{\partial u}{\partial
t}}\left(t,x\right)+{\operatorname{\mathop{}\\!\mathbin{\bigtriangleup}}u}\left(t,x\right)+{\left|u\right|^{\frac{4}{N-2}}u}\left(t,x\right)=0,$
(1.9)

and the $\dot{H}^{1}$-critical nonlinear wave (NLW) equation,

${\frac{\partial^{2}u}{\partial
t^{2}}}\left(t,x\right)+{\operatorname{\mathop{}\\!\mathbin{\bigtriangleup}}u}\left(t,x\right)+{\left|u\right|^{\frac{4}{N-2}}u}\left(t,x\right)=0,$
(1.10)

the nondegeneracy of the ground state solution $W$ to the corresponding
elliptic equation

$-{\operatorname{\mathop{}\\!\mathbin{\bigtriangleup}}u}\left(t,x\right)-{\left|u\right|^{\frac{4}{N-2}}u}\left(t,x\right)=0,$
(1.11)

is crucial in the construction of blow-up solutions to the equations (1.9) and
(1.10) (see, for instance, [6, 7, 8, 10, 9, 13]). With the help of Theorem
1.1, we are able to construct blow-up solutions to the $\dot{H}^{1}$-critical
nonlinear Schrödinger equation with Hartree terms,

$i{\frac{\partial u}{\partial
t}}\left(t,x\right)+{\operatorname{\mathop{}\\!\mathbin{\bigtriangleup}}u}\left(t,x\right)+\int_{\operatorname{\mathbb{R}}^{6}}\frac{\left|{u}\left(y\right)\right|^{2}}{\left|x-y\right|^{4}}\mathrm{d}y\,{u}\left(x\right)=0,\quad
x\in\operatorname{\mathbb{R}}^{6},$

which will be considered in our subsequent work.

The nondegeneracy of the ground state solution to the semilinear elliptic
equation (1.11) is also necessary in the construction of multi-bump solutions
of the equation (1.11), see, for example [3, 4, 5, 15, 14]. We hope Theorem
1.1 can be used to construct multi-bump solutions to (1.1).

The paper is organized as follows. In Section 2, we give some notations and
review several lemmas which will be frequently used in the remainder of the
paper. In Section 3, we mainly prove Theorem 1.1.

##  2\. Notation and useful lemmas

### Notation and conventions

As usual, we use $\mathbb{S}^{5}$ to denote $5$-dimensional unit sphere in
$6$-dimensional Euclidean space $\operatorname{\mathbb{R}}^{6}$,

$\mathbb{S}^{5}=\\{\,x=\left(x_{1},x_{2},\cdots,x_{6}\right)\in\operatorname{\mathbb{R}}^{6}\;|\;\left|x\right|^{2}=\sum_{j=1}^{6}{x_{j}^{2}}=1\,\\}.$

For any $x,y\in\operatorname{\mathbb{R}}^{6}$ with
$\left|x\right|\neq\left|y\right|$ , let us denote,

$\displaystyle x\vee
y:=\begin{cases}x&\quad\text{if}\quad\left|x\right|>\left|y\right|,\\\
y&\quad\text{if}\quad\left|x\right|<\left|y\right|,\end{cases}\quad\text{and}\quad
x\wedge y:=\begin{cases}y&\quad\text{if}\quad\left|x\right|>\left|y\right|,\\\
x&\quad\text{if}\quad\left|x\right|<\left|y\right|.\end{cases}$ (2.1)

An elementary calculation implies that,

$\left|x\vee
y\right|=\max\left\\{\left|x\right|,\,\left|y\right|\right\\},\quad\text{and}\quad\left|x\wedge
y\right|=\min\left\\{\left|x\right|,\,\left|y\right|\right\\}.$ (2.2)

We use $L^{2}\left(\operatorname{\mathbb{R}}^{6}\right)$ denote the real
Hilbert space of measurable functions $f$ on $\operatorname{\mathbb{R}}^{6}$
with the inner product,

$\left({f},g\right):=\int_{\operatorname{\mathbb{R}}^{6}}{f}\left(x\right){g}\left(x\right)\mathrm{d}x.$

We shall also use $L^{2}\left(\mathbb{S}^{5}\right)$ denote the space of
measurable functions $f$ on $\mathbb{S}^{5}$ for which

$\int_{\mathbb{S}^{5}}\left|{f}\left(x\right)\right|^{2}\mathrm{d}{\bm{\sigma}}\left(x\right)$

is finite, where $\mathrm{d}\bm{\sigma}$ is the surface area measure. With a
slight abuse of notation of notation, we write both ${f}\left(x\right)$ and
${f}\left(\left|x\right|\right)$ for radial functions $f$ on
$\operatorname{\mathbb{R}}^{6}$. Moreover, we will use
$L^{2}\left(\left(0,+\infty\right),r^{5}\right)$ denote the Hilbert space with
measurable functions $f$ on $\left(0,+\infty\right)$ with the inner product,

$\langle{f},g\rangle:=\int_{0}^{+\infty}{f}\left(r\right){g}\left(r\right)r^{5}\mathrm{d}r.$

Next, we recall some well-known results related to spherical harmonics. We use
$\operatorname{\mathcal{H}}_{k}$ to denote the space of spherical harmonics of
degree $k$ (i.e. the restrictions to $\mathbb{S}^{5}$ of real, homogeneous
harmonic polynomials of degree $k$). In fact, the dimension of
$\operatorname{\mathcal{H}}_{k}$ is[2, 17] is

$\dim{\operatorname{\mathcal{H}}_{k}}=\alpha_{k}:=\begin{cases}1,&\quad\text{if}\quad
k=0,\\\ 6,&\quad\text{if}\quad k=1,\\\
\binom{k+5}{k}-\binom{k+3}{k-2}&\quad\text{if}\quad k\geq 2.\end{cases}$

We use ${Y_{k,j}}\left(1\leq j\leq\alpha_{k}\right)$ to denote an orthogonal
basis for $\operatorname{\mathcal{H}}_{k}$, i.e.

$\int_{\mathbb{S}^{5}}{Y_{k,i}}\left(\xi\right){Y_{k,j}}\left(\xi\right)\mathrm{d}{\bm{\sigma}}\left(\xi\right)=\pi^{3}\begin{cases}1,&\text{if
}i=j,\\\ 0,&\text{if }i\neq j.\end{cases}$ (2.3)

Specially,
$\operatorname{\mathcal{H}}_{0}=\operatorname{span}\left\\{1\right\\}$, and

$\operatorname{\mathcal{H}}_{1}=\operatorname{span}\\{\,\frac{x_{j}}{\left|x\right|}\;|\;1\leq
j\leq 6\,\\},$ (2.4)

therefore, we may set ${Y_{1,j}}\left(\xi\right)=\sqrt{6}\xi_{j}$ with
$\xi_{j}=\frac{x_{j}}{\left|x\right|}$. Moreover, the space
$\operatorname{\mathcal{H}}_{k}$ coincides with the eigenspace of the
eigenvalue $-k\left(k+4\right)$ for the Laplace–Beltrami operator
$\operatorname{\mathop{}\\!\mathbin{\bigtriangleup}}_{\mathbb{S}^{5}}$ on
$\mathbb{S}^{5}$, i.e.

$\operatorname{\mathop{}\\!\mathbin{\bigtriangleup}}_{\mathbb{S}^{5}}Y=-k\left(k+4\right)Y,\quad\text{for
any}\quad Y\in\operatorname{\mathcal{H}}_{k}.$ (2.5)

An elementary calculation implies that,

$\int_{\mathbb{S}^{5}}{Y_{k,i}}\left(\xi\right){Y_{l,j}}\left(\xi\right)\mathrm{d}\xi=0,\quad\text{for
all }1\leq k\neq l<\infty,~{}~{}1\leq i\leq\alpha_{k},~{}~{}1\leq
j\leq\alpha_{l}.$ (2.6)

Now, for any $f\in L^{2}\left(\operatorname{\mathbb{R}}^{6}\right)$, let us
denote

${f_{k,j}}\left(r\right):=\frac{1}{\pi^{3}}\int_{\mathbb{S}^{5}}{f}\left(r\xi\right){Y_{k,j}}\left(\xi\right)\mathrm{d}{\bm{\sigma}}\left(\xi\right),$
(2.7)

then we have the following direct sum decomposition (e.g. see [17]),

$\displaystyle{f}\left(x\right)=\sum_{k=0}^{\infty}\sum_{j=1}^{\alpha_{k}}{f_{k,j}}\left(\left|x\right|\right){Y_{k,j}}\left(\frac{x}{\left|x\right|}\right).$
(2.8)

Moreover, the following identity holds,

$\displaystyle\int_{\operatorname{\mathbb{R}}^{6}}\left|{f}\left(x\right)\right|^{2}\mathrm{d}x=\pi^{3}\sum_{k=0}^{\infty}\sum_{j=1}^{\alpha_{k}}\int_{0}^{+\infty}\left|{\varphi_{k,j}}\left(r\right)\right|^{2}r^{5}\mathrm{d}r.$
(2.9)

In order to deal with the Newtonian potential appeared in (1.1), we need the
following lemma, which related to Gegenbauer functions.

######  Lemma  ([2, Lemma 1.2.3, Theorem1.2.6]).

For any $r\in\left(-1,\,1\right)$, $t\in\left[-1,\,1\right]$, we have,

$\frac{1}{\left(1-2rt+r^{2}\right)^{2}}=\sum_{k=0}^{\infty}{C^{(2)}_{k}}\left(t\right)r^{k},$
(2.10)

where ${C^{(2)}_{k}}\left(t\right)$ is the Gegenbauer polynomial of degree $k$
associated with $2$ (see [2, p.418]). Moreover, the following addition formula
holds,

${P^{2}_{k}}\left(\xi\cdot\eta\right)=\frac{2}{k+2}\sum_{j=1}^{\alpha_{k}}{Y_{k,j}}\left(\xi\right){Y_{k,j}}\left(\eta\right),\quad\text{for
any}\quad\xi,~{}\eta\in\mathbb{S}^{5}.$ (2.11)

Now, we are able to give an explicit expression of the Newtonian potential in
$\operatorname{\mathbb{R}}^{6}$, by making use of spherical harmonics.

######  Lemma .

For any $x$, $y\in\operatorname{\mathbb{R}}^{6}\setminus\\{0\\}$ with
$\left|x\right|\neq\left|y\right|$, we have,

$\displaystyle\frac{1}{\left|x-y\right|^{4}}=\sum_{k=0}^{\infty}\frac{k}{k+2}\frac{\left|x\wedge
y\right|^{k}}{\left|x\vee
y\right|^{k+4}}\sum_{m=1}^{\alpha_{k}}{Y_{k,j}}\left(\frac{x}{\left|x\right|}\right){Y_{k,j}}\left(\frac{y}{\left|y\right|}\right),$
(2.12)

###### Proof.

First, for any $x$, $y\in\operatorname{\mathbb{R}}^{6}\setminus\\{0\\}$
satisfying $\left|x\right|\neq\left|y\right|$, we have,

$\displaystyle\left|x-y\right|^{2}=\left|x\right|^{2}-2x\cdot
y+\left|y\right|^{2}.$ (2.13)

By making use of (2.1) and (2.2), we have, from (2.13),

$\displaystyle\left|x-y\right|^{2}=\left|x\vee y\right|^{2}-2\left|x\vee
y\right|\left|x\wedge
y\right|\frac{x}{\left|x\right|}\cdot\frac{y}{\left|y\right|}+\left|x\wedge
y\right|^{2}.$ (2.14)

Therefore, by (2.14), we obtain that,

$\displaystyle\frac{1}{\left|x-y\right|^{4}}=\frac{1}{\left|x\vee
y\right|^{4}}\frac{1}{\left(1-2\frac{\left|x\wedge y\right|}{\left|x\vee
y\right|}\frac{x}{\left|x\right|}\cdot\frac{y}{\left|y\right|}+\left(\frac{\left|x\wedge
y\right|}{\left|x\vee y\right|}\right)^{2}\right)^{2}}.$ (2.15)

Now, by taking $r=\frac{\left|x\wedge y\right|}{\left|x\vee y\right|}$ in
(2.10), then taking $\xi=\frac{x}{\left|x\right|}$ and
$\eta=\frac{y}{\left|y\right|}$ in (2.11), The formula (2.12) follows directly
from (2.15). This ends the proof of Section 2. ∎

######  Remark 2.1.

The spherical harmonics expansion for the Newtonian potential in
$\operatorname{\mathbb{R}}^{n}(3\leq n\leq 4)$ and
$\operatorname{\mathbb{R}}^{4}$ is well-known, (see, for example, [1, 11]).

The following proposition plays an important role in the nondegenerate
analysis of positive solutions to (1.1).

######  Proposition .

Let $\Phi$ be defined by (1.7). For any $f\in
L^{2}\left(\operatorname{\mathbb{R}}^{6}\right)$, we have,

$\displaystyle{\Phi
f}\left(x\right)=\sum_{k=0}^{\infty}\sum_{j=1}^{\alpha_{k}}\int_{0}^{+\infty}{\operatorname{\mathcal{K}}_{k}}\left(r,\left|x\right|\right){f_{k,j}}\left(r\right)\mathrm{d}r{Y_{k,j}}\left(\frac{x}{\left|x\right|}\right),$
(2.16)

where ${f_{k,j}}\left(r\right)$ is defined by (2.7), and

${\operatorname{\mathcal{K}}_{k}}\left(r,\left|x\right|\right)=\frac{2\pi^{3}}{k+2}\begin{cases}\frac{r^{k+5}}{\left|x\right|^{k+4}},&\quad\text{if}\quad
r<\left|x\right|,\\\ \frac{\left|x\right|^{k}}{r^{k-1}},&\quad\text{if}\quad
r>\left|x\right|.\end{cases}$ (2.17)

###### Proof.

On the one hand, in view of Section 2,

$\displaystyle\frac{1}{\left|x-y\right|^{4}}=\sum_{k=0}^{\infty}\frac{k}{k+2}\frac{\left|x\wedge
y\right|^{k}}{\left|x\vee
y\right|^{k+4}}\sum_{m=1}^{\alpha_{k}}{Y_{k,j}}\left(\frac{x}{\left|x\right|}\right){Y_{k,j}}\left(\frac{y}{\left|y\right|}\right),$
(2.18)

where $x$, $y\in\operatorname{\mathbb{R}}^{6}\setminus\\{0\\}$ satisfy
$\left|x\right|\neq\left|y\right|$.

On the other hand, by (2.8), we have

${f_{k,j}}\left(r\right)=\frac{1}{\pi^{3}}\int_{\mathbb{S}^{5}}{f}\left(r\xi\right){Y_{k,j}}\left(\xi\right)\mathrm{d}{\bm{\sigma}}\left(\xi\right),$
(2.19)

where ${f_{k,j}}\left(r\right)$ is defined by (2.7).

Combining with (2.18) and (2.19), we have,

$\displaystyle\int_{\operatorname{\mathbb{R}}^{6}}\frac{{f}\left(y\right)}{\left|x-y\right|^{4}}\mathrm{d}y$
$\displaystyle=$
$\displaystyle\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\sum_{j=1}^{\alpha_{k}}\sum_{m=1}^{\alpha_{l}}\int_{\operatorname{\mathbb{R}}^{6}}\frac{k}{k+2}\frac{\left|x\wedge
y\right|^{k}}{\left|x\vee
y\right|^{k+4}}{Y_{k,j}}\left(\frac{x}{\left|x\right|}\right){Y_{k,j}}\left(\frac{y}{\left|y\right|}\right){f_{l,m}}\left(\left|y\right|\right){Y_{l,m}}\left(\frac{y}{\left|y\right|}\right)\mathrm{d}y.$
(2.20)

By the definition of $x\wedge y$ and $x\vee y$ (see (2.1)), and (2.2), from
(2.20), we find,

$\displaystyle\int_{\operatorname{\mathbb{R}}^{6}}\frac{{f}\left(y\right)}{\left|x-y\right|^{4}}\mathrm{d}y$
$\displaystyle=$
$\displaystyle\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\sum_{m=1}^{\alpha_{k}}\sum_{j=1}^{\alpha_{k}}\frac{k}{k+2}\int_{\left|y\right|<\left|x\right|}\frac{\left|y\right|^{k}}{\left|x\right|^{k+4}}{Y_{k,j}}\left(\frac{x}{\left|x\right|}\right){Y_{k,j}}\left(\frac{y}{\left|y\right|}\right){f_{l,m}}\left(\left|y\right|\right){Y_{l,m}}\left(\frac{y}{\left|y\right|}\right)\mathrm{d}y$
$\displaystyle+\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\sum_{m=1}^{\alpha_{k}}\sum_{j=1}^{\alpha_{k}}\frac{k}{k+2}\int_{\left|y\right|>\left|x\right|}\frac{\left|x\right|^{k}}{\left|y\right|^{k+4}}{Y_{k,j}}\left(\frac{x}{\left|x\right|}\right){Y_{k,j}}\left(\frac{y}{\left|y\right|}\right){f_{l,m}}\left(\left|y\right|\right){Y_{l,m}}\left(\frac{y}{\left|y\right|}\right)\mathrm{d}y.$
(2.21)

With the change of variables $r=\left|x\right|$ and
$\xi=\frac{x}{\left|x\right|}$ and using (2.3), (2.6), (2.21) becomes

$\displaystyle\int_{\operatorname{\mathbb{R}}^{6}}\frac{{f}\left(y\right)}{\left|x-y\right|^{4}}\mathrm{d}y$
$\displaystyle=$
$\displaystyle\sum_{k=0}^{\infty}\sum_{j=1}^{\alpha_{k}}\frac{k}{k+2}{Y_{k,j}}\left(\frac{x}{\left|x\right|}\right)\left(\int_{0}^{\left|x\right|}\pi^{3}\frac{r^{k+5}}{\left|x\right|^{k+4}}{f_{k,j}}\left(r\right)\mathrm{d}r+\int_{\left|x\right|}^{+\infty}\pi^{3}\frac{\left|x\right|^{k}}{r^{k-1}}{f_{k,j}}\left(r\right)\mathrm{d}r\right),$

which means that (2.16) holds with
${\operatorname{\mathcal{K}}_{k}}\left(r,\left|x\right|\right)$ defined by
(2.17). ∎

As a consequence of Section 2, we obtain Newton’s theorem (see, for instance,
[12, Theorem 9.7]).

######  Corollary .

For any radial $f\in L^{2}\left(\operatorname{\mathbb{R}}^{6}\right)$, ${\Phi
f}\left(x\right)$ is radial. Moreover, we have,

${\Phi
f}\left(x\right)=\int_{0}^{+\infty}{\operatorname{\mathcal{K}}_{0}}\left(r,\left|x\right|\right){f}\left(r\right)\mathrm{d}r,$
(2.22)

where ${\operatorname{\mathcal{K}}_{0}}\left(r,\left|x\right|\right)$ is
defined by (2.17) with $k=0$. Specially, the follows identity holds,

$\displaystyle{\Phi}\left(\omega^{2}\right)\left(x\right)=2\pi^{\nicefrac{{3}}{{2}}}{\omega}\left(x\right),$
(2.23)
$\displaystyle-\operatorname{\mathop{}\\!\mathbin{\bigtriangleup}}{\omega}\left(x\right)-2\pi^{\nicefrac{{3}}{{2}}}{\omega}\left(x\right)=0.$
(2.24)

###### Proof.

Since $f$ is radial, by (2.7), we have

${f_{0,1}}\left(r\right)={f}\left(r\right),\quad\text{and}\quad{f_{k,j}}\left(r\right)\equiv
0,\quad\text{for all}\quad k\geq 1~{}\text{and}~{}1\leq j\leq\alpha_{k}.$

Therefore, (2.22) follows from (2.16). The identity (2.23) follows from
elementary calculations and (2.22) by taking
${f}\left(r\right)={\omega^{2}}\left(r\right)$. Moreover, by inserting (2.23)
into (1.4) with $\lambda=1$ and $z=0$, we obtain that (2.24) holds. ∎

The next theorem concerns the decomposition of the operator $L$ defined by
(1.6) in terms of spherical harmonics.

######  Theorem 2.2.

Let $L$ be defined by (1.6). For any $f\in
L^{2}\left(\operatorname{\mathbb{R}}^{6}\right)$, we have,

$\displaystyle{Lf}\left(x\right)=\sum_{k=0}^{\infty}\sum_{j=1}^{\alpha_{k}}{\operatorname{\mathcal{L}}_{k}f_{k,j}}\left(\left|x\right|\right){Y_{k,j}}\left(\frac{x}{\left|x\right|}\right),$
(2.25)

where $f_{k,j}$ is defined by (2.7), and

$\displaystyle{\operatorname{\mathcal{L}}_{k}f}\left(r\right)=$
$\displaystyle-{f^{\prime\prime}}\left(r\right)-\frac{5}{r}{f^{\prime}}\left(r\right)+\frac{k\left(k+4\right)}{r^{2}}{f}\left(r\right)$
$\displaystyle-2\pi^{\nicefrac{{3}}{{2}}}{\omega}\left(r\right){f}\left(r\right)-2{\omega}\left(r\right)\int_{0}^{+\infty}{\operatorname{\mathcal{K}}_{k}}\left(t,r\right){\omega}\left(t\right){f}\left(t\right)\mathrm{d}t,$
(2.26)

with ${\operatorname{\mathcal{K}}_{k}}$ defined by (2.17).

###### Proof.

First, by (2.8) and (2.7), for any $f\in
L^{2}\left(\operatorname{\mathbb{R}}^{6}\right)$, we have,

$\displaystyle{f}\left(x\right)=\sum_{k=0}^{\infty}\sum_{j=1}^{\alpha_{k}}{f_{k,j}}\left(\left|x\right|\right){Y_{k,j}}\left(\frac{x}{\left|x\right|}\right),$
(2.27)

where

${f_{k,j}}\left(r\right)=\frac{1}{\pi^{3}}\int_{\mathbb{S}^{5}}{f}\left(r\xi\right){Y_{k,j}}\left(\xi\right)\mathrm{d}{\bm{\sigma}}\left(\xi\right).$
(2.28)

Using the fact that,

$\operatorname{\mathop{}\\!\mathbin{\bigtriangleup}}\left({f_{k,j}}\left(\left|x\right|\right){Y_{k,j}}\left(\frac{x}{\left|x\right|}\right)\right)={f_{k,j}^{\prime\prime}}\left(\left|x\right|\right)+\frac{5}{\left|x\right|}{f_{k,j}^{\prime}}\left(\left|x\right|\right)+\frac{1}{\left|x\right|^{2}}\operatorname{\mathop{}\\!\mathbin{\bigtriangleup}}_{\mathbb{S}^{5}}{Y_{k,j}}\left(\frac{x}{\left|x\right|}\right),$

and (2.5), we obtain that,

$\displaystyle-\operatorname{\mathop{}\\!\mathbin{\bigtriangleup}}{f}\left(x\right)=\sum_{k=0}^{\infty}\sum_{j=1}^{\alpha_{k}}\left(-{f_{k,j}^{\prime\prime}}\left(\left|x\right|\right)-\frac{5}{\left|x\right|}{f_{k,j}^{\prime}}\left(\left|x\right|\right)+\frac{k\left(k+4\right)}{\left|x\right|^{2}}{f_{k,j}}\left(\left|x\right|\right)\right){Y_{k,j}}\left(\frac{x}{\left|x\right|}\right).$
(2.29)

Next, since $\omega$ is radial (see (1.3)), by Section 2, we have,

$\displaystyle-{\Phi}\left(\omega^{2}\right)\left(x\right){f}\left(x\right)=-\sum_{k=0}^{\infty}\sum_{j=1}^{\alpha_{k}}2\pi^{\nicefrac{{3}}{{2}}}{\omega}\left(\left|x\right|\right){f_{k,j}}\left(\left|x\right|\right){Y_{k,j}}\left(\frac{x}{\left|x\right|}\right).$
(2.30)

Now, by (2), we have,

$\displaystyle-2{\Phi}\left(\omega
f\right)\left(x\right){\omega}\left(x\right)=-2\sum_{k=0}^{\infty}\sum_{j=1}^{\alpha_{k}}\int_{0}^{+\infty}{\operatorname{\mathcal{K}}_{k}}\left(r,\left|x\right|\right){\left(\omega
f\right)_{k,j}}\left(r\right)\mathrm{d}r{Y_{k,j}}\left(\frac{x}{\left|x\right|}\right){\omega}\left(\left|x\right|\right).$
(2.31)

Since $\omega$ is radial, it follows from (2.7) that,

${\left(\omega
f\right)_{k,j}}\left(r\right)={\omega}\left(r\right){{f}_{k,j}}\left(r\right),$

which, together with (2.31), implies that,

$\displaystyle-2{\Phi}\left(\omega
f\right)\left(x\right){\omega}\left(x\right)=-2\sum_{k=0}^{\infty}\sum_{j=1}^{\alpha_{k}}\int_{0}^{+\infty}{\operatorname{\mathcal{K}}_{k}}\left(r,\left|x\right|\right){\omega}\left(r\right){{f}_{k,j}}\left(r\right)\mathrm{d}r{Y_{k,j}}\left(\frac{x}{\left|x\right|}\right){\omega}\left(\left|x\right|\right).$
(2.32)

Combining (2.29), (2.30) and (2.32), we obtain that (2.25) holds with
$\operatorname{\mathcal{L}}_{k}$ defined by (2.26). ∎

######  Lemma .

Let $\operatorname{\mathcal{L}}_{k}$ be defined by (2.26). For any $f\geq 0$
with $f\neq 0$, we have,

$\displaystyle\langle{\operatorname{\mathcal{L}}_{k}f},f\rangle>\langle{\operatorname{\mathcal{L}}_{k-1}f},f\rangle.$
(2.33)

###### Proof.

By an elementary computation, we have,

$\displaystyle\langle{\operatorname{\mathcal{L}}_{k}f},f\rangle-\langle{\operatorname{\mathcal{L}}_{k-1}f},f\rangle$
$\displaystyle=$
$\displaystyle\int_{0}^{+\infty}\left(k\left(k+4\right)-\left(k-1\right)\left(k+3\right)\right){f^{2}}\left(r\right)r^{3}\mathrm{d}r$
$\displaystyle+2\pi^{3}\int_{0}^{+\infty}\frac{{\omega}\left(r\right){f}\left(r\right)}{r^{k-1}}\int_{0}^{r}\left(r-t\right){\omega}\left(t\right)t^{k+4}{f}\left(t\right)\mathrm{d}t\mathrm{d}r$
$\displaystyle+2\pi^{3}\int_{0}^{+\infty}{{\omega}\left(r\right){f}\left(r\right)}{r^{k+4}}\int_{r}^{+\infty}\left(r-t\right)\frac{{\omega}\left(t\right){f}\left(t\right)}{t^{k-1}}\mathrm{d}t\mathrm{d}r,$

which, combining with $f\geq 0$ and $f\neq 0$ shows that, (2.33) holds. ∎

The following lemma shows that the operator $L$, defined by (1.6), is
nonnegative under suitable orthogonal condition.

######  Lemma  ([16]).

Let $L$ be defined by (1.6) and $\omega$ be defined by (1.3). If $f\in
H^{1}\left(\operatorname{\mathbb{R}}^{6}\right)$ satisfies $\left({\nabla
f},\nabla\omega\right)=0$, then $\left({Lf},f\right)\geq 0$.

As a consequence of Section 2, using (2.24), we obtain the following results,
which will be important in our analysis.

######  Corollary .

Let $L$ be defined by (1.6) and $\omega$ be defined by (1.3). If $f\in
L^{2}\left(\operatorname{\mathbb{R}}^{6}\right)$ satisfies $Lf\in
L^{2}\left(\operatorname{\mathbb{R}}^{6}\right)$ and
$\left({f},\omega^{2}\right)=0$, then we have $\left({Lf},f\right)\geq 0$.

######  Corollary .

Let $L$ be defined by (1.6), $Y\in\operatorname{\mathcal{H}}_{k}$ for some
$k\geq 1$, and $g\in L^{2}\left(\left(0,+\infty\right),r^{5}\right)$. If
${f}\left(x\right)={g}\left(\left|x\right|\right){Y}\left(\frac{x}{\left|x\right|}\right)$
satisfies $Lf\in L^{2}\left(\operatorname{\mathbb{R}}^{6}\right)$, then we
have $\left({Lf},f\right)\geq 0$.

###### Proof.

First, since $\omega^{2}$ is radial, by changing the variables
$r=\left|x\right|$ and $\xi=\frac{x}{\left|x\right|}$, we have,

$\int_{\operatorname{\mathbb{R}}^{6}}{f}\left(x\right){\omega^{2}}\left(x\right)\mathrm{d}x=\int_{0}^{+\infty}{f}\left(r\right){\omega^{2}}\left(r\right)r^{5}\mathrm{d}r\int_{\mathbb{S}^{5}}{Y}\left(\xi\right)\mathrm{d}{\bm{\sigma}}\left(\xi\right),$

which, combining (2.3) and (2.6), implies that
$\left({f},\omega^{2}\right)=0$. Therefore, by Section 2, we have
$\left({Lf},f\right)\geq 0$. ∎

##  3\. Proof of main results

In this section, we prepare the proof of Theorem 1.1 by means of several
propositions. The following proposition shows that the kernel of the operator
(1.6) restricted to the radial functions is
$\operatorname{span}\left\\{\Lambda\omega\right\\}$.

######  Proposition .

Let $\operatorname{\mathcal{L}}_{0}$ be defined by (2.26) with $k=0$. If
$\varphi\in L^{2}\left(\left(0,+\infty\right),r^{5}\right)$ with
$\varphi\not\equiv 0$ satisfies $\operatorname{\mathcal{L}}_{0}\varphi=0$,
then there exists $\eta\in\operatorname{\mathbb{R}}$, such that

${\varphi}\left(r\right)=\eta{\Lambda\omega}\left(r\right),$ (3.1)

where $\Lambda\omega$ is defined by (1.5).

###### Proof.

First of all, for any $\varphi\in
L^{2}\left(\left(0,+\infty\right),r^{5}\right)$ satisfies
$\operatorname{\mathcal{L}}_{0}\varphi=0$, a bootstrap argument implies that
$\varphi$ is smooth and ${\varphi^{\prime}}\left(0\right)=0$.

Now, we cliam that,

${\operatorname{\mathcal{L}}_{0}\Lambda\omega}\left(r\right)=0.$ (3.2)

Indeed, for all $\lambda>0$,
${\omega_{\lambda}}\left(x\right):=\lambda^{2}{\omega}\left(\lambda x\right)$
satisfy (1.1), i.e.

$\displaystyle-\operatorname{\mathop{}\\!\mathbin{\bigtriangleup}}{\omega_{\lambda}}\left(x\right)-\int_{\operatorname{\mathbb{R}}^{6}}\frac{\left|{\omega_{\lambda}}\left(y\right)\right|^{2}}{\left|x-y\right|^{4}}\mathrm{d}y\,{\omega_{\lambda}}\left(x\right)=0,\quad
x\in\operatorname{\mathbb{R}}^{6},$ (3.3)

by differentiation equation (3.3) with respect to $\lambda$, we obtain that,

$\displaystyle{L\Lambda\omega}\left(x\right)=0,\quad
x\in\operatorname{\mathbb{R}}^{6},$

where $L$ is defined by (1.6). Moreover, since ${\Lambda\omega}\left(x\right)$
is radial, it follows from (2.7) that,

${\left(\Lambda\omega\right)_{0,1}}\left(r\right)={\Lambda\omega}\left(r\right),\quad\text{and}\quad{\left(\Lambda\omega\right)_{k,j}}\left(r\right)=0,\quad\text{for
all~{}~{}}k\geq 1\text{~{}~{}and~{}~{}}1\leq j\leq\alpha_{k},$

which, together with Theorem 2.2 implies that (3.2) holds.

Next, let us rewirte ${\operatorname{\mathcal{L}}_{0}\varphi}\left(r\right)$
as

$\displaystyle{\operatorname{\mathcal{L}}_{0}\varphi}\left(r\right)={\operatorname{\mathfrak{L}}_{0}\varphi}\left(r\right)-2\pi^{3}{\omega}\left(r\right)\int_{0}^{+\infty}{\varphi}\left(t\right){\omega}\left(t\right)t\mathrm{d}t,$
(3.4)

where

$\displaystyle{\operatorname{\mathfrak{L}}_{0}\varphi}\left(r\right)=$
$\displaystyle-{\varphi^{\prime\prime}}\left(r\right)-\frac{5}{r}{\varphi^{\prime}}\left(r\right)-2\pi^{\nicefrac{{3}}{{2}}}{\omega}\left(r\right){\varphi}\left(r\right)-2\pi^{3}\int_{0}^{r}\left(\frac{t^{4}}{r^{4}}-1\right){\varphi}\left(t\right){\omega}\left(t\right)t\mathrm{d}t.$
(3.5)

Therefore, for any $\varphi\in L^{2}\left(\left(0,+\infty\right),r^{5}\right)$
with $\varphi\not\equiv 0$ satisfying
$\operatorname{\mathcal{L}}_{0}\varphi=0$, by (3.4) and (3.5), we have,

$\displaystyle{\operatorname{\mathfrak{L}}_{0}\varphi}\left(r\right)=2\pi^{3}{\omega}\left(r\right)\int_{0}^{+\infty}{\varphi}\left(t\right){\omega}\left(t\right)t\mathrm{d}t,$
(3.6)

and specially,

$\displaystyle{\operatorname{\mathfrak{L}}_{0}\Lambda\omega}\left(r\right)=2\pi^{3}{\omega}\left(r\right)\int_{0}^{+\infty}{\Lambda\omega}\left(t\right){\omega}\left(t\right)t\mathrm{d}t.$
(3.7)

By setting

$\displaystyle{\phi}\left(r\right)={\varphi}\left(r\right)-\frac{\int_{0}^{+\infty}{\varphi}\left(t\right){\omega}\left(t\right)t\mathrm{d}t}{\int_{0}^{+\infty}{\Lambda\omega}\left(t\right){\omega}\left(t\right)t\mathrm{d}t}{\Lambda\omega}\left(r\right),$

we have $\phi\in L^{2}\left(\left(0,+\infty\right),r^{5}\right)$ is smooth and
${\phi^{\prime}}\left(0\right)=0$, Moreover, an elementary calculation implies
that,

${\operatorname{\mathfrak{L}}_{0}\phi}\left(r\right)=0.$

Now, by Appendix A, we have

${\phi}\left(r\right)=0\quad\text{for all}\quad r\in\left[0,+\infty\right),$

which implies that (3.1) holds with
$\eta=\frac{\int_{0}^{+\infty}{\varphi}\left(t\right){\omega}\left(t\right)t\mathrm{d}t}{\int_{0}^{+\infty}{\Lambda\omega}\left(t\right){\omega}\left(t\right)t\mathrm{d}t}$.
This ends the proof of Section 3.

∎

######  Proposition .

Let $\operatorname{\mathcal{L}}_{k}$ be defeind by (2.26). For each $k\geq 1$,
the operator $\operatorname{\mathcal{L}}_{k}$ is bounded below and essentially
self-adjoint on $C_{0}^{\infty}\left(0,+\infty\right)\subset
L^{2}\left(\left(0,+\infty\right),r^{5}\mathrm{d}r\right)$. Moreover, for each
$k\geq 1$, the operator $\operatorname{\mathcal{L}}_{k}$ enjoys the Perron-
Frobenius property, i.e. if

$\lambda_{k}^{0}=\inf\\{\,\langle{\operatorname{\mathcal{L}}_{k}f},f\rangle\;|\;\int_{0}^{+\infty}\left|{f}\left(r\right)\right|^{2}r^{5}\mathrm{d}r=1\,\\}$

is attained, then the lowest eigenvalue $\lambda_{k}^{0}$ of the operator
$\operatorname{\mathcal{L}}_{k}$ is simple, and the corresponding
eigenfunction ${\chi_{k}^{0}}\left(r\right)$ dose not change sign on
$\left(0,+\infty\right)$.

###### Proof.

The proof of Section 3 is almost identical to that of [11, Lemma 7], therefore
we omit the details. ∎

######  Proposition .

Let $\operatorname{\mathcal{L}}_{1}$ be defined by (2.26) with $k=1$. If
$\psi\in L^{2}\left(\left(0,+\infty\right),r^{5}\right)$ with $\psi\not\equiv
0$ satisfies $\operatorname{\mathcal{L}}_{1}\psi=0$, then there exists
$\beta\in\operatorname{\mathbb{R}}$, such that

${\psi}\left(r\right)=\beta{\omega^{\prime}}\left(r\right).$ (3.8)

Moreover,

$\lambda_{1}^{0}=\inf\\{\,\langle{\operatorname{\mathcal{L}}_{1}f},f\rangle\;|\;\int_{0}^{+\infty}\left|{f}\left(r\right)\right|^{2}r^{5}\mathrm{d}r=1\,\\}=0.$
(3.9)

###### Proof.

First, by differentiation (1.4) with respect to $z$, and taking $\lambda=1$,
$z=0$, we obtain,

${L\frac{\partial\omega}{\partial x_{j}}}\left(x\right)=0,\qquad 1\leq j\leq
6.$ (3.10)

Since $\omega$ is radial, by (2.4), we have,

${\frac{\partial\omega}{\partial
x_{j}}}\left(x\right)=6{\omega^{\prime}}\left(\left|x\right|\right){Y_{1,j}}\left(\frac{x_{j}}{\left|x\right|}\right),\qquad
1\leq j\leq 6,$ (3.11)

which, together with (2.7) implies that,

${\omega^{\prime}_{k,j}}\left(r\right)=\begin{cases}0,&k=0,j=1,\\\
{\omega^{\prime}}\left(r\right),&k=1,1\leq j\leq 6,\\\ 0,&k\geq 2,1\leq
j\leq\alpha_{k}.\end{cases}$ (3.12)

By Theorem 2.2, using (3.10), (3.11) and (3.12), we get,

${\operatorname{\mathcal{L}}_{1}\omega^{\prime}}\left(r\right)=0,$

which implies that, $0$ is an eigenvalue of the operator
$\operatorname{\mathcal{L}}_{1}$ with the eigenfunction
${\omega^{\prime}}\left(r\right)$. Moreover, since
${\omega^{\prime}}\left(r\right)<0$ for all $r\in\left(0,+\infty\right)$, by
Section 3,

$0=\inf\\{\,\left({\operatorname{\mathcal{L}}_{1}f},f\right)\;|\;\int_{0}^{+\infty}\left|{f}\left(r\right)\right|^{2}r^{5}\mathrm{d}r=1\,\\}$

is the lowest eigenvalue of the operator $\operatorname{\mathcal{L}}_{1}$, and
any function $\psi\in L^{2}\left(\left(0,+\infty\right),r^{5}\right)$
satisfying ${\operatorname{\mathcal{L}}_{1}\psi}\left(r\right)=0$ must belongs
to the set
$\\{\,\beta\omega^{\prime}\;|\;\beta\in\operatorname{\mathbb{R}}\,\\}$. This
ends the proof of Section 3. ∎

######  Proposition .

Let $k\geq 2$ and $\operatorname{\mathcal{L}}_{k}$ be defined by (2.26). For
any $\varrho\in L^{2}\left(\left(0,+\infty\right),r^{5}\right)$ satisfying
$\operatorname{\mathcal{L}}_{1}\varrho=0$, we have $\varrho\equiv 0$.

###### Proof.

We argue by contradiction. For any $\varrho\in
L^{2}\left(\left(0,+\infty\right),r^{5}\right)$, by letting

${f_{\rho,k}}\left(x\right)={\varphi}\left(\left|x\right|\right){Y_{k,1}}\left(\frac{x}{\left|x\right|}\right),$

we have

$f_{\rho,k}\in
L^{2}\left(\operatorname{\mathbb{R}}^{6}\right),\quad\text{and}\quad\left({f_{\rho,k}},\omega^{2}\right)=0,$

which together with Section 2, implies that,

$\langle{\operatorname{\mathcal{L}}_{1}\rho},\rho\rangle=\left({Lf_{\rho,k}},f_{\rho,k}\right)\geq
0,$

therefore,

$\lambda_{k}^{0}=\inf\\{\,\langle{\operatorname{\mathcal{L}}_{k}\rho},\rho\rangle\;|\;\int_{0}^{+\infty}\left|{\rho}\left(r\right)\right|^{2}r^{5}\mathrm{d}r=1\,\\}\geq
0.$

If $\lambda_{k}^{0}=0$ is attained or $\lambda_{k}^{0}>0$, then for any
$\rho\in L^{2}\left(\left(0,+\infty\right),r^{5}\right)$ with $\rho\neq 0$, we
have $\left({\operatorname{\mathcal{L}}_{k}\rho},\rho\right)>0$, which
contradicts with $\operatorname{\mathcal{L}}_{k}\rho=0$.

If $\lambda_{k}^{0}=0$ is attained, then by Section 3, the lowest eigenvalue
$0$ of the operator $\operatorname{\mathcal{L}}_{k}$ is simple, and the
corresponding eigenfunction $\varrho$ with
$\int_{0}^{+\infty}\left|{\rho}\left(r\right)\right|r^{5}\mathrm{d}r=1$ does
not change sign on $\left(0,+\infty\right)$. Without loss of generality, we
may assume that ${\varrho}\left(r\right)>0$ for all
$r\in\left(0,+\infty\right)$. By (2), we obtain that,

$\langle{\operatorname{\mathcal{L}}_{1}\varrho},\varrho\rangle<\langle{\operatorname{\mathcal{L}}_{k}\varrho},\varrho\rangle=0,$

which contradicts with (3.9). This ends the proof of Section 3. ∎

###### Proof of Theorem 1.1.

First, by (2.7), (2.8), and Theorem 2.2, we have,

$f\in
L^{2}\left(\operatorname{\mathbb{R}}^{6}\right)\quad\text{satisfies}\quad{Lf}=0,$

if and only if

${\operatorname{\mathcal{L}}_{k}f_{k,j}}\left(r\right)=0,\quad\text{for
all}\quad k\geq 0\quad\text{and}\quad 1\leq j\leq\alpha_{k},$

where $f_{k,j}\in L^{2}\left(\left(0,+\infty\right),r^{5}\right)$ is defined
by (2.7), and

$\displaystyle{f}\left(x\right)=\sum_{k=0}^{\infty}\sum_{j=1}^{\alpha_{k}}{f_{k,j}}\left(\left|x\right|\right){Y_{k,j}}\left(\frac{x}{\left|x\right|}\right).$

Next, by Section 3, Section 3 and Section 3, there exist real numbers $\eta$
and $\beta_{j}$, ($1\leq j\leq 6$) such that,

${f}\left(x\right)=\eta{\Lambda\omega}\left(x\right)+\sum_{j=1}^{6}\beta_{j}{\omega^{\prime}}\left(\left|x\right|\right){\frac{x_{j}}{\left|x\right|}},$

i.e.

$f\in\operatorname{span}\left\\{\Lambda\omega,~{}\frac{\partial\omega}{\partial{x_{1}}},~{}\frac{\partial\omega}{\partial{x_{2}}},~{}\frac{\partial\omega}{\partial{x_{3}}},~{}\frac{\partial\omega}{\partial{x_{4}}},~{}\frac{\partial\omega}{\partial{x_{5}}},~{}\frac{\partial\omega}{\partial{x_{6}}}\right\\}.$

This ends the proof of Theorem 1.1. ∎

##  Appendix A Properties of the linear operator
$\operatorname{\mathfrak{L}}_{0}$

######  Proposition .

Let $\operatorname{\mathfrak{L}}_{0}$ be define by (3.5). If $\phi\in
C^{2}\left[0,+\infty\right)$ with ${\phi}\left(0\right)\neq 0$ and
${\phi^{\prime}}\left(0\right)=0$ satisfies
${\operatorname{\mathfrak{L}}_{0}\phi}\left(r\right)=0$, then
${\phi}\left(r\right)$ does not change sign and

${\left|{\phi}\left(r\right)\right|}>\frac{\left|{\phi}\left(0\right)\right|}{4},\quad\text{for
all}\quad r\in\left[0,+\infty\right).$ (A.1)

###### Proof.

First, by setting

${\tilde{\phi}}\left(r\right)=\frac{2{\omega}\left(0\right)}{{\phi}\left(0\right)}{\phi}\left(r\right),\quad\text{and
}\quad{\lambda}\left(r\right)=\frac{{\tilde{\phi}}\left(r\right)}{{\omega}\left(r\right)}.$
(A.2)

we obtain that $\tilde{\phi}$ satisfies the linear equaion
${\operatorname{\mathfrak{L}}_{0}\tilde{\phi}}\left(r\right)=0$ as well.
Moreover, since $\phi\in C^{2}\left[0,+\infty\right)$ and $\omega\in
C^{\infty}\left[0,+\infty\right)$, by (A.2), we have $\lambda\in
C^{2}\left[0,+\infty\right)$ with ${\lambda}\left(0\right)>1$. Let us define

$\displaystyle
r^{\ast}=\sup\\{\,r>0\;|\;{\lambda}\left(t\right)>1,\quad\text{for all }\quad
t\in\left[0,r\right]\,\\}.$

By the definition of $r^{\ast}$, it is obvious that,

${\lambda^{\prime}}\left(r^{\ast}\right)\leq 0.$ (A.3)

We claim that

$r^{\ast}=+\infty.$ (A.4)

For this, we argue by contradiction, assuming that $r^{\ast}<+\infty$ and
obtaining a contradiction with the definition of $r^{\ast}$. By noting that
$\omega$ satisfies

$-{\omega^{\prime\prime}}\left(r\right)-\frac{5}{r}{\omega^{\prime}}\left(r\right)-2\pi^{\nicefrac{{3}}{{2}}}{\omega^{2}}\left(r\right)=0,$

we have,

${\operatorname{\mathfrak{L}}_{0}\omega}\left(r\right)=-2\pi^{3}{\omega}\left(r\right)\int_{0}^{r}\left(\frac{t^{4}}{r^{4}}-1\right){\omega}\left(t\right){\omega}\left(t\right)t\mathrm{d}t.$
(A.5)

By combining ${\operatorname{\mathfrak{L}}_{0}\tilde{\phi}}\left(r\right)=0$
with (A.5), we obtain that,

$\displaystyle
2\pi^{3}{\tilde{\phi}}\left(r\right){\omega}\left(r\right)\int_{0}^{r}\left(\frac{t^{4}}{r^{4}}-1\right){\omega}\left(t\right){\omega}\left(t\right)t\mathrm{d}t.$
$\displaystyle=$
$\displaystyle-{\omega}\left(r\right){\tilde{\phi}^{\prime\prime}}\left(r\right)-\frac{5}{r}{\omega}\left(r\right){\tilde{\phi}^{\prime}}\left(r\right)+{\tilde{\phi}}\left(r\right){\omega^{\prime\prime}}\left(r\right)+\frac{5}{r}{\tilde{\phi}}\left(r\right){\omega^{\prime}}\left(r\right)$
$\displaystyle-2\pi^{3}{\omega^{2}}\left(r\right)\int_{0}^{r}\left(\frac{t^{4}}{r^{4}}-1\right){\tilde{\phi}}\left(t\right){\omega}\left(t\right)t\mathrm{d}t+2\pi^{3}{\tilde{\phi}}\left(r\right){\omega}\left(r\right)\int_{0}^{r}\left(\frac{t^{4}}{r^{4}}-1\right){\omega^{2}}\left(t\right)t\mathrm{d}t.$

Moreove, it is elementary to check that,

$\displaystyle\left[r^{5}\left({\tilde{\phi}^{\prime}}\left(r\right){\omega}\left(r\right)-{\omega^{\prime}}\left(r\right){\tilde{\phi}}\left(r\right)\right)\right]^{\prime}=-2\pi^{3}r^{5}{\omega^{2}}\left(r\right)\int_{0}^{r}\left(\frac{t^{4}}{r^{4}}-1\right){\tilde{\phi}}\left(t\right){\omega}\left(t\right)t\mathrm{d}t,$
(A.6)

which implies that,

$\displaystyle\left[r^{5}{\omega^{2}}\left(r\right){\lambda^{\prime}}\left(r\right)\right]^{\prime}=2\pi^{3}r{\omega^{2}}\left(r\right)\int_{0}^{r}\left({r^{4}}-{t^{4}}\right){\tilde{\phi}}\left(t\right){\omega}\left(t\right)t\mathrm{d}t.$
(A.7)

By integrating (A.7) from $0$ to $r^{\ast}$, we obtain that,

$\displaystyle{\lambda^{\prime}}\left(r^{\ast}\right)=$
$\displaystyle\frac{2\pi^{3}}{{r^{\ast}}^{5}{\omega^{2}}\left(r^{\ast}\right)}\int_{0}^{r^{\ast}}s{\omega^{2}}\left(s\right)\int_{0}^{s}\left({s^{4}}-{t^{4}}\right){\tilde{\phi}}\left(t\right){\omega}\left(t\right)t\mathrm{d}t\mathrm{d}s$
$\displaystyle>$
$\displaystyle\frac{2\pi^{3}}{{r^{\ast}}^{5}{\omega^{2}}\left(r^{\ast}\right)}\int_{0}^{r^{\ast}}s{\omega^{2}}\left(s\right)\int_{0}^{s}\left({s^{4}}-{t^{4}}\right){\omega^{2}}\left(t\right)t\mathrm{d}t\mathrm{d}s.$

Using the fact ${\omega}\left(r\right)>0$, we have,
${\lambda^{\prime}}\left(r^{\ast}\right)>0$, which contradicts with (A.3).
Therefore (A.4) holds.

Now, by integrating (A.7) from $0$ to $r$ and an elementary calculation, we
obtain that,

$\displaystyle{\lambda}\left(r\right)-{\lambda}\left(0\right)>\frac{9}{5}r^{4}+\frac{12}{5}r^{2}-\frac{12}{5}{\log}\left(1+r^{2}\right),$

which, together with ${\lambda}\left(0\right)>1$, implies that,
${\lambda}\left(r\right)>r^{4}+1\geq\frac{1}{2}\left(1+r^{2}\right)^{2}$. By
(1.3) and (A.2), we obtain that
$\frac{{\phi}\left(r\right)}{{\phi}\left(0\right)}>\frac{1}{4}$. Therefore,
(A.1) holds. ∎

As a consequence of Appendix A, we have the following corollary.

######  Corollary .

Let $\operatorname{\mathfrak{L}}_{0}$ be define by (3.5). If $\phi\in
L^{2}\left(\left(0,+\infty\right),r^{5}\right)\cap
C^{2}\left(0,+\infty\right)$ satisfies
${\operatorname{\mathfrak{L}}_{0}\phi}\left(r\right)=0$ with
${\phi^{\prime}}\left(0\right)=0$, then ${\phi}\left(r\right)=0$ for all
$r\in\left[0,+\infty\right)$.

### Acknowledgements.

G. Xu was supported by National Key Research and Development Program of China
(No. 2020YFA0712900) and by NSFC (No. 11831004). X. Tang was supported by NSFC
(No. 12001284).

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