**On Positive Quadratic Hyponormality of a Unilateral Weighted Shift with Recursively Generated by Five Weights**

#### **Chunji Li <sup>1</sup> and Cheon Seoung Ryoo 2,\***


Received: 16 January 2019; Accepted: 8 February 2019; Published: 25 February 2019

**Abstract:** Let 1 < *a* < *b* < *c* < *d* and *α*ˆ[5] := 1, <sup>√</sup>*a*, √*b*, <sup>√</sup>*c*, √*d* ∧ be a weighted sequence that is recursively generated by five weights 1, <sup>√</sup>*a*, √*b*, <sup>√</sup>*c*, <sup>√</sup>*d*. In this paper, we give sufficient conditions for the positive quadratic hyponormalities of *Wα*(*x*) and *Wα*(*y*,*x*), with *α* (*x*) : <sup>√</sup>*x*, *<sup>α</sup>*ˆ[5] and *α* (*y*, *x*) : √*y*, <sup>√</sup>*x*, *<sup>α</sup>*ˆ[5].

**Keywords:** positively quadratically hyponormal; quadratically hyponormal; unilateral weighted shift; recursively generated

**MSC:** 47B37; 47B20

#### **1. Introduction**

Let H be a separable, infinite dimensional, complex Hilbert space, and let L(H) be the algebra of all bounded linear operators on H. An operator *T* in L(H) is said to be *normal* if *T*∗*T* = *TT*∗, *hyponormal* if *<sup>T</sup>*∗*<sup>T</sup>* <sup>≥</sup> *TT*∗, and *subnormal* if *<sup>T</sup>* <sup>=</sup> *<sup>N</sup>*|H , where *<sup>N</sup>* is normal on some Hilbert space *<sup>K</sup>* ⊇ H. For *A*, *B* ∈ L(H), let [*A*, *B*] := *AB* − *BA*. We say that an *n*-tuple *T* = (*T*1, ... , *Tn*) of operators in L(H) is *hyponormal* if the operator matrix ([*T*∗ *<sup>j</sup>* , *Ti*])*<sup>n</sup> <sup>i</sup>*,*j*=<sup>1</sup> is positive on the direct sum of *n* copies of H. For arbitrary positive integer *<sup>k</sup>*, *<sup>T</sup>* ∈ L(H) is (strongly) *<sup>k</sup>*-*hyponormal* if (*I*, *<sup>T</sup>*, ... , *<sup>T</sup>k*) is hyponormal. It is well known that *T* is subnormal if and only if *T* is ∞-hyponormal. An operator *T* in L(H) is said to be *weakly n*-*hyponormal* if *p*(*T*) is hyponormal for any polynomial *p* with degree less than or equal to *n*. An operator *T* is *polynomially hyponormal* if *p*(*T*) is hyponormal for every polynomial *p*. In particular, the weak two-hyponormality (or weak three-hyponormality) is referred to as quadratical hyponormality (or cubical hyponormality, resp.) and has been considered in detail in [1–9].

Let {*en*}<sup>∞</sup> *<sup>n</sup>*=<sup>0</sup> be the canonical orthonormal basis for Hilbert space *l* <sup>2</sup> (Z+),and let *<sup>α</sup>* :<sup>=</sup> {*αn*}<sup>∞</sup> *<sup>n</sup>*=<sup>0</sup> be a bounded sequence of positive numbers. Let *W<sup>α</sup>* be a unilateral weighted shift defined by *Wαen* := *αnen*+<sup>1</sup> (*n* ≥ 0). It is well known that *W<sup>α</sup>* is hyponormal if and only if *α<sup>n</sup>* ≤ *αn*+<sup>1</sup> (*n* ≥ 0). The moments of *W<sup>α</sup>* are usually defined by *γ*<sup>0</sup> := 1, *γ<sup>i</sup>* := *α*<sup>2</sup> <sup>0</sup> ··· *<sup>α</sup>*<sup>2</sup> *<sup>i</sup>*−<sup>1</sup> (*<sup>i</sup>* <sup>≥</sup> <sup>1</sup>). It is well known that *<sup>W</sup><sup>α</sup>* is subnormal if and only if there exists a Borel probability measure *μ* supported in 0, *Wα*<sup>2</sup> , with *Wα*<sup>2</sup> <sup>∈</sup> supp *<sup>μ</sup>*, such that [10] *γ<sup>n</sup>* = \$ *t <sup>n</sup>*d*<sup>μ</sup>* (*t*) (∀*<sup>n</sup>* ≥ <sup>0</sup>). It follows from [11] (Theorem 4) that *<sup>W</sup><sup>α</sup>* is subnormal if and only if for every *k* ≥ 1 and every *n* ≥ 0, the Hankel matrix:

$$A\left(n,k\right) := \begin{bmatrix} \gamma\_n & \gamma\_{n+1} & \gamma\_{n+2} & \cdots & \gamma\_{n+k} \\ \gamma\_{n+1} & \gamma\_{n+2} & \gamma\_{n+3} & \cdots & \gamma\_{n+k+1} \\ \gamma\_{n+2} & \gamma\_{n+3} & \gamma\_{n+4} & \cdots & \gamma\_{n+k+2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \gamma\_{n+k} & \gamma\_{n+k+1} & \gamma\_{n+k+2} & \cdots & \gamma\_{n+2k} \end{bmatrix} \ge 0.1$$

A weighted shift *W<sup>α</sup>* is said to be *recursively generated* if there exists *i* ≥ 1 and *ϕ* = (*ϕ*0,..., *ϕi*−1) ∈ C*<sup>i</sup>* such that:

$$
\gamma\_n = \varphi\_{i-1}\gamma\_{n-1} + \dots + \varphi\_0\gamma\_{n-i} \quad (n \ge i), i
$$

where *γ<sup>n</sup>* is the moment of *Wα*, i.e., *γ*<sup>0</sup> := 1, *γ<sup>i</sup>* := *α*<sup>2</sup> <sup>0</sup> ··· *<sup>α</sup>*<sup>2</sup> *<sup>i</sup>*−<sup>1</sup> (*<sup>i</sup>* <sup>≥</sup> <sup>1</sup>), equivalently,

$$\kappa\_n^2 = \wp\_{i-1} + \frac{\wp\_{i-2}}{a\_{n-1}^2} + \dots + \frac{\wp\_0}{a\_{n-1}^2 \cdot \dots \cdot a\_{n-i+1}^2} \quad (n \ge i) \dots$$

Given an initial segment of weights *α* : *α*0, ... , *α*2*<sup>k</sup>* (*k* ≥ 0), there is a canonical procedure to generate a sequence (denoted *α*ˆ) in such a way that *Wα*<sup>ˆ</sup> is a recursively-generated shift having *α* as an initial segment of weight. In particular, given an initial segment of weights *α* : <sup>√</sup>*a*, √*b*, <sup>√</sup>*<sup>c</sup>* with <sup>0</sup> <sup>&</sup>lt; *<sup>a</sup>* <sup>&</sup>lt; *<sup>b</sup>* <sup>&</sup>lt; *<sup>c</sup>*, we obtain *<sup>ϕ</sup>*<sup>0</sup> <sup>=</sup> <sup>−</sup>*ab*(*c*−*b*) *<sup>b</sup>*−*<sup>a</sup>* and *<sup>ϕ</sup>*<sup>1</sup> <sup>=</sup> *<sup>b</sup>*(*c*−*a*) *<sup>b</sup>*−*<sup>a</sup>* .

In [12,13], Curto-Putinar proved that there exists an operator that is polynomially hyponormal, but not two-hyponormal. Although the existence of a weighted shift, which is polynomially hyponormal, but not subnormal, was established in [12,13], a concrete example of such weighted shifts has not been found yet. Recently, the authors in [14] proved that the subnormality is equivalent to the polynomial hyponormality for recursively-weighted shift *Wα* with *α* : <sup>√</sup>*x*, <sup>√</sup>*a*, √*b*, √*c* ∧ . Based on this, in this paper, we have to consider the weighted shift operator with five generated elements.

The organization of this paper is as follows. In Section 2, we recall some terminology and notations concerning the quadratic hyponormality and positive quadratic hyponormality of unilateral weighted shifts *Wα*. In Section 3, we give some results on the unilateral weighted shifts with recursively generated by five weights *α*ˆ[5] := 1, <sup>√</sup>*a*, √*b*, <sup>√</sup>*c*, √*d* ∧ (1 < *a* < *b* < *c* < *d*). In Section 4, we consider positive quadratic hyponormalities of *W<sup>α</sup>* with weights *α* : <sup>√</sup>*x*, *<sup>α</sup>*ˆ[5] and *<sup>α</sup>* : √*y*, <sup>√</sup>*x*, *<sup>α</sup>*ˆ[5]. In Section 5, we give more results on the positive quadratic hyponormality for any unilateral weighted shift *Wα*. In Section 6, we present the conclusions.

#### **2. Preliminaries and Notations**

Recall that a weighted shift *W<sup>α</sup>* is *quadratically hyponormal* if *W<sup>α</sup>* + *sW*<sup>2</sup> *<sup>α</sup>* is hyponormal for any *<sup>s</sup>* <sup>∈</sup> <sup>C</sup> [2], i.e., *<sup>D</sup>*(*s*) := [(*W<sup>α</sup>* <sup>+</sup> *sW*<sup>2</sup> *<sup>α</sup>* )∗, *W<sup>α</sup>* + *sW*<sup>2</sup> *<sup>α</sup>* ] <sup>≥</sup> 0, for any *<sup>s</sup>* <sup>∈</sup> <sup>C</sup>. Let {*ei*}<sup>∞</sup> *<sup>i</sup>*=<sup>0</sup> be an orthonormal basis for H, and let *Pn* be the orthogonal projection on ∨*<sup>n</sup> <sup>i</sup>*=0{*ei*}. For *<sup>s</sup>* <sup>∈</sup> <sup>C</sup>, we let:

$$\begin{array}{rcl} D\_{\mathfrak{n}}(\mathbf{s}) & = & P\_{\mathfrak{n}}[(\mathcal{W}\_{\mathfrak{n}} + \mathrm{s}\mathcal{W}\_{\mathfrak{n}}^{2})^{\*}, \mathcal{W}\_{\mathfrak{n}} + \mathrm{s}\mathcal{W}\_{\mathfrak{n}}^{2}]P\_{\mathfrak{n}} \\ & & \left[ \begin{array}{ccccc} q\_{0} & r\_{0} & 0 & \cdots & 0 & 0 \\ \overline{r\_{0}} & q\_{1} & r\_{1} & \cdots & 0 & 0 \\ 0 & \overline{r\_{1}} & q\_{2} & \ddots & 0 & 0 \\ \vdots & \vdots & \ddots & \ddots & \ddots & \vdots \\ \vdots & \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & q\_{n-1} & r\_{n-1} \\ 0 & 0 & 0 & \cdots & \overline{r\_{n-1}} & q\_{n} \end{array} \right] \end{array}$$

where:

$$\begin{array}{lcl}\boldsymbol{q}\_{k} & := \boldsymbol{u}\_{k} + \left|\boldsymbol{s}\right|^{2} \boldsymbol{v}\_{k} & \boldsymbol{r}\_{k} := \boldsymbol{w}\_{k} \bar{\mathtt{s}},\\\boldsymbol{u}\_{k} & := \boldsymbol{a}\_{k}^{2} - \boldsymbol{a}\_{k-1}^{2} \quad \boldsymbol{v}\_{k} := \boldsymbol{a}\_{k}^{2} \boldsymbol{a}\_{k+1}^{2} - \boldsymbol{a}\_{k-2}^{2} \boldsymbol{a}\_{k-1}^{2},\\\boldsymbol{w}\_{k} & := \boldsymbol{a}\_{k}^{2} (\boldsymbol{a}\_{k+1}^{2} - \boldsymbol{a}\_{k-1}^{2})^{2}, \text{ for } k \ge 0,\end{array}$$

and *<sup>α</sup>*−<sup>1</sup> <sup>=</sup> *<sup>α</sup>*−<sup>2</sup> :<sup>=</sup> 0. Hence, *<sup>W</sup><sup>α</sup>* is quadratically hyponormal if and only if *Dn*(*s*) <sup>≥</sup> 0 for every *<sup>s</sup>* <sup>∈</sup> <sup>C</sup> and every *n* ≥ 0. Hence, we consider *dn* (·) := det *Dn* (·), which is a polynomial in *t* := |*s*| <sup>2</sup> of degree *n* + 1, with Maclaurin expansion *dn* (*t*) := ∑*n*+<sup>1</sup> *<sup>i</sup>*=<sup>0</sup> *c* (*n*, *i*)*t i* . It is easy to find the following recursive relations [2]:

$$\begin{cases} \begin{aligned} d\_0\left(t\right) &= q\_{0\prime} \\ d\_1\left(t\right) &= q\_0 q\_1 - \left|r\_0\right|^2 \\ d\_{n+2}\left(t\right) &= q\_{n+2} d\_{n+1}\left(t\right) - \left|r\_{n+1}\right|^2 d\_n\left(t\right) \quad \left(n \ge 0\right) \end{aligned} \end{cases}$$

Furthermore, we can obtain the following:

$$\begin{aligned} \mathcal{c}\left(0,0\right) &= \quad u\_{0\prime} \; \mathcal{c}\left(0,1\right) = v\_{0\prime} \\ \mathcal{c}\left(1,0\right) &= \quad u\_1 u\_{0\prime} \; \mathcal{c}\left(1,1\right) = u\_1 v\_0 + u\_0 v\_1 - w\_{0\prime} \; \mathcal{c}\left(1,2\right) = v\_1 v\_{0\prime} \end{aligned}$$

and:

$$\begin{aligned} \mathcal{L}\left(n+2,i\right) &= \begin{array}{c} u\_{n+2}c\left(n+1,i\right) + v\_{n+2}c\left(n+1,i-1\right) - w\_{n+1}c\left(n,i-1\right) \\ \end{array} \\ &\quad \left(n \ge 0, \text{ and } 0 \le i \le n+1\right). \end{aligned}$$

In particular, for any *n* ≥ 0, we have:

$$c\left(n,0\right) = \mu\_0 \mu\_1 \cdots \mu\_n, \ c\left(n, n+1\right) = \upsilon\_0 \upsilon\_1 \cdots \upsilon\_n.$$

Furthermore, we can obtain the following results.

**Lemma 1.** *Let ρ* := *v*<sup>2</sup> (*u*0*v*<sup>1</sup> − *w*0) + *v*<sup>0</sup> (*u*1*v*<sup>2</sup> − *w*1). *Then, for any n* ≥ 4, *we have:*

$$\begin{aligned} \mathcal{L}\left(n,n\right) &= \begin{aligned} u\_n \mathfrak{c}\left(n-1,n\right) + \left(u\_{n-1}v\_n - w\_{n-1}\right)\mathfrak{c}\left(n-2,n-1\right) \\ &+ \sum\_{i=1}^{n-3} v\_n v\_{n-1} \cdot \cdots \upsilon\_{i+3} \left(u\_{i+1}v\_{i+2} - w\_{i+1}\right)\mathfrak{c}\left(i,i+1\right) + v\_n v\_{n-1} \cdot \cdots \upsilon\_3 \rho. \end{aligned} $$

**Lemma 2.** *Let τ* := *u*<sup>0</sup> (*u*1*v*<sup>2</sup> − *w*1). *Then, for any n* ≥ 4, *we have:*

$$\begin{aligned} \mathcal{L}\left(n, n-1\right) &= \begin{aligned} u\_{il}\boldsymbol{c}\left(n-1, n-1\right) + \left(u\_{n-1}v\_{n} - w\_{n-1}\right)\boldsymbol{c}\left(n-2, n-2\right) \\ &+ \sum\_{i=1}^{n-3} v\_{n}v\_{n-1} \cdot \boldsymbol{\cdot} \boldsymbol{v}\_{i+3} \left(u\_{i+1}v\_{i+2} - w\_{i+1}\right)\boldsymbol{c}\left(i, i\right) + v\_{n}v\_{n-1} \cdot \boldsymbol{\cdot} \boldsymbol{v}\_{3}\boldsymbol{\tau} .\end{aligned} \end{aligned}$$

**Lemma 3.** *For any n* ≥ 5 *and* 0 ≤ *i* ≤ *n* − 2, *we have:*

$$\begin{split} \mathcal{c}\left(n,i\right) &= \quad \mu\_{\mathbb{R}}\mathcal{c}\left(n-1,i\right) + \left(\mu\_{n-1}v\_{\mathbb{R}} - w\_{n-1}\right)\mathcal{c}\left(n-2,i-1\right) \\ &\quad + \sum\_{j=1}^{n-3} v\_{\mathbb{R}}v\_{n-1} \cdot \cdots \upsilon\_{j+3} \left(u\_{j+1}v\_{j+2} - w\_{j+1}\right)\mathcal{c}\left(j,j+i-n+1\right) \\ &\quad + \upsilon\_{\mathbb{R}}\upsilon\_{n-1} \cdot \cdots \upsilon\_{5}\mathcal{c}\left(i-n+5,0\right) \left(u\_{i-n+6}v\_{i-n+7} - w\_{i-n+6}\right) .\end{split}$$

To detect the positivity of *dn* (*t*), we need the following concept.

**Definition 1.** *Let α* : *α*0, *α*1, ... *be a positive weight sequence. We say that W<sup>α</sup> is positively quadratically hyponormal if c* (*n*, *i*) ≥ 0 *for all n*, *i* ≥ 0 *with* 0 ≤ *i* ≤ *n* + 1, *and c* (*n*, *n* + 1) > 0 *for all n* ≥ 0 *[2].*

Positive quadratic hyponormality implies quadratic hyponormality, but the converse is false [15]. In addition, the authors in [15] showed that the positive quadratic hyponormality is equivalent to the quadratic hyponormality for recursively-generated weighted shift *W<sup>α</sup>* with *α* : <sup>√</sup>*x*, <sup>√</sup>*a*, √*b*, √*c* ∧ (here, 0 < *x* ≤ *a* < *b* < *c*).

#### **3. Recursive Relation of** *Wα*ˆ[5]

Given the initial segment of weights *<sup>α</sup>* : 1, <sup>√</sup>*a*, √*b*, <sup>√</sup>*c*, <sup>√</sup>*<sup>d</sup>* with 1 <sup>&</sup>lt; *<sup>a</sup>* <sup>&</sup>lt; *<sup>b</sup>* <sup>&</sup>lt; *<sup>c</sup>* <sup>&</sup>lt; *<sup>d</sup>*, we obtain the moments:

$$
\gamma\_0 = \gamma\_1 = 1, \quad \gamma\_2 = a, \ \gamma\_3 = ab, \ \gamma\_4 = abc, \ \gamma\_5 = abcd.
$$

Let:

$$V\_0 = \begin{pmatrix} \gamma\_0 \\ \gamma\_1 \\ \gamma\_2 \end{pmatrix}, \quad V\_1 = \begin{pmatrix} \gamma\_1 \\ \gamma\_2 \\ \gamma\_3 \end{pmatrix}, \quad V\_2 = \begin{pmatrix} \gamma\_2 \\ \gamma\_3 \\ \gamma\_4 \end{pmatrix},$$

and we assume that *V*0, *V*1, *V*<sup>2</sup> are linearly independent, i.e.,

$$\det\left(V\_{0\prime}V\_{1\prime}V\_2\right) = \det A(0,2) \neq 0.$$

Then, there exist three nonzero numbers *ϕ*0, *ϕ*1, *ϕ*2, such that:

$$
\begin{pmatrix} \gamma\_3\\\gamma\_4\\\gamma\_5 \end{pmatrix} = \wp\_0 \begin{pmatrix} \gamma\_0\\\gamma\_1\\\gamma\_2 \end{pmatrix} + \wp\_1 \begin{pmatrix} \gamma\_1\\\gamma\_2\\\gamma\_3 \end{pmatrix} + \wp\_2 \begin{pmatrix} \gamma\_2\\\gamma\_3\\\gamma\_4 \end{pmatrix}.
$$

A straightforward calculation shows that:

$$\begin{array}{rcl}\varphi\_0 &=& \frac{ab(ab^2 - 2abc + bc^2 + acd - bcd)}{a^2 - 2ab + ab^2 + bc - abc},\\\varphi\_1 &=& \frac{-ab(ab - ac - bc + bc^2 + cd - bcd)}{a^2 - 2ab + ab^2 + bc - abc},\\\varphi\_2 &=& \frac{b(a^2 - ab - ac + abc + cd - acd)}{a^2 - 2ab + ab^2 + bc - abc}.\end{array}$$

Thus:

$$
\gamma\_{n+1} = \varphi\_0 \gamma\_{n-2} + \varphi\_1 \gamma\_{n-1} + \varphi\_2 \gamma\_n \qquad \qquad (n \ge 5),
$$

i.e.,

$$\alpha\_n^2 = \varphi\_2 + \frac{\varphi\_1}{a\_{n-1}^2} + \frac{\varphi\_0}{a\_{n-1}^2 a\_{n-2}^2} \tag{1}$$

By (1), we can obtain a recursively-generated weighted shift, and we set it as *α*ˆ[5]. In this case, we call the weighted shift operator *Wα*ˆ[5] with rank three.

**Proposition 1.** *Wα*ˆ[5] *with rank three is subnormal if and only if: (1)* 1 < *a* < *b*, *(2) c* > *<sup>a</sup> b* (*b*−1) 2 *<sup>a</sup>*−<sup>1</sup> <sup>+</sup> <sup>1</sup> , *(3) d* > *<sup>b</sup> c* (*c*−*a*) 2 *<sup>b</sup>*−*<sup>a</sup>* <sup>+</sup> *<sup>a</sup>* .

**Proof.** See [16], Example 3.6.

**Proposition 2.** *If Wα*ˆ[5] *with rank three is subnormal, then ϕ*<sup>0</sup> > 0, *ϕ*<sup>1</sup> < 0 *and ϕ*<sup>2</sup> > 0.

**Proof.** By Proposition 1, we know that:

$$\begin{aligned} F\_{1} &: \quad = ab^2 - 2ab + bc + a^2 - abc < 0, \\ F\_{2} &: \quad = ab^2 + bc^2 - 2abc + acd - bcd < 0. \end{aligned}$$

Thus, *ϕ*<sup>0</sup> > 0. Since:

$$ab - ac - bc + bc^2 + c\left(1 - b\right)d < \left(c - b\right)\frac{F\_1}{b - a} < 0\_\prime$$

and:

 $a^2 - ab - ac + abc + c$   $(1 - a)d < (c - a)\frac{F\_1}{b - a} < 0$ .

we have *ϕ*<sup>1</sup> < 0 and *ϕ*<sup>2</sup> > 0. The proof is complete.

**Proposition 3.** *Let un*−<sup>1</sup> *un* = *β<sup>n</sup>* (*n* ≥ 5). *Then:*

$$\beta\_n = -\frac{a\_{n-1}^2 a\_{n-2}^2 a\_{n-3}^2}{\varphi\_1 a\_{n-3}^2 + \varphi\_0 + \varphi\_0 \beta\_{n-1}}.\tag{2}$$

**Proof.** Since:

$$\begin{aligned} u\_n &=& -a\_n^2 - a\_{n-1}^2 \\ &=& -\frac{q\_1}{a\_{n-1}^2 a\_{n-2}^2} u\_{n-1} - \frac{q\_0}{a\_{n-1}^2 a\_{n-2}^2 a\_{n-3}^2} \left( u\_{n-1} + u\_{n-2} \right) \\ &=& -\left( \frac{q\_1}{a\_{n-1}^2 a\_{n-2}^2} + \frac{q\_0}{a\_{n-1}^2 a\_{n-2}^2 a\_{n-3}^2} \right) u\_{n-1} - \frac{q\_0}{a\_{n-1}^2 a\_{n-2}^2 a\_{n-3}^2} u\_{n-2}, \end{aligned}$$

so:

$$\begin{split} 1 &=& -\left(\frac{\varphi\_1}{a\_{n-1}^2 a\_{n-2}^2} + \frac{\varphi\_0}{a\_{n-1}^2 a\_{n-2}^2 a\_{n-3}^2}\right) \frac{u\_{n-1}}{u\_n} - \frac{\varphi\_0}{a\_{n-1}^2 a\_{n-2}^2 a\_{n-3}^2} \frac{u\_{n-2}}{u\_n} \\ &=& -\left(\frac{\varphi\_1}{a\_{n-1}^2 a\_{n-2}^2} + \frac{\varphi\_0}{a\_{n-1}^2 a\_{n-2}^2 a\_{n-3}^2} + \frac{\varphi\_0}{a\_{n-1}^2 a\_{n-2}^2 a\_{n-3}^2} \frac{u\_{n-2}}{u\_{n-3}}\right) \frac{u\_{n-1}}{u\_n}. \end{split}$$

Thus, we have:

$$\begin{split} \beta\_{n} &=& -\frac{1}{\left(\frac{q\_{1}}{a\_{n-1}^{2}a\_{n-2}^{2}} + \frac{q\_{0}}{a\_{n-1}^{2}a\_{n-2}^{2}a\_{n-3}^{2}} + \frac{q\_{0}}{a\_{n-1}^{2}a\_{n-2}^{2}a\_{n-3}^{2}}\beta\_{n-1}\right)} \\ &=& -\frac{a\_{n-1}^{2}a\_{n-2}^{2}a\_{n-3}^{2}}{\varphi\_{1}a\_{n-3}^{2} + \varphi\_{0} + \varphi\_{0}\beta\_{n-1}}. \end{split}$$

Thus, we have our conclusion.

Since lim*n*→<sup>∞</sup> *<sup>α</sup>*<sup>2</sup> *<sup>n</sup>* <sup>=</sup> *<sup>L</sup>*2, we let lim*<sup>n</sup>*→<sup>∞</sup> *<sup>β</sup><sup>n</sup>* :<sup>=</sup> *<sup>β</sup>*∗, and by (2), we have *<sup>β</sup>*<sup>∗</sup> <sup>=</sup> <sup>−</sup> *<sup>L</sup>*<sup>6</sup> *<sup>ϕ</sup>*1*L*2+*ϕ*0+*ϕ*0*β*<sup>∗</sup> ; hence:

$$\beta^\* = -\frac{1}{2} - \frac{\varrho\_1}{2\varrho\_0} L^2 + \frac{1}{2\varrho\_0} \sqrt{\left(\varrho\_1^2 - 4\varrho\_0\varrho\_2\right) L^4 - 2\varrho\varrho\varrho\_1 L^2 - 3\varrho\_0^2}. \tag{3}$$

#### **4. Main Results**

First, we give the following result (cf. [11], Corollary 5).

**Proposition 4.** *Let W<sup>α</sup> be any unilateral weighted shift. Then, W<sup>α</sup> is two-hyponormal if and only if θ<sup>k</sup>* := *ukvk*<sup>+</sup><sup>1</sup> <sup>−</sup> *wk* <sup>≥</sup> 0, <sup>∀</sup>*<sup>k</sup>* <sup>∈</sup> <sup>N</sup>.

It is well-known that if *Wα* is two-hyponormal or positively quadratically hyponormal, then *Wα* is quadratically hyponormal. By Proposition 4 and Lemma 1∼3, we have the following result.

**Theorem 1.** *Let Wα be any unilateral weighted shift. If Wα is 2-hyponormal, then Wα is positively quadratically hyponormal.*

*4.1. The Positive Quadratic Hyponormality of Wα*(*x*)

Let *α* (*x*) : <sup>√</sup>*x*, *<sup>α</sup>*ˆ[5] with 0 <sup>&</sup>lt; *<sup>x</sup>* <sup>≤</sup> 1, and we consider the following (*<sup>n</sup>* <sup>+</sup> <sup>1</sup>) <sup>×</sup> (*<sup>n</sup>* <sup>+</sup> <sup>1</sup>) matrix:

$$D\_{\mathtt{n}} = \begin{bmatrix} u\_0 + v\_0 t & \sqrt{w\_0 t} & 0 & \cdots & 0 \\ \sqrt{w\_0 t} & u\_1 + v\_1 t & \sqrt{w\_1 t} & \cdots & 0 \\ 0 & \sqrt{w\_1 t} & u\_2 + v\_2 t & \ddots & 0 \\ \vdots & \vdots & \ddots & \ddots & \sqrt{w\_{\mathtt{n-1}} t} \\ 0 & 0 & \cdots & \sqrt{w\_{\mathtt{n-1}} t} & u\_\mathtt{n} + v\_\mathtt{n} t \end{bmatrix} . \tag{4}$$

Let *dn* = det *Dn* = ∑*n*+<sup>1</sup> *<sup>i</sup>*=<sup>0</sup> *c* (*n*, *i*)*t i* . Then:

**Lemma 4.** *c*(*n*, *i*) ≥ 0, *n* = 0, 1, 2, *and* 0 ≤ *i* ≤ *n* + 1*.*

**Proof.** In fact, *c*(1, 1) = *x* (*a* − *x*) > 0, *c*(2, 1) = *ax* (1 − *x*) (*b* − 1) > 0, and:

$$\begin{aligned} x(2,2) &=& ax \left( \left( ab - 1 \right) - \left( b - 1 \right) x \right) \\ &>& ax \left( \left( ab - a \right) - \left( b - 1 \right) x \right) \\ &=& ax \left( a - x \right) \left( b - 1 \right) > 0. \end{aligned}$$

Thus, we have our conclusion.

**Lemma 5.** *Assume that θ<sup>k</sup>* := *ukvk*<sup>+</sup><sup>1</sup> − *wk* ≥ 0 *for k* ≥ 2. *Then, for n* ≥ 3, 0 ≤ *i* ≤ *n* + 1, *we have:*

$$\mathcal{L}(n,i) \ge u\_n \mathcal{c}(n-1,i) + \upsilon\_n \cdot \dots \cdot \upsilon\_3 [\upsilon\_2 \mathcal{c}(1,i-n+1) - w\_1 \mathcal{c}(0,i-n+1)].$$

**Proof.** For *n* = 3, 0 ≤ *i* ≤ 4,

$$\begin{aligned} \mathcal{L}c(3,i) &=& \mathfrak{u}\_3 c(2,i) + v\_3 c(2,i-1) - w\_2 c(1,i-1) \\ &=& \mathfrak{u}\_3 c(2,i) + (\mathfrak{u}\_2 \mathbf{v}\_3 - \mathbf{w}\_2) c(1,i-1) + v\_3 [v\_2 c(1,i-2) - w\_1 c(0,i-2)] \\ &\ge& \mathfrak{u}\_3 c(2,i) + v\_3 [v\_2 c(1,i-2) - w\_1 c(0,i-2)]. \end{aligned}$$

By the inductive hypothesis, we have our result.

Thus, if *θ<sup>k</sup>* := *ukvk*<sup>+</sup><sup>1</sup> − *wk* ≥ 0 for *k* ≥ 2, then by Lemma 1∼3 and Lemma 5, for *n* ≥ 3, we have:

$$
\mathcal{L}(n,i) \ge \begin{cases}
\upsilon\_n \cdot \cdot \upsilon\_2 c(1,2), & \text{for} \quad i = n+1, \\
\upsilon\_n c(n-1,n) + \upsilon\_n \cdot \cdot \upsilon\_3 \rho\_\prime & \text{for} \quad i = n, \\
\upsilon\_n c(n-1,n-1) + \upsilon\_n \cdot \cdot \upsilon\_3 \tau\_\prime & \text{for} \quad i = n-1, \\
\upsilon\_n c(n-1,i)\_\prime & \text{for} \quad 0 \le i \le n-2, \\
\end{cases}
$$

where *ρ* = *ax* (*b* − 1) (*a* − *x*) > 0, *τ* = *x* (*a* (*b* − *a*) − (*ab* − 2*a* + 1) *x*). Therefore, when *n* ≥ 3, we have:

$$\begin{cases} \ c(n, n+1) > 0, \\ c(n, n) > u\_n c(n-1, n) + v\_n \cdot \cdots \cdot v\_3 \rho \ge 0, \\ c(n, i) \ge u\_n \cdot \cdots \cdot u\_{i+2} c(i+1, i) \ (n \ge 3, 0 \le i \le n-2). \end{cases}$$

To complete our analysis of the coefficients *c*(*n*, *i*), it suffices to determine the values of *x* for which *c*(*n*, *n* − 1) ≥ 0 (*n* ≥ 3).

**Lemma 6.** *c*(*n*, *n* − 1) ≥ 0 (*for any n* ≥ 3), *if c*(3, 2) ≥ 0, *c*(4, 3) ≥ 0 *and An* := *v*0*v*1*v*2*unun*−<sup>1</sup> + *unvn*−1*ρ* + *vnvn*−1*τ* ≥ 0 (*n* ≥ 5).

**Proof.** For *n* ≥ 4, by Lemma 2, we have:

$$\begin{aligned} x(n, n-1) &\geq \quad u\_n c(n-1, n-1) + v\_n \cdot \cdot v\_3 \tau \\ &\geq \quad u\_n [u\_{n-1} c(n-2, n-1) + v\_{n-1} \cdot \cdot \cdot v\_3 \rho] + v\_n \cdot \cdot v\_3 \tau \end{aligned}$$

and since *c*(*n* − 2, *n* − 1) = *vn*−<sup>2</sup> ··· *v*0, we get:

$$\mathcal{L}(n, n-1) \ge u\_n(u\_{n-1}v\_{n-2}\cdots v\_0 + v\_{n-1}\cdots v\_3\rho) + v\_n\cdots v\_3\tau.$$

If *n* ≥ 5, we can factor *vn*−<sup>2</sup> ··· *v*<sup>3</sup> to get:

$$\begin{aligned} \mathcal{L}(\boldsymbol{v}, \boldsymbol{n}-1) &\geq & \boldsymbol{v}\_{\boldsymbol{n}-2} \cdots \boldsymbol{v}\_{3} (\boldsymbol{v}\_{0} \boldsymbol{v}\_{1} \boldsymbol{v}\_{2} \boldsymbol{u}\_{\boldsymbol{n}} \boldsymbol{u}\_{\boldsymbol{n}-1} + \boldsymbol{u}\_{\boldsymbol{n}} \boldsymbol{v}\_{\boldsymbol{n}-1} \boldsymbol{\rho} + \boldsymbol{v}\_{\boldsymbol{n}} \boldsymbol{v}\_{\boldsymbol{n}-1} \boldsymbol{r}) \\ &= & \boldsymbol{v}\_{\boldsymbol{n}-2} \cdots \boldsymbol{v}\_{3} A\_{\boldsymbol{n}}. \end{aligned}$$

Hence, we have our result.

Let:

$$\mathfrak{x}\_n := \sup \{ \mathfrak{x} : c(n, n-1) \ge 0 \text{ in } \mathcal{W}\_a \} \quad (n \ge 3).$$

By direct computations, we have:

$$\begin{array}{llll}\texttt{x}\_{3} &=& \frac{a\theta\_{2} + a\left(b - a\right)v\_{3} + a\left(ab - 1\right)u\_{3}}{\theta\_{2} + a\left(b - 1\right)u\_{3} + \left(-2a + ab + 1\right)v\_{3}},\\\texttt{x}\_{4} &=& \frac{a\left(ab - 1\right)\theta\_{3} + a\left(u\_{4} + v\_{4}\right)\theta\_{2} + \left(a^{2}\left(b - 1\right)u\_{4} + v\_{4}a\left(b - a\right)\right)v\_{3} + a^{2}bu\_{3}u\_{4}}{a\left(b - 1\right)\theta\_{3} + v\_{4}\theta\_{2} + \left(\left(ab - 2a + 1\right)v\_{4} + u\_{4}a\left(b - 1\right)\right)v\_{3} + au\_{3}u\_{4}}.\end{array}$$

For *n* ≥ 5, a calculation using the specific form of *v*0, *v*1, *v*2, *ρ* and *τ* shows that:

$$\begin{aligned} A\_n &= -\left[a^2 b u\_n u\_{n-1} + a^2 (b-1) u\_n v\_{n-1} + a(b-a) v\_n v\_{n-1}\right. \\ &\quad - \left(a u\_n u\_{n-1} + a(b-1) u\_n v\_{n-1} + (ab+1-2a) v\_n v\_{n-1}\right) x] x\_n \end{aligned}$$

it follows that:

$$\mathbf{x}\_{n} = \frac{a^2 b u\_n u\_{n-1} + a^2 (b-1) u\_n v\_{n-1} + a (b-a) v\_n v\_{n-1}}{a u\_n u\_{n-1} + a (b-1) u\_n v\_{n-1} + (ab+1-2a) v\_n v\_{n-1}}.$$

Let *zn* := *vn un* (*un* = 0). Then, for *n* ≥ 5,

$$\mathbf{x}\_n = \frac{a^2 b + a^2 (b - 1) z\_{n-1} + a (b - a) z\_n z\_{n-1}}{a + a(b - 1) z\_{n-1} + (ab + 1 - 2a) z\_n z\_{n-1}}.\tag{5}$$

**Lemma 7.** lim*n*→<sup>∞</sup> *zn* = *K* := (1 + *β*∗) *<sup>ϕ</sup>*<sup>2</sup> <sup>−</sup> *<sup>ϕ</sup>*<sup>0</sup> *<sup>L</sup>*<sup>4</sup> *β*<sup>∗</sup> , *where β*∗ *as in (3)*.

**Proof.** Since:

$$
\alpha\_{n+1}^2 = \varphi\_2 + \frac{\varphi\_1}{\alpha\_n^2} + \frac{\varphi\_0}{\alpha\_n^2 \alpha\_{n-1}^2} \quad (n \ge 5),
$$

from which it follows that:

$$
\alpha\_n^2 \alpha\_{n+1}^2 = \varphi\_2 \alpha\_n^2 + \varphi\_1 + \frac{\varphi\_0}{\alpha\_{n-1}^2}.
$$

Thus:

$$
\omega\_n = \varrho\_2(u\_n + u\_{n-1}) - \frac{\varrho\_0(u\_{n-1} + u\_{n-2})}{a\_{n-1}^2 a\_{n-3}^2}. \tag{6}
$$

Thus, we have (for *n* ≥ 5):

$$z\_{\rm tr} = \wp\_2(1 + \beta\_{\rm tr}) - \frac{\wp\_0}{\alpha\_{\rm n-1}^2 \alpha\_{\rm n-3}^2} \beta\_{\rm n} (1 + \beta\_{\rm n-1}) \dots$$

Since *α*<sup>2</sup> *<sup>n</sup>* → *<sup>L</sup>*2, we have:

$$\lim\_{n \to \infty} z\_n = (1 + \beta^\*) \left( \wp\_2 - \frac{\wp\_0}{L^4} \beta^\* \right) \dots$$

Thus, we have our conclusion.

Let:

$$f(z,w) := \frac{a^2b + a^2(b-1)z + a(b-a)zw}{a + a(b-1)z + (ab+1-2a)zw}.$$

By Lemma 7 and the fact in [2] (p. 399), if *zn* is increasing, then we know that {*xn*}*n*≥<sup>5</sup> in (5) is decreasing and inf*n*≥<sup>5</sup> *xn* = *f*(*K*, *K*). Thus, we have the following result.

**Theorem 2.** *Assume that Wα*ˆ[5] *with rank three is subnormal. Let α* (*x*) : <sup>√</sup>*x*, *<sup>α</sup>*ˆ[5], *and let:*

> *h*+ <sup>2</sup> := sup{*x* : *Wα*(*x*) *be positively quadratically hyponormal*}.

*If zn* := *vn un* (*n* ≥ 5) *is increasing, then:*

$$h\_2^+ \ge \min\left\{1, \mathbf{x}\_{3\prime}, \mathbf{x}\_{4\prime} \frac{a^2b + a^2(b-1)K + a(b-a)K^2}{a + a(b-1)K + (ab+1-2a)K^2}\right\} \ge $$

*where:*

$$\begin{array}{llll}\mathbf{x}\_{3} &=& \frac{a\theta\_{2} + a\left(b - a\right)\upsilon\_{3} + a\left(ab - 1\right)\upsilon\_{3}}{\theta\_{2} + a\left(b - 1\right)\upsilon\_{3} + \left(-2a + ab + 1\right)\upsilon\_{3}},\\\mathbf{x}\_{4} &=& \frac{a\left(ab - 1\right)\theta\_{3} + a\left(\mu\_{4} + \upsilon\_{4}\right)\theta\_{2} + \left(a^{2}\left(b - 1\right)\mu\_{4} + a\left(b - a\right)\upsilon\_{4}\right)\upsilon\_{3} + a^{2}b\upsilon\_{3}\mu\_{4}}{a\left(b - 1\right)\theta\_{3} + \upsilon\_{4}\theta\_{2} + \left(a\left(b - 1\right)\upsilon\_{4} + \left(ab - 2a + 1\right)\upsilon\_{4}\right)\upsilon\_{3} + a\upsilon\_{3}\mu\_{4}},\\\mathbf{K}\_{\perp} &=& \left(1 + \boldsymbol{\theta}^{\*}\right)\left(\boldsymbol{\varrho}\_{2} - \frac{\boldsymbol{\varrho}\boldsymbol{\varrho}\_{0}}{L^{4}}\boldsymbol{\delta}^{\*}\right). \end{array}$$

**Remark 1.** *By (6), we know that if β<sup>n</sup> is increasing, then so is zn*. *Hence, our problems are as follows.*

**Problem 1.** *Let <sup>α</sup>* :<sup>=</sup> {*αn*}<sup>∞</sup> *<sup>n</sup>*=<sup>0</sup> *be any unilateral weighted sequence. If W<sup>α</sup> is subnormal, is β<sup>n</sup> increasing or not? In particular, what is the answer for subnormal Wα*ˆ[5] *with rank three?*

**Example 1.** *Let α* (*x*) : <sup>√</sup>*x*, 1, <sup>√</sup>2, <sup>√</sup>3, <sup>√</sup>4, <sup>√</sup><sup>5</sup> ∧ . *Then, <sup>ϕ</sup>*<sup>0</sup> = 6, *<sup>ϕ</sup>*<sup>1</sup> = −18, *<sup>ϕ</sup>*<sup>2</sup> = 9, *<sup>L</sup>*<sup>2</sup> ≈ 6.2899, *and <sup>K</sup>* <sup>≈</sup> 32.118, *<sup>x</sup>*<sup>3</sup> <sup>=</sup> <sup>17</sup> <sup>18</sup> <sup>≈</sup> 0.94444, *<sup>x</sup>*<sup>4</sup> <sup>=</sup> <sup>226</sup> <sup>249</sup> <sup>≈</sup> 0.90763, *<sup>f</sup>*(*K*, *<sup>K</sup>*) <sup>≈</sup> 0.77512. *We obtain <sup>h</sup>*<sup>+</sup> <sup>2</sup> - 0.77512. *That is, if* 0 < *x* 0.77512, *then Wα*(*x*) *is positively quadratically hyponormal. Numerically, we know that β<sup>n</sup> and zn are all increasing. See the following Table 1.*


**Table 1.** Numerical data for *βn* and *zn* in Example 1.

#### *4.2. The Positive Quadratic Hyponormality of Wα*(*y*,*x*)

Let *α* (*y*, *x*) : √*y*, <sup>√</sup>*x*, *<sup>α</sup>*ˆ[5]. We also consider the matrix as in (4), and let *dn* <sup>=</sup> det *Dn* <sup>=</sup> ∑*n*+<sup>1</sup> *<sup>i</sup>*=<sup>0</sup> *c* (*n*, *i*)*t i* . Then:

$$\begin{aligned} \mathcal{c}\left(1,1\right) &=& \text{xy}\left(1-y\right)>0, \\ \mathcal{c}\left(2,1\right) &=& \mathcal{y}\left(\mathbf{x}-\mathbf{y}\right)\left(a-\mathbf{x}\right)>0, \\ \mathcal{c}\left(2,2\right) &=& \text{xy}\left(a-ay+\mathbf{x}y-\mathbf{x}^2\right) \ge \mathbf{x}\mathbf{y}\left(1-y\right)\left(a-\mathbf{x}\right)>0, \end{aligned}$$

and:

$$\begin{aligned} \mathcal{L}\left(3,1\right) &\geq \quad \mathfrak{u}\_3 \mathfrak{c}\left(2,1\right) > 0, \\ \mathcal{L}\left(3,2\right) &= \quad \mathfrak{y}\left(\left(\mathbf{x}-\boldsymbol{a}\right)\left(\mathbf{x}^2 - 2\mathbf{x} + \boldsymbol{a}\boldsymbol{b}\right)\mathfrak{y} + \mathfrak{x}\left(\left(1-\boldsymbol{a}\right)\mathbf{x}^2 + \boldsymbol{a}\left(1-\boldsymbol{b}\right)\mathbf{x} + \boldsymbol{a}\left(\boldsymbol{a}\boldsymbol{b}-1\right)\right)\mathfrak{y}\right), \\ \mathcal{L}\left(3,3\right) &\geq \quad \mathfrak{u}\_3\mathfrak{c}\left(2,3\right) + \mathfrak{v}\_3\mathfrak{\rho} > 0. \end{aligned}$$

Since:

$$\begin{array}{rcl} \rho &=& \text{xy} \left(1 - y\right) \left(a - x\right) > 0, \\ \tau &=& y \left(\left(2x - x^2 - a\right) y + x \left(a - 1\right)\right), \end{array}$$

we can similarly show that *c*(*n*, *n* − 1) ≥ 0 for all *n* ≥ 3, if *c*(3, 2) ≥ 0, *c*(4, 3) ≥ 0 and *<sup>y</sup>* <sup>≤</sup> *<sup>x</sup>*((*a*−1)*K*2+(*a*−*x*)*K*+*ax*) (*a*−2*x*+*x*2)*K*2+(*a*−*x*)*xK*+*x*<sup>3</sup> , where *<sup>K</sup>* = (<sup>1</sup> <sup>+</sup> *<sup>β</sup>*∗) *<sup>ϕ</sup>*<sup>2</sup> <sup>−</sup> *<sup>ϕ</sup>*<sup>0</sup> *<sup>L</sup>*<sup>4</sup> *β*<sup>∗</sup> . Thus, we have the following result.

\*\*Theorem 3.\*\* Let  $a\left(\underline{y},\mathbf{x}\right) : \sqrt{\underline{y}}$ ,  $\sqrt{\mathbf{x}}$ ,  $\hbar \left[\underline{s}\right]$ .  $\operatorname{If } z\_{n} := \frac{\underline{v}\_{n}}{\underline{u}\_{n}}$  ( $n \ge 5$ ) is increasing, and:  $\mathbf{x}(1)$   $0 < \mathbf{x} \le \min\left\{ a - \sqrt{a\left(a - 1\right)}, \frac{ab(c - 1) - \sqrt{ab(a - 1)(a^{2}b + bc^{2} - ab - ac + 2a^{2} - a^{3} - abc)}}{a(a - 1) + b(c - a)} \right\}$ .

*(2)* 0 < *y* ≤ min {*x*, *f*<sup>1</sup> (*x*), *f*<sup>2</sup> (*x*), *f*<sup>3</sup> (*x*)} , *where:*

$$\begin{array}{rcl} f\_1(\mathbf{x}) &=& \frac{\mathbf{x}\left(a\left(ab-1\right) + a\left(1-b\right)\mathbf{x} - \left(a-1\right)\mathbf{x}^2\right)}{\left(a-\mathbf{x}\right)\left(ab-2\mathbf{x} + \mathbf{x}^2\right)},\\ f\_2\begin{pmatrix} x\\ \end{pmatrix} &=& -\frac{\mathbf{x}p\_1\left(\mathbf{x}\right)}{p\_2\left(\mathbf{x}\right)},\\ \end{array} \\\ \begin{split} f\_1(\mathbf{x}) &=& \left(a^2b + bc - a^2 - abc\right)\mathbf{x}^2 + \left(a^2b - 2ab + a^2 - ab^2c + abc\right)\mathbf{x} \\ &+ \left(a^2b^2 - 2a^3b + 2a^2b - cab\right),\\ p\_2\begin{pmatrix} x\\ \end{pmatrix} &=& \left(bc - ab - a + a^2\right)\mathbf{x}^3 + \left(a - a^2b + 3ab - 2bc - abc\right)\mathbf{x}^2\\ &+ \left(a^3b - 3a^2b - a^2 + 6ab^2 + 2cab\right)\mathbf{x} + a^2b\left(a - bc\right),\\ f\_3\begin{pmatrix} x\\ \end{pmatrix} &=& \frac{\mathbf{x}\left(\left(a - 1\right)K^2 + \left(a - \mathbf{x}\right)K + ax\right)}{\left(a - 2\mathbf{x} + \mathbf{x}^2\right)K^2 + \left(a - \mathbf{x}\right)\mathbf{x}K + \mathbf{x}^3},\ K = \left(1 + \boldsymbol{\beta}^\*\right)\left(\mathbf{p}\_2 - \frac{q\boldsymbol{p}\_0}{\mathbf{I}^4}\boldsymbol{\beta}^\*\right),\end{split}$$

*then Wα*(*y*,*x*) *is positively quadratically hyponormal.*

**Example 2.** *Let <sup>a</sup>* <sup>=</sup> 2, *<sup>b</sup>* <sup>=</sup> 3, *<sup>c</sup>* <sup>=</sup> 4, *<sup>d</sup>* <sup>=</sup> 5. *If* <sup>0</sup> <sup>&</sup>lt; *<sup>x</sup>* <sup>≤</sup> <sup>2</sup> <sup>−</sup> <sup>√</sup><sup>2</sup> <sup>≈</sup> 0.585 79, *<sup>y</sup>* <sup>≤</sup> *<sup>x</sup>*(*K*2+(2−*x*)*K*+2*x*) (*x*2−2*x*+2)*K*2+(2*x*−*x*2)*K*+*x*<sup>3</sup> *with K* ≈ 32.118, *then Wα*(*y*,*x*) *is positively quadratically hyponormal. See the following Figure 1.*

**Figure 1.** A subset of the region of positive quadratic hyponormality of *Wα*(*y*,*x*) in Example 2.

#### **5. More Results**

From the above discussions, we obtain the following criteria for any unilateral weighted shifts.

**Proposition 5.** *Let α* (*x*) : <sup>√</sup>*x*, 1, <sup>√</sup>*a*, √*b*, <sup>√</sup>*c*, <sup>√</sup>*d*, <sup>√</sup>*α*5, <sup>√</sup>*α*6, ... , *and <sup>α</sup>* : 1, <sup>√</sup>*a*, √*b*, <sup>√</sup>*c*, <sup>√</sup>*d*, <sup>√</sup>*α*5, <sup>√</sup>*α*6, ... *be subnormal weighted shifts. Let:*

> *h*+ <sup>2</sup> := sup{*x* : *Wα*(*x*) *be positively quadratically hyponormal*}.

*If zn* = *vn un* (*n* ≥ 5) *is increasing and zn* → *K (as n* → ∞*), then:*

$$h\_2^+ \ge \min\left\{1, \mathbf{x}\_{3\prime}, \mathbf{x}\_{4\prime} \frac{a^2b + a^2(b-1)K + a(b-a)K^2}{a + a(b-1)K + (ab+1-2a)K^2}\right\} \mathbf{x}$$

*where:*

$$\begin{array}{llll}\texttt{x}\_{3} &=& \frac{a\theta\_{2} + a\left(b - a\right)\upsilon\_{3} + a\left(ab - 1\right)\upsilon\_{3}}{\theta\_{2} + a\left(b - 1\right)\upsilon\_{3} + \left(-2a + ab + 1\right)\upsilon\_{3}},\\\texttt{x}\_{4} &=& \frac{a\left(ab - 1\right)\theta\_{3} + a\left(\mathsf{u}\_{4} + \mathsf{v}\_{4}\right)\theta\_{2} + \left(a^{2}\left(b - 1\right)\mathsf{u}\_{4} + a\left(b - a\right)\upsilon\_{4}\right)\upsilon\_{3} + a^{2}\mathsf{b}\mathsf{u}\_{3}\mathsf{u}\_{4}}{a\left(b - 1\right)\theta\_{3} + \mathsf{v}\_{4}\theta\_{2} + \left(a\left(b - 1\right)\mathsf{u}\_{4} + \left(ab - 2a + 1\right)\upsilon\_{4}\right)\upsilon\_{3} + a\mathsf{u}\_{3}\mathsf{u}\_{4}}.\end{array}$$

By Proposition 5, we can have the following results, but we omit the concrete computations.

.

\*\*Example 3.\*\* (1) Let  $\mathfrak{a}\left(\mathbf{x}\right) : \sqrt{\mathfrak{x}}$ ,  $\sqrt{\frac{2}{3}} \sqrt{\frac{3}{4}} \sqrt{\frac{4}{5}} \dots$ . Then,  $h\_{2}^{+} = \frac{2}{3}$  (cf. 111, Proposition 7).
(2) Let  $\mathfrak{a}\left(\mathbf{x}\right) : \sqrt{\mathfrak{x}}$ ,  $\sqrt{\frac{5}{8}} \sqrt{\frac{3}{4}} \sqrt{\frac{4}{5}} \sqrt{\frac{4}{5}} \times$ . Then,  $h\_{2}^{+} = \frac{1945}{3136}$  (cf. 1151, Theorem 3.7).
(3) Let  $\mathfrak{a}\left(\mathbf{x}\right) : \sqrt{\mathfrak{x}}$ ,  $\mathbf{1}\_{\prime} \left(\sqrt{2}, \sqrt{2.1}, \sqrt{12.1}\right)^{\wedge}$ . Then,  $h\_{2}^{+} \underset{\mathbf{x}}{\stackrel{\scriptstyle\rightarrow}{\rightarrow}} 0.16682.$ 
(4) Let  $\mathfrak{a}\left(\mathbf{x}\right) : \sqrt{\mathfrak{x}}$ ,  $\sqrt{\frac{n}{n+1} \cdot \frac{1}{2} \cdot \frac{2^{n+1} - 1}{2^n - 1}}$ . Then,  $h\_{2}^{+} = \frac{3}{4}$  (cf. 1161, Example 3.4).

#### **6. Conclusions**

In this work, we study a weighed shift operator for which the weights are recursively generated by five weights. We give sufficient conditions of the positive quadratic hyponormalities. Next, it is worth studying the cubic hyponormality, semi-weak *k*-hyponormalities, and so on.

**Author Contributions:** All authors contributed equally in writing this article. All authors read and approved the final manuscript.

**Funding:** This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MEST) (No. 2017R1A2B4006092).

**Conflicts of Interest:** The authors declare no conflict of interest.

#### **References**


c 2019 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).

### *Article* **Ground State Solutions for Fractional Choquard Equations with Potential Vanishing at Infinity**

**Huxiao Luo 1, Shengjun Li 2,\* and Chunji Li <sup>3</sup>**


Received: 8 December 2018; Accepted: 1 February 2019; Published: 5 February 2019

**Abstract:** In this paper, we study a class of nonlinear Choquard equation driven by the fractional Laplacian. When the potential function vanishes at infinity, we obtain the existence of a ground state solution for the fractional Choquard equation by using a non-Nehari manifold method. Moreover, in the zero mass case, we obtain a nontrivial solution by using a perturbation method. The results improve upon those in Alves, Figueiredo, and Yang (2015) and Shen, Gao, and Yang (2016).

**Keywords:** variational methods; fractional Choquard equation; ground state solution; vanishing potential

**MSC:** 35J50; 58E30

#### **1. Introduction**

In this paper, we deal with the following nonlocal equation:

$$\begin{cases} (-\Delta)^s u + V(\mathbf{x})u = \left( \int\_{\mathbb{R}^N} \frac{Q(y)F(u(y))}{|\mathbf{x} - \mathbf{y}|^\mu} dy \right) Q(\mathbf{x}) f(u), \text{ in } \mathbb{R}^N, \\\ u \in D^{s, 2}(\mathbb{R}^N), \end{cases} \tag{1}$$

where *<sup>N</sup>* <sup>≥</sup> 3, 0 <sup>&</sup>lt; *<sup>s</sup>* <sup>&</sup>lt; 1, 0 <sup>&</sup>lt; *<sup>μ</sup>* <sup>&</sup>lt; *<sup>N</sup>*, *<sup>V</sup>* <sup>∈</sup> *<sup>C</sup>*(R*N*, [0, <sup>∞</sup>)), *<sup>Q</sup>* <sup>∈</sup> *<sup>C</sup>*(R*N*,(0, <sup>∞</sup>)), *<sup>f</sup>* <sup>∈</sup> *<sup>C</sup>*(R, <sup>R</sup>) and *F*(*t*) = \$ *<sup>t</sup>* <sup>0</sup> *<sup>f</sup>*(*s*)*ds*. The fractional Laplacian (−Δ)*<sup>s</sup>* is defined as

$$(-\Delta)^s u(\mathbf{x}) = \mathbb{C}\_{N, \mathfrak{s}} P.V. \int\_{\mathbb{R}^N} \frac{u(\mathbf{x}) - u(y)}{|\mathbf{x} - y|^{N + 2s}} dy, \quad \mu \in \mathcal{S}(\mathbb{R}^N),$$

where P.V. denotes the *principal value* of the singular integral, <sup>S</sup>(R*N*) is the Schwartz space of rapidly decaying *C*<sup>∞</sup> functions in R*N*, and

$$\mathbb{C}\_{N,s} = \frac{2^{2s}s\Gamma(N+s)}{\pi^{N/2}\Gamma(1-s)}.$$

(−Δ)*<sup>s</sup>* is a pseudo-differential operator, and can be equivalently defined via Fourier transform as

$$\mathcal{A}^{\mathbb{P}}[( - \Delta)^{s}u](\xi) = |\xi|^{2s} \mathcal{A}^{\mathbb{P}}[u](\xi), \ u \in \mathcal{S}(\mathbb{R}^{N}),$$

*Mathematics* **2019**, *7*, 151; doi:10.3390/math7020151 www.mdpi.com/journal/mathematics

where F is the Fourier transform, that is,

$$\mathcal{R}^{\mathbb{P}}[u](\xi) = \frac{1}{(2\pi)^{\frac{N}{2}}} \int\_{\mathbb{R}^N} e^{-i\frac{\pi}{\xi}\cdot x} u(x) dx, \quad u \in \mathcal{S}(\mathbb{R}^N).$$

The fractional Laplace operator (−Δ)*<sup>s</sup>* is the infinitesimal generator of Lévy stable diffusion processes, and appears in several areas such as the thin obstacle problem, anomalous diffusion, optimization, finance, phase transitions, crystal dislocation, multiple scattering, and materials science, see [1–5] and their references.

Recently, a great deal of work has been devoted to the study of the Choquard equations, see [6–14] and their references. For instance, Alves, Cassani, Tarsi, and Yang [7] studied the following singularly perturbed nonlocal Schrödinger equation:

$$-\varepsilon^2 \Delta u + V(\mathbf{x})u = \varepsilon^{\mu - 2} \left[ \frac{1}{|\mathbf{x}|^{\mu}} \* F(\mu) \right] f(\mu), \text{ in } \mathbb{R}^2 \text{ } \mu$$

where 0 < *μ* < 2 and *ε* is a positive parameter, the nonlinearity *f* has critical exponential growth in the sense of Trudinger–Moser. By using variational methods, the authors established the existence and concentration of solutions for the above equation.

In [6], Alves, Figueiredo and Yang studied the following Choquard equation:

$$\begin{cases} -\Delta u + V(\mathbf{x})u = (\frac{1}{|\mathbf{x}|^{\mu}} \* F(u))f(u), & \text{in } \mathbb{R}^{N} . \\\ u \in H^{1}(\mathbb{R}^{N}). \end{cases} \tag{2}$$

Under the assumption *V*(*x*) → 0 as |*x*| → ∞, the authors obtained a nontrivial solution for (2) by using a penalization method.

In the physical case *N* = 3, *μ* = 1, *V*(*x*) = 1 and *F*(*t*) = *<sup>t</sup>* 2 <sup>2</sup> , (2) is also known as the stationary Hartree equation [15]. It dates back to the description of the quantum mechanics of a polaron at rest by Pekar in 1954 [16]. In 1976, Choquard used (2) to describe an electron trapped in its own hole, in a certain approximation to the Hartree–Fock theory of one-component plasma [11]. In 1996, Penrose proposed (2) as a model of self-gravitating matter, in a programme in which quantum state reduction is understood as a gravitational phenomenon [15].

In addition, there is little literature on the fractional Choquard equations. Frank and Lenzmann [17] established the uniqueness and radial symmetry of ground state solutions for the following equation:

$$(-\Delta)^{\frac{1}{2}}u + u = (|x|^{-1} \* |u|^2)u, \text{ in } \mathbb{R}^N.$$

D'Avenia, Siciliano, and Squassina [18] obtained the existence, regularity, symmetry, and asymptotic of the solutions for the nonlocal problem

$$(-\Delta)^s \mu + \omega \mu = (|\mathfrak{x}|^{-\mu} \* |\mathfrak{u}|^p) |\mathfrak{u}|^{p-2} \mathfrak{u}\_\prime \text{ in } \mathbb{R}^N.$$

In [19], Shen, Gao, and Yang studied the following fractional Choquard equation:

$$(-\Delta)^{s}\mu + \mu = (|x|^{-\mu} \* F(\mu))f(\mu), \text{ in } \mathbb{R}^{N}, \tag{3}$$

where *N* ≥ 3,*s* ∈ (0, 1), and *μ* ∈ (0, *N*). Under the general Berestycki–Lions-type conditions [20], the authors obtained the existence and regularity of ground states for (3). The authors also established the Pohozaev identity for ˘ (3):

$$\frac{N-2s}{2} \int\_{\mathbb{R}^N} |( - \Delta)^{\frac{s}{2}} u|^2 dx + \frac{N}{2} \int\_{\mathbb{R}^N} u^2 dx = \frac{2N - \mu}{2} \int\_{\mathbb{R}^N} (|x|^{-\mu} \* F(u)) F(u) dx.$$

Motivated by the above works, in the first part of this article, we study the ground state solution for (1). We assume

(I) *<sup>V</sup>*(*x*), *<sup>Q</sup>*(*x*) <sup>&</sup>gt; 0 for all *<sup>x</sup>* <sup>∈</sup> <sup>R</sup>*N*, *<sup>V</sup>* <sup>∈</sup> *<sup>C</sup>*(R*N*, <sup>R</sup>) and *<sup>Q</sup>* <sup>∈</sup> *<sup>C</sup>*(R*N*, <sup>R</sup>) <sup>∩</sup> *<sup>L</sup>*∞(R*N*, <sup>R</sup>);

(II) if {*An*} ⊂ <sup>R</sup>*<sup>N</sup>* is a sequence of Borel sets such that meas{*An*} ≤ *<sup>δ</sup>* for all *<sup>n</sup>* and some *<sup>δ</sup>* <sup>&</sup>gt; 0, then

$$\lim\_{r \to \infty} \int\_{A\_n \cap B\_r^c(0)} [Q(x)]^{\frac{2N}{2N-\mu}} dx = 0 \text{ uniformly in } n \in \mathbb{N};$$

(III) one of the below conditions occurs:

$$\frac{Q}{V} \in L^{\infty}(\mathbb{R}^{N}),$$
 
$$(4)$$

or there exists *p* ∈ (2, 2<sup>∗</sup> *<sup>s</sup>* ) such that

$$\frac{[Q(\mathbf{x})]^{\frac{2N}{2N-\mu}}}{[V(\mathbf{x})]^{\frac{2\mu}{2\mu}-\frac{\mu}{2}}} \to 0, \quad |\mathbf{x}| \to \infty,\tag{5}$$

where 2∗ *<sup>s</sup>* <sup>=</sup> <sup>2</sup>*<sup>N</sup> <sup>N</sup>*−2*<sup>s</sup>* is the fractional critical exponent; 2*N*−*μ*

$$\text{(F1)} \quad F(t) = o(|t|^{\frac{2N-\mu}{N}}) \text{ as } t \to 0 \text{ if (4) holds; or } F(t) = o(|t|^{\frac{p(2N-\mu)}{2N}}) \text{ as } t \to 0 \text{ if (5) holds; or } F(t) = o(|t|^{\frac{2N-\mu}{N}}) $$


It is necessary for us to point out that the original of assumptions (I)–(III) come from [21–23]. The assumptions can be used to prove that the work space *E* is compactly embedded into the weighted Lebesgue space *L<sup>q</sup> <sup>K</sup>*(R*N*), see Section <sup>2</sup> and Lemma 1.

Now, we can state the first result of this article.

**Theorem 1.** *Suppose that* (*I*),(*I I*),(*III*) *and (F1)–(F4) hold. Then* (1) *has a ground state solution.*

**Remark 1.** *Since the Nehari-type monotonicity condition for f is not satisfied, the Nehari manifold method used in [24] no longer works in our setting. To prove Theorem 2, we use the non-Nehari manifold method developed by Tang [25], which relies on finding a minimizing sequence outside the Nehari manifold by using the diagonal method (see Lemma 8).*

In the second part of this article, we consider the following fractional Choquard equation with zero mass case:

$$\begin{cases} \left( -\Delta \right)^{s} \mu = \left( \frac{1}{|x|^{\overline{\mu}}} \* F(\mu) \right) f(\mu), & \text{in } \mathbb{R}^{N} \\\ \mu \in D^{s,2}(\mathbb{R}^{N}), \end{cases} \tag{6}$$

where *<sup>N</sup>* <sup>≥</sup> 3, 0 <sup>&</sup>lt; *<sup>s</sup>* <sup>&</sup>lt; 1, 0 <sup>&</sup>lt; *<sup>μ</sup>* <sup>&</sup>lt; min{*N*, 4*s*}. The homogeneous fractional Sobolev space *<sup>D</sup>s*,2(R*N*), also denoted by *H*˙ *<sup>s</sup>*(R*N*), can be characterized as the space

$$D^{s,2}(\mathbb{R}^N) = \left\{ \mu \in L^{2^\*\_s}(\mathbb{R}^N) : \int\_{\mathbb{R}^N} \int\_{\mathbb{R}^N} \frac{|u(\mathbf{x}) - u(y)|^2}{|\mathbf{x} - y|^{N + 2s}} d\mathbf{x} dy < +\infty \right\}.$$

*<sup>f</sup>* <sup>∈</sup> *<sup>C</sup>*(R, <sup>R</sup>) satisfy the following Berestycki–Lions-type condition [19,20]:

(*F*5) *F* is not trivial, that is, *F* ≡ 0;

(*F*6) there exists *<sup>C</sup>* <sup>&</sup>gt; 0 such that for every *<sup>t</sup>* <sup>∈</sup> <sup>R</sup>,

$$|tf(t)| \le C|t|^{\frac{2N-p}{N-2s}};$$

(*F*7)

$$\lim\_{t \to 0} \frac{F(t)}{|t|^2} = \lim\_{t \to \infty} \frac{F(t)}{|t|^{\frac{2N-\mu}{N-2s}}} = 0.$$

The second result of this paper is as follows.

**Theorem 2.** *Suppose that f satisfies (F5)–(F7). Then* (6) *has a nontrivial solution.*

**Remark 2.** *Notice that the method used in [13] is no longer applicable for* (6)*, because it relies heavily on the constant potentials. In the zero mass case, we use the perturbation method and the Pohozaev identity established ˘ in [19] to overcome this difficulty.*

In this article, we make use of the following notation:


#### **2. Ground State Solutions for** (1)

Set

$$D^{s,2}(\mathbb{R}^N) := \left\{ \mu \in L^{2^\*\_s}(\mathbb{R}^N) : \int\_{\mathbb{R}^N} \int\_{\mathbb{R}^N} \frac{|u(\mathbf{x}) - u(y)|^2}{|\mathbf{x} - y|^{N + 2s}} d\mathbf{x} dy < +\infty \right\},$$

endowed with the Gagliardo (semi)norm

$$\mathbb{E}[\boldsymbol{\mu}] := \left( \int\_{\mathbb{R}^N} \int\_{\mathbb{R}^N} \frac{|\boldsymbol{\mu}(\boldsymbol{x}) - \boldsymbol{\mu}(\boldsymbol{y})|^2}{|\boldsymbol{x} - \boldsymbol{y}|^{N + 2s}} d\boldsymbol{x} d\boldsymbol{y} \right)^{1/2}.$$

From [5], we have the following identity:

$$\left[\left[\mu\right]\right]^2 = \int\_{\mathbb{R}^N} |(-\Delta)^{\frac{s}{2}}\mu|^2 d\mathbf{x} = \int\_{\mathbb{R}^N} |\xi|^{2s} |\mathcal{J}^\mathbf{e}[\mu](\xi)|^2 d\xi.$$

From [26], *Ds*,2(R*N*) is continuously embedded into *L*2<sup>∗</sup> *<sup>s</sup>* (R*N*). Then, we can define the best constant *S* > 0 as

$$S := \sup\_{\boldsymbol{\mu} \in D^{s,2}(\mathbb{R}^N)} \frac{\left(\int\_{\mathbb{R}^N} |\boldsymbol{\mu}|^{2^\*\_s} d\boldsymbol{x}\right)^{\frac{2}{2^\*\_s}}}{\int\_{\mathbb{R}^N} |( - \boldsymbol{\Delta})^{\frac{s}{2}} \boldsymbol{u}|^2 d\boldsymbol{x}}.$$

Let

$$E := \left\{ \mu \in D^{s,2}(\mathbb{R}^N) : \int\_{\mathbb{R}^N} V(x) \mu^2 dx < +\infty \right\}.$$

Under the assumptions (I)–(III), following the idea of ([21], Proposition 2.1) or ([22], Proposition 2.2), we can prove that the Hilbert space *E* endowed with scalar product and norm

$$\mathbb{P}(u,v) = \int\_{\mathbb{R}^N} [( - \Delta)^{\frac{\mathsf{H}}{2}} u ( - \Delta)^{\frac{\mathsf{H}}{2}} v + V(\mathsf{x})uv] dx, \ \|u\| = \left( \int\_{\mathbb{R}^N} [|( - \Delta)^{\frac{\mathsf{H}}{2}} u|^2 + V(\mathsf{x})u^2] dx \right)^{\frac{1}{2}}$$

is compactly embedded into the weighted space *L<sup>q</sup> <sup>K</sup>*(R*N*) for every *<sup>q</sup>* <sup>∈</sup> (2, 2<sup>∗</sup> *<sup>s</sup>* ), where *K*(*x*) := [*Q*(*x*)]2*N*/(2*N*−*μ*) and

$$L\_K^q(\mathbb{R}^N) := \left\{ u : \text{meas}\{u\} < \infty \text{ and } \int\_{\mathbb{R}^N} K(x) |u|^q dx < \infty \right\}, \quad \forall q \ge 2.$$

**Lemma 1.** *Assume that (I)–(III) hold. If* (*K*1) *holds, E is compactly embedded in L<sup>q</sup> <sup>K</sup>*(R*N*) *for all <sup>q</sup>* <sup>∈</sup> (2, 2<sup>∗</sup> *<sup>s</sup>* )*. If* (*K*2) *holds, E is compactly embedded in L<sup>p</sup> <sup>K</sup>*(R*N*)*.*

**Proof.** If (*K*1) holds, then

$$\frac{K(\boldsymbol{x})}{V(\boldsymbol{x})} = \frac{Q(\boldsymbol{x})}{V(\boldsymbol{x})} [Q(\boldsymbol{x})]^{\frac{\mu}{2N-\mu}} \in L^{\infty}(\mathbb{R}^N).$$

Given *ε* > 0 and fixed *q* ∈ (2, 2<sup>∗</sup> *<sup>s</sup>* ), there exist 0 < *t*<sup>0</sup> < *t*<sup>1</sup> and *C* > 0 such that

$$\mathcal{K}(\mathbf{x})|t|^q \le \varepsilon \mathbb{C}(V(\mathbf{x})|t|^2 + |t|^{2^\*\_s}) + \mathcal{K}(\mathbf{x})\chi\_{\left[t\_0, t\_1\right]}(|t|)|t|^{2^\*\_s} \ \forall t \in \mathbb{R}.$$

Hence,

$$\int\_{B\_r^c(0)} \mathbb{K}(\mathbf{x}) |u|^q d\mathbf{x} \le \varepsilon \mathbb{C} \mathcal{W}(\mathbf{u}) + \mathbb{C} \mathcal{K}(\mathbf{x}) \int\_{A \cap B\_r^c(0)} \mathbb{K}(\mathbf{x}) d\mathbf{x} \,\,\,\forall \mathbf{u} \in E,\tag{7}$$

where

$$\mathcal{W}(\mathfrak{u}) = \int\_{\mathbb{R}^N} V(\mathfrak{x}) |\mathfrak{u}|^2 d\mathfrak{x} + \int\_{\mathbb{R}^N} |\mathfrak{u}|^{2^\*\_s} d\mathfrak{x}$$

and

$$A = \{ \mathfrak{x} \in \mathbb{R}^N : s\_0 \le |\mathfrak{u}(\mathfrak{x})| \le s\_1 \}.$$

Let {*vn*} be a sequence such that *vn v* in *E*, then there exists a constant *M*<sup>1</sup> > 0 such that

$$\int\_{\mathbb{R}^N} [|( -\Delta)^{\frac{s}{2}} \boldsymbol{\upsilon}\_{\boldsymbol{\Pi}}|^2 + V(\boldsymbol{x}) |\boldsymbol{\upsilon}\_{\boldsymbol{\Pi}}|^2] d\boldsymbol{x} \le M\_1 \text{ and } \int\_{\mathbb{R}^N} |\boldsymbol{\upsilon}\_{\boldsymbol{\Pi}}|^{2^\*\_s} d\boldsymbol{x} \le M\_1 \text{ } \forall \boldsymbol{n} \in \mathbb{N}\_\*$$

which implies that {*W*(*vn*)} is bounded. On the other hand, setting

$$A\_n = \{ \mathbf{x} \in \mathbb{R}^N : \mathbf{s}\_0 \le |v\_n(\mathbf{x})| \le s\_1 \},$$

we have

$$s\_0^{2^\*\_s} |A\_n| \le \int\_{A\_n} |v\_n|^{2^\*\_s} dx \le M\_1 \,\,\forall n \in \mathbb{N}$$

and so sup *<sup>n</sup>*∈<sup>N</sup> |*An*| < +∞. Therefore, from (*I I*), there is *r* > 0 such that

$$\int\_{A\_{\mathbb{R}} \cap B\_{r}^{c}(0)} \mathbb{K}(\mathbf{x}) d\mathbf{x} < \frac{\varepsilon}{s\_{1}^{2^{\*}\_{s}}} \,\,\forall n \in \mathbb{N}.\tag{8}$$

Combining (7) and (8), we have

$$\int\_{B\_r^\varepsilon(0)} \mathcal{K}(\mathbf{x}) |\boldsymbol{\upsilon}\_n|^q d\mathbf{x} < \varepsilon \mathcal{C} M\_1 + s\_1^{2^\*\_s} \int\_{F\_R \cap B\_r^\varepsilon(0)} \mathcal{K}(\mathbf{x}) d\mathbf{x} < (\mathcal{C} M\_1 + 1)\varepsilon \,\,\forall n \in \mathbb{N}.\tag{9}$$

By *q* ∈ (2, 2<sup>∗</sup> *<sup>s</sup>* ), we have from Sobolev embeddings that

$$\lim\_{n \to +\infty} \int\_{B\_r(0)} \mathcal{K}(\mathbf{x}) |v\_{\mathcal{U}}|^q d\mathbf{x} = \int\_{B\_r(0)} \mathcal{K}(\mathbf{x}) |v|^q d\mathbf{x}.\tag{10}$$

Combining (9) and (10), we have

$$\lim\_{n \to +\infty} \int\_{\mathbb{R}^N} \mathcal{K}(\mathfrak{x}) |v\_n|^q d\mathfrak{x} = \int\_{\mathbb{R}^N} \mathcal{K}(\mathfrak{x}) |v|^q d\mathfrak{x}\_{\mathcal{I}}$$

which yields

$$
\upsilon\_n \to \upsilon \text{ in } L^q\_K(\mathbb{R}^N) \text{ } \forall q \in (\mathfrak{L}, \mathfrak{L}^\*\_s) \text{ }.
$$

Next, we suppose that (*K*2) holds. For each *<sup>x</sup>* <sup>∈</sup> <sup>R</sup>*<sup>N</sup>* fixed, we observe that the function

$$\lg(t) = V(\mathbf{x})t^{2-p} + t^{2^\*\_s - p} \quad \forall t > 0$$

has *CpV*(*x*) 2∗ *s* −*p* 2∗ *<sup>s</sup>* <sup>−</sup><sup>2</sup> as its minimum value, where

$$\mathbb{C}\_{p} = \left(\frac{p-2}{2\_s^\*-p}\right)^{\frac{2-p}{2\_s^\*-2}} + \left(\frac{p-2}{2\_s^\*-p}\right)^{\frac{2\_s^\*-p}{2\_s^\*-2}}.$$

Hence

$$\mathcal{C}\_{\mathcal{V}}V(\mathbf{x})^{\frac{2^\*-p}{2^\*-2}} \le V(\mathbf{x})t^{2-p} + t^{2^\*\_s-p} \,\,\forall \mathbf{x} \in \mathbb{R}^N \text{ and } t > 0.$$

Combining this inequality with (*K*2), given *ε* ∈ (0, *Cp*), there exists *r* > 0 large enough such that

$$K(\mathbf{x})|t|^p \le \varepsilon (V(\mathbf{x})|t|^2 + |t|^{2^\*\_s}) \ \forall t \in \mathbb{R} \text{ and } |\mathbf{x}| \ge r\_\prime$$

leading to

$$\int\_{B\_r^\varepsilon(0)} \mathbb{K}(\mathbf{x}) |u|^p d\mathbf{x} \le \varepsilon \int\_{B\_r^\varepsilon(0)} (V(\mathbf{x}) |u|^2 + |u|^{2^\*\_s}) d\mathbf{x} \,\,\forall u \in E.$$

Let {*vn*} be a sequence such that *vn v* in *E*, then there exists a constant *M*<sup>2</sup> > 0 such that

$$\int\_{\mathbb{R}^N} V(\mathbf{x}) |\upsilon\_n|^2 d\mathbf{x} \le M\_2 \text{ and } \int\_{\mathbb{R}^N} |\upsilon\_n|^{2^\*\_s} d\mathbf{x} \le M\_2 \text{ } \forall n \in \mathbb{N}\_s$$

and so,

$$\int\_{B\_r^c(0)} \mathbb{K}(x) |v\_n|^p dx \le 2\varepsilon M\_2 \quad \forall n \in \mathbb{N}.\tag{11}$$

Since *p* ∈ (2, 2<sup>∗</sup> *<sup>s</sup>* ) and *K* is a continuous function, we have

$$\lim\_{n \to +\infty} \int\_{B\_r^\varepsilon(0)} \mathcal{K}(\mathbf{x}) |v\_n|^p d\mathbf{x} = \int\_{B\_r^\varepsilon(0)} \mathcal{K}(\mathbf{x}) |v|^p d\mathbf{x}.\tag{12}$$

From (11) and (12), we have

$$\lim\_{n \to +\infty} \int\_{\mathbb{R}^N} K(\mathbf{x}) |v\_n|^p d\mathbf{x} = \int\_{\mathbb{R}^N} K(\mathbf{x}) |v|^p d\mathbf{x}.$$

Therefore

$$
v\_n \to v \text{ in } L^p\_K(\mathbb{R}^N).
$$

**Lemma 2.** *(Hardy–Littlewood–Sobolev inequality, see [26]). Let* <sup>1</sup> <sup>&</sup>lt; *<sup>r</sup>*, *<sup>t</sup>* <sup>&</sup>lt; <sup>∞</sup>*, and <sup>μ</sup>* <sup>∈</sup> (0, *<sup>N</sup>*) *with* <sup>1</sup> *<sup>r</sup>* <sup>+</sup> <sup>1</sup> *<sup>t</sup>* = <sup>2</sup> <sup>−</sup> *<sup>μ</sup> <sup>N</sup> . If <sup>φ</sup>* <sup>∈</sup> *<sup>L</sup>r*(R*N*) *and <sup>ψ</sup>* <sup>∈</sup> *<sup>L</sup><sup>t</sup>* (R*N*)*, then there exists a constant C*(*N*, *μ*,*r*, *t*) > 0*, such that*

$$\int\_{\mathbb{R}^N} \int\_{\mathbb{R}^N} \frac{\phi(x)\psi(y)}{|x-y|^\mu} dx dy \le C(N, \mu, r, t) \|\phi\|\_r \|\psi\|\_{t=0}$$

**Lemma 3.** *Assume that (I)–(III) and* (*F*1)*–*(*F*3) *hold. Then for u* ∈ *E*

$$\left| \int\_{\mathbb{R}^N} \int\_{\mathbb{R}^N} \frac{\mathcal{Q}(\mathbf{x}) \mathcal{Q}(\mathbf{y}) F(\mathbf{u}(\mathbf{x})) F(\mathbf{u}(\mathbf{y}))}{|\mathbf{x} - \mathbf{y}|^\mu} d\mathbf{x} d\mathbf{y} \right| < +\infty,\tag{13}$$

*and there exists a constant C*<sup>1</sup> > 0 *such that*

$$\left| \int\_{\mathbb{R}^N} \int\_{\mathbb{R}^N} \frac{\mathcal{Q}(\mathbf{x}) \mathcal{Q}(\mathbf{y}) F(\mathbf{u}(\mathbf{x})) f(\mathbf{u}(\mathbf{y})) \mathbf{v}(\mathbf{y})}{|\mathbf{x} - \mathbf{y}|^\mu} d\mathbf{x} d\mathbf{y} \right| < \mathcal{C}\_1 \|\mathbf{v}\|\_{\prime} \quad \forall \mathbf{v} \in E. \tag{14}$$

*Furthermore, let* {*un*} ⊂ *E be a sequence such that un u in E, then*

$$\lim\_{n \to \infty} \int\_{\mathbb{R}^N} \int\_{\mathbb{R}^N} \frac{\mathcal{Q}(\mathbf{x}) \mathcal{Q}(\mathbf{y}) [F(u\_{\mathrm{il}}(\mathbf{x})) F(u\_{\mathrm{il}}(\mathbf{y})) - F(u(\mathbf{x})) F(u(\mathbf{y}))]}{|\mathbf{x} - \mathbf{y}|^\mu} d\mathbf{x} d\mathbf{y} = 0 \tag{15}$$

*and*

$$\lim\_{n \to \infty} \int\_{\mathbb{R}^N} \int\_{\mathbb{R}^N} \frac{\mathcal{Q}(\mathbf{x}) \mathcal{Q}(\mathbf{y}) F(u\_n(\mathbf{x})) f(u\_n(\mathbf{y})) [u\_n(\mathbf{y}) - u(\mathbf{y})]}{|\mathbf{x} - \mathbf{y}|^\mu} d\mathbf{x} d\mathbf{y} = 0. \tag{16}$$

**Proof.** Set

$$
\beta = \begin{cases}
\ 2, \text{ if } (K1) \text{ holds,} \\
\ p, \text{ if } (K2) \text{ holds.}
\end{cases}
$$

By (*F*1),(*F*2), Lemma 2, Hölder inequality and Sobolev inequality, we have

$$\begin{split} \int\_{\mathbb{R}^{N}} K(\mathbf{x}) |F(\mathbf{u})|^{\frac{2N}{2N-\mu}} d\mathbf{x} &\leq \mathsf{C}\_{1} \int\_{\mathbb{R}^{N}} K(\mathbf{x}) \left[ |\mathbf{u}|^{\frac{\beta(2N-\mu)}{2N}} + |\mathbf{u}|^{\frac{2N-\mu}{N-2\mu}} \right]^{\frac{2N}{2N-\mu}} d\mathbf{x} \\ &\leq \mathsf{C}\_{2} \int\_{\mathbb{R}^{N}} K(\mathbf{x}) |\mathbf{u}|^{\beta} d\mathbf{x} + \mathsf{C}\_{2} \int\_{\mathbb{R}^{N}} |\mathbf{u}|^{2^{\*}\_{s}} d\mathbf{x} \\ &\leq \mathsf{C}\_{3} (\|\mathbf{u}\|^{\beta} + [\mathbf{u}]^{2^{\*}\_{s}}), \ \forall \mathbf{u} \in E \end{split} \tag{17}$$

and

$$\begin{split} \int\_{\mathbb{R}^{N}} K(\mathbf{x}) |f(\boldsymbol{u}) \boldsymbol{v}|^{\frac{2N}{2N-p}} d\boldsymbol{x} &\leq \mathsf{C}\_{1} \int\_{\mathbb{R}^{N}} K(\mathbf{x}) \left[ |\boldsymbol{u}|^{\frac{\theta(2N-p)-2N}{2N}} + |\boldsymbol{u}|^{\frac{N-p+2\mu}{N-2\mu}} \right]^{\frac{2N}{2N-p}} |\boldsymbol{v}|^{\frac{2N}{2N-p}} d\boldsymbol{x} \\ &\leq \mathsf{C}\_{4} \int\_{\mathbb{R}^{N}} [K(\mathbf{x})]^{\frac{\theta(2N-p)-2N}{\theta(2N-p)}} |\boldsymbol{u}|^{\frac{\theta(2N-p)-2N}{2N-p}} [K(\mathbf{x})]^{\frac{2N}{\theta(2N-p)}} |\boldsymbol{v}|^{\frac{2N}{2N-p}} d\boldsymbol{x} \\ &+ \mathsf{C} \mathfrak{s} \int\_{\mathbb{R}^{N}} |\boldsymbol{u}|^{\frac{2N(N+2\mu-p)}{(N-2\mu)(2N-p)}} |\boldsymbol{v}|^{\frac{2N}{2N-p}} d\boldsymbol{x} \\ &\leq \mathsf{C}\_{6} \left[ \left\| \boldsymbol{u} \right\|^{\frac{\theta(2N-p)-2N}{2N-p}} + \left\| \boldsymbol{u} \right\|^{\frac{2N(N+2\mu-p)}{(N-2\mu)(2N-p)}} \right] \|\boldsymbol{v}\|^{\frac{2N}{2N-p}}, \ \forall \boldsymbol{u}, \boldsymbol{v} \in E. \end{split} \tag{18}$$

Applying Lemma 2 and (17), we have

$$\begin{aligned} & \left| \int\_{\mathbb{R}^N} \int\_{\mathbb{R}^N} \frac{Q(\mathbf{x}) Q(\mathbf{y}) F(\mathbf{u}(\mathbf{x})) F(\mathbf{u}(\mathbf{y}))}{|\mathbf{x} - \mathbf{y}|^\mu} d\mathbf{x} d\mathbf{y} \right| \\ & \leq C \gamma \left[ \int\_{\mathbb{R}^N} K(\mathbf{x}) |F(\mathbf{u})|^{\frac{2N}{N-\mu}} d\mathbf{x} \right]^{\frac{2N-\mu}{N}} \\ & \leq C \varsigma \left[ \|\boldsymbol{u}\|^{\frac{\beta(2N-\mu)}{N}} + \|\boldsymbol{u}\|^{\frac{2(2N-\mu)}{N-2\mu}} \right], \ \forall \boldsymbol{u} \in E, \end{aligned} \tag{19}$$

which yields (13) holds. Similarly, we have

$$\begin{split} & \left| \int\_{\mathbb{R}^N} \int\_{\mathbb{R}^N} \frac{\mathcal{Q}(\mathbf{x}) \mathcal{Q}(\mathbf{y}) F(\mathbf{u}(\mathbf{x})) f(\mathbf{u}(\mathbf{y})) \mathbf{v}(\mathbf{y})}{|\mathbf{x} - \mathbf{y}|^\mu} d\mathbf{x} d\mathbf{y} \right| \\ & \leq \mathsf{C}\_{\theta} \left[ \int\_{\mathbb{R}^N} K(\mathbf{x}) |F(\mathbf{u})|^{\frac{2N}{2N-\mu}} d\mathbf{x} \right]^{\frac{2N-\mu}{2N}} \left[ \int\_{\mathbb{R}^N} K(\mathbf{x}) |f(\mathbf{u}) \mathbf{v}|^{\frac{2N}{2N-\mu}} d\mathbf{x} \right]^{\frac{2N-\mu}{2N}}, \ \forall \mathbf{u}, \mathbf{v} \in E, \end{split} \tag{20}$$

which, together with (17) and (18), implies that (14) holds.

Similar to ([21], Lemma 2), by (*F*2),(*F*3), and Lemma 2, we have

$$\lim\_{n \to \infty} \int\_{\mathbb{R}^N} K(\mathbf{x}) |\mathcal{F}(u\_n) - \mathcal{F}(u)|^{\frac{2N}{2N-p}} d\mathbf{x} = 0,\\ \lim\_{n \to \infty} \int\_{\mathbb{R}^N} \mathcal{K}(\mathbf{x}) |f(u\_n)|^{\frac{2N}{2N-p}} |u\_n - u|^{\frac{2N}{2N-p}} d\mathbf{x} = 0. \tag{21}$$

Combining (18), (20), and (21), we deduce that (15) and (16) hold.

The energy functional <sup>Φ</sup> : *<sup>E</sup>* → <sup>R</sup> given by

$$\Phi(u) := \frac{1}{2} \int\_{\mathbb{R}^N} |( - \Delta)^{\frac{\nu}{2}} u|^2 dx + \frac{1}{2} \int\_{\mathbb{R}^N} V(x) |u|^2 dx - \frac{1}{2} \int\_{\mathbb{R}^N} \int\_{\mathbb{R}^N} \frac{Q(x) Q(y) F(u(x)) F(u(y))}{|x - y|^\mu} dx dy. \tag{22}$$

By Lemmas 2 and 3, Φ is well-defined and belongs to *C*1-class. Moreover, we have

$$\begin{split} \langle \Phi'(\mathsf{u}), \mathsf{v} \rangle &= \int\_{\mathbb{R}^N} (-\Delta)^{\frac{\mathsf{s}}{2}} \mathsf{u}(-\Delta)^{\frac{\mathsf{s}}{2}} \mathsf{v} d\mathsf{x} + \int\_{\mathbb{R}^N} V(\mathsf{x}) \mathsf{u} \mathsf{v} d\mathsf{x} \\ &- \int\_{\mathbb{R}^N} \int\_{\mathbb{R}^N} \frac{Q(\mathsf{x}) Q(\mathsf{y}) F(\mathsf{u}(\mathsf{x})) f(\mathsf{u}(\mathsf{y})) \mathsf{v}(\mathsf{y})}{|\mathsf{x} - \mathsf{y}|^\mu} d\mathsf{x} d\mathsf{y}, \ \forall \mathsf{u}, \mathsf{v} \in \mathsf{E}. \end{split} \tag{23}$$

**Lemma 4.** *Assume that* (*F*1)*–*(*F*3) *hold. Then, for all t* <sup>≥</sup> <sup>0</sup> *and <sup>τ</sup>*1, *<sup>τ</sup>*<sup>2</sup> <sup>∈</sup> <sup>R</sup>*,*

$$l(t, \tau\_1, \tau\_2) := F(t\tau\_1)F(t\tau\_2) - F(\tau\_1)F(\tau\_2) + \frac{1 - t^2}{2} [F(\tau\_1)f(\tau\_2)\tau\_2 + F(\tau\_2)f(\tau\_1)\tau\_1] \ge 0. \tag{24}$$

**Proof.** Firstly, it follows from (*F*1) that *f*(0) = 0. By (*F*3), we have

$$f(\tau) \ge 0, \,\forall \tau \ge 0; \,\, f(\tau) \le 0, \,\,\forall \tau \le 0; \,\, F(\tau) \ge 0, \,\,\forall \tau \in \mathbb{R}$$

and

$$f(\tau)\tau \ge \int\_0^\tau f(t)dt = F(\tau), \ \forall \tau \in \mathbb{R}.\tag{25}$$

It is easy to verify that (24) holds for *t* = 0. For *τ* = 0, we have from (25) that

$$\left[\frac{F(\tau)}{\tau}\right]' = \frac{f(\tau)\tau - F(\tau)}{\tau^2} \ge 0. \tag{26}$$

For every *<sup>τ</sup>*1, *<sup>τ</sup>*<sup>2</sup> <sup>∈</sup> <sup>R</sup>, we deduce from (*F*3) and (26) that

$$\begin{aligned} &\frac{d}{dt}l(t,\tau\_1,\tau\_2) \\ &= \tau\_1\tau\_2 t \left[ \frac{F(t\tau\_1)}{t\tau\_1} f(t\tau\_2) + \frac{F(t\tau\_2)}{t\tau\_2} f(t\tau\_1) - \frac{F(\tau\_1)}{\tau\_1} f(\tau\_2) - \frac{F(\tau\_2)}{\tau\_2} f(\tau\_1) \right], \\ &\left\{ \begin{array}{l} \ge 0, \ t \ge 1, \\ \le 0, \ 0 < t < 1, \end{array} \right. \end{aligned}$$

which implies that *<sup>l</sup>*(*t*, *<sup>τ</sup>*1, *<sup>τ</sup>*2) <sup>≥</sup> *<sup>l</sup>*(1, *<sup>τ</sup>*1, *<sup>τ</sup>*2) = 0 for all *<sup>t</sup>* <sup>&</sup>gt; 0 and *<sup>τ</sup>*1, *<sup>τ</sup>*<sup>2</sup> <sup>∈</sup> <sup>R</sup>.

**Lemma 5.** *Assume that* (*I*)*–*(*III*) *and* (*F*1)*–*(*F*4) *hold. Then*

$$
\Phi(u) \ge \Phi(tu) + \frac{1 - t^2}{2} \langle \Phi'(u), u \rangle, \ \forall u \in E, \ t \ge 0. \tag{27}
$$

**Proof.** By (22), (23), and (24), we have

$$\begin{split} & \left| \Phi(u) - \Phi(tu) - \frac{1 - t^2}{2} \langle \Phi'(u), u \rangle \right| \\ &= \frac{1}{2} \int\_{\mathbb{R}^N} \int\_{\mathbb{R}^N} \frac{1}{|\mathbf{x} - \mathbf{y}|^\mu} \left[ F(tu(\mathbf{x})) F(tu(\mathbf{y})) - F(u(\mathbf{x})) F(u(\mathbf{y})) \right. \\ & \quad + \frac{1 - t^2}{2} \left( F(u(\mathbf{x})) f(u(\mathbf{y})) u(\mathbf{y}) + F(u(\mathbf{y})) f(u(\mathbf{x})) u(\mathbf{x}) \right) \right] d\mathbf{x} dy \\ &= \frac{1}{2} \int\_{\mathbb{R}^N} \int\_{\mathbb{R}^N} \frac{l(t, u(\mathbf{x}), u(\mathbf{y}))}{|\mathbf{x} - \mathbf{y}|^\mu} d\mathbf{x} dy \\ &\ge 0, \quad \forall u \in E, \ t \ge 0. \end{split}$$

**Corollary 1.** *Assume that* (*I*)*–*(*III*) *and* (*F*1)*–*(*F*4) *hold. Let*

$$\mathcal{N} := \{ \mathfrak{u} \in E \mid \{ 0 \} : \langle \Phi'(\mathfrak{u}), \mathfrak{u} \rangle = 0 \}.$$

*Then*

$$\Phi(u) = \max\_{t \ge 0} \Phi(tu), \ \forall u \in \mathcal{N}.$$

**Lemma 6.** *Assume that* (*I*)*–*(*III*) *and* (*F*1)*–*(*F*4) *hold. Then, for any u* ∈ *E* \ {0}*, there exists tu* > 0 *such that tuu* ∈ N *.*

**Proof.** Let *u* ∈ *E* \ {0} be fixed. Define a function *ζ*(*t*) := Φ(*tu*) on (0, ∞). By (22) and (23), we have

$$\mathcal{L}'(t) = 0 \Longleftrightarrow t \|u\|^2 - \int\_{\mathbb{R}^N} \int\_{\mathbb{R}^N} \frac{Q(\mathbf{x}) Q(\mathbf{y}) F(tu(\mathbf{x})) F(tu(\mathbf{y}))) f(tu(\mathbf{y})) u(\mathbf{y})}{|\mathbf{x} - \mathbf{y}|^\mu} d\mathbf{x} dy = 0, \quad \Longleftrightarrow t u \in \mathcal{N}.$$

By (19), we have for *u* ∈ *E*

$$\Phi(u) \ge \begin{cases} \frac{1}{2} \|u\| - \mathbb{C}\_8 \left[ \|u\| \frac{4N - 2\mu}{N} + \|u\| \frac{4N - 2\mu}{N - 2s} \right], & \text{if (K1) holds,} \\\frac{1}{2} \|u\| - \mathbb{C}\_8 \left[ \|u\| \frac{2pN - p\mu}{N} + \|u\| \frac{4N - 2\mu}{N - 2s} \right], & \text{if (K2) holds,} \end{cases} \tag{28}$$

which implies that there exists *ρ*<sup>0</sup> > 0 such that

$$\delta\_0 := \inf\_{\|u\| = \rho\_0} \Phi(u) > 0. \tag{29}$$

Therefore, lim*t*→<sup>0</sup> *ζ*(*t*) = 0 and *ζ*(*t*) > 0 for small *t* > 0. By (*F*4), for *t* large, we have

$$\mathcal{L}(t) = \frac{t^2}{2} \left[ \left\| u \right\|^2 - \frac{1}{2} \int\_{\mathbb{R}^N} \int\_{\mathbb{R}^N} \frac{\mathcal{Q}(\mathbf{x}) \mathcal{F}(tu(\mathbf{x}))}{|tu(\mathbf{x})|} \frac{\mathcal{Q}(\mathbf{y}) \mathcal{F}(tu(\mathbf{y}))}{|tu(\mathbf{y})|} \frac{|u(\mathbf{x})u(\mathbf{y})|}{|\mathbf{x} - \mathbf{y}|^\mu} d\mathbf{x} d\mathbf{y} \right] < 0. \tag{30}$$

Therefore max *<sup>t</sup>*∈[0,∞) *ζ*(*t*) is achieved at some *tu* > 0 so that *ζ* (*tu*) = 0 and *tuu* ∈ N .

**Lemma 7.** *Assume that* (*I*)*–*(*III*) *and* (*F*1)*–*(*F*4) *hold. Then*

$$\inf\_{u \in \mathcal{N}} \Phi(u) := c = \inf\_{u \in E(\{0\})} \max\_{t \ge 0} \Phi(tu) > 0.$$

**Proof.** Corollary 1 and Lemma 6 imply that

$$\mathcal{c} = \inf\_{u \in E \backslash \{0\}} \max\_{t \ge 0} \Phi(tu).$$

By (22) and (29),

$$\mathcal{L} \ge \inf\_{\mu \in E(\{0\})} \Phi \left( \frac{\rho\_0}{\|\mu\|} \mu \right) = \inf\_{\|\mu\| = \rho\_0} \Phi(\mu) > 0.$$

Next, we will seek a Cerami sequence for Φ outside N by using the diagonal method, which is used in [25,27,28].

**Lemma 8.** *Assume that* (*I*)*–*(*III*) *and* (*F*1)*–*(*F*4) *hold. Then there exist* {*un*} ⊂ *E and c*<sup>∗</sup> ∈ (0, *c*] *such that*

$$\Phi(u\_n) \to \mathfrak{c}^\*, \quad (1 + \|u\_n\|) \|\Phi'(u\_n)\| \to 0,\tag{31}$$

*as n* → ∞*.*

N

**Proof.** For *c* = inf Φ, we can choose a sequence {*vk*}⊂N such that

$$c \le \Phi(v\_k) < c + \frac{1}{k'} \quad k \in \mathbb{N}.\tag{32}$$

By (29) and (30), it is easy to verify that Φ(0) = 0, Φ(*Tvk*) < 0 when *T* is large enough, and Φ(*u*) ≥ *δ*<sup>0</sup> > 0 when *u* = *ρ*0. Therefore, from Mountain Pass Lemma ([29]), there is a sequence {*un*,*k*} such that

$$\Phi(u\_{k,n}) \to c\_k \in [\delta\_{0\prime} \sup\_{t \in [0,1]} \Phi(t\nu\_k)], \quad (1 + \|u\_{k,n}\|) \|\Phi^\prime(u\_{k,n})\| \to 0, \quad k \in \mathbb{N}.\tag{33}$$

By Corollary 1 and {*vk*}⊂N , we have

$$
\Phi(t\upsilon\_k) \le \Phi(\upsilon\_k), \quad \forall \, t \ge 0. \tag{34}
$$

It follows from (34) that Φ(*vk*) = sup *t*∈[0,1] Φ(*tvk*). Hence, by (32)–(34), we have

$$\Phi(w\_{k,n}) \to \mathfrak{c}\_k \in \left[\delta\_{0\prime}\mathfrak{c} + \frac{1}{k}\right), \quad (1 + \|\boldsymbol{u}\_{k,n}\|) \|\Phi^\prime(\boldsymbol{u}\_{k,n})\| \to 0, \quad k \in \mathbb{N}.$$

Then, we can choose {*nk*} ⊂ <sup>N</sup> such that

$$\Phi(u\_{k,n\_k}) \in \left[\delta\_0, c + \frac{1}{k}\right), \quad (1 + ||u\_{k,n\_k}||) \|\Phi'(u\_{k,n\_k})\| < \frac{1}{k'}, \quad k \in \mathbb{N}.$$

Let *uk* = *uk*,*nk* , *<sup>k</sup>* <sup>∈</sup> <sup>N</sup>. Therefore, up to a subsequence, we have

$$\Phi(\mathfrak{u}\_n) \to \mathfrak{c}^\* \in [\delta\_{0\prime}\mathfrak{c}], \quad (1 + ||\mathfrak{u}\_n||) ||\Phi^\prime(\mathfrak{u}\_n)|| \to 0.$$

**Lemma 9.** *Assume that* (*I*)*–*(*III*) *and* (*F*1)*–*(*F*4) *hold. Then, the sequence* {*un*} *satisfying* (31) *is bounded in E.*

**Proof.** Arguing by contradiction, suppose that *un* → <sup>∞</sup>. Let *vn* <sup>=</sup> *un un* , then *vn* <sup>=</sup> 1. Passing to a subsequence, we have *vn v* in *E*. There are two possible cases: (i). *v* = 0; (ii) *v* = 0.

Case (i) *v* = 0. In this case

$$\begin{cases} \int\_{\mathbb{R}^N} \frac{\mathbb{Q}(\mathbf{x}) \mathbb{Q}(\mathbf{y}) F(2\sqrt{c^\*+1} \mathbf{v}\_n(\mathbf{x})) F(2\sqrt{c^\*+1} \mathbf{v}\_n(\mathbf{y}))}{|\mathbf{x} - \mathbf{y}|^\mu} d\mathbf{x} d\mathbf{y} \Big|\_{\mathbf{x}} \\ \quad \le C\_1 \left[ \int\_{\mathbb{R}^N} K(\mathbf{x}) |F(2\sqrt{c^\*+1} \mathbf{v}\_n(\mathbf{x}))|^{\frac{2N}{2N-\mu}} d\mathbf{x} \right]^{\frac{2N-\mu}{N}} \\ \quad = o(1). \end{cases} \tag{35}$$

Combining (27), (31), and (35), we have

$$\begin{aligned} \varepsilon^\* + o(1) &= \Phi(u\_n) \\ &\ge \Phi\left(\frac{2\sqrt{\varepsilon^\* + 1}}{||u\_n||} u\_n\right) + \frac{1 - \left(\frac{2\sqrt{\varepsilon^\* + 1}}{||u\_n||}\right)^2}{2} \langle \Phi'(u\_n), u\_n \rangle \\ &= \Phi(2\sqrt{\varepsilon^\* + 1}v\_n) + o(1) \\ &= 2(\varepsilon^\* + 1) + o(1), \end{aligned}$$

which is a contradiction.

Case (ii) *<sup>v</sup>* <sup>=</sup> 0. In this case, since <sup>|</sup>*un*<sup>|</sup> <sup>=</sup> <sup>|</sup>*vn*|*un* and *un*/*un* → *<sup>v</sup>* a.e. in <sup>R</sup>*N*, we have lim*n*→<sup>∞</sup> <sup>|</sup>*un*(*x*)<sup>|</sup> <sup>=</sup> <sup>∞</sup> for *<sup>x</sup>* ∈ {*<sup>y</sup>* <sup>∈</sup> <sup>R</sup>*<sup>N</sup>* : *<sup>v</sup>*(*x*) <sup>=</sup> <sup>0</sup>}. Hence, it follows from (22), (31), (*F*4), and Fatou's lemma that

$$\begin{split} 0 &= \lim\_{n \to \infty} \frac{c^\* + o(1)}{||u\_n||^2} = \lim\_{n \to \infty} \frac{\Phi(u\_n)}{||u\_n||^2} \\ &= \frac{1}{2} - \frac{1}{2} \lim\_{n \to \infty} \int\_{\mathbb{R}^N} \int\_{\mathbb{R}^N} \frac{Q(\mathbf{x}) F(u\_n(\mathbf{x}))}{|u\_n(\mathbf{x})|} \frac{Q(\mathbf{y}) F(u\_n(\mathbf{y}))}{|u\_n(\mathbf{y})|} \frac{|v\_n(\mathbf{x}) v\_n(\mathbf{y})|}{|\mathbf{x} - \mathbf{y}|^\mu} d\mathbf{x} d\mathbf{y} \\ &\leq \frac{1}{2} - \frac{1}{2} \int\_{\mathbb{R}^N} \int\_{\mathbb{R}^N} \lim\_{n \to \infty} \int \frac{Q(\mathbf{x} + k\_n) F(u\_n(\mathbf{x}))}{|u\_n(\mathbf{x})|} \frac{Q(\mathbf{y} + k\_n) F(u\_n(\mathbf{y}))}{|u\_n(\mathbf{y})|} \frac{|v\_n(\mathbf{x}) v\_n(\mathbf{y})|}{|\mathbf{x} - \mathbf{y}|^\mu} d\mathbf{x} d\mathbf{y} \\ &= -\infty. \end{split}$$

This contradiction shows that {*un*} is bounded in *E*.

**Proof of Theorem 1.** In view of Lemmas 8 and 9, there exists a bounded sequence {*un*} ⊂ *E* such that (31) holds. Passing to a subsequence, we have *un u* in *E*. Thus, it follows from (22), (23), (31), and Lemma 3 that

$$\|\|u\_{\mathbb{R}} - u\|\|^2 = \langle \Phi'(u\_{\mathbb{R}}), u\_{\mathbb{R}} - u \rangle + \int\_{\mathbb{R}^N} \int\_{\mathbb{R}^N} \frac{Q(\mathbf{x}) Q(\mathbf{y}) F(u\_{\mathbb{R}}(\mathbf{x})) f(u\_{\mathbb{R}}(\mathbf{y})) [u\_{\mathbb{R}}(\mathbf{y}) - u(\mathbf{y})]}{|\mathbf{x} - \mathbf{y}|^\mu} = o(1),$$

which implies that Φ (*u*) = 0 and Φ(*u*) = *c*<sup>∗</sup> ∈ (0, *c*]. Moreover, since *u* ∈ N , we have Φ(*u*) ≥ *c*. Hence, *u* ∈ *E* is a ground state solution for (1) with Φ(*u*) = *c* > 0.

#### **3. Zero Mass Case**

In this section, we consider the zero mass case, and give the proof of Theorem 2. In the following, we suppose that (*F*5)–(*F*7) and *<sup>μ</sup>* <sup>&</sup>lt; <sup>4</sup>*<sup>s</sup>* hold. Fix *<sup>q</sup>* <sup>∈</sup> (2, <sup>2</sup>*N*−*<sup>μ</sup> <sup>N</sup>*−2*<sup>s</sup>* ), by (*F*7), for every <sup>&</sup>gt; 0 there is *<sup>C</sup>* <sup>&</sup>gt; <sup>0</sup> such that

$$\epsilon \left| f(t)t \right| \le \epsilon (\left| t \right|^2 + \left| t \right|^{\frac{2\lambda - \mu}{N - 2s}}) + \mathbb{C}\_{\mathfrak{c}} \left| t \right|^q, \left| F(t) \right| \le \epsilon (\left| t \right|^2 + \left| t \right|^{\frac{2\lambda - \mu}{N - 2s}}) + \mathbb{C}\_{\mathfrak{c}} \left| t \right|^q, \ \forall t \in \mathbb{R}. \tag{36}$$

To find nontrivial solutions for (6), we study the approximating problem

$$\begin{cases} \left( -\Delta \right)^{s} u + \varepsilon u = \left( \frac{1}{|x|^{\overline{\mathbb{T}}}} \* F(u) \right) f(u), \text{ in } \mathbb{R}^{N}, \\\ u \in H^{s}(\mathbb{R}^{N}), \end{cases} \tag{37}$$

where *ε* ≥ 0 is a small parameter. The energy functional associated to (37) is

$$\Phi\_{\varepsilon}(u) = \frac{1}{2} \int\_{\mathbb{R}^N} [|( - \Delta)^{\frac{\varepsilon}{2}} u|^2 + \varepsilon u^2] dx - \frac{1}{2} \int\_{\mathbb{R}^N} \int\_{\mathbb{R}^N} \frac{F(u(x)) F(u(y))}{|x - y|^\mu} dx dy. \tag{38}$$

By using (*F*5)–(*F*7) and Lemma 2, it is easy to check that <sup>Φ</sup><sup>0</sup> <sup>∈</sup> *<sup>C</sup>*1(*Ds*,2(R*N*), <sup>R</sup>) and <sup>Φ</sup>*<sup>ε</sup>* <sup>∈</sup> *<sup>C</sup>*1(*Hs*(R*N*), <sup>R</sup>) for every *<sup>ε</sup>* <sup>&</sup>gt; 0. Moreover, for every *<sup>ε</sup>* <sup>≥</sup> 0,

$$\langle \Phi'\_{\varepsilon}(u), v \rangle = \int\_{\mathbb{R}^{N}} [(-\Delta)^{\frac{\mathfrak{s}}{2}} u(-\Delta)^{\frac{\mathfrak{s}}{2}} v + \varepsilon uv] dx - \frac{1}{2} \int\_{\mathbb{R}^{N}} \int\_{\mathbb{R}^{N}} \frac{F(u(x)) f(u(y)) v(y)}{|x - y|^{\mu}} dx dy. \tag{39}$$

In view of ([19], Proposition 2), for every *ε* > 0, any critical point *u* of Φ*<sup>ε</sup>* in *Hs*(R*N*) satisfies the following Pohozaev identity ˘

$$\begin{split} \mathcal{P}\_{\varepsilon}(u) := \frac{N - 2\varepsilon}{2} \int\_{\mathbb{R}^{N}} |( - \Delta)^{\frac{\varepsilon}{2}} u |^{2} dx + \frac{N}{2} \varepsilon \int\_{\mathbb{R}^{N}} |u|^{2} dx - \frac{2N - \mu}{2} \int\_{\mathbb{R}^{N}} \int\_{\mathbb{R}^{N}} \frac{\mathcal{F}(u(x)) \mathcal{F}(u(y))}{|x - y|^{\mu}} dx dy \\ = 0. \end{split} \tag{40}$$

For every *ε* > 0, let

$$\begin{aligned} \mathcal{M}\_{\mathfrak{c}} &:= \{ u \in H^s(\mathbb{R}^N) \mid \{ 0 \} : \Phi\_{\mathfrak{c}}'(u) = 0 \}, \\ \Gamma\_{\mathfrak{c}} &:= \{ \gamma \in \mathbb{C}([0, 1], H^s(\mathbb{R}^N)) : \gamma(0) = 0, \Phi\_{\mathfrak{c}}(\gamma(1)) < 0 \}, \\ \mathfrak{c}\_{\mathfrak{c}} &:= \inf\_{\gamma \in \Gamma\_{\mathfrak{c}}} \max\_{t \in [0, 1]} \Phi\_{\mathfrak{c}}(\gamma(t)). \end{aligned}$$

**Lemma 10.** *For every <sup>ε</sup>* <sup>&</sup>gt; <sup>0</sup>*,* (37) *has a ground state solution <sup>u</sup><sup>ε</sup>* <sup>∈</sup> *<sup>H</sup>s*(R*N*) *such that* <sup>0</sup> <sup>&</sup>lt; <sup>Φ</sup>*ε*(*uε*) = inf M*ε* Φ*<sup>ε</sup>* = *cε. Moreover, there exists a constant K*<sup>0</sup> > 0 *independent of ε such that c<sup>ε</sup>* ≤ *K*<sup>0</sup> *for all ε* ∈ (0, 1]*.*

**Proof.** In view of ([19], Theorem 1.3), under the assumption (*F*5)–(*F*7), for every *ε* > 0, (37) has a ground state solution *<sup>u</sup><sup>ε</sup>* <sup>∈</sup> *<sup>H</sup>s*(R*N*) such that 0 <sup>&</sup>lt; <sup>Φ</sup>*ε*(*uε*) = inf M*ε* Φ*<sup>ε</sup>* = *cε*. Let *γ* ∈ Γ1, since Φ*ε*(*u*) ≤ Φ1(*u*) for *<sup>u</sup>* <sup>∈</sup> *<sup>H</sup>s*(R*N*) and *<sup>ε</sup>* <sup>∈</sup> (0, 1], we have *<sup>γ</sup>* <sup>∈</sup> <sup>Γ</sup>*<sup>ε</sup>* for *<sup>ε</sup>* <sup>∈</sup> (0, 1], and so

$$c\_{\varepsilon} \le \max\_{t \in [0,1]} \Phi\_{\varepsilon}(\gamma(t)) = \Phi\_{\varepsilon}(\gamma(t\_{\varepsilon})) \le \Phi\_{1}(\gamma(t\_{\varepsilon})) \le \max\_{t \in [0,1]} \Phi\_{1}(\gamma(t)) := \mathbb{K}\_{0}, \ \forall \varepsilon \in (0,1].$$

where *t<sup>ε</sup>* ∈ (0, 1).

**Lemma 11.** *There exists a constant K*<sup>1</sup> > 0 *independent of ε such that*

$$\left[\boldsymbol{u}\_{\ell}\right] \ge \mathbb{K}\_{1}, \ \forall \boldsymbol{u}\_{\ell} \in \mathcal{M}\_{\ell}.\tag{41}$$

**Proof.** Since Φ *<sup>ε</sup>*(*uε*), *uε* = 0 for *u<sup>ε</sup>* ∈ M*ε*, from (*F*6), (39), and Sobolev inequality, we have

$$\begin{split} \left[\left[u\_{\varepsilon}\right]^{2}\right] &= \int\_{\mathbb{R}^{N}} \left|(-\Delta)^{\frac{\mu}{2}}u\_{\varepsilon}\right|^{2} dx \leq \int\_{\mathbb{R}^{N}} \left|(-\Delta)^{\frac{\mu}{2}}u\_{\varepsilon}\right|^{2} + \varepsilon u\_{\varepsilon}^{2} \big| dx \\ &= \int\_{\mathbb{R}^{N}} \int\_{\mathbb{R}^{N}} \frac{F(u\_{\varepsilon}(x))f(u\_{\varepsilon}(y))u\_{\varepsilon}(y)}{|x-y|^{\mu}} dx dy \\ &\leq \mathbb{C}\_{1} \left(\int\_{\mathbb{R}^{N}} \left|F(u\_{\varepsilon})\right|^{\frac{2N}{2N-\mu}} dx\right)^{\frac{2N-\mu}{2N}} \left(\int\_{\mathbb{R}^{N}} \left|f(u\_{\varepsilon})u\_{\varepsilon}\right|^{\frac{2N}{2N-\mu}} dx\right)^{\frac{2N-\mu}{2N}} \\ &\leq \mathbb{C}\_{2} \left(\int\_{\mathbb{R}^{N}} \left|u\_{\varepsilon}\right|^{\frac{2N}{N-2\mu}} dx\right)^{\frac{2N-\mu}{N}} \\ &\leq \mathbb{C}\_{2} S^{\frac{2N-\mu}{N-2\mu}} \left[u\_{\varepsilon}\right]^{\frac{2(2N-\mu)}{N-2\mu}}, \ \forall u\_{\varepsilon} \in \mathcal{M}\_{\varepsilon}, \end{split}$$

which, together with (2*N* − *μ*)/(*N* − 2*s*) > 1, implies that (41) holds.

The following lemma is a version of Lions' concentration-compactness Lemma for fractional Laplacian.

**Lemma 12.** *([18]) Assume* {*un*} *is a bounded sequence in Hs*(R*N*)*, which satisfies*

$$\lim\_{n \to +\infty} \sup\_{y \in \mathbb{R}^N} \int\_{B\_1(y)} |u\_n(x)|^2 dx = 0.$$

*Then un* <sup>→</sup> <sup>0</sup> *in Lq*(R*N*) *for q* <sup>∈</sup> (2, 2<sup>∗</sup> *s* )*.*

**Proof of Theorem 2.** We choose a sequence {*εn*} ⊂ (0, 1] such that *ε<sup>n</sup>* 0. In view of Lemma 10, there exists a sequence {*uε<sup>n</sup>* }⊂M*ε<sup>n</sup>* such that 0 < Φ*ε<sup>n</sup>* (*uε<sup>n</sup>* ) = inf M*εn* Φ*ε<sup>n</sup>* = *cε<sup>n</sup>* ≤ *K*0. For simplicity, we use *un* instead of *<sup>u</sup>ε<sup>n</sup>* . Now, we prove that {*un*} is bounded in *<sup>D</sup>s*,2(R*N*). Since <sup>P</sup>*ε<sup>n</sup>* (*un*) = 0 for *un* ∈ M*ε<sup>n</sup>* , it follows from (38) and (40) that

$$\begin{split} K\_0 \ge \mathfrak{c}\_{\mathfrak{c}\_n} &= \Phi\_{\mathfrak{c}\_n}(u\_n) - \frac{1}{2N - \mu} \mathcal{P}\_{\mathfrak{c}\_n}(u\_n) \\ &= \left[ \frac{1}{2} - \frac{N - 2\mathfrak{s}}{2(2N - \mu)} \right] [u\_n]^2 + \left[ \frac{1}{2} - \frac{N}{2(2N - \mu)} \right] \varepsilon\_n \| u\_n \|\_2^2. \end{split} \tag{42}$$

Thus, {*un*} is bounded in *<sup>D</sup>s*,2(R*N*) and *<sup>L</sup>*2(R*N*). If

$$\delta := \lim\_{n \to \infty} \sup\_{y \in \mathbb{R}^N} \int\_{B\_1(y)} |u\_n|^2 d\mathfrak{x} = 0.$$

Then, by Lemma 12, for *<sup>q</sup>* <sup>∈</sup> (2, <sup>2</sup>*N*−*<sup>μ</sup> <sup>N</sup>*−2*<sup>s</sup>* ), we have

$$\int\_{\mathbb{R}^{\mathbb{N}}} |u\_n|^{\frac{4N}{2N-p}} d\mathfrak{x} \to 0, \quad \int\_{\mathbb{R}^N} |u\_n|^{\frac{2Nq}{2N-p}} d\mathfrak{x} \to 0.$$

Therefore, by (36) and Sobolev embedding for *Ds*,2(R*N*), for every > 0 there exists *C* > 0 such that

$$\begin{split} \left| \int\_{\mathbb{R}^{N}} |\boldsymbol{F}(\boldsymbol{u}\_{\boldsymbol{n}})|^{\frac{2\lambda}{2N-\mu}} d\boldsymbol{x} \right| &\leq \varepsilon \left[ \int\_{\mathbb{R}^{N}} \left( |\boldsymbol{u}\_{\boldsymbol{n}}|^{\frac{4\lambda}{2N-\mu}} + |\boldsymbol{u}\_{\boldsymbol{n}}|^{2^{\*}\_{s}} \right) d\boldsymbol{x} \right] + \mathbb{C}\_{\mathfrak{c}} \int\_{\mathbb{R}^{N}} |\boldsymbol{u}\_{\boldsymbol{n}}|^{\frac{2Nq}{2N-\mu}} d\boldsymbol{x} \\ &\leq \varepsilon \mathsf{C} + o(1). \end{split}$$

By the arbitrariness of , we get

$$\int\_{\mathbb{R}^N} |F(u\_n)|^{\frac{2N}{2N-\mu}} dx \to 0. \tag{43}$$

Combining (36), (43), and Lemma 2, we have

$$\begin{split} & \left| \int\_{\mathbb{R}^N} \int\_{\mathbb{R}^N} \frac{F(u\_n(\mathbf{x})) f(u\_n(\mathbf{y})) u\_n(\mathbf{y})}{|\mathbf{x} - \mathbf{y}|^\mu} d\mathbf{x} d\mathbf{y} \right| \\ & \leq C\_1 \left( \int\_{\mathbb{R}^N} |F(u\_n)|^{\frac{2N}{2N-\mu}} d\mathbf{x} \right)^{\frac{2N-\mu}{2N}} \left( \int\_{\mathbb{R}^N} |f(u\_n) u\_n|^{\frac{2N}{2N-\mu}} d\mathbf{x} \right)^{\frac{2N-\mu}{2N}} \\ & = o(1). \end{split} \tag{44}$$

.

Notice that {*un*} is bounded in *<sup>L</sup>*2(R*N*), we have from (44) and *un* ∈ M*ε<sup>n</sup>* that [*un*] <sup>2</sup> = *o*(1). This contradicts (41). Thus, we get *<sup>δ</sup>* <sup>&</sup>gt; 0. Passing to a subsequence, there exists a sequence {*yn*} ⊂ <sup>R</sup>*<sup>N</sup>* such that

$$\int\_{B\_{1+\sqrt{N}}(y\_n)} |u\_n|^2 d\mathbf{x} > \frac{\delta}{2}$$

Let *u*˜*n*(*x*) = *un*(*x* + *yn*). Then

$$
\Phi\_{\ell\_n}'(\vec{u}\_{\hbar}) = 0,\ \Phi\_{\ell\_n}(\vec{u}\_{\hbar}) = c\_{\ell\_n}
$$

$$
\int\_{B\_{1+\sqrt{N}}(0)} |\tilde{u}\_{\hbar}|^2 d\mathbf{x} > \frac{\delta}{2}.\tag{45}
$$

and

Passing to a subsequence, we have *<sup>u</sup>*˜*<sup>n</sup> <sup>u</sup>*<sup>0</sup> in *<sup>D</sup>s*,2(R*N*). Clearly, (45) implies that *<sup>u</sup>*<sup>0</sup> <sup>=</sup> 0. By the standard argument, *<sup>u</sup>*<sup>0</sup> <sup>∈</sup> *<sup>D</sup>s*,2(R*N*) is a nontrivial solution for (6).

#### **4. Conclusions**

In this work, we study a class of nonlinear Choquard equation driven by the fractional Laplacian. When potential function vanishes at infinity and the Nehari-type monotonicity condition for the nonlinearity is not satisfied, we prove that the fractional Choquard equation has a ground state solution by using the non-Nehari manifold method. Unlike the Nehari manifold method, the main idea of our approach lies in finding a minimizing sequence for the energy functional outside the Nehari manifold by using the diagonal method. Moreover, by using a perturbation method, we obtain a nontrivial solution in the zero mass case.

**Author Contributions:** All authors contributed equally in writing this article. All authors read and approved the final manuscript.

**Funding:** This work is supported by Hainan Natural Science Foundation (Grant No.118MS002 and No.117005), the National Natural Science Foundation of China (Grant No.11861028 and No.11461016 ), Young Foundation of Hainan University (Grant No.hdkyxj201718).

**Conflicts of Interest:** The authors declare no conflict of interest.

#### **References**



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