## 1 Introduction and main results

In this article, we shall assume that the reader is familiar with the fundamental results and the standard notations of the Nevanlinna theory (e.g., see [13]). In addition, we will use the notations λ(f) to denote the exponent of convergence of zero sequences of meromorphic function f(z); σ(f) to denote the order of f(z). We say that meromorphic functions f and g share a finite value a CM when f - a and g - a have the same zeros with the same multiplicities. For a non-zero constant c, the forward difference ${\Delta }_{c}f\left(z\right)=f\left(z+c\right)-f\left(z\right)$, ${\Delta }_{c}^{n+1}f\left(z\right)={\Delta }_{c}^{n}f\left(z+c\right)-{\Delta }_{c}^{n}f\left(z\right)$, n = 1, 2,.... In general, we use the notation C to denote the field of complex numbers.

Currently, there has been an increasing interest in studying difference equations in the complex plane. Halburd and Korhonen [4, 5] established a version of Nevanlinna theory based on difference operators. Ishizaki and Yanagihara [6] developed a version of Wiman-Valiron theory for difference equations of entire functions of small growth.

Recently, Liu and Yang [7] establish a counterpart result to the Brück conjecture [8] valid for transcendental entire function for which σ(f) < 1. The result is stated as follows.

Theorem A. Let f be a transcendental entire function such that σ(f) < 1. If f and ${\Delta }_{c}^{n}f$ share a finite value a CM, n is a positive integer, and c is a fixed constant, then

$\frac{{\Delta }_{c}^{n}f-a}{f-a}=\tau$

for some non-zero constant τ.

Heittokangas et al. [9], prove the following result which is a shifted analogue of Brück conjecture valid for meromorphic functions.

Theorem B. Let f be a meromorphic function of order of growth σ(f) < 2, and let cC. If f(z) and f(z + c) share the values aC andCM, then

$\frac{f\left(z+c\right)-a}{f\left(z\right)-a}=\tau$

for some constant τ.

Here, we also study the shift analogue of Brück conjecture, and obtain the results as follows.

Theorem 1.1. Let f(z) be a non-constant entire function, σ(f) < 1 or 1 < σ(f) < 2 and λ(f) < σ(f) = σ. Set L1(f) = a n (z) f(z + n) + a n -1(z) f(z + n - 1) +... + a1(z) f(z + 1) + a0(z) f(z), where a j (z)(0 ≤ jn) are entire functions with a n (z)a0(z) ≢ 0. Suppose that if σ(f) < 1, then max{σ(a j )} = α < 1, and if 1 < σ(f) < 2, then max{σ(a j )} = α < σ - 1. If f and L1(f) share 0 CM, then

${L}_{1}\left(f\right)=cf,$

where c is a non-zero constant.

Theorem 1.2. Let f(z) be a non-constant entire function, 2 < σ(f) <and λ(f) < σ(f). Set L2(f) = a n (z) f(z + n) + a n -1(z) f(z + n - 1) +... + a1(z) f(z + 1) + ezf(z), a j (z)(1 ≤ jn) are entire functions with σ(a j ) < 1 and a n (z) ≢ 0. If f and L2(f ) share 0 CM, then

${L}_{2}\left(f\right)=h\left(z\right)f,$

where h(z) is an entire function of order no less than 1.

Theorem 1.3. Let f(z) be a non-constant entire function, σ(f) < 1 or 1 < σ(f) < 2, λ(f) < σ(f). Set L3(f) = a n (z) f(z + n) + a n- 1(z) f(z + n - 1) + ... + a1(z) f(z + 1) + a0(z) f(z), a j (z)(0 ≤ jn) are polynomials and a n (z) ≢ 0. If f and L3(f ) share a polynomial P(z) CM, then

${L}_{3}\left(f\right)-p\left(z\right)=c\left(f\left(z\right)-p\left(z\right)\right),$

where c is a non-zero constant.

Theorem 1.4. Let f(z) be a non-constant entire function, σ(f) < 1 or 1 < σ(f) < 2, λ(f) < σ(f). Set a(z) is an entire function with σ(a) < 1. If f and a(z)f(z + n) share a polynomial P(z) CM, then

$a\left(z\right)f\left(z+n\right)-p\left(z\right)=c\left(f\left(z\right)-p\left(z\right)\right),$

where c is a non-zero constant.

The method of the article is partly from [10].

## 2 Preliminary lemmas

Lemma 2.1. [11] Let f(z) be a meromorphic function with σ(f) = η < ∞. Then for any given ε > 0, there is a set E1 ⊂ (1, +∞) that has finite logarithmic measure, such that

$|f\left(z\right)|\phantom{\rule{2.77695pt}{0ex}}\le \text{exp}\left\{{r}^{\eta +\epsilon }\right\},$

holds for |z| = r ∉ [0, 1] ∪ E1, r → ∞.

Applying Lemma 2.1 to $\frac{1}{f}$, it is easy to see that for any given ε > 0, there is a set E2 ⊂ (1, ∞) of finite logarithmic measure, such that

$\text{exp}\left\{-{r}^{\eta +\in }\right\}\le \phantom{\rule{2.77695pt}{0ex}}|f\left(z\right)|\phantom{\rule{2.77695pt}{0ex}}\le \text{exp}\left\{{r}^{\eta +\epsilon }\right\},$

holds for |z| = r ∉ [0, 1] ∪ E2, r → ∞.

Lemma 2.2. [11] Let

$Q\left(z\right)={b}_{n}{z}^{n}+{b}_{n-1}{z}^{n-1}+\cdots +{b}_{0},$

where n is a positive integer and ${b}_{n}={\alpha }_{n}{e}^{i{\theta }_{n}},{\alpha }_{n}>0,{\theta }_{n}\in \left[0,2\pi \right)$. For any given $\epsilon \left(0<\epsilon <\frac{\pi }{4n}\right)$, we introduce 2n open sectors

${S}_{j}:-{\theta }_{n}+\left(2j-1\right)\frac{\pi }{2n}+\epsilon <\theta <-{\theta }_{n}+\left(2j+1\right)\frac{\pi }{2n}-\epsilon \left(j=0,1,\dots ,2n-1\right).$

Then there exists a positive number R = R(ε) such that for |z| = r > R,

$Re\left\{Q\left(z\right)\right\}>{\alpha }_{n}\left(1-\epsilon \right)\text{sin}\left(n\epsilon \right){r}^{n}$

if zS j where j is even; while

$Re\left\{Q\left(z\right)\right\}<-{\alpha }_{n}\left(1-\epsilon \right)\text{sin}\left(n\epsilon \right){r}^{n}$

if zS j where j is odd.

Now for any given θ ∈ [0, 2π), if $\theta \ne -\frac{{\theta }_{n}}{n}+\left(2j-1\right)\frac{\pi }{2n}$, (j = 0, 1,..., 2n - 1), then we take ε sufficiently small, there is some S j , j ∈ {0, 1,...,2n - 1} such that θS j .

Lemma 2.3. [12] Let f(z) be a meromorphic function of order σ = σ(f) < ∞, and let λ' and λ'' be, respectively, the exponent of convergence of the zeros and poles of f. Then for any given ε > 0, there exists a set E ⊂ (1, ∞) of |z| = r of finite logarithmic measure, so that

$2\pi i{n}_{z,\eta }+\text{log}\frac{f\left(z+\eta \right)}{f\left(z\right)}=\eta \frac{{f}^{\prime }\left(z\right)}{f\left(z\right)}+O\left({r}^{\beta +\epsilon }\right),$

or equivalently,

$\frac{f\left(z+\eta \right)}{f\left(z\right)}={e}^{\eta \frac{{f}^{\prime }\left(z\right)}{f\left(z\right)}+O\left({r}^{\beta +\epsilon }\right)},$

holds for rE ∪ [0, 1], where n z,η is an integer depending on both z and η, β = max{σ - 2, 2λ - 2} if λ < 1 and β = max{σ - 2, λ - 1} if λ ≥ 1 and λ = max{λ', λ''} .

Lemma 2.4. [2] Let f(z) be an entire function of order σ, then

$\sigma =\underset{r\to \infty }{\text{lim}\text{sup}}\frac{\text{log}\nu \left(r\right)}{\text{log}r}$

where ν(r) be the central index of f.

Lemma 2.5. [2, 13, 14] Let f be a transcendental entire function, let $0<\delta <\frac{1}{4}$ and z be such that|z| = r and that

$|f\left(z\right)|>M\left(r,g\right)\nu {\left(r,g\right)}^{-\frac{1}{4}+\delta }$

holds. Then there exists a set FR+ of finite logarithmic measure, i.e., ${\int }_{F}\frac{dt}{t}<\infty$, such that

$\frac{{f}^{\left(m\right)}\left(z\right)}{f\left(z\right)}={\left(\frac{\nu \left(r,f\right)}{z}\right)}^{m}\left(1+o\left(1\right)\right)$

holds for all m ≥ 0 and all rF.

Lemma 2.6. [10] Let f(z) be a transcendental entire function, σ(f) = σ <, and G = {ω1, ω2,..., ω n }, and a set E ⊂ (1, ∞) having logarithmic measure lmE <. Then there is a positive number $B\left(\frac{3}{4}\le B\le 1\right)$, a point range $\left\{{z}_{k}={r}_{k}{e}^{i{\omega }_{k}}\right\}$ such that |f(z k )| ≥ BM(r k , f ), ω k ∈ [0, 2π), lim k →∞ ω k = ω0 ∈ [0, 2π), r k E ∪ [0, 1], r k → ∞, for any given ε > 0, we have

${r}_{k}^{\sigma -\epsilon }<\nu \left({r}_{k},f\right)<{r}_{k}^{\sigma +\epsilon }.$

## 3 Proof of Theorem 1.1

Under the hypothesis of Theorem 1.1, see [3], it is easy to get that

$\frac{{L}_{1}\left(f\right)}{f}={e}^{Q\left(z\right)},$
(3.1)

where Q(z) is an entire function. If σ(f) < 1, we get Q(z) is a constant. Then Theorem 1.1 holds. Next, we suppose that 1 < σ(f) < 2 and λ(f) < σ(f) = σ. We divide this into two cases (Q(z) is a constant or a polynomial with deg Q = 1) to prove.

Case (1): Q(z) is a constant. Then Theorem 1.1 holds.

Case (2): deg Q = 1. By Lemma 2.3 and λ(f) < σ(f) = σ, for any given $0<\epsilon <\text{min}\left\{\frac{\sigma -1}{2},\frac{1-\alpha }{2},\frac{\sigma -\lambda \left(f\right)}{2},\frac{\sigma -1-\alpha }{2}\right\}$, there exists a set E1 ⊂ (1, ∞) of |z| = r of finite logarithmic measure, so that

$\frac{f\left(z+j\right)}{f\left(z\right)}=\text{exp}\left\{j\frac{{f}^{\prime }\left(z\right)}{f\left(z\right)}+o\left({r}^{\sigma \left(f\right)-1-\epsilon }\right)\right\},j=1,2,\dots ,n$
(3.2)

holds for rE1 ∪ 0[1].

By Lemma 2.5, there exists a set E2 ⊂ (0, ∞) of finite logarithmic measure, such that

$\frac{{f}^{\prime }\left(z\right)}{f\left(z\right)}=\left(1+o\left(1\right)\right)\frac{\nu \left(r,f\right)}{z},$
(3.3)

holds for |z| = rE2 ∪ [0, 1], where z is chosen as in Lemma 2.5.

By Lemma 2.1, for any given ε > 0, there exists a set E3 ⊂ (1, ∞) that has finite logarithmic measure such that

$\text{exp}\left\{-{r}^{\alpha +\epsilon }\right\}\le \phantom{\rule{2.77695pt}{0ex}}|{a}_{j}\left(z\right)|\phantom{\rule{2.77695pt}{0ex}}\le \text{exp}\left\{{r}^{\alpha +\epsilon }\right\}\left(j=0,1,\dots ,n\right)$
(3.4)

holds for |z| = r ∉ [0, 1] ∪ E3, r → ∞.

Set E = E1E2E3 and $G=\left\{-\frac{{\phi }_{n}}{n}+\left(2j-1\right)\frac{\pi }{2n}|j=0,1\right\}\cup \left\{\frac{\pi }{2},\frac{3\pi }{2}\right\}$. By Lemma 2.6, there exist a positive number $B\in \left[\frac{3}{4},1\right]$, a point range $\left\{{z}_{k}={r}_{k}{e}^{i{\theta }_{k}}\right\}$ such that |f(z k )| ≥ BM (r k , f], θ k ∈ [0, 2π), lim k →∞ θ k = θ0 ∈ [0, 2π) \ G, r k E ∪ [0, 1], r k → ∞, for any given ε > 0, as r k → ∞, we have

${r}_{k}^{\sigma \left(f\right)-\epsilon }<\nu \left({r}_{k},f\right)<{r}_{k}^{\sigma \left(f\right)+\epsilon }$
(3.5)

By (3.1)-(3.3), we have that

${a}_{n}\mathrm{exp}\left\{n\left(1+o\left(1\right)\right)\frac{\nu \left({r}_{k},f\right)}{{z}_{k}}\right\}+\cdots +{a}_{1}\mathrm{exp}\left\{\left(1+o\left(1\right)\right)\frac{\nu \left({r}_{k},f\right)}{{z}_{k}}\right\}+{a}_{0}={e}^{Q\left(z\right)}\right\}$
(3.6)

Let $Q\left(z\right)=\tau {e}^{i{\theta }_{1}}z+{b}_{0}$, τ > 0, θ1 ∈ [0, 2π). By Lemma 2.4, there are two opened angles for above ε,

${S}_{j}:-{\theta }_{1}+\left(2j-1\right)\frac{\pi }{2}+\epsilon <\theta <-{\theta }_{1}+\left(2j+1\right)\frac{\pi }{2}+\epsilon \left(j=0,1\right)$

For the above θ0, there are two cases: (i) θ0S0; (ii) θ0S1.

Case (i). θ0S1. Since S j is an opened set and lim k →∞ θ k = θ0, there is a K > 0 such that θ k S j when k > K. By Lemma 2.2, we have

$Re\left\{Q\left({r}_{k}{e}^{i{\theta }_{k}}\right)\right\}<-\eta {r}_{k},$
(3.7)

where η = η(1 - ε) sin(ε) > 0. By Lemma 2.2, if Rez k > ζr k (0 < ζ ≤ 1). By (3.4)-(3.7), we have

$\begin{array}{c}\text{exp}\left\{{r}_{k}^{\sigma \left(f\right)-1-\epsilon }-{r}_{k}^{\alpha +\epsilon }\right\}\\ \le \left|{a}_{n}\text{exp}\left\{n\left(1+o\left(1\right)\right)\frac{\nu \left({r}_{k},f\right)}{{z}_{k}}\right\}\right|\\ \le 3\left|{a}_{n}\text{exp}\left\{n\left(1+o\left(1\right)\right)\frac{\nu \left({r}_{k},f\right)}{{z}_{k}}\right\}+\cdots +{a}_{1}\text{exp}\left\{\left(1+o\left(1\right)\right)\frac{\nu \left({r}_{k},f\right)}{{z}_{k}}\right\}+{a}_{0}\right|\\ =3\left|{e}^{Q\left(z\right)}\right|\le 3{e}^{-\eta {r}_{k}},\end{array}$
(3.8)

which contradicts that 0 < σ(f) - 1 - α - ε.

If Rez k < - ζr k (0 < ζ ≤ 1), By (3.4)-(3.7), we have

$\begin{array}{c}1\le \left|\frac{{a}_{n}}{{a}_{0}}\text{exp}\left\{n\left(1+o\left(1\right)\right)\frac{\nu \left({r}_{k},f\right)}{{z}_{k}}\right\}+\cdots +\frac{{a}_{1}}{{a}_{0}}\text{exp}\left\{\left(1+o\left(1\right)\right)\frac{\nu \left({r}_{k},f\right)}{{z}_{k}}\right\}\right|+\left|\frac{{e}^{Q\left(z\right)}}{{a}_{0}}\right|\\ \le 2n\text{exp}\left\{-\eta {r}_{k}^{\sigma \left(f\right)-1+\epsilon }+2{r}_{k}^{\alpha +\epsilon }\right\}+{e}^{-\eta {r}_{k}}\text{exp}\phantom{\rule{2.77695pt}{0ex}}\left\{{r}_{k}^{\alpha +\epsilon }\right\},\end{array}$
(3.9)

which implies that 1 < 0, r → ∞, a contradiction.

Case (ii). θ0S0. Since S0 is an opened set and lim k →∞ θ k = θ0, there is K > 0 such that θ k S j when k > K. By Lemma 2.2, we have

$Re\left\{Q\left({r}_{k}{e}^{i{\theta }_{k}}\right)\right\}>\eta {r}_{k},$
(3.10)

where η = τ(1 - ε) sin(ε) > 0. By (3.4)-(3.6), (3.9), we obtain

$\begin{array}{c}\left(n+1\right)\text{exp}\left\{n{r}_{k}^{\sigma \left(f\right)-1+\epsilon }+{r}_{k}^{\alpha +\epsilon }\right\}\\ \ge \phantom{\rule{2.77695pt}{0ex}}|{a}_{n}\text{exp}\left\{n\left(1+o\left(1\right)\right)\frac{\nu \left({r}_{k},f\right)}{{z}_{k}}\right\}+\cdots +{a}_{1}\text{exp}\left\{\left(1+o\left(1\right)\right)\frac{\nu \left({r}_{k},f\right)}{{z}_{k}}\right\}+{a}_{0}|\\ =\phantom{\rule{2.77695pt}{0ex}}|{e}^{Q\left(z\right)}|\phantom{\rule{2.77695pt}{0ex}}\ge {e}^{\eta {r}_{k}}.\end{array}$
(3.11)

From (3.11), we get that σ(f) ≥ 2, a contradiction. Theorem 1.1 is thus proved.

## 4 Proof of Theorem 1.2

Under the hypothesis of Theorem 1.2, see [3], it is easy to get that

$\frac{{L}_{2}\left(f\right)}{f}={e}^{Q\left(z\right)},$
(4.1)

where Q(z) is an entire function. For Q(z), we discuss the following two cases.

Case (1): Q(z) is a polynomial with deg Q = n ≥ 1. Then Theorem 1.2 is proved.

Case (2): Q(z) is a constant. Using the similar reasoning as in the proof of Theorem 1.1, we get that

${a}_{n}\text{exp}\left\{n\left(1+o\left(1\right)\right)\frac{\nu \left({r}_{k},f\right)}{{z}_{k}}\right\}+\cdots +{a}_{1}\text{exp}\left\{\left(1+o\left(1\right)\right)\frac{\nu \left({r}_{k},f\right)}{{z}_{k}}\right\}+a=-{e}^{{z}_{k}},$
(4.2)

where a is some non-zero constant.

If Rez k < -ηr k (η ∈ (0, 1]), By (3.4), (3.5), (4.2), we have

$\begin{array}{c}|a|\phantom{\rule{2.77695pt}{0ex}}\le \left|{a}_{n}\text{exp}\left\{n\left(1+o\left(1\right)\right)\frac{\nu \left({r}_{k},f\right)}{{z}_{k}}\right\}+\cdots +{a}_{1}\text{exp}\left\{\left(1+o\left(1\right)\right)\frac{\nu \left({r}_{k},f\right)}{{z}_{k}}\right\}\right|+|\text{exp}\left\{{z}_{k}\right\}|\\ \le \text{exp}\left\{-\eta {r}_{k}\right\}+n\text{exp}\left\{-\eta {r}_{k}^{\sigma \left(f\right)-1+\epsilon }+2{r}_{k}^{\alpha +\epsilon }\right\},\end{array}$
(4.3)

which is impossible.

If Rez k > ηr k (η ∈ (0, 1]), By (3.4), (3.5) and (4.2), we get

$\begin{array}{c}\text{exp}\left\{\eta {r}_{k}^{\sigma \left(f\right)-1-\epsilon }\right\}<\text{exp}\left\{n\frac{\nu \left({r}_{k},f\right)}{{z}_{k}}-{r}_{k}^{\alpha +\epsilon }\right\}\\ \le 2\left|{a}_{n}\text{exp}\left\{n\left(1+o\left(1\right)\right)\frac{\nu \left({r}_{k},f\right)}{{z}_{k}}\right\}+\cdots +{a}_{1}\text{exp}\left\{\left(1+o\left(1\right)\right)\frac{\nu \left({r}_{k},f\right)}{{z}_{k}}\right\}+a\right|\\ =2|-\text{exp}\left\{{z}_{k}\right\}|\phantom{\rule{2.77695pt}{0ex}}\le 2\text{exp}\left\{{r}_{k}\right\},\end{array}$
(4.4)

which contradicts that σ(f) > 2. This completes the proof of Theorem 1.2.

## 5 Proof of Theorem 1.3

Since f and L3(f) share P CM, we get

$\frac{{L}_{3}\left(f\right)}{f}={e}^{Q\left(z\right)},$
(5.1)

where Q(z) is an entire function. If σ(f) < 1, we get Q(z) is a constant. Then Theorem 1.3 holds. Next, we suppose that 1 < σ(f) < 2 and λ(f) < σ(f) = σ. Set F(z) = f(z) - P(z), then σ(F) = σ(f). Substituting F(z) = f(z) - p(z) into (5.1), we obtain

$\frac{{a}_{n}\left(z\right)F\left(z+n\right)+{a}_{n-1}\left(z\right)F\left(z+n-1\right)+\cdots +{a}_{1}\left(z\right)F\left(z+1\right)}{F\left(z\right)}+{a}_{0}\left(z\right)+\frac{b\left(z\right)}{F\left(z\right)}={e}^{Q\left(z\right)},$
(5.2)

where b(z) = a n (z)P(z + n) + ... + a1(z)P (z + 1) + a0(z)p(z) is a polynomial. We discuss the following two cases.

Case 1. Q(z) is a complex constant. Then Theorem 1.3 holds.

Case 2. Q(z) is a polynomial with deg Q = 1. By Lemma 2.3 and λ(f) < σ(f) = σ, for any given $0<\epsilon <\text{min}\left\{\frac{\sigma -1}{2},\frac{1-\alpha }{2},\frac{\sigma -\lambda \left(f\right)}{2},\frac{\sigma -1-\alpha }{2}\right\}$, there exists a set E1 ⊂ (1, ∞) of |z| = r of finite logarithmic measure, so that

$\frac{f\left(z+j\right)}{f\left(z\right)}=\text{exp}\left\{j\frac{{f}^{\prime }\left(z\right)}{f\left(z\right)}+o\left({r}^{\sigma \left(f\right)-1-\epsilon }\right)\right\},j=1,2,\dots ,n$
(5.3)

holds for rE1 ∪ [0, 1].

By Lemma 2.5, there exists a set E2 ⊂ (0, ∞) of finite logarithmic measure, such that

$\frac{{f}^{\prime }\left(z\right)}{f\left(z\right)}=\left(1+o\left(1\right)\right)\frac{\nu \left(r,f\right)}{z},$
(5.4)

holds for |z| = rE2 ∪ [0, 1], where z is chosen as in Lemma 2.5.

Set E = E1E2 and $G=\left\{-\frac{{\phi }_{n}}{n}+\left(2j-1\right)\frac{\pi }{2n}|j=0,1\right\}\cup \left\{\frac{\pi }{2},\frac{3\pi }{2}\right\}$. By Lemma 2.6, there exist a positive number $B\in \left[\frac{3}{4},1\right]$, a point range $\left\{{z}_{k}={r}_{k}{e}^{i{\theta }_{k}}\right\}$ such that | f (z k )| ≥ BM(r k , f), θ k ∈ [0, 2π), lim k →∞θ k = θ0 ∈ [0, 2π) \ G, r k E ∪ 0[1], r k → ∞, for any given ε > 0, as r k → ∞, we have

${r}_{k}^{\sigma \left(f\right)-\epsilon }<\nu \left({r}_{k},f\right)<{r}_{k}^{\sigma \left(f\right)+\epsilon }.$
(5.5)

Since F is a transcendental entire function and |f(z k )| ≥ BM (r k , f), we obtain

$\frac{b\left({z}_{k}\right)}{F\left({z}_{k}\right)}\to 0,\left({r}_{k}\to \infty \right).$
(5.6)

By (5.2)-(5.6), we have that

${a}_{n}\text{exp}\left\{n\left(1+o\left(1\right)\right)\frac{\nu \left({r}_{k},f\right)}{{z}_{k}}\right\}+\cdots +{a}_{1}\text{exp}\left\{\left(1+o\left(1\right)\right)\frac{\nu \left({r}_{k},f\right)}{{z}_{k}}\right\}+{a}_{0}+o\left(1\right)={e}^{Q\left(z\right)}.$
(5.7)

Using similar proof as in proof of Theorem 1.1, we can get a contradiction. Hence, Theorem 1.3 holds.

## 6 Proof of Theorem 1.4

Using similar proof as in proof of Theorem 1.1, we can get Theorem 1.4 holds.

## Author's contributions

YL completed the main part of this article, YL, XQ and HX corrected the main theorems. All authors read and approved the final manuscript.