Theory and Modern Applications

# Some results on difference polynomials sharing values

## 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 [1â€“3]). In addition, we will use the notations Î»(f) to denote the exponent of convergence of zero sequences of meromorphic function f(z); Ïƒ(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 ${\mathrm{Î”}}_{c}f\left(z\right)=f\left(z+c\right)-f\left(z\right)$, ${\mathrm{Î”}}_{c}^{n+1}f\left(z\right)={\mathrm{Î”}}_{c}^{n}f\left(z+c\right)-{\mathrm{Î”}}_{c}^{n}f\left(z\right)$, 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 BrÃ¼ck conjecture [8] valid for transcendental entire function for which Ïƒ(f) < 1. The result is stated as follows.

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

$\frac{{\mathrm{Î”}}_{c}^{n}f-a}{f-a}=\mathrm{Ï„}$

for some non-zero constant Ï„.

Heittokangas et al. [9], prove the following result which is a shifted analogue of BrÃ¼ck conjecture valid for meromorphic functions.

Theorem B. Let f be a meromorphic function of order of growth Ïƒ(f) < 2, and let c âˆˆ C. If f(z) and f(z + c) share the values a âˆˆ C and âˆž CM, then

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

for some constant Ï„.

Here, we also study the shift analogue of BrÃ¼ck conjecture, and obtain the results as follows.

Theorem 1.1. Let f(z) be a non-constant entire function, Ïƒ(f) < 1 or 1 < Ïƒ(f) < 2 and Î»(f) < Ïƒ(f) = Ïƒ. 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 â‰¤ j â‰¤ n) are entire functions with a n (z)a0(z) â‰¢ 0. Suppose that if Ïƒ(f) < 1, then max{Ïƒ(a j )} = Î± < 1, and if 1 < Ïƒ(f) < 2, then max{Ïƒ(a j )} = Î± < Ïƒ - 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 < Ïƒ(f) < âˆž and Î»(f) < Ïƒ(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 â‰¤ j â‰¤ n) are entire functions with Ïƒ(a j ) < 1 and a n (z) â‰¢ 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, Ïƒ(f) < 1 or 1 < Ïƒ(f) < 2, Î»(f) < Ïƒ(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 â‰¤ j â‰¤ n) are polynomials and a n (z) â‰¢ 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, Ïƒ(f) < 1 or 1 < Ïƒ(f) < 2, Î»(f) < Ïƒ(f). Set a(z) is an entire function with Ïƒ(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 Ïƒ(f) = Î· < âˆž. Then for any given Îµ > 0, there is a set E1 âŠ‚ (1, +âˆž) that has finite logarithmic measure, such that

$|f\left(z\right)|\phantom{\rule{2.77695pt}{0ex}}â‰¤\text{exp}\left\{{r}^{\mathrm{Î·}+\mathrm{Îµ}}\right\},$

holds for |z| = r âˆ‰ [0, 1] âˆª E1, r â†’ âˆž.

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

$\text{exp}\left\{-{r}^{\mathrm{Î·}+âˆˆ}\right\}â‰¤\phantom{\rule{2.77695pt}{0ex}}|f\left(z\right)|\phantom{\rule{2.77695pt}{0ex}}â‰¤\text{exp}\left\{{r}^{\mathrm{Î·}+\mathrm{Îµ}}\right\},$

holds for |z| = r âˆ‰ [0, 1] âˆª E2, r â†’ âˆž.

Lemma 2.2. [11] Let

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

where n is a positive integer and ${b}_{n}={\mathrm{Î±}}_{n}{e}^{i{\mathrm{Î¸}}_{n}},{\mathrm{Î±}}_{n}>0,{\mathrm{Î¸}}_{n}âˆˆ\left[0,2\mathrm{Ï€}\right)$. For any given $\mathrm{Îµ}\left(0<\mathrm{Îµ}<\frac{\mathrm{Ï€}}{4n}\right)$, we introduce 2n open sectors

${S}_{j}:-{\mathrm{Î¸}}_{n}+\left(2j-1\right)\frac{\mathrm{Ï€}}{2n}+\mathrm{Îµ}<\mathrm{Î¸}<-{\mathrm{Î¸}}_{n}+\left(2j+1\right)\frac{\mathrm{Ï€}}{2n}-\mathrm{Îµ}\left(j=0,1,â€¦,2n-1\right).$

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

$Re\left\{Q\left(z\right)\right\}>{\mathrm{Î±}}_{n}\left(1-\mathrm{Îµ}\right)\text{sin}\left(n\mathrm{Îµ}\right){r}^{n}$

if z âˆˆ S j where j is even; while

$Re\left\{Q\left(z\right)\right\}<-{\mathrm{Î±}}_{n}\left(1-\mathrm{Îµ}\right)\text{sin}\left(n\mathrm{Îµ}\right){r}^{n}$

if z âˆˆ S j where j is odd.

Now for any given Î¸ âˆˆ [0, 2Ï€), if , (j = 0, 1,..., 2n - 1), then we take Îµ sufficiently small, there is some S j , j âˆˆ {0, 1,...,2n - 1} such that Î¸ âˆˆ S j .

Lemma 2.3. [12] Let f(z) be a meromorphic function of order Ïƒ = Ïƒ(f) < âˆž, and let Î»' and Î»'' be, respectively, the exponent of convergence of the zeros and poles of f. Then for any given Îµ > 0, there exists a set E âŠ‚ (1, âˆž) of |z| = r of finite logarithmic measure, so that

$2\mathrm{Ï€}i{n}_{z,\mathrm{Î·}}+\text{log}\frac{f\left(z+\mathrm{Î·}\right)}{f\left(z\right)}=\mathrm{Î·}\frac{{f}^{\mathrm{â€²}}\left(z\right)}{f\left(z\right)}+O\left({r}^{\mathrm{Î²}+\mathrm{Îµ}}\right),$

or equivalently,

$\frac{f\left(z+\mathrm{Î·}\right)}{f\left(z\right)}={e}^{\mathrm{Î·}\frac{{f}^{\mathrm{â€²}}\left(z\right)}{f\left(z\right)}+O\left({r}^{\mathrm{Î²}+\mathrm{Îµ}}\right)},$

holds for r âˆ‰ E âˆª [0, 1], where n z,Î· is an integer depending on both z and Î·, Î² = max{Ïƒ - 2, 2Î» - 2} if Î» < 1 and Î² = max{Ïƒ - 2, Î» - 1} if Î» â‰¥ 1 and Î» = max{Î»', Î»''} .

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

$\mathrm{Ïƒ}=\underset{râ†’\mathrm{âˆž}}{\text{lim}\text{sup}}\frac{\text{log}\mathrm{Î½}\left(r\right)}{\text{log}r}$

where Î½(r) be the central index of f.

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

$|f\left(z\right)|>M\left(r,g\right)\mathrm{Î½}{\left(r,g\right)}^{-\frac{1}{4}+\mathrm{Î´}}$

holds. Then there exists a set F âŠ‚ R+ of finite logarithmic measure, i.e., ${âˆ«}_{F}\frac{dt}{t}<\mathrm{âˆž}$, such that

$\frac{{f}^{\left(m\right)}\left(z\right)}{f\left(z\right)}={\left(\frac{\mathrm{Î½}\left(r,f\right)}{z}\right)}^{m}\left(1+o\left(1\right)\right)$

holds for all m â‰¥ 0 and all r âˆ‰ F.

Lemma 2.6. [10] Let f(z) be a transcendental entire function, Ïƒ(f) = Ïƒ < âˆž, and G = {Ï‰1, Ï‰2,..., Ï‰ n }, and a set E âŠ‚ (1, âˆž) having logarithmic measure lmE < âˆž. Then there is a positive number $B\left(\frac{3}{4}â‰¤Bâ‰¤1\right)$, a point range $\left\{{z}_{k}={r}_{k}{e}^{i{\mathrm{Ï‰}}_{k}}\right\}$ such that |f(z k )| â‰¥ BM(r k , f ), Ï‰ k âˆˆ [0, 2Ï€), lim k â†’âˆž Ï‰ k = Ï‰0 âˆˆ [0, 2Ï€), r k âˆ‰ E âˆª [0, 1], r k â†’ âˆž, for any given Îµ > 0, we have

${r}_{k}^{\mathrm{Ïƒ}-\mathrm{Îµ}}<\mathrm{Î½}\left({r}_{k},f\right)<{r}_{k}^{\mathrm{Ïƒ}+\mathrm{Îµ}}.$

## 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 Ïƒ(f) < 1, we get Q(z) is a constant. Then Theorem 1.1 holds. Next, we suppose that 1 < Ïƒ(f) < 2 and Î»(f) < Ïƒ(f) = Ïƒ. 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 Î»(f) < Ïƒ(f) = Ïƒ, for any given $0<\mathrm{Îµ}<\text{min}\left\{\frac{\mathrm{Ïƒ}-1}{2},\frac{1-\mathrm{Î±}}{2},\frac{\mathrm{Ïƒ}-\mathrm{Î»}\left(f\right)}{2},\frac{\mathrm{Ïƒ}-1-\mathrm{Î±}}{2}\right\}$, there exists a set E1 âŠ‚ (1, âˆž) of |z| = r of finite logarithmic measure, so that

$\frac{f\left(z+j\right)}{f\left(z\right)}=\text{exp}\left\{j\frac{{f}^{\mathrm{â€²}}\left(z\right)}{f\left(z\right)}+o\left({r}^{\mathrm{Ïƒ}\left(f\right)-1-\mathrm{Îµ}}\right)\right\},j=1,2,â€¦,n$
(3.2)

holds for r âˆ‰ E1 âˆª 0[1].

By Lemma 2.5, there exists a set E2 âŠ‚ (0, âˆž) of finite logarithmic measure, such that

$\frac{{f}^{\mathrm{â€²}}\left(z\right)}{f\left(z\right)}=\left(1+o\left(1\right)\right)\frac{\mathrm{Î½}\left(r,f\right)}{z},$
(3.3)

holds for |z| = r âˆ‰ E2 âˆª [0, 1], where z is chosen as in Lemma 2.5.

By Lemma 2.1, for any given Îµ > 0, there exists a set E3 âŠ‚ (1, âˆž) that has finite logarithmic measure such that

$\text{exp}\left\{-{r}^{\mathrm{Î±}+\mathrm{Îµ}}\right\}â‰¤\phantom{\rule{2.77695pt}{0ex}}|{a}_{j}\left(z\right)|\phantom{\rule{2.77695pt}{0ex}}â‰¤\text{exp}\left\{{r}^{\mathrm{Î±}+\mathrm{Îµ}}\right\}\left(j=0,1,â€¦,n\right)$
(3.4)

holds for |z| = r âˆ‰ [0, 1] âˆª E3, r â†’ âˆž.

Set E = E1âˆªE2âˆªE3 and $G=\left\{-\frac{{\mathrm{Ï†}}_{n}}{n}+\left(2j-1\right)\frac{\mathrm{Ï€}}{2n}|j=0,1\right\}âˆª\left\{\frac{\mathrm{Ï€}}{2},\frac{3\mathrm{Ï€}}{2}\right\}$. By Lemma 2.6, there exist a positive number $Bâˆˆ\left[\frac{3}{4},1\right]$, a point range $\left\{{z}_{k}={r}_{k}{e}^{i{\mathrm{Î¸}}_{k}}\right\}$ such that |f(z k )| â‰¥ BM (r k , f], Î¸ k âˆˆ [0, 2Ï€), lim k â†’âˆž Î¸ k = Î¸0 âˆˆ [0, 2Ï€) \ G, r k âˆ‰ E âˆª [0, 1], r k â†’ âˆž, for any given Îµ > 0, as r k â†’ âˆž, we have

${r}_{k}^{\mathrm{Ïƒ}\left(f\right)-\mathrm{Îµ}}<\mathrm{Î½}\left({r}_{k},f\right)<{r}_{k}^{\mathrm{Ïƒ}\left(f\right)+\mathrm{Îµ}}$
(3.5)

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

${a}_{n}\mathrm{exp}\left\{n\left(1+o\left(1\right)\right)\frac{\mathrm{Î½}\left({r}_{k},f\right)}{{z}_{k}}\right\}+â‹¯+{a}_{1}\mathrm{exp}\left\{\left(1+o\left(1\right)\right)\frac{\mathrm{Î½}\left({r}_{k},f\right)}{{z}_{k}}\right\}+{a}_{0}={e}^{Q\left(z\right)}\right\}$
(3.6)

Let $Q\left(z\right)=\mathrm{Ï„}{e}^{i{\mathrm{Î¸}}_{1}}z+{b}_{0}$, Ï„ > 0, Î¸1 âˆˆ [0, 2Ï€). By Lemma 2.4, there are two opened angles for above Îµ,

${S}_{j}:-{\mathrm{Î¸}}_{1}+\left(2j-1\right)\frac{\mathrm{Ï€}}{2}+\mathrm{Îµ}<\mathrm{Î¸}<-{\mathrm{Î¸}}_{1}+\left(2j+1\right)\frac{\mathrm{Ï€}}{2}+\mathrm{Îµ}\left(j=0,1\right)$

For the above Î¸0, there are two cases: (i) Î¸0 âˆˆ S0; (ii) Î¸0 âˆˆ S1.

Case (i). Î¸0 âˆˆ S1. Since S j is an opened set and lim k â†’âˆž Î¸ k = Î¸0, there is a K > 0 such that Î¸ k âˆˆ S j when k > K. By Lemma 2.2, we have

$Re\left\{Q\left({r}_{k}{e}^{i{\mathrm{Î¸}}_{k}}\right)\right\}<-\mathrm{Î·}{r}_{k},$
(3.7)

where Î· = Î·(1 - Îµ) sin(Îµ) > 0. By Lemma 2.2, if Rez k > Î¶r k (0 < Î¶ â‰¤ 1). By (3.4)-(3.7), we have

$\begin{array}{c}\text{exp}\left\{{r}_{k}^{\mathrm{Ïƒ}\left(f\right)-1-\mathrm{Îµ}}-{r}_{k}^{\mathrm{Î±}+\mathrm{Îµ}}\right\}\\ â‰¤\left|{a}_{n}\text{exp}\left\{n\left(1+o\left(1\right)\right)\frac{\mathrm{Î½}\left({r}_{k},f\right)}{{z}_{k}}\right\}\right|\\ â‰¤3\left|{a}_{n}\text{exp}\left\{n\left(1+o\left(1\right)\right)\frac{\mathrm{Î½}\left({r}_{k},f\right)}{{z}_{k}}\right\}+â‹¯+{a}_{1}\text{exp}\left\{\left(1+o\left(1\right)\right)\frac{\mathrm{Î½}\left({r}_{k},f\right)}{{z}_{k}}\right\}+{a}_{0}\right|\\ =3\left|{e}^{Q\left(z\right)}\right|â‰¤3{e}^{-\mathrm{Î·}{r}_{k}},\end{array}$
(3.8)

which contradicts that 0 < Ïƒ(f) - 1 - Î± - Îµ.

If Rez k < - Î¶r k (0 < Î¶ â‰¤ 1), By (3.4)-(3.7), we have

$\begin{array}{c}1â‰¤\left|\frac{{a}_{n}}{{a}_{0}}\text{exp}\left\{n\left(1+o\left(1\right)\right)\frac{\mathrm{Î½}\left({r}_{k},f\right)}{{z}_{k}}\right\}+â‹¯+\frac{{a}_{1}}{{a}_{0}}\text{exp}\left\{\left(1+o\left(1\right)\right)\frac{\mathrm{Î½}\left({r}_{k},f\right)}{{z}_{k}}\right\}\right|+\left|\frac{{e}^{Q\left(z\right)}}{{a}_{0}}\right|\\ â‰¤2n\text{exp}\left\{-\mathrm{Î·}{r}_{k}^{\mathrm{Ïƒ}\left(f\right)-1+\mathrm{Îµ}}+2{r}_{k}^{\mathrm{Î±}+\mathrm{Îµ}}\right\}+{e}^{-\mathrm{Î·}{r}_{k}}\text{exp}\phantom{\rule{2.77695pt}{0ex}}\left\{{r}_{k}^{\mathrm{Î±}+\mathrm{Îµ}}\right\},\end{array}$
(3.9)

which implies that 1 < 0, r â†’ âˆž, a contradiction.

Case (ii). Î¸0 âˆˆ S0. Since S0 is an opened set and lim k â†’âˆž Î¸ k = Î¸0, there is K > 0 such that Î¸ k âˆˆ S j when k > K. By Lemma 2.2, we have

$Re\left\{Q\left({r}_{k}{e}^{i{\mathrm{Î¸}}_{k}}\right)\right\}>\mathrm{Î·}{r}_{k},$
(3.10)

where Î· = Ï„(1 - Îµ) sin(Îµ) > 0. By (3.4)-(3.6), (3.9), we obtain

$\begin{array}{c}\left(n+1\right)\text{exp}\left\{n{r}_{k}^{\mathrm{Ïƒ}\left(f\right)-1+\mathrm{Îµ}}+{r}_{k}^{\mathrm{Î±}+\mathrm{Îµ}}\right\}\\ â‰¥\phantom{\rule{2.77695pt}{0ex}}|{a}_{n}\text{exp}\left\{n\left(1+o\left(1\right)\right)\frac{\mathrm{Î½}\left({r}_{k},f\right)}{{z}_{k}}\right\}+â‹¯+{a}_{1}\text{exp}\left\{\left(1+o\left(1\right)\right)\frac{\mathrm{Î½}\left({r}_{k},f\right)}{{z}_{k}}\right\}+{a}_{0}|\\ =\phantom{\rule{2.77695pt}{0ex}}|{e}^{Q\left(z\right)}|\phantom{\rule{2.77695pt}{0ex}}â‰¥{e}^{\mathrm{Î·}{r}_{k}}.\end{array}$
(3.11)

From (3.11), we get that Ïƒ(f) â‰¥ 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 â‰¥ 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

${a}_{n}\text{exp}\left\{n\left(1+o\left(1\right)\right)\frac{\mathrm{Î½}\left({r}_{k},f\right)}{{z}_{k}}\right\}+â‹¯+{a}_{1}\text{exp}\left\{\left(1+o\left(1\right)\right)\frac{\mathrm{Î½}\left({r}_{k},f\right)}{{z}_{k}}\right\}+a=-{e}^{{z}_{k}},$
(4.2)

where a is some non-zero constant.

If Rez k < -Î·r k (Î· âˆˆ (0, 1]), By (3.4), (3.5), (4.2), we have

$\begin{array}{c}|a|\phantom{\rule{2.77695pt}{0ex}}â‰¤\left|{a}_{n}\text{exp}\left\{n\left(1+o\left(1\right)\right)\frac{\mathrm{Î½}\left({r}_{k},f\right)}{{z}_{k}}\right\}+â‹¯+{a}_{1}\text{exp}\left\{\left(1+o\left(1\right)\right)\frac{\mathrm{Î½}\left({r}_{k},f\right)}{{z}_{k}}\right\}\right|+|\text{exp}\left\{{z}_{k}\right\}|\\ â‰¤\text{exp}\left\{-\mathrm{Î·}{r}_{k}\right\}+n\text{exp}\left\{-\mathrm{Î·}{r}_{k}^{\mathrm{Ïƒ}\left(f\right)-1+\mathrm{Îµ}}+2{r}_{k}^{\mathrm{Î±}+\mathrm{Îµ}}\right\},\end{array}$
(4.3)

which is impossible.

If Rez k > Î·r k (Î· âˆˆ (0, 1]), By (3.4), (3.5) and (4.2), we get

$\begin{array}{c}\text{exp}\left\{\mathrm{Î·}{r}_{k}^{\mathrm{Ïƒ}\left(f\right)-1-\mathrm{Îµ}}\right\}<\text{exp}\left\{n\frac{\mathrm{Î½}\left({r}_{k},f\right)}{{z}_{k}}-{r}_{k}^{\mathrm{Î±}+\mathrm{Îµ}}\right\}\\ â‰¤2\left|{a}_{n}\text{exp}\left\{n\left(1+o\left(1\right)\right)\frac{\mathrm{Î½}\left({r}_{k},f\right)}{{z}_{k}}\right\}+â‹¯+{a}_{1}\text{exp}\left\{\left(1+o\left(1\right)\right)\frac{\mathrm{Î½}\left({r}_{k},f\right)}{{z}_{k}}\right\}+a\right|\\ =2|-\text{exp}\left\{{z}_{k}\right\}|\phantom{\rule{2.77695pt}{0ex}}â‰¤2\text{exp}\left\{{r}_{k}\right\},\end{array}$
(4.4)

which contradicts that Ïƒ(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 Ïƒ(f) < 1, we get Q(z) is a constant. Then Theorem 1.3 holds. Next, we suppose that 1 < Ïƒ(f) < 2 and Î»(f) < Ïƒ(f) = Ïƒ. Set F(z) = f(z) - P(z), then Ïƒ(F) = Ïƒ(f). Substituting F(z) = f(z) - p(z) into (5.1), we obtain

$\frac{{a}_{n}\left(z\right)F\left(z+n\right)+{a}_{n-1}\left(z\right)F\left(z+n-1\right)+â‹¯+{a}_{1}\left(z\right)F\left(z+1\right)}{F\left(z\right)}+{a}_{0}\left(z\right)+\frac{b\left(z\right)}{F\left(z\right)}={e}^{Q\left(z\right)},$
(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 Î»(f) < Ïƒ(f) = Ïƒ, for any given $0<\mathrm{Îµ}<\text{min}\left\{\frac{\mathrm{Ïƒ}-1}{2},\frac{1-\mathrm{Î±}}{2},\frac{\mathrm{Ïƒ}-\mathrm{Î»}\left(f\right)}{2},\frac{\mathrm{Ïƒ}-1-\mathrm{Î±}}{2}\right\}$, there exists a set E1 âŠ‚ (1, âˆž) of |z| = r of finite logarithmic measure, so that

$\frac{f\left(z+j\right)}{f\left(z\right)}=\text{exp}\left\{j\frac{{f}^{\mathrm{â€²}}\left(z\right)}{f\left(z\right)}+o\left({r}^{\mathrm{Ïƒ}\left(f\right)-1-\mathrm{Îµ}}\right)\right\},j=1,2,â€¦,n$
(5.3)

holds for r âˆ‰ E1 âˆª [0, 1].

By Lemma 2.5, there exists a set E2 âŠ‚ (0, âˆž) of finite logarithmic measure, such that

$\frac{{f}^{\mathrm{â€²}}\left(z\right)}{f\left(z\right)}=\left(1+o\left(1\right)\right)\frac{\mathrm{Î½}\left(r,f\right)}{z},$
(5.4)

holds for |z| = r âˆ‰ E2 âˆª [0, 1], where z is chosen as in Lemma 2.5.

Set E = E1 âˆª E2 and $G=\left\{-\frac{{\mathrm{Ï†}}_{n}}{n}+\left(2j-1\right)\frac{\mathrm{Ï€}}{2n}|j=0,1\right\}âˆª\left\{\frac{\mathrm{Ï€}}{2},\frac{3\mathrm{Ï€}}{2}\right\}$. By Lemma 2.6, there exist a positive number $Bâˆˆ\left[\frac{3}{4},1\right]$, a point range $\left\{{z}_{k}={r}_{k}{e}^{i{\mathrm{Î¸}}_{k}}\right\}$ such that | f (z k )| â‰¥ BM(r k , f), Î¸ k âˆˆ [0, 2Ï€), lim k â†’âˆžÎ¸ k = Î¸0 âˆˆ [0, 2Ï€) \ G, r k âˆ‰ E âˆª 0[1], r k â†’ âˆž, for any given Îµ > 0, as r k â†’ âˆž, we have

${r}_{k}^{\mathrm{Ïƒ}\left(f\right)-\mathrm{Îµ}}<\mathrm{Î½}\left({r}_{k},f\right)<{r}_{k}^{\mathrm{Ïƒ}\left(f\right)+\mathrm{Îµ}}.$
(5.5)

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

$\frac{b\left({z}_{k}\right)}{F\left({z}_{k}\right)}â†’0,\left({r}_{k}â†’\mathrm{âˆž}\right).$
(5.6)

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

${a}_{n}\text{exp}\left\{n\left(1+o\left(1\right)\right)\frac{\mathrm{Î½}\left({r}_{k},f\right)}{{z}_{k}}\right\}+â‹¯+{a}_{1}\text{exp}\left\{\left(1+o\left(1\right)\right)\frac{\mathrm{Î½}\left({r}_{k},f\right)}{{z}_{k}}\right\}+{a}_{0}+o\left(1\right)={e}^{Q\left(z\right)}.$
(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.

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## Acknowledgements

The authors thank the referee for his/her valuable suggestions to improve the present article. This research was partly supported by the NNSF of China (No. 11171184), the NSF of Shangdong Province, China (No. Z2008A01) and Shandong University graduate student independent innovation fund (yzc11024).

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Correspondence to XiaoGuang Qi.

### Competing interests

The authors declare that they have no competing interests.

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Liu, Y., Qi, X. & Yi, H. Some results on difference polynomials sharing values. Adv Differ Equ 2012, 1 (2012). https://doi.org/10.1186/1687-1847-2012-1