Infinite growth of solutions of second order complex differential equations with meromorphic coefficients

1.   Introduction and main results

Throughout this paper, we assume that the reader is familar with the fundamental results and the standard notations of Nevanlinna's value distribution theory (see ^[9,14,28]). In addition, we use $\rho(f)$ and $\mu(f)$ to denote the order and lower order of a meromorphic function $f(z)$ respectively, which are defined as

The second order linear differential equation

where $A(z)$ and $B(z)$ are meromorphic functions, is the focus of this paper. To begin, we look for the conditions of coefficients that guarantee that every non-trivial meromorphic solution of Eq (1.1) has infinite order. Every non-trivial solution of Eq (1.1) must be an entire function, if $A(z)$ and $B(z)$ are entire functions, as is widely known. When $A(z)$ and $B(z)$ are entire functions, a lot of progress has been made, see Gundersen ^[5], Hellerstein, Miles and Rossi ^[10], and Ozawa ^[19]. The following is a summary of their work.

Theorem A. Suppose that $A(z)$ and $B(z)$ are entire functions satisfying any one of the following additional hypotheses:

(1) $\rho(A) < \rho(B)$, see ^[5];

(2) $A(z)$ is a polynomial and $B(z)$ is transcendental, see ^[10];

(3) $\rho(B) < \rho(A)\leq \frac{1}{2}$, see ^[19].

Then, every non-trivial solution $f$ of Eq (1.1) is of infinite order.

One may ask a question based on Theorem A.

Question 1: If $\rho(A) = \rho(B)$, or if $\rho(A) > \rho(B)$ and $\rho(A) > \frac{1}{2}$, is every non-trivial solution $f$ of Eq (1.1) is of infinite order?

In general, the answer to Question 1 is negative.

Example 1.1. Let $Q(z)$ be any non-constant polynomial, let $B(z)\not\equiv 0$ be any entire function with $\rho(B) < \deg(Q)$, let $f$ be any antiderivative of $e^{Q(z)}$ that satisfies $\rho(f) = \deg(Q)$, and set $A(z) = -Q'-B(z)fe^{-Q}$. Then $\rho(B) < \rho(A) = \deg(Q) = \rho(f)$, and $f''+A(z)f'+B(z)f = 0$. This shows that it is possible to have a finite order non-trivial solution $f$ of Eq (1.1) where $\rho(B) < \rho(A)$ and $\rho(A)$ may be any positive integer.

Example 1.2 Let $Q(z)$ be any non-constant polynomial, let $A(z)\not\equiv 0$ be any entire function, and set $B(z) = -Q''-(Q')^2-A(z)Q'$. Then $\rho(A) = \rho(B)$ and it can be verifies that $f(z) = e^{Q(z)}$ satisfies the equation $f''+A(z)f'+B(z)f = 0$. This shows that it is possible to have a finite order non-trivial solution $f$ of Eq (1.1) where $\rho(A) = \rho(B)$.

In some special cases, however, an entire solution of Eq (1.1) can have infinite order, see, for example, ^{[3,13,15,16,22,23]}. Gundersen ^[8] took into account a special case in which the coefficient $A(z)$ of Eq (1.1) is an exponential function.

Theorem B. ^[8] Let $A(z) = e^{-z}$ and $B(z)$ is a transcendental entire function with order $\rho(B)\ne 1$. Then every non-trivial solution $f$ of Eq (1.1) has infinite order.

When $\rho(B) = 1$, the entire solution of Eq (1.1) may be finite order, according to Theorem B. What conditions can guarantee that every non-trivial solution $f$ of Eq (1.1) has infinite order if $\rho(B) = 1$? Chen ^[3] considered this question and proved the following result.

Theorem C. ^[3] Suppose that $A_j(z)(j = 0, 1)$ are entire functions with $\rho(A_j) < \infty$, $a$ and $b$ are complex constants with $ab\ne 0$ and $a = cb(c > 1)$. Let $A(z) = A_1(z)e^{az}$, $B(z) = A_2(z)e^{bz}$. Then every non-trivial solution $f$ of Eq (1.1) has infinite order.

Remark 1.1. Because $\rho(A) = 1$ and $\rho(B)\ne 1$ in Theorem B, $\rho(A) > \rho(B)$ and $\rho(A) > \frac{1}{2}$ can occur, whereas $\rho(A) = \rho(B) = 1$ in Theorem C. As a result, Theorems B and C partially answer Question 1 as well.

Question 1 was recently studied by several scholars who assumed that $A(z)$ is a nontrivial solution of a second order differential equation. We have the following collection theorem.

Theorem D. Let $A(z)$ be a nontrivial solution of $w''+P(z)w = 0$, where $P(z)$ is a nonconstant polynomials with $\deg(P) = n$, and satisfying any one of the following additional hypotheses:

(1) $\rho(B) < \frac{1}{2}$, see ^[24];

(2) $B(z)$ is an entire function with Fabry gaps, see ^[16]; Here, an entire function $f(z) = \sum\limits_{n = 0}^{\infty}a_nz^{\lambda_n}$ is said to have Fabry gaps if $\frac{\lambda_n}{n}\to\infty\mbox{ as }n\to\infty$.

(3) $T(r, B)\sim \log M(r, B)$ as $r\rightarrow \infty$ outside a set of finite logarithmic measure such that $\rho(A)\ne \rho(B)$, see ^[29]. Then every nontrivial solution of Eq (1.1) is of infinite order.

Remark 1.2. We know that $\rho(A) = \frac{n+2}{2}$ based on Theorem D's assumptions. Clearly, Theorem D partially answers Question 1 because $\rho(A) = \rho(B)$ or $\rho(A) > \rho(B)$ with $\rho(A) > \frac{1}{2}$ can occur if $B(z)$ meets one of the conditions (1)-(3) in Theorem D. We also find that the proof of three cases of Theorem D is based on the observation that if $A(z)$ is a nontrivial solution of $w''+P(z)w = 0$, then the whole plane may be divided into $n+2$ sectors $S_j(j = 0, 1, \cdots, n+1)$ in which $A(z)$ either blows up or decays to zero exponentially.

We will continue to study Question 1 in this paper. We consider a more general case in which $A(z)$ in some angular domains either blows up or decays to zero rapidly. The coefficients of Eq (1.1) in particular, are meromorphic rather than entire.

Theorem 1.1. Let $\{\phi_k\}$ be a finite set of real numbers satisfying $\phi_1 < \phi_2 < ... < \phi_{2n} < \phi_{2n+1}$ with $\phi_{2n+1} = \phi_1+2\pi$, and set

Suppose that $A(z)$ and $B(z)$ are meromorphic functions such that for some constant $\alpha\geq 0$ and a set $H\subset[0, 2\pi)$ of linear measure zero,

as $z\rightarrow \infty$ in $\arg z\in(\phi_{2k-1}, \phi_{2k})\backslash H$ for $k = 1, ..., n$, and where $B(z)$ is transcendental with a deficient value $\infty$ and

Then every non-trivial meromorphic solution $f$ of Eq (1.1) has infinite order.

Remark 1.3. We can use a specific example to illustrate our point. If $A(z) = e^{P(z)}$, where $P(z)$ is a polynomial with $\deg (P) = n$, and $B(z)$ is an entire function with $\mu(B) < n$, then the example meets Theorem 1.1's criteria.

Corollary 1.1. Suppose that $A(z) = h_1(z)e^{p_1(z)}+h_2(z)e^{p_2(z)}+p_3(z)$, where $p_3(z)$ is a polynomial, $p_i(z) = a_iz^n+...(i = 1, 2)$ are two non-constant polynomials of degree $n$ with $\arg a_1-\arg a_2\ne \pm \pi$, and $h_i(z)(i = 1, 2)$ are meromorphic functions of order less than $n$. Let $B(z)$ be given as Theorem 1.1. Then every non-trivial solution $f$ of Eq (1.1) has infinite order.

Remark 1.4. For $p_1(z)$ and $p_2(z)$, from remark 2.1, there exist $\Omega_k(p_1)$ and $\Omega_k(p_2)$ such that when $k$ is odd, $\delta(p_1, \theta) < 0$ if $\theta\in \Omega_k(p_1)$ and $\delta(p_2, \theta) < 0$ if $\theta\in \Omega_k(p_2)$. Since $\arg a_1-\arg a_2\ne \pm \pi$, $\Omega_k(p_1)\cap\Omega_k(p_2)\ne\emptyset$. Then we can redivide the plane into $2n$ open angles $S_j(j = 0, 1, ..., 2n-1)$ such that for $\theta\in S_j$, $\delta(p_1, \theta) < 0$ and $\delta(p_2, \theta) < 0$ if $j$ is odd, while $\delta(p_1, \theta)\geq 0$ or $\delta(p_2, \theta)\geq 0$ if $j$ is even. Suppose that $\deg(p_3) = m$. By Lemma 2.3, we know $|A(re^{i\theta})| = O(r^m)$ in $S_j$ outside a set of linear measure zero when $j$ is odd. By Theorem 1.1, Corollary 1.1 holds.

Corollary 1.2. Suppose $A(z) = h_1(z)e^{p_1(z)}+h_2(z)e^{p_2(z)}+p_3(z)$, where $p_3$ is a polynomial, $p_i(i = 1, 2)$ are two non-constant polynomials with $\deg(p_1)\ne\deg(p_2)$, and $h_i(z)(i = 1, 2)$ are meromorphic functions of order less than $\deg(p_i)$. Let $B(z)$ be given as Theorem 1.1. Then every non-trivial solution $f$ of Eq (1.1) has infinite order.

Remark 1.5. Let $p_1(z) = a_nz^n+...+a_0$, $p_2(z) = b_tz^t+...+b_0$. Similarly as remark 1.4, we also can redivide the plane into $2s$ open angles $S_j(j = 0, 1, ..., 2s-1)$ where $s$ may be different from $n$ and $t$, and depends on $\deg(p_i)(i = 1, 2)$, $a_n$ and $b_t$, such that for $\theta\in S_j$, $\delta(p_1, \theta) < 0$ and $\delta(p_2, \theta) < 0$ if $j$ is odd; while $j$ is even, $\delta(p_1, \theta)\geq 0$ or $\delta(p_2, \theta)\geq 0$. Then Corollary 1.2 holds.

Next, we consider the lower bound estimate of the measure of the angular domain, in which the radial order of every non-trivial solution of (1.1) is infinite. The following notations and notions are provided to further illustrate our concerns.

Assuming $0\leq\alpha < \beta\leq 2\pi$, we denote

and use $\overline{\Omega}(\alpha, \beta)$ to denote the closure of $\Omega(\alpha, \beta)$. Similarly, we denote

Suppose that $f(z)$ is an analytic function in $\overline{\Omega}(\alpha, \beta)$, we use $\rho_{\alpha, \beta}$ to denote the order of growth of $f$ on $\Omega(\alpha, \beta)$, that is

where $M(r, \Omega(\alpha, \beta), f) = \sup_{\alpha\leq\theta\leq\beta}|f(re^{i\theta})|$. Furthermore, we denote the radial order of $f(z)$ by

In recent years, some progress in the angular distribution of entire solutions of linear differential equations have been obtained by several researchers, see ^{[12,20,21,24,25,26,27]}. In particular, Huang and Wang ^[11] considered the following question. We know that for a transcendental entire function $f(z)$ of infinite order, at least one ray $\arg z = \theta$ exists with the radial order $\rho_\theta(f) = \infty$. However, for any angular domain $\Omega(\alpha, \beta)$, $\rho(f) = \infty$ cannot ensure $\rho_{\alpha, \beta}(f) = \infty$, $f(z) = e^{e^z}$, for example, satisfies $\rho(f) = \infty$, whereas $\rho_{\pi/2, 3\pi/2}(f) = 0$. Motivated by this fact, we raise an interesting question: how wide are such $\Omega(\alpha, \beta)$ with $\rho_{\alpha, \beta}(f) = \infty$? Let

Clearly, $I(f)$ is closed and measurable. In 2015, Huang and Wang ^[11] proved the following result.

Theorem E. Suppose that $A(z)$, $B(z)$ are entire function with $\rho(A) < \mu(B)$. If $f(z)$ is a non-trivial solution of Eq (1.1), then $mes I(f)\geq\min\{2\pi, \pi/\mu(B)\}$.

Inspired by Theorem E, we consider the lower bound of measure of the set of infinite radial order solutions of Eq (1.1). As a result, we have the following.

Theorem 1.2. Let $A(z)$ be an entire function of finite order having a finite Borel exceptional value, and $B(z)$ be a transcendental entire function with $\mu(B) < \rho(A)$. Then, every non-trivial solution $f$ of Eq (1.1) is of infinite order and satisfies $\mathit{\mbox{mes}}(I(f))\geq \min\{k_1, k_2\}$, where $k_1: = \pi$, $k_2: = \begin{cases} \frac{[\rho(A)/\mu(B)]}{2}\frac{\pi}{\rho(A)}, & \mathit{\mbox{if}} [\frac{\rho(A)}{\mu(B)}] \mathit{\mbox{is even}}\\ \frac{\pi}{\mu(B)}-\frac{[\rho(A)/\mu(B)]+1}{2}\frac{\pi}{\rho(A)}, & \mathit{\mbox{if}} [\frac{\rho(A)}{\mu(B)}] \mathit{\mbox{is odd}}. \end{cases}$

Remark 1.6. In Theorem 1.2, we use $[x]$ to denote the largest integer not exceeding the real number $x$, for example, $[1.2] = 1$.

Remark 1.7. Under the assumptions of Corollary 1.1, we know every non-trivial solution $f$ of Eq (1.1) is an entire function with infinite order. Hence, from the proof of Theorem 1.2, we obtain that $\mbox{mes}(I(f))\geq k_1$ if $\mu(B)\in [0, \frac{1}{2})$, while $\mbox{mes}(I(f))\geq k_2$ if $\mu(B)\in [\frac{1}{2}, \rho(A))$.

2.   Preliminary lemmas

Lemma 2.1. ^[17] Let $f$ be an entire function of finite order having a finite Borel exceptional value $c$. Then

where $h(z)$ is an entire function with $\rho(h) < \rho(f)$, $Q(z)$ is a polynomial of degree $deg(Q) = \rho(f)$.

Lemma 2.2. ^[6] Let $g(z)$ be a transcendental meromorphic function of order $\rho(g) = \rho < \infty$. Then there exists a set $E\subset[0, 2\pi)$ of linear measure zero such that, for any $\theta\in[0, 2\pi)\backslash E$, there exists a positive constant $R_0 = R_0(\theta) > 1$ such that, for all $z$ satisfying $\arg z = \theta$ and $|z| > R_0$,

where $k = 1, 2$ and $\varepsilon$ is given positive real constant.

Remark 2.1. For the polynomial $p(z)$ with degree $n$, set $p(z) = (\alpha+i\beta)z^n+p_{n-1}(z)$ with $\alpha, \beta$ real, and denote $\delta(p, \theta): = \alpha\cos n\theta-\beta\cos n\theta$. The rays

satisfying $\delta(p, \theta_k) = 0$ can split the complex domain into $2n$ equal angles. We denote these angle domains as

And we know that for $\theta\in\Omega_k(p)$, $\delta(p, \theta) > 0$ if $k$ is even, while $\delta(p, \theta) < 0$ if $k$ is odd.

Lemma 2.3. ^[4,Lemma 1.20] Let $p(z)$ be a polynomial of degree $n\geq 1$. Suppose that $h(z)\not\equiv 0$ is a meromorphic function and $\rho(h) < n$. Consider the function $g(z): = h(z)e^{p(z)}$, there exists a set $H_1\subset[0, 2\pi)$ of linear measure zero, for every $\theta\in[0, 2\pi)\backslash(H_1\cup H_2)$ and $r > r_0(\theta) > 0$, we have

(1) If $\delta(p, \theta) > 0$,

holds here;

(1) If $\delta(p, \theta) < 0$,

holds here, where $H_2 = \{\theta:\delta(p, \theta) = 0, 0\leq\theta < 2\pi\}$ is a finite set.

The following lemma due to Markushevich ^[18].

Lemma 2.4. ^[18] Suppose $f(z) = e^{P(z)}$, where $P(z)$ is a polynomial of degree $n$:

We know that $f(z)$ is a function of order $n$. Let $b_n = \alpha_ne^{i\theta_n}$, $\alpha_n > 0$, $\theta_n\in[0, 2\pi)$ and $z = re^{i\theta}$, we now divide the plane into $2n$ equal open angles

where $j = 0, 1, 2, ..., 2n-1$. We also introduce $2n$ closed angles

where $0 < \varepsilon < \pi/2$ and $S_j(\varepsilon)\subset S_{j}$. Then there exists a positive number $R = R(\varepsilon)$ such that for $|z| = r > R$,

if $\arg z\in S_j(\varepsilon)$ when $j$ is even; While

if $\arg z\in S_j(\varepsilon)$ when $j$ is odd.

Remark 2.2. Clearly, for any $\arg z\in S_j$, we always find an $\varepsilon$, $\varepsilon\in(0, \pi/2)$ and a positive number $R = R(\varepsilon)$ such that $\arg z\in S_j(\varepsilon)$, and for $|z| = r > R$,

if $j$ is even; While

if $j$ is odd.

Lemma 2.5. ^[1] Suppose that $g(z)$ is an entire function with $\mu(g)\in [0, 1)$. Then, for every $\alpha\in (\mu(g), 1)$, there exists a set $E\subset [0, \infty)$ such that $\overline{\log dens}E\geq 1-\frac{\mu(g)}{\alpha}$, where $E = \{r\in [0, \infty):m(r) > M(r)\cos\pi\alpha\}$, $m(r) = \inf_{|z| = r}\log |g(z)|$ and $M(r) = \sup_{|z| = r}\log |g(z)|$.

Lemma 2.6. ^[22] Suppose that $f$ is an entire function and lower order $\mu(f)\in [\frac{1}{2}, +\infty)$. Then, there exists an domain $\Omega(\alpha, \beta)$, where $\alpha, \beta$ satisfy $\beta-\alpha\geq\frac{\pi}{\mu(f)}$ and $0\leq\alpha\leq\beta\leq 2\pi$, such that

for all $\psi\in(\alpha, \beta)$.

Lemma 2.7. ^[7] Let $z = r\exp(i\psi), r_0+1 < r$ and $\alpha\leq\psi\leq\beta$, where $0 < \beta-\alpha\leq2\pi$. Suppose that $n(\geq 2)$ is an integer, and that $g(z)$ is analytic in $\Omega(r_0, \alpha, \beta)$ with $\rho_{\alpha, \beta} < \infty$. Choose $\alpha < \alpha_1 < \beta_1 < \beta$. Then, for every $\varepsilon\in (0, \frac{\beta_j-\alpha_j}{2})(j = 1, 2, ..., n-1)$ outside a set of linear measure zero with

there exist $K > 0$ and $M > 0$ only depending $g$, $\varepsilon_1, ..., \varepsilon_{n-1}$ and $\Omega(\alpha_{n-1}, \beta_{n-1})$, and not depend on $z$ such that

and

for all $z\in\Omega(\alpha_{n-1}, \beta_{n-1})$ outside an $R$-set $D$, where $k = \pi/(\beta-\alpha)$ and $k_{\varepsilon_j} = \pi/(\beta_j-\alpha_{j}(j = 1, 2, ..., n-1)).$

Lemma 2.8. ^[2] Let $f(z)$ be a transcendental meromorphic function of finite lower order $\mu$, and have one deficient value $a$. Let $\Lambda(r)$ be a positive function with $\Lambda(r) = o(T(r, f))$ as $r\rightarrow \infty$. Then for any fixed sequence of Pólya peaks $\{r_n\}$ of order $\mu$, we have

where $D_{\Lambda}(r, a)$ is defined by

and for finite $a$,

3.   Proof of Theorem 1.1

Suppose on the contrary to the assertion that there exists a non-trivial solution $f$ with $\rho(f) < \infty$. We aim for a contradiction. Let $\{\phi_k\}$ be a finite collection of real numbers satisfying $\phi_1 < \phi_2 < ... < \phi_{2n} < \phi_{2n+1}$ with $\phi_{2n+1} = \phi_1+2\pi$, and set

Suppose that $A(z)$ is a meromorphic function such that for some constant $\alpha\geq 0$ and a set $H\subset[0, 2\pi)$ of linear measure zero,

as $z\rightarrow \infty$ in $\arg z\in(\phi_{2k-1}, \phi_{2k})\backslash H$ for $k = 1, ..., n$. From Eq (1.1), we get the following inequality

From Lemma 2.2, we can obtain that there exists a set $E^*\subset[0, 2\pi)$ of linear measure zero and a positive constant $R_0 > 1$ such that

for any $\theta\in [0, 2\pi)\backslash E^*$ and $r > R_0$.

Let $B(z)$ be a transcendental meromorphic function having a deficient value $\infty$ and $\mu(B) < 4\arcsin\sqrt{\frac{\delta(\infty, B)}{2}}/v$. Then we have

Next, we define

It is clear that $\Lambda(r) = o(T(r, B))$ and $\Lambda(r)/\log r\rightarrow \infty$ as $r\rightarrow \infty$ since $B(z)$ is a transcendental meromorphic function. Applying Lemma 2.8 to $B(z)$ gives the existence of the Pólya peaks $\{r_j\}$ of order $\mu(B)$ such that for sufficiently large $j$,

where $D_{\Lambda}(r_j, \infty)$ is defined by

From (3.4), there exists at least one sector $(\phi_{2k-1}, \phi_{2k})$ such that

where $k = 1, ..., n$.

Let $F_j: = (\phi_{2k-1}, \phi_{2k})\cap D_{\Lambda}\left(r_{j}, \infty\right)$. On the one hand, for any $\theta\in F_j$ and sufficiently large $j$, we have

Since $\Lambda(r)/\log r\rightarrow \infty$ as $r\rightarrow \infty$,

Substituting (3.1) and (3.3) into (3.2), for any $\theta\in F_j\backslash (E^*\cup H)$ and sufficiently large $j$, we have

where $\alpha\geq 0$ is a constant. Coupling (3.5)and (3.6) yields a contradiction. Thus, every non-trivial solution $f$ of Eq (1.1) is of infinite order.

4.   Proof of Theorem 1.2

We assume the contrary to the assertion that $m(I(f)) < k$, where $k = k_1$ or $k_2$. Then $t: = k-\operatorname{mes} I(f) > 0$. Our goal is to obtain a contradiction. Since $I(f)$ is closed, $H: = (0, 2\pi)\backslash I(f)$ is open. So it consists of at most countably many open intervals. We can choose finitely many open intervals $I_i = (\alpha_i, \beta_i)(i = 1, 2, ..., m)$ in $H$ such that

It easy to see that $I_i\cap I(f) = \emptyset$, and hence $\rho_{\alpha_i, \beta_i}(f) < \infty$ for each $i = 1, 2, ..., m$. Apply Lemma 2.7 to $f$, for sufficiently small $\xi > 0$, there exist two constants $M > 0$ and $K > 0$ such that

for all $z\in\cup_{i = 1}^{m}\Omega(r_j, \alpha_i+\xi, \beta_i-\xi)$ outside an $R$-set $D$.

Suppose that $A(z)$ is an entire function of finite order having a finite Borel exceptional value $a$. By Lemma 2.1, we have

where $h(z)$ is an entire function with $\rho(h) < \rho(A)$, $Q(z)$ is a polynomial of degree $deg(Q) = \rho(A)$. Let $Q(z) = b_dz^d+b_{d-1}z^{d-1}+...+b_0$, where $d$ is a positive integer and $b_d = \alpha_de^{i\theta_d}$, $\alpha_d > 0$ and $\theta_d\in[0, 2\pi)$. We now divide the plane into $2d$ equal open angles

where $j = 0, 1, 2, ..., 2d-1$. We also introduce $2d$ closed angles

where $0 < \varepsilon < \pi/4$ and $S_j(\varepsilon)\subset S_{j}$. According to Lemma 2.4, for $z = re^{i\theta}$, if $\theta\in S_j(\varepsilon)$ and $j$ is odd, there exists a positive number $R(\varepsilon)$, such that

for $r > R(\varepsilon)$. Then for any given $\varepsilon\in(0, \pi/4)$, $\rho(h) < d = \rho(A)$ and $\theta\in S_j(\varepsilon)$($j$ is odd), we have

for all sufficiently large $r$, where $C(\varepsilon)$ is a positive only constant depending on $\alpha_d$, $\varepsilon$ and $\rho(A)$. Therefore, for any $\theta\in S_j$($j$ is odd), we can find an $\varepsilon$ and a positive constant $C(\varepsilon)$ such that $\theta\in S_j(\varepsilon)$ and

for all sufficiently large $r$. From Eq (1.1), we have

In the following, we consider three cases.

Case 1. Suppose $\mu(B) = 0$. Then $k = k_1$. Define

Clearly, $\mbox{mes }S = \pi$. Thus,

Hence

Then we can conclude that exists at least one open interval $I_{i0} = (\alpha_0, \beta_0)$ such that

and there exists at least one sector $S_j$($j$ is odd) such that

We can find an $\varepsilon > 0$ such that $mes(S_j(\varepsilon)\cap (\alpha_0, \beta_0)) > 0$. So $D(\varepsilon): = S_j(\varepsilon)\cap (\alpha_0+\xi, \beta_0-\xi)\ne\emptyset$, where $0 < \xi < mes(S_j(\varepsilon)\cap (\alpha_0, \beta_0))/4$. Applying Lemma 2.5 to $B(z)$, we can choose $\alpha = \frac{1}{4}$ and there is a set $E\subset [0, \infty)$ such that $\overline{\log dens}E = 1$ such that, for all $r\in E$,

where $M(r, B) = \sup_{|z| = r} |B(z)|$. By substituting (4.5), (4.3) and (4.2) into (4.4), for any $\theta\in D(\varepsilon)$ and $r\in E$ outside an $R$-set $D$, we have

for all sufficiently large $r$. Since $B(z)$ is a transcendental entire function, we know that

From (4.6) and (4.7), we can deduce a contradiction. Therefore, $mes(I(f))\geq k_1$.

Case 2. Suppose $0 < \mu(B) < \frac{1}{2}$. So $k = k_1$. Similarly as in Case 1, we have a sector $D(\varepsilon): = S_j(\varepsilon)\cap (\alpha_0+\xi, \beta_0-\xi)\ne\emptyset.$ According to Lemma 2.5, for any given $\alpha\in(\mu(B), \frac{1}{2})$, there exists a set $E\subset [0, \infty)$ such that $\overline{\log dens}E\geq 1-\frac{\mu(B)}{\alpha}$, where $E = \{r\in [0, \infty):m(r) > M(r)\cos\pi\alpha\}$, $m(r) = \inf_{|z| = r}\log |B(z)|$ and $M(r) = \sup_{|z| = r}\log |B(z)|$. Thus, there exists a constant $R_1 > 0$ such that, for arbitrarily small $\eta > 0$ and all $r\in E\backslash [0, R_1]$,

Taking (4.8), (4.3) and (4.2) into (4.4), for any $\theta\in D(\varepsilon)$ and $r\in E\backslash[0, R_1]$ outside an $R$-set $D$, we deduce

for all sufficiently large $r$. This is a contradiction. Therefore, $mes(I(f))\geq k_1$.

Case 3. Suppose $\frac{1}{2}\leq\mu(B) < \rho(A)$. So $k = k_2$. By using Lemma 2.6 to $B(z)$, we can get a sector $\Omega(\alpha, \beta)$ with $\beta-\alpha\geq\frac{\pi}{\mu(B)} > \frac{\pi}{\rho(A)} = \frac{\pi}{d}$, which satisfies

for all $\theta\in(\alpha, \beta)$. That is, there exist a sequence $\{r_n\}$ such that, for arbitrarily small $\eta > 0$,

Let

Clearly, $\mbox{mes}(G)\geq k_2$. By using the similar method in Case 1, we have at least one open interval $I_{i1} = (\alpha_1, \beta_1)$ such that

Then there exists at least one sector $S_j$($j$ is odd) such that

We can find an $\varepsilon > 0$ such that $\mbox{mes }(S_j(\varepsilon)\cap (\alpha_1, \beta_1)) > 0$. So $F(\varepsilon): = S_j(\varepsilon)\cap (\alpha_1+\xi, \beta_1-\xi)\ne\emptyset.$ Substituting (4.10), (4.3) and (4.2) into (4.4), for any $\arg z = \theta\in F(\varepsilon)$ and $\{r_n\}$ outside an $R$-set $D$, we have

for all sufficiently large $n$. A contradiction. Therefore, $mes(I(f))\geq k_2$.

Conflict of interest

All authors declare no conflicts of interest in this paper.