On the Heins Theorem

TL;DR

This paper investigates pairs of entire functions of finite exponential type with bounded minimum modulus, revealing they must be bounded on certain rotating half-planes and providing bounds for these regions.

Contribution

It extends understanding of the behavior of entire functions of finite exponential type, specifically characterizing their boundedness on rotating half-planes.

Findings

Functions must be bounded on some rotating half-planes

Derived bounds for rotation functions of these half-planes

Clarified limitations of Heins Theorem extension

Abstract

It is known that the famous Heins Theorem (also known as the de Branges Lemma) about the minimum of two entire functions of minimal type does not extend to functions of finite exponential type. We study in detail pairs of entire functions $f, g$ of finite exponential type satisfying $\sup_{z\in\mathbb{C}}\min\{|f(z)|,|g(z)|\}<\infty.$ It turns out that $f$ and $g$ have to be bounded on some rotating half-planes. We also obtain very close upper and lower bounds for possible rotation functions of these half-planes.

Equations

$$\underset{z\in \mathbb{C}}{\sup }\min \{|f(z)|,|g(z)|\}\le 1,$$

$$|f(z)|\le 1,|g(z)|>1,z\in C\backslash B,$$

$$|f(z)|>1,|g(z)|\le 1,z\in (-C)\backslash B.$$

$${\Omega }_s=\{r{e}^{i\alpha }\mid r>0,s(r)<\alpha <s(r)+\pi \}.$$

$$\underset{0}{\overset{\infty }{\int }}{h}^{\prime }(x{)}^{2}dx<\infty$$

$$\underset{0}{\overset{\infty }{\int }}{h}^{\prime }(x{)}^{2}dx=\infty \text{and}\underset{0}{\overset{\infty }{\int }}|h^{\prime \prime }(x)|dx<\infty .$$

$$\underset{r\to \infty }{\mathrm{limsup}}\frac{s(r)}{\sqrt{\log r}}>0,$$

$$V_u=\overline{\{z\mid u(z)>0\}},$$

$$m_u(R)=\sup \{|I|/(2\pi ):I\text{ is an open interval},R{e}^{ix}\in V_u,x\in I\},$$

$${\eta }_u(s)=\frac{1}{m_u(e^s)}.$$

$$\underset{0}{\overset{2\pi }{\int }}{u}^2\left(e^{\tau }{e}^{i\theta }\right)d\theta \ge C\underset{0}{\overset{\tau }{\int }}\exp \left(\underset{0}{\overset{{\tau }^{\prime }}{\int }}{\eta }_u(s)ds\right)d{\tau }^{\prime }.$$

$$\begin{gathered} \underset{0}{\overset{2\pi }{\int }}q(\tau )u^2(e^{\tau }{e}^{i\theta })+v^2(e^{\tau }{e}^{i\theta })d\theta \ge \\ C\underset{0}{\overset{\tau }{\int }}q(\tau )\exp \left(\underset{0}{\overset{{\tau }^{\prime }}{\int }}{\eta }_u(s)ds\right)+\exp \left(\underset{0}{\overset{{\tau }^{\prime }}{\int }}{\eta }_v(s)ds\right)d{\tau }^{\prime }. \end{gathered}$$

$$\begin{gathered} \underset{0}{\overset{2\pi }{\int }}q(\tau )u^2(e^{\tau }{e}^{i\theta })+v^2(e^{\tau }{e}^{i\theta })d\theta \ge \\ 2C\sqrt{q(\tau )}\underset{0}{\overset{\tau }{\int }}\exp \left(\underset{0}{\overset{{\tau }^{\prime }}{\int }}\frac{{\eta }_u(s)+{\eta }_v(s)}{2}ds\right)d{\tau }^{\prime }. \end{gathered}$$

$$m_u(e^s)+m_v(e^s)\le 1.$$

$$\underset{0}{\overset{2\pi }{\int }}q(\tau )u^2(e^{\tau }{e}^{i\theta })+v^2(e^{\tau }{e}^{i\theta })d\theta \ge C\sqrt{q(\tau )}(e^{2\tau }-1).$$

$$\underset{0}{\overset{2\pi }{\int }}{u}^2(e^{\tau }{e}^{i\theta })+v^2(e^{\tau }{e}^{i\theta })d\theta \ge 2C\underset{0}{\overset{\tau }{\int }}\exp \left(\underset{0}{\overset{{\tau }^{\prime }}{\int }}\frac{{\eta }_u(s)+{\eta }_v(s)}{2}ds\right)d{\tau }^{\prime }.$$

$$\underset{0}{\overset{\tau }{\int }}\left(\frac{{\eta }_u(s)+{\eta }_v(s)}{2}\right)ds>2\tau +C_3,\tau >{\tau }_0.$$

$$\begin{gathered} \underset{0}{\overset{2\pi }{\int }}{u}^2(e^{\tau }{e}^{i\theta })+v^2(e^{\tau }{e}^{i\theta })d\theta \ge \\ 2C\underset{{\tau }_0}{\overset{\tau }{\int }}\exp \left(2{\tau }^{\prime }+C_3\right)d{\tau }^{\prime }=C{e}^{C_3}(e^{2\tau }-e^{2{\tau }_0}). \end{gathered}$$

$$\frac{1}{x}+\frac{1}{y}=\frac{4}{(x+y)-\frac{(x+y-2x{)}^2}{x+y}}>\frac{4}{1-{\epsilon }^2},$$

$$S=\{z=u+iv\mid {\phi }_{-}(u)<v<{\phi }_{+}(u),u>M\},-\infty \le M<+\infty$$

$$\pi \underset{u_1}{\overset{u_2}{\int }}\frac{du}{\theta (u)}\le x_2-x_1+4\pi ,$$

$$x_2-x_1\le \pi \underset{u_1}{\overset{u_2}{\int }}\frac{1+{\psi }^{\prime 2}(u)}{\theta (u)}du+\frac{\pi }{12}\underset{u_1}{\overset{u_2}{\int }}\frac{{\theta }^{\prime 2}(u)}{\theta (u)}du+8\pi \left(1+\frac{4}{3}{m}^2\right),$$

$$x_2-x_1\ge \pi \underset{u_1}{\overset{u_2}{\int }}\frac{1+{\psi }^{\prime 2}(u)}{\theta (u)}du-\frac{\pi }{4}\underset{u_1}{\overset{u_2}{\int }}\frac{{\theta }^{\prime 2}(u)}{\theta (u)}du+o(1),$$

$$v=f_y(u)=\psi (u)+\frac{\theta (u)}{\pi }y+o(\theta (u)),u\to +\infty .$$

$$x_2-x_1\le \pi \underset{u_1}{\overset{u_2}{\int }}\frac{du}{\theta (u)}+\underset{u_1}{\overset{u_2}{\int }}\frac{{\psi }^{\prime }(u{)}^2}{\theta (u)}du+\frac{\pi }{12}\underset{u_1}{\overset{u_2}{\int }}\frac{{\theta }^{\prime }(u{)}^2}{\theta (u)}du+O(1)=I_1+I_2+I_3+O(1).$$

$$x_2\le u_2-u_1+x_1+O(1)=u_2+O(1).$$

$$f(z)=\frac{1}{2\pi i}\underset{\partial \Omega }{\int }\frac{g(w)}{z-w}dw.$$

$$\Omega =\left\{r{e}^{i\alpha }\mid s(r)+\pi +\frac{1}{{\log }^{2}r+1}<\alpha <s(r)+2\pi -\frac{1}{{\log }^{2}r+1},r>10\right\},$$

$$|\arg (z)-\arg (w)|\ge \frac{1}{{\log }^2|z|+1}-\delta \underset{|z|-\delta <r<|z|+\delta }{\max }{s}^{\prime }(r).$$

$$f(z)=g(z)+\frac{1}{2\pi i}\underset{\partial \Omega }{\int }\frac{g(w)}{z-w}dw.$$

Full text

2010 Mathematics Subject Classification: Primary 30D15; Secondary 30E20, 31A15.

Key words and phrases: growth of entire functions, harmonic measure, extremal length

The work was supported by the Russian Science Foundation grant 17-11-01064.

1. Introduction and main results

If two entire functions $f$ and $g$ of minimal exponential type are such that

[TABLE]

then at least one of them must be constant. This beautiful result, known to some experts as the de Branges Lemma, was proved by Maurice Heins in 1959 (see [7, Theorem 5.1]). It plays a crucial role in the proof of de Branges’s Ordering Theorem [4, Theorem 35] and has many applications in function theory and spectral theory of differential operators (see e.g. [2, 3]). First of all we prove a small refinement of Heins’s Theorem.

Theorem 1.1**.**

Let $f,g$ be entire functions of minimal and finite exponential type, respectively, which satisfy (1.1). Then either $f$ or $g$ is constant.

This theorem obviously fails in the case of both functions having finite exponential type (consider $f(z)=e^{z},g(z)=e^{-z}$). Nevertheless something can be said about domains where these functions are small. The main aim of this paper is to study geometric properties of such domains.

A careful analysis of the proof of Heins’s Theorem shows that the sets $\{|f(z)|\leq 1\}$ and $\{|g(z)|\leq 1\}$ have to be close to half-planes.

Theorem 1.2**.**

Let $f,g$ be nonconstant entire functions of finite exponential type which satisfy (1.1). Then for any $\varepsilon>0$ there exists a set $E_{\varepsilon}$ of finite measure such that for $\tau\in\mathbb{R}_{+}\backslash E_{\varepsilon}$ the sets $\{|f(z)|\leq 1\}$ and $\{|g(z)|\leq 1\}$ intersect circle of radius $e^{\tau}$ centered at the origin in two opposite semicircles modulo a set of the circle of angular measure at most $\varepsilon$, i.e. there exists a semicircle $C\subset\{|z|=e^{\tau}\}$ and a set $B$ of angular measure at most $\varepsilon$ such that

[TABLE]

[TABLE]

In view of this theorem it is reasonable to ask whether orientation of these semicircles can change when the radius tends to infinity. We are able to answer this question in an affirmative way by constructing an entire function of finite exponential type which is bounded in some rotating half-plane, and we also give very close upper and lower bounds on possible rotation functions of this half-plane.

Definition 1**.**

Let $s:\mathbb{R}_{+}\to\mathbb{R}$ be a non-decreasing continuous function. We will say that $\Omega_{s}$ is a rotating half-plane with rotation function $s$ if**

[TABLE]

Note that the interior of the complement of a rotating half-plane $\Omega_{s}$ is a rotating half-plane with rotation function $s(r)+\pi$. Hence, if an entire function $f$ is bounded on a rotating half-plane $\Omega_{s}$, then $\sup_{z\in\mathbb{C}}\min\{|f(z)|,|f(-z)|\}<\infty$.

Now we are ready to formulate the main results of the paper.

Put $h(r)=s(e^{r})$. To avoid inessential technicalities we will always assume that $s\in C^{2}(\mathbb{R}_{+})$.

Theorem 1.3**.**

For every rotating half-plane $\Omega_{s}$ such that $h^{\prime}(x)\to 0$ as $x\to+\infty$ and

[TABLE]

there exists a nonconstant entire function of finite exponential type bounded in $\Omega_{s}$.

It turns out that under additional regularity of the function $h$ condition (1.2) is also necessary for the existence of an entire function of finite exponential type bounded in $\Omega_{s}$.

Theorem 1.4**.**

Let $\Omega_{s}$ be a rotating half-plane with rotation function $s$ and let $f$ be an entire function of finite exponential type bounded on $\Omega_{s}$. Assume that $s$ is such that

[TABLE]

Then $f$ is constant.

In particular, if the function $h(x)$ is concave, then these two theorems give us that condition (1.2) is necessary and sufficient (note that $h^{\prime}(x)\geq 0$ since we assume that $s(r)$ is increasing).

Proofs of Theorems 1.3, 1.4 are based on the estimates for conformal mappings between infinite strips from [10].

Interestingly, we can deduce a version of Theorem 1.4 without assuming any regularity of the function $s$ using estimates for harmonic measure in simply connected domains.

Theorem 1.5**.**

Let $\Omega_{s}$ be a rotating half-plane with rotation function $s$ and let $f$ be an entire function of finite exponential type bounded on $\Omega_{s}$. If the function $s$ is such that

[TABLE]

then $f$ is constant.

Acknowledgments

The author would like to thank Yurii Belov and Alexander Borichev for many helpful discussions. The author also would like to thank Mikhail Sodin for pointing out the paper of Heins [7] where the aforementioned result about two functions of minimal type was proved. The author is grateful to the anonymous referees for numerous remarks which substantially improved the exposition, and for the suggestion to look into the paper of Warschawski [10] which allowed the extension and simplification of the proof of Theorem 1.3 and the proof of Theorem 1.4.

2. Functions of minimal and finite exponential type

In this section we prove Theorems 1.1 and 1.2.

2.1. Functions of minimal exponential type

For the proof of Theorem 1.1 we need the following variant of Lemma 27 from [9] (see also [6, Theorem 8.1]).

Lemma 2.1**.**

Let $u:\mathbb{C}\to\mathbb{R}$ be a continuous subharmonic function satisfying the following conditions:

1. (i)

$u(z)\geq 0$* and $u(0)>0$,* 2. (ii)

$u$* is smooth in some neighbourhood of [math],* 3. (iii)

$u$* is not constant in any neighbourhood of [math].*

Put

[TABLE]

[TABLE]

[TABLE]

Then there exists $C=C(u)>0$ such that for all $\tau>0$

[TABLE]

We tacitly assume that $m_{u}(R)=\infty$, $\eta_{u}(\log R)=0$ when $R\mathbb{T}\subset V_{u}$.

Proof of Theorem 1.1.

Suppose that neither $f$ nor $g$ is constant. Put ${f_{1}(z)=f(z)-f(0)}$, $g_{1}(z)=g(z)-g(0)$. Since $f$ and $g$ are nonconstant, $f_{1}$ and $g_{1}$ are not identically zero and so for some $k,n\in\mathbb{N}$ and $C_{1}>0$ the functions $f_{2}(z)=C_{1}f_{1}(z)/z^{k}$ and $g_{2}(z)=C_{1}g_{1}(z)/z^{n}$ are entire and $|f_{2}(0)|,|g_{2}(0)|>1$.

Since $\min\{|f(z)|,|g(z)|\}\leq 1$, for some $C_{2}>1$ we have $\min\{|f_{2}(z)|,|g_{2}(z)|\}\leq\frac{C_{2}}{|z|},|z|>1$. So, if we consider $f_{3}(z)=f_{2}(C_{2}z)$ and $g_{3}(z)=g_{2}(C_{2}z)$ then they will satisfy $\min\{|f_{3}(z)|,|g_{3}(z)|\}\leq\frac{1}{|z|}<1,$ $|z|>1$. On the other hand, $|f_{3}(0)|,|g_{3}(0)|>1$. It is easy to see from the maximum principle that neither $f_{3}$ nor $g_{3}$ is constant.

Put $u(z)=\max\{0,\log|f_{3}(z)|\}$ and $v(z)=\max\{0,\log|g_{3}(z)|\}$. Now we apply Lemma 2.1 for $u$ and $v$.

We consider some function $q:\mathbb{R}_{+}\to\mathbb{R}_{+}$ which we choose later and sum estimates (2.1) for $u$ and $v$ with weights $q(\tau)$ and $1$.

[TABLE]

Applying the inequality between the arithmetic and geometric means to the right-hand side, we get

[TABLE]

We claim that $\eta_{u}(s)+\eta_{v}(s)\geq 4$ for all $s>0$.

From the definition of $u$ and $v$ we have $|f_{3}(z)|\geq 1,z\in V_{u}$ and ${|g_{3}(z)|\geq 1,z\in V_{v}}$. Let $I,J\subset\mathbb{R}$ be intervals such that $e^{s}e^{ix}\in V_{u}$, $x\in I$ and $e^{s}e^{ix}\in V_{v}$, $x\in J$.

Since $\min\{|f_{3}(e^{s}e^{ix})|,|g_{3}(e^{s}e^{ix})|\}<1$, $x\in\mathbb{R}$ either $|I|+|J|\leq 2\pi$ or one of these intervals has length more than $2\pi$. But if, for example, $|I|>2\pi$, then $|g_{3}(e^{s}e^{ix})|<1$, $x\in\mathbb{R}$. This contradicts the maximum principle. Hence, $|I|+|J|\leq 2\pi$. Taking the supremum over $I$ and $J$ we get

[TABLE]

The claim now follows from the inequality between the arithmetic and harmonic means.

From $\eta_{u}(s)+\eta_{v}(s)\geq 4$ and (2.2) we get

[TABLE]

Now, since $f_{3}$ is of minimal exponential type and $g_{3}$ is of finite exponential type, we have $u(e^{\tau}e^{i\theta})=o(e^{\tau})$ and $v(e^{\tau}e^{i\theta})=O(e^{\tau})$ and so we can choose $q$ such that $q(\tau)\to\infty$ as $\tau\to\infty$ and $q(\tau)u^{2}(e^{\tau}e^{i\theta})=O(e^{2\tau})$. But for this choice of $q$ the left-hand side of (2.4) is $O(e^{2\tau})$ while right-hand side is not. We arrive at a contradiction.∎

2.2. Functions of finite exponential type

We fix two entire functions $f$ and $g$ of finite exponential type satisfying (1.1).

Lemma 2.2**.**

We have $\int\limits_{0}^{\infty}\left(\frac{\eta_{u}(s)+\eta_{v}(s)}{2}-2\right)ds<\infty$ .

Proof.

We consider estimate (2.2) with $q(\tau)=1$

[TABLE]

Suppose that $\int\limits_{0}^{\infty}\left(\frac{\eta_{u}(s)+\eta_{v}(s)}{2}-2\right)ds=\infty$. Since $\frac{\eta_{u}(s)+\eta_{v}(s)}{2}-2\geq 0$ (see the proof of Theorem 1.1), for any $C_{3}>0$ there exists $\tau_{0}>0$ such that

[TABLE]

Thus, for $\tau>\tau_{0}$ we have

[TABLE]

The left hand side of (2.6) is bounded by $ce^{2\tau}$ for some $c=c(f,g)>0$. We can choose $C_{3}$ such that $c<Ce^{C_{3}}$. We arrive at a contradiction for big enough $\tau$.

∎

Lemma 2.3**.**

If $x,y>0$, $x+y\leq 1$ and $|x-\frac{1}{2}|>\varepsilon$ then $\frac{1}{x}+\frac{1}{y}>4+\delta$ for some $\delta=\delta(\varepsilon)>0$.

Proof.

If $1-x-y>\varepsilon$, then $\frac{1}{x}+\frac{1}{y}\geq\frac{4}{x+y}>\frac{4}{1-\varepsilon}$. Otherwise

[TABLE]

Hence, if $\delta=\min\{\frac{4}{1-\varepsilon^{2}}-4,\frac{4}{1-\varepsilon}-4\}$ we get the desired estimate. ∎

Lemma 2.4**.**

For any $\varepsilon>0$ there exists $E_{\varepsilon}\subset\mathbb{R}_{+}$, $|E_{\varepsilon}|<\infty$ such that for any $\tau\in\mathbb{R}_{+}\setminus E_{\varepsilon}$ the set $\{z:|z|=e^{\tau},|f_{3}(z)|\geq 1\}$ contains an arc of angular measure at least $\pi-\varepsilon$.

Proof.

We know that $m_{u}(e^{\tau})+m_{v}(e^{\tau})\leq 1$ (see (2.3)). From Lemmas 2.2 and 2.3 it follows that for all $\tau>0$ except for the set of finite measure $m_{u}(e^{\tau})>\frac{1}{2}-\frac{\varepsilon}{2\pi}$. By definition, this means that $V_{u}\bigcap\{z:|z|=e^{\tau}\}$ contains an arc of angular measure at least $\pi-\varepsilon$. ∎

Lemma 2.5**.**

For any $\varepsilon>0$ there exists $E_{\varepsilon}\subset\mathbb{R}_{+}$, $|E_{\varepsilon}|<\infty$ such that for any $\tau\in\mathbb{R}_{+}\setminus E_{\varepsilon}$ the set $\{|z|=e^{\tau},|f(z)|>1\}$ contains an arc of angular measure at least $\pi-\varepsilon$.

Proof.

We consider only those $\tau>\log C_{2}$ for which the statement of Lemma 2.4 is fulfilled with $\tau^{\prime}=\tau-\log C_{2}$. If $|f(e^{\tau}e^{i\theta})|\leq 1$ then $|f_{3}(e^{\tau^{\prime}}e^{i\theta})|<1$ and so $\{\theta:|f_{3}(e^{\tau^{\prime}}e^{i\theta})|\geq 1\}\subset\{\theta:|f(e^{\tau}e^{i\theta})|>1\}$. From Lemma 2.4 we get the result. ∎

Now we are ready to prove Theorem 1.2. We will prove a slightly more general result.

Theorem 2.6**.**

For any $\varepsilon>0$ there exists $E_{\varepsilon}\subset\mathbb{R}_{+}$, $|E_{\varepsilon}|<\infty$, such that for any $\tau\in\mathbb{R}_{+}\backslash E_{\varepsilon}$ there are two disjoint arcs $I$, $J$ of angular measure at least $\pi-\varepsilon$ on the circle $\{|z|=e^{\tau}\}$ such that $|f(z)|>1$ and $|g(z)|\leq 1$ on $I$ and $|g(z)|>1$ and $|f(z)|\leq 1$ on $J$.

Proof.

In view of Lemma 2.5 we consider only those $\tau>0$ for which there exist arcs $I$, $J$ of angular measure at least $\pi-\varepsilon$ such that for $|f(z)|>1$, $z\in I$ and for $|g(z)|>1$, $z\in J$. From $\min\{|f(z)|,|g(z)|\}\leq 1$ we get that $I\cap J=\varnothing$ and moreover $|f(z)|\leq 1,z\in J$ and $|g(z)|\leq 1,z\in I$. ∎

3. Rotating half-planes

In this section we prove Theorems 1.3, 1.4 and 1.5.

3.1. Proofs of Theorems 1.3 and 1.4

For continuous functions ${\varphi_{-}(u)<\varphi_{+}(u)}$ we will call the set

[TABLE]

a semi-infinite strip bounded by the curves $v=\varphi_{+}(u),v=\varphi_{-}(u),u=M$. Let $\theta(u)=\varphi_{+}(u)-\varphi_{-}(u)$ and $\psi(u)=\frac{1}{2}(\varphi_{+}(u)+\varphi_{-}(u))$.

We will denote by $Z(w)=X(w)+iY(w)$ a conformal mapping from $S$ to $S_{0}=\{x+iy\mid|y|<\frac{\pi}{2}\}$ such that $\Re w$ and $X(w)$ tends to $+\infty$ simultaneously. The function $W(z)=U(z)+iV(z)$ will denote the inverse of $Z(w)$.

Now we are ready to formulate results from [10] that we will use in the proofs of Theorems 1.3 and 1.4.

Theorem 3.1**.**

[10, Theorem III a]** Let $w_{1}=u_{1}+iv_{1},w_{2}=u_{2}+iv_{2}$ and $u_{1}\leq u_{2}$. Then

[TABLE]

where $x_{k}=X(w_{k})$.

Theorem 3.2**.**

[10, Theorem IV a]** If $|\varphi_{\pm}^{\prime}(u)|\leq m$, then for some $x_{0}$, for $x_{0}\leq x_{1}\leq x_{2}$ and $|y_{1,2}|<\frac{\pi}{2}$ we have the bound

[TABLE]

where $u_{k}=U(x_{k}+iy_{k})$.

Theorem 3.3**.**

[10, Theorem VI]** If $|\varphi_{\pm}^{\prime}(u)|\to 0,u\to+\infty$ and ${|\varphi_{\pm}^{\prime\prime}(u)|\in L^{1}(\mathbb{R})}$, then for $w_{1}=u_{1}+iv_{1}$ and $w_{2}=u_{2}+iv_{2}$, $u_{1}<u_{2}$, we have

[TABLE]

where $x_{k}=X(w_{k})$ and $o(1)$ is uniform as $u_{1}\to+\infty$.

Theorem 3.4**.**

[10, Theorem X iii]** If $|\varphi_{\pm}^{\prime}(u)|\to 0,u\to+\infty$, then for every $y,|y|<\frac{\pi}{2}$, the line $\Lambda_{y}=\{z\mid\Im z=y\}$ is mapped by $Z$ onto a curve $L_{y}$ which for big enough $u$ is represented by the equation

[TABLE]

Proof of Theorem 1.3.

We will construct a function $f$ in two steps. First of all, we will construct an unbounded analytic function $g:\bar{\Omega}\to\mathbb{C}$ of finite exponential type which is bounded on $\partial\Omega$ for some suitable domain $\Omega$. Then we will construct function $f$ as a Cauchy integral of $g$ over the boundary of $\Omega$ and show that $f$ is bounded in $\Omega_{s}$.

Put $\varphi_{-}(u)=h(u)+\pi+\frac{1}{u^{2}+1},\varphi_{+}(u)=h(u)+2\pi-\frac{1}{u^{2}+1}$. Note that $\theta(u)=\pi-\frac{2}{u^{2}+1}$ and $\psi(u)=h(u)+\frac{3\pi}{2}$.

Let $x_{0}=x_{1}\leq x_{2}$ and consider estimate (3.2)

[TABLE]

Direct computation shows that $I_{1}+I_{3}=u_{2}-u_{1}+O(1)$. Since $\int h^{\prime}(u)^{2}du<\infty$ and $\theta(u)\geq\pi-2$, we have $I_{2}=O(1)$. Therefore we get

[TABLE]

Put $e(z)=\exp(\exp(z)),z\in\bar{S_{0}}$. Note that $e$ is bounded on $\partial S_{0}$ and unbounded on $S_{0}$. Consider the function $b(w)=e(Z(w)),w\in\bar{S}$. This function is bounded on $\partial S$ and from the estimate (3.5) we get ${|b(u+iv)|\leq\exp(C\exp(u))}$.

Put $\Omega=\{re^{i\alpha}\mid\varphi_{-}(\log r)<\alpha<\varphi_{+}(\log r),r>10\}$ and consider the function $g(z)=\frac{b(\log z)}{z^{2}},z\in\bar{\Omega}$. We have $|g(z)|=O\left(\frac{1}{|z|^{2}}\right),z\in\partial\Omega$ and $|g(z)|\leq\exp(C|z|),z\in\Omega$.

For $z\in\mathbb{C}\backslash\bar{\Omega}$ put

[TABLE]

Since $|\varphi_{\pm}^{\prime}|$ are bounded, the intersection of $\partial\Omega$ with the annulus ${\{n<|z|<n+1\}}$ has length $O(1)$. Therefore this integral converges absolutely.

Note that by shifting contour from $\partial\Omega$ to $(\partial\Omega\cap\mathbb{C}\backslash R\mathbb{D})\cup(\partial R\mathbb{D}\cap\Omega)$ we can analytically extend $f$ to the disk $R\mathbb{D}$. Since $R$ can be chosen arbitrary large the function $f$ has an analytic extension to the whole plane. Moreover, choosing $R=2|z|$ we obtain $|f(z)|\leq 10\max\limits_{|w|=R}|g(w)|+O(1)$. Since ${|g(w)|\leq\exp(C|w|)}$, the function $f$ is of finite exponential type. It remains to prove that $f$ is nonconstant and bounded in $\Omega_{s}$.

First of all we will prove that $\operatorname{dist}(\Omega,\Omega_{s})=\varepsilon>0$. Note that

[TABLE]

therefore $\Omega\cap\Omega_{s}=\varnothing$.

Suppose that $z\in\Omega,w\in\Omega_{s},|z-w|<\delta$. Then $||z|-|w||<\delta$. Hence,

[TABLE]

From the boundedness of $h^{\prime}(u)$ it follows that $s^{\prime}(r)=O\left(\frac{1}{r}\right)$. Hence for small enough $\delta$ the arguments of $z$ and $w$ differ by at least $\frac{1}{2(\log^{2}|z|+1)}$. Thus $|z-w|\geq\frac{|z|}{20(\log^{2}|z|+1)}>\delta$ for small enough $\delta$. We arrive at a contradiction.

Therefore for $z\in\Omega_{s}$, $|f(z)|\leq\frac{1}{2\pi\varepsilon}\int\limits_{\partial\Omega}|g(w)||dw|=C<\infty$.

To prove that $f$ is nonconstant we first note the following identity for $z\in\Omega$:

[TABLE]

Indeed, when the contour passes through the point $z$ the residue at $z$ adds to the integral which can be easily seen to be equal to $g(z)$.

Consider the line $\Lambda_{0}=\mathbb{R}\subset S_{0}$. From Theorem 3.4 under the mapping $z\mapsto\exp(W(z))$ it is mapped onto the curve

[TABLE]

Reasoning as before we can see that $dist(\Gamma,\partial\Omega)>0$. Therefore the integral in the (3.6) is $O(1)$.

From Theorem 3.1 and definition of $e(z)$ we can see that ${|g(z)|\geq\frac{\exp(c|z|)}{|z|^{2}}},z\in\Gamma$, for some $c>0$. So $g(z)$ is unbounded on $\Gamma$. Therefore $f(z)$ is unbounded on $\Gamma$ as well and thus is nonconstant. ∎

Remark 3.5**.**

It seems plausible that Theorem 1.3 can be proved assuming only boundedness of $h^{\prime}(x)$ instead of $h^{\prime}(x)\to 0$, by modifying methods from [10]. The only step where we used the more restrictive hypothesis is when we proved that $f$ is nonconstant and there we had a very big degree of freedom.**

We will now proof a slightly weaker form of Theorem 1.4.

Theorem 3.6**.**

Let $\Omega_{s}$ be a rotating half-plane with rotation function $s$ and let $f$ be an entire function of finite exponential type bounded on $\Omega_{s}$. Assume that $s$ does not satisfy (1.2), $h^{\prime}(x)\to 0$ as $x\to\infty$ and

[TABLE]

Then $f$ is constant.

Proof.

Let $S=\{u+iv\mid h(u)+\pi<v<h(u)+2\pi\}$, that is $\theta(u)=\pi$, $\psi(u)=h(u)+\frac{3\pi}{2}$.

The function $g(z)=f(\exp(W(z)))$ is analytic on $S_{0}$ and bounded on $\partial S_{0}$. Fix some big $u_{1}$ and consider estimate (3.3),

[TABLE]

Since the function $f$ is of finite exponential type, we have

[TABLE]

Note that since the map $W$ is conformal and sends infinity to infinity, we have $U(x+iy)\to+\infty$ as $x\to+\infty$. Therefore $|g(x+iy)|=o(\exp(c\exp(x)))$ for every $c>0$ as $x\to+\infty$. Thus, the function $b(z)=g(\log z),{\Re z\geq 0}$, is bounded on $\{\Re z=0\}$ and $b(z)=o(\exp(c|z|)),|z|\to\infty$. From the Phragmén-Lindelöf Theorem (see, e.g., [8, Lecture 6, Theorem 3]) it follows that $b$ is bounded. Therefore $f$ is bounded on $\mathbb{C}\backslash\bar{\Omega}_{s}$. Since by assumption $f$ is bounded on $\Omega_{s}$ we get that $f$ is constant by Liouville’s Theorem. ∎

Now we will derive Theorem 1.4 from Theorem 3.6 and Theorem 1.5 (the proof of which we postpone to the next section).

Proof of Theorem 1.4.

Since $h^{\prime\prime}(u)\in L^{1}(\mathbb{R})$, there exists $\lim_{u\to\infty}h^{\prime}(u)=H$. If $H=0$, then we may apply Theorem 3.6 to conclude that $f$ is constant.

If $H\neq 0$ then we have $h(u)=(H+o(1))u\neq o(\sqrt{u})$. Therefore we can apply Theorem 1.5 and again conclude that $f$ is constant. ∎

3.2. Proof of Theorem 1.5

We will need the following estimate for harmonic measure in bounded simply connected domains:

Theorem 3.7** ([5, Theorem 5.3]).**

Let $G$ be a simply connected domain bounded by a rectifiable Jordan curve and let $E\subset\partial G$ be an arc. If $\sigma$ is any curve connecting point $z_{0}\in G$ and $\partial G\backslash E$ which lies entirely inside of $G$, then

[TABLE]

where $\omega(z_{0},E,G)$ is the harmonic measure at $z_{0}$ of $E$ in $G$ and ${\lambda(\sigma,E,G)=\frac{dist(\sigma,E)^{2}}{A(G)}}$, where $A(G)$ is the area of $G$ and $dist(\sigma,E)$ is the infimum of the lengths of paths which connect points on $\sigma$ with $E$ and lie entirely inside of $G$.

In Theorem 5.3 from [5] we can vary not only $\sigma$ but also the metric $\rho$. Theorem 3.7 is a special case with $\rho(z)=1$.

Put $h(x)=s(\exp(x))+\pi$, ${V=\{a+bi\mid a>0,h(a)<b<h(a)+\pi\}}$. The function $g(z)=f(e^{z})$ is defined on $\bar{V}$, bounded on $\partial V$ and satisfies the estimate $|g(a+bi)|\leq\exp(A+C\exp(a))$ for some ${A,C>0}$. Without loss of generality we may assume that ${|g(z)|\leq 1}$ on $\partial V$. Put ${G_{t}=V\cap\{\Re z<t\}}$, ${E_{t}=[t+h(t)i,t+(h(t)+\pi)i]}$ for some ${t\in\mathbb{R}}$ and consider function ${u(z)=C\exp(t)\omega(z,E_{t},G_{t})}$. Obviously, we have ${A+u(z)\geq\log|g(z)|,\ z\in\partial G_{t}}$. Since $\log|g(z)|$ is a subharmonic function, this estimate holds for $z\in G_{t}$ as well. Thus, if for all $z_{0}=a_{0}+b_{0}i\in V$ we can prove that $u(z_{0})\to 0,\ t\to\infty$, then we would have $|g(z_{0})|\leq\exp(A)$, from which it follows that $f$ is bounded on $\mathbb{C}\backslash\Omega_{s}$. Since $f$ is also bounded on $\Omega_{s}$, by Liouville’s Theorem $f$ is constant. So it remains to prove that $u(z_{0})\to 0,\ t\to\infty$.

Put $\sigma=[a_{0}+h(a_{0})i,a+bi]$. By the assumption of Theorem 1.5, there exists a sequence of numbers $t_{k}$ tending to infinity such that $h(t_{k})\geq c\sqrt{t_{k}}$ for some $c>0$. Moreover, we may assume that $\frac{h(t_{k+1})}{2}>h(t_{k})+\pi+\frac{c^{2}}{8}$ and that $t_{1}>a_{0}$. We have

[TABLE]

where $n+1$ is the number of $t_{k}$’s not greater than $t$. Using (3.7) we get

[TABLE]

Obviously, when $n$ tends to infinity the right-hand side of (3.8) tends to [math], and since $n\to\infty$ as $t\to\infty$ we have that $u(z_{0})\to 0$ as $t\to\infty$.

∎