Abstract
We study the behavior of the minimum modulus of analytic functions in the unit disc in terms of \(\rho _\infty \)-order, which is the limit of the orders of \(L_p\)-norms of \(\log |f(re^{i\theta })|\) over the circle as \(p\rightarrow \infty \). This concept coincides with the usual order of the maximum modulus function if the order is greater than one. New results are obtained for analytic functions of order smaller than 1.
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1 Introduction and Main Results
Let \(\mathbb {D}_R=\{z\in \mathbb {C}:|z|<R\}\), \(0<R\le \infty \), and \(\mathbb {D}=\mathbb {D}_1\). For an analytic function \(f\) on \(\mathbb {D}_R\), we define the minimum modulus
and the maximum modulus
Interplay between \(\mu (r,f)\) and \(M(r,f)\) has been studied in a large number of papers. In the case of entire functions, i.e., \(R=\infty \), a survey of results up to 1989 can be found in Hayman’s book ([16], Chap. 6]).
The orders of the growth of an analytic function \(f\) in \(\mathbb {D}_\infty \), and in \(\mathbb {D}\), respectively, are defined as
For entire functions of order \(\rho [f]\le 1\), there are a lot of sharp results on the behavior such as \(\cos \pi \rho \)-theorem ([1, 16]).
Theorem
([1]) Suppose that \(0\le \rho <\alpha <1\). If \(f\) is an entire function of order \(\rho \) and \(f(z)\not \equiv const\) then
where
One of the most interesting open problems for entire functions of order greater than 1 is to find the asymptotic behavior of the minimum modulus with respect to the maximum modulus, especially for values of \(\rho [f]\) close to 1 ([14, 15]). The most precise results concerning the minimum modulus of entire and subharmonic functions of order zero can be found in [2–4, 11–13].
For analytic functions in the unit disc \(\mathbb {D}\) the situation, in a certain sense, is the opposite. Known results are much weaker in accuracy than the statements of the \(\cos \pi \rho \)-theorem type. Moreover, these results mainly concern analytic functions with \(\rho _M[f]\ge 1\).
We start with an old result of M. Heins.
Theorem A
[18] If \(f(z)\) is analytic in \(\mathbb {D}\), \(f(z)\not \equiv const\), \(f(z)\) is bounded in \(\mathbb {D}\), then there exist a constant \(K>0\) and a sequence \((r_n)\), \(r_n\nearrow 1\) such that
For the function \(f(z)=\exp (\frac{1}{z-1})\), we have
Thus, inequality (1) is sharp in the class of bounded analytic functions in the unit disc. A description of exceptional sets for the relation \((1-|z|)\log |B(z)| \rightarrow 0\), \(|z| \nearrow 1\), where \(B\) is a Blaschke product, has been very recently obtained in [17].
In the general case, we have the following theorem of C.N. Linden.
Theorem B
[20] Let \(f(z)\) be an analytic function, \(f(z)\not \equiv const\) in \(\mathbb {D}\), \(\rho =\rho _M[f]>1\), then there is a constant \(K(\rho )\) such that
for some sequence of number \((r_n),\,r_n\nearrow 1.\)
The following theorem plays a key role for the estimates of minimum modulus.
Theorem C
[20] Let \(f\) be analytic in \(\mathbb {D}\), and suppose that \(\frac{1}{2}\le \alpha <1\). Then, there exists \(R_0=R_0(\alpha )\in (0,1)\) such that for arbitrary \(R\in [R_0,1)\), there is a set \(E_R\subset [R^2,R(R+\frac{1}{16}(1-R))]\) of measure at least \(\frac{1}{32}R(1-R)\) such that
\(r\in E_R\), \(C=C(\alpha ,R_0)>0\).
Such an approach for analytic functions \(f\) of order \(\rho _M[f]<1\) allows us to get the following results.
Theorem D
[22] Let \(0\le \rho _M[f]<1\), \(f(z)\) be an analytic in \(\mathbb {D}\), \(f(0)=1\) and
Then, there exist \(R_1\in (0,1)\) depending on \(\rho _M\) and \(K=K(A,R_1)\) such that, if \(R\in (R_1,1)\), then the interval \((R,\frac{1}{2}(1+R))\) contains a set of values \(r\) of measure at least \(\frac{1}{4}(1-R)\) such that
Let \((a_n)\) be a sequence of zeros of analytic function \(f\) in \(\mathbb {D}\). For this sequence, we define
Theorem E
[22] Let \(f(z)\) be analytic in \(\mathbb {D}\), \(f(z)\not \equiv const\), \(\rho _M[f]<1\). If there are \(r_0\in (0,1)\) and a constant \(B\) such that
then there exist \(K>0, L\ge \frac{1}{4}\), \(\rho _0\in (0,1)\) such that, if \(R\in (\rho _0,1)\), then the interval \((R,\frac{1}{2}(1+R))\) contains a set \(r\) of measure at least \(L(1-R)\) such that
A characteristic feature of Theorems D and E is that their conclusions do not depend on the corresponding value of order \(\rho _M[f]\in [0,1]\). It appears that, in this case, the value \(\rho _M[f]\) does not allow the behavior of the minimum modulus in terms of conditions on zeros of \(f\) to be described more precisely. The aim of this paper is to correct this defect. Note that, some classes of bounded analytic functions satisfying the inequality
were found in [5].
For an analytic function \(f(z)\), \(z\in \mathbb {D}\), \(f\not \equiv 0\) and \(p\ge 1\), we define
We write
We define the order \(\rho _\infty [f]\) of the function \(f\) as
The limit exists since \(\rho _p\) is a non-decreasing function in \(p\) ([24]). This quantity appeared for the first time in a work of Linden [23], who proved that \(\rho _M[f]=\rho _\infty [f]\) provided that \(\rho _M[f]>1\), and \(\rho _M[f]\le \rho _\infty [f]\), but he did not study the classes of functions defined by the order \(\rho _\infty [f]\) when \(\rho _\infty [f]<1\). Applications of this concept to factorization of analytic functions in \(\mathbb {D}\), and logarithmic derivative estimates can be found in [6, 9].
Let a sequence \((a_n)\) in \(\mathbb {D}\) satisfy the condition
Consider the canonical product, \(s\in \mathbb {N}\),
where \(E(w,0)=1-w\),
is the Weierstrass primary factor, and \(A_n(z)=\frac{1-|a_n|^2}{1-\bar{a}_n z}\). The function \(P(z)\) is analytic in the unit disc with the zero sequence \((a_n)\) provided that (2) holds. We note that if \(s=0\), we have \(P_0(z)=CB(z),\) where \(C=\prod _n |a_n|\),
is the Blaschke product corresponding to the sequence \((a_n)\) provided that \(\sum _{n} (1-|a_n|)<\infty \). We define
Let \(E\subset [0,1)\) be a measurable set. The upper density of \(E\) is defined by
where \(\lambda _1(E\cap [r,1))\) denotes the Lebesgue measure of \(E\cap [r,1)\). Theorem 1 describes the minimum modulus of canonical products of genus \(s\in \mathbb {N}\).
Theorem 1
Given a sequence \((a_n)\) in \(\mathbb {D}\), suppose that \(n_z\left( \frac{1-|z|}{2}\right) \le \left( \frac{1}{1-|z|}\right) ^\beta \), for some \(\beta >0\), and all \(z\in \mathbb {D}{\setminus }\mathbb {D}_{r_0},\, 0\le r_0<1\), and let \(P(z)\) be the canonical product of genus \(s\ge [\beta ]+1\) with zeros \((a_n)\). Then, for arbitrary \(K_1, K_2>1\), there exist a constant \(C\in (0, \frac{2}{3}]\) and a set \(F\subset [0,1)\) such that
and
where \(D_1(F)\le C\).
Remark 1
An example from [20], Theorem 6] shows that for all \(\beta \ge 1\) there exists an analytic function in \(\mathbb {D}\) satisfying the conditions of Theorem 1, and of order \(\rho _M[f]=\beta \) such that
holds for all \(r\in [0,1)\) and some constant \(K_3(\beta )>0\).
In the general case, we need the following factorization theorem.
Theorem F
[6] Let \(f\) be an analytic function in \(\mathbb {D}\), and of finite order \(\rho _\infty [f]\). Then,
where \(P(z)\) is a canonical product of form (3) displaying the zeros of \(f\), \(p\) is a non-negative integer, \(g\) is non-zero and both \(P\) and \(g\) are analytic, and \(\rho _\infty [P]\le \rho _\infty [f]\), \(\rho _\infty [g]\le \rho _\infty [f]\).
Let \(u(z)\) be a harmonic function in \(\mathbb {D}\). We then define
Denote \(M_\infty (r,u)=\max \{|u(z)|: |z|=r\}\). The following statement is of some independent interest. It gives another way to compute \(\rho _\infty \)-order of an analytic function without zeros.
Proposition 1
Let \(u(z)\) be a harmonic function in \(\mathbb {D}\). Then, we have
The main result of this paper is the following.
Theorem 2
Let \(f\) be analytic in \(\mathbb {D}\), \(\rho _\infty [f]=\rho \), \(\rho <+\infty \). Then, for arbitrary \(\varepsilon >0\), there exists \(C\in (0,1)\), and a set \(F\subset [0,1)\), such that
for \(r\in [0,1){\setminus } F\), \(D_1(F)\le C\).
Remark 2
Theorem 2 gives us substantially new information when \(\rho _\infty [f]<1\).
Remark 3
The function \(g(z)=\exp \{-\frac{1}{(1-z)^\alpha }\log \frac{1}{1-z}\},\, g(0)=1\), shows that \(\varepsilon \) in the inequality (5) cannot be omitted.
Some generalizations of Theorems 1 and 2 are considered in Sect. 3.
2 Proof of the Main Results
The following lemma is important in our investigation.
Lemma 1
[21] For a given value \(\theta \) let \(S_{m,k}\) denote the region
where \(k\) and \(m\) are integers such that \(k>0\) and \(-2^{k-1}\le m\le 2^{k-1}-1\). Let \(k_0\) be a positive integer and \(\beta >0\). Suppose that there are a finite number of points \(a_n\) in \(\{|z|<1-2^{-k_0}\}\) and that for some value \(\theta \) such that \(0\le \theta <2\pi \) there are at most \(C 2^{k\beta }\) points \(a_n\) in each region (6) for \(k\ge k_0\). Then, if \(s\) is an integer greater than \(\beta \) the function \(P\) defined by (3) is analytic in \(\mathbb {D}\) and
where \(K\) depends on \(s,\beta ,C\).
Proof of Theorem 1
Without loss of generality, we may assume that \(r_0=0\). Otherwise,
where \(C_{r_0}=\max \left\{ 1, n_0\left( \frac{1+r_0}{2}\right) \right\} \), because \(n_z\left( \frac{1-|z|}{2}\right) \le n_0\left( \frac{1+r_0}{2}\right) \) if \(|z|\le r_0\). Since each \(S_{m,k}\) can be covered by a uniformly bounded number of discs of the form \(\{\zeta : |\zeta -z|<\frac{1-|z|}{2}\}\) with the centers in \(S_{m,k}\), the assumptions of Lemma 1 are satisfied. Hence, the inequality \(n_z\left( \frac{1-|z|}{2}\right) \le \left( \frac{1}{1-|z|}\right) ^\beta \) holds in \(\mathbb {D}\) and we have from Lemma 1
We denote \(r_N=1-\bigl (\frac{3}{4}\bigr )^{N}\), and use the following lemma. \(\square \)
Lemma 2
[15] Suppose that \(r>0,\; h>0\) and that for \(|z|=r\) we have \(n_z(h)\le n_0\). Then there exist a set \(\fancyscript{E}\subset [r,r+\frac{h}{2}]\) having measure at least \(\frac{1}{4}h\) such that, for \(R\) in \(\fancyscript{E}\) and \(|z|=R\),
where \(A\) is an absolute constant.
We apply this lemma with \(r=r_N\), \(h= \frac{1}{2} (1-r_N)=\frac{1}{2} \left( \frac{3}{4}\right) ^{N}\), and \(n_0=\left( \frac{4}{3}\right) ^{N\beta } \). Since \(r_N+ \frac{h}{2}=r_{N+1}\), there exists a set \(\fancyscript{E}_N\subset \{R: 1- r_N \le R\le 1-r_{N+1}\}\) such that \(\lambda _1(\fancyscript{E}_N)\ge \frac{1}{8} \bigl (\frac{3}{4}\bigr )^{N}\), and for \(R\in \fancyscript{E}_N\) and \(|z|=R\) we obtain
We define \(F=\bigcup _{N=1}^\infty F_N\) where \(F_N=[1-r_N;1-r_{N+1}]{\setminus } \fancyscript{E}_N\). Let us prove that
holds, where \(0<C \le \frac{2}{3}\). Note that,
Consider two cases:
-
i)
Let \(r\in [1-\bigl (\frac{3}{4}\bigr )^{k}, 1-\frac{7}{6}\cdot \bigl (\frac{3}{4}\bigr )^{k+1}]\), \(k\in \mathbb {N}\). Then,
$$\begin{aligned} \lambda _1([r,1)\cap F)\le & {} \sum \limits _{N=k}^\infty \lambda _1\Bigl (\Bigl [1-\Bigl (\frac{3}{4}\Bigr )^{N},1-\Bigl (\frac{3}{4}\Bigr )^{N+1}\Bigr ]\cap F_N\Bigr )=\sum \limits _{N=k}^\infty \lambda _1(F_N)\le \\\le & {} \sum \limits _{N=k}^\infty \frac{1}{8} \left( \frac{3}{4}\right) ^{N} = \frac{1}{2} \left( \frac{3}{4}\right) ^{k}. \end{aligned}$$Since \(r\le 1- \frac{7}{6} \cdot \bigl (\frac{3}{4}\bigr )^{k+1}\), we obtain \(\bigl (\frac{3}{4}\bigr )^{k}\le \frac{8}{7}(1-r)\). Hence,
$$\begin{aligned} \lambda _1([r,1)\cap F)\le \frac{4}{7}(1-r). \end{aligned}$$ -
ii)
Let \(r\in [1- \frac{7}{6} \cdot \bigl (\frac{3}{4}\bigr )^{k+1}, 1-\bigl (\frac{3}{4}\bigr )^{k+1}]\). Thus,
$$\begin{aligned} \lambda _1([r,1)\cap F)\le & {} \sum \limits _{N=k+1}^\infty \lambda _1(F_N)+ \frac{1}{6} \cdot \left( \frac{3}{4}\right) ^{k+1}\\\le & {} \sum \limits _{N=k+1}^\infty \frac{1}{8} \left( \frac{3}{4}\right) ^{N}+ \frac{1}{6} \cdot \left( \frac{3}{4}\right) ^{k+1} = \frac{2}{3} \cdot \left( \frac{3}{4}\right) ^{k+1}\le \frac{2}{3} (1-r). \end{aligned}$$Therefore, we have proved that \(\lambda _1([r,1)\cap F)\le \frac{2}{3}(1-r)\) holds for all \(r<1\) sufficiently close to 1. Thus, \(D_1(F)\le C\le \frac{2}{3}\).
To complete the proof of Theorem 1 we need the following lemma.
Lemma 3
[7] For arbitrary \(\delta \in (0,1)\) and arbitrary \(z\in \mathbb {D}\), the inequality
holds, where \(C_1\) is a positive constant depending on \(s\) and \(\delta \).
From Lemmas 3 and 1, and (2) we get
where \(K>1\) is an arbitrary constant. Theorem 1 is proved.
Proof of Proposition 1
Since \(|u(re^{i\theta })|\le M_\infty (r,u)\), we have \(m_p(r,u)\le M_\infty (r,u)\). It implies
Let \(P(z,w)= \mathrm {Re}\frac{w+z}{w-z}\) be the Poisson kernel. The Poisson formula together with Hölder’s inequality yields
where \(\frac{1}{p}+\frac{1}{q}=1,\; 0<r<R<1,\; 0\le \theta <2\pi .\) \(\square \)
For the estimate of the Poisson kernel, we use the following lemma
Lemma 4
[10] If \(a>0\) and \(R=\frac{1}{2}(1+r)\) then
Therefore,
Putting \(R=\frac{1}{2}(1+r)\), we get
Thus, as \(p\rightarrow \infty \) we obtain
Hence, from arbitrariness of \(\varepsilon >0\) it follows that \(\rho ^*\le \rho _\infty [u]\), and finally
Proof of Theorem 2
By Theorem F, we have
where \(P(z)\) is a canonical product of form (3) displaying the zeros of \(f\), \(p\) is a non-negative integer, \(g\) is non-zero, both \(P\) and \(g\) are analytic, and \(\rho _\infty [P]\le \rho \), \(\rho _\infty [g]\le \rho \). Further,
Note that, by [6], Thm. 1.4]
Applying Theorem 1, with \(\beta =\rho +\varepsilon \) we get
where \(D_1(F)\le C\).
We set \(u=\log |g(z)|\), \(u\) is harmonic in \(\mathbb {D}\). Since \(\rho _M[g]\le \rho _\infty [g]\le \rho \), applying Proposition 1, we deduce
which is equivalent to
Thus, we have
Theorem 2 is proved. \(\square \)
3 Generalizations
Remark 4
One can replace the condition \(n_z\left( \frac{1-|z|}{2}\right) \le \left( \frac{1}{1-|z|}\right) ^\beta \) in Theorem 1 by a more general one of the form \(n_z\left( \frac{1-|z|}{2}\right) \le \psi \left( \frac{1}{1-|z|}\right) ,\) where \(\psi : [1,+\infty )\rightarrow \mathbb {R}_+\) is a non-decreasing function such that \(\psi (2x)=O(\psi (x)),\, x\rightarrow \infty .\) Then, one should replace the factor \(\frac{1}{(1-r)^\beta }\log \frac{1}{1-r}\) by \(\tilde{\psi }\left( \frac{1}{1-r}\right) \log \frac{1}{1-r} \) in the conclusion of Theorem 1, where \(\tilde{\psi }(x)=\int _1^x \frac{\psi (t)}{t}dt\) (see [7], for details).
If we have additional information on the factors in the factorization formula (4) we can state more, using the same method.
Theorem 3
Let \(f\) be an analytic function in \(\mathbb {D}\) of the form (4), \(\psi \) is a positive non-decreasing function such that \(\psi (2x)=O(\psi (x))\) on \([1,+\infty ).\) If
and the counting functions of the zeros of \(f\) satisfy
then
where \(\tilde{\psi }(x)=\int _1^x \frac{\psi (t)}{t}dt.\)
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We would like to thank the anonymous referee for careful reading of the manuscript and valuable remarks.
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Communicated by James K. Langley.
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Chyzhykov, I., Kravets, M. On the Minimum Modulus of Analytic Functions of Moderate Growth in the Unit Disc. Comput. Methods Funct. Theory 16, 53–64 (2016). https://doi.org/10.1007/s40315-015-0118-y
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DOI: https://doi.org/10.1007/s40315-015-0118-y
Keywords
- Analytic function
- Minimum modulus
- Order of growth
- Factorization
- Zero distribution
- Canonical product
- Harmonic function