Jointly maximal products in weighted growth spaces

0.13 MB·English

by  Janne Gröhn

JOINTLY MAXIMAL PRODUCTS IN WEIGHTED GROWTH
SPACES
2
1
0 JANNEGRO¨HN,JOSE´ A´NGEL PELA´EZ, ANDJOUNI RA¨TTYA¨
2

t
c Abstract. Itisshownthatforanynon-decreasing,continuousandunbounded
O doubling function ω on [0,1), there exist two analytic infinite products f
0
and f1 such that the asymptotic relation |f0(z)|+|f1(z)| ≍ ω(|z|) is satis-
1
1 fije=d f0o,r1aslaltizsfyinTt(hre,fujn)it≍dliosgc.ω(Irt),isasalrso→sh1o−w,natnhdathebnoctehgfiuvnecetixoanmspfljesfoorf
] analytic functions for which the Nevanlinna characteristic admits the regu-
V lar slow growth induced by ω. The applied construction produces two recent
C results concerning regularly growing analytic functions in the unit disc.
.
h
t
a
m
1. Introduction and results
[

1 Let H(D) denote the algebra of all analytic functions in the unit disc D of the
v complex plane C. To consider the growth and the zero distribution of functions
8 in H(D), we use the following classical notation. The non-integrated counting
1
3 function n(r,f,0) counts the zeros of f in {z ∈ C : |z| ≤ r} according to mul-
3 tiplicities. Quantities M (r,f), M (r,f), where 0 < p < ∞, N(r,f,a), where
∞ p
.
0 a ∈ C, and T(r,f) denote the maximum modulus of f, the Lp-mean of f, the
1
integrated counting function of a-points of f and the Nevanlinna characteristic
2
1 of f, respectively. We also employ the notation a ≍ b, which is equivalent to the
: conditions a . b and b . a, where the former means that there exists a constant
v
i C > 0 such that a ≤ Cb, and the latter is defined analogously.
X
Let ω : [0,1) → (0,∞) be non-decreasing, continuous and unbounded. Such a
r
a function ω is said to be doubling, if there exists a constant B > 1 such that
(1) ω(1−r/2) ≤ Bω(1−r), 0 < r ≤ 1.
Thefollowingresultshowsthatforanydoublingfunctionω thereexisttwojointly
maximal products in the sense that the sum of their moduli behaves asymptoti-
cally as ω(|z|) in D.
Date: October12, 2012.
2010 Mathematics Subject Classification. Primary 30J99.
Key words and phrases. Doublingfunction and infiniteproduct and zero distribution.
This research was supported in part by the Ram´on y Cajal program of MICINN (Spain),
Ministerio de Educaci´on y Ciencia, Spain, (MTM2011-25502), from La Junta de Andaluc´ıa,
(FQM210) and(P09-FQM-4468), MICINN-Spainref. MTM2011-26538, EuropeanNetworking
Programme HCAA of theEuropean Science Foundation and theFinnish Cultural Foundation.
1
2 JANNEGRO¨HN,JOSE´ A´NGELPELA´EZ,ANDJOUNIRA¨TTYA¨
Theorem 1. Let ω :[0,1) → (0,∞) be doubling. Then, there exist f ,f ∈ H(D)
0 1
such that
(2) |f (z)|+|f (z)| ≍ ω(|z|), z ∈D,
0 1
where both functions f for j = 0,1 satisfy n(r,f ,0) = O (1−r)−1 , as r → 1−.
j j
Moreover,
(cid:0) (cid:1)
(3) M (r,f ) ≍ ω(r), r → 1−,
p j
for all 0 < p ≤ ∞, and
T(r,f )≍ N(r,f ,a) ≍ log ω(r), r → 1−,
j j
for all a ∈ C.
Theasymptoticrelation (2)reproducesarecentresult[1,Lemma1]concerning
regularly growing analytic functions in D. The proof of [1, Lemma 1] rests upon
the use of lacunary series, which have been the key tool to solve similar prob-
lems in the existing literature. The pioneering result [7, Proposition 5.4], which
concerns (2) for ω(r) = 1/(1−r), have been a source of inspiration for several
authors. For example, [4, Theorem 1.2] proves (2) for ω(r) = log 1/(1−r) . It
is well known that (2) has many applications in the operator theory; for details,
(cid:0) (cid:1)
we refer to [1].
The main advantage of our self-contained and constructive proof of Theorem 1
is that the zero distribution of the products f and f is explicit. These products
0 1
aresimilar tothoseappliedtostudythezerodistributionoffunctionsinweighted
Bergman spaces [6, Section 3]. Note also that our argument gives an alternative
way to prove [6, Theorem 3.15], whose original proof is based on certain lacunary
series. In fact, Theorem 1 generalizes [6, Theorem 3.15] to doubling functions.
The current state of the development concerning complex linear differential
equation f′′+A(z)f = 0 in D allows us to deduce a significant amount of infor-
mation on solutions f, whenever we can analyze the coefficient A in detail. If
we take A to be one of the functions f and f in the Theorem 1, then we get
0 1
an important and intriguing family of examples of such differential equations.
These particular equations are way too complicated to be solved explicitly, but
the growth and the oscillation of their solutions are well understood due to the
asymptotic properties satisfied by the coefficient A. To be brief with regards to
this matter, we settle to mention two cases in the recent literature of which the
first one concerns polynomial regular functions. This class of regularly growing
analytic functions in D arises naturally in the theory of ODEs [3]. In the sense
of linear differential equations, polynomial regular functions play a similar role
in the unit disc as polynomials do in the complex plane. For a more general
example, see [2].
2. Proof of Theorem 1
The proof of Theorem 1 is divided in several steps. The point of departure is
the construction of the infinite products f ,f ∈ H(D), which is followed by a
0 1
JOINTLY MAXIMAL PRODUCTS IN WEIGHTED GROWTH SPACES 3
discussionoftheirgrowth. Finally,weconsidertheassertedasymptoticproperties
of the products f and f .
0 1
2.1. Construction of the products. Before going into the details of the con-
struction, we note the following lemmas on doubling functions.
Lemma 2. Let ω : [0,1) → (0,∞) be doubling. If B > 1 is the constant in (1),
then
α
1−r
(4) ω(t) ≤ C ω(r), 0 ≤ r ≤ t < 1,
1−t
(cid:18) (cid:19)
where C = max Bω(1/2)/ω(0),B2 and α = log B.
2
Conversely, it(cid:8)is obvious that, if ω(cid:9):[0,1) → (0,∞) is non-decreasing, continu-
ous, unbounded and it satisfies (4) for some C > 1 and α > 0, then ω must be
doubling.
Proof of Lemma 2. Since ω is doubling, (1) implies ω(t) ≤ Bω(2t − 1) for all
t ∈ [2−1,1). Moreover, if 0 ≤ r ≤ t < 1, then there exist unique constants
j,k ∈ N∪{0},j ≥ k,suchthatt ∈ 1−2−j,1−2−j−1 andr ∈ 1−2−k,1−2−k−1 .
If k = 0, then
(cid:2) (cid:1) (cid:2) (cid:1)
Bjω(1/2) Bω(1/2) 1−r log2B
ω(t) ≤ Bjω 2j(t−1)+1 ≤ ω(r)≤ ω(r),
ω(0) ω(0) 1−t
(cid:18) (cid:19)
(cid:0) (cid:1)
while if k > 0, then
log B
1−r 2
ω(t) ≤ Bj−k+1ω 2j−k+1(t−1)+1 ≤ Bj−k+1ω(r)≤ B2 ω(r).
1−t
(cid:18) (cid:19)
The assertion follo(cid:0)ws. (cid:1) (cid:3)
The second lemma introduces a sequence of natural numbers depending on
the growth of the doubling function ω. This sequence is the foundation of our
construction.
Lemma 3. Let ω : [0,1) → (0,∞) be doubling. Then, there exist a sequence
{n }∞ of natural numbers, real constants λ and µ, and a constant d ∈ (0,1)
k k=1
such that the sequence {a }∞ , defined by
k k=1
ω 1−1/n
a = k+2 , k ∈N,
k
ω 1−1/n
(cid:0) k (cid:1)
satisfies
(cid:0) (cid:1)
loga n
(5) 1 < λ ≤ a ≤ µ < ∞, k+1 < d k+1, k ∈ N.
k
loga n
k k
Proof. Let α > 0 and C > 1 be the constants ensured by Lemma 2. Now, let γ
be a sufficiently large real constant such that
2γ +α+log C 1 2γ/α
(6) 2γ−αC−1 > 1, 2 < −1 .
2γ −α−log C 21/α C1/α
2 !
4 JANNEGRO¨HN,JOSE´ A´NGELPELA´EZ,ANDJOUNIRA¨TTYA¨
Taket = 1/2, anddefinethesequence{t }∞ inductivelybyω(t )/ω(t )= 2γ
1 k k=1 k+1 k
for k ∈ N. Let n = floor (1−t )−1 , where floor(x) = max{n ∈ N : n ≤ x}.
k k
By means of Lemma 2, and the estimates 2−1 ≤ n (1−t ) ≤ 1, we may define
(cid:0) (cid:1) k k
1 < λ < µ < ∞ by
Cω(t ) ω(t ) ω(t )
a ≤ k+2 ≤ 2αC k+2 k+1 = 22γ+αC = µ,
k α
n (1−t ) ω(t ) ω(t ) ω(t )
k k k k+1 k
and
(cid:0) (cid:1)
α
n (1−t ) ω(t ) 1 ω(t ) ω(t )
a ≥ k+2 k+2 k+2 ≥ k+2 k+1 = 22γ−αC−1 = λ,
k Cω(t ) 2αC ω(t ) ω(t )
(cid:0) k(cid:1) k+1 k
since these inequalities hold for all k ∈ N. By Lemma 2 we conclude
n 1 1−t
k+1 k
> −1 (1−t )> −1
k
n 1−t 1−t
(7) k (cid:18) k+1 (cid:19) k+1
1 ω(t ) 1/α 2γ/α
≥ k+1 −1 = −1, k ∈ N,
C1/α ω(t ) C1/α
(cid:18) k (cid:19)
and further by (6), we have
loga logµ 2γ +α+log C 1 2γ/α 1 n
k+1 ≤ = 2 < −1 < k+1.
loga logλ 2γ −α−log C 21/α C1/α 21/α n
k 2 ! k
This confirms the last inequality in (5) for d = 2−1/α. (cid:3)
Let {n }∞ be the sequence ensured by Lemma 3, and define
k k=1
f (z) = ∞ 1+a2k+jzn2k+j, z ∈ D, j =0,1.
j 1+a−1 zn2k+j
k=1 2k+j
Y
Evidently both functions f belong to H(D), since all factors are bounded func-
j
tions in D, and according to (5) the sum
∞ 1+a2k+jzn2k+j −1 ≤ ∞ a2k+j −a−2k1+j |z|n2k+j ≤ (1+µ) ∞ |z|n2k+j
(cid:12)1+a−1 zn2k+j (cid:12) 1−a−1
Xk=1(cid:12) 2k+j (cid:12) Xk=1 2k+j Xk=1
converg(cid:12)(cid:12)es uniformly on com(cid:12)(cid:12)pact subsets of D.
(cid:12) (cid:12)
2.2. Growth estimates for the maximum modulus of the products. To
estimate the growth of fj for j = 0,1 we define r2m+j = e−1/n2m+j for m ∈ N,
and write
m a−1 +zn2k+j ∞ 1+a zn2(m+k)+j
2k+j 2(m+k)+j
(8) |f (z)| = a .
j (cid:12) 2k+j1+a−1 zn2k+j(cid:12)(cid:12) 1+a−1 zn2(m+k)+j(cid:12)
(cid:12)kY=1 2k+j (cid:12)(cid:12)kY=1 2(m+k)+j (cid:12)
First, we prove t(cid:12)(cid:12)hat the infinite subproduc(cid:12)(cid:12)t(cid:12)(cid:12)in (8) is bounded in D. To(cid:12)(cid:12)this end,
(cid:12) (cid:12)(cid:12) (cid:12)
let τ = 2γ/αC−1/α−1 be the lower bound in (7). According to (6) we know that
τ > 1, and
n n n
(9) 2(m+k)+j = 2(m+k)+j ··· 2m+j+1 ≥ τ2k, k,m ∈N.
n n n
2m+j 2(m+k)+j−1 2m+j
JOINTLY MAXIMAL PRODUCTS IN WEIGHTED GROWTH SPACES 5
Since h (x) = (y+x)/(1+yx) is increasing on [0,1) for each y ∈ [0,1), we obtain
1
1+a zn2(m+k)+j a−1 +zn2(m+k)+j
2(m+k)+j 2(m+k)+j
= a
(cid:12)1+a−1 zn2(m+k)+j(cid:12) 2(m+k)+j(cid:12)1+a−1 zn2(m+k)+j(cid:12)
(cid:12) 2(m+k)+j (cid:12) (cid:12) 2(m+k)+j (cid:12)
(cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12)a−1 +|z|n2(m+k)+j(cid:12)(cid:12)
(cid:12) (cid:12) ≤ a (cid:12) 2(m+k)+j (cid:12)
(10) 2(m+k)+j 1+a−1 |z|n2(m+k)+j
2(m+k)+j
n2(m+k)+j
1+a 1 n2m+j
< 2(m+k)+j e
n2(m+k)+j
1+a−1 (cid:0)1(cid:1) n2m+j
2(m+k)+j e
for |z| < r and k,m ∈ N. Moreover, since h (x(cid:0),y(cid:1)) = (1+xy)/(1+x−1y) is
2m+j 2
increasinginbothvariables,providedthatx > 1and0 ≤ y < 1,estimates(5),(9)
and (10) imply
∞ 1+a zn2(m+k)+j ∞ 1+µ 1 τ2k
(11) 2(m+k)+j < e ≤ C⋆ < ∞,
(cid:12)(cid:12)kY=11+a−2(1m+k)+jzn2(m+k)+j(cid:12)(cid:12) kY=11+µ−1(cid:0) (cid:1)1e τ2k
(cid:12) (cid:12)
for |z| <(cid:12)(cid:12)r2m+j and m ∈ N, where C(cid:12)(cid:12)⋆ > 0 is a const(cid:0)an(cid:1)t independent of m ∈ N.
Second, we proceed to derive an upper estimate for the maximum modulus of f .
j
By means of (5), (8), (11) and the inequality 1−x ≤ e−x for x ≥ 0, we get
m ω 1−1/n ω 1−1/n
|f (z)| < C⋆ a = C⋆ 2(m+1)+j ≤ C⋆µ 2m+j
j 2k+j
ω 1−1/n ω 1−1/n
k=1 (cid:0) 2+j (cid:1) (cid:0) 2+j (cid:1)
Y
(12) ≤ C⋆µ ω(r2m+j) , |(cid:0)z| < r ,(cid:1) m ∈ N. (cid:0) (cid:1)
2m+j
ω 1−1/n
2+j
If |z| ≥ r , then r(cid:0) ≤(cid:1)|z| < r for some m ∈ N \{1}. Note that
2+j 2(m−1)+j 2m+j
by (9) there exists t ∈ N such that n > 2n for all m ∈ N. Since
2(m+t)+j 2m+j
e−x ≤ 1−x/2 for 0≤ x ≤ 1, we conclude
r ≤ 1−(2n )−1 < 1−1/n , m ∈ N.
2m+j 2m+j 2(m+t)+j
Then (5), (12) and the inequality 1−x ≤ e−x for 0 ≤ x ≤ 1, give
ω 1−1/n ω 1−1/n
|f (z)| < C⋆µ 2(m+t)+j ≤ C⋆µ2+t 2(m−1)+j
j
ω 1−1/n ω 1−1/n
(cid:0) 2+j (cid:1) (cid:0) 2+j (cid:1)
≤ C⋆µ2+t ω(cid:0) r2(m−1)+j(cid:1) ≤ C⋆µ2+t ω(cid:0)|z| . (cid:1)
ω 1−1/n ω 1−1/n
(cid:0) 2+j(cid:1) (cid:0) (cid:1)2+j
Consequently, the maximu(cid:0)m modulus(cid:1)of f satisfies(cid:0) (cid:1)
j
(13) M (r,f )= max f (z) . ω(r), 0≤ r < 1.
∞ j j
|z|=r
(cid:12) (cid:12)
(cid:12) (cid:12)
6 JANNEGRO¨HN,JOSE´ A´NGELPELA´EZ,ANDJOUNIRA¨TTYA¨
2.3. Growth estimates for the minimum modulus of the products. The
following discussion shows that the difference between the maximum modulus
and the minimum modulus of f for j = 0,1 is small in a large subset of the
j
unit disc. Define E = ∞ I , where I is the closed interval whose
j m=1 2m+j 2m+j
endpoints are
S
minI = a−n−2m1+j 1−δ a−n−2(1m+1)+j δ, m ∈ N,
2m+j 2m+j 2(m+1)+j
(cid:18) (cid:19) (cid:18) (cid:19)
and
maxI = a−n−2m1+j δnn2m2m++1+jj a−n−2(1m+1)+j 1−δn2nm2m++1+jj , m ∈ N.
2m+j 2m+j 2(m+1)+j
(cid:18) (cid:19) (cid:18) (cid:19)
Here 0 < δ < 1 is a sufficiently small constant, which is to be determined later.
According to(5) all elements inthesequence a−1/nm ∞ belongtotheinterval
m m=1
(0,1), this sequence is strictly increasing, and it converges to 1, as m → ∞.
(cid:8) (cid:9)
Moreover, I ⊂ a−1/n2m+j,a−1/n2(m+1)+j for all m ∈ N. First, we prove that
2m+j 2m+j 2(m+1)+j
the infinite subproduct in (8) is uniformly bounded away from zero for |z| ∈ E .
(cid:0) (cid:1) j
If |z| ∈ I2m+j, then |z|n2(m+k)+j < a−2(1m+k)+j for all k ∈ N, and therefore
1+a zn2(m+k)+j a−1 −|z|n2(m+k)+j
2(m+k)+j 2(m+k)+j
≥ a
(cid:12)1+a−1 zn2(m+k)+j(cid:12) 2(m+k)+j 1−a−1 |z|n2(m+k)+j
(14) (cid:12) 2(m+k)+j (cid:12) 2(m+k)+j
(cid:12)(cid:12) (cid:12)(cid:12) 1−a2(m+k)+j|z|n2(m+k)+j
(cid:12) (cid:12) =
1−a−1 |z|n2(m+k)+j
2(m+k)+j
for |z| ∈ I and k, m ∈ N. Since h (x,y) = (1−xy)/(1−x−1y) is decreasing
2m+j 3
in both variables, when x > 1 and 0 ≤ y < 1, estimates (5), (9) and (14) imply
that there exists a constant C∗ > 0, independent of m ∈ N, such that
∞ 1+a zn2(m+k)+j
2(m+k)+j
(cid:12) 1+a−1 zn2(m+k)+j(cid:12)
(cid:12)kY=1 2(m+k)+j (cid:12)
(cid:12)(cid:12)(cid:12) 1−a a−(cid:12)(cid:12)(cid:12)δ a−nn22(mm++11)++jj(cid:18)1−δn2nm2m++1+jj(cid:19) nn22(mm++k1+)+jj
∞ 2(m+k)+j 2m+j 2(m+1)+j
!
(15) ≥
kY=11−a−1 a−δ a−nn22(mm++11)++jj(cid:18)1−δn2nm2m++1+jj(cid:19) nn22(mm++k1+)+jj
2(m+k)+j 2m+j 2(m+1)+j
!
∞ 1−µ λ−δ τ2k−1
≥ ≥ C∗, |z| ∈ I , m ∈ N.
k=11−µ−(cid:0)1 λ−(cid:1)δ τ2k−1 2m+j
Y
(cid:0) (cid:1)
JOINTLY MAXIMAL PRODUCTS IN WEIGHTED GROWTH SPACES 7
Second, we proceed to estimate the minimum modulus of f on E . Note that
j j
the last inequality in (5) implies
− 1 (1−δ) − 1 δ n2k+j
a |z|n2k+j ≥ a a n2m+j a n2(m+1)+j
2k+j 2k+j 2m+j 2(m+1)+j
!
a a
2k+j 2k+j
= ≥
(16) ann22mk++jj(1−δ)an2(nm2k++1j)+jδ ad2k2(+mj−k)(1−δ)ad2k2(+mj+1−k)δ
2m+j 2(m+1)+j
a
≥ 2k+j = aδ(1−d) ≥ λδ(1−d) > 1
a1−δ adδ 2k+j
2k+j 2k+j
for |z| ∈ I2m+j when 1 ≤ k ≤ m, and m ∈ N; in particular, |z|n2k+j > a−2k1+j.
Moreover, choose t ∈ N sufficiently large such that 1−λ−1 ≥ τ−2t. Since h (x) =
4
1−(1−a)x−1−a1/x ≥ 0 for all x ∈ [1,∞), provided that a ∈ (0,1), by applying
(5) and (9), we obtain
|z| ≤ a−1/n2(m+1)+j ≤ λ−1/n2(m+1)+j ≤ 1− 1−λ−1 n−1
2(m+1)+j 2(m+1)+j
(17)
≤ 1−τ−2tn−1 ≤ 1−1/n (cid:0) , |z(cid:1)| ∈ I , m ∈ N.
2(m+1)+j 2(m+1+t)+j 2m+j
Therefore (8), (15) and (17) yield
m a−1 +zn2k+j
|f (z)| ≥ C∗ a 2k+j
j 2k+j(cid:12)1+a−1 zn2k+j(cid:12)
kY=1 (cid:12) 2k+j (cid:12)
ω 1−1/n(cid:12)(cid:12) m a(cid:12)(cid:12)−1 +zn2k+j
= C∗ (cid:12)2(m+1)+j (cid:12)2k+j
(18) (cid:0)ω 1−1/n2+j (cid:1) kY=1(cid:12)(cid:12)1+a−2k1+jzn2k+j(cid:12)(cid:12)
ω 1(cid:0)−1/n (cid:1) m(cid:12)(cid:12) |z|n2k+j −a−1(cid:12)(cid:12)
≥ C∗ 2(m+1+t)+j (cid:12) 2k+(cid:12)j
(cid:0)µtω 1−1/n2+j (cid:1) k=11−a−2k1+j|z|n2k+j
Y
ω(cid:0)(|z|) (cid:1)m |z|n2k+j −a−1
≥ C∗ 2k+j
µtω 1−1/n2+j k=11−a−2k1+j|z|n2k+j
Y
(cid:0) (cid:1)
for |z| ∈ I and m ∈ N. For our purposes, it suffices to show that the product
2m+j
in the last line of (18) is uniformly bounded away from zero for |z| ∈ E . To
j
simplify computations, we prove that the reciprocal of this product is uniformly
bounded for such values of z. Now, since logx ≤ x−1 for x ≥ 1, we have
m 1−a−1 |z|n2k+j m 1−a−1 |z|n2k+j
2k+j 2k+j
= exp log
|z|n2k+j −a−1 |z|n2k+j −a−1 !
k=1 2k+j k=1 2k+j
Y X
m 1−|z|n2k+j
≤ exp (1+µ)
a2k+j|z|n2k+j −1!
k=1
X
8 JANNEGRO¨HN,JOSE´ A´NGELPELA´EZ,ANDJOUNIRA¨TTYA¨
for |z| ∈ I and m ∈ N. By means of (9), (16), and the estimate 1−e−x ≤ x
2m+j
for x ≥0, we conclude
m 1−a−1 |z|n2k+j 1+µ m
2k+j ≤ exp (1−|z|n2k+j)
|z|n2k+j −a−1 λδ(1−d) −1 !
k=1 2k+j k=1
Y X
m
1+µ
≤ exp 1−a−n2k+j/n2m+j
λδ(1−d) −1 2m+j !
Xk=1(cid:16) (cid:17)
m
1+µ n
2k+j
≤ exp logµ
λδ(1−d) −1 n2m+j!
k=1
X
∞
1+µ 1
≤ exp logµ
λδ(1−d) −1 τk
!
k=0
X
for |z| ∈ I and m ∈ N, which gives the desired uniform lower bound for the
2m+j
product in the last line of (18). Hence, by (13) and (18), we get
∞
(19) |f (z)| ≍ ω(|z|), |z| ∈ E = I .
j j 2m+j
m=1
[
2.4. The covering property of the sets where the products are maximal.
It remains to prove that the sets E and E induce a covering of [minI ,1). Note
0 1 2
that the closed intervals {I }∞ are pairwise disjoint, which is also true for
2m m=1
{I }∞ . Consequently, it is sufficient to show that
2m+1 m=1
(20) minI ≤ maxI , minI ≤ maxI , m ∈ N.
2m+1 2m 2(m+1) 2m+1
We proceed to prove the first inequality in (20). By the definition of I , the
2m+j
first inequality in (20) is equivalent to
− 1−δ − δ − δ − 1 1−δ n2m
(21) a n2m+1a n2m+3 ≤ a n2m+1a n2m+2(cid:16) n2m+1(cid:17), m ∈ N.
2m+1 2m+3 2m 2m+2
By taking the logarithm to the base a on the both sides of (21), and then
2m
solving the resulting inequality with respect to δ, we conclude that the first
inequality in (21) is valid if and only if δ ≤ T(m) for all m ∈ N, where
− 1 − 1
log a n2m+2 −log a n2m+1
T(m) = a2m 2m+2 a2m 2m+1 .
− 1 − 1 − n2m
1 −log a n2m+1 +log a n2m+3 +log a n2m+1n2m+2
n2m+1 a2m 2m+1 a2m 2m+3 a2m 2m+2
Note that the denominator of T(m) can be written in the form
− 1 − 1
log a n2m+3 −log a n2m+1
a2m 2m+3 a2m 2m+1
n − 1 − 1
+ 2m log a n2m+2 −log a n2m > 0, m ∈ N,
n a2m 2m+2 a2m 2m
2m+1 (cid:18) (cid:19)
JOINTLY MAXIMAL PRODUCTS IN WEIGHTED GROWTH SPACES 9
and hence T(m) is strictly positive for all m ∈ N. By means of (5) we get
− 1
(d−1)log a n2m+1 (1−d)log a
T(m)≥ a2m 2m+1 = a2m 2m+1, m ∈N.
− 1 1+log a
1 −log a n2m+1 a2m 2m+1
n2m+1 a2m 2m+1
This implies that, if
(1−d)log λ
µ
(22) 0 < δ ≤ ,
1+log µ
λ
then the first inequality in (20) is satisfied for all m ∈ N. The second inequality
in (21) follows by a similar argument, and the choice (22) for δ is again adequate.
We conclude that
(23) |f (z)|+|f (z)| ≍ ω(|z|)
0 1
for |z| ≥ minI . Finally, since a−1/nm ∞ is strictly increasing, (23) holds also
2 m m=1
for |z| ≤ minI , and hence f and f are analytic functions satisfying (2).
2 0 1
(cid:8) (cid:9)
2.5. Asymptotic properties of the products. Product f for j = 0,1 has
j
exactly n simple zeros on the each circle z : |z| = s , where s =
2m+j 2m+j 2m+j
a−1/n2m+j for m ∈N. Therefore, we obtain
2m+j (cid:8) (cid:9)
m m
1
n ≤ n(s ,f ,0) = n = n
2m+j 2m+j j 2k+j 2m+j n2m+j
Xk=1 Xk=1 n2k+j
∞
1
≤ n . n , m ∈ N,
2m+j τk 2m+j
k=0
X
by (9). By applying the estimates 1−x < logx−1 < 2(1−x), which are valid
for 4−1 < x < 1, it follows that n(s ,f ,0) ≍ (1−s )−1 for all m ∈ N.
2m+j j 2m+j
Consequently,
n(r,f ,0) = O (1−r)−1 , r → 1−.
j
Now we observe that E ∪E = [minI ,1), so it is not possible that d(E ) =
0 1 (cid:0) 2 (cid:1) 0
d(E )= 0, where
1
m F ∩[r,1)
d(F) = liminf
r→1− (cid:0) 1−r (cid:1)
is the lower density of the set F ⊂ [0,1), and where m denotes the Lebesgue
measure. Consequently, forsomej = 0,1wehaved(E )> 0, whichtogether with
j
the nature of the sets E , implies that d(E )> 0 for both j = 0,1. Consequently,
j j
(19) holds outside a set E⋆ = [0,1)\E , which satisfies
j j
m E⋆∩[r,1)
d(E⋆)= limsup j < 1, j = 0,1.
j r→1− (cid:0) 1−r (cid:1)
So, (19), [5, Lemma 4.3] and Lemma 2 yield M (r,f ) ≍ w(r), as r → 1−, where
p j
the constants in the asymptotic relation are independent of 0 < p ≤ ∞. This
proves (3). On the other hand, for any a ∈ C, Jensen’s formula and (19) imply
10 JANNEGRO¨HN,JOSE´ A´NGELPELA´EZ,ANDJOUNIRA¨TTYA¨
that N(r,f ,a) ≍ log ω(r) for r ∈ [0,1)\E⋆. The fact that the same estimate
j 0
holds also without the exceptional set E⋆ follows again from [5, Lemma 4.3] and
0
Lemma 2. Furthermore, log ω(r) ≍ N(r,f ,0) . T(r,f ) ≤ logM (r,f ) ≍
j j ∞ j
log ω(r), as r → 1−, again by Jensen’s formula. This completes the proof of
Theorem 1.
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E-mail address: [email protected]