Wiman’s type inequality for some double power series

УДК 517.55

A. O. Kuryliak, L. O. Shapovalovska, O. B. Skaskiv

WIMAN’S TYPE INEQUALITY FOR SOME DOUBLE POWER SERIES

A. O. Kuryliak, L. O. Shapovalovska, O. B. Skaskiv.Wiman’s type inequality for some double power series, Mat. Stud.39_{(2013), 134–141.}

In this paper we prove some analogue of Wiman’s inequality for analytic functionsf(z₁, z₂) in the domainT_={z∈C2_{:|z1|<1, |z2|<+∞}. The obtained inequality is sharp.}

А. О. Куриляк, Л. О. Шаповаловска, О. Б. Скаскив. Неравенство типа Вимана для некоторых двойных степенных рядов // Мат. Студiї. – 2013. – Т.39, №2. – C.134–141.

Доказывается аналог неравенства Вимана для функцийf(z₁, z₂),аналитических в об-ластиT_={z∈C2_{: |z1|<1, |z2|<+∞}. Полученное неравенство точное.}

1. Introduction. Let f be analytic function in the disc _{z: _|z_| < R_}, 0 < R _≤ +_∞,

represented by the power series

f(z) = +∞

X

n=0

anzn. (1)

Forr _∈(0, R)we denote Mf(r) = max{|f(z)|: |z|=r}, µf(r) = max{|an|rn: n ≥0}.

It is well known ([1, p. 9], [2, p. 28]) that for each nonconstant entire function f(z) and every ε >0 there exists a setE(ε, f)_⊂[1,+_∞)such that Wiman’s inequality

Mf(r)≤µf(r)(lnµf(r))1/2+ε

holds for all r _∈ [1,+_∞)_\E(ε, f), where the set E(ε, f) has finite logarithmic measure, i.e.

R

E(ε,f)

dr

r <+∞.

Let R = 1, i.e. f(z) be an analytic function in the unit disc D_{={z: |z| <1}. For such}

a functionf(z)and everyδ >0there exists a setEf(δ)⊂(0,1)of finite logarithmic measure

on(0,1), i.e.

Z

Ef(δ) dr

1₋r <+∞,

such that for all r_∈(0,1)_\Ef(δ)the inequality

Mf(r)≤

µf(r)

(1₋r)1+δ ln

1/2+δ µf(r)

1₋r

holds ([3]). Similar inequality for analytic function in the unit disc one can find in [4].

2010Mathematics Subject Classification:30B20, 30D20.

Keywords:maximum modulus; maximal term; double power series; Wiman’s type inequality.

c

Also in [3] it is noted that for the function g(z) = P+∞

n=1exp{nε}zn, ε∈(0,1) we have

lim

r→1−0

Mg(r) µg(r)

1−_r ln

1/2µg(r)

1−_r

≥C > 0.

Some analogues of Wiman’s inequality for entire functions of several complex variables one can find in [5]–[11].

The aim of this paper is to prove some analogues of Wiman’s inequality for analytic functions represented by the series

f(z) =f(z1, z2) = +∞

X

n+m=0

anmz1nz2m (2)

with the domain of convergence T_{={z ∈}C2_{: |z}

1|<1, z2 ∈C}.

By_A2_{we denote the class of analytic functions of form (2) with the domain of convergence}

T _and ∂

∂z2f(z1, z2)6≡0in

T_.

2. Wiman’s type inequality for analytic functions in T_{.Forr = (r1, r2)∈T:=[0,1)×}

[0,+_∞) and function f _{∈ A}2 _{we denote}

△r={(t1, t2)∈T: t1 > r1, t2 > r2}, Mf(r) = max{|f(z)|: |z1| ≤r1,|z2| ≤r2},

µf(r) = max{|anm|rn1r2m: (n, m)∈Z2+}, Mf(r) =

+∞

X

n+m=0

|anm|r1nr2m.

Let Df(r) = (Dij)be a 2×2 matrix such that

Dij =ri

∂ ∂ri

rj

∂ ∂rj

lnM_f(r)

=∂i∂jlnMf(r), ∂i =ri

∂ ∂ri

, i, j _{∈ {}1,2_}.

The following statement can be proved by verbatim repetition of the proof of Theorem 3.1 from [5].

Theorem 1. Letf _{∈ A}2_{. There exists an absolute constantC}

0 such that

M_{f(r)≤C0µf(r)(det(Df(r) +I))}1/2_,

where I is the identity2_×2 matrix.

We say that E _⊂T is set ofasymptotically finite logarithmic measure on T if there exists

r0 ∈T such that

νln(E_{∩ △}r0):=

Z Z

E∩△_r

0

dr1dr2 (1₋r1)r2

<+_∞,

i.e. the set E_{∩ △}r0 is a set of finite logarithmic measure on T.

Lemma 1. Letδ > 0 and let h: R2

+→ R+ be an increasing function on each variable such

that

+∞

Z

1 +∞

Z

1

du1du2

h(u1, u2)

Then there exists a set E _⊂T of asymptotically finite logarithmic measure such that for all

r_∈T_\E the inequalities

det(Df(r) +I)≤

1 1₋r1 ·

h ∂

∂r1

lnM_{f(r), r2} ∂ ∂r2

lnM_{f(r) + lnr2, (3)}

∂ ∂r1

lnM_f(r)≤ 1

1₋r1 ·

hlnM_{f(r),lnr2, (4)}

∂ ∂r2

lnM_f(r)≤ 1

r2(1₋r1)δ(lnMf(r))

1+δ ₍₅₎

hold.

Proof. LetE1 ⊂T be a set for which inequality (3) does not hold. Now we prove thatE1 is a set of asymptotically finite logarithmic measure. Sincerj_∂r∂_jlnMf(r) is increasing to +∞

forj _{∈ {}1,2_} asr1 →1−0, r2 →+∞, there existsr0 ∈T such that r01 > 12, r 0

2 >1and for allr _{∈ △}r0 we have

rj

∂ ∂rj

lnM_{f(r) + lnrj >1.}

Then

νln(E1∩ △r0) =

Z Z

E1∩△_r0

dr1dr2 (1₋r1)r2 ≤

Z Z

E1∩△_r0

det(Df(r) +I)(1−r1)dr1dr2

h ∂

∂r1 ln

M_{f(r), r2} ∂

∂r2 ln

M_{f(r) + lnr2(1−r1)r2} ≤

≤

Z Z

E1∩△_r0

1

r1r2 ·

det(Df(r) +I)dr1dr2

hr1_∂r∂1 ln

M_{f(r) + lnr1, r2} ∂

∂r2 ln

M_{f(r) + lnr2} .

Let U: T _→ R2

+ be a mapping such that U = (u1(r), u2(r)) and uj(r) = rj_∂r∂_j lnMf(r) +

lnrj, j ∈ {1,2}, r= (r1, r2).Then for i, j ∈ {1,2} we obtain

∂ui

∂ri

= ∂

∂ri

ri

∂ ∂ri

lnM_{f(r) + lnri=} 1 ri

∂i∂ilnMf(r) +

1

ri,

∂ui

∂rj

= ∂

∂rj

ri

∂ ∂ri

lnM_{f(r) + lnri}

= 1

rj

∂i∂jlnMf(r), i6=j.

So, the Jacobian

J1:=

D(u1, u2)

D(r1, r2) =

∂u1

∂r1

∂u1

∂r2

∂u2

∂r1

∂u2

∂r2

= det(Df(r) +I)

1

r1r2

.

Therefore,

νln(E1∩ △r0)≤

Z Z

U(E1∩△_r0)

du1du2

h(u1, u2) ≤ +∞

Z

1 +∞

Z

1

du1du2

h(u1, u2)

<+_∞.

Suppose that E2 ⊂T is a set for which inequality (4) does not hold. Then we can choose

r0 _{∈T so that lnM}

f(r0)>1 and r02 > e.

νln(E2∩ △r0) =

Z Z

E2∩△_r0

dr1dr2 (1₋r1)r2 ≤

Z Z

E2∩△_r0

∂ ∂r1 ln

M_{f(r)·(1−r1)dr1dr2}

Consider the mapping V : T _→ R2

+, where V = (v1(r), v2(r)) and v1 = lnMf(r), v2 = lnr2, r= (r1, r2). So,

J2:=D(v1, v2)

D(r1, r2) =

∂ ∂r1 ln

M_f(r) ∂

∂r2 ln

M_f(r)

0 _r1₂

= 1

r2·

∂ ∂r1

lnM_f(r).

Therefore

νln(E2∩ △r0)≤

Z Z

E2∩△_r0

∂ ∂r1 ln

M_f(r)dr1dr2

h(lnM_f(r),lnr2)r2 = Z Z

V(E2∩△_r0)

du1du2

h(u1, u2) ≤ +∞

Z

1 +∞

Z

1

du1du2

h(u1u2)

<+_∞.

Let E3 ⊂T be a set for which inequality (5) does not hold. Then

νln(E3∩ △r0)≤

Z Z

E3∩△_r0

∂ ∂r2 ln

M_f(r)r2dr1dr2

1 (1−_r1)δ ln

1+δ_M

f(r)(1−r1)r2

.

Letr0be such thatlnMf(r0)>1.Define the mappingW: T →T,whereW = (w1(r), w2(r))

and w1 =r1, w2 = lnMf(r), r= (r1, r2). So,

J3:=D(w1, w2)

D(r1, r2) =

1 0

∂ ∂r1 ln

M_f(r) ∂

∂r2 ln

M_f(r)

= ∂

∂r2

lnM_f(r).

Therefore,

νln(E3∩ △r0)≤

Z Z

W(E3∩△_r0)

du1du2 (1₋u1)1−δu1+2 δ

≤

1

Z

0

du1 (1₋u1)1−δ ·

+∞

Z

1

du2

u1+₂ δ <+∞.

It remains to remark that the set E = S3

j=1Ej is also a set of asymptotically finite logarithmic measure inT.

Theorem 2. Letf _{∈ A}2_{. For every δ >0 there exists a setE =E(δ, f)⊂T of}

asymptoti-cally finite logarithmic measure such that for all r _∈T_\E we obtain

Mf(r)≤

µf(r)

(1₋r1)1+δ ln

1+δµf(r)

1₋r1

·ln1/2+δr2. (6)

Proof. Let E′

and E0 be the exceptional sets from Theorem 1 and Lemma 1, respectively. Then for E =E′

∪E0 and h(r1, r2) = (r1r2)1+δ, δ∈(0,1), we get for all r∈T\E

M_{f(r)≤C0µf(r)(det(Df(r) +I))}1/2 _≤

≤C0µf(r)

1 1₋r1 ·

h ∂

∂r1

lnM_{f(r), r2} ∂ ∂r2

lnM_{f(r) + lnr2} !1/2

≤

≤C0µf(r)

1 1₋r1

∂

∂r1

lnM_f(r)1+δ_r2 ∂ ∂r2

lnM_{f(r) + lnr2}1+δ !1/2

Hence by Lemma 1 there exists r0 _{∈T such that for all r∈ △}

r0\E we get

M_{f(r)≤C0µf(r)} 1

(1₋r1)2

lnM_f(r)·lnr2

(1+δ)2 1

(1₋r1)δ(lnMf(r))

1+δ_{+ lnr}

2

1+δ !1/2

<

< µf(r)

(1₋r1)1+δln

(1+δ)2

M_{f(r) ln}1/2+2δ_{r2. (7)}

Using inequality (7) we obtain

lnM_f(r)≤ln µf(r)

(1₋r1)1+δ + (1 +δ)

2_{ln lnM}

f(r) +

1

2+ 2δ

ln lnr2 ≤

≤ln µ 1+δ f (r)

(1₋r1)1+δ

+ 8 ln lnM_f(r).

Therefore,lnM_{f(r)≤2 ln}µf(r)

1−r1.Finally for all r ∈ △r

0\E we have

Mf(r)≤Mf(r)≤

µf(r)

(1₋r1)1+δ 2 ln

µf(r)

1₋r1

!1+2δ+δ2

ln1/2+2δr2 <

< µf(r)

(1₋r1)1+δ1 ln

µf(r)

1₋r1

!1+δ1

ln1/2+δ1

r2,

where δ1 >2(δ+δ2).

3. Some examples. From Theorem 2 we conclude, that for every δ >0 the set

E =nr _∈T: Mf(r)>

µf(r)

(1₋r1)1+δ ln

1+δµf(r)

1₋r1

ln1/2+δr2

o

has asymptotically finite logarithmic measure.

Remark that exponent 1 by ₁−1r1 in inequality (6) cannot be replaced with a smaller

number. We suppose that one can replace this exponent withε_∈(0,1).Consider the function

f(z) = f(z1, z2) =

γ(z2) 1₋z1

,

where γ(z2) is an entire function such thatlnµγ(r2)< r2,r2 →+∞. Then

Mf(r) =

Mγ(r2)

1₋r1

, µf(r) = µγ(r2).

Denote r′

(ε)_∈(0,1) such that for all r1 > r′(ε) we have

1

(1₋r1)1−_ε −ln

2 1 1₋r1

> e.

Thus,

B′

=nr_∈[r′

(ε),1)_×[e,+_∞) : 1

(1₋r1)1−ε −

ln2 1 1₋r1

> r₂2o_⊂

⊂nr_∈[r′(ε),1)_×[e,+_∞) : 1 1₋r1

> 1

(1₋r1)ε

ln 1 1₋r1

+r2

p

lnr2

o

⊂nr_∈[r′(ε),1)_×[e,+_∞) : µγ(r2) 1₋r1

> µγ(r2)

(1₋r1)ε ln

µγ(r2)

1₋r1

p

lnr2

o

⊂

⊂nr_∈[r′

(ε),1)_×[e,+_∞) : Mf(r1, r2)>

µf(r1, r2) (1₋r1)ε

lnµf(r1, r2) 1₋r1

p

lnr2

o

=B

Denote R = q₍₁−_r11)1−ε −ln

2 1

1−_r1. Then for νln-measure of the set B is estimated as

follows

νln(B) =

Z Z

B

dr1dr2 (1₋r1)r2 ≥

Z Z

B′

dr1dr2 (1₋r1)r2 ≥

1

Z

r′_(ε) R

Z

1

dr2

r2

! dr1 1₋r1 ≥

= 1 2

1

Z

r′_(ε)

ln 1

(1₋r1)1−ε −

ln2 1 1₋r1

! dr1 1₋r1 ≥

1 2

1

Z

r′_(ε)

dr1 1₋r1

= +_∞.

We remark, that none of the exponents 1in (6) can be replaced with a smaller number. Let us consider the function

f(z1, z2) = +∞

X

n=0

zn

1

n! +∞

X

m=1

emεz2m =ϕ(z1)·ψ(z2), ε∈(0,1).

Remark that for this function there exists r0 ∈T such that for someδ >0 we have

n

r _{∈ △}r0: M_f(r)>

µf(r)

(1₋r1)1+δ

ln1+δµf(r) 1₋r1

ln1/2+δr2

o

⊂

⊂nr _{∈ △}r0: M_f(r)>

µf(r)

(1₋r1)1+δln

1+2δµf(r)

1₋r1

o .

For the analytic functionϕ(z1)there exist constantsC′

1(ε)∈(0,1)and r1 ∈(r10,1)such that for all r1 ≥r10 we get

C₁′(ε)µϕ(r1) 1₋r1 ≤

Mϕ(r1)

p

lnMϕ(r1) ≤

1

C′

1(ε)

µϕ(r1)

1₋r1

. (8)

It follows from inequality (8), that for r1 ≥ r01 and some constant C1(ε) < C

′

1(ε) we obtain

Mϕ(r1)≥C1(ε)

µϕ(r1) 1₋r1

ln1/2µϕ(r1) 1₋r1

, (9)

and for the entire function ψ(z2) we obtain for every ε >0 and r2 ∈(r02,+∞)

Mψ(r2)≥(

√

2π₋ε)µψ(r2)ln

1 2 µ

ψ(r2). (10)

So, Mf(r1, r2) = Mϕ(r1)Mψ(r2)and for r ∈(r10; 1)×(r02,+∞) we get

Mf(r)≥(

√

2π₋ε)C1(ε)µf(r) 1₋r1

lnµϕ(r1) 1₋r1 ·

lnµψ(r2)

!12

Consider the positive increasing functionsg1(r1) = ln(µϕ(r1)/(1−r1))andg2(r2) = lnµψ(r2).

We define

A =nr _∈T: g1(r1)·g2(r2)> 1

18(g1(r1) +g2(r2)) 2o

⊃nr_∈T: 1 2 <

g1(r1)

g2(r2)

<2o=E∗.

Then, for r_∈A we get

g1(r1)g2(r2)≥

g2 2(r2)

2 = 1

18(g2(r2) + 2g2(r2)) 2

≥ ₁₈1 (g1(r1) +g2(r2))2.

Moreover, there exists the inverse functiong−1

2 : R+ →(r0,1), which is also increasing. For r1 ∈(r10,1)we define r

∗

1 and r

∗

2 such that

r1∗ =g

−₁

2

g

1(r1) 2

, r2∗ =g

−₁

2 (2g1(r1)).

Therefore, from inequality (11) we have for all r_∈E∗

Mf(r)≥

(√2π₋ε)C1(ε) 18

µf(r)

1₋r1

lnµϕ(r1) 1₋r1

+ lnµψ(r2)

!

= (

√

2π₋ε)C1(ε) 18

µf(r)

1₋r1

ln µf(r) 1₋r1

.

It remains to prove that set E∗

is a set of infinite asymptotically logarithmic measure.

µ(E∗

) =

Z Z

△

r0 1

∩E∗

dr1dr2 (1₋r1)r2

= 1 Z r1 0 r∗ 2 Z r∗ 1

dr1dr2 (1₋r1)r2

= = 1 Z r1 0

(lnr∗

2 −lnr

∗

1)

dr1 1₋r1

= 1

Z

r1 0

lng−1

2 (2g1(r1))−lng2−1

g1(r1)

2

!

dr1 1₋r1

.

Since lim

r→+∞

Mψ(r)

µψ(r) ln1/2µψ(r) = √

2π and g2(r) = lnµψ(r), we get

r₋ln(√2π+ε)₋ 1

2lng2(r)< g2(r)< r−ln(

√

2π₋ε)₋ 1

2lng2(r), r→+∞. (12)

From (12) we have

g−1

2 (r)> r, (13)

g2(r)> r₋ln(√2π+ε)₋1

2lng2(r)> r−ln(

√

2π+ε)₋ 1 2lnr >

2 3r,

g2−1(r)< 3

2r, r→+∞. (14)

Using (13) and (14) we finally obtain

νln(E

∗ )_≥ 1 Z r1 0

ln(2g1(r1))−ln

1

2· 3 2g1(r1)

!

dr1 1₋r1

= 1 Z r1 0 ln8 3 · dr1 1₋r1

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Ivan Franko National University of Lviv kurylyak88@gmail.com

shap.ludmila@gmail.com matstud@franko.lviv.ua