Generalization of some extremal problems on non-overlapping domains with free poles

U N I V E R S I T A T I S M A R I A E C U R I E - S K Ł O D O W S K A L U B L I N – P O L O N I A

VOL. LXVII, NO. 1, 2013 SECTIO A 11–22

IRYNA V. DENEGA

Generalization of some extremal problems on non-overlapping domains with free poles

Abstract. Some results related to extremal problems with free poles on radial systems are generalized. They are obtained by applying the known methods of geometric function theory of complex variable. Sufficiently good numerical results for γ are obtained.

1. Introduction. In geometric function theory of complex variable ex- tremal problems on non-overlapping domains form the well-known clas- sic direction. In the paper [1] Lavrent’ev posed and solved a problem of maximizing the product of conformal radii of two non-overlapping simply connected domains. Topics connected with the study of problems on non- overlapping domains was developed in papers [1]–[21]. This paper summa- rizes some results obtained in [5], [2].

Let N, R be the set of natural and real numbers, respectively, C be the complex plane, C = C ∪ {∞} be the one point compactification and R⁺= (0, ∞).

Let r(B, a) be an inner radius of a domain B ⊂ C with respect to a point a ∈ B (see [6], [13], [3]) and χ(t) = ¹₂(t + t⁻¹).

Let n ∈ N. A set of points An :=a_k∈ C : k = 1, n is called n-radial system if |a_k| ∈ R⁺, k = 1, n, and 0 = arg a₁ < arg a₂< . . . < arg a_n< 2π.

2010 Mathematics Subject Classification. 30C75.

Key words and phrases. Extremal problems on non-overlapping domains, inner radius, n-radial system of points, separating transformation.

Denote

P_k(A_n) := {w : arg a_k < arg w < arg a_k+1}, θ_k:= arg a_k, an+1:= a₁, θn+1:= 2π, α_k:= 1

πarga_k+1

a_k , αn+1:= α₁, k = 1, n.

This work is based on application of separating transformation developed in [4]–[6]. For specific use of this method we consider a special system of conformal mappings. By ζ = π_k(w) = −i e^−iθkw
¹

αk, k = 1, n we denote unique branch of multi-valued analytic function π_k(w) performing univalent and conformal mapping P_k(An) onto the right half plane Re ζ > 0.

For an arbitrary n-radial system of points A_n = {a_k} and γ ∈ R⁺ we assume that

L^(γ)(A_n) :=

n

Y

k=1

 χ

  

a_k a_k+1

1 2αk

1−¹₂γα²_k

·

n

Y

k=1

|a_k|^{1+14γ(αk+αk−1)},

L⁽⁰⁾(A_n) :=

n

Y

k=1

 χ

  

a_k a_k+1

1 2αk



· |a_k|.

The class of n-radial systems of points for which L^(γ)(A_n) = 1 (L⁽⁰⁾(A_n) = 1) automatically includes all systems with n different points located on the unit circle.

The main purpose of this work is to obtain exact upper estimates for the functionals

(1) Jn(γ) = r^γ(B0, 0)

n

Y

k=1

r (B_k, a_k) ,

I_n(γ) = [r (B₀, 0) r (B∞, ∞)]^γ

n

Y

k=1

r (B_k, a_k) ,

where γ ∈ R⁺, A_n = {ak}ⁿ_k=1 is n-radial system of points, a₀ = 0, and {B_k}ⁿ_k=0 is a system of non-overlapping domains (i.e. B_p∩ B_j = ∅ if p 6= j) such that a_k∈ B_k, k = 0, n.

2. Main results.

Theorem 1. Let n ∈ N, n ≥ 2 and γ ∈ (0, 1]. Then for any n-radial system of points A_n = {a_k}ⁿ_k=1 such that L^(γ)(An) = 1 and any system of non-overlapping domains B_k, a_k ∈ B_k⊂ C, k = 1, n, a0= 0 ∈ B0 we have the inequality

(2) Jn(γ) ≤ 4^n+nγγ^γnnⁿ (n²− γ)^n+nγ

 n −√ γ n +√

γ

2√ γ

.

Equality in (2) is achieved when a_k and B_k, k = 0, n are, respectively, poles and circular domains of the quadratic differential

Q(w)dw² = −(n²− γ)wⁿ+ γ w²(wⁿ− 1)² dw².

Theorem 2. Let n ∈ N, n ≥ 2 and γ = ¹₂. Then for any n-radial system of points A_n = {a_k}ⁿ_k=1 such that L⁽⁰⁾(A_n) = 1 and any system of non- overlapping domains B_k, B₀, B_∞, a_k∈ B_k ⊂ C, k = 1, n, a0 = 0 ∈ B₀ ⊂ C, a∞= ∞ ∈ B∞⊂ C we have the inequality

(3) [r (B0, 0) r (B∞, ∞)]¹²

n

Y

k=1

r (Bk, ak) ≤ 2²ⁿ⁺ⁿ¹ (n²− 2)ⁿ¹⁺ⁿ²

n −√ 2 n +√

2

!

√ 2

. Equality in (3) is achieved, when a_k and B_k are, respectively, poles and circular domains of the quadratic differential

Q(w)dw² = −w²ⁿ+ wⁿ(2n²− 2) + 1 w²(wⁿ− 1)² dw².

Theorem 3. Let γ ∈ (0; γ₂⁰], γ₂⁰ = 1.1, ε = 0.25. Then for any 2-radial system of points A₂= {ak}²_k=1 such that L^(γ)(A2) = 1, 1 − ε < |ak| < 1 + ε, k = 1, 2 and any system of non-overlapping domains {B_k}²_k=0, a_k ∈ B_k, k = 0, 2, a₀ = 0 ∈ B₀, we have the inequality

r^γ(B₀, 0)

2

Y

k=1

r(B_k, a_k) ≤ r^γ(D₀, 0)

2

Y

k=1

r (D_k, d_k) ,

where D_k, d_k, k = 0, 2, d0 = 0, are circular domains and poles of the quadratic differential

Q(w)dw² = −(4 − γ)w²+ γ w²(w²− 1)² dw².

Proof of Theorem 1. We use the method due to Bakhtin [2]–[3] and prop- erties of separating transformation (see [4]–[7], [3], [8]). We make separating transformation of domains {B_k}ⁿ_k=1. Suppose

P_k:= P_k(A_n) := {w ∈ C\{0} : θk< arg w < θ_k+1}.

Consider the introduced system of functions ζ = π_k(w) = −i e^−iθkw
¹

αk, k = 1, n.

Let Ω⁽¹⁾_k , k = 1, n be a domain of the plane Cζ obtained by combining the connected component π_k(B_kT P_k) containing a point π_k(a_k), with its symmetrical reflection with respect to the imaginary axis.

By Ω⁽²⁾_k , k = 1, n, we denote the domain of the plane Cζ, obtained by combining the connected component π_k(B_k+1T P_k), containing the point

π_k(a_k+1), with its symmetrical reflection with respect to the imaginary axis, Bn+1 := B₁, π_n(an+1) := πn(a1).

Besides, by Ω⁽⁰⁾_k we denote the domain of Cζ, obtained by combining the connected component π_k(B0T P_k), containing the point ζ = 0, with its symmetrical reflection with respect to the imaginary axis. Denote π_k(a_k) :

= ω_k⁽¹⁾, π_k(ak+1) := ω_k⁽²⁾, k = 1, n, π_n(an+1) := ω⁽²⁾n . The definition of π_k implies that

|π_k(w) − ω⁽¹⁾_k | ∼ 1

α_k|a_k|^{αk1 −1}· |w − a_k|, w → a_k, w ∈ P_k,

|π_k(w) − ω_k⁽²⁾| ∼ 1 αk

|a_k+1|^{αk1 −1}· |w − a_k+1|, w → a_k+1, w ∈ P_k,

|π_k(w)| ∼ |w|

1

αk, w → 0, w ∈ P_k.

Then, using results of papers [4]–[7], [3], we obtain the inequalities

(4) r (B_k, a_k) ≤





 r

Ω⁽¹⁾_k , ω_k⁽¹⁾

· r

Ω⁽²⁾_k , ω_k⁽²⁾

1

αk|a_k|^{αk1 −1}·_α¹

k−1|a_k|

1 αk−1−1







1 2

,

k = 1, n, Ω⁽²⁾₀ := Ω⁽²⁾_n , ω₀⁽²⁾ := ω_n⁽²⁾,

(5) r (B0, 0) ≤

" _n Y

k=1

r^α2k

 Ω⁽⁰⁾_k , 0



#¹

2

.

Repeating arguments given in [3] in the proof of Theorem 5.2.1 and taking into account introduced sets of domains {P_k}ⁿ_k=1, functions {π_k}ⁿ_k=1 and numbers {θ_k}ⁿ_k=1, we obtain the following inequality for the investigated functional (1):

(6)

Jn(γ) ≤

n

Y

k=1

h r

 Ω⁽⁰⁾_k , 0

i

α2k 2 γ

·

n

Y

k=1





 r



Ω⁽²⁾_k−1, ω⁽²⁾_k−1

 r



Ω⁽¹⁾_k , ω⁽¹⁾_k



1 αk−1·α_k|a_k|

1 αk−1−1

· |a_k|^{αk1 −1}







1 2

=

n

Y

k=1

αk·

n

Y

k=1

|a_k|

|a_ka_k+1|^2αk1

×

" _n Y

k=1

r^γα2k

Ω⁽⁰⁾_k , 0Yⁿ

k=1

r

Ω⁽¹⁾_k , ω_k⁽¹⁾ r

Ω⁽²⁾_k , ω⁽²⁾_k 

#¹₂ . Expression in parentheses of the last formula in (6) is a product of the func- tional r^β2



Ω⁽⁰⁾_k , 0 r

Ω⁽¹⁾_k , ω_k⁽¹⁾ r

Ω⁽²⁾_k , ω_k⁽²⁾

on triples of domains

 Ω⁽⁰⁾_k , Ω⁽¹⁾_k , Ω⁽²⁾_k 

of the plane Cζ.

It is known [15] that the functional

Y₃(σ₁, σ₂, σ₃) = r^σ1(D₁, d₁) · r^σ2(D₂, d₂) · r^σ3(D₃, d₃)

|d₁− d₂|^{σ1+σ2−σ3}· |d₁− d₃|^{σ1−σ2+σ3}· |d₂− d₃|^{−σ1+σ2+σ3}, σ_k ∈ R⁺, d_k ∈ D_k ⊂ C, Dk∩ D_p = ∅, k = 1, 2, 3, p = 1, 2, 3, k 6= p, is invariant under all conformal automorphisms of the complex plane C.

With this relation in mind, the following estimate holds:

J_n(γ) ≤

n

Y

k=1

α_k

!

·

n

Y

k=1

|a_k|

|a_kak+1|

1 2αk

×







n

Y

k=1

r^γα2k

 Ω⁽⁰⁾_k , 0



· r

Ω⁽¹⁾_k , ω_k⁽¹⁾



· r

Ω⁽²⁾_k , ω_k⁽²⁾



|ω_k⁽¹⁾· ω⁽²⁾_k |^γα2k|ω⁽¹⁾_k − ω_k⁽²⁾|^2−γα2k







1 2

×

" _n Y

k=1

|ω⁽¹⁾_k · ω_k⁽²⁾|^γα2k|ω_k⁽¹⁾− ω⁽²⁾_k |^2−γα2k

#¹₂ .

Note that |ω⁽¹⁾_k | = |a_k|

1

αk, |ω_k⁽²⁾| = |a_k+1|

1

αk, |ω_k⁽¹⁾ − ω_k⁽²⁾| = |a_k|

1 αk +

|a_k+1|

1 αk.

Taking into account these equalities, we obtain J_n(γ) ≤

n

Y

k=1

α_k

!

·

n

Y

k=1

|a_k|

|a_kak+1|

1 2αk

×

n

Y

k=1

|ω_k⁽¹⁾− ω_k⁽²⁾|

! _n Y

k=1

|ω_k⁽¹⁾· ω⁽²⁾_k |

|ω_k⁽¹⁾− ω⁽²⁾_k |

!

γα2k 2

×







n

Y

k=1

r^γα2k

Ω⁽⁰⁾_k , 0

· r

Ω⁽¹⁾_k , ω⁽¹⁾_k 

· r

Ω⁽²⁾_k , ω_k⁽²⁾

|ω_k⁽¹⁾· ω_k⁽²⁾|^γα2k|ω⁽¹⁾_k − ω⁽²⁾_k |^2−γα2k







1 2

= 2ⁿ·

n

Y

k=1

α_k

!

·

n

Y

k=1

χ    

a_k ak+1

1 2αk

!

|a_k|

× 2⁻

γ 2

n

P

k=1

αk

" _n Y

k=1

χ    

a_k a_k+1

1 2αk

!#⁻

γα2k

2 n

Y

k=1

a_k+1 a_k

!

γα2k 2

×







n

Y

k=1

r^γα2k

 Ω⁽⁰⁾_k , 0



· r

Ω⁽¹⁾_k , ω⁽¹⁾_k



· r

Ω⁽²⁾_k , ω_k⁽²⁾



|ω_k⁽¹⁾· ω_k⁽²⁾|^γα2k|ω⁽¹⁾_k − ω⁽²⁾_k |^2−γα2k







1 2

= 2

n−^γ₂

n

P

k=1

α²_k

·

n

Y

k=1

αk

!

·

n

Y

k=1

 χ

  

ak

a_k+1   

1 2αk

1−^γα2₂^k

×

n

Y

k=1

|a_k|^{1+14γ(αk+αk−1)}

×







n

Y

k=1

r^γα2k

 Ω⁽⁰⁾_k , 0



· r

Ω⁽¹⁾_k , ω_k⁽¹⁾



· r

Ω⁽²⁾_k , ω⁽²⁾_k



|ω⁽¹⁾_k · ω_k⁽²⁾|^γα2k|ω_k⁽¹⁾− ω_k⁽²⁾|^2−γα2k







1 2

= 2

n−^γ₂

n

P

k=1

α²_k

·

n

Y

k=1

αk

!

· L^(γ)(An)

×







n

Y

k=1

r^γα2k

Ω⁽⁰⁾_k , 0

· r

Ω⁽¹⁾_k , ω_k⁽¹⁾

· r

Ω⁽²⁾_k , ω⁽²⁾_k 

|ω⁽¹⁾_k · ω_k⁽²⁾|^γα2k|ω_k⁽¹⁾− ω_k⁽²⁾|^2−γα2k







1 2

.

For each k =1, n we can easily define conformal automorphism ζ = T_k(z) of complex numbers of the plane C such that Tk(0) = 0, Tk

 ω^(s)_k



= (−1)^s· i, G^(q)_k := T_k

Ω^(q)_k



, k =1, n, s = 1, 2, q = 0, 1, 2.

Then







n

Y

k=1

r^γα2k

Ω⁽⁰⁾_k , 0

· r

Ω⁽¹⁾_k , ω⁽¹⁾_k 

· r

Ω⁽²⁾_k , ω_k⁽²⁾

|ω_k⁽¹⁾· ω⁽²⁾_k |^γα2k|ω⁽¹⁾_k − ω_k⁽²⁾|^2−γα2k







1 2

=







n

Y

k=1

r^α2kγ

 G⁽⁰⁾_k , 0



· r

G⁽¹⁾_k , −i



· r G⁽²⁾_k , i

 2^2−γα2k







1 2

.

Then using results of [3], [15], we obtain

Jn(γ) ≤ 2

n−^γ₂

n

P

k=1

α²_k

·

n

Y

k=1

αk

!

· L^(γ)(An)

×

n

Y

k=1





 r^α2kγ

G⁽⁰⁾_k , 0

· r

G⁽¹⁾_k , −i

· r

G⁽²⁾_k , i 2^2−γα2k







1 2

= 2ⁿ⁻

γ 2

n

P

k=1

α²_k n

Y

k=1

α_k

!

· L^(γ)(A_n) · 2

−n+^γ

2 n

P

k=1

α²_k

×

" _n Y

k=1

r^α2kγ

 G⁽⁰⁾_k , 0



· r

G⁽¹⁾_k , −i



· r G⁽²⁾_k , i



#¹

2

≤

n

Y

k=1

α_k

!

· L^(γ)(A_n) ·

" _n Y

k=1

r^α2kγ

G⁽⁰⁾_k , 0

· r

G⁽¹⁾_k , −i

· r

G⁽²⁾_k , i

#¹₂ .

Following the paper [15], we have

(7) J_n(γ) ≤ γ⁻ⁿ²

" _n Y

k=1

Ψ (β_k)

#1/2

,

where Ψ(β) = 2^β2+6· β^β2+2(2 − β)^{−12(2−β)2} · (2 + β)^−12(2+β)2, β ∈ [0, 2].

Similarly to [5], we consider the next extremal problem:

n

Y

k=1

Ψ(β_k) → sup;

n

X

k=1

β_k= 2√

γ, β_k= α_k√

γ, 0 < β_k≤ 2.

Necessary conditions have the form Ψ⁰(β_k)

Ψ(βk) = −λ

n

Q

k=1

Ψ(β_k)

, k = 1, n.

We will show that all β_kare equal. We investigate behavior of the function F (β) = ^Ψ_Ψ(β)^0(β) = 2β ln(2β)+_β²+(2−β) ln(2−β)−(2+β) ln(2+β) on interval β ∈ [0, 2]. It is strictly decreasing on the interval (0; β0], β0 ∈ (1.32; 1.33) and increasing on [β₀; 2). Then we use the method of the proof of Theorem 4 [5] and obtain that unique solution of the extremal problem is the point (_n², . . . ,²_n). Estimates (4), (5), (7) yield inequality (2) of Theorem 1. The case of equality is verified directly and Theorem 1 is proved.  Dubinin proved Theorem 1 if γ = 1 and for any distinct points a_k that lie on the unit circle and any non-overlapping domains B_k (see [5], [8]).

Proof of Theorem 2. We retain all notation for separating transforma- tion of domains introduced in the proof of Theorem 1 for domains B_k, k = 0, n. By Ω^(∞)_k we denote the domain of plane Cζ, obtained by com- bining the connected component π_k(B∞T E_k) containing the point ζ = ∞ with its symmetrical reflection with respect to the imaginary axis. The family n

Ω^(∞)_k on k=1

is a result of separating transformation of an arbitrary domain B∞, ∞ ∈ B∞⊂ C with respect to the families {Pk}ⁿ_k=1and {π_k}ⁿ_k=1 at the point ζ = ∞.

By Theorem 2 in [5] we have

(8) r(B∞, ∞) ≤

" _n Y

k=1

r^α2k



Ω^(∞)_k , ∞



#¹₂ . Using (4), (5), (8), we obtain

[r (B0, 0) r (B∞, ∞)]¹²

n

Y

k=1

r (B_k, a_k) ≤ 2ⁿ·

n

Y

k=1

α_k

!

· L⁽⁰⁾(An)

×







n

Y

k=1

r^α2kγ

Ω⁽⁰⁾_k , 0

· r^α2kγ

Ω^(∞)_k , ∞

· r

Ω⁽¹⁾_k , ω_k⁽¹⁾

· r

Ω⁽²⁾_k , ω_k⁽²⁾ 

ω⁽¹⁾_k − ω_k⁽²⁾  

2







1 2

.

Theorem 6 in [5] gives

(9) [r (B₀, 0) r (B∞, ∞)]^α2kγ·r (B₁, a₁) r (B₂, a₂)

|a₁− a₂|² ≤ Ψ(β)

= β^2β2 · |1 − β|^−(1−β)2 · (1 + β)^−(1+β)2, 0 < β ≤

√ 2.

Inequality (9) was obtained by Dubinin using the results of Kolbina [15].

Similarly to [5], we consider the extremal problem:

n

Y

k=1

Ψ(β_k) → sup;

n

X

k=1

β_k=√ 2.

We introduce a function F (β) = ^Ψ_Ψ(β)^0(β). Calculations show that this function is decreasing on the interval (0; β₀] and increases on [β0;√

2), 0.85 < β₀ < 1. Further, as in the proof of Theorem 6 in [5], we verify that the unique solution of the extremal problem is the point (

√ 2 n , . . . ,

√ 2 n ).

Estimates (4), (5), (8), (9) yield the inequality (3) of Theorem 2. The case of equality is verified directly. Theorem 2 is proved.  Proof of Theorem 3. The proof is based on application of separating transformation, developed in details in [6]. According to the conditions of Theorem 3, a₀= 0, 1 − ε < |ak| < 1 + ε, k = 1, 2. Assume

0 = arg a₁< arg a₂ < 2π.

Let α₁ := _π¹ · (arg a₂− arg a₁), α₂ := _π¹· (2π − arg a₂), P_k := {w : arg a_k<

arg w < arg a_k+1}, k = 1, 2, arg a₃ = 2π, P₀ := P₂, P₃ := P₁.

The family of two symmetrical domains {D⁽¹⁾_k ; D⁽²⁾_k−1} with respect to the imaginary axis is called a result of separating transformation of the domain B_k.

Further, as in Theorem 5.2.1 in [3], using the separating transformation we obtain

(10) J₂(γ) ≤ L^(γ)(A₂)

" ₂ Y

k=1

α²_kr^γ·α2k(D₀, 0) · r(D₁, 1) · r(D₂, −1)

#¹₂ . We will prove that for α₀ ≥ ^√2_γ, α₀ = max{α1, α2} extremal configura- tions different than those referred in Theorem 3 do not exist. For this we find a value of the functional (see Theorem 5.2.3 in [3])

J₂⁰(γ) = r^γ(D0, 0) ·

2

Q

k=1

r(D_k, d_k) = 4 · ^(γ)

γ 2

(^1−γ₄)^2+γ2 ·

1−

√γ

2

1+

√γ 2

2√ γ

if γ = 1.1. We have that J₂⁰(1.1) ≈ 0.8315.

Denote r(B₀, 0) = r0, r(B₁, a1) = r1, r(B₂, a2) = r2. The Lavrent’ev’s theorem [17] gives r₀r₁ < |a₁|², r₀r₂ < |a₂|², r₀²r₁r₂ < |a₁|²|a₂|² =⇒ r₁r₂ <

|a₁|²|a₁|² r²₀ .

Then r^γ(B₀, a₀) ·

2

Q

k=1

r(B_k, a_k) = r₀^γ · Q²

k=1

r(B_k, a_k) ≤ r₀^γ · ^|a1|2·|a2|2

r²₀ ≤ J₂⁰(γ) ⇒ r₀ ≥ _|a

1|²·|a₂|² J₂⁰(γ)

_2−γ¹

. If r₀ ≥ _|a

1|²·|a₂|² J₂⁰(γ)

_2−γ¹

, then the extremal configurations do not exist. Consider the case r₀ ≤_|a

1|²·|a₂|² J₂⁰(γ)

_2−γ¹ . J₂(γ) ≤ r₀^γ|a₁− a₂|² = r^γ₀

(|a₁| − |a₂|)²+ 4|a₁| · |a₂| sin²(2 − α₀)π 2



≤ |a₁| · |a₂| J₂⁰(γ)

_2−γ^γ

·

(|a₁| − |a₂|)²+ 4|a₁| · |a₂| sin²(2 − α₀)π 2

≤ J₂⁰(1.1).

Substituting ε = 0.25, γ = 1.1, n = 2, |a₁| = 1 − ε, |a₂| = 1 + ε, J₂⁰(1.1) = 0.8315 in the last inequality, we obtain

(11) (1 + ε)^2−1,11.1



4ε²+ 4(1 + ε) sin²



2 − 2

√1.1

 π 2



≤ J₂⁰(1.1)^1+2−1.11.1 . Performing calculations of right and left sides of the inequality (11), we have 0.6085 < 0.6663. From this it follows that if ε = 0.25, then inequality (11) is true. Hence J = ^J2(1.1)

J₂⁰(1.1) ≤ ^0.7316_0.8315 = 0.8798 < 1, i.e. if α₀ > ^√2_γ, J2(γ) < J₂⁰(γ) then the maximum value of the functional J2(γ) for such domains is not attained. Then α₀ ≤ ^√2_γ and we can apply inequality (10).

Using a result obtained in the proof of Theorem 4 in [5], we can write the following inequality

J₂(γ) ≤ ^√1_γ ·

 ₂ Q

k=1

2^σk2+6· σ_k^σk2+2· (2 − σ_k)^{−12(2−σk)2}(2 + σ_k)^{−12(2+σk)2}

¹₂ ,

where σ_k=√

γ · α_k. Consider the function

Ψ(σ) = 2^σ2+6· σ^σ2+2· (2 − σ)^{−12(2−σ)2}(2 + σ)^−12(2+σ)2,

σ ∈ [0, 2] and we will conduct detailed investigation of its graph on this interval (see Fig.1).

Ψ(σ) is logarithmically convex on interval [0; x₀], x₀ ≈ 1.32. On [0; x₁], x1 ≈ 1.05 the function increases from Ψ(0) = 0 to Ψ(x₁) ≈ 0.9115, and it decreases on interval [x₁; x₂], x₂ ≈ 1.6049 to Ψ(x₂) ≈ 0.86, and on [x₂; 2] it increases to Ψ(2) = 1. x₃ ≈ 1.9, Ψ(x₃) = Ψ(x1) ≈ 0.9115.

Figure 1.

Using equality σ₁ + σ₂ = 2√

γ, we will prove that Ψ(σ₁) · Ψ(σ₂) ≤ (Ψ(x1))² ≈ 0.8308. For σ ∈ [0; x₀] we make appropriate conclusion from the logarithmic convexity of the function Ψ(σ). For σ₂ ∈ [x₀; x₃] from properties of the graph of the function Ψ(σ), we have Ψ(σ₂) ≤ Ψ(x1) and Ψ(σ₁) ≤ Ψ(x₁) and thus Ψ(σ₁) · Ψ(σ₂) ≤ (Ψ(x₁))².

If σ₂ ∈ [x₃; 2] then Ψ(σ₂) < Ψ(2) = 1, Ψ(σ₁) < Ψ(0, 2)  0.4, and hence Ψ(σ1) · Ψ(σ2) < 0, 4 < (Ψ(x1))². So, J₂(γ) ≤ J₂⁰(γ). Inequality (11) is true

and Theorem 3 is proved. 

Acknowledgement. The author is grateful to Prof. Bakhtin for suggesting problems and useful discussions.

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Iryna V. Denega

Department of Complex Analysis and Potential Theory

Institute of Mathematics of National Academy of Sciences of Ukraine Tereshchenkivska St. 3

01601 Kyiv Ukraine

e-mail: iradenega@yandex.ru Received September 20, 2011