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Complex Anal. Oper. Theory
https://doi.org/10.1007/s11785-018-0770-0
Complex Analysis
and Operator Theory
Bavrin’s Type Factorization of the Temljakov Operator
for Holomorphic Functions in Circular Domains of Cn
Renata Długosz1,2·Piotr Liczberski2·
Edyta Trybucka3
Received: 11 July 2017 / Accepted: 18 January 2018
© The Author(s) 2018
Abstract The paper concerns investigations of holomorphic functions of several com-
plex variables with a factorization of their Temljakov transform. Firstly, there were
considered some inclusions between the families CG,MG,NG,RG,VGof such holo-
morphic functions on complete n-circular domain Gof Cnin some papers of Bavrin,
Fukui, Higuchi, Michiwaki. A motivation of our investigations is a condensation of the
mentioned inclusions by some new families of Bavrin’s type. Hence we consider some
families Kk
G,k≥2,of holomorphic functions f:G→C,f(0)=1,defined also
by a factorization of Lfonto factors from CGand MG.We present some interesting
properties and extremal problems on Kk
G.
Keywords Holomorphic functions on n-circular domains in Cn·Minkowski
function ·Estimates of homogeneous polynomials of Taylor series ·Temljakov
operator ·Bavrin’s families of functions
Communicated by David Shoikhet.
BPiotr Liczberski
piotr.liczberski@p.lodz.pl
Renata Długosz
renata.dlugosz@p.lodz.pl
Edyta Trybucka
eles@ur.edu.pl
1Centre of Mathematics and Physics, Lodz University of Technology, Al. Politechniki 11,
90-924 Lodz, Poland
2Institute of Mathematics, Lodz University of Technology, Wólcza´nska 215, 93-005 Lodz, Poland
3Faculty of Mathematics and Natural Sciences, University of Rzeszów, Prof. St. Pigonia 1,
35-310 Rzeszow, Poland
R. Długosz et al.
Mathematics Subject Classification 32A30 ·30C45
1 Introduction
We say that a domain G⊂Cn, is complete n-circular if zλ=(z1λ1,...,znλn)∈G
for each z=(z1,...,zn)∈Gand every λ=(λ1,...,λ
n)∈Un, where Uis the unit
disc {ζ∈C:|ζ|<1}.FromnowbyGwill be denoted a bounded complete n-circular
domain in Cn,n≥2.By HGlet us denote the space of all holomorphic functions
f:G−→ Cand by HG(1)the collection of all f∈HG, normalized by f(0)=1.
Many authors (cf., eg., [1,2,5–7,11,18,19,23]) considered some Bavrin’s subfam-
ilies XGof the family HG(1). In the definitions of these families XGthemainrole
play the families CG(α), α ∈[0,1),
CG(α) ={f∈HG(1):Re f(z)>α,z∈G}
and the following invertible Temljakov [24] linear operator L:HG−→ HG
Lf(z)=f(z)+Df(z)(z), z∈G,
where Df(z)is the Fréchet derivative of fat the point z.By a Bavrin’s family XG
we mean a collection of functions f∈HG(1)whose the Temljakov transform Lf
has a functional factorization Lf=p·g, where p∈CG≡CG(0)and gis from a
fixed subfamily of HG(1). Below, we recall the factorizations which define a few well
known Bavrin’s families XG,like as
VG:Lf=p·1,p∈CG,
MG:Lf=p·f,p∈CG,
NG:Lf=p·LL f,p∈CG,
RG:Lf=p·Lϕ, ϕ ∈NG,p∈CG.
It is known that functions of these families were used to construct biholomorphic
mappings in Cn(cf., eg., [10,13,20]). Let us note that the above families have geometric
interpretation, in particular the functions f∈MGmap biholomorphically some
planar intersections Sof Gonto starlike domains in C,(see [1]). It is very important,
because the starlikeness plays a central role in many different subjects of geometry
and topology and in particular, in geometric function theory.
Let us recall also that Bavrin showed the inclusions NGRG,VGRGand
pointed that the first of them can be complete to the following double inclusion NG
MGRG. Thus, it is natural to ask whether is possible to do the same in the case
of the second above inclusion. In the paper [12] the authors defined a family K−
G,
which satisfies the inclusion VGK−
GRG. An adequate definition of K−
Ghas the
form: A function f∈HG(1)belongs to K−
Gif its Temljakov transform Lfhas the
factorization
Bavrin’s Type Factorization of the Temljakov Operator for…
Lf(z)=p(z)·h(z)·h(−z), z∈G,h∈MG1
2,p∈CG,
where the family MG(α), α ∈[0,1), is defined similarly as MG,but in this case
p∈CG(α).
In the present paper we consider Bavrin’s type families Kk
G,k≥2(K2
G=K−
G)sepa-
rating also the families VG,RG,i.e., satisfying the inclusions VGKk
GRG,k≥2.
The formal definition of such family has the following form.
A function f∈HG(1)belongs to Kk
Gif there exist a function p∈CGand a function
h∈MG(k−1
k)such that the Temljakov transform Lfof fhas the factorization
Lf(z)=p(z)·
k−1
l=0
h(εlz), z∈G,(1.1)
where ε=εk=exp 2πi
kis a generator of the cyclic group of kth roots of unity.
Let us observe that Kk
G,k≥2 are nonempty families. Indeed, the function f=1
belongs to Kk
G,because it satisfies the factorization (1.1) with p=1∈CGand
h=1∈MG(k−1
k)).
In the future, we will use a characterization of the family Kk
Gby a notion of (j,k)-
symmetry, which is connected with a functional decomposition with respect to the
above group.
Let us observe that bounded complete n-circular domains Gare k-symmetric sets
for k=2,3,..., that is εG=G.For j=0,1,...,k−1 we define the collections
Fj,k(G)of functions (j,k)-symmetrical, i.e., all functions f:G→Csuch that
f(εz)=εjf(z),z∈G.
If n=1 and G=U,then we write Fj,k(U).
The mentioned functional decomposition appears in the following result from [14].
Theorem A For every function f :G→Cthere exists exactly one sequence of func-
tions f j,k∈Fj,k(G), j=0,1,...,k−1,such that
f=
k−1
j=0
fj,k.
Moreover,
fj,k(z)=1
k
k−1
l=0
ε−jl fεlz,z∈G.
The functions fj,k,which are uniquely determined by the above decomposition,
will be called (j,k)-symmetrical components of the function f.Some interesting
applications of the above partition may also be found in [15,16] and [17].
R. Długosz et al.
2 Results
Now we can present a characterization of f∈Kk
G,simpler than (1.1).
Theorem 1 A function f ∈HG(1)belongs to the family Kk
G,k≥2if and only if
there exists a function g ∈MG∩F0,k(G)and a function p ∈CGsuch that
Lf=p·g.(2.1)
Proof Let f∈Kk
G.Then there exists p∈CGand h∈MG(k−1
k)such that
Lf(z)=p(z)·g(z),z∈G,
where
g(z)=
k−1
l=0
h(εlz), z∈G.
It is obvious that g∈F0,k(G). We show that g∈MG.To do it, using the differen-
tiation product rule and the form of the operator L,we have at z∈G
Lg(z)
g(z)=1+Dg(z)(z)
g(z)=1+
k−1
l=0
Dh(εlz)(εlz)
h(εlz)=1−k+
k−1
l=0
Lh(εlz)
h(εlz).
Hence and by the fact that h∈MG(k−1
k), we obtain that Re Lg(z)
g(z)>1−k+kk−1
k=0.
Thus g∈MG.
Now, let us suppose that fsatisfies the equality (2.1), with a p∈CGand a g∈
MG∩F0,k(G). Let us put h(z)=(g(z))1
k,z∈G,with the power function taking
the value 1 at the point 1.Since g(z)= 0 (see [1]), the function his holomorphic. It
remains to show that h∈MG(k−1
k)and the equality (1.1) is fulfilled. To this end we
compute step by step
Re Lh(z)
h(z)=Re L(g(z))1
k
(g(z))1
k
=1+1
kRe (g(z))1
k−1Dg(z)(z)
(g(z))1
k
=1+1
kRe Dg(z)(z)
g(z)=k−1
k+1
kRe Lg(z)
g(z)>k−1
k.
The formula (1.1) follows from the definition of the function h. Indeed,
g(z)=(h(z))k=
k−1
l=0
h(εlz), z∈G,
because h∈F0,k(G).
The proof is complete.
Bavrin’s Type Factorization of the Temljakov Operator for…
Now we consider an extremal problem for f∈Kk
G.More precisely, we look for
some estimates for G-balances of m-homogeneous polynomias Qf,mof its unique
power series expansion
f(z)=1+
∞
m=1
Qf,m(z), z∈G.(2.2)
In our considerations the Minkowski function
μG(z)=inf {t>0:1
tz∈G},z∈Cn,
will be very useful. This function gives a possibility to redefine the domain Gand its
boundary ∂Gas follows:
G={z∈Cn:μG(z)<1},∂G={z∈Cn:μG(z)=1}.
The notion of G-balance of m-homogeneous polynomial Qm:Cn→C,m∈N∪{0},
was defined in [3] as the quantity
μG(Qm)=sup
w∈Cn\{0}
|Qm(w)|
(μG(w))m=sup
v∈∂G
|Qm(v)|=sup
u∈G
|Qm(U)|.
The G-balance μG(Qm)generalizes the norm Qmof the polynomial Qmand if G
is convex,then μG(Qm)reduces to Qm,because
|Qm(w)|≤μG(Qm)(μG(w))m,w ∈Cn
and for bounded convex complete n-circular domains Galso μG(w) =||w|| (see, e.g.,
[21]).
We present the announced estimates of G-balances μG(Qf,m)of m-homogeneous
polynomials Qf,mfrom the Taylor series of f∈Mk
Gin the following theorem.
Theorem 2 If the expansion of the function f ∈Kk
G,k≥2,into a series of m-
homogenous polynomials Q f,mhas the form (2.2), then for the G-balances μG(Qf,m)
of polynomials Q f,mthe following sharp estimate hold:
μG(Qf,m)≤⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
2
m
m
k−1
p=11+2
pk for m =k,2k,3k,...
2
m+1
m
k
p=11+2
pk for remaining m ∈N
,
where qmeans the integral part of the number q.We use a standard convention that
the product
l2
l=l1
alis equal to 1for l2<l1.
R. Długosz et al.
Proof Let f∈Kk
Gbe arbitrarily fixed. Then, by Theorem 1, the factorization (2.1)
holds with a function p∈CGof the form
p(z)=1+
∞
ν=1
Qp,ν (z), z∈G
and a function g∈MG∩F0,k(G)of the form
g(z)=1+
∞
ν=1
Qg,kν(z), z∈G.(2.3)
From the above, by the series expansion of Lf
Lf(z)=1+
∞
m=1
QLf,m(z)=1+
∞
m=1
(m+1)Qf,m(z), z∈G
and by the equalities Qf,0=Qp,0=Qg,0=1,we obtain the recursive formula for
m∈N
(m+1)Qf,m(z)=
m
k
l=0
Qg,kl(z)Qp,m−kl (z), z∈G.
Hence
(m+1)Qf,m(z)≤
m
k
l=0Qg,kl(z)Qp,m−kl(z),z∈G.(2.4)
Since
Qp,ν (z)≤2,ν∈N,z∈G,(2.5)
(see [1]) we need some bounds for Qg,kμ(z). We show that for g∈MG∩F0,k(G)
and μ∈Nthere hold the inequalities
Qg,kμ(z)≤2
kμ
μ−1
ν=11+2
kν,z∈G.(2.6)
For this purpose let us observe that for each z∈G,the function
G(ζ ) =ζg(ζ z), ζ ∈U
belongs to the family S∗∩F1,k(U)of (1,k)-symmetric univalent starlike mappings
(in the unit disc U)and its Taylor series has the form
G(ζ ) =ζ+
∞
μ=1
bkμ+1ζkμ+1=1+
∞
μ=1
Qg,kμ(z)ζ kμ+1,ζ ∈U.
Bavrin’s Type Factorization of the Temljakov Operator for…
Thus, in view of the estimates [25] of the coefficients of functions from S∗∩F1,k(U)
we get the announced bounds (2.6).
In two next parts of the proof we use also the fact [4] that for every k,s∈N\{1}
there holds the identity:
1+2
k+
s
l=2
2
lk
l−1
ν=11+2
νk=
s
ν=11+2
νk.(2.7)
Now, we will estimate the quantities Qf,m(z),z∈G,using all the conditions (2.4)-
(2.7).
First let us assume that m=ks,where s∈N.Since Qp,m−kl(z)=1forl=s,we
get from (2.4) that
(m+1)Qf,m(z)≤Qg,ks(z)+2
s−1
l=0Qg,kl(z),z∈G.
Thus for z∈G,inviewof(2.6) and (2.7),
(m+1)Qf,m(z)≤2
sk
s−1
ν=11+2
νk+21+2
k+
s−1
l=2
2
lk
l−1
ν=11+2
νk
=−2
sk
s−1
ν=11+2
νk+21+2
k+
s
l=2
2
lk
l−1
ν=11+2
νk
≤−2
sk
s−1
ν=11+2
νk+2
s
ν=11+2
νk
=2(sk +1)
sk
s−1
ν=11+2
νk.
Hence, for m=k,2k,3k,...
Qf,m(z)≤2
m
m
k−1
ν=11+2
νk,z∈G.
Now let us consider the case m=ks+r,where s∈N∪{0}andr∈{1,2,...,k−1}.
In this case we apply in (2.4) the inequality Qp,m−kl(z)≤2,l=0,...,s=m
k,
which follows from estimates (2.5), because m−kl >0. Thus, in view of (2.6) and
(2.7) we get step by step
R. Długosz et al.
(m+1)Qf,m(z)≤2
m
k
l=0Qg,kl(z)≤2⎡
⎣1+2
k+
m
k
l=2
2
lk
l−1
ν=11+2
νk⎤
⎦
≤2
m
k
ν=11+2
νk.
Summing up the results of both cases we get
Qf,m(z)≤⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
2
m
m
k−1
ν=11+2
νkfor m=k,2k,3k,...
2
m+1
m
k
ν=11+2
νkfor remaining m∈N
,z∈G
and consequently
sup
z∈GQf,m(z)≤⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
2
m
m
k−1
ν=11+2
νkfor m=k,2k,3k,...
2
m+1
m
k
ν=11+2
νkfor remaining m∈N
.
These inequalities and the definition of G-balances μG(Qf,m)of m-homogeneous
polynomials imply the estimates from the statement of the theorem.
Now, we will show the sharpness of the above estimates.
For the linear functional I=μG(J)−1J,with
J(z)=
n
l=1
zl,z=(z1,...,zn)∈Cn,
let us denote by Zan analytic set G∩I−1{0}and let Im(z)=(Iz)m,z∈G,m∈
N∪{0}.The equalities in our estimates are achieved for the following function f∈
Kk
G,k≥2,
f(z)=⎧
⎪
⎨
⎪
⎩
k−1
l=0
Il−1(z)
(1−Ik(z))2
k
−1
I(z)−
k−1
l=3
l−2
lIl−1(z)H(2
k,l
k,l+k
k,Ik(z)) for z∈GZ
1for z ∈Z
,
(2.8)
where H(a,b,c,ζ) :U→Cis a hypergeometric function
H(a,b,c,ζ) =
∞
ν=0
(a)ν(b)ν
(c)ν
ζν
ν!,ζ ∈U,
Bavrin’s Type Factorization of the Temljakov Operator for…
defined by Pochhamer symbols (a)ν,(b)ν,(c)ν:
(a)ν=a(a+1)...(a+ν−1), ν ∈N
1,ν=0,
and the branch of the power function x2
ktakes the value 1 at the point x=1.In the
case k=2,3 we use a standard convention that the sum
k−1
l=3
l−2
lIl−1(z)H2
k,l
k,l+k
k,Ik(z),z∈G
is equal to zero, if the superscript of the sum is smaller than the subscript.
In the paper [4], it was proven that the above function gives the equalities in the
bounds from the statement of the theorem. It remains to show that f∈Kk
Gfor k≥2.
To do it, let us observe that as shown in [4]
Lf(z)=1+I(z)
1−I(z)
1
1−Ik(z)2
k
,z∈G.
This implies, in view of Theorem 1, the relation f∈Kk
G,because the functions
p(z)=1+I(z)
1−I(z),g(z)=1
1−Ik(z)2
k
,z∈G
belong to CGand to MG∩F0,k(G), respectively.
We use the estimates of G-balances μG(Qf,m)of polynomials Qf,mto solve the
mentioned separation problem for the families VG,Kk
G,RG. We prove the following
theorem:
Theorem 3 For every k ≥2there holds the double inclusion
VGKk
GRG.
Proof We start with the inclusion VG⊂Kk
G.To do it, let us assume that f∈VG,then
Lf∈CG.Putting p=Lfand h=1,we obtain the factorization (1.1) with p∈CG
and g=1∈MG∩F0,k(G). Hence f∈Kk
G.It remains to show the relation VG= Kk
G.
To do it, let us observe that for f∈VGthere hold the sharp estimates μG(Qf,m)≤
2
m+1,m∈N(cf., eg., [1]), while for f∈Kk
Gthe sharp estimates μG(Qf,m)≤
B(m)(Theorem 2.), with the obvious bound B(m)> 2
m+1,m∈N{1}.Hence, the
extremal function f∈Kk
Gdoes not belong to VG.
Now we prove that Kk
G⊂RG.To this end, let us suppose that f∈Kk
G.Then there
exist functions p∈CG,g∈MG∩F0,k(G)such that Lf=p·g.Denoting ϕ=L−1g,
we have that ϕ∈NG(by the Aleksander type theorem [1]) and Lf=pLϕ. Thus
R. Długosz et al.
f∈RG. It remains to show the relation Kk
G= RG.For this purpose, let us observe
that in the above estimates μG(Qf,m)≤B(m), m∈N,we have B(m)≤1,m∈N
(see below), while for f∈RGthere hold the sharp estimates μG(Qf,m)≤m+1(see
for instance [1]). Therefore, the extremal function f∈RGdoes not belong to Kk
G.
To complete the proof, we show that B(m)≤1,m∈N.To do it, we consider two
cases, according to the partition m=ks +r,r∈{0,1,...,k−1},from the proof of
Theorem 2.
1. Let us suppose that r=0.Then, if s=m
k=1,we see that the superscript s−1
of the first product in Theorem 2is smaller than its subscript 1.Hence, we replace
the referred product by 1 and consequently, we get μG(Qf,m)≤2
m≤1,because
m=k≥2. Next, if s≥2,then from Theorem 2, by the inequality 1+2
νk≤ν+1
ν,ν ∈
N,k∈N{1},we obtain
μG(Qf,m)≤2
m
s−1
ν=1
ν+1
ν≤2
ms=2
k≤1.
2. Let us suppose that r∈{1,...,k−1}.Then, if s=m
k=0,we see that the
superscript of the second product in Theorem 2 is smaller than its subscript. Hence we
replace the referred product by 1 and consequently, we get μG(Qf,m)≤2
m+1≤1,
because m≤k−1.Next, if s=m
k≥1,then similarly as in step 1,we obtain
μG(Qf,m)≤2
m+1
s
ν=1
ν+1
ν≤2
m+1(s+1)≤2(s+1)
ks +2≤2(s+1)
2s+2≤1.
Now, we give a growth theorem for f∈Kk
Gand its Temljakov transform Lf.
Theorem 4 For functions f ∈Kk
Gthere follow the following sharp estimates
1−r
1+r
1
1+rk2
k
≤|Lf(z)|≤1+r
1−r
1
1−rk2
k
,r=μG(z)∈[0,1),
(2.9)
1
r
r
0
1−
1+
1
1+k2
k
d≤|f(z)|≤1
r
r
0
1+
1−
1
1−k2
k
d,
r=μG(z)∈[0,1). (2.10)
Proof First, let us observe that the above estimates are true for z=0(in(2.10)the
values at r=0,of the left and right hand sides, mean the limit if r→0+). Thus, in
the sequel we will assume that z∈G{0}.We start with the estimates (2.9). Since
f∈Kk
G,there exist a function p∈CGand a function g∈MG∩F0,k(G)such that
the factorization (2.1) holds. Therefore, we show for such functions gthe following
inequalities
Bavrin’s Type Factorization of the Temljakov Operator for…
1
1+rk2
k
≤|g(z)|≤1
1−rk2
k
,r=μG(z)∈(0,1).
To this aim, let us fix arbitrarily a point z∈Gsuch that μG(z)=r∈(0,1)and let us
consider the function
G(ζ ) =ζg(ζ z
μG(z)), ζ ∈U.
Then Gis (1,k)-symmetric, holomorphic, normalized and satisfies the condition
Re ζG(ζ )
G(ζ ) =Re
Lg(ζ z
μG(z))
g(ζ z
μG(z))>0,ζ ∈U.
Hence G∈S∗∩F1,k(U)and by [9, Thm. 2.2.13]
|ζ|
1+|ζ|k2
k
≤|G(ζ )|≤|ζ|
1−|ζ|k2
k
,ζ ∈U.
Putting ζ=μG(z)in the above we obtain, by the definition of the function G,the
announced inequality.
On the other hand, there hold for p∈CGthe following estimates [1]
1−r
1+r≤|p(z)|≤1+r
1−r,r=μG(z)∈(0,1),
Using the estimates of |p(z)|and |g(z)|we get the estimates (2.9). The sharpness of
the upper bounds (2.9) confirms the function given by (2.8). Indeed, for r∈(0,1)and
function f∈Kk
Ggiven by (2.8), we get
Lf(z)=1+r
1−r
1
1−rk2
k
at points z∈G,μ
G(z)=r∈(0,1)such that I(z)=r(this condition is fulfilled by
the points z=rz∗,where z∗∈∂Gand I(z∗)=1).
The sharpness of the lower bounds (2.9) can be proven in a similar way.
Now, we prove the estimates (2.10). To obtain the upper bound (2.10), we use the
proved above upper bound (2.9) and the fact that the Temljakov operator Lis invertible
and
L−1u(z)=
1
0
u(tz)dt,u∈HG,z∈G.
R. Długosz et al.
Indeed, we have for f∈Kk
Gand z∈G,μ
G(z)=r∈(0,1),
|f(z)|=L−1Lf(z)=
1
0
L(tz)dt
≤
1
0
1+rt
(1−rt)1−(rt)k2
k
dt
=1
r
r
0
1+
(1−) 1−k2
k
d.
To prove the lower bound (2.10) let us consider the function
F(ζ ) =ζfζz
μG(z),ζ ∈U,
with arbitrarily fixed f∈Kk
Gand z∈G,μ
G(z)=r∈(0,1). Since
F(ζ ) =Lfζz
μG(z),ζ ∈U,
we get, by Theorem 1, that there exist functions g∈MG∩F0,k(G)and p∈CGsuch
that the factorization (2.1) is true. Thus
F(ζ ) =P(ζ ) ·G(ζ), ζ ∈U,
where for ζ∈U
G(ζ ) =ζgζz
μG(z),P(ζ ) =pζz
μG(z).
Moreover, G∈S∗∩F1,k(U)( see the proof of the estimates (2.9)) and P:U→
C,P(0)=1,is a holomorphic function with a positive real part. Therefore, F
belongs to a subclass K(k)(considered in [22] and for k=2in[8]) of the class of
close-to-convex functions. Hence, Fis univalent in the disc U.
On the other hand, by the lower bound (2.9), we have that
|F(ζ )|≥ 1−|ζ|
1+|ζ|
1
1+|ζ|k2
k
,
because r=μGζz
μG(z)=|ζ|.Now we show that
|F(ζ )|≥
r
0
1−
1+
1
1+k2
k
d, |ζ|=r∈(0,1).
Bavrin’s Type Factorization of the Temljakov Operator for…
To this aim, it is sufficient to show that it holds for the nearest point F(ζ0)from
zero (|ζ0|=r∈(0,1)), otherwise, we have |F(ζ )|≥|F(ζ0)|,|ζ|=r.Since Fis
univalent in the disc U,the original image of the line segment 0,F(ζ0)is a piece of
arc F−10,F(ζ0)in the disc rU.Thus
|F(ζ0)|=
0,F(ζ0)
|dw|=
F−10,F(ζ0)F(ζ )|dζ
≥
r
0
1−
1+
1
1+k2
k
d, r∈(0,1)).
Thus, by the definition of F, we get
ζfζz
μG(z)
≥
r
0
1−
1+
1
1+k2
k
d, |ζ|=r∈(0,1).
Hence, putting ζ=μG(z)=r∈(0,1), we have the lower bound (2.10).
Finally, let us note that we obtain the equalities in the inequalities (2.10)forthe
function (2.8) in adequate points z∈G.
We close the paper with a sufficient condition guaranteeing that a function f∈
HG(1)belongs to Kk
G.We formulate it in the term of G-balances of m-honogeous
polynomials in developments of functions from HG(1).
Theorem 5 Let f ∈HG(1)has the form (2.2). If there exists a function g ∈MG∩
F0,k(G)of the form (2.3)such that
∞
m=1
(m+1)μGQf,m+
∞
m=1
μGQg,mk≤1,
then f ∈Kk
G.
Proof Since g, as a function from MGomits zero [1], we will prove that
Re Lf(z)
g(z)>0,z∈G.
To do it, we compute step by step
R. Długosz et al.
|Lf(z)−g(z)|−|Lf(z)+g(z)|
=
∞
m=1
(m+1)Qf,m(z)−
∞
m=1
Qg,mk (z)
−
2+
∞
m=1
(m+1)Qf,m(z)
+
∞
m=1
Qg,mk (z)
≤2∞
m=1
(m+1)Qg,mk (z)+
∞
m=1Qg,mk (z)−1
≤2∞
m=1
(m+1)μGQf,m+
∞
m=1
μGQg,mk−1≤0.
Thus
Lf(z)
g(z)−1
≤
Lf(z)
g(z)+1
,z∈G
and hence
Re Lf(z)
g(z)≥0,z∈G.
This gives the mentioned inequality by a maximum principle for pluriharmonic
functions of several complex variables. Putting p(z)=Lf(z)
g(z),z∈G,we obtain
that the transform Lfhas the factorization (1) with g∈MG∩F0,k(G)and
p∈CG.Consequently, f∈Kk
G.
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