Techniques of the differential subordination for domains bounded by conic sections

PII. S0161171203302212 http://ijmms.hindawi.com © Hindawi Publishing Corp.

TECHNIQUES OF THE DIFFERENTIAL SUBORDINATION

FOR DOMAINS BOUNDED BY CONIC SECTIONS

STANISŁAWA KANAS

Received 24 February 2003

We solve the problem of finding the largest domainDfor which, under givenψ andq, the differential subordinationψ(p(z), zp(z))∈D⇒p(z)≺q(z), where Dandq(ᐁ)are regions bounded by conic sections, is satisfied. The shape of the domainDis described by the shape ofq(ᐁ). Also, we find the best dominant of the differential subordinationp(z)+(zp(z)/(βp(z)+γ))≺pk(z), when the functionpk(k∈[0,∞))maps the unit disk onto a conical domain contained in a right half-plane. Various applications in the theory of univalent functions are also given.

2000 Mathematics Subject Classification: 30C45, 34A25, 33E05, 30C35.

1. Introduction. Fork∈[0,∞)define the domainΩkas follows:

Ωk=u+iv:u2> k2(u−1)2+k2v2. (1.1)

For fixedk the above domain represents the conic region bounded, succes-sively, by the imaginary axis (k=0), the right branch of a hyperbola (0< k <1), a parabola (k=1), and an ellipse (k >1). Also we note that, for no choice of parameterk (k >1),Ωkreduces to a disk. We present below the univalent func-tions mapping the unit diskᐁontoΩk, denoted bypk, such thatpk(0)=1 and

p_k(0) >0:

pk(z)=

                            

1+z

1−z, fork=0,

1+₁₋2_k2sinh2A(k)arctanh√z, fork∈(0,1),

1+_π22log 21+

√ z

1−√z, fork=1,

1+ 2 k2₋₁sin

2 π 2᏷(t)Ᏺ

√_z √

t, t

, fork >1,

(1.2)

whereA(k)=(2/π )arccosk, Ᏺ(w, t)is the Legendre elliptic integral of the first kind,

Ᏺ(w, t)=

w

0

dx √

The functionp0is well known; an explicit form ofp1is due to Rønning [16] and independently Ma and Minda [11]; the remaining cases belong to Kanas [5] and Kanas and Wi´sniowska [8].

We also note that the function pk(z)=p(k, z) is continuous as regards

k, k∈[0,∞)and has real coefficients fork∈[0,∞)(cf. [4,5]).

Letᏼ denote the well-known Carathéodory class of functions, analytic in the unit disk with positive real part there, so that ᏼ= {p :panalytic inᐁ,

p(0)=1, Rep(z) >0}. Denote byᏼ(pk)the subclass of the classᏼconsisting of all functionspsuch thatp(ᐁ)⊂pk(ᐁ).

Suppose that functionsgand Gare analytic in the unit discᐁ. The func-tiongis said to besubordinatetoG, writteng≺G(org(z)≺G(z), z∈ᐁ), ifGis univalent inᐁ, g(0)=G(0), andg(ᐁ)⊂G(ᐁ). If the functiong(z)= ψ(p(z), zp(z))with ψ, pbeing analytic and having appropriate normaliza-tion, then the subordination is called thefirst-order differential subordination. The theory of differential subordinations was developed by Miller and Mocanu in their numerous works (cf., e.g., [12,13,14]) and by Eenigenburg et al. [1] and has wide applications in the theory of univalent functions.

We briefly recall one of the problems that characterizes this theory. For more details the reader should consult [13,14]. Assume thatpis analytic inᐁand such thatp(0)=aand p(z)≡a. Also, let ψ:C2 _{Cbe analytic and such} thatψ(p(0),0)=a. Givenψand univalent functionq,q(0)=p(0), find the “largest” domainD, such thata∈D, and the relation

ψp(z), zp(z)∈D ⇒p(z)≺q(z) (1.4)

holds. The second important problem of this theory is finding the “smallest” functionq, whenDandψare given, and (1.4) holds.

Now, we recall some results from the theory of differential subordinations (cf., e.g., [12,13,14]) useful for further investigation (byᐁrwe denote the open disk centred at the origin and with the radiusr).

Lemma1.1(see [13]). Letfbe analytic inᐁ, and letgbe analytic and univa-lent onᐁ¯, withf (0)=g(0). Iffis not subordinate tog, then there exist points z0∈ᐁ,ζ0∈∂ᐁ, andm≥1for which

fᐁ|z₀|⊂gᐁ|z0|

, fz0

=gζ0

, z0f

z0

=mζ0g

ζ0

. (1.5)

Lemma1.2(see [13]). Letψ:C2_{→Cbe an analytic function in a domain}

D⊂C2_{, and letp(z)=a+p}

nzn+···be analytic inᐁ, such that

wherehis analytic and univalent inᐁwithψ(p(0),0)=h(0). If

ψhζ0

, mζ0h

ζ0

∉h(ᐁ) whenm≥n, z∈ᐁ, ζ0=1, (1.7)

thenp≺hinᐁ.

The univalent functionqis called adominant of the solution of the differential subordination

ψp(z), zp(z)≺h(z), (1.8)

or more simply adominant if p≺q for allp satisfying (1.8). A dominant ˜q

that satisfies ˜q≺qfor all dominantsqof (1.8) is said to bethe best dominant of (1.8). (Note that the best dominant is unique up to the rotation ofᐁ.)

Lemma1.3(see [1,12,14]). Letβandγbe complex constants, and lethbe convex(univalent)inᐁ, withRe(βh(z)+γ) >0. Also, suppose that the differ-ential equation

q(z)+ zq(z)

βq(z)+γ =h(z) (1.9)

has a univalent solution q, which satisfies q≺ h. If p is analytic in ᐁ with p(0)=h(0)and satisfies

p(z)+_βp(z)zp(z)_+γ ≺h(z), (1.10)

thenp≺q≺hinᐁ, andqis the best dominant of (1.10).

Lemma1.4(see [14, page 97]). Letβ, γ∈C,β≠0, andRe(β+γ) >0. Letq be analytic inᐁ, withq(0)=h(0)=1and (1.9) is satisfied. If Re(βh(z)+γ) >0 andhis convex inᐁ, then the solutionqof (1.10) is univalent andqis the best dominant of (1.10).

In fact, the above lemmas appear in a more general form, but the above form is sufficient for our considerations. Lemmas 1.3 and 1.4 characterize properties of subordination, known as the subordination of the Briot-Bouquet type, described in detail by Miller and Mocanu in [12,14] and by Eenigenburg et al. in [1].

Numerous subordination results concerning various choices of functionq

by parabolas and hyperbolas, were considered although the other forms of the functionψ seem to be more useful. For instance, when we putp(z)= zf(z)/f (z)andψ(p, zp)=p+zp/p, then the relationψ(p(z), zp(z))∈ D⇒p(z)≺q(z), for various choices of functions q and D, gives inclusion relations between subclasses of convex and subclasses of starlike functions. Hence, in this case, the functionψin the formψ(p, zp)=p+γzpis useless. Our principal goal in this paper is to develop the theory of the first-order dif-ferential subordinations towards subordinations in conic domains. Such kind of domains is closely related tok-uniformly convex andk-starlike functions (cf. [7,8,9]). We employ geometric techniques to obtain certain differential in-clusions when the functionsψandq(orψand a domainD) are given. Mainly, we discuss the caseψ(p, zp)=p+zp/p and the functionhmapping the unit disk onto a domain bounded by a parabola.

InSection 2, we solve the problem of finding the largest domainDfor that, under givenψandq, the condition (1.4) is satisfied. The shape of the domain

Dis described by the shapeq(ᐁ). InSection 3, we present the best dominant of the differential subordination (1.10). Various applications in the theory of univalent functions are given inSection 4.

2. Subordination results related to conic domains. Consider first the re-gion bounded by a parabolau=v2_{/2+1/2 that is the domain p}

1(ᐁ) with

p1given by (1.2). Next, we consider the family described by the equalityu=

v2_{/2+(2a+1)/2, a <1/2, that consists of parabolas with the vertex atw=}

(2a+1)/2, symmetric about the real axis (our previous case corresponds to

a=0). The family of domains containing point 1 inside and bounded by those parabolas may be characterized as

Ᏸa=w: Re(w−a) >|w−1−a|, (2.1)

or equivalently

Ᏸa=w=u+iv: 2u > v2_{+2a+1. (2.2)}

Note thatᏰ0=Ω1.

Theorem2.1. Letpbe a function analytic in the unit disk such thatp(0)=1. Also, leta <1/2. If

p(z)+zp_p(z)(z)∈Ᏸa, (2.3)

then

p(z)≺1+_π22log 21+√z

Proof. Suppose, on the contrary, thatp≺p1. Then, byLemma 1.1, there exist pointsz0∈ᐁ, ζ0∈∂ᐁ, ζ0≠1, andm≥1 such that p(z0)=p1(ζ0),

p(|z| < |z0|) ⊂ p1(ᐁ), and z0p(z0) = mζ0p1(ζ0). Setting p1(z) = 1+

(2/π2_)log2_{q(z)withq(z)=(1+}√_z)/(1−√_{z), q(0)=1, we obtain}

zp₁(z) p1(z) =

4

π2

zq(z) q(z)

logq(z)

1+2/π2_log2_q(z). (2.5)

Note that forζ0=eiθ,θ∈(0,2π ), we haveq(ζ0)=icot(θ/4)=ix, x >0, andζ0q(ζ0)/q(ζ0)=iy=i(x+1/x)/4. Thus

pz0

=p1

ζ0

=1+_π22log

2_(ix)=1 2+

2

π2log 2_x+i2

πlogx, z0pz0

pz0

=mζ0p1

ζ0 p1 ζ0

=4miy_π2

logx+i(π /2)

(1/2)+2/π2_log2_{x+i(2/π )logx}

=4my_π2

 

(1/π )logx−(π /4)+ilogx

(3/2)+2/π2_log2_x

(1/2)+2/π2_log2_x2_+4/π2_log2_x

  .

(2.6)

From the above we have

Re

pz0

+z0p

z0

pz0

=1

2+ 2

π2log

2_x−my π /4−(1/π )log2x log2x+π /4+(1/π )log2x2

,

Im

pz0

+z0p

z0

pz0

=logx

 

_π2+my 3/2+

2/π2_log2_x log2x+π /4+(1/π )log2x2

  .

(2.7)

Observe, by the definition ofᏰa, that we will be led to a contradiction if we show that

Im2

pz0

+z0p

z0

pz0

−2 Re

pz0

+z0p

z0

pz0

which, after necessary reductions, is equivalent to

m2_y2_log2_x

3/2+2/π2_log2

x2

log2x+π /4+(1/π )log2x22

+4m π ylog

2

x 3/2+

2/π2_log2

x

log2x+π /4+(1/π )log2x2

+2my π /4−(1/π )log

2_x log2x+π /4+(1/π )log2x2

+2a≥0.

(2.9)

Sett=log2x, t≥0. Then the above inequality can be rewritten as

m2y2t

3/2+2/π2_t2

t+π /4+(1/π )t22

+my

8/π3_t2_{+(4/π )t+π /2}

t+π /4+(1/π )t2 +2a≥0.

(2.10)

Sincem≥1, “coefficients” ofy2 _{andy are both nonnegative, andy=(x+} 1/x)/4=(e√t_+e−√t_)/4=[cosh√_{t]/2, condition (2.10) will be fulfilled if}

cosh2t t

3/2+2/π2_t2

t+π /4+(1/π )t22

+2 cosht

8/π3_t2_{+(4/π )t+(π /2)}

t+π /4+(1/π )t2 +8a≥0.

(2.11)

Observe now that

cosh2t t

3/2+2/π2_t2

t+π /4+(1/π )t22

+2 cosht

8/π3_t2_{+(4/π )t+π /2}

t+π /4+(1/π )t2 +8a

≥2

8/π3_t2_{+(4/π )t+π /2}

t+π /4+(1/π )t2 +8a

=16 π

1− t

t+π /4+(1/π )t2

+8a

≥ 8 π+8a.

(2.12)

inequality follows. Hence

Im2

pz0+

z0p

z0

pz0

−2 Re

pz0+

z0p

z0

pz0

+2a+1≥0 (2.13)

if and only if 8/π+8a≥0, that is, whena≥a0= −1/π≈ −0.318309 that leads to the contradiction. The proof is completed.

Now, we extend the result of Theorem 2.1 to the case when p(z)+ zp(z)/(βp(z)+γ)∈Ᏸa implies p≺p1. Since the method of the proof is the same as inTheorem 2.1, we omitted the details.

Theorem2.2. Letβ, γ∈Rbe such thatβ > γ≥0,β≠0,a <1/2, and let p∈ᏼ. If

p(z)+_βp(z)zp(z)_+γ ∈Ᏸa, (2.14)

then

p(z)≺1+_π2₂log21+

√ z

1−√z, (2.15)

whereᏰa is given by (2.1) and

a≥ − 2 π β+

2

π2r

t0=:a1, (2.16)

where

r (t)=

2(2β+γ)/πt+π γ(β+2γ)/2β

4β2_/π2_{t+(β+2γ)/2+2β/π}2_t2,

t0=

π2_{2ββ(β+2γ)β}2_−γ2_−γ(β+γ) 4β(2β+γ) .

(2.17)

Reasoning in the other direction we consider the function

Qb(z)=1+

2(1−b) π2 log

21+

√ z

1−√z, b <1, (2.18)

that maps the unit disk analytically and univalently onto the parabolic region

Ωb= {u+iv: 2u > v2/(1−b)+1+b}having a vertex atw=(1+b)/2. When

b grows, the parabola becomes narrower until it degenerates forb=1. The functionQbwas defined by Rønning in [15]. Define now the family of domains, having 1 inside and creating fromΩbby a translation through the vector(a,0) as follows:

Ᏸa,b=w=u+iv: 2(u−a) > v

2

1−b+1+b

Reasoning along the same lines as inTheorem 2.1, in this case, we obtain the following theorem.

Theorem2.3. Leta < (1−b)/2, b <1, and letp be a function analytic in the unit disk such thatp(0)=1. If

p(z)+zp(z)

p(z) ∈Ᏸa,b, (2.20)

thenp(z)≺Qb(z), whereᏰa,bis given by (2.19),Qb(z)is given by (2.18), and

a≥ −_π2+2(1_π−2b)r

t0

=:a2, (2.21)

where

r (t)=

2(2−b)/πt+π b(1+b)/2(1−b)

4(1−b)2_/π4_t2_{+2(1−b)(3+b)/π}2_t+(1+b)2_/4,

t0=

π2_{(1−2b)1−b}2_−b(1+b) 2(2−b)(1−b) .

(2.22)

3. Problem of the best dominant. The second problem of the theory of differential subordinations, mentioned in the introduction, is the problem of finding the “smallest” functionqsuch that (1.9) and (1.10) implyp≺q≺h. This problem was posed by Miller and Mocanu in [13], repeated among others in [14] and completely solved with reference to Briot-Bouquet differential sub-ordination, although, an explicit form of the best dominant was found only for the functions mapping the unit disk onto a half-plane or a disk (given in term of hypergeometric functions). In this section, employing Miller and Mocanu re-sults, we provide the form of the best dominant of Briot-Bouquet differential subordination in the case when the functionh≡pk.

In this instance we have the following proposition.

Proposition3.1. Let0≤k <∞. Also, letβ, γ∈Cbe such thatβ≠0and Re(βk/(k+1)+γ) >0. Ifpis analytic inᐁ,p(0)=1,psatisfies

p(z)+_βp(z)zp(z)_+γ ≺pk(z), (3.1)

andqis an analytic solution of

q(z)+ zq(z)

βq(z)+γ =pk(z), (3.2)

thenqis univalent,p≺q≺pk, andqis the best dominant of (3.1).

the solutionqis the best dominant of the subordination (3.1). Such a solution is given by

q(z)=

zγ+β_exp z 0

pk(t)−1

/tdtβ

β 0zuβ+γ−1

exp 0u

pk(x)−1/xdxβdu

−γ_β, (3.3)

or equivalently q(z)=  β 1 0

tβ+γ−1exp tz

z

pk(u)−1

u du β dt   −1

−γ_β (3.4)

(cf. [12,14]). Thus, by making use ofLemma 1.3, we havep≺q≺pkwithq being the best dominant.

In particular, forβ=1 andγ=0, one obtains

q(z)=

!

zexp z

0

pk(t)−1

t dt "!z 0 exp u 0

pk(x)−1

x dx

du

"−1

, (3.5) or equivalently q(z)= 1 0 exp tz z

pk(u)−1

u du

dt

−1

. (3.6)

In the case whenk >1 we may approximate the domain, being an elliptical domain, by a circular domain. An ellipse must be contained in the disk which is tangent to the ellipse at points common with the real axis. The case of subordi-nations related to disks is completely solved in the literature. The first attempt was given by Jakubowski and Kami´nski [3], but a final result is due to Miller and Mocanu (cf., e.g., [14, page 110]). Since the ellipse intersects the real axis at pointsk/(k+1)andk/(k−1), then such a circleK(S, R)has the center at

S=k2_/(k2_{−1)and a radiusR=k/(k}2_{−1)and contains the pointz=1 inside.} Then the functionϕ:ᐁ→K(S, R)has a formϕ(z)=k/(k−z), and therefore Proposition 3.1withβ=1 andγ=0 yields the following proposition.

Proposition3.2. Let1< k <∞and letpbe analytic inᐁ,p(0)=1, andp satisfies (3.1). Thenp(z)≺z/[(z−k)log(1−z/k)]and

Rep(z) > 1

(k+1)log(1+1/k). (3.7)

Proof. Sincepsatisfies (3.1) and for each fixedk,pk(z)≺k/(k−z), then

p(z)+zp(z)/p(z)≺k/(k−z)inᐁ. In view of (3.6), insertingϕ(z)=k/(k−z)

in place ofpk, we obtainp(z)≺q(z)=z/[(z−k)log(1−z/k)]. Then we have

Remark3.3. Making use ofProposition 3.2, for instance, for the casek=2, one obtains that if p(z)+zp(z)/p(z)≺p2(z)=1/(1−z/2)(it is obvious that then Re(p(z)+zp(z)p(z)) >2/3) implies Rep(z) >1/[3 log(3/2)]≈

0.813. . . .

4. Applications. Functionspk and the classᏼ(pk)were constructed with reference to the subclass of the class of univalent functions, calledk -uniform-ly convex, and denotedk-ᐁᏯᐂ(cf. [7,8,9]), so that every domain appeared in Section 2is strictly related with the classᏼ(pk)and, obviously, withk-ᐁᏯᐂ. It should be mentioned that the classk-ᐁᏯᐂwas defined pure geometrically as the class of such functions that map every circular arc, with the center

ζ, |ζ| ≤k, (0≤k <∞)contained inᐁ, onto a convex arc. With reference to the classk-ᐁᏯᐂ, Kanas and Wi´sniowska defined the classk-᏿᐀= {g∈Ᏼ:g= zf, f∈k-ᐁᏯᐂ}, called the class ofk-starlike functions. HereᏴdenotes the class of functions analytic in the unit disk and such thatf (0)=f(0)−1=0. One of the most important cases ofk-ᐁᏯᐂfunctions is the uniformly convex functions case which corresponds to the choicek=1. In this case, condition (1.1), when setting u+iv =1+zf(z)/f(z), z∈ᐁ, means that the range of the expression 1+zf(z)/f(z)is the region bounded by a parabolau= v2_/2+1/2.

Theorem4.1. Letf∈Ᏼand let1+zf(z)/f(z)∈Ᏸa, whereᏰa given in (2.1) anda≥ −1/π. Then

zf(z) f (z) ≺1+

2

π2log 21+√z

1−√z, (4.1)

or equivalentlyf∈1-᏿᐀.

Proof. Assume that 1+zf(z)/f(z)∈Ᏸa. Settingp(z)=zf(z)/f (z), we havep(0)=1, and the above condition can be rewritten as

p(z)+zp(z)

p(z) ∈Ᏸa (4.2)

inᐁ. Now, applyingTheorem 2.1, we conclude the assertion.

ApplyingTheorem 2.2and proving similarly as above, we obtain the follow-ing theorem.

Theorem4.2. Letβ, γ∈R,β≠0,β > γ≥0, and letf∈Ᏼ. If

1+zf(z)/f(z)−zf(z)/f (z)

β+γf (z)/zf_(z) ∈Ᏸa, (4.3)

whereᏰa is given in (2.1) andasatisfies (2.16), thenf∈1-᏿᐀.

Theorem4.3. Leta <1/2, andb <1. Also, letf∈Ᏼand1+zf(z)/f(z)∈

Ᏸa,b, whereᏰa,b is given by (2.19) anda≥a1 (the valuea1is described by (2.21)). Thenzf(z)/f (z)≺Qb(z)withQbgiven by (2.18).

Conversely to the presented applications we may assume that

f∈k-ᐁᏯᐂ⇐⇒1+zf_f(z)(z)≺pk(z), (4.4)

and from this, by making use of results ofSection 3, we conclude the contain-ment results aboutk-᏿᐀.

On applyingProposition 3.1(forβ=1 andγ=0) andProposition 3.2, we have the following relations between the classesk-ᐁᏯᐂandk-᏿᐀below.

Proposition4.4. Let0≤k <∞. Also, letf∈k-ᐁᏯᐂ. Thenzf(z)/f (z)≺ q(z)≺pk(z), whereqis given by (3.6).

Proposition4.5. Let1< k <∞and letf∈k-ᐁᏯᐂ. Thenzf(z)/f (z)≺ z/[(z−k)log(1−z/k)]and

Rezf(z)

f (z) >

1

(k+1)log(1+1/k). (4.5)

We presented applications of results of Sections3and 4only fork-ᐁᏯᐂ andk-᏿᐀functions. However, by settingp(z)=f (z)/z,p(z)=f(z),p(z)=

2f (z)/z−1, p(z)=2zf(z)/f (z)−1, p(z)=2f(z)−1, and so forth, in the propositions and the theorems of the mentioned sections, we may obtain many more results related to conical domains.

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Stanisława Kanas: Department of Mathematics, Rzeszów University of Technology, W. Pola 2, Pl-35-959 Rzeszów, Poland