On a local version of Jack’s lemma
Mamoru Nunokawa, Janusz Sok´o l
Abstract
The purpose of this paper is to provide a result which concerns with the boundary behavior of analytic functions. It may be a local version of the well known Jack’s lemma when we change the function normalization at the origin.
1 Introduction
Let H denote the class of analytic functions in the unit disk D ={z ∈ C:
|z|<1}. Let A(p) denote the class of all functions analytic in the unit diskD which have the form
f(z) =zp+
∞
X
n=1
ap+nzp+n, z∈D, (1) wherepis positive integer. In this section we develop a key lemma that forms the groundwork for many of the results. It is a local version of the following lemma, well known as the Jack’s lemma.
LEMMA 1.1. [1] Let w(z) be non-constant and analytic function in the unit disc D with w(0) = 0. If |w(z)| attains its maximum value on the disc
|z| ≤r at the point z0,|z0|=r, thenz0w0(z0) =kw(z0)andk≥1.
The Jack’s lemma has found several of the applications and generalizations in the theory of differential subordinations, see for instance [2], [3] and [4]. In this paper we generalize the following Nunokawa’s lemma, [5], see also [6] for its angle version.
Key Words: convex, starlike, analytic functions, univalent functions, Jack’s lemma.
2010 Mathematics Subject Classification: Primary 30C45, Secondary 30C80.
Received: 01.03.2018 Accepted: 24.09.2018.
101
LEMMA 1.2. Let p be analytic function in |z| < 1, with p(0) = 1. If there exists a point z0, |z0| < 1, such that Re{p(z)} > 0 for |z| <|z0| and p(z0) =±iafor somea >0, then we have
z0p0(z0)
p(z0) = 2ikarg{p(z0)}
π , arg{p(z0)}=±π 2 for somek≥(a+a−1)/2≥1.
LEMMA 1.3. Let p(z) = 1 +P∞
n=1cnzn be analytic in D with p(z) 6= 0 therein. If there exists a pointz1,0<|z1|<1and the sectorSδ(z1), for which max{z∈Sδ(z1) :|p(z)|}=|p(z1)| (2) wherez1=|z1|eiθ1
Sδ(z1) ={reiθ: 0≤r≤ |z1|, |θ−θ1| ≤δ}, then we have
z1p0(z1)
p(z1) ∈R, z1p0(z1)
p(z1) ≥0, (3)
moreover
Re
1 + z1p00(z1) p0(z1)
≥z1p0(z1)
p(z1) ≥0. (4)
- 6
Re Im
z-plane
rz1
r1
Fig.1. z-plane.
- 6
Re w-plane Im
w=p(z) rp(z1)
r 1
Fig.2. w-plane.
Proof. From the hypothesis, we can have the above pictures, Fig. 1. and Fig. 2. Then it follows that
zp0(z)
p(z) =d log|p(z)|+id arg{p(z)}
idθ =d arg{p(z)}
dθ −i 1
|p(z)|
d|p(z)|
dθ , (5)
where z moves on the arc z = |z1|eiθ and θ1−δ ≤ θ ≤ θ1+δ. From the hypothesis, we have also
d|p(z)|
dθ
z=z1
= 0 (6)
and from geometrical observation, we have d arg{p(z)}
dθ
z=z1
≥0. (7)
It completes the proof of (3). To prove (4) let us put q(z) = zp0(z)
p(z) , q(0) = 0. (8)
From the hypothesis,q(z) is analytic inDand q(z)6= 0, z∈Sδ(z1).
Then it follows that
q(z) = zp0(z)
p(z) = d arg{p(z)}
dθ −i 1
|p(z)|
d|p(z)|
dθ ,
wherez=|z1|eiθ and θ1−δ≤θ≤θ1+δ. Then, from the above picture, we
have d|p(z)|
dθ ≥0, θ1−δ≤θ≤θ1
and d|p(z)|
dθ ≤0, θ1≤θ≤θ1+δ.
Therefore, we have
Im{q(z)}) < 0 for θ1−δ≤θ≤θ1, Im{q(z)}) = 0 for θ=θ1,
Im{q(z)}) > 0 for θ1≤θ≤θ1+δ.
This shows that d arg{q(z)}
dθ
z=z1
= Re
zq0(z) q(z)
z=z1
= Re
1 +zp00(z)
p0(z) −zp0(z) p(z)
z=z1
≥ 0
This shows that 1 +Re
1 +z1p00(z1) p0(z1)
≥Re
z1p0(z1) p(z1)
= z1p0(z1) p(z1) . It completes the proof of (4).
RemarkThe results of Lemma 1.3 and Theorem 2.1 below, hold to be correct not only for the case|p(z)|and|f(z)|take its local maximum value at the point z=z0 in the domain|z| ≤ |z0| but at the pointz1 in the subsetSδ(z1)⊂D. It is an improvement of the known results from [1] and [4]. Lemma 1.3 is applicable for the pointsz=αand not forz=β, Fig. 3.
- 6
u iv
r
α rα
r α
rα rα
rα rα rβ
rβ
rβ rβ
r 1
Fig.3. p(|z| ≤ |z1|).
2 Applications
THEOREM 2.1. Let f(z) = zp+P∞
n=p+1anzn, 1 ≤ p, be analytic and p-valent in D. If there exists a point z1,0 <|z1|<1 and the sector Sδ(z1), for which
max{z∈Sδ(z1) :|f(z)|}=|f(z1)|, (9) wherez1=|z1|eiθ1 and
Sδ(z1) ={reiθ: 0≤r≤ |z1|, |θ−θ1|< δ}, then we have
z1f0(z1)
f(z1) ∈R, z1f0(z1)
f(z1) ≥p, (10)
moreover Re
1 + z1f00(z1) f0(z1)
≥Re
z1f0(z1) f0(z1)
= z1f0(z1)
f0(z1) ≥p. (11) Proof. For the proof of (10), let us put
p(z) = f(z)
zp , p(0) = 1.
From the hypothesis, we have thatp(z) is analytic in D and p(z) 6= 0 in D sincef(z) isp-valent inD. Then it follows that|p(z)|takes its maximum value at the pointz=z1 in the sector Sδ(z1). Therefore, applying Lemma 1.3, we have
z1p0(z1)
p(z1) = Re
z1p0(z1) p(z1)
= z1f0(z1) f(z1) −p
= Re
z1f0(z1) f(z1)
−p
≥ 0.
It completes the proof of (10).
For the proof of (11), let us put q(z) =zf0(z)
pf(z), q(0) = 1.
From the hypothesis, and from (10),q(z) is analytic inDand z1f0(z1)
f(z1) ≥p2>0.
Applying Lemma 1.3, we have z1q0(z1)
q(z1) = Re
1 + z1f00(z1)
f0(z1) −z1f0(z1) f(z1)
= Re
1 + z1f00(z1) f0(z1)
−Re
z1f0(z1) f(z1)
≥ 0.
this shows that 1 +Re
z1f00(z1) f0(z1)
≥Re
z1f0(z1) f(z1)
= z1f0(z1) f(z1) ≥0.
It completes the proof of (11).
LEMMA 2.2. Let p(z) = 1 +P∞
n=1cnzn be analytic in D with p(z) 6= 0 with p(z)6= 0 therein. If there exists a pointz1, 0<|z1|<1 and the sector Sδ(z1), for which
min{|z| ≤r <1 :|p(z)|}=|p(z1)| (12) where|z1|=r <1. Then we have
z1p0(z1)
p(z1) ∈R, z1p0(z1)
p(z1) ≤0, (13)
moreover
Re
1 + z1p00(z1) p0(z1)
≤z1p0(z1)
p(z1) ≤0. (14)
Proof. then we have z1p0(z1)
p(z1) = d logp(z) d logz
z=z
1
= d log|p(z)|+id arg{p(z)}
idϕ
z=z
1
= d arg{p(z)}
dϕ − i
|p(z)|
d|p(z)|
dϕ z=z
1
= d arg{p(z)}
dϕ z=z
1
≤ 0, (15)
because of (12). This gives (13). For the proof of (14) consider d logzp0(z)
p(z)
d log{z} =
d log
zp0(z) p(z)
idθ − i
idθ 1
|p(z)|
d|p(z)|
dθ
= −d dθ
1
|p(z)|
d|p(z)|
dθ
−i d dθ
d arg{p(z)}
dθ
= 1
|p(z)|2
d|p(z)|
dθ 2
− 1
|p(z)|
d2|p(z)|
dθ2
−id2arg{p(z)}
dθ2
= 1 +zp00(z)
p0(z) −zp0(z) p(z) ,
wherez=reiθand 0≤θ≤2π. If we putz=z1, then we have 1 + z1p00(z1)
p0(z1) −z1p0(z1) p(z1)
= 1
|p(z)|2
d|p(z)|
dθ 2
z=z1
− 1
|p(z)|
d2|p(z)|
dθ2
z=z1
−i
d2arg{p(z)}
dθ2
z=z1
= − 1
|p(z)|
d2|p(z)|
dθ2
z=z1
−i
d2arg{p(z)}
dθ2
z=z1
because of (12). Therefore, Re
1 + z1p00(z1)
p0(z1) −z1p0(z1) p(z1)
= − 1
|p(z)|
d2|p(z)|
dθ2
z=z1
≤ 0
because |p(z)| attains its minimum value at z = z1, and from the known geometric property, we have
d2|p(z)|
dθ2
z=z1
≥0.
It completes the proof of (14).
Applying Lemma 2.2 and the same method as in the proof of Theorem 2.1 we can proof the following theorem.
THEOREM 2.3. Let f(z) = zp+P∞
n=p+1anzn, 1 ≤ p, be analytic and p-valent in D. If there exists a point z1,0 <|z1|<1 and the sector Sδ(z1), for which
max{z∈Sδ(z1) :|f(z)|}=|f(z1)|, (16) wherez1=|z1|eiθ1 and
Sδ(z1) ={reiθ: 0≤r≤ |z1|, |θ−θ1|< δ}, then we have
z1f0(z1)
f(z1) ∈R, z1f0(z1)
f(z1) ≤p, (17)
moreover Re
1 + z1f00(z1) f0(z1)
≤Re
z1f0(z1) f0(z1)
= z1f0(z1)
f0(z1) ≤p. (18) For some related results we refer to [7, 8, 9].
References
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3(1971) 469–474.
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[3] S. S. Miller, P. T. Mocanu, On some classes of first order differential subordinations and univalent functions, Michigan Math. J. 32(1985) 185–
195.
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Japan Acad. Ser. A 68(6)(1992) 152–153.
[6] M. Nunokawa, On the order of strongly starlikeness of strongly convex functions, Proc. Japan Acad. Ser. A 69(7)(1993) 234–237.
[7] M. Nunokawa, J. Sok´o l, On some geometric properties of multivalent functions, Journal of Inequalities and Applications, 2015, 2015:300.
[8] M. Nunokawa, J. Sok´o l, On some differential subordinations, Studia Sci- ent. Math. Hungarica 54(4)(2017) 1–10.
[9] M. Nunokawa, J. Sok´o l, N. E. Cho, Some applications of Nunokawa’s lemma, Bull. Malaysian Math. Sci. Soc. 40(4)(2017) 1791–1800.
Mamoru NUNOKAWA, University of Gunma,
Chuou-Ward, Chiba, 260-0808, Japan.
Email: mamoru−[email protected] Janusz SOK ´O L,
Faculty of Mathematics and Natural Sciences, University of Rzesz´ow,
ul. Prof. Pigonia 1, 35-310 Rzesz´ow, Poland.
Email: [email protected]
