Conditions for one direction convexity and starlikeness
Nunokawa et al. Journal of Inequalities and Applications (2016) 2016:90
DOI 10.1186/s13660-016-1034-z
RESEARCH
Open Access
Conditions for one direction convexity
and starlikeness
Mamoru Nunokawa1 , Oh Sang Kwon2 , Young Jae Sim2 , Ji Hyang Park3 and Nak Eun Cho3*
*
Correspondence:
Department of Applied
Mathematics, Pukyong National
University, Busan, 48513, Korea
Full list of author information is
available at the end of the article
3
Abstract
We investigate several sufficient conditions on a function to be convex in one
direction or starlike in one direction.
MSC: Primary 30C45
Keywords: analytic functions; univalent functions; convex functions in one
direction; starlike functions in one direction
1 Introduction
Let H denote the class of functions analytic in the unit disk D := {z ∈ C : |z| < }, and denote
by A the class of analytic functions in H that are normalized by f () = = f () – . Also,
let S denote the subclass of A composed of functions that are univalent in D.
We say that a function f is starlike in one direction if f it maps |z| = r for every r near
onto a contour C that is cut by a straight-line passing through the origin in two and no
more than two points. Robertson [] found the following sufficient condition for starlikeness in one direction.
Lemma Let f (z) be analytic in |z| ≤ r, and f (z) = in < |z| ≤ r. Further, let f () = .
Suppose that
π
zf (z)
R
f (z) dθ < π,
z = ρeiθ , for every ρ ≤ r.
Then, for every ρ ≤ r, f (z) maps |z| = ρ onto a curve that is starlike in one direction.
A function is said to be convex in one direction in |z| < r (r > ) if the function maps
|z| = ρ < r for every ρ near r into a contour that may be cut by every straight-line parallel
to this direction in no more than two points. It is known (see []) that if f ∈ A and zf (z) is
starlike in one direction, then f (z) is convex in one direction and belongs to S . Therefore,
we can obtain the following lemma (see also [–]).
Lemma Let f (z) = z+
that
∞
n= an z
n
be analytic for |z| ≤ and f (z) = on |z| = r < . Suppose
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Nunokawa et al. Journal of Inequalities and Applications (2016) 2016:90
π
+ R zf (z) dθ < π,
f (z)
Page 2 of 6
z = reiθ , for every r < .
Then f (z) is convex in one direction, and hence f (z) is univalent in |z| ≤ .
We may refer to [–] for more sufficient conditions on analytic functions to be convex
in one direction.
In the present paper, we investigate several sufficient conditions on functions in A to
be convex in one direction using various methods. Also, we find sufficient conditions for
starlikeness in one direction.
2 Main results
Theorem Let f (z) ∈ A and suppose that
+ R zf (z) < R + z
f (z)
–z
(z ∈ D).
()
Then f (z) is convex in one direction, and hence f (z) is univalent in D.
Proof Let ≤ r < . From hypothesis () we have
π
+ R zf (z) dθ
(z)
f
+z
R
<
dθ
–z
|z|=r
π
– r
dθ
=
– r cos θ + r
= π.
Therefore, by Lemma , f (z) is convex in one direction in D.
Example Consider the function f : D → C defined by f (z) = z/( – z). Then we have
+
zf (z) + z
=
f (z)
–z
(z ∈ D).
Moreocer, we can easily check that condition () holds for the function f . Therefore, by
Theorem the function f is convex in one direction and univalent in D.
Theorem Let f (z) ∈ A and suppose that
+ zf (z) ≤ + z
f (z)
– z
(z ∈ D).
()
Then f (z) is convex in one direction in |z| < r = . · · · , where r is the root of the
equation
+r
π
–r
+r
+ log
–r
= π.
()
Nunokawa et al. Journal of Inequalities and Applications (2016) 2016:90
Page 3 of 6
Proof Let ≤ r < . From inequality () we have
π
+ R zf (z) dθ
f (z)
π
zf (z)
≤
+ f (z) dθ
π
+ z
≤
– z dθ
π
– r
r sin θ
+
≤
– r cos θ + r – r cos θ + r dθ
+r
+r
+ log
.
≤ π
–r
–r
Define the function g : [, ) → R by
+r
g(r) = π
–r
+r
+ log
.
–r
Then g() = π and g(r) → ∞ as r → – . Also, we have that the function g is increasing
on [, ) since
g (r) =
π
+
>
( – r) – r
for all r ∈ [, ). Therefore, there exists a unique root r in [, ) such that g(r) = π . Hence,
we have
π
+ R zf (z) dθ < π
f (z)
for |z| < r . It follows from Lemma that f (z) is convex in one direction in |z| < r .
Theorem Let f (z) ∈ A and suppose that
R + zf (z) – < R + z +
f (z)
–z
(z ∈ D).
Then f (z) is convex in one direction, and hence f (z) is univalent in D.
Proof Let ≤ r < . From hypothesis () we have
R + zf (z) – dθ
f (z)
|z|=r
R + zf (z) – dθ
≤
f (z)
|z|=r
+z
<
+
dθ
R
–z
|z|=r
= π.
()
Nunokawa et al. Journal of Inequalities and Applications (2016) 2016:90
Page 4 of 6
Therefore, we have
R + zf (z) dθ < π
f (z)
|z|=r
for |z| = r < . This shows that f (z) is convex in one direction in D.
Corollary Let f (z) ∈ A and suppose that
R + zf (z) < R + z + (z ∈ D).
f (z)
–z
Then f (z) is convex in one direction, and hence f (z) is univalent in D.
Theorem Let f (z) ∈ A and suppose that
√
+ zf (z) <
f (z)
(z ∈ D).
()
Then f (z) is convex in one direction, and hence f (z) is univalent in D.
Proof Let ≤ r < . From () we have
+ zf (z) dθ < π.
(z)
f
|z|=r
()
Note that
zf (z)
zf (z)
zf (z)
+
+
dθ = π
+
dθ =
f (z)
f (z)
f (z)
|z|=r
|z|=r
()
zf (z)
+
dθ = π.
f (z)
|z|=r
()
and
Therefore, from (), (), and () we have
zf (z)
+R
dθ
f (z)
|z|=r
zf (z)
zf (z)
zf (z)
=
+
dθ
+ +
+ +
|z|=r
f (z)
f (z)
f (z)
< π.
()
Hence, applying the Cauchy-Schwarz inequality and (), we get
+ R zf (z) dθ ≤
f (z)
|z|=r
π
|z|=r
+ R zf (z) dθ < π.
f (z)
This completes the proof of Theorem .
Nunokawa et al. Journal of Inequalities and Applications (2016) 2016:90
Page 5 of 6
Example Consider the function f : D → C defined by
√
√
z+
f (z) =
–
(z ∈ D).
√
Then we have
√
√
√
+ zf (z) = √z + (...truncated)
