Sections of univalent harmonic mappings

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by  Saminathan Ponnusamy

SECTIONS OF UNIVALENT HARMONIC MAPPINGS
7 SAMINATHAN PONNUSAMY, ANBAREESWARAN SAIRAM KALIRAJ,
1
AND VICTOR V. STARKOV
0
2
n Abstract. In this article, we determine the radius of univalence of sections of
a normalized univalent harmonic mappings for which the range is convex (resp.
J starlike, close-to-convex, convex in one direction). Our result on the radius of

1 univalenceofsections (f)issharpespeciallywhenthecorrespondingmappings
n,n
2
have convex range. In this case, each section s (f) is univalent in the disk of
n,n

radius 1/4 for all n 2, which may be compared with classicalresult of Szego¨ on
] ≥
V conformal mappings.
C
.
h
t
a 1. Introduction and main results
m
Since confirmation of the Bieberbach conjecture by Louis de Branges [3] on the
[
class , of all normalized univalent analytic functions φ defined in the unit disk
1 D = Sz : z < 1 , one of the open problems about the class is that of determining
v
{ | | } S
1 the precise value of r such that all sections s (φ) of φ are univalent in z < r .
n n n
4 Here we say that φ is normalized if φ(0) = 0 = φ′(0) 1. Also, let | |
0 −
6 n
0 (1) s (φ)(z) = a zk
. n k
1
k=1
0 X
whenever
7
1 ∞
: (2) φ(z) = a zk.
v k
i
X Xk=1
In [24], Szeg¨o proved that the section/partial sum s (φ) of φ is univalent in
r n
a ∈ S
z < 1/4 for all n 2. The constant 1/4 is sharp as the second section of the Koebe
| | ≥
function k(z) = z/(1 z)2 suggests. In [20], Robertson proved that the section s (k)
n
is starlike in the disk−z < 1 3n−1logn for n 5, and that the number 3 cannot be
| | − ≥
replaced by a smaller constant. Later in the year 1991, Bshouty and Hengartner [4]
showed that the Koebe function is not extremal for the problem of determining the
radius of univalency of the partial sums of functions in . At this time the best
S
known result is due to Jenkins [11] who proved that s (φ) is univalent in z < r
n n
| |
for φ , where the radius of univalence r is at least 1 (4logn log(4logn))/n
n
∈ S − −
for n 8. For related investigations on this topic, see the recent articles [14,16] and
≥
the references therein. More interestingly, as investigated recently in [12,13], our
Date: January 24, 2017.
2010 Mathematics Subject Classification. Primary: 30C45;Secondary: 31A05, 30C55, 32E30.
Key words and phrases. Harmonic univalent, starlike, close-to-convex and convex mappings,
convex in one direction, partial sums .
File: PonSaiStarkov˙Partia˙Sum7˙2015˙˙IndagMath.tex,printed: 2017-1-24,1.34.
1
2 S.Ponnusamy,A.Sairam Kaliraj, and V.V.Starkov
main aim in this article is to consider the analogous problem for univalent harmonic
mappings in the unit disk, since harmonic mappings have interesting links with
geometric function theory, minimal surfaces and locally quasiconformal mappings.
Every harmonic mapping f in a simply connected domain can be written as
f = h+g, where h and g are analytic. In particular, we consider the class of all
complex-valued harmonic functions f = h+g in D normalized by h(0) = g(0H) = 0 =
h′(0) 1. We call h and g, the analytic and the co-analytic parts of f, respectively,
−
and obviously they have the following power series representation
∞ ∞
(3) h(z) = z + a zk and g(z) = b zk, z D.
k k
∈
k=2 k=1
X X
Throughout the discussion we shall use this representation. Since the Jacobian J
f
of f = h + g is J (z) = h′(z) 2 g′(z) 2, we say that f is sense-preserving in
f
D if J (z) > 0 in D. Let| |de−no|te the| class of all sense-preserving harmonic
f H
S
univalent mappings f and set 0 = f : f (0) = 0 . For many basic
∈ H SH { ∈ SH z }
results on univalent harmonic mappings, we refer to the monograph of Duren [7] and
also[15]. Harmonicmappingstechniques havebeenusedtostudyandsolvefluidflow
problems (see [1]). In particular, the study of univalent harmonic functions having
special geometric property such as convexity, starlikeness, and close-to-convexity
arises naturally while dealing with planar fluid dynamics problems. For example, in
[1, Theorem 4.5], Aleman et al. considered a fluid flow problem on a convex domain
Ω satisfying an interesting geometric property. In view of results from [17,19], one
0
obtains that harmonic mappings, z H (z)+G (z) such that ReH′(z) > G′(z)
7→ 0 0 0 | 0 |
on the convex domain Ω , considered by the authors in [1, Theorem 4.5] are indeed
0
close-to-convex in Ω .
0
Another reasons in studying the sections of harmonic mappings is that approx-
imation of real valued harmonic functions by harmonic polynomials attracted the
attention of mathematicians (see [25]) as it has many advantages. For example, a
harmonic function has its maximum and minimum values on the boundary of the
regions of consideration. Because planar harmonic mappings f = h+g defined on
D have series representation as in (3), sections of f can be thought of as an approx-
imation of f by complex-valued harmonic polynomials and thus, approximation of
univalent harmonic mappings by univalent harmonic polynomials might lead to new
applications in fluid flow problems, in particular. Until recently, much is not known
about the univalence of sections of univalent harmonic mappings. In 2013, Li and
Ponnusamy [12,13] initiated the study on this topic by considering certain classes
of univalent harmonic mappings. However, the harmonic analog of these results are
not known in the literature even for well-known geometric subclasses of 0, namely,
the classes 0 are 0 , ∗0, and 0 mapping D onto, respectively, convSexH, starlike,
and close-toS-HconvexKdHomSaHins, justCHas , ∗, and are the subclasses of mapping
D onto these respective domains. AtKthSis place,Cit is worth to recall thSat general
theorems on convolutions [22] (see also [6, p. 256, 273]) allow one to conclude that
s (φ) isconvex, starlike, or close-to-convex inthe disk z < 1 3n−1logn, forn 5,
n
whenever φ is convex (resp. starlike or close-to-co|n|vex) −in D. ≥
∈ S
Sections of univalent harmonic mappings 3
Another interesting geometric subclass of 0 which attracted function theorists is
the class of univalent harmonic functions f CfoHr which f(D) is convex in a direction
α. Recall that a domain D C is called convex in the direction α (0 α < π) if the
⊂ ≤
intersection of D with each line parallel to the line through 0 and eiα is connected
(or empty). See, for example, [5,9,21]. Now, we recall the class 0( ) introduced
SH S
in [18], where
0( ) = h+g 0 : h+eiθg for some θ R
SH S ∈ SH ∈ S ∈
and as in [18], let ((cid:8)) = f = f +bf : f 0 ( ) and b D(cid:9) . One of the
SH S 0 0 0 ∈ SH S ∈
conjectures stated in [18] reads as follows.
(cid:8) (cid:9)
Conjecture 1. 0 = 0( ). That is, for every function f = h + g 0 , there
exists at least oneSHθ RSHsuSch that h+eiθg . ∈ SH
∈ ∈ S
In[18], itwasalsoremarked thatthetruthofthisconjectureverifies thecoefficient
conjecture of Clunie and Sheil-Small for f = h+g 0 , namely,
∈ SH
(2n+1)(n+1) (2n 1)(n 1)
an , a−n − − , and an a−n n
| | ≤ 6 | | ≤ 6 | |−| | ≤
(cid:12) (cid:12)
for each n 2, where bn = a−n. Conjecture 1 remains open. Th(cid:12)e bound a−(cid:12)2 1/2
≥ | | ≤
is well-known and sharp which follows from the classical Schwarz lemma. However,
the conjectured bounds of Clunie and Sheil-Small have been verified for a number
of subclasses of 0, namely, for ∗0, 0 , 0 ( ) and the class of harmonic mappings
SH SH CH SH S
convex in one direction. More recently, Starkov [23] established a criteria for func-
tions belonging to the class 0( ) and as a consequence, several examples including
SH S
harmonic univalent polynomials are also obtained for a given f .
H
∈ S
For f = h+g 0 with power series representation as in (3), the sections/partial
∈ SH
sums s (f) of f are defined as
n,m
s (f)(z) = s (h)(z)+s (g)(z)
n,m n m
where n 1 and m 2. However, the special case m = n 2 seems interesting in
≥ ≥ ≥
its own merit. We now state our main results.
Theorem 1. Let f = h + g 0 with series representation as in (3). Suppose
that f belongs to any one of t∈heSfHollowing geometric subclasses of 0 : ∗0, 0 ,
SH SH CH
0 ( ) or the class of harmonic mappings convex in one direction. Then the section
SH S
s (f) is univalent in the disk z < r . Here r is the unique positive root of
n,m n,m n,m
| |
the equation ψ(n,m,r) = 0, where
3 6
1 1 r 1 r
(4) ψ(n,m,r) = − 1 − R T ,
n m
12r 1+r − 1+r − −
" #
(cid:18) (cid:19) (cid:18) (cid:19)
with
∞ ∞
k(k +1)(2k +1)
(5) R = A rk−1, T = A rk−1, where A = ,
n k m −k k
− 6
k=n+1 k=m+1
X X
4 S.Ponnusamy,A.Sairam Kaliraj, and V.V.Starkov
In particular, every section s (f)(z) is univalent in the disk z < r , where
n,n n,n
| |
(7logn 4log(logn))
r > rL := 1 − for n 15.
n,n n,n − n ≥
Moreover, r rL, where l = min n,m 15.
n,m ≥ l,l { } ≥
For functions in the convex family 0 of harmonic mappings, we have the follow-
KH
ing interesting result which may compared with the original conjecture for functions
in .
S
Theorem 2. Let f = h + g 0 with series representation as in (3). Then the
∈ KH
section s (f) is univalent in the disk z < r , where r is the unique positive
n,m n,m n,m
| |
root of the equation µ(n,m,r) = 0. Here
∞ ∞
1 r k(k +1) k(k 1)
(6) µ(n,m,r) = − rk−1 − rk−1 .
(1+r)3 − 2 − 2
k=n+1(cid:20) (cid:21) k=m+1(cid:20) (cid:21)
X X
In particular, for n 5, and θ R, the harmonic function
≥ ∈
s (f;θ)(z) = s (h)(z)+eiθs (g)(z)
n,n n n
is univalent and close-to-convex in the disk z < 1 3n−1logn. Moreover, we have
| | −
r 1 (4logl 2log(logl))/l, where l = min n,m 7.
n,m
≥ − − { } ≥
It is worth to remark that if f 0 , then we actually prove that for n 5,
s (f) is stable harmonic close-to-c∈onKveHx (see [10]) in z < 1 3n−1logn. ≥
n,n
| | −
The paper is organized as follows. In Section 2, we recall certain known results
which are crucial in the proof of our main theorems. In Section 3, we present the
proofs of Theorems 1 and 2, and as a consequence, we state a couple of corollaries.
2. Useful Lemmas
Now, we recall some results that are needed for the proofs of our main results.
The following result due to Bazilevich [2] gives the necessary and sufficient condition
for a normalized analytic function to be univalent in D.
Theorem A.An analytic function φ defined in D and determined by (2) is univalent
in D if and only if for each z D and each t [0,π/2],
∈ ∈
∞
φ(reiη) φ(reiψ) sinkt
(7) − := a zk−1 = 0,
reiη reiψ k sint 6
− k=1
X
where t = (η ψ)/2, z = rei(η+ψ)/2 and sinkt = k.
− sint t=0
Recently, Starkov [23] generalized this res(cid:12)ult to the class of normalized sense-
(cid:12)
preserving harmonic mappings in the following form.
Theorem B. A sense-preserving harmonic function f = h + g defined in D de-
termined by (3) is univalent in D if and only if for each z D 0 and each
∈ \ { }
Sections of univalent harmonic mappings 5
t (0,π/2],
∈
∞
f(reiη) f(reiψ) sinkt
(8) − := (a zk b zk) = 0,
reiη reiψ k − k sint 6
− k=1(cid:20) (cid:21)
X
where t = (η ψ)/2 and z = rei(η+ψ)/2.
−
The following two point distortion theorem of Graf et al. [8] plays a crucial role
in the proof of our main results.
Lemma C. If f = h+g 0 , r (0,1), t,ψ R, then
∈ SH ∈ ∈
f(reit) f(reiψ) 1 1 r α 1 r 2α
− − 1 − ,
reit reiψ ≥ 4αr 1+r − 1+r
" #
(cid:12) − (cid:12) (cid:18) (cid:19) (cid:18) (cid:19)
(cid:12) (cid:12)
where α = sup(cid:12) h′′(0) /2. (cid:12)
(cid:12)f∈SH | | (cid:12)
Finally, we recall the following well-known identities which are also easy to derive.
Lemma 1. The following identities are true for 0 < r < 1:
∞
rn
(i) krk−1 = [1+n(1 r)].
(1 r)2 −
k=n+1 −
X∞
rn
(ii) k2rk−1 = [2+(2n 1)(1 r)+n2(1 r)2].
(1 r)3 − − −
k=n+1 −
X∞
rn
(iii) k3rk−1 = [6+(6n 6)(1 r)+(3n2 3n+1)(1 r)2+n3(1 r)3].
(1 r)4 − − − − −
k=n+1 −
X
Proof. The identity (i) is obvious. To obtain (ii) we may multiply (i) by r and then
differentiate it with respect to r. The proof of case (iii) is similar. So, we omit its
(cid:3)
proof.
3. Partial sums of Univalent Harmonic Mappings
3.1. Proof of Theorem 1. Suppose that f = h+g belongs to either ∗0 or 0 or
SH CH
0 ( ) or to the class of harmonic mappings convex in one direction, where h,g are
SH S
given by the power series (3) with b = 0. Set F (z) = f(rz)/r for 0 < r < 1. Then
1 r
F (z) 0 and
r ∈ SH ∞ ∞
F (z) = z + a rk−1zk + b rk−1zk.
r k k
k=2 k=2
X X
Evidently, finding the largest radius of univalence of s (f)(z) is equivalent to
n,m
finding the largest value r such that s (F )(z) is univalent in D. From Theorem
n,m r
B, it is clear that s (F )(z) is univalent in D if and only if s (F )(z) is sense-
n,m r n,m r
preserving in D and the associated section P (z) has the property that
n,m,r
M
sinkt
P (z) := (a′zk b′zk) = 0, for all z D 0 , and t (0,π/2],
n,m,r k − k sint 6 ∈ \{ } ∈
k=1(cid:20) (cid:21)
X
6 S.Ponnusamy,A.Sairam Kaliraj, and V.V.Starkov
where M = max n,m , l = min n,m , a′ = a rk−1, b′ = b rk−1 for all k l,
{ } { } k k k k ≤
a rk−1 for all k > l if M = n,
a′ = k
k 0 for all k > l if M > n,
(cid:26)
and
b rk−1 for all k > l if M = m,
b′ = k
k 0 for all k > l if M > m.
(cid:26)
Setting t = (η ψ)/2, z = ρei(η+ψ)/2 D in (8) and from the univalency of F for
r
− ∈
0 < r < 1, we get that
∞ 3 6
sinkt 1 1 r 1 r
(9) (a zk b zk)rk−1 − 1 − .
k k
− sint ≥ 12r 1+r − 1+r
(cid:12) (cid:12) " #
(cid:12)Xk=1(cid:20) (cid:21)(cid:12) (cid:18) (cid:19) (cid:18) (cid:19)
(cid:12) (cid:12)
In order(cid:12)to find a lower bound for P (cid:12) (z) , we need to find an upper bound for
n,m,r
(cid:12) | (cid:12) |
∞ ∞
sinkt sinkt
R (z) = a rk−1zk (b rk−1zk) .
n,m,r k k
| | sint − sint
(cid:12) (cid:12)
(cid:12)k=Xn+1(cid:20) (cid:21) k=Xm+1(cid:20) (cid:21)(cid:12)
(cid:12) (cid:12)
By the assumption on(cid:12)f, it follows that (see for instance [7,18]) (cid:12)
(cid:12) (cid:12)
(k +1)(2k+1) (k 1)(2k 1)
a and b − − , for all k 2,
k k
| | ≤ 6 | | ≤ 6 ≥
and hence
∞ ∞
k(k +1)(2k +1) k(k 1)(2k 1)
R (z) rk−1 + − − rk−1
n,m,r
| | ≤ 6 6
k=n+1(cid:20) (cid:21) k=m+1(cid:20) (cid:21)
X X
(10) = R +T (see (5)).
n m
From (9) and (10), we get that
3 6
1 1 r 1 r
P (z) − 1 − R T = ψ(n,m,r).
n,m,r n m
| | ≥ 12r 1+r − 1+r − −
" #
(cid:18) (cid:19) (cid:18) (cid:19)
The inequality P (z) > 0 holds for all z D 0 , whenever ψ(n,m,r) > 0,
n,m,r
| | ∈ \ { }
where ψ(n,m,r) is defined by (4). This gives that ψ(n,m,r) > 0 for all r
∈
(0,r ), where r is the positive root of the equation ψ(n,m,r) = 0 which lies
n,m n,m
in the interval (0,1). In order to complete the proof, we have to show that s (f)
n,m
is locally univalent in z < r . However, s (f) = s (h) + s (g) is locally
n,m n,m n m
| |
univalent in z < r if and only if the analytic functions s (h) + eiθs (g) is
n,m n m
locally unival|en|t in z < r for every θ R. That is, we have to show that
n,m
s (h′)(z)+eiθs |(g|′)(z) = 0 for all z <∈r and θ R. It is easy to see that
n−1 m−1 n,m
6 | | ∈
∞ ∞
s (h′)(z)+eiθs (g′)(z) = (h′(z)+eiθg′(z)) ka zk−1 +eiθ kb zk−1 .
n−1 m−1 k k
−
!
k=n+1 k=m+1
X X
From the hypothesis, it is clear that f = h+g belongs to either 0 or 0 (S) (see
CH SH
for instance, [18]). As the affine spanning of 0 as well as 0 (S) are linear invariant
CH SH
Sections of univalent harmonic mappings 7
families [18],
h′′(0)
α := sup | | = 3.
2
f∈CHSSH(S)
Therefore, fromawell knownresult onlinearinvariant familyofharmonicmappings,
it follows that (see [7, p. 99])
(1 r)2
′ ′
h(z) g (z) − for z = r .
| |−| | ≥ (1+r)4 | |
Moreover, for 0 < r < 1 (see (9)), we see that
(1 r)2 1 1 r 3 1 r 6 1 1 r 3 1 r 6
min − , − 1 − = − 1 −
(1+r)4 12r 1+r − 1+r 12r 1+r − 1+r
( " #) " #
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
and thus,
(1 r)2
sn−1(h′)(z)+eiθsm−1(g′)(z) ≥ (1−+r)4 −Rn −Tm
(cid:12) (cid:12)
3 6
(cid:12) (cid:12) 1 1 r 1 r
− 1 − R T
n m
≥ 12r 1+r − 1+r − −
" #
(cid:18) (cid:19) (cid:18) (cid:19)
0,
≥
where thelast inequality here gives the condition z < r . This observation proves
n,m
| |
that s (f) is univalent in the disk z < r and the proof of the first part of the
n,m n,m
| |
theorem is complete.
From the above discussion, it is apparent that r r , where l = min n,m
n,m l,l
≥ { } ≥
2. Next, we need to consider the special case m = n and determine the lower bound
for r with certain restriction on n. In this case, the sufficient condition (4) for
n,n
the univalence of s (f) reduces to
n,n
(1 r)3(3+10r2 +3r4)
(11) ψ(n,n,r) = − (R +T ) 0,
3(1+r)9 − n n ≥
where R +T determined from (5) may be rearranged in a convenient form as
n n
∞ ∞
1 2
R +T = krk−1+ k3rk−1
n n
3 3
k=n+1 k=n+1
X X
rn[12+12(n 1)(1 r)+3(2n2 2n+1)(1 r)2 +(2n3 +n)(1 r)3]
= − − − − − .
3(1 r)4
−
As lim (R +T ) = 0, it is clear that the radius of univalence r 1. Setting
n→∞ n n n,n
→
r = 1 α /n, where α = o(n), we see that (4) holds, whenever 0 < t(α ,n) < 1,
n n n
−
where
e−xn7 x 9 [12n4 +12(n 1)xn3 +3(2n2 2n+1)x2n2 +(2n3 +n)x3n]
t(x,n) = 2 − − .
x7 − n 16n4 32n3x+28n2x2 12nx3 +3x4
(cid:16) (cid:17) − −
From the definition of t(α ,n), it is clear that eαnt(α ,n) = O(n7) and hence α
n n n
can be chosen to be 7logn alog(logn) for some positive real number a. However,
−
8 S.Ponnusamy,A.Sairam Kaliraj, and V.V.Starkov
computations shows that
64
lim t(α ,n) = > 0 for a = 4, and lim t(α ,n) = for a > 4.
n n
n→∞ 2401 n→∞ ∞
Therefore, we may set γ = 7logn 4log(logn).
n
−
It is easy to see that 1 γ /n > 0 for all n 15. For n 15, we shall prove
n
− ≥ ≥
that r > 1 γ /n. For n 15, it is sufficient to prove that t(x,n) is a decreasing
n,n n
− ≥
function in x, whenever
γ x n, 0 < t(γ ,n) < 1 and t(n,n) > 0.
n n
≤ ≤
In order to do this, we first differentiate t(x,n) with respect to x and obtain that
t′(x,n) = q (x,n)q (x,n), where
1 2
(2n x)8e−x
q (x,n) = − −
1 x8[16n4 32n3x+28n2x2 12nx3 +3x4]2
− −
and
q (x,n) = 2688n7 +2688(n 3)n6x+3(448n7 2368n6 +3648n5 x7)x2
2
− − −
+64n4(7n3 48n2 +137n 132)x3 +16n2(59n3 128n2 +178n 75)x5
− − − −
+2n2(32n5 80n4(6+x)+1672n3 4n2(774+13x3)+2040n 3x5)x4
− − −
+2n(88n4 240n3 +434n2 390n+81)x6
− −
+2n(78n2 98n+57)x7 +6(6n3 2n2 +6n 1)x8.
− − −
From the definition of q (x,n), it is clear that q (x,n) < 0 for all γ x n,
1 1 n
where n 15. To conclude t′(x,n) < 0, we need to prove that q (x,n) >≤0 fo≤r all
2
≥
n 15 and γ x n. From the grouping of terms in q (x,n), one can see that
n 2
≥ ≤ ≤
q (x,n) > 0 for n 15, and x (0,n/3]. Next, we show that q (x,n) > 0 for all
2 2
≥ ∈
x [n/3,n]. To do this, for any fixed n 15, we set x = n/k, where k [1,3].
∈ ≥ ∈
Then, q (x,n) reduces to
2
n7(1 2k)2 n8 n9 n10 n11
Q(k,n) = − Q (k)+ Q (k)+ Q (k)+ Q (k)+ Q (k),
k6 1 k8 2 k9 3 k8 4 k9 5
where
Q (k) = 6(112k4 224k3 +204k2 92k +27),
1
− −
Q (k) = 2(1344k7 3552k6 +4384k5 3096k4+1424k3 390k2 +57k 3),
2
− − − −
Q (k) = 1344k7 3072k6+3344k5 2048k4 +868k3 196k2 +36k 3,
3
− − − −
Q (k) = 4(112k5 240k4 +236k3 120k2 +39k +3) and
4
− −
Q (k) = 2(32k5 80k4 +88k3 52k2 +18k 3).
5
− − −
A computation shows that Q (k) has no real root. Moreover, the only real roots
1
of Q (k), Q (k), Q (k) and Q (k) are 0.104153, 0.143187, 0.0630667 and 0.5,
2 3 4 5
−
respectively. Therefore, for 1 j 5, Q (k) will have same sign for all k [1,3].
j
≤ ≤ ∈
As Q (1) > 0 for 1 j 5, we conclude that Q (k) > 0 for all k [1,3]. This
j j
≤ ≤ ∈
shows that Q(k,n) > 0 for all k [1,3] and hence, q (x,n) > 0 for all x [n/3,n].
2
∈ ∈
Since
e−n
t(n,n) = (2n3 +6n2 +7n+3) > 0 for all n N,
3 ∈
Sections of univalent harmonic mappings 9
Value of n Value of r
n,n
2 0.108193
3 0.147197
4 0.182263
5 0.214025
10 0.337088
50 0.675001
100 0.788521
287 0.900122
Table 1. Values of r for certain values of n
n,n
from the fact that t′(x,n) < 0, we infer that t(x,n) is a positive and decreasing
function of x in the interval (0,n], for each n 15. To complete the proof, we
≥
have to show that t(γ ,n) < 1 for all n 15. By making use of upper bounds of
n
≥
γ /n for various values of n, it is easy to see that t(γ ,n) < 1 for n 73. A direct
n n
≥
computation using mathematica shows that t(γ ,n) < 1 for 15 n < 73 also. The
n
≤ (cid:3)
proof is complete.
Corollary 1. Let f 0 satisfies the hypothesis of Theorem 1. Then s (f)(z) is
∈ SH n,n
univalent in the disk
(i) z < 1/4, whenever n 7,
| | ≥
(ii) z < 1/2, whenever n 22,
| | ≥
(ii) z < 3/4, whenever n 78.
| | ≥
The bound for the radius of univalence r of s (f) for certain values of n are
n,n n,n
listed in Table 1.
The following shearing theorem due to Clunie and Sheil-Small is needed for the
proof of Theorem 2.
Theorem D. [5, Theorem 5.3] A locally univalent harmonic function f = h+g in
D is a univalent mapping of D onto a domain convex in the direction θ if and only
if φ = h ei2θg is a conformal univalent mapping of D onto a domain convex in
θ
−
the direction θ.
The proof of Theorem 2 is similar to the proof in Theorem 1 with the help of the
corresponding coefficients inequalities and the sharp lower bound for the two point
distortion theorem (Lemma 2) for f 0 .
∈ KH
Lemma 2. If f = h+g 0 , r (0,1), t,ψ R, then
∈ KH ∈ ∈
f(reit) f(reiψ) 1 r
− − .
reit reiψ ≥ (1+r)3
(cid:12) − (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
Proof. For every pair of po(cid:12)ints reit and reiψ(cid:12)in D, we can find a θ R such that
∈
(h(reit) h(reiψ))+(g(reit) h(reiψ)) = (h(reit) h(reiψ))+eiθ(g(reit) h(reiψ)).
− − − −
10 S.Ponnusamy,A.Sairam Kaliraj, and V.V.Starkov
Since f 0 , f is convex in every direction and hence, by Lemma D, the function
h ei2θg∈isKuHnivalent in D for every θ R. The desired conclusion follows from the
− ∈
two point distortion theorem for univalent analytic functions (see [6, Corollary 7,
(cid:3)
p. 127]).
3.2. Proof of Theorem 2. Let f = h+g 0 . Then the Taylor coefficients of h
∈ KH
and g satisfy the inequality (see [5,7])
n+1 n 1
(12) a and b − for all n 2.
n n
| | ≤ 2 | | ≤ 2 ≥
Following the proof of Theorem 1 with the same notation and (12), we get that
∞ ∞
k(k +1) k(k 1)
R (z) rk−1 + − rk−1 = R +T .
n,m,r n m
| | ≤ 2 2
k=n+1(cid:20) (cid:21) k=m+1(cid:20) (cid:21)
X X
From Lemma 2, we obtain that
1 r
P (z) − R T .
| n,m,r | ≥ (1+r)3 − n − m
The inequality P (z) > 0 holds for all z D 0 , whenever µ(n,m,r) > 0,
n,m,r
| | ∈ \ { }
where µ(n,m,r) is defined by (6). However, µ(n,m,r) > 0 for all r (0,r ),
n,m
∈
where r is the unique positive root of the equation (6) which is less than 1. Since
n,m
the affine span of 0 is a linear invariant family and α = sup h′′(0) /2 = 2, we
KH f∈KH | |
have (see [7, p. 99])
(1 r)
′ ′
h(z) g (z) − for z = r.
| |−| | ≥ (1+r)3 | |
Following the proof technique of Theorem 1, we conclude that s (f) is locally
n,m
univalent in z < r . Now, let us first consider the special case m = n. In this
n,m
| |
case, (6) reduces to
∞
1 r
µ(n,n,r) = − k2rk−1.
(1+r)3 −
k=n+1
X
From Lemma 1, the expression for µ(n,n,r) simplifies to
1 r rn[2+(2n 1)(1 r)+n2(1 r)2]
(13) µ(n,n,r) = − − − − .
(1+r)3 − (1 r)3
−
For 0 < r < 1, µ(n,n,r) > 0 if and only if ψ(n,r) > 0, where
ψ(n,r) = (1 r)4 [2+(2n 1)(1 r)+n2(1 r)2](1+r)3rn.
− − − − −
From the continuity of ψ(n,r) and from the fact that ψ(n,0) > 0 and ψ(n,1) < 0
, it is evident that there exists a real number r > 0 such that ψ(n,r) > 0 for all
n
r (0,r ) and ψ(n,r ) = 0. To complete the proof, we need to find the lower bound
n n
∈
for r for large values of n. By letting r = 1 α /n in ψ(n,r) and making use of
n n
−