CC BY
81DOI: 10.15393/j3.art.2014.2729 Issues of Analysis. Vol. 3(21), No. 2, 2014
V. V. STARKQY
UNIVALENCE OF HARMONIC FUNCTIONS, PROBLEM OF PONNUSAMY AND SAIRAM, AND CONSTRUCTIONS OF UNIVALENT POLYNOMIALS1
Abstract. The criteria of the univalence of a harmonic mapping is obtained in this paper. Particularly, it permits to formulate the hypothesis of the harmonic function classes equality S% = S% (S) (hypothesis of Ponnusamy and Sairam), in analytic form. The method of construction of the univalent harmonic polynomials with desired properties, according to a given harmonic function, is obtained by means of the univalence criteria.
Key words: harmonic functions, criteria of the univalence, harmonic univalent polynomials.
2010 Mathematical Subject Classification: 30C55, 30C10, 31A05.
1. Introduction. Let S be the class of all analytic and univalent functions of the form f (z) = anzn in the disk A = {z E C : \z\ < 1}.
n=2
The problem of determining a necessary coefficient condition in this class was set up by Bieberbach [1]. It consisted of validity of the inequality \an\ < n for each n E N (equality for Koebe function k(z) = z/(1 — z)2 and its rotations ke(z) = e-%ek(ze%e)). The Bieberbach hypothesis contributed largely to the origin and development of a great number of ideas and methods in complex analysis. The full solution of this problem was finally given by de Branges [2].
The theory of univalent harmonic functions began its active development since the eighties of the previous century. Here the main object
1 This work was supported by RFBR (project No 14-01-92692 and No 14-01-00510) and by Program of Strategic Development of Petrozavodsk State University.
© Starkov V. V., 2014
of research is the class S h of harmonic univalent and sense-preserving functions f in A given by f (z) = h(z) + g(z), where
h(z) = ^^ anzn and g(z) = ^^ a-nzn
n=1 n=1
with a1 = 1 (see, for example, [3]). The class Sh is an analog of the class S. For obtaining information about functions from Sh it is often convenient to have such information about functions from subclass SH C C Sh, where SH = {f G Sh : a-1 = 0}. This circumstance explains the interest in studying the class SH. In [4] Clunie and Sheil-Small formulated
the following conjecture (the problem about coefficients in S0): for each 0 and „ natur„i number „ g n the ;-----„^
function f G S0 and a natural number n G N, the inequalities
. . (2n +1)(n + 1) , (2n - 1)(n - 1) , ,
\an -^-J-, \a-n\ < --^-J-, \an - a-n\ < n (1)
are true. A great deal of papers are dedicated to this conjecture. Particularly, in the paper of Ponnusamy and Sairam Kaliraj [5] this conjecture, together with some other results were proved for the subclass
S0 (S) = {f = h + g G S0 : h + e^g G S for some 0 G R}
of the class S0. Besides, in this paper the authors formulated hypothesis about the equality of these classes, namely, S0 (S) = S0, the proof of which would permit to obtain the full solution of the coefficients problems (1) of Clunie and Sheil-Small. In the present paper criteria of univalency of harmonic functions (Theorem 1) is obtained. With the aid of this, a criteria for functions belonging to the class S0(S) (Theorem 2) is obtained and several examples are performed. Theorem 3 permits to construct harmonic univalent polynomials for a given f G S .
2. Univalence criteria and hypothesis of Ponnusamy and Sairam Kaliraj. The univalence criteria of arbitrary harmonic function in A of the form
f (z) = Y,(anZn + a-nzn ) (2)
will be obtained analogously to Bazilevich's [6] univalence criteria for analytic functions:
Theorem A. [6] An analytic function f (z) = ^^ anzn in A is univalent
n=1
in A if and only if for each z E A and each t E [0, n/2],
£ anS^z-i =0
n=1
sin t
i sin nt\ sin t
= n.
t=0
Theorem 1. Harmonic sense-preserving function in A, determined by the formula (2), is univalent in A if and only if for each z E A \ {0} and each t E (0,n/2],
£
n=1
(anz a—nz )
sin nt sin t
= 0.
(3)
Proof. Let f E Sh■ Then, for z1,z2 (z1 = z2) from A, we have
f (z2 ) - f (zi)
z2 - zi
= 0.
Particularly, for z1 = re"0i = rei02 = z2, r E (0,1), 9k E R, this is equivalent to
f (rei02) - f (rei01) =
rei02 _rei0i
OO /
= £ rn—1 (a,
n=1 ^
ein02 _ein0i
ei02 — e"
e—in02 _e—in0i
+a—n
)
= 0.
Without loss of generality, we may assume that 91 < 92 < 91 + n. Let
t = E (0, n/2] and 9 = E R.
22
Then
ein02 _ein0i
ei02 — ei0i
and
= ei(n — 1) 92
=e
92 — 91
+£i ein 2 — e— 2
■92—9i
=e
e 2 —e 2
e—in02 _e—in0i
„int „—int i(n — 1)0 e — e
eit e—it
ei02 — ei0i
en-e
2—9i
8n-8
2 —9i
gin02 _ gindi \ _ Q — i^1
/ gin02 _ çinBl \
V eiQ2 - eiQi )
e
i&2 _ ei^i
„int „—int = e—i(n — 1)0 e - e
"T" (-e—2id ).
eit e
it
Hence (4) may be represented as
E
n=1
(anv'
l—iei(n—l)e - a_nTn
—ie—i(n—i)û e—2ie )sin nt
sin t
which is equivalent to
= 0
E
(a n Z a—nz
i sin nt
sin t
= 0,
(5)
where 2 = rel° E A, 0 = arg2, t E (0,^/2]. But (5) ^^ (3). Let us note, that (5) is fulfilled for 2 = 0 as well, because \a\ \ > \a-1 \, since f is sense-preserving.
Let us next prove the inverse proposition. Suppose a harmonic function (2) is sense-preserving in A and condition (3) is fulfilled. According to the accepted designations this is equivalent to fulfilling of the condition (4), i.e.
f (Z2) - f (zi)
Z2 - Zi
= 0 Vzi = rei01 = rei= Z2, r E (0,1), 6k E R.
Thus f is univalent on any circle {z E C : \z\ = r}.
The local univalence of the function f implies that df (Ar) C f (dAr), where Ar = {z E C : \z\ < r}. Then the assumption that f is not univalent (but locally univalent) in A implies the existence of a disk Ar (a disk of f univalence), in which f is univalent, but on c)Ar there exist points zi = z2 such that f (zi) = f (z2). This contradicts the univalence of f on the circle {z E C : \z\ = R}. This contradiction proves Theorem 1. □
The following theorem represents the criteria of belonging of a function to the class SH (S).
Theorem 2. Let f E SH. Define
A = A(z, t) = > a.
n=i
sin nt
n
sin t
z
B = B (z,t) =
^ .
_sin nt r
y a_n z
sin t
n=i
and
E = {(z,t) E A x (0, n/2] : \A(z,t)\ = \B(z,t)\}.
Then f E S°H (S) if and only if there exists a 0 E [0, 2n) such that
A(z, t) = -e^B(z, t) V (z, t) E E.
Proof. Let f = h + g E SH (S) and it be determined by (2). According to the definition, SH (S) 9 f if and only if there exists a 0 E [0, 2n) such that h + e1^g E S. Theorem A implies that
^ . t
/ (an +
e aa—n) . z = 0, for z E A \ {0}, t E (0, n/2], sin t
n=1
^ A(z,t) = -e^B(z,t) V (z,t) E E. This completes the proof. □
Corollary 1. Let f E SH and E be the set defined in Theorem 2. If E = then f E S°H(S).
, x ^—sin nt n
Remark 1. Denote q(z) = > -z . Then
sin t
sin t
n=1
L ) L ) f
Hence the statement of Theorem 2 may be represented by means of these integrals.
Example. For fixed R E (0,1), consider the harmonic function f = h + g in A, where
h(z) = 7-and g(z) = kz2, k E R.
(1 — Rz )2
By means of Theorem 1 we determine for which values of k and R, f E SH • Applying Theorem 2, let us show that for these values of parameters, the function f E SH (S)•
Firstly we need to determine, for which values of the parameters, the function f is sense-preserving in A. The condition of sense-preservation means the validity for any z E A, the inequality
W | — \g' | =
1 + Rz
(1 — Rz)3
— |2kz| > 0
which holds if
mm
\z\=r
1 + Rz
(1 - Rz)3
1 - Rr (1 + Rr)3
> 2\k\r, V r e [0,1).
Since the function equality gives
1 - Rr r(1 + Rr)3
decreases with respect to r, the latter in-
2\k\ <
1R
(6)
(1 + R)3'
The values of parameters for which the function f is sense-preserving in A is determined from (6). Further, let the condition (6) be valid. According to Theorem 1, f is univalent if and only if A(z,t) = B(z,t) (designations from Theorem 2) in (A \ {0}) x (0,^/2]. Let us show that the equality A(z,t) = B(z,t) is not possible. We see that
A(z,t) = J]
nR
eint _ e-int
n-l^ne e
n=l 1
2i sin t
2i sin t
ze
it
ze
it
(1 - Rzeit)2 (1 - Rze-it)2
z(1 - R2z2)
(1 - 2Rz cos t + R2z2)2
and
B(z,t) = 2kz2 cos t. Now we show that for t E (0, n/2) and z G A \ {0}, the equation
z (1 - R2z )
(1 - 2Rz cos t + R2z2)2
= 2kz2 cost
(7)
has no solution. It is suffices to show that in (7) the absolute values of the left hand and right hand sides are not equal. If
\2k\ = then for some t,
1 R2 z
22
z cos t(1 - 2Rcos tz + R2z2)2
= L(z,t),
\2k\> min L(z, t) = min L(z,t) = L(-1,t) =
0<\z\<1 \z\ = 1
= 1 — R2 1 — R
= cost(1 + 2Rcost + R2)2 > (1 + R)3
for values of t under consideration. Thus a contradiction with (6) is obtained. Therefore, if the condition (6) is fulfilled, then all functions f from this example are univalent. As shown above, \A(z,t)\ = \B(z, t)\, in (A \{0}) x (0, n/2] and therefore the set E defined in Theorem 2 is empty. Hence, for parameters values satisfying the inequality (6), f E SH (S).
3. Univalent harmonic polynomials are functions of the following form P = h + g, where h and g are classic polynomials in z. Generally speaking, there is not much information about univalent harmonic polynomials than about other functions from Sh (here we speak about univalence in A) (see [7]).
For example, in a survey paper [8] the authors note: "Finding a method of constructing sense-preserving univalent harmonic polynomials is another important problem". In the analytic case, Bazilevich [6] proposed a method of construction of univalent polynomials, associated with a given function from S. Further, his idea has been transferred to harmonic case in Theorem 3. Moreover, unlike with analytic case, the proof will be constructive. Thus, Theorem 3 gives an opportunity to construct harmonic univalent polynomials of sufficiently high power for any function f E SH.
Lemma 1. If f e S°H,r e (0,1), e R, then
f (relt) — f (rei*)
re.it_rei^
def
>
1 — r i 1 — r\a 1 /1 — r\
V1 + r/ [ V 1 + r)
4ar V 1 + r
2a
where a(= ordSH) = supfeSn \a2\.
Proof. If f = h + g e SH, then the linear invariance of the class S h (see [9]) implies that
f ( H+a ) — f (a)
F(z)= ht + M I |2) e Sh Va e A. h' (a)(1 — \a\2)
Let
a = rel*
z+a ' 1 + az
it
= re , i. e. z =
it
re — a 1 — areit
From the affine invariance (see. [9]) of the class S h , it follows that the function
^(z) Hence,
F(z) - a-1 F(z) 1 -\a-i\2
e SH, where a-1 = -7— (0) =
dF
dz
g'(a)
h'(a)
\F(z)|(1 + |a—i|) > \F(z) - a-!F(z)\ > \0(z)\(1 - |a-i\2),
so that \F(z)\ > \0(z)\(1 - |a-1|). Since f E S°H, then |k(0) = 0, i.e. g'(0) = 0. Since f is sense-preserving in A, we have \g'(z)/h'(z) \ < 1 in A. Therefore, according to Schwarz's lemma, \a-1 \ = \g'(a)/h'(a)| < r and
\F(z)\ > \0(z)\(1 - r). For any function 0 = H + G E SH and any z G A, one has [9]:
(8)
1
2a
1
(1 _ |zi)a-1 < \^(z)\ and -^^ < \H'(z)\,
(1 + \z\)a+1
from which with regard to (8) we obtain
1
\f (re ) - f (re*)\> —
1
(1 - r)(1 - r2)\h'(re^)\ >
>
2a
1 - r f 1-r\a I" /1 - \z\ \
V1 + r) [ V 1+ \z \ )
where
z=
r(eit - e^) 1 _ r2ei(t-4>)
1_<w
Denote by w = eis = ei(t-^) and Z = ~--—• Then
w =
1 - Z
1 - r2 Z
, 0 <\z\ = r\Z\<
1 - r2
2r
1+r
w
-, \1 - w\ =
\Z\(1 - r2) \1 - r2Z \
Represent (9) as follows
(9)
f (reit) - f (re^)
reit_rei^
>
2a
1 - r\Z\
1 - r (1-r\a L /1 - r\Z \ N
V1 + r) L v 1 + r\z\J
\1 - r2Z \ r\Z\(1 - r2)
>
a
a-
>
2a\l+r) I V 1 + x)
2\
1 — rx x(1 + r)
x = r\C\ G [0, 2r/(1 + r2)]. Define
u(x) = —
x
1
(—x)
V1 + x)
0 2r '
, x G 0, 1 + r2
We show that u is decreasing on (0, 2r/(1 + r2)). Then,
a—1
x2u'(x) < 0
(—x)
V1 + x)
2ax
(1 + x)2
1 +
V1 + x)
2ax +1 — x2 < (1 — x2)(1±x^ ^^ u1(x) < u2 (x),
where
(10)
0
(11)
u1 (x)=ln(1 + 2ax — x2) and u2(x) = a(ln(1 + x) — ln(1 — x)) + ln(1 — x2). Note that ui (0) = 0 = u2(0) and
, , . 2(a — x) 2(a — x) . , .
ui (x) = -—^--/ = u2 (x),
1K ' 1 + 2ax — x 1 — x2 2V h
since (see [9]) 3 < a < 48,9. This proves the validity of (11) and hence u(x) is decreasing on (0, 2r/(1 + r2)) and hence
min u(x) = u . _
xE[0,2r/(1+r2 )] V 1 + r2
( — ) V 1 + r2)
1 + r2 2r
1
(1 + r)
2a
Then from (10) we obtain the desired inequality in Lemma 1. The proof of the lemma is complete. □
In Lemma 2 below the estimates of coefficients of functions from Sjj will be obtained. The estimates are not exact, but they are sufficient to achieve the goal which we set up in this section (see Theorem 3).
Lemma 2. If a = ordS#, S° 3 f with series expansion (2), then the following inequality is true:
(2e2)a
\a±n\ < na, n G N.
2a
a-
a
a
Proof. Sheil-Small [9] proved that for f G S°, the inequality
lf (z)!< è
G+S )
- 1
, z G A.
Hence for n G N and r G (0,1),
\a-n\ =
f (z )dz
2mJ| , |=r zn+1
<
2arn
■0(r) 2a '
The same estimate is also true for \an\, n G N. In order to find a minimum of the right hand side of inequality we find the point of minimum of the function ln -0(r):
(ln -0(r))' =
2a n
2a
1 - r2
--= 0 ^^ r2 + — r - 1=0.
n
Therefore, r0 = \Ja2/n2 + 1 — a/n is the point of minimum and
\an\ < 0) for all integer n,
2a
where
^(r0 ) =
=( ^+n)" [2a (i+/ n2 n
Using inequality V1 + x < 1 + \fx, x > 0, we obtain
*(ro) < Q ' (
1+n
n
n 2a
2a
/ a\a
(2+a) =
= <2nr(2n + 1) (1 + 2n)
i/(2a)'
2a
Introduce
tf (y) = y ln 1 + -
y
('+a-
(12)
1
1
r
a
Then supy>0 ^(y) = 1, since limy^+o ^(y) < 1, ^(y) = 1,
and, if tf(y) has maximum on the interval (0, œ) at the point y0, then *'(yo) = 0. This gives
lnfl + -M = 1
V yoj
yo) 1 + yo
which implies that
Hence from (12) we have
^(yo ) =
^(ro) < (2n)ae2a and \a±n\ <
yo
1 + yo
< 1.
2\a
(2e2 ) 2a
n
and the proof of Lemma 2 is complete. □
Theorem 3. Let f E SH and have the series expansion (1), a = = ordSH, s = [a + 2], where [.] denotes the integer part of a number. Let m E N, e E (0, (2e2)a/(2a)) and
r
(
1
2ae
(2e2 m)a
l/r
,1
(13)
let N E N be so large that
2 1 - r
(a(2 + ln 2) + ln(s!) - (s + 1) ln \ ln r\ - ln(1 - r) - a ln
\ ln r\ ln
1
(1 + r)
2a
ln N ln r
+ 2 < N, m < N, <
N - 2[a + 2]
Then the harmonic polynomial
N
N
P (z) = J2(Ck zk + c-k zk ) = Y^(ak rk zk + a-k rk zk )
ak k k k k
k=1 k=1
is univalent in A and moreover \a±k — c±k\ < e for all k = 1,... ,m.
Proof. Let us note that 2ae/(2e2)a < 1 for the indicated values e. Hence 2ae
x < 1 for k = 1,... ,m.
(2e2k)a ' '
The function ((x) = (1 — xa(2ae)/(2e2)a)x decreases on (0,1], since d ln ((x)
--- < 0 for x E (0,1]. Therefore, ((1/k) < ((1/m) for each k =
dx
= 1,... ,m, and (13) implies the inequality ((1/k) < ((1/m) < r. Hence
(1 — rk)(2e k) < e for k = 1,... ,m. 2a
From these inequalities and Lemma 2 we obtain inequalities for the first coefficients c±k = rka±k of the polynomial P(z):
\a±k — c±k\ = \a±k — rka±k\ < e, k = 1,...,m.
Let us verify, by the method of mathematical induction, that
\ sinnt\ < nsint V t E (0,^/2] and V n E N.
The inequality is true for n = 1. Assume that the inequality is true for (n — 1). Then
\ sinnt\ < \ sin(n — 1)t\ cost + \ cos(n — 1)t\ sint <
< sin t[(n — 1) cos t + \ cos(n — 1)t\] < n sin t.
Let us show under the hypothesis of the theorem that the polynomial P(z) is univalent in A. For z E A, let us estimate the remainder Rn of the series. We have
\Rn \ = £ (anzn — a-nzn )r^ J]
£-' Q1T1 T rv -'
n=N+1
sin n^ _ (2e2)a
sin t
n ' r
a
n=N+1
according to Lemma 2. By hypotheses, N\ ln r\ > 2s ln N > a + 1, and
+ , to ), the function T(x) =
a +1
therefore, N > —--. But for x E . .
\ ln r\ \ ln r\
= xa+1 rx decreases. Hence with the increasing values of n, the terms of
a+1 n
the series ^^ na+1 rn decreases and therefore,
n=N
V na+1 rn < I T(x)dx < I xsrxdx.
n=N+1 N N
oo
Integrating by parts successfully s times, we obtain
Ns rN sN s-1rN s(s — 1)Ns-2 r
Ts-N
s-2„ N
xsrxdx =--+
N \ ln r\ \ ln r\ 2
+
\ lnr\3
+ —+
s!r
N
\ lnr\s+1
Let us represent the condition N \ ln r\ > 2s ln N of the theorem as r N Ns < < 1. Hence rNNj < rN/2 for any j = 1,..., s. Taking into account of these inequalities, we have the estimate
xsrxdx <
N
<
sir
N/2
\ ln r\
s+1
I lnr|s I lnr|s 1
+ L,—^ +... +
\ ln r\ , rN/2
Hence
si (s — 1)i ..... 1i
(2e2)a rN/2-1sl
\RN \ < -"pj-p+Y.
a \ lnr\s+1
From (14), Theorem 1 and Lemma 1 we obtain
+ rN
„N/2-1si
<
\ ln r\
s+1 •
(14)
N
E
n=1
(anzn — a-n zn)r
sin nt
>
E
n=1
(anzn — a-nan )rn
sin t sin nt
>
sin t
>
4a 1 + r
1 — r i 1 — r\a 1 /1 — r\
\1 + ^ V1 + r)
2a
— \Rn\ >
(2e2)a rN/2-1 si
a \ ln r\ s+1
= Q.
If now Q > 0, then the univalency of the polynomial P follows from Theorem 1. The latter inequality is equivalent to
2 1 — r
(a(2 + ln 2) + ln(s!) — (s + 1) ln \ ln r\ — ln(1 — r) — a ln
\ ln r\
1+r
ln
1
(1 + r)
2a
+ 2 <N,
which proves Theorem 3. □
The proof of Theorem 3 implies the following result.
DO
DO
Corollary 2. Let f G S0 and f have the expansion (1), a = ordSH, r G (0,1) and 0 = m G Z. Then the m-th coefficient cm = amrm of the univalent function
f (rz) = Y,(cnzn + — zn )
n=1
may be replaced by cm + A, where A G C, and
|A| <
1 + r /
1 — r /1 — r\ i /1 — r\
V1 + ^ V 1 + r)
2a
without loss of univalence. Proof. According to Lemma 1,
E
n=1
(an zn — a-n zn)rn
sin nt sin t
>
4a
1 — r i 1 — r \ " 1 /1 — r\
V1 + ^ V 1 + r)
2a
By Theorem 1, the condition of the univalence of the new function, obtained by the variation of the m-th coefficient cm gives
E
n=1
(an zn — a-n zn)rn
sin nt sin t
+ aAp
\m\
sin mt sin t
> 0, z G A \ 0, p = Izl,
t G (0,^/2], where |a| = 1. It is fulfilled provided
A
sin mt
sin t
<
4a 1 + r
1 — r /1 — r \ a 1 /1 — r\
V1 + ^ V 1 + r)
2a
Hence it is certainly fulfilled if the inequality concerning |A|, from the statement of Lemma, is true. □
Acknowledgment. I would like to thank the reviewers for reading this article carefully and making valuable comments.
References
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[3] Duren P. Harmonic mappings in the plane. Cambridge, 2004, 214 p.
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Received September 1, 2014.
Petrozavodsk State University,
Lenin Avenue, 33, 185910 Petrozavodsk, Russia.
E-mail: [email protected]
