Re[zf"(z)/f`(z) + 1] > 0, z £ U => Re[zf`(z)/f(z)} >\, z

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PROCEEDINGS
OF THE
AMERICAN MATHEMATICAL
Volume
99, Number
3, March
SOCIETY
1987
MARX-STROHHÄCKER DIFFERENTIAL
SUBORDINATION SYSTEMS
SANFORD S. MILLER AND PETRU T. MOCANU
ABSTRACT. Let f(z) = z + aiz2 + ■■■be analytic in the unit disc U and let
k(z) = z/(l — 2). The classic Marx-Strohhäcker
result, that a convex (univalent) function / is starlike of order ^, can be written in terms of differential
subordinations
as
zf"(z)/f'(z)
< zk"(z)/k'(z)
=* zf'(z)/f(z)
< zk'(z)/k(z).
The authors determine general conditions on k for which this relation holds.
They also determine a different set of general conditions on k for which
zf'(z)lf(z)
Finally,
differential
< zk'(z)/k(z)
subordinations
=* }(z)/z < k(z)/z.
with starlike
superordinate
functions
are
considered.
1. Introduction.
Let / be analytic in the unit disc U — {z: [z\ < 1}. The
function /, with /'(0) 7^ 0, is convex (univalent) if and only if Re[zf"(z)/f'(z)
+ l] >
0 in U. The function /, with /(0) = 0 and /'(O) ^ 0, is starlike (univalent) if and
only if Re[zf'(z)/f(z)}
> 0 in U. If, in addition, Re[z/'(z)//(,z)]
> ± then / is said
to be starlike of order | [1, Vol. 1, p. 137].
The classic result of Marx [3] and Strohhäcker
is starlike of order ^, that is
(1)
Re[zf"(z)/f'(z)
[10] asserts that a convex function
+ 1] > 0, z £ U => Re[zf'(z)/f(z)} >\,
z£U.
We can simplify this relation by expressing it in terms of subordination.
If F and
G are analytic in U, then F is subordinate to G, written F < G or F(z) < G(z), if
G is univalent, F(0) = G(0) and F(U) C G(U).
If we let k(z) = z/(l —z), then (1) can be rewritten as
(2)
zf"(z)/f'(z)
< zk"(z)/k'(z)
=> zf'(z)/f(z)
< zk'(z)/k(z).
This Marx-Strohhäcker differential subordination system of first type holds for other
functions k; in [5, p. 194] and in [7, Corollary 2.3] it is shown that (2) also holds
when k(z) = z/(l — z)2 is the Koebe function. In §2 of this article we determine
general conditions on k for which (2) holds.
If k is the Koebe function, then Marx [3] and Strohhäcker [10] also showed that
(3)_
zf'(z)/f(z)
< zk'(z)/k(z)
=> f(z)/z < k(z)/z.
Received by the editors December 30, 1985.
1980 Mathematics Subject Classification (1985 Revision).
Primary 30C80; Secondary 30C45,
34A20, 34A40.
Key words and phrases.
Differential subordination,
superordinate,
convex function, starlike
function, univalent function.
The first author acknowledges
support received from the International
Research Exchange
Board through its exchange program with the Romanian National Council of Science and Technology.
©1987
American
0002-9939/87
527
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528
S. S. MILLER AND P. T. MOCANU
§3 deals with general first order differential subordinations of the form ip(p(z),zp'(z))
< h(z), where h is a starlike function. In addition, as a special application of these
results we determine conditions on k for which the Marx-Strohhäcker
differential
subodination system of second type, as given in (3), holds.
We close this section with two lemmas that will be used in subsequent sections.
LEMMA 1. Let F be analytic in U and let G be analytic and univalent on U,
with F(0) = G(0). If F is not subordinate to G, then there exist points zq £ U and
Ço€ dU, and an m > 1 for which F(\z\ < \z0\) C G(U),
(i) F(z0) = G(ço), and
(n)z0F'{z0) = mçi)G'{Ço)The proof of a more general form of this lemma may be found in [4, Lemma 1].
The second lemma deals with the concept of a subordination chain. A function
L(z, t), z £ U, t > 0, is a subordination chain if L(-,t) is analytic and univalent in
U for all t > 0, L(z,-) is continuously differentiable on [0, oo) for all z £U, and
L(z,ti) < L(z,t2) when 0 < £i < t2.
LEMMA 2 [8, p. 159]. The function L(z,t) = ai(t)z-\-,
with ai(t) ^ 0 for
all t > 0, is a subordination chain if and only if Re[z(dL/dz)/(dL/dt)}
> 0 for
z £U and t > 0.
2. Marx-Strohhäcker
differential
subordination
systems
of first type.
THEOREM 1. Let q be univalent in U, with q(0) = 1. Set Q(z) = zq'(z)/q(z),
h(z) — q(z) + Q(z) and suppose that
(i) Q is starlike in U (logç is convex in U), and
(ii) Re[zh'(z)/Q(z)[ = Re[q(z) + zQ'(z)/Q(z)\ >Q,z£U.
If B is an analytic function
(4)
on U such that
B(z)<q(z)
+ zq'(z)/q(z)
then the analytic solution p of the differential
(5)
zp'(z) + B(z)p(z) = 1
= h(z),
equation
(p(0) = 1)
satisfies p -< l/q.
PROOF.
Conditions
(i) and (ii) imply that the function
h is univalent
(close-
to-convex) [1, Vol. 2, p. 2], h(0) = 1, and hence (4) is well defined with 7?(0) = 1.
If we let b(z) = /02[77(í) — l]í_1cíí, then (5) has an analytic solution given by
p(z) = j* eb^ at)zeb<-z). Condition (i) implies that q(z) ^ 0 in U. Hence the
function l/q is analytic and univalent in U.
Without loss of generality we can assume q is univalent and q(z) ^ 0 on U. If not,
then we can replace q, B, and p by qr(z) = q(rz), Br(z) = B(rz), and pr(z) = p(rz)
respectively, where 0 < r < 1. These new functions satisfy the conditions of the
theorem on U. We would then prove pr < l/qr, and by letting r —►1~ we obtain
P < !/<?•
The function
(6)
L(z, t) = h(z) + tQ(z) = q(z) + (1 + t)zq'(z)/q(z)
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MARX-STROHHACKER
is analytic
z£U.
DIFFERENTIAL
SUBORDINATION
in U for all t > 0, and it is continuously
SYSTEMS
differentiable
Since Q'(0) = q'(0) ^ 0 we have dL/dz(0,t)
529
on [0, oo) for all
= <?'(0)(2+1) ¿ 0 for t > 0.
From (ii) we obtain
dL I dL
Z~dz~ lk
Re
Re
MHi +t)«™
Q(z)
Q(z)
for z £ U and t > 0. Hence by Lemma 2, L(z, t) is a subordination chain and so
we have L(z,ti) < L(z,t2) for 0 < ii < t2. From (6) we have L(z,0) = /i(z) and
so we obtain
(7)
L(c,t)#h(U)
for |ç| = 1 and t > 0.
Now assume that p -^ l/q. From Lemma 1 there exist points zq £U and ço G dU,
and an m > 1 such that p(z0) = l/q(Ço) and z0p'(zo) = -mçoL/(co)/q2(oj).
Using
these results together with (5) and (6) we obtain
or
\
1
zop'(zo)
P(zo)
p(zo)
ß(zo) = -t—t-i~^-
,
s
mÇo9'(ç0)
= <?to +-t-t—
<7Ko)
,,
,>
= Mfo, m - 1 .
By using (7) we deduce that 7?(zo) ^ /i(í7), which contradicts (4). Hence p -< 1/g,
completing the proof of the theorem.
We now turn our attention to considering the Marx-Strohhäcker
differential subordination system of first type as given in (2). If we let p(z) = zf'(z)/f(z)
and
q(z) = zk'(z)/k(z)
then (2) becomes
p(z) + zp'(z)/p(z)
< q(z) + zq'(z)/q(z)
=» p(z) < q(z).
This differential subordination result was considered in [5, Theorem 3]. Before using
this general result to prove (2), however, we have to first prove that zf'(z)/f(z)
is
analytic in U, a condition which does not follow directly from the left side of (2).
THEOREM 2. Let q satisfy the conditions of Theorem 1 and let
(8)
k(z) = 2exp /
Jo
[q(t) - l]i_1 dt.
If f(z) = z + a2z2 + ■■■ is analytic in U, and
(9)
zf"(z)/f'(z)
then zf'(z)/f(z)
< zk"(z)/k'(z),
is analytic in U and zf'(z)/f(z)
-< zk'(z)/k(z).
PROOF. From (8) we obtain
. .
zk'(z)
q(z) =
y
k(z)
,
and
,. ,
, ,
zq'(z)
t
zk"(z)
h(z) = q(z) + -—^r = 1 +
w
^ '
q(z)
k'(z) '
Thus zk'(z)/k(z)
is univalent, and by (ii) of Theorem 1 we see that zk"(z)/k'(z)
is also univalent.
Since condition (9) implies f'(z) ^ 0, the function B(z) =
1 + zf"(z)/f'(z)
is analytic in U, and satisfies B < h. For this particular B
equation (5) has the analytic solution p(z) = f(z)/zf'(z).
Since all the conditions
of Theorem 1 are satisfied, we have p < l/q. Since l/q ^ 0 this implies that p / 0,
and so l/p(z) — zf'(z)/f(z)
is analytic in U. In addition, from p ■< l/q and l/q ^ 0
we obtain 1/p < q, which yields the desired conclusion zf'(z)/f(z)
-< zk'(z)/k(z).
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S. S. MILLER AND P. T. MOCANU
530
EXAMPLES, (a) The function q(z) = [1 + (1 - 2a)z]/(l
satisfies (i) and (ii) of Theorem
z/(l-z)2^^
and
1. From Theorem
zf"(z)
zk"(a,z)
f'(z)
< k'(a,z)
zf'(z)
f(z)
- z), with 0 < a < 1,
2 we obtain k(z) — k(a,z)
—
_ 2z(l - a)(2 + (1 - 2a)z)
:
(l-*)(l + (l-2a)«)
zk'(a,z) _ l + {l-2a)z
k(a,z)
1 —z
The cases a = ^ and a — 0 correspond to the functions z/(l — z) and z/(l — z)2
respectively, which were mentioned in §1.
(b) The function q(z) — z/(l — e~z), which is a convex (univalent) function [6,
p. 70] satisfies q(0) = 1 and
Q(z) = zq'(z)/q(z)
= 1 + z/(l -ez)
= l-
q(~z).
Since q is convex we conclude that Q is also convex. By using (1) we obtain
RezQ'(z)/Q(z)
> A, which implies condition (i) of Theorem 1. Condition (ii) also
holds since
Re[q(z) + zQ'(z)/Q(z)\ > l/(e - 1) + 1/2 > 0.
Hence by Theorem
2 we obtain k(z) = ez — 1 and
zf"(z)/f'(z)
< zk"(z)/k'(z) = z
=> zf'(z)/f(z)
-< zk'(z)/k(z) = z/(l - e~*).
(c) The function q(z) = [(1 + z)/(l - z)]ß, with 0 < ß < 1, satisfies (i) and (ii)
of Theorem 1 and by Theorem 2 we have
1+ z
f'(z)
. z2 + 2ßz-l
1-z2
1+ z
1-z
zf'(z)
f(z)
*
(d) Employing the function q(z) = eXz in Theorem 1 we obtain Q(z) = Xz and
q(z) + zQ'(z)/Q(z)
= eXz + 1. Condition (i) is satisfied and condition (ii) will
be satisfied if Re[eA2: + 1] > 0. A simple calculation shows that this condition
holds for |A| < rp. = 2.183..., where ro = yo/SÏNyo and yo is the smallest root
of COS y + exp(yCOTy) = 0. Hence by Theorem 2, for |A| < r0 = 2.183..., we
obtain k(z) = zexp JQz(eXt—l)t_1 dt and
zf"(z)/f'(z)
< zk"(z)/k'(z)
=> zf'(z)/f(z)
= eXz+ Xz - 1
-< zk'(z)/k(z)
= eXz.
(e) The function q(z) = 1 4- Xz, \X\ < 1, satisfies the conditions
and 2 and we obtain k(z) = zeXz and
*/"(*) . £^M _ A^(2+ Xz)
f'(z)
k'(z)
(1 + Xz)
z_f(zl
f(z)
of Theorems
1
z_k^l
* k(z)
+
•
In the special case A = 1 this simplifies to
zf"(z)/f'(z)
< z(2 + z)/(l + z)^
\zf'(z)/f(z)
This is an improvement of the result \f"(z)/f'(z)\
proved in [7, Theorem 3] and the result \zf"(z)/f'(z)
- 1| < 1.
< 3/2 => \zf'(z)/f(z)
- 1| < 1
+ l\ <2=> \zf'(z)/f(z)-l\
<
1 proved in [5, Theorem 5].
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MARX-STROHHACKER
3. Differential
DIFFERENTIAL
subordinations
with
SUBORDINATION
starlike
THEOREM 3. Let H(z) = z + a2z2 -\-be
a function
SYSTEMS
superordinate
function.
starlike in U and let i¡j: C2 -* C be
of the form iß(r, s) as w(r) ■sa with a > 0. In addition,
(10)
i>(q(z),zq'(z))
= w(q(z))
531
■(zq'(z))a
suppose that
= H(z)
has a univalent solution q. If p is analytic in U, with p(0) = q(0), then
(11)
w(p(z)) ■(zp'(z)r
< H(z) =>p(z) < q(z).
PROOF. If p -/i,q then by Lemma 1 there exist points zo £U and Ço& dU, and
an m > 1 such that p(zo) — g(Ço) and zop'(zo) = rafoç''((.o)■ Using these results in
(10) we obtain
w(p(zo))
■(z0p'(z0))a
= w(q(ç0))
■(mç0q'(io))a
= maH(ç0).
Since H(U) is starlike and ma > 1 we obtain w(p(z0)) ■(zop'(zo))a & H(U). This
contradicts (10) and so we must have p < q.
Note that (10) requires t/>(<j(0),0) = 0. The special case ip(r,s) = r1~asa, with
0 < a < 1, satisfies Theorem 3, with p(0) = g(0), and was considered in [4, p. 169].
For the purposes of this paper, Theorem 3 is most interesting when tp(r,s) = s/r.
Applying the theorem in this case we obtain
THEOREM 4. Let H(z) = z + a2z2 + ■• ■ be starlike in U and suppose that
(12)
Q(z) = exp /
Jo
is univalent.
(13)
H(t)t~xdt
If p is analytic in U, with p(0) = 1, then
zp'(z)/p(z) < zQ'(z)/Q(z) = H(z) => p(z) < Q(z).
COROLLARY 4.1.
If H is a convex (univalent) and ReH(z)
given in (12) is univalent
> —1, then Q as
and (13) holds.
PROOF. From (1) and (12) we obtain
Re[l + zQ"(z)/Q'(z)] = Re[77(¿) + zH'(z)/H(z)\
> -1 + £ = -¿,
which shows that Q is univalent [9, Corollary 3].
If we set 77(a) = q(z) — 1 and p(z) = f(z)/z we obtain
COROLLARY4.2.
Let q be convex with Req(z) > 0 and let
k(z) = zexp /
Jo
[q(t) - l]t~l dt.
U f{z) — z + a2z2 + ■■■is analytic in U, then
zf'(z)/f(z)
< zk'(z)/k(z) => f(z)/z < k(z)/z.
REMARKS. 1. This relation is precisely the one that the Koebe function satisfied
in (3) of §1. The corollary lists sufficient conditions on k to ensure that the MarxStrohhäcker differential subordination system of second type holds.
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532
S. S. MILLER AND P. T. MOCANU
2. The fonctions q used in Examples (a), (b), (c), and (e) also satisfy the conditions of the corollary, whereas q in (d) does not for |A| > 1. As a result, for
Example (a) we have
zf"(z)
zk"(a,z)
f'(z)
' k'(a,z)
zf'(z)
f(z
where k(a,z) = z/(l - z)2'1"^
For Example (b) we have
zf"(z)/f'(z)<z
f(z)
<Zk'(a,z)
k'(a,z)
,k(a,z)
and 0 < a < 1.
=►zf(z)/f(z)<z/(l-e
■■)=> f(z)/z
< (ez - l)/z,
and for Example (c) we have the extension
zf'(z)
f{z)
f(z)
1+ z
1-z
<
< exp
1+ t
i
rldt
/"
Jo
for 0 < ß < 1. In the special case ß = \ this simplifies to
zf'(z)
f(z)
1/2
<
f(z)
1
2exp(SENT1z)
i + (i-z2y/2
QU
From the last corollary we know that Re(l + zQ"(z)/Q'(z))
> —\ and that Q
is univalent. To show that the function Q is also convex it is sufficient to show
Re(l + zQ"(z)/Q'(z))
1 + zQ"(z)/Q'(z)
> 0 for |z| = 1. Since
= [(1 + z)/(l - z)]1'2 - 1 + ¿[(1 - z)"1 + (1 - z2)"1/2],
if we let z = elt, 0 < t < n, we obtain
(1 + z)/(l - z) = iCOT(i/2),
1 - z2 = -2¿e¿tSINí,
and
Re[l + zQ"(z)/Q'(z)\ = (\ COT(t/2))1/2 - | + [COS(í/2) + SIN(í/2)]/4(SINí)1/2.
If we let x = (COT(i/2))1/2
> 0, we obtain
Re[l + zQ"(z)/Q'(z)\ = (5x2 - 3^2x + l)/4\^i
> 0.
We close this section by considering another differential subordination having a
starlike superordinate
function. In [4, p. 170] and in [2, p. 192] it is shown that if
h is convex and p is analytic in U, then
p(z) + zp'(z) <h(z)
=> p(z) < z'1
/
Jo
h(t)dt.
We can replace the superordinate
function h with a function of the form
[(1 + z)/(l — z)]a. For a > 1 this function is not convex, but is starlike with
respect to 1.
THEOREM 5. Let ß0 = 1.21872... be the solution of ß-K = 3tt/2 - TAN-1/?
and let a = a(ß) = ß + 2TAN-1 ß/n for 0 < ß < ßQ- If p is analytic in U, with
p(0) = 1, then
p(z) + zp'(z) <
1
p(z) <
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1 + z 1/3
MARX-STROHHACKER
DIFFERENTIAL
SUBORDINATION
SYSTEMS
533
PROOF. Note that ßu < 3tt/2 - TAN"1 ß and ß < 3 - a for 0 < ß < ß0. If we
let h(z) = [(1 + z)/(l - z)]a and q(z) = [(1 + z)/(l - z)}ß then h(U) and q(U) are
domains given by the sectors |arg/i| < q7t/2 and |argg| < ßir/2 respectively. We
need to show that p < q. If this is not true, then there exist points zo £ U and
ço £ dll such that p(zo) = q(Ço) and p(|z| < |zo|) C q(U). We need to consider
separately the case p(zo) ^ 0 which corresponds to a point on one of the rays on the
sector q(U), and the case p(zo) = 0 which corresponds to the corner of the sector.
Case (i). If p(zo) ¥" 0 tnen çb ^ ±1. If we let ri = (1 + Ço)/(l - Ço) and "se
Lemma 1 we obtain
p(zo) + zop'(zo) = q(Ço)+ mçor/to) = (n)ß ~ (ri)ß-xmß(l
+ r2)¡2
= (ri)0[l + imß(l + r2)/2r\,
and
arg(p(z0) + zop'(zo)) = ßir/2 + TAN"1 (m/3(r + l/r)/2).
Since TAN-
is an increasing
function and m > 1, by considering
the cases r > 0
and r < 0 we obtain
ßn/2 + TAN-1/3 < |arg(p(z0) + z0p'(z0))\ < ßix/2 + tt/2.
Using the fact that ßn < 3tt/2 - TAN-1 ß for 0 < ß < ß0 and the definition of a
we obtain
qtt/2 < |arg(p(z0) + z0p'(z0))|
< 2tt - aw/2.
This implies that p(zo) + ^oP'(^o) hes outside the sector h(U), and since this contradicts the hypothesis of the theorem we must have p < q.
Case (ii). The case p(zo) = 0 can occur only if ß > 1, since ß < 1 implies
that the sector angle of q(U) is less than 7r and p(|z| = |zo|) cannot pass through
such a corner without itself having a corner. If p'(zo) / 0, then the smallest
possible value of arg(zop'(2o)) is given by (27r — ßir/2) — tt/2, which occurs when
p(|z| = |zo|) is tangent to the lower ray of the sector q(U). The largest possible
value of arg(zop'(zo)) is given by /?7r/2 + 7r/2, which occurs when p(|z| = |zo|) is
tangent to the upper ray of the sector q(U). Using these limitations and the fact
that ß < 3 —a we obtain
Q7T/2 < arg(p(z0)
+ zp'(z0))
< 2n - an/2,
which gives the same contradiction as Case (i). The subcase p'(zo) = 0 also yields
a contradiction since p(zo) + zop'(zo) —0^. h(U).
REMARKS. If we take ß = 1 in Theorem 5 we obtain a(l) = | and
p(z) + zp'(z) <
1 + Z 3/2
1-z
p(z) <
1
In this case the angle of the sector h(U) is 270°. The case with the largest angle
corresponds to Qo = a(ßo) — o(1.218...)
= 1.781 — In this case the angle of the
sector h(U) is 320.62...°, while the angle of sector q(U) is 219.36...°.
References
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2. D. J. Hallenbeck
Univalent functions.
and S. Ruscheweyh,
Vols. 1, 2, Mariner, Tampa, Florida, 1983.
Subordination
by convex functions,
Proc.
Math. Soc. 52 (1975), 191-195.
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
Amer.
S. S. MILLER AND P. T. MOCANU
534
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Current
of Mathematics,
Babe§-Bolyai University,
State
University
3400 Cluj-Napoca,
of
Roma-
address of P. T. Mocanu)
address
(S. S. Miller):
Department
of Mathematics,
University
California 94720
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of California,
Berkeley,
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