J o u r n a l of

Mathematics

and Applications

No 32, pp 109-116 (2010)

On coefficients’ estimates of concave univalent functions

Paweł Zaprawa

Submitted by: Jan Stankiewicz

Abstract: In the class Co(p) , p ∈ (0, 1) of univalent concave functions with a pole at p we find some estimates on the third coefficient as well as the residuum of a function, while its second coefficient is fixed

AMS Subject Classification: Primary 30C45; Secondary 30C80 Key Words and Phrases: covering domain, Koebe domain, convex in the direction of imaginary axis, real coefficients

Concave functions

Let a univalent meromorphic function map the unit disk Δ onto the complement (with respect to \overline{ℂ}) of a convex set. Such a function is usually called concave. If a concave function f has a simple pole at z = p, p ∈ (0, 1) and is normalized by f(0) = 0, f′(0) = 1 then we write f ∈ Co(p). Each function f from Co(p) can be expanded into a series

f(z) = z + ∑_{n=2}^{∞} a_{n}z^{n} , |z| < p  (1)

or

f(z) = ∑_{n=−1}^{∞} b_{n}(z − p)^{n} , |z − p| < 1 − p .  (2)

The problem of variability of n-th coefficient of functions in the class Co(p), as well as in the class S of univalent analytic functions, was considered by many authors. In the most stimulating paper Livingstone [4] conjectured that

|a_{n} − \frac{1 − p^{2n+2}}{p^{n−1}(1 − p^{4})}| ≤ \frac{p^{2}(1 − p^{2n−2})}{p^{n−1}(1 − p^{4})} .

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This conjecture was proved in 2007 by Avkhadiev and Wirths [3]. It is worth to recall a few papers connected with this conjecture, and more generally with the problem of coefficients. They were published by Miller [5], Avhadiev, Pommerenke, Wirths [1, 2], Wirths [7, 8] and others.

In [5] Miller proved the necessary and sufficient condition for a function f to belong to Co(p).

Theorem A. Let f be univalent meromorphic function. Then f ∈ Co(p) if and only if

Re \{\frac{2zf′(z)}{f(z) − f(ζ)} − \frac{z + ζ}{z − ζ} + \frac{z + p}{z − p} − \frac{1 + pz}{1 − pz} \} < 0 for z, ζ ∈ Δ .  (3)

Instead of (3) one can write

P_{f} ∈ P ,  (4)

where

P_{f}(z) ≡ −\frac{2zf′(z)}{f(z) − f(ζ)} + \frac{z + ζ}{z − ζ} − \frac{z + p}{z − p} + \frac{1 + pz}{1 − pz}  (5)

and P stands for the class of analytic functions p, normalized by p(0) = 1 and satisfying Re p(z) > 0 , z ∈ Δ.

Two first coefficients of the decomposition of P_{f} into power series around the origin are equal to

p_{1} = 2 (\frac{1}{f(ζ)} − \frac{1}{ζ} + \frac{1 + p^{2}}{p} )  (6)

and

p_{2} = 2 (\frac{1}{f(ζ)^{2}} − \frac{1}{ζ^{2}} + \frac{2a_{2}(f)}{f(ζ)} + \frac{1 + p^{4}}{p^{2}} ) .  (7)

Using the well known bounds on the coefficients of functions from P we get

|\frac{1}{f(ζ)} − \frac{1}{ζ} + \frac{1 + p^{2}}{p} | ≤ 1 ,  (8)

see [5], and

|\frac{1}{f(ζ)^{2}} − \frac{1}{ζ^{2}} + \frac{2a_{2}(f)}{f(ζ)} + \frac{1 + p^{4}}{p^{2}} | ≤ 1 .  (9)

Both estimates, (8) and (9), are valid for f ∈ Co(p) , p ∈ (0, 1) and ζ ∈ Δ \ {0}.

Coefficients a_{n}

One can use (8) and (9) to estimate coefficients of f(z) = z + ∑_{n=2}^{∞} a_{n}z^{n} ∈ Co(p). Taking the limit in (8) as ζ tends to 0 we get

|a_{2}(f) − \frac{1 + p^{2}}{p} | ≤ 1 .  (10)

Unfortunatelly this estimate is not sharp and it is rather week.

This idea applied to (9) leads to On coefficients’ estimates of concave univalent functions 111 Theorem 1. If f ∈ Co(p) , p ∈ (0, 1) has expansion (1) then

|2a_{3}(f) − a_{2}(f)^{2} − \frac{1 + p^{4}}{p^{2}} | ≤ 1 .  (11)

Proof. For ζ ∈ Δ \ {0} and f ∈ Co(p) there is

\frac{1}{f(ζ)^{2}} − \frac{1}{ζ^{2}} + \frac{2a_{2}}{f(ζ)} = \frac{(a_{2}^{2} − 2a_{3})ζ^{2} + (4a_{2}a_{3} + 2a_{4})ζ^{3} + …}{ζ^{2}(1 + a_{2}ζ + …)^{2}} .

Hence

lim_{ζ→0} (\frac{1}{f(ζ)^{2}} − \frac{1}{ζ^{2}} + \frac{2a_{2}}{f(ζ)} + \frac{1 + p^{4}}{p^{2}} ) = a_{2}^{2} − 2a_{3} + \frac{1 + p^{4}}{p^{2}} .

For f ∈ Co(p) with expansion (1) Miller proved [5] that

|a_{3}(f) − \frac{1 + p^{2}}{p} a_{2}(f) + 1| ≤ \frac{2p}{3(1 + p^{2})} .  (12)

This result is partially improved by the following theorem. Theorem 2. If f ∈ Co(p) , p ∈ (0, 1) has expansion (1) then

|a_{3}(f) − \frac{1 + p^{2}}{p} a_{2}(f) + 1| ≤ \frac{1}{2} (1 − |a_{2}(f) − \frac{1 + p^{2}}{p} |^{2}) .  (13)

Proof. It is known that for any p ∈ P of the form p(z) = 1 + ∑_{j=1}^{∞} p_{j}z^{j} there is

|2p_{2} − p_{1}^{2}| ≤ 4 − |p_{1}|^{2} .  (14)

Let f ∈ Co(p) , p ∈ (0, 1) and let P_{f} be of the form (5). Since the coefficients of P_{f} are given by (6) and (7), the inequality (14) leads to

|\frac{1}{f(ζ)^{2}} − \frac{1}{ζ^{2}} + \frac{2a_{2}(f)}{f(ζ)} + \frac{1 + p^{4}}{p^{2}} − (\frac{1}{f(ζ)} − \frac{1}{ζ} + \frac{1 + p^{2}}{p} )^{2}| ≤

1 − |\frac{1}{f(ζ)} − \frac{1}{ζ} + \frac{1 + p^{2}}{p} |^{2} . (15)

As ζ tends to 0 we obtain

|a_{2}^{2} − 2a_{3} + \frac{1 + p^{4}}{p^{2}} − (−a_{2} + \frac{1 + p^{2}}{p} )^{2}| ≤ 1 − |−a_{2} + \frac{1 + p^{2}}{p} |^{2} .

Hence the assertion follows.

Since (14) is stronger then |p_{2}| ≤ 2, the estimate (13) is better then (11). Moreover, observe that if

|a_{2}(f) − \frac{1 + p^{2}}{p} | > \sqrt{1 − \frac{4}{3} \frac{p}{1 + p^{2}}}  (16)

then estimate established in Theorem 2 is better then the one given by (12).

The region of variability of the second coefficient of f ∈ Co(p) coincides with (see [5, 3])

|a_{2}(f) − \frac{1 + p^{2}}{p} + \frac{p}{1 + p^{2}} | ≤ \frac{p}{1 + p^{2}} .  (17)

Hence (16) does not hold for any a_{2}(f) while p ∈ (0, p_{0}), where p_{0} = 0.425 … is the unique solution of \frac{6p}{1+p^{2}} = \sqrt{10} − 1 in (0, 1). On the other hand, for p ∈ [p_{0}, 1) there are such functions f ∈ Co(p) with a_{2}(f) satisfying (16).

Residuum

Let f ∈ Co(p) , p ∈ (0, 1) and let b denote the residuum of f at the point p. The bounds of coefficients of P_{f} defined by (5), and consequently the inequalities (8) and (9), are useful in determining the set of variability of the residuum of f at p.

In [8] Wirths found that

|b + \frac{p^{2}}{1 − p^{4}} | ≤ \frac{p^{4}}{1 − p^{4}}  (18)

and he proved that the equality held for

f_{λ}(z) = \frac{z − \frac{p}{1+p^{2}} (1 + λ)z^{2}}{(1 − z/p)(1 − zp)} , |λ| = 1 .  (19)

It is possible to modify the proof of Wirths’ theorem taking into account the bound on p_{2} instead of p_{1}, where p_{1}, p_{2} are given by (6) and (7). This enable us to estimate the residuum of f at p while the coefficient a_{2}(f) is fixed.

Theorem 3. If f ∈ Co(p) , p ∈ (0, 1) has expansion (1) and b is the residuum of f at p then

|b + \frac{4p^{3}a_{2}(f)}{4 − p^{6}(1 + p^{2})^{2}} | ≤ \frac{2p^{6}(1 + p^{2})|a_{2}(f)|}{4 − p^{6}(1 + p^{2})^{2}} .  (20)

Proof. Let f ∈ Co(p) , p ∈ (0, 1) and let

ω(z) ≡ (\frac{1}{f(z)} + a_{2})^{2} − \frac{1}{z^{2}} + \frac{1 + p^{4}}{p^{2}} − a_{2}^{2} .  (21)

From (9) it follows that |ω(z)| ≤ 1 , z ∈ Δ \ {p}. Moreover, ω has the holomorphic extension onto Δ by assumption ω(p) = p^{2}.

By the Schwarz Lemma (see for example [6])

|ω′(p)| ≤ \frac{1 − |ω(p)|^{2}}{1 − p^{2}} ,  (22)

hence

|ω′(p)| ≤ 1 + p^{2} .  (23)

From (21) and the expansion (2) we get

ω′(z) = −2\frac{f′(z)}{f(z)^{2}} (\frac{1}{f(z)} + a_{2}) + \frac{2}{z^{3}} .  (24)

and

lim_{z→p} \frac{f′(z)}{f(z)^{2}} = lim_{z→p} \frac{\frac{−b}{(z−p)^{2}} + b_{1} + 2b_{2}(z − p) + …}{\frac{b^{2}}{(z−p)^{2}} [1 + \frac{b_{0}}{b} (z − p) + …]^{2}} = −\frac{1}{b .}  (25)

Combining it with (23) and (24) we have

|\frac{2a_{2}}{b} + \frac{2}{p^{3}} | ≤ 1 + p^{2}

which is equivalent to (20).

Taking (14) instead of |p_{2}| ≤ 2 we can obtain

Theorem 4. If f ∈ Co(p) , p ∈ (0, p_{0}) has expansion (1) and b is the residuum of f at p then

|b + \frac{4p^{3}(1 − p^{2})}{4(1 − p^{2})^{2} − p^{6}} (a_{2}(f) − p)| ≤ \frac{2p^{6}}{4(1 − p^{2})^{2} − p^{6}} |a_{2}(f) − p| .  (26)

The number p_{0} is the only solution of p^{3} + 2p^{2} − 2 = 0 in (0, 1).

Proof. Let f ∈ Co(p) , p ∈ (0, p_{0}). Then (15) holds (compare the proof of Theorem 2). Assume

ω(z) ≡ \frac{\frac{1}{f(ζ)^{2}} − \frac{1}{ζ^{2}} + \frac{2a_{2}(f)}{f(ζ)} + \frac{1+p^{4}}{p^{2}} − (\frac{1}{f(ζ)} − \frac{1}{ζ} + \frac{1+p^{2}}{p} )^{2}}{1 − (\frac{1}{f(ζ)} − \frac{1}{ζ} + \frac{1+p^{2}}{p} )^{2}} .  (27)

In view of (8) the function ω is properly defined for z ∈ Δ \ {p}, and it can be extended to the holomorphic function in whole Δ by ω(p) = 0. Moreover, by (15)

|ω(z)| ≤ 1, z ∈ Δ .

After not complicated calculus we have

ω′(p) = \frac{2}{1 − p^{2}} (\frac{1}{p^{3}} − \frac{1}{p} + \frac{1}{b} (a_{2} − p))  (28)

and consequently from (22) there is

|\frac{1}{b}(a_{2} − p) + \frac{1 − p^{2}}{p^{3}} | ≤ \frac{1}{2} .  (29)

We shall prove that the above is valid for p ∈ (0, p_{0}) only. Indeed, according to (17)

|(a_{2} − p) − \frac{1}{p(1 + p^{2})}| ≤ \frac{p}{1 + p^{2}} , p ∈ (0, 1) ,

hence Re(a_{2} − p) > 0. This and (18) mean that Re \frac{1}{b}(a_{2} − p) < 0. But from (29) it follows that if \frac{1}{2} − \frac{1−p^{2}}{p^{3}} ≥ 0 then the real part of \frac{1}{b}(a_{2} − p) could be greater then 0, which is impossible. Hence (29) is valid only for p ∈ (0, p_{0}).

Ihe inequality (29) for these p is eventually equivalent to

|b + \frac{4p^{3}(1 − p^{2})}{4(1 − p^{2})^{2} − p^{6}} (a_{2}(f) − p)| ≤ \frac{2p^{6}}{4(1 − p^{2})^{2} − p^{6}} |a_{2}(f) − p| .

This completes the proof.

Figures.

Figure 1: The comparision of the results for p = 3/5, a_{2} = \frac{1+p^{4}}{p(1+p^{2})}

a) Thm 1 (solid line) and the result of Miller (dashed line)

b) Thm 2 (solid line) and the result of Miller (dashed line) .

On coefficients’ estimates of concave univalent functions 115 we

References

[1] Avkhadiev F.G., Pommerenke Ch., Wirths K.J., On the coefficients of concave univalent functions. Math. Nachr. 271, 3-9 (2004).

[2] Avkhadiev F.G., Pommerenke Ch., Wirths K.J., Sharp inequalities for the coefficients of concave schlicht functions. Comment. Math. Helv. 81, 801-807 (2006).

[3] Avkhadiev F.G., Wirths K.J., A proof of the Livingstone conjecture. Forum Math. 19, 149-157 (2007).

[4] Livingstone A.E., Convex meromorphic functions. Ann. Pol. Math. 59, No.3, 275-291 (1994).

[5] Miller J., Convex and starlike meromorphic functions. Proc. Am. Math. Soc. 80, 607-613 (1980).

[6] Privalov I.I., Introduction to the Theory of Functions of a Complex Variable, GITTL, Moscow-Leningrad, 1948.

[7] Wirths K.J., A proof of the Livingston conjecture for the fourth and the fifth coefficient of concave univalent functions. Ann. Pol. Math. 83, 87-93 (2004).

[8] Wirths K.J., On the residuum of concave univalent functions. Serdica Math. J. 32, 209-214 (2006).

Paweł Zaprawa

email: p.zaprawa@pollub.pl

Departament of Applied Mathematics, Lublin University of Technology, Nadbystrzycka 38D, 20-618 Lublin, Poland Received 18.09.2009 Figure 2: The comparision of the results for p = 2/5, a_{2} = \frac{1+p^{4}}{p(1+p^{2})}

a) Thm 3 (solid line) and the result of Wirths (dashed line)

b) Thm 4 (solid line) and the result of Wirths (dashed line).

Figure 3: The comparision of the results for p = 1/2, a_{2} = \frac{1+p^{2}+p^{4}}{p(1+p^{2})}

a) Thm 3 (solid line) and the result of Wirths (dashed line)

b) Thm 4 (solid line) and the result of Wirths (dashed line).

Figure 4: The comparision of the results for p = 3/5, a_{2} = \frac{1+p^{2}}{p}

a) Thm 3 (solid line) and the result of Wirths (dashed line)

b) Thm 4 (solid line) and the result of Wirths (dashed line).