The norm of pre-Schwarzian derivative on subclasses of bi-univalent functions

which are analytic in the open unit disk ∆ = {z ∈ C : |z| < 1} and normalized by the conditions f (0) = 0 and f ′(0) = 1. An analytic function in a domain D is said to be univalent in D if it does not take the same value twice i.e, f (z1) , f (z2) for all pairs of distinct points z1 and z2 in D. The Koebe one-quarter theorem et al [3] ensures that the image of ∆ under every univalent function f ∈ A contains the disk with the center at origin and of the radius 1/4. Thus, every univalent function f ∈ A has an inverse f −1 : f (∆)→ ∆, satisfying f −1( f (z)) = z, (z ∈ ∆) and


Introduction and definitions
Let A be the class of functions f of the form f (z) = z + ∞ n=2 a n z n (1.1) which are analytic in the open unit disk ∆ = {z ∈ C : |z| < 1} and normalized by the conditions f (0) = 0 and f (0) = 1. An analytic function in a domain D is said to be univalent in D if it does not take the same value twice i.e, f (z 1 ) f (z 2 ) for all pairs of distinct points z 1 and z 2 in D. The Koebe one-quarter theorem et al [3] ensures that the image of ∆ under every univalent function f ∈ A contains the disk with the center at origin and of the radius 1/4. Thus, every univalent function f ∈ A has an inverse f −1 : f (∆) → ∆, satisfying f −1 ( f (z)) = z, (z ∈ ∆) and Moreover, it is easy to see that the inverse function has the series expansion of the form which implies that f −1 is analytic. The derivative of f −1 (see pp. 1038 [4]) is given by .
A function f ∈ A is said to be bi-univalent in ∆ if both f and f −1 are univalent in ∆. We denote the class of bi-univalent functions by σ.(see [2]) The function f in class A is said to be starlike of order where z ∈ ∆. We denote the class of starlike functions of order α by S * (α). The function f of the form (1) is said to be bi-starlike function of order α where 0 ≤ α < 1 if each of the following conditions are satisfied where f ∈ σ, g = f −1 and w = f (z). We denote the class of bi-starlike functions of order α by S * σ (α) (see [5]). If f and g are analytic functions in ∆, we say that f is subordinate to g, written as f ≺ g, if there exists a Schwarz function w analytic in ∆, with w(0) = 0 and |w(z)| < 1 (z ∈ ∆), such that f (z) = g (w(z)). In particular, when g is univalent then the above definition reduces to f (0) = 0 and f (∆) ⊆ g(∆). The pre-Schwarzian derivative of f is denoted by and its norm is given by This norm have a significant meaning in the theory of Teichmuller spaces. For a univalent function f it is well known that ||T f || < 6. This is the best possible estimation. Defining two subclasses for bi-univalent functions as follows In this paper, we shall give the best norm estimation for the classes Proof. Since f ∈ S * σ [A, B], let us assume that Using the definition of subordination, there exists a Schwarz function φ : ∆ → ∆ with φ (0) = 0 and |φ (z) | < 1, such that Hence, h(z) becomes By logarithmic differentiation of (3), we get .
Above equation gives us the pre-Schwarzian derivative of f , i.e, Setting φ (z) = id ∆ (as φ belongs to the class of Schwarz functions and φ (z) ≺ z on ∆) and rearranging the terms, we get Taking the supremum value both sides in the unit disc, the above equation becomes As −1 ≤ B, we get (1 − |z|) ≤ (1 + B |z|), therefore the above inequality becomes The above inequality gives us the norm of pre-Schwarzian derivative of f , denoted by ||T f ||. To estimate the upper bound of ||T f || in the unit disc ∆, z must lead to 1 and therefore Finally we get For the second part of the proof, let us assume that where z = f −1 (w) = g(w). By definition of subordination, there exists a Schwarz function φ : ∆ → ∆ with φ (0) = 0 and |φ (z) | < 1, such that Since f ∈ σ , both f and f −1 are analytic and univalent in ∆. The derivative of f −1 is given by Therefore, k(z) can be expressed as Taking logarithmic differentiation of (5), we get Bφ (z)) .
Setting φ (z) = id ∆ (as φ belongs to the class of Schwarz functions and φ (z) ≺ z on ∆), we have .
Following the previous steps and using ( For upper bound of ||T f ||, z must lead to 1, i.e, . Combining (4) and (6), the proof is complete.
Proof. Since f ∈ V * σ [A, B], let us assume that Therefore, there exists a Schwarz function φ : ∆ → ∆ with φ (0) = 0 and |φ (z) | < 1 such that Therefore, k(z) can be expressed as Setting φ (z) = id ∆ , the pre-Schwarzian derivative of f becomes Following the similar steps as in the first part of this theorem, we get Finally, . (2.7) Combining (8) and (9), the proof is complete.