Maximal Area Integral Problem for Certain Class of Univalent Analytic Functions

Abstract

One of the classical problems concerns the class of analytic functions f on the open unit disk |z| < 1 which have finite Dirichlet integral ∆(1, f ), where ∆(r, f ) = |z|<r |f ′ (z)| 2 dxdy (0 < r ≤ 1). The class S * (A, B) of normalized functions f analytic in |z| < 1 and satisfies the subordination condition zf ′ (z)/f (z) ≺ (1 + Az)/(1 + Bz) in |z| < 1 and for some −1 ≤ B ≤ 0, A ∈ C with A = B, has been studied extensively. In this paper, we solve the extremal problem of determining the value of max f ∈S * (A,B) ∆(r, z/f ) as a function of r. This settles the question raised by Ponnusamy and Wirths in [11]. One of the particular cases includes solution to a conjecture of Yamashita which was settled recently by Obradović et. al [9].

Some inequalities for a certain subclass of starlike functions

“…Over the years, the definition of a certain subclass of analytic functions by using the subordination relation has been investigated by many works including (for example) [5], [7], [8], [11], [12], [15] and [17]. We now recall from [16], the following definition which is used from subordination.…”

Integral Means and Dirichlet Integral for Certain Classes of Analytic Functions

“…In 2013, Yamashita's conjecture was settled in a more general setting for functions in the class S * (α) by Obradović et al [14]. In 2014, Ponnusamy and Wirths [18], Obradović et al [15] and Sahoo and Sharma [21] discussed the maximum area problem for functions of type z/ f (z) when f belongs to certain subclasses of the class S. Moreover, recently, Ponnusamy et al [17] solved the same problem for the class S * (A, B), where −1 ≤ B < A ≤ 1.…”

“…12[17, Theorems 2.1 and 2.3]. Let f ∈ S * 1 (A, B, 1) = S * (A, B) for some −1 ≤ B ≤ 0, A B and A ∈ C. Then, for0 < r ≤ 1, max f ∈S * (A,B) ∆ r, z f (z) = E A,B (r), |A| 2 r 2 ) for B = 0.The maximum is attained by the rotation of the function k A,B (z) defined by(1.4).Finally, if we choose A = (b 2 − a 2 + a)/b, B = (1 − a)/b with a + b ≥ 1, a ∈ [b, 1 + b] in Corollary 2.12, then we obtain the result of Ponnusamy et al [17, Corollary 2.7].…”