The spherical metric and univalent harmonic mappings

Theorem 1. If f = h + g ∈S0 H, then |a2| ≤20.9197. The proof of Theorem 1 is presented in Section 3. It requires several other basic results which will be discussed in Section 2. For f = h + g ∈S0 H given by (1.1), it was shown in [3] that the analytic part h of f lies in Hardy spaces Hp for some small p > 0, namely, 0 < p < (2α0 + 2)−2 with α0 = supSH |a2|. Thus, determining sharp estimate for |a2| is an important problem. On the other hand, while the bound in Theorem 1 may not be sharp, it is

Corollary 2, a new estimate on the second coefficient is obtained for functions belonging to a certain class of conformal mappings. In Section 4, specifically in Theorem 4, we show that for a K–quasiconformal univalent harmonic map f = h + g (that is, |fz| ≤α|fz| a.e. on D, where α = (K −1)/(K + 1) with K ≥1), the spherical area As(h) is dominated by

Lemma 1. Suppose that Q(z) = P∞ n=1 Anzn is given by (2.1) and f(z) = P∞ n=1 anzn analytic in D satisfy f(z) ≺Q(z) for z ∈D. Then |an| ≤An for n ≥1.

Lemma 2. [16] If f ∈F, D = f(D) and a = d(0, ∂D), then f(z) ≺aQ(z). If D is a region in the complex plane, denote by Dc its complement C\D. As an immediate consequence of Lemma 2, here is a result which reveals an important geometric fact.

Corollary 1. (Compare with [15, Theorem II]) Suppose that h ∈F, h(z) = P∞ n=1 anzn, h(D) = D, d(0, ∂D) = |a|, a ∈∂D, and Q(z) = P∞ n=1 Anzn is given by (2.1). Then |an| ≤|a|An for n ≥1. An important question to ask is whether the coefficient estimate is sharp. At least in

Theorem 2 in the next section, we present better estimates for a2 and a3. 3. Coefficient estimates for hyperbolic conformal maps Here is our first basic result which gives better estimates for a2 and a3 than the estimates given by Corollary 1.

Theorem 2. Suppose that h ∈F, h(z) = z + P∞ n=2 anzn, h(D) = D, and d(0, ∂D) = |a| for some a ∈∂D. Then a2 and a3 satisfy the following inequalities: (3.1) 1 16 ≤|a|, |a2| ≤16|a| + 1 2|a|, and (3.2) |a3| ≤704|a|. If in addition D is hyperbolic, then |a| < 1.

Corollary 2. If h ∈F and h(z) = z + P∞ n=2 anzn, then |a2| ≤16.5 and |a3| ≤704. Now, we are in a position to formulate an important general result. First note that if h(z) = P∞ n=1 anzn is analytic on D, h(D) = D and d(0, ∂D) = |a| for some a ∈∂D, then the function z (h(z) −a) /(−a) belongs to F. Furthermore, we remark that z(h(z) −a) is zero only at 0.

Theorem 3. Suppose that h is conformal in D, h(D) = D is hyperbolic and d(0, ∂D) = |a|, where a ∈∂D and h(0) = h′(0) −1 = 0. Then 1 16.5 ≤|a| < 1.

Lemma 3. Suppose that h(z) = z + a2z2 + · · · is analytic in D and misses the disk D(c, r) := {z : |z−c| < r} which touches the boundary ∂h(D). Then the function Ψ defined by Ψ(z) = c−h(z) c misses the disk D(0, ρ), where ρ = r/|c| > 1/16, and |a2| < 20.9197|c|.

Theorem 4. Let f = h+g be a sense-preserving univalent harmonic mapping in D given by (1.1) with the dilatation ϕ. Suppose that α = ||ϕ||∞< 1 and D = h(D) is hyperbolic. Then the spherical area of the covering surface of D satisfies 1 −α2 4 As(D) ≤As(Ω) := As(Ω, f) ≤4π, where f(D) = Ωand As(D) = ZZ D |h′(z)|2 dA (1 + |h(z)|2)2.

Theorem 5. Let f = h + g ∈S0 H, f(D) = Ω, h(D) = D is hyperbolic, and a ∈∂D be the nearest point to the origin 0. Then (5.4) 1 16 ≤d(0, ∂D) ≤1 and (5.5) 1 16(1 −|µ(z)|) ≤d(f(z), ∂Ω) ≤2d(h(z), ∂D).