Hyperbolically convex functions

Theorem 1. Let Ω⊂D be a hyperbolically k-convex region with C2 boundary. Then κD(z, ∂Ω) ≥k for all z ∈∂Ω∩D. P r o o f. Since both hyperbolic curvature and hyperbolic k-convexity are invariant under the group Aut(D), it is enough to show that κD(0, ∂Ω) ≥k if 0 ∈∂Ω. First, we consider the case k > 0. Assume that κD(0, ∂Ω) < k. Then κe(0, ∂Ω)=κD(0, ∂Ω)<k. Let L be the circular arc tangent to ∂Ωat 0 with euclidean curvature k such that Ω∩{z : |z| < ε} lies inside L for sufficiently small ε > 0. Let ∆b

Theorem 2. Let f ∈Kh(α). Then either f(D) contains the closed disk  w : |w| ≤ α 1 + √ 1 −α2  or else f is a rotation of kα, that is, f(z) = eiθkα(e−iθz) for some real θ. P r o o f. Let c be a point in D \ f(D) which is closest to the origin. Then F(z) = f(z) −c 1 −cf(z) is also hyperbolically convex and 0 ̸∈F(D). So F(D) must lie in some half-disk of D which has a diameter L′ as part of its boundary. Let Ωbe

Lemma 1. Suppose Ω⊂D is hyperbolically convex in D. Then for c ∈ Ω∩D and ϱ > 0, Ω∩DD(c, ϱ) is hyperbolically convex in Ω. P r o o f. Since the hyperbolic distance and hyperbolic convexity are both invariant under Aut(D), we may assume that c = 0. Now we prove that the set Ω∩DD(0, ϱ) = Ω∩{z : |z| < tanh(ϱ)} is hyperbolically convex in Ω. For each z ∈Ω, the half-open line segment (0, z] lies in Ωbecause Ω is hyperbolically convex. Consequently, Ωis starlike with respect to 0. By

Proposition 2 of [10], we see that Ω∩{z : |z| < tanh(ϱ)} is hyperbolically convex in Ω.

Theorem 3. Suppose f is holomorphic and locally univalent in D with f(D) ⊂D. Then the following are equivalent. (i) f is hyperbolically convex.

Lemma 1 implies that Ω∩DD(βeiθ, artanh(β)) is hyperbolically convex in Ωfor all real θ. Note that DD(βeiθ, artanh(β)) =  w :

Theorem 1 we deduce that κΩ(0, γθ) ≥0 for all real θ. We may now conclude that

Theorem 4. Suppose f is holomorphic and locally univalent in D with f(D) ⊂D. Then the following are equivalent. (i) f is hyperbolically convex. (ii) For each a ∈D the function Fa(z) = f  z + a 1 + az  −f(a) 1 −f(a)f  z + a 1 + az  is starlike with respect to the origin in D.

Theorem 3 characterizes hyperbolically convex functions, it is not a sharp inequality. We now establish the sharp version of Theorem 3(ii) by using a technique of Wirths [16].

Theorem 5. Let f be hyperbolically convex in D. Then, for z ∈D, |Dh2f(z)/(2Dh1f(z))| ≤1 −|Dh1f(z)|2. P r o o f. By considering f(rz) and letting r tend to 1 if necessary, we may assume that f is holomorphic and univalent on D with |f(z)| ≤R < 1 (z ∈D). Define Hf(z) = |Dh1f(z)|2 1 −|Dh2f(z)/(2Dh1f(z))|. The desired inequality is the same as Hf(z) ≤1. Under our assumptions, Hf(z) tends to 0 as |z| tends to 1. If the inequality Hf(z) ≤1 were not valid, then there would exist a ∈D such that Hf(a) > 1

Theorem 6. Suppose f is hyperbolically convex in D and f ̸∈Aut(D). Then for any path γ in D from a to b, |Dh1f(a)| p 1 −|Dh1f(a)|2 exp[−2 lengthh(γ)] ≤ |Dh1f(b)| p 1 −|Dh1f(b)|2 ≤ |Dh1f(a)| p 1 −|Dh1f(a)|2 exp[2 lengthh(γ)], where lengthh(γ) is the hyperbolic length of the path γ. P r o o f. It suffices to establish the upper bound because the lower bound then follows by interchanging the roles of a and b. Let γ : z = z(s), 0 ≤s ≤

Theorem 5 gives (d/ds)|Dh1f(z)| |Dh1f(z)|(1 −|Dh1f(z)|2) ≤2, since f ̸∈Aut(D). By integrating this differential inequality over the interval [0, L], we obtain log |Dh1f(b)| p 1 −|Dh1f(a)|2 |Dh1f(a)| p 1 −|Dh1f(b)|2 ≤2L. This gives the upper bound in the theorem. Corollary. For f ∈Kh(α), we have the sharp bounds α(1 −|z|) p

Theorem 7. Let f ∈Kh(α). Then we have sharp estimates −kα(−|z|) ≤|f(z)| ≤kα(|z|). P r o o f. To prove the upper bound, fix any a ∈D, set γ = [0, a] and Γ = f ◦γ. Then from the Corollary to Theorem 6, dD(0, f(a)) ≤R Γ |dw| 1 −|w|2 = R γ |f ′(z)||dz| 1 −|f(z)|2 = R γ |Dh1f(z)| |dz| 1 −|z|2

Theorem 8. For k ∈[0, 4] the radius of hyperbolic k-convexity for the family SB(α) is the unique root r(α, k) in the interval (0, 1] of the polynomial pα,k(r) = r4 −(2 + αk)r3 + (8α + 2αk −6)r2 −(2 + αk)r + 1. The root r(α, k) is an increasing function of α which satisfies the sharp bounds 2 − √ 3 ≤r(α, k) ≤  1 if 0 ≤k ≤2,  k − √ k2 −4 

Theorem 9. Suppose f : D →D is locally univalent. If eαh(f) < 1, then f ∈Aut(D). For f ̸∈Aut(D), eαh(f) ≥1 with equality if f is hyperbolically convex. P r o o f. Let α = eαh(f). We begin by showing that for z ∈D, (6) |f ′(0)| 1 −|f(0)|2 · (1 −|z|)α−1 (1 + |z|)α+1 ≤ |f ′(z)| 1 −|f(z)|2 . Let z = z(s), 0 ≤s ≤L, be a parametrization of the radial segment γ = [0, z] by hyperbolic arc length. Then as in the proof of Theorem 6 we obtain (d/ds)|Dh1f(z)| |Dh1f(z)| = Re

Theorem 10. Suppose f : D →D is locally univalent and f ̸∈Aut(D). Then ϱh(f) = ϱ if and only if eαh(f) = coth(2ϱ). In particular, Ch(ϱ) = eF(coth(2ϱ)). P r o o f. A geometric proof completely parallel to that of [6, Theorem 3] is easily constructed. Note that the condition eαh(f) ≥1 is needed in the proof. Corollary. If f : D →D is locally univalent and eαh(f) = 1, then f is hyperbolically convex.

Theorem 11. Suppose f : D →D is locally univalent. Then kh(f) = 2eαh(f). P r o o f. A geometric proof parallel to [8, Theorem 9] is readily given. 7. Open problems. For the family Kh(α) it would be interesting to obtain the sharp upper bound on |f (n)(0)| for n ≥3 and on the Schwarzian derivative. For α=1/ √ 2 we note that k1/ √ 2 cannot be the extremal function for the sharp upper bound on |f (3)(0)| because k(3) 1/ √ 2(0) = 0. What are sharp upper and lower bounds on |f ′(z)| for the family Kh