Univalent harmonic mappings II

Lemma 1. Let T(x) = 1 \ 0  a(1 + t)2 (1 + xt + t2)2 + b(1 −t)2 (1 −xt + t2)2  dt (2.1) and S(x) =

Lemma 2. Let P(z) be in P and Q(x) = a 1 \ 0 1 −t2 (1 + xt + t2)2 Re P(t) dt (2.3) + b 1 \ 0 1 −t2 (1 −xt + t2)2 Re P(−t) dt where a < 0 < b, a + b ≥0 and −2 < x < 2.

Lemma 3. With the same hypotheses as in Lemma 2 and with a and b fixed we have c1 ≤c ≤c2 where c1 and c2 are given in Lemma 1. The range for c is sharp in the sense that for each c, c1 ≤c ≤c2, there exists P(z) in P such that the corresponding Q given by (2.3) satisfies Q(c) = 0.

Theorem 1. If f is a member of F(a, b), then f is harmonic, sense- preserving and univalent in U. Moreover, f(U) is convex in the direction of the real axis and f(U) ⊂Ω(a, b). P r o o f. Let f = h + g = Re F + i Re G; then F(z) = A z \ 0 (1 −ζ)2P(ζ) (1 + cζ + ζ2)2 dζ and G(z) = −iAz 1 + cz + t2 . Since

Theorem 2. SH(U, Ω(a, b)) ⊂F(a, b). P r o o f. Let f be a member of SH(U, Ω(a, b)) and f = h + g. Since Ω(a, b) is convex in the direction of the real axis, by a result of Clunie and

Lemma 4. F(a, b) is closed. P r o o f. Let fn be a sequence in F(a, b) with fn converging to f uniformly

Theorem 3. SH(U, Ω(a, b)) = F(a, b). P r o o f. Let f(z) have the form (3.1) where (3.2) is satisfied and let rn be a sequence with 0 < rn < 1 and lim rn = 1. Let Pn(z) = P(rnz) and denote by fn(z) the function obtained from (3.1) and (3.2) by replacing P(z) with Pn(z). Let cn be the value of c satisfying (3.2) when P is replaced by Pn. We claim that fn is a member of SH(U, Ω(a, b)). To see this let An = b  Re 1 \ 0 (1 −ζ2)Pn(ζ) (1 + cnζ + ζ2)2 dζ, Fn(z) = An

Lemma 5. Let f ∈F(−b, b) and be odd. If f(z) = h(z) + g(z), then both h and g are odd. P r o o f. Since f(−z) = −f(z), we have h(z)+g(z) = −(h(−z)+g(−z)). Thus h(z) + h(−z) = −g(z) + g(−z). It follows that h(z) + h(−z) and h(z) + h(−z) are both analytic in U. Thus h(z) + h(−z) is constant. Since its value is 0 at z = 0, we have h(z) = −h(−z). Similarly, g(z) is odd.

Lemma 6. If f ∈F(−b, b) and f is odd then in the representation (3.1), P(z) is even and c = 0.

Lemma 7. Let f ∈F(−b, b) with representation (3.1). If P(z) is even, then c = 0 and f is odd. P r o o f. If P(z) is even, then Q(x) defined by (2.3), with a = −b, satisfies Q(0) = − 1 \ 0 (1 −t2) Re P(t) (1 + t2)2 dt + 1 \ 0 (1 −t2) Re P(−t) (1 + t2)2

Theorem 4. If f ∈G(−b, b), then (4.5) 4b π ≤a1 ≤8b π and the inequalities are sharp. P r o o f. Since P ∈P and P is even, (1 −|z|2)/(1 + |z|2) ≤Re P(z) ≤ (1 + |z|2)/(1 −|z|2). Thus π 8 = 1 \ 0 (1 −t2)2 (1 + t2)3 dt ≤

Theorem 5. Let f(z) = h(z) + g(z) be in G(−b, b) and suppose h(z) = ∞ X n=0 a2n+1z2n+1 and g(z) = ∞ X n=1 b2n+1z2n+1. Then |a2n+1| ≤(n + 1)2 2n + 1 |a1|,

Theorem 6. Let f(z) = h(z) + g(z) be a member of G(−b, b). Then for |z| = r < 1, (4.14) |a1|(1 −r2) (1 + r2)3 ≤|fz(z)| ≤|a1|(1 + r2) (1 −r2)3 and the inequalities are sharp.