Univalent functions with half-integral coefficients

Theorem 1.1. Suppose that E ⊂C is uniformly discrete. Then S(E) consists of finitely many functions. For instance, S( 1 N Z) is a finite set for every natural number N, where 1 N Z = {n/N : n ∈Z}. Note also that the ring O of integers in an imaginary quadratic number field is uniformly discrete. Therefore, we obtain finiteness also for the case E = 1 cO for a non-zero element c in O. In these cases, we can say more. Indeed, the following remarkable result is a special case of Salem’s theorem [14, Th

Theorem 1.2. Suppose that all the coefficients an of a function f in S are half-integers. Then f is either one of the nine functions in Theorem B or else one of the following twelve functions: z ± z2 2 , z(2 ± z) 2(1 ± z), z(2 ± z2) 2(1 ± z2), z(2 ± z) 2(1 −z2), z(2 ± z) 2(1 ± z)2, z(2 ± z + z2) 2(1 ± z + z2).

Lemma 2.1. Let E be a uniformly discrete subset of C with bound r0 and let f(z) = z + a2z2 + a3z3 + · · · and g(z) = z + A2z2 + A3z3 + · · · be functions in the class T (E). We write 1/f(z) = 1/z + b0 + b1z + · · · . Suppose that an = An for n = 2, . . . , N and that (2.1) 2 s 1 −PN−2 n=1 n|bn|2 N −1 < r0. Then f = g.

Lemma 2.2. A function f ∈A is univalent on D if and only if its Grunsky matrix Gf(n) of order n is positive semi-definite for every n ≥1. Prawitz’s inequality, which is an extension of Gronwall’s inequality, is also useful as a univalence criterion. See [10] for details.

Lemma 2.3 (Prawitz’s inequality). Let f ∈S and [z/f(z)]α = 1 −P∞ n=1 σnzn. Then ∞ X n=1 (n −α)|σn|2 ≤α for every α > 0. It is elementary, but not easy by hand, to compute the Grunsky matrices for a given function. However, by using a suitable computer software, we can check positivity of Gf(n) rigorously for a specific f and a small enough n. We collect useful formulae to compute these coefficients in Appendix. 3. Properties of the twelve functions As part of the proof of Theorem 1.2, we check univ

Lemma 2.1 guarantees that a function f ∈S( 1 2Z) with f(z) = z +a2z2 +· · ·+aNzN +· · · is uniquely determined. 4.1. Case when a2 = 3/2. We start with the case when a2 = 3/2. Then we have |b1| = |a3 −9/4| ≤1, which is equivalent to 5/4 ≤a3 ≤13/4. Therefore, we have only the possibilities that a3 = 3/2, 2, 5/2. We will show that a3 must be 2 in this case. Suppose that a3 = 3/2. Then b1 = 3/4 and b2 = −a4 + 9/8. Since |b2| ≤ p 7/16 · 2 < 0.47, a4 = 1, 3/2. Assume first that a4 = 1. Then b2 = 1/8, b

Lemma 2.1 to conclude that f = f6 in this case. Finally, we suppose that a4 = 0. Then b2 = −1/8 and b3 = −a5 + 1/16. Since |b3| ≤ p 29/32 · 3 < 0.55, we have a5 = 0, 1/2. If a5 = 1/2, we have b3 = −7/16 and b4 = −a6 + 15/32. Since |b4| ≤ p 85/256 · 4 < 0.29, we have a6 = 1/2. Then b4 = −1/32 and b5 = −a7 + 9/64. Since |b5| ≤ p 21/64 · 5 < 0.26, we have a7 = 0. Then b5 = −9/64 and b6 = −a8 −17/128. Since |b6| ≤ p 939/4096 · 6 < 0.2, we have a8 = 0. Then b6 = −17/128 and b7 = −a9 + 89/256. Since |

Theorem 1.2. Let f(z) = z + a2z2 + a3z3 + · · · be in A and 1/f(z) = 1/z + b0 + b1z + b2z2 + · · · . We first note that the coefficients bn are computed in terms of an’s recursively by the formula (5.1) bn−1 = −an+1 − n X k=2 akbn−k, n ≥1. In particular, we have b0 = −a2, b1 = −a3 + a2 2, b2 = −a4 + 2a2a3 −a3