ON POSITIVITY OF TAYLOR COEFFICIENTS OF CONFORMAL MAPS

TOSHIYUKI SUGAWA

Abstract. We provide an approach to the proof of positivity of the Taylor coefficients for a given conformal map of the unit disk onto a plane domain. This short note is a summary of the joint work [2] with Stanisława Kanas.

1. Introduction

If a univalent function f(z) = a_{0} +a_{1}z +a_{2}z^{2} +⋯ in the unit disk 𝔻 = {z ∈ ℂ; |z| < 1} has non-negative Taylor coefficients about the origin, namely, a_{k} ≥ 0 for all k ≥ 0, various sharp estimates can easily be deduced. For example, one can show the sharp inequalities

|f(z) − a_{0} − a_{1}z − ⋯ − a_{k}z^{k}| ≤ f(|z|) − a_{0} − a_{1}|z| − ⋯ − a_{k}|z|^{k}

and

|f^{(k)}(z)| ≤ f^{(k)}(|z|)

for k = 0, 1, 2, …. Note that this sort of inequalities are, in general, not easy to establish.

As one immediately sees, a necessary condition for a univalent function f to have nonnegative Taylor coefficients is that the image domain Ω = f(𝔻) is symmetric in the real axis. Under the assumption of this symmetric property, however, it seems to be difficult to give a sufficient condition for non-negativity of the coefficients in terms of the shape of Ω. For instance, the convexity of Ω is not sufficient. In fact, for a constant 0 < c < 1, the function

f(z) = \frac{z}{1 + cz} = z − cz^{2} + c^{2}z^{3} − c^{3}z^{4} + ⋯

maps 𝔻 univalently onto a disk but has a negative coefficient. (In general, when f(z) has non-negative Taylor coefficients, the function f̂(z) = −f(−z) has a negative coefficient unless f is an odd function.)

In this note, we will explain one approach to show positivity of the Taylor coefficients of a specific conformal map of the interior of a conic section.

2. Conformal mappings onto domains bounded by conic sections

For k ∈ [0, ∞), we set

Ω_{k} = {u + iv ∈ ℂ; u^{2} > k^{2}(u − 1)^{2} + k^{2}v^{2}, u > 0}.

Date: February 13, 2004.

Note that 1 ∈ Ω_{k} for all k. Ω_{0} is nothing but the right half plane. When 0 < k < 1, Ω_{k} is the unbounded domain enclosed by the right half of the hyperbola

(\frac{u + k^{2}/(1 − k^{2})}{k/(1 − k^{2})})^{2} − \frac{v^{2}}{1/(1 − k^{2})} = 1

with focus at 1. Ω_{1} becomes the unbounded domain enclosed by the parabola

v^{2} = 2u − 1

with focus at 1. When k > 1, the domain Ω_{k} is the interior of the ellipse

(\frac{u − k^{2}/(k^{2} − 1)}{k/(k^{2} − 1)})^{2} + \frac{v^{2}}{1/(k^{2} − 1)} = 1

with focus at 1. For every k, the domain Ω_{k} is convex and symmetric in the real axis. Note also that Ω_{k_{1}} ⊃ Ω_{k_{2}} if 0 ≤ k_{1} ≤ k_{2}.

Kanas and Wiśniowska [3] treated the family Ω_{k} in their study of k-uniformly convex functions and gave the explicit formulae for the conformal homeomorphisms p_{k} : 𝔻 → Ω_{k} determined by p_{k}(0) = 1 and p′_{k}(0) > 0. Here, an analytic function f(z) in the unit disk with f(0) = 0, f′(0) = 1 is called k-uniformly convex if the function 1 + zf′′(z)/f′(z) maps the unit disk analytically into Ω_{k}. A function is 1-uniformly convex precisely when it is uniformly convex (see [4]).

In order to state their result, we prepare some notation. Let K(z, t) and K(t) be the normal and complete elliptic integrals, respectively, i.e.,

K(z, t) = ∫_{0}^{z} \frac{dx}{\sqrt{(1 − x^{2})(1 − t^{2}x^{2})}}

and K(t) = K(1, t). The quantity

μ(t) = \frac{πK(\sqrt{1 − t^{2}})}{2K(t)}

is known as the modulus of the Groetszch ring 𝔻 \ [0, t] for 0 < t < 1. Note that μ(t) is a strictly decreasing smooth function. For details, see [1].

Proposition 1 (Kanas-Wiśniowska [3]). The conformal map p_{k} : 𝔻 → Ω_{k} with p_{k}(0) = 1 and p′_{k}(0) > 0 is given by

p_{k}(z) = \begin{cases}(1 + z)/(1 − z) & if k = 0, \\ (1 − k^{2})^{−1} cosh[C_{k} log(1 + \sqrt{z})/(1 − \sqrt{z})] − k^{2}/(1 − k^{2}) & if 0 < k < 1, \\ 1 + (2/π^{2})[log(1 + \sqrt{z})/(1 − \sqrt{z})]^{2} & if k = 1, \\ (k^{2} − 1)^{−1} sin[C_{k}K((z/\sqrt{t} − 1)/(1 − \sqrt{t}z), t)] + k^{2}/(k^{2} − 1) & if 1 < k, \end{cases}

where C_{k} = (2/π) arccos k for 0 < k < 1 and C_{k} = π/2K(t) and t ∈ (0, 1) is chosen so that k = cosh(μ(t)/2) for k > 1.

3. Main Results

For each k ∈ [0, ∞), we write

p_{k}(z) = 1 + A_{1}(k)z + A_{2}(k)z^{2} + ⋯

for the conformal mapping p_{k} of 𝔻 onto Ω_{k} with p_{k}(0) = 1 and p′_{k}(0) > 0. Since Ω_{k} lies in the right half-plane, Carathéodory’s theorem yields that |A_{n}(k)| ≤ 2 holds for each n ≥ 1 and k ∈ [0, ∞). Our main result is the following.

Theorem 2. A_{n}(k) > 0 for all n ≥ 1 and k ∈ [0, +∞).

Since p_{0}(z) = 1 + 2z + 2z^{2} + 2z^{3} + ⋯ and

p_{1}(z) = 1 + \frac{2}{π^{2}} (z + \frac{z^{2}}{3} + \frac{z^{3}}{5} + ⋯)^{2} ,

the assertion of the theorem is trivial for k = 0 and k = 1. When 0 < k < 1, the assertion is also trivial because the function cosh has the non-negative Taylor coefficients.

In what follows, we consider the cases when k > 1. Due to complexity of the representation of p_{k} given above for k > 1, we try to simplify it.

We now consider the conformal mapping J of 𝔻 onto ℂ̂ \ [−1, 1] defined by f(z) = (z + z^{−1})/2. Since

J(e^{−s+it}) = cosh s cos t − i sinh s sin t,

the circle |z| = e^{−s} is mapped by J onto the ellipse E_{s} given by

(\frac{u}{cosh s})^{2} + (\frac{v}{sinh s})^{2} = 1

for s > 0 and the radial segment (0, e^{it}) is mapped by J into the component H_{t} of the hyperbola given by

(\frac{u}{cos t})^{2} − (\frac{v}{sin t})^{2} = 1, u cos t > 0,

for t ∈ ℝ with (2/π)t ∉ ℤ.

Let T_{n} be the Chebyshev polynomial of degree n, i.e., T_{n}(cos θ) = cos(nθ). Then it is well known that the n-fold mapping z ↦ z^{n} is conjugate under J to T_{n}, in other words,

J(z^{n}) = T_{n}(J(z))

holds in |z| < 1. In particular, one can see that the ellipse E_{s} is mapped by T_{n} onto E_{ns} and that the hyperbola H_{t} is mapped by T_{n} onto H_{nt}.

Applying the above argument to T_{2}(w) = 2w^{2} − 1, we obtain the following.

Lemma 3. The Chebyshev polynomial T_{2}(w) = 2w^{2} − 1 maps the domain bounded by H_{t} and H_{π−t} onto the connected component of ℂ \ H_{2t} containing −1. Also, T_{2} maps the domain bounded by the ellipse E_{s} onto the domain bounded by E_{2s}.

On the basis of the above lemma, we can obtain another representation of p_{k}.

Theorem 4. For k > 0, the function p_{k} is written by p_{k}(z) = 1 + Q_{k}(\sqrt{z})^{2}, where

Q_{k}(z) = \begin{cases}\sqrt{\frac{2}{1−k^{2}}} sinh(C_{k} arctanhz) & if 0 < k < 1, \\ \sqrt{\frac{1}{2π^{2}}} arctanhz & if k = 1, \\ \sqrt{\frac{2}{k^{2}−1}} sin (C′_{k}K(z/\sqrt{s}, s)) & if 1 < k. \end{cases}

Here, C_{k} = (2/π) arccos k when 0 < k < 1, and s ∈ (0, 1) is chosen so that k = cosh μ(s) and C′_{k} = (π/2)/K(s) when k > 1.

Furthermore, the function Q_{k} is odd and maps the unit disk conformally onto the domain D_{k} = {x + iy : (k − 1)x^{2} + (k + 1)y^{2} < 1}.

Note that D_{k} is the inside of a hyperbola when k < 1 and D_{k} is the interior of an ellipse when k > 1. When k = 1, the domain D_{k} becomes the parallel strip −1/\sqrt{2} < Im z < 1/\sqrt{2}. Also note that D_{k} is invariant under the involution z ↦ −z.

4. Rough idea of the proof

We indicate here how to deduce Theorem 2. A detailed exposition will appear in [2]. In order to prove positivity of the Taylor coefficients of p_{k}, it is enough to show that of

Q_{k} thanks to Theorem 4. Though the assertion is trivial in the case when 0 < k < 1, we first treat this case in order to highlight an idea of the present method. When 0 < k < 1, one can check that w = Q_{k}(z) satisfies the linear differential equation

(1 − z^{2})^{2}w′′ − 2z(1 − z^{2})w′ − C_{k}^{2}w = 0 (1)

in 𝔻.

Lemma 5. Let Q(z) be an analytic solution of (1) in 𝔻 with Q(0) = 0 and Q′(0) > 0. Then Q has Taylor expansion in the form Q(z) = ∑_{n=0}^{∞} B_{n}z^{2n+1} and the coefficients satisfy the inequalities

(2n + 1)B_{n} − (2n − 1)B_{n−1} > 0 and B_{n} > 0 (2)

for each n ≥ 1.

Proof. By the linear differential equation (1), one obtains the recursive formula for coefficients

(2n + 2)(2n + 3)B_{n+1} − {2(2n + 1)^{2} + C_{k}^{2}}B_{n} + 2n(2n − 1)B_{n−1} = 0

for n ≥ 0, here we have set B_{−1} = 0. We now suppose that the assertion is true up to n. Then, by the above formula, we get

(2n + 2){(2n + 3)B_{n+1} − (2n + 1)B_{n}}
={2(2n + 1)^{2} − (2n + 2)(2n + 1) + C_{k}^{2}}B_{n} − 2n(2n − 1)B_{n−1}
(3)  ≥{2(2n + 1)^{2} − (2n + 2)(2n + 1)}B_{n} − 2n(2n − 1)B_{n−1}
=2n{(2n + 1)B_{n} − (2n − 1)B_{n−1}} > 0

Therefore, the assertion is also true for n + 1. By induction, the proof is done.

In the case when k > 1, the function w = Q_{k}(z) satisfies the similar differential equation

(1 − sz^{2})(1 − z^{2}/s)w′′ − 2z((s + s^{−1})/2 − z^{2})w′ + \frac{C′_{k}^{2}}{s} w = 0

in 𝔻, where s ∈ (0, 1) is chosen so that k = cosh μ(s) and C′_{k} = π/2K(s). Note that Q_{k}(z) satisfies Q_{k}(0) = 0 and Q′_{k}(0) > 0.

The above two differential equations can also be unified into the form

(1 − 2Mz^{2} + z^{4})w′′ − 2z(M − z^{2})w′ − cw = 0, (4)

where M = 1 and c = C_{k}^{2} for 0 < k < 1 and M = (s + s^{−1})/2 ≥ 1 and c = −C′_{k}^{2}/s = −π^{2}/4sK(s)^{2} for k > 1. Let w = Q(z) be the solution of the equation with the initial condition Q(0) = 0 and Q′(0) = 1. In the same way as above, one obtains the relations for the coefficients of Q(z) = ∑_{n=0}^{∞} B_{n}z^{2n+1} :

(2n + 2)(2n + 3)B_{n+1} − {2M(2n + 1)^{2} + c}B_{n} + 2n(2n − 1)B_{n−1} = 0 (5)

for n ≥ 0, where we also have set B_{−1} = 0.

In the case when k > 1, however, the above argument breaks down at the inequality (3) because now c < 0. In fact, the coefficients B_{n} tend rapidly to 0 as n → ∞, therefore, some renormalization techniques are required in this case. See [2] for the details.

References

1. G. D. Anderson, M. K. Vamanamurthy, and M. K. Vuorinen, Conformal Invariants, Inequalities, and Quasiconformal Maps, Wiley-Interscience, 1997.

2. S. Kanas and T. Sugawa, Conformal representations of domains bounded by conic sections and related classes of Carathéodory functions, in preparation.

3. S. Kanas and A. Wiśniowska, Conic regions and k-uniform convexity, J. Comp. Appl. Math. 105 (1999), 327–336.

4. F. Rønning, Uniformly convex functions and a corresponding class of starlike functions, Proc. Amer. Math. Soc. 118 (1993), 189–196.

Department of Mathematics, Graduate School of Science, Hiroshima University, Higashi-Hiroshima, 839-8526 Japan

  E-mail address: sugawa@math.sci.hiroshima-u.ac.jp