Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 31, 2006, 329–348

ON CONFORMAL REPRESENTATIONS OF THE INTERIOR OF AN ELLIPSE

Stanisława Kanas and Toshiyuki Sugawa

Rzeszów University of Technology, Department of Mathematics W. Pola 2, PL-35-959 Rzeszów, Poland; skanas@prz.rzeszow.pl Hiroshima University, Department of Mathematics, Graduate School of Science Higashi-Hiroshima, 839-8526 Japan; sugawa@math.sci.hiroshima-u.ac.jp

Abstract. We consider the conformal mappings f and g of the unit disk onto the inside of an ellipse with foci at ±1 so that f(0) = 0, f′(0) > 0, g(0) = −1 and g′(0) > 0. The main purpose of this article is to show positivity of the Taylor coefficients of f and g about the origin. To this end, we use a special relation between f and g and the fact that f satisfies a second-order linear ODE. Some applications are given to the class of k -uniformly convex functions.

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, various sharp estimates can be easily deduced. For example, one can show the sharp inequalities

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

and

(1.2)  |f^{(k)}(z)| ≤ f^{(k)}(|z|), |z| < 1,

for k = 0, 1, 2, ….

As one immediately sees, necessary conditions for a univalent function f to have non-negative Taylor coefficients about the origin are that f(0) ≥ 0, f′(0) > 0 and that the image domain Ω = f(D) is symmetric in the real axis. Note that these conditions imply that the relation \overline{f(z̄)} = f(z) holds, and hence, all coefficients are real. It is also necessary that the farthest point of ∂Ω from f(0) is the right-most point of \overline{Ω} ∩ R. Conversely, under these conditions, however, it seems to be difficult to give a sufficient geometric condition for positivity of the Taylor coefficients. For instance, convexity of Ω is not sufficient. In fact, for constants 0 < c < 1 and N < α < N + 1 with cα ≤ 1 and N being a positive integer, the function

2000 Mathematics Subject Classification: Primary 30C20; Secondary 30C45, 33E05.

The second author was partially supported by the Ministry of Education, Grant-in-Aid for Encouragement of Young Scientists (B), 14740100.

f(z) = (1 + cz)^{α} = ∑_{n=0}^{∞} \binom{α}{n}(cz)^{n}

is univalent in D and has convex image because

Re(1 + \frac{zf′′(z)}{f′(z)}) = 1 + (α − 1) Re \frac{cz}{1 + cz} > 1 − (α − 1)\frac{c}{1 − c} ≥ 0.

Since

f^{(k)}(z) = c^{k}α(α − 1) ⋯ (α − k + 1)(1 + cz)^{α−k},

we observe that (1.2) is fulfilled by k = 0, 1, …, N but not by k = N + 1. Note that one can deduce (1.1) for k from (1.2) for k + 1 by repeated integrations.

In this paper, we show non-negativity of the Taylor coefficients of specific conformal mappings of the unit disk onto an ellipse. Let E_{ξ} be the ellipse given by

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

and let D_{ξ} be the interior of E_{ξ} for ξ > 0. Note that E_{ξ} has foci at 1 and −1 and that an arbitrary ellipse is similar to E_{ξ} for some ξ. We prove the following two results.

Theorem 1.1. Let f_{ξ} be the conformal mapping of the unit disk onto the interior D_{ξ} of the ellipse E_{ξ} determined by f_{ξ}(0) = 0 and f′_{ξ}(0) > 0. Then f_{ξ} has positive odd Taylor coefficients about the origin.

Theorem 1.2. Let g_{ξ} be the conformal mapping of the unit disk onto the interior D_{ξ} of the ellipse E_{ξ} determined by g_{ξ}(0) = −1 and g′_{ξ}(0) > 0. Then g_{ξ} has positive Taylor coefficients about the origin except for the first one.

Since D_{ξ} is invariant under the rotation by angle π about the origin, f_{ξ} is an odd univalent function and is of the form

f_{ξ}(z) = a(z + A_{1}z^{3} + A_{2}z^{5} + ⋯), a = f′_{ξ}(0) > 0.

In particular, there exists a univalent function φ with φ(0) = 0, φ′(0) > 0 such that (f_{ξ}(z))^{2} = φ(z^{2}) (see [4]). Then

φ(z) = a^{2}z ∑_{n=0}^{∞}( ∑_{m=0}^{n} A_{m}A_{n−m})z^{n},

where we set A_{0} = 1. Therefore, Theorem 1.1 yields that φ has positive Taylor coefficients about the origin except for the first one. We will show the relation

2φ = g_{2ξ} + 1

in Section 2 (see Theorem 2.1). In this way, Theorem 1.2 follows from Theorem 1.1.

An explicit form of f_{ξ} was first given by Schwarz as early as in 1869 and is well known nowadays. There is, however, less awareness of that f_{ξ} satisfies a secondorder homogeneous linear ordinary differential equation (see Section 3). Using this ODE, we obtain linear recurrence relations between three successive coefficients A_{n−1} , A_{n} and A_{n+1} . It is still difficult to show positivity of the coefficients. The final stroke will be made by a theory of continued fractions, which will be presented in Section 4.

We apply Theorem 1.2 to the study of k -uniformly convex functions introduced by the first author and Wiśniowska [8]. Indeed, the present article grew out of a part of the first author’s habilitation [6] which summarizes the study of that class. See Section 5 for details.

Acknowledgements. The authors are grateful to the referee for careful reading of the manuscript and helpful suggestions.

2. Conformal representation of the interior of an ellipse

We begin with introduction of special functions involving elliptic integrals. Let K(z, t) and K(t) be the normal and complete elliptic integrals of the first kind, 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) for 0 < t < 1. Note that Jacobi’s elliptic function sn( · , t) is defined as the inverse function of K( · , t) with sn(0, t) = 0, where our notation K(z, t) and sn(w, t) may not agree with the traditional one. It is well known that K( · , t) maps the upper half plane conformally onto the rectangle with vertices at ±K(t) and ±K(t)+iK(t′), where t′ = \sqrt{1 − t^{2}} (see, for instance, [11, Chapter VI, Section 3]). Since the interval [−1, 1] is mapped to the interval [−K(t), K(t)], the function K( · , t) can be continued analytically to the slit domain C \ ((−∞, −1] ∪ [1, +∞)) by the Schwarz reflection principle. In what follows, the function K( · , t) will be understood in this way.

The quantity

μ(t) = \frac{π}{2} · \frac{K(t′)}{K(t)}, t′ = \sqrt{1 − t^{2}} ,

is known as the modulus of the Grötzsch ring D \ [0, t] for 0 < t < 1. Note that μ(t) decreases from +∞ to 0 when t moves from 0 to 1. For details, see [3].

We are now in a position to present an explicit expression of the function f_{ξ}. Choose a number s ∈ (0, 1) so that μ(s) = 2ξ. Then the formula

(2.1)  f_{ξ}(z) = sin[\frac{π}{2K(s)}K(z/\sqrt{s} , s)]

can be deduced. Note that the inverse function is given by

z = \sqrt{s} sn((2K(s)/π) arcsin w, s)

as is shown by [11, p. 296, (51)]^{1} (see also [13]).

Let us give an outline of the proof of (2.1) for the reader’s convenience. Recall that the function K(z, s) maps the upper half-plane conformally onto the rectangle \{u + iv : −K(s) < u < K(s), 0 < v < K(s′)\}, where s′ = \sqrt{1 − s^{2}} . Since

K(1/(sz̄), s) = \overline{K(z, s)} + iK(s′)

holds, the upper half of the disk |z| < 1/\sqrt{s} is mapped conformally onto the rectangle {u+iv : −K(s) < u < K(s), 0 < v < K(s′)/2}. Therefore, the function (π/(2K(s)))K(z/\sqrt{s} , s) maps the upper half of the unit disk onto the rectangle R = {u + iv : −π/2 < u < π/2, 0 < v < ξ}. On the other hand, since

(2.2)  sin(x + iy) = sin x cosh y + i cos x sinh y,

the function sin z maps R onto the upper half of D_{ξ}. In this way, we see that the function sin((π/(2K(s)))K(z/\sqrt{s} , s)) maps the upper half of the unit disk onto the upper half of D_{ξ}. By the Schwarz reflection principle, we obtain the expression in (2.1).

Since f_{ξ}(−\sqrt{s} ) = −1, the function g_{ξ} can be expressed by

g_{ξ}(z) = f_{ξ}(\frac{z − \sqrt{s}}{1 − \sqrt{s} z} ).

This formula is, however, not convenient to compute the Taylor coefficients of g_{ξ} about the origin. This is a motivation of deduction of the following formula.

Theorem 2.1. For ξ > 0, the relation g_{2ξ}(z) = 2(f_{ξ}(\sqrt{z} ))^{2} − 1 holds for |z| < 1.

As was indicated, Theorem 1.2 immediately follows from Theorem 1.1 by means of this identity. See also Theorem 5.1 for a related result.

The representations f_{ξ} and g_{ξ} also give explicit values of the hyperbolic density of the domain D_{ξ}. Recall that the hyperbolic density ϱ_{D} of a simply connected domain D with #(C \ D) ≥ 2 is defined by ϱ_{D}(w_{0}) = 1/|f′(0)| for a conformal mapping f of the unit disk D onto D with f(0) = w_{0} .

1 We remark that there is a confusion on p. 296 of Nehari’s book. Since c = πK′/(2K) in his notation, the norm q = e^{−K′/K} should be given by e^{−2c/π} instead of e^{−2c}.

Corollary 2.2. Let ξ = μ(s)/2 for s ∈ (0, 1). Then

ϱ_{D_{ξ}}(0) = \frac{2\sqrt{s}}{π} K(s) and ϱ_{D_{2ξ}}(−1) = ϱ_{D_{2ξ}}(1) = \frac{2s}{π^{2}} (K(s))^{2}.

Proof. Since f′_{ξ}(0) = π/(2\sqrt{s} K(s)) and g′_{2ξ}(0) = 2(f′_{ξ}(0))^{2} we obtain the required relations.

In order to prove Theorem 2.1, we recall some facts about Chebyshev polynomials. We first consider the conformal mapping J of D onto Ĉ \ [−1, 1] defined by J(z) = (z + z^{−1})/2. Since

J(e^{−ξ+iη}) = cosh ξ cos η − i sinh ξ sin η,

the circle |z| = e^{−ξ} is mapped by J onto the ellipse E_{ξ} for ξ > 0 and the radial segment (0, e^{iη}) is mapped by J into the branch H_{η} of a hyperbola given by

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

for η ∈ R with (2/π)η ∉ Z. Note that these conic sections have the common foci at −1 and 1.

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 to T_{n} under J, in other words,

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

holds for |z| < 1. In particular, one can see that the ellipse E_{ξ} is mapped by T_{n} onto E_{nξ} in an n-to-one fashion and that the branch H_{η} of a hyperbola is mapped by T_{n} bijectively to H_{nη}.

Applying the above argument to T_{2}(w) = 2w^{2} − 1, we obtain the following. We recall that D_{ξ} is the interior of the ellipse E_{ξ}.

Lemma 2.3. The Chebyshev polynomial T_{2}(w) = 2w^{2} − 1 maps D_{ξ} onto D_{2ξ}. Also, T_{2} maps the domain bounded by H_{η} and H_{π−η} onto the connected component of C \ H_{2η} containing −1. Both are two-sheeted branched covering projections.

On the basis of the above lemma, we can prove Theorem 2.1.

Proof of Theorem 2.1. By Lemma 2.3, the composed function T_{2} ∘ f_{ξ} is a two-sheeted covering projection of D onto D_{2ξ} which sends the origin to the focus −1 of E_{2ξ}. Since T_{2}∘f_{ξ} is even, the function g(z) = (T_{2}∘f_{ξ})(\sqrt{z} ) is single-valued and analytic in D. By construction, g is conformal and satisfies g′(0) > 0, and therefore, g = g_{2ξ}. Thus the theorem has been proved.

The same reasoning yields a relation between conformal mappings onto domains bounded by hyperbolas. Let F_{η} be the conformal mapping of D onto the domain bounded by H_{η} and H_{π−η} which are given by (2.3) with F_{η}(0) = 0 and F′_{η}(0) > 0 for 0 < η < π/2. We also let G_{η} be the conformal mapping of D onto the left component of C \ H_{η} with G_{η}(0) = −1, G′_{η}(0) > 0 for 0 < η < π, η ≠ π/2. We define G_{π/2} as the limit of G_{η} as η → π/2, that is, G_{π/2}(z) = (z − 1)/(z + 1). Then we obtain the following.

Proposition 2.4. Let η ∈ (0, π/2). Then G_{2η}(z) = 2(F_{η}(\sqrt{z} ))^{2} − 1 for |z| < 1.

In view of (2.2) we see that the function sin z maps the parallel strip | Re z| < a conformally onto the domain {u + iv : (u/ sin a)^{2} − (v/ cos a)^{2} < 1} for 0 < a < π/2. Noting that the function arctan z maps the unit disk onto the strip | Re w| < π/2, we have the expression

(2.4)  F_{η}(z) = sin((2 − \frac{4η}{π} ) arctan z) for 0 < η < π/2.

3. A linear ODE satisfied by the conformal representation f_{ξ}

It is a noteworthy fact that the conformal representations f_{ξ} and F_{η}, which are given in (2.1) and (2.4), respectively, satisfy simple second-order linear ordinary differential equations (ODE).

Indeed, if we write b = 2 − 4η/π, we have F′_{η}(z) = b(1 + z^{2})^{−1} cos(b arctan z). Differentiating both sides of (1 + z^{2})F′_{η}(z) = b cos(b arctan z), we see that w = F_{η} satisfies the differential equation

(1 + z^{2})^{2}w′′ + 2z(1 + z^{2})w′ + b^{2}w = 0.

Similarly, one can check that the function w = f_{ξ}(z) satisfies the differential equation

(3.1)  (1 − 2Mz^{2} + z^{4})w′′ − 2z(M − z^{2})w′ + cw = 0

in D, where M = (s + s^{−1})/2, c = π^{2}/(4s(K(s))^{2}) and s ∈ (0, 1) is chosen so that μ(s) = 2ξ.

Let w = f(z) be the solution to the differential equation (3.1) with the initial conditions f(0) = 0 and f′(0) = 1. Note that f_{ξ} can be written in the form f_{ξ} = f′_{ξ}(0)f, and hence, positivity of the Taylor coefficients of f_{ξ} is equivalent to that of f.

As was seen in the introduction, f(z) has the Taylor expansion of the form

f(z) = ∑_{n=0}^{∞} A_{n}z^{2n+1},

with A_{0} = 1. Substituting the above expansion to the equation (3.1), we obtain the following recurrence relations for the coefficients A_{n}:

(3.2) (2n + 2)(2n + 3)A_{n+1} − {2M(2n + 1)^{2} − c}A_{n} + 2n(2n − 1)A_{n−1} = 0

for n ≥ 0; here we have set A_{−1} = 0.

Since the image f(D) is bounded by an ellipse and, in particular, convex, |A_{n}| ≤ 1 holds for every n ≥ 1 (see [4, p. 45]). As was explained in the introduction, all the coefficients A_{n} are real. However, we have no a priori information about the sign of A_{n}.

We take a closer look at the recurrence formula (3.2). We now transform the sequence by

B_{n} = (2n + 1)A_{n}.

Then B_{−1} = 0, B_{0} = 1 and the relation (3.2) turns to

(3.3)  (n + 1)B_{n+1} − \{M(2n + 1) − \frac{c}{2(2n + 1)}\}B_{n} + nB_{n−1} = 0

for n ≥ 0. We further set

E_{n} = \frac{B_{n}}{B_{n−1}}

for n ≥ 0. Here, we adopt the convention E_{n} = ∞ when B_{n−1} happens to be zero. Thus, for instance, E_{0} = ∞. By dividing both sides of (3.3) by (n + 1)B_{n}, we obtain

E_{n+1} = \frac{M(2n + 1)}{n + 1} − \frac{c}{(2n + 1)(2n + 2)} − \frac{n}{n + 1} · \frac{1}{E_{n}} .

By letting

(3.4)  p_{n} = \frac{n}{n + 1}, q_{n} = \frac{M(2n − 1)}{n} − \frac{c}{2n(2n − 1)},

the last relation can be rewritten in the form

(3.5)  E_{n+1} = q_{n+1} − \frac{p_{n}}{E_{n}} , n = 0, 1, 2, ….

We recall that the constants M and c are given by

M = \frac{1}{2}(s + \frac{1}{s}) and c = \frac{π^{2}}{4s(K(s))^{2}} .

We remark that c = (f′_{ξ}(0))^{2} = 1/(ϱ_{D_{ξ}}(0))^{2} (cf. Corollary 2.2). For a later use, we give estimates of c.

Lemma 3.1. The quantity c = π^{2}/(4s(K(s))^{2}) satisfies the double inequality

\frac{1}{s} − s ≤ c ≤ \frac{1}{s} − \frac{s}{2}.

Proof. It is easily checked that the required inequality is equivalent to

(3.6)  \frac{π}{2\sqrt{1 − s^{2}/2}} ≤ K(s) ≤ \frac{π}{2\sqrt{1 − s^{2}}}

for 0 < s < 1. First, using the inequality \sqrt{1 − s^{2}x^{2}} ≥ \sqrt{1 − s^{2}} for 0 < x < 1, we obtain

K(s) = ∫_{0}^{1} \frac{dx}{\sqrt{(1 − x^{2})(1 − s^{2}x^{2})}} ≤ \frac{1}{\sqrt{1 − s^{2}}} ∫_{0}^{1} \frac{dx}{\sqrt{1 − x^{2}}} = \frac{π}{2\sqrt{1 − s^{2}}} .

To show the other part, we need another technique. We first express K(s) in the form

K(s) = ∫_{0}^{π/2} \frac{dθ}{\sqrt{1 − s^{2} sin^{2} θ}} = ∫_{0}^{π/4}(\frac{1}{\sqrt{1 − s^{2} sin^{2} θ}} + \frac{1}{\sqrt{1 − s^{2} cos^{2} θ}} ) dθ.

Note here the inequality

(1 − s^{2} sin^{2} θ)(1 − s^{2} cos^{2} θ) = 1 − s^{2} + \frac{s^{4}}{4} sin^{2}(2θ) ≤ (1 − \frac{s^{2}}{2} )^{2}.

We now use the inequality 1/X + 1/Y ≥ 2/\sqrt{XY} for X, Y > 0 to deduce

K(s) ≥ ∫_{0}^{π/4} \frac{2}{\sqrt[4]{(1 − s^{2} sin^{2} θ)(1 − s^{2} cos^{2} θ)}} dθ ≥ \frac{π}{2\sqrt{1 − s^{2}/2}} .

Remark. Matti Vuorinen told us that the inequalities

\frac{2}{1 + \sqrt{1 − s^{2}}} ≤ \frac{2}{π} K(s) ≤ \frac{1}{\sqrt[4]{1 − s^{2}}} ,

which are better than (3.6), are known (see [3, 4.6(3)]). See also [2, Section 7.4] for a different kind of inequalities and related references. The authors thank him for the above information.

Let us explain the difficulty of the recurrence relations (3.5). First note that p_{n} → 1 and q_{n} → s + s^{−1} as n → ∞. Thus, one can think that the dynamical system E_{n+1} = q_{n+1} − p_{n+1}/E_{n} converges to the autonomous system E_{n+1}^{∘} = s + s^{−1} − 1/E_{n}^{∘} as n → ∞. As is easily observed, the linear fractional transformation f(x) = s + s^{−1} − x^{−1} has the attracting fixed point x = s^{−1} and the repelling fixed point x = s. On the other hand, in reality, a numerical computation suggests that E_{n} → s as n → ∞. Therefore, usual methods of approximation and even induction arguments seem to fail to show positivity of E_{n}. Therefore, we have to take a different approach.

By (3.5), we can express E_{n} in terms of a continued fraction:

E_{n} = q_{n} − \frac{p_{n−1}}{q_{n−1} − \frac{p_{n−2}}{... − \frac{p_{1}}{q_{1}}}} = q_{n} − \frac{p_{n−1}}{q_{n−1}} − \frac{p_{n−2}}{q_{n−2}} − ⋯ − \frac{p_{1}}{q_{1}} .

We define the double sequence q_{m,n} for 1 ≤ m ≤ n by induction of n − m. Fix a positive integer n. First we set

q_{n,n} = q_{n}.

Suppose that q_{n,n}, q_{n−1,n}, …, q_{m+1,n} have already been defined for 1 ≤ m < n. Then, we set

(3.7)  q_{m,n} = q_{m} − \frac{p_{m}}{q_{m+1,n}} .

In this way, we can define q_{n,n}, …, q_{1,n}. Then we can restate positivity of E_{n} in terms of q_{m,n}.

Lemma 3.2. Let n be a positive integer. Then E_{m} > 0 holds for each m with 1 ≤ m ≤ n if and only if q_{m,n} > 0 for each m with 1 ≤ m ≤ n.

Proof. We first assume that q_{m,n} > 0 for all 1 ≤ m ≤ n. Then q_{1,n} > 0 implies

E_{1} = q_{1} > \frac{p_{1}}{q_{2,n}} .

In particular, E_{1} > 0 (though this is implied by Lemma 3.1). Since q_{2,n} > 0 by assumption, we obtain

q_{2,n} = q_{2} − \frac{p_{2}}{q_{3,n}} > \frac{p_{1}}{E_{1}} ,

which is equivalent to

E_{2} = q_{2} − \frac{p_{1}}{E_{1}} > \frac{p_{2}}{q_{3,n}} .

In particular, we observe E_{2} > 0. We now use q_{3,n} > 0 to see

q_{3,n} = q_{3} − \frac{p_{3}}{q_{4,n}} > \frac{p_{2}}{E_{2}} .

We repeat this procedure to get finally

q_{n,n} = q_{n} > \frac{p_{n−1}}{E_{n−1}} ,

which yields

E_{n} = q_{n} − \frac{p_{n−1}}{E_{n−1}} > 0.

The converse can be seen by tracing back the above.

At this stage, we collect some elementary properties of q_{m,n}. When we regard q_{m,n} as a function of s in (0, 1), we sometimes write q_{m,n}(s) to indicate the argument s. We also write q_{m,n}(1) = lim_{s→1} q_{m,n}(s) if the limit exists. In particular, we have q_{m}(1) = q_{m,m}(1) = 2 − 1/m because c = c(s) → 0 when s → 1. We first prepare the following lemma.

Lemma 3.3. Let m be a positive integer. The function q_{m}(s) is positive in 0 < s < 1 and the inequality q_{m}(s) ≥ q_{m}(1) holds for 0 < s ≤ (2m − 2)/(2m − 1).

Remark. If the inequality q_{m}(s) ≥ q_{m}(1) held for all 0 < s < 1, the proof of positivity of E_{n} would be simpler. Unfortunately, this is not the case.

Proof. By Lemma 3.1, we have c < s^{−1} − s/2. Thus, we obtain

q_{m} ≥ q_{1} = \frac{s + s^{−1}}{2} − \frac{c}{2} > \frac{3s}{4} > 0.

It is easily verified that the condition q_{m}(s) ≥ q_{m}(1) = 2−1/m is equivalent to (2/π)(1−s)K(s) ≥ 1/(2m−1). Since K(s) ≥ π/2, the condition s ≤ 1−1/(2m−1) is enough to ensure the inequality q_{m}(s) ≥ q_{m}(1).

The following result is readily shown by (reverse) induction on m.

Lemma 3.4. Let a_{1}, …, a_{n}; x_{1}, …, x_{n} and x′_{1}, …, x′_{n} be positive numbers with x_{m} ≤ x′_{m} for m = 1, …, n. Define y_{m} by reverse induction: y_{n} = x_{n} and y_{m} = x_{m} − a_{m+1}/y_{m+1} for m < n. Similarly, set y′_{n} = x′_{n} and y′_{m} = x′_{m} − a_{m+1}/y′_{m+1} for m < n. Further suppose that y_{m} > 0 for all 1 ≤ m ≤ n. Then y_{m} ≤ y′_{m} and, in particular, y′_{m} > 0 for all m.

With the aid of the above lemma, we can now show the following.

Lemma 3.5. Let m and n be integers with 2 ≤ m ≤ n. Then the quantity q_{m,n} = q_{m,n}(s) is positive for 0 < s ≤ (2m − 2)/(2m − 1).

Proof. We first show that q_{m,n}(1) > 0 for 1 ≤ m ≤ n. In view of Lemma 3.2, it is enough to see that E_{n}(1) is positive for each n ≥ 1. The solution w = f(z) with f(0) = 0, f′(0) = 1 to the equation (3.1) corresponding to the case when s = 1 is nothing but the function arctanh z. Clearly, this has positive odd Taylor coefficients about the origin, and therefore, the inequality E_{n}(1) > 0 follows. Lemmas 3.3 and 3.4 now yield the inequality q_{m,n}(s) ≥ q_{m,n}(1) > 0 for 0 < s ≤ (2m − 2)/(2m − 1) and for 2 ≤ m ≤ n.

The next simple fact will be a key to the proof of Theorem 1.1.

Lemma 3.6. Let n and n_{0} be integers with 2 ≤ n_{0} ≤ n. Suppose that q_{m,n} > 0 holds for every m with n_{0} ≤ m ≤ n. Then q_{m,n−1} > q_{m,n} holds for every m with n_{0} − 1 ≤ m ≤ n − 1.

Proof. We shall show q_{m,n−1} > q_{m,n} by reverse induction on m. For m = n − 1, the inequality holds because q_{n−1,n} = q_{n−1,n−1} − p_{n−1}/q_{n}. We now assume that q_{m,n−1} > q_{m,n} holds for some m with n_{0} ≤ m ≤ n − 1. Note now that q_{m,n} > 0 by assumption. Since

q_{m−1,n−1} − q_{m−1,n} = −\frac{p_{m−1}}{q_{m,n−1}} + \frac{p_{m−1}}{q_{m,n}} = \frac{p_{m−1}(q_{m,n−1} − q_{m,n})}{q_{m,n−1}q_{m,n}},

we obtain q_{m−1,n−1} > q_{m−1,n}. This procedure can be continued up to m − 1 = n_{0} − 1.

At this stage, we can show that lim_{n→∞} q_{m,n} always exists.

Proposition 3.7. For each m ≥ 1, the sequence q_{m,n} has a limit in Ĉ when n → ∞.

Proof. Since q_{m,n} and q_{m+1,n} are related by a Möbius transformation described in (3.7), if q_{m,n} has a limit for some m then q_{m,n} does for all m. By Lemma 3.5, for a fixed s, there exists an integer N such that q_{m,n} > 0 whenever N ≤ m ≤ n. Lemma 3.6 now implies that q_{m−1,n} is monotone decreasing with respect to n. In particular, q_{m−1,n} has a limit as n → ∞ for m ≥ N − 1.

We denote by q_{m,∞} the limit of q_{m,n} as n → ∞. In order to find a value of q_{m,∞}, we employ the general theory of continued fractions, which will be explained in the next section.

4. A continued fraction approach

In order to apply the general theory of continued fractions to our problem, we recall some notions and results in the theory based on the work of L. Jacobsen (Lorentzen) and W. J. Thron [5].

Let {T_{n}} be a sequence of Möbius maps. The sequence is said to be restrained if there exist sequences {u_{n}} and {v_{n}} of points in the Riemann sphere Ĉ = C ∪ {∞} such that

lim inf_{n→∞} d(u_{n}, v_{n}) > 0

and that

lim_{n→∞} d(T_{n}(u_{n}), T_{n}(v_{n})) = 0,

where d(z, w) denotes the chordal distance between z and w, namely, d(z, w) = |z − w|/\sqrt{(1 + |z|^{2})(1 + |w|^{2})} . Note that the asymptotic behaviour of \{T_{n}(u_{n})\} is unique in the sense that d(T_{n}(u_{n}), T_{n}(u′_{n})) → 0 for any other pair of sequences {u′_{n}} and {v′_{n}} satisfying lim inf d(u′_{n}, v′_{n}) > 0 and lim d(T_{n}(u′_{n}), T_{n}(v′_{n})) = 0 (see [5, Theorem 2.1]). Sometimes we say that {T_{n}} is restrained with {u_{n}} when we want to indicate the associated sequence. A sequence {w_{n}} is said to be exceptional with respect to the restrained sequence {T_{n}} with {u_{n}} if

lim sup_{n→∞} d(T_{n}(w_{n}), T_{n}(u_{n})) > 0.

Among several interesting results in [5], the following will be made use of in the present paper.

Lemma 4.1 (Proposition 2.4 in [5]). Let {T_{n}} be a sequence of Möbius maps which is restrained with {u_{n}}. Suppose that lim inf_{n→∞} d(T_{n}(u_{n}), ∞) > 0. Then for an exceptional sequence {w_{n}}

lim inf_{n→∞} d(w_{n}, T_{n}^{−1}(∞)) = 0.

We now return to our problem. Let

R_{m}(z) = q_{m} − \frac{p_{m}}{z} , S_{m} = R_{1} ∘ ⋯ ∘ R_{m}, and T_{m} = S_{m}^{−1}

for m ≥ 1, where p_{m} and q_{m} are given by (3.4). Then, by definition,

q_{m,n} = (R_{m} ∘ ⋯ ∘ R_{n})(∞).

In particular,

(4.1)  q_{1,n} = S_{n}(∞) = S_{n+1}(0).

Also, by noting the relation R_{m}^{−1}(w) = p_{m}/(q_{m} − w), we observe

(4.2)  T_{n}(0) = \frac{p_{n}}{E_{n}} .

We now claim that our {T_{n}} is restrained. More concretely, we show the following.

Lemma 4.2. For a fixed s ∈ (0, 1), there exists a non-empty open interval I = I(s) in R such that T_{n}(x) → s for every x ∈ I.

Proof. Since q_{n} → s + s^{−1} (> 2) as n → ∞, one can take an integer N so that q_{n} > 2 for all n ≥ N. Let α_{n} = 1/\sqrt{p_{n}} = \sqrt{(n + 1)/n} and choose t_{n} ∈ (0, 1) for n ≥ N so that α_{n}q_{n} = t_{n} + 1/t_{n}, and thus,

(4.3)  \frac{1}{t_{n}} = \frac{α_{n}q_{n} + \sqrt{(α_{n}q_{n})^{2} − 4}}{2} .

We now investigate the asymptotic behaviour of t_{n} as n → ∞. Since

α_{n} = 1 + \frac{1}{2n} − \frac{1}{8n^{2}} + \frac{1}{16n^{3}} + O(n^{−4})

as n → ∞, we have

α_{n}q_{n} = s + \frac{1}{s} − \frac{2c + 3(s + s^{−1})}{8} · \frac{1}{n^{2}} + \frac{s + s^{−1} − 2c}{8} · \frac{1}{n^{3}} + O(n^{−4})

and

(4.4)  α_{n+1}q_{n+1} − α_{n}q_{n} = \frac{2c + 3(s + s^{−1})}{4} · \frac{1}{n^{3}} + O(n^{−4})

as n → ∞. In particular, α_{n+1}q_{n+1} > α_{n}q_{n} and hence t_{n+1} < t_{n} for sufficiently large n. It is obvious that t_{n} → s as n → ∞. Also, by (4.3) and (4.4), we obtain 1/t_{n+1} − 1/t_{n} = O(n^{−3}). If we set s_{n} = t_{n}/α_{n}, then

\frac{1}{s_{n−1}} − \frac{1}{s_{n}} = \frac{α_{n−1} − α_{n}}{t_{n−1}} + α_{n}(\frac{1}{t_{n−1}} − \frac{1}{t_{n}} ) = \frac{1}{2sn^{2}} + o(n^{−2})

as n → ∞. In particular, s_{n−1} < s_{n} for sufficiently large n. We replace N by a larger number if necessary so that t_{n+1} < t_{n} and s_{n−1} < s_{n} hold for all n ≥ N. In particular, we have t_{n} ≤ τ for n ≥ N, where we set τ = t_{N}(< 1). Since 1/α_{n} − s_{n} → 1 − s(> 0) as n → ∞, we may further assume that the inequality

(4.5)  τ(\frac{1}{α_{n}} − s_{n}) < \frac{1}{α_{n+1}} − s_{n+1}

holds for each n ≥ N. In the rest of the proof, we consider only integers n with n ≥ N.

It is easy to check that s_{n} is a fixed point of the map U_{n} = R_{n}^{−1} . Furthermore,

U_{n}(x) − U_{n}(s_{n}) = \frac{p_{n}(x − s_{n})}{(q_{n} − x)(q_{n} − s_{n})} = \frac{s_{n}}{q_{n} − x}(x − s_{n}).

When s_{n} < x < 1/α_{n}, one has 0 < s_{n}/(q_{n}−x) = t_{n}/(α_{n}q_{n}−α_{n}x) ≤ τ/(2−1) = τ, and therefore

(4.6)  0 < U_{n}(x) − U_{n}(s_{n}) ≤ τ(x − s_{n}).

For x_{N} ∈ R, we set x_{n} = (U_{n} ∘ ⋯ ∘ U_{N+1})(x_{N}) for n > N. Then, by (4.6) and (4.5), if s_{n} < x_{n} < 1/α_{n} we have

\begin{aligned}0 & < x_{n+1} − s_{n+1} = U_{n+1}(x_{n}) − U_{n+1}(s_{n+1}) ≤ τ(x_{n} − s_{n+1}) < τ(x_{n} − s_{n}) \\ & < τ(\frac{1}{α_{n}} − s_{n}) < \frac{1}{α_{n+1}} − s_{n+1}, \end{aligned}

and therefore, s_{n+1} < x_{n+1} < 1/α_{n+1} . By induction, we have s_{n} < x_{n} < 1/α_{n} for all n ≥ N if we take x_{N} from (s_{N}, 1/α_{N}). Therefore, the interval I = S_{N}((s_{N}, 1/α_{N})) works for the assertion.

If we choose a pair of distinct points x_{0} and x′_{0} from the interval I in the last lemma, then d(T_{n}(x_{0}), T_{n}(x′_{0})) → d(s, s) = 0 as n → ∞. In particular, {T_{n}} is a restrained sequence with constant sequence {x_{0}}. With the aid of this fact, we can now show the following.

Lemma 4.3. We have q_{1,∞} = 0.

Proof. Fix a number x_{0} ∈ I. Then T_{n}(x_{0}) → s as n → ∞. In particular, d(T_{n}(x_{0}), ∞) → d(s, ∞) > 0. Next we observe the relation T_{n}^{−1}(∞) = S_{n}(∞) = q_{1,n} by (4.1). Suppose now that the constant sequence {0} were not exceptional with respect to {T_{n}}. That would mean T_{n}(0) → s when n → ∞. Now we recall the relation (4.2). Since E_{n} = B_{n}/B_{n−1} and B_{n} = (2n + 1)A_{n}, we would have

lim_{n→∞} \frac{A_{n}}{A_{n−1}} = lim_{n→∞} \frac{B_{n}}{p_{n}B_{n−1}} = lim_{n→∞} \frac{1}{T_{n}(0)} = \frac{1}{s} > 1,

which would violate the boundedness of the sequence {A_{n}}. Thus we have concluded that the constant sequence {0} is exceptional with respect to {T_{n}}. Lemma 4.1 now yields

lim inf_{n→∞} d(0, q_{1,n}) = lim inf_{n→∞} d(0, T_{n}^{−1}(∞)) = 0,

which implies that 0 is a limit point of the convergent sequence {q_{1,n}} (see Proposition 3.7). Thus q_{1,∞} = lim_{n→∞} q_{1,n} must be 0.

Proof of Theorem 1.1. We now show the inequality q_{m,n} > 0 for all m ≤ n. This implies E_{n} > 0 for all n ≥ 1 by Lemma 3.2, and thus Theorem 1.1 follows.

For each s ∈ (0, 1), by Lemma 3.5, we see that there exists an integer N ≥ 1 such that q_{m,n} > 0 for any pair of integers m, n with N ≤ m ≤ n. We denote by N(s) the minimum of such numbers N for s ∈ (0, 1). Lemma 3.5 implies also that N(s) ≤ m for s ∈ (0, (2m−2)/(2m−1)]. In particular, N(s) ≤ 2 for s ∈ (0, 2/3]. If N(s) ≤ 2, then Lemma 3.6 gives us the information q_{1,n} > q_{1,n+1} for n ≥ 1. Since q_{1,n} → 0 by Lemma 4.3, we now conclude that q_{1,n} > 0 for n ≥ 1. Thus N(s) must be 1 in this case. In particular, N(s) = 1 for s ∈ (0, 2/3].

Suppose that N(s) > 2 for some s. Let s_{0} be the infimum of the set {s ∈ (0, 1) : N(s) > 2}. As we observed above, N(s) = 1 for s < s_{0} , namely, q_{m,n}(s) > 0 for all 1 ≤ m ≤ n, and thus 0 < E_{n}(s) < +∞, n ≥ 1, for s < s_{0} .

Since N(s) is locally bounded in 0 < s < 1, there exist an integer N > 2 and a decreasing sequence {s_{k}} such that N(s_{k}) = N and s_{k} → s_{0} as k → ∞. Since q_{N,n}(s_{k}) > 0 for n ≥ N, Lemma 3.6 yields q_{N−1,n−1}(s_{k}) > q_{N−1,n}(s_{k}) for n ≥ N. In particular, we obtain q_{N−1,n−1}(s_{k}) > q_{N−1,∞}(s_{k}) for n ≥ N. By minimality of N(s_{k}), we see that −∞ ≤ q_{N−1,∞}(s_{k}) < 0. On the other hand, q_{N−1,∞}(s) ≥ 0 for s < s_{0} . By continuity, we obtain q_{N−1,∞}(s_{0}) = ∞ or 0. Since

S_{N−2}(q_{N−1,∞}) = S_{N−2}( lim_{n→∞} q_{N−1,n}) = lim_{n→∞} S_{N−2}(q_{N−1,n}) = lim_{n→∞} q_{1,n} = 0,

we see that q_{N−1,∞} = T_{N−2}(0) = p_{N−2}/E_{N−2} . So E_{N−2}(s_{0}) = 0 or E_{N−2}(s_{0}) = ∞.

Suppose first that E_{N−2}(s_{0}) = 0. Now E_{N−2}(s) approaches 0 from the right as s → s0^{−}. On the other hand, E_{N−1}(s) > 0 implies

q_{N−1}(s) > \frac{p_{N−1}}{E_{N−2}(s)}

for s < s_{0} . Therefore, as s → s0^{−} the right-hand side goes to ∞, which forces q_{N−1}(s_{0}) to be infinity. Thus we have reached a contradiction.

We next suppose that E_{N−2}(s_{0}) = ∞. In this case, E_{N−3}(s_{1}) = 0 by (3.5). Then the same argument as above leads to a contradiction.

At any event, we get a contradiction. Thus the possibility that N(s) > 2 for some 0 < s < 1 has been ruled out. Therefore N(s) = 1, namely, q_{m,n}(s) > 0 for 1 ≤ m ≤ n.

Numerical experiments suggest the following conjectures, which seem to be difficult to prove by simple induction arguments. Recall that E_{n} = E_{n}(s) is defined as (2n + 1)A_{n}/(2n − 1)A_{n−1} in Section 3.

Conjecture 4.4.

(i) E_{n}(s) is increasing in 0 < s < 1 for each n ≥ 1. (ii) E_{n}(s) monotonically increases to s as n → ∞. (iii) E_{n}(s)/s increases from (2n + 1)/(2n + 2) to 1 when s moves from 0 to 1. (iv) Let a ∈ D_{ξ} ∩ (−∞, 0). The conformal map f of the unit disk D onto D_{ξ}

determined by f(0) = a and f′(0) > 0 has positive Taylor coefficients except for the first one.

Repeated use of (iii) would yield the inequality

\frac{(2n − 1)!!}{(2n)!!} · \frac{s^{n}}{n + 1} ≤ A_{n}(s) ≤ \frac{s^{n}}{2n + 1}

for each n ≥ 1 and for each s ∈ (0, 1). Note that Stirling’s formula implies

\frac{(2n − 1)!!}{(2n)!!} = \frac{(2n)!}{2^{2n}(n!)^{2}} ∼ \frac{1}{\sqrt{πn}} as n → ∞.

5. Applications to k -uniformly convex functions

We consider the domain

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

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

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

with foci at 1 and −(1 + k^{2})/(1 − k^{2}). When k = 1, the domain 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{(k^{2} − 1)u − k^{2}}{k} )^{2} + (\frac{(k^{2} − 1)v}{\sqrt{k^{2} − 1}} )^{2} = 1

with foci at 1 and (k^{2} + 1)/(k^{2} − 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} and that Ω_{k} converges to Ω_{k_{0}} in the sense of Carathéodory when k → k_{0} .

An analytic function f in the unit disk D normalized by f(0) = 0 and f′(0) = 1 is called k -uniformly convex if 1 + zf′′(z)/f′(z) ∈ Ω_{k} for z ∈ D. This concept was introduced and studied by the first author and Wiśniowska [8], [7]. Clearly, 0-uniformly convex functions are exactly same as convex functions. Moreover, uniformly convex functions introduced by Goodman are characterized as 1-uniformly convex functions (see [10] and [12]).

Let P_{k} be the conformal mapping of D onto Ω_{k} determined by the conditions P_{k}(0) = 1 and P′_{k}(0) > 0. They gave a concrete expression for P_{k}.

Theorem A (Kanas–Wiśniowska [8]). 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{matrix}(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{matrix}

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.

By Theorem 2.1, we can obtain another expression which is more convenient to compute the Taylor expansion about the origin.

Theorem 5.1. For k ≤ 0, the conformal map P_{k} of the unit disk onto the domain Ω_{k} with P_{k}(0) = 1 and P′_{k}(0) > 0 is expressed as

(5.2)  P_{k}(z) = 1 + (Q_{k}(\sqrt{z} ))^{2},

where

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

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

Furthermore, the function Q_{k} is odd and maps the unit disk conformally onto the domain W_{k} = {u + iv : (k − 1)u^{2} + (k + 1)v^{2} < 1}.

This result was presented in the first author’s habilitation [6, Theorem 2.2.2] with a proof slightly different from the one here.

It is easily checked that W_{k} is the inside of a hyperbola when k < 1 and W_{k} is the interior of an ellipse when k > 1. When k = 1, the domain W_{k} becomes the parallel strip −1/\sqrt{2} < Im w < 1/\sqrt{2} . Note that W_{k} is invariant under the involution w ↦ −w. The relation between t and s in Theorems A and 5.1 is given by s = 2\sqrt{t} /(1 + t) (see [11, p. 293, (43)]).

The reader might expect that the functions P_{k} could be expressed in a unified way for all 0 < k < ∞ by introducing another kind of special functions. It is, however, hopeless to do that because we discarded the left half of the “interior” of the hyperbola when k became less than 1 (see also that the forms of corresponding differential equations are different). Though we can prove Theorem 5.1 by using Theorem A, we give an independent proof so that the present article be selfcontained as far as possible.

Proof of Theorem 5.1. First let k > 1. Choose ξ > 0 so that cosh(2ξ) = k, namely, k = cosh(μ(s)). Since the similarity

L(z) = (z + k^{2})/(k^{2} − 1) = 1 + (z + 1)/(k^{2} − 1)

maps −1 to 1 and 1 to (k^{2}+1)/(k^{2}−1), respectively, the image L(D_{2ξ}) coincides with Ω_{k}. Thus P_{k} = L ∘ g_{2ξ} = 1 + (g_{2ξ} + 1)/(k^{2} − 1). By Theorem 1.2, we obtain the relation P_{k}(z) = 1 + 2(f_{ξ}(\sqrt{z} ))^{2}/(k^{2} − 1). Hence, we conclude that P_{k}(z) = 1 + (Q_{k}(\sqrt{z} ))^{2} .

The case when 0 < k < 1 can be treated as above. Indeed, take a number η ∈ (0, π/4) so that k = cos 2η and let L(z) = (z + k^{2})/(k^{2} − 1). Then the similarity L maps the left component of C \ H_{π−2η} onto Ω_{k}. Noting L′(−1) < 0, we obtain the relation P_{k}(z) = L(G_{π−2η}(−z)) for z ∈ D. Proposition 2.4 and formula (2.4) now yield

P_{k}(z) = L(2(F_{π/2−η}(i\sqrt{z} ))^{2} − 1) = L(2 sin((4η/π) arctan(i\sqrt{z} ))^{2} − 1)
= L(−2 sinh((4η/π) arctanh(\sqrt{z} ))^{2} − 1) = 1 + (Q_{k}(\sqrt{z} ))^{2}.

When k = 1, the expression is obtained as the limiting case when k → 1.

As an immediate corollary, we have another characterization of k -uniformly convex functions (cf. [6, Theorem 3.8.1]).

Theorem 5.2. For a function f locally univalent and analytic in the unit disk D and normalized so that f(0) = 0, f′(0) = 1, let g be the analytic branch of log f′ determined by g(0) = 0. The function f is k -uniformly convex if and only if \sqrt{zg′(z)} ∈ W_{k} for each z ∈ D.

Proof. Set p(z) = 1 + zf′′(z)/f′(z). It is enough to show that p(D) ⊂ Ω_{k}. By assumption, there exists a point w ∈ W_{k} such that zg′(z) = w^{2} for each z ∈ D. Choose a point ζ ∈ D so that Q_{k}(ζ) = w. Then, p(z) = 1 + zg′(z) = 1 + (Q_{k}(ζ))^{2} = P_{k}(ζ^{2}) by Theorem 5.1, which implies p(z) ∈ Ω_{k}.

The next result was also claimed in [6, Theorem 2.3.3]. However, the proof for the case k > 1 contained a serious error. In order to correct it, the present joint research was initiated.

Theorem 5.3. The conformal representation P_{k} of Ω_{k} with P_{k}(0) = 1, P′_{k}(0) > 0 has positive Taylor coefficients about the origin for each k ≥ 0.

Proof. The assertion for 0 ≤ k ≤ 1 can be deduced by Theorem 5.1 because sinh z and arctanh z both have positive odd Taylor coefficients about the origin. The assertion for k > 1 follows from Theorem 1.2 since P_{k} = 1+(g_{2ξ} +1)/(k^{2} −1) as observed in the proof of Theorem 5.1.

We remark that for 0 < k < 1, the function w = Q_{k}(z) satisfies the linear ODE

(5.3)  (1 − z^{2})^{2}w′′ − 2z(1 − z^{2})w′ − C_{k}^{2}w = 0

in D. By using this, one can also show positivity of the Taylor coefficients of P_{k} about the origin. Note that this differential equation is a special case of the Legendre equation (see, for example, [1, Chapter 8]).

An analytic function p in the unit disk is called a Carathéodory function if p(0) = 1 and if p has positive real part, in other words, Re p(z) > 0 holds for |z| < 1. The class of Carathéodory functions will be denoted by 𝒫.

For two analytic functions f and g in the unit disk D, we say that f is subordinate to g and denote it by f ≺ g if there exists an analytic map ω: D → D such that f = g ∘ ω and ω(0) = 0. For each Carathéodory function q, we define the subclass

𝒫(q) = {p ∈ 𝒫 : p ≺ q}

of 𝒫. For instance, a normalized analytic function f in the unit disk is k - uniformly convex if and only if the function 1 + zf′′(z)/f′(z) belongs to 𝒫(P_{k}).

If q has some nice properties, then elements of 𝒫(q) are dominated by q in various ways.

Proposition 5.4. Suppose that a Carathéodory function q is convex univalent and has non-negative Taylor coefficients about the origin. Then each element p of 𝒫(q) satisfies

q(−|z|) ≤ Re p(z) ≤ |p(z)| ≤ q(|z|)

for |z| < 1.

Proof. We write p_{r}(z) = p(rz) for 0 < r < 1. The Lindelöf principle says that p ≺ q implies p_{r} ≺ q_{r} for all 0 < r < 1. Therefore, for a fixed r ∈ [0, 1), we obtain

min_{|z|=r} Re q(z) ≤ Re p(z_{0}) ≤ |p(z_{0})| ≤ max_{|z|=r} |q(z)|

for any z_{0} with |z_{0}| = r. Notice now that q is symmetric, i.e., \overline{q(z)} = q(z̄), because q has non-negative (thus real) coefficients. Since q_{r} is convex (see the proof of Theorem 2.1 in [4]) and symmetric and Re q_{r} > 0, the relation

min_{|z|=r} Re q(z) = q(−r)

can be deduced. On the other hand, the non-negativity of the coefficients means that q can be expressed in the form

q(z) = 1 + ∑_{n=1}^{∞} a_{n}z^{n}, |z| < 1,

for some a_{n} ≥ 0. Therefore,

|q(z)| ≤ 1 + ∑_{n=1}^{∞} a_{n}|z|^{n} = q(|z|),

and thus,

max_{|z|=r} |q(z)| = q(r)

follows.

This, together with Theorem 5.3, implies the following.

Corollary 5.5. Let 0 ≤ k < ∞ and p ∈ 𝒫(P_{k}). Then the inequalities

Re p(z) ≤ |p(z)| ≤ \{ \begin{matrix}1 + \frac{2}{1 − k^{2}} sinh^{2}(\frac{2 arctanh \sqrt{r}}{π arccos k} ) & for k ∈ [0, 1), \\ 1 + \frac{8}{π^{2}} arctanh^{2} \sqrt{r} & for k = 1, \\ 1 + \frac{2}{k^{2} − 1} sin^{2}(\frac{π}{2K(t)}K(\sqrt{r/t} , t)) & for k > 1, \end{matrix}

and

|p(z)| ≥ Re p(z) ≥ \{ \begin{matrix}1 − \frac{2}{1 − k^{2}} sin^{2}(\frac{2 arctan \sqrt{r}}{π arccos k} ) & for k ∈ [0, 1), \\ 1 − \frac{8}{π^{2}} arctan^{2} \sqrt{r} & for k = 1, \\ 1 − \frac{2}{k^{2} − 1} sinh^{2}(\frac{π}{2K(t)}K̃(\sqrt{r/t} , t)) & for k > 1, \end{matrix}

348 Stanisława Kanas and Toshiyuki Sugawa

hold for z ∈ D with |z| = r. In particular, the inequalities

\frac{k}{k + 1} < Re p(z) ≤ |p(z)|

hold for all |z| < 1 and all k > 0 and |p(z)| < k/(k − 1) holds for all |z| < 1 and k > 1.

In the above, we set

K̃(r, t) = −iK(ir, t) = ∫_{0}^{r} \frac{dx}{\sqrt{(1 + x^{2})(1 + t^{2}x^{2})}} .

It is expected that the positivity theorem (Theorem 5.1) has more applications to the class of k -uniformly convex functions. For instance, it gives an improvement of the bound for the lengths of the images of the unit circle under k -uniformly starlike functions (see [9, Theorem 3.5]).

References

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[3] Anderson, G. D., M. K. Vamanamurthy, and M. Vuorinen: Topics in special functions. - Rep. Univ. Jyväskylä 83, 2001, 5–26.

[4] Duren, P. L.: Univalent Functions. - Springer-Verlag, 1983.

[5] Jacobsen, L., and W. J. Thron: Limiting structures for sequences of linear fractional transformations. - Proc. Amer. Math. Soc. 99, 1987, 141–146.

[6] Kanas, S.: Generalized notion of starlikeness and convexity. - Rzeszów University of Technology, 2003.

[7] Kanas, S., and A. Wiśniowska: Conic regions and k -uniform convexity II. - Zeszyty Nauk. Politech. Rzeszowskiej Mat. 170(22), 1998, 65–78.

[8] Kanas, S., and A. Wiśniowska: Conic regions and k -uniform convexity. - J. Comput. Appl. Math. 105, 1999, 327–336.

[9] Kanas, S., and A. Wiśniowska: Conic domains and starlike functions. - Rev. Roumaine Math. Pures Appl. 45, 2000, 647–657.

[10] Ma, W., and D. Minda: Uniformly convex functions. - Ann. Polon. Math. 57, 1992, 165–175.

[11] Nehari, Z.: Conformal Mappings. - McGraw-Hill, New York, 1952.

[12] Rønning, F.: Uniformly convex functions and a corresponding class of starlike functions. - Proc. Amer. Math. Soc. 118, 1993, 189–196.

[13] Szegö, G.: Conformal mapping of the interior of an ellipse onto a circle. - Amer. Math. Monthly 57, 1950, 474–478.

Received 29 March 2005