Geometric properties of

functions with small Schwarzian

derivatives

Toshiyuki Sugawa

Department of Mathematics, Graduate School of Science,

Hiroshima University, Higashi-Hiroshima, 739-8526 Japan

sugawa@math.sci.hiroshima-u.ac.jp
http://sugawa@cajpn.org/

July 29, 2003, Chiang Mai, Thailand

Joint work with Ji-A Kim at POSTECH, Korea Schwarzian derivative.

For a non-constant meromorphic function f in a plane domain, the Schwarzian derivative S_{f} of f is defined by

S_{f} = (\frac{f′′}{f′})′ − \frac{1}{2} (\frac{f′′}{f′})^{2} .

S_{f} is holomorphic at z_{0} ⇔ f is locally univalent at z_{0}.

Fact: S_{f} = 0 if and only if f is (a restriction of) a Möbius map. Moreover, S_{L∘f} = S_{f} for any Möbius transformation L.

⟹ Schwarzian derivative measures deviation of the function from Möbius maps.

Univalence criteria.

Theorem 1 (Nehari ∼ 1949) If f is univalent meromorphic in the unit disk 𝔻, then

|S_{f}(z)| ≤ 6(1 − |z|^{2})^{−2}.

Conversely, if a meromorphic function f in 𝔻 satisfies

|S_{f}(z)| ≤ 2(1 − |z|^{2})^{−2},

then f is univalent in 𝔻. The numbers 6 and 2 are sharp.

• The Koebe function K(z) = z/(1 − z)^{2} satisfies

S_{K}(z) = \frac{−6}{(1 − z^{2})^{2}} .

• The function L(z) = (1/2) log (1 + z)/(1 − z) which maps 𝔻 onto the paralell strip |Im w| < π/2 satisfies

S_{L}(z) = \frac{2}{(1 − z^{2})^{2}} .

Theorem 2 (Nehari 1949, Pokornyi 1951) If f satisfies one of the following conditions in 𝔻, then f is univalent in 𝔻 :

|S_{f}(z)| ≤ \frac{π^{2}}{2},
|S_{f}(z)| ≤ 4(1 − |z|^{2})^{−1}.

These numbers are sharp.

• Extremal functions are given, respectively, by

tan \frac{πz}{2} and \frac{z}{2(1 − z^{2})} + \frac{1}{4} log \frac{1 + z}{1 − z} .

Based on these results, a more general univalence criteria were deduced by Avkhadiev and et al.

Connection with a linear ODE.

For a given holomorphic function φ in the unit disk 𝔻, we can construct a locally univalent meromorphic function f so that S_{f} = φ in 𝔻. Indeed, let y_{0} and y_{1} be the analytic solutions to the ODE

2y′′ + φy = 0

in 𝔻 with the initial conditions

y_{0}(0) = 1, y_{1}(0) = 0,
y′_{0}(0) = 0, y′_{1}(0) = 1.

Note: the Wronskian is identically 1 :

y_{0}y′_{1} − y′_{0}y_{1} ≡ 1.

Then the quotient f = y_{1}/y_{0} is a desired one, because the logarithmic derivative of

f′ = \frac{y_{0}y′_{1} − y′_{0}y_{1}}{y_{0}^{2}} = \frac{1}{y_{0}^{2}}

yields

\frac{f′′}{f′} = −\frac{2y′_{0}}{y_{0}} .

Hence,

S_{f} = (−\frac{2y′_{0}}{y_{0}} )′ − \frac{1}{2} (−\frac{2y′_{0}}{y_{0}} )^{2}

= − (\frac{2y′′_{0}}{y_{0}} ) + 2 (\frac{y′_{0}}{y_{0}} )^{2} − 2 (\frac{y′_{0}}{y_{0}} )^{2}

= φ.

Note: the constructed function f satisfies f(0) = 0, f′(0) = 1 and f′′(0) = 0. Third condition can easily be missed!

Normalizations.

M: the set of meromorphic functions f in the unit disk 𝔻 with f(0) = 0, f′(0) = 1. For a complex number c, set

M(c) = {f ∈ M : f′′(0) = 2c}.

Fact:for φ and for c ∈ ℂ, there is the unique function f = f_{φ,c} in M(c) for which S_{f} = φ holds.

Indeed, such an f can be given by

f_{φ,c} = \frac{y_{1}}{y_{0} − cy_{1}} = \frac{f_{φ,0}}{1 − cf_{φ,0}} .

Note: f_{φ,c}(z) = z + cz^{2} + · · · .

Omitted values. Set K(φ) = {c ∈ ℂ : 1/c ∉ f_{φ,0}(𝔻)}. Note that K(φ) is always compact.

• f_{φ,c} is pole-free (i.e., analytic) ⇔ c ∈ K(φ).

• |c| ≤ 2 for each c ∈ K(φ) if f_{φ,0} is univalent meromorphic.

Weight functions.

A(x), 0 ≤ x < 1, is called a weight function if it is locally Lipschitz, non-decreasing, and positive.

Example: A(x) = C(1 − x^{2})^{−μ}.

Let U_{0}, U_{1}, V_{0} and V_{1} be the functions on [0, 1) determined by

2U_{0}′′ = AU_{0}, U_{0}(0) = 1, U_{0}′(0) = 0,
2U_{1}′′ = AU_{1}, U_{1}(0) = 0, U_{1}′(0) = 1,
2V_{0}′′ = −AV_{0}, V_{0}(0) = 1, V_{0}′(0) = 0,
2V_{1}′′ = −AV_{1}, V_{1}(0) = 0, V_{1}′(0) = 1.

When we need to indicate the weight function A, we write, for example, U_{0}(x, A) = U_{0}(x).

• U_{0} > 0 and U_{0}′ > 0 hold on the interval [0, 1) for any weight function A.

A general univalence criterion

Theorem 3 (Nehari 1954) If

(i) A(x)(1−x^{2})^{2} is non-increasing in 0 ≤ x < 1, (ii) V_{0}(x, A) is positive for 0 ≤ x < 1, then the condition |S_{f}(z)| ≤ A(|z|) for a function f ∈ M implies univalence of f in 𝔻.

Examples: For A(x) = π^{2}/2, one has V_{0}(x) = cos (πx/2). For A(x) = 4(1−x^{2})^{−1}, one has V_{0}(x) = 1−x^{2}. For A(x) = 2(1−x^{2})^{−2}, one has V_{0}(x) = √1 − x^{2} .

Problem: What geometric properties can we say about those functions with prescribed growth for the Schwarzian derivatives?

Starlike and convex functions.

f ∈ M is called starlike if f is univalent analytic and the image f(𝔻) is starlike with respect to the origin, in other words,

Re \frac{zf′(z)}{f(z)} > 0, |z| < 1.

f ∈ M is called convex if f is univalent analytic and the image f(𝔻) is convex, in other words,

Re (1 + \frac{zf′′(z)}{f′(z)} ) > 0, |z| < 1.

For a constant α ∈ [0, 1), a function f ∈ M is called starlike of order α if

Re \frac{zf′(z)}{f(z)} > α, |z| < 1.

• Strohhäcker theorem: a convex function is starlike of order 1/2.

• Starlike functions of order 1/2 play an important role in the theory of convolution (Hadamard product).

• These properties are not preserved by postcomposition of Möbius maps unlike univalence.

Starlikeness Theorem.

Theorem 4 Let A be a weight function and c be a complex number. Suppose

2 ∫_{0}^{1} U_{0}′(x)U_{1}(x)dx + |c|U_{1}(1)^{2} ≤ 1.

If a function f ∈ M(c) satisfies |S_{f}(z)| ≤ A(|z|) in |z| < 1, then f is starlike of order 1/2.

As the special case when A is a positive constant and c = 0, we obtain Corollary 5 Let C_{0} = 2β_{0}^{2} ≈ 2.37036, where β_{0} is the unique positive root of the equation sinh (2β) = 4β. If a function f ∈ M(0) satisfies the inequality |S_{f}(z)| ≤ C_{0} in |z| < 1, then f is a starlike function of order 1/2. The constant C_{0} is sharp.

Remarks:

Gabriel (1955) proved that |S_{f}(z)| ≤ C′_{0} implies starlikeness of f ∈ M(0), where C′_{0} = 2β′_{0}^{2} ≈ 2.71707 and β′_{0} is the unique root of the equation 2β = tan β in 0 < β < π/2.

On the other hand, Chiang (1994) showed that C′_{0} cannot be replaced by a larger number than C′′_{0} = (ξ^{2} + η^{2})/2 ≈ 4.6351, where ξ and η are the smallest positive roots of the equations ξ tan ξ = −1 and η tanh η = 1.

By some experiments, it is likely that C′′_{0} is the best possible constant.

Convexity Theorem.

Theorem 6 Let A be a weight function and c be a complex number. Suppose that the functions V_{0} and V_{1} satisfy the inequalities

V_{0}(x) − |c|V_{1}(x) > 0, 0 ≤ x < 1,

and

V_{0}(1) + 2V_{0}′(1) ≥ |c|(V_{1}(1) + 2V_{1}′(1)).

If a function f ∈ M(c) satisfies |S_{f}(z)| ≤ A(|z|) in |z| < 1, then f is convex.

As a corollary, we obtain an improvement of a result of Chiang (1994).

Corollary 7 Let C_{1} = 2β_{1}^{2} ≈ 0.853526, where β_{1} is the unique root of the equation 2β tan β = 1 in 0 < β < π/2. If a function f ∈ M(0) satisfies the inequality |S_{f}(z)| ≤ C_{1} in |z| < 1, then f is a convex function.

Remarks:

The constant C_{1} above is not sharp. More precisely, C_{1} is the sharp constant for which |S_{f}(z)| ≤ C_{1} implies the inequality |zf′′(z)/f′(z)| < 1 in |z| < 1.

As Chiang (1994) showed, the constant C_{1} cannot be replaced by a larger number than C′_{1} = 2β′_{1}^{2} ≈ 1.19105, where β′_{1} is the unique positive root of the equation β tanh β = 1/2.

Growth theorems.

Our main theorems are based on some growth theorems for solutions to the ODE introduced earlier.

Lemma 8 Let A be a weight function and suppose that |φ(z)| ≤ A(|z|) in |z| < 1. The solutions y_{0} and y_{1} to the differential equation 2y′′ + φy = 0 in 𝔻 with the initial conditions y_{0}(0) = 1, y′_{0}(0) = 0, y_{1}(0) = 0, y′_{1}(0) = 1 then satisfy the inequalities

Ṽ_{0}(|z|, A) ≤|y_{0}(z)| ≤ U_{0}(|z|, A),
|y′_{0}(z)| ≤ U_{0}′(|z|, A),
Ṽ_{1}(|z|, A) ≤|y_{1}(z)| ≤ U_{1}(|z|, A),
|y′_{1}(z)| ≤ U_{1}′(|z|, A)

for z ∈ 𝔻, where Ṽ (x) = V (x) for 0 ≤ x < x_{0} and Ṽ (x) = 0 for x ≥ x_{0} and x_{0} is the smallest positive zero of V (x) (if there is no such zero, set x_{0} = 1).

Lemma 9 Under the same hypothesis as in the previous lemma, let y_{2} = y_{0} − cy_{1}, where c is a complex constant for which the function V_{2} = V_{0} − |c|V_{1} is positive on (0, 1). Then the inequality

|\frac{y′_{2}(z)}{y_{2}(z)} | ≤ −\frac{V_{2}′(|z|)}{V_{2}(|z|)}

holds for every z ∈ 𝔻.

Idea of proof:

For a fixed ζ ∈ ∂𝔻, we set w(t) = y′_{2}(tζ)/y_{2}(tζ) and v(t) = −V′_{2}(t)/V_{2}(t). Then, the function w satisfies the Riccati equation

w′ = −\frac{φ}{2} − w^{2}.

Hence, the function u(t) = |w(t)| satisfies the differential inequality

u′ ≤ |w′| ≤ \frac{A}{2} + u^{2}.

Similarly, the function v satisfies v′ = A/2+v^{2}. Use a comparison theorem!

A comparison theorem.

The following is a specialized comparison theorem for the present situation.

Lemma 10 (cf. Walter, ”ODE”, p. 96) Let A be a non-negative continuous function on [0, 1) and set P w = w′ − A/2 − w^{2}. If absolutely continuous real-valued functions u, v on [0, 1) satisfy the inequalities

(a) P u ≤ P v a.e. in [0, 1) and

(b) u(0) ≤ v(0), then u ≤ v holds in [0, 1).

Proof of Starlikeness Theorem. Let f = y_{1}/y_{2}, where y_{2} = y_{0} − cy_{1}. Then the quantity p(z) = zf′(z)/f(z) satisfies

\frac{1}{p(z)} = \frac{y_{1}(z)y_{2}(z)}{z}
= ∫_{0}^{1} (y_{1}y_{2})′(tz)dt
= 1 + 2 ∫_{0}^{1} y_{1}(tz)y′_{2}(tz)dt.

We now use the growth theorem to get

|\frac{1}{p(z)} − 1| ≤ 2 ∫_{0}^{1} U_{1}(t|z|)U_{2}′(t|z|)dt
≤ 2 ∫_{0}^{1} U_{1}(t)U_{2}′(t)dt

= 2 ∫_{0}^{1} U_{1}(t)U_{0}′(t)dt + |c|U_{1}(1)^{2}.

We now conclude that |1/p(z) − 1| < 1, which is equivalent to Re p(z) > 1/2.

Proof of Convexity Theorem.

Use the same notation as in the previous proof. Further we set V_{2} = V_{0} − |c|V_{1}. Then, since f′ = y_{2}^{−2},

1 + \frac{zf′′(z)}{f′(z)} = 1 − 2 \frac{y′_{2}(z)}{y_{2}(z)} .

By the second growth lemma,

|2 \frac{y′_{2}(z)}{y_{2}(z)} | ≤ −2 \frac{V_{2}′(|z|)}{V_{2}(|z|)} < −2 \frac{V_{2}′(1)}{V_{2}(1)}

The last term is certainly not greater than 1. Therefore, Re (1 + zf′′(z)/f′(z)) > 0.

Computations.

• The simplest case is when A is a positive constant. If we write A = 2β^{2}, where β is a positive number, then

U_{0}(x) = cosh (βx),
U_{1}(x) = sinh (βx)/β,
V_{0}(x) = cos (βx),
V_{1}(x) = sin(βx)/β.

• For A(x) = C(1 − x^{2})^{−2}, where the constant C is allowed to be negative for convenience. If we write C = 2(4α^{2} − 1), then

U_{0}(x) = √1 − x^{2} cosh [α log (\frac{1 + x}{1 − x} )] ,
U_{1}(x) = \frac{√1 − x^{2}}{2α} sinh [α log (\frac{1 + x}{1 − x} )] .

• For A(x) = C(1 − x^{2})^{−1} with positive constant C = (1 − α^{2})/2,

U_{0}(x) = F (−\frac{1+α}{4} , −\frac{1−α}{4}; \frac{1}{2}; x^{2})
U_{1}(x) = x F (\frac{1+α}{4} , \frac{1−α}{4}; \frac{3}{2}; x^{2}),

where F (a, b; c; x) stands for the hypergeometric function.

As a corollary, we get

Corollary 11 Let C_{2} = (1 + β_{2}^{2})/2 ≈ 1.52444, where β_{2} is the unique positive root of the equation

∫_{0}^{1} x^{2}F (\frac{3+iβ}{4} , \frac{3−iβ}{4} ; \frac{3}{2}; x^{2})F (\frac{1+iβ}{4} , \frac{1−iβ}{4} ; \frac{3}{2}; x^{2})dx

= \frac{2}{1 + β^{2}} .

If a function f ∈ M(0) satisfies the inequality |S_{f}(z)| ≤ C_{2}/(1 − |z|^{2}) in |z| < 1, then f is a starlike function of order 1/2. The constant C_{2} is sharp.

Final remark.

The Schwarzian radius of convexity (cf. Chiang 1994) must be zero unless we impose some restriction on the second coefficient c = f′′(0)/2.

Proposition 12 Let φ be analytic in the unit disk. Suppose that f_{φ,0} is univalent and that f_{φ,c} is convex for every c ∈ K(φ). Then, φ = 0.

This follows from the more geometric assertion.

Proposition 13 Let D be a proper subdomain of the complex plane ℂ. Suppose that L(D) is convex for each Möbius transformation L such that L^{−1}(∞) ∉ D. Then D is a disk or a halfplane.