Univalence Criteria and Analogs of the John Constant

Theorem 1.3. $$\frac{\pi}{3} = 1.04719 \dots < \delta_0 < \frac{5\pi}{7} = 2.24399 \dots$$

Theorem 1.4. $$\frac{7\pi}{25} = 0.87964 \dots < \delta_1 < \frac{5\pi}{7} = 2.24399 \dots$$ We remark that the above results are not optimal. Indeed, more elaborative numerical computations would yield slightly better bounds as will be suggested at the end of Section 2. In order to obtain a lower bound, we need a univalence criterion due to Becker [1] with numerical computations as we will explain in Section 2. On the other hand, to give an upper bound, we should construct a non-univalent function satisfying the condition in Problems 1.1 or 1.2. The function $F_a \in \mathcal{A}$ determined by the differential equation (1.1) $$\frac{zF_a'(z)}{F_a(z)} = \left(\frac{1-iz}{1+iz}\right)^{ai}$$ is a candidate for an extremal one, where a is a positive constant and i is the imaginary unit $\sqrt{-1}$ . As will be seen later, $L(F_a)/l(F_a) = e^{\pi a}$ . We will give a detailed account on this function and provide the upper bound in the above theorems in Section 3. The proof is involved with matrices of large order. Therefore, we made use of Mathematica 8.0 to carry out symbolic computations. The most interesting problem is to determine the values of $\delta_0$ and $\delta_1$ . However, this seems to the authors very hard. We end the section with a couple of open questions, which may be easier to solve. Let $a$ be the supremum of the numbers a such that $F_a$ is univalent on $\mathbb{D}$ . Likewise let $a$ be the infimum of the numbers of a such that $F_a$ is not univalent on $\mathbb{D}$ . Obviously, $\delta_0 \leq \pi a_ \leq \pi a$ . In the proof of the above theorems, we indeed show that $a_* < 5/7$ . - (1) Is it true that $a_ = a$ ? - (2) Is it true that $\delta_0 = \pi a_*$ ? - (3) Is it true that $\delta_0 = \delta_1$ ? Acknowledgement. The authors would like to thank the referee for careful reading of the manuscript and for suggestions which helped us to improve the exposition.

Lemma 2.1. Let f be a non-constant analytic function on $\mathbb{D}$ . If $$(1 - |z|^2) \left| \frac{zf''(z)}{f'(z)} \right| \le 1, \quad z \in \mathbb{D},$$ then f is univalent on $\mathbb{D}$ . Sometimes, it is more convenient to consider the pre-Schwarzian norm $$||f|| = \sup_{z \in \mathbb{D}} (1 - |z|^2) \left| \frac{f''(z)}{f'(z)} \right|$$ because it has several nice properties (see [8] for example). By Becker's theorem above, we see that the condition $||f|| \le 1$ implies univalence of f on $\mathbb{D}$ . We used this norm to deduce the estimate $\pi/6 \le \delta_0$ . In this note, however, we do use the original form (Lemma 2.1) of Becker's theorem to improve the estimate. For a non-negative number c, we consider the quantity $$\Phi(c) = \sup_{0 < r < 1} \Big\{ r + c(1 - r^2) \operatorname{arctanh} r \Big\} = c \sup_{0 < r < 1} \Big\{ c^{-1}r + (1 - r^2) \operatorname{arctanh} r \Big\}.$$ It is easy to see that $\Phi(c)$ is non-decreasing in c and that $c^{-1}\Phi(c)$ is non-increasing in c. In terms of this function, we will prove the following technical lemma which yields lower bounds for $\delta_0$ and $\delta_1$ as corollaries.

Lemma 2.2. Let $f \in A$ . If $L(f)/l(f) < +\infty$ and if the inequality $$\frac{2}{\pi}\Phi(L(f))\log\frac{L(f)}{l(f)} \le 1$$ holds, then f is univalent on $\mathbb{D}$ . The lemma immediately yields the following results.

Corollary 2.3. Let $\delta > 0$ be given. If (2.2) $$\frac{2\delta}{\pi}\Phi(e^{\delta/2}) \le 1,$$ then $\delta < \delta_0$ . If $$(2.3) \frac{2\delta}{\pi} \Phi(e^{\delta}) \le 1,$$ then $\delta \leq \delta_1$ . To show the corollary, we first assume (2.2) and consider a function $f \in \mathcal{A}$ satisfying $e^{-\delta/2} < |zf'(z)/f(z)| < e^{\delta/2}$ . Then $L(f) \le e^{\delta/2}$ and $\log L(f)/l(f) \le \delta$ so that $$\frac{2}{\pi}\Phi(L(f))\log\frac{L(f)}{l(f)} \le \frac{2\delta}{\pi}\Phi(e^{\delta/2}) \le 1.$$ We now apply Lemma 2.2 to conclude univalence of f. Secondly, we assume (2.3) and consider a function $f \in \mathcal{A}$ satisfying $L(f) \leq e^{\delta} l(f)$ . Then $L(f) \leq e^{\delta}$ and the conclusion follows similarly. Let us prepare for the proof of Lemma 2.2. We note that the function $\arctan z = \frac{1}{2i} \log \frac{1+iz}{1-iz}$ maps the unit disk $\mathbb D$ conformally onto the vertical parallel strip $|\operatorname{Re} w| < \pi/4$ . Therefore, for a constant a>0, the function (2.4) $$Q_a(z) = \exp(2a \arctan z) = \left(\frac{1 - iz}{1 + iz}\right)^{ai}$$ is the universal covering projection of $\mathbb{D}$ onto the annulus $e^{-\pi a/2} < |w| < e^{\pi a/2}$ . We note that the function $Q_a$ satisfies $Q_a(0) = 1$ and $$\frac{Q_a'(z)}{Q_a(z)} = \frac{2a}{1+z^2}.$$ Proof of Lemma 2.2. Let p(z)=zf'(z)/f(z) for a function $f\in\mathcal{A}$ . If p is a constant, then f is clearly univalent. We can thus assume that p is not a constant so that l(f)<1< L(f). Let $\delta=\log L(f)/l(f)<+\infty$ and $m=\sqrt{L(f)l(f)}$ . We consider the universal covering map $Q=mQ_a$ of $\mathbb D$ onto the annulus $W=\{w:l(f)<|w|< L(f)\}=\{w:me^{-\delta/2}<|w|< me^{\delta/2}\}$ , where $Q_a$ is given in (2.4) with $a=\delta/\pi$ . Note that $p(\mathbb D)\subset W$ by assumption. Since the real interval (-1,1) is mapped onto (l(f),L(f)) by Q, we can choose an $\alpha\in(-1,1)$ so that $Q(\alpha)=1$ . Then, $P=Q\circ T$ is a universal covering map of $\mathbb D$ onto W with P(0)=1, where $T(z)=(z+\alpha)/(1+\alpha z)$ . Since $P:\mathbb D\to W$ is a covering map, we can take a lift $\omega$ of p with respect to P so that $\omega(0)=0$ and $p=P\circ\omega$ . We write $w=\omega(z)$ . Note here that the Schwarz lemma implies $|w|\leq |z|$ . We now have $$\frac{zf''(z)}{f'(z)} = \frac{zp'(z)}{p(z)} + p(z) - 1 = \frac{z\omega'(z)P'(w)}{P(w)} + P(w) - 1.$$ Set $\tau = T(w) \in \mathbb{D}$ . Since T is a hyperbolic isometry of $\mathbb{D}$ , one has the relation $(1 - |w|^2)|T'(w)| = 1 - |\tau|^2$ . Therefore, by using (2.1), we have $$(1 - |z|^2) \left| \frac{\omega'(z)P'(w)}{P(w)} \right| \le (1 - |w|^2) \left| \frac{Q'(\tau)T'(w)}{Q(\tau)} \right|$$ $$= (1 - |\tau|^2) \left| \frac{Q'(\tau)}{Q(\tau)} \right|$$ $$= \frac{2a(1 - |\tau|^2)}{|1 + \tau^2|}$$ $$\le 2a.$$ Let $\gamma$ be the image of the line segment (0, w) under the Möbius mapping T. Then, $$P(w) - 1 = \int_0^w P'(t)dt = \int_0^w Q'(T(t))T'(t)dt = \int_\gamma Q'(u)du = \int_\gamma \frac{2aQ(u)}{1 + u^2}du.$$ Since $|Q(u)| \leq L(f)$ , we obtain $$|P(w) - 1| \le 2aL(f) \int_{\gamma} \frac{|du|}{1 - |u|^2} = 2aL(f) \int_{0}^{w} \frac{|du|}{1 - |u|^2}$$ $$= 2aL(f)d(0, w) \le 2aL(f)\operatorname{arctanh}|z|.$$ Therefore, $$(2.5) (1-|z|^2) \left| \frac{zf''(z)}{f'(z)} \right| \le 2a|z| + 2aL(f)(1-|z|^2) \operatorname{arctanh}|z|.$$ Hence, $$\sup_{z\in\mathbb{D}} (1-|z|^2) \left| \frac{zf''(z)}{f'(z)} \right| \le 2a\Phi(L(f)) = \frac{2\delta}{\pi} \Phi(L(f)).$$ Lemma 2.1 now implies the required assertion. The above method also gives a norm estimate of the pre-Schwarzian derivative. Though we do not use it in this note, we record it for the possible future reference.

Proposition 2.4. Suppose that $L(f)/l(f) < +\infty$ for a function $f \in A$ . Then the pre-Schwarzian norm of f is estimated as $$||f|| \le \frac{2}{\pi} (1 + L(f)) \log \frac{L(f)}{l(f)}.$$

Lemma 2.5. Let c > 1. If a number $x_1 \in (0,1)$ satisfies the inequality $x_1$ arctanh $x_1 < \frac{1+c}{2c}$ , then $\Phi(c) < H(x_1,c)$ . Proof. Let $g(x) = x + c(1 - x^2)\operatorname{arctanh} x$ . Then $g'(x) = 1 + c - 2cx \operatorname{arctanh} x$ . Since $x \operatorname{arctanh} x$ (strictly) increases from 0 to $+\infty$ when x moves from 0 to 1, there exists a unique zero $x_0 \in (0,1)$ of g'(x) so that g'(x) > 0 in $0 < x < x_0$ and g'(x) < 0 in $x_0 < x < 1$ . Note here that the assumption implies that $0 < x_1 < x_0$ . We see now that g(x) takes its maximum at $x = x_0$ and therefore, we have $$\Phi(c) = g(x_0) = \frac{1-c}{2}x_0 + \frac{1+c}{2}x_0^{-1} = H(x_0, c).$$ Since $H_x(x,c) = (1-c)/2 - (1+c)/(2x^2) < 0$ , the function H(x,c) is decreasing in x > 0 for a fixed c > 1. Hence, $x_1 < x_0$ implies $H(x_0,c) < H(x_1,c)$ , which proves the assertion. Proof of Theorem 1.3. Let $\delta = \pi/3$ and set $c = e^{\delta/2} = e^{\pi/6}$ . If we take $x_1 = 17/22$ , then $$\frac{1+c}{2c} - x_1 \operatorname{arctanh} x_1 = \frac{1+e^{-\pi/6}}{2} - \frac{17}{44} \log \frac{39}{5} = 0.00255 \dots > 0.$$ Therefore, Lemma 2.5 yields $$\frac{2\delta}{\pi}\Phi(e^{\delta/2}) = \frac{2}{3}\Phi(c) < \frac{2}{3}H(x_1, c) = \frac{773 + 195e^{\pi/6}}{1122} = 0.982\dots < 1.$$ We now apply Corollary 2.3 to obtain $\pi/3 < \delta_0$ . Proof of Theorem 1.4. We will proceed in the same line as above. Let $\delta = 7\pi/25$ and set $c = e^{\delta}$ . We take $x_1 = 20/27$ and have $$\frac{1+c}{2c} - x_1 \operatorname{arctanh} x_1 = \frac{1+e^{-7\pi/25}}{2} - \frac{10}{27} \log \frac{47}{7} = 0.00219... > 0.$$ Lemma 2.5 now implies $$\frac{2\delta}{\pi}\Phi(e^{\delta}) = \frac{14}{25}\Phi(c) < \frac{14}{25}H(x_1, c) = \frac{7903 + 2303e^{7\pi/25}}{13500} = 0.9965 \dots < 1.$$ We again apply Corollary 2.3 to obtain $7\pi/25 < \delta_1$ . Remark. We can slightly improve Theorems 1.3 and 1.4 by changing the choice of $\delta$ and $x_1$ in the above proofs. For instance, concerning Theorem 1.3, we can take $(\delta, x_1) = (\frac{22\pi}{65}, \frac{17}{22}), (\frac{87\pi}{257}, \frac{2765}{3578})$ , to have lower bounds $22\pi/65 = 1.06330...$ and $87\pi/257 = 1.06349...$ , respectively, for $\delta_0$ . Numerical computations with Mathematica 8 suggest that the solution to the equation $\frac{2\delta}{\pi}\Phi(e^{\delta/2})=1$ is about $\delta=1.0635213$ . Therefore, it seems that we would obtain at most this value as a lower bound for $\delta_0$ by the above method. Similarly, concerning Theorem 1.4, we can take $(\delta, x_1) = (\frac{25\pi}{89}, \frac{622}{839}), (\frac{127\pi}{452}, \frac{321}{433})$ , to have lower bounds $25\pi/89 = 0.882469...$ and $127\pi/452 = 0.882704...$ , respectively, for $\delta_1$ . We see that the numerical solution to the equation $\frac{2\delta}{\pi}\Phi(e^{\delta}) = 1$ is about $\delta = 0.8827139$ . Therefore, the above method seems to give only a lower bound of $\delta_1$ not better than this value.

Lemma 3.1. Let $f \in A$ and $\{c_{j,k}\}$ be its Grunsky coefficients. If f is univalent on |z| < 1 then $$\sum_{m=1}^{\infty} m \left| \sum_{k=1}^{n} c_{m,k} t_k \right|^2 \le \sum_{m=1}^{n} \frac{|t_m|^2}{m}$$ holds for arbitrary $n \geq 1$ and $t_1, \ldots, t_n \in \mathbb{C}$ . We remark that the Grunsky coefficients are usually defined for the function $g(\zeta) = 1/f(1/\zeta)$ . This change affects only the coefficients $c_{j,0} = c_{0,j}$ , which do not involve the Grunsky inequalities. See [5] for more information. From Lemma 3.1, the inequality (3.2) $$\sum_{m=1}^{n} m \left| \sum_{k=1}^{n} c_{m,k} t_k \right|^2 \le \sum_{m=1}^{n} \frac{|t_m|^2}{m}$$ follows for every n and $t_1, \ldots, t_n \in \mathbb{C}$ . This implies that the Hermitian matrix $G_f(n) = (\gamma_{j,k}^{(n)})$ of order n is positive semi-definite; in other words, $\mathbf{t}G_f(n)\mathbf{t}^* \geq 0$ for any $\mathbf{t} = (t_1, \ldots, t_n) \in \mathbb{C}^n$ , where $$\gamma_{j,k}^{(n)} = \frac{\delta_{j,k}}{j} - \sum_{m=1}^{n} m c_{m,j} \overline{c_{m,k}} \qquad (1 \le j, k \le n),$$ $\delta_{j,k}$ means Kronecker's delta and $\mathbf{t}^*$ is the conjugate transpose of $\mathbf{t}$ as a matrix. Letting $t_k = \delta_{j,k}$ in (3.2), we have $\sum_{m=1}^n m |c_{m,j}|^2 \le 1/j$ for $j \le n$ , which implies $|c_{m,j}| \le 1/\sqrt{mj} \le 1$ for $m, j \ge 1$ . This guarantees that the series expansion in (3.1) is convergent in |z| < 1, |w| < 1, and therefore, that f is univalent on $\mathbb{D}$ . We shall call $G_f(n)$ the Grunsky matrix of order n. We have observed the following assertion.

Corollary 3.2. A function $f \in \mathcal{A}$ is univalent on $\mathbb{D}$ if and only if its Grunsky matrix $G_f(n)$ of order n is positive semi-definite for every $n \geq 1$ . In order to compute the Grunsky coefficients of $F_a(z)$ , it is convenient to have recursion formulae for relavant coefficients. The following elementary lemma gives a recursion formula for exponentiation.

Lemma 3.3. Let $g(z) = b_1 z + b_2 z^2 + \cdots$ be a given function analytic around z = 0 and let $h(z) = e^{g(z)} = c_0 + c_1 z + c_2 z^2 + \cdots$ . Then $c_n$ can be computed recursively by $c_0 = 1$ and $$c_n = \frac{1}{n} \sum_{k=0}^{n-1} (n-k)b_{n-k}c_k \quad (n \ge 1).$$

Lemma 3.4. The Grunsky coefficients $c_{j,k}$ of a function $f(z) = z + a_2 z^2 + \cdots$ in A satisfy the recursion formulae (3.3) $$c_{j,k} = \sum_{l=1}^{k-1} \frac{l}{k} a_{k-l} c_{j+1,l} - \sum_{m=1}^{j} a_{m+1} c_{j-m,k} - \frac{a_{j+k+1}}{k}$$ for $j \ge 0$ and $k \ge 1$ .