Univalence Criteria and Analogs of the John Constant

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. <span id="page-3-3"></span>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 <span id="page-3-1"></span><span id="page-3-0"></span> $$\frac{2}{\pi}\Phi(L(f))\log\frac{L(f)}{l(f)} \le \frac{2\delta}{\pi}\Phi(e^{\delta/2}) \le 1.$$ 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 <span id="page-3-2"></span> $$\frac{Q_a'(z)}{Q_a(z)} = \frac{2a}{1+z^2}.$$ 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, <span id="page-4-0"></span> $$\sup_{z\in\mathbb{D}} (1-|z|^2) \left| \frac{zf''(z)}{f'(z)} \right| \le 2a\Phi(L(f)) = \frac{2\delta}{\pi} \Phi(L(f)).$$ 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 <span id="page-6-2"></span> $$\gamma_{j,k}^{(n)} = \frac{\delta_{j,k}}{j} - \sum_{m=1}^{n} m c_{m,j} \overline{c_{m,k}} \qquad (1 \le j, k \le n),$$ $\delta_{j,k}$ 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 $k \ge 1$ .