Coefficient Behavior of a Class of Meromorphic Functions

Theorem 1. Suppose k > 2 and $f \in \Lambda_k$ is given by (2). Then $$\omega = \lim_{r \to 1} (1 - r)^{k/2-1} M(r, f')$$ exists, is finite, and equals 0 unless $s_2$ is of the form $z/(1-ze^{-i\theta})^2$ . If $\omega > 0$ , Received September 9, 1974 and in revised form, February 6, 1975. This research was supported in part by NSF Grant GP-43774. there exists $\theta$ such that $$\omega = \lim_{r \to 1} (1 - r)^{k/2 - 1} |f'(re^{i\theta})|.$$ The proofs of the analogous results for $V_k$ and $K(\beta)$ depend on the linear invariance (in the sense of Pommerenke) of these two classes. Since $\Lambda_k$ is not a linear invariant family, Theorem 1 requires a different method of proof. An application of the major-minor arc technique of W. K. Hayman yields the following result.

Theorem 2. Suppose k > 2 and $f \in \Lambda_k$ is given by (1). Then $$\lim_{n\to\infty} \frac{|a_n|}{n^{(k/2)-3}} = \frac{\omega}{\Gamma((k/2)-1)}.$$ Define $F_k \in \Lambda_k$ by $$(4) \quad F_{k}'(z) = -\frac{1}{z^{2}} \frac{\left(1+z^{2}-2z\frac{k-2}{k+1}\right)^{(k+2)/4}}{\left(1-z\right)^{k/2-1}},$$ and set $$F_k(z) = z^{-1} + A_0 + \sum_{n=1}^{\infty} A_n z^n.$$

Theorem 3. Suppose k > 2, $f \in \Lambda_k$ is given by (1), and $F_k$ is as above. Then $$\lim_{n\to\infty} |a_n|/|A_n|$$ exists, is at most 1, and equals 1 if and only if $f(z) = e^{i\theta} F_k(e^{i\theta}z)$ for some $\theta$ . We note that for fixed $f \in \Lambda_k$ , there exists n(f), depending only on f, such that $|a_n| \leq |A_n|$ for $n \geq n(f)$ . This result is clearly false for k = 2, since then $F_2(z) = z^{-1} + A_0 + z$ (and $A_n = 0$ for $n \geq 2$ ). It is also interesting to note that although $F_k$ is not the solution (for all n) of the problem of determining max $\{|a_n|: f \in \Lambda_k\}$ , Theorem 3 shows that the coefficients of $F_k$ do in fact eventually dominate the coefficients of any fixed $f \in \Lambda_k$ . In other words, $F_k$ is the unique solution to the asymptotic coefficient problem. 2. Proof of Theorem 1. If $f \in \Lambda_k$ is given by (2) with $s_2$ not of the form $z/(1-ze^{i\theta})^2$ , then [6] there exists d < 2 such that $s_2(re^{i\theta}) = O(1)(1-r)^{-d}$ . It follows immediately, since k > 2, that $\omega = 0$ . We now assume $s_2(z) = z/(1-ze^{i\theta})^2$ for some $\theta$ . For notational ease we assume $\theta = 0$ . Set $$\omega_1 = \limsup_{r \to 1} (1 - r)^{k/2-1} M(r, f')$$ $$\omega_2 = \liminf_{r \to 1} (1 - r)^{k/2-1} M(r, f'),$$ and note that $0 \le \omega_2 \le \omega_1 < \infty$ . The remainder of the proof will be divided into a sequence of lemmas. We first state a definition: a sequence $\{z_n\}_1^{\infty}$ , with $|z_n| < |$ , $\lim_{n\to\infty} z_n = 1$ , is said to approach 1 strictly tangentially if, given any Stolz angle S with vertex 1, there exists N(S) such that $z_n \notin S$ for $n \ge N(S)$ .

Lemma 1. If $\{z_n\}_{1}^{\infty}$ approaches 1 strictly tangentially, then $$\lim_{n\to\infty} (1-|z_n|)^{k/2-1}|f'(z_n)| = 0.$$

Lemma 2. If $\omega_1 > 0$ , then $\omega_1 = \omega_2$ .

Lemma 3. Suppose k > 2 and $$\omega = \lim_{r \to 1} (1 - r)^{k/2 - 1} |f'(r)| > 0.$$ Then given $\delta > 0$ , there exists $C(\delta) > 0$ and $r(\delta) < 1$ such that $$\int_{E} |g'(re^{i\theta})| d\theta < \frac{\delta}{(1-r)^{k/2-1}}$$ for $$r(\delta) < r < 1$$ , where $E = \{\theta : C(\delta)(1-r) \leq |\theta| \leq \pi\}$ .

Lemma 4. Suppose $$k>2$$ , $f\in \Lambda_k$ , $\omega>0$ . For $n\geq 2$ , set $r_n=1-1/n$ , $\omega_n=(k/2-1)s_1(r_n)^{-(k+2)/4}$ , $g_n{}'(z)=\omega_n(1-z)^{-k/2}$ . Put $$I_n = \{\theta : 0 \le |\theta| \le c(1 - r_n)\}.$$ Then with $z = r_n e^{i\theta}$ , $g'(z)/g_n'(z) \to 1$ uniformly for $\theta \in I_n$ , as $n \to \infty$ .

Lemma 5. Suppose $f \in \Lambda_k(k > 2)$ is given by (2). Assume that $$\omega = \lim_{r \to 1} (1 - r)^{k/2 - 1} |f'(r)| > 0,$$ and suppose that $s_1$ in (2) is given by $$s_1(z) = z \exp \left\{ - \int_{-\pi}^{\pi} \log (1 - ze^{-it}) d\alpha(t) \right\}$$ where $\alpha$ is increasing on $[-\pi, \pi]$ with $$\int_{-\pi}^{\pi} d\alpha(t) = 2, \int_{-\pi}^{\pi} e^{-it} d\alpha(t) = 2(k-2)/(k+2).$$ Then $$\omega^{\frac{-4}{k+2}} = \exp\left\{-\int_{-\pi}^{\pi} \log|1 - e^{-it}| d\alpha(t)\right\}.$$

Lemma 6. With k > 2, $N \ge 2$ , $X = (x_1, ..., x_N)$ , and $A = (a_1, ..., a_N)$ , set $$h(X, A) = \prod_{j=1}^{N} (1 - x_j)^{n_j/2}.$$ Let $\epsilon > 0$ be given. Suppose that $X^ = (x_1^, \dots, x_N^)$ and $A^ = (a_1^, \dots, a_N^)$ are such that $h(X^, A^) = \max h(X, A)$ , where the maximum is taken subject to the conditions $-1 \le x_j \le 1$ and $a_j \ge 0$ $(1 \le j \le N)$ , $$|x_1 - (k-2)/(k+2)| \ge \epsilon$$ , $a_1 \ge \epsilon$ and $$\sum_{j=1}^{N} a_j = 2, \quad \sum_{j=1}^{N} a_j x_j = 2(k-2)/(k+2).$$ Then there exists $\epsilon_1 > 0$ depending only on $\epsilon$ and k (in particular, $\epsilon_1$ is independent of N) such that $h(X^, A^) \leq 4/(k+2) - \epsilon_1$ . The proof of Lemma 6 consists of a tedious but straightforward application of the Lagrange multiplier theorem. Since the arguments required are similar in nature to the argument given in [5, Lemma 3.2], we omit the details. The fact that $\epsilon_1$ is independent of N follows from the fact that, in the course of applying the Lagrange multiplier theorem, one shows that at the maximum point $(X^, A^)$ , the variables $x_j^*$ , $1 \le j \le N$ , can assume at most four distinct values. We now complete the proof of Theorem 3. First note that $F_k$ has a representation of the form (2) with $S_2(z) = z/(1-z)^2$ and $$S_1(z) = z \exp \left\{ - \int_{-\pi}^{\pi} \log (1 - ze^{-it}) d\alpha(t) \right\},$$ where $\alpha$ is a step function having jumps of magnitude 1 at each of the points $t = \pm \delta$ ; here $\delta = \arccos(k-2)/(k+2)$ . Suppose that $f_1 \in \Lambda_k$ , $f_1 \neq F_k$ . We shall show $\omega(f_1) < \omega^*$ . Clearly we may assume $$0 < \omega(f_1) = \lim_{r \to 1} (1 - r)^{k/2 - 1} |f_1'(r)|;$$ we suppose that $s_1$ and $s_2$ correspond to $f_1$ via (2). Since $\omega(f_1) > 0$ , we have $s_2(z) = z/(1-z)^2$ , and since $f_1 \neq F_k$ , we have $s_1 \neq S_1$ . Hence $\alpha_1$ , the starlike integrator of $s_1$ , is not the same as $\alpha$ , the starlike integrator for $S_1$ . If $\alpha_1$ were to concentrate all its mass at $t = \pm \delta$ , the condition $$\int_{-\pi}^{\pi} e^{-it} d\alpha_1(t) = 2(k-2)/(k+2)$$ would imply $\alpha_1 = \alpha$ . Therefore we can choose $\eta > 0$ (depending only on $\alpha_1$ and hence only on $f_1$ ) such that $$(7) \quad \int_{E} d\alpha_{1}(t) \geq \eta > 0,$$ where $E = \{t \in [-\pi, \pi] : |t - \delta| \ge \eta \text{ or } |t + \delta| \ge \eta\}.$ In view of (7) and Lemma 5, we choose a sequence $\{\mu_N\}_{2}^{\infty}$ of step functions with the following properties: $\mu_N$ has at most N discontinuities, $$\int_{-\pi}^{\pi} d\mu_N = 2, \int_{-\pi}^{\pi} e^{-it} d\mu_N = 2(k-2)/(k+2),$$ $$\int_{E} d\mu_N \ge \eta/2 > 0, \text{ and } \lim_{N \to \infty} \omega_N = \omega(f_1),$$ where $$\omega_N^{-4/(k+2)} = \exp \left\{ - \int_{-\pi}^{\pi} \log |1 - e^{-it}| d\mu_N(t) \right\}.$$ Denote by $\{a_j\}_{1}^{N}$ and $\{t_j\}_{1}^{N}$ respectively the magnitudes and positions of the jumps of $\mu_N$ . It follows that $$\frac{1}{2} \omega_N^{4/(k+2)} = \prod_{j=1}^N (1 - x_j)^{a_j/2}$$ where $$x_j = \cos t_j$$ , $\sum_{i=1}^{N} a_j = 2$ , and $\sum_{j=1}^{N} a_j x_j = 2(k-2)/(k+2)$ . It follows from (7) that there exists $\epsilon > 0$ depending only on $f_1$ such that $a_1 \ge \epsilon$ , $|x_1 - (k-2)/(k+2)| \ge \epsilon$ (relabel variables if necessary). By Lemma 6, there exists $\epsilon_1 > 0$ depending only on $f_1$ such that $\omega_N \le (8/(k+2) - 2 \epsilon_1)^{(k+2)/4}$ . Therefore $$\omega(f_1) = \lim_{N \to \infty} \omega_N < (8/(k+2))^{(k+2)/4} = \omega^*.$$ This completes the proof of Theorem 3.