On a maximal outer area problem for a class of meromorphic univalent fuctions

Theorem 1. The solution to the extremal problem $$\max_{f \in S_{\mathcal{D}}(a)} \overline{A}(f)$$ is given by (3) provided that the residue a belongs to the disk A of values determined by the inequality (4) $$\frac{|a|p^2}{|a+p^2|} \ge (1+p)^2$$ 2 Proof. Case I: a = -p . In this case, $$F(z;p,-p^2) = \frac{pz}{p-z}.$$ p - z Since cu = 0 for k > 1 , the formula for the outer area (1) gives 4 2 2 2 A(F) = Tip /(I - p ) • Moreover, since any other function in S (-p ) must have at least one non-zero coefficient a, , any other function in this class must also have a strictly smaller outer area. Finally, since F is linear-fractional, it must map U onto the exterior of a certain 2 2 2 disk. This disk is centered at -2p/(l - p ) and has radius p /(I - p ). 2 Case II: a ^ -p . Since (3) is meromorphic and satisfies the appropriate normalisations, it remains to show that it is univalent on U . Since F can be written in the form $$F(z;p,a) = \frac{a}{z-p} + \left(\frac{2a}{p} + p\right) + \left(1 + \frac{a}{p^2}\right) (z-p)$$ It is sufficient to show that, under the assumed condition, the function G(w) = A/w + Bw is univalent on the set U = {w:w = C, - p, X, e V} , where A=a and $B=1+a/p^2$ . We note first that $\left|w_1w_2\right|<(1+p)^2$ for any $w_1,w_2\in U_p$ . Suppose now that G is not univalent; that is, that there exists $w_1,w_2\in U_p$ , $w_1\neq w_2$ , such that $$G(w_1) = \frac{A}{w_1} + Bw_1 = \frac{A}{w_2} + Bw_2 = G(w_2)$$ . Then a little arithmetic shows that $$(1+p)^2 > |w_1w_2| = \left|\frac{A}{B}\right| = \frac{|a|p^2}{|a+p^2|},$$ contrary to assumption. Hence, (4) implies that F(z;p,a) is univalent on U and thereby belongs to $S_p(a)$ . Since (4) may be rewritten in the form $$\left|\frac{1}{a} + \frac{1}{p^2}\right| \le \frac{1}{\left(1+p\right)^2}$$ it is clear that the reciprocals $\{1/a\}$ described by this condition belong to a disk; consequently, the set of residues $\{a\}$ do as well. In explicit form, we may write $$\Delta_p = \{s: |s + (1 + p)^2 \delta_p | \le p^2 \delta_p \},$$ where $$\delta_p = \frac{p^2 (1+p)^2}{(1+2p)(1+2p+2p^2)} .$$ This concludes the proof of the theorem. For the sake of completeness, we now briefly discuss the mapping properties of F(z;p,a). Since F can be rewritten in the form $$F(z;p,a) = \frac{2a}{p} + p$$ $$+ 2\left[\frac{a}{p^2} + 1\right] \left(\frac{z-p}{2} + \frac{1}{2} \left(\frac{ap^2}{a+p^2}\right) \left(\frac{1}{z-p}\right)\right],$$ it is easily seen to be the successive composition of three maps: (1) $$\zeta = C(z) = z - p$$ , (2) $\xi = D(\zeta) = \zeta/2 + \beta/\zeta$ , $\beta = \frac{1}{2} \left[ \frac{\alpha p^2}{\alpha + p^2} \right]$ , and (3) $$w = E(\xi) = (2a/p + p) + 2(1 + a/p^2)\xi$$ . Maps C and E are linear. Map D is discussed in [1, p. 64-67]; it univalently takes either the interior or the exterior of the circle $\{|\zeta| = |\sqrt{\beta}|\}$ onto the plane, cut from $-\sqrt{2\beta}$ to $+\sqrt{2\beta}$ .  Since $$\left| \sqrt{2\beta} \right| = \left| \frac{ap^2}{a+p^2} \right|^{1/2} \ge (1+p)$$ , the set $C(U)$ is contained entirely in the interior of the circle $\{|\zeta|=|\sqrt{2\beta}|\}$ . The set D(C(U)) is the plane, except for a compact set of positive area, which contains the interval $[-\sqrt{2\beta}]$ , $+\sqrt{2\beta}$ . Map E stretches and translates the set D(C(U)). If strict inequality holds in (4), then the set F(U;p,a) will have an analytic boundary. If equality holds in (4), then F(U;p,a) will have an analytic boundary except for one point; the tear-shaped exterior in this case is sometimes called a <u>Joukowski aerofoil</u> and is obtained whenever $\alpha \in \partial \Delta$ .

Theorem 2. Let g(z) be an extremal function for the problem $$\max_{f \in S_p(a)} \overline{A}(f)$$ and assume that $ext\{g(U)\} \neq \phi$ . Then $$\langle g,Q \circ g \rangle = \lim_{r \to 1} \left[ \frac{-1}{2\pi i} \int_{g(|z|=r)} \overline{Q(w)} dw \right] = 0$$ for every Q satisfying the conditions of the Variational Lemma. Proof. If g(z) has an expansion of the form $$g(z) = \frac{\alpha}{z - p} + \sum_{k=0}^{\infty} \alpha_k z^k$$ and Q(g(z)) has an expansion of the form $$Q(g(z)) = \sum_{k=0}^{\infty} q_k z^k$$ then $g_{\pm}(z)$ will have an expansion of the form $$g_t(z) = \frac{\alpha}{z-p} + \alpha_0 + \alpha_1 z + \sum_{k=0}^{\infty} (\alpha_k + t q_k) z^k + O(t^2)$$ . Since g(z) is an extremal function for the maximal problem, we must have $\overline{A}(g) \geq \overline{A}(g_t)$ for every complex t with $|t| < 1/K_h$ . It follows from (1) that $$\sum_{k=0}^{\infty} k |\alpha_k + tq_k|^2 + O(|t|^2) \ge \sum_{k=0}^{\infty} k |\alpha_k|^2$$ Equivalently, we have $$\sum_{k=0}^{\infty} 2k \operatorname{Re} \left( \frac{\overline{t}}{|t|} \alpha_{k} \overline{q}_{k} \right) + O(|t|^{2}) \geq 0$$ for all t , $\left|t\right|$ < $1/K_h$ ; but this last inequality can hold only if $$\sum_{k=0}^{\infty} k \alpha_{k} \overline{q}_{k} = 0.$$ The proof is now complete, since this last equality is equivalent to the result in the statement of the theorem, as we have shown earlier in this section. As a specific example, we now show that the extremal function g(z) and the functions $f_c^N(z)$ , $N\geq 2$ , defined within the Variational Lemma above, are orthogonal in the space $L_p$ with respect to the Hermitian product <, >.

Corollary 2.1. Suppose that $$g(z) = \frac{a}{z - p} + \sum_{k=0}^{\infty} \alpha_k z_k$$ is an extremal function for the problem $$\max_{f \in S_{\mathcal{D}}(a)} \overline{A}(f)$$ and that $ext\{g(U)\} \neq \phi$ . If we set $$f_c(z) = \frac{cg(z)}{(c + g(z))} = z + \sum_{k=0}^{\infty} a_k(c) z^k$$ and $$f_c^N(z) = z^N + \sum_{N=1}^{\infty} a_k^{(N)}(c) z^k$$ , then $$\langle g, f_c^N \rangle = \sum_{N}^{\infty} k \alpha_k \overline{\alpha_k^{(N)}(c)} = 0$$ for every $N \ge 2$ and every c such that $-c \in ext\{g(U)\}$ . Proof. Choose $$Q(w) = c^2 \left( \frac{cw}{c + w} \right)^N . \qquad (N \ge 2)$$ Introducing the change of variable w = ct/(c - t) , we have $$\frac{1}{2\pi i} \int_{g(|z|=r)} \overline{Q(w)} dw = \frac{\overline{c^2}}{2\pi i} \int_{g(|z|=r)} \overline{(cw)}^N dw$$ $$= \frac{|c|^4}{2\pi i} \int_{f_c(|z|=r)} \overline{t^N} \frac{dt}{(c-t)^2}$$ $$= \frac{\overline{c^2}}{2\pi i} \int_{|z|=r} \overline{f_c^N(z)} g'(z) dz$$ $$= \overline{c^2} \int_{N} k\alpha_k \overline{a_k^{(N)}(c)} r^{2k} .$$ Now observe that since $f_c(z)$ is bounded, $f_c^N(z) \in L_p$ . The corollary now follows from Theorem 2 by letting r tend to one. Another Corollary to Theorem 2 concerning the coefficients of extremal functions is of interest.

Corollary 2.2. Suppose that $$g(z) = \frac{\alpha}{z - p} + \sum_{k=0}^{\infty} \alpha_k z^k$$ is an extremal function for the problem $$\begin{array}{cc} max & \overline{A}(f) \\ f \in S_p(a) \end{array}$$ and that $ext\{g(U)\} \neq \emptyset$ . Then either $\alpha_k = 0$ for all $k \geq 2$ or $\alpha_k \neq 0$ for infinitely many $k \geq 2$ . Proof. Suppose that there exists a K such that $\alpha_K \neq 0$ and a, = 0 for all k > K . Applying the equality in Corollary 2.1, with N = K , we get $$0 = \sum_{k=1}^{\infty} k \alpha_{k} \overline{a_{k}^{(K)}(c)} = K \alpha_{k} \overline{a_{k}^{(K)}(c)}$$ (K) But aK (c) = 1 . Thus we have obtained a contradiction.