Area and length maxima for univalent functions

Theorem 1. We have $$\max_{f \in S} \triangle(r, 1/F_f) = 2\pi r^2 (r^2 + 2) \quad \text{for} \quad 0 < r \leqslant 1.$$ For each r, 0 < r ^ 1, the maximum is attained only by Kg 's. PROOF: Given / E S, we can apply the area theorem [3, p.29] to $$f(1/z)^{-1} = z - a_2 + \sum_{n=1}^{\infty} b_n z^{-n} \quad (|z| > 1)$$ to obtain $$(2) \sum_{n=1}^{\infty} n |b_n|^2 \leqslant 1$$ Since $$1/F_f(z) = 1 - a_2 z + \sum_{n=1}^{\infty} b_n z^{n+1}, \quad z \in D$$ it follows from (2), together with |a2| ^2 , that $$\begin{align} \pi^{-1} \triangle (r,1/F_f) &= |a_2|^2 \ r^2 + 2r^4 \sum_{n=1}^\infty 2^{-1} (n+1) \, |b_n|^2 \ r^{2n-2} \ &\leqslant 4r^2 + 2r^4 \sum_{n=1}^\infty n \, |b_n|^2 \leqslant 2r^2 \big(r^2 + 2\big). \end{split}$$ Since A(r, 1/Fxa) = 2irr² (r2 +2), we now have the identity. If the maximum is attained by / , then |a21 =2 , so that / = Kg for some 0. U It follows that A(l, 1/Ff) ^ 6n. This shows that each function / G 5 is the quotient of two functions, z and 1/Ff(z), both of which are bounded and Dirichletfinite in D\ see estimate (6) for the bound |1/.F/| ^ 4. Each / G 5 maps {\z\ — r} onto a curve of length $$L(r,f) = r \int_0^{2\pi} \left| f' \big( r e^{it} \big) \right| dt \qquad (0 < r < 1).$$ It is known that, for 0 < r < 1, (3) $$2^{-1}\pi r(1+r)(1-r)^{-2} < L(r,K) \leqslant \sup_{f \in S} L(r,f);$$ see [2, Theorem 2] and [3, p.39]. Now, as another application of the de Branges theorem we have (4) $$\max_{f \in S} \triangle(r, f) = \triangle(r, K) = \pi r^2 (r^4 + 4r^2 + 1) (1 - r^2)^{-4},$$ for 0 < r < 1. The maximum is attained only by Kg's. We improve (3) in

Theorem 2. For $0 < \tau < 1$ we have (5) $$2\pi r (r^4 + 4r^2 + 1)^{1/2} (1 - r^2)^{-2} \leqslant L(r, K) \leqslant \sup_{f \in S} L(r, f).$$ PROOF: This is a consequence of the expression of $\triangle(r,K)$ in (4), without appealing to the expression of L(r,K) in terms of elliptic integrals (see [2]). We only apply to K the isoperimetric inequality: $$\triangle(r,f) \leqslant \pi \{L(r,f)/(2\pi)\}^2 \text{ for } f \in S,$$ which says that, of all rectifiable Jordan curves with the given perimeter L(r, f), (0 < r < 1), the circle has interior of maximum area. Since $$\inf_{0 < r < 1} (r^4 + 4r^2 + 1)^{1/2} (1 + r)^{-3} = \sqrt{6}/8 > 1/4,$$ estimate (5) is better than (3). We recall that $$L(r,f) \leqslant 2\pi r (1-r)^{-2} \equiv \gamma(r)$$ for $f \in S$ and 0 < r < 1 [3, p.40]. Estimate (5) now yields $$\left(\sqrt{6}/4\right)\gamma(r)\leqslant \sup_{f\in S}L(r,\,f)\leqslant \gamma(r).$$ Note that $\gamma(r)$ is the length of the boundary circle of $\delta_r = \{|z| < r(1-r)^{-2}\}$ . We recall the growth theorem for $f \in S$ : (6) $$(1+|z|)^{-2} \leqslant |F_f(z)| \leqslant (1-|z|)^{-2}, \quad z \in D;$$ see [3, p.33]. The image $f(\{|z| < r\})(f \in S)$ is contained in the disc $\delta_r$ with area $\pi r^2(1-r)^{-4}$ and $$\triangle(r,f)/\{\pi r^2(1-r)^{-4}\}$$ is at most: $$(r^4 + 4r^2 + 1)(1+r)^{-4}, \quad 0 < r < 1,$$ which decreases from 1 to 3/8 as r increases from 0 to 1. Therefore, one may say that the upper estimate of (6) becomes "worse" as r increases because $f(\{|z| < r\})$ occupies only a small part of $\delta_r$ in area. We next assume that $F_f$ is nonconstant. The Riemann surface $\Phi_r$ ( $\Phi_r^*$ , respectively) which is the image of $\{|z| < r\}$ by $F_f$ (1/ $F_f$ , respectively), by (6), has projection contained in the disc with centre 0 and radius $(1-r)^{-2}$ $((1+r)^2$ , respectively). The "sheet-number" of the covering surface $\Phi_r$ ( $\Phi_r^*$ , respectively) over this disc: $$\triangle(r, F_f)/\{\pi(1-r)^{-4}\}(\triangle(r, 1/F_f)/\{\pi(1+r)^4\}, \text{ respectively})$$ is at most $2r^2(r^2+2)(1+r)^{-4}$ which increases from 0 to 3/8 as r increases from 0 to 1. In this sense (6) yields little information on the distribution of the values of $F_f(z)$ $(1/F_f(z), respectively)$ , for |z| < r. Let C be the family of all $f \in S$ such that f(D) is convex. With the aid of the coefficient estimate [3, p.45, Corollary] we have $$\max_{f \in C} \triangle(r, F_f) = \pi r^2 (1-r)^{-2}, \qquad 0 < r < 1.$$ For each r, 0 < r < 1, the maximum is attained only by $J_{\theta}(z) \equiv z/(1 - e^{i\theta}z)$ , $z \in D$ , $\theta$ real. A natural conjecture is that $$\max_{f \in C} \triangle(r, 1/F_f) = \pi r^2, \qquad 0 < r \leqslant 1,$$ where the maximum is attained only by $J_{\theta}$ 's. REMARK. If $a_2 = 0$ in (1) for $f \in S$ , then $F'_f(0) = a_2 = 0$ , so that $F_f$ is not univalent in any disc with centre 0. To consider the case $a_2 \neq 0$ , we first note that the function $$\varphi(x) = -\log(1-x^2) + (3x^2-2x^4)(1-x^2)^{-2}$$ increases from 0 to $+\infty$ as z increases from 0 to 1. Therefore there exists a number $R \equiv R(a_2)$ , 0 < R < 1, such that $\varphi(R) = |a_2|^2$ . We shall show that $F_f$ is univalent in $\{|z| < R(a_2)\}$ . The expression for $1/F_f$ in the proof of Theorem 1 shows that $$g(z) \equiv \{1 - 1/F_f(z)\}/a_2 = z - a_2^{-1} \sum_{n=2}^{\infty} b_{n-1} z^n \text{ in } D.$$ The Schwarz inequality, together with (2), yields that $$|a_{2}|^{-1} \sum_{n=2}^{\infty} n |b_{n-1}| R^{n-1}$$ $$\leq |a_{2}|^{-1} \left\{ \sum_{n=2}^{\infty} (n-1) |b_{n-1}|^{2} \right\}^{1/2} \left\{ \sum_{n=2}^{\infty} n^{2} (n-1)^{-1} (R^{2})^{n-1} \right\}^{1/2}$$ $$\leq |a_{2}|^{-1} \varphi(R)^{1/2} = 1.$$ By [3, p.73, Problem 24 (b)] we have that R~¹ g{Rz) is univalent and starlike in D. Thus g, and hence Ff, are univalent in {\z\ < R} as we wished. We note that the image of {|z| < R} under 1/Ff is starlike with respect to 1 also. Unfortunately we cannot claim that R(a.2) is sharp. In fact, for the Koebe function K with 02 = 2 we have $$R(2)=0.6823\ldots\ldots,$$ while FK is univalent in D. Finally, since \a2\ < 2 for / € S, we have R(a2) < -R(2).