Maximizing the generalized Fekete-Szego functional over a class of hyperbolically convex functions

Lemma 2.1. For $$f = \alpha z + a_2 z^2 + a_3 z^3 + \dots \in H$$ , $\Re a_3 \le \alpha (1 - \alpha^2)(2 - \alpha^2)$ Proof. From the aforementioned authors we know that if $h(z) = c_1 z + c_2 z^2 + \cdots$ is a conformal map from $\mathbb D$ into $\mathbb D$ , then for $\lambda \in \mathbb R$ we have $\lambda^2 \log |c_1| + 2\lambda \Re \frac{c_2}{c_1} + \Re \left[ \frac{c_3}{c_1} - \frac{c_2^2}{c_1^2} \right] + |c_1|^2 \le 1$ . If $f(z) = \alpha z + a_2 z^2 + a_3 z^3 + \cdots$ and $\lambda = 0$ , then $\Re \left[ \frac{a_3}{\alpha} - \frac{a_2^2}{\alpha^2} \right] + |\alpha|^2 \le 1$ . So, $\alpha \Re a_3 \le \alpha^2 (1 - \alpha^2) + a_2^2 \le \alpha^2 (1 - \alpha^2) + \alpha^2 (1 - \alpha^2)^2 = \alpha^2 (1 - \alpha^2)(2 - \alpha^2)$ . Hence, $\Re a_3 \le \alpha (1 - \alpha^2)(2 - \alpha^2)$ Unfortunately, we have no evidence to believe our result is sharp. In 1999, Ma and Minda found that hyperbolically convex functions obeyed the following inequality: $\left|\frac{a_3}{\alpha} - \left(\frac{a_2}{\alpha}\right)^2\right| + \frac{1}{2}\left|\frac{a_2}{\alpha}\right|^2 \le \frac{1}{2}$ . Using the reverse triangle inequality, we can arrive at $|a_3| \le \frac{1}{2}\alpha^5 - \alpha^3 + \alpha$ . A result from Mejía and Pommerenke ([10]) will improve one of these and will be of some use to us. They showed that for hyperbolically convex functions of the form $f(z) = \alpha z + a_2 z^2 + \cdots$ , $\Re\left\{z\frac{f'(z)}{f(z)}\right\} > \frac{1}{2}$ . That is, these functions are starlike of order $\frac{1}{2}$ . As a consequence, $|a_2| < 1$ and $|a_3| < 1$ . These results will prove useful particularly when we examine specific coefficients by doing computations with the aid of computer programs. Another result of hyperbolically convex functions can be shown using the Beiberbach conjecture.

Lemma 2.2. If $f(z) = \alpha z + a_2 z^2 + \cdots \in H$ is rotated so that $a_2$ is real, then $2\alpha - a_2 \ge 0$ .

Lemma 4.1. Lt(f) is rotationally invariant. Proof. We wish to show Lt(f) = Lt(e ⁻iθf(e iθz)) for θ ∈ [0, 2π). Recall that f(z) = αz + a2z ² + a3z ³ + · · · . Letting θ ∈ [0, 2π) be given, we have $$e^{-i\theta}f(e^{i\theta}z) = e^{-i\theta}[\alpha e^{i\theta}z + a_2 e^{2i\theta}z^2 + a_3 e^{3i\theta}z^3 + \cdots]$$ $$= \alpha z + a_2 e^{i\theta}z^2 + a_3 e^{2i\theta}z^3 + \cdots.$$ Thus, $$\mathcal{L}_{t}(e^{-i\theta}f(e^{i\theta}z)) = |ta_{3}e^{2i\theta} + (a_{2}e^{i\theta})^{2}|$$ $$= |ta_{3}e^{2i\theta} + a_{2}^{2}e^{2i\theta}|$$ $$= |e^{2i\theta}| \cdot |ta_{3} + a_{2}^{2}|$$ $$= \mathcal{L}_{t}(f).$$ Recall that the Julia variation is given by $$f_{\epsilon}(z) = f(z) + \frac{\epsilon z f'(z)}{2\pi} \int_{\Gamma} \frac{1+\xi z}{1-\xi z} d\Psi(\xi) + o(\epsilon),$$ where $\Psi$ is dependent upon $\xi$ . Using a Taylor series expansion about z=0, we get $$\frac{1+\xi z}{1-\xi z} = 1 + 2\xi z + 2\xi^2 z^2 + 2\xi^3 z^3 + \cdots$$ Also, $zf'(z) = \alpha z + 2a_2z^2 + 3a_3z^3 + \cdots$ . Multiplying the last two equations: $$zf'(z) \cdot \frac{1+\xi z}{1-\xi z} = \alpha z + 2a_2 z^2 + 3a_3 z^3 + \cdots$$ $$+2\alpha \xi z^2 + 4a_2 \xi z^3 + \cdots$$ $$+2\alpha \xi^2 z^3 + \cdots$$ $$= \alpha z + (2a_2 + 2\alpha \xi) z^2 + (3a_3 + 4a_2 \xi + 2\alpha \xi^2) z^3 + \cdots$$ Substituting this into the Julia variational formula, we see that $$f_{\epsilon}(z) = \alpha z + a_{2}z^{2} + a_{3}z^{3} + \cdots + \frac{\epsilon}{2\pi} \int_{\Gamma} \{\alpha z + (2a_{2} + 2\alpha\xi)z^{2} + (3a_{3} + 4a_{2}\xi + 2\alpha\xi^{2})z^{3} + \cdots \} d\psi + o(\epsilon) = \left[\alpha + \frac{\epsilon}{2\pi} \alpha \int_{\Gamma} d\psi\right] z + \left[a_{2} + \frac{\epsilon}{2\pi} \int_{\Gamma} (2a_{2} + 2\alpha\xi) d\psi\right] z^{2} + \left[a_{3} + \frac{\epsilon}{2\pi} \int_{\Gamma} (3a_{3} + 4a_{2}\xi + 2\alpha\xi^{2}) d\psi\right] z^{3} + \cdots + o(\epsilon).$$ Applying the functional to $f_{\epsilon}(z)$ , we get $$\mathcal{L}_{t}(f_{\epsilon}(z)) = \left| t \left[ a_{3} + \frac{\epsilon}{2\pi} \int_{\Gamma} (3a_{3} + 4a_{2}\xi + 2\alpha\xi^{2}) \ d\psi \right] + \left[ a_{2} + \frac{\epsilon}{2\pi} \int_{\Gamma} (2a_{2} + 2\alpha\xi) \ d\psi \right]^{2} + o(\epsilon) \right|$$ $$= \left| t a_{3} + a_{2}^{2} + \frac{\epsilon}{2\pi} \int_{\Gamma} [t(3a_{3} + 4a_{2}\xi + 2\alpha\xi^{2}) + (4a_{2}^{2} + 4a_{2}\alpha\xi)] \ d\psi + o(\epsilon) \right|.$$ We wish to find $\frac{\partial}{\partial \epsilon}[\mathcal{L}_t(f_{\epsilon})]\big|_{\epsilon=0}$ as this will be zero for an extremal function f. To this end, let $F(\epsilon) = ta_3 + a_2^2 + \frac{\epsilon}{2\pi} \int_{\Gamma} [t(3a_3 + 4a_2\xi + 2\alpha\xi^2) + (4a_2^2 + 4a_2\alpha\xi)] d\psi + o(\epsilon)$ , with $o(\epsilon) = \epsilon^2 g(\epsilon)$ for some $g \in C^1(\epsilon)$ . We can then express the derivative, using $|F(\epsilon)| = \exp{\Re(\log F(\epsilon))}$ , as $$\frac{\partial}{\partial \epsilon} \mathcal{L}_t(f_{\epsilon}) = \exp\{\Re(\log F(\epsilon))\}\Re\left\{\frac{F'(\epsilon)}{F(\epsilon)}\right\},\,$$ where $$F'(\epsilon) = \frac{1}{2\pi} \int_{\Gamma} [t(3a_3 + 4a_2\xi + 2\alpha\xi^2) + 4a_2(a_2 + \alpha\xi)] d\psi + 2\epsilon g(\epsilon) + \epsilon^2 g'(\epsilon).$$ Evaluating F and F' at $\epsilon = 0$ , we get $F(0) = ta_3 + a_2^2$ and $F'(0) = \frac{1}{2\pi} \int_{\Gamma} K(\xi) d\psi$ , where $K(\xi) = t(3a_3 + 4a_2\xi + 2\alpha\xi^2) + 4a_2(a_2 + \alpha\xi)$ . Thus, $$\frac{\partial}{\partial \epsilon} \mathcal{L}_t(f_{\epsilon})\big|_{\epsilon=0} = \frac{\exp\{\Re(\log(ta_3 + a_2^2))\}\Re\{\frac{1}{2\pi}\int_{\Gamma}K(\xi) d\psi\}}{ta_3 + a_2^2}$$ $$= \frac{|ta_3 + a_2^2|}{ta_3 + a_2^2}\Re\left\{\frac{1}{2\pi}\int_{\Gamma}K(\xi) d\psi\right\}$$ $$= \frac{1}{2\pi}\int_{\Gamma}\Re\{K(\xi)\} d\psi$$ (after rotating so that the functional is real). It is the kernel of this integral that we will be concerned with.

Lemma 4.1. (The Step Down Lemma) If f is extremal, then it can have at most two proper sides Proof. Suppose that f is extremal and f(D) has exactly four sides. If the image under K(ξ) of the pre-image of one of the sides is entirely in the left or right half-plane, then we can increase the value of Lt(f²) and hence, f is not extremal. Therefore, all images of proper sides under K(ξ) must intersect the imaginary axis at least once. Let Γ be one of the sides having one endpoint in the left half-plane and the other in the right half-plane. We must have at least one such side as there are four sides and our kernel crosses the axis at most four times. Let z⁰ be a vertex whose image under K(ξ) lies in the left half-plane. Apply the side adding variation at the vertex z0, making sure to keep the entire image of the new side entirely in the left half-plane. Therefore, $$\frac{\partial}{\partial \epsilon} \mathcal{L}_t(f_{\epsilon})\big|_{\epsilon=0} = \Re\{K(e^{i\theta})\} \left( \int_{\gamma} d\psi + \int_{\gamma_k} d\psi \right) > 0$$ where γ is our new side. Hence, Lt(f²) > Lt(f), which is a contradiction to f being extremal. Therefore, f(D) cannot have exactly four proper sides. If we suppose that our extremal has exactly three proper sides, then we can repeat all of our arguments. This is because we must still have at least one proper side with endpoints on opposite sides of the imaginary axis as the kernel has at most four values that are zero and we have three sides crossing. Therefore, f(D) cannot have exactly three sides. Thus, n ≤ 2. It should be noted that, in general, this is the best possible bound on the number of sides of the extremal. This can be seen if we let t = 4 and α = 0.7. The value of the functional applied to the one-sided map is approximately 0.156. However, there is a two-sided map (having −1, e iγ, and e ⁻iγ, where γ ≈ 0.5889π, as points of intersection with ∂D) for which the functional has a greater value: approximately 0.530. However, there are restrictions which we can impose that would allow us to reduce the number of proper sides to one. We have shown that the extremal for L^t cannot be a function whose image of D has three or more sides. To help us find conditions under which we may further reduce the problem, we will create a library of functions having two or fewer proper sides. For the most part, we will be concerned with those mappings which are symmetric about the real axis, since they will have real coefficients. With these maps, we will be able to produce images of the kernel with the hope that this will give us insight into the correct conditions.

Lemma 6.1. For symmetric maps in $H^2$ having non-zero second coefficient, the image of $\partial \mathbb{D}$ under the kernel $K(\xi) = 2t\alpha\xi^2 + 4a_2(t+\alpha)\xi + 3ta_3 + 4a_2^2$ is (Euclidean) convex for $t \in \left[ -\frac{\alpha a_2}{2\alpha + a_2}, \frac{\alpha a_2}{2\alpha - a_2} \right]$ . Proof. The image of $z + az^n$ is (Euclidean) convex if and only if $|a| \leq \frac{1}{n^2}$ . Our kernel is of the form $K(z) = Az^2 + Bz + C$ with $C \in \mathbb{R}$ (symmetric mappings). Since C has no affect on the convexity of $K(\partial \mathbb{D})$ , we can define $\widetilde{K} = Az^2 + Bz$ . Then, since $B \in \mathbb{R} \setminus \{0\}$ , we can factor it out and define a new function. Let $\frac{1}{B}\widetilde{K} = K^ = \frac{A}{B}z^2 + z$ . Then $K$ is convex if and only if $\left|\frac{A}{B}\right| \leq \frac{1}{4}$ . Therefore, the image of $\partial \mathbb{D}$ under $K(\xi) = 2t\alpha\xi^2 + 4a_2(t+\alpha)\xi + (3ta_3 + 4a_2^2)$ is (Euclidean) convex provided $\left|\frac{2t\alpha}{4a_2(t+\alpha)}\right| \leq \frac{1}{4}$ . There are two possibilities to consider. Case 1: $t \in \mathbb{R}^+$ . In this case, $\frac{2t\alpha}{4a_2(t+\alpha)}$ is positive, so we can rewrite our inequality as $t\alpha \leq \frac{1}{2}a_2(t+\alpha)$ . Collecting t on one side, we have $t(2\alpha - a_2) \leq a_2\alpha$ . By lemma 2.2, we know that $2\alpha - a_2 \ge 0$ . Thus, we have the right-hand side of our interval. Case 2: t=-p, for $p\in\mathbb{R}^+$ . Replacing t with -p, we can achieve $2p\alpha\leq a_2|\alpha-p|$ . If $\alpha-p>0$ , then we have $2p\alpha\leq a_2(\alpha-p)$ . After collecting terms and dividing we get $p\leq \frac{\alpha a_2}{2\alpha+a_2}$ . Replacing p=-t, we get the left-hand side of our interval. If, on the other hand, $\alpha-p<0$ , then $2p\alpha\leq a_2(p-\alpha)$ . Again, collecting terms and dividing we get $\frac{\alpha a_2}{2\alpha-a_2}\leq t$ . This, again, is the left-hand side of our interval. $\square$ As an illustration, figure 6.4 shows a kernel of $g_{\alpha}$ , $\alpha = \frac{1}{5}$ ( $a_2 \approx 0.03996$ and $a_3 \approx 0.0222$ ), with a t-value (t = -0.018) in its interval of convexity: [-0.01817, 0.0222].  Figure 6.4: A kernel in its interval of convexity This result has a perfect cross over into the non-symmetric mappings.

Theorem 6.1. For $f(z) = \alpha z + a_2 z^2 + a_3 z^3 + \cdots \in H^2$ $(a_2 \neq 0)$ the image of $\partial \mathbb{D}$ under the kernel $K(\xi) = 2t\alpha \xi^2 + 4a_2(t+\alpha)\xi + 3ta_3 + 4a_2^2$ is (Euclidean) convex for $t \in \left[ -\frac{\alpha a_2}{2\alpha + a_2}, \frac{\alpha a_2}{2\alpha - a_2} \right]$ .

Theorem 6.1. Suppose f ∈ H is such that a² 6= 0 and f is not the half-plane mapping. Then there exists a range of t such that f is not extremal for Lt, in particular, for t in the interval of convexity: $$t \in \left[ -\frac{\alpha a_2}{2\alpha + a_2}, \frac{\alpha a_2}{2\alpha - a_2} \right].$$ As an example, figure 6.5 shows the kernel of h^α with α = 0.42 and b = 0.5 (a² ≈ 0.2004 and a³ ≈ 0.1959). t = 0.13, which is in the interval of convexity [−0.8089, 0.1316]  Figure 6.5: h^α in its interval of convexity Proof. Assume f has non-zero second coefficient, it is not the half-plane map, and it is extremal over the interval. As in the proof of lemma 6.1, the kernel is not only (Euclidean) convex, but univalent. Since our kernel is quadratic and its image is (Euclidean) convex, it can cross the imaginary axis no more than two times. Therefore, we have three possibilities for how our sides can cross the imaginary axis. Case 1: One side lies in the right half-plane. In this case, we can apply the variation that pushes out this side. Hence, $\epsilon > 0$ and the real part of this side is positive. Therefore, such a variation would increase the functional. This would contradict f being extremal. Case 2: One side lies in the left half-plane. In this case, we can apply the variation that pulls in this side. Hence, $\epsilon < 0$ and the real part of this side is negative. Therefore, such a variation would increase the functional. This would contradict f being extremal. Case 3: Both sides cross the imaginary axis. If both sides cross the imaginary axis, then one side must have an endpoint in the left half-plane. We can apply the side-adding variation (making sure to keep the new side in the left-half plane). This would increase the functional implying the extremal has more than two sides; contradicting the step-down lemma. Therefore, in the interval of convexity, the half-plane mapping is extremal. Recall that $f_{\alpha}(z) = \alpha z + \alpha \left(\frac{\alpha^2 + \cos(2\tilde{\theta})}{3}\right) z^3 + \cdots$ . This is a map which has a zero second coefficient. We will show that the kernel of this map crosses the imaginary axis. The kernel applied to this map is $K(t,\theta) = 2t\alpha e^{2i\theta} + 3t\alpha \left(\frac{\alpha^2 + \cos(2\tilde{\theta})}{3}\right)$ and we can let $f(\theta) = \Re\{K(t,\theta)\} = 2t\alpha\cos(2\theta) + 3ta_3$ . We can use calculus optimization techniques to find the minimum and maximum and show that this case cannot be eliminated for all $t \neq 0$ . Since $f(\theta) = 2t\alpha\cos(2\theta) + 3ta_3$ , $f'(\theta) = -4t\alpha\sin(2\theta)$ . $f'(\theta) = 0$ for $\theta = 0$ , $\frac{\pi}{2}$ , $\pi$ , $\frac{3\pi}{2}$ , and $2\pi$ . For t > 0, we have $f'(\theta) < 0$ on $(0, \frac{\pi}{2})$ and $(\pi, \frac{3\pi}{2})$ and $f'(\theta) > 0$ on $(\frac{\pi}{2}, \pi)$ and $(\pi, \frac{3\pi}{2})$ . Thus, we have a minimum of $f(\frac{\pi}{2}) = f(\frac{3\pi}{2}) = -2t\alpha + 3ta_3$ and a maximum f(π) = f(0) = f(2π) = 2tα + 3ta3. For t < 0, interchange f ⁰ > 0 with f ⁰ < 0 and the maximums and minimums. We claim that the kernel will always cross the imaginary axis for the baseball map. Let t be positive. Then the minimum is −2tα + 3ta³ = −2tα + tα(α ² + cos 2θe) = −tα(2 − α ² − cos 2θe) = −tα[(1 − α 2 ) + (1 − cos 2θe)] < 0. The maximum is 2tα + 3ta³ = 2tα(α ² + cos 2θe)) = tα(2 + α ² + cos 2θe) > 0. Thus, it must cross the imaginary axis for every t except zero. In the following three figures, we have we have plotted the kernel of f^α with α ≈ 0.7284 and θe = π 8 , for t = −4 (figure 6.6), t = 0.01 (figure 6.7), and t = 6 (figure 6.8).  Figure 6.6: Kernel of the "baseball map", I  Figure 6.7: Kernel of the "baseball map", II As expected, the kernels are all seem (Euclidean) convex (as the equation of the  Figure 6.8: Kernel of the "baseball map", III kernel is a circle) and all cross the imaginary axis. However, they are not convex as the images wrap around twice. Therefore, we cannot relax our condition of a² 6= 0 in theorem 6.1.