Star-center points of univalent functions

Lemma 1. The set of all star center points of a function in S0 is convex. PROOF. Let g e S?, zx, z2 e D such that g{zx), g(z2) belong to Sg. We show that the segment [g(zx), g(z2)] is contained in S . Suppose [g(zx), g(z2)\ £ Sg and let w e (g(zx), g{z2)) be such that w £ Sg. Since g(zi), g(z2) are s.c.p of g(D) we have w e g(D). By the hypothesis on w there is z0 e D such that [g(z0), w] <f. g(D). Observe that if the points g(z0), g(zx), g(z2) are collinear then there is nothing to prove. Otherwise there is w{ e (g(z0), w) such that w{ \$ g(D). We have [#(z,), g(zQ)] c g(D) because g(z{) e Sg and g(z0) e g(D). Let w2 be the intersection of the segment [g(z{), g(z0)] and the straight line determined by the points g(z2) and w{. These two sets intersect because w{ is an interior point of the triangle {g(z0), g{zx), g(z2)} . We have w2 e g(D). Since g(z2) € Sg it follows that w{ e g(D) which contradicts wx £ g(D). Hence S is convex.$

Lemma 2. Let f e <9\Q, ^:Z)->5 ° be a univalent analytic function such that <^(0) = 0, £(£>) = S°, and let z0, zx be complex numbers such that \zo\ < |zj| = r < 1. Then the segment [/(z,), £(z0)] is contained in f(Dr), where 7)r = {z: \z\ < r}.* PROOF. For £(z0) = 0 the lemma is known [2, page 220]. Let p and 6 be two real numbers such that 0 < p < I, -n < 6 < n, pe'e zx = zQ. Put <j>(z) = tf{z) + (1 - t)£(pe'e z), zeD, 0 < t < 1. Clearly <P is analytic in D, 0(0) = /(0) = 0, and for each z the point \$(pe'e z) is a s.c.p of f{D). Hence <1> is subordinate to / , so O(z) = f(cp{z)), where <p is analytic in D, and $|\varphi(z)| \leq |z|$ . We have $$\begin{split} & \Phi(z_1) = t f(z_1) + (1-t) \xi(\rho e^{i\theta} z_1) = t f(z_1) + (1-t) \xi(z_0) = f(\varphi(z_1)) \\ & \text{and } |\varphi(z_1) \leq |z_1| \text{. Hence } \Phi(z_1) \in f(\overline{D}_r) \text{.} \end{split}$$$

Lemma 3. Let n > 2 be an integer. Given $1/2 \le x \le 1$ and integers $1 \le p \le q$ , define $$F_{q,p} = (p-x)(p+1-x)\cdots(q-x).$$ Then (1) $$-n!n + nF_{n,2}(x) + 2x[F_{n,3}(x) + 2!2F_{n,4}(x) + \cdots + (n-2)!(n-2)F_{n,n}(x) + (n-1)!(n-1)] \le 0.$$ PROOF. We proceed by induction on n. Observe that (1) holds for n = 3. We assume that it holds for n and we prove that it holds for n+1. It suffices to show that the left-hand side of (1) is nonincreasing in n, for each fixed $x \in [1/2, 1]$ , or equivalently $$(n+1)F_{n+1,2}(x) - nF_{n,2}(x) + 2xn!n + 2x(n-x)$$ $$\cdot [F_{n,3}(x) + 2!2F_{n,4}(x) + \dots + (n-1)!(n-1)]$$ $$\leq (n+1)!(n+1) - n!n.$$ Now by the induction hypothesis we have $$2x[F_{n,3}(x)+2!2F_{n,4}(x)+\cdots+(n-1)!(n-1)] \le n!n-nF_{n,2}(x).$$ Hence (2) will hold if the following holds: (3) $$(n+1)F_{n+1,2}(x) - nF_{n,2}(x) + 2xn!n + (n-x)(n!n - nF_{n,2}(x))$$ $$\leq (n+1)!(n+1) - n!n.$$ This is equivalent to (4) $$F_{n+1,2}(x) + n!nx - (n+1)! \le 0.$$ To prove (4) we proceed as follows. We put $\Phi(x) = F_{n+1,2}(x) + n!nx - (n+1)!$ and we claim that the derivative $\Phi'(x)$ is nonnegative for $1/2 \le x \le 1$ . If this is proven it will mean that $\Phi(x)$ is nondecreasing so that its maximum value will be taken for x = 1. But since $\Phi(1) = 0$ (4) will be proven. We show that (5) $$\Phi'(x) = n!n + F'_{n+1,2}(x) \ge 0, \qquad 1/2 \le x \le 1.$$ Observe that from the definition of F p{x) it follows that (6) $$F'_{n+1,2}(x) = -F_{n+1,2}(x) \cdot \sum_{k=2}^{n+1} \frac{1}{k-x},$$ so (5) can be written (7) $$n!n - F_{n+1,2}(x) \cdot \sum_{k=2}^{n+1} \frac{1}{k-x} \ge 0.$$ Since 1/2 < x < 1, to prove (7) it suffices to show (8) $$n!n - F_{n+1,2}\left(\frac{1}{2}\right) \cdot \sum_{k=2}^{n+1} \frac{1}{k - \frac{1}{2}} \ge 0.$$ We again proceed by induction on n. It is easily seen that (8) holds for n = 3 . Assume that it holds for n . To show that (8) holds for n + l we prove that the left-hand side of (8) is nondecreasing in n , or that $$(9) \qquad (n+1)!(n+1) - n!n \ge F_{n+1,2}\left(\frac{1}{2}\right) \left[\left(n + \frac{1}{2}\right) \cdot \sum_{k=2}^{n+1} \frac{1}{k - \frac{1}{2}} + 1\right].$$ If in (9) the expression Fn+l 2(j) is replaced by $$n!n / \sum_{k=2}^{n+1} \frac{1}{k-\frac{1}{2}}$$ (10) $$(n+1)!(n+1) - n!n \ge \left[ \frac{n!n}{\sum_{k=2}^{n+1} \frac{1}{k - \frac{1}{2}}} \right] \cdot \left[ \left( n + \frac{1}{2} \right) \cdot \sum_{k=2}^{n+1} \frac{1}{k - \frac{1}{2}} + 1 \right].$$ Since (8) holds, it follows that (9) is true if (10) holds. But (10) is equivalent to (11) $$\frac{n}{n+2} \le \sum_{k=1}^{n} \frac{1}{2k+1}, \qquad n \ge 3,$$ which is easily seen to be true by induction. It follows that (9) holds, and this proves the lemma.

Theorem 1. Let f(z) = z + Yl7=2a nz " be a function in <9\0 and let a be a point of S°f. Then* - (i) 0 < n(f, a) < 1. - (ii)//>(/, a) = 1 then \an\ < 1, n = l,2,.... - (iii) n(f, a) = 1 if and only if S°f = f(D). - (iv) Ifn(f, a) < 1 then \an\ < An(f, a) + R^o) = Mn(f, a), n>2, \) = l + (n-l)f\ⁿ k=2{k-l)/(k-o), a= l/(l+n(f, a)), and* $$\begin{split} R_n(\sigma) &= \frac{-n!n}{\prod_{k=2}^{n+1} (k-\sigma)} + \frac{n}{n+1-\sigma} + \frac{2\sigma}{n+1-\sigma} \\ & \cdot \left[ \frac{1}{2-\sigma} + \frac{2!2}{(2-\sigma)(3-\sigma)} + \cdots \right. \\ & \left. + \frac{(n-2)!(n-2)}{(2-\sigma)\cdots(n-1-\sigma)} + \frac{(n-1)!(n-1)}{(2-\sigma)\cdots(n-\sigma)} \right]. \end{split}$$ (v) $$|a_n| \le B(f, n), n \ge 2$$ , where $B(f, n) = \inf_{\alpha \in S_f^0} (M_n(f, \alpha))$ . PROOF. Put g = g~ l where ga is the function denned in (12). Then g: D -• 5° is analytic in D and has the following properties: $$(a')$$ $g(0) = \alpha$ , $g'(0) = 1/g'_{\alpha}(\alpha) > 0$ ; (b') g is one-to-one; $$(c') g(D) = S_f^0.$$ Let ^Q(0) = p . Then fi e D and g(fi) = 0. (13) $$G(z) = g\left(\frac{z+\beta}{1+\overline{\beta}z}\right), \qquad z \in D.$$ The function G: D -» 51 ? is analytic in Z> and has the following properties: (a") $$G(0) = g(\beta) = 0$$ ; $G'(0) = g'(\beta)(1 - |\beta|^2) = (1 - |\beta|^2)/g'_{\alpha}(0) = (1 - |g_{\alpha}(0)|^2)/g'_{\alpha}(0)$ ; (b") $G$ is one-to-one; $$(c'') G(D) = S_f^0.$$ Clearly G is subordinate to / . It follows that (14) $$G(z) = f(\omega(z))$$ where co is analytic on D and |to(z)| < \z\. Put G{z) = Y^Li bnz" , z e D. We have since, G'(0) does not vanish, that (15) $$0 < b_1 = G'(0) = \omega'(0) = [1 - |g_{\alpha}(0)|^2] / g_{\alpha}'(0) = \mu(f, \alpha) \le 1.$$ This proves assertion (1) of Theorem 1. The function G{z)/bl = Y^Li^n/^i)2 " belongs to the class i ^ and maps D onto the region (1/2^)5° = {w/bl: w e S®} which is convex since 5° is convex. It follows that (16) $$|b_n/b_1| \le 1$$ , $n = 1, 2, ....$ Observe that co is univalent in D because the composition of two univalent functions is univalent. Summarizing the properties of co, we have: - (i) co is univalent in D; - (ii) co{D) c D; - (iii) co(0) = 0; - (iv) 0 < &/(0) = ^ < 1. If in addition we had co{D) = D then we would have co{z) = z, co'(0) = bl = 1 and it would follow from (14) and (16) that G(z) = f{z) so that an = bn, \an\ < 1. This proves assertion (ii) of Theorem 1. Next assume that co{D) is a proper subset of D. Then it follows from the condition for equality in Schwarz's lemma that <y'(0) < 1. The above imply (i) $$\omega(D) = D$$ if and only if $\omega'(0) = 1$ , (17) (ii) if $$\omega'(0) < 1$$ then $0 < b_1 < 1$ , $$(iii) |b_n| \le |b_1| \le 1,$$ and assertion (iii) of Theorem 1 follows from (17)(i). Let z, zQ e D such that \zo\ < \z\ = r < 1. Put G(z0) = w e S°f, f{z) - w - Re'x , z — re'e . It follows from Lemma 2 that w is a s.c.p of f(Dr). Therefore $$\frac{\partial}{\partial \theta} \arg[f(z) - w] = \frac{\partial \tau}{\partial \theta} \ge 0.$$ We have $$\log[f(z) - w] = \log R + i\tau$$ so $$\operatorname{Im}\left[\frac{\partial}{\partial \theta} \log(f(z) - w)\right] \ge 0$$ In view of $$\frac{\partial}{\partial \theta} = ire^{i\theta} \frac{d}{dz} = iz \frac{d}{dz}$$ we get $$\text{Re}[zf'(z)/(f(z)-G(z_0))] \ge 0.$$ The last inequality holds for all z, z0 in D for which \z\ > \zo\. Therefore if X is a real number such that 0 < X < 1, we have $$\operatorname{Re}[zf'(z)/(f(z)-G(-\lambda z))] \geq 0, \qquad z \in D.$$ Put (18) $$F(z) = [zf'(z)/(f(z) - G(-\lambda z))] = \sum_{n=0}^{\infty} c_n z^n, \qquad z \in D.$$ It is easily seen that F is analytic in D and that c0 = 1/(1 + bxX). Due to the inequality $$\operatorname{Re} F(z) \geq 0, \qquad z \in D$$ we have $$|c_n| \le 2c_0 = \frac{2}{1 + b_1 \lambda}.$$ From (18) we get $$zf'(z) = \sum_{n=1}^{\infty} na_n z^n = \sum_{n=1}^{\infty} [a_n - b_n(-\lambda)^n] z^n \cdot \sum_{n=0}^{\infty} c_n z^n.$$ The last equation gives $$na_n = \sum_{k=1}^n [a_k - (-\lambda)^k b_k] c_{n-k}, \qquad n = 1, 2, ...,$$ or (20) $$(n - c_0)a_n = \sum_{k=1}^{n-1} a_k c_{n-k} - \sum_{k=1}^n (-\lambda)^k b_k c_{n-k}.$$ If we set X = 0 then (20) and (19) provide the well known inequality \an\ < n, n = 2, 3, ... . From (20) we obtain, on account of (16) and (19), $$\begin{aligned} |a_n| &\leq \frac{2c_0}{n-c_0} \sum_{k=1}^{n-1} |a_k| + \frac{1}{n-c_0} \sum_{k=1}^{n} \lambda^k |c_{n-k}| |b_k| \\ &\leq \frac{2c_0}{n-c_0} \sum_{k=1}^{n-1} |a_k| + \frac{\lambda^n b_1 c_0}{n-c_0} + \frac{1}{n-c_0} \sum_{k=1}^{n-1} 2b_1 c_0 \lambda^k. \end{aligned}$$ Now if we let $\lambda \to 1$ we get, since $b_1 \sigma = 1 - \sigma$ , that (21) $$|a_n| \le \frac{2\sigma}{n-\sigma} \sum_{k=1}^{n-1} |a_k| + \frac{(1-\sigma)(2n-1)}{n-\sigma}, \qquad n \ge 2.$$ From (21) we deduce that for $n \ge 2$ , (22) $$|a_n| \le A_n(f, \alpha) + R_{n-1}(\sigma) = M_n(f, \alpha) \le A_n(f, \alpha).$$ The last part of (22) follows immediately from Lemma 3, since $R_n(\sigma)$ is nonpositive for $n \ge 1$ and $1/2 \le \sigma \le 1$ . To prove the first part of (22) we proceed by induction on n. It is easily seen that for n = 2, 3, (21) provides $$\begin{split} |a_2| & \leq 1 + \frac{1}{2 - \sigma} = A_2(f, \alpha) + R_1(\sigma) = A_2(f, \alpha), \\ |a_3| & \leq 1 + \frac{2!2}{(2 - \sigma)(3 - \sigma)} = A_3(f, \alpha) + R_2(\sigma) = A_3(f, \alpha), \end{split}$$ because $R_1(\sigma)=R_2(\sigma)=0$ , which proves that (22) holds for n=2,3. Assume that (22) holds for n. We get from (21), after some calculations, that $$|a_{n+1}| \le \frac{2\sigma}{n+1-\sigma} \sum_{k=1}^{n} |a_k| + \frac{(1-\sigma)(2n+1)}{n+1-\sigma}$$ $$\le A_{n+1}(f,\alpha) + R_n(\sigma) = M_{n+1}(f,\alpha).$$ It follows that (22) holds for n + 1. This proves assertion (iv) of Theorem 1, while assertion (v) is obvious. The theorem is proved. REMARK. If in (19) and (16) equality holds for n=2,3,4 then for $c_1=c_2=c_3=2\sigma$ , $b_2=b_4=-b_1$ , $b_3=b_1$ , $\lambda=1$ , it is easily checked that (22) is sharp for $n\leq 4$ . Indeed we find $$\begin{aligned} a_2 &= 1 + \frac{1}{2 - \sigma} \,, \qquad a_3 &= 1 + \frac{4}{(2 - \sigma)(3 - \sigma)} \,, \\ a_4 &= 1 + \frac{18}{(2 - \sigma)(3 - \sigma)(4 - \sigma)} + \frac{\sigma^2 - \sigma}{(2 - \sigma)(3 - \sigma)(4 - \sigma)} . \end{aligned}$$ However the sharpness of (22) for all n remains open. We make the following conjecture. Conjecture. Let $f \in \mathcal{S}_0$ , $\alpha \in S_f^0$ . Then (\*) $$|a_n| \leq A_n(f, \alpha) + R_{n-1}(\sigma) + H_n(\sigma), \qquad n \geq 2,$$ where $$H_n(\sigma) = \sum_{k=3}^{n-2} \left[ R_k(\sigma) (2\sigma)^{n-k-1} \middle/ \prod_{p=5}^{n+3-k} (p-\sigma) \right] ,$$ for $n \ge 5$ and $H_n(\sigma) = 0$ for n < 5. Furthermore, if equality holds in (16) and (19) and if $$c_n = 2\sigma$$ , $b_{2q} = -b_1$ , $b_{2q-1} = b_1$ , $n = 1, 2, ..., q = 1, 2, ...$ then for the $a_n$ obtained from (20), (\*) is sharp.

Theorem 2. Let $f_1$ , $f_2$ be functions in $\mathcal{S}_0$ . Let $B(f_1, n)$ , $B(f_2, n)$ be the corresponding bounds on the Taylor coefficients of $f_1$ and $f_2$ respectively, as these are defined in Theorem 1(v). Suppose $S_{f_1}^0 \subset S_{f_2}^0$ . Then (23) $$B(f_2, n) \leq B(f_1, n).$$ PROOF. Let $\alpha \in S_{f_1}^0$ . Let $G_1$ be the function obtained from $f_1$ exactly the same way as G was obtained from f in (13). Similarly, since $\alpha$ also belongs to $S_{f_2}^0$ , let $G_2$ be the function obtained from $f_2$ . We have $$G_1(D) = S_{f_1}^0 \subset S_{f_2}^0 = G_2(D), \qquad G_1(0) = G_2(0) = 0,$$ and both $G_1$ and $G_2$ are regular and univalent in D. It follows that $G_1$ is subordinate to $G_2$ , so $G_1(z) = G_2(\varphi(z))$ , where $\varphi$ is analytic in D and $|\varphi(z)| \leq |z|$ . We have $G_1'(z) = G_2'(\varphi(z))\varphi'(z)$ , or $$G_1'(0) = \mu(f_1, \alpha) = G_2'(0)\varphi'(0) = \mu(f_2, \alpha)\varphi'(0).$$ Since $|\varphi'(0)| \le 1$ we have (24) $$\mu(f_1, \alpha) \leq \mu(f_2, \alpha).$$ Put $$\sigma_1 = \frac{1}{1 + \mu(f_1, \alpha)}, \qquad \sigma_2 = \frac{1}{1 + \mu(f_2, \alpha)}.$$ We have from (24) that $$\sigma_1 \geq \sigma_2.$$ Now the function $M_n(f, \alpha) = A_n(f, \alpha) + R_{n-1}(\sigma)$ , defined in the statement of Theorem 1, can be written as follows $$\begin{split} M_n(f,\alpha) &= 1 + \frac{n-1}{n-\sigma} + \frac{2\sigma}{n-\sigma} \\ &\cdot \left[ \frac{1}{2-\sigma} + \frac{2!2}{(2-\sigma)(3-\sigma)} + \dots + \frac{(n-2)!(n-2)}{(2-\sigma)\dots(n-1-\sigma)} \right]. \end{split}$$ It is easily seen that the derivative of $M_n(f,\alpha)$ with respect to $\sigma$ is nonnegative, which implies that $M_n(f,\alpha)$ is an increasing function of $\sigma$ . It follows that $$(26) M_n(f_1, \alpha) \ge M_n(f_2, \alpha).$$ By taking the infinum of the left side of (26) for $\alpha \in S_{f_1}^0$ and of the right side for $\alpha \in S_{f_2}^0$ , we get (23) because $S_{f_1}^0 \subset S_{f_2}^0$ . This proves the theorem.