A radius of convexity problem

Lemma 1. Let o)(s) = a zⁿ + a +.z + be regular in A and < 1 . Then |e | 5 1 $$|\omega(z)| \le r^n ((r+|c_n|)/(1+|c_n|r)) , |z| = r .$$ Proof. Let w (s) = a)(s)/3n - c + ... . Then w (2) is regular and satisfies |io (s) | S 1 with |c | 5 1 . Define (2.2) $$\omega_2(z) = \left(\omega_1(z) - c_n\right) / \left(1 - \overline{c}_n \omega_1(z)\right).$$ Then u? (0) = 0 and |u2 (3)| 5 1 . Thus, by Schwarz's lemma [5], $$|\omega_2(z)| \leq |z|$$ . But, from (2.2), $$\omega_{1}(z) = (\omega_{2}(z) + c_{n}) / (1 + \bar{c}_{n} \omega_{2}(z))$$ Therefore $$\begin{split} |\omega(z)/z^n| &= |\omega_1(z)| = |(\omega_2(z) + c_n)/(1 + \overline{c}_n \omega_2(z))| \\ &\leq (|\omega_2(z)| + |c_n|)/(1 + |c_n||\omega_2(z)|) \\ &\leq (r + |c_n|)/(1 + |c_n|r) \; . \end{split}$$ Now the result immediately follows.

Lemma 2. Jf p(a) = 1 t J z" t V &,s^ is regular and has k=n+l Re p(3 ) > 0 for z € A , then, for \z\ = r , (2.3) Re $$p(z) \ge \left(2 + b_n r - b_n r^n - 2r^{n+1}\right) / \left(2 + b_n r + b_n r^n + 2r^{n+1}\right)$$ $$|p(z)| \le \left(2 + b_n r + b_n r^n + 2r^{n+1}\right) / \left(2 + b_n r - b_n r^n - 2r^{n+1}\right)$$ where 0 Sf) 5 2 . 5ot?z the estimates are sharp. Proof. Since Re p(s) > 0 and p(0) = 1 , the function p(z) is subordinate to the function (1+z)/(1-z). Therefore there exists a function h(z) which is regular for $z\in\Delta$ with h(0)=0 and |h(z)|<1 such that (2.5) $$p(z) = (1+h(z))/(1-h(z)) = 1 + b_n z^n + \sum_{k=n+1}^{\infty} b_k z^k.$$ A direct computation gives $h(z) = (b_n/2)z^n + \dots$ . Since h(z) satisfies the conditions of Lemma 1, applying it we have $$(2.6) |h(z)| \leq r^{n} ((r + \frac{1}{2}b_{n}) / (1 + \frac{1}{2}b_{n}r)) = r^{n} ((2r + b_{n}) / (2 + b_{n}r)).$$ Another direct computation shows that $$(2.7) \quad \text{Re } p(z) = \left(1 - \left|h(z)\right|^2\right) / \left(\left|1 - h(z)\right|^2\right) \ge \left(1 - \left|h(z)\right|\right) / \left(1 + \left|h(z)\right|\right).$$ Since the right hand side of (2.7) is monotone decreasing with respect to |h(z)|, applying (2.6) to (2.7), we get (2.3) immediately. To prove (2.4) consider the function $$q(z) = (p(-z))^{-1}.$$ Since q(z) satisfies the hypothesis of Lemma 2 we have $$1/|p(z)| = |q(-z)| \ge \text{Re } q(z) \ge \left(2 + b_n r - b_n r^n - 2r^{n+1}\right) / \left(2 + b_n r + b_n r^n + 2r^{n+1}\right)$$ which gives (2.4). It is easy to see that the estimate in (2.3) is sharp for the function $p(z) = \left(2 + b_n z - b_n z^n - 2z^{n+1}\right) / \left(2 + b_n z + b_n z^n + 2z^{n+1}\right) \quad \text{while the estimate in (2.4)}$ is sharp for the function $p(z) = \left(2 + b_n z + b_n z^n + 2z^{n+1}\right) / \left(2 + b_n z - b_n z^n - 2z^{n+1}\right)$ for each $b_n$ , $0 \le b_n \le 2$ .

Lemma 3. If $$p(z) = 1 + b_n z^n + \sum_{k=n+1}^{\infty} b_k z^k$$ is regular and has Re $p(z) > 0$ for $z \in \Delta$ , then, for $|z| = r$ , $$(2.8) |z(p'(z)/p(z))| \le 2r^n \cdot \left[ \left[ 2nb_n + \left( \frac{1}{4}n + \frac{1}{4} + (n-1)b_n^2 \right) r + 2nb_n r^2 \right] / \left( \left( 2 + b_n r \right)^2 - r^{2n} \left( 2r + b_n \right)^2 \right) \right]$$ where $b_n \ge 0$ . Proof. As in Lemma 2, there exists a function h(z) regular in $\Delta$ with |h(z)| < 1 such that $$p(z) = (1+h(z))/(1-h(z))$$ . Furthermore, since h(0)=0, |h(z)|<1 and h(z) has the power series expansion $h(z)=\left(b_n/2\right)z^n+\ldots$ , there exists a function $\phi(z)$ regular in $\Delta$ with $|\phi(z)|<1$ such that $h(z)=z^n\phi(z)$ . Using this, we obtain $$(2.9) p(z) = (1+z^n\phi(z))/(1-z^n\phi(z)) .$$ Taking the logarithmic derivatives on both sides of equation (2.9), we have $$p'(z)/p(z) = \left(2\left\{nz^{n-1}\phi(z) + z^{n}\phi'(z)\right\}\right) / \left(1 - z^{2n}\left\{\phi(z)\right\}^{2}\right).$$ Thus $$(2.10) |p'(z)/p(z)| \leq 2((nr^{n-1}|\phi(z)|+r^n|\phi'(z)|)/(1-r^{2n}|\phi(z)|^2)).$$ But it is well known [5] that (2.11) $$|\phi'(z)| \leq (1-|\phi(z)|^2)/(1-|z|^2)$$ . Since the right hand side of (2.10) is monotone increasing with respect to $|\phi'(z)|$ , by substituting (2.11) into (2.10), we obtain (2.12) $$|p'(z)/p(z)|$$ $$\leq \left(2/\left(1-r^{2}\right)\right)\left[\left(nr^{n-1}\left(1-r^{2}\right)\left|\phi(z)\right|+r^{n}\left(1-\left|\phi(z)\right|^{2}\right)\right)/\left(1-r^{2n}\left|\phi(z)\right|^{2}\right)\right]\ .$$ It is easy to check that the expression in the square bracket in (2.12) is monotone increasing with respect to $|\phi(z)|$ . Furthermore, $$\left|\phi(z)\right| = \left|h(z)/z^n\right| \le \left(2r + b_n\right)/\left(2 + b_n r\right) .$$ Hence $$\begin{split} & \left| p'(z)/p(z) \right| \\ & \leq 2 \left[ \left( n \left( 1 - r^2 \right) r^{n-1} \left( \frac{2r + b_n}{2 + b_n r} \right) + r^n \left\{ 1 - \left( \frac{2r + b_n}{2 + b_n r} \right)^2 \right\} \right] / \left( \left( 1 - r^2 \right) \left\{ 1 - r^{2n} \left( \frac{2r + b_n}{2 + b_n r} \right)^2 \right\} \right] \right] \\ & = 2 r^{n-1} \left[ \left( 2n b_n + \left( 4 + 4n + (n-1) b_n^2 \right) r + 2n b_n r^2 \right) / \left( \left( 2 + b_n r \right)^2 - r^{2n} \left( 2r + b_n \right)^2 \right) \right] \end{split}$$ which completes the