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Results & Lemmas (10)

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Lemma 1 Lemma 1. If, then there exists an such that implies that.
Lemma 1. If $f(z) = z + \sum_{n=2}^{\infty} b_n z^n \notin LS$ , then there exists an $\varepsilon > 0$ such that $g \in F_{\varepsilon}(f)$ implies that $g \notin LS$ .
Theorem 1 · coeff Theorem 1. If with, then there exists an such that implies that.
Theorem 1. If $f(z) = z - \sum_{n=2}^{\infty} |a_n| z^n$ with $\sum_{n=2}^{\infty} n |a_n| > 1$ , then there exists an $\varepsilon > 0$ such that $g \in F_{\varepsilon}(f)$ implies that $g \notin LS$ .
Theorem 2 · coeff Theorem 2. If, then there exist real numbers such that.
Theorem 2. If $\sum_{n=2}^{N} n |a_n| > 1$ , then there exist real numbers $\alpha_2, \dots, \alpha_N$ $(-\pi < \alpha_j \leq \pi)$ such that $$f(z)=z+\sum\limits_{n=2}^{N}a_{n}e^{i\alpha_{n}}z^{n}+\sum\limits_{n=N+1}^{\infty}a_{n}z^{n}\notin LS$$ .
Corollary 1 · coeff Corollary 1. If, then there exist real numbers and an such that for each, Proof. The corollary is established upon applying Lemma 1 to…
Corollary 1. If $\sum_{n=2}^{N} n|a_n| > 1$ , then there exist real numbers $\alpha_2, \dots, \alpha_N$ $(-\pi < \alpha_j \leq \pi)$ and an $\varepsilon > 0$ such that for each $\lambda, -\varepsilon \leq \lambda \leq \varepsilon$ , $$f_{\lambda}\!(z)=z\,+\,e^{i\lambda}\!\!\left(\sum\limits_{n=2}^{N}a_{n}e^{i\alpha_{n}}z^{n}\,+\,\sum\limits_{n=N+1}^{\infty}a_{n}z^{n}\right)\!\notin\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!$$ Proof. The corollary is established upon applying Lemma 1 to Theorem 2.
Corollary 2 Corollary 2. If, then there exist real numbers and such that for each,
Corollary 2. If $\sum_{k=2}^{N} n_k |a_{n_k}| > 1$ , then there exist real numbers $\alpha_2, \dots, \alpha_N \ (-\pi < \alpha_j \leq \pi)$ and $\varepsilon > 0$ such that for each $\lambda, -\varepsilon \leq \lambda \leq \varepsilon$ , $$f_{\lambda}\!(z) = z \,+\, e^{i\lambda}\!\!\left(\sum\limits_{k=2}^N a_{n_k}\!e^{i\alpha_k}z^{n_k} + \sum\limits_{\substack{n=2\n\neq n}}^\infty a_nz^n\right)\!\notin\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!$$
Lemma 2 Lemma 2. For there exist real numbers such that.
Lemma 2. For $n \geq 3$ there exist real numbers $\alpha_1, \alpha_2, \dots, \alpha_n$ such that $|\sum_{k=1}^n e^{i\alpha_k} z^k| < n \ (|z| \leq 1)$ .
Theorem 3 Theorem 3. (i) If, then for some. (ii) If N > 3, then we can find such that and for all.
Theorem 3. (i) If $\sum_{k=2}^{3} n_k |a_{n_k}| > 1$ , then $f_{\lambda}(z) = z + e^{i\lambda} (a_{n_2} z^{n_2} + a_{n_2} z^{n_3}) \notin LS$ for some $\lambda$ . (ii) If N > 3, then we can find $\{a_{n_k} | k = 2, 3, \dots, N\}$ such that $\sum_{k=2}^{N} n_k |a_{n_k}| > 1$ and $f_{\lambda}(z) = z + e^{i\lambda} \sum_{k=2}^{N} a_{n_k} z^{n_k} \in S$ for all $\lambda$ .
Lemma 3 · radius Lemma 3. Set and. For, the values for (zg'(z)/g(z)),, lie in a disk centered at a(r) and having radius d(r).
Lemma 3. Set $a(r)=1/(1-r^2)-\varepsilon^2r^2/(1-\varepsilon^2r^2)$ and $d(r)=r/(1-r^2)+\varepsilon r/(1-\varepsilon^2r^2)$ . For $g\in G_{\varepsilon}$ , the values for (zg'(z)/g(z)), $|z|\leq r$ , lie in a disk centered at a(r) and having radius d(r).
Lemma 4 Lemma 4. A function defined by (7) is starlike if and only if.
Lemma 4. A function $g \in G_{\varepsilon}$ defined by (7) is starlike if and only if $\varepsilon \leq 1/3$ .
Theorem 4 · radius Theorem 4. Suppose,, and, with a(r) and d(r) defined in Lemma 3. Then h *f is starlike in a disk, where is the first positive root of…
Theorem 4. Suppose $h(z) = z + e^{i\lambda} \sum_{n=2}^{\infty} z^n \in H_{\varepsilon}$ , $\varepsilon \leq 1/3$ , and $f \in F$ , with a(r) and d(r) defined in Lemma 3. Then h\*f is starlike in a disk $|z| < r_0$ , where $r_0$ is the first positive root of $$a(r) - \frac{d(r)}{\sqrt{(1-\varepsilon^2 r^2)(1-r^2)} - \varepsilon r^2} = 0.$$ Proof. In view of (5), $$(hf)(z) = \int_X h\frac{z}{1-xz}d\mu(x) = \int_X \frac{h(xz)}{x}d\mu(x)$$ , $|x|=1$ . By Lemma 4, the kernel functions are all starlike. An application of Lemma 3 to Lemma A shows that h \* f is starlike in a disk whose radius is at least as large as the first positive root of $$a(r) - d(r) \sec b(r) = 0,$$ where $$b(r) = \max_{|z|=r} \left\{ \arg \frac{g(z)}{z} \middle| g \in G_{\epsilon} \right\}$$ . Since $$\begin{align} b(r) & \leq \max_{|z|=r} |\arg\left(1+\varepsilon z\right)| + \max_{|z|=r} |\arg\left(1-z\right)| \ & = \sin^{-1}(\varepsilon r) + \sin^{-1}(r) = t(r) \; , \end{split}$$ $$(11) \qquad \sec b(r) \leq \sec t(r) = \frac{1}{\cos t(r)} = \frac{1}{\sqrt{(1-\varepsilon^2 r^2)(1-r^2)}-\varepsilon r^2} \; .$$ A substitution of the right-hand side of (11) into (10) yields the desired result. REMARK. If $\varepsilon=0$ then (9) reduces to $1/(1-r^2)-r/((1-r^2)^{3/2})=0$ , whose smallest positive root is $1/\sqrt{2}$ . This coincides with a sharp result of MacGregor [4].
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