Certain classes of univalent functions with negative coefficients II

Theorem 1. A function $$f(z) = z - \sum_{n=2}^{\infty} |a_n| z^n$$ is in $P^*(\alpha, \beta)$ if and only if $$\sum_{n=2}^{\infty} n(1+\beta) |a_n| \le 2\beta(1-\alpha) .$$ This result is sharp. Proof. Let |z| = 1. Then $$\begin{split} \left| f'(z) - 1 \right| - \beta \left| f'(z) + (1 - 2\alpha) \right| &= \left| -\sum_{n=2}^{\infty} n |a_n| z^{n-1} \right| - \beta \left| 2(1 - \alpha) - \sum_{n=2}^{\infty} n |a_n| z^{n-1} \right| \\ &\leq \sum_{n=2}^{\infty} n (1 + \beta) |a_n| - 2\beta (1 - \alpha) \\ &\leq 0 \text{, by hypothesis.} \end{split}$$ Hence, by the maximum modulus theorem, $f \in P^*(\alpha, \beta)$ . For the converse, assume that $$\left| \frac{f'(z) - 1}{f'(z) + (1 - 2\alpha)} \right| = \left| \left( -\sum_{n=2}^{\infty} n |a_n| z^{n-1} \right) / \left( 2(1 - \alpha) - \sum_{n=2}^{\infty} n |a_n| z^{n-1} \right) \right| < \beta \quad (|z| < 1) .$$ Since $|Re(z)| \le |z|$ for all z, we have (2.1) $$\operatorname{Re}\left\{\left(\sum_{n=2}^{\infty} n |a_n| z^{n-1}\right) / \left(2(1-\alpha) - \sum_{n=2}^{\infty} n |a_n| z^{n-1}\right)\right\} < \beta.$$ Choose values of z on the real axis so that f'(z) is real. Upon clearing the denominator in (2.1) and letting $z \to 1$ through real values, we obtain $$\sum_{n=2}^{\infty} n |a_n| \leq 2\beta(1-\alpha) - \beta \sum_{n=2}^{\infty} n |a_n|.$$ This gives the required condition. The result is sharp, the extremal function being $$f(z) = z - [2\beta(1-\alpha)/n(1+\beta)]z^n$$

Theorem 2. If f € P\(a, g) , then* (3.1) $$r - \frac{\beta(1-\alpha)}{1+\beta} r^2 \le |f(z)| \le r + \frac{\beta(1-\alpha)}{1+\beta} r^2 \quad (|z| = r) ,$$ and (3.2) $$1 - \frac{2\beta(1-\alpha)}{1+\beta} r \le |f'(z)| \le 1 + \frac{2\beta(1-\alpha)}{1+\beta} r \quad (|z| = r).$$ Proof. In view of Theorem 1, we have $$\sum_{n=2}^{\infty} |a_n| \le \frac{\beta(1-\alpha)}{1+\beta} .$$ Hence $$|f(z)| \le r + r^2 \sum_{n=2}^{\infty} |a_n|$$ $$\le r + \frac{\beta(1-\alpha)}{1+\beta} r^2$$ and $$|f(z)| \ge r - r^2 \sum_{n=2}^{\infty} |a_n|$$ $$\ge r - \frac{\beta(1-\alpha)}{1+\beta} r^2.$$ Thus (3.1) follows. Also $$|f'(z)| \le 1 + r \sum_{n=2}^{\infty} n |a_n|$$ $$\le 1 + \frac{2\beta(1-\alpha)}{1+\beta} r$$ and $$|f'(z)| \ge 1 - \frac{2\beta(1-\alpha)}{1+\beta} r$$ This completes the proof of the theorem. REMARK. The bounds in (3.1) and (3.2) are sharp since the equalities are attained for the function $$f(z) = z - \frac{\beta(1-\alpha)}{1+\beta} z^2 \quad (z = \pm r)$$ .

Theorem 3. Let f Z P\*(a, 3) . Then the disc \z\ < 1 is mapped onto a domain that contains the disc \w\ < P . The result is sharp with extremal • function z ~ ~ / z .

Theorem 4. If f € C\*(a, B) , then f i ^[3^ , 3^ ) • This result is sharp with the extremal function -L+p $$f(z) = z - [\beta(1-\alpha)/1+\beta(3-2\alpha)]z^2$$ . Proof. In view of Theorem 1, it is sufficient to show that $$\sum n(2-\beta)|a_n| \leq \frac{2(1-\alpha)}{3-2\alpha}.$$ But / € C\*(a, B) implies, [/], $$\sum_{n=2}^{\infty} n\{(n-1) + \beta(n+1-2\alpha)\} |a_n| \le 2\beta(1-\alpha) .$$ Also, we note that $$\frac{n\{(n-1)+\beta(n+1-2\alpha)\}}{2\beta(1-\alpha)} \geq \frac{n(2-\beta)(3-2\alpha)}{2(1-\alpha)},$$ since $$(n-1) + (n+1-2\alpha) \ge \beta(2-\beta)(3-2\alpha) \quad (n = 2, 3, 4, ...)$$ Hence the result follows.

Theorem 5. If f t. P\*(a, &) , then f is convex in the disc \z\ < r = r(a, 3) , where $$r(\alpha, \beta) = \inf_{n} \left(\frac{1+\beta}{2\beta n(1-\alpha)}\right)^{1/(n-1)}$$ $(n = 2, 3, \ldots)$ This result is sharp, the extremal function being of the form $$f(z) = z - [2\beta(1-\alpha)/n(1+\beta)]z^n$$ . Proof. It suffices to show that $\left|\frac{zf''}{f'}\right| \le 1$ for $|z| \le 1$ . First, we note that $$\left|\frac{zf''}{f''}\right| \leq \left(\sum_{n=2}^{\infty} n(n-1)\left|a_n\right|\left|z\right|^{n-1}\right) / \left(1 - \sum_{n=2}^{\infty} n\left|a_n\right|\left|z\right|^{n-1}\right).$$ Thus the result follows if $$\sum_{n=2}^{\infty} n(n-1) |a_n| |z|^{n-1} \le 1 - \sum_{n=2}^{\infty} n |a_n| |z|^{n-1}$$ which is equivalent to $$\sum_{n=2}^{\infty} n^2 |a_n| |z|^{n-1} \le 1.$$ But by Theorem 1, $$\sum_{n=2}^{\infty} n(1+\beta) |a_n| \le 2\beta(1-\alpha) .$$ Hence f is convex if $$n^2 |z|^{n-1} \le \frac{n(1+\beta)}{2\beta(1-\alpha)}, \quad n = 2, 3, \ldots;$$ that is, $$|z| \le \left(\frac{1+\beta}{2\beta n(1-\alpha)}\right)^{1/(n-1)}, \quad n = 2, 3, \ldots$$ This completes the proof. 4. In this section we assert that the class $P^*(\alpha, \beta)$ is closed under 'arithmetic mean' and 'convex linear combinations'.

Theorem 6. If $$f(z) = z - \sum_{n=2}^{\infty} |a_n| z^n$$ and $$g(z) = z - \sum_{n=2}^{\infty} |b_n| z^n$$ are in $P^*(\alpha, \beta)$ , then $$h(z) = z - \frac{1}{2} \sum_{n=2}^{\infty} |a_n + b_n| z^n$$ is also in $P^*(\alpha, \beta)$ .

Theorem 7. Let $$f_1(z) = z$$ , $f_n(z) = \frac{2\beta(1-\alpha)}{n(1+\beta)} z^n$ $(n = 2, 3, ...)$ . Then $f \in P^*(\alpha, \beta)$ if and only if it can be expressed in the form $$f(z) = \sum_{n=1}^{\infty} \lambda_n f_n(z) ,$$ where $\lambda_n \geq 0$ , and $$\sum_{n=1}^{\infty} \lambda_n = 1.$$ Proofs of Theorems 6 and 7 follow on the lines of the proofs of Theorems 8 and 9 in [1]. The details are omitted.