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Abstract
In this article, we determine the Rogosinski radii for certain subclasses of close-to-convex functions defined on open unit disc $\mathbb{D}= \{z \in \mathbb{C}: |z| < 1\}$. Furthermore, we establish improved versions of the classical Bohr inequality and the Bohr-Rogosinski inequality pertaining to these subclasses. We demonstrate that all results derived in the study are sharp.
Results & Lemmas (12)
Each result is auto-extracted from the paper via OCR - the math is shown verbatim and may carry occasional transcription artifacts. Click to expand the full statement.
Theorem 1.1 · radius
Theorem 1.1. [15] Suppose that and. Then where is the positive root of the equation. The radius is the best possible. Moreover, where is…
Theorem 1.1. [15] Suppose that $f: \mathbb{D} \to \mathbb{D}$ and $f(z) = \sum_{n=0}^{\infty} a_n z^n$ . Then
$$|f(z)| + \sum_{n=N}^{\infty} |a_n r^n| \le 1 \quad for |z| = r \le R_N,$$
where $R_N$ is the positive root of the equation $2(1+r)r^N - (1-r)^2 = 0$ . The radius $R_N$ is the best possible. Moreover,
$$|f(z)|^2 + \sum_{n=N}^{\infty} |a_n r^n| \le 1 \quad for |z| = r \le R'_N,$$
where $R'_N$ is the positive root of the equation $(1+r)r^N - (1-r)^2 = 0$ . The radius $R'_N$ is the best possible.
For the results on Bohr-Rogosinski inequality, we refer to the articles [6, 14, 19, 20]. In the present work, we extend this line of research by determining the sharp Bohr–Rogosinski radius for the classes C1, C<sup>2</sup> and C<sup>3</sup> together with improved versions of the classical Bohr's inequality . In section 2, section 3, and section 4, we find the sharp Bohr–Rogosinski radius and its various versions for the classes C1, C<sup>2</sup> and C<sup>3</sup> respectively.
Theorem 2.4 · coeff
Theorem 2.4. Suppose that, then (2.8) for and, where is the solution of
Theorem 2.4. Suppose that $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{C}_1$ , then
$$|z| + \sum_{n=2}^{\infty} |a_n z^n| + \sum_{n=2}^{\infty} |a_n|^p |z|^{np} \le d(f(0), \partial f(\mathbb{D}))$$
(2.8)
for $|z| \le r_p$ and $p \ge 1$ , where $r_p$ is the solution of
$$(1-r)\sum_{n=2}^{\infty} \left(2 - \frac{1}{n}\right)^p r^{pn} - \left(1 - 3r + r\log 2 - \log(2 - 2r) + r\log(1 - r)\right) = 0.$$
Theorem 2.6 · radius
Theorem 2.6. Suppose that and is an integer, then (2.11) for, where is the solution of The result is sharp.
Theorem 2.6. Suppose that $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{C}_1$ and $N \geq 2$ is an integer, then
$$|f(z)| + \sum_{n=N}^{\infty} |a_n z^n| \le d(f(0), \partial f(\mathbb{D}))$$
(2.11)
for $|z| \leq r_{1,N}$ , where $r_{1,N}$ is the solution of
$$\frac{2r}{1-r} + \log(1-r) + \sum_{n=N}^{\infty} \left(2 - \frac{1}{n}\right) r^n - (1 - \log 2) = 0.$$
The result is sharp.
Lemma 3.1 · coeff
Lemma 3.1. [22] If, then for all. Equality in all cases is obtained for f(z) = z/(1-z). Using the above coefficient bound, growth and…
Lemma 3.1. [22] If $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in C_2$ , then
$$|a_n| \le 1 \tag{3.1}$$
$$\frac{r}{1+r} \le |f(z)| \le \frac{r}{1-r} \tag{3.2}$$
$$\frac{1}{(1+r)^2} \le |f'(z)| \le \frac{1}{(1-r)^2} \tag{3.3}$$
for all $|z| \le r$ . Equality in all cases is obtained for f(z) = z/(1-z).
Using the above coefficient bound, growth and distortion results we obtain the following Bohr-type inequalities for the class $C_2$ .
Theorem 3.2 · radius
Theorem 3.2. Suppose that, then (3.4) for, where is the solution of The result is sharp.
Theorem 3.2. Suppose that $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{C}_2$ , then
$$|f(z)| + |f'(z)||z| + \sum_{n=2}^{\infty} |a_n z^n| \le d(f(0), \partial f(\mathbb{D}))$$
(3.4)
for $|z| \le r_{21} \simeq 0.173417$ , where $r_{21}$ is the solution of
$$1 - 6r + r^2 + 2r^3 = 0.$$
The result is sharp.
Theorem 3.5 · radius
Theorem 3.5. Suppose that and is an integer. Then (3.10) for, where is the solution of The result is sharp.
Theorem 3.5. Suppose that $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{C}_2$ and $N \geq 2$ is an integer. Then
$$|f(z)| + \sum_{n=N}^{\infty} |a_n z^n| \le d(f(0), \partial f(\mathbb{D}))$$
(3.10)
for $|z| \leq r_{1,N}$ , where $r_{1,N}$ is the solution of
$$3r + 2r^N - 1 = 0.$$
The result is sharp.
Theorem 3.6 · radius
Theorem 3.6. Suppose that, then (3.12) for, where is the solution of
Theorem 3.6. Suppose that $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{C}_2$ , then
$$|f(z)|^2 + \sum_{n=N}^{\infty} |a_n z^n| \le d(f(0), \partial f(\mathbb{D}))$$
(3.12)
for $|z| \leq r_{2,N}$ , where $r_{2,N}$ is the solution of
$$1 - 2r - r^2 - 2r^N + 2r^{1+N} = 0. (3.13)$$
Lemma 4.1 · coeff
Lemma 4.1. [22] If, then This result is sharp, with equality for Lemma 4.2. [22] If then and (4.3) for all and equality holds in all cases…
Lemma 4.1. [22] If $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in C_3$ , then
$$|a_n| \le \frac{2}{3} + \frac{1}{3n^2}. (4.1)$$
This result is sharp, with equality for
$$f(z) = \frac{2z}{3(1-z)} - \frac{1}{3} \int_0^z \frac{\log(1-\zeta)}{\zeta} d\zeta.$$
Lemma 4.2. [22] If $f \in \mathcal{C}_3$ then
$$\frac{2r}{3(1+r)} + \frac{1}{3} \int_0^r \frac{\log(1+t)}{t} dt \le |f(z)| \le \frac{2r}{3(1-r)} - \frac{1}{3} \int_0^r \frac{\log(1-t)}{t} dt \tag{4.2}$$
and
$$\frac{2}{3(1+r)^2} + \frac{\log(1+r)}{3r} \le |f'(z)| \le \frac{2}{3(1-r)^2} - \frac{\log(1-r)}{3r}$$
(4.3)
for all $(0 < |z| \le r)$ and equality holds in all cases for the extremal function
$$f(z) = \frac{2z}{3(1-z)} - \frac{1}{3} \int_0^z \frac{\log(1-\zeta)}{\zeta} d\zeta.$$
Using the above coefficient bound, growth and distortion results we obtain the following Bohr-type inequalities for the class $C_3$ .
Theorem 4.3 · radius
Theorem 4.3. Suppose that, then (4.4) for, where is the solution of The radius is sharp.
Theorem 4.3. Suppose that $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{C}_3$ , then
$$|f(z)| + |f'(z)||z| + \sum_{n=2}^{\infty} |a_n z^n| \le d(f(0), \partial f(\mathbb{D}))$$
(4.4)
for $|z| \leq r_{31}$ , where $r_{31}$ is the solution of
$$\frac{2(2-r^2)r}{3(1-r)^2} - \frac{1}{3} \int_0^r \frac{\log(1-t)}{t} dt - \frac{1}{3} \log(1-r) + \sum_{n=2}^\infty \frac{r^n}{3n^2} - \left(\frac{1}{3} + \frac{\pi^2}{36}\right) = 0.$$
The radius is sharp.
Theorem 4.4 · coeff
Theorem 4.4. Suppose that, then (4.10) П for and, where is the root of the equation
Theorem 4.4. Suppose that $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{C}_3$ , then
$$|z| + \sum_{n=2}^{\infty} |a_n z^n| + \sum_{n=2}^{\infty} |a_n|^p |z|^{np} \le d(f(0), \partial f(\mathbb{D}))$$
(4.10)
П
for $|z| \leq R_p$ and $p \geq 1$ , where $R_p \in (0, 1/2)$ is the root of the equation
$$\frac{(3-r)r}{3(1-r)} + \sum_{n=2}^{\infty} \frac{r^n}{3n^2} + \sum_{n=2}^{\infty} \left(\frac{2}{3} + \frac{1}{3n^2}\right)^p r^{np} - \frac{1}{36}\left(12 + \pi^2\right) = 0. \tag{4.11}$$
Theorem 4.5 · radius
Theorem 4.5. Suppose that, then for, where is the solution of The result is sharp.
Theorem 4.5. Suppose that $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{C}_3$ , then
$$|f(z)| + \sum_{n=N}^{\infty} |a_n z^n| \le d(f(0), \partial f(\mathbb{D}))$$
$$(4.14)$$
for $|z| \leq R_{1,N}$ , where $R_{1,N}$ is the solution of
$$\sum_{n=N}^{\infty} \frac{r^n}{3n^2} - \frac{1}{3} \int_0^r \frac{\log(1-t)}{t} dt - \frac{12 + \pi^2 - 36r - \pi^2 r - 24r^N}{36(1-r)} = 0.$$
The result is sharp.
Theorem 4.6 · radius
Theorem 4.6. Suppose that, then (4.15) for, where is the solution of Proof. It follows from Lemma 4.1 and Lemma 4.2 that for |z| ≤ r.…
Theorem 4.6. Suppose that $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{C}_3$ , then
$$|f(z)|^2 + \sum_{n=N}^{\infty} |a_n z^n| \le d(f(0), \partial f(\mathbb{D}))$$
(4.15)
for $|z| \leq R_{2,N}$ , where $R_{2,N}$ is the solution of
$$\left(\frac{2r}{3(1-r)} - \frac{1}{3} \int_0^r \frac{\log(1-t)}{t} dt\right)^2 + \sum_{n=1}^{\infty} \frac{r^n}{3n^2} + \frac{24r^N + (12+\pi^2)r - 12 - \pi^2}{36(1-r)} = 0.$$
Proof. It follows from Lemma 4.1 and Lemma 4.2 that
$$|f(z)|^2 + \sum_{n=N}^{\infty} |a_n z^n| \le \left(\frac{2r}{3(1-r)} - \frac{1}{3} \int_0^r \frac{\log(1-t)}{t} dt\right)^2 + \sum_{n=N}^{\infty} \left(\frac{2}{3} + \frac{1}{3n^2}\right) r^n$$
$$= \left(\frac{2r}{3(1-r)} - \frac{1}{3} \int_0^r \frac{\log(1-t)}{t} dt\right)^2 + \sum_{n=N}^{\infty} \frac{r^n}{3n^2} + \frac{2r^N}{3(1-r)}$$
for |z| ≤ r.
Define a function S : [0, 1) → R by
$$S(r) = \left(\frac{2r}{3(1-r)} - \frac{1}{3} \int_0^r \frac{\log(1-t)}{t} dt\right)^2 + \sum_{n=N}^{\infty} \left(\frac{2}{3} + \frac{1}{3n^2}\right) r^n - \frac{1}{3} - \frac{1}{3} \int_0^1 \frac{\log(1+t)}{t} dt.$$
Note that S(0) = −1/3 − π <sup>2</sup>/36. Also, an easy calculation yields
$$S(1/2) = \left(\frac{2}{3} - \frac{1}{3} \int_{0}^{1/2} \frac{\log(1-t)}{t} dt\right)^{2} + \sum_{n=N}^{\infty} \frac{1}{3n^{2}2^{n}} + \frac{2^{2-N}}{3} - \frac{1}{36} \left(12 + \pi^{2}\right)$$
$$= \frac{\left(24 + \pi^{2} - 6(\log 2)^{2}\right)^{2}}{1296} + \sum_{n=N}^{\infty} \frac{1}{3n^{2}2^{n}} + \frac{2^{2-N}}{3} - \frac{1}{36} \left(12 + \pi^{2}\right)$$
$$= \frac{\left(24 + \pi^{2} - 6(\log 2)^{2}\right)^{2}}{1296} - \frac{1}{36} \left(12 + \pi^{2}\right) + \sum_{n=N}^{\infty} \frac{1}{3n^{2}2^{n}} + \frac{2^{2-N}}{3}$$
$$= \frac{144 + 12\pi^{2} + \pi^{4} - 288(\log 2)^{2} - 12\pi^{2}(\log 2)^{2} + 36(\log 2)^{4}}{1296} + \sum_{n=N}^{\infty} \frac{1}{3n^{2}2^{n}} + \frac{2^{2-N}}{3}$$
$$> 0.$$
In view of the above, S(0)S(1/2) < 0, and by the Intermediate Value Theorem there exists a root ζ ∈ (0, 1/2) such that S(r) ≤ 0 for r ≤ ζ. Furthermore, Since
$$S'(r) = 2\left(\frac{2r}{3(1-r)} - \frac{1}{3} \int_0^r \frac{\log(1-t)}{t} dt\right) \left(\frac{2}{3(1-r)^2} - \frac{\log(1-r)}{3r}\right) + \frac{2r^{-1+N}(N+r-Nr)}{3(1-r)^2} + \sum_{n=N}^{\infty} \frac{r^{n-1}}{3n} > 0,$$
there exists the unique R2,N ∈ (0, 1/2) such that the inequality (4.15) holds for r ≤ R2,N , and satisfying the equation S(R2,N ) = 0.
The result is sharp for the function
$$f(z) = \frac{2z}{3(1-z)} - \frac{1}{3} \int_0^z \frac{\log(1-\zeta)}{\zeta} d\zeta.$$
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