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Abstract

In this paper using $q$ calculus operator we obtain some sufficient conditions on $f_1$ and $f_2$ so that their linear combination $% f=tf_{1}+(1-t)f_{2},\ t\in \left[ 0,1\right] $, is univalent and convex in the direction of the real axis. Some examples are also illustrated to support our main results.

Results & Lemmas (7)

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 Theorem 1. [3] A locally univalent harmonic function in is a univalent mapping of real axis onto a domain convex in the direction of real…
Theorem 1. [3] A locally univalent harmonic function $f = h + \overline{g}$ in $\mathbb{D}$ is a univalent mapping of real axis $\mathbb{D}$ onto a domain convex in the direction of real axis if and only if h - g is a analytic univalent mapping of $\mathbb{D}$ onto a domain convex in the direction of real axis. In this paper using quantum approach we find the new harmonic functions. Such that, if we have an analytic function of the form Example 1. $$h(z) - g(z) = z - \frac{1}{2}z^2$$ and $\omega_q(z) = \frac{\partial_q(g(z))}{\partial_q(h(z))} = \frac{[2]_q}{2}z^2$ $$\partial_q(h(z)) - \partial_q(g(z)) = 1 - \frac{[2]_q}{2}z \qquad \partial_q(g(z)) = \frac{[2]_q}{2}z\partial_q(h(z))$$ we get $$\partial_a(h(z)) = 1$$ which on q-integration and normalization gives $$h(z) = z$$ $g(z) = \frac{[2]_q}{4}z^2$ Thus the mapping $f = h + \bar{g}$ given by $$f = z - \frac{[2]_q}{4} \overline{z^2}$$ On $q \to 1^-$ , we get the original function $f = z - \frac{1}{2}\overline{z^2}$
Lemma 1 Lemma 1. [ ](#page-6-16) Let Ω ⊂ C be a domain convex in the direction of the real axis. Also let p be a real-valued continuous function in…
Lemma 1. [\[3\]](#page-6-16) Let Ω ⊂ C be a domain convex in the direction of the real axis. Also let p be a real-valued continuous function in Ω. Then the mapping ω 7→ ω + p(ω) is univalent in Ω if and only if it is locally univalent. If it is univalent, then its range is convex in the direction of the real axis.
Lemma 2 Lemma 2. [ ](#page-6-17)Let f be analytic function in D with f(0) = 0 and f ′ (0) 6= 0. Suppose also that If then f is convex in the…
Lemma 2. [\[18\]](#page-6-17)Let f be analytic function in D with f(0) = 0 and f ′ (0) 6= 0. Suppose also that $$\varphi(z) = \frac{z}{(1 + ze^{i\theta})(1 + ze^{-i\theta})} \quad (\theta \in \mathbb{R}; z \in \mathbb{D}).$$ If $$\Re\left(\frac{zf'(z)}{\varphi(z)}\right) > 0 \quad (z \in \mathbb{D}),$$ then f is convex in the direction of real axis. Dorff and Rolf [\[5\]](#page-6-15) applied another way of constructing a univalent harmonic map by taking two suitable harmonic maps f<sup>1</sup> and f<sup>2</sup> with same dilatations, whose linear combination f<sup>3</sup> = tf<sup>1</sup> + (1 − t)f2, t ∈ [0, 1] is univalent and convex in the direction of the imaginary axis. Wang et al. [\[26\]](#page-6-18) derived several sufficient conditions on harmonic univalent functions f<sup>1</sup> and f<sup>2</sup> so that their linear combination f = tf<sup>1</sup> + (1 − t)f2, t ∈ [0, 1], is univalent and convex in the direction of the real axis. More results on the linear combination f of f<sup>1</sup> and f<sup>2</sup> may also be found in [\[6,](#page-6-19) [13,](#page-6-20) [21,](#page-6-21) [23,](#page-6-22) [24,](#page-6-23) [26\]](#page-6-18) etc. (also see the references cited in these). In this paper using quantum approach we find sufficient conditions on f<sup>1</sup> and f<sup>2</sup> so that their linear combination f = tf<sup>1</sup> + (1 − t)f2, t ∈ [0, 1], is univalent and convex in the direction of the real axis. Some examples are also illustrated to support our main results.
Lemma 3 Lemma 3. Let the function f: D → C be an analytic function with f(0) = 0 and ∂q(f(0)) 6= 0. Suppose that (2.1) If the function f satisfy…
Lemma 3. Let the function f : D → C be an analytic function with f(0) = 0 and ∂q(f(0)) 6= 0. Suppose that $$\lim_{q \to 1^{-}} \varphi_{q}(z) = \varphi(z) = \frac{z}{(1 + ze^{i\theta})(1 + ze^{-i\theta})} \quad (\theta \in \mathbb{R}; z \in \mathbb{D}).$$ (2.1) If the function f satisfy $$\Re e\left(\frac{z\partial_q(f(z))}{\varphi_q(z)}\right) > 0 \quad (z \in \mathbb{D}),$$ (2.2) then the function f is convex in the direction of real axis.
Theorem 2 Theorem 2. Let for with (2.3) If and satisfy the condition for some function given by (2.2), then, is univalent and convex in the direction…
Theorem 2. Let for $j = 1, 2, f_j = h_j + \overline{g_j} \in S_{\mathcal{H}}$ with $$F_j = h_j(z) - g_j(z).$$ (2.3) If $\omega_{q_1}(z) = \omega_{q_2}(z)$ and satisfy the condition $\Re\left(\frac{z\partial_q(F_j(z))}{\varphi_q(z)}\right) > 0$ for some function $\varphi_q(z)$ given by (2.2), then $f_3 = tf_1 + (1-t)f_2$ , $t \in [0,1]$ is univalent and convex in the direction of real axis.
Corollary 1 Corollary 1. Let for with the condition (2.3). If and satisfy the condition for some function given by (2.2). Then such that. is univalent…
Corollary 1. Let for $j=1,2,...,n, f_j=h_j+\overline{g_j}\in S_{\mathcal{H}}$ with the condition (2.3). If $\omega_{q_1}(z)=\omega_{q_2}(z)=...=\omega_{q_n}(z)$ and satisfy the condition $\Re\left(\frac{z\partial_q(F_J(z))}{\varphi_q(z)}\right)>0$ for some function $\varphi_q(z)$ given by (2.2). Then $F=\sum_{j=1}^n t_j f_j,\ t_j\in [0,1]$ such that $\sum_{j=1}^n t_j=1$ . is univalent and convex in the direction of real axis.
Theorem 3 Theorem 3. Let for be convex in the direction of the real axis. If, then is convex in the direction of real axis.
Theorem 3. Let for $j=1,2, f_j=h_j+\overline{g_j}\in S_{\mathcal{H}}$ be convex in the direction of the real axis. If $\Re\left\{\left(1-\omega_{q_1}\overline{\omega_{q_2}}\right)\partial_q(h_1(z)\overline{\partial_q(h_2(z)}\right\}\geq 0$ , then $f_3=tf_1+(1-t)f_2,\ t\in[0,1]$ is convex in the direction of real axis.

Definitions (1)

Def 2 Definition 2. Harmonic q right half-plane mappings: If we have an analytic function of the form and. Then we get which on q-integration and…
Definition 2. Harmonic q right half-plane mappings: If we have an analytic function of the form $$h(z) + g(z) = \frac{z}{1 - z}$$ and $\omega_q(z) = \frac{z^2 - 2z + qz}{1 - qz}$ . Then $$\partial_q(h(z)) + \partial_q(g(z)) = \frac{1}{(1-z)(1-qz)}$$ we get $$\partial_q(h(z)) = \frac{1}{(1-z)^3}$$ which on q-integration and normalization gives $$h(z) = \sum_{n=0}^{\infty} \frac{(n+1)(n+2)}{2[n+1]_q} z^{n+1}$$ Using this we get $$g(z) = \sum_{n=0}^{\infty} \frac{2[n+1]_q - (n+1)(n+2)}{2[n+1]_q} z^{n+1}$$ Thus h and g of the form $$h(z) = \sum_{n=0}^{\infty} \frac{(n+1)(n+2)}{2[n+1]_q} z^{n+1} \qquad g(z) = \sum_{n=0}^{\infty} \frac{2[n+1]_q - (n+1)(n+2)}{2[n+1]_q} z^{n+1}$$ On q → 1 <sup>−</sup>, we get the mapping defined in [\(1.1\)](#page-2-0).
Function classes studied:

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