A hybrid forecasting algorithm based on SVR and wavelet decomposition

Timotheos Paraskevopoulos; Peter N Posch; Timotheos Paraskevopoulos; Peter N Posch

doi:10.3934/QFE.2018.3.525

Quantitative Finance and Economics

2018, Volume 2, Issue 3: 525-553. doi: 10.3934/QFE.2018.3.525

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Research article

A hybrid forecasting algorithm based on SVR and wavelet decomposition

TU Dortmund University, Finance, Otto-Hahn-Str. 6, 44227 Dortmund, Germany

Received: 13 November 2017 Accepted: 12 February 2018 Published: 23 July 2018
JEL Codes: C63, C53

We present a forecasting algorithm based on support vector regression emphasizing the practical benefits of wavelets for financial time series. We utilize an e ective de-noising algorithm based on wavelets feasible under the assumption that the data is generated by a systematic pattern plus random noise. The learning algorithm focuses solely on the time frequency components, instead of the full time series, leading to a more general approach. Our findings propose how machine learning can be useful for data science applications in combination with signal processing methods. The timefrequency decomposition enables the learning algorithm to solely focus on periodical components that are beneficial to the forecasting power as we drop features with low explanatory power. The proposed integration of feature selection and parameter optimization in a single optimization step enable the proposed algorithm to be scaled for a variety of applications. Applying the algorithm to real life financial data shows wavelet decompositions based on the Daubechie and Coiflet basis functions to deliver the best results for the classification task.

Keywords:

Citation: Timotheos Paraskevopoulos, Peter N Posch. A hybrid forecasting algorithm based on SVR and wavelet decomposition[J]. Quantitative Finance and Economics, 2018, 2(3): 525-553. doi: 10.3934/QFE.2018.3.525

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Abstract

1. Introduction and definitions

To understand in a clear way the notions used in our main results, we need to add here some basic literature of Geometric function theory. For this we start first with the notation $\mathcal{A}$ which denotes the class of holomorphic or analytic functions in the region $\mathfrak{D} = \left \{ z\in \mathbb{C} :\left \vert z\right \vert < 1\right \}$ and if a function $g\in \mathcal{A}$ , then the relations $g\left(0\right) = g^{\prime }\left(0\right) -1 = 0$ must hold. Also, all univalent functions will be in a subfamily $\mathcal{S}$ of $\mathcal{A}$ . Next we consider to define the idea of subordinations between analytic functions $g_1$ and $g_2,$ indicated by $g_1\left(z\right) \prec g_2\left(z\right),$ as; the functions $g_1, g_2\in \mathcal{A}$ are connected by the relation of subordination, if there exists an analytic function $w$ with the restrictions $w\left(0\right) = 0$ and $\left \vert w\left(z\right) \right \vert < 1$ such that $g_1(z) = g_2(w(z)).$ Moreover, if the function $g_2\in \mathcal{S}$ in $\mathfrak{D},$ then we obtain:

$\begin{equation*} g_1(z)\prec g_2(z)\Leftrightarrow \left[ g_1(0) = g_2(0)\ \ \text{& }\ g_1(\mathfrak{D} )\subset g_1(\mathfrak{D}\mathbb{})\right] . \end{equation*}$

In 1992, Ma and Minda ^[16] considered a holomorphic function $\varphi$ normalized by the conditions $\varphi (0) = 1$ and $\varphi ^{\prime }(0) > 0$ with $Re\varphi > 0$ in $\mathfrak{D}$ . The function $\varphi$ maps the disc $\mathfrak{D}$ onto region which is star-shaped about $1$ and symmetric along the real axis. In particular, the function $\varphi (z) = (1+Az)/(1+Bz), \ (-1\leq B < A\leq 1)$ maps $\mathfrak{D}$ onto the disc on the right-half plane with centre on the real axis and diameter end points $\frac{1-A}{1-B}$ and $\frac{1+A}{1+B}$ . This interesting familiar function is named as Janowski function ^[10]. The image of the function $\varphi (z) = \sqrt{1+z}$ shows that the image domain is bounded by the right-half of the Bernoulli lemniscate given by $|w^{2}-1| < 1, \$ ^[25]. The function $\varphi (z) = 1+\frac{4}{3}z+\frac{2}{3}z^{2}$ maps $\mathfrak{D}$ into the image set bounded by the cardioid given by $(9x^{2}+9y^{2}-18x+5)^{2}-16(9x^{2}+9y^{2}-6x+1) = 0,$ ^[21] and further studied in ^[23]. The function $\varphi (z) = 1+\sin z$ was examined by Cho and his coauthors in ^[3] while $\varphi (z) = e^{z}$ is recently studied in ^[17] and ^[24]. Further, by choosing particular $\varphi, \$ several subclasses of starlike functions have been studied. See the details in ^{[2,4,5,11,12,14,19]}.

Recently, Ali et al. ^[1] have obtained sufficient conditions on $\alpha$ such that

$\begin{equation*} \begin{array}{llll} 1+zg^{\prime }(z)/g^{n}(z)\prec \sqrt{1+z} & \Rightarrow & g\left( z\right) \prec \sqrt{1+z}, & \text{for }n = 0,1,2. \end{array} \end{equation*}$

Similar implications have been studied by various authors, for example see the works of Halim and Omar ^[6], Haq et al ^[7], Kumar et al ^[13,15], Paprocki and Sokól ^[18], Raza et al ^[20] and Sharma et al ^[22].

In 1994, Hayman ^[8] studied multivalent ( $p$ -valent) functions which is a generalization of univalent functions and is defined as: an analytic function $g$ in an arbitrary domain $\mathcal{D}\subset \mathbb{C}$ is said to be $p$ -valent, if for every complex number $\omega$ , the equation $g(z) = \omega$ has maximum $p$ roots in $\mathcal{D}$ and for a complex number $\omega _{0}$ the equation $g(z) = \omega _{0}$ has exactly $p$ roots in $\mathcal{D}$ . Let $\mathcal{A}_{p}$ $\left(p\in \mathbb{N} = \left \{ 1, 2, \ldots \right \} \right)$ denote the class of functions, say $g\in \mathcal{A}_{p}$ , that are multivalent holomorphic in the unit disc $\mathfrak{D}$ and which have the following series expansion:

$\begin{equation} g(z) = z^{p}+\sum \limits_{k = p+1}^{\infty }a_{k}z^{k},\text{ }\left( z\in \mathfrak{D}\right) . \end{equation}$

(1.1)

Using the idea of multivalent functions, we now introduce the class $\mathcal{SL}_{p}^{\ast }$ of multivalent starlike functions associated with lemniscate of Bernoulli and as given below:

$\begin{equation*} \mathcal{SL}_{p}^{\ast } = \left \{ g(z)\in \mathcal{A}_{p}:\frac{zg^{\prime }\left( z\right) }{pg(z)}\prec \sqrt{1+z},\text{ }\left( z\in \mathfrak{D} \right) \right \} . \end{equation*}$

In this article, we determine conditions on $\alpha$ such that for each

$\begin{equation*} \begin{array}{lllll} 1+\alpha \frac{z^{2+p\left( j-1\right) }g^{\prime }\left( z\right) }{ pg^{j}\left( z\right) },\text{ for each }j = 0,1,2,3, \end{array} \end{equation*}$

are subordinated to Janowski functions implies $\frac{g\left(z\right) }{ z^{p}}\prec \sqrt{1+z}, \ \left(z\in \mathfrak{D}\right)$ . These results are then utilized to show that $g$ are in the class $\mathcal{SL}_{p}^{\ast }.$

1.1. Lemma

Let $w$ be analytic non-constant function in $\mathfrak{D}$ with $w\left(0\right) = 0.$ If

$\begin{equation*} \left \vert w\left( z_{0}\right) \right \vert = \max \left \{ \left \vert w\left( z\right) \right \vert ,\ \ \ \left \vert z\right \vert \leq \left \vert z_{0}\right \vert \right \} ,\ \ z\in \mathfrak{D}\text{,} \end{equation*}$

then there exists a real number $m\ \left(m\geq 1\right)$ such that $z_{0}w^{\prime }\left(z_{0}\right) = mw\left(z_{0}\right).$

This Lemma is known as Jack's Lemma and it has been proved in ^[9].

2. Main results

2.1. Theorem

Let $g\in \mathcal{A}_{p}$ and satisfying

$\begin{equation*} 1+\frac{\alpha z^{1-p}g^{\prime }\left( z\right) }{p}\prec \frac{1+Az}{1+Bz}, \end{equation*}$

with the restriction on $\alpha$ is

$\begin{equation} \left \vert \alpha \right \vert \geq \frac{2^{\frac{3}{2}}p\left( A-B\right) }{1-\left \vert B\right \vert -4p\left( 1+\left \vert B\right \vert \right) } . \end{equation}$

(2.1)

Then

$\begin{equation*} \frac{g\left( z\right) }{z^{p}}\prec \sqrt{1+z}. \end{equation*}$

Proof

Let us define a function

$\begin{equation} p\left( z\right) = 1+\frac{\alpha z^{1-p}g^{\prime }\left( z\right) }{p}, \end{equation}$

(2.2)

where the function $p$ is analytic in $\mathfrak{D}$ with $p(0) = 1.$ Also consider

$\begin{equation} \frac{g\left( z\right) }{z^{p}} = \sqrt{1+w\left( z\right) }. \end{equation}$

(2.3)

Now to prove our result we will only require to prove that $\left \vert w\left(z\right) \right \vert < 1.$ Logarithmically differentiating $\left(2.3\right)$ and then using $\left(2.2\right),$ we get

$\begin{equation*} p\left( z\right) = 1+\frac{\alpha zw^{\prime }\left( z\right) }{2p\sqrt{ 1+w\left( z\right) }}+\alpha \sqrt{1+w\left( z\right) }, \end{equation*}$

and so

$\begin{eqnarray*} \left \vert \frac{p\left( z\right) -1}{A-Bp\left( z\right) }\right \vert & = &\left \vert \frac{\frac{\alpha zw^{\prime }\left( z\right) }{2p\sqrt{ 1+w\left( z\right) }}+\alpha \sqrt{1+w\left( z\right) }}{A-B\left( 1+\frac{ \alpha zw^{\prime }\left( z\right) }{2p\sqrt{1+w\left( z\right) }}+\alpha \sqrt{1+w\left( z\right) }\right) }\right \vert \\ & = &\left \vert \frac{\alpha zw^{\prime }\left( z\right) +2p\alpha \left( 1+w\left( z\right) \right) }{2p\left( A-B\right) \sqrt{1+w\left( z\right) } -B\left( \alpha zw^{\prime }\left( z\right) +2p\alpha \left( 1+w\left( z\right) \right) \right) }\right \vert . \end{eqnarray*}$

Now, we suppose that a point $z_{0}\in \mathfrak{D}$ occurs such that

$\begin{equation*} \underset{\left \vert z\right \vert \leq \left \vert z_{0}\right \vert }{ \max }\left \vert w\left( z\right) \right \vert = \left \vert w\left( z_{0}\right) \right \vert = 1. \end{equation*}$

Also by Lemma 1.1, a number $m\geq 1$ exists with $z_{0}w^{\prime }\left(z_{0}\right) = mw\left(z_{0}\right)$ . In addition, we also suppose that $w\left(z_{0}\right) = e^{i\theta }$ for $\theta \in \left[ -\pi, \pi \right].$ Then we have

$\begin{eqnarray*} \left \vert \frac{p\left( z_{0}\right) -1}{A-Bp\left( z_{0}\right) }\right \vert & = &\left \vert \frac{\alpha mw\left( z_{0}\right) -2p\alpha \left( 1+w\left( z_{0}\right) \right) }{2p\left( A-B\right) \sqrt{1+w\left( z_{0}\right) }-B\left( \alpha mw\left( z_{0}\right) +2p\alpha \left( 1+w\left( z_{0}\right) \right) \right) }\right \vert , \\ &\geq &\frac{\left \vert \alpha \right \vert m-2p\left \vert \alpha \right \vert \left( \left \vert 1+e^{i\theta }\right \vert \right) }{2p\left( A-B\right) \sqrt{\left \vert 1+e^{i\theta }\right \vert }+\left \vert B\right \vert \left( \left \vert \alpha \right \vert m+2p\alpha \left \vert 1+e^{i\theta }\right \vert \right) }, \\ &\geq &\frac{\left \vert \alpha \right \vert m-4p\left \vert \alpha \right \vert }{2^{\frac{3}{2}}p\left( A-B\right) +\left \vert B\right \vert \left \vert \alpha \right \vert \left( m+4p\right) }. \end{eqnarray*}$

Now if

$\begin{equation*} \phi \left( m\right) = \frac{\left \vert \alpha \right \vert \left( m-4p\right) }{2^{\frac{3}{2}}p\left( A-B\right) +\left \vert B\right \vert \left \vert \alpha \right \vert \left( m+4p\right) }, \end{equation*}$

then

$\begin{equation*} \phi ^{\prime }\left( m\right) = \frac{2^{\frac{3}{2}}p\left( A-B\right) \left \vert \alpha \right \vert +8\left \vert \alpha \right \vert ^{2}p\left \vert B\right \vert }{\left( 2^{\frac{3}{2}}p\left( A-B\right) +\left \vert B\right \vert \left \vert \alpha \right \vert \left( m+4p\right) \right) ^{2} } \gt 0, \end{equation*}$

which illustrates that the function $\phi \left(m\right)$ is increasing and hence $\phi \left(m\right) \geq \phi \left(1\right)$ for $m\geq 1,$ so

$\begin{equation*} \left \vert \frac{p\left( z_{0}\right) -1}{A-Bp\left( z_{0}\right) }\right \vert \geq \frac{\left \vert \alpha \right \vert \left( 1-4p\right) }{2^{ \frac{3}{2}}p\left( A-B\right) +\left \vert B\right \vert \left \vert \alpha \right \vert \left( 1+4p\right) }. \end{equation*}$

Now, by using $\left(2.1\right),$ we have

$\begin{equation*} \left \vert \frac{p\left( z_{0}\right) -1}{A-Bp\left( z_{0}\right) }\right \vert \geq 1 \end{equation*}$

which contradicts the fact that $\begin{array}{c} p\left(z\right) \prec \frac{1+Az}{1+Bz}. \end{array}$ Thus $\left \vert w\left(z\right) \right \vert < 1$ and so we get the desired result.

Taking $g\left(z\right) = \frac{z^{p+1}f^{\prime }\left(z\right) }{pf(z)}$ in the last result, we obtain the following Corollary:

2.2. Corollary

Let $f\in \mathcal{A}_{p}$ and satisfying

$\begin{equation} 1+\frac{\alpha zf^{\prime }\left( z\right) }{p^{2}f\left( z\right) }\left( p+1+\frac{zf^{\prime \prime }\left( z\right) }{f^{\prime }\left( z\right) }- \frac{zf^{\prime }\left( z\right) }{f\left( z\right) }\right) \prec \frac{ 1+Az}{1+Bz}, \end{equation}$

(2.4)

with the condition on $\alpha$ is

$\begin{equation*} \left \vert \alpha \right \vert \geq \frac{2^{\frac{3}{2}}p\left( A-B\right) }{1-\left \vert B\right \vert -4p\left( 1+\left \vert B\right \vert \right) } . \end{equation*}$

Then $f\in \mathcal{SL}_{p}^{\ast }.$

2.3. Theorem

If $g\in \mathcal{A}_{p}$ such that

$\begin{equation} 1+\frac{\alpha }{p}\frac{zg^{\prime }\left( z\right) }{g\left( z\right) } \prec \frac{1+Az}{1+Bz}, \end{equation}$

(2.5)

with

$\begin{equation} \left \vert \alpha \right \vert \geq \frac{8p\left( A-B\right) }{1-\left \vert B\right \vert -4p\left( 1+\left \vert B\right \vert \right) }, \end{equation}$

(2.6)

then

$\begin{equation*} \frac{g\left( z\right) }{z^{p}}\prec \sqrt{1+z}. \end{equation*}$

Proof

Let us choose a function $p$ by

$\begin{equation*} p\left( z\right) = 1+\alpha \frac{zg^{\prime }\left( z\right) }{pg\left( z\right) }, \end{equation*}$

in such a way that $p$ is analytic in $\mathfrak{D}$ with $p(0) = 1.$ Also consider

$\begin{equation*} \frac{g\left( z\right) }{z^{p}} = \sqrt{1+w\left( z\right) }. \end{equation*}$

Using some simple calculations, we obtain

$\begin{equation*} p\left( z\right) = 1+\frac{\alpha zw^{\prime }\left( z\right) }{2p\left( 1+w\left( z\right) \right) }+\alpha , \end{equation*}$

and so

$\begin{eqnarray*} \left \vert \frac{p\left( z\right) -1}{A-Bp\left( z\right) }\right \vert & = &\left \vert \frac{\frac{\alpha zw^{\prime }\left( z\right) }{2p\left( 1+w\left( z\right) \right) }+\alpha }{A-B\left( 1+\frac{\alpha zw^{\prime }\left( z\right) }{2p\left( 1+w\left( z\right) \right) }+\alpha \right) } \right \vert \\ & = &\left \vert \frac{\alpha zw^{\prime }\left( z\right) +2p\alpha \left( 1+w\left( z\right) \right) }{2p\left( A-B\right) \left( 1+w\left( z\right) \right) -B\left( \alpha zw^{\prime }\left( z\right) +2p\alpha \left( 1+w\left( z\right) \right) \right) }\right \vert . \end{eqnarray*}$

Let a point $z_{0}\in \mathfrak{D}$ exists in such a way

Then, by virtue of Lemma 1.1, a number $m\geq 1$ occurs such that $\ z_{0}w^{\prime }\left(z_{0}\right) = mw\left(z_{0}\right)$ . In addition, we set $w\left(z_{0}\right) = e^{i\theta },$ so we have

$\begin{eqnarray*} \left \vert \frac{p\left( z_{0}\right) -1}{A-Bp\left( z_{0}\right) }\right \vert & = &\left \vert \frac{\alpha mw\left( z_{0}\right) +2p\alpha \left( 1+w\left( z_{0}\right) \right) }{2p\left( A-B\right) \left( 1+w\left( z_{0}\right) \right) -B\left( \alpha mw\left( z_{0}\right) +2p\alpha \left( 1+w\left( z_{0}\right) \right) \right) }\right \vert , \\ &\geq &\frac{\left \vert \alpha \right \vert m-2p\left \vert \alpha \right \vert \left \vert 1+e^{i\theta }\right \vert }{2\left( A-B\right) \left \vert 1+e^{i\theta }\right \vert +\left \vert B\right \vert \left \vert \alpha \right \vert m+2p\left \vert B\right \vert \left \vert \alpha \right \vert \left \vert 1+e^{i\theta }\right \vert }, \\ & = &\frac{\left \vert \alpha \right \vert m-2p\left \vert \alpha \right \vert \sqrt{2+2\cos \theta }}{2\left( 2\left( A-B\right) +\left \vert B\right \vert \left \vert \alpha \right \vert \right) p\sqrt{2+2\cos \theta }+\left \vert B\right \vert \left \vert \alpha \right \vert m}, \\ &\geq &\frac{\left \vert \alpha \right \vert \left( m-4p\right) }{4p\left( 2\left( A-B\right) +\left \vert B\right \vert \left \vert \alpha \right \vert \right) +\left \vert B\right \vert \left \vert \alpha \right \vert m}. \end{eqnarray*}$

Now let

$\begin{equation*} \phi \left( m\right) = \frac{\left \vert \alpha \right \vert \left( m-4p\right) }{4p\left( 2\left( A-B\right) +\left \vert B\right \vert \left \vert \alpha \right \vert \right) +\left \vert B\right \vert \left \vert \alpha \right \vert m}, \end{equation*}$

it implies

$\begin{equation*} \phi ^{\prime }\left( m\right) = \frac{\left \vert \alpha \right \vert 8p\left( \left( A-B\right) +\left \vert \alpha \right \vert \left \vert B\right \vert \right) }{\left( 4p\left( 2\left( A-B\right) +\left \vert B\right \vert \left \vert \alpha \right \vert \right) +\left \vert B\right \vert \left \vert \alpha \right \vert m\right) ^{2}} \gt 0, \end{equation*}$

which illustrates that the function $\phi \left(m\right)$ is increasing and so $\phi \left(m\right) \geq \phi \left(1\right)$ for $m\geq 1,$ hence

$\begin{equation*} \left \vert \frac{p\left( z_{0}\right) -1}{A-Bp\left( z_{0}\right) }\right \vert \geq \frac{\left \vert \alpha \right \vert \left( 1-4p\right) }{ 4p\left( 2\left( A-B\right) +\left \vert B\right \vert \left \vert \alpha \right \vert \right) +\left \vert B\right \vert \left \vert \alpha \right \vert }. \end{equation*}$

Now, by using $\left(2.6\right),$ we have

$\begin{equation*} \left \vert \frac{p\left( z_{0}\right) -1}{A-Bp\left( z_{0}\right) }\right \vert \geq 1, \end{equation*}$

which contradicts $\left(2.5\right).$ Thus $\left \vert w\left(z\right) \right \vert < 1$ and so the desired proof is completed.

Putting $g\left(z\right) = \frac{z^{p+1}f^{\prime }\left(z\right) }{pf(z)}$ in last Theorem, we get the following Corollary:

2.4. Corollary

If $f\in \mathcal{A}_{p}$ and satisfying

$\begin{equation*} 1+\frac{\alpha }{p}\left( p+1+\frac{zf^{\prime \prime }\left( z\right) }{ f^{\prime }\left( z\right) }-\frac{zf^{\prime }\left( z\right) }{f\left( z\right) }\right) \prec \frac{1+Az}{1+Bz}, \end{equation*}$

with

$\begin{equation*} \left \vert \alpha \right \vert \geq \frac{8p\left( A-B\right) }{1-\left \vert B\right \vert -4p\left( 1+\left \vert B\right \vert \right) }, \end{equation*}$

then $f\in \mathcal{SL}_{p}^{\ast }.$

2.5. Theorem

If $g\in \mathcal{A}_{p}$ and satisfy the subordination relation

$\begin{equation} 1+\alpha \frac{z^{1-p}g^{\prime }\left( z\right) }{p\left( g\left( z\right) \right) ^{2}}\prec \frac{1+Az}{1+Bz}, \end{equation}$

(2.7)

with the condition on $\alpha$

$\begin{equation} \left \vert \alpha \right \vert \geq \frac{2^{\frac{5}{2}}p\left( A-B\right) }{1-\left \vert B\right \vert -4p\left( 1+\left \vert B\right \vert \right) } \end{equation}$

(2.8)

is true, then

$\begin{equation*} \frac{g\left( z\right) }{z^{p}}\prec \sqrt{1+z}. \end{equation*}$

Proof

Let us define a function

$\begin{equation*} p\left( z\right) = 1+\alpha \frac{z^{1-p}g^{\prime }\left( z\right) }{p\left( g\left( z\right) \right) ^{2}}. \end{equation*}$

Then $p$ is analytic in $\mathfrak{D}$ with $p(0) = 1.$ Also let us consider

$\begin{equation*} \frac{g\left( z\right) }{z^{p}} = \sqrt{1+w\left( z\right) }. \end{equation*}$

Using some simplification, we obtain

$\begin{equation*} p\left( z\right) = 1+\frac{\alpha zw^{\prime }\left( z\right) }{2p\left( 1+w\left( z\right) \right) ^{\frac{3}{2}}}+\frac{\alpha }{\sqrt{1+w\left( z\right) }}, \end{equation*}$

and so

$\begin{eqnarray*} \left \vert \frac{p\left( z\right) -1}{A-Bp\left( z\right) }\right \vert & = &\left \vert \frac{\frac{\alpha zw^{\prime }\left( z\right) }{2p\left( 1+w\left( z\right) \right) ^{\frac{3}{2}}}+\frac{\alpha }{\sqrt{1+w\left( z\right) }}}{A-B\left( 1+\frac{\alpha zw^{\prime }\left( z\right) }{2p\left( 1+w\left( z\right) \right) ^{\frac{3}{2}}}+\frac{\alpha }{\sqrt{1+w\left( z\right) }}\right) }\right \vert \\ & = &\left \vert \frac{\alpha zw^{\prime }\left( z\right) +2p\alpha \left( 1+w\left( z\right) \right) }{2p\left( A-B\right) \left( 1+w\left( z\right) \right) ^{\frac{3}{2}}-B\alpha zw^{\prime }\left( z\right) -2p\alpha B\left( 1+w\left( z\right) \right) }\right \vert . \end{eqnarray*}$

Let us choose a point $z_{0}\in \mathfrak{D}$ such a way that

Then, by the consequences of Lemma 1.1, a number $m\geq 1$ occurs such that $z_{0}w^{\prime }\left(z_{0}\right) = mw\left(z_{0}\right)$ and also put $w\left(z_{0}\right) = e^{i\theta },$ for $\theta \in \left[ -\pi, \pi \right],$ we have

$\begin{eqnarray*} \left \vert \frac{p\left( z_{0}\right) -1}{A-Bp\left( z_{0}\right) }\right \vert & = &\left \vert \frac{\alpha mw\left( z_{0}\right) +2p\alpha \left( 1+w\left( z_{0}\right) \right) }{2p\left( A-B\right) \left( 1+w\left( z_{0}\right) \right) ^{\frac{3}{2}}-B\alpha mw\left( z_{0}\right) -2p\alpha B\left( 1+w\left( z_{0}\right) \right) }\right \vert , \\ &\geq &\frac{\left \vert \alpha \right \vert m-2p\left \vert \alpha \right \vert \left \vert 1+e^{i\theta }\right \vert }{2p\left( A-B\right) \left \vert 1+e^{i\theta }\right \vert ^{\frac{3}{2}}+\left \vert B\right \vert \left \vert \alpha \right \vert m+2p\left \vert \alpha \right \vert \left \vert B\right \vert \left \vert 1+e^{i\theta }\right \vert }, \\ & = &\frac{\left \vert \alpha \right \vert m-4p\left \vert \alpha \right \vert }{2^{\frac{5}{2}}p\left( A-B\right) +\left \vert B\right \vert \left \vert \alpha \right \vert m+4p\left \vert \alpha \right \vert \left \vert B\right \vert }, \\ &\geq &\frac{\left \vert \alpha \right \vert \left( m-4p\right) }{2^{\frac{5 }{2}}p\left( A-B\right) +\left \vert B\right \vert \left \vert \alpha \right \vert m+4p\left \vert \alpha \right \vert \left \vert B\right \vert } = \phi \left( m\right) \text{ (say).} \end{eqnarray*}$

Then

$\begin{equation*} \phi ^{\prime }\left( m\right) = \frac{2^{\frac{5}{2}}p\left( A-B\right) +8\left \vert \alpha \right \vert ^{2}\left \vert B\right \vert p}{\left( 2^{ \frac{5}{2}}p\left( A-B\right) +B\left \vert \alpha \right \vert m+4p\left \vert \alpha \right \vert B\right) ^{2}} \gt 0, \end{equation*}$

which demonstrates that the function $\phi \left(m\right)$ is increasing and thus $\phi \left(m\right) \geq \phi \left(1\right)$ for $m\geq 1,$ hence

$\begin{equation*} \left \vert \frac{p\left( z_{0}\right) -1}{A-Bp\left( z_{0}\right) }\right \vert \geq \frac{\left \vert \alpha \right \vert \left( 1-4p\right) }{2^{ \frac{5}{2}}p\left( A-B\right) +\left \vert B\right \vert \left \vert \alpha \right \vert +4p\left \vert \alpha \right \vert \left \vert B\right \vert }. \end{equation*}$

Now, using $\left(2.8\right),$ we have

$\begin{equation*} \left \vert \frac{p\left( z_{0}\right) -1}{A-Bp\left( z_{0}\right) }\right \vert \geq 1, \end{equation*}$

which contradicts $\left(2.7\right)$ . Thus $\left \vert w\left(z\right) \right \vert < 1$ and so we get the required proof.

If we set $g\left(z\right) = \frac{z^{p+1}f^{\prime }\left(z\right) }{pf(z)}$ in last theorem, we easily have the following Corollary:

2.6. Corollary

Assume that

$\begin{equation*} \left \vert \alpha \right \vert \geq \frac{2^{\frac{5}{2}}p\left( A-B\right) }{1-\left \vert B\right \vert -4p\left( 1+\left \vert B\right \vert \right) } , \end{equation*}$

and if $f\in \mathcal{A}_{p}$ satisfy

$\begin{equation*} 1+\frac{\alpha f\left( z\right) }{z^{2p+1}f^{\prime }\left( z\right) }\left( p+1+\frac{zf^{\prime \prime }\left( z\right) }{f^{\prime }\left( z\right) }- \frac{zf^{\prime }\left( z\right) }{f\left( z\right) }\right) \prec \frac{ 1+Az}{1+Bz}, \end{equation*}$

then $f\in \mathcal{SL}_{p}^{\ast }.$

2.7. Theorem

If $g\in \mathcal{A}_{p}$ satisfy the subordination

$\begin{equation*} 1+\alpha \frac{z^{1-2p}g^{\prime }\left( z\right) }{p\left( g\left( z\right) \right) ^{3}}\prec \frac{1+Az}{1+Bz}, \end{equation*}$

with restriction on $\alpha$ is

$\begin{equation} \left \vert \alpha \right \vert \geq \frac{8p\left( A-B\right) }{1-\left \vert B\right \vert -4p\left( 1+\left \vert B\right \vert \right) }. \end{equation}$

(2.9)

then

$\begin{equation*} \frac{g\left( z\right) }{z^{p}}\prec \sqrt{1+z}. \end{equation*}$

Proof. Let us define a function

$\begin{equation*} p\left( z\right) = 1+\alpha \frac{z^{1-2p}g^{\prime }\left( z\right) }{ p\left( g\left( z\right) \right) ^{3}}, \end{equation*}$

where $p$ is analytic in $\mathfrak{D}$ with $p(0)-1 = 0.$ Also let

$\begin{equation*} \frac{g\left( z\right) }{z^{p}} = \sqrt{1+w\left( z\right) }. \end{equation*}$

Using some simple calculations, we obtain

$\begin{equation*} p\left( z\right) = 1+\frac{\alpha zw^{\prime }\left( z\right) }{2p\left( 1+w\left( z\right) \right) ^{2}}+\frac{\alpha }{1+w\left( z\right) }, \end{equation*}$

and so

$\begin{eqnarray*} \left \vert \frac{p\left( z\right) -1}{A-Bp\left( z\right) }\right \vert & = &\left \vert \frac{\frac{\alpha zw^{\prime }\left( z\right) }{2p\left( 1+w\left( z\right) \right) ^{2}}+\frac{\alpha }{1+w\left( z\right) }}{ A-B\left( 1+\frac{\alpha zw^{\prime }\left( z\right) }{2p\left( 1+w\left( z\right) \right) ^{2}}+\frac{\alpha }{1+w\left( z\right) }\right) }\right \vert \\ & = &\left \vert \frac{\alpha zw^{\prime }\left( z\right) +2p\alpha \left( 1+w\left( z\right) \right) }{2p\left( A-B\right) \left( 1+w\left( z\right) \right) ^{2}-B\alpha zw^{\prime }\left( z\right) -2p\alpha B\left( 1+w\left( z\right) \right) }\right \vert . \end{eqnarray*}$

Let us pick a point $z_{0}\in \mathfrak{D}$ in such a way that

Then, by using Lemma 1.1 $,$ a number $m\geq 1$ exists such that $z_{0}w^{\prime }\left(z_{0}\right) = mw\left(z_{0}\right) \$ and put $w\left(z_{0}\right) = e^{i\theta },$ for $\theta \in \left[ -\pi, \pi \right]$ , we have

$\begin{eqnarray*} \left \vert \frac{p\left( z_{0}\right) -1}{A-Bp\left( z_{0}\right) }\right \vert & = &\left \vert \frac{\alpha mw\left( z_{0}\right) +2p\alpha \left( 1+w\left( z_{0}\right) \right) }{2p\left( A-B\right) \left( 1+w\left( z_{0}\right) \right) ^{2}-B\alpha mw\left( z_{0}\right) -2p\alpha B\left( 1+w\left( z_{0}\right) \right) }\right \vert \\ &\geq &\frac{\left \vert \alpha \right \vert m-2p\left \vert \alpha \right \vert \left \vert 1+e^{i\theta }\right \vert }{2p\left( A-B\right) \left \vert 1+e^{i\theta }\right \vert ^{2}+\left \vert B\right \vert \left \vert \alpha \right \vert m+2p\left \vert \alpha \right \vert \left \vert B\right \vert \left \vert 1+e^{i\theta }\right \vert } \\ & = &\frac{\left \vert \alpha \right \vert m-2p\left \vert \alpha \right \vert \sqrt{2+2\cos \theta }}{2p\left( A-B\right) \left( \sqrt{2+2\cos \theta } \right) ^{2}+\left \vert B\right \vert \left \vert \alpha \right \vert m+2p\left \vert \alpha \right \vert \left \vert B\right \vert \sqrt{2+2\cos \theta }} \\ &\geq &\frac{\left \vert \alpha \right \vert \left( m-4p\right) }{8p\left( A-B\right) +\left \vert B\right \vert \left \vert \alpha \right \vert m+4p\left \vert \alpha \right \vert \left \vert B\right \vert }, \end{eqnarray*}$

Now let

$\begin{equation*} \phi \left( m\right) = \frac{\left \vert \alpha \right \vert \left( m-4p\right) }{8p\left( A-B\right) +\left \vert B\right \vert \left \vert \alpha \right \vert m+4p\left \vert \alpha \right \vert \left \vert B\right \vert }, \end{equation*}$

then

$\begin{equation*} \phi ^{\prime }\left( m\right) = \frac{8p\left \vert \alpha \right \vert \left( A-B\right) +8\left \vert \alpha \right \vert ^{2}\left \vert B\right \vert p}{\left( 8p\left( A-B\right) +\left \vert B\right \vert \left \vert \alpha \right \vert m+4p\left \vert \alpha \right \vert \left \vert B\right \vert \right) ^{2}} \gt 0 \end{equation*}$

which shows that $\phi \left(m\right)$ is an increasing function and hence it will have its minimum value at $m = 1$ , so

$\begin{equation*} \left \vert \frac{p\left( z_{0}\right) -1}{A-Bp\left( z_{0}\right) }\right \vert \geq \frac{\left \vert \alpha \right \vert \left( 1-4p\right) }{ 8p\left( A-B\right) +\left \vert B\right \vert \left \vert \alpha \right \vert +4p\left \vert \alpha \right \vert \left \vert B\right \vert }. \end{equation*}$

Using $\left(2.9\right),$ we easily obtain

$\begin{equation*} \left \vert \frac{p\left( z_{0}\right) -1}{A-Bp\left( z_{0}\right) }\right \vert \geq 1, \end{equation*}$

which is a contradiction to the fact that $p\left(z\right) \prec \frac{1+Az }{1+Bz},$ and so $\left \vert w\left(z\right) \right \vert < 1.$ Hence we get the desired result.

If we put $g\left(z\right) = \frac{z^{p+1}f^{\prime }\left(z\right) }{pf(z)}$ in last Theorem, we achieve the following result:

2.8. Corollary

If $f\in \mathcal{A}_{p}$ and satisfy the condition

$\begin{equation*} \left \vert \alpha \right \vert \geq \frac{8p\left( A-B\right) }{1-\left \vert B\right \vert -4p\left( 1+\left \vert B\right \vert \right) }, \end{equation*}$

and

$\begin{equation*} 1+\alpha \frac{p\left( f\left( z\right) \right) ^{2}}{z^{3p+2}\left( f^{\prime }\left( z\right) \right) ^{2}}\left( p+1+\frac{zf^{\prime \prime }\left( z\right) }{f^{\prime }\left( z\right) }-\frac{zf^{\prime }\left( z\right) }{f\left( z\right) }\right) \prec \frac{1+Az}{1+Bz}, \end{equation*}$

then $f\in \mathcal{SL}_{p}^{\ast }.$

Conflict of interest

All authors declare no conflict of interest in this paper.

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