# Leveraging Multimodal Features and Item-level User Feedback for Bundle
Construction

Yunshan Ma  National University of Singapore<EMAIL_ADDRESS>,  Xiaohao
Liu  University of Chinese Academy of Sciences<EMAIL_ADDRESS>,
Yinwei Wei  Monash University<EMAIL_ADDRESS>,  Zhulin Tao
Communication University of China<EMAIL_ADDRESS>,  Xiang Wang
University of Science and Technology of China<EMAIL_ADDRESS>and
Tat-Seng Chua  National University of Singapore<EMAIL_ADDRESS>
###### Abstract.

Automatic bundle construction is a crucial prerequisite step in various
bundle-aware online services. Previous approaches are mostly designed to model
the bundling strategy of existing bundles. However, it is hard to acquire
large-scale well-curated bundle dataset, especially for those platforms that
have not offered bundle services before. Even for platforms with mature bundle
services, there are still many items that are included in few or even zero
bundles, which give rise to sparsity and cold-start challenges in the bundle
construction models. To tackle these issues, we target at leveraging
multimodal features, item-level user feedback signals, and the bundle
composition information, to achieve a comprehensive formulation of bundle
construction. Nevertheless, such formulation poses two new technical
challenges: 1) how to learn effective representations by optimally unifying
multiple features, and 2) how to address the problems of modality missing,
noise, and sparsity problems induced by the incomplete query bundles. In this
work, to address these technical challenges, we propose a Contrastive
Learning-enhanced Hierarchical Encoder method (CLHE). Specifically, we use
self-attention modules to combine the multimodal and multi-item features, and
then leverage both item- and bundle-level contrastive learning to enhance the
representation learning, thus to counter the modality missing, noise, and
sparsity problems. Extensive experiments on four datasets in two application
domains demonstrate that our method outperforms a list of SOTA methods. The
code and dataset are available at https://github.com/Xiaohao-Liu/CLHE.

Bundle Construction, Multimodal Modeling, Contrastive Learning

††conference: March 4–8, 2023; Mérida, México;

##  1\. Introduction

Product bundling has been a popular and effective marketing strategy, tracing
back from ancient commercial times and persisting through to the rapidly
growing e-commerce and online services today. By combining a set of individual
items into a bundle, both the sellers (or service providers) and consumers can
benefit a lot from multiple aspects, including the reduced cost of packaging,
shipment, and installation, to promoting sales of old or new items by
combining them with some popular or essential items with discounts. To
implement product bundling, the first and foremost step is constructing
bundles from individual items, _aka._ bundle construction, which is
traditionally carried out by human experts. However, the explosive growth of
item sets poses significant challenges to such high-cost manual approaches.
Hence, automatic approaches to bundle construction are imperative and have
garnered more and more attention in recent years.

Figure 1. The motivations of leveraging multimodal features and item-level
user feedback for bundle construction.

By analyzing prior studies, we find that they mostly build the bundles based
on the co-occurrence relationship of items in existing training bundles.
However, there are two key problems that have not been well studied: 1)
previous approaches heavily rely on large-scale high-quality bundle dataset
for training, and 2) they cannot properly handle the sparsity and cold-start
issues. First, most previous bundle construction methods require high-quality
supervision signals from a large set of well-curated bundles. However, there
is a dilemma in such an approach especially for platforms that have not
offered bundle service before or have just deployed bundle service for a short
period of time, it is difficult for such platforms to collect sufficient
bundle data for training. Second, even for platforms with mature bundle
services, the situation is far from ideal due to the various cold-start
problems. On the one hand, there are quite a number of items that are only
involved in a few bundles, consequently, it is challenging to obtain
informative representations for these sparse items to construct new bundles.
Worse still, there many new items, which haven’t been part of previous bundles
while are continuously pushed online, and how to swiftly bundle these cold-
start items with existing warm items is crucial for platforms to promote new
products and keep sustained growth.

Addressing these challenges, instead of seeking any silver bullet, we are more
keen on practical solutions that make full use of the large amount of easy-to-
access resources: multimodal features and item-level user feedback. The
motivation behind this solution is that these data are well aligned to diverse
bundling strategies.  First, multimodal features, such as text, image, and
audio, contain rich semantic information that is helpful to find either
similar or compatible items and form bundles, as shown in Figure 1. More
importantly, most items, even those sparse and newly introduced items, usually
have one or multiple such features. A plethora of previous efforts, such as
personalized recommendation (Wei et al., 2019), have demonstrated the efficacy
of multimodal features in handling sparse and cold-start items. Second, item-
level user feedback information endows precious crowd-sourcing knowledge that
is crucial to bundle construction. Intuitively, the items that users
frequently co-interact with are strong candidates for bundling. More
importantly, a large amount of such user feedback signals are available even
to platforms that do not offer bundle services. Compared with previous works
(Guan et al., 2021), we pioneer the integration of multimodal features and
item-level user feedback for bundle construction.

Given the outlined motivations, we aim to leverage both multimodal features
and item-level user feedback, along with the existing bundles, to develop a
comprehensive model for bundle construction. However, it is non-trivial to
design a model to capture all three types of information and achieve optimal
bundle construction performance. First, how to learn effective representations
in each modality and well capture the cooperative association among the three
modalities is a key challenge. Second, some items might not be associated with
user feedback or affiliated to bundles comprehensively, thus the so-called
modality-missing issue may degrade the modeling capability. What’s more,
during the inference stage of bundle construction, we usually need to provide
several seed items as a partial bundle to initiate the construction process.
However, the incompleteness of the partial bundle imposes noise and sparsity
challenges to the bundle representation learning, which will impede the bundle
construction performance.

In this work, to address the aforementioned challenges, we propose a
Contrastive Learning-enhanced Hierarchical Encoder (CLHE) for bundle
construction. In order to obtain the representations of items, we make use of
the recently proposed large-scale multimodal foundation models (_i.e.,_ BLIP
(Li et al., 2022b) and CLAP (Wu et al., 2022)) to extract the multimodal
features of items. Concurrently, we pre-train a collaborative filtering
(CF)-based model (_i.e.,_ LightGCN (He et al., 2020a)) to obtain the items’
representations that preserve the user feedback information. Then, we employ a
hierarchical encoder to learn the bundle representation, where the self-
attention mechanism is devised to expertly fuse multimodal information and
multi-item representations. To tackle the modality missing problem and the
sparsity/noise issues induced by the incomplete partial bundle, we employ two
levels of contrastive learning (Wu et al., 2021; Ma et al., 2022), _i.e.,_
item-level and bundle-level, to fully take advantage of the self-supervision
signals. We conduct experiments on four datasets from two domains, and the
results demonstrate that our method outperforms multiple leading methods.
Various ablation and modal studies further justify the effectiveness of key
modules and demonstrate multiple crucial properties of our proposed model. We
summarize the key contributions of this work as follows:

  * •

We introduce a pioneering approach to bundle construction by holistically
combining multimodal features, item-level user feedback, and existing bundles.
This integration addresses prevailing challenges such as data insufficiency
and the cold-start problem.

  * •

We highlight multiple technical challenges of this new formulation and propose
a novel method of CLHE to tackle them.

  * •

Our method outperforms various leading methods on four datasets from two
application domains with different settings, and further diverse studies
demonstrate various merits of our method.

##  2\. Methodology

We first formally define the problem of bundle construction by considering all
three types of data. Then we describe the details of our proposed method CLHE
(as shown in Figure 2).

###  2.1. Problem Formulation

Given a set of items $\mathcal{I}=\\{i_{1},i_{2},\cdots,i_{N}\\}$, each item
has a textual input $t_{i}$, which can be its title, description, or metadata,
and a media input $m_{i}$, which can be an image, audio, or video of the item.
In addition, for the items that have been online for a while, we have
collected some item-level user feedback data, which is denoted as a user-item
interaction matrix $\mathbf{X}_{M\times
N}=\\{x_{ui}|u\in{\mathcal{U}},i\in{\mathcal{I}}\\}$, where
$\mathcal{U}=\\{u_{1},u_{2},\cdots,u_{M}\\}$ is the user set. We define a
bundle as a set of items, denoted as $b=\\{i_{1},i_{2},\cdots,i_{n}\\}$, where
$n=\lvert b\rvert$ is the size of the bundle. Given a partial bundle
$b_{s}\subset b$ (_i.e.,_ a set of seed items), where $\lvert
b_{s}\rvert<\lvert b\rvert$, the bundle construction model targets at predict
the missing items $i\in b\setminus b_{s}$. We have a set of known bundles for
training, denoted as $\mathcal{B}=\\{b_{1},b_{2},\cdots,b_{O}\\}$ and a set of
unseen bundles for testing, denoted as
$\mathcal{\bar{B}}=\\{b_{O+1},b_{O+2},\cdots,b_{O+\bar{O}}\\}$, where $O$ is
the number of training bundles and $\bar{O}$ is the number of testing bundles.
We would like to train a model based on the training set $\mathcal{B}$, for an
unseen bundle $b\in\mathcal{\bar{B}}$, when given a few seed items
$\bar{b}_{s}$, _aka._ the partial bundle, the model can predict the missing
items $b\setminus b_{s}$ thus to construct the entire bundle.

Figure 2. The overall framework of our proposed method CLHE, which consists of
two main components: hierarchical encoder (_aka._ multimodal feature
extraction, item and bundle representation learning) and contrastive learning.

###  2.2. Hierarchical Encoder

We utilize a hierarchical encoder for multimodal bundle representation.
Initially, we extract multimodal features using multimodal foundation models,
while concurrently pre-training a CF-based model to capture item-level user
feedback. Subsequently, a self-attention encoder is introduced to integrate
these multimodal features, resulting in a fused item representation. Another
self-attention encoder then aggregates these representations, producing a
comprehensive bundle representation.

####  2.2.1. Item Representation Learning

We first detail the feature extraction process and then present the self-
attention encoder.

Multimodal Feature Extraction. We seek large-scale multimodal foundation
models to extract the textual and media features of items. Compared with
previous uni-modal feature extractors, such as in Computer Vision (CV) (He et
al., 2016; Liu et al., 2021a), Natural Language Processing (NLP) (Devlin et
al., 2019; Liu et al., 2019), or audio (Kong et al., 2020; Chen et al., 2022),
multimodal foundation models are more powerful to capture the multimodal
semantics of the input data, which have demonstrated to be effective in
transferring or generalizing to various downstream tasks. Concretely, for
image data, we use the BLIP (Li et al., 2022b) model to extract both textual
and visual features. For audio data, we use the CLAP (Wu et al., 2022) model
to extract textual and audio features. After the feature extraction, we obtain
the textual feature $\mathbf{t}_{i}\in\mathbb{R}^{768}$ and media feature
$\textbf{m}_{i}\in\mathbb{R}^{768}$. Given their shared representation space,
we perform a simple average pooling over them, resulting in the content
feature of the item, denoted as
$\mathbf{c}_{i}=\text{average}(\mathbf{t}_{i},\mathbf{m}_{i})$.

Item-level User Feedback Feature Extraction. We employ the well-performing CF-
based model, _i.e.,_ LightGCN (He et al., 2020a), to obtain item
representations from user feedback. Specifically, we devise a bipartite graph
based on the user-item interaction matrix, then train a LightGCN 111Other CF-
based models can also be used. model over the bipartite graph, denoted as:

(1) $\left\\{\begin{aligned}
\mathbf{p}_{u}^{(k)}&=\sum_{i\in\mathcal{N}_{u}}{\frac{1}{\sqrt{|\mathcal{N}_{u}|}\sqrt{|\mathcal{N}_{i}|}}\mathbf{p}^{(k-1)}_{i}},\\\
\mathbf{p}_{i}^{(k)}&=\sum_{u\in\mathcal{N}_{i}}{\frac{1}{\sqrt{|\mathcal{N}_{i}|}\sqrt{|\mathcal{N}_{u}|}}\mathbf{p}^{(k-1)}_{u}},\end{aligned}\right.$

where $\mathbf{p}_{u}^{(k)},\mathbf{p}_{i}^{(k)}\in\mathbb{R}^{d}$ are
embeddings for user $u$ and item $i$ at the $k$-th layer, and $d$ is the
dimensionality of the hidden representation; $\mathcal{N}_{u}$ and
$\mathcal{N}_{i}$ are the neighbors of the user $u$ and item $i$ in the user-
item interaction graph. We only make use of the item representation
$\mathbf{p}_{i}$, which captures the item-level user feedback information. It
is tailored by aggregating the item representations over $K$ layers’
propagation, denoted as:

(2) $\mathbf{p}_{i}=\frac{1}{K}\sum^{K}_{k=0}{\mathbf{p}^{(k)}_{i}}.$

ID Embedding Initialization. We also initialize an id embedding
$\mathbf{v}_{i}\in\mathbb{R}^{d}$ for each item to capture its bundle-item
affiliation patterns. Please note that for those items (both during training
and testing) that do not have user feedback features, we copy the content
feature to its user feedback feature slot. Analogously, for the cold-start
item that do not have an id embedding, we copy its corresponding content
feature to take the slot.

Modality Fusion via Self-attention. Given the three types of features, _i.e.,_
$\mathbf{c}_{i}$, $\mathbf{p}_{i}$, and $\mathbf{v}_{i}$, we first apply a
feature transformation layer to project the multimodal and user-feedback
features into the same latent space with the id embeddings, then we
concatenate all the three features into a feature matrix
$\mathbf{F}_{i}\in\mathbb{R}^{3\times d}$, denoted as:

(3)
$\mathbf{F}_{i}=\text{concat}(\mathbf{c}_{i}\mathbf{W}_{c},\mathbf{p}_{i}\mathbf{W}_{p},\mathbf{v}_{i}),$

where $\mathbf{W}_{c}\in\mathbb{R}^{768\times d}$ and
$\mathbf{W}_{p}\in\mathbb{R}^{768\times d}$ are the transformation matrices
for multimodal and user-feedback features, respectively;
$\text{concat}(\cdot)$ is the concatenation function. Then, we devise a self-
attention layer to model the correlations of multiple features, denoted as:

(4) $\left\\{\begin{aligned}
\mathbf{A}_{i}^{(l)}&=\frac{1}{\sqrt{d}}\mathbf{\hat{F}}_{i}^{(l-1)}\mathbf{W}_{I}^{K}\bigl{(}\mathbf{\hat{F}}_{i}^{(l-1)}\mathbf{W}_{I}^{Q}\bigr{)}^{\intercal},\\\
\mathbf{\tilde{F}}_{i}^{(l)}&=\text{softmax}\bigl{(}\mathbf{A}_{i}^{(l)}\bigr{)}\mathbf{\hat{F}}_{i}^{l-1},\end{aligned}\right.$

where $\mathbf{W}_{I}^{K}\in\mathbb{R}^{d\times d}$ and
$\mathbf{W}_{I}^{Q}\in\mathbb{R}^{d\times d}$ are the trainable parameters for
this item-level encoder to project the input feature embeddings into the key
and value spaces; $\mathbf{\hat{F}}_{i}^{(l)}\in\mathbb{R}^{3\times d}$ is the
hidden feature representations in the intermediate layer $l$, and
$\mathbf{\hat{F}}_{i}^{(0)}=\mathbf{F}_{i}$; $\text{softmax}(\cdot)$ is the
softmax function and $\mathbf{\tilde{F}}_{i}^{L}$ denotes the features’
representations after $L$ layers of self-attention. We then average the
multiple features to obtain the item representation
$\mathbf{f}_{i}\in\mathbb{R}^{d}$ after multimodal fusion, formally defined
as:

(5)
$\mathbf{f}_{i}=\text{average}\bigl{(}{\mathbf{\tilde{F}}_{i}^{(L)}}\bigr{)}.$

####  2.2.2. Bundle Representation Learning

After obtaining the item representation, we build a second self-attention
module to learn the representation of the given partial bundle. For a certain
partial bundle $b_{s}$, its representation $\mathbf{e}_{b_{s}}$ 222For
simplicity, we omit the subscript $s$ in $b_{s}$ and just use $b$ and
$\mathbf{e}_{b}$ to represent the partial bundle and its representation if
there is no ambiguity. is learned by:

(6) $\left\\{\begin{aligned}
\mathbf{A}_{b}^{(z)}&=\frac{1}{\sqrt{d}}\mathbf{\hat{E}}_{b}^{(z-1)}\mathbf{W}_{B}^{K}\bigl{(}\mathbf{\hat{E}}_{b}^{(z-1)}\mathbf{W}_{B}^{Q}\bigr{)}^{\intercal},\\\
\mathbf{\tilde{E}}_{b}^{(z)}&=\text{softmax}\bigl{(}\mathbf{A}_{b}^{(z)}\bigr{)}\mathbf{\hat{E}}_{b}^{z-1},\end{aligned}\right.$

where $\mathbf{W}_{B}^{K}\in\mathbb{R}^{d\times d}$ and
$\mathbf{W}_{B}^{Q}\in\mathbb{R}^{d\times d}$ are the trainable parameters in
the bundle-level to project the input item embeddings into the key and value
spaces; $\mathbf{\hat{E}}_{b}^{(z)}\in\mathbb{R}^{\lvert b\rvert\times d}$ is
the hidden representations in the middle layer $z$, and
$\mathbf{\hat{E}}_{b}^{(0)}=\text{concat}(\\{\mathbf{f}_{i}\\}_{i\in b})$;
$\mathbf{\tilde{E}}_{b}^{Z}$ denotes the features’ representations after $Z$
layers of self-attention. We then average the multiple features to obtain the
item representation $\mathbf{e}_{b}$ after multimodal fusion, formally defined
as:

(7)
$\mathbf{e}_{b}=\text{average}\bigl{(}{\mathbf{\tilde{E}}_{b}^{(Z)}}\bigr{)}.$

Table 1. The statistics of the four datasets on two different domains.

Dataset | #U | #I | #B | #B-I | #U-I | #Avg.I/B | #Avg.B/I | #Avg.I/U | #Avg.U/I
---|---|---|---|---|---|---|---|---|---
POG | 17,449 | 48,676 | 20,000 | 72,224 | 237,519 | 3.61 | 1.48 | 13.61 | 4.88
POG_dense | 2,311,431 | 31,217 | 29,686 | 105,775 | 6,345,137 | 3.56 | 3.39 | 2.75 | 203.26
Spotify | 118,994 | 254,155 | 20,000 | 1,268,716 | 36,244,806 | 63.44 | 4.99 | 304.59 | 142.61
Spotify_sparse | 118,899 | 213,325 | 12,486 | 549,900 | 32,890,315 | 44.04 | 2.58 | 276.62 | 154.18

###  2.3. Contrastive Learning

Even though the hierarchical encoder can well attain the correlations among
multiple features and multiple items, it still suffers from noise, sparsity,
or even cold-start problems in both item and bundle levels. Specifically, at
the item level, the items that have fewer user feedbacks or are involved in
fewer bundles during training may also be prone to deteriorate
representations, which is the so-called sparsity issue. Even worse, some cold-
start items may have never interacted with any users or been included in any
bundles before, therefore, the cold-start problem will severely deteriorate
the representation quality. Second, at the bundle level, the partial bundle’s
representation is susceptible to noise and sparsity issues. Instead of a
complete bundle that is sufficient to depict all the functionalities or
properties of the bundle, the given partial bundle only encompasses some of
the items. Consequently, the bundle representation may be biased due to the
arbitrary seed items.

To tackle these problems, we aim to harness contrastive learning over both
item and bundle levels to mine the self-supervision signals. Recently,
contrastive learning has achieved great success in various tasks, including CV
(Chen et al., 2020), NLP (Gao et al., 2021), and recommender systems (Wu et
al., 2021). The main idea is to first corrupt the original data and generate
some augmented views for the same data point, and then leverage an InfoNCE
loss to pull close the representations across multiple augmented views for the
same data point, while pushing away the representations of different data
points. Therefore, the representations could be more robust to combat noise
and sparsity.

####  2.3.1. Item-level Contrastive Learning

For each item $i$, we tailor its representation $\mathbf{f}_{i}$ in Equation
5. We leverage various data augmentations to generate the augmented view
$\mathbf{f}_{i}^{\prime}$. The item-level data augmentation methods we used
include: 1) No Augmentation (NA) (Ma et al., 2022): just use the original
representation as the augmented feature without any augmentation; 2) Feature
Noise (FN) (Yu et al., 2022b): add a small-scaled random noise vector to the
item’s features; 3) Feature Dropout (FD) (Wu et al., 2021): randomly dropout
some values over the feature vectors; and 4) Modality Dropout (MD): dropout
the whole feature of a randomly selected modality on a randomly selected item.
Then, we use the InfoNCE (Ma et al., 2022) to generate the item-level
contrastive loss, denoted as:

(8)
$\mathcal{L}^{C}_{I}=\frac{1}{|\mathcal{I}|}\sum_{i\in\mathcal{I}}{-\text{log}\frac{\text{exp}({\text{cos}(\mathbf{f}_{i},\mathbf{f}_{i}^{\prime})/\tau})}{\sum_{v\in\mathcal{I}}{\text{exp}({\text{cos}(\mathbf{f}_{i},\mathbf{f}_{v}^{\prime})/\tau})}}},$

where $\text{cos}(\cdot)$ is the cosine similarity, and $\tau$ is the
temperature.

####  2.3.2. Bundle-level Contrastive Learning

For each bundle $b$ and its original representation $\mathbf{e}_{b}$, we also
implement various data augmentations to generate an augmented view
$\mathbf{e}_{b}^{\prime}$. The data augmentation methods we leveraged include:
1) Item Dropout (ID): randomly dropout some items in the bundle; and 2) Item
Replacement (IR): randomly select some items in the bundle and replace them
with some other items that have not appear in the bundle. Following on, the
bundle-level contrastive loss is tailored by:

(9)
$\mathcal{L}^{C}_{B}=\frac{1}{|\mathcal{B}|}\sum_{b\in\mathcal{B}}{-\text{log}\frac{\text{exp}({\text{cos}(\mathbf{e}_{b},\mathbf{e}_{b}^{\prime})/\tau})}{\sum_{v\in\mathcal{B}}{\text{exp}({\text{cos}(\mathbf{e}_{b},\mathbf{e}_{v}^{\prime})/\tau})}}}.$

Table 2. The overall performance of our CLHE and baselines. ”%Improv.” denotes
the relative improvement over the strongest baseline. The best baselines are
underlined.

Models | POG | POG_dense | Spotify | Spotify_sparse
---|---|---|---|---
Rec@20 | NDCG@20 | Rec@20 | NDCG@20 | Rec@20 | NDCG@20 | Rec@20 | NDCG@20
MultiDAE | 0.0119 | 0.0063 | 0.3213 | 0.2179 | 0.0578 | 0.0944 | 0.0506 | 0.0574
MultiVAE | 0.0196 | 0.0104 | 0.3221 | 0.2086 | 0.0400 | 0.0656 | 0.0325 | 0.0365
Bi-LSTM | 0.0170 | 0.0097 | 0.2932 | 0.1745 | 0.0833 | 0.1486 | 0.0645 | 0.0822
Hypergraph | 0.0207 | 0.0111 | 0.3063 | 0.2256 | 0.0572 | 0.0941 | 0.0529 | 0.0590
Tranformer | 0.0215 | 0.0114 | 0.3525 | 0.2527 | 0.0875 | 0.1460 | 0.0768 | 0.0902
TranformerCL | 0.0202 | 0.0134 | 0.3170 | 0.2374 | 0.1014 | 0.1696 | 0.0874 | 0.1062
CLHE (ours) | 0.0284 | 0.0193 | 0.3811 | 0.2773 | 0.1081 | 0.1806 | 0.0980 | 0.1212
%Improv. | 32.45 | 44.03 | 8.13 | 9.71 | 6.61 | 6.49 | 12.12 | 14.15

###  2.4. Prediction and Optimization

After obtain the partial bundle representation $\mathbf{e}_{b_{s}}$ and the
item representations $\mathbf{f}_{i}$, we leverage the inner-product function
to induce the score $\hat{y}_{b_{s},i}$ that indicates the possibility of item
$i$ being included into bundle $b$ to make it complete, defined as:

(10) $\hat{y}_{b_{s},i}=\mathbf{e}_{b_{s}}\mathbf{f}_{i}^{\intercal}.$

To optimize our model, we follow the previous approaches (Wu et al., 2016;
Liang et al., 2018) and leverage the negative log-likelihood loss, therefore,
the loss for bundle $b$ is denoted as:

(11)
$\mathcal{L}_{b}=\frac{1}{|\mathcal{I}|}\sum_{i\in\mathcal{I}}{-y_{b_{s},i}\text{log}\pi_{b_{s}}(\hat{y}_{b_{s},i})},$

where $\pi(\cdot)$ is the softmax function which produces the probabilities
over the entire items. In collaboration with the contrastive loss and
regularization, we have the final loss, denoted as:

(12)
$\mathcal{L}=\frac{1}{\lvert\mathcal{B}\rvert}\sum_{b\in\mathcal{B}}{\mathcal{L}_{b}}+\alpha_{1}\mathcal{L}_{I}^{C}+\alpha_{2}\mathcal{L}_{B}^{C}+\beta{\|\mathbf{\Theta}\rVert}_{2}^{2},$

where ${\alpha}_{1}$, ${\alpha}_{2}$ and $\beta$ are hyper-parameters to
balance different loss terms, ${\|\mathbf{\Theta}\rVert}_{2}^{2}$ is the L2
regularization term, and ${\|\mathbf{\Theta}\rVert}$ denotes all the trainable
parameters in our model.

##  3\. Experiments

We evaluate our proposed methods on two application domains of product
bundling, _i.e.,_ fashion outfit and music playlist. We are particularly
interested in answering the research questions as follow:

  * •

RQ1:  Does the proposed CLHE method beat the leading methods?

  * •

RQ2:  Are the key modules, _i.e.,_ hierarchical transformer and contrastive
learning, effective?

  * •

RQ3:  How does out method work in countering the problems of cold-start items,
modality missing, noise and sparsity of the partial bundle? How the detailed
configurations affect its performance and how about the computation
complexity?

###  3.1. Experimental Settings

There are various application scenarios that are suitable for product
bundling, such as e-commerce, travel package, meal, _etc._ , each of which has
one or multiple public datasets. However, only datasets that include all the
multimodal item features, user feedback data, and bundle data can be used to
evaluate our method. Therefore, we choose two representative domains, _i.e.,_
fashion outfit and music playlist. We use the POG (Chen et al., 2019a) for
fashion outfit. For the music playlist, we use the Spotify (Chen et al., 2018)
dataset for the bundle-item affiliations, and we acquire the user feedback
data from the Last.fm dataset (Bertin-Mahieux et al., 2011). Since the average
bundle size is quite small in POG (it makes sense for fashion outfit), we re-
sample a second version POG_dense which has denser user feedback connections
for each item. In contrast, the average bundle size in Spotify dataset is
large, thus we sample a sparser version Spotify_sparse, which has smaller
average bundle size. To be noted, we keep the integrity of all the bundles in
all the versions, which means we do not corrupt any bundles during the
sampling. For each dataset, we randomly split all the bundles into
training/validation/testing set with the ratio of 8:1:1. The statistics of the
datasets are shown in Table 1. We use the popular ranking protocols of
Recall@K and NDCG@K as the evaluation metric, where K=20.

####  3.1.1. Compared Methods

Due to the new formulation of our work, there are no previous works that have
exactly same setting with ours. Therefore, we pick several leading methods and
adapt them to our settings. For fair comparison, all the baseline methods use
all the three types of extracted features that are same with our method. In
addition, they all use the same negative log-likelihood loss function.

  * •

MultiDAE (Wu et al., 2016) is an auto-encoder model which uses an average
pooling to aggregate the items’ representations to get the bundle
representation.

  * •

MultiVAE (Liang et al., 2018) is an variational auto-encoder model which
employ the variational inference on top of the MultiDAE method.

  * •

Bi-LSTM (Han et al., 2017) treats each bundle as a sequence and uses bi-
directional LSTM to learn the bundle representation.

  * •

Hypergraph (Yu et al., 2022a) formulates each bundle a hyper-graph and devises
a GCN model to learn the bundle representation.

  * •

Transformer (Brosh et al., 2022; Wei et al., 2023b) tailors a transformer to
capture the item interactions and generate the bundle representation.

  * •

TransformerCL is the version that we add bundle-level contrastive loss to the
above Transformer model.

####  3.1.2. Hyper-parameter Settings

The embedding and hidden representation size is 64, and we use Xavier (Glorot
and Bengio, 2010) initialization, batch size 2048, and Adam optimizer (Kingma
and Ba, 2014). We find the optimal hyper-parameter setting by adopting grid
search. Wherein, learning rate is searched in range of $\\{10^{-2},2\times
10^{-2},10^{-3},2\times 10^{-3},10^{-4},2\times 10^{-4}\\}$ and $\beta$ is
tuned in range of $\\{10^{-3},10^{-4},10^{-5},10^{-6}\\}$. In most cases, the
optimal value of learning rate is $10^{-3}$ and the one of $\beta$ is
$10^{-5}$. According to the contrastive learning, we search
${\alpha}_{1},{\alpha}_{2}$ and $\tau$ in range of $\\{0.1,0.2,0.5,1,2\\}$ and
$\\{0.1,0.2,0.5,1,2,5\\}$, respectively. Besides, we dropout features and
modalities in augmentation step randomly with the ratio in range of
$\\{0,0.1,0.2,0.5\\}$ and add noise with a weight in range of
$\\{0.01,0.02,0.05,0.1\\}$. We search the number of propagation layers $K,L,Z$
from $\\{1,2,3\\}$. For the baselines, we follow the designs in their articles
to achieve the best performance. Certainly, we keep the same settings to
ensure a fair comparison.

###  3.2. Overall Performance Comparison (RQ1)

Table 2 shows the overall performance comparison between our model CLHE and
the baseline methods. We have the following observations. First, our method
beats all the baselines on all the datasets, demonstrating the competitive
performance of our model. Second, over the baselines, Transformer and
TransformerCL achieve the best performance, showing that the self-attention
mechanism and contrastive learning can well preserve the correlations among
items within the bundle, thus yielding good bundle representations. Third,
comparing the results between different versions of dataset, we find that: 1)
the performance on POG_dense is much larger than that on POG due to denser
user-item interactions, demonstrating that user feedback information is quite
helpful to the performance; 2) the performance of Spotify_sparse is relatively
smaller than that on Spotify since the sparser bundle-item affiliation data,
justifying our hypothesis that large-scale and high-quality bundle dataset is
vital to bundle construction. Finally, we have an interesting observation that
the performance improvements on the four datasets is negatively correlated
with ”#Avr.B/I”, as shown in Table 1, in another word, in scenarios that items
are included in fewer bundles (_i.e.,_ the dataset include more sparse items),
our method performs even better. This phenomenon further justifies the
advantage of our method in countering the issue of sparse items.

Table 3. Ablation study of the hierarchical encoder and contrastive learning
(the performance is NDCG@20).

Settings | POG | POG_dense | Spotify | Spotify_sparse
---|---|---|---|---
CLHE | 0.0193 | 0.2773 | 0.1806 | 0.1212
w/o user feedback | 0.0168 | 0.2733 | 0.1695 | 0.1174
SelfAtt. | w/o item | 0.0168 | 0.2551 | 0.1334 | 0.0822
w/o bundle | 0.0127 | 0.2141 | 0.1785 | 0.1210
w/o both | 0.0034 | 0.20647 | 0.0418 | 0.0334
CL | w/o item | 0.0171 | 0.2742 | 0.1735 | 0.1203
w/o bundle | 0.0176 | 0.2598 | 0.1740 | 0.1123
w/o both | 0.0178 | 0.2662 | 0.1730 | 0.1084

###  3.3. Ablation Study of Key Modules (RQ2)

To further evaluate the effectiveness of the key modules of our model, we
conduct a list of ablation studies and the results are shown in Table 3. First
and foremost, we aim to justify the effectiveness of the user feedback
features. Thereby, we remove the user feedback features from our model
(_i.e.,_ remove $\mathbf{p}_{i}$ from $\mathbf{f}_{i}$) and build an ablated
version of model, _i.e.,_ w/o user feedback. According to the result in Table
3, after removing user feedback features, the performance reduces clearly,
verifying that user feedback feature is significant for bundle construction.
Second, we would like to evaluate whether each component of the hierarchical
encoder is useful. We progressively remove the two self-attention modules from
our model and replace them with an vanilla average pooling, thus yielding
three ablated models, _i.e.,_ w/o item, w/o bundle, and w/o both. The results
in Table 3 show that the removal of either self-attention modules causes
performance drop. These results further verify the efficacy of our self-
attention-based hierarchical encoder framework. Third, to justify the
contribution of contrastive learning, we progressively remove the two levels
of contrastive loss, thus generating three ablations, _i.e.,_ w/o item, w/o
bundle, and w/o both. Table 3 depicts the results, which demonstrate the both
contrastive losses are helpful, especially on the sparser version of datasets.

###  3.4. Model Study (RQ3)

To explicate more details and various properties of our method, we further
conduct a list of model studies.

Table 4. The overall performance (NDCG@20) of our CLHE and baselines on the
warm setting. ”%Improv.” denotes the relative improvement over the strongest
baseline.

Models | POG | POG_dense | Spotify | Spotify_sparse
---|---|---|---|---
MultiDAE | 0.0338 | 0.2725 | 0.1527 | 0.0839
MultiVAE | 0.0424 | 0.2871 | 0.0900 | 0.0384
Bi-LSTM | 0.0145 | 0.1997 | 0.0946 | 0.0281
Hypergraph | 0.0393 | 0.2860 | 0.0822 | 0.0401
Tranformer | 0.0520 | 0.2969 | 0.1837 | 0.1199
TranformerCL | 0.0280 | 0.2747 | 0.1766 | 0.1152
CLHE (ours) | 0.0554 | 0.3218 | 0.1846 | 0.1245
%Improv. | 6.53 | 8.38 | 0.48 | 3.81

####  3.4.1. Cold-start Items

One of the main challenges for bundle construction is cold-start items that
have never been included in previous bundles. It is difficult to directly
evaluate the methods solely based on cold-start items since there are few
testing bundles where both the input and result partial bundles purely consist
of cold-start items. Nevertheless, we come up with an alternative way to
indirectly test how these methods perform against cold-start items.
Specifically, we remove all the cold-start items and just keep the warm items
in the testing set, _i.e.,_ the warm setting. We test our method and all the
baseline models on this warm setting. The results shown in Table 4 illustrate
that: 1) the performance of all the models on the warm setting are much better
than that of the warm-cold hybrid setting (the normal setting as shown in
Table 2), exhibiting that the existence of cold-start items significantly
deteriorate the performance; and 2) the performance gap between CLHE and the
strongest baseline in the hybrid setting is obviously much larger than that on
the warm setting, implying that our method’s strength in dealing with cold-
start items.

Figure 3. Performance analysis with varying rates of sparsity and noise in the
partial bundle.

####  3.4.2. Sparsity and Noise in Bundle

Another merit of our approach is that the contrastive learning is able to
counter the sparsity and noise issue when the input partial bundle is
incomplete. To elicit this property, we corrupt the testing dataset to make
the input partial bundle sparse and noisy. Specifically, we randomly remove
certain portion of items from the input partial bundle to make them sparser.
To make the partial bundle more noisy, we randomly sample some items from the
whole item set and add them to the bundle. Then we test our model and the
model without both levels of contrastive loss, and the performance curves are
shown in Figure 3, where the x-axis is the ratio of bundle size after
corruption compared with the original bundle, and the ratio=1 corresponds to
the original clean data. From this figure, we can derive the conclusion that:
1) with the sparsity and noise degree increasing, both our method and
baselines’ performance drops; 2) our method still outperforms baselines even
under quite significant sparsity or noise rate, such as removing 50% seed
items or adding 50% more noisy items; and 3) the contrastive loss in our model
is able to combat the parse and noise bundle issue to some extent.

Table 5. Study of data augmentations (NDCG@20).

| Setting | POG | POG_dense | Spotify | Spotify_sparse
---|---|---|---|---|---
Item | NA | 0.0148 | 0.2428 | 0.1712 | 0.1202
FN | 0.0193 | 0.2753 | 0.1735 | 0.1172
FD | 0.0178 | 0.2763 | 0.1718 | 0.1200
MD | 0.0162 | 0.2773 | 0.1806 | 0.1212
Bundle | ID | 0.0184 | 0.2773 | 0.1791 | 0.1212
IR | 0.0193 | 0.2750 | 0.1806 | 0.1185

####  3.4.3. Data Augmentations

Data augmentation is the crux to contrastive learning. We search over multiple
different data augmentation strategies at both item- and bundle-level
contrastive learning, in order to find the best-performing setting. In Table
5, we present the performance of CLHE under various data augmentations at both
item- or bundle-level. Overall speaking, data augmentation methods may affect
the performance and proper selection is important for good results.

Figure 4. Computational complexity analysis.

####  3.4.4. Computational Complexity

Self-attention calculates every pair of instances in a set, _i.e.,_ features
of an item or items of a bundle, thus it usually suffers from high
computational complexity. We record the time used for every training epoch and
the time used from the beginning of training till convergence, and the records
of our method and two baselines, _i.e.,_ MultiDAE and Transformer, are
illustrated in Figure 4. The bar chart reveals that on the one hand, our
method is computationally heavy since it takes the longest time for each
training epoch; on the other hand, our method takes the least training time to
reach convergence on three datasets. In conclusion, our method is effective
and efficient during training, while the inherent complexity induced by
hierarchical self-attention may impose the inference slower. We argue that
various self-attention acceleration approaches could be considered in
practice, which is out of the scope of this work.

Figure 5. Illustration of similarity a) between each feature and the whole
item representation; and b) between each item and the whole bundle
representation. The size of the circles is positively correlated with its
corresponding cosine similarity.

####  3.4.5. Case Study

We would like to further illustrate some cases to portrait how the
hierarchical encoder learn the association of multimodal features and the
multiple items’ representations. Specifically, for both item- and bundle-level
self-attention modules, we take the last layer’s output representation as each
feature’s (item’s) representation, and calculate the cosine similarity score
with the whole item (bundle). We cherry pick some example items and bundles,
as shown in Figure 5. The results of feature-item similarity exhibit that the
three type of features could play distinctive roles in different items,
showing the importance of all the three types of features. For the item-bundle
similarity results, we can find that items do not equally contribute to their
affiliated bundles, thus it is crucial to model the bundle composition
patterns. Here we just intuitively illustrate some hints about the bundle
representation learning, more sophisticate analysis, such as feature pair or
item pair co-effects, is left for future work.

Figure 6. Hyper-parameter analysis.

####  3.4.6. Hyper-parameter Analysis

We also present the model’s performance change _w.r.t._ the key hyper-
parameters, _i.e.,_ the temperature $\tau$ in contrastive loss and the weights
${\alpha}_{1},{\alpha}_{2}$ for the two contrastive losses. The curves in
Figure 6 reveal that the model is still sensitive to these hyper-parameters,
and proper tuning is required to achieve optimal performance.

##  4\. Related Work

We review the literature about bundles, including: 1) bundle recommendation
and construction, and 2) bundle representation learning.

###  4.1. Bundle Recommendation and Construction

Product bundling is a mature marketing strategy that has been applied in
various application scenarios, including fashion outfit (Chen et al., 2019a;
Li et al., 2020), e-commerce (Sun et al., 2022), online music playlist (Cao et
al., 2017; Chen et al., 2018), online games (Deng et al., 2020), travel
package (Lim et al., 2018), meal (Li et al., 2022a), and _etc._. Personalized
bundle recommendation (Chen et al., 2019b; Chang et al., 2021; Ma et al.,
2022) is the pioneering work that first focuses on bundle-oriented problems in
the data science community. Soon after that, researchers realize that just
picking from predefined bundles cannot satisfy people’s diverse and
personalized needs. Thereby, the task of personalized bundle generation (Bai
et al., 2019; Chang et al., 2023; Deng et al., 2021; Wei et al., 2022a; He et
al., 2020b; Ding et al., 2023) is naturally proposed where the model aims to
automatically generate a bundle from a large set of items catering to a given
user. It has to simultaneously deal with both users’ personalization and item-
item compatibility patterns, where the user-item interaction is specifically
utilized for personalization modeling. In this paper, we only focus on bundle
construction, which is committed to generate more bundles to enrich the bundle
catalog for the platform. In addition, most of the bundle-oriented research in
general domain still falls into the id-based paradigm, where very few domains,
such as the fashion domain, have explored multimodality. We extend the
multimodal learning to one more domain of music playlist. Moreover, we also
leverage user feedback to multimodal bundle construction.

###  4.2. Bundle Representation Learning

Bundle representation learning is the crux of all the bundle-oriented
problems. Initial studies (Rendle et al., 2010) treat a bundle as a special
type of item and just use the bundle id to represent it. Naturally and
reasonably, people get to consider the encapsulated items within a bundle to
generate more detailed representation. The simplest method is performing
average pooling over the included items (Wu et al., 2016). Later on,
sequential models, such as Bi-LSTM (Han et al., 2017), are utilized to capture
the relations between two consecutive items. However, the items within a
bundle are not ordered essentially, and sequential models cannot well capture
all the pair-wise correlations. To address the limitation, attention models
(Chen et al., 2019b; Lin et al., 2020; He et al., 2020b), Transformer (Liu et
al., 2021b; Li et al., 2022c; Brosh et al., 2022; Wei et al., 2023b; Sarkar et
al., 2023; Vaswani et al., 2017) and graph neural networks (GNNs) (Chang et
al., 2021; Ren et al., 2023; Zhao et al., 2022; Yu et al., 2022a; Ding et al.,
2021) are leveraged to model not only every pair of items within a bundle, but
also the higher-order relations by stacking multiple layers.

Even though many efforts have been paid to the item correlation learning to
achieve good bundle representation, the multimodal information has been less
explored. Multimodal information, such as textual, visual, or knowledge graph
information of items, demonstrates to be effective in general recommendation
(Wei et al., 2019, 2023a, 2022b). In the fashion domain, visual and textual
features have been extensively investigated for pairwise mix-and-match (Yang
et al., 2019; Guan et al., 2021) or outfit compatibility modeling (Cui et al.,
2019; Song et al., 2023). However, these works have not been extended to other
domains, such as music playlist, where the audio modality has been rarely
studied in the bundle recommendation or construction problem. More
importantly, we argue that the user-item interaction information, which is
widely utilized in the personalized recommendation problem, can serve as an
additional modality in bundle construction. Sun _et al._ (Sun et al., 2022)
leverage a pre-trained CF model to obtain item representation to enhance the
bundle completion task, while they have not fully justify the rationale and
motivation. To the best of our knowledge, none of the previous works put
together all the user-item interaction, bundle-item affiliation, and item
content information for bundle construction.

##  5\. Conclusion and Future Work

In this work, we systematically study the problem of bundle construction and
define a more comprehensive formulation by considering all the three types of
data, _i.e.,_ multimodal features, item-level user feedback data, and existing
bundles. Based on this formulation, we highlight two challenges: 1) how to
learn expressive bundle representations given multiple features; and 2) how to
counter the modality missing, noise, and sparity problem. To tackle these
challenges, we propose a novel method of Contrastive Learning-enhanced
Hierarchical Encoder (CLHE) for bundle construction. Our method beats a list
of leading methods on four datasets of two application domains. Extensive
ablation and model studies justify the effectiveness of the key modules.

Despite the great performance that has been achieved by this work, there is
still large space to be explored for bundle construction. First, the current
evaluation setting is a little bit rigid and inflexible, it is interesting to
extend it to more flexible setting to align with real applications. For
example, given arbitrary number of seed items, the model is asked to construct
the bundle. Second, some of the feature extractors are pre-trained and fixed,
_i.e.,_ the multimodal feature extraction and user-item interaction models. Is
it possible to optimize these feature extractors in an end-to-end fashion thus
the extracted features would be more aligned to the bundle construction task?
Finally, this work just targets at unpersonalized bundle construction. It is
an interesting and natural direction to push forward this work to personalized
bundle construction.

## acknowledgement

This research is supported by NExT Research Center, National Natural Science
Foundation of China (9227010114), and the University Synergy Innovation
Program of Anhui Province (GXXT-2022-040).

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# On certain analytic functions defined by differential inequality

Prachi Prajna Dash‡, Jugal Kishore Prajapat† ‡Department of Mathematics,
Central University of Rajasthan, Bandarsindri, Kishangarh-305817, Dist.-
Ajmer, Rajasthan, India<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

For the family of analytic functions $f(z)$ in the open unit disk $\mathbb{D}$
with $f(0)=f^{\prime}(0)-1=0$, satisfying the differential equation

$zf^{\prime}(z)-f(z)=\dfrac{1}{2}z^{2}\phi(z),\quad|\phi(z)|\leq 1,$

we obtain radii of convexity, starlikeness, and close-to-convexity of partial
sums of $f(z)$. We also study the generalization of this family having the
form

$zf^{\prime}(z)-f(z)=\lambda z^{2}\phi(z),\quad|\phi(z)|\leq 1,$

where $\lambda>0,$ and obtain some useful properties of these functions.

###### Key words and phrases:

Univalent functions, Convex functions, Starlike functions, Close-to-Convex
functions, Partial sums, Radius of convexity, Radius of Starlikeness, Radius
of close-to-convexity.

###### 2020 Mathematics Subject Classification:

30C45

##  1\. Introduction

Let $\mathcal{A}$ denote the family of analytic functions in the open unit
disk $\mathbb{D}=\\{z\in\mathbb{C}:|z|<1\\}$ satisfying
$f(0)=0=f^{\prime}(0)-1$, and let $\mathcal{S}$ be the subclass of univalent
functions in $\mathcal{A}$. Let $E\subset\mathcal{A}$ and $P$ denote the
property of the image of $f(z)\in E$. The largest $\gamma_{f}$ such that
$f(z)\in E$ has a property $P$ in each disk
$\mathbb{D}_{r}=\\{z\in\mathbb{C}:\,|z|<r\\}$ for $0<r\leq\gamma_{f}$ is known
as the radius of the property $P$ of $f(z)$. The number
$\gamma:=\mbox{inf}\\{\gamma_{f}:\;f(z)\in E\\}$ is the radius of the property
$P$ of the class $E$. If $\mathcal{G}$ is the class of all $f(z)\in E$
characterized by the property $P$, then $\gamma$ is called the
$\mathcal{G}$-radius of the class $E$.

A function $f(z)\in\mathcal{A}$ is said to be convex if it maps $\mathbb{D}$
conformally onto a convex domain. A function $f(z)\in\mathcal{A}$ is convex if
and only if $\Re\left(1+zf^{\prime\prime}(z)/f^{\prime}(z)\right)>0$ for all
$z$ in $\mathbb{D}$. The class of these functions is denoted by $\mathcal{K}.$
A function $f(z)\in\mathcal{A}$ is said to be starlike if it maps $\mathbb{D}$
conformally onto a starlike domain. A function $f(z)\in\mathcal{A}$ is
starlike if and only if $\Re\left(zf^{\prime}(z)/f(z)\right)>0$ for all $z$ in
$\mathbb{D}$. The class of these functions is denoted by $\mathcal{S}^{*}$.
Note that $\mathcal{K}\subset\mathcal{S}^{*}\subset\mathcal{S}$. An analytic
function $f(z)\in\mathcal{A}$ is close-to-convex in $\mathbb{D}$ if
$\mathbb{C}\backslash f(\mathbb{D})$ can be represented as a union of non-
crossing half lines. The class of these functions is denoted by $\mathcal{C}$.
For every convex function $g(z)$, a function $f(z)\in\mathcal{C}$ satisfies
the inequality $\Re\left(f^{\prime}(z)/g^{\prime}(z)\right)>0$ for $z$ in
$\mathbb{D}$.

A function $f(z)$ is subordinate to the function $g$, written as $f(z)\prec
g(z)$, if there exists a Schwarz function $w(z)$, with $w(0)=0$ and
$|w(z)|\leq|z|$, such that $f(z)=g(w(z))$ for all $z\in\mathbb{D}$. If
$g(z)\in\mathcal{S}$, then the following relation:

$f(z)\prec
g(z)\quad\Longleftrightarrow\quad[f(0)=g(0)\;\;\mbox{and}\;\;f(\mathbb{D})\subset
g(\mathbb{D})],$

holds true. The convolution (or Hadamard product) of two analytic functions
$f(z)=\sum_{k=0}^{\infty}a_{k}z^{k}$ and $g(z)=\sum_{k=0}^{\infty}b_{k}z^{k}$
is defined by

$(f*g)(z)=\sum_{k=0}^{\infty}a_{k}\,b_{k}\,z^{k}.$

The convolution has the algebraic properties of ordinary multiplication.
Rusheweyh and Sheil-Small [9] proved an important conjecture known as the
Pólya-Schoenberg Conjecture that the class $\mathcal{K}$ is preserved under
convolution. They also proved that the classes $\mathcal{S^{*}}$ and
$\mathcal{C}$ are closed under convolution with the class $\mathcal{K}$. Many
other convolution problems were studied by Rusheweyh in [8] and have found
many applications in various fields. Finding classes of univalent functions
that are preserved under convolution and studying their properties is an
important research problem. Recently, Peng and Zhong [11] proved that
$f(z)\in\mathcal{A}$ that satisfy the differential equation

(1.1) $zf^{\prime}(z)-f(z)=\dfrac{1}{2}z^{2}\phi(z),\quad|\phi(z)|\leq 1,$

are closed under convolution. Let $\Omega$ be the subset of $\mathcal{A}$ that
consists of all functions satisfying (1.1). Note that the functions
$f\in\Omega$ satisfy the inequality

(1.2) $|zf^{\prime}(z)-f(z)|<\dfrac{1}{2},\qquad z\in\mathbb{D}.$

The functions in $\Omega$ are univalent and starlike in $\mathbb{D}$ (see [11,
4]). Obradović and Peng [5] gave certain sufficient conditions for functions
belonging to the class $\Omega$. Also, Wani and Swaminathan [1] obtained
certain sufficient conditions and subordination properties for functions
belonging to the class $\Omega$.

To obtain an example of a function in $\Omega$, let $f_{\mu}(z)\in\mathcal{A}$
such that

$zf^{\prime}_{\mu}(z)-f_{\mu}(z)=\dfrac{1}{2}z^{2}\phi_{\mu}(z),$

where $\phi_{\mu}(z)=\dfrac{\mu+z}{1+\mu z},\;\mu\in[-1,1]$. Then

$zf^{\prime}_{\mu}(z)-f_{\mu}(z)=\dfrac{1}{2}z^{2}\left(\mu+(1-\mu^{2})\sum_{k=1}^{\infty}(-\mu)^{k-1}\,z^{k}\right),$

now comparing the coefficients on both sides, we obtain that
$f_{\mu}(z)\in\Omega$, where

(1.3)
$f_{\mu}(z)=z+\dfrac{\mu}{2}z^{2}+\dfrac{1-\mu^{2}}{2}\sum_{k=1}^{\infty}\dfrac{(-\mu)^{k-1}}{k+1}\,z^{k+2},\quad\mu\in[-1,1].$

For $\lambda>0$, let $\Omega_{\lambda}$ denote the set of functions in
$\mathcal{A}$ satisfying the differential equation

(1.4) $zf^{\prime}(z)-f(z)=\lambda z^{2}\phi(z),\quad|\phi(z)|\leq 1.$

Note that the functions $f(z)\in\Omega_{\lambda}$ satisfy the inequality

(1.5) $|zf^{\prime}(z)-f(z)|<\lambda,\qquad z\in\mathbb{D}.$

The class $\Omega_{\lambda}$ is non-empty as the identity function satisfies
the inequality (1.5). Note that $\Omega_{1/2}=\Omega$ and
$\Omega_{\lambda_{1}}\subseteq\Omega_{\lambda_{2}}$ when
$\lambda_{1}\leq\lambda_{2}$. Consequently, the functions in
$\Omega_{\lambda}$ are univalent and starlike in $\mathbb{D}$ for
$\lambda\in\left(0,1/2\right]$. We observe that the function

(1.6) $f(z)=z+\dfrac{\lambda z^{2}}{2}+\dfrac{\lambda
z^{3}}{4},\qquad\lambda>0,\,z\in\mathbb{D},$

belongs to the class $\Omega_{\lambda}$.

A function $f(z)\in\mathcal{A}$ has the power series representation

(1.7)
$f(z)=z+\sum_{k=2}^{\infty}a_{k}\,z^{k},\qquad\left(a_{k}=\dfrac{f^{(k)}(z)}{k!},\;k=2,\,3,\dots,\;z\in\mathbb{D}\right),$

and the $n$-th partial sum of $f(z)$ is the polynomial

(1.8) $s_{n}(z;f)=z+\sum_{k=2}^{n}a_{k}\,z^{k}.$

In 1928, Szegö demonstrated that $s_{n}(z;f)$ for $f(z)\in\mathcal{S}$ are
univalent in the disk $\mathbb{D}_{1/4}$, and this radius cannot be extended
to a larger value [3]. Later, in 1941, Robertson [6] established that
$s_{n}(z;f)$ remains univalent in the disk $|z|<1-(3\log n)/n$ for all $n\geq
5$. It is true that the Koebe function is extremal in many subclasses of
$\mathcal{S}$, but Bshouty and Hergartner [2] showed in 1991 that it is not
extremal for the radius of univalence of $s_{n}(z;f)$ for
$f(z)\in\mathcal{S}$. This result was resolved using the convolution theorem
by Ruscheweyh and others [9] for various geometric subclasses of
$\mathcal{S}$. However, determining the exact largest radius of univalence of
$s_{n}(z;f)$ for $f(z)\in\mathcal{S}$ remains an open problem (see [7, page
246]).

We shall use the following results.

######  Lemma 1.1.

[11, 4] If $f(z)=z+\sum_{k=2}^{\infty}a_{k}z^{k}\in\Omega$, then

  1. (a)

$|z|-\dfrac{|z|^{2}}{2}\leq|f(z)|\leq|z|+\dfrac{|z|^{2}}{2}$;

  2. (b)

$1-|z|\,\leq\,|f^{\prime}(z)|\,\leq\,1+|z|$;

  3. (c)

$|a_{k}|\,\leq\,\dfrac{1}{2(k-1)},\;k\geq 2$.

For each $z\in\mathbb{D}\setminus\\{0\\},$ equality occurs in (a) and (b) if
and only if $f(z)=z+\dfrac{1}{2}e^{i\psi}z^{2},$ where $\psi\in\mathbb{R}$,
and equality holds for (c) if $f(z)=z+\dfrac{1}{2(k-1)}z^{k}$, where
$k=2,3,\dots$.

######  Lemma 1.2.

[1] If $f(z)\in\Omega$, then

$\bigg{|}\dfrac{zf^{\prime}(z)}{f(z)}-1\bigg{|}\leq\dfrac{r}{2-r},\quad|z|=r<1,$

for each $z\in\mathbb{D}$.

######  Lemma 1.3.

[10] Let $q(z)=\sum_{n=1}^{\infty}b_{n}z^{n}\in\mathcal{K}$. If
$p(z)=\sum_{n=1}^{\infty}a_{n}z^{n}$ is analytic in $\mathbb{D}$ and satisfies
the subordination $p(z)\prec q(z)$, then $|a_{n}|\leq|b_{1}|$ where
$n\in\mathbb{N}$.

######  Lemma 1.4.

Let $f(z)=z+\sum_{k=2}^{\infty}a_{k}z^{k}\in\Omega$ and
$\rho_{n}(z;f)=\sum_{k=n+1}^{\infty}a_{k}z^{k}$. Then, for $n\geq 2$ and
$|z|=r<1$,

(1.9) $\displaystyle|\rho_{n}(z;f)|$
$\displaystyle\leq\,-\,\dfrac{r}{2}\,\mbox{ln}(1-r)-\dfrac{r^{2}}{2},$ (1.10)
$\displaystyle|\rho^{\prime}_{n}(z;f)|$
$\displaystyle\leq\,\dfrac{2r^{2}-r}{2(1-r)}-\dfrac{\mbox{ln}(1-r)}{2},$
(1.11) $\displaystyle|z\rho^{\prime\prime}_{n}(z;f)|$
$\displaystyle\leq\,\dfrac{r^{2}(3-2r)}{2(1-r)^{2}}.$

###### Proof.

Using Lemma 1.1$(c)$, we obtain

$\displaystyle|\rho_{n}(z;f)|$
$\displaystyle\leq\;\dfrac{1}{2}\sum^{\infty}_{k=1}\dfrac{r^{k+n}}{k+n-1}\,\leq\,\dfrac{1}{2}\sum^{\infty}_{k=1}\dfrac{r^{k+2}}{k+1}=-\dfrac{r}{2}\ln(1-r)-\dfrac{r^{2}}{2},$
$\displaystyle|\rho^{\prime}_{n}(z;f)|$
$\displaystyle\leq\;\dfrac{1}{2}\sum_{k=1}^{\infty}\dfrac{(k+n)\,r^{k+n-1}}{k+n-1}\leq\dfrac{1}{2}\left(\sum_{k=1}^{\infty}r^{k+1}+\sum_{k=1}^{\infty}\dfrac{r^{k+1}}{k+1}\right)$
$\displaystyle=\;\dfrac{2r^{2}-r}{2(1-r)}-\dfrac{\ln(1-r)}{2},\quad\text{and}$
$\displaystyle|z\rho^{\prime\prime}_{n}(z;f)|$
$\displaystyle\leq\;\dfrac{1}{2}\sum_{k=1}^{\infty}\left(k+n\right)r^{k+n-1}\,\leq\,\dfrac{1}{2}\sum_{k=1}^{\infty}(k+2)r^{k+1}=\dfrac{r^{2}(3-2r)}{2(1-r)^{2}}.$

∎

##  2\. Radius properties of the Class $\Omega$

The functions in $\Omega$ are convex in $\mathbb{D}_{1/2}$ [11, Theorem 3.4].
However, the partial sums of $f(z)\in\Omega$ need not be convex in
$\mathbb{D}_{1/2}$. For example, if
$s_{3}(z;f)=z+\dfrac{\mu}{2}z^{2}+\dfrac{1-\mu^{2}}{4}z^{3}$ of $f_{\mu}$
given by (1.3), we obtain

$\displaystyle\Re\left(1+\dfrac{z\,s_{3}^{\prime\prime}(z;f)}{s_{3}^{\prime}(z;f)}\right)\geq
1-\left|\dfrac{\mu z+\dfrac{3}{2}(1-\mu^{2})z^{2}}{1+\mu
z+\dfrac{3}{4}(1-\mu^{2})z^{2}}\right|$ $\displaystyle\geq
1-\dfrac{|\mu||z|+\dfrac{3}{2}(1-\mu^{2})|z|^{2}}{1-|\mu||z|-\dfrac{3}{4}(1-\mu^{2})|z|^{2}}$
$\displaystyle=\dfrac{1-2|\mu||z|-\dfrac{9}{4}(1-\mu^{2})|z|^{2}}{1-|\mu||z|-\dfrac{3}{4}(1-\mu^{2})|z|^{2}}.$

Note that $1-|\mu||z|-\dfrac{3}{4}(1-\mu^{2})|z|^{2}>0$ in $|z|<0.964$ if
$\mu=0.9$, but
$\Re\left(1+\dfrac{z\,s_{3}^{\prime\prime}(z;f)}{s_{3}^{\prime}(z;f)}\right)\not>0$
for $\mu=0.9$ and $|z|\geq 0.4969$. Hence, $s_{3}(z;f)$ is not convex in
$\mathbb{D}_{1/2}$. In the next result, we obtain the radius of convexity for
the partial sums of $f\in\Omega$.

######  Theorem 2.1.

Let $f(z)=z+\sum_{k=2}^{\infty}a_{k}z^{k}\in\Omega$, and $r_{\mathcal{K}}$
denote the positive root of

(2.1) $(1-r)^{2}\ln(1-r)+2-7r+4r^{2}=0.$

Then $s_{n}(z;f)$ is convex in the disk $|z|<r_{\mathcal{K}}$.

###### Proof.

By using Lemma 1.1$(c)$, we obtain

$\Re\left(1+\dfrac{zs_{2}^{\prime\prime}(z;f)}{s_{2}^{\prime}(z;f)}\right)\geq
1-\dfrac{2|a_{2}||z|}{1-2|a_{2}||z|}\geq 1-\dfrac{|z|}{1-|z|},$

which is positive for $|z|<1/2$. Thus, $s_{2}(z;f)$ is convex in the disk
$|z|<1/2:=r_{1}$. Further, using Lemma 1.1$(c)$, we obtain

$\displaystyle\Re\left(1+\dfrac{zs_{3}^{\prime\prime}(z;f)}{s_{3}^{\prime}(z;f)}\right)$
$\displaystyle=\Re\left(\dfrac{1+4a_{2}z+9a_{3}z^{2}}{1+2a_{2}z+3a_{3}z^{2}}\right)=2-\Re\left(\dfrac{1-3a_{3}z^{2}}{1+2a_{2}z+3a_{3}z^{2}}\right)$
$\displaystyle\geq
2-\left|\dfrac{1-3a_{3}z^{2}}{1+2a_{2}z+3a_{3}z^{2}}\right|\geq
2-\dfrac{1+3|a_{3}||z|^{2}}{1-2|a_{2}||z|-3|a_{3}||z|^{2}}$ $\displaystyle\geq
2-\dfrac{4+3|z|^{2}}{4-4|z|-3|z|^{2}}=\dfrac{4-8|z|-9|z|^{2}}{4-4|z|-3|z|^{2}}\;,$

which is positive for $|z|<\min\\{r_{2},r_{2.1}\\}=r_{2}$, where $r_{2}\approx
0.3568$ is the unique root of $9r^{2}+8r-4=0$ and $r_{2.1}\approx 0.6667$ is
the unique root of $3r^{2}+4r-4=0$ in $(0,1)$. Thus, $s_{3}(z;f)$ is convex in
the disk $|z|<r_{2}$.

Notice that the radii are decreasing as $n$ increases. Next, we consider the
general case $n\geq 2$. To prove that $s_{n}(z;f)$ is convex in
$|z|<r_{\mathcal{K}},$ it is sufficient to prove that for
$0<|z|<r_{\mathcal{K}},$ $s^{\prime}_{n}(z;f)\neq 0$ and

(2.2)
$\displaystyle\displaystyle{\Re\left(1+\dfrac{zs_{n}^{\prime\prime}(z;f)}{s_{n}^{\prime}(z;f)}\right)}$
$\displaystyle=\Re\left(1+\dfrac{zf^{\prime\prime}(z)}{f^{\prime}(z)}+\dfrac{\dfrac{zf^{\prime\prime}(z)}{f^{\prime}(z)}\rho^{\prime}_{n}(z;f)-z\rho^{\prime\prime}_{n}(z;f)}{f^{\prime}(z)-\rho^{\prime}_{n}(z;f)}\right)$
$\displaystyle\geq\Re\left(1+\dfrac{zf^{\prime\prime}(z)}{f^{\prime}(z)}\right)-\dfrac{\left|\dfrac{zf^{\prime\prime}(z)}{f^{\prime}(z)}\right||\rho^{\prime}_{n}(z;f)|+|z\rho^{\prime\prime}_{n}(z;f)|}{|f^{\prime}(z)|-|\rho^{\prime}_{n}(z;f)|}>0,$

where $\rho_{n}(z;f)=\sum^{\infty}_{k=n+1}a_{k}z^{k}$. If $f\in\Omega$, then
for $r\leq\dfrac{\sqrt{5}-1}{2},$

(2.3)
$\left|\dfrac{zf^{\prime\prime}(z)}{f^{\prime}(z)}\right|\leq\dfrac{r}{1-r}$

(see [11, Theorem 3.4]). Hence,

$\Re\left(1+\dfrac{zf^{\prime\prime}(z)}{f^{\prime}(z)}\right)\geq\dfrac{1-2r}{1-r},\qquad|z|\leq\dfrac{\sqrt{5}-1}{2}.$

To prove the inequality (2.2), it is sufficient to show that

(2.4) $|\rho^{\prime}_{n}(z;f)|<|f^{\prime}(z)|,\qquad|z|<r_{\mathcal{K}}.$

and

(2.5)
$\dfrac{\Big{|}\dfrac{zf^{\prime\prime}(z)}{f^{\prime}(z)}\Big{|}|\rho^{\prime}_{n}(z;f)|+|z\rho^{\prime\prime}_{n}(z;f)|}{|f^{\prime}(z)|-|\rho^{\prime}_{n}(z;f)|}\leq\dfrac{1-2r}{1-r},\qquad|z|=r<r_{\mathcal{K}},$

where $r_{\mathcal{K}}$ is the radius of convexity of $s_{n}(z;f)$.

Note that $\dfrac{\rho^{\prime}_{n}(z;f)}{f^{\prime}(z)}$ is analytic in
$\mathbb{D}$ since $f^{\prime}(z)\neq 0$ (by local mapping theorem and as
$f\in\mathcal{S}$) for $0<|z|<1$. Further, by the use of the maximum modulus
theorem, if (2.4) holds for $|z|=r_{\mathcal{K}},$ then the same inequality
holds for $0<|z|<r_{\mathcal{K}}$ when $s^{\prime}_{n}(z;f)\neq 0$ for
$0<|z|<r_{\mathcal{K}}$. Consequently,
$1+\dfrac{zs^{\prime\prime}_{n}(z;f)}{s^{\prime}_{n}(z;f)}$ is analytic on
$|z|\leq r_{\mathcal{K}}$, and if
$\Re\left(1+\dfrac{zs^{\prime\prime}_{n}(z;f)}{s^{\prime}_{n}(z;f)}\right)\geq
0$ on $|z|=r_{\mathcal{K}}$, then $s_{n}(z;f)$ is convex in
$|z|<r_{\mathcal{K}}$ by the maximum modulus theorem. But the inequality
$\Re\left(1+\dfrac{zs^{\prime\prime}_{n}(z;f)}{s^{\prime}_{n}(z;f)}\right)\geq
0$ on $|z|=r_{\mathcal{K}}$ follows from (2.5). If $|z|=r$, then by using
Lemma 1.4 and Lemma 1.1$(b)$, we obtain

$|f^{\prime}(z)|-|\rho^{\prime}_{n}(z;f)|\geq
1-r-\left(\dfrac{2r^{2}-r}{2(1-r)}-\dfrac{\ln(1-r)}{2}\right)=\dfrac{2-3r}{2(1-r)}+\dfrac{\ln(1-r)}{2},$

which is positive when $|z|<r_{3},$ where $r_{3}\approx 0.5471.$ Hence (2.4)
holds when $r<r_{3}\approx 0.5471.$

Now by using inequalities (1.10), (1.11), (2.3) and Lemma 1.1$(b)$, we obtain

$\displaystyle\dfrac{1-2r}{1-r}-\dfrac{\left|\dfrac{zf^{\prime\prime}(z)}{f^{\prime}(z)}\right||\rho^{\prime}_{n}(z;f)|+|z\rho^{\prime\prime}_{n}(z;f)|}{|f^{\prime}(z)|-|\rho^{\prime}_{n}(z;f)|}$
$\displaystyle\geq\dfrac{1-2r}{1-r}-\dfrac{\dfrac{r}{1-r}\left|\rho^{\prime}_{n}(z;f)\right|+\left|z\rho^{\prime\prime}_{n}(z;f)\right|}{1-r-\left|\rho^{\prime}_{n}(z;f)\right|}$
$\displaystyle\geq\dfrac{1-2r}{1-r}-\dfrac{\dfrac{r}{(1-r)}\left(\dfrac{2r^{2}-r}{2(1-r)}-\dfrac{\ln(1-r)}{2}\right)+\dfrac{r^{2}(3-2r)}{2(1-r)^{2}}}{1-r-\dfrac{2r^{2}-r}{2(1-r)}+\dfrac{\ln(1-r)}{2}}$
$\displaystyle=\dfrac{1-2r}{1-r}-\dfrac{2r^{2}-r(1-r)\ln(1-r)}{(1-r)(2-3r+(1-r)\ln(1-r))},$

which is positive for $|z|<r_{4},$ where $r_{4}\approx 0.3181$ is the unique
root of (2.1) in $(0,(\sqrt{5}-1)/2)$. Thus $s_{n}(z;f),\,n\geq 2$ is convex
in the disk $|z|<\min\\{r_{1},r_{2},r_{3},r_{4}\\}=r_{4}:=r_{\mathcal{K}}$. ∎

The functions in $\Omega$ are starlike in $\mathbb{D}$ [11, Theorem 3.1], but
the partial sums of $f(z)\in\Omega$ may not be starlike in $\mathbb{D}$. For
example, if we consider $s_{3}(z;f)$ of $f_{\mu}$ given by (1.3), then we
obtain

$\displaystyle\Re\left(\dfrac{zs^{\prime}_{3}(z;f)}{s_{3}(z;f)}\right)\geq
1-\left|\dfrac{2\mu z^{2}+2(1-\mu^{2})z^{3}}{4z+2\mu
z^{2}+(1-\mu^{2})z^{3}}\right|$ $\displaystyle\geq
1-\dfrac{2|\mu||z|+2(1-\mu^{2})|z|^{2}}{4-2|\mu||z|-(1-\mu^{2})|z|^{2}}$
$\displaystyle=\dfrac{4-4|\mu||z|-3(1-\mu^{2})|z|^{2}}{4-2|\mu||z|-(1-\mu^{2})|z|^{2}}.$

Note that $4-2|\mu||z|-(1-\mu^{2})|z|^{2}>0$ in $\mathbb{D}$ if $\mu=0.7$, but
$\Re\left(\dfrac{zs_{3}^{\prime}(z;f)}{s_{3}(z;f)}\right)\not>0$ for $\mu=0.7$
and $|z|\geq 0.9428$. Hence, $s_{3}(z;f)$ is not starlike in $\mathbb{D}$. In
the following result, we obtain the radius of starlikeness for the partial
sums of $f\in\Omega$.

######  Theorem 2.2.

Let $f(z)=z+\sum_{k=2}^{\infty}a_{k}z^{k}\in\Omega$, and $r_{\mathcal{S}^{*}}$
denote the positive root of

(2.6) $3(1-r)(2-r)\ln(1-r)+4-4r-3r^{2}+2r^{3}=0.$

Then $s_{n}(z;f)$ is starlike in the disk $|z|<r_{\mathcal{S}^{*}}$.

###### Proof.

By using Lemma 1.1$(c)$, we obtain

$\Re\left(\dfrac{zs_{2}^{\prime}(z;f)}{s_{2}(z;f)}\right)=\Re\left(\dfrac{z+2a_{2}z^{2}}{z+a_{2}z^{2}}\right)\geq
1-\dfrac{|a_{2}||z|^{2}}{|z|-|a_{2}||z|^{2}}\geq\dfrac{2(1-|z|)}{2-|z|},$

which is positive for $|z|<1$. Thus, $s_{2}(z;f)$ is startlike in
$|z|<1:=r_{5}$. Further, using Lemma 1.1$(c)$, we obtain

$\Re\left(\dfrac{zs_{3}^{\prime}(z;f)}{s_{3}(z;f)}\right)=\Re\left(\dfrac{z+2a_{2}z^{2}+3a_{3}z^{3}}{z+a_{2}z^{2}+a_{3}z^{3}}\right)\geq\dfrac{4-4|z|-3|z|^{2}}{4-2|z|-|z|^{2}},$

which is positive for $|z|<r_{6},$ where $r_{6}\approx 0.6667$ is the unique
root of $3r^{2}+4r-4=0$ in $(0,1)$. Thus, $s_{3}(z;f)$ is starlike in
$|z|<r_{6}$.

Notice that the radii are decreasing as $n$ increases. Next, we consider the
general case $n\geq 2$. To prove that $s_{n}(z;f)$ is starlike in
$|z|<r_{\mathcal{S}^{*}},$ it is sufficient to prove that for
$0<|z|<r_{\mathcal{S}^{*}},$ $s_{n}(z;f)\neq 0$ and

(2.7) $\displaystyle\Re\left(\dfrac{zs^{\prime}_{n}(z;f)}{s_{n}(z;f)}\right)$
$\displaystyle=\Re\left(\dfrac{zf^{\prime}(z)}{f(z)}+\dfrac{\displaystyle{\dfrac{zf^{\prime}(z)}{f(z)}\rho_{n}(z;f)-z\rho^{\prime}_{n}(z;f)}}{f(z)-\rho_{n}(z;f)}\right)$
$\displaystyle\geq\Re\left(\dfrac{zf^{\prime}(z)}{f(z)}\right)-\dfrac{\left|\dfrac{zf^{\prime}(z)}{f(z)}\right||\rho_{n}(z;f)|+|z\rho^{\prime}_{n}(z;f)|}{|f(z)|-|\rho_{n}(z;f)|}>0.$

If $f\in\Omega$, then Lemma 1.2 shows that

(2.8)
$\Re\left(\dfrac{zf^{\prime}(z)}{f(z)}\right)\geq\dfrac{2(1-r)}{2-r},\qquad|z|=r<1.$

To prove the inequality (2.7), it is sufficient to show that

(2.9) $|\rho_{n}(z;f)|<|f(z)|,\quad|z|<r_{\mathcal{S}^{*}},$

and

(2.10)
$\dfrac{\displaystyle{\left|\dfrac{zf^{\prime}(z)}{f(z)}\right||\rho_{n}(z;f)|+|z\rho^{\prime}_{n}(z;f)|}}{|f(z)|-|\rho_{n}(z;f)|}\leq\dfrac{2(1-r)}{2-r},\qquad|z|=r<r_{\mathcal{S}^{*}},$

where $r_{\mathcal{S}^{*}}$ is the radius of starlikeness of $s_{n}(z;f)$.

Note that $\dfrac{\rho_{n}(z;f)}{f(z)}$ is analytic in $\mathbb{D}$ since
$f(z)\neq 0$ for $0<|z|<1$. Further, by the use of the maximum modulus
theorem, if (2.9) holds for $|z|=r_{\mathcal{S}^{*}},$ then the same
inequality holds for $0<|z|<r_{\mathcal{S}^{*}}$ when $s_{n}(z;f)\neq 0$ for
$0<|z|<r_{\mathcal{S}^{*}}$. Consequently,
$\dfrac{zs^{\prime}_{n}(z;f)}{s_{n}(z;f)}$ is analytic on $|z|\leq
r_{\mathcal{S}^{*}}$ and if
$\Re\left(\dfrac{zs^{\prime}_{n}(z;f)}{s_{n}(z;f)}\right)\geq 0$ on
$|z|=r_{\mathcal{S}^{*}}$, then $s_{n}(z;f)$ is starlike in
$|z|<r_{\mathcal{S}^{*}}$ by the maximum modulus theorem. But the inequality
$\Re\left(\dfrac{zs^{\prime}_{n}(z;f)}{s_{n}(z;f)}\right)\geq 0$ on
$|z|=r_{\mathcal{S}^{*}}$ follows by using (2.8) and (2.10) in (2.7). Using
(1.9) and Lemma 1.1$(a)$, we obtain

$|f(z)|-|\rho_{n}(z;f)|\geq
r-\dfrac{r^{2}}{2}+\dfrac{r}{2}\left(\ln(1-r)+r\right),$

which is positive when $|z|<r_{7},$ where $r_{7}\approx 0.8647.$ Hence (2.9)
holds when $|z|<r_{7}.$

By using the inequalities (1.9), (1.10), (2.8), (2.9), and (2.10), we obtained

$\displaystyle\Re\left(\dfrac{zs^{\prime}_{n}(z;f)}{s_{n}(z;f)}\right)$
$\displaystyle>\dfrac{2(1-r)}{2-r}-\dfrac{\left|\dfrac{zf^{\prime}(z)}{f(z)}\right||\rho_{n}(z;f)|+|z\rho^{\prime}_{n}(z;f)|}{|f(z)|-|\rho_{n}(z;f)|}$
$\displaystyle\geq\dfrac{2(1-r)}{2-r}-\dfrac{\dfrac{2}{2-r}\left(-\dfrac{r\ln(1-r)}{2}-\dfrac{r^{2}}{2}\right)+\dfrac{2r^{3}-r^{2}}{2(1-r)}-\dfrac{r\ln(1-r)}{2}}{\displaystyle{r-\dfrac{r^{2}}{2}+\dfrac{r^{2}}{2}+\dfrac{r\ln(1-r)}{2}}}$
$\displaystyle=\dfrac{2(1-r)}{2-r}+\dfrac{\dfrac{1}{2-r}\left(r\ln(1-r)+r^{2}\right)-\dfrac{2r^{3}-r^{2}}{2(1-r)}+\dfrac{r\ln(1-r)}{2}}{r+\dfrac{r\ln(1-r)}{2}},$

which is positive for $|z|<r_{8},$ where $r_{8}\approx 0.4899$ is the unique
root of (2.6). Thus $s_{n}(z;f),\,n\geq 2$ is starlike in the disk
$|z|<\min\\{r_{5},r_{6},r_{7},r_{8}\\}=r_{8}:=r_{\mathcal{S}^{*}}$. ∎

The functions in $\Omega$ are close-to-convex in $\mathbb{D}$ [4, Lemma 3.2],
but the partial sums of $f(z)\in\Omega$ may not be close-to-convex in
$\mathbb{D}$. For example, if we consider $s_{3}(z;f)$ of $f_{\mu}$ given by
(1.3), we obtain

$\Re\left(s^{\prime}_{3}(z;f)\right)\geq 1-\left|\mu
z+\dfrac{3}{4}(1-\mu^{2})z^{2}\right|\geq
1-|\mu||z|-\dfrac{3}{4}(1-\mu^{2})|z|^{2},$

is positive when $\mu=0.7$ and $|z|<0.9428$. Thus,
$\Re\left(s^{\prime}_{3}(z;f)\right)\not>0$ for $\mu=0.7$ and $|z|\geq
0.9428$. Hence, $s_{3}(z;f)$ is not close-to-convex in $\mathbb{D}$. In the
next result, we obtain the radius of close-to-convexity for the partial sums
of $f\in\Omega$.

######  Theorem 2.3.

Let $f(z)=z+\sum_{k=2}^{\infty}a_{k}z^{k}\in\Omega$, and $r_{\mathcal{S}}$
denote the positive root of

(2.11) $(1-r)\ln(1-r)+2-3r=0.$

Then $s_{n}(z;f)$ is close-to-convex in the disk $|z|<r_{\mathcal{S}}$.

###### Proof.

Using Lemma 1.1$(c)$, we have

$\Re\left(s_{2}^{\prime}(z;f)\right)\geq 1-2|a_{2}||z|\geq 1-|z|,$

which is positive for $|z|<1$. Thus, $s_{2}(z,f)$ is close-to-convex in the
disk $|z|<1:=r_{9}$. Further, using Lemma 1.1$(c)$, we have

$\Re\left(s_{3}^{\prime}(z;f)\right)\geq 1-2|a_{2}||z|-3|a_{3}||z|^{2}\geq
1-|z|-\dfrac{3}{4}|z|^{2},$

which is positive for $|z|<r_{10},$ where $r_{10}\approx 0.66667$ is the
unique root of $3r^{2}+4r-4=0$ in $(0,1)$. Thus, $s_{3}(z;f)$ is close-to-
convex in the disk $|z|<r_{10}$.

Next, we consider the general case $n\geq 2$. Following the proof of [4, Lemma
3.2], we have

(2.12) $\Re(f^{\prime}(z))\geq 1-r.$

Using (1.10) and (2.12), we obtain

$\displaystyle\Re(s^{\prime}_{n}(z;f))$
$\displaystyle\geq\Re(f^{\prime}(z))-\Re(\rho^{\prime}_{n}(z;f))$
$\displaystyle\geq 1-r-\left(\dfrac{2r^{2}-r-(1-r)\ln(1-r)}{2(1-r)}\right)$
$\displaystyle=\dfrac{2(1-r)^{2}-2r^{2}+r+(1-r)\ln(1-r)}{2(1-r)}$
$\displaystyle=\dfrac{2-3r+(1-r)\ln(1-r)}{2(1-r)},$

which is positive for $|z|<r_{11},$ where $r_{11}\approx 0.5471$ is the unique
root of (2.11). Thus, $s_{n}(z;f),\,n\geq 2$ is close-to-convex in the disk
$|z|<\min\\{r_{9},r_{10},r_{11}\\}=r_{11}:=r_{\mathcal{S}}$. ∎

##  3\. Inequalities on the partial sums of the Class $\Omega$

If $f(z)\in\Omega$, then following the proof of Theorem 2.3, we observe that
the radius of close-to-convexity is decreasing as $n$ is increasing, but it
remains constant for all $n\geq 12$, which can be seen in Figure 1 of $n$ and
approximate value of radius of disk in which
$\Re(s^{\prime}_{n}(z;f))>0,\;f(z)\in\Omega$. This section is devoted to
establishing this fact.

Figure 1. Graph between $n$ and approximate value of radius of disk in which
$\Re(s^{\prime}_{n}(z;f))>0$.

######  Theorem 3.1.

Let $f(z)\in\Omega$. Then, for each $n\geq 2$, we have

(3.1)
$\bigg{|}\dfrac{s_{n}^{\prime}(z,f)}{f^{\prime}(z)}-1\bigg{|}\leq|z|^{n}\left(\dfrac{n+1}{2n}+A_{n}\dfrac{|z|}{1-|z|}\right),\quad|z|=r<1,$

where $A_{n}=\dfrac{\sqrt{2r-r^{2}}}{2(1-r)}(n+1+\ln(n-1)+\gamma)$ and
$\gamma\approx 0.57722$ is the Euler-Mascheroni constant.

###### Proof.

Let $f(z)\in\Omega$. Since $f(z)\in\Omega$, it is univalent in $\mathbb{D}$,
and so $f^{\prime}(z)$ is non-vanishing in $\mathbb{D}$. Hence,
$1/f^{\prime}(z)$ can be represented in the form

$\dfrac{1}{f^{\prime}(z)}=1+d_{1}z+d_{2}z^{2}+\dots\;,$

where $d_{1}=-2a_{2}$, $d_{2}=4a_{2}^{2}-3a_{3}$ and so on, for some complex
coefficients $d_{n}$, $n\geq 1$.
Now considering the identity
$f^{\prime}(z)\displaystyle{\dfrac{1}{f^{\prime}(z)}}\equiv 1$, we have

$(1+2a_{2}z+3a_{3}z^{2}+\dots)(1+d_{1}z+d_{2}z^{2}+\dots)=1,\quad
z\in\mathbb{D}.$

From this, we get

(3.2) $ma_{m}+\sum_{k=1}^{m-1}(m-k)a_{m-k}d_{k}=0,$

where $m\geq 2$ and $a_{1}=1$.
Now using (3.2) and the series representation of the partial sum $s_{n}(z,f)$,
we get

(3.3) $\displaystyle\dfrac{s_{n}^{\prime}(z,f)}{f^{\prime}(z)}$
$\displaystyle=(1+2a_{2}z+\dots+na_{n}z^{n-1})(1+d_{1}z+d_{2}z^{2}+\dots)$
$\displaystyle\equiv 1+c_{n}z^{n}+c_{n+1}z^{n+1}+\dots\;,$

where

(3.4) $\displaystyle c_{n}$
$\displaystyle=na_{n}d_{1}+(n-1)a_{n-1}d_{2}+\dots+2a_{2}d_{n-1}+d_{n},$
$\displaystyle c_{n+1}$
$\displaystyle=na_{n}d_{2}+(n-1)a_{n-1}d_{3}+\dots+2a_{2}d_{n}+d_{n+1},$
$\displaystyle c_{n+2}$
$\displaystyle=na_{n}d_{3}+(n-1)a_{n-1}d_{4}+\dots+2a_{2}d_{n+1}+d_{n+2},$

and so on.
Putting $m=n+1$ in (3.2) and comparing with (3.4), we get,

$c_{n}=-(n+1)a_{n+1}.$

Now using Lemma 1.1$(c)$, we get

(3.5) $|c_{n}|\leq\dfrac{n+1}{2n},\qquad n\geq 1.$

Let $M$ be a constant such that

(3.6) $\left|\dfrac{1}{f^{\prime}(z)}\right|\leq M,\qquad z\in\mathbb{D}.$

Then, using Lemma 1.1$(b)$, we obtain that $M\geq 1$. Now using (3.6), for
$z=re^{i\theta}$, we get

$\displaystyle\dfrac{1}{2\pi}\int_{0}^{2\pi}\bigg{|}\dfrac{1}{f^{\prime}(re^{i\theta})}\bigg{|}^{2}d\theta$
$\displaystyle=\dfrac{1}{2\pi}\int_{0}^{2\pi}|1+d_{1}re^{i\theta}+d_{2}r^{2}e^{2i\theta}+\dots|^{2}d\theta$
(3.7) $\displaystyle=1+\sum_{k=1}^{\infty}|d_{k}|^{2}r^{2k}\leq M^{2}.$

Allowing $r\rightarrow 1^{-}$, we obtain

$\sum_{k=1}^{\infty}|d_{k}|^{2}\leq M^{2}-1.$

From this, we conclude that

(3.8) $|d_{k}|\leq\sqrt{M^{2}-1},$

for each $k\in\mathbb{N}$.

Observing the series $c_{n+1}$, $c_{n+2}$, and so on, for $m\geq n+1$, we
obtain

$c_{m}=na_{n}d_{m-n+1}+(n-1)a_{n-1}d_{m-n+2}+\dots+2a_{2}d_{m-1}+d_{m}.$

Using Lemma 1.1$(c)$ and (3.8), we get for $m\geq n+1$,

(3.9) $\displaystyle|c_{m}|$
$\displaystyle\leq\sum_{k=2}^{n}\dfrac{k}{2(k-1)}|d_{m-k+1}|+|d_{m}|$
$\displaystyle\leq\left(1+\dfrac{1}{2}\sum_{k=2}^{n}\dfrac{k}{k-1}\right)\sqrt{M^{2}-1}$
$\displaystyle=\left(1+\dfrac{1}{2}\sum_{k=2}^{n}\left(1+\dfrac{1}{k-1}\right)\right)\sqrt{M^{2}-1}$
$\displaystyle=\left(1+\dfrac{n-1}{2}+\dfrac{1}{2}\sum_{k=1}^{n-1}\dfrac{1}{k}\right)\sqrt{M^{2}-1}$
$\displaystyle\approx(n+1+\ln(n-1)+\gamma)\;\dfrac{\sqrt{M^{2}-1}}{2},$

where $\gamma\approx 0.57722$ is the Euler-Mascheroni constant. Further using
(3.6), for $m\geq n+1$, we get

(3.10)
$|c_{m}|\leq\dfrac{\sqrt{2r-r^{2}}}{2(1-r)}(n+1+\ln(n-1)+\gamma):=A_{n}.$

Now using (3.5) and (3.10) in (3.3), we get

(3.11)
$\displaystyle\bigg{|}\dfrac{s_{n}^{\prime}(z,f)}{f^{\prime}(z)}-1\bigg{|}$
$\displaystyle=|c_{n}z^{n}+c_{n+1}z^{n+1}+\dots|$
$\displaystyle\leq|c_{n}||z^{n}|+|c_{n+1}||z|^{n+1}+\dots$
$\displaystyle\leq|z|^{n}\left(\dfrac{n+1}{2n}+A_{n}|z|+A_{n}|z|^{2}+\dots\right)$
$\displaystyle=|z|^{n}\left(\dfrac{n+1}{2n}+A_{n}\dfrac{|z|}{1-|z|}\right),$

for $|z|=r<1$. This completes the proof of the theorem. ∎

######  Theorem 3.2.

Let $f(z)\in\Omega$. Then, for each $r\in(0,1)$ and $n\geq 2$, we have

(3.12)
$\bigg{|}\dfrac{s_{n}^{\prime}(z,f)}{f^{\prime}(z)}-1\bigg{|}\leq|z|^{n}\left(\dfrac{n+1}{2n}+B_{n}\dfrac{|z|}{r-|z|}\right),\quad|z|<r,$

where $B_{n}=\dfrac{\sqrt{2r-r^{2}}}{2(1-r)r^{n}}(n+1+\ln(n-1)+\gamma)$.

###### Proof.

Let $f(z)\in\Omega$. Following the proof of previous Theorem 3.1 and using
(3.9), we have

$\displaystyle|c_{m}|$
$\displaystyle\leq\sum_{k=2}^{n}\dfrac{k}{2(k-1)}|d_{m-k+1}|+|d_{m}|,$ (3.13)
$\displaystyle\leq\sum_{k=2}^{n}\left(\dfrac{k}{2(k-1)r^{m-k+1}}|d_{m-k+1}|\right)+\dfrac{1}{r^{m}}|d_{m}|,$

for any arbitrary fixed $r\in(0,1)$ and $m\geq n+1$. From (3), we get

(3.14) $|d_{k}|r^{k}\leq\sqrt{M^{2}-1},$

for each $k\in\mathbb{N}$. Now using the inequality (3.14) in (3), we get

$\displaystyle|c_{m}|$
$\displaystyle\leq\left(1+\dfrac{1}{2}\sum_{k=2}^{n}\dfrac{k}{k-1}\right)\dfrac{\sqrt{M^{2}-1}}{r^{m}}$
$\displaystyle=\left(1+\dfrac{n-1}{2}+\dfrac{1}{2}\sum_{k=1}^{n-1}\dfrac{1}{k}\right)\dfrac{\sqrt{M^{2}-1}}{r^{m}}$
$\displaystyle\approx(n+1+\ln(n-1)+\gamma)\dfrac{\sqrt{M^{2}-1}}{2r^{m}},$

for $r\in(0,1)$ and $\gamma\approx 0.57722$. Now using (3.6) here, for $m\geq
n+1$ we have

(3.15)
$|c_{m}|\leq\dfrac{1}{r^{m}}\left(\dfrac{\sqrt{2r-r^{2}}}{2(1-r)}(n+1+\ln(n-1)+\gamma)\right):=\dfrac{1}{r^{m}}D_{n}.$

Now using (3.5) and (3.15) in (3.3), we get

$\displaystyle\bigg{|}\dfrac{s_{n}^{\prime}(z,f)}{f^{\prime}(z)}-1\bigg{|}$
$\displaystyle\leq|c_{n}||z^{n}|+|c_{n+1}||z|^{n+1}+\dots$
$\displaystyle\leq|z|^{n}\left(\dfrac{n+1}{2n}+\dfrac{D_{n}}{r^{n}}\left(\dfrac{|z|}{r}+\dfrac{|z|^{2}}{r^{2}}\dots\right)\right)$
$\displaystyle=|z|^{n}\left(\dfrac{n+1}{2n}+\dfrac{D_{n}}{r^{n}}\dfrac{|z|/r}{1-|z|/r}\right)$
$\displaystyle=|z|^{n}\left(\dfrac{n+1}{2n}+B_{n}\dfrac{|z|}{r-|z|}\right),\quad|z|<r,$

where $B_{n}=\dfrac{\sqrt{2r-r^{2}}}{2(1-r)r^{n}}(n+1+\ln(n-1)+\gamma)$ and
$\gamma\approx 0.57722$ is the Euler-Mascheroni constant. This completes the
proof of the theorem. ∎

######  Theorem 3.3.

Let $f(z)\in\Omega$. Then $\Re(s^{\prime}_{n}(z,f))>0$ in $|z|\leq r_{0}$ for
$n\geq 12$, where $r_{0}=0.547$.

###### Proof.

If $f(z)\in\Omega$, then in view of [11, Equation 3.6],

$f^{\prime}(z)=1+\dfrac{1}{2}z\phi(z)+\dfrac{1}{2}z\int_{0}^{1}\phi(zt)dt,$

where $|\phi(z)|\leq 1$. Hence

$|f^{\prime}(z)-1|=\left|\dfrac{1}{2}\phi(z)+\dfrac{1}{2}z\int_{0}^{1}\phi(zt)dt\right|<|z|.$

Therefore, using the maximum modulus principle for $|z|\leq r_{0}=0.547$, we
obtain

(3.16) $\underset{|z|=r_{0}}{\max}|\arg
f^{\prime}(z)|\leq\sin^{-1}(r_{0})<33.16^{\circ}.$

Now taking $r=0.99$ in inequality (3.12), we get

$\bigg{|}\dfrac{s_{n}^{\prime}(z;f)}{f^{\prime}(z)}-1\bigg{|}\leq
C_{n},\quad|z|<r_{0}.$

where,

$C_{n}=(0.547)^{n}\left(\dfrac{n+1}{2n}+\dfrac{61.735}{(0.99)^{n}}(n+1+\ln(n-1)+\gamma)\right).$

Now, using the maximum modulus principle, we obtain

(3.17)
$\underset{|z|=r_{0}}{\max}\bigg{|}\arg\dfrac{s_{n}^{\prime}(z;f)}{f^{\prime}(z)}\bigg{|}\leq\sin^{-1}(C_{n}),\quad|z|<r_{0}.$

Using (3.16) and (3.17)

$|\arg s^{\prime}_{n}(z;f)|\leq|\arg
f^{\prime}(z)|+\bigg{|}\arg\dfrac{s_{n}^{\prime}(z;f)}{f^{\prime}(z)}\bigg{|}<33.16^{\circ}+\sin^{-1}(C_{n})<\dfrac{\pi}{2}.$

This inequality holds true if $\sin^{-1}(C_{n})\leq 56.84^{\circ}$. Now the
result follows, as the last inequality holds true for $n\geq 12$. ∎

##  4\. Properties of the Class $\Omega_{\lambda}$

The functions in class $\Omega_{\lambda}$ are starlike in $\mathbb{D}$ when
$\lambda\in\left(0,1/2\right]$. If $f(z)$ has the form (1.6), then

$\Re\left(\dfrac{zf^{\prime}(z)}{f(z)}\right)\geq 1-\left|\dfrac{2\lambda
z+2\lambda z^{2}}{4+2\lambda z+\lambda z^{2}}\right|\geq
1-\;\dfrac{2\lambda|z|+2\lambda|z|^{2}}{4-2\lambda|z|-\lambda|z|^{2}}>1-\;\dfrac{4\lambda}{4-3\lambda}\geq
0,\quad(z\in\mathbb{D}),$

whenever $\lambda\leq 4/7$. That is, $f$ is not starlike when $\lambda>4/7$.
Thus, all functions in $\Omega_{\lambda}$ are not starlike when $\lambda>1/2$.
Therefore, in the next result, we obtain a subdisk of $\mathbb{D}$ where all
the functions in class $\Omega_{\lambda}$ are starlike when $\lambda>1/2$.

######  Theorem 4.1.

The radius of starlikeness of $\Omega_{\lambda}$ is $\dfrac{1}{2\lambda}$.
Further, if $f(z)\in\Omega_{\lambda}$, then

(4.1) $\displaystyle|z|-\lambda|z|^{2}\leq$
$\displaystyle\;|f(z)|\leq\;|z|+\lambda|z|^{2},$ (4.2) $\displaystyle
1-2\lambda|z|\leq$ $\displaystyle\;|f^{\prime}(z)|\leq\;1+2\lambda|z|.$

Equality occurs in both inequalities if and only if $f(z)=z+\lambda\eta z^{2}$
for $z\in\mathbb{D}$ and $z\neq 0$, where $|\eta|=1$.

###### Proof.

If $f(z)\in\Omega_{\lambda}$, then from (1.4) we obtain

(4.3) $zf^{\prime}(z)-f(z)=\lambda z^{2}\phi(z),$

where $\phi(z)$ is analytic in $\mathbb{D}$ such that $|\phi(z)|\leq 1$.
Equation (4.3) is equivalent to

(4.4) $f(z)=z+\lambda z\int_{0}^{z}\phi(\zeta)d\zeta=z+\lambda
z^{2}\int_{0}^{1}\phi(zt)dt.$

Using the bound of $\phi$ for $z\in\mathbb{D}$ and setting the inequality to
both sides of $|f(z)|$, we get the estimate of (4.1). Further,

(4.5) $f^{\prime}(z)=1+\lambda z\phi(z)+\lambda z\int_{0}^{1}\phi(zt)dt.$

Now using the bound of $\phi$ for $z\in\mathbb{D}$ and setting the inequality
to both sides of $|f^{\prime}(z)|$, we get the estimate of (4.2). The equality
in both estimates occurs if and only if $|\int_{0}^{1}\phi(zt)dt|=1$. It
follows that $|\phi(zt)|\equiv 1$ for $0\leq t\leq 1$. Thus
$\phi(z)\equiv\eta\;(|\eta|=1)$ by the maximum modulus theorem, and so
$f(z)=z+\lambda\eta z^{2}$ where $|\eta|=1$.

Moreover, using the lower bound of (4.1), we obtain

$\bigg{|}\dfrac{zf^{\prime}(z)}{f(z)}-1\bigg{|}\leq\dfrac{\lambda|z|^{2}}{|z|-\lambda|z|^{2}}<1$

if and only if $|z|<\dfrac{1}{2\lambda}$. Thus, $f$ is starlike in the disk
$|z|<\dfrac{1}{2\lambda}$. ∎

The functions in class $\Omega_{\lambda}$ are univalent in $\mathbb{D}$ and
convex in $\mathbb{D}_{1/2}$ when $\lambda\in\left(0,1/2\right]$. If $f(z)$
has the form (1.6), then

$\Re(f^{\prime}(z))\geq 1-\left|\lambda
z+\dfrac{3\lambda}{4}z^{2}\right|>1-\lambda-\dfrac{3\lambda}{4}\geq
0,\quad(z\in\mathbb{D}),$

whenever $\lambda\leq 4/7$. That is, $f$ is not univalent when $\lambda>4/7$.
Thus, all functions in $\Omega_{\lambda}$ are not univalent when
$\lambda>1/2$. Similarly, if $f(z)$ has the form (1.6), then

$\displaystyle\Re\left(1+\dfrac{zf^{\prime\prime}(z)}{f^{\prime}(z)}\right)$
$\displaystyle\geq 1-\left|\dfrac{\lambda
z+\dfrac{3\lambda}{2}z^{2}}{1+\lambda z+\dfrac{3\lambda}{4}z^{2}}\right|\geq
1-\dfrac{\lambda|z|+\dfrac{3\lambda}{2}|z|^{2}}{1-\lambda|z|-\dfrac{3\lambda}{4}|z|^{2}}$
$\displaystyle>1-\dfrac{\lambda+\dfrac{3\lambda}{2}}{1-\lambda-\dfrac{3\lambda}{4}}\;\geq\;0,\qquad(z\in\mathbb{D}),$

whenever $\lambda\leq 4/17$. That is, $f$ is not convex whenever
$\lambda>4/17$. Thus, all functions in $\Omega_{\lambda}$ are not convex when
$\lambda>1/2$. Therefore, we obtain subdisks of $\mathbb{D}$ and
$\mathbb{D}_{1/2}$ where the functions in class $\Omega_{\lambda}$ are
univalent and convex, respectively, when $\lambda>1/2$.

######  Theorem 4.2.

Let $f(z)\in\Omega_{\lambda}$. Then $f$ is univalent in
$|z|<\dfrac{1}{2\lambda}$ and convex in $|z|<\dfrac{1}{4\lambda}$.

###### Proof.

Let $f\in\Omega_{\lambda}$. Using (4.5), we obtain

(4.6) $|f^{\prime}(z)-1|\leq 2\lambda|z|<1.$

if $|z|<\dfrac{1}{2\lambda}$. Thus, $f$ is univalent in
$|z|<\dfrac{1}{2\lambda}$. Again, using (4.5), we obtain

(4.7) $f^{\prime\prime}(z)=\lambda\phi(z)+\lambda(z\phi(z))^{\prime}.$

Now using [7, Dieudonné’s Lemma (page 198)] for $z\phi(z)$, we obtain

(4.8)
$|(z\phi(z))^{\prime}|\leq|\phi(z)|+\dfrac{|z|^{2}-|z\phi(z)|^{2}}{|z|(1-|z|^{2})}.$

Using (4.8) in (4.7) and the bound of $\phi(z)$, we obtain

(4.9)
$|f^{\prime\prime}(z)|\leq\lambda|\phi(z)|+\lambda\left(|\phi(z)|+\dfrac{|z|^{2}-|z\phi(z)|^{2}}{|z|(1-|z|^{2})}\right)\leq
2\lambda.$

Now using the lower bound of (4.2) and using (4.9), we obtain

$\left|\dfrac{zf^{\prime\prime}(z)}{f^{\prime}(z)}\right|\leq\dfrac{2\lambda|z|}{1-2\lambda|z|}<1,$

if $|z|<\dfrac{1}{4\lambda}$. Thus, $f$ is convex in
$|z|<\dfrac{1}{4\lambda}$. ∎

######  Theorem 4.3.

Let $f(z)\in\mathcal{A}$. If $\,|f^{\prime\prime}(z)|\leq 2\lambda\,$ then
$\,f\in\Omega_{\lambda}$. The number $2\lambda$ is the best possible.

###### Proof.

Let $|f^{\prime\prime}(z)|\leq 2\lambda$. If $g(z)=zf^{\prime}(z)-f(z)$, then
$g^{\prime}(z)=zf^{\prime\prime}(z)=2\lambda z\phi(z)$, where $\,|\phi(z)|\leq
1$. Note that

$g(z)=2\lambda\int_{0}^{z}\zeta\phi(\zeta)d\zeta=2\lambda
z^{2}\int_{0}^{1}t\phi(zt)dt.$

Therefore,

$|g(z)|=\left|2\lambda
z^{2}\int_{0}^{1}t\phi(zt)dt\right|<2\lambda\int_{0}^{1}tdt=\lambda,\qquad
z\in\mathbb{D}.$

This implies $f(z)\in\Omega_{\lambda}$.

Further, let $|f^{\prime\prime}(z)|\leq\eta$ and $\eta>2\lambda$. Considering
$f(z)=z+\dfrac{1}{2}\eta z^{2}$, we get $|f^{\prime\prime}(z)|\leq\eta$. But
$f(Z)$ is not univalent in $\mathbb{D}_{1/2\lambda}$ as $f^{\prime}(z)$
vanishes at $\;-\,\dfrac{1}{\eta}\in\mathbb{D}_{1/2\lambda}$. Therefore, the
number $2\lambda$ is best possible. ∎

######  Theorem 4.4.

Let $f(z)\in\mathcal{A}$. If

(4.10) $|z^{2}f^{\prime\prime}(z)+zf^{\prime}(z)-f(z)|\leq 3\lambda,$

then $f(z)\in\Omega_{\lambda}$. The number $3\lambda$ is best possible.

###### Proof.

Let $f(z)\in\mathcal{A}$ satisfy the inequality (4.10). Then

$(z^{2}f^{\prime}(z)-zf(z))^{\prime}=3\lambda z^{2}\phi(z),$

where $\phi$ is analytic in $\mathbb{D}$ such that $|\phi(z)|\leq 1$. Thus,

$z^{2}f^{\prime}(z)-zf(z)=3\lambda\int_{0}^{z}\zeta^{2}\phi(\zeta)d\zeta=3\lambda
z^{3}\int_{0}^{1}t^{2}\phi(zt)dt,$

consequently,

$|zf^{\prime}(z)-f(z)|=\left|3\lambda
z^{2}\int_{0}^{1}t^{2}\phi(zt)dt\right|<3\lambda\int_{0}^{1}t^{2}dt=\lambda,\quad
z\in\mathbb{D}.$

This implies $f(z)\in\Omega_{\lambda}$. Let
$|z^{2}f^{\prime\prime}(z)+zf^{\prime}(z)-f(z)|\leq\eta$ and $\eta>3\lambda$.
Considering $f(z)=z+\dfrac{1}{3}\eta z^{2}$, we obtain
$|z^{2}f^{\prime\prime}(z)+zf^{\prime}(z)-f(z)|=|\eta
z^{2}|\leq\eta,\,z\in\mathbb{D}$. But $f(z)$ is not univalent in
$\mathbb{D}_{1/2\lambda}$ as $f^{\prime}(z)$ vanishes at
$\;-\,\dfrac{3}{2\eta}\in\mathbb{D}_{1/2\lambda}$. Thus, the number $3\lambda$
is best possible. ∎

######  Theorem 4.5.

Let $f_{1}(z)$, $f_{2}(z)\in\Omega_{\lambda}$. Then $(f_{1}\ast
f_{2})(z)\in\Omega_{\lambda}$, if $\lambda\leq 1$.

###### Proof.

Let $f_{1}(z)$, $f_{2}(z)\in\Omega_{\lambda}$. Then

$f_{1}(z)=z+\lambda\int_{0}^{1}z^{2}\phi_{1}(zt)dt,\qquad
f_{2}(z)=z+\lambda\int_{0}^{1}z^{2}\phi_{2}(zt)dt,$

where $|\phi_{1}(z)|\leq 1$ and $|\phi_{2}(z)|\leq 1$. Now

$\displaystyle f_{1}(z)\ast f_{2}(z)$
$\displaystyle=\left(z+\lambda\int_{0}^{1}z^{2}\phi_{1}(zt)dt\right)\ast
f_{2}(z)$
$\displaystyle=z+\lambda\int_{0}^{1}\bigg{(}z^{2}\phi_{1}(zt)\bigg{)}\ast\left(z+\lambda\int_{0}^{1}z^{2}\phi_{2}(z\zeta)d\zeta\right)dt$
$\displaystyle=z+\lambda^{2}\int_{0}^{1}\bigg{(}\int_{0}^{1}z^{2}[\phi_{1}(zt)]\ast[\phi_{2}(z\zeta)]d\zeta\bigg{)}dt$
$\displaystyle=z+\lambda\int_{0}^{1}z^{2}\phi_{0}(zt)dt,$

where
$\phi_{0}(z)=\lambda\displaystyle{\int_{0}^{1}[\phi_{1}(z)]\ast[\phi_{2}(z\zeta)]d\zeta}$.
Now, using [11, Lemma 3.6], we obtain

$|\phi_{0}(z)|\leq\lambda\int_{0}^{1}|\phi_{1}(z)\ast\phi_{2}(z\zeta)|d\zeta\leq\lambda.$

The result follows if $\lambda\leq 1$. ∎

######  Theorem 4.6.

Let $f(z)=z+\sum_{k=2}^{\infty}a_{k}z^{k}\in\mathcal{A}$. If

(4.11) $\sum_{k=2}^{\infty}(k-1)|a_{k}|<\lambda,$

then $f(z)\in\Omega_{\lambda}$. In particular, for $f(z)\in\Omega_{\lambda}$,
$|a_{k}|\leq\dfrac{\lambda}{k-1}$ for all $k\geq 2$.

###### Proof.

Using (4.11) we obtain

$|zf^{\prime}(z)-f(z)|=\bigg{|}\sum_{k=2}^{\infty}ka_{k}z^{k}-\sum_{k=2}^{\infty}a_{k}z^{k}\bigg{|}\leq\sum_{k=2}^{\infty}(k-1)|a_{k}|<\lambda.$

Hence $f(z)\in\Omega_{\lambda}$. Now, for $f(z)\in\Omega_{\lambda}$

$zf^{\prime}(z)-f(z)=\sum_{k=2}^{\infty}(k-1)a_{k}z^{k}\prec\lambda
z^{2}\prec\lambda z,\quad z\in\mathbb{D}.$

Since $\lambda z$ is convex univalent in $\mathbb{D}$, we obtain the
particular result using Lemma 1.3. ∎

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