# MadDog:
A Web-based System for Acronym Identification and Disambiguation

Amir Pouran Ben Veyseh1, Franck Dernoncourt2,
Walter Chang2, and Thien Huu Nguyen1
1 Department of Computer and Information Science, University of Oregon,
Eugene, OR 97403, USA
2 Adobe Research, San Jose, CA, USA
<EMAIL_ADDRESS>
<EMAIL_ADDRESS>
###### Abstract

Acronyms and abbreviations are the short-form of longer phrases and they are
ubiquitously employed in various types of writing. Despite their usefulness to
save space in writing and reader’s time in reading, they also provide
challenges for understanding the text especially if the acronym is not defined
in the text or if it is used far from its definition in long texts. To
alleviate this issue, there are considerable efforts both from the research
community and software developers to build systems for identifying acronyms
and finding their correct meanings in the text. However, none of the existing
works provide a unified solution capable of processing acronyms in various
domains and to be publicly available. Thus, we provide the first web-based
acronym identification and disambiguation system which can process acronyms
from various domains including scientific, biomedical, and general domains.
The web-based system is publicly available at http://iq.cs.uoregon.edu:5000
and a demo video is available at https://youtu.be/IkSh7LqI42M. The system
source code is also available at https://github.com/amirveyseh/MadDog.

##  1 Introduction

Textual contents such as books, articles, reports, and web-blogs in various
domains are replete with phrases that are commonly used by people in that
field. In order to save space in text writing and also facilitate
communication among people who are already familiar with these phrases, the
shorthanded form of long phrases, known as acronyms and abbreviations, are
frequently used. However, the use of acronyms could also introduce challenges
to understand the text, especially for newcomers. More specifically, two types
of challenges might hinder reading text with acronyms: 1) In long documents,
e.g., a book chapter, an acronym might be defined somewhere in the text and
used several times throughout the document. For someone who is not familiar
with the definition of the acronym and interested in reading a part of the
document, it might be time-consuming to find the definition of the acronym in
the document. To solve this problem, an automatic acronym identification tool
is required whose goal is to find all acronyms and their definitions that are
locally provided in the same document. 2) Some of the acronyms might not be
even defined in the document itself. These acronyms are commonly used by
writers in a specific domain. To find the correct meaning of them, a reader
must look-up the acronym in a dictionary of acronyms. However, due to the
shorter length of acronyms compared to their long-form, multiple phrases might
be shortened with the same acronym, thereby, they will be ambiguous. In these
cases, a deep understanding of the domain is required to recognize the correct
meaning of the acronym among all possible long-forms. To solve this issue, a
system capable of disambiguating an acronym based on its context is necessary.

Each of the aforementioned problems, i.e., acronym identification (AI) and
acronym disambiguation (AD), has been extensively studied by the research
community or software developers. One of the methods which are widely used in
acronym identification research is proposed by Schwartz and Hearst (2002).
This is a rule-based model that utilizes character-match between acronym
letters and their context to find the acronym and its long-form in text.
Later, some feature-based models have been also used for acronym
identification Kuo et al. (2009); Liu et al. (2017). In addition, some of the
existing software employs regular expressions for acronym identification in
the biomedical domain Gooch (2011). Acronym disambiguation is also approached
with feature-based models Wang et al. (2016) or more advanced deep learning
methods Wu et al. (2015); Ciosici et al. (2019). The majority of deep models
employ word embeddings to compute the similarity between the candidate long-
form and the acronym context. In addition to the existing research for AD,
there is some web-based software that employ dictionary look-up to expand an
acronym to its long-form ABBREX (2018). Note that the methods based on
dictionary look-up are not able to disambiguate the acronym if it has multiple
meanings.

Despite the progress made on the AI and AD task in the last two decades, there
are some limitations in the prior works that prevent achieving a functional
system to be used in practice. More specifically, considering the research on
the AD task, all of the prior works employ a small-size dataset covering a few
hundred to a few thousand long-forms in a specific domain. Therefore, the
models trained in these works are not capable to expand all acronyms of a
domain or acronyms in other domains other than the one used in the training
set. Although in the recent work Wen et al. (2020), authors proposed a big
dataset for acronym disambiguation in the medical domain with more than 14
million samples, it is still limited to a specific domain (i.e., medical
domain). Another limitation in prior works is that they do not provide a
unified system capable of performing both tasks in various domains and to be
publicly available. To our knowledge, the only exiting web-based system for AI
and AD is proposed by Ciosici and Assent (2018). For acronym identification,
this system employs the rule-based model introduced by Schwartz and Hearst
(2002). To handle corner cases, they add extra rules in addition to Schwartz’s
rules in their system. Unfortunately, they do not provide detailed information
about these corner cases and extra rules or any evaluation to assess the
performance of the model. For acronym disambiguation, they resort to a
statistical model in which a pre-computed vector representation for each
candidate long-form is employed to compute the similarity between candidate
long-form with the context of the ambiguous acronym represented using another
vector. However, there are two limitations with this approach: first, the pre-
computed long-form vectors are obtained via only Wikipedia, thus limiting this
system to the general domain and incapable of disambiguating acronyms in other
domains such as scientific papers or biomedical texts; Second, the AD model
based on the pre-computed vectors is a statistical model and is not benefiting
from the advanced deep architectures, thereby it might have inferior
performance compared to a deep AD model.

To address the shortcomings and limitations of the prior research works or
systems for AI and AD, in this work, we introduce a web-based system for
acronym identification and disambiguation that is capable of recognizing and
expanding acronyms in multiple domains including general (e.g., Wikipedia
articles), scientific (e.g., computer science papers), biomedical (e.g.,
Medline abstracts), or financial (e.g., financial discussions in Reddit). More
specifically, we first propose a rule-based model for acronym identification
by extending the set of rules proposed by Schwartz and Hearst (2002). We
empirically show that the proposed model outperforms both the previous rule-
based model and also the existing state-of-the-art deep learning models for
acronym identification on the recent benchmark dataset SciAI Veyseh et al.
(2020). Next, we use a large dataset created from corpora in various domains
for acronym disambiguation to train a deep model for this task. Specifically,
we employ a sequential deep model to encode the context of the ambiguous
acronym and solve the AD task using a feed-forward multi-class classifier. We
also evaluate the performance of the proposed acronym disambiguation model on
the recent benchmark dataset SciAD Veyseh et al. (2020).

To summarize, our contributions are:

  * •

The first web-based multi-domain acronym identification and disambiguation
system

  * •

Extensive evaluation of the proposed model on the two benchmark datasets SciAI
and SciAD

##  2 System Description

The proposed system is a web-based system consisting of two major components:
1) Acronym Identification which consists of a set of prioritized rules to
recognize the mentions of acronyms and their long-forms in the text; 2)
Acronym Expansion which involves a dictionary look-up to expand acronyms with
only one possible long-form and a pre-trained deep learning model to predict
the long-form of an ambiguous acronym using its context. The system takes as
input a piece of text and returns the text with highlighted acronyms in which
the user can click on the acronyms and their long-form will be shown in a pop-
up window. The acronym glossary extracted from the text is also shown at the
end of the text. Note that users can also enable/disable the acronym expansion
component. This section studies the details of the aforementioned components.

###  2.1 Acronym Identification

Acronym Identification aims to find the mentions of acronyms and their long-
forms in text. This is the first stage in the proposed system to identify the
acronyms and their immediate definitions. Generally, this task is modeled as a
sequence labeling problem. In our system, we employ a rule-based model to
extract acronyms and their meanings from a given text. In particular, the
proposed AI model is a collection of rules mainly inspired by the rule
introduced in Schwartz and Hearst (2002). More specifically, the following
rules are employed in the proposed AI model:

  * •

Acronym Detector: This rule identifies all acronyms in text, regardless of
having an immediate definition or not. Specifically, all words that at least
60% of their characters are upper-cased letters and the number of their
characters is between 2 and 10 are recognized as an acronym (i.e., short-
form).

  * •

Bounded Schwartz’s: Similar to Schwartz and Hearst (2002), we look for
immediate definitions of detected acronyms if they follow one of the templates
long-form (short-form) or short-form (long-form). In particular, considering
the first template, we take the $min(|A|+5,2*|A|)$ words, where $|A|$ is the
number of characters in the acronym, that appear immediately before the
parentheses as the candidate long-form111Note that we use the same candidate
long-form in other rules too. Then, a sub-sequence of the candidate long-form
that some of its characters could form the acronym is selected as the long-
form. However, despite the original Schwartz’s rule that does not restrict the
first and last word of the long-form to be used in the acronym, we enforce
this restriction. This modification could fix erroneous long-form detection by
Schwartz’s rule. For instance, in the phrase User-guided Social Media Crawling
method (USMC), the modified rule identifies the long-form User-guided Social
Media Crawling, excluding the leading word method.

  * •

Character Match: While the Bounded Schwartz’ rule could identify the majority
of the long-forms, it might also introduce some noisy meanings. For instance,
in the phrase Analyzing Avatar Boundary Matching (AABM), the Bounded
Schwartz’s rule identifies Avatar Boundary Matching as the long-form of AABM,
missing the starting word Analyzing. To solve this issue and increase the
model’s accuracy, we also employ a character match rule that assesses if the
initials of the words in the candidate long-form could form the acronym. In
the given example, it identifies the full phrase Analyzing Avatar Boundary
Matching as the long-form. Since this rule is more restricted and it has
higher precision than Bounded Schwartz’s rule, in our system, it has a higher
priority than the Bounded Schwartz’s rule.

  * •

Initial Capitals: One issue with the proposed Character Matching rule is that
if there is a word in the long-form that is not used in the acronym, the rule
fails to correctly identify the long-form. For instance, in the phrase
Analysis of Avatar Boundary Matching (AABM) the Character Matching rule fails
due to the existence of the word of. To mitigate this issue, we propose
another high-precision rule, Initial Capitals. In this rule, if the
concatenation of the initials of the words of the candidate long-form which
are upper-cased could form the acronym, the candidate is selected as the
expanded form of the acronym. This rule has the highest priority in our
system.

In addition to the mentioned general rules, we also add some other rules to
handle the special cases (e.g., acronyms with a hyphen, roman numbers,
definitions provided in some templates (e..g, CNN stands for convolution
neural network)).

Acronym | Long-form | Rule
---|---|---
AABM | Analyzing Avatar Boundary Matching | Character Match
ABBREX | Abbreviation Expander | Bounded Schwartz’s
AD | acronym disambiguation | Character Match
AI | Acronym identification | Character Match
BADREX | Biomedical Abbreviations using | Bounded Schwartz’s
| Dynamic Regular Expressions |
BiLSTM | Bi - directional Long ShortTerm Memory | Bounded Schwartz’s
DOG | Diverse acrOnym Glossary | Bounded Schwartz’s
MAD | Massive Acronym Disambiguation | Capital Initials
MF | most frequent | Character Match
USMC | User - guided Social Media Crawling | Capital Initials

Table 1: The acronym glossary extracted from the text of this paper using
MadDog.

In the web-based system, the user could enter the text and the system
recognizes both acronyms without any definition in text and also acronyms that
are locally defined with their identified long-forms. Users could also click
on each detected acronym to see its definition in a pop-up window. Also, a
glossary of detected acronyms and their long-forms is shown at the bottom of
the page. A screenshot of the output of the system is shown in Figure 1.
Moreover, Table 1 shows the glossary extracted from the text of this paper
using the rule-based component of the system. In section 3 we compare the
performance of the proposed rule-based model with the existing state-of-the-
art models for AI Veyseh et al. (2020).

Figure 1: A screenshot of acronym identification by MadDog. It identifies all
acronyms and their local-long forms. This interface highlights the detected
acronyms and by clicking on them, a pop-up window shows the recognized meaning
of the acronym.

###  2.2 Acronym Expansion

Although the proposed rule-based model is effective to recognize locally
defined acronyms, it might not be able to expand acronyms that don’t have any
immediate definition in the text itself. To alleviate this issue and expand
acronyms even without local definition, two resources are required: 1) A
dictionary that provides the list of possible expansion for a given acronym,
2) A model to exploit the context of the given acronym and choose the most
likely expansion for a given acronym. For the acronym dictionary, we employ
the glossary obtained by exploiting our proposed rule-based AI model on
corpora in various domains (i.e., Wikipedia, Arxiv papers, Reddit submissions,
Medline abstracts, and PMC OA subset). The obtained glossary, named as Diverse
acrOnym Glossary (DOG), contains 426,389 unique acronyms and 3,781,739 unique
long-forms. Note that the previously available web-based acronym
disambiguation system Ciosici and Assent (2018) employed only Wikipedia
corpus, therefore, it covers limited domains and acronyms compared to our
system.

Model | Acronym | Long Form |
---|---|---|---
| P | R | F1 | P | R | F1 | Macro F1
NOA | 80.31 | 18.08 | 29.51 | 88.97 | 14.01 | 24.20 | 26.85
ADE | 79.28 | 86.13 | 82.57 | 98.36 | 57.34 | 72.45 | 79.37
UAD | 86.11 | 91.48 | 88.72 | 96.51 | 64.38 | 77.24 | 84.09
BIOADI | 83.11 | 87.21 | 85.11 | 90.43 | 73.79 | 77.49 | 82.35
LNCRF | 84.51 | 90.45 | 87.37 | 95.13 | 69.18 | 80.10 | 83.73
LSTM-CRF | 88.58 | 86.93 | 87.75 | 85.33 | 85.38 | 85.36 | 86.55
MadDog | 89.98 | 87.56 | 88.75 | 96.45 | 79.53 | 87.18 | 88.12

Table 2: Performance of models for acronym identification (AI)

In DOG, the average number of long-forms per acronym is 6.9 and 81,372
ambiguous acronyms exist. Due to this ambiguity, a simple dictionary look-up
is not sufficient for acronym expansion in the web-based system that uses DOG
to expand acronyms with non-local definitions. In order to tackle this
problem, we propose to train a supervised model in which the input is the text
and the position of the ambiguous acronym in it and the model predicts the
correct long-form among all possible candidates. To train this model, we use
an automatically labeled dataset obtained by extracting samples from large
corpora for each long-form in DOG. This dataset contains 46 million records
and we call it the Massive Acronym Disambiguation (MAD) dataset. To split the
dataset into train/dev/test splits, we use 80% of samples of each long-form
for training, 10% for the development set, and 10% for the test set. It is
noteworthy that to facilitate training, before splitting the dataset into
train/dev/test splits, we first create chunks of size 100,000 samples in which
all samples of an acronym are assigned to the same chunk. Since each acronym
appears only in one chuck, we train a separate acronym disambiguation model
for each chunk. During inference, we first identify which chuck the ambiguous
acronym belongs to, then, we use the corresponding model to predict the
expanded form of the acronym.

In this work, we use a deep sequential model to be trained on the MAD dataset
for acronym disambiguation. More specifically, given the input text
$T=[w_{1},w_{2},\ldots,w_{n}]$ with the ambiguous acronym $w_{a}$, we first
represent each word using the corresponding GloVe embedding, i.e.,
$X=[x_{1},x_{2},\ldots,x_{n}]$. Afterward, the vectors $X$ are consumed by a
Bi-directional Long Short-Term Memory network (BiLSTM) to encode the
sequential order of the words. Next, we take the hidden states of the BiLSTM
neurons, i.e., $H=[h_{1},h_{2},\ldots,h_{n}]$, and compute the text
representation by computing the max-pool of the vectors $H$, i.e.,
$\bar{h}=MAX\\_POOL(h_{1},h_{2},\ldots,h_{n})$. Finally, the concatenation of
the text representation, i.e., $\bar{h}$, and the acronym representation,
i.e., $h_{a}$, is fed into a 2-layer feed-forward neural network whose final
layer dimension is equal to the total number of long-forms in the dataset
(i.e., dataset chunks explained above).

Figure 2: Sorted list of candidate long-forms along with their scores for the
acronym AFD in the sentence After 1991, the presidential system of government
by Act of Parliament was abolished, and by October 1994, the AFD was
integrated into the Prime Minister’s Office and concurrently the combined
armed forces authority was transferred to this government body.

In the proposed system, the long-form of acronyms predicted by the acronym
disambiguation model is presented in the glossary at the end of the page (See
Figure 1). Moreover, by clicking on the acronym word in text, a pop-up window
shows the model’s prediction and also the sorted list of other candidate long-
forms for the selected acronym. An example is shown in Figure 2. In the
provided example, the system correctly predicts Gross Domestic Production as
the long-form of the ambiguous acronym GDP. We name the proposed acronym
identification and disambiguation system as MAdDog.

##  3 Evaluation

This section provides more insight into the performance of the proposed
acronym identification and disambiguation models. To evaluate the performance
of the models in comparison with other state-of-the-art AI and AD models, we
report the performance of the proposed models on SciAI and SciAD benchmark
datasets Veyseh et al. (2020). We also compare the performance of the proposed
model with the baselines provided in the recent work Veyseh et al. (2020).
More specifically, on SciAI, we compare our model with rule-based models NOA
Charbonnier and Wartena (2018), ADE Li et al. (2018) and UAD Ciosici et al.
(2019); and also the feature-based models BIOADI Kuo et al. (2009) and LNCRF
Liu et al. (2017); and finally the SOTA deep model LSTM-CRF Veyseh et al.
(2020). For evaluation metrics, following prior work, we report precision,
recall, and F1 score for the acronym and long-form prediction and also their
macro-averaged F1 score. The results are shown in Table 2. This table shows
that our model outperforms both rule-based and more advanced feature-based or
deep learning models. More interestingly, while the proposed model has
comparable precision with the existing rule-based models, it enjoys higher
recall.

Model | P | R | F1
---|---|---|---
MF | 89.03 | 42.2 | 57.26
ADE | 86.74 | 43.25 | 57.72
NOA | 78.14 | 35.06 | 48.40
UAD | 89.01 | 70.08 | 78.37
BEM | 86.75 | 35.94 | 50.82
DECBAE | 88.67 | 74.32 | 80.86
GAD | 89.27 | 76.66 | 81.90
MadDog | 92.27 | 85.01 | 88.49

Table 3: Performance of models for acronym disambiguation (AD)

To assess the performance of the proposed acronym disambiguation model, we
evaluate its performance on the benchmark dataset SciAD Veyseh et al. (2020)
and compare it with the existing state-of-the-art models. Specifically, we
compare the model with non-deep learning models including most frequent (MF)
meaning Veyseh et al. (2020), feature-based model (i.e., ADE Li et al.
(2018)), and deep learning models including NOA Charbonnier and Wartena
(2018), UAD Ciosici et al. (2019), BEM Zettlemoyer and Blevins (2020), DECBAE
Jin et al. (2019) and GAD Veyseh et al. (2020). The results are shown in Table
3. This table demonstrates the effectiveness of the proposed model compared
with the baselines. Our hypothesis for the higher performance of the proposed
model is the massive number of training examples for all acronyms which
results in low generalization error.

##  4 Related Work

Acronym identification (AI) and acronym disambiguation (AD) are two well-known
tasks with several prior works in the past two decades. For AI, both rule-
based models Park and Byrd (2001); Wren and Garner (2002); Schwartz and Hearst
(2002); Adar (2004); Nadeau and Turney (2005); Ao and Takagi (2005); Kirchhoff
and Turner (2016) and supervised feature-based or deep learning models Kuo et
al. (2009); Liu et al. (2017); Veyseh et al. (2020); Pouran Ben Veyseh et al.
(2021) are utilized. Due to the higher accuracy of rule-based models, they are
dominantly used in the majority of the related works, especially to
automatically create acronym dictionary Ciosici et al. (2019); Li et al.
(2018); Charbonnier and Wartena (2018). However, the existing works prepare a
small-size dictionary in a specific domain. In contrast, in this work, we
first improve the existing rules for acronym identification, then, we use a
diverse acronym glossary in our system. For acronym disambiguation, prior
works employ either feature-based models Wang et al. (2016); Li et al. (2018)
or deep learning methods Wu et al. (2015); Antunes and Matos (2017);
Charbonnier and Wartena (2018); Ciosici et al. (2019); Pouran Ben Veyseh et
al. (2021). In this work, we also employ a sequential deep learning model for
AD. However, unlike prior work that proposes an acronym disambiguation model
for a specific domain and limited acronyms, our proposed model covers more
acronyms and it is able to expand an acronym in various domains.

Another common limitation of the existing research-based models for AI and AD
is that they do not provide any publicly available system that could be
quickly incorporated into a text-processing application. Although there is
some software for acronym identification such as expanding Biomedical
Abbreviations using Dynamic Regular Expressions (BADREX) Gooch (2011) or
Abbreviation Expander (ABBREX) ABBREX (2018), unfortunately, they are
incapable of acronym disambiguation. To our knowledge, the most similar work
to ours is proposed by Ciosici and Assent (2018). Specifically, similar to our
work, this web-based system is able to identify and expand acronym in text. A
rule-based model is employed for AI and this model is also used to create a
dictionary of acronyms. For AD, unlike our work that trains a deep model, they
use word embedding similarity to predict the most likely expansion. However,
there are some limitations to this previous system. Firstly, it is restricted
to the general domain (i.e., Wikipedia) and it covers a limited number of
acronyms. Second, it does not provide any analysis and evaluations of the
performance of the proposed model. Lastly, it is not publicly available
anymore. The proposed MadDog system could be useful for many downstream
applications including definition extraction Pouran Ben Veyseh et al. (2020a);
Spala et al. (2020, 2019), information extraction Pouran Ben Veyseh et al.
(2019, 2020b, 2020c) or question answering Perez et al. (2020)

##  5 System Deployment

MadDog is purely written in Python 3 and could be run as a FLASK Grinberg
(2018) server. For text toknization, it employs SpaCy 2 Honnibal and Montani
(2017). Also, the trained acronym expansion model requires PyTorch 1.7 and 64
GB of disk space. Note that all acronyms with their long-forms are encoded in
the trained model so they can perform both the dictionary look-up operation
and the disambiguation task. Moreover, the trained models could be loaded both
on GPU and CPU.

##  6 Conclusion

In this work, we propose a new web-based system for acronym identification and
disambiguation. For AI, we employ a refined set of rules which is shown to be
more effective than the previous rule-based and deep learning models.
Moreover, using a massive acronym disambiguation dataset with more than 46
million records in various domains, we train a supervised model for acronym
disambiguation. The experiments on the existing benchmark datasets reveal the
efficacy of the proposed AD model.

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††thanks: 2020 Mathematics Subject Classification: primary 39B52, secondary
39B32
Key words and phrases: Semigroups, multiplicative function, sine addition law.

#  A Kannappan-sine addition law on semigroups

Jafar Ahmed1 Ajebbar Omar2 and Elqorachi Elhoucien1 1 Ibn Zohr University,
Faculty of sciences, Department of mathematics, Agadir, Morocco  2 Sultan
Moulay Slimane University, Multidisciplinary Faculty, Department of
mathematics and computer science, Beni Mellal, Morocco
<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

Let $S$ be a semigroup and $z_{0}$ a fixed element in $S.$ We determine the
complex-valued solutions of the following Kannappan-sine addition law
$f(xyz_{0})=f(x)g(y)+f(y)g(x),x,y\in S.$

__

##  1 Introduction

Let $S$ be a semigroup. Let $\mathbb{C}$ be a field of complex numbers. The
sine addition law for unknown functions $f,g:S\longrightarrow\mathbb{C}$ is
the functional equation

$f(xy)=f(x)g(y)+f(y)g(x),\hskip 11.38092ptx,y\in S,$ (1.1)

The functional equation (1.1) has been solved on groups and semigroups
generated by their squares (see for example [2], [6] and [9]). The solutions
of the functional equation (1.1) on a general semigroup were given recently by
Ebanks [4, Theorem 3.5], [5, Theorem 2.1 ] and [3, Theorem 3.1]. Moreover, the
continuous solutions are also known on topological groups and semigroups. For
additional discussions of this functional equation and their history, see [1,
Chapter 13], [11, Chapter 4] and their references.
In this paper we consider the following Kannappan-sine addition law (1.1),
namely

$f(xyz_{0})=f(x)g(y)+f(y)g(x),\;x,y\in S,$ (1.2)

where $S$ is a semigroup and $z_{0}$ is a fixed element in $S$.
On a monoid with identity element $e$ the functional equation (1.2) becomes
$g(e)f(xy)+f(e)g(xy)=f(x)g(y)+f(y)g(x)$. So, if $f(e)=0$ or $g(e)=0$, the last
functional equation is the sine addition law, which was solved recently by
Ebanks on semigroups [3, 4]. When $f(e)=\beta\neq 0$ and $g(e)=\alpha\neq 0$
we can easily verify that $f,g$ satisfy the following cosine addition law
$2\bigl{[}\dfrac{f}{\beta}+\dfrac{g}{\alpha}](xy)=[\dfrac{f}{\beta}+\dfrac{g}{\alpha}](x)[\dfrac{f}{\beta}+\dfrac{g}{\alpha}](y)-[\dfrac{-f}{\beta}+\dfrac{g}{\alpha}](x)[\dfrac{-f}{\beta}+\dfrac{g}{\alpha}](y)$,
and from [5] we get the solutions of the functional equation (1.2) on monoids.
It should be noted that the traitement above on monoids are used in [12,
Corollary 2.4] to solve equation (1.2) on groups.
We notice also that, when $z_{0}$ is in the centre of $S$, then equation (1.2)
can be written as follows $f(x\ast y)=f(x)g(y)+f(y)g(x)$, where $x\ast
y=xyz_{0}$ is an associative law on $S$, and then the functional equations
(1.1) and (1.2) coincide, and as mentioned above their solutions are known, so
the contributions of the present paper to the theory of trigonometric
functional equations in the non-abelian case.
The special case of (1.2) in which $g=\dfrac{1}{2}f$, is the functional
equation

$f(xyz_{0})=f(x)f(y),\hskip 11.38092ptx,y\in S,$ (1.3)

which has been solved on semigroups by Stetkær [10, Proposition 16]. He proved
that any non-zero solution $f:S\longrightarrow\mathbb{C}$ of (1.3) has the
form $f=\chi(z_{0})\chi,$ where $\chi$ is exponential function on $S$ such
that $\chi(z_{0})\neq 0$. We shall use his result in the present paper.
The result by Kannappan [7] and its generalization (see for example [8],
[10]), has been an inspiration by its treatment of similar functional
equations on groups.
Our main goal is to extend works in the literature in which $S$ is a monoid or
a group, by showing that existence of an identity element or a group inversion
are not needed, because we present proofs that work for general semigroups. In
this context we obtain in Theorem 4.4 explicit formulas for the solutions of
(1.2).
Our main contributions to the knowledge about (1.2) are the following:
1\. We extend the setting from groups to semigroups.
2\. We relate the solutions of (1.2) to those of (1.1) and (1.3).
We also show that the functional equation

$f(xyz_{0})=f(x)g(y)-f(y)g(x),\hskip 11.38092ptx,y\in S$ (1.4)

has only trivial solutions on a semigroup $S$, with $z_{0}$ is a fixed element
in $S$.

##  2 Set up, notations and terminology

Throughout this paper we enforce the set up below.
$S$ is a semigroup (a set with an associative composition rule) and $z_{0}$ is
a fixed element in $S$.

If $X$ is a topological space we denote by $C(X)$ the algebra of continuous
functions from $X$ to the field of complex numbers $\mathbb{C}$. Let
$\mathbb{C}^{*}=\setminus\\{0\\}$.

A map $A:S\longrightarrow\mathbb{C}$ is said to be additive if

$A(xy)=A(x)+A(y)$ for all $x,y\in S$,

and a function $\chi:S\longrightarrow\mathbb{C}$ is multiplicative if

$\chi(xy)=\chi(x)\chi(y)$ for all $x,y\in S.$

If $\chi$ is multiplicative and $\chi\neq 0$ then we call $\chi$ an
exponential. For a multiplicative function $\chi:S\longrightarrow\mathbb{C}$,
we define the null space $I_{\chi}$ by

$I_{\chi}:=\\{x\in S\hskip 2.84544pt|\hskip 2.84544pt\chi(x)=0\\}$.

Then $I_{\chi}$ is either empty or a proper subset of $S$ and $I_{\chi}$ is a
two sided ideal in S if not empty and $S\setminus I_{\chi}$ is a subsemigroup
of $S$.

A function $f:S\longrightarrow\mathbb{C}$ is said to be central if
$f(xy)=f(yx)$ for all $x,y\in S$. A function $f$ is said to be abelian if
$f(x_{1}x_{2},...,x_{n})=f(x_{\sigma(1)}x_{\sigma(2)},...,x_{\sigma(n)})$ for
all $x_{1},x_{2},...,x_{n}\in S$, all permutations $\sigma$ of $n$ elements
and all $n\in\mathbb{N}$.
Note that additive and multiplicative functions are abelian.

For any subset $T\subseteq S$ let $T^{2}:=\\{xy\hskip 2.84544pt|\hskip
2.84544ptx,y\in T\\}$ and for any fixed element $z_{0}$ in S we let
$T^{2}z_{0}:=\\{xyz_{0}\hskip 2.84544pt|\hskip 2.84544ptx,y\in T\\}$.

To characterize and describe some solutions of our functional equations, we
use the partition of the null space $I_{\chi}$ into the disjoint union
$I_{\chi}=P_{\chi}\cup(I_{\chi}\setminus P_{\chi})$, with

$P_{\chi}:=\\{p\in I_{\chi}\setminus I_{\chi}^{2}\hskip 2.84544pt|\hskip
2.84544ptup,pv,upv\in I_{\chi}\setminus I_{\chi}^{2}\hskip
2.84544pt\text{for}\hskip 2.84544pt\text{all}\hskip 2.84544ptu,v\in S\setminus
I_{\chi}\\}.$

For more discussions about the partition of the null space $I_{\chi}$ we refer
to [4], [5] and [3]. The following lemma will be used later.

######  Lemma 2.1.

$p\in P_{\chi}\implies up,pv,upv\in P_{\chi}$ for all $u,v\in S\setminus
I_{\chi}\ $.

###### Proof.

Follows directly from the definition of $P_{\chi}$. ∎

##  3 Preparatory works

In this section we prove some useful lemmas that will be used in the proof of
our main results.

######  Lemma 3.1.

Let $S$ be a semigroup, $n\in\textit{N},$ and
$\chi_{1},\chi_{2},...\chi_{n}:S\longrightarrow\mathbb{C}$ be exponential
functions. Then
(a) $\\{\chi_{1},\chi_{2},...,\chi_{n}\\}$ is linearly independent.
(b) If $A:S\setminus I_{\chi}\longrightarrow\mathbb{C}$ is a non-zero additive
function and $\chi$ is exponential on $S$, then the set $\\{\chi A,\chi\\}$ is
linearly independent on $S\setminus I_{\chi}$.

###### Proof.

(a): See [11, Theorem 3.18]. (b): See [2, Lemma 4.4]. ∎

To obtain some of the solutions of the functional equation (1.2) we need the
solutions of the following functional equation

$f(xyz_{0})=f(x)+f(y),\hskip 8.5359ptx,y\in S.$ (3.1)

Lemma 3.2 describes the solutions of the functional equation (3.1) on
semigroups.

######  Lemma 3.2.

Let $S$ be a semigroup. The solutions of the functional equation (3.1) are the
function of the form $f=A+A(z_{0})$, where $A:S\longmapsto\mathbb{C}$ is
additive.

###### Proof.

Let $x,y,z\in S$ be arbitrary. Using (3.1) we find that

$f((xy)z_{0}^{2}z_{0})=f(xy)+f(z_{0}^{2}).$

Similarly we find that

$f((xyz_{0})z_{0}z_{0})=f(xyz_{0})+f(z_{0})=f(x)+f(y)+f(z_{0}).$

Comparing the previous equations we get that

$f(xy)=f(x)+f(y)+[f(z_{0})-f(z_{0}^{2})].$

Therefore, we see that $A:=f+[f(z_{0})-f(z_{0}^{2})]$ is additive. Now, using
(3.1) and the fact that $A$ is additive we obtain that

$\begin{split}0&=f(xyz_{0})-f(x)-f(y)\\\
&=A(xyz_{0})-[f(z_{0})-f(z_{0}^{2})]-(A(x)-[f(z_{0})-f(z_{0}^{2})])\\\
&\quad-(A(y)-[f(z_{0})-f(z_{0}^{2})])\\\
&=A(z_{0})+[f(z_{0})-f(z_{0}^{2})].\end{split}$

Thus $A(z_{0})=-[f(z_{0})-f(z_{0}^{2})]$, and hence $f=A+A(z_{0})$. The
converse statement is trivial. This completes the proof of Lemma 3.2. ∎

##  4 Solutions of the Kannappan-sine addition law on semigroups

In this section we solve the Kannappan-sine addition law (1.2) on semigroups.
The following lemmas will be used in the proof of Theorem 4.4.

######  Lemma 4.1.

Let $S$ be a semigroup, and let $f,g:S\longmapsto\mathbb{C}$ be a solution of
the functional equation (1.2). Then we have the following.
(i) If $f(z_{0})=0$ then

$g(z_{0}^{2})f(xy)=g(z_{0})[f(x)g(y)+f(y)g(x)]-f(z_{0}^{2})g(xy),\hskip
8.5359ptx,y\in S.$ (4.1)

(ii) If $f(z_{0})\neq 0$ then there exists $\gamma\in\mathbb{C}$ such that

$g(xyz_{0})=g(x)g(y)+\gamma^{2}f(x)f(y),\hskip 8.5359ptx,y\in S.$ (4.2)

###### Proof.

(i) Let $x,y,z\in$ $S$ be arbitrary. Using (1.2) and the fact that
$f(z_{0})=0$ we have

$f((xyz_{0})z_{0}z_{0})=f(xyz_{0})g(z_{0})+f(z_{0})g(xyz_{0})=g(z_{0})[f(x)g(y)+f(y)g(x)]$

and

$f((xyz_{0})z_{0}z_{0})=f((xy)z_{0}^{2}z_{0})=f(xy)g(z_{0}^{2})+f(z_{0}^{2})g(xy).$

Then we deduce that

$f(xy)g(z_{0}^{2})+f(z_{0}^{2})g(xy)=g(z_{0})[f(x)g(y)+f(y)g(x)],\hskip
8.5359ptx,y\in S.$

This proves (i).
(ii) Let $x,y\in S$ be arbitrary. By using (1.2) we get

$\begin{split}f((xyz_{0})z_{0}z_{0})&=f(xyz_{0})g(z_{0})+f(z_{0})g(xyz_{0})\\\
&=g(z_{0})[f(x)g(y)+f(y)g(x)]+f(z_{0})g(xyz_{0})\\\
&=g(z_{0})f(x)g(y)+g(z_{0})f(y)g(x)+f(z_{0})g(xyz_{0}).\end{split}$

Similarly, we find

$\begin{split}f(x(yz_{0}^{2})z_{0})&=f(x)g(yz_{0}^{2})+f(yz_{0}^{2})g(x)\\\
&=f(x)g(yz_{0}^{2})+g(x)[f(y)g(z_{0})+f(z_{0})g(y)]\\\
&=f(x)g(yz_{0}^{2})+g(z_{0})g(x)f(y)+f(z_{0})g(x)g(y).\end{split}$

Then the associativity of the operation in $S$ implies

$f(z_{0})[g(xyz_{0})-g(x)g(y)]=f(x)[g(yz_{0}^{2})-g(y)g(z_{0})].$ (4.3)

Since $f(z_{0})\neq 0$, dividing equation (4.3) by $f(z_{0})$ we get that

$g(xyz_{0})=g(x)g(y)+f(x)\phi(y),$ (4.4)

where

$\phi(y):=\dfrac{g(yz_{0}^{2})-g(y)g(z_{0})}{f(z_{0})},\hskip 5.69046pty\in
S.$

Now, substituting (4.4) come back into (4.3) we get

$f(z_{0})[f(x)\phi(y)]=f(x)[f(y)\phi(z_{0})].$

By putting $x=z_{0}$ in the last equation, we deduce that

$\phi(y):=cf(y),$ where $c:=\dfrac{\phi(z_{0})}{f(z_{0})}$.

Choosing $\gamma\in\mathbb{C}$ such that $\gamma^{2}=c$, equation (4.4)
becomes

$g(xyz_{0})=g(x)g(y)+\gamma^{2}f(x)f(y).$

This proves part (ii) and completes the proof of Lemma 4.1. ∎

######  Lemma 4.2.

Let $S$ be a semigroup, and let $f,g:S\longmapsto\mathbb{C}$ be a solution of
the functional equation (1.2) such that $f\neq 0$ and $f(z_{0})=0$. Then we
have the following.
(i) If $f(z_{0}^{2})\neq 0$ then $g(z_{0}^{2})=0$.
(ii) If $\\{f,g\\}$ is linearly independent then the following holds

(a) $g(z_{0})\neq 0.$

(b) If $f(z_{0}^{2})=0$ then $g(z_{0}^{2})\neq 0$.

###### Proof.

(i) Assume that $f(z_{0})=0$ and letting $x=y=$ $z_{0}$ in (1.2) we get that

$f(z_{0}^{3})=f(z_{0})g(z_{0})+f(z_{0})g(z_{0})=0.$

Similarly, letting $x=z_{0}^{2}$ and $y=z_{0}^{2}$ and then $x=z_{0}$ and
$y=z_{0}^{3}$ in (1.2) yields

$\begin{split}f(z_{0}^{5})&=2f(z_{0}^{2})g(z_{0}^{2})\\\
&=f(z_{0})g(z_{0}^{3})+f(z_{0}^{3})g(z_{0})=0.\end{split}$

Since $f(z_{0}^{2})\neq 0,$ we deduce that $g(z_{0}^{2})=0$.
(ii)- (a) Assume that $g(z_{0})=0$. Making the substitutions $(x,yz_{0})$ and
$(xy,z_{0})$ in (1.2) we get respectively that

$f(xyz_{0}z_{0})=f(x)g(yz_{0})+f(yz_{0})g(x)=f(xy)g(z_{0})+f(z_{0})g(xy)=0,$

which implies that

$f(x)g(yz_{0})+f(yz_{0})g(x)=0,\;x,y\in S.$

Since $f$ and $g$ are linearly independent we get in particular that
$f(yz_{0})=0$ for all $y\in S$. Now, by (1.2) we have

$f(x)g(y)=-f(y)g(x),\hskip 8.5359ptx,y\in S.$ (4.5)

Applying [11, Exercice 1.1(b)] we get $f=0$ or $g=0$, which is a contradiction
because $f$ and $g$ are linearly independent. So $g(z_{0})\neq 0$.
(ii)-(b) Assume, for the sake of contradiction, that $g(z_{0}^{2})=0$. Since
$f(z_{0})=0$, and $f(z_{0}^{2})=0$ by hypothesis, equation (4.1) reduces to

$g(z_{0})f(x)g(y)+g(z_{0})f(y)g(x)=0,$ $x,y\in S,$

which can be written as follows

$f(x)g(y)=-f(y)g(x),$ $x,y\in S,$

since $g(z_{0})\neq 0$ by (b)-(i). Then $f=0$, which contradicts the fact that
$f$ and $g$ are linearly independent. So $g(z_{0}^{2})\neq 0$. This completes
the proof of Lemma 4.2.∎

In the following Proposition we solve a special case of Kannappan-sine
addition law on semigroups, namely

$f(xyz_{0})=\chi(z_{0})f(x)\chi(y)+\chi(z_{0})f(y)\chi(x),\;x,y\in S,$ (4.6)

where $\chi$ is an exponential function on $S$. Proposition 4.3 is a
generalization of Ebanks’s work on the special case of the sine addition law
on monoids $(z_{0}=e)$ [4, Theorem 3.1].

######  Proposition 4.3.

Let $S$ be a semigroup and $\chi:S\longmapsto\mathbb{C}$ be an exponential
function such that $\chi(z_{0})\neq 0$. If $f:S\longmapsto\mathbb{C}$ is a
solution of functional equation (4.6), then

$f(x)=\begin{cases}\chi(x)(A(x)+A(z_{0}))\hskip 14.22636ptfor\hskip
8.5359ptx\in S\backslash I_{\chi}\\\ \rho(x)\hskip 79.6678ptfor\hskip
8.5359ptx\in P_{\chi}\\\ 0\hskip 93.89418ptfor\hskip 11.38092ptx\in
I_{\chi}\setminus P_{\chi},\end{cases}$ (4.7)

where $A:S\setminus$ $I_{\chi}\longrightarrow\mathbb{C}$ is additive and
$\rho:P_{\chi}\longrightarrow\mathbb{C}$ is the restriction of $f$ to
$P_{\chi}$.
In addition, $f$ is abelian and satisfies the following conditions:
(I) $f(xy)=f(yx)=0$ for all $x\in I_{\chi}\setminus P_{\chi}$ and $y\in
S\setminus I_{\chi}$.
(II) If $x\in\\{up,pv,upv\\}$ with $p\in P_{\chi}$ and $u,v\in S\setminus
I_{\chi}$, then $x\in P_{\chi}$ and we have respectively
$\rho(x)=\rho(p)\chi(u)$, $\rho(x)=\rho(p)\chi(v)$ or
$\rho(x)=\rho(p)\chi(uv)$.
Conversely, the function f of the form (4.7) define a solution of (4.6).
Moreover, if $S$ is a topological semigroup and $f\in C(S)$, then $\chi\in
C(S),A\in C(S\setminus I_{\chi})$ and $\rho\in C(P_{\chi})$.

###### Proof.

If $I_{\chi}=\emptyset$ then $\chi$ vanishes nowhere, so the functional
equation (4.6) reduces to

$\dfrac{f(xyz_{0})}{\chi(xyz_{0})}=\dfrac{f(x)}{\chi(x)}+\dfrac{f(y)}{\chi(y)},\hskip
11.38092ptx,y\in S.$ (4.8)

According to Lemma 3.2 we get that $f/\chi=:A+A(z_{0})$ where
$A:S\longrightarrow\mathbb{C}$ is additive. Then we deduce that
$f=\chi(A+A(z_{0}))$. This is a special case of (4.7) (here the middle and the
bottom lines of (4.7) are disappearing). So from now on we may assume that
$I_{\chi}\neq\emptyset$. For all $x,y\in S\setminus I_{\chi}$ we have
$\chi(xyz_{0})=\chi(x)\chi(y)\chi(z_{0})\neq 0$, because $\chi(z_{0})\neq 0$.
Therefore, dividing equation (4.6) by $\chi(xyz_{0})$ we get that

$\dfrac{f(xyz_{0})}{\chi(xyz_{0})}=\dfrac{f(x)}{\chi(x)}+\dfrac{f(y)}{\chi(y)},\hskip
11.38092ptx,y\in S\setminus I_{\chi}.$

By Lemma 3.2 we deduce that $f/\chi=:A+A(z_{0})$, where $A:S\setminus
I_{\chi}\longrightarrow\mathbb{C}$ is additive, then $f=\chi(A+A(z_{0}))$ and
we have the top line of (4.7).

Let $x,y\in I_{\chi}$. Using (4.6) we find that

$f((xy)(z_{0})z_{0})=f(xy)\chi(z_{0})\chi(z_{0})+f(z_{0})\chi(xy)\chi(z_{0})=f(xy)\chi(z_{0})^{2}$
(4.9)

and

$f(x(yz_{0})z_{0})=f(x)\chi(y)\chi(z_{0})\chi(z_{0})+f(yz_{0})\chi(x)\chi(z_{0})=0,$
(4.10)

since $\chi(xy)=\chi(x)=\chi(y)=0$. By using (4.9), (4.10) and the
associativity of the operation in $S$ we have $f(xy)\chi(z_{0})^{2}=0,$ which
gives $f(xy)=0$ since $\chi(z_{0})\neq 0$. This proves the case $x\in
I_{\chi}^{2}$ of the third line of formula (4.7).

Next suppose $x\in I_{\chi}\setminus(I_{\chi}^{2}\cup P_{\chi})$. Then there
exist $u,v\in S\setminus I_{\chi}$ such that one of $ux,vx$ or $uxv\in
I_{\chi}^{2}$. For $ux\in I_{\chi}^{2}$, by using (4.6) again we have

$\begin{split}f(z_{0}(ux)z_{0})&=f(z_{0})\chi(u)\chi(x)\chi(z_{0})+f(ux)\chi(z_{0})\chi(z_{0})=0\\\
&=f((z_{0}u)(x)z_{0})=f(z_{0}u)\chi(x)\chi(z_{0})+f(x)\chi(z_{0}u)\chi(z_{0})\\\
&=f(x)\chi(u)\chi(z_{0})^{2},\end{split}$

because $f(ux)=\chi(x)=0$. It follows that $f(x)\chi(u)\chi(z_{0})^{2}=0$ and
hence $f(x)=0$ since $\chi(u)\neq 0$ and $\chi(z_{0})\neq 0$. The case $xv\in
I_{\chi}^{2}$ can be treated similarly. For the case $uxv\in I_{\chi}^{2}$, we
have

$\begin{split}f((uxv)(z_{0})z_{0})&=f(uxv)\chi(z_{0})\chi(z_{0})+f(z_{0})\chi(uv)\chi(x)\chi(z_{0})=0\\\
&=f((u)(xvz_{0})z_{0})=f(u)\chi(xvz_{0})\chi(z_{0})+f(xvz_{0})\chi(u)\chi(z_{0})\\\
&=f(u)\chi(vz_{0})\chi(x)\chi(z_{0})+\chi(uz_{0})[f(x)\chi(v)\chi(z_{0})\\\
&\quad+f(v)\chi(x)\chi(z_{0})]\\\ &=f(x)\chi(uv)\chi(z_{0})^{2},\end{split}$

because $f(uxv)=\chi(x)=0$. Consequently $f(x)=0$ since $\chi(uv)\neq 0$ and
$\chi(z_{0})\neq 0$. Then the proof of the third line of (4.7) is completed.
For the bottom line of (4.7) there is nothing left to prove.

For the condition (I), let $x\in I_{\chi}\setminus P_{\chi}$ and $y\in
S\setminus I_{\chi}$. Using (4.6) we have

$\begin{split}f((xy)z_{0}z_{0})&=f(xy)\chi(z_{0})\chi(z_{0})+f(z_{0})\chi(xy)\chi(z_{0})=f(xy)\chi(z_{0})^{2}\\\
&=f((x)(yz_{0})z_{0})=f(x)\chi(yz_{0})\chi(z_{0})+f(yz_{0})\chi(x)\chi(z_{0})\\\
&=0,\end{split}$

which gives $f(xy)=0$, since $f(x)=0$ and $\chi(z_{0})\neq 0$. The
verification of $f(yx)=0$ is similar.

For the condition (II), we begin by the case $x=up$ with $p\in P_{\chi}$ and
$u\in S\setminus I_{\chi}$. Then by (4.6) we have

$\begin{split}f(xz_{0}^{2}z_{0})&=f(x)\chi(z_{0}^{2})\chi(z_{0})+f(z_{0}^{2})\chi(x)\chi(z_{0})=f(x)\chi(z_{0})^{3}\\\
&=f((u)(pz_{0}^{2})z_{0})=f(u)\chi(pz_{0}^{2})\chi(z_{0})+f(pz_{0}^{2})\chi(u)\chi(z_{0})\\\
&=f(u)\chi(p)\chi(z_{0})^{3}+\chi(u)\chi(z_{0})[f(p)\chi(z_{0})\chi(z_{0})\\\
&\quad+f(z_{0})\chi(p)\chi(z_{0})]\\\
&=\rho(p)\chi(u)\chi(z_{0})^{3},\end{split}$

because $\chi(p)=0$. Therefore $f(x)=\rho(p)\chi(u)$ since $\chi(z_{0})\neq
0$. In addition we have for any $z,t\in S\setminus I_{\chi}$, that
$zx=zup=(zu)p,\;xt=upt=(up)t,\;zxt=zupt=(zu)pt\in I_{\chi}\setminus
I_{\chi}^{2}$, so $x\in P_{\chi}$ and $\rho(x)=\rho(p)\chi(u)$. The case
$x=pv$ can be treated similarly. For the case $x=upv$ with $p\in P_{\chi}$ and
$u,v\in S\setminus I_{\chi}$, we have

$\begin{split}f(x(z_{0}^{2})z_{0})&=f(x)\chi(z_{0}^{2})\chi(z_{0})+f(z_{0}^{2})\chi(x)\chi(z_{0})=f(x)\chi(z_{0})^{3}\\\
&=f((u)(pvz_{0}^{2})z_{0})=f(u)\chi(pvz_{0}^{2})\chi(z_{0})+f(pvz_{0}^{2})\chi(u)\chi(z_{0})\\\
&=f(u)\chi(pv)\chi(z_{0})^{3}+\chi(u)\chi(z_{0})[f(p)\chi(vz_{0})\chi(z_{0})\\\
&\quad+f(vz_{0})\chi(p)\chi(z_{0})]\\\
&=\rho(p)\chi(u)\chi(v)\chi(z_{0})^{3},\end{split}$

because $\chi(p)=0$. Then $f(x)=\rho(p)\chi(u)\chi(v)$ since $\chi(z_{0})\neq
0$. In addition we have also for any $z,t\in S\setminus I_{\chi}$, that
$zx=zupv=(zu)pv,xt=up(vt)=(upv)t,zxt=zupvt=(zu)p(vt)\in I_{\chi}\setminus
I_{\chi}^{2}$, then $x\in P_{\chi}$. Therefore $\rho(x)=\rho(p)\chi(uv)$ and
this completes the proof of condition (II) of Proposition (4.3).

Conversely, suppose that $f:S\longrightarrow\mathbb{C}$ satisfies (4.7),
$A:S\setminus I_{\chi}\longrightarrow\mathbb{C}$ is additive,
$\rho:P_{\chi}\longrightarrow\mathbb{C}$ and conditions (I) and (II) hold. We
shall prove that $f$ satisfy (3.2). The case $x,y\in S\setminus I_{\chi}$ is
straightforward so we omit it. For $x\in I_{\chi}\setminus P_{\chi}$ and $y\in
S\setminus I_{\chi}$ we have $yz_{0}\in S\setminus I_{\chi}$. By condition (I)
we have

$f(xyz_{0})=f(x(yz_{0}))=0=f(x)\chi(y)\chi(z_{0})+f(y)\chi(x)\chi(z_{0}),$

since $f(x)=\chi(x)=0$. The case $y\in I_{\chi\setminus}P_{\chi}$ and $x\in
S\setminus I_{\chi}$ can be treated similarly.

For $x\in P_{\chi}$ and $y\in S\setminus I_{\chi}$ we have $yz_{0}\in
S\setminus I_{\chi}$ and then by Lemma 2.1 we get $xyz_{0}\in P_{\chi}$, then

$\begin{split}f(xyz_{0})&=\rho(x(yz_{0}))=\rho(x)\chi(yz_{0})=\rho(x)\chi(y)\chi(z_{0})\\\
&=f(x)\chi(y)\chi(z_{0})+f(y)\chi(x)\chi(z_{0}),\end{split}$

since $\chi(x)=0$. The case $y\in P_{\chi}$ and $x\in S\setminus I_{\chi}$
also can be treated similarly.

Now, for $x\in I_{\chi}$ , $y\in I_{\chi}$ we have $xyz_{0}=x(yz_{0})\in
I_{\chi}^{2}\subset I_{\chi}\setminus P_{\chi}$. Then

$f(xyz_{0})=0=f(x)\chi(y)\chi(z_{0})+f(y)\chi(x)\chi(z_{0})$

since $\chi(x)=\chi(y)=0$. Which proves (4.6).

Next we prove that $f$ is abelian. By computing $f((xyu)z_{0}z_{0})$ in two
different ways, and using (4.6) we have for all $x,y,u\in S$, by

$\begin{split}f((xyu)z_{0}z_{0})&=f(xyu)\chi(z_{0})\chi(z_{0})+f(z_{0})\chi(xyu)\chi(z_{0})\\\
&=f(x)\chi(yuz_{0})\chi(z_{0})+f(yuz_{0})\chi(x)\chi(z_{0})\\\
&=\chi(x)\chi(z_{0})[f(y)\chi(u)\chi(z_{0})+f(u)\chi(y)\chi(z_{0})]\\\
&\quad+f(x)\chi(yuz_{0})\chi(z_{0})\\\
&=\chi(x)\chi(z_{0})[f(u)\chi(y)\chi(z_{0})+f(y)\chi(u)\chi(z_{0})]\\\
&\quad+f(x)\chi(uyz_{0})\chi(z_{0})\\\
&=f(x)\chi(uyz_{0})\chi(z_{0})+f(uyz_{0})\chi(x)\chi(z_{0})\\\
&=f(xuyz_{0}z_{0})\\\
&=f(xuy)\chi(z_{0})\chi(z_{0})+f(z_{0})\chi(xuy)\chi(z_{0}),\end{split}$

then we deduce that
$f(xyu)\chi(z_{0})\chi(z_{0})=f(xuy)\chi(z_{0})\chi(z_{0})$ which implies
$f(xyu)=f(xuy)$ for all $x,y,u\in S$, since $\chi(z_{0})\neq 0$. Next we have

$\begin{split}f((xy)z_{0}^{2}z_{0})&=f(xy)\chi(z_{0}^{2})\chi(z_{0})+f(z_{0}^{2})\chi(xy)\chi(z_{0})\\\
&=f(xyz_{0})\chi(z_{0})\chi(z_{0})+f(z_{0})\chi(xyz_{0})\chi(z_{0})\\\
&=f(yxz_{0})\chi(z_{0})\chi(z_{0})+f(z_{0})\chi(yxz_{0})\chi(z_{0})\\\
&=f((yx)z_{0}^{2}z_{0})\\\
&=f(yx)\chi(z_{0}^{2})\chi(z_{0})+f(z_{0}^{2})\chi(yx)\chi(z_{0},)\\\
\end{split}$

which implies that

$f(xy)\chi(z_{0}^{2})\chi(z_{0})=f(yx)\chi(z_{0}^{2})\chi(z_{0}),$

and therefore we obtain $f(xy)=f(yx)$ for all $x,y\in S$. This proves the
centrality of $f$. Thus $f$ is abelian.

Finally, suppose that $S$ is a topological semigroup. Putting $y=a$ in (4.6)
such that $f(a)\neq 0$ (because $f\neq 0$ by assumption) we get that

$\chi(x)=\frac{f(xaz_{0})-f(x)\chi(a)\chi(z_{0})}{f(a)\chi(z_{0})}$

since $\chi(z_{0})\neq 0$. This gives the continuity of $\chi$. Therefore
$A:=f/\chi-A(z_{0})$ is continuous on $S\setminus I_{\chi}$ since $\chi\neq 0$
on $S\setminus I_{\chi}$. Here also $\rho$ is continuous by restriction. This
completes the proof of Proposition 4.3. ∎

Now, we are ready to describe the solutions of the functional equation (1.2)
on semigroups.

######  Theorem 4.4.

The solutions $f,g:S\longrightarrow\mathbb{C}$ of the functional equation
(1.2) are the following pairs of functions
(1) $f=0$ and $g$ arbitrary.
(2) There exist a constant $b\in\mathbb{C}^{*}$ and an exponential function
$\chi$ on $S$ with $\chi(z_{0})\neq 0$ such that

$f=\dfrac{\chi(z_{0})}{2b}\chi\hskip 8.5359ptand\hskip
8.5359ptg=\dfrac{\chi(z_{0})}{2}\chi.$

(3) There exist a constant $\delta\in\mathbb{C}^{*}$ and two different
exponentials functions $\chi_{1}$ and $\chi_{2}$ on $S$ with
$\chi_{2}(z_{0})=-\chi_{1}(z_{0})\neq 0$ such that

$f=\delta(\chi_{1}+\chi_{2})\hskip 8.5359ptand\hskip
8.5359ptg=\dfrac{\chi_{1}(z_{0})}{2}(\chi_{1}-\chi_{2}).$

(4) There exist a constant $c\in\mathbb{C}^{*}$ and two different exponentials
functions $\chi_{1}$ and $\chi_{2}$ on $S$ with
$\chi_{2}(z_{0})=\chi_{1}(z_{0})\neq 0$ such that

$f=c(\chi_{1}-\chi_{2})\hskip 8.5359ptand\hskip
8.5359ptg=\dfrac{\chi_{1}(z_{0})}{2}(\chi_{1}+\chi_{2}).$

(5) There exist a constant $\gamma\in\mathbb{C}^{*}$ and two different
exponentials functions $\chi_{1}$ and $\chi_{2}$ on $S$ with
$\chi_{1}(z_{0})\neq 0$, $\chi_{2}(z_{0})\neq 0$ and
$\chi_{1}(z_{0})\chi_{1}\neq\chi_{2}(z_{0})\chi_{2}$ such that

$f=\dfrac{\chi_{1}(z_{0})\chi_{1}-\chi_{2}(z_{0})\chi_{2}}{2\gamma}\hskip
8.5359ptand\hskip
8.5359ptg=\dfrac{\chi_{1}(z_{0})\chi_{1}+\chi_{2}(z_{0})\chi_{2}}{2}.$

(6) $S\neq S^{2}z_{0}$ and we have

$g=0\;and\;f(x)=\begin{cases}f_{z_{0}}(x)\;for\;x\in S\setminus S^{2}z_{0}\\\
0\;for\;x\in S^{2}z_{0},\end{cases}$

where $f_{z_{0}}:S\setminus S^{2}z_{0}\longrightarrow\mathbb{C}$ is an
arbitrary non-zero function.
(7) There exist an exponential function $\chi$ on $S$, an additive function
$A:S\setminus$$I_{\chi}\longrightarrow\mathbb{C}$ and a function
$\rho:P_{\chi}\longrightarrow\mathbb{C}$ with $\rho$ is the restriction of $f$
to $P_{\chi}$ and $\chi(z_{0})\neq 0$ such that

$g=\chi(z_{0})\chi\;and\;f(x)=\begin{cases}\chi(x)(A(x)+A(z_{0}))\hskip
14.22636ptfor\hskip 8.5359ptx\in S\backslash I_{\chi}\\\ \rho(x)\hskip
85.35826ptfor\hskip 8.5359ptx\in P_{\chi}\\\ 0\hskip 99.58464ptfor\hskip
8.5359ptx\in I_{\chi}\setminus P_{\chi}.\end{cases}$

In addition, in the case (7) $f$ satisfies the following conditions:
(I) $f(xy)=f(yx)=0$ for all $x\in I_{\chi}\setminus P_{\chi}$ and $y\in
S\setminus I_{\chi}$.
(II) If $x\in\\{up,pv,upv\\}$ with $p\in P_{\chi}$ and $u,v\in S\setminus
I_{\chi}$, then $x\in P_{\chi}$ and we have respectively
$\rho(x)=\rho(p)\chi(u)$, $\rho(x)=\rho(p)\chi(v)$ or
$\rho(x)=\rho(p)\chi(uv)$.

Moreover if $S$ is a topological semigroup and $f\in C(S)$, then
$g,\chi,\chi_{1},\chi_{2}\in C(S),A\in C(S\setminus I_{\chi})$, $f_{z_{0}}\in
C(S\setminus S^{2}z_{0})$ and $\rho\in C(P_{\chi})$.
Note that $f$ and $g$ are abelian in cases (2)-(5) and (7).

###### Proof.

It is easy to check that if $f=0$, then $g$ can be chosen arbitrary, so we
have the solution family (1). From now on we may assume that $f\neq 0$.
If $f$ and $g$ are linearly dependent, then there exists a constant
$b\in\mathbb{C}$ such that $g=bf$ and therefore equation (1.2) becomes
$f(xyz_{0})=2bf(x)f(y),\;x,y\in S.$
If $b=0$ then $g=0$ and $f(xyz_{0})=0$ for all $x,y\in S$, so necessarily we
have $S\neq S^{2}z_{0}$, because $f\neq 0$, and then we have the solution
family (6) for an arbitrary non-zero function $f_{z_{0}}:S\setminus
S^{2}z_{0}\longrightarrow\mathbb{C}$. If $b\neq 0$, then by [10, Proposition
16] we infer that there exists a non-zero multiplicative function $\chi$ on
$S$ such that $2bf=:\chi(z_{0})\chi$, which gives

$f=\dfrac{\chi(z_{0})}{2b}\chi$ and $g=\dfrac{\chi(z_{0})}{2}\chi$,

and this proves that $(f,g)$ belongs to the solution family (2).
For the rest of the proof we assume that $f$ and $g$ are linearly independent.
Then, according to Lemma 4.2 (ii)-(a), we get that $g(z_{0})\neq 0$.

To complete the proof we split it into two cases according to whether
$f(z_{0})=0$ or $f(z_{0})\neq 0$.

Case 1. Suppose $f(z_{0})=0$. From Lemma 4.1 we get that $f$ and $g$ satisfy
the following functional equation

$g(z_{0}^{2})f(xy)=g(z_{0})f(x)g(y)+g(z_{0})f(y)g(x)-f(z_{0}^{2})g(xy),\;x,y\in
S,$ (4.11)

Subcase 1.1: Suppose $f(z_{0}^{2})=0$, then by lemma 4.2 (ii)-(b) we get that
$g(z_{0}^{2})\neq 0$. Therefore equation (4.11) can be rewritten as follows

$f(xy)=\alpha f(x)g(y)+\alpha f(y)g(x),\;x,y\in S,$

with $\alpha:=\dfrac{g(z_{0})}{g(z_{0}^{2})}\neq 0$. Then the pair $(f,\alpha
g)$ satisfies the sine addition law (1.1). According to [11, Theorem 4.1 (b)]
we infer that there exist two multiplicative functions $\chi_{1}$ and
$\chi_{2}$ on $S$ such that $\alpha g=\dfrac{\chi_{1}+\chi_{2}}{2}.$
(i) For the case $\chi_{1}\neq\chi_{2}$, there exists a constant
$c\in\mathbb{C}^{\ast}$ such that

$f=c(\chi_{1}-\chi_{2})$ and $g=\dfrac{\chi_{1}+\chi_{2}}{2\alpha}.$

Furthermore, if $\chi_{1}=0$ or $\chi_{2}=0$, we get $f,g$ linearly dependent,
which contradict the assumption that $f$ and $g$ are linearly independent, so
$\chi_{1}$ and $\chi_{2}$ are exponentials.
By using (1.2) we get

$\begin{split}0&=f(xyz_{0})-f(x)g(y)-f(y)g(x)\\\
&=c(\chi_{1}-\chi_{2})(xyz_{0})-c(\chi_{1}-\chi_{2})(x)\left(\dfrac{\chi_{1}+\chi_{2}}{2\alpha}\right)(y)\\\
&\quad-c(\chi_{1}-\chi_{2})(y)\left(\dfrac{\chi_{1}+\chi_{2}}{2\alpha}\right)(x)\\\
&=\left(c\chi_{1}(z_{0})-\dfrac{c}{\alpha}\right)\chi_{1}(xy)-\left(c\chi_{2}(z_{0})-\dfrac{c}{\alpha}\right)\chi_{2}(xy),\end{split}$

for all $x,y\in S$. Since $\chi_{1}$ and $\chi_{2}$ are different we get by
Lemma 3.1 (a) that

$c\chi_{1}(z_{0})-\dfrac{c}{\alpha}=0$ and
$c\chi_{2}(z_{0})-\dfrac{c}{\alpha}=0,$

which gives $\chi_{1}(z_{0})=\chi_{2}(z_{0})=\dfrac{1}{\alpha}$, since $c\neq
0$. Then $g$ becomes $g=\chi_{1}(z_{0})(\chi_{1}+\chi_{2})/2$. This gives that
the solution is in part (4). (ii) For the case $\chi_{1}=\chi_{2}=:\chi$ with
$\chi$ exponential, we have $\alpha g=\chi$ and $f$ has the form (3) of [3,
Theorem 3.1]. Since $0\neq\alpha g(z_{0})=\chi(z_{0})$ in the present Subcase
1.1, we get that $z_{0}\in S\setminus I_{\chi}$. Now for all $x,y$ $\in
S\setminus I_{\chi}$ we have $xyz_{0}\in S\setminus I_{\chi}$ and by using
(1.2) we get that

$\begin{split}0&=f(xyz_{0})-f(x)g(y)-f(y)g(x)\\\
&=A(xyz_{0})\chi(xyz_{0})-\dfrac{A(x)\chi(xy)}{\alpha}-\dfrac{A(y)\chi(xy)}{\alpha}\\\
&=A(xy)\chi(xy)\left(\chi(z_{0})-\dfrac{1}{\alpha}\right)+A(z_{0})\chi(z_{0})\chi(xy).\end{split}$

If $A\neq 0$, then by Lemma 3.1.(b) we get that

$\chi(z_{0})-\dfrac{1}{\alpha}=0$ and $A(z_{0})\chi(z_{0})=0.$

Then we deduce that $\chi(z_{0})=\dfrac{1}{\alpha}\neq 0$ and therefore
$A(z_{0})=0$. So we are in solution family (7).
If $A=0,$ then $f=A\chi=0$ on $S\setminus I_{\chi}$ and necessarily we have
$\rho\neq 0$ because $f\neq 0$. Next, let $x\in P_{\chi}$ and $y\in S\setminus
I_{\chi}$, then $xyz_{0}\in P_{\chi}$ by Lemma 2.1. By using (1.2) and Lemma
2.1 we get that

$\rho(x)\chi(y)\chi(z_{0})=\rho(xyz_{0})=f(xyz_{0})=f(x)g(y)+f(y)g(x)=\dfrac{1}{\alpha}\rho(x)\chi(y),$

since $f=0$ on $S\setminus I_{\chi}$ and $\chi(x)=0$. This implies that
$\chi(z_{0})=1/\alpha$ because $\rho\neq 0$ and $\chi(y)\neq 0$. So we have
the special case of the solution family (7).

Subcase 1.2: Suppose $f(z_{0}^{2})\neq 0$. From lemma 4.2 (i) we read that
$g(z_{0}^{2})=0$. Therefore Eq.(4.11) yields

$g(xy)=\beta g(x)f(y)+\beta g(y)f(x),\;x,y\in S,$

with $\beta:=\dfrac{g(z_{0})}{f(z_{0}^{2})}\neq 0$ since $g(z_{0})\neq 0$.
This means that the pair $(g,\beta f)$ satisfy the sine addition formula (1.1)
with $g\neq 0$ since $f$ and $g$ are linearly independent. Applying [11,
Theorem 4.1 (b)] we infer that there exist two multiplicative functions
$\chi_{1}$ and $\chi_{2}$ on $S$ such that: $\beta
f=\dfrac{\chi_{1}+\chi_{2}}{2}.$
(i) In the case $\chi_{1}\neq\chi_{2}$, there exists a constant
$c\in\mathbb{C}^{*}$ such that

$g=c(\chi_{1}-\chi_{2})$ and $f=\dfrac{\chi_{1}+\chi_{2}}{2\beta}.$

Furthermore, $\chi_{1}$ and $\chi_{2}$ are exponentials, because $f$ and $g$
are linearly independent. A small computation based on (1.2) shows that

$\left(\dfrac{c}{\beta}-\dfrac{\chi_{1}(z_{0})}{2\beta}\right)\chi_{1}(xy)-\left(\dfrac{c}{\beta}+\dfrac{\chi_{2}(z_{0})}{2\beta}\right)\chi_{2}(xy)=0.$

Since $\chi_{1}\neq\chi_{2}$ we get by Lemma 3.1 (a) that
$\chi_{1}(z_{0})=-\chi_{2}(z_{0})=2c$. So we are in the part (3) of our
statement, with $\delta:=1/2\beta$. (ii) In the case $\chi_{1}=\chi_{2}:=\chi$
with $\chi$ exponential, we have $\beta f=\chi$ and $g$ has the form (3) of
[3, Theorem 3.1]. Since $0=\beta f(z_{0})=\chi(z_{0})$ by the hypothesis of
this case 1, we have necessarily $z_{0}\in I_{\chi}$. On the other hand since
$g(z_{0})\neq 0$ we get that $z_{0}\in P_{\chi}$, because $g=0$ on
$I_{\chi}\setminus P_{\chi}$ and $I_{\chi}=P_{\chi}\cup(I_{\chi}\setminus
P_{\chi})$. Now let $x,y\in S\setminus I_{\chi}$, then by Lemma 2.1 we get
$xyz_{0}\in P_{\chi}$. Therefore by using (1.2) we obtain

$\begin{split}0&=f(xyz_{0})-f(x)g(y)-f(y)g(x)\\\
&=\rho(xyz_{0})-\dfrac{1}{\beta}\chi(x)A(y)\chi(y)-\dfrac{1}{\beta}\chi(y)A(x)\chi(x)\\\
&=\rho(z_{0})\chi(xy)-\dfrac{1}{\beta}A(y)\chi(xy)-\dfrac{1}{\beta}A(x)\chi(xy)A(x)\\\
&=\rho(z_{0})\chi(xy)-\dfrac{1}{\beta}\chi(xy)A(xy),\end{split}$

which gives $\rho(z_{0})=\dfrac{1}{\beta}A(xy)$. Therefore $A=0$ and
$\rho(z_{0})=0$ which is a contradiction, because $\rho(z_{0})=g(z_{0})\neq 0$
by Lemma 4.2 (ii)-(a). So this case is excluded.

Case 2. Suppose $f(z_{0})\neq 0$. Multiplying (1.2) by $\pm\gamma$ where
$\gamma\in\mathbb{C}$ is the constant in the identity (4.2) and adding the
result to the identity (4.2) we get

$(g+\gamma f)(xyz_{0})=(g+\gamma f)(x)(g+\gamma f)(y)$

and

$(g-\gamma f)(xyz_{0})=(g-\gamma f)(x)(g-\gamma f)(y).$

According to [10, Proposition 16], we infer that there exist two exponentials
functions $\chi_{1}$ and $\chi_{2}$ on $S$ such that

$\chi_{1}(z_{0})\chi_{1}:=g+\gamma f\hskip 8.5359ptand\hskip
8.5359pt\chi_{2}(z_{0})\chi_{2}:=g-\gamma f,$

with $\chi_{1}(z_{0})\neq 0$ and $\chi_{2}(z_{0})\neq 0$, because $g-\gamma
f\neq 0$ and $g+\gamma f\neq 0$ since $f$ and $g$ are linearly independent.
Then we get

$g=\dfrac{\chi_{1}(z_{0})\chi_{1}+\chi_{2}(z_{0})\chi_{2}}{2}.$

If $\chi_{1}(z_{0})\chi_{1}\neq\chi_{2}(z_{0})\chi_{2}$, then $2\gamma
f=\chi_{1}(z_{0})\chi_{1}-\chi_{2}(z_{0})\chi_{2}$ and $\gamma\neq 0$.
Therefore

$f=\dfrac{\chi_{1}(z_{0})\chi_{1}-\chi_{2}(z_{0})\chi_{2}}{2\gamma}$.

This solution falls in solution family (5).
If $\chi_{1}(z_{0})\chi_{1}=\chi_{2}(z_{0})\chi_{2}$, then we get by Lemma 3.1
(a) that $\chi_{1}=\chi_{2}$, since $\chi_{1}(z_{0})\neq 0$ and
$\chi_{2}(z_{0})\neq 0$. This allows us to define $\chi:=\chi_{1}=\chi_{2}$
with $\chi$ exponential such that $\chi(z_{0})\neq 0$, and we obtain
$g=\chi(z_{0})\chi$. Therefore Eq.(1.2) becomes

$f(xyz_{0})=f(x)\chi(z_{0})\chi(y)+f(y)\chi(z_{0})\chi(x).$ (4.12)

This equation was solved in Proposition 4.3. The solutions in this case occur
in part (7).

That $f$ and $g$ are abelian in the cases (2)-(5) can be derived from the
solutions formulas above and the case (7) is established in Proposition 4.3.

Conversely it is easy to check that the formulas of $f$ and $g$ listed in
Theorem 4.4 define solutions of (1.2).

Finally, suppose that $S$ is a topological semigroup. The continuity of cases
(1)-(5) follows from [11, Theorem 3.18] since $f\neq 0$, $c\neq 0,b\neq
0,\delta\neq 0,\chi(z_{0})\neq 0,\chi_{1}(z_{0})\neq 0$, $\chi_{2}(z_{0})\neq
0$ and also $\chi_{1}$ and $\chi_{2}$ are different. For the case (6),
$f_{z_{0}}$ is continuous by restriction. The case (7) follows directly from
Proposition 4.3. ∎

In the following Proposition we show that the functional equation (1.4) has
only trivial solutions on semigroup.

######  Proposition 4.5.

The solutions $f,g:S\longrightarrow\mathbb{C}$ of the functional equation
(1.4) are the following pairs of functions listed below
(1) $f=0$ and $g$ arbitrary
(2) $S\neq S^{2}z_{0}$ and there exists a constant $c\in\mathbb{C}$ such that

$g=cf\hskip 11.38092ptand\hskip 11.38092ptf(x)=\begin{cases}f_{z_{0}}(x)\hskip
17.07182ptfor\hskip 8.5359ptx\in S\setminus S^{2}z_{0}\\\ 0\hskip
42.67912ptfor\hskip 8.5359ptx\in S^{2}z_{0},\end{cases}$

where $f_{z_{0}}:S\setminus S^{2}z_{0}\longrightarrow\mathbb{C}$ is an
arbitrary non-zero function.

###### Proof.

If $f=0$ then $g$ can be chosen arbitrary. So from now we may assume that
$f\neq 0$. If $f$ and $g$ are linearly dependent, then there exists a constant
$c\in\mathbb{C}$ such that $g=cf$. Therefore Eq.(1.4) becomes $f(xyz_{0})=0$
for all $x,y\in S$, so necessarily we have $S\neq S^{2}z_{0}$, because $f\neq
0$. Then we get the solution family (2).
If $f$ and $g$ are linearly independent. By interchanging $x$ and $y$ in (1.4)
we get

$f(xyz_{0})=-f(yxz_{0})$, $x,y\in S$.

Using this we obtain

$f(xyzz_{0})=-f(zxyz_{0})=f(yzxz_{0})=-f(xyzz_{0})$, $x,y,z\in S,$

which gives $f(xyzz_{0})=0$ for all $x,y,z\in S$. Therefore by using (1.4) we
get that

$f(xyzz_{0})=f(xy)g(z)-f(z)g(xy)=0,$ for all $x,y,z\in S,$

which implies

$f(xy)g(z)=f(z)g(xy),\;x,y,z\in S.$ (4.13)

If there exist $x_{0},y_{0}\in S$ such that $f(x_{0}y_{0})\neq 0$, then (4.13)
with $(x,y)=(x_{0},y_{0})$ contradicts that $f$ and $g$ are linearly
independent. Thus $f=0$ on $S^{2}$. In particular $f(xyz_{0})=f((xy)z_{0})=0$,
so by (1.4) $f(x)g(y)=f(y)g(x)$ for all $x,y\in S$ which contradicts that $f$
and $g$ are linearly independent. This completes the proof of Proposition 4.5.
∎

## References

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