# Large Transformers are Better EEG Learners

Bingxin WANG  Department of Mathematics, The Hong Kong University of Science
and Technology, Clear Water Bay, Hong Kong SAR, China  Xiaowen Fu  Department
of Mathematics, The Hong Kong University of Science and Technology, Clear
Water Bay, Hong Kong SAR, China  Yuan LAN  Department of Mathematics, The Hong
Kong University of Science and Technology, Clear Water Bay, Hong Kong SAR,
China  Luchan ZHANG  Yang Xiang  Department of Mathematics, The Hong Kong
University of Science and Technology, Clear Water Bay, Hong Kong SAR, China

###### Abstract

Pre-trained large transformer models have achieved remarkable performance in
the fields of natural language processing and computer vision. Since the
magnitude of available labeled electroencephalogram (EEG) data is much lower
than that of text and image data, it is difficult for transformer models pre-
trained from EEG to be developed as large as GPT-4 100T to fully unleash the
potential of this architecture. In this paper, we show that transformers pre-
trained from images as well as text can be directly fine-tuned for EEG-based
prediction tasks. We design AdaCE, plug-and-play Adapters for Converting EEG
data into image as well as text forms, to fine-tune pre-trained vision and
language transformers. The proposed AdaCE module is highly effective for fine-
tuning pre-trained transformers while achieving state-of-the-art performance
on diverse EEG-based prediction tasks. For example, AdaCE on the pre-trained
Swin-Transformer achieves 99.6%, an absolute improvement of 9.2%, on the EEG-
decoding task of human activity recognition (UCI HAR). Furthermore, we
empirically show that applying the proposed AdaCE to fine-tune larger pre-
trained models can achieve better performance on EEG-based predicting tasks,
indicating the potential of our adapters for even larger transformers. The
plug-and-play AdaCE module can be applied to fine-tuning most of the popular
pre-trained transformers on many other time-series data with multiple
channels, not limited to EEG data and the models we use. Our code will be
available at https://github.com/wangbxj1234/AdaCE.

##  1 Introduction

Electroencephalogram, or EEG, a non-invasive way to measure brain activity,
contains a wealth of information about the functioning of the brain, which
makes it an attractive source of data for deep learning applications. The
development of deep learning in EEG feature analysis has shown great promise
for new diagnostic and therapeutic tools in the field of neurological
disorders. EEG signals are usually measured by placing electrodes on the scalp
and represented as a two-dimensional matrix or graph, where time is one
dimension and the location of electrodes is another. Most EEG-decoding
approach researchers regard multiple channels as the spatial dimension of EEG
data, corresponding to the temporal dimension.

Large language models (LLMs), such as GPT [1] [2] [3] [4], have achieved
remarkable results in various traditional natural language processing tasks.
[5] combined the pre-trained BERT-like model [6] with their local attention
modules for the EEG-To-Text decoding and sentence sentiment classification
tasks, in which they treated EEG sequences as encoded sentences. However,
regarding a sequence of multi-channel EEG data as a sentence may neglect the
spatial dimension of EEG signals, which limits the performance of this
approach in prediction tasks. Most existing EEG decoding methods construct
special CNN or transformer modules to extract features from the spatial
dimension, abandoning which could lead to a significant drop in model
performance [7] [8] [9] [10].

We are committed to designing a solution that feeds the complete information
of EEG as a two-dimensional matrix into the pre-trained model without
requiring any additional feature extraction neural networks, which will
introduce additional parameters to be trained. During the fine-tuning process,
the gradients of these parameters need to be backpropagated throughout the
entire model, which will significantly increase the computational cost.

[11] showed that common pre-trained models get low intrinsic dimensions, which
leads to their better generalization performance when fine-tuned on
traditional tasks. Could the powerful generalization ability of pre-trained
vision transformers help us complete the EEG prediction tasks? To validate
this idea, in this paper, we design the EEG-to-Image adapter for converting
multi-channel or long single-channel EEG data into image data. We also design
an EEG-to-Text adapter for converting moderate single-channel EEG data into
text, which effectively solves the overflow issue of regarding EEG as input
text for LLMs. These two adapters are collectively referred to as AdaCE,
Adapters for Converting EEG data into image and text forms, as shown in Figure
1. The proposed AdaCE achieves state-of-the-art performance on diverse EEG-
based prediction or classification tasks. We also show that applying the
proposed AdaCE to fine-tune larger transformers can achieve even better
performance on these tasks. Taking Epilepsy Seizure Recognition [12] as an
example, we show that the fine-tuned GPT-2 model [2] achieves 98.7% on the
epilepsy seizure prediction task, and that fine-tuned language transformers of
the same architecture get an average improvement of 1.3% when the number of
parameters changes from about 100 million to 300 million.

The main contributions of our paper are as follows:

  * •

We show that large models (LMs) pre-trained from images as well as text can be
directly fine-tuned for EEG-based prediction tasks without introducing extra
parameters to be trained. This provides a new application scenario for LMs.

  * •

We propose the AdaCE-to-Image, adapter for converting multi-channel EEG into
images, which can directly feed the complete information of two-dimensional
EEG data into the pre-trained vision transformers.

  * •

We also design the AdaCE-to-Text, adapter for converting moderate single-
channel EEG data into text, which effectively solves the overflow issue of
regarding EEG as input text for LLMs.

  * •

The proposed AdaCE module preserves the feature information of each original
data point to the maximum extent in two different scenarios, achieving state-
of-the-art results in several EEG prediction tasks.

  * •

Furthermore, we empirically show that applying the proposed AdaCE to fine-tune
larger pre-trained models can achieve better performance on EEG-based
predicting tasks, indicating the potential of our adapters for even larger
transformers.

  * •

The plug-and-play AdaCE module could be applied to fine-tuning most of the
popular pre-trained large models on other time-series data with multiple
channels.

Figure 1: Framework: Adapters to convert EEG into images and text for fine-
tuning pre-trained large transformers.

##  2 Related Works

###  2.1 Transformer Combined with CNN

Whether long-range correlations are essential for decoding EEG data depends on
the specific EEG data and decoding task. In some cases, long-range correlation
can be important for accurately decoding EEG signals, and the attention-based
transformer architecture [13] will be a more suitable choice for decoding EEG
data to have better capability to extract long-range dependencies.

[9] used the combined Transformer+CNN model, in which they expected the
transformer module to pre-extract temporal and spatial features by calculating
attention across time and space separately. While [10] applied CNN+transformer
architecture, using the CNN module to pre-extract temporal and spatial
features by calculating convolution across time and space separately. Most
existing transformer+CNN methods EEG-Transformers would face performance
degradation without the pre-extraction module [7] [8] [9] [10].

However, adding the attention or CNN pre-extraction module to the pre-trained
large transformer will introduce additional parameters to be trained. During
the fine-tuning process, the gradients of these parameters need to be
backpropagated throughout the entire model, which will significantly increase
the computational cost.

###  2.2 Self-supervised Learning

Considering the cost of labeling and the limited number of labeled EEG data,
[8] proposed the time-series self-supervised paradigm by conducting different
temporal and contextual data augmentation for contrast. Soon after they also
proposed a Class-Aware semi-supervised learning paradigm [14]. Their method
has made a significant contribution towards addressing the problem of limited
availability of labeled EEG data, which is much lower in magnitude compared to
that of text and image data. However, the performance by measuring absolute
prediction accuracy from Contrastive Learning ( [8] [14]) on diagnostic tasks
is not that good.

How to get better performance with limited training data? GPT 2’s authors [2]
demonstrated that large language models could perform many traditional
language tasks without any parameter or architecture modification, a
capability referred to as "zero-shot" learning. Given that large language
models have already exhibited the remarkable ability to perform "zero-shot"
learning, the issue of limited training samples is no longer a major
bottleneck. For further exploring LM’s potential, We fine-tuned pre-trained
transformers on diverse EEG datasets to pursue better performance than that
from the "zero-shot" learning setting.

###  2.3 Synthesize EEG by GPT

Figure 2: AdaCE-to-Image: Adapter for Converting EEG into images to maintain
the complete feature information.

[15] and [16] used GPT by synthesizing EEG signal for data augmentation.
However, synthetic signals may not necessarily be reliable. When training
models for future medical diagnosis, it is preferable to use real data. [2]
and [3] have provided evidence that large language models possess the ability
to perform few-shot or even zero-shot learning in conventional natural
language processing tasks. Therefore, we did not introduce any synthetic
biological signals machine-generated by LLMs in our experiments.

###  2.4 BERT-like Transformers on EEG Dataset

[5] proposed to combine the pre-trained BART [6] with their local attention
modules for EEG-To-Text decoding as well as sentence sentiment classification
tasks. They assume that the brain has already encoded the data, treating each
EEG sequence as an encoded sentence. However, regarding a sequence of multi-
channel EEG data as a sentence may neglect the spatial dimension of EEG
signals, which limits the performance of this approach in prediction tasks.

Moreover, the local attention modules will introduce additional parameters
that need to be trained. As [5] placed them before the pre-trained language
model, the gradients of these parameters must be backpropagated throughout the
entire architecture. The pre-extraction modules will significantly increase
the computational cost when we switch from BART to a larger model.

[17] and [18] both trained BERT-like transformers on their EEG dataset, which
was the beginning of pre-training transformer models on EEG datasets. Although
it is widely known that GPT-3 achieved better performance than BERT with the
help of a larger number of parameters, pre-trained transformers based on EEG
datasets can’t go larger since the magnitude of publicly available labeled EEG
data is much lower than that of text and image data.

Unlike the approach taken by [5] in which multi-channel EEG sequences are
treated as encoded sentences, [18] consider each EEG channel as a sentence and
process multiple channels as a whole text. Although preserving all the
information points, their approach transforms the parallel relationships
between different channel data into contextual relationships, which may
overlook the parallel nature of data between different channels.

Since the input sequence’s length of large language transformers is limited
due to the quadratic complexity of attention calculation, connecting multiple
channels as one text input is usually faced with the overflow issue.

##  3 AdaCE: Adapters for Converting EEG Data into Images as well as Text

EEG signals are usually measured by placing electrodes on the scalp and
represented as a two-dimensional matrix or graph. The magnitude and sign of
numbers in EEG data indicate the polarity and strength of the electrical
activity of the brain neurons: A positive value generally indicates that the
neuron is firing, while a negative value indicates inhibited; the magnitude of
the data indicates how strongly the neuron fires or is inhibited. Deep
learning models can be used to analyze EEG signals and extract meaningful
information for downstream tasks.

Fine-tuning pre-trained large models may be a promising solution for higher
prediction accuracy, but these vision or language transformers were not
designed for decoding EEG. The existing EEG-to-TEXT methods [5] [18] neglected
the original information in either spatial dimension or parallel relationship
among different channels. When it comes to fine-tuning larger transformers, we
also need to address the extra computational cost from local modules and
overflow issues caused by long time series.

In this section, we introduce our solution for fine-tuning pre-trained large
transformers on EEG datasets. We design AdaCE, plug-and-play Adapters for
converting EEG data into image as well as text forms, to fine-tune pre-trained
vision and language transformers.

We first propose the EEG-to-Image adapter for converting multi-channel EEG
into images, which can directly feed the complete information of two-
dimensional EEG data into the pre-trained vision transformers. Apart from
this, we also design the EEG-to-Text adapter for converting moderate single-
channel EEG data into text, which effectively solves the overflow issue of
regarding EEG as input text for LLMs.

###  3.1 AdaCE-to-Image

Figure 3: AdaCE-to-Text: Adapter for Converting EEG into text while solving
the overflow issue.

To maintain the complete feature information of two-dimensional EEG data, we
propose to convert multi-channel EEG into images, as shown in Figure 2.

The first step is to construct a dimension of three corresponding to the RGB
pattern required by the pre-trained vision transformer, and we propose two
schemes for different conditions:

  * •

In the case where the size of EEG’s spatial dimension is a multiple of three,
we decompose the spatial dimension into two new dimensions, with the first
newly generated dimension having a dimensionality of three.

  * •

In other cases, we apply interpolation on the spatial dimension to make its
dimensionality multiple of three and apply the same type of decomposition.

In the second step, we apply reshape operation and bilinear interpolation on
the last two dimensions to align their dimensionalities with the input height
and width of the pre-trained vision transformers. The interpolation can be
either upsampling or downsampling, depending on the size of the vector
obtained in the first step.

The proposed adapter can also be used to convert single-channel EEG into
images. In this case, we modify the first step by inserting a new dimension of
size one before the temporal dimension and replicate data along the newly
generated dimension to three for alignment with the RGB pattern.

We empirically show that the proposed EEG-to-Image adapter is highly effective
for fine-tuning large vision transformers on diverse EEG-based prediction
tasks while achieving state-of-the-art performance.

###  3.2 AdaCE-to-Text

For EEG data with a moderate number of samples and a single channel, we
propose another adapter for EEG-to-Text, which effectively solves the overflow
issue of regarding EEG as input text for LLMs.

When we treat EEG data as text, the most intuitive idea is to connect all the
numerical points from the same instance to formulate a text, where each
sampling point at a timestamp should be mapped as a word in the language.
However, the input sequence’s lengths of the existing pre-trained language
transformers are limited due to the quadratic complexity of attention
calculation. Under these conditions, we aim to retain the essential
information of each sampling point while using the minimum number of digits
when converting the EEG signals into numerical sequences.

MinMaxScaler or other normalization method would lead to too many decimal
digits. We propose to concatenate the raw integer data directly into text
without the normalization step. For some datasets, the provided data may
already be normalized as decimal values. In such cases, we amplify the values
by a factor of $\alpha$ and round them to the nearest integer.

Another issue is that the number of EEG’s temporal dimension can be larger
than the pre-trained model’s input size. Automatic truncation will lead to
severe information loss. Since EEG data is often densely sampled over a short
period of time (such as a few seconds), its adjacent sampling points on the
timestamp exhibit obvious continuity characteristics. Given this feature, we
downsample data using non-overlapping sliding windows to the target language
transformer’s maximum input length, as shown in Figure 3. We do not choose the
1D CNN layer with finite kernel size since we believe that using sliding
windows better preserves the original information of the EEG data for tuning
pre-trained transformers.

The experimental results confirmed the effectiveness of the proposed adapter
for EEG-to-Text. We fine-tuned GPT-2-medium 355M with AdaCE on the Epileptic
Seizure Dataset [12] to state-of-the-art performance.

However, it should be noted that based on our experience, when the generated
text exceeds three times the acceptable input length of the pre-trained model,
using non-overlapping sliding windows for downsampling may result in
significant information loss. In such cases, the EEG-to-Image adapter can be
used to process this single-channel EEG data.

###  3.3 Fine-tune Pre-trained Transformers on Converted Dataset

Despite the varying structures of different pre-trained large models, the
principle of using most multi-layer transformers for prediction tasks remains
the same. We follow the fine-tuning approach of [1]. Suppose we train K
instances at a time, the fine-tuning objective is to maximize the following
function:

$L=\sum_{k=1}^{K}\log\left(H_{c}\left(z^{n}_{k}\right)\right),$ (1)

where $z^{n}$ is the last vector of the transformer’s output sequence, and
$H_{c}$ is the classification head consisting of a linear layer followed by a
softmax layer.

##  4 Experiments

We load the pre-trained transformer models for image or sequence
classification from huggingface [19] and evaluate our method on diverse EEG
decoding tasks: UCI HAR [20], Epileptic Seizure Recognition [12], and Sleep-
EDF [21]. Following the setting in [8], we split the datasets into 60%, 20%,
and 20% for training, validating, and testing. We fine-tune the pre-trained
large transformers using Huggingface Trainer [19] with AdamW optimizer,
initial learning rate of 5e-5, training batch size per device of 16, and
gradient accumulation steps of 4. We report the prediction accuracy and macro-
averaged f1 scores. Each experiment we conduct is trained within 20 epochs,
and further training does not significantly improve the performance. For
experiments involving the conversion of EEG data into text, we set the
hyperparameter $\alpha$ to 1000. We conduct fine-tuning with PyTorch
1.13.0+cu116 and Huggingface transformers 4.31.0 on two NVIDIA Quadro RTX 8000
GPUs.

###  4.1 UCI HAR Dataset

The Human Activity Recognition Dataset [20] was collected from 30 subjects
performing six different activities (Walking, Walking Upstairs, Walking
Downstairs, Sitting, Standing, Laying), consisting of inertial sensor data
that was collected using a smartphone carried by the subjects. The inside EEG
data has nine channels: three for acceleration signals of X, Y, and Z; three
for body acceleration obtained by subtracting gravity from the total
acceleration of X, Y, and Z; three for the angular velocity vector of X, Y,
and Z. There are 7352 pieces of EEG data, each piece containing 9*128 data
points.

Figure 4: Visualization: Display of some images generated from the UCI HAR
Dataset.

Method | Accuracy | Macro-F1
---|---|---
SSL-ECG [22] | $65.3$ | $63.8$
SimCLR [23] | $81.0$ | $80.2$
TS-TCC [8] | $90.4$ | $90.4$
AdaCE-to-Image (28M) | $98.2$ | $98.1$
AdaCE-to-Image (87M) | $\mathbf{99.6}$ | $\mathbf{99.5}$

Table 1: Comparisons of our AdaCE-to-Image with those of the previous state-
of-the-art transformer-based methods on UCI HAR. Figure 5: Training Process
Analysis of AdaCE-to-Image on UCI HAR Dataset. The baseline refers to TS-TCC
[8].

Since this EEG dataset has multiple channels, we choose to fine-tune pre-
trained vision transformers with the proposed EEG-to-Image adapter. The pre-
trained models we choose are Swin-Transformer [24] base model (87M) pre-
trained on ImageNet-21k [25] and Swin-Transformer tiny model (28M) pre-trained
on ImageNet-1k.

Table 1 presents the results of our AdaCE-to-Image method, as well as the
results of previous state-of-the-art transformer-based methods tested by [8].
The proposed AdaCE-to-Image method noticeably surpasses the previous state-of-
the-art transformer-based methods, achieving an improvement of +9.2% in
classification accuracy for AdaCE 87M (99.6%) over TS-TCC (90.4%), and an
improvement of +7.8% in classification accuracy for AdaCE 28M (98.2%) over TS-
TCC (90.4%). Table 1 also shows that AdaCE performs much better than the
earlier models [23] and [22].

Through an epoch-wise training process analysis presented in Figure 5, we
demonstrate that AdaCE-to-Image on diverse pre-trained transformers
outperforms the previous state-of-the-art transformer-based method within only
5 epochs. This highlights the efficiency and universality of the AdaCE-to-
Image method.

###  4.2 Epileptic Seizure Prediction

The Epileptic Seizure Dataset [12] consists of 500 files, with each file
representing a single subject. Each file is a recording of brain activity for
23.6 seconds. The corresponding time series was sampled into 4097 data points.
Every 4097 data points were divided and shuffled into 23 chunks, and each
chunk contains 178 data points for 1 second. There are 23 x 500 = 11500 pieces
of information.

Since the temporal dimensionality of this single-channel dataset is below
three times the input limit of the pre-trained language transformers, we
conduct the experiments with the proposed AdaCE-to-Text method. The pre-
trained models we choose are GPT-2 of 124M and GPT-2 of 355M [2].

Method | Accuracy | Macro-F1
---|---|---
SSL-ECG [22] | $93.7$ | $89.2$
SimCLR [23] | $96.0$ | $93.5$
TS-TCC [8] | $97.2$ | $95.5$
AdaCE-to-Text (124M) | $97.4$ | $96.5$
AdaCE-to-Text (355M) | $\mathbf{98.7}$ | $\mathbf{97.9}$

Table 2: Comparisons of our AdaCE-to-Text with those of the previous state-of-
the-art transformer-based methods on Epilepsy Seizure Prediction.

Table 2 presents the results of our AdaCE-to-Text method, as well as the
results of previous state-of-the-art transformer-based methods tested by [8].
Compared with the previous state-of-the-art transformer-based methods, the
proposed AdaCE-to-Text method achieves better accuracy without CNN pre-
extraction.

We also show that applying AdaCE to a larger pre-trained model achieves a
better performance: +1.3% for GPT-2 355M (98.7%) over GPT-2 124M (97.4%) by
prediction accuracy. Inspired by this, we conduct comparative experiments for
fine-tuning pre-trained models of different sizes and include them in the
ablation study at the end of this section.

The Sleep-EDF Dataset [21] contains 197 whole-night PolySomnoGraphic sleep
recordings, containing EEG, EOG, chin EMG, and event markers. We follow [7],
using the first 20 subjects’ records out of 78 to construct the single-channel
EEG train dataset of 42308 pieces of information, with each piece of
information containing 1*3000 data points.

###  4.3 Sleep-EDF

We conduct the experiments using the proposed AdaCE-to-Image method. The pre-
trained models we choose are Swin-Transformer-v2 [26] tiny model of 28M pre-
trained on ImageNet-1k and Swin-Transformer-v2 base model of 110M pre-trained
on ImageNet-21k [25]. Similar to the two datasets discussed above, the
proposed method achieves state-of-the-art performance, and applying the
proposed AdaCE to fine-tune larger pre-trained models can achieve better
performance, as shown in Table 3.

If we apply the EEG-to-Text method on Sleep-EDF Dataset, the generated text
exceeds three times the acceptable input length of the largest pre-trained
model we use, even using non-overlapping sliding windows for downsampling may
result in significant information loss. To demonstrate this, we conduct
comparative experiments for fine-tuning pre-trained models with different
approaches and include them in the ablation study at the end of this section.

Method | Accuracy | Macro-F1
---|---|---
SSL-ECG [22] | $74.6$ | $65.4$
SimCLR [23] | $78.9$ | $68.6$
TS-TCC [8] | $83.0$ | $73.6$
AttnSleep [7] | $84.4$ | $\mathbf{78.1}$
AdaCE-to-Image (28M) | $83.4$ | $74.5$
AdaCE-to-Image (110M) | $\mathbf{85.3}$ | $76.4$

Table 3: Comparisons of our AdaCE-to-Image with those of the previous state-of-the-art transformer-based methods on SLEEP-EDF. Approach (Pre-trained Model) | HAR | Sleep-EDF
---|---|---
| Accuracy | Macro-F1 | Accuracy | Macro-F1
AdaCE-to-Image (DeiT 86M [27]) | $99.5$ | $99.3$ | $80.7$ | $70.3$
AdaCE-to-Image (Swin-Transformer 87M [24]) | $\mathbf{99.6}$ | $\mathbf{99.5}$ | $\mathbf{84.7}$ | $\mathbf{76.2}$
AdaCE-to-Image (ViT 87M [28]) | $99.5$ | $99.2$ | $79.9$ | $68.7$
AdaCE-to-Text (BERT 110M [29]) |  |  | $77.0$ | $65.9$
AdaCE-to-Text (GPT-2 124M [2]) |  |  | $76.6$ | $67.6$
EEG-to-Text (BERT 110M [29]) | $83.3$ | $84.6$ | $71.1$ | $60.8$
EEG-to-Text (GPT-2 124M [2]) | $85.5$ | $86.6$ | $72.0$ | $60.3$
Table 4: Comparisons of the proposed AdaCE-to-Image, AdaCE-to-Text and the
previous EEG-to-text method on UCI HAR and SLEEP-EDF. Note: The blank cells in
the table for HAR indicate that we do not design AdaCE-to-Text method for
multi-channel EEG data.

###  4.4 Ablation Study

We conduct ablation studies on the UCI HAR Dataset and the Sleep-EDF Dataset
to demonstrate the effectiveness of AdaCE.

We first use the conventional method of processing each sequence of EEG data
as a text input, and fine-tune BERT and GPT-2 separately. Then we apply the
proposed AdaCE-to-Text method on the two pre-trained models separately. For
further comparison, we also convert the EEG into images with the proposed
AdaCE-to-Image and fine-tune three different pre-trained vision transformers
of similar sizes.

Table 4 demonstrates that the AdaCE-to-Image method performs best on the two
datasets. Table 4 also shows that using the proposed AdaCE-to-Text method
outperforms the previous EEG-to-Text method on the Sleep-EDF Dataset, but is
inferior to the proposed AdaCE-to-Image method. The latter performance gap is
particularly evident on the UCI HAR Dataset, which further highlights the
effectiveness of the proposed EEG-to-Image adapter for processing multi-
channel EEG signals.

It can also be observed from Table 4 that for the same EEG transformation
method, pre-trained transformers with similar number of trainable parameters
but different structures have similar performance on the EEG-decoding tasks.

We then compare the performance of pre-trained models with different sizes on
the UCI HAR Dataset, as shown in Table 5. We use the proposed adapters to
convert the EEG data into images at resolution $224\times 224$, RGB pattern.
Subsequently, we employ six distinct vision transformers, comprising three
types, each with a smaller and a larger variant. Throughout the fine-tuning
process of different pre-trained models, we maintain consistent settings. From
Table 5 we can find that fine-tuning a larger pre-trained model of the same
type leads to a higher accuracy, indicating that our method has the potential
to be applied to larger transformers.

Pre-trained Model | Accuracy | Macro-F1
---|---|---
DeiT (22M) [27] | $98.7$ | $97.6$
Swin-Transformer (28M) [24] | $98.2$ | $98.1$
ViT (22M) [28] | $98.8$ | $98.0$
DeiT (86M) [27] | $99.5$ | $99.3$
Swin-Transformer (87M) [24] | $\mathbf{99.6}$ | $\mathbf{99.5}$
ViT (87M) [28] | $99.5$ | $99.2$

Table 5: Comparisons of fine-tuning pre-trained vision transformers in
different sizes on UCI HAR with our AdaCE-to-Image.

##  5 Conclusions and Future Work

Decoding medical diagnosis-related electroencephalograms is a new application
scenario for LMs, we have shown that large transformers pre-trained from the
images as well as text can be directly fine-tuned for EEG-based prediction
tasks without introducing extra parameters to be trained. Specifically, We
have proposed adapters for converting EEG data into image and text formats for
LMs respectively. The proposed AdaCE-to-Image enables input of EEG data into
pre-trained vision transformers, and the AdaCE-to-Text solves the overflow
issue when regarding EEG as input text for LLMs.

The proposed AdaCE module preserves the feature information of each original
data point to the maximum extent, achieving state-of-the-art results in
several EEG prediction tasks. Furthermore, we empirically show that applying
the proposed AdaCE to fine-tune larger pre-trained models can achieve better
performance on EEG-based predicting tasks, indicating the potential of our
adapters for even larger transformers.

The plug-and-play AdaCE module can be applied to fine-tuning most of the
popular pre-trained transformers on many other time-series data with multiple
channels, not limited to EEG data and the models in our paper.

As mentioned above, the Sleep-EDF Dataset also provides ECG
(electrocardiogram) and EMG (electromyogram) records, both are also important
medical data for diagnosing health conditions and various diseases of the
body. AdaCE can also be applied to fine-tuning pre-trained transformers on ECG
and EMG data, since they are also multi-channel time-series data measuring the
human body’s electrical activities.

Moreover, we can also fine-tune pre-trained mobile-friendly transformers [30]
with AdaCE, to provide a new technical route for the development of portable
disease detectors.

Fine-tuning larger transformers on EEG datasets can also be conducted with
parameter-efficient methods (for instance, LoRA [31]), which we leave to
future work.

## Acknowledgments

This work was partially supported by the Project of Hetao Shenzhen-HKUST
Innovation Cooperation Zone HZQB-KCZYB-2020083.

## References

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On The Problem Of Entire Functions That Share a Small Function With Their
Difference Operators

Abdallah EL FARISSI1, Zinelâabidine LATREUCH2,  Benharrat BELAÏDI2 and Asim
ASIRI3

1Department of Mathematics and Informatics,

Faculty of Exact Sciences,

University of Bechar-(Algeria)

<EMAIL_ADDRESS>
2Department of Mathematics

Laboratory of Pure and Applied Mathematics

University of Mostaganem (UMAB)

B. P. 227 Mostaganem-(Algeria)

<EMAIL_ADDRESS>
<EMAIL_ADDRESS>
3Department of Mathematics,

Faculty of Science, King Abdulaziz University,

P.O.Box 80203, Jeddah 21589, Saudi Arabia

<EMAIL_ADDRESS>
Abstract. In this paper, we study uniqueness problems for an entire function
that shares small functions of finite order with their difference operators.
In particular, we give a generalization of results in $[2,3,13]$.

2010 Mathematics Subject Classification:30D35, 39A32.

Key words: Uniqueness, Entire functions, Difference operators.

##  1 Introduction and Main Results

In this paper, by meromorphic functions we will always mean meromorphic
functions in the complex plane. In what follows, we assume that the reader is
familiar with the fundamental results and the standard notations of the
Nevanlinna’s value distribution theory of meromorphic functions
$(\left[9\right],$ $\left[11\right],$ $\left[17\right])$. In addition, we will
use $\rho\left(f\right)$ to denote the order of growth of $f$ and
$\lambda\left(f\right)$ to denote the exponent of convergence of zeros of $f$,
we say that a meromorphic function $\varphi\left(z\right)$ is a small function
of $f\left(z\right)$ if $T\left(r,\varphi\right)=S\left(r,f\right),$ where
$S\left(r,f\right)=o\left(T\left(r,f\right)\right),$ as $r\rightarrow\infty$
outside of a possible exceptional set of finite logarithmic measure, we use
$S\left(f\right)$ to denote the family of all small functions with respect to
$f\left(z\right)$. For a meromorphic function $f\left(z\right),$ we define its
shift by $f_{c}\left(z\right)=f\left(z+c\right)$ and its difference operators
by

$\Delta_{c}f\left(z\right)=f\left(z+c\right)-f\left(z\right),\text{ \ \
}\Delta_{c}^{n}f\left(z\right)=\Delta_{c}^{n-1}\left(\Delta_{c}f\left(z\right)\right),\text{
}n\in\mathbb{N},\text{ }n\geq 2.$

In particular, $\Delta_{c}^{n}f\left(z\right)=\Delta^{n}f\left(z\right)$ for
the case $c=1.$

Let $f$ and $g$ be two meromorphic functions and let $a$ be a finite nonzero
value. We say that $f$ and $g$ share the value $a$ CM provided that $f-a$ and
$g-a$ have the same zeros counting multiplicities. Similarly, we say that $f$
and $g$ share $a$ IM provided that $f-a$ and $g-a$ have the same zeros
ignoring multiplicities. It is well-known that if $f$ and $g$ share four
distinct values CM, then $f$ is a Möbius transformation of $g$. In
$\left[15\right],$ Rubel and Yang proved that if an entire function $f$ shares
two distinct complex numbers CM with its derivative $f^{\prime}$, then
$f\equiv f^{\prime}$ . In 1986, Jank et al. $\left(\text{see
}\left[10\right]\right)$ proved that for a nonconstant meromorphic function
$f$, if $f$, $f^{\prime}$ and $f^{\prime\prime}$ share a finite nonzero value
CM, then $f^{\prime}\equiv f$ . This result suggests the following question:

Question 1 $\left[17\right]$ Let $f$ be a nonconstant meromorphic function,
let $a$ be a finite nonzero constant, and let $n$ and $m$ $\left(n<m\right)$
be positive integers. If $f$, $f^{\left(n\right)}$ and $f^{\left(m\right)}$
share $a$ CM, then can we get the result $f^{\left(n\right)}\equiv f$?

The following example $\left(\text{see }\left[18\right]\right)$ shows that the
answer to the above question is, in general, negative. Let $n$ and $m$ be
positive integers satisfying $m>n+1$, and let $b$ be a constant satisfying
$b^{n}=b^{m}\neq 1$. Set $a=b^{n}$ and $f\left(z\right)=e^{bz}+a-1$. Then $f$,
$f^{\left(n\right)}$ and $f^{\left(m\right)}$ share the value $a$ CM, and
$f^{\left(n\right)}\not\equiv f$. However, when $f$ is an entire function of
finite order and $m=n+1$, the answer to Question 1 is still positive. In fact,
P. Li and C. C. Yang proved the following:

Theorem A $\left[14\right]$ Let $f$ be a nonconstant entire function, let $a$
be a finite nonzero constant, and let $n$ be a positive integer. If $f$,
$f^{\left(n\right)}$ and $f^{\left(n+1\right)}$ share the value $a$ CM, then
$f\equiv f^{\prime}.$

Recently several papers have focussed on the Nevanlinna theory with respect to
difference operators see, e.g. $[1],$ $[5],\left[7\right],\left[8\right]$.
Many authors started to investigate the uniqueness of meromorphic functions
sharing values with their shifts or difference operators. In
$\left[2,3\right],$ B. Chen et al. proved a difference analogue of result of
Jank et al. and obtained the following results:

Theorem B $\left[2\right]$ Let $f\left(z\right)$ be a nonconstant entire
function of finite order, and let $a\left(z\right)\left(\not\equiv 0\right)\in
S\left(f\right)$ be a periodic entire function with period $c$. If
$f\left(z\right),$ $\Delta_{c}f\left(z\right)$ and
$\Delta_{c}^{2}f\left(z\right)$ share $a\left(z\right)$ CM, then
$\Delta_{c}f\equiv\Delta_{c}^{2}f.$

Theorem C $\left[3\right]$ Let $f\left(z\right)$ be a nonconstant entire
function of finite order, and let $a\left(z\right)\left(\not\equiv 0\right)\in
S\left(f\right)$ be a periodic entire function with period $c$. If
$f\left(z\right),$ $\Delta_{c}f\left(z\right)$ and
$\Delta_{c}^{n}f\left(z\right)$ $\left(n\geq 2\right)$ share $a\left(z\right)$
CM, then $\Delta_{c}f\equiv\Delta_{c}^{n}f.$

Theorem D $\left[3\right]$ Let $f\left(z\right)$ be a nonconstant entire
function of finite order. If $f\left(z\right),$ $\Delta_{c}f\left(z\right)$
and $\Delta_{c}^{n}f\left(z\right)$ share $0$ CM, then
$\Delta_{c}^{n}f\left(z\right)=C\Delta_{c}f\left(z\right),$ where $C$ is a
nonzero constant.

Recently in $\left[13\right],$ Z. Latreuch et al. proved the following
results:

Theorem E $\left[13\right]$ Let $f\left(z\right)$ be a nonconstant entire
function of finite order, and let $a\left(z\right)\left(\not\equiv 0\right)\in
S\left(f\right)$ be a periodic entire function with period $c$. If
$f\left(z\right)$, $\Delta_{c}^{n}f\left(z\right)$ and
$\Delta_{c}^{n+1}f\left(z\right)$ $\left(n\geq 1\right)$ share
$a\left(z\right)$ CM, then
$\Delta_{c}^{n+1}f\left(z\right)\equiv\Delta_{c}^{n}f\left(z\right).$

Theorem F $\left[13\right]$ Let $f\left(z\right)$ be a nonconstant entire
function of finite order. If $f\left(z\right),$
$\Delta_{c}^{n}f\left(z\right)$ and $\Delta_{c}^{n+1}f\left(z\right)$ share
$0$ CM, then $\Delta_{c}^{n+1}f\left(z\right)=C\Delta_{c}^{n}f\left(z\right),$
where $C$ is a nonzero constant.

For the case $n=1,$ A. El Farissi and others gave the following improvement.

Theorem G $\left[6\right]$ Let $f\left(z\right)$ be a non-periodic entire
function of finite order, and let $a\left(z\right)\left(\not\equiv 0\right)\in
S\left(f\right)$ be a periodic entire function with period $c$. If
$f\left(z\right),$ $\Delta_{c}f\left(z\right)$ and
$\Delta_{c}^{2}f\left(z\right)$ share $a\left(z\right)$ CM, then
$\Delta_{c}f\left(z\right)\equiv f\left(z\right).$

Remark 1.1 The Theorem G is essentially known in [6]. For the convenience of
readers, we give his proof in the Lemma 2.4.

It is naturally now to ask the following question: Under the hypotheses of
Theorem E, can we get the result $\Delta_{c}f\left(z\right)\equiv
f\left(z\right)$? The aim of this paper is to answer this question and to give
a difference analogue of result of P. Li and C. C. Yang in $\left[14\right]$.
In fact we obtain the following results:

Theorem 1.1 Let $f\left(z\right)$ be a nonconstant entire function of finite
order such that $\Delta_{c}^{n}f\left(z\right)\not\equiv 0$, and let
$a\left(z\right)\left(\not\equiv 0\right)\in S\left(f\right)$ be a periodic
entire function with period $c$. If $f\left(z\right)$,
$\Delta_{c}^{n}f\left(z\right)$ and $\Delta_{c}^{n+1}f\left(z\right)$
$\left(n\geq 1\right)$ share $a\left(z\right)$ CM, then
$\Delta_{c}f\left(z\right)\equiv f\left(z\right).$

Remark 1.2 The condition $\Delta_{c}^{n}f\left(z\right)\not\equiv 0$ is
necessary. Let’s take for example the entire function
$f\left(z\right)=1+e^{2\pi iz}$ and $c=a=1,$ then $f-a$ and
$\Delta^{n}f-a=\Delta^{n+1}f-a=-1$ have the same zeros but $\Delta f\neq f.$
On the other hand, under the conditions of Theorem 1.1
$\Delta_{c}^{n}f\left(z\right)\not\equiv 0$ can not be a periodic entire
function because
$\Delta_{c}^{n+1}f\left(z\right)\equiv\Delta_{c}^{n}f\left(z\right)$ (Theorem
E, $\left[13\right]$).

Example 1.1 Let $f\left(z\right)=e^{z\ln 2}$ and $c=1.$ Then, for any
$a\in\mathbb{C},$ we notice that $f\left(z\right),$
$\Delta_{c}^{n}f\left(z\right)$ and $\Delta_{c}^{n+1}f\left(z\right)$ share
$a$ CM for all $n\in\mathbb{N}$ and we can easily see that
$\Delta_{c}f\left(z\right)\equiv f\left(z\right).$ This example satisfies
Theorem 1.1.

Theorem 1.2 Let $f\left(z\right)$ be a nonconstant entire function of finite
order such that $\Delta_{c}^{n}f\left(z\right)\not\equiv 0$, and let
$a\left(z\right),$ $b\left(z\right)\left(\not\equiv 0\right)\in
S\left(f\right)$ such that $b\left(z\right)$ is a periodic entire function
with period $c$ and $\Delta_{c}^{m}a\left(z\right)\equiv 0$ $\left(1\leq m\leq
n\right)$. If $f\left(z\right)-a\left(z\right),$
$\Delta_{c}^{n}f\left(z\right)-b\left(z\right)$ and
$\Delta_{c}^{n+1}f\left(z\right)-b\left(z\right)$ share $0$ CM, then
$\Delta_{c}f\left(z\right)\equiv
f\left(z\right)+b\left(z\right)+\Delta_{c}a\left(z\right)-a\left(z\right).$

Remark 1.3 The condition $b\left(z\right)\not\equiv 0$ is necessary in the
proof of Theorem 1.2, for the case $b\left(z\right)\equiv 0,$ please see
Theorem 1.4.

Remark 1.4 The condition $\Delta_{c}^{m}a\left(z\right)\equiv 0$ in Theorem
1.2 is more general than the condition ”periodic entire function of period
$c$”.

For the case $m=1,$ we deduce the following result.

Corollary 1.1 Let $f\left(z\right)$ be a nonconstant entire function of finite
order such that $\Delta_{c}^{n}f\left(z\right)\not\equiv 0$, and let
$a\left(z\right),$ $b\left(z\right)\left(\not\equiv 0\right)\in
S\left(f\right)$ be periodic entire functions with period $c$. If
$f\left(z\right)-a\left(z\right),$
$\Delta_{c}^{n}f\left(z\right)-b\left(z\right)$ and
$\Delta_{c}^{n+1}f\left(z\right)-b\left(z\right)$ share $0$ CM, then
$\Delta_{c}f\left(z\right)\equiv
f\left(z\right)+b\left(z\right)-a\left(z\right).$

Example 1.2 Let $f\left(z\right)=e^{z\ln 2}-2,$ $a=-1$ and $b=1.$ It is clear
that $f\left(z\right)-a,$ $\Delta^{n}f\left(z\right)-b$ and
$\Delta^{n+1}f\left(z\right)-b$ share $0$ CM. Here, we also get $\Delta
f\left(z\right)=f\left(z\right)+b-a.$

Example 1.3 Let $f\left(z\right)=e^{z\ln 2}+z^{3}-1,$ $a\left(z\right)=z^{3}$
and $b=1.$ It is clear that $f\left(z\right)-z^{3},$
$\Delta^{4}f\left(z\right)-1$ and $\Delta^{5}f\left(z\right)-1$ share $0$ CM.
On the other hand, we can verify that $\Delta
f\left(z\right)=f\left(z\right)+1+\Delta z^{3}-z^{3}$ which satisfies Theorem
1.2.

Theorem 1.3 Let $f\left(z\right)$ be a nonconstant entire function of finite
order such that $\Delta_{c}^{n}f\left(z\right)\not\equiv 0$. If
$f\left(z\right),$ $\Delta_{c}^{n}f\left(z\right)$ and
$\Delta_{c}^{n+1}f\left(z\right)$ share $0$ CM, then
$\Delta_{c}f\left(z\right)\equiv Cf\left(z\right),$ where $C$ is a nonzero
constant.

Example 1.4 Let $f\left(z\right)=e^{az}$ and $c=1$ where $a\neq 2k\pi i$
$\left(k\in\mathbb{Z}\right),$ it is clear that
$\Delta_{c}^{n}f\left(z\right)=\left(e^{a}-1\right)^{n}e^{az}$ for any integer
$n\geq 1.$ So, $f\left(z\right),$ $\Delta_{c}^{n}f\left(z\right)$ and
$\Delta_{c}^{n+1}f\left(z\right)$ share $0$ CM for all $n\in\mathbb{N}$ and we
can easily see that $\Delta_{c}f\left(z\right)\equiv Cf\left(z\right)$ where
$C=e^{a}-1.$ This example satisfies Theorem 1.3.

Corollary 1.2 Let $f\left(z\right)$ be a nonconstant entire function of finite
order such that $f\left(z\right),$ $\Delta_{c}^{n}f\left(z\right)$
$\left(\not\equiv 0\right)$ and $\Delta_{c}^{n+1}f\left(z\right)$ $\left(n\geq
1\right)$ share $0$ CM. If there exists a point $z_{0}$ and an integer $m\geq
1$ such that $\Delta_{c}^{m}f\left(z_{0}\right)=f\left(z_{0}\right)\neq 0,$
then $\Delta_{c}^{m}f\left(z\right)\equiv f\left(z\right).$

By combining Theorem 1.2 and Theorem 1.3 we can prove the following result.

Theorem 1.4 Let $f\left(z\right)$ be a nonconstant entire function of finite
order such that $\Delta_{c}^{n}f\left(z\right)\not\equiv 0$, and let
$a\left(z\right)\in S\left(f\right)$ such that
$\Delta_{c}^{m}a\left(z\right)\equiv 0$ $\left(1\leq m\leq n\right)$. If
$f\left(z\right)-a\left(z\right),$ $\Delta_{c}^{n}f\left(z\right)$ and
$\Delta_{c}^{n+1}f\left(z\right)$ share $0$ CM, then
$\Delta_{c}f\left(z\right)\equiv
Cf\left(z\right)+\Delta_{c}a\left(z\right)-a\left(z\right),$ where $C$ is a
nonzero constant.

##  2 Some lemmas

Lemma 2.1 $\left[5\right]$ Let $\eta_{1},\eta_{2}$ be two arbitrary complex
numbers such that $\eta_{1}\neq\eta_{2}$ and let $f\left(z\right)$ be a finite
order meromorphic function. Let $\sigma$ be the order of $f\left(z\right)$,
then for each $\varepsilon>0,$ we have

$m\left(r,\frac{f\left(z+\eta_{1}\right)}{f\left(z+\eta_{2}\right)}\right)=O\left(r^{\sigma-1+\varepsilon}\right).$

By combining Theorem 1.4 in $\left[4\right]$ and Theorem 2.2 in
$\left[12\right],$ we can prove the following lemma.

Lemma 2.2 Let
$a_{0}\left(z\right),a_{1}\left(z\right),\cdots,a_{n}\left(z\right)\left(\not\equiv
0\right),$ $F\left(z\right)\left(\not\equiv 0\right)$ be finite order
meromorphic functions, $c_{k}$ ($k=0,\cdots,n$) be constants, unequal to each
other. If $f$ is a finite order meromorphic solution of the equation

$a_{n}\left(z\right)f\left(z+c_{n}\right)+\cdots+a_{1}\left(z\right)f\left(z+c_{1}\right)+a_{0}\left(z\right)f\left(z+c_{0}\right)=F\left(z\right)$
(2.1)

with

$\max\left\\{\rho\left(a_{i}\right),\left(i=0,\cdots,n\right),\rho\left(F\right)\right\\}<\rho\left(f\right),$

then $\lambda\left(f\right)=\rho\left(f\right).$

Proof. By $\left(2.1\right)$ we have

$\frac{1}{f\left(z+c_{0}\right)}=\frac{1}{F}\left(a_{n}\frac{f\left(z+c_{n}\right)}{f\left(z+c_{0}\right)}+\cdots+a_{1}\frac{f\left(z+c_{1}\right)}{f\left(z+c_{0}\right)}+a_{0}\right).$
(2.2)

Set $\max\left\\{\rho\left(a_{j}\right)\text{
}\left(j=0,\cdots,n\right),\rho\left(F\right)\right\\}=\beta<\rho\left(f\right)=\rho.$
Then, for any given $\varepsilon$
$\left(0<\varepsilon<\frac{\rho-\beta}{2}\right)$, we have

$\overset{n}{\underset{j=0}{\sum}}T\left(r,a_{j}\right)+T\left(r,F\right)\leq\left(n+2\right)\exp\left\\{r^{\beta+\varepsilon}\right\\}=o\left(1\right)\exp\left\\{r^{\rho-\varepsilon}\right\\}.$
(2.3)

By $\left(2.2\right)$, $\left(2.3\right)$ and Lemma 2.1, we obtain

$T\left(r,f\right)=T\left(r,\frac{1}{f}\right)+O\left(1\right)=m\left(r,\frac{1}{f}\right)+N\left(r,\frac{1}{f}\right)+O\left(1\right)$
$\leq
N\left(r,\frac{1}{f}\right)+m\left(r,\frac{1}{F}\right)+\overset{n}{\underset{j=0}{\sum}}m\left(r,a_{j}\right)+\overset{n}{\underset{j=1}{\sum}}m\left(r,\frac{f\left(z+c_{j}\right)}{f\left(z+c_{0}\right)}\right)+O\left(1\right)$
$\leq
N\left(r,\frac{1}{f}\right)+T\left(r,\frac{1}{F}\right)+\overset{n}{\underset{j=0}{\sum}}T\left(r,a_{j}\right)+\overset{n}{\underset{j=1}{\sum}}m\left(r,\frac{f\left(z+c_{j}\right)}{f\left(z+c_{0}\right)}\right)+O\left(1\right)$
$\leq
N\left(r,\frac{1}{f}\right)+O\left(r^{\rho-1+\varepsilon}\right)+o\left(1\right)\exp\left\\{r^{\rho-\varepsilon}\right\\}.$
(2.4)

By $\left(2.4\right)$, we obtain that
$\rho\left(f\right)\leq\lambda\left(f\right)$ and since
$\lambda\left(f\right)\leq\rho\left(f\right)$ for every meromorphic function,
we deduce that $\lambda\left(f\right)=\rho\left(f\right).$

Remark 2.1 Recently, Shun-Zhou Wu and Xiu-Min Zheng (see $\left[16\right]$)
obtained Lemma 2.2 by using a different proof.

Lemma 2.3 $\left[17\right]$ Suppose $f_{j}\left(z\right)$ $(j=1,2,\cdots,n+1)$
and $g_{j}\left(z\right)$ $(j=1,2,\cdots,n)$ $(n\geq 1)$ are entire functions
satisfying the following conditions:

$\left(\text{i}\right)$
$\overset{n}{\underset{j=1}{\sum}}f_{j}\left(z\right)e^{g_{j}\left(z\right)}\equiv
f_{n+1}\left(z\right);$

$\left(\text{ii}\right)$ The order of $f_{j}\left(z\right)$ is less than the
order of $e^{g_{k}\left(z\right)}$ for $1\leq j\leq n+1,$ $1\leq k\leq n.$ And
furthermore, the order of $f_{j}\left(z\right)$ is less than the order of
$e^{g_{h}\left(z\right)-g_{k}\left(z\right)}$ for $n\geq 2$ and $1\leq j\leq
n+1,$ $1\leq h<k\leq n.$

Then $f_{j}\left(z\right)\equiv 0,$ $\left(j=1,2,\cdots n+1\right).$

Lemma 2.4 $\left[6\right]$ Let $f\left(z\right)$ be a non-periodic entire
function of finite order, and let $a\left(z\right)\left(\not\equiv 0\right)\in
S\left(f\right)$ be a periodic entire function with period $c$. If
$f\left(z\right),$ $\Delta_{c}f\left(z\right)$ and
$\Delta_{c}^{2}f\left(z\right)$ share $a\left(z\right)$ CM, then
$\Delta_{c}f\left(z\right)\equiv f\left(z\right).$

Proof. Suppose that $\Delta_{c}f\left(z\right)\not\equiv f\left(z\right).$
Since $f,$ $\Delta_{c}f$ and $\Delta_{c}^{2}f$ share $a\left(z\right)$ CM,
then we have

$\frac{\Delta_{c}f\left(z\right)-a\left(z\right)}{f\left(z\right)-a\left(z\right)}=e^{P\left(z\right)}$

and

$\frac{\Delta_{c}^{2}f\left(z\right)-a\left(z\right)}{f\left(z\right)-a\left(z\right)}=e^{Q\left(z\right)}$

where $P$ $\left(e^{P}\not\equiv 1\right)$ and $Q$ are polynomials. By using
Theorem B, we obtain that $\Delta_{c}^{2}f\equiv\Delta_{c}f$, which means that

$\alpha\left(z\right)=\Delta_{c}f\left(z\right)-f\left(z\right)$ (2.5)

is entire periodic function of period $c.$ By $\left(2.5\right)$ we have

$\Delta_{c}f\left(z\right)-a\left(z\right)=f\left(z\right)-a\left(z\right)+\alpha\left(z\right),$

then

$\frac{\Delta_{c}f\left(z\right)-a\left(z\right)}{f\left(z\right)-a\left(z\right)}=1+\frac{\alpha\left(z\right)}{f\left(z\right)-a\left(z\right)}=e^{P\left(z\right)},$

which is equivalent to

$f\left(z\right)-a\left(z\right)=\frac{\alpha\left(z\right)}{e^{P\left(z\right)}-1}.$
(2.6)

Since $\alpha\left(z\right)$ and $a\left(z\right)$ are periodic functions of
period $c$, then we have

$\Delta_{c}f\left(z\right)=\alpha\left(z\right)\Delta_{c}\left(\frac{1}{e^{P\left(z\right)}-1}\right)$
(2.7)

and

$\Delta_{c}^{2}f\left(z\right)=\alpha\left(z\right)\Delta_{c}^{2}\left(\frac{1}{e^{P\left(z\right)}-1}\right).$
(2.8)

We have the two following subcases:

$\left(\text{i}\right)$ If $P\equiv K$ $\left(K\neq 2k\pi i,\text{
}K\in\mathbb{Z}\right)$, then by $\left(2.7\right)$ we have
$\Delta_{c}f\left(z\right)=0.$ On the other hand, by using $\left(2.5\right),$
$\left(2.6\right)$ and $\Delta_{c}f\left(z\right)=0,$ we deduce that

$f\left(z\right)-a\left(z\right)=\frac{-f\left(z\right)}{e^{K}-1},\text{
}K\in\mathbb{C}-\left\\{2k\pi i,k\in\mathbb{Z}\right\\}.$

So,

$f\left(z\right)=\frac{e^{K}-1}{e^{K}}a\left(z\right).$

Hence

$T\left(r,f\right)=S\left(r,f\right),$

which is a contradiction.

$\left(\text{ii}\right)$ If $P$ is nonconstant and since
$\Delta_{c}^{2}f\left(z\right)=\Delta_{c}f\left(z\right),$ then

$e^{P_{c}\left(z\right)+P\left(z\right)}-3e^{P_{2c}\left(z\right)+P\left(z\right)}+2e^{P_{2c}\left(z\right)+P_{c}\left(z\right)}+e^{P_{2c}\left(z\right)}-3e^{P_{c}\left(z\right)}+2e^{P\left(z\right)}=0$

which is equivalent to

$e^{P_{c}\left(z\right)}+\left(2e^{\Delta_{c}P\left(z\right)}-3\right)e^{P_{2c}\left(z\right)}=-e^{\Delta_{c}P_{c}\left(z\right)+\Delta_{c}P\left(z\right)}+3e^{\Delta_{c}P\left(z\right)}-2.$
(2.9)

Since $\deg\Delta_{c}P=\deg P-1,$ then we have

$\rho\left(e^{P_{c}}+\left(2e^{\Delta_{c}P}-3\right)e^{P_{2c}}\right)=\rho\left(-e^{\Delta_{c}P_{c}+\Delta_{c}P}+3e^{\Delta_{c}P}-2\right)\leq\deg
P-1.$ (2.10)

On the other hand,

$\rho\left(e^{P_{c}}+\left(2e^{\Delta_{c}P}-3\right)e^{P_{2c}}\right)=\rho\left(e^{P_{c}}\right)=\deg
P$ (2.11)

because if we have the contrary

$\rho\left(e^{P_{c}}+\left(2e^{\Delta_{c}P}-3\right)e^{P_{2c}}\right)<\rho\left(e^{P_{c}}\right),$

we obtain the following contradiction

$\deg
P=\rho\left(\frac{e^{P_{c}}+\left(2e^{\Delta_{c}P}-3\right)e^{P_{2c}}}{e^{P_{c}}}\right)=\rho\left(1+\left(2e^{\Delta_{c}P}-3\right)e^{\Delta
P_{c}}\right)\leq\deg P-1.$

By using $\left(2.10\right)$ and $\left(2.11\right),$ we obtain

$\deg P\leq\deg P-1$

which is a contradiction. This leads to
$\Delta_{c}f\left(z\right)=f\left(z\right)$. Thus, the proof of Lemma 2.4 is
completed.

##  3 Proof of the Theorems and Corollary

Proof of the Theorem 1.1. Obviously, suppose that
$\Delta_{c}f\left(z\right)\not\equiv f\left(z\right)$. By using Theorem E, we
have

$\frac{\Delta_{c}^{n}f\left(z\right)-a\left(z\right)}{f\left(z\right)-a\left(z\right)}=e^{P\left(z\right)}$
(3.1)

and

$\frac{\Delta_{c}^{n+1}f\left(z\right)-a\left(z\right)}{f\left(z\right)-a\left(z\right)}=e^{P\left(z\right)},$
(3.2)

where $P$ $\left(e^{P}\not\equiv 1\right)$ is polynomial. Dividing the proof
of Theorem 1.1 into two cases:

Case 1. $P$ is a nonconstant polynomial. Setting now
$g\left(z\right)=f\left(z\right)-a\left(z\right).$ Then, we have from
$\left(3.1\right)$ and $\left(3.2\right)$

$\Delta_{c}^{n}g\left(z\right)=e^{P\left(z\right)}g\left(z\right)+a\left(z\right)$
(3.3)

and

$\Delta_{c}^{n+1}g\left(z\right)=e^{P\left(z\right)}g\left(z\right)+a\left(z\right).$
(3.4)

By $\left(3.3\right)$ and $\left(3.4\right),$ we have

$g_{c}\left(z\right)=2e^{P-P_{c}}g\left(z\right)+a\left(z\right)e^{-P_{c}}.$

Using the principle of mathematical induction, we obtain

$g_{ic}\left(z\right)=2^{i}e^{P-P_{ic}}g\left(z\right)+a\left(z\right)\left(2^{i}-1\right)e^{-P_{ic}},\text{
}i\geq 1.$ (3.5)

Now, we can rewrite $\left(3.3\right)$ as

$\Delta_{c}^{n}g\left(z\right)=\overset{n}{\underset{i=1}{\sum}}C_{n}^{i}\left(-1\right)^{n-i}\left(2^{i}e^{P-P_{ic}}g\left(z\right)+a\left(z\right)\left(2^{i}-1\right)e^{-P_{ic}}\right)$
$+\left(-1\right)^{n}g\left(z\right)=e^{P}g\left(z\right)+a\left(z\right),$

which implies

$\left(\overset{n}{\underset{i=0}{\sum}}C_{n}^{i}\left(-1\right)^{n-i}2^{i}e^{P-P_{ic}}-e^{P}\right)g\left(z\right)$
$+a\left(z\right)\left(\overset{n}{\underset{i=0}{\sum}}C_{n}^{i}\left(-1\right)^{n-i}\left(2^{i}-1\right)e^{-P_{ic}}-1\right)=0.$

Hence

$A_{n}\left(z\right)g\left(z\right)+B_{n}\left(z\right)=0,$ (3.6)

where

$A_{n}\left(z\right)=\overset{n}{\underset{i=0}{\sum}}C_{n}^{i}\left(-1\right)^{n-i}2^{i}e^{P-P_{ic}}-e^{P}$

and

$B_{n}\left(z\right)=a\left(z\right)\left(\overset{n}{\underset{i=0}{\sum}}C_{n}^{i}\left(-1\right)^{n-i}\left(2^{i}-1\right)e^{-P_{ic}}-1\right).$

By the same method, we can rewrite $\left(3.4\right)$ as

$A_{n+1}\left(z\right)g\left(z\right)+B_{n+1}\left(z\right)=0,$ (3.7)

where

$A_{n+1}\left(z\right)=\overset{n+1}{\underset{i=0}{\sum}}C_{n+1}^{i}\left(-1\right)^{n+1-i}2^{i}e^{P-P_{ic}}-e^{P}$

and

$B_{n+1}\left(z\right)=a\left(z\right)\left(\overset{n+1}{\underset{i=0}{\sum}}C_{n+1}^{i}\left(-1\right)^{n+1-i}\left(2^{i}-1\right)e^{-P_{ic}}-1\right).$

We can see easily from the equations $\left(3.6\right)$ and $\left(3.7\right)$
that

$h\left(z\right)=A_{n}\left(z\right)B_{n+1}\left(z\right)-A_{n+1}\left(z\right)B_{n}\left(z\right)\equiv
0.$ (3.8)

On the other hand, we remark that

$e^{P}B_{n}\left(z\right)=a\left(z\right)e^{P}\left(\overset{n}{\underset{i=0}{\sum}}C_{n}^{i}\left(-1\right)^{n-i}2^{i}e^{-P_{ic}}-\overset{n}{\underset{i=0}{\sum}}C_{n}^{i}\left(-1\right)^{n-i}e^{-P_{ic}}-1\right)$
$=a\left(z\right)e^{P}\left(\overset{n}{\underset{i=0}{\sum}}C_{n}^{i}\left(-1\right)^{n-i}2^{i}e^{-P_{ic}}-1-\Delta_{c}^{n}\left(e^{-P}\right)\right)$
$=a\left(z\right)\left(A_{n}\left(z\right)-e^{P}\Delta_{c}^{n}\left(e^{-P}\right)\right).$

Then

$B_{n}\left(z\right)=a\left(z\right)\left(e^{-P}A_{n}\left(z\right)-\Delta_{c}^{n}\left(e^{-P}\right)\right).$
(3.9)

By the same method, we obtain

$B_{n+1}\left(z\right)=a\left(z\right)\left(e^{-P}A_{n+1}\left(z\right)-\Delta_{c}^{n+1}\left(e^{-P}\right)\right).$
(3.10)

Return now to the equation $\left(3.8\right)$, by using $\left(3.9\right)$ and
$\left(3.10\right),$ we get

$h\left(z\right)=A_{n}\left(z\right)B_{n+1}\left(z\right)-A_{n+1}\left(z\right)B_{n}\left(z\right)$
$=A_{n}\left(z\right)\left[a\left(z\right)\left(e^{-P}A_{n+1}\left(z\right)-\Delta_{c}^{n+1}\left(e^{-P}\right)\right)\right]$
$-A_{n+1}\left(z\right)\left[a\left(z\right)\left(e^{-P}A_{n}\left(z\right)-\Delta_{c}^{n}\left(e^{-P}\right)\right)\right]$
$=a\left(z\right)\left[A_{n+1}\left(z\right)\Delta_{c}^{n}\left(e^{-P}\right)-A_{n}\left(z\right)\Delta_{c}^{n+1}\left(e^{-P}\right)\right]\equiv
0.$

Hence

$A_{n+1}\left(z\right)\Delta_{c}^{n}\left(e^{-P}\right)-A_{n}\left(z\right)\Delta_{c}^{n+1}\left(e^{-P}\right)\equiv
0.$

Therefore

$\Delta_{c}^{n}\left(e^{-P}\right)\left(\overset{n+1}{\underset{i=0}{\sum}}C_{n+1}^{i}\left(-1\right)^{n+1-i}2^{i}e^{-P_{ic}}-1\right)$
$-\Delta_{c}^{n+1}\left(e^{-P}\right)\left(\overset{n}{\underset{i=0}{\sum}}C_{n}^{i}\left(-1\right)^{n-i}2^{i}e^{-P_{ic}}-1\right)=0.$

Thus

$\displaystyle\Delta_{c}^{n}\left(e^{-P}\right)\overset{n+1}{\underset{i=0}{\sum}}C_{n+1}^{i}\left(-1\right)^{n+1-i}2^{i}e^{-P_{i}c}-\Delta_{c}^{n+1}\left(e^{-P}\right)\overset{n}{\underset{i=0}{\sum}}C_{n}^{i}\left(-1\right)^{n-i}2^{i}e^{-P_{ic}}$
$\displaystyle=$
$\displaystyle\Delta_{c}^{n}\left(e^{-P}\right)-\Delta_{c}^{n+1}\left(e^{-P}\right)=\Delta_{c}^{n}\left(2e^{-P}-e^{-P_{c}}\right).$

Then

$\overset{n}{\underset{i=0}{\sum}}\left(\Delta_{c}^{n}\left(e^{-P}\right)C_{n+1}^{i}\left(-1\right)^{n+1-i}-\Delta_{c}^{n+1}\left(e^{-P}\right)C_{n}^{i}\left(-1\right)^{n-i}\right)2^{i}e^{-P_{ic}}$
$+\Delta_{c}^{n}\left(e^{-P}\right)2^{n+1}e^{-P_{\left(n+1\right)c}}=\Delta_{c}^{n}\left(2e^{-P}-e^{-P_{c}}\right),$

which yields

$\overset{n}{\underset{i=0}{\sum}}\left(\Delta_{c}^{n}\left(e^{-P}\right)C_{n+1}^{i}+\Delta_{c}^{n+1}\left(e^{-P}\right)C_{n}^{i}\right)\left(-1\right)^{n+1-i}2^{i}e^{P_{\left(n+1\right)c}-P_{ic}}$
$+\Delta_{c}^{n}\left(e^{-P}\right)2^{n+1}=e^{P_{\left(n+1\right)c}}\Delta_{c}^{n}\left(2e^{-P}-e^{-P_{c}}\right).$
(3.11)

Let us denote

$\alpha_{i}\left(z\right)=\left(-1\right)^{n+1-i}2^{i}e^{P_{\left(n+1\right)c}-P_{ic}},\text{
}i=0,\cdots,n$

and

$\alpha_{n+1}\left(z\right)=e^{P_{\left(n+1\right)c}}\Delta_{c}^{n}\left(2e^{-P}-e^{-P_{c}}\right).$

It is clear that $\rho\left(\alpha_{i}\right)\leq\deg P-1$ for all
$i=0,2,\cdots,n+1.$ The equation $\left(3.11\right)$ will be

$\overset{n}{\underset{i=0}{\sum}}\left(\Delta_{c}^{n}\left(e^{-P}\right)C_{n+1}^{i}+\Delta_{c}^{n+1}\left(e^{-P}\right)C_{n}^{i}\right)\alpha_{i}\left(z\right)+\Delta_{c}^{n}\left(e^{-P}\right)2^{n+1}$
$=\left(\overset{n}{\underset{i=0}{\sum}}C_{n+1}^{i}\alpha_{i}\left(z\right)+2^{n+1}\right)\Delta_{c}^{n}\left(e^{-P}\right)$
$+\left(\overset{n}{\underset{i=0}{\sum}}C_{n}^{i}\alpha_{i}\left(z\right)\right)\Delta_{c}^{n+1}\left(e^{-P}\right)=\alpha_{n+1}\left(z\right).$
(3.12)

For convenience, we denote by $M\left(z\right)$ and $N\left(z\right)$ the
following

$M\left(z\right)=\overset{n}{\underset{i=0}{\sum}}C_{n+1}^{i}\alpha_{i}\left(z\right)+2^{n+1},\text{
}N\left(z\right)=\overset{n}{\underset{i=0}{\sum}}C_{n}^{i}\alpha_{i}\left(z\right).$

Equation $\left(3.12\right)$ is equivalent to

$M\left(z\right)\overset{n}{\underset{i=0}{\sum}}C_{n}^{i}\left(-1\right)^{n-i}e^{-P_{ic}}+N\left(z\right)\overset{n+1}{\underset{i=0}{\sum}}C_{n+1}^{i}\left(-1\right)^{n+1-i}e^{-P_{ic}}$
$=\overset{n}{\underset{i=0}{\sum}}\left(C_{n}^{i}M\left(z\right)-C_{n+1}^{i}N\left(z\right)\right)\left(-1\right)^{n-i}e^{-P_{ic}}+N\left(z\right)e^{-P_{\left(n+1\right)c}}=\alpha_{n+1}\left(z\right).$
(3.13)

As conclusion, we can say that $\left(3.13\right)$ can be written as follow

$a_{n+1}\left(z\right)e^{-P\left(z+\left(n+1\right)c\right)}+a_{n}\left(z\right)e^{-P\left(z+nc\right)}+\cdots+a_{0}\left(z\right)e^{-P\left(z\right)}=\alpha_{n+1}\left(z\right),$
(3.14)

where $a_{0}\left(z\right),\cdots,a_{n+1}\left(z\right)$ and
$\alpha_{n+1}\left(z\right)$ are entire functions. We distingue the following
two subcases.

$\left(\text{i}\right)$ If $\deg P>1,$ then we have

$\max\left\\{\rho\left(a_{i}\right)\text{
}\left(i=0,\cdots,n+1\right),\rho\left(\alpha_{n+1}\right)\right\\}<\deg P.$
(3.15)

In order to prove that $\alpha_{n+1}\left(z\right)\not\equiv 0,$ it suffices
to show that $\Delta_{c}^{n}\left(2e^{-P}-e^{-P_{c}}\right)\not\equiv 0.$
Suppose the contrary. Thus

$\overset{n}{\underset{i=0}{\sum}}C_{n}^{i}\left(-1\right)^{n-i}\left(2e^{-P_{ic}}-e^{-P_{\left(i+1\right)c}}\right)\equiv
0.$ (3.16)

The equation $\left(3.16\right)$ can be written as

$\overset{n+1}{\underset{i=0}{\sum}}b_{i}e^{-P_{ic}}\equiv 0,$

where

$b_{i}=\left\\{\begin{array}[]{c}2\left(-1\right)^{n},\text{ if }i=0\\\
\left(2C_{n}^{i}+C_{n}^{i-1}\right)\left(-1\right)^{n-i},\text{ if }1\leq
i\leq n\\\ -1,\text{ if }i=n+1.\end{array}\right.$

Since $\deg P=m>1,$ then for any two integers $j$ and $k$ such that $0\leq
j<k\leq n+1,$ we have

$\rho\left(e^{-P_{kc}+P_{jc}}\right)=\deg P-1.$

It’s clear now that all the conditions of Lemma 2.3 are satisfied. So, by
Lemma 2.3 we obtain $b_{i}\equiv 0$ for all $i=0,...,n+1,$ which is
impossible. Then, $\alpha_{n+1}\left(z\right)\not\equiv 0.$ By Lemma 2.2,
$\left(3.14\right)$ and $\left(3.15\right),$ we deduce that
$\lambda\left(e^{P}\right)=\deg P>1,$ which is a contradiction.

$\left(\text{ii}\right)$ $\deg P=1.$ Suppose now that $P\left(z\right)=\mu
z+\eta$ $\left(\mu\neq 0\right).$ Assume that
$\alpha_{n+1}\left(z\right)\equiv 0.$ It easy to see that

$\Delta_{c}^{n}\left(2e^{-P}-e^{-P_{c}}\right)=\left(2-e^{-\mu
c}\right)\Delta_{c}^{n}\left(e^{-P}\right).$

In the following two subcases, we prove that both of $\left(2-e^{-\mu
c}\right)$ and $\Delta_{c}^{n}\left(e^{-P}\right)$ are not vanishing.

(A) Suppose that $2=e^{-\mu c}.$ Then for any integer $i$, we have $e^{-i\mu
c}=2^{i}$ and $e^{-P_{ic}}=2^{i}e^{-P},$ applying that on the equation
$\left(3.6\right)$, we get

$A_{n}\left(z\right)=\overset{n}{\underset{i=0}{\sum}}C_{n}^{i}\left(-1\right)^{n-i}2^{i}e^{-i\mu
c}-e^{P}=3^{n}-e^{P},$

and

$B_{n}\left(z\right)=a\left(z\right)\left(\overset{n}{\underset{i=0}{\sum}}C_{n}^{i}\left(-1\right)^{n-i}\left(2^{i}-1\right)e^{-P_{ic}}-1\right)$
$=a\left(z\right)\left(\overset{n}{\underset{i=0}{\sum}}C_{n}^{i}\left(-1\right)^{n-i}\left(4^{i}-2^{i}\right)e^{-P}-1\right)=a\left(z\right)\left(\left(3^{n}-1\right)e^{-P}-1\right).$

Then

$\left(3^{n}-e^{P}\right)g\left(z\right)+a\left(z\right)\left(\left(3^{n}-1\right)e^{-P}-1\right)=0,$

which is equivalent to

$g\left(z\right)=a\left(z\right)\frac{e^{P}-\left(3^{n}-1\right)}{e^{P}\left(3^{n}-e^{P}\right)}.$
(3.17)

By the same arguing as before and the equation $\left(3.7\right),$ we obtain

$g\left(z\right)=a\left(z\right)\frac{e^{P}-\left(3^{n+1}-1\right)}{e^{P}\left(3^{n+1}-e^{P}\right)},$

which contradicts $\left(3.17\right)$.

(B) Suppose now that $\Delta_{c}^{n}\left(e^{-P}\right)\equiv 0.$ Thus

$\Delta_{c}^{n}\left(e^{-P}\right)=\overset{n}{\underset{i=0}{\sum}}C_{n}^{i}\left(-1\right)^{n-i}e^{-\mu\left(z+ic\right)-\eta}=e^{-P}\overset{n}{\underset{i=0}{\sum}}C_{n}^{i}\left(-1\right)^{n-i}e^{-\mu
ic}$ $=e^{-P}\left(e^{-\mu c}-1\right)^{n}.$

This together with $\Delta_{c}^{n}e^{-P}\equiv 0$ gives $\left(e^{-\mu
c}-1\right)^{n}\equiv 0,$ which yields $e^{\mu c}\equiv 1.$ Therefore, for any
$j\in\mathbb{Z}$

$e^{P\left(z+jc\right)}=e^{\mu z+\mu jc+\eta}=\left(e^{\mu
c}\right)^{j}e^{P\left(z\right)}=e^{P\left(z\right)}.$ (3.18)

On the other hand, we have from $\left(3.1\right)$

$\Delta_{c}^{n}f\left(z\right)=e^{P\left(z\right)}\left(f\left(z\right)-a\left(z\right)\right)+a\left(z\right).$
(3.19)

By $\left(3.18\right)$ and $\left(3.19\right),$ we have

$\Delta_{c}^{n+1}f\left(z\right)=e^{P\left(z\right)}\Delta_{c}f\left(z\right)$
(3.20)

Combining $\left(3.2\right)$ and $\left(3.20\right),$ we obtain

$\Delta_{c}f\left(z\right)=\left(f\left(z\right)-a\left(z\right)\right)+a\left(z\right)e^{-P\left(z\right)}$

which means that
$\Delta_{c}^{n+1}f\left(z\right)=\Delta_{c}^{n}f\left(z\right)$ for all $n\geq
1.$ Therefore,  $f\left(z\right),$ $\Delta_{c}f\left(z\right)$ and
$\Delta_{c}^{2}f\left(z\right)$ share $a\left(z\right)$ CM and by Lemma 2.4 we
obtain $\Delta_{c}f\left(z\right)=f\left(z\right),$ which contradicts the
hypothesis. Then $\Delta_{c}^{n}\left(e^{-P}\right)\not\equiv 0.$ From the
subcases (A) and (B), we can deduce that $\alpha_{n+1}\left(z\right)\not\equiv
0.$ It is clear that

$\max\left\\{\rho\left(a_{i}\right),\rho\left(\alpha_{n+1}\right),i=0,...,n+1\right\\}<\deg
P=1.$

By using Lemma 2.2, we obtain $\lambda\left(e^{P}\right)=\deg P=1,$ which is a
contradiction, and $P$ must be a constant.

Case 2. $P\left(z\right)\equiv K$ , $K\in\mathbb{C}-\left\\{2k\pi
i,k\in\mathbb{Z}\right\\}.$ We have from $\left(3.1\right)$

$\Delta_{c}^{n}f\left(z\right)=e^{K}\left(f\left(z\right)-a\left(z\right)\right)+a\left(z\right).$

Hence

$\Delta_{c}^{n+1}f\left(z\right)=e^{K}\Delta_{c}f\left(z\right).$ (3.21)

Combining $\left(3.2\right)$ and $\left(3.21\right),$ we obtain

$\Delta_{c}f\left(z\right)=\left(f\left(z\right)-a\left(z\right)\right)+a\left(z\right)e^{-K}$

which means that
$\Delta_{c}^{n+1}f\left(z\right)=\Delta_{c}^{n}f\left(z\right)$ for all $n\geq
1.$ Therefore,  $f\left(z\right),$ $\Delta_{c}f\left(z\right)$ and
$\Delta_{c}^{2}f\left(z\right)$ share $a\left(z\right)$ CM and by Lemma 2.4 we
obtain $\Delta_{c}f\left(z\right)=f\left(z\right),$ which contradicts the
hypothesis. Then $e^{P}\equiv 1$ and the proof of Theorem 1.1 is completed.

Proof of the Theorem 1.2. Setting
$g\left(z\right)=f\left(z\right)+b\left(z\right)-a\left(z\right).$ Since
$\Delta_{c}^{m}a\left(z\right)\equiv 0$ $\left(1\leq m\leq n\right),$ then we
can remark that

$g\left(z\right)-b\left(z\right)=f\left(z\right)-a\left(z\right),$
$\Delta_{c}^{n}g\left(z\right)-b\left(z\right)=\Delta_{c}^{n}f\left(z\right)-b\left(z\right)$

and

$\Delta_{c}^{n+1}g\left(z\right)-b\left(z\right)=\Delta_{c}^{n}f\left(z\right)-b\left(z\right),\text{
}n\geq 2.$

Since $f\left(z\right)-a\left(z\right),$
$\Delta_{c}^{n}f\left(z\right)-b\left(z\right)$ and
$\Delta_{c}^{n+1}f\left(z\right)-b\left(z\right)$ share $0$ CM, then
$g\left(z\right),$ $\Delta_{c}^{n}g\left(z\right)$ and
$\Delta_{c}^{n+1}g\left(z\right)$ share $b\left(z\right)$ CM. By using Theorem
1.1, we deduce that $\Delta_{c}g\left(z\right)\equiv g\left(z\right),$ which
leads to $\Delta_{c}f\left(z\right)\equiv
f\left(z\right)+b\left(z\right)+\Delta_{c}a\left(z\right)-a\left(z\right)$ and
the proof of Theorem 1.2 is completed.

Proof of the Theorem 1.3. Note that $f\left(z\right)$ is a nonconstant entire
function of finite order. Since $f\left(z\right),$
$\Delta_{c}^{n}f\left(z\right)$ and $\Delta_{c}^{n+1}f\left(z\right)$ share
$0$ CM, then it is known by Theorem F that
$\Delta_{c}^{n+1}f\left(z\right)=C\Delta_{c}^{n}f\left(z\right),$ where $C$ is
a nonzero constant. Then we have

$\frac{\Delta_{c}^{n}f\left(z\right)}{f\left(z\right)}=e^{P\left(z\right)}$
(3.22)

and

$\frac{\Delta_{c}^{n+1}f\left(z\right)}{f\left(z\right)}=Ce^{P\left(z\right)},$
(3.23)

where $P$ is a polynomial. By $\left(3.22\right)$ and $\left(3.23\right)$ we
obtain

$f_{ic}\left(z\right)=\left(C+1\right)^{i}e^{P-P_{ic}}f\left(z\right).$ (3.24)

Then

$\Delta_{c}^{n}f\left(z\right)=\left(\overset{n}{\underset{i=0}{\sum}}C_{n}^{i}\left(-1\right)^{n-i}\left(C+1\right)^{i}e^{P-P_{ic}}\right)f\left(z\right)=e^{P\left(z\right)}f\left(z\right).$
(3.25)

The equality $\left(3.25\right)$ leads to $\deg P=0.$ Hence
$P\left(z\right)-P_{ic}\left(z\right)\equiv 0$ and $\left(3.25\right)$ will be

$\overset{n}{\underset{i=0}{\sum}}C_{n}^{i}\left(-1\right)^{n-i}\left(C+1\right)^{i}=C^{n}=e^{P\left(z\right)}.$
(3.26)

By $\left(3.22\right),\left(3.23\right)$ and $\left(3.26\right)$ we deduce

$\Delta_{c}^{n}f\left(z\right)=C^{n}f\left(z\right)$

and

$\Delta_{c}^{n+1}f\left(z\right)=C^{n+1}f\left(z\right).$

Then

$\Delta_{c}^{n+1}f\left(z\right)=\Delta_{c}\left(\Delta_{c}^{n}f\left(z\right)\right)=\Delta_{c}\left(C^{n}f\left(z\right)\right)=C^{n}\Delta_{c}f\left(z\right)=C^{n+1}f\left(z\right),$

which implies $\Delta_{c}f\left(z\right)=Cf\left(z\right).$ Thus, the proof of
Theorem 1.3 is completed.

Proof of Corollary 1.2. By Theorem 1.3 we have
$\Delta_{c}f\left(z\right)=Cf\left(z\right),$ where $C$ is a nonzero constant.
Then

$\Delta_{c}^{m}f\left(z\right)=C\Delta_{c}^{m-1}f\left(z\right)=C^{m}f\left(z\right),\text{
}m\geq 1.$ (3.27)

On the other hand, for $z_{0}\in\mathbb{C}$ we have

$\Delta_{c}^{m}f\left(z_{0}\right)=f\left(z_{0}\right).$ (3.28)

By $\left(3.27\right)$ and $\left(3.28\right)$ we deduce that $C^{m}=1.$ Hence
$\Delta_{c}^{m}f\left(z\right)=f\left(z\right)$.

Proof of the Theorem 1.4. Setting
$g\left(z\right)=f\left(z\right)-a\left(z\right),$ we can remark that

$g\left(z\right)=f\left(z\right)-a\left(z\right),$
$\Delta_{c}^{n}g\left(z\right)=\Delta_{c}^{n}f\left(z\right)-b\left(z\right)$

and

$\Delta_{c}^{n+1}g\left(z\right)=\Delta_{c}^{n}f\left(z\right)-b\left(z\right),\text{
}n\geq 2.$

Since $f\left(z\right)-a\left(z\right),$
$\Delta_{c}^{n}f\left(z\right)-b\left(z\right)$ and
$\Delta_{c}^{n+1}f\left(z\right)-b\left(z\right)$ share $0$ CM, then
$g\left(z\right),$ $\Delta_{c}^{n}g\left(z\right)$ and
$\Delta_{c}^{n+1}g\left(z\right)$ share $0$ CM. By using Theorem 1.3, we
deduce that $\Delta_{c}g\left(z\right)\equiv Cg\left(z\right),$ where $C$ is a
nonzero constant, which leads to $\Delta_{c}f\left(z\right)\equiv
Cf\left(z\right)+\Delta_{c}a\left(z\right)-a\left(z\right)$ and the proof of
Theorem 1.4 is completed.

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