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# Automated Statement Extraction from Press Briefings

Jüri Keller111Technische Hochschule Köln, Claudiusstraße 1, 50678 Köln,
Germany<EMAIL_ADDRESS>Meik Bittkowski222Science Media Center
Germany, Schloss-Wolfsbrunnenweg 33, 69118 Heidelberg, Germany
<EMAIL_ADDRESS>Philipp Schaer333Technische Hochschule Köln,
Claudiusstraße 1, 50678 Köln, Germany<EMAIL_ADDRESS>
###### Abstract

Scientific press briefings are a valuable information source. They consist of
alternating expert speeches, questions from the audience and their answers.
Therefore, they can contribute to scientific and fact-based media coverage.
Even though press briefings are highly informative, extracting statements
relevant to individual journalistic tasks is challenging and time-consuming.
To support this task, an automated statement extraction system is proposed.
Claims are used as the main feature to identify statements in press briefing
transcripts. The statement extraction task is formulated as a four-step
procedure. First, the press briefings are split into sentences and passages,
then claim sentences are identified through sequence classification.
Subsequently, topics are detected, and the sentences are filtered to improve
the coherence and assess the length of the statements.

The results indicate that claim detection can be used to identify statements
in press briefings. While many statements can be extracted automatically with
this system, they are not always as coherent as needed to be understood
without context and may need further review by knowledgeable persons.

###### keywords:

Computational Journalism Claim Detection Data Mining Natural Language
Processing

##  1 Introduction

Scientific press briefings are a valuable instrument in scientific
communication. They consist of alternating expert speeches and answers to
questions from the audience, which let scientists and journalists immediately
and jointly address the information needs of journalists. The SMC press
briefings444https://www.sciencemediacenter.de/alle-angebote/alle-
angebote/?tx_solr%5Bfilter%5D%5B0%5D=type%3APress+Briefing used in this work,
typically begin with an introduction of the participating experts followed by
moderated questions from journalists and answers from invited experts.
Additionally, the key results are concluded in closing statements. Press
briefings can directly contribute to a more fact-based media coverage by
connecting journalists and scientists. Even though press briefings are highly
informative, filtering this information and extracting relevant statements
remains challenging and time-consuming due to the high entropy and domain-
specific language used.

In this context, the research question will be answered: To what extent can
claim detection be used to extract statements from scientific press briefings?

To approach this domain-specific problem and automatically identify relevant
statements from press briefings, a pipeline is proposed, relying on claims as
a central element of a statement. A transformer based language model
[chanGermanNextLanguage2020] is used to classify claim sentences, which are
subsequently filtered and clustered to improve the relevance and coherence of
the emerging statements.

For example, a sentence like ”We do not yet know what a long covid course of
Omicron infection looks like.” is considered a complete claim, while the
sentence ”If I only have mild symptoms, it doesn’t mean I won’t have problems
in the long run.” is only considered an incomplete claim as it can not be
understood without context.

The main contribution of this work is the application of state-of-the-art
methods in claim detection to develop a system that extracts statements from
transcripts of scientific press briefings. Additionally, a novel dataset of 53
transcribed German press briefings is created and partially annotated. The
dataset and implementation are made publicly available via GitHub555Press
Briefing Claim Dataset at GitHub:
https://github.com/jueri/press_briefing_claim_dataset666Statement Extractor at
GitHub: https://github.com/jueri/statement_extractor and Zenodo
[keller_juri_2022_6047551].

##  2 Related Work

Automated extraction of statements from press briefings occurs at the
intersection of journalism and computer science. While the premise and context
emerge from journalism, the technical methods originate from computer science.

Various definitions of a claim exist [daxenbergerWhatEssenceClaim2017,
liebeckWhatAirportMining2016]. Generically, a claim indicates the author’s
position related to a main concept (topic) of a sentence
[levyUnsupervisedCorpuswideClaim2017]. Therefore, claims are indicated as the
main feature of a statement. Generically, a claim is a short phrase people can
have different opinions on [liebeckWhatAirportMining2016,
lippiArgumentationMiningState2016]. Since claims are conceptually complex, it
is often difficult to determine what a claim is and what not
[lippiContextindependentClaimDetection2015]. To clarify, this work refers to a
claim as a sentence asserting something on the main concept.

Claim detection is an information extraction task often part of a data
processing pipeline solving more complex problems. Automatically detecting
claims in texts has a variety of applications in fields like decision-making,
argument mining, fact-checking, or document processing
[levyArgumentativeContentSearch2018, daxenbergerWhatEssenceClaim2017]. As
diverse as the applications for claim detection are, so are the domains it is
applied in, such as for example political, debate, legal, and the web
[lippiArgumentationMiningState2016].

Besides binary classification, different claim types or the part of the
sentence that is a claim can be identified
[konstantinovskiyAutomatedFactcheckingDeveloping2018,
lippiArgumentationMiningState2016]. Levy et al.
[levyContextDependentClaim2014] first formally defined the task of automated
claim detection for computational argumentation while working on automatic
argument mining [levyContextDependentClaim2014,
slonimAutonomousDebatingSystem2021]. The systems developed in the following
years can be categorized by their ability to detect claims without further
context or the scope of application, e.g., a corpus of documents or a single
one [lippiContextindependentClaimDetection2015,
daxenbergerWhatEssenceClaim2017]. Chakrabarty et al. and Daxenberger et al.
provide more extensive overviews of different claim detection models and
datasets [chakrabartyIMHOFinetuningImproves2019,
daxenbergerWhatEssenceClaim2017].

##  3 Statement Detection System

Figure 1 visualizes the pipeline approach to extract statements from press
briefings. The press briefing needs to be initially segmented into passages
(1). Then, at the system’s core, a classification model is used to identify
sentences containing claims (2). Further, the main concept of the press
briefing and the sentences are detected (3). This information is used in the
last step to filter out the sentences that are not or incomplete claims (4).

Figure 1: Schematically visualization of the pipeline approach.

1\. Segmentation: The press briefings are split by a sentence tokenizer, and
neighboring and coherent sentences are combined into statements of multiple
sentences. Therefore, the sentences are represented in vector space through
spaCy [SPACY] and neighboring sentences are clustered by similarity with the
TextSplit [alemiTextSegmentationBased2015] algorithm. Starting with a list of
vector representations of sentences, TextSplit calculates a score for each
segment. By modifying the segment boundaries, the score changes. The optimal
segmentation is determined by minimizing the aggregated segment scores
[alemiTextSegmentationBased2015, modyFindingBestPart2020].

2\. Claim Detection: A German language model [chanGermanNextLanguage2020]
based on the BERT [devlinBERTPretrainingDeep2019] architecture is fine-tuned
for the task of sentence classification. The model’s hyperparameters are tuned
to achieve the best results on available training data. The German BERT
version GBERT is used as the foundational model for the classifier
[chanGermanNextLanguage2020]. This model is fine-tuned as a claim sentence
classifier using a dataset of 3000 sentences created for this purpose from the
publicly available SMC press
briefings777https://www.sciencemediacenter.de/alle-angebote. The dataset
contains a total of 53 press briefings from a time span of over four years. It
consists of 24,897 sentences with an average length of 17.31 tokens.

3\. Main Concept Detection: Based on the intuitive assertion that a claim
sentence with a topic similar to the overall topic of the press briefing the
sentence originates from is most relevant, the topics of both the sentences
and the overall press briefings (i.e., the title and the introduction text)
are detected. Therefore, Wikipedia articles are used as representations for
the topic of a sentence or the whole press briefing. All Wikipedia articles
related to tokens in the sentences and the title and the introduction texts
are detected. The sentence is considered more relevant if a sentence can be
linked to the same Wikipedia articles as the title or introduction text. For
this task, the wikification APIs Dandelion888https://dandelion.eu/ and
TagMe999https://sobigdata.d4science.org/web/tagme/tagme-help are used
[ferraginaTAGMEOntheflyAnnotation2010].

4\. Sentence Filtering: Two techniques are applied to filter out claim
sentences with a low topical similarity between the main concept of the
sentence and the overall main concept to improve the coherence and relevance.
The similarity is measured by two approaches based on embeddings or Wikipedia
concepts. Both approaches create a similarity score that can be used for
filtering with a minimum threshold.

The first approach calculates similarity by creating a vector representation
from the title of the press briefing and the individual sentences and then
measuring the cosine similarity between both. The vector representations are
created by combining spaCy [SPACY] word embeddings. The second approach is
based on the Wikipedia concepts detected as described previously. The
confidence scores of shared Wikipedia concepts between the title or
introduction text and the individual sentence are summed to create a score of
topical relatedness.

##  4 Experimental Evaluation

Three experiments were conducted to evaluate the system’s ability to extract
statements from press briefings. Therefore, three new press briefings with 799
sentences from different categories were annotated as ground truth.

Claim Detection: For hyperparameter optimization, models with different
configurations were trained for six epochs. The optimal number of epochs was
determined in conjunction with the learning rate by comparing the evaluation
loss rates after each epoch. Learning rates, ranging from 0.00005 to 0.01,
were tested. The best results could be achieved with a learning rate of 1e-5
and three epochs. For the final system, a model with an F1 Score of 0.89, a
precision of 0.92 and a recall of 0.86 was used. This model assigned slightly
more false positive than false negative labels, but since the human in the
loop can decide on the misclassified sentences, this behavior is preferred.

To evaluate the claim detection component of the system, claim sentences
classified with minimal confidence scores of 0.7, 0.8 and 0.9 are analyzed
independently. Since the confidence score limits the number of sentences
considered as claims by the model, a score small enough to allow claim
sentences to be classified and high enough to exclude non-claim sentences
leads to the best results. The overall best results for detecting claims
independent of their completeness were achieved at a confidence of 0.7 with an
F1 score of 0.68, a precision of 0.89 and a recall of 0.55. By investigating
individual results for the claim types, complete claims (that can stand on
their own) and incomplete claims (that can not), it can be assessed if the
filtering and sentence clustering methods improve the coherence. In total, a
maximum of 167 of 224 incomplete claims and 102 of 152 complete claims claim
sentences could be identified with various system configurations. With the
claim detection model and minimum confidence of 0.8, the highest F1 score of
0.47 could be achieved. Raising the confidence threshold to 0.9 results in a
higher precision of 0.43. Decreasing the score to 0.7 leads to a higher recall
of 0.67.

Statement Filtering: Based on the results of the first experiments, the claim
sentences with a confidence score of 0.8 are further filtered as this
threshold achieved the best F1 score. To exclude incoherent statements, the
similarity between the sentence’s main concept and the press briefing’s main
concept is calculated. The main concepts of the press briefing based on
embeddings, title wikification and introduction wikification are evaluated
individually. For each method, a minimum similarity is chosen by investigating
the similarity distribution for bends. With the introduction wikification
based filtering, a precision of 0.43 was achieved. The title wikification
method reaches a precision of 0.456, and the embedding method exceeds this
score with the best precision of 0.463. All results are presented in table 1.

Table 1: Results for detected complete claims. The first three columns hold the results for different confidence thresholds of the claim detection model. The last three columns contain the results of a system using a claim detection model threshold of 0.8 and the different main concept methods. The best results are highlighted in bold. Confidence | 0.9 | 0.8 | 0.7 |  0.8 embedding |  0.8 w. title |  0.8 w. intro
---|---|---|---|---|---|---
F1 | 0.466 | 0.473 | 0.450 | 0.481 | 0.339 | 0.341
Precision | 0.426 | 0.378 | 0.339 | 0.463 | 0.456 | 0.430
Recall | 0.513 | 0.632 | 0.671 | 0.500 | 0.270 | 0.283

Sentence Clustering: The last experiment evaluates the sentence clustering
method. Claim sentences with a confidence of at least 0.9 are enlarged to
statements of multiple sentences. The statements are then assessed according
to their coherence. To measure if the larger statements created by the
similarity-based sentence clustering are more coherent, the complete claim and
incomplete claim ratio for all methods are compared. With 53 % complete
claims, the coherence of the statements extracted exceeds all other methods
and the baseline. 46 % of the statements extracted with the claim detection
module and a minimum confidence score of 0.9 are complete statements. The
embedding method of the main concept module extracts 50 % complete claims.

##  5 Discussion and Conclusion

The analysis shows that the press briefings are very argumentative content
full of claims. While most claims are incomplete and can only be understood
with further information, 46 % can stand alone. The sentences extracted by the
proposed system are highly likely to be claims, although not necessarily
complete claims. This finding is supported by the low false-positive rate of
the system considering all claims. Missing coherence is the main error of the
statements extracted.

Compared to the initial results of the claim detection model with an accuracy
of 0.89, the results achieved in the experiments are noticeably lower. This
may be caused by the dataset used for training, which only contains complete
claims and no claims and should be improved in future work. By adding a filter
based on main concepts, some incoherent claim sentences could be successfully
excluded, and the precision improved. A smaller confidence threshold of the
initial claim detection module could improve these results further by
providing more claim sentences in general. The embedding-based main concept
method provided the best results. The Wikification-based methods come to an
extent as some topics can only be expressed poorly by Wikipedia concepts, and
other Wikipedia concepts may not exist or be captured by the model.

Assessing statements of multiple sentences, some statements previously
classified as incomplete claims can be understood without additional
information through the added neighboring sentences. By systematically adding
sentences, more coherent statements could be created. However, this method
sacrifices the conciseness of statements since they get longer.

In summary, this work investigates how statements can be mined from scientific
press briefings. A dataset is created from publicly available press briefings
and used to fine-tune a claim detection model. Additional methods are tested
to improve the quality of the extracted statements. More context is added to
the statements, and the statements are filtered based on their topic. The
results are compared to a gold standard to evaluate the system’s performance.

The results show that the system can differentiate between claims of any type
and no claim sentences with high precision (0.89), but only half of the claims
can be recalled (0.55). Considering only complete claim sentences, the
system’s overall performance decreases (F1 0.47), but relatively more complete
claims can be identified. These results emphasize the system’s difficulty
differentiating between complete and incomplete claim sentences. The resulting
statements gain coherence by adding more context. Similarly, incomplete claim
sentences can be filtered out by low topical similarity, which improves the
results slightly. The evaluations show that the presence of a claim can be
used as an indicator to detect statements. However, the results leave room for
improvement in many aspects. Especially the coherence of the extracted
statements needs to be improved to extract statements that can stand on their
own.


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#  Local moduli of continuity for permanental processes that are zero at zero.

Michael B. Marcus Jay Rosen Research of Jay Rosen was partially supported by
grants from the Simons Foundation.

###### Abstract

Let $u(s,t)$ be a continuous potential density of a symmetric Lévy process or
diffusion with state space $T$ killed at $T_{0}$, the first hitting time of
$0$, or at $\lambda\wedge T_{0}$, where $\lambda$ is an independent
exponential time. Let

$f(t)=\int_{T}u(t,v)\,d\mu(v),$

where $\mu$ is a finite positive measure on $T$. Let
$X_{\alpha}=\\{X_{\alpha}(t),t\in T\\}$ be an $\alpha-$permanental process
with kernel

$v(s,t)=u(s,t)+f(t).$

Then when $\lim_{t\to 0}u(t,t)=0$,

$\limsup_{t\downarrow 0}\frac{X_{\alpha}(t)}{u(t,t)\log\log 1/t}\geq
1,\qquad\text{a.s.}$

and

$\limsup_{t\downarrow 0}\frac{X_{\alpha}(t)}{u(t,t)\log\log 1/t}\leq
1+C_{u,h},\qquad\text{a.s.}$

where $C_{u,\mu}\leq|\mu|$ is a constant that depends on both $u$ and $\mu$,
which is given explicitly, and is different in the different examples.

00footnotetext:  Key words and phrases: permanental processes with non-
symmetric kernels, law of the iterated logarithm at zero 00footnotetext:  AMS
2020 subject classification: 60E07, 60G15, 60G17, 60G99, 60J25

##  1 Introduction

Let $K=\\{K(s,t),s,t\in T\\}$ be a kernel with the property that for all
${\mathbf{t}_{n}}=(t_{1},\ldots,t_{n})$ in $T^{n}$, the matrix $\mathbf{{\cal
K}(t_{n})}=\\{K(t_{i},t_{j}),i,j\in[1,n]\\}$ determines an $n-$dimensional
random variable random variable $(X_{\alpha}(t_{1}),\ldots,X_{\alpha}(t_{n}))$
with Laplace transform,

$E\left(e^{-\sum_{i=1}^{n}s_{i}X_{\alpha}(t_{i})}\right)=\frac{1}{|I+\mathbf{{\cal
K}(t_{n})}S(\mathbf{s_{n}})|^{\alpha}},$ (1.1)

where $S(\mathbf{s_{n}})$ is a diagonal matrix with positive entries
$\mathbf{s_{n}}=(s_{1},\ldots,s_{n})$, and $\alpha>0$. We refer to
$\mathbf{{\cal K}(t_{n})}$ as the kernel of
$(X_{\alpha}(t_{1}),\ldots,X_{\alpha}(t_{n}))$. It follows from the Kolmogorov
Extension Theorem that $\\{K(s,t),s,t\in T\\}$ determines a stochastic process
which we denote by $X_{\alpha}=\\{X_{\alpha}(t),t\in T\\}$ and refer to as an
$\alpha-$permanental process with kernel $K$.

There are well known examples of $\alpha-$permanental processes when
$\alpha=k/2$. Let $\eta=\\{\eta(t);t\in T\\}$ be a mean zero Gaussian process
with covariance $U=\\{U(s,t),s,t\in T\\}.$ Let $\\{\eta_{i};i=1,\ldots,k\\}$
be independent copies of $\eta$. Define,

$Y_{k/2}(t)=\sum_{i=1}^{k}\frac{\eta^{2}_{i}(t)}{2},\qquad t\in T.$ (1.2)

The positive stochastic process $\\{Y_{k/2}(t),t\in T\\}$ is referred to as a
chi–square process of order $k$ with covariance $U$. It is well known that for
$\\{t_{1},\ldots,t_{n}\\}\subset T$

$E\left(e^{-\sum_{j=1}^{n}s_{j}Y_{k/2}(t_{j})}\right)=\frac{1}{|I+\overline{U}(\mathbf{t_{n}})S(\mathbf{s_{n}})|^{k/2}},$
(1.3)

where $\overline{U}(\mathbf{t_{n}})$ is the $n\times n$ matrix,
$\\{U(t_{j},t_{k});1\leq j,k\leq n\\}$. Therefore $Y_{k/2}=\\{Y_{k/2}(t);t\in
T\\}$ is a $k/2$–permanental process with kernel $U$.

It is important to note that whereas (1.3) holds for all integers $k$ whenever
$U$ is the covariance of a Gaussian process, it may or may not hold when $k/2$
is replaced by some $\alpha>0$ that is not a multiple of 1/2. (The question of
when (1.3) holds for all $\alpha>0$ was raised by Paul Lévy when he asked when
does a Gaussian process have infinitely divisible squares. This was answered
by Griffiths and Bapat. An extensive treatment of these results is given in
[3, Chapter 13].) Nevertheless, it follows from Eisenbaum and Kaspi, [1,
Theorem 3.1] that if $U$ is the potential density of a transient Markov
process on $T$ which is finite for all $s,t\in T$, then there exist
$\alpha-$permanental processes $\\{Y_{\alpha}(t),t\in T\\}$ with kernel $U$
for all $\alpha>0$. I.e., (1.3) holds with $k/2$ replaced by $\alpha$ for all
$\alpha>0.$

It is at this point that this becomes interesting because in the
Eisenbaum–Kaspi theorem the kernel $U$ is not necessarily symmetric, or
equivalent to a symmetric matrix; (see [5]). When it is not, the corresponding
permanental processes are not squares of a Gaussian process. They are a new
kind of positive stochastic process defined by kernels that need not be
symmetric. It is intriguing to study sample path properties of these
processes.

In [6] we consider symmetric Lévy processes and diffusions $Y$, that are
killed at the end of an independent exponential time or the first time they
hit 0, with potential densities $u(x,y)$ with respect to some $\sigma$-finite
measure on $T\subseteq R^{1}$. Let $f$ be a finite excessive function for $Y$.
Set

$u_{f}(x,y)=u(x,y)+f(y),\qquad x,y\in T.$ (1.4)

Under general smoothness conditions on $f$, $u(x,y)$ and points $d\in T$, laws
of the iterated logarithm are found for $X_{k/2}=\\{X_{k/2}(t),t\in T\\}$, a
$k/2-$permanental process with kernel $\\{u_{f}(x,y),x,y\in T\\}$, of the
following form: For all integers $k\geq 1$,

$\limsup_{x\to
0}\frac{|X_{k/2}(d+x)-X_{k/2}(d)|}{\left(2\sigma^{2}\left(x\right)\log\log
1/x\right)^{1/2}}=\left(2X_{k/2}(d)\right)^{1/2},\qquad a.s.,$ (1.5)

where,

$\sigma^{2}(x)=u(d+x,d+x)+u(x,x)-2u(d+x,x).$ (1.6)

In this paper we consider the question of finding the local modulus of
continuity for $X_{\alpha}$ at 0 when $X_{\alpha}(0)=0$. This, it seems, is
difficult, and, in general, we can only get different upper and lower bounds.
On the other hand our results hold for $X_{\alpha}$ for all $\alpha>0$.

We consider the following situation: Let $T$ a be locally compact set with a
countable base. Let $Y=(\Omega,{\cal F}_{t},Y_{t},\theta_{t},P^{x})$ be a
transient Borel right process with state space $T$, and continuous strictly
positive potential densities $u(x,y)$ with respect to some $\sigma$–finite
measure $m$ on $T$. Let $\mu$ be a finite positive measure on $T$ with mass
$|\mu|=\mu(T)$. We call the function

$f(x)=\int_{T}u(y,x)\,d\mu(y)$ (1.7)

a left potential for $Y$.

The following is [4, Theorem 6.1].

###### Theorem 1.1

Let $Y$ be a transient Borel right process with state space $T$, as described
above. Then for any left potential $f$ for $Y$, there exists a transient Borel
right process $\widetilde{Y}\\!=\\!(\Omega,{\cal
F}_{t},\widetilde{Y}_{t},\theta_{t},P^{x})$ with state space
$S=T\cup\\{\ast\\}$, where $\ast$ is an isolated point, such that
$\widetilde{Y}$ has potential densities,

$\displaystyle\widetilde{u}(x,y)=u(x,y)+f(y),\hskip 14.45377ptx,y\in T$ (1.8)
$\displaystyle\widetilde{u}(\ast,y)=f(y),\hskip 14.45377pt\mbox{ and }\hskip
14.45377pt\widetilde{u}(x,\ast)=\widetilde{u}(\ast,\ast)=1,$

with respect to the measure $\widetilde{m}$ on $S$ which is equal to $m$ on
$T$ and assigns a unit mass to $\ast$.

Let $\\{Z_{t},t\in S\\}$ be the $\alpha$ permanental process with kernel
$\widetilde{u}(x,y)$. We have the following general lower bound:

###### Theorem 1.2

Let $Y$ be a strongly symmetric transient Borel right process with state space
$(0,1)\subseteq T\subseteq R^{1}$ and continuous strictly positive potential
densities $u(x,y)$ with respect to some $\sigma$–finite measure $m$ on $T$, as
in Theorem 1.1. Assume that $\lim_{t\downarrow 0}u(t,t)=0.$ Then

$\limsup_{t\to 0}\frac{Z_{t}}{u(t,t)\log\log 1/t}\geq 1,\qquad a.s.$ (1.9)

###### Remark 1.1

In Examples 1.1–1.4, $T=R^{1}-\\{0\\}$ or $(0,\infty)$ and

$\widetilde{u}(x,y)=u(x,y)+f(y),\hskip 14.45377ptx,y\in(0,1]$ (1.10)

has a continuous extension to $x,y\in[0,1]$. We continue to denote this
extension by $\widetilde{u}(x,y)$. It follows from the proof of [4, Lemma 6.2]
that this extension is the kernel of an $\alpha-$permanental process.

We use [4, Theorem 1.2] to get upper bounds. We repeat it here for the
convenience of the reader.

###### Theorem 1.3

Let $X_{\alpha}=\\{X_{\alpha}(t),t\in[0,1]\\}$ be an $\alpha-$permanental
process with kernel $\\{\widetilde{u}(s,t),s,t\in[0,1]\\}$ with
$\widetilde{u}(0,0)=0$. Define the function,

$\widetilde{\sigma}(s,t)=\left(\widetilde{u}(s,s)+\widetilde{u}(t,t)-2(\widetilde{u}(s,t)\widetilde{u}(t,s))^{1/2}\right)^{1/2}.$
(1.11)

Let $\varphi$ be a bounded increasing function which is regularly varying at
zero with positive index such that for all $s,t\in[0,1]$,

$\widetilde{\sigma}(s,t)\leq\varphi(|s-t|),$ (1.12)

If, in addition, $\varphi^{2}(h)=O(\widetilde{u}(h,h))$ at $0$, then

$\limsup_{t\downarrow 0}\frac{X_{\alpha}(t)}{\widetilde{u}(t,t)\log\log
1/t}\leq 1,\qquad\text{a.s.}$ (1.13)

Note that when

$\widetilde{u}(s,t)=u(s,t)+f(t),$ (1.14)

the denominator in (1.13) is larger than the denominator in (1.9).

Let us now consider Theorem 1.3 with $\widetilde{u}(s,t)$ as in (1.10) and

$f(t):=f_{u,\mu}(t)=\int_{T}u(t,v)\,d\mu(v),$ (1.15)

for some positive finite measure $\mu$. By [3, Lemma 3.4.3], $u(t,v)\leq
u(t,t)\wedge u(v,v)$ so that

$f(t)\leq u(t,t)|\mu|$ (1.16)

In this case we can write (1.13) as,

$\limsup_{t\downarrow 0}\frac{X_{\alpha}(t)}{u(t,t)\log\log 1/t}\leq
1+|\mu|,\qquad\text{a.s.}$ (1.17)

However, in the examples we consider, (1.17) is generally too large. In the
examples we consider we show that,

$f(t)\sim C_{u,\mu}u(t,t),\qquad\text{as $t\downarrow 0$},$ (1.18)

where $0<C_{u,\mu}\leq|\mu|$ is a constant that depends on both $u$ and $\mu$.
In this case (1.13) can be written as,

$\limsup_{t\downarrow 0}\frac{X_{\alpha}(t)}{u(t,t)\log\log 1/t}\leq
1+C_{u,h},\qquad\text{a.s.}$ (1.19)

We consider four examples in which $u(s,t)$ are continuous potential densities
of strongly symmetric Borel right processes killed the first time they hit 0.
In the first two examples the strongly symmetric Borel right processes are
Lévy processes, or exponentially killed Lévy processes, killed when they hit
zero.

###### Example 1.1

Let $Z=\\{Z_{t};t\in R^{1}\\}$ be a real valued symmetric Lévy process with
characteristic exponent $\psi$, i.e.,

$E\left(e^{i\lambda Z_{t}}\right)=e^{-t\psi(\lambda)},\qquad\forall\,t\geq 0,$
(1.20)

where $\psi(\lambda)$ is an even function and,

$\frac{1}{1+\psi(\lambda)}\in L^{1}(R^{1}).$ (1.21)

In this example we consider processes $Z$ for which $0$ is recurrent.

Let $Z^{\prime}=\\{Z^{\prime}(t);t\in R^{1}\\}$ be a strongly symmetric Borel
right process with state space $T=R^{1}-\\{0\\}$ obtained by killing $Z$ the
first time it hits $0$. It follows from [3, Theorem 4.2.4] that the process
$Z^{\prime}$ has continuous potential densities with respect to Lebesgue
measure restricted to $T$, which we denote by $\Phi=\\{\Phi(x,y),x,y\in T\\}$,
where

$\Phi(x,y)=\frac{(\sigma^{0})^{2}(x)}{2}+\frac{(\sigma^{0})^{2}(y)}{2}-\frac{(\sigma^{0})^{2}(x-y)}{2},\qquad
x,y\in R^{1}-\\{0\\},$ (1.22)

and

$(\sigma^{0})^{2}(x)=\frac{2}{\pi}\int_{0}^{\infty}\frac{1-\cos\lambda
x}{\psi(\lambda)}\,d\lambda.$ (1.23)

Note that,

$\Phi(x,x)=(\sigma^{0})^{2}(x).$ (1.24)

As we point out in Remark 1.1 there exists an $\alpha-$permanental process
$\\{X_{\alpha}(x),x\in[0,1]\\}$ with kernel,

$v(x,y)=\Phi(x,y)+f_{\Phi,\mu}(y),$ (1.25)

where

$f_{\Phi,\mu}(y)=\int_{T}\Phi(x,y)\,d\mu(x)$ (1.26)

and $|\mu|<\infty$.

The next theorem is an application of Theorem 1.3 with $\widetilde{u}=v$.

###### Theorem 1.4

Let $\\{X_{\alpha}(x),x\in[0,1]\\}$ be an $\alpha-$permanental process with
kernel $v(x,y)$. When $(\sigma^{0})(t)\in C^{1}(0,\infty)$ is a regularly
varying function at zero with positive index, $X_{\alpha}(t)$ is continuous on
$[0,\delta]$. Furthermore, when

$\lim_{t\downarrow 0}\frac{t}{(\sigma^{0})^{2}(t)}=0,$ (1.27)
$\limsup_{t\downarrow 0}\frac{X_{\alpha}(t)}{(\sigma^{0})^{2}(t)\log\log
1/t}\leq 1+\frac{|\mu|}{2},\qquad a.s.,$ (1.28)

and when

$\lim_{t\downarrow 0}\frac{t}{(\sigma^{0})^{2}(t)}=\frac{1}{C},\qquad C>0,$
(1.29) $\limsup_{t\downarrow 0}\frac{X_{\alpha}(t)}{Ct\log\log
1/t}\leq\left(1+\frac{|\mu|}{2}\right)+\frac{1}{2C}\int_{-\infty}^{\infty}((\sigma^{0})^{2}(v))^{\prime}\,d\mu(v),\quad
a.s.$ (1.30)

We give conditions when $(\sigma^{0})(t)\in C^{1}(0,\infty)$ in Remark 7.2.

###### Remark 1.2

We show in [6, Lemma 4.1] that either,

$\lim_{t\downarrow
0}\frac{t}{(\sigma^{0})^{2}(t)}=0,\qquad\text{or}\qquad\lim_{t\downarrow
0}\frac{t}{(\sigma^{0})^{2}(t)}=C,$ (1.31)

for some constant $C>0.$

Note also that when $\mu$ is a symmetric measure, since
$((\sigma^{0})^{2}(v))^{\prime}$ is an odd function, the integral in (1.30) is
equal to 0. If $\mu$ is concentrated on thee negative half–line it is
negative.

###### Example 1.2

Let $Z_{\beta}=\\{Z_{\beta}\left(t\right);t\in R^{1}\\}$ be a symmetric Markov
process obtained by killing $Z$ at the end of an independent exponential time
with mean $1/\beta$, $\beta>0$. The $\beta-$potential density of $Z$ with
respect to Lebesgue measure, which is the same as the $0-$potential density of
$Z_{\beta}$ with respect to Lebesgue measure, is

$u^{\beta}(x,y)=u^{\beta}(x-y)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{\cos\lambda(x-y)}{\beta+\psi(\lambda)}\,d\lambda,\qquad
x,y\in R^{1},$ (1.32)

where $\psi$ is an even function; see e.g., [3, (4.84)]. Let

$\displaystyle(\sigma^{\beta})^{2}(x-y))$ $\displaystyle=$ $\displaystyle
u^{\beta}(x,x)+u^{\beta}(y,y)-2u^{\beta}(x,y)$ (1.33) $\displaystyle=$
$\displaystyle\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{1-\cos\lambda(x-y)}{\beta+\psi(\lambda)}\,d\lambda,\qquad
x,y\in R^{1}.$ (1.34)

Let

$f_{u^{\beta},\mu}(y)=\int_{-\infty}^{\infty}u^{\beta}(x,y)\,d\mu(x),$ (1.35)

where $\mu$ is a finite positive measure with $\mu(\\{0\\})=0$.

Let $Z^{\prime}_{\beta}=\\{Z^{\prime}_{\beta}(t);t\in R^{1}\\}$ be a strongly
symmetric Borel right process with state space $T=R^{1}-\\{0\\}$ obtained by
killing $Z_{\beta}$ the first time it hits $0$. The process
$Z_{\beta}^{\prime}$ has continuous potential densities with respect to
Lebesgue measure restricted to $T$, which we denote by
$v^{\beta}=\\{v^{\beta}(x,y),x,y\in T\\}$, where

$\displaystyle v^{\beta}(x,y)$ $\displaystyle=$ $\displaystyle
u^{\beta}(x,y)-\frac{u^{\beta}(x,0)u^{\beta}(0,y)}{u^{\beta}(0,0)}$
$\displaystyle=$ $\displaystyle
u^{\beta}(x-y)-\frac{u^{\beta}(x)u^{\beta}(y)}{u^{\beta}(0)};$

see [3, (4.165)].

It follows from Remark 1.1 that there exists an $\alpha-$permanental process
$\\{X_{\alpha,\beta}(x),x\in[0,1]\\}$ with kernel

$v(x,y)=v^{\beta}(x,y)+f_{v^{\beta},\,\mu}(y),$ (1.37)

where,

$f_{v^{\beta},\mu}(y)=\int_{T}v^{\beta}(x,y)\,d\mu(x)$ (1.38)

for $\mu$ as given above. Since $\mu(\\{0\\})=0$ we can write (1.38) as

$f_{v^{\beta},\mu}(y)=\int_{-\infty}^{\infty}v^{\beta}(x,y)\,d\mu(x)$ (1.39)

The next theorem is an application of Theorem 1.3 with $\widetilde{u}=v$. The
reader should note that in (1.41) and (1.43) we use $f_{u^{\beta},\mu}$.

###### Theorem 1.5

Let $\\{X_{\alpha}(x),x\in[0,1]\\}$ be an $\alpha-$permanental process with
kernel $v(x,y)$. When $(\sigma^{\beta})^{2}(t)\in C^{1}(T)$ and
$|(\sigma^{\beta})^{2}(t)^{\prime}|$ is bounded on
$(-\infty,-\Delta)\cup(\Delta,\infty)$ for all $\Delta>0$ and
$(\sigma^{\beta})^{2}(t)$ is regularly varying at zero with positive index,
$X_{\alpha,\beta}(t)$ is continuous on $[0,\delta]$, for some $\delta>0$.
Furthermore, when

$\lim_{t\downarrow 0}\frac{t}{(\sigma^{\beta})^{2}(t)}=0,$ (1.40)
$\limsup_{t\downarrow 0}\frac{X_{\alpha}(t)}{(\sigma^{\beta})^{2}(t)\log\log
1/t}\leq 1+\frac{f_{u^{\beta},\mu}(0)}{2u^{\beta}(0)},\qquad a.s.,$ (1.41)

and when

$\lim_{t\downarrow 0}\frac{t}{(\sigma^{\beta})^{2}(t)}=\frac{1}{C},\qquad
C>0,$ (1.42) $\limsup_{t\downarrow 0}\frac{X_{\alpha}(t)}{Ct\log\log 1/t}\leq
1+\frac{f_{u^{\beta},\mu}(0)}{2u^{\beta}(0)}+\frac{1}{2C}\int_{-\infty}^{\infty}((\sigma^{\beta})^{2}(v))^{\prime}\,d\mu(v),\quad
a.s.$ (1.43)

In Remark 7.2 we give examples of Lévy processes that satisfy the hypotheses
of this theorem. Also we show in [6, Lemma 8.1] that for all $\beta\geq 0$,

$\lim_{t\downarrow
0}\frac{(\sigma^{\beta})^{2}(t)}{(\sigma^{0})^{2}(t)}=1,\qquad{a.s}$ (1.44)

so we can replace $(\sigma^{\beta})^{2}$ by $(\sigma^{0})^{2}$ in (1.41).

###### Example 1.3

Let $\mathcal{Z}$ be a diffusion in $R^{1}$ that is regular and without traps
and is symmetric with respect to a $\sigma$–finite measure $m$, called the
speed measure, which is absolutely continuous with respect to Lebesgue
measure. (A diffusion is regular and without traps when
$P^{x}\left(T_{y}<\infty\right)>0,\forall x,y\in R^{1}$.) We consider
diffusions $\mathcal{Z}$ with generators of the form

$\frac{1}{2}a^{2}(x)\frac{d^{2}}{dx^{2}}+c(x)\frac{d}{dx},$ (1.45)

where $a^{2}(x)$ and $c(x)\in C\left(R^{1}\right)$ and $a(x)\neq 0$ for any
$x\in R^{1}$. For details see [8, Section 7.3] and [2, Chapter 16].

Let $s\in C^{2}(R^{1})$ be a scale function for $\cal Z$. The function $s$ is
strictly increasing and is unique up to linear transformation so that we take
$s(0)=0$. Assume that $\lim_{x\to\infty}s(x)=\infty$.

Let $\overline{Z}$ be the process with state space $T=(0,\infty)$ that is
obtained by starting $\cal Z$ in $T$ and then killing it the first time it
hits 0. The potential density of $\overline{Z}$ with respect to the speed
measure $m$ is,

$u_{T_{0}}(x,y)=s(x)\wedge s(y),\qquad x,y>0,$ (1.46)

where $s\in C^{2}(0,\infty)$ is strictly positive and strictly increasing.

By Remark 1.1 there exists an $\alpha-$ permanental process
$\\{X_{\alpha}(x),x\in[0,1]\\}$ with kernel

$v(x,y)=u_{T_{0}}(x,y)+f_{u_{T_{0}},\mu}(y),$ (1.47)

where

$f_{u_{T_{0}},\mu}(y)=\int_{T}u_{T_{0}}(x,y)\,d\mu(x)$ (1.48)

and $|\mu|<\infty$.

The next theorem is an application of Theorem 1.3 where $\widetilde{u}=v$.

###### Theorem 1.6

Let $\\{X_{\alpha}(t),t\in[0,1]\\}$ be an $\alpha-$permanental process with
kernel $v(x,y)$. Then $X_{\alpha}(t)$ is continuous on $[0,\delta]$ and,

$\limsup_{t\downarrow 0}\frac{X_{\alpha}(t)}{s(t)\log\log 1/t}\leq
1+|\mu|,\qquad a.s.$ (1.49)

(Note that we can replace $s(t)$ by $s^{\prime}(0)t$ in (1.49).)

###### Example 1.4

Let $Z$ be $\cal Z$ killed at the end of an independent exponential time with
mean $1/\beta$. The potential density of $Z$ with respect to the speed measure
$m$ is

$\widetilde{u}^{\beta}(x,y)=\left\\{\begin{array}[]{cc}p_{\beta}(x)q_{\beta}(y),&\quad
x\leq y\\\ q_{\beta}(x)p_{\beta}(y),&\quad y\leq x\end{array}\right.,$ (1.50)

where $p_{\beta},q_{\beta}\in C^{2}(R^{1})$ are positive and $p_{\beta}$ is
strictly increasing and $q_{\beta}$ is strictly decreasing. (For the
dependence of $p_{\beta}$ and $q_{\beta}$ on $\beta$ see [3, (4.14)].)

Let $Z^{\prime}$ be the process with state space $T=(0,\infty)$ that is
obtained by starting $Z$ in $T$ and then killing it the first time it hits 0.
The potential density of $Z^{\prime}$ is

$\widetilde{v}^{\beta}(x,y)=\widetilde{u}^{\beta}(x,y)-\frac{\widetilde{u}^{\beta}(x,0)\widetilde{u}^{\beta}(0,y)}{\widetilde{u}^{\beta}(0,0)},\qquad
x,y>0,$ (1.51)

i.e.,

$\widetilde{v}^{\beta}(x,y)=\left\\{\begin{array}[]{cc}p_{\beta}(x)q_{\beta}(y)-\displaystyle\frac{p_{\beta}(0)}{q_{\beta}(0)}q_{\beta}(x)q_{\beta}(y),&\quad
0<x\leq y\\\ \\\
q_{\beta}(x)p_{\beta}(y)-\displaystyle\frac{p_{\beta}(0)}{q_{\beta}(0)}q_{\beta}(x)q_{\beta}(y),&\quad
0<y\leq x.\end{array}\right.$ (1.52)

As we point out in Remark 1.1 there exists an $\alpha-$permanental process
$\\{X_{\alpha}(x),x\in[0,1]\\}$ with kernel

$v(x,y)=\widetilde{v}^{\beta}(x,y)+f_{\widetilde{v}^{\beta},\mu}(y),$ (1.53)

where

$f_{\widetilde{v}^{\beta},\mu}(y)=\int_{T}\widetilde{v}^{\beta}(x,y)\,d\mu(x),$
(1.54)

and $|\mu|<\infty$.

The next theorem is an application of Theorem 1.3 where $\widetilde{u}=v$.

###### Theorem 1.7

Let $\\{X_{\alpha}(t),t\in[0,1]\\}$ be an $\alpha-$permanental process with
kernel $v(x,y)$. Then $X_{\alpha}(t)$ is continuous on $[0,\delta]$ and,

$\limsup_{t\downarrow 0}\frac{X_{\alpha}(t)}{\rho_{\beta}(0)t\log\log
1/t}\leq\left(1+\int_{0}^{\infty}\frac{q_{\beta}(y)}{q_{\beta}(0)}\,d\mu(y)\right),\qquad
a.s.$ (1.55)

where,

$\rho_{\beta}(0)={p_{\beta}^{\prime}(0)}{q_{\beta}(0)}-{q_{\beta}^{\prime}(0)}{p_{\beta}(0)}.$
(1.56)

##  2 Proof of Theorems 1.2

We make several observations that are used in the proof of Theorem 1.2. Let
$U$ be a non-singular $n\times n$ matrix. Let $U^{-1}$ denote the inverse of
$U$ and $U^{j,k}$ denote the elements of $U^{-1}$. Let $U_{f}$ be the
$(n+1)\times(n+1)$ matrix

$\displaystyle U_{f}$ $\displaystyle=$ $\displaystyle\left(\begin{array}[]{
cccc }1&f(1)&\ldots&f(n)\\\ 1&U_{1,1}+f(1)&\ldots&U_{1,n}+f(n)\\\
\vdots&\vdots&\ddots&\vdots\\\
1&U_{n,1}+f(1)&\ldots&U_{n,n+}+f(n)\end{array}\right).$ (2.5)

where $f$ is any function. One can check that

$U_{f}^{-1}=\left(\begin{array}[]{ cccc
}1+\rho&-\sum_{i=1}^{n}f(i)U^{i,1}&\dots&-\sum_{i=1}^{n}f(i)U^{i,n}\\\
-\sum_{j=1}^{n}U^{1,j}&U^{1,1}&\dots&U^{1,n}\\\ \vdots&\vdots&\ddots&\vdots\\\
-\sum_{j=1}^{n}U^{n,j}&U^{n,1}&\dots&U^{n,n}\end{array}\right),$ (2.6)

where

$\rho=\sum_{j=1}^{n}\sum_{i=1}^{n}f(i)U^{i,j}.$ (2.7)

We note the following remarkable property of $U_{f}$ that we use,

$(U_{f}^{-1})^{i,i}=U^{i-1,i-1},\qquad i=2,\ldots,n+1.$ (2.8)

Let $\xi_{\alpha}$ be a gamma random variable with probability density
function,

$\frac{x^{\alpha-1}e^{-x}}{\Gamma(\alpha)}.$ (2.9)

The following Theorem is [7, Theorem 1.1]:

###### Theorem 2.1

Let $X=(X_{0},X_{1},\ldots,X_{n})$ be an $\alpha-$permanental vector with non-
singular kernel $K$. Let $A=K^{-1}$ and denote the diagonal entries of $A$ by
$(a_{0},a_{1},\ldots,a_{n})$. Then there exists a coupling between $X$ and an
$n-$tuple $(\xi^{(0)}_{\alpha},\xi^{(1)}_{\alpha},\ldots,\xi_{\alpha}^{(n)})$
of independent identically distributed. copies of $\xi_{\alpha}$ such that,

$X\geq\left(a_{0}^{-1}\xi^{(0)}_{\alpha},a_{1}^{-1}\xi^{(1)}_{\alpha},\ldots,a_{n}^{-1}\xi_{\alpha}^{(n)}\right),\qquad
a.s.$ (2.10)

Proof of Theorem 1.2 Let $\\{t_{j}=\theta^{j}\\}_{j=1}^{n}$ and let
$X=(X(*),X(t_{1}),\ldots,X(t_{n})$ be an $\alpha-$permanental vector with non-
singular kernel $K$

$\displaystyle K$ $\displaystyle=$ $\displaystyle\left(\begin{array}[]{ cccc
}1&f(t_{1})&\ldots&f(t_{n})\\\
1&u(t_{1},t_{1})+f(t_{1})&\ldots&u(t_{1},t_{n})+f(t_{n})\\\
\vdots&\vdots&\ddots&\vdots\\\
1&u(t_{n},t_{1})+f(t_{1})&\ldots&u(t_{n},t_{n})+f(t_{n})\end{array}\right).$
(2.15)

where $f$ is given in (1.7). $A=K^{-1}$ is an $M$ matrix. Denote its diagonal
entries by $(a_{0},a_{1},\ldots,a_{n})$. Let ${\cal
M}=\\{u(t_{i},t_{j})\\}_{i,j=1}^{n}$. It follows from (2.8) that

$a_{i}={\cal M}^{i,i},\qquad i=1,\ldots,n.$ (2.16)

Let $\overline{\cal M}$ be the matrix with elements,

$\overline{\cal M}_{i,j}=\frac{{\cal M}_{i,j}}{{\cal M}^{1/2}_{i,i}{\cal
M}^{1/2}_{j,j}},\qquad i,j=1,\ldots,n.$ (2.17)

We have,

$\overline{\cal M}=D{\cal M}D,$ (2.18)

where $D$ is a diagonal matrix with diagonal elements $\\{1/{\cal
M}^{1/2}_{i,i}\\}_{i=1}^{n}$. Therefore,

${\cal M}^{-1}=D\overline{\cal M}^{-1}D.$ (2.19)

Let $A={\cal M}^{-1}$ and $\overline{A}=\overline{\cal M}^{-1}$. We have,

$a_{i}={\cal M}^{i,i}=A_{i,i}=\left({\cal
M}^{-1}\right)_{i,i}=\left(D\overline{\cal
M}^{-1}D\right)_{i,i}=\frac{\overline{A}_{i,i}}{{\cal M}_{i,i}},\qquad
i=1,\ldots,n.$ (2.20)

We see by [7, Lemma 5.1] that,

$\overline{A}_{i,i}\leq\frac{1}{\overline{\cal M}_{i,i}-\max_{j\neq
i}\overline{\cal M}_{i,j}}.$ (2.21)

Note that it follows from the symmetry of $Y$ and [3, Lemma 3.4.3] that when
$i\neq j$,

$\overline{\cal M}_{i,j}\leq\frac{{\cal M}^{1/2}_{i,i}}{{\cal
M}^{1/2}_{j,j}}\wedge\frac{{\cal M}^{1/2}_{j,j}}{{\cal
M}^{1/2}_{i,i}}\leq\theta.$ (2.22)

Therefore, since $\overline{\cal M}_{i,i}=1$,

$\overline{A}_{i,i}\leq\frac{1}{1-\theta},$ (2.23)

and by (2.20),

$a_{i}\leq\frac{1+2\theta}{{\cal
M}_{i,i}}=\frac{1+2\theta}{u(\theta^{i},\theta^{i})},\qquad i=1,\ldots,n.$
(2.24)

It follows from Theorem 2.1 that

$\left(X_{t_{1}},X_{t_{2}},\cdots,X_{t_{n}}\right)\geq\left(a_{1}^{-1}\xi^{(1)}_{\alpha},a_{2}^{-1}\xi^{(2)}_{\alpha},\ldots,a_{n}^{-1}\xi^{(n)}_{\alpha},\right)$
(2.25)

so that

$\left(a_{1}X_{t_{1}},a_{2}X_{t_{2}},\cdots,a_{n}X_{t_{n}}\right)\geq\left(\xi^{(1)}_{\alpha},\xi^{(2)}_{\alpha},\ldots,\xi^{(n)}_{\alpha}\right).$
(2.26)

By (2.24) we see that,

$\left(\frac{X_{t_{1}}}{u(t_{1},t_{1})},\frac{X_{t_{2}}}{u(t_{2},t_{2})},\cdots,\frac{X_{t_{n}}}{u(t_{n},t_{n})}\right)\geq\frac{1}{1+2\theta}\left(\xi^{(1)}_{\alpha},\xi^{(2)}_{\alpha},\ldots,\xi^{(n)}_{\alpha}\right).$
(2.27)

Consequently,

$\max_{2\leq i\leq n}\frac{X_{\theta^{i}}}{u(\theta^{i},\theta^{i})\log
i}\geq\frac{1}{1+2\theta}\max_{2\leq i\leq n}\frac{\xi^{(i)}_{\alpha}}{\log
i},\qquad a.s.$ (2.28)

It now follows from [4, Lemma 7.2] that for any $\alpha$ and $\epsilon>0$
there exists an $l=l(\epsilon)$ such that,

$P\left(\max_{l\leq i\leq n}\frac{X_{\theta^{i}}}{u(\theta^{i},\theta^{i})\log
i}>\frac{1-\epsilon}{1+2\theta}\right)\geq 1-\frac{l+1}{n+1}.$ (2.29)

Now let $\\{Z_{t},t\in S\\}$ be an $\alpha-$permanental process with kernel
$\widetilde{u}(x,y)$ in (1.8). It follows from (2.29) that,

$P\left(\max_{l\leq i\leq n}\frac{Z_{\theta^{i}}}{u(\theta^{i},\theta^{i})\log
i}>\frac{1-\epsilon}{1+2\theta}\right)\geq 1-\frac{l+1}{n+1}.$ (2.30)

Since we can take $\theta$ arbitrarily small we get (1.9).

The next lemma breaks down the condition in (1.12) and is used in everything
that follows to show that (1.12) is satisfied.

###### Lemma 2.1

Let $\widetilde{u}$ be as given in (1.8) and assume that $u$ symmetric. Let
$\widetilde{\sigma}$ be as given in (1.11). If

$u(s,s)+u(t,t)-2u(s,t)\leq Cu(|t-s|,|t-s|),$ (2.31)

and

$\big{|}f(s)-f(t)\big{|}\leq Cu(|t-s|,|t-s|)$ (2.32)

for all $s,t\in[0,\delta]$, then

$\widetilde{\sigma}^{2}(s,t)\leq 2Cu(|t-s|,|t-s|),$ (2.33)

for all $s,t\in[0,\delta]$.

Proof Consider $\widetilde{\sigma}^{2}(s,t)$. Using the fact that $u(s,t)$ is
symmetric we can write this as,

$\displaystyle\widetilde{\sigma}^{2}(s,t)$ $\displaystyle=$
$\displaystyle\widetilde{u}(s,s)+\widetilde{u}(t,t)-\widetilde{u}(s,t)-\widetilde{u}(t,s)$
$\displaystyle\qquad+\left(\widetilde{u}^{1/2}(s,t)-\widetilde{u}^{1/2}(t,s)\right)^{2}$
$\displaystyle=$ $\displaystyle
u(s,s)+u(t,t)-2u(s,t)+\left(\widetilde{u}^{1/2}(s,t)-\widetilde{u}^{1/2}(t,s)\right)^{2}.$

Once again, using the fact that $u(s,t)$ is symmetric we have,

$\left(\widetilde{u}^{1/2}(s,t)-\widetilde{u}^{1/2}(t,s)\right)^{2}\leq\big{|}\widetilde{u}(s,t)-\widetilde{u}(t,s)\big{|}\leq\big{|}f(s)-f(t)\big{|}.$
(2.35)

Now using (2.31) and (2.32) to bound the last line of (2) we get (2.33).

We use Lemma 2.1 to get a more usable version of Theorem 1.3.

###### Theorem 2.2

Let $X_{\alpha}=\\{X_{\alpha}(t),t\in[0,1]\\}$ be an $\alpha-$permanental
process with kernel $\\{\widetilde{u}(s,t),s,t\in[0,1]\\}$ where
$\widetilde{u}(0,0)=0$. If $u(v,v)$ is regularly varying at zero with positive
index and (2.31) and (2.32) hold then

$\limsup_{t\downarrow 0}\frac{X_{\alpha}(t)}{(u(t,t)+f(t))\log\log 1/t}\leq
1,\qquad\text{a.s.}$ (2.36)

Proof If $u(v,v)$ is regularly varying at zero with positive index it is
asymptotic to an increasing function at zero. Denote this function by
$\widehat{\varphi}^{2}(v)$. It follows from (2.33) that

$\widetilde{\sigma}^{2}(s,t)\leq C^{\prime}\widehat{\varphi}(|t-s|),\qquad$
(2.37)

and, since $\widehat{\varphi}^{2}(v)=O(u(v,v))$ at 0, we have
$\widehat{\varphi}^{2}(v)=O(\widetilde{u}(v,v))$ at zero. Consequently, (2.36)
follows from (1.13).

We proceed to prove Theorems 1.4–1.7 by establishing (2.31) and (2.32).

##  3 Proof of Theorem 1.4

By definition,

$\Phi(s,s)+\Phi(t,t)-2\Phi(s,t)=\Phi(|t-s|,|t-s|)=(\sigma^{0})^{2}(t-s),$
(3.1)

first for all $s,t\in R^{1}-\\{0\\}$. Since $\Phi(s,t)$ is continuous for all
$s,t\in R^{1}$ we take it to be defined and continuous on $R^{1}$. This gives
(2.31).

We have the following remarkable regularity property of $(\sigma^{0})^{2}$
that is proved at the end of this section.

###### Lemma 3.1

$|(\sigma^{0})^{2}(y)-(\sigma^{0})^{2}(x-y)|\leq(\sigma^{0})^{2}(x),\qquad
x,y\in R^{1},$ (3.2)

Using this we show that,

$|f_{\Phi,\mu}(s)-f_{\Phi,\mu}(t)|\leq(\sigma^{0})^{2}(t-s)|\mu|,\quad s,t\in
R^{1}.$ (3.3)

It follows from Lemma 3.1 that

$|(\sigma^{0})^{2}(s-v)-(\sigma^{0})^{2}(t-v)|\leq(\sigma^{0})^{2}(t-s).$
(3.4)

Therefore,

$\displaystyle|f_{\Phi,\mu}(s)-f_{\Phi,\mu}(t)|\leq\int_{T}|\Phi(s,v)-\Phi(t,v)|\,d\mu(v)$
$\displaystyle\leq\frac{1}{2}\int_{T}\left(|(\sigma^{0})^{2}(t)-(\sigma^{0})^{2}(s)|+|(\sigma^{0})^{2}(t-v)-(\sigma^{0})^{2}(s-v)|\right)d\mu(v)$
$\displaystyle\leq\frac{1}{2}\int_{T}(\sigma^{0})^{2}(t-s)d\mu(v)=(\sigma^{0})^{2}(t-s)|\mu|.$

The relationships in (3.1) and (3.3) and Theorem 2.2 show that

$\limsup_{t\downarrow
0}\frac{X_{\alpha}(t)}{\left((\sigma^{0})^{2}(t)+f_{\Phi,\mu}(t)\right)\log\log
1/t}\leq 1,\qquad a.s.$ (3.6)

We now get an estimate in the form of (1.18) so that we can write (3.6) in the
form of (1.19). Using Lemma 3.1 we have,

$\displaystyle f_{\Phi,\mu}(t)$ $\displaystyle=$
$\displaystyle\int_{T}\Phi(t,v)d\mu(v)=\frac{1}{2}\int_{T}\left((\sigma^{0})^{2}(t)+(\sigma^{0})^{2}(v)-(\sigma^{0})^{2}(t-v)\right)d\mu(v)$
(3.7) $\displaystyle=$
$\displaystyle\frac{(\sigma^{0})^{2}(t)}{2}{}|\mu|+\frac{1}{2}\int_{T}\left((\sigma^{0})^{2}(v)-(\sigma^{0})^{2}(t-v)\right)d\mu(v).$

(Here we use the fact that $(\sigma^{0})^{2}$ is an even function and we
extend it to $T$.) Consequently

$\frac{f_{\Phi,\mu}(t)}{(\sigma^{0})^{2}(t)}=\frac{|\mu|}{2}+\frac{1}{2}\int_{T}\frac{\left((\sigma^{0})^{2}(v)-(\sigma^{0})^{2}(v-t)\right)}{(\sigma^{0})^{2}(t)}\,d\mu(v).$
(3.8)

By (3.4)

$\frac{|(\sigma^{0})^{2}(v)-(\sigma^{0})^{2}(v-t)|}{(\sigma^{0})^{2}(t)}\leq
1.$ (3.9)

Therefore, using the Dominated Convergence Theorem we see that,

$\lim_{t\downarrow
0}\frac{f_{\Phi,\mu}(t)}{(\sigma^{0})^{2}(t)}={\frac{|\mu|}{2}}+\frac{1}{2}\int_{T}\lim_{t\downarrow
0}\frac{\left((\sigma^{0})^{2}(v)-(\sigma^{0})^{2}(v-t)\right)}{(\sigma^{0})^{2}(t)}\,d\mu(v).$
(3.10)

Note that,

$\displaystyle\lim_{t\downarrow
0}\frac{(\sigma^{0})^{2}(v)-(\sigma^{0})^{2}(v-t)}{(\sigma^{0})^{2}(t)}$
$\displaystyle=$ $\displaystyle\lim_{t\downarrow
0}\frac{(\sigma^{0})^{2}(v)-(\sigma^{0})^{2}(v-t)}{t}\,{\frac{t}{(\sigma^{0})^{2}(t)}}$
(3.11) $\displaystyle=$
$\displaystyle((\sigma^{0})^{2}(v))^{\prime}\lim_{t\downarrow
0}\frac{t}{(\sigma^{0})^{2}(t)}.$

Consequently, the integral in (3.10) is equal to

$\frac{1}{2}{}\int_{T}((\sigma^{0})^{2}(v))^{\prime}\lim_{t\downarrow
0}\frac{t}{(\sigma^{0})^{2}(t)}\,d\mu(v).$ (3.12)

Using this and (3.10) we get (1.28) when (1.27) holds and (1.30) when (3.19)
holds.

Proof of Lemma 3.1 Since $\Phi(x,y)$ is a continuous potential density of a
strongly symmetric Borel right process with state space $T=R^{1}-\\{0\\}$ it
follows from [3, Lemma 3.4.3] that

$\Phi(x,y)\leq\Phi(x,x)\wedge\Phi(y,y)=(\sigma^{0})^{2}(x)\wedge(\sigma^{0})^{2}(y),\quad
x,y\in R^{1}-\\{0\\}.$ (3.13)

Therefore, it follows from this and (1.22) that,

$(\sigma^{0})^{2}(x)+(\sigma^{0})^{2}(y)-(\sigma^{0})^{2}(x-y)\leq
2(\sigma^{0})^{2}(x),\quad x,y\in R^{1}-\\{0\\},$ (3.14)

Since $\sigma^{0}$ is continuous on $R^{1}$ we have,

$(\sigma^{0})^{2}(x)+(\sigma^{0})^{2}(y)-(\sigma^{0})^{2}(x-y)\leq
2(\sigma^{0})^{2}(x),\quad x,y\in R^{1}.$ (3.15)

Or equivalently,

$(\sigma^{0})^{2}(y)-(\sigma^{0})^{2}(x-y)\leq(\sigma^{0})^{2}(x),\qquad
x,y\in R^{1}.$ (3.16)

Furthermore, since $\Phi(x,y)\geq 0$ it follows from (1.22) that

$(\sigma^{0})^{2}(x-y)-(\sigma^{0})^{2}(y)\leq(\sigma^{0})^{2}(x).$ (3.17)

Combining this with (3.16) we have (3.2).

###### Remark 3.1

It is interesting to note that since $((\sigma^{\beta})^{2})^{\prime}(t)$ is
an odd function whenever $\mu$ is a symmetric measure the integral in (1.30)
is equal to 0 and we get

$\limsup_{t\downarrow 0}\frac{X_{\alpha}(t)}{Ct\log\log 1/t}\leq
1+\frac{|\mu|}{2},\qquad a.s.$ (3.18)

###### Example 3.1

Suppose that $\mu$ is concentrated on the negative half line and

${(\sigma^{0})^{2}(t)}={C|t|}.$ (3.19)

Then

$\frac{1}{2C}\int_{-\infty}^{\infty}((\sigma^{0})^{2}(v))^{\prime}\,d\mu(v)=-\frac{|\mu|}{2}.$
(3.20)

Therefore, it follows from (1.30) and Theorem 1.2 that,

$\limsup_{t\downarrow 0}\frac{X_{\alpha}(t)}{Ct\log\log 1/t}=1,\qquad a.s.$
(3.21)

##  4 Proof of Theorem 1.5

We have,

$\displaystyle(\sigma^{\beta})^{2}(x-y))$ $\displaystyle=$ $\displaystyle
u^{\beta}(x,x)+u^{\beta}(y,y)-2u^{\beta}(x,y)$ $\displaystyle=$ $\displaystyle
2\left(u^{\beta}(0)-u^{\beta}(x-y)\right).$

Define,

$(\sigma_{v}^{\beta})^{2}(x,y))=v^{\beta}(x,x)+v^{\beta}(y,y)-2v^{\beta}(x,y).$
(4.2)

Using (1.2) we see that,

$(\sigma_{v}^{\beta})^{2}(x,y)\leq(\sigma^{\beta})^{2}(x-y).$ (4.3)

In addition,

$\displaystyle v^{\beta}(x,x)$ $\displaystyle=$ $\displaystyle
u^{\beta}(0)-\frac{(u^{\beta}(x))^{2}}{u^{\beta}(0)}=\frac{(u^{\beta}(0))^{2}-(u^{\beta}(x))^{2}}{u^{\beta}(0)}$
$\displaystyle=$
$\displaystyle\frac{(\sigma^{\beta})^{2}(x)}{2u^{\beta}(0)}\left(u^{\beta}(0)+u^{\beta}(x)\right),$

so that,

$v^{\beta}(x,x)\sim{(\sigma^{\beta})^{2}(x)},\qquad\text{as $x\downarrow 0.$}$
(4.5)

It follows from (4.3) and (4) that,

$(\sigma_{v}^{\beta})^{2}(x,y)\leq Cv^{\beta}(x-y,x-y),\qquad\text{for all
$x,y\in[0,\delta].$}$ (4.6)

This gives (2.31) for this example.

To obtain (2.32) we use following lemma which is given in [3, Lemma 7.4.2]:

###### Lemma 4.1

Let $\\{G(t),t\in R^{1}\\}$ be a mean zero stationary Gaussian process with
continuous covariance $u(s,t)=u(t-s)$, that is the potential density of a
strongly symmetric Borel right process. Set,

$\sigma^{2}(t)=E(G(t)-G(0))^{2}=2\left(u(0)-u(t)\right).$ (4.7)

Then,

$|\sigma^{2}(t)-\sigma^{2}(s)|\leq\frac{2u(0)}{u(t)+u(s)}\sigma^{2}(t-s).$
(4.8)

Since $\\{u^{\beta}(x,y)$, $x,y\in R^{1}\\}$ is the potential density of a
strongly symmetric Borel right process with state pace $R^{1}$ it follows from
[3, Lemma 3.3.3] that it is positive definite and therefore it is the
covariance of a stationary Gaussian process. Applying (4.8) to $u^{\beta}$ and
$\sigma^{\beta}$ we have,

$\Big{|}(\sigma^{\beta})^{2}(t)-(\sigma^{\beta})^{2}(s)\Big{|}\leq\frac{2u^{\beta}(0)}{u^{\beta}(t)+u^{\beta}(s)}(\sigma^{\beta})^{2}(t-s).$
(4.9)

The potential of $Z_{\beta}$ is given in (1.35). We take $\mu(\\{0\\})=0$ so
that potential of $Z^{\prime}_{\beta}$ is

$f_{v^{\beta},\mu}(y)=\int_{-\infty}^{\infty}v^{\beta}(z,y)\,d\mu(z).$ (4.10)

Consequently,

$\big{|}f_{v^{\beta},\mu}(x)-f_{v^{\beta},\mu}(y)\big{|}\leq{\big{|}f_{u^{\beta},\mu}(x)-f_{u^{\beta},\mu}(y)\big{|}}+\frac{f_{u^{\beta},\mu}(0)}{2u^{\beta}(0)}\big{|}(\sigma^{\beta})^{2}(x)-(\sigma^{\beta})^{2}(y)\big{|}.$
(4.11)

It follows from (4.9) that,

$\frac{f_{u^{\beta},\mu}(0)}{2u^{\beta}(0)}\big{|}(\sigma^{\beta})^{2}(x)-(\sigma^{\beta})^{2}(y)\big{|}\leq\frac{f_{u^{\beta},\mu}(0)}{(u^{\beta}(x)+u^{\beta}(y))}(\sigma^{\beta})^{2}(x-y).$
(4.12)

Since $u^{\beta}(x)$ is continuous, $u^{\beta}(x)>0$, for all $x$ sufficiently
small. Therefore, for $\Delta$ sufficiently small when $x,y$ in $[0,\Delta]$
we have,

$\frac{f_{u^{\beta},\mu}(0)}{2u^{\beta}(0)}\big{|}(\sigma^{\beta})^{2}(x)-(\sigma^{\beta})^{2}(y)\big{|}\leq
C_{\Delta}(\sigma^{\beta})^{2}(x-y),$ (4.13)

for some finite constant $C_{\Delta}.$

We now consider the first term after the equal sign in (4.11). We have,

$\displaystyle|f_{u^{\beta},\mu}(x)-f_{u^{\beta},\mu}(y)|$ (4.14)
$\displaystyle\qquad=\bigg{|}\int_{-\infty}^{\infty}u^{\beta}(z-x)\,d\mu(z)-\int_{-\infty}^{\infty}u^{\beta}(z-y)\,d\mu(z)\bigg{|}$
$\displaystyle\qquad=\bigg{|}\int_{-\infty}^{\infty}\left(\left(u^{\beta}(0)-u^{\beta}(z-y)\right)-\left(u^{\beta}(0)-u^{\beta}(z-x)\right)\right)\,d\mu(z)\bigg{|}$
$\displaystyle\qquad\leq\frac{1}{2}\int_{-\infty}^{\infty}\Big{|}(\sigma^{\beta})^{2}(z-y)-(\sigma^{\beta})^{2}(z-x)\Big{|}\,d\mu(z).$

It follows from (4.9) that for $\Delta$ sufficiently small,

$\left(\int_{0}^{\Delta}+\int_{-\Delta}^{0}\right)\left((\sigma^{\beta})^{2}(z-y)-(\sigma^{\beta})^{2}(z-x)\right)\,d\mu(z)\leq
C^{\prime}_{\Delta}(\sigma^{\beta})^{2}(x-y),$ (4.15)

for some finite constant $C^{\prime}_{\Delta}.$

In addition

$\displaystyle
H_{\Delta}(x-y):=\left(\int_{-\infty}^{-\Delta}+\int_{\Delta}^{\infty}\right)\left((\sigma^{\beta})^{2}(z-y)-(\sigma^{\beta})^{2}(z-x)\right)\,d\mu(z)$
$\displaystyle\qquad\leq\max_{|z|\geq\Delta}(\sigma^{\beta})^{2}(z)^{\prime}\mu(\\{(-\infty,-\Delta)\cup\mu(\Delta,\infty)\\})|x-y|,$
(4.16)

and,

$\frac{H_{\Delta}(x-y)}{(\sigma^{\beta})^{2}(x-y)}\leq\max_{|z|\geq\Delta}(\sigma^{\beta})^{2}(z)^{\prime}\mu(\\{(-\infty,-\Delta)\cup\mu(\Delta,\infty)\\})\frac{|x-y|}{(\sigma^{\beta})^{2}(x-y)}.$
(4.17)

Using this (4.15), (4.13) and (1.31), we get,

$\lim_{x,y\downarrow
0}\frac{\big{|}f_{v^{\beta},\mu}(x)-f_{v^{\beta},\mu}(y)\big{|}}{(\sigma^{\beta})^{2}(x-y)}\leq
C,$

for some constant $C$. Together with (4.5) this gives us the version of (2.32)
we need for this theorem.

We now get an estimate for $f_{v^{\beta},\mu}$ at 0.

$\displaystyle f_{v^{\beta},\mu}(x)$ $\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}u^{\beta}(x,y)\,d\mu(y)-\frac{u^{\beta}(0,x)}{u^{\beta}(0,0)}\int_{-\infty}^{\infty}u^{\beta}(0,y)\,d\mu(y)$
$\displaystyle=$
$\displaystyle{f_{u^{\beta},\mu}(x)}-\frac{u^{\beta}(0,x)f_{u^{\beta},\mu}(0)}{u^{\beta}(0,0)}$
$\displaystyle=$
$\displaystyle{f_{u^{\beta},\mu}(x)}-{f_{u^{\beta},\mu}(0)}+\frac{f_{u^{\beta},\mu}(0)}{u^{\beta}(0,0)}\left(u^{\beta}(0,0)-u^{\beta}(0,x)\right)$
$\displaystyle=$
$\displaystyle{f_{u^{\beta},\mu}(x)-f_{u^{\beta},\mu}(0)}+\frac{(\sigma^{\beta})^{2}(x)f_{u^{\beta},\mu}(0)}{2u^{\beta}(0,0)}.$

In addition,

$\displaystyle f_{u^{\beta},\mu}(x)-f_{u^{\beta},\mu}(0)$ (4.19)
$\displaystyle\qquad=\int_{-\infty}^{\infty}u^{\beta}(z-x)\,d\mu(z)-\int_{-\infty}^{\infty}u^{\beta}(z)\,d\mu(z)$
$\displaystyle\qquad=\int_{-\infty}^{\infty}\left(u^{\beta}(0)-u^{\beta}(z)\right)-\left(u^{\beta}(0)-u^{\beta}(z-x)\right)\,d\mu(z)$
$\displaystyle\qquad=\frac{1}{2}\int_{-\infty}^{\infty}\left((\sigma^{\beta})^{2}(z)-(\sigma^{\beta})^{2}(z-x)\right)\,d\mu(z).$

For $x\in[0,\Delta]$ it follows from (4.9) that,

$\displaystyle 2I_{1}(x)$ $\displaystyle:=$
$\displaystyle\int_{-\Delta}^{\Delta}\left((\sigma^{\beta})^{2}(z)-(\sigma^{\beta})^{2}(z-x)\right)\,d\mu(z)$
$\displaystyle\leq$
$\displaystyle(\sigma^{\beta})^{2}(x)\int_{-\Delta}^{\Delta}\frac{2u^{\beta}(0)}{u^{\beta}(z-x)+u^{\beta}(z)}\,d\mu(z).$

Consequently for $\Delta$ sufficiently small,

$\frac{I_{1}(x)}{(\sigma^{\beta})^{2}(x)}\leq
C_{\Delta}\mu([-\Delta,\Delta]),$ (4.21)

for some $C_{\Delta}<\infty$ with $\lim_{\Delta\to 0}C_{\Delta}=2$.

Now consider

$2I_{2}(x):=\left(\int_{\Delta}^{\infty}+\int_{-\infty}^{-\Delta}\right)\left((\sigma^{\beta})^{2}(z)-(\sigma^{\beta})^{2}(z-x)\right)\,d\mu(z).$
(4.22)

Note that for $x\in(0,\delta)$ where $0<\delta\leq\Delta$,

$(\sigma^{\beta})^{2}(z)-(\sigma^{\beta})^{2}(z-x)=((\sigma^{\beta})^{2})^{\prime}(z^{*}(z,x))x$
(4.23)

for some $z^{*}(z,x)\in[z-x,z]$. Therefore,

$\frac{|I_{2}(x)|}{(\sigma^{\beta})^{2}(x)}=\frac{x}{(\sigma^{\beta})^{2}(x)}\frac{1}{2}\left(\int_{\Delta}^{\infty}+\int_{-\infty}^{-\Delta}\right)((\sigma^{\beta})^{2})^{\prime}(z^{*}(z,x))\,d\mu(z).$
(4.24)

Since $|(\sigma^{\beta})^{2})^{\prime}(z^{*}(z,,x))|$ is bounded on the range
of integration we can use the Dominated Convergence Theorem to get,

$\lim_{x\downarrow
0}\frac{|I_{2}(x)|}{(\sigma^{\beta})^{2}(x)}=\lim_{x\downarrow
0}\frac{x}{(\sigma^{\beta})^{2}(x)}\frac{1}{2}\left(\int_{\Delta}^{\infty}+\int_{-\infty}^{-\Delta}\right)((\sigma^{\beta})^{2})^{\prime}(z)\,d\mu(z).$
(4.25)

When (1.40) holds this is equal to 0. Using this (4) and (4.21) we see that in
this case,

$\lim_{x\downarrow 0}\frac{f_{v^{\beta},\mu}(x)}{(\sigma^{\beta})^{2}(x)}\leq
C_{\Delta}\mu([-\Delta,\Delta])+\frac{f_{u^{\beta},\mu}(0)}{2u^{\beta}(0,0)}.$
(4.26)

Since this holds for all $\Delta>0$ and $\lim_{\Delta\to 0}C_{\Delta}=2$ we
get (1.41).

When (1.42) holds the right hand side of (4.25) is equal to,

$\frac{1}{2C}\left(\int_{\Delta}^{\infty}+\int_{-\infty}^{-\Delta}\right)((\sigma^{\beta})^{2})^{\prime}(z)\,d\mu(z),$
(4.27)

which in this case gives,

$\lim_{x\downarrow
0}\frac{f_{v^{\beta},\mu}(x)}{(\sigma^{\beta})^{2}(x)}\leq\frac{1}{2C}\int_{-\infty}^{\infty}((\sigma^{\beta})^{2})^{\prime}(z)\,d\mu(z)+\frac{f_{u^{\beta},\mu}(0)}{2u^{\beta}(0,0)}.$
(4.28)

Using this we get (1.43).

###### Remark 4.1

When (1.42) holds the integral in (4.28) is finite, since it, and the
hypothesis that $(\sigma^{\beta})^{2}(t)\in C^{1}(T)$ and
$|(\sigma^{\beta})^{2}(t)^{\prime}|$ is bounded on
$(-\infty,-\Delta)\cup(\Delta,\infty)$ for all $\Delta>0$ implies that
$|((\sigma^{\beta})^{2})^{\prime}(t)|$ is bounded on $R^{1}.$

It is interesting to note that since $((\sigma^{\beta})^{2})^{\prime}(t)$ is
an odd function whenever $\mu$ is a symmetric measure the integral in (4.28)
is equal to 0 and we get

$\limsup_{t\downarrow 0}\frac{X_{\alpha}(t)}{Ct\log\log 1/t}\leq
1+\frac{f_{u^{\beta},\mu}(0)}{2u^{\beta}(0)},\qquad a.s.$ (4.29)

###### Example 4.1

In [6, 1.25] we point out that for $\beta>0$,

$u^{\beta}_{\gamma}(x-y):=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{\cos\lambda(x-y)}{\beta+\gamma\lambda^{2}}\,d\lambda=\frac{1}{2}\frac{e^{-({\beta/\gamma})^{1/2}\,|x-y|}}{(\beta\gamma)^{1/2}},\qquad
x,y\in R^{1}.$ (4.30)

In this example we simply take $\beta=K$ and $\gamma=1/K$ so that,

$u_{\gamma}^{\beta}(x-y)=\frac{1}{2}\exp^{-K|x-y|}.$ (4.31)

and

$(\sigma_{\gamma}^{\beta})^{2}(x)=1-\exp^{-K|x|}.$ (4.32)

Suppose that $\mu$ is symmetric. Then

$f_{u_{\gamma}^{\beta},\mu}(0)=\frac{1}{2}\int_{-\infty}^{\infty}e^{-K|y|}\,\,d\mu(y)=\int_{0}^{\infty}e^{-Ky}\,d\mu(y),$
(4.33)

so that

$\frac{f_{u_{\gamma}^{\beta},\mu}(0)}{2u_{\gamma}^{\beta}(0)}=\int_{0}^{\infty}e^{-Ky}\,d\mu(y).$
(4.34)

In addition,

$\lim_{x\downarrow 0}\frac{(\sigma_{\gamma}^{\beta})^{2}(x)}{x}={K}.$ (4.35)

Therefore, it follows from (4.29) that,

$\limsup_{t\downarrow 0}\frac{X_{\alpha,\beta}(t)}{Kt\log\log
1/t}\leq\left(1+\int_{0}^{\infty}e^{-Ky}\,d\mu(y)\right),\qquad a.s.$ (4.36)

Suppose that $\mu$ is concentrated on the negative half line. We evaluate the
two terms on the right in (4.28) for $u^{\beta}_{\gamma}$ in (4.31). Note that
in this case $C$ in (4.28) is equal to $K$. We have,

$\displaystyle f_{u_{\gamma}^{\beta},\mu}(0)$ $\displaystyle=$
$\displaystyle\int_{-\infty}^{0}u_{\gamma}^{\beta}(-y)\,d\mu(y)$ (4.37)
$\displaystyle=$
$\displaystyle\frac{1}{2}\int_{-\infty}^{0}\exp^{-K|-y|}d\mu(y)=\frac{1}{2}\int_{-\infty}^{0}\exp^{Ky}d\mu(y).$

Since $2u_{\gamma}^{\beta}(0)=1$ we get,

$\frac{f_{u_{\gamma}^{\beta},\mu}(0)}{2u_{\gamma}^{\beta}(0)}=\frac{1}{2}\int_{-\infty}^{0}\exp^{Ky}d\mu(y).$
(4.38)

In addition,

$\displaystyle\frac{1}{2K}\int_{-\infty}^{0}((\sigma^{\beta})^{2})^{\prime}(z)\,d\mu(z)=\frac{1}{2K}\int_{-\infty}^{0}(1-\exp^{Kz})^{\prime}\,d\mu(z)$
(4.39) $\displaystyle=-\frac{1}{2}\int_{-\infty}^{0}\exp^{Kz}d\mu(z).$

Therefore, it follows from (4.29) and Theorem 1.2 that,

$\limsup_{t\downarrow 0}\frac{X_{\alpha,\beta}(t)}{Kt\log\log 1/t}=1,\qquad
a.s.$ (4.40)

##  5 Proof of Theorem 1.6

It is easy to check that,

$u_{T_{0}}(x,x)+u_{T_{0}}(y,y)-2u_{T_{0}}(x,y)=|s(x)-s(y)|,$ (5.1)

and that

$|u_{T_{0}}(x,v)-u_{T_{0}}(y,v)|\leq|s(x)-s(y)|.$ (5.2)

Therefore,

$|f_{u_{T_{0}},\mu}(x)-f_{u_{T_{0}},\mu}(y)|\leq|s(x)-s(y)|\,\,|\mu|.$ (5.3)

The condition that $s\in C^{2}(0,\infty)$ shows that the criteria in Lemma 2.1
are satisfied. The next lemma completes the proof of this theorem.

###### Lemma 5.1

$f_{u_{T_{0}},\mu}(x)\sim s(x)|\mu|,\qquad\text{as $x\downarrow 0$}.$ (5.4)

Proof We have,

$f_{u_{T_{0}},\mu}(x)=s(x)\int_{x}^{\infty}\,d\mu(y)+\int^{x}_{0}s(y)\,d\mu(y).$
(5.5)

Therefore,

$\frac{f_{u_{T_{0}},\mu}(x)}{s(x)}=\int_{x}^{\infty}\,d\mu(y)+\int^{x}_{0}\frac{s(y)}{s(x)}\,d\mu(y).$
(5.6)

Since $s(y)$ is increasing and $\mu$ is supported on $(0,\infty)$,

$\int^{x}_{0}\frac{s(y)}{s(x)}\,d\mu(y)\leq\int^{x}_{0}\,d\mu(y),$ (5.7)

Using this in (5.6) we get (5.4).

The proof is completed as in the proofs of Theorems 1.5 and 1.6.

##  6 Proof of Theorem 1.7

Let

$(\widetilde{\sigma}^{\beta})^{2}(x,y))=\widetilde{u}^{\beta}(x,x)+\widetilde{u}^{\beta}(y,y)-2\widetilde{u}^{\beta}(x,y),$
(6.1)

and define,

$(\widetilde{\sigma}_{v}^{\beta})^{2}(x,y))=\widetilde{v}^{\beta}(x,x)+\widetilde{v}^{\beta}(y,y)-2\widetilde{v}^{\beta}(x,y).$
(6.2)

It follows from (1.52) that

$\widetilde{v}^{\beta}(x,y)=\widetilde{u}^{\beta}(x,y)-\frac{p_{\beta}(0)}{q_{\beta}(0)}q_{\beta}(x)q_{\beta}(y).$
(6.3)

Consequently,

$(\widetilde{\sigma}_{v}^{\beta})^{2}(x,y)=(\widetilde{\sigma}^{\beta})^{2}(x,y)-\frac{p_{\beta}(0)}{q_{\beta}(0)}\left(q_{\beta}(x)-q_{\beta}(y)\right)^{2}.$
(6.4)

We show in [6, Lemma 4.9] that

$(\widetilde{\sigma}^{\beta})^{2}(x,y)\sim\rho_{\beta}(0)|x-y|,\qquad\text{as
$|x-y|\to 0$},$ (6.5)

where,

$\rho_{\beta}(0)=p_{\beta}^{\prime}(0)q_{\beta}(0)-q_{\beta}^{\prime}(0)p_{\beta}(0).$
(6.6)

Since,

$|q_{\beta}(x)-q_{\beta}(y)|\sim q_{\beta}^{\prime}(0)|x-y|,\qquad\text{as
$x,y\downarrow 0$},$ (6.7)

it follows by (6.4) and (6.5) that,

$(\widetilde{\sigma}_{v}^{\beta})^{2}(x,y)\sim\rho_{\beta}(0)|x-y|,\qquad\text{as
$x,y\downarrow 0$}.$ (6.8)

By (1.52)

$\widetilde{v}^{\beta}(x,x)=\left(p_{\beta}(x)-\displaystyle\frac{p_{\beta}(0)}{q_{\beta}(0)}q_{\beta}(x)\right)q_{\beta}(x)$
(6.9)

Since,

$p_{\beta}(x)=p_{\beta}(0)+p_{\beta}^{\prime}(0)x+o(x)\quad\text{and}\quad
q_{\beta}(x)=q_{\beta}(0)+q_{\beta}^{\prime}(0)x+o(x),$ (6.10)

as $x\downarrow 0$, we see that,

$\displaystyle\widetilde{v}^{\beta}(x,x)$ $\displaystyle=$
$\displaystyle\left(\frac{p_{\beta}(x)}{p_{\beta}(0)}-\frac{q_{\beta}(x)}{q_{\beta}(0)}\right)p_{\beta}(0)q_{\beta}(x)$
(6.11) $\displaystyle=$
$\displaystyle\left(\frac{p_{\beta}^{\prime}(0)}{p_{\beta}(0)}-\frac{q_{\beta}^{\prime}(0)}{q_{\beta}(0)}\right)p_{\beta}(0)q_{\beta}(x)x+o(x)$
$\displaystyle=$ $\displaystyle{\rho_{\beta}(0)x}+o(x),\qquad\text{as
$x\downarrow 0$ }.$

Consequently,

$\widetilde{v}^{\beta}(x-y,x-y)\sim\rho_{\beta}(0)|x-y|,\quad\text{as
$x,y\downarrow 0$},$ (6.12)

and, by (6.8)

$(\widetilde{\sigma}_{v}^{\beta})^{2}(x,y)\leq
C\widetilde{v}^{\beta}(x-y,x-y),\quad\text{as $x,y\downarrow 0$},$ (6.13)

for some constant $C$. This is the condition in (2.31).

Let

$f_{\widetilde{u}^{\beta},\mu}(x)=\int_{0}^{\infty}\widetilde{u}^{\beta}(x,y)\,d\mu(y)$
(6.14)

and

$f_{\widetilde{v}^{\beta},\mu}(x)=\int_{0}^{\infty}\widetilde{v}^{\beta}(x,y)\,d\mu(y).$
(6.15)

Using (6.3) we see that

$f_{\widetilde{v}^{\beta},\mu}(x)=f_{\widetilde{u}^{\beta},\mu}(x)-\frac{p_{\beta}(0)}{q_{\beta}(0)}q_{\beta}(x)\int_{0}^{\infty}q_{\beta}(y)\,d\mu(y).$
(6.16)

Note that,

$f_{\widetilde{u}^{\beta},\mu}(x)=p_{\beta}(x)\int_{x}^{\infty}q_{\beta}(y)\,d\mu(y)+q_{\beta}(x)\int_{0}^{x}p_{\beta}(y)\,d\mu(y)$
(6.17)

and,

$f^{\prime}_{\widetilde{u}^{\beta},\mu}(x)=p_{\beta}^{\prime}(x)\int_{x}^{\infty}q_{\beta}(y)\,d\mu(y)+q_{\beta}^{\prime}(x)\int_{0}^{y}p_{\beta}(y)\,d\mu(y).$
(6.18)

This shows that $f_{\widetilde{u}^{\beta},\mu}\in C^{1}([0,\infty))$ and we
see from (6.16) that for $x,y\in[0,\delta]$,

$\big{|}f_{\widetilde{v}^{\beta},\mu}(x)-f_{\widetilde{v}^{\beta},\mu}(y)\big{|}\leq\sup_{s\in[0,\delta]}f_{\widetilde{u}^{\beta},\mu}^{\prime}(s)|x-y|+\frac{p_{\beta}(0)}{q_{\beta}(0)}\big{|}q_{\beta}(x)-q_{\beta}(y)\big{|}\int_{0}^{\infty}q_{\beta}(z)\,d\mu(z).$
(6.19)

Using (6.7) we see that,

$\big{|}f_{\widetilde{v}^{\beta},\mu}(x)-f_{\widetilde{v}^{\beta},\mu}(y)\big{|}=O(|x-y|).$
(6.20)

This is the condition in (2.32).

We now consider the permanental process with kernel

$v^{\beta}(t,t)=\widetilde{v}^{\beta}(t,t)+f_{\widetilde{v}^{\beta},\mu}(t).$
(6.21)

The behavior of $\widetilde{v}^{\beta}(t,t)$ at 0 is given in (6.12). We now
consider the behavior of $f_{\widetilde{v}^{\beta},\mu}$ at 0.

###### Lemma 6.1

$f_{\widetilde{v}^{\beta},\mu}(x)={x\,\rho_{\beta}(0)}\,\int_{0}^{\infty}\frac{q_{\beta}(y)}{q_{\beta}(0)}\,d\mu(y)+o(x),\qquad\text{as
$x\downarrow 0$}.$ (6.22)

Proof Rewrite (1.52) as,

$\widetilde{v}^{\beta}(x,y)=\left\\{\begin{array}[]{cc}\left(\displaystyle\frac{p_{\beta}(x)}{p_{\beta}(0)}-\frac{q_{\beta}(x)}{q_{\beta}(0)}\right)p_{\beta}(0)q_{\beta}(y),&\quad
0<x\leq y\\\
\left(\displaystyle\frac{p_{\beta}(y)}{p_{\beta}(0)}-\frac{q_{\beta}(y)}{q_{\beta}(0)}\right)p_{\beta}(0)q_{\beta}(x),&\quad
0<y\leq x.\end{array}\right.$ (6.23)

We have,

$\displaystyle f_{\widetilde{v}^{\beta},\mu}(x)$ $\displaystyle=$
$\displaystyle\int^{\infty}_{x}\widetilde{v}^{\beta}(x,y)\,d\mu(y)+\int^{x}_{0}\widetilde{v}^{\beta}(x,y)\,d\mu(y)$
$\displaystyle=$
$\displaystyle\left(\frac{p_{\beta}(x)}{p_{\beta}(0)}-\frac{q_{\beta}(x)}{q_{\beta}(0)}\right)p_{\beta}(0)\int_{x}^{\infty}q_{\beta}(y)\,d\mu(y)$
$\displaystyle\qquad+p_{\beta}(0)q_{\beta}(x)\int_{0}^{x}\left(\displaystyle\frac{p_{\beta}(y)}{p_{\beta}(0)}-\frac{q_{\beta}(y)}{q_{\beta}(0)}\right)\,d\mu(y)$
$\displaystyle:=$ $\displaystyle I_{1}+I_{2}.$

It follows as in (6.11) that,

$\displaystyle\frac{p_{\beta}(x)}{p_{\beta}(0)}-\frac{q_{\beta}(x)}{q_{\beta}(0)}$
$\displaystyle=$
$\displaystyle\left(\frac{p_{\beta}^{\prime}(0)}{p_{\beta}(0)}-\frac{q_{\beta}^{\prime}(0)}{q_{\beta}(0)}\right)x+o(x),\qquad\text{as
$x\downarrow 0$ }$ $\displaystyle=$
$\displaystyle\frac{\rho_{\beta}(0)x}{p_{\beta}(0)q_{\beta}(0)}+o(x),\qquad\text{as
$x\downarrow 0$ }.$

Using this and (6) we see that,

$I_{1}(x)=x{\rho_{\beta}(0)}\int_{0}^{\infty}\frac{q_{\beta}(y)}{q_{\beta}(0)}\,d\mu(y)+o(x),\qquad\text{as
$x\downarrow 0$. }$ (6.26)

We complete the proof by showing that $I_{2}(x)=o(x)$ as $x\downarrow 0$.
Using (6) we have,

$\displaystyle I_{2}(x)$ $\displaystyle=$ $\displaystyle
q_{\beta}(0)p_{\beta}(x)\int_{0}^{x}\left(\frac{p_{\beta}(y)}{p_{\beta}(0)}-\frac{q_{\beta}(y)}{q_{\beta}(0)}\right)\,d\mu(y)$
$\displaystyle=$ $\displaystyle
q_{\beta}(0)p_{\beta}(x)\int_{0}^{x}\left(\frac{\rho_{\beta}(0)y}{p_{\beta}(0)q_{\beta}(0)}+o(y)\right)\,d\mu(y)$
$\displaystyle=$
$\displaystyle\frac{p_{\beta}(x)\rho_{\beta}(0)}{p_{\beta}(0)}\int_{0}^{x}y\,d\mu(y)+q_{\beta}(0)p(x)\int_{0}^{x}o(y)\,d\mu(y).$

The line above is,

$\leq\frac{p(x)\rho_{\beta}(0)x}{p_{\beta}(0)}\int_{0}^{x}\,d\mu(y)+q_{\beta}(0)p_{\beta}(x)o(x)\int_{0}^{x}\,d\mu(y)=o(x),\quad\text{as
$x\downarrow 0$},$ (6.28)

since $\mu$ is supported on $(0,\infty)$.

The proof is completed as in the proofs of Theorems 1.5 and 1.6.

###### Example 6.1

In (1.50) we can take,

$p_{\beta}(x)=e^{Kx},\qquad\text{and}\qquad q_{\beta}(y)=\frac{1}{2}e^{-Ky};$
(6.29)

see [6, Example 9.3], in which case,

$\widetilde{u}^{\beta}(x,y)=\frac{1}{2}e^{-K|x-y|}.$ (6.30)

It is easy to see that in this case,

$\rho_{\beta}(0)=K.$ (6.31)

Consequently, (1.55) gives,

$\limsup_{t\downarrow 0}\frac{X_{\alpha}(t)}{Kt\log\log
1/t}=\left(1+\int_{0}^{\infty}e^{-Ky}d\mu(y)\right),\qquad a.s.,$ (6.32)

as in (4.36).

##  7 Additional observations

###### Remark 7.1

In Theorems 1.4 and 1.5 we require that $(\sigma^{0})^{2}$ and
$(\sigma^{\beta})^{2}$ are regularly varying at 0 with positive index. We show
in [6, Section 8] that these conditions are realized in many cases.

###### Remark 7.2

We give a simple criteria for the conditions that $(\sigma^{\beta})^{2}(t)\in
C^{1}(0,\infty)$ and $|(\sigma^{\beta})^{2}(x)^{\prime}|$ is bounded on
$[\delta,\infty)$ for all $\delta>0$.

###### Lemma 7.1

For all $\beta\geq 0$, if

$\sup_{0\leq\lambda<\infty}\frac{\lambda|\psi^{\prime}(\lambda)|}{\beta+\psi(\lambda)}<\infty,$
(7.1)

$(\sigma^{\beta})^{2}(x)\in C^{1}(0,\infty)$ and

$|(\sigma^{\beta})^{2}(x)^{\prime}|\leq
C\frac{(\sigma^{\beta})^{2}(x)}{x},\qquad\forall\,x\in(0,\infty),$ (7.2)

for some constant $C$.

Proof The proof is a modification of [6, Lemma 8.4] which gives sharper
estimates for $(\sigma^{0})^{\prime}(t)$ under more restrictive conditions on
$\psi(\lambda)$ . When $x\neq 0$,

$(\sigma^{\beta})^{2}(x)=\frac{4}{\pi}\int_{0}^{\infty}\frac{\sin^{2}\lambda
x/2}{\beta+\psi(\lambda)}\,d\lambda=\frac{4}{\pi
x}\int_{0}^{\infty}\frac{\sin^{2}t/2}{\beta+\psi(t/x)}\,dt,$ (7.3)

and,

$\displaystyle(\sigma^{\beta})^{2}(x)^{\prime}$ $\displaystyle=$
$\displaystyle-\frac{(\sigma^{\beta})^{2}(x)}{x}+\frac{4}{\pi
x}\frac{d}{dx}\left(\int_{0}^{\infty}\sin^{2}t/2\frac{1}{\beta+\psi(t/x)}\,dt\right)$
(7.4) $\displaystyle:=$
$\displaystyle-\frac{(\sigma^{\beta})^{2}(x)}{x}+\nu_{\beta}(x).$

We have,

$\nu_{\beta}(x)=\frac{4}{\pi
x^{2}}\int_{0}^{\infty}\frac{\sin^{2}t/2}{\beta+\psi(t/x)}\frac{(t/x)\psi^{\prime}(t/x)}{\beta+\psi(t/x)}\,dt.$
(7.5)

(See [6, Lemma 8.3] for the justification of differentiating under the
integral sign.) Consequently, when

$C:=\sup_{0\leq\lambda<\infty}\bigg{|}\frac{\lambda\psi^{\prime}(\lambda)}{\beta+\psi(\lambda)}\bigg{|}<\infty,$
(7.6) $\nu_{\beta}(x)\leq\frac{C(\sigma^{\beta})^{2}(x)}{x}.$ (7.7)

The inequalities in (7.2) follow from (7.4) and (7.7).

Note that the condition in (7.1) is satisfied for $\beta>0$ when
$\psi(\lambda)$ is a normalized regularly varying function at infinity, i.e.
for some $0<p\leq 2$,

$\psi(\lambda)=C\lambda^{p}\exp\left(\int_{1}^{\lambda}\frac{\epsilon(u)}{u}\,du\right),$
(7.8)

where $\lim_{u\to\infty}\epsilon(u)=0$. When $\beta=0$ wecan also require that
$\psi(\lambda)$ is a normalized regularly varying function at 0.

###### Remark 7.3

There are kernels for which the upper and lower bounds of the local moduli are
the same.

###### Theorem 7.1

Let $Y$ be a strongly symmetric transient Borel right process with state space
$T=R^{1}-\\{0\\}$ or $(0,\infty)$ and continuous strictly positive potential
densities $u(x,y)$ with respect to some $\sigma$–finite measure $m$ on $T$,
where $\lim_{t\downarrow 0}u(t,t)=0.$

Let $\\{X_{\alpha,1}(t),t\in[0,1]\\}$ and $\\{X_{\alpha,2}(t),t\in[0,1]\\}$ be
an $\alpha-$permanental processes with kernels

$v_{1}(s,t)=u(s,t),\qquad\text{and }\qquad v_{2}(s,t)=u(s,t)+g(s)f(t),$ (7.9)

respectively, where $f$ is given in (1.7) and

$g(s)=\int_{T}u(s,v)\,d\nu(v)$ (7.10)

for some positive finite measure $\nu$. If $u(v,v)$ is regularly varying at
zero with positive index and

$u(s,s)+u(t,t)-2u(s,t)\leq Cu(|t-s|,|t-s|).$ (7.11)

then

$\limsup_{t\downarrow 0}\frac{X_{\alpha,1}(t)}{u(t,t)\log\log 1/t}=1.$ (7.12)

If, in addition,

$\big{|}f(s)-f(t)\big{|}\leq
Cu(|t-s|,|t-s|)\quad\text{and}\quad\big{|}g(s)-g(t)\big{|}\leq
Cu(|t-s|,|t-s|)$ (7.13)

for some constant $C$ for all $s,t\in[0,\delta]$, for some $\delta>0$, then
(7.12) holds for $X_{\alpha,2}$. (The existence of $X_{\alpha,2}$ follows from
[6, Theorem 1.1].)

Proof of Theorem 7.1 When the kernel is $v_{1}(s,t)$ the lower bound follows
from Theorem 1.2 and the upper bound follows from Theorem 2.2.

Suppose the kernel is $v_{2}(s,t)$. Consider $\widetilde{\sigma}^{2}(s,t)$.
Using (7.9) and the fact that $u(s,t)$ is symmetric we can write this as,

$\displaystyle\widetilde{\sigma}^{2}(s,t)$ $\displaystyle=$ $\displaystyle
v_{2}(s,s)+v_{2}(t,t)-v_{2}(s,t)-v_{2}(t,s)$
$\displaystyle\qquad+\left(v_{2}^{1/2}(s,t)-v_{2}^{1/2}(t,s)\right)^{2}$
$\displaystyle=$ $\displaystyle u(s,s)+u(t,t)-2u(s,t)+(f(s)-f(t))(g(s)-g(t))$
$\displaystyle\qquad+\left(v_{2}^{1/2}(s,t)-v_{2}^{1/2}(t,s)\right)^{2}.$

Using (7.11) and (7.13) we see that,

$\displaystyle u(s,s)+u(t,t)-2u(s,t)+(f(s)-f(t))(g(s)-g(t))$ (7.15)
$\displaystyle\qquad\leq C^{\prime}u(|t-s|,|t-s|),$

for some constant $C^{\prime}$, for all $s,t\in[0,\delta]$. Using the fact
that $u(s,t)$ is symmetric we have,

$\displaystyle\left(v_{2}^{1/2}(s,t)-v_{2}^{1/2}(t,s)\right)^{2}\leq\big{|}v_{2}(s,t)-v_{2}(t,s)\big{|}$
(7.16)
$\displaystyle\qquad=\big{|}f(s)g(t)-f(t)g(s)\big{|}=\big{|}f(s)(g(t)-g(s))-(f(t)-f(s))g(s)\big{|}$
$\displaystyle\qquad\leq
f(s)\big{|}g(s)-g(t)\big{|}+g(s)\big{|}f(s)-f(t)\big{|}.$

Consequently,

$\sigma^{2}(s,t)\leq
C^{\prime}u(|t-s|,|t-s|)+2\left(f(s)\big{|}g(s)-g(t)\big{|}+g(t)\big{|}f(s)-f(t)\big{|}\right),$
(7.17)

and by (1.16),

$\displaystyle f(s)\big{|}g(s)-g(t)\big{|}+g(t)\big{|}f(s)-f(t)\big{|}$ (7.18)
$\displaystyle\qquad\leq|\mu|u(s,s)\big{|}g(s)-g(t)\big{|}+|\nu|u(t,t)\big{|}f(s)-f(t)\big{|}.$

It now follows by a minor generalization of Theorem 2.2 that,

$\limsup_{t\downarrow
0}\frac{X_{\alpha,2}(t)}{(u(t,t)+C^{\prime}u^{2}(t,t))\log\log 1/t}\leq
1,\qquad\text{a.s.},$ (7.19)

for some constant $C^{\prime}$. This implies that

$\limsup_{t\downarrow 0}\frac{X_{\alpha,2}(t)}{u(t,t)\log\log 1/t}\leq
1,\qquad\text{a.s.}$ (7.20)

The lower bound for $X_{\alpha,2}$ follows from a minor modification of the
proof of Theorem 1.2. Let $U_{f,g}$ be a non-singular $n\times n$ matrix,

$\displaystyle\left(\begin{array}[]{ cccc }1&f(1)&\ldots&f(n)\\\
g(1)&U_{1,1}+g(1)f(1)&\ldots&U_{1,n}+g(1)f(n)\\\
\vdots&\vdots&\ddots&\vdots\\\
g(n)&U_{n,1}+g(n)f(1)&\ldots&U_{n,n}+g(n)f(n)\end{array}\right).$ (7.25)

where $g$ is any function.

Let $U_{f,g}^{-1}$ denote the inverse of $U_{f,g}$ and $U_{f,g}^{j,k}$ denote
the elements of $U_{f,g}^{-1}$. One can check that,

$U_{f,g}^{-1}=\left(\begin{array}[]{ cccc
}1+\rho^{\prime}&-\sum_{i=1}^{n}f(i)U^{i,1}&\dots&-\sum_{i=1}^{n}f(i)U^{i,n}\\\
-\sum_{j=1}^{n}g(1)U^{1,j}&U^{1,1}&\dots&U^{1,n}\\\
\vdots&\vdots&\ddots&\vdots\\\
-\sum_{j=1}^{n}g(n)U^{n,j}&U^{n,1}&\dots&U^{n,n}\end{array}\right),$ (7.26)

where

$\rho^{\prime}=\sum_{j=1}^{n}\sum_{i=1}^{n}g(j)f(i)U^{i,j}.$ (7.27)

As in (2.8) we have

$U_{f,g}^{i,i}=U^{i-1,i-1},\qquad i=2,\ldots,n+1.$ (7.28)

The proof now proceeds like the proof of Theorem 1.2.

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