text:
Daya Is Looking For A 'Bad Girl?' Sign Us Up!
by Emily Defoor
Emily Defoor
Shungudzo Releases Video for 'To be me'
In 2016, Daya blew the world away and took the industry by storm with her hit single 'Sit Still, Look Pretty,' won a Grammy for the song, 'Don't Let Me Down' with The Chainsmokers, and released her debut album. After this, she took a critical step back to focus on personal growth and channel that into the art she wanted to create, on her terms and in her words.
Now, she's back! To kick off her 2021 music career, she's dropped a bombshell single titled, 'Bad Girl.' Daya uses the song to confront her sexuality and identity as a bisexual woman, the desires that have propelled and shaped her, and the stereotype of the term "Bad Girl" itself. It's a pretty fierce track, produced by Charlie Puth and Andrew Goldstein, but that's not our favorite thing about it! You can tell in Daya's voice how much this song means to her. She's having fun performing it, and we love to see the personal growth she HAS gone through over the last few years.
Not only do we love that Daya is living her best life on this track, but it's got indulgent, hypnotic beats that we can't seem to stop dancing to, along with dynamic vocals that make you feel good and some pretty intoxicating visuals. The music video shows Daya entering a bar where all eyes are on her, but she doesn't focus on any of the guys swooning over her, just the mysterious and striking lady across the bar. They lock eyes, and she subtly invites Daya to follow her out of the bar. We wish this was a movie so we could see how it plays out, but at least we can watch this video on repeat! Oh, and did we mention Daya's fire outfit choices? HOLY SMOKES! YAS QUEEN! Work it!
To me, the lyric of 'bad girl' can be used in an almost ironic way to play out the 'problem child' or archetypal rebel, flipping the script so that a 'bad girl' can be anyone who is confident and asserts themselves boldly in the world and doesn't necessarily have to adhere to the stereotypical bad girl 'look,' but rather can appear anywhere on the binary from leather jackets & boots to red lips and dresses and pearls, in the case of the song's music video.
Daya about 'Bad Girl'
We know this song will put Daya back on everyone's radar, and you can bet that we'll keep promoting her until she is appreciated the way she deserves! This song is just a small sneak peek of what's to come as she has a new EP coming out later this year, but until then, stream the song everywhere you go! It's time for Daya world domination!
What do you think of 'Bad Girl?' Are you glad Daya is back at it and making badass music? Are you stoked for her new EP? Let us know in the comments below or by tweeting us @TheHoneyPOP! You can also hit us up on Facebook and Instagram!
STAYC Makes Their Japanese Debut With 'POPPY'
Check out these other articles to quench your pop-culture thirst!
TO LEARN MORE ABOUT DAYA:
Featured Image Source: Clyde Munroe
Emily Defoor February 5, 2021


Our spiritual growth is similar to the patterns of our physical growth—it happens slowly over time so that we sometimes hardly notice it. Many times it's not until someone brings it up that we realize how much we've grown. To grow spiritually doesn't happen without some effort on our part in seeking God in prayer, Bible study, and fellowship. Our faith in God grows as we learn to trust Him even in the cloudy days of disappointment, despair, or uneasiness. What are the top 7 signs of growing closer to Christ?
Sign #1) You are less angry or irritated when things don't go your way.
The closer we get to Christ, the less control our flesh has on us. We learn to recognize the sins of selfishness that results in outbursts of anger or irritation. It's not that we don't ever get angry, it's rather we begin to harness our sinful nature and see the broader picture of how God uses adversity to help us grow up spiritually.
Sign #2) You are more likely to pray first before calling on anyone else.
"But I pray to you, Lord, in the time of your favor; in your great love, O God, answer me with your sure salvation" (Psalm 69:13).
As a new Christian, you may have called on your spiritual mentor regularly when faced with the enormity of keeping the faith. However, after a time—you're more compelled to fall on your knees to beseech your heavenly Father first. Our spiritual growth is marked by our willingness to trust God in prayer before we discuss our situations with others.
Sign #3) You are less anxious and worried over troubles.
"Be still before the Lord and wait patiently for him; do not fret when people succeed in their ways, when they carry out their wicked schemes" (Psalm 37:7).
A worrisome spirit can be tormenting for a believer. Either we completely trust God or not. When we've experienced God's deliverance, miracles, and intervention in the past, we learn to trust in His faithfulness and refuse to let anxiousness dominate our thoughts. Over time we learn the discipline of being still before God instead of being busy with worry.
Sign #4) You are more concerned about your own motives and integrity.
Spiritual maturity is most evident in those who take the time to reflect on their motives and actions in all situations. You've grown closer to Christ when you immediately respond to the conviction of the Holy Spirit. When we press the magnifying glass of judgment on ourselves, we are less judgmental and harsh towards other people.
Sign #5) You are less doubtful about God's control over your life.
"Lord, you are my God; I will exalt you and praise your name, for in perfect faithfulness you have done wonderful things, things planned long ago" (Isaiah 25:1).
No matter where we are in the faith—a new or mature believer, we will have days where we are perplexed by sudden events. These can be the most fertile times of doubt due to the hardship of the situation. However as we grow in our relationship with Jesus, we learn to accept that nothing can touch us that hasn't first cleared His divine purpose and plan for our lives.
Sign #6) You are more engaged in God's word.
"I will never forget your precepts, for by them you have preserved my life" (Psalm 119:93).
Unfortunately some believers' growth has been stunted due to their lack of engagement with the Bible. Yet for those who've tasted of God's goodness, revelation, and faithfulness through His word—it becomes as necessary as our natural food. We cling to the knowledge that God is not man that He should lie—He will keep His word.
Sign #7) You are less connected to the things of this world.
"Do not conform to the pattern of this world, but be transformed by the renewing of your mind. Then you will be able to test and approve what God's will is—his good, pleasing and perfect will" (Romans 12:2).
Many of us began our relationship with God with one foot still connected with the world. However over time a choice had to be made—either God or the world because we can't live a successful life being double minded. Our relationship with God becomes bonded and sealed with commitment to Him and a desire for purity.
Our continued faithfulness to God on a daily basis will consistently bring us closer to Christ. In our spiritual walk there will be testing times when we may question our faith in God or even God's faithfulness to us. However, when there is a solid relationship built with Christ—these fleeting moments of doubt will be overwhelmed by our faith in a loving God.
Crystal McDowell is a writer, speaker, and teacher with a passion to encourage believers to know and grow in their relationship with Jesus Christ through practical application of God's word.
Crystal and her husband, Marshall, raise their five children in the Bloomington, IL. As a freelance writer and editor for over 13 years, Crystal has published numerous Christian curriculums for Sunday School and VBS as well as many articles dealing with marriage, motherhood, and relationships. She's a weekly staff writer for www.whatchristianswanttoknow.com as well as writing her own blog at http://crystalmcdowellspeaks.blogspot.com.


text:
Q: Standard Matrix vs. Identity Matrix Is there a difference between identity and standard matrices? Or are they are just different notations to distinguish between regular matrices and linear transformations?

A: The $n\times n$ identity matrix is always $I=[\delta_{ij}]$ where $\delta_{ij}=\begin{cases} 1 & \text{if } i=j \\ 0 & \text{otherwise}\end{cases}$. For example the 3x3 identity matrix is $I=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
The standard matrix of a linear transformation $T:\mathbb{R}^n \to \mathbb{R}^m$ is an $m \times n$ matrix $A$ with the property that for all vectors $x$ evaluating the linear transformation is the same as multiplying by the matrix. That is $T(x)=Ax$.
So it is possible for an identity matrix to be a standard matrix, but it's not always the case.

redpajama_set_name: RedPajamaStackExchange


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text:
\section{Introduction}
\setcounter{equation}{0}
In this paper we present expansions of the form
\begin{equation}
\label{La_Fourche}
{\mathbf K}(t,s)=\sum_{n=1}^\infty K^n(t,s)K_n(t,s)
\end{equation}
where $K(t,s)$ and $K_n(t,s),\, n=1,2,\ldots$ are positive kernels
on the real line, for a family of kernels ${\mathbf K}(t,s)$ of the
the form
\begin{equation}
\label{Maubert_Mutualite}
{\mathbf K}(t,s)=\varphi(r(t)+r(s))-\varphi(r(t-s))
\end{equation}
with appropriate choices of functions $r$ and $\varphi$. The
functions $r$ we consider are real--valued and such that
$r(0)=0$, but it is
better to assume them complex--valued and with
a possibly non zero $r(0)$ at this stage of the
exposition. The case where $\varphi(x)=x$, corresponds to a class
of functions introduced by J. von Neumann and I. Schoenberg in
the late 1930s (see \cite{MR1503439}, \cite{MR0004644}), and M.G.
Krein  in the 1940s (see \cite{MR0012176} and the 1954 paper
\cite{krein_1954}; Krein's papers are reprinted in
\cite{MR1321817}). This class plays an important role both in analysis and
in the theory of stochastic processes. To recall the role of the
functions $r$ such that the kernel
\begin{equation}
\label{Duroc_Vaneau}
K_r(t,s)=r(t)+r(s)^*-r(t-s)-r(0),
\end{equation}
is positive on the real line, we first give a definition: A
{\em helical arc} is a continuous map $t\mapsto \xi_t$ from the real
line into a metric space (with metric $d$) such that
$d(\xi_t,\xi_s)$ is a function of $t-s$:
\begin{equation}
\label{160407} d(\xi_t,\xi_s)=
\rho(t-s).
\end{equation}
When the metric space is a {\sl real} Hilbert space,
it is a consequence
of a result of K. Menger (see \cite{menger}) that a function
$\rho$ satisfies \eqref{160407} if and only if the kernel
\[
K_{\rho^2}(t,s)=\rho^2(t)+\rho^2(s)-\rho^2(t-s)
\]
is positive on the real line; see
\cite[Lemma 1, p. 229]{MR0004644}. Using this fact,
von Neumann and Schoenberg
give in \cite[Theorem 1, p. 229]{MR0004644}  an integral
representation of the functions $\rho$ corresponding to a helical arc
with values in a real Hilbert space. These are the
functions $\rho$ such that:
\begin{equation}
\label{eureka} \rho^2(t)=\int_0^\infty
\frac{\sin^2tu}{u^2}d\sigma(u),
\end{equation}
where the positive real measure $\sigma$ is such that
\[
\int_1^\infty\frac{d\sigma(u)}{u^2}<\infty.
\]
We note the formula (see \cite[(1.6) p. 229]{MR0004644})

\[
\rho^2(t)+\rho^2(s)-\rho^2(t-s)=\int_0^\infty\frac{
(1-\cos2tu)(1-\cos2su)+\sin2tu\sin 2su}{2u^2}d\sigma(u)
\]
and hence the kernel $\rho^2(t)+\rho^2(s)-\rho^2(t-s)$, which is
of the form  \eqref{Duroc_Vaneau}, is positive on the real line.
The result of von Neumann and Schoenberg expresses in particular
that the converse is true: if $r$ is an even real-valued function
such that $r(0)=0$ and $K_r(t,s)$ is positive on the real line, then
$r$ is
of the form \eqref{eureka}.\\

The motivation of von Neumann and Schoenberg
comes from imbedding problems of metric spaces into Hilbert
spaces; we will not address this question here, but refer the
interested reader to
\cite[pp.
81--84]{MR86b:43001} for further information.\\

In \cite{MR0012176}, Krein considers helical arcs defined on an
open interval $(a,b)\subset{\mathbb R}$, with values in a {\sl
complex} Hilbert space. Condition \eqref{160407} is replaced by
the condition on the inner product, that is, by the condition that the
function
\[
B(t,s)=\langle \xi_{s+v}-\xi_v,\xi_{t+v}-\xi_v\rangle_{\mathcal
H}\]
does not depend on $v$. Krein states that a function $B(t,s)$
corresponds to a helical arc defined on the real line if and only
if it is of the form
\begin{equation}
B(t,s)=\int_{\mathbb R}
\frac{e^{itu}-1}{u}\frac{e^{-isu}-1}{u}dm(u)
\end{equation}
for an appropriate increasing function $m$; see \cite[Theorem
4]{MR0012176} and \cite[p. 115]{MR1321817}. One can then take
\[
\xi_t(u)=e^{itu}\quad{\rm and}\quad {\mathcal H}= {\mathbf
L}_2\left(\frac{dm(u)}{u^2}\right).
\]

In particular,
$B(t,s)$ defines a positive kernel on ${\mathbb R}$. Setting
\[
g(t)=i\gamma t+\int_{\mathbb
R}\left\{e^{itu}-1-\frac{itu}{u^2+1}\right\}\frac{dm(u)}{u^2}
\]
one has
\[
B(t,s)=g(t-s)-g(s)-g(s)^*.
\]
It follows (see also the paper \cite[equations (11.16) and
(11.17), p. 402]{KL_jot} of M.G. Krein and H. Langer and the
preprint \cite{denisov}) that a function $r(t)$, satisfying to
$r(-t)=r(t)^*$, is such that the kernel $K_r(t,s)$ in
\eqref{Duroc_Vaneau} is positive on the real line if and only if
it can be written as
\begin{equation}
\label{babylone}
r(t)=r_0+i\gamma t-\int_{\mathbb
R}\left\{e^{itu}-1-\frac{itu}{u^2+1}\right\}\frac{dm(u)}{u^2}
\end{equation}
where $r_0=r(0)$ and $\gamma$ are real numbers and where $m$ is a
positive measure on ${\mathbb R}$ such that
$$\int_{\mathbb
R}\frac{dm(u)}{u^2+1}<\infty.$$
Furthermore, equation \eqref{babylone} implies the representation
of the kernel
\begin{equation}
\label{marseille}
r(t)+r(s)^*-r(t-s)-r(0)=\int_{\mathbb
R}\frac{e^{itu}-1}{u}\frac{e^{-isu}-1}{u}dm(u)
\end{equation}
as an inner product.
See \cite[equation (11.17) p. 402]{KL_jot}.\\

Since $m$ is positive and hence real valued, we have:
\begin{equation}
\begin{split}
{\rm Re}\left\{r(t)\right\}&=\int_{\mathbb R} \dfrac{1-\cos(tu)}{u^2}dm(u)\\
{\rm Im}\left\{r(t)\right\}&=\int_{\mathbb R}
\left\{-\sin(tu)+\dfrac{tu}{u^2+1}\right\}dm(u)
\end{split}
\end{equation}
In particular, when the measure is even and $r(0)=0$, one has
\[
r(t)=\int_{\mathbb R} \dfrac{1-\cos(tu)}{u^2}dm(u).\]
With the equality $1-\cos 2u=2\sin^2 u$, and a change of variables,
one then recognizes formula \eqref{eureka} of von
Neumann and Schoenberg. Note also that the
kernel $K_r(t,s)$ takes the simpler form
\begin{equation}
\label{mathilde_est_revenue}
K_r(t,s)=r(t)+r(s)-r(t-s).
\end{equation}

The choice where $dm(u)=\frac{1}{\pi}|u|^{1-2H}du$ with
$H\in(0,1)$, corresponds to covariance function of the fractional
Brownian motion
with Hurst parameter $H$. Indeed, we have the following computation:
\[
\int_0^\infty\dfrac{1-\cos(tu)}{u^2}|u|^{1-2H}
du=\int_0^\infty\dfrac{1-\cos
y}{y^2t^{-2}} y^{1-2H} t^{-(1-2H)}\dfrac{dy}{t}\\
=c({1-2H})t^{2H},
\]
where
\[
c({1-2H})=\int_0^\infty (1-\cos
y)y^{-1-2H}dy=\dfrac{1}{(1-2H){2H}}\int_0^\infty y^{1-2H}\cos y
dy,
\]
where we have assumed that $H\not=1/2$, and where we have
integrated twice by parts. The last integral is computed by
integrating the function
\[f(z)=e^{iz}e^{(1-2H) \ln z},\]
where $\ln z$ is the principal value of the logarithm in ${\mathbb
C}\setminus{\mathbb R}_-$, along an appropriate contour and is
found equal to
\[
\Gamma(2-2H)\cos(H\pi),
\]
where $\Gamma$ denotes Euler's Gamma function.
Thus, \(r(t)=\dfrac{V_H}{2}|t|^{2H}\), where
\[
V_H=\dfrac{\Gamma(2-2H)\cos(H\pi)}{\pi (1-2H)H}.
\]

The function
\begin{equation}
\label{ioanna}
K_{H,\alpha}(t,s)=(|t|^{2H}+|s|^{2H})^\alpha-|t-s|^{2H\alpha},\
\end{equation}
with
\[
0<\alpha\le 1\quad{\rm and}\quad 0<H< 1,
\]
is a special case of \eqref{Maubert_Mutualite} with
$r(t)=|t|^{2H}$ and $\varphi(x)=x^\alpha$. It was introduced in
\cite[Exercise 2.12 (h) p. 79]{MR86b:43001}, is positive in
${\mathbb R}$, and, by Kolmogorov's theorem, may assume the role
of the covariance function of a zero mean Gaussian stochastic
process. This process was introduced and first studied by Houdr\'e
and Villa in \cite{MR2037165}, who coined for it the term {\sl
bi--fractional Brownian motion}. The classical Brownian motion
corresponds to the choice $\alpha=1$ and $H=1/2$ while the choice
$\alpha=1$ and $H<1$ ($H\not =1/2$) corresponds, up to a
multiplicative constant factor, to the covariance function of the
fractional Brownian motion. For $\alpha<1$, one can view this
kernel as a generalization of the covariance function of the
fractional Brownian motion. It has been recently the topic of a
number of
papers; see \cite{MR2037165}, \cite{russo_tudor}.\\

It is well known that the reproducing kernel space, associated
with the covariance function of the Brownian motion, is the
Sobolev space of functions absolutely continuous on the real line
and with square summable derivatives; see \cite[p. 25]{Malliavin},
\cite{MR99f:60082}. The reproducing kernel Hilbert space
associated with the covariance function of the fractional Brownian
motion has also been characterized in a number of places; see for
instance \cite[Theorem 4.1 p. 948]{fbm88}.\\

We note that Schoenberg \cite[(4), p. 788]{MR1503370} also considers the
multidimensional kernel case.\\

We now turn to the outline of the paper. The paper consists of
five sections besides the introduction. Sections 2 and 3 are of a
review nature. The main results appear in Sections 4,5 and 6. In
Section 2 we recall some of M.G. Krein's results on positive
kernels of the form \eqref{Duroc_Vaneau} and of the associated
reproducing kernel Hilbert spaces. These results have been
rediscovered within the setting of the fractional Brownian motion
a number of times without reference to Krein's work. In the third
section we recall several facts on integrals and products of
positive kernels and of the associated reproducing Hilbert kernel
spaces. Section 4 contains two proofs of the positivity of the
kernel of the fractional Brownian motion. The first is valid for
any $H\in(0,1)$ while the second one is valid only for
$H\in(0,1/2)$.
In both instances, the positivity of
the kernel is proved by exhibiting it as a sum of positive
kernels of the form \eqref{La_Fourche}; see formulas
\eqref{total-recall-nice-1991} and \eqref{ohohoh} below. In
Section 5 we consider the case of the bi-fractional Brownian
motion and associate to its covariance function an expansion of
the form \eqref{La_Fourche}. In Section 6 we study some nonlinear
transforms associated with the expansions \eqref{La_Fourche}.\\

A note on terminology: Throughout we use the terms
positive kernels, positive functions, covariance kernels or
covariance functions interchangeably.

\section{The reproducing kernel Hilbert space ${\mathcal H}(K_r)$}
\setcounter{equation}{0}
Recall that a function $K(t,s)$ defined on a set $\Omega$ is said
to be positive if it is hermitian:
\[K(t,s)=K(s,t)^*,\quad \forall t,s\in \Omega,\]
and if, for every integer $n$ and every choice of points
$t_1,\ldots, t_n\in\Omega$, the $n\times n$ matrix with $\ell j$
entry $K(t_\ell,t_j)$ is nonnegative. In preparation for the
forthcoming sections (see in particular formula \eqref{St
Philippe du Roule, ligne 9} below), one also may recall that sums
and products of positive functions are still positive functions.
The Aronszajn--Moore theorem states that a function $K(t,s)$ is
positive on a set $\Omega$ if and only if it can be written as
\begin{equation}
\label{zxcvbnm}
K(t,s)=\langle f_s,f_t\rangle_{\mathcal C},
\end{equation}
where ${\mathcal C}$ is a auxiliary Hilbert space and where
$s\mapsto f_s$ is a function from $\Omega$ into ${\mathcal C}$.
The space ${\mathcal C}$ can be chosen to be the reproducing
kernel Hilbert space associated with $K(t,s)$. This is a Hilbert
space of functions on $\Omega$, denoted by ${\mathcal H}(K)$,
uniquely determined by the following two conditions: $(a)$ For
every $s\in\Omega$ the function
\[K_s\,\,:\,\, u\mapsto K(u,s)\]
belongs to ${\mathcal H}(K)$, and: $(b)$ for every $s\in\Omega$
and $F\in{\mathcal H}(K)$
\begin{equation}
\langle F, K_s\rangle_{{\mathcal H}(K)}=F(s). \label{rk_pr}
\end{equation}

Setting $f_s=K_s$ in \eqref{rk_pr}, leads to the factorization
\eqref{zxcvbnm}.\\

If $K$is positive on $\Omega$ so is
the function $mK$, where $m$ is a strictly positive function. We
recall that
\begin{equation}
\label{poisse}
\|F\|^2_{{\mathcal H}(K)}=
m\|F\|^2_{{\mathcal H}(mK)}.
\end{equation}
For future reference we now recall the
characterization of the reproducing kernel space when a
representation \eqref{zxcvbnm} is available.
\begin{La}
\label{Vaugirard}
Let $K(t,s)$ be of the form \eqref{zxcvbnm} and let ${\mathcal
C}(f)$ be the closed linear span in ${\mathcal C}$ of the functions
$f_s$ where $s$ runs throughout $\Omega$. Then the reproducing kernel
Hilbert space with reproducing kernel $K(t,s)$ is the set of
functions of the form
\[
F(t)=\langle x, f_t\rangle_{\mathcal C},\quad x\in{\mathcal C}(f)
\]
with norm
\[
\|F\|_{{\mathcal H}(K)}=\|x\|_{\mathcal C}.
\]
\end{La}

The next theorem characterizes the reproducing kernel Hilbert
space associated with $K_r(t,s)$. It is a direct consequence of
formula \eqref{marseille} and Lemma \ref{Vaugirard}.

\begin{Tm}
\label{totoche}
The reproducing kernel Hilbert space ${\mathcal H}(K_r)$
associated with $K_r(t,s)$ consists of functions of the form
\begin{equation}
{\mathcal H}(K_r)=\left\{\, F\,:\, F(t)=\int_{\mathbb
R}\dfrac{e^{itu}-1}{u}f(u)dm(u),\right\}
\label{St-Michel}
\end{equation}
where $f$ is in the closed linear span in ${\mathbf L}_2(dm)$ of
the functions
\begin{equation}
\chi_s(u)=\frac{e^{isu}-1}{u},\quad s\in{\mathbb R},
\label{haifa}
\end{equation}
with the norm
\begin{equation}
\label{napoleon} \|F\|_{{\mathcal H}(K_r)}=\|f\|_{{\mathbf
L}_2(dm)}.
\end{equation}
\label{attention-les-yeux}
\end{Tm}

Theorem \ref{totoche} is proved through different methods
in \cite[Theorem 4.1 p.
948]{fbm88} for $dm(u)=\frac{1}{\pi}|u|^{1-2H}du$ with
$H\in(1/2,1)$. See also \cite{MR1280932},
\cite{MR2145497}. We note that in the case $H\in(1/2,1)$ the
covariance kernel can also be expressed in terms of double
integrals as follows: one has
\[
|t|^{2H}+|s|^{2H}-|t-s|^{2H}=2H(2H-1)\int_0^t\int_0^s
|u-v|^{2H-2}dudv
\]
and the associated reproducing kernel Hilbert space can be
expressed in terms of the preHilbert space of functions $f$ such
that
\[
\iint_{{\mathbb R}_+^2}f(u)f(v)|u-v|^{2H-2}dudv<\infty,
\]
See \cite{MR1801485}, \cite{duncan1}.
This last space is not complete; see \cite[p. 270]{MR1790083}.\\

{\bf Remarks:}\\
$(a)$ In \eqref{St-Michel}, one can take $f\in{\mathbf L}_2(m)$,
rather than in the stated closed linear span; one then looses in
general the uniqueness of $f$ in the representation of $F$ and one
has to change \eqref{napoleon} to $\inf \|f\|_{{\mathbf
L}_2(m)}$, where the $\inf$ is over all $f$ corresponding to the
given function
$F$.\\
$(b)$ An important case is when $dm$ is absolutely continuous with
respect to Lebesgue measure and when moreover its derivative
satisfies
\[
\int_{\mathbb R}\dfrac{\ln m^\prime(u)}{1+u^2}du>-\infty
\]
Then $m^\prime$ admits an outer factorization $m^\prime=|h|^2$;
see \cite{Dmk}.\\
$(c)$ For any $m$ such that the space
${\mathbf L}_2(dm)$ is infinite dimensional and any $T>0$, the
closed linear span of the functions $\chi_t$ (defined by
\eqref{haifa}) for $|t|\le T$ has a special structure. It is a
reproducing kernel space of {\it entire functions} of the kind
introduced by L. de Branges; see \cite{MR38:6386}, \cite{dbbook},
\cite{dym-70}, \cite{Dmk}. Its reproducing kernel is of the form
\[
\frac{a(z)a(w)^*-b(z)b(w)^*}{-i(z-w^*)}
\]
where the functions $a$ and $b$ are entire functions of finite
exponential type. In the case of the fractional Brownian motion,
these are the homogeneous de Branges spaces of entire functions
(see \cite[pp. 184-189]{dbbook}). The functions $a$ and $b$ are
then expressed in terms of Bessel functions, a fact already
observed (in a slightly
different language) by Krein in \cite{krein_1954}. See also
\cite{MR2145497}. We refer to \cite{ad1} for more information
on the reproducing kernel spaces of the kind introduced by
de Branges and Rovnyak.\\
$(d)$ To prove that the kernels $K_{H,1}$ are positive, one can
also use the theory of negative kernels (see \cite{MR86b:43001}).
Another proof can be found in \cite[p. 209]{Mallat-francais}. The
associated reproducing kernel Hilbert space is given in for
instance in \cite{MR0242247} (with different representations of the kernels
corresponding to  $H<1/2$ and $H>1/2$; see
\cite[p. 136]{MR0242247}), in \cite[p. 319]{berlinet}, and can
also be deduced from the model for the fractional Brownian motion
given in e.g. \cite[p. 50]{Lifshits95}, \cite[\S 7.2 p.
318]{MR1280932}, \cite{benassi_cohen}. A characterization as
operator range can be deduced from \cite[Lemma 3.1 p. 182]{DS}).
When $H=1/2$, it is well known that the associated reproducing
kernel Hilbert space is the Sobolev space of functions absolutely
continuous on the real line and with square summable derivatives;
see \cite[p.
25]{Malliavin}, \cite{MR99f:60082}.\\

We now study a relation between the functions
$f$ and $F$ of \eqref{St-Michel}. We restrict ourselves to the case
where $m$ is
absolutely continuous with respect to Lebesgue measure, and
resort to the theory of distributions, as we now explain. Denote
by ${\mathcal S}$ the space of rapidly decreasing functions. These
are the infinitely differentiable functions $\sigma(u)$ such that
for every choice of $(k,r)\in{\mathbb N}^2$,
\[
\sup_{u\in{\mathbb R}}|(1+u^2)^r\sigma^{(k)}(u)|<\infty.
\]
We refer the reader to \cite{Schwartz66} and \cite{Treves67} for
more information on this space and in particular on its topology.
We denote by ${\mathcal S}^\prime$ the set of continuous linear
functionals on ${\mathcal S}$. The elements of ${\mathcal
S}^\prime$ are called {\it tempered distributions}.

\begin{Tm}
A function $F$ of the form \eqref{St-Michel}, defines an element
in ${\mathcal S}^\prime$ whose derivative is the Fourier
transform of the distribution
\begin{equation}
\label{Chevaleret} T_{m^\prime f}\sigma=\int_{\mathbb
R}\sigma(u)m^\prime(u)f(u)du.
\end{equation}
\end{Tm}

{\bf Proof:} We first note that the function $m^\prime f$ indeed
defines an element in ${\mathcal S}^\prime$ via the formula
\eqref{Chevaleret} since, for every $\sigma\in{\mathcal S}$, the integral
\[
\begin{split}
\int_{\mathbb R}m^\prime(u)f(u){\sigma}(u)du&=\int_{\mathbb
R}\dfrac{m^\prime(u)}{(1+u^2)^\ell}f(u)(1+u^2)^\ell{\sigma}(u)du,
\end{split}
\]
exists,
where $\ell\in{\mathbb N}$ is such that $(1+u^2)^\ell \sigma(u)$ is
bounded on the real line.
Using Fubini's theorem and integration by parts, we have for
${\sigma}\in{\mathcal S}$
\[
\begin{split}
D_F{\sigma}&=\int_{\mathbb R}{\sigma}^{\prime}(t)F(t)dt\\
&=\int_{\mathbb R}{\sigma}^\prime(t)\left(\int_{\mathbb
R}\dfrac{e^{itu}-1}{u}m^\prime(u)f(u)du\right)dt\\
&=\int_{\mathbb R}\left(\int_{\mathbb
R}{\sigma}^\prime(t)\dfrac{e^{itu}-1}{u}dt\right)m^\prime(u)f(u)du\\
&=-\int_{\mathbb R}\widetilde{{\sigma}}(u)m^\prime(u)f(u)du,
\end{split}
\]
where we denote by $\widetilde{\sigma}$ the Fourier transform
\[
\widetilde{\sigma}(w)=\int_{\mathbb R}\sigma(t)e^{itw}dt.
\]

This last equality expresses the fact that
$\widetilde{D_F}=T_{m^\prime f}$, and that the distributional
derivative of $F$ is the distributional Fourier transform of
$m^\prime f$.
\mbox{}\qed\mbox{}\\

Let us now restrict the above results to $dm(u)=du$. Then, we note
that
\[
\int_0^\infty\dfrac{1-\cos(tu)}{u^2}\,du=\dfrac{\pi|t|}{2}
\]
(as is seen by integrating the function
$f(z)=\frac{1-e^{itz}}{z^2}$ around an appropriate contour), and
so
\[
\begin{split}
\dfrac{1}{2}\left\{|t|+|s|-|t-s|\right\}&=\dfrac{1}{\pi}\int_0^\infty
\dfrac{\cos((t-s)u)-\cos(tu)-\cos(su)+1}{u^2}\,du\\
&={\rm Re}~\dfrac{1}{2\pi}\int_{\mathbb
R}\dfrac{e^{i(t-s)u}-e^{itu}-e^{-isu}+1}{u^2}du\\
&=\dfrac{1}{2\pi}\int_{\mathbb
R}\dfrac{e^{i(t-s)u}-e^{itu}-e^{-isu}+1}{u^2}du\\
&=\dfrac{1}{2\pi}\int_{\mathbb R}
\dfrac{e^{itu}-1}{u}\dfrac{e^{-isu}-1}{u}\,du\\
&=\frac{1}{2\pi}\langle \chi_t,\chi_s\rangle_{{\mathbf L}_2(du)}.
\end{split}
\]
The closed linear span in ${\mathbf L}_2(du)$ of the functions
$\chi_t$ ($t\in{\mathbb R}$) is equal to ${\mathbf L}_2(du)$, and
the reproducing kernel Hilbert space associated with the kernel
$\frac{1}{2}\left\{|t|+|s|-|t-s|\right\}$ is equal to the set of
functions $F$ of the form
\[
F(t)=\int_{\mathbb R}\dfrac{e^{itu}-1}{u}f(u)du,
\]
where $f\in{\mathbf L}_2(du)$, with the norm
$\|F\|=\frac{1}{\sqrt{2\pi}}
\|f\|_{{\mathbf L}_2(du)}$. This space is known to be the
Sobolev space. We regenerate this result as follows. For
$f\in{\mathbf L}_1(du)\cap{\mathbf L}_2(du)$, Fubini's theorem
shows that
\begin{equation}
\label{2.5}
F(t)=\int_{\mathbb R}\dfrac{e^{itu}-1}{u}f(u)du
=\int_0^t \left(\int_{\mathbb R}e^{ius}f(u)du\right)ds
=\int_0^t\widetilde{f}(s)ds,
\end{equation}
and
\[
\|F\|=\frac{1}{\sqrt{2\pi}}\|f\|_{{\mathbf L}_2({\mathbb R})}=
\|\widetilde{f}\|_{{\mathbf L}_2({\mathbb R})}.
\]
The result for general $f\in{\mathbf L}_2(du)$
follows by approximation. If we restrict $t,s\in[0,T]$ for some
$T>0$, then $f$ belongs to the closed linear span in ${\mathbf
  L}_2({\mathbb R})$ of the functions $\chi_s$, $s\in[-t,0]$. Thus
$f(t)=\int_0^T e^{-iut}g(u)du$ for some
$g\in{\mathbf L}_2([0,T])$,
and we have
\[F(t)=\int_0^t\left(\int_{\mathbb
  R}e^{isx}\left(\int_0^Te^{-ixu}g(u)du\right)
dx\right)ds=\int_0^tg(u)du.\]

\section{Integrals and products of positive functions}
\label{clementine}
\setcounter{equation}{0}
In this section we briefly review a number of results related to
reproducing kernel Hilbert spaces. The results are stated for
functions positive on the real line because this is the setting
we consider, but are valid for functions positive on any set. The
first result is in Schwartz's paper \cite[p. 170]{schwartz}. We
outline the proof in our special case for completeness.

\begin{Tm}
Let $k_u(t,s)=k(t,s,u)$ be a continuous function on ${\mathbb
R}^2\times{\mathbb R}_+$ such that for every $u\in{\mathbb R}_+$
the kernel $k(t,s,u)$ is positive on ${\mathbb R}$, and let $q$ be a
function positive and increasing on ${\mathbb R}_+$. Assume that
$\int_0^\infty k(t,t,u)dq(u)<\infty$.
Then the function
\begin{equation}
\mathbf k(t,s)=\int_0^\infty k(t,s,u)dq(u)
\label{Temple, ligne 3}
\end{equation}
is positive on the real line and
its reproducing kernel Hilbert space
is the Hilbert space $\int_0^\infty{\mathcal
H}(k_u)dq(u)$.
\end{Tm}
{\bf Proof:} First we note that the function
\eqref{Temple, ligne 3} is well defined. Indeed, from the
positivity of $k_u$ we get that
\[
|k(t,s,u)|^2\le k(t,t,u)k(s,s,u)
\]
and hence by Cauchy-Schwarz inequality we have
\[
\begin{split}
|\int_0^\infty k(t,s,u)dq(u)|&\le\int_0^\infty
k(t,t,u)^{1/2}k(s,s,u)^{1/2}dq(u)\\
&\le\left(\int_0^\infty
k(t,t,u)dq(u)\right)^{1/2}\left(\int_0^\infty
k(s,s,u)dq(u)\right)^{1/2}<\infty.
\end{split}
\]
Consider now the vector space ${\mathcal M}(\mathbf k)$ spanned by the
finite linear combinations of the functions ${\mathbf k}(t,s)$,
with inner product
\[
\langle {\mathbf k}(\cdot, w),{\mathbf k}(\cdot, v)
\rangle_{{\mathcal M}({\mathbf k})}={\mathbf k}(v,w).
\]
An element $F\in{\mathcal M}({\mathbf k})$ can thus be written as
a finite sum
\[
F(t)=\sum_{\ell=1}^m \mathbf k(t,s_\ell)c_\ell=\int_0^\infty
\left(\sum_{\ell=1}^m k(t,s_\ell,u)c_\ell\right)dq(u),\quad
c_\ell\in{\mathbb C},\]
with norm
\[
\|F\|^2_{{\mathcal M}({\mathbf k})}=\int_0^\infty\|
\sum_{\ell=1}^m k(\cdot,s_\ell,u)c_\ell\|^2_{{\mathcal
H}(k_u)}dq(u).
\]
The space ${\mathcal H}({\mathbf k})$ is on the one hand the
completion of ${\mathcal M}({\mathbf k})$ with respect to this
inner product. By \cite[\S5, p. 146]{MR0352996}, this is also the
description of the space $\int_0^\infty {\mathcal H}(k_u)dq(u)$.
By uniqueness of the completion we have
the result. \qed\mbox{}\\

Elements of the space $\int_0^\infty {\mathcal H}(k_u)dq(u)$ can be
described as follows (see \cite[p. 170]{schwartz}, \cite[pp. 77 and
96]{saitoh}): A function $F$ belongs to ${\mathcal
H}({\mathbf k})$ if and only if it can be written as
\begin{equation}
\label{La_Muette} F(t)=\int_0^\infty x(t,u)dq(u)
\end{equation}
where for every $u$, the function $x_u\,:\,t\mapsto
x(t,u)\in{\mathcal H}(k_u)$ and the map $u\mapsto x_u$ is weakly
measurable. Moreover,
\begin{equation}
\label{perlette} \|F\|_{{\mathcal H}({\mathbf
k})}^2=\inf\int_0^\infty \|x_u\|^2_{{\mathcal H}(k_u)}dq(u)
\end{equation}
where the infimum is over all the representations of a given $F$
in \eqref{La_Muette}.\\

We refer the reader to \cite{MR0044536}, \cite{MR0056611},
\cite{MR0029101} and \cite[pp. 139--161]{MR0352996} for more
information on integrals of Hilbert spaces.\\

As a special case we have:

\begin{Tm}
\label{Chevaleret, ligne 6} Let $k_n(t,s)$, $n=1,2,\ldots$ be
positive functions on ${\mathbb R}$ and let
\[
\mathbf k(t,s)=\sum_{n=1}^\infty k_n(t,s).
\]
Assume that $\mathbf k(t,t)<\infty$ for all $t\in{\mathbb R}$.
 Then
\begin{equation}
\label{helene}
{\mathcal H}(\mathbf k)=\sum_{n=1}^\infty{\mathcal H}(k_n)
\end{equation}
in the following sense: ${\mathcal H}(\mathbf k)$ consists of
the functions which can be written as
\begin{equation}
\label{Rue du Bac} f(t)=\sum_{n=1}^\infty f_n(t),
\end{equation}
where $f_n\in{\mathcal H}(k_n)$ and
\begin{equation}
\sum_{n=1}^\infty\|f_n\|^2_{{\mathcal H}(k_n)}<\infty.
\label{Rue_du_Bac2}
\end{equation}
Moreover, we have:
\begin{equation}
\label{quartier_latin}
\|f\|_{{\mathcal H}(\mathbf
k)}^2=\inf\sum_{n=1}^\infty\|f_n\|^2_{{\mathcal H}(k_n)},
\end{equation}
where the $\inf$ is over all the possible decompositions
\eqref{Rue du Bac} satisfying \eqref{Rue_du_Bac2}.
\end{Tm}

We note that the infimum is in fact a minimum and is achieved for
the choice
\[
f_n=i_ni_n^*f
\]
where $i_n$ is the inclusion map $f\mapsto f$ from ${\mathcal
H}(k_n)$ into ${\mathcal H}(\mathbf k)$. See \cite{aron}, \cite{saitoh}.\\

The sum \eqref{helene} is in general not direct, and the spaces
are called complementary; see \cite{dbjfa} for more information on
complementation theory.\\

The following auxiliary result will be instrumental in the
following sections. Note that the result is stated in a form
convenient for our present purpose.

\begin{Tm}
\label{rosemary}
 Let $k_n(t,s,u), n=1,2,\ldots$ be a sequence of continuous functions
defined on ${\mathbb R}^2\times {\mathbb R}_+$ and assume that
for every $u\in{\mathbb R}_+$ and every positive integer $n$ the
function $k_n(t,s,u)$ is positive on ${\mathbb R}$. Assume that
\[
\sum_{n=1}^\infty k_n(t,s,u)
\]
is continuous on ${\mathbb R}^2\times {\mathbb R}_+$ and that for
every $t\in{\mathbb R}$
\[
\int_0^\infty\left(\sum_{n=1}^\infty k_n(t,t,u)\right)du<\infty
\]
Then,
\[
\int_0^\infty \left(\sum_{n=1}^\infty k_n(t,s,u)\right)du=
\sum_{n=1}^\infty\int_0^\infty k_n(t,s,u)du.
\]
\end{Tm}
{\bf Proof:} For every $u\in{\mathbb R}_+$ and $n\in{\mathbb
N}^*$ we have
\[
|k_n(t,s,u)|^2\le k_n(t,t,u)k_n(s,s,u),\]
and so,
\[
\begin{split}
|\sum_{n=1}^\infty k_n(t,s,u)|&\le\sum_{n=1}^\infty
 \sqrt{k_n(t,t,u)}\sqrt{k_n(s,s,u)}\\
 &\le\left(\sum_{n=1}^\infty k_n(t,t,u)\right)^{1/2}\left(\sum_{n=1}^\infty
 k_n(s,s,u)\right)^{1/2}
\end{split}
\]
and the function $u\mapsto \sum_{n=1}^\infty k_n(t,s,u)$ is absolutely
summable for every choice of $t,s\in{\mathbb R}$.
The result is then a direct consequence of the dominated
convergence theorem.\mbox{}\qed\mbox{}\\

The next theorem is classical; see for instance
\cite{aron},\cite{saitoh}.
\begin{Tm}
Let $k_1$ and $k_2$ be two functions positive on the real line.
Then the function $k_1k_2$ is also positive on the real line.
Assume moreover that the functions $k_1$ and $k_2$ are continuous.
Then the Hilbert spaces ${\mathcal H}(k_1)$ and ${\mathcal
H}(k_2)$ are separable. The space ${\mathcal H}(k_1k_2)$ consists
of the restrictions on the diagonal of the elements of tensor
product ${\mathcal H}(k_1)\otimes{\mathcal H}(k_2)$. Let
$(f_n)_{n=0,\ldots}$ and $(g_n)_{n=0,\ldots}$ be Hilbert space
basis of the spaces ${\mathcal H}(k_1)$ and ${\mathcal H}(k_2)$
respectively. Then ${\mathcal H}(k_1k_2)$ consists of the
functions of the form
\[
F(t)=\sum_{n,m=0}^\infty c_{n,m}f_n(t)g_m(t)\]
with
\[
\|F\|^2_{{\mathcal H}(k_1k_2)}=\sum_{n,m=0}^\infty |c_{n,m}|^2.
\]
\end{Tm}
From this result we obtain that $f_1f_2\in{\mathcal H}(k_1k_2)$ where
$f_\ell\in{\mathcal H}(k_\ell)$, $\ell=1,2$, and
\begin{equation}
\label{saitoh_1}
\|f_1f_2\|_{{\mathcal H}(k_1k_2)}\le \|f_1\|_{{\mathcal
H}(k_1)}\|f_2\|_{{\mathcal H}(k_2)}\quad\mbox{{\rm and that}}\quad
\|f^n\|^2_{{\mathcal H}(k^n)}\le \|f\|^{2n}_{{\mathcal H}(k)}
\end{equation}
where $k$ is a positive kernel. See \cite[p 243]{saitoh2} for the
second inequality.
\section{Positivity of the kernels $K_{H,1}(t,s)$}
\label{Avenue Foch}
\setcounter{equation}{0}
In  this section we present two decompositions
of the form \eqref{La_Fourche} of the kernels
$K_{H,1}(t,s)$, defined in \eqref{ioanna}. The first is valid
for any $H\in(0,1)$ while the second is valid only for
$H\in(0,1/2)$.

\begin{Tm} Let $0<H<1$. \\
$(a)$ It holds that
\begin{equation}
\label{total-recall-nice-1991}
\begin{split}
\frac{\Gamma(1-H)}{2H}
(|t|^{2H}+|s|^{2H}-|t-s|^{2H})&=\\
&\hspace{-2.5cm} =\int_0^\infty
\frac{(1-e^{-u^2t^2})(1-e^{-u^2s^2})}{u^{1+{2H}}}du+\\
&\hspace{-2cm}+
 \sum_{n=1}^\infty
\frac{2^{n-1}\Gamma(n-H)}{n!}\frac{t^ns^n}{(t^2+s^2)^{n-H}}.
\end{split}
\end{equation}
$(b)$A function $F$ belongs to ${\mathcal
H}(\frac{2H}{\Gamma(1-H)}K_{H,1})$ if and only if it can be
written as
\[
F(t)=\sum_{n=0}^\infty F_n(t)
\]
with
\[
F_0(t)=\int_0^\infty \frac{(1-e^{-u^2t^2})x_0(u)}{u^{1+2H}}du,
\]
where $x_0$ belongs to the closed linear span of the functions
$1-e^{-u^2s^2}$ ($s\in{\mathbb R}$) in ${\mathbf
L}_2(1/u^{1+{2H}})$, and where for every $n\ge 0$,
\[
F_n(t)=t^n\int_0^\infty e^{-u^2t^2}x_n(u)u^{2n-1-2H}du,\] where
$x_n$ belongs to the closed linear span of the functions
$e^{-u^2s^2}$ ($s\in{\mathbb R}$) in ${\mathbf
L}_2(u^{2n-1-{2H}})$. For any such decomposition,
\begin{equation}
\label{Porte d'Orleans, ligne 4}
\|F\|^2\le\sum_{n=0}^\infty\frac{n!}{2^n}\|x_n\|^2_{{\mathbf
L}^2(u^{2n-1-2H})},
\end{equation}
and there is a unique decomposition for which equality hold in
\eqref{Porte d'Orleans, ligne 4}.
\end{Tm}
{\bf Proof:} The proof is based on an idea of I. Schoenberg. We
use the equality (see \cite[equation (8) p. 526]{MR1501980})
\begin{equation}
 |t|^{2H}=\dfrac{\int_0^\infty (1-e^{-u^2t^2})u^{-1-{2H}}du}
{\int_0^\infty (1-e^{-u^2})u^{-1-{2H}}du}. \label{schoenberg}
\end{equation}
The equality itself is proved by using the change of variable
$u\mapsto u|t|$ in the integral
\[\int_0^\infty
(1-e^{-u^2t^2})u^{-1-{2H}}du,\quad 0<H<1.\]
See \cite[p. 526]{MR1501980}. Moreover, integration by part leads
to
\[
\int_0^\infty (1-e^{-u^2})u^{-1-{2H}}du=\dfrac{\Gamma(1-H)}{2H}.
\]
From \eqref{schoenberg} it follows that
\begin{equation}
\label{total-recall}
\begin{split}
\left(\frac{\Gamma(1-H)}{2H}\right)
(|t|^{2H}+|s|^{2H}-|t-s|^{2H})&=\\
&\hspace{-2.5cm} =\int_0^\infty
\frac{(1-e^{-u^2t^2})(1-e^{-u^2s^2})}{u^{1+{2H}}}du+\\
&\hspace{-2cm}+
\int_0^\infty\frac{e^{-u^2t^2}(e^{2u^2ts}-1)e^{-u^2s^2}}{u^{1+{2H}}}du.
\end{split}
\end{equation}
The second term is a decomposition of the form \eqref{Temple,
ligne 3} with
\[
k(t,s,u)=e^{-u^2t^2}(e^{2u^2ts}-1)e^{-u^2s^2}\quad{\rm and}\quad
q^\prime(u)=\frac{1}{u^{1+{2H}}}.
\]
The positivity of $K_{H,1}(t,s)$ follows by using the fact that
sums and products of positive functions are still positive. We
note that equation \eqref{total-recall} proves that
$K_{H,1}(t,s)$ is a covariance function, but it does not provide
a {\it model} for it (in the sense given in \cite[p.
41]{Lifshits95}). Still, one can be somewhat more explicit by
using formulas \eqref{La_Muette} and \eqref{perlette}: A function
$F$ belongs to the reproducing kernel Hilbert space with
reproducing kernel
\[
M(t,s)=
\int_0^\infty\frac{e^{-u^2t^2}(e^{2u^2ts}-1)e^{-u^2s^2}}{u^{1+{2H}}}du
\]
if and only if it can be written as
\[
F(t)=\int_0^\infty \frac{f(t,u)}{u^{1+2H}}du
\]
where for every $u$ the function $f(t,u)$ belongs to the
reproducing kernel Hilbert space ${\mathcal H}(N_u)$ with reproducing
kernel
\[
N_u(t,s)= e^{-u^2t^2}(e^{2u^2ts}-1)e^{-u^2s^2}
\]
and
\[
\int_0^\infty\frac{\|f(t,\cdot)\|^2_{{{\mathcal H}(N_u)}}}{{u^{1+{2H}}}}
du<\infty.
\]
Next, and using Theorem \ref{rosemary}, write:
\[
\begin{split}
\int_0^\infty\frac{e^{-u^2t^2}(e^{2u^2ts}-1)e^{-u^2s^2}}{u^{1+{2H}}}du&=
\sum_{n=1}^\infty \frac{2^nt^ns^n}{n!}\int_0^\infty
e^{-u^2(t^2+s^2)}u^{2n-1-2H}du\\
&= \sum_{n=1}^\infty
\frac{2^{n-1}\Gamma(n-H)}{n!}\frac{t^ns^n}{(t^2+s^2)^{n-H}},
\end{split}
\]
which recovers \eqref{total-recall-nice-1991}.
\mbox{}\qed\mbox{}\\

For instance, for a given real number $s$ and for
\[
F(t)=\frac{2H}{\Gamma(1-H)}K_{H,1}(t,s)
\]
we have \( x_0(u)=(1-e^{-u^2s^2})\) and for $n\ge 1$,
\[
x_n(u)=\frac{2^n}{n!}s^ne^{-u^2s^2}.\]
For this choice of $x_0,x_1,\ldots$ equality holds in
\eqref{Porte d'Orleans, ligne 4}.\\

The second approach is also related to Schoenberg's work. Let
$dm(u)$ be a positive measure on $[0,\infty)$ subject to the
condition
\[
\int_1^\infty dm(u)<\infty.
\]
(Note that the Lebesgue measure does not satisfy this last
requirement.)\\

The function
\begin{equation}
\label{new_r}
r(t)=\int_0^\infty (1-e^{-u|t|})dm(u)
\end{equation}
is then defined for all real $t$'s. Such functions $r(t)$ have
been introduced in \cite[Definition 2, p. 825]{MR1503439}.
Functions of the form $\sqrt{r(t^2)}$ are exactly the functions
such that a separable real Hilbert space ${\mathcal H}$ (with
norm denoted by $\|\cdot\|$), endowed with the metric $F(\|x\|)$,
can be isometrically imbedded in ${\mathcal H}$. See
\cite[Theorem 6, p. 828]{MR1503439}. In particular, $r$ is of the
form \eqref{eureka}.

\begin{Tm}
Let $r(t)$ be of the form \eqref{new_r}. Then the kernel
$K_r(t,s)$:
\[
K_r(t,s)=r(t)+r(s)-r(t-s).
\]
is positive on the real line. Moreover, with
$K_{1,1}(t,s)=|t|+|s|-|t-s| \stackrel{\rm def.}{=}K(t,s)$,
\begin{equation}
\label{St Philippe du Roule, ligne 9}
\begin{split}
K_r(t,s) &=\int_0^\infty(1-e^{-u|t|})(1-e^{-u|s|})dm(u)+\\
&\hspace{5mm}+\sum_{n=1}^\infty\dfrac{K^n(t,s)}{n!}
\left(\int_0^\infty u^ne^{-u(|t|+|s|)}dm(u)\right).
\end{split}
\end{equation}
\end{Tm}
{\bf Proof:} Using Theorem \ref{rosemary} we have:
\[
\begin{split}
K_r(t,s)&=\int_0^\infty \left\{
1-e^{-u|t|}+1-e^{-u|s|}+e^{-u|t-s|}-1\right\}
m(u) du\\
&=\int_0^\infty(1-e^{-u|t|})(1-e^{-u|s|})dm(u)+ \int_0^\infty
\left\{e^{-u|t-s|}-e^{-u(|t|+|s|)}\right\}dm(u)\\
&=\int_0^\infty(1-e^{-u|t|})(1-e^{-u|s|})dm(u)+\\
&\hspace{5mm}+ \int_0^\infty e^{-u|t|}
(e^{u(|t|+|s|-|t-s|)}-1)e^{-u|s|}dm(u).
\end{split}
\]
which proves the positivity of the kernel since for every $u\ge
0$, one has
\[
e^{u(|t|+|s|-|t-s|)}-1=\sum_{1}^\infty
\dfrac{u^n}{n!}(|t|+|s|-|t-s|)^n
\]

\mbox{}\qed\\

{\bf Remarks:}\\
$(a)$ The function
\begin{equation}
\label{segolene} \Gamma(z,\mu)=\int_0^\infty u^{z-1}e^{-\mu
u}dm(u)
\end{equation}
is the generalization for any $m(u)$ of the classical Gamma
function. Furthermore, the decomposition \eqref{St Philippe du
Roule, ligne 9}
is conducive in applications with regard to non--linear
transforms. See Section \ref{NL}.\\
$(b)$ The formula \eqref{St Philippe du Roule, ligne 9} allows to
characterize the associated reproducing kernel Hilbert space.
\\
$c)$ In view of the result of von Neumann and Schoenberg
mentioned in the introduction, and as we already noticed above,
any function of the form
\eqref{new_r} is of the form \eqref{eureka}.\\

In the case where
\[
dm(u)=\dfrac{e^{-u}}{u}du
\]
we have $r(t)=\ln(1+|t|)$. This follows from the formula in
\cite[Corollary 2.10]{MR86b:43001}:
\[
\ln (1+z)=\int_0^\infty(1-e^{-uz})\dfrac{e^{-u}du}{u},\quad {\rm
Re}~z\ge 0.
\]
It follows that the function
\[
q(t,s)=\ln(1+|t|)+\ln(1+|s|)-\ln(1+|t-s|)
\]
is positive on ${\mathbb R}$.\\

The choice
\[
dm(u)=\dfrac{2H}{\Gamma(1-2H)} \dfrac{1}{u^{1+2H}}du
\]
corresponds to the covariance function of the
fractional Brownian motion for $H\in(0,1/2)$
since (see \cite[Corollary 2.10 p. 78]{MR86b:43001}),
\[
z^{2H}=\dfrac{2H}{\Gamma(1-2H)}\int_0^\infty
(1-e^{-uz})\dfrac{du}{u^{1+2H}},\]
where $H\in(0,1/2)$ and ${\rm Re}~z\ge 0$. We check this formula
for positive $t$ as follows. The change of variable $v=tu$ leads
to
\[
\int_0^\infty (1-e^{-uz})\dfrac{du}{u^{1+2H}}=t^{2H}\int_0^\infty
(1-e^{-u})\dfrac{du}{u^{1+2H}}=\frac{\Gamma(1-2H)}{2H}
\]
where the last equality is obtained by integration by parts.\\

Specializing \eqref{St Philippe du Roule, ligne 9} to this case
leads to:

\begin{Tm} Let $H\in(0,1/2)$. Then:
\begin{equation}
\label{ohohoh}
\begin{split}
|t|^{2H}+|s|^{2H}-|t-s|^{2H}
&=2H(|t|+|s|-|t-s|)+\\
&\hspace{5mm}+\frac{2H}{\Gamma(1-2H)}\left\{
\int_0^\infty\frac{(1-e^{-u|t|})(1-e^{-u|s|})}{u^{1+2H}}du+\right.\\
&\hspace{3cm}+\left.
\sum_{n=2}^\infty\dfrac{K^n(t,s)\Gamma(n-2H)}{n!(|t|+|s|)^{n-2H}}\right\}.
\end{split}
\end{equation}
In particular, the space ${\mathcal H}(K_{H,1})$ contains
contractively a copy of ${\mathcal H}(K_{1,1})$.
\label{surprise}
\end{Tm}
{\bf Proof:} Indeed, using Theorem \ref{rosemary} we have:
\begin{equation*}
\begin{split}
|t|^{2H}+|s|^{2H}-|t-s|^{2H}&=\frac{2H}{\Gamma(1-2H)}\left\{
\int_0^\infty\frac{(1-e^{-u|t|})(1-e^{-u|s|})}{u^{1+2H}}du+\right.\\
&\hspace{3cm}+\left.
\sum_{n=1}^\infty\dfrac{K^n(t,s)\Gamma(n-2H)}{n!(|t|+|s|)^{n-2H}}
\right\}\\
&=2H(|t|+s|-|t-s|)+\\
&\hspace{5mm}+\frac{2H}{\Gamma(1-2H)}\left\{
\int_0^\infty\frac{(1-e^{-u|t|})(1-e^{-u|s|})}{u^{1+2H}}du+\right.\\
&\hspace{3cm}+\left.
\sum_{n=2}^\infty\dfrac{K^n(t,s)\Gamma(n-2H)}{n!(|t|+|s|)^{n-2H}}\right\}.
\end{split}
\end{equation*}
\mbox{}\qed\mbox{}\\

{\bf Remark:} Formula \eqref{total-recall-nice-1991} is valid for
any $0<H<1$ and we see that terms of the form $\Gamma(n-H)$
appear in it ($n\ge 1$). On the other hand, formula
\eqref{ohohoh} is valid only for $0<H<1/2$, with terms of the form
$\Gamma(n-2H)$.

\section{A general family of positive functions}
\setcounter{equation}{0}
\begin{Tm}
Let $r(t)$ be a real valued function such that the kernel
$K_r(t,s)$, defined by \eqref{mathilde_est_revenue}, is positive
on the real line, and let $\varphi$ is of the form \eqref{new_r}.
Then:
\begin{equation}
\label{jezabel} \varphi(r(t)+r(s))-\varphi(r(t-s)) =
\sum_{n=1}^\infty\dfrac{K_r(t,s)^n}{n!}\Gamma(n, r(t)+r(s))
\end{equation}
where $\Gamma(z,\mu)$ has been defined in \eqref{segolene}.
In particular the function
\begin{equation}
\label{Bercy}
V(t,s)=\varphi(r(t)+r(s))-\varphi(r(t-s))
\end{equation}
is positive on the real line.
\end{Tm}

{\bf Proof:} With $\varphi$ of the form \eqref{new_r}, one has

\begin{equation*}
\begin{split}
\varphi(r(t)+r(s))-\varphi(r(t-s))&=\int_0^\infty \left\{
e^{-ur(t-s)}-e^{-u(r(t)+r(s))}\right\}dm(u)\\
&=\int_0^\infty e^{-ur(t)})\left\{
e^{u(r(t)+r(s)-r(t-s))}-1\right\}e^{-ur(s)}dm(u)\\
&= \sum_{n=1}^\infty\dfrac{K_r(t,s)^n}{n!}\Gamma(n, r(t)+r(s)).
\end{split}
\end{equation*}
This allows to conclude the proof since the functions
$K_r(t,s)^n$ and $\Gamma(n, r(t)+r(s))$ are positive on the real
line, and so is
their product. \mbox{}\qed\mbox{}\\

We now consider the case $\varphi(z)=z^\alpha$ and obtain the
following analogue of Theorem \ref{surprise}:

\begin{Tm} Let $\alpha\in(0,1)$. Then:
\begin{equation}
\label{Monceau, ligne 2}
\begin{split}
(r(t)+r(s))^\alpha-r(t-s)^\alpha&= \alpha
K_r(t,s)+\frac{\alpha}{\Gamma(1-\alpha)}
\sum_{n=2}^\infty\frac{K_r(t,s)^n\Gamma(n-\alpha)}{n!(r(t)+r(s))^{n-\alpha}}.
\end{split}
\end{equation}
In particular, the space with reproducing kernel
$(r(t)+r(s))^\alpha-r(t-s)^\alpha$ contains contractively a copy
of ${\mathcal H}(K_r)$.
\end{Tm}
{\bf Proof:} We have already recalled that
\[
z^\alpha=\frac{\alpha}{\Gamma(1-\alpha)}\int_0^\infty(1-e^{-uz})
\frac{du}{u^{1+\alpha}}.
\]
Using the preceding arguments we therefore obtain:

\begin{equation*}
\begin{split}
(r(t)+r(s))^\alpha-r(t-s)^\alpha&=
\frac{\alpha}{\Gamma(1-\alpha)}\sum_{n=1}^\infty\frac{K_r(t,s)^n}{n!}
\int_0^\infty u^ne^{-u(r(t)+r(s))}\frac{du}{u^{1+\alpha}}\\
&=
\frac{\alpha}{\Gamma(1-\alpha)}\sum_{n=1}^\infty\frac{K_r(t,s)^n}{n!}
\dfrac{1}{(r(t)+r(s))^{n-\alpha}}\Gamma(n-\alpha)\\
&=\alpha K_r(t,s)+\frac{\alpha}{\Gamma(1-\alpha)}
\sum_{n=2}^\infty\frac{K_r(t,s)^n\Gamma(n-\alpha)}{n!(r(t)+r(s))^{n-\alpha}}.
\end{split}
\end{equation*}
The last claim of the theorem follows from
\eqref{Monceau, ligne 2} and Theorem \ref{Chevaleret, ligne 6}
\mbox{}\qed\mbox{}\\

{\bf Remark:} Formula \eqref{Monceau, ligne 2} is still valid for
$\alpha=1$ since the function $\Gamma(z)$ has a pole at the
origin.

\section{Associated non-linear transforms}
\setcounter{equation}{0}
\label{NL}

We now associate a non-linear transform (in the sense of Saitoh;
see \cite[Appendix 2, p. 243] {saitoh2}) to the decomposition
\eqref{jezabel}. We begin with a general result.

\begin{Tm}
Let $K$ and $K_n,\, n=1,2,\ldots$ be positive
kernels on the real line and define
\[
{\mathbf K}(t,s)=\sum_{n=1}^\infty K^n(t,s)K_n(t,s).
\]
Assume that ${\mathbf K}(t,t)$ converges for all
$t\in{\mathbb R}$. For every choice of $t_0\in{\mathbb R}$, the
map which to $f\in{\mathcal H}(K)$ associates the function
\[
\varphi(f)(t)=\sum_{n=1}^\infty K_n(t,t_0)f(t)^n
\]
maps ${\mathcal H}(K)$ into ${\mathcal H}({\mathbf K})$.
Furthermore,
\begin{equation}
\label{rue_des_francs_bourgeois}
\|\varphi(f)\|_{{\mathcal H}({\mathbf K})}^2
\le\sum_{n=1}^\infty
 K_n(t_0,t_0)\|f\|^{2n}_{{\mathcal H}(K)}.
\end{equation}
\end{Tm}

{\bf Proof:} Using \eqref{quartier_latin} and the two inequalities
in \eqref{saitoh_1} one after the other, we have
\[
\begin{split}
\|\varphi(f)\|_{{\mathcal H}({\mathbf K})}^2&\le\sum_{n=1}^\infty
\|K_n(\cdot, t_0)f(\cdot)^n\|^2_{{\mathcal H}(K_nK^n)}\\
&\le \sum_{n=1}^\infty\|K_n(\cdot, t_0)\|^2_{{\mathcal H}(K_n)}
\|f^n\|^2_{{\mathcal H}(K^n)}\\
&\le\sum_{n=1}^\infty
 K_n(t_0,t_0)\|f\|^{2n}_{{\mathcal H}(K)}
\end{split}
\]
since
\[
\|K_n(\cdot, t_0)\|^2_{{\mathcal H}(K_n)}=K_n(t_0,t_0).\]
\mbox{}\qed\mbox{}\\

As a consequence we have:

\begin{Tm} Let $t_0\in{\mathbb R}\setminus\{0\}$, and
let $\psi$ be the map which to $f\in{\mathcal H}(K_r)$ associates
the function
\begin{equation}
\label{lydia}
\psi(f)(t)=\sum_{n=1}^\infty\frac{\Gamma(n,r(t)+r(t_0))}{n!}f(t)^n.
\end{equation}
Then $\psi(f)$ belongs to the reproducing kernel Hilbert space
${\mathcal H}(K_{\psi,r})$ with reproducing kernel
$K_{\psi,r}(t,s)=\psi(r(t)+r(s))-\psi(r(t-s))$ and we have
\begin{equation}
\label{asie} \|\psi(f)\|_{{\mathcal
H}(K_{\psi,r})}^2\le\sum_{n=1}^\infty
\frac{\Gamma(n,2r(t_0))}{n!}\|f\|_{{\mathcal H}(K_r)}^{2n}
\end{equation}
\end{Tm}
Indeed, the kernel $\frac{\Gamma(n, r(t)+r(s))}{n!}$
is positive on the real line, and
\[
\left\|\frac{\Gamma(n, r(\cdot)+r(t_0))}{n!}\right\|^2_{{\mathcal H}\left(
\frac{\Gamma(n, r(t)+r(s))}{n!}\right)}=
\frac{\Gamma(n,2r(t_0))}{n!};
\]
See formula \eqref{poisse} if need be.\\

Similar theorems can be stated for the decompositions \eqref{St
Philippe du Roule, ligne 9} and \eqref{ohohoh}. The case
\eqref{ohohoh} will not be pursued here. To present the results
pertaining to the case \eqref{St Philippe du Roule, ligne 9}, we
make a number of additional assumptions. First we restrict
ourselves to positive $t$ and $s$. The kernel $(|t|+|s|-|t-s|)^n$
then takes the form
\[
K_{1,1}^n(t,s)=(|t|+|s|-|t-s|)^n=2^n\max(t^n, s^n),\]
and the corresponding reproducing kernel Hilbert space is equal to the
set of functions of the form
\[
F(t)=\int_0^{t^n} f(u)du, \quad f\in{\mathbf L}_2({\mathbb R}),\]
with norm (recall \eqref{poisse})
\[
\|F\|^2_{{\mathcal H}(K_{1,1}^n)}=
\frac{
\int_0^\infty |f|^2(u)du}{2^n}=\frac{\int_0^\infty
\frac{|F^\prime(u)|^2}{nu^{n-1}}du}{2^n}.\]
See \cite{MR1423330}, \cite{saitoh2}.
We will also assume that for some $t_0\ge 0$
\begin{equation}
\label{edf_gdf}
\sum_{n=1}^\infty \frac{\int_0^\infty u^ne^{-2ut_0} dm(u)}{2^n}<\infty.
\end{equation}

We have:

\begin{Tm} Assume that \eqref{edf_gdf} holds.
The map which to $F\in{\mathcal H}(K_{1,1})$ associates
\[
\psi(F)(t)=\sum_{n=1}^\infty\left(\int_0^{t^n}F^\prime(u)du\right)
\left (\int_0^\infty u^ne^{-u(t+t_0)}dm(u)\right)
\]
sends ${\mathcal H}(K_{1,1})$ into ${\mathcal H}(K_r)$ and we have:
\[
\|\psi(F)\|^2_{{\mathcal H}(K_r)}\le\|F^\prime\|^2_{{\mathbf
    L}_2({\mathbb R}^+)}\left\{
\sum_{n=1}^\infty \frac{\int_0^\infty u^ne^{-2ut_0} dm(u)}{2^n}\right\}.\]
\end{Tm}

In the case of the fractional Brownian motion, the above theorem
becomes (recall that $K_{H,1}(t,s)$ is defined by \eqref{ioanna}):

\begin{Tm} Let $0<H<1/2$ and let $t_0>1/2$.
The map which to $F\in{\mathcal H}(K_{1,1})$ associates the
\[
\psi(F)(t)=\frac{2H}{\Gamma(1-2H)}
\sum_{n=2}^\infty\left(\int_0^{t^n}F^\prime(u)du\right)
\frac{\Gamma(n-2H)}{n!(t+t_0)^{n-2H}}
\]
sends ${\mathcal H}(K_{1,1})$ into ${\mathcal H}(K_{H,1})$ and we have:
\begin{equation}
\label{totototo}
\|\psi(F)\|^2_{{\mathcal
H}(K_r)}\le\|F^\prime\|^2_{{\mathbf
    L}_2({\mathbb R}^+)}
\sum_{n=2}^\infty \frac{\Gamma(n-2H)}{2^nn!(2t_0)^{n-2H}}.
\end{equation}
\end{Tm}

We note that the series \eqref{totototo} converges  for $t_0>1/2$
since $\Gamma(n)=(n-1)!$.

redpajama_set_name: RedPajamaArXiv


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8.1. Claims Against Customer. TalentWall will defend, at its own expense, any claim, suit or action against you brought by a third party to the extent that such claim, suit or action is based upon an allegation that the Software infringes any intellectual property rights of such third party (each, "Customer Claim"), and TalentWall shall indemnify and hold you harmless from and against those losses, liabilities, costs and expenses (including reasonable attorneys' fees and costs) that are attributable to such Customer Claim or those costs and damages agreed to in a monetary settlement of such Customer Claim. The foregoing obligations are conditioned on you: (a) promptly notifying TalentWall in writing of such Customer Claim; (b) giving TalentWall sole control of the defense thereof and any related settlement negotiations; and (c) reasonably cooperating and, at TalentWall's request and expense, assisting in such defense. In the event that the use of the Software is enjoined, TalentWall shall, at its option and at its own expense either (a) procure the right for you to continue using the Software, (b) replace the Software with a non-infringing but functionally equivalent product, (c) modify the Software so it becomes non-infringing or (d) terminate these Terms and refund the amounts you paid for System Access that relate to the period during which the Software was not usable by you. Notwithstanding the foregoing, TalentWall will have no obligation under this Section 8.1 or otherwise with respect to any infringement claim based upon: (1) any use of the System in violation of these Terms; (2) any use of the System in combination with products, equipment, software, or data not supplied or approved by TalentWall if such infringement would have been avoided without the combination with such other products, equipment, software or data; or (3) any modification of the System by any person other than TalentWall or its authorized agents or subcontractors. This Section 8.1 states TalentWall's entire liability and your sole and exclusive remedy for infringement claims or actions.
8.2. Claims Against TalentWall. You will defend, at your own expense, any claim, suit or action against TalentWall brought by a third party to the extent that such claim, suit or action is based upon your or TalentWall's use of any Customer Data in accordance with these Terms ("TalentWall Claim"), and you shall indemnify and hold TalentWall harmless from and against those losses, liabilities, costs and expenses (including reasonable attorneys' fees and costs) attributable to such TalentWall Claim or those costs and damages agreed to in a monetary settlement of such TalentWall Claim. The foregoing obligations are conditioned on TalentWall: (a) promptly notifying you in writing of such TalentWall Claim; (b) giving you sole control of the defense thereof and any related settlement negotiations; and (c) cooperating and, at your request and expense, assisting in such defense. Notwithstanding the foregoing, you will have no obligation under this Section 8.2 or otherwise with respect to any TalentWall Claim to the extent based upon any use of the Customer Data by TalentWall in violation of these Terms.
9. Limitation of Liability. In no event will EITHER PARTY be liable for any consequential, indirect, EXEmplary, special, or incidental damages, OR FOR any lost data, lost profits OR costs of procurement of substitute goods or services, arising from or relating to these Terms, however caused and under any theory of liability (including negligence), even if SUCH PARTY has been advised of the possibility of such damages. EACH PARTY's total cumulative liability in connection with these Terms and the System, whether in contract or tort or otherwise, will not exceed the amount OF FEES PAID OR PAYABLE BY CUSTOMER TO TALENTWALL UNDER THESE Terms IN THE 12 MONTHS PRIOR TO THE EVENT GIVING RISE TO THE LIABILITY. the limitation of liabilities set forth in this section 9 shall not apply to the obligations under Section 8.
10. Confidentiality.
10.1. Definition. "Confidential Information" means all information disclosed by one party ("Discloser") to the other party ("Recipient") relating to or disclosed in the course of the performance of these Terms. Confidential Information includes information that is marked or identified as confidential and, if not marked or identified as confidential, information that should reasonably have been understood by Recipient to be proprietary and confidential to Discloser or to a third party, because of legends or other markings, the circumstances of disclosure or the nature of the information itself. The Software, System and Documentation shall be considered TalentWall's Confidential Information, notwithstanding any failure to mark or identify it as such.
10.2. Protection. Recipient will not use any Confidential Information of the Discloser for any purpose not expressly permitted by these Terms, and will disclose Confidential Information only to the employees or, subject to Section 3, individual independent contractors of Recipient who have a need to know such Confidential Information for purposes of these Terms and who are under a duty of confidentiality no less restrictive than Recipient's duty hereunder. Recipient will protect Confidential Information from unauthorized use, access, or disclosure in the same manner as Recipient protects its own confidential or proprietary information of a similar nature and with no less than reasonable care.
10.3. Exceptions. Recipient's obligations under Section 10.2 above with respect to any Confidential Information of Discloser will terminate if and when Recipient can document that such information: (a) was already lawfully known to Recipient at the time of disclosure by Discloser; (b) is disclosed to Recipient by a third party who had the right to make such disclosure without any confidentiality restrictions; (c) is, or through no fault of Recipient has become, generally available to the public; or (d) is independently developed by Recipient without access to or use of the Confidential Information. In addition, Recipient may disclose Confidential Information to the extent that such disclosure is required by law or by the order of a court or similar judicial or administrative body, provided that Recipient notifies Discloser of such required disclosure in writing prior to making such disclosure and cooperates with Discloser, at Discloser's reasonable request and expense, in any lawful action to contest or limit the scope of such required disclosure.
10.4. Protection of Customer Data. Without limiting the above, TalentWall will maintain appropriate administrative, physical, and technical safeguards for protection of the security, confidentiality and integrity of the Customer Data. TalentWall will not (a) modify the Customer Data except as updated by the Customer in the System, (b) disclose the Customer Data except as compelled by law in accordance with Section 10.3 or as expressly permitted in writing by you, or (c) access the Customer Data except to provide the Services or prevent or address service or technical problems, or at your request in connection with customer support matters.
11. General Provisions.
11.1. Assignment. Neither party may assign or transfer, by operation of law or otherwise, these Terms or any of its rights under these Terms to any third party without the other party's prior written consent, such consent shall not be unreasonably withheld or delayed; except that either party may assign these Terms without consent by operation of law or otherwise to any successor to its business or assets to which these Terms relates, whether by merger, sale of assets, sale of stock, reorganization or otherwise. Any attempted assignment or transfer in violation of the foregoing will be null and void. These Terms shall be binding upon and inure to the benefit of the parties hereto and their respective successors and permitted assigns, and shall not confer any rights or remedies upon any person or entity not a party hereto.
11.2. Force Majeure. Except for any payment obligations, neither party shall be liable hereunder by reason of any failure or delay in the performance of its obligations hereunder as a result of any cause which is beyond the reasonable control of such party.
11.3. Notices.
(a) Notice to TalentWall. All notices, consents, and approvals to TalentWall under these Terms must be delivered in writing by electronic mail to legal-notices@talentwall.io and will be effective upon receipt.
(b) Notice to Customer. All notices, consents, and approvals to you under these Terms must be delivered in writing by electronic mail, courier, or certified or registered mail (postage prepaid and return receipt requested) to you at the address that TalentWall has on file for you and will be effective upon receipt. You may update your address on file by notifying TalentWall of such address in accordance with Section 11.3(a).
11.4. Governing Law. These Terms will be governed by and interpreted in accordance with the laws of the State of Colorado without reference to its choice of law rules.
11.5. Remedies. Except as otherwise expressly provided in these Terms, the parties' rights and remedies under these Terms are cumulative. Each party acknowledges and agrees that any actual or threatened breach of Sections 3 or 10 will constitute immediate, irreparable harm to the non-breaching party for which monetary damages would be an inadequate remedy, that injunctive relief is an appropriate remedy for such breach, and that if granted, the breaching party agrees to waive any bond that would otherwise be required. If any legal action is brought by a party to enforce these Terms, the prevailing party will be entitled to receive its attorneys' fees, court costs, and other collection expenses, in addition to any other relief it may receive from the non-prevailing party.
11.6. Relationship of the Parties. The parties acknowledge that TalentWall is an independent contractor of you, and its employees are not your employees. Nothing in these Terms or any order document will be construed as creating a partnership, joint venture, or agency relationship between the parties, or as authorizing either party to act as an agent for the other or to enter contracts on behalf of the other. Nothing in these Terms is intended to confer any rights or remedies on any other person or entity, which is not a party to these Terms.
11.7. Waivers. Any waiver or failure to enforce any provision of these Terms on one occasion will not be deemed a waiver of any other provision or of such provision on any other occasion. Any waiver must be in writing and signed by the party entitled to the benefit of the right being waived. Unless otherwise stated in the waiver, any waiver applies only to the specific circumstance for which the waiver is given and not to any subsequent circumstance involving the same or any other right.
11.8. Severability. If any provision of these Terms is held by a court of competent jurisdiction to be unenforceable, such provision will be changed and interpreted to accomplish the objectives of such provision to the greatest extent possible under applicable law and the remaining provisions of these Terms will continue in full force and effect.
11.9. Entire Agreement. These Terms, including any order document and any exhibits or attachments thereto, constitute the entire agreement between the parties regarding the subject hereof and supersedes all prior or contemporaneous agreements, understandings, and communication, whether written or oral. This Agreement sets forth the general terms and conditions applicable to all Services provided by TalentWall to you under any specific terms and conditions that may be set forth in the applicable order document. By executing these Terms, the parties agree that the provision and receipt of Services are expressly conditioned on the acceptance of the agreed upon terms in these Terms. No other terms apply. No terms and conditions proposed by either party shall be binding on the other party unless accepted in writing by both parties, and each party hereby objects to and rejects all terms and conditions not so accepted. To the extent of any conflict between the provisions of these Terms and the provisions of any applicable order document, the provisions of the applicable order document shall govern. In addition, to the extent of any conflict or inconsistency between the terms and conditions of these Terms and your purchase order, the terms and conditions of these Terms will control. These Terms will not be modified except by a subsequently dated written amendment signed on behalf of TalentWall and you by duly authorized representatives.

redpajama_set_name: RedPajamaCommonCrawl


text:
WorldPhotos
WN.com Search Results for: Kristofer and Hivju
photo: Creative Commons / Marco Verch
Netflix's 'The Witcher' filming continues despite Henry Cavill's injury on set
DNA India | 14 Dec 2020
Filming on the sets of the super hit Netflix drama series 'The Witcher' continues on Sunday without its lead star Henry Cavill who is recovering from an injury sustained during production at Arborfield studios, London. According to Deadline, Cavill who is still recovering from what is being described as a minor leg muscle injury will be...
Netflix - Web Series
photo: AP / Jorge Saenz
Search for a COVID-19 vaccine heats up in China, US
Indian Express | 15 Apr 2020
Three potential COVID-19 vaccines are making fast progress in early-stage testing in volunteers in China and the U.S., but it's still a long road to prove if they'll really work. China's CanSino Biologics has begun the second phase of testing its vaccine candidate, China's Ministry of Science and Technology said Tuesday. In...
wikipedia:COVID-19 vaccine
photo: Creative Commons / Ewen Roberts
Lost actor Daniel Dae Kim tests positive for coronavirus
Dawn | 20 Mar 2020
Actor Daniel Dae Kim said on Thursday he had tested positive for the coronavirus. Kim, speaking from his home in Hawaii, announced the news in an Instagram post and video. "Yesterday I was diagnosed with COVID-19, the...
Wikipedia: Daniel Dae Kim
photo: AP / Joel Ryan
Jared Leto Hadn't Heard About The Coronavirus But Now He Has
Newsweek | 17 Mar 2020
Actor Jared Leto has recently discovered the latest news about the coronavirus outbreak, after being isolated during his meditative trip of self-discovery. In a real-life twisted take of Washington Irving's short story, "Rip Van Winkle," Leto has awakened to the latest news of how America has dramatically changed. Celebrities Who Have Tested...
Wikipedia: Jared Leto
photo: AP / Jordan Strauss/Invision
'Game of Thrones' star Kristofer Hivju tests positive for Coronavirus
One India | 17 Mar 2020
Oslo, Mar 17: "Game of Thrones" star Kristofer Hivju has revealed that he tested positive for the coronavirus. The Norwegian actor, who played Tormund Giantsbane in HBO's epic fantasy series, took to Instagram on Monday to share his diagnosis. ......
wikipedia:Kristofer Hivju
photo: Creative Commons / Gage Skidmore
'Game of Thrones' slashes its way to historic 32 Emmy Award nominations
The Press Enterprise | 16 Jul 2019
By LYNN ELBER LOS ANGELES — HBO's "Game of Thrones" has slashed its way to a record-setting 32 Emmy nominations Tuesday for its eighth and final season. The bloodthirsty saga's total eclipsed the all-time series record of 27 nods earned by "NYPD Blue" in 1994. The showing by "Game of Thrones" led HBO...
Emmy Nominations
Wikipedia: Game of Thrones
Game of Thrones prequel will not feature Targaryens and is not based on Fire and ...
The Independent | 20 Nov 2018
With George RR Martin's latest Westeros-based novel almost on book-seller's shelves, fans have been wondering whether Fire and Blood could form the basis of the upcoming Game of Thrones spin-off show. The new book offers an extensive history of one of its main families, House Targaryen, and begins 300 years before the tale told in A Song...
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text:
Home Sports Stars Female Sports Stars Katie Zelem Height, Weight, Age, Body Statistics
Katie Zelem Height, Weight, Age, Body Statistics
Katie Zelem Quick Info
Zodiac Sign Aquarius
Katie Zelem is an English professional soccer player and TV personality. She is best known as the captain and midfielder for club Manchester United in the FA Women's Super League starting in 2018. Between 2013 and 2017, she played for Liverpool, and between 2017 and 2018 for Juventus.
Katie Leigh Zelem
Katie Zelem in an Instagram selfie as seen in January 2019 (Katie Zelem / Instagram)
Oldham, Greater Manchester, England, United Kingdom
Manchester, Greater Manchester, England, United Kingdom
Professional Soccer Player, TV Personality
Father – Alan Zelem (Former Professional Goalkeeper for Macclesfield Town F.C.)
Mother – Gillian Zelem
Siblings – None
Others – Peter Zelem (Uncle) (Former Professional Soccer Player), Ross Zelem (Cousin) (Played cricket for Lancashire and soccer for Oldham schools), Louise Zelem (Cousin)
Katie Zelem is represented by –
KIN Partners (Talent Agent) in London, England, United Kingdom
Next Level Collective (Talent Agent) in the United Kingdom
5 – Juventus
10 – Manchester United
20 – Liverpool
Playing Position
5 ft 4 in or 162.5 cm
Katie Zelem in an Instagram post in March 2020 (Katie Zelem / Instagram)
Girlfriend / Boyfriend / Spouse
Katie Zelem has dated –
Jessica Sigsworth – Rumor
Dark Brown (Natural)
Katie Zelem as seen in February 2019 (James Boyes / Flickr / CC BY-2.0)
Dimpled smile
Freckled face
Thin nose
Katie Zelem has endorsed brands such as Adidas.
She also worked with the RMCH Charity that benefitted Royal Manchester Children's Hospital in 2019.
Katie Zelem Favorite Things
Midfielder – Fara Williams
Professional Soccer Player – Wayne Rooney
Source – Wikipedia, Women's Competitions
Katie Zelem in an Instagram post in September 2019 (Katie Zelem / Instagram)
Katie Zelem Facts
She started playing soccer at the age of 6, greatly influenced by her dad.
Her parents enrolled her in the Manchester United Girls' Centre of Excellence academy at the age of 8.
She stayed at the Manchester United's academy for 8 years, between 2004 and 2013, and only left because there was no senior team she could graduate into.
She broke her collarbone on April 11, 2014, which stopped her from playing for Liverpool at the start of the 2014 FA WSL season.
Her dad always came to every game she played in, at home or away, and sat at the same spot in the crowd. When her mom booked a holiday, he demanded she postpones it so he can watch Katie play.
At the end of each game, her dad gives her what she calls a "run-through" or a "scout report" on how she and others performed.
Katie was nominated for the title of the "WSL Development Player of the Year" alongside Natasha Baptiste and Carla Humphrey in 2014. Natasha won the title but Katie was named Liverpool's "Young Player of the Year".
In late 2014 and early 2015, she was compared to a well-known soccer player, Frank Lampard, by Liverpool's coach Matt Beard. He also predicted her a great career ahead.
She is inseparable with her teammates, Martha Harris, with whom she played in Liverpool, and Jessica Sigsworth, with whom she played in Manchester United.
Katie was the 1st ever player to win the Manchester United Women's "Player of the Month" award in March 2019. At the end of that season, she also won the "Player of the Year" award.
She was named Manchester United's women club captain ahead of the 2019-2020 season after Alex Greenwood signed a contract with Olympique Lyonnais and departed.
She stated that if people asked some of the girls in Manchester United, they'd say she was quite vocal and has always been. Katie added she was "chewing their ear off" because she wanted them to improve and become the best players they can be.
Katie used to be the "dressing room DJ" until she took the mantle of the club captain. Her "replacements" were Millie Turner and Mollie Green, who also love to listen to dance and R&B as Katie does.
She thinks Leah Galton and Jessica Sigsworth are the fastest players in her Manchester United squad. On the other hand, Lauren James is the most skillful player, according to her.
Katie is the slowest at getting ready among her club teammates.
Featured Image by Katie Zelem / Instagram
Katie Zelem
Professional Soccer Player
VMN Author 2235
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CARLI I-Share library patrons have easy access to a consortial collection that ranks among the world's largest research libraries.
For a collated list, please review the Weekly highlights of the Alma and Primo VE project page.
CARLI is now seeking volunteers to serve on eight advisory committees. Volunteers chosen will serve a three-year term, July 1, 2019 through June 30, 2022. Faculty and staff members from all CARLI member libraries, with all levels of experience, are encouraged to apply.


Tomas Forsberg, better kent bi his stage name "Quorthon" (Februar 17, 1966, Hägersten, Stockholm – Juin 3, 2004, Hässelby, Stockholm), wis a Swadish hivy metal muisicker. He wis the foonder an sangwriter o the pioneerin Swadish black metal baund Bathory. He is an aa creditit wi creatin the Viking metal genre. He composed the muisic an wrote the leerics on aw o Bathory's albums.


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What is "beasley"
n. (surname)
Beasley, TX -- U.S. city in Texas
Population (2000): 590
Housing Units (2000): 232
Land area (2000): 1.003548 sq. miles (2.599176 sq. km)
Water area (2000): 0.000000 sq. miles (0.000000 sq. km)
Total area (2000): 1.003548 sq. miles (2.599176 sq. km)
FIPS code: 06272
Located within: Texas (TX), FIPS 48
Location: 29.496435 N, 95.917297 W
ZIP Codes (1990): 77417
Note: some ZIP codes may be omitted esp. for suburbs.
Headwords:
Beasley, TX
Beasley may refer to:
Beasley (Tampa)
Beasley is a neighborhood within the city limits of Tampa, Florida. The estimated population stands at 122. The neighborhood is part of the East Tampa region and is located within District Five of the Tampa City Council.
Beasley (surname)
Beasley is a surname. Notable people with the surname include:
Aaron Beasley, a United States-born National Football League cornerback
Allyce Beasley, a United States-born actress
Bruce Beasley (American sculptor), a United States sculptor
Cecil A. Beasley (1876-1959), American lawyer and politician from Alabama.
Cheri Beasley, a United States-born judge
Christine Beasley, a British-born dame and the Chief Nursing Officer of England
DaMarcus Beasley, a United States-born soccer player; brother of Jamar
David Beasley, a United States-born politician who served as the Governor of South Carolina
Debra Lafave, formerly Debra Beasley, a United States-born former teacher and registered sex offender
Eileen Beasley, Welsh language campaigner
Fred Beasley, a United States-born National Football League cornerback
Henry Beasley, Indian-born British officer and bridge player
Jamar Beasley, a United States-born soccer player; brother of DaMarcus
Jack Beasley, an Australian-born politician
Jere Beasley, a United States-born politician and Lieutenant Governor of Alambama
Jerome Beasley, a United States-born basketball player
John Beasley (actor), a United States-born actor
John Beasley (basketball), a United States-born basketball player
John Beasley (football player), a United States-born football player and color commentator
John Beasley (musician), a United States-born composer, pianist, producer, and arranger
Larry Beasley, a Canadian-born urban planner
Marco Beasley, a Neapolitan-British tenor and musicologist
Marsha Beasley, a United States-born competitive rifle shooter and coach
Michael Beasley, a United States-born basketball player
Pat Beasley, an English professional footballer and coach
Pierce Beasley, or Piaras Béaslaí, a Liverpool-born Irish Republican politician
Richard Beasley (politician), a United States-born, Canadian-residing businessman, politician, and soldier
Richard Beasley (author), an Australian author
Richard Beasley, convicted of luring three men to their deaths via a Craigslist ad in 2011
Richard Lee Beasley, an American politician from South Carolina
Sheree Beasley, An Australian murder victim
Simon Beasley, an Australian rules footballer
Thomas W. Beasley, American lawyer, political activist, businessman; co-founder of Corrections Corporation of America
Tom Beasley (born 1954), American football player
Usage examples of "beasley".
I could hear the weather report on TV, which made me think of a friend of mine, Warren Beasley, who used to be a weatherman.
Warren Beasley, the controversial radio personality, was probably the last person to talk to the popular young television executive.
This is Warren Beasley at the White House in Washington saying this is Warren Beasley at the White House in Washington.
Beasley served them tea and scones, and gave them a blanket to spread on the grass.
Beasley did not like such men or their connivings, but Arbigost whispered, "Why pay the auctioneer an unnecessary commission?
Some were in favor of consigning you and your vessel to perdition, but Prophet Beasley thought you might be worthy of the kingdom of God.
Roy was surrounded -- and shot from behind Beasley -- by that four-flush Riggs!
Beasley, do you remember last fall when you met Snake Anson with his gang up in the woods--and hired him to make off with me?
Beasley, do you remember last fall when you met Snake Anson with his gang up in the woods -- and hired him to make off with me?
And the Beasley mentioned he was one of the two biggest ranchers and sheep-raisers of the White Mountain ranges.
And the Beasley mentioned -- he was one of the two biggest ranchers and sheep-raisers of the White Mountain ranges.
He an' Al was pards once, an' now Beasley has most folks here believin' he got the short end of thet deal.
words rhyming with beasley, words from word "beasley", words starting with "b", words starting with "be", words starting with "bea", words starting with "beas", words starting with "beasl", words ending with "y", words ending with "ey", words ending with "ley", words ending with "sley", words containing "e", words containing "ea", words containing "eas", words containing "easl",


A live question and answer session was hosted by CZ where a lot of questions were answered. Binance's CEO stated that Binance would 'definitely' join hands with Ripple with the intention of boosting its XRP-powered solution xRapid. The CEO cleared that this step is in the pipeline and is for the distant future. The two companies are yet to work out the details of the project. The company is focussed on launching BinanceDEX and Binance Chain.
In order to accelerate xRapis'd liquidity, Ripple joined hands with different crypto exchanges like Coin ph, Bitso, Bittrex and Bitstamp. Such changes enable financial institutions to convert local currency into XRP while enabling you to send digital assets to countries allowing immediate conversion to fiat.


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Wirtsbach ist ein Gewässername:

 Wirtsbach (Goldbach), rechter Zufluss des Goldbachs bei Appensee, Gemeinde Frankenhardt, Landkreis Schwäbisch Hall, Baden-Württemberg
 Wirtsbach (Starnberger See), südöstlicher Zufluss des Starnberger Sees in Sankt Heinrich, Gemeinde Münsing, Landkreis Bad Tölz-Wolfratshausen, Bayern

redpajama_set_name: RedPajamaWikipedia


text:
WWE SmackDown Viewership Prelim Numbers Up Big From Last Week's Show
SpoilerTV.com reports that last Friday's episode of WWE SmackDown was able to draw a total average of 2.441 million viewers, with a rating of 0.5 in the key 18-49 demographic. This is up from last week's 2.213 million viewers.
In terms of hour-by-hour breakdown, the first hour of WWE SmackDown drew 2,441,000 viewers, while the second hour of WWE SmackDown would remain at 2,411,000 viewers.
Prelim rating for last night's Smackdown on Fox was 2,441,000.
I project around 2,590,000 for the final based on an estimated +6.2% adjustment. Final will be reported Tuesday.
Read more (ad-free, no paywall): https://t.co/aY96MPC14b pic.twitter.com/C5jXkEj87U
— Brandon Thurston (@BrandonThurston) December 31, 2022