text:
\section{Introduction and problem formulation}\label{Intro section}

In this paper, we study the minimization of energy functionals of the form
\begin{equation}\label{energy functional eq}
J_\gamma[\phi]=\frac\gamma2\int_0^T\phi(t)^2\dif t+\frac12\int_0^T\int_0^TG(|t-s|)\phi(s)\phi(t)\dif s\dif t,\qquad \phi\in L^2[0,T],
\end{equation}
where $\gamma\ge0$, $T>0$, and $G:(0,\infty)\to[0,\infty)$ is a continuous and nonconstant  function satisfying
\begin{equation}\label{eq: weakly singular G def}
\int_0^TG(t)\dif t<\infty\qquad\text{for all $T>0$.}
\end{equation}
 Problems of this type have a long history. An early reference is \cite{Hilbert}, where the minimization and maximization of $J_0[\phi]$ is studied under the constraint $\int_0^T\phi(t)^2\,dt=1$ if  $G$ is of positive type in the sense that
\begin{equation}\label{G of positive type}
\frac12\int_0^T\int_0^TG(|t-s|)\phi(t)\phi(s)\dif t\dif s\ge0\quad\text{for all $\phi\in L^2[0,T]$ and all $T>0$.}
\end{equation}
  In potential theory, one usually takes $\gamma=0$ and considers the minimization of
$$J_0[\mu]=\frac12\int\int G(|t-s|)\,\mu(ds)\,\mu(dt),
$$
over Borel probability measures $\mu$ supported on a given compact set $K\subset[0,T]$. If a minimizing measure $\mu^*$ exists, it is the capacitary measure for $K$, and $1/J_0[\mu^*]$ is the capacity of $K$; see, e.g., \cite{choquet}.  Note that the requirement that $\mu$ is a probability measure corresponds to the infinitely many convex constraints   $\mu(K)=1$ and $\mu(A)\ge0$ for every Borel set $A\subset K$.
It was proved in \cite{Gatheral2012} that, for convex and nonincreasing  $G$, the latter constrained minimization problem  can be replaced by the much simpler minimization of $J_0[\mu]$ over all finite signed Borel measures  $\mu$ on $K$ that have finite total variation and satisfy the single linear constraint $\mu(K)=1$. This observation enabled in particular an approach to compute $\mu^*$ for $K=[0,T]$ by means of singular control \citep{Alfonsi2013}. Here, we will instead exploit the fact that, for $\gamma>0$, minimizers of $J_\gamma[\phi]$ under the constraint $\int_0^T\phi(t)\dif t =1$ can be characterized as the solution of the following Fredholm integral equation of second kind,
\begin{equation}\label{Fredholm eq}
\gamma\phi(t)+\int_0^TG(|t-s|)\phi(s)\dif s=\sigma\qquad\text{for a.e.~$t\in[0,T]$,}
\end{equation}
where the constant $\sigma$ is equal to the minimal energy (see Proposition~\ref{Fredholm prop}).

In this paper, we focus  on the qualitative properties of minimizers. For instance, explicit computations or numerical simulations reveal that minimizers of $J_\gamma$ are often convex functions of $t\in[0,T]$ with a minimum at $T/2$. In addition, it is easy to see that every solution $\phi$ must be symmetric around $T/2$, i.e., $\phi(t)=\phi(T-t)$. These two facts are reminiscent of the celebrated Riesz rearrangement inequality, which states that for decreasing $G$,
\begin{equation}\label{eq:Riesz}
\int_0^T\int_0^TG(|t-s|)f(s)g(t)\dif s\dif t\le \int_0^T\int_0^TG(|t-s|)f^*(s)g^*(t)\dif s\dif t,
\end{equation}
where $f^*$ and $g^*$ are the symmetric decreasing rearrangements of the nonnegative functions $f$ and $g$; see \cite{Riesz}. Although a lower bound in \eqref{eq:Riesz} is generally not available, it would be tempting to conjecture that minimizers of $J_\gamma$ are equal to their symmetric increasing rearrangements. This conjecture, however, cannot be true in general since the choice $G(t)=(1- t)^+$  provides a counterexample; see  Example~\ref{capped linear example} and Figure 1. So the following question arises:
\begin{equation}
\begin{split}
&\text{\emph{For which kernels $G$ is the minimizer $\phi$, respectively the solution of \eqref{Fredholm eq},}}\\
&\text{\emph{convex with a minimum at $T/2$?}}
\end{split}\tag{$*$}
\end{equation}
Our main result shows that this is the case whenever $G$ is completely monotone. As a matter of fact, we will actually prove a much stronger result: If $G$ is completely monotone, then $\phi$ is \emph{symmetrically totally monotone} in the sense that it is analytic in $(0,T)$ and its power series development around $T/2$ is of the form  $\phi(t)=\sum_{n=0}^\infty a_{2n}(t-T/2)^{2n}$  for coefficients $a_{2n}\ge0$.

Problems such as the minimization of $J_\gamma$ or the solution of Fredholm integral equations \eqref{Fredholm eq} have a large number of applications, for instance in machine learning; see, e.g., \cite{Chen2002}.  \cite{Gatheral2012} and \cite{Alfonsi2013}, on the other hand, were motivated by the problem of optimal portfolio liquidation in financial markets. There, the solution $\phi$ corresponds to an optimal trading rate for liquidating a large initial position of shares during the time interval $[0,T]$. Since the position is large, its liquidation affects asset prices in an unfavorable way, which creates additional execution costs. The temporal evolution of this price impact can be described by means of a kernel $G$, for which some empirical studies suggest a behavior of the form $G(t)\sim t^{-\alpha}$ for some $\alpha\in(0,1)$; see, e.g., \cite{Gatheral2010}.
Assumption \eqref{G of positive type} is reasonable in this context: it excludes the existence of price manipulation strategies that generate profit through their own price impact \citep{Huberman2004, Gatheral2010}.
The term$\frac\gamma2\int_0^T\varphi(t)^2\dif t$ can be interpreted as costs arising from \lq slippage\rq\ or temporary price impact as in \cite{Almgren}. In this  financial context, the question $(*)$ was asked by J.~Gatheral, and the possible convexity of the optimal portfolio liquidation strategy $\phi$ has the practical significance that it matches the empirically observed U-shape of the daily distribution of market liquidity. That is, if the answer to $(*)$ is \lq Yes\rq\ and the liquidation horizon is one trading day, as it  is often the case, then the optimal liquidation strategy $\phi$ involves fast trading toward the beginning and end of the trading day when liquidity is high and slower trading when liquidity is low.

This paper is organized as follows. In Section~\ref{results section}, we present our main results and a few explicit examples. All proofs are given in Section~\ref{proofs section}.

\section{Main results}\label{results section}

For simplicity, we will assume henceforth that
\begin{equation}\label{G assumption}
G:(0,\infty)\to[0,\infty)\quad\text{is  continuous, nonincreasing, nonconstant, and satisfies \eqref{eq: weakly singular G def}.}
\end{equation}
For $\gamma> 0$ and $T>0$, we consider the following variational problem,
\begin{equation}\label{eq: J gamma}
\text{minimize}\quad J_\gamma[\phi]=\frac\gamma2\int_0^T\phi(t)^2\dif t+\frac12\int_0^T\int_0^TG(|t-s|)\phi(t)\phi(s)\dif t\dif s\quad\text{over $\phi\in\Phi_1$,}
\end{equation}
where $\Phi_1$ consists of all  functions $\phi\in L^2[0,T]$ that satisfy the linear constraint $\int_0^T\phi(t)\dif t=1$ and for which the double integral on the right is well-defined and finite. For $\gamma=0$ we consider the following problem,
\begin{equation}\label{eq: J 0}
\text{minimize}\quad J_0[\mu]=\frac12\int_{[0,T]}\int_{[0,T]} G(|t-s|)\,\mu(dt)\,\mu(ds)\quad\text{over $\mu\in\Phi_0$,}
\end{equation}
where we put $G(0):=G(0+)\in(0,\infty]$ and  where $\Phi_0$ consists of all  signed Borel measures $\mu$ on $[0,T]$ that satisfy $\mu([0,T])=1$ and whose total variation measure $|\mu|$ is finite and such that
$$\int_{[0,T]}\int_{[0,T]} G(|t-s|)\,|\mu|(dt)\,|\mu|(ds)<\infty.
$$

For $\gamma>0$, standard Hilbert space arguments easily yield  the existence and uniqueness of minimizers to \eqref{eq: J gamma} if $G$ satisfies \eqref{G of positive type}. For $\gamma=0$, however, the existence of a minimizer  for \eqref{eq: J 0} is nontrivial even if $G$ is bounded and satisfies \eqref{G of positive type}. Indeed, it was shown in  \cite{Gatheral2012} that minimizers do not exist  for a large class of kernels for which $G(|\cdot|)$ is analytic, such as for $G(t)=e^{-t^2}$ or $G(t)=1/(1+t^2)$, despite the fact that these kernels are of positive type \eqref{G of positive type}. But it was shown in Theorem 2.24 of \cite{Gatheral2012} that \eqref{eq: J 0} admits a unique minimizer $\mu^*\in\Phi_0$ provided that $G$  is convex and satisfies \eqref{G assumption}. It was shown moreover that convexity guarantees that $\mu^*$ is a probability measure. The following proposition extends this latter result to the case $\gamma>0$. Its proof also provides an alternative proof for the existence of minimizers of  \eqref{eq: J 0}. Note that every convex, nonincreasing, and nonnegative function $G$ is of positive type in the sense of \eqref{G of positive type} due to Equation \eqref{eq: Jgamma Fourier rep} below.\footnote{This fact relies on the well-known result that $G(|\cdot|)$ is positive definite in the sense of Bochner  for every bounded, convex, and nonincreasing function $G:[0,\infty)\to[0,\infty)$. This latter result is often attributed to \cite{Polya49}, although it is also an easy consequence of \cite{Young}.}

\begin{proposition}\label{GSS prop}Suppose that $\gamma>0$, $T>0$, and  $G$ satisfies \eqref{G assumption}. If  $G$ is convex on  $(0,T]$, then the unique minimizer of \eqref{eq: J gamma} is a probability density.
\end{proposition}

The nonnegativity of minimizers to  \eqref{eq: J gamma} and \eqref{eq: J 0}, which only involve a one-dimensional linear constraint, yields the solutions to the minimization of the functional $J_\gamma$ over probability measures or probability densities.  The latter problem is of interest in many applications (see, e.g., \cite{Gatheral2012} and \cite{Alfonsi2013}).

The following proposition links the minimizer of $J_\gamma$ for $\gamma>0$ to the solution of a Fredholm integral equation of second kind with constant free term.

\begin{proposition}\label{Fredholm prop}Suppose that $\gamma>0$, $T>0$, and  $G$ satisfies \eqref{G assumption}. For a function $\phi\in\Phi_1$, the following conditions are equivalent.
\begin{enumerate}[label={\rm({\alph*})}]
\item $\phi$ solves \eqref{eq: J gamma}.
\item There exists a constant $\sigma$  such that $\phi$ solves \eqref{Fredholm eq}.
\end{enumerate}
In this case, the constant $\sigma$ from {\rm(b)} is equal to $2J_\gamma[\phi]=2\inf_{\psi\in\Phi_1}J_\gamma[\psi] $ and is strictly positive.
\end{proposition}

Now we prepare for the statement of our main result. Let $\tau\in(0,\infty]$.  Recall that a function $f:(0,\tau)\to\mathbb{R}$ is called \emph{completely monotone on $(0,\tau)$} if $f$ admits derivatives of all orders throughout $(0,\tau)$ and if
$(-1)^nf^{(n)}(x)\ge0$ for all $x\in (0,\tau)$ and $n=0,1,\dots$. According to Bernstein's theorem, completely monotone functions on $(0,\infty)$ are a special case, as they can be represented as the Laplace transforms of positive Radon measures on $[0,\infty)$. This representation may fail if $\tau<\infty$. A simple example is the function $f(t)=e^t+e^{T-t}$ for $T>0$, which is completely monotone on $(0,T/2)$  but not on  $(0,T)$. This function, however, belongs to the following class.

\begin{definition}A function $f:(0,T)\to\mathbb{R}$ is called \emph{symmetrically totally monotone} if it is analytic in $(0,T)$ and its power series development around $T/2$ is of the form
$$f(x)=\sum_{n=0}^\infty a_{2n}(x-T/2)^{2n}
$$
for coefficients $a_{2n}\ge0$.
\end{definition}

This terminology is motivated by the fact that  any symmetrically totally monotone function $f$ on $(0,T)$ is symmetric in the sense that $f(x)=f(T-x)$, completely monotone on $(0,T/2)$, and absolutely monotone on $(T/2,T)$ (i.e., $f^{(n)}(x)\ge0$ for $x\in(T/2,T)$ and $n=0,1,\dots$). In particular, every symmetrically totally monotone function on $(0,T)$ is convex and has a minimum at $T/2$.

\begin{theorem}\label{th:completemonotonicity}
	Suppose that  that $T>0$ and that $G:(0,\infty)\to[0,\infty)$ is completely monotone, nonconstant, and satisfies  \eqref{eq: weakly singular G def}.
	\begin{enumerate}[label={\rm({\alph*})}]
	\item For $\gamma>0$, the unique minimizer of \eqref{eq: J gamma} is symmetrically totally monotone.
	\item For $\gamma=0$, the restriction  to $(0,T)$ of the unique minimizer $\mu^*$ of \eqref{eq: J 0} admits a symmetrically totally monotone Lebesgue density.
	\end{enumerate}
\end{theorem}

\begin{remark} Theorem~\ref{th:completemonotonicity}
answers our initial question $(*)$ by providing a sufficient criterion on $G$ that guarantees that solutions of \eqref{Fredholm eq} are convex with a minimum at $T/2$. It makes sense, however, to ask whether our condition of complete monotonicity can perhaps be replaced by $n$-monotonicity for some $n\ge2$. Recall that a function $f$ on $(0,\infty)$ is called $n$-monotone if $(-1)^kf^{(k)}$ is nonnegative, nonincreasing, and convex for $k=0,1,\dots, n-2$. According to \cite{Williamson}, any such function $f$ can be represented in the form
$$f(t)=\int ((1-\rho t)^+)^{n-1}\,\mu(\dif \rho)
$$
for some Radon measure $\mu$ on $(0,\infty)$.
We know from  \cite{Gatheral2012}  that 2-monotonicity of $G$ is a sufficient condition for the nonnegativity of $\varphi$. It is thus tempting to conjecture that 4-monotonicity of $G$ is sufficient for the convexity of $\varphi$. Unfortunately, however, our numerical analysis provided in Figure~\ref{comp mon figure} suggests that this conjecture is not true: there are nonconvex solutions to \eqref{Fredholm eq} for $G(t)==((1-10 t)^+)^4$ and $G(t)=((1-10 t)^+)^5$.
\end{remark}

\begin{remark}One may wonder why in Theorem~\ref{th:completemonotonicity},   $G$ is defined on the entire interval $(0,\infty)$, although only its values on $(0,T]$ are relevant for our problems \eqref{eq: J gamma} and  \eqref{eq: J 0}. The reason is that our proof of Theorem~\ref{th:completemonotonicity} makes heavy use of Bernstein's theorem that gives a one-to-one correspondence between the completely monotone functions on $[0,\infty)$ and the Laplace transforms of nonnegative finite Borel measures on $[0,\infty)$. This correspondence may fail if $G$ is only defined on a finite interval.  An example is the function $\arcsin(1-t)$, which is completely monotone on $(0,1)$. \end{remark}

For $\gamma=0$, the unique minimizer $\mu^*$ has strictly positive point masses in $0$ and $T$ as soon as both $G(0+)$ and $G'(0+)$ are finite and $G$ is convex in addition to \eqref{G assumption} \citep[Theorem 2.23]{Gatheral2012}. If, however,   $G(0+)=\infty$, then we must have $\mu^*(\{0\})=\mu^*(\{T\})=0$, and so $\mu^*$ will be absolutely continuous with respect to the Lebesgue measure on all of $[0,T]$. We presently do not know what happens if $G(0+)<\infty$ and $|G'(0+)|=\infty$.

\begin{example}[Exponential kernel]Consider a completely monotone  kernel of the form
$$
	G(t) = \sum_{k=1}^n a_k e^{-\sqrt{b_k} t}
$$
for  coefficients $a_1, a_2, \dots, a_n > 0$ and $b_n > b_{n-1} > \dots > b_1 > 0$.  We will show in Section~\ref{exponentialkernels} that the unique solution of
\eqref{Fredholm eq} is of the form
$$\phi(t)=z_0+\sum_{i=1}^nz_i\big(e^{\sqrt{c_i}t}+e^{\sqrt{c_i}(T-t)}\big)
$$
where $z_i\ge0$ and the coefficients $c_i$ are equal to the eigenvalues of the matrix $M$ from \eqref{eq:M matrix} and satisfy $c_n>b_n>c_{n-1}>b_{n-1}>\cdots c_1>b_1>0$. This function $\phi$ is clearly symmetrically totally monotone.
In the special case $n=1$ with $G(t)=e^{-\sqrt b t}$, we have $c=b+\frac2\gamma \sqrt{b}$ and a direct calculation   yields that
	$$
		\phi(t) = \frac{ \sigma b}{\gamma c} \Big( 1 + \frac{2 ( e^{\sqrt c t} + e^{\sqrt c (T-t)} )}{\gamma\big(e^{\sqrt c T} (\sqrt b+\sqrt c) + \sqrt b-\sqrt c\big)} \Big),
		\qquad t \in [0,T],
		$$
	where the constant $\sigma>0$ is as in \eqref{Fredholm eq}. For solving \eqref{eq: J gamma}, $\sigma$ can be determined through the condition $\int_0^T\phi(t)\dif t=1$.
\end{example}

The following two examples illustrate that the assertions of Proposition~\ref{GSS prop} and Theorem~\ref{th:completemonotonicity} need no longer be true if the corresponding hypotheses are not satisfied. More precisely, the following Example~\ref{capped linear example} shows that the minimizer $\phi$ need not be convex even if $G$ is convex and nonincreasing, and Example~\ref{trig example} illustrates that $\phi$ can become negative if $G$ is merely of positive type and not convex.

\begin{example}[Capped linear kernel]\label{capped linear example}Consider the convex nonincreasing kernel $G(t)=(1- t)^+$ and the equation
\begin{equation}
\label{eq:cappedlinear}
	\gamma \phi(t) + \int_0^{T} \big( 1- |t-s| \big)^+ \phi(s) \dif s = \sigma,
\end{equation}
where we assume for simplicity that $T=n\in\mathbb{N}$. For $i = 1, \dots, n,$ define
$
	\lambda_i \coloneqq 2 \big( 1- \cos \big( \frac{i\pi}{n+1}\big) \big)
$
and
$
	b_i \coloneqq \sqrt{\lambda_i / \gamma}.
$
Let $B \coloneqq \diag(b_1, \dots, b_n),$
$$
	Q \coloneqq \Big( \sin \Big( \frac{ij\pi}{n+1} \Big) \Big)_{i,j = 1, \dots, n}
$$
and $E(t) \coloneqq \diag\big( e^{b_1 t}, \dots, e^{b_n t}\big).$
Furthermore, define ${\bm \sigma} \coloneqq (\sigma, \dots, \sigma) \in \mathbb{R}^n$,
denote by $I$ the $n$-dimensional identity matrix,  let $J \coloneqq \diag(1, -1, 1, \dots, \pm 1) \in \mathbb{R}^{n\times n}$,
and put $K \coloneqq I +
( \delta_{ j,n-i} )_{i,j = 1, \dots, n}$. For the solution $\phi$ of \eqref{eq:cappedlinear}, as provided by Propositions~\ref{GSS prop} and~\ref{Fredholm prop}, define $\phi_i(t) \coloneqq \phi(t+i-1)$ for $t \in [0,1], i = 1, \dots, n.$
We will prove in Section~\ref{capped linear section} that the functions $\phi_1,\dots, \phi_n$ are given by
\begin{equation}\label{capped linear formula}
\big( \phi_1(t), \dots, \phi_n(t) \big)^\top = Q \big( E(t) + E(1-t) J \big) a,
	\qquad t \in [0,1],
\end{equation}
where
$$
	a \coloneqq \Big( \gamma Q (E(1) + J) + KQ \big( (E(1)-I)(J-I) + B(E(1)-J)\big) B^{-2} \Big)^{-1} {\bm \sigma}.
$$
See Figure 1 for an illustration.
\end{example}

\begin{example}[Trigonometric kernel]\label{trig example}Let $G(t)=\cos(\rho t)$ for a constant $\rho>0$. It is well known that $G$ is a positive definite function and hence satisfies \eqref{G of positive type}, but it is of course not convex. One easily verifies that the solution $\phi$ of \eqref{Fredholm eq}  is given by
$$\phi(t)=\frac\sigma\gamma\bigg(1-\frac{2\tan(\rho T/2)\big(\cos(\rho t)+\cos(\rho(T-t)\big)}{\rho(2\gamma+T)+\sin(\rho T)} \bigg).
$$
This function  clearly takes negative values; see  Figure 2.
\end{example}

\begin{minipage}[b]{8.5cm}
\includegraphics[width=8.5cm]{Plots/capped_linear_n=11_gamma=0,01.pdf}\\
Figure 1: Solution $\phi$ of \eqref{eq:cappedlinear} for $G(t)=(1-t)^+$, $\gamma=0.01$, and $T=11$. Although $\phi$ is positive, it is not convex.
\end{minipage}\qquad
\begin{minipage}[b]{8.5cm}
\vskip-1cm\includegraphics[width=8.5cm]{Plots/trigonometric_rho=0,5_T=1.pdf}\vspace{10pt}
Figure 2: Solution $\phi$ of \eqref{Fredholm eq}  for $G(t)=\cos (t/2)$, $\gamma=0.001$, and $T=1$.\\ $ $
\end{minipage}
\addtocounter{figure}{2}
\begin{figure}[h]
\centering
\includegraphics[width=8.5cm]{Plots/CappedQuarticDecay9.pdf}\qquad
\includegraphics[width=8.5cm]{Plots/CappedQuintDecay9.pdf}
\caption{Solutions $\varphi$ of \eqref{Fredholm eq}  for $G(t)=((1-10 t)^+)^4$ (left) and $G(t)=((1-10 t)^+)^5$ with $T=1$ and $\gamma=0.001$.  }
\label{comp mon figure}
\end{figure}

\section{Auxiliary results on symmetrically totally monotone functions}

Let us introduce the notation $\Delta_hf(x):=f(x+h)-f(x)$ for a function $f$. We will say that $f$ is symmetric around $T/2$ if $f(x)=f(T-x)$.

\begin{lemma}\label{symmetric totally mon lemma}For an analytic function $f:(0,T)\to\mathbb{R}$, the following conditions are equivalent.
\begin{enumerate}[label={\rm({\alph*})}]

\item $f$ is symmetrically totally monotone.
\item $f$ is  symmetric around $T/2$, completely monotone on $(0,T/2)$, and absolutely monotone on $(T/2,T)$.
\item $f$ is  symmetric around $T/2$ and
$$\Delta_h^nf(x)=\sum_{k=0}^n(-1)^{n-k}{n\choose k}f(x+kh)\ge0
$$
for $x> T/2$, $n=1,2,\dots$,  and $h>0$ with $x+nh<T$.
\end{enumerate}\end{lemma}

\begin{proof} The implication (a)$\Rightarrow$(c) is straightforward. The proof of  (c)$\Rightarrow$(b) relies on results by \cite{Bernstein1914}. It was proved there (p.~451) that a function $f$ satisfying condition (c) (without necessarily being analytic \emph{a priori}) is absolutely monotone on $(T/2,T)$ and admits an analytic continuation $\tilde f$ to all of $(0,T)$. By analyticity, $\tilde f$ must coincide with $f$ on $(0,T)$. The symmetry of $f$ now implies that $f$ is completely monotone on $(0,T/2)$.  To show (b)$\Rightarrow$(a), note that complete monotonicity in $(0,T/2)$ together with absolute monotonicity in $(T/2,T)$ implies that $f^{(2n)}(T/2)\ge0$ and $f^{(2n+1)}(T/2)=0$ for $n=0,1,\dots$. Thus, developing $f$ into a power series around $T/2$ gives (a).
\end{proof}

The function $\arcsin(|1-x|)$ shows that the condition of analyticity in Lemma~\ref{symmetric totally mon lemma} cannot simply be dropped.
The following lemma shows in particular that the class of symmetrically totally monotone functions is closed under pointwise convergence.

\begin{lemma}\label{symm tot mon pointwise lemma}Suppose that $(f_n)$ is a sequence of symmetrically totally monotone functions on $(0,T)$ and $D$ is a dense subset of $(0,T)$ such that for each $x\in D$ the limit $\lim_nf_n(x)$ exists and is finite. Then the limit of $f_n(x)$ exists for every $x\in(0,T)$ and the function $f(x):=\lim_nf_n(x)$ is symmetrically totally monotone. Moreover,  $f_n\to f$ uniformly on compact subsets of $(0,T)$ and the coefficients in the power series development $f_n(x)=\sum_{k=0}^\infty a_k^{(n)}(x-T/2)^k$ converge to those in the corresponding development of $f$.\end{lemma}

\begin{proof}Clearly, every function $f_n$ is convex, so Theorem 10.8 from \cite{Rockafellar} yields the existence of the limit $f(x):=\lim_nf_n(x)$ for every $x\in (0,T)$, the uniform convergence $f_n\to f$ on compact subsets of $(0,T)$, and the fact that $f$ is convex and, hence, continuous on $(0,T)$. Moreover, we clearly have $\Delta^k_hf_n(x)\to\Delta^k_hf(x)$ for every $x\in(T/2,T)$, $k=0,1,2,\dots$, and $h>0$ with $x+kh<T$.  The result from  \cite{Bernstein1914} quoted in the proof of Lemma~\ref{symmetric totally mon lemma} implies that $f$ is analytic on $(T/2,T)$ and can be extended to an analytic function $\tilde f$ on $(0,T)$. We will show below that $\tilde f$ is symmetrically totally monotone. Then the symmetry of the functions $f_n$,  $f$, and $\tilde f$ and the continuity of $f$ and $\tilde f$  will imply that $f=\tilde f$ on all of $(0,T)$, and the result will be proved.

Now we prove that $\tilde f$ is symmetrically totally monotone. To this end, let $f_n(x)=\sum_{k=0}^\infty a_k^{(n)}(x-T/2)^k$ and $\tilde f(x)=\sum_{k=0}^\infty \tilde a_k(x-T/2)^k$ denote the power series developments of $f_n$ and $\tilde f$ around $T/2$.  In a first step, we note that $a_0^{(n)}=f_n(T/2)\to\tilde f(T/2)=\tilde a_0$, according to the convergence established in the preceding paragraph. Next, consider the functions $f_{n,1}(x)=\sum_{k=0}^\infty a_{k+1}^{(n)}|x-T/2|^k$. Since $a_{k}^{(n)}\ge0$ and $a_{2k+1}^{(n)}=0$, these functions are convex and we have
$$f_{n,1}(x)=\sgn(x-T/2)\sum_{k=0}^\infty a_{k+1}^{(n)}(x-T/2)^k=\frac{f_n(x)-a_0^{(n)}}{|x-T/2|}.$$
Therefore,  these functions converge pointwise on $(0,T/2)\cup (T/2,T)$ to $\tilde f_1(x)=(\tilde f(x)-\tilde a_0)/|x-T/2|$. Using once again Theorem 10.8 from \cite{Rockafellar}, we conclude that $\tilde f_1$ has a continuous and convex extension to all of $(0,T)$ and that $f_{n,1}\to\tilde f_1$ locally uniformly. It follows that $a_1^{(n)}=f_{n,1}(T/2)\to \tilde f_1(T/2)=\tilde a_1$. Next, by considering the convex functions $f_{n,2}(x)=\sum_{k=0}^\infty a_{k+2}^{(n)}(x-T/2)^k$, we conclude in the same way that $a^{(n)}_2\to \tilde a_2$. Iterating this argument further yields that $a^{(n)}_k\to \tilde a_k$  for all $k$, and hence that $\tilde a_k\ge0$ and $\tilde a_{2k}=0$  for all $k$. Therefore, $\tilde f$ is indeed symmetrically totally monotone.
\end{proof}

\begin{lemma}\label{weak convergence lemma}Let $\mathscr{M}$ denote the class of all nonnegative finite Borel measures on $[0,T]$ whose restrictions to $(0,T)$ admit a symmetrically totally monotone Lebesgue density. Then $\mathscr{M}$ is closed with respect to weak convergence of  measures.
\end{lemma}

\begin{proof}Let $(\mu_n)_{n=1,2,\dots}$ be a sequence of  measures in $\mathscr{M}$ such that $\mu_n$ converges weakly to a finite measure $\mu_0$ on $[0,T]$ and denote by $F_n(x)=\mu_n([0,x])$ the corresponding distribution functions. Then $F_n(x)\to F_0(x)$ for all continuity points of $F_0$ and hence on a dense subset of $(0,T)$. By assumption, $F_n$ is the integral of an absolutely monotone function on $(T/2,T)$ and thus absolutely monotone there itself. In particular, $F_n$ is convex on $[T/2,T]$. Since, moreover, $F_n(x)=F_n(T)-F_n(T-x)$ for $x\in[0,T/2]$, it is concave on $[0,T/2]$. By arguing as in the proof of Lemma~\ref{symm tot mon pointwise lemma}, we thus conclude that $F_n(x)\to F_0(x)$ for all $x\in(0,T)$, that $F_0$ is absolutely monotone on $(T/2,T)$, and that $F_0$ is analytic on $(0,T)$ with a symmetrically totally monotone  derivative there. \end{proof}

\begin{lemma}\label{weakly closed lemma}The class of all symmetrically totally monotone functions in $L^2[0,T]$ is weakly closed in $L^2[0,T]$.
\end{lemma}

\begin{proof}Let $\mathscr{S}$ denote the class of all symmetrically totally monotone functions in $L^2[0,T]$. If a sequence $(f_n)$ in $\mathscr{S}$ converges in $L^2[0,T]$ to a function $f$, then $f\ge0$ and the finite measures $f_n(x)\dif x$ converge weakly to the finite measure $f(x)\dif x$. Hence, $f\in\mathscr{S}$ by Lemma~\ref{weak convergence lemma}. Hence, the convex cone $\mathscr{S}$ is closed in $L^2[0,T]$ and thus also weakly closed.\end{proof}

\section{Proofs of the results from Section~\ref{results section}}\label{proofs section}

\subsection{Preliminaries}

Let us start by recalling some facts from \cite{Gatheral2012} on convex, nonincreasing, and nonconstant kernels $G:(0,\infty)\to[0,\infty)$ satisfying the condition \eqref{eq: weakly singular G def}. By Lemma 4.1 in \cite{Gatheral2012}, there exists a positive Radon measure $\eta$ on $(0,\infty)$ such that
\begin{equation}
\int_{(0,\infty)}\min\{y, y^2\}\,\eta(dy)<\infty
\end{equation}
and
\begin{equation}\label{psietaRelationEq}
G(x)=G(\infty-)+\int_{(0,\infty)}(y-x)^+\,\eta(dy)\quad\text{for $x>0$.}
\end{equation}
The Fourier transform of a Radon measure $\mu$ on $\mathbb{R}$ for which $\mu([-x,x])$ grows at  most polynomially in $x$ can be defined through
$$\widehat \mu(f)=\int\widehat f\,d\mu,\qquad f\in\mathscr{S}(\mathbb{R}),
$$
where $\mathscr{S}(\mathbb{R})$ is the usual  Schwartz space   of rapidly decreasing $C^\infty$-functions and
$\widehat f(z)=\int e^{izx}f(x)\dif x$
is the Fourier transform of $f$ (in the convention of \cite{Gatheral2012}). With this definition, it was shown in Lemma 4.2 of \cite{Gatheral2012} that $G(|\cdot|)$ can be represented as the Fourier transform of the positive Radon measure
$$\nu(dx)={G}(\infty-)\delta_0(dx)+g(x)\,dx,
$$
on $\mathbb{R}$, where the density $g$ is given by
\begin{equation}\label{eq:g function}
g(x)=\frac1\pi\int_{(0,\infty)}\frac{1-\cos xy}{x^2}\,\eta(dy)
\end{equation}
and the measure $\eta$ is in \eqref{psietaRelationEq}.
Now let $\mu$ be any signed Borel measure     on $[0,T]$  whose total variation measure $|\mu|$ is finite and such that
$\int \int G(|t-s|)\,|\mu|(dt)\,|\mu|(ds)<\infty$.
Proposition 4.5 in \cite{Gatheral2012} then shows that
\begin{equation}\label{eq: J0 Fourier rep}
J_0[\mu]=\frac12\int|\widehat\mu(z)|^2\,\nu(dz).
\end{equation}
It therefore follows from Plancherel's theorem that for $\gamma>0$,
\begin{equation}\label{eq: Jgamma Fourier rep}
J_\gamma[\phi]=\frac\gamma2\int|\widehat\phi(z)|^2\,dz+\frac12\int|\widehat\phi(z)|^2\,\nu(dz),\quad\phi\in\Phi_1.
\end{equation}
As a matter of fact, the preceding identity extends to the space $L^2_G[0,T]$ of all functions $\phi\in L^2[0,T]$ for which $J_\gamma[\phi]$ is finite. It is clear from \eqref{eq: Jgamma Fourier rep}
 and Minkowski's inequality that $L^2_G[0,T]$ is a vector space.

\subsection{Proof of Proposition~~\ref{GSS prop}}

For the proof of  Proposition~\ref{GSS prop}, we will need two auxiliary lemmas. For $n\in\mathbb{N}$, we let $\Phi_1^{(n)}$ denote the set of all $\phi \in L^2[0,T]$
	that satisfy $\int_0^T \phi(t) \dif t = 1$ and that are constant
	on all intervals of the form $[t_k, t_{k+1})$, where $t_k = k2^{-n}T$ for $k=0,\dots, 2^n$.
	Any such $\phi$ is thus of the form
	\begin{equation}
	\label{alpha in Xn}
		\phi = \sum_{k=0}^{2^n-1} \phi_k \Ind{[t_k, t_{k+1})}
	\end{equation}
	for certain real coefficients $\phi_k$ that sum up to  $2^n/T$. In particular, $\phi$ belongs to $L^\infty[0,T]$ and hence to $\Phi_1$. We need the following simple lemma.

	\begin{lemma}
	\label{n quadratic form lemma}
		Suppose that $G$ satisfies \eqref{G assumption}. For $\phi \in \Phi_1^{(n)}$ of the form \eqref{alpha in Xn}, we have
		$$
			 J_\gamma[\phi] = \sum_{i,j=0}^{2^n}\phi_i\phi_jG_n(|t_i-t_j|)
		$$
		where
		\begin{align*}
			G_n(0) &= \gamma 2^{-(n+1)}  T  + 2^{-2n+1}  T^2  \int_0^1 G(2^{-n} T s) \, (1-s) \dif s, \\
			G_n(t) &= 2^{-2n} T^2  \int_{-1}^1G(t+2^{-n} T s) \, (1-|s|) \dif s
			\qquad\text{for $t \ge 2^{-n}T$,}
		\end{align*}
		and $G_n(t)$ linearly interpolated between $G_n(0)$ and $G_n(2^{-n}T)$ for $ t \in (0,2^{-n}T)$.
	\end{lemma}
	\begin{proof}
		We  have
		$$
			\frac{\gamma}{2} \int_0^T\phi(t)^2 \dif t
			= \frac{\gamma}{2} \sum_{i=0}^{2^n-1} \phi_i^2(t_{i+1}-t_i)
			= \gamma 2^{-(n+1)}  T  \sum_{i=0}^{2^n-1} \phi_i^2.
		$$
		Moreover,
		\begin{align*}
			\frac{1}{2} \int_0^T \int_0^T G(|t-s|) \phi(t) \phi(s) \dif s \dif t
			&=\frac{1}{2} \sum_{i,j=0}^{2^n-1}\phi_i \phi_j \int_{t_i}^{t_{i+1}} \int_{t_j}^{t_{j+1}} G(|t-s|) \dif s \dif t.
		\end{align*}
		For $i<j$, we have
		\begin{align*}
			\int_{t_i}^{t_{i+1}} \int_{t_j}^{t_{j+1}} G(|t-s|) \dif s \dif t
			&= 2^{-2n}  T^2  \int_0^1 \int_0^1 G(t_j-t_i+2^{-n} T (s-r)) \dif s \dif r
			= G_n(t_j-t_i).
		\end{align*}
		For $i=j$, we have
		\begin{align*}
			\int_{t_i}^{t_{i+1}} \int_{t_i}^{t_{i+1}} G(|t-s|) \dif s \dif t
						&= 2^{-2n+1} T^2  \int_0^1 G(2^{-n} T s)(1-s) \dif s.
		\end{align*}
		This determines the values of $G$ in the points $t_k$ for $k = 0,\dots, 2^n$.
		The values of $G_n(t)$ for all other $t$ do actually not matter for the representation of $J_\gamma[\phi]$,
		and hence can be chosen arbitrarily, for instance, as in the statement of the lemma.
	\end{proof}

Note  that the function $G_n$ need not be convex if $G$ is convex.
	This can be seen by taking, for instance, $n=0$,  $T=1$, $G(t)=(3-t)^+$,  and $\gamma$ small. However, we have the following result.

	\begin{lemma}\label{convex Gn lemma}If $\gamma>0$ and the convex function $G$ satisfies \eqref{G assumption}, $G(0+)<\infty$, and has a finite right-hand derivative $G'_+(0)$ at zero, then there exists $n_0\in\mathbb N$ such that the function $G_n$ defined in Lemma~\ref{n quadratic form lemma} is convex for each $n\ge n_0$.\end{lemma}

\begin{proof}On $[0,2^{-n}T],$ the function~$G_n$ is linear.
	On $[2^{-n}T,\infty)$, it is a mixture
	of the convex, nonincreasing, and nonnegative functions $G(\cdot+2^{-n}Ts)$ for $-1 \le s \le 1$.
	Hence, $G_n$ also has these properties on $[2^{-n}T,\infty)$.	We conclude that~$G_n$ is convex if and only if the  left-hand derivative~$G_{n,-}'(2^{-n}T)$ of~$G_n$ in~$2^{-n}T$
	is smaller than or equal to the right-hand derivative~$G_{n,+}'(2^{-n}T).$
	The convexity of $G$ and dominated convergence imply that
	\begin{align*}
	G_{n,+}'(2^{-n}T)&=\lim_{t\downarrow 0}\frac{G_n(t+2^{-n}T)-G_n(2^{-n}T)}{t}\\
		&=2^{-2n}T^2\int_{-1}^1G'_+(2^{-n}T(1+s))(1-|s|)\dif s\\
		&\ge 2^{-2n }T^2G'_+(0).
	\end{align*}
	On the other hand, since $G$ is nonnegative and nonincreasing,
	\begin{align*}
	G_{n,-}'(2^{-n}T)&=\frac{G_n(2^{-n}T)-G_n(0)}{2^{-n }T}\\
	&=-\frac\gamma 2+2^{-n}T\int_{-1}^1\big(G(2^{-n}T(1+s))-G(2^{-n}T|s|)\big)(1-|s|)\dif s\\
	&\le -\frac\gamma 2+2^{-n}TG(0+)
	\end{align*}
	Thus, if $n$ is sufficiently large, then the term $\gamma/2$ becomes dominant and ensures the convexity of $G_n$.
\end{proof}

	\begin{proof}[Proof of Proposition~\ref{GSS prop}]
 The uniqueness of minimizers follows immediately from the fact that $J_\gamma$ is strictly convex by \eqref{eq: Jgamma Fourier rep}.
To show the existence of a nonnegative minimizer, we  consider  first  the case in which both $G(0+)$ and the right-hand derivative $G'_+(0)$ are finite. When letting $G(0):=G(0+)$, the function $G(|\cdot|)$ is  a bounded and continuous function on $\mathbb{R}$.

	Now consider the problem of minimizing $J_\gamma[\phi]$ over $\phi\in \Phi_1^{(n)}$.
	By Lemma~\ref{n quadratic form lemma}, this problem is equivalent to the minimization of the quadratic form
	$\sum_{i,j=0}^{2^n} \phi_i \phi_j G_n(|t_i-t_j|)$ over $\phi_0, \dots, \phi_{2^n} \in \mathbb{R}$ that sum up to $2^n/T$.  The fact that $G_n$ is convex, nonincreasing, nonnegative, and nonconstant implies that the matrix with entries
	$ G_n(|t_i-t_j|)$ is positive definite due to \eqref{eq: J0 Fourier rep}. Thus, our minimization problem has a unique minimizer as soon as $n \ge n_0$.
	Moreover, by Theorem~1 in \cite{ASS}, together with Lemmas~\ref{n quadratic form lemma} and~\ref{convex Gn lemma}, all components $\phi_k$ of this minimizer will be nonnegative.
	Thus, also the problem of minimizing $J_\gamma[\phi]$ over $\phi \in \Phi_1^{(n)}$ has a unique minimizer $\phi^{(n)}$,
	which is nonnegative, as soon as $n\ge n_0$.

	Next, since $\Phi_1^{(n)}\subset\Phi_1^{(n+1)}$, we have $J_\gamma[\phi^{(n_0)}] \ge J_\gamma[\phi^{(n)}]$ for all $n\ge n_0$.
	Since moreover $J_\gamma[\phi] \ge \frac{\gamma}{2}\|\phi\|^2$ due to \eqref{eq: J0 Fourier rep},
	we get that the $L^2$-norms $\|\phi^{(n)}\|$ are uniformly bounded for all $n \ge n_0$. By passing to a subsequence if necessary, we may therefore assume that
	the sequence $(\phi^{(n)})_{n\ge n_0}$ converges weakly in $L^2[0,T]$ to some nonnegative limit $\phi^*$.

	We claim that $\phi^*$ is the  minimizer of $J_\gamma[\phi]$
	over $\phi \in \Phi_1$. To see this, let us assume by way of contradiction that $\phi^*$ is not the minimizer. Then there exists another function $\hat\phi\in\Phi_1$ with $J_\gamma[\hat\phi]<J_\gamma[\phi^*]$. By $\hat \phi_n$ we denote the conditional expectation
	of $\hat\phi$ with respect to the $\sigma$-field on $[0,T]$ generated by the intervals $[t_0,t_1),\dots, [t_{2^n-1},t_{2^n})$
	and under the (normalized) Lebesgue measure. Then $\hat\phi_n $ belongs to $\Phi_1^{(n)}$,
	and so $J_\gamma[\hat\phi_n] \ge J_\gamma[\phi^{(n)}]$.
	By martingale convergence we have $\hat\phi_n\to\hat\phi$ in $L^2[0,T]$. 	The fact that $G(|\cdot|)$ is bounded and continuous   gives $J_\gamma[\hat\phi_n] \to J_\gamma[\hat\phi]$. On the other hand,  the map $\phi \mapsto J_\gamma[ \phi ]$ is weakly lower semicontinuous, and so
	$$J_\gamma[\phi^*] \le \liminf_{n\uparrow\infty} J_\gamma[\phi^{(n)}]\le\lim_{n\uparrow\infty} J_\gamma[\hat\phi_n] =J_\gamma[\hat\phi],
	$$
which is a contraction. Therefore, the nonnegative function $\phi^*$ is indeed the minimizer.

Let us now consider the case $G(0+)=\infty$. As in the proof of Theorem 2.24 of \cite{Gatheral2012}, we can consider approximations $G^{(n)}$ of $G$ defined through the measures
\begin{equation}\label{eq:etan}
\eta_n(dy)=\Ind{(1/n,\infty)}(y)\,\eta(dy)
\end{equation}
in \eqref{psietaRelationEq}. These functions $G^{(n)}$ are then continuous, nonincreasing, nonnegative, convex and they satisfy $G^{(n)}(0+)<\infty$ and $(G^{(n)})'_+(0)>-\infty$. They correspond to functions $g_n$ defined as in \eqref{eq:g function} and  energy functionals $J_\gamma^{(n)}$ satisfying \eqref{eq: Jgamma Fourier rep} for $\nu_n(dx)=G(\infty-)\, \delta_0(dx)+g_n(x)\dif x$.
Then let $(\psi_n)_{n=1,2,\dots}$ be a minimizing sequence for $J_\gamma$ in $\Phi_1$. Since $g(x)\ge g_n(x)$, we have $J_\gamma[\psi_n]\ge J_\gamma^{(n)}[\psi_n]$.  For each $n$, we take moreover a minimizer $\phi_n$ of $J^{(n)}_\gamma$ in $\Phi_1$. Since $G^{(n)}(0+)<\infty$ and $(G^{(n)})'_+(0)>-\infty$, we already know from the first part of this proof that $\phi_n\ge0$. Moreover, we have  $J_\gamma[\psi_n]\ge J_\gamma^{(n)}[\psi_n]\ge J_\gamma^{(n)}[\phi_n] $. This implies that $\gamma\int_0^T\phi_n(t)^2\dif t$ is uniformly bounded in $n$, and so, after passing to a subsequence if necessary, we may assume that the sequence $(\phi_n)$ converges weakly in $L^2[0,T]$ to a function $\phi\in L^2[0,T]$, which must also be nonnegative.  Due to the compactness of $[0,T]$, it follows that the Fourier transforms $\widehat\phi_n$ converge pointwise to $\widehat\phi$.  Since moreover  $g_n$ increases pointwise to the function $g$  from \eqref{eq:g function}, we get
\begin{align*}
\inf_{\psi\in\Phi_1}J_\gamma[\psi]&=\lim_{n\uparrow\infty}J_\gamma[\psi_n]\ge\liminf_{n\uparrow\infty}J_\gamma^{(n)}[\phi_n]\\&=\liminf_{n\uparrow\infty}\bigg(\frac\gamma2\int|\widehat\phi_n(z)|^2\,dz+G^{(n)}(\infty-)|\widehat\phi_n(0)|^2+\frac12\int|\widehat\phi_n(z)|^2g_n(z)\dif z\bigg)\\
&\ge J_\gamma[\phi],\end{align*}
where we have used Fatou's lemma in the final step. This shows that the function $\phi$ is the desired nonnegative minimizer.\end{proof}

\subsection{Proof of Proposition~\ref{Fredholm prop}}

 For $f,g\in L^2_G[0,T]$, we can define the symmetric bilinear form
$$J_\gamma[f,g]:=\frac12\big(J_\gamma[f+g]-J_\gamma[f]-J_\gamma[g]\big).
$$
Now suppose $\phi$ solves \eqref{eq: J gamma}. We take a nonzero function $\psi\in   L^2_G[0,T]$ such that $\int_0^T\psi(t)\dif t=0$ and $\alpha\in\mathbb{R}$. Then $\phi+\alpha\psi\in\Phi_1$ and
$$J_\gamma[\phi+\alpha\psi]=J_\gamma[\phi]+\alpha^2 J_\gamma[\psi]+2\alpha J_\gamma[\phi,\psi].
$$
The optimality of $\phi$ implies that the right-hand side is minimized at $\alpha=0$, which implies that $J_\gamma[\phi,\psi]=0$. Thus, $\gamma\phi(t)+\int_0^TG(|t-s|)\phi(s)\dif s$ must be orthogonal to every $\psi\in   L^2_G[0,T]$ with  $\int_0^T\psi(t)\dif t=0$, which gives \eqref{Fredholm eq}.

Conversely,  \eqref{Fredholm eq} implies that $J_\gamma[\phi,\psi]=0$ for every $\psi\in   L^2_G[0,T]$ with  $\int_0^T\psi(t)\dif t=0$. For every $\tilde\phi\in\Phi_1$ and $\psi:=\tilde\phi-\phi$,
$$J_\gamma[\tilde\phi]=J_\gamma[\phi]+J_\gamma[\psi]+2J_\gamma[\phi,\psi]\ge J_\gamma[\phi],
$$
and so $\phi$ solves \eqref{eq: J gamma}. Finally, it is clear that $J_\gamma[\phi]=\sigma/2$ if $\phi$ solves \eqref{Fredholm eq}.
\qed

\subsection{Proof of Theorem~\ref{th:completemonotonicity}}

\subsubsection{Proof of Theorem~\ref{th:completemonotonicity} for exponential kernels}
\label{exponentialkernels}

Assume first that $G$ is an {\em exponential kernel (of order $n$)},
i.e., there are $a_1, a_2, \dots, a_n > 0$ and $b_n > b_{n-1} > \dots > b_1 > 0$ such that
\begin{equation}
\label{eq:exponentialkernel}
	G(t) = \sum_{k=1}^n a_k e^{-\sqrt{b_k} t}.
\end{equation}
Clearly, any such $G$ is completely monotone and satisfies \eqref{eq: weakly singular G def}.
Let $\phi$ be the unique minimizer of \eqref{eq: J gamma}.
By Proposition~\ref{Fredholm prop} there is a $\sigma > 0$ such that $\phi$ solves \eqref{Fredholm eq}.
By scaling $\phi$ and $G$, we may assume without loss of generality that $\sigma=\gamma$.

All matrices considered in this proof are $n$-dimensional square matrices,
and all vectors $n$-dimensional column vectors.
We denote the diagonal matrix with $x_1, \dots, x_n$ on its main diagonal as
$\diag(x_i)_{i=1,\dots,n}$, and say that a matrix is a {\em positive diagonal matrix} if it is diagonal and all diagonal entries are positive.

Let $A \coloneqq \diag(a_i)_{i = 1, \dots, n}$ and $B \coloneqq \diag(b_i)_{i = 1, \dots, n}.$
Define the function $\psi = (\psi_1, \psi_2, \dots, \psi_n)$ via
$$
	\psi_k(t) \coloneqq a_k \int_{0}^{T} e^{-\sqrt{b_k} |t-s|} \phi(s) \dif s,
		\qquad t \in [0,T],
		\quad k =1, 2, \dots, n.
$$
Then
\begin{equation}
\label{eq:phipsi}
	\phi = 1 - \lambda \sum_k \psi_k = 1 - \lambda\, {\bm 1}^{\top} \psi,
\end{equation}
where $\lambda \coloneqq 1/\gamma$ and ${\bm 1} \coloneqq (1, 1, \dots, 1) \in \mathbb{R}^n.$

Let us first give an outline of the proof:
\begin{enumerate}[label={\bf {\arabic*.}},topsep=0pt,itemsep=-1ex,partopsep=1ex,parsep=1ex]
	\item Show that $\psi$ solves a system of $n$ ordinary differential equations $\psi'' = M\psi - 2 A B^{1/2} {\bm 1}$ with boundary conditions
		$\psi(0) = \psi(T)$ and $\psi'(0) = B^{1/2} \psi(0).$
		Here, $M$ is a nonsingular matrix.
	\item Show that $M$ has $n$ distinct, real eigenvalues $c_n > c_{n-1} > \dots > c_1 > 0.$
		Let $C \coloneqq \diag(c_i)_{i = 1, \dots, n}.$
		Obtain an eigendecomposition $M = Q C Q^{-1},$
		where $Q$ is a nonsingular matrix.
	\item Conclude with 1. that
		\begin{equation}\label{eq:phi}
			\phi(t) =  d\, \big( 1 + 2\lambda\, {\bm 1}^{\top} \big( e^{M^{1/2} t} + e^{M^{1/2} (T-t)} \big) N^{-1} {\bm 1} \big),
		\end{equation}
		where $d > 0$ and $N$ is a nonsingular matrix.
	\item Use the eigendecomposition of $M$ to rewrite \eqref{eq:phi} as
		\begin{equation}\label{eq:phi2}
			\phi(t) = d\, \big( 1 +  {\bm 1}^{\top} E(t) \tilde{N}^{-1} {\bm 1} \big).
		\end{equation}
		Here, $\tilde{N}$ is a nonsingular matrix. The matrices $E(t)$ are positive diagonal matrices, and each diagonal entry of the mapping $t \mapsto E(t)$ is symmetrically totally monotone.
	\item Decompose $\tilde{N}^{-1} = \tilde{N}_1 (\tilde{N}_2 + \tilde{N}_3)^{-1} \tilde{N}_4$ such that
		$\tilde{N}_1$ and $\tilde{N}_3$ are positive diagonal matrices,
		$\tilde{N}_2$ is positive definite,
		and all off-diagonal entries of $(\tilde{N}_2 + \tilde{N}_3)^{-1}$ are nonpositive.
	\item Show that all entries of $\tilde{N}_2^{-1} \tilde{N}_4\, {\bm 1}$ are nonnegative.
		Show that this implies that all entries of $\tilde{N}^{-1}{\bm 1} = \tilde{N}_1 (\tilde{N}_2 + \tilde{N}_3)^{-1} \tilde{N}_4\, {\bm 1}$
		are nonnegative.
	\item Conclude with \eqref{eq:phi2} and Step {\bf 6} that $\phi$ is symmetrically totally monotone.
\end{enumerate}

\medskip
\medskip

{\bf 1.} Recall that $A = \diag(a_i)_{i = 1, \dots, n}$ and $B = \diag(b_i)_{i = 1, \dots, n}$ are positive diagonal matrices,
and that $b_n > b_{n-1} > \dots > b_1.$
Notice that ${\bm 1}\one^{\top}$ is the matrix containing all ones.
Define
\begin{equation}\label{eq:M matrix}
	M
		\coloneqq B + 2 \lambda\, A B^{1/2} {\bm 1}\one^{\top}
		= \begin{pmatrix}
			b_1 + 2 \lambda a_1 \sqrt{b_1} & 2\lambda a_1 \sqrt{b_1} & \cdots & 2\lambda a_1 \sqrt{b_1} \\
			2\lambda a_2 \sqrt{b_2} & b_2 + 2\lambda a_2 \sqrt{b_2} & \cdots & 2\lambda a_2 \sqrt{b_2} \\
			\vdots & \vdots & \ddots & \vdots \\
			2\lambda a_n \sqrt{b_n} & 2\lambda a_n \sqrt{b_n} & \cdots & b_n + 2\lambda a_n \sqrt{b_n}
		\end{pmatrix}.
\end{equation}

\begin{enumerate}[label={\bf 1.{\arabic*}}, wide, labelwidth=!, labelindent=0pt]
	\item
	{\em $\psi$ solves the  system of $n$ ordinary differential equations $\psi'' = M\psi - 2 A B^{1/2} {\bm 1}.$}\\
		Let $t \in [0,T]$ and $k=1, 2, \dots, n.$
		Differentiating and plugging in from \eqref{eq:phipsi} shows
		\begin{align*}
			\psi_k''(t)
				&= a_k \sqrt{b_k} \, \frac{\dif}{\dif t}\Big[ - \int_0^t e^{-\sqrt{b_k} (t-s)} \phi(s) \dif s + \int_t^T e^{-\sqrt{b_k}(s-t)} \phi(s) \dif s \Big]\\
			&= a_k \sqrt{b_k} \, \Big(\sqrt{b_k} \int_0^T e^{-\sqrt{b_k}|t-s|} \phi(s) \dif s - 2 \phi(t) \Big)\\
			&= b_k \psi_k(t) - 2 a_k \sqrt{b_k} \phi(t)\\
			&= b_k \psi_k(t) - 2 a_k \sqrt{b_k} \Big( 1 - \lambda \sum_l \psi_l(t) \Big).
		\end{align*}
		We conclude
		$
			\psi''
				= (B + 2\lambda\, A B^{1/2} {\bm 1} {\bm 1}^{\top}) \psi - 2 A B^{1/2} {\bm 1} = M \psi - 2 A B^{1/2} {\bm 1}.
		$
	\item\label{step:boundary1}
	{\em $\psi(0) = \psi(T)$.}\\
		Recall that $\phi(t) = \phi(T-t)$ for all $t \in [0,T].$
		Let $t \in [0,T]$ and $k = 1, 2, \dots, n.$
		Integration by substitution shows
		\begin{align*}
			\psi_k(t)
				&= a_k \int_0^T e^{-\sqrt{b_k}|t-s|} \phi(s) \dif s\\
			&= a_k \int_0^T e^{-\sqrt{b_k} |(T-t) - (T-s)|} \phi(T-s) \dif s\\
			&= a_k \int_0^T e^{-\sqrt{b_k} |(T-t) - s| } \phi(s) \dif s\\
			&= \psi_k(T-t).
		\end{align*}
		In particular, $\psi_k(0) = \psi_k(T).$
	\item\label{step:boundary2}
	{\em $\psi'(0) = B^{1/2} \psi(0).$}\\
		Let $k = 1, 2, \dots, n.$
		Then
		\begin{align*}
			\psi_k'(0)
				&= a_k \sqrt{b_k} \Big[ - \int_0^t e^{-\sqrt{b_k} (t-s)} \phi(s) \dif s + \int_t^T e^{-\sqrt{b_k}(s-t)} \phi(s) \dif s \Big]_{t=0}\\
			&= a_k \sqrt{b_k} \int_0^T e^{-\sqrt{b_k} s} \phi(s) \dif s\\
			&= \sqrt{b_k} \, \psi_k(0).
		\end{align*}
\end{enumerate}

\medskip

\begin{enumerate}[label={\bf 2.{\arabic*}}, wide, labelwidth=!, labelindent=0pt]
	\item\label{step:eigenvalues}
	{\em $M$ has $n$ distinct, real eigenvalues $c_1, c_2, \dots, c_n$ that satisfy $c_n > b_n > c_{n-1} > b_{n-1} > \dots > c_1 > b_1 > 0.$}\\
		Let
		$v \coloneqq 2\lambda (a_1 \sqrt{b_1}, a_2 \sqrt{b_2}, \dots, a_n \sqrt{b_n}) \in \mathbb{R}^n$
		and $x \in [0,\infty) \setminus \{b_1, b_2, \dots, b_n\}.$
		 The matrix $xI-M$ is the sum of a diagonal matrix and the outer product $v{\bm 1}^\top$.
		Hence
		\begin{align*}
			\det(x I - M)
				&=\det(x I - B -v {\bm 1}^{\top})\\
			&= \big( 1 - v^{\top} (xI-B)^{-1} {\bm 1} \big) \det(x I - B)\\
			&= \Big( 1 - 2\lambda \sum_k \frac{a_k \sqrt{b_k}}{x - b_k} \Big) \prod_k (x - b_k).
		\end{align*}
		The following argument is due to \cite{Terrell2017}.
		Define $f \colon [0,\infty) \setminus \{b_1, b_2, \dots, b_n\} \to \mathbb{R}$ via
		$$
			f(x)
				\coloneqq 1 - 2\lambda \sum_k \frac{a_k \sqrt{b_k}}{x - b_k}.
		$$
		Let $k = 1, 2, \dots, n-1.$
		Then $f$ is continuous on $(b_k, b_{k+1})$, with
		$$
			\lim_{x \searrow b_k} f(x)
				= - \infty
			\qquad \text{ and } \qquad
			\lim_{x \nearrow b_{k+1}} f(x)
				= + \infty.
		$$
		We conclude that $f$ has a root $c_k \in (b_k, b_{k+1})$.
		Furthermore,
		$$
			\lim_{x \searrow b_n} f(x)
				= - \infty
			\qquad \text{ and } \qquad
			\lim_{x \nearrow +\infty} f(x)
				= 1,
		$$
		showing that $f$ has another root $c_n \in (b_n, +\infty).$
		Since $\det(c_k I - M) = 0$ for $k=1, 2, \dots, n,$ each $c_k$ is an eigenvalue of $M$.
	\item\label{step:eigenvectors}
	{\em If $c$ is an eigenvalue of $M$, then
		$$
			\Big( \frac{a_1 \sqrt{b_1}}{c-b_1} , \frac{a_2 \sqrt{b_2}}{c-b_2}, \dots, \frac{a_n \sqrt{b_n}}{c-b_n}\Big)
		$$
		is a corresponding eigenvector.}\\
		Let $c \in [0,\infty) \setminus \{b_1, \dots, b_n\}$ be an eigenvalue of $M$,
		and $v = (v_1, v_2, \dots, v_n)$ a corresponding eigenvector.
		The definition $Mv = cv$ translates into the following system of equations:
		$$
			b_k v_k + 2\lambda a_k \sqrt{b_k} \sum_l v_l
				= c v_k,
				\qquad k = 1, 2, \dots, n.
		$$
		It must be true that $\sum_l v_l \neq 0.$
		Otherwise, $b_k v_k = c v_k$ for all $k=1, 2, \dots, n.$
		Since $c \notin \{b_1, b_2, \dots, b_n\}$ (see Step~\ref{step:eigenvalues}),
		this implies $v = 0,$ which contradicts the definition of an eigenvector.
		Hence we may set ${\bm 1}^{\top} v = \sum_l v_l = \frac1{2\lambda}$ without loss of generality.
		We obtain
		$$
			v
				= \Big( \frac{a_1 \sqrt{b_1}}{c - b_1}, \frac{a_2 \sqrt{b_2}}{c - b_2}, \dots, \frac{a_n \sqrt{b_n}}{c - b_n} \Big).
		$$
\end{enumerate}

Let $c_n > c_{n-1} > \dots > c_1 > 0$ be the eigenvalues of $M$.
Define $C \coloneqq \diag(c_i)_{i = 1, \dots, n},$
\begin{equation}
\label{eq:Q}
	\tilde{Q} \coloneqq \Big(\frac1{c_j - b_i}\Big)_{i,j = 1, 2, \dots, n}
	\qquad\text{ and }\qquad
	Q \coloneqq A B^{1/2} \tilde{Q}.
\end{equation}

\begin{enumerate}[label={\bf 2.{\arabic*}}, resume, wide, labelwidth=!, labelindent=0pt]
	\item\label{step:eigendecomposition}
	{\em $M = Q C Q^{-1}.$}\\
		By Step~\ref{step:eigenvectors}, the columns of $Q$ are eigenvectors corresponding to the eigenvalues $c_1, c_2, \dots, c_n$.
		Eigenvectors corresponding to different eigenvalues are linearly independent, hence $Q$ is nonsingular.
		We obtain the eigendecomposition
		$
			M
				= Q C Q^{-1}.
		$
	\item\label{step:colsumsQ}
	{\em ${\bm 1}^{\top} Q = \frac1{2\lambda} {\bm 1}^{\top}$.}\\
		This follows from Step~\ref{step:eigenvectors},
		where we assumed that each eigenvector contained in $Q$ sums to $\frac1{2\lambda}.$
\end{enumerate}

\medskip

{\bf 3.}
Define
$$
	d
		\coloneqq \Big( 1 + 2\lambda \sum_k \frac{a_k}{\sqrt{b_k}} \Big)^{-1} > 0.
$$
We let
$
	M^{1/2} \coloneqq Q \diag(\sqrt{c_i})_{i = 1, \dots, n}\, Q^{-1}
$
and denote by
$
	e^{M^{1/2} T} = Q \diag(e^{\sqrt{c_i} T})_{i = 1, \dots, n}\, Q^{-1}
$
the matrix exponential of $M^{1/2} T.$
Define
$$
	N
		\coloneqq A^{-1} \Big( M^{1/2} \big( e^{M^{1/2} T} - I \big) + B^{1/2} \big( e^{M^{1/2} T} + I \big) \Big),
$$
where $I$ denotes the identity matrix.

The general solution to the system of $n$ ordinary differential equations
$
	f''
		= M f - 2 A B^{1/2} {\bm 1}
$
is
$$
	f(t)
		= e^{M^{1/2} t} x_0 + e^{M^{1/2} (T-t)} x_1 + 2 d A B^{-1/2} {\bm 1},
		\qquad t \in [0,T],
$$
for $x_0, x_1 \in \mathbb{R}^n.$
To see this, let $t \in [0,T]$ and $x_0, x_1 \in \mathbb{R}^n.$
Writing $d = 1 / (1 + 2\lambda\, {\bm 1}^{\top} A B^{-1/2} {\bm 1})$ shows
\begin{align*}
	d\, M A B^{-1/2} {\bm 1}
		&= d\, \big( A B^{1/2} {\bm 1} + 2\lambda\, A B^{1/2} {\bm 1}\,{\bm 1}^{\top} A B^{-1/2} {\bm 1} \big)\\
	&= d\, \Big( 1 + \frac1d - 1 \Big) A B^{1/2}{\bm 1}\\
	&= A B^{1/2}{\bm 1}.
\end{align*}
Therefore,
\begin{align*}
	f''(t)
		&= M \big( e^{M^{1/2} t} x_0 + e^{M^{1/2} (T-t)} x_1 \big)\\
	&= M f(t) - 2 d\, M A B^{-1/2} {\bm 1}\\
	&= M f(t) - 2 A B^{1/2} {\bm 1}.
\end{align*}

It remains to choose $x_0$ and $x_1$ in such a way that the boundary conditions from Steps~\ref{step:boundary1} and~\ref{step:boundary2}
are satisfied.
First,
$
	f(0) - f(T)
		= ( e^{M^{1/2} T} - I ) (x_1 - x_0).
$
By Step~\ref{step:eigendecomposition},
\begin{align*}
	e^{M^{1/2} T} - I
		&= Q \diag\big(e^{\sqrt{c_i} T}\big)_{i = 1, \dots, n}\, Q^{-1} - I\\
	&= Q \diag\big(e^{\sqrt{c_i} T} - 1\big)_{i = 1, \dots, n}\, Q^{-1}
\end{align*}
is nonsingular.
Hence $f(0) = f(T)$ if and only if $x_0 = x_1.$
Set $x_0 = x_1.$
Second,
\begin{align*}
	f'(0) - B^{1/2} f(0)
		&= \big( M^{1/2} \big( I - e^{M^{1/2} T} \big) - B^{1/2} \big( I + e^{M^{1/2} T} \big) \big) x_0 - 2d A {\bm 1}\\
	&= - A ( N x_0 + 2d\, {\bm 1}).
\end{align*}
We show in Step~\ref{step:Nnonsingular} that $N$ is nonsingular.
Hence, $f'(0) = B^{1/2} f(0)$ if and only if $x_0 = -2d\, N^{-1} {\bm 1}.$
We conclude
\begin{align*}
	\psi(t)
		&= e^{M^{1/2} t} x_0 + e^{M^{1/2} (T-t)} x_1 + 2d\, A B^{-1/2} {\bm 1}\\
	&= \big( e^{M^{1/2} t} + e^{M^{1/2} (T-t)} \big) x_0 + 2d\, A B^{-1/2} {\bm 1}\\
	&= 2d\, \big( A B^{-1/2} - \big( e^{M^{1/2} t} + e^{M^{1/2} (T-t)} \big) N^{-1} \big) {\bm 1}
\end{align*}
for all $t \in [0,T].$
Notice that
$$
	1 - 2d \lambda\, {\bm 1}^{\top} A B^{-1/2} {\bm 1}
		= 1 - d \Big( \frac1d - 1 \Big)
		= d,
$$
so
\begin{align*}
	\phi(t)
		&= 1 - \lambda\, {\bm 1}^{\top} \psi(t)\\
	&= 1 - 2d \lambda\, {\bm 1}^{\top} A B^{-1/2} {\bm 1} + 2d \lambda\, {\bm 1}^{\top} \big( e^{M^{1/2} t} + e^{M^{1/2} (T-t)} \big) N^{-1} {\bm 1}\\
	&= d\, \big( 1 + 2\lambda\, {\bm 1}^{\top} \big( e^{M^{1/2} t} + e^{M^{1/2} (T-t)} \big) N^{-1} {\bm 1} \big)
\end{align*}
for all $t \in [0,T].$

\medskip

{\bf 4.}
Define
$$
	E(t)
		\coloneqq \diag \Big( \frac{e^{\sqrt{c_i} t} + e^{\sqrt{c_i} (T-t)} }{e^{\sqrt{c_i}T} - 1}\Big)_{i = 1, \dots, n,}
		\qquad t \in [0,T],
$$
and
$$
	\tilde{N}
		\coloneqq A^{-1} \big( Q C^{1/2} + B^{1/2} Q E(T) \big).
$$
$E(t)$ is a positive diagonal matrix for all $t \in [0,T]$ and thus nonsingular. The diagonal entries of the mapping $t \mapsto E(t)$ are symmetrically totally monotone.
Using Step~\ref{step:eigendecomposition}, we obtain
\begin{align*}
	N &= A^{-1} \big( Q C^{1/2} \diag \big( e^{\sqrt{c_i} T} - 1\big)_{i = 1, \dots, n}\, Q^{-1}
		+ B^{1/2} Q \diag \big( e^{\sqrt{c_i} T} + 1\big)_{i = 1, \dots, n}\, Q^{-1} \big)\\
	&= A^{-1} \big( Q C^{1/2} + B^{1/2} Q E(T) \big) \diag \big( e^{\sqrt{c_i} T} - 1 \big)_{i = 1, \dots, n}\, Q^{-1}\\
	&= \tilde{N} \diag \big(e^{\sqrt{c_i} T} - 1 \big)_{i = 1, \dots, n}\, Q^{-1}.
\end{align*}
Hence $N$ is nonsingular if and only if $\tilde{N}$ is nonsingular.
This, in combination with Steps~\ref{step:eigendecomposition},~\ref{step:colsumsQ} and {\bf 3}, shows
\begin{align*}
	\phi(t)
		&= d\, \big( 1 + 2\lambda\, {\bm 1}^{\top} \big( e^{M^{1/2} t} + e^{M^{1/2} (T-t)} \big) N^{-1} {\bm 1} \big)\\
	&= d\, \big( 1 + 2\lambda\, {\bm 1}^{\top} Q \diag\big( e^{\sqrt{c_i} t} + e^{\sqrt{c_i} (T-t)} \big)_{i = 1, \dots, n}\, Q^{-1} N^{-1} {\bm 1} \big)\\
	&= d\, \big( 1 +  {\bm 1}^{\top} E(t) \tilde{N}^{-1} {\bm 1} \big)
\end{align*}
for all $t \in [0,T].$

\medskip
{\bf 5.}
Define the real-valued functions
$$
	\beta(x) \coloneqq \prod_l (x-b_l),
	\qquad
	\gamma(x) \coloneqq \prod_l (x-c_l).
$$
Let
$$
	D_1 \coloneqq \diag \Big( \frac{\beta(c_i)}{\gamma'(c_i)}\Big)_{i = 1, \dots, n,}
	\qquad\text{and}\qquad
	D_2 \coloneqq \diag \Big( \hspace{-.1cm}-\hspace{-.1cm}  \frac{\gamma(b_i)}{\beta'(b_i)}\Big)_{i = 1, \dots, n.}
$$
We show in Step~\ref{step:D1D2positive} that $D_1$ and $D_2$ are positive diagonal matrices.
In particular, they are nonsingular.

\begin{enumerate}[label={\bf 5.{\arabic*}}, wide, labelwidth=!, labelindent=0pt]
	\item\label{step:Qinverse}
	{\em $\tilde{Q}^{-1} = D_1 \tilde{Q}^{\top} D_2$ and $\tilde{Q}^{-1} {\bm 1} = D_1 {\bm 1}.$}\\
		The matrix $-\tilde{Q}$ as defined in \eqref{eq:Q} is known as a Cauchy matrix.
		Both results are due to \cite{Schechter1959}.
	\item\label{step:D1D2positive}
	{\em $D_1$ and $D_2$ are positive diagonal matrices.}\\
		Let $k = 1, 2, \dots, n.$
		Then
		$$
			\frac{\beta(c_i)}{\gamma'(c_i)}
				= \frac{\prod_l (c_i - b_l)}{\sum_m \prod_{l \neq m} (c_i - c_l)}
				= \frac{\prod_l (c_i - b_l)}{\prod_{l \neq i} (c_i - c_l)}
				= (c_i - b_i) \prod_{l \neq i} \frac{c_i - b_l}{c_i - c_l}.
		$$
		Recall from Step~\ref{step:eigenvalues} that $c_i > b_i$, and that $c_i > b_l$ if and only if $c_i > c_l$
		for all $l = 1, 2, \dots, n.$
		Similarly,
		$$
			- \frac{\gamma(b_i)}{\beta'(c_i)} = (c_i - b_i) \prod_{l \neq i} \frac{b_i - c_l}{b_i - b_l} > 0.
		$$
\end{enumerate}

Define
$$
	\tilde{N}_1 \coloneqq C^{-1/2},
	\quad
	\tilde{N}_2 \coloneqq \tilde{Q}^{\top} D_2 B^{-1/2} \tilde{Q},
	\quad
	\tilde{N}_3 \coloneqq D_1^{-1} E(T) C^{-1/2},
	\quad
	\tilde{N}_4 \coloneqq \tilde{Q}^{\top} D_2 B^{-1}.
$$
All four matrices are nonsingular (see Steps~\ref{step:Qinverse} and~\ref{step:D1D2positive} in particular).
We show in Step~\ref{step:Nnonsingular} that $\tilde{N}_2 + \tilde{N}_3$ is nonsingular.

\begin{enumerate}[label={\bf 5.{\arabic*}}, resume, wide, labelwidth=!, labelindent=0pt]
	\item
	{\em $\tilde{N}^{-1} = \tilde{N}_1 (\tilde{N}_2 + \tilde{N}_3)^{-1} \tilde{N}_4.$}\\
	By definition, $\tilde{Q} = A^{-1} B^{-1/2} Q.$
	Using Step~\ref{step:Qinverse}:
	\begin{align*}
		\tilde{N}_4^{-1} (\tilde{N}_2 + \tilde{N}_3) \tilde{N}_1^{-1}
			&= B D_2^{-1} \tilde{Q}^{-T} \big( \tilde{Q}^{\top} D_2 B^{-1/2} \tilde{Q} + D_1^{-1} E(T) C^{-1/2} \big) C^{1/2} \\
			&= \big( B^{1/2} \tilde{Q} C^{1/2} + B (D_1 \tilde{Q}^{\top} D_2)^{-1} E(T) \big) \\
			&= \big( A^{-1} Q C^{1/2} + B \tilde{Q} E(T) \big) \\
			&= A^{-1} \big( Q C^{1/2} + B^{1/2} Q E(T) \big) \\
			&= \tilde{N}.
	\end{align*}
	\item\label{step:N1N3positive,N2posdef}
	{\em $\tilde{N}_1$ and $\tilde{N}_3$ are positive diagonal matrices, and $\tilde{N}_2$ is positive definite.}\\
		$B, C$ and $E(T)$ are positive diagonal matrices.
		By Step~\ref{step:D1D2positive},
		the same is true for $D_1$ and $D_2$.
		Hence $D_2 B^{-1/2}$ is positive definite.
		Since $\tilde{Q}$ is nonsingular (see Step~\ref{step:Qinverse}),
		$\tilde{N}_2 = \tilde{Q}^{\top} D_2 B^{-1/2} \tilde{Q}$ is also positive definite.
	\item\label{step:Nnonsingular}
	{\em $\tilde{N}_2+\tilde{N}_3, \tilde{N}$ and $N$ are nonsingular.}\\
		It follows from Step~\ref{step:N1N3positive,N2posdef} that $\tilde{N}_2 + \tilde{N}_3$ is positive definite and hence nonsingular.
		Since $\tilde{N}_1$ and $\tilde{N}_3$ are nonsingular,
		$\tilde{N}$ is nonsingular.
		We have shown in Step {\bf 4} that $N$ is nonsingular if (and only if) $\tilde{N}$ is nonsingular.
\end{enumerate}

A square matrix is called a {\em $Z$-matrix} if all its off-diagonal entries are nonpositive.
Given that some matrix $U$ is a nonsingular $Z$-matrix, the following two conditions are equivalent:
\begin{enumerate}[label=(M{\arabic*})]
	\item\label{M1}
		There exists a positive diagonal matrix $V$ such that $UV + VU^{\top}$ is positive definite.
	\item\label{M2}
		$U$ is nonsingular and all entries of $U^{-1}$ are nonnegative.
\end{enumerate}
In this case, $U$ is called an {\em $M$-matrix}.
In particular, condition (M1) implies that every positive definite $Z$-matrix is an $M$-matrix.
See Theorem~2.3 in \cite{Berman1994} for proofs and further equivalent characterizations of $M$-matrices.

\begin{enumerate}[label={\bf 5.{\arabic*}}, wide, labelwidth=!, labelindent=0pt]
\setcounter{enumi}{5}
	\item\label{step:N2nonpositive}
	{\em $\tilde{N}_2^{-1}$ is a $Z$-matrix.}\\
		With Step~\ref{step:Qinverse}, we obtain
		$$
			\tilde{N}_2^{-1}
				= \tilde{Q}^{-1} B^{1/2} D_2^{-1} \tilde{Q}^{-T}
				= D_1 \tilde{Q}^{\top} D_2 B^{1/2} \tilde{Q} D_1.
		$$
		$D_1$ is a positive diagonal matrix, so it suffices to show that all off-diagonal entries of
		$
			\tilde{Q}^{\top} D_2 B^{1/2} \tilde{Q}
		$
		are nonpositive.
		Fix $i,j \in \{1, 2, \dots, n\}$ such that $i\neq j.$
		Define
		$$
			\alpha
				\coloneqq \big( \tilde{Q}^{\top} D_2 B^{1/2} \tilde{Q} \big)_{ij}
				= - \sum_k \frac{\sqrt{b_k}\, \gamma(b_k)}{(b_k-c_i) (b_k-c_j) \beta'(b_k)}
				= - \sum_k \frac{\sqrt{b_k} \prod_{l \neq i,j} (b_k - c_l)}{\prod_{l \neq k} (b_k - b_l)}.
		$$
		The following argument is due to \cite{Petrov2017}.
		Define $f \colon [0,\infty) \to \mathbb{R},$
		$$
			f(x) \coloneqq - \sqrt{x} \prod_{l \neq i,j} (x-c_l).
		$$
		There are positive constants $z_0, z_1, \dots, z_{n-2}$ such that
		$$
			f(x) = - \sum_{k=0}^{n-2} (-1)^{n-2-k} z_k\, x^{k+1/2}
				= \sum_{k=0}^{n-2} (-1)^{n-1-k} z_k\, x^{k+1/2}.
		$$
		Differentiating $n-1$ times yields
		$$
			f^{(n-1)}(x)
				= \sum_{k=0}^{n-2} (-1)^{n-1-k} z_k\, x^{k - n + 3/2} \prod_{l=0}^{n-2} (k+1/2-l).
		$$
		For $k=0, 1, \dots, n-2,$ the factor $k+1/2-l$ is positive if $l=0, 1, \dots, k$ and negative if $l=k+1, k+2, \dots, n-2.$
		Hence
		\begin{align*}
			(-1)^{n-1-k} \prod_{l=0}^{n-2} ( k+1/2-l )
				&= (-1)^{n-1-k} (-1)^{ n-2-(k+1)+1 } \prod_{l=0}^{n-2} |k+1/2-l|\\
			&= - \prod_{l=0}^{n-2} |k+1/2-l| < 0.
		\end{align*}
		We conclude that $f^{(n-1)}(x) < 0$ for all $x > 0.$

		The Lagrange polynomial interpolation $p$ of $f$ in the points $b_1, b_2, \dots, b_n$ is
		$$
			p(x)
				= \sum_k f(b_k) \prod_{l \neq k} \frac{x-b_l}{b_k - b_l}
				= \Big( - \sum_k \frac{ \sqrt{b_k} \prod_{l \neq i,j} (b_k - c_l) }{ \prod_{l \neq k} (b_k - b_l) } \Big) x^{n-1} + q(x)
				= \alpha \,x^{n-1} + q(x)
		$$
		for some polynomial $q$ of degree at most $n-2$.
		The interpolation is exact for $x = b_1, b_2, \dots, b_n$.
		By Rolle's theorem, there is an $x_0 > 0$ such that
		$f^{(n-1)}(x_0) = p^{(n-1)}(x_0)$ \cite[Chapter~1]{Milne2000}.
		Hence
		$$
			0 > f^{(n-1)}(x_0)
				= p^{(n-1)} (x_0)
				= (n-1)! \, \alpha,
		$$
		showing that
		$
			\big( \tilde{Q}^{\top} D_2 B^{1/2} \tilde{Q} \big)_{ij}
		$
		is nonpositive if $i \neq j.$

	\item\label{step:N2invMmatrix}

	{\em $\tilde{N}_2^{-1}$ is a nonsingular $M$-matrix.}\\
		We have shown in Step~\ref{step:N2nonpositive} that $\tilde{N}_2^{-1}$ is a $Z$-matrix.
		Since $\tilde{N}_2$ is positive definite by Step~\ref{step:N1N3positive,N2posdef},
		$\tilde{N}_2^{-1}$ is positive definite as well.
		Hence $\tilde{N}_2^{-1}$ is a nonsingular $M$-matrix by condition~\ref{M1}.

	\item\label{step:N2invN3inv}
	{\em All entries of $(\tilde{N}_2^{-1} + \tilde{N}_3^{-1})^{-1}$ are nonnegative.}\\
		As a positive diagonal matrix, $\tilde{N}_3^{-1}$ is positive definite and a nonsingular $M$-matrix.
		The sum of positive definite $Z$-matrices is again a positive definite $Z$-matrix.
		Hence $\tilde{N}_2^{-1} + \tilde{N}_3^{-1}$ is a positive definite $Z$-matrix (see Step~\ref{step:N2invMmatrix});
		and therefore an $M$-matrix.
		By condition~\ref{M2}, all entries of $(\tilde{N}_2^{-1} + \tilde{N}_3^{-1})^{-1}$ are nonnegative.

	\item\label{step:N2N3inv}
	{\em All off-diagonal entries of $(\tilde{N}_2 + \tilde{N}_3)^{-1}$ are nonpositive.}\\
		By the Woodbury matrix identity,
		$$
			(\tilde{N}_2 + \tilde{N}_3)^{-1}
				= \tilde{N}_3^{-1} - \tilde{N}_3^{-1} (\tilde{N}_2^{-1} + \tilde{N}_3^{-1})^{-1} \tilde{N}_3^{-1}.
		$$
		Recall that $\tilde{N}_3^{-1}$ is a positive diagonal matrix.
		It follows from Step~\ref{step:N2invN3inv} that all off-diagonal entries of
		$\tilde{N}_3^{-1} (\tilde{N}_2^{-1} + \tilde{N}_3^{-1})^{-1} \tilde{N}_3^{-1}$
		are nonnegative.
\end{enumerate}

\medskip
\begin{enumerate}[label={\bf 6.{\arabic*}}, wide, labelwidth=!, labelindent=0pt]
	\item\label{step:N2invN4nonnegative}
	{\em All entries of $\tilde{N}_2^{-1} \tilde{N}_4\, {\bm 1}$ are nonnegative.}\\
		Using Step~\ref{step:Qinverse}, we obtain
		\begin{align*}
			\tilde{N}_2^{-1} \tilde{N}_4\, {\bm 1}
				&= \tilde{Q}^{-1} B^{1/2} D_2^{-1} \tilde{Q}^{-T} \tilde{Q}^{\top} D_2 B^{-1} {\bm 1} \\
			&= D_1 \tilde{Q}^{\top} D_2 B^{-1/2} \, {\bm 1} \\
			&= D_1 \tilde{Q}^{\top} D_2 B^{-1/2} \tilde{Q} \, \tilde{Q}^{-1} {\bm 1} \\
			&= D_1 \tilde{N}_2 D_1 {\bm 1}.
		\end{align*}
		We have shown in Step~\ref{step:N2invMmatrix} that $\tilde{N}_2^{-1}$ is a nonsingular $M$-matrix.
		Hence all entries of $\tilde{N}_2$ are nonnegative by condition~\ref{M2}.
		The same is true for $D_1$ by Step~\ref{step:D1D2positive}.
	\item\label{step:Ntilde232nonnegative}
	{\em All entries of $(\tilde{N}_2 + \tilde{N}_3)^{-1} \tilde{N}_2$ are nonnegative.}\\
		Define $U \coloneqq (\tilde{N}_2 + \tilde{N}_3)^{-1} \tilde{N}_2$.
		Writing
		$$
			U = I - (\tilde{N}_2 + \tilde{N}_3)^{-1} \tilde{N}_3
		$$
		shows that all off-diagonal entries of $U$ are nonnegative (see Steps~\ref{step:N1N3positive,N2posdef} and~\ref{step:N2N3inv}).

		We now use the following result about positive definite matrices:
		If two matrices $U$ and $V$ are positive definite, then $U-V$ is positive definite if and only if $V^{-1} - U^{-1}$ is positive definite
		\cite[Corollary~7.7.4]{Horn2013}.
		The matrices $(\tilde{N}_2 + \tilde{N}_3), \tilde{N}_3$ and $(\tilde{N}_2 + \tilde{N}_3) - \tilde{N}_3 = \tilde{N}_2$ are positive definite
		(see Step~\ref{step:N1N3positive,N2posdef}).
		Hence
		$$
			\tilde{N}_3^{-1} - (\tilde{N}_2 + \tilde{N}_3)^{-1} = U \tilde{N}_3^{-1}
		$$
		is positive definite.
		All entries on the main diagonal of a positive definite matrix are nonnegative.
		Therefore, all entries on $U$'s main diagonal are nonnegative.
	\item\label{step:Ntildenonnegative}
	{\em All entries of $\tilde{N}^{-1} {\bm 1}$ are nonnegative.}\\
		All entries of $\tilde{N}_1, (\tilde{N}_2 + \tilde{N}_3)^{-1} \tilde{N}_2$
		and $\tilde{N}_2^{-1} \tilde{N}_4\, {\bm 1}$ are nonnegative
		(see Steps~\ref{step:N2invN4nonnegative} and~\ref{step:Ntilde232nonnegative}).
		Hence all entries of the product
		$$
			\tilde{N}_1 (\tilde{N}_2 + \tilde{N}_3)^{-1} \tilde{N}_2 \tilde{N}_2^{-1} \tilde{N}_4\, {\bm 1}
				= \tilde{N}_1 (\tilde{N}_2 + \tilde{N}_3)^{-1} \tilde{N}_4\, {\bm 1}
				= \tilde{N}^{-1} {\bm 1}
		$$
		are nonnegative.
\end{enumerate}

\medskip
{\bf 7.}
Conclude with Steps {\bf 4} and~\ref{step:Ntildenonnegative} that there are $z_1, z_2, \dots, z_n \ge 0$ such that
\begin{align*}
	\phi(t) &= d\, \big( 1 +  {\bm 1}^{\top} E(t) \tilde{N}^{-1} {\bm 1} \big)
		= d \Big( 1 + \sum_{i=1}^n z_i \big( e^{\sqrt{c_i} t} + e^{\sqrt{c_i}(T-t)} \big) \Big), \qquad t \in [0,T].
\end{align*}
This function is clearly  symmetrically totally monotone.\qed

\subsubsection{Proof of Theorem~\ref{th:completemonotonicity} for  $\bm{G}$ arbitrary and $\bm{\gamma>0}$}

Let $G:(0,\infty)\to[0,\infty)$ be a nonconstant and  completely monotone kernel. We assume first that  $G(0+)<\infty$. Then we may assume  without loss of generality that $G(0):=G(0+)=1$. By Bernstein's theorem,
there exists a Borel probability measure $\mu$ on $[0,\infty)$ such that $G$ is equal to the Laplace transform of $\mu$.  Since the set of finite convex combinations of Dirac measures  is dense in the set of all Borel probability measures on $[0,\infty)$ with respect to weak convergence, there exists a corresponding sequence $(\mu_n)_{n=1,2,\dots}$ that converges weakly to $\mu$. Clearly, the corresponding Laplace transforms,
$$G_n(t)=\int_{[0,\infty)}e^{-tx}\,\mu_n(dx)\qquad\text{for $t\ge0$ and $n=1,2,\dots$}
$$
are all exponential kernels of type \eqref{eq:exponentialkernel}. The weak convergence $\mu_n\to\mu$ implies that $G_n(t)\to G(t)$ for all $t\ge0$. By slight abuse of notation, let us write
$$J_\gamma^{(n)}[\phi]:=\frac\gamma2\int_0^T\phi(t)^2\dif t+\int_0^T\int_0^TG_n(|t-s|)\phi(s)\phi(t)\dif s\dif t$$
 for every $\gamma\ge0$
 and $\phi\in L^2[0,T]$. Then
\begin{align*}
\big|J_0[\phi]-J_0^{(n)}[\phi]\big|&\le \|\phi\|_{L^2[0,T]} \int_0^T   \bigg(\int_0^T\big(G(|t-s|)- G_n(|t-s|)\big)^2\dif s\bigg)^{1/2}|\phi(t)|\dif t\\
&\le 2\sqrt{T}\|\phi\|^2_{L^2[0,T]} \|G-G_n\|_{L^2[0,T]} .
\end{align*}
Since $ \|G-G_n\|_{L^2[0,T]}\to0$ by dominated convergence, we conclude that $J_0^{(n)}[\phi]\to J_0[\phi]$ uniformly in functions $\phi$ from any bounded subset of $ L^2[0,T]$.

 For each $n$, let $\phi_n$ be the minimizer in $\Phi_1$ of the energy functional $J^{(n)}_\gamma$. By Section~\ref{exponentialkernels}, each function $\phi_n$ is symmetrically totally monotone. Since the function $f\equiv 1/T$ belongs to $\Phi_1$, one sees that there exists a constant $C$ such that $\|\phi_n\|_{L^2[0,T]}\le C$. By passing to a subsequence if necessary, we may therefore assume without loss of generality that the sequence $(\phi_n)_{n\in\mathbb{N}}$ converges weakly in $L^2[0,T]$ to a limiting function $\tilde \phi$, which by Lemma~\ref{weakly closed lemma} admits a symmetrically totally monotone version. Let $\phi$ be the minimizer of $J_\gamma$. Then $J^{(n)}_\gamma[\phi]\ge J_\gamma^{(n)}[\phi_n]$ for each $n$. Hence, the uniform convergence of $J_\gamma^{(n)}$ yields that
\begin{align*}
J_\gamma[\phi]&=\lim_{n\uparrow\infty}J^{(n)}_\gamma[\phi]\ge \liminf_{n\uparrow\infty}J^{(n)}_\gamma[\phi_n]=\liminf_{n\uparrow\infty}J_\gamma[\phi_n]\ge J_\gamma[\tilde\phi],
\end{align*}
where the latter inequality follows from the weak lower semicontinuity of $J_\gamma$. This shows that $\phi=\tilde\phi$ and concludes the proof for $G(0+)<\infty$.

If $G$ is weakly singular and satisfies $G(0+)=\infty$, we use its approximation as in \eqref{eq:etan} by kernels $G_n$ with $G_n(0+)<\infty$. As in the final part of the proof of Proposition~\ref{GSS prop} one sees that the symmetrically totally monotone minimizers for $G_n$ converge weakly in $L^2[0,T]$ to the minimizer for $G$. Thus, this latter minimizer is also symmetrically totally monotone by Lemma~\ref{weakly closed lemma}.

\subsubsection{Proof of Theorem~\ref{th:completemonotonicity} for  $\bm{\gamma=0}$}

Let  $\mu^*$ be the minimizer of $J_0$ as provided by Theorem 2.24 of \cite{Gatheral2012}. We approximate  $\mu^*$ in the weak topology by probability measures of the form $\mu^*_n(dx)=\psi_n(x)\dif x$, where each $\psi_n$ is a bounded nonnegative function on $[0,T]$ satisfying $\int_0^T\psi_n(x)\dif x=1$. Then we choose a sequence $\gamma_n\downarrow0$ that is such that $\gamma_n\int_0^T\psi_n(x)^2\dif x\to0$. Then it follows from \eqref{eq: Jgamma Fourier rep} that $J_{\gamma_n}[\psi_n]\to J_0[\mu^*]$.

Next, we let $\phi_n$ be the minimizer of $J_{\gamma_n}$ in $\Phi_{1}$. By passing to a subsequence if necessary, we may assume that the probability measures $\mu_n(dx)=\phi_n(x)\dif x$ on $[0,T]$ converge weakly to a probability measure $\mu$ on $[0,T]$. By Lemma~\ref{weak convergence lemma}, the restriction of $\mu$ to $(0,T)$ is absolutely continuous with respect to the Lebesgue measure and admits a symmetrically totally monotone density. Finally, we claim that $\mu=\mu^*$. Indeed, by \eqref{eq: J0 Fourier rep} and Fatou's lemma, $J_0$ is lower semicontinuous with respect to weak convergence of measures, and hence
\begin{align*}
J_0[\mu]&\le\liminf_{n\uparrow\infty}J_0[\mu_n]\le \liminf_{n\uparrow\infty}J_{\gamma_n}[\phi_n]\le \liminf_{n\uparrow\infty}J_{\gamma_n}[\psi_n]=J_0[\mu^*],
\end{align*}
and so the uniqueness of the minimizer yields $\mu=\mu^*$.\qed

\subsection{Proof of the formula from Example~\ref{capped linear example}}\label{capped linear section}

To prove the representation \eqref{capped linear formula}
, note first that
$$
		\int_0^{n} \big(1 - |t-s| \big)^+ \phi(s) \dif s =
		\begin{cases}
			\int_0^t (1-t+s) \phi(s) \dif s + \int_t^{t+1} (1+t-s) \phi(s) \dif s,
				& t \in [0,1], \\
			\int_{t-1}^t (1-t+s) \phi(s) \dif s + \int_t^{t+1} (1+t-s) \phi(s) \dif s,
				& t \in [1,n-1], \\
			\int_{t-1}^t (1-t+s) \phi(s) \dif s + \int_t^{n} (1+t-s) \phi(s) \dif s,
				& t \in [n-1,n].
		\end{cases}
$$
Differentiating this identity twice and replacing $\phi$ with $\phi_1, \dots, \phi_n$ yields
\begin{align*}
	\gamma \phi''_1(t) &= 2 \phi_1(t) - \phi_2(t), \\
	\gamma \phi''_i(t) &= 2 \phi_i(t) - \phi_{i-1}(t) - \phi_{i+1}(t), \qquad i = 2, \dots, n-1, \\
	\gamma \phi''_n(t) &= 2 \phi_n(t) - \phi_{n-1}(t).
\end{align*}

Hence $f \coloneqq (\phi_1, \dots, \phi_n)$ solves the following $n$-dimensional system of ordinary differential equations on $[0,1]:$
$$
	f'' = \frac1\gamma
	\begin{pmatrix}
		2 & -1 & \dots & 0 & 0\\
		-1 & 2 & \dots & 0 & 0 \\
		\vdots & \vdots & \vdots & \ddots & \vdots \\
		0 & 0 & \dots & 2 & -1 \\
		0 & 0 & \dots & -1 & 2 \end{pmatrix} f.
$$
Let $M_n$ denote the preceding triangular matrix, denote by $\lambda_1, \dots, \lambda_n$ its eigenvalues, and let  $Q$ contain the corresponding eigenvectors as columns.
Then
$$
	f(t) = Q \big( E(t) x_0 + E(1-t) x_1 \big)
$$
for some vectors $x_0, x_1 \in \mathbb{R}^n.$

Define $m \coloneqq \lceil n/2 \rceil.$
Let $I_m, J_m, 0_m$ denote the $m$-dimensional identity matrix, reverse identity matrix, and zero matrix, respectively.
The symmetry of $\phi$ implies that $\phi_i(t) = \phi_{n+1-i}(1-t)$, and so
\begin{equation}
\label{eq:cappedlinear_symmetry}
	\begin{bmatrix} I_m & 0_m \end{bmatrix} Q \big( E(t) x_0 + E(1-t) x_1 \big)
	= \begin{bmatrix} 0_m & J_m \end{bmatrix} Q \big( E(1-t) x_0 + E(t) x_1 \big),
	\quad t \in [0,T].
\end{equation}
Since
$$
	\sin\Big( \frac{(n+1-i) j \pi}{n+1} \Big)
	= \sin(j \pi) \cos\Big( \frac{ij\pi}{n+1} \Big) -  \cos(j \pi) \sin\Big( \frac{ij\pi}{n+1} \Big)
	= (-1)^{j+1} \sin\Big( \frac{ij\pi}{n+1} \Big)
$$
for all $i \in \{1, \dots, m\}$ and $j \in \{1, \dots, n\},$ it holds that $\begin{bmatrix} 0_m & J_m \end{bmatrix} Q = \begin{bmatrix} I_m & 0_m \end{bmatrix} Q J.$
Notice that $J^{-1} = J.$
Hence \eqref{eq:cappedlinear_symmetry} is satisfied if and only if $x_1 = J x_0.$

Let $t = i \in \{1, \dots, n-1\}.$
Then the symmetry of $\phi$ shows that
\begin{align*}
	\sigma &= \gamma \phi(i) + \int_{i-1}^i (1-i+s) \phi(s) \dif s + \int_i^{i+1} (1+i-s) \phi(s) \dif s \\
	&= \gamma \phi_i(1) + \int_{i-1}^i (1-i+s) \phi_i(s-i+1) \dif s + \int_i^{i+1} (1+i-s) \phi_{n-i}(1+i-s) \dif s \\
	&= \gamma \phi_i(1) + \int_0^1 s \big( \phi_i(s) + \phi_{n-i}(s) \big) \dif s.
\end{align*}
Similar arguments yield $\sigma = \gamma \phi_n(1) + \int_0^1 s \phi_n(s) \dif s.$

A straightforward calculation shows
$
	\int_0^1 s f(s) \dif s = Q \big( (E(1)-I)(J-I) + B(E(1)-J)\big) B^{-2}.
$

Hence
\begin{equation}
\label{eq:cappedlinear_x0}
	 {\bm\sigma} = \Big(\gamma Q \big( E(1) + J \big) + K Q \big( (E(1)-I)(J-I) + B(E(1)-J)\big) B^{-2} \Big) x_0.
\end{equation}
Existence and uniqueness of a solution   to \eqref{eq:cappedlinear} imply that \eqref{eq:cappedlinear_x0} can be uniquely solved for $x_0$.

redpajama_set_name: RedPajamaArXiv


When and where does the volunteering take place: This role is based in Oxford with an immediate start until November 5th 2017.
What the roles involves: The events manager will assist the Foundation's Executive Director in managing GCRF's UK-based events, most notably the Global China Research Conference to be held in November.
Why you should sign up: This is a great opportunity to get valuable event management skills and international experience. GCRF is a global organisation with ties and operations in North America and East Asia. Volunteers will be trained in event management skills and receive excellent networking opportunities. In addition, they will be exposed to significant knowledge about China's role and impact in the world, gaining Asia capacity, which is very valuable to future professional pursuits.
Skills required: Knowledge of the Microsoft Office Suite will be valuable for this role as well as an ability to communicate with individuals from different backgrounds. You do not need past experience.
You will receive training in event management, fundraising and marketing.
Out of city travel expenses will be reimbursed as will the cost of printing and purchasing event supplies.


The Sino French Water Development Company, a 50% subsidiary of Suez Environment, and Sichuan Dayi Water Company have formed a new joint venture, Shuangliu Dayi Environmental Company Limited.
Under a 25-year strategic partnership, Sino French Water will hold 65% of the units in the new company and will jointly manage and operate eight wastewater treatment plants located in rural areas in Shuangliu County with Shuangliu Dayi.
The partnership will also involve the construction, management and operation of the Shuangliu Maojiawan wastewater treatment plant, which will have an ultimate capacity of 250,000 m3 per day.
The plant will primarily serve industrial and commercial customers and the residents of the new Tianfu Urban Area, which is expanding rapidly.
The projects as a whole are expected to generate revenues of around €156 million for the joint venture over the 25-year period. More broadly, the aim of the 25-year partnership is to broaden the cooperation between both parties on new Dayi Water projects in Sichuan Province.
China's 12th Five-Year Plan identifies Tianfu as one of five key development areas in South-West China.


Welcome to WRITING LITE #2. I'm still talking about suspense, but this time I'm starting at the beginning. I talk about the power of that first line, how to approach it, and give you some of my all-time favorite examples. It is a little longer than I anticipated (at three and a half minutes) but hang in there. And, as always, let me know what you think.


text:
Home » Resources » Calendar/Events » CalGreen And Energy Codes Workshop, August 2, Fresno, CA
CalGreen And Energy Codes Workshop, August 2, Fresno, CA
Start time : 10:00 AM
End time : 04:00 PM
Event Host : greenTechnology
Event Location : Radisson Hotel Fresno Conference Center, 1055 Van Ness Ave, Room: El Captain, Fresno, CA, United States
In 2008, California became the first state in the nation to include mandatory green building requirements in its building code. This groundbreaking step meant that every structure built in the state – whether a home, school, commercial building or other structure – would have to meet guidelines for energy and water efficiency, low emission flooring and building materials and more. These requirements have also extended to building renovation.
Over the ensuing decade, CALGreen has continued to raise the bar in keeping with California's commitment to reducing greenhouse gas emissions and promoting building health. The latest updates to the code will become mandatory in January 2020.
California's Energy Code, which CALGreen defers to in regard to building efficiency standards, is the strictest code of its kind in the nation. Years ago, it was announced that when the current revisions became mandatory in January 2020, all new residential construction will be required to meet Zero Net Energy requirements. The 2019 updates address this efficiency goal.
Anyone working in the building sector can benefit from knowing what's coming as soon as possible, whether to ensure that projects in the pipeline take new changes into account, to prepare for future plan review or inspections, or to ensure that current renovations and upgrades take the latest standards for building efficiency and health into account.
Since the first publication of CALGreen, Green Technology has organized training related to code requirements, reaching thousands of building professionals at locations throughout the state.
We're now offering two new seminars that will address current changes. They will also provide opportunities for attendees to gain insight into current priorities for buildings in California and where code may be headed in the future and to ask questions regarding common challenges in code compliance.
The seminars can be attended individually, or together. See schedule below
CALGreen Seminar
Registration: $115
Energy Code Seminar
Registration for both seminars (includes lunch): $255
Host Company/Organization Name
greenTechnology
http://www.rsvpbook.com/event.php?472613
Energy Code, Green Building Standards
California Green Building
Reply/Leave a Comment (You must be logged in to leave a comment)
Not a Member Yet? Register and Join the Community | Log in
Best of Greenbuild: Building the First LEED Platinum City Panel and Webinar, July 17, Jacksonville, FL
Location: Jacksonville
Home Energy Score 2-Day Boot Camp, July 17-18, San Diego, CA
From Blueprint to Bill: NYC's Building Emissions Law Webinar, July 17, 9:30-10:30am
Stanley Marketplace Green Building Tour, July 17, 2:30-4:00pm, Aurora, CO
Presinar: Reimagining Building Codes: How to Green All Buildings, July 17, 4:30-6:00pm, Boston, MA
High-Performance Building Assemblies Course, July 18, Boston, MA
Webinar; A Path to Net Zero- Driving Energy Efficiency in Smart Buildings, July 18
LEED Tour - South Bend Fire Station #4, July 18, 3:00-4:00pm, South Bend, IN
Location: South Bend
CALGreen and Energy Code Seminar, July 19, 10:00-4:00pm, Irvine, CA
Location: Irvine
Rain Barrel Workshop, July 20, Englewood, FL 10:00am - 11:00pm
Location: Englewood
Every organization and everyone can submit to Rate It Green's green building calendar! Simply click register, verify your email address, and create a username and password. You can then decide if you'd like to engage more fully as a community member, but you'll be able to post events.

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Western Backlash Rises Over Zimbabwe Vote
Published: 6 August 2013
Zimbabwe prepares for a tough 2019
Australia calls up Zimbabwe-born all-rounder
Mphoko's foot in the mouth gaffes baffle many
Zimbabwe deep in limbo
Zimbabwe: UN boss bags award
Macheso appeals for divine intervention
Zimbabwe woman sends Mugabe nude picture via Whatsapp
Tuku's wife wife dismisses hubby's rumours of ill-health
Harare Man refuses to pay maintanance until wife returns underwear
RBZ demands Tsvangirai pay back $1.5m loan
Who will you vote for in 2013 Elections?
Zimbabwe's election is receiving wide criticism after longtime President Robert Mugabe swept the vote in an election the opposition says was rigged.
Australia is calling for a rerun of the poll, and the United States and Britain have said they do not think the results were credible. In neighboring South Africa, Zimbabwean expatriates are reacting to the results with disbelief.
On Monday, the opposition Movement for Democratic Change said dozens of its members were arrested a day after the electoral commission announced that longtime President Robert Mugabe trounced challenger Morgan Tsvangirai, winning 61 percent of the vote.
Those reports follow strong expressions of disapproval from the top diplomats of Australia, the United States and Britain. Zimbabwe's largest poll observer mission also said the poll was "seriously compromised" by multiple irregularities, including problems with voter rolls, vote tampering and voter intimidation.
But the African Union and the Southern African Development Community observation missions considered the poll relatively free and fair, though neither has presented a full report. Rugare Gumbo, a spokesman for Mugabe's ZANU-PF party, said the party won honestly.
"These people have to accept that their biggest problem was that they didn't have a program to offer to the people," Gumbo said. "They were complacent. As ZANU-PF, we went into the field quietly like old typical guerrilla fighters. We started restructuring the party from the bottom, while these people [opposition parties] were complaining or [creating their party] constitution, we were organizing the party, and that's why we came up on top."
South African President Jacob Zuma sent Mugabe his "profound congratulations" and said the observers had found the poll to be "an expression of the will of the people."
That opinion is not shared by many Zimbabwean expatriates in South Africa. Mugabe has few fans among the Zimbabwean community here, which is believed to number two million people. Many of them say he is the reason they fled here and claimed political asylum.
Sox Chikohwero is one of them. The opposition party activist he fled after he was arrested and tortured in 2002. He said Zimbabweans of all persuasions are questioning the results.
"The elections are not free and fair," Chikohwero said. "They've tried to force Morgan [Tsvangirai] to accept the results. Morgan has refused, because he's not talking for himself, he's talking for the people. And even people in ZANU-PF are surprised where this figure came from. You saw [ZANU-PF chairman] Simon Khaya Moyo commenting after winning so resoundingly, we know how ZANU-PF boasts about winning, but Simon Khaya Moyo was saying, no one is a winner in this election, which means he is also surprised and shocked about the level of winning. The type of rigging, I think they overdid it," he said.
Chikohwero's concerns are echoed by top Western officials.
U.S. Secretary of State John Kerry said he did not believe the outcome "represent[s] a credible expression of the will of the Zimbabwean people." British Foreign Secretary William Hague said he had "grave concerns." Australian Foreign Minister Bob Carr is calling for a complete rerun of the poll.
Mugabe was forced into a power-sharing government with Tsvangirai after elections in 2008 turned violent. That coalition has now come to a bitter end.
One thing does appear to be sure: Mugabe, who is now 89 years old, is not a man who is easily defeated. He has ruled Zimbabwe since its independence in 1980 and says he is still going strong.
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Home Health News
p27: A Marker for Breast Cancer?
TOPICS:CancerHarvard Stem Cell InstituteHarvard UniversityStem Cells
By Joseph Caputo, Harvard Gazette August 12, 2013
Graphical Abstract. Credit: Sibgat Choudhury, et al., doi: 10.1016/j.stem.2013.05.004
Researchers are in the final stages of testing p27, a mammary gland progenitor marker that they believe could lead to the commercial development of a clinical test for breast cancer risk.
An international scientific collaborative led by the Harvard Stem Cell Institute's Kornelia Polyak has discovered why women who give birth in their early 20s are less likely to develop breast cancer eventually than women who don't, triggering a search for a way to confer this protective state on all women.
The researchers are now testing p27, a mammary gland progenitor marker, on tissue samples collected from thousands of women over decades — women whose medical histories have been followed extremely closely — to see if it is an accurate breast cancer predictor in a large population. If the hypothesis is confirmed, which appears likely within a few months, Polyak says the commercial development of a clinical test for breast cancer risk would follow.
In a paper just published in the journal Cell Stem Cell, the researchers describe how a full-term pregnancy when a woman is in her early 20s reduces the relative number and proliferative capacity of mammary gland progenitors — cells that have the ability to divide into milk-producing cells — making them less likely to acquire mutations that lead to cancer.
By comparing numerous breast tissue samples, the scientists found that women at high risk for breast cancer, such as those who inherit a mutated BRCA1 or BRCA2 gene, have higher-than-average numbers of mammary gland progenitors. In general, women who carried a child to full term had the lowest populations of mammary gland progenitors, even when compared with cancer-free women who had never been pregnant. In addition, in women who gave birth relatively early but later developed breast cancer, the number of mammary gland progenitors was again observed to be higher than average.
"The reason we are excited about this research is that we can use a progenitor cell census to determine who's at particularly high risk for breast cancer," said Polyak, a Harvard Stem Cell Institute principal faculty member and a Harvard Medical School professor at the Dana-Farber Cancer Institute. "We could use this strategy to decrease cancer risk because we know what regulates the proliferation of these cells, and we could deplete them from the breast."
Research shows that two trends are contributing to an increase in the number of breast cancer diagnoses, a rise in obesity and the ever-increasing number of women postponing childbearing. The scientists' long-range goal is to develop a treatment that would mimic the protective effects of early childbearing.
The research, which took five years to complete, began with conversations between Polyak and Saraswati Sukumar, a professor at Johns Hopkins University School of Medicine. The two scientists formed collaborations with clinicians at cancer centers that see large numbers of high-risk women, in order to obtain breast tissue samples. They also worked with genomics experts and bioinformaticians to analyze gene expression in different breast cell types. At times, Polyak and Sukumar had trouble gaining cooperation for the study, which is unique in the breast cancer field for its focus on risk prediction and prevention.
"In general, people who study cancer always want to focus on treating the cancer. But, in reality, preventing cancer can have the biggest impact on cancer-associated morbidity and mortality," Polyak said. "I think the mentality has to change because breast cancer affects so many women, and even though many of them are not dying of breast cancer, there's a significant personal and societal burden."
Contributors to the research included members of the Dana-Farber Cancer Institute, Brigham and Women's Hospital, Beth Israel Deaconess Medical Center, Harvard Medical School, Harvard School of Public Health, John Hopkins University School of Medicine, Coimbra University Hospital, Thomson Reuters Healthcare & Science, NICTA Victoria Research Laboratory, University of Melbourne, University of Oslo, Baker IDI Heart and Diabetes Institute, Baylor-Charles A. Sammons Cancer Center, St. Vincent's Institute, USC Norris Comprehensive Cancer Center, UCSF Helen Diller Family Comprehensive Cancer Center, and Peter MacCallum Cancer Center.
The main supporter of this research was the Avon Foundation, with additional contributions by the National Cancer Institute, the Susan G. Komen Foundation, the Terri Brodeur Breast Cancer Foundation, the U.S. Army Congressionally Directed Medical Research Program, the Victorian Breast Cancer Research Consortium, the St. Vincent's Hospital Melbourne Research Endowment Fund, the Victorian Government's OIS Program, the Program for Advanced Medical Education, and the Cellex Foundation.
Publication: Sibgat Choudhury, et al., "Molecular Profiling of Human Mammary Gland Links Breast Cancer Risk to a p27+ Cell Population with Progenitor Characteristics," Cell Stem Cell, Volume 13, Issue 1, 117-130, 2013; doi: 10.1016/j.stem.2013.05.004
Image: Sibgat Choudhury, et al., doi: 10.1016/j.stem.2013.05.004
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M. Ødegaard
Odegaard for the Ballon d'Or? Arsenal's midfield maestro backed to challenge for top prize by club legend Lauren
Peter McVitie
19:21 SAST 2023/01/25
Arsenal captain Martin Odegaard has the potential to win the Ballon d'Or and have a nine-figure valuation, according to club legend Lauren.
Odegaard key for Premier League leaders
Has scored eight goals in top flight
Lauren says he could be one of world's best
WHAT HAPPENED? Odegaard, 24, has been a vital player for Mikel Arteta's side as they have dominated the Premier League this season. The Norway international is their top scorer in the division with eight goals and has registered five assists. Former Arsenal 'Invincible' and Cameroon hero Lauren believes the midfielder can go on to challenge for the world's top individual prize, and reckons he will soon be worth around £200 million ($247m).
WHAT THEY SAID: "Odegaard was not a cheap player, but he has been great and still has more to show. If he continues to improve, he can be a £100m or £200m player in the future," Lauren told FairBettingSites. "I can see Odegaard compete for the top individual awards such as the Ballon d'Or, the performances of the team are helping him to grow. In the future, if Arsenal win the Premier League, then perform well in the Champions League next season, I believe Odegaard will be among the best in the world."
THE BIGGER PICTURE: Odegaard, who joined Arsenal from Real Madrid in 2021, has helped Mikel Arteta's team build a five-point lead at the top of the Premier League this season. Meanwhile, they are also in contention to win the FA Cup and Europa League this term.
IN TWO PHOTOS:
WHAT NEXT FOR ARSENAL? Odegaard and his team-mates will hope to lead the Gunners into the fifth round of the FA Cup when they face Manchester City on Friday.
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"I Would Figure Out a Way to Get By" | Woman Becomes Israel's First Doctor with Cerebral Palsy
By Steven Briggs August 20, 2021
Becoming a doctor is hard enough. But Dr. Hodaya Oliel faced additional challenges to become Israel's first doctor with cerebral palsy. Now that she finally has her degree, she's opening up about what it was like to grow up with CP and how it informs her practice as a medical professional.
If you ask Oliel whether she feels like she's making history, she will say NO.
"I don't see myself that way. I'm just a regular doctor, as I've dreamt of becoming since I was a small child," Oliel says. "It's just that my difficulty is very visible, whereas the challenges lots of other people have to overcome might not be so obvious."
She's currently a resident in the Pediatrics department at the Shamir Medical Center but it was her condition that inspired her to go into healthcare.
After high school, she completed her national service at Asif, the school at the hospital where she spent most of her time as a kid.
"I worked mainly with CP patients," Oliel explains. "When I had been hospitalized there as a kid, I was encouraged to join in on the activities with the kids who learned in that school. That's why it was so meaningful for me to do my national service there."
She took her psychometric exam and got into the Technion – Israel Institute of Technology, but the idea of going to college felt intimidating to Oliel.
"It was a very challenging transition," she recalls. "Up until then, I'd lived at home with my parents, and all of a sudden I was far away, without any help. Granted, I was a very independent person, but there were so many things I didn't know how to do yet, or that took me a really long time to do, and each day I would encounter a new difficulty, and who knows if there was going to be somebody nearby who could help me. In short, that first year was horrible, but I knew that I had to push on, and that I would figure out a way to get by."
She learned how to live on her own while pursuing a degree in medicine, but her chosen path felt like anything but a guarantee.
"I wanted more than anything else in the world to become a pediatrician. That was the only thing I could think about. I didn't always have someone to guide or support me, but I knew that if I was just a little more stubborn, and stuck it out through the tough times, I would succeed," she says.
When she finished her college studies and began her residency, she posted a message on Facebook that quickly went viral.
"I'm sending hugs to everyone who is in my position and thinks they don't stand a chance. You do, but that doesn't mean things will be easy. It just means that you need to keep breathing, give yourself daily boosts of confidence and believe that tomorrow might be a better day," she wrote in 2019.
"Throughout my years in college, I'd been uploading posts, but this particular post caught the eye of [journalist] Sivan Rahav-Meir, who invited me to come for an interview with her. This was my first step toward becoming well-known," Oliel says. "I never imagined so many people would be interested in my story."
Later that year, she was chosen to light the torch at the country's 71st Independence Day ceremony, which thrusted her into the spotlight.
Growing Up with CP
Oliel was born prematurely at just 27 weeks. She spent three months in the NICU and was diagnosed with cerebral palsy when she was six months old. She went on to lose some motor function in her legs but didn't suffer any cognitive impairment.
But those early days in the hospital made an impression on young Oliel.
"Every time I had to undergo surgery when I was a child or teen, I would say to myself, 'I have to keep coming back to the hospital because God wants me to know what it's like to be a child who is hospitalized.' It was never easy, but I remember being so curious about everything I saw, even the operating room. These experiences are what spurred me to succeed in high school, and while I was studying for my psychometric exam. I didn't make any backup plans for if I didn't succeed. That was not an option."
She often felt ostracized around the other kids when she was in elementary school.
"To make things even worse, I had to have an aide follow me around all day at school, so you can imagine how all the other kids reacted to that. They looked at me like I was an alien. I was extraordinarily lucky, however, that my siblings were so supportive and would help me boost my self-esteem. I just couldn't wait for elementary school to end so I could move to a new school and make a fresh start. I made a pledge to myself that I would do everything in my power to have a great social life at my new school."
She soon decided to forgo the aide and have her dad walk her to school.
"When I began walking with a crutch, I was finally able to walk on my own, which was amazing," she added.
Oliel will never forget what it's like to grow up with CP.
With medical school behind her, she says she plans on specializing in pediatric neurology, so she can help children and families struggling with the same condition.
She gets lots of messages from parents and individuals living with CP that she's inspired since going public with her condition.
"Ever since I participated in the Independence Day ceremony, the number has risen dramatically. I don't always have time to respond, since my hours in the hospital working as a resident are so long, but I do try to write back to everyone. But seeing how many people I can help makes me feel like it was worth it all the times I experienced hardship."
Steven Briggs
Steven Briggs is a healthcare writer for Scrubs Magazine, hailing from Brooklyn, NY. With both of his parents working in the healthcare industry, Steven writes about the various issues and concerns facing the industry today.
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