text:
\begin{document}

\begin{frontmatter}
\title{Penalization of Wasserstein barycenters for $\varphi$-exponential distributions}
\author{S. Kum}
\address[S. Kum]{Department of Mathematics Education, Chungbuk National University, Cheongju 28644, Republic of Korea}
\author{M. H. Duong}
\address[M. H. Duong]{School of Mathematics, University of Birmingham, B15 2TT, UK. \textit{Corresponding author. Email}: h.duong@bham.ac.uk}
\author{Y. Lim}
\address[Y. Lim]{Department of Mathematics, Sungkyunkwan University, Suwon 440-746, Republic of Korea}
\author{S. Yun}
\address[S. Yun]{Department of Mathematics Education, Sungkyunkwan University,  Seoul 03063, Republic of Korea}
\begin{abstract}
In this paper we study the penalization of barycenters in the
Wasserstein space for $\varphi$-exponential
distributions. We obtain an explicit characterization of the
barycenter in terms of the variances of the measures generalizing existing results for Gaussian measures. We then develop a
gradient projection method for the computation of the barycenter establishing a Lipstchitz continuity for the gradient function. We also numerically show the influence of parameters and stability of the algorithm under small perturbation of data.
\end{abstract}

\end{frontmatter}
\section{Introduction}
\subsection{Penalization of barycenters in the Wasserstein space} In this paper we are interested in \textit{the penalization of barycenters in the Wasserstein space}, which is a minimization problem of the form
\begin{equation}
\label{eq: penalization}
\min_{\mu \in \A}\sum_{i=1}^n\frac{1}{2}\lambda_i W_2^2(\mu,\mu_i)+\gamma F(\mu),
\end{equation}
where $\A$ is a subset of $\P_2(\R^d)$, which is the Wasserstein
space of probability measures on $\R^d$ with finite second moments;
$\{\mu_i\}_{i=1}^{n}$ are $n$ given probability measures in $\A$;
$W_2$ is the $L^2$-Wasserstein distance between two probability
measures in $\P_2(\R^d)$ (cf. Section \ref{sec: preliminary}),
and $F:\P_2(\R^d)\rightarrow \R$ is an entropy
functional. Finally $\gamma\geq 0$ is a given
regularization/penalization parameter; $\lambda_1,\ldots, \lambda_n$
are given non-negative numbers (weights) satisfying
$\sum_{i=1}^n\lambda_i=1$.

\subsection{Literature review} Problem \eqref{eq: penalization}
 for $\gamma=0$ has been studied intensively in the literature.
 It was first studied by Knott and Smith \cite{KnottSmith1984} for Gaussian measures.
 In \cite{Agueh2011}, Agueh and Carlier studied the general case proving,
 among other things, the existence and uniqueness of a minimizer provided that one of $\mu_i's$
  vanishes on small sets (i.e., sets whose Hausdorff dimension is at most $d-1$). The minimizer
  is called the barycenter of the measures $\mu_i$ with weights $\lambda_i$ extending a classical
  characterization of the Euclidean barycenter. The article \cite{Agueh2011} has sparked off many
   research activities from both theoretical and computational aspects over the last years.
    Wasserstein barycenters in different settings, such as over Riemannian manifolds  and over discrete data,
     have been investigated \cite{KimPass2017,AnderesBorgwardtMiller2016}. Connections between Wasserstein barycenters
     and optimal transports have been explored \cite{Pass2012, KimPass2015}. Several computational methods for the computation
      of the barycenter have been developed \cite{CutiriDoucet2014, Alvarez2016, KumYun2018,PeyreCuturi2019}. Recently Wasserstein
      barycenters has found many applications in statistics, image processing and machine learning
       \cite{Rabin2011, MallastoFeragen2017, Srivastava2018}. We refer the reader to the mentioned papers and references therein
       for a more detailed account of the topic.

The case $\gamma>0$ has been studied in the recent paper
 \cite{BigotCazellesPapadakis2019} where the existence,
 uniqueness and stability of a minimizer, which is called the penalized barycenter,
  has been established. The regularization parameter $\gamma$ was proved to provide smooth
barycenters especially when the input probability measures are
irregular which is useful for data analysis
\cite{BCP2018,Redko2019}. In addition, the penalized barycenter
problem also resembles the discretization formulation of Wasserstein
gradient flows for dissipative evolution equations
\cite{JKO1998,AGS08,Carlier2017} and the fractional heat equation
\cite{DuongJin2019} at a given time step where $\{\mu_i\}$ represent
discretized solutions at the previous steps and $\gamma$ is
proportional to the time-step parameter.

Gaussian measures play an important role in the study of Wasserstein barycenter problem since in this case an useful characterization of the barycenter exists \cite{Agueh2011,Bhatia2018} which gives rise to efficient computational algorithms such as the fixed point approach \cite{Alvarez2016} and the gradient projection method \cite{KumYun2018}. Our aim in this paper is to seek for a large class of probability measures so that the penalized barycenter can be explicitly characterized and computed similarly to the case of Gaussian measures.  We will study the penalization problem
\eqref{eq: penalization} for an important classes of probability measures, namely $\varphi$-exponential measures,
  where the entropy functional is the Tsallis entropy functional respectively. The class of $\varphi$-exponential measures significantly enlarges that of Gaussian measures and containing also $q$-Gaussian measures as special cases, cf. Section \ref{sec: intro varphi-exponential} below. To state our main results, we briefly recall the definition of $\varphi$-exponential measures; more detailed will be given in Section \ref{sec: preliminary}.

\subsection{$\varphi$-exponential distributions}
\label{sec: intro varphi-exponential}
Let $\varphi$
 be an increasing, positive, continuous function on $(0,\infty)$, the $\varphi$-logarithmic is defined by \cite{Takatsu2013}
\begin{equation}
\label{eq: varphi log}
\ln_{\varphi}(t):=\int_1^t\frac{1}{\varphi(s)}\,ds,
\end{equation}
which is increasing, concave and $C^1$ on $(0,\infty)$. Let $l_\varphi$ and $L_\varphi$ be respectively the infimum and the supremum of $\ln_\varphi$, that is
\begin{equation}
l_\varphi:=\inf\limits_{t>0}\ln_\varphi(t)=\lim\limits_{t\downarrow 0} \ln_{\varphi}(t)\in [-\infty,0), \quad L_\varphi:=\sup\limits_{t>0}\ln_\varphi(t)=\lim\limits_{t\uparrow \infty} \ln_{\varphi}(t)\in (0,+\infty).
\end{equation}
The function $\ln_\varphi$ has the inverse function, which is called the $\varphi$-exponential function, and is defined on $(l_\varphi,L_\varphi)$. This inverse function can be extended to the whole $\R$ as
\begin{equation}
\label{eq: varphi exp}
\exp_{\varphi}(s):=\begin{cases}
0\qquad\text{for}~~s\leq l_\varphi,\\
\ln_\varphi^{-1}(s)\qquad\text{for}~~ s\in(l_\varphi,L_\varphi),\\
\infty\qquad\text{for}~~ s\geq L_\varphi,
\end{cases}
\end{equation}
which is $C^1$ on $(l_\varphi,L_\varphi)$.

Let ${\Bbb S} (d,\R)$ (${\Bbb S}(d,\R)_+$) be the set of symmetric (positive definite, respectively) matrices of order $d$. Let $v\in
\mathbb{R}^d$ be a given vector and $V\in {\Bbb H}_+$ be a given
symmetric positive definite matrix. The $\varphi$-exponential
measure with mean $v$ and covariance matrix $V$ is the probability
measure on $\mathbb{R}^d$ with Lebesgue density
\begin{equation}
\label{eq: gvarphi}
g_\varphi(v,V)(x):=\exp_{\varphi}(\lambda_\varphi(I_d)-c_\varphi(I_d)|x-v|_V^2)\Big(\det(V)\Big)^{-\frac{1}{2}},
\end{equation}
where $|x|_V^2:=\langle x, V^{-1}x\rangle$, $\lambda_\varphi$ and $c_\varphi$ are
 continuous functions on ${\Bbb S}(d,\R)_+$ playing the role of normalization constants. Two important examples of $\varphi$-exponential
measures include Gaussian measures and $q$-Gaussian measures
corresponding to $\varphi(s)=s$ and $\varphi(s)=s^q$ respectively.
The $\varphi$-exponential measures play an important role in
statistical physics, information geometry and in the analysis of
nonlinear diffusion equations \cite{Otto2001,Ohara2009,Takatsu2012,
Takatsu2013}. More information about $\varphi$-exponential measures
will be reviewed in Section \ref{sec: preliminary}.
\subsection{Main results of the paper}
As already mentioned, in this paper we study the penalization problem \eqref{eq: penalization} for Gaussian measures and $\varphi$-exponential measures, where the entropy functional is the (negative) Boltzmann entropy functional and the Tsallis entropy functional respectively. Main results of the present paper are explicit characterizations of the minimizer of \eqref{eq: penalization} and properties of the objective functions that can be summarized as follows.
\begin{theorem}
\label{thm: main theorem 1}
Suppose $\mu_i$ are either Gaussian measures or $q$-Gaussian measures or $\varphi$-exponential measures (in this case, $\gamma=0$) with mean zero. Then the minimization problem \eqref{eq: penalization} has a unique minimizer whose variance solves the nonlinear matrix equation \eqref{eq: optimal equation Gaussian} or \eqref{eq: optimal equation q-Gaussian} or \eqref{eq: optimal equation varphi} respectively. Furthermore, the objective function is strictly convex.
\end{theorem}
\begin{theorem}
\label{thm: main theorem 2}
The gradient function of the objective function is Lipschitz continuous.
\end{theorem}
Theorem \ref{thm: main theorem 1} summarizes Theorem \ref{thm: Gaussian} (for Gaussian measures), Theorem \ref{thm: q-Gaussian} (for $q$-Gaussian measures) and Theorem \ref{thm: varphi} (for general $\varphi$-exponential measures). Theorem \ref{thm: main theorem 2} summarizes Theorem \ref{thm: Lipschitz continuity} (for Gaussian measures)
and Theorem \ref{thm: Lipschitz continuity q-Gaussians} (for $q$-Gaussian measures).

The key to the analysis of the present paper is that  the spaces of $\varphi$-exponential measures
and  Gaussian measures are isometric in the sense of Wasserstein
geometry \cite{Takatsu2012, Takatsu2013}, that is
$$W_2(g_\varphi(v,V),g_\varphi(u, U))=W_2(\N(v, V),\N(u,U)),$$
where $\N(v, V)$ denotes a Gaussian measure with mean $v$ and covariance matrix $V$.
Therefore, since the Wassertein distance between Gaussian measures
can be computed explicitly, the objective functional in \eqref{eq:
penalization} can also be computed explicitly in terms of the
variances and \eqref{eq: penalization} becomes a minimization
problem over the space of symmetric positive definite matrices. We
then prove the strict convexity of the objective function and the
existence of solutions to the optimality equation using matrix
analysis tools as in \cite{Bhatia2018}. Theorems \ref{thm:
Gaussian},   \ref{thm: q-Gaussian} and \ref{thm: varphi} establish
the existence and uniqueness of a minimizer and provide an explicit
characterization of the minimizer in terms of nonlinear matrix
equations for the variance generalizing the characterization of the
Wasserstein barycenter for Gaussian measures in
\cite{Agueh2011,Bhatia2018} to the penalized Wasserstein barycenter
for Gaussian measures and $\varphi$-exponential measures. Theorem
\ref{thm: Lipschitz continuity} and Theorem \ref{thm: Lipschitz
continuity q-Gaussians} prove the Lipschitz continuity of the
gradient of the objective function providing an explicit upper bound
for the Lipschitz constant generalizing the results of
\cite{KumYun2018} for the barycenter for Gaussian measures to our
setting. We also perform numerical experiments to show the affect of the parameter $q$ and a stability property of the algorithm under small perturbation of the data, cf. Section \ref{sec: numerical experiments}.

\subsection{Organization of the paper}
The rest of the paper is organized as follows. In Section \ref{sec: preliminary} we review relevant knowledge that will be used in subsequent sections on the Wasserstein metric and the Wasserstein geometry of Gaussian and $\varphi$-exponential distributions. Then we study the penalization of barycenters for Gaussian measures in Section \ref{sec: Gaussians} and extend these results to $q$-Gaussian and $\varphi$-exponential measures in Section \ref{sec: q-Gaussians} and Section \ref{sec: vaphi expo measures}. In Section \ref{sec: GPM} we describe a gradient projection method for the computation of the minimizer and prove that the gradient function is Lipschitz continuous. Finally, in Section \ref{sec: numerical experiments}, we numerically show affect of parameters to the minimizer and stability of the algorithm under small perturbation of data.
\section{Wasserstein metric, Gaussian measures and $\varphi$-exponential measures}
\label{sec: preliminary}
In this section, we summarize relevant knowledge that will be used in subsequent sections on the Wasserstein metric and the Wasserstein geometry of Gaussian and $\varphi$-exponential distributions.
\subsection{Wasserstein metric}
We recall that $\P_2(\R^d)$ is the space of probability measures in $\R^d$ with finite second moments, that is
\begin{equation}
\label{eq: P2} \P_2(\R^d):=\left\{\mu:\R^d\rightarrow
(0,\infty)~\text{measurable
and}~\int_{\R^d}|x|^2\mu(dx)<\infty\right\}.
\end{equation}
Let $\mu$ and $\nu$ be two probability measures belonging to $\P^2(\R^d)$. The $L^2$-Wasserstein distance, $W_2(\mu,\nu)$, between $\mu$ and $\nu$ is defined via
\begin{equation}
\label{eq: W2}
W^2_2(\mu,\nu):=\inf_{\gamma\in \Gamma(\mu,\nu)}\int_{\R^d\times\R^d}|x-y|^2\,d\gamma(x,y),
\end{equation}
where $\Gamma(\mu,\nu)$ denotes the set of transport plans between $\mu$ and $\nu$, i.e., the set of all probability measures on $\R^d\times \R^d$ having $\mu$ and $\nu$ as the first and the second marginals respectively. More precisely,
$$
\Gamma(\mu,\nu):=\{\gamma\in \P(\R^d\times \R^d): \gamma(A\times \R^d)=\mu(A)~\text{and}~ \gamma(\R^d\times A)=\nu(A)\},
$$
for all Borel measurable sets $A\subset\R^d$. It has been proved that, under rather general conditions (e.g., when $\mu$ and $\nu$ are absolutely continuous with respect to the Lesbegue measure), an optimal transport plan in \eqref{eq: W2} uniquely exists and is of the form $\gamma=[\mathrm{id}\times\nabla\psi]_\#\mu$ for some convex function $\psi$ where $\#$ denotes the push forward \cite{Brenier1991, GangboMcCann1996}.

The Wasserstein distance is an instance of a Monge-Kantorevich optimal transportation cost functional and plays a key role in many branches of mathematics such as optimal transportation, partial differential equations, geometric analysis and has been found many applications in other fields such as economics, statistical physics and recently in machine learning. We refer the reader to the celebrated monograph \cite{Villani2003} for a great exposition of the topic.

We now consider two important classes of probability measures, namely Gaussian measures and $\varphi$-exponential measures, for which there is an explicit expression for the Wasserstein distance between two members of the same class. Although Gaussian measures are special cases of $\varphi$-exponential measures, but we consider them separately since many proofs for the former are much simplified than those for the latter.
\subsection{Wasserstein distance of Gaussian measures}
The Wasserstein distance between two Gaussian measures is well-known \cite{GivensShortt1984}, see also e.g., \cite{Takatsu2012}:
\begin{equation}
\label{eq: W2-Gaussians}
W_2(\N(\mu,U),\N(\nu,V))^2=|\mu-\nu|^2+\mathrm{tr}U+\mathrm{tr}V-2\mathrm{tr}\sqrt{V^\frac{1}{2}U V^\frac{1}{2}}.
\end{equation}
Furthermore, $[\mathrm{id}\times \nabla \mathcal{T}]_\#\N(\mu,U)$ is the optimal plan between them, where
\begin{equation}
\label{eq: optimal plan Gaussian}
\mathcal{T}(x)=\frac{1}{2}\langle x-\mu, T(x-\mu)\rangle+\langle x,\nu\rangle, \quad T=U^\frac{1}{2}\Big(U^\frac{1}{2} V U^\frac{1}{2}\Big)^{-\frac{1}{2}}U^\frac{1}{2}.
\end{equation}
\subsection{The entropy of Gaussian measures}
The (negative) Boltzmann entropy functional of a probability measure is defined by
\begin{equation}
\label{eq: entropy}
F(\mu):=\int \mu \log \mu.
\end{equation}
Using Gaussian integral, the (negative) Boltzmann entropy of a Gaussian measure can be computed explicitly \cite[Theorem 9.4.1]{CoverThomas1991}:
\begin{equation}
\label{eq: entropy of Gaussian measures}
F(\N(\mu, U))=-\frac{d}{2}\ln(2\pi e)-\frac{1}{2}\ln\det(U).
\end{equation}
We now consider the second class of probability measures: $\varphi$-exponential measures.
\subsection{$\varphi$-exponential measures and Wassertein distance}

We recall that for a given increasing, positive and continuous function $\varphi$ on $(0,\infty)$, the $\varphi$-logarithmic function and the $\varphi$-exponential function are respectively defined in \eqref{eq: varphi log} and \eqref{eq: varphi exp}. Two important classes of $\varphi$-exponential functions are:
\begin{enumerate}[(i)]
\item  $\varphi(s)=s$: the $\varphi$-logarithmic function and the $\varphi$-exponential function become the traditional logarithmic and exponential functions: $\ln_\varphi(t)=\ln (t),~~\exp_{\varphi}(t)=\exp(t)$.
\item $\varphi(s)=s^q$ for some $q>0$: the $\varphi$-logarithmic function and the $\varphi$-exponential function become the $q$-logarithmic and $q$-exponential functions respectively
\begin{equation}
\label{eq: q-exp}
\ln_{\varphi}(t)=\log_q(t)=\frac{t^{1-q}-1}{1-q} \quad \text{for}~ t>0,\quad \exp_{\varphi}(t)=\exp_q(t)=\Big( 1+(1-q)t\Big)_+^\frac{1}{1-q},
\end{equation}
where $[x]_+=\max\{0,x\}$ and by convention $0^a\colonequals\infty$. The $q$-logarithmic function satisfies the following property
\begin{equation}
\label{eq: q-log properties}
\ln_q(xy)=\ln_q(x)+\ln_q(y)+(1-q)\ln_q(x)\ln_q(y).
\end{equation}
\end{enumerate}
\begin{definition} For any $a\in\R$, we define $\O(a)$ to be the set of all increasing, continuous function $\varphi$ on $(0,\infty)$ such that $\max\{\delta_\varphi,\delta^\varphi\}<a$ where
$$
\delta_\varphi:=\inf\Big\{\delta\in \R\Big\vert \lim\limits_{s\downarrow 0}\frac{s^{1+\delta}}{\varphi(s)}~\text{exists}\Big\}, \quad  \delta^\varphi:=\inf\Big\{\delta\in \R\Big\vert \lim\limits_{s\uparrow \infty}\frac{s^{1+\delta}}{\varphi(s)}=\infty\Big\}.
$$
\end{definition}
It is proved in \cite[Proposition
3.2]{Takatsu2013} that for any $\varphi\in \O(2/(d+2))$ there exist continuous functions $\lambda_\varphi$ and
$c_\varphi$ on ${\Bbb S}(d,\R)_+$ such that (cf. \eqref{eq: gvarphi} in the Introduction)

\begin{equation}
\label{eq: varphi exp distr}
g_\varphi(v,V)(x):=\exp_{\varphi}(\lambda_\varphi(I_d)-c_\varphi(I_d)|x-v|_V^2)\,\Big(\det(V)\Big)^{-\frac{1}{2}},
\end{equation}
where $|x|_V^2:=\langle x, V^{-1}x \rangle$, is a probability density on $\R^d$ with mean $v$ and
 covariance matrix $V$, which is called a $\varphi$-exponential
 distribution. Note that, in the above expression, $\lambda_\varphi$ and
$c_\varphi$ are enough to define only at the identity matrix $I_d$,
not on all ${\Bbb S}(d,\R)_+$. We define the space of all $\varphi$-exponential distribution measures by

\begin{equation}
\G_{\varphi}:=\Big\{G_\varphi(v,V):=g_{\varphi}(v,V)\L^d\big\vert (v,V)\in \R^d\times {\Bbb S}(d,\R)_+\Big\}.
\end{equation}
Above $\L^d$ is the Lesbesgue measure on $\R^d$. Two important cases:
\begin{enumerate}[(i)]
\item $\varphi=s$, $\G_\varphi$ reduces to the class of Gaussian measures with mean $v$ and covariance matrix $V$.
\item In the case $\varphi=s^q$, $\G_{\varphi}$ becomes the class of all $q$-Gaussian measures
$$\G_q=\Big\{G_q(v,V)\big\vert (v,V)\in
\R^d\times {\Bbb S}(d,\R)_+\Big\}$$ where
\begin{equation*}
G_q(v,V)=C_0(q,d)(\det V)^{-\frac{1}{2}}\exp_q\Big(-\frac{1}{2}C_1(q,d)\langle x-v,V^{-1}(x-v)\rangle\Big)\L^d,
\end{equation*}
and $C_0(q,d), C_1(q,d)$ are given by
\begin{align*}
& C_1(q,d)=\frac{2}{2+(d+2)(1-q)},\\
&C_0(q,d)=
\begin{cases} \frac{\Gamma\Big(\frac{2-q}{1-q}+\frac{d}{2}\Big)}{\Gamma\Big(\frac{2-q}{1-q}\Big)} \Big(\frac{(1-q)C_1(q,d)}{2\pi}\Big)^\frac{d}{2}& \text{if $0<q<1$,}
\\ \frac{\Gamma\Big(\frac{1}{q-1}\Big)}{\Gamma\Big(\frac{1}{q-1}-\frac{d}{2}\Big)}\Big(\frac{(q-1)C_1(q,d)}{2\pi}\Big)^\frac{d}{2}& \text{if $1<q<\frac{d+4}{d+2}$.}
\end{cases}
\end{align*}
Note that when $q=1$, $C_1(q,d)=1$. Thus Gaussian measures are special cases of $q$-Gaussian measures.
\end{enumerate}
The $\varphi$-exponential measures play an important role in statistical physics, information geometry and in the analysis of nonlinear diffusion equations \cite{Otto2001,Ohara2009,Takatsu2012, Takatsu2013}. We refer to \cite{Ohara2009,Takatsu2012, Duong2015} for further details on $q$-Gaussian measures, $\varphi$-exponential measures and and their properties.

The following result explains why $q$-Gaussian measures and $\varphi$-exponential measures are special. It will play a key role in the analysis of this paper.
\begin{proposition}
\label{prop: q-Gaussian properties}
The following statements hold \cite{Takatsu2012,Takatsu2013}
\begin{enumerate}
\item For any $q\in (0,1)\cup \Big(1,\frac{d+4}{d+2}\Big)$, the space of $q$-Gaussian measures is convex and isometric to the space of Gaussian measures with respect to the Wasserstein metric.
\item For any $\varphi\in\O(2/(d+2))$ with $d\geq 2$, the space $\G_\varphi$ is convex and isometric to the space of Gaussian measures with respect to the Wasserstein metric.
\item Let $G_\varphi(\nu, V)$ and $G_\varphi(\mu, U)$ be two $\varphi$-exponential distributions. Then~ $[\mathrm{id}\times\nabla \mathcal{T}]_{\#}G_\varphi(\mu,U)$, where $\mathcal{T}$ is defined in \eqref{eq: optimal plan Gaussian}, is the optimal plan in the definition of $W^2_\varphi(G_q(\nu, V), G_\varphi(\mu, U)$.
\item We have
\begin{align} \label{eq: W2 formula}
W_2(G_\varphi(\mu,U),G_\varphi(\nu,V))^2&=W_2(G_q(\mu,U),G_q(\nu,V))^2\notag
\\&=W_2(\N(\mu,U),\N(\nu,V))^2\notag\\
&=|\mu-\nu|^2+\mathrm{tr}U+\mathrm{tr}V-2\mathrm{tr}\sqrt{V^\frac{1}{2}U
V^\frac{1}{2}}.
\end{align}
\end{enumerate}
\end{proposition}
\subsection{The Tsallis entropy of a $q$-Gaussian measure}
The Tsallis entropy is defined by
\begin{equation}
\label{eq: Tsalli}
F_q(\mu):=\frac{1}{1-q}\int_{\R^d} \mu(x)\ln_q\mu(x)\,dx.
\end{equation}
The Tsallis entropy of a $q$-Gaussian can also be computed explicitly using the property \eqref{eq: q-log properties} and similar computations as in the Gaussian case.
\begin{lemma}
\label{lem: Tsallis entropy of q-Gaussian}
It holds that  \cite{Duong2015}
\begin{equation}
F_q(G_q(\mu, U))=-\frac{d}{2}C_1(q,d)+\Big[1-(1-q)\frac{d}{2}C_1(q,d)\Big]\ln_{q}\frac{C_0(q,d)}{(\det U)^\frac{1}{2}}.
\end{equation}
\end{lemma}
\section{Penalization of barycenters for Gaussian measures}
\label{sec: Gaussians}
In this section we study the following penalization of barycenters in the space of Gaussian measures
\begin{equation}
\label{eq: penalization Gaussian}
\min_{\mu\in\N}\sum_{i=1}^n\frac{1}{2}\lambda_i W_2^2(\mu,\mu_i)+\gamma F(\mu),
\end{equation}
where $F$ the (negative) Boltzmann entropy functional of a
probability measure defined in \eqref{eq: entropy} and $\gamma>0$ is a regularization parameter.

We assume that $\mu_i\sim \N(0,A_i)$ and seek for a Gaussian
minimizer $\mu\sim \N(0,X)$. We note that we consider here Gaussian measures with zero mean just for simplicity. The main results of the paper can be easily extended to the case of non-zero mean. From now on, we equip ${\Bbb S}(d,\R)$ with the
Frobenius inner product $\langle X,Y\rangle :=\tr(XY)$. The
Frobenius norm is defined by
$\|X\|_F=\Big(\tr(X^2)\Big)^\frac{1}{2}$. For
$X,Y\in {\Bbb S}(d,\R),$ we write $X\leq Y$ if $Y-X$ is positive
semidefinite, and $X<Y$ if $Y-X$ is positive definite. Note that
$X\leq Y$ if and only if $\langle x,Xx\rangle\leq \langle
x,Yx\rangle$ for all $x\in {\Bbb C}^N.$ We denote $[X,Y]$ by the
L\"owner order interval $[X,Y]:=\{Z:X\leq Z\leq Y\}.$
\begin{theorem}
\label{thm: Gaussian}
Assume that $\mu_i\sim \N(0,A_i)$. The penalization of barycenters problem \eqref{eq: penalization} has a unique solution $\mu\sim \N(0,X)$ where the covariance matrix $X$ solves the following nonlinear matrix equation
\begin{equation}
\label{eq: optimal equation Gaussian}
X-\gamma I=\sum_{i=1}^n \lambda_i (X^\frac{1}{2} A_i X^{1/2})^\frac{1}{2}.
\end{equation}
In particular, in the scalar case ($d=1$), we obtain
\begin{equation}
\label{eq: 1D}
X=\frac{\left[\sum_{i=1}^n\lambda_i a_i^\frac{1}{2}+\Big(\big(\sum_{i=1}^n\lambda_i a_i^\frac{1}{2}\big)^2+4\gamma\Big)^\frac{1}{2}\right]^2}{4}.
\end{equation}
\end{theorem}
Before proving this theorem, we show the existence of solutions to equation \eqref{eq: optimal equation Gaussian}.
\begin{lemma}
\label{lem: existence Gaussian}
Equation \eqref{eq: optimal equation Gaussian}
 has a positive definite solution.
\end{lemma}
\begin{proof}
 Pick
$0<\alpha_0<\beta_0$ so that $\alpha_0 I\leq A_i\leq \beta_0 I$ for
all $i=1,\dots, n.$ Set
$$
\alpha_*:=\left(\frac{\sqrt{\alpha_0}+\sqrt{\alpha_0+4\gamma}}{2}\right)^2,
\qquad \beta_*:=\left(\frac{\sqrt{\beta_0}+\sqrt{\beta_0+4\gamma}}{2}\right)^2.
$$

Then for $\alpha_* I\leq X\leq \beta_* I,$
$$\alpha_0 X\leq X^{1/2}A_iX^{1/2}\leq \beta_0 I, \qquad i=1,\dots,
n$$ and hence
$$\sqrt{\alpha_0}\sqrt{\alpha_*} I\leq \sqrt{\alpha_0}X^{1/2}\leq
(X^{1/2}A_iX^{1/2})^{1/2}\leq \sqrt{\beta_0}X^{1/2}\leq
\sqrt{\beta_0}\sqrt{\beta_*}I.$$ By definition of $\alpha_*$ and
$\beta_*$,
\begin{eqnarray*}
\alpha_* I&=&\sqrt{\alpha_0}\sqrt{\alpha_*}I+\gamma
I\leq\sum_{i=1}^n\lambda_i(X^{1/2}A_iX^{1/2})^{1/2}+\gamma I\\
&\leq&\sqrt{\beta_0}\sqrt{\beta_*}I+\gamma I=\beta_* I
\end{eqnarray*}
for every $X\in [\alpha_* I,\beta_* I]:=\{Z: \alpha_* I\leq Z\leq
\beta_* I\}.$ This shows that the map
$$
f(X):=\sum_{i=1}^n\lambda_i(X^{1/2}A_iX^{1/2})^{1/2}+\gamma I
$$
is a continuous self map on the L\"owner order interval $[\alpha_*
I,\beta_* I].$ By Brouwer's fixed point theorem, it has a fixed
point.
\end{proof}
We are now ready to prove Theorem \ref{thm: Gaussian}
\begin{proof}[Proof of Theorem \ref{thm: Gaussian}]
According to \eqref{eq: W2-Gaussians} and \eqref{eq: entropy of Gaussian measures} we have
\begin{align*}
W_2^2(\mu_i, \mu)&=\tr X+\tr A_i-2\tr\Big(A_i^\frac{1}{2}XA_i^\frac{1}{2}\Big)^\frac{1}{2},
\\ F(\mu)&=-\frac{d}{2}\ln(2\pi e)-\frac{1}{2}\ln(\det X).
\end{align*}
Thus we can write \eqref{eq: penalization} as a minimization problem in the space of positive definite matrices
\begin{equation}
\label{eq: penlization2}
\min_{X\in \H} \frac{1}{2} f(X)
\end{equation}
where
\begin{align}
\label{eq: f}
f(X)&:=\sum_{i=1}^n \lambda_i\tr A_i+\sum_{i=1}^n \lambda_i \tr\Big(X-2\big(A_i^\frac{1}{2}X A_i^\frac{1}{2}\big)^\frac{1}{2}\Big)
-\gamma\ln \det(X)-\gamma d \ln(2\pi e)\notag
\\&:=f_1(X)+\gamma f_2(X),
\end{align}
where
\begin{align*}
&f_1(X)=\sum_{i=1}^n \lambda_i\tr A_i+\sum_{i=1}^n \lambda_i \tr\Big(X-2\big(A_i^\frac{1}{2}X A_i^\frac{1}{2}\big)^\frac{1}{2}\Big),
\\& f_2(X)=-\ln \det(X)-d \ln(2\pi e).
\end{align*}
It has been proved \cite{Bhatia2018} that
\begin{enumerate}[(i)]
\item $X\mapsto f_1(X)$ is strictly convex,
\item $Df_1(X)(Y)=\tr\Big(I-\sum_{i=1}^n \lambda_i (A_i\sharp X^{-1})\Big) Y$,
\end{enumerate}
where $A\sharp B$ denotes the  geometric mean between $A$ and $B$ defined by
\begin{equation}
\label{eq: geometric mean}
A\sharp B=A^{1/2} (A^{-1/2}BA^{-1/2})^{1/2}A^{1/2},
\end{equation}
which is symmetric in $A$ and $B$. According to \cite[Proof of
Theorem 8, Chapter 10]{Lax2007} $X\mapsto
-\ln\det(X)$ is strictly convex. Using Jacobi's formula for the
derivative of the determinant and the chain rule, we get
\begin{equation}
Df_2(X)(Y)=-\frac{d}{dt}\ln\det(X+\varepsilon Y)\Big\vert_{t=0}=-\frac{1}{\det X}\cdot\det X \cdot\tr (X^{-1}Y)=-\tr (X^{-1}Y).
\end{equation}
It follows that $X\mapsto f(X)$ is strictly convex. Furthermore, we have
\begin{equation}
Df(X)(Y)=\tr\Big(I-\gamma X^{-1}-\sum_{i=1}^n \lambda_i (A_i\sharp X^{-1})\Big) Y,
\end{equation}
From this we deduce that
\begin{equation}
\label{eq: nabla f} \nabla f(X)=I-\gamma X^{-1}-\sum_{i=1}^n
\lambda_i (A_i\sharp X^{-1}),
\end{equation}
where the gradient is with respect to the Frobenius inner product. Hence $\nabla f(X)=0$ if and only if
\begin{equation*}
I-\gamma X^{-1}=\sum_{i=1}^n \lambda_i (A_i\# X^{-1}).
\end{equation*}
Using the definition \eqref{eq: geometric mean} of the geometric mean, the above equation can be written as
\begin{equation*}
 X-\gamma I=\sum_{i=1}^n \lambda_i (X^\frac{1}{2} A_i
 X^{1/2})^\frac{1}{2},
\end{equation*}
which is equation \eqref{eq: optimal equation Gaussian}.  By Lemma \ref{lem: existence Gaussian} this equation has a positive definite solution. This together with the strict convexity of $f$ imply that $f$ has a unique minimizer which is a Gaussian measure $\N(0,X)$ where $X$ solves \eqref{eq: optimal equation Gaussian}. In the one dimensional case this equation reads
$$
X-\gamma=\sqrt{X}\sum_{i=1}^n\lambda_i\sqrt{a_i},
$$
which results in
$$
X=\frac{\left[\sum_{i=1}^n\lambda_i a_i^\frac{1}{2}+\Big(\big(\sum_{i=1}^n\lambda_i a_i^\frac{1}{2}\big)^2+4\gamma\Big)^\frac{1}{2}\right]^2}{4}.
$$
This completes the proof of the theorem.
\end{proof}
\section{Penalization of barycenters for $q$-Gaussian measures}
\label{sec: q-Gaussians}
In this section we study the following penalization of barycenters in the space of $q$-Gaussian measures
\begin{equation}
\label{eq: penalization q-Gaussian}
\min_{\mu\in\G_q}\sum_{i=1}^n\frac{1}{2}\lambda_i W_2^2(\mu,\mu_i)+\gamma F_q(\mu),
\end{equation} where $F_q$ the Tsallis entropy functional defined by
\begin{equation}
\label{eq: Tsallis entropy}
F_q(\mu):=\int \mu \log_q \mu.
\end{equation}
We assume that $\mu_i\sim G_q(0,A_i)$ and seek for a Gaussian minimizer $\mu\sim G_q(0,X)$.
\begin{theorem}
\label{thm: q-Gaussian}
Assume that $\mu_i\sim G_q(0,A_i)$. Suppose that $\alpha I\leq A_i \leq \beta I$ for all $i=1,\ldots, n$.
The penalization of barycenters problem \eqref{eq: penalization q-Gaussian} has a unique solution $\mu\sim G_q(0,X)$ for all $\gamma\geq 0$ if either $0< q \leq 1$ or $1<q\leq 1+\frac{2\alpha^2}{d \beta^2}$ and for $\gamma$ sufficiently small if $1+\frac{2\alpha^2}{d \beta^2}<q<\frac{d+4}{d+2}$. The covariance matrix $X$ solves the following nonlinear matrix equation
\begin{equation}
\label{eq: optimal equation q-Gaussian}
X-\gamma m(q,d)(\det X)^{\frac{q-1}{2}} I=\sum_{i=1}^n \lambda_i \Big(X^\frac{1}{2}A_i X^\frac{1}{2}\Big)^\frac{1}{2},
\end{equation}
where $m(q,d)$ is defined by
\begin{equation}
\label{eq: m}
m(q,d):=\frac{2(2-q)C_0(q,d)^{1-q}}{2+(d+2)(1-q)}.
\end{equation}
\end{theorem}
The following proposition shows that equation \eqref{eq: optimal equation q-Gaussian} possesses a positive definite solution.
\begin{proposition}
\label{prop: existence solution q Gaussians}
 Equation \eqref{eq: optimal equation q-Gaussian} has a positive definite solution.
\end{proposition}
\begin{proof} Similarly as the proof of Lemma \ref{lem: existence Gaussian} we will also apply Brouwer's fixed point theorem. We will show that
$$
\psi(X):=\sum_{i=1}^n \lambda_i(X^{1/2}A_iX^{1/2})^{1/2}+\gamma m(q,d)(\det X)^{\frac{q-1}{2}} I
$$
has a fixed point which is a positive definite matrix. Due to the appearance of the second term on the left-hand side of \eqref{eq: optimal equation q-Gaussian} the proof of this proposition is significantly involved than that of Lemma \ref{lem: existence Gaussian}. Suppose that $\alpha_0 I\leq A_i \leq \beta_0 I$ for all $i=1,\ldots, n$. Then similarly as in the proof of Lemma \ref{lem: existence Gaussian}, for $\alpha_* I\leq X\leq \beta_* I$ (with $\alpha_*, \beta_*$ chosen later), we have
$$
\sqrt{\alpha_0}\sqrt{\alpha_*} I\leq \sqrt{\alpha_0}X^{1/2}\leq
(X^{1/2}A_iX^{1/2})^{1/2}\leq \sqrt{\beta_0}X^{1/2}\leq
\sqrt{\beta_0}\sqrt{\beta_*}I,\quad i=1,\ldots, n,
$$
so that
$$
\sqrt{\alpha_0}\sqrt{\alpha_*} I\leq (X^{1/2}A_iX^{1/2})^{1/2}\leq \sqrt{\beta_0}\sqrt{\beta_*}I.
$$
Multiplying this inequality with $\lambda_i$ then adding them together, noting that $\sum\lambda_i=1$, we obtain
\begin{equation*}
\sqrt{\alpha_0}\sqrt{\alpha_*} I\leq \sum_{i=1}^n \lambda_i(X^{1/2}A_iX^{1/2})^{1/2}\leq \sqrt{\beta_0}\sqrt{\beta_*}I,
\end{equation*}
from which it follows that
\begin{multline}
\label{eq: estimate1}
\sqrt{\alpha_0}\sqrt{\alpha_*} I+\gamma m(q,d)(\det X)^{\frac{q-1}{2}} I\leq \sum_{i=1}^n \lambda_i(X^{1/2}A_iX^{1/2})^{1/2}+\gamma m(q,d)(\det X)^{\frac{q-1}{2}} I\\\leq \sqrt{\beta_0}\sqrt{\beta_*}I+\gamma m(q,d)(\det X)^{\frac{q-1}{2}} I.
\end{multline}
To continue we consider two cases.
\vskip 1pc

Case 1: $1<q<\frac{d+4}{d+2}$.
It follows from \eqref{eq: estimate1} that
\begin{multline}
\label{eq: estimate2}
\sqrt{\alpha_0}\sqrt{\alpha_*} I+\gamma m(q,d)\alpha_*^{\frac{d(q-1)}{2}} I\leq
\sqrt{\alpha_0}\sqrt{\alpha_*} I+\gamma m(q,d)(\det X)^{\frac{q-1}{2}} I\\
\leq \gamma m(q,d)(\det X)^{\frac{q-1}{2}} I+\sum_{i=1}^n \lambda_i(X^{1/2}A_iX^{1/2})^{1/2}\\\leq \sqrt{\beta_0}\sqrt{\beta_*}I+\gamma m(q,d)(\det X)^{\frac{q-1}{2}} I\leq \sqrt{\beta_0}\sqrt{\beta_*}I+\gamma m(q,d)\beta_*^{\frac{d(q-1)}{2}} I
\end{multline}
Since $1<q<\frac{d+4}{d+2}$, we have $0<(q-1)d<\frac{2d}{d+2}<2$.
\vskip 1pc
Case 1.1: $d(q-1)\leq 1$. Consider the following equation
$$
g_1(t):=t^{1-\frac{q(d-1)}{2}}-\sqrt{\alpha_0}t^{\frac{1-d(q-1)}{2}}-\gamma m(q,d)=0.
$$
We have $\lim\limits_{t\rightarrow 0}g_1(t)=-\gamma m(q,d)<0$ and $\lim\limits_{t\rightarrow +\infty}g_1(t)=+\infty$. Since $g_1$ is continuous, it follows that there exists $\alpha_*\in (0,\infty)$ such that $g_1(\alpha_*)=0$, that is
$$
\alpha_*^{1-\frac{q(d-1)}{2}}=\sqrt{\alpha_0}\alpha*^{\frac{1-d(q-1)}{2}}+\gamma m(q,d), \quad\text{i.e.,}\quad \alpha_*=\sqrt{\alpha_0}\sqrt{\alpha_*}+\gamma m(q,d)\alpha_*^{\frac{d(q-1)}{2}}.
$$
Similarly by considering the function $g_2(t):=t^{1-\frac{q(d-1)}{2}}-\sqrt{\beta_0}t^{\frac{1-d(q-1)}{2}}-\gamma m(q,d)$, we deduce that there exists $\beta_*\in (0,\infty)$ such that
$$
\beta_*=\sqrt{\beta_0}\sqrt{\beta_*}+\gamma m(q,d)\beta_*^{\frac{d(q-1)}{2}}.
$$
\vskip 1pc
Case 1.2: $d(q-1)>1$. Using the same argument as in the previous case for
$$
g_3(t)=t^{1/2}-\sqrt{\alpha_0}-\gamma m(q,d)t^{\frac{d(q-1)-1}{2}}~\text{and}~g_4(t)=t^{1/2}-\sqrt{\beta_0}-\gamma m(q,d)t^{\frac{d(q-1)-1}{2}}
$$
we can show that there exist $\alpha_*, \beta_*\in (0,\infty)$ such that
$$
\alpha_*=\sqrt{\alpha_0}\sqrt{\alpha_*}+\gamma m(q,d)\alpha_*^{\frac{d(q-1)}{2}}~\text{and}~\beta_*=\sqrt{\beta_0}\sqrt{\beta_*}+\gamma m(q,d)\beta_*^{\frac{d(q-1)}{2}}.
$$
\vskip 1pc
Therefore in both Cases 1.1 and 1.2, there exist  $\alpha_*, \beta_*\in (0,\infty)$ such that
$$
\alpha_*=\sqrt{\alpha_0}\sqrt{\alpha_*}+\gamma m(q,d)\alpha_*^{\frac{d(q-1)}{2}}~\text{and}~\beta_*=\sqrt{\beta_0}\sqrt{\beta_*}+\gamma m(q,d)\beta_*^{\frac{d(q-1)}{2}}.
$$
Substituting these quantities into \eqref{eq: estimate2} we obtain
\begin{multline}
\alpha_* I=\sqrt{\alpha_0}\sqrt{\alpha_*} I+\gamma m(q,d)\alpha_*^{\frac{d(q-1)}{2}} I \\\leq \gamma m(q,d)(\det X)^{\frac{q-1}{2}} I+\sum_{i=1}^n \lambda_i(X^{1/2}A_iX^{1/2})^{1/2} \\
\leq \sqrt{\beta_0}\sqrt{\beta_*}I+\gamma m(q,d)\beta_*^{\frac{d(q-1)}{2}} I=\beta_* I.
\end{multline}
Thus $\alpha_* I \leq \psi(X)\leq \beta_* I$.  By Brouwer's fixed point theorem, $\psi(X)$ has a fixed point in $[\alpha_* I, \beta_* I]$ as desired.
\vskip 1pc

Case 2. $0<q<1$.
\vskip 1pc

It follows from \eqref{eq: estimate1} that
\begin{multline}
\label{eq: estimate3}
\sqrt{\alpha_0}\sqrt{\alpha_*} I+\gamma m(q,d)\beta_*^{\frac{d(q-1)}{2}} I\leq
\sqrt{\alpha_0}\sqrt{\alpha_*} I+\gamma m(q,d)(\det X)^{\frac{q-1}{2}} I\\
\leq \gamma m(q,d)(\det X)^{\frac{q-1}{2}} I+\sum_{i=1}^n \lambda_i(X^{1/2}A_iX^{1/2})^{1/2}\\\leq \sqrt{\beta_0}\sqrt{\beta_*}I+\gamma m(q,d)(\det X)^{\frac{q-1}{2}} I\leq \sqrt{\beta_0}\sqrt{\beta_*}I+\gamma m(q,d)\alpha_*^{\frac{d(q-1)}{2}} I
\end{multline}
Next we will show that following system has positive solutions $0<\alpha_*<\beta_*<\infty$:
\begin{equation}
\label{eq: system1}
\begin{cases}
\alpha_*=\sqrt{\alpha_0}\sqrt{\alpha_*}+\gamma m(q,d)\beta_*^{\frac{d(q-1)}{2}} \\
\beta_*=\sqrt{\beta_0}\sqrt{\beta_*}+\gamma m(q,d)\alpha_*^{\frac{d(q-1)}{2}}.
\end{cases}
\end{equation}
Define $f: (0,\infty)^2\rightarrow (0,\infty)^2$ by
$$
f\left(\begin{pmatrix}
x\\y
\end{pmatrix}\right)=\begin{pmatrix}
\sqrt{\alpha_0}\sqrt{x}+\gamma m(q,d)y^{\frac{d(q-1)}{2}}\\
\sqrt{\beta_0}\sqrt{y}+\gamma m(q,d)x^{\frac{d(q-1)}{2}}
\end{pmatrix}
$$

Set
\begin{align*}
&a_*=\left(\frac{\sqrt{\alpha_0}+\sqrt{\alpha_0+4\gamma m(q,d)\beta_0^{(q-1)d/2}}}{2}\right)^2,
\\& b_*=\left(\frac{\sqrt{\beta_0}+\sqrt{\beta_0+4\gamma m(q,d)\alpha_0^{(q-1)d/2}}}{2}\right)^2.
\end{align*}
Thus $a_*$ and $b_*$ satisfy
$$
a_*=\sqrt{\alpha_0}\sqrt{a_*}+\gamma m(q,d)\beta_0^{(q-1)d/2},\quad b_*=\sqrt{\beta_0}\sqrt{b_*}+\gamma m(q,d)\alpha_0^{(q-1)d/2}.
$$
We now show that $f: [\alpha_0, a_*]\times [\beta_0, b_*]\rightarrow [\alpha_0, a_*]\times [\beta_0, b_*]$.
In fact, consider $\alpha_0\leq x\leq a_*$ and $\beta_0\leq y\leq b_*$. We have
\begin{align*}
&\alpha_0\leq \sqrt{\alpha_0}\sqrt{x}\leq \sqrt{\alpha_0}\sqrt{x}+\gamma m(q,d)y^{\frac{d(q-1)}{2}}\leq \sqrt{\alpha_0}\sqrt{x}+\gamma m(q,d)\beta_0^{\frac{d(q-1)}{2}}=a_*,\\
&\beta_0\leq \sqrt{\beta_0}\sqrt{y}\leq \sqrt{\beta_0}\sqrt{y}+\gamma m(q,d)x^{\frac{d(q-1)}{2}}\leq \sqrt{\beta_0}\sqrt{y}+\gamma m(q,d)\alpha_0^{\frac{d(q-1)}{2}}=b_*.\\
\end{align*}
Thus $f((x,y)^T)\in [\alpha_0, a_*]\times [\beta_0, b_*]$. By Brouwer's fixed point theorem, $f$ has a fixed point in $[\alpha_0, a_*]\times [\beta_0, b_*]$, which means that system \eqref{eq: system1} has a positive solution $(\alpha_*,\beta_*)$. Using this solution in \eqref{eq: estimate3} we obtain
\begin{multline}
\alpha_* I=\sqrt{\alpha_0}\sqrt{\alpha_*} I+\gamma m(q,d)\beta_*^{\frac{d(q-1)}{2}} I \\ \leq \gamma m(q,d)(\det X)^{\frac{q-1}{2}} I+\sum_{i=1}^n \lambda_i(X^{1/2}A_iX^{1/2})^{1/2}\\ \leq  \sqrt{\beta_0}\sqrt{\beta_*}I+\gamma m(q,d)\alpha_*^{\frac{d(q-1)}{2}} I=\beta_* I.
\end{multline}
Hence by Brouwer's fixed point theorem again, $\psi$ has a fixed point in $[\alpha_* I, \beta_* I]$ as desired. This finishes the proof of the proposition.
\end{proof}
Next we will show that the functional that we wish to mimimize in \eqref{eq: penalization q-Gaussian} is strictly convex under rather general conditions. According to Proposition \ref{prop: q-Gaussian properties} and Lemma \ref{lem: Tsallis entropy of q-Gaussian} we have
\begin{align*}
W_2^2(\mu_i, \mu)&=\tr X+\tr A_i-2\tr\Big(A_i^\frac{1}{2}XA_i^\frac{1}{2}\Big)^\frac{1}{2},
\\ F_q(\mu)&=-\frac{d}{2}C_1(q,d)+\Big[1-(1-q)\frac{d}{2}C_1(q,d)\Big]\ln_{q}\frac{C_0(q,d)}{(\det U)^\frac{1}{2}}.
\end{align*}
Therefore the minimization problem \eqref{eq: penalization q-Gaussian} can be written as
\begin{equation}
\label{eq: reformulation of penal. q Gaussian}
\min_{X\in\H}\frac{1}{2} g(X)
\end{equation}
where
\begin{align}
\label{eq: g1}
g(X)&=\sum_{i=1}^n\lambda_i\tr A_i+\sum_{i=1}^n \lambda_i\tr\Big(X-2(A_i^\frac{1}{2}XA_i^\frac{1}{2})^\frac{1}{2}\Big)
\\&\qquad+\gamma \Big[2-(1-q)d C_1(q,d)\Big]\ln_{q}\frac{C_0(q,d)}{(\det U)^\frac{1}{2}}-\gamma d C_1(q,d)\notag
\\&=f_1(X)+\gamma \Big[2-(1-q)d C_1(q,d)\Big]\ln_{q}\frac{C_0(q,d)}{(\det U)^\frac{1}{2}}-\gamma d C_1(q,d),
\end{align}
with $f_1(X)=\sum_{i=1}^n \lambda_i\tr A_i+\sum_{i=1}^n \lambda_i\tr\Big(X-2(A_i^\frac{1}{2}XA_i^\frac{1}{2})^\frac{1}{2}\Big)$, which appeared in \eqref{eq: f}.
Note that by definition of the $q$-logarithmic function we have
\begin{equation*}
\ln_{q}\frac{C_0(q,d)}{(\det U)^\frac{1}{2}}=\frac{1}{1-q}\left[C_0(q,d)^{1-q}(\det U)^{-\frac{1-q}{2}}-1\right].
\end{equation*}
Using explicit formula of $C_1(q,d)$ we get
\begin{align*}
2-(1-q)d C_1(q,d)&=2-(1-q)d\frac{2}{2+(d+2)(1-q)}
\\&=\frac{4(2-q)}{2+(d+2)(1-q)}.
\end{align*}
Substituting these expressions into \eqref{eq: g1} we get
\begin{align}
\label{eq: g2}
g(X)&=f_1(X)+\frac{4\gamma(2-q)C_0(q,d)^{1-q}}{(2+(d+2)(1-q))(1-q)}(\det X)^{-\frac{1-q}{2}}\nonumber
\\&\qquad-\frac{4(2-q)}{(1-q)(2+(d+2)(1-q))}-\gamma d C_1(q,d).
\end{align}
The following proposition studies the convexity of $g$.
\begin{proposition}
\label{prop: convexity}
Suppose that $\alpha I\leq A_i, X,\leq \beta I$ for all $i=1,\ldots, n$. The functional $g$ given in \eqref{eq: g2} is strictly convex for all $\gamma\geq 0$ when one of the following condition holds
\begin{enumerate}
\item $0<q<1$,
\item $1<q\leq 1+\frac{2\alpha^2}{d \beta^2}$.
\end{enumerate}
The second condition is fulfilled if $ \beta^2\leq \frac{d+2}{d}\alpha^2$. In addition, if $1+\frac{2\alpha^2}{d \beta^2}<q<\frac{d+4}{d+2}$, then $g$ is strictly convex for $0\leq\gamma<\gamma_0$ where
$$
\gamma_0=\frac{1}{2}\frac{\alpha^{1/2}}{\beta^{3/2}}\frac{1}{\frac{1}{\beta^2}-\frac{(q-1)d}{2\alpha^2}}\frac{1}{m(q,d)}\frac{1}{\beta^{d(q-1)/2}}.
$$
\end{proposition}
\begin{proof} We consider two cases.

\noindent{\bf Case 1.} $1< q <\frac{d+4}{d+2}.$

Let $k(X):=\frac{4\gamma(2-q)C_0(q,d)^{1-q}}{(2+(d+2)(1-q))(1-q)}(\det X)^{\frac{q-1}{2}}$. Let $h(X):= (\det X)^\frac{q-1}{2}$. Similarly as in the proof of Theorem \ref{thm: Gaussian}, using again Jacobi's formula for the derivative of the determinant and the chain rule, we get
\begin{equation*}
Dh(X)(Y)=\frac{q-1}{2}(\det X)^{\frac{q-3}{2}} \cdot \det(X)\cdot\tr (X^{-1} Y)=\frac{q-1}{2}(\det X)^{\frac{q-1}{2}}\tr (X^{-1} Y).
\end{equation*}
Therefore, using the definition of $m(q,d)$, we have
\begin{equation}
\label{eq: derivative k}
\nabla k(X) = -\gamma m(q,d)(\det X)^{\frac{q-1}{2}}X^{-1}=-\gamma m(q,d) h(X)X^{-1}.
\end{equation}
In the computations below the linear operator $P(X)$ is defined to be $P(X)Y = XYX$. This operator is called the quadratic representation in the literature. By the Leibniz rule, we get
\begin{align*}
\nabla^2 k(X)(H) &= D(\nabla k)(X)(H)\\
& = -\gamma m(q,d)[D h(X)(H) X^{-1} + h(X)(-P(X^{-1}))(H)]\\
& = -\gamma m(q,d)[\langle \nabla h(X), H\rangle X^{-1} - h(X)X^{-1}H X^{-1}]\\
& = -\gamma m(q,d)\left[\left\langle \frac{q-1}{2}(\det X)^{\frac{q-1}{2}}X^{-1},  H\right\rangle X^{-1}- (\det X)^{\frac{q-1}{2}}X^{-1}H X^{-1}\right]\\
& = -\gamma m(q,d)(\det X)^{\frac{q-1}{2}}\left[\left\langle \frac{q-1}{2}X^{-1}, H\right\rangle X^{-1}- X^{-1}H X^{-1}\right].
\end{align*}
Thus
\begin{align}
\langle\nabla^2 k(X)(H), H\rangle &=-\gamma m(q,d)(\det X)^{\frac{q-1}{2}}\left[\frac{q-1}{2}\left\langle X^{-1}, H\right\rangle^2-\langle X^{-1}H, X^{-1}H\rangle\right]\\
&=-\gamma m(q,d)(\det X)^{\frac{q-1}{2}}\left[\frac{q-1}{2}\tr^2 (X^{-1}H)-\| X^{-1}H\|^2\right].
\end{align}
\noindent Furthermore, according to \cite{Bhatia2018}, for $\alpha I \le A_i, X \le \beta I$, we have
$$
\langle\nabla^2 f_1 (X)(H), H\rangle\geq \frac{1}{2}\frac{\alpha^{1/2}}{\beta^{3/2}}\|H\|^2.
$$
Thus we get
\begin{align*}
\langle\nabla^2 g(X)(H), H\rangle &= \langle\nabla^2 f_1 (X)(H), H\rangle + \langle\nabla^2 k(X)(H), H\rangle\\
&\ge -\gamma m(q,d)(\det X)^{\frac{q-1}{2}}\left[\frac{q-1}{2}\tr^2 (X^{-1}H)-\| X^{-1}H\|^2\right] + \frac{1}{2}\frac{\alpha^{1/2}}{\beta^{3/2}}\|H\|^2\\
&= \gamma m(q,d)(\det X)^{\frac{q-1}{2}}\left[\langle P(X^{-1})H, H\rangle - \frac{q-1}{2}\tr^2 (X^{-1}H)\right] + \frac{1}{2}\frac{\alpha^{1/2}}{\beta^{3/2}}\|H\|^2\\
& \ge \gamma m(q,d)(\det X)^{\frac{q-1}{2}}\left[\frac{1}{\beta^2}\|H\|^2 - \frac{q-1}{2}\|X^{-1}\|^2\|H\|^2\right] + \frac{1}{2}\frac{\alpha^{1/2}}{\beta^{3/2}}\|H\|^2\\
& = \left\{\gamma m(q,d)(\det X)^{\frac{q-1}{2}}\left[\frac{1}{\beta^2} - \frac{q-1}{2}\|X^{-1}\|^2\right] + \frac{1}{2}\frac{\alpha^{1/2}}{\beta^{3/2}}\right\}\|H\|^2\\
& \ge\left\{\gamma m(q,d)(\det X)^{\frac{q-1}{2}}\left[\frac{1}{\beta^2} - \frac{q-1}{2}\frac{d}{\alpha^2}\right] + \frac{1}{2}\frac{\alpha^{1/2}}{\beta^{3/2}}\right\}\|H\|^2\\
& \ge \left\{\gamma m(q,d)(\det X)^{\frac{q-1}{2}}\left[\frac{1}{\beta^2} - \frac{q-1}{2}\frac{d}{\alpha^2}\right] + \frac{1}{2}\frac{\alpha^{1/2}}{\beta^{3/2}}\right\}\|H\|^2.
\end{align*}
From this estimate, we deduce the following cases
\begin{enumerate}[(i)]
\item If
$$
1<q\leq 1+\frac{2\alpha^2}{d \beta^2},
$$
thus $\frac{1}{\beta^2}-\frac{q-1}{2}\frac{d}{\alpha^2}>0$, which implies that the Hessian of $g$ is positive for all $\gamma$. Note that the above condition is fulfilled if $\alpha$ and $\beta$ satisfy  $ \beta^2\leq \frac{d+2}{d}\alpha^2$. In fact, we have
$$
q<1+\frac{2}{d+2}\leq 1+\frac{2\alpha^2}{d\beta^2},
$$
\item If
$$
1+\frac{2\alpha^2}{d \beta^2}<q<\frac{d+4}{d+2}.
$$
then for
$$
\gamma<\frac{1}{2}\frac{\alpha^{1/2}}{\beta^{3/2}}\frac{1}{\frac{1}{\beta^2}-\frac{(q-1)d}{2\alpha^2}}\frac{1}{m(q,d)}\frac{1}{\beta^{d(q-1)/2}}
$$
the Hessian of $g$ is positive since
\begin{multline}
\gamma<\frac{1}{2}\frac{\alpha^{1/2}}{\beta^{3/2}}\frac{1}{\frac{1}{\beta^2}-\frac{(q-1)d}{2\alpha^2}}\frac{1}{m(q,d)}\frac{1}{\beta^{d(q-1)/2}} \\ \leq \frac{1}{2}\frac{\alpha^{1/2}}{\beta^{3/2}}\frac{1}{\frac{1}{\beta^2}-\frac{(q-1)d}{2\alpha^2}}\frac{1}{m(q,d)}\frac{1}{(\det X)^{(q-1)/2}}
\end{multline}
\end{enumerate}
\vskip 1pc
\noindent{\bf Case 2.} $0< q <1.$
Similarly, we obtain
\begin{align*}
\langle\nabla^2 k(X)(H), H\rangle &=\gamma m(q,d)(\det X)^{\frac{q-1}{2}}\left[\frac{1-q}{2}\left\langle X^{-1}, H\right\rangle^2+\langle P(X^{-1})H, H\rangle\right]\\
&\ge\gamma m(q,d)(\det X)^{\frac{q-1}{2}}{\frac{1}{\lambda^2_{\max}(X)}}\|H\|^2.
\end{align*}
Hence the Hessian of $g$ is always positive definite in this case.
\end{proof}
We are now ready to proof Theorem \ref{thm: q-Gaussian}.
\begin{proof}[Proof of Theorem \ref{thm: q-Gaussian}]
Suppose that the hypothesis of the statement of Theorem \ref{thm: q-Gaussian} is satisfied, that is either (i) $0<q\leq 1$ or (ii) $1<q\leq 1+\frac{2\alpha^2}{d \beta^2}$ or (iii) $1+\frac{2\alpha^2}{d \beta^2}<q<\frac{d+4}{d+2}$. Suppose that $\gamma$ is sufficiently small in the last case; in the other cases it can be arbitrarily positive. As has been shown in the paragraph before Proposition \ref{prop: convexity}, the minimization problem \eqref{eq: penalization q-Gaussian} can be written as
\begin{equation*}
\min_{X\in\H}\frac{1}{2} g(X),
\end{equation*}
where $g(X)$ is given in \eqref{eq: g2}
$$
g(X)=f_1(X)+k(X)-\frac{4(2-q)}{(1-q)(2+(d+2)(1-q))}-\gamma d C_1(q,d).
$$
By Proposition \ref{prop: convexity}, $X\mapsto g(X)$ is strictly convex.
Now we compute the derivative of $g(X)$. We have
\begin{equation}
\label{eq: derivative 1}
\nabla g(X)= \nabla f_1(X)+ \nabla k(X),
\end{equation}
According to the proof of Theorem \ref{thm: Gaussian} we have
\begin{equation*}
\nabla f_1(X)=I-\sum_{i=1}^n \lambda_i (A_i\sharp X^{-1}).
\end{equation*}
By \eqref{eq: derivative k}, we have
$$
\nabla k(X)=-\gamma m(q,d) (\det X)^{\frac{q-1}{2}} X^{-1}
$$
Substituting these computations into \eqref{eq: derivative 1} we obtain
\begin{equation}
\nabla g(X)=\Big(I-\sum_{i=1}^n \lambda_i (A_i\sharp X^{-1})\Big)-\gamma m(q,d)(\det X)^{\frac{q-1}{2}}X^{-1}.
\end{equation}
Thus $\nabla g(X)=0$ if and only if
$$
I-\gamma m(q,d)(\det X)^{\frac{q-1}{2}}X^{-1}=\sum_{i=1}^n \lambda_i (A_i\sharp X^{-1}),
$$
which, by using the definition of the geometric mean \eqref{eq: geometric mean}, is equivalent to
$$
X-\gamma m(q,d)(\det X)^{\frac{q-1}{2}} I=\sum_{i=1}^n \lambda_i \Big(X^\frac{1}{2}A_i X^\frac{1}{2}\Big)^\frac{1}{2}.
$$
This is precisely equation \eqref{eq: optimal equation q-Gaussian}. By Proposition \ref{prop: existence solution q Gaussians}, it has a positive definite solution. This, together with the strictly convexity of $g$, guarantees the existence and uniqueness of a minimizer of $g$. We complete the proof of the theorem.
\end{proof}

\section{Barycenters for $\varphi$-exponential measures}
\label{sec: vaphi expo measures}
In this section we consider the following barycenter problem in the space of $\varphi$-exponential measures:
\begin{equation}
\label{eq: baryceter varphi}
\min_{\mu\in\G_\varphi}\frac{\lambda_i}{2}W_2^2(\mu,\mu_i).
\end{equation}
In contrast to the Gaussian and q-Gaussian measures, we are not aware of an explicit for the entropy for a general $\varphi$-exponential measure. Therefore, in the above formulation we do not include the penalization term. The main result of this section is the following theorem that states that the equation determining the barycenter for $\varphi$-exponential measures is the same as that of for Gaussian-measures.
\begin{theorem}
\label{thm: varphi}
Let $\varphi\in \O(2/(d+2))$ with $d\geq 2$. Assume that $\mu_i\sim G_{\varphi}(0,A_i)$. The penalization of barycenters problem \eqref{eq: penalization} has a unique solution $\mu\sim G_\varphi(0,X)$ where the covariance matrix $X$ solves the following nonlinear matrix equation
\begin{equation}
\label{eq: optimal equation varphi}
X=\sum_{i=1}^n \lambda_i (X^\frac{1}{2} A_i X^{1/2})^\frac{1}{2}.
\end{equation}
In particular, for $n=2$, $X$ is given explicitly by
\begin{equation}
\label{eq: optimal n=2}
X=\lambda_1^2 A_1+ \lambda_2^2 A_2+\lambda_1 \lambda_2 \Big[(A_1A_2)^\frac{1}{2}+(A_2A_1)^\frac{1}{2}\Big].
\end{equation}
\end{theorem}
\begin{proof}
This theorem is a direct consequence of Proposition \ref{prop: q-Gaussian properties} and \cite[Theorem 6.1]{Agueh2011} or \cite[Theorem 8]{Bhatia2018}. In fact, similarly as in the proof of \ref{thm: q-Gaussian}, by using \eqref{eq: W2 formula} we can write \eqref{eq: baryceter varphi} as
$$
\min_{X\in \H}\frac{1}{2}f_1(X)
$$
where $f_1(x)=\sum_{i=1}^n \lambda_i\tr(A_i)+\sum_{i=1}^n\lambda_i\tr\Big(X-2(A_i^\frac{1}{2}XA_i^\frac{1}{2})\Big)^\frac{1}{2}$. Then the statement can be proved exactly as  \cite[Theorem 6.1]{Agueh2011} or \cite[Theorem 8]{Bhatia2018}, see also computations in the proof of Theorems \ref{thm: Gaussian} and \ref{thm: q-Gaussian} when $\gamma=0$.
Explicit formula \eqref{eq: optimal n=2} for the minimizer for the case $n=2$ is given in \cite[Eq. (63)]{Bhatia2018}.
\end{proof}

\section{Gradient projection method}
\label{sec: GPM}
In this section, we describe a gradient projection method for the computation of the minimizer to the penalization problems (\ref{eq: penlization2}) and (\ref{eq: reformulation of penal. q Gaussian}), and analyze its convergence properties.

First, we formally describe the algorithmic procedure for the gradient projection method (GPM) below.
\begin{algorithm}\label{A:alg1}
	\caption{GPM}
	\begin{algorithmic}
		\STATE Choose $X^0\in \Pi$.  Initialize $k=0$. Update $X^{(k+1)}$ from
		$X^{(k)}$ by the following template:
		\begin{description}
			\item[Step 1.] Find $\bar{X}^{(k)}=[X^{(k)}-\nabla \psi(X^{(k)})]^+$,
			\item[Step 2.] Select a stepsize $t^{(k)}$,
			\item[Step 3.] $X^{(k+1)}=X^{(k)}+t^{(k)}(\bar{X}^{(k)}-X^{(k)})$.
		\end{description}

		Here $[\cdot]^+$ denotes the projection on the set $\Pi:=[\hat{\alpha}I,\hat{\beta}I]$.
	\end{algorithmic}
\end{algorithm}

The stepsize is selected by Armijo rule along the feasible direction \cite{Ber99}. It is described in the below.

\vspace{4mm} \centerline{\fbox{\parbox{5in}{ Let $t^{(k)}$ be the largest element of
			$\{\xi^j\}_{j=0,1,...}$ satisfying
			\begin{equation}\label{armijo}
			\psi(X^{(k)}+t^{(k)} D^{(k)}) \le \psi(X^{(k)}) - \sigma t^{(k)}
			\inprod{\nabla \psi(X^{(k)})}{D^{(k)}},
			\end{equation}
			where $0<\xi<1$, $0<\sigma<1$, and
			$D^{(k)}=\bar{X}^{(k)}-X^{(k)}$. }}}

Note that $\psi=f$ for the penalization problem (\ref{eq: penlization2}) and $\psi=g$ for the penalization problem (\ref{eq: reformulation of penal. q Gaussian}). The projection of the matrix $S\in {\mathcal S}^d$, where ${\mathcal S}^d$ is the set of $d\times d$ symmetric matrices, onto the set $\Pi$ is to find the solution of the following minimization problem
$$
\min_{X\in \Pi}\;\norm{X-S}_F.
$$
The solution of the above problem is
$$
[S]^+=U{\rm Diag}(\min(\max(\hat{\alpha},\lambda_1),\hat{\beta}),\dots,\min(\max(\hat{\alpha},\lambda_d),\hat{\beta})U^T,
$$
where $\lambda_1\ge\cdots\ge\lambda_d$ are the eigenvalues of $S$ and $U$ is a corresponding orthogonal matrix of eigenvalues of $S$.

Now, we establish the global convergence of GPM. For the proof, we refer to
\cite[Proposition 2.3.1]{Ber99}.

\begin{theorem}\label{thm1}
	Let $\{X^{(k)}\}$ be the sequence generated by GPM with $t^{(k)}$ chosen by Armijo rule
	along the feasible direction. Then
	every limit point of $\{X^{(k)}\}$ is stationary.
\end{theorem}

In the following subsections, we show the Lipschitz continuity of the gradient function of the penalization problems. In this case, we can use a constant stepsize for
the gradient projection method. That is, $t^{(k)}=\frac{1}{L}$ where $L$ is a Lipschitz constant. Then we have
\begin{equation}\label{con}
X^{(k+1)}=X^{(k)}+\frac{1}{L}(\bar{X}^{(k)}-X^{(k)}).
\end{equation}

\subsection{Penalization of barycenters for Gaussian measures}
We recall that the unique minimizer of the minimization problem \eqref{eq: penalization Gaussian} in the space of Gaussian measures satisfies the following nonlinear matrix equation $\nabla f(X)=0$ where
\begin{equation}
\nabla f(X)=I-\sum_{i=  1}^n\lambda_i (A_i\sharp X^{-1})-\gamma X^{-1}=:F_1(X)-\gamma F_2 (X).
\end{equation}

We establish the following theorem for the Lipschitz continuity of the gradient function.
\begin{theorem}
\label{thm: Lipschitz continuity}
Suppose that $A_i\in [\alpha I, \beta I]$ for all $i=1,\ldots, n$. Then for $\alpha I\leq X\neq Y\leq \beta I$ we have
\begin{equation*}
\frac{\|\nabla f(X)-\nabla f(Y)\|_F}{\|X-Y\|_F}\leq \frac{\beta^2}{2\alpha^3}+\frac{\gamma}{\alpha^2}.
\end{equation*}
\end{theorem}
\begin{proof}
According to \cite[Proof of Theorem 3.1]{KumYun2018} we have
\begin{align*}
&\frac{\|F_1(X)-F_1(Y)\|_F}{\|X-Y\|_F}\leq \frac{\beta^2}{2\alpha^3}~~\text{and}~~\frac{\|F_2(X)-F_2(Y)\|_F}{\|X-Y\|_F}\leq \frac{1}{\alpha^2}.
\end{align*}
Therefore we get
\begin{align}
\frac{\|\nabla f(X)-\nabla f(Y)\|_F}{\|X-Y\|_F}&\leq \frac{\|F_1(X)-F_1(Y)\|_F+\gamma \|F_2(X)-F_2(Y)\|_F}{\|X-Y\|_F} \leq\frac{\beta^2}{2\alpha^3}+\frac{\gamma}{\alpha^2}.
\end{align}
\end{proof}

\subsection{Penalization of barycenters for $q$-Gaussian measures}
We recall that the unique minimizer of the minimization problem \eqref{eq: penalization q-Gaussian} in the space of $q$-Gaussian measures solves the nonlinear matrix equation $\nabla g(X)$=0 where
\begin{equation}
\label{eq: nabla g(X)}
\nabla g(X)=\Big(I-\sum_{i=1}^n \lambda_i (A_i\sharp X^{-1})\Big)-\gamma m(q,d)(\det X)^{\frac{q-1}{2}}X^{-1}=: F_1(X)-\gamma m(q,d)\tilde h(X),
\end{equation}
where $F_1(X)=\Big(I-\sum_{i=1}^n \lambda_i (A_i\sharp X^{-1})\Big)$ as in the previous section and $\tilde h(X) =(\det X)^{\frac{q-1}{2}}X^{-1}=h(X)X^{-1}$.
The following main theorem of this section proves the Lipschitz continuity of $\nabla g$.
\begin{theorem}
\label{thm: Lipschitz continuity q-Gaussians}
Suppose that $A_i\in [\alpha I, \beta I]$ for all $i=1,\ldots, n$. Then for  $\alpha I\leq X\neq Y\leq \beta     I$, we have
\begin{equation}
\frac{\|\nabla g(X)-\nabla g(Y)\|_F}{\|X-Y\|_F}\leq \begin{cases}
\frac{\beta^2}{2\alpha^3}+\frac{\gamma}{\alpha^2}+\frac{\gamma m(q,d)}{\alpha^2}\cdot \beta^{\frac{q-1}{2}d} \bigg(1+ \frac{q-1}{2}d\bigg),\quad \text{if}~~1< q <\frac{d+4}{d+2},\\
\\\frac{\beta^2}{2\alpha^3}+\frac{\gamma}{\alpha^2}+\gamma m(q,d)\alpha^{-2 +\frac{q-1}{2}d} \bigg(1+ \frac{1-q}{2} d\bigg),\quad \text{if}~~0< q <1.
\end{cases}
\end{equation}
\end{theorem}
\begin{proof}
Let $\alpha I\le X,\, Y \le \beta I$. According to the proof of Theorem \ref{thm: Lipschitz continuity}, we have
\begin{equation}
\label{eq: F1}
\frac{\|F_1(X)-F_1(Y)\|_F}{\|X-Y\|_F}\leq \frac{\beta^2}{2\alpha^3}+\frac{\gamma}{\alpha^2}.
\end{equation}
It remains to study the Lipschitz continuity of $\tilde h(X) =(\det X)^{\frac{q-1}{2}}X^{-1}=h(X)X^{-1}$.
\vskip 1pc
\noindent{\bf Case 1.} $1< q <\frac{d+4}{d+2}.$

First, we have
\begin{align*}
|h(X) - h(Y)| &= \big|\exp (\ln (\det X)^{\frac{q-1}{2}})- \exp (\ln (\det Y)^{\frac{q-1}{2}})\big| \\
&= e^{\theta}\, \big|\ln (\det X)^{\frac{q-1}{2}}-\ln (\det Y)^{\frac{q-1}{2}}\big|\\
&\le \beta^{\frac{q-1}{2}d}\, \big|\ln (\det X)^{\frac{q-1}{2}}-\ln (\det Y)^{\frac{q-1}{2}}\big|\\
& =   \frac{q-1}{2}\cdot\beta^{\frac{q-1}{2}d}\,|\ln \det X-\ln \det Y|\\
& \le  \frac{q-1}{2}\cdot\beta^{\frac{q-1}{2}d} \bigg(\max_{\alpha I\le X \le \beta I} \|X^{-1}\|\bigg) \|X-Y\|\\
& \le \frac{q-1}{2}\cdot\beta^{\frac{q-1}{2}d}\cdot \frac{\sqrt{d}}{\alpha}  \|X-Y\|
\end{align*}
where $ \ln \alpha^{\frac{q-1}{2}d}\le \theta \le \ln \beta^{\frac{q-1}{2}d}$ because $\ln \alpha^{\frac{q-1}{2}d}\le \ln (\det X)^{\frac{q-1}{2}} \le \ln \beta^{\frac{q-1}{2}d}$.
The second equality and inequality are derived from the mean value theorem. Moreover, we get
\begin{align}
\label{eq: tilde h1}
\| \tilde h(X) - \tilde h(Y)\| & = \| h(X) (X^{-1} -Y^{-1}) + (h(X) - h(Y)) Y^{-1}\|\notag\\
&\le h(X) \|X^{-1} -Y^{-1}\| + |h(X) - h(Y)|\, \|Y^{-1}\|\notag \\
&\le \bigg(\max_{\alpha I\le X \le \beta I} h(X) \bigg)\cdot \frac{1}{\alpha^2} \| X - Y\|\notag\\
&+ \bigg(\max_{\alpha I\le Y \le \beta I} \|Y^{-1}\|\bigg)\cdot \frac{q-1}{2}\cdot\beta^{\frac{q-1}{2}d}\cdot \frac{\sqrt{d}}{\alpha}  \|X-Y\| \notag \\
&= \bigg(\beta^{\frac{q-1}{2}d}\cdot \frac{1}{\alpha^2} + \frac{\sqrt{d}}{\alpha}\cdot\frac{q-1}{2}\cdot\beta^{\frac{q-1}{2}d}\cdot \frac{\sqrt{d}}{\alpha}\bigg)  \|X-Y\|\notag \\
& = \frac{1}{\alpha^2}\cdot \beta^{\frac{q-1}{2}d} \bigg(1+ \frac{q-1}{2}d\bigg)  \|X-Y\|
\end{align}
where the second inequality comes from [KY18, Proof of Theorem 3.1].
\vskip 1pc
\noindent{\bf Case 2.} $0< q <1.$
Similarly, we obtain
$$
|h(X) - h(Y)| \le \frac{1-q}{2}\cdot\alpha^{\frac{q-1}{2}d}\cdot \frac{\sqrt{d}}{\alpha}  \|X-Y\|.
$$
Hence
\begin{align}
\label{eq: tilde h2}
\| \tilde h(X) - \tilde h(Y)\| & \le \bigg(\max_{\alpha I\le X \le \beta I} h(X) \bigg)\cdot \frac{1}{\alpha^2} \| X - Y\|\notag\\
&+ \bigg(\max_{\alpha I\le Y \le \beta I} \|Y^{-1}\|\bigg)\cdot \frac{1-q}{2}\cdot\alpha^{\frac{q-1}{2}d}\cdot \frac{\sqrt{d}}{\alpha}  \|X-Y\| \notag \\
& = \bigg(\alpha^{\frac{q-1}{2}d}\cdot \frac{1}{\alpha^2} + \frac{\sqrt{d}}{\alpha}\cdot\frac{1-q}{2}\cdot\alpha^{\frac{q-1}{2}d}\cdot \frac{\sqrt{d}}{\alpha}\bigg)  \|X-Y\| \notag \\
& = \alpha^{-2 +\frac{q-1}{2}d} \bigg(1+ \frac{1-q}{2} d\bigg) \|X-Y\|.
\end{align}
Substituting the estimates \eqref{eq: F1}, \eqref{eq: tilde h1} and \eqref{eq: tilde h2} back into \eqref{eq: nabla g(X)} we obtain the desired inequality.
\end{proof}

\section{Numerical Experiments}
\label{sec: numerical experiments}
In this section, we numercally observe how the solution is effected as $q\to 1$. To see this, we report numerical results of a gradient projection method applied for the penalization of barycenters for $q$-Gaussian measures on $n$ randomly generated matrices of the size $d\times d$. The random matrices we use for our test are generated by matlab code as follows:
$$
\begin{array}{rl}
for & i=1:n \\
& [Q,~]=qr(randn(d)); \\
& A_i=Q*diag({\rm eiglb}+{\rm eigub}*rand(d,1))*Q';
\end{array}
$$
The eigenvalues of generated matrices are randomly distributed in the interval\\ $[{\rm eiglb},{\rm eiglb}+{\rm eigub}]$. In our experiments, we set $n=100$, $d=10$ if $q<1$ and $n=50$, $d=5$ if $q>1$. And we set ${\rm eiglb}=0.1$ and ${\rm eigub}=9.9$.

We set $\xi=0.5$, $\sigma=0.1$, $\hat{\alpha}=10^{-5}$, and $\hat{\beta}=10^5$ for GPM in our experiment.
All runs are performed on a Laptop with Intel Core i7-10510U CPU (2.30GHz)
and 16GB Memory, running 64-bit windows 10 and MATLAB (Version 9.8).
Throughout the experiments, we choose the initial iterate to be $X^0 = I$ and stop the algorithm when $\norm{D^{(k)}}_F\le 10^{-8}$.

\begin{table}[htb!]
	\caption{Test results of the value $\|X_{0.5}-X_q\|_F/$ where $X_{0.5}$ is the final estimated solution of the model (\ref{eq: reformulation of penal. q Gaussian}) with $q=0.5$ and $x_q$ is that with various given $q$ less than $1$ on $5$ random data sets.} \label{t1}
	\begin{center}

			\begin{tabular}{| c | c| c| c| c| c|} \hline
				\mc{1}{|c|}{q} &\mc{5}{c|}{difference when $\gamma=1$}\\[3pt] \hline
				0.6 & 0.00502 & 0.00503 &0.00521  & 0.00481  & 0.00497 \\[3pt]
				0.7 & 0.04672 & 0.04679 & 0.04761 & 0.04572  & 0.04649 \\[3pt]
				0.8 & 0.39716 & 0.39688 & 0.39667 & 0.39682 & 0.39653 \\[3pt]
				0.9 & 3.14528 & 3.13602 & 3.07209 & 3.21587 & 3.15235  \\[3pt]
				0.99 & 10.24065 & 10.19128 & 9.81731 & 10.67508 & 10.29661 \\[3pt] \hline
				\mc{1}{|c|}{q} &\mc{5}{c|}{difference when $\gamma=0.1$}\\[3pt] \hline
				0.6 & 0.000501 & 0.000503 & 0.000520 & 0.000481 & 0.000497 \\[3pt]
				0.7 & 0.00466 & 0.00467 & 0.00475  & 0.00456 & 0.00464 \\[3pt]
				0.8 & 0.03947  & 0.03944 & 0.03941 & 0.03944 & 0.03940 \\[3pt]
				0.9 & 0.33191 & 0.33097 & 0.32457 & 0.33896 & 0.33259 \\[3pt]
				0.99 & 2.08373 & 2.08373 & 1.99968 & 2.16978 & 2.09474 \\[3pt] \hline
				\mc{1}{|c|}{q} &\mc{5}{c|}{difference when $\gamma=0.01$}\\[3pt] \hline
				0.6 & 0.0000501 & 0.0000502 & 0.0000519 & 0.0000481 & 0.0000497 \\[3pt]
				0.7 & 0.00047 & 0.00047 & 0.00047 & 0.00046 & 0.00046 \\[3pt]
				0.8 & 0.00394 & 0.00394 & 0.00394 & 0.00394 & 0.00394 \\[3pt]
				0.9 & 0.03337 & 0.03327 & 0.03263 & 0.03407 & 0.03343 \\[3pt]
				0.99 & 0.22964 & 0.22856 & 0.22042 & 0.23908 & 0.23085 \\[3pt] \hline
		\end{tabular}
	\end{center}
\end{table}

\begin{table}[htb!]
	\caption{Test results of the value $\|X_{1.2}-X_q\|_F/$ where $X_{1.2}$ is the final estimated solution of the model (\ref{eq: reformulation of penal. q Gaussian}) with $q=1.2$ and $X_q$ is that with various given $q$ greater than $1$ on $5$ random data sets.} \label{t2}
	\begin{center}

		\begin{tabular}{| c | c| c| c| c| c|} \hline
			\mc{1}{|c|}{q} &\mc{5}{c|}{difference when $\gamma=0.1$}\\[3pt] \hline
			1.2 & 1.11865 & 1.05088 & 1.01739 & 1.16269 & 1.13214 \\[3pt]
			1.1 & 3.54827 & 3.29863 & 3.17136  & 3.71393 & 3.60257 \\[3pt]
			1.01 & 5.49360 & 5.08760 & 4.87969 & 5.76457 & 5.58347 \\[3pt] \hline
			\mc{1}{|c|}{q} &\mc{5}{c|}{difference when $\gamma=0.01$}\\[3pt] \hline
			1.2 & 1.05337 & 0.95910 & 0.91030 & 1.11727 & 1.07516 \\[3pt]
			1.1 & 2.09875 & 1.90800 & 1.80934 & 2.22836 & 2.14300 \\[3pt]
			1.01 & 2.46054 & 2.23918 & 2.12473 & 2.61087 & 2.51182 \\[3pt] \hline
		\end{tabular}
	\end{center}
\end{table}

We report in Table \ref{t1} our numerical results, showing the Frobenius norm of the difference between the final estimated solution of the model (\ref{eq: reformulation of penal. q Gaussian}) with $q=0.5$ and that with various given $q$ less than $1$. In Table \ref{t2}, the difference between the final estimated solution of the model (\ref{eq: reformulation of penal. q Gaussian}) with $q=1.25$ and that with various given $q$ greater than $1$ is reported. From Tables \ref{t1}-\ref{t2}, we see that the difference is increasing as $q$ goes to $1$ and the bigger the penalty parameter $\gamma$ is, the bigger the difference is.

In the next experiemnt, we investigate stability properties for the model (\ref{eq: reformulation of penal. q Gaussian}).
We perburb the given data, $A_i$ as follows:
$$
B_i=A_i+\epsilon I\quad i=1,\dots,n
$$

\begin{table}[htb!]
	\caption{Test results of the value $\|X_B-X_A\|_F/\epsilon$ where $X_A$ is the final estimated solution of the model (\ref{eq: reformulation of penal. q Gaussian}) with data $A_i$ and $X_B$ is that with the perturbed data $B_i$ on $5$ random data sets when $q<1$.} \label{t3}
\begin{center}

		{\scriptsize
		\begin{tabular}{| c | c| c| c| c| c| c| c| c| c| c| c| c| c| c| c|} \hline
			\mc{1}{|c|}{q} &\mc{5}{c|}{ $\gamma=1$ and $\epsilon=10^{-2}$}&\mc{5}{c|}{ $\gamma=1$ and $\epsilon=10^{-3}$}&\mc{5}{c|}{ $\gamma=1$ and $\epsilon=10^{-5}$}\\[3pt] \hline
			0.6 & 3.90 & 3.79 &	3.88 &	3.83 &	3.84 & 3.91 & 3.79 &	3.88 & 3.84 & 3.85 & 3.91 & 3.79 &	3.89 & 3.84 & 3.85\\[3pt]
			0.7 & 3.90 & 3.79 &	3.88 &	3.84 &	3.85 & 3.91 & 3.79 &	3.89 & 3.84 & 3.86 & 3.91 & 3.80 &	3.89 & 3.84 & 3.86\\[3pt]
			0.8 & 3.90 & 3.79 &	3.88 &	3.83 &	3.84 & 3.91 & 3.79 &	3.88 & 3.87 & 3.85 & 3.91 & 3.79 &	3.88 & 3.84 & 3.85\\[3pt]
			0.9 & 3.45 & 3.35 &	3.42 &	3.40 &	3.40 & 3.45 & 3.35 &	3.43 & 3.40 & 3.41 & 3.46 & 3.35 &	3.43 & 3.40 & 3.41 \\[3pt]
			0.99 & 1.19 & 1.15 & 1.18 &	1.17 &	1.17 & 1.19 & 1.15 & 1.18 &	1.17 & 1.17 & 1.19 & 1.15 & 1.18 &	1.17 & 1.17\\[3pt] \hline
			\mc{1}{|c|}{} &\mc{5}{c|}{ $\gamma=0.1$ and $\epsilon=10^{-2}$}&\mc{5}{c|}{ $\gamma=0.1$ and $\epsilon=10^{-3}$}&\mc{5}{c|}{ $\gamma=0.1$ and $\epsilon=10^{-5}$}\\[3pt] \hline
		    0.6 & 3.90 & 3.79 &	3.88 & 3.83 & 3.84 & 3.90 & 3.79 &	3.88 & 3.84 & 3.85 & 3.91 & 3.79 &	3.88 & 3.84 & 3.85\\[3pt]
		    0.7 & 3.90 & 3.79 &	3.88 & 3.83 & 3.84 & 3.91 & 3.79 &	3.88 & 3.84 & 3.85 & 3.91 & 3.79 &	3.88 & 3.84 & 3.85\\[3pt]
		    0.8 & 3.90 & 3.79 &	3.88 & 3.83 & 3.84 & 3.90 & 3.79 &	3.88 & 3.84 & 3.85 & 3.91 & 3.79 &	3.88 & 3.84 & 3.85\\[3pt]
		    0.9 & 3.85 & 3.74 &	3.83 & 3.79 & 3.80 & 3.86 & 3.75 &	3.84 & 3.79 & 3.80 & 3.86 & 3.75 &	3.84 & 3.79 & 3.81\\[3pt]
		    0.99 & 3.36 & 3.27 & 3.34 &	3.30 & 3.31 & 3.37 & 3.27 & 3.35 & 3.31 & 3.32 & 3.37 & 3.27 & 3.35 &	3.31 & 3.32\\[3pt] \hline
			\mc{1}{|c|}{} &\mc{5}{c|}{ $\gamma=0.01$ and $\epsilon=10^{-2}$}&\mc{5}{c|}{ $\gamma=0.01$ and $\epsilon=10^{-3}$}&\mc{5}{c|}{ $\gamma=0.01$ and $\epsilon=10^{-5}$}\\[3pt] \hline
			0.6 & 3.90 & 3.79 &	3.88 & 3.83 & 3.84 & 3.90 & 3.79 &	3.88 & 3.84 & 3.85 & 3.91 & 3.79 &	3.88 & 3.84 & 3.85\\[3pt]
			0.7 & 3.90 & 3.79 &	3.88 & 3.83 & 3.84 & 3.90 & 3.79 &	3.88 & 3.84 & 3.85 & 3.91 & 3.79 &	3.88 & 3.84 & 3.85\\[3pt]
			0.8 & 3.90 & 3.79 &	3.88 & 3.83 & 3.84 & 3.90 & 3.79 &	3.88 & 3.84 & 3.85 & 3.91 & 3.79 &	3.88 & 3.84 & 3.85\\[3pt]
			0.9 & 3.89 & 3.78 &	3.87 & 3.83 & 3.84 & 3.90 & 3.79 &	3.88 & 3.83 & 3.84 & 3.90 & 3.79 &	3.88 & 3.83 & 3.85 \\[3pt]
			0.99 & 3.84 & 3.73 & 3.82 &	3.77 & 3.78 & 3.85 & 3.73 & 3.82 &	3.78 & 3.79 & 3.85 & 3.74 & 3.82 &	3.78 & 3.79\\[3pt] \hline
		\end{tabular}}
	\end{center}
\end{table}

\begin{table}[htb!]
	\caption{Test results of the value $\|X_B-X_A\|_F/\epsilon$ where $X_A$ is the final estimated solution of the model (\ref{eq: reformulation of penal. q Gaussian}) with data $A_i$ and $X_B$ is that with the perturbed data $B_i$ on $5$ random data sets when $q>1$.} \label{t4}
	\begin{center}

		{\scriptsize
			\begin{tabular}{| c | c| c| c| c| c| c| c| c| c| c| c| c| c| c| c|} \hline
				\mc{1}{|c|}{} &\mc{5}{c|}{ $\gamma=0.1$ and $\epsilon=10^{-2}$}&\mc{5}{c|}{ $\gamma=0.1$ and $\epsilon=10^{-3}$}&\mc{5}{c|}{ $\gamma=0.1$ and $\epsilon=10^{-5}$}\\[3pt] \hline
				1.2 & 0.77 & 0.77 & 0.82 & 0.74 & 0.76 & 0.77 & 0.77 & 0.82 & 0.74 & 0.76 & 0.77 & 0.77 & 0.82 & 0.74 & 0.76 \\[3pt]
				1.1 & 1.54 & 1.53 &	1.61 & 1.50 & 1.53 & 1.54 & 1.53 &	1.61 & 1.50 & 1.53 & 1.54 & 1.53 &	1.61 & 1.50 & 1.53 \\[3pt]
				1.01 & 2.21 & 2.18 & 2.28 &	2.16 & 2.20 & 2.21 & 2.18 & 2.28 &	2.17 & 2.20 & 2.21 & 2.18 & 2.28 &	2.17 & 2.21 \\[3pt] \hline
				\mc{1}{|c|}{} &\mc{5}{c|}{ $\gamma=0.01$ and $\epsilon=10^{-2}$}&\mc{5}{c|}{ $\gamma=0.01$ and $\epsilon=10^{-3}$}&\mc{5}{c|}{ $\gamma=0.01$ and $\epsilon=10^{-5}$}\\[3pt] \hline
				1.2 & 2.13 & 2.11 &	2.22 & 2.08 & 2.12 & 2.13 & 2.12 &	2.22 & 2.08 & 2.12 & 2.13 & 2.12 &	2.22 & 2.08 & 2.12 \\[3pt]
				1.1 & 2.55 & 2.52 &	2.63 & 2.49 & 2.54 & 2.55 & 2.52 &	2.64 & 2.50 & 2.54 & 2.55 & 2.52 &	2.64 & 2.50 & 2.54 \\[3pt]
				1.01 & 2.68 & 2.64 & 2.76 &	2.62 & 2.67 & 2.68 & 2.65 & 2.77 &	2.63 & 2.67 & 2.68 & 2.65 & 2.77 &	2.63 & 2.67 \\[3pt] \hline
		\end{tabular}}
	\end{center}
\end{table}

From Tables \ref{t3}-\ref{t4}, we can observe that $\|X_B-X_A\|_F\le 4\epsilon$, where $X_A$ is the final estimated solution of the model (\ref{eq: reformulation of penal. q Gaussian}) with data $A_i$ and $X_B$ is that with the perturbed data $B_i$, for all the cases. The value $\|X_B-X_A\|_F/\epsilon$ tends to reduce if the penalty parameter $\gamma$ and $q$ are getting large.

\bibliographystyle{abbrv}


TITLE: Proving that the series with the general term $u_n = \int_{0}^{\frac{\pi}{2}} \sin^n(x)dx$ diverges.
QUESTION [6 upvotes]: $$u_n = \int_{0}^{\frac{\pi}{2}} \sin^n(x)dx$$ for $ n \in \mathbb{N}^*$.

Prove that $(u_n)$ is convergent toward $0$.
Prove that the series with the general term $(-1)^n u_n$ converges.
Prove that $$\sum_{n=0}^{\infty} (-1)^n u_n = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sin(x)}dx$$
Compute $$\int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sin(x)}dx$$
Hint: You can start by proving $$\int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sin(x)}dx = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \cos(x)}dx$$
Prove that the series with the general term $u_n$ diverges.
Hint: You can start by proving $$u_n \geq \frac{1}{n + 1}$$

For $ 0 \leq x \leq \frac{\pi}{2}$ we have:  $0 \leq u_n \leq \frac{\pi}{2}$ wich I can't derive the convergence from it. How can I prove it is convergent toward 0?

I am stuck with question 4. and 5, I could not see how to prove the hints.

REPLY [0 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
u_{n} & \equiv \int_{0}^{\pi/2}\sin^{n}\pars{x}\,\dd x =
\int_{0}^{\pi/2}\cos^{n}\pars{x}\,\dd x =
\int_{0}^{\pi/2}\exp\pars{n\ln\pars{\cos\pars{x}}}\,\dd x
\end{align}

With
  Laplace Method:

\begin{align}
&u_{n}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
\int_{0}^{\infty}\expo{-nx^{2}/2}\pars{1 - {nx^{4} \over 12}}\,\dd x
\\[5mm] \implies &\
\bbx{u_{n} \equiv
\int_{0}^{\pi/2}\sin^{n}\pars{x}\,\dd x \sim
\bbox[10px,#ffd]{\pars{{1 \over n^{1/2}} - {1 \over 4n^{3/2}}}\root{\pi \over 2}}
\quad\mbox{as}\quad n \to \infty}
\end{align}


TITLE: Visualizing Lattice Products
QUESTION [6 upvotes]: Partial orders can be thought of as a set $S$, and a partial relation on that set, $\le$. If a greatest lower bound and least upper bound exist for any subset of $S$, the poset defines a lattice.
The sum operator can be defined on lattices as follows:
$$
L_1 + L_2+...+ L_n = \{ (i, x_i)\ \vert \ x_i \in L_i \backslash \{\bot, \top \}\} \cup \{ \bot, \top\}
$$
where $\top$ and $\bot$ are the trivial greatest and least elements, respectively, when dealing with lattices over finite sets. This can be very elegantly visualized as the lattices lined up, with both top and bottom joined at a single point, respectively.
I now have a question in two parts.
1) Is there is an easy way to visually understand the Lattice product, as defined by:
$$
L_1 \times L_2\times\ ...\times\ L_n = \{(x_1, x_2, ..., x_n)\ \vert \ x_i \in L_i\}
$$
2) According to defintions, $\le$ and g.u.b and l.u.b are defined point-wise on lattice products.
Computing $\le$ pointwise makes sense to me, but pointwise bounds has my head creaking. Could someone provide some intuition behind this, or otherwise help my understanding?

REPLY [6 votes]: Whether or not this is easy depends on $n$ and the lattices involved.
For $n\leq 2$, the answer is yes, (especially if the lattices are bounded) there is an easy way to visualize direct product of two lattices (see explanation below).  For $n=3$, as long as the lattices are not too complicated, it's fairly easy to visualize the direct product.  For larger $n$, unless you are very good at thinking in $n>3$ spatial dimensions, it seems hard to draw a good Hasse diagram for the direct product of lattices by hand.  Of course, there are computer programs which do it for you, like the UA Calculator, and Ralph Freese (author of UACalc) has a paper about automated lattice drawing available here.
I'll explain how to visualize direct products of lattices in the easy cases I mentioned above.
Let's start with the smallest non-trivial example.  Let $L_1$ be the two element lattice with universe $\{x_0, x_1\}$, and  $x_0 < x_1$.  We sometimes say $L_1 \cong \mathbf 2$.  Let $L_2$ be another two element lattice with universe $\{y_0, y_1\}$, and $y_0<y_1$. Then $L_1 \times L_2 \cong \mathbf 2 \times \mathbf 2$ is just the lattice whose Hasse diagram looks like a diamond.  The top is the element $(x_1, y_1)$.  The bottom is $(x_0, y_0)$.  The other elements $(x_0, y_1)$ and $(x_1, y_0)$ are incomparable.
Now lets look at a slightly more complicated example.  Consider $L_1 \times L_2 \times L_3$, where $L_3 \cong \mathbf 3$ is the three element chain.  The Hasse diagram in this case is the lattice on the left in the figure below.  You can see in this case how the figure is just the diamond (that represents $L_1 \times L_2$) crossed with the three element chain (so there are three diamonds, as you move up and to the right in the diagram).
More generally.  Take two arbitrary bounded lattices $L_1$ and $L_2$.  You can visualize the direct product $L_1 \times L_2$ just as we did with the $\mathrm 2 \times \mathrm 2$ example, except now there may be other points $(x,y)$ in the middle of the diagram, where $x\in L_1$ and $y\in L_2$.

As for the second part of your question, point-wise simply means coordinate-wise; i.e. $(x_1, \dots, x_n) \leq (y_1,\dots, y_n)$ iff $x_i \leq y_i$ for each $i$.  It should be clear that this is the order being used in the figures above.


TITLE: Euclidean algorithm for ordinals
QUESTION [3 upvotes]: I am trying to prove the Euclidean algorithm for ordinals. More specifically:
For any ordinals $\alpha, \beta$ where $\beta >0$, there are unique $\gamma, \delta$ such that $\alpha = \beta\gamma+\delta$ and $\delta<\beta$.
For now, I have difficulties. For example, I am not sure what to do when, $\alpha$ is a limit ordinal and $\beta$ is finite (a natural number).. since "diminishing" multiples of $\beta$ from $\alpha$ will not help.. and I don't really have an action of division...
Can anyone help? maybe give me a clue or a direction??
Thanks!
Shir

REPLY [7 votes]: Hints: Let $\xi$ be the least ordinal such that $\alpha\le\beta\cdot\xi.$ If $\alpha=\beta\cdot\xi,$ you're done. Suppose that $\alpha<\beta\cdot\xi.$ Show that $\xi$ cannot be a limit ordinal (by our choice), and cannot be $0$, let $\gamma=\sup\xi,$ and proceed to prove existence.
Uniqueness isn't too tricky.


text:
\section{Properties of the $g_d$ function}
\label{sec:g-d-property}

In this section, we generalized the $g_d$ function and make some observations about its properties.  We then give the proof of Lemma~\ref{lem-main} in the next section.

From Theorem \ref{def-g_d},
we can see that it is very easy to compute $g_d(\iterator)$ if we write $\iterator$ in base $p$.
It will be convenient to consider a more general class of functions in other bases.
\begin{defn}\label{def-g_d,q}
Fix $d\in \mathbb{Z}$ and $q\in \mathbb{Z}_{>0}$.
Let $\func{d}{q}:\mathbb{Z}\rightarrow \mathbb{Z}$ be a function defined as
$\func{q}{d}(\iterator)=0$ when $d<0$ or $\iterator=0$,
$\func{q}{d}(1)=1$ when $d\geq 0$,
$\func{d}{q}(q^r)=\sum_{i=0}^{q-1}\func{d-i}{q}(q^{r-1})$,
and for other cases,
let $r$ be the largest integer so that $\iterator\geq q^r$,
then
\[
\func{d}{q}(\iterator)=\func{d}{q}(q^r)+\func{d-1}{q}(\iterator-q^r)
\]
\end{defn}

We can verify that $g_d(\iterator)$ defined in last section is the same as $\func{d}{p}(\iterator)$.
\begin{proposition}\label{prop-expand}
$
g_d(\iterator)=\func{d}{p}(\iterator).
$
\end{proposition}
\begin{proof}
When $d<0$ or $\iterator=0$,
by definition of $g_d$, we have
$g_d(\iterator)=0$.
Also,
when $d\geq 0$,
by Theorem~\ref{thm-g_d-base}, we have
$g_d(1)=|\monomial{p}{d}{0}|=1$.

Next, we claim that $g_d(p^r)=\sum_{i=0}^{p-1}g_{d-i}(p^{r-1})$.
Since $g_{d}(p^r)=|\monomial{p}{d}{r}|$,
we can split all monomials over $x_1,\cdots,x_r$ of degree at most $d$ by their degree in $x_1$.
That is,
\[
\monomial{q}{d}{r}=\bigcup_{i=0}^{q-1} x_1^i\cdot \monomial{q}{d-i}{r-1},
\]
which precisely yields the claim.

Finally,
for input $\iterator$ not of the above forms,
let $r$ be the largest integer so that $\iterator\geq q^r$,
then Theorem \ref{def-g_d} implies that
\[
g_d(\iterator)-g_{d-1}(\iterator-p^r)=g_d(p^r).
\]
So, $g_d(\iterator)$ satisfies all the requirements for $\func{d}{p}(\iterator)$.
Notice that for fixed $p$,
Definition \ref{def-g_d,q} uniquely determines a function over $\mathbb{Z}$,
so we have $g_d(\iterator)=\func{d}{p}(\iterator)$.
\end{proof}

We prove the following important property of the $\func{d}{q}$ function that
we will repeatedly apply later.

\begin{proposition}\label{prop-main}
For $d>d'$ and $q,a> 0$,
\[
\func{d}{q}(a)-\func{d}{q}(a-1)\geq \func{d'}{q}(a)-\func{d'}{q}(a-1).
\]
\end{proposition}

\begin{proof}
We first write down $\func{d}{q}(a+1)-\func{d}{q}(a)$ as a function of $d$.
We represent $a$ and $a-1$, respectively, in base $q$ as
\[
a=\sum_{i=0}^r a_i\cdot q^i \quad\mbox{ and } \quad  a-1=\sum_{i=0}^r b_i\cdot q^i.
\]
With this representation,
by Definition \ref{def-g_d,q}
we can explicitly compute $$\func{d}{q}(a)=\sum_{i=0}^{r}\func{d-\sum_{j=i+1}^{r}a_j}{q}(a_i\cdot q^r)$$
Let $i_0$ be the largest integer so that $a_{i_0}\neq b_{i_0}$.
Notice that
\[
\func{d}{q}(a)-\func{d}{q}(a-1)=\sum_{i=0}^{r}\func{d-\sum_{j=i+1}^{r}a_j}{q}(a_i\cdot q^r)-\sum_{i=0}^{r}\func{d-\sum_{j=i+1}^{r}b_j}{q}(b_i\cdot q^r)
\]
When $i>i_0$, $a_i=b_i$,
so we have $\func{d-\sum_{j=i+1}^{r}a_j}{q}(a_i\cdot q^r)=\func{d-\sum_{j=i+1}^{r}b_j}{q}(b_i\cdot q^r)$.
Hence
\begin{equation}\label{equation-g_d,q-main}
\func{d}{q}(a)-\func{d}{q}(a-1)=\sum_{i=0}^{i_0}\func{d-\sum_{j=i+1}^{r}a_j}{q}(a_i\cdot q^r)-\sum_{i=0}^{i_0}\func{d-\sum_{j=i+1}^{r}b_j}{q}(b_i\cdot q^r)\tag{*}
\end{equation}

We consider two cases based on the value of $i_0$.

\bigskip\noindent
\underline{\textsc{Case  $i_0=0$:}}   In this case $b_{0}=a_{0}-1$,
and \eqref{equation-g_d,q-main} is simply
\[
\begin{split}
\func{d}{q}(a)-\func{d}{q}(a-1)
&=\func{d-\sum_{j=1}^r a_j}{q}(a_0)-\func{d-\sum_{j=1}^r a_j}{q}(a_0-1)\\
&=g_{d+1-\sum_{j=0}^r a_j}(1).
\end{split}
\]
Notice that $\func{d+1-\sum_{j=0}^r a_j}{q}(1)=\begin{cases}0 &\mbox{if }d+1<\sum_{j=0}^r a_j,\\ 1&\mbox{otherwise};\end{cases}$
so this is an increasing function of $d$.
Hence we have
$
\func{d}{q}(a)-\func{d}{q}(a-1)\geq \func{d'}{q}(a)-\func{d'}{q}(a-1)
$
as required.

\bigskip\noindent\underline{\textsc{Case $i_0>0$:}}
Then, it must be the case that $a_i=0, b_i=q-1$ for $0\leq i<i_0$.
Therefore, using the fact that $\func{d}{q}(0)=0$ for all $d$,
\eqref{equation-g_d,q-main} is
\[
\begin{split}
\func{d}{q}(a)-\func{d}{q}(a-1)
=&\func{d-\sum_{j=i_0+1}^r a_j}{q}(a_{i_0}\cdot q^{i_0})-\func{d-\sum_{j=i_0+1}^r b_j}{q}((a_{i_0}-1)\cdot q^{i_0})\\
&-\sum_{i=0}^{i_0-1}\func{d-\sum_{j=i+1}^{r}b_j}{q}((q-1)\cdot q^i)\\
=&\func{d+1-\sum_{j=i_0}^r a_j}{q}(q^{i_0})-\sum_{i=0}^{i_0-1}\func{d-\sum_{j=i+1}^r b_j}{q}((q-1)\cdot q^i)
\end{split}
\]

Define $h(i)=(\sum_{j=i_0}^r a_j)-1+(q-1)\cdot i$.
Then, we can verify that for $i\leq i_0-1$,
$$\sum_{j=i+1}^r b_j=(\sum_{j=i_0+1}^r a_j)+a_{i_0}-1+(q-1)(i_0-1-i)=h(i_0-1-i).$$

We claim that
$\displaystyle
\func{d-h(i)}{q}(q^{i_0-i})-\func{d-h(i)}{q}((q-1)\cdot q^{i_0-1-i})=\func{d-h(i+1)}{q}(q^{i_0-i-1}):
$

\medskip\noindent
Indeed, by Definition \ref{def-g_d,q},
\[
\func{d-h(i)}{q}((q-1)\cdot q^{i_0-i-1})=\sum_{j=0}^{p-2}\func{d-h(i)-j}{q}(q^{i_0-i-1}).
\]
By Proposition \ref{prop-expand},
we have
\[
\func{d-h(i)}{q}(q^{i_0-i})=\sum_{j=0}^{p-1}\func{d-h(i)-j}{q}(q^{i_0-i-1}).
\]
Comparing the two summations,
and using the fact $h(i)+q-1=h(i+1)$ we get the claim.

\medskip\noindent
So,
we can repeatedly apply the claim to get
\[
\begin{split}
\func{d}{q}(a)-\func{d}{q}(a-1)&=\func{d-h(0)}{q}(q^{i_0})-\sum_{i=0}^{i_0-1}\func{d-h(i)}{q}((q-1)\cdot q^{i_0-i-1})\\
&=\func{d-h(1)}{q}(q^{i_0-1})-\sum_{i=1}^{i_0-1}\func{d-h(i)}{q}((q-1)\cdot q^{i_0-i-1})\\
&=\cdots\\
&=\func{d-h(i_0-1)}{q}(q^{0})-\func{d-h(i_0-1)}{q}((q-1)\cdot q^{0})\\
&=\func{d+1-i_0\cdot (q-1)-\sum_{j=i_0}^r a_j}{q}(1).
\end{split}
\]
This is increasing in $d$,
so we obtain the required inequality.
\end{proof}

By telescoping Lemma~\ref{prop-main},
we have
\begin{corollary}\label{prop-lifting}
For $a>b$, $d>d'$, for any $q>0$,
we have
\[
\func{d}{q}(a)-\func{d}{q}(b)\geq \func{d'}{q}(a)-\func{d'}{q}(b).
\]
\end{corollary}

Moreover,
if we set $b=0$,
then we can see that for fixed $a$,
$\func{d}{q}(a)$ is monotone in $d$.
\begin{corollary}\label{prop-monotone}
For $d>d'$, for any $q>0$,
we have
$
\func{d}{q}(a)\geq \func{d'}{q}(a).
$
\end{corollary}

config: arxiv


text:
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\DeclareMathOperator{\tr}{tr}
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\newcommand{\om}{\omega }

\newtheorem{Assumption}{Assumption}
\newtheorem{lemma}{Lemma}[section]
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text:
\begin{document}

\maketitle

\begin{abstract}
In this paper we characterize the elements of $PSL_2(\mathbb{Z})$,
as a subgroup of Thompson group $T$, in the language of reduced
tree pair diagrams and in terms of piecewise linear maps as well.
Actually, we construct the reduced tree pair diagram for every
element of $PSL_2(\mathbb{Z})$ in normal form. This allows us to
estimate the length of the elements of $PSL_2(\mathbb{Z})$ through
the number of carets of their reduced tree pair diagrams and, as a
consequence, to prove that $PSL_2(\mathbb{Z})$ is a non distorted
subgroup of $T$. In particular, we find non-distorted free non
abelian subgroups of $T$.
\end{abstract}

\textbf{MSC classsification:} 20F65.

\textbf{Keywords:} Thompson's group, $PSL_2(\mathbb{Z})$,
distortion, free group, free subgroups, rooted binary trees,
Minkowski question mark function, piecewise projective
homeomorphisms.

\section{Introduction}

\input{1Intro}

\section{$PSL(2,\mathbb{Z})$ as a subgroup of Thompson's group $T$}

\input{2PPSL}

\section{Tree diagram characterization of $PSL_2(\mathbb{Z})$}

\input{3Arbres}

\section{Piecewise linear characterization of $PSL_2(\mathbb{Z})$}

\input{4Pendents}

\bibliographystyle{amsplain}
\bibliography{referencies}

\end{document}


TITLE: Proof of supremum
QUESTION [1 upvotes]: Let E be a nonempty bounded subset of R. Let λ ∈ R and define
λE = {λx : x ∈ E}.
Prove that
sup(λE) = max{λ sup E, λ inf E}.
Hint: Substantial partial credit will be given if you prove that if λ ≥ 0, then sup(λE) = λ sup E.
Answer: I have gotten so far as to say that we know that the Sup(λE) exists since this is a nonempty bounded subset. I tried by starting to prove the hint, but I got stuck. Does anyone have any suggestions? Thanks

REPLY [1 votes]: The trick is to use an $\epsilon$-argument.
I will show one side of the equality for $\lambda > 0$. It suffices to show that for each given $\epsilon >0$
$$\lambda \sup( E) \leq \sup(\lambda E) +\epsilon.$$
From the definition of $\sup$, there exists $x\in E$ such that
$$\sup (E) -x \leq \frac{\epsilon}{\lambda}$$
$$\lambda \sup (E) - \lambda x \leq \epsilon$$
therefore
$$\lambda \sup (E) \leq \lambda x + \epsilon \leq \sup (\lambda E) + \epsilon.$$


TITLE: conditional probability and expectation of random variables
QUESTION [1 upvotes]: It is given that $E[Y/A]$ where $A = {X \in C}, C \in B(R) $; Want to indicate how to compute  $E[Y/A]$. Here $X$ and $Y$ are random variables with joint density $f(x,y)$. Any ideas will be very helpful on this problem.
I started doing as follows;
$E[Y/A] = E[Y/X \in C] $.
I don't know how to continue from here.
Thank you.

REPLY [1 votes]: The conditional expectation is $$\mathbb{E}[Y \mid X \in C] = \dfrac{\displaystyle \int_{y\in \mathbb{R}}\int_{x \in C} y f(x,y) \, dx\, dy}{\displaystyle \int_{y\in \mathbb{R}}\int_{x \in C} f(x,y) \, dx\, dy}$$  though as Ian says there is an issue if the denominator is zero.


TITLE: Find normal closure of $\mathbb Q\left( (1+i)\sqrt[4]{5} \right)/\mathbb Q$
QUESTION [3 upvotes]: I've been solving problems from my Galois Theory course, and I've had problems finishing this one. It says:

Being $a=(1+i)\sqrt[4] 5$, find the normal closures of these field
extensions:

$\mathbb Q(a)/\mathbb Q(i\sqrt 5)$
$\mathbb Q(i\sqrt 5)/\mathbb Q$
$\mathbb Q(a)/\mathbb Q$

This is the work I've done so far:
The first I tried to find where the degrees of the extensions involved. It was easy to notice $X^2+5$ is the minimal polynomial of $i\sqrt 5$ over $\mathbb Q$, so then the extension $\mathbb Q(i\sqrt 5)/\mathbb Q$ has degree 2. Now, since it has degree two, then $\mathbb Q(i\sqrt5)$ is splitting field of $X^2+5$ over $\mathbb Q$, hence $\mathbb Q(i\sqrt5)/\mathbb Q$ is a normal extension and then it's normal closure is $\mathbb Q(i\sqrt5)$ itself.
Then, it was easy to see that $a$ is root of $X^4+20$, and it's irreducible for Eisenstein so $[\mathbb Q(a):\mathbb Q]=4$. Now, given that $a^2=2i\sqrt5\in\mathbb Q(a)$, it's clear that $\mathbb Q(i\sqrt5)\subset\mathbb Q(a)$, and the degree of $\mathbb Q(a)/\mathbb Q(i\sqrt 5)$ can only be 1 or 2 (because of transitivity), but it's clearly not $1$ because $a\notin\mathbb Q(i\sqrt5)$, so it's degree is 2, hence it's a normal extension so it's normal closure is $\mathbb Q(a)$ itself.
The last part, to be said, finding normal closure of $\mathbb Q(a)/\mathbb Q$ is where I'm not sure. What I did is first notice that, since $a$ is a complex numer, and $X^4+20$ is a biquadratic polynomial with rational coefficients, the it's 4 different roots are $a,-a,\bar a,-\bar a$. So the normal closure of $\mathbb Q(a)/\mathbb Q$ is $\mathbb Q(\pm a,\pm\bar a)$, but I want this expression to be reduced as much as possible. The first reduction is obviously removing the negative terms, to be said,
$$\mathbb Q(\pm a,\pm\bar a)=\mathbb Q(a,\bar a).$$
And here is where I don't know how to end. I'm not sure if $\mathbb Q(a,\bar a)=\mathbb Q(a)$ or not. I tried to find a way to make $\bar a$ out of combinations of elements from $\mathbb Q(a)$ but had no success, and something tells me $\bar a\notin\mathbb Q(a)$, but I don't know how to prove it. If this was true, then the normal closure of $\mathbb Q(a)/\mathbb Q$ would be $\mathbb Q(a,\bar a)$. If not, then it would be a normal extension and the normal closure would be $\mathbb Q(a)$ itself, but I don't think this is the case.
Is what I did correct? How can I finish the last part? If it's true, how can I prove $\bar a\notin\mathbb Q(a)$? If not, how can I prove the opposite? Any help will be appreciated, thanks in advance.

REPLY [2 votes]: As you guessed, $\bar{a}$ is not in $\mathbb{Q}(a)$. If this is true, then
$$
i = -\frac{\bar{a}}{a} \in \mathbb{Q}(a), \sqrt[4]{5} = \frac{a}{1+i}\in \mathbb{Q}(a)
$$
so $\mathbb{Q}(a)\supseteq \mathbb{Q}(i, \sqrt[4]{5}) = K$. However, the field $K$ has degree 8 over $\mathbb{Q}$, so this is impossible.


TITLE: A dense set within another dense set (nested)
QUESTION [1 upvotes]: Suppose
1. The subset $A$ is dense within the space $X$.
2. The subset $B$ is also dense within $A$.
Can I say the subset $B$ is dense within the space $X$?
For example, $X$ could be $L^1(\mathbb{R})$, $A$ could be the set of integrable simple functions, $B$ could be the set of step functions.

REPLY [1 votes]: Let $C$ be a closed subspace of $X$ which contains $B$, $C\cap A$ is a closed subspace of $A$ which contains $B$, since $B$ is dense in $A$, and $B\subset  C\cap A$, we deduce that $C\cap A=A$ thus $C$ contains $A$ and is closed, since $A$ is dense, $C=X$. done.