text:
\section{Introduction}

We consider the following multi-marginal entropy-transport problem
\begin{equation}\label{ETIntro}
I_{\varepsilon}[\rho] = \inf_{\gamma\in\Pi^{\mathrm{sym}}_N(\rho)}( C_0[\gamma] + \varepsilon E[\gamma]),
\end{equation}
where $C_0[\gamma] = \int_{X^N} c\,\d\gamma$ is the transportation cost related to a cost function $c$,  $E[\gamma]$ is the entropy, and $\varepsilon \ge 0$ is a parameter, see Section \ref{sec:entropic} for details.
We consider the setting where $(X,d,\mathfrak m)$ is a Polish measure space and $\rho\mathfrak m \in \mathcal{P}^{ac}(X)$ is an absolutely continuous probability measure with respect to the reference measure $\mathfrak m$. An element $\gamma \in \Pi^{\mathrm{sym}}_N(\rho)$ is called a symmetric coupling (or transport plan), that is, a symmetric probability measure in $X^N$ having all marginals equal to $\rho\mathfrak m$.

We are interested in a class of repulsive cost functions $c\colon X^N\to\mathbb{R}\cup\lbrace +\infty\rbrace$ of the form
\[
c(x_1,\ldots, x_N)=\sum_{1\le i<j\le N}f(d(x_i,x_j)),\quad \text{ for all } \, (x_1,\ldots,x_N)\in X^N.
\]
We assume $f\colon ]0,\infty[\to \mathbb{R}$ to be a continuous and decreasing function that approaches $+\infty$ if $d(x_i,x_j)\to 0$. Among the examples of such cost functions we have the Coulomb cost $f(z) = 1/\vert z\vert$, the Riesz cost  $f(z) = 1/\vert z\vert^s, n \geq s\geq \max\lbrace n-2, 0\rbrace$ (in $\mathbb{R}^n$) and the logarithmic cost $f(z) = -\log(\vert z\vert)$. We observe that when $\varepsilon=0$, this entropy-transport problem  reduces to the classical multi-marginal optimal transport problem with repulsive costs \cite{BuDePGor,CFK,CFK17,DMaGerNen}.

The motivation of this paper comes from both theory and numerics. For repulsive cost functions, the entropy term in \eqref{ETIntro} plays a role of a \textit{regularizer} to compute numerically a solution $\gamma$ of the multi-marginal optimal transport problem $I_{0}[\rho]$, see \cite{BenCaNen}. Numerical experiments suggest that when the regularization parameter $\varepsilon$ goes to $0$, the minimizer $\gamma_{\varepsilon}$ converges to a minimizer of $I_{0}[\rho]$ having minimal entropy among the minimizer of $I_{0}[\rho]$.

From a theoretical viewpoint, this type of a functional has direct relevance in Density Functional Theory.  By choosing carefully the parameter $\varepsilon$, the functional \eqref{ETIntro} provides a lower bound for the Hohenberg-Kohn functional in Density Functional Theory \cite{NenThesis, DMaGerNenGorSei}. This is an immediate consequence of the Log-Sobolev Inequality.

The entropy-transport problem has appeared previously in the literature in the attractive case, in particular when $c(x_1,x_2) = d(x_1,x_2)^2$. We mention briefly below some of the connections of the entropy-transport with other fields and point out the relevance in the Coulomb case. \medskip

\noindent
\textbf{Brief comments on some applications of the entropy-transport}

\noindent
\textit{Optimal Transport and Sinkhorn algorithm:}  The entropy-transport \eqref{ETIntro} was introduced by M. Cuturi \cite{CutSin} in order to compute numerically the optimal transport plan for the distance squared cost in the $2$-marginals case via the Sinkorn algorithm. Due to its reasonable computational cost, it has been applied to a wide range of problems in various research areas, including Information Theory, Computer Graphics, Statistical Inference, Machine Learning, and Mean-Field Games. The entropic regularization method was also considered in the (attractive) multi-marginal case in the so-called \textit{barycenter problem} introduced by M. Agueh and G. Carlier \cite{AguCar} (see also \cite{CarDuvPeySch}) and in numerical methods in the time discretization of Brenier's relaxed formulation of the incompressible Euler equation \cite{BenCarNenEuler}. For a thorough presentation of the computational aspects we refer to M. Cuturi and G. Peyr\' e's book \cite{CutPeyBook}.  \medskip

\noindent
\textit{Second-order Calculus on $RCD$ spaces:} N. Gigli and L. Tamanini \cite{GigTam} studied the entropic-transport problem on a class of metric spaces with (Riemannian) Ricci curvature bounded from below (2-marginals case, $c(x_1,x_2) = d(x_1,x_2)^2$). The entropic regularization procedure was crucial for  establishing a second-order differential structure in that setting.  \medskip

\noindent
\textit{Schr\"odinger Problem:} In 1926, E. Schr\"odinger introduced the (linear) Schr\"odinger equations describing the non-relativistic evolution of a single particle in an electric field with potential energy and also established an equivalence between such equations and a system of diffusion equations \cite{Schrodinger31}. Roughly speaking, the variational problem (see \eqref{ETIntro} with $X = C([0,1],\mathbb{R}^d)$ and $N=2$)
arises in the Schr\"odinger manuscript while studying the limit $k\to\infty$ $(N=2)$ of the empirical measures associated to the evolution of $k$ i.i.d. Brownian motions. We refer the reader to C. L\'eonard survey \cite{LeoSurvey} for technical details and historical notes. \medskip

\noindent
\textit{Lower bound on the Hohenberg-Kohn functional in Density Functional Theory:} This is the particular case where the entropy-transport problem with Coulomb cost comes into play. It has been shown in \cite{NenThesis, DMaGerNenGorSei} that the functional \eqref{ETIntro} provides a lower bound for computing the ground state energy of the Hohenberg-Kohn functional \cite{BuDePGor, CFK, CFK17, DMaGerNen,Lieb83}. Below we give a brief description of the result. Notice that in this context $X = \mathbb{R}^d$ and $\mathfrak m$ is the Lebesgue measure on $\mathbb{R}^d$.

Assume that $\gamma \in \Pi_N(\rho)$ such that $\sqrt{\gamma} \in H^1(\mathbb{R}^{dN})$. This is the case, for example, when $\gamma(x_1,\dots,x_N) = \vert \psi(x_1,\dots,x_N)\vert^2$, where $\psi \in H^1(\mathbb{R}^{dN})$ is a ground-state wave function solving the $N$-electron Schr\"odinger Equation (see \cite{CFK, CFK17,DMaGerNen, DMaGerNenGorSei} for details). Then, we can define the Hohenberg-Kohn functional by
\[
\tilde{F}_{\hbar}^{HK}[\rho] = \inf_{\gamma\in \Pi_N(\rho), \sqrt{\gamma} \in H^1(\mathbb{R}^{dN})}\bigg\lbrace \dfrac{\hbar^2}{2}\int_{\mathbb{R}^{dN}}\vert \nabla \sqrt{\gamma}\vert^2 dx_1\dots dx_N + \int_{\mathbb{R}^{dN}}\sum_{1\leq i<j\leq N}\dfrac{1}{\vert x_i-x_j\vert}d\gamma \bigg\rbrace.
\]

Now, as a consequence of the logarithmic Sobolev inequality for the Lebesgue measure \cite{GozLeo}, the following result holds: if $\rho\mathcal{L}^d\in\mathcal{P}(\mathbb{R}^d)$ and $\sqrt{\gamma}\in H^1(\mathbb{R}^{dN})$ then
\[
C_0[\rho]\leq C_{\varepsilon}[\rho] \leq  \tilde{F}_{\hbar}^{HK}[\rho], \quad \text{ with } \varepsilon = \pi\hbar^2/2.
\]

\subsection*{Example of optimal entropy couplings}
\label{sec:examples}

Let us present some computational examples of minimizers of $I_\varepsilon[\rho]$ illustrating the role of the parameter $\varepsilon$.
Before this, we recall a result on the characterization of minimizers in the one-dimensional case \cite{CoDePDMa}. In particular, according to it the minimizer of $I_0[\rho]$
is  concentrated on finitely many graphs and thus singular with respect to the product reference measure.

\begin{theorem}[\cite{CoDePDMa}]\label{teo:1DN} Let $\mu \in \mathcal{P}(\mathbb{R})$ be an absolutely continuous probability measure and $f\colon\mathbb{R}\to \mathbb{R}$ strictly convex, bounded from below and non-increasing function. Then there exists a unique optimal symmetric plan $\gamma \in \Gamma^{\mathrm{sym}} (\mu)$ that solves
$$ \min_{ \gamma \in \Pi_N^{\mathrm{sym}}(\mu) } \int_{\mathbb{R}^{N} } \sum_{1 \leq i < j \leq N}  f(|x_j-x_i|) \, \d  \gamma. $$
Moreover, this plan is induced by an optimal cyclical map $T$, that is, $\gamma_{\mathrm{sym}}=\left(\gamma_T\right)^S$, where $\gamma_T=(\mathrm{id},T,T^{(2)} , \ldots, T^{(N-1)})_{\sharp} \mu$. An explicit optimal cyclical map is
$$ T(x) =\begin{cases}  F_{\mu}^{-1} (F_{\mu}(x) + 1/N) \qquad & \text{ if }F_{\mu}(x) \leq (N-1)/N \\ F_{\mu}^{-1} ( F_{\mu}(x) +1 - 1/N ) & \text{ otherwise.} \end{cases}$$
Here $F_{\mu}(x)=\mu ( -\infty , x]$ is the distribution function of $\mu$, and $F_{\mu}^{-1}$ is its lower semicontinuous left inverse.
\end{theorem}

\subsubsection*{One-dimensional Entropic-Transport with Coulomb cost and a Gaussian measure}
Let $\rho$ be the normal distribution on the real line with zero mean and standard deviation $\sigma = 5$.
We compute numerically the solution of the entropic-transport problem with Coulomb cost in the real line using the Sinkorn algorithm \cite{CutSin}.
Notice that by Theorem \ref{teo:1DN}, we know that the minimizer of  $I_0[\rho]$ is concentrated on a graph.
See Figure \ref{fig} for an illustration of the computational results. Our code is based on the Python implementation available at POT library \cite{POT}.

\begin{figure}[h]
	\centering
	\includegraphics[ scale=0.3]{gaussian.jpg}
	\caption{The dependence of the minimizer of the entropic-transport problem \eqref{ETIntro} on the entropic parameter $\varepsilon$ for the one-dimensional Coulomb cost, $N=2$ and $\rho \sim N(0,5)$. The pictures show part of the support of the optimal coupling $\gamma_{\varepsilon}$ around the origin. From the left to right: $\varepsilon = 10^4,10^{-2},10^{-3},10^{-4},10^{-5}.$ }
	\label{fig}
\end{figure}

\subsection*{Organization of the paper} In Section \ref{sec:entropic} we introduce the setting and study sufficient conditions for the existence of minimizers for the entropy-transport problem \eqref{ETIntro}. Section \ref{sec:gammaconvergence} is devoted to the $\Gamma-$convergence proof of the entropic-transport functional $C_{\varepsilon}[\gamma]$ to the multi-marginal optimal transport with repulsive costs $C_0[\gamma]$. In Section \ref{sec:duality}, we study the Kantorovich duality for the entropic-transport problem.

\subsection*{Strategy of the main proof and some technical remarks} The main result of this paper is Theorem \ref{gammaconv:maintheorem}, in which we prove the $\Gamma$-convergence of the entropic-regularized functional $C_{\varepsilon}[\gamma]$ to $C_0[\gamma]$. The technical difficulty on dealing with the $\Gamma$-convergence comes from the fact that while for the entropic part $E[\gamma]$ the minimizer $\gamma$ tends to be \textit{as spread as possible} with respect to $\mathfrak m$, for the cost $C_{0}[\gamma]$ a minimizer can be very singular and have infinite entropy.

We divide the proof in two parts. The part (I), the $\liminf-$inequality, follows basically from the lower-semicontinuity of the costs $C_0[\gamma]$ and $C_{\varepsilon}[\gamma]$ - which are obtained from the assumption $\rho \log \rho \in L_\mathfrak m^1(X)$ on the marginal measure $\rho\mathfrak m$, giving a lower bound on the entropy. The part (II), the $\limsup-$inequality, is more involved.  In Section \ref{sec:blockapproximation}, we construct a block approximation $\gamma'_n$ for a coupling $\gamma$ with $C_{0}[\gamma] <+\infty$. Such a construction is done in several steps, since we need to construct a competitor $\gamma'_n$ such that $E[\gamma'_n] <\infty$ and $\gamma'_n \in \Pi^{\mathrm{sym}}_N(\rho)$. The main idea and the rigorous construction is done at section \ref{sec:blockapproximation}.

Futhermore, we point out that our construction can deal with the case when the space $X$ is a domain in $\mathbb{R}^d$, answering a question raised in \cite{BenCarNenEuler}. There the $\Gamma$-convergence was proven using convolutions; an approach that does not seem to be easy to implement for domains, or in general metric spaces.\medskip

\noindent
\emph{Related works:} A proof of the $\Gamma$-convergence of \eqref{ETIntro} to the Monge-Kantorovich problem for $c(x,y) = d(x,y)^p$ first appeared in \cite{LeoGamma,Mik04} via probabilistic methods. In \cite{CarDuvPeySch}, G. Carlier, V. Duval, G. Peyr\' e and B. Schmitzer provided an alternative and more analytical proof carrying out a similar block approximation procedure for the two-marginal squared distance cost in the Euclidean space and the Wasserstein Barycenter.

\section{The entropy-regularized repulsive costs}\label{sec:entropic}

Let $(X,d)$ be a Polish space and $\mathfrak m$ be a reference measure on $X$. We denote by $\mathcal{P}(X)$  the set of Borel probability measures on $X$, and $\mathcal{P}^{ac}(X)$ the set of Borel probability measures on $X$ that are absolutely continuous with respect to $\mathfrak m$.  We denote by $\mathfrak m_{N}$ the product measure $\mathfrak m\otimes \mathfrak m\otimes\cdots\otimes \mathfrak m$. This is the reference measure we use on the product space $X^N$. On $X^N$ we use the sup-metric, which we denote by $d_N$.

The class of cost functions $c\colon X^N\to \mathbb{R}\cup\{+\infty\}$ of our interest is given by functions of the form
\[
c(x_1,\ldots, x_N)=\sum_{1\le i<j\le N}f(d(x_i,x_j)),\quad \text{ for all }(x_1,\ldots,x_N)\in X^N,
\]
where $f:[0,\infty[\to\mathbb{R}\cup\{+\infty\}$ satisfies the following conditions
\begin{align*}
&f|_{]0,\infty[}\text{ is continuous, decreasing ,and }\tag{F1}\\
&\lim_{t\to 0+}f(t)=+\infty\,.\tag{F2}
\end{align*}
Above and from now on, we denote by $(x_1,\ldots, x_N)$ points in $X^N$, so $x_i\in X$ for each $i$.

We denote by
\[
\Pi_N(\rho)=\left\{\gamma\in \mathcal{P}(X^N)~|~\mathtt{pr}^i_\sharp \gamma=\rho\mathfrak m\text{ for all }i\in \{1,\ldots,N\}\right\}
\]
the set of \textit{couplings} or \textit{transport plans}, where $\mathtt{pr}^i$ is the projection
\[
\mathtt{pr}^i(x_1,\ldots,x_i,\ldots,x_N)=x_i~~~\text{for all }(x_1,\ldots, x_i,\ldots ,x_N)\in X^N\,.
\]
For the definition of the set of symmetric couplings $\Pi^{\mathrm{sym}}_N(\rho)$, see Definition \ref{def:symmeasures}.
We define the functional $C_0[\gamma]$ to be the cost related to the coupling $\gamma$
\[C_0[\gamma]=\int_{X^N}c(x_1,\ldots,x_N)\,\d \gamma(x_1,\ldots, x_N)\,.\]
Given $\varepsilon\ge 0$, we denote by $C_\varepsilon[\gamma]$ the entropy-regularized cost
\begin{equation}\label{eq:entropicfunctional}
 C_\varepsilon[\gamma]=C_0[\gamma]+\varepsilon E[\gamma],\quad \text{ for all }\gamma\in \Pi^{\mathrm{sym}}_N(\rho),
\end{equation}
where the entropy $E[\gamma]:\mathcal{P}(X^N)\to \mathbb{R}\cup\lbrace -\infty,+\infty\rbrace$ is defined as

\begin{equation}\label{eq:entropy}
E[\gamma]=\begin{cases}\int_{X^N}\rho_\gamma\log\rho_\gamma\,\d \mathfrak m_{N}&\text{ if }\gamma\ll \mathfrak m_{N}\\
+\infty&\text{ otherwise}\end{cases}\,.
\end{equation}

The notation $\rho_\gamma$ stands for the Radon-Nikodym derivative of $\gamma$ with respect to the reference measure $\mathfrak m_{N}$ and $\gamma \ll \mathfrak m_{N}$ means that $\gamma$ is absolutely continuous with respect to the reference measure $\mathfrak m_{N}$.
 Let $\rho \mathfrak m\in \mathcal{P}^{ac}(X)$. In this paper we are interested in the following infimum
\begin{equation}\label{eq:inf}
I_\varepsilon[\rho \mathfrak m]:=\inf_{\gamma\in\Pi^{\mathrm{sym}}_N(\rho)} C_\varepsilon[\gamma]\,.
\end{equation}

In order to guarantee the lower semicontinuity for $C_\varepsilon$, we will assume $\rho \log \rho \in L_\mathfrak m^1(X)$. This will take care of the entropy part $E[\cdot]$ of the cost. In order to establish the lower semicontinuity for the functional $C_0[\cdot]$, we assume that the measure $\rho$ satisfies
 the following two conditions:
\begin{align*}
&\lim_{r\to 0}\sup_{x\in X}\rho (B(x,r))<\frac{1}{N(N-1)^2}~~~\text{and}\tag{A}
\\
&\int_{X\setminus B(o,r_0)}f\left(2d(x,o)\right)\,\d\rho(x)> - \infty~~~\text{for some }o\in X\,.\tag{B}
\end{align*}Above we have, by an abuse of notation, denoted the measure $\rho \mathfrak m$ by only the density $\rho$; we will use the same abbreviation in the rest of the paper if there is no risk of confusion. The Condition $(B)$ is a similar assumption than requiring, in the case of the quadratic cost, that the marginal measures have finite second moments. The Condition $(A)$ guarantees that the cost is finite.

If we endow the spaces $\mathcal{P}(X^N)$ and $\mathcal{P}(X)$ with $w^\ast$-topology then, by Prokhorov's theorem, any subset of $\mathcal{P}(X)$ (or $\mathcal{P}(X^N)$) is tight if and only if it is relatively compact. \begin{remark}[Entropy-transport seen as a Kullback-Leibler divergence]\label{rm:kl}

If $\mu$ and $\nu$ are measures on a set $X$, the Kullback-Leibler divergence of $\mu$ with respect to $\nu$ is defined as
\[
\operatorname{KL}\nolimits(\mu|\nu)=\begin{cases}
\int_X\log\left(\frac{\d \mu}{\d\nu}\right)\d\mu&\text{ if }\mu\ll \nu\\
+\infty&\text{ otherwise}\end{cases}\,.\]
Now, if both measures $\mu$ and $\nu$ are absolutely continuous with respect to some reference measure $R$ of the space $X$ with densities $\rho_\mu$ and $\rho_\nu$, respectively,  we can write:
\[
\operatorname{KL}\nolimits(\mu|\nu)=\begin{cases}
\int_X\rho_\mu\log\left(\frac{\rho_\mu}{\rho_\nu}\right)\d\mu&\text{ if }\mu\ll \nu\\
+\infty&\text{ otherwise}\end{cases}\,.\]
Considering the entropy-regularized MOT problem, we see that the cost functional $C_{\varepsilon}[\gamma]$ can be alternatively written as the Kullback-Leibler divergence between $\gamma$ and a \textit{kernel} $\kappa$ defined below

\[
C_{\varepsilon}[\gamma]= \varepsilon\operatorname{KL}\nolimits(\gamma|\kappa) = \varepsilon\int_{X^N}\rho_{\gamma} \ln\bigg(\dfrac{\rho_{\gamma}}{\rho_{\kappa}}\bigg)\,\d\mathfrak m_{N},
\]
where $\kappa= e^{-c/\varepsilon}\mathfrak m_N$.
\end{remark}

For the most part, in this paper we have chosen to consider as a reference measure the measure $\mathfrak m_N$. However, as the following lemma shows, we could also assume the reference measure to be $(\rho\mathfrak m)^{\otimes N}$ since the minimizers of the entropy-regularized MOT problem (\ref{eq:inf}) do not depend on the choice of the reference measure, at least if there exists a minimizer with finite cost. To state the lemma, let us introduce the notation of relative entropy: for each reference measure $R$ of a Polish space $Y$, and for each $\gamma\in \mathcal{P}(Y)$, we denote by $E(\gamma|R)$ the relative entropy of $\gamma$ with respect to $R$, defined as
\[E(\gamma|R)=
\begin{cases}
\int_Y\log\left(\frac{\d\gamma}{\d R}\right)\,\d\gamma&\text{ if }\gamma \ll R\\
+\infty&\text{ otherwise}\end{cases}\,.\]
Now we may consider two, a priori different, entropy-regularized MOT problems: the one introduced in (\ref{eq:inf})
\begin{equation}\label{eq:usual}
I_\epsilon(\rho\mathfrak m)=\inf_{\gamma\in\Pi^{\mathrm{sym}}_N(\rho)} C_\varepsilon[\gamma]=:I_\epsilon(\rho\mathfrak m|\mathfrak m)\,,
\end{equation}
and the problem with the reference measure chosen to be $(\rho\mathfrak m)^{\otimes N}$
\begin{equation}\label{eq:rhoasreference}
I_\epsilon(\rho\mathfrak m|\rho\mathfrak m):=\inf_{\gamma\in\Pi^{\mathrm{sym}}_N(\rho)}\left(C_0[\gamma]+E(\gamma|(\rho \mathfrak m)^{\otimes N})\right)\,.
\end{equation}
The folowing Lemma \ref{lm:changeofreference} is used only to go from the compact to the general case in the duality Theorem \ref{kanto:dualthm}. The proof in \cite{DMaGer19}
can be directly applied here to prove Lemma \ref{lm:changeofreference}.
\begin{lemma}\label{lm:changeofreference}
Let $(X,d,\mathfrak m)$ be a Polish measure space, $\rho\mathfrak m\in \mathcal{P}(X)$ a measure satisfying (A) and (B), and $c$ a cost function satisfying  (F1) and (F2). Now for all $\epsilon > 0$ we have
\begin{equation}\label{eq:change}
I_\epsilon(\rho\mathfrak m|\mathfrak m)=I_\epsilon(\rho\mathfrak m|\rho\mathfrak m)+N\epsilon \operatorname{KL}\nolimits(\rho\mathfrak m|\mathfrak m)=I_\epsilon(\rho\mathfrak m|\rho\mathfrak m)+N\epsilon \int_X\rho\log\rho\d\mathfrak m\,.
\end{equation}Moreover, whenever at least one side of the equality above is finite, the problems (\ref{eq:usual}) and (\ref{eq:rhoasreference}) have the same minimizers.
\end{lemma}

\subsection{Some properties of the entropy functional

Let us start by noting that the minimum of the entropy is attained by the product measure and that its value is not $-\infty$.

\begin{proposition}\label{ent:min}
Let $(X,d, \mathfrak m)$ be a Polish metric measure space,
and let $\rho\mathfrak m \in \mathcal{P}^{ac}(X)$ with $\rho\log\rho \in L_\mathfrak m^1(X)$ .
Then
\[
\min_{\gamma\in\Pi^{\mathrm{sym}}_N(\rho)}E[\gamma] = \int_{X^N}\bigg(\otimes^N_{i=1}\rho\bigg)\log\bigg(\otimes^N_{i=1}\rho\bigg) \,\d\mathfrak m_{N} = N \int_X \rho\log\rho\,\d\mathfrak m > -\infty.
\]
\end{proposition}
\begin{proof}
As we will see, the minimality is an immediate consequence of Jensen's inequality. Let $\gamma \in \Pi(\rho)$. Then
\begin{align*}
E[\gamma] & = \int_{X^N}\rho_{\gamma}\log(\rho_\gamma)\,\d\mathfrak m_{N} =
\int_{X^N}\frac{\rho_{\gamma}}{\otimes^N_{i=1}\rho}\left(\log\left(\frac{\rho_{\gamma}}{\otimes^N_{i=1}\rho}\right) + \log\left(\otimes^N_{i=1}\rho\right)\right)\otimes^N_{i=1}\rho\,\d\mathfrak m_{N} \\
&\ge \bigg(\int_{X^N}\rho_{\gamma}\,\d\mathfrak m_{N}\bigg)\log\bigg(\int_{X^N}\rho_{\gamma}\,\d\mathfrak m_{N}\bigg) +
\int_{X^N}\rho_{\gamma}\log\left(\otimes^N_{i=1}\rho\right)\,\d\mathfrak m_{N} \\
& = 0 + E[\otimes^N_{i=1}\rho]. \qedhere
\end{align*}
\end{proof}

Using Proposition \ref{ent:min} we immediately get the lower semicontinuity of the entropy functional by representing the entropy as relative entropy against the probability measure $\otimes_{i=1}^N(\rho\mathfrak m)$. See for instance \cite[Lemma 4.1]{Sturm} for the lower semicontinuity of the entropy when the reference measure is finite.

\begin{corollary}\label{ent:lsc}
Let $(X,d, \mathfrak m)$ be a Polish metric measure space,
and let $\rho\mathfrak m \in \mathcal{P}^{ac}(X)$ with $\rho\log\rho \in L_\mathfrak m^1(X)$.
Then $E[\cdot]$ is lower semicontinuous in the set $\Pi^{\mathrm{sym}}_N(\rho)$.
\end{corollary}

Now we are ready to prove the existence of the  minimizers for entropy-regularized MOT:

\begin{proposition}\label{prop:minexists}
Let $(X,d,\mathfrak m)$ be a Polish metric measure space. Assume that the measure $\rho \mathfrak m\in \mathcal{P}^{ac}(X)$ satisfies $\rho\log\rho \in L_\mathfrak m^1(X)$ along with Conditions (A) and (B). Assume that $c\colon X^N\to\mathbb{R}\cup\lbrace +\infty\rbrace$ satisfies the conditions $(F1)$ and $(F2)$. Then, for each $\varepsilon\ge 0$, there exists a minimizer $\gamma\in \Pi^{\mathrm{sym}}_N(\rho)$ for the entropic-regularized cost $C_{\varepsilon}[\gamma]$.
\end{proposition}

\begin{proof}
We notice that the set $\Pi^{\mathrm{sym}}_N(\rho)$ is compact in the $w^\ast$-topology \cite{Kel}.
 The functional $E$ is lower semicontinuous by Corollary \ref{ent:lsc}, and in our setting
 the lower semicontinuity of $C_0$ is proven as a part of the proof of \cite[Proposition 3.1]{GerKauRaj17}.
Since for each $\varepsilon\ge 0$ the functional $C_\varepsilon$ is convex, we conclude that it has a minimizer in the set $\Pi^{\mathrm{sym}}_N(\rho)$\end{proof}

\subsection{Some properties of the coupling cost $C_0[\gamma]$}Notice that in this section $\Pi_N(\rho)$ denotes the set of couplings in $X^N$ (not necessarily symmetric). Moreover, we need to assume extra hypothesis on the probability measure $\rho$ in order to guaranteee that $C_0[\gamma]$ is bounded from below for a $\gamma\in\Pi_N(\rho)$ (e.g. $f(z) = -\log(\vert z\vert)$).

 The next theorem from \cite{BuChaDeP} (see also \cite[Theorem 3.2]{GerKauRaj17}) states that for measures $\rho$ satisfying the assumptions (A) and (B) there exists $\overline{\alpha}>0$ for which the support of any optimal plan is concentrated away from the set
\[D_{\alpha}:=\{(x_1,\ldots,x_N)\in X^N~|~d(x_i,x_j)<\alpha\text{ for some }i\ne j\}\,.\]

\begin{theorem}{\cite[Theorem 3.2]{BuChaDeP}}\label{ent:supportoutofD}
Let $(X,d)$, $\rho$, $f$, $c$ be such that the measure $\rho$ satisfies (A) and (B) and the function $f$ satisfies (F1) and (F2).  Let $\gamma$ be a minimizer of
\[
I_0[\rho] =  \min_{\gamma\in\Pi_N(\rho)}\int_{X^N} c(x_1,\dots,x_N)\,\d\gamma(x_1,\dots,x_N).
\]
Let us fix $0<\beta<1$ such that
\[\sup_{x\in X}\rho (B(x,\beta)) <\frac{1}{N(N-1)^2}.\]
Then, we have for all
\begin{equation}\label{eq:alphaneedstobe}
\alpha<f^{-1}\left(\frac{N^2(N-1)}{2}f(\beta)\right)\end{equation}
the inclusion
\begin{equation}\label{eq:claim}
\mathop{\rm spt}\nolimits (\gamma)\subset X^N\setminus D_\alpha,\end{equation}
where $f^{-1}$ stands for the left-inverse of $f$.
\end{theorem}

Next we observe that one can restrict the problem $\min_{\gamma\in\Pi(\rho)}C_0[\gamma]$  to the class of symmetric couplings in $X^N$ having all the marginals equal to $\rho$.

\begin{definition}[Symmetric measures]\label{def:symmeasures}
A measure $\gamma \in \mathcal{P}(X^N)$ is symmetric if
\[
\int_{X^N}\phi(x_1,\dots,x_N)\,\d\gamma = \int_{X^N} \phi(\overline{\sigma}(x_1,\dots,x_N))\,\d\gamma, \text{ for all } \phi \in  \mathcal{C}(X^N)
\]
 for all permutations $\overline{\sigma}$ of the $N$ symbols $(x_1,\ldots, x_N)$. We denote by $\mathcal{P}^{\mathrm{sym}}(X^N)$ the set of symmetric probability measures in $X^N$, and notice that $\Pi^{\mathrm{sym}}_N(\rho) :=  \Pi_N(\rho)\cap \mathcal{P}^{\mathrm{sym}}(X^N)$.
\end{definition}

Let us also introduce the notation for symmetrized measures. If $\gamma$ is a Borel measure on $X^N$, we denote by $\gamma^S$ the symmetrized measure
\[\gamma^S:=\frac{1}{N!}\sum_{\sigma\in \mathcal{S}_N}\sigma_\sharp\gamma\,,\]
where $\mathcal{S}_N$ is the set of permutations of the $\{1,\ldots,N\}$ koordinates $(x_1,\ldots, x_N)$.
The following result follows immediately.

\begin{proposition}\label{ent:c0symplans}
Under the hypothesis of Proposition \ref{ent:supportoutofD}, we have that
\begin{equation}\label{eq:MKsymMK}
\min_{ \gamma \in \Pi_N(\rho) } \int_{X^{N} } c(x_1,\ldots, x_N) \, \d  \gamma = \min_{ \gamma \in \Pi_N^{\mathrm{sym}}(\rho) } \int_{X^{N} } c(x_1,\ldots, x_N) \, \d  \gamma.
\end{equation}
\end{proposition}
\vspace{5mm}

\section{The $\Gamma$-convergence of Entropic-regularized cost}\label{sec:gammaconvergence}

Now let us turn to the $\Gamma$-convergence.
 From now on,  $(\tau_n)_{n\in\mathbb{N}}$ is any sequence of positive real numbers decreasing to zero. Let us introduce the following functionals: for each $n\in\mathbb{N}$
\[\mathcal{C}_n:\mathcal{P}^{\mathrm{sym}}(X^N)\to\mathbb{R}\cup\{+\infty\},~~ \mathcal{C}_n[\gamma]=\begin{cases}C_{\tau_n}[\gamma]&\text{ if }\gamma\in \Pi_N(\rho)\\+\infty&\text{ otherwise}\end{cases}\]
and
\[\mathcal{C}:\mathcal{P}^{\mathrm{sym}}(X^N)\to\mathbb{R}\cup\{+\infty\},~~ \mathcal{C}[\gamma]=\begin{cases}C[\gamma]&\text{ if }\gamma\in \Pi_N(\rho)\\+\infty&\text{ otherwise}\end{cases}.\]

The goal of this section is to prove that the sequence $(\mathcal{C}_{n\in\mathbb{N}})$ $\Gamma$-converges to $\mathcal{C}$ in the space $\mathcal{P}^{\mathrm{sym}}(X^N)$.

\begin{theorem}\label{gammaconv:maintheorem}
Let $(X,d,\mathfrak m)$ be a Polish metric measure space. Let $\rho \in \mathcal{P}^{ac}(X)$ with $\rho\log\rho \in L_\mathfrak m^1(X)$ satisfying (A) and (B).
Then the sequence $(\mathcal{C}_n)$ $\Gamma$-converges to $\mathcal{C}$ in the space $\mathcal{P}^{\mathrm{sym}}(X^N)$.
\end{theorem}

Let us fix $\gamma\in \mathcal{P}^{\mathrm{sym}}(X^N)$. We need to show that
\begin{align*}
&\text{For each sequence }(\gamma_n)_{n\in\mathbb{N}}\text{ that converges to }\gamma\\
&\text{we have }\liminf_{n\to\infty} \mathcal{C}_n[\gamma_n]\ge C[\gamma]\text{, and }\tag{I}\\
&\text{There exists a sequence }(\gamma_n)_{n\in\mathbb{N}}\text{ that converges to }\gamma\text{ and }\\
&\limsup_{n\to\infty }\mathcal{C}_n[\gamma_n]\le C[\gamma]\,.\tag{II}\end{align*}

The proof of Theorem \ref{gammaconv:maintheorem} is divided into two parts. The proof of the first part, the liminf-inequality (I), is short and is established in the next subsection. The remainder of this section is then divided into subsections in which the second part, the limsup-inequality (II) is proven.

\subsection{Proof of condition (I)} We fix a sequence $(\gamma_n)_{n\in\mathbb{N}}$ that converges to $\gamma$. If $\gamma\notin \Pi_N(\rho)$, then since the set $\Pi_N(\rho)$ is compact, for large indices we also have $\gamma_n\notin \Pi_N(\rho)$, so both sides of inequality (I) are $+\infty$, and we are done. Hence we may assume that $\gamma$ and $\gamma_n$'s are elements of the set $\Pi_N(\rho)$. Since now $\gamma_n\in \Pi_N(\rho)$, the claim (I) follows from the lower-semicontinuity of $\gamma\mapsto \int c\,\d\gamma$ and from the entropy lower bound shown in Proposition \ref{ent:min}.

\subsection{Constructing an approximation of the coupling $\gamma$}\label{sec:blockapproximation}

First of all, we need to construct an approximation of $\gamma$ only in the case where $C_0[\gamma] < \infty$: if this is not the case then any sequence $(\gamma_n)$ converging to $\gamma$ can be used to prove  Condition (II). The idea of the construction is to redefine a large part of $\gamma$ to be a product measure on finitely many Borel sets with small diameter. In order not to increase the cost by too much, the Borel sets we are using have to be far away from the diagonal compared to the diameter of the sets. We call the part of the measure defined in this way \emph{the core part} of the approximation. For the rest of the measure, we take another finite combination of product measures. However, this time the sets do not need to have small (or even bounded) diameter, but just small measure. This part will be called \emph{the remainder part} of the approximation.

We start the construction by taking out a small part of $\gamma$ that will later be used to deal with the remainder part of the approximation. For this we take a sequence of radii defined as $r_n = 1/n$. Since $C[\gamma] < \infty$, there exists a point $x =(x_1,\dots, x_N)\in \mathop{\rm spt}\nolimits(\gamma)$ with
\[
x_i \ne x_j \quad \text{ if }1 \le i <  j \le N.
\]
Moreover, since $\gamma\in \Pi_N(\rho)$ and $\rho$ satisfies (A), we have
\begin{align*}
\gamma(\{(y_1,\dots, & y_N)  \in X^N  \,| \, y_i \ne x_j \text{ for all }i,j\}) \\
& \ge 1 - \sum_{i\ne j}  \gamma(\{(y_1,\dots, y_N) \in X^N \,| \, y_i = x_j \}) \\
& = 1 - \sum_{i\ne j}  \rho(X \setminus \{x_j\}) \ge 1 - N(N-1) \frac{1}{N(N-1)^2} > 0.
\end{align*}

Thus, using again $C[\gamma] < \infty$, there exists another point $x'=(x_{N+1},\dots, x_{2N}) \in \mathop{\rm spt}\nolimits(\gamma)$, so that
\[
x_i \ne x_j \quad \text{ if }1 \le i <  j \le 2N.
\]
From now on, we consider $x,x'$ fixed. Therefore, for $n \in \mathbb{N}$ sufficiently large we have
\begin{equation}
d(x_i,x_j)>r_n \quad \text{ if }1 \le i <  j \le 2N.\label{eq:farenough}
\end{equation}

Let us denote by
\[
B_n:=B(x,\tfrac{r_n}{10})\quad\text{and}\quad B_n':=B(x',\tfrac{r_n}{10})
\]
the balls around $x$ and $x'$ with radii $r_n/10$ in the sup-metric of the product space. So,
\[y=(y_1,\ldots,y_N)\in B_n\]
 if and only if
\[d(x_i,y_i)<\tfrac{r_n}{10}\text{ for all }i\in \{1,\ldots, N\}\,,\]
and analogously for $B_n'$ with the relevant index modifications.

Let us now define
\begin{align*}
\gamma_{B_n}=\left(\frac{\gamma\restr{B_n} }{\gamma(B_n)} \right)^S\quad\text{and}\quad
\gamma_{B_n'}=\left(\frac{\gamma\restr{B_n'} }{\gamma(B_n')} \right)^S\,.
\end{align*}
Observe that $\gamma_{B_n}$ and $\gamma_{B_n'}$ are symmetric probability measures.
Since the marginals of a symmetric measure are the same, we may denote by $\rho_{B_n}$ the marginal of $\gamma_{B_n}$ and similarly by $\rho_{B_n'}$ the marginal of $\gamma_{B_n'}$.
Let us further denote $\tilde B_n:=\mathop{\rm spt}\nolimits\gamma_{B_n}$, $\tilde B_n':=\mathop{\rm spt}\nolimits\gamma_{B_n'}$ and
\begin{equation}\label{eq:epsilondefinition}
 \varepsilon_n := \frac1N\min\left\{\gamma(\tilde B_n),\gamma(\tilde B_n'),r_n,\frac{r_n}{f(2r_n/5)}\right\}\,.
\end{equation}

We then define a measure
\begin{align*}
\gamma_{0,n}:= & \gamma\restr{X^n\setminus (\tilde B_n\cup \tilde B_n')}
+\frac{\gamma(\tilde B_n)-\varepsilon_n}{\gamma(\tilde B_n)}\gamma\restr{\tilde B_n}+\frac{\gamma(\tilde B_n')-\varepsilon_n}{\gamma(\tilde B_n')}\gamma\restr{\tilde B_n'}.
\end{align*}
The idea behind the measure $\gamma_{0,n}$ is that we have chopped of a small part of the measure around the points $x$ and $x'$ (symmetrically) for later use. Since we are working with a singular cost, we still need to take out a small neighbourhood of the diagonals before approximating by product measures. We do this now.

We fix a compact $K_n\subset X$ such that
\begin{equation}\label{eq:largeCompact}
\gamma_{0,n}(X^N\setminus K_n^N)<\frac{\varepsilon_n}2
\end{equation}
and take a small enough $\delta_n \in (0,r_n)$ so that
\begin{equation}\label{eq:smalldiagonal}
  \gamma_{0,n}(D_{\delta_n}) < \frac{\varepsilon_n}2.
\end{equation}
Using $K_n$ and  $\delta_n$ we then define
\begin{equation}\label{gammaconv:gamma2n}
\gamma_{1,n}:=\gamma_{0,n}|_{K_n^N\setminus D_{\delta_n}}.
\end{equation}
The measure $\gamma_{1,n}$ is now the core part of the measure that we approximate.
We  denote by $\rho_{1,n}$ the marginals of the symmetric measure $\gamma_{1,n}$.

Let us then approximate the measure $\gamma_{1,n}$. We take $\lambda_n \in (0,\delta_n/n)$
so that
\begin{equation}\label{eq:lambdachoice}
 |f(r)-f(s)| < \varepsilon_n \qquad \text{for all }r,s \in [\delta_n/2,2\mathop{\rm diam}\nolimits(K_n)]\text{ with }|r-s| \le 2\lambda_n.
\end{equation}
Such $\lambda_n$ exists by the uniform continuity of $f$ on the compact set$[\delta_n/2,2\mathop{\rm diam}\nolimits(K_n)]$.
Since the set $K_n$ is compact, we may fix a finite Borel partition $\{B_n^i\}_{i=1}^{M_n}$ of the set $\mathop{\rm spt}\nolimits(\rho_{1,n})$ such that
\begin{align*}
&\mathop{\rm diam}\nolimits(B_n^i)<\lambda_n\quad\text{and}\quad
0<\rho_{1,n}(B_n^i)<\varepsilon_n\text{ for all }i\in \{1,\ldots, M_n\}\,.\end{align*}
We are now ready to define the core part approximants $\gamma_{1,n}^a$ as
\begin{equation}
\label{gammaconv:gamma2na}
\gamma_{1,n}^a=\sum_{(k_1,\ldots,k_N)\in M_n^N}\frac{\gamma_{1,n}(B_n^{k_1}\times\cdots\times B_n^{k_N})}{\rho_{1,n}(B_n^{k_1})\cdots\rho_{1,n}(B_n^{k_N})}\rho_{1,n}|_{B_n^{k_1}}\otimes\cdots\otimes\rho_{1,n}|_{B_n^{k_N}}\,.
\end{equation}

Now let us handle the main part of the remainder of the measure, namely the measure
\[
\gamma_{2,n}:=\gamma_{0,n}|_{D_{\delta_n}\cup (X^N\setminus K_n^N)}\,.
\]
Because $\gamma_{0,n}$ and the set where we restrict it is symmetric, also $\gamma_{2,n}$ is.
We may thus denote its marginals by $\rho_{2,n}$.

In order to determine which part of the remaining marginal measure should be coupled where, we define a partition $\{A_{i,n}\}_{i=1}^N$ of the space $X$ by setting, for all $i\in \{1,\ldots, N-1\}$
\[A_{i,n}:=\{y\in X~|~d(x_i,y)\le \tfrac{r_n}{2}\}\,,\]
and
\[A_{N,n}:=X\setminus \bigcup_{i=1}^{N-1}A_{i,n}\,.\]
Condition \eqref{eq:farenough} guarantees that the sets $A_{i,n}$ are pairwise disjoint.

 Now we approximate $\gamma_{2,n}$ by the measure
\[\gamma_{2,n}^a:=N\left(\sum_{i=1}^N\eta_{n,i}\right)^S\,,\]
 where for all $i$ the measure $\eta_{n,i}$ is the product
\[
\eta_{n,i} := \left(\bigotimes_{k=1}^{i-1}\frac{\rho_{B_n}\restr{B(x_k,r_n/10)}}{\rho_{B_n}(B(x_k,r_n/10))}\right)
\otimes \rho_{2,n}\restr{A_{i,n}}
\otimes \left(\bigotimes_{k=i+1}^N\frac{\rho_{B_n}\restr{B(x_k,r_n/10)}}{\rho_{B_n}(B(x_k,r_n/10))}\right).
\]
By the definition of the sets $A_{i,n}$, for every $(y_1,\dots,y_N) \in \mathop{\rm spt}\nolimits(\gamma_{2,n}^a)$ we have for each $i \ne j$
\begin{equation}\label{eq:remainderoffdiagonal}
d(y_i,y_j) \ge |d(y_i,x_j) - d(x_j,y_j)| > \frac{r_n}{2}-\frac{r_n}{10} = \frac{2r_n}{5},
\end{equation}
where we have assumed (which we can do without loss of generality) that $y_j \in \overline{B}(x_j,r_n/10)$.

What we have done using the measure $\gamma_{2,n}^a$ is that we have coupled the marginals of the measure $\gamma_{2,n}$ with some suitable parts of the marginals of the reserved  measure that was taken out around the point $x$. In this way we have used unevenly the marginals of this reserved part. To handle the rest of the reserved part of the measure around the point $x$, we now use the reserved measure around the point $x'$. So, we need to redefine the coupling for the part of the marginal given by
\[
\rho_{3,n} := (\texttt{pr}^1)_\sharp\frac{\varepsilon_n}{\gamma(\tilde B_n)}\gamma\restr{\tilde B_n}  +\rho_{2,n} - (\texttt{pr}^1)_\sharp\gamma_{2,n}^a.
\]
We define it as
\[
\gamma_{3,n}^a := \left(\sum_{i=1}^N\phi_{n,i}  \right)^S,
\]
where each $\phi_{n,i}$ is defined as
\[
\phi_{n,i} := \left(\bigotimes_{k=N+1}^{N+i-1}\frac{\rho_{B_n}\restr{B(x_k,r_n/10)}}{\rho_{B_n}(B(x_k,r_n/10))}\right)
\otimes \rho_{3,n}
\otimes \left(\bigotimes_{k=N+i+1}^{2N}\frac{\rho_{B_n}\restr{B(x_k,r_n/10)}}{\rho_{B_n}(B(x_k,r_n/10))}\right).
\]
Since $\mathop{\rm spt}\nolimits(\rho_{3,n}) \subset \mathop{\rm spt}\nolimits(\rho_{B_n})$, we have that for every $(y_1,\dots,y_N) \in \mathop{\rm spt}\nolimits(\gamma_{3,n}^a)$ and each $i \ne j$
\begin{equation}\label{eq:remainderoffdiagonal2}
d(y_i,y_j) \ge |d(x_{k(i)},x_{k(j)})- d(y_i,x_{k(i}) - d(x_{k(j)},y_j)| > {r_n}-\frac{r_n}{10}-\frac{r_n}{10} = \frac{4r_n}{5},
\end{equation}
where $k(i)\ne k(j)$ are the indices for which $y_j \in \overline{B}(x_{k(j)},r_n/10)$ and $y_i \in \overline{B}(x_{k(i)},r_n/10)$.

What remains is the part of the measure around $x'$ that was not used for $\gamma_{3,n}^a$. Since $\gamma_{3,n}^a$ used the marginals from this part of the reserved measure evenly, we may simply couple the rest by a measure
\[
\gamma_{4,n}^a := b\left(\bigotimes_{k=N+1}^{2N}\frac{\rho_{B_n}\restr{B(x_k,r_n/10)}}{\rho_{B_n}(B(x_k,r_n/10))}\right)^S,
\]
with $b$ being the correct scaling constant.
Similarly as for the previous remainder part, we have that for every $(y_1,\dots,y_N) \in \mathop{\rm spt}\nolimits(\gamma_{4,n}^a)$ and each $i \ne j$ the inequality \eqref{eq:remainderoffdiagonal2} holds.

Now we are ready to define the full approximation as
\[
\gamma_n' = \gamma_{1,n}^a + \gamma_{2,n}^a +  \gamma_{3,n}^a +  \gamma_{4,n}^a.
\]
By construction $\gamma_n' \in \Pi^\mathrm{sym}(\rho)$.

\subsection{Narrow convergence of the approximations}

Let us now prove that the sequence $(\gamma_n')_n$ narrowly converges to $\gamma$. We could argue this by using the Wasserstein distance. However, let us do it here directly using the definition of narrow convergence.

\begin{lemma}\label{lma:weakconvergence}
The sequence $(\gamma_n')_n$ narrowly converges to $\gamma$.
\end{lemma}
\begin{proof}

Let $\varphi \in C_b(X^N)$ and $\varepsilon > 0$.
We need an index $N_0\in\mathbb{N}$ such that
\begin{equation}\label{eq:close}
\left\lvert\int_{X^N}\varphi\d\gamma-\int_{X^N}\varphi d\gamma_n'\right\rvert<\varepsilon\text{ for all }n\ge N_0\,.
\end{equation}
Let us denote $M:=\sup_{x\in X^N}|\varphi(x)|$; we may assume that $M>0$. Since $\rho$ is inner regular, we can fix a compact set $K\subset X$ such that
\[\rho(X\setminus K)<\frac{\varepsilon}{12NM}\,.\]
Since $\gamma\in \Pi_N(\rho)$, we now have
\[\gamma(X^N\setminus K^N)<\frac{\varepsilon}{12M}\,.\]
The function $\varphi$, when restricted to $K^N$, is uniformly continuous. Hence there exists $\delta > 0$ so that
\begin{equation}\label{eq:uniform}
|\varphi(x)-\varphi(y)| < \frac{\varepsilon}{12}\text{ for all }x,y\in K^N\text{ for which }d_N(x,y)<\delta\,.
\end{equation}
Now let $N_0 \in \mathbb{N}$ be so large that

\[
  \sqrt N\lambda_n < \delta \qquad \text{and} \qquad 6M\varepsilon_n < \frac{\varepsilon}{6}~~~\text{for all }n\ge N_0\,. \]

Let us show that this choice of $N_0$ fulfills Condition (\ref{eq:close}). First we note that for all $n\ge N_0$ we have
\begin{align}
\left\lvert \int_{X^N}\varphi\,\d\gamma-\int_{X^N}\varphi\,\d\gamma_n'\right\rvert
&\le  \left\lvert\int_{X^N}\varphi\,\d\gamma-\int_{X^N}\varphi\,\d\gamma_{1,n}\right\rvert+\left\lvert\int_{X^N}\varphi\,\d\gamma_{1,n}-\int_{X^N}\varphi\,\d\gamma_n'\right\rvert\nonumber\\
&\le \left\lvert\int_{X^N}\varphi\,\d\gamma-\int_{X^N}\varphi\,\d\gamma_{1,n}\right\rvert+\left\lvert\int_{X^N}\varphi\,\d\gamma_{1,n}-\int_{X^N}\varphi\,\d\gamma_{1,n}^a\right\rvert\nonumber\\
&\quad+\left\lvert\int_{X^N}\varphi\,\d(\gamma_{2,n}^a+\gamma_{3,n}^a+\gamma_{4,n}^a)\right\rvert\nonumber\\
&<\left\lvert\int_{X^N}\varphi\,\d\gamma_{1,n}-\int_{X^N}\varphi\,\d\gamma_{1,n}^a\right\rvert+\frac{\varepsilon}{6}\,,\label{eq:tailsestimated}\end{align}
where in the last inequality we have used the following facts:
$\gamma(X^N)-\gamma_{1,n}(X^N)<3\varepsilon_n$ for all $n$, and
for the remainder part of the measure $\gamma_n'$ we have
\[
 (\gamma_{2,n}^a + \gamma_{3,n}^a + \gamma_{4,n}^a)(X^N) < 3\varepsilon_n~~~\text{ for all }n\in\mathbb{N}.\]

It remains to show that for all $n\ge N_0$ we have
\begin{equation}\label{eq:core}
 \left|\int_{X^N} \varphi \,\d\gamma_{1,n} - \int_{X^N} \varphi \,\d\gamma_{1,n}^a \right| < \frac{5\varepsilon}{6}\,.
\end{equation}
We first estimate the integrals in the set $K^N$.
Let us fix, for each $(k_1,\ldots, k_N)\in M_n^N$ for which the set
\[(B_n^{k_1}\times\cdots\times B_n^{k_N})\cap K^N\]
is nonempty, an element
\[z_{k_1,\ldots,k_N}\in(B_n^{k_1}\times\cdots\times B_n^{k_N})\cap K^N\,.\]
Now we have, for a fixed $(k_1,\ldots, k_N)$, denoting for simplicity
\[z_0:=z_{k_1,\ldots, k_N}, Q=B_n^{k_1}\times\cdots\times B_n^{k_N},~\gamma=\gamma_{1,n},~\gamma_a=\gamma_{1,n}^a\,,\]
\begin{align*}
\bigg\lvert\int_{Q\cap K^N}&\varphi(z)\,\d\gamma-\int_{Q\cap K^N}\varphi(z)\,\d\gamma_a\bigg\rvert\\
&\le\left\lvert\int_{Q\cap K^N}\varphi(z)\,\d\gamma-\varphi(z_0)\gamma(Q\cap K^N)\right\rvert
+\left\lvert\varphi(z_0)\gamma_a(Q\cap K^N)-\int_{Q\cap K^N}\varphi(z)\,\d\gamma_a\right\rvert\\
&\quad+\left\lvert\varphi(z_0)\gamma(Q\cap K^N)-\varphi(z_0)\gamma_a(Q\cap K^N)\right\rvert\\
&\le\int_{Q\cap K^N}|\varphi(z)-\varphi(z_0)|\,\d\gamma
+\int_{Q\cap K^N}|\varphi(z_0)-\varphi(z)|\,\d\gamma_a\\
&\quad+M|\gamma(Q\cap K^N)-\gamma_a(Q\cap K^N)|\\
&\overset{a)}{<}\gamma(Q\cap K^N)\cdot\frac{\varepsilon}{12}+\gamma_a(Q\cap K^N)\cdot\frac{\varepsilon}{12}
+M|\gamma(Q\cap K^N)-\gamma_a(Q\cap K^N)|\\
&=\gamma(Q\cap K^N)\cdot\frac{\varepsilon}{12}+\gamma_a(Q\cap K^N)\cdot\frac{\varepsilon}{12}\\
&\quad+M|\gamma(Q\cap K^N)-\gamma(Q)+\gamma_a(Q)-\gamma_a(Q\cap K^N)|\\
&\le \gamma(Q\cap K^N)\cdot\frac{\varepsilon}{12}+\gamma_a(Q\cap K^N)\cdot\frac{\varepsilon}{12}+M\gamma(Q\setminus K^N)+M\gamma_a(Q\setminus K^N)\,,
\end{align*}
where in a) we have used Condition (\ref{eq:uniform}), and in b) the fact that the measures $\gamma$ and $\gamma_a$ coincide on 'cubes' $Q$.
Summing the estimate above over all cubes $Q=B_{k_1}\times\cdots\times B_{k_N}$, $(k_1,\ldots, k_N)\in M_n^N$, gives
\begin{align}
 \bigg|\int_{K^N} \varphi \,\d\gamma_{1,n} &- \int_{K^N} \varphi \,\d\gamma_{1,n}^a \bigg|\nonumber\\
&< \gamma_{1,n}(K^N)\cdot\frac{\varepsilon}{12}+\gamma_a(K^N)\cdot\frac{\varepsilon}{12}+M\gamma(X^N\setminus K^N)+M\gamma_a(X^N\setminus K^N)\nonumber\\
&\overset{a)}{<}\frac{\varepsilon}{12}+\frac{\varepsilon}{12}+\frac{\varepsilon}{12}+\frac{\varepsilon}{12}=\frac{\varepsilon}{3}\,.\label{eq:inKN}
\end{align}
In inequality a) we have used the fact that $\rho(X\setminus K)<\frac{\epsilon}{12MN}$ and, since the marginals of $\gamma_{1,n}$ and $\gamma_{1,n}^a$ are restrictions of $\rho$, we can bound both $\gamma_{1,n}(X^N\setminus K^N)$ and $\gamma_{1,n}^a(X^N\setminus K^N)$ by $\frac{\varepsilon}{12M}$. For the same reason, we have
\begin{equation}\label{eq:outofKN}
\left\lvert\int_{X^N\setminus K^N}\varphi\,\d\gamma_{1,n}-\int_{X^N\setminus K^N}\varphi\,\d\gamma_{1,n}^a\right\rvert <2M\cdot\frac{\varepsilon}{12M}=\frac{\varepsilon}{6}\,.
\end{equation}
Combining estimates (\ref{eq:inKN}) and (\ref{eq:outofKN}) gives
\[\left\lvert\int_{X^N}\varphi\,\d\gamma_{1,n}-\int_{X^N}\varphi\,\d\gamma_{1,n}^a\right\rvert<\frac{\varepsilon}{3}+\frac{\varepsilon}{6}=\frac{3\varepsilon}{6}<\frac{5\varepsilon}{6}\,,\]
proving Condition (\ref{eq:core}).
\end{proof}

\subsection{Convergence of the cost functional}

 In order to prove the $\Gamma$-limsup inequality (II), we need the cost $C_0[\cdot]$ to converge along the approximating sequence $\gamma_n$. We prove this in the following lemma.

\begin{lemma}\label{lma:costconvergence}
We have $C_0[\gamma_n'] \to C_0[\gamma]$ as $n \to \infty$.
\end{lemma}

\begin{proof}
 Let us first consider the remainder part. Recall that for all $n \in \mathbb{N}$ we have
 \[
  (\gamma_n'-\gamma_{1,n}^a)(X^N) =  (\gamma_{2,n}^a + \gamma_{3,n}^a + \gamma_{4,n}^a)(X^N) < 3\varepsilon_n.
 \]
 Thus, using the lower bounds \eqref{eq:remainderoffdiagonal} and \eqref{eq:remainderoffdiagonal2} for distances in the support of the remainder part, and the definition \eqref{eq:epsilondefinition} of $\varepsilon_n$, we get
 \begin{equation}\label{eq:remaindercost}
  \int_{X^N} c\,\d (\gamma_n'-\gamma_{1,n}^a) \le \frac{N(N-1)}{2}f\left(\frac{2r_n}{5}\right) 3\varepsilon_n \le \frac{3N(N-1)}{2}r_n \to 0
 \end{equation}
 as $n \to \infty$.
 By continuity of the integral, we get
 \begin{equation}\label{eq:remaindercostorginalpart}
  \int_{X^N} c\,\d (\gamma -\gamma_{1,n}) \to 0
 \end{equation}
 as $n \to \infty$.

 Let us now estimate the core part of the approximation. By the construction \eqref{gammaconv:gamma2na} of $\gamma_{1,n}^a$
 and the choice \eqref{eq:lambdachoice} of $\lambda_n$, we have
 \begin{equation}\label{eq:corecost}
  \left| \int_{X^N} c\,\d\gamma_{1,n}^a - \int_{X^N} c\,\d\gamma_{1,n}\right| \le
   \int_{X^N} \frac{N(N-1)}{2} \varepsilon_n\,\d\gamma_{1,n} < \frac{N(N-1)}{2} \varepsilon_n.
 \end{equation}

 Combining the above estimate with \eqref{eq:remaindercost}, \eqref{eq:remaindercostorginalpart} and \eqref{eq:corecost} we get
 \begin{align*}
  \left|C_0[\gamma_n'] - C_0[\gamma] \right| \le & \left| \int_{X^N} c\,\d\gamma_{1,n}^a - \int_{X^N} c\,\d\gamma_{1,n}\right | +
  \left|\int_{X^N} c\,\d (\gamma_n'-\gamma_{1,n}^a) \right|\\
  & + \left|\int_{X^N} c\,\d (\gamma -\gamma_{1,n}) \right| \to 0
 \end{align*}
 as $n \to \infty$.
\end{proof}

\subsection{Finiteness of the entropy for the approximations}

Next we show that the entropy is finite for the approximating sequence. Notice that, in order to prove (II), we do not need a better estimate on the entropy.

\begin{lemma}\label{lma:finiteentropy}
 For each $n\in \mathbb{N}$ we have $E[\gamma_n'] < \infty$.
\end{lemma}
\begin{proof}
 In order to see the finiteness of the entropy, it suffices to notice that each $\gamma_n'$ is a sum of finitely many measures $(\tilde\gamma_{n,k})_{k=1}^{N_n}$ each of which
 is of the form $\tilde\gamma_{n,k} = \tilde\rho_1^k\mathfrak m \otimes \cdots \otimes \tilde\rho_N^k\mathfrak m$ with $\tilde\rho_i^k \ll \rho$ and $\frac{\d\tilde\rho_i^k}{\d\rho} \le 1$.
Indeed, by Proposition \ref{ent:min}, the entropy is always bounded from below, and so we can make a crude estimate:
 \begin{align*}
  E[\gamma_n'] &= \int_{X^N} \log\left(\sum_{k=1}^{N_n}\frac{\d\tilde\gamma_{n,k}}{\d\mathfrak m}\right)\,\d\left(\sum_{k=1}^{N_n} \tilde\gamma_{n,k}\right) \\
  &\le \log(N_n) + \sum_{k=1}^{N_n}
  \int_{X^N}\log\left(\frac{\d\tilde\gamma_{n,k}}{\d\mathfrak m}\right)\,\d \tilde\gamma_{n,k} < \infty.\qedhere
 \end{align*}
 \end{proof}

\subsection{Proof of condition (II)}

We are now ready to prove the $\Gamma$-$\limsup$ inequality (II). By Lemma \ref{lma:weakconvergence} we already know that $(\gamma_n')_n$ converges to $\gamma$.
However, $\mathcal C_n[\gamma_n']$ need not converge to $\mathcal C[\gamma]$. This can be solved by making the convergence of $(\gamma_n')_n$ slower by repeating always the same measure for sufficiently (but finitely) many times before moving to the next one. We define $k(n)$ for every $n \in \mathbb{N}$ as
\[
k(n) = \min\left(n,\max\left(1, \sup\left\{k\in \mathbb{N}\,|\,\sqrt{\tau_n} E[\gamma_j'] < 1 \text{ for all }j \le k\right\}\right)\right).
\]
By definition, $1 \le k(n) \le n$. Moreover, since for every $j\in\mathbb{N}$ we have $E[\gamma_j'] < \infty$ by Lemma \ref{lma:finiteentropy} and  $\tau_n \to 0$ by definition, we have that $k(n) \to \infty$ as $n\to \infty$. Thus, defining $\gamma_n = \gamma_{k(n)}'$, for large enough $n\in \mathbb{N}$ we have
\[
\mathcal C_n[\gamma_n] = C_0[\gamma_{k(n)}'] + \tau_nE[\gamma_{k(n)}'] < C_0[\gamma_{k(n)}'] + \sqrt{\tau_n}.
\]
Recalling that by Lemma \ref{lma:costconvergence} we have $C_0[\gamma_{k(n)}'] \to C_0[\gamma]$, we conclude the proof.

In Proposition (\ref{prop:minexists}) the existence of a minimizer for the entropy-regularized cost was established. Now that we know that measures $\gamma$ for which $C_0(\gamma)<\infty$ can be approximated by measures with not only finite costs but also finite entropy, we can say more:
\begin{corollary}
Let $(X,d,\mathfrak m)$ be a Polish metric measure space. Assume that $\rho \mathfrak m\in \mathcal{P}^{ac}(X)$ satisfies $\rho\log\rho \in L_\mathfrak m^1(X)$ and Conditions (A) and (B). Assume that $c\colon X^N\to\mathbb{R}\cup\lbrace +\infty\rbrace$ satisfies  Conditions $(F1)$ and $(F2)$. Then, for each $\varepsilon> 0$, there exists a unique minimizer $\gamma\in \Pi^{\mathrm{sym}}_N(\rho)$ for the entropic-regularized cost $C_{\varepsilon}[\gamma]$, and this minimizer has a finite cost.
\end{corollary}
\begin{proof}
Our marginal measure satisfies Conditions (A) and (B), so there exists a measure $\gamma\in\Pi^{\mathrm{sym}}_N(\rho)$ that minimizes $C_0$ with $C_0[\gamma]<\infty$. It must be noted that this measure can have infinite entropy. However, because of the approximation result presented in the proof of Condition (II) above, we get the existence of a measure $\gamma'\in \Pi^{\mathrm{sym}}_N(\rho)$ such that $C_\epsilon[\gamma']<\infty$. The uniqueness claim now follows, since the functional $\gamma\mapsto C_\epsilon[\gamma]$ is strictly convex for $\epsilon>0$.
\end{proof}

\section{Entropic-Kantorovich Duality for Coulomb-type costs}\label{sec:duality}

We start by recalling the classical Fenchel-Rockafellar Theorem and we refer to the I. Ekeland and R. T\' emam's book \cite[Theorem 4.2]{EkeTem} for a more complete presentation and references.

\begin{theorem}[Fenchel-Rockafellar]\label{dual:FenRocthm}
Let $\mathcal X$ and $\mathcal Y$ be Banach spaces and $A:\mathcal X \to \mathcal Y$ be linear and continuous. Let $F:\mathcal X\to \mathbb{R}\cup \lbrace +\infty\rbrace$ and $G:\mathcal Y\to\mathbb{R}\cup\lbrace +\infty \rbrace$ be proper and convex functions. Then
$$ \inf \big\lbrace F[x] + G[Ax] ~\big|~ x \in \mathcal X\big\rbrace  = \sup \big\lbrace -F^{*}[-A^* \gamma] - G^{*}[\gamma] ~\big|~ \gamma\in \mathcal Y^* \big\rbrace $$
where $A^*:\mathcal Y^*\to \mathcal X^{*}$ denotes the adjoint operator of A.
\end{theorem}

Next we prove the Entropic-Kantorovich duality for the problem \eqref{ent:min}.

\begin{theorem}[Entropic Duality for repulsive costs]\label{kanto:dualthm}
Let $(X,d,\mathfrak m)$ be a Polish measure space. Suppose $\rho\mathfrak m\in \mathcal{P}(X)$ such that $(A)$ and $(B)$ hold and $\rho\log\rho\in L_m^1(X)$, and $c\colon X^N\to[0,\infty]$ is a cost function
\[
c(x_1,\ldots, x_N)=\sum_{1\le i<j\le N}f(d(x_i,x_j)),\quad \text{ for all } \, (x_1,\ldots,x_N)\in X^N,
\]
where $f\colon [0,+\infty[\to [0,\infty]$ is a function satisfying $(F1)$ and $(F2)$. Then, for $\varepsilon> 0$, the duality holds
\begin{align*}
\min_{\gamma\in\Pi(\rho)}C_\varepsilon[\gamma]&=
\sup_{u_i\in C_b(X)}\left\{\sum_{i=1}^N\int_Xu_i\d\rho\mathfrak m-\varepsilon\int_{X^N}\exp\left(\frac{u_1\oplus\cdots\oplus u_N-c}{\varepsilon}\right)\,\d \mathfrak m_{N}\right\}+\varepsilon\\
&=\sup_{u\in C_b(X)}\left\{N\int_Xu\d\rho\mathfrak m-\varepsilon\int_{X^N}\exp\left(\frac{u\oplus\cdots\oplus u-c}{\varepsilon}\right)\,\d \mathfrak m_{N}\right\}+\varepsilon\,,
\end{align*}
where $v_1\oplus\dots \oplus v_N$ denotes the operator $(v_1\oplus\dots \oplus v_N)(x_1,\dots, x_N) = v_1(x_1)+ \dots + v_N(x_N)$.

\end{theorem}
\begin{proof}
First let us assume that $X$ is a compact space. We denote by $\mathcal X = (C_b(X))^N$ and $\mathcal Y=C_b(X^N)$, where $C_b(X)$ is the space of continuous and bounded functions on $X$, and similarly for $X^N$.
By Riesz representation theorem, the space $\mathcal Y$ is dual to the space $\mathcal M(X^N)$ of signed regular Borel measures on $X^N$.
Thus we may define the Legendre-Fenchel transform $G^{*}$ of a functional $G\colon \mathcal Y\to \mathbb{R}\cup\lbrace +\infty\rbrace$ by

\[
G^{*}\colon\mathcal M(X^N)\to\mathbb{R}\cup\lbrace +\infty\rbrace, \quad G^{*}[\pi] = \sup_{\psi \in C_b(X^N)} \bigg\lbrace \int_{X^N}\psi \,\d\pi - G[\psi] \bigg\rbrace.
\]

We apply Fenchel-Rockafellar Theorem \ref{dual:FenRocthm} to the functionals

\[
F\colon (C_b(X))^N\to\mathbb{R}\cup\{+\infty\}\,,~~(u_1,\ldots, u_N)\mapsto -\sum_{i=1}^N\int_Xu_i\,\d\rho\mathfrak m
\]
and
\[
G\colon C_b(X^N)\to\mathbb{R}\cup\{+\infty\}\,,~~\psi\mapsto \varepsilon\int_{X^N}e^{\tfrac{1}{\varepsilon}(\psi-c)}\,\d \mathfrak m_{N}\,,
\]
and to the operator
\[A\colon (C_b(X))^N\to C_b(X^N)\,,~~(u_1,\ldots, u_N)\mapsto u_1\oplus\cdots\oplus u_N\,.\]

Now $F$ and $G$ are proper and convex functionals and $A$ is a linear and continuous operator. So we may apply Fenchel-Rockafellar duality to get
\[\inf\{F[x]+G[Ax]~|~x\in \mathcal X\}=\sup\{-G^*[\gamma]-F^*[-A^\dag \gamma]~|~\gamma\in \mathcal Y^*\}\,.\]
This gives (since for every set $S$ we have $\inf(S)=-\sup(-S)$ and $\sup S=-\inf (-S)$)
\[\inf\{G^*[\gamma]+F^*[-A^\dag \gamma]~|~\gamma\in \mathcal Y^*\}=\sup\{-F[x]-G[Ax]~|~x\in \mathcal X\}\,.\]
It remains to show that the above expression has exactly the form of our duality claim. The claim that the right-hand sides correspond to each other follows immediately from our choices of $\mathcal X$, $F$, and $G$.  So it remains to show that
\begin{equation}\label{eq:frclaim}
\inf\{G^*[\gamma]+F^*[-A^\dag \gamma]~|~\gamma\in \mathcal Y^*\}=C_\varepsilon[\gamma]-\varepsilon\,.
\end{equation}

To prove it, let $\gamma \in \mathcal M(X^N)$. Now we have
\begin{align*}
F^*[-A^*\gamma] &= \sup \bigg\lbrace \int_{X^N}\sum^N_{i=1}u_i(x_i)d\gamma - \sum^N_{i=1}\int_{X}u_i(x_i)\,\d\rho\mathfrak m(x_i) ~\bigg|~ (u_1,\dots,u_N)\in C_b(X)^N \bigg\rbrace \\
&=\begin{cases}0 &\text{ if }\gamma\in \Pi_N(\rho)\\
+\infty&\text{ otherwise}\end{cases} \,.
\end{align*}
\noindent
Let us then compute $G^*[\gamma]$:
\[G^*[\gamma]=\sup_{\psi \in C_b(X^N)} \left\{ \int_{X^N}\psi \,\d\gamma - \varepsilon\int_{X^N}e^{\tfrac{1}{\varepsilon}(\psi-c)}\,\d \mathfrak m_{N} \right\}\,.\]
If $\gamma$ is not absolutely continuous with respect to $\mathfrak m_{N}$ we have $G^*[\gamma]=+\infty$. If $\gamma\ll \mathfrak m_{N}$,
then the supremum (that appears in the definition of $G^{*}[\gamma]$) is realized at $\psi=\varepsilon\log\rho_\gamma+c$; this holds also if the function $\rho_\gamma$ is not continuous since it can be approximated by a sequence of continuous functions. Thus we get for $\gamma \ll \mathfrak m_{N}$
\begin{align*}
G^*[\gamma]&=\int_{X^N}\left(\rho_\gamma\cdot\psi -\varepsilon e^{\frac{1}{\varepsilon}(\psi-c)}\right)\,\d \mathfrak m_{N}\\
&=\int_{X^N}(\varepsilon\rho_\gamma\log\rho_\gamma+c\rho_\gamma-\varepsilon\rho_\gamma)\,\d \mathfrak m_{N}\,.
\end{align*}
Hence, if $\gamma\in \Pi_N(\rho)$, we have
\[G^*[\gamma]=C_0[\gamma]+\varepsilon E[\gamma]-\varepsilon\,.\]
This concludes the duality proof when $X$ is a compact space.\\

\noindent
\textbf{The noncompact case: }Due to Lemma \ref{lm:changeofreference}, it suffices to prove the claim in the case where the reference measure is $\rho\mathfrak m$ instead of $\mathfrak m$; the finiteness of the measure $\rho\mathfrak m$ now gives access to inner regularity and to the approximability by compact sets. We will for simplicity denote $\rho :=\rho\mathfrak m$. \\\\ The claim is
\[\min_{\gamma\in \Pi_N(\rho)}C_\varepsilon[\gamma]=\sup_{u\in C_b(X)}\left\{N\int_Xu\d\rho-\varepsilon\int_{X^N}\exp\left(\frac{u\oplus\cdots\oplus u-c}{\varepsilon}\right)\,\d \rho^{\otimes N}\right\}+\varepsilon\,.\]

For simplicity, let us denote
\[D_\rho(u):=\left\{N\int_Xu\d\rho-\varepsilon\int_{X^N}\exp\left(\frac{u\oplus\cdots\oplus u-c}{\varepsilon}\right)\,\d \rho^{\otimes N}\right\}+\varepsilon~~~\text{for all  }u\in C_b(X)\,.\]
We may assume that $\sup_{u\in C_b(X)}D_\rho(u)>-\infty$; indeed, since we can test with  the function $u\equiv 0$, this always holds for cost functions that are bounded from below.

Let us make, in the notation of the primal functional, the dependence on the reference measure explicit by the notation $\gamma\mapsto C_\epsilon(\gamma |\mu)$ when the  reference measure on the space $X$ is $\mu$. Thus the original notation $\gamma\mapsto C_\epsilon[\gamma]$ corresponds to $\gamma\mapsto C_\epsilon(\gamma |\mathfrak m)$.

Since the measures $\rho$ and $\gamma$ are inner regular, there exists a sequence $(K_n)_{n\in\mathbb{N}}$ of compact subsets of $X$ such that
\[\rho(K_n)\to \rho(X)\qquad\text{and}\qquad\gamma(K_n^N)\to \gamma(X^N)\,.\]
Let us denote $\gamma_n:=\frac{1}{\gamma(K_n^N)}\gamma\restr{K_n^N}$ and $\rho_n:=\frac{1}{\rho(K_n)}\rho\restr{K_n}$. Let us also denote by $\gamma_n^{\text{min}}$ the minimizer of the problem $I_\epsilon(\rho_n)$. Since $\gamma$ is the minimizer of the problem $I_\epsilon(\rho)$ and since (due to the absolute continuity of the integral and the continuity of the function $t\mapsto t\log t$)
\begin{equation}\label{eq:abscontinuity}
\lim_{n\to\infty}|C_\epsilon(\gamma_n|\rho_n)- C_\epsilon(\gamma|\rho)|=0\,,
\end{equation}
we have
\begin{equation}\label{eq:minapproaches}
\lim_{n\to\infty}|C_\epsilon(\gamma_n|\rho_n)-C_\epsilon(\gamma_n^{\text{min}}|\rho_n)|=0\,.
\end{equation}
By the duality result proven above for compact spaces, we have for all $n \in\mathbb{N}$
\[\sup_{u\in C_b(K_n)}D_{\rho_n}(u)=C_\epsilon(\gamma_n^{\text{min}}|\rho_n)\,.\]
Again, due to the absolute continuity of the integral, we have \[\lim_{n\to\infty}\left\lvert\sup_{u\in C_b(K_n)}D_{\rho_n}(u)-\sup_{u\in C_b(X)}D_\rho(u)\right\rvert=0\,.\]
Putting these conditions together, we get for all $n\in \mathbb{N}$
\begin{align*}
&\left|C_\epsilon(\gamma|\rho)-\sup_{u\in C_b(X)}D_\rho(u)\right|\\
&\le \left|C_\epsilon(\gamma|\rho)-C_\epsilon(\gamma_n|\rho_n)\right|+\left|C_\epsilon(\gamma_n|\rho_n)-C_\epsilon(\gamma_n^{\text{min}}|\rho_n)\right|\\
&\quad +\left|C_\epsilon(\gamma_n^{\text{min}}|\rho_n)-\sup_{u\in C_b(K_n)}D_{\rho_n}(u)\right|+\left|\sup_{u\in C_b(K_n)}D_{\rho_n}(u)-\sup_{u\in C_b(X)}D_{\rho}(u)\right|\,.
\end{align*}
The claim follows by letting $n\to\infty$.
\end{proof}

\begin{remark}
Notice that we always have
\[
\sup_{u_i\in C_b(X)}\left\{\sum_{i=1}^N\int_Xu_i\,\d\rho-\varepsilon\int_{X^N}\exp\left(\frac{u_1\oplus\cdots\oplus u_N-c}{\varepsilon}\right)\,\d \mathfrak m_{N}\right\}+\varepsilon
\]
\[
\geq \sup_{u\in C_b(X)}\left\{N\int_Xu\,\d\rho-\varepsilon\int_{X^N}\exp\left(\frac{u\oplus\cdots\oplus u-c}{\varepsilon}\right)\,\d \mathfrak m_{N}\right\}+\varepsilon.
\]

We show that the equality actually holds by simply choosing a potential $u(x) = (u_1(x)+u_2(x)+\cdots + u_N(x))/N$.
\end{remark}

\bibliographystyle{siam}

redpajama_set_name: RedPajamaArXiv


Grab a paper bag, you're gonna need something to breathe into in a hot second.
Director Paul Feig, who gifted us with "The Heat" and "Bridesmaids," sent a tweet that we hope is what we think it is. See, Feig tweeted a composite featuring pictures of Melissa McCarthy (Feig's collaborator in both "The Heat" and "Bridesmaids," as well as the upcoming "Spy"), Kristen Wiig ("Bridesmaids") and current "Saturday Night Live" stars Kate McKinnon and Leslie Jones.
Pardon the capslock, but when you need it, you need it: IS THIS THE CAST OF THE ALL-FEMALE "GHOSTBUSTERS" FEIG IS DIRECTING?
Of course, Feig didn't say anything with this picture, because that would just be too easy, wouldn't it?
According to The Hollywood Reporter. the four actresses are nearing deals to star in the reboot.
A few hours later, Feig tweeted the movie's release date: July 22, 2016.
Wiig and McCarthy have both been long rumored for parts in the movie, and McKinnon and Jones, both standouts on "SNL" (please give McKinnon a chance to use her "Weekend Update" Olya Pavlotsky character, please), would be fantastic finishing pieces to the puzzle.
On top of all of this, we can take this as yet another sign that Bill Murray just knows things the rest of us don't. Back in September, "Ghostbusters" original cast member Murray was asked which women should don the iconic jumpsuits for Feig's reboot.
"Melissa [McCarthy] would be a spectacular Ghostbuster. And Kristen Wiig is so funny — God, she's funny!" he said.
Later, Wiig was told of his recommendation. "I will do whatever Bill Murray tells me to," she said.


text:
News of Record
UDTV Magazine
Sierra Lifestyles Magazine
47° Fog
Sonora, CA (95370)
Considerable cloudiness. Low 33F. Winds light and variable.
Sonora's (2) Emily Van Noord takes a shot at goal during the CIF NorCal Regional Girls Water Polo game between the Sonora High School Wildcats and Tamalpais High School Red Tailed Hawks at Independence High School on November 23, 2019 in San Jose, CA.
Sam Stringer Photography
Sonora's Head Coach Matt Personius looks on as his team competes during the CIF NorCal Regional Girls Water Polo game between the Sonora High School Wildcats and Tamalpais High School Red Tailed Hawks at Independence High School on November 23, 2019 in San Jose, CA.
Sonora's (1) Goalie Liberty Wheeler tries to block a ball during the CIF NorCal Regional Girls Water Polo game between the Sonora High School Wildcats and Tamalpais High School Red Tailed Hawks at Independence High School on November 23, 2019 in San Jose, CA.
Sonora's (8) Megan Slater in action during the CIF NorCal Regional Girls Water Polo game between the Sonora High School Wildcats and Tamalpais High School Red Tailed Hawks at Independence High School on November 23, 2019 in San Jose, CA
Sonora's (2) Emily Van Noord competes during the CIF NorCal Regional Girls Water Polo game between the Sonora High School Wildcats and Tamalpais High School Red Tailed Hawks at Independence High School on November 23, 2019 in San Jose, CA. (
The Sonora Wild Cats team huddles up after losing a close game during the CIF NorCal Regional Girls Water Polo game between the Sonora High School Wildcats and Tamalpais High School Red Tailed Hawks at Independence High School on November 23, 2019 in San Jose, CA.
Sonora girls water polo team falls to Tamalpais
Guy McCarthy
The girls varsity water polo team from Sonora High School fought to the end and fell 15-14 to Tamalpais in a thrilling NorCal Division III Championship final Saturday morning in San Jose.
The game was tied at the end of each of the first three quarters, 1-1, 5-5, and then 13-13. Sonora held the lead three times in the second period and fought back from a three-goal deficit in the third period.
Trailing 6-9 with 5 minutes 20 seconds left in the third period, the Wildcats battled back with scores by Emily Van Noord, Megan Slater, Van Noord again, and Slater twice more to make it 12-12 with 34 seconds left in the third.
Van Noord scored again to make it 13-12. That was the last time Sonora led in the match. The Tamalpais Hawks answered to tie it again at 13-13 to end the third period. The Hawks added two more goals to open scoring in the fourth period.
Slater scored the last goal of the match with 4 minutes 4 seconds left in the fourth period to make it Tamalpais 15 Sonora 14.
Other goal scorers Saturday for Sonora included Kayla Landman and Sofia Alomia.
"It was a crazy game," Sonora girls varsity head coach Matt Personius said after the match. "I don't think you could ask for a better championship final. Both teams had a lot of parity."
The outcome of the match came down to small differences, who made a critical error at a critical time, who finished with a critical shot when their team needed it, Personius said. Little things made a difference.
"I'm very happy with how our team played today," Personius said. "I want to give shout outs to Megan Salter and Emily Van Noord, who were both unstoppable. The only ones who could stop them were themselves. The opposing defenders couldn't stop them.
"I don't have one single complaint about my team's performance today. I could not be more proud of the way they played and performed and comported themselves," Personius said.
In the moments immediately after the match ended, it was a mix of emotions for the Sonora Wildcats. Personius said his players had sadness about the finality of the match.
"When you win, you don't think about the finality of it so much," Personius said. "I think as soon as it was over I think the girls had a sense of sorrow, a sense of loss for the opportunity to stay together and be together as a team, and real disappointment at feeling like this is a game they might well have won. To be honest, I think we all felt we were good enough to win this game on most days. It didn't go that way this day."
Personius paid tribute to the seniors who played their final match for Sonora on Saturday. They are Taetum Arnett, Brielle Estwin, Kayla Landman, Megan Ronneburg, Emily Van Noord, and goalkeeper Liberty Wheeler.
The NorCal Division III Championship final was played at Independence High School in San Jose.
Tamalpais High School from Mill Valley in Marin County are CIF NorCal Division III champions. The Sonora Wildcats are runners-up in the regional Northern California final. It was the farthest the Wildcats had ever advanced in CIF NorCal competition.
The 2019 Wildcats end their season as three-time CIF Sac-Joaquin Section Division III champions, having secured the section title by defeating Kimball 16-7 a week ago in Roseville.
Contact Guy McCarthy at gmccarthy@uniondemocrat.com or 588-4585. Follow him on Twitter at @GuyMcCarthy.
Tamalpais High School
Sonora High School
Independence High School In San Jose
Storm brings rain to foothills, snow down to 2,000 feet
Documents offer new details on investigation into death of Waterford toddler
Jumpers 10th in the state
Investigator: kerosene poured outside bathroom minutes before fatal house fire
Man on felony probation arrested for tampering with vehicles, public intoxication
KFC and Wendy's construction moving along
Death notices for Thursday, Jan. 16, 2020
Linoberg park takes shape
Diane Anderson released from jail
Sonora man identified as a firefighter arrested for underage sex crimes
"Karl, get the kids," mother, trapped in a house fire, screamed
Police searching for man accused of robbing Walmart at gunpoint
Family fights for justice in toddler death
Two bodies found inside houseboat at Lake McClure
Christina Karlsen died of smoke inhalation
Death notices for Saturday, Jan. 11, 2010
Death notice for Tuesday, Jan. 14, 2020
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text:
User Manual Extract
Orthopaedic Product News Article
Ross Wark Medical Ltd
Orthopaedic Product News Article June 2018 Breaking the mould in cast removal
Consultant orthopaedic surgeon David Ross encountered many problems when removing casts with oscillating cast saws, so he designed and developed the CasterpillarTM, a completely new device for cast removal, which would be quieter and safer for both the patient and the operator
The majority of upper and lower limb fractures are still treated by application of a cast. Plaster of Paris is often used for the initial cast and this is relatively easy to remove using plaster shears or scissors, cutting the padding of a simple back-slab. Lightweight synthetic casting materials have been developed for longer-term fracture immobilisation and are removed after the fracture has healed, usually after several weeks.
These synthetic casts have many advantages, they are lighter and stronger than Plaster of Paris and have good wear and water resistance. Due to the strength of synthetic casts they currently have to be removed using a high-speed oscillating saw. The problems associated with oscillating cast saws are well documented; they are noisy and can be terrifying, particularly for young children. The noise prevents good communication between the clinician and the patient and can be so loud that it affects other patients in adjacent rooms or cubicles.
Small quadrant shaped saw blades fitted to small saws designed for paediatric use can also become hot very quickly; larger circular blades on standard saws allow more heat diffusion and can be rotated to bring a relatively cooler area into use. Linear skin burns caused by a hot saw blade, most likely to occur when the saw is used by an inexperienced operator, are not common but distressing and probably medico-legally indefensible. A child can become very distressed when having a cast removed and it may not be possible for the clinician to distinguish between the anxiety caused by the appearance and noise of the saw, and actual pain.
The hands of the person operating the high speed oscillating saw are likely to be exposed to a significant level of vibration. Vibration exposure has been linked to hand arm vibration syndrome (HAVS or vibration white finger) and carpal tunnel syndrome. The dust produced by the saw has to be collected by a large and noisy vacuum extractor as this dust may be harmful if inhaled. All of these problems are well recorded in the medical scientific literature.
While working as an orthopaedic surgeon in Stirling, I contacted Scottish Enterprise who introduced me to George Miller, an electronics engineer. We formed Ross Wark Medical Ltd, to design and develop a completely new device for cast removal, which would be quieter and safer for both the patient and the operator. "Smart Prototyping" techniques, including the use of CAD-CAM (computer aided design and manufacture) and 3D printing were employed, and in 2014 CasterpillarTM was launched.
This cast cutter was a major innovation: a powered shear rather than a saw. Casterpillar is quiet, produces little vibration and virtually no dust. It simultaneously cuts through the synthetic cast material, the stockinet and under-cast padding and is therefore probably quicker at removing a cast than a cast saw. Casterpillar is quiet and does not look like a circular saw, so is unlikely to produce an anxiety reaction in patients. Casterpillar is likely to be more economical than a cast saw as regular replacement of blades, vacuum filters and tubing will not be required. Since 2015 a number of the first version of Casterpillar have been in use in hospitals in Scotland and England, with very positive feedback.
Since retiring from clinical practice, I have been able to concentrate on the further development of Casterpillar, which now has UK, US, Canadian and German patents. With help from Business Gateway we were able to apply successfully for funding from Scottish Enterprise to improve the Casterpillar Cast Cutter. Electronic technological advances allowed the use of a smaller, lighter, but more powerful motor and we were able to simplify the control switches and low voltage power supply unit.
A second generation Casterpillar Cast Cutter has been developed which is lighter, easier to use and has a better cutting performance.
We do not anticipate that Casterpillar will be able to replace every cast saw; cast saws will still be required for very thick casts, such as a long-leg cast for a large adult. However, we hope that most patients, particularly children and the frail elderly, will benefit from having casts removed quietly and safely by a Casterpillar.
The full article can be down loaded by clicking on the link below....
OPN_185_proof-David May 18.pdf
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Full service on site catering from extensive & diverse preplanned menus or customized to your individual needs & desires. Please call for information & quotes for your event.
Planning a Wedding, Anniversary Party, Hail and Farewell, Prom, Birthday Party or a Company Party? Come in, tour the venue to see our beautiful private rooms of all sizes great for small parties and large. Your personal experienced banquet director will guide you through the event planning process making your event a success using one or all of a variety of banquet options. You may choose one or all of nine intimate dining and meeting rooms.
Only a few minutes drive west of Colorado Springs, Briarhurst Manor is located in the tiny valley of Manitou Springs. The stately Victorian manor house was built in 1872 by the founder of Manitou Springs, Dr. William Bell of London, England. Our restaurant seats up to 500 guests and features casual yet elegant dining in nine intimate rooms, an extensive menu and wine list complete with five course dinners featuring domestic and game meats, fresh seafood, pastas, poultry, specialty salads, vegetarian, and handcrafted desserts. The finely grained pink sandstone Tudor manor displays the architecture and landscaping of an English country house complete with the bubbling Fountain Creek passing through the estate. There is also an unrivaled view of Pike's Peak.
It was great customer service.
Excellent food and atmosphere. Beautiful and romantic setting.
The chef and waitress were amazing. There was nothing we could have asked for that they wouldn't do. Just an overall amazing night.
Overall a very good experience and would definitely go back.
Amazing evening! Food and service were exceptional. Will go back and take friends!
This is one of those places that you don't want to let too many people know that a dining gem is around because you're afraid that the place will get packed and reservations will be hard to get!
We had not dined here for a good number of years as we live out of town. We were fortunate to spend an evening in Manitou Springs and said, "Let's go to the Briarwood".
Dinner was WONDERFUL! From the graceful manor house to the gentleman who greeted us, Christian who waited on us, the food was INCREDIBLE right along with Christian who gave us his expert advice from cocktails to wine pairing to dinner entree selections and finally, that unbelievable dessert! My husband appreciated the 15 year old scotch instead of the normal 12 year that most restaurants have. It was smoother and more fragrant. The appetizer of escargot melted in your mouth. The salad was a nice fresh take on the usual. The boar and roasted vegetables were so tender and flavorful...yum! The dessert was so decadent that my husband is still talking about it! Just a wonderful, wonderful eating and dining experience! Shhh...don't tell anyone!
Our waiter was so kind and attentive! And the food was great, especially the soup & filet!
Once again, the preparation and presentation was excellent! The chefs are so confident (And rightly so) in their skills and personal pallete, you will find neither salt nor pepper on your table. If you try your meal as prepared, without adding anything to it, you may be delightfully surprised at how flavorful it is.
Excellent service and food. We had an enjoyable evening.
The five course meal turns an everyday dinner into a special occasion. The lobster bisque is the best in the city. I had the duck which had a perfect seat, and my wife had the boar which was rich and delicious. I was disappointed we missed the murder mystery night so we'll have to make sure to keep an eye open for the next one because that sounds like a blast.
Food was wonderful, service was great, and the bill was nowhere near as high as I expected.
This place was elegant to say the least. The ambiance and faint music while sitting in a cozy nook made the experience all that much more romantic. Our coats were checked at the front. Our server Christian was well versed on the history of the Manor. He also knew everything about the meal and how it is prepared. I had the Elk and stuffed sausage. The flavor was acute with every bite. We will be going back for date night more often.
Excellent food and ambiance but expect to pay a C note per person if you get drinks and the five course meal. A great spot for that occasional spurge. I will return.
The place is really neat, lot of history. We were sat and you could hear a pin drop. People were whispering at their tables because it was so quiet, awkwardly quiet, no music or anything. They finally had a violinist play a few songs, but that only lasted about an hour. My salad was good, the lobster bisque was good, but the entrees and dessert were sub par. The waiter pushed the $18 add on to all of our meals saying it was more cost effective. He was very nice, but service was extremely slow. The tables around us had their orders and drinks messed up the whole time we were there. Not sure if it's new management, new staff or what, but we paid a lot of money for an awful meal and experience. I heard it wasn't always this way so hopefully they get their stuff together.
Food defiantly just was not worth the money that you pay.
This a a very romantic place for a nice quiet dinner. I am not sure what others have said about the dress code but there were plenty of people dressed casually, blue jeans etc. The food was great, the service was amazing and learning the history was cool too. We will definitely eat there again but probably only for special occasions. The price is at the top end.
As usual, we were excited about the evening as we have visited the Manor before. As expected, we we greeted warmly and sat in a very quiet corner among several couples enjoying the evening as well. Our server was delightfully attentive so we expected the true Briarhurst hospitality we had experienced several times before. We started with the Bisque, which was and has always, been incredible. In addition, we ordered the cheese plate as an appetizer, with the Wild Boar and Fillet as entrees, and finishing with the cheesecake and creme brulee. Now, I understand that serving times may have been a little longer due to the occasion and ambiance being portrayed, but they were longer than expected. Up to this point, all is well until we see a couple with a baby being sat by us. Now, don't get us wrong. We have kids and we love to take them out to eat as well but... it's a Special Valentine's Day Dinner specifically advertised for couples. The entire reason for coming to this venue was to have COUPLE time WITHOUT the kids. If I had wanted Family style surroundings for this special day, I'd have gone to Texas Roadhouse! And, for the record, by no means is the couple to be faulted. They were simply looking for a rightfully deserved moment for themselves, as well! However, this changed the mood and celebrating became more of a dash to finish our dinner before having to experience the inevitable natural baby things. Unfortunately, the fillet was overdone, med-well over the med-rare requested, so the wife and I ate the Boar while we waited. When her dish arrived, the meat was cooked well enough, but the root vegetables were raw. We didn't have much luck with Creme Brulee either as they forgot to fire the candy sugar crust and gave us wet custard instead.
When I spoke with Ken Healey, GM, regarding the experience, he did offer to make it up but as we are not in the area, it was not possible. A month later, I'm still waiting for some semblance of an alternate "make it up".
We have been coming here for years and this time was horrible! It took 3 hours for us to even get to the 4th course and a table of 4 got their 2nd entree before us who had reservations 30 min after us! Complained to the manager and he gave us excuse after excuse and we asked for our entree and dessert to go! He gave us a discount on the bill which didn't make up for our ruined valentines dinner!
The service was great. Food was amazing. People were friendly. Building was fantastic.
Probably the best restaurant in town. Finest dining. Beautiful dining rooms. Pretty perfect service.
Excellent food in a romantic environment, service, food, and staff was great!
Delicious food. The team was even able to work with my Celiac disease and I had a great dining experience.
dining room was cold. Was informed that the lobster appetizer was not available at 6:30 pm; restaurant opens at 5:00 pm; then after inquiring with our server, was told that the appetizer could be served for $2.00 extra because the menu's were changing the next day. I found the bread service to be a joke. The duck entree was wonderful, however the skin on the duck should be crispy not mushy. Had to send it back for further cooking.
We have visited Briarhurst Manor a number of times now and have always had a wonderful experience. The restaurant has such amazing character and charm with a number of unique dining rooms. We just love the place. Feels like a little bit of English Manor House tucked into Manitou Springs. Great staff! They are friendly, professional, but not stuffy.
We have been coming here for years on our anniversary with the exception of this year everything was great. This year was very unacceptable. My wife's mushrooms were not washed and she ended up with a mouth full of sand. The Crème brûlée was cold with and the crust was soggy. Overall this was the worst experience we have ever had here. I'm not sure if there has been a change in management or what but if things do t get fixed we will not be returning.
We have been here before for special occasions with a Groupon. This time, went for my birthday and decided to get the Steak and Lobster special for $19.95. We decided to do the 5 course, which added $18. Well worth it. Very good food, the steak was rare as asked for, and I recommend the creme brulee for dessert. If you aren't really hungry, just get the Steak and Lobster for $19.95. Plenty of food!! Our waiter was wonderful, attentive and thought of everything. Great place for a romantic evening.
We had a wonderful Christmas Eve dinner at Briarhust!
We went there for our 25th wedding anniversary because we wanted something extra special. We were not disappointed. The food was excellent, probably the best Wellington I've ever had, and we loved the desserts, especially the s'mores bomb. My husband's filet was also perfect. The service was wonderful, and the ambiance was terrific. We will definitely go back for other special occasions, although it's a little pricey for every day. And we would recommend Briarhurst for anyone who wants a memorable meal.
Food is very good, although portions are small. The furniture (tables, chairs, etc.) need some attention. Continues to be charming, however without some top to bottom attention, the aging facilities will eventually cancel out the notable ambiance.
The Briarhurst hits another grand slam! Everything including the service from Tyler was simply impeccable. Jaime (the head host) has this place humming. The chef is so creative. I hate salmon and he had it in the amuse bouche which I thoroughly enjoyed. The steak Diane was once again incredible and we'll be back for more soon!
Great ambiance and service! Food was tasty and provided a fantastic sample of Rocky Mountain cuisine. Food presentation was beautiful and I was really impressed by how healthy many of the dishes were. A perfect place to bring visitors to get a great taste of CO in a gorgeous, unique space.
We had a fabulous five course meal at Briarhurst Manor. The presentation of the food was beautiful and the service outstanding. It's a great place for a special occasion or romantic date. I'd highly recommend!
Wonderful experience. Great food and atmosphere.
We recently moved back to Colorado and the Briarhurst is just as good as it was years ago. The food was fantastic and very wonderful. Our server Corbin was excellent and very knowledgeable about the menu and the wine list. I recommend this restaurant to everyone that is looking for a sophisticated restaurant with an excellent menu, wonderful ambience, and impeccable service. The experience is well worth the cost.
I decided to come here Saturday for my boyfriends birthday by the suggestion of a friend. She told me the food was outstanding and it would be perfect for the date. But that was completely wrong, the food was terrible, the waiter was snobby about us not knowing the correct way to place our silverware to indicate we were done eating. The only good thing about this restaurant was the bread. The house special was elk that night and was recromended. It was bland and dry. I was really disappointed after seeing the great reviews. Maybe it was an off night. Never going the spend that much money for the food again.


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Jack Ward -MI 50, M55
24 July 2020 /in Vale /by anareoffice
Vale Jack Ward –MI 50, M55.
Jack Ward died on Saturday 14th September 2013 aged 92. He would have been 93 on the 1st October.
How to describe Jack…he was quite simply a delight, an absolute gentleman, an astute observer of himself and others. He had a great love of organ music, and appreciation of art and conversation.
Jack had a 'fairly longstanding fascination with the stories of the Antarctic and the explorers and the strangeness of it'…but never imagined he would go South. However in 1949, after seeing several advertisements for the Heard and Macquarie expeditions he '..rang the Antarctic Division and an appointment was made to interview you. In my case, it was in the back of a taxi with Phil Law, who asked a few questions and then said, I think you can assure yourself that you will be going…'
After Macquarie Island Jack worked for several years as a ships radio operator out of England then '…Phil Law wrote me a letter saying he was compiling the membership of the second expedition and he invited me to join. I was tossing up and decided I was enjoying things in England so much I would like another year there and also I was a bit scared whether I was up to the rigors of a mainland expedition so I asked him could he put it off for another year which he kindly agreed to do and I was put on the second year at Mawson…'
On his return from Mawson Jack became the librarian at RMIT for 25 years, before retiring at the age of 60. His reasoning was '… I come from an extremely short lived family, my father was dead at 56, my brother at 52 my sisters at 54 I think my oldest sister made 60 so I decided that if I retired at 60 I would be lucky to get 5 years retirement…'
All those who have known him over the years will be so pleased Jack got that bit very wrong…
There will not be a funeral service as per Jacks wishes.
From Anare Tas Newsletter 19 Sept 2013
https://www.anareclub.org/wp-content/uploads/2020/07/ANARE-Club-logo-Web-1.png 0 0 anareoffice https://www.anareclub.org/wp-content/uploads/2020/07/ANARE-Club-logo-Web-1.png anareoffice2020-07-24 22:22:112020-07-24 22:22:11Jack Ward -MI 50, M55
Peter Neal -M 69 Jannes Keuken- Davis 1959, Anare Club Treasurer


text:
What is spoken text?
Spoken texts include oral stories, interviews, dialogues, monologues (e.g. a welcome to country speech, a presentation to the class), phone conversations, discussions, role plays, or any other piece of spoken language. This is an opportunity for language development.
What are the criteria to have good spoken text?
Answer. Answer: These criteria, pronunciation, vocabulary, accuracy, communication, interaction and fluency are all markers of a student's overall speaking abilities.
What are the basic elements of spoken texts?
Basic structural elements: TEI distinguishes seven structural elements in spoken texts: utterances, pauses, vocal, kinesic, events, writing and shifts.
What are the 3 basic elements of spoken language?
Answer. The three elements of a spoken language are words, grammar, and speech/writing.
What's a good narrative essay?
In a narrative essay, the writer tells a story about a real-life experience. Everyone enjoys a good story—especially one that captures the imagination. However, the narrative essay goes further. In it, the writer places a personal experience within the context of a larger theme, such as a lesson learned.
What is language and its types?
A language is a structured system of communication used by humans, including speech (spoken language), gestures (sign language) and writing. Most languages have a writing system composed of glyphs to inscribe the original sound or gesture and its meaning. The scientific study of language is called linguistics.
What are text elements?
Literary Text: A story about people, animals, or events that is made up by an author.
Characters: The people or animals in a story.
Setting: The time and place, or when and where, a story happens.
Events: The things that happen in a story.
Sequence: The order in which events happen.
What is the difference between spoken and written text?
→ Spoken text is more informal and longer than written text, and uses more words and more sentences. → The first person (I, me, we, etc.) is more acceptable in speech than in most academic writing, and signposting words like 'also', 'next', and 'for example' are used more often than in written work.
How do you write an audio essay?
GUIDE – Writing Process for Audio Essays
STEP 1: ROUGH DRAFT. If you'd like to produce an audio essay, the first step is to write a rough draft the same way you would for any other kind of essay project.
STEP 2: TRIAL RECORDING. After you make further revisions to the transcript, it's time to make a trial recording.
STEP 3: FULL DRAFT.
STEP 4: FINAL VERSION.
What is a personal narrative essay?
A personal narrative essay is about a personal experience, so you should write it in the first person. A personal narrative is a story about yourself, and great personal narrative essay topics include experiences you've had, people you know, your reactions to books or other writing, and many other options.
What is an example of language?
The definition of language is speech or other forms of communication. An example of language is words spoken. An example of language is words read in a book. (countable) A form of communication using words either spoken or gestured with the hands and structured with grammar, often with a writing system.
What is language in your own words?
Language, a system of conventional spoken, manual (signed), or written symbols by means of which human beings, as members of a social group and participants in its culture, express themselves.
What is narration example?
In writing or speech, narration is the process of recounting a sequence of events, real or imagined. It's also called storytelling. For example, if a story is being told by someone insane, lying, or deluded, such as in Edgar Allen Poe's "The Tell-Tale Heart," that narrator would be deemed unreliable.
What is narrative paragraph and examples?
A narrative paragraph tells a story. Something happens first, second, third, etc. That's a good assignment because it allows you to write in the first person and the details in this kind of paragraph are likely to be very vivid. Here's an example: I remember when I first arrived in the United States.
What are the five elements of spoken language?
Linguists have identified five basic components (phonology, morphology, syntax, semantics, and pragmatics) found across languages.
What are the 5 main elements?
Everything in nature is made up of five basic elements: earth, water, fire, air, and space.
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