text:
\begin{document}

\title[Fubini-Tonelli type theorem]{Fubini-Tonelli type theorem for non product measures  in a product space}
\author{ Jorge Salazar}

\address{DMAT, Universidade de \'Evora, \'Evora - Portugal}
\email{salazar@uevora.pt}

\subjclass{ }
\keywords{}

\date{}

\begin{abstract}
I prove a theorem about iterated integrals for non-product measures in a product space. The first task is to show the existence of a family of measures on the second space, indexed by the points on of the first space (outside a negligible set), such that integrating the measures on the index against the first marginal gives back the original measure (see Theorem \ref{teo}). At the end, I give a simple application in Optimal Transport.
\end{abstract}
\maketitle

\section{Introduction}
\noindent
The Fubini-Tonelli theorem states that the integral of a function defined on a product space, against a measure which is a product of  measures on the factor spaces, can be obtained  by iterated  integration, i.e. integrating one variable  (against its marginal measure) at the time.
\vskip 2mm
\noindent
 If a measure on a product space is not a product measure, is it still possible to decompose the measure and evaluate the integral using iterated integration? To better understand the problem, imaging we are dealing with a measure $ \zeta  $ which is absolutely continuous with respect to the product measure $ \mu\otimes\nu $, i.e. there is a function $ \delta: X\times Y\rightarrow \left[ 0, \infty\right] $, such that for all measurable set $C\subseteq X\times Y$,
 \[ \zeta\left( C\right) =\int_{C} \delta(x,y)\,\mu\otimes\nu\left( dx, dy \right) . \]
 \vskip 2mm
 \noindent
 Then, using the classical Fubini-Tonelli theorem, we can decompose this integral into
 \[ \zeta\left( C\right) =\int_X\left( \int_{C_x} \delta(x,y)\, \nu\left( dy \right) \right)  \mu\left( dx \right) ,\]
 where  $C_x :=\left\lbrace y\in Y;\, \left( x,y\right)\in C \right\rbrace  $
  is the slice of $C$ at $X$.
 Accordingly, $\nu$ decomposes into the measures
 \[ \nu_x(dy):=\delta(x,y)\, \nu\left( dy \right) , \]
 which integrates against $\mu$ to give  $\zeta$. Symbolically,
 \[ \zeta\left( dx, dy \right) =\nu_x(dy)\, \mu\left( dx \right) . \]
  With this decomposition the order of integration is not interchangeable, since the first measure depends on the second variable.
 To interchange the order of integration, we must decompose  $\mu$ in a similar way,
  \[ \mu_y(dx):=\delta(x,y)\, \mu\left( dx \right) , \]
  which integrates against $\nu$ to give  $\zeta$. Symbolically,
  \[ \zeta\left( dx, dy \right) =\mu_y(dx)\, \nu\left( dy \right) . \]

\noindent
 In this paper,  the existence of this kind of decomposition is established for arbitrary  Borel probability measures on
 the product of two complete, separable,  locally compact,
 metric spaces  (see Theorem \ref{teo}). I restricted myself  to the case of probability measures to simplify the discourse, although the results stay valid for $\sigma-$finite measures.
\vskip 2mm
\noindent
 In the best of my knowledge, there is nothing of the kind in the literature on foundations of measure theory that describes similar results. Nonetheless, this question is natural and I believe it may provide a useful calculation and/or analytical tool as much as the classical Fubini-Tonelli theorem does.
 \vskip 2mm
 \noindent
 Optimal Transport, for example,  deals with fixed marginal probability measures and a minimal cost is seek  among all the couplings of the given marginal probabilities, i.e. among all the probabilities on the product space, such that the marginal measures are the ones given.
 It would be a nice research project to look for a new characterization of the optimal transport plans in terms of the measures along the ``fibers'', obtained from  the  decomposition described in Theorem \ref{teo}.
In section \ref{OT}, I give a simple application of Theorem \ref{teo}, showing that a pair of competitive price functions, whose integral with respect to some transference plan matches the transport cost, are conjugate to each other almost surely. This complements the Kantorovich duality theorem on the nature, regarding convexity/concavity, of pair of competitive prices maximizing the profit. See Villani's book \cite{vill}, page 70, for a very detailed discussion of the Kantorovich theorem.
  \vskip 2mm
  \noindent
Another interesting project is the application of Theorem \ref{teo} to the study of measures on the Tangent bundle of  Riemannian manifolds. Indeed, the local charts of the tangent bundle are Cartesian products of Euclidean  open sets. Using local charts,   we can transport the measure to this product to be decomposed and then sent back the family of measures fiber-wise. In the literature, the measures on tangent bundles are a kind of product measures, as is the volume  obtained from the  Sasaki \cite{sasaki}  metric, or the measure on the unit sphere on the tangent space, integrated against the volume element of the base manifold. I believe Theorem \ref{teo}  is a tool that could help  exploring general integration on tangent bundles.

\section{Main theorem}

{\thm\label{teo} Let $X\times Y$ be the product of two complete, separable,  locally compact,
	metric spaces. We equip $X$, $Y$, and $X\times Y$  with their Borel $\sigma-$algebras, denoted by $\mathfrak{B}_X$, $\mathfrak{B}_Y$, and $\mathfrak{B}_{X\times Y}$ respectively.
	\vskip 3pt
	\noindent
	Let $\zeta$ be a  probability measure on $\mathfrak{B}_{X\times Y}$, and denote $\mu$ and $\nu$  the marginal probabilities on $\mathfrak{B}_X$, $\mathfrak{B}_Y$ respectively. i.e.
	\[ \forall A\in \mathfrak{B}_X, \ \mu(A)= \zeta(A\times Y) \]
	and
	\[ \forall B\in \mathfrak{B}_Y, \ \nu(B)= \zeta(X\times B) .\]
\vskip 3pt
\noindent
Then, outside an exceptional $\mu-$negligible set $E_1 \in \mathfrak{B}_X$  ($\mu\left( E_1\right) = 0$), for all
	$ x \in X\setminus E_1 $, there is a measure $\nu_x$ defined on $\mathfrak{B}_Y$, such that for all $C\in \mathfrak{B}_{X\times Y}$, the function
\begin{equation}\label{meas0}
	 x \in X\setminus E_1 \longrightarrow \nu_x\left( C_x \right)  ,
	\end{equation}
where $ C_x = C\cap\left(  \left\lbrace x \right\rbrace\times Y \right)   $,	is $\mathfrak{B}_X-$measurable and
\begin{equation}\label{int0}
	 \zeta(C)= \int_X\nu_x\left(C_x \right) \mu(dx).
	\end{equation}
	In particular,
	\[ \forall B\in \mathfrak{B}_Y, \ \nu(B)= \int_{X} \nu_x\left(B \right) \mu(dx)  ,\]
Moreover, for all positive $\mathfrak{B}_{X\times Y}-$measurable function,  $f: X\times Y\rightarrow \mathbb{R}$,
\begin{equation}\label{meas0f}
x \in X\setminus E_1 \longrightarrow \int_Y f(x,y)\, \nu_x\left(dy\right)
\end{equation}
 is $\mathfrak{B}_X-$measurable   and
 \begin{equation}\label{int0f}
	\int_{X\times Y} f(x,y)\, \zeta\left(dx,dy\right)= \int_X \left( \int_Y f(x,y)\, \nu_x\left(dy\right)  \right)  \mu(dx).
	\end{equation}

\vskip 3mm
\noindent
Likewise, there is a  $\nu-$negligible set $E_2  \in \mathfrak{B}_Y$, such that   for every $\,  y\in Y\setminus E_2 $
  there is a measure $\mu_y$  on $\mathfrak{B}_X$,  such that for all $C\in \mathfrak{B}_{X\times Y}$, the function
\[ y\in Y\setminus E_2 \longrightarrow \mu_y\left(C_y \right)  ,\]
where $ C_y = C\cap\left(  X \times\left\lbrace y \right\rbrace   \right)   $,	 is $\mathfrak{B}_{ Y}-$measurable and
\[  \zeta(C)= \int_{X} \mu_y\left(C_y \right)  \mu(dx). \]
In particular,
\[ \forall A\in \mathfrak{B}_X, \ \mu(A)=\int_{Y} \mu_y\left(A \right) \nu(dx)  .\]
Moreover, for all positive $\mathfrak{B}_{X\times Y}-$measurable function,  $f: X\times Y\rightarrow \mathbb{R}$,
\[ y\in Y\setminus E_2 \rightarrow \int_{X} f(x,y)\, \mu_y\left(dx\right)  \] is  $\mathfrak{B}_{Y}-$measurable  and
\[  \int_{X\times Y} f(x,y)\, \zeta \left(dx, dy\right) = \int_{Y} \left(  \int_{X} f(x,y)\, \mu_y \left(dx\right) \right)  \nu(dy). \]

\vskip 3mm
\noindent
As a consequence, given a $\mathfrak{B}_{X\times Y}-$measurable function,  $f: X\times Y\rightarrow \mathbb{R}$, the following affirmations are equivalent

\vskip 2mm

\begin{enumerate}
	\item $f: X\times Y\rightarrow \mathbb{R}$ is $\zeta-$integrable.
\vskip 3mm
\item $ x \in X\setminus E_1 \rightarrow \int_Y \left| f(x,y)\right| \, \nu_x\left( dy\right)   $
is $\mu-$integrable.
\vskip 3mm
\item $ y\in Y\setminus E_2 \rightarrow \int_{X}\left| f(x,y)\right| \, \mu_y\left( dx\right) $  is $\nu-$integrable.
\end{enumerate}
\vskip 2mm\noindent
And
\[ \int_{X\times Y} f(x,y)\, \zeta\left( dx, dy\right)= \int_X \left( \int_Y f(x,y)\, \nu_x\left( dy\right)  \right)  \mu\left( dx\right) . \]
\[  \qquad \qquad \qquad \qquad \qquad = \int_{Y} \left(  \int_{X} f(x,y)\, \mu_y \left( dx\right) \right)  \nu\left(  dy\right) . \]
}

\section{Proof of Theorem \ref{teo}}

\noindent
\emph{Note about the notation}: We will use $x$ and $y$ to denote generic points in $X$ and $Y$ respectively. In this way,  $B_r\left( x\right) $ automatically refers to a ball in $X$, of center $x$ and and radius $r$, while $B_r\left( y\right)$ represents a ball in $Y$ (different space, different metric).
\vskip 2mm
\noindent
The proof of Theorem \ref{teo} will be given in several steps.

\subsection{Definition of $ \mathit{l}_x $}

Let  $\mathcal{Y} $ be a  dense subset of $Y$. Denote by $ \mathcal{B} $  the set of open balls $B_r\left( y\right) $ with center $y\in \mathcal{Y} $  and radius  $ r\in\mathbb{Q} $. i.e.
\begin{equation}\label{balls}
\mathcal{B}:= \left\lbrace B_r \left( y\right) ;\, y\in\mathcal{Y},\ r\in \mathbb{Q} \right\rbrace .
\end{equation}

\noindent
Consider also the complement of the closed balls,
\[  \mathcal{B}_{\mathrm{c}} :=\left\lbrace Y\setminus \overline{B}_r\left( y\right) ; {B}_r\left( y\right) \in\mathcal{B} \right\rbrace .  \]

\noindent
Finally, let $ \mathcal{L} $ be the set of finite unions of finite intersections of elements of $ \mathcal{B}\cup \mathcal{B}_{\mathrm{c}} $ (note that $\emptyset\in \mathcal{L}$).
\vskip 2mm
\noindent
For each $O\in \mathcal{L}$, define the measure
\[ A\in \mathfrak{B}_X\rightarrow \mu_O \left( A\right) =\zeta\left( A\times O\right)  .\]

\noindent
Since  $\mu_O \left( A\right) \le \mu\left( A\right) $,   $\mu_O $ is absolutely continuous with respect to $ \mu $. By  Radon-Nikodym's Theorem, there is a density function $ \frac{d\mu_O}{d\mu} $, defined $\mu-$almost surely, such that $\mu_O $ is represented as an integral of this density against $\mu$.
\vskip 2mm
\noindent
To obtain a common exceptional $\mu-$negligible set outside which $ \frac{d\mu_O}{d\mu} $
is well defined (by a formula) for all $ O\in \mathcal{L} $, we choose the version of $ \frac{d\mu_O}{ d\mu} $ given by the limit of the quotient of balls. To avoid talking about measurability issues, we fix once and for all a sequence $\rho_k$  decreasing to $0$. Given $ O\in \mathcal{L} $,  define
\[ \overline{\mathit{l}}_x\left( O\right) :=\limsup_{k\rightarrow \infty} \frac{\mu_O \left( B_{\rho_k}(x)\right) }{\mu \left( B_{\rho_k}(x)\right) },\]
and \[ \underline{\mathit{l}}_x\left( O\right) :=\liminf_{k\rightarrow \infty} \frac{\mu_O \left( B_{\rho_k}(x)\right) }{\mu \left( B_{\rho_k}(x)\right) },\]
\vskip 2mm
\noindent
It is well known, by a generalization of Lebesgue differentiation theorem (see for example Federer \cite{fed}, section 2.9),
 that $ \overline{\mathit{l}}_x $ and $ \underline{\mathit{l}}_x $  are versions of
 $ \frac{d\mu_O}{d\mu} $.
i.e. For all $  A\in \mathfrak{B}_X$,
\begin{equation}\label{int}
\mu_O \left( A\right) =\int_A  \overline{\mathit{l}}_x\left( O\right) \mu\left( dx\right)  =\int_A  \underline{\mathit{l}}_x \left( O\right) \mu\left( dx\right) .
\end{equation}
In particular,
$$ \overline{\mathit{l}}_x\left( O\right) = \underline{\mathit{l}}_x\left( O\right) ,\  \mu-\mathrm{a.s.}.$$
\noindent
Let $E_O$ be the exceptional set where the limit does not exist. i.e.
$$E_O:=\left\lbrace x\in X;\ \overline{\mathit{l}}_x\left( O\right) -\underline{\mathit{l}}_x \left( O\right) >0\right\rbrace \in \mathfrak{B}_X$$
 \noindent
Put
\[ E:=\bigcup_{O\in \mathcal{L} } E_O. \]

\noindent
Since  $\mu\left( E_O\right) =0$ for all $ O\in \mathcal{L} $, and $\mathcal{L}$ is numerable, we  have $$\mu\left( E\right) =0 .$$
 For  $O\in \mathcal{L}$, and   $x\in X\setminus E$, put
 $$ \mathit{l}_x\left( O\right) = \overline{\mathit{l}}_x \left( O\right) =\underline{\mathit{l}}_x \left( O\right) .$$

\vskip 2pt
\noindent
Recapitulating, for all $O\in \mathcal{L}$ and every $ A\in \mathfrak{B}_Y $, by (\ref{int}) we have
\begin{equation}\label{int1}
 \zeta\left( A\times O\right)  =\int_A  {\mathit{l}}_x\left( O\right) \mu\left( dx\right)  .
\end{equation}

\subsection{Outer measure $ \nu^*_x $}

Changing the standpoint, we  fix   $x\in X\setminus E$ and consider the set function
$$ O\in \mathcal{L}\rightarrow \mathit{l}_x\left( O\right) .$$
For future reference, observe that $ \mathit{l}_x $ has the following properties:
For all $ O$ and  $ \tilde{O}\in \mathcal{L} $,
\vskip 3pt
\noindent
\emph{Finite additivity}
\begin{equation}\label{add}
 \mathit{l}_x\left(  O\right) +\mathit{l}_x\left(  \tilde{O}\right) = \mathit{l}_x\left(  O\cup \tilde{O}\right) + \mathit{l}_x\left(  O\cap \tilde{O}\right) ,
\end{equation}
\vskip 2pt
\noindent
\emph{Finite subadditivity}
	\begin{equation}\label{subadd}
\mathit{l}_x\left(  O\cup \tilde{O}\right)\le	\mathit{l}_x\left(  O\right) +\mathit{l}_x\left(  \tilde{O}\right)
	\end{equation}
\vskip 2pt
\noindent
 \emph{Monotonicity}
\begin{equation}\label{mono}
   O\subseteq \tilde{O} \, \Rightarrow \, \mathit{l}_x\left(  O\right) \le \mathit{l}_x\left(  \tilde{O}\right)
\end{equation}
\vskip 3pt
\noindent
We need a measure on $\mathfrak{B}_Y$, capable of fulfilling the role of  $\mathit{l}_x$ in equation (\ref{int1}). Let us start by defining the outer measure
\begin{equation} \label{outer}
C\subseteq Y\rightarrow \nu^*_x\left( C\right) :=\inf\sum_{i=1}^{\infty}\mathit{l}_x\left( O_i\right) ,
\end{equation}
where the infimum is taken over all the covers $ \left\lbrace O_i \right\rbrace _{i\in \mathbb{N}} \subseteq \mathcal{L} $ of $C$. i.e.
\[  C\subseteq \bigcup_{i=1}^\infty O_i  \, ,\ \mathrm{and}\ \forall i\in \mathbb{N}, \,  O_i\in  \mathcal{L}\]

\noindent
It is well known that $\nu^*_x$, restricted to the set of $\nu^*_x-$measurable sets,  is a measure (denoted $\nu_x$). What we need to prove are: Firstly, that every Borel subset of $Y$ (i.e. in $\mathfrak{B}_Y$) is $\nu^*_x-$measurable and secondly that the integration property (\ref{int1}) is preserved when $\mathit{l}_x$ is replaced by $\nu_x$ (and therefore valid for any set in $\mathfrak{B}_Y$).
\vskip 2mm
\noindent
 Unfortunately, $\mathit{l}_x$ is not countably subadditive, as a  result,  $\nu^*_x$  is not  the extension of $\mathit{l}_x$, hardening  our task  a little bit. Indeed, for all $ O\in  \mathcal{L} $, we clearly have $ \nu^*_x(O)\le \mathit{l}_x(O)$, but the inverse inequality may fail, as the following example shows.
\vskip 3mm
\noindent
\emph{Example}: Let $X=Y=\left[ 0,1\right] $. Let $\mu=\nu$ be the Lebesgue measure on $ \left[ 0,1\right]  $ and $\zeta$ the normalized length on the diagonal $\left\lbrace \left(x,x \right) ;\, x\in \left[ 0,1\right] \right\rbrace $.
\vskip 3mm
\noindent
Observe that for every $ x\in \left] 0,1\right[  \, $ and $\rho_k$ small enough,
\[  \frac{\mu_{\left[ 0,x\right[} \left( B_{\rho_k}(x)\right) }{\mu \left( B_{\rho_k}(x)\right) }=\frac{1}{2}.\]

\noindent
So, $ \mathit{l}_x  \left( \left[ 0,x\right[\right) =   1/2 $, while $ \nu_x^*  \left( \left[ 0,x\right[\right) =   0 $. In fact, we can cover $ \left[ 0,x\right[ $ with a sequence of intervals $ \left[ 0,x_n\right[ $, where $x_n \in \mathbb{Q}$ increases to $x$. Each one of the intervals $ \left[ 0,x_n\right[ $ verify
\[  \frac{\mu_{\left[ 0,x_n\right[} \left( B_{\rho_k}(x)\right) }{\mu \left( B_{\rho_k}(x)\right) }=0, \]
for all  $\rho_k$ small enough. Therefore, for all $n\in \mathbb{N}$,
$ \mathit{l}_x  \left( \left[ 0,x_n\right[\right) =0 $ and
\[ \nu_x^*  \left( \left[ 0,x\right[\right) \le \sum_{i=1}^\infty  \mathit{l}_x  \left( \left[ 0,x_n\right[\right) =   0 .\]

\vskip 3mm
\noindent
\subsection{Borel subsets of $Y$ are $\nu^*_x-$measurable} Let's prove first that any open ball $B_r(y)\in\mathcal{L}$ is $\nu^*_x-$measurable. To this end, fix $C\subseteq Y$ and a cover  $ \left\lbrace O_i\right\rbrace_{i\in\mathbb{N}}  \subseteq \mathcal{L}$ of $C$. We must  show that
\begin{equation}\label{measurable}
\nu^*_x \left( C \cap B_r(y) \right) + \nu^*_x \left( C\setminus B_r(y)\right) \le \sum_{i=1}^\infty \mathit{l}_x\left( O_i\right).
\end{equation}
\noindent
Since $ O_i\cap B_r(y)  \in \mathcal{L}$ and  $ \left\lbrace O_i\cap B_r(y)  \right\rbrace_{i\in\mathbb{N}} $  is a covering of $ C\cap B_r(y) $,
\begin{equation}\label{uno}
 \nu^*_x\left( C\cap B_r(y)\right) \le \sum_{i=1}^\infty  \mathit{l}_x\left( O_i\cap B_r(y)  \right)    .
\end{equation}
\noindent
Now, let $\alpha_{i}<1$, $\alpha_{i}\in\mathbb{Q}$. Then,  $O_i\setminus \overline{B}_{ \alpha_{i} r}(y)\in \mathcal{L}$ and  $ \left\lbrace O_i\setminus \overline{B}_{ \alpha_{k_i} r}(y) \right\rbrace_{i\in\mathbb{N}}  $ is a covering of $C\setminus B_r(y) $. So,
  \begin{equation}\label{dos}
 \nu^*_x\left( C\setminus B_r(y)\right) \le \sum_{i=1}^\infty  \mathit{l}_x\left(   O_i\setminus \overline{B}_{ \alpha_i r}(y)  \right) .
 \end{equation}
 By (\ref{add}) and (\ref{mono}), we have

\begin{equation}\label{tres}
 \mathit{l}_x\left( O_i\cap B_r(y)  \right)  +   \mathit{l}_x\left(   O_i\setminus \overline{B}_{ \alpha r}(y)  \right) \le  \mathit{l}_x\left( O_i \right)  +   \mathit{l}_x \left( B_r(y) \setminus \overline{B}_{ \alpha_i r}(y)  \right).
 \end{equation}
\vskip 2mm
\noindent
Adding (\ref{uno}) and (\ref{dos}), and using (\ref{tres}), we obtain
\[\nu^*_x\left( C\cap B_r(y)\right) +\nu^*_x\left( C\setminus B_r(y)\right) \]
\[\le \sum_{i=1}^\infty  \mathit{l}_x\left( O_i \right) +\sum_{i=1}^\infty   \mathit{l}_x\left( B_r(y)\setminus \overline{B}_{ \alpha_i r}(y)  \right) .\]
\vskip 2mm
\noindent
The result follows if we can make
the second sum as small as we want.
Unfortunately, for a fixed $x\in E$, we might fail to do so, even though
\begin{equation}\label{empty}
\bigcap_{i=1}^\infty B_r(y) \setminus \overline{B}_{ \alpha_i r}(y) =\emptyset ,
\end{equation}
for any sequence  $\alpha_i\rightarrow 1$.
In fact,  we can not switch the limits in
\[ \lim_{i\rightarrow \infty}\mathit{l}_x\left( B_r(y)\setminus \overline{B}_{ \alpha_i r}(y)  \right) = \lim_{i\rightarrow \infty}\lim_{k\rightarrow \infty} \frac{\zeta \left( B_{\rho_k}\left( x\right) \times \left(   B_r(y)\setminus \overline{B}_{ \alpha_i r}(y)  \right) \right) }{\mu\left(  B_{\rho_k}\left( x\right) \right)} .\]
So, we need to look back at what happens for $x$ variable and check whether we can solve the problem by  throwing  away a few more points (meaning to enlarge $E$).
\vskip 2mm
\noindent

By (\ref{int1}) and ($\ref{empty}$), and $\gamma < 1$,
\[ \int_X \mathit{l}_x\left( B_r(y) \setminus \overline{B}_{ \gamma r}(y)\right)  \mu\left( dx\right) =\nu\left(  B_r(y) \setminus \overline{B}_{ \gamma r}(y)\right) \longrightarrow_{\gamma \nearrow 1} 0. \]
Therefore,
\begin{equation}\label{null-a.e.}
\mathit{l}_x\left( B_r(y) \setminus \overline{B}_{ \gamma r}(y)\right)  \longrightarrow_{\gamma \nearrow 1} 0,\ \mu-\mathrm{a.s.}
\end{equation}
\vskip 2mm
\noindent
Now, fix an increasing sequence  $ \left\lbrace \gamma_j\right\rbrace_{j\in\mathbb{N}}  \subseteq  \mathbb{Q}$, $ \gamma_j \rightarrow 1$. Define, for all $y\in\mathcal{Y}$ and $r\in\mathbb{Q}$,
\[ {E}_{r,y}:=\left\lbrace x\in X\setminus E ;\ \liminf_{j\rightarrow \infty}\mathit{l}_x\left( B_r(y) \setminus \overline{B}_{  \gamma_j r}(y)\right)  > 0 \right\rbrace  .\]
By (\ref{null-a.e.}), $\mu\left( {E}_{r,y}\right) =0 $. Since
\[   {E}_1:= E\cup\bigcup_{r\in\mathbb{Q} ,\, y\in\mathcal{Y} }{E}_{r,y} \]
is a countable union of sets of $\mu-$measure 0, we  have $$\mu\left( {E}_1\right) =0 . $$
For all  $x\in X\setminus E_1  $, and every $\epsilon>0$, we can choose a subsequence $ \left\lbrace \alpha_i\right\rbrace_{i\in\mathbb{N}}  $ of $ \left\lbrace \gamma_j\right\rbrace_{k\in\mathbb{N}}  $, such that
\[ \sum_{i=1}^\infty   \mathit{l}_x\left( B_r(y)\setminus \overline{B}_{ \alpha_i r}(y)  \right) < \epsilon .\]
This completes the proof of (\ref{measurable}).
\vskip 2mm
\noindent
Consequently, for all  $x \in X\setminus   E_1$, $\nu_x$ is a measure defined at least in the $\sigma-$field generated by $\mathcal{B}$, i.e.  $\mathfrak{B}_Y$.

\subsection{Measurability and integrability  for compact sets}
Our task now is to prove that given $B\in \mathfrak{B}_Y$, the function
\begin{equation}\label{meas}
 x\in X\setminus  E_1 \longrightarrow \nu_x\left( B\right)
\end{equation}
is measurable and, for all $A\in \mathfrak{B}_X$
\begin{equation}\label{int2}
\zeta\left( A\times B\right) =\int_A  \nu_x\left( B\right) \mu\left( dx\right)   .
\end{equation}
\vskip 2pt
\noindent
Let's consider first a finite intersection of closed, compact balls
$$ \overline{B}=\overline{B}_{r_1}(y_1)
\cap\cdots\cap \overline{B}_{r_n}(y_n) ,$$
with $y_1,\cdots,y_n\in\mathcal{Y}$  and $ r_1,\cdots,r_n\in\mathbb{Q} $.
 By the measurability of $\mathit{l}_x$ and (\ref{int1}), the properties (\ref{meas}) and (\ref{int2}) are proven at once if we show
\begin{equation}\label{l_x}
\nu_x \left( \overline{B} \right) =1-\mathit{l}_x\left( Y\setminus \overline{B}\right) ,\ \mu-\mathrm{a.s.}
\end{equation}
(We use $\mathit{l}_x\left( Y\setminus \overline{B}\right) $ just because $\mathit{l}_x $ is not defined for $ \overline{B}$.)
\vskip 2pt
\noindent
Let $O_1,\, O_2,  \cdots, O_m\subseteq \mathcal{L}$ be a covering of
$ \overline{B} $. We can assume the covering is finite, since $\overline{B}$ is compact and the sets in $\mathcal{L}$ are open.
\vskip 3pt
\noindent
Clearly, for all $ x\in X\setminus E $,
\[ \lim_{k\rightarrow \infty} \frac{\zeta\left( {B}_{\rho_k}(x)\times \overline{B}\right) }{\mu\left( {B}_{\rho_k}(x)\right) }=\lim_{k\rightarrow \infty} \left( 1-\frac{\zeta\left( {B}_{\rho_k}(x)\times \left( Y\setminus \overline{B} \right)\right)  }{\mu\left( {B}_{\rho_k} (x)\right) }\right) =1-\mathit{l}_x\left( Y\setminus \overline{B}\right) . \]
Since the covering is finite, by  (\ref{subadd}),
 \[ \lim_{k\rightarrow \infty}  \frac{\zeta\left( {B}_{\rho_k}(x)\times \overline{B}\right) }{\mu\left( {B}_{\rho_k}(x)\right) }\le \mathit{l}_x\left( \bigcup_{i=1}^{n} O_i\right) \le \mathit{l}_x\left( O_1\right) + \cdots +   \mathit{l}_x\left( O_m\right).  \]
Then,
\[ 1-\mathit{l}_x\left( Y\setminus \overline{B}\right) \le  \nu_x \left( \overline{B} \right) .\]
\vskip 2pt
\noindent
On the other hand, given $ \eta >1 $, $ \eta\in\mathbb{Q} $, and denoting
\[ B_\eta={B}_{\eta r_1}(y_1)
\cap\cdots\cap {B}_{\eta r_n}(y_n) , \]
we have
\[ \nu_x \left( \overline{B}\right) \le \mathit{l}_x\left(  {B}_\eta \right) .\]
Consequently, taking any sequence $\eta_k\searrow 1$, $ \eta_k \in\mathbb{Q} $, we have
\[ 1-\mathit{l}_x\left( Y\setminus \overline{B}\right) \le  \nu_x \left( \overline{B}\right)  \le \lim_{k\rightarrow\infty}\mathit{l}_x\left(  {B}_{\eta_k}\right) .\]
\vskip 2pt
\noindent
Since the functions at the left and at the  right of the above inequalities are $\mathfrak{B}_{X}-$measurable, and equal between them $\mu-$almost-surely, using (\ref{int1}), we have proved (\ref{l_x})  and, a fortiori,  (\ref{meas})  and (\ref{int2}), at least for a finite intersection of compact balls.

\subsection{Measurability and integrability  for Borel sets}

 Let $\mathcal{M}$ be the collection  of sets  $B\in \mathfrak{B}_Y$  actually verifying (\ref{meas})  and  (\ref{int2}).
 \vskip 3pt
 \noindent
 Clearly, $\emptyset\in \mathcal{M} $ and $ Y\setminus B\in \mathcal{M} $, whenever  $B \in \mathcal{M}  $.
\vskip 3pt
\noindent
Now, take a disjoint sequence   $B_1,B_2,\, \cdots \, \in \mathcal{M}$  ($B_i\cap B_j= \emptyset $,   $ i\neq j $).
Since, for all $x\in X\setminus  E_1  $, $\nu_x$ is a measure on $\mathfrak{B}_Y$,
\begin{equation}\label{suma}
 \nu_x\left( \bigcup_{i=1}^{\infty} B_i\right) = \sum_{i=1}^{\infty}\nu_x\left(  B_i\right) .
\end{equation}
By  (\ref{meas}), $ x\rightarrow  \nu_x\left( \bigcup_{i=1}^{\infty} B_i\right)  $ is $\mathfrak{B}_{X}-$measurable, being a countable sum of $\mathfrak{B}_{X}-$measurable functions.

\vskip 2pt
\noindent
By (\ref{suma}), the monotone convergence theorem, and  (\ref{int2}) (remember $B_i \in \mathcal{M}$, for all  $i\in \mathbb{N}$), given $ A\in\mathfrak{B}_X $,
\[ \int_A  \nu_x\left( \bigcup_{i=1}^{\infty} B_i\right) \mu\left( dx\right) = \int_A  \sum_{i=1}^{\infty}\nu_x\left(  B_i\right) \mu\left( dx\right)   \]
\[ \ \quad  \qquad  \qquad  \qquad  \qquad = \sum_{i=1}^{\infty} \int_A  \nu_x\left(  B_i\right)  \mu\left( dx\right)  \]
\[     \qquad   \quad  \qquad  \qquad = \sum_{i=1}^{\infty}   \zeta\left( A\times B_i\right) \]
\[ \qquad \qquad  \qquad  \qquad = \zeta \left( \bigcup_{i=1}^{\infty} A\times B_i\right) . \]
\[ \qquad \qquad  \qquad  \qquad = \zeta \left( A\times\bigcup_{i=1}^{\infty}  B_i\right) . \]
 Then,
 \[ \bigcup_{i=1}^{\infty} B_i\in \mathcal{M}. \]
Since $\mathcal{M}$ contains all the finite intersections of compact balls with center in $\mathcal{Y}$ and rational radius, by the $\pi-\lambda$ theorem (see \cite{Bi}, page 36),
 $$ \mathcal{M}=\mathfrak{B}_Y. $$

\subsection{Proof of the theorem} We have proven so far (\ref{meas0}) and (\ref{int0})   for sets of the form $C=A\times B$, with $ A\in \mathfrak{B}_X $  and $ B\in \mathfrak{B}_Y $.
\vskip 2mm
\noindent
Using a similar argument as before, let $\tilde{\mathcal{M}}$ denote the collection  of sets  $C\in \mathfrak{B}_{X\times Y} $  verifying (\ref{meas0})  and  (\ref{int0}). We readily see that $\emptyset\in \mathcal{M} $.

\vskip 2mm
\noindent
Now, let $C\in\tilde{\mathcal{M}} $. Since $\left( \left( X\times Y\right) \setminus  C\right)_x = Y  \setminus  C_x $, the function
\[ x\rightarrow \nu_x\left(\left( \left( X\times Y\right) \setminus  C\right) _x \right)  = 1- \nu_x\left( C_x\right) \]
 is $ \mathfrak{B}_{X}-$measurable, and
\[ \int_X \nu_x\left(\left( \left( X\times Y\right) \setminus  C\right)_x \right)  \mu\left( dx\right)  = 1- \int_X \nu_x\left(C_x \right) \mu\left( dx \right)   \]
\[ \ \qquad  \qquad  \qquad  \qquad = 1-\zeta\left(  C\right)\]
\[\  \qquad  \qquad  \qquad  \qquad  \qquad  \ \quad = \zeta\left( \left( X\times Y\right) \setminus  C\right). \]

\noindent
Then, for all $C\in\tilde{\mathcal{M}} $, we have $\left( X\times Y\right) \setminus  C \in\tilde{\mathcal{M}} $.
\vskip 3pt
\noindent
Finally,  let  $C_1,C_2,\, \cdots \, \in \tilde{\mathcal{M}} $ a sequence of disjoint sets ($C_i\cap C_j= \emptyset $, for all $ i\neq j $).
Since
$\nu_x$ is a measure on $\mathfrak{B}_Y$,
\begin{equation}\label{suma1}
\nu_x\left(  \left( \bigcup_{i=1}^{\infty} C_i\right)_x\,\right) = \nu_x\left( \bigcup_{i=1}^{\infty} \left( C_i\right) _x\right) = \sum_{i=1}^{\infty}\nu_x\left(  \left( C_i\right)_x \right)  .
\end{equation}
Then, $  x\rightarrow  \nu_x\left(  \left( \bigcup_{i=1}^{\infty} C_i\right)_x \,\right)  $ is $\mathfrak{B}_{X}-$measurable, being a sum of $\mathfrak{B}_{X}-$measurable functions, since $C_i \in \tilde{\mathcal{M}}$,  for all $i\in \mathbb{N}$.

\vskip 3pt\noindent
By (\ref{suma1}), the monotone convergence theorem, and  (\ref{int0})  ($C_i \in \tilde{\mathcal{M}}$),
\[ \int_X  \nu_x\left(  \left( \bigcup_{i=1}^{\infty} C_i\right)_x\,\right) \mu\left( dx \right) = \int_X  \sum_{i=1}^{\infty}\nu_x\left(  \left( C_i\right) _x\right) \mu\left( dx \right) \ \quad   \quad   \]
\[ \ \quad  \qquad  \qquad  \qquad  \qquad = \sum_{i=1}^{\infty} \int_X  \nu_x\left(  \left( C_i\right) _x\right)  \mu\left( dx \right) \]
\[  \qquad  \qquad = \sum_{i=1}^{\infty}   \zeta\left(  C_i\right) \]
\[ \  \qquad  \qquad = \zeta \left( \bigcup_{i=1}^{\infty}  C_i\right)  \]
Then,
\[ \bigcup_{i=1}^{\infty} C_i\in \tilde{\mathcal{M}}. \]
\vskip 2mm
\noindent
Since
\[ \left\lbrace A\times B;\, A\in \mathfrak{B}_X ,\, B\in \mathfrak{B}_Y \right\rbrace  \subseteq \tilde{\mathcal{M}} ,\]

by the $\pi-\lambda$ theorem,
\[ \tilde{\mathcal{M}} = {\mathfrak{B}_{X\times Y}}  .\]
\vskip 3mm
\noindent
The remaining of the proof is standard. Assume $f :{X\times Y}\rightarrow \mathbb{R}$ is a  positive, $\mathfrak{B}_{X\times Y}-$measurable functions. Then, $f$ can be approached by an increasing sequence of simple, $\mathfrak{B}_{X\times Y}-$measurable functions  (linear combinations of characteristic functions)
\[ f_k(x,y)=\sum_{i=1}^{n_k} \lambda_{k,i}C_{k,i},\]
where $\lambda_{k,i}\in\mathbb{R}$ and  $C_{k,i} \in \mathfrak{B}_{X \times Y}$ ($i=1,\cdots, n_k $, $ k\in  \mathbb{N}$).
\vskip 3mm
\noindent
By linearity, for each $ f_k$, properties (\ref{meas0f}) and (\ref{int0f}) are a direct consequence of (\ref{meas0}) and (\ref{int0}). Passing to the limit as $k\rightarrow \infty$, equations (\ref{meas0f}) and (\ref{int0f}) are conserved, therefore valid  for any positive function like $f$.
\vskip 3mm
\noindent
To establish the integrability equivalence, we apply the preceding result to the positive and negative parts of the given function. In this way,  the proof of Theorem \ref{teo} is complete.

\section{An aplication to Optimal Transport}  \label{OT}

\noindent
In this section we use Theorem \ref{teo} to show that a pair of competitive price functions, whose integral with respect to some transference plan equals the transport cost, are  conjugate to each other almost surely, complementing  Kantorovich's duality theorem on the nature of a pairs of competitive prices maximizing the profit. See Villani's book \cite{vill}, page 70, for a very detailed discussion on  Kantorovich's theorem, in particular
 Theorem 5.1, part \textit{(ii)}, item \textit{(d)}.  For this result, we do not need lower semicontinuity  of the cost and other assumptions used to prove Kantorovich's theorem, so we state the following  lemma in its simplest form, using the notations in  Theorem \ref{teo}, by the way.

{\lem\label{ot} Let $ c : X \times Y \rightarrow \mathbb{R} \cup \left\lbrace +\infty\right\rbrace $ be a ${\mathfrak{B}_{X\times Y}}-$measurable,
	cost function, and
	$ \psi :X\rightarrow \mathbb{R}\cup \left\lbrace +\infty\right\rbrace  \ \mathrm{and}\ \phi :Y\rightarrow \mathbb{R}\cup \left\lbrace -\infty\right\rbrace $
	be a pair of  competitive prices, i.e.
	\[ \forall \, \left( x,y\right) \in X\times Y, \ \phi \left( y\right) -\psi\left( x\right) \le c\left( x,y\right) . \]
Let  $\pi$  be a probability measure on ${\mathfrak{B}_{X\times Y}}$, with  marginal measures $\mu$ on ${\mathfrak{B}_X}$ and $\nu$ on ${\mathfrak{B}_Y}$. (i.e.  $\pi$   is a transference plan between $\mu$ and $\nu$.)
Assume  that $ \phi -\psi$ and $ c $ are $\pi-$integrable and
\[  \phi \left( y\right) -\psi\left( x\right) = c\left( x,y\right) , \ \pi-\mathrm{a.s.} . \]
Then
\begin{equation}\label{intes01}
\psi\left( x\right) = \sup_{y\in Y}\left(\phi \left( y\right)  - c\left( x,y\right) \right)  , \ \mu-\mathrm{a.s.}
\end{equation}
and
\begin{equation}\label{intes02}
 \phi \left( y\right) = \inf_{x\in X}\left( \psi \left( y\right)  + c\left( x,y\right)\right) , \ \nu-\mathrm{a.s.}
 \end{equation}
}
\vskip 2mm
\noindent
\emph{Proof}: Since $ \psi+ c-\phi = 0 $, $\pi-$a.s.,
\[\int_{X\times Y} \left( \psi(x)+ c(x,y)-\phi(y) \right)  \pi\left(dx,dy\right)= 0. \]
Using the decomposition  given by Theorem \ref{teo}, equation (\ref{int0f}),
\begin{equation}\label{nullint}
\int_X \left( \int_Y \left( \psi(x)+ c(x,y)-\phi(y) \right)  \nu_x\left(dy\right)  \right)  \mu(dx)=0.
\end{equation}
Since $ \psi+ c-\phi \ge 0 $,
\begin{equation}\label{sup}
\psi(x)\ge \sup_{y\in Y}  \left(\phi(y) - c(x,y) \right)
\end{equation}
and, for all $x$ where it is defined, the function
\[ x\longrightarrow \int_Y \left( \psi(x)+ c(x,y)-\phi(y) \right)  \nu_x\left(dy\right)
 \]
is nonnegative. By (\ref{nullint}),
\[  \int_Y \left( \psi(x)+ c(x,y)-\phi(y) \right)  \nu_x\left(dy\right)  = 0,\ \mu-\mathrm{a.s.} \]
Then,
 \begin{equation}\label{intas}
\psi(x)= \int_Y \left(\phi(y) - c(x,y)\right)  \nu_x\left(dy\right)\le \sup_{y\in Y}  \left(\phi(y) - c(x,y) \right) ,\ \mu-\mathrm{a.s.}
\end{equation}
Combining (\ref{sup}) and (\ref{intas}), we obtain (\ref{intes01}). Equation (\ref{intes02}) is validated in a similar way.

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\begin{document}

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\title{A complete conformal metric of preassigned negative Gaussian
curvature for a punctured hyperbolic Riemann surface}

\markboth{Rukmini Dey}{Preassigned negative Gaussian curvature}

\author{RUKMINI DEY}

\address{Department of Mathematics, Indian Institute of Technology,
Kanpur~208~016, India\\
\noindent E-mail: rukmini@iitk.ac.in}

\volume{114}

\mon{May}

\parts{2}

\Date{MS received 1 December 2003; revised 14 February 2004}

\begin{abstract}
Let $h$ be a complete metric of Gaussian curvature $K_0$ on a punctured
Riemann surface of genus $g \geq 1$ (or the sphere with at least three
punctures). Given a smooth negative function $K$ with $K=K_0$ in
neighbourhoods of the punctures we prove that there exists a metric
conformal to $h$ which attains this function as its Gaussian curvature
for the punctured Riemann surface. We do so by minimizing an appropriate
functional using elementary analysis.
\end{abstract}

\keyword{Punctured Riemann surfaces; prescribed curvature.}

\maketitle

\section{Introduction}

Let $\Sigma$ be a punctured hyperbolic Riemann surface, namely a Riemann
surface of genus $g \geq 1$ with at least one puncture (or the sphere
with at least three punctures). Let $\d s^2 = h\, \d z \otimes \d
\bar{z} $ be a complete metric on $\Sigma$ with $K_0= K_0(z, \bar{z})$
its Gaussian curvature, negative near the punctures. Example of an
initial metric is given in \S4.

In this paper we prove the result that given an {\it arbitrary} smooth
negative function $K$, with $K=K_0$ in neighbourhoods of the punctures,
there exists a metric conformal to $h$ which attains this function as
its Gaussian curvature\footnote{Note that there are minimal surfaces
with ends, topologically equivalent to punctured Riemann surfaces, which
have negative curvature everywhere \cite{JM,O,X}.}. Our proof is
elementary, using Hodge theory, i.e., the existence of the Green's
operator for the Laplacian on a compact manifold with boundary. This
proof is a generalization of the proof given in an earlier paper that
for a compact Riemann surface of genus $g>1$ any negative function $K$
is attained as the Gaussian curvature of a metric conformal to the given
one~\cite{D}.

In~\cite{HT} Appendix A, there is a theorem.

\begin{theorem}[\!]\hskip -.5pc {\bf (Hulin--Troyanov).}\ \
Let $S^{\prime}$ be a non-compact Riemann surface of finite type. Assume
$S^{\prime}$ is neither conformally equivalent to ${\CC}$ nor
${\CC}^{*}$. Let $K\hbox{\rm :}\ S^{\prime} \rightarrow {\RR}$ be any bounded
locally H\"{o}lder-continuous function. Then{\rm ,} there exists a conformal
metric $g$ on $S^{\prime}$ with curvature $K$.
\end{theorem}

However, such metrics are not usually complete. If we restrict ourselves
to complete metrics but forgo having an example in each conformal class,
there is a result by Burago (proof is to be found in \cite{HT})
generalizing Kazdan and Warner that any function negative\newpage \noindent on an open set
with non-positive value somewhere near the punctures is achieved as a
prescribed Gaussian curvature \cite{Bu,HT,KW2}.

\begin{theorem}[\!]\hskip -.4pc {\bf (Burago--Kazhdan--Warner).}\ \
Let $K$ be a smooth function on an open surface $S^{\prime}$ of finite
type. The following conditions are necessary and sufficient for $K$
to be the curvature of some complete metric of $S^{\prime}${\rm :}

\begin{enumerate}
\renewcommand{\labelenumi}{\rm (\roman{enumi})}
\leftskip .1pc
\item $\lim \inf_{x \rightarrow \infty}\ K(x) \leq 0$ at each end of
$S^{\prime}${\rm ;}

\item when Euler characteristic of $S^{\prime} <0${\rm ,} assume also{\rm ,}
\hbox{$\inf K <0$}{\rm ;} when Euler characteristic of $S^{\prime} = 0${\rm ,}
assume also \hbox{$\inf K <0$} or $K \equiv 0$.
\end{enumerate}
\end{theorem}

The proof of this theorem is also given in \cite{HT}. However they do
not find a metric in every conformal class which has this property.

Our proof is very different and our result is new in the sense that we
show that if the initial metric has negative Gaussian curvature near the
punctures and given any negative function with the same value as the
initial one near the punctures, there exists a metric in the conformal
class of the initial metric which has this function as its Gaussian
curvature. Moreover, the behaviour of the initial metric near the
punctures do not change (the conformal equivalence factor
$\hbox{e}^{\sigma}$ is 1 near the punctures). Thus if we started with a
complete metric, completeness is preserved. If the {\it initial} metric
has Gaussian curvature the function $K_0$, negative somewhere, such
that $\lim_{x \rightarrow \infty} K_0 = f_0 <0 $ (i.e. its value is
$f_0$ near the punctures) then by Burago--Kazhdan--Warner's result we
would have a complete metric which achieves $K_0$ as its Gaussian
curvature. Let $K <0$ be another function such that $\lim_{x \rightarrow
\infty} K = f_0 $ then our result shows that there exists a metric
conformal to the previous one such that $K$ is achieved as its Gaussian
curvature. This indicates that there could be many conformal classes
which contain a metric with $K$ as its Gaussian curvature. In the last
section we give an example of the initial metric which has negative
Gaussian curvature near the punctures.

By Thereom 2.1a, we also have the result that any arbitrary negative
function $K$ is allowed as the Gaussian curvature of an unpunctured
Riemann surface of genus $g \geq 1$ with at least one boundary component
or a sphere with at least three boundary components.

For prescribed curvature on surfaces with conical singularities, see
~\cite{T}.

Let $p_i$, $i=1,\ldots,n$ be the punctures on $\Sigma$. Let us choose a
disc $D_i$ about the point $p_i$ with coordinates $ (r_i, \theta_i) $
such that $D_i = \{ (r_i, \theta_i)$\,: $0< r_i < 1,\ 0< \theta_i \leq 2
\pi \}$ with the point $p_i$ corresponds to $r_i=0$. No other puncture
lies in this neighbourhood. Let $M = \Sigma - \cup_{i=1}^n D_i$. Thus
$M$ is a compact manifold with boundary $\partial M = \cup_{i=1}^n
\partial D_i$. We assume $K_0$ is negative on $\partial M$. In \S4 we
show by an example that this is not a stringent condition on the metric
$h$. Also $M$ is of finite volume with respect to the metric $h$ and
$K_0$ is smooth on $M$. $K_0$ may have singularities at the punctures.
Let $K <0 $ be a given function on $M$ and $K = K_0 $ on $\cup_{i=1}^n
D_i$. On $M$ we show that there exists a metric conformal to $h$ which
has $K$ as its Gaussian curvature. We show this by minimizing the
functional
\begin{equation*}
S[\sigma] = \int_M (K(\sigma)-K)^2 \e^{2\sigma}\d\mu
\end{equation*}
over functions in class $T$ where $T= \{ \sigma \in
C^{2}(M,\RR)|\sigma|_{\partial M} =\partial_{\nu} \sigma|_{\partial M} =
\Delta \sigma|_{\partial M} =0 \}^{-}$ where $-$ denotes the closure in
$W^{2,2}$. Here $K(\sigma)$ stands for the Gaussian curvature of the
metric $\e^{\sigma} h$, and $\d\mu = \frac{\sqrt{-1}}{2} h\ \d z
\wedge\d\bar{z}$ is the area form for the metric $h$. Using the Sobolev
embedding theorem for compact manifolds with boundary we show that
$S[\sigma]$ takes its absolute minimum, zero, on $C^{\infty}(M)$ which
corresponds to a metric on $M$ of negative curvature $K$. Then we extend
the minimizer $\sigma$ by zero on $D_i$ and thus near the punctures the
metric $\e^{\sigma} h\, \d z \otimes \d \bar{z} $ and the curvature $K$
remain $ h\, \d z \otimes \d \bar{z}$ and $K_0$ respectively.

\section{The main theorem}

\subsection{\it All notations are as in {\rm \S1}}

The functional $S[\sigma] = \int_M
(K(\sigma) - K)^2 \e^{2\sigma} \d \mu$ is non-negative on $T$, so that
its infimum
\begin{equation*}
S_0 = \inf\{S[\sigma],\,\sigma \in T\}
\end{equation*}
exists and is non-negative. Let $\{\sigma_n\}_{n=1}^{\infty}\subset T$
be a corresponding minimizing sequence,
\begin{equation*}
\lim\limits_{n\rightarrow\infty} S[\sigma_n] =S_0.
\end{equation*}
Our main result is the following\vspace{.2pc}

\begin{theorem}[\!]
{\rm (a)} Let $M = \Sigma - \cup_{i=1}^n D_i$ be a compact Riemann
surface with boundary{\rm ,} with at least three boundary components if
genus $g=0$. Let $h\, \d z \wedge \d \bar{z}$ be its initial metric and
$K_0$ be its initial curvature such that $K_0 <0$ on $\cup_{i=1}^n D_i$.
Let $K <0$ be an arbitrary negative function with $K = K_0$ on
$\cup_{i=1}^n D_i$ {\rm (}i.e.{\rm ,} the curvature is left intact near the
punctures{\rm )}. The infimum $S_0 =0$ is attained at a unique
$\sigma\in C^{\infty}(M,\RR)${\rm ,} i.e.{\rm ,} the minimizing sequence
$\{\sigma_n\}$ contains a subsequence that converges in strong $W^{2,2}$
to a unique $\sigma\in C^{\infty} (M,\RR)$ such that $S[\sigma]=S_0 =
0$. The corresponding metric $\e^{\sigma}h\,\d z\otimes\d\bar{z}$ is a
metric on $M$ of preassigned negative curvature $K$. {\rm (b)} On
$\cup_{i=1}^n D_i${\rm ,} $\sigma$ is extended as\break zero.\vspace{.3pc}
\end{theorem}

\subsection{\it Uniform bounds}

Since $\{\sigma_n\}$ is a minimizing sequence, we have the obvious
inequality
\begin{equation}\label{1}
S[\sigma_n]=\int_{M}(K_n-K)^2\e^{2\sigma_n}\d\mu =\int_M \left(K_0 -
\frac{1}{2}\Delta_h\sigma_n -K\e^{\sigma_n}\right)^2 \d\mu\leq m
\end{equation}
for some $m>0$, where we denoted by $K_n$ the Gaussian curvature
$K(\sigma_n)$ of the metric $\e^{\sigma_n}h$ and by $K_0$ that of the
metric $h$, and used that
\begin{equation*}
K_n=\e^{-\sigma_n}\left(K_0 - \frac{1}{2} \Delta_h \sigma_n\right).
\end{equation*}
\begin{note}
Here $\Delta_h = 4h^{-1} ({\del^2}/{\del z\del\bar{z}})$
stands for the Laplacian defined by the metric $h$ on $M$.\vspace{.2pc}
\end{note}

\setcounter{defin}{1}
\begin{lemma}
There exists a constant $C_1$  such that{\rm ,} uniformly in $n${\rm ,}
\begin{equation*}
\int_M (\Delta_h\sigma_n)^2 \d\mu <C_1.
\end{equation*}
\end{lemma}
\vspace{.3pc}

\begin{proof}
By Minkowski inequality, and using \eqref{1}, we get
\begin{align*}
\left[\! \int_M \left(\! -\frac{1}{2} \Delta_h\sigma_n - K\e^{\sigma_n}\!\right)^2 \d \mu\!\right]^{1/2}\! &\leq
\left[\int_M \left(K_0 - \frac{1}{2}\Delta_h\sigma_n -
K\e^{\sigma_n}\right)^2 \d \mu\right]^{1/2}\\
&\quad\, + \left[\int_M (K_0)^2 \d \mu\right ]^{1/2}\leq m^{1/2}+c=C,
\end{align*}
so that
\begin{equation}\label{2}
\frac{1}{4}\int_M (\Delta_h\sigma_n)^2 \d\mu+ \int_M K^2 \e^{2\sigma_n}\d\mu +
\int_M \Delta_h\sigma_n \e^{\sigma_n} K \d\mu \leq C^2.
\end{equation}

We will show that
\begin{equation}
\int_M K^2 \e^{2\sigma_n}\d\mu + \int_M \Delta_h\sigma_n \e^{\sigma_n} K \d \mu
= B_{n1} + B_{n2} + B_{n3},
\end{equation}
where $B_{n1} \geq 0$, $B_{n2} \geq 0$, $|B_{n3}| \leq 3 D^2$ where $D^2$ is
a constant independent of $n$.

From (2.3) and (2.2) the result  will follow since we will have
\begin{align*}
C^2 + 3D^2 &\geq C^2 - B_{n3} \geq \frac{1}{4}\int_M (\Delta_h\sigma_n)^2 \d\mu
+ B_{n1} + B_{n2}\\[.2pc]
&\geq \frac{1}{4}\int_M (\Delta_h\sigma_n)^2 \d\mu.
\end{align*}
Just renaming the constants, we will have the result.

Integrating by parts we get
\begin{align*}
 \int_M \Delta_h\sigma_n \e^{\sigma_n} K \d \mu &= - \int_M
|\partial_{z} \sigma_n|^2 \e^{\sigma_n} K \d \mu -  \int_M (\partial_z
\sigma_n) (\partial_{\bar{z}} K) \e^{\sigma_n} \d \mu\\[.2pc]
&\quad\, -\int_{\partial M} (\partial_{\nu} \sigma_n) K \e^{\sigma_n} \d \mu\\[.2pc]
&=  \int_M |\partial_{z} \sigma_n|^2 \e^{\sigma_n} |K|\d \mu -  \int_M (\partial_z \sigma_n) g |K|\ \e^{\sigma_n} \d \mu
\end{align*}
since $K$ is negative, $\partial_{\nu} \sigma_{n} |_{\partial M} = 0$ and
where we define $g= {\partial_{\bar{z}} K}/{|K|}$.\vspace{.2pc}

Let $M = \Omega_{n1} \cup \Omega_{n2} \cup \Omega_{n3}$, a disjoint union of
sets defined as follows:\vspace{.2pc}

On $\Omega_{n1}$, (1) $|\partial_{z} \sigma_n |> |g|$.\vspace{.4pc}

On $\Omega_{n2}$, (2) $|\partial_{z} \sigma_n |\leq |g|$ and
$ |K|\ \e^{\sigma_n} > |g|^2$.\vspace{.4pc}

On $\Omega_{n3}$, (3) $|\partial_{z} \sigma_n |\leq |g|$ and
$ |K|\ \e^{\sigma_n} \leq |g|^2$.\vspace{.4pc}

Let $B_{ni} = \int_{\Omega_{ni}} K^2 \e^{2\sigma_n}\d\mu + \int_{\Omega_{ni}}
\Delta_h\sigma_n \e^{\sigma_n} K \d \mu,\ i=1,2,3.$\vspace{.3pc}

We will show that $B_{n1} \geq 0$.
\begin{align*}
B_{n1} &= \int_{\Omega_{n1}} K^2 \e^{2\sigma_n} \d \mu + \int_{\Omega_{n1}} |\partial_{z} \sigma_n|^2 |K|\ \e^{\sigma_n} \d \mu
- \int_{\Omega_{n1}} (\partial_{z} \sigma_n) g |K|\ \e^{\sigma_n} \d \mu\\[.2pc]
&\geq  \int_{\Omega_{n1}} K^2 \e^{2\sigma_n} \d \mu + \int_{\Omega_{n1}} |\partial_{z} \sigma_n|^2 |K|\ \e^{\sigma_n} \d \mu
- \int_{\Omega_{n1}} |\partial_{z} \sigma_n| |g| |K|\ \e^{\sigma_n} \d \mu
\end{align*}
\begin{align*}
&=\int_{\Omega_{n1}} K^2 \e^{2\sigma_n} \d \mu + \int_{\Omega_{n1}} |\partial_{z} \sigma_n| |K|\ \e^{\sigma_n}(|\partial_z \sigma_n| - |g|) \d\mu\\[.2pc]
&\geq 0
\end{align*}
by (1) in the definition of $\Omega_{n1}$.

Next we shall show that $B_{n2} \geq 0$.
\begin{align*}
B_{n2} &= \int_{\Omega_{n2}} K^2 \e^{2\sigma_n} \d \mu + \int_{\Omega_{n2}} |\partial_{z} \sigma_n|^2 |K|\ \e^{\sigma_n} \d \mu - \int_{\Omega_{n2}} (\partial_{z} \sigma_n) g |K|\ \e^{\sigma_n} \d \mu\\[.2pc]
&\geq  \int_{\Omega_{n2}} K^2 \e^{2\sigma_n} \d \mu + \int_{\Omega_{n2}} |\partial_{z} \sigma_n|^2 |K|\ \e^{\sigma_n} \d \mu - \int_{\Omega_{n2}} |\partial_{z} \sigma_n| |g| |K|\ \e^{\sigma_n} \d \mu\\[.2pc]
&=\int_{\Omega_{n2}} |K|\ \e^{\sigma_n} (|K|\ \e^{\sigma_n} - |\partial_{z} \sigma_n| |g|) \d \mu + \int_{\Omega_{n2}} |\partial_z \sigma_n|^2 \e^{\sigma_n} |K| \d \mu\\[.2pc]
&\geq \int_{\Omega_{n2}} |K|\ \e^{\sigma_n} (|K|\ \e^{\sigma_n} - |g|^2) \d \mu +
\int_{\Omega_{n2}} |\partial_z \sigma_n|^2 \e^{\sigma_n} |K| \d \mu\\[.2pc]
&\geq 0,
\end{align*}
by using the two conditions (2) defining $\Omega_{n2}$.

Next we shall show that $B_{n3}$ is uniformly bounded.
\begin{align*}
|B_{n3}| &\leq \int_{\Omega_{n3}} K^2 \e^{2\sigma_n} \d \mu +
\int_{\Omega_{n3}} |\partial_{z} \sigma_n|^2 |K|\ \e^{\sigma_n} \d \mu +
\int_{\Omega_{n3}} |\partial_{z} \sigma_n| |g| |K|\ \e^{\sigma_n} \d
\mu\\[.2pc]
&\leq 3 D^2,
\end{align*}
where $D^2 = \max |g|^4 \mu(M)$, where $\mu(M)$ is the volume of $M$.
This follows from the two conditions $|K|\ \e^{\sigma_n} \leq |g|^2$ and
$ |\partial_{z} \sigma_n| \leq |g|$ on $\Omega_{n3}$. $D^2$ is a finite
constant ($\max |g|^4$ is finite since $K$ is non-zero and the volume of
$M$ is finite), independent of $n$. Thus the result follows.\hfill$\Box$\vspace{.3pc}
\end{proof}

\subsection{\it Pointwise convergence of $\sigma_n$}

\setcounter{defin}{2}
\begin{lemma}
$\{ \sigma_n \}$ is uniformly bounded in $W^{2,2}(M)$.
\end{lemma}

\begin{proof}
Let us recall that there is a Green's function  on the compact manifold
$M$ with boundary such that if $u \in C^{2}(M)$ then
\begin{equation*}
-u(x) = \int \int_{M} G(x,z) \Delta u(z) \d \mu(z) +
\int_{\partial M} \frac{\partial G}{\partial \nu}(x, w) u (w) \d l
\end{equation*}
(see for e.g.~\cite{C}, p.~174).  In particular if $u|_{\partial M} = 0$
then
\begin{equation*}
-u(x)= \int \int_{M} G(x,z) \Delta u (z) \d \mu(z).
\end{equation*}
Thus there exists a Green's operator $G$ such that $\Delta G u = -u$ and
$G \Delta u = - u$ (see ~\cite{C}, p.~177). Let $u = \sigma_n.$ Since
$\sigma_n |_{\partial M} =0$, $G \Delta \sigma_n = -\sigma_n$.

Next we will show that since by Lemma~2.2, $\Delta \sigma_n$ is uniformly
bounded in $L^2(M)$, $\sigma_n$ is uniformly bounded in $W^{2,2}(M)$.

Let $\tau_n = \Delta \sigma_n$. We know that $\tau_n|_{\partial M}=0$
and is uniformly bounded in $L^2(M)$, i.e., $\| \tau_n \|_2 < C, \forall
n$. Then $\tau_n$ can be written as $\tau_n = \sum_{i=1}^{\infty} c_n^i
\phi_i $ where $\phi_i$ is a solution to the equation $\Delta_h \phi +
\lambda_i \phi = 0$, with $\phi|_{\partial M} = 0$, (\cite{C},
pp.~8--9). $\{\phi_i\}_{i=1}^{\infty}$ is an orthonormal sequence of
eigenfunctions of the Laplacian. In fact since we are in the Dirichlet
case, the least eigenvalue $\lambda_1 >0$. We will show that since $\{
\tau_n \} $ is uniformly bounded in $L^2(M)$, $\{ G\tau_n \}$ is
uniformly bounded in $W^{2,2}(M)$. $G \tau_n = \sum_{i=1}^{\infty} c_n^i
G \phi_i$. Since $\phi_i|_{\partial M} = 0$, $G \Delta \phi_i = -
\phi_i$. Since $\Delta_h \phi_i = -\lambda_i \phi_i$, $G \phi_i =
{\phi_i}/{\lambda_i}$. Thus $G\tau_n = \sum_{i=1}^{\infty}
({c_n^i}/{\lambda_i}) \phi_i$. Since $\lambda_i \rightarrow \infty$ and
since $\| \tau_n \|_2 = (\sum_{i=1}^{\infty} c_n^{i2})^{1/2} < C$, $\| G
\tau_n \|_2 = (\sum_{i=1}^{\infty} ({c_n^i}/{\lambda_i})^2)^{1/2} < C,$
$\forall n$ if $\lambda_1 >1$. If $\lambda_1 <1$, $\| G \tau_n \|^2_2
\leq ({C^2}/{\lambda_1^2})$ $\forall n$. Thus $\|G \tau_n\| \leq C_2\
\forall n$ where
\begin{align*}
C_2 &= C\quad \hbox{if}\ \ \lambda_1 >1\\
&= \frac{C}{\lambda_1}\quad \hbox{if}\ \ \lambda_1 <1.
\end{align*}

Next we show that the first and second derivative norms of $G \tau_n$
are uniformly bounded in $L^2(M)$. The second derivative is defined as
$\nabla ^2 \tau = \nabla^1 \nabla \tau$ where $\nabla \tau $ is the
usual covariant derivative of $f$ taking values in $T^*M$ and $\nabla^1$
is the covariant derivative of sections of $T^*M$.

First we show that the first derivative norm is uniformly bounded. For
this we observe that the norm of the Laplacian is bounded uniformly.
This is because $\Delta_h G \tau_n = -\tau_n $ and $\tau_n$ is uniformly
bounded in $L^2(M)$. Thus $\| \Delta_h G \tau_n \|_2 < C$.

To show that  $\| \nabla G \tau_n \|^2_2 <C^2$ we note that since $G\tau_n$
is zero on $\partial M$,
\begin{align*}
\int_M | \nabla G \tau_n |^2 \d \mu &= - \int_M G\tau_n \Delta_h G \tau_n\\[.2pc]
& \leq \| G \tau_n \|_2 \| \Delta G \tau_n \|_2\\[.2pc]
& \leq C_2 C.
\end{align*}

Next we show that the second derivative norm $\| \nabla^2 G \tau_n \|_2
< C_1$, $\forall n$. The norm of the Laplacian is bounded in $L^2(M)$.
Using Weitzenbock formula we show that then the second derivative norm
is uniformly bounded. By Weitzenbock formula ~\cite{B}, $\Delta_h =
\nabla^* \nabla + \tau(K_0) $ where $\nabla^*$ is the adjoint of the
covariant derivative. Note that $\nabla^* \nabla = - \nabla^{1} \nabla =
- \nabla^2$ (\cite{B}, p.~52). $\tau(K_0)$ is some continuous function
of the curvature $K_0$. In other words, $\Delta_h (G\tau_n) = - \nabla^2
(G\tau_n) + \tau(K_0) (G \tau_n)$. Thus by Minkowski's inequality,
\begin{align*}
\| \nabla^2 (G \tau_n) \|_2 &\leq \| \Delta_h (G \tau_n) \|_2 +
\| \tau(K_0) G \tau_n \|_2\\[.2pc]
& \leq C + \min_M |\tau(K_0)| \| G \tau_n \|_2\\[.2pc]
& \leq C + \min_M |\tau(K_0)| C_2\\[.2pc]
& \leq C_1.
\end{align*}
Since $M$ is a compact manifold with boundary, $\min_M |\tau(K_0)|$ is bounded.

Thus $G\tau_n $ is uniformly bounded in $W^{2,2}$ since $\| G \tau_n\|_2$,
$\| \nabla (G \tau_n)\|_2$, $\| \nabla^2 (G \tau_n)\|_2 $ are uniformly bounded.
Since $\tau_n = \Delta_h \sigma_n$ and $G \Delta_h \sigma_n
= \sigma_n$, $\sigma_n$ are uniformly bounded in $W^{2,2}$.\hfill $\Box$\vspace{.4pc}
\end{proof}

Now we can formulate the main result of this subsection.

\newpage

\begin{proposition}$\left.\right.$\vspace{.5pc}

\noindent The sequence $ \{\sigma_n \}_{n=1}^{\infty}$
contains a subsequence $ \{ \sigma_{l_n}\}_{n=1}^{\infty}$ with
the following properties{\rm :}

\begin{enumerate}
\renewcommand{\labelenumi}{\rm (\alph{enumi})}
\item The sequences $\{ \sigma_{l_n} \}_{n=1}^{\infty}$ and
$\{\e^{\sigma_{l_n}}\}$ converge in $C^0(M)$ topology to continuous
functions $\sigma$ and $\e^{\sigma}$ respectively. Moreover{\rm ,}
$\sigma \in W^{2,2}(M)$.

\item The subsequence $\{\Delta_h \sigma_{l_n} \}$ converges
weakly in $L^2$ to $f\doteq \Delta_h^{\rm distr} \tilde{\sigma}$ -- a
distribution Laplacian of $\sigma$.

\item Passing to this subsequence $\{\sigma_{l_n} \}${\rm ,} the
following limits exist{\rm :}
\begin{align*}
\hskip -.55cm &\lim\limits_{n\rightarrow\infty}\|\Delta_h
\sigma_{l_n}\|_2 =\| \Delta_h^{\rm distr} \sigma \|_2,\\
\hskip -.55cm &\lim\limits_{n\rightarrow\infty}S[\sigma_{l_n}]
=S_0=\int_M \left( K_0-\frac{1}{2} \Delta_h^{\rm distr} \sigma -K
\e^{\sigma} \right)^2 \d\mu.
\end{align*}
In fact{\rm ,} the convergence in {\rm (b)} is strong in $L^2$.
\end{enumerate}
\end{proposition}

\begin{proof}
Part (a) follows from the Sobolev embedding theorem and Rellich
lemma for compact manifolds with boundary, since, for $\dim M =
2$, the space $W^{2,2}(M)$ is compactly embedded into $C^0(M)$
(see, e.g.~\cite{A}). Therefore the sequence $\{\sigma_n\}$,
which  according to Lemma~2.3, is uniformly bounded in
$W^{2,2}(M)$, contains a convergent subsequence in $C^0(M)$.
Passing to this subsequence $\{ \sigma_{l_n} \}$ we can assume
that there exists a function $\sigma \in C^0(M)$ such that
\begin{equation*}
\lim\limits_{n\rightarrow\infty}\sigma_{l_n}=\sigma.
\end{equation*}
Since $\sigma_n$'s are uniformly bounded in a Hilbert space
$W^{2,2} (M)$, they weakly converge to $s\in W^{2,2}(M)$ (after
passing to a subsequence if necessary). The uniform limit
coincides with $s$  so that $\sigma=s\in W^{2,2}(M)$.
\end{proof}

In order to prove (b), set $\psi_n=\Delta_h \sigma_{l_n}$ and
observe that, according to part (a) of Lemma~2.2, the sequence
$\{\psi_n\}$ is bounded in $L^2$. Therefore, passing to a
subsequence, if necessary, there exists $f\in L^2(M)$ such that
\begin{equation*}
\lim\limits_{n\rightarrow\infty}\int_M\psi_n g=\int_M fg
\end{equation*}
for all $g\in L^2(M)$. In particular, considering $g\in
C^{\infty}(M)$, this implies $f =\Delta_h^{\rm distr} \sigma$.

In order to prove (c) we use the following lemma.\vspace{.2pc}

\begin{lemma}
If a sequence $\{\psi_n\}$ converges to $f\in L^2$ in the weak
topology{\rm ,} then
\begin{equation*}
\lim\limits_{n\rightarrow\infty}\|\psi_n\|\geq\|f\|.
\end{equation*}
Further $\lim_{n \rightarrow \infty} \| \psi_n \| = \|f \| $ iff there is
strong convergence.\vspace{.2pc}
\end{lemma}

\begin{proof}
The lemma follows from considering the following inequality:
\begin{equation*}
\lim\limits_{n \rightarrow \infty} \int ( \psi_n - f )^2 \d \mu \geq 0.
\end{equation*}

$\left.\right.$\vspace{-2.6pc}

\hfill$\Box$\vspace{1.4pc}

To continue with the proof of the proposition, suppose $\lim_{n\rightarrow
\infty}\|\psi_n\|>\|f\|$. Using the definition of the functional, we have
\begin{align*}
S[\sigma_n] &= \int_M \left(K_0 - \frac{1}{2}\Delta_h \sigma_n
-K\e^{\sigma_n}\right)^2 \d\mu\\[.2pc]
&= \frac{1}{4}\|\psi_n\|^2+\|K_0
-K\e^{\sigma_n}\|^2 - \int_M \psi_n(K_0-K\e^{\sigma_n})\d\mu.
\end{align*}
From parts (a) and (b) it follows that the sequence $S[\sigma_n]$
converges to $S_0$ and
\begin{align*}
S_0 &=\lim\limits_{n\rightarrow\infty} S[\sigma_n]\\[.2pc]
&=\lim\limits_{n\rightarrow \infty}\frac{1}{4}\|\psi_n\|^2+\|K_0-K\e^{\sigma}\|^2-
 \int_M f(K_0-K\e^{\sigma})\d\mu\\[.2pc]
&\geq\frac{1}{4}\|f\|^2+\|K_0-K \e^{\sigma}\|^2-\int_M f(K_0-K
\e^{\sigma}) \d\mu\\[.2pc]
&=\bigg\|-\frac{1}{2}f+K_0-K \e^{\sigma}\bigg\|^2.
\end{align*}

We will show that this is an equality since the inequality contradicts
the fact that $\{\sigma_n\}$ was a minimizing sequence. That is we can
construct a sequence $\{\tau \} \in C^{\infty}(M)$ such that $S[\tau]$
gets as close to $\big\|-\frac{1}{2}f+K_0-K\e^{\sigma}\big\|^2$ as we like.

Namely, for any $\epsilon>0$ we can construct, by the density of
$C^{\infty}$ in $W^{2,2}$, a function $\tau\in C^{\infty}(M)$
approximating $\sigma \in W^{2,2}$ such that
$\|\Delta_h\tau-f\|<\epsilon$ and $\|4(\e^{\tau}- \e^{\sigma})\|<
\epsilon/2$. Since
\begin{equation*}
S_{\tau} = \lim\limits_{n \rightarrow \infty}
S[\tau]=\bigg\|-\frac{1}{2}\Delta_h\tau+K_0 -K \e^{\tau}\bigg\|^2,
\end{equation*}
we have\vspace{.2pc}
\begin{equation*}
\bigg\| \sqrt{S_{\tau}}-\bigg \|-\frac{1}{2}f+K_0-K \e^{\sigma} \bigg\| \leq
\bigg\|\frac{1}{2}(f-\Delta_h\tau)-K(\e^{\tau}-
\e^{\sigma})\bigg\|\leq\epsilon.
\end{equation*}

$\left.\right.$\vspace{-1.1pc}

\noindent Now setting $\delta=\sqrt{S_0}-\|-\frac{1}{2}f+K_0-K
\e^{\sigma}\|>0$ and choosing $\epsilon <\delta/2$, and using
$\sqrt{S_{\tau}}\leq\|-\frac{1}{2}f+K_0-K \e^{\sigma}\|+ \epsilon$
we get, $\sqrt{S_{\tau}}<\sqrt{S_0}-\frac{\delta}{2}$ -- a
contradiction, since $S_0$ is the infimum of the functional.

Thus, $\lim_{n\rightarrow\infty}\|\Delta_h \sigma_n\|=\|f\|$, so
that, in fact, by Lemma~2.5, the convergence is in the strong
$L^2$ topology. This proves part (c).\hfill$\Box$
\end{proof}

\section{Smoothness and uniqueness}

Here we complete the proof of the main Theorem~2.1 by showing that\vspace{.2pc}

\setcounter{equation}{0}
\setcounter{defin}{0}
\begin{proposition}$\left.\right.$
\begin{enumerate}
\renewcommand{\labelenumi}{\rm (\alph{enumi})}
\item The minimizing function $\sigma\in C^0(M)$ is smooth
and unique and corresponds to a metric of  negative curvature $K$.

\item $\sigma$  extends to $M - \cup_{i=1}^{n} p_i$ by zero.
\end{enumerate}\vspace{-1.5pc}
\end{proposition}

\pagebreak

\begin{proof}
(a) Let $b=(K_0-\frac{1}{2}\Delta_h^{\rm
distr}\sigma-K\e^{\sigma})\in L^2(M)$; note that $b=0$ on the boundary
$\partial M$, since $\sigma|_{\partial M} = 0$ and by redefining
$\Delta^{\rm distr}_h \sigma$ on the boundary, we can take $\Delta^{\rm
distr}_h \sigma|_{\partial M} = 0$ and we know that $K|_{\partial M} =
K_0$. According to Proposition~2.4, (c) $S_0= \int_M b^2\d\mu$.
Set $G(t)=S(\sigma+t\beta)-S_0$, where $\beta\in \{ f \in C^{\infty}(M)
| \partial_{\nu} f|_{\partial M} = 0, f|_{\partial M} =0 \}$. $G(t)$ for
fixed $\beta$ is smooth, $G(0)=0$ and $G(t)\geq 0$ for all $t$.
Therefore,
\begin{equation*}
\frac{\d G}{\d t}\bigg|_{t=0}=0.
\end{equation*}
A simple calculation yields
\begin{equation*}
\frac{\d G}{\d t}\bigg|_{t=0}=\int_M(-b\Delta_h\beta-2K\e^{\sigma}b\beta) \d\mu.
\end{equation*}
Now, $\int_M b \Delta_h \beta\ \d \mu - \int_M \beta \Delta_h b\ \d \mu
= \int_{\partial M} b \partial_{\nu} \beta\ \d l - \int_{\partial M}
\beta \partial_{\nu} b\ \d l =0$. Thus $b\in L^2(M)$ satisfies, in a
distributional sense, the following equation:
\begin{equation} \label{eq}
 -\Delta_h b-2K\e^{\sigma} b=0.
\end{equation}

First, we will show that $b=0$ is the only weak $L^2$ solution to
eq.~\eqref{eq}. Indeed, by elliptic regularity $b$ is smooth, so that multiplying
\eqref{eq} by $b$ and integrating over $M$ using the Stokes formula, we get
\begin{equation*}
\int_M \nabla b \cdot \nabla b -\int_{\partial M} \partial_{\nu} b \cdot
b - \int_M 2 K b^2\e^{\sigma}\d\mu=0.
\end{equation*}
The second term drops since we redefined $\Delta^{\rm distr} \sigma$ to be zero on the boundary and had $K=K_0$ on the boundary, thus   $b=0$ on $\partial M$. Recalling
that $K<0$ on $M$, all the remaining terms are positive or zero,
$b=0$ everywhere. Thus we have shown that $S_0=0$.

Secondly, equation $b=0$ for the minimizing function $\sigma\in
C^0(M)$ reads
\begin{equation} \label{boot}
\frac{1}{2}\Delta_h^{\rm distr}\sigma=K_{0}-K\e^{\sigma}\in C^{0}(M).
\end{equation}
Therefore, $\Delta_h^{\rm distr}\sigma$ belongs to $L^p(M)$ so that
$\sigma\in W^{2,p}$ for all $p$. By the Sobolev embedding theorem
it follows that $\sigma\in C^{1,\alpha}(M)$ for some
$0<\alpha<1$. Therefore, the right-hand side of eq.~\eqref{boot}
actually belongs to the space $C^{1,\alpha}(M)$, and
therefore $\sigma\in C^{3,\alpha}(M)$ and so on. This kind of
bootstrapping argument shows that $\sigma$ is smooth.

Equation $b \equiv 0$ satisfied by $\sigma$ now translates to
$K(\sigma) \equiv K$, where $K(\sigma)$ is the Gaussian curvature of the
metric ${\rm e}^{\sigma}h\ \d z \otimes \d \bar{z}$, $\sigma \in C^{\infty}(M)$.

Next to show uniqueness, let $\eta$ be another solution in class $T$. Then
$\eta$ satisfies
\begin{equation} \label{boot}
\frac{1}{2}\Delta_h^{\rm distr}\eta=K_{0}-K\e^{\eta}\in C^{0}(M)
\end{equation}
so that
\begin{equation*}
\Delta_h (\sigma - \eta) = - 2 K (\e^{\sigma} - \e^{\eta}).
\end{equation*}
Multiplying this equation by $\sigma - \eta$ and remembering that $\sigma$
and $\eta$ and their normal derivatives vanish on the boundary, we get
\begin{equation*}
- \int_M \d \zeta \wedge * \d \zeta = \int_M -2K (\sigma - \eta)(\e^{\sigma} -
\e^{\eta}) \d \mu,
\end{equation*}
where we set $\zeta = \sigma - \eta$. Since $-2K(\sigma - \eta) (\e^{\sigma} -
\e^{\eta}) \geq 0$, we conclude that $\d \zeta = 0$ and in fact $\zeta = 0$.

(b) $\sigma$ is in class $T$, i.e. $\sigma|_{\partial M} = \partial_{\nu}
\sigma |_{\partial M} = \Delta_h \sigma |_{\partial M} = 0$. Supposing we
minimize over the class of functions all of whose derivatives are zero at
$\partial M$. Let $\tau$ be the smooth function to which this minimizing
sequence converge to. By uniqueness of the solution, $\tau = \sigma$.
We can extend $\tau$ by zero on the discs $\cup_{i=1}^{n} D_i$.
Thus we can extend $\sigma$ by zero as well. \hfill $\Box$
\end{proof}

\section{An example of the initial metric}\vspace{.2pc}

Let $\Sigma - \{ p_i \}_{i=1}^{n}$ be the Riemann surface with
punctures. Let $D_i= \{ (r_i, \theta_i): 0< r_i < 1 \}$ be a disc about
$p_i$ such that $p_i$ is the only puncture on $D_i$. Recall $M = \Sigma
- \cup_{i=1}^{n} D_i $. $\partial M = \cup_{i=1}^{n} \partial D_i =
\cup_{i=1}^{n} \{ (r_i, \theta_i): r_i = 1 \}$. Let $D_i^{\prime} = \{
(r_i, \theta_i): 0 <r_i < 1.5 \} $ and $D_i^{\prime \prime} = \{ (r_i,
\theta_i): 0< r_i < 2 \}$. Let $f_i\ \d z \otimes \d \bar{z}$ be a
metric on $D_i^{\prime \prime}- \ p_i$. Let $\tilde{h}\ \d z \otimes \d
\bar{z}$ be a metric on $M$. We wish to construct a metric which will
interpolate between the metric on $M$ and the metric on $D_i^{\prime
\prime} $, $i=1,\ldots,n$. Let
\begin{align*}
\rho_i &= 1\quad \hbox{on}\  \{ 0 \leq r_i \leq 1.5 \}\\
       &= 0\quad \hbox{on}\  \{ r_i \geq 2 \}\\
&\quad\ \, 0 < \rho < 1 \quad \hbox{on}\ \{ 1.5 < r_i < 2 \}\\
\intertext{and}
\tilde{\rho} &= 1\quad \hbox{on}\ \Sigma - \cup_{i=1}^{n} D_i^{\prime \prime}\\
&= 0 \quad \hbox{on}\ D_i^{\prime}\ \forall i\\
& \quad\ \, 0 <  \tilde{\rho} < 1 \quad \hbox{on}\ \{ 1.5 < r_i <2 \}.
\end{align*}

Let
\begin{equation*}
h\ \d z \otimes \d \bar{z} = \sum\limits_{i=1}^{n} \rho_i f_i\ \d z \otimes \d \bar{z} + \tilde{\rho} \tilde{h}\ \d z \otimes \d \bar{z}.
\end{equation*}
This is a metric because $h$ is positive everywhere. On $D_i^{\prime}$,
$h\ \d z \otimes \d \bar{z} = f_i\ \d z \otimes \d \bar{z}$ and on
$\Sigma - \cup_{i=1}^{n} D_i^{\prime \prime}$, $h\ \d z \otimes d
\bar{z} = \tilde{h}\ \d z \otimes \d \bar{z}$.

We can choose $f_i$ such that the Gaussian curvature $K_0$ of $h\ \d z
\otimes \d \bar{z}$ is negative on $\partial M$. One such example would
be to choose $f_i\ \d z \otimes \d \bar{z} $ on the disc
$D_i^{\prime}-p_i $\ \,to be\ \,$ -\log(\frac{r_i}{4}) [(\d r_i)^2 + r_i^2 (\d
\theta_i)^2 ] $, where $(r_i, \theta_i )$ are standard polar coordinates
on the disc. This metric has a true singularity at the puncture. $K_0 =
\frac{1}{2 \sqrt{EG}}
\left[\left(\frac{E_{\theta_i}}{\sqrt{EG}}\right)_{\theta_i} +
\left(\frac{G_{r_i}}{\sqrt{EG}}\right)_{r_i}\right]$ where $E=
-\log\left(\frac{r_i}{4}\right)$, $G= -r_i^2
\log\left(\frac{r_i}{4}\right)$ \cite{DoC}. On $D_i$, $K_0 = \frac{1}{2
r_i^2 \left(\log (r_{i}/4)\right)^3}$ and therefore
negative. In particular, on $\partial M = \cup_{i=1}^{n} \{(r_i,
\theta_i): r_i=1 \}$, $K_0$ is negative.

\section*{Acknowledgement}\vspace{.2pc}

The author would like to thank V~Pati and E~K~Narayanan for very useful
discussions. The author would also like to thank the referee for his
comments.

config: arxiv


TITLE: Asymptotic distribution of $\lambda_1$ under the $z$-measure for partitions
QUESTION [11 upvotes]: The following question about $z$-measures on Young diagrams came up in some ongoing work with Ofir Gorodetsky. I recall the background and then state our question below in the box.
For parameters $z$ and $z'$, define a measure on $\lambda \vdash n$ with the weights
$$
M_{n,z,z'}(\lambda):= \frac{(\dim \lambda)^2}{n!\, (z z')_n} \prod_{\square \in \lambda} (z + c(\square))(z' + c(\square)).
$$
Here $\dim \lambda$ is the number of standard Young tableaux of shape $\lambda$, the product is over all squares $\square$ in the diagram of $\lambda$, and $c(\square) := \mathrm{column}(\square) - \mathrm{row}(\square)$ is the content of $\square$. $(t)_n := t (t+1)\cdots (t + n-1)$. Definitions of the terminology used here can be found in e.g. vol. II of Stanley's Enumerative Combinatorics. Kerov discovered the remarkable fact that these weights induce a probability measure:
$$
\sum_{\lambda \vdash n} M_{n, z, z'}(\lambda) = 1.
$$
(At any rate they sum to $1$; they are a probability measure if all of them are non-negative.) Some of the proofs of this fact can be found for instance here or in section 8 here.
Let $\lambda^{(z,z',n)}$ be a random partition of $n$ according to these weights. We are interested in the limiting distribution of $\lambda_1$ under these weights in the case that $z = z' = \alpha > 0$. For $c \in [0,1]$, define
$$
F_\alpha(c) := \lim_{n\rightarrow\infty}\mathbb{P}(\lambda_1^{(\alpha, \alpha, n)} \leq cn) = \lim_{n\rightarrow\infty}\sum_{\substack{\lambda \vdash n \\ \lambda_1 \leq cn}} M_{n,z,z'}(\lambda).
$$
This limit is known to exist (and in fact converges to the distribution of $\alpha_1$ of the Thoma complex under the spectral z-measure; see this paper of Borodin). There are a variety of formulas characterizing the spectral z-measure in that paper (e.g. Theorem 2.2.1 for correlation functions of the Thoma point process), but we are interested in positivity of $F_\alpha(c)$ and have not been successful in applying these formulas to this. More exactly, we are interested in the following question:

Question: Is it true for each $\alpha \in (0,1)$, that $F_\alpha(c) > 0$ for all $c > 0$?

We are also interested in the behavior of $F_\alpha(c)$ when $\alpha > 1$ (we really understand matters completely only when $\alpha$ is a positive integer), and any information relevant would be interesting, but the question above is the one that is really pressing.
Below are some approximate graphs (generated from $M_{n,\alpha,\alpha}$ with $n=45$) of $F_\alpha(c)$ for $\alpha = .01, .1, .3, .5,$ and $.7$. The graph for $\alpha = .01$ is at the top and $\alpha = .7$ is on the bottom. $c$ of course is the horizontal axis.

(The apparent changes in convexity for $c$ near $1$ in the bottom two graphs are just a remnant of polynomial interpolation and should be ignored.)

REPLY [4 votes]: A positive answer to Brad and mine's question is the main result of a recent paper of Olshanski, titled "The Topological Support of the z-Measures on the Thoma Simplex" (link). His proof is based on a previous work of his, as well as a recent paper of Korotkikh.
Interestingly, this question has (conjecturally) a simple number-theoretical interpretation, in terms of the distribution of the $z$th divisor function in short intervals and in arithmetic progressions. See Conjecture 4 in Brad and mine's preprint.


TITLE: Suppose $f$ is a one-one function from $X$ to $Y$ and $f^{-1}(U)\subset f^{-1}(V)$, where $U$,and $V$ be the subsets of $Y$,then$U\subset V$?
QUESTION [0 upvotes]: Can someone prove this. In general it is not true i.e. If $f^{-1}(U)\subset f^{-1}(V)$, then it is not always true that $U\subset V$ in $Y$.

REPLY [0 votes]: Here's my suggested answer, written in haste, so read it critically:
$f$ is a bijection $f \colon X \rightarrow Y$, so $g :=f^{-1}$ is a well defined bijection $g \colon Y \rightarrow X$, with $g(f^{-1}(U)) = U$ and $g(f^{-1}(V)) = V$. $f^{-1}(U) \subset f^{-1}(V)$ means that $\forall \ x \in f^{-1}(U) \colon \ x \in f^{-1}(V)$, and $\forall \ y \in U \exists \ x \in f^{-1}(U) \colon y = g(x)$. From this one can conclude that $\forall y \in U: y \in V$, ie. $U \subset V$


TITLE: How prove this $\sin{t}=t-\frac{\sin{(\theta t)}}{2}t^2$
QUESTION [0 upvotes]: show that:there exsit $\theta\in(0,1)$,such that

$$\sin{t}=t-\dfrac{\sin{(\theta t)}}{2}t^2\tag{1}$$

the book say this is Taylor formula
But I kown that Taylor formula, there exsit $\xi$ such that
$$\sin{t}=t-\dfrac{-cos{\xi}}{3!}t^3$$
How about prove $(1)$? Thank you

REPLY [2 votes]: Taylor's formula in general says

Let $f \colon [x_0, x_0+t] \to \mathbb R$ be $(n+1)$-times differentiable. Then there is a $\theta \in [0,1]$ such that
  $$ f(x_0+t) = \sum_{k=0}^n \frac{1}{k!}f^{(k)}(x_0)t^k + \frac{1}{(k+1)!}f^{(k+1)}(x_0+\theta t)t^{k+1} $$

Apply this with $k=1$ and $x_0 = 0$ here.


TITLE: Trigonometric identity problem
QUESTION [0 upvotes]: $$\frac{\sin(2x)}{2 \sin (x)}-\frac{\cos(2x)}{\cos(x)+\sin(x)}=\sin(x).$$
I got it on a test and want an answer. I always hit a dead end with the identities I learned.

REPLY [1 votes]: $\sin(2x) = 2\sin(x)\cos(x)$, and $\cos(2x) = \cos^{2}(x) - \sin^{2}(x)$, which factors as $(\cos(x) - \sin(x))(\cos(x) + \sin(x))$. That turns your identity into $$ \frac{2\sin(x)\cos(x)}{2\sin(x)} - \frac{\cos^{2}(x) - \sin^{2}(x)}{\cos(x) + \sin(x)} = \cos(x) - (\cos(x) - \sin(x)) = \sin(x).$$