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\begin{document}

\title{SOME SHARP INEQUALITIES RELATED TO MOSER-TRUDINGER-ONOFRI INEQUALITY}
\author{Suyu Li}
\date{April 2014}

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{ %\large
UNIVERSITY OF OKLAHOMA
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GRADUATE COLLEGE
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SOME SHARP INEQUALITIES RELATED TO
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MOSER-TRUDINGER-ONOFRI INEQUALITY
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A DISSERTATION
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SUBMITTED TO THE GRADUATE FACULTY
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in partial fulfillment of the requirements for the
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Degree of
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DOCTOR OF PHILOSOPHY
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By
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SUYU LI
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Norman, Oklahoma
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2014
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SOME SHARP INEQUALITIES RELATED TO MOSER-TRUDINGER-ONOFRI INEQUALITY
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A DISSERTATION APPROVED FOR THE
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DEPARTMENT OF MATHEMATICS
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BY
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\  \  \  \  \  \  \  & Dr. Meijun Zhu, Chair\\
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  & Dr. Yiqi Luo \\
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 & Dr. John Albert\\
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 & Dr. Murad \"{O}zayd{\i}n \\
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   & Dr. Nikola Petrov  \\
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\copyright \    Copyright by SUYU LI \   2014

All rights reserved.
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DEDICATION
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To my parents Fuliang Li and Guiling Wang.
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To my husband Zhaowen and our children Jerry and Jasper.
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Words cannot express how much I love you all.
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\chapter*{Acknowledgements}

\ \ \ \ \ I would like to express my deep appreciation and gratitude to my PhD advisor, Dr. Meijun Zhu, for his guidance, encouragement and patience throughout my graduate studies. I am grateful to have him as my advisor.

I am also grateful to Dr. Albert, Dr. Luo, Dr, \"{O}zayd{\i}n and Dr. Petrov. I thank them for serving on my committee and spending time reviewing this work. I also thank the OU Mathematics department for the financial support and the teaching and guidance of the faculty, stuff and students.

Last, but not least, I would like to thank my dear husband and best friend Zhaowen for his love and support during all these years, and for the wonderful sons we have been raising. My parents, parents-in-law and my sister receive my love for their continuous support. Without any one of them, this dissertation would not have been possible.


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%\begin{abstract}
\chapter*{Abstract}

In this dissertation, we focus on the study of sharp inequalities of Moser-Trudinger-Onofri type. We first establish the analog Bliss and Hardy inequalities with sharp constant involving exponential weight function. One special case of the inequalities (for $n=2$ ) leads to a direct proof of Onofri inequality on $S^2$. Then we
establish the sharp trace inequalities on any smooth bounded simply connected domain in $\mathbb{R}^2$.
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\chapter{Introduction}
The sharp Sobolev inequality on $\mathbb{R}^n$ ($n \ge 3$) was
obtained  by Aubin \cite{AU}  and Talenti \cite{TA}, respectively,
in 1976: for all $u(x)$ satisfying
$u\in L^{2n/(n-2)}(\mathbb{R}^n)$ and $\nabla u\in L^{2}(\mathbb{R}^n)$
\begin{equation}
 (\int_{\mathbb{R}^n} |u|^{\frac{2n}{n-2}} )^{\frac{n-2}n}  \le S_1  \int_{\mathbb{R}^n}
|\nabla u|^2, \label{0-1}
\end{equation}
where  $1/S_1=\pi n(n-2)(\Gamma(n/2)/\Gamma(n))^{2/n}$ is called
the best constant of Sobolev inequality. Extremal functions were
also given by them. By even reflection, One can have the sharp Sobolev inequality on the upper half space:
$$(\int_{\mathbb{R}^n_+} |u|^{\frac{2n}{n-2}} )^{\frac{n-2}n}  \le 2^{\frac2n}S_1  \int_{\mathbb{R}^n_+}
|\nabla u|^2$$
for all $\ u\ $ satisfying $\ u\in L^{2n/(n-2)}(\mathbb{R}^n_+)\ $ and $\nabla u\in L^{2}(\mathbb{R}^n_+),$
where $\mathbb{R}^n_+=\{(x_1,..., x_n) \ |  \ x_n>0\}$ is the upper half space.
And (\ref{0-1}) also implies the following
local sharp Sobolev inequality: on any smooth bounded domain
$\Omega \subset \mathbb{R}^n$,
$$
(\int_{\Omega} |u|^{\frac{2n}{n-2}} )^{\frac{n-2}n}  \le S_1
\int_{\Omega} |\nabla u|^2,  \ \ \ \forall \ \ u \in
H^1_0(\Omega).$$ Based on this local  inequality, Aubin [2] was
able to establish the sharp type Sobolev inequality on any compact
Riemmanian manifolds via the partition of unity: if $M$ is  an
n-dimensional compact Riemmanian manifold ($n \ge 3$), then for
every $\epsilon>0$, there is a constant $C=C(\epsilon)>0$, such
that
\begin{equation}
(\int_M |u|^{\frac {2n}{n-2}} dv_g)^{\frac{n-2}n} \le
(S_1+\epsilon) \int_M |\nabla_g u|^2 dv_g +C(\epsilon) \int_M u^2
dv_g. \label{0-2}
\end{equation}
Inequality (\ref{0-2}) is successfully used by Aubin  to settle
down the Yamabe problem for high dimensional ($n \ge 5$) and
non-conformally flat manifolds. Later, Yamabe problem was
completely solved through the work of Schoen \cite{Sch}.

It is certainly an interesting mathematical problem to study the
behavior of $C(\epsilon)$ as $\epsilon$ tends to zero. In fact
it was conjectured by Aubin in 1976 that on any
smooth n-dimensional compact Riemannian manifold without boundary
$(M, g)$ ($n\ge 3$),  there is a constant $C(M,g)$ such that
$\forall u \in H^1(M)$
\begin{equation}
(\int_M |u|^{\frac {2n}{n-2}} dv_g)^{\frac{n-2}n} \le S_1 \int_M
|\nabla_g u|^2 dv_g +C(M,g) \int_M u^2 dv_g. \label{0-3}
\end{equation}

Aubin's conjecture was later proved by Hebey and Vaugon \cite{HV}
in 1996.

\medskip

On the upper half space $\mathbb{R}^n_+$, the  sharp trace inequality  was proved
by  P. L. Lions \cite{LP}:
$$
 (\int_{\partial \mathbb{R}^n_+} |u|^{\frac{2(n-1)}{n-2}} )^{\frac{n-2}{n-1}}\le S
 \int_{\mathbb{R}^n_+}
|\nabla u|^2
$$
for all $u$ satisfying $u\in L^{2(n-1)/(n-2)}({\partial
\mathbb{R}^n_+})$ and $\nabla u \in L^2(\mathbb{R}^n_+),$ where
$S=2/(n-2) \cdot (2\pi^{n/2}/(\Gamma(n/2)))^{1/(1-n)}$ is called
the best constant of trace inequality. Later the extremal
functions were found by  Escobar \cite{Esc} and Beckner \cite{Bec}
independently.

Stimulated by the work of Hebey and Vaugon \cite{HV},  Y.Y. Li and
M. Zhu \cite{LYYZ} established  the corresponding sharp trace
inequality on any compact Riemannian manifold with boundary.
Namely, they showed that for any smooth $n-$dimensional compact
Riemannian manifold $(M,g) $ ( $n\ge 3$) with smooth boundary,
there exists a constant $A(M,g)>0$, such that $ \forall u \in
H^1(M)$
\begin{equation}
(\int_{\partial M} |u|^{\frac{2(n-1)}{n-2}}ds_g)^{\frac{n-2}{n-1}}
\le  S \int_M |\nabla_g u|^2 dv_g + A(M,g) \int_{\partial M
}u^2ds_g, \label{0-4}
\end{equation}
where $dv_g$ denotes the volume form of $(M,g)$  and $ds_g$
denotes the induced volume form on $\partial M.$ Using the same method, they \cite{LYYZ2}
established a sharp Sobolev inequality on any bounded domain with smooth boundary:
For any smooth bounded open set $\Omega\subset\mathbb{R}^n (n\ge3),$  there
exists some constant $A(\Omega)>0$  such that
 \begin{equation}
(\int_\Omega |u|^{\frac{2n}{n-2}})^{\frac{n-2}{n}}
\le 2^{\frac2n} S_1 \int_\Omega |\nabla u|^2  + A(\Omega) \int_{\partial \Omega
}u^2 \ \ \ \forall u \in H^1(\Omega). \label{0-5}
\end{equation}

\medskip

The two dimensional case is quite different.

\medskip

Let $\Omega$ be a smooth bounded domain in $\mathbb{R}^2$.  It is
well-known that $L^\infty(\Omega)$ is not a subset of
$H_0^{1}(\Omega)$ even if $\Omega$ is a two dimensional ball.
Nevertheless,  it was shown by Trudinger \cite{Tru} that functions
in $H^{1}_0(\Omega)$ are actually in the exponential class.

\medskip

\noindent{\bf Theorem A} ({\bf Trudinger's inequality}) \ {\it Let
$\Omega \subset \mathbb{R}^2$ be a smooth bounded domain. There are  two
positive constants $\beta_0$ and  $C_1$ ($C_1$ depending on the area
of $\Omega$) such that if $u\in H_0^{1}(\Omega)$ and $||\nabla u||_{L^2} \le 1$,
then
\begin{equation} \int_{\Omega} e^{\beta u^{2} } dx\le C_1
 \label{0-6}
\end{equation}
for all $\beta \le \beta_0$. Moreover, if $\{ u_i\}_{i-1}^\infty$ is bounded in $W_0^{1, n}(\Omega)$,
then up to a subsequence of $i$, $u_i\rightharpoonup u_0$ in $W_0^{1, n}(\Omega)$
and $\exp(nu_i)\rightarrow\exp(nu_0)$ in $L^1$ norm.}

\medskip


The key observation in Trudinger's proof is that in the following
embedding inequality
\begin{equation}
\bigg\{\frac 1{|\Omega|}\int_\Omega |u|^{p} dx \bigg\}^{1/p} \le
c(p) \bigg\{ \int_\Omega |\nabla u|^2 dx \bigg\}^{1/2},
\label{0-7}
\end{equation}
the coefficient satisfies $c(p) \le C p^{1-1/2}$. Adams in \cite{Ad} pointed out that one
can obtain the similar results involving higher order derivatives.


The best constant $\beta_0$  in Theorem A  was later found by
Moser \cite{MO}, $\beta_0= n \omega_{n-1}^{\frac1{n-1}}$
and $\omega_{n-1}$ is the $(n-1)-$dimensional surface of the unit sphere.
Carleson and Chang \cite{CC} state that there is an extremal
function which realizes the equality in Theorem A when
$\beta=\beta_0$ and $\Omega$ is a unit ball in $\mathbb{R}^n$.
For $n=3$, Onofri derived the following inequality \cite{On} (See also
\cite{HO} and \cite{OS}):

\medskip

\noindent{\bf Theorem B} ({\bf Onofri inequality}) \ {\it Let
$(S^2, g_0)$ be the standard unit sphere in $\mathbb{R}^{3}$. For any $u
\in W^{1,2}(S^2)$,
$$\ln (\frac 1{4 \pi} \int_{S^2}e^{2u}dx) \le \frac 1{4 \pi}
\int_{S^2 } (|\nabla u|^2+2u)dx. $$ The equality holds if and only
if the curvature under metric $e^{2u}g_0$ is constant.}

\medskip

Onofri proved this inequality in his study of the volume element
in string theory integrals (see also Hong \cite{HO} for an
analytic approach). It was re-discovered by Osgood, Phillips and
Sarnak in their proof of the $C^\infty$-compactness of isospectral
metrics on compact surfaces. As another important application,
Onofri inequality is used to derive the lower bound
 for the Liouville energy on a topological two-sphere, which can be used to derive
 the global existence of
 Ricci flow and Calabi flow
on 2-sphere, see, for example, Chow \cite{CHO} and Chen \cite{Che}.

The proofs of Theorem  B   by Onofri  and others rely on the
highly nontrivial sharp inequalities of Moser mentioned above
\cite{MO}. This makes their proofs more difficult to digest. In
\cite{LZ}, J. Li and M. Zhu gave a more direct and
simpler proof of Theorem B, which does not depend on Moser's sharp
inequality, but based on the following sharp local inequality:
Let $\Omega \subset \mathbb{R}^2$ be a bounded smooth domain, $\Omega^*$ be
the ball in $\mathbb{R}^2$ that has the same area as $\Omega$, and denote
$$
D_{a,b}(\Omega)=\{f(y): \ f(y)-b\in H^1_0(\Omega) , \ \int_{\Omega}
e^{2f}dy=a\},
$$
where $a (>0)$ and $b$ are two constants. Then
\begin{equation}
\inf_{w \in D_{a,b}(\Omega)}  \int_{\Omega}|\nabla w|^2 d x\ge 4
\pi \cdot (\ln \frac{a e^{-2b}}{\pi r^2} +\frac {\pi r^2}{a
e^{-2b}}-1),\label{JZ}
\end{equation}
where $r$ is the radius of $\Omega^*$.

Comparing the proof of sharp Sobolev inequality (based on the
Bliss Lemma) with that of Onofri inequality, we believe that there
are undiscovered calculus inequalities, which turn out to be the
main theorem of our paper\cite{LSZ}. We include this in Chapter 2.
One of the inequalities we derived is the following:
\begin{theorem}{\label{main}}
(1). Let $n>1$ be given.  For any nonnegative function $u\in
C^1[0, +\infty)$ with $u(0)=0$ and $\int_0^\infty e^{nu-nr} dr
=a>1/n$, we have
\begin{align}\label{anaBliss}
\int_0^\infty|u_r|^n dr &\geq \left(\frac{n}{n-1}\right)^{n-1}
\bigg\{\ln (na) - \int_{1/na}^1 \frac 1{t}(1-
 (1-t)^{n-[n]})\ d t\\&- \sum_{i=1}^{[n]-1} \frac {(na-1)^{n-i}}{(n-i)(na)^{n-i}}
\bigg\},
\nonumber
\end{align}
 where $[n]$ is the integer part of $n$. The
equality in (\ref{anaBliss}) holds if and only if $$u(r)=\ln
\frac{ na (na-1)^{-1}}{(na-1)^{-1}+e^{-nr/(n-1)}}.$$

(2). For any non-negative function $u\in C^1[0, +\infty)$ with
$u(0)=0$,
\begin{equation}
\ln \int_0^{+\infty} \frac {e^{u}}{e^{r}}dr \le
 \int_0^{+\infty}|u_r| dr,\label {n=1}
\end{equation}
the equality holds if and only if $u(r)=0$.
\end{theorem}
As a consequence of such sharp local inequality (\ref{JZ}), one can obtain a sharp
inequality on any smooth bounded domain $\Omega \subset
\mathbb{R}^2$:
\begin{equation}
\int_\Omega e^{2u} \le meas(\Omega) \cdot e \cdot exp\{\frac
1{4\pi} \int_\Omega |\nabla u|^2\}, \ \ \ \ \forall u\in
H^1_0(\Omega). \label{0-8}
\end{equation}
Also, for any smooth Riemann surface $(M, g),$ given $\epsilon>0,$
there are $r_0>0$ and $C(\epsilon)>0$ such that if $\Omega\subset B(x_0, r_0)$
for some point $x_0\in M$, then
\begin{equation}
\int_\Omega e^w dV_g \le C(\epsilon)\exp\{(\frac1{16\pi}+\epsilon)\int_\Omega |\nabla_g w|^2 dV_g\},
\ \ \ \ \forall w\in H^1_0(\Omega). \label{0-9}
\end{equation}

From inequalities (\ref{0-8}) and (\ref{0-9}), X. Chen and M. Zhu \cite{CZ} derived that
on Riemann surface $(M, g),$ given $\epsilon>0,$ there are  constants $C_1=C_1(\epsilon)$  and
$C_2=C_2(\epsilon)$ such that
$$
\int_{M} e^{w}dV_g \le C_1 \exp\{(\frac 1{16 \pi}+\epsilon)
  \int_{M}|\nabla_g w|^2 d V_g+C_2\int_M w^2\}, \ \ \ \ \ \forall  w\in H^1(M).
$$
They followed the strategy to prove Hebey and Vaugon's inequality
(\ref{0-3}), without using the uniformization theorem and Onofri inequality,
showed that on any
topological two sphere $(M, g)$, there is a constant $C(M, g)>0,$
such that, for all $u\in H^1(M),$
\begin{equation}
\int_M  (|\nabla_g u|^2+2 R_g u) dV_{g}-\ln\int_M e^u dV_g\ge
C(M^2, g), \label{0-10}
\end{equation}
where $R_g$ is twice the Gaussian curvature with respect to metric
$g$.


\medskip

Comparing at the high dimensional sharp trace inequalities, it is quite natural to seek two-dimensional sharp trace inequality. In Chapter 3 we will derive such inequality on simply connected domain in $\mathbb{R}^2$.
\begin{theorem}{\label{main2}} Let $\Omega$ be a  simply connected domain in $\mathbb{R}^2$
with bounded Geodesic curvature $K_g$.
Then there exists a constant $C(\Omega)$ such that for any $u\in
H^1(\Omega)$,
$$\ln\int_{\partial\Omega} e^u dS \le \frac 1{4 \pi}
  \int_{\Omega}  |\nabla u|^2 d y + \frac 1{2 \pi} \int_{\partial\Omega} K_g \cdot udS + C(\Omega). $$
\end{theorem}

\chapter{A Hardy Inequality and Its Applications}
The classical Hardy inequality says that for any non-negative
function $f(x)$ on $ [0, +\infty)$, if $F(x)=\int_0^x f(t) dt$,
then
$$
\int_0^\infty \left(\frac Fx\right)^k dx \le \left(\frac
k{k-1}\right)^k\int_0^\infty f^k dx,$$
where $k>1$ is a given parameter. See, for example, Inequality 327 in  the book by Hardy,
Littlewood and Polya \cite{HLP}. It is important to note that the
constant $( \frac k{k-1})^k$ is the optimal one and the equality
in the inequality never holds.  Using H\"older inequality, Hardy
and Littlewood \cite{HL} were able to derive that
$$\int_0^\infty \frac{F^l}{x^{l-\alpha}}dx \le \left(\frac k{k-1} \right)^k
\left(\int_0^\infty f^k dx\right)^{\frac l k},$$ where $l\ge k$
and $\alpha=l/k-1$.  It was quite clear to them that the constant
is not optimal for $l>k$. Though they guessed what is the best
constant, it was later  proved by Bliss, who obtained  nowadays
the famous Bliss Lemma (see the interesting papers \cite{HL, BL}):

\medskip

\noindent {\bf Bliss Lemma}: {\it Let $k,\ l$ be constants, such
that $l>k>1$, and let $f(x)$ be a   non-negative measurable
function in the interval $0 \le x<\infty$, such that the integral $J=\int_0^\infty f^k dx$ is finite. Then the integral $y=\int_0^x f dx$ is finite for every $x$ and
\begin{equation} I=\int_0^\infty
\frac{y^l}{x^{l-\alpha}} dx \le C_{b} J^{l/k},\label{bl-1}
\end{equation}
where $$\alpha=\frac lk -1, \ \ \ \ C_{b}=\frac
1{l-\alpha-1}\bigg[\frac{\alpha \Gamma(l/\alpha)}{\Gamma(1/\alpha)
\Gamma((l-1)/\alpha)} \bigg]^\alpha.$$ The equality in
(\ref{bl-1}) holds if and only if $f(x)=c/(1+d
x^\alpha)^{(\alpha+1)/\alpha}$ for some positive constants $c,\
d$. }

\medskip

The Bliss Lemma  later (after more than forty years) became a
crucial ingredient in the proof of  sharp Sobolev inequality by
Aubin \cite{AU}, and Talenti \cite{TA}, respectively. The latter
inequality has played an essential role in the resolution of the
Yamabe problem, which mainly concerns about finding a canonical
metric with constant scalar curvature on compact manifolds with
dimension higher than or equal to three.
Comparing the proof of sharp Sobolev inequality (based on the
Bliss Lemma) with the prove of Onofri inequality in \cite{LZ}
directly from Trudinger's inequality,
we believe that there are undiscovered calculus inequalities.

In this chapter, we establish the following
analog Bliss and Hardy inequalities with sharp
constant involving exponential weight function. One special case
of the inequalities (for $n=2$)  leads to a direct proof of Onofri
inequality on $S^2$.

\medskip

\noindent{\bf Theorem 1.1.}\ {\it (1). Let $n>1$ be given.  For any nonnegative function $u\in
C^1[0, +\infty)$ with $u(0)=0$ and $\int_0^\infty e^{nu-nr} dr
=a>1/n$, we have$$
\begin{array}{rll}
\int_0^\infty|u_r|^n dr &\geq \left(\frac{n}{n-1}\right)^{n-1}
\bigg\{\ln (na) - \int_{1/na}^1 \frac 1{t}(1-
 (1-t)^{n-[n]})\ d t\\&- \sum_{i=1}^{[n]-1} \frac {(na-1)^{n-i}}{(n-i)(na)^{n-i}}
\bigg\},
\nonumber
\end{array}$$
 where $[n]$ is the integer part of $n$. The
equality in (\ref{anaBliss}) holds if and only if $$u(r)=\ln
\frac{ na (na-1)^{-1}}{(na-1)^{-1}+e^{-nr/(n-1)}}.$$

(2). For any non-negative function $u\in C^1[0, +\infty)$ with
$u(0)=0$,
$$
\ln \int_0^{+\infty} \frac {e^{u}}{e^{r}}dr \le
 \int_0^{+\infty}|u_r| dr,
$$
the equality holds if and only if $u(r)=0$.}

Theorem \ref{main} immediately yields the following sharp
inequality:

\medskip

\begin{corollary}\ Let $n>1$ be given.  For any nonnegative function $u\in C^1[0, +\infty)$ with $u(0)=0$
\begin{equation}
\ln \bigg\{n \int_0^{+\infty} \frac {e^{nu}}{e^{nr}}dr\bigg\} \le
\left(\frac{n-1}n\right)^{n-1} \int_0^{+\infty}|u_r|^n dr +C_n,
\label{hardyet}
\end{equation}
where the constant $C_n$ is given by
$$C_n= \int_{0}^1 \frac 1{t}(1-
 (1-t)^{n-[n]})
\ d t+ \sum_{i=1}^{[n]-1} \frac {1}{(n-i)},$$ and $[n]$ is the
integer part of $n$. Both constants $(\frac{n-1}n)^{n-1}$ and
$C_n$ are optimal, and the equality never holds.
\end{corollary}

We first prove  inequality (\ref{anaBliss}) with a larger
coefficient in Sec. 2.1 (Proposition \ref{rough-1} below). The
argument  is elementary and simple. It needs to be pointed out
that for $n>1$ being an integer, Theorem \ref{main} can be read
out from Theorem 1.3 in \cite{LZ}. For general positive constant,
it seems impossible to prove Theorem \ref{main} from that theorem,
rather, Theorem \ref{main} provides an alternative proof of that
theorem (Corollary \ref{LZcor} in this paper). Recall the original
proof of Theorem 1.3 in \cite{LZ} does rely on Trudinger's
inequality. Quite interestingly, we also recall that Moser
\cite{MO} used a similar argument to give a very simple proof of
the improved Trundinger's inequality (with best constant):

\medskip

\begin{corollary} {\bf (Weak Moser's Inequality).}
Let $\Omega \subset \mathbb{R}^n$ (for $n \ge 2$) be a smooth bounded
domain. For any $\ \beta <n \omega_{n-1}^{1/(n-1)} \ $, there is a
constant $C(\Omega,\beta)$ depending on the volume of $\Omega$ and
$\beta$, such that for all $ \ u\in W^{1, n}_0(\Omega) \ $ with
$\int_\Omega |\nabla u|^n dx \le 1$,
$$\int_\Omega e^{\beta u^{\frac n{n-1}}} dx \le C(\Omega,\beta).$$
\label{TM-1}
\end{corollary}
Here and throughout this chapter, we use $\omega_n$ for the volume
of unit sphere $S^n$ in $\mathbb{R}^{n+1}$.  This result is slightly weaker
than Moser's inequality since it does not include the case of
$\beta= n \omega_{n-1}^{1/(n-1)}$. It seems that one needs the
argument due to Moser \cite{MO}, or Carleson and Chang \cite{CC}
to cover this extremal case.

In Sec. 2.2 we will show how to improve the rough inequality
(Proposition \ref{rough-1}) and complete the proof of  Theorem \ref{main}.
One particular reason that we can achieve this (but not
for Moser's inequality) is that we can classify all extremal
functions.

 As Bliss Lemma yields a sharp Sobolev inequality, in Sec. 2.3 we will
show that  Theorem \ref{main} can be used to give a more direct
proof of the Onofri inequality (thus without even using Trudinger's inequality).
In fact, let $B_r(0) \subset \mathbb{R}^n$ (now $n$ is an integer greater
than or equal to two) be a
 ball in $\mathbb{R}^n$ with radius $r$ centered at the origin, and
$$
D_{a,b}(B_r(0))=\bigg\{f(y)\ : \ f(y)-b\in W_0^{1,n}(B_r(0)), \ \
\int_{B_r(0)} e^{nf}dy=a\bigg\},
$$
where $a $ is  a constant  satisfying $ a> \frac {\omega_{n-1} r^n
e^{nb}}n$. We will show that Theorem \ref{main} yields:

\begin{corollary} {\bf (Local Sharp Inequality for $n=2$).}
$$\inf_{w \in D_{a,b}(B_r)}  \int_{B_r}|\nabla w|^2 d y= 4
\pi \cdot \left(\ln \frac{a e^{-2b} }{\pi r^2} +\frac {\pi r^2}{a
e^{-2b} }-1\right).
$$
\label{LZcor1}
\end{corollary}
It is known now that this corollary implies
 Onofri inequality on $S^2$, see \cite{LZ}. For the readers' convenience, we include a complete proof of
the Onofri inequality in Sec. 2.3.

 In Sec. 2.3, we shall also discuss the applications of Theorem \ref{main} to
other  geometric problems.
\section{Rough inequality}
We shall establish two elementary calculus inequalities ((\ref{mm1})
and (\ref{mm2})) in this section. The first one will be used to
prove Theorem \ref{main}, and the second one will be used to derive Corollary
\ref{TM-1}.

\begin{proposition}
(1). Let $n> 1$  and $\beta_{0}>(\frac{n-1}n)^{\frac{n-1}n}$. There
is a constant $c_1(\beta_{0})$, such that for any $u(r) \in C^1[0,
+\infty)$ satisfying $ u(0)=0$,
\begin{equation}
\ln \int_0^\infty e^{n(u-r)} dr \le \beta_{0}^n \int_0^\infty|u_r|^n
dr+c_1(\beta_{0}). \label{mm1}
\end{equation}

(2). For $u(r) \in C^1[0, +\infty)$ satisfying $ u(0)=0$,
$$
\ln \int_0^\infty \frac{e^u}{e^r} dr \le \int_0^\infty|u_r| dr.$$
Equality holds if and only if $u(r)=0$. \label{rough-1}
\end{proposition}

\medskip

For $n \ge 2$ we will improve the inequality by variational method in next section.

\begin{proof}
Let  $u(r) $ be any  function in $C^1[0, +\infty)$ satisfying $
u(0)=0$.  We have
$$u(r) \le \int_0^\infty |u_r| dr,$$
thus
$$
\int_0^\infty \frac{e^u}{e^r} dr \le \exp\{\int_0^\infty|u_r|
dr\},$$ which yields
$$
\ln \int_0^\infty \frac{e^u}{e^r} dr \le \int_0^\infty|u_r| dr.$$ It
is easy to see that the equalities in the above inequalities hold if
and only if $u_r=0$, thus $u(r)=0$.

Now, for  given $n>1$ and positive parameter $\beta >0$, we have
\begin{align*}
u(r)& =\int_0^r u_r dr \le (\int_0^r|u_r|^n dr)^{1/n} \cdot
r^{\frac{n-1}n}\\
&\le \frac {\beta^n \int_0^r|u_r|^n dr}{n}+\frac{\beta^{-\frac
n{n-1}} r}{n/(n-1)}.
\end{align*}
Thus
\begin{align}\label{rough}
\int_0^\infty \frac{e^{nu}}{e^{nr}} dr &\le \int_0^\infty
\frac{\exp\{ \beta^n \int_0^\infty|u_r|^n dr+(n-1)\beta^{-\frac
n{n-1}}r\}}{e^{nr}} dr \nonumber\\
&=\exp \{ \beta^n \int_0^\infty|u_r|^n dr\} \cdot \int_0^\infty
e^{[(n-1)\beta^{-\frac n{n-1}}-n]r} dr. \nonumber
\end{align}
If we choose
\begin{equation}
\beta=\beta_{0}>(\frac{n-1}n)^{\frac{n-1}n},\label{2-1}
\end{equation}
then
$$
\int_0^\infty e^{[(n-1)\beta_{0}^{-\frac n{n-1}}-n]r}
dr=c(\beta_{0})$$ is a finite number depending  on $\beta_{0}$. It
follows that
$$\ln \int_0^\infty e^{n(u-r)} dr \le  \beta_{0}^n \int_0^\infty|u_r|^n
dr+c_1(\beta_{0})
$$
for  $c_1(\beta_{0})=\ln c(\beta_{0})$.

\end{proof}

It is obvious in the above proof that $c(\beta_{0}), c_1(\beta_{0})
\to +\infty$ as $\beta_{0}\to (\frac{n-1}n)^{\frac{n-1}n}.$ We need
another argument to derive the main theorem.

\begin{remark}\label{rem2-1}
From (\ref{mm1}) we can see that for $\beta_{0}$ satisfying
(\ref{2-1}),
\begin{align}\label{rem2}
\int_R^\infty \frac{e^{nu}}{e^{nr}} dr &\le \exp \{ \beta_{0}^n
\int_0^\infty|u_r|^n dr\} \cdot \int_R^\infty
e^{[(n-1)\beta_{0}^{-\frac n{n-1}}-n]r} dr\\
&=o_R(1)\exp \{ \beta_{0}^n \int_0^\infty|u_r|^n dr\}, \nonumber
\end{align}
where $o_R(1)\to 0$ as $R\to \infty$.
\end{remark}

We now compare this with Moser's proof of Trudinger's inequality
\begin{lemma} For $n>1$, $a>0$  and $\beta<n a^{\frac 1{1-n}}$, there is a constant
$C_{\beta,a}$ depending only on $\beta$ and $a$, such that for any
nonnegative function $u \in C^1[0, +\infty)$ with $u(0)=0$ and
$\int_0^\infty|u_r|^n dr\le a$, \begin{equation}\int_0^\infty
\frac{e^{\beta u^{\frac n{n-1}}}}{e^{nr}} dr \le C_{\beta,a}.
\label{mm2}
\end{equation}
\label{propM-1}
\end{lemma}
\begin{proof}
For given $n>1$  we have
$$
u(r) =\int_0^r u_r dr \le (\int_0^r|u_r|^n dr)^{1/n} \cdot
r^{\frac{n-1}n} \le  a^{\frac 1n} r^{\frac{n-1}n} . $$ Thus for any
positive parameter $\tau
>0$,
\begin{equation}\label{roughM-1}
\int_0^\infty \frac{e^{\tau  u^{\frac n{n-1}}}}{e^{nr}} dr \le
\int_0^\infty \exp\{\tau a^{\frac 1{n-1}}-n\}r  dr. \end{equation}
The right hand side of the above inequality is bounded if we choose
$\tau=\beta<n a^{\frac 1{1-n}}$.
\end{proof}

Based on Lemma \ref{propM-1}, one can verify Corollary \ref{TM-1} as
follows.

Due to the rearrangement and rescaling, we only need to prove
Corollary \ref{TM-1} when $\Omega=B_1(0)$ and $u\in C_0^1(B_1(0))$
is radially symmetric and nonnegative.

From $\int_{B_1} |\nabla u|^n dx\le 1$, we know that (let $r=-\ln
s$)
\begin{align*}
1 \ge \int_{B_1}  |\nabla u|^n dx =\omega_{n-1} \int_0^1 |u_s|^n
s^{n-1} ds= \omega_{n-1} \int_0^\infty |u_r|^n dr.
\end{align*}
Also,
\begin{align*}
\int_{B_1} e^{\beta u^{\frac n{n-1}}} dx=\omega_{n-1} \int_0^1
e^{\beta u^{\frac n{n-1}}} s^{n-1} ds=\omega_{n-1} \int_0^\infty
\frac{e^{\beta u^{\frac n{n-1}}}}{e^{nr}} dr.
\end{align*}
One immediately has Corollary \ref{TM-1} by  using Lemma
\ref{propM-1} with $a=\omega_{n-1}^{-1}.$

This result is slightly weaker than Moser's inequality since it does
not include the case of $\beta= n \omega_{n-1}^{1/(n-1)}$. It seems
that one needs the argument due to Moser \cite{MO}, or Carleson and
Chang \cite{CC} to cover this extremal case.
\section{Sharp Inequality}
We shall prove Theorem \ref{main} in this section. Since the case of $n=1$
has been settled by Proposition \ref{rough-1}, we will focus on the
case of $n>1$. For given $a>0$, define
\begin{equation}\label{set-1}
D_a^n:=\{ u(r)\in W^{1,n}(\mathbb{R}^+) \ : \ u(0)=0, \ \int_0^\infty
\exp\{nu-nr\} dr=a\}.
\end{equation}

\begin{lemma}\label{sharp}
There is a $v \in D_a^n$ such that
$$
\int_0^\infty |v_r|^n dr =\inf_{u \in D_a^n} \int_0^\infty |u_r|^n
dr:=\xi. $$
\end{lemma}

\begin{proof}
Let $\{v^i\}$ be a  minimizing sequence of $\inf_{u \in D_a^n}
\int_0^\infty |u_r|^n dr$. Then
$$v^i \rightharpoonup v \ \ \ \mbox{in} \ W^{1,n}(\mathbb{R}^+), \ \ \mbox{and}
\ \ \ \int_0^\infty |v_r|^n dr \le \underline{\lim}_{i\to \infty}
 \int_0^\infty |v^i_r|^n dr=\xi $$ for some $v\in W^{1,n}(\mathbb{R}^+).$
We  need to verify $v \in D_a^n.$

First, from (\ref{rem2}), we know that for $w= v^i,$ or $v$:
$$\int_R^\infty \frac{e^{nw}}{e^{nr}} dr =o_R(1).$$ On the other
hand,  it follows from  the embedding $H^1(0, R)\hookrightarrow
C^{0, 1/2}(0, R)$ and Arzela-Ascoli lemma  that
$$
\lim_{ i \to \infty} \int_0^R \exp\{nv^i -nr\} dr = \int_0^R
\exp\{nv -nr\} dr.$$ Letting $i, \ R\to \infty$, we have
$\int_0^\infty \exp\{nv -nr\} dr=a$, that is $v \in D_a^n.$
\end{proof}

We now begin the proof of Theorem \ref{main}.
\begin{proof}  We only need to consider
the case of $n>1$ for nontrivial nonnegative functions. For $a>1/n$,
let $v$ be the minimizer of $\inf_{u \in D_a^n}\int_0^\infty |u_r|^n
dr$. It is easy to see that $v_r \ge 0$.  So it satisfies the
following Euler-Lagrange equation:
\begin{equation}\label{equ-1}
v_r^{n-2} v_{rr}=-\tau e^{nv -nr}, \ \ \ \ v(0)=0 \end{equation} for
some $\tau>0$.
Though it is not obvious how to obtain the general
solution from the uniqueness of the ordinary differential equation
since $v_r$ could be zero,  one can follow the argument given by
Carleson and Chang \cite{CC} to show that the general
solution to (\ref{equ-1}) is given by
$$ v(r)=\ln \frac {1}{\lambda_0 + e^{-nr/(n-1)}}-\frac 1n \ln \frac{\tau }
{ (\frac{n}{n-1})^n \lambda_0} ,$$ where $ \lambda_0 $ is a positive
constant and
$\tau=\frac{(\frac{n}{n-1})^n\lambda_0}{(\lambda_0+1)^n}. $ Thus
\begin{equation}\label{sol}  v(r)=\ln \frac
{\lambda_0+1}{\lambda_0 + e^{-nr/(n-1)}}. \end{equation} Since
$a=\int_0^\infty e^{nv-nr} \ d r ,$ we have
\begin{align*}
a&=\int_0^\infty\left(\frac{\lambda_0+1}{\lambda_0+e^{-nr/(n-1)}}\right)^n
e^{-nr}\ d r\\
&=\int_0^1 \left(\frac{\lambda_0+1}{\lambda_0+s^{n/(n-1)}}\right)^n
s^n (\frac1{s}) \ d s \\
&=(\lambda_0+1)^n  \int_0^1
\frac{s^{n-1}}{(\lambda_0+s^{n/(n-1)})^n} \ d s\\
&=(\lambda_0+1)^n \frac {s^n}{n\lambda_0 (\lambda_0 +
s^{n/(n-1)})^{n-1}} \bigr|^{1}_{s=0}\\
&=\frac{\lambda_0+1}{n\lambda_0} ,\\
\end{align*}
That is
\begin{equation}\label{a-1} \lambda_0=\frac1{na-1} .
\end{equation}
We compute
\begin{align*}
\int_0^\infty |v_r|^n dr &= \int_0^\infty \left| \frac{\frac{n}{n-1}
e^{-nr/(n-1)}}{\lambda_0+e^{-nr/(n-1)}}\right|^n \ d r \\
&= (\frac{n}{n-1})^n \int_0^\infty
\left(\frac{e^{-nr/(n-1)}}{\lambda_0+e^{-nr/(n-1)}}\right)^n \ d r\\
&=(\frac{n}{n-1})^{n-1} \int_0^{1/\lambda_0} \frac{t^{n-1}}{(1+t)^n} \ d t\\
&=(\frac{n}{n-1})^{n-1} \int_1^{1/\lambda_0+1}
\frac{(\tau-1)^{n-1}}{\tau^n} \ d \tau\\
&=-(\frac{n}{n-1})^{n-1} \int_1^{1/\lambda_0+1}(\tau -1)(1-\frac
1 {\tau})^{n-2} \ d \frac 1{\tau}\\
& =(\frac{n}{n-1})^{n-1} \int_{{\lambda_0}/(\lambda_0+1)}^1 (\frac 1
t-1)(1-t)^{n-2} \ d t.
\end{align*}
Using (\ref{a-1}) we have: If $n \in  $ {\bf N}$,$
\begin{align*}
 \int_0^\infty |v_r|^n \ d r
&=(\frac{n}{n-1})^{n-1} \cdot
\{ -\int_{{\lambda_0}/(\lambda_0+1)}^1 (1-t)^{n-2} \ d t \\
&- \cdots -
 \int_{{\lambda_0}/(\lambda_0+1)}^1 (1-t) \ d t - \int_{{\lambda_0}/(\lambda_0+1)}^1 (1-\frac 1 {t}) \ d t
 \}\\
&=(\frac{n}{n-1})^{n-1} \{ \ln \frac {\lambda_0+1}{\lambda_0}
-\frac1{\lambda_0+1}-\sum_{i=1}^{n-2} \frac {(\frac
1{\lambda_0+1})^{n-i}}{n-i} \} \\
&=(\frac{n}{n-1})^{n-1} \{\ln (na) - \sum_{i=1}^{n-1} \frac
{(na-1)^{n-i}}{(n-i)(na)^{n-i}} \};
\end{align*}
For general $n>1$, we have
\begin{align*}
 \int_0^\infty |v_r|^n \ d r &=(\frac{n}{n-1})^{n-1} \cdot
\{ -\int_{{\lambda_0}/(\lambda_0+1)}^1 (1-t)^{n-2} \ d t \\
&- \cdots -
 \int_{{\lambda_0}/(\lambda_0+1)}^1 (1-t)^{n-[n]} \ d t + \int_{{\lambda_0}/(\lambda_0+1)}^1
 \frac 1{t}(1-t)^{n-[n]} \ d t
 \}\\
&=( \frac n{n-1} )^{n-1} \{\int_{\lambda_0/(\lambda_0+1)}^1 \frac
1{t}(1-t)^{n-[n]} \ d t - \sum_{i=1}^{[n]-1} \frac {( \frac
1{\lambda_0+1})^{n-i}}{n-i} \} \\
&= ( \frac n{n-1} )^{n-1} \{\int_{\lambda_0/(\lambda_0+1)}^1 \frac
1{t}+ \int_{\lambda_0/(\lambda_0+1)}^1 \frac 1{t}((1-t)^{n-[n]}-1)
\ d t \\
&- \sum_{i=1}^{[n]-1} \frac {( \frac
1{\lambda_0+1})^{n-i}}{n-i} \}\\
 &= (\frac{n}{n-1})^{n-1} \{\ln (na) -  \int_{\lambda_0/(\lambda_0+1)}^1 \frac 1{t}(1-
 (1-t)^{n-[n]})
\ d t\\
&- \sum_{i=1}^{[n]-1} \frac {(na-1)^{n-i}}{(n-i)(na)^{n-i}} \},
\end{align*}
where $[n]$ is the integer part of $n$. Let $a \to \infty$, then
$\lambda_0 \to 0$ by (\ref{a-1}).  We know that $C_n$ is optimal.
The proof  is completed.
\end{proof}

\begin{remark}
For negative function $u$, we can certainly improve the
inequalities. In particular, similar argument will yield Theorem 1.3
(ii) in \cite{LZ} for integer $n>1$.
\end{remark}

\section{Applications}
We first show that Theorem \ref{main} implies:

\medskip

\noindent{\bf Corollary \ref{LZcor1}} {\bf (Local sharp inequality for $n=2$)}
$$\inf_{w \in D_{a,b}(B_r)}  \int_{B_r}|\nabla w|^2 d y= 4
\pi \cdot (\ln \frac{a e^{-2b} }{\pi r^2} +\frac {\pi r^2}{a e^{-2b}
}-1).
$$

\medskip

\begin{proof}
Let $v\in D^2_\alpha$ (recalling
the notation in (\ref{set-1})). We have, from  Theorem \ref{main}, that
\begin{equation}
\inf_{v \in D_\alpha^2} \int_0^\infty |v_r|^2 \ d r \geq 2 \{\ln
(2\alpha) +\frac{1}{2 \alpha } -1 \}, \label{n=2}
\end{equation}
where $\int_0^\infty e^{2v-2r}dr=\alpha$. For $w\in D_{a,0}(B_1(0))$,
$$
\int_{B_1}  |\nabla w|^2 dx =2\pi \int_0^1 |w_s|^n s ds= 2 \pi
\int_0^\infty |w_r|^2 dr.
$$
and
$$
\int_{B_1} e^{2w } dx=2 \pi \int_0^1 e^{2w} s ds=2 \pi \int_0^\infty
\frac{e^{2 w}}{e^{2r}} dr.
$$
Combing with (\ref{n=2}), we have
$$
\inf_{w \in D_{a,0}(B_1)} \int_{B_1}  |\nabla w|^2 dx = 4 \pi \cdot
(\ln \frac{\int_{B_1} e^{2w } dx}{\pi} +\frac {\pi}{\int_{B_1} e^{2w
} dx}-1).$$ After rescaling and shifting, we get Corollary
\ref{LZcor1}.
\end{proof}

\medskip

In the same spirit, from (\ref{hardyet}) we easily obtain
\begin{corollary}Let $u\in C^1(B_1)$ be a nonnegative
function satisfying $u=0$ on $\partial B_1$
$$
\ln \frac{n \int_{B_1} e^{nu}}{\omega_{n-1}} < (\frac{n-1}n)^{n-1}
\omega_n^{-1} \int_{B_1} |\nabla u|^n+F(1).$$
 where
$$F(1) = 1+\frac 12 +\cdots+\frac 1{n-1}.$$ \label{LZcor}
\end{corollary}

The fact that the strict inequality holds on a bounded domain
coincides with the one that the strict sharp Sobolev inequality
holds on a bounded domain. Corollary \ref{LZcor} was first proved in
\cite{LZ} using Trudinger's inequality. The proof presented here
does not rely on Trudinger's inequality. Inequality in Corollary
\ref{LZcor} was refereed as {\it local sharp inequality} in
\cite{LZ}, which is  easily adapted  for manifolds. See related
topics in Chen and Zhu \cite{CZ}.

\medskip

Finally, we shall show that one can prove the Onofri inequality
using Corollary \ref{LZcor1}.

Due to the rearrangement, we only need to prove Onofri inequality
for $u\in C^1(S^2)$ which
 depends only on $x_3$  and
is monotonically decreasing in $x_3$. Also, we can assume that
$u(x_3)\mid_{x_3=1}=0$ (otherwise, we replace $u(x)$ by
$u(x)-u(1))$. We can approximate $u$ by  a sequence of functions
$u_i \in C^1(S^2)$ such that $u_i(x)=u_i(x_3)$ is monotonically
decreasing in $x_3$, and $u_i(x)=0$ in the geodesic ball
$B_{1/i}(N)$ of the north pole $N$ for $i\in {\bf N}$. Denote
$S^2_i:=S^2 \setminus B_{1/i}(N)$.

Let $\Phi$: $x \in S^2 \to y \in \mathbb{R}^2$ be a stereographic projection
given by
$$
x_i=\frac {2 y_i}{1+|y|^2}, \ \ \ \ \ \ for \ \ \ i=1, 2;
$$
and
$$
x_{3} =\frac{|y|^2-1}{|y|^2+1}.
$$
Denote
$$
g_0=\sum_{i=1}^{3}dx_i^2=(\frac{2}{1+|y|^2})^2 dy^2:=e^{2
\varphi(y)} d y^2.
$$
Thus
$$\varphi(y)=\ln\frac  2{1+|y|^2}.
$$
It is easy to check that $\varphi(y)$ satisfies
\begin{equation}
-\Delta \varphi=e^{2\varphi } \ \ \ \ \ in  \ \ \ \mathbb{R}^2. \label{2.4}
\end{equation}
Let  $\Phi(S^2_i)=B_{R_i}$. It is obvious that $R_i \to +\infty$ as
$i\to +\infty$.  For
$$
w_i(y)=u_i(x)+\varphi(y)=u_i(\Phi^{-1}(y))+\varphi(y),$$ we have
$$
\int_{B_{R_i}} e^{2 w_i(y)} dy= \int_{S^2_i} e^{2u_i}dx:=a_i,
$$
and \begin{align*}
&\int_{B_{R_i}} |\nabla w_i|^2dy\\
=&\int_{B_{R_i}}|\nabla (u_i\circ \Phi^{-1})|^2 dy+2
\int_{B_{R_i}}\nabla (u_i\circ \Phi^{-1}) \cdot \nabla
\varphi dy +\int_{B_{R_i}}|\nabla \varphi|^2d y\\
=& \int_{S^2_i} |\nabla u_i|^2 dx+2 \int_{S^2_i} u_i
dx+\int_{B_{R_i}}|\nabla \varphi|^2dy,
\end{align*}
where we use the fact that $ \varphi $ satisfies (\ref{2.4}). Since
$w_i(y)=\ln\frac 2{1+R_i^2}$  on $\partial B_{R_i}$, it follows from
Corollary \ref{LZcor1} that
$$
\int_{B_{R_i}} |\nabla w_i|^2 dy \ge 4 \pi (\ln \frac{a_i \cdot
(\frac {1+R_i^2}2)^2}{\pi R_i^2} +\frac {\pi R_i^2} {a_i \cdot
(\frac {1+R_i^2}2)^2} -1).
$$
Also, one can check that
$$
\int_{B_{R_i}}|\nabla \varphi|^2dy=4\pi\left(\ln(1+R_i^2)+\frac
1{1+R_i^2}-1\right).$$ We conclude
\begin{align*}
&\int_{S^2_i} |\nabla u_i|^2 dx+2 \int_{S^2_i} u_i dx \\
\ge &4 \pi\left(\ln \frac{a_i \cdot \left(\frac {1+R_i^2}2\right)^2}{\pi
R_i^2} +\frac {\pi R_i^2} {a_i \cdot \left(\frac {1+R_i^2}2\right)^2} -1\right)
-4\pi\left(\ln(1+R_i^2)+\frac 1{1+R_i^2}-1\right)\\
=&4 \pi \left(\ln \frac{a_i \cdot ( {1+R_i^2})}{4\pi R_i^2} +\frac
{4\pi R_i^2} {a_i \cdot ({1+R_i^2})^2} -\frac 1{1+R_i^2}\right).
\end{align*}
Sending $i\to +\infty$, we have
$$
\int_{S^2} |\nabla u|^2 dx+2 \int_{S^2} u dx \ge 4 \pi \left( \ln
\frac 1{4\pi} \int_{S^2} e^{2u}dx\right).
$$

\medskip

\chapter{ Two Dimensional Sharp Trace Inequalities}
The main purpose of this chapter is to derive a global sharp trace
inequality on two dimensional simply connected domain.

\medskip

\noindent{\bf Theorem 1.2.}\ \ {\it Let $\Omega$ be a  simply
connected domain in $\mathbb{R}^2$ with bounded Geo
-desic curvature $K_g$.
Then there exists a constant $C(\Omega)$ such that for any $u\in
H^1(\Omega)$,
$$\ln\int_{\partial\Omega} e^u dS \le \frac 1{4 \pi}
  \int_{\Omega}  |\nabla u|^2 d y + \frac 1{2 \pi} \int_{\partial\Omega} K_g \cdot udS + C(\Omega). $$}
We start with a local sharp inequality,
which allows us to obtain an $\epsilon-$level sharp trace
inequality in Theorem \ref{theorem1}. We then define a functional and study the
behavior  of the minimizing sequence as the parameter $\epsilon$ goes to
$0$. It will lead to the
result in Theorem \ref{main2}.

\section{Sharp local trace inequality}
Let $\Phi : y \in \mathbb{R}^2_+ \to x \in B_1$ be a conformal mapping given
by $$y_1=\frac {-2x_2}{(x_1-1)^2+x_2^2} \ \ \mbox{and }\ \  y_2=\frac
{1-x_1^2-x_2^2}{(x_1-1)^2+x_2^2},$$
where $B_1$ is the unit disk in $\mathbb{R}^2$ centered at the origin.

\noindent Define $$D = \{ (y_1, y_2)\in \mathbb{R}^2_+: y_1^2+(y_2+1)^2 < 2\}$$ and
$$\Omega = \{(x_1, x_2)\in B_1 : (x_1-1)^2+x_2^2 > 2\}.$$
Then $\Phi(D)=\Omega \ \ \mbox{and}\ \ \Phi(\Gamma_2)=\Gamma_1$ ,
where $$\Gamma_1=\{(x_1, x_2)\in B_1 : (x_1-1)^2+x_2^2 = 2\}$$ and
$$\Gamma_2=\{ (y_1, y_2)\in \mathbb{R}^2_+ : y_1^2+(y_2+1)^2 = 2\}.$$
\begin{figure}
\begin{center}
\includegraphics[scale=0.8]{mapping.eps}
\end{center}
\end{figure}
Denote $$g_x = \sum_{i=1}^2 dx_i \wedge
dx_i = \frac 4{(y_1^2+(y_2+1)^2)^2}\sum_{i=1}^2 dy_i \wedge
dy_i:=e^{2\varphi}g_y.$$ Thus $$\varphi(y)=\ln \frac
2{y_1^2+(y_2+1)^2}.$$
It is easy to check that $\varphi$ satisfies
$$
\left\{
\begin{array}{rll}
-\Delta_y \varphi &= 0 \ \ &\mbox{in} \ \ \mathbb{R}^2_+\\
\frac {\partial \varphi}{\partial\nu} &= e^\varphi \ \ &\mbox{on} \ \
\partial \mathbb{R}^2_+.
\end{array}
\right.
$$
For $u \in H^1(D)$, we define
$$ {\tilde u}(y)=
\left\{
\begin{array}{rll}
&u(y) \ \ &\mbox{when} \ \ y\in D\\ 
&\varphi(y)\ \ &\mbox{when} \ \ y\in \mathbb{R}_+^2\backslash D.
\end{array}
\right.
$$
Let
$h(x)=\tilde{u}(\phi^{-1}(x))-\varphi(\phi^{-1}(x)).$
There is a well-known inequality

\medskip

\noindent{\bf Theorem C} ({\bf Lebedev-Milin inequality})\ \ {\it Let $B_1$ be unit disk in $\mathbb{R}^2$ centered at the origin. Then
for any $u\in H^1(B_1)$
$$ \ln\int_{\partial
B_1}e^{u}\frac{d\theta}{2\pi}\leq\int_{B_1}|\nabla u|^2\frac{dx}{4\pi}+\int_{\partial B_1}u\frac{d\theta}{2\pi}.$$}

Applying Lebedev-Milin inequality, we have $$\ln \int_{\partial B_1} e^h \frac {d
\theta}{2\pi}\le \frac 14 \int_{B_1} |\nabla h|^2 \frac {dx}\pi+
\int_{\partial B_1}h \frac {d\theta}{2\pi}.$$ Since $$\ln
\int_{\partial B_1} e^h \frac {d \theta}{2\pi}=\ln\int_ {\partial
B_1}e^{{\tilde u}\circ\Phi^{-1}-\varphi\circ\Phi^{-1}}\frac
{d\theta}{2\pi}=\ln(\int_\mathbb{R} e^{\tilde u} \frac {dy_1}{2\pi})$$
and
\begin{align*}
&\ \ \ \frac 14 \int_{B_1} |\nabla h|^2 \frac {dx}\pi+
\int_{\partial B_1}h \frac {d\theta}{2\pi}\\&=\frac 1{4\pi}
\int_{B_1} |\nabla_x ({\tilde u}\circ\Phi^{-1}-\varphi\circ\Phi^{-1})|^2
dx+\frac
1{2\pi}\int_{\partial B_1}({\tilde u}\circ\Phi^{-1}-\varphi\circ\Phi^{-1})d\theta\\
&=\frac 1{4\pi}[\int_\Omega|\nabla_x({\tilde u}\circ\Phi^{-1})|^2
dx+\int_\Omega|\nabla_x(\varphi\circ\Phi^{-1})|^2 dx\\
&\ \ \
-2\int_\Omega\nabla_x({\tilde u}\circ\Phi^{-1})\cdot\nabla_x(\varphi\circ\Phi^{-1})dx]+\frac
1{2\pi} \int_{-1}^1({\tilde u}-\varphi)e^\varphi dy_1\\
&=\frac 1{4\pi}[\int_D |\nabla_y {\tilde u}|^2 dy+\int_D |\nabla_y \varphi|^2
dy+2\int_\Omega
{\tilde u}\circ\Phi^{-1}\cdot\Delta_x(\varphi\circ\Phi^{-1})dx\\&\ \ \
-2\int_{\partial \Omega} {\tilde u}\circ\Phi^{-1}\frac
{\partial(\varphi\circ\Phi^{-1})}{\partial\nu}dS_x]+\frac 1{2\pi}
\int_{-1}^1({\tilde u(y_1,0)}-\varphi(y_1,0))e^\varphi(y_1,0) dy_1\\
&=\frac 1{4\pi}\int_D |\nabla u|^2 dy+\frac 1{4\pi}\int_D |\nabla
\varphi|^2 dy- \frac 1{2\pi}\int_{-1}^1 \varphi(y_1,0) e^{\varphi(y_1,0)} dy_1\\
&\le\frac 1{4\pi}\int_D |\nabla u|^2 dy+C, \ \ \mbox{for some constant C},
\end{align*}
we obtain a Lebedev-Milin type inequality on the special domain $D=\{ (y_1, y_2)\in \mathbb{R}^2_+: y_1^2+(y_2+1)^2 < 2\},$
\begin{equation}
\ln\int_{-1}^1 e^u \frac {dy_1}{2\pi}\le\frac 1{4\pi}\int_D |\nabla u|^2 dy+C, \ \ \ \forall  u\in H^1(D).\label{Ltype}
\end{equation}
This inequality will be used to obtain $\epsilon-$ level sharp inequality on general domains.
\section{Global $\epsilon-$ level sharp trace inequality}
\begin{lemma}\ \  Assume $\Omega$ is a smooth bounded domain in $\mathbb{R}^2$. For any given
$\epsilon>0$, there are  constants $C_1=C_1(\epsilon) \ and \ C_2=C_2(\epsilon)$ such that
$$
\ln\int_{\partial\Omega} e^{u}dS \le (\frac 1{4 \pi}+\epsilon)
  \int_{\Omega}|\nabla u|^2 d y+C_1\int_{\Omega} u^2\ dy +C_2 ,\ \ \ \ \ \forall  u\in H^1(\Omega).
$$
\label{lem3-1}
\end{lemma}
\begin{proof}\ \ Step 1. For $\overline{y}\in \partial\Omega$, we
first straighten the boundary $\partial\Omega$ near $\overline{y}$.
Without loss of generality, we may assume that $\overline{y}$ is the coordinate origin and
$y_{2}$-axis is the inner normal to $\partial\Omega$ at
$\overline{y}$. Therefore, there exists a smooth function
$\varphi(y_1)\ge 0$ defined for $|y_1|\le \alpha$ for some small
$\alpha$, satisfying $\varphi(0)=0$ and $\varphi'(0)=0$, such that
$\partial\Omega$ is given by $(y_1, \varphi(y_1))$ in a neighborhood
of $\overline{y}$, written as $\partial\Omega(\overline{y},
\alpha)=(y_1, \varphi(y_1))$.
Let $ D_\alpha=\{(x_1, x_2)\in \mathbb{R}^2_+  \ :  x_1^2+(x_2+\alpha)^2\le
2\alpha^2\}.$
Define $y\ :=\phi(x)=(y_1, y_2)$ by
$$
\left\{
\begin{array}{rll}
&y_1 = x_1\\
&y_2 = x_2+\varphi(x_1).
\end{array}
\right.
$$
Since the differential map of $\phi$ satisfies $D\phi(0)$=Identity,
we know that $\phi$ has the inverse in a neighborhood of $\{0\}$.
If we denote $x=\psi(y):=\phi^{-1}(y)$ in a neighborhood of
$\overline{y}$, then
$$
\left\{
\begin{array}{rll}
&x_1 = y_1\\
&x_2 = y_2-\varphi(y_1).
\end{array}
\right.
$$
Thus
\begin{align*}
&\\\int_{\partial\Omega(\overline{y},\alpha)} e^{u(y)} d S_y &=\int_{-\alpha}^\alpha
e^{u\circ\Phi(x_1,0)}\sqrt {1+\varphi\prime(x_1)^2}dx_1 \\
&\le\int_{-\alpha}^\alpha e^{u\circ\Phi(x_1,0)}(1+\varphi\prime(x_1)^2) d x_1.
\end{align*}
When $\alpha$ is small enough, we have
\begin{equation}
\int_{\partial\Omega(\overline{y},\alpha)} e^{u(y)} d S_y
\le\int_{-\alpha}^\alpha e^{u\circ\Phi(x_1,0)}(1+c_0^2\alpha^2) d x_1 ,\ \ \mbox{for some } c_0>0 .\label{1}
\end{equation}
Define $v(z)\:=u(\alpha z)$, then $v\in H^1(D).$
We have
$$
\ln\int_{-1}^1 e^v \frac {dz_1}{2\pi}\le\frac 1{4\pi}\int_D |\nabla v(z)|^2 dz+C,
$$
where we use the fact that $ v $ satisfies (\ref{Ltype}).
We know
$$
\ln\int_{-1}^1 e^v \frac {dz_1}{2\pi}=\ln\int_{-1}^1 e^{u(\alpha z)} \frac {dz_1}{2\pi}\\
=\ln\int_{-\alpha}^{\alpha} e^{u(y)} \frac {dy_1}{2\pi}-\ln\alpha
$$
and
$$\frac 1{4\pi}\int_D |\nabla v(z)|^2 dz=\frac 1{4\pi}\int_D |\nabla_z u(\alpha z)|^2 dz
=\frac 1{4\pi}\int_{D_\alpha} |\nabla_y u(y)|^2 dy.
$$
Hence \begin{equation}
\ln\int_{-\alpha}^{\alpha} e^{u(y)} \frac {dy_1}{2\pi}\le\frac 1{4\pi}\int_{D_\alpha} |\nabla_y u(y)|^2 dy+\ln\alpha+c.\label{2}
\end{equation}
Applying inequalities (\ref{1}) and (\ref{2}), we can derive
\begin{align*}
&\\\int_{\partial\Omega(\overline{y},\alpha)} e^{u(y)} d S_y
&\le\int_{-\alpha}^\alpha e^{u\circ\Phi(x_1,0)}(1+c_0^2\alpha^2) d x_1\\
&\le e^c(1+c_0^2\alpha^2)2\pi\alpha\cdot\exp\{\frac 1{4\pi}\int_{D_\alpha} |\nabla_x (u\circ\Phi(x))|^2 dx\}\\
&=e^c(1+c_0^2\alpha^2)2\pi\alpha\cdot\exp\{\frac 1{4\pi}\int_{\Phi (D_\alpha)} |\nabla_x u(y)|^2 dy\}.
\end{align*}
By the definition of $\Phi$, we have
\begin{align*}
&\\|\nabla_x u(y)|^2&=|(\frac{\partial(u\circ\Phi)}{\partial x_1},\frac{\partial(u\circ\Phi)}{\partial x_2})|^2\\
&=|\nabla_y u(y)|^2+2\frac{\partial u}{\partial y_1}\frac{\partial u}{\partial y_2}\frac{d\varphi}{dx_1}+(\frac{\partial u}{\partial y_2})^2(\frac{d\varphi}{dx_1})^2\\
&\le|\nabla_y u(y)|^2[1+\frac{d\varphi}{dx_1}+(\frac{d\varphi}{dx_1})^2]\\
&\le|\nabla_y u(y)|^2(1+c_0\alpha+c_0^2\alpha^2).
\end{align*}
Therefore
$$
\ln\int_{\partial\Omega(\overline{y},\alpha)} e^{u(y)} d S_y
\le\frac{1+c_0\alpha+c_0^2\alpha^2}{4\pi}\int_{\Phi (D_\alpha)} |\nabla_y u(y)|^2 dy
+c+\ln[2\pi\alpha(1+c_0^2\alpha^2)].
$$
For any given $\epsilon_0>0$, choose $\alpha$ small enough
such that $\frac{c_0\alpha+c_0^2\alpha^2}{4\pi}<\epsilon_0$, then
\begin{equation}
\ln\int_{\partial\Omega(\overline{y},\alpha)} e^u d S_y
\le(\frac1{4\pi}+\epsilon_0)\int_{\Phi (D_\alpha)} |\nabla_y u|^2 dy+c(\epsilon_0).\label{4}
\end{equation}
Step 2. Let $\alpha$ be the constant chosen above.
Since $\Omega$ is a bounded domain in $\mathbb{R}^2$, we can find
finitely many points $\overline{y_i}\in\partial\Omega$,  where $0<\alpha_i<\alpha$, and
corresponding neighborhoods  $\partial\Omega(\overline{y_i},\frac{\alpha_i}2)$ such that
$\partial\Omega\subset\cup_{i=1}^{i=N}\partial\Omega(\overline{y_i},\frac{\alpha_i}2)$.
Let $\psi_i$ be smooth cutoff functions subordinate to this covering, satisfying
$$ \psi_i(y)=
\left\{
\begin{array}{rll}
&1 \ \ \mbox{when} \ \ y\in\Phi(D_{\frac{\alpha_i}2}(\overline{y_i}))\\
&0 \ \ \mbox{when} \ \ y\notin\Phi(D_{\alpha_i}(\overline{y_i})),
\end{array}
\right.
$$
and $0\le\psi_i^2\le1$ , $|\nabla\psi_i|<c$ some constant c.

According to (\ref{4}),
\begin{align*}
\int_{\partial\Omega} e^{u(y)} d S_y
\le&\displaystyle\sum^N_{i=1}\int_{\partial\Omega(\overline{y_i},\frac{\alpha_i}2)} e^{u(y)} d S_y\\
\le&\displaystyle\sum^N_{i=1}\int_{\partial\Omega(\overline{y_i},\alpha)} e^{u\cdot\psi_i} d S_y\\
\le&\displaystyle\sum^N_{i=1}\exp\{(\frac1{4\pi}+\epsilon_0)
\int_{\Phi (D_\alpha(\overline{y_i}))} |\nabla_y (u\cdot\psi_i)|^2 dy+c(\epsilon_0)\}\\
=&\displaystyle\sum^N_{i=1}\exp\{(\frac1{4\pi}+\epsilon_0)
\int_{\Phi (D_\alpha(\overline{y_i}))} |(\nabla_y u)\psi_i|^2\\
&+2(\nabla_y u)\psi_i\cdot u\nabla_y\psi_i+|u\nabla_y\psi_i|^2 dy+c(\epsilon_0)\}\\
\le&\displaystyle\sum^N_{i=1}\exp\{(\frac1{4\pi}+\epsilon_0)
\int_{\Phi (D_\alpha(\overline{y_i}))} |\nabla_y u|^2 \psi_i^2+ u^2|\nabla_y\psi_i|^2 \\
&+\epsilon_0|\nabla_y u|^2 \psi_i^2+\frac{u^2|\nabla_y\psi_i|^2}{\epsilon_0} dy+c(\epsilon_0)\}\\
\le&\displaystyle\sum^N_{i=1}\exp\{(\frac1{4\pi}+\epsilon_0)
\int_{\Phi (D_\alpha(\overline{y_i}))} (1+\epsilon_0)|\nabla_y u|^2\\
&+c(1+\frac1{\epsilon_0}) u^2] dy+c(\epsilon_0)\}\\
\le& N\cdot\exp\{(\frac1{4\pi}+\epsilon_0)\int_{\Omega} (1+\epsilon_0)|\nabla_y u|^2+c(1+\frac1{\epsilon_0})u^2 dy+c(\epsilon_0)\}.
\end{align*}
It follows that
\begin{align*}
 &\ln\int_{\partial\Omega} e^{u(y)} d S_y\\
\le&[\frac1{4\pi}+\epsilon_0(1+\frac1{4\pi}+\epsilon_0)]\int_{\Omega}|\nabla_y u|^2 dy\\
&+c(\frac1{4\pi}+\epsilon_0)(1+\epsilon_0)\int_{\Omega} u^2dy+\ln N+c(\epsilon_0).
\end{align*}
Hence, for any given $\epsilon>0$, choose $\epsilon_0>0$  small enough such that
$\epsilon_0(1+\frac1{4\pi}+\epsilon_0)<\epsilon,$  there exist constants $C_1(\epsilon)$ and $C_2(\epsilon)$ so that
$$
\ln\int_{\partial\Omega} e^{u(y)} d S_y\le(\frac1{4\pi}+\epsilon)\int_{\Omega}|\nabla_y u|^2 dy+C_1\int_{\Omega} u^2dy+C_2.
$$
\end{proof}
We are now ready to establish the following
  $\epsilon-$level inequality.
\begin{theorem}\ \  Let $\Omega$ be a  simply connected domain in $\mathbb{R}^2$
 with bounded Geodesic curvature $K_g$. Then for any given $\epsilon>0$,
 there is a constant $C_3=C_3(\epsilon)$ such that for any $u\in H^1(\Omega)$ with
$\int_{\partial\Omega} K_g \cdot u dS_g =0$,
\begin{equation}
\ln\int_{\partial\Omega} e^u dS \le (\frac 1{4 \pi}+\epsilon)
  \int_{\Omega}  |\nabla u|^2 d y + C_3. \label{rough-2}
\end{equation} \label{theorem1}
\end{theorem}
To prove this theorem, we need the following Poincar\'e-Sobolev type
inequality.
\begin{lemma}\ \  Assume $\Omega$ is a  simply connected domain in $\mathbb{R}^2$
 with bounded Geodesic curvature $K_g$.
 Then for any $p\ge 1$, there is a constant $c_p$ such that for any
$u \in H^1(\Omega)$ with  $\int_{\partial\Omega} K_g \cdot udS
=0$, $$ (\int_\Omega |u|^p)^{2/p}dy \le c_p \int_\Omega |\nabla u|^2 dy.$$  \label{lemma3-2}
\end{lemma}
\begin{proof}\ \  We prove it by contradiction. Suppose this is
not true, then there exists a sequence of functions $\{u_n\}$ in
$H^1(\Omega)$ with  $\int_{\partial\Omega} K_g \cdot u_ndS =0$
such that
$$
(\int_\Omega |u_n|^p)^{2/p}\ge n \int_\Omega |\nabla u_n|^2.$$ Let
$$ v_n=\frac {u_n}{(\int_\Omega |u_n|^p)^{1/p}}.$$ Then
$||v_n||_{L^p(\Omega)}=1$,\  and by assumption \
$||\nabla v_n||_{L^2(\Omega)}=\frac{||\nabla u_n||_{L^2(\Omega)}}{||u_n||_{L^p(\Omega)}}
\le\frac1{\sqrt n}.$
Thus$$
||v_n||_{H^1(\Omega)}\le \sqrt 2, \ \ \mbox{and} \
\ ||\nabla v_n||_{L^2(\Omega)} \to 0 \ \ \mbox{as} \ \ n \to
\infty.$$
Therefore, $v_n \rightharpoonup v_0$ in $H^1(\Omega)$ and $v_0$
is a constant in $\Omega$, say $v_0=c$ in $\Omega$. Let $u_0=v_0-c$
in $\overline{\Omega}$, then $||u_0||_{H^1(\Omega)}= 0$. By trace
inequality $||u_0||_{L^p(\partial\Omega)}\le
C||u_0||_{H^1(\Omega)}$, $u_0=0$  on $\partial\Omega$ almost everywhere and $v_0=c$ in
$\overline\Omega$. By the compact embedding, we know that $v_n \to
v_0$ in $L^q(\Omega)$ for any $q<\infty$, thus
$||v_0||_{L^p(\Omega)}=1$. Hence $c \neq 0$.

On the other hand, by the compact embedding $W^{1,p}(\Omega)\hookrightarrow\hookrightarrow L^q(\partial\Omega)$, we know that $v_n \to v_0$ in
$L^q(\partial\Omega)$, thus $\int_{\partial\Omega} K_g \cdot v_0 dS = 0$. Since $\int_{\partial\Omega} K_g dS = 2\pi$, we
conclude that $v_0=0$ on $\partial\Omega$. A contradiction.
\end{proof}
We now return to the proof of Theorem \ref{theorem1}.
\begin{proof}\ \ Due to Lemma 1 in
\cite{aubin} %on page 586
, we only need to prove the theorem for
functions in $C^1(\Omega)$ with only non-degenerate critical points.
Let $u(x) $ be such a function  with $\int_{\partial\Omega} K_g
\cdot u dS=0$. For small enough $\eta>0$, there exists $a_\eta$
such that
$$Vol \{ x \in \Omega \ : \ u(x)>a_\eta\} =\eta.$$\label{2-2}
Using Lemma \ref{lem3-1} we have  for any $\epsilon_1>0$, there are
constants $C_1(\epsilon_1)$ and $ C_2(\epsilon_1)$ such that
\begin{align*}
 &\ln\int_{\partial\Omega} e^u dS  \\
\le &a_\eta +\ln\int_{\partial\Omega} e^{(u-a_\eta)} dS \\
\le &a_\eta + \ln\int_{\partial\Omega} e^{(u-a_\eta)_+ }dS\\
\le &a_\eta + (\frac 1{4 \pi}+\epsilon_1)
  \int_{\Omega}|\nabla (u-a_\eta)_+|^2 d x\\
  &+C_1(\epsilon_1)\int_\Omega |(u-a_\eta)_+|^2 dx + C_2(\epsilon_1),
\end{align*}
where $f_+(x)=\max \{ f(x),  0\}.$
We know that
$$ a_\eta \cdot \eta \le \int_{\{x\in \Omega \ : u(x)>a_\eta\}} udx \le
\int_\Omega|u| dx \le c_1^{1/2}(\int_\Omega|\nabla u|^2)^{1/2},
$$
thus
$$
a_\eta \le \eta ||\nabla u||_{L^2(\Omega)}^2 +\frac {c_1}{4\eta^3}.
$$
Also, applying Lemma \ref{lemma3-2}, we obtain
\begin{align*}
\int_\Omega|(u-a_\eta)_+|^2 dx = & \int_{\{x\in \Omega \ :
u(x)>a_\eta\}}|(u-a_\eta)_+|^2 dx \\
\le & (\int_{\{x\in \Omega \ : u(x)>a_\eta\}}|(u-a_\eta)_+|^4
dx)^{1/2} \cdot \eta^{1/2} \\
\le &  (\int_{\Omega}|u|^4
dx)^{1/2} \cdot \eta^{1/2}\\
\le& c_4||\nabla u||_{L^2(\Omega)}^2 \cdot \eta^{1/2}.
\end{align*}
Therefore,
\begin{align*}
 &\ln\int_{\partial\Omega} e^u dS  \\
 \le & a_\eta + (\frac 1{4\pi}+\epsilon_1)\int_{\Omega}|\nabla (u-a_\eta)_+|^2 d x
 + C_1(\epsilon_1)\int_\Omega |(u-a_\eta)_+|^2 dx + C_2(\epsilon_1) \\
\le & (\frac 1{4\pi}+\epsilon_1+\eta+C_1(\epsilon_1)\cdot c_4\cdot
\eta^{1/2})||\nabla u||_{L^2(\Omega)}^2+\frac
c{4\eta^3}+C_2(\epsilon_1).
\end{align*}
Choosing $\epsilon_1$ and $\eta$ sufficiently small such that
$
\eta+\epsilon_1+C_1 \cdot c_4 \cdot \eta^{1/2} <\epsilon, $ we
obtain inequality (\ref{rough-2}).
\end{proof}
Notice that Theorem \ref{theorem1} can also be stated as the following.

\medskip

\noindent{\bf Theorem \ref{theorem1}$^\prime$}\ \ Let $\Omega$  be a  simply connected domain in $\mathbb{R}^2$
 with bounded Geodesic curvature $K_g$. Then for any given $\epsilon>0$,
 there is a constant $C_3=C_3(\epsilon)$ such that for any $v\in H^1(\Omega)$,
\begin{equation}
\ln\int_{\partial\Omega} e^v dS \le (\frac 1{4 \pi}+\epsilon)
  \int_{\Omega}  |\nabla v|^2 d y +\frac1{2\pi}\int_{\partial\Omega}K_g\cdot v dS+ C_3. \label{rough-2'}
\end{equation}
\begin{proof}\ \ Let $$u=v-\frac{\int_{\partial\Omega}K_g v dS}{\int_{\partial\Omega}K_g dS},$$
then $$\int_{\partial\Omega}K_g u dS = 0.$$
According to Theorem \ref{theorem1},
we obtain inequality (\ref{rough-2'}).
\end{proof}

\section{Global sharp trace inequality}
In this section, we always assume that $\Omega\subset\mathbb{R}^2$ is  simply
connected.

For any $\epsilon>0$, we define a functional
$$ I_\epsilon(u)=\frac 18 \int_\Omega |\nabla u|^2+\frac{1 -\epsilon}{ 4 }\int_{\partial\Omega} K_g u
-\frac{\pi(1-\epsilon)}{2}\ln \int_{\partial\Omega} e^u,$$ \label{4-1}  and denote
$$
E_\epsilon=\inf_{u \in H^1(\Omega), \ \int_{\partial\Omega} K_g u dS
=0} I_\epsilon (u).
$$
It is easy to see that $I_\epsilon(u+C)=I_\epsilon(u)$ for any
constant $C$. Due to Theorem \ref{theorem1} we know that
$E_\epsilon$ is a finite number. Moreover, we can show
\begin{lemma}\ \ $E_\epsilon$ is achieved by a function $u_\epsilon \in
H^1(\Omega)$. \label{lemma3-1}
\end{lemma}
\begin{proof}\ \ Let $\{u_n\}$ be a minimizing sequence with
$\int_{\partial\Omega} K_g u_n dS=0$. Choose a small positive number
$\epsilon_1$ such that
$$ \frac 18-\epsilon_1 > \frac{\pi(1-\epsilon_1)}{2}\cdot(\frac 1{4 \pi}+\epsilon_1).$$
Applying Theorem \ref{theorem1} we have
\begin{align*}
E_\epsilon \leftarrow I_\epsilon(u_n)&= \frac 18 \int_\Omega |\nabla u_n|^2 -\frac {\pi(1-\epsilon_1)}2 \ln\int_{\partial\Omega} e^{u_n}\\
&=(\frac 18-\epsilon_1) \int_\Omega |\nabla u_n|^2 -\frac {\pi(1-\epsilon_1)}2 \ln\int_{\partial\Omega} e^{u_n}+\epsilon_1 \int_\Omega |\nabla u_n|^2 \\
&\ge(\frac 18-\epsilon_1) \int_\Omega |\nabla u_n|^2 \\
&\ \ - \frac {\pi(1-\epsilon_1)}2\left[ (\frac 1{4 \pi}+\epsilon_1)
  \int_{\Omega}  |\nabla u_n|^2 d y + C_3(\epsilon_1) \right]+\epsilon_1 \int_\Omega |\nabla u_n|^2\\
&\ge C(\epsilon_1)+\epsilon_1 \int_\Omega |\nabla u_n|^2,
\end{align*}
thus $||\nabla u_n||_{L^2(\Omega)}<C.$ It follows from Lemma
\ref{lemma3-2} that $||u_n||_{H^1(\Omega)}<C.$ Further, it implies
that $u_n \rightharpoonup u_\epsilon$ in $H^1(\Omega)$ and $e^{\beta
u_n} \rightarrow e^{\beta u_\epsilon}$ in $L^1(\Omega)$ for any
positive number $\beta$ by Trudinger's inequality. Thus, $||u_\epsilon||_{H^1(\Omega)}^2
\le\underline{\lim}_{n \to \infty} ||u_n||_{H^1(\Omega)}^2.$

\medskip

For any \ $1<p<2,$
$$
\begin{array}{rll}
\int_\Omega |\nabla e^{\beta u_n}|^p &= \int_\Omega |e^{\beta
u_n}\beta\nabla u_n|^p \\ &\le \beta^p (\int_\Omega e^{\frac
{2p\beta}{2-p}u_n})^{ (2-p)/2}(\int_\Omega |\nabla u_n|^2)^{p/2}
\\ &\le c(\int_\Omega |\nabla u_n|^2 )^{p/2}
\\ &\le C,
\end{array}
$$
thus $\{e^{\beta u_n}\}$ is a bounded sequence in $W^{1,
p}(\Omega).$ Therefore, for any $\beta>0$ and $1<p<2$,  by the compact embedding,
$e^{\beta u_n}\rightarrow e^{\beta u_\epsilon}$ in $L^p(\partial\Omega)$ upto
subsequence. Hence $e^{\beta u_n}\rightarrow e^{\beta u_\epsilon}$
in $L^1(\partial\Omega)$ for any $\beta > 0$. So, $K_g u_n
\rightarrow K_g u_\epsilon$ and $\int_{\partial\Omega} K_g
u_\epsilon = 0.$
\medskip
It follows that
$$u_\epsilon\in\{u\in H^1(\Omega) \mid \int_{\partial\Omega} K_g\cdot u=0 \}$$
and
$$\underline{\lim}_{n \to \infty}I_\epsilon (u_n)\ge I_\epsilon(u_\epsilon).$$
Thus $u_\epsilon$ is a minimizer.
\end{proof}
To prove Theorem \ref{main2}, we only need to show that
$E_\epsilon$ is uniformly bounded from below as $\epsilon \to 0$.
Following the similar arguments given in \cite{CZ}, \cite{DJLW1}, \cite{DJLW} and \cite{NZ},
we argue by contradiction. Suppose that $E_\epsilon$ is not bounded
from below, that is, up to a subsequence (due to the nature of the
proof, for convenience we will not distinguish subsequence
$\{\epsilon_i\}$ and the original sequence $\{\epsilon\}$  in this
section),
\begin{equation}
\lim_{\epsilon \to 0} E_\epsilon = -\infty.\label{3-4}
\end{equation}

Let $$ v_\epsilon= u_\epsilon-\ln\int_{\partial\Omega}
e^{u_\epsilon},$$
 then $v_\epsilon$ is also a
minimizer, which satisfies the following Euler-Lagrange equations
\begin{gather}
\left\{
\begin{array}{rll}
-\Delta v_\epsilon &= 0 \ \ &\mbox{in} \ \ \Omega\\
\frac {\partial v_\epsilon}{\partial\nu} + (1-\epsilon)K_g &=
2\pi(1-\epsilon)e^{v_\epsilon} \ \ &\mbox{on} \ \ \partial\Omega.
\end{array}
\right.
 \label{3-3}
\end{gather}
Let $\lambda_\epsilon=v_\epsilon(x_\epsilon):=\max_{x \in \overline{\Omega}} v_\epsilon(x),$
then $\lambda_\epsilon=\max_{x \in {\partial\Omega}} v_\epsilon(x).$ We first claim
\begin{lemma}
$$\lim_{\epsilon \to 0} \lambda_\epsilon=+\infty.$$
 \label{lem4-2}
 \end{lemma}
\begin{proof}\ \ Let $$v_\epsilon^a=\frac{\int_{\partial\Omega}v_\epsilon dS}{\int_{\partial\Omega}dS}.$$
If $\lambda_\epsilon<C$,
then $e^{v_\epsilon}<C$. From (\ref{3-3}), we have
$$
0=\int_\Omega -v_\epsilon\Delta v_\epsilon
=\int_\Omega|\nabla v_\epsilon|^2-\int_{\partial\Omega}v_\epsilon\frac{\partial v_\epsilon}{\partial\nu},
$$
and
$$
\int_\Omega|\nabla (v_\epsilon-v_\epsilon^a)|^2=\int_{\partial\Omega}\frac{\partial v_\epsilon}{\partial\nu}v_\epsilon
=\int_{\partial\Omega}[-(1-\epsilon)K_g+2\pi (1-\epsilon)e^{v_\epsilon}]v_\epsilon.
$$
$\|\nabla(v_\epsilon-v_\epsilon^a)\|_{L^2(\Omega)}<C $ and  $\|v_\epsilon-v_\epsilon^a\|_{L^2(\Omega)}<C .$ Since $\|\Delta(v_\epsilon-v_\epsilon^a)\|_{L^\infty(\Omega)}<C$, $v_\epsilon-v_\epsilon^a \in C^\alpha(\Omega).$ Also,
$v_\epsilon-v_\epsilon^a \in C^\alpha(\partial\Omega)$
since $\|\frac{\partial(v_\epsilon-v_\epsilon^a)}{\partial\nu}\|_{L^\infty(\partial\Omega)}<C.$

\medskip
If $v_\epsilon^a$ is bounded, then $v_\epsilon=(v_\epsilon-v_\epsilon^a)+v_\epsilon^a$ is bounded,
 thus $E_\epsilon$ is bounded from below, this contradicts our assumption
 $\lim_{\epsilon\to 0} E_\epsilon=-\infty.$

\medskip
If  $v_\epsilon^a \to -\infty$ is unbounded, then $v_\epsilon-v_\epsilon^a \to w,$ where $w$ satisfies
$$
\left\{
\begin{array}{rll}
-\Delta w &= 0 \ \ &\mbox{in} \ \ \Omega\\
\frac{\partial w}{\partial\nu} &= -K_g \ \ &\mbox{on} \ \ \partial\Omega.
\end{array}
\right.
$$
Thus, $0=\int_\Omega -\Delta w
=\int_{\partial\Omega}\frac{\partial w}{\partial\nu} = \int_{\partial\Omega}-K_g = -2\pi,$ a contradiction.

\medskip
Hence $\lim_{\epsilon\to 0} \lambda_\epsilon=+\infty.$
\end{proof}

\medskip

Assume that, up to a subsequence, $x_\epsilon \to \bar x\in\partial\Omega$.
  In a neighborhood of  $\bar x$, we choose
a normal coordinate system, and define
$$
\varphi_\epsilon(x)=v_\epsilon(\tau_\epsilon^{-1}x+x_\epsilon)-\ln \tau_\epsilon,$$
where $$\tau_\epsilon:=e^{\lambda_\epsilon}.$$
For any given $R>0$, if $\epsilon$ is sufficiently small, $\varphi_\epsilon$
satisfies
$$
\left\{
\begin{array}{rll}
-\Delta\varphi_\epsilon &= 0 \ \ &\mbox{in} \ \ B_{2R}(0, -\frac{R}{\sqrt2})_+\\
\frac {\partial \varphi_\epsilon}{\partial\nu} + \frac{1-\epsilon}{\tau_\epsilon}K_g &=
2\pi(1-\epsilon)e^{\varphi_\epsilon} \ \ &\mbox{on} \ \ \Gamma_2,
\end{array}
\right.
$$
where $\Gamma_2:=\{(y_1, 0)| -\frac{\sqrt7}{2}R\le y_1\le \frac{\sqrt7}{2}R\}$
and $B_{2R}(0, -\frac{R}{\sqrt2})_+:= \{(y_1, y_2)\in\mathbb{R}_+^2 | y_1^2+(y_2+\frac{R}{\sqrt2})^2 < (2R)^2\}.$
The behavior of $v_\epsilon$ in
a tiny neighborhood of $\bar x$ can be described by the behavior
of $\varphi_\epsilon$ in a large set $B_{2R}(0, -\frac{R}{\sqrt2})_+ $.
\begin{lemma}\ \ For a fixed $R>0$, there is a constant $C(R)$ such that
$$
|\varphi_\epsilon(x)| <C(R) \ \ \ \ \forall \ \ x \in \overline {B_{R}(0, -\frac{R}{\sqrt2})_+} .
$$
\label{lem3-3}
\end{lemma}
\begin{proof}\ \ Let $\varphi_\epsilon^{(1)}$ be the unique solution of
$$
\left\{
\begin{array}{rll}
-\Delta\varphi_\epsilon^{(1)} &= 0 \ \ &\mbox{in} \ \ B_{2R}(0, -\frac{R}{\sqrt2})_+\\
\varphi_\epsilon^{(1)} &= 0 \ \ &\mbox{on} \ \ \Gamma_1\\
\frac {\partial \varphi_\epsilon^{(1)}}{\partial\nu} + \frac{1-\epsilon}{\tau_\epsilon}K_g &=
2\pi(1-\epsilon)e^{\varphi_\epsilon} \ \ &\mbox{on} \ \ \Gamma_2,
\end{array}
\right.
$$
where $ \Gamma_1:= \partial B_{2R}(0, -\frac{R}{\sqrt2})_+\cap\mathbb{R}_+^2. $

Since $e^{\varphi_\epsilon}\le 1$,
we have  $|\varphi_\epsilon^{(1)}|< C_1(R)$  for all $x \in \overline{B_R(0, -\frac{R}{\sqrt2})_+}. $

Let
$$
\varphi_\epsilon^{(2)}(x) = \varphi_\epsilon^{(2)}(x_1, x_2):=
\left \{
\begin{array}{rll}
&\varphi_\epsilon(x)-\varphi_\epsilon^{(1)}(x) \ \ &\mbox{if} \ \ x_2\ge 0\\
&\varphi_\epsilon^{(2)}(x_1, -x_2) \ \ &\mbox{if} \ \ x_2<0 ,
\end{array}
\right.
$$
then
$$
\left \{
\begin{array}{rll}
-\Delta\varphi_\epsilon^{(2)} &= 0 \ \ &\mbox{in} \ \ B_{2R}(0, -\frac{R}{\sqrt2})_+ \cup B_{2R}(0, +\frac{R}{\sqrt2})_-\\
\varphi_\epsilon^{(2)} &= 0 \ \ &\mbox{on}\ \ \Gamma_1\\
\frac{\partial\varphi_\epsilon^{(2)}}{\partial\nu} &= 0 \ \ &\mbox{on}\ \ \Gamma_2,
\end{array}
\right.
$$
where $B_{2R}(0, +\frac{R}{\sqrt2})_-:=\{(x_1, x_2)\in\mathbb{R}^2 : x_1^2+(x_2-\frac{R}{\sqrt2})^2<4R^2 \ and \ x_2<0\}.$

By weak solution, we check that
$$-\Delta\varphi_\epsilon^{(2)} = 0  \ \ in \ \
 B_{2R}(0, -\frac{R}{\sqrt2})_+ \cup B_{2R}(0, +\frac{R}{\sqrt2})_-\cup\Gamma_2;$$
$\forall x \in \Gamma_2, $ choose $\delta$ small enough such that
$$B_{\delta}(x)\subset \left [B_{2R}(0, -\frac{R}{\sqrt2})_+ \cup B_{2R}(0, +\frac{R}{\sqrt2})_-\cup\Gamma_2\right].$$
Let $\psi$ be a test function with $0$ on $\partial B_{\delta}(x)$, then
\begin{align*}
&\\-\int_{B_\delta(x)}\psi\Delta\varphi_\epsilon^{(2)}&=\int_{B_\delta(x)}\nabla\psi\nabla\varphi_\epsilon^{(2)}\\
&=\int_{B_\delta(x)_+}\nabla\psi\nabla\varphi_\epsilon^{(2)}+\int_{B_\delta(x)_-}\nabla\psi\nabla\varphi_\epsilon^{(2)}\\
&=-\int_{B_\delta(x)_+}\psi\Delta\varphi_\epsilon^{(2)}+\int_{\partial(B_\delta(x)_+)}\frac{\partial\varphi_\epsilon^{(2)}}{\partial\nu}\psi\\
&-\int_{B_\delta(x)_-}\psi\Delta\varphi_\epsilon^{(2)}+\int_{\partial(B_\delta(x)_-)}\frac{\partial\varphi_\epsilon^{(2)}}{\partial\nu}\psi\\
&=0.
\end{align*}
Thus $-\Delta\varphi_\epsilon^{(2)} = 0 $  in $B_{2R}(0, -\frac{R}{\sqrt2})_+ \cup B_{2R}(0, +\frac{R}{\sqrt2})_-\cup\Gamma_2 $.
Also, $ \varphi_\epsilon^{(2)}(x)$  is bounded from above by $C_1(R)$.
Applying Harnack inequality to $2C_1(R)-\varphi_\epsilon^{(2)}$, we have
$$\frac1{C}\le\frac{2C_1(R)-\varphi_\epsilon^{(2)}(x)}{2C_1(R)-\varphi_\epsilon^{(2)}(0)}\le C \ \ \ \ \
\forall x\in\overline{B_R(0,-\frac{R}{\sqrt2})_+\cup B_R(0,+\frac{R}{\sqrt2})_-}.$$
This and the bound of $\varphi_\epsilon^{(1)}$ yield Lemma \ref{lem3-3}.
\end{proof}
Since $\varphi_\epsilon$ is uniformly bounded in any fixed $\overline{B_R(0, -\frac{R}{\sqrt2})_+}$,
 based on the standard elliptic estimates,  we have
\begin{gather}
\varphi_\epsilon \to \varphi_0 \ \ \ \ \mbox{in} \
C^{1,\alpha}(\overline{B_R(0, -\frac{R}{\sqrt2})_+}),\label{e-1}
\end{gather}
for some $\alpha \in (0,1)$,  where $\varphi_0$ satisfies
$$
\left\{
\begin{array}{rll}
-\Delta\varphi_0 &= 0 \ \ &\mbox{in} \ \ \mathbb{R}_+^2\\
\frac{\partial\varphi_0}{\partial\nu} &= 2\pi e^{\varphi_0} \ \ &\mbox{on} \ \ \partial(\mathbb{R}_+^2),
\end{array}
\right.
$$
and
\begin{align*}
\int_{\partial\mathbb{R}_+^2} e^{2\varphi_0}&=\lim_{R \to \infty}\overline{\lim_{\epsilon \to 0}}\int_{\partial B_R(0, -\frac{R}{\sqrt2})_+\cap\partial\mathbb{R}^2_+}e^{2\varphi_\epsilon}\\
&=\lim_{R \to \infty}\overline{\lim_{\epsilon \to 0}}\int_{\partial B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})_+\cap\partial\mathbb{R}^2_+}e^{2v_\epsilon}\\
&\le[\overline{\lim_{\epsilon \to 0}}\int_{\partial\Omega}e^{v_\epsilon}]^2\\
&\le 1.
\end{align*}
From a uniqueness theorem for harmonic functions on upper-half plane in \cite{Ou} we know that
\begin{equation}
\varphi_0(x)=\varphi_0(x_1, x_2)=\ln\frac{2\tilde x_2}{(x_1-\tilde x_1)^2+(x_2+\tilde x_2)^2}-\ln(2\pi),\label{3-7}
\end{equation}
where $\tilde x_1$ is any real number and $\tilde x_2$ is any positive number, are fundamental solutions.


Away from the singular point $\bar x$, we have the following global estimate.
\begin{lemma}\ \ For any compact domain $K \subset \subset \bar\Omega\setminus\{\bar x\}$,
there is a constant $C(K)$ such that $$||v_\epsilon - v_\epsilon^a||_{L^\infty(K)} <C(K). $$ \label{lem3-4}
\end{lemma}
In order to prove Lemma \ref{lem3-4}, we need the following inequality.
\begin{lemma}\ \  Assume $K\subset\Omega$ is a bounded domain in $\mathbb{R}^2$ and $u$ is a solution of
$$ \left \{
\begin{array}{rll} -\Delta u & =0 \ \ &\mbox {in} \ \ K\\
u&=0  \ \ &\mbox{on} \ \ \partial K\cap\Omega\\
\frac{\partial u}{\partial\nu}&=f  \ \ &\mbox{on} \ \ \partial K\cap\partial\Omega.
\end{array}
\right.
$$
If $f \in L^1(\partial K\cap\partial\Omega)$, then for every $0<\delta< 1$, there is
a constant $C(K, \delta )$ such that
$$ \int_K \exp\left\{\frac{\pi(1-\delta)|u(y)|}{||f||_{L^1(\partial K\cap\partial\Omega)}}\right\}\le C(K, \delta).
$$\label{lem3-7}
\end{lemma}
\begin{proof}\ \ The proof is similar to Theorem 1. in \cite{BM} by Brezis and Merle.

Let $R=diamK$ such that $K\subset B_R^+$ for some upper-half ball of radius $R$ in $\mathbb{R}^2$.
Extend $f$ to be zero outside $K$.
Let $$\tilde u(y)=\frac1{2\pi} \int_{\partial\mathbb{R}^2_+\cap\partial B_R^+} (\log \frac{2R}{|x-y|}+\log \frac{2R}{|x-\bar y|})|f(x)| dx,\ y\in \overline{B_R^+},$$
where $\bar y$ is the reflection point of $y$ about the horizontal axis.
A direct computation yields
$$ \left \{
\begin{array}{rll} -\Delta \tilde u & =0\ \ &\mbox {in} \ \ B_R^+\\
\frac{\partial\tilde u}{\partial\nu}&=|f| \ \ &\mbox{on} \ \ \partial B_R^+\cap\partial\mathbb{R}^2_+.
\end{array}
\right.
$$
Note that $\tilde u(y)\ge 0\ \ \mbox{for}\ \ y\in \overline{B_R^+},$ and $$ \left \{
\begin{array}{rll} -\Delta (\tilde u \pm u) & =0\ \ &\mbox {in} \ \ B_R^+\\
\frac{\partial(\tilde u \pm u)}{\partial\nu}& \ge  0 \ \ &\mbox{on} \ \ \partial B_R^+\cap\partial\mathbb{R}^2_+.
\end{array}
\right.
$$
It follows from the maximum principle that $|u| \le \tilde u$ on $K$.
Thus
$$\int_K \exp\{\frac{\pi(1-\delta)|u(y)|}{||f||_{L^1(\partial K\cap\partial\Omega)}}\} dy \le\int_{B_R^+} \exp\{\frac{\pi(1-\delta)\tilde
u(y)}{||f||_{L^1(\partial K\cap\partial\Omega)}}\} dy .$$
We estimate the right-hand side of the above inequality using Jensen's inequality
$$F(\int w(x)\varphi(x) dx)\le \int w(x)F(\varphi(x)) dx$$
with $F(t) = \exp t$, $w(x) = \frac{|f(x)|}{\|f\|_{L^1(\partial K\cap\partial\Omega)}}$
and $\varphi(x) = \frac{1-\delta}{2}(\log\frac{2R}{|x-y|}+\log\frac{2R}{|x-\bar y|}).$
We have
\begin{align*}
&\int_{B_R^+} \exp\{\frac{\pi(1-\delta)\tilde u(y)}{||f||_{L^1(\partial K\cap\partial\Omega)}}\} dy \\
\le & \int_{B_R^+}\int_{\partial B_R^+\cap\partial\mathbb{R}^2_+}
\frac{|f(x)|}{\|f\|_{L^1(\partial K\cap\partial\Omega)}}
\exp\{ \frac{1-\delta}{2}(\log\frac{2R}{|x-y|}+\log\frac{2R}{|x-\bar y|})dx dy \}\\
= &\int_{\partial B_R^+\cap\partial\mathbb{R}^2_+}
\frac{|f(x)|}{\|f\|_{L^1(\partial K\cap\partial\Omega)}}
\int_{B_R^+}(\frac{2R}{|x-y|})^{\frac{1-\delta}{2}}(\frac{2R}{|x-\bar y|})^{\frac{1-\delta}{2}}dydx\\
\le& \frac{2^{2-\delta}}{\delta}(diam K)^2=C(K, \delta).
\end{align*}
\end{proof}
Now we are ready to prove Lemma \ref{lem3-4}.
\begin{proof}\ \ Let $K \subset\subset \Omega\setminus \{\bar x\}$ 
  and  $v_\epsilon^1$ be a solution of
\begin{gather} \left\{
\begin{array}{rll}
-\Delta v_\epsilon^1 &=0 \ \ &\mbox{in} \ \ K\\
v_\epsilon^1&=0 \ \ &\mbox{on} \ \ \partial K\cap\Omega\\
\frac{\partial v_\epsilon^1}{\partial\nu} \ &=2\pi(1-\epsilon)e^{v_\epsilon}\ \ &\mbox{on} \ \ \partial K\cap\partial\Omega.
\end{array}
\right. \label{3-21}
\end{gather}
From Lemma \ref{lem3-7}, we know that $e^{|v_\epsilon^1|} \in
L^p(K)$ for some $1<p<2 $, thus $||v_\epsilon^1||_{L^{p}(K)} \le C.$

Let $v_\epsilon^2= v_\epsilon -v_\epsilon^1-v_\epsilon^a,$  then
$$ \left\{
\begin{array}{rll}
-\Delta v_\epsilon^2 &=0 \ \ &\mbox{in} \ \ K\\
\frac{\partial v_\epsilon^2}{\partial\nu} + (1-\epsilon)K_g \ &=0 \ \ &\mbox{on} \ \ \partial K\cap\partial\Omega.
\end{array}
\right.
$$
We have, from the $W^{2, p}$ estimates,
\begin{gather}
\begin{array}{rll}
||v_\epsilon^2||_{L^\infty(K)} & \le C\{||(1-\epsilon)K_g||_{W^{1-\frac1p, p}(\partial K\cap\partial\Omega)}+||v_\epsilon^2||_{L^{p}(K)}\}\\
& \le C(||v_\epsilon-v_\epsilon^a||_{L^{p}(K)}+||v_\epsilon^1|_{L^{p}(K)})\\
&\le C(||\nabla v_\epsilon||_{L^p(K)}+||v_\epsilon^1|_{L^{p}(K_2)}).
\end{array}\label{3-22}
\end{gather}

Note that  for any $1< p<2$,
\begin{equation}
||\nabla v_\epsilon||_{L^p(\Omega)}<C_q.\label{e-2}
\end{equation} This is because for any
$\varphi \in W^{1,p/(p-1)}(\Omega)$ with  $\int_\Omega \varphi =0$ and
$||\varphi ||_{W^{1, p/(p-1)}(\Omega)}=1$ (thus $\varphi \in L^\infty(\Omega)$),
\begin{align*}
&|\int_\Omega \nabla v_\epsilon \nabla \varphi|\\
=&|-\int_\Omega \Delta v_\epsilon
\varphi + \int_{\partial\Omega} \frac{\partial v_\epsilon}{\partial\nu}\varphi| \\ =&|\int_{\partial\Omega}(2\pi(1-\epsilon)e^{v_\epsilon}-(1-\epsilon)K_g)\varphi|\\
\le&C ||\varphi ||_{L^\frac{p}{p-1}(\partial\Omega)}\\
\le&C ||\varphi ||_{W^{1, \frac{p}{p-1}}(\Omega)}\\
\le& C.
\end{align*}
We derive from the above and (\ref{3-4}) that
\begin{gather}
\begin{array}{rll}
2 \pi v_\epsilon^a = & \int_{\partial\Omega} K_g v_\epsilon- \int_{\partial\Omega} K_g(v_\epsilon-v_\epsilon^a) \\
\le & \int_{\partial\Omega} K_g v_\epsilon -C \to -\infty. \end{array}\label{e-3}\end{gather}
From (\ref{e-2}) and (\ref{3-22}) we have $||v_\epsilon^2 ||_{L^\infty(K)}\le C$.
Thus
$$
\int_{K} e^{pv_\epsilon}  \le \int_{K} e^{pv_\epsilon^a} \cdot
e^{pv_\epsilon^1}
\cdot e^{pv_\epsilon^2}\le C,
$$
we then know from (\ref{3-21}) that $||v_\epsilon^1||_{L^\infty(K)}\le
C$, thus $||v_\epsilon-v_\epsilon^a||_{L^\infty(K)}\le C.$
\end{proof}
From Lemma \ref{lem3-4}, we conclude, via the
standard elliptic estimates, that
\begin{equation}
 v_\epsilon - v_\epsilon^a \to G(x, \bar x) \ \ \ in \ \ C^{1,\alpha}(K)  \label{3-10}
\end{equation}
for any compact
domain $K$ in $\bar\Omega\setminus\{\bar x\},$ where $G(x, \bar x)$  satisfies
\begin{gather} \left\{
\begin{array}{rll}
-\Delta G &=0 \ \ &\mbox{in}\ \ \Omega\\
\frac{\partial G}{\partial \nu}+ K_g &=2\pi \delta_{\bar
x} \ \ &\mbox{on} \ \ \partial\Omega   \\
\int_{\partial\Omega} G &=0.
\end{array}
\right.
 \label{3-12}
\end{gather}
Thus
\begin{equation} G(x, \bar x)=-\ln |x-\bar x|+A(\bar x)+o(1), \label{3-13}
\end{equation}
where $A(\bar x)$ is a constant depending on the location of
$\bar x$, $o(1) \to 0$ as $|x| \to 0$.

To complete the proof of the main theorem, we also need a lower
bound for $v_\epsilon$ away from the singular point $\bar x$.
For convenience, we choose a normal coordinate system centered at $x_\epsilon$.
\begin{lemma}\ \ For any fixed $R>0$, let $r_\epsilon=\tau_\epsilon^{-1}R$. Then
\begin{gather}
 v_\epsilon(x) \ge G(x, x_\epsilon)- \ln \frac{2\pi }{(\sqrt{2}-1)} -A(\bar x)+o_\epsilon(1)
 \ \ \ \ \ \ \  \forall x
\in \Omega\setminus B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt{2}}), \label{3-16}
\end{gather}
where $o_\epsilon(1)\to 0$ as $\epsilon \to 0$. \label{lem3-5}
\end{lemma}
\begin{proof}\ \ On $\partial B_{r_\epsilon}( 0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega$, we define
$C^*_\epsilon:=(v_\epsilon-G) |_{|x-(0, -\frac{r_\epsilon}{\sqrt2})|=r_\epsilon}.$
From (\ref{e-1}) and (\ref{3-7}), we know that
$$C^*_\epsilon= \lambda_\epsilon + \ln \frac{\sqrt2}{2\pi R} + \ln |x|-A(\bar x)+o_\epsilon(1)
\ge -\ln\frac{2\pi}{\sqrt2-1}-A(\bar x)+o_\epsilon(1) .
$$
Let $$ \tilde K_g(x)=\left\{
\begin{array}{rll}
&K_g(x)\ \ &\mbox {if} \ \ K_g(x)\le 0\\
&\phi(x) K_g(x)\ \ &\mbox{if} \ \ K_g(x)>0,
\end{array}
\right.
$$
where $0\le \phi \le 1$ is a measurable function such that
$\int_{\partial\Omega} \tilde K_g =0.$
Let $h$ satisfy
$$\left\{
\begin{array}{rll}
-\Delta h &=0 \ \ &\mbox{in} \ \ \Omega\\
\frac{\partial h}{\partial \nu}+ \tilde K_g &=0 \ \ &\mbox{on} \ \ \partial\Omega.
\end{array}
\right.
$$
In $\Omega\setminus B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2}),$
we  consider $v_\epsilon(x) -G(x, x_\epsilon)-C^*_\epsilon+ \epsilon h(x)$ .
Since $$-\Delta (v_\epsilon(x) -G(x, x_\epsilon)-C^*_\epsilon+ \epsilon h(x))=0
\ \ \mbox{in} \ \ \Omega\setminus B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})$$
and $$\frac{\partial}{\partial\nu}(v_\epsilon(x) -G(x, x_\epsilon)-C^*_\epsilon
+ \epsilon h(x))\ge 0\ \ \mbox{on} \ \ \partial(\Omega\setminus B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})),$$
we have (\ref{3-16}) via the maximum principle.
\end{proof}
We are now ready to complete the proof of  the main theorem.

\medskip

We need to estimate $E_\epsilon=I_\epsilon(v_\epsilon)$. For any fixed  small
$\delta>0$, we assume that $\epsilon$ is sufficiently small so that
$\delta>r_\epsilon$. Then
\begin{align*}
&\int_\Omega|\nabla v_\epsilon|^2\\
=&\int_{\Omega\setminus B_\delta(0, -\frac{r_\epsilon}{\sqrt{2}})}|\nabla v_\epsilon|^2
+\int_{\Omega\bigcap B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\setminus
B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt{2}})} |\nabla v_\epsilon|^2
+\int_{\Omega\bigcap B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})} |\nabla v_\epsilon|^2 \\
:=& {\mathbf I}_1+{\mathbf I}_2+{\mathbf I}_3.
\end{align*}
To estimate ${\mathbf I}_3$, we first let $\tilde x_1=0$ and $\tilde x_2=\frac{R}{\sqrt2}$,
then from (\ref{e-1}) and (\ref{3-7})
\begin{gather}
\begin{array}{rll}
{\mathbf I}_3
=&\int_{B_R(0, -\frac R{\sqrt2})_+}|\nabla\varphi_0|^2dx+o_\epsilon(1)\\
=&\int_{B_R(0, -\frac R{\sqrt2})_+}\left|\nabla
\left(\ln\frac{\sqrt2 R}{x_1^2+(x_2+\frac R{\sqrt2})^2}-\ln 2\pi\right)\right|^2dx+o_\epsilon(1)\\
=&\int_{B_R(0, -\frac R{\sqrt2})_+}\frac 4{x_1^2+(x_2+\frac R{\sqrt2})^2}+o_\epsilon(1)\\
=&8\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_{\frac{R}{\sqrt2\sin\theta}}^R \frac 1{r^2}r drd\theta+o_\epsilon(1)\\
=&2\pi\ln\sqrt2 +c +o_\epsilon(1).
\end{array}
\label{3-17}
\end{gather}

From (\ref{3-10}) and (\ref{3-12}), we obtain
\begin{gather}
\begin{array}{rll}
 {\mathbf I}_1=& \int_{\Omega\setminus B_\delta(0,-\frac{r_\epsilon}{\sqrt2})}|\nabla v_\epsilon|^2\\
=&\int_{\Omega\setminus B_\delta(0,-\frac{r_\epsilon}{\sqrt2})}|\nabla G|^2+o_\epsilon(1) \\
=& - \int_{\Omega\setminus B_\delta(0,-\frac{r_\epsilon}{\sqrt2})} G \Delta G
+\int_{(\partial\Omega)\setminus B_\delta(0,-\frac{r_\epsilon}{\sqrt2})} G\frac{\partial G}{\partial\nu}\\
&-\int_{(\partial B_\delta(0,-\frac{r_\epsilon}{\sqrt2}))\bigcap\Omega} G\frac{\partial G}{\partial\nu}+o_\epsilon(1)\\
=&-\int_{(\partial\Omega)\setminus B_\delta(0,-\frac{r_\epsilon}{\sqrt2})} G K_g
-\int_{(\partial B_\delta(0,-\frac{r_\epsilon}{\sqrt2}))\bigcap\Omega} G\frac{\partial G}{\partial\nu}+o_\epsilon(1).
\end{array}
\label{3-18}
\end{gather}
To estimate ${\mathbf I}_2$, we first use (\ref{3-3}) to get
\begin{align*}
{\mathbf I}_2=& \int_{\Omega\bigcap (B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\setminus
B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt{2}}))} |\nabla v_\epsilon|^2\\
=&-\int_{\Omega\bigcap (B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\setminus
B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt{2}}))}\Delta v_\epsilon\cdot v_\epsilon+
\int_{\partial({\Omega\bigcap (B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\setminus
B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt{2}}))})}
v_\epsilon\frac{\partial v_\epsilon}{\partial\nu}\\
=&\int_{\partial B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega}
v_\epsilon\frac{\partial v_\epsilon}{\partial\nu}
-\int_{\partial B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega}
v_\epsilon\frac{\partial v_\epsilon}{\partial\nu}\\
&+ \int_{(B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\setminus
B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt{2}})) \bigcap \partial\Omega}
v_\epsilon\frac{\partial v_\epsilon}{\partial\nu}\\
=&\int_{\partial B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega}
v_\epsilon\frac{\partial v_\epsilon}{\partial\nu}
-\int_{\partial B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega}
v_\epsilon\frac{\partial v_\epsilon}{\partial\nu}\\
&+\int_{(B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\setminus
B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt{2}}))) \bigcap \partial\Omega}
(-(1-\epsilon)K_gv_\epsilon + 2\pi(1-\epsilon)v_\epsilon e^{v_\epsilon})
\end{align*}
Applying  Lemma \ref{lem3-5}, we have
\begin{align*}%$$\begin{array}{rll}
&\int_{(B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\setminus
B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt{2}})) \bigcap \partial\Omega}
2\pi(1-\epsilon)v_\epsilon e^{v_\epsilon}\\
\ge &2\pi(1-\epsilon)\int_{(B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\setminus
B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt{2}})) \bigcap \partial\Omega}
G e^{v_\epsilon}+o_\epsilon(1)+o_R(1)
\end{align*}%\end{array}$$
where  $o_\epsilon(1)\to 0$ as $\epsilon \to 0$ and  $o_R(1)\to 0$ as $R \to
+\infty$.

From (\ref{3-3}), (\ref{3-10}) and (\ref{3-12})
we obtain
\begin{align*}
&2\pi(1-\epsilon)\int_{(B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\setminus
B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt{2}})) \bigcap \partial\Omega}
G e^{v_\epsilon}\\
=&\int_{(B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\setminus
B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt{2}})) \bigcap \partial\Omega}
G \frac{\partial v_\epsilon}{\partial\nu}+ \int_{(B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\setminus
B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt{2}})) \bigcap \partial\Omega}(1-\epsilon)K_g G\\
=&\int_{\partial((B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\setminus
B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2}))\bigcap\Omega)} G \frac{\partial v_\epsilon}{\partial\nu}
- \int_{\partial B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega} G \frac{\partial v_\epsilon}{\partial\nu}\\
 &+ \int_{\partial B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega} G \frac{\partial v_\epsilon}{\partial\nu} +o_\epsilon(1)+o_\delta(1)\\
 =&\int_{\partial B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega} v_\epsilon \frac{\partial G}{\partial\nu} - \int_{\partial B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega} v_\epsilon \frac{\partial G}{\partial\nu} + \int_{(B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\setminus
B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt{2}})) \bigcap \partial\Omega}v_\epsilon \frac{\partial G}{\partial\nu}\\
 &- \int_{\partial B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega} G \frac{\partial v_\epsilon}{\partial\nu}
 + \int_{\partial B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega} G \frac{\partial v_\epsilon}{\partial\nu} +o_\epsilon(1)+o_\delta(1)\\
 =& \int_{\partial B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega} v_\epsilon^a \frac{\partial G}{\partial\nu}- \int_{\partial B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega} v_\epsilon \frac{\partial G}{\partial\nu} + \int_{(B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\setminus
B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt{2}})) \bigcap \partial\Omega}v_\epsilon \frac{\partial G}{\partial\nu}\\
& + \int_{\partial B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega} G \frac{\partial v_\epsilon}{\partial\nu} +o_\epsilon(1) + o_\delta(1)\\
 =&\int_{\partial B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega} v_\epsilon^a \frac{\partial G}{\partial\nu} - \int_{\partial B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega} v_\epsilon \frac{\partial G}{\partial\nu}+ \int_{\partial B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega} G \frac{\partial v_\epsilon}{\partial\nu}\\
 &- \int_{(B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\setminus B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt{2}})) \bigcap \partial\Omega}
 v_\epsilon K_g +o_\epsilon(1)+ o_\delta(1)
\end{align*}
Thus
\begin{align*}
{\mathbf I}_2\ge &-(2-\epsilon)\int_{(B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\setminus
B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt{2}})) \bigcap \partial\Omega}K_g v_\epsilon
+\int_{(\partial B_\delta(0, -\frac{r_\epsilon}{\sqrt2}))\bigcap\Omega}
v_\epsilon\frac{\partial v_\epsilon}{\partial\nu}\\
&-\int_{(\partial B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2}))\bigcap\Omega}
v_\epsilon\frac{\partial v_\epsilon}{\partial\nu}+ \int_{\partial B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega} v_\epsilon^a \frac{\partial G}{\partial\nu}\\
 &- \int_{\partial B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega} v_\epsilon \frac{\partial G}{\partial\nu}+ \int_{\partial B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega} G \frac{\partial v_\epsilon}{\partial\nu} +o_\epsilon(1)+ o_\delta(1)+o_R(1)
\end{align*}
We now estimate the boundary term in the right hand side of the
above inequality. From (\ref{e-1}), (\ref{3-7}) and Lemma
\ref{lem3-5}, we know that
$$
-\int_{\partial B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega}  \frac{\partial v_\epsilon}{\partial\nu} (v_\epsilon-G)\ge \pi(-A(\bar x)- \ln\frac{2\pi }{\sqrt2-1})+o_\epsilon(1),
$$
and
\begin{align*}
 &-\int_{\partial B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega} v_\epsilon \frac{\partial G}{\partial\nu}\\
&=-\lambda_\epsilon \int_{\partial B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega} \frac{\partial
G}{\partial\nu}-(\ln \frac{\sqrt2}{R}) \int_{\partial B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega} \frac{\partial
G}{\partial\nu}+o_\epsilon(1)+o_R(1)\\
&=-\lambda_\epsilon \int_{\partial\Omega\setminus B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})}\frac{\partial G}{\partial\nu}
-(\ln \frac{\sqrt2}{R}) \int_{\partial B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})} \frac{\partial G}{\partial\nu}+o_\epsilon(1)+o_R(1)\\
 &\ge\lambda_\epsilon \int_{\partial\Omega\setminus B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})}K_g+(\ln \frac{\sqrt2}{R})(\frac{\pi}{2}+\frac{\pi}{2(\sqrt2-1)}) +o_\epsilon(1)+o_R(1).
\end{align*}
From (\ref{3-12}) we have
$$
v^a_\epsilon \cdot \int_{\partial B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega}\frac{\partial
G}{\partial\nu} = v^a_\epsilon \int_{ \partial\Omega\setminus B_\delta(0, -\frac{r_\epsilon}{\sqrt2})} \frac{\partial G}{\partial\nu}= - v^a_\epsilon \int_{ \partial\Omega\setminus B_\delta(0, -\frac{r_\epsilon}{\sqrt2})} K_g,
$$
and
\begin{align*}
&v^a_\epsilon \cdot \int_{\partial B_\delta(0,\frac{r_\epsilon}{\sqrt2})\bigcap\Omega}
\frac{\partial v_\epsilon}{\partial\nu}\\
&=v^a_\epsilon \cdot \int_{(\partial\Omega)\setminus B_\delta(0,\frac{r_\epsilon}{\sqrt2})}
\frac{\partial v_\epsilon}{\partial\nu}\\
&=v^a_\epsilon \int_{(\partial\Omega)\setminus B_\delta(0,\frac{r_\epsilon}{\sqrt2})}
( -(1-\epsilon)K_g + 2\pi(1-\epsilon) e^{v_\epsilon})\\
&= -(1-\epsilon)v^a_\epsilon \int_{(\partial\Omega)\setminus B_\delta(0,\frac{r_\epsilon}{\sqrt2})}K_g
+ v_\epsilon^a e^{v_\epsilon^a}\int_{(\partial\Omega)\setminus B_\delta(0,\frac{r_\epsilon}{\sqrt2})}
2\pi(1-\epsilon) e^{v_\epsilon-v_\epsilon^a}\\
&= -(1-\epsilon)v^a_\epsilon \int_{(\partial\Omega)\setminus B_\delta(0,\frac{r_\epsilon}{\sqrt2})}K_g + o_\epsilon(1).
\end{align*}
From (\ref{3-10}),
\begin{align*}
&\int_{(B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\setminus
B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt{2}})) \bigcap \partial\Omega}K_g v_\epsilon\\
&=v^a_\epsilon \int_{(B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\setminus
B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt{2}})) \bigcap \partial\Omega}K_g+\int_{(B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\setminus
B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt{2}})) \bigcap \partial\Omega}K_g ({v_\epsilon}-v^a_\epsilon)\\
&=v^a_\epsilon \int_{(B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\setminus
B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt{2}})) \bigcap \partial\Omega}K_g +o_\delta(1)\\
&=2\pi v^a_\epsilon -v^a_\epsilon \int_{\partial\Omega\setminus B_\delta(0, -\frac{r_\epsilon}{\sqrt2})}K_g
+ v^a_\epsilon \int_{\partial\Omega\bigcap B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})}K_g +o_\delta(1)
\end{align*}
We conclude
\begin{gather}
\begin{array}{rll}{\mathbf I}_2\ge &-2\pi(2-\epsilon)v^a_\epsilon
+ (2-\epsilon)v^a_\epsilon \int_{\partial\Omega\bigcap B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})}K_g
+\pi(-A(\bar x)- \ln\frac{2\pi R}{(\sqrt2-1)R})\\
&+\lambda_\epsilon( 2\pi - \int_{B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})\bigcap \partial\Omega} K_g)
+ \int_{\partial B_\delta(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\Omega} G \frac{\partial G}{\partial\nu}\\
&+(\ln \frac{\sqrt2}{R})(\frac{\pi}{2}+\frac{\pi}{2(\sqrt2-1)})+o_\epsilon(1)+o_R(1)+o_\delta(1)
\end{array}
\label{3-19}
\end{gather}
We have, from (\ref{3-17})-(\ref{3-19}), that
\begin{align*}
I_\epsilon(v_\epsilon)\ge &-\frac{\pi}{8} A(\bar x)-\frac1{8}\ln\frac{2\pi}{\sqrt2-1}+\frac{\pi\ln\sqrt2}{4} +\lambda(\frac{\pi}{4}-\frac1{8}\int_{B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\partial\Omega} K_g)\\
&-\frac{\pi}{4}\epsilon v_\epsilon^a+\frac{1-\epsilon}{4}\int_{B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})\bigcap\partial\Omega} K_g v_\epsilon^a
-\frac18\int_{(\partial\Omega)\setminus B_\delta(0,-\frac{r_\epsilon}{\sqrt2})} G K_g\\
&+\frac{1-\epsilon}{4}\int_{\partial\Omega} K_g (v_\epsilon- v_\epsilon^a)+\frac{c}8
+o_\epsilon(1)+o_R(1)+o_\delta(1).
\end{align*}
From  (\ref{e-3}), we know that $v^a_\epsilon \to -\infty$. Also, since
$\lambda_\epsilon \cdot  vol(B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})) =o_\epsilon(1)$, we know from Lemma
\ref{lem3-5}, that $
 v^a_\epsilon \cdot vol(B_{r_\epsilon}(0, -\frac{r_\epsilon}{\sqrt2})) =o_\epsilon(1)$.  We finally have
 $$
  \lim_{\epsilon \to 0} E_\epsilon \ge -\frac{\pi}{8} A(\bar x)-\frac1{8}\ln\frac{2\pi}{\sqrt2-1}+\frac{\pi\ln\sqrt2}{4} +C.
 $$
for some constant $C$. Thus Theorem \ref{main2} is proved.

\chapter{Conclusion and Future Work}
In this research, we study the inequalities related to Moser-Trudinger-Onofri inequality. We first prove an analog Hardy inequality with sharp constant involving exponential weight function in chapter 2. When $n=2$, this inequality allows us to give a direct and simple proof of Onofri inequality on $S^2$. In chapter 3, we start with Lebedev-Milin inequality and derive the sharp local trace inequality
$$\ln\int_{-1}^1 e^u \frac {dy_1}{2\pi}\le\frac 1{4\pi}\int_D |\nabla u|^2 dy+C\ \ \forall  u\in H^1(D),$$ where $D=\{ (y_1, y_2)\in \mathbb{R}^2_+: y_1^2+(y_2+1)^2 < 2\}.$ This inequality and partition of unity are used to establish the following sharp inequality on general bounded domain $\Omega\in\mathbb{R}^2$,
$$
\ln\int_{\partial\Omega} e^{u}dS \le (\frac 1{4 \pi}+\epsilon)
  \int_{\Omega}|\nabla u|^2 d y+C_1\int_{\Omega} u^2\ dy +C_2  \ \ \ \ \forall  u\in H^1(\Omega).
$$
This inequality is followed by the $\epsilon-$level sharp trace inequality on simply connected domain $\Omega\in\mathbb{R}^2$,
$$\ln\int_{\partial\Omega} e^u dS \le (\frac 1{4 \pi}+\epsilon)
  \int_{\Omega}  |\nabla u|^2 d y + C_3  \ \ \forall  u\in H^1(\Omega),\ \ \mbox{with} \int _{\partial\Omega} K_g\cdot u dS_g=0.$$
We then define a functional and study the behavior of the minimizing sequence as the parameter $\epsilon$ goes to zero. It leads to the sharp trace inequality on two-dimensional simply connected domain,
$$\ln\int_{\partial\Omega} e^u dS \le \frac 1{4 \pi}
  \int_{\Omega}  |\nabla u|^2 d y + \frac 1{2 \pi} \int_{\partial\Omega} K_g \cdot udS + C(\Omega) \ \ \forall  u\in H^1(\Omega).$$

It is quite natural to ask whether one can adapt this trace inequality for
two-dimensional compact manifold with boundary. I plan to continue my research
on extending such trace inequality to two-dimensional compact manifold with boundary.


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