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| − | \begin{document}
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| − | \title{Lecture 2}
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| − | \newcommand{\e}{\epsilon}
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| − | \newcommand{\ra}{\rightarrow}
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| − | \newcommand{\g}{\mathfrak{g}}
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| − | \newcommand{\M}{\mathcal{M}}
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| − | \newcommand{\al}{\alpha}
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| − | \newcommand{\PR}{\mathcal{P}}
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| − | \newcommand{\Q}{\mathbb{Q}}
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| − | \newcommand{\G}{\mathcal{G}}
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| − | \newcommand{\F}{\mathcal{F}}
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| − | \newcommand{\C}{\mathbb{C}}
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| − | \newcommand{\R}{\mathbb{R}}
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| − | \newcommand{\A}{\mathcal{A}}
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| − | \newcommand{\B}{\mathcal{B}}
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| − | \newcommand{\dvol}{\text{ d\textit{vol}}}
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| − | \newtheorem*{thm}{Proposition}
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| − | \newtheorem*{thm1}{Claim}
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| − | \maketitle
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| − | \section{Sobolev Embedding}
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| − |
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| − | Let $X$ be smooth manifold of dimension $n$, $\dvol$, let $E$ be a
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| − | vector bundle over $X$, and let $A$ be a connection on $E$. WE
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| − | define the Sobolev Space $L^p_{k,A}(X,E)$ to be the completion of
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| − | $C^\infty(X,E)$ with respect to the norm
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| − |
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| − | $$||s||^p_{p,k,A}=\sum_{i=0}^k\int_{X}|\nabla_A\overset{i \text{
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| − | times}}{\circ\cdots\circ}\nabla_A s|^p\dvol$$
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| − |
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| − | If no vector bundle is specified, then the trivial bundle, i.e. the
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| − | set of smooth functions on $X$, is implied. Other notations for
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| − | this space are $W^p_{k,A}$ and in the case $p=2$, $H_{k,A}$. We
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| − | would like to know when $L^p_k(X)=subset L^q_l(X)$ holds. Define
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| − | the conformal weight of $L^p_k$, to be $k-n/p$. Then for $f\in
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| − | L^p_k$, $||f(tx)||_{p,k}\simeq t^{k-n/p}||f(x)||$, so for such an
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| − | inclusion to hold, we must have $k-n/p\leq l-n/q$. If $X$ is
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| − | compact, then this, and the condition that $k\geq l$ are sufficient
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| − | for $L^p_k(X)=subset L^q_l(X)$. Furthermore, Rellich's theorem
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| − | tells us that if both inequalities are strict, this inclusion is a
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| − | compact embedding,
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| − |
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| − | Let $C^{l,\al}(X)$ be the space of functions with $l$ derivatives,
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| − | such that the $l^{th}$ derivative is Holder continuous with exponent
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| − | $\al$. If $0\al<1$, then $L^p_k(X)=subset C^{l,\al}\iff k\geq l$,
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| − | and $k-n/p\geq l+\al$. The case $\al=0$ is much more subtle. For
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| − | example, if $\Sigma$ is a Riemann surface, $L^2_1(\Sigma)\not=subset
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| − | C^0(\Sigma)$, and if $X$ is a 4-manifold, $L^4_1(X)\not=subset
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| − | C^0(X)$.
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| − |
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| − | \section{Sobolev Multiplication}
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| − | Multiplication gives a map $L^p_k\times L^q_l\ra L^r_m$ if
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| − | $$\text{a) }(k-n/p)+(l-n/q)>m-n/r\text{ and }k,l\geq m,\text{
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| − | or}$$
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| − | $$\text{b) }k-n/p>0\text{ and }k-n/p,l-n/q\geq m-n/r$$
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| − |
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| − | The proofs of these two results follow from the embedding theorem,
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| − | and Holder's inequality.
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| − |
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| − | \section{Differentiation on Banach Spaces}
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| − |
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| − | Let $X,Y$ be Banach spaces, $U=subset X$ an open subset, and $f:U\ra
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| − | Y$ a map. We say that $f$ is differentiable at $x\in U$ if there is
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| − | $C>0$ and a bounded linear operator $A:X\ra Y$ such that
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| − | $||f(x+h)-f(x)-A\dot h||_Y\leq C||h||^2_X$ (this exponent 2 here is
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| − | not quite the correct definition, but it will do for our purposes).
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| − | In this case, we write $d_x f$ for $A$, and call this the derivative
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| − | of $f$ at $x$.
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| − |
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| − | We say that $f$ is continuously differentiable if $x\mapsto A_x:U\ra
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| − | \text{Hom}(X,Y)$ is a continuous map. We say that $f$ is twice
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| − | differentiable at $x$ is it is continuously differentiable at $x$,
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| − | and the map $x'\mapsto d_{x'}f$ is differentiable at $x$. Iterating
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| − | this definition allows us to define higher derivatives. In
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| − | particular, $f$ is $C^\infty$ if it has derivatives of all orders.
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| − |
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| − | It turns out that if $(f,g)\mapsto \mu(f,g)=f\cdot g:L^p_k\times
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| − | L^q_l\ra L^r_m$ is continuous, then it is $C^\infty$. In fact, if
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| − | we let $\delta$ denote a small variation, and letting
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| − | $d_{\mu(f,g)}(\delta f,\delta g)=f\delta g+g\delta f$, we have
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| − |
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| − | $$||\mu(f+\delta f,g+\delta g)-\mu(f,g)-d_{\mu(f,g,)}(\delta
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| − | f,\delta g)||_{r.m}$$
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| − |
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| − | $$=||\delta f\cdot\delta g||_{r,m}\leq C(||\delta f||_{p,k}||\delta
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| − | g||_{l,q})^2$$
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| − |
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| − | where the last inequality follows from the multiplication theorem.
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| − | The derivative $d_{\mu(f,g)}(\delta f,\delta g)$ only depends on the
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| − | product $f\cdot g$, so the higher derivatives all vanish.
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| − |
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| − | \section{Application to the Space of Connections}
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| − |
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| − | Let $\pi:P\ra X$ be a principal $G$-bundle, and let $\g$ be the Lie
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| − | algebra of $G$. Let $A$ denote a connection on $P$,
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| − | $F_A\in\Omega^2(X,\text{ad }P)$ its curvature. If, $\tau$ is a
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| − | local trivialization, we let $a^\tau$ denote the one-form associated
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| − | to the connection, $A\mapsto F_A$ corresponds to $a^\tau\mapsto
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| − | da^\tau+[a^\tau\wedge a^\tau]$. Recall the following definition:
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| − | $$\A^p_k=\{\text{connections }A|\text{ if }\tau:\pi^{-1}(U)\ra
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| − | U\times P\text{ is a }C^\infty\text{ trivialization such that if }$$
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| − | $$\phi\in C^\infty_0(U)\text{ we have } a_\tau\in\Omega^1(U,\g),
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| − | \phi a_\tau\in L^p_k(U,T^pU\otimes\g)\}$$
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| − |
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| − | Let $\F:\A^p_k\ra L^q_l(X,\Lambda^2\otimes\text{ad }P)$ be the map
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| − | sending a connection to its curvature. We would like to know when
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| − | $\F$ is continuous. Clearly, $a^\tau\mapsto da^\tau$ is continuous
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| − | into $L^p_{k-1}$. From the multiplication theorem, $a^\tau\mapsto
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| − | [a^\tau\wedge a^\tau]$ is continuous when
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| − | $(k-n/p)+(k-n/p)>(k-1)-n/p$, i.e. $k+1\geq n/p$. In fact, in this
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| − | case it is $C^\infty$. For example, for $n=4$, $(p,k)=(2,1)$,
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| − | $(q,l)=(4,0)$, $a^\tau\mapsto [a^\tau\wedge a^\tau]$ is continuous.
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| − |
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| − | We specialize to the the situation where $G=O(n),U(n),Sp(n)$, amd
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| − | where $P$ is the frame bundle of some vector bundle $E$. A smooth
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| − | gauge transformation $u$ is a smooth bundle endomorphism of $E$ such
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| − | that $u$ is an orthogonal (resp. unitary or symplectic) isomorphism
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| − | on each fiber. In other words, $u\in C^\infty(X,\text{End}(E))$,
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| − | and $u^*u=1$, where $u^*$ denotes the transpose if $G=O(n)$, the
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| − | conjugate transpose if $G=U(n)$, etc. We define
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| − | $$\G^p_k=\{u\in L^p_k(X,\text{End}(E))|u^*u=1\text{ almost everywhere}\}$$
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| − |
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| − | We would like $\G^p_k$ to be a group, and ideally, a Banach Lie
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| − | group.
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| − |
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| − | \begin{thm}
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| − | If $(k-n/p)>0$, then $\G^p_k$ is a Banach Lie group, and
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| − | $L^p_k(X,\text{End}(E))$ is a Banach space and algebra
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| − | \end{thm}
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| − |
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| − | \begin{proof}
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| − | Define $\Phi:L^p_k(X,\text{End}(E))\ra L^p_k(X,\text{End}(E))$ by
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| − | $\Phi(s)=s^*s$. Then $\Phi$ is a $C^\infty$ map, and
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| − | $d_s\Phi(t)=s^*t+t^*s$. Using $s^*s=1$, once easily sees that
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| − | $d_s\Phi(sw/2)=w$, so that $d_s\Phi$ is surjective for all
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| − | $s\in\G^p_k$. It is also clear that
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| − | $\text{ker}(d_s\Phi)=\{sr|r=-r^*\}$. These two facts allows us to
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| − | apply the implicit function theorem to conclude that
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| − | $\Phi^{-1}(1)=subset\G^p_k$ is a Banach Lie group.
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| − | \end{proof}
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| − |
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| − | We would next like to determine when $\G^p_k$ acts on $\A^q_l$. The
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| − | multiplication theorem implies that $L^p_{k-1}$ is an $L^p_k$ module
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| − | if $(k-n/p)$>0. The argument at the end of section 3 shows that
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| − | this multiplication is smooth, so that $\G^p_k\times\A^p_{k-1}\ra
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| − | \A^p_{k-1}$ is a smooth.
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| − |
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| − | Let $\B^p_{k-1}=\A^p_{k-1}/\G^p_k$. We want to know when this is a
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| − | Hausdorff space. Recall the following criterion: If $G$ is a
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| − | topological group, that acts continuously on a topological space
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| − | $X$, then $X/G$ is Hausdorff if and only if the graph
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| − | $\Gamma=\{(x,xg)|x\in X,g\in G\}$ is closed in $X\times X$.
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| − |
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| − | \begin{thm1}
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| − | If $(k-n/p)>0$, then $\B^p_{k-1}$ is Hausdorff
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| − | \end{thm1}
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| − |
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| − | \begin{proof}
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| − | Suppose we have a sequence $(A_i,u_iA_i)$ that converges to $(A,A')$
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| − | in $L^p_{k-1}$. Then $a^\tau_i$ converges to $a^\tau$ in $L^p_k$,
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| − | and $u^{-1}_idu_i+u^{-1}_ia^\tau_iu_i\ra a'^\tau$. Rearranging this
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| − | gives $du_i=u_ia'^\tau-a^\tau_iu_i$. Suppose $k=1$, $p=n+\e$. Then
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| − | since $a^\tau_i$ converges, and $u_i$ are all in $\G^p_k$, this
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| − | implies that $du_i$ is uniformly bounded. If we pass to a
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| − | subsequence where $u_i$ converges, say to $u$, in $C^0$, then $du_i$
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| − | converges in $L^{n+\e}$, so that $u_i$ converges to $u$ in
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| − | $L^{n+\e}_1$. For $k>1$, one applies a bootstrap argument.
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| − | \end{proof}
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| − | \end{document}
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