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\leftheadtext{A degree theory for BMO maps}
\rightheadtext{H. Brezis and L. Nirenberg}
\topmatter
\jourlogo{\eightpoint Selecta Mathematica, {\bf 1} (1995), p. 197-263}
\medskip
\medskip
\title  Degree theory and BMO;\\
PART I:  Compact Manifolds without Boundaries\endtitle
\author  Ha\"im Brezis$^{(1)}$
and
Louis Nirenberg$^{(2)}$\endauthor
\affil  $^{(1)}$Universit\'e P. et Marie Curie, Paris, France\\
and Rutgers University, New Brunswick, New Jersey\\
$^{(2)}$New York University, Courant Institute, New York\endaffil

\address  $^{(1)}$Analyse Num\'erique\endgraf
Universit\'e P. et M. Curie\endgraf
4, Pl. Jussieu\endgraf
75252 Paris, Cedex 05 France\endgraf
and\endgraf
Rutgers University\endgraf
Department of Mathematics\endgraf
Hill Center, Busch Campus,\endgraf
New Brunswick, N.J. 08903\endgraf
\endaddress
\email   brezis\@ann.jussieu.fr;  brezis\@math.rutgers.edu;\endemail
\address  $^{(2)}$Courant Institute\endgraf
New York University\endgraf
251 Mercer Street,\endgraf New York, NY 10012\endaddress
\email  nirenl\@cims.nyu.edu\endemail
\endtopmatter
\document
\noindent
{\bf  Section I.0.  Introduction}
\medskip
In this paper we consider degree theory for mappings $u$ from one
compact smooth $n$-dimensional manifold $X$ to a connected compact
smooth manifold $Y$ of the same dimension.  These are manifolds
without boundary and which are oriented.
\medskip
The classical degree counts the ``number of times'' $Y$ is covered by
$u(X)$, taking into account algebraic multiplicity.  For instance, if
$u\in C^1$ and $y\in Y$ is a regular value of the map $u$,\; i.e., $u^{-1}(y)$
consists of a finite number of points $x_1,\ldots, x_k$ at each of
which the Jacobian of the map, $J_u$, in terms of local coordinates
(with the given orientation), is nonsingular, then
$$
\deg(u,X,y) = \sum_j \text{sgn}\det J_u (x_j).
$$
A basic fact is that this degree is independent of the choice of the
regular value $y$, and we then denote this degree by $\deg(u,X,Y)$.
\medskip
Degree extends to continuous maps $u$ from $X$ to $Y$ because of the
fundamental fact that if $u, v\in C^1(X,Y)$, and are close in the $C^0$
topology, then they have the same degree.  Degree theory is often
defined directly for continuous maps via the action of the map on $n$th
degree homology.
\medskip
One of the important properties of degree is that if it is not zero
then the map is onto $Y$.  Another basic fact is that the degree is
invariant under continuous deformation of the map (homotopy).
\medskip
For a $C^1$-map there is an integral formula for the degree.  Namely,
if $\mu$ is a smooth $n$-form on $Y$ then
$$
\deg(u,X,Y) \int_Y \mu = \int_X \mu\circ u. \tag 0.1
$$
(see e.g., L.~Nirenberg~[1]).  This may be expressed using local
coordinates by
$$
\int_X f(u) \det J_u(x) dx_1 \wedge\ldots\wedge dx_n,
$$
if $\mu = f(y) dy_1 \wedge\ldots\wedge dy_n$.  In particular, if $X$
and $Y$ are Riemannian manifolds, then
$$
\deg(u,X,Y) = \frac{1}{\text{vol}(Y)} \int_X \det J_u (x) d\sigma(x)
\tag 0.2
$$
where $d\sigma$ is the volume element on $X$, and $J_u$ is computed using
geodesic normal coordinates at $x$ and geodesic normal coordinates at
$u(x)$.
\medskip
Specializing further, consider $X = \partial\Omega$, where $\Omega$ is
a smooth bounded domain in $\Bbb R^{n+1}$, and $Y = S^n$.  Consider
$u\in C^1(X,Y)$ and let $\tilde u$ be any $C^1$-extension inside
$\Omega$ with values in $\Bbb R^{n+1}$.  There is another integral
formula for the degree of $u$:
$$
\deg (u,\partial\Omega, S^n) = \frac{1}{|B|} \int_\Omega \det
J_{\tilde u} dx_1 \ldots dx_{n+1}, \tag 0.3
$$
where $|B|$ is the volume of the unit ball $B$ in $\Bbb R^{n+1}$.
Since $\det J_{\tilde u}$ is a divergence expression, using Green's
theorem, one easily obtains the equality of the two integral formulas.
\medskip
Formulas (0.2), (0.3) suggest the possibility of extending degree
theory to another class of maps---which need not be
continuous---namely, maps in appropriate Sobolev spaces.  This was
done in the 80's:

(a)  In connection with their proof of the existence of
``large'' harmonic maps, H.~Brezis and J.~M.~Coron~[1] (see also
H.~Brezis~[2]) were led to consider degree for $H^1$ maps from $S^2$
to $S^2$.  This degree is given by the integral on the right hand side
of (0.2).  To prove that this integral is an integer relies on the
fact that smooth maps from $S^2$ to $S^2$ are dense in $H^1(S^2, S^2)$
(see R.~Schoen and K.~Uhlenbeck~[1]).
\smallskip
(b)  Motivated by a question concerning the Ginzburg-Landau
equation (see Boutet de Monvel-Berthier, Georgescu and Purice [1]),
L.~Boutet de Monvel and O.~Gabber introduced a degree for maps $u\in
H^{1/2}(S^1, S^1)$.  It is the familiar case of (0.2), namely the
``change in argument''
$$
\deg(u,S^1,S^1) = \frac{1}{2\pi i} \int_{S^1} \frac{du}{u} =
\frac{1}{2\pi i} \int_{S^1} \bar u du. \tag 0.4
$$
Using the duality between $H^{1/2}$ and $H^{-1/2}$ one sees that this
is well defined.  (The degree may also be expressed in terms of the
Fourier coefficients of $u$; see Section I.5.  It is then transparent
that degree makes sense for $u\in H^{1/2}$.)  That the expression (0.4)
(or the analogue in terms of the Fourier coefficients) is an integer
for $u\in H^{1/2}$ is proved by approximation, as above.
Alternatively, one may extend $u\in H^{1/2} (S^1, S^1)$ to $\tilde{u}
\in H^1(B,\Bbb R^2)$, and then use formula (0.3).  This $H^{1/2}$
degree is also used in Bethuel, Brezis and H\'elein [1].
\medskip
The natural generalization of (a) is to maps in the Sobolev space
$W^{1,n}(X,Y)$, while (b) extends degree to maps in
$W^{\frac{n}{n+1},n+1} (\partial\Omega,S^n)$---a space slightly larger
than $W^{1,n}$.  These are borderline spaces:  embedding into
continuous functions just fails.
\medskip
In connection with degree for $H^{1/2}(S^1,S^1)$, L.~Boutet de Monvel
and O.~Gabber made the interesting observation that the notion of
degree for maps from $S^1$ to $S^1$ makes sense for maps in the class
VMO:  the closure in the BMO($=$ bounded mean oscillation) topology of
smooth maps.  However they did not establish the basic properties of
VMO degree, such as stability under homotopy within VMO, surjectivity
if $\deg\not=0$, etc.  The VMO degree is not defined by an integral
formula; it is defined via approximation.  More precisely, they
pointed out that if $u\in\text{VMO}(S^1,S^1)$ and
$$
\bar u_\vare(\theta) = \frac{1}{2\vare}
\int^{\theta+\vare}_{\theta-\vare} u(s) ds,
$$
then $|\bar u_\vare(\theta)|\rightarrow 1$ uniformly in
$\theta$---despite the fact that $u$ need not be continuous.  Then,
for $\vare$ small, 
$$
u_\vare(\theta) = \frac{\bar u_\vare(\theta)}{|\bar u_\vare(\theta)|}
$$
has a well defined degree, which is independent of $\vare$.  This is
their definition of the degree.
\medskip
In this paper we develop this concept for maps between $n$-dimensional
manifolds $X,Y$, and establish its basic properties.  The degree is
defined via approximation, in the BMO topology, by smooth maps from
$X$ to $Y$.
\medskip

A natural related question is:  Are smooth maps from $X$
to $Y$ dense in $W^{1,p} (X,Y)$, with $1\leq p < \infty$?  Here $X$
and $Y$ might have different dimensions.  For $p\geq \dim X$ the
answer is always yes.  For $p < \dim X$ the answer was given by
F.~Bethuel [1]:  a necessary and sufficient condition for density is
that $\Pi_{[p]} (Y) = 0$, where $\Pi$ denotes the homotopy class and
$[p]$ is the integral part of $p$.
\medskip
We now describe the organization of the paper.
\medskip
In Section I.1 we recall the notion of BMO maps in Euclidean spaces
and describe its extension to maps between manifolds.  For this
purpose it is convenient to put a Riemannian metric on $X$ and to
embed $Y$ smoothly into some $\Bbb R^N$.  However, the notion of
BMO$(X,Y)$ is independent of the particular metric or embedding---as
will be the degree.
\medskip
The BMO (semi) norm of a map $u$ from $X$ to $\Bbb R^N$ is
$$
\|u\|_{BMO} = \sup\Sb x\in X\\\vare < r_0\endSb \Mint_{B_\vare(x)}
|u(y) - \bar u_\vare(x)| d\sigma(y) \tag 0.5
$$
where
$$
\bar u_\vare(x) = \Mint_{B_\vare(x)} u d\sigma. \tag 0.6
$$
Here, for any $x\in X$, $B_\vare(x)$ is the geodesic ball centred at
$x$ with radius $\vare < r_0$, the injectivity radius of $X$ and
$\dsize\Mint_A u$ denotes the average of $u$ in a set $A$.  A very
convenient equivalent (semi) norm is
$$
\|u\|_\star = \sup\Sb x\in X\\ \vare < r_0\endSb
\Mint_{B_\vare(x)}\;\Mint_{B_\vare(x)} |u(y) - u(z)| d\sigma(y)
d\sigma(z). \tag 0.7
$$
(Incidentally, (0.7) suggests a notion of BMO for maps into a general metric
space $Y$.  Such maps enjoy some of the basic properties of BMO maps,
e.g. the John-Nirenberg~[1] inequality holds---via the usual proof.)
\medskip
The space VMO($=$ vanishing mean oscillation) is the completion of
smooth maps in the BMO norm.  This space was introduced by Sarason~[1]
who established a useful characterization (see Lemma 3).  In the same
section we present some of the properties of VMO maps, such as the
effect of left composition by a Lipschitz map $F$ (see Lemma $2'$ and
the more general Lemma A.7 in Appendix A).  The map $\Psi : u\mapsto
F\circ u$, for $u\in \text{BMO}(X,\Bbb R^N)$, {\it need not} be
continuous in its dependence on $u$---as a map from BMO to BMO---but
it is continuous at every $u\in\text{VMO}$ (see Lemma A.8 and Remark
A.1).  Lemma 4 gives a characterization of compact sets in VMO---an
adaptation of Arzel\`a-Ascoli to VMO.
\medskip
The proofs of many technical statements are given in Appendix A.
\medskip
Section I.2 takes up various examples of BMO and VMO maps.  In
addition to continuous maps, VMO contains all the ``borderline''
Sobolev spaces $W^{s,p}$ for $1 < p, s p = n$.
\medskip
The degree for VMO maps is defined in Section I.3.  The first main
result, Theorem 1, deals with its stability under perturbation in VMO:
given $u\in\text{VMO}(X,Y)$, there exists $\delta$ depending on $u$
such that, for $v\in\text{VMO}(X,Y)$ with $\|u - v\|_{\text{BMO}} <
\delta$, it has the same degree as $u$; this implies in turn the
invariance of degree under homotopy within VMO.  
\medskip
Surprisingly the $\delta$ really depends on $u$ (see Lemma 6).  This
is in contrast to the standard perturbation of continuous maps; there
the $\delta$ is uniform.  We point out in Remark 7 that the degree can
also be defined for $u$ in BMO$(X,Y)$ provided $u$ is ``close'' to
VMO.
\medskip
In Section I.4 we carry over standard properties of degree to VMO.
For example, we prove that if $\deg u\not=0$, then $u$ is ``onto''
$Y$.  This is more subtle than for the continuous case because $u$ may
be changed on a set of measure zero.  We are led to a notion of
``essential range'' of $u$ which is independent of the choice of
representatives in the class of equivalent maps.
\medskip
The formulas (0.2), (0.3) extend when $u$ is in some appropriate
``borderline'' Sobolev space (see Properties 4 and 5).
\medskip
In Section I.5 we take up a natural question concerning maps from $X$
to $Y$, not necessarily of the same dimension.  BMO$(X,Y)$---as well
as $L^p(X,Y), 1\leq p\leq \infty$---is arcwise connected, while
VMO$(X,Y)$ has components which are simply the closures of the components of
$C^0$ maps.
\medskip
Section I.6 deals with a question first considered by R.~Coifman and
Y.~Meyer [1], namely the possibility of lifting a map
$u\in\text{BMO}(X,S^1)$ to BMO$(X,\Bbb R)$.  Theorem 3 asserts that this
can be done with VMO if and only if $u$ is homotopic to a constant
within VMO.  In Theorem 4, which is directly related to a result in
Coifman and Meyer [1], we show that any $u\in\text{BMO}(X,S^1)$
with small BMO norm may be written as $u = e^{i\varphi}$ with
$\varphi\in\text{BMO}(X,\Bbb R)$ and $\|\varphi\|_{\text{BMO}} \leq 4
\|u\|_{\text{BMO}}$.  The proofs are quite technical.  They make use
of the John-Nirenberg inequality; various forms of this inequality for
manifolds are presented in Appendix B.
\medskip
Of course degree theory extends to maps on domains or manifolds with
boundary.  In Part II we will consider this situation for VMO maps.  A
new feature is that VMO maps in a domain need not have a trace on the
boundary.  This makes the theory more delicate.
\medskip
The plan of the paper is the following:
\medskip
\roster
\item"{I.1.}"  BMO and VMO
\smallskip
\item"{I.2.}"  Some examples of BMO and VMO functions
\smallskip
\item"{I.3.}"  Degree for VMO maps
\smallskip
\item"{I.4.}"  Some properties of degree
\smallskip
\item"{I.5.}"  Further comments
\smallskip
\item"{I.6.}"  Lifting of VMO maps
\endroster
\smallskip
Appendix A.  Some useful estimates on BMO, et al.
\smallskip
Appendix B.  John-Nirenberg inequality on manifolds, et al.

\medskip
We wish to express our thanks to a number of colleagues for
interesting discussions and encouragement:  L.~Boutet de Monvel,
F.~Browder, S.~Chanillo, G. David, H.~Furstenberg, I.~M.~Gelfand, A.~Granas,
Y.~Meyer and P.~Mironescu, with special thanks to P.~Jones.
\bigskip
\noindent
{\bf Section I.1.  BMO and VMO}
\medskip
Let $X$ be a smooth $n$-dimensional compact manifold without boundary.
In this section we recall the notion of BMO and VMO functions and maps
defined on $X$ and we state some of their properties.  There is much
literature on BMO, but mainly defined in Euclidean space; e.g., E.~Stein~[1] where many references may be found.  People have worked with BMO
on some manifolds, but the subject is mainly folklore to people in the
field.

\definition{Definition of BMO for real functions on $X$}  
\medskip
We first put a smooth Riemannian metric on $X$.  (Later we shall show
that the notion of BMO is independent of the choice of metric.)
Consider a real function $f$ in $L^1(X)$, using the measure associated
to the metric.  Set
$$
\|f\|_{BMO}=\sup\Sb \varepsilon < r_0\\x\in X\endSb
\Mint_{B_\varepsilon(x)} |f(y) - \bar f_\varepsilon(x)|d\sigma(y), \tag
1
$$
where $r_0 = r_0(X)$, the injectivity radius of $X$ (see
e.g., M.~P.~do~Carmo~[1]), $\sigma$ is the element of volume on $X$, $B_\varepsilon (x)$
denotes the geodesic ball in $X$ of radius $\varepsilon < r_0$,
centered at $x$, and
$$
\bar f_\varepsilon(x) = \Mint_{B_\varepsilon(x)} f(z)d\sigma(z).
$$

As usual, $\dsize\Mint_A f = \dfrac{1}{|A|} \int_A f$ denotes the average of
$f$ on $A$.   BMO$(X,\Bbb R)$---often denoted by BMO---consists of those functions with
$\|f\|_{\text{BMO}} < \infty$.  For these, (1) defines a norm on BMO---modulo
constants, (see E.~Stein [1])---and BMO is complete under this norm.
\enddefinition
\medskip
Clearly
$$
\align
\Mint_{B_\vare (x)} \vert f(y) - \bar f_\vare(x)| d\sigma(y)&\leq
\Mint_{B_\vare(x)} \Mint_{B_\vare (x)} |f(y) - f(z) | d\sigma(z)
d\sigma(y)\\
   &= \Mint_{B_\vare(x)} \Mint_{B_\vare (x)} |f(y) - \bar f_\vare (x)
+ \bar f_\vare(x) - f(z)|d\sigma(z) d\sigma(y)\\
   &\leq 2 \Mint_{B_\vare (x)} |f(y) - \bar f_\vare(x)|d\sigma(y).
\endalign
$$
Consequently, the following is an equivalent norm on BMO:
$$
\|f\|_\star = \operatornamewithlimits{sup}\Sb \vare < r_0\\ x\in
X\endSb \Mint_{B_\vare(x)} \Mint_{B_\vare(x)} |f(y) - f(z)| d\sigma(y)
d\sigma(z);\tag{$1^{'}$}
$$
in fact,
$$
\|f\|_{BMO}\leq \|f\|_{{}_\star} \leq 2\|f\|_{BMO}.
\tag{$1^{''}$}
$$
\medskip

A first simple but useful property is

\proclaim{Lemma 1}  There exists a constant $C$, depending on $X$ (and
the metric) such that for every $ f\in BMO$,
$$
\|f\|_{L^1}\leq C\|f\|_{BMO} + |\int_X f|.
$$
\endproclaim
This is proved in the Appendix;  see Lemma A.1.

\remark{\bf Remark 1}  If we replace $r_0$ by any positive
$r_1 < r_0$ we get a new norm $\|f\|_1$.  The two
norms are equivalent.
\medskip
Indeed, $\|f\|_1\leq \|f\|_{BMO} =: \|f\|_0$.  Conversely, if
$r_1 \leq \varepsilon < r_0$ then
$$
\align
\Mint_{B_\varepsilon(x)} |f - \bar f_\varepsilon (x)|&\leq
2\Mint_{B_\varepsilon(x)} |f|\\
    &\leq \frac{2}{|B_{r_1}(x)|} \int_X |f|\leq C \int_X
|f|.\endalign
$$
We now use Lemma 1.  Since we may suppose $\int_X f = 0$, we obtain the desired conclusion
$$
\|f\|_1 \leq C \|f\|_0.
$$
\medskip
If one changes the Riemannian metric on $X$ one obtains an equivalent
BMO norm.  More generally, if $X_1$ and $X_2$ are two smooth compact
Riemannian manifolds of dimension $n$, without boundary, and $\varphi:
X_1\rightarrow X_2$ is a $C^1$ diffeomorphism, then $f\in\text{BMO}(X_2)$
implies that $f\circ\varphi\in\text{BMO}(X_1)$ and
$$
\|f\circ\varphi\|_{BMO(X_1)}\leq C \|f\|_{BMO(X_2)}
$$
(see Lemma A.10 in the Appendix; a more general form, where $\varphi$
is only quasi-conformal and $X_1 = X_2 = \Bbb R^n$, was proved by
H.~M.~Reimann~[1]).
\medskip

If $\Omega$ is an open subset of $X$, we set 
$$
\|f\|_{BMO(\Omega)} =\text{
the sup in (1) taken over all balls $B_\varepsilon(x)$ in $\Omega$,
with $\varepsilon < r_0$}.
$$
\medskip
An $L^1$ map $u : X\rightarrow\Bbb R^N$ belongs to BMO$(X,\Bbb R^N)$
provided each component of $u$ is in BMO.  As norm, we use the
definition (1), except that the absolute value refers to the Euclidean
norm in $\Bbb R^N$.
\endremark
\definition{Definition of BMO maps into a manifold}
\medskip
  Let $Y$ be a
compact manifold without boundary which we always take to be smoothly
embedded in some $\Bbb R^N$.  We say that a map $u$ belongs to
BMO$(X,Y)$, if $u\in\text{BMO}(X,\Bbb R^N)$ and $u(x)\in Y$ a.e.
\enddefinition

\medskip
\proclaim{Claim}  The notion of BMO$(X,Y)$ is independent of the
metric on $X$ and of the embedding of $Y$.
\endproclaim

  That it does not depend on the metric in $X$ follows from a previous consideration.  To verify
its independence of the embedding of $Y$, we use the following


\proclaim{Lemma 2}  Let $F$ be a Lipschitz map from $\Bbb R^N$ into
$\Bbb R^D$ and let $u\in BMO(X,\Bbb R^N)$.  Then $F \circ u$ is in
$BMO(X,\Bbb R^D)$ and
$$
\|F\circ u\|_{BMO}\leq 2\|F\|_{Lip} \|u\|_{BMO}.
$$
\endproclaim

\noindent
This follows immediately from $(1^{''})$; see also the more general Lemma A.2.
\medskip

\demo{Proof of Claim}  Suppose $\varphi_1$ and $\varphi_2$ are smooth
embeddings of $Y$ into $\Bbb R^N$ and $\Bbb R^D$.  Then $\eta =
\varphi_2\circ\varphi_1^{-1}$ is a smooth map of $\varphi_1(Y)$ onto
$\varphi_2(Y)$.  Let $F$ be a Lipschitz continuous extension of
$\eta$ as a map from $\Bbb R^N$ to $\Bbb R^D$.  Using Lemma 2 we see
that the BMO norm using $\varphi_2$, is bounded by a constant times
the BMO norm using $\varphi_1$.  Thus the norms are equivalent.
\medskip
Having chosen a Riemannian metric on $X$, and a smooth embedding of
$Y$ in some $\Bbb R^N$, BMO$(X,Y)$ is equipped with a metric
$$
d(u,v) = \|u - v\|_{BMO(X,\Bbb R^N)}.
$$
A different choice of the Riemannian metric on $X$ and of the
embedding of $Y$ in some $\Bbb R^D$ yields an equivalent metric.
Thus it makes sense to say that a sequence of maps $u_j: X\rightarrow
Y$ converges to $u$ in BMO$(X,Y)$, independently of the choice of
metric on $X$ and embedding of $Y$.
\enddemo

\medskip
In view of $(1^{'})$ there is a natural notion of BMO of a map from $X$
into any metric space $Y$, namely
$$
\|u\|_{{}_\star} = \operatornamewithlimits{sup}\Sb \vare < r_0\\x\in
X\endSb \Mint_{B_\vare(x)} \Mint_{B_\vare(x)} \text{dist}(u(y), u(z))
d\sigma(y) d\sigma(z).
$$
\medskip
It is clear that $C^0(X) \subset\text{BMO}(X)$; other examples of BMO
functions will be given later (see Section I.2).  In particular, the examples show that
smooth functions are not dense in BMO.  It is therefore natural to
introduce the following definition (see D.~Sarason~[1]).
\medskip
\definition{Definition of VMO functions and maps}
\medskip
  VMO is the
completion of smooth functions in the BMO norm, namely, a real
function $f$ on $X$ belongs to VMO$(X,\Bbb R)$ if $f\in\text{BMO}(X,\Bbb
R)$, and there is a sequence $(f_j)$ of smooth functions such that
$\|f_j - f\|_{BMO} \rightarrow 0$.  In view of Lemma 1 we may also
suppose that $\|f_j - f\|_{L^1} \rightarrow 0$.
\enddefinition

\medskip
VMO is equiped with the BMO norm.
\medskip
\remark{\bf Remark 2}  In the definition of VMO$(X,\Bbb R)$ one could
use continuous $f_j$ instead of smooth $f_j$.  This follows easily from two
facts:
\roster
\item"{(i)}"  $C^\infty(X,\Bbb R)$ is dense in $C^0(X,\Bbb R)$,

\item"{(ii)}"  $C^0(X,\Bbb R)\subset\text{BMO}(X,\Bbb R)$ and
$$\|f\|_{BMO} \leq 2|f|_{C^0}.
$$
\endroster
\endremark

\medskip
Similarly, one defines VMO$(X,\Bbb R^N)$.  Furthermore,  a map $u : X\rightarrow\Bbb
R^N$ belongs to VMO$(X,Y)$ if $u\in \text{VMO}(X,\Bbb R^N)$ and $u(x)\in Y$
a.e. $x\in X$.
\medskip
As above, VMO$(X,Y)$ is independent of the Riemannian metric on $X$.
The fact that it is also independent of the choice of embedding of $Y$
in some Euclidean space follows from a variant of Lemma 2.  In view
of this fact, unless we say otherwise, from now on we fix a Riemannian
metric on $X$, and an embedding of $Y$ in $\Bbb R^N$.  The variant of
Lemma 2 is:

\proclaim{Lemma $2'$}  Let $F$ be a Lipschitz map of $\Bbb R^N$ into
$\Bbb R^D$ and let $u\in \text{VMO}(X,\Bbb R^N)$.  Then $F \circ u\in
\text{VMO}(X,\Bbb R^D)$.
\endproclaim

For proof, see the more general Lemma A.7.  A natural way to prove the lemma would be
to take a sequence of smooth maps $u_j$ tending to $u$ in BMO$\,\cap\, L^1$ and to
show that $F\circ u_j\rightarrow F\circ u$ in BMO.  Indeed,
this method of proof works, but it is more delicate than it would
appear.  In fact $u\mapsto F\circ u$ is not continuous in BMO$\,\cap\, L^1$; it
is however continuous at points $u$ in VMO.  See Lemma A.8 and Remark A.1.  The
proof of Lemma $2'$ that we present is different; it relies on Sarason's
characterization of VMO:
\medskip
\proclaim{Lemma 3 (D.~Sarason [1])}  $u\in \text{VMO}(X,\Bbb R^N)$ iff $u\in
\text{BMO}(X,\Bbb R^N)$ and
$$
\lim_{\varepsilon\rightarrow 0} \Mint_{B_\varepsilon(x)} |u -\bar
u_\varepsilon (x)| = 0\text{  uniformly in $x\in X$}. \tag 2
$$
\endproclaim

Again, in view of $(1')$, property (2) is equivalent to
$$
\lim_{\vare\rightarrow 0} \Mint_{B_\vare(x)} \Mint_{B_\vare(x)} |u(y) -
u(z)| = 0\quad\text{uniformly in $x\in X$}. \tag{$2'$}
$$
\medskip
The implication:  VMO$\,\Rightarrow\, (2)$ is easy.  Indeed, given $\delta
> 0$, there is a\newline $v\in C^0 (X,\Bbb R^N)$ such that
$$
\|u - v\|_{BMO} < \delta/2.
$$
\medskip
Write
$$
\align
\Mint_{B_\varepsilon(x)} |u - \bar u_\varepsilon(x)|&\leq \Mint_{B_\varepsilon(x)} |(u - v)
- (\bar u_\varepsilon(x) - \bar v_\varepsilon(x))| + \Mint_{B_\varepsilon(x)} |v - \bar
v_\varepsilon(x)|\\
    &\leq \frac\delta2 + \frac\delta2\endalign
$$
provided $\varepsilon$ is sufficiently small (depending on $v$).
Property (2) follows easily.
\medskip
The converse implication is more delicate.  It is in fact a
consequence of a more general form of Lemma 3, Lemma $3'$ below.  First
some notation:
\medskip

For $u\in \text{BMO}(X,\Bbb  R^N)$ and $0 < a < r_0(X)$, set
$$
\align
M_a = M_a(u) &= \sup\Sb \varepsilon\leq a\\x\in X\endSb
\Mint_{B_\varepsilon(x)} |u - \bar u_\varepsilon(x)|\leq \|u\|_{BMO}\\
M_0 &= M_0 (u) = \lim_{a\searrow 0} M_a(u).
\endalign
$$
\medskip

\proclaim{Lemma $3'$ (D.~Sarason [1])}  There is a constant $A$
depending only on $X$  (and choice of Riemannian metric), such that, if $u\in BMO(X,\Bbb R^N)$, then
$$
M_0(u)\leq \text{ dist}  (u,VMO(X,\Bbb R^N))\leq A\, M_0(u). \tag 3
$$
Here distance is measured using the BMO norm.  More precisely,
$$
 \|u - \bar
u_\varepsilon\|_{BMO} \leq A\; M_\vare(u)\quad\forall \vare <
r_0,\quad\forall u\in\text{ BMO}(X,\Bbb R^N). \tag 4
$$
\endproclaim

This is Lemma A.5.
\medskip
Note that (2) says that $M_0(u) = 0$; hence (3) yields the
implication $\Leftarrow$ in Lemma 3.  In addition we have
\medskip
\proclaim{Corollary 1}  For any $u\in VMO(X,\Bbb R^N)$, 
$$
\|\bar u_\varepsilon - u\|_{BMO} \longrightarrow 0,\; \bar
u_\varepsilon\longrightarrow u\text{  in $L^1$ as
$\varepsilon\rightarrow 0$}. \tag 5
$$
\endproclaim

The first assertion follows from (3) and (4); the last assertion is
well known.  Another consequence of (4) is:
\medskip
\proclaim{Corollary 2}  There is a constant $\widetilde A$ depending only
on $X$ (and choice of metric), such that
$$
\|\bar u_\varepsilon\|_{BMO} \leq \widetilde A \|u\|_{BMO}\; \;
\forall\vare < r_0,\quad\forall u\in BMO(X,\Bbb R^N). \tag 6
$$
\endproclaim

If $u\in C^0(X,Y)$, the map $x\mapsto\bar u_\varepsilon(x)$ maps $X$
into $\Bbb R^N$, but not into $Y$.  However, for $\varepsilon$ small
it lies close to $Y$.  (This is clear since $\bar
u_\varepsilon\rightarrow u$ uniformly.)  Surprisingly the same is true
for $u\in\text{VMO}(X,Y)$---even though $u$ need not be continuous.  Indeed
we have
$$
\text{dist} (\bar u_\varepsilon(x), Y)\leq \Mint_{B_\varepsilon (x)}
|u(y) - \bar u_\varepsilon(x)|\leq M_\vare (u). \tag 7
$$
and $M_\vare(u)$ tends to $0$, by (2).
\medskip

\remark{\bf Remark 3}  Note that (7) holds if $Y$ is any closed set in
$\Bbb R^N$---not just a smooth manifold.
\endremark
\smallskip
This fact is at the heart of the paper, because it allows us to
project $\bar u_\varepsilon$, for $\varepsilon$ small, onto its
nearest point in $Y$.
\medskip
As pointed out in the introduction, the role of VMO maps in
conjunction with (7) was first observed by L.~Boutet de Monvel and
O.~Gabber.
\medskip
Denote by $P$ the projection operator in $\Bbb R^N$ to the nearest point on
$Y$ (this is well defined in a tubular neighbourhood of $Y$).  For
$\vare$ less than some $\vare_0$,
$$
u_\varepsilon(x) = P\bar u_\varepsilon(x)\quad\text{is well defined}. \tag 8
$$
\medskip
Here is one more
\medskip
\proclaim{Corollary 3}  There is a constant $C$ depending only on $X$
such that for any\newline $a < r_0, \vare\leq \vare_0$,
$$
M_a(u_\vare)\leq C\left(M_a(u) + M_\vare (u)\right)\quad\forall u\in VMO(X,Y).
$$
\endproclaim

\demo{Proof}  We have
$$
M_a(u_\vare)\leq C M_a(\bar u_\vare)
$$
since $P$ is Lipschitz continuous.  On the other hand,
$$
\align
M_a(\bar u_\vare)&\leq M_a(\bar u_\vare - u) + M_a(u)\\
     &\leq A M_\vare (u) + M_a(u)
\endalign
$$
by (4).  $\quad\hfill\qed$
\medskip
 Presumably, in the assertion of the corollary, the term $C M_\vare(u)$
could be omitted.  This is clear in Euclidean space.   \qed
\enddemo

\medskip
We will be considering families of maps in VMO$(X,Y)$.  Let ${\scr F}\subset
\text{VMO}(X,Y)$ be a collection of maps. For each individual map
$u\in {\scr F}$ we
have
$$
\lim_{\vare\rightarrow 0} \text{ dist}(\bar u_\varepsilon(x), Y) = 0
\tag 9
$$
uniformly in $x\in X$, but this does not hold uniformly with respect
to the map $u$.  However, if $\scr F$ is a compact subset of
VMO$(X,Y)$, then (9) holds uniformly in $x\in X$ and $u\in {\scr F}$. 
This is an immediate consequence of (7) and the following:

\proclaim{Lemma 4  (Characterization of compact sets in VMO)}  Assume $\scr F$ is a compact subset of VMO$(X,\Bbb
R^N)$.  Then 
$$
\underset \varepsilon\rightarrow 0\to\lim  M_\vare (u) =
0\quad\text{holds uniformly in $u\in \scr F$}.\tag 10
$$
Conversely, if $\scr F$ is any collection of maps in VMO$(X,\Bbb R^N)$
such that (10) holds, then $\scr F$ is contained in a compact subset
of VMO$(X,\Bbb R^N)$.
\endproclaim

Proof of the first assertion (the second, which is more delicate, is
proved in the Appendix; see Lemma A.16):
\medskip

Given any $\delta > 0$ we may cover $\scr F$ by a finite number of balls
$$
{\scr F} \subset \underset i=1\to{\overset k\to\bigcup} B_{\delta/2} (v_i)
$$
(where $B$ refers to balls for the BMO norm).
\medskip
For each $i$ there is some $\vare_i > 0$ such that $\forall
\vare <  \vare_i$,
$$
M_\vare(v_i) < \delta/2. \tag 11
$$
Set $\vare_0 = \underset 1\leq i\leq k\to\min \vare_i$.  Given $v\in
{\scr F}$, there is
some $i$ such that
$$
\|v - v_i\|_{BMO} < \delta/2.
$$
Then $\forall \vare < \vare_0$
$$
\align
M_\vare(v)\leq M_\vare(v - v_i) + M_\vare(v_i)&\leq \|v -
v_i\|_{\text{BMO}} + M_\vare(v_i)\\
    &\leq (\delta/2) + (\delta/2)\quad\text{by (11)}.  \qed
\endalign
$$
%$\quad\hfill\qed$
\medskip
Some further consequences of Lemma $3'$ and Corollary 1 are:

\proclaim{Corollary 4}  Given $u\in\text{VMO}(X,Y)$,
$$
\|u_\vare - u\|_{BMO} \longrightarrow 0,\; u_\vare\longrightarrow
u\text{  a.e. as $\vare\rightarrow 0$}.
$$
\endproclaim

\demo{Proof}  We have
$$
\align
\|u - P\bar u_\vare \|_{BMO} &\leq \|u - \bar u_\vare\|_{BMO} + \|\bar u_\vare
- P \bar u_\vare\|_{BMO}\\
    &\leq \|u - \bar u_\vare \|_{BMO} + 2 |\bar u_\vare - P \bar
u_\vare |_{C^0}\\
    &\leq \|u - \bar u_\vare \|_{BMO} + 2 \sup_x \text{dist}(\bar
u_\vare(x), Y)\\
    &\longrightarrow 0\quad\text{as $\vare\rightarrow 0$},\endalign
$$ 
by (5) and (9).  \hfill\qed
\enddemo
\medskip
\proclaim{Corollary 5}  Given $u\in \text{ VMO}(X,Y)$, there exists a
sequence $u_j\in C^\infty(X,Y)$ such that $u_j\rightarrow u$ in BMO
and a.e.
\endproclaim

\demo{Proof}  Since $C^\infty(X,Y)$ is dense in $C^0(X,Y)$ the result
follows with the aid of Corollary 4.
\enddemo

\bigskip
\noindent
{\bf Section I.2.  Some examples of BMO and VMO functions}
\medskip
As we have said in Remark 2, continuous functions $f$ on $X$ belong to
VMO and
$$
\|f\|_{BMO} \leq 2 |f|_{C^0}. \tag 12
$$
\medskip
A less obvious class consists of functions in Sobolev spaces
corresponding to limiting cases---where the embedding is into $L^p$
for every $p < \infty$, but not into $L^\infty$.
\medskip
\example{Example 1}  $W^{1,n}(X)\subset\text{VMO}(X)$ with continuous
embedding.
\endexample

\demo{Proof}  We first prove that $W^{1,n}(X) \subset\text{BMO}(X)$, with
continuous embedding.  By Poincar\'e's inequality---which even holds
on a manifold---we have, for $\vare < r_0$,
$$
\int_{B_\vare(x)} |u - \bar u_\vare(x)|\leq C\vare \int_{B_\vare (x)}
|\nabla u|.
$$
Hence
$$
\int_{B_\vare(x)} |u - \bar u_\vare (x)|\leq C
\vare^n\left(\int_{B_\vare(x)} |\nabla u|^n\right)^{1/n}
$$
and thus
$$
\Mint_{B_\vare(x)} |u - \bar u_\vare(x)|\leq C\left(\int_{B_\vare(x)}
|\nabla u|^n\right)^{1/n}, \tag 13
$$
which implies the desired conclusion.  The embedding in VMO now follows easily from (13) and Lemma 3.
\enddemo

\medskip
More generally, we have:

\example{Example 2}  $W^{s,p}(X)\subset \text{VMO}(X)$ in the limiting case
of the Sobolev embedding, i.e., $s p = n,\; 0 < s < n$ ($s$ may or may
not be an integer).\footnote{For the definition and general properties
of fractional Sobolev spaces, see e.g., R.~A.~Adams~[1], Chapter VII.}
\endexample

\demo{Proof}  We distinguish two cases:\enddemo
\medskip
\noindent
{\bf Case 1:}  $s\geq 1$.  Then, by the Sobolev embedding, (see e.g.,
R.~A.~Adams~[1], Theorem 7.57),
$$
W^{s,p}(X)\subset W^{1,n}(X)
$$
and the conclusion follows from Example 1.
\medskip
\noindent
{\bf Case 2:}  $0 < s < 1$.  Recall that
$$
W^{s,p}(X) = \left\{u;\;\int_X\; \int_X \frac{|u(x) -
u(y)|^p}{|\text{dist}(x,y)|^{sp + n}} < \infty\right\}
$$
and in our case $sp = n$, so that
$$
W^{s,p}(X) = \left\{u;\;\; \int_X\; \int_X\;\frac{|u(x) -
u(y)|^p}{|\text{dist}(x,y)|^{2n}} < \infty\right\}.
$$
As before, we first prove that
$$
W^{s,p}\subset\text{BMO}\quad\text{with continuous injection}.\tag 14
$$
\medskip

To prove (14), we compute for $\vare\leq r_0$,
$$
\align
 \Mint_{B_\vare(x)}\Mint_{B_\vare(x)} |u(y) - u(z)|
    &\leq \Mint_{B_\vare(x)} \Mint_{B_\vare(x)} \frac{|u(y) - u(z)|}
{|\text{dist}(y, z)|^{\frac{2n}{p}}}\; C \vare^{2n/p}\\
   &\leq C  \int_{B_\vare (x)} \int_{B_\vare (x)} \frac{|u(y) -
u(z)|^p}{|\text{dist} (y,z)|^{2n}},
\endalign
$$
which yields (14).  It then follows, as above, that $W^{s,p}\subset\text{VMO}$.
\medskip
Next we present some specific functions in BMO or VMO.  The functions
are defined on some bounded domain $\Omega$ in $\Bbb R^n$ containing
the origin; $\Omega$ may be considered as an open subset of a compact
manifold $X$.

\medskip
\example{Example 3}  The function $\log|x|$ belongs to BMO$(\Omega)$,
for any $n$ (see F.~John and L.~Nirenberg [1] or E.~Stein[1], Chapter IV,
Section I.1.2).  However, $\log |x|$ is not in VMO.  To see this,
observe that
$$
I = \Mint_{B_\vare (0)} \bigg\vert\log |y| - \Mint_{B_\vare (0)} \log|x|\bigg\vert =
\Mint_{B_1(0)}\bigg\vert \log|y| - \Mint_{B_1(0)} \log|x|\bigg\vert,
$$
and thus $I$ does not tend to zero as $\vare \rightarrow 0$.
\endexample

\medskip
\example{Example 4}  The function $f(x) = \log
\big\vert\log|x|\big\vert$ is in VMO$(\Omega)$.  An easy way to verify
this is to observe that $f$ belongs to $W^{1,n}(\Omega)$, when $n\geq
2$, because
$$
|\nabla f|\leq \frac{C}{|x|\big\vert\log|x|\big\vert}.
$$
For $n = 1$,  $f$ is the trace on $\Bbb R$ of the function
$\log\big\vert\log|x|\big\vert$ in $\Bbb R^2$---which belongs to
$W^{1,2} = H^1$ in any bounded region on $\Bbb R^2$.  Consequently its
trace belongs to $H^{1/2}(\Omega) = W^{\frac12,2} (\Omega)$.  By
Example 2, this function is then in VMO.
\endexample

\medskip
Applying Lemma $2'$ we see that the functions $\exp(i \log|\log|x||)$ or
$\sin(\log|\log |x||)$ also belong to VMO.
\medskip
\example{Example 5}  The function $f(x) = \big\vert\log|x|\big\vert^\alpha$, for $0 <
\alpha < 1$ is in VMO$(\Omega)$.
\endexample

\medskip
\demo{Proof}  Observe first that $f\in W^{1,n}_{\text{loc}}$ in case $n
> 1/(1-\alpha)$.  If $n \leq 1/(1-\alpha)$, fix an integer $m >
1/(1-\alpha)$.  Then the function $f(x)$ belongs to
$W^{1,m}_{\text{loc}}(\Bbb R^m)$.  Consequently its trace on $\Bbb
R^{m-1}$ belongs to $W^{1-\frac1m, m}_{\text{loc}} (\Bbb R^{m-1})$.
Continuing to take traces, we find that $f\in
W^{\frac{n}{m},m}_{\text{loc}}(\Bbb R^n)$.  Again by Example 2, $f\in\text{
VMO}(\Omega)$.
\enddemo

We conclude this section with a particular, but useful sequence of VMO functions in
$\Bbb R$.


\example{Example 6}  In $\Bbb R$, consider the sequence
$$
f_j(x) = \cases  \hphantom{- {}}1&\quad\text{if $|x|\leq 1/j^2$}\\
      -1 - \dfrac{\log|x|}{\log j}&\quad\text{if $\dfrac{1}{j^2}\leq
|x|\leq \dfrac1j$}\\
  \hphantom{- {}}0&\quad\text{if $|x|\geq \dfrac1j$}.\endcases
$$
Then
$$
\|f_j\|_{H^{1/2}} \longrightarrow 0\quad\text{as $j\rightarrow\infty$}.
$$
In particular, $\| f_j\|_{BMO}\rightarrow 0$.
\endexample

\demo{Proof}  Consider the sequence for $x\in \Bbb R^2$.  One easily
verifies that
$$
\|f_j\|_{H^1(\Bbb R^2)}\longrightarrow 0.
$$
The desired result then follows by taking trace.
\enddemo

The reader may prefer a direct argument showing that
$\|f_j\|_{\text{BMO}}\rightarrow 0$, an argument which does not rely on
trace.  Here is one:  $f_j$ may be written as
$$
f_j(x) = \min\left\{1,\max\{0, -1 - \frac{\log|x|}{\log j}\}\right\}.
$$
\medskip
Since $F(t) = \max\{0,t\}$ is Lipschitz with Lipschitz constant $1$,
it follows from Lemma 2 that
$$
\align
\bigg\Vert\max\{0, - 1 - \frac{\log|x|}{\log j}\}\bigg\Vert_{BMO} &\leq \frac{2}{\log
j}  \big\Vert\log|x|\big\Vert_{BMO}\\
   &\leq\frac{C}{\log j}.
\endalign
$$
Similarly,
$$
\|f_j\|_{BMO} \leq \frac{C}{\log j}.
$$

This argument shows that for any $n, f_j$ as defined above, in $\Bbb
R^n$, satisfies
$$
\|f_j\|_{BMO} \leq \frac{C}{\log j}.\quad\hfill\qed
$$

\remark{\bf Remark 4}  In Example 6 it seems natural to replace $f_j$ by a
simpler sequence of functions, in which $f_j$ is linear on
$(\frac{1}{j^2}, \frac1j)$.  However, the reader may verify that, then
$\|f_j\|_{\text{BMO}}$ does not tend to zero.  Our sequence $(f_j)$ is the kind of sequence which is commonly used to
prove that in two dimensions, a point has zero capacity.
\endremark

\bigskip
\noindent
{\bf Section I.3.  Degree for VMO maps}
\medskip
This is our main topic.  We consider VMO maps from $X$ to $Y$.  Here,
$X$ and $Y$ are smooth $n$-dimensional compact manifolds without
boundaries---{\it which we now assume to be oriented manifolds}.  We shall
define a degree for such maps and show that it has some of the usual
properties of a degree.
\medskip
We first put a Riemannian metric on $X$ and consider $Y$ as smoothly embedded in
some $\Bbb R^N$.  Recall that for a $C^1$ mapping $u : X\rightarrow Y$,
$$
\deg u = \frac{1}{\text{vol } Y} \int_X \det J_u(x),
$$
where $J_u(x)$ is the Jacobian at $x$ of the map $u$ computed in terms of
geodesic normal coordinates at $x$ and at $u(x)$.  This integral
clearly makes sense for a map in $W^{1,n}(X,Y)$.  We shall prove later
that for such a map, this expression is indeed an integer. This fact suggests
that degree theory, which extends to continuous maps, 
extends also to maps in $W^{1,n}(X,Y)$.  In an attempt to find a general class of maps
including both of these, the natural candidate seems to be the class
VMO.
\medskip
We now proceed to define the degree for a VMO map $u: X\rightarrow Y$.
\medskip
\definition{Definition}  Let $u\in \text{VMO}(X,Y)$.  For $0 < \vare$ small,
recall
$$
\bar u_\vare(x) = \Mint_{B_\vare(x)} u\quad\text{and $u_\vare(x) =
P\bar u_\vare(x)$}.
$$
Define
$$
\deg(u,X,Y) = \deg(u_\vare, X,Y)
$$
for $\vare$ small.  We claim that this is independent of $\vare$.
Indeed for $\vare$ small, since $u_\vare$ is continuous,
$\deg(u_\vare, X,Y)$ is defined.  Furthermore, using the deformation
$u_{t\vare + (1-t)\vare'}$, for $\vare, \vare'$ small, $0\leq t\leq
1$, we see that $\deg u_\vare = \deg u_{\vare '}$.
\enddefinition

In principle, $\deg(u,X,Y)$ depends on the choices of metric on $X$
and of the embedding of $Y$.  We shall see soon that it is independent of
these choices.  We first establish an important fact about this degree,
namely, that it is stable under perturbation in VMO:

\proclaim{Theorem 1}  Let $u\in \text{VMO}(X,Y)$.  Then there exists
$\delta > 0$ depending on $u$, such that if $v\in \text{VMO}(X,Y)$ and
$$
d(u,v) < \delta,
$$
then
$$
\deg v = \deg u. \tag 15
$$
\endproclaim

Recall that $d(u,v)$ refers to the metric induced by the norm of
BMO$(X,\Bbb R^N)$ once an embedding of $Y$ has been chosen.  Easy
consequences of Theorem 1 are

\proclaim{Corollary 6}  Let $H_t(\cdot)$ be a one-parameter family of
VMO maps from $X$ to $Y$, depending continuously in the BMO topology,
on the parameter $t$.  Then
$$
\deg H_t(\cdot)\quad\text{is independent of $t$}.
$$
\endproclaim

\proclaim{Corollary 7}  $\deg(u,X,Y)$ is independent of the choices of
Riemannian metric on $X$ and of the embedding of $Y$.
\endproclaim

\demo{Proof}  Suppose we have another metric on $X$ and a smooth
embedding of $Y$ in some $\Bbb R^D$.  We then obtain another family
$\tilde u_\vare$ mapping $X\rightarrow Y$.  There is a corresponding
degree $\tilde d$.   By Corollary 4, $\tilde u_\vare\rightarrow u$ in
BMO$(X,Y)$ as $\vare\rightarrow 0$.  Applying Theorem 1 we obtain the
desired conclusion.
\enddemo

In the proof of Theorem 1 we shall use

\proclaim{Lemma 5}  Consider $u\in \text{VMO}(X,Y)$.  Suppose that for some
constant vector $\xi\in\Bbb R^N$,
$$
u(x) + \xi \in Y\quad\text{a.e.}
$$
Then
$$
\deg(u + \xi) = \deg u.
$$
\endproclaim

\demo{Proof of Lemma 5}  We need only consider $\xi\not=0$, and, in
fact, in this case we will also prove that both degrees are zero.  We
may suppose $\xi = (\xi_1, 0, 0)$, $\xi_1 > 0$ and also that $0\in Y$, and that $y_1 \leq 0\;\;\forall y\in Y$.  Then
$$
u_1(x)\leq - \xi_1,\quad\text{a.e. on $X$}.
$$
Consequently for $\vare < r_0$, the first component of $\bar u_\vare(x)
\leq - \xi_1$.  It follows that for $\vare$ small,  the first component
of $P\bar u_\vare(x) = u_\vare (x)$ is less than $-\xi_1 /2$.  This
implies that the image of $u_\vare$ does not cover $Y$, and so $\deg
u_\vare = \deg u = 0$.

Reversing the roles of $u$ and $u + \xi$, we conclude that $\deg(u + \xi) = 0$.  \qed

\enddemo

\demo{Proof of Theorem 1}  Suppose the assertion of the theorem is
false.  Then there exists a sequence $(v_j)$ such that
$$
\|v_j - u\|_{BMO} \longrightarrow 0\quad\text{and $|\deg v_j - \deg
u|\geq 1$}.
$$
Since the $(v_j)$ are compact in VMO, we know by Lemma 4 and (7) that
$$
\text{dist}(\bar v_{j,\vare} (x), Y)\rightarrow 0\quad\text{as
$\vare\rightarrow 0$}
$$
uniformly in $j$ and in $x\in X$.  Hence there exists $\vare_0 > 0$, such
that
$$
v_{j,\vare} = P \bar v_{j,\vare}
$$
is well defined for all $j$ and all $\vare\leq \vare_0$.  By
definition,
$$
\deg v_j = \deg v_{j,\vare} \quad\forall j, \quad\forall \vare\leq
\vare_0. \tag 16
$$
Fix some $\vare < \vare_0$.  Set
$$
\xi_j = \Mint_X(v_j - u).
$$
By Lemma 1,
$$
\int_X |v_j - u - \xi_j|\leq C \|v_j - u\|_{BMO} \rightarrow 0.
$$
\medskip
For a subsequence, we may assume $\xi_j$ converges to some vector $\xi$, since
$Y$ is bounded.  Hence $v_j\rightarrow u + \xi$ in $L^1$ and a.e.
Therefore $u + \xi\in Y$ a.e. and also
$$
\bar v_{j,\vare}\rightarrow \bar u_\vare + \xi \quad\text{uniformly on
$X$ as $j\rightarrow \infty$},
$$
---recall that $\vare$ is fixed.  Hence $v_{j, \vare} \rightarrow (u +
\xi)_\vare$ as $j\rightarrow\infty$ uniformly on $X$.  For $j$ large
it follows that $\deg v_{j, \vare} = \deg (u + \xi)_\vare$ and
consequently
$$
\deg v_j = \deg (u + \xi)
$$
by (16) and by definition of $\deg(u + \xi)$.  By Lemma 5 the proof is
complete.
\enddemo

\remark{\bf Remark 5}  We have defined $\deg u$ with the aid of particular
approximations of $u$ by continuous functions tending to $u$ in BMO,
namely the $u_\varepsilon$ (see Corollary 4), and we set $\deg u = \deg
u_\vare$.  The preceding theorem shows that we could have used any
approximation by continuous maps, tending to $u$ in BMO.  Theorem 1 is
somewhat subtle for various reasons:
\medskip

1)  BMO convergence is weaker than uniform convergence but stronger than
any $L^p$, $p<\infty$ (modulo constants).  Degree is not preserved
under small perturbations in $L^p$, $p < \infty$.  For example, the
following maps $u_j$ of $S^1$ to $S^1$, have degree one, but their $L^p$
limit is a constant---and thus has degree zero:
$$
u_j(\theta) = e^{i\varphi_j(\theta)}
$$
where $\varphi_j(\theta) = 0$ on $(0, 2\pi - \frac1j)$ and $\varphi_j$
goes linearly from $0$ to $2\pi$ on $[2\pi - \frac1j, 2\pi]$.
\endremark

\medskip
2)  Recall that when working with the $C^0$ norm, there is a {\it uniform}
$\delta > 0$ such that
$$
|u - v|_{C^0} < \delta\Longrightarrow \deg(u,X,Y) = \deg(v,X,Y).
$$
For then, $u$ is easily deformed to $v$ via
$$
P(t v + (1 - t) u).
$$

Surprisingly, in Theorem 1, the $\delta$ really depends on $u$.  Here
is an example for $X = Y = S^1$ showing that if $u$ and $v$ are maps of
$S^1$ to $S^1$ which are close even in $H^{1/2}$, they need not have
the same degree.

\proclaim{Lemma 6}  Given $\vare > 0$ there are two smooth maps, $u,v$
of $S^1$ to $S^1$ with
$$
\|u - v\|_{H^{1/2}} < \vare\tag 17
$$
such that
$$
\deg u = 0\quad\text{and $\deg v = 1$}.\tag 18
$$
\endproclaim

\demo{Proof}  
\medskip
\noindent
{\bf Step 1.}  We first construct  $u,v\in C^0(S^1,S^1)$ with $u - v$
in $H^{1/2}$, satisfying (17) and (18).  Recall that there is a continuous function $\rho$ defined on $\Bbb R$
with support, in $[\pi - \delta, \pi + \delta], \rho > 0$ in $(\pi -
\delta, \pi + \delta)$, $\rho$ symmetric about $\pi$ nondecreasing on
$(\pi - \delta, \pi), \rho(\pi) = 2$, and such that
$$
\|\rho\|_{H^{1/2}} < \vare.
$$
Here $\delta$ depends on $\vare$ (see Example 6 in Section I.2).
\medskip
Using $\rho$, we construct $u$ and $v$ of the form
$$
u = e^{if},\quad v = e^{i(f+g)}
$$
on $[0,2\pi]$ such that
$$
f(0) = f(2\pi), \quad g(2\pi) - g(0) = 2\pi.
$$
Thus we will have $\deg u = 0, \deg v = 1$.
\enddemo

\medskip
We first define $g$ as a continuous nondecreasing function on
$(0,2\pi)$, with
$$
g(\theta) = \cases  0&\quad\text{on $[0,\pi - \delta]$}\\
  2\pi &\quad\text{on $[\pi + \delta, 2\pi]$},
\endcases
$$
and such that
$$
|e^{ig(\theta)} - 1| = \rho(\theta)\quad\text{on $[0,2\pi]$}.
$$
This defines $g$ in a unique manner.
\medskip
Next we define $f$ on $(\pi - \delta, \pi + \delta)$ as $f = -\text{arg}
(e^{ig} - 1)$.  Note that
$$
\align &f(\theta)\rightarrow - \frac{\pi}{2}\quad\text{as
$\theta\searrow (\pi - \delta)$},\\
  &f(\theta)\rightarrow - \frac{3 \pi}{2}\quad\text{as
$\theta\nearrow (\pi + \delta)$}.
\endalign
$$
We then extend $f$ to $[0,2\pi]$ continuously so that $f(0) =
f(2\pi)$.  Any such extension will do.
\medskip
The point is that 
$$
v - u = e^{if} (e^{ig} - 1)\equiv\rho.
$$
This is clear on $(\pi - \delta, \pi + \delta)$, and even
clearer outside.
\medskip
\noindent
{\bf Step 2.}  We may approximate $v$ in $C^0$ by smooth functions,
and may approximate $u - v$ by smooth functions in the $C^0\cap
H^{1/2}$ topology.  The sum of these approximates is an approximation
of $u$ in $C^0$.  \qed

\remark{\bf Remark 6}  By a slight modification we may even construct two
smooth maps $u,v$ of $S^1$ to $S^1$ with
$$
\|u - v\|_{H^{1/2}} < \vare
$$
such that
$$
\deg u = 0\quad\text{and $\deg v = k$}
$$
(for any given integer $k$ and any given $\vare > 0$).
\endremark

\remark{\bf Remark 7}  We have defined the degree for VMO maps of $X$ to
$Y$.  The degree can, in fact, also be defined for $u\in \text{BMO}(X,Y)$,
with $u$ ``close'' to VMO.  More precisely, there is a $\delta > 0$
such that if $u\in \text{BMO}(X,Y)$, and
$$
\text{dist$(u, VMO(X,Y)):= \underset v\in C^0(X,Y)\to\inf d(u,v) <
\delta$},
$$
then $u$ has a well defined degree.  The distance $d$, and hence the
number $\delta$, depend on a particular choice of Riemannian metric on
$X$ and on the embedding of $Y$.
\endremark

\bigskip
\noindent
{\bf Section I.4.  Some properties of degree}
\medskip
The setting is the same as in the previous section.  We consider VMO
maps from $X$ to $Y$ and show that standard properties of degree
carry over.  Here are some:
\medskip
\noindent
{\bf Property 1}.  If $\deg u\not=0$ then
$$
\text{ess} R(u) = Y.
$$

Here, $\text{ess}R(u)$, the essential range of $u$, has to be explained.
\medskip

The notion of the range of a measurable map is not well defined, since
the map may always be modified on a set of measure zero, keeping the
new map in the same equivalence class.  It is important to introduce a
notion of the range of $u$ which is independent of the choice of
representative in the class of equivalent maps.
\medskip

\definition{Definition $(\text{ess} R$)}  The essential range of a map $u$,
$\text{ess}R(u)$, is the smallest closed set $\Sigma$ in $Y$ such that
$$
u(x)\in\Sigma \text{  a.e.}
$$
\enddefinition

\proclaim{Claim}  This is well defined.
\endproclaim

\demo{Proof}  Let $(\Sigma_\alpha)_{\alpha\in\Lambda}$ be the family of
all closed sets $(\Sigma_\alpha)$ in $Y$, such that $\forall \alpha$,
$$
u(x) \in \Sigma_\alpha\text{  a.e.}
$$
Set $\Sigma = \underset \alpha\in\Lambda\to\bigcap \Sigma_\alpha$.  We
assert that
$$
u(x)\in\Sigma\text{  a.e.}
$$
This follows easily from the general fact that there is a {\it
countable} subset $J\subset\Lambda$ such that
$$
\Sigma = \underset \alpha\in J\to\bigcap \Sigma_\alpha.
$$
\medskip
To see this, let $O_\alpha = \Sigma^c_\alpha$\,; $O = \underset
\alpha\in\Lambda\to\bigcup O_\alpha = \Sigma^c$.  The open set $O$ may
be written as a countable union of increasing compact subsets, $K_i, i
= 1,2,\ldots\;$.  Each $K_i$ is covered by a finite number of the
$O_\alpha$.  Hence $O$ is the countable union of these.  $\Sigma$
is the intersection of their complements.  \qed
\medskip

The notion of essential range for a complex-valued measurable function
$f$ is commonly used in the theory of Banach algebras (see e.g.
R.~G.~Douglas~[1]).  There it is defined as the set of all
$\lambda\in\Bbb C$ for which $\{x\in X; |f(x) - \lambda| <
\varepsilon\}$ has positive measure for every $\varepsilon > 0$.  It
is easy to see that this notion is equivalent to our definition when
$\Bbb C$ is replaced by $Y$.
\enddemo
\medskip
\demo{\it Proof of Property 1}  We argue by contradiction.  Suppose
$\text{ess$R(u)$ omits a point $y_0$}$.  Then, for some $r > 0$,
$$
\text{ess}R(u)\cap B_r (y_0) = \phi.
$$
Setting
$$
\Sigma = Y\backslash B_r(y_0),
$$
we clearly have $u(x)\in\Sigma$ a.e.  Since $u\in$\,VMO,
$$
\underset \vare\rightarrow0\to\lim \Mint_{B_\vare(x)} |u -\bar
u_\vare (x)| = 0\text{\quad uniformly in $x$}.
$$
Therefore
$$
\text{dist}(\bar u_\vare(x),\Sigma)\rightarrow 0\quad\text{uniformly
in $x$},
$$
which implies
$$
\text{dist}(u_\vare(x), \Sigma)\rightarrow 0\text{\quad uniformly in
$x$}.
$$
Consequently, for $\vare$ small, $\deg u_\vare = 0$, contradicting the
assumption that\newline $\deg u_\vare = \deg u\not= 0$.
\medskip
\noindent
{\bf Property 2 (Hopf).}  If $u,v\in\,$VMO$(S^n,S^n)$ and have the same
degree, then they are homotopic within VMO$(S^n, S^n)$.
\enddemo
\medskip

\demo{Proof}  By our construction, $\deg u_\vare = \deg u = \deg v =
\deg v_\vare$.  The well known result of Hopf says that $u_\vare$ and
$v_\vare$ are homotopic within $C^0(S^n,S^n)$ and therefore within
VMO$(S^n,S^n)$.  On the other hand, $u_\vare$ is homotopic to $u$
within VMO$(S^n,S^n)$ (via $u_{t\vare}$).
\enddemo

\noindent
{\bf Property 3 (Borsuk).}  Let $U$ and $V$ be symmetric open bounded
neighbourhoods of the origin in $\Bbb R^n$, with smooth boundaries
$\partial U,\partial V$---each of which is connected.  Let
$u\in$\,VMO$(\partial U,\partial V)$ be an odd map.  Then $\deg u$ is
odd.

\demo{Proof}  One may simply apply Borsuk's theorem to $u_\vare$ which
is also an odd map.
\enddemo

\medskip
\noindent
{\bf Property 4.}  We return to the setting of Section I.3.  Let $u\in
W^{1,n}(X,Y)$ so that $u\in$\,VMO$(X,Y)$---see Example 1 in Section
I.2.  Then
$$
\deg u \,\int_Y \mu = \int_X \mu\circ u \tag 19
$$
where $\mu$ is any smooth $n$-form on $Y$.
\medskip
Observe that in local coordinates, the integrand involves the
determinant of the Jacobian of the map, and hence is integrable.
Formula (19) is well known for smooth maps (see e.g., L.~Nirenberg~[1]).
\medskip
The proof relies on the following:

\proclaim{Lemma 7}  Given $u\in W^{1,n}(X,Y)$, there is sequence
$(u_j)$ of smooth maps from $X$ to $Y$, converging to $u$ in
$W^{1,n}$.
\endproclaim

  Lemma 7 follows R.~Schoen and K.~Uhlenbeck [1], and is proved in
the Appendix---see Lemma A.11.
\medskip
\remark{\bf Remark 8}  In general, if $p < n$, smooth maps from $X$ to
$Y$ are not dense in $W^{1,p} (X,Y)$.  However, F.~Bethuel [1] has
shown that they are dense iff the homotopy group $\pi_{[p]}(Y)$ is
zero.
\endremark

\medskip
Assuming the lemma we give the
\medskip
\demo{\it Proof of Property 4}  Let $(u_j)$ be the sequence of Lemma
7.  Then, by convergence in $W^{1,n}$,
$$
\int_X \mu\circ u_j\longrightarrow \int_X \mu\circ u.
$$
On the other hand $u_j\rightarrow u$ in BMO, by Example 1 in Section
I.2, and so, by Theorem 1, $\deg u_j = \deg u$ for $j$ large.
\enddemo

\noindent
{\bf Property 5.}  Let $\Omega$ be a bounded domain in $\Bbb R^n$ with
smooth connected boundary $\partial\Omega$.  Suppose $u\in
W^{1,n}(\Omega,\Bbb R^n)$ (so its trace is defined on $\partial\Omega$)
and suppose that
$$
u : \partial\Omega\rightarrow S^{n-1}.
$$
Then
$$
\deg(u_{\vert_{\partial\Omega}}, \partial \Omega, S^{n-1}) =
\frac{1}{|B_1|} \int_\Omega \det J_u. \tag 20
$$
\medskip
Formula (20) is a well known formula for the degree in case $u$ is
smooth.  In the more general situation, the right hand side makes
sense because $u\in W^{1,n}$.  The left hand side makes sense because
$u_{\vert_{ \partial\Omega}}$ belongs to
$W^{1-\frac1n, n}(\partial\Omega)$ which, by Example 2 in Section I.2,
is contained in VMO$(\partial\Omega)$.
\medskip
\proclaim{Corollary 8}  Let $\Omega$ be a bounded domain in $\Bbb
R^n$ with smooth connected boundary $\partial\Omega$.  If $u\in
W^{1,n} (\Omega,S^{n-1})$ then
$$
\deg (u_{\vert_{\partial\Omega}}, \partial\Omega, S^{n-1}) = 0.
$$
\endproclaim
Returning to Property 5, we shall prove (20), as expected, via
approximation; namely, using the following---which we formulate more
generally (see Lemma A.13):

\proclaim{Lemma 8}  Let $\Omega \subset \Bbb R^n$ be a domain with
smooth connected boundary $\partial\Omega$.  Let $Y$ be a compact
manifold without boundary, smoothly embedded in $\Bbb R^N$.  Let $u\in
W^{1,n}(\Omega, \Bbb R^N)$ such that
$$
u(\partial\Omega)\subset Y.
$$
Then there is a sequence $(u_j)$ of smooth maps of $\overline\Omega$
into $\Bbb R^N$ such that
$$
u_j(\partial\Omega)\subset Y,\quad\forall j, \text{ and $u_j\rightarrow u$
in $W^{1,n}$}.
$$
\endproclaim

Assertion (20) is a simple consequence of Lemma 8, for the
$u_{j\vert\partial\Omega}$ converges to $u_{\vert
\partial\Omega}$ in $W^{1-\frac1n,n}(\partial\Omega)$ and hence in
BMO.  Thus
$$
\deg(u_{j\vert\partial\Omega}, \partial\Omega,Y)\longrightarrow
\deg(u_{\vert\partial\Omega}, \partial\Omega, Y)
$$
by Theorem 1.  Furthermore, the integrals on the right of (20) for the
$u_j$ tend to that for $u$.
\medskip
We use Corollary 8 to prove a stronger result:

\proclaim{Theorem 2}  Let $\Omega$ and $Z$ be smooth bounded domains
in $\Bbb R^n$ with $\partial\Omega$ and $\partial Z$ connected.  Let
$u\in W^{1,n}(\Omega,\Bbb R^n)$ be such that
$$
u(\partial\Omega)\subset \partial Z
$$
and
$$
\deg(u_{\vert_{\partial\Omega}}, \partial\Omega,\partial Z)\not=0.
\tag 21
$$
Then
$$
\text{ess}R(u)\supset \overline Z.
$$
\endproclaim

\demo{Proof}  Since ess$R(u)$ is closed it suffices to show that
ess$R(u)\supset Z$.  Suppose not, i.e., suppose that some $z_0\in Z$ is
not in ess$R(u)$.  Then a closed ball $\overline B_r(z_0)$ lies in
$Z$ and is disjoint from ess$R(u)$.  Let $P$ be the nearest point
projection onto $\overline B_r(z_0)$.  Set
$$
v = P\circ u.
$$
Clearly $v\in W^{1,n}(\Omega,\Bbb R^n)$ and $v(\Omega)\subset
S_r(z_0)$, the sphere of radius $r$ centered at $z_0$.
\medskip
\noindent
Hence, by Corollary 8,
$$
\deg(v_{\vert_{\partial\Omega}}, \partial\Omega, S_r) = 0. \tag 22
$$
\medskip
Next we claim that
$$
\deg(u_{\vert\partial\Omega}, \partial\Omega,\partial Z) =
\deg(v_{\vert\partial\Omega}, \partial\Omega, S_r). \tag 23
$$
The conclusion of the theorem is then an immediate consequence of
(21--23).
\enddemo

\demo{Proof of (23)}  We reduce it to the smooth case.  Namely, by
Lemma 8, with $Y = \partial Z$, we know that there is a sequence $(u_j)$ of
smooth maps from $\overline\Omega$ into $\Bbb R^n$ such that
$u_j(\partial\Omega)\subset\partial Z$, and $u_j\rightarrow u$ in
$W^{1,n}$.  Since $u_{j|\partial\Omega} \rightarrow
u_{|\partial\Omega}$ in $W^{1 - \frac1n, n} (\partial\Omega)$ it also
converges in VMO$(\partial\Omega)$.  Set $v_j = P u_j$.  Applying
Lemma A.8 we find that $v_{j|\partial\Omega} \rightarrow
v_{|\partial\Omega}$ in VMO$(\partial\Omega)$.  By Theorem 1 and Example 2 of Section I.2, for $j$ large,
$$
\deg(u_{j\vert_{\partial\Omega}},\partial\Omega,\partial Z) =
\deg(u_{\vert_{\partial\Omega}}, \partial\Omega,\partial Z)
$$
and
$$
\deg(v_{j \vert_{\partial\Omega}}, \partial\Omega, S_r) =
\deg(v_{\vert\partial\Omega}, \partial\Omega, S_r).
$$
However, it is well known that
$$
\deg(u_{j|\partial\Omega}, \partial\Omega,\partial Z) =
\deg(u_j,\Omega, z_0)
$$
and
$$
\deg (v_{j\vert\partial\Omega}, \partial\Omega, S_r) =
\deg(v_j,\Omega,z_0).
$$
Finally, the two degrees on the right hand sides are the same by the
following homotopy
$$
H_t(x) = t u_j(x) + (1-t) v_j(x),\quad t\in [0,1].
$$
This completes the proof of (23).    \qed
\enddemo

\medskip
\remark{\bf Remark 9}  The assumption that $u$ belongs to $W^{1,n}$ in
Theorem 2 is sharp in that it may not be replaced by $u\in W^{1,p}$
with $p < n$---even if $u$ is smooth near $\partial\Omega$, so that
the degree of $u_{\vert\partial\Omega}$ makes sense.  Namely, for
$n\geq 2$ and $\Omega = B_1(0)$ the map $u(x) = x/|x|$ is in
$W^{1,p}(\Omega,\Bbb R^n)$ for any $p < n$; moreover
$u_{\vert\partial\Omega} = \text{Id}$ and so has degree one, but the
conclusion of Theorem 2 does not hold since ess$R(u) = S^{n-1}$.
\endremark

\medskip
The reader may ask if in Theorem 2, the condition $u\in
W^{1,n}(\Omega,\Bbb R^n)$ may be replaced by $u\in$ VMO$(\Omega,\Bbb
R^n)$.  This is a delicate issue, since maps in VMO$(\Omega)$ do not
in general admit a trace on $\partial\Omega$.  We will be led in Part II,
Section 3, to the notion of a special class of maps in VMO$(\Omega)$
admitting a trace on $\partial\Omega$, which belongs again to
VMO$(\partial \Omega)$.  For such a class, which includes
$W^{1,n}(\Omega)$, we will have a generalisation of Theorem 2 (in Part
II, Section 4).
\bigskip
\noindent
{\bf Section I.5.  Further comments}
\medskip
\noindent
{\bf 1.}  One may discuss BMO and VMO maps from compact $X$ to compact $Y$
even if their dimensions are different, say maps of $S^n$ to $S^k$.
The space of continuous maps from $X$ to $Y$ decomposes naturally into
its components $\Cal C_i$ namely, maps $u$ and $v$ are in the same component
if there is a homotopy within $C^0(X,Y)$ of $u$ to $v$. Similarly, the space of VMO maps from
$X$ to $Y$ also decomposes into components via homotopy.
There are two natural notions of homotopy for maps in VMO:
\medskip
a)  Two maps $u,v\in$ VMO$(X,Y)$ are said to be homotopic
within VMO\,$\cap\, L^1$ if there is some deformation $H\in C([0,1]$,
VMO\,$\cap\, L^1)$ such that $H(0)=u$ and $H(1)=v$.
\medskip
b)  Two maps $u,v\in$ VMO$(X,Y)$ are said to be homotopic
within VMO if there is some deformation $H\in C([0,1]$, VMO) such that
$H(0)=u$ and $H(1)= v$.
\medskip
Clearly, the first notion is stronger and, in general, it is {\bf
strictly stronger} (see Remark A.6).  Surprisingly, the two notions
are {\bf equivalent} in the special case where $Y = S^k,
k\geq 1$, (see Lemma A.23).
\medskip
Homotopy classes of VMO$(X,Y)$ in the sense of definition
a)---homotopy within\newline VMO\,$\cap\,L^1$---are in one-to-one
correspondence with the homotopy classes of
$C^0(X,Y)$.  They are simply the closures of the above
$\Cal C_i$ in VMO\,$\cap\, L^1$ (see Lemma A.21).
\medskip
In contrast, the spaces $L^p (X,Y)$, $\;1\leq p\leq \infty$, and
BMO$(X,Y)$ are arcwise connected.  It suffices to prove this for
$p=\infty$.  We sketch a proof.
\medskip
\noindent
{\bf Step 1.}  Denote by PC$(X,Y)$ the set of measurable maps from $X$
to $Y$ taking on only a finite number of values in $Y$.  Given any
measurable map $f:X\rightarrow Y$ and any $\vare > 0$, there exists
$g\in \text{PC}(X,Y)$ such that $\|f - g\|_{L^\infty} < \vare$.
\medskip
\noindent
{\bf Step 2.}  For  $t\in [0,1]$, $tf(x) + (1-t)g(x)$ lies within $\vare$ of
$Y$.  Hence $h_t(x) = P(tf(x) + (1-t) g(x))$ lies in $Y$ and connects
$g$ to $f$ continuously in $t$, within $L^\infty(X,Y)$.
\medskip
\noindent
{\bf Step 3.}  Given $g_0,g_1\in \text{PC}(X,Y)$, they may be
connected by a continuous arc within $L^\infty(X,Y)$.  Namely, we may
always assume that $g_0$ and $g_1$ have the form
$$
g_0 = \Sigma \chi_{\omega_i} a_i,\;\; g_1 = \Sigma \chi_{\omega_i} b_i
$$
for some finite partition $(\omega_i)$ of $X$, with $a_i,b_i\in Y$.  For
each $i$ let $\varphi_i(t), 0\leq t\leq 1$, be a continuous arc in $Y$
connecting $a_i$ to $b_i$.  Then the maps
$$
g_t(x) = \Sigma \chi_{\omega_i} (x)\varphi_i(t)\quad 0\leq t\leq 1
$$
connect $g_0$ to $g_1$.
\medskip
A continuous map from $X$ to $Y$ naturally induces a map from homology
in $X$ to homology in $Y$.  The same is true for a VMO map $u$---via
approximation by $u_\vare$.
\medskip
\noindent
{\bf 2.}  A.~Granas pointed out that several authors have
previously considered fixed point properties, and degree theory, for
some classes of maps which are not continuous (see O.~H.~Hamilton [1],
J.~Stallings [1], H.~A.~De Kleine and J.~E.~Girolo [1]).  A class which plays an
essential role in their considerations is one introduced by J.~Nash [1]
called ``connectivity maps''.  By this is meant that the graph of such
a map $f$ over every connected subset of $X$, is a connected set.
H.~A.~De~Kleine and J.~E.~ Girolo [1] developed a degree theory for a somewhat more
general class (``almost continuous'' maps).  However, their degree is a
collection of integers.  We do not see how our degree is related to
theirs.
\medskip
We point out, however, that if $u\in \text{VMO}(X,Y)$ and $A$ is a
connected subset of $X$ then $\Sigma = \text{ess}R(u_{\vert A})$ is
connected.  This may be seen as follows:  If $\Sigma$ is not connected
then $\Sigma = \Sigma_1\cup\Sigma_2$ where $\Sigma_1,\Sigma_2$ are
nonempty disjoint closed sets.  Recall that since $u\in$ VMO,
$$
\text{dist}(u_\vare(A), \Sigma)\longrightarrow\text{  as
$\vare\longrightarrow 0$}.
$$
On the other hand, $u_\vare(A)$ is connected.  Therefore for $\vare$
small, dist $u_\vare(A)$ to either $\Sigma_1$ or $\Sigma_2$ is $<
\frac\delta2 = \frac12\text{dist}(\Sigma_1,\Sigma_2)$.  Consequently,
using a sequence $\vare_i\rightarrow 0$, we conclude that
ess$R(u_{\vert A})$ is contained either in $\Sigma_1$ {\bf or} in $\Sigma_2$.
Impossible.
\medskip
\noindent
{\bf 3.}  Our degree theory holds in particular for a map $u\in
H^{1/2}(S^1,S^1)$.  As remarked earlier, L.~Boutet~de Monvel
and O.~Gabber previously
defined a degree for such maps, given by
$$
\deg u = \frac{1}{2\pi i} \int^{2\pi}_0 \bar u \,du. \tag 24
$$
This integral makes sense since $\bar u\in H^{1/2}$ and $\dot u\in
H^{-1/2}$.  That $u$ is in $H^{1/2}$ may be expressed in
terms of its Fourier coefficients.  If
$$
u = \sum^\infty_{-\infty} a_j e^{ij\theta}
$$
then
$$
\|u\|^2_{H^{1/2}} = \sum^\infty_{-\infty} |j|\; |a_j|^2.
$$
\medskip
I.~M.~Gelfand raised the question:  what is $\deg u$ in terms of its
Fourier coefficients?  More generally, for maps $u: S^n\rightarrow
S^n$, what is $\deg u$ in terms of its expansion coefficients in
spherical harmonics?  It is easily seen from (24) that
$$
\deg u = \sum^\infty_{-\infty} j|a_j|^2. \tag 25
$$
From the fact that $|u(\theta)| = 1$ it is not a priori clear that the
right hand side of (25) is an integer.  The condition that $|u(\theta)| =
1$ on $S^1$ is equivalent to
$$
\align
&\sum |a_j|^2 = 1\\
  &\sum^\infty_{j=-\infty} \bar a_j a_{j+k} = 0\quad\text{for all
integers $k\not= 0$}.
\endalign
$$
\medskip
For a continuous map (or VMO map) $u : S^1\rightarrow S^1$, its
Fourier coefficients are defined.  But the series in (25) need not be
absolutely convergent.  
\bigskip


\noindent
{\bf Open Problem:}  What summation process makes it summable so that
(25) holds?
\medskip
When working with maps from $S^n$ to $S^n$ one would use the formula
that
$$
\deg u = \frac{1}{|S^n|} \int_{S^n} \det(u, u_{x_1},\ldots, u_{x_n})
d\sigma
$$
computed using normal geodesic coordinates $(x_1,\ldots,x_n)$.  If one
represents $u$ via spherical harmonics, this leads to some complicated
expressions.
\bigskip
\noindent
{\bf 4.}  Recently, M.~Giaquinta, G.~Modica and J.~Soucek~[1]
introduced a notion of degree for rectifiable currents and for
approximately differentiable maps with Jacobian determinant in $L^1$
(see also a related work by M.~Esteban and S.~M\"uller [1]).  We do
not know if it is related to our degree.
\medskip
\noindent
{\bf 5.}  To every function $\varphi\in L^\infty(S^1,\Bbb C)$
corresponds a Toeplitz operator $T_\varphi$ in the Hardy space
$H^2$; see e.g. R.~G.~Douglas [1], Chapter 7.  When $\varphi\in
C^0(S^1,\Bbb C)$, $T_\varphi$ is Fredholm if and only if
$|\varphi|\geq \alpha > 0$\,; moreover
$$
\text{ind}(T_\varphi) = - \deg(\frac{\varphi}{|\varphi|}, S^1, S^1).
$$
A similar result holds assuming only $\varphi\in L^\infty(S^1,\Bbb C) \cap
VMO(S^1,\Bbb C)$ ; see Theorem 7.36 in Douglas [1] (where it is stated
in different terms).  We will return to this topic in Part II.
\bigskip
\noindent
{\bf Section I.6.  Lifting of BMO maps}
\medskip
\leftheadtext{A degree theory for BMO maps}
\rightheadtext{H. Brezis and L. Nirenberg}
This section is largely inspired by an interesting result in
R.~Coifman and Y.~Meyer [1].
\medskip
One form of their result asserts that there exist constants $\delta,
C > 0$ such that every $u\in\text{BMO}((0,1), S^1)$ with
$$
\|u\|_{BMO} < \delta \tag 26
$$
may be lifted as
$$
\cases  u = e^{i\varphi},\; \varphi\in {BMO}((0,1),\Bbb R)\\
  \|\varphi\|_{BMO} \leq C \|u\|_{BMO}.
\endcases
\tag 27
$$

We present variants of this result, in that we replace the interval
$(0,1)$ by our compact $n$-dimensional manifold $X$ without boundary.
In addition, we will see that (26) is not needed when working with VMO,
as opposed to BMO.
\medskip
Here is a first result:
\medskip
\proclaim{Theorem 3}  Any $u\in\text{VMO}(X,S^1)$ which is
homotopic within VMO$(X,S^1)$ to a constant map may be uniquely
written as
$$
u = e^{i\varphi}\text{  with  $\varphi\in VMO(X, \Bbb R),\;
0\leq \Mint_X \varphi < 2\pi$}.  \tag 28
$$
Furthermore, the map $u\mapsto \varphi$ is continuous from VMO$\,\cap\,
L^1$ (respectively VMO) into VMO$\,\cap\, L^1$ (respectively VMO).
\endproclaim

\remark{\bf Remark 10}  (i)  Note that we give no estimate of
$\|\varphi\|_{\text{BMO}}$.  (ii)  The converse of Theorem 3 also
holds, namely, any $u$ of the form in (28) is homotopic to a constant
within VMO---because of Lemma A.8, via the homotopy $e^{it\varphi},
0\leq t\leq 1$.  (iii)  As a consequence of Theorem 3, and Property 2
in Section I.4, we may assert that any map $u\in\text{VMO}(S^1, S^1)$
with degree zero, may be written as $e^{i\varphi}$ with
$\varphi\in\text{VMO}(S^1,\Bbb R)$, and conversely.  More generally,
any $u\in \text{VMO}(S^1,S^1)$ may be written as
$$
u(\theta) = e^{i k\theta + i\varphi(\theta)}
$$
where $k = \deg u$, and $\varphi\in\text{VMO}(S^1,\Bbb R)$.  This is
easily seen by considering $e^{-ik\theta} u(\theta)$.  (iv)  If $\pi_1(X) = 0$, then {\it every} map $u\in\text{VMO}(X,S^1)$ may
be written as in (28).  This is a consequence of the corresponding
fact for continuous maps:  one repeats the argument used in proving
Theorem 3.
\endremark

A variant which is closer to the result above of Coifman-Meyer is

\proclaim{Theorem 4}  There exists $\delta > 0$ (depending only on
$X$), such that if $u\in\text{BMO}(X,S^1)$ and
$$
\|u\|_{BMO} \leq \delta, \tag 29
$$
then
$$
u = e^{i\varphi}\text{  with $\varphi\in BMO(X,\Bbb R)$},
$$
and
$$
\|\varphi\|_{BMO} \leq 4\|u\|_{BMO}. \tag 30
$$
\endproclaim

The central idea in the proof of Theorem 3 and 4 is the same.  We
proceed as follows:  (i)  We approximate $u$ by our $u_\vare$ (using
averaging, $\bar u_\vare$, and projection on $S^1$);\; (ii)  We
lift $u_\vare$ as $u_\vare = e^{i\varphi_\vare}$, and derive estimates
for $\varphi_\vare$---in case of Theorem 4 we prove (30) for
$\varphi_\vare$; (iii)  Finally, we show that $\varphi_\vare$
converges as $\vare\rightarrow 0$ to the desired function $\varphi\in
\text{BMO}(X,\Bbb R)$.  
\medskip
Our proof of Step (ii) is very different from that in R.~Coifman and
Y.~Meyer~[1].  It relies on the John-Nirenberg inequality, in the form
described in Appendix B, while they used estimates of commutators.  We
have been informed that L.~Carleson, in a personal communication to
Y.~Meyer in 1979, also proved Step (ii) using the John-Nirenberg
inequality rather than commutators. 
\medskip
There is a result which includes Theorem 3 and part of Theorem 4; it
involves\newline $u\in\text{BMO}(X,S^1)$ with small distance to
VMO$(X,S^1)$---see Theorem 5.
\medskip  
\demo{Proof of Theorem 3}  Since $u\in\text{VMO}(X,S^1)$, there
exists some $\vare_0 > 0$ such that for every $\vare \leq \vare_0$,
$u_\vare = P\bar u_\vare$ is well defined, and converges to $u$
in BMO$\; \cap\; L^1$ as $\vare\rightarrow 0$.  In what follows, we always
take $\vare < \vare_0$, and sometimes restrict $\vare$ further.
\enddemo

\noindent
{\bf Step 1.}  There is some $\vare_1\leq \vare_0$ such that for every
$\vare\leq \vare_1$, $u_\vare$ may be written as
$$
u_\vare = e^{i\varphi_\vare}\tag 31
$$
with $\varphi_\vare\in C^0(X,\Bbb R)$ and
$$
0 \leq \Mint_X \varphi_\vare < 2 \pi. \tag 32
$$
\medskip
\demo{Proof}  We rely on Lemma A.23 according to which $u$ is
homotopic to a constant within VMO\,$\cap\, L^1$.  Denote the homotopy by
$H(t), 0 \leq t\leq 1$, with $H(0) = u$, $H(1) = $ a constant.  Since
$H([0,1])$ is a compact set in VMO, by Lemma 4, there is some
$\vare_1\leq \vare_0$ such that for every $\vare\leq \vare_1,
H(t)_\vare$ is well defined.  It yields a homotopy of $u_\vare$ to a
constant within $C^0(X,S^1)$.  Here we use the fact that $t\mapsto H(t)$
is continuous in $L^1$.  We are therefore reduced to the classical
continuous case, yielding (31).  By adding an appropriate integral
multiple of $2 \pi$ to $\varphi_\vare$, we may achieve (32).
\medskip
  From now on, $\vare\leq \vare_1$.
\enddemo

\medskip
\noindent
{\bf Step 2.}  There is a constant $\overline C$ depending only on $X$
such that
$$
M_t(\varphi_\vare)\leq 2 M_t(u_\vare) + \overline C M^2_t(
\varphi_\vare)\quad\forall t < r_0,\quad \forall \vare\leq \vare_1.
\tag 33
$$
\medskip
\demo{Proof}  Since
$$
|e^{it} - 1 - it|\leq \frac12 t^2\quad\forall t\in\Bbb R, \tag 34
$$
one easily finds that, for $\alpha, \beta\in\Bbb R$,
$$
|\alpha - \beta|\leq |e^{i\alpha} - e^{i\beta}| + \frac12 (\alpha -
\beta)^2. \tag 35
$$
\enddemo

To verify (33) we recall that
$$
M_t(\varphi_\vare) \leq \sup\Sb x\in X\\ r\leq t\endSb \Mint_{B_r(x)}
\Mint_{B_r(x)} |\varphi_\vare(y) - \varphi_\vare (z)|.
$$
\medskip
\noindent
Applying (35), with $\alpha = \varphi_\vare(y)$, $\beta =
\varphi_\vare(z)$ we find, as in $(1'')$, that
$$
M_t(\varphi_\vare) \leq 2 M_t(u_\vare) + \frac 12 \sup\Sb x\in X\\ r
\leq t\endSb \Mint_{B_r(x)} \Mint_{B_r(x)} |\varphi_\vare(y) -
\varphi_\vare (z)|^2. \tag 36
$$
We now use Lemma B.4 to see that the last term in (36) is
$$
\leq \overline C M^2_t (\varphi_\vare);
$$
(33) is proved.
\bigskip
\noindent
{\bf Step 3.}  There is some $a > 0$ such that
$$
M_t(\varphi_\vare)\leq 4 M_t(u_\vare)\quad\forall t\leq a,\quad\forall
\vare\leq \vare_1. \tag 37
$$
\medskip
\demo{Proof}  Since for $\vare\leq \vare_1$, the family $(u_\vare)$ is
compact in VMO, there is some $a > 0$, by Lemma 4, such that
$$
M_a(u_\vare)\leq \frac{1}{9\overline C}\quad\forall \vare\leq \vare_1.
$$
We now claim that for such $a$, $M_t(\varphi_\vare) <
\dfrac{1}{2\overline C}\;\;\forall t\leq a$---which yields (37) via (33).
\medskip
Indeed, since $M_t(\varphi_\vare)$ tends to zero as $t\rightarrow 0$,
we see by (33), that $M_t(\varphi_\vare) < \dfrac{1}{2\overline C}$ in
some neighbourhood of $t = 0$.  If the claim were false, then since
$M_t(\varphi_\vare)$ is continuous in $t$---see Lemma A.15---there
would be a first value of $t\leq a$ such that
$$
M_t(\varphi_\vare) = \frac{1}{2\overline C}.
$$
But then by (33), for that $t$,
$$
\frac{1}{2\overline C}= M_t(\varphi_\vare) \leq 4 M_t(u_\vare) \leq
\frac{4}{9\overline C}.
$$
Impossible.
\medskip  %new addition
\noindent
{\bf Step 4. Existence of $\boldsymbol\varphi$.}   We make use of an
elementary, but very useful, observation of G.~David (which simplifies
our original presentation).
\enddemo

\medskip
\proclaim{Lemma 9}  Let $f\in C^0((0,\alpha), \Bbb R)$, for some
$\alpha > 0$, and assume that $\underset \vare\rightarrow 0\to{\lim}
e^{if(\vare)}$ exists.  Then $\underset \vare\rightarrow 0\to{\lim}
f(\vare)$ exists.
\endproclaim

The proof of Lemma 9 is obvious and relies on the connectedness of the
range of $f$.
\medskip
Since $u_\vare\in C^0((0,\vare_1) \times X,S^1)$ we may lift it
as $u_\vare = e^{i\varphi_\vare}$ with $\varphi_\vare \in
C^0((0,\vare_1) \times X,\Bbb R)$.  Recall that $\underset
\vare\rightarrow 0\to{\lim}\, u_\vare(x)$ exists at every Lebesgue
point $x$ of $u$.  By Lemma 9, $\varphi(x) = \underset
\vare\rightarrow 0\to{\lim} \varphi_\vare(x)$ exists at every such
$x$, hence a.e. on $X$.  Moreover $u = e^{i\varphi}$.
\medskip
We now prove that $\varphi\in \text{VMO}$.  From (37) we deduce that,
for every $x\in X$,
$$
\Mint_{B_r(x)}\, \Mint_{B_r(x)} |\varphi_\vare(y) - \varphi_\vare (z)
|\leq 8 M_r (u_\vare)\quad\forall r \leq a,\quad\forall \vare\leq
\vare_1.
$$
Recall (see Corollary 3) that
$$
M_r(u_\vare) \leq C(M_r(u) + M_\vare (u)).
$$
We may then pass to limit as $\vare\rightarrow 0$, using Fatou's
lemma, and conclude that
$$
\Mint_{B_r(x)}\, \Mint_{B_r(x)} |\varphi(y) - \varphi(z)|\leq 8 C
M_r(u)\quad\forall r \leq a.
$$
It follows, by Sarason's Lemma 3, that $\varphi\in \text{VMO}$.
%++++++++++++++++++++++++++++
\medskip
\noindent
{\bf Step 5. Uniqueness.}  Suppose $\varphi_1$ and $\varphi_2$ are
solutions of (28).  Then $\eta = \dfrac{1}{2\pi} (\varphi_1 -
\varphi_2)\in\Bbb Z$ a.e.  On the other hand, since
$\eta\in\text{VMO}$,  ess $R(\eta)$ is connected, see Item 2 in
Section I.5.  Hence ess$R(\eta)$ is reduced to a point, i.e., $\eta$
is constant; by (28), $\eta = 0$.
\medskip
\noindent
{\bf Step 6.  Claim:}  There exists $\alpha > 0$ depending only on $X$
such if $a < r_0$ and
$$
M_a(u) \leq \alpha,
$$
then
$$
M_a(\varphi)\leq 4 M_a(u).
$$
Here $u$ is as in the theorem, and $\varphi$ is the unique solution of
(28).
\medskip
\demo{Proof}  We may take $\alpha = 1/(9\overline C)$ of Step 3 and
repeat the arguments of Steps 2 and 3, deleting $\vare$ everywhere.
\enddemo

\medskip
\noindent
{\bf Step 7.  Continuous dependence of $\bold u \mapsto \boldsymbol\varphi$.}
\medskip
At first, let $(u_j)$ be a sequence converging in VMO to $u$, each
homotopic to a constant.  We have the corresponding $\varphi_j$.  By
Lemma 4, we know that there exists some $a_0$ such that---for $\alpha$
in Step 6---
$$
M_{a_0} (u_j)\leq \alpha\quad\forall j.
$$
Hence for $a\leq a_0$,
$$
M_a(\varphi_j)\leq 4 M_a(u_j)\quad\forall j. \tag 38
$$
Since $M_a(u_j)\rightarrow 0$ as $a\rightarrow 0$, uniformly in $j$,
the same is true for $M_a(\varphi_j)$.  Consequently by Lemma 4,
again, the $(\varphi_j)$ lie in a compact set in VMO.  A subsequence,
still called $(\varphi_j)$, converges in VMO to $\psi$.
\medskip
In view of the fact that $0\leq \dsize\Mint_X \varphi_j\leq 2\pi$ we
may assume that
$$
\Mint_X (\varphi_j - \psi) \rightarrow \ell.
$$
By Lemma 1, $\varphi_j\rightarrow \psi + \ell = :\varphi$ in $L^1$.
For a further subsequence, still denoted $\varphi_j$,
$\varphi_j\rightarrow \varphi$ a.e.
\medskip
\noindent
{\bf 1.}  Continuity from VMO$\,\cap\,L^1$ into VMO$\,\cap\,L^1$.  We suppose,
then, in addition that $u_j\rightarrow u$ in $L^1$.  Consequently
$$
u = e^{i\varphi}.
$$
Convergence of the full sequence $(\varphi_j)$ follows from the
uniqueness of $\varphi$.
\medskip
\noindent
{\bf 2.}  Continuity from VMO to VMO.  We established above that for a
subsequence
$$
\varphi_{j_k} \rightarrow\varphi\quad\text{in $VMO\,\cap\,L^1$}.
$$
\medskip
We can no longer infer that $u = e^{i\varphi}$; we can only say that
$$
u = e^{i\varphi} + c
$$
for some constant $c$.  We need only consider the case $c\not= 0$.  In
this case, $u$ takes its values in $S^1\cap (S^1 + c)$, which consists
of one or two points.  Since ess$R(u)$ is connected (see paragraph 2 in
Section I.5), $u$ must be a constant, thus also $\varphi$.  Since
$\varphi_j\rightarrow\varphi$ in BMO it follows that
$\|\varphi_j\|_{BMO}\rightarrow 0$.  Continuity is proved.  \qed
\medskip
We turn now to the

\demo{Proof of Theorem 4}  Given $u\in \text{BMO}(X,Y)$, recall that
$\bar u_\vare$ is defined for every $\vare < r_0$; by (7),
$$
\text{dist}(\bar u_\vare(x), Y)\leq M_\vare(u)\leq
\|u\|_{BMO},\quad\forall x\in X.
$$
Thus if $\|u\|_{BMO} \leq $ some small $\delta$ (depending only
on $X$), we may define $u_\vare = P\bar u_\vare$ for every $\vare
< r_0$.
\enddemo

Now consider $u\in \text{BMO}(X,S^1)$ with $\|u\|_{BMO}
\leq\delta$.  We will show that for $\delta$ sufficiently small, $u =
e^{i\varphi}$, with $\|\varphi\|_{BMO} \leq 4
\|u\|_{\text{BMO}}$.
\medskip
\noindent
{\bf Step 1.}  According to Lemma A.18 there is a $\delta$ depending
only on $X$ such that if\newline $\|u\|_{BMO} \leq\delta$, then for each
$\vare < r_0$, $u_\vare$ is homotopic (within $C^0(X,S^1)$) to a
constant.  Hence we lift $u_\vare$ and write it as
$$
u_\vare = e^{i\varphi_\vare},\quad\varphi_\vare\in C^0(X,\Bbb R), \tag
39
$$
with
$$
0\leq \Mint_X \varphi_\vare < 2\pi. \tag 40
$$
\medskip
By Step 2 in the proof of Theorem 3, we have
$$
M_t(\varphi_\vare)\leq 2 M_t(u_\vare) + \overline C
M^2_t(\varphi_\vare)\quad\forall t < r_0,\quad\forall \vare < r_0.
\tag 41
$$
Now
$$
M_t(u_\vare)\leq \|u_\vare\|_{BMO} \leq C\|\bar
u_\vare\|_{BMO}
$$
by Lemma 2, for some constant $C$.
\medskip
Using Corollary 2 we conclude that
$$
M_t(u_\vare)\leq C^\star\|u\|_{BMO} \leq C^\star
\delta\quad\forall t, \vare < r_0. \tag 42
$$

Next, arguing as in Step 3, we find that if
$$
C^\star \delta < \frac{1}{9\overline C}, \tag 43
$$
then
$$
M_t(\varphi_\vare)\leq 4 M_t(u_\vare)\quad\forall t, \vare < r_0. \tag
44
$$ 
In particular, by (42),
$$
\|\varphi_\vare\|_{BMO} \leq 4 C^\star \|u\|_{BMO} \leq
4 C^\star\delta. \tag 45
$$
\medskip
\noindent
{\bf Step 2.  Existence of $\boldsymbol\varphi$.}  We see, as in the
proof of Theorem 3 (Step 4), that\newline $\varphi(x) = \underset
\vare\rightarrow 0\to{\lim} \varphi_\vare(x)$ exists a.e., and that $u
= e^{i\varphi}$.  From (45) we deduce that, for every $B_r(x)$,
$$
\Mint_{B_r(x)}\, \Mint_{B_r(x)} |\varphi_\vare(y) -
\varphi_\vare(z)|\leq 8 C^* \|u\|_{BMO}. \tag 46
$$
\medskip
Using Fatou's lemma we conclude that $\varphi\in \text{BMO}$ and
$$
\|\varphi\|_{BMO} \leq 8 C^* \|u\|_{BMO}. \tag 47
$$
\medskip
Finally, we get rid of the factor $C^\star$ in (47).  Namely, as in
Step 2 of the proof of Theorem 3, applied to $u$ and $\varphi$, we find
$$
\align
\|\varphi\|_{BMO} &\leq 2\|u\|_{BMO} + \overline
C\|\varphi\|^2_{BMO}\\
   &\leq 2 \|u\|_{BMO} + 8 C^\star\overline C\delta
\|\varphi\|_{BMO}
\endalign
$$
by (47).  Assuming $16 C^* \overline C \delta \leq 1$ we obtain the desired conclusion.   \qed
\medskip

\remark{\bf Remark 11}  The reader may think that the space $S^1$
plays a special role in Theorems 3 and 4.  However this is not the
case.  One considers, in addition to the target space $Y$, a covering
space $Z$.  For $Y = S^1, Z$ is $\Bbb R$.
\medskip
We denote by $F$ the covering map of $Z$ to $Y$, i.e., $F$ is onto,
and every point in $Z$ has a neighbourhood $U$ such that $F$ is a
diffeomorphism of $U$ onto $F(U)$.
\endremark
\medskip

The proof of Theorem 3 extends to give the following:
\medskip
\proclaim{Theorem $3'$}  Any $u$ in VMO$(X,Y)$ which is homotopic
within VMO$(X,Y)\cap L^1$ to a constant map may be written
as
$$
u = F\circ \varphi\quad\text{for some $\varphi\in {VMO}(X,Z)$}.
$$
\endproclaim

In the proof the following inequality replaces (35); here we use a
Riemannian metric on $Y$ (and its lift to $Z$):  For $\alpha,
\beta\in Z$,
$$
\text{dist}(\alpha,\beta)\leq \text{dist}(F(\alpha), F(\beta)) + C
\text{dist}^2(\alpha,\beta).
$$
\medskip
Similarly, one has an extension of Theorem 4:
\medskip
\proclaim{Theorem $4'$}  There exists $\delta$ depending on $X,Y$ and
$Z$ such that if $u\in \text{BMO}(X,Y)$ and $\|u\|_{\text{BMO}}\leq
\delta$, then $u$ may be lifted to a map $\varphi \in
\text{BMO}(X,Z)$, i.e., $u = F\circ\varphi$ such that
$$
\|\varphi\|_{BMO} \leq 4 \|u\|_{BMO}.
$$
\endproclaim

\remark{\bf Remark 12}  We have carried out lifting for VMO or BMO
maps.  Can one do the same for Sobolev maps, say $u\in W^{s,p}(X,S^1)
$?  Some partial results are known (see F.~Bethuel and X.~Zheng [1],
F.~Demengel [1], P.~Mironescu in H.~Brezis [1]), also for $X$ with
boundary.  If $X$ is a bounded domain in $\Bbb R^n$ then the answer is
positive in the following cases:  (a)  $sp > n$ (by Sobolev embedding)
and (b) $s=1, p=2$, in any dimension.  However if $s = 1$ and $p < 2$,
the answer is sometimes negative.
\endremark

\remark{\bf Remark 13}  We now present a lifting result related to
both Theorems 3 and 4.  In doing so we consider the class $\scr C_\delta$
of maps $u\in\text{BMO}(X,S^1)$ such that
$$
d_0(u) = \text{dist}(u, {VMO}(X,\Bbb R^2)) \leq\delta\tag 48
$$
and such that there is continuous deformation $h(t,x)$,
$$
\cases  h\in C([0,1], BMO(X,S^1)\,\cap\, L^1)\text{ satisfying}\\
   h(0) = u, h(1) = \text{constant and}\\
d_0(h(t))\leq \delta\quad\forall t\in [0,1].\endcases  \tag 49
$$
\endremark

Recall that $d_0$ is equivalent to $M_0$ (see Lemma $3'$); in fact
$M_0 \leq d_0\leq A M_0$.
\medskip
\proclaim{Theorem 5}  There exist $\delta, C > 0$ depending only on
$X$ (and a Riemannian metric on it) such that for every $u\in
\scr C_\delta$, there is a lifting $\varphi\in\text{BMO}(X,\Bbb R)$ of $u$,
i.e.,
$$
u = e^{i\varphi}, \tag 50
$$
satisfying
$$
d_0(\varphi)\leq C d_0(u). \tag 51
$$
Here, $d_0(\varphi) = \text{dist}(\varphi,\text{VMO}(X,\Bbb R))$.
\endproclaim
We do not include the proof here.  It follows the lines of that of 
Theorem 4, but there are some additional technical points which the
reader is spared.
\bigskip
%+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
\noindent
{\bf APPENDIX A.  Some useful estimates on BMO, et al.}
\medskip
We present some simple facts about BMO and Sobolev maps on manifolds.
These are well known to people working on the subject, but are not all
easily found in the literature.  Unless stated otherwise, $X$ is
always assumed to be a connected compact Riemannian manifold without
boundary.
\medskip
\proclaim{Lemma A.1}  There exists a constant $C$ depending on $X$,
such that for every\newline $u\in \text{BMO}(X,\Bbb R)$,
$$
\|u\|_{L^1(X)} \leq C\|u\|_{BMO} + \big\vert\int_X u\big\vert.
\tag{A.1}
$$
\endproclaim

\demo{Proof}  We may suppose that $\dsize\int_X u = 0$.  We argue by
contradiction;  suppose there is no such $C$.  Then there is a
sequence $u_j$ with $\dsize\int_X u_j = 0$, such that
$$
\|u_j\|_{BMO} \rightarrow 0\quad\text{and \quad $\|u_j\|_{L^1} = 1$}.
\tag{A.2}
$$
Cover $X$ by a finite number of balls $B_i = B_{r_0/2}(x_i)$.  It
follows from the definition of BMO that a subsequence of the $u_j$
converges to a constant a.e., and in $L^1$, on each $B_i$.
Necessarily, the constant is the same for all $B_i$.  Since $\dsize\int_X
u_j = 0$, the constant must be zero.  This contradicts the second
assertion in (A.2).  \hfill\qed
\enddemo
\medskip

\proclaim{Lemma A.2}  Let $u\in \text{BMO} (X,\Bbb R^N)$ and let $F$
be a uniformly continuous mapping from $\Bbb R^N$ to $\Bbb R^D$.  Then
$F\circ u\in\text{BMO}(X,\Bbb R^D)$.
\endproclaim

We shall make use of
\medskip
\proclaim{Lemma A.3}  Let $F$ be a uniformly continuous map  of $\Bbb
R^N$ to $\Bbb R^D$.  Then $F$ has a concave modulus of continuity
$\omega$.
\endproclaim

\demo{Proof}  For $t > 0$, set
$$
\alpha(t) = \underset |a-b|< t\to\sup |F(a) - F(b)|.
$$
It is easy to verify that $\forall t_0 > 0,\;\;\exists A = A(t_0)$ such that
$$
\alpha(t)\leq At\quad\text{for $t\geq t_0$}.
$$
($A$ depends on the function $F$).  Hence $\alpha(t)\leq A t + \alpha(t_0)\;\;
\forall t \geq 0$.  Thus we may introduce the concave hull $\omega(t)$
of $\alpha(t)$, namely the least concave function $\geq\alpha$.  Since
$\omega(t)\leq A(t_0)t + \alpha(t_0)$ it follows that
$$
\underset t\searrow 0\to\lim \omega(t)\leq \alpha(t_0),
$$
and consequently this limit must be zero.\quad\hfill\qed
\enddemo

\demo{Proof of Lemma A.2}  We use $(1')$ and $(1'')$.  For any $B_\vare(x)$ in $X$, $\vare <
r_0$,
$$
\align
\Mint_{B_\vare(x)}\Mint_{B_\vare(x)} \vert F(u(y)) - F(u(z))\vert\
&\leq \Mint_{B_\vare (x)}\Mint_{B_\vare(x)} \omega(|u(y) - 
u(z)|)\\
&\leq \omega\left(\Mint_{B_\vare(x)}\Mint_{B_\vare(x)}  |u(y) -  u_(z)|\right)
\endalign
$$
by the concavity of $\omega$.  Hence, by $(1'')$,
$$
\|F\circ u\|_{BMO} \leq \|F\circ u\|_\star\leq
\omega(\|u\|_\star)\leq  \omega\left(2\|u\|_{BMO}\right).
$$
\enddemo

We shall often make use of the following simple

\proclaim{Lemma A.4}  Given two measurable sets $A\subset B$ in a
measure space, for any integrable function $f$
$$
\left\vert\Mint_A f - \Mint_B f\right\vert \leq \Mint_A \left\vert f -
\Mint_B f\right\vert\leq \frac{|B|}{|A|} \Mint_B \left\vert f -
\Mint_B f\right\vert\tag{A.3}
$$
$$
\Mint_A \left\vert f - \Mint_A f\right\vert,\;\; \Mint_B\left\vert f - \Mint_A
f\right\vert \leq 2 \frac{|B|}{|A|} \Mint_B\left\vert f - \Mint_B
f\right\vert. \tag{A.4}
$$
\endproclaim

We often refer to this as Lemma A-B.
\medskip
\demo{Proof}  Inequality (A.3) is obvious.  The first term in (A.4) is
bounded by
$$
\Mint_A\left\vert f - \Mint_B f\right\vert + \left\vert\Mint_B f - \Mint_A
f\right\vert\leq 2\frac{|B|}{|A|} \Mint_B \left\vert f - \Mint_B
f\right\vert,
$$
by (A.3).  Finally, the second term in (A.4) is bounded by
$$
\Mint_B\left\vert f - \Mint_B f\right\vert + \left\vert\Mint_B f - \Mint_A
f\right\vert \leq \left( 1 + \frac{|B|}{|A|}\right) \Mint_B \left\vert f -
\Mint_B f\right\vert.  \qed
$$
\enddemo
%++++texed to here 12/8/94.
\medskip
We now restate and prove Sarason's Lemma 3'.  Recall that for $u\in
\text{BMO}(X,\Bbb R^N)$ and $0 < a < r_0(X)$
$$
\align
M_a &= M_a(u) = \sup\Sb \vare\leq a\\ x\in X\endSb \Mint_{B_\vare(x)}
|u - \bar u_\vare(x)|\leq \|u\|_{BMO},\\
  M_0 &= M_0(u) = \underset a\searrow 0\to\lim\; M_a(u).
\endalign
$$
\medskip
\proclaim{Lemma A.5}  There is a constant $A$ depending on $X$ (and
its metric) such that if\newline $u\in \text{BMO}(X,\Bbb R^N)$, then
$$
M_0(u)\leq \text{dist$(u, VMO(X,\Bbb R^N))\leq A M_0(u)$}.\tag{A.5}
$$
In fact
$$
\|u - \bar
u_\vare\|_{BMO} \leq A M_\varepsilon(u)\quad\forall \varepsilon
< r_0. \tag{A.6}
$$
\endproclaim

\demo{Proof}  To prove the first inequality in (A.5), note that
$\forall u, v\in\text{BMO$(X,\Bbb R^N)$}$,
$$
M_a(u)\leq M_a(v) + M_a(u-v) \quad\forall a\in [0, r_0).
$$
For $v\in C^0(X,\Bbb R^N)$, $M_0(v) = 0$ and thus
$$
M_0(u)\leq M_0(u - v)\leq \|u - v\|_{BMO}.
$$
This yields the first inequality in (A.5).
\enddemo
\medskip

The other inequality in (A.5) is an obvious consequence of (A.6).  The
proof of (A.6) relies on the
following simple
\medskip
\proclaim{Lemma A.6}  There is a number $B$ depending only on $X$ such
that for any given numbers $\vare,\delta,\;0 < \vare\leq \delta <
r_0$, any ball $B_\delta(x)$ in $X$ may be covered by a finite number
of balls $B_\vare(x_i)$, $x_i\in B_\delta(x), i = 1,\ldots,K$, such
that dist$(x_i,x_j) \geq \vare$ for $i\not= j$, and
$$
\sum^K_1 |B_\vare (x_i)|\leq B|B_\delta(x)|. \tag{A.7}
$$
\endproclaim

\demo{Proof of Lemma A.6}  Let
$B_{\vare/2} (x_i), i = 1,\ldots,K$, be a maximal collection of
disjoint balls with centres $x_i$ in $B_\delta(x)$.  From the
maximality, it follows easily that
$$
\cup  B_\vare(x_i)\supset B_\delta(x).
$$
Since
$$
B_{\vare/2} (x_i)\subset B_{\delta + \vare/2} (x)\subset
B_{2\delta}(x),
$$
$$
\sum |B_{\vare/2} (x_i)|\leq |B_{2\delta}(x)|.
$$
Therefore
$$
\align
\sum |B_\vare(x_i)| &\leq C\sum_i |B_{\vare/2} (x_i)|\\
   &\leq C|B_{2\delta} (x)|\leq C|B_\delta (x)|. \qed
\endalign
$$
\enddemo
\medskip
We now return to the 
\smallskip
%++++++++++++++++++++new material(B.Mastrian)
\demo{Proof of (A.6)}  We first claim that
$$
|\bar u_\vare (y) - \bar u_\vare(z)| \leq C M_\vare (u)\;\;\forall
\vare < r_0,\;\;\forall y,z\in X\;\;\text{with $d(y,z) <
\vare/2$},\tag{A.8}
$$
where $C$ depends only on $X$.
\medskip
Indeed, by Lemma A-B (i.e., Lemma A.4), the following inequalities
hold:
$$
|\bar u_{\vare/2} (y) - \bar u_{\vare} (z)| \leq C M_\vare(u),
\tag{A.9}
$$
$$
|\bar u_{\vare/2} (x) - \bar u_\vare (x)|\leq C M_\vare(u)\quad\forall
\vare < r_0,\quad\forall x\in X. \tag{A.10}
$$
Their combination yields (A.8).
\medskip
For $\vare < r_0$ we have to estimate $\|u - \bar
u_\vare\|_{BMO}$; more precisely we want to show that for any
$B_\delta(x)\subset X$, $\delta < r_0$,
$$
I = \Mint_{B_\delta(x)} \bigg\vert (u - \bar u_\vare) -
\Mint_{B_\delta(x)} (u - \bar u_\vare)\bigg\vert \leq A M_\vare(u).
\tag{A.11}
$$
We distinguish two cases.
\medskip
\noindent
{\bf Case (i):}  $\delta < \vare/4$.  We have
$$
\align
I &\leq \Mint_{B_\delta(x)} |u - \Mint_{B_\delta(x)} u| +
\Mint_{B_\delta(x)} \Mint_{B_\delta(x)} |\bar u_\vare (y) - \bar
u_\vare (z)|\\
 &\leq M_\delta(u) + C M_\vare(u)\quad\text{by (A.8)},\\
 &\leq C M_\vare(u)\qquad\text{\quad since $\delta < \varepsilon$}.
\endalign
$$
\enddemo
\medskip
\noindent
{\bf Case (ii):}  $\delta \geq \vare/4$.  We now use the covering of
$B_\delta(x)$ by $B_{\vare/2} (x_i)$, $i = 1,\ldots, K$ given by Lemma
A.6.  Then
$$
I\leq 2 \sum^K_1 \frac{1}{|B_\delta(x)|} \int_{B_{\vare/2} (x_i)} |u -
\bar u_\vare|,
$$
so that
$$
\align
I&\leq \frac{2}{|B_\delta(x)|} \sum^K_1 \int_{B_{\vare/2} (x_i)} \left[
|u(y) - \bar u_{\vare/2} (x_i)| + |\bar u_{\vare/2} (x_i) - \bar
u_\vare(x_i)| + |\bar u_\vare(x_i) - \bar u_\vare(y)|\right]\\
   &\leq \frac{C}{|B_\delta(x)|} M_\vare(u) \sum^K_1 |B_{\vare/2}
(x_i)|\qquad\text{by (A.10) and (A.8)}.
\endalign
$$
Using Lemma A.6 we conclude that
$$
I \leq C M_\vare (u). \quad\hfill \qed
$$
\medskip
A simple consequence of (A.8) is the following:  For any two points
$y,z$ in $X$, with $d = \text{dist}(y,z)\geq \vare/2$,
$$
|\bar u_\vare (y) - \bar u_\vare (z)|\leq
C\frac{\text{dist}(y,z)}{\vare} M_\vare(u), \tag{$A.12$}
$$
where $C$ depends only on $X$.  Indeed there is a chain of points
$y,y_1,y_2,\ldots,y_{k-1}, z$, with $k=1+[\frac{2d}{\vare}]$, such
that the distance of any two successive ones is bounded by $\vare/2$.
Adding the corresponding inequalities (A.8) we obtain (A.12).   \quad\hfill\qed
\medskip
Returning to Lemma A.2  we know that $F\circ u$ is in BMO whenever
$u\in$ BMO and $F$ is uniformly continuous.  The same holds in VMO:
\medskip
\proclaim{Lemma A.7}  Let $u\in\text{VMO}(X,\Bbb R^N)$ and let $F$ be
a uniformly continuous map from $\Bbb R^N$ to $\Bbb R^D$.  Then
$$
F\circ u\in {VMO}(X,\Bbb R^D).
$$
\endproclaim

\demo{Proof}  Recall---see the proof of Lemma A.2---that
$$
\Mint_{B_\vare(x)}\Mint_{B_\vare(x)} \left\vert F(u(y) - F(u(z))
\right\vert \leq \omega\left(\Mint_{B_\vare(x)}\Mint_{B_\vare(x)}  |u(y) -  u(z)|\right).
$$
Since the right hand side goes to zero as $\vare\rightarrow 0$, we
conclude by Sarason's characterization of VMO that $F\circ u\in
\text{VMO}(X,\Bbb R^D)$.$\quad\hfill\qed$
\medskip
We now take up a more delicate property, namely the continuity of the
map $u\mapsto F(u)$.  The map $F$ induces a map $\Psi$ from BMO(VMO)
into BMO(VMO).  We show that $\Psi$ is continuous at every point
$u\in$ VMO.  We show also that $\Psi$ need {\bf not} be continuous
outside VMO.
\medskip

\proclaim{Lemma A.8}  Let $F$ be a uniformly continuous map from $\Bbb
R^N$ to $\Bbb R^D$.  Then $\Psi$ is continuous in the BMO$\; \cap\;  L^1$
topology at every point $u$ in VMO$(X,\Bbb R^N)$.
\endproclaim

\demo{Proof}  Let $u$ be in VMO$(X,\Bbb R^N)$.  Given $\vare$ we shall
show that there exists $\tau > 0$, depending on $\vare$ and $u$,
such that if $v\in BMO(X,\Bbb R^N)$ and $\|v\|_{BMO} +
\|v\|_{L^1} < \tau$, then
$$
\|F(u+v) - F(u)\|_{BMO} < \vare,
$$
i.e., we show that for any ball $B_\delta(x)$ in $X$, $\delta < r_0$,
$$
\align
J = \Mint_{B_\delta(x)}\Mint_{B_\delta(x)}  \vert F\left(u(y)+v(y)\right)
&- F(u(y))\\
 &- \left(F(u(z) + v(z)) - F(u(z))\right)\vert < \vare. \tag{$A.13$}
\endalign
$$
\medskip

For $\xi\in X$, $p\in\Bbb R^N$, set
$$
G(\xi,p) = F(u(\xi) + p) - F(u(\xi)).
$$
Given a measurable set $A$ in $X$, we wish to estimate
$$
I: = \Mint_A \Mint_A\left\vert G(y, v(y)) - G(z,v(z))\right\vert dz dy.
\tag{A.14}
$$
\medskip
We make use of the concave modulus of continuity $\omega$ of $F$ of
Lemma A.3 and establish two estimates for $I$:
$$
I\leq 2\omega\left(\Mint_A |v|\right)\tag{A.15}
$$
and
$$
I\leq 2\omega \left(2\Mint_A |u - \Mint_A u|\right) + \omega\left(2\Mint_A |v -
\Mint_A v|\right). \tag{A.16}
$$
\medskip
Since $G(\xi,p)\leq \omega(|p|)$, we have
$$
\align
I\leq 2\Mint_A |G(y,v(y))| &\leq 2\Mint_A \omega( |v(y)|)\\
  &\leq 2\omega\left(\Mint_A |v|\right)
\endalign
$$
by concavity of $\omega$.  We have proved (A.15). 
\medskip
To prove (A.16), we rely on an obvious inequality:
$$
|G(\xi,p) - G(\eta,q)|\leq 2\omega(|u(\xi) - u(\eta)|) +
\omega(|p-q|). \tag{A.17}
$$
\enddemo
\medskip
\noindent
Using (A.17) in (A.14) we find
$$
\align
I&\leq \Mint_A \Mint_A \left[ 2 \omega(|u(y) - u(z)|) + \omega(|v(y) -
v(z)|)\right] dz dy\\
  &\leq 2 \omega\left(\Mint_A\Mint_A |u(y) - u(z)|\right) +
\omega\left(\Mint_A\Mint_A |v(y) - v(z)|\right).
\endalign
$$
This yields (A.16) (as in $(1^{\prime\prime})$).
\medskip
Recall now that $\vare$ is fixed and we wish to find $\tau$.  Since
$u\in$ VMO, there exists $\delta_0 < r_0$, such that for every $\delta\leq
\delta_0$ and every $x\in X$,
$$
2\omega\left(2\Mint_{B_\delta(x)} \left\vert u - \Mint_{B_\delta(x)} u\right\vert \right) < \frac{\vare}{2}.
$$
\medskip
Hence, taking $A = B_\delta(x)$, we see from (A.16) that
$$
I\leq \frac{\vare}{2} + \omega\left(2\Mint_{B_\delta(x)}\left\vert v -
\Mint_{B_\delta(x)} v\right\vert \right)\leq \frac{\vare}{2} +
\omega(2\|v\|_{BMO}). \tag{A.18}
$$
\medskip
Now we explain how to choose $\tau$.  We first require that
$$
\omega(2\tau)\leq \vare/2, \tag{A.19}
$$
so that, by (A.18), $I\leq \vare$ whenever $\delta \leq \delta_0$.  It
remains to consider the case where $\delta > \delta_0$.  Here we use
(A.15) to conclude that
$$
I \leq 2\omega\left(\frac{\tau}{|B_{\delta_0}(x)|}\right)\leq
2\omega(\alpha\tau)
$$
where $\alpha = \underset x\in X\to{\operatorname{Sup}}
\frac{1}{|B_{\delta_0} (x)|}$.  Choosing $\tau > 0$ such that
$2\omega(\alpha\tau) < \vare$, we obtain the desired conclusion,
$I\leq \vare$, in both cases $(\delta\leq \delta_0$ and $\delta >
\delta_0)$.
\medskip
\remark{\bf Remark A.1}  The following example shows that the map
$\Psi$ of Lemma A.8 need {\bf not} be continuous at a point $u$ in
BMO$(X,\Bbb R^N)$.  Here we take $X = \Bbb R$,---it is not compact,
but our functions will all have support in $[-1,1]$.
\endremark
\medskip
Let $\sigma$ be a positive function in VMO with support in $[-1,1]$,
even in $x$, decreasing and continuous on $(0,1)$ and
$$
\operatornamewithlimits{\lim}_{x\to 0} \sigma(x) = \infty.
$$
Consider
$$
u(x) = \cases  \hphantom{- {}}0\quad\text{for $x < 0$}\\
     -1\quad\text{for $0 < x < \frac12$}\\
    \hphantom{- {}}0\quad\text{for $x > \frac12$}.
\endcases
$$
\medskip
Set $F(s) = s^+$, so that $u^+ = F(u)\equiv 0$.  $F$ is clearly
Lipschitz but we claim that for $v_j = \sigma/j$,
$$
\|F(u + v_j) - F(u)\|_{BMO} \geq \frac14\quad\text{for $j$
large}. \tag{A.20}
$$
To see this, observe that there is a unique $\delta_j , 0 < \delta_j <
1$, such that $\sigma(\delta_j) = j$.  Then
$$
F(u+v_j) = \cases  v_j&\quad\text{for $x < 0$},\\
    v_j-1&\quad\text{for $0 < x < \delta_j$},\\
    0&\quad\text{for $x > \delta_j$}.
\endcases
$$
Set
$$
J = \Mint^{\delta_j}_{-\delta_j} \left\vert F(u+v_j) -
\Mint^{\delta_j}_{- \delta_j}F(u + v_j) \right\vert.
$$
One checks that
$$
\Mint^{\delta_j}_{-\delta_j} F(u + v_j) = \frac1j \Mint^{\delta_j}_0
\sigma - \frac12 = \Mint^{\delta_j}_{-\delta_j} v_j - \frac12.
$$
Hence
$$
\align
J &= \Mint^{\delta_j}_{-\delta_j} \left\vert v_j + u -
\Mint^{\delta_j}_{-\delta_j} v_j + \frac12\right\vert\\
  &\geq \Mint^{\delta_j}_{-\delta_j} |u + \frac12 | -
\|v_j\|_{BMO}\\
  &= \frac12 - \|v_j\|_{BMO}
\endalign
$$
which yields (A.20).
\medskip
By a small modification of the $F$ above, we may prove that $\|F(u+v_j) -
F(u)\|_{\text{BMO}} \geq \frac18$ for $j$ large---for a {\it smooth}
function $F$ for which the Lipschitz constant is $1$.  Namely, take
$F$ to be any smooth nondecreasing function on $\Bbb R$ with $\dot
F\leq 1$ such that 
$$
F(s) = \cases  \hphantom{- {}}s\quad\text{for $s > 0$}\\
  - \dfrac18\quad\text{for $s < -1$}.
\endcases
$$
\medskip
Later we shall have need of an extension of Lemma A--B---where some
averages are taken with respect to a positive weight function $w$
satisfying
$$
\frac{1}{C_0} \leq w\leq C_0,
$$
for some constant $C_0 > 0$.  For any measurable set $\Sigma$, and
integrable function $f$ on $\Sigma$ we denote
$$
\Mint_{\Sigma,w} f := \left(\int_\Sigma w\right)^{-1} \int_\Sigma fw.
\tag{A.21}
$$
\medskip
\proclaim{Lemma A.9}  Let $A\subset B$ be measurable sets in $X$.
Then
$$
\Mint_B \left\vert f - \Mint_{A,w} f\right\vert \leq 4 C^2_0
\frac{|B|}{|A|}  \Mint_B \left\vert f - \Mint_B f\right\vert,
\tag{A.22}
$$
and
$$
\Mint_{A,w} |f - \Mint_{A,w}f |\leq 4 C_0^2 \frac{|B|}{|A|} \Mint_B
\left\vert f - \Mint_B f\right\vert.\tag{A.23}
$$
\endproclaim

\demo{Proof}  Using Lemma A-B repeatedly we find
$$
\align
\Mint_B \left\vert f - \Mint_{A,w} f\right\vert &\leq
\Mint_B\left\vert f - \Mint_A f\right\vert + \left\vert\Mint_A f -
\Mint_{A,w} f\right\vert\\
   &\leq 2 \frac{|B|}{|A|} \Mint_B \left\vert f - \Mint_B f\right\vert
+ \left\vert (\int_A w)^{-1} \int_A \left(f - \Mint_A f\right) w \right\vert\\
   &\leq 2\frac{|B|}{|A|} \Mint_B \left\vert f - \Mint_B f\right\vert +
C^2_0 \Mint_A \left\vert f - \Mint_A f\right\vert\\
   &\leq 2\frac{|B|}{|A|} (1 + C^2_0) \Mint_B \left\vert f - \Mint_B
f\right\vert.
\endalign
$$
This proves (A.22).
\medskip
Turning to (A.23), we have
$$
\align
\Mint_{A,w} \left\vert f - \Mint_{A,w} f\right\vert &\leq \Mint_{A,w}
\left\vert f - \Mint_A f\right\vert + \left\vert \Mint_A f -
\Mint_{A,w} f\right\vert\\
    &\leq C^2_0 \Mint_A \left\vert f - \Mint_A f\right\vert + C^2_0
\Mint_A \left\vert f - \Mint_A f\right\vert\\
\intertext{as above,}
    &\leq 4 C^2_0 \frac{|B|}{|A|} \Mint_B \left\vert f - \Mint_B
f\right\vert.          \qquad\hfill\qed\endalign
$$
\enddemo

\medskip
Next we establish the fact that the BMO notion is invariant under
$C^1$ diffeomorphism.  It is a simple but essential fact.
\medskip

\proclaim{Lemma A.10}  Let $\varphi: X_1\rightarrow X_2$ be a $C^1$
diffeomorphism of a smooth compact $n$-dimensional Riemannian manifold
without boundary onto another $X_2$.  If $f\in \text{BMO}(X_2)$ then
$f\circ\varphi\in \text{BMO}(X_1)$ and
$$
\|f\circ\varphi\|_{BMO(X_1)} \leq C \|f\|_{\text{BMO}(X_2)}.
\tag{A.24}
$$
Here $C$ depends only on the Riemannian manifolds $X_1,X_2$.
\endproclaim

\demo{Proof}  It is not difficult to verify that there are constants $\vare_0,
K > 0$ depending only on $X_1$ and $X_2$ such that for every $x\in X_1$,
and every $\vare < \vare_0$
$$
B_{\vare/K} (\varphi (x)) \subset \varphi (B_\vare(x))
\subset B_{\vare K} (\varphi (x)). \tag{A.25}
$$
Here $\vare_0$ is less than the injectivity radius of $X_1$ and $\vare_0K$ is
less than that of $X_2$.
\enddemo

\medskip
For $f\in \text{BMO}(X_2)$, set $g = f\circ \varphi$.  To prove that
$g\in \text{BMO}(X_1)$ it suffices, by Remark 1 in Section I.1, to
show that $\forall \xi\in X_1,\quad\forall \vare < \vare_0$
$$
\operatornamewithlimits{\sup}\Sb \vare < \vare_0\\ \xi\in X_1\endSb
\Mint_{B_\vare(\xi)} |g - \bar g_\vare(\xi)|\leq C\|f\|_{BMO}.
\tag{A.26}
$$
We proceed to estimate
$$
J = \Mint_{B_\vare(\xi)} |g - \bar g_\vare (\xi)|
$$
by changing variables; we require $\vare < \vare_0$.  Setting $A_\vare
= \varphi (B_\vare(\xi))$ we see easily that
$$
J = \frac{1}{|B_\vare(\xi)|} \int_{A_\vare} |f - \bar g_\vare (\xi)| w
= \Mint_{A_{\vare,w}} |f - \bar g_\vare (\xi)|,
$$
where $w$ is a smooth positive function obtained by the change of
variables.
\medskip
On the other hand
$$
\bar g_\vare(\xi) = \Mint_{B_\vare(\xi)} g = \frac{1}{|B_\vare(\xi)|}
\int_{A_\vare} f w d\eta = \Mint_{A_{\vare,w}} f
$$
since
$$
\int_{A_\vare} w = |B_\vare(\xi)|.
$$
Using Lemma A.9 and (A.25) we obtain the desired conclusion.\quad\hfill\qed
\medskip
Here is another proof of Lemma A.10.  In view of $(1')$ and $(1'')$ we
estimate
$$
\align
I &= \Mint_{B_\vare(x)} \Mint_{B_\vare(x)} |f(\varphi(y)) -
f(\varphi(z))|\\
  &\leq \frac{C}{|B_\vare(x)|^2} \int_{\varphi(B_\vare(x))}
\int_{\varphi(B_\vare(x))} |f(\eta) - f(\xi)|\\
    &\leq C\, \Mint_{B_{\vare K}(\varphi (x))} \Mint_{B_{\vare K}
(\varphi (x))} |f(\eta) - f(\xi)|,\quad\text{by (A.25)},\\
    &\leq C \|f\|_{BMO}. \endalign  
$$
$\quad\hfill\qed$
\medskip

Next we take up the proofs of two approximation lemmas of Section I.1:
Lemmas 7 and 8.  We restate them.
\medskip
Let $X,Y$ be our usual compact connected 
manifolds without boundaries.  $X$ has a Riemannian metric, dim$X = n$, and $Y$ is
smoothly embedded in $\Bbb R^N$.  (Here, $Y$ need not have the same
dimension as $X$.)
\medskip
\proclaim{Lemma A.11}  Given $u\in W^{1,n}(X,Y)$ i.e., $u\in
W^{1,n}(X,\Bbb R^N)$ and $u(x)\in Y$ a.e., there exists a sequence
$(u^j)$ of smooth maps from $X$ to $Y$ tending to $u$ in $W^{1,n}$.
\endproclaim

\demo{Proof}  We first construct a sequence $(u^j) \subset W^{1,n}
(X,Y)\cap C^0(X,Y)$ tending to $u$ in $W^{1,n}(X,\Bbb R^N)$.  Cover
$X$ by a finite number of balls $B_{r_0/4}(a_i) = B_i$.  Let $\zeta_i$
be a subordinate partition of unity on $X$.  Set
$u_i = \zeta_i u$; clearly $u_i\in W^{1,n}(X,\Bbb R^N)$.  Let
$\varphi$ be the smooth diffeomorphism (given by geodesic normal
coordinates) from the Euclidean ball $B_{r_0} (0)$ onto $B_{r_0}
(a_i), \varphi(0) = a_i$.  Denote by $v_i$ the transplant of $u_i$ to
$B_{r_0}(0)$, i.e.,
$$
v_i(\xi) = u_i(\varphi(\xi))\quad\text{for $\xi\in B_{r_0}(0)$}.
$$
$v_i$ has its support in $B_{r_0/4}(0)$ and belongs to $W^{1,n}$.
Denote by $\bar v_{i,\vare}(\xi)$ the Euclidean average of $v_i$ in
$B_\vare(\xi)$
$$
\bar v_{i,\vare}(\xi) = \Mint_{B_\vare(\xi)} v_i.
$$
It has support in $B_{r_0/2} (0)$ if $\vare < r_0/4$, which we always
assume.  It is well known that $\bar v_{i,\vare}\rightarrow v_i$ in
$W^{1,n}(\Bbb R^n)$.  Carrying back $\bar v_{i,\vare}$ to $X$, we set
$$
u_{i,\vare}(x) = \bar v_{i,\vare} (\varphi^{-1}(x)).
$$
It has support in $B_{r_0/2} (a_i)$, and converges to $u_i$ in
$W^{1,n}(X,\Bbb R^N)$.  Set
$$
u_\vare = \sum_i u_{i,\vare}. \tag{A.27}
$$
Clearly $u_\vare\in C^0(X,\Bbb R^N) \cap W^{1,n}$ and
$u_\vare\rightarrow u$ in $W^{1,n}$ as $\vare\rightarrow 0$.
\enddemo

\proclaim{Claim} 
$$
\text{dist}(u_\vare(x), Y)\rightarrow 0\quad\text{as $\vare\rightarrow
0$, uniformly in $x\in X$}. \tag{A.28}
$$
\endproclaim

The main ingredient in proving the claim is the assertion that
for every $i$,
$$
J_\vare(x) := \Mint_{B_\vare(x)} |u_i(y) - u_{i,\vare} (x)| d\sigma (y)
\rightarrow 0\quad\text{uniformly in $x\in X$}. \tag{A.29}
$$
\medskip
Assuming (A.29), the claim follows easily because
$$
\sum_i \Mint_{B_\vare(x)} |u_i(y) - u_{i,\vare}(x)|
d\sigma(y) \rightarrow 0.
$$
The left hand side majorizes
$$
\Mint_{B_\vare(x)} |u(y) - u_\vare(x)| d\sigma(y)\geq
\text{dist}(u_\vare(x), Y).
$$
Then (A.28) holds.
\medskip

\demo{Proof of (A.29)}  Note that $J_\vare(x) = 0$ if $x\notin
B_{r_0/2}(a_i)$.
\medskip
We may write
$$
u_{i,\vare}(x) = \frac{1}{|B_\vare(\xi)|} \int_{A_\vare(x)} u_i w
$$
where $A_\vare(x) = \varphi(B_\vare(\xi))\quad\text{and $\xi =
\varphi^{-1}(x)$}$.
 \medskip
Here $w$ is a smooth function coming from the change of variables, and
$$
\frac{1}{C_0} \leq w\leq C_0,
$$
with $C_0$ depending only on $X$.  Recall that
$$
\int_{A_\vare(x)} w = |B_\vare(\xi)|,
$$
so that, using the notation (A.21),
$$
u_{i,\vare}(x) = \Mint_{A_\vare,w} u_i. \tag{A.30}
$$
\medskip
As in the proof of Lemma A.10, there are constants $0 < \vare_1$ and $K
> 1$, such that
$$
B_{\vare/K}(x)\subset A_\vare(x)\subset B_{\vare K}(x)\quad\forall
x\in B_{r_0/2} (a_i),\quad\forall \vare < \vare_1,
$$
and we require that $\vare_1 K < r_0/4$.
\medskip
Returning to $J_\vare$, we have
$$
\align
J_\vare(x) &= \Mint_{B_\vare(x)} \left\vert u_i(y) -
\Mint_{A_\vare(x),w} u_i\right\vert\\
   &\leq C \Mint_{B_{\vare K}(x)} \left\vert u_i - \Mint_{A_\vare
(x),w} u_i\right\vert.
\endalign
$$
Applying (A.22) of Lemma A.9, we find
$$
J_\vare(x) \leq C\Mint_{B_{\vare K}(x)}\left\vert u_i -
\Mint_{B_{\vare K}(x)} u_i\right\vert.
$$
By Example 1 of Section I.2, $J_\vare(x)\rightarrow 0$ as
$\vare\rightarrow 0$ uniformly in $x\in X$.
\medskip
To summarize, we have a family $u_\vare\in W^{1,n}(X,\Bbb 
R^N)\cap C^0(X,\Bbb R^N)$ such that as $\vare\rightarrow 0$,
$u_\vare\rightarrow u$ in $W^{1,n}$ and dist$(u_\vare(x),
Y)\rightarrow 0$ uniformly.
\medskip
Using the projection $P$ onto closest point in $Y$, the functions
$$
\widetilde u_\vare = P u_\vare
$$
take values in $Y$ and tend to $u$ in $W^{1,n}$, since in general, as
is well known, $u\mapsto F(u)$ is continuous in $W^{1,p}$ if $F$ is
Lipschitz and smooth.  Letting $\vare\rightarrow 0$ through a sequence
$\vare_j$, the functions
$$
\widetilde u^j = \widetilde u_{\vare_j}
$$
belong to $C^0(X,Y)\cap W^{1,n}(X,Y)$ and tend to $u$ in $W^{1,n}$.
\medskip
For each fixed $j$ there is a smooth map $u^j$ from $X$ to $Y$ with
$$
\|u^j - \widetilde u^j\|_{W^{1,n}} + |u^j - \widetilde u^j|_{C^0} \leq
\frac1j
$$
---by standard smoothing and projection on $Y$.\quad\hfill\qed
\enddemo
\medskip
\remark{\bf Remark A.2}  R.~Schoen and K.~Uhlenbeck [1] use a slightly
different, but natural,\newline approach in their proof (for $n=2$).  Namely,
they embed $X$ in some $\Bbb R^M$, and extend $u$ to a tubular
neighbourhood of $X$ as constant on normals to $X$.  They then mollify
the above extension $\widetilde u$.
\endremark
\medskip
\remark{\bf Remark A.3}  In Lemma A.11 if $u$ is merely in
$W^{1,p}(X,Y)$, $1\leq p < n$ and also in VMO$(X,Y)$, then there
is a sequence $(u^j)$ of smooth maps of $X$ into $Y$, tending to $u$
in $W^{1,p}$ and in BMO.  This is proved in essentially the same way
as the lemma, using in addition Lemma B.8 below.  The latter is used
to ensure that $u_i = \zeta_i u\in$\,VMO.
\endremark
\medskip
Essentially the same argument as in the proof of Lemma A.11 yields the
following more general form:
\medskip
\proclaim{Lemma A.12}  Assume $u\in W^{s,p} (X,Y)$ with $sp = n$, $0 <
s < n$ ($s$ may or may not be an integer).  Then there exists a
sequence $(u^j)$ of smooth maps from $X$ to $Y$ tending to $u$ in
$W^{s,p}$.
\endproclaim

We use here Example 2 of Section I.2 instead of Example 1, and
standard properties of $W^{s,p}$.  A special case of Lemma A.12 occurs
in F.~Bethuel [2].
\medskip
\proclaim{Lemma A.13}  Assume $\Omega \subset\Bbb R^n$ is a smooth
bounded domain with connected boundary $\partial\Omega$.  Let $u\in
W^{1,n}(\Omega,\Bbb R^N)$ be such that
$$
u(\partial\Omega)\subset Y
$$
where $Y$ is, as usual, a compact connected manifold smoothly embedded
in $\Bbb R^N$.  Then there exists a sequence $(u_j)$ of smooth maps
from $\overline\Omega$ to $\Bbb R^N$ such that
$$
u_j(\partial\Omega)\subset Y\quad\forall j
$$
and
$$
u_j\rightarrow u\quad\text{in $W^{1,n}(\Omega,\Bbb R^N)$}.
$$
\endproclaim

\demo{Proof}  Set $\varphi = u_{|\partial\Omega}$, so that $\varphi\in
W^{1 - \frac1n,n} (\partial\Omega,Y)$.  Applying Lemma A.12 with $X =
\partial\Omega, s = 1 - \frac1n, p = n$ (note that dim$X = n-1$), we
obtain a sequence $(\varphi^j)$ of smooth maps from $\partial\Omega$
to $Y$ such that
$$
\varphi^j\rightarrow\varphi\quad\text{in $W^{1 - \frac1n, n}
(\partial\Omega,\Bbb R^N)$}.
$$
Let $v_j$ be the harmonic extension of $\varphi^j$ in $\Omega$.  It is
well known that
$$
v_j\rightarrow v\quad\text{in $W^{1,n}(\Omega,\Bbb R^N)$},
$$
where $v$ is the harmonic extension of $\varphi$ in $\Omega$.
\medskip
Since $u - v\in W^{1,n}_0 (\Omega,\Bbb R^N)$ there is a sequence
$(w_j)$ in $C^\infty_c (\Omega,\Bbb R^N)$ such that
$$
w_j\rightarrow (u - v)\quad\text{in $W^{1,n} (\Omega,\Bbb R^N)$}.
$$
The sequence $u_j = v_j + w_j$ has the required properties.\quad\hfill\qed
\enddemo

\medskip
\remark{\bf Remark A.4}  The same argument as in the proof of Lemma
A.13 shows that if\newline $u\in W^{1,p}(\Omega,\Bbb R^N)$ with $1\leq p < n$
and
$$
u(\partial \Omega) \subset Y
$$
with $u_{|\partial\Omega} \in \text{VMO} (\partial\Omega,Y)$, then
there exists a sequence $(u_j)$ of smooth maps from $\overline \Omega$
to $\Bbb R^N$ such that $u_j (\partial\Omega)\subset Y\quad\forall j$
and $u_j\rightarrow u$ in $W^{1,p}$.
\medskip
If we do not make the assumption that $u_{|\partial\Omega} \in
\text{VMO} (\partial\Omega,Y)$ then the conclusion may fail.  Here is
an example.  Let $\Omega = B_1\subset \Bbb R^3$, let $Y = S^1$ and
take $p=2$.  We use coordinates $x = (x_1, x_2, x_3) = (x', x_3)$.
Consider the map
$$
\varphi(x) = \frac{x'}{|x'|}\text{\quad defined on $\partial\Omega$,
with values in $S^1$}.
$$
It is smooth except at the north and south poles.  Near there, it
belongs to $W^{1,q}(\partial\Omega)$ for any $q < 2$; in particular,
it belongs to $H^{1/2} (\partial\Omega,\Bbb R^2)$---by Sobolev.  Hence
$\varphi$ admits an extension $u$ into $\Omega$, belonging to
$H^1(\Omega,\Bbb R^2)$.  Suppose there is a sequence of smooth maps $u_j\rightarrow u$
in $H^1(\Omega,\Bbb R^2)$ and $u_j: \partial\Omega\rightarrow S^1$.  Then
on some disc$D=\{x_3 = $ constant\}$\,\cap\,\Omega$,\newline $u_j\rightarrow u$ in
$H^1(D)$, and therefore $u\rightarrow u$ in $H^{1/2}(\partial D)$.
\medskip
By Theorem 1,
$$
\text{deg}(u_j,\partial D, S^1) = \deg(u,\partial D,
S^1)\quad\text{for $j$ large}.
$$
Clearly the right hand side equals $1$.  On the other hand the left
hand side is zero because $\partial D$ is the boundary of a spherical
cap on $\partial\Omega$, which is mapped by $u_j$ into $S^1$.
\medskip
The argument is related to one of R.~Schoen and K.~Uhlenbeck [1], in
which they prove that the map $x/|x|$ of $\Omega$ above into $S^2$,
cannot be $H^1$--approximated by smooth maps into $S^2$.
\endremark
%++++++++++++++++++++++++new part 1/5/95
\medskip
Returning to the function $M_t$ defined before Lemma $3'$, we prove a
useful fact:

\proclaim{Lemma A.14}  There is a constant $A$ {\it depending only on} $X$, such that for\newline $u\in\text{BMO}(X,\Bbb R^N)$,
$$
M_{2t} (u) \leq A\, M_t(u)\quad\forall t < r_0/2.
$$
\endproclaim

\demo{Proof}  In view of $(1')$, $(1'')$ we estimate, for $t\leq
\delta\leq 2t$,
$$
I = \Mint_{B_\delta(x)} \Mint_{B_\delta(x)} |u(y) - u(z)|.
$$
Using Lemma A.6, with $\vare = t$, we cover $B_\delta(x)$ by balls
$B_t(x_i)$, with dist$(x_i, x_j)\geq t$ for $i\not= j$, and
$$
\sum |B_t (x_i)|\leq C|B_\delta(x)|.
$$
Thus
$$
I\leq \frac{1}{|B_\delta(x)|^2} \left[ \sum_{i\not=j} \int_{B_t(x_i)}
\int_{B_t(x_j)} |u(y) - u(z)| + \sum_i \int_{B_t(x_i)} \int_{B_t(x_i)}
|u(y) - u(z)|\right].
$$
The last sum is bounded by
$$
2M_t(u) \sum_i|B_t(x_i)|^2 \leq C M_t(u)|B_\delta(x)|^2
$$
while the first sum is bounded by
$$
\align
\sum_{i\not=j} \int_{B_t(x_i)} \int_{B_t(x_j)}  \bigg[|u(y) -
\bar u_t(x_i)| + |\bar u_t(x_i) - \bar u_t(x_j)|
   + |\bar u_t(x_j) - u(z)|\bigg]\\
\leq 2 M_t(u)\sum_{i\not=j} |
B_t(x_i)|\; |B_t(x_j)| + \sum_{i\not=j} |\bar u_t(x_i) -
\bar u_t(x_j)|\; |B_t(x_i)|\; |B_t(x_j)|.
\endalign
$$
\medskip
\noindent
Since dist$(x_i, x_j) \leq 2\delta\leq 4t$, we find using
$(A.12)$ that the expression above is bounded by
$$
C M_t(u) \sum_{i\not=j} |B_t(x_i)|\; |B_t(x_j)|\leq C M_t(u)
|B_\delta(x)|^2.
$$

Thus we conclude that
$$
I \leq C\; M_t(u).   \qquad\hfill\qed
$$
\enddemo

\proclaim{Lemma A.15}  For any $u\in \text{BMO}(X,\Bbb R^N)$ the
function $t\mapsto M_t(u)$ is continuous in $[0,r_0)$.
\endproclaim

The proof is left to the reader.
\medskip

We now turn to the proof of Lemma 4 (the characterization of compact
sets in VMO), which we restate

\proclaim{Lemma A.16}  A set $\Cal F$ in VMO$(X,\Bbb R^N)$ is
relatively compact if and only if
$$
\lim_{\vare\to 0} M_\vare(u)\quad\text{holds uniformly in $u\in \Cal
F$}.
$$
\endproclaim

After the statement of Lemma 4 we proved $\Rightarrow$.  Now, we prove
$\Leftarrow$.
\medskip
It suffices to show that for any given $\delta > 0$, $\Cal F$ may be
covered by a finite number of balls in BMO of radius $\delta$.  We may
assume that
$$
\int_X u = 0\quad\forall u\in\Cal F.
$$
We denote by $\bar{\bar{u}}_\vare$ the $\vare$-averaging
iterated twice.  Applying Lemma $3'$ repeatedly we find
$$
\align
\|u - \bar{\bar{u}}_\vare\|_{BMO} &\leq \|u - \bar u_\vare
\|_{BMO} + \| \bar u_\vare - \bar{\bar{u}}_\vare\|_{BMO}\\
   &\leq A\, M_\vare(u) + A\, M_\vare (\bar u_\vare)\\
  &\leq A\, M_\vare (u) + A\, M_\vare(\bar u_\vare - u) + A\,
M_\vare(u)\\
  &\leq 2A\, M_\vare(u) + A \|\bar u_\vare - u\|_{BMO}
\leq (2A + A^2)\, M_\vare(u).
\endalign
$$
By our hypothesis there exists $\vare$, $0 < \vare < r_0$ such that
$$
\|u - \bar{\bar{u}}_\vare \|_{BMO} \leq 3
A\,M_\vare(u) \leq \delta/2\quad\forall u\in\Cal F.
$$
{\it Fix} this $\vare$.  In view of Remark 1,
$$
\|u\|_{BMO} \leq C\quad\forall u\in \Cal F
$$
and then by Lemma 1,
$$
\|u\|_{L^1}\leq C\quad\forall u\in\Cal F.
$$
(Here, and in what follows, all the constants $C$ depend on $\vare$,
which has been fixed.)  Consequently,
$$
\|\bar u_\vare\|_{L^\infty} +
\|\bar{\bar{u}}_\vare\|_{L^\infty} \leq C\quad\forall
u\in\Cal F. \tag{A.31}
$$
\medskip

We now claim that the family $(\bar{\bar{u}}_\vare)$,
$u\in\Cal F$, satisfies the conditions of the Arzela-Ascoli theorem;
more precisely, there is a constant $C$ such that
$$
J = |\bar{\bar{u}}_\vare(x) - \bar{\bar
{u}}_\vare(y)|\leq C\; \text{dist}(x,y)\quad\forall x,y\in X,\quad\forall
u\in \Cal F. \tag{A.32}
$$


\demo{Proof of (A.32)}  We distinguish two cases:
\medskip
\noindent
{\bf Case 1:}  dist$(x,y) \geq \vare/100$.  In this case
$$
J\leq 2 \|\bar{\bar{u}}_\vare \|_{L^\infty} \leq C.
$$
\medskip
\noindent
{\bf Case 2:}  dist$(x,y) < \vare /100$.  Then
$$
\left\vert \frac{1}{|B_\vare(x)|} - \frac{1}{|B_\vare(y)|}\right\vert
\leq C\; \text{dist}(x,y).
$$
Thus
$$
J\leq \int_{B_\vare(x)} |\bar u_\vare (z)|
\bigg\vert\frac{1}{|B_\vare(x)|} - \frac{1}{|B_\vare(y)|}\bigg\vert +
\frac{1}{|B_\vare(y)|} \bigg\vert \int_{B_\vare(x)} \bar
u_\vare(z) - \int_{B_\vare(y)} \bar u_\vare(z)\bigg\vert.
$$
We have only to estimate the last term, $K$.  Let $S$ be the symmetric
difference of $B_\vare(x)$ and $B_\vare(y)$, i.e., $S = (B_\vare(x)\cup
B_\vare(y))\backslash (B_\vare(x)\cap B_\vare(y))$.  Clearly,
$$
S\subset \left(B_{\vare+\text{dist}(x,y)} (x)\backslash
B_\vare(x)\right) \cup \left(B_{\vare +\text{dist}(x,y)} (y)\backslash
B_\vare(y)\right).
$$
Hence
$$
|S|\leq C \text{ dist}(x,y)
$$
and therefore
$$
K \leq C |S|\, \|\bar u_\vare \|_{L^\infty} \leq C\text{
dist}(x,y)\quad\text{by (A.31)}.
$$
The desired inequality (A.32) follows by combining this with the
earlier estimate.
\medskip
Returning to the proof of Lemma A.16, we may now assert that the
family $(\bar{\bar{u}}_\vare)$, $u\in\Cal F$, is relatively
compact in $C^0(X,\Bbb R^N)$ and thus in BMO$(X,\Bbb R^N)$.  We may
cover the family $(\bar{\bar{u}}_\vare)$ by a finite number
of balls in BMO$(X,\Bbb R^N)$ of radius $\delta/2$.  The concentric
balls of radius $\delta$ then cover $\Cal F$. $\quad\hfill\qed$
\medskip
We mention a simple application of Lemma 3 to the sequence of truncates.  Given a real-valued function $f$ on $X$ and an integer $k$, set
$$
f^k(x) = \cases k&\qquad\text{if $f(x) \geq k$}\\
   f(x)&\qquad\text{if $-k < f(x) < k$}\\
   -k&\qquad\text{if $f(x)\leq -k$}.
\endcases
$$
\enddemo
\medskip
\proclaim{Lemma A.17}  For every $f\in \text{VMO}(X,\Bbb R)$ the
sequence $(f^k)$ converges to $f$ in BMO.
\endproclaim
\demo{Proof}  Given any $\delta > 0$ we will show that there exists
$k_0$ such that
$$
\|f - f^k\|_{BMO} < \delta\quad\text{for $k > k_0$}.
$$
Consider any $B_\vare (x)$ in $X$.  We have
$$
\Mint_{B_\vare (x)} \,\Mint_{B_\vare (x)}\,|f^k(y) - f^k(z)| \leq
\Mint_{B_\vare(x)}\, \Mint_{B_\vare(x)} \,|f(y) - f(z)|\qquad\forall
k.
$$
It follows, with the aid of Lemma 3, that there exists $\vare_0 > 0$,
such that
$$
\Mint_{B_\vare(x)} |(f - f^k) - \Mint_{B_\vare(x)} (f - f^k)| <
\delta\quad\text{for $\vare < \vare_0,\quad\forall k$}.
$$
Since $f^k\rightarrow f$ in $L^1$, the same inequality holds for
$\vare \geq \vare_0$ provided $k\geq k_0$, for some $k_0$ depending on
$\vare_0$.
\enddemo
\medskip

\remark{\bf Remark A.5}  If $f\in\text{BMO}(X,\Bbb R)$ then $(f^k)$ need
{\it not} converge to $f$ in BMO.  (It is easy to construct an example
using $f(x) = \log|x|$; recall that this $f$ belongs to BMO, but not
VMO---see Example 3 in Section I.2).
\endremark

\medskip
We now present various results concerning homotopy properties for BMO
and VMO maps.  They are used in the proofs of Theorems 3 and 4, as well
as in paragraph 1 of Section I.5.  Let $X,Y$ be our usual compact
connected manifolds with $X$ Riemannian ($X$ and $Y$ need not have the
same dimension.)
\proclaim{Lemma A.18}  There exists $\delta >0$ (depending on $X,Y$)
such that for every $u\in$ BMO$(X,Y$) with $\| u \|_{BMO}
<\delta$ and every $\varepsilon \in (0,r_0), u_\varepsilon$ is
homotopic within $C^0(X,Y)$ to a constant.
\endproclaim

\demo{Proof}  First observe that
$$
\text{dist}\left(\bar u_\varepsilon(x),Y\right)\leq \|u\|_{BMO}\quad \forall
x\in X,\quad \forall\varepsilon\in(0,r_0)\tag{A.33}
$$
and thus $u_\varepsilon=P\bar u_\vare$ is well defined for every
$\varepsilon\in(0,r_0)$, provided $\|u\|_{BMO}\leq \delta <
\delta_0$, with $\delta_0$ sufficiently small.  Moreover, $u_\vare$ is
homotopic to $u_{\vare'}$ within $C^0(X,Y)$ for every
$\varepsilon,\varepsilon' \in(0,r_0)$, using the deformation
$u_{t\varepsilon+(1-t)\varepsilon'}, 0\leq t\leq 1$.
\medskip
{\it Fix}  any $\varepsilon\in (0,r_0)$ - for example
$\varepsilon=r_0/2$.  We have, by Lemma A.1,
$$
|\bar u_\varepsilon(x) - \Mint_X u|\leq{1\over |B_\varepsilon(x)|}\int_X
|u-\Mint_X u|\leq C\|u\|_{BMO};\tag{A.34}
$$
here $C$ also depends on $\varepsilon$, but $\varepsilon$ has been
fixed and we do not stress the $\varepsilon$-dependence.  Combining
(A.33) and (A.34) we obtain 
$$
\text{dist}\left((1-t)\bar u_\varepsilon(x) + t\Mint_X u,\; Y\right)
\leq \left(C+1\right)\|u\|_{BMO}\quad \forall x\in X,\quad\forall t\in
[0,1].
$$
Thus
$$
P\left( (1-t)\bar u_\varepsilon + t\Mint_X u \right) , \, 0\leq
t\leq 1,
$$
is well defined provided $\|u\|_{BMO}\leq \delta <\delta_1$
with $\delta_1$ sufficiently small.  Hence $u_\varepsilon$ is
homotopic within $C^0(X,Y)$ to a constant, via $t\in[0,1]$.
\enddemo  
\proclaim{Lemma A.19}  Given $u\in \text{VMO}(X,Y)$ there exists
$\delta=\delta(u) >0$ such that every\newline $v\in \text{VMO}(X,Y)$
satisfying
$$
\|v-u\|_{BMO} + \|v-u\|_{L^1}<\delta
$$
is homotopic to $u$ within VMO$\,\cap\, L^1$.  Moreover $\delta$ is
uniform when $u$ lies in a compact subset $\Cal F$ of VMO$\,\cap\,L^1$.
\endproclaim
\demo{Proof}  We argue by contradiction and assume that there is a
sequence $(u_j)$ in VMO$(X, Y)$ such that $u_j\to u$ in VMO$\,\cap\,L^1$ and
each $u_j$ is not homotopic to $u$ within VMO$\,\cap\, L^1$.
\medskip
In view of Lemma 4 and (7) we know that there is some
$\varepsilon_0 > 0$ such that $u_{j,\varepsilon}$ is well defined for
every $j$ and every $\varepsilon < \varepsilon_0$.  {\it Fix} any
$\varepsilon < \varepsilon_0$.  Since $u_j\to u$ in $L^1$, we deduce
that $u_{j,\varepsilon} \to u_\varepsilon$ uniformly, as $j\to\infty$.
In particular, for $j$ large, $u_{j,\varepsilon}$ is homotopic to
$u_\varepsilon$ within $C^0(X,Y)$---and thus within VMO$\,\cap\, L^1$.
\medskip
On the other hand, $u_\varepsilon$ is homotopic to $u$ within  VMO$\,\cap\, L^1$ (through $u_{t\varepsilon}$, by Corollary 4), and similarly
$u_{j,\varepsilon}$ is homotopic to $u_j$ within VMO$\,\cap\, L^1$.
Therefore $u_j$ is homotopic to $u$ within VMO$\,\cap\, L^1$. A
contradiction.
\medskip
The fact that $\delta$ is uniform when $u\in \Cal F$ is easy to
establish by contradiction.  If not, there would exist equences $(u_j)$
in $\Cal F$ and $(v_j)$ in VMO such that
$$
\|v_j -u_j\|_{BMO} + \|v_j -u_j\|_{L^1} \to 0
$$  
and, for each $j, u_j$ is not homotopic to $v_j$ within VMO$\,\cap
\,L^1$.
\medskip
Since $\Cal F$ is compact we may assume, for a subsequence, that
$u_j\to u$ and $v_j \to u$ in VMO$\,\cap\, L^1$.  From the first
assertion in the lemma we deduce that $u_j$ and $v_j$ are homotopic to
$u$ within VMO$\,\cap\, L^1$, for $j$ large.  A contradiction.\quad\hfill\qed
\enddemo
\proclaim{Lemma A.20}  Assume $u,v\in C^0(X,Y)$ are homotopic within
VMO$\,\cap\,L^1$. Then they are homotopic within $C^0(X,Y)$.
\endproclaim
\demo{Proof}  Let $H(t)$ be a homotopy connecting $u$ and $v$ within
VMO$\,\cap\, L^1$.  Since $\Cal F=H ([0,1])$ is compact in VMO, we know
by Lemma 4 that $H(t)_\varepsilon$ is well defined for all
$\varepsilon < \varepsilon_0$ and all $t\in [0,1]$.  Fix any
$\varepsilon<\varepsilon_0$.  Since $H\in C([0,1]),L^1)$ we deduce
that $H(t)_\varepsilon$ is a homotopy connecting $u_\varepsilon$ to
$v_\varepsilon$ within $C^0(X,Y)$.  On the other hand, $u_\varepsilon$
is homotopic to $u$ within $C^0(X,Y)$ (via $u_{t\varepsilon}$), and
similarly for $v$ and $v_\varepsilon$.  Thus, $u$ and $v$ are
homotopic within $C^0(X, Y)$.
\enddemo
\medskip
Given a homotopy class $\Cal C$ in $C^0(X,Y)$ we denote by $\overline{\Cal
C}$ its closure in the VMO$\,\cap\, L^1$ topology.
\medskip
\proclaim{Lemma A.21}  If $u,v\in \overline{\Cal C}$, then $u$ is homotopic
to $v$ within VMO$\,\cap\, L^1$.
Conversely, if $u,v\in \text{VMO}(X,Y)$ are homotopic within VMO$\,\cap\, L^1$,
then there exists a unique homotopy class $\Cal C$ in
$C^0(X, Y)$ such
that $u,v\in \overline{\Cal C}$.
\endproclaim
\demo{Proof}  The first assertion is clear from Lemma A.19.  We turn
to the proof of the converse.  Let $(u_j)$ be a sequence in
$C^0(X,Y)$
such that $u_j\to u$ in VMO$\,\cap\, L^1$ (we may for example take
$u_\varepsilon$ with $\varepsilon = 1/j$).  Similarly, let $(v_j)$ be a
sequence in $C^0(X,Y)$ such that $v_j\to v$ in VMO$\,\cap\,
L^1$.  Applying Lemma A.19 we see that $u_j$ is homotopic to $u$
within VMO$\,\cap\, L^1$ for all $j\geq N$.  Similarly, $v_k$ is
homotopic to $v$ within VMO$\,\cap\, L^1$ for all $k\geq N$. Hence $u_j$
is homotopic to $v_k$ for all $j,k\geq N$, within VMO$\,\cap\, L^1$.  We
deduce from Lemma A.20 that $u_j$ and $v_k$ are also homotopic within
$C^0(X,Y)$.  Consequently there is a homotopy class $\Cal
C$ in $C^0(X,Y)$ such that $u_j,v_j\in \Cal C \quad\forall j\geq
N$.  Thus $u,v\in \bar\Cal C$.
\medskip
Finally we prove the uniqueness of $\Cal C$.  It suffices to show
that if $\Cal C_1$ and $C_2$ are two homotopy classes in
$C^0(X,Y)$ such that $\overline{\Cal C}_1\cap \overline{ \Cal C}_2\neq
\phi$, then $\Cal C_1 = \Cal C_2$.  Let $u\in \overline{\Cal C}_1 \cap
\overline{\Cal C}_2$ and let $(u_j)\subset \Cal C_1, (v_j)\subset \Cal C_2$
be sequences such that $u_j\to u$ in VMO$\,\cap\, L^1, v_j\to u$ in VMO$\,\cap\, L^1$.  In view of Lemma A.19, we may assume that $u_j$ and $v_j$
are homotopic to $u$ within\newline VMO$\,\cap\, L^1$ for all $j$. By
Lemma A.20 we know that $u_j$ and $v_j$ are homotopic within
$C^0(X,Y)$, i.e., $\Cal C_1 =\Cal C_2$.  \quad\hfill\qed
\enddemo
\medskip
In what follows we consider the special case where $Y =
S^k,k\geq 1$, and we show that some of the properties concerning
homotopy can be improved.  One suprising fact is that the notion of
``homotopy within VMO'' is {\bf equivalent} to the notion of
``homotopy within VMO$\,\cap\, L^1$'' (see Lemma A.23).
\medskip
Throughout the rest of Appendix A we take $Y = S^k,k\geq
1$.  A basic ingredient is

\proclaim{Lemma A.22}  Let $u\in$ VMO$(X,Y)$ be such that
for some constant $c\neq 0,\, u+c\in Y$ a.e.  Then $u$ is
homotopic to a constant within VMO$\cap L^1$.  In particular, $u$ is
homotopic to $(u+c)$ within VMO$\,\cap\, L^1$.
\endproclaim

\demo{Proof}  Set $\Sigma = Y \cap (Y - c)$; since
$\Sigma\neq Y$, it is contractible to a point in $Y$, i.e., there is a
continuous map $h(t,\sigma):[0,1] \times \Sigma\to Y$ such that  $h(0,\sigma)=\sigma \;\;\forall \sigma \in \Sigma$
and $h(1,\sigma)$ is a constant.
\medskip 
Set
$$
H(t,x) = h(t,u(x)).
$$
It is easy to verify (using Lemma A.7) that 
$$H\in C([0,1], {VMO} \cap L^1);
$$
moreover $H(0,x) = u(x)$ and $H(1,x)$ is a constant.
\enddemo
\medskip
Next, an improvement of Lemma A.19.

\proclaim{Lemma A.19$'$}  Given $u\in\text{VMO}(X,Y)$
there exists $\delta=\delta(u) > 0$ such that every\newline $v\in$ VMO
$(X,Y)$ satisfying
$$
\|v-u\|_{BMO}< \delta
$$
is homotopic to $u$ within VMO$\,\cap\, L^1$.  Moreover $\delta$ is
uniform when $u$ lies in a compact subset $\Cal F$ of VMO.
\endproclaim

\demo{Proof}  We argue by contradiction and assume that there is a
sequence $(u_j)$ in VMO$(X,Y)$ such that $u_j\to u$ in VMO
and each $u_j$ is not homotopic to $u$ within VMO$\,\cap\, L^1$.
\enddemo
\medskip
Set
$$c_j = \Mint_X(u_j-u).
$$
Passing to a subsequence we may assume that $c_j\to c$.  Then, by
Lemma 1, $u_j\to u+c$ in $L^1$.

In view of Lemma 4 and (7) we know that there is some
$\varepsilon_0>0$ such that $u_{j,\varepsilon}$ is well defined for
every $j$ and every $\varepsilon<\varepsilon_0$.  {\it Fix}  any
$\varepsilon<\varepsilon_0$.  Since $u_j\to (u+c)$ in $L^1$ we deduce
that $u_{j,\varepsilon}\to (u+c)_\varepsilon$ uniformly as $j\to
\infty$.
\medskip
In particular, for $j$ large, $u_{j,\varepsilon}$ is homotopic to
$(u+c)_\varepsilon$ within $C^0(X,Y)$---and thus within
VMO$\,\cap\, L^1$.
\medskip
On the other hand, $(u+c)_\varepsilon$ is homotopic to $(u+c)$ within
VMO$\cap L^1$ and similarly $u_{j,\varepsilon}$ is homotopic to $u_j$
within VMO$\cap L^1$.  Therefore $u_j$ is homotopic to $(u+c)$ within
\newline VMO$\,\cap\, L^1$ for $j$ large.
\medskip
Finally, we apply Lemma A.22 to assert that $(u+c)$ is homotopic to
$u$ within\newline VMO$\cap L^1$ (this is also true when $c=0$!).  Hence
$u_j$ is homotopic to $u$ within VMO$\,\cap\, L^1$ for $j$ large.  A
contradiction.
\medskip
The fact that $\delta$ is uniform when $u\in \Cal F,$ a compact subset
of VMO, is derived as in the proof of Lemma A.19.$\qquad\hfill\qed$

\proclaim{Lemma A.23}  Assume $u,v\in$ VMO$(X,Y)$ are
homotopic within VMO.  Then $u,v$ are homotopic within VMO$\,\cap\, L^1$.
\endproclaim  
\demo{Proof}  Let $H(t)$ be a homotopy connecting $u$ to $v$ within
VMO.  Let $\Cal F=H([0,1])$, so that $\Cal F$ is compact subset of
VMO. Let $\delta$ be as in Lemma A.$19'$ (relative to $\Cal F$).
There is a chain $u=u_0,u_1, \dots, u_k,u_{k+1} = v$ in $\Cal F$ such
that $\|u_{i+1} - u_i\|_{\text{BMO}} <\delta$ for $i=0,1,\dots,k$.
Thus $u_{i+1}$ is homotopic to $u_i$ within VMO$\,\cap\, L^1$ for
$i=0,1,\dots,k$.  Consequently $v$ is homotopic to $u$ within VMO$\,\cap\, L^1$.
\enddemo

\remark{\bf Remark A.6}  The conclusion of Lemma A.23 may fail for a
general manifold $Y$.  Consider, for example, a manifold
$Y$ lying in $\Bbb R^3$ diffeomorphic to a $2$-$d$ torus $T^2$.
Assume that $Y$ contains a circle $\Sigma$ contractible to a
point in $Y$ and another circle $\Sigma^\prime= \Sigma +c$ (for
some constant $c$) such that $\Sigma^\prime$ in {\bf not} contractible
to a point in $Y$.  Let $X=\Sigma$; the map $u(x)=x$ is
homotopic to a constant within  VMO$\,\cap\,L^1$ and the map $v(x)= x+c$
(viewed as a map from $X$ into $Y$) is not homotopic to a
constant within VMO$\,\cap\,L^1$ (by Lemma A.20). On the other hand $u$
and $v$ are clearly homotopic---in fact they are the same---within VMO.
\endremark
\proclaim{Lemma A.24}  Let $\Cal C$ be a homotopy class in
$C^0(X,Y)$.  Then its closure $\overline{\Cal C}$ in VMO$\,\cap\,
L^1$ coincides with its closure $\widetilde{\Cal C}$ in VMO.
\endproclaim 
\demo{Proof}  Clearly $\overline{\Cal C}\subset \widetilde{\Cal C}$.  To prove the
reverse inclusion, consider some $u\in \widetilde{\Cal C}$.  There is a
sequence $(u_j)$ in $\Cal C$ such that $u_j\to u$ in VMO.  By Lemma
A.$19'$, $u_j$ is homotopic to $u$ within VMO$\,\cap\, L^1$ for $j$
sufficiently large.  Applying Lemma A.21 we conclude that $u\in \bar
\Cal C$.
\enddemo
\bigskip

\enddemo
\subhead  APPENDIX B.  John-Nirenberg inequality on manifolds, et
al\endsubhead
\medskip

We begin by stating of the John-Nirenberg inequality on a cube $Q_0$
in $\Bbb R^n$ (with edges parallel to the axes).  It is inequality
$(3)''$ in John-Nirenberg [1]:
\medskip
There exist $\beta, A > 0$ depending only on $n$ such that if
$u\in BMO(Q_0,\Bbb R^N)$ and\newline $\|u\|_{BMO(Q_0)} \leq 1$, then
$$
\int_{Q_0} \left( e^{\beta |u - \bar{u}_{Q_0}|} - 1\right)\leq A
\int_{Q_0} |u - \bar{u}_{Q_0}| \tag{B.1}
$$
where $\bar{u}_{Q_0} = \dsize\Mint_{Q_0} u$.
\medskip
Here, $\|\,\,\|_{BMO(Q_0)}$ refers to the sup in (1) taken over
all parallel subcubes rather than balls.
\medskip
An immediate consequence is: for $p$ an integer $\geq 2$,
$$
\int_{Q_0} |u - \bar{u}_{Q_0}|^p \leq \frac{A}{\beta^p} p!
\int_{Q_0} |u - \bar{u}_{Q_0}|.
$$
The scaled version of this is
$$
\align
\int_{Q_0} |u - \bar{u}_{Q_0}|^p &\leq \frac{Ap!}{\beta^p}
\big\|u\big\|^{p-1}_{BMO(Q_0)} \int_{Q_0} |u - \bar{u}_{Q_0}|\\
   &\leq C^p p! \big\|u\big\|^{p-1}_{BMO(Q_0)}\int_{Q_0} | u - \bar{u}_{Q_0}|.\tag{B.2} \endalign
$$
with $C$ depending only on $n$.  It follows directly that, for a
different $C$, depending only on $n$, 
$$
\int_{Q_0} \int_{Q_0} |u(y) - u(z)|^p \leq C^p p!
\big\|u\big\|^{p-1}_{BMO(Q_0)} \int_{Q_0} \int_{Q_0} |u(y) - u(z)|.
\tag{B.3}
$$
\medskip
We wish next to present corresponding inequalities on a compact
manifold $X$, without boundary.  Here is a form of (B.3) on $X$.
\medskip
\proclaim{Lemma B.1}  There exists a constant $A$ depending only on
$X$, such that $\forall t < r_0/\sqrt{n}$,\newline $\forall x\in X$,
$$
\int_{B_t(x)} \int_{B_t(x)} |u(y) - u(z)|^p \leq A^p p!\;M^{p-1}_t (u)
\int_{B_{kt}(x)} \int_{B_{kt}(x)} |u(y) - u(z)| \tag{B.4}
$$
where $k = \sqrt{n}$.
\endproclaim

\demo{Proof}  We use geodesic normal coordinates in $B_t(x)$.  Then
one easily sees that
$$
I = \int_{B_t(x)} \int_{B_t(x)} |u(y) - u(z)|^p \leq C \int_Q \int_Q
|\tilde u (y) - \tilde u(z)|^p
$$
where in the right hand side $Q$ is a cube centred at the origin (in
our coordinate patch) with side length $2t$, and $\tilde u(y)$
represents the transplanted function.  Here $C$ comes from the change
of variables, and depends only on $X$.  By (B.3),
$$
I\leq C^p p! \big\|\tilde u\big\|^{p-1}_{BMO(Q)}\, \int_Q \int_Q
|\tilde u(y) - \tilde u(z)|. \tag{B.5}
$$
\enddemo

\proclaim{Claim}  There is a constant $C$ depending only on $X$ such
that
$$
\|\tilde u\|_{BMO(Q)} \leq C\, M_t(u). \tag{B.6}
$$
\endproclaim

Assuming that (B.6) holds, the proof of (B.4) is easily completed,
since $Q$ is contained in the transplant of $B_{kt}(x)$ so that
$$
\int_Q \int_Q |\tilde u(y) - \tilde u(z) |\leq C\, \int_{B_{kt}(x)}
\int_{B_{kt}(x)} |u(y) - u(z)|.
$$

\demo{Proof of (B.6)}  We have to estimate for any parallel subcube
$\widetilde Q$ of $Q$, centred at $\xi$, of side length $2\tau$,
$$
J = \Mint_{\widetilde Q} \Mint_{\widetilde Q} |\tilde u(y) - \tilde
u(z)|
$$
in terms of $M_t(u)$.  $\widetilde Q$ is contained in a ball
$B_{k\tau} (\xi)$.  Its transplant to $X$ is contained in a ball
$B_{K\tau}(\tilde x)$ and contains $B_{\tau/K}(\tilde x)$ where
$\tilde x$ is  the transplant of $\xi$ and $K$ is a constant depending
only on $X$ (see Proof of Lemma A.10).  Hence
$$
J\leq C \Mint_{B_{K\tau}(\tilde x)} \Mint_{B_{K\tau}(\tilde x)} |u(y) -
u(z)| \leq 2 C\, M_{K\tau}(u). \tag{B.7}
$$
where $C$ depends only on $X$.  
\medskip
By Lemma A.14 applied a number of times, we find that
$$
J\leq C\, M_\tau(u).
$$
Inserting this in (B.7) we obtain (B.6).
\medskip
Lemma B.1 is proved.
\enddemo
\medskip
The next result is a more global form of Lemma B.1.
\medskip
\proclaim{Lemma B.2}  For every $t\leq r_0/\sqrt{n}$, and for every
integer $p > 1$, there is a finite number of balls $B_t(x_i)$ in $X$,
$i = 1,\ldots, m,$ depending on $t$, such that
$$
\int_X |u|^p \leq A^p p! M_t^{p-1} (u) \int_X |u| + 2^p \sum_i
\frac{1}{|B_t(x_i)|^{p-1}} \bigg\vert \int_{B_t(x_i)} u\bigg\vert^p
\tag{B.8}
$$
where $A$ depends only on $X$.
\endproclaim
\medskip
Before proving Lemma B.2 we present some more civilized corollaries.
\medskip
\proclaim{Lemma B.3}  There is a constant $A$ depending only on $X$
such that for $p > 1$, an integer,
$$
\int_X \big\vert u - \Mint_X u\big\vert^p \leq A^p p! \big\|u\big\|^{p-1}_{BMO} \big\|u -
\Mint_X u\big\|_{L^1}.
$$
\endproclaim

\demo{Proof}  We may assume $\int_X u = 0$.  In (B.8) we choose $t =
r_0/\sqrt{n}$.  Then we have
$$
\int_X |u|^p \leq A^p p! \|u\|^{p-1}_{BMO} \big\|u\big\|_{L^1} + A^p \big\|u\big\|^p_{L^1}.
$$
The desired conclusion follows with the aid of Lemma 1.
\enddemo
\medskip
Another consequence is

\proclaim{Lemma B.4}
$$
\sup_{x\in X} \Mint_{B_t(x)} \Mint_{B_t(x)} |u(y) - u(z)|^p \leq A^p p!
M_t^p(u),\quad\forall t < r_0.
$$
\endproclaim

\demo{Proof}  For $t \leq r_0/\sqrt{n}$, the claim follows immediately
from Lemma B.1 with the aid of Lemma A.14.  Suppose $r_0/\sqrt{n} < t <
r_0$; we may assume that $\int_X u = 0$.  Then the desired result
follows easily from Lemma B.3.                                       
\enddemo
\medskip
Another simple consequence of Lemma B.3 is

\proclaim{Lemma B.5}  Given $\theta > 0$, there is a number $\beta$
depending on $\theta$ and on $X$ such that if $\|u\|_{\text{BMO}} \leq
1$, then
$$
\int_X \left(e^{\beta|u - \bar{u}|} - 1\right) \leq \theta \int_X |u -
\bar{u}|
$$
where $\bar{u} = \dsize\Mint_X u$.
\endproclaim

We point out that Lemmas B.3 and B.5 may be proved in a more direct
fashion, not via Lemma B.2.  Namely, one starts by proving, directly,
Lemma B.5 using (B.1) locally and summing, as we did in the proof of
Lemma B.1.  Lemma B.3 follows from Lemma B.5.
\medskip

We now turn to the proof of Lemma B.2.  We shall make use of the
following covering lemma:

\proclaim{Lemma B.6}  Given $t, 0 < t < r_0$ and $k > 1$, there is a
covering of $X$ by a finite number of balls $B_t(x_i)$, $i =
1,\ldots,m = m(t)$ with the property that every $y\in X$ belongs to at
most $\mu$ balls $B_{kt} (x_i)$, where $\mu$ depends only on $k$ and
$X$. $\boldsymbol\mu$ {\bf is independent of} $\bold t$.
\endproclaim

\demo{Proof}  Consider a maximal family of disjoint balls in $X$ of
radius $t/2: B_{t/2} (x_i)$,\newline $i = 1,\ldots,m$.  Clearly the $B_t(x_i)$
cover $X$.  Suppose $y\in X$ belongs to $\mu$ of the $B_{kt}(x_i)$,
say for $i = 1,\ldots,\mu$.  Since dist$(y,x_i) \leq k t,\; i =
1,\ldots,\mu$, it follows that
$$
B_{t/2} (x_i) \subset B_{(k+\frac12)t} (y), \quad i = 1,\ldots,\mu.
$$
Since the balls $B_{t/2} (x_i)$ are disjoint, we find on adding their
measures, that
$$
\sum^\mu_{i=1} |B_{t/2} (x_i)|\leq |B_{(k+\frac12)t}(y)|.
$$
Using the fact that
$$
\alpha r^n\leq |B_r(x)|\leq \beta r^n\quad\forall r,
$$
for some positive constants $\alpha,\beta$ depending only on $X$, we deduce a
bound for $\mu$ which depends only on $k$ and $X$.
\enddemo
\medskip
Now the
\smallskip
\demo{Proof of Lemma B.2}  Observe first that for any ball $B = B_t
(x)$,
$$
\align
\left(\Mint_B |u|^p\right)^{1/p} &\leq  \left(\Mint_B |u - \bar{u}_t(x)|^p\right)^{1/p} + |\bar{u}_t(x)|\\
   &\leq \left[ \Mint_B \Mint_B |u(y) - u(z)|^p\right]^{1/p} +
|\bar{u}_t(x)|
\endalign
$$
by the triangle inequality.  Hence
$$
\int_B |u|^p \leq \frac{2^p}{|B|} \int_B\; \int_B |u(y) - u(z)|^p +
\frac{2^p}{|B|^{p-1}} \bigg|\int_B u\bigg|^p.
$$
By Lemma B.1, we find, with a different $A$:
$$
\align
\int_B |u|^p &\leq \frac{A^p p!}{|B|} M^{p-1}_t (u)
\int_{B_{kt}(x)}\;\int_{B_{kt}(x)} |u(y) - u(z)| +
\frac{2^p}{|B|^{p-1}} \bigg\vert\int_B u\bigg\vert^p\\
   &\leq A^p p! M^{p-1}_t (u) \int_{B_{kt}(x)} |u| + \frac{2^p}{|B|^{p-1}}
\bigg\vert \int_B u \bigg\vert^p.
\endalign
$$
\medskip

Using the preceding covering lemma, and this last inequality, with $B
= B_t(x_i)$, and summing, we obtain the desired conclusion.
$\quad\hfill\qed$
\enddemo
\medskip
\proclaim{Lemma B.7}  There are constant $\beta,C$ depending only on
$X$ such that for any measurable set $A\subset X$, and every
$u\in\text{BMO}(X,\Bbb R^N)$
$$
\beta \Mint_A |u|\leq \|u\|_{BMO} \left(C + \log
\frac{|X|}{|A|}\right) + \beta \bigg\vert\Mint_X u\bigg\vert.
$$
\endproclaim

\demo{Proof}  We may assume $\int_X u = 0$ and $\|u\|_{BMO}\leq
1$.  Recall Young's inequality:  For\newline $t > 0, \alpha \geq 1$,
$$
\alpha t \leq e^t + \alpha \log \alpha - \alpha. \tag{B.9}
$$
\medskip
We apply this with
$$
\alpha = \frac{|X|}{|A|}\text{ and  $t = \beta |u(x)|$}
$$
with $\beta$ as in Lemma B.5, where we take $\theta = 1$.   Integrating
the inequality over $A$ we find 
$$
\beta |X| \Mint_A |u| \leq \int_A e^{\beta|u|} + |X| \log
\frac{|X|}{|A|} - |X|.
$$
Hence
$$
\beta \Mint_A |u|\leq \Mint_X (e^{\beta|u|} - 1) + \log
\frac{|X|}{|A|}.
$$
The desired conclusion follows with the aid of Lemmas B.5 and 1.
$\quad\hfill\qed$
\enddemo

Next, a lemma on the effect of multiplying a BMO function by some
function.
\medskip
\proclaim{Lemma B.8}  Let $a$ be a Lipschitz function on $X$ and let
$f$ be in BMO$(X)$ (respectively VMO). Then $af$ is in BMO (respectively
VMO), and
$$
\|af\|_{BMO} \leq C \left( \big|a\big|_{C^0} + \|a\|_{\text{Lip}}\right)
\|f\|_{BMO} + \|a\|_{BMO} \bigg\vert\Mint_X f\bigg\vert.
$$
where $C$ depends only on $X$.
\endproclaim

\demo{Proof}  We may assume that $\int_X f = 0$.  We then have to
evaluate
$$
\align
J &= \Mint_{B_\vare(x)} \Mint_{B_\vare(x)} |a(y) f(y) - a(z) f(z)|\\
   &\leq \Mint_{B_\vare(x)}\,\Mint_{B_\vare(x)} |(a(y) - a(z)) f(y) + a(z) (f(y) - f(z))|\\
   &\leq 2\vare \|a\|_{\text{Lip}} \Mint_{B_\vare (x)} |f| +
2|a|_{C^0} \|f\|_{BMO}.
\endalign
$$
\medskip
Using Lemma B.7, with $A = B_\vare(x)$, the desired estimate follows.
The VMO assertion follows easily from Lemma 3.
\enddemo
\medskip
\remark{\bf Remark B.1}  Note that in Lemma B.8, instead of assuming
that $a$ is Lipschitz continuous, we could have assumed that $a$ is
H\"older continuous or even merely that
$$
|a(x) - a(y)| \leq \frac{C}{1 + |\log\text{ dist}(x,y)|}\quad\forall
x\not= y.
$$
D.~Stegenga [1] has obtained necessary and sufficient conditions for a
function $a$ to be a multiplier preserving BMO.
\endremark

\medskip
We conclude this Appendix with a lemma asserting that if $u\in\text{BMO}$, then $\bar u_\vare$ is ``almost'' Lipschitz.

\medskip
\proclaim{Lemma B.9}  For $u\in \text{BMO}(X,\Bbb R^N)$ and $\vare <
r_0$, there is a constant $C_\vare$ such that
$$
J = |\bar{u}_\vare(x) - \bar{u}_\vare(y)|\leq C_\vare
\|u\|_{BMO} \text{dist}(x,y) \left(1 + \log
\frac{\text{diam }X}{\text{dist}(x,y)}\right).
$$
\endproclaim

\demo{Proof}  We may suppose that $\int_X u = 0$ and
$\|u\|_{\text{BMO}} = 1$.  Then by Lemma 1, $\int_X |u|\leq C$.
\medskip
\noindent
{\bf Case 1.}  dist$(x,y) \geq \vare/100$.  In this case
$$
\align
J&\leq \frac{1}{|B_\vare(x)|} \int_X |u| + \frac{1}{|B_\vare(y)|}
\int_X |u| \leq C_\vare.
\endalign
$$
\medskip
\noindent
{\bf Case 2.}  dist$(x,y) < \vare/100$.  Then
$$
\bigg\vert \frac{1}{|B_\vare(x)|} - \frac{1}{|B_\vare(y)|} \bigg\vert
\leq C_\vare \text{ dist}(x,y).
$$
Thus
$$
J\leq \int_{B_\vare(x)} |u(z)| \bigg\vert\frac{1}{|B_\vare(x)|} -
\frac{1}{|B_\vare (y)|}\bigg\vert + \frac{1}{|B_\vare(y)|} \bigg\vert
\int_{B_\vare(x)} u(z) - \int_{B_\vare(y)} u(z)\bigg\vert.
$$
As in the proof of Lemma A.16 we introduce the symmetric difference
$S$ of $B_\vare(x)$ and $B_\vare(y)$ and we have
$$
|S|\leq C_\vare \text{dist}(x,y).
$$
\medskip
Applying Lemma B.7 we see that
$$
\big\vert \int_S u(z)\big\vert \leq C|S| \left(C + \log \frac{|X|}{|S|}\right).
$$
\medskip
The desired inequality follows by combining this with the earlier
estimate.
\enddemo.
\medskip
\noindent
{\bf Acknowledgment.}  The second author was partly supported by
grants NSF-DMS 9114456 and ARO-DAAL-03-92-G-0143.
%=end of Appendix B.

\bigskip

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