\magnification=\magstep1
\input amstex
\documentstyle{amsppt}
\loadbold
\nologo
\pagewidth{6.5truein}
\pageheight{8.5truein}
\def\BR{\Bbb R}
\def\varep{\varepsilon}
\def\Mint{\diagup\hskip-.50truecm\int}
\def\mint{\diagup\hskip-.38truecm\int}
\topmatter
\title  Degree and Sobolev Spaces\endtitle
\leftheadtext{Degree and Sobolev Spaces}
\author  Ha\"im Brezis$^{(1)(2)}$, Yanyan Li$^{(2)}$, Petru
Mironescu$^{(3)}$ and Louis Nirenberg$^{(4)}$\endauthor
\dedicatory  Dedicated to Jurgen Moser in friendship and admiration
\enddedicatory
\address  $(1)$ Institut Universitaire de France and Analyse Numerique,
Universit\'e P. et M. Curie, 4 pl. Jussieu, 75252 Paris Cedex 05,
France;\endgraf
$(2)$  Department of Mathematics, Rutgers University,
Piscataway, NJ 08854,USA;\endgraf
$(3)$  Departement de Mathematiques,
universit\'e Paris-Sud, Bat. 425, 91405 Orsay Cedex, France;\endgraf
 $(4)$  Courant Institute, New York University, 251 Mercer St., New York, NY
10012, USA
\endaddress
\thanks  Acknowledgments:  The first author (H.B.) is partially
supported by a European Grant ERB FMRX CT98 0201. The second author (Y.L.) is partially
supported by NSF Grant DMS-9706887 and a Rutgers University Research
Council grant.  Part of this work was done when the third author
(P.M.) was visiting Rutgers University; he thanks the Mathematics
Department for its invitation and hospitality.  The visit of the
fourth author (L.N.) to Paris was supported by the Institut
Universitaire de France and he thanks it for its hospitality.
\endthanks
\endtopmatter
\document
\subhead  Introduction\endsubhead
\medskip
J. Rubinstein and P. Sternberg established in [9] the following
result.  Let $\Omega$ be a solid 3-dimensional torus, i.e., $\Omega =
S^1\times\Lambda$ where $\Lambda$ is the unit disc in $\BR^2$.  Let
$u\in H^1(\Omega;S^1)$.  For a.e. $\lambda\in\Lambda$ the map
$$
x\in S^1\mapsto u(x,\lambda)\in S^1
$$
belongs to $H^1(S^1,S^1)$; thus it is continuous and has a degree.
Conclusion:
$$
\deg (u(\;\cdot\;,\lambda))\quad\text{is independent of $\lambda$}.
$$
\medskip
This result is somewhat surprising because $H^1$ functions in 3-d
need not be continuous, and not even in VMO.  If $\Omega$ were a 2-d
annulus, $\Omega = S^1\times (0,1)$, instead of a 3-d torus the conclusion
would still be surprising; however, in this case one can give a
straightforward proof via the $H^{1/2} (S^1,S^1)$ degree theory of
L. Boutet de Monvel and O. Gabber (see [4] and also [5]).  Indeed, by standard trace theory, the map
$$
\lambda\in (0,1)\mapsto u(\;\cdot\;,\lambda) \in H^{1/2} (S^1,S^1)
$$
is continuous.  We recall that any map $\varphi\in H^{1/2} (S^1,S^1)$
has a degree which depends continuously on the $H^{1/2}$ norm.
Therefore $\deg (u(\;\cdot\;,\lambda))$ is well-defined for {\it every}
$\lambda\in (0,1)$ and is independent of $\lambda$.
\medskip
By contrast, in 3-d, there is no similar argument since a general $H^1$
function does not have trace on every line.
\medskip
In this paper we first give, in Section 1, a direct generalization with
simple proof.  We then present in Section 2 a still more general
result which holds in fractional Sobolev spaces.
\medskip
\subhead Section 1\endsubhead
\medskip

  Let $X$ and $Y$ be compact, oriented, $n$-dimensional smooth
manifolds without boundary, $Y$ is connected.  Let $\Lambda$ be a
domain in $\BR^k$.  Let $u$ be a map
from $\Omega = X\times\Lambda$ into $Y$ which belongs to $W^{1,n+1}
(\Omega,Y)$, i.e., if $Y$ is smoothly embedded in some $\BR^N$ then $u$
is a map from $\Omega$ into $\BR^N$ having each component in
$W^{1,n+1}$ and such that $u(x,\lambda)\in Y$ a.e. in $\Omega$.  For
a.e. $\lambda\in\Lambda$, $u(\;\cdot\;,\lambda)$ belongs to $W^{1,n+1}
(X,Y)$; so it is continuous and therefore $\deg (u(\;\cdot\;,\lambda))$ is
well defined.

\proclaim{Theorem 1}  $\deg (u(\;\cdot\;,\lambda))$ is independent of
$\lambda$ and we call it simply \hbox{deg $u$}.  Moreover,\linebreak $\text{deg }u$ is stable under convergence in $W^{1,n}(\Omega)$,
i.e., if a sequence $(u_j)$ in $W^{1,n+1}(\Omega)$ converges in the
$W^{1,n}$ norm to some $u\in W^{1,n+1}(\Omega)$, then deg $u_j =
\text{ deg }u$ for sufficiently large $j$.
\endproclaim

\remark{Remark 1}  The result need not hold if $u$ is merely in
$W^{1,p}(\Omega,Y)$ with $p < n+1$.  Here is an example.  Let $X = Y
= S^{n}$ and let $\Omega = (0,2)^k$
$$
u(x,\lambda) = \frac{x-\lambda_1 e_1}{|x - \lambda_1 e_1|}
$$
It is easily seen that $u\in W^{1,p}(\Omega,Y)$ for every $p < n+1$.
On the other hand\linebreak $\deg(u(\;\cdot\;,\lambda)) = 0$ for $\lambda_1 > 1$ and
$\deg (u(\;\cdot\;,\lambda)) = 1$ for $\lambda_1 < 1$.
\medskip
The proof of the theorem uses a standard representation of the degree
of a $C^1$ map $\varphi$ as an integral.  If $\omega$ is a smooth $n$-form on
$Y$ with
$$
\int_Y\omega = 1
$$
then
$$
\deg\varphi = \int_X \omega\circ \varphi
$$
where $\omega\circ\varphi$ is the pull-back of $\omega$ by $\varphi$
(see e.g. [8]).  That formula still holds (by density) if $\varphi\in
C^0\cap W^{1,n}$.  (In fact, it suffices that $\varphi\in W^{1,n}$
since it is then in VMO and VMO-maps have a degree, see [5]).
\endremark

\demo{Proof of Theorem 1}  It is convenient to work with a special
form $\omega$ on $Y$ having small support.  For then we can use one
fixed local coordinate system near a point.  Assume $0\in
Y\subset\BR^N$; we may also choose the embedding $e$ of $Y$ into $\BR^N$
in such a way that, in a neighborhood $V$ of $0$ in $\BR^N$,
$$
e(Y) = \left\{y;\;\;y^{n+1} = y^{n+2} = y^N = 0\right\}.
$$
Let $\zeta$ be a smooth function with support in $V$ such that
$$
\int \zeta(y^1,y^2,\ldots y^n, 0,\ldots 0) dy^1\ldots dy^n = 1.
$$
We consider the $n$-form $\tilde\omega$ on $\BR^N$,
$$
\tilde\omega = \zeta(y^1,y^2,\ldots,y^N) dy^1\wedge\cdots\wedge dy^n
$$
and take as $w$ on $Y$, the pull back of $\tilde\omega$ under $e$; in
our local coordinates it has the form
$$
\omega =\zeta(y^1,y^2,\ldots y^n,0,\ldots,0)dy^1\wedge\cdots\wedge dy^n
$$
and thus $\int_Y \omega = 1$.
\enddemo

We have to prove that
$$
\int_X \omega\circ u (\;\cdot\;,\lambda) \tag 1
$$
is independent of $\lambda$, a.e. in $\Lambda$.  The natural argument
would be to differentiate the integral with respect to the parameter
$\lambda$.  However, the integrand in (1) already involves first order
derivatives of $u$ and $\lambda$-differentiation introduces
second-order derivatives.  We get rid of these by integration by
parts.  To carry this out we use approximation by smooth functions.

Let $u_\varep$ be a family of smooth maps from $\overline\Omega$ into $\BR^N$
converging, as $\varep\rightarrow 0$, to $u$ in $W^{1,n+1}$.  Note
that, in general, the $u_\varep$'s do not map into $Y$ (and not even into
a neighborhood of $Y$, see [2]).  Set
$$
\psi_\varep(\lambda) = \int_X \tilde\omega\circ u_\varep
(\;\cdot\;,\lambda)
$$
and differentiate $\psi_\varep$ with respect to one of the
$\lambda$'s, still denoted by $\lambda$.

We find, with $u^i_{\varep\lambda} = \frac{\partial}{\partial\lambda}
u^i_\varep$,
$$
\align
\frac{\partial}{\partial\lambda} \psi_\varep(\lambda) &= \int_X
\sum^N_1 \frac{\partial\zeta}{\partial y^j} (u_\varep)
u^j_{\varep\lambda} du^1_\varep\wedge\ldots\wedge d u^n_\varep\\
&+\int_X \zeta(u_\varep) du^1_{\varep\lambda}\wedge d
u^2_\varep\wedge\ldots\wedge du^n_\varep\tag 2\\
&+\cdots+\int_X \zeta(u_\varep) du^1_\varep\wedge\cdots\wedge
du^{n-1}_\varep \wedge d u^n_{\varep\lambda}.
\endalign
$$
Now
$$
\align
\int_X \zeta(u_\varep) du^1_{\varep\lambda} \wedge d u^2_\varep
\wedge\cdots\wedge du^n_\varep & = \int_X d\big[\zeta(u_\varep)
u^1_{\varep\lambda} d u^2_\varep\wedge\cdots\wedge d u^n_\varep\big]\\
&- \int_X \sum^N_1 u^1_{\varep\lambda}\frac{\partial\zeta}{\partial y^j}
(u_\varep) du^j_\varep\wedge d u^2_\varep\wedge\cdots\wedge
du^n_\varep\\
&= - \int_X u^1_{\varep\lambda} \frac{\partial\zeta}{\partial y^1}
(u_\varep) du^1_\varep\wedge du^2_\varep\wedge\cdots\wedge d
u^n_\varep\\
&- \int_X \sum^N_{n+1} u^1_{\varep\lambda}
\frac{\partial\zeta}{\partial y^j} (u_\varep) d u^j_\varep \wedge
du^2_\varep\wedge\cdots\wedge du^n_\varep.
\endalign
$$
Similar expressions hold for the term after this one in (2).  Inserting
these expressions into (2), we find
$$
\align
&\frac{\partial}{\partial\lambda} \psi_\varep(\lambda) = \int_X
\sum^N_{n+1} \frac{\partial\zeta}{\partial y^j} (u_\varep)
u^j_{\varep\lambda} du^1_\varep\wedge\cdots\wedge du^n_\varep\\
&\qquad - \int_X \sum^N_{n+1} \frac{\partial\zeta}{\partial y^j} (u_\varep)
\biggl[u^1_{\varep\lambda} du^j_\varep\wedge du^2_\varep\wedge\cdots\wedge
du^n_\varep\tag 3\\
&\qquad + u^2_{\varep\lambda} du^1_\varep\wedge du^j_\varep\wedge
du^3_\varep\wedge\cdots\wedge d u^n_\varep + \cdots+
u^n_{\varep\lambda} d u^1_\varep\wedge\cdots\wedge du^{n-1}_{\varep}
\wedge d u^j_\varep\biggr].
\endalign
$$
\medskip
Next we claim that, as $\varepsilon\rightarrow 0$,
$$
\int_{\Omega} |\frac{\partial\psi_\varep}{\partial\lambda}|\rightarrow
0 .\tag 4
$$
Indeed by (3) we have
$$
|\frac{\partial\psi_{\varep(\lambda)}}{\partial\lambda}|\leq C
\sum^N_{n+1} \int_X \left|\frac{\partial\zeta}{\partial y^j}
(u_\varep)\right| |D u^j_\varep| |D u_\varep|^n
$$
where $D$ denotes the full gradient (in $x$ and $\lambda$).
\medskip
%texed and corrected to here.
Thus
$$
\int_\Lambda |\frac{\partial\psi_\varep}{\partial\lambda} |\leq C
\sum^N_{n+1} \left[ \int_\Omega |\frac{\partial\zeta}{\partial y^j}
(u_\varep)|^{n+1} |D u^j_\varep|^{n+1}\right]^{1/(n+1)} \|u_\varep\|^n_{W^{1,n+1}}(\Omega).
$$
Since $u_\varep\rightarrow u$ in $W^{1,n+1} (\Omega)$ we have
%corrected by YYLi to here
$$
\int_\Lambda |\frac{\partial\psi_\varep}{\partial\lambda}|\leq C
\sum^N_{n+1}\left[\int_\Omega |\frac{\partial\zeta}{\partial y^j}
(u_\varep)|^{n+1} |D u^j_\varep|^{n+1}\right]^{1/(n+1)}.
$$
Next observe that (passing to a subsequence)
$$
|\frac{\partial\zeta}{\partial y^j} (u_\varep)||D u^j_\varep|\rightarrow
|\frac{\partial\zeta}{\partial y^j} (u)| |D u^j|\quad\text{a.e. on
$\Omega$}
$$
and, for $j>n$, $\frac{\partial\zeta}{\partial y^j}(u) Du^j = 0$ a.e. on $\Omega$
(since on the set $\{(x,\lambda); u(x,\lambda)\in V\}, u^j = 0$ for $j
= n+1,\ldots,N$ and hence $D u^j = 0$).  On the other hand (passing
to a subsequence)  we may assume that $|D u^j_\varep|$ is bounded by a
fixed function in $L^{n+1}(\Omega)$ and hence, by dominated
convergence,
$$
\int_\Omega |\frac{\partial\zeta}{\partial y^j} (u_\varep)|^{n+1} |D
u^j_\varep|^{n+1}\rightarrow 0,\quad\text{for $j > n$},
$$
which yields (4).
\medskip

Finally we claim that
$$
\psi_\varep(\lambda)\rightarrow \psi(\lambda) = \int_X
\tilde\omega\circ u(\;\cdot\;,\lambda) = \int_X \omega\circ
u(\;\cdot\;,\lambda)\quad\text{in $L^{(n+1)/n}(\Lambda)$}\tag 5
$$
Indeed the integrand in $\psi_\varep$ can be estimated pointwise by
$$
|\tilde\omega\circ u_\varep|\leq C |D u_\varep|^n
$$
and thus (passing to a subsequence)
$$
|\tilde\omega\circ u_\varep - \tilde\omega\circ u|\leq f
$$
where $f$ is a fixed function in $L^{(n+1)/n}(\Omega)$.
\medskip
Therefore
$$
|\psi_\varep(\lambda) - \psi(\lambda)|^{(n+1)/n}\leq C \int_X
|f(x,\lambda)|^{(n+1)/n} dx
$$
and the right-hand side is a fixed function in $L^1(\Lambda)$.  The
claim (5) follows, again by dominated convergence, since
$\psi_\varep(\lambda)\rightarrow\psi(\lambda)$ a.e.
\medskip
Combining (4) and (5) we see that $\psi\in W^{1,1}(\Lambda)$ and
$$
\frac{\partial\psi}{\partial\lambda}= 0.
$$
 Hence $\psi$ is independent of $\lambda$.
\medskip
The last assertion in the theorem, i.e., stability of degree under
$W^{1,n}$ convergence follows easily from the formula
$$
\text{deg }u = \Mint_\Lambda \int_X \omega\circ u
$$
and the fact that the integrand in the right-hand side involves
$n$-products of derivatives of $u$.


\medskip
\remark{Remark 2} The above computation for computing the
$\lambda$-derivative of a pull back can be expressed globally, and
more succinctly, in terms of differential forms.
\medskip
Namely, consider $X$ and $\Lambda$ as above and a smooth map $u$ from
$\Omega = X\times\Lambda$ into an oriented manifold $Z$ (in the case
above, $Z = \BR^N$).  Let $\tilde\omega$ be a smooth $n$-form on $Z$.
The $\lambda$-derivative of the pullback $\tilde\omega \circ
u(\;\cdot\;,\lambda)$ is simply
$$
\partial_\lambda \tilde\omega\circ u(\;\cdot\;,\lambda) = d A + B \tag
6
$$
where $A$ and $B$ are $(n-1)$ and $n$-forms respectively on $X$.  They
are expressed using the tangent vector $u_\lambda$ which is defined at
points of $Z$ in the image of $u(\;\cdot\;,\lambda)$.  $A$ and $B$ are
given by
$$
A = (\tilde\omega\; \lrcorner \;u_\lambda)\circ u \tag 7
$$
$$
B = (d\tilde\omega\;\lrcorner\; u_\lambda)\circ u. \tag 8
$$
Here $u_\lambda$ is the $\lambda$-derivative (say with respect to one
of the $\lambda$ coordinates) of $u$.   The symbol $\lrcorner$ denotes
contraction of a differential form and a vector (see [7]).
\endremark

Formula (6) holds for a smooth map.  It still holds for maps in
$W^{1,n+1}$, provided one interprets (6) in the distribution sense.
In fact, the coefficients of $\tilde\omega\circ u$ and of $A$ are $n$
products of functions in $L^{n+1}$, the coefficients of $B$ are $n+1$
products of functions in $L^{n+1}$.  To justify (6) in such generality one smoothes $u$ by
$u_\varep$ as above, mapping however into a high dimension Euclidean
space in which $Z$ is embedded.
\medskip
In the special case that $\text{dim }Z = \text{dim }X$ (for example if
$Z = Y$ as above) then $d\tilde\omega = 0$ and thus $B = 0$.  
\smallskip
\noindent
{\it Warning:}  The reader might think that in this case (6) holds
with $B = 0$ assuming only that $u\in W^{1,n}$.  This is not true as the
counterexample in Remark 1 shows.
\smallskip
When using degree theory one often considers $X$, the domain space
with a boundary, $Y$ connected and open.  One wishes to compute the degree of
$u:  X\rightarrow Y$ at some point $y\in Y$ which is not in the image
of the boundary (if the map $u\in C(\overline X)$,  Theorem 1 easily
extends to such a situation).  Here is one form of such a result:
\smallskip
Let $X$ be an open subset of an $n$-dimensional smooth oriented
manifold $\widetilde X$ with $\overline X$, the closure of $X$, compact in
$\widetilde X$ and $\partial X$ smooth.  Let $Y$ be an open oriented,
connected, $n$-dimensional smooth Riemannian manifold.  Let $\Lambda$
be a domain in $\BR^k$.  Let $u$ be a map from $\Omega =
X\times\Lambda$ into $Y$ which belongs to $W^{1,n+1} (\Omega,Y)$.  For
a.e. $\lambda\in\Lambda$, $u(\;\cdot\;, \lambda)$ belongs to
$W^{1,n+1}(X,Y)$ so it is continuous in $\overline X$.  Assume that
$y\in Y$ is such that for some $\delta > 0$ and for a.e. $\lambda$ as
above,
$$
\text{dist }(y,u(\partial X,\lambda)) \geq \delta
$$
Then the degree of $u$ at $y$, $\deg (u(\;\cdot\;,\lambda), X,y)$ is
well defined.

\proclaim{Theorem $\bold 1^\prime$}  $\deg (u (\;\cdot\;,\lambda),X,y)$ is
independent of $\lambda$.
\endproclaim

The proof is just the same as that of Theorem 1.  We may suppose that $y$
is the origin in $\BR^N$ and that $Y$ near $0$ is flat.  Then we take
the forms $\tilde\omega$ and $w$ as above, with supp $\omega$ lying in
a $\delta/2$ neighborhood (with respect to the metric on $Y$) of $y$.  Then proceed as before. 
%texed to here 7/16. I can't find a Section 1

\medskip
\subhead  Section 2\endsubhead
\medskip
 Let $X,Y$ and $\Lambda$ be as in Theorem 1 and let $u$ be a map
from $\Omega = X\times\Lambda$ into $Y$ which belongs to
$W^{s,p}(\Omega,Y)$ with $s > 0$ and $1 < p < \infty$.  Recall
that for a.e. $\lambda\in\Lambda$, $u(\;\cdot\;,\lambda)$ belongs to
$W^{s,p}(X,Y)$.  This is clear if $s$ is an integer; when $0 < s < 1$ such
property is an easy consequence of the equivalence of two $W^{s,p}$
norms in $\BR^m$:
$$
\|f\|^p_{W^{s,p}} = \|f\|^p_{L^p} + \int_{\BR^m} \int_{\BR^m}
\frac{|f(x) - f(y)|^p}{|x - y|^{m+sp}} dx dy
$$
and
$$
\||f\||^p_{W^{s,p}} = \|f\|^{p}_{L^p} + \sum^m_{i=1} \int^1_0
\int_{\BR^m} \frac{|f(x+te_i) - f(x)|^p}{t^{sp}} dx dt
$$
(see e.g. Adams [1], p. 208-214 or Triebel [10]).  The case of a
general $s > 0$ follows easily.
\medskip
Assuming further that
$$
sp\geq n+1 \tag 9
$$
we find that for a.e. $\lambda\in\Lambda$, $u(\;\cdot\;,\lambda)\in
W^{s,p} (X,Y) \subset C^0 (X,Y)$; therefore $\deg
(u(\;\cdot\;,\lambda))$ is well defined for a.e. $\lambda\in\Lambda$.
\medskip
\proclaim{Theorem 2}  Assume that $u\in W^{s,p}(\Omega,Y)$ and that
(9) holds.  Then
$$
\deg (u(\;\cdot\;,\lambda))\text{\quad is independent of $\lambda$}.
$$
Moreover, this degree is stable under convergence in any $W^{s^\prime,
p^\prime}$ norm provided $s^\prime p^\prime\geq n$.

\endproclaim


\demo{Proof}  We may assume that $0 < s\leq 1$ and the case $s > 1$ is
handled with minor modifications.  In this generality there is no integral representation
for the degree and the argument is quite different from the proof of
Theorem 1.  Clearly it suffices to prove that the degree is locally
constant a.e.  Hence it suffices to consider the case where $\Lambda =
(0,1)^k$.  We assume first that $k = 1$ and the general case will be
done by reduction to $k = 1$ as in Bethuel and Demengel [3] (Lemma A.1).
\enddemo

\noindent
{\bf Case where $\boldsymbol\Lambda = (\bold 0,\bold 1)$}.  By the
standard trace theory a map $u\in W^{s,p} (X\times (0,1), Y)$ can be identified with a map
$u\in C([0,1], W^{s-1/p,p} (X,Y)$.  Since $(s - 1/p)p = sp - 1\geq n$,
$W^{s-1/p,p}(X)\subset\text{ VMO }(X)$ (see e.g. [5]); we also recall that
there is a degree theory on VMO$(X,Y)$ and that this degree is stable
under small VMO perturbation.  In this case
$\deg (u(\;\cdot\;,\lambda))$ is well defined for {\it every}
$\lambda\in [0,1]$ and it is independent of $\lambda$.
\medskip
\newpage

\noindent
{\bf Case where $\boldsymbol\Lambda = ({\bold 0,\bold 1})^{\bold k}$}.  We start with two lemmas:

\medskip
\proclaim{Lemma 1}  The map $\lambda\mapsto\deg (u(\;\cdot\;,\lambda))
= \psi(\lambda)$ is measurable.
\endproclaim

\demo{Proof}  Consider a sequence $(u_j)$ of smooth functions on
$X\times\overline\Lambda\rightarrow\BR^N$ ($Y$ is embedded in $\BR^N$)
such that
$$
u_j\rightarrow u\quad\text{in $W^{s,p}(X\times\overline\Lambda)$}\
$$
Passing to a subsequence (and using the equivalence of norms mentioned
above) we may assume that for a.e. $\lambda\in\Lambda$
$$
u_j(\;\cdot\;,\lambda)\rightarrow u(\;\cdot\;,\lambda)\quad\text{in
$W^{s,p}(X)$}.
$$
In particular for a.e. $\lambda\in\Lambda$,
$$
u_j(\;\cdot\;,\lambda)\rightarrow u(\;\cdot\;,\lambda)\quad\text{uniformly in $X$}. \tag 10
$$
Let $\delta > 0$ be sufficiently small so that in the closed
$\delta$-neighborhood $N_\delta(Y)$ of $Y$ in $\BR^N$ the projection $P_Y$ onto $Y$ is well defined.
\enddemo
\medskip
For every $j = 1,2,\ldots$ and every $\lambda\in\overline\Lambda$ set
$$
\gamma_j(\lambda)= \operatornamewithlimits{Sup}_{x\in X}
\text{dist }(u_j(x,\lambda), Y),
$$
(so that each $\gamma_j$ is continuous--even Lipschitz--in $\lambda$) and
$$
\psi_j(\lambda) = \cases  \deg(P_Y(u_j(\;\cdot\;,\lambda))
\frac{(\delta - \gamma_j(\lambda))}{\delta}&\quad\text{if
$\gamma_j(\lambda)\leq\delta$}\\
0&\quad\text{if $\gamma_j(\lambda) > \delta$}.
\endcases
$$
In view of (10) it is clear that
$$
\psi_j (\lambda)\rightarrow \psi(\lambda),\quad\text{as
$j\rightarrow\infty$, a.e. in $\lambda\in\Lambda$}.
$$
On the other hand, it is easy to check that for every $j$, the
function $\lambda\mapsto\psi_j(\lambda)$ is continuous on
$\overline\Lambda$.  Thus $\psi$ is measurable on $\Lambda$.
\medskip
The second lemma is purely measure theoretical.
\medskip

\proclaim{Lemma 2}  Let $\Lambda = (0,1)^k$ and let $\psi$ be a
measurable function on $\Lambda$ such that for each $1\leq i\leq n$
and for a.e. $(\lambda_1,\ldots\lambda_{i-1},\lambda_{i+1},\ldots\lambda_k)$
in $(0,1)^{k-1}$, the function
$$
a\in (0,1)\mapsto\psi(\lambda_1,\ldots\lambda_{i-1}, a,
\lambda_{i+1},\ldots\lambda_k)
$$
is constant a.e. on $(0,1)$.  Then $\psi$ is constant a.e. on $\Lambda$.
\endproclaim

\demo{Proof}  We may always assume that $\psi$ is also bounded (and thus
integrable) since otherwise we may replace $\psi$ by Arctan $\psi$.
By the triangle inequality, with
$$
\lambda = (\lambda_1,\ldots\lambda_k)\quad\text{and $\mu =
(\mu_1,\ldots\mu_k)$},
$$
we have
$$
\align
|\psi(\lambda) &- \psi(\mu)|\leq
|\psi(\lambda_1,\lambda_2,\ldots\lambda_{k-1}, \lambda_k) -
\psi(\lambda_1,\lambda_2,\ldots\lambda_{k-1}, \mu_k)|\\
&+ |\psi(\lambda_1,\lambda_2,\ldots\lambda_{k-1}, \mu_k) -
\psi(\lambda_1,\lambda_2,\ldots\mu_{k-1},\mu_k)|\\
&+\cdots+ |\psi(\lambda_1,\mu_2,\ldots\mu_{k-1},\mu_k) -
\psi(\mu_1,\mu_2,\ldots\mu_{k-1},\mu_k)|
\endalign
$$
It follows from the assumption that
$$
\int_{(0,1)^k} \int_{(0,1)^k} |\psi(\lambda) - \psi(\mu) |d\lambda
d\mu = 0.
$$
Consequently, $\psi(\lambda) - \psi(\mu) = 0$ a.e. on $(0,1)^k\times
(0,1)^k$ which implies that $\psi(\lambda)$ is constant a.e. on
$(0,1)^k$.
\enddemo
\medskip
We now return to the proof of the theorem and, in view of the $\Lambda
= (0,1)$ case, apply Lemma 2 to $\psi(\lambda) = \text{ deg }(u(\cdot\,,\lambda))$ to conclude that deg $(u(\cdot\,, \lambda))$ is
constant a.e. in $(0,1)^k$.
\medskip
To establish the stability under $W^{s^\prime,p^\prime}$ convergence with $s^\prime
p^\prime\geq n$ we argue as follows.  Consider a sequence $(u_j)$ in
$W^{s,p}$ converging in the $W^{s^\prime,p^\prime}$ norm to some $u \in W^{s,p}$ with $sp\geq
n+1$.  As in Lemma 1, passing to a subsequence we may assume that, for
a.e. $\lambda\in\Lambda$,
$$
u_j(\cdot\,, \lambda)\rightarrow u (\cdot\,, \lambda)\quad\text{in
$W^{s^\prime,p^\prime}(X)$}.
$$
Since $s^\prime p^\prime \geq n$, $W^{s^\prime,p^\prime}$ is contained in
VMO, and we may infer from the result of [5] that, for
a.e. $\lambda\in\Lambda$,
$$
\text{deg }(u_j(\cdot\,,\lambda))\rightarrow \text{ deg }(u(\cdot\,,\lambda)).
$$
The conclusion follows by picking any $\lambda$ outside a countable
union of sets of measure zero.  The uniqueness of the limit implies
the convergence of the full sequence.

\medskip
\remark{Remark 3}  The above argument extends to the case where $X$
and $Y$ need not have the same dimension, and degree is replaced by
homotopy classes.  More precisely we have
\endremark

\proclaim{Theorem $\bold 2^\prime$}  Assume that $u\in W^{s,p}
(\Omega,Y)$ and that (9) holds with $n = \dim X$, then there is a
homotopy class $\Cal C$ in $C^0(X,Y)$ such that
$$
u(\;\cdot\;,\lambda)\in\Cal C\quad\text{for a.e. $\lambda\in\Lambda$}.
$$ 
\endproclaim

\demo{Proof}  When $\Lambda = (0,1)$ we may invoke Lemma A.20 in [5]
to assert that two continuous maps which are homotopic within VMO are
also homotopic in $C^0(X,Y)$.
\enddemo

In the general case we denote by $(\Cal C_k)$, $k = 1,2,\ldots$, the
homotopy classes of $C^0(X,Y)$ (the connected components of $C^0(X,Y)$
are countable since $C^0(X,Y)$ is separable).  For every $v\in
C^0(X,Y)$ we set
$$
\deg v = k\quad\text{provided $v\in\Cal C_k$}.
$$
and the above argument remains unchanged.
\medskip
We conclude with a similar question in the VMO framework.  Let $X,Y$
and $\Lambda$ be as in Section 1 and let
$$
u\in \text{VMO }(\Omega,Y),
$$
\bigskip
\noindent
{\bf  Open Problem:}  Is it true that for a.e. $\lambda\in\Lambda$
$$
u(\;\cdot\;,\lambda)\in\text{ VMO $(X,Y)$}
$$
and if so, is deg $(u(\;\cdot\;,\lambda))$ constant a.e. in $\Lambda$?
\medskip\
This question is also related to a question of H. Amann and result in
[6] (p.332-333).

\bigskip
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\enddocument

