for any r > 0 and so a suitable subsequence converges pointwise. The
integrands are also uniformly integrable. Thus the Vitali convergence theorem can be
applied to each of the integrals in the above sum and obtain that for a suitable
subsequence, e
∑ ∑m ∑p ∫ ∂ (ψ a )( )∑m ∂x
= ----jI- R −j1ε (u) --kεA1ldu + e(ε)
I k=1j=1 Rj(Uj) ∂xk l=1 ∂ul
where A_{1l} is the 1l^{th} cofactor for the determinant
( )
∂--xkε,xi1ε⋅⋅⋅xin−1ε-
∂(u1,⋅⋅⋅,un)
which is determined by a particular I. I am suppressing the ε for the sake of notation.
Then the above reduces to
∑ ∑p ∫ ∑n ∑m ∂-(ψjaI) ( −1 ) ∂xkε
= R (U ) A1l ∂xk Rjε (u ) ∂ul du +e (ε)
I j=1 j∫j l=1 k=1
∑ ∑p n∑ -∂-( −1)
= Rj(Uj)A1l∂ul ψjaI ∘R jε (u)du+ e (ε) (28.4.13)
I j=1 l=1
(Note l goes up to n not m.) Recall R_{j}
(Uj)
is relatively open in ℝ_{≤}^{n}. Consider the
integral where l > 1. Integrate first with respect to u_{l}. In this case the boundary term
vanishes because of ψ_{j} and you get
∫
− A (ψ a ∘ R−1)(u)du (28.4.14)
Rj(Uj) 1l,l j I jε
(28.4.14)
Next consider the case where l = 1. Integrating first with respect to u_{1}, the term reduces
to
where R_{j}V_{j} is an open set in ℝ^{n−1} consisting of
{(u ,⋅⋅⋅,u ) ∈ ℝn −1 : (0,u ,⋅⋅⋅,u ) ∈ R (U )}
2 n 2 n j j
and du_{1} represents du_{2}du_{3}
⋅⋅⋅
du_{n} on R_{j}V_{j} for short. Thus V_{j} is just the part
of ∂Ω which is in U_{j} and the mappings S_{j}^{−1} given on R_{j}V_{j} = R_{j}
(Uj ∩ ∂Ω)
by
−1 −1
S j (u2,⋅⋅⋅,un) ≡ R j (0,u2,⋅⋅⋅,un)
are such that
{(Sj,Vj)}
is an atlas for ∂Ω. Then if 28.4.14 and 28.4.15 are placed in
28.4.13, then it follows from Lemma 28.3.1 that this reduces to
p ∫
∑ ∑ −1
RjVj ψjaI ∘R jε (0,u2,⋅⋅⋅,un)A11du1 + e(ε)
I j=1
Now as before, there exists a subsequence, still denoted as ε such that each ∂x_{sε}∕∂u_{r}
converges pointwise to ∂x_{s}∕∂u_{r} and then using that these are bounded in every L^{p},
one can use the Vitali convergence theorem to pass to a limit obtaining finally
∑ ∑p ∫
ψjaI ∘R −j1(0,u2,⋅⋅⋅,un)A11du1
I j=1 RjVj
∑ ∑p ∫
= ψjaI ∘S−j1 (u2,⋅⋅⋅,un )A11du1
I j=1 SjVj
This is done by using a partition of unity which has the property that ψ_{j} equals 1 on K
which forces all the other ψ_{k} to equal zero there. Using the same trick involving a
judicious choice of the partition of unity, ∫dω is also equal to
∑ ∫ ( )
aI ∘S −1(v2,⋅⋅⋅,vn) ∂-xi1 ⋅⋅⋅xin−1-(0,v2,⋅⋅⋅,vn)dv1
I Si(Vj∩Vj) i ∂(v2,⋅⋅⋅,vn)
Similarly if A is an open connected subset of S_{i}
(Vj ∩Vj)
whose measure zero
boundary separates ℝ^{n} into two components, and K is a compact subset of
S_{i}^{−1}
(A )
, containing spta_{I} for all I,∫dω equals each of 28.4.18 and 28.4.17
below.
∫ ( )
∑ a ∘S −1(v ,⋅⋅⋅,v ) ∂-xi1 ⋅⋅⋅xin−1-(0,v ,⋅⋅⋅,v )dv (28.4.17)
I A I i 2 n ∂(v2,⋅⋅⋅,vn) 2 n 1
(28.4.17)
∑ ∫ ( )
aI ∘S −j1(u2,⋅⋅⋅,un) ∂-xi1 ⋅⋅⋅xin−1-du1 (28.4.18)
I Sj∘S−i1(A) ∂ (u2,⋅⋅⋅,un )
∑ ∫ ( )
aI ∘S− 1(u2,⋅⋅⋅,un) ∂-xi1 ⋅⋅⋅xin−1 ⋅
I Sj∘S−i1(A ) j ∂(u2,⋅⋅⋅,un)
(1 − d(u ,A,S ∘S−1))du = 0
1 j i 1
Now by invariance of domain, it follows S_{j}∘S_{i}^{−1}
(A )
is an open connected set contained
in a single component of
(Sj ∘S− 1(∂A))
i
^{C} and so the above degree is constant on
S_{j}∘ S_{i}^{−1}
(A)
. If this degree is not 1 then it follows that for any choice of the a_{I} having
compact support in S_{i}^{−1}
(A )
,
∑ ∫ −1 ∂(xi1 ⋅⋅⋅xin−1)
−1 aI ∘ Sj (u2,⋅⋅⋅,un) ∂-(u,⋅⋅⋅,u-)-du1 = 0 (28.4.19)
I Sj∘Si (A) 2 n
(28.4.19)
Next let I always denote an increasing list of indices. Note that S_{j}∘ S_{i}^{−1} maps the
open set A to an open set which therefore has positive Lebesgue measure. It follows from
the area formula that
( n×m m ×n )
( ( −1)) | ◜-(--(◞◟−1---)◝)◜--◞−◟1--◝|
det D Sj ∘ Si = det( D Sj Si (u) DS i (u)) (28.4.20)
(28.4.20)
must be nonzero on a set of positive measure. It follows that at least some
( )
∂-xi1 ⋅⋅⋅xin−1-
∂(u2,⋅⋅⋅,un)
must be nonzero since by the Binet Cauchy theorem, the above determinant in 28.4.20 is
the sum of products of these multiplied by other determinants which come from
deleting corresponding columns in the matrix for D
( ( ))
Sj S−i1 (u )
. It follows
that
∑ ( ( ))2
∂-xi1 ⋅⋅⋅xin−1
I ∂(u2,⋅⋅⋅,un)
is positive on a set of positive measure. Let lim_{p→∞}a_{Ip}∘ S_{j}^{−1} =
∂(xi1⋅⋅⋅xin−1)
∂(u2,⋅⋅⋅,un)
in
L^{2}
( −1 )
Sj ∘S i (A)
for each I =
(i1,⋅⋅⋅,in−1)
. Replacing a_{I}∘ S_{j}^{−1} with a_{Ip}∘ S_{j}^{−1} in
28.4.19 and passing to the limit, it follows