First is an easy result about approximation of continuous functions with smooth
ones.
Theorem 16.1.1Let Ω be a bounded open set in ℝ^{n}and let f ∈ C_{c}
(ℝn)
.Then there exists g ∈ C_{c}^{∞}
(ℝn )
with
||g− f||
_{∞}< ε.
Proof:Form g ≡ f ∗ψ_{n} for a mollifier ψ_{n}. This will approximate f uniformly on Ω, and
will be in C_{c}^{∞}
(ℝn)
. ■
Using the Weierstrass approximation theorem, you could also get g to equal a function
from G described in the development of the Fourier transform for all x ∈ Ω. Simply apply
Theorem 21.3.9 to the functions in G.
Applying this result to the components of a vector valued function yields the following
corollary.
Corollary 16.1.2If f ∈ C
(-- )
Ω;ℝn
for Ω a bounded subset of ℝ^{n}, then for all ε > 0,there exists g ∈ C^{∞}
(-- )
Ω;ℝn
such that
||g − f||∞ < ε.
The following lemma is a wonderful application of the Vitali covering theorem.
Lemma 16.1.3Let h be differentiable on U. If T ⊆ U and m_{p}
(T)
= 0, thenm_{p}
(h(T ))
= 0.
Proof:For k ∈ ℕ
Tk ≡ {x ∈ T : ||Dh (x)|| < k}
and let ε > 0 be given. Since T_{k} is a subset of a set of measure zero, it is measurable, but we
don’t need to pay much attention to this fact. Now by outer regularity, there exists an open
set V , containing T_{k} which is contained in U such that m_{p}
(V )
< ε. Let x ∈ T_{k}. Then by
differentiability,
h(x+ v ) = h(x)+ Dh (x)v + o(v)
and so there exist arbitrarily small r_{x}< 1 such that B
(x,5rx)
⊆ V and whenever
||v||
≤ 5r_{x},
||o(v )||
<
15
||v||
. Thus
h (B (x,5rx)) ⊆ Dh (x) (B (0,5rx)) +h (x)+ B (0,rx) ⊆ B (0,k5rx)+
+B (0,rx)+ h (x) ⊆ B (h (x) ,(5k +1)rx) ⊆ B(h (x ),6krx)
From the Vitali covering theorem, there exists a countable disjoint sequence of these balls,
{B(xi,ri)}
_{i=1}^{∞} such that
{B (xi,5ri)}
_{i=1}^{∞} =
{ ^ }
Bi
_{i=1}^{∞} covers T_{k} Then letting m_{p}
denote the outer measure determined by m_{p},
The following is Sard’s lemma. It is important in defining the degree.
Lemma 16.1.4(Sard) Let U be an open set in ℝ^{p}and let h : U → ℝ^{p}be differentiable.Let
Z ≡ {x ∈ U : detDh (x) = 0}.
Then m_{p}
(h(Z ))
= 0.
Proof: For convenience, assume the balls in the following argument come from
||⋅||
_{∞}.
First note that Z is a Borel set because h is continuous and so the component functions of the
Jacobian matrix are each Borel measurable. Hence the determinant is also Borel
measurable.
Suppose that U is a bounded open set. Let ε > 0 be given. Also let V ⊇ Z with V ⊆ U
open, and
mp (Z )+ ε > mp (V) .
Now let x ∈ Z. Then since h is differentiable at x, there exists δ_{x}> 0 such that if r < δ_{x},
then B
(x,r)
⊆ V and also o
(v)
< η
∥v∥
for
∥v∥
< r. Thus
h(x+B (0,r)) = h(B (x,r)) ⊆ h (x)+ Dh (x)(B (0,r))+ B (0,rη), η < 1.
Regard Dh
(x)
as an n × n matrix, the matrix of the linear transformation Dh
(x)
with
respect to the usual coordinates. Since x ∈ Z, it follows that there exists an invertible matrix
A such that ADh
(x)
is in row reduced echelon form with a row of zeros on the bottom.
Therefore,
mp (A (h (B(x,r)))) ≤ mp (ADh (x)(B (0,r))+ AB (0,rη)) (16.1)
(16.1)
The diameter of ADh
(x)
(B (0,r))
is no larger than
||A||
||Dh (x)||
2r and it lies in
ℝ^{p−1}×
{0}
. The diameter of AB
(0,rη)
is no more than
||A ||
(2rη)
.Therefore, the measure of
the right side in 16.1 is no more than
p−1
[(||A||||Dh (x)||2r+ ||A||(2η))r] (rη)
≤ C (||A||,||Dh (x)||)(2r)pη
That is,
p
mp (A (h(B (x,r)))) ≤ C(||A ||,||Dh (x)||)(2r) η
Hence from the change of variables formula for linear maps,