As an application of the inverse function theorem is a simple proof of the important
invariance of domain theorem which says that continuous and one to one functions defined on
an open set in ℝ^{n} with values in ℝ^{n} take open sets to open sets. You know that this is true
for functions of one variable because a one to one continuous function must be either strictly
increasing or strictly decreasing. This will be used when considering orientations of curves
later. However, the n dimensional version isn’t at all obvious but is just as important if you
want to consider manifolds with boundary for example. The need for this theorem occurs in
many other places as well in addition to being extremely interesting for its own
sake.
Lemma 6.10.1Let f be continuous and map B
(p,r)
⊆ ℝ^{n}to ℝ^{n}. Suppose that for allx ∈B
(p,r)
,
|f (x)− x| < εr
Then it follows that
f (B-(p,-r)) ⊇ B (p,(1 − ε)r)
Proof:This is from the Brouwer fixed point theorem, Corollary 4.3.2. Consider for
y ∈ B
(p,(1 − ε)r)
h (x ) ≡ x− f (x) +y
Then h is continuous and for x ∈B
(p,r)
,
|h(x) − p | = |x − f (x)+ y− p | < εr+ |y − p| < εr+ (1− ε)r = r
Hence h :B
(p,r)
→B
(p,r)
and so it has a fixed point x by Corollary 4.3.2. Thus
x− f (x)+ y = x
so f
(x)
= y. ■
The notation
∥f∥
_{K} means sup_{x∈K}
|f (x )|
.
Lemma 6.10.2Let K be a compact set in ℝ^{n}and let g : K → ℝ^{n}be continuous,z ∈ K isfixed. Let δ > 0. Then there exists a polynomial q (each component a polynomial) suchthat
∥q− g∥K < δ, q(z) = g (z), Dq (z)− 1 exists
Proof:By the Weierstrass approximation theorem, Theorem 3.12.7, (apply
this theorem to the algebra of real polynomials) there exists a polynomial
ˆq
such
that
∥ˆq− g ∥ < δ
K 3
Then define for y ∈ K
q (y ) ≡ ˆq(y)+ g (z) −qˆ(z)
Then
q (z) = ˆq(z)+ g(z)− ˆq (z) = g(z)
Also
|q(y) − g (y)| ≤ |(ˆq(y)+ g (z)− ˆq (z))− g (y)|
≤ |ˆq(y)− g (y)|+ |g(z)− ˆq (z)| < 2δ
3
(Actually, since g
(z)
=
ˆq
(z)
, we could put δ∕3 on the right.) and so since y was
arbitrary,
∥q− g∥ ≤ 2δ-< δ
K 3
If Dq
(z)
^{−1} exists, then this is what is wanted. If not, let
0 < η < {|λ | : λ is an eigenvalue of Dq (z),λ ⁄= 0}
Then if η is small enough, q
(y)
could be replaced with q
(y)
+ η
(y − z)
and the above
inequality would be preserved along with q
(z)
= g
(z)
but now Dq
(z)
would have
no zero eigenvalues and would therefore be invertible. Simply use the modified q.
■
Lemma 6.10.3Let f :B
(p,r)
→ ℝ^{n}where the ball is also in ℝ^{n}. Let f be one to one, fcontinuous. Then there exists δ > 0 such that
(------)
f B (p,r) ⊇ B (f (p ),δ) .
In other words, f
(p)
is an interior point of f
(------)
B (p,r)
.
Proof: Since f
( )
B-(p,r)
is compact, it follows that f^{−1} : f
( )
B-(p,r)
→B
(p,r)
is
continuous. By Lemma 6.10.2, there exists a polynomial q : f
( )
q∘ f B-(p,r) ⊇ B (p,(1− ε)r) = B(q (f (p)),(1− ε)r)
So do q^{−1‘} to both sides. Then you would have f
(------)
B (p,r)
⊇ q^{−1}
(B (q(f (p)),(1− ε)r))
,
and it becomes a question whether the set on the right contains an open set B
(f (p),δ)
and
whether g^{−1} exists. This is where the inverse function theorem is used. One reduces
(1− ε)
r
to make this happen.
PICT
By the inverse function theorem, there is an open set containing f
(p)
denoted as W such
that on W, q is one to one and q and its inverse, defined on an open set V = q
(W )
both map
open sets to open sets.
PICT
By the construction, p = q
(f (p))
∈ V and so if η is small enough, it follows
that
(------)
B (p,η) = B (q (f (p)),η) ⊆ B (p,(1− ε)r)∩ V ⊆ q∘ f B (p,r) .
and q,q^{−1} both map open sets to open sets. Thus q^{−1}
(B (q(f (p)),η))
is an open set
containing f
(p )
and this open set is contained in f
(------)
B (p,r)
. Hence if δ is small
enough,
(------)
f B (p,r) ⊇ B (f (p),δ) ■
With this lemma, the invariance of domain theorem comes right away. This remarkable
theorem states that if f : U → ℝ^{n} for U an open set in ℝ^{n} and if f is one to one and
continuous, then f
(U)
is also an open set in ℝ^{n}.
Theorem 6.10.4Let U be an open set in ℝ^{n}and let f : U → ℝ^{n}be onetoone and continuous. Then f