⊆ ℝ^{n} where U is open and f is one to one and
continuous, then f
(U)
is also open. To do this, we first prove the following lemma. I found
something like this on the web. I liked it a lot because it shows how the Brouwer fixed point
theorem implies the invariance of domain. The other ways I know about involve
degree theory or some sort of algebraic topology. There is a proof of this in the
chapter on degree theory, but if it is all you want, here is a proof from Brouwer
fixed point theorem. This is a very important theorem in the study of manifolds
with boundary. It is also used in a crucial way in the reduction theorem for the
center manifold in ordinary differential equations in addition to being extremely
interesting in its own right. Compare to the inverse function theorem where one uses
differentiability to obtain that a mapping is open. This gets it done with just one to one and
continuous.
First I will give an informal argument to identify the main ideas. It is actually a valid
proof but leaves out details. Sometimes, I think that too many details obscure the main ideas
and so I am trying to give this informal argument hoping that if the details are too much, the
main idea will still be available.
Lemma 9.12.1Let B be a closed ball in ℝ^{n}centered at a which has radius r. Letf : B → ℝ^{n}. Then f
(a)
is an interior point of f
(B )
.
Proof:Since f
(B )
is compact and f is one to one, f^{−1} is continuous on f
(B )
. See
Problem 18 on Page 764. Use Tietze extension theorem on components of f^{−1} or some such
thing to obtain g : ℝ^{n}→ ℝ^{n} such that g is continuous and equals f^{−1} on f
(B)
.
Then multiply by a suitable truncation function to get g uniformly continuous on
ℝ^{n}.
Suppose f
(a )
is not an interior point of f
(B )
. Then there exists c_{k}→ f
(a)
but c_{k}
∕∈
f
(B )
.
In the picture, let C_{k} be a sphere whose radius is
2 |ck − f (a)|
PICT
Let ĝ_{k} be C^{1} and let it satisfy
∥ˆgk− g∥ ≡ max |ˆg(y)− g(y)| < εk
f(B)∪D y∈f(B)∪D
ε_{k} is very small, ε_{k}→ 0. How small will be considered later. Here D is a large closed disk
which contains all of the spheres C_{k} considered above. The idea is to have a large compact set
which includes everything of interest below.
To get ĝ_{k},you could use the Weierstrass approximation theorem, Theorem 3.12.7. An
easier way involving convolution will be presented in the next chapter. Also let a
∕∈
ĝ_{k}
(Ck )
.
This is no problem. C_{k} has measure zero and so ĝ
(Ck)
also has measure zero thanks to the
assumption that ĝ is C^{1} and Lemma 9.6.1. Therefore, you could simply add a small enough
nonzero vector to ĝ to preserve the above inequality of ĝ and g so that ĝ
(Ck )
no longer
contains a. That is, replace ĝ with ĝ + a − b where
|a − b|
is very small but b
∈∕
ĝ
(Ck )
.
There is a set Σ_{k} consisting of that part of f
(B )
which is outside of the sphere C_{k} in the
picture along with the sphere C_{k} itself. By construction, ĝ_{k} misses a on C_{k}. As to the other
part of Σ_{k}, g misses a on this part, because f is one to one and so f^{−1} is also. Now
we will squash the part of f
(B)
inside C_{k} onto C_{k} while leaving the rest of f
(B )
unchanged.
Let Φ_{k} be defined on f
(B )
( 2|ck − f (a)| )
Φk(y) ≡ max -----------,1 (y− ck)+ ck
|y − ck|
This Φ_{k} squishes the part of f
(B )
inside C_{k} to C_{k} and leaves the rest of f
(B )
unchanged.
Thus
Φk : f (B) → f (B )∩ [y : |y − ck| ≥ 2|ck − f (a)|]∪ Ck
a compact set. Now
∥ˆgk ∘ Φk − g∥
_{f(B)
}→ 0 and g misses a on the part of f
(B )
outside of C_{k}. In the above, we chose ĝ_{k} so close to g that it also misses a on the
part of f
(B)
which is outside of C_{k}. Then by construction, ĝ_{k} misses a on C_{k}
and so in fact ĝ_{k}∘ Φ_{k} misses a on f
(B)
. Now consider a + x −ĝ_{k}
(Φk (f (x)))
for
x ∈ B.
|a + x− ˆg (Φ (f (x)))− a| = |x− ˆg (Φ(f (x)))|
k k
= |g (f (x )) − ˆgk (Φk(f (x)))|
For f
(x)
outside of C_{k}, we could have chosen ĝ_{k} such that
∥g− ˆgk∥
_{f(B)
}<
r2
and
this was indeed done. When f
(x )
is inside C_{k}, then eventually, for large k, both
g
(f (x ))
,ĝ_{k}
(Φk (f (x)))
are close to g
(f (a))
. To see this,
|ˆgk (Φk (f (x)))− g (Φk(f (x)))|+ |g(Φk(f (x)))− g(f (x))|
≤ ∥ˆg − g∥ + |g (Φ (f (x)))− g(f (x))|
k f(B )∪D k
the last term being small for large k, and so for large k,x →a + x −ĝ_{k}
(Φk (f (x)))
maps
B to B and so by Brouwer fixed point theorem, it has a fixed point and hence
ĝ_{k}
(Φk (f (x)))
= a contrary to what was argued above. Hence, f
(a)
must be an interior point
after all. ■
Now here is the same lemma with the details.
Lemma 9.12.2Let B be a closed ball in ℝ^{n}centered at a which has radius r. Letf : B → ℝ^{n}. Then f
(a)
is an interior point of f
(B )
.
Proof:Since f
(B )
is compact and f is one to one, f^{−1} is continuous on f
(B )
. See
Problem 18 on Page 764. Use Tietze extension theorem on components of f^{−1} or some
such thing to obtain g : ℝ^{n}→ ℝ^{n} such that g is continuous and equals f^{−1} on
f
(B )
.
Suppose f
(a)
is not an interior point of f
(B)
. Then for every ε > 0 there exists
c_{ε}∈ B
(f (a),ε)
∖ f
(B )
. So fix ε small and refer to c_{ε} as c. ε will be so small that
r-
|g (y )− g(f (a))| < 10 for y ∈ B (f (a),4ε)
There is δ > 0 such that if
|x− ˆx|
< δ, then
|f (x)− f (ˆx)| < ε (9.25)
(9.25)
Let ĝ be C^{1} and on f
(B )
, let it satisfy
( )
∥ˆg − g∥ ≡ max |ˆg(y)− g(y)| < min -r, δ
f(B) y∈f(B) 10 2
To get ĝ_{k},you could use the Weierstrass approximation theorem, Theorem 3.12.7.
An easier way involving convolution will be presented in the next chapter. Also
let
a ∕∈ ˆg(∂B (c,2ε)).
This is no problem. ∂B
(a,2ε)
has measure zero and so ĝ
(∂B (c,ε))
also has measure zero
thanks to the assumption that ĝ is C^{1} and Lemma 9.6.1. Therefore, you could simply add a
small enough nonzero vector to ĝ to preserve the above inequality of ĝ and g so that
ĝ
(∂B (c,2ε))
no longer contains a. That is, replace ĝ with ĝ + a − b where
|a− b|
is very
small but b
∈∕
ĝ
(∂B (c,2ε))
. A summary of the rest of the argument is contained in the
following picture in which the sphere has radius 2ε.
PICT
There is a set Σ consisting of that part of f
(B )
which is outside of the sphere C in the
picture along with the sphere C itself. By construction, ĝ misses a on C. As to the other part
of Σ, that g misses a on this part, follows from the assumption that f is one to one and so f^{−1}
is also. Then ĝ missing a follows from ĝ being close enough to g. We define a continuous
mapping Φ which maps f
(B )
to this set Σ. This map squishes that part of f
(B )
which
is inside C onto C and does nothing to the part of f
(B)
which is outside of C.
This is where c not in f
(B )
but close to f
(B)
is used. Then we argue that ĝ∘ Φ
is continuous and close to g and misses a. It will be close to g and ĝ because of
the above assumption that everything inside C is close to f
(a)
and g and ĝ are
continuous. This will yield an easy contradiction from a use of the Brouwer fixed point
theorem.
Now let Φ be defined on f
(B)
( )
Φ (y) ≡ max --2ε--,1 (y− c) +c
|y − c|
If
|y − c|
≥ 2ε, then Φ
(y )
= y. If
|y − c|
< 2ε, then Φ
(y)
=
|y2ε−c|
(y− c)
+ c and
so
|Φ(y)− c| = 2ε |y-−-c|= 2ε
|y − c|
Note that this function is well defined because c
∕∈
f
(B )
. Thus
Φ : f (B) → f (B) ∩[y : |y − c| ≥ 2ε]∪ ∂B (c,2ε) ≡ Σ
a compact set. Now the interesting thing about this set Σ is this. For y ∈ Σ,ĝ