This is an interesting theorem which holds in arbitrary normal topological spaces. However, I am
specializing to ℝ^{p} to keep the emphasis on that which is most familiar. The presentation depends on
Lemma 2.2.46.
Lemma 2.3.1Let H,K be two nonempty disjoint closed subsets of ℝ^{p}. Then there exists acontinuous function, g : ℝ^{p}→
[− 1∕3,1∕3]
such that g
(H)
= −1∕3, g
(K )
= 1∕3,g
(ℝp)
⊆
[− 1∕3,1∕3]
.
Proof: Let
dist(x,H )
f (x) ≡--------------------.
dist(x,H )+ dist(x,K)
The denominator is never equal to zero because if dist
(x,H )
= 0, then x ∈ H because H is closed. (To see
this, pick h_{k}∈ B
(x,1∕k )
∩ H. Then h_{k}→ x and since H is closed, x ∈ H.) Similarly, if dist
(x,K )
= 0,
then x ∈ K and so the denominator is never zero as claimed. Hence f is continuous and from its definition,
f = 0 on H and f = 1 on K. Now let g
(x)
≡
2
3
( 1)
f (x)− 2
. Then g has the desired properties.
■
Definition 2.3.2For f : M ⊆ ℝ^{p}→ ℝ, define
∥f∥
_{M}as sup
{|f (x)| : x ∈ M }
. This is justnotation. I am not claiming this is a norm.
Lemma 2.3.3Suppose M is a closed set in ℝ^{p}and suppose f : M →
[− 1,1]
is continuous at everypoint of M. Then there exists a function, g which is defined and continuous on all of ℝ^{p}such that
||f − g||
_{M}<
2
3
,g
p
(ℝ )
⊆
[− 1∕3,1∕3]
.
Proof: Let H = f^{−1}
([− 1,− 1∕3])
,K = f^{−1}
([1∕3,1])
. Thus H and K are disjoint closed subsets of M.
Suppose first H,K are both nonempty. Then by Lemma 2.3.1 there exists g such that g is a continuous
function defined on all of ℝ^{p} and g
(H )
= −1∕3, g
(K )
= 1∕3, and g
(ℝp)
⊆
[− 1∕3,1∕3]
. It follows
||f − g||
_{M}< 2∕3. If H = ∅, then f has all its values in
[− 1∕3,1]
and so letting g ≡ 1∕3, the desired
condition is obtained. If K = ∅, let g ≡−1∕3. ■
Lemma 2.3.4Suppose M is a closed set in ℝ^{p}and suppose f : M →
[− 1,1]
is continuous atevery point of M. Then there exists a function g which is defined and continuous on all of ℝ^{p}suchthat g = f on M and g has its values in
[− 1,1]
.
Proof: Using Lemma 2.3.3, let g_{1} be such that g_{1}
can play the role of f in the first step of the proof. Therefore,
there exists g_{m+1} defined and continuous on all of ℝ^{p} such that its values are in
[− 1∕3,1∕3]
and
∥ ( ) ∥
∥∥(3)m m∑ (2)i− 1 ∥∥ 2
∥∥ 2 f − 3 gi − gm+1∥∥ ≤ 3.
i=1 M
Hence
∥( ) ∥
∥∥ ∑m ( 2)i−1 ( 2)m ∥∥ ( 2)m+1
∥∥ f − 3 gi − 3 gm+1∥∥ ≤ 3 .
i=1 M
so the Weierstrass M test applies and shows convergence is uniform. Therefore g must be continuous by
Theorem ??. The estimate 2.14 implies f = g on M. ■
The following is the Tietze extension theorem.
Theorem 2.3.5Let M be a closed nonempty subset of ℝ^{p}and let f : M →
[a,b]
be continuous atevery point of M. Then there exists a function, g continuous on all of ℝ^{p}which coincides with f onM such that g
p
(ℝ )
⊆
[a,b]
.
Proof: Let f_{1}
(x)
= 1 +
b2−a
(f (x )− b)
. Then f_{1} satisfies the conditions of Lemma 2.3.4 and
so there exists g_{1} : ℝ^{p}→
[− 1,1]
such that g is continuous on ℝ^{p} and equals f_{1} on M. Let
g