This is an interesting theorem which holds in arbitrary normal topological spaces. In
particular it holds in metric space and this is the context in which it will be discussed. First,
here is a useful lemma.
Lemma 3.12.1Let X be a metric space and let S be a nonempty subset ofX.
dist(x,S ) ≡ inf{d(x,z) : z ∈ S }
Then
|dist(x,S) − dist(y,S)| ≤ d(x,y).
Proof:Say dist
(x,S )
≥dist
(y,S)
. Then letting ε > 0 be given, there exists z ∈ S such
that
≤ d (x,z) − (d(y,z)− ε) ≤ d (x,y)+ d(y,z)− d(y,z)+ ε = d (x,y )+ ε
Since ε is arbitrary,
|dist(x,S )− dist(y,S)| ≤ d(x,y)
It is similar if dist
(x,S)
<dist
(y,S)
. ■
Then this shows that x →dist
(x,S)
is a continuous real valued function.
Lemma 3.12.2Let H,K be two nonempty disjoint closed subsets of X. Thenthere exists a continuous function, g : X →
[− 1∕3,1∕3]
such that g
(H)
= −1∕3,g
(K )
= 1∕3,g
(X )
⊆
[− 1∕3,1∕3]
.
Proof: Let
f (x ) ≡------dist(x,H-)-----.
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)
≡
23
( )
f (x)− 12
. Then g has the desired properties.
■
Definition 3.12.3For f : M ⊆ X → ℝ, define
∥f∥
_{M}as
sup
{|f (x )| : x ∈ M }
. This is just notation. I am not claiming this is a norm.
Lemma 3.12.4Suppose M is a closed set in X and suppose f : M →
[− 1,1]
iscontinuous at every point of M. Then there exists a function, g which is defined andcontinuous on all of X such that
||f − g||
_{M}<
23
,g
(X)
⊆
[− 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 3.12.2 there exists g
such that g is a continuous function defined on all of X and g
(H)
= −1∕3, g
(K )
= 1∕3, and
g
(X )
⊆
[− 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 3.12.5Suppose M is a closed set in X and suppose f : M →
[− 1,1]
iscontinuous at every point of M. Then there exists a function g which is defined andcontinuous on all of X such that g = f on M and g has its values in
[− 1,1]
.
Proof: Using Lemma 3.12.4, let g_{1} be such that g_{1}
||∞ ( )i− 1 || m ( )i−1
|g(x)| ≤ ||∑ 2 g (x)||≤ ∑ 2 1 ≤ 1
|i=1 3 i | i=1 3 3
and
||( )i− 1 || ( )i−1
|| 2 gi(x)||≤ 2 1
| 3 | 3 3
so the Weierstrass M test applies and shows convergence is uniform. Therefore
g must be continuous by Theorem 2.9.3. The estimate 3.28 implies f = g on M.
■
The following is the Tietze extension theorem.
Theorem 3.12.6Let M be a closed nonempty subset of X and let f : M →
[a,b]
be continuous at every point of M. Then there exists a function, g continuous onall of X which coincides with f on M such that g
(X )
⊆
[a,b]
.
Proof: Let f_{1}
(x)
= 1 +
b2−a
(f (x)− b)
. Then f_{1} satisfies the conditions of Lemma 3.12.5
and so there exists g_{1} : X →
[− 1,1]
such that g is continuous on X and equals f_{1} on M. Let
g
(x)
=
(g1(x)− 1)
( )
b−2a
+ b. This works. ■
With the Tietze extension theorem, here is a better version of the Weierstrass
approximation theorem.
Theorem 3.12.7Let K be a closed and bounded subset of ℝ^{n}and let f : K → ℝ becontinuous. Then there exists a sequence of polynomials
{p }
m
such that
lim (sup{|f (x)− pm (x )| : x ∈ K }) = 0.
m →∞
In other words, the sequence of polynomials converges uniformly to f on K.
Proof: By the Tietze extension theorem, there exists an extension of f to a continuous
function g defined on all ℝ^{n} such that g = f on K. Now since K is bounded, there exist
intervals,
[ak,bk]
such that
∏n
K ⊆ [ak,bk] = R
k=1
Then by the Weierstrass approximation theorem, Theorem 3.11.5 there exists a sequence of
polynomials
{pm }
converging uniformly to g on R. Therefore, this sequence of
polynomials converges uniformly to g = f on K as well. This proves the theorem.
■
By considering the real and imaginary parts of a function which has values in ℂ one can
generalize the above theorem.
Corollary 3.12.8Let K be a closed and bounded subset of ℝ^{n}and let f : K → F becontinuous. Then there exists a sequence of polynomials
{pm}
such that
lim (sup{|f (x)− pm (x )| : x ∈ K }) = 0.
m →∞
In other words, the sequence of polynomials converges uniformly to f on K.
More generally, the function f could have values in ℝ^{n}. There is no change in the proof.
You just use norm symbols rather than absolute values and nothing at all changes
in the theorem where the function is defined on a rectangle. Then you apply the
Tietze extension theorem to each component in the case the function has values in
ℝ^{n}.
class=”left” align=”middle”(V,W) As A Vector Space3.13. CONNECTEDNESS IN
NORMED LINEAR SPACE