5.7. A GENERALIZATION 121

that if |x− y| < δ , then | f ′ (x)− f ′ (y)| < ε and let m be so large that 1/m < δ/2. Then if∣∣x− km

∣∣< δ/2, it follows that∣∣x− xk,m

∣∣< δ and so

∣∣ f ′ (x)− f ′(xk,m)∣∣= ∣∣∣∣∣ f ′ (x)− f

( k+1m

)− f

( km

)1/m

∣∣∣∣∣< ε.

Now as before, letting M ≥ | f ′ (x)| for all x,

∣∣p′m (x)− f ′ (x)∣∣≤ m−1

∑k=0

(m−1

k

)xk (1− x)m−1−k ∣∣ f ′ (xk,m

)− f ′ (x)

∣∣≤∑{

x:|x− km |< δ

2

}(

m−1k

)xk (1− x)m−1−k

ε +Mm−1

∑k=0

(m−1

k

)4(k−mx)2

m2δ2 xk (1− x)m−1−k

≤ ε +4M14

m1

m2δ2 = ε +M

1

mδ2 < 2ε

whenever m is large enough. Thus this proves uniform convergence. ■There is a more general version of the Weierstrass theorem which is easy to get. It

depends on the Tietze extension theorem, a wonderful little result which is interesting forits own sake.

5.7 A GeneralizationThis is an interesting theorem which holds in arbitrary normal topological spaces. In par-ticular it holds in metric space and this is the context in which it will be discussed. First,review Lemma 3.12.1.

Lemma 5.7.1 Let H,K be two nonempty disjoint closed subsets of X . Then there existsa 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 hk ∈ B(x,1/k)∩H.Then hk → x and since H is closed, x ∈ H.) Similarly, if dist(x,K) = 0, then x ∈ K andso 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

(f (x)− 1

2

). Then g has the desired

properties. ■

Definition 5.7.2 For f : M ⊆ X→R, let ∥ f∥M ≡ sup{| f (x)| : x ∈M} . This is justnotation. I am not claiming this is a norm.

Lemma 5.7.3 Suppose M is a closed set in X and suppose f : M→ [−1,1] is continuousat every point of M. Then there exists a function, g which is defined and continuous on allof X such that ∥ f −g∥M ≤ 2

3 , g(X)⊆ [−1/3,1/3] . If X is a normed vector space,and f isodd, meaning that M is symmetric (x ∈M if and only if −x ∈M) and f (−x) = − f (x) .Then we can assume g is also odd.