5.8. A GENERALIZATION 135

Definition 5.8.2 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 5.8.3 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.

Proof: Let H = f−1 ([−1,−1/3]) ,K = f−1 ([1/3,1]) . Thus H and K are disjoint closedsubsets of M. Suppose first H,K are both nonempty. Then by Lemma 5.8.1 there exists gsuch 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 = /0, then f has all its values in[−1/3,1] and so letting g≡ 1/3, the desired condition is obtained. If K = /0, let g≡−1/3.If both H,K = /0, let g = 0.

When M is symmetric and f is odd, g(x)≡ 13

dist(x,H)−dist(x,K)dist(x,H)+dist(x,K) . When x ∈H this gives

13−dist(x,K)dist(x,K) = − 1

3 . Then x ∈ K, this gives 13

dist(x,H)dist(x,H) =

13 . Also g(H) = −1/3, f (H) ⊆

[−1,−1/3] so for x ∈ H, |g(x)− f (x)| ≤ 23 . It is similar for x ∈ K. If x is in neither H

nor K, then g(x) ∈ [−1/3,1/3] and so is f (x) . Thus ∥ f −g∥M ≤ 23 . Now by assumption,

since f is odd, H =−K. It is clear that g is odd because

g(−x) =13

dist(−x,H)−dist(−x,K)

dist(−x,H)+dist(−x,K)=

13

dist(−x,−K)−dist(−x,−H)

dist(−x,−K)+dist(−x,−H)

=13

dist(x,K)−dist(x,H)

dist(x,K)+dist(x,H)=−g(x) . ■

Lemma 5.8.4 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 g = f on M and g has its values in [−1,1] . If X is a normed linear spaceand f is odd, then we can also assume g is odd.

Proof: Using Lemma 5.8.3, let g1 be such that g1 (X)⊆ [−1/3,1/3] and ∥ f −g1∥M ≤23 . Suppose g1, · · · ,gm have been chosen such that g j (X)⊆ [−1/3,1/3] and∥∥∥∥∥ f −

m

∑i=1

(23

)i−1

gi

∥∥∥∥∥M

<

(23

)m

. (5.2)

This has been done for m = 1. Then∥∥∥( 3

2

)m(

f −∑mi=1( 2

3

)i−1gi

)∥∥∥M≤ 1 and so

(32

)m(

f −m

∑i=1

(23

)i−1

gi

)

can play the role of f in the first step of the proof. Therefore, there exists gm+1 defined andcontinuous on all of X such that its values are in [−1/3,1/3] and∥∥∥∥∥

(32

)m(

f −m

∑i=1

(23

)i−1

gi

)−gm+1

∥∥∥∥∥M

≤ 23.