10.5. TIETZE EXTENSION THEOREM 267

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 10.5.2 For f : M ⊆ X → R, let ∥ f∥M ≡ sup{| f (x)| : x ∈M} . This is just no-tation. I am not claiming this is a norm.

Lemma 10.5.3 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 10.5.1 there existsg 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 = /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 10.5.4 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 10.5.3, 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

. (10.12)

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

)

10.5. TIETZE EXTENSION THEOREM 267Proof: Let f(x) = worst: The denominator is never equal to zero becauseif dist (x,H) = 0, then x € H because H is closed. (To see this, pick hy € B(w,1/k) NH.Then hy — x and since H is closed, x € H.) Similarly, if dist (a,K) = 0, then w € K andso the denominator is never zero as claimed. Hence f is continuous and from its definition,f =0 on H and f = 1 0n K. Now let g(x) = 3 (f(x) —3).- Then g has the desiredproperties.Definition 10.5.2 For f: MCX OR, let ||f ||, = sup {|f (x)|: 2 €M}. This is just no-tation. Iam not claiming this is a norm.Lemma 10.5.3 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\ly < z, g(X) C [-1/3,1/3]. If X is anormed vector space,and f isodd, meaning that M is symmetric (x € M if and only if —x € M) and f (—x) = —f (a).Then we can assume g is also odd.Proof: Let H = f~' ([-1,—1/3]),K =f! ({1/3, 1]) . Thus H and K are disjoint closedsubsets of M. Suppose first H,K are both nonempty. Then by Lemma 10.5.1 there existsg such that g is a continuous function defined on all of X and g(H) = —1/3, g(K) = 1/3,and g(X) C [-1/3, 1/3]. It follows ||,f — g||,, < 2/3. If H =9, 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 = 90, let g =0.When M is symmetric and f is odd, g(a) = } ae= 3 dist(a H)tdist(@ K)* When z € H this givesSey = —}- Then a € K, this gives 53527) = 4. Also g(H) = —1/3, f(H) C[—1,—1/3] so for x € H,|g(a) — f(ax)| < 2. It is similar for x € K. If x is in neither Hnor K, then g(a) € [-1/3, 1/3] and so is f (a). Thus || f — g||y < 4. Now by assumption,since f is odd, H = —K. It is clear that g is odd becauseI dist(—ax,H) —dist(—x,K) _ 1 dist(—x,—K) — dist(—a, —H)3 dist(—a,H)+dist(—a,K) 3 dist(—w,—K)+dist (—a2,—H)1 dist (a, K) — dist (a,H)= = = . |_|3 dist(w,K) +dist(a,H) 8")g(-") =Lemma 10.5.4 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 10.5.3, let g; be such that g) (X) C [—1/3, 1/3] and || f — gi||y <3. Suppose g1,--+ ,8m have been chosen such that g;(X) C [—1/3, 1/3] andf- (5) 8i <(5) ; (10.12)EG) ol <GThis has been done for m = 1. Then | (3)” (¢-m 1 (3)' ‘8i)i=3) (-£() «)M| <1 and soM