84 CHAPTER 3. STONE WEIERSTRASS APPROXIMATION THEOREM

•(⃗0,2)

(⃗0,1)•

p

θ(p)

Rn

This map θ is one to one onto Sp \{(⃗

0,2)}

. More precisely, if you have (⃗a,an+1)

on Sp \(⃗

0,2)

to get θ−1 (⃗a,an+1) , you form the line from

(⃗0,2)

through this point and

see where it hits Rp. The line is(⃗

0,2)+ t((⃗a,an+1)−

(⃗0,2))

and it hits Rp when 2+

t (an+1−2) = 0 which is when t = 22−an+1

. Thus θ−1 (⃗a,an+1) =

(2⃗a

2−an+1,0). From this

formula, it is clear that θ−1 is continuous and one to one. It is also onto because if x⃗ ∈ Rp

you can take the line from(⃗

0,2)

to (⃗x,0) and where it intersects Sp is the point which iswanted. It is also easy to see from this that θ is continuous. Indeed, suppose x⃗k → x⃗ inRp. Does it follow that θ (⃗xk)→ θ (⃗x)? We know that {⃗xk} is bounded since it converges.Therefore, there is an open ball, B

((⃗0,2),r)

such that θ (⃗xk) ∈ Sp \B((⃗

0,2),r)≡ K a

compact set. If θ (⃗xk) fails to converge to θ (⃗x) , then there is a subsequence, still denotedas θ (⃗xk) such that θ (⃗xk)→ y ∈ K where y ̸= θ (⃗x) . But then, the continuity of θ

−1 impliesxk→ θ

−1 (y) and so θ−1 (y) = x which implies y = θ (x) , a contradiction. Thus both θ and

θ−1 are continuous, one to one and onto mappings between Rp and Sp \

{(⃗0,2)}

.

Theorem 3.3.3 Let A be an algebra of functions of C0 (X ,R) which separates thepoints of the closed set X ⊆Rp and annihilates no point of X. Then A is dense in C0 (X ;R).

Proof: Ã denote all finite linear combinations of the form{n

∑i=1

ci f̃i + c0 : f ∈A , ci ∈ R

}

where for f ∈C0 (X ;R) ,

f̃ (x)≡

{f(θ−1 (x)

)if x ∈ θ (X)

0 if x =(⃗

0,2) .

Then à is obviously an algebra of functions in C (Sp;R). It separates points because thisis true of A . Similarly, it annihilates no point because of the inclusion of c0 an arbitraryelement of R in the definition of à above. Therefore from Theorem 3.2.4, à is dense inC (Sp;R) . Letting f ∈C0 (X ;R) , it follows f̃ ∈C (Sp;R) . It is clearly continuous on θ (X) .

What about at(⃗

0,2)

? If you have xn→(⃗

0,2), then

∣∣θ−1 (xn)∣∣→ ∞ and therefore, since

f ∈C0, f(θ−1 (xn)

)≡ f̃ (xn)→ 0≡ f̃

((⃗0,2))

and so indeed f̃ is in C (Sp;R) as claimed.

Thus there exists a sequence {hn} ⊆ Ã such that hn converges uniformly to f̃ . Now hn is

84 CHAPTER 3. STONE WEIERSTRASS APPROXIMATION THEOREMThis map @ is one to one onto S? \ { (6.2) \. More precisely, if you have (4, dan+1)on S? \ (6.2) to get 9! (@,an41), you form the line from (0.2) through this point andsee where it hits R?. The line is (6,2) +t (a, Anti) — (6.2) ) and it hits R? when 2+oh | 2 -l(> 2a :t (an41 —2) = 0 which is when t = T<a,n, Thus 6 (G,dn41) = (24..0) . From thisformula, it is clear that @~! is continuous and one to one. It is also onto because if ¥ € R?you can take the line from (6. 2) to (¥,0) and where it intersects S? is the point which iswanted. It is also easy to see from this that @ is continuous. Indeed, suppose x, — X inR?. Does it follow that 6 (%;,) —> 6 ()? We know that {x;,} is bounded since it converges.Therefore, there is an open ball, B ((6.2) .r) such that 6 (x;,) € S?\B ( (6.2) .r) =Kacompact set. If 6 (x;) fails to converge to 6 (%), then there is a subsequence, still denotedas 0 (X;) such that 0 (x,) + y € K where y 4 6 (X) . But then, the continuity of @~' impliesxz > 0~!(y) and so 6”! (y) =x which implies y = 6 (x), a contradiction. Thus both @ and6~' are continuous, one to one and onto mappings between R? and S? \ { (6, 2) \ .Theorem 3.3.3 Let & be an algebra of functions of Co (X ,IR) which separates thepoints of the closed set X CR? and annihilates no point of X. Then & is dense in Co (X;R).Proof: ra denote all finite linear combinations of the formn ~Vcifitco: fed, cq ERi=lwhere for f € Co (X;R),~ . { f(@-' (x) ifxe (xX)fx) =| Oifx= (6.2)Then ./ is obviously an algebra of functions in C(S?;R). It separates points because thisis true of </. Similarly, it annihilates no point because of the inclusion of co an arbitraryelement of R in the definition of & above. Therefore from Theorem 3.2.4, & is dense inC(S?;R). Letting f € Co (X;R), it follows f € C(S?;R). It is clearly continuous on 6 (X).What about at (0.2) ? If you have x, — (6,2) , then |o-! (xn)| —> oo and therefore, sincefECQ,f (07! (Xn)) = f(x») 20= F((6.2)) and so indeed f is in C ($?;IR) as claimed.Thus there exists a sequence {h,,} C & such that h, converges uniformly to f. Now hy, is