82 CHAPTER 3. STONE WEIERSTRASS APPROXIMATION THEOREM

This proves the lemma. Now continue the proof of Theorem 3.2.4.First note that A satisfies the same axioms as A but in addition to these axioms, A

is closed. The closure of A is taken with respect to the usual norm on C (A), ∥ f∥∞≡

max{| f (x)| : x ∈ A} . Thus A consists, by definition, of all functions in A along withall uniform limits of these functions. Suppose f ∈ A and suppose M is large enoughthat ∥ f∥

∞< M. Using Corollary 3.2.2, let pn be a sequence of polynomials such that

∥pn−|·∥|∞→ 0, pn (0) = 0. It follows that pn ◦ f ∈A and so | f | ∈A whenever f ∈A .Also note that

max( f ,g) =| f −g|+( f +g)

2,min( f ,g) =

( f +g)−| f −g|2

.

Therefore, this shows that if f ,g ∈ A then max( f ,g) , min( f ,g) ∈ A . By induction, iffi, i = 1,2, · · · ,m are in A then

max( fi, i = 1,2, · · · ,m) , min( fi, i = 1,2, · · · ,m) ∈A .

Now let h ∈ C (A;R) and let x ∈ A. Use Lemma 3.2.5 to obtain fxy, a function of Awhich agrees with h at x and y. Letting ε > 0, there exists an open set U (y) containing ysuch that

fxy (z)> h(z)− ε if z ∈U(y).

Since A is compact, let U (y1) , · · · ,U (yl) cover A. Let

fx ≡max(

fxy1 , fxy2 , · · · , fxyl

).

Then fx ∈A and fx (z)> h(z)−ε for all z∈ A and fx (x) = h(x). This implies that for eachx ∈ A there exists an open set V (x) containing x such that for z ∈ V (x), fx (z) < h(z)+ ε.Let V (x1) , · · · ,V (xm) cover A and let f ≡ min( fx1 , · · · , fxm).Therefore, f (z) < h(z)+ ε

for all z ∈ A and since fx (z)> h(z)−ε for all z ∈ A, it follows f (z)> h(z)−ε also and so| f (z)−h(z)|< ε for all z. Since ε is arbitrary, this shows h ∈A and proves A =C (A;R).■

3.3 The Case of a Closed Set in Rp

You can extend this theory to the case where A = X a closed set. More generally, this isdone with a locally compact Hausdorff space but this kind of space has not been consideredhere.

Definition 3.3.1 Let X be a closed set in Rp. C0 (X) denotes the space of real orcomplex valued continuous functions defined on X with the property that if f ∈C0 (X) , thenfor each ε > 0 there exists a compact set K such that | f (x)|< ε for all x ∈ X \K. Define

∥ f∥∞= sup{| f (x)| : x ∈ X}.

These functions are said to vanish at infinity.

Lemma 3.3.2 For X a closed set inRp with the above norm, C0 (X) is a complete space,meaning that every Cauchy sequence converges.