What is so special about polynomials that allows them to uniformly approximate continuous functions? Can we find other families of nice functions that uniformly approximate continuous functions? We now wish to generalizeTheorem 3.32to general metric spaces and to generalize to collections other than polynomials.
Definition 3.39 (Algebra). LetA be a collection of real valued functions all defined on some set E. We say A is an algebra if it is closed under pointwise addition, scalar multiplication, and pointwise multiplication, that is, given f, g ∈A and λ ∈ R, then f + g ∈ A , λ · f ∈ A , and f · g ∈ A .
Definition 3.40. Let A be a collection of real valued functions all defined on some set E. We say A is uniformly closed if f ∈A whenever fn∈A for n ∈ N, and fn → f uniformly on E.
Given this collectionA , we define the uniform closure B of A as
B = {f : fn∈A for all n ∈ N such that fn→ f uniformly}, that is,B is the collection of uniform limit points of sequences in A .
Let A be the set of polynomials ddefined on [a, b]. By Stone-Weierstrass, the uniform closure of A is C ([a, b]). This means
(a) If there exists a sequence of polynomials (Pn)n∈N such that Pn → f uniformly on [a, b], then f is continuous on [a, b].
(b) For f ∈C ([a, b]), there exists a sequence of polynomials (Pn)n∈Nin A such that Pn→ f uniformly on [a, b].
Theorem 3.41. Let A be an algebra of bounded real-valued functions on a compact set X and B be the uniform closure of A . Then, B is a uniformly closed algebra. In other words, the uniform closure of an algebra is still an algebra.
Proof. Let f, g ∈B and λ ∈ R. Then, there are uniformly convergent sequences (fn)n∈Nand (gn)n∈N in A such that fn → f and gn∈ g uniformly. Since A is an algebra, so fn+ λ · gn∈A for n ∈ N. Now,
d∞(fn+ λ · gn, f + λ · g) = sup x∈X |(fn(x) + λ · gn(x)) − (f (x) + λ · g(x))| ≤ sup x∈X |fn(x) − f (x)| + λ · sup x∈X |gn(x) − g(x)| = d∞(fn, f ) + λ · d∞(gn, g).
Since d∞(fn, f ) → 0 and d∞(gn, g) → 0, then
d∞(fn+ λ · gn, f + λ · g) → 0.
Therefore, f + λ · g ∈B, which means that B is closed under addition and scalar multiplication.
Since f and g are defined on X, which is compact, so there exists an M1, M2∈ R such that |fn(x)| < M1 and |gn(x)| < M2 for all n ∈ N and x ∈ X. Fix ε > 0. Since fn→ f and gn → g uniformly, so there exists N1, N2 ∈ N such that |fn(x) − f (x)| < ε
2M2 for all n ≥ N1 and |gn(x) − g(x)| < ε
2M1 for all n ≥ N2. Let N = max{N1, N2}. Then, for n ≥ N and x ∈ X, we have
|fn(x) · gn(x) − f (x) · g(x)| = |fn(x) · gn(x) − fn(x) · g(x) + fn(x) · g(x) − f (x) · g(x)| ≤ |fn(x)| · |gn(x) − g(x)| + |g(x)| · |fn(x) − f (x)| < M1· ε 2M1 + M2· ε 2M2 = ε.
So, fn· gn→ f · g uniformly on X. As A is an algebra, so fn· gn ∈A for n ∈ N, and this proves that the product f · g is in the uniform closure ofA . Thus, B is closed under pointwise multiplication. So, B is an algebra.
Let (fn)n∈Nbe an uniformly convergent sequence of functions defined on X fromB. There is a sequence of functions (gn)n∈Nsuch that |fn(x) − gn(x)| < 1
n. If fn→ f , then gn → f as well, so by the definition of uniform closureB, so f ∈ B and B is uniformly closed.
Definition 3.42. LetA be a family of functions on a set E. We say that A separates points on E if for every x1, x2∈ E such that x16= x2, there exists f ∈A such that f(x1) 6= f (x2), andA vanishes at no point of E if for every x ∈ E, there exists g ∈A such that g(x) 6= 0.
Example 3.43 (Polynomials).
(a) The algebraA = {P : polynomials on R} is separating and vanishing nowhere.
(b) The algebra A1 = {P : polynomials on R such that P (x) = P (−x) for x ∈ R} is not separating. Consider x1= 1 and x2= −1, for every P ∈A1, we have P (1) = P (−1).
(c) The algebraA2 = {P : polynomials on R such that P (−x) = −P (x) for x ∈ R} is vanishing some- where. For every P ∈ A2, we have P (x) + P (−x) = 0, so P (0) = 0, which means A2 vanishes at x = 0.
Let X be a compact metric space. By the Extreme Value Theorem, the function f : X → R is bounded. Thus, the supremum norm d∞(f, g) = sup
x∈X
|f (x) − g(x)| is a genuine metric on C (X). Let (fn)n∈N be a sequence of functions inC (X). Then, we say (fn) uniformly approximates a function f ∈C (X) means that d∞(fn, f ) → 0 as n → ∞. In this setting, the Stone-Weierstrass Approximation Theorem says that every function f ∈C ([a, b]) is the uniform limit of some sequence (pn) of polynomials on [a, b], that is, the set of polynomials is dense.
Theorem 3.44 (Stone-Weierstrass). Let A be an algebra of functions f : K → R where K is compact. If A separates points on K and vanishes at no point of K, then the uniform closure B of A is the set of continuous functions on K, that is,C (K).
The real part of Theorem 3.44 is the statement that a real sub-algebra ofC (K) that separates points and vanishes nowhere is dense inC (K). In light ofTheorem 3.41, we aim to prove that if A ⊆ C (K) is a uniformly closed algebra that separates points and vanishes nowhere, thenA = C (K). To do that, we need two more lemmas first.
Lemma 3.45. Suppose A is an algebra of functions on E that separates points on E and vanishes at no point of E. Suppose x1, x2∈ E but x1 6= x2 and c1, c2 ∈ R. Then, there exists a function f ∈ A such that f (x1) = c1 and f (x2) = c2.
Proof. By assumption, there exists g, h, k : E → R and g, h, k ∈ A such that g(x1) 6= g(x2), h(x1) 6= 0, and k(x2) 6= 0.
Let
u = g · k − g(x1) · k and v = g · h − g(x2) · h and u, v ∈A . Note that u(x1) = 0 = v(x2). Then,
u(x2) = (g(x2) − g(x1)) · k(x2) 6= 0 and v(x1) 6= 0. Now, define
f = c1· v v(x1)
+ c2· u u(x2) . Then, f ∈A , and we conclude that f(x1) = c1 and f (x2) = c2.
Lemma 3.46. Let K be a compact metric space, and let A ⊆ C (K) be a uniformly closed algebra. Then, for each f ∈A , |f| ∈ A as well. Consequently, if f, g ∈ A , then max{f, g} and min{f, g} are in A as well. Proof. Let M = sup
x∈K
|f (x)|. Fix an ε > 0. By Stone-Weierstrass Approximation Theorem, there exists a sequence of polynomials (Pn)n∈N such that Pn → |y| uniformly on [−M, M ]. In particular, for ε > 0, there exists c1, . . . , cn∈ R such that
n X i=1 ci· yi− |y| < ε
for −M ≤ y ≤ M . SinceA is an algebra, then the function g = n X i=1 ci· fi is in A . Then, |g(x) − |f (x)|| = n X i=1 ci· (f (x))i− |f (x)| < ε
for x ∈ K. Consider a sequence (εn) such that εn → 0, and let gn(x) ∈A be the corresponding function such that |gn− |f || < ε, then gn→ |f | uniformly on K and gn ∈A for n ∈ N. Since A is uniformly closed, so |f | ∈A . We note that max{f, g} =f + g 2 + |f − g| 2 and min{f, g} = f + g 2 − |f − g| 2 .
By the definition of algebra and prior paragraph, we conclude that max{f, g} ∈A and min{f, g} ∈ A as well. Further, by induction, for f1, . . . , fn ∈A , then max
1≤i≤nfi, min1≤i≤nfi∈A .
Lecture 21 Wednesday February 24 We are finally in a position to complete the proof for a more general version of Stone-Weierstrass Theorem
(in a compactness setting).
Proof of Theorem 3.44. Fix a function f ∈C (K). Fix a point x ∈ K. Since A ⊆ B and A separates points and vanishes at no points of K, then for any point y ∈ K, by Lemma 3.45, there exists a function hy such that
hy(x) = f (x) and hy(y) = f (y).
Fix an ε > 0. By continuity of hy ∈B, there exists an open neighborhood Vy ∈ K such that y ∈ Vy and hy(t) > f (y) −
ε
2 and f (t) < f (y) + ε
2 for all t ∈ Vy. In particular, hy(t) > f (t) − ε for all t ∈ Vy. Now, K is compact, and {Vy : y ∈ K} covers K, so there exists y1, . . . , yn such that K ⊆Sni=1Vyi. Define
gx= max{hy1, hy2, . . . , hyn}.
By (induction on) Lemma 3.46, gx ∈ B. By contruction, gx(t) ≥ hyi(t) > f (t) − ε for all t ∈ K. The function hy satisfies hy(x) = f (x), so gx(x) = f (x) for x ∈ K.
Since gx and f are continuous, there is an open neighborhood Vxsuch that x ∈ Vx and gx(u) < f (x) + ε for all u ∈ Vx. (Consider r = gx− f , then r(x) = 0 < ε. Then, there is a neighborhood Vx of x such that r(t) < ε for t ∈ Vx, so gx(t) − f (t) < ε for t ∈ Vx. Thus, gx(t) < f (t) + ε for all t ∈ Vx.) Let {Vx: x ∈ K} covers K. By compactness of K, there exists x1, . . . , xm∈ K such that K ⊆Smi=1Vxi. Now, define
h = min{gx1, gx2, . . . , gxm}.
Again by (induction on) Lemma 3.46, h ∈ B. Thus, h(t) > f(t) − ε for t ∈ K. By construction, since K ⊆Smi=1Vxi, there is 1 ≤ i ≤ m such that t ∈ Vxi, so
h(t) ≤ gxi(t) < f (t) + ε.
Therefore, f (t) − ε < h(t) < f (t) + ε for t ∈ K. We conclude that for every ε > 0, there is a function h ∈B such that |f (t) − h(t)| < ε for all t ∈ K.
Let εn= 1
n. We can construct a sequence of functions fn∈B such that |fn(t) − f (t)| < εn=
1 n
for t ∈ K. So, fn→ f uniformly on K. SinceB is uniformly closed, so f ∈ B, which means C (K) ⊆ B. By assumption,B is the uniform closure of a family of continuous functions on compact set K, then B ⊆ C (K) since the uniform limit of continuous function is continuous. Hence,C (K) = B.