There is a profound generalization of the Weierstrass approximation theorem due to Stone. It
has to be one of the most elegant things available.
Definition 21.3.1A is an algebraof functions if A is a vector space and ifwhenever f,g ∈A then fg ∈A.
To begin with assume that the field of scalars is ℝ. This will be generalized later. Theorem
3.10.2 implies the following corollary. See Corollary 3.10.3.
Corollary 21.3.2The polynomials are dense in C
([a,b])
.
Here is another approach to proving this theorem. It is the original approach used by
Weierstrass. Let m ∈ ℕ and consider c_{m} such that
whenever m is large enough. Hence, for large enough m,
sup |pm (x)− f (x)| ≤ (1+ 4M )ε
x∈[−1,1]
Since ε is arbitrary, this shows that the functions p_{m} converge uniformly to f on
[− 1,1]
. However, p_{m} is actually a polynomial. To see this, change the variables and
obtain
∫ x+1
pm (x) = f (t)ϕm (x− t)dt
x−1
which will be a polynomial. To see this, note that a typical term is of the form
∫
x+1 k
x− 1 f (t) a(x− t) dt
which is clearly a polynomial in x. This proves Corollary 21.3.2 in case
[a,b]
=
[− 1,1]
. In the
general case, there is a linear one to one onto map l :
[− 1,1]
→
[a,b]
.
b−-a-
l(t) = 2 (t+ 1)+ a
Then if f ∈ C
([a,b])
,f ∘ l ∈ C
([− 1,1])
. Hence there is a polynomial p such that
max |f ∘l(t)− p(t)| < ε
t∈[−1,1]
Then letting t = l^{−1}
(x)
=
2(x−a)
-b−a-
− 1, for x ∈
[a,b]
,
max ||f (x )− p(l− 1(x ))|| < ε
x∈[a,b]
but x → p
(l−1(x))
is a polynomial. This gives an independent proof of that corollary.
The next result is the key to the profound generalization of the Weierstrass theorem due
to Stone in which an interval will be replaced by a compact or locally compact set
and polynomials will be replaced with elements of an algebra satisfying certain
axioms.
Corollary 21.3.3On the interval
[− M, M ]
, there exist polynomials p_{n}suchthat
pn(0) = 0
and
lim ||pn − |⋅|||∞ = 0.
n→∞
recall that
∥f ∥
_{∞}≡ sup_{t∈[−M,M ]
}
|f (t)|
.
Proof:By Corollary 21.3.2 there exists a sequence of polynomials,
{˜p }
n
such that
˜p
_{n}→
|⋅|
uniformly. Then let p_{n}
(t)
≡
˜p
_{n}
(t)
−
p˜
_{n}
(0)
. ■
Definition 21.3.4An algebra of functions, A defined on A, annihilates nopoint of A if for all x ∈ A, there exists g ∈A such that g
(x)
≠0. The algebra separatespoints if whenever x_{1}≠x_{2}, then there exists g ∈A such that g
(x1)
≠g
(x2)
.
The following generalization is known as the Stone Weierstrass approximation
theorem.
Theorem 21.3.5Let A be a compact topological space and let A⊆ C
(A;ℝ )
be an algebra of functions which separates points and annihilates no point. Then A isdense in C
(A; ℝ)
.
Proof:First here is a lemma.
Lemma 21.3.6Let c_{1}and c_{2}be two real numbers and let x_{1}≠x_{2}be two points of A.Then there exists a function f_{x1x2}such that
fx1x2 (x1) = c1,fx1x2 (x2) = c2.
Proof of the lemma:Let g ∈A satisfy
g(x1) ⁄= g(x2).
Such a g exists because the algebra separates points. Since the algebra annihilates no point,
there exist functions h and k such that
h(x1) ⁄= 0,k(x2) ⁄= 0.
Then let
u ≡ gh− g(x2)h,v ≡ gk− g(x1)k.
It follows that u
(x1)
≠0 and u
(x2)
= 0 while v
(x2)
≠0 and v
(x1)
= 0. Let
fx1x2 ≡ -c1u-+ --c2v-.
u(x1) v (x2)
This proves the lemma. Now continue the proof of Theorem 21.3.5.
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 }.
Suppose f ∈A and suppose M is large enough that
||f||∞ < M.
Using Corollary 21.3.3, let p_{n} be a sequence of polynomials such that
||pn − |⋅|||∞ → 0,pn(0) = 0.
It follows that p_{n}∘ f ∈A and so
|f|
∈A whenever f ∈A. Also note that
max (f,g) = |f −-g|+-(f-+g)
2
(f +-g)−-|f −-g|
min (f,g) = 2 .
Therefore, this shows that if f,g ∈A then
--
max (f,g), min(f,g) ∈ A.
By induction, if f_{i},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 21.3.6 to obtain f_{xy}, a function of A
which agrees with h at x and y. Letting ε > 0, there exists an open set U
(y)
containing y
such that
fxy(z) > h (z) − ε if z ∈ U (y).
Since A is compact, let U
(y1)
,
⋅⋅⋅
,U
(yl)
cover A. Let
f ≡ max (f ,f ,⋅⋅⋅,f ).
x xy1 xy2 xyl
Then f_{x}∈A and
fx(z) > h(z)− ε
for all z ∈ A and f_{x}
(x)
= h
(x)
. This implies that for each x ∈ 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 f_{x}
(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