Lecture #22
MATH 321: Real Variables II
University of British Columbia
Lecture #22:
Instructor:
Scribe:
March 3, 2008
Dr. Joel Feldman
Peter Wong
Theorem. (Weierstraß) If f : [a, b] → C is continuous, then there is a sequence of polynomials { Pn }n∈N such that
sup |f (x) − Pn (x)| → 0
as
n → ∞.
x∈[a,b]
Proof.
Reduction #1: It suffices to consider [a, b] = [0, 1].
Reduction #2: It suffices to consider the case in which f (0) = f (1) = 0.
R1
Define Qn (x) = cn (1 − x2 )n , where cn = { −1 (1 − x2 )n }−1 . Qn obeys
(1) Qn (x) ≥ 0 on [−1, 1],
R1
(2) −1 Qn (x) dx = 1,
(3) 0 < δ ≤ |x| ≤ 1 =⇒ 0 ≤ Qn (x) ≤ cn (1 − δ 2 ) ≤
√
n(1 − δ 2 )n .
Define for 0 ≤ x ≤ 1,
Pn (x) =
=
Z
1
f (x + t)Qn (t) dt,
−1
Z 1+x∈[1,2]
−1+x∈[−1,0]
setting s = x + t
f (s)Qn (s − x) ds =
Z
1
f (s)
0
P 2n
Qn (s − x)
| {z }
m=0 (polynomial
ds
in S)xm
is obviously a polynomial in x, which is real if f : [0, 1] → R.
Let ε > 0. Since f is continuous on [0, 1],
(i) f is bounded: ∃M such that |f (x)| ≤ M on [0, 1],
(ii) f is uniformly continuous: ∃δ > 0 such that |f (x) − f (y)| <
ε
2
if |x − y| < δ.
Then for all 0 ≤ x ≤ 1,
|f (x) − Pn (x)| = f (x)
≤
≤
<
=
Z
1
Z−1
ε
2
ε
2
|
1
−1
Qn (t) dt − Pn (x) =
Z
1
−1
[f (x) − f (t − x)]Qn (t) dt
|f (x) − f (t + x)|Qn (t) dt
|t|<δ
Z
Z
|f (x) − f (t + x)| Qn (t) dt +
{z
}
|
|t|<δ
Z
δ≤|t|≤1
ε/2
√
Qn (t) dt +2 · 2M · n(1 − δ 2 )n
{z
≤1
}
√
+ 4M n(1 − δ 2 )n < ε
√
as long as n is big enough so that 4M n(1 − δ 2 )n < 2ε .
|f (x) − f (t + x)| Qn (t)
|
{z
} | {z }
2M
√
n(1−δ 2 )n
dt
2
MATH 321: Lecture #22
Idea: Generalize f : [a, b] → C continuous to f : K → C continuous, where K is any compact metric space.
Problem: If x ∈ K, then x2 , x3 , . . . are not defined (i.e., there is no concept of polynomials on K.)
Notation: Let
• K be a compact metric space,
• F = { K : R →| f bounded } with the metric d(f, g) = supx∈K |f (x) − g(x)|,
• C = { K : R →| f continuous } with the metric d(f, g) = supx∈K |f (x) − g(x)|,
• A ⊂ C as approximators.
Definition. A is an algebra if
(1) f, g ∈ A =⇒ f + g ∈ A,
(2) f, g ∈ A =⇒ f g ∈ A,
(3) f ∈ A, c ∈ R =⇒ cf ∈ A.
Conjecture. Let K be a compact metric space and C = { f : K → R | f continuous } and let A ⊂ C be an algebra.
Then A = C. ( A is the set of all uniform limits of sequences in A.)
To search for conter-examples, we use the following approach:
If (1) every p ∈ A has property X,
(2) for every pn has property X, limpn has the property X, and
(3) there exists an f ∈ C which does not have the property X,
then the conjecture fails.
Exercise. Find as many such properties X as possible.
Lecture #23
MATH 321: Real Variables II
University of British Columbia
Lecture #23:
Instructor:
Scribe:
March 5, 2008
Dr. Joel Feldman
Peter Wong
Notation: Let
• K = a compact metric space,
• F = { K : R →| f bounded } with the metric d(f, g) = supx∈K |f (x) − g(x)|,
• C = { K : R →| f continuous } with the same metric d,
• A ⊂ C the set of approximators.
Definition. A is an algebra if
(1) f, g ∈ A =⇒ f + g ∈ A,
(2) f, g ∈ A =⇒ f g ∈ A,
(3) c ∈ R, f ∈ A =⇒ cf ∈ A.
Definition.
A = the closure of A
= the set of all limits of sequences in A
Theorem. If A ⊂ F is an algebra, then A is a closed algebra.
Proof. A is closed since any closure is closed. A is an algebra since
f, g ∈ A =⇒ ∃ sequences { fn } , { gn } ⊂ A such that f = lim fn , g = lim gn (uniformly)


f + g = lim fn + lim gn = lim (fn + gn ) ∈ A



n→∞
n→∞
n→∞




f g = lim fn · lim gn = lim (fn gn ) ∈ A
=⇒
n→∞
n→∞
n→∞






 cf = c lim fn = lim (cfn ) ∈ A

n→∞
n→∞
Conjecture. If A ⊂ C is an algebra. Then A = C.
Conclusion: False:
(1) If ∃ x ∈ K and A ⊂ { f : K → R | f (x) = 0 }, then A ⊂ { f : K → R | f (x) = 0 }
(2) If ∃ x1 , x2 ∈ K, x1 6= x2 and A ⊂ { f : K → R | f (x1 ) = f (x2 ) }, then A ⊂ { f : K → R | f (x1 ) = f (x2 ) }
Theorem. (Stone, 1937) If K is a compact metric space and A ⊂ C = { f : K → R | f continuous } obeys
(1) A is an algebra,
(2) A vanishes nowhere (i.e., ∀x ∈ K, ∃f ∈ A with f (x) 6= 0.)
(3) A separates points (i.e., x1 , x2 ∈ K, x1 6= x2 implies ∃f ∈ A with f (x1 ) 6= f (x2 ).)
Then A = C.
2
MATH 321: Lecture #23
Proof.
Step 1: Show that f ∈ A =⇒ |f | ∈ A.
Proof. Let ε > 0. It suffieces to find g ∈ A such that ∀x ∈ K, |g(x) − |f (x)|| < ε. Let a = supx∈K |f (x)|.
By the Weierstraß Theorem:
n
X
∃c0 , . . . , cn ∈ R such that
j=0
cj y j − |y| < 2ε , ∀|y| ≤ a
=⇒ |c0 | < ε/2 (Set y = 0)
=⇒
n
X
cj y j − |y| <
n
X
cj f (x)j − |y| < ε, ∀x ∈ K
j=1
=⇒
j=1
ε
2
+
ε
2
= ε, ∀|y| ≤ a
Step 2: If f1 , f2 ∈ A, then the pointwise max(f1 , f2 ), min(f1 , f2 ) ∈ A, where
(
(
f1 (x), if f1 (x) ≥ f2 (x)
f1 (x),
max(f1 , f2 )(x) =
and min(f1 , f2 )(x) =
f2 (x), if f1 (x) ≤ f2 (x)
f2 (x),
if f1 (x) ≤ f2 (x)
if f1 (x) ≥ f2 (x)
Proof.
max(f1 , f2 ) =
|f1 − f2 |
f1 + f2
+
2
2
and
min(f1 , f2 ) =
f1 + f2
|f1 − f2 |
−
2
2
Step 3: Show that if f1 , . . . , fn ∈ A, then max{f1 , . . . , fn }, min f {f1 , . . . , fn } ∈ A.
Proof. Apply the proof of Step 2 inductively.
Step 4: Let x1 , x2 ∈ K, x1 6= x2 . Let c1 , c2 ∈ R. Then there is an f ∈ A with f (x1 ) = c1 and f (x2 ) = c2 .
Proof. Since A separates points, ∃g ∈ A with g(x1 ) 6= g(x2 ) and A vanishes nowehere ∃h ∈ A with
h(x1 ) 6= 0, and ∃k ∈ A with k(x2 ) 6= 0,
g(x) − g(x2 ) h(x)
g(x) − g(x1 ) k(x)
f (x) = c1
+c2
g(x1 ) − g(x2 ) h(x1 )
g(x2 ) − g(x1 ) k(x2 )
| 
| 
{z
}
{z
}
=

1

0
at x = x1
at x = x2
=

0

1
at x = x1
at x = x2
Lecture #24
MATH 321: Real Variables II
University of British Columbia
Lecture #24:
Instructor:
Scribe:
March 7, 2008
Dr. Joel Feldman
Peter Wong
Theorem. (Stone, 1937) If K is a compact metric space and A ⊂ C = { f : K → R | f continuous } with
d(f, g) = supx∈K |f (x) − g(x)|, A ⊂ C is an algebra which vanishes nowhere and separates points, then A = C.
Proof.
Step 1: f ∈ A =⇒ |f | ∈ A
Step 2: f1 , f2 ∈ A =⇒ max(f1 , f2 ), min(f1 , f2 ) ∈ A, pointwise-speaking
Step 3: f1 , . . . , fn ∈ A =⇒ max{f1 , . . . , fn }, min f {f1 , . . . , fn } ∈ A
Step 4: x1 , x2 ∈ K, x1 6= x2 , c1 , c2 ∈ R =⇒ ∃ f ∈ A with f (x1 ) = c1 and f (x2 ) = c2
Step 5: Let f ∈ C. Let x ∈ K. Let ε > 0. Then ∃ gx ∈ A (gx does not mean
∂g
∂x )
that obeys
(1) gx (x) = f (x)
(2) gx (t) > f (t) − ε, for all t ∈ K.
Proof. By Step 4, for each y ∈ K, there is an hy ∈ A obeying hy (x) = f (x), hy (y) = f (y). Since hy − f
is continuous and hy (y) − f (y) = 0, there is an open set Jy with y ∈ Jk with hy − f (t) > −ε for all
t ∈ Jy (i.e., hy (t) > f (t) − ε.) { Jy }y∈K is an open cover of the compact set K =⇒ there is a finite
subcover Jy1 , . . . , Jym . Then gx (t) = max { hy1 (t), . . . , hym (t) } works. We check that
(a) hy1 , . . . , hym ∈ A ⊂ A =⇒ gx ∈ A. (By Step 3.)
(b) gx (x) = max { f (x), . . . f (x) } = f (x).
(c) gx ( |{z}
t ) = max{hy1 (t), . . . , hy1001 (t), . . . , hym (t)}
| {z }
∈Jy1001
>f (t)−ε
Step 6: (Final Step) Let f ∈ C. Let ε > 0. Then ∃ g ∈ A such that d(f, g) < ε.
Proof. By Step 5, for each x ∈ K, ∃ gx ∈ A that obeys (1) and (2). So gx − f is continuous and
gx (x) − f (x) = 0 =⇒ ∃ an open set Kx obeying x ∈ Kx and gx (t) − f (t) < ε for all t ∈ Kx . Thus,
{ Kx }x∈K is an open cover of K, which is compact =⇒ there is a finite subcover Kx1 , . . . , Kxn . Taking
g(t) = min{ gx1 , . . . , gxn } ∈ A
|{z}
|{z}
∈A
∈A
works since
(a) g(t) = min{ gx1 (t) , . . . , gxn (t) } > f (t) − ε
| {z }
| {z }
>f (t)−ε
>f (t)−ε
(b) g( |{z}
t ) = min{ gx1 (t) , . . . , gx1001 (t), . . . , gxn (t) } > f (t) + ε
| {z }
| {z }
| {z }
∈Kx1001
>f (t)−ε
f (t)+ε
>f (t)−ε
Theorem. (Generalized Stone-Weierstraß for Complex numbers)
If K is a compact metric space, C = { f : K → C | f continuous } with the metric d(f, g) = supx∈K |f (t) − g(t)|,
A ⊂ C is an algebra in C which vanishes nowhere, separates points, and is self-adjoint (i.e., f ∈ A =⇒ f ∗ ∈ A,
where f ∗ is the complex conjugate of f . See Problem Set 7 Question 2.) Then A = C.
2
MATH 321: Lecture #24
Proof. Set AR = { f ∈ A | f is real-valued }. Since AR is a real algebra which vanishes nowhere and spearates point.
By Step 4, there is for each x1 6= x2 , and f ∈ A which f (x1 ) = 1 and f (x2 ) = 0. By Stone-Weierstraß for the
reals, A = { f : K → R } f continuous. For any f ∈ C,
f (x) =
f (x) − f ∗ (x)
f (x) + f ∗ (x)
+i
∈A
2
2i
|
{z
} |
{z
}
=<(f ) ∈ AR ⊂ A
==(f ) ∈ AR ⊂ A