MATH 321: Real Variables II
Lecture #19
University of British Columbia
Lecture #19:
Instructor:
Scribe:
February 25, 2008
Dr. Joel Feldman
Peter Wong
Theorem. (Arzelà-Ascoli) If K is a compact metric space and { fn }n∈N is a sequence of continuous functions
on K that are
(H1) pointwise-bounded, and
(H2) equicontinuous,
then
(a) the sequence is uniformly bounded, and
(b) it contains a uniformly convergent subsequence.
Step 4 (of 4): Proof of (b).
• By Lemma 1, there is a countable dense subset { p` }`∈N in K.
• By Lemma 2, there is a subsequence { fnk }k∈N such that limk→∞ fnk (p` ) exists for each ` ∈ N.
• Choose { fnk } as our subsequence. Rename fnk to gk . We shall prove the { gk } converges
uniformly by verifying
∀ε > 0, ∃N ∈ N such that ∀p ∈ K, i, j ≥ N =⇒ |gi (p) − gj (p)| < ε.
Let ε > 0.
• By equicontinuity (H2), ∃δ such that if dK (p, p0 ) < δ then |gk (p) − gk (p0 )| < ε/3.
• Now, we look for a finite number of p` ’s, p`1 , . . . , p`r such that every p ∈ K is less than a distance
of δ away from at least one of p`1 , . . . , p`r . Since K is compact, K is totally bounded, which
means that: Given any δ > 0, K is covered by finitely many open balls of radius δ/2. That is,
there are finitely many points q1 , . . . , qr in K such that each p ∈ K is less than a distance of
δ/2 away from at least one of q1 , . . . , qr . By the density of { p` }, for each 1 ≤ m ≤ r, there is a
p`m within a distance of δ/2 away from qm . Thus, { p`1 , . . . , p`r } work.
• For each 1 ≤ m ≤ r, { gk (p`m ) } converges. This means that
∃Nm
such that
i, j ≥ Nm =⇒ |gi (p`m ) − gj (p`m )| <
ε
.
3
• Choose N = max 1 ≤ m ≤ rNm . Then
i, j ≥ N =⇒ |gi (p) − gj (p)| ≤ |gi (p) − gi (p`m )| + |gi (p`m ) − gj (p`m )| + |gj (p`m ) − gj (p)|
<
ε/3
+
ε/3
+
ε/3
=ε
(equicontinuity)
(convergence of p`m )
(equicontinuity)
with m chosen so that dk (p, p`m ) < δ.
Corollary. Let a < b, and c ∈ [a, b]. Let, for each n ∈ N, fn : [a, b] → R be a diferentiable function. Assume that
(i) M = sup{ |fn (c)| : n ∈ N } < ∞
(ii) M 0 = sup{ |fn0 (x)| : x ∈ [a, b] n ∈ N } < ∞
Then { fn } has a uniformly convergent subsequence.
2
MATH 321: Lecture #19
Proof. Apply Arzelà-Ascoli with K = [a, b], which is compact.
Step (1) Verification of equicontinuity (H2): Let ε > 0. Choose δ = ε/M 0 . Then x, y ∈ [a, b] obey
|x − y| < δ =⇒ |fn (x) − fn (y)| = |fn0 (c̃)(x − y)| ≤ M 0 |x − y| < M 0 δ = M ε/M 0 = ε.
Step (2) Verification of Pointwise boundedness (H1): Let x ∈ [a, b]. Then
|fn (x)| ≤ |fn (c)| + |fn (x) − fn (c)| ≤ M + |fn0 (c̃)(x − c)| ≤ M + M 0 |x − c| = Φ(x) ≤ M + M 0 (b − a)
Notes: No lecture on this Wednesday – Lecture #20 will be a midterm exam session.
MATH 321: Real Variables II
Lecture #21
University of British Columbia
Lecture #21:
Instructor:
Scribe:
February 29, 2008
Dr. Joel Feldman
Peter Wong
Notes: No lecture on Wednesday – Lecture #20 was a midterm exam session.
Third Issue: How to approximate some possibly complicated functions by some simple functions?
Theorem. (Weierstraß, 1885) If f : [a, b] → C is continuous, then there exists a sequence of polynomials { Pn }n∈N
that converges uniformly to f . Further, if f is real-valued, then the Pn ’s may also be taken as real-valued.
Idea of the proof:
Z
f (x) =
f (t)δ(x − t) dt
Of course, f cannot be so naı̈vely defined by a Dirac delta function this way. What weRwant to do is to find some
polynomials Qn which approximated δ(t) very well. To do this, we will define Pn (x) = f (t)Qn (x − t) dt.
Proof. Reduction #1: It suffices to consider [a, b] = [0, 1]. Otherwise, define h : [0, 1] → C by h(y) = f ((b−a)y+a).
(Check: h(0) = f (a) and h(1) = f (b))
x−a
x − a sup |h(y) − Pn (y)| = sup h
− Pn
b−a
b−a y∈[0,1]
x∈[a,b]
Reduction #2: It suffices to consider the case in which f (0) = 0, f (1) = 0. Otherwise, we should write
f (x) = f (0) + x [f (1) − f (0)] + {f (x) − f (0) − x [f (1) − f (0)]}
Let g(x) = {f (x) − f (0) − x [f (1) − f (0)]}. Notice that g obeys g(0) = 0 and g(1) = 0. Also, f (0) + x [f (1) − f (0)]
is already a polynomial.
R1
Definition of Qn (x)(≈ δ(x)): Qn (x) = cn (1 − x2 )n , where cn is chosen such that −1 Qn (x) dx = 1 (i.e., cn =
R1
{ −1 (1 − x2 )n }−1 .) Also, notice that


= 1, for x = 0,
(1 − x2 )n = 0, for x = ±1,


< 1, for 0 < |x| < 1.
b
b
−1
0
Properties of Qn (x) for −1 ≤ x ≤ 1:
1
(1) Qn (x) ≥ 0
(2) decreases as |x| increases
R1
(3) −1 Qn (x) dx = 1
√
(4) 0 < δ ≤ |x| ≤ 1 =⇒ Qn (x) ≤ Qn (δ) = cn (1 − δ 2 ) ≤ n(1 − δ 2 )n → 0 as n → ∞, since
Z 1
Z 1
2 n
1
(1 − x ) dx = 2
(1 − x2 )n dx
cn ≥
0
−1
≥2
Z
≥2
Z
0
√
1/ n
√
1/ n
0
≥
and since (1 − y)n = 1 − ny +
(1 − x2 )n dx
h
(1 − nx ) dx = 2 x −
2
nx3
3
√1
n
1 d2
2! d2 y (1
− y)n y=c = 1 − ny +
1
2! n(n
√
i1/ n
=
0
4 √1
3 n
− 1)(1 − c)n−2 y 2 ≥ 1 − ny.
2
MATH 321: Lecture #21
Definition of Pn (x): Extend f (x) to −∞ < x < ∞ by setting f (x) = 0 for x ∈ [0, 1]. Define
Pn (x) =
=
=
Z
1
f (x + t)Qn (t) dt,
−1
Z 1+x∈[1,2]
setting s = x + t
f (s)Qn (s − x) ds
−1+x∈[−1,0]
Z 1
f (s)Qn (s − x) ds
0
where Qn is now a polynomial in s and x. Notice that Pn is a real-valued polynomial if f is real-valued.
Midterm Results and Comment: x̄ =
18.2
30
#2(a) If you wrote |fn (xn ) − f (x)| ≤ |fn (xn ) − fn (x)| + |fn (x) − f (x)|, you ran into a wall. See solution.