Lecture #16
MATH 321: Real Variables II
University of British Columbia
Lecture #16:
Instructor:
Scribe:
February 11, 2008
Dr. Joel Feldman
Peter Wong
Remark. (Not part of this course)
1. Reference for Arzelà’s theorem: Tom Apostol, Mathematical Analysis. pp.405-406
2. (Lebesgue dominated convergence theorem) Assume
• µ is a measure on R
• for each n ∈ N, fn : R → R is measurable
• g : R → R is measurable
• for each x ∈ R, limn→∞ fn (x) = f (x) exists
• |fn (x)| ≤ g(x) for all x ∈ R
R
• g 0 (x) dµ(x) < ∞
Then f is measurable and limn→∞
Special case:
R
fn (x) dµ(x) =
R
f (x) dµ(x).
• µ is a Lebesgue measure on [a, b]
• fn is Riemann integrable
• g(x) = M for some constant M
• for each x ∈ R, limn→∞ fn (x) = f (x) exists
Convergence and Differentiation
Question: If fn is differentiable and converges to f , then is it true that f 0 exists and lim fn0 = f 0 ?
Counterexamples.
1. Problem Set 5 Question 2(h): fn differentiable and converges uniformly to f , fn0 converges pointwise to g, but
f 0 does not exists.
2. Problem Set 5 Question 2(i): fn differentiable and converges uniformly to a differentiable f , fn0 converges
pointwise to g, but f 0 6= g.
Theorem. If
• for each n ∈ N, fn : [a, b] → R differentiable
• { fn0 }n∈N converges uniformly on [a, b]
• for some x0 ∈ [a, b], { fn (x0 ) }n∈N converges (to prevent, say, fn (x) = n, which diverges).
Then
(a) { fn }n∈N converge uniformly on [a, b]. Call the limit f (x)
(b) f (x) is differentiable on [a, b], and f 0 (x) = limn→∞ fn0 (x).
2
MATH 321: Lecture #16
Proof.
(a) Use the Cauchy test
|fn (x) − fm (x)| ≤ |fn (x) − fm (x) − fn (x0 ) + fm (x0 )| + |fn (x0 ) − fm (x0 )|
= |(fn − fm )(x) − (fn − fm )(x0 )| + |fn (x0 ) − fm (x0 )|
= |(fn − fm )0 (c)(x − x0 )| + |fn (x0 ) − fm (x0 )| for some c between x and x0
0
≤ kfn0 − fm
k∞ (b − a) + |fn (x0 ) − fm (x0 )|
Let ε > 0 Then
∃N1
such that n, m ≥ N1 =⇒ |fn (x0 ) − fm (x0 )| <
∃N2
0
such that n, m ≥ N2 =⇒ kfn0 − fm
k∞ <
=⇒
(b)
ε
2
ε
2(b−a)
n, m ≥ max{N1 , N2 } =⇒ kfn − fm k∞ < ε.
Step 1: Fix x ∈ [a, b]. Define
 0
fn (x),
φn (t) =
 fn (x)−fn (t)
x−t
if
t=x
φ(t) =
, otherwise

0
(x),
limm→∞ fm
 f (x)−f (t)
x−t
,
if t = x
otherwise
Since fn ’s are differentiable, φn (t)’s must be continuous. Thus, it suffices to prove that φ(t) is
continuous:
lim φ(t) = φ(x)
=⇒
t→x
lim
t→x
f (x) − f (t)
0
= lim fm
(x)
m→∞
x−t
It suffices to prove that φn ’s converge uniformly to φ:
 1
 |x−t| |fn (x) − fn (t) − f (x) + f (t)|,
|φn (t) − φ(t)| =
 0
0
|fn (x) − limm→∞ fm
(x)|,
which does exist
if t 6= x
if t = x
For t 6= x
|φn (t) − φ(t)| =
≤
1
|x−t| |(fn − fm )(x) − (fn − fm )(t)|
0
0
0
1
|x−t| lim sup kfn − fm k∞ |x − t| ≤ lim sup kfn
m→∞
m→∞
0
− fm
k∞
Lecture #17
MATH 321: Real Variables II
University of British Columbia
Lecture #17:
Instructor:
Scribe:
February 13, 2008
Dr. Joel Feldman
Peter Wong
Proof (Continued):
Step 2: We know that f is differentiable and f 0 (x) = limn→∞ fn0 (x). Fix any x ∈ [a, b]. Define
(
fn (x)−fn (t)
, for t 6= x
x−t
φn (t) =
fn0 (x),
for t = x
• φn (t) is continuous at t = x because limt→x φn (t) = φn (x)
• φn (t) is continuous at t 6= x because the denominator x − t 6= 0
so φn (t) is continuous everywhere for all n ∈ N. Define
(
f (x)−f (t)
,
x−t
φ(t) =
limn→∞ fn0 (x),
for x 6= t
for x = t
• Need to show φ(t) is continuous: limx→t φ(t) = φ(x)
• This means that:
lim φ(t) = lim fn0 (x)
x→t
0
n→∞
f (x) = lim fn0 (x)
n→∞
Let ε > 0. We are assuming { fn0 } converges uniformly; that is,
0
(t)| < ε, for all t ∈ [a, b]
∃N such that n, m ≥ N =⇒ |fn0 (t) − fm
Now show that for all t ∈ [a, b], n, m ≥ N =⇒ |φn (t) − φ(t)| ≤ ε. For t = x,
0
(x)| ≤ ε
|φn (x) − φ(x)| = lim |fn0 (x) − fm
m→∞
For t 6= x,
fn (x) − fn (t) f (x) − f (t) |φn (x) − φ(x)| = −
x−t
x−t 1
|fn (x) − fn (t) − fm (x) + fm (t)|
= lim
m→∞ |x − t|
1
= lim
|(fn − fm )(x) − (fn − fm )(t)|
m→∞ |x − t|
1
= lim
|(fn − fm )0 (c)||x − t|
m→∞ |x − t|
0
= lim |fn0 (c) − fm
(c)| ≤ ε
when m ≥ N
m→∞
Theorem. (Arzelà-Ascoli, See Rudin pp.154-158) Let K be a compact metric space and C(K) be the space
of real (or complex) valued continuous functions on K. If { fn } is a sequence in C(K) obeying
(H1) for each p ∈ K, there is a ϕ(p) > 0 such that |fn (p)| < ϕ(p) for all n ∈ N (then the sequence { fn } is called
pointwise-bounded), and
2
MATH 321: Lecture #17
(H2) ∀ε > 0, ∃δ > 0 such that for all n ∈ N, dK (p, p0 ) < δ
=⇒
|fn (p) − fn (p0 )| < ε.
Then
(a) there is an M > 0 such that |fn (p)| < M for all p ∈ K and all n ∈ N
(b) the sequence { fn } has a uniformly convergent subsequence
Proof.
Step 1 (of 4):
Lemma 1. There is a countable subset { p` | ` ∈ N } ⊂ K that is dense in K.
Proof of Lemma 1: K is compact =⇒ K is totally bounded, that is, ∀ε > 0, K can be covered
by a finite union of balls of radius ε. In particular, for each m ∈ N, there are a finite number
1
of one of
of points pm,1 , pm,2 , . . . , pm,nm such that every point in K is within a distance of M
pm,1 , pm,2 , . . . , pm,nm . We can then choose
{ p1 , p2 , . . . } = { p1,1 , . . . , p1,n1 , p2,1 , . . . , p2,n2 , . . . }
|
{z
} |
{z
}
m=1
m=2
Lecture #18
MATH 321: Real Variables II
University of British Columbia
Lecture #18:
Instructor:
Scribe:
February 15, 2008
Dr. Joel Feldman
Peter Wong
Step 1 (of 4): Lemma 1. There is a countable subset { p` | ` ∈ N } ⊂ K that is dense in K.
Proof of Lemma 1: Done last class.
Step 2 (of 4): Lemma 2. Let { fn }n∈N be any pointwise bounded sequence of functions on the countable set
{ p` }`∈N . Then there is a subsequence { fnk }k∈N of { fn }n∈N that converges at every p` , ` ∈ N.
(i.e., limk→∞ fnk (p` ) exists for each ` ∈ N.)
Proof.
• Concentrate on p1 . By hypothesis, { fn (p1 ) }n∈N is bounded
=⇒
∃M1
such that
|fn (p1 )| ≤ M1
for alln ∈ N
Since { y ∈ R | |y| ≤ M1 } is compact,
there is a convergent subsequence fnj (p1 ) j∈N
=⇒
Rename nj to N1 (j).
• Concentrate on p2 . By hypothesis,
=⇒
∃M2
fN1 (j) (p2 )
n∈N
is bounded
such that |fN1 (j) (p2 )| ≤ M2
for allj ∈ N
Since { y ∈ R | |y| ≤ M2 } is compact,
=⇒
there is a convergent subsequence fNj (jk ) (p2 ) k∈N
Rename N1 (jk ) to N2 (k).
• Repeat: For each ` ∈ N, we construct a subsequence fN` (m) m∈N of fN`−1 (p) p∈N and
hence of { fn }n∈N for which fN` (m) (p` ) m∈N converges.
• Choose the subsequence in the conclusion of this Lemma to be fNn (n) n∈N . This subse
quence converges at every p` because, for each ` ∈ N, fNn (n) (p` ) n≥` is a subsequence
of fn` (j) (p` ) j∈N , because for n ≥ `, Nn (n) is in the range of Nn , R(Nn ), and R(Nn ) ⊂
R(Nn−1 ) ⊂ · · · ⊂ R(N` ).
Step 3 (of 4): Proof of Arzelà-Ascoli (a), i.e.,
(H1) Pointwise bounded
+
(H2) equicontinuity
=⇒
(a) uniformly bounded.
Proof. We have to find an M ≥ 0 such that |fn (p)| ≤ M for all n ∈ N and p ∈ K. Apply
equicontinuity with ε = 1. So
∃δ
such that |fn (p) − fn (q)| < 1
for all n ∈ N,
p, q ∈ K
with
dK (p, q) < δ
Since K is compact, there are finitely many points p1 , . . . , pr in K such that every p ∈ K is within
a distance of δ of at least one of the points. For all n ∈ N, p ∈ K,
|fn (p)| ≤ |fn (p) − fn (pi )| + |fn (pi )| < 1 + φ(pi )
≤ 1 + max φ(pi ) = M
1≤i≤r
with pi chosen such that dK (p, pi ) < δ
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MATH 321: Lecture #18
Step 4 (of 4): Proof of Arzelà-Ascoli (b)
(H1) Pointwise bounded
+
(H2) equicontinuity
=⇒ (b) there is a subsequence which converges uniformly.
Proof.
• Apply Lemma 1 to give a countable dense subset { p` }`∈N of K.
• Apply Lemma 2 to give a subsequence fnj j∈N which converges at each p` , ` ∈ N.