Lecture #10
MATH 321: Real Variables II
University of British Columbia
Lecture #10:
Instructor:
Scribe:
January 28, 2008
Dr. Joel Feldman
Peter Wong
Please refer to the handout product.pdf – Products of Riemann Integrable Functions
Proof. (Continued) Set
A = { 1 ≤ i ≤ n | Mi − mi < δ },
Control of
Control of
P
P
∗
i∈A (Mi
− m∗i )∆αi :
i ∈ A =⇒ Mi − mi < δ
∗
i∈B (Mi
B = { 1 ≤ i ≤ n | M i − mi ≥ δ }
uniform
=⇒
continuity
Mi∗ − m∗i ≤ ε0 =⇒
X
(Mi∗ − m∗i )∆αi ≤ ε0 [α(b) − α(a)]
i∈A
− m∗i )∆αi :
Mi∗ − m∗i =
[ϕ(f (x)) − ϕ(f (y))] <≤ 2Mϕ
sup
since |ϕ(z)| ≤ Mϕ , ∀|z| ≤ M
xi−1 ≤x,y≤xi
Recall that
f ∈ R(α) =⇒ U (Pη , f, α) − L(Pη , f, α) =
n
X
(Mi − mi )∆i < η
i=1
=⇒
X
(Mi − mi )∆αi < η
i∈B
=⇒
X
δ∆αi < η =⇒
i∈B
X
∆αi <
i∈B
η
δ
End game:
n
X
(Mi∗ − m∗i )∆αi =
i=1
if we choose (3) ε0 =
X
(Mi∗ − m∗i )∆αi +
i∈A
ε
2[α(b)−α(a)]
X
(Mi∗ − m∗i )∆αi < ε0 [α(b) − α(a)] + 2Mϕ
i∈B
and (4) η =
η
<ε
δ
εδ
4Mϕ
Functions of Bounded Variation (See variation.pdf on the Web)
Motivation. If β is monotone increasing and f is continuous, then f ∈ R(β). If γ is monotone increasing, then
f ∈ R(β). By linearity, we must have f ∈ R(β − γ). Our question is: Given a function α, can it be written as
a difference of two increasing functions? We shall show that α = β − γ with β, γ increasing if and only if α is of
bounded variation.
Definition. Let a < b. Let α : [a, b] → R.
(a) α is of bounded variation if there exists an M > 0 such that for a partition P = { x0 , . . . , xn } of [a, b], we
have
n
n
X
X
|(∆α)i | < M
|α(xi ) − α(xi−1 )| =
i=1
i=1
(b) If so, the (total) variation of α on [a, b] is
sup
P
X
i∈P
|∆αi | = Vα (a, b)
2
MATH 321: Lecture #10
Examples.
(a) If α is increasing
X
|∆αi | =
P
If α is decreasing
X
|∆αi | =
P
X
∆αi = α(b) − α(a) = Vα (a, b).
P
X
−∆αi = −[α(b) − α(a)] = α(a) − α(b) = Vα (a, b).
P
(b) If α is differentiable with |α0 (x)| ≤ M
X
X
X
X
|α(xi ) − α(xi−1 )| =
|α0 (yi )(xi − xi−1 )| ≤
|α0 (yi )|(xi − xi−1 ) =
M (xi − xi−1 ) = M (b − a)
P
P
P
P
(c) Define the function α : [0, 1] → R by
α(x) =
(
x cos πx , x 6= 0
0,
x=0
xn−2
xn = 1
xn−1 =
For each n ∈ N, choose
1 1 1 1
1
,..., , , , ,1 ,
Pn = 0,
2n
5 4 3 2
1
2
α(Pn ) =
1
1 1 1 1
0,
, . . . , − , , − , , −1
2n
5 4 3 2
Lecture #11
MATH 321: Real Variables II
University of British Columbia
Lecture #11:
Instructor:
Scribe:
January 30, 2008
Dr. Joel Feldman
Peter Wong
Last time, we started the discussion of how a function might not be of bounded/finite variation. . .
(c) Define the function α : [0, 1] → R by
α(x) =
(
x cos πx , x 6= 0
0,
x=0
xn−2
xn = 1
xn−1 =
For each n ∈ N, choose
1 1 1 1
1
Pn = 0, , . . . , , , , , 1 ,
2n
5 4 3 2
1
2
α(Pn ) =
1
1 1 1 1
0, , . . . , − , , − , , −1
2n
5 4 3 2
so that we have
N
X
i=1
Since
cos(2n−1)π
−
0
|α(xi ) − α(xi−1 )| = cos(2nπ)
+ 2n−1 −
2n
P∞
1
m=2 m
cos(2nπ) 2n
+ cos(2n−2)π
−
2n−2
cos(2n−1)π 2n−1
1
−1
1
1 −1 = 2n
− 0 + 2n−1
− 2n
− 2n−1
+ 2n−2
+ · · · + −1 − 12 1 1
1
1
1 1 +
+
+
+
= 2n
2n−1
2n
2n−2
2n−1 + · · · + 1 + 2
h
i
1
1
= 2 2n
+ 2n−1
+ · · · + 21 + 1
+ · · · + −1 − 21 PN
diverges, for any M > 0, ∃Pn such that i=1 |α(xi ) − α(xi−1 )| > M .
Now, we can prove some properties of bounded variation. For our notational convenience, we let
X
P
|∆i α| :=
n
X
|α(xi ) − α(xi−1 )|,
for P = {x0 , . . . , xn }.
i=1
Theorem.
(1) If α, β : [a, b] → R are of bounded variation and c, d ∈ R, then
cα + dβ is of bounded variation,
and
Vcα+dβ (a, b) ≤ |c|Vα (a, b) + |d|Vβ (a, b).
Proof. For any partition P = {x0 , . . . , xn },
X
X
X
|∆i (cα + dβ)| ≤ |c|
|∆i α| + |d|
|∆i β| ≤ |c|Vα (a, b) + |d|Vβ (a, b)
P
P
P
(2) If α : [a, b] → R is of bounded variation on [a, b] and [c, d] ⊂ [a, b], then
α is of bounded variation on [c, d],
and
Vα (c, d) ≤ Vα (a, b).
2
MATH 321: Lecture #11
Proof. If P is a partition of [c, d], we have
X
X
|∆i α| ≤ Vα (a, b)
|∆i α| ≤
P
P ∪[a,b]
(3) If α : [a, b] → R is of bounded variation and c ∈ (a, b), then Vα (a, b) = Vα (a, c) + Vα (c, b).
Proof. (≤) We know from the triangle inequality that
|∆i α| = |α(xi ) − α(xi−1 )| ≤ |α(xi ) − α(c)| + |α(c) − α(xi−1 )|
So,
X
|∆i α| ≤
P
X
|∆i α| ≤
X
|∆i α| +
|∆i α| ≤ Vα (a, c) + Vα (c, b)
(P ∪{c})∩[c,b]
(P ∪{c})∩[a,c]
P ∪{c}
X
(≥) Let ε > 0. Choose
P1 of [a, c] such that
X
|∆i α| ≥ Vα (a, c) − ε
P1
P2 of [c, b] such that
X
|∆i α| ≥ Vα (c, b) − ε
P2
Then
X
|∆i α| =
P ∪{c}
X
|∆i α| =
P1 ∪P2
=⇒
∀ε > 0,
X
|∆i α| +
P1
X
|∆i α| ≥ Vα (a, c) + Vα (c, b) − 2ε
P2
Vα (a, b) ≥ Vα (a, c) + Vα (c, b) − 2ε
Hence, Vα (a, b) ≥ Vα (a, c) + Vα (c, b).
(4) If α : [a, b] → R is of bounded variation, then the functions
V (x) = Vα (a, x),
and
V (x) − α(x)
Proof. Let a ≤ x1 ≤ x2 ≤ b. Then by (2), V (x1 ) = Vα (a, x1 ) ≤ Vα (a, x2 ) = V (x2 ). By (3),
[V (x2 ) − α(x2 )] − [V (x1 ) − α(x1 )] = Vα (x1 , x2 ) − [α(x2 ) − α(x1 )].
Following from
|α(x2 ) − α(x1 )| =
X
|∆i α| ≤ Vα (x1 , x2 ),
{x1 ,x2 }
we have V (x2 ) − α(x2 ) ≥ V (x1 ) − α(x1 )
are both increasing on [a, b].
(5) The function α : [a, b] → R is of bounded variation iff it is the difference of two increasing functions.
Proof. (=⇒) If α is of bounded variation, then we can express α as the difference of two increasing functions
as follows:
α(x) = Vα (a, x) − [Vα (a, x) − α(x)].
(⇐=) α = β − γ for some increasing β, γ =⇒ β, γ are of bounded variation by an earlier example. Then by
(1), α is of bounded variation.
Examples. See the notes variation.pdf on the web for details.
)
(
f continuous
=⇒ f ∈ R(α) on [a, b]
α is of bounded variation
)
(
Z b
Z b
f is of bounded variation
=⇒ f ∈ R(α) on [a, b] with
f dα = f (b)α(b) − f (a)α(a) −
α df
α continuous
a
a
Lecture #12
MATH 321: Real Variables II
University of British Columbia
Lecture #12:
Instructor:
Scribe:
February 1, 2008
Dr. Joel Feldman
Peter Wong
Sequences and Series of Function (Rudin §7)
The issues:
(1) “limn→∞ fn = f ” has a number of different meanings. How are they related?
(2) Suppose you know
(a) limn→∞ fn (x) = f (x)
(b) each fn (x) has some property (e.g. continuity, differentiability, or integrability) Can you conclude that f
has the same property?
(3) Suppose you need to work with a really ugly function f (x). Can you approximate f (x) to arbitrarily good
accuracy by some really nice function? E.g.: polynomials, trigonometric polynomials (Fourier series)
Definition. Let a < b. Let I be some interval or domain ((a, b),[a, b),(a, b], or [a, b]). Let, for each n ∈ N,
fn : I → R or C. Let f : I → R or C. The sequence { fn }n∈N is said to converge to f
(a) pointwise ⇐⇒ for each x in the domain D, limn→∞ fn (x) = f (x)
(b) uniformly (or in L∞ ) ⇐⇒ limn→∞ supx∈D |fn (x) − f (x)| = 0
Rb
(c) in the mean (or in L2 , denoted f = limn→∞ fn ) ⇐⇒ limn→∞ kfn − f k2 = 0, where kfn − f k2 = [ a |fn (x) −
1
f (x)|2 dx] 2 (Engineering notation: rms)
Rb
1
(d) in Lp , 1 ≤ p ≤ ∞ ⇐⇒ limn→∞ kfn − f kp = 0, where kfn − f kp = [ a |fn (x) − f (x)|p dx] p
More generally, if d(f, g) is any metric on some set of functions of ineterest, like
C([a, b]) = { f : [a, b] → R | f is continuous }
you say { fn } converges to f with respect to the metric d if limn→∞ d(fn , f ) = 0. For example,
(
supx∈[a,b] |f (x) − g(x)| = kf − gk∞ , for uniform convergence
Rb
d(f, g) =
1
kf − gkp = [ a |fn (x) − f (x)|p dx] p , for Lp convergence
Claim. Uniform convergence =⇒ pointwise convergence.
Proof. Note the |fn (x) − f (x)| ≤ kfn − f k∞ . Since uniform convergence requires that kfn − f k∞ goes to zero as
n → ∞, |fn (x) − f (x)| is forced to zero, resulting in pointwise convergence.
Claim. Uniform convergence =⇒ Lp convergence if a, b are finite.
Proof. Since
kfn − f kp =
"Z
b
# p1
|fn (x) − fn |p dx
a
"
≤ kfn − f kp∞
Z
a
b
# p1
1 dx
1
= kfn − f k∞ (b − a) p ,
the right hand side approaches zero as n → ∞, provided that b − a is finite. This forces the left hand side to
zero.
2
MATH 321: Lecture #12
Unfortunately, the good news ends here.
Counterexamples.
1. Uniform convergence does NOT imply Lp convergence if a or b is infinity. (See Problem Set 5 # 2(c))
2. Pointwise convergence does NOT imply uniform or Lp convergence. (See Problem Set 5 # 2(a))
3. Lp convergence does NOT imply pointwise or uniform convergence. (See Problem Set 5 # 2(b))
Theorem. (Tests for uniform convergence)
(a) (Weierstraß M-test) If
1. for each n ∈ N, fn : E → R or C,
2. |fn (x)| ≤ Mn for x ∈ E, n ∈ N, and
P∞
3.
n=1 Mn < ∞,
P∞
then n=1 fn (x) converges uniformly.
Pn
Pn
Proof. Step 1: Guess the limit. P
For each x ∈ E, { Fn (x)
= m=1 fm (x) }, we know m=1 fm (x) converges
P
n
∞
absolutely by comparison with
m=1 fm (x) converges to, say, F (x). More on this
n=1 Mn , so we know
next week.