MA244 - Analysis III
Anthony Manning
October - December 2007
Contents
1 Analysis I & II, Uniform Continuity and Continuity by Open
Sets
2
2 The Integral for Step Functions
12
3 The Integral for Regulated Functions
17
4 Indefinite Integral and the Fundamental Theorem of Calculus 21
5 Normed Vector Spaces
24
6 Closed, Convergent, Cauchy, Completeness, Contractions
31
7 Pointwise Convergence and its Disadvantages
34
8 Uniform Convergence: its Advantages for Integrals and Continuity
36
9 Uniform Convergence: its Advantages for Differentiability
38
10 A Space Filling Curve
40
11 Series of Functions
43
12 Power Series
48
13 A Solution for an ODE
53
Typeset by Daniel Manson.
1
Chapter 1
Analysis I & II, Uniform
Continuity and Continuity
by Open Sets
Definition 1 1 A real sequence is a function a : N → R usually written (an )∞
n=1
when an := a(n)
Lemma 1 The set of real sequences forms a real vector space, V say, under
the operations:
(a + b)(n) := an + bn
(λa)(n) := λan
Proof
Check the axioms.
Definition 2 (xn ) ∈ V is said to be convergent if ∃l ∈ R s.t ∀ > 0 ∃N ∈ N
s.t. ∀n ≥ N |xn − l| < . We can write N as N () to emphasise that it depends
on . limn→∞ xn means l.
Proposition 2 The subset U of V consisting of convergent sequences is a
vector subspace of V . lim : U → R, (xn )∞
n=1 7→ limn→∞ xn is a linear map.
Proof xn → l, yn → m as n → ∞ ⇒ λxn + µyn → λl + µm as m → ∞
was proved in Analysis I.
Theorem 3 (Bolzano-Weierstrass) Any bounded real sequence has a convergent subsequence.
1 Lecture
03-10-07 starts here
2
Definition 3 Fix A ⊂ R and a function f : A → R. f is said to be continuous
at c ∈ A if ∀ > 0 ∃δ = δc () > 0 s.t. |x − c| < δ and x ∈ A ⇒ |f (x) − f (c)| < .
f is continuous if it is continuous at each c ∈ A.
Definition 30 (equivalent to Definition 3) f is continuous if xn → l in A ⇒
f (xn ) → f (l) in R.
Theorem 4 Fix a < b in R and f : [a, b] → R. If f is continuous then it is
bounded above (and below), thus ∃k s.t. ∀x ∈ [a, b] |f (x)| ≤ k.
Proof Suppose not. Then ∀n ∈ N ∃xn ∈ [a, b] s.t. f (xn ) > n. By Theorem
3 (xn ) has a convergent subsequence, (xnk )∞
k=1 with xnk → l, say, as k → ∞ in
[a, b]. By Definition 30 f (xnk ) → f (l) ∈ R as k → ∞ contradicting f (xnk ) >
nk → ∞ as k → ∞.
Question
Where does the proof break down for:
(0, 1) → R x 7→
1
x
R → R x 7→ x2
Those domains are not closed intervals.
Definition 4
D ⊂ R then:
2
If f : A → R, g : C → R are functions and B ⊂ A ⊂ C ⊂ R
(i) the restriction of f to B is f |B : B → R is the function for which ∀x ∈
B (f |B )(x) = f (x)
+
D
+
|{z
B
}
|
{z}
A
|
C
(ii) if g|A = f we say g is an extension of f to C.
2 Lecture
04-10-07 starts here
3
|
(iii) f −1 (D) := {a ∈ A : f (a) ∈ D} is called the pre-image of D, although
there is no function f −1 since f is not injective.
Example 1 cos : R → R
O
1+
+
+
−2π
−π
+
0
+
π
/
2π
−1+
differentiable with cos0 = − sin so continuous
cos−1 ((−1, 1)) = R \ {nπ : n ∈ Z} =
[
(nπ, (n + 1)π)
n∈Z
cos−1
1 1
− ,
2 2
[
=
n∈Z
nπ +
π
2π
, nπ +
3
3
−1
((6, 7)) = ∅
[
π
π
cos−1 ((0, 7)) =
2nπ − , 2nπ +
2
2
cos
n∈Z
cos ((−10, 10)) = [−1, 1]
Definition 5
r, x + r) ⊂ U .
A subset U ⊂ R is said to be open if ∀x ∈ U ∃r > 0 s.t. (x −
Remark Hence an open interval (a, b) := {t ∈ R : a < t < b}, where
a, b ∈ R ∪ {−∞, ∞} is open (use r = min{x − a, b − x}) and so is ∅ and R.
(
a
×
x
(
x−r
)
x+r
any union of open intervals is open because x ∈ ∪i∈I (ai , bi ) ⇒ ∃j ∈ I s.t. x ∈
(aj , bj ) ⇒ putting r = min{x − aj , bj − x}, (x − r, x + r) ⊂ (aj , bj ) ⊂ ∪i∈I (ai , bi ).
Theorem 5 f : R → R is continuous (Definition 3) ⇒ ∀U ⊂ R open f −1 (U )
is open in R.
4
Proof ⇒ Suppose f : R → R is continuous and U ⊂ R is open. Take
x ∈ f −1 (U ); then f (x) ∈ U so, for some > 0 (f (x) − , f (x) + ) ⊂ U .
Since f is continuous at x then there is δ > 0 s.t. t ∈ (x − δ, x + δ) ⇒ f (t) ∈
(f (x) − , f (x) + ) ⊂ U . So (x − δ, x + δ) ⊂ f −1 (U ), which is open.
⇐ Take x ∈ R, > 0. Then f −1 (f (x) − , f (x) + ) is open and contains x.
Thus ∃δ > 0 s.t. (x − δ, x + δ) ⊂ f −1 ((f (x) − , f (x) + )). So t ∈ (x − δ, x + δ) ⇒
f (t) ∈ (f (x) − , f (x) + ) and f is continuous.
Corollary 6
3
If f, g : R → R are continuous then so is f ◦ g.
Proof
_
g −1 (f −1 (U ))
= (f ◦ g)−1 (U )
_
g
?
"
U
"
f
?
f −1 (U )
5
ff ◦ g
f −1 (U ) is open since f is continuous. (f ◦ g)−1 is open since g is continuous.
So f ◦ g continuous.
More on this in Metric Spaces and when we study normed spaces.
Proposition 7 The set of functions f : [a, b] → R is a real vector space, W
say. The bounded functions B[a, b] and the continuous functions C 0 [a, b] are
vector subspaces.
C 0 [a, b] ⊂ B[a, b] ⊂ W
Proof f, g : [a, b] → R continuous ⇒ ∀λ, µ ∈ R λf + µg : [a, b] → R x 7→
λf (x) + µg(x) is continuous by Analysis II. If ∀x ∈ [a, b] |f (x)| ≤ k, |g(x)| ≤ l
then ∀λ, µ ∈ R ∀x ∈ [a, b] |(λf + µg)(x)| = |λf (x) + µg(x)| ≤ |λ||f (x)| +
|µ||g(x)| ≤ |λ|k + |µ|l. C 0 [a, b] ⊂ B[a, b] by Theorem 4.
Example
2 sin x + 3x2 is a linear combination of sin and x 7→ x2 .
Remark Polynomials form a vector subspace of C 0 [a, b] and {xk : k ∈ N ∪
{0}} spans with ∀n ∈ N {xk : 0 ≤ k ≤ n} is independent so dim{polynomials} ≥
n.
Definition 6 For h ∈ B[a, b] define its supremum norm khk∞ by khk∞ :=
sup{|h(x)| : x ∈ [a, b]}. This represents the ‘size’ of the function h, as |y| does
for y ∈ R.
3 Lecture
05-10-07 starts here
5
Example
ksin |[−100,100] k∞ = 1 but ksin |[0, π6 ] k∞ =
Proposition 8
1
2
in C 0 [0, π6 ] ⊂ B[0, π6 ]
For h1 , h2 ∈ B[a, b], k·k∞ satisfies:
(i) kh1 k∞ ≥ 0 and kh1 k∞ = 0 iff h1 = 0
(ii) ∀λ ∈ R kλh1 k∞ = |λ|kh1 k∞
(iii) kh1 + h2 k∞ ≤ kh1 k∞ + kh2 k∞
Proof (i) ∀x ∈ [a, b] |h1 (x)| ≥ 0 so supx∈[a,b] |h1 (x)| ≥ 0. supx∈[a,b] |h1 (x)| =
0 ⇒ ∀x ∈ [a, b] h1 (x) = 0. k0k∞ = supx∈[a,b] |0| = 0 where the 0 is the constantzero function.
(ii) kλh1 k∞ = supx∈[a,b] |λh1 (x)| = |λ| supx |h1 (x)| = |λ|kh1 k∞
(iii)4
kh1 + h2 k∞
:=
sup{|h1 (x) + h2 (x)| : a ≤ x ≤ b}
≤
sup{|h1 (x)| + |h2 (x)| : a ≤ x ≤ b}
≤
sup{|h1 (x)| : a ≤ x ≤ b} + sup{|h2 (x)| : a ≤ x ≤ b}
=
kh1 k∞ + kh2 k∞
Definition 7 If f, g : [a, b] → R are bounded (so f, g ∈ B[a, b]) in particular if
they are continuous we define the uniform distance between them by d(f, g) :=
sup{|f (x) − g(x)| : a ≤ x ≤ b} = kf − gk∞ .
Proposition 9
d satisfies:
(i) ∀f, g d(f, g) ≥ 0 and d(f, g) = 0 iff f = g.
(ii) ∀f, g ∈ B[a, b] d(f, g) = d(g, f ).
(iii) ∀f, g, h ∈ B[a, b] d(f, h) ≤ d(f, g) + d(g, h).
Proof
λ = −1.
(i) and (iii) from Proposition 8. (ii) from Proposition 8(ii) with
Remark Proposition 9 gives the standard properties of a metric (on any
set) while Proposition 8 gives the standard properties of a norm, giving distance
between two vectors in a vector space. (More on those in Metric Spaces and
Differentiation next term).
4 Lecture
10-10-07 starts here
6
Remark If f, g ∈ B[a, b] and d(f, g) = kf − gk∞ = t then ∀x ∈ [a, b]
f (x) − t ≤ g(x) ≤ f (x) + t so the graph of g lies between f − t and f + t.
1
For example, f (x) = x2 on [0, 1], g(x) = x2 + 10
sin(2πx). So we have
1
kf − gk∞ = 10 .
O
...
...
.
.
.
.
..
..
.
...
.
.
..
...
...
...
.
.
.
.
...
...
....
...
.
.
.
.
.
.
..
..
.....
+ ....
+ /
0
1
Definition 8 A ⊂ R, f : A → R is said to be uniformly continuous if ∀ > 0
∃δ = δ() > 0 s.t. (x, y ∈ A and |x − y| < δ) ⇒ |f (x) − f (y)| < .
Note ‘Uniform’ means consistently along the domain. This is not the
same as continuity where ∀c ∈ A ∀ > 0 ∃δ = δc () > 0 s.t. y ∈ A and
|c − y| < δ ⇒ |f (y) − f (c)| < , but uniform continuity ⇒ f continuous (just
put δc () = δ()).
Example 2 f : (0, 1) → R, f (x) = x1 is continuous but not uniformly continu1
1
| < δ but |f (δ) − f ( 1 +1
)| = 1 ≮ .
ous, because for = 1 any δ > 0 has |δ − 1 +1
δ
Theorem 10
ous.
Example
δ
Suppose [a, b] → R is continuous. Then it is uniformly continu-
See Example 3 after proof of Theorem 10.
Proof 5 Suppose, if possible, that f : [a, b] → R is continuous but not
uniformly continuous. Then ∃0 > 0 s.t. ∀δ > 0 ∃x, y ∈ A s.t. |x − y| < δ and
|f (x) − f (y)| ≥ 0 . For n = 1, 2, 3, . . . consider δ = n1 . Then ∃xn , yn ∈ A with
|xn − yn | < n1 and |f (xn ) − f (yn )| ≥ 0 .
The sequence, (xn ) on [a, b] is bounded. By Theorem 3, it has a convergent
subsequence (xnk ) with xnk → u, say, as k → ∞. Note that u ∈ [a, b].
By Definition 30 , f (xnk ) → f (u) as k → ∞ and f (ynk ) → f (u) so 0 ≤
|f (xnk ) − f (ynk )| → |f (u) − f (u)| = 0 contradiction.
Question
(0, 1)?
5 Lecture
Where does the argument break down for Example 2,
11-10-07 starts here
7
1
x
on
√
Example 3 6 f : (0, 1) → R, f (x) = x is uniformly continuous because
f = g|(0,1) where g : [0, 1] → R is continuous (so by Theorem 10 is uniformly
continuous).
Definition 9 ϕ : [a, b] → R is called a step function if there is a finite set
P ⊂ (a, b) s.t. ϕ is constant on each open interval of (a, b) \ P .
×
×
×
×
×
+ ++ +
6
Z
a
66V--- F
66-- P
+
b
If P and Q are two such ‘partitions’ we say Q is a refinement of P if P ⊂ Q.
P ∪ Q is called the common refinement of P and Q.
Number the points of P in order so that a = p0 < p1 < . . . < pk−1 < pk = b
with P = {p1 , . . . , pk−1 }, k = #P + 1 , and let cj be the value of ϕ|(pj−1 ,pj ) .
×
× ×
c2+
c3+
×
c1+
+ + + +
a
p1 p2
b
Note ϕ takes at most 2k + 1 values {cj : 1 ≤ j ≤ k} ∪ {ϕ(pj ) : 0 ≤ j ≤ k}
so ϕ ∈ B[a, b].
6 Example
3 was given before the proof of Theorem 10, at the end of the previous lecture
8
Proposition 11 Let f : [a, b] → R be continuous and let > 0. Then there is
a step function ϕ : [a, b] → R with kf − ϕk∞ ≤ O
~
~~
~
~~
~~
~
~~
~~
~
~~
~~
+
+
a
/
b
Proof By Theorem 10, f is uniformly continuous. Take δ = δ() > 0
s.t. x, y ∈ [a, b], |x − y| < δ ⇒ |f (x) − f (y)| < . Take k with b−a
k < δ, and
:=
put pj = a + j b−a
for
0
≤
j
≤
k.
Put
ϕ(p
)
f
(p
)
for
0
≤
j ≤ k and
j
j
k
∀x ∈ (pj−1 , pj ), ϕ(x) := f (pj ) 1 ≤ j ≤ k. Then ∀x ∈ [a, b] ∃j ∈ {0, . . . , k}
s.t. x ∈ (pj−1 , pj ] and |f (x) − ϕ(x)| = |f (x) − f (pj )| < (since |x − pj | < δ).
So kf − ϕk∞ = sup|f (x) − ϕ(x)| ≤ .
Remark 7 ∀n ∈ N take ϕn with kf − ϕn k∞ ≤ n1 . Then (ϕn )∞
n=1 is a
sequence in B[a, b] and kf − ϕn k∞ → 0 as n → ∞.
We shall want to say that (ϕn ) converges to f in (B[a, b], k·k∞ ).
Definition 10 A ⊂ R f : A → R is said to be differentiable with derivative
(c)
→ g(c) as x → c in A.
g : A → R if ∀c ∈ A f (x)−f
x−c
0
g is usually written f . f is said to be C 1 (or of class C 1 ) if also f 0 : A → R
is continuous.
There is a sense (Brownian motion) in which almost all continuous functions
are not differentiable.
Example 4 f2 : R → R, f2 (x) := x cos x1 , f2 (0) := 0
O
/
7 Lecture
12-10-07 starts here
9
is continuous (at 0 this is because |f2 (x)| ≤ x).
f (x)−f (0)
= cos x1 takes values ±1 for x arbitrarily near 0. So f is not
x−0
differentiable at 0.
Proposition 12 The set C 1 [a, b] of C 1 functions f : [a, b] → R is a vector
subspace of C 0 [a, b] and the function D : C 1 [a, b] → C 0 [a, b] given by D(f ) = f 0
is linear.
Proof f differentiable ⇒ f continuous by Analysis II. D(f ) ∈ C 0 [a, b].
D(λf + µg) = λD(f ) + µD(g) by Analysis II.
Example 5 f3 : R → R, f3 (x) := x2 cos x1 f3 (0) := 0
O
/
has f30 (x) = 2x cos x1 + sin x1 x 6= 0,
f30 (0)
=
0
x2 cos x1 − 0
x→0
x−0
1
= lim x cos
x→0
x
=
lim
but as x → 0, f30 (x) − sin x1 → 0 but f30 (x) does not converge as x → 0 f3 is not
C 1.
Theorem 13 (Mean Value Theorem) Suppose f : [a, b] → R is continuous and
f |(a,b) is differentiable. Then ∃c ∈ (a, b) s.t. f (b) − f (a) = (b − a)f 0 (c). This
10
does not need f 0 continuous, and does not say where c is.
O
×
×
×
+
+
a
Slogan
Proof
c
+
/
b
The MVT relates changes in f to values of f 0 .
See Analysis II.
Note The MVT and Taylor’s Theorem (nth MVT) are developed further
in Differentiation.
11
Chapter 2
The Integral for Step
Functions
The mid-ordinate rule Given f : [a, b] → R and n ∈ N put h =
ar := a + rh
b−a
n
and
f
a4 a5 a6 a7
b
+ + + + + + + + +
a
a1 a2 a3
Consider total area of rectangles
An =
n
X
i=1
hf (xr )
1
where xr = a + (r − )h
2
Rb
We might hope to use limn→∞ An for a f but we don’t know that (An )∞
n=1
converges. And if it does, will lim An have the nice properties we expect of
Rb
f?
a
12
Pn
Example 6 1 f1 , f2 : [0, 1] → R,P
f1 (x) = x2 . Show that An = r=1 f (ar ) (ar − ar−1 ) →
n
r
1
1
2
r=1 r = 6 n(n + 1)(2n + 1).
3 as n → ∞ with ar = n , using
(
2 x∈Q
f2 =
1 x∈
/Q
f1
f2
+
xr
R1
Pn
f2 each xr ∈ Q so f2 (xr ) = 2 and Ar = 1 n1 × 2 → 2 as n → ∞.
√
√
R 2
Pn √
However, for 0 2 f2 . each xr ∈
/ Q so f2 (xr ) = 1 and An = 1 n1 22 × 1 = 22 ,
suggesting that
For
0
√
Z
2
2
Z
f2 +
0
1
√
2
2
√
√
Z 1
2
2
f2 =
+ (1 −
) = 1 6= 2 =
f2
2
2
0
This is most unsatisfying and appears to be connected to f2 not being continuous.
Books often use (upper and lower) ‘Riemann Sums’. Replacing f (xr ) by
sup{f (x) : ar−1 ≤ x ≤ ar } and not insisting that the intervals are of equal
length. We use step and regulated functions (see e.g. Walker).
Problem 1 Let f : [a, b] → R be Ra bounded function (and a ≤ b in R). At
x
least for nice f , ∀x ∈ [a, b] define a f ∈ R representing the ‘area under the
graph’ of f |[a,x] and satisfying:
Rx
Rb
Rb
(i) a f + x f = a f
Rx
Rx
Rx
(ii) a (λf + µg) = λ a f + µ a g
Rb
(iii) a 1 = b − a
Problem 2 Is there, at least for nice f , an antiderivative, i.e a differentiable
function F : [a, b] → R with F 0 = f ?
Example 5 suggests that f need not be continuous.
1 Lecture
17-10-07 starts here
13
Problem 3 Understand the relation between Problems 1 and 2. ‘Area under
the graph’ is not obviously related to the slope of the tangent of F .
We focus on this general theory and not much on formulae for an antiderivative (the antiderivative of a familiar function may not be familiar). Thus x 7→ xn
n+1
has antiderivative x 7→ xn+1 unless n = −1 when a new function log is needed.
Arc length along an ellipse gives elliptic functions. Area is easily defined for
rectangles so first consider step functions.
Proposition 14 The set of step functions ϕ : [a, b] → R is a vector subspace,
S[a, b] say, of B[a, b].
Proof ϕ is bounded since it takes at most 2k + 1 values. Suppose λ, µ ∈ R
and ϕ, ψ ∈ S[a, b] are constant on the open intervals of partitions P and Q
respectively. Then ϕ and ψ and so λϕ + µψ are constant on the refinement of
these partitions P ∪ Q. So λϕ + µψ : [a, b] → R is a step function.
Definition 11 2 Let ϕ : [a, b] → R be a step function constant on the open
intervals of a partition P = {p1 , . . . , pk−1 } and ∀j ∈ {1, . . . , k} ∀x ∈ (pj−1 , pj ),
Rb
Pk
ϕ(x) = cj . Define a ϕ = j=1 cj (pj − pj−1 ).
Note
This ignores ϕ(pj ) If cj < 0 the contribution is negative.
Lemma 15
partition P .
For a step function ϕ : [a, b] → R,
Rb
a
ϕ is independent of the
Proof If P and Q are partitions for which ϕ is constant on their open
Rb
intervals, it suffices to show a ϕ is the same for P and P ∪ Q = R (because
then also the same for Q and R). When comparing P and R it suffices to add
points to P one at a time. Thus consider P and P ∪ {r} where pj−1 < r < pj ,
pj−1 < x < r ⇒ ϕ(x) = cj and r < x < pj ⇒ ϕ(x) = cjP
. cj (r−pj )+cj (pj −r) =
cj (pj − pj−1 ). The other summands are unchanged at j6=i cj (pj − pj−1 ) for P
Rb
and for P ∪ {r} so a P is the same using P and P ∪ {r}.
Next we establish the properties of the integral of step functions. Let
ϕ : [a, b] → R be a step function and let a ≤ u < v < w ≤ b. Then ϕ|[u,w]
is a step function using the partition P ∩ (u, w).
Proposition 16
(Additivity)
Z w
Z
v
ϕ=
u
2 Lecture
Z
ϕ+
u
18-10-07 starts here
14
w
ϕ
v
Proof Take partitions u = p0 < p1 < . . . < pk = v = q0 < q1 < . . . < pl =
w so that ϕ is constant at cj on (pj−1 , pj ) and at c0j on (qj−1 , qj ). By Lemma
15:
Z
w
ϕ=
u
k
X
cj (pj − pj−1 ) +
j=1
l
X
Z
v
cj (qj − qj−1 ) =
Z
ϕ+
u
j=1
w
ϕ
v
(Linearity) Let ϕ, ψ : [a, b] → R be step functions and λ, µ ∈
Proposition 17
R. Then
Z
b
Z
(λϕ + µψ) = λ
a
Thus I : S[a, b] → R, I(ϕ) :=
b
Z
ϕ+µ
a
Rb
a
b
ψ
a
ϕ is linear.
Proof As in Proposition 14, let R = P ∪ Q be a partition such that
ϕ, ψ, λϕ + µψ are each constant on the open intervals R. Proposition 16 reduces the proof to the case that ϕ, ψ and λϕ + µψ are constant functions, where
it is obvious.
ϕ = cj on (pj−1 , pj ) and ψ = c0j on (pj−1 , pj ) so
Z
pj
λϕ + µψ = (λcj + µc0j )(pj − pj − 1)
pj−1
Proposition 18 (Fundamental Theorem of Calculus for Step Functions) Let
ϕ : [a, b] → R be a step function and ϕ|(pj−1 ,pj ) = cj for a = p0 < p1 < . . . <
Rx
pk = b. Then Φ : [a, b] → R, Φ(x) := a ϕ is differentiable on ∪kj=1 (pj−1 , pj )
and ∀x ∈ (pj−1 , pj ), Φ0 (x) = ϕ(x).
Proof
3
For 1 ≤ i ≤ k ∀x ∈ (pj−1 , pj )
Z x
Φ(x) =
φ
a
Z pj−1
Z
=
φ+
a
Z
x
φ
pj−1
pj−1
φ + cj (x − pj−1 )
=
a
=
constant + cj x
so Φ is differentiable at x with Φ0 (x) = cj = φ(x).
3 Lecture
19-10-07 starts here
15
Example
If
(
2
φ(x) =
−1
if 0 ≤ x ≤ 1
if 1 < x ≤ 3
then
Z
Φ(x) =
0
x
(
2x
if 0 ≤ x ≤ 1
φ=
2(1 − 0) + (−1)(x − 1) = 3 − x if 1 < x ≤ 3
(1) Φ is differentiable on [0, 1) ∪ (1, 3] with Φ0 = φ there.
(2) Φ is continuous but not differentiable at 1.
(3) Changing φ(1) does not change Φ or (1) or (2).
Proposition 19 (Bounds for the integral) Let φ : [a, b] → R be a step function,
Rb
satisfying ∀x ∈ [a, b], m ≤ φ(x) ≤ M . Then m(b − a) ≤ a φ ≤ M (b − a) and
Rb
∀φ ∈ S[a, b], | a φ| ≤ kφk∞ (b − a).
Proof Let a = p0 < p1 . . . < pk = b be a partition s.t. ∀j ∈ [1, . . . , k] ∀x ∈
(pj−1 , pj ) φ(x) = cj then m ≤ cj ≤ M and m(pj − pj−1 ) ≤ cj (pj − pj−1 ) ≤
M (pj − pj−1 ).
Rb
Adding these inequalities for 1 ≤ j ≤ k gives m(b − a) ≤ a φ ≤ M (b − a)
as required.
Put m = −kφk∞ = − sup {|φ(x)| : x ∈ [a, b]}. Thus ∀x ∈ [a, b] m ≤ φ(x) ≤
Rb
M so by the first part −kφk∞ (b − a) ≤ a φ ≤ kφk∞ (b − a).
16
Chapter 3
The Integral for Regulated
Functions
Definition 12 The function f : [a, b] → R is said to be regulated if ∀ >
0 ∃φ ∈ S[a, b] s.t. kφ − f k∞ < . Equivalently f is regulated if there is a
sequence (φn )∞
n=1 in S[a, b] with kφn − f k∞ → 0 as n → ∞ because
⇒ ∀n ∈ N choose φn with kφn −f k ≤ n1 ⇐ ∀ > 0 pick n > 1 and use φn for φ.
If kφn − f k∞ → 0 as n → ∞ we shall say that the sequence (φn ) converges
uniformly to f in S[a, b].
Note Any regulated function is bounded since, with φ ∈ S[a, b] and kφ −
f k∞ < 1 we have ∀x ∈ [a, b], φ(x) − 1 ≤ f (x) ≤ φ(x) + 1. inf φ(x) − 1 ≤ f (x) ≤
sup φ(x) + 1.
Any step function, and by Proposition 11, any continuous function on [a, b]
is regulated.
1
O
...
..
..
.. ... ..... . ..
f +
.. ..
... .. ....
.
.
f
.
.
... .
.. .... . .. f −
.
.
.
...
... . ...
+ /
b
+
a
Rb
Definition 13 For a regulated function f : [a, b] → R we define a f to be
Rb
limn→∞ a ϕn where (ϕn ) is a sequence of step functions converging uniformly
to f . This needs Theorem 20:
1 Lecture
24-10-07 starts here
17
Theorem 20 Let f : [a, b] → R be a regulated function and (ϕn )∞
n=1 be a seRb
∞
quence of step functions converging uniformly to f . Then ( a ϕn )n=1 converges
in R. If (ψn )∞
of step functions converging uniformly to
n=1R is another sequence
Rb
b
f then limn→∞ a ψn = limn→∞ a ϕn .
Proof
∀ > 0 ∃N1 () s.t. ∀n > N1 () kϕn − f k∞ < ; then
kϕm − ϕn k∞
= kϕm − f + f − ϕn k∞
≤
by Proposition 8
kϕm − f k∞ + kϕn − f k∞
≤ 2
so by Proposition 19
Z
|
b
Z
ϕm −
a
b
Z b
ϕn | = | (ϕm − ϕn )| ≤ (b − a)2
a
a
Rb
Thus ( a ϕn ) is a Cauchy sequence and converges in R.
Given (ψn ) define (ωn )∞
n and ω2n−1 := ϕn . Since (ωn )
n=1 by ω2n := ψR
b
converges uniformly to f the first part says ( a ωn ) converges. So
Z
lim
n→∞
b
Z
ϕn = lim
a
n→∞
b
Z
ωn = lim
n→∞
a
b
ψn
a
We now show that the integral of regulated functions has good properties as
for step functions:
Property 21 (Additivity) Let a ≤ u ≤ v ≤ wR≤ b and
R vletf : R[ wb] → R be a
w
regulated function. Then f |[u,w] is regulated and u f = u f + v f .
Proof Choose a sequence of step functions (ϕn : [a, b] → R)∞
n=1 converging
uniformly to f . (kϕn − f k∞ → 0 as n → ∞.)
kϕn |[u,w] − f[u,w] k∞ ≤ kϕn − f k∞
so (ϕn |[u,w] )∞
n=1 is a sequence of step functions converging uniformly to f |[u,w]
as n → ∞, which is therefore regulated. The same applies to f |[u,v] and f |[v,w] .
Hence, by Proposition 16
Z v
Z w
Z w
Z w ϕn +
ϕn
f := lim
ϕn = lim
n→∞ u
n→∞
u
u
v
Z v
Z w
(by Analysis I) = lim
ϕn + lim
ϕn
n→∞
n→∞ u
Z v uZ w
=
f+
f
u
18
v
Proposition 22 (Linearity)2 Let f, g : [a, b] → R be a regulated function and
Rb
Rb
let µ, λ ∈ R. Then λf + µg : [a, b] → R is regulated and a (λf + µg) = λ a f +
µba g
The set R[a, b] say, of regulated functions on [a, b] is a vector space (subspace
Rb
of B[a, b]) and I : R[a, b] → R, I(f ) := a f is linear.
Proof Choose sequences (ϕn ) and (ψn ) in S[a, b] with kϕn − f k∞ → 0 and
kψn − gk∞ → 0 as n → ∞. Then by Proposition 14 λϕn + µψn ∈ S[a, b] ∀n ∈ N.
k(λϕm + µψn ) − (λf + µg)k∞
= kλ(ϕn − f ) + µ(ψn − g)k∞
≤
|λ|kϕn − f k∞ + |µ|kψn − gk∞ → 0 as n → ∞
∴ (λf + µg) is regulated (by (λϕn + µψn )) and hence R[a, b] is a vector space
Z
b
(λf + µg) :=
and
b
Z
a
lim
(λϕn + µψn )
n→∞
a
b
Z
(by Proposition 17)
=
lim
λ
n→∞
λ lim
n→∞
Z
=
λ
so I is linear.
ψn
a
b
Z
ϕn + µ lim
n→∞
a
b
Z
f +µ
a
!
b
ϕn + µ
a
Z
=
Z
b
ψn
a
b
g
a
Proposition 23 (Bounds) If f : [a, b] → R is regulated and ∀x ∈ [a, b] m ≤
Rb
Rb
f (x) ≤ M then m(b − a) ≤ a f ≤ M (b − a). Hence ∀f ∈ R[a, b] | a f | ≤
(b − a)kf k∞ .
Proof Let (ϕn )∞
n=1 be a sequence in S[a, b] with kϕn −f k∞ → 0 as n → ∞.
Replace any values of ϕn > M by M and < m by m; this cannot increase
Rb
kϕn − f k∞ . By Proposition 19, ∀n m(b − a) ≤ a ϕn ≤ M (b − a). Letting
2 Lecture
25-10-07 starts here
19
n → ∞ we obtain m(b − a) ≤
R
f ≤ M (b − a).
M+
+
+
a
b
m+
Also −kf k∞ (b − a) ≤
f (x) ≤ kf k∞ .
Rb
a
f ≤ kf k∞ (b − a) follows since ∀x ∈ [a, b] − kf k∞ ≤
20
Chapter 4
Indefinite Integral and the
Fundamental Theorem of
Calculus
Definition 14 Let f : [a, b] → R be regulated. Define its indefinite integral
F : [a, b] → R by
Z x
F (x) :=
f
notice F (a) = 0
a
Theorem 24 The Indefinite integral F : [a, b] → R of a regulated function
f : [a, b] → R is uniformly continuous.
Ry
Proof For a ≤ x ≤ y ≤ b, |F (y) − F (x)| = | x f | by Proposition 21.
Ry
| x f | ≤ kf k∞ |y − x| by Proposition 23. So ∀ > 0 |y − x| < kf k∞ x, y ∈ [a, b]
⇒ |F (y) − F (x)| ≤ kf k∞ kf k∞ = ∴ uniform continuous.
Theorem 25 1 Let f : [a, b] → R be regulated and suppose f is continuous at
some c ∈ [a, b]. Then the indefinite integral F : [a, b] → R of f is differentiable
at c with F 0 (c) = f (c).
Proof ∀ > 0 ∃δ > 0 s.t. c − δ < x < c + δ and x ∈ [a, b] ⇒ f (c) − <
f (x) < f (c) + .
R c+h
Thus if 0 < h < δ ⇒ (f (c) − )h < c f = F (c + h) − F (c) < (f (c) + )h ⇒
(c)
(c)
| F (c+h)−F
− f (c)| < . And similarly for −δ < h < 0. Thus F (c+h)−F
→
h
h
f (c) as h → 0, as required.
1 Lecture
26-10-07 starts here
21
Corollary 26 (First form of the Fundamental Theorem of Calculus). Suppose
f : [a, b] → R is continuous. Then:
(1) ∃ a differentiable function g : [a, b] → R with g 0 = f .
(2) Rif h : [a, b] is differentiable with h0 = f then ∃k s.t. ∀x ∈ [a, b] h(x) =
x
f + k.
a
Rx
Proof (1) g(x) := a f will do by Theorem 25.
(2) (h − g)0 = 0 so h − g constant by Mean Value Theorem (Theorem
13).
Theorem 27 (Second form of Fundamental Theorem of Calculus) Let f : [a, b] →
R be a regulated function and suppose there is a differentiable g : [a, b] → R with
Rb
g 0 = f . Then a f = g(b) − g(a).
Notes
(1) If f is continuous, this follows from Corollary 26.
(2) Putting x for b in Theorem 27 and ∀x ∈ [a, b] gives F (x) :=
is differentiable (since g is differentiable).
Rx
a
f (= g(x) − g(a))
Proof Fix any > 0 and choose ϕ ∈ S[a, b] with kϕ − f k∞ < . Take
a partition a = p0 < p1 < . . . < pk−1 < pk = b with ∀i ∈ {1, 2, . . . k}, ∀x ∈
(pi−1 , pi ) ϕ(x) = ci . The Mean Value Theorem (Theorem 13) for g|[pj−1 ,pj ] gives
xj ∈ (pj−1 , pj ) with g(pj ) − g(pj−1 ) = g 0 (xj )(pj − pj−1 ) = f (xj )(pj − pj−1 ).
kϕ − f k∞ < gives cj − < f (xj ) < cj + . Adding (cj − )(pj − pj−1 ) <
g(pj ) − g(pj−1 ) < (cj + )(pj − pj−1 ) gives
Z b
Z b
ϕ − (b − a) < g(b) − g(a) <
ϕ + (b − a)
(4.1)
a
a
Rb
Rb
Rb
Rb
by Proposition 23 | a ϕ− a f | = | a (ϕ−f )| < (b−a). So |g(b)−g(a)− a f | <
Rb
2(b − a) as |g(b) − g(a) − a ϕ| < (b − a) by (4.1). This holds for any > 0 so
Rb
g(b) − g(a) = a f .
.
Rb
Rb
Notation 2 If f is given by f (s) = cos(2s) then a f may be written a f (t)dt
Rb
or a cos(2u)du.
n+1
For example ∀n ∈ N fn (x) := xn , gn (x) := xn+1 satisfy gn0 = fn on R. If
a < b in R then fn |[a,b] is differentiable so continuous so regulated and Theorem
h n+1 ib
Rb
Rb
27 (FTC2) gives a fn = gn (b)−gn (a) and one might write a xn dx = xn+1 =
a
bn+1 −an+1
.
n+1
2 Lecture
31-10-07 starts here
22
Corollary 28 (Integration by parts) If F, G : [a, b] → R are differentiable and
f := F 0 and g := G0 are regulated then
Z b
Z b
F g = F (b)G(b) − F (a)G(a) −
fG
a
1
Proof H := F G is differentiable (by Analysis II) with H 0 = f G + F g ∈
R[a, b] (by Qn5 on Sheet 4) so by Theorem 27:
Z b
Z b
Z b
0
F (b)G(b) − F (a)G(a) = H(b) − H(a) =
H =
fG +
Fg
a
a
a
Corollary 29 (Integration by Substitution) Suppose that G : [c, d] → R is
differentiable. G0 ∈ C 0 [c, d] and ∀x ∈ (c, d) G0 (x) > 0. Put a := G(c), b := G(d).
0
(Then by Analysis II G : [c, d] → [a, b] is
a bijection). If f ∈ C [a, b] then
R
Rd
Rb
d
f = c (f ◦ G)G0 = c f (G(t))G0 (t)dt .
a
Rx
Proof Consider F : [a, b] → R F (x) := a f and define H : [c, d] → R
by H(t) := F (G(t)). Then F is differentiable by Corollary 26 (FTC1) and so
H = F ◦ G is also differentiable by Analysis II with H 0 = (F 0 ◦ G)G0 . Now
Theorem 27 says
Z d
Z d
f (G(t))G0 (t)dt =
H0
c
c
= H(d) − H(c)
= F (b) − F (a)
Z b
=
f
a
Example 1
O
1_
b+
0
+
_
arcsin b
π
2
G = sin. 0 < b ≤ 1 has G0 > 0 on (0, π2 ). f (x) :=
Z
b
Z
f=
0
0
b
√
1
dx =
1 − x2
Z
0
arcsin b
cos(t)
q
1 − sin2 (t)
23
/
√ 1
1−x2
Z
dt =
arcsin b
1dt = arcsin b
0
Chapter 5
Normed Vector Spaces
Definition 15 Let V be a real (or complex) vector space. A norm on V is a
function V → R, written k·k : V → R. v 7→ kvk satisfying:
(i) ∀v ∈ V kvk ≥ 0, kvk = 0 ⇔ v = 0v
(ii) ∀λ ∈ R (or C ) ∀v ∈ V kλvk = |λ|kvk.
(iii) ∀v, v 0 ∈ V kv + v 0 k ≤ kvk + kv 0 k
e.g By Proposition 8, the supremum norm k·k∞ on B[a, b] is a norm.
e.g. |·| is a norm on R.
1
A normed vector space is a pair (V, k·k) when k·k is a norm on the vector
space V .
Proposition 30 (Norms on Rn ) On Rn = {x = (x1 , . . . , xn ) : xj ∈ R 1 ≤ j ≤ n}
kxk1 :=
n
X
|xj |
j=1
v
uX
u n 2
xj
kxk2 := t
j=1
kxk∞ := max |xj |
1≤j≤n
are norms.
Remark These are the cases p = 1, 2, limp→∞ of

 p1
n
X
kxkp :=  |xj |p 
j=1
which is a norm for 1 ≤ p < ∞, but proof needs Minkowski’s inequality.
1 Lecture
01-11-07 starts here
24
Pn
Proof (i) and (ii) are easily checked, and taking j=1 or max1≤j≤n of
|xj +x0j | ≤ |xj |+|x0j | gives (iii) for k·k1 and k·k∞ . It remains to show kx+yk2 ≤
kxk2 + kyk2

n
X

2
aj bj 

= 
j=1
n
X

a2j  
≤ 
n
X

a2j  
j=1
so
n
X
aj bj
b2j  −
n
X
X
(aj bk − ak bj )2
1≤j≤k≤n

b2j 
j=1
v
v
u n
uX
u n 2 uX
≤ t
aj × t
b2j
j=1
kx + yk22

j=1
j=1

n
X
j=1
=
n
X
(5.1)
j=1
(xj + yj )2
j=1
=
n
X
xj (xj + yj ) +
j=1
≤
qX
n
X
yj (xj + yj )
j=1
x2j
qX
(xj + yj )2 +
qX
yj2
qX
(xj + yj )2
by (5.1)
= kxk2 kx + yk2 + kyk2 kx + yk2
so
kx + yk2
≤
kxk2 + kyk2
Definition 16 The norms k·ka , k·kb on a vector space V are said to be equivalent if ∃k ∈ R s.t. ∀v ∈ V k −1 kvka ≤ kvkb ≤ kkvka . This is clearly an
equivalence relation on the set of norms on V
Lemma 31 ∀n ∈ N the norms k·k1 , k·k2 and k·k∞ on Rn are equivalent.
O
(−1, 1)
(1, 1)
_? X
•
•
m f  ??? Q I
??
u
ullll kxk∞ = 1

?
?


??

gO5OO
??

OOO
??


?? - O


?? &
kxk2 = 1

??

?

?
/
?
??


& ??
??
 mm kxk1 = 1
??
mmmmm

m ??
vmmm
??
5

?

??
?


?

?
I

Q X ???  m u
f
_

•
•
(−1, −1)
(1, −1)
25
Proof
∀x ∈ Rn kxk∞ ≤ kxk2 ≤ kxk1 ≤ nkxk∞
yXX
X
Xadd |xj | ≤ max1≤j≤n |xj |
consider (x1 , 0, 0) triangle
inequality for k·k
2
Theorem 32
Proof
etc.) so
n
Any norm on R is equivalent to k·k∞ .
x = (x1 , . . . , xn ) =
kxk
Pn
i=1
= k
≤
xi ei (where e1 = (1, 0, . . . , 0) e2 = (0, 1, 0, . . . , 0)
n
X
xi ei k
i=1
n
X
|xi |kei k
j=1
≤
n
X
!
kej k
kxk∞
(5.2)
i=1
2
It remains to show that 0 < k := inf {kxk : kxk∞ = 1} = inf
n
kxk
kxk∞
kxk
kλxk
since ∀λ ∈ R \ {0} kxk
= kλxk
.
∞
∞
k ∞
Suppose not and take (x )k=1 in Rn with kxk k∞ = 1 and kxk k < k1 .
O
(−1, 1)
(1, 1)
•
•
/
(−1, −1)
•
? ? ??
y
? ?
? ?
•
(1, −1)
The cube has 2n faces so take j in {1, . . . , n} and a subsequence s.t. all terms
have xkj = 1 (or −1).
Now take a subsequence of this for which xk1 converges (to y1 ) by Theorem 3, then a subsequence of this for which xk2 converges (to y2 ) ,. . . , then
2 Lecture
02-11-07 starts here
26
o
: x ∈ Rn \ {0}
a subsequence of this for which xkn converges (to yn ). Now yj = 1 (or −1).
k
n and kyk > 0. Then 0 < kyk = kx
So y = (y1 , . . . , yn ) 6= 0RP
+ y − xk k ≤
n
1
k
k
k
kx k + ky − x k < k + ( 1 kei k)ky − x k∞ by equation 5.2.
=
n
X
1
+( kei k) max |yi − xki |
1≤i≤n
k
|{z}
1
|
{z
}
→0
→0
This contradiction completes the proof.
Definition 17 For x in a normed vector space (V, k·kV ) and δ > 0 define the
open ball centre x radius δ as B(x, δ) = B(x, δ, k·k) := {y ∈ V : ky − xkV < δ}.
U ⊂ V is said to be an open subset of if ∀x ∈ U ∃δ s.t. B(x, δ) ⊂ U .
Lemma 33 B(x, δ) is an open subset of (V, k·kV ). B(x, δ) = x + δB(0V , 1).
Proof For y ∈ B(x, δ) kz − ykV < δ − ky − xkV ⇒ kz − xkV ≤ kz − ykV +
ky − xkV < δ
O
z
?
nnn B(x, δ)
x ? vnnynn
?
/
Thus B(y, δ − ky − xkV ) ⊂ B(x, δ) and δ − ky − xkV > 0.
kvk ∈ B(0, 1) ⇔ kvkV < 1 ⇔ kδvk < δ ⇔ k(x + δv) − xkV < δ ⇔ x + δv ∈
B(x, δ).
Definition 18 Let (V, k·kV ) and (W, k·kW ) be normed vector spaces. Then
a map f : V → W is said to be (k·kV , k·kW )-continuous or just continuous at
x ∈ V if ∀ > 0 ∃δ = δx () > 0 s.t. kx − ykV < δ ⇒ kf (x) − f (y)kW < (or
equivalently ∀ > 0 ∃δ > 0 s.t. f (B(x, δ)) ⊂ B(f (x), )).
3
If δ can be chosen depending on but not on x we say f is uniformly
continuous.
Lemma 34 If k·ka and k·kb are equivalent norms on V then U is open in
(V, k·ka ) ⇔ U is open in (V, k·kb ).
3 Lecture
07-11-07 starts here
27
Proof Take k ∈ R s.t. ∀v ∈ V k −1 kvka ≤ kvkb ≤ kkvka . Then B(x, kδ , k·kb ) :=
{y ∈ V : kx−ykb < kδ } ⊂ {y ∈ V : kx−yka < δ} := B(x, δ, k·ka ). ⇒ If U is open
in (V, k·ka ) then ∀x ∈ U ∃δ > 0 s.t. B(x, δ, k·ka ) ⊂ U and then B(x, kδ , k·kb ) ⊂ U
so U is open in (V, k·kb ). Proof ⇐ is similar.
Proposition 35
The set U of those subsets of V open in (V, k·kV ) satisfies:
(i) ∅, V ∈ U
(ii) U1 , U2 ∈ U ⇒ U1 ∩ U2 ∈ U
S
(iii) ∀i ∈ I Ui ∈ U ⇒ i∈I Ui ∈ U
Proof (i) For ∅ there is nothing to prove. ∀x ∈ V B(x, 1, k·kV ) ⊂ V .
(ii) x ∈ U1 ∩ U2 ⇒ ∃δ1 , δ2 > 0 s.t. B(x, δ1 , k·kV ) ⊂ U1 B(x, δ2 , k·kV ) ⊂ U2
but now B(x,
Smin(δ1 , δ2 ), k·kV ) ⊂ U1 ∩ U2 .
(iii)
x
∈
i∈I Ui ⇒ ∃j ∈ I s.t. x ∈ Uj ⇒ ∃δ > 0 s.t. B(x, δ, k·kV ) ⊂ Uj ⊂
S
i∈I Ui .
Remark U is called the topology of (V, k·kV ), more in Metric Spaces.
Look up norm in book on differential calculus (QA304 in the library).
Lemma 36
B(x, δx , k·kV ).
U ∈ V is open in (V, k·kV ) ⇔ it is a union of open balls
Proof ⇐ by Proposition 35 (iii) and Lemma 33.
S ⇒ Take U open ∀x ∈ U ∃δxS> 0 s.t. B(x, δx , k·kV ) ⊂ U . Then U ⊂
x∈U B(x, δx , k·kV ) ⊂ U . So U =
x∈U B(x, δx , k·kV ).
Compare with Definition 5 in (R, |·|) (x − r, x + r) = B(x, r, |·|)
Proposition 37 a) f : V → W is (k·kV , k·kW )-continuous ⇔ ∀U open in
(W, k·kW ), f −1 (U ) is open in (V, k·kV )
(b) If f : V → W is (k·kV , k·kW )-continuous then it is also continuous if we
replace k·kV or k·kW by an equivalent norm.
Proof 4 ⇒ Given U open in (W, k· kV ) and x ∈ f −1 (U ) take f (x) >
0 s.t. B(f (x), f (x) , k·kW ) ⊂ U . Since f is continuous at x ∃δx > 0 s.t.
f (B(x, δx , k·kV )) ⊂ B(f (x), f (x) , k·kW ) ⊂ U . Thus B(x, δx , k·kV ) ⊂ f −1 (U ),
so f −1 (U ) is open.
⇐ Given x ∈ V and > 0. B(f (x), , k·kW ) is open in (W, k·kW ) by Lemma
33, so f −1 (B(f (x), , k·kW )) is open in (V, k·kV ). x ∈ f −1 (B(f (x), , k·kW ))
so ∃δ > 0 s.t. B(x, δ, k·kV ) ⊂ f −1 (B(f (x), , k·kW )). Thus f (B(x, δ, k·kW )) ⊂
B(f (x), , k·kW ) and so f is continuous at x.
4 Lecture
08-11-07 starts here
28
Replacing k·kV or k·kW by an equivalent norm does not change which subsets
of V or W are open. Now use the bit above.
Proposition 38 Let (V, k·kV ) , (W, k·kW ) be normed vector spaces. Let T :
V → W be linear. Then the following are equivalent:
(i) T is continuous at 0V .
(ii) T is continuous.
(iii) {kT vkW : kV kV < 1} is bounded. (In which case we say T is bounded.)
T
O
O
$
V
W
/
?



max




/
Proof (i) ⇒ (ii) Assume T continuous at 0V . ∀ > 0 ∃ δ > 0 s.t.
T (B(0V , δ, k·kV )) ⊂ B(0W , , k·kW ). Then y ∈ B(v, δ, k·kV ) ⇔ ky − vkV < δ
⇒ kT y − T vkW = kT (y − v)kW < ⇔ T (y) ∈ B(T (v), , k·kW )
Thus T is continuous at any v ∈ V . (T is uniformly continuous).
(ii) ⇒ (i) is obvious.
(i) ⇒ (iii) Assume T is continuous at 0V so for = 1 ∃ δ > 0 s.t. T (B(0V , δ, k·kV )) ⊂
B(0W , 1, k·kW ). kV kV ≤ 1 ⇒ for 0 < t < δ tkT (v)kW = kT (tv)kW < 1 ⇒
kt(v)kW ≤ 1δ so {kT vkW : kvkV < 1} is bounded.
(iii) ⇒ (i) If k := sup{kT vkW : kvkV ≤ 1} < ∞ then kvkV < k ⇒ k k vkV <
1 ⇒ k k T (v)kW ≤ k ⇒ kT (v)kW ≤ . Hence T is continuous at 0.
Proposition 39 5 V, W normed vector spaces L(V, W ) set of bounded linear
maps V → W . Then L(V, W ) is a vector space and kT k = sup{kT vkW : kvkV ≤
1} is a norm on it.
5 Lecture
09-11-07 starts here
29
Remark L(Rn , Rn ) is isomorphic to the space of m × n real matrices. All
of these are bounded.
Proof
λ, µ ∈ R S, T ∈ L(V, W ) ⇒ λS + µT : V → W which is in L(V, W ).
kλS + µT k
=
sup{k(λS + µT )(v)kW : kvkV ≤ 1}
≤ sup{|λ| · kSvkW + |µ|kT vkW : kvkV ≤ 1}
≤
|λ| sup{kSvkW : kvkV ≤ 1} + |µ| sup{kT vkW : kvkV ≤ 1}
= |λ| · kSk + |µ| · kT k
which is finite
so kS + T k is linear and bounded.
and
kS + T k
≤
kSk + kT k
kλSk
=
sup{kλSvkW : kvkV ≤ 1}
= |λ| sup{kSvkW : kvkV ≤ 1}
= |λ| · kSk
kT k ≥ 0 and kT k = 0 iff T = 0.
kT k = sup{kT vkW : kvk ≤ 1} = 0 ⇔ kT vkW = 0 ∀v, kvkV ≤ 1
But k·kW is a norm so ⇔ T = 0 ∴ kT k := sup{kT vkW : kvkV ≤ 1} is the
‘operator norm’ on L(V, W ).
30
Chapter 6
Closed, Convergent,
Cauchy, Completeness,
Contractions
Definition 19
V \ Z is open.
Let (V, k·kV ) be a normed space, Z ⊂ V is said to be closed iff
Example In (R, |·|), [a, b] is closed because R \ [a, b] = (−∞, a) ∪ (b, ∞),
which is open.
Proposition 40 f : V → W is (k·kV , k·kW )-continuous ⇔ every closed subset
Z ⊂ (W, k·kW ) has f −1 (Z) closed in (V, k·kV ).
Proof f −1 (W \ Z) = V \ f −1 (Z) so this follows from the formulation for
open subsets shown in Proposition 37 to be equivalent to continuity.
Definition 20 In a normed space (V, k·kV ) we say a sequence (xn ) converges
to a in (V, k·kV ) if ∀ > 0 ∃ N () s.t. ∀n ≥ N , kxn − akV < .
Proposition 41 Suppose Z closed in (V, k·kV ) and ∀n ∈ N, xn ∈ Z. If xn → a
then a ∈ Z.
Proof If a ∈ V \ Z (which is open) then ∃r > 0 s.t. ky − akv < r ⇒ y ∈
V \Z ⇒y ∈
/ Z which contradicts kxN (r) − ak < r.
Definition 21 1 A sequence (xn )∞
n=1 in a normed space (V, k·kV ) is said to be
Cauchy if ∀ > 0, ∃ M () s.t. ∀m, n ≥ M, kxm − xn kV < .
1 Lecture
14-11-07 starts here
31
Lemma 42 If xn → a then the sequence is Cauchy.
Proof If > 0 then corresponding to 2 there is N ( 2 ) s.t. ∀n ≥ N, kxn −
akV < 2 . Then m, n ≥ N ( 2 ) ⇒ kxm − xn kV ≤ kxm − akV + kxn − akV <
2 + 2 = .
Definition 22 The normed space (V, k·kV ) is said to be complete if every
Cauchy sequence in (V, k·kV ) converges (to some a ∈ V ). A complete normed
space is called a Banach space.
Proposition 43
Rq with any of its equivalent norms is a Banach space.
Proof Let (xn ) be a Cauchy sequence in (Rq , k·k∞ ) and xn = (xn1 , . . . , xnq ).
Then for 1 ≤ j ≤ q
n
m
n
|xm
j − xj | ≤ max |xj − xj | =: kxm − xn k∞
1≤j≤q
so (xnj )∞
n=1 is a Cauchy sequence in (R, |·|) and converges (by Analysis I) to
some aj ∈ R.
q
Then (xn )∞
n=1 → a = (a1 , . . . , aq ) in (R , k·k∞ ).
Theorem 44 (Contraction Mapping Theorem). Let (V, k·kV ) be a Banach
space, ∅ =
6 Z ⊂ V a closed subset 0 < k < 1 and f : Z → Z a function satisfying
x, y ∈ Z ⇒ kf (x) − f (y)kV ≤ kkx − ykV
Then there exists a unique z ∈ Z s.t. f (z) = z. Moreover, ∀x ∈ Z |f n (x)| → z
as n → ∞ when we define f 2 (x) = f (f (x)), f n (x) = f (f n−1 (x)).
Example
Z = V = R, f (x) = kx + c
O
o
ooo
_o oo
o
oo
ooo
o
o
ooo
co_o o
o
o
ooo
_
/
z
f (z) _
c
has f (z) = z if z = 1−k
. f (0) = c, f 2 (0) = f (c) = kc + c = c(1 + k).
c
f n (0) = c(1 + k + k 2 + . . . + k n−1 ) → 1−k
32
Proof
f (z1 ) = z1 f (z2 ) = z2
⇒ kz1 − z2 kV = kf (z1 ) − f (z2 )kV ≤ kkz1 − z2 kV
⇒ kz1 − z2 kV = 0
⇒ z1 = z2
For x ∈ Z
kf n (x) − f n+1 (x)kV ≤ kkf n−1 (x) − f n (x)kV ≤ . . . ≤ k n kx − f (x)kV
so by triangle inequality m > n
⇒ kf n (x) − f m (x)kV
≤
kf n (x) − f n+1 (x)kV + . . . + kf m−1 (x) − f m (x)kV
≤ (k n + k n−1 + . . . + k m−1 )kx − f (x)kV
kn
kx − f (x)kV → 0 as n → ∞
≤
1−k
so (f n (x)) is Cauchy so converges (to some z ∈ V and since Z is closed, z ∈ Z
by Proposition 41).
V
, ∃N s.t. ∀n ≥ N kf n (x)−zkV < .
If kf (z)−zkv > 0 then for = kf (z)−zk
3
N
N +1
This kf (x) − zkV < and kf
(x) − f (z)kV ≤ k < . So kz − f (z)kV ≤
kf N +1 (x) − zkV + kf N +1 (x) − f (z)kV < + = 23 kf (z) − zkV which shows that
f (z) = z.
2
2 Lecture
15-11-07 starts here
33
Chapter 7
Pointwise Convergence and
its Disadvantages
Definition 23 Let A ⊂ R and ∀n ∈ N let fn : A → R be a function. We
say (fn ) converges pointwise to some f : A → R if ∀x ∈ A (fn (x)) converges
in R to f (x). More specifically, ∀x ∈ A, ∀ > 0 ∃N = Nx () s.t n ≥ Nx () ⇒
|fn (x) − f (x)| < .
Example 1 fn : [−1, 1] → R, fn (x) = x2n−1
O



f1 


f99





_ /
_



−1
+1











f : [−1, 1] → R


if x = 1
1
f (x) = 0
if − 1 < x < 1


−1 if x = −1
34
(fn ) converges pointwise to f . Nx () is very large when x is near ±1. fn is
(differentiable and so) continuous. But f is not continuous at ±1
Example 2 A = [0, 1]

1
2

if 0 ≤ x ≤ 2n
2n x
1
2
f (x) = 2n − 2n x if 2n ≤ x ≤ n1


0
if n1 ≤ x ≤ 1
O
,,
,,,
,,
,, fn
,,
,,
,,
_
_, _ _ _ _ _ _ _ _
1
1
1
2n
/
n
fn converges pointwise to f . f (x) := 0 for 0 ≤ x ≤ 1. Since fn (0) = 0∀n and
for 0 < x ≤ 1 fn (x) = 0 for n > x1 .
Z
1
fn =
0
Z
6=
1
1
→ as n → ∞
2
2
1
Z
f=
0
1
0=0
0
Disadvantages:
1 The pointwise limit of a sequence of continuous functions need not be
continuous.
2 The integral of the pointwise limit of a sequence of functions need not be
the limit of their integrals.
35
Chapter 8
Uniform Convergence: its
Advantages for Integrals
and Continuity
Definition 24 Let A ∈ R. We say that a sequence (fn ) of functions fn : A →
R converges uniformly to the function f : A → R if ∀ > 0∃N = N () s.t
(n ≥ N, x ∈ A) ⇒ |fn (x) − f (x)| < . Equivalently kfn − f k∞ → 0 as n → ∞
where kfn − f k∞ := supx∈A |fn (x) − f (x)|.
In pointwise convergence N can depend on x as well as . In uniform convergence N depends on but works for all x.
1
Uniform convergence implies pointwise convergence. (Just use N () for
each Nx ().)
Examples 1 and 2 show pointwise convergence does not imply uniform convergence.
We say (fn ) is uniformly Cauchy if ∀ > 0 ∃M = M () s.t.
(m, n ≥ M ()) ⇒ |fm (x) − fn (x)| < (8.1)
Or ∀ > 0 m, n ≥ M ⇒ kfm − fn k∞ ≤ . In this case ∀x ∈ A (fn (x)) is a
Cauchy sequence in R so has a limit in R, call that f (x) which defines f : A → R
Let n → ∞ in equation 8.1 to get ∀m ≥ M () |fm (x)−f (x)| ≤ so fn converges
uniformly to f : A → R.
Theorem 45 Suppose that fn : [a, b] → R is a sequence of regulated functions
Rb
Rb
and that (fn ) → f uniformly as n → ∞. Then f is regulated and ( a fn ) → a f
as n → ∞ in R. Hence (R[a, b], k·k∞ ) is a Banach Space.
Slogan
1 Lecture
Uniform convergence makes integral converge.
20-11-07 starts here
36
Find fn within
Idea
2
of f and a step function within
2
of fn .
Given > 0 choose N = N ( 2 ) s.t.
Proof
(n ≥ N, x ∈ [a, b]) ⇒ |fn (x) − f (x)| <
2
(8.2)
fN is regulated so choose ϕ ∈ S[a, b] s.t. kϕ − fN k∞ < 2 . Then kf − ϕk∞ ≤
kf − fN k∞ + kfN − ϕk∞ < , so f is regulated.
By equation 8.2 and Propositions 22 and 23:
Z
n≥N ⇒|
b
Z
fn −
a
Hence (
Rb
a
fn ) →
Rb
a
f.
a
b
Z b
f | = | (fn − f )| < (b − a)
2
a
Theorem 46 Let A ⊂ R and let (fn ) be a sequence of continuous functions
fn : A → R that converges uniformly to f : A → R. Then f is continuous. Hence
(C 0 [a, b], k·k∞ ) is a Banach space.
Slogan
tinuous.
Uniform convergence for continuous functions make the limit con-
Idea To prove f is continuous at c we estimate |f (x) − f (c)| approximate
f (x) and f (c) to within 3 by fN (x) and fN (c) for fixed N . Then control
|fN (x) − fN (c)| by continuity of fN at c.
Proof Fix any c ∈ A. Fix any > 0 and use uniform convergence to
choose N = N ( 3 ) s.t. (n ≥ N, t ∈ A) ⇒ |fn (t) − f (t)| < 3 . Use continuity of
fN to give δ > 0 s.t. (|c − x| < δ and x ∈ A) implies
|f (x) − f (c)| ≤ |f (x) − fN (x)| + |fN (x) − fN (c)| + |fN (c) − f (c)| <
37
+ + =
3 3 3
Chapter 9
Uniform Convergence: its
Advantages for
Differentiability
Example 3
1
f, fn : R → R. fn (x) :=
1
n
sin(nx). f (x) := 0
O
f1
/
f
f2
Then kfn −f k∞ = n1 → 0 as n → ∞, fn → f uniformly. Each fn is differentiable
and fn0 (x) = cos(nx). (fn0 (0)) = 1 → 1 6= f 0 (0). (fn0 (π)) = (−1)n and so does
not converge. Although f is differentiable, we do not have that (fn0 (x)) → f 0 (x),
even at x = 0 or x = π.
1 Lecture
21-11-07 starts here
38
Example 4 f, fn : R → R, fn (x) =
q
x2 + n1 , f (x) = |x|.
f
??

??


??

??

??

??


??

??

??


??

??

?? 
?
∀x ∈ R
with
|fn (x)−f (x)| ≤ |fn (0)−f (0)| =
fn0 (x) = q
→ 0 as n → ∞, fn is differentiable


−1 if x < 0
→ 0
if x = 0


1
if x > 0
x
x2
+
√1
n
1
n
so fn → f uniformly but f is not differentiable.
Theorem 47 Let fn : [a, b] → R be a sequence of C 1 functions that converges
pointwise to a function f : [a, b] → R. Suppose (fn0 ) converges uniformly to
g : [a, b] → R (which is continuous by Theorem 46). Then g is C 1 and f 0 = g.
Idea
Proof
says:
The antiderivative of lim fn0 is lim fn + constant.
Fix x ∈ [a, b]. Since (fn0 ) converges uniformly to g, Theorem 45
Z
a
x
Z
fn0 →
x
g as n → ∞
a
Rx
Now a fn0 = fn (x) − fn (a) by FTC2 (Theorem 27) and fn (x) − fn (a) → f (x) −
Rx
f (a) by the pointwise convergence. Thus ∀x ∈ [a, b] f (x) − f (a) = a g and
FTC1 (Corollary 26, swap f and g) says that f is differentiable with derivative
g. Since g is continuous, f is C 1 .
In Example 4 above, Theorem 47 says (fn0 ) cannot converge uniformly. Check
this directly.
39
Chapter 10
A Space Filling Curve
A path in R2 is a continuous function β : [a, b] → R2 . Using β(R.|·|) →
(R2 , k·k∞ ) it is easy to show β is continuous iff β1 and β2 are continuous (where
β(t) = ((β1 (t), β2 (t)) ∈ R2 . If ∀n ∈ Nαn : [0, 1] → R2 is continuous and (αn )
is uniformly Cauchy then one can prove as in Theorem 46 that β[0, 1] → R2 is
continuous where ∀t ∈ [0, 1], β(t) = limn→∞ αn (t).
Theorem 48 Let ∆ denote the region enclosed by this equilateral triangle in
R2 .
O
1
2
√
3_
2
222
22
22
22
22
22
_
_
1
1
0
/
2
Then there is a path β : [0, 1] → R2 with β([0, 1]) = ∆.
Proof
1 Lecture
1
Let α0 (t) :=
(
(t, √t3 )
√ )
(t, 1−t
3
0≤t≤
1
2
1
2
≤t≤1
22-11-07 starts here
40
∆ is the union of four
1
2
size triangles ∆0 , ∆1 , ∆2 , ∆3 as shown:
∆
rrL3LLL& L
r
8
r
LL
rLrLLr
Lf LL
L∆1
∆
L
O
rL0L
∆2 LLL& LL
8 rr LL& LL
r
r
L
L
rr
and α1 maps [0, 14 ] into ∆0 , and [ 14 , 12 ] into ∆1 , and [ 12 , 34 ] into ∆2 , and [ 34 , 1]
into ∆3 . With
1
1
α1 (t) := α0 (4t)
0≤t≤
2
4
1
1 1√
1
3
α1 (t) := α0 (4t − 2) + ( ,
3)
≤t≤
2
4 4
2
4
1
1
and for ∆1 , ∆3 , α1 is 2 α0 (4t − 1) and 2 α0 (4t − 3) reflected and translated to
the new starting point.
For n ≥ 2, αn maps
[0, 41 ] into ∆0 as 12 αn−1 (4t)
[ 14 , 12 ] into ∆1 as 21 αn−1 (4t − 1) reflected and translated
1
[ 21 , 34 ] into ∆2 as 21 αn−1 (4t − 2) + ( 41 , 4√
)
3
[ 34 , 1] into ∆3 as 12 αn−1 (4t − 3) reflected and translated
α2
_
0
_
1
rLL
LrLrrL8f L& L
L LLL& O LLL&
r
8
r
L
rMrMMf qx qLMMMf
Mq
M ) L q
rL
qMM O LL
LrLrrL8f L& LqMqMqM8f M& MLLLf Lr&x Lr
L M Lrr
L
O LLL& O LLL& r8 rLLL&
L
r
L
L
Lrr
L
rrr8 & L
Thus αn maps each of the 4n sub intervals of [0, 1] of length 4−n into one of the
4n equilateral triangles with side 2−n . αn+1 , αn+2 , . . . each maps each of the 4n
intervals of length 4−n into the same one of the 4n triangles of side 2−n as αn
does. Hence ∀ > 0 m ≥ n ⇒ kαm − αn k∞ := sup0≤t≤1 kαm (t) − αn (t)k∞ ≤
2−n < , provided n ≥ − log2 . Hence the sequence (αn )∞
n=0 is uniformly
Cauchy and as in Theorem 46 its limit β : [0, 1] → R2 is continuous.
Now for any z ∈ ∆ let ∆1 > ∆2 > ∆3 > . . . denote the triangles of sides
−1 −2 −3
2 , 2 , 2 , . . . to which z belongs. (Make a choice if z lies on a boundary.)
∀n ∈ N let [an , an + 4−n ] denote the interval with αn ([an , an + 4−n ]) ⊂ ∆n .
41
For m ≥ n we have αm ([an , an + 4−n ]) ⊂ ∆n . ∆n is closed in (R2 , k·k∞ ) so
∀t ∈ [an , an + 4−n ] β(t) := limm→∞ αm (t) ∈ ∆n by Proposition 41.
Now a := supm∈N am has
{a} =
∞
\
[am , am + 4−m ]
m=0
and β(a) ∈
∞
\
∆n = {z}
n=1
Hence β([0, 1]) = ∆.
Note
Area
S∞
n=0
αn ([0, 1]) = 0 so
S∞
42
n=0
αn ([0, 1]) 6= ∆.
Chapter 11
Series of Functions
1
Many functions are not elementary. (i.e. not given in terms of polynomials
exp, log, sin, cos). For example:
1 Elliptic functions (related to arc length on an ellipse)
P∞
2 Riemann zeta functions ζ : (1, ∞) → R ζ(x) := n=1 n1x
R∞
Rb
3 Gamma function Γ : (1, ∞) → R Γ(x) = 0 tx−1 e−t dt (= limb→∞ 0 )
R∞
x−2 −t
Integrating by parts gives Γ(x) = [−tx−1 e−t ]∞
e dt =
0 + 0 (x − 1)t
(x − 1)Γ(x − 1)
R∞
Γ(1) = 0 e−t dt = [−e−t ]∞
0 = 1 so Γ(n + 1) = n! so we might define
x! = Γ(x + 1)
Rx
4 Dilogarithm function. Li2 (x) := − 0 t−1 log(t−1)dt
Li2 : (−1, 1) → R
For these we shall have to use approximations by elementary functions
e.g.
P∞power series is approximated by polynomials , a trigonometric series
r=1 ar sin(rx) is a limit of differentiable functions (see MA250 PDEs).
Taylor series If f : R →
is ∀n ∈ N n-times differentiable at 0 define the
PR
r
∞
1
Taylor series T (f, x)(0) = r=0 f (r) (0) xr! = f (0) + xf 0 (0) + x2 f 00 2!
+ ...
Q1 For which x does this series converge?
Q2 If the series converges, is the sum = f (x)?
1 Lecture
23-11-07 starts here
43
1
f (x) = e− x2
Example 1
limx→0
f
(r−1)
(x)−f
x
(r−1)
(0)
(x 6= 0), f (0) = 0 Check that ∀r ∈ N f (r) (0) :=
=0
O
1_
/
T (f, x) =
P
r
r
0 xr! which converges to 0 not to f (x).
Example 2 f (x) := log(1 + x), f (r) (0) = (−1)r (r − 1)!, r > 0
T (f, x)
=
∞
X
(−1)r−1
r=1
= x−
xr
r
x3
x4
x2
+
−
+ ···
2
3
4
which converges if −1 < x ≤ 1. IfPit is permissible to differentiate
this
P∞
∞
term by term we get the derivative is r=1 (−1)(r−1) xr−1 = q=0 (−1)q xq =
Rx 1
P∞
P∞ r
r
(1 + x)−1 and then r=1 (−1)r−1 xr = 0 1+t
dt. Let g(x) = r=1 xr2 which
converges for |x| ≤ 1. Can we differentiate term by term to get g 0 (x) =
P∞ xr−1
= − x1 log(1 − x) if above was true. If so then Theorem 27 gives
r=1 Rr
x
g(x) = 0 −t−1 log(1 − t)dt =: Li2 (x).
2
If ∀r ∈ P
N fr : A → R is a function, we form
P the partial sums sn : A →
n
R sn (x) :=
f
(x).
We
say
that
series
r=1 r
r fr converges uniformly (or
pointwise to a function f : A → R if (sn )∞
converges
uniformly (or pointwise)
n=1
to f . We get immediately:
Theorem 45’ If A = [a, b] ∀r ∈ N fr : [a, b] → R is regulated and (sn ) → f
Rb
Pn R b
uniformly then f is regulated and ( r=1 a fr ) → a γ.
Proof
Apply Theorem 45 to the (sn ) sequence.
Theorem 46’ If ∀r ∈ N fr : A → R is continuous and (sn ) → f uniformly
then f : A → R is continuous.
2 Lecture
28-11-07 starts here
44
Theorem 47’ If A = [a, b] ∀r ∈ N fr : [a, b] → R is C 1 (sn ) → f pointwise
and s0n converges uniformly to some g : [a, b] → R then f is C 1 and f 0 = g.
Theorem 49 (The Weierstrass M-test for uniform convergence of a series) Let
fr : A → P
R be functions. Suppose that there
Pn are Mr ∈ R ∀x ∈ A |fr (x)| ≤ Mr
and that Mr converges. Then (sn ) = ( r=1 fr ) converges uniformly (to some
function f A → R).
Pn
Proof fn :=
r=1 Mr converges so (fn ) is a Cauchy sequence in R so
∀ > 0 ∃M () s.t. m ≥ n ≥ M () ⇒ |fm − fn | < . We shall show that (sn ) is
uniformly Cauchy. For all x ∈ A:
|sm (x) − sn (x)| = |
m
X
fr (x)| ≤
r=n+1
m
X
|fr (x)| ≤
r=n+1
m
X
Mr < r=n+1
if m ≥ n ≥ M (). P
n
P∞Thus we have r=1 fr (x) converges uniformly to f where ∀x ∈ A f (x) :=
r=1 fr (x)
Corollary 50 If the series
P
P
|ar | and |br | converge then the Fourier series:
∞
a0 X
+
(ar cos(rx) + br sin(rx))
2
r=1
converges uniformly. (and so the limit function R → R is continuous by Theorem
46’).
Proof ∀x ∈ R |ar cos(rx) + br sin(rx)| ≤ |ar | + |br | =: Mr . Now the
convergence is uniform by Theorem 49.
P∞
Proposition 51 (The Riemann Zeta Function)PThe series r=1 r1x converges
∞
pointwise for x > 1 and ζ : (1, ∞) → R, ζ(x) := r=1 r1x is continuous.
Proof
Fix a > 1. ∀r ∈ N
n
X
1
a
r
r=2
Z
n
1
dt
a
t
1
1−a n
t
=
1−a 1
≤
=
≤
1
n1−a
−
a−1 a−1
1
a−1
45
O
+
+
+
1
2
n−1
+
n
/t
P
If Mr = r1a then
Mr converges
∀x ∈ [a, ∞) | r1x | ≤ Mr so by the WeierP∞ and
1
strass M-test (Theorem 49), r=1 rx converges uniformly on [a, ∞). By Theorem 46’, ζ [a,∞) is continuous.
3
In particular ζ is continuous at each point of (a, ∞). Since this holds
∀a > 1 we have ζ is continuous on (1, ∞)
P∞
Note On (1, ∞) it is not true that r=1 r1x converges uniformly because,
P2n
P2n
as x → 1 |s2n (x) − sn (x)| = r=n+1 r1x → r=n+1 1r > 12 so it is not true that
∀n > some N ∀x ∈ (1, ∞) |s2n (x) − sn (x)| < 41
Theorem 52 (A nowhere differentiable function) There exists a continuous
function f : R → R s.t. ∀x ∈ R f is not differentiable at x.
Proof
Define ϕ : R → R by ϕ(x) = |x| for |x| ≤ 1 and ∀x ∈ R
ϕ(x) = ϕ(x + 2)
(11.1)
|ϕ(s) − ϕ(t)| ≤ |s − t|
(11.2)
Note that ∀s, t ∈ R
and ϕ is uniformly continuous (put δ = ).
O
//
/ /
/
//
//
///
//
//
//
//
//
//
ϕ
//
//
//
// // // _/ _
_/
_
_/
/
−2
−1
0
1
2
n
r
P
P∞
Define f : R → R by f (x) := n=0 43 ϕ(4n x). Since Mr = 43 has
Mr
convergent and ∀t|ϕ(t)| ≤ 1 the series converges uniformly to f by Theorem 49.
By Theorem 46’ f is continuous.
3 Lecture
29-11-07 starts here
46
Fix x ∈ R and m ∈ N. Put δm = ± 12 4−m , where the sign is chosen so that
there is no integer in the interval of length 12 from 4m x to 4m (x + δm ).
n
n
Consider γn : = ϕ(4 (x+δδmm))−ϕ(4 x) , which contributes to the slope of the
chord from (x, f (x)) to (x + δm , f (x + δm )). In n > m then 4n δm is an even
integer and γn = 0 by equation 11.1.
For 0 ≤ n < m, equation 11.2 gives |γm | ≤ 4n . Since |γm | = 4m we have:
∞ n
X
f (x + δm ) − f (x) 3
= γn δm
4
n=0 m
n
X
3
γn = 4
n=0
n m−1
m X 3
γn = 3 +
4
n=0
≥ 3m −
m−1
X
3n
n=0
=
As m → ∞ δm → 0 and we can see that
R
.
47
1 m
(3 + 1)
2
f (x+δm )−f (x)
δm
cannot tend to a limit in
Chapter 12
Power Series
Now apply the preceding theory to power series of the form
P∞
r=0
ar xr (ar ∈ R).
P∞
r
Theorem 53 1 Suppose that the power
r=0 ar x converges for x =
P∞ series
r
x0 6= 0 and that 0 < bP
< |x0 |. Then r=0 ar x converges uniformly on [−b, b]
∞
and the derived series r=0 rar xr−1 converges uniformly on [−b, b].
Idea
Use
b
|x0 |
< 1 and M-test.
P∞
r
r
Proof
r=0 ar x0 converges so ar x0 → 0 as r → ∞ and the terms are
bounded. Choose k ∈ R s.t. rP≥ 0 ⇒ |ar xr0 | ≤ k. If |x| ≤ b then
∞
k
|axr | ≤ |ar br | ≤ k( |xb0 | )r := Mr .
0 Mr converges with sum 1− b , so the
|x0 |
P∞
r
M-test
(Theorem
49)
says
that
a
x
converges
uniformly
on
[−b,
r
r=0
P∞ b]. Also
P
kr( |xb0 | )r−1 converges (by the ratio test) so the M-test says that r=0 rar xr−1
converges uniformly on [−b, b].
Remark The same proof works with ar , x, x0 ∈ C and 0 < b < |x0 |. Both
series converge uniformly
on the disk |x| ≤ b.
P
R := sup{|x0 | :
ar xr converges} ∈ [0, ∞] defines
radius of convergence
P the
of the series. R satisfies z ∈ C and|z| < R ⇒
ar z r converges and z ∈ C
and|z| > R ⇒ {|ar z r | : r ≥ 0} is unbounded.
P∞
Theorem 45 Let R > 0 and let r=0 ar xr be a real power series that converges pointwise on (−R, R) ⊂ R to f : (−R,
→ R. Then f is continuous and
PR)
∞
differentiable with ∀x ∈ (−R, R) f 0 (x) = r=1 rar xr−1 . If −R < c < d < R
Rd
P∞ ar r+1
(d
− cr+1 ).
then f |[c,d] is continuous and c f = r=0 r+1
1 Lecture
30-11-07 starts here
48
P
ar xr0 converges so, by Theorem 53,
P∞Proofr Take 0 < b < |x0 | < R. Then
r=0 ar x converges uniformly on [−b, b]. By Theorem 46’ f |[−b,b]
S is continuous;
in particular f is continuous at each x ∈P(−b, b). (−R, R) = 0<b<R (−b, b) so
∞
f : (−R, R) → R is continuous . Also r=1 rar xr−1 converges uniformly on
[−b, b] by Theorem 53 and, by Theorem 47’, f is C 1 on (−b, b) with f 0 (x) =
P
∞
r−1
so also on (−R, R). The integral result comes from Theorem 45’
r=1 rar x
and
!
Z d
Z d X
∞
∞
X
ar
dr+1 − cr+1
f=
ar xr dx =
r
+
1
c
c
r=0
r=0
Remark By applying Theorem 54 repeatedly we find that f : (−R, R) → R
is infinitely differentiable,
with ∀r ∈ N f (r) (0) = r! · ar . Hence the Taylor series
P∞
r
r
of f is T (f, x) = r=0 r!·a
r! x which is f . (If f is given by a power series then
1
its Taylor series is that power series which does converge to f. Hence e− x2 is
not given by any power series.)
Definition 25
2
Define the following functions by power series:
exp : C → C
cosh : C → C
sinh : C → C
cos : C → C
exp(x) :=
cosh(x) :=
sinh(x) :=
cos(x) :=
∞
X
xn
n!
n=0
∞
X
x2n
(2n)!
n=0
∞
X
x2n+1
(2n + 1)!
n=0
∞
X
(−1)n
n=0
sin : C → C
sin(x) :=
but log : (0, ∞) → R
∞
X
x2n
(2n)!
x2n+1
(2n + 1)!
n=0
Z x
1
dt
log(x) :=
1 t
(−1)n
Proposition 55 The radius of convergence for each of the series for exp, cosh, sinh, cos, sin
is ∞.
2 Lecture
05-12-07 starts here
49
Proof
For cos ∀x ∈ C:
2r+2
|
x
(−1)r+1 (2r+2)!
|x|2
→ 0 as r → ∞
(2r + 1)(2r + 2)
|=
x2r
(−1)r (2r)!
so ∀x this converges absolutely by the ratio test,
Corollary 56 On R, exp0 = exp, cosh0 = sinh, sinh0 = cosh, cos0 =
− sin, sin0 = cos. Note: true also on C
Proof
By Theorem 45 within (−R, R) = R
exp0 (x) =
∞
X
rxr−1
r!
r=1
sin0 (x) =
∞
X
∞
X
xq
q=0
q!
= exp(x) (putting q = r − 1)
∞
(2r + 1)
r=0
Others similarly.
=
X
x2r
x2r
(−1)r =
(−1)r
= cos(x)
(2r + 1)!
(2r)!
r=0
P
P
Recall that a series
aP
|aj | < ∞, in which
j is absolutely convergent if
case any rearrangement of
aj converges with the same sum and a product:
X X X
aj bk
aj
bk =
j,k
Proposition 57
∀x ∈ C
cosh(x) =
1
(exp(x) + exp(−x))
2
1
(exp(x) − exp(x))
2
cosh(ix) = cos(x)
sinh(x) =
sinh(ix) = i sin(x)
exp(ix) = cos(x) + i sin(x)
∀x, y ∈ C
exp(x + y) = exp(x). exp(y)
cos(x + y) = cos(x) cos(y) − sin(x) sin(y)
sin(x + y) = sin(x) cos(y) + cos(x) sin(y)
(cos(x))2 + (sin(x))2 = 1
50
Proof
Rearranging 12 (exp(x) − exp(−x))
∞
∞
1 X xr
xr
1X
=
−
(−1)r
2 r=0 r!
2 r=0
r!
gives
X xr
= sinh(x)
r!
r odd
exp(x + y)
=
=
=
∞
X
(x + y)r
r=0
∞ X
r
X
r!
xr−q y q
(r − q)! q!
r=0 q=0
! ∞
!
∞
X yq
X
xp
p!
q!
q=0
p=0
putting p = r − q gives
=
exp(x). exp(y)
1 = cos(0) = cos(x − x) = cos(x) cos(−x) − sin(x) sin(−x)
= cos(x) cos(x) + sin(x) sin(x)
Proposition 58
There is π > 0 in R with x ∈ C
sin(x + π) = − sin(x)
cos(x + π) = − cos(x)
and so
sin(x + 2π) = sin(x)
cos(x + 2π) = cos(x)
Proof 3 sin0 (0) = cos(0) = 1 > 0, and sin(0) = 0 so sin(x) > 0 for all
sufficiently small x > 0. For 0 < x < 12:
sin x < x −
since
−
x11
x13
+
<0
11!
13!
and
sin 4 < 4 −
3 Lecture
x3
x5
x7
x9
+
−
+
3!
5!
7!
9!
etc.
43
45
47
49
+
−
+
3!
5!
7!
9!
06-12-07 starts here
51
sin 4 < −0.6617 < 0
By IVT (Analysis II) sin takes value 0 in (0, 4). Define π as the least positive x
with sin x = 0.
Then cos0 = − sin < 0 on (0, π) so MVT says cos |[0,π] is strictly decreasing.
By Proposition 57 cos2 (π) + sin2 (π) = 1 so cos(π) = ±1 and now we see that
cos(π) = −1.
Hence
and
sin(x + π) = sin(x) cos(π) + sin(π) cos(x) = − sin(x)
cos(x + π) = cos(x) cos(π) − sin(x) sin(π) = − cos(x) Notes ∀r ∈ N ∀x ∈ C, sin(rx) and cos(rx) are (2π)-periodic functions
as are any finite P
linear combination of these and with x ∈ R, any pointwise
convergent series (ar cos(rx) + br sin(rx))
sin00 = − sin < 0 on (0, π).
52
Chapter 13
A Solution for an ODE
Theorem 59 If for M ∈ R F : P = [x0 − a0 , x0 + a0 ] × [y0 − b0 , y0 + b0 ] → R
satisfies ∀(x, y), (x0 , y 0 ) ∈ P , |F (x, y) − F (x0 , y 0 )| ≤ M (|x − x0 | + |y − y 0 |) then
there is a ∈ (0, a0 ) and a unique C 1 function g : [x0 − a, a0 + a] → R with
g(x0 ) = y0 and ∀x ∈ [x0 − a, x0 + a], g 0 (x) = F (x, g(x)).
2
Y ^ m ~ t
O Z k z tv _
y0 _ _ l ho r _  { Y_
iK
| | z
U a]lm v  v _
O
k yv Y e] p _
_
_
x0 −a x
x0 +a
0
Proof ∀(x, y) ∈ P |F (x, y)| ≤ |F (x0 , y0 )| + M (a0 + b0 ) =: K. Take a ∈
(0, a0 ) s.t. aM < 1 and aK < b0 (so as not to go out of the top of P ) and
put Z = {h ∈ C 0 [x0 − a, x0 + a] : h([x0 − a, x − 0 + a]) ⊂ [y0 − aK, y0 + aK]}.
(C 0 [x0 − a, x0 + a], k·k∞ ) is a Banach space by Theorem 46, and Z is a closed
subset (because h ∈ C 0 [x0 − a, x0 + a] and (t, h(t)) ∈ [x0 − a, x0 + a] × (y0 +
aK, ∞) ⇒ B(h, h(t) − (y0 + aK)) ∩ Z = ∅).
Rx
1
Define G : Z → Z so that (G(h))(x) := y0 + x0 F (t, h(t))dt for x ∈ [x0 −
a, x0 + a]. Note that ∀t ∈ [x0 − a, x0 + a] (t, h(t)) ∈ P so |F (t, h(t))| ≤ K and
by Proposition 23 (bounds) and Theorem 24 (indefinite integral is uniformly
continuous), G(h) ∈ Z. If h1 , h2 ∈ Z then t ∈ [x0 − a, x0 + a] ⇒ |h1 (t) − h2 (t)| ≤
kh1 − h2 k∞ ⇒ |F (t, h1 (t)) − F (t, h2 (t))| ≤ M kh1 − hR2 k∞ .
x
Thus ∀x ∈ [x0 −a, x0 +a] |G(h1 )(x)−G(h2 )(x)| = | x0 F (t, h1 (t))−F (t, h2 (t))dt| ≤
aM kh1 −h2 k∞ so kG(h1 )−G(h2 )k∞ ≤ aM kh1 −h2 k∞ . Since aM < 1 G : Z → Z
is a contraction and by CMT (Theorem 44) has
R x a unique fixed point g ∈ Z.
∀x ∈ [x0 − a, x0 + a] g(x) = (G(g))(x) = y0 + x0 F (t, g(t))dt, t 7→ (t, g(t)) 7→
1 Lecture
07-12-07 starts here
53
F (t, g(t)) is a composition of continuous functions so continuous, so by Corollary
26 (FTC1) ∀x ∈ [x0 − a, x0 + a] g 0 (x) = F (x, g(x)) as required.
If h : [x0 − a, x0 + a] → R is C 1 with
R xh(x0 ) = y0 and ∀x ∈ [x0 − a, x0 + a]
0
h (x) = F (x, h(x)) then |h(x) − y0 | = | x0 F (t, h(t))dt| ≤ aK (and h ∈ C 0 ) so
Rx
Rx
h ∈ Z and (G(h))(x) = y0 + x0 F (t, h(t))dt = y0 + x0 h0 (t)dt = y0 + h(x) −
h(x0 ) = h(x), so G(h) = h and by CMT (Theorem 44) h = g.
Example
F (x, y) = 2xy, (x0 , y0 ) = (0, 1) P = [−1, 1] × [0, 2]
FDQ
F LL R
F R
F LL R
F R
4 LR
L_

_ nm {{
_
_ m {
_ m {
_ m {
_ mW l
|F (x, y)−F (x0 , y 0 )| = |2xy −2x0 y 0 | = 2|xy −xy 0 +xy 0 −x0 y 0 | ≤ 2|x||y −y 0 |+2|x−
x0 ||y 0 | ≤ 2|y − y 0 | + 4|x − x0 | so take M = 4. Also sup|F (x, y)| = 4 =: K. Any
a ∈ (0, 14 ) has aM < 1 and aK < b0 = 1. Try ∀x h0 (x) = 1, and hn = Gn (h0 ).
Z x
Z x
h1 (x) = G(h0 )(x) = 1 +
F (t, h0 (t))dt = 1 +
2tdt = 1 + x2
0
Z
h2 (x) = G(h1 )(x) = 1 +
0
x
Z
F (t, h1 (t))dt = 1 +
0
0
x
2t(1 + t2 )dt = 1 + x2 +
x4
2
By induction
x4
x6
x2n
+
+ ... +
2!
3!
n!
Then, by CMT (Theorem 44) hn → g uniformly on [−a, a]. But hn → (x →
7
exp(x2 )). Note that g(x) = exp(x2 ) gives g 0 (x) = 2x exp(x2 ) = F (x, g(x)).
hn (x) = 1 + x2 +
54