CHAPTER 9
Sequences of Functions
1. Pointwise Convergence
We have accumulated much experience working with sequences of numbers.
The next level of complexity is sequences of functions. This chapter explores
several ways that sequences of functions can converge to another function. The
basic starting point is contained in the following definitions.
D EFINITION 9.1. Suppose S Ω R and for each n 2 N there is a function f n :
S ! R. The collection { f n : n 2 N} is a sequence of functions defined on S.
For each fixed x 2 S, f n (x) is a sequence of numbers, and it makes sense to
ask whether this sequence converges. If f n (x) converges for each x 2 S, a new
function f : S ! R is defined by
f (x) = lim f n (x).
n!1
The function f is called the pointwise limit of the sequence f n , or, equivalently, it
S
is said f n converges pointwise to f . This is abbreviated f n °! f , or simply f n ! f ,
if the domain is clear from the context.
E XAMPLE 9.1. Let
Then f n ! f where
8
x <0
>
<0,
f n (x) = x n , 0 ∑ x < 1 .
>
:
1,
x ∏1
f (x) =
(
0, x < 1
1, x ∏ 1
.
(See Figure 1.) This example shows that a pointwise limit of continuous functions
need not be continuous.
E XAMPLE 9.2. For each n 2 N, define f n : R ! R by
nx
f n (x) =
.
1 + n2 x2
(See Figure 2.) Clearly, each f n is an odd function and lim|x|!1 f n (x) = 0. A bit of
calculus shows that f n (1/n) = 1/2 and f n (°1/n) = °1/2 are the extreme values of
f n . Finally, if x 6= 0,
Ø nx Ø Ø nx Ø ØØ 1 ØØ
Ø
Ø Ø
Ø
Ø
| f n (x)| = Ø
Ø < Ø 2 2 Ø = ØØ
2
2
1+n x
n x
nx Ø
9-1
9-2
CHAPTER 9. SEQUENCES OF FUNCTIONS
1.0
0.8
0.6
0.4
0.2
0.2
0.4
0.6
0.8
1.0
F IGURE 1. The first ten functions from the sequence of Example 9.1.
0.4
0.2
!3
!2
1
!1
2
3
!0.2
!0.4
F IGURE 2. The first four functions from the sequence of Example 9.2.
implies f n ! 0. This example shows that functions can remain bounded away
from 0 and still converge pointwise to 0.
E XAMPLE 9.3. Define f n : R ! R by
8 2n+4
1
3
x ° 2n+3 ,
< x < 2n+2
>
<2
2n+1
3
f n (x) = °22n+4 x + 2n+4 , 2n+2
∑ x < 21n
>
:
0,
otherwise
To figure out what this looks like, it might help to look at Figure 3.
The graph of f n is a piecewise linear function supported on [1/2n+1 , 1/2n ] and
the area under the isosceles triangle of the graph over this interval is 1. Therefore,
R1
0 f n = 1 for all n.
If x > 0, then whenever x > 1/2n , we have f n (x) = 0. From this it follows that
f n ! 0.
The lesson to be learned from this example is that it may not be true that
R1
R1
limn!1 0 f n = 0 limn!1 f n .
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Pointwise and Uniform Convergence
9-3
64
32
16
8
1 1
16 8
1
4
1
2
1
F IGURE 3. The first four functions f n ! 0 from the sequence of Exam-
ple 9.3.
1/2
1/4
1/8
1
1
F IGURE 4. The first ten functions of the sequence f n ! |x| from Ex-
ample 9.4.
E XAMPLE 9.4. Define f n : R ! R by
(
n 2
1
x + 2n
, |x| ∑
f n (x) = 2
|x|,
|x| >
1
n
1
n
.
(See Figure 4.) The parabolic section in the center was chosen so f n (±1/n) = 1/n
and f n0 (±1/n) = ±1. This splices the sections together at (±1/n, ±1/n) so f n is
differentiable everywhere. It’s clear f n ! |x|, which is not differentiable at 0.
This example shows that the limit of differentiable functions need not be
differentiable.
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9-4
CHAPTER 9. SEQUENCES OF FUNCTIONS
The examples given above show that continuity, integrability and differentiability are not preserved in the pointwise limit of a sequence of functions. To
have any hope of preserving these properties, a stronger form of convergence is
needed.
2. Uniform Convergence
D EFINITION 9.2. The sequence f n : S ! R converges uniformly to f : S ! R
on S, if for each " > 0 there is an N 2 N so that whenever n ∏ N and x 2 S, then
| f n (x) ° f (x)| < ".
S
In this case, we write f n ‚
f , or simply f n ‚ f , if the set S is clear from the
context.
f(x) + ε
f(x)
fn(x)
f(x
a
ε
b
F IGURE 5. | f n (x) ° f (x)| < " on [a, b], as in Definition 9.2.
The difference between pointwise and uniform convergence is that with
pointwise convergence, the convergence of f n to f can vary in speed at each
point of S. With uniform convergence, the speed of convergence is roughly the
same all across S. Uniform convergence is a stronger condition to place on the
sequence f n than pointwise convergence in the sense of the following theorem.
S
S
T HEOREM 9.3. If f n ‚
f , then f n °! f .
P ROOF. Let x 0 2 S and " > 0. There is an N 2 N such that when n ∏ N , then
| f (x) ° f n (x)| < " for all x 2 S. In particular, | f (x 0 ) ° f n (x 0 )| < " when n ∏ N . This
shows f n (x 0 ) ! f (x 0 ). Since x 0 2 S is arbitrary, it follows that f n ! f .
⇤
The first three examples given above show the converse to Theorem 9.3
is false. There is, however, one interesting and useful case in which a partial
converse is true.
S
D EFINITION 9.4. If f n °! f and f n (x) " f (x) for all x 2 S, then f n increases to f
S
on S. If f n °! f and f n (x) # f (x) for all x 2 S, then f n decreases to f on S. In either
case, f n is said to converge to f monotonically.
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2. UNIFORM CONVERGENCE
9-5
The functions of Example 9.4 decrease to |x|. Notice that in this case, the convergence is also happens to be uniform. The following theorem shows Example
9.4 to be an instance of a more general phenomenon.
T HEOREM 9.5 (Dini’s Theorem). If
(a)
(b)
(c)
(d)
S is compact,
S
f n °! f monotonically,
f n 2 C (S) for all n 2 N, and
f 2 C (S),
then f n ‚ f .
P ROOF. There is no loss of generality in assuming f n # f , for otherwise we
consider ° f n and ° f . With this assumption, if g n = f n ° f , then g n is a sequence
of continuous functions decreasing to 0. It suffices to show g n ‚ 0.
To do so, let " > 0. Using continuity and pointwise convergence, for each
x 2 S find an open set G x containing x and an N x 2 N such that g Nx (y) < " for all
y 2 G x . Notice that the monotonicity condition guarantees g n (y) < " for every
y 2 G x and n ∏ N x .
The collection {G x : x 2 S} is an open cover for S, so it must contain a finite
subcover {G xi : 1 ∑ i ∑ n}. Let N = max{N xi : 1 ∑ i ∑ n} and choose m ∏ N . If
x 2 S, then x 2 G xi for some i , and 0 ∑ g m (x) ∑ g N (x) ∑ g Ni (x) < ". It follows that
g n ‚ 0.
⇤
1
fn (x) = xn
1/2
2
1/n
1
F IGURE 6. This shows a typical function from the sequence of Example 9.5.
E XAMPLE 9.5. Let f n (x) = x n for n 2 N, then f n decreases to 0 on [0, 1). If 0 <
a < 1 Dini’s Theorem shows f n ‚ 0 on the compact interval [0, a]. On the whole
interval [0, 1), f n (x) > 1/2 when 2°1/n < x < 1, so f n is not uniformly convergent.
(Why doesn’t this violate Dini’s Theorem?)
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9-6
CHAPTER 9. SEQUENCES OF FUNCTIONS
3. Metric Properties of Uniform Convergence
If S Ω R, let B (S) = { f : S ! R : f is bounded}. For f 2 B (S), define k f kS =
lub {| f (x)| : x 2 S}. (It is abbreviated to k f k, if the domain S is clear from the
context.) Apparently, k f k ∏ 0, k f k = 0 () f ¥ 0 and, if g 2 B (S), then k f ° g k =
kg ° f k. Moreover, if h 2 B (S), then
k f ° g k = lub {| f (x) ° g (x)| : x 2 S}
∑ lub {| f (x) ° h(x)| + |h(x) ° g (x)| : x 2 S}
∑ lub {| f (x) ° h(x)| : x 2 S} + lub {|h(x) ° g (x)| : x 2 S}
= k f ° hk + kh ° g k
Combining all this, it follows that k f ° g k is a metric1 on B (S).
The definition of uniform convergence implies that for a sequence of bounded
functions f n : S ! R,
f n ‚ f () k f n ° f k ! 0.
Because of this, the metric k f ° g k is often called the uniform metric or the
sup-metric. Many ideas developed using the metric properties of R can be carried over into this setting. In particular, there is a Cauchy criterion for uniform
convergence.
D EFINITION 9.6. Let S Ω R. A sequence of functions f n : S ! R is a Cauchy
sequence under the uniform metric, if given " > 0, there is an N 2 N such that
when m, n ∏ N , then k f n ° f m k < ".
T HEOREM 9.7. Let f n 2 B (S). There is a function f 2 B (S) such that f n ‚ f iff
f n is a Cauchy sequence in B (S).
P ROOF. ()) Let f n ‚ f and " > 0. There is an N 2 N such that n ∏ N implies
k f n ° f k < "/2. If m ∏ N and n ∏ N , then
k fm ° fn k ∑ k fm ° f k + k f ° fn k <
" "
+ ="
2 2
shows f n is a Cauchy sequence.
(() Suppose f n is a Cauchy sequence in B (S) and " > 0. Choose N 2 N so
that when k f m ° f n k < " whenever m ∏ N and n ∏ N . In particular, for a fixed
x 0 2 S and m, n ∏ N , | f m (x 0 ) ° f n (x 0 )| ∑ k f m ° f n k < " shows the sequence f n (x 0 )
is a Cauchy sequence in R and therefore converges. Since x 0 is an arbitrary point
of S, this defines an f : S ! R such that f n ! f .
Finally, if m, n ∏ N and x 2 S the fact that | f n (x) ° f m (x)| < " gives
| f n (x) ° f (x)| = lim | f n (x) ° f m (x)| ∑ ".
m!1
This shows that when n ∏ N , then k f n ° f k ∑ ". We conclude that f 2 B (S) and
fn ‚ f .
⇤
1Definition 2.11
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4. SERIES OF FUNCTIONS
9-7
A collection of functions S is said to be complete under uniform convergence,
if every Cauchy sequence in S converges to a function in S . Theorem 9.7 shows
B (S) is complete under uniform convergence. We’ll see several other collections
of functions that are complete under uniform convergence.
E XAMPLE 9.6. For S Ω R let L(S) be all the functions f : S ! R such that
f (x) = mx + b for some constants m and b. In particular, let f n be a Cauchy
sequence in L([0, 1]). Theorem 9.7 shows there is an f : [0, 1] ! R such that
f n ‚ f . In order to show L([0, 1]) is complete, it suffices to show f 2 L([0, 1]).
To do this, let f n (x) = m n x + b n for each n. Then f n (0) = b n ! f (0) and
Given any x 2 [0, 1],
m n = f n (1) ° b n ! f (1) ° f (0).
f n (x) ° (( f (1) ° f (0))x + f (0)) = m n x + b n ° (( f (1) ° f (0))x + f (0))
= (m n ° ( f (1) ° f (0)))x + b n ° f (0) ! 0.
This shows f (x) = ( f (1) ° f (0))x + f (0) 2 L([0, 1]) and therefore L([0, 1]) is complete.
E XAMPLE 9.7. Let P = {p(x) : p is a polynomial}. The sequence of polynomiP
als p n (x) = nk=0 x k /k! increases to e x on [0, a] for any a > 0, so Dini’s Theorem
shows p n ‚ e x on [0, a]. But, e x › P , so P is not complete. (See Exercise 7.7.25.)
4. Series of Functions
The definitions of pointwise and uniform convergence are extended in the
P
natural way to series of functions. If 1
f is a series of functions defined on a
k=1 k
set S, then the series converges pointwise or uniformly, depending on whether
P
the sequence of partial sums, s n = nk=1 f k converges pointwise or uniformly,
respectively. It is absolutely convergent or absolutely uniformly convergent, if
P1
n=1 | f n | is convergent or uniformly convergent on S, respectively.
The following theorem is obvious and its proof is left to the reader.
P
P
T HEOREM 9.8. Let 1
defined on S. If 1
n=1 f n be a series of functions
n=1 f n
P1
is absolutely convergent, then it is convergent. If n=1 f n is absolutely uniformly
convergent, then it is uniformly convergent.
The following theorem is a restatement of Theorem 9.5 for series.
P
T HEOREM 9.9. If 1
conn=1 f n is a series of nonnegative continuous functions
P
verging pointwise to a continuous function on a compact set S, then 1
n=1 f n
converges uniformly on S.
A simple, but powerful technique for showing uniform convergence of series
is the following.
T HEOREM 9.10 (Weierstrass M-Test). If f n : S ! R is a sequence of functions
and M n is a sequence nonnegative numbers such that k f n kS ∑ M n for all n 2 N
P
P1
and 1
n=1 M n converges, then n=1 f n is absolutely uniformly convergent.
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9-8
CHAPTER 9. SEQUENCES OF FUNCTIONS
P
P ROOF. Let " > 0 and s n be the sequence of partial sums of 1
n=1 | f n |. Using
the Cauchy criterion for convergence of a series, choose an N 2 N such that when
P
n > m ∏ N , then nk=m M k < ". So,
ks n ° s m k = k
n
X
k=m+1
fk k ∑
n
X
k=m+1
k fk k ∑
n
X
k=m
M k < ".
This shows s n is a Cauchy sequence and must converge according to Theorem 9.7.
⇤
n
E XAMPLE 9.8. Let a > 0 and M n = a /n!. Since
M n+1
a
lim
= lim
= 0,
n!1 M n
n!1 n + 1
P1
the Ratio Test shows n=0 M n converges. When x 2 [°a, a],
Ø nØ
Ø x Ø an
Ø Ø∑
Ø n! Ø n! .
P
n
The Weierstrass M-Test now implies 1
n=0 x /n! converges absolutely uniformly
on [°a, a] for any a > 0. (See Exercise 9.4.)
5. Continuity and Uniform Convergence
T HEOREM 9.11. If f n : S ! R is such that each f n is continuous at x 0 and
S
fn ‚
f , then f is continuous at x 0 .
P ROOF. Let " > 0. Since f n ‚ f , there is an N 2 N such that whenever n ∏ N
and x 2 S, then | f n (x) ° f (x)| < "/3. Because f N is continuous at x 0 , there is a
± > 0 such that x 2 (x 0 ° ±, x 0 + ±) \ S implies | f N (x) ° f N (x 0 )| < "/3. Using these
two estimates, it follows that when x 2 (x 0 ° ±, x 0 + ±) \ S,
| f (x) ° f (x 0 )| = | f (x) ° f N (x) + f N (x) ° f N (x 0 ) + f N (x 0 ) ° f (x 0 )|
∑ | f (x) ° f N (x)| + | f N (x) ° f N (x 0 )| + | f N (x 0 ) ° f (x 0 )|
< "/3 + "/3 + "/3 = ".
⇤
Therefore, f is continuous at x 0 .
The following corollary is immediate from Theorem 9.11.
C OROLLARY 9.12. If f n is a sequence of continuous functions converging uniformly to f on S, then f is continuous.
Example 9.1 shows that continuity is not preserved under pointwise convergence. Corollary 9.12 establishes that if S is compact, then C (S) is complete under
the uniform metric.
The fact that C ([a, b]) is closed under uniform convergence is often useful
because, given a “bad” function f 2 C ([a, b]), it’s often possible to find a sequence
f n of “good” functions in C ([a, b]) converging uniformly to f . Following is the
most widely used theorem of this type.
T HEOREM 9.13 (Weierstrass Approximation Theorem). If f 2 C ([a, b]), then
there is a sequence of polynomials p n ‚ f .
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5. CONTINUITY AND UNIFORM CONVERGENCE
9-9
To prove this theorem, we first need a lemma.
≥R
¥°1
1
L EMMA 9.14. For n 2 N let c n = °1 (1 ° t 2 )n d t
and
k n (t ) =
(
c n (1 ° t 2 )n , |t | ∑ 1
0,
|t | > 1
.
(See Figure 7.) Then
(a) k n (t ) ∏ 0 for all t 2 R and n 2 N;
R1
(b) °1 k n = 1 for all n 2 N; and,
(c) if 0 < ± < 1, then k n ‚ 0 on (°1, °±] [ [±, 1).
1.2
1.0
0.8
0.6
0.4
0.2
!1.0
!0.5
0.5
1.0
F IGURE 7. Here are the graphs of k n (t ) for n = 1, 2, 3, 4, 5.
P ROOF. Parts (a) and (b) follow easily from the definition of k n .
To prove (c) first note that
µ
∂
Z1
Z1/pn
2
1 n
2 n
1=
kn ∏
c
(1
°
t
)
d
t
∏
c
1
°
.
p
n
n
p
n
n
°1
°1/ n
°
¢n
p
Since 1 ° n1 " 1e , it follows there is an Æ > 0 such that c n < Æ n.2 Letting ± 2 (0, 1)
and ± ∑ t ∑ 1,
p
k n (t ) ∑ k n (±) ∑ Æ n(1 ° ±2 )n ! 0
by L’Hospital’s Rule. Since k n is an even function, this establishes (c).
⇤
2Repeated application of integration by parts shows
n + 1/2 n ° 1/2 n ° 3/2
3/2
°(n + 3/2)
£
£
£···£
=p
.
n
n °1
n °2
1
º°(n + 1)
p
With the aid of Stirling’s formula, it can be shown c n º 0.565 n.
cn =
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9-10
CHAPTER 9. SEQUENCES OF FUNCTIONS
A sequence of functions satisfying conditions such as those in Lemma 9.14 is
called a convolution kernel or a Dirac sequence.3 Several such kernels play a key
role in the study of Fourier series, as we will see in Theorems 10.5 and 10.13. The
one defined above is called the Landau kernel.4
We now turn to the proof of the theorem.
P ROOF. There is no generality lost in assuming [a, b] = [0, 1], for otherwise we
consider the linear change of variables g (x) = f ((b ° a)x + a). Similarly, we can
assume f (0) = f (1) = 0, for otherwise we consider g (x) = f (x) ° (( f (1) ° f (0))x °
f (0), which is a polynomial added to f . We can further assume f (x) = 0 when
x › [0, 1].
Set
Z1
(82)
p n (x) =
f (x + t )k n (t ) d t .
°1
To see p n is a polynomial, change variables in the integral using u = x + t to arrive
at
Zx+1
Z1
p n (x) =
f (u)k n (u ° x) d u =
f (u)k n (x ° u) d u,
x°1
0
because f (x) = 0 when x › [0, 1]. Notice that k n (x ° u) is a polynomial in u
with coefficients being polynomials in x, so integrating f (u)k n (x ° u) yields a
polynomial in x. (Just try it for a small value of n and a simple function f !)
Use (82) and Lemma 9.14(b) to see for ± 2 (0, 1) that
Ø Z1
Ø
Ø
Ø
(83) |p n (x) ° f (x)| = ØØ
f (x + t )k n (t ) d t ° f (x)ØØ
°1
Ø Z1
Ø
Ø
Ø
Ø
= Ø ( f (x + t ) ° f (x))k n (t ) d t ØØ
°1
=
Z±
°±
∑
Z1
°1
| f (x + t ) ° f (x)|k n (t ) d t
| f (x + t ) ° f (x)|k n (t ) d t +
Z
±<|t |∑1
| f (x + t ) ° f (x)|k n (t ) d t .
We’ll handle each of the final integrals in turn.
Let " > 0 and use the uniform continuity of f to choose a ± 2 (0, 1) such that
when |t | < ±, then | f (x + t ) ° f (x)| < "/2. Then, using Lemma 9.14(b) again,
Z±
Z
" ±
"
(84)
| f (x + t ) ° f (x)|k n (t ) d t <
k n (t ) d t <
2 °±
2
°±
3Given two functions f and g defined on R, the convolution of f and g is the integral
f ? g (x) =
Z1
°1
f (t )g (x ° t ) d t .
The term convolution kernel is used because such kernels typically replace g in the convolution
given above, as can be seen in the proof of the Weierstrass approximation theorem.
4
It was investigated by the German mathematician Edmund Landau (1877–1938).
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5. CONTINUITY AND UNIFORM CONVERGENCE
9-11
According to Lemma 9.14(c), there is an N 2 N so that when n ∏ N and |t | ∏ ±,
"
then k n (t ) < 8(k f k+1)(1°±)
. Using this, it follows that
(85)
Z
±<|t |∑1
=
| f (x + t ) ° f (x)|k n (t ) d t
Z°±
°1
| f (x + t ) ° f (x)|k n (t ) d t +
∑ 2k f k
< 2k f k
Z°±
°1
Z1
±
| f (x + t ) ° f (x)|k n (t ) d t
k n (t ) d t + 2k f k
Z1
±
k n (t ) d t
"
"
"
(1 ° ±) + 2k f k
(1 ° ±) =
8(k f k + 1)(1 ° ±)
8(k f k + 1)(1 ° ±)
2
Combining (84) and (85), it follows from (83) that |p n (x)° f (x)| < " for all x 2 [0, 1]
and p n ‚ f .
⇤
C OROLLARY 9.15. If f 2 C ([a, b]) and " > 0, then there is a polynomial p such
that k f ° pk[a,b] < ".
The theorems of this section can also be used to construct some striking
examples of functions with unwelcome behavior. Following is perhaps the most
famous.
E XAMPLE 9.9. There is a continuous f : R ! R that is differentiable nowhere.
P ROOF. Thinking of the canonical example of a continuous function that
fails to be differentiable at a point—the absolute value function—we start with a
“sawtooth” function. (See Figure 8.)
(
x ° 2n,
2n ∑ x < 2n + 1, n 2 Z
s 0 (x) =
2n + 2 ° x, 2n + 1 ∑ x < 2n + 2, n 2 Z
Notice that s 0 is continuous and periodic with period 2 and maximum value 1.
Compress it both vertically and horizontally:
µ ∂n
°
¢
3
s n (x) =
s 0 4 n x , n 2 N.
4
Each s n is continuous and periodic with period p n = 2/4n and ks n k = (3/4)n .
1.0
0.8
0.6
0.4
0.2
0.5
1.0
1.5
2.0
2.5
3.0
3.5
F IGURE 8. s 0 , s 1 and s 2 from Example 9.9.
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9-12
CHAPTER 9. SEQUENCES OF FUNCTIONS
3.0
2.5
2.0
1.5
1.0
0.5
0.5
1.0
1.5
2.0
F IGURE 9. The nowhere differentiable function f from Example 9.9.
Finally, the desired function is
f (x) =
1
X
n=0
s n (x).
Since ks n k = (3/4)n , the Weierstrass M -test implies the series defining f is uniformly convergent and Corollary 9.12 shows f is continuous on R. We will show
f is differentiable nowhere.
Let x 2 R, m 2 N and h m = 1/(2 · 4m ).
If n > m, then h m /p n = 4n°m°1 2 N, so s n (x ± h m ) ° s n (x) = 0 and
(86)
m s (x ± h ) ° s (x)
f (x ± h m ) ° f (x) X
m
k
k
=
.
±h m
±h
m
k=0
On the other hand, if n < m, then a worst-case estimate is that
Ø
Ø µ ∂n µ ∂
Ø s n (x ± h m ) ° s n (x) Ø
Ø
Ø ∑ 3 / 1 = 3n .
Ø
Ø
hm
4
4n
This gives
(87)
Ø
Ø
Ø
Øm°1
Ø m°1 Ø
Ø X s k (x ± h m ) ° s k (x) Ø X ØØ s k (x ± h m ) ° s k (x) ØØ
Ø
Ø∑
Ø
Ø k=0
Ø k=0 Ø
±h m
±h m
∑
3m ° 1 3m
<
.
3°1
2
Since s m is linear on intervals of length 4°m = 2 · h m with slope ±3m on those
linear segments, at least one of the following is true:
Ø
Ø
Ø
Ø
Ø s m (x + h m ) ° s(x) Ø
Ø s m (x ° h m ) ° s(x) Ø
m
Ø
Ø
Ø
Ø = 3m .
(88)
Ø
Ø = 3 or Ø
Ø
hm
°h m
Suppose the first of these is true. The argument is essentially the same in the
second case.
July 13, 2016
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6. INTEGRATION AND UNIFORM CONVERGENCE
9-13
Using (86), (87) and (88), the following estimate ensues
Ø
Ø
Ø Ø1
Ø
Ø f (x + h m ) ° f (x) Ø ØØ X
s
(x
+
h
)
°
s
(x)
Ø
m
k
k
Ø
Ø=Ø
Ø
Ø
Ø
Ø
Ø
hm
hm
k=0
Ø
Ø
ØX
m s (x + h ) ° s (x) Ø
Ø
Ø
m
k
k
=Ø
Ø
Øk=0
Ø
hm
Ø
Ø m°1 Ø
Ø
Ø s m (x + h m ) ° s m (x) Ø X Ø s k (x ± h m ) ° s k (x) Ø
Ø
Ø
Ø
Ø
∏Ø
Ø°
Ø
Ø
hm
±h m
k=0
> 3m °
3m 3m
=
.
2
2
⇤
Since 3m /2 ! 1, it is apparent f 0 (x) does not exist.
There are many other constructions of nowhere differentiable continuous
functions. The first was published by Weierstrass [20] in 1872, although it was
known in the folklore sense among mathematicians earlier than this. (There is
an English translation of Weierstrass’ paper in [10].) In fact, it is now known in a
technical sense that the “typical” continuous function is nowhere differentiable
[4].
6. Integration and Uniform Convergence
One of the recurring questions with integrals is when it is true that
lim
n!1
Z
fn =
Z
lim f n .
n!1
This is often referred to as “passing the limit through the integral.” At some
point in her career, any student of advanced analysis or probability theory will
be tempted to just blithely pass the limit through. But functions such as those of
Example 9.3 show that some care is needed. A common criterion for doing so is
uniform convergence.
Rb
T HEOREM 9.16. If f n : [a, b] ! R such that a f n exists for each n and f n ‚ f
on [a, b], then
Zb
Zb
f = lim
fn
a
n!1 a
.
P ROOF. Some care must be taken in this proof, because there are actually
two things to prove. Before the equality can be shown, it must be proved that f is
integrable.
July 13, 2016
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9-14
CHAPTER 9. SEQUENCES OF FUNCTIONS
To show that f is integrable, let " > 0 and N 2 N such that k f ° f N k < "/3(b°a).
If P 2 part ([a, b]), then
n
n
X
X
°
¢
°
¢
|R f , P, x k§ ° R f N , P, x k§ | = |
f (x k§ )|I k | °
f N (x k§ )|I k ||
(89)
k=1
=|
∑
N
X
k=1
N
X
k=1
k=1
( f (x k§ ) ° f N (x k§ ))|I k ||
| f (x k§ ) ° f N (x k§ )||I k |
n
X
"
|I k |
3(b ° a)) k=1
"
=
3
<
According to Theorem 8.10, there is a P 2 part ([a, b]) such that whenever
P ø Q 1 and P ø Q 2 , then
°
¢
°
¢
"
(90)
|R f N ,Q 1 , x k§ ° R f N ,Q 2 , y k§ | < .
3
Combining (89) and (90) yields
Ø °
¢
°
¢Ø
ØR f ,Q 1 , x § ° R f ,Q 2 , y § Ø
k
k
Ø °
¢
°
¢
°
¢
= ØR f ,Q 1 , x k§ ° R f N ,Q 1 , x k§ + R f N ,Q 1 , x k§
°
¢
°
¢
°
¢Ø
°R f N ,Q 1 , x k§ + R f N ,Q 2 , y k§ ° R f ,Q 2 , y k§ Ø
Ø °
¢
°
¢Ø Ø °
¢
°
¢Ø
∑ ØR f ,Q 1 , x k§ ° R f N ,Q 1 , x k§ Ø + ØR f N ,Q 1 , x k§ ° R f N ,Q 1 , x k§ Ø
Ø °
¢
°
¢Ø
+ ØR f N ,Q 2 , y k§ ° R f ,Q 2 , y k§ Ø
" " "
< + + ="
3 3 3
Another application of Theorem 8.10 shows that f is integrable.
Finally, when n ∏ N ,
ØZb
Ø Zb
Zb Ø ØZb
Ø
Ø Ø
Ø
"
"
Ø
Ø
Ø
f°
f n Ø = Ø ( f ° f n )ØØ <
= <"
Ø
3
a
a
a
a 3(b ° a)
Rb
Rb
shows that a f n ! a f .
⇤
P1
C OROLLARY 9.17. If n=1 f n is a series of integrable functions converging
uniformly on [a, b], then
Zb X
1
1 Zb
X
fn =
fn
a n=1
n=1 a
E XAMPLE 9.10. It was shown in Example 4.2 that the geometric series
1
X
n=0
July 13, 2016
tn =
1
,
1°t
°1 < t < 1.
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7. DIFFERENTIATION AND UNIFORM CONVERGENCE
9-15
In Exercise 9.3, you are asked to prove this convergence is uniform on any compact subset of (°1, 1). Substituting °t for t in the above formula, it follows that
1
X
1
(°t )n ‚
1
+
t
n=0
on [0, x], when 0 < x < 1. Corollary 9.17 implies
Zx
1 Zx
X
dt
ln(1 + x) =
=
(°t )n d t = x ° x 2 + x 3 ° x 4 + · · · .
1
+
t
0
n=0 0
The same argument works when °1 < x < 0, so
when x 2 (°1, 1).
ln(1 + x) = x ° x 2 + x 3 ° x 4 + · · ·
Combining Theorem 9.16 with Dini’s Theorem, gives the following.
C OROLLARY 9.18. If f n is a sequence of continuous functions converging monoRb
Rb
tonically to a continuous function f on [a, b], then a f n ! a f .
7. Differentiation and Uniform Convergence
The relationship between uniform convergence and differentiation is somewhat more complex than those we’ve already examined. First, because there are
two sequences involved, f n and f n0 , either of which may converge or diverge at a
point; and second, because differentiation is more “delicate” than continuity or
integration.
Example 9.4 is an explicit example of a sequence of differentiable functions
converging uniformly to a function which is not differentiable at a point. The
derivatives of the functions from that example converge pointwise to a function
that is not a derivative. The Weierstrass Approximation Theorem and Example 9.9
push this to the extreme by showing the existence of a sequence of polynomials
converging uniformly to a continuous nowhere differentiable function.
The following theorem starts to shed some light on the situation.
T HEOREM 9.19. If f n is a sequence of derivatives defined on [a, b] and f n ‚ f ,
then f is a derivative.
P ROOF. For each n, let F n be an antiderivative of f n . By considering F n (x) °
F n (a), if necessary, there is no generality lost with the assumption that F n (a) = 0.
Let " > 0. There is an N 2 N such that
"
m, n ∏ N =) k f m ° f n k <
.
b°a
If x 2 [a, b] and m, n ∏ N , then the Mean Value Theorem and the assumption that
F m (a) = F n (a) = 0 yield a c 2 [a, b] such that
(91)
|F m (x) ° F n (x)| = |(F m (x) ° F n (x)) ° (F m (a) ° F n (a))|
= | f m (c) ° f n (c)| |x ° a| ∑ k f m ° f n k(b ° a) < ".
This shows F n is a Cauchy sequence in C ([a, b]) and there is an F 2 C ([a, b]) with
Fn ‚ F .
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9-16
CHAPTER 9. SEQUENCES OF FUNCTIONS
It suffices to show F 0 = f . To do this, several estimates are established.
Let M 2 N so that
"
m, n ∏ M =) k f m ° f n k < .
3
Notice this implies
"
(92)
k f ° f n k ∑ , 8n ∏ M .
3
For such m, n ∏ M and x, y 2 [a, b] with x 6= y, another application of the
Mean Value Theorem gives
Ø
Ø
Ø F n (x) ° F n (y) F m (x) ° F m (y) Ø
Ø
Ø
°
Ø
Ø
x°y
x°y
Ø
1 ØØ
=
(F n (x) ° F m (x)) ° (F n (y) ° F m (y))Ø
|x ° y|
Ø
1 ØØ
"
=
f n (c) ° f m (c)Ø |x ° y| ∑ k f n ° f m k < .
|x ° y|
3
Letting m ! 1, it follows that
Ø
Ø
Ø F n (x) ° F n (y) F (x) ° F (y) Ø "
Ø
Ø ∑ , 8n ∏ M .
(93)
°
Ø
Ø 3
x°y
x°y
Fix n ∏ M and x 2 [a, b]. Since F n0 (x) = f n (x), there is a ± > 0 so that
Ø
Ø
Ø F n (x) ° F n (y)
Ø "
Ø
(94)
° f n (x)ØØ < , 8y 2 (x ° ±, x + ±) \ {x}.
Ø
x°y
3
Finally, using (93), (94) and (92), we see
Ø
Ø
Ø F (x) ° F (y)
Ø
Ø
° f (x)ØØ
Ø
x°y
Ø
Ø F (x) ° F (y) F n (x) ° F n (y)
= ØØ
°
x°y
x°y
Ø
Ø
F n (x) ° F n (y)
+
° f n (x) + f n (x) ° f (x)ØØ
x°y
Ø
Ø
Ø F (x) ° F (y) F n (x) ° F n (y) Ø
Ø
∑ ØØ
°
Ø
x°y
x°y
Ø
Ø
Ø F n (x) ° F n (y)
Ø Ø
Ø
Ø
+Ø
° f n (x)ØØ + Ø f n (x) ° f (x)Ø
x°y
" " "
< + + = ".
3 3 3
This establishes that
F (x) ° F (y)
lim
= f (x),
y!x
x°y
as desired.
⇤
C OROLLARY 9.20. If G n 2 C ([a, b]) is a sequence such that G n0 ‚ g and G n (x 0 )
converges for some x 0 2 [a, b], then G n ‚ G where G 0 = g .
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8. POWER SERIES
9-17
P ROOF. Suppose G n (x 0 ) ! Æ. For each n choose an antiderivative F n of g n
such that F n (a) = 0. Theorem 9.19 shows g is a derivative and an argument
similar to that in the proof of Theorem 9.19 shows F n ‚ F on [a, b], where F 0 = g .
Since F n0 °G n0 = 0, Corollary (7.16) shows G n (x) = F n (x)+(G n (x 0 )°F n (x 0 )). Define
G(x) = F (x) + (Æ ° F (x 0 )).
Let " > 0 and x 2 [a, b]. There is an N 2 N such that
"
"
n ∏ N =) kF n ° F k < and |G n (x 0 ) ° Æ| < .
3
3
If n ∏ N ,
|G n (x) °G(x)| = |F n (x) + (G n (x 0 ) ° F n (x 0 )) ° (F (x) + (Æ ° F (x 0 )))|
∑ |F n (x) ° F (x)| + |G n (x 0 ) ° Æ| + |F n (x 0 ) ° F (x 0 )|
" " "
< + +
3 3 3
="
This shows G n ‚ G on [a, b] where G 0 = F 0 = g .
⇤
C OROLLARY 9.21. If f n is a sequence of differentiable functions defined on [a, b]
P
P
such that 1
f (x ) exists for some x 0 2 [a, b] and 1
f 0 converges uniformly,
k=1 k 0
k=1 k
then
√
!0
1
1
X
X
f =
f0
k=1
k=1
⇤
P ROOF. Left as an exercise.
E XAMPLE 9.11. Let a > 0 and f n (x) = x n /n!. Note that f n0 = f n°1 for n 2 N.
P
0
Example 9.8 shows 1
n=0 f n (x) is uniformly convergent on [°a, a]. Corollary 9.21
shows
µ 1 n ∂0 1 n
X x
X x
(95)
=
.
n=0 n!
n=0 n!
on [°a, a]. Since a is an arbitrary positive constant, (95) is seen to hold on all of
R.
P
n
If f (x) = 1
n=0 x /n!, then the argument given above implies the initial value
problem
(
f 0 (x) = f (x)
f (0) = 1
As is well-known, the unique solution to this problem is f (x) = e x . Therefore,
ex =
1 xn
X
.
n=0 n!
8. Power Series
8.1. The Radius and Interval of Convergence. One place where uniform
convergence plays a key role is with power series. Recall the definition.
July 13, 2016
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9-18
CHAPTER 9. SEQUENCES OF FUNCTIONS
D EFINITION 9.22. A power series is a function of the form
1
X
(96)
f (x) =
a n (x ° c)n .
n=0
Members of the sequence a n are the coefficients of the series. The domain of f is
the set of all x at which the series converges. The constant c is called the center of
the series.
To determine the domain of (96), let x 2 R \ {c} and use the root test to see the
series converges when
lim sup |a n (x ° c)n |1/n = |x ° c| lim sup |a n |1/n < 1
and diverges when
|x ° c| lim sup |a n |1/n > 1.
If r lim sup |a n |1/n ∑ 1 for some r ∏ 0, then these inequalities imply (96) is absolutely convergent when |x ° c| < r . In other words, if
R = lub {r : r lim sup |a n |1/n < 1},
(97)
then the domain of (96) is an interval of radius R centered at c. The root test gives
no information about convergence when |x ° c| = R. This R is called the radius of
convergence of the power series. Assuming R > 0, the open interval centered at c
with radius R is called the interval of convergence. It may be different from the
domain of the series because the series may converge at one endpoint or both
endpoints of the interval of convergence.
The ratio test can also be used to determine the radius of convergence, but,
as shown in (31), it will not work as often as the root test. When it does,
Ø
Ø
Ø a n+1 Ø
Ø
Ø < 1}.
(98)
R = lub {r : r lim Ø
Ø
n!1 a
n
This is usually easier to compute than (97), and both will give the same value for
R.
E XAMPLE 9.12. Calling to mind Example 4.2, it is apparent the geometric
P
n
power series 1
n=0 x has center 0, radius of convergence 1 and domain (°1, 1).
P
n
n
E XAMPLE 9.13. For the power series 1
n=1 2 (x + 2) /n, we compute
µ n ∂1/n
2
1
lim sup
= 2 =) R = .
n
2
Since the series diverges when x = °2 ± 12 , it follows that the interval of convergence is (°5/2, °3/2).
P
n
E XAMPLE 9.14. The power series 1
n=1 x /n has interval of convergence
(°1, 1) and domain [°1, 1). Notice it is not absolutely convergent when x = °1.
P
n
2
E XAMPLE 9.15. The power series 1
n=1 x /n has interval of convergence
(°1, 1), domain [°1, 1] and is absolutely convergent on its whole domain.
The preceding is summarized in the following theorem.
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8. POWER SERIES
9-19
T HEOREM 9.23. Let the power series be as in (96) and R be given by either (97)
or (98).
(a) If R = 0, then the domain of the series is {c}.
(b) If R > 0 the series converges absolutely at x when |c ° x| < R and
diverges at x when |c°x| > R. In the case when R = 1, the series converges
everywhere.
(c) If R 2 (0, 1), then the series may converge at none, one or both of c ° R
and c + R.
8.2. Uniform Convergence of Power Series. The partial sums of a power
series are a sequence of polynomials converging pointwise on the domain of the
series. As has been seen, pointwise convergence is not enough to say much about
the behavior of the power series. The following theorem opens the door to a lot
more.
T HEOREM 9.24. A power series converges absolutely and uniformly on compact
subsets of its interval of convergence.
P ROOF. There is no generality lost in assuming the series has the form of (96)
with c = 0. Let the radius of convergence be R > 0 and K be a compact subset of
(°R, R) with Æ = lub {|x| : x 2 K }. Choose r 2 (Æ, R). If x 2 K , then |a n x n | < |a n r n |
P
P1
n
n
for n 2 N. Since 1
n=0 |a n r | converges, the Weierstrass M -test shows n=0 a n x
is absolutely and uniformly convergent on K .
⇤
The following two corollaries are immediate consequences of Corollary 9.12
and Theorem 9.16, respectively.
C OROLLARY 9.25. A power series is continuous on its interval of convergence.
C OROLLARY 9.26. If [a, b] is an interval contained in the interval of converP
n
gence for the power series 1
n=0 a n (x ° c) , then
Zb X
1
a n=0
n
a n (x ° c) =
1
X
n=0
an
Zb
a
(x ° c)n .
E XAMPLE 9.16. Define
f (x) =
(
sin x
x ,
1,
x 6= 0
x =0
.
Since limx!0 f (x) = 1, f is continuous everywhere. Suppose we want
an accuracy of five decimal places.
If x 6= 0,
f (x) =
July 13, 2016
Rº
0
f with
1 (°1)n+1
1
X
1 X
(°1)n 2n
x 2n°1 =
x
x n=1 (2n ° 1)!
n=0 (2n + 1)!
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9-20
CHAPTER 9. SEQUENCES OF FUNCTIONS
The latter series converges to f everywhere. Corollary 9.26 implies
Zº
0
(99)
Zº µ X
1
∂
(°1)n 2n
f (x) d x =
x
dx
0 n=0 (2n + 1)!
Zº
1
X
(°1)n
=
x 2n d x
(2n
+
1)!
0
n=0
1
X
(°1)n
=
º2n+1
(2n
+
1)(2n
+
1)!
n=0
The latter series satisfies the Alternating Series Test. Since º1 5/(15 £ 15!) º 1.5 £
10°6 , Corollary 4.20 shows
Zº
0
f (x) d x º
(°1)n
º2n+1 º 1.85194
(2n
+
1)(2n
+
1)!
n=0
6
X
The next question is: What about differentiability?
Notice that the continuity of the exponential function and L’Hospital’s Rule
give
µ
∂
µ
∂
ln n
ln n
1/n
lim n
= lim exp
= exp lim
= exp(0) = 1.
n!1
n!1
n!1 n
n
Therefore, for any sequence a n ,
(100)
lim sup(na n )1/n = lim sup n 1/n a n1/n = lim sup a n1/n .
P
n
Now, suppose the power series 1
n=0 a n x has a nontrivial interval of convergence, I . Formally differentiating the power series term-by-term gives a new
P
n°1
power series 1
. According to (100) and Theorem 9.23, the term-byn=1 na n x
term differentiated series has the same interval of convergence as the original.
Its partial sums are the derivatives of the partial sums of the original series and
Theorem 9.24 guarantees they converge uniformly on any compact subset of I .
Corollary 9.21 shows
1
1 d
1
X
X
d X
an x n =
an x n =
na n x n°1 , 8x 2 I .
d x n=0
d
x
n=0
n=1
This process can be continued inductively to obtain the same results for all higher
order derivatives. We have proved the following theorem.
P
n
T HEOREM 9.27. If f (x) = 1
n=0 a n (x ° c) is a power series with nontrivial
interval of convergence, I , then f is differentiable to all orders on I with
(101)
f (m) (x) =
1
X
n!
a n (x ° c)n°m .
n=m (n ° m)!
Moreover, the differentiated series has I as its interval of convergence.
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8. POWER SERIES
9-21
P
n
8.3. Taylor Series. Suppose f (x) = 1
n=0 a n x has I = (°R, R) as its interval
of convergence for some R > 0. According to Theorem 9.27,
f (m) (0) =
m!
f (m) (0)
a m =) a m =
, 8m 2 !.
(m ° m)!
m!
Therefore,
f (x) =
1 f (n) (0)
X
x n , 8x 2 I .
n!
n=0
This is a remarkable result! It shows that the values of f on I are completely
determined by its values on any neighborhood of 0. This is summarized in the
following theorem.
T HEOREM 9.28. If a power series f (x) =
of convergence I , then
(102)
f (x) =
P1
n=0 a n (x °c)
n
has nontrivial interval
1 f (n) (c)
X
(x ° c)n , 8x 2 I .
n!
n=0
The series (102) is called the Taylor series5 for f centered at c. The Taylor
series can be formally defined for any function that has derivatives of all orders at
c, but, as Example 7.9 shows, there is no guarantee it will converge to the function
anywhere except at c. Taylor’s Theorem 7.18 can be used to examine the question
of pointwise convergence. If f can be represented by a power series on an open
interval I , then f is said to be analytic on I .
8.4. The Endpoints of the Interval of Convergence. We have seen that at
the endpoints of its interval of convergence a power series may diverge or even
absolutely converge. A natural question when it does converge is the following:
What is the relationship between the value at the endpoint and the values inside
the interval of convergence?
T HEOREM 9.29 (Abel). A power series is continuous on its domain.
P
n
P ROOF. Let f (x) = 1
n=0 (x ° c) have radius of convergence R and interval of
convergence I . If I = {0}, the theorem is vacuously true from Definition 6.9. If
I = R, the theorem follows from Corollary 9.25. So, assume R 2 (0, 1). It must be
shown that if f converges at an endpoint of I = (c °R, c +R), then f is continuous
at that endpoint.
It can be assumed c = 0 and R = 1. There is no loss of generality with either
of these assumptions because otherwise just replace f (x) with f ((x + c)/R). The
theorem will be proved for Æ = c + R since the other case is proved similarly.
5When c = 0, it is often called the Maclaurin series for f .
July 13, 2016
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9-22
CHAPTER 9. SEQUENCES OF FUNCTIONS
Pn
Set s = f (1), s °1 = 0 and s n =
n
X
k=0
ak x k =
=
k=0
n
X
k=0
n
X
k=0
a k for n 2 !. For |x| < 1,
(s k ° s k°1 )x k
sk x k °
= sn x n +
n
X
k=1
n°1
X
k=0
s k°1 x k
sk x k ° x
= s n x n + (1 ° x)
n°1
X
n°1
X
sk x k
k=0
sk x k
k=0
When n ! 1, since s n is bounded and |x| < 1,
(103)
Since (1 ° x)
(104)
f (x) = (1 ° x)
P1
n=0 x
n
1
X
sk x k .
k=0
= 1, (103) implies
Ø
Ø
Ø
Ø
1
X
Ø
kØ
| f (x) ° s| = Ø(1 ° x)
(s k ° s)x Ø .
Ø
Ø
k=0
Let " > 0. Choose N 2 N such that whenever n ∏ N , then |s n ° s| < "/2. Choose
± 2 (0, 1) so
N
X
±
|s k ° s| < "/2.
k=0
Suppose x is such that 1 ° ± < x < 1. With these choices, (104) becomes
Ø
Ø Ø
Ø
Ø
Ø Ø
Ø
N
1
X
X
Ø
kØ Ø
kØ
| f (x) ° s| ∑ Ø(1 ° x)
(s k ° s)x Ø + Ø(1 ° x)
(s k ° s)x Ø
Ø
Ø Ø
Ø
k=0
k=N +1
Ø
Ø
Ø " "
N
1
X
X
" ØØ
Ø
<±
|s k ° s| + Ø(1 ° x)
xk Ø < + = "
Ø
Ø 2 2
2
k=0
k=N +1
It has been shown that limx"1 f (x) = f (1), so 1 2 C ( f ).
⇤
Here is an example showing the power of these techniques.
E XAMPLE 9.17. The series
1
X
n=0
(°1)n x 2n =
1
1 + x2
has (°1, 1) as its interval of convergence. If 0 ∑ |x| < 1, then Corollary 9.17 justifies
Zx
Zx X
1
1 (°1)n
X
dt
n 2n
arctan(x) =
=
(°1)
t
d
t
=
x 2n+1 .
2
1
+
t
2n
+
1
0
0 n=0
n=0
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8. POWER SERIES
9-23
This series for the arctangent converges by the alternating series test when x = 1,
so Theorem 9.29 implies
1 (°1)n
X
º
= lim arctan(x) = arctan(1) = .
x"1
2n
+
1
4
n=0
(105)
A bit of rearranging gives the formula
µ
∂
1 1 1
º = 4 1° + ° +··· ,
3 5 7
which is known as Gregory’s series for º.
Finally, Abel’s theorem opens up an interesting idea for the summation of
P
series. Suppose 1
n=0 a n is a series. The Abel sum of this series is
A
1
X
n=0
a n = lim
1
X
x"1 n=0
Consider the following example.
an x n .
E XAMPLE 9.18. Let a n = (°1)n so
1
X
an = 1 ° 1 + 1 ° 1 + 1 ° 1 + · · ·
n=0
diverges. But,
A
1
X
n=0
a n = lim
1
X
x"1 n=0
(°x)n = lim
x"1
1
1
= .
1+x 2
This shows the Abel sum of a series may exist when the ordinary sum does
not. Abel’s theorem guarantees when both exist they are the same.
Abel summation is one of many different summation methods used in areas
such as harmonic analysis. (For another see Exercise 4.4.25.)
P
T HEOREM 9.30 (Tauber). If 1
n=0 a n is a series satisfying
then
(a) na n ! 0 and
P
(b) A 1
n=0 a n = A,
P1
n=0 a n =A.
P
P ROOF. Let s n = nk=0 a k . For x 2 (0, 1) and n 2 N,
Ø
Ø Ø
Ø
Ø
Ø ØX
Ø
1
n
n
1
X
X
X
Ø
kØ Ø
k
kØ
ak x Ø = Ø
ak °
ak x °
ak x Ø
Øs n °
Ø
Ø Øk=0
Ø
k=0
k=0
k=n+1
Ø
Ø
ØX
Ø
n
1
X
Ø
Ø
=Ø
a k (1 ° x k ) °
ak x k Ø
Øk=0
Ø
k=n+1
Ø
Ø
ØX
Ø
n
1
X
Ø
k°1
kØ
=Ø
a k (1 ° x)(1 + x + · · · + x
)°
ak x Ø
Øk=0
Ø
k=n+1
(106)
July 13, 2016
∑ (1 ° x)
n
X
k=0
k|a k | +
1
X
k=n+1
|a k |x k .
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9-24
CHAPTER 9. SEQUENCES OF FUNCTIONS
Let " > 0. According to (a) and Exercise 3.3.21, there is an N 2 N such that
(107)
n ∏ N =) n|a n | <
n
"
1 X
"
and
k|a k | < .
2
n k=0
2
Let n ∏ N and 1 ° 1/n < x < 1. Using the right term in (107),
(108)
(1 ° x)
n
X
k=0
k|a k | <
Using the left term in (107) gives
1
X
k=n+1
|a k |x k <
n
1 X
"
k|a k | < .
n k=0
2
1
X
" k
x
k=n+1 2k
" x n+1
2n 1 ° x
"
< .
2
Combining (107) and (107) with (106) shows
Ø
Ø
Ø
Ø
1
X
Ø
Ø
a k x k Ø < ".
Øs n °
Ø
Ø
k=0
(109)
<
⇤
Assumption (b) implies s n ! A.
9. Exercises
9.1. If f n (x) = nx(1 ° x)n for 0 ∑ x ∑ 1, then show f n converges pointwise, but
not uniformly on [0, 1].
9.2. Show sinn x converges uniformly on [0, a] for all a 2 (0, º/2). Does sinn x
converge uniformly on [0, º/2)?
P
9.3. Show that x n converges uniformly on [°r, r ] when 0 < r < 1, but not on
(°1, 1).
P
n
9.4. Prove 1
n=0 x /n! does not converge uniformly on R.
9.5. The series
1 cos nx
X
nx
n=0 e
is uniformly convergent on any set of the form [a, 1) with a > 0.
9.6. A sequence of functions f n : S ! R is uniformly bounded on S if there is an
M > 0 such that k f n kS ∑ M for all n 2 N. Prove that if f n is uniformly convergent
on S and each f n is bounded on S, then the sequence f n is uniformly bounded
on S.
9.7. Let S Ω R and c 2 R. If f n : S ! R is a Cauchy sequence, then so is c f n .
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9. EXERCISES
9-25
9.8. If S Ω R and f n , g n : S ! R are Cauchy sequences, then so is f n + g n .
9.9. Let S Ω R. If f n , g n : S ! R are uniformly bounded Cauchy sequences, then
so is f n g n .
9.10. Prove or give a counterexample: If f n is a sequence of monotone functions
converging pointwise to a continuous function f , then f n ‚ f .
9.11. Prove or give a counterexample: If f n : [a, b] ! R is a sequence of monotone
functions converging pointwise to a continuous function f , then f n ‚ f .
9.12. Prove there is a sequence of polynomials on [a, b] converging uniformly to
a nowhere differentiable function.
9.13. Prove Corollary 9.21.
(°1)n
. (This is the Madhava-Leibniz series which
n
n=0 3 (2n + 1)
was used in the fourteenth century to compute º to 11 decimal places.)
1
p X
9.14. Prove º = 2 3
9.15. If exp(x) =
9.16. Is
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1 xn
X
, then
n=0 n!
d
dx
exp(x) = exp(x) for all x 2 R.
1
X
1
Abel convergent?
n
ln
n
n=2
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