Sequences and series of functions - Draft Notes
J.D.S. Jones
October 2000
1
Introduction
At this stage we have a good mathematical understanding of the four fundamental infinite processes in analysis: convergence, continuity, differentiation,
and integration. We have precise mathematical definitions of these concepts
which are suitable for rigorous proofs and computations. We have a body
of results, ranging from general theorems to specific examples, which we can
use in many different ways. Now, we turn our attention to applying this
theory to functions.
The elementary functions of analysis are polynomials, the circular functions (sin, cos, tan, arcsin, arccos, arctan), the exponential functions (ex and
the logarithm function log(x)), together with all the functions one can construct using algebraic operations, including composition, and forming quotients, from these functions. If we take an elementary function and differentiate it, we get another elementary function. On the other hand, if we start
with an elementary function and integrate it, we could easily end up with a
non-elementary function. Here are some examples.
Example 1.1. Consider the integral
Z x
dt
p
.
(1 − t2 )(1 − kt2 )
0
This integral is very closely related to (but is not the same as) the integral
which gives the arc-length of the ellipse. It is an example of an elliptic
integral; it is usually called the Jacobi elliptic integral of modulus k.
If we can set k = 0 in the above integral we get
Z x
dt
p
= arcsin(x).
(1 − t2 )
0
1
This suggest we should really study the inverse function to the function
defined by the above integral.
The Jacobi elliptic function function of modulus k is the function snk (x)
defined by the equation
Z snk (x)
dt
p
= x.
2
(1 − t )(1 − kt2 )
0
The function snk arises when you solve the equation of motion of a simple
pendulum (without the assumption that the angle of displacement is small).
Example 1.2. Another example is the dilogarithm function Li2 (x). This is
the function defined by the integral
Z x
log(1 − t)
Li2 (x) = −
dt.
t
0
This integral occurs in many areas of modern mathematics. For example,
it arises in the calculation of volumes in 3 dimensional non-euclidean (or
hyperbolic) geometry. It also arises in quantum field theory and in the branch
of algebra known as algebraic K-theory.
Example 1.3. The Gamma function Γ(x) is the function defined by the
integral
Z ∞
Γ(x)
tx−1 e−t dt.
0
It is a very good exercise to use integration by parts to establish the following
functional equation for the Gamma function
Γ(x) = (x − 1)Γ(x − 1).
It is also easy to check that
Γ(1) = 1
and so it follows that
Γ(n) = (n − 1)!.
This function therefore gives meaning to x! when x is a real (or for that
matter complex) number.
The three functions defined above are defined by integrating elementary
functions but they are not elementary functions. We will not prove that they
are not elementary functions here but we will discuss this point later in these
notes. Here is another example of a function which is not an elementary
function.
2
Example 1.4. One of the most famous functions in mathematics is the Riemann ζ-function. To describe the ζ-function recall that for any positive
number a and any real number x the number ax is defined by the formula
ax = ex log(a) .
If we fix x with x > 1 then the series
∞
X
1
nx
n=1
converges. The limit defines a function
ζ : (1, ∞) → R.
This is the Riemann ζ-function and it is of major (mega) importance in
number theory. Perhaps the most famous unsolved problem in mathematics
is the celebrated Riemann hypothesis which concerns the behaviour of the ζfunction. We will discuss the ζ-function in some detail later in these notes and
give the precise statement of the Riemann hypothesis. For the moment, let
us simply ask the following question: what methods do we have for studying
this function? For example: is the ζ-function continuous? Is it differentiable?
If it is differentiable what is its derivative?
Exercise 1.5 (Challenge). In fact, the Riemann ζ-function is differentiable.
Can you get a formula for the derivative of the ζ-function by some kind of
formal calculations? Here, formal calculations means you should not worry
too much about any convergence questions when you do your calculations. It
might help to consult a book on number theory, for example “???” by Hardy
and Wright. If you get an answer, then as you go through these notes, you
should see if the theorems proved and the techniques developed help you to
give a rigorous justification of your answer.
One basic line of thought in the analysis of functions is this. We have some
class of functions we understand in detail. Now we try to understand a more
general function f by expressing it in terms of these elementary functions.
And then we try to deduce properties of the function f from this expression.
1.1
Taylor series
For example, we can take the Taylor series of f : R → R
∞
X
f (n) (0)
n=1
n!
3
xn .
To write down this series we need to assume that f is infinitely differentiable
at 0 and this gives a restriction on the class of functions which have a Taylor
series. But before we can use the Taylor series of f to say anything about
f we need to answer the following question: in what sense does the Taylor
series “represent” the function f ? In this example we are trying to express
the function f in terms of polynomials. Let us phrase this more formally.
Let Tn (f ; x) be the polynomial
n
X
f (n) (0)
Tn (f ; x) =
r=1
n!
xn .
This is the n-th Taylor polynomial of f . Does Tn (f ; x) converge to f (x) as
n → ∞?
We can draw interesting conclusions from the fact that Tn (f ; x) converges
to f (x). For example it is not difficult to compute the Taylor series of log(1+
x): it is
∞
X
xn
x2 x3 x4
(−1)n−1
=x−
+
−
+ ··· .
n
2
3
4
n=0
Assuming
log(1 + x) = x −
x2 x3 x4
+
−
+ ···
2
3
4
we can substitute x = 1 to get
log(2) = 1 −
1 1 1
+ − + ··· .
2 3 4
This gives us the sum of the Leibniz series
∞
X
(−1)( n − 1)
n
n=1
.
However, it is not always the case that the n-th Taylor polynomial of f
converges to f as the following example shows.
Exercise 1.6. Let f : R → R be the function defined by
( 1
e− x2 if 6= 0,
f (x) =
0
if x = 0.
Show that f (n) (0) = 0 for all n. It follows that the Taylor expansion of
f is zero. So in this example it is definitely not true that the n-th Taylor
polynomial converges to the function.
4
Exercise 1.7. (Challenge)
Calculate by any means you can, the Taylor series of the functions
Z x
dt
p
2
(1 − t )(1 − k 2 t2 )
0
snk (x)
Li2 (x)
Γ(x)
ζ(x).
You should not worry too much about convergence if you try this Exercise
but you should try to decide if your answer is sensible. If you get an answer,
then as you go through these notes, you should see if the theorems proved
and the techniques developed help you to give a rigorous justification of your
answer.
1.2
Fourier series
Another example where we expand a function in terms of elementary functions is given by Fourier series. The Fourier series of a function f : [0, 2π] → R
is the series
∞
a0 X
+
(an cos nx + bn sin nx)
2
n=1
where
Z
1 2π
f (x) cos nxdx
an =
π 0
Z
1 2π
bn =
f (x) sin nxdx
π 0
First we should think about whether the Fourier series of f exists. This
means that we must think about the following problem: give conditions on
the function f which ensures that the integrals
Z
Z
1 2π
1 2π
f (x) cos nxdx,
f (x) sin nxdx
π 0
π 0
exist. These integrals certainly exist if f is a regulated function and in these
notes we will only consider the Fourier series of regulated functions. If we
want Fourier series of more general functions then we must use a more general
theory of integration.
5
1.3
Term-by-term calculations
Here is an example of the kind of argument we want analyse in detail. You
may well have come across the following proof that
d x
e = ex .
dx
The function ex is given by the series
1+x+
x2 x3
xr
+
+ ··· +
+ ....
2!
3!
r!
If we differentiate this series term-by-term we get the series
0+1+x+
x2 x3
xr
+
+ ··· +
+ ....
2!
3!
r!
This is the same series—therefore
d x
e = ex .
dx
Is this a proof? The reason for asking whether this is a proof is not just
pedantry—if it is a proof, it is a very good one. And it is a good one because
the idea behind it can be used in many other examples where the answer is
not obvious.
As a particular example, consider the series
x+
xr
x2 x3
+
+
·
·
·
+
+ ....
22
32
r2
What function does this series represent? The simplest way to answer this
question is to differentiate the series term-by-term. So suppose
f (x) = x +
x2 x3
xr
+
+
·
·
·
+
+ ....
22
32
r2
Then, it should be the case that
df
x x2
xr−1
(x) = 1 + +
+ ··· +
+ ....
dx
2
3
r
Now comparing this with the Taylor series of log(1 + x) we see that
−
log(1 − x)
x x2
xr−1
=1+ +
+ ··· +
+ ....
x
2
3
r
6
Therefore
and so
d
log(1 − x)
f (x) = −
dx
x
Z x
log(1 − t)
f (x) =
dt
t
0
and therefore,
f (x) = Li2 (x).
1.4
The programme
The most obvious way of studying non-elementary functions like the examples
above is to approximate the functions by elementary functions and take a
limit. For example, both Taylor series and Fourier series fit into this strategy.
But we do as usual have to deal with a pretty basic theoretical question.
• What exactly do we mean by approximate and take a limit? In more
precise terms we must say precisely what we mean by saying that a
sequence of functions fn converges to a function f .
We will then concentrate on three basic questions.
• Suppse the sequence of functions fn converges to the function f . Suppose in addition that each fn is continuous. Is it true that f is continuous?
• Suppose the sequence of functions fn converges to the function f . Suppose in addition that each fn is differentiable. Is it true that f is
differentiable and that
lim fn0 (x) = f 0 (x)?
n→∞
In other words when can we differentiate term-by-term and get the
right answer?
• Suppose the sequence of functions fn converges to the function f . Suppose in addition that each fn is regulated (= integrable). Is it true that
f is regulated and that
Z
Z
lim
fn (x) = f (x)?
n→∞
In other words when can we integrate term-by-term and get the right
answer?
7
Note that these questions involve interchanging two limiting processes. For
example the second asks for hypotheses under which we can assert
fn (x + h) − fn (x)
fn (x + h) − fn (x)
lim lim
= lim lim
.
n→∞ h→0
h→0 n→∞
h
h
Questions involving interchanging two limiting processes lead to some of the
most subtle techniques in Analysis.
The final point to make is that there are many essentially different notions
of convergence when dealing with sequences of functions. In this part of the
notes we will deal in detail with two, pointwise convergence and uniform
convergence.
2
Pointwise convergence
The most obvious notion of convergence when dealing with sequences of
function is pointwise convergence.
Definition 2.1. Let A be a subset of R and let fn : A → R be a sequence of
functions. Then fn converges pointwise to a function f : A → R if for each
x ∈ A the sequence fn (x) converges to f (x). That is, given x ∈ A
lim fn (x) = f (x).
n→∞
We can express this definition in terms of and N as follows. Choose
x ∈ A and choose > 0. Then there exists an integer Nx, , which depends
on x and , such that
n ≥ Nx, =⇒ |fn (x) − f (x)| < .
We will now give examples to show that if we assume only that fn converges pointwise to f then if each fn is continuous we cannot deduce that
fn is conitnuous, if each fn is differentiable then we cannot deduce that f is
differentiable Rand if each
fn is integrable we cannot deduce that the limit of
R
the sequence fn is f .
Pointwise convergence is a very important notion. But these examples
show that by itself it is not good enough to deal with the fundamental questions we want to ask. The conclusion is that pointwise convergence needs
to be strengthened to give a useful concept of convergence of sequences of
functions.
8
Example 2.2. Consider the sequence of functions fn : [−1, 1] → R defined
by


if 1/n ≤ x ≤ 1,
1
fn (x) = nx if −1/n ≤ x ≤ 1/n,


−1 if −1 ≤ x ≤ −1/n.
diagram
Fix x with 0 < x ≤ 1. Then
1/n < x =⇒ fn (x) = 1.
So, if we choose an integer N with 1/N < x then if n ≥ N the sequence
fn (x) is constant; therefore
lim fn (x) = 1.
n→∞
Since fn (0) = 0 for all n it follows that
lim fn (0) = 0
n→∞
If we fix x with −1 ≤ x < 0 then a simple modification of the argument for
the case 0 < x ≤ 1 will show that
lim fn (x) = −1.
n→∞
Therefore, this sequence converges pointwise to


if 0 < x ≤ 1,
1
f (x) = 0
if x = 0,


−1 if −1 ≤ x < 0.
diagram
Notice that the limit function f is not continuous at 0. However, each fn
is continuous. This is an example of a sequence of continuous functions
which converges pointwise to a discontinuous function. Therefore if f is the
pointwise limit of a sequence of continuous functions, we cannot conclude
that f is continuous.
9
Example 2.3. Perhaps the problem with the previous example is that while
each of the functions fn is continuous it is not the case that each fn is
differentiable. Indeed fn is not differentiable at the points −1/n and 1/n:
these are the corners of the graph of fn . And, as n goes to ∞ these points
converge to 0 which is where the pointwise limit f fails to be continuous.
However, consider sequence fn : [−1, 1] → R defined by


if 1/n ≤ x ≤ 1,
1
fn (x) = sin(nπx/2) if −1/n ≤ x ≤ 1/n,


−1
if −1 ≤ x ≤ −1/n.
diagram
This is just a smoothed out version of the previous sequence: instead of
joining (−1/n, −1) to (1/n, 1) by a straight line we have joined them up by
(a speeded up) sine curve.
In particular the arguments of the previous example will work just as well
to prove that the pointwise limit of the sequence fn is the function


if 0 < x ≤ 1,
1
f (x) = 0
if x = 0,


−1 if −1 ≤ x < 0.
diagram
The limit function f is not continuous at 0 and therefore not differentiable
at 0. However, each fn is differentiable, indeed each fn is infinitely differentiable. Therefore we cannot conclude that the pointwise limit of a sequence of
differentiable functions is differentiable. It also shows that we do not improve
matters by improving the functions—the problem lies not in the functions
but in the way they converge.
Example 2.4. Now consider the sequence of functions fn : [0, 1] → R defined by

2

if 0 ≤ x ≤ 1/2n,
2n x
2
fn (x) = 2n − 2n x if 1/2n ≤ x ≤ 1/n,


0
if 1/n ≤ x ≤ 1.
10
diagram
The graph of fn is an isoceles triangle with base 1/n and height n. Each fn
is continuous, and therefore integrable, and
Z 1
1
fn = .
2
0
This answer 1/2 is the area of an isoceles triangle with base 1/n and height
n.
Fix x ∈ [0, 1]. Then for all but a finite number of n the sequence fn (x) is
zero and therefore
lim fn (x) = 0.
n→∞
R1
Thus fn converges pointwise to the zero function but 0 fn does not converge
to 0.
Therefore if f is the pointwise limit of a sequence of continuous functions,
we cannot conclude that the integral of f is the limit of the integrals of the
fn .
Example 2.5. Consider the sequence of functions fn : [−1, 1] → R defined
by
r
1
fn (x) = x2 + .
n
diagram
It is not difficult to prove that
(
x
if x ≥ 0,
lim fn (x) = |x| =
n→∞
−x if x ≤ 0.
In this example each fn is differentiable, indeed infinitely differentiable, but
the pointwise limit |x| is not: its graph has a corner at x = 0.
3
Uniform convergence
We will now introduce a second notion of convergence for sequences of functions—
-uniform convergence.
11
Definition 3.1. Let A be a subset of R and let fn : A → R be a sequence
of functions. The sequence fn converges uniformly to a function f : A → R
if: given > 0 there exists an integer N , (which depends on ), such that
n ≥ N =⇒ |fn (x) − f (x)| < for all x ∈ A.
Let us compare this definition with the definition of pointwise convergence. Fix > 0. The sequence fn : A → R converges pointwise to a
function f : A → R if:
• given x ∈ A there exists an integer Nx , which depends on x, such that
n ≥ Nx =⇒ |fn (x) − f (x)| < .
Thus for pointwise convergence we need to start by fixing x ∈ A and then
we must find an integer Nx which depends on x
n ≥ Nx =⇒ |fn (x) − f (x)| < .
The sequence fn : A → R converges uniformly to a function f : A → R if:
• there is an integer N such that
n ≥ N =⇒ |fn (x) − f (x)| < for all x ∈ A.
Thus, for uniform convergence we must be able to choose one N which works
for all x.
At an intuitive level, uniform convergence means the graphs of the functions fn converge to the graph of f in the sense that given > 0 there exists
N such that if n ≥ N the vertical distance between the graph of f and the
graph of fn is never more than .
diagram
This intuition is formalised in the following lemma.
Lemma 3.2. Let fn : A → R be a sequence of functions which converges
pointwise to a function f : A → R. Suppose in addition that for each integer
n, the function f − fn : A → R is bounded. Then fn converges uniformly to
f : A → R if and only if the sequence of real numbers
sup |f (x) − fn (x)|
x∈A
converges to zero.
12
The reason for the condition that, for all n the function f − fn is bounded
is to ensure that the supremum
sup |f (x) − fn (x)|
x∈A
exists.
Thus pointwise convergence means
lim |f (x) − fn (x)| = 0.
n→∞
Provided that for all n the function f − fn is bounded, uniform convergence
means that in addition
lim sup |f (x) − fn (x)| = 0.
n→∞
x∈A
This lemma helps to clarify the difference between pointwise and uniform
convergence. It shows how uniform convergence is a strengthening of pointwise convergence. It also gives a method for doing computations.
Proof of Lemma ??. First suppose that fn converges to f uniformly. Given
> 0 choose N such that
n ≥ N =⇒ |fn (x) − f (x)| < /2,
for all x ∈ A.
Then, since |fn (x) − f (x)| < /2 for all x ∈ A, it follows that
sup |fn (x) − f (x)| ≤ /2.
x∈A
This shows that
n ≥ N =⇒ sup |fn (x) − f (x)| ≤ /2 < ,
x∈A
and therefore
lim
n→∞
sup |f (x) − fn (x)| = 0.
x∈A
On the other hand, suppose
lim sup |f (x) − fn (x)| = 0.
n→∞
x∈A
Thus, given > 0 there exists N such that
n ≥ N =⇒ sup |fn (x) − f (x)| < .
x∈A
13
However, from the definition of the supremum
|fn (x) − f (x)| ≤ sup |fn (x) − f (x)|,
for all x ∈ A,
x∈A
and so
n ≥ N =⇒ |fn (x) − f (x)| < ,
for all x ∈ A.
This proves that fn converges uniformly to f .
Here is another useful method for proving a sequence of functions does
not converge uniformly.
Lemma 3.3. Let fn : A → R be a sequence of functions. Suppose that the
sequence fn converges pointwise to a function f : A → R. Suppose there is a
real number k > 0 and for each n there is a point pn ∈ A such that
|f (pn ) − fn (pn )| ≥ k.
Then fn does not converge uniformly to f .
Proof. Choose < k in the definition of uniform convergence. Choose any
integer N > 0. Then if n ≥ N we know there is a point pn such that
|f (pn ) − fn (pn )| ≥ k > .
Therefore it is not true that
n ≥ N =⇒ |fn (x) − f (x)| < ,
for all x ∈ A.
and so we cannot have uniform convergence.
We will now work through these ideas in the first example of the previuos
section.
Example 3.4. Consider the sequence of functions fn : [−1, 1] → R defined
by


if 1/n ≤ x ≤ 1,
1
fn (x) = nx if −1/n ≤ x ≤ 1/n,


−1 if −1 ≤ x ≤ −1/n.
diagram
14
This sequence converges pointwise to


if 0 < x ≤ 1,
1
f (x) = 0
if x = 0,


−1 if −1 ≤ x < 0.
diagram
This was proved in the previous section.
The proof uses the following argument. Fix x with 0 < x ≤ 1. Then
1/n < x =⇒ fn (x) = 1.
Therefore fn (x) = 1 for all but a finite number of n and so
lim fn (x) = 1.
n→∞
Let us express this in terms of -N . Fix x with 0 < x ≤ 1. Given > 0
choose Nx such that
1
< x.
Nx
If n ≥ Nx then 1/n ≤ 1/Nx < x and so fn (x) = 1. Therefore,
n ≥ Nx =⇒ |fn (x) − 1| = 0 < ,
and this proves that if 0 < x ≤ 1 then
lim fn (x) = 1.
n→∞
If x = 0 then fn (x) = 0 for all n and f (0) = 0; therefore
lim fn (0) = 0.
n→∞
If −1 ≤ x < 0, then a simple modification of the argument for the case
0 < x ≤ 1 will show that
lim fn (x) = −1.
n→∞
In all cases we have shown that
lim fn (x) = f (x)
n→∞
15
and so fn converges to f pointwise.
However, notice that this method of choosing Nx will not produce an
integer N which works for all x. For example if 0 < x ≤ 1 we choose Nx so
that
1
< x.
Nx
For this to work for all x we would need an integer N such that
1
< x, for all x with 0 < x ≤ 1.
N
However, there is no such integer.
Although this is very convincing, it is not quite a proof that fn does not
converge uniformly to f : there could be a much more clever way of choosing
Nx . To get a proof we use the above lemma.
First note that in this example fn is bounded for each n and f is also
bounded. Therefore we can apply the lemma. To use the lemma we do not
need to calculate
sup |f (x) − fn (x)|
x∈[−1,1]
exactly; we only need to prove it does not converge to zero. (Actually it is
not difficult to calculate this supremum—it is 1 and you should try to prove
this once you have understood the proof that the sequence does not converge
uniformly.) Now if you look at a graph of fn you will see straightaway that
there is a point pn in the region −1/n ≤ x ≤ 1/n such that
|f (pn ) − fn (pn )| = 1/2.
Indeed take
pn = 1/2n.
Now since
sup |f (x) − fn (x)| ≥ |f (pn ) − fn (pn )| = 1/2
x∈[−1,1]
this proves that for any n
sup |f (x) − fn (x)| ≥ 1/2.
x∈[−1,1]
In particular this shows that
lim
n→∞
!
sup |f (x) − fn (x)|
6= 0.
x∈[−1,1]
Thus, using the lemma, this proves that fn does not converge uniformly to
f.
16
The discussion of this example may seem like labouring the obvious, but
it is really important to understand the notion of uniform convergence. One
of the best ways of doing this is to understand, from many different points
of view, how uniform convergence works in examples.
As usual let A be a subset of R. Suppose we have a sequence of functions
fn : A → R which converge pointwise to a function f : A → R. Suppose,
in addition, that this sequence fn converges uniformly to f . We will now
prove that we can indeed conclude that if each fn is continuous then f is
continuous.
Theorem 3.5. Let A be a subset of R and let fn : A → R be a sequence of
functions which converges uniformly to f : A → R. If each fn is continuous
then f is continuous.
Proof. To prove the limit function f : A → R is continuous we must estimate
|f (x) − f (y)|. Since fn (x) converges to f (x) we can approximate f (x) by
fn (x), and f (y) by fn (y). Now we can control |fn (x) − fn (y)| by using the
continuity of f . This line of thought suggests we should use the three step
estimate
|f (x) − f (y)| = |f (x) − fn (x) + fn (x) − fn (y) + fn (y) − f (y)|
≤ |f (x) − fn (x)| + |fn (x) − fn (y)| + |fn (y) − f (y)|
and so an “/3 proof”. The way uniform convergence comes in to the proof is
that it shows that it is possible choose some large integer N such that fN (x)
and fN (y) approximate f (x) and f (y) to the “same order of magnitude”. To
make the notion “same order of magnitude” precise is exactly where comes
in.
The formal proof goes as follows. Given > 0 choose N = N such that
n ≥ N =⇒ |fn (x) − f (x)| < /3 for all x ∈ A.
This integer N exists since fn converges uniformly to f . We will use the
function fN to give the required approximations: we will approximate f (x)
by fN (x) and f (y) by fN (y).
Now fix x. Then since fN is continuous at x we can choose δx such that
|x − y| < δx =⇒ |fN (x) − fN (y)| < /3.
Now suppose |x − y| < δx . Then
|f (x) − f (y)| = |f (x) − fN (x) + fN (x) − fN (y) + fN (y) − f (y)|
≤ |f (x) − fN (x)| + |fN (x) − fN (y)| + |fN (y) − f (y)|
< /3 + /3 + /3
= .
17
This shows that f is continuous at x.
Next, we will prove that if fn converges uniformly to f then the integrals
of fn converge to f .
Theorem 3.6. Let fn : [a, b] → R be a sequence of functions which converges
uniformly to f : [a, b] → R. If each fn is regulated then f is regulated and
lim
n→∞
Z
b
fn =
a
Z
b
f.
a
Proof. To prove that f is regulated we must prove that f can be approximated up to “any order of magnitude” by a step function. To do this, we use
the obvious argument: approximate f by one of the functions fn and then
approximate fn by a step function. The two step approximation leads to an
“/2 proof”.
Given > 0 choose N = N such that
n≥N
=⇒
|f (x) − fn (x)| < /2 for all x ∈ [a, b].
This is where uniform convergence is used. Now choose a step function
φ : [a, b] → R such that
|fN (x) − φ(x)| < /2 for all x ∈ [a, b].
This is where the hypothesis that fN is regulated is used. Now
|f (x) − φ(x)| = |f (x) − fN (x) + fN (x) − φ(x)|
≤ |f (x) − fN (x)| + |fN (x) − φ(x)|
< /2 + /2
=
and since this inequality is valid for all x ∈ [a, b] this proves that f is regulated.
Rb
Rb
Now we must prove that a fn converges to a f . Given > 0 choose
N = N such that
n≥N
=⇒
|f (x) − fn (x)| < for all x ∈ [a, b].
¿From the properties of the integral of regulated functions, it follows that
Z b
Z b
f (x) −
fn (x) < (b − a)
n ≥ N =⇒ a
a
18
Since this inequality is valid for all > 0 it follows that the sequence
converges and
Z b
Z b
lim
fn =
f.
n→∞
a
Rb
a
fn
a
So far we have proved that continuity and integration interact well with
uniform convergence. The situation with differentiation is not so straightforward.
Example 3.7. Show that Example ?? gives an example of a sequence of
differentiable functions fn : [−1, 1] → R which converges uniformly to a
function f : [−1, 1] → R which is not differentiable at every point of [−1, 1].
However, assuming that the sequence of differentiable functions fn converges uniformly to f , then Theorem ??, which concerns integration and
uniform convergence, and the fundamental theorem of calculus guarantee
that something useful is true about the differentiability of the limit function
f.
Theorem 3.8. Let fn : [a, b] → R be a sequence of differentiable functions
which converges pointwise to a function f : [a, b] → R. Suppose, in addition,
that fn0 : [a, b] → R converges uniformly to a continuous function g : [a, b] →
R. Then f is differentiable and
f 0 (x) = lim fn0 (x).
n→∞
Proof. Fix x ∈ [a, b]. Since fn0 converges uniformly to g, Theorem ?? shows
that
Z x
Z x
g = lim
fn0 .
a
n→∞
a
Now using the fundamental theorem of calculus it follows that
Z x
fn0 = fn (x) − fn (a).
a
Since fn0 converges pointwise to f we know that
lim (fn (x) − fn (a)) = f (x) − f (a).
n→∞
Putting these three equations together gives
Z x
g = f (x) − f (a).
a
19
Now g is continuous so it follows that the indefinite integral
tiable and its derivative is g. Therefore
Z x
f (x) =
g + f (a)
Rx
a
g is differen-
a
is differentiable and
f 0 (x) = g(x) = lim fn0 (x).
n→∞
Notice how this argument is nothing more or less that making precise the
idea that Theorem ??, and the fundamental theorem of calculus guarantee
that something useful is true about the differentiability of the limit function
f.
Next we turn to the notion of a uniformly Cauchy sequence. As you
should expect, the role of the uniform Cauchy condition is that it gives a
way of proving that a sequence of functions converges uniformly without
having to know what the limit is.
Definition 3.9. Let A be a subset of R and let fn : A → R be a sequence
of functions. Then fn is a uniformly Cauchy sequence if: given > 0 there
is an integer N , which depends on , such that
n, m ≥ N =⇒ |fn (x) − fm (x)| < ,
for all x ∈ A.
This definition is immediately followed by the theorem that a sequence
of functions fn : A → R is a uniformly Cauchy sequence if and only if it
converges uniformly to a function f : A → R.
Theorem 3.10. Suppose fn : A → R is a sequence of functions which converges uniformly to a function f : A → R. Then the sequence fn is a uniformly Cauchy sequence.
Conversely, suppose that fn : A → R is a uniformly Cauchy sequence.
Then there is a function f : A → R such that the sequence fn converges
uniformly to f .
Proof. The proof of the first part involves the two-step estimate
|fn (x) − fm (x)| = |fn (x) − f (x) + f (x) − fm (x)|
≤ |fn (x) − f (x)| + |f (x) − fm (x)|
and therefore an “/2 proof”. Given > 0, choose N such that
n ≥ N =⇒ |fn (x) − f (x)| < /2,
20
for all x ∈ A.
Such an N exists since fn converges uniformly to f . Now, if n, m ≥ N then
|fn (x) − fm (x)| ≤ |fn (x) − f (x) + f (x) − fm (x)|
< /2 + /2
=
This inequality is true for all x ∈ A and so this proves that fn is a uniformly
Cauchy sequence.
To prove the second part, note that since fn is a uniformly Cauchy sequence, it follows that for each x ∈ A the sequence fn (x) is a Cauchy sequence
of real numbers; therefore it converges. Define f : A → R by
f (x) = lim fn (x).
n→∞
Thus fn converges pointwise to f . We must now show that, in addition, fn
converges uniformly to f . So given > 0 choose N such that
n ≥ N =⇒ |fn (x) − fn+k (x)| < /2,
for all x ∈ A and for all integers k ≥ 0. This is where we use the fact that
the sequence fn is a uniformly Cauchy sequence. Now let k tend to infinity
in this inequality. Since
lim fn+k (x) = f (x)
k→∞
it follows that
n ≥ N =⇒ |fn (x) − f (x)| ≤ /2 < ,
for all x ∈ A. This shows that fn converges uniformly to f .
4
Series of functions
Now let us turn our attention to series of functions
∞
X
fr .
r=1
Given functions fr : A → R let sn : A → R be the function
sn = f1 + f2 + · · · + fn =
n
X
r=1
Thus sn is the n-th partial sum of the series
21
P
fr .
fr .
Definition 4.1. Let fr : A → R be functions. The series
∞
X
fr
r=1
converges uniformly to a function f : A → R if the sequence partial sums
sn =
n
X
fr
r=1
converges uniformly to f .
Now there are versions of the theorems concerning uniform convergence
and continuity, uniform convergence and integration, uniform convergence
and differentiation for series of functions. The statements are given below
and the proofs are simple deductions from the corresponding statements for
sequences.
Theorem 4.2. Let A be a subset
P of R and let fr : A → R be continuous
functions. Suppose the series ∞
r=1 fr converges uniformly to a function f :
A → R. Then f is continuous.
Proof. ThePhypothesis that each fr is continuous guarantees that the partial
sum sn = nr=1 fr is continuous. Now, by definition, sn converges uniformly
to f . So by Theorem ?? it follows that f is continuous.
Theorem
4.3. Let fr : [a, b] → R be regulated functions. Suppose the series
P∞
f
converges
uniformly to a function f : [a, b] → R. Then f is regulated,
r=1 r
the series of real numbers
∞ Z b
X
fr
r=1
converges, and
∞ Z
X
r=1
a
b
fr =
a
Z
b
f.
a
Proof. ThePhypothesis that each fr is regulated guarantees that the partial
sum sn = nr=1 fr is regulated. Now, by definition, sn converges uniformly
to f . So the result follows from Theorem ??.
Theorem 4.4. Let fr : [a, b] → R be differentiable functions. Suppose
P∞
•
r=1 fr converges pointwise to a function f : [a, b] → R;
22
•
P∞
r=1
fr0 converges uniformly to a continuous function.
Then f is differentiable, and
f 0 (x) =
∞
X
fr0 (x).
r=1
Proof. The hypotheses guarantee that we can apply Theorem ?? to the series
of partial sums.
Next we give a general test for uniform convergence of series of functions—
the Weierstrass M -test. It is perhaps the single most useful test for uniform
convergence of series.
Theorem 4.5 (The Weierstrass M -test). Let fr : A → R be functions.
Suppose there are real numbers Mr ≥ 0 such that
• |fr (x)| ≤ Mr for all x ∈ A;
P∞
•
r=1 Mr converges.
P
Then ∞
r=1 fr converges uniformly to a function f : A → R.
Proof. Fix x ∈ A. Then since
|fr (x)| ≤ Mr
P
P∞
and ∞
r=1 Mr converges, it follows, from the comparison test, that
r=1 fr (x)
converges. Thus we can define f : A → R by
f (x) =
∞
X
fr (x)
r=1
P
and ∞
r=1 fr converges pointwise to f . We must go on to prove that the
series converges uniformly.
Given integers m and n, consider the partial sums
sn =
n
X
fr ,
sm =
r=1
m
X
r=1
We must estimate the difference
|sn (x) − sm (x)|.
23
fr .
To do this suppose, without loss of generality, that n ≥ m. Then
n
X
|sn (x) − sm (x)| = fr (x)
r=m+1
≤
≤
n
X
|fr (x)|
r=m+1
n
X
Mr
r=m+1
Now since
P∞
r=1
Mr converges it follows that the partial sums
n
X
tn =
Mr
r=1
form a Cauchy sequence and this will give us control over
n
X
Mr = tn − tm .
r=m+1
Fix > 0. Choose N such that
n ≥ m ≥ N =⇒
n
X
Mr < .
r=m+1
Then, using the above estimate, it follows that
n ≥ m ≥ N =⇒ |sn (x) − sm (x)| < for all x ∈ A. Therefore the P
partial sums sn form a uniformly Cauchy
sequence and so it follows that ∞
r=1 fr converges uniformly.
5
Power series
We will now apply the preceding theory to power series, that is series of the
form
∞
X
an x n
n=0
where the coefficients an are real numbers.
24
Theorem 5.1. Suppose that for some x0 ∈ R with x0 6= 0 the power series
P
an xn0 converges; then
P
•
an xn converges for all x with −|x0 | < x < |x0 |;
• the limit
f (x) =
X
an x n
defines a continuous function
f : (−|x0 |, |x0 |) → R;
• the derived series
P
nan xn−1 converges for all x with −|x0 | < x < |x0 |;
• the function f is differentiable and
X
f 0 (x) =
nan xn−1 , for all x ∈ (−|x0 |, |x0 |);
• the function f is regulated on any closed interval [a, b] with −|x0 | <
a ≤ b < |x0 |;
• if −|x0 | < a ≤ b < |x0 | then
Z
a
b
f=
∞
X
an bn+1
n=0
n+1
−
∞
X
an an+1
n=0
n+1
.
This theorem is the main theorem on power series. It is a very important,
and practical, theorem and it is used time and time again to justify termby-term calculations with power series. It is also quite a striking theorem:
if we can prove that a power series converges at some point then the theorem guarantees convergence in an interval. Not only that but the theorem
guarantees that the limit is continuous, differentiable, and integrable.
Proof of Theorem ??. We are assuming that the power series converges at
some point x0 6= 0. The first step is to use the Weierstrass M -test to prove
that the series converges uniformly on any interval of the form [−b, b] with
0 < b < |x0 |.
We must estimate
the terms |an xn |. First let us consider the terms |an x0 |n .
P
Since the series an xn0 converges we know that the terms an xn0 must converge
to zero:
lim an xn0 = 0
n→∞
25
In particular it follows that the sequence |an xn0 | is bounded. Choose M such
that
|an xn0 | ≤ M, for all n.
Suppose |x| ≤ b < |x0 |, then
|an xn | = |an ||x|n
n
|x|
= |an ||x0 |
|x0 |
n
b
n
≤ |an ||x0 |
|x0 |
n
b
≤M
|x0 |
n
Now take
Mn = M
b
|x0 |
n
in the Weierstrass M -test. The above estimate shows that
|an xn | ≤ Mn ,
for all x ∈ [−b, b].
P
The series (b/|x0 |)n is a geometric series with ration b/|x0 | < 1 and so it
converges. This shows that the series
n
∞
∞ X
X
b
Mn = M
|x0 |
n=0
n=0
converges. Thus
P we ncan apply the Weierstrass M -test, Theorem ??, and
conclude that
an x converges uniformly
on [−b, b].
P
n
It now follows that the series
an x defines a continuous function
f : [−b, b] → R,
f (x) =
∞
X
an x n .
n=0
But this true for any closed interval [−b, b] ⊂ (−|x0 |, |x0 |). It follows that
the series defines a continuous function
f : (−|x0 |, |x0 |) → R.
To show that f is defined and continuous at some point x ∈ (−|x0 |, |x0 |)
simply choose some b with
x ∈ [−b, b] ⊂ (−|x0 |, |x0 |).
26
Next consider the derived series
∞
X
nan xn−1 .
n=1
We will use the same method of estimating the terms to show that this derived
series converges uniformly on any interval of the form [−b, b] ⊂ (−|x0 |, |x0 |).
If x ∈ [−b, b] then
|nan xn | = n|an ||x|n−1
n−1
|x|
= n|an ||x0 |
|x0 |
n−1
b
n
≤ n|an ||x0 |
|x0 |
n−1
b
≤ Mn
.
|x0 |
n−1
Now take
Mn = M n
b
|x0 |
n−1
in the Weierstrass M -test. The above estimate shows that
|nan xn−1 | ≤ Mn , for all x ∈ [−b, b].
P
Since b/|x0 | < 1, the series
n(b/|x0 |)n−1 converges—this is one of the
standard series from first-year analysis. Therefore the series
∞
X
n
∞ X
b
Mn = M
|x0 |
n=0
n=0
converges and we
P can apply the Weierstrass M -test, Theorem ??, to conclude
that the series nan xn−1 converges uniformly
P onn[−b, b] Now combining this
with Theorem ?? shows that the series
nan x converges uniformly to a
continuous function on [−b, b].
So far, we have proved that given any b with 0 < b < |x0 |
P∞
n
•
n=0 an x converges uniformly, and in particular pointwise, on [−b, b];
P∞
n−1
•
converges uniformly to a continuous function on [−b, b].
n=0 nan x
•
27
¿From this, Theorem ?? shows that the limit function f : [−b, b] → R is
differentiable and
0
f (x) =
∞
X
aan xn ,
for all x ∈ [−b, b].
n=1
But since this is valid for any b with 0 < b < |x0 | it follows that f :
(−|x0 |, |x0 |) → R is differentiable and
f 0 (x) =
∞
X
nan xn ,
for all x ∈ (−|x0 |, |x0 |).
n=0
We are left to prove that the limit function
f : (−|x0 |, |x0 |) → R
is regulated, and its integral is given by term-by-term integration of the
defining
power series. But this follows immediately from the fact that the
P∞
series n=0 an xn on any interval of the form [−b, b] ⊂ (−|x0 |, |x0 |) together
with Theorem ??. Of course, we need as input the fact that each functions
an xn is regulated and the formula (given by the fundamental theorem of
calculus) for its integral.
Notice that the theorem can be applied to the derived series of a given
power series, and to the derived series of the derived series, and so on. This
line of thought leads to the following corollary.
Corollary
5.2. Suppose that for some x0 ∈ R with x0 6= 0 the power series
P
an xn0 converges. Then the function
f : (−|x0 |, |x0 |) → R
defined by
f (x) =
X
an x n
is infinitely differentiable and
f (n) (0) =
an
.
n!
Thus, the Taylor expansion of f around the origin is given by
X
f (x) =
an x n
Furthermore, the Taylor series of f converges for all x ∈ (−|x0 |, |x0 |).
28
Proof. This corollary follows by a straightforward argument from the main
theorem on power series Theorem ??.
The main theorem on power series has many applications. And it is
very important to understand how to apply this theorem shows to justify
term-by-term calculations with power series.
Example 5.3. Consider the exponential series
∞
X
xn
n=0
n!
.
Using the ratio test it is straightforward to show that this series converges
for all x ∈ R. Theorem ?? shows that the function
x
e =
∞
X
xn
n=0
n!
is:
• continuous, differentiable, and integrable;
•
•
d x
e = ex
dx
Z
b
ex dx = eb − ea .
a
Therefore, it justifies all the properties of the exponential function.
Example 5.4. Let us consider the series
f (x) =
∞
X
xn
n=1
n
.
This series converges when x = −1 by the Liebniz alternating series test.
Therefore, by Theorem ??, it converges provided |x| < 1. Notice how this
is a very delicate application of the theorem: the series does not converge if
x = 1. This shows that the hypothesis of Theorem ?? cannot be weakened:
the series converges for −1 ≤ x < 1 but it does not converge outside this
range.
29
How can we identify the function f : (−1, 1) → R defined by this power
series? Term-by-term differentiation of this power series gives
∞
X
0
f (x) =
xn
n=0
=
1
.
1−x
Theorem ?? shows that
d
1
f (x) =
, for all x ∈ (−1, 1).
dx
1−x
The fundamental theorem of calculus now shows that
f (x) = − log(1 − x) + A,
for all x ∈ (−1, 1)
where A is a constant. Now, if we check values when x = 0 we conclude that
A = 0. This shows that
∞
X
xn
− log(1 − x) =
, for all x ∈ (−1, 1).
n
n=1
What happens if |x| ≥ 1? For example, if x = −1, it gives the formula
− log 2 =
∞
X
(−1)n
n=1
n
.
This formula is correct: in its more usual form it gives
∞
X
(−1)n−1
log(2) =
n
n=1
=1−
1 1 1
+ − + ....
2 3 4
This answer is correct.
As another example, when x = 1 this series expansion gives the formula
∞
X
1
− log 0 =
.
n
n=1
P
The series
1/n diverges to ∞ and so, given the best will in the world, we
seem to be able to conclude that
log 0 = −∞.
Intuitively, this seems valid. What is more, we know from first-year analysis
that
lim log(1 − x) = −∞.
x→1
30
5.1
The Riemann ζ-function
We will apply the Weierstrass M -test is to the Riemann ζ-function ζ :
(0, ∞) → R is continuous. First we must do the analysis required to prove
that this function exists! Then, we will use the Weierstrass M -test to prove
that the ζ-function is infinitely differentiable.
Let a be a real number with a > 0. Then ax is defined by
ax = ex log(a) ,
and x 7→ ax defines a function R → R. In particular, if n is a positive integer
and s is a real number, we can form ns . Now consider the series
∞
X
1
.
s
n
n=1
Theorem 5.5. The series
∞
X
1
.
s
n
n=1
converges provided s > 1.
Proof. Fix s > 1. We want to estimate the partial sums
n
X
1
An =
.
s
r
r=1
To do this, we will consider the area under the graph y = 1/xs in other words
the integral
Z
dx
.
xs
The idea of the proof is explained in the following diagram.
diagram
In this diagram, the area of the shaded region is
n
X
1
An =
.
rs
r=1
Clearly this area is less that
intn0
dx
.
xs
31
But xs goes to ∞ as x → 0 so there could be a problem with this integral. It
is not particularly difficult to deal with this problem directly, but it is much
easier to avoid it by the following argument.
Consider
n
X
1
Bn =
;
rs
r=2
This is the area of the shaded region in the following diagram.
diagram
It is straightforward to show that
Bn ≤
Z
1
n
dx
.
xs
This is done by constructing a step function φn : [1, n] → R such that
φn (x) <
and
1
,
xs
Z
for all x ∈ [1, n]
n
φn = Bn .
1
The diagram immediately suggests the definition of φn ; the details are left
to the reader. Since
φn (x) <
1
,
xs
it follows that
Bn =
Z
for all x ∈ [1, n]
n
φn ≤
1
Z
1
n
dx
xs
as required. Now we evaluate the integral:
1−s n
Z n
dx
x
=
s
1−s 1
1 x
n1−s
1
=
−
1−s 1−s
and this gives the inequality
Bn ≤
n1−s
1
−
.
1−s 1−s
32
Now Bn is an increasing sequence, we have
Bn ≤ Cn
where
n1−s
1
Cn =
−
1−s 1−s
and, since 1 − s < 0, it follows that Cn converges: in fact
lim Cn = −
n→∞
1
1
=
.
1−s
s−1
Therefore Bn converges and
lim Bn ≤
n→∞
1
.
s−1
Therefore
An = 1 + Bn
converges and
lim An ≤ 1 +
n→∞
1
s
=
.
s−1
s−1
Now define the Riemann ζ-function
ζ : (1, ∞) → R
by
∞
X
1
ζ(x) =
.
x
n
n=1
Theorem 5.6. This function ζ : (1, ∞) → R is continuous.
Proof. Fix c ∈ (1, ∞) and choose a closed interval [a, b] ⊂ (1, ∞) with c ∈
[a, b]. To prove that ζ is continuous at c it is enough to prove that ζ : [a, b] →
R is continuous. Now for any x ∈ [a, b]
nx ≥ na ,
1
1
< a.
x
n
n
So take
Mn =
33
1
na
in the Weierstrasss M -text. Then
1
≤ Mn ,
nx
and
∞
X
for all x ∈ [a, b],
Mn =
n=1
converges. Therefore
∞
X
1
na
n=1
∞
X
1
nx
n=1
converges uniformly. Since each of the functions 1/nx is continuous it follows
that the limit function, which is the Riemann ζ-function, is a continuous
function [a, b] → R.
Exercise 5.7. Prove that the Riemann ζ-function ζ : [a, b] → R is differentiable. Indeed prove that it is infinitely differentiable.
6
A space filling curve
34