When you understand sequences and series of numbers it is easy to consider sequences and
series of functions.
Definition 6.9.1A sequence of functions is a map defined on ℕ or some set ofintegers larger than or equal to a given integer, m which has values which are functions.It is written in the form
{fn}
_{n=m}^{∞}where f_{n}is a function. It is assumed also that thedomain of all these functions is the same.
In the above, where do the functions have values? Are they real valued functions? Are
they complex valued functions? Are they functions which have values in ℝ^{n}? It turns
out it does not matter very much and the same definition holds. However, if you
like, you can think of them as having values in F. This is the main case of interest
here.
Example 6.9.2Suppose f_{n}
(x)
= x^{n}for x ∈
[0,1]
. Here is a graph of the functionsf
(x )
= x,x^{2},x^{3},x^{4},x^{5}.
PICT
Definition 6.9.3Let
{fn}
be a sequence of functions. Then the sequenceconverges pointwise to a function f if for all x ∈ D, the domain of the functions in thesequence,
f (x) = lim fn(x)
n→ ∞
This is always the definition regardless of where the f_{n}have their values.
Thus you consider for each x ∈ D the sequence of numbers
{fn(x)}
and if this sequence
converges for each x ∈ D, the thing it converges to is called f
= 1.
Therefore, this sequence of functions converges pointwise to the function f
(x)
given by
f
(x )
= 0 if 0 ≤ x < 1 and f
(1)
= 1. However, given small ε > 0, and n, there is always some
x such that
|f (x) − fn (x)|
> ε. Just pick x less than 1 but close to 1. Then f
(x )
= 0 but
f_{n}
(x )
will be close to 1.
Pointwise convergence is a very inferior thing but sometimes it is all you can
get. It’s undesirability is illustrated by Example 6.9.4. The limit function is not
continuous although each f_{n} is continuous. Now here is another example of a sequence of
functions.
Example 6.9.5Let f_{n}
(x)
=
1
n
sin
( 2 )
n x
.
In this example,
|fn(x)|
≤
1n
so this function is close to 0 for all x at once provided n is
large enough. There is a difference between the two examples just given. They both involve
pointwise convergence, but in the second example, the pointwise convergence happens for all
x at once. In this example, you have uniform convergence. Here is a picture of the first
four of these graphed on
[− π,π]
.
PICT
Definition 6.9.6Let
{fn}
be a sequence of functions defined on D. Then
{fn}
issaid to converge uniformly to f if it convergespointwise to f and for every ε > 0 thereexists N such that for all n ≥ N
sup |f (x)− fn(x)| < ε
x∈D
The following picture illustrates the above definition.
PICT
The dotted lines define sort of a tube centered about the graph of f and the graph of the
function f_{n} fits in this tube.
The reason uniform convergence is desirable is that it drags continuity along with it and
imparts this property to the limit function.
Theorem 6.9.7Let
{fn}
be a sequence offunctions defined on D which arecontinuous at z and suppose this sequence converges uniformly to f. Then f is alsocontinuous at z. If each f_{n}is uniformly continuous on D, then f is also uniformlycontinuous on D.
Proof: Let ε > 0 be given and pick z ∈ D. By uniform convergence, there exists N such
that if n > N, then for all x ∈ D,
|f (x)− fn(x)| < ε∕3. (6.1)
(6.1)
Pick such an n. By assumption, f_{n} is continuous at z. Therefore, there exists δ > 0 such that
if
be a sequence of functions defined on D. Then thesequence is said to be uniformly Cauchyif for every ε > 0 there exists N such that wheneverm,n ≥ N,
sup|fm(x)− fn (x)| < ε
x∈D
Then the following theorem follows easily.
Theorem 6.9.9Let
{fn}
be a uniformlyCauchy sequence of F valued functionsdefined on D. Then there exists f defined on D such that
{fn}
converges uniformly tof.
Proof: For each x ∈ D,
{fn(x)}
is a Cauchy sequence. Therefore, it converges to some
number because of completeness of F. (Recall that completeness is the same as saying every
Cauchy sequence converges.) Denote by f
(x)
this number. Let ε > 0 be given and let N be
such that if n,m ≥ N,
|fm (x )− fn(x)| < ε∕2
for all x ∈ D. Then for any x ∈ D, pick n ≥ N and it follows from Theorem 4.4.11
As before, once you understand sequences, it is no problem to consider series.
Definition 6.9.12Let
{fn}
be a sequence of functions defined on D. Then
( )
∞∑ ∑n
fk (x) ≡ nl→im∞ fk(x) (6.2)
k=1 k=1
(6.2)
whenever the limit exists. Thus there is a new function denoted by
∞∑
fk (6.3)
k=1
(6.3)
and its value at x is given by the limit of the sequence of partial sums in 6.2. If for all x ∈ D,the limit in 6.2exists, then 6.3is said to converge pointwise.∑_{k=1}^{∞}f_{k}is said to convergeuniformly on D if the sequence of partial sums,
{ }
∑n
fk
k=1
converges uniformly.If the indices for the functions start at some other value than 1, youmake the obvious modification to the above definition as was done earlier with series ofnumbers.
Theorem 6.9.13Let
{fn}
be a sequence of functions defined on D. The series∑_{k=1}^{∞}f_{k}converges pointwise if and only if for each ε > 0 and x ∈ D, there exists N_{ε,x}which may depend on x as well as ε such that when q > p ≥ N_{ε,x},
| |
||∑q ||
|| fk (x)||< ε
|k=p |
The series∑_{k=1}^{∞}f_{k}convergesuniformly on D if for every ε > 0 there exists N_{ε}such that ifq > p ≥ N_{ε}then
|| q ||
sup||∑ fk(x)||< ε (6.4)
x∈D||k=p ||
(6.4)
Proof: The first part follows from Theorem 5.1.7. The second part follows from observing
the condition is equivalent to the sequence of partial sums forming a uniformly Cauchy
sequence and then by Theorem 6.9.11, these partial sums converge uniformly to a function
which is the definition of ∑_{k=1}^{∞}f_{k}. ■
Is there an easy way to recognize when 6.4 happens? Yes, there is. It is called the
Weierstrass M test.
Theorem 6.9.14Let
{fn}
be a sequence of functions defined on D. Supposethere exists M_{n}such that sup
{|fn(x)| : x ∈ D }
< M_{n}and∑_{n=1}^{∞}M_{n}converges.Then∑_{n=1}^{∞}f_{n}converges uniformly on D.
whenever m is large enough because of the assumption that ∑_{n=1}^{∞}M_{n} converges. Therefore,
the sequence of partial sums is uniformly Cauchy on D and therefore, converges uniformly to
∑_{k=1}^{∞}f_{k} on D. ■
Theorem 6.9.15If
{fn}
is a sequence of functions defined on D which arecontinuous at z and∑_{k=1}^{∞}f_{k}converges uniformly, then the function∑_{k=1}^{∞}f_{k}mustalso be continuous at z.
Proof: This follows from Theorem 6.9.7 applied to the sequence of partial sums of the
above series which is assumed to converge uniformly to the function ∑_{k=1}^{∞}f_{k}.■