2.2.5 Convergence of Functions
There are two kinds of convergence for a sequence of functions described in the next definition, pointwise
convergence and uniform convergence. Of the two, uniform convergence is far better and tends to be the
kind of convergence most encountered in complex analysis. Pointwise convergence is more often encounted
in real analysis and necessitates much more difficult theorems.
Definition 2.2.30 Let S ⊆ ℂp and let fn : S → ℂq for n = 1,2,
is said to converge
pointwise to f on S if for all x ∈ S,
for each x. The sequence is said to converge uniformly to f on S if
is denoted as
∞ or just
is called the uniform norm.
To illustrate the difference in the two types of convergence, here is a standard example.
Example 2.2.31 Let
Also let fn
≡ xn for x ∈
. Then fn converges pointwise to f on
but does not converge
uniformly to f on
Note how the target function is not continuous although each function in the sequence is. The next
theorem shows that this kind of loss of continuity never occurs when you have uniform convergence. The
theorem holds generally when S ⊆ X a normed linear space and f,fn have values in Y another normed
linear space. You should fill in the details to be sure you understand this. You simply replace
for an appropriate norm.
Theorem 2.2.32 Let fn : S → ℂq be continuous and let fn converge uniformly to f on S. Then if
fn is continuous at x ∈ S, it follows that f is also continuous at x.
Proof: Let ε > 0 be given. Let N be such that if n ≥ N, then
Pick such an n. Then by continuity of fn at x, there exists δ > 0 such that if
< δ,y ∈ S,
is continuous at x
as claimed. ■
Definition 2.2.33 Let fn : S → ℂq. Then fn is said to be uniformly Cauchy if for every ε > 0, there
exists N such that if m,n > ε, then if m,n ≥ N, then
satisfies the axioms of a norm.
Consider the triangle inequality.
As to the axiom about scalars,
It is clear that
then clearly f
for each x
and so f
Theorem 2.2.35 Let fn : S → ℂq be bounded functions:supx∈S
Cn < ∞. Then there
exists bounded f
: S → ℂq such that
if and only if
is uniformly Cauchy.
Proof: ⇐First suppose
is uniformly Cauchy. Then for each
and it follows that for each x,
is a Cauchy sequence. By completeness of
this converges. Let
be that to which it converges. Now pick
such that for m,n ≥ N,
Then in 2.12,
so, since x is arbitrary, it follows each fn is bounded since
is bounded above by
Also, for n ≥ N,
Now, take the limit as m →∞ on the right to obtain that for all x,
if n ≥ N
so, since ε
is arbitrary, this shows that limn→∞
⇒Conversely, if there exists bounded f : S → ℝq to which
converges uniformly, why is
By assumption, there is N
such that if n ≥ N,
Then if m,n ≥ N,
it follows that for
any x ∈ S,
is uniformly Cauchy.
Now here is an example of an infinite dimensional space which is also complete.
Example 2.2.36 Denote by BC
the bounded continuous functions defined on S with values
in ℂq. Then this is a complex vector space with norm
defined above. It is also complete.
It is obvious that this is a complex vector space. Indeed, it is a subspace of the set of functions having
values in ℂq and it is clear that the given set of functions is closed with respect to the vector space
operations. It was explained above in Observation 2.2.34 that this uniform norm really is a norm. It
remains to verify completeness. Suppose then that
is a Cauchy sequence. By Theorem
exists a bounded function f
such that fn
converges to f
uniformly. Then by Theorem 2.2.32
it follows that f
is also continuous. Since every Cauchy sequence converges, this says that BC
is complete. Of course
this all works if
is a subset of a normed linear space and ℂq
is replaced with Y
a complete normed linear
Example 2.2.37 Let K be a compact set and consider C
with the uniform norm. This is
a complete normed linear space.
If you have a Cauchy sequence, then the functions are automatically bounded thanks to the extreme
value theorem. Thus, by the above example, the Cauchy sequence converges uniformly to a continuous
One can consider convergence of infinite series the same way as done in calculus.
Definition 2.2.38 The symbol ∑
provided this limit
exists. This is called pointwise convergence of the infinite sum. Thus the infinite sum means the limit
of the sequence of partial sums. The infinite sum is said to converge uniformly if the sequence of
paritial sums converges uniformly.
Note how this theorem includes the case of ∑
k=1∞ak as a special case. Here the ak don’t depend on
The following theorem is very useful. It tells how to recognize that an infinite sum is converging or
converging uniformly. First is a little lemma which reviews standard calculus.
Lemma 2.2.39 Suppose Mk ≥ 0 and ∑
k=1∞Mk converges. Then
Proof: By assumption, there is N such that if m ≥ N, then if n > m,
Then letting n →∞, one can pass to a limit and conclude that
It follows that for m > N,∑
k=m∞Mk < ε. The part about passing to a limit follows from the fact that
k=m+1nMk is an increasing sequence which is bounded above by ∑
k=1∞Mk. Therefore, it
converges by completeness of ℝ. ■
Theorem 2.2.40 For x ∈ S, if ∑
< ∞, then ∑
converges pointwise. If
there exists Mk such that Mk ≥
for all x ∈ S, then ∑
Proof: Let m < n. Then
whenever m is large enough due to the assumption that ∑
. Thus the partial sums are a
Cauchy sequence and so the series converges pointwise.
If Mk ≥
x ∈ S,
then for M
Thus, taking sup
and so the partial sums are uniformly Cauchy sequence. Hence they converge uniformly to what is defined
x ∈ S
The latter part of this theorem is called the Weierstrass M test. As a very interesting application,
consider the question of nowhere differentiable functions.
Now consider the following description of a function. The following is the graph of the function on
The height of the function is 1/2 and the slope of the rising line is 1 while the slope of the falling line is −
Now extend this function to the whole real line to make it periodic of period 1. This means
x ∈ ℝ
and n ∈ ℤ
, the integers. In other words to find the graph of f
simply slide the graph of
a distance of 1 to get the same tent shaped thing on
this way. The following picture illustrates what a piece of the graph of this function looks like. Some might
call it an infinite sawtooth.
Letting Mk =
an application of the Weierstrass M
test, Theorem 2.2.40
continuous. This is because each function in the sum is continuous and the series converges uniformly on ℝ
However, this function is nowhere differentiable. This is shown next.
Let δm = ±
where we assume
That of interest will be m →∞
If you take k > m,
The absolute value of the last term in the sum is
and we choose the sign of δm such that both 4m
are in some interval [k∕
which is certainly possible because the distance between these two points is 1/4 and such half open
intervals include all of ℝ
. Thus, since f
has slope ±
1 on the interval just mentioned,
As to the other terms, 0 ≤ f
2 and so
Since δm → 0 as m →∞, g′
does not exist because the difference quotients are not bounded.
This proves the following theorem.
Theorem 2.2.41 There exists a function defined on ℝ which is continuous and bounded but fails
to have a derivative at any point.
Proof: It only remains to verify that the function just constructed is bounded. However,
Note that you could consider
to get a function which is continuous, has values between 0 and
which has no derivative.