2Series of functions
IB Analysis II
2.1 Convergence of series
Recall that in Analysis I, we studied the convergence of a series of numbers.
Here we will look at a series of functions. The definitions are almost exactly the
same.
Definition (Convergence of series). Let g
n
; E → R be a sequence of functions.
Then we say the series
P
∞
n=1
g
n
converges at a point
x ∈ E
if the sequence of
partial sums
f
n
=
n
X
j=1
g
j
converges at x. The series converges uniformly if f
n
converges uniformly.
Definition (Absolute convergence).
P
g
n
converges absolutely at a point
x ∈ E
if
P
|g
n
| converges at x.
P
g
n
converges absolutely uniformly if
P
|g
n
| converges uniformly.
Proposition. Let
g
n
:
E → R
. If
P
g
n
converges absolutely uniformly, then
P
g
n
converges uniformly.
Proof.
Again, we don’t have a candidate for the limit. So we use the Cauchy
criterion.
Let
f
n
=
n
P
j=1
g
j
and
h
n
(
x
) =
n
P
j=1
|g
j
|
be the partial sums. Then for
n > m
,
we have
|f
n
(x) − f
m
(x)| =
n
X
j=m+1
g
j
(x)
≤
n
X
j=m+1
|g
j
(x)| = |h
n
(x) − h
m
(x)|.
By hypothesis, we have
sup
x∈E
|h
n
(x) − h
m
(x)| → 0 as n, m → ∞.
So we get
sup
x∈E
|f
n
(x) − f
m
(x)| → 0 as n, m → ∞.
So the result follows from the Cauchy criteria.
It is important to remember that uniform convergence plus absolute pointwise
convergence does not imply absolute uniform convergence.
Example. Consider the series
∞
X
n=1
(−1)
n
n
x
n
.
This converges absolutely for every
x ∈
[0
,
1) since it is bounded by the geometric
series. In fact, it converges uniformly on [0
,
1) (see example sheet). However,
this does not converge absolutely uniformly on [0, 1).
We can consider the difference in partial sums
n
X
j=m
(−1)
j
j
x
j
=
n
X
j=m
1
j
|x|
j
≥
1
m
+
1
m + 1
+ ··· +
1
n
|x|
n
.
For each
N
, we can make this difference large enough by picking a really large
n, and then making x close enough to 1. So the supremum is unbounded.
Theorem (Weierstrass M-test). Let
g
n
:
E → R
be a sequence of functions.
Suppose there is some sequence M
n
such that for all n, we have
sup
x∈E
|g
n
(x)| ≤ M
n
.
If
P
M
n
converges, then
P
g
n
converges absolutely uniformly.
This is in fact a very easy result, and we could as well reproduce the proof
whenever we need it. However, this pattern of proving absolute uniform conver-
gence is so common that we prove it as a test.
Proof. Let f
n
=
n
P
j=1
|g
j
| be the partial sums. Then for n > m, we have
|f
n
(x) − f
m
(x)| =
n
X
j=m+1
|g
j
(x)| ≤
n
X
j=m+1
M
j
.
Taking supremum, we have
sup |f
n
(x) − f
m
(x)| ≤
n
X
j=m+1
M
j
→ 0 as n, m → ∞.
So done by the Cauchy criterion.