Recall that a sequence of functions
with common domain
is said to be pointwise convergent if for all
and for all
there exists an
such that if
then:
Also recall that a sequence of functions
with common domain
is said to be uniformly convergent if for all
there exists an
such that if
then for all
we have that:
We will now extend the concept of pointwise convergence and uniform convergence to series of functions.
Definition: Let
be a sequence of functions with common domain
. The corresponding series
is said to be Pointwise Convergent to the sum function
if the corresponding sequence of partial sums
(where
) is pointwise convergent to
.
For example, consider the following sequence of functions defined on the interval
:
We now that this series converges pointwise for all
since the result series
is simply a geometric series to the sum function
.
Definition: Let
be a sequence of functions with common domain
. The corresponding series
is said to be Uniformly Convergent to the sum function
if the corresponding sequence of partial sums is uniformly convergent to
.
The geometric series given above actually converges uniformly on
, though, showing this with the current definition of uniform convergence of series of functions is laborious. We will soon develop methods to determine whether a series of functions converges uniformly or not without having to brute-force apply the definition for uniform convergence for the sequence of partial sums.
Cauchy's Uniform Convergence Criterion for Series of Functions
If we have a sequence of functions
with common domain
then the corresponding series of functions
is said to be uniformly convergent if the corresponding sequence of partial sums
is a uniformly convergent sequence of functions.
We will now look at a nice theorem known as Cauchy's uniform convergence criterion for series of functions.
Theorem 1: Let
be a sequence of real-valued functions with common domain
. Then
is uniformly convergent on
if and only if for all
there exists an
such that if
and for all
we have that
for all
.
- Proof:
Suppose that
is uniformly convergent to some limit function
on
. Let
denote the sequence of partial sums for this series. Then we must have that
uniformly on
. So, for
there exists an
such that if
and for all
we have that:
- For any
let
. Then
and so:
Suppose that for all
there exists an
such that if
and for all
we have that:
- Let
. Assume without loss of generality that
and that
for some
. Then from above we see that for all
:
- So
converges uniformly by the Cauchy uniform convergence criterion for sequences of functions. So
converges uniformly on
. ![{\displaystyle \blacksquare }](https://wikimedia.org/api/rest_v1/media/math/render/svg/8733090f2d787d03101c3e16dc3f6404f0e7dd4c)
The Weierstrass M-Test for Uniform Convergence of Series of Functions
Recall that if
is a sequence of real-valued functions with common domain
, then we say that the corresponding series of functions
is uniformly convergent if the sequence of partial sums
is a uniformly convergent sequence.
We will now look at a very nice and relatively simply test to determine uniform convergence of a series of real-valued functions called the Weierstrass M-test.
Theorem 1: Let
be a sequence of real-valued functions with common domain
, and let
be a sequence of nonnegative real numbers such that
for each
and for all
. If
converges then
uniformly converges on
.
- Proof: Suppose that there exists a sequence of nonnegative real numbers
such that for all
and for all
we have that:
- Furthermore, suppose that
converges to some
,
. Then we have that for all
:
- So the
converges for each
by the comparison test. ![{\displaystyle \blacksquare }](https://wikimedia.org/api/rest_v1/media/math/render/svg/8733090f2d787d03101c3e16dc3f6404f0e7dd4c)
Weierstrass Example
The series expansion of the exponential function can be shown to be uniformly convergent on any bounded subset
using the Weierstrass M-test.
Theorem (Weierstrass M-test). Let
be a sequence of functions
and let
be a sequence of positive real numbers such that
for all
and
If
converges, then
converges uniformly on
.
The complex exponential function can be expressed as the series:
![{\displaystyle \sum _{n=0}^{\infty }{\frac {z^{n}}{n!}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/12a6f8ee1a41e61276321f7b74826a6e83622ba3)
Any bounded subset is a subset of some disc
of radius
centered on the origin in the complex plane. The Weierstrass M-test requires us to find an upper bound
on the terms of the series, with
independent of the position in the disc:
![{\displaystyle \left|{\frac {z^{n}}{n!}}\right|\leq M_{n},\forall z\in D_{R}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b43c37ca88b4757759c4b7f188c09fa0f42e6def)
To do this, we notice
![{\displaystyle \left|{\frac {z^{n}}{n!}}\right|\leq {\frac {|z|^{n}}{n!}}\leq {\frac {R^{n}}{n!}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d4847cb43da241c67dbbce88c27a3ea55628a034)
and take
If
is convergent, then the M-test asserts that the original series is uniformly convergent.
The ratio test can be used here:
![{\displaystyle \lim _{n\to \infty }{\frac {M_{n+1}}{M_{n}}}=\lim _{n\to \infty }{\frac {R^{n+1}}{R^{n}}}{\frac {n!}{(n+1)!}}=\lim _{n\to \infty }{\frac {R}{n+1}}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/60bd4234373cc86f39105fcc41c4d8d14e9fd036)
which means the series over
is convergent. Thus the original series converges uniformly for all
and since
, the series is also uniformly convergent on
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