1Uniform convergence

IB Analysis II



1 Uniform convergence
In IA Analysis I, we understood what it means for a sequence of real numbers
to converge. Suppose instead we have sequence of functions. In general, let
E
be any set (not necessarily a subset of
R
), and
f
n
:
E R
for
n
= 1
,
2
, ···
be
a sequence of functions. What does it mean for
f
n
to converge to some other
function f : E R?
We want this notion of convergence to have properties similar to that of
convergence of numbers. For example, a constant sequence
f
n
=
f
has to
converge to
f
, and convergence should not be affected if we change finitely many
terms. It should also act nicely with products and sums.
An obvious first attempt would be to define it in terms of the convergence of
numbers.
Definition (Pointwise convergence). The sequence
f
n
converges pointwise to
f
if
f(x) = lim
n→∞
f
n
(x)
for all x.
This is an easy definition that is simple to check, and has the usual properties
of convergence. However, there is a problem. Ideally, We want to deduce
properties of
f
from properties of
f
n
. For example, it would be great if continuity
of all
f
n
implies continuity of
f
, and similarly for integrability and values of
derivatives and integrals. However, it turns out we cannot. The notion of
pointwise convergence is too weak. We will look at many examples where
f
fails
to preserve the properties of f
n
.
Example. Let
f
n
: [
1
,
1]
R
be defined by
f
n
(
x
) =
x
1/(2n+1)
. These are all
continuous, but the pointwise limit function is
f
n
(x) f(x) =
1 0 < x 1
0 x = 0
1 1 x < 0
,
which is not continuous.
x
y
Less excitingly, we can let f
n
be given by the following graph:
x
y
1
n
1
n
which converges to the same function as above.
Example. Let
f
n
: [0
,
1]
R
be the piecewise linear function formed by joining
(0, 0), (
1
n
, n), (
2
n
, 0) and (1, 0).
x
y
0
2
n
1
n
n
The pointwise limit of this function is f
n
(x) f(x) = 0. However, we have
Z
a
0
f
n
(x) dx = 1 for all n;
Z
1
0
f(x) dx = 0.
So the limit of the integral is not the integral of the limit.
Example. Let f
n
: [0, 1] R be defined as
f
n
(x) =
(
1 n!x Z
0 otherwise
Since
f
n
has finitely many discontinuities, it is Riemann integrable. However,
the limit is
f
n
(x) f(x) =
(
1 x Q
0 x ∈ Q
which is not integrable. So integrability of a function is not preserved by pointwise
limits.
This suggests that we need a stronger notion of convergence. Of course, we
don’t want this notion to be too strong. For example, we could define
f
n
f
to mean
f
n
=
f
for all sufficiently large
n
”, then any property common to
f
n
is obviously inherited by the limit. However, this is clearly silly since only the
most trivial sequences would converge.
Hence we want to find a middle ground between the two cases a notion of
convergence that is sufficiently strong to preserve most interesting properties,
without being too trivial. To do so, we can examine what went wrong in the
examples above. In the last example, even though our sequence
f
n
does indeed
tends pointwise to
f
, different points converge at different rates to
f
. For
example, at
x
= 1, we already have
f
1
(1) =
f
(1) = 1. However, at
x
= (100!)
1
,
f
99
(
x
) = 0 while
f
(
x
) = 1. No matter how large
n
is, we can still find some
x
where
f
n
(
x
) differs a lot from
f
(
x
). In other words, if we are given pointwise
convergence, there is no guarantee that for very large
n
,
f
n
will “look like”
f
,
since there might be some points for which
f
n
has not started to move towards
f.
Hence, what we need is for
f
n
to converge to
f
at the same pace. This is
known as uniform convergence.
Definition (Uniform convergence). A sequence of functions
f
n
:
E R
con-
verges uniformly to f if
(ε)(N)(x)(n > N) |f
n
(x) f(x)| < ε.
Alternatively, we can say
(ε)(N)(n > N) sup
xE
|f
n
(x) f(x)| < ε.
Note that similar to pointwise convergence, the definition does not require
E
to be a subset of
R
. It could as well be the set
{Winnie, Piglet, Tigger}
. However,
many of our theorems about uniform convergence will require
E
to be a subset
of R, or else we cannot sensibly integrate or differentiate our function.
We can compare this definition with the definition of pointwise convergence:
(ε)(x)(N)(n > N) |f
n
(x) f(x)| < ε.
The only difference is in where there (
x
) sits, and this is what makes all the
difference. Uniform convergence requires that there is an
N
that works for every
x
, while pointwise convergence just requires that for each
x
, we can find an
N
that works.
It should be clear from definition that if
f
n
f
uniformly, then
f
n
f
pointwise. We will show that the converse is false:
Example. Again consider our first example, where
f
n
: [
1
,
1]
R
is defined
by f
n
(x) = x
1/(2n+1)
. If the uniform limit existed, then it must be given by
f
n
(x) f(x) =
1 0 < x 1
0 x = 1
1 1 x < 0
,
since uniform convergence implies pointwise convergence.
We will show that we don’t have uniform convergence. Pick
ε
=
1
4
. Then
for each
n
,
x
= 2
(2n+1)
will have
f
n
(
x
) =
1
2
,
f
(
x
) = 1. So there is some
x
such
that |f
n
(x) f(x)| > ε. So f
n
→ f uniformly.
Example. Let
f
n
:
R R
be defined by
f
n
(
x
) =
x
n
. Then
f
n
(
x
)
f
(
x
) = 0
pointwise. However, this convergence is not uniform in
R
since
|f
n
(
x
)
f
(
x
)
|
=
|x|
n
, and this can be arbitrarily large for any n.
However, if we restrict
f
n
to a bounded domain, then the convergence is
uniform. Let the domain be [a, a] for some positive, finite a. Then
sup |f
n
(x) f(x)| =
|x|
n
a
n
.
So given ε, pick N such that N >
a
ε
, and we are done.
Recall that for sequences of normal numbers, we have normal convergence and
Cauchy convergence, which we proved to be the same. Then clearly pointwise
convergence and pointwise Cauchy convergence of functions are equivalent. We
will now look into the case of uniform convergence.
Definition (Uniformly Cauchy sequence). A sequence
f
n
:
E R
of functions
is uniformly Cauchy if
(ε > 0)(N)(m, n > N) sup
xE
|f
n
(x) f
m
(x)| < ε.
Our first theorem will be that uniform Cauchy convergence and uniform
convergence are equivalent.
Theorem. Let
f
n
:
E R
be a sequence of functions. Then (
f
n
) converges
uniformly if and only if (f
n
) is uniformly Cauchy.
Proof.
First suppose that
f
n
f
uniformly. Given
ε
, we know that there is
some N such that
(n > N) sup
xE
|f
n
(x) f(x)| <
ε
2
.
Then if n, m > N, x E we have
|f
n
(x) f
m
(x)| |f
n
(x) f(x)|+ |f
m
(x) f(x)| < ε.
So done.
Now suppose (
f
n
) is uniformly Cauchy. Then (
f
n
(
x
)) is Cauchy for all
x
. So
it converges. Let
f(x) = lim
n→∞
f
n
(x).
We want to show that
f
n
f
uniformly. Given
ε >
0, choose
N
such that
whenever
n, m > N
,
x E
, we have
|f
n
(
x
)
f
m
(
x
)
| <
ε
2
. Letting
m
,
f
m
(x) f(x). So we have |f
n
(x) f(x)|
ε
2
< ε. So done.
This is an important result. If we are given a concrete sequence of functions,
then the usual way to show it converges is to compute the pointwise limit and
then prove that the convergence is uniform. However, if we are dealing with
sequences of functions in general, this is less likely to work. Instead, it is often
much easier to show that a sequence of functions is uniformly convergent by
showing it is uniformly Cauchy.
We now move on to show that uniform convergence tends to preserve proper-
ties of functions.
Theorem (Uniform convergence and continuity). Let
E R
,
x E
and
f
n
, f
:
E R
. Suppose
f
n
f
uniformly, and
f
n
are continuous at
x
for all
n
.
Then f is also continuous at x.
In particular, if
f
n
are continuous everywhere, then
f
is continuous every-
where.
This can be concisely phrased as “the uniform limit of continuous functions
is continuous”.
Proof. Let ε > 0. Choose N such that for all n N , we have
sup
yE
|f
n
(y) f(y)| < ε.
Since f
N
is continuous at x, there is some δ such that
|x y| < δ |f
N
(x) f
N
(y)| < ε.
Then for each y such that |x y| < δ, we have
|f(x) f(y)| |f(x) f
N
(x)| + |f
N
(x) f
N
(y)|+ |f
N
(y) f(y)| < 3ε.
Theorem (Uniform convergence and integrals). Let
f
n
, f
: [
a, b
]
R
be
Riemann integrable, with f
n
f uniformly. Then
Z
b
a
f
n
(t) dt
Z
b
a
f(t) dt.
Proof. We have
Z
b
a
f
n
(t) dt
Z
b
a
f(t) dt
=
Z
b
a
f
n
(t) f(t) dt
Z
b
a
|f
n
(t) f(t)| dt
sup
t[a,b]
|f
n
(t) f(t)|(b a)
0 as n .
This is really the easy part. What we would also want to prove is that if
f
n
is integrable,
f
n
f
uniformly, then
f
is integrable. This is indeed true, but we
will not prove it yet. We will come to this later on at the part where we talk a
lot about integrability.
So far so good. However, the relationship between uniform convergence and
differentiability is more subtle. The uniform limit of differentiable functions
need not be differentiable. Even if it were, the limit of the derivative is not
necessarily the same as the derivative of the limit, even if we just want pointwise
convergence of the derivative.
Example. Let f
n
, f : [1, 1] R be defined by
f
n
(x) = |x|
1+1/n
, f(x) = |x|.
Then f
n
f uniformly (exercise).
Each
f
n
is differentiable this is obvious at
x
= 0, and at
x
= 0, the
derivative is
f
n
(0) = lim
x0
f
n
(x) f
n
(0)
x
= lim
x0
sgn(x)|x|
1/n
= 0
However, the limit f is not differentiable at x = 0.
Example. Let
f
n
(x) =
sin nx
n
for all x R. Then
sup
xR
|f
n
(x)|
1
n
0.
So f
n
f = 0 uniformly in R. However, the derivative is
f
n
(x) =
n cos nx,
which does not converge to f
= 0, e.g. at x = 0.
Hence, for differentiability to play nice, we need a condition even stronger
than uniform convergence.
Theorem. Let
f
n
: [
a, b
]
R
be a sequence of functions differentiable on [
a, b
]
(at the end points
a
,
b
, this means that the one-sided derivatives exist). Suppose
the following holds:
(i) For some c [a, b], f
n
(c) converges.
(ii) The sequence of derivatives (f
n
) converges uniformly on [a, b].
Then (
f
n
) converges uniformly on [
a, b
], and if
f
=
lim f
n
, then
f
is differentiable
with derivative f
(x) = lim f
n
(x).
Note that we do not assume that
f
n
are continuous or even Riemann inte-
grable. If they are, then the proof is much easier!
Proof.
If we are given a specific sequence of functions and are asked to prove
that they converge uniformly, we usually take the pointwise limit and show that
the convergence is uniform. However, given a general function, this is usually not
helpful. Instead, we can use the Cauchy criterion by showing that the sequence
is uniformly Cauchy.
We want to find an
N
such that
n, m > N
implies
sup |f
n
f
m
| < ε
. We
want to relate this to the derivatives. We might want to use the fundamental
theorem of algebra for this. However, we don’t know that the derivative is
integrable! So instead, we go for the mean value theorem.
Fix x [a, b]. We apply the mean value theorem to f
n
f
m
to get
(f
n
f
m
)(x) (f
n
f
m
)(c) = (x c)(f
n
f
m
)(t)
for some t (x, c).
Taking the supremum and rearranging terms, we obtain
sup
x[a,b]
|f
n
(x) f
m
(x)| |f
n
(c) f
m
(c)| + (b a) sup
t[a,b]
|f
n
(t) f
m
(t)|.
So given any
ε
, since
f
n
and
f
n
(
c
) converge and are hence Cauchy, there is some
N such that for any n, m N ,
sup
t[a,b]
|f
n
(t) f
m
(t)| < ε, |f
n
(c) f
m
(c)| < ε.
Hence we obtain
n, m N sup
x[a,b]
|f
n
(x) f
m
(x)| < (1 + b a)ε.
So by the Cauchy criterion, we know that
f
n
converges uniformly. Let
f
=
lim f
n
.
Now we have to check differentiability. Let
f
n
h
. For any fixed
y
[
a, b
],
define
g
n
(x) =
(
f
n
(x)f
n
(y)
xy
x = y
f
n
(y) x = y
Then by definition,
f
n
is differentiable at
y
iff
g
n
is continuous at
y
. Also, define
g(x) =
(
f(x)f(y)
xy
x = y
h(y) x = y
Then
f
is differentiable with derivative
h
at
y
iff
g
is continuous at
y
. However,
we know that
g
n
g
pointwise on [
a, b
], and we know that
g
n
are all continuous.
To conclude that
g
is continuous, we have to show that the convergence is
uniform. To show that
g
n
converges uniformly, we rely on the Cauchy criterion
and the mean value theorem.
For x = y, we know that
g
n
(x) g
m
(x) =
(f
n
f
m
)(x) (f
n
f
m
)(y)
x y
= (f
n
f
m
)(t)
for some
t
[
x, y
]. This also holds for
x
=
y
, since
g
n
(
y
)
g
m
(
y
) =
f
n
(
y
)
f
m
(
y
)
by definition.
Let
ε >
0. Since
f
converges uniformly, there is some
N
such that for all
x = y, n, m > N , we have
|g
n
(x) g
m
(x)| sup |f
n
f
m
| < ε.
So
n, m N sup
[a,b]
|g
n
g
m
| < ε,
i.e.
g
n
converges uniformly. Hence the limit function
g
is continuous, in particular
at x = y. So f is differentiable at y and f
(y) = h(y) = lim f
n
(y).
If we assume additionally that
f
n
are continuous, then there is an easy proof
of this theorem. By the fundamental theorem of calculus, we have
f
n
(x) = f
n
(c) +
Z
x
c
f
n
(t) dt. ()
Then we get that
sup
[a,b]
|f
n
(x) f
m
(x)| |f
n
(c) f
m
(c)| + sup
x[a,b]
Z
x
c
(f
n
(t) f
m
(t)) dt
|f
n
(c) f
m
(c)| + (b a) sup
t[a,b]
|f
n
(t) f
m
(t)|
< ε
for sufficiently large n, m > N.
So by the Cauchy criterion,
f
n
f
uniformly for some function
f
: [
a, b
]
R
.
Since the
f
n
are continuous,
h
=
lim
n→∞
f
n
is continuous and hence integrable.
Taking the limit of (), we get
f(x) = f(c) +
Z
x
c
h(t) dt.
Then the fundamental theorem of calculus says that
f
is differentiable and
f
(x) = h(x) = lim f
n
(x). So done.
Finally, we have a small proposition that can come handy.
Proposition.
(i)
Let
f
n
, g
n
:
E R
, be sequences, and
f
n
f
,
g
n
g
uniformly on
E
.
Then for any a, b R, af
n
+ bg
n
af + bg uniformly.
(ii)
Let
f
n
f
uniformly, and let
g
:
E R
is bounded. Then
gf
n
:
E R
converges uniformly to gf.
Proof.
(i) Easy exercise.
(ii) Say |g(x)| < M for all x E. Then
|(gf
n
)(x) (gf)(x)| M|f
n
(x) f(x)|.
So
sup
E
|gf
n
gf| M sup
E
|f
n
f| 0.
Note that (ii) is false without assuming boundedness. An easy example is to
take
f
n
=
1
n
,
x R
, and
g
(
x
) =
x
. Then
f
n
0 uniformly, but (
gf
n
)(
x
) =
x
n
does not converge uniformly to 0.