IB Analysis II - Uniform convergence

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

be any set (not necessarily a subset of

), and

E → R

for

= 1

, ···

a sequence of functions. What does it mean for

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

has to

converge to

, 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

converges pointwise to

f(x) = lim

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

from properties of

. For example, it would be great if continuity

of all

implies continuity of

, 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

fails

to preserve the properties of f

Example. Let

: [

−

→ R

be defined by

(

) =

1/(2n+1)

. These are all

continuous, but the pointwise limit function is

(x) → f(x) =











1 0 < x ≤ 1

0 x = 0

−1 −1 ≤ x < 0

which is not continuous.

Less excitingly, we can let f

be given by the following graph:

−

which converges to the same function as above.

Example. Let

: [0

→ R

be the piecewise linear function formed by joining

(0, 0), (

, n), (

, 0) and (1, 0).

The pointwise limit of this function is f

(x) → f(x) = 0. However, we have

(x) dx = 1 for all n;

f(x) dx = 0.

So the limit of the integral is not the integral of the limit.

Example. Let f

: [0, 1] → R be defined as

(x) =

(

1 n!x ∈ Z

0 otherwise

Since

has finitely many discontinuities, it is Riemann integrable. However,

the limit is

(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

to mean “

for all sufficiently large

”, then any property common to

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

does indeed

tends pointwise to

, different points converge at different rates to

. For

example, at

= 1, we already have

(1) =

(1) = 1. However, at

= (100!)

−1

(

) = 0 while

(

) = 1. No matter how large

is, we can still find some

where

(

) differs a lot from

(

). In other words, if we are given pointwise

convergence, there is no guarantee that for very large

will “look like”

since there might be some points for which

has not started to move towards

Hence, what we need is for

to converge to

at the same pace. This is

known as uniform convergence.

Definition (Uniform convergence). A sequence of functions

E → R

con-

verges uniformly to f if

(∀ε)(∃N)(∀x)(∀n > N) |f

(x) −f(x)| < ε.

Alternatively, we can say

(∀ε)(∃N)(∀n > N) sup

x∈E

(x) −f(x)| < ε.

Note that similar to pointwise convergence, the definition does not require

to be a subset of

. It could as well be the set

{Winnie, Piglet, Tigger}

. However,

many of our theorems about uniform convergence will require

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

(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

that works for every

, while pointwise convergence just requires that for each

, we can find an

that works.

It should be clear from definition that if

→ f

uniformly, then

→ f

pointwise. We will show that the converse is false:

Example. Again consider our first example, where

: [

−

→ R

is defined

by f

(x) = x

1/(2n+1)

. If the uniform limit existed, then it must be given by

(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

. Then

for each

= 2

−(2n+1)

will have

(

) =

(

) = 1. So there is some

such

that |f

(x) −f(x)| > ε. So f

→ f uniformly.

Example. Let

R → R

be defined by

(

) =

. Then

(

)

→ f

(

) = 0

pointwise. However, this convergence is not uniform in

since

(

)

− f

(

)

|x|

, and this can be arbitrarily large for any n.

However, if we restrict

to a bounded domain, then the convergence is

uniform. Let the domain be [−a, a] for some positive, finite a. Then

sup |f

(x) −f(x)| =

|x|

≤

So given ε, pick N such that N >

, 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

E → R

of functions

is uniformly Cauchy if

(∀ε > 0)(∃N)(∀m, n > N) sup

x∈E

(x) −f

(x)| < ε.

Our first theorem will be that uniform Cauchy convergence and uniform

convergence are equivalent.

Theorem. Let

E → R

be a sequence of functions. Then (

) converges

uniformly if and only if (f

) is uniformly Cauchy.

Proof.

First suppose that

→ f

uniformly. Given

, we know that there is

some N such that

(∀n > N) sup

x∈E

(x) −f(x)| <

Then if n, m > N, x ∈ E we have

(x) −f

(x)| ≤ |f

(x) −f(x)|+ |f

(x) − f(x)| < ε.

So done.

Now suppose (

) is uniformly Cauchy. Then (

(

)) is Cauchy for all

. So

it converges. Let

f(x) = lim

n→∞

(x).

We want to show that

→ f

uniformly. Given

ε >

0, choose

such that

whenever

n, m > N

x ∈ E

, we have

(

)

− f

(

)

| <

. Letting

m → ∞

(x) → f(x). So we have |f

(x) − f(x)| ≤

< 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

E → R

. Suppose

→ f

uniformly, and

are continuous at

for all

Then f is also continuous at x.

In particular, if

are continuous everywhere, then

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

y∈E

(y) − f(y)| < ε.

Since f

is continuous at x, there is some δ such that

|x − y| < δ ⇒ |f

(x) − f

(y)| < ε.

Then for each y such that |x − y| < δ, we have

|f(x) − f(y)| ≤ |f(x) − f

(x)| + |f

(x) − f

(y)|+ |f

(y) − f(y)| < 3ε.

Theorem (Uniform convergence and integrals). Let

, f

: [

a, b

]

→ R

Riemann integrable, with f

→ f uniformly. Then

(t) dt →

f(t) dt.

Proof. We have



(t) dt −

f(t) dt



(t) − f(t) dt



≤

(t) − f(t)| dt

≤ sup

t∈[a,b]

(t) − f(t)|(b − a)

→ 0 as n → ∞.

This is really the easy part. What we would also want to prove is that if

is integrable,

→ f

uniformly, then

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

, f : [−1, 1] → R be defined by

(x) = |x|

1+1/n

, f(x) = |x|.

Then f

→ f uniformly (exercise).

Each

is differentiable — this is obvious at

x 

= 0, and at

= 0, the

derivative is

′

(0) = lim

x→0

(x) − f

(0)

= lim

x→0

sgn(x)|x|

1/n

= 0

However, the limit f is not differentiable at x = 0.

Example. Let

(x) =

sin nx

√

for all x ∈ R. Then

sup

x∈R

(x)| ≤

√

→ 0.

So f

→ f = 0 uniformly in R. However, the derivative is

′

(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

: [

a, b

]

→ R

be a sequence of functions differentiable on [

a, b

]

(at the end points

, this means that the one-sided derivatives exist). Suppose

the following holds:

(i) For some c ∈ [a, b], f

(ii) The sequence of derivatives (f

′

) converges uniformly on [a, b].

Then (

) converges uniformly on [

a, b

], and if

lim f

, then

is differentiable

with derivative f

′

(x) = lim f

′

(x).

Note that we do not assume that

′

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

such that

n, m > N

implies

sup |f

− f

| < 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

− f

to get

− f

)(x) − (f

− f

)(c) = (x − c)(f

′

− f

′

)(t)

for some t ∈ (x, c).

Taking the supremum and rearranging terms, we obtain

sup

x∈[a,b]

(x) − f

(x)| ≤ |f

(c)| + (b − a) sup

t∈[a,b]

′

(t) − f

′

(t)|.

So given any

, since

′

and

(

) converge and are hence Cauchy, there is some

N such that for any n, m ≥ N ,

sup

t∈[a,b]

′

(t) − f

′

(t)| < ε, |f

(c)| < ε.

Hence we obtain

n, m ≥ N ⇒ sup

x∈[a,b]

(x) − f

(x)| < (1 + b − a)ε.

So by the Cauchy criterion, we know that

converges uniformly. Let

lim f

Now we have to check differentiability. Let

′

→ h

. For any fixed

y ∈

[

a, b

define

(x) =

(

(x)−f

(y)

x−y

x = y

′

(y) x = y

Then by definition,

is differentiable at

iff

is continuous at

. Also, define

g(x) =

(

f(x)−f(y)

x−y

x = y

h(y) x = y

Then

is differentiable with derivative

iff

is continuous at

. However,

we know that

→ g

pointwise on [

a, b

], and we know that

are all continuous.

To conclude that

is continuous, we have to show that the convergence is

uniform. To show that

converges uniformly, we rely on the Cauchy criterion

and the mean value theorem.

For x = y, we know that

(x) − g

(x) =

− f

)(x) − (f

− f

)(y)

x − y

= (f

′

− f

′

)(t)

for some

t ∈

[

x, y

]. This also holds for

, since

(

)

−g

(

) =

′

(

)

−f

′

(

)

by definition.

Let

ε >

0. Since

′

converges uniformly, there is some

such that for all

x = y, n, m > N , we have

(x) − g

(x)| ≤ sup |f

′

− f

′

| < ε.

n, m ≥ N ⇒ sup

[a,b]

− g

| < ε,

i.e.

converges uniformly. Hence the limit function

is continuous, in particular

at x = y. So f is differentiable at y and f

′

(y) = h(y) = lim f

′

(y).

If we assume additionally that

′

are continuous, then there is an easy proof

of this theorem. By the fundamental theorem of calculus, we have

(x) = f

′

(t) dt. (∗)

Then we get that

sup

[a,b]

(x) − f

(x)| ≤ |f

(c)| + sup

x∈[a,b]



′

(t) − f

′

(t)) dt



≤ |f

(c)| + (b − a) sup

t∈[a,b]

′

(t) − f

′

(t)|

< ε

for sufficiently large n, m > N.

So by the Cauchy criterion,

→ f

uniformly for some function

: [

a, b

]

→ R

Since the

′

are continuous,

lim

n→∞

′

is continuous and hence integrable.

Taking the limit of (∗), we get

f(x) = f(c) +

h(t) dt.

Then the fundamental theorem of calculus says that

is differentiable and

′

(x) = h(x) = lim f

′

(x). So done.

Finally, we have a small proposition that can come handy.

Proposition.

(i)

Let

, g

E → R

, be sequences, and

→ f

→ g

uniformly on

Then for any a, b ∈ R, af

+ bg

→ af + bg uniformly.

(ii)

Let

→ f

uniformly, and let

E → R

is bounded. Then

E → R

converges uniformly to gf.

Proof.

(i) Easy exercise.

(ii) Say |g(x)| < M for all x ∈ E. Then

|(gf

)(x) − (gf)(x)| ≤ M|f

(x) − f(x)|.

sup

|gf

− gf| ≤ M sup

− f| → 0.

Note that (ii) is false without assuming boundedness. An easy example is to

take

x ∈ R

, and

(

) =

. Then

→

0 uniformly, but (

)(

) =

does not converge uniformly to 0.