Sequences of functions Pointwise and Uniform Convergence

[Pages:5]MATH 401 - NOTES

Sequences of functions Pointwise and Uniform Convergence

Fall 2005

Previously, we have studied sequences of real numbers. Now we discuss the topic of sequences of real valued functions. A sequence of functions {fn} is a list of functions (f1, f2, . . .) such that each fn maps a given subset D of R into R.

I. Pointwise convergence

Definition. Let D be a subset of R and let {fn} be a sequence of functions defined on D. We say that {fn} converges pointwise on D if

lim fn(x) exists for each point x in D.

n

This means that lim fn(x) is a real number that depends only on x. n

If {fn} is pointwise convergent then the function defined by f (x) = lim fn(x), n

for every x in D, is called the pointwise limit of the sequence {fn}.

Example 1. Let {fn} be the sequence of functions on R defined by fn(x) = nx. This sequence does not converge pointwise on R because lim fn(x) =

n

for any x > 0.

Example 2. Let {fn} be the sequence of functions on R defined by fn(x) = x/n. This sequence converges pointwise to the zero function on R.

Example 3. Consider the sequence {fn} of functions defined by

nx + x2 fn(x) = n2 Show that {fn} converges pointwise.

for all x in R.

Solution: For every real number x, we have:

x x2

lim fn(x)

n

=

lim

n

n

+

n2

=

x

1 lim n n

+ x2

1

lim

n

n2

=0+0=0

1

Thus, {fn} converges pointwise to the zero function on R.

Example 4. Consider the sequence {fn} of functions defined by

sin(nx + 3)

fn(x) =

n+1

for all x in R.

Show that {fn} converges pointwise. Solution: For every x in R, we have

-1 sin(nx + 3) 1

n+1

n+1

n+1

Moreover,

1

lim

= 0.

n n + 1

Applying the squeeze theorem for sequences, we obtain that

lim fn(x) = 0 for all x in R.

n

Therefore, {fn} converges pointwise to the function f 0 on R.

Example 5. Consider the sequence {fn} of functions defined by fn(x) = n2xn for 0 x 1. Determine whether {fn} is pointwise convergent.

Solution: First of all, observe that fn(0) = 0 for every n in N. So the sequence {fn(0)} is constant and converges to zero. Now suppose 0 < x < 1 then n2xn = n2en ln(x). But ln(x) < 0 when 0 < x < 1, it follows that

lim fn(x) = 0 for 0 < x < 1

n

Finally, fn(1) = n2 for all n. So, lim fn(1) = . Therefore, {fn} is not n

pointwise convergent on [0, 1].

Example 6. Let {fn} be the sequence of functions defined by fn(x) = cosn(x) for -/2 x /2. Discuss the pointwise convergence of the sequence.

Solution: For -/2 x < 0 and for 0 < x /2, we have

0 cos(x) < 1.

2

It follows that

lim (cos(x))n = 0 for x = 0.

n

Moreover, since fn(0) = 1 for all n in N, one gets lim fn(0) = 1. Therefore, n

{fn} converges pointwise to the function f defined by

f (x) =

0

if

-

2

x

<

0

or

0

<

x

2

1 if

x=0

Example 7. Consider the sequence of functions defined by

fn(x) = nx(1 - x)n on [0, 1]. Show that {fn} converges pointwise to the zero function.

Solution: Note that fn(0) = fn(1) = 0, for all n N. Now suppose 0 < x < 1, then

lim fn(x) = lim nxen ln(1-x) = x lim nen ln(1-x) = 0

n

n

n

because ln(1 - x) < 0 when 0 < x < 1. Therefore, the given sequence converges pointwise to zero.

Example 8. Let {fn} be the sequence of functions on R defined by

fn(x) =

n3

if

0

<

x

1 n

1 otherwise

Show that {fn} converges pointwise to the constant function f = 1 on R.

Solution: For any x in R there is a natural number N such that x does not belong to the interval (0, 1/N ). The intervals (0, 1/n) get smaller as n . Therefore, fn(x) = 1 for all n > N . Hence,

lim fn(x) = 1 for all x.

n

The formal definition of pointwise convergence

Let D be a subset of R and let {fn} be a sequence of real valued functions defined on D. Then {fn} converges pointwise to f if given any x in D and given any > 0, there exists a natural number N = N (x, ) such that

|fn(x) - f (x)| < for every n > N.

Note: The notation N = N (x, ) means that the natural number N depends on the choice of x and .

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I. Uniform convergence

Definition. Let D be a subset of R and let {fn} be a sequence of real valued functions defined on D. Then {fn} converges uniformly to f if given any > 0, there exists a natural number N = N () such that

|fn(x) - f (x)| < for every n > N and for every x in D.

Note: In the above definition the natural number N depends only on . Therefore, uniform convergence implies pointwise convergence. But the converse is false as we can see from the following counter-example.

Example 9. Let {fn} be the sequence of functions on (0, ) defined by

nx

fn(x)

=

1

. + n2x2

This function converges pointwise to zero. Indeed, (1 + n2x2) n2x2 as n gets larger and larger. So,

nx 1 1

lim fn(x)

n

=

lim

n

n2x2

=

x

lim

n

n

=

0.

But for any < 1/2, we have

1

11

fn n - f n

= - 0 > . 2

Hence {fn} is not uniformly convergent.

Theorem. Let D be a subset of R and let {fn} be a sequence of continuous functions on D which converges uniformly to f on D. Then f is continuous on D

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Homework

Problem 1. Let {fn} be the sequence of functions on [0, 1] defined by fn(x) = nx(1-x4)n. Show that {fn} converges pointwise. Find its pointwise limit.

Problem 2.

Is

the

sequence of

functions

on

[0,

1)

defined by

fn(x)

=

(1

-

x)

1 n

pointwise

convergent? Justify your answer.

Problem 3. Consider the sequence {fn} of functions defined by

n + cos(nx) fn(x) = 2n + 1

for all x in R.

Show that {fn} is pointwise convergent. Find its pointwise limit.

Problem 4. Consider the sequence {fn} of functions defined on [0, ] by fn(x) = sinn(x). Show that {fn} converges pointwise. Find its pointwise limit. Using the above theorem, show that {fn} is not uniformly convergent.

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