8/7/2019 M401-notes1

1/5

MATH 401 - NOTES

Sequences of functionsPointwise and Uniform ConvergenceFall 2005

Previously, we have studied sequences of real numbers . Now we discussthe topic of sequences of real valued functions . A sequence of functions {f n }is a list of functions (f 1, f 2, . . .) such that each f n maps a given subset D of R into R .

I. Pointwise convergence

Denition. Let D be a subset of R and let {f n }be a sequence of functionsdened on D . We say that {f n }converges pointwise on D if lim

n f n (x) exists for each point x in D.

This means that limn

f n (x) is a real number that depends only on x.

If {f n }is pointwise convergent then the function dened by f (x) = limn f n (x),for every x in D, is called the pointwise limit of the sequence {f n }.Example 1. Let {f n }be the sequence of functions on R dened by f n (x) =nx . This sequence does not converge pointwise on R because limn f n (x) = for any x > 0.Example 2.Let {f n }be the sequence of functions on R dened by f n (x) = x/n . Thissequence converges pointwise to the zero function on R .Example 3. Consider the sequence {f n }of functions dened by

f n (x) =nx + x2

n2for all x in R .

Show that {f n }converges pointwise.Solution: For every real number x, we have:

limn

f n (x) = limn

xn

+x2

n2= x lim

n

1n

+ x2 limn

1n2

= 0 + 0 = 0

1

8/7/2019 M401-notes1

2/5

Thus, {f n }converges pointwise to the zero function on R .Example 4. Consider the sequence {f n }of functions dened by

f n (x) =sin(nx + 3)n + 1 for all x in R .

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

1n + 1 sin(nx + 3)n + 1

1n + 1

Moreover,lim

n

1n + 1 = 0 .

Applying the squeeze theorem for sequences, we obtain that

limn

f n (x) = 0 for all x in R .

Therefore, {f n }converges pointwise to the function f 0 on R .Example 5. Consider the sequence {f n }of functions dened by f n (x) =n

2x

n

for 0 x 1. Determine whether {f n

}is pointwise convergent.Solution: First of all, observe that f n (0) = 0 for every n in N . So thesequence {f n (0)}is constant and converges to zero. Now suppose 0 < x < 1then n2x n = n2en ln( x ) . But ln( x) < 0 when 0 < x < 1, it follows that

limn

f n (x) = 0 for 0 < x < 1

Finally, f n (1) = n2 for all n. So, limn

f n (1) = . Therefore, {f n }is notpointwise convergent on [0 , 1].

Example 6. Let {f n

}be the sequence of functions dened by f n (x) = cos

n

(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

8/7/2019 M401-notes1

3/5

It follows that

limn (cos(x))n

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

n f n (0) = 1. Therefore,

{f n }converges pointwise to the function f dened byf (x) = 0 if

2 x < 0 or 0 < x

21 if x = 0

Example 7. Consider the sequence of functions dened by

f n (x) = nx (1 x)n on [0, 1].

Show that {f n }converges pointwise to the zero function.Solution: Note that f n (0) = f n (1) = 0, for all n

N . Now suppose0 < x < 1, then

limn

f n (x) = limn

nxe n ln(1 x ) = x limn

ne n ln(1 x ) = 0

because ln(1 x) < 0 when 0 < x < 1. Therefore, the given sequence con-verges pointwise to zero.Example 8. Let {f n }be the sequence of functions on R dened by

f n (x) = n3 if 0 < x 1n1 otherwise

Show that {f n }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 doesnot belong to the interval (0 , 1/N ). The intervals (0 , 1/n ) get smaller asn . Therefore, f n (x) = 1 for all n > N . Hence,

limn

f n (x) = 1 for all x.

The formal denition of pointwise convergenceLet D be a subset of R and let {f n }be a sequence of real valued functionsdened on D. Then

{f n

}converges pointwise to f if given any x in D and

given any > 0, there exists a natural number N = N (x, ) such that

|f n (x) f (x)| < for every n > N.Note: The notation N = N (x, ) means that the natural number N dependson the choice of x and .

3

8/7/2019 M401-notes1

4/5

I. Uniform convergence

Denition. Let D be a subset of R and let {f n }be a sequence of realvalued functions dened on D. Then {f n }converges uniformly to f if givenany > 0, there exists a natural number N = N () such that

|f n (x) f (x)| < for every n > N and for every x in D.Note: In the above denition the natural number N depends only on .Therefore, uniform convergence implies pointwise convergence. But the con-verse is false as we can see from the following counter-example.

Example 9. Let

{f n

}be the sequence of functions on (0,

) dened by

f n (x) =nx

1 + n2x2.

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

gets larger and larger. So,

limn

f n (x) = limn

nxn2x2

=1x

limn

1n

= 0 .

But for any < 1/ 2, we have

f n1n f

1n =

12 0 > .

Hence {f n }is not uniformly convergent.

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

4

8/7/2019 M401-notes1

5/5

Homework

Problem 1.Let {f n }be the sequence of functions on [0, 1] dened by f n (x) = nx (1x4)n .Show that {f n }converges pointwise. Find its pointwise limit.Problem 2.Is the sequence of functions on [0, 1) dened by f n (x) = (1 x)

1

n pointwiseconvergent? Justify your answer.

Problem 3.Consider the sequence

{f n

}of functions dened by

f n (x) =n + cos( nx )

2n + 1for all x in R .

Show that {f n }is pointwise convergent. Find its pointwise limit.Problem 4.Consider the sequence {f n }of functions dened on [0, ] by f n (x) = sin

n (x).Show that {f n }converges pointwise. Find its pointwise limit. Using theabove theorem, show that {f n }is not uniformly convergent.

5