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
n2
Show that {fn } converges pointwise.
for all x in R.
fn (x) =
Solution: For every real number x, we have:
x x2
1
1
2
lim fn (x) = lim + 2 = x lim
+x
lim
=0+0=0
n→∞
n→∞ n
n→∞ n
n→∞ n2
n
1
Thus, {fn } converges pointwise to the zero function on R.
Example 4. Consider the sequence {fn } of functions defined by
fn (x) =
sin(nx + 3)
√
n+1
for all x in R.
Show that {fn } converges pointwise.
Solution: For every x in R, we have
√
sin(nx + 3)
1
−1
≤ √
≤√
n+1
n+1
n+1
Moreover,
1
= 0.
n→∞
n+1
Applying the squeeze theorem for sequences, we obtain that
lim √
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) =
n2 xn 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 n2 xn = n2 en 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 −π/2 ≤ x ≤ π/2. Discuss the pointwise convergence of the sequence.
Solution: For −π/2 ≤ x < 0 and for 0 < x ≤ π/2, we have
0 ≤ cos(x) < 1.
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It follows that
lim (cos(x))n = 0
n→∞
for x 6= 0.
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
0 if − π2 ≤ x < 0 or 0 < x ≤
f (x) =
1 if
x=0
π
2
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
3
n
if
0 < x ≤ n1
fn (x) =
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 ε > 0, there exists a natural number N = N (x, ε) such that
|fn (x) − f (x)| < ε for every n > N.
Note: The notation N = N (x, ε) means that the natural number N depends
on the choice of x and ε.
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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 ε > 0, there exists a natural number N = N (ε) such that
|fn (x) − f (x)| < ε for every n > N
and for every x in D.
Note: In the above definition the natural number N depends only on ε.
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
fn (x) =
nx
.
1 + n2 x2
This function converges pointwise to zero. Indeed, (1 + n2 x2 ) ∼ n2 x2 as n
gets larger and larger. So,
1
1
nx
=
lim
= 0.
n→∞ n2 x2
x n→∞ n
lim fn (x) = lim
n→∞
But for any ε < 1/2, we have
fn 1 − f 1 = 1 − 0 > ε.
n
n 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.
1
Is the sequence of functions on [0, 1) defined by fn (x) = (1 − x) n pointwise
convergent? Justify your answer.
Problem 3.
Consider the sequence {fn } of functions defined by
fn (x) =
n + cos(nx)
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, π] 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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