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
n→∞lim fn(x) exists for each point x in D.
This means that lim
n→∞fn(x) is a real number that depends only on x.
If {fn} is pointwise convergent then the function defined by f (x) = lim
n→∞fn(x), 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
n→∞fn(x) = ∞ 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 fn(x) = nx + x2
n2 for all x in R.
Show that {fn} converges pointwise.
Solution: For every real number x, we have:
n→∞lim fn(x) = lim
n→∞
x n +x2
n2 = x
n→∞lim 1 n
+ x2
n→∞lim 1 n2
= 0 + 0 = 0
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
√−1
n + 1 ≤ sin(nx + 3)
√n + 1 ≤ 1
√n + 1 Moreover,
n→∞lim
√ 1
n + 1 = 0.
Applying the squeeze theorem for sequences, we obtain that
n→∞lim fn(x) = 0 for all x in R.
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
n→∞lim fn(x) = 0 for 0 < x < 1 Finally, fn(1) = n2 for all n. So, lim
n→∞fn(1) = ∞. Therefore, {fn} is not 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.
It follows that
n→∞lim(cos(x))n= 0 for x 6= 0.
Moreover, since fn(0) = 1 for all n in N, one gets lim
n→∞fn(0) = 1. Therefore, {fn} converges pointwise to the function f defined by
f (x) = 0 if −π2 ≤ x < 0 or 0 < x ≤ π2
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
n→∞lim fn(x) = lim
n→∞nxen ln(1−x) = x lim
n→∞nen 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 {fn} be the sequence of functions on R defined by fn(x) = n3 if 0 < x ≤ n1
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,
n→∞lim fn(x) = 1 for all x.
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.
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 con- verse 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 + n2x2.
This function converges pointwise to zero. Indeed, (1 + n2x2) ∼ n2x2 as n gets larger and larger. So,
n→∞lim fn(x) = lim
n→∞
nx n2x2 = 1
x lim
n→∞
1 n = 0.
But for any ε < 1/2, we have
fn 1 n
− f 1 n
= 1
2− 0 > ε.
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
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)n1 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.
