Math 318 HW #1 Solutions
1. Exercise 6.2.3. Consider the sequence of functions
x
hn (x) =
1 + xn
over the domain [0, ∞).
(a) Find the pointwise limit of (hn ) on [0, ∞).
Answer. Let h be the function


x
h(x) = 1/2


0
if 0 ≤ x < 1
if x = 1
if x > 1.
Then the claim is that hn → h pointwise. To see this, let > 0. There are three cases
to consider:
ln 0 ≤ x < 1: If x = 0, then hn (x) = 0 for all n. Otherwise, let N > ln
x , which is equivalent
N
+1
to assuming that x
< . Then, for any n ≥ N , we have that
|hn (x) − h(x)| =
x
−xn+1
xn+1
−
x
=
=
< xn+1 ≤ xN +1 < .
1 + xn
1 + xn
1 + xn
x = 1: Clearly, hn (1) = 1/2 for all n, so
|hn (1) − h(1)| = 0 < for any n.
x > 1: Let N >
ln(1/)
ln x
+ 1, which is equivalent to having
|hn (x) − h(x)| =
1
xN −1
< . Then for any n ≥ N ,
x
1
1
x
x
−0 =
< n = n−1 ≤ N −1 < .
n
n
1+x
1+x
x
x
x
In all three cases, there exists N ∈ N such that n ≥ N implies that
|hn (x) − h(x)| < .
Since the choice of > 0 was arbitrary, we conclude that hn → h pointwise.
(b) Explain how we know that the convergence cannot be uniform on [0, ∞).
Answer. Each hn is continuous, whereas the function h is clearly not continuous at x = 1.
If the convergence were uniform, then the function h would have to be continuous by
Theorem 6.2.6. Since that isn’t the case, we know the convergence cannot be uniform.
(c) Choose a smaller set over which the convergence is uniform and supply an argument to
show that this is indeed the case.
Answer. Pick r > 1. Then I claim that hn → h uniformly on [r, ∞). To see this, first
note that h(x) = 0 on [r, ∞). Now, let > 0 and choose N > ln(1/)
ln r + 1. Then for all
n ≥ N we know that
|hn (x) − 0| =
x
1
1
1
< n−1 ≤ n−1 ≤ N −1 < .
n
1+x
x
r
r
Since the choice of > 0 was arbitrary, we conclude that hn → 0 uniformly on [r, ∞).
1
2. Exercise 6.2.6. Using the Cauchy Criterion for convergent sequences of real numbers (Theorem
2.6.4), supply a proof for Theorem 6.2.5. (First, define a candidate for f (x), and then argue
that fn → f uniformly.)
Proof. (⇒) Suppose (fn ) converges uniformly on A to some function f . Let > 0. Then, by
definition of uniform convergence, there exists N ∈ N such that n ≥ N implies that
|fn (x) − f (x)| < /2
for any x ∈ A.
In particular, if n, m ≥ N , then
|fn (x) − fm (x)| = |fn (x) − f (x) + f (x) − fm (x)| ≤ |fn (x) − f (x)| + |f (x) − fm (x)|
by the triangle inequality, but then the fact that n, m ≥ N implies that the right hand side
is less than /2 + /2 = . In other words, n, m ≥ N implies that
|fn (x) − fm (x)| < for any x ∈ A, which was the desired conclusion.
(⇐) Suppose, on the other hand, that for every > 0 there exists N ∈ N such that |fn (x) −
fm (x)| < for all m, n ≥ N and all x ∈ A.
Then for each x ∈ A the sequence of real numbers (fn (x)) is Cauchy and, hence, converges
to some number limn→∞ fn (x). This is true for all x ∈ A, so we can define the function
f (x) = lim fn (x);
n→∞
the claim is that fn → f uniformly.
Let > 0. Then there exists N ∈ N such that n, m ≥ N implies that |fn (x) − fm (x)| < /2 for
all x ∈ A. Moreover, for any x ∈ A, since the sequence (fn (x)) converges to the number f (x),
we know that there exists Nx ∈ N such that Nx ≥ N and k ≥ Nx implies |fk (x) − f (x)| < /2.
Hence, if n ≥ N , we see that
|fn (x) − f (x)| = |fn (x) − fNx (x) + fNx (x) − f (x)| ≤ |fn (x) − fNx (x)| + |fNx (x) − f (x)|
by the triangle inequality. The first term on the right hand side is less than /2 since n, Nx ≥
N , and the second is less than /2 since Nx ≥ Nx . Therefore, we see that if n ≥ N , then
|fn (x) − f (x)| < /2 + /2 = for all x ∈ A. Since our choice of > 0 was arbitrary, we can conclude that fn → f uniformly
on A.
3. Exercise 6.2.8. Decide which of the following conjectures are true and which are false. Supply
a proof for those that are valid and a counterexample for each one that is not.
2
(a) If fn → f pointwise on a compact set K, then fn → f uniformly on K.
Answer. False. We’ve already seen that the sequence given by fn (x) = xn converges
pointwise but not uniformly on the compact set [0, 1] to the function
(
0 if x < 1
f (x) =
1 if x = 1.
(b) If fn → f uniformly on A and g is a bounded function on A, then fn g → f g uniformly
on A.
Answer. True. Suppose |g(x)| ≤ M for all x ∈ A. Let > 0. Then by hypothesis there
exists N ∈ N such that n ≥ N implies that |fn (x) − f (x)| < /M for all x ∈ A. Then
n ≥ N implies that
|fn (x)g(x) − f (x)g(x)| = |g(x)||fn (x) − f (x)| ≤ M |fn (x) − f (x)| < M /M = for all x ∈ A. Since the choice of > 0 was arbitrary, we conclude that fn g → f g
uniformly on A.
(c) If fn → f uniformly on A, and if each fn is bounded on A, then f must also be bounded.
Answer. True. Since fn → f uniformly, there exists N ∈ N such that n ≥ N implies
that |fn (x) − f (x)| < 1 for all x ∈ A. If M is an upper bound for the function fN
(meaning that |fN (x)| < M for all x ∈ A), then |fN (x) − f (x)| < 1 implies that
|f (x)| < |fN (x)| + 1 ≤ M + 1
for all x ∈ A, so we see that f is bounded.
(d) If fn → f uniformly on a set A, and if fn → f uniformly on a set B, then fn → f
uniformly on A ∪ B.
Answer. True. Let > 0. Then fn → f on A implies that there exists N1 ∈ N such that
n ≥ N1 implies that
|fn (x) − f (x)| < for all x ∈ A. Also, fn → f uniformly on B implies that there exists N2 ∈ N such that
n ≥ N2 implies that
|fn (x) − f (x)| < for all x ∈ B. Therefore, if N = max{N1 , N2 }, then n ≥ N implies that
|fn (x) − f (x)| < for all x ∈ A ∪ B.
(e) If fn → f uniformly on an interval, and if each fn is increasing, then f is also increasing.
Answer. True. Suppose x and y are in the interval with x < y. Then, for each n,
fn (x) ≤ fn (y). Therefore,
f (x) = lim fn (x) ≤ lim fn (y) = f (y),
n→∞
n→∞
where the inequality follows from the Order Limit Theorem.
3
(f) Repeat conjecture (e) assuming only pointwise convergence.
Answer. It’s still true. Nowhere in the proof did I use uniform convergence.
4. Exercise 6.2.13 (Cantor Function). Review the construction of the Cantor set C ⊆ [0, 1] from
Section 3.1. This exercise makes use of results and notation from this discussion.
(a) Define f0 (x) = x for all x ∈ [0, 1]. Now, let


(3/2)x
f1 (x) = 1/2


(3/2)x − 1/2
for 0 ≤ x ≤ 1/3
for 1/3 < x < 2/3
for 2/3 ≤ x ≤ 1.
Sketch f0 and f1 over [0, 1] and observe that f1 is continuous, increasing, and constant
on the middle third (1/3, 2/3) = [0, 1]\C1 .
Answer. f1 is clearly continuous away from 1/3 and 2/3. At 1/3 we have that
lim f1 (x) =
x→1/3−
lim (3/2)x = 1/2 = f1 (1/3)
x→1/3−
and
lim f1 (x) =
x→1/3+
lim 1/2 = 1/2 = f1 (1/3),
x→1/3+
so indeed f1 is continuous at 1/3. A similar argument shows that f1 is continuous at
2/3.
Since each of the three pieces of f1 are increasing functions, the fact that f1 is continuous
implies that, in fact, f1 is increasing. Finally, f1 is defined to be constant on (1/3, 2/3).
(b) Construct f2 by imitating this process of flattening out the middle third of each nonconstant segment of f1 . Specifically, let


for 0 ≤ x ≤ 1/3
(1/2)f1 (3x)
f2 (x) = f1 (x)
for 1/3 < x < 2/3


(1/2)f1 (3x − 2) + 1/2 for 2/3 ≤ x ≤ 1.
If we continue this process, show that the resulting sequence (fn ) converges uniformly
on [0, 1].
Proof. My goal is to show by induction that for all n, |fn+1 (x) − fn (x)| ≤ 6·21 n for all
x ∈ [0, 1]. If this is true, then a suitable modification of the solution to Midterm #1,
Problem 5 from last semester will yield the desired result.
Base Case: Clearly,


if 0 ≤ x ≤ 1/3
|(3/2)x − x|
|f1 (x) − f0 (x)| = |1/2 − x|
if 1/3 < x < 2/3


|(3/2)x − 1/2 − x| if 2/3 ≤ x ≤ 1


if 0 ≤ x ≤ 1/3
|x/2|
= |1/2 − x|
if 1/3 < x < 2/3


|x/2 − 1/2| if 2/3 ≤ x ≤ 1
≤ 1/6,
4
as desired.
Inductive Hypothesis: Suppose |fn (x) − fn−1 (x)| ≤ 6·21n−1 . Then


|(1/2)fn (3x) − (1/2)fn−1 (3x)|
|fn+1 (x) − fn (x)| = |fn (x) − fn−1 (x)|


|(1/2)fn (3x − 2) + 1/2 − ((1/2)fn−1 (3x − 2) + 1/2)|


if 0 ≤ x ≤ 1/3
1/2|fn (3x) − fn−1 (3x)|
= 0
if 1/3 < x < 2/3


1/2|fn (3x − 2) − fn−1 (3x − 2) if 2/3 ≤ x ≤ 1
1
1
≤ ·
2 6 · 2n−1
1
=
.
6 · 2n
Therefore, we conclude, by induction, that |fn+1 (x) − fn (x)| ≤
result follows.
1
6·2n ,
if 0 ≤ x ≤ 1/3
if 1/3 < x < 2/3
if 2/3 ≤ x ≤ 1
and the desired
(c) Let f = lim fn . Prove that f is a continuous, increasing function on [0, 1] with f (0) = 0
and f (1) = 1 that satisfies f 0 (x) = 0 for all x in the open set [0, 1]\C. Recall that the
“length” of the Cantor set C is 0. Somehow, f manages to increase from 0 to 1 while
remaining constant on a set of “length 1”.
Proof. Since each fn is continuous and since fn → f uniformly, Theorem 6.2.6 implies
that f is continuous. Since each fn is increasing, Problem 3(e) above implies that f is
increasing. Since fn (0) = 0 and fn (1) = 1 for all n, it must be the case that f (0) = 0
and f (1) = 1.
Finally, the fact that f 0 (x) = 0 for all x ∈ [0, 1]\C is a corollary of the following claim:
Claim: If I is an interval contained in [0, 1]\C, then f is constant on I.
Proof. Since each Cn is defined by deleting the “middle thirds” of the intervals that
make up Cn−1 and since I is connected, it must be the case that I is contained in the
“middle third” of some interval in CN for some N ∈ N. Then since fN +1 is defined to
be constant on the middle thirds of the intervals making up CN , we know that fN +1 is
constant on an interval containing I.
Moreover, for all n ≥ N + 1,
fn (x) = fn−1 (x) = · · · = fN +1 (x)
for all x ∈ I, so we see that fn is constant on I. In turn, this implies that f is constant
on I.
5. Exercise 6.2.15. A sequence of functions (fn ) defined on a set E ⊆ R is called equicontinuous
if for every > 0 there exists a δ > 0 such that |fn (x)−fn (y)| < for all n ∈ N and |x−y| < δ
in E.
5
(a) What is the difference between saying that a sequence of functions (fn ) is equicontinuous
and just asserting that each fn in the sequence is individually uniformly continuous?
Answer. If each fn is uniformly continuous, then, for each n, if > 0 then there exists
δ > 0 such that |x − y| < δ implies that |fn (x) − fn (y)| < , which looks similar to
the definition of equicontinuity. However, in the above situation, the number δ depends
not only on , but also potentially on n. On the other hand, to satisfy the definition of
equicontinuity, δ could only depend on .
Stated another way, for a sequence of functions to be equicontinuous, it must be the
case that not only is each function in the sequence uniformly continuous, but that, for
a given > 0, the same choice of δ > 0 works for all the functions in the sequence.
(b) Give a qualitative explanation for why the sequence gn (x) = xn is not equicontinuous
on [0, 1]. Is each gn uniformly continuous on [0, 1]?
Answer. Each gn is uniformly continuous on [0, 1] (since each gn is continuous and [0, 1]
is compact), but the sequence is not equicontinuous. Heuristically, this is because the
maximum slope of the graph of gn gets arbitrarily large as n gets large. The larger
the maximum slope of gn , the closer together I can pick two numbers x and y while
maintaining some fixed distance between gn (x) and gn (y).
(More rigorously, for any δ > 0, if |x − 1| < δ and n >
be bigger than 1/2.)
6
ln(1/2)
ln x ,
then |gn (x) − gn (1)| will