MAT 337 Problem Set 4 Sample Solutions
Ian Zwiers  March 25, 2009
This document is for demonstration purposes; it is not a marking key.
5.5(I)
Let
f
be a continuous function on
(0
,
1]
. Show that
f
is uniformly continuous if and only if
lim
x
→
0
+
f
(
x
)
exists.
Note:
We assume that
f
: (0
,
1]
7→
R
M
, as that matches Deﬁnition 5.5.1. It is essential that
the range lives inside a complete space.
(
⇒
) Let
f
be uniformly continuous. Suppose lim
x
→
0
+
f
(
x
) does not exist. Then
∃
sequence
a
n
>
0 with lim
n
→∞
a
n
= 0 such that lim
n
→∞
f
(
a
n
) does not exist.
Let
± >
0 be arbitrary. By uniform continuity,
∃
δ >
0 such that

f
(
x
)

f
(
y
)

< ±
whenever

x

y

< δ
(for
x,y
∈
(0
,
1]). Since lim
n
→∞
a
n
exists, it is Cauchy, so there exists
N
such
that

a
n

a
m

< δ
for all
n,m > N
. But then

f
(
a
n
)

f
(
a
m
)

< ±
for all
n,m > N
, which
proves that (
f
(
a
n
))
∞
n
=1
is Cauchy. By the completeness of
R
M
, lim
n
→∞
f
(
a
n
) exists. By
contradiction, our supposition is false and lim
x
→
0
+
f
(
x
) exists.
(
⇐
) Let lim
x
→
0
+
f
(
x
) exist. Then we may deﬁne
g
: [0
,
1]
7→
R
M
as an extension of
f
:
g
(
y
) =
(
lim
x
→
0
+
f
(
x
)
if
y
= 0
f
(
y
)
if
y
∈
(0
,
1]
For
y
= 0,
g
is continuous by deﬁnition. For
y
∈
(0
,
1],
g
is continuous since
f
is continuous.
That is,
g
is continuous on [0
,
1], which is a compact set. By Theorem 5.5.9,
g
is uniformly
continuous.
Given
± >
0, there exists
δ >
0 such that

g
(
x
)

g
(
y
)

< ±
whenever
x,y
∈
[0
,
1] with

x

y

< δ
. For
x,y
∈
(0
,
1] with

x

y

< δ
,

f
(
x
)

f
(
y
)

=

g
(
x
)

g
(
y
)

< ±
, which proves
that
f
is uniformly continuous.
5.6(H) (a)
Show that a continuous function on
(
∞
,
+
∞
)
cannot take every real value exactly
twice.
Suppose that