Math 3110 Homework 9 Solutions
Exercise 10.1.4
We can use the triangle inequality and that

cos(
x
)
 ≤
1 for all
x
to get
k
n
=1
a
n
cos(
nx
)
≤
k
n
=1

a
n
cos(
nx
)
 ≤
k
n
=1

a
n

,
so the function is bounded by
k
n
=1

a
n

.
Exercise 10.1.9
(a)
True. If
f
is bounded, then there is some
M
such that

f
(
x
)
 ≤
M
for all
x
. In particular,
for every
y
, if
x
=
g
(
y
), we get

f
(
g
(
y
))

=

f
(
x
)
 ≤
M
, so
f
(
g
(
x
)) is bounded.
(b)
False. Consider
f
(
x
) = sin(
x
) and
g
(
x
) =
1
x
if
x
= 0 and
g
(0) = 0. Pick
M >
0. If
x
=
1
M
,
then since sin(
x
)
< x
, we get
g
(
f
(
x
)) =
1
sin(
x
)
>
1
x
=
M
, so
M
is not an upper bound for
g
(
f
(
x
)). Since this holds for all
M >
0,
g
(
f
(
x
)) is not bounded above.
Exercise 10.3.2
Let the period of
f
be
c
. We are given that there is some
q
such that
f
(
x
) is bounded for all
x
≥
q
. Then there is some
M
such that

f
(
x
)
 ≤
M
for all
x
≥
q
. So for any
x
, since
f
(
x
)
has period
c
,

f
(
x
)

=
f
x
+
q

x
c
c
≤
M,
because
x
+
q

x
c
c
≥
x
+
q

x
c
c
=
q.
This means
f
(
x
) is bounded by
M
for all
x
.
1
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Problem 102
As the suggestion recommends, define
S
by
S
=
{
a <
1

f
(
x
) is constant on [0
, a
)
}
.
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 '08
 RAMAKRISHNA
 Math, Order theory, Metric space, upper bound, sup

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