This implies that
y
=
y
0
and
k
=
k
0
because the diﬀerence of the frac
tional parts has to equal zero for the expression to equal zero, uniqueness
follows.
Exercise 16.1.2
(a) If
f
∈
C(R/Z; C)
,
then
f
is bounded( i.e., there exists a real number
M >
0 such that

f
(
x
)
 ≤
M
for all
x
∈
R
.
Proof
Let us consider the closed interval [0
,
1], extending the result to the general
case follows immediately. For
±
= 1, we can choose a
δ
s.t

x

y
 ≤
δ
implies

f
(
x
)

f
(
y
)
 ≤
1
.
Now, we have to choose a suitable lower bound for
δ
, this
can be done by selecting an integer
z
s.t
δ
≥
1
z
. Next, we substitute
x
k
=
k
z
for
k
∈ {
1
,...z

1
}
.
For a suitable
x
∈
[0
,
1], there exists
k
∈ {
1
,...,z

1
}
such that

x

x
k
 ≤
1
z
≤
δ
. All that remains to show is that we can select
a suitable bound for
f
(
x
), this can be done as follows:

f
(
x
)
 ≤ 
f
(
x
)

f
(
x
k
)

+

f
(
x
k
)
 ≤
1 + max
1
≤
k
≤
z

1

f
(
x
k
)

(4)
As we can see, the upper bound is clearly independent of
x
. The claim
follows.
Exercise 16.1.2
(b)If
f,g
∈
C(R/Z;C)
, then the functions
f
+
g
,
f

g
, and
fg
are also in
C(R/Z;C)
. Also, if
c
is any complex number, then the function
cf
is also
in
C(R/Z;C)
.
Proof
Let
f,g
∈
C(R/Z;C)
be continuous and
x
∈
R/Z
. Now, consider
y
∈
R/Z
.
Then
f
(
x
) +
g
(
x
)

f
(
y
)

g
(
y
) = (
f
(
x
)

f
(
y
)) + (
g
(
x
)

g
(
y
))
(5)
Let
± >
0 be arbitrary, we can choose
δ
1
>
0 such that for all

x

y

< δ
1

(
f
(
x
)

f
(
y
))

<
±
2
,
(6)
similarly, we choose a
δ
2
>
0 such that for all

x

y

< δ
2
2