3
FOURIER ANALYSIS
52
3.
y
=
(
sin
1
x
x
6
= 0
c
x
= 0
is not of bounded variation on any interval containing 0.
Notation.
Let
V
f
(
a, b
) = sup
P
P

f
(
x
k
)

f
(
x
k

1
)

denote the
total variation of
f
over [
a, b
]
.
Proposition.
If
f
is of bounded variation, then
f
is bounded.
Proof.
Consider the partition
{
a, x, b
}
.
We have that

f
(
x
)

f
(
a
)

+

f
(
b
)

f
(
x
)

V
f
(
a, b
).
Thus

f
(
x
)


f
(
a
)

+
V
f
(
a, b
).
Proposition.
If
f
and
g
are of bounded variation, then so are
f
±
g
.
Proof.
Exercise – Use triangle inequality.
Proposition.
Let
f
: [
a, b
]
!
R
be of bounded variation.
Then for
x
2
[
a, b
], put
V
(
x
) =
V
f
(
a, x
) (note that
V
(
a
) = 0), and let
D
(
x
) =
V
(
x
)

f
(
x
). Then
V
and
D
are both increasing functions.
Proof.
Let
a
x < y
b
. We first prove that
V
f
(
a, y
) =
V
f
(
a, x
) +
V
f
(
x, y
). Take a partition
a
=
x
0
< x
1
<
· · ·
<
x
n
=
x
=
y
0
< y
1
<
· · ·
< y
m
=
y
. Then
V
f
(
a, y
)
≥
n
X
k
=1

f
(
x
k
)

f
(
x
k

1
)

+
m
X
j
=1

f
(
y
j
)

f
(
y
j

1
)

This is true for all partitions of [
a, x
] and [
x, y
], so
V
f
(
a, y
)
≥
V
f
(
a, x
) +
V
f
(
x, y
). Now take any partition
P
of [
a, y
].
Let
P
0
be the refined partition that includes
x
. Notice that
X
P
Δ
f
X
P
0
Δ
f
V
f
(
a, x
) +
V
f
(
x, y
)
Since this is true for any partition
P
,
V
f
(
a, y
)
V
f
(
a, x
) +
V
f
(
x, y
).
Now consider
V
(
y
)

V
(
x
) =
V
f
(
a, y
)

V
f
(
a, x
) =
V
f
(
x, y
)
≥
0, and thus
V
(
y
)
≥
V
(
x
) when
y
≥
x
. Also
D
(
y
)

D
(
x
) =
V
(
y
)

f
(
y
)

[
V
(
x
)

f
(
x
)] =
V
f
(
x, y
)

[
f
(
y
)

f
(
x
)]
≥

f
(
y
)

f
(
x
)


[
f
(
y
)

f
(
x
)]
≥
0
so
D
(
y
)
≥
D
(
x
) whenever
y
≥
x
.
Corollary.
f
is of bounded variation if and only if
f
is a di
↵
erence of increasing functions.
Proof.
If
f
is of bounded variation, then
f
=
V

D
where
V, D
are increasing. Conversely if
f
=
g

h
where
g
and
h
are increasing functions, then
g
and
h
are of bounded variation, so
f
=
g

h
is of bounded variation.
Corollary.
If
f
is of bounded variation, then
σ
n
f
(
t
0
) converges.
Proof.
Write
f
=
V

D
where
V
and
D
are increasing.
V
and
D
only have jump discontinuities, so
f
only has jump
discontinuities. Thus
f
(
t
+
0
) and
f
(
t

0
) exist. Thus
σ
n
f
(
t
0
)
!
f
(
t
+
0
)+
f
(
t

0
)
2
by Fejer’s theorem.
Note.
In fact, if
f
is continuous then
V
and
D
are continuous.