Ex2-2

**Solution to Chapter 2, Exercise 2**

Theorem 2.1 (i)

If

f

{\displaystyle f}

is of bounded variation on

[
a
,
b
]

{\displaystyle [a, b]}

, then

f

{\displaystyle f}

 is bounded on

[
a
,
b
]

{\displaystyle [a, b]}

.

Proof

It can be proved by taking contrapositive. We need to consider two cases. Case 1: Both

f
(
a
)

<
∞

{\displaystyle |f(a)|<\infty }

 and

f
(
b
)

<
∞

{\displaystyle |f(b)|<\infty }

; Case 2:

f
(
a
)

=
∞

{\displaystyle |f(a)|=\infty }

 or

f
(
b
)

=
∞

{\displaystyle |f(b)|=\infty }

.

Case1 : Both

f
(
a
)

<
∞

{\displaystyle |f(a)|<\infty }

 and

f
(
b
)

<
∞

{\displaystyle |f(b)|<\infty }

.

Assume that

f

{\displaystyle f}

 is unbounded on

[
a
,
b
]

{\displaystyle [a, b]}

, then there exists a sequence

{

c

j

}
⊂
[
a
,
b
]

{\displaystyle \{c\_{j}\}\subset [a,b]}

 such that

f
(

c

j

)

>
j

{\displaystyle |f(c\_{j})|>j}

. Now consider a partition

Γ

j

=
{
a
,

c

j

,
b
}

{\displaystyle \Gamma \_{j}=\{a,c\_{j},b\}}

. Then we approximate the variation by

S

Γ

j

=

∑

# k
1

2

f
(

x

k

)
−
f
(

x

k
−
1

)

=
f
(

c

j

)
−
f
(
a
)
f
(
b
)
−
f
(

c

j

)

=
f
(

c

j

)
−
f
(
a
)
f
(

c

j

)
−
f
(
b
)

≥
f
(

c

j

)
−
f
(
a
)
f
(

c

j

)
−
f
(
b
)

≥
2
f
(

c

j

)
−
f
(
a
)
−
f
(
b
)

≥
2
j
−
f
(
a
)
−
f
(
b
)
.

{\displaystyle {\begin{aligned}S\_{\Gamma \_{j}}&=\sum \limits \_{k=1}^{2}\vert f(x\_{k})-f(x\_{k-1})\vert \\&=\vert f(c\_{j})-f(a)\vert +\vert f(b)-f(c\_{j})\vert \\&=\vert f(c\_{j})-f(a)\vert +\vert f(c\_{j})-f(b)\vert \\&\geq \vert f(c\_{j})\vert -\vert f(a)\vert +\vert f(c\_{j})\vert -\vert f(b)\vert \\&\geq 2\vert f(c\_{j})\vert -\vert f(a)\vert -\vert f(b)\vert \\&\geq 2j-\vert f(a)\vert -\vert f(b)\vert .\end{aligned}}}

Next we take the sup to find the variation

V
(
f
,
[
a
,
b
]
)

=

sup

Γ

S

Γ

≥

sup

j

S

Γ

j

≥

sup

j

{
2
j
−
f
(
a
)
−
f
(
b
)
}
.

{\displaystyle {\begin{aligned}V(f,[a,b])&=\sup \_{\Gamma }S\_{\Gamma }\\&\geq \sup \_{j}S\_{\Gamma \_{j}}\\&\geq \sup \_{j}\{2j-\vert f(a)\vert -\vert f(b)\vert \}.\end{aligned}}}

Since

j

{\displaystyle \displaystyle j}

 can be arbitrarily large, so

sup

j

{
2
j
−
f
(
a
)
−
f
(
b
)
}

{\displaystyle \sup \_{j}\{2j-\vert f(a)\vert -\vert f(b)\vert \}}

 is unbounded. Consequently,

V
(
f
,
[
a
,
b
]
)

{\displaystyle V(f,[a,b])}

 is unbounded. Under our assumption,

f

{\displaystyle f}

 is not of bounded variation.

Case 2:

f
(
a
)

=
∞

{\displaystyle |f(a)|=\infty }

 or

f
(
b
)

=
∞

{\displaystyle |f(b)|=\infty }

.

Similarly, there exists a sequence

{

c

j

}
⊂
[
a
,
b
]

{\displaystyle \{c\_{j}\}\subset [a,b]}

 such that

f
(

c

j

)

>
j

{\displaystyle |f(c\_{j})|>j}

 since f is unbounded. We choose a partition

# Γ
{
a
,

c

j

,
d
,
b
}

{\displaystyle \Gamma =\{a,c\_{j},d,b\}}

, where

d
∈
(
a
,
b
)

{\displaystyle d\in (a,b)}

 satisfies that

f
(
d
)

<
∞

{\displaystyle |f(d)|<\infty }

.

Then, we calculate the variation by

S

Γ

j

=

∑

# k
1

3

f
(

x

k

)
−
f
(

x

k
−
1

)

=
f
(

c

j

)
−
f
(
a
)
f
(
d
)
−
f
(

c

j

)
f
(
b
)
−
f
(
d
)

≥
f
(
d
)
−
f
(

c

j

)

≥
f
(

c

j

)
−
f
(
d
)

{\displaystyle {\begin{aligned}S\_{\Gamma \_{j}}&=\sum \limits \_{k=1}^{3}\vert f(x\_{k})-f(x\_{k-1})\vert \\&=\vert f(c\_{j})-f(a)\vert +\vert f(d)-f(c\_{j})\vert +\vert f(b)-f(d)\vert \\&\geq \vert f(d)-f(c\_{j})\vert \\&\geq \vert f(c\_{j})\vert -\vert f(d)\vert \end{aligned}}}

Next we take the sup to find the variation

V
(
f
,
[
a
,
b
]
)

=

sup

Γ

S

Γ

≥

sup

j

S

Γ

j

≥

sup

j

(
f
(

c

j

)
−
f
(
d
)
)
.

{\displaystyle {\begin{aligned}V(f,[a,b])&=\sup \_{\Gamma }S\_{\Gamma }\\&\geq \sup \_{j}S\_{\Gamma \_{j}}\\&\geq \sup \_{j}(\vert f(c\_{j})\vert -\vert f(d)\vert ).\end{aligned}}}

Since

j

{\displaystyle \displaystyle j}

 can be arbitrarily large,

V
(
f
,
[
a
,
b
]
)

{\displaystyle V(f,[a,b])}

 is unbounded in this case, which means

f

{\displaystyle f}

 is not of bounded variation.

Another proof of Theorem 2.1 (i)

Let

V
(
f
;
[
a
,
b
]
# )

sup

Γ

S

Γ

≤
M

{\displaystyle V(f;[a,b])=\sup \_{\Gamma }S\_{\Gamma }\leq M}

. Let

x
∈
(
a
,
b
)

{\displaystyle x\in(a,b)}

.

Consider the partition

# Γ
{
a
,
x
,
b
}

{\displaystyle \Gamma =\{a,x,b\}}

.

Then

f
(
x
)
−
f
(
a
)

≤

f
(
x
)
−
f
(
a
)

f
(
b
)
−
f
(
x
)

=

S

Γ

≤

sup

Γ

S

Γ

≤
M

.

This implies

f
(
a
)
−
M
≤
f
(
x
)
≤
f
(
a
)
M

{\displaystyle f(a)-M\leq f(x)\leq f(a)+M}

, so

f

{\displaystyle f}

 is bounded.

Proof of Theorem 2.1 (ii)

Suppose

f

{\displaystyle f}

and

g

{\displaystyle g}

are of bounded variation on

[
a
,
b
]

{\displaystyle [a, b]}

and

# Γ
{

x
0

,

x
1

,
.
.
.
,

x
m

}

{\displaystyle \Gamma=\{x\_0,x\_1,...,x\_m\}}

is a partition of

[
a
,
b
]

{\displaystyle [a, b]}

. Since

S

Γ

[
c
f
;
a
,
b
]

=

∑

# i
1

m

c
f
(

x

i

)
−
c
f
(

x

i
−
1

)

=

c

∑

# i
1

m

f
(

x

i

)
−
f
(

x

i
−
1

)

=

c

S

Γ

[
f
;
a
,
b
]
,

we have

V
[
c
f
;
a
,
b
]

=

sup

Γ

S

Γ

[
c
f
;
a
,
b
]

=

c

V
[
f
;
a
,
b
]
.

{\displaystyle {\begin{aligned}V[cf;a,b]&=\sup \_{\Gamma }{S\_{\Gamma }[cf;a,b]}\\&=\left|c\right|V[f;a,b].\end{aligned}}}

Therefore, if

V
[
f
;
a
,
b
]
<
∞

{\displaystyle V[f; a , b ] < +\infty
}

, then

V
[
c
f
;
a
,
b
]
<
∞

{\displaystyle V[cf;a,b]<+\infty }

.
Observe that

S

Γ

[
f
g
;
a
,
b
]

=

∑

# i
1

m

f
(

x

i

)
g
(

x

i

)
−
f
(

x

i
−
1

)
−
g
(

x

i
−
1

)

≤

∑

# i
1

m

f
(

x

i

)
−
f
(

x

i
−
1

)

∑

# i
1

m

g
(

x

i

)
−
g
(

x

i
−
1

)

=

S

Γ

[
f
;
a
,
b
]

S

Γ

[
g
;
a
,
b
]

≤
V
[
f
;
a
,
b
]
V
[
g
;
a
,
b
]
.

Thus,

V
[
f
g
;
a
,
b
# ]

sup

Γ

S

Γ

[
f
g
;
a
,
b
]

≤
V
[
f
;
a
,
b
]
V
[
g
;
a
,
b
]
.

{\displaystyle V[f+g;a,b]=\sup \_{\Gamma }{S\_{\Gamma }[f+g;a,b]}\leq V[f;a,b]+V[g;a,b].}

Therefore,

V
[
f
g
;
a
,
b
]
<
∞

{\displaystyle V[f+g;a,b]<+\infty }

as well.
Since

f

{\displaystyle f}

and

g

{\displaystyle g}

are of bounded variation, by Theorem 2.1(i) they are bounded as well. For some

M
,
N
>
0

{\displaystyle M,N>0}

,

f
(
x
)

≤
M

{\displaystyle |f(x)|\leq M}

 and

g
(
x
)

≤
N

{\displaystyle |g(x)|\leq N}

for all

x

{\displaystyle x}

. Then

S

Γ

[
f
g
;
a
,
b
]

=

∑

# i
1

m

f
(

x

i

)
g
(

x

i

)
−
f
(

x

i
−
1

)
g
(

x

i
−
1

)

=

∑

# i
1

m

f
(

x

i

)
g
(

x

i

)
−
f
(

x

i

)
g
(

x

i
−
1

)
f
(

x

i

)
g
(

x

i
−
1

)
−
f
(

x

i
−
1

)
g
(

x

i
−
1

)

≤

∑

# i
1

m

f
(

x

i

)
g
(

x

i

)
−
f
(

x

i

)
g
(

x

i
−
1

)

∑

# i
1

m

f
(

x

i

)
g
(

x

i
−
1

)
−
f
(

x

i
−
1

)
g
(

x

i
−
1

)

=

∑

# i
1

m

f
(

x

i

)

g
(

x

i

)
−
g
(

x

i
−
1

)

∑

# i
1

m

g
(

x

i
−
1

)

f
(

x

i

)
−
f
(

x

i
−
1

)

≤
M

∑

# i
1

m

g
(

x

i

)
−
g
(

x

i
−
1

)

N

∑

# i
1

m

f
(

x

i

)
−
f
(

x

i
−
1

)

=
M

S

Γ

[
g
;
a
,
b
]
N

S

Γ

[
f
;
a
,
b
]

≤
M
V
[
g
;
a
,
b
]
N
V
[
f
;
a
,
b
]

Therefore,

V
[
f
g
;
a
,
b
]
≤
M
V
[
g
;
a
,
b
]
N
V
[
f
;
a
,
b
]
<
∞

{\displaystyle V[fg;a,b]\leq MV[g;a,b]+NV[f;a,b]<+\infty }

.
Suppose

∃
ϵ
>
0

{\displaystyle \exists \epsilon > 0 }

 such that

g
(
x
)

≥
ϵ

{\displaystyle |g(x)|\geq \epsilon }

 for all

x
∈
[
a
,
b
]

{\displaystyle x \in [a, b]}

. Let

# M
1

/

ϵ

{\displaystyle M=1/\epsilon }

. Then

1

g
(
x
)

≤

1
ϵ

=
M
.

{\displaystyle \left|{\frac {1}{g(x)}}\right|\leq {\frac {1}{\epsilon }}=M.}

It suffices to show that

1

/

g

{\displaystyle 1/g}

is of bounded variation. Then the result will follow since

f

/

g

{\displaystyle f/g}

is the product of

f

{\displaystyle f}

and

1

/

g

{\displaystyle 1/g}

. Notice that

S

Γ

[
1

/

g
;
a
,
b
]

=

∑

# i
1

m

1

g
(

x

i

)

−

1

g
(

x

i
−
1

)

=

∑

# i
1

m

g
(

x

i
−
1

)
−
g
(

x

i

)

g
(

x

i

)
g
(

x

i
−
1

)

≤

∑

# i
1

m

M

2

g
(

x

i

)
−
g
(

x

i
−
1

)

=

M

2

S

Γ

[
g
;
a
,
b
]

≤

M

2

V
[
g
;
a
,
b
]
.

Therefore,

V
[
1

/

g
;
a
,
b
]
≤

M

2

V
[
g
;
a
,
b
]
<
∞

{\displaystyle V[1/g;a,b]\leq M^{2}V[g;a,b]<+\infty }

.