Ex2-4

**Chapter 2 Question 4**

*Question: Let

{

f
k

}

{\displaystyle \{ f\_k \}}

 be a sequence of functions of bounded variation on

[
a
,
b
]

{\displaystyle [a, b]}

. If

V
[

f

k

;
a
,
b
]
≤
M
<
∞

{\displaystyle V[f\_{k};a,b]\leq M<+\infty }

 for all

k

{\displaystyle k}

 and if

f
k

→
f

{\displaystyle f\_k \to f}

 pointwise on

[
a
,
b
]

{\displaystyle [a, b]}

, show that

f

{\displaystyle f}

 is of bounded variation and that

V
[
f
;
a
,
b
]
≤
M

{\displaystyle V[f; a , b ] \leq M
}

. Give an example of a convergent sequence of functions of bounded variation whose limit is not of bounded variation.*

Proof: Given

ε
>
0

{\displaystyle \varepsilon > 0}

, suppose

# Γ
{

x
0

,

x
1

,
…
,

x
n

}

{\displaystyle \Gamma=\{x\_0,x\_1,\ldots,x\_n\}}

 is any partition of

[
a
,
b
]

{\displaystyle [a, b]}

, then for each

# i
0
,
1
,
…
,
n

{\displaystyle i=0,1,\ldots ,n}

we can choose a natural number

K
i

{\displaystyle K\_i}

such that

f

k

(

x

i

)
−
f
(

x

i

)

<

ε

2
n

∀
k
≥

K

i

.

{\displaystyle |f\_{k}(x\_{i})-f(x\_{i})|<{\frac {\varepsilon }{2n}}\quad \forall k\geq K\_{i}.}

(We will see why there is

2
n

{\displaystyle 2n}

 inside the fraction later)

By letting

# K
max
{

K

0

,

K

1

,
…
,

K

n

}

{\displaystyle K=\max\{K\_{0},K\_{1},\ldots ,K\_{n}\}}

, we have

f

k

(

x

i

)
−
f
(

x

i

)

<

ε

2
n

∀
# i
0
,
1
,
…
,
n
,
∀
k
≥
K
.

{\displaystyle |f\_{k}(x\_{i})-f(x\_{i})|<{\frac {\varepsilon }{2n}}\quad \forall i=0,1,\ldots ,n,\forall k\geq K.}

We find that

S

Γ

[
f
]

=

∑

# i
1

n

f
(

x

i

)
−
f
(

x

i
−
1

)

≤

∑

# i
1

n

{

f

K

(

x

i

)
−

f

K

(

x

i
−
1

)

f

K

(

x

i

)
−
f
(

x

i

)

f

K

(

x

i
−
1

)
−
f
(

x

i
−
1

)

}

<

∑

# i
1

n

{

f

K

(

x

i

)
−

f

K

(

x

i
−
1

)

ε
n

}

=

S

Γ

[

f

K

]
ε

≤
M
ε
.

Since

ε

{\displaystyle \varepsilon}

 is arbitrary, we conclude

S

Γ

[
f
]
≤
M

{\displaystyle S\_{\Gamma }[f]\leq M}

 for each partition

Γ

{\displaystyle \Gamma}

 of

[
a
,
b
]

{\displaystyle [a, b]}

, thus the proof is complete.

## What if the variations do not bounded above uniformly?

Define a sequence

{

f

k

}

# n
1

∞

{\displaystyle \{f\_{k}\}\_{n=1}^{\infty }}

 of functions on

[
0
,
1
]

{\displaystyle [0,1]}

 such that

f

k

(
x
# )

{

1

 if

0
<
x
≤

1

2

k

,

0

 otherwise

.

{\displaystyle f\_{k}(x)={\begin{cases}1&{\text{ if }}0<x\leq {\frac {1}{2^{k}}},\\0&{\text{ otherwise}}.\end{cases}}}

We also define a sequence

{

g

n

}

# n
1

∞

{\displaystyle \{g\_{n}\}\_{n=1}^{\infty }}

 on

[
0
,
1
]

{\displaystyle [0,1]}

 such that

g

n

(
x

# k
1

n

f

k

(
x
)
.

{\displaystyle g\_{n}(x)=\sum \_{k=1}^{n}f\_{k}(x).}

Clearly, each function

f
k

{\displaystyle f\_k}

 is of bounded variation with variation

V
[

f

k

;
0
,
1
# ]
2

{\displaystyle V[f\_{k};0,1]=2}

, so each

g
n

{\displaystyle g\_n}

 being a finite sum of functions of bounded variation, is also of bounded variation. One can check that

V
[

g

n

;
0
,
1
# ]
2
n

{\displaystyle V[g\_{n};0,1]=2n}

 for all

n

{\displaystyle n}

. Now

{

g

n

}

# n
1

∞

{\displaystyle \{g\_{n}\}\_{n=1}^{\infty }}

 is a sequence of functions of bounded variation. We prove this sequence has a pointwise convergence to a function

g

{\displaystyle g}

. It is clear that

g
(
0
# )
0

{\displaystyle g(0) = 0}

 since

g

n

(
0
# )
0

{\displaystyle g\_{n}(0)=0}

 for all

n

{\displaystyle n}

. For

x
∈
(
0
,
1
]

{\displaystyle x\in(0,1]}

 we can choose nonnegative integer

N

{\displaystyle N}

 such that

1

2

N
1

<
x
≤

1

2

N

{\displaystyle {\frac {1}{2^{N+1}}}<x\leq {\frac {1}{2^{N}}}}

, hence

g
(
x
# )

lim

n
→
∞

g

n

(
x

# k
1

N

f

k

(
x
# )
N

{\displaystyle g(x)=\lim \_{n\to \infty }g\_{n}(x)=\sum \_{k=1}^{N}f\_{k}(x)=N}

since

f
k

(
x
# )
0

{\displaystyle f\_k(x) = 0}

 for

k
>
N

{\displaystyle k > N}

.
Note that for each positive integer

N

{\displaystyle N}

, there is always a positive number

x

{\displaystyle x}

 such that

x
≤
1

/

2

N

{\displaystyle x\leq 1/2^{N}}

, therefore

g

{\displaystyle g}

 is unbounded and therefore it is not a function of bounded variation by Theorem 2.1(i).