Ex2-27

If

f

{\displaystyle f}

 is an even function on

[
−
1
,
1
]

{\displaystyle [-1, 1]}

, verify the formulas:

(i)

V
[
f
;
−
1
,
1
# ]
2

V

P

[
f
;
−
1
,
1
]
;

{\displaystyle V[f;-1,1]=2V\_{P}[f;-1,1];}

(ii)

V
[
f
;
0
,
1
# ]

V

P

[
f
;
−
1
,
1
]
;

{\displaystyle V[f;0,1]=V\_{P}[f;-1,1];}

(iii)

V

P

[
f
;
0
,
1
# ]

V

N

[
f
;
−
1
,
0
]
.

{\displaystyle V\_{P}[f;0,1]=V\_{N}[f;-1,0].}

f

{\displaystyle f}

 is an even function on

[
−
1
,
1
]

{\displaystyle [-1, 1]}

 implies that it is symmetric to the y-axis. By the definition of even functions,

f
(
x
# )
f
(
−
x
)

for all

x
∈
[
0
,
1
]
.

{\displaystyle f(x)=f(-x)~{\text{for all}}~x\in [0,1].}

This implies that

f
(
−
1
# )
f
(
1
)

{\displaystyle f(-1)=f(1) }

.

Consider the case when

V

{\displaystyle V}

 is finite. Then, by Theorem 2.6,

V

P

{\displaystyle
V\_{P}
}

 and

V

N

{\displaystyle V\_{N}}

 is finite. Moreover,

V

P

V

N

=
V

{\displaystyle V\_{P}+V\_{N}=V}

 and

V

P

−

V

N

=
f
(
1
)
−
f
(
−
1
# )
0

{\displaystyle V\_{P}-V\_{N}=f(1)-f(-1)=0}

.

(i)

V

P

[
f
;
−
1
,
1
]

=

1
2

[
V
[
f
;
−
1
,
1
]
f
(
1
)
−
f
(
−
1
)
]

=

1
2

V
[
f
;
−
1
,
1
]

⟹

V
[
f
;
−
1
,
1
]

=
2

V

P

[
f
;
−
1
,
1
]
.

{\displaystyle {\begin{aligned}V\_{P}[f;-1,1]&={\frac {1}{2}}[V[f;-1,1]+f(1)-f(-1)]\\&={\frac {1}{2}}V[f;-1,1]\\\implies V[f;-1,1]&=2V\_{P}[f;-1,1].\end{aligned}}}

(ii) Since the graph remains unchanged after reflection about the *y*-axis, we have

V
[
f
;
−
1
,
0
# ]
V
[
f
;
0
,
1
]

{\displaystyle V[f;-1,0]=V[f;0,1]}

, thus

V
[
f
;
−
1
,
1
]

=
V
[
f
;
−
1
,
0
]
V
[
f
;
0
,
1
]

⟹

V
[
f
;
−
1
,
1
]

=
2
V
[
f
;
−
1
,
0
# ]
2
V
[
f
;
0
,
1
]
.

{\displaystyle {\begin{aligned}V[f;-1,1]&=V[f;-1,0]+V[f;0,1]\\\implies V[f;-1,1]&=2V[f;-1,0]=2V[f;0,1].\end{aligned}}}

Combining with the result of (i),

2

V

P

[
f
;
−
1
,
1
# ]
V
[
f
:
−
1
,
1
# ]
2
V
[
f
;
0
,
1
]

⟹

V
[
f
;
0
,
1
# ]

V

P

[
f
;
−
1
,
1
]
.

{\displaystyle {\begin{aligned}2V\_{P}[f;-1,1]=V[f:-1,1]=2V[f;0,1]\\\implies V[f;0,1]=V\_{P}[f;-1,1].\end{aligned}}}

(iii)

```
The idea of this proof is more or less right, but it is too vague. Someone should edit it to make the calculations more explicit.

```

Given

f

{\displaystyle f}

 an even function. Consider a partition

# Γ
{
−
# 1
−

x

m

,
−

x

m
−
1

,
…
−

x

1

,

x

0

=
0
,

x

1

,
…
,

x

m
−
1

,

x

m

=
1
}

{\displaystyle \Gamma =\{-1=-x\_{m},-x\_{m-1},\ldots -x\_{1},x\_{0}=0,x\_{1},\ldots ,x\_{m-1},x\_{m}=1\}}

. Choose

x

i

,

x

i
1

∈
Γ

{\displaystyle x\_{i},x\_{i+1}\in \Gamma }

, where

i
∈
[
0
,
m
−
1
]

{\displaystyle i\in [0,m-1]}

. If

f

{\displaystyle f}

 is increasing (decreasing) in the interval

[
−

x

i
1

,
−

x

i

]

{\displaystyle [-x\_{i+1},-x\_{i}]}

, where

{
−

x

i
1

,
−

x

i

}
∈
[
−
1
,
0
]

{\displaystyle \{-x\_{i+1},-x\_{i}\}\in [-1,0]}

, then

f

{\displaystyle f}

 is decreasing (increasing) in the interval

[

x

i
1

,

x

i

]

{\displaystyle [x\_{i+1},x\_{i}]}

, where

{

x

i
1

,

x

i

}
∈
[
0
,
1
]

{\displaystyle \{x\_{i+1},x\_{i}\}\in [0,1]}

. Moreover, they are with the same variation. Thus, by reviewing the definition of

V

P

{\displaystyle
V\_{P}
}

 and

V

N

{\displaystyle V\_{N}}

,

V

P

[
f
;
0
,
1
# ]

V

N

[
f
;
−
1
,
0
]
.

{\displaystyle V\_{P}[f;0,1]=V\_{N}[f;-1,0].}