Ex2-9

**Solution to Chapter 2, Exercise 9**

*Question: Let

C

{\displaystyle C}

be a curve with parametric equations

# x
ϕ
(
t
)

{\displaystyle x=\phi(t)}

and

# y
ψ
(
t
)

{\displaystyle y=\psi(t)}

,

a
≤
t
≤
b

{\displaystyle a \leq t \leq b}

.*

*(a) If

ϕ

{\displaystyle \phi}

and

ψ

{\displaystyle \psi}

 are of bounded variation and continuous, show that

# L

lim

Γ

→
0

l
(
Γ
)

{\displaystyle L=\lim\_{|\Gamma|\to 0}l(\Gamma)}

.*

*(b) If

ϕ

{\displaystyle \phi}

and

ψ

{\displaystyle \psi}

 are continuously differentiable, show that

# L

∫
a
b

(
[

ϕ
′

(
t
)

]
2

[

ψ
′

(
t
)

]
2

)

1

/

2

d
t

{\displaystyle L=\int\_a^b([\phi'(t)]^2+[\psi'(t)]^2)^{1/2}dt}

.*

Proof:

(a) If

ϕ

{\displaystyle \phi}

and

ψ

{\displaystyle \psi}

are of bounded variation, we know that

C

{\displaystyle C}

is rectifiable. We want to prove that for any

M
<
L

{\displaystyle M<L}

, there exists

δ
>
0

{\displaystyle \delta > 0}

such that if

Γ

<
δ

{\displaystyle |\Gamma|<\delta}

, then

l
(
Γ
)
>
M

{\displaystyle l(\Gamma)>M}

. Choose

μ
>
0

{\displaystyle \mu>0}

so that

M
μ
<
L

{\displaystyle M+\mu<L}

. Then, choose a partition

Γ
¯

{\displaystyle \bar{\Gamma}}

of

[
a
,
b
]

{\displaystyle [a, b]}

such that

L
(

Γ
¯

)
>
M
μ

{\displaystyle L(\bar{\Gamma})>M+\mu}

 and let

Γ
¯

=
{

t
0

¯

,

t
1

¯

,
.
.
.
,

t
k

¯

}

{\displaystyle \bar{\Gamma}=\{\bar{t\_0},\bar{t\_1},...,\bar{t\_k}\}}

.

ϕ

{\displaystyle \phi}

and

ψ

{\displaystyle \psi}

 are uniformly continuous since they are continuous on a compact set. Then there exists

δ
′

>
0

{\displaystyle \delta' > 0}

such that if

s
−
t

<

δ
′

{\displaystyle |s-t|<\delta'}

, then

(i)

ϕ
(
s
)
−
ϕ
(
t
)

<

μ

8

(
k
1
)

{\displaystyle |\phi(s)-\phi(t)|<\frac{\mu}{\sqrt{8}(k+1)}}

(ii)

ψ
(
s
)
−
ψ
(
t
)

<

μ

8

(
k
1
)

{\displaystyle |\psi(s)-\psi(t)|<\frac{\mu}{\sqrt{8}(k+1)}}

for all

s
,
t
∈
[
a
,
b
]

{\displaystyle s,t\in[a,b]}

.

Let

# Γ
{

t
0

,

t
1

,
.
.
.
,

t
m

}

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

be any partition that satisfies

(iii)

Γ

<

δ
′

{\displaystyle |\Gamma|<\delta'}

(iv)

Γ

<
min
{

t
i

¯

−

t

i
−
1

¯

1
≤
i
≤
m
}

{\displaystyle |\Gamma|<\min\{\overline{t\_i}-\overline{t\_{i-1}}\vert1\leq i \leq m\}}

We have

l
(
Γ
)

=

∑

# i
1

m

(
[
ϕ
(

t
i

)
−
ϕ
(

t

i
−
1

)

]
2

[
ψ
(

t
i

)
−
ψ
(

t

i
−
1

)

]
2

)

1

/

2

=

∑
′

∑
″

(1)

{\displaystyle
\begin{align}
l(\Gamma)&=\sum\_{i=1}^m ([\phi(t\_i)-\phi (t\_{i-1})]^2+[\psi(t\_i)-\psi (t\_{i-1})]^2)^{1/2}\\
&=\sum' + \sum ''~~~~~~~~~\text{(1)}
\end{align}
}

where

∑
′

{\displaystyle \sum '}

 is extended over all

i

{\displaystyle i}

such that

(

t

i
−
1

,

t
i

)

{\displaystyle (t\_{i-1},t\_i)}

does not contain some

t
j

¯

{\displaystyle \bar{t\_j}}

while

∑
″

{\displaystyle \sum ''}

is extended over all

i

{\displaystyle i}

such that

(

t

i
−
1

,

t
i

)

{\displaystyle (t\_{i-1},t\_i)}

contains some

t
j

¯

{\displaystyle \bar{t\_j}}

. Notice that any

(

t

i
−
1

,

t
i

)

{\displaystyle (t\_{i-1},t\_i)}

 can contain at most one

t
j

¯

{\displaystyle \bar{t\_j}}

 by (iv), and the number of terms of

∑
″

{\displaystyle \sum ''}

is at most

k
1

{\displaystyle k+1}

.

Now we consider the partition

Γ
∪

Γ
¯

{\displaystyle \Gamma \cup \bar{\Gamma}}

.

l
(
Γ
∪

Γ
¯

)

{\displaystyle l(\Gamma \cup \bar{\Gamma})}

can be written as

l
(
Γ
∪

Γ
¯

# )

∑
′

∑
‴

{\displaystyle
l(\Gamma \cup \bar{\Gamma})=\sum' + \sum '''
}

where

∑
′

{\displaystyle \sum '}

is the same as before while

∑
‴

{\displaystyle \sum '''}

is obtained from

∑
″

{\displaystyle \sum ''}

by replacing each

(
[
ϕ
(

t
i

)
−
ϕ
(

t

i
−
1

)

]
2

[
ψ
(

t
i

)
−
ψ
(

t

i
−
1

)

]
2

)

1

/

2

{\displaystyle ([\phi(t\_i)-\phi (t\_{i-1})]^2+[\psi(t\_i)-\psi (t\_{i-1})]^2)^{1/2}}

term by

(
[
ϕ
(

t
i

)
−
ϕ
(

t

j

¯

)

]
2

[
ψ
(

t
i

)
−
ψ
(

t

j

¯

)

]
2

)

1

/

2

(
[
ϕ
(

t
j

¯

)
−
ϕ
(

t

i
−
1

)

]
2

[
ψ
(

t
j

¯

)
−
ψ
(

t

i
−
1

)

]
2

)

1

/

2

{\displaystyle ([\phi(t\_i)-\phi (\bar{t\_{j}})]^2+[\psi(t\_i)-\psi (\bar{t\_{j}})]^2)^{1/2} + ([\phi(\bar{t\_j})-\phi (t\_{i-1})]^2+[\psi(\bar{t\_j})-\psi (t\_{i-1})]^2)^{1/2}}

.

Then each term in

∑
″

{\displaystyle \sum ''}

will be smaller than or equal to those in

∑
‴

{\displaystyle \sum '''}

and hence

∑
‴

≥

∑
″

{\displaystyle \sum ''' \geq \sum ''}

. Thus,

l
(
Γ
∪

Γ
¯

)
≥
l
(
Γ
)

{\displaystyle l(\Gamma \cup \bar{\Gamma}) \geq l(\Gamma)}

. Similarly, we can show that

l
(
Γ
∪

Γ
¯

)
≥
l
(

Γ
¯

)

{\displaystyle l(\Gamma \cup \bar{\Gamma}) \geq l(\bar{\Gamma})}

.

∑
‴

{\displaystyle \sum '''}

has at most

2
(
k
1
)

{\displaystyle 2(k+1)}

 terms and by

(
i
)
,
(
i
i
)
,
(
i
i
i
)

{\displaystyle (i),(ii),(iii)}

, we have

∑
‴

<

(

μ
2

8
(
k
1

)
2

μ
2

8
(
k
1

)
2

)

1

/

2

2
(
k
1
)

=

(

μ
2

4
(
k
1

)
2

)

1

/

2

2
(
k
1
)

=
μ
.

{\displaystyle
\begin{align}
\sum '''&< \left( \frac{\mu^2}{8(k+1)^2}+\frac{\mu^2}{8(k+1)^2} \right)^{1/2}2(k+1)\\
&=\left( \frac{\mu^2}{4(k+1)^2} \right)^{1/2}2(k+1)\\
&=\mu.
\end{align}
}

Since

l
(
Γ
∪

Γ
¯

# )

∑
′

∑
‴

{\displaystyle
l(\Gamma \cup \bar{\Gamma})=\sum' + \sum '''
}

, we have

∑
′

=
l
(
Γ
∪

Γ
¯

)
−

∑
‴

>
l
(
Γ
∪

Γ
¯

)
−
μ
.

{\displaystyle
\begin{align}
\sum ' &= l(\Gamma \cup \bar{\Gamma}) - \sum '''\\
&> l(\Gamma \cup \bar{\Gamma}) - \mu.
\end{align}
}

Recall that

Γ
¯

{\displaystyle \bar{\Gamma}}

 has been chosen so that

l
(

Γ
¯

)
>
M
μ

{\displaystyle l(\bar{\Gamma})>M+\mu}

. Then

l
(
Γ
)

=

∑
′

∑
″

≥

∑
′

>
l
(
Γ
∪

Γ
¯

)
−
μ

≥
l
(

Γ
¯

)
−
μ

>
M
μ
−
μ

=
M

{\displaystyle
\begin{align}
l(\Gamma) &= \sum ' + \sum ''\\
&\geq \sum '\\
&> l(\Gamma \cup \bar{\Gamma}) - \mu\\
&\geq l(\bar{\Gamma}) - \mu\\
&> M+\mu-\mu\\
&=M
\end{align}
}

Take

# δ
min
{

δ
′

,
min
{

t
i

¯

−

t

i
−
1

¯

}
}

{\displaystyle \delta=\min\{ \delta',\min\{\overline{t\_i}-\overline{t\_{i-1}}\}\}}

, then for any

Γ

{\displaystyle \Gamma}

 with

Γ

<
δ

{\displaystyle |\Gamma|<\delta}

, we will have

l
(
Γ
)
>
M

{\displaystyle l(\Gamma)>M}

.

(b) Let

# Γ
{

t
0

,

t
1

,
.
.
.
,

t
m

}

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

. By Mean Value Theorem,

ϕ
(

t
i

)
−
ϕ
(

t

i
−
1

)

=

ϕ
′

(

c
i

)
(

t
i

−

t

i
−
1

)

ψ
(

t
i

)
−
ψ
(

t

i
−
1

)

=

ψ
′

(

d
i

)
(

t
i

−

t

i
−
1

)

{\displaystyle
\begin{align}
\phi(t\_i)-\phi(t\_{i-1})&=\phi'(c\_i)(t\_i-t\_{i-1})\\
\psi(t\_i)-\psi(t\_{i-1})&=\psi'(d\_i)(t\_i-t\_{i-1})
\end{align}
}

for some

c
i

{\displaystyle c\_i}

and

d
i

{\displaystyle d\_i}

 in

(

t

i
−
1

,

t
i

)

{\displaystyle (t\_{i-1},t\_i)}

. Then

l
(
Γ
)

=

∑

# i
1

m

(
[
ϕ
(

t
i

)
−
ϕ
(

t

i
−
1

)

]
2

[
ψ
(

t
i

)
−
ψ
(

t

i
−
1

)

]
2

)

1

/

2

=

∑

# i
1

m

(
[

ϕ
′

(

c
i

)
(

t
i

−

t

i
−
1

)

]
2

[

ψ
′

(

d
i

)
(

t
i

−

t

i
−
1

)

]
2

)

1

/

2

=

∑

# i
1

m

(
[

ϕ
′

(

c
i

)

]
2

[

ψ
′

(

d
i

)

]
2

)

1

/

2

(

t
i

−

t

i
−
1

)

{\displaystyle
\begin{align}
l(\Gamma)&=\sum\_{i=1}^m ([\phi(t\_i)-\phi (t\_{i-1})]^2+[\psi(t\_i)-\psi (t\_{i-1})]^2)^{1/2}\\
&=\sum\_{i=1}^m ([\phi'(c\_i)(t\_i-t\_{i-1})]^2+[\psi'(d\_i)(t\_i-t\_{i-1})]^2)^{1/2}\\
&=\sum\_{i=1}^m ([\phi'(c\_i)]^2+[\psi'(d\_i)]^2)^{1/2}(t\_i-t\_{i-1})
\end{align}
}

If we take the limit as

Γ

→
0

{\displaystyle |\Gamma| \to 0}

on both sides, the left hand side converges to

L

{\displaystyle L}

 by part (a) while the limit of the right hand side is the integral

∫
a
b

(
[

ϕ
′

(
t
)

]
2

[

ψ
′

(
t
)

]
2

)

1

/

2

d
t

{\displaystyle
\int\_a^b ([\phi'(t)]^2+[\psi'(t)]^2)^{1/2}~dt
}