proof of estimating theorem of contour integral

proof of estimating theorem of contour integral

WLOG consider

g

⁢

(

t

)

:

ℝ

→

ℂ

a parameterization of the

γ

curve along which the

integral

is evaluated with

|

g

′

⁢

(

t

)

|

=

1

. This amounts to a canonical parameterization and is always possible.
Since the integral is independent of re-parameterization

1

1

apart from a possible sign change due to exchange of

orientation

of the path

the result will be completely general.

With this in mind, the

contour integral

can be explicitly written as

∫

γ

f

⁢

(

z

)

⁢

𝑑

z

=

∫

0

L

f

⁢

(

g

⁢

(

t

)

)

⁢

g

′

⁢

(

t

)

⁢

𝑑

t

(1)

where

L

is the

arc length

of the curve

γ

.

Consider the set of all

continuous functions

[

0

,

L

]

→

ℂ

as a

vector space

2

2

axioms are trivial to verify

, we can define an

inner product

in it via

⟨

f

,

g

⟩

=

∫

0

L

f

⁢

(

t

)

⁢

g

¯

⁢

(

t

)

⁢

𝑑

t

(2)

The axioms are easy to verify:

•

⟨

k

1

⁢

a

1

+

k

2

⁢

a

2

,

a

3

⟩

=

∫

0

L

(

k

1

⁢

a

1

⁢

(

t

)

+

k

2

⁢

a

2

⁢

(

t

)

)

⁢

a

3

¯

⁢

(

t

)

⁢

𝑑

t

=

k

1

⁢

⟨

a

1

,

a

3

⟩

+

k

2

⁢

⟨

a

2

,

a

3

⟩

•

⟨

a

,

b

⟩

=

∫

0

L

a

⁢

(

t

)

⁢

b

¯

⁢

(

t

)

⁢

𝑑

t

=

∫

0

L

b

⁢

(

t

)

⁢

a

¯

⁢

(

t

)

¯

⁢

𝑑

t

=

∫

0

L

b

⁢

(

t

)

⁢

a

¯

⁢

(

t

)

⁢

𝑑

t

¯

=

⟨

b

,

a

⟩

¯

•

⟨

a

,

a

⟩

=

∫

0

L

a

⁢

(

t

)

⁢

a

¯

⁢

(

t

)

⁢

𝑑

t

=

∫

0

L

|

a

⁢

(

t

)

|

2

⁢

𝑑

t

≥

0

since the integrand is a non-negative (real)

function

, and

0

iff

|

a

|

2

=

0

everywhere in the

interval

, that is:

⟨

a

,

a

⟩

=

0

⇔

a

=

0

With all this in mind, equation

1

can be written as

∫

γ

f

⁢

(

z

)

⁢

𝑑

z

=

⟨

f

∘

g

,

g

¯

′

⟩

(3)

Where by definition

∥

f

∥

=

<

f

,

f

>

is the norm associated with the inner product defined previously.

Using

Cauchy-Schwarz inequality

we can write that

|

⟨

f

∘

g

,

g

¯

′

⟩

|

≤

∥

f

∘

g

∥

⁢

∥

g

¯

′

∥

(4)

But since by assumption the parameterization

g

is canonic,

∥

g

¯

′

∥

=

∥

g

′

∥

=

∫

0

L

1

⁢

𝑑

t

=

L

.

On the other hand

∥

f

∘

g

∥

=

∫

0

L

f

⁢

(

g

⁢

(

t

)

)

⁢

f

¯

⁢

(

g

⁢

(

t

)

)

⁢

𝑑

t

≤

∫

0

L

M

2

⁢

𝑑

t

=

M

⁢

L

, where

|

f

⁢

(

g

⁢

(

t

)

)

|

≤

M

for every point on

γ

.

The previous paragraphs imply that

|

∫

γ

f

⁢

(

z

)

⁢

𝑑

z

|

≤

M

⁢

L

(5)

which is the result we aimed to prove.

Cauchy-Schwarz inequality says more, it also says that

|

⟨

a

,

b

⟩

|

=

∥

a

∥

⁢

∥

b

∥

⇔

a

=

λ

⁢

b

where

λ

is a constant.

So if

|

⟨

f

∘

g

,

g

¯

′

⟩

=

∥

f

∘

g

∥

∥

g

¯

′

∥

then

f

∘

g

=

λ

⁢

g

¯

′

, where

λ

∈

ℂ

is a constant.
If

g

is a canonical parameterization

|

g

′

|

=

1

and we get the absolute

modulus

|

λ

|

=

|

f

∘

g

|

(which must be constant) and all that remains is to find the phase of

λ

which must also be constant.

Title

proof of estimating theorem of contour integral

Canonical name

ProofOfEstimatingTheoremOfContourIntegral

Date of creation

2013-03-22 15:46:02

Last modified on

2013-03-22 15:46:02

Owner

cvalente (11260)

Last modified by

cvalente (11260)

Numerical id

22

Author

cvalente (11260)

Entry type

Proof

Classification

msc 30E20

Classification

msc 30A99

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