proof of Goursat’s theorem

proof of Goursat’s theorem

We argue by

contradiction

. Set

η

=

∮

∂

⁡

R

f

⁢

(

z

)

⁢

𝑑

z

,

and suppose that

η

≠

0

. Divide

R

into four

congruent

rectangles

R

1

,

R

2

,

R

3

,

R

4

(see Figure 1), and set

η

i

=

∮

∂

⁡

R

i

f

⁢

(

z

)

⁢

𝑑

z

.

Figure 1: subdivision of the rectangle contour.

Now subdivide each of the four sub-rectangles, to get 16 congruent
sub-sub-rectangles

R

i

1

⁢

i

2

,

i

1

,

i

2

=

1

⁢

…

⁢

4

, and then continue

ad
infinitum

to obtain a

sequence

of nested families of rectangles

R

i

1

⁢

…

⁢

i

k

, with

η

i

1

⁢

…

⁢

i

k

the values of

f

⁢

(

z

)

integrated along the corresponding contour.

Orienting the boundary of

R

and all the sub-rectangles in the usual counter-clockwise fashion
we have

η

=

η

1

+

η

2

+

η

3

+

η

4

,

and more generally

η

i

1

⁢

…

⁢

i

k

=

η

i

1

⁢

…

⁢

i

k

⁢

1

+

η

i

1

⁢

…

⁢

i

k

⁢

2

+

η

i

1

⁢

…

⁢

i

k

⁢

3

+

η

i

1

⁢

…

⁢

i

k

⁢

4

.

In as much as the

integrals

along oppositely oriented

line
segments

cancel, the contributions from the interior

segments

cancel,
and that is why the right-hand side reduces to the integrals along the
segments at the boundary of the

composite

rectangle.

Let

j

1

∈

{

1

,

2

,

3

,

4

}

be such that

|

η

j

1

|

is the

maximum

of

|

η

i

|

,

i

=

1

,

…

,

4

. By the

triangle inequality

we have

|

η

1

|

+

|

η

2

|

+

|

η

3

|

+

|

η

4

|

≥

|

η

|

,

and hence

|

η

j

1

|

≥

1

/

4

⁢

|

η

|

.

Continuing inductively, let

j

k

+

1

be such that

|

η

j

1

⁢

…

⁢

j

k

⁢

j

k

+

1

|

is the maximum of

|

η

j

1

⁢

…

⁢

j

k

⁢

i

|

,

i

=

1

,

…

,

4

. We then have

|

η

j

1

⁢

…

⁢

j

k

⁢

j

k

+

1

|

≥

4

-

(

k

+

1

)

⁢

|

η

|

.

(1)

Now the sequence of nested rectangles

R

j

1

⁢

…

⁢

j

k

converges

to some point

z

0

∈

R

; more formally

{

z

0

}

=

⋂

k

=

1

∞

R

j

1

⁢

…

⁢

j

k

.

The

derivative

f

′

⁢

(

z

0

)

is assumed to exist, and hence for every

ϵ

>

0

there
exists a

k

sufficiently large, so that for all

z

∈

R

j

1

⁢

…

⁢

j

k

we have

|

f

⁢

(

z

)

-

f

′

⁢

(

z

0

)

⁢

(

z

-

z

0

)

|

≤

ϵ

⁢

|

z

-

z

0

|

.

Now we make use of the following.

Lemma 1

Let

Q

⊂

C

be a rectangle, let

a

,

b

∈

C

, and let

f

⁢

(

z

)

be a

continuous

,

complex

valued

function

defined and

bounded

in a domain containing

Q

. Then,

∮

∂

⁡

Q

(

a

⁢

z

+

b

)

⁢

𝑑

z

=

0

|

∮

∂

⁡

Q

f

⁢

(

z

)

|

≤

M

⁢

P

,

where

M

is an

upper bound

for

|

f

⁢

(

z

)

|

and where

P

is the length
of

∂

⁡

Q

.

The first of these assertions follows by the

Fundamental Theorem of
Calculus

; after all the function

a

⁢

z

+

b

has an anti-derivative. The
second assertion follows from the fact that the

absolute value

of an
integral is smaller than the integral of the absolute value of the

integrand

— a standard result in integration theory.

Using the Lemma and the fact that the perimeter of a rectangle is

greater

than its diameter we infer that for every

ϵ

>

0

there
exists a

k

sufficiently large that

η

j

1

⁢

…

⁢

j

k

=

|

∮

∂

⁡

R

j

1

⁢

…

⁢

j

k

f

⁢

(

z

)

⁢

𝑑

z

|

≤

ϵ

⁢

|

∂

⁡

R

j

1

⁢

…

⁢

j

k

|

2

=

4

-

k

⁢

|

∂

⁡

R

|

2

⁢

ϵ

.

where

|

∂

⁡

R

|

denotes the
length of perimeter of the rectangle

R

. This contradicts the
earlier estimate (

1

). Therefore

η

=

0

.

Title

proof of Goursat’s

theorem

Canonical

name

ProofOfGoursatsTheorem

Date of creation

2013-03-22 12:54:37

Last modified on

2013-03-22 12:54:37

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

13

Author

rmilson (146)

Entry type

Proof

Classification

msc 30E20

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