proof of identity theorem of holomorphic functions

proof of identity theorem of holomorphic functions

Since

z

0

∈

D

, there exists an

ϵ

0

>

0

the closed disk of radius

ϵ

about

z

0

is

contained

in

D

. Furthermore, both

f

1

and

f

2

are

analytic

inside this disk. Since

z

0

is a

limit point

, there must exist a

sequence

x

k

k

=

1

∞

of distinct points of

s

which

converges

to

z

0

. We may further assume that

|

x

k

-

z

0

|

<

ϵ

0

for every

k

.

By the

theorem

on the

radius of convergence

of a

complex function

, the

Taylor series

of

f

1

and

f

2

about

z

0

have radii of

convergence

greater

than or equal to

ϵ

0

. Hence, if we can show that the Taylor series of the two

functions

at

z

0

ar equal, we will have shown that

f

1

⁢

(

z

)

=

g

1

⁢

(

z

)

whenever

z

<

ϵ

.

The

n

-th

coefficient

of the Taylor series of a function is constructed from the

n

-th

derivative

of the function. The

n

-th derivative may be expressed
as a limit of

n

-th divided differences

f

(

n

)

⁢

(

z

0

)

=

lim

y

1

,

…

⁢

y

n

→

z

0

⁡

Δ

n

⁢

f

⁢

(

y

1

,

…

,

y

n

)

Δ

n

⁢

(

y

1

,

…

,

y

n

)

Suppose we choose the points at which to compute the divided differences as points of the sequence

x

i

. Then we have

f

(

n

)

⁢

(

z

0

)

=

lim

m

→

∞

⁡

Δ

n

⁢

f

⁢

(

x

m

+

1

,

…

,

x

m

+

n

)

Δ

n

⁢

(

x

m

+

1

,

…

,

x

m

+

n

)

Since

f

1

⁢

(

x

i

)

=

f

2

⁢

(

x

i

)

, it follows that

f

1

(

n

)

⁢

(

z

0

)

=

f

2

(

n

)

⁢

(

z

0

)

for all

n

and hence

f

1

⁢

(

z

)

=

f

2

⁢

(

z

)

when

|

z

-

z

-

0

|

<

ϵ

0

.

If

D

happens to be a circle centred about

z

0

, we are done. If not, let

z

1

be any point of

D

such that

|

z

1

-

z

0

|

≥

ϵ

. Since every

connected

open subset

of the plane is

arcwise connected

, there exists an arc

C

with

endpoints

z

1

and

z

0

.

Define the function

M

:

D

→

ℝ

as follows

M

⁢

(

z

)

=

sup

⁡

{

r

∣

|

z

-

w

|

<

r

⇒

w

∈

D

}

∩

[

0

,

1

]

Because

D

is open, it follows that

0

<

M

⁢

(

z

)

≤

1

for all

z

∈

D

.

We will now show that

M

is

continuous

. Let

w

1

and

w

2

be any two distinct points of

D

. If

M

⁢

(

w

1

)

>

|

w

1

-

w

2

|

, then a disk of radius

M

⁢

(

w

1

)

-

|

w

1

-

w

2

|

about

w

2

will be contained in the disk of radius

M

⁢

(

w

1

)

about

w

1

. Hence, by the definition of

M

, it will follow that

M

⁢

(

w

2

)

≥

M

⁢

(

w

1

)

-

|

w

1

-

w

2

|

. Therefore, for any two points

w

1

and

w

2

, it is the case that

|

M

⁢

(

w

1

)

-

M

⁢

(

w

2

)

|

≤

|

w

1

-

w

2

|

, which implies that

M

is continuous.

Since

M

is continuous and the arc

C

is

compact

,

M

attains a

minimum

value

m

on

C

. Let

μ

>

0

be chosen smaller strictly less than both

m

/

2

and

ϵ

0

. Consider the set of all open disks of radius

μ

centred about ponts of

C

. By the way

μ

was selected, each of these disks lies inside

D

. Since

C

is compact a finite subset of these disks will serve to cover

D

. In other words, there exsits a

finite set

of points

y

1

,

y

2

,

…

⁢

y

n

such that, if

z

∈

C

, then

|

z

-

y

j

|

<

μ

for some

j

∈

{

1

,

2

,

…

,

n

}

. We may assume that the

y

j

’s are ordered so that, as one traverses

C

from

z

0

to

z

1

, one encounters

y

j

before one encounters

y

j

+

1

. This imples that

|

y

j

-

y

j

+

1

|

<

μ

. Without loss of generality, we may assume that

y

1

=

z

0

and

y

n

=

z

1

.

We shall now show that

f

1

⁢

(

z

)

=

f

2

⁢

(

z

)

when

|

z

-

y

j

|

<

μ

for all

j

by

induction

. From our definitions it follows that

f

1

⁢

(

z

)

=

f

2

⁢

(

z

)

when

|

z

-

y

1

|

<

μ

. Next, we shall now show that if

f

1

⁢

(

z

)

=

f

2

⁢

(

z

)

when

|

z

-

y

j

|

≤

m

/

2

, then

f

1

⁢

(

z

)

=

f

2

⁢

(

z

)

when

|

z

-

y

j

+

i

|

≤

m

/

2

. Since

|

y

j

-

y

j

+

1

|

<

μ

, there exists a point

w

∈

C

and a

constant

ϵ

>

0

such that

|

w

-

z

|

<

ϵ

implies

|

z

-

y

j

|

≤

μ

and

|

z

-

y

j

+

i

|

≤

μ

. By the

induction hypothesis

,

f

1

⁢

(

z

)

=

f

2

⁢

(

z

)

when

|

z

-

y

j

|

<

μ

. Consider a disk of radius

m

about

w

. By the definition of

m

, this disk lies inside

D

and, by what we have already shown,

f

1

⁢

(

z

)

=

f

2

⁢

(

z

)

when

|

z

-

w

|

≤

m

. Since

|

w

-

y

j

+

1

|

<

μ

<

m

/

2

, it follows from the

triangle inequality

that

f

1

⁢

(

z

)

=

f

2

⁢

(

z

)

when

|

z

-

y

j

+

1

|

<

μ

.

In particular, the

proposition

just proven implies that

f

1

⁢

(

z

1

)

=

f

1

⁢

(

z

1

)

since

z

1

=

y

n

. This means that we have shown that

f

1

⁢

(

z

)

=

f

2

⁢

(

z

)

for all

z

∈

D

.

Title

proof of identity theorem of holomorphic functions

Canonical name

ProofOfIdentityTheoremOfHolomorphicFunctions

Date of creation

2013-03-22 14:40:41

Last modified on

2013-03-22 14:40:41

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

12

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 30A99

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