proof of Bohr-Mollerup theorem

proof of Bohr-Mollerup theorem

To show that the

gamma function

is

logarithmically convex

, we can examine the product representation:

Γ

⁢

(

x

)

=

1

x

⁢

e

-

γ

⁢

x

⁢

∏

n

=

1

∞

n

x

+

n

⁢

e

-

x

/

n

Since this product

converges absolutely

for

x

>

0

, we can take the

logarithm

term-by-term to obtain

log

⁡

Γ

⁢

(

x

)

=

-

log

⁡

x

-

γ

⁢

x

-

∑

n

=

1

∞

log

⁡

(

n

x

+

n

)

-

x

n

It is justified to

differentiate

this series twice because the series of

derivatives

is absolutely and

uniformly convergent

.

d

2

d

⁢

x

2

⁢

log

⁡

Γ

⁢

(

x

)

=

1

x

2

+

∑

n

=

1

∞

1

(

x

+

n

)

2

=

∑

n

=

0

∞

1

(

x

+

n

)

2

Since every term in this series is positive,

Γ

is logarithmically convex. Furthermore, note that since each term is

monotonically decreasing

,

log

⁡

Γ

is a

decreasing function

of

x

. If

x

>

m

for some integer

m

, then we can bound the series term-by-term to obtain

d

2

d

⁢

x

2

⁢

log

⁡

Γ

⁢

(

x

)

<

∑

n

=

0

∞

1

(

m

+

n

)

2

=

∑

n

=

m

∞

1

n

2

Therefore, as

x

→

∞

,

d

2

⁢

Γ

/

d

⁢

x

2

→

0

.

Next, let

f

satisfy the hypotheses of the

Bohr-Mollerup theorem

. Consider the

function

g

defined as

e

g

⁢

(

x

)

=

f

⁢

(

x

)

/

Γ

⁢

(

x

)

. By hypothesis 3,

g

⁢

(

1

)

=

0

. By hypothesis 2,

e

g

⁢

(

x

+

1

)

=

e

g

⁢

(

x

)

, so

g

⁢

(

x

+

1

)

=

g

⁢

(

x

)

. In other ,

g

is periodic.

Suppose that

g

is not constant. Then there must exist points

x

0

and

x

1

on the real axis such that

g

⁢

(

x

0

)

≠

g

⁢

(

x

1

)

. Suppose that

g

⁢

(

x

1

)

>

g

⁢

(

x

0

)

for definiteness. Since

g

is periodic with period 1, we may assume without loss of generality that

x

0

<

x

1

<

x

0

+

1

. Let

D

2

denote the second

divided difference

of

g

:

D

2

=

Δ

2

⁢

(

g

;

x

0

,

x

1

,

x

0

+

1

)

=

-

g

⁢

(

x

0

)

x

0

-

x

1

+

g

⁢

(

x

1

)

(

x

1

-

x

0

)

⁢

(

x

1

-

x

0

-

1

)

-

g

⁢

(

x

0

+

1

)

x

0

-

x

1

+

1

By our assumptions,

D

2

<

0

. By linearity,

D

2

=

Δ

2

⁢

(

log

⁡

f

;

x

0

,

x

1

,

x

0

+

1

)

-

Δ

2

⁢

(

log

⁡

Γ

;

x

0

,

x

1

,

x

0

+

1

)

By periodicity, we have

D

2

=

Δ

2

⁢

(

log

⁡

f

;

x

0

+

n

,

x

1

+

n

,

x

0

+

n

+

1

)

-

Δ

2

⁢

(

log

⁡

Γ

;

x

0

+

n

,

x

1

+

n

,

x

0

+

n

+

1

)

for every integer

n

>

0

. However,

|

Δ

2

⁢

(

log

⁡

Γ

;

x

0

+

n

,

x

1

+

n

,

x

0

+

n

+

1

)

|

<

max

x

0

+

n

≤

x

≤

x

0

+

1

⁡

d

2

d

⁢

x

2

⁢

log

⁡

Γ

⁢

(

x

)

As

n

→

∞

, the right hand side approaches zero. Hence, by choosing

n

sufficiently large, we can make the left-hand side smaller than

|

D

2

|

/

2

. For such an

n

,

Δ

2

⁢

(

log

⁡

f

;

x

0

+

n

,

x

1

+

n

,

x

0

+

n

+

1

)

<

0

However, this contradicts hypothesis 1. Therefore,

g

must be constant. Since

g

⁢

(

0

)

=

0

,

g

⁢

(

x

)

=

0

for all

x

, which implies that

e

0

=

f

⁢

(

x

)

/

Γ

⁢

(

x

)

. In other words,

f

⁢

(

x

)

=

Γ

⁢

(

x

)

as desired.

Title

proof of Bohr-Mollerup theorem

Canonical name

ProofOfBohrMollerupTheorem

Date of creation

2013-03-22 14:53:39

Last modified on

2013-03-22 14:53:39

Owner

Andrea Ambrosio (7332)

Last modified by

Andrea Ambrosio (7332)

Numerical id

18

Author

Andrea Ambrosio (7332)

Entry type

Proof

Classification

msc 33B15

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