elementary results about multiplicative functions and convolution

elementary results about multiplicative functions and convolution

One of the most important

elementary results about multiplicative functions and convolution

(

http://planetmath.org/DirichletConvolution

) is:

Theorem

.

If

f

,

g

, and

h

are

arithmetic functions

with

f

*

g

=

h

and at least two of them are

multiplicative

, then all of them are multiplicative.

The above

theorem

will be proven in two separate parts.

Lemma 1

.

If

f

and

g

are multiplicative, then so is

f

*

g

.

Proof.

Note that

(

f

*

g

)

⁢

(

1

)

=

f

⁢

(

1

)

⁢

g

⁢

(

1

)

=

1

⋅

1

=

1

since

f

and

g

are multiplicative.

Let

a

,

b

∈

ℕ

with

gcd

⁡

(

a

,

b

)

=

1

. Then any

divisor

d

of

a

⁢

b

can be uniquely factored as

d

=

d

1

⁢

d

2

, where

d

1

divides

a

and

d

2

divides

b

. When a divisor

d

of

a

⁢

b

is factored in this manner, the fact that

gcd

⁡

(

a

,

b

)

=

1

implies that

gcd

⁡

(

d

1

,

d

2

)

=

1

and

gcd

⁡

(

a

d

1

,

b

d

2

)

=

1

. Thus,

(

f

*

g

)

⁢

(

a

⁢

b

)

=

∑

d

|

a

b

f

⁢

(

d

)

⁢

g

⁢

(

a

⁢

b

d

)

=

∑

d

1

|

a

d

2

|

b

f

⁢

(

d

1

⁢

d

2

)

⁢

g

⁢

(

a

⁢

b

d

1

⁢

d

2

)

=

∑

d

1

|

a

d

2

|

b

f

⁢

(

d

1

)

⁢

f

⁢

(

d

2

)

⁢

g

⁢

(

a

d

1

)

⁢

g

⁢

(

b

d

2

)

=

(

∑

d

1

|

a

f

⁢

(

d

1

)

⁢

g

⁢

(

a

d

1

)

)

⁢

(

∑

d

2

|

b

f

⁢

(

d

2

)

⁢

g

⁢

(

b

d

2

)

)

=

(

f

*

g

)

⁢

(

a

)

⁢

(

f

*

g

)

⁢

(

b

)

.

∎

Lemma 2

.

If

f

is an arithmetic function and

g

and

h

are

multiplicative functions

with

f

*

g

=

h

, then

f

is multiplicative.

Proof.

Let

a

,

b

∈

ℕ

with

gcd

⁡

(

a

,

b

)

=

1

.

Induction

will be used on

a

⁢

b

to establish that

f

⁢

(

a

⁢

b

)

=

f

⁢

(

a

)

⁢

f

⁢

(

b

)

.

If

a

⁢

b

=

1

, then

a

=

b

=

1

. Note that

f

⁢

(

1

)

=

f

⁢

(

1

)

⋅

1

=

f

⁢

(

1

)

⁢

g

⁢

(

1

)

=

(

f

*

g

)

⁢

(

1

)

=

h

⁢

(

1

)

=

1

. Thus,

f

⁢

(

a

⁢

b

)

=

f

⁢

(

1

)

=

1

=

1

⋅

1

=

f

⁢

(

a

)

⁢

f

⁢

(

b

)

.

Now let

a

⁢

b

be an arbitrary

natural number

. The

induction hypothesis

yields that, if

d

1

divides

a

,

d

2

divides

b

, and

d

1

⁢

d

2

<

a

⁢

b

, then

f

⁢

(

d

1

⁢

d

2

)

=

f

⁢

(

d

1

)

⁢

f

⁢

(

d

2

)

. Thus,

h

⁢

(

a

)

⁢

h

⁢

(

b

)

=

h

⁢

(

a

⁢

b

)

=

(

f

*

g

)

⁢

(

a

⁢

b

)

=

∑

d

|

a

b

f

⁢

(

d

)

⁢

g

⁢

(

a

⁢

b

d

)

=

f

⁢

(

a

⁢

b

)

⁢

g

⁢

(

1

)

+

∑

d

|

a

b

d

<

a

⁢

b

f

⁢

(

d

)

⁢

g

⁢

(

a

⁢

b

d

)

=

f

⁢

(

a

⁢

b

)

⋅

1

+

∑

d

1

|

a

d

2

|

b

d

1

⁢

d

2

<

a

⁢

b

f

⁢

(

d

1

⁢

d

2

)

⁢

g

⁢

(

a

⁢

b

d

1

⁢

d

2

)

=

f

⁢

(

a

⁢

b

)

+

∑

d

1

|

a

d

2

|

b

d

1

⁢

d

2

<

a

⁢

b

f

⁢

(

d

1

)

⁢

f

⁢

(

d

2

)

⁢

g

⁢

(

a

d

1

)

⁢

g

⁢

(

b

d

2

)

=

f

⁢

(

a

⁢

b

)

-

f

⁢

(

a

)

⁢

f

⁢

(

b

)

⁢

g

⁢

(

1

)

⁢

g

⁢

(

1

)

+

∑

d

1

|

a

d

2

|

b

f

⁢

(

d

1

)

⁢

f

⁢

(

d

2

)

⁢

g

⁢

(

a

d

1

)

⁢

g

⁢

(

b

d

2

)

=

f

⁢

(

a

⁢

b

)

-

f

⁢

(

a

)

⁢

f

⁢

(

b

)

⋅

1

⋅

1

+

(

∑

d

1

|

a

f

⁢

(

d

1

)

⁢

g

⁢

(

a

d

1

)

)

⁢

(

∑

d

2

|

b

f

⁢

(

d

2

)

⁢

g

⁢

(

b

d

2

)

)

=

f

⁢

(

a

⁢

b

)

-

f

⁢

(

a

)

⁢

f

⁢

(

b

)

+

(

f

*

g

)

⁢

(

a

)

⁢

(

f

*

g

)

⁢

(

b

)

=

f

⁢

(

a

⁢

b

)

-

f

⁢

(

a

)

⁢

f

⁢

(

b

)

+

h

⁢

(

a

)

⁢

h

⁢

(

b

)

.

It follows that

f

⁢

(

a

⁢

b

)

=

f

⁢

(

a

)

⁢

f

⁢

(

b

)

.
∎

The theorem follows from these two lemmas and the fact that

convolution

is

commutative

.

The theorem has an

obvious

corollary.

Corollary

.

If

f

is multiplicative, then so is its convolution inverse.

Proof.

Let

f

be multiplicative. Since

f

⁢

(

1

)

≠

0

,

f

has a convolution inverse

f

-

1

. (See

convolution inverses for arithmetic functions

for more details.) Since

f

*

f

-

1

=

ε

, where

ε

denotes the convolution identity function, and both

f

and

ε

are multiplicative, the theorem yields that

f

-

1

is multiplicative.
∎

Title

elementary results about multiplicative functions and convolution

Canonical

name

ElementaryResultsAboutMultiplicativeFunctionsAndConvolution

Date of creation

2013-03-22 16:07:37

Last modified on

2013-03-22 16:07:37

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

16

Author

Wkbj79 (1863)

Entry type

Theorem

Classification

msc 11A25

Related topic

ConvolutionInversesForArithmeticFunctions

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