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IJMMS 29:11 (2002) 633–639 PII. S0161171202012498

http://ijmms.hindawi.com © Hindawi Publishing Corp.

CHARACTERIZING COMPLETELY MULTIPLICATIVE FUNCTIONS BY GENERALIZED MÖBIUS

FUNCTIONS

VICHIAN LAOHAKOSOL, NITTIYA PABHAPOTE,

and NALINEE WECHWIRIYAKUL

Received 23 March 2001 and in revised form 15 August 2001

Using the generalized Möbius functions, µα, first introduced by Hsu (1995), two charac- terizations of completely multiplicative functions are given; save a minor condition they read (µαf)−1 = µ−αf and fα = µ−αf . 2000 Mathematics Subject Classification: 11A25.

1. Introduction. Hsu [6], see also Brown et al. [3], introduced a very interesting

arithmetic function

µα(n)= ∏ p|n

( α

νp(n)

) (−1)νp(n), (1.1)

where α∈R, and n=∏p primepνp(n) denotes the prime factorization of n. This function is called the generalized Möbius function because µ1 = µ, the well-

known Möbius function. Note that µ0 = I, the identity function with respect to Dirichlet convolution, µ−1 = ζ, the arithmetic zeta function and µα+β = µα∗µβ; α, β being real numbers. Recall that an arithmetic function f is said to be completely multiplicative if f(1)≠ 0 and f(mn)= f(m)f(n) for allm andn. As a tool to characterize completely multiplicative functions, Apostol [1] or Apostol [2, Problem 28(b), page 49], it is known

that for a multiplicative function f , f is completely multiplicative if and only if

(µf)−1 = µ−1f = µ−1f . (1.2)

Our first objective is to extend this result to µα.

Theorem 1.1. Let f be a nonzero multiplicative function and α a nonzero real number. Then f is completely multiplicative if and only if

( µαf

)−1 = µ−αf . (1.3) In another direction, Haukkanen [5] proved that if f is a completely multiplicative

function and α a real number, then fα = µ−αf . Here and throughout, all powers refer to Dirichlet convolution; namely, for positive integral α, define fα := f ∗···∗f (α times) and for real α, define fα = Exp(αLogf), where Exp and Log are Rearick’s operators [9]. Our second objective is to establish the converse of this result. There

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634 VICHIAN LAOHAKOSOL ET AL.

is an additional hypothesis, referred to as condition (NE) which appears frequently.

By condition (NE), we refer to the condition that: if α is a negative even integer, then assume that f(p−α−1)= f(p)−α−1 for each prime p.

Theorem 1.2. Let f be a nonzero multiplicative function and α ∈ R−{0,1}. As- suming condition (NE), if fα = µ−αf , then f is completely multiplicative.

Because of the different nature of the methods, the proof of Theorem 1.2 is divided

into two cases, namely, α ∈ Z and α �∈ Z. As applications of Theorem 1.2, we deduce an extension of Corollary 3.2 in [11] and a modified extension of [7, Theorem 4.1(i)].

2. Proof of Theorem 1.1. If f is completely multiplicative, then (µαf)−1 = µ−αf follows easily from Haukkanen’s theorem [5]. To prove the other implication, it suffices

to show that f(pk)= f(p)k for each prime p and nonnegative integer k. This is trivial for k= 0,1. Assuming f(pj)= f(p)j for j = 0,1, . . . ,k−1, we proceed by induction to settle the case j = k > 1. From hypothesis, we get

µαf ∗µ−αf = I. (2.1)

Thus

0= I(pk)= ∑ i+j=k

µ−α ( pi ) f ( pi ) µα ( pj ) f ( pj )

= (−1)k ∑ i+j=k

(−α i

)( α j

) f ( pi ) f ( pj ) .

(2.2)

Simplifying and using induction hypothesis, we get

− [( α+k−1

k

) +(−1)k

( α k

)] f ( pk )= k−1∑

j=1

[ (−1)j

( α j

)( α+k−j−1

k−j

)] f(p)k. (2.3)

From Riordan [10, identity (5), page 8], the coefficient of f(p)k on the right-hand side is equal to

0− [ (−1)0

( α 0

)( α+k−1

k

) +(−1)k

( α k

)( α+k−k−1

k−k

)]

=− [( α+k−1

k

) +(−1)k

( α k

)] ≠ 0

(2.4)

and the desired result follows.

Remark 2.1. (1) To prove the “only if” part of Theorem 1.1, instead of using

Haukkanen’s result, a direct proof based on [1, Theorem 4(a)] can be done as follows:

if f is completely multiplicative, then (µαf)∗(µ−αf)= (µα∗µ−α)f = µ0f = If = I.

CHARACTERIZING COMPLETELY MULTIPLICATIVE FUNCTIONS . . . 635

(2) To prove the “if” part of Theorem 1.1, instead of using [10, identity (5)], a self-

contained proof can be done as follows: from (1+z)α ·(1+z)−α = 1 we infer that, for k > 1,

∑ i+j=k

(−α i

)( α j

) = 0 (2.5)

which implies

(−α k

) + ( α k

) =−

[k−1∑ i=1

(−α i

)( α k−i

)] . (2.6)

Thus,

0= ∑ i+j=k

(−α i

)( α j

) f ( pi ) f ( pj )

= [(−α

k

) + ( α k

)] f ( pk )+

[k−1∑ i=1

(−α i

)( α k−i

)] f(p)k

(2.7)

implies f(pk)= f(p)k.

3. Proof of Theorem 1.2. The proof of Theorem 1.2 is much more involved and we

treat the integral and nonintegral cases separately. This is because the former can be

settled using only elementary binomial identities, while the proof of the latter, which

is also valid for integral α, makes use of Rearick logarithmic operator, which deems nonelementary to us.

Proposition 3.1. Let f be a nonzero multiplicative function and r a positive integer ≥ 2. If f r = µ−r f , then f is completely multiplicative.

Proof. Since f is multiplicative, it is enough to show that

f ( pk )= f(p)k, (3.1)

where p is a prime and k a nonnegative integer. This clearly holds for k= 0,1. As an induction hypothesis, assume this holds for 0,1, . . . ,k−1 (≥ 1).

From

( µ−r f

)( pk )= f r (pk), (3.2)

we get, using induction hypothesis,

(−r k

) (−1)kf (pk)= ∑

j1+···+jr=k f ( pj1

) f ( pj2

)···f (pjr )

= rf (pk)+f(p)k ∑ j1+···+jr=k

all ji≠k

1. (3.3)

636 VICHIAN LAOHAKOSOL ET AL.

Simplifying, we arrive at

[( k+r −1 r −1

) −r

]( f ( pk )−f(p)k)= 0. (3.4)

Since r ≥ 2, then (k+r−1r−1 )−r ≠ 0, and we have the result. Remark 3.2. The case r = 1 is excluded for µ−1f = ζf is always equal to f , and

so the assumption is empty. The case r = 0 is excluded because I = f 0 = µ0f = If holds for any arithmetic function f with f(1)= 1.

Proposition 3.3. Let f be a nonzero multiplicative function and −α= r a positive integer. Assuming condition (NE), if f−r = µrf , then f is completely multiplicative.

Proof. As in Proposition 3.1, we show by induction that f(pk)= f(p)k for prime p and nonnegative integer k, noting that it holds trivially for k = 0,1. The main as- sumption of the theorem gives

µrf ∗f r = I. (3.5)

We have, for k≥ 2,

0= I(pk)= ∑ i+j1+···+jr=k

( µrf

)( pi ) f ( pj1

)···f (pjr ). (3.6)

Using induction hypothesis and [10, identity (5)], the right-hand expression is

∑ j1+···+jr=k

f ( pj1

)···f (pjr )+f(p)k k−1∑ i=1 (−1)i

( r i

) ∑ j1+···+jr=k−i

1+(µrf )(pk)

= rf (pk)+f(p)k [( k+r −1 r −1

) −r

] +f(p)k

k−1∑ i=1 (−1)i

( r i

)( k−i+r −1 r −1

)

+(−1)k ( r k

) f ( pk )=

[ r +(−1)k

( r k

)]( f ( pk )−f(p)k).

(3.7)

For positive integers r and k (≥ 2), observe that r + (−1)k(rk) = 0 if and only if k= r −1 and k is odd. The conclusion hence follows.

Remark 3.4. In the case of r being a positive even integer, without an additional as- sumption on f(pr−1), Proposition 3.3 fails to hold as seen from the following example.

Take r = 4. For each prime p, set

f(1)= f(p)= f (p2)= 1, f (p3)= 0 (3.8) and for k≥ 4, define f(pk) by the relation (µ4f ∗f 4)(pk)= I(pk)= 0.

Define other values of f by multiplicativity, namely,

f ( pa11 ···pakk

)= f (pa11 )···f (pakk ); (3.9)

CHARACTERIZING COMPLETELY MULTIPLICATIVE FUNCTIONS . . . 637

pi prime, ai nonnegative integer. This particular function satisfies µ4f = f−4 and is multiplicative, but not completely multiplicative.

Now for the case of nonintegral index, we need one more auxiliary result. For more

details about the Rearick logarithm, see [8, 9].

Lemma 3.5. Let f be an arithmetic function, p a prime, k a positive integer, and let Log denote the Rearick logarithmic operator defined by

Logf(1)= logf(1),

Logf(n)= 1 logn

∑ d|n f(d)f−1

( n d

) logd (n > 1). (3.10)

If f(1)= 1, f(pi)= f(p)i (i= 1,2, . . . ,k−1), then

(Logf) ( pi )= f(p)i

i (i= 1,2, . . . ,k−1). (3.11)

Proof. From hypothesis, we have

f(1)= f−1(1)= 1, f−1(p)=−f(p), (3.12)

and so

Logf(1)= 0, Logf(p)= f(p). (3.13)

Next,

Logf ( p2 )= 1

logp2 [ f ( p2 ) logp2+f(p)f−1(p) logp

] = 1

2 f(p)2. (3.14)

Now proceed by induction noting, as in the l

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