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Master’s Thesis
Continuous Bernoulli Distribution and p-adic Periodic Zeta Function
Sunmin Gwak
Department of Mathematical Sciences
Ulsan National Institute of Science and Technology
2021
Continuous Bernoulli Distribution and p-adic Periodic Zeta Function
Sunmin Gwak
Department of Mathematical Sciences
Ulsan National Institute of Science and Technology
Abstract
Thep-adic DirichletL-function or the Kubota-Leopoldtp-adicL-function is ap-adic version of DirichletL-function. It is defined by theMellin-Mazur transformof the Bernoulli distribution µk on Z×p for an integerk≥1. In this thesis, we extend the Bernoulli distribution with integer parameters to a continuous version i.e., a Bernoulli number with a continuous parameter. Also, we generalize the p-adic Dirichlet zeta functions that is defined only for Dirichlet characters to thep-adic periodic zeta functions that is defined for arbitrary periodic functions with values in Cp.
Contents
1 Introduction . . . 1
2 Preliminaries . . . 3
2.1 Bernoulli number . . . 3
2.2 p-adic numbers . . . 3
2.3 Hensel’s lemma . . . 5
2.4 Teichüller character . . . 6
3 p-adic integration . . . 7
4 p-adic interpolation . . . 10
5 Continuous Bernoulli Distribution andp-adic periodic zeta function . . . 17
6 p-adic periodic zeta function . . . 25
References . . . 27
Acknowledgements . . . 30
1 Introduction
Thep-adic DirichletL-function or the Kubota-Leopoldtp-adicL-function is a p-adic version of DirichletL-function. In other words, it is ap-adic analytic function onZp that interpolates the special values of DirichletL-functions at the non-positive integers. It possesses several properties that the complex DirichletL-functions enjoy.
The importance ofp-adic Dirichlet L-functions comes from several roles played by the func- tion in the Iwasawa theory. For example, the vanishing of µ-invariant of the p-adic Dirich- let L-function conjectured by Iwasawa in 1960 and had been an open problem until Ferrero- Washington [1] finally resolved in 1979. Relevant problems and generalizations can be found in [2], [3], [6], [7], [4], [14], [15], [16], and [17].
The p-adic Dirichlet L-function is defined by the Mellin-Mazur transform of the Bernoulli distribution µk on Z×p for an integer k ≥ 1 (See [5], [8], [9], and [18] for other constructions).
One can show that the Bernoulli distribution satisfies an additive property on the disjoint union of compact-open subsets of Zp, which comes from the properties of the Bernoulli polynomial.
Also, using Mazur’s treatment, one can normalize the Bernoulli distribution in order to get a p-adically bounded distribution, i.e., ap-adic measure, sayµk,αfor ap-adic integerα. A famous identity present a relation between the measures µ1,α andµk,α.
The Mellin-Mazur transform of a measure is the integration of the continuous character x 7→ (x/ω(x))sχ(x) for the Teichmüller character ω and a Dirichlet character χ on Z×p, with respect to the measure. This amounts to the complex version, namely that the Dirichlet L- function is a Mellin transform of the continuous character y7→ ys with respect to the measure e−2πy dyy on the group R×>0.
In this thesis, we will generalize the previous discussion to the case of p-adic zeta functions for arbitrary periodic functions with values in Cp (see section 6). For this generalization, we will first extend the Bernoulli distribution with integer parameters to a continuous version, i.e., a Bernoulli number with a continuous parameter (see section 5). By a Bernoulli number with a continuous parameter, we mean a p-adic analytic functions that interpolate modified Bernoulli numbers.
Following the previous construction in a similar way, we also define a distribution µs,λ onZp
for a continuous parameters∈Zp and a periodic function λ(see Definition 5.1). We also adopt the treatment of Mazur in order to obtain another distribution µs,λ,κ,α that depends onp-adic integerss,α, and two periodic functionsλ and κ by normalizing. In fact, we present a precise condition on the periodic functions λand κ in order for µs,λ,κ,α to be p-adically bounded (see Definition 5.2 and Theorem 5.4 (1)).
By extending the previous result on the relation betweenµ1,αandµk,α, we obtain a relation between µs,λ,κ,α and µ1,λ,κ,α. In particular, it can be shown that a limit lims→0s−1dµs,λ,κ,α of measures, is again a measure (see Theorem 5.4 (2)) and it also can be written in terms ofx/ω(x) and µ1,λ,κ,α. This result is not visible when one considers only the Bernoulli distribution with
the integer parameters. In addition, the same discussion applies to distributions Bs and Ds
defined by p-adic analytic functions (see Definition 5.5).
2 Preliminaries
This section describes the materials that will be necessary in the latter part of thesis. The main reference is [12]. Please refer to the reference for more explanation.
2.1 Bernoulli number
In this section, we present basic definitions on the Bernoulli number that will be used in latter part of thesis.
Definition 2.1. Define the Bernoulli numbers, denoted Bn as the n-th coefficient of a power series
t et−1 =
∞
X
n=0
Bntn n! .
From the series, one can get B2k+1= 0 for k≥1. Also, it is already known that B0 = 1, B1=−1
2, B2 = 1
6, B4 =− 1
30, B6= 1 42, ...
The following definition will play an important role in this thesis.
Definition 2.2. Define the n-th Bernoulli polynomial as
Bn(x) =
n
X
k=0
n k
Bkxn−k.
Since
ext=
∞
X
i=0
(xt)i i! ,
one can easily get a generating function as follows, text
et−1 =
∞
X
k=0
Bk(x)tk
k! . (1)
2.2 p-adic numbers
Let us review basic definitions of p-adic numbers. For this, we review first on a general theory on the completion of a metric space.
Definition 2.3. Let k·k be a norm on a field F. Define a metric on field F by setting the distance between x and y as d(x, y) =kx−yk.
Definition 2.4. A metric space is a setF with a metricdthat assigns a nonnegative real number d(x, y) for every x, y∈F satisfying the following:
(1) d(x, y)≥0 and d(x, y) = 0⇐⇒x=y, (2) d(x, y) =d(x, y),
(3) d(x, y) +d(y, z)≥d(x, z). (The triangle inequality)
Definition 2.5. Let F be a field and let d be a metric on F. A completion of a metric space (F, d) is a pair consisting of a complete metric space (F∗, d∗) and an isometry φ:F →F∗ such that φ[F] is dense in F∗.
Let F be a field with norm k·k. Also, let R be the set of all Cauchy sequences {an}∞n=1 with respect to the norm k·kwhen an∈F. Define addition and multiplication of sequences as follows:
{an}∞n=1+{bn}∞n=1 ={an+bn}∞n=1 {an}∞n=1× {bn}∞n=1={anbn}∞n=1.
Then it is a tedious task to show that(R,+,×)is a commutative ring. Anull Cauchy sequence is defined as a sequence that converges to zero. Letmbe the set of null Cauchy sequences. Then the set m is a subset of R and it can be easily verified that it is a maximal ideal of R. Thus R/mis a field.
The field F can be embedded inR with the map a7→(a, a, ...) and obviously(a, a, ...) is a Cauchy sequence. Then the field F can be considered as a subfield of R/m. Thus R/mis the completion of F with respect to k·k.
From now on, we will see a specific metric spaceQ with the following norm:
Definition 2.6. Let p be a prime number and x a rational number x 6= 0. Express x as x=pvp(x)y wherey is a rational number coprime top andvp(x) is an integer. Define thep-adic absolute value of x, denoted |x|p as
|x|p:=p−vp(x). When x= 0, set |0|p= 0.
One can show with no difficulty that the distance function d(x, y) =|x−y|p satisfies three properties in Definition 2.4. In particular, it can be easily checked that the triangle inequality comes from the following inequality:
|x+y|p≤max(|x|p,|y|p). (2) Furthermore, the equality in (2) holds when|x|p6=|y|p.
Let F = Q with the usual absolute value. Then by above construction, one can get the field of real numbers. Now, consider that F =Q withp-adic norm |·|p. Then by completingQ with respect to the new distance, one can also get a new field that is called the field of p-adic numbers, denoted byQp.
Definition 2.7. Define the set ofp-adic integers as follows:
Zp ={x∈Qp :|x|p≤1}.
Remark 2.1.
1. Zp is a subring of Qp.
2. It can be shown thatZis dense in Zp.
3. One can write elements of Zp as formal power series as follows:
∞
X
n=0
bnpn,0≤bn≤p−1.
Now, consider the function f(x) = ax when a is a fixed number. Consider in the case of complex analysis, one can easily compute the derivative of f(x) i.e., f0(x) = (loga)ax. In the p-adic case, although the proof is not easy, one can get a similar result. Then by this result, one can get the following formula,
logpa= lim
n→∞
apn−1 pn .
This formula can be interpreted as the p-adic derivative of ax at x = 0. So, the conclusion is that(apn−1)/pn behave like the logarithm.
We will introduce here the N-th partial sum of the p-adic expansion ofα denoted by(α)N. Suppose thatα∈Zp. Let (α)N be the rational integer that congruent to α(modpN). If
α=a0+a1p+a2p2+· · ·, then
(α)N =a0+a1p+· · ·+an−1pN−1. 2.3 Hensel’s lemma
We introduce well-known Hensel’s lemma here to explain the Teichmüller character in Section 2.4.
Theorem 2.1 (Hensel’s lemma). Let f(x)∈ Zp[x]be a polynomial that coefficients are in Zp. Suppose that f(x1) ≡ 0 (modp) and f0(x1) 6≡ 0 (modp). Then there exists a unique p-adic integer y such that f(y) = 0 and y≡x1(modp).
Proof. We prove inductively that if
f(x)≡0 (modpn)
has a solutionan then it can be lifted uniquely toan+1(modpn+1)such that f(an+1)≡0 (modpn+1)
and an+1 ≡an(modpn). Indeed, if we write an+1 =pnt+an, then we get f(pnt+an)≡0 (modpn+1).
And writef(x) =P
icixi, then we obtain f(pnt+an) =X
i
ci(an+pnt)i
≡X
i
ci(ain+ipntai−1n ) (modpn+1)
≡f(an) +pntf0(an) (modpn+1).
Now, we need to solve fortin the congruence
pntf0(an) +f(an)≡0 (modpn+1).
Sincef(an)≡0(modpn) this reduces to
tf0(an)≡ −(f(an)/pn) (modp)
which has a unique solution t(modp) because an ≡a1 (modp) and f0(a1)6≡0 (modp). Also, the sequence {an}∞n=1 is a Cauchy sequence, and limit of this sequence is the required solution.
Since an+1 is a unique lifting (mod pn+1) of an (modpn), the uniqueness of the limit solution is clear. This completes the proof of the theorem.
2.4 Teichüller character
Now let the polynomialf(x) =xm−1withm|p−1. We already know that (Z/pZ)×is a cyclic group of orderp−1. Sof(x)≡0(modp)hasmdistinct solutions. Consider the case whenp >2 andm=p−1. By Hensel’s lemma, for each1≤i≤p−1, there exists a numberω(i)∈Zp such thatω(i)≡i(modp) and ω(i)p−1= 1. These ω(i) are called the Teichmüller representatives of the residue classes mod p. Now, consider the map
(Z/pZ)×→Zp
given by x 7→ ω(x). Then this map defines the Teichmüller character. And the Teichmüller character is a multiplicative character. ω(x) is defined by ω(x mod p) for x ∈ Z×p. For every x∈Z×p, one can writex uniquely asx=ω(x)hxiwhen hxi ∈1 +pZp. For p6= 2, we have
ω(x) = lim
n→∞xpn,
(By using induction, it is easy to verify that a ≡b (mod p) implies apn ≡bpn (mod pn+1), so that)
ω(x) =ω(x)pn = xpn hxipn and the denominator will go to 1 whenn→ ∞.
3 p-adic integration
Note that in the p-adic topology, all sets (the author of [12] calls them “intervals") of the form a+pNZp
are both closed and open. The Riemann integral can be built over these sets. Suppose thatX is a compact-open subset ofQp (such as an “interval", Zp or Z×p). Then a p-adic distributionµon X can be regarded as aQp-linear functional on theQp-vector space of locally constant functions onX. Recall that a functionf is locally constant if for everyx∈Xthere exists a neighbourhood U such that f is constant in U. Letcbe a constant andχU be the characteristic function ofU. Then any locally constant function can be expressed in a linear combination of functions of the formcχU. Now, one can write µ(U) for µ(χU). Let us present another definition:
Definition 3.1. Ap-adic distribution µon X is an additive map from the set of compact-open sets in X to Qp. That is, ifU ⊆X is a disjoint union of compact-open setsU1, U2, ..., Un, then
µ(U) =µ(U1) +· · ·+µ(Un).
A typical “interval" in Qp is of the form
a+pnZp ={x∈Qp :|x−a|p≤p−n}.
A basis of open sets forQp is the collection of such sets. So, ap-adic distribution is determined by its values on such “intervals". This idea is explained in the following theorem, which is the fundamental property of distributions on Qp.
Theorem 3.1. Every map µ from the set of “intervals" of X to Qp satisfying
µ(a+pnZp) =
p−1
X
b=0
µ(a+bpn+pn+1Zp) (3) whena+pnZp⊆X, extends uniquely to a p-adic distribution onX.
Proof. Every compact-open setsU ⊆X can be expressed as a finite disjoint union of intervals, i.e., U =S
iIi. So, one can write
µ(U) =X
i
µ(Ii).
Now, one have to check that well-definedness of this. Assume that U = S
iIi and U = S
iJi, whenIi andJi are “intervals". Then one can refine these decompositions toS
ijIij when theIij are intervals. Thus the result is verified and uniqueness is also easy to verified.
In the p-adic integration, the locally constant function is used like the step function of the Riemann integration. By above theorem,R
xf dµcan be defined using locally constant functions.
First, when Ui are compact-open sets, one can writef as a linear combination of characteristic functionsχUi, i.e.,
f =X
i
ciχUi.
ThenR
xf dµcan be defined as follows, Z
X
f dµ=X
i
ciµ(Ui).
Example 3.1. Recall that for each positive integer k, Bk(x) is the k-th Bernoulli polynomial.
The Bernoulli distribution is defined as
µk(a+pnZp) =pn(k−1)Bk(a/pn).
It is necessary to check that (3) holds, i.e., it is necessary to show that pn(k−1)Bk(a/pn) =
p−1
X
b=0
p(n+1)(k−1)Bk((a+bpn)/pn+1)
holds. Now multiply p−n(k−1) on both sides of this equation. Then the identity to be proved follows from
Bk(px) =pk−1
p−1
X
b=0
Bk(x+b
p) (4)
which is easily deduced from the power series generating function for the Bernoulli polynomials as follows. Recall the equation (1):
∞
X
k=0
Bk(x)tk
k! = text et−1. Then one can get
∞
X
k=0
Bk(px)tk
k!= tepxt et−1.
On the other hand, the right hand side of the equation (4) is the coefficient of tk in the power series expansion of
pk−1
p−1
X
b=0
te(x+b/p)t
et−1 = pk−1text et−1
p−1
X
b=0
ebt/p= pk−1text
et−1 · et−1
et/p−1 = pk(t/p)epx(t/p) e(t/p)−1 ,
and the coefficient oftk on the right hand side isBk(px). So one can conclude thatµkis ap-adic distribution.
Now, we introduce here a p-adic measure to definep-adic integration.
Definition 3.2. A distribution µof X is a p-adic measure on X if
|µ(U)|p≤B where B is a constant and for all compact-open U ⊆X.
We are ready to give the p-adic version of the Riemann integration:
Theorem 3.2. Let µ be a p-adic measure on X and let f :X → Qp be a continous function.
Then the “Riemann sums"
SN := X
0≤a<pN a+pNZp⊆X
f(xa,N)µ(a+pNZp)
(where the sum is taken over all afor which a+pNZp ⊆X andxa,N is chosen in the “interval"
a+pNZp) converge to a limit inQp as N → ∞. This limit is independent of the{xa,N}.
Proof. First, it is necessary to check that the sequence {SN} is Cauchy sequence. Let X be a finite union of intervals. Choose large enoughN so that every intervala+pNZp is contained in X or disjoint from X. Suppose thatM > N. Let˜abe the least non-negative residue of amod pN. Then, since µholds additivity, SN can be rewritten as
SN = X
0≤a<pM a+pMZp⊆X
f(xa,N˜ )µ(a+pMZp).
Let > 0. Since X is compact and f is a continuous function, f is uniformly continuous.
Then for large enoughN,|f(x)−f(y)|p≤whenx≡y(modpN). Thus, we have
|SN −SM|p =
X
0≤a<pM a+pMZp⊆X
(f(x˜a,N)−f(xa,M))µ(a+pMZp) p
≤max
a
|f(xa,N˜ )−f(xa,M)|p·|µ(a+pMZp|p
≤B,
since x˜a,N ≡xa,M (modpN). So, one can conclude that the “Riemann sums" have a limit. Also it is easy to see that the limit does not depend on the choice of the {xa,N}. The theorem is verified.
Assume f is a function that satisfies the hypothesis of the above theorem. Then denote the limit of the Riemann sums by the symbol
Z
X
f dµ.
Now, choosexa,N =a. Then the above integral is equivalent to the following limit
N→∞lim X
0≤a<pN
f(a)µ(a+pNZp).
Suppose that µ is a distribution and α ∈ Qp. Then αµ is again a distribution. Now, let α∈Z×p, and we define µ0 asµ0(U) =µ(αU). Then also µ0 is again a distribution.
4 p-adic interpolation
Most of materials in this section comes from [12].
Given a sequence of integers {ak}∞k=1, one can ask when we get a continuous function f : Zp → Qp such that f(k) = ak. Since Zp is compact, such a continuous function is uniformly continuous and bounded. Thus, the following condition is a necessary condition: for each m, there exist an integerN =N(m) such that
k≡k0(modpN)⇒ak ≡ak0(modpm). (5) That is, if k and k0 are close p-adically, thenak and ak0 are also close p-adically. Conversely, whenever (5) is hold, and
sup
n
|an|p
is bounded, then one can get a continuous function f :Zp →Qp, and the function is as follows.
Letni be a sequence of integers tending to x for x∈Zp. Now define f(x) = lim
i→∞f(ni).
The limit exists by (5). In addition, ifa0i is any other sequence tending to x, one can easily see that
i→∞lim f(ni) = lim
i→∞f(n0i)
so that the function is well-defined. Therefore, the interpolation problem can be reduced to checking (5). These observations can be stated as follows:
Theorem 4.1. Assume that a sequence of numbersf(k), fork= 0,1,2, ...isp-adically bounded and satisfies the following condition. For each natural number m, there exist a natural number N =N(m) such that
k≡k0(modpN)⇒f(k)≡f(k0)(modpm).
Then, there exist a continuous function f˜:Zp→Zp such that f˜(k) =f(k).
As an application of the above theorem, notice that if n≡1(modp), the functionf(k) =nk can be p-adically interpolated since the condition (5) is easily verified. In fact, f(s) =ns is a p-adically analytic function for s ∈ Cp satisfying |s|p< p
p−2
p−1. For odd primes p, this region is larger than the unit disc.
Recall that the classical theorem of Weierstrass: a continuous function on a closed interval can be uniformly approximated by polynomials. Thep-adic version of this theorem is Mahler’s theorem. This theorem says that for any continuous function f :Zp → Qp, there are numbers ck inQp satisfying
f(x) =
∞
X
k=0
x k
ck.
Therefore the sequence of partial sums of this series gives the polynomial approximations for f(x).
Now, begin with some observations from combinatorial analysis. The proof of Mahler’s theorem will follow a treatment due to Bojanic.
Suppose that the p-adic expansions ofnand kare as follows:
n=a0+a1p+a2p2+· · · k=b0+b1p+b2p2+· · ·. Then it is easy to see that for primep,
n k
≡ a0
b0 a1
b1 a2
b2
· · ·(modp).
Especially,
pn k
≡0(modp) for 1≤k≤pn−1.
Recall the binomial inversion formula:
bn=
n
X
k=0
n k
ak (6)
if and only if
an=
n
X
k=0
n k
(−1)n−kbk. (7)
Now fix a positive integer m, and let ak =
(−1)k if k=m 0 otherwise.
Then the above binomial inversion formula gives the following relation:
n
X
k=0
(−1)k n
k k
m
=
(−1)m if n=m 0 otherwise.
(8)
Suppose thatf :Zp →Qp is continuous, and let an(f) =
n
X
k=0
n k
(−1)n−kf(k). (9)
Then one can get
f(n) =
n
X
k=0
n k
ak(f)
by (6). This suggests how to solve the interpolation problem. Given a sequencef(0), f(1), f(2), ..., definean(f) by (9) and let
f(x) :=
∞
X
k=0
x k
ak(f). (10)
If one can prove that this is a p-adically convergent series, it is done. Observe that x
n
=
x(x−1)· · ·(x−n+ 1)/n! if n≥1
1 if n= 0
is a continuous function of ap-adic variable and takes integer values forx∈Z. So, by continuity,
x n
p ≤1
for allx∈Zp. The main idea in Mahler’s theorem is to show that|ak(f)|p→0 ask→ ∞ when (5) is holds.
Now, new operators∆n is defined as follows,
∆nf(x) =
n
X
k=0
n k
(−1)n−kf(x+k).
So we get
∆f(x) =f(x+ 1)−f(x),
∆2f(x) =f(x+ 2)−2f(x+ 1) +f(x),
· · ·.
And it is clear that∆n is a linear operator. The following lemmas will be used:
Lemma 4.2.
∆nf(x) =
m
X
j=0
m j
∆n+jf(x−m).
Proof. To prove this lemma, it is enough to show that f(x) =
m
X
j=0
m j
∆jf(x−m).
Then the result can be derived by applying ∆n to both sides of the above equation. But
m
X
j=0
m j
∆jf(x−m) =
m
X
j=0
m j
j X
k=0
j k
(−1)j−kf(x−m+k)
=
m
X
k=0
(−1)kf(x−m+k)
m
X
j=0
m j
j k
(−1)j.
By (8), the inner sum is zero except for k=m. Wherek=m, it is(−1)m. The result is now clear.
Lemma 4.3. For any sequencef(0), f(1), f(2), f(3), ..., define an(f) :=
n
X
k=0
(−1)n−k n
k
f(k).
Then,
m
X
j=0
m j
an+j(f) =
n
X
k=0
(−1)n−k n
k
f(k+m).
Proof. Ifm= 0, the result is clear. By Lemma 4.2,
∆nf(m) =
m
X
j=0
m j
∆n+jf(0).
But
∆nf(m) =
n
X
k=0
n k
(−1)n−kf(m+k) and
∆nf(0) =an(f).
Then the result is now immediate.
Theorem 4.4 (Mahler, 1961). Assume that f :Zp→Qp is continuous function. Let an(f) =
n
X
k=0
(−1)n−k n
k
f(k).
Thenan(f)→0 as n→ ∞. Thus the series
∞
X
k=0
x k
ak(f) converges uniformly inZp. Moreover,
f(x) =
∞
X
k=0
x k
ak(f).
Proof. SinceZp is compact andf is continuous, it is uniformly continuous. So, for any positive integer s, there is a positive integer tsuch that
|x−y|p≤p−t⇒ |f(x)−f(y)|p≤p−s, wherex, y∈Zp. In particular, for k= 0,1,2, ...,
|f(k+pt)−f(k)|p≤p−s.
Sincef is continuous onZp and Zp is compact, it is bounded there. Without loss of generality, let |f(x)|p≤1 for all x ∈Zp. Then, |an(f)|p≤1 for all n. By Lemma 4.3, and the formula for an given there, one can get
an+pt(f) =−
pt−1
X
j=1
pt j
an+j(f) +
n
X
k=0
(−1)n−k n
k
{f(k+pt)−f(k)}.
In first sum of the above formula, p divides each of the binomial coefficients. So, one can get the following inequality:
|an+pt(f)|p≤ max
1≤j<pt{p−1|an+j(f)|p, p−s}
≤p−1 for n≥pt
since |an(f)|p≤1. Now if nis replaced byn+pt in the penultimate inequality, one can get the following inequality:
|an(f)|p≤p−2 for n≥2pt. And if one repeat the argument(r−1)times, one can get
|an(f)|p≤p−s for n≥spt. So, by above argument,an(f)→0 asn→ ∞. Therefore, the series
∞
X
k=0
x k
ak(f)
converges to f(x) since both are continuous and equal on a dense set and
x k
p ≤1 for x∈Zp.
Remark 4.1. This is thep-adic analogue of the classical theorem of Weierstrass that any contin- uous function on[−1,1]can be uniformly approximated by polynomials.
Corollary 4.5. Let {an} be any sequence in Cp. Define bn:=
n
X
k=0
(−1)n−k n
k
ak.
Then there exists a continuous function f :Zp → Cp such that f(n) =an if and only if bn →0 as n→ ∞.
Proof. This is clear from Mahler’s theorem.
Now consider the convergence problem for x ∈Cp. When certain conditions holds, one can prove that the above series extends to an analytic function for|x|p< RwhereR >1. To achieve such a result, one can follow Washington’s approach.
Lemma 4.6. Let Pi(x) =P∞
n=0an,ixn be a sequence of power series that converge in some fixed subset D of Cp. Assume that for each n, an,i → an,0 as i → ∞. And suppose that for each x∈D and >0 there exist an n0 =n0(x, ) such that
X
n≥n0
an,ixn <
uniformly in i. Thenlimi→∞Pi(x) =P0(x) for all x∈D.
Proof. Let >0 and x ∈D be given. And choosen0 as the above. Then for sufficiently large i, one can get
|P0(x)−Pi(x)|≤max
n<n0
{,|an,0−an,i|p|x|np}=.
Theorem 4.7. Suppose r < p−1/(p−1) <1 and f(x) =
∞
X
n=0
x n
an
for x ∈ Cp when |an|p≤ M rn for some M. Then, f(x) can be written as a power series. And the radius of convergence of that power series is at leastR = (rp1/(p−1))−1.
Proof. First, consider the partial sums Pi(x) =X
n≤i
x n
an=X
n≤i
an,ixn.
Then
an,i=an
integer
n! +an+1
integer
(n+ 1)! +· · ·. Since ordpj!≤j/(p−1), one can get
|an,i|p≤max
j≥n|aj/j!|p≤max
j≥n M rjpj/(p−1)≤M R−n. And also
|an,i−an,i+k|p=
ai+1integer
(i+ 1)! +· · ·+ai+kinteger (i+k)!
p
≤M R−(i+1)→0
as i → ∞. So, by the above, one can conclude that {an,i}∞i=1 is a Cauchy sequence. Let an,0 = limi→∞an,i. Then|an,0|p≤M R−n. Now let
P0(x) =
∞
X
n=0
an,0xn.
Then for x ∈Cp, P0 converges where|x|p< R. Since thePi are polynomials, they converge in the following region,
D={x∈Cp:|x|p< R}.
Furthermore,
X
n≥n0
an,ixn
≤max
n≥n0M R−n|x|np→0
as n0 → ∞uniformly in i for |x|p< R. Therefore for |x|p< R, |P0(x)−Pi(x)|p≤ uniformly.
So, the result is now clear.
Remark 4.2. Since the above Ris bigger than 1, i.e.,R >1, the series in the above theorem has a wider radius of convergence.
By Mahler’s theorem, when (5) is hold, f(x) =
∞
X
n=0
x n
af(n)
is a continuous function onZp. Also, suppose that the growth condition of Theorem 4.7 holds in the sequence an=f(n). Then one can get a p-adic analytic continuation off(x) with a wider domain. As we show below in Theorem 4.8, one can sharpen the Theorem 4.7.
Theorem 4.8. Let f :Zp →Cp be a continuous function with the Mahler series f(x) =
∞
X
k=0
x k
ck.
Then, f is the restriction of an analytic power series
∞
X
n=0
anxn
with an→0 if and only if|ck/k!|p→0 ask→ ∞.
Proof. The natural basis of the vector space of polynomials:
1, x, x2, ....
But one can also use the basis
1,(x)1,(x)2, ...,
where(x)k=x(x−1)...(x−k+ 1). Let s(n, k)be the Stirling numbers of the first kind. Then one can get the formulas:
(x)n=
n
X
k=0
s(n, k)xk.
Also, let S(n, k) be the Stirling numbers of the second kind. Then one can get xn=
n
X
k=0
S(n, k)(x)k.
The Stirling numbers of the second kind interpret combinatorically the number of partitions of a set of n elements into k blocks. In other words, S(n, k) interpret the number of equivalence relations on a set of n elements with k equivalence classes. Therefore, if one express f(x) = P
nanxn, one can also express it asP
kbk(x)k. And it is already known that the Stirling numbers are integers. So, one can get
sup
n
|an|p= sup
k
|bk|p. Butck =k!bk. So the result follows immediately.
Suppose thatn≡1(modp). Then one can get ns= (1 + (n−1))s=
∞
X
k=0
s k
(n−1)k.
And we can see that if we apply Theorem 4.7, then it gives us a p-adic continuation ofns.
5 Continuous Bernoulli Distribution and p-adic periodic zeta func- tion
The modified Bernoulli numbers Qk−1ω−k(a)Bk(a/Q), k ≥ 0 for p|Q,1 ≤ a < Q can be interpolated by an analytic function on Zp. Indeed the analytic function B(s, a, Q) in s ∈ Zp
defined by
B(s, a, Q) := hais Q
∞
X
n=0
s n
Q a
n
Bn (11)
interpolates the numbers
B(k, a, Q) = hais Q
∞
X
n=0
s n
Q a
n
Bn=Qk−1ω(a)−kBk a
Q
.
Definition 5.1. Regarding λ as a periodic function with the period N pm, for s∈Zp we define a function µs,λ on basic open subsets of Z×p such that
µs,λ(a+pmZp) = X
r≡a(pm) 0≤r<N pm
λ(r)B(s, r, N pm). (12)
Proposition 5.1. Let s = k ∈ Z≥1, and λ = ωk. Then µk,ωk = µk is the k-th Bernoulli distribution.
Proof. We omit this proof.
Proposition 5.2. Let λk(n) =ξnωk(n) for an N-th root of unity ξ, then µk,λk(a+pmZp) =
T d dT
k−1 T(a)m Tpm−1−1
T=ξ.
In particular, µ1,λ1(a+pmZp) = ξξpm(a)m−1 is a measure onZp, which is considered in [9] to deduce several properties of p-adic Dirichlet L-function.
Proof. By the above definitions, we have
µk,λk(a+pmZp) = (pmN)k−1 X
r≡a(pm) 0≤r<N pm−1
ξrBk r N pm
. (13)
SinceBk(x) = d
dt
k tetx et−1
t=0, the above formula (13) is equal to (N pm)k−1
N−1
X
q=0
ξpmq+aBkpmq+a N pm
= (N pm)k−1X
q
ξpmq+ad dt
kte(
pmq+a N pm )t
et−1 t=0
= (N pm)k−1ξad dt
k
t
N−1
X
q=0
ξpmqe(pmq+aN pm )t
et−1
= (N pm)k−1ξad dt
k−1
tepmNa tN−1X
q=0
ξpmqeqtN
et−1
= (N pm)k−1ξa d
dt k−1
epmNa t N−1X
q=0
ξpmqeqtN
et−1
t=0. (14) LetT =epmN1 t. Then we have dtd = pm1N ·TdTd , and the formula (14) is equal to
(N pm)k−1ξa 1
pmNT d dT
k−1
Ta NX−1
k=0
ξpmqTpmq
TpmN −1 T=1
= T d
dT
k−1(ξT)a(ξTTpm)pmNξpm−1−1
TpmN −1
= T d
dT
k−1 (ξT)a (ξT)pm−1
T=1. (15) Now let S = ξT. Then the above formula (15) is equal to
S d
dS
k−1 Sa Spm−1
S=ξ. Thus we conclude that
µk,λk(a+pmZp) =
T d dT
k−1 T(a)m Tpm−1
T=ξ
holds.
Now we will show that µ1,λ1(a+pmZp) is a measure on Zp. Since ξpm−1 6= 0, we have ξpm ∈ {ξ, ξ2, ξ3, . . . , ξN−1}. So, |ξpm −1|p∈ {|ξ −1|p,|ξ2 −1|p, . . . ,|ξN−1 −1|p}. Now let B = min
1≤i≤N−1(|ξi −1|p) > 0. Then we have 1
|ξpm−1|p ≤ B. And since |ξ(a)m|p= 1, we get
ξ(a)m ξpm−1
p
≤B. Therefore, we conclude thatµ1,λ1(a+pmZp) is a measure on Zp. For general λ, we have
Proposition 5.3. The function µs,λ becomes a distribution on Z×p. Proof. Let us consider the sum
p−1
X
q=0
µs,λ(a+pmq+pm+1Zp) =
p−1
X
q=0
X
r=a+pmq(pm+1) 1≤r<N pm+1
λ(r)B(s, r, N pm+1). (16)
Setting r =N pml+twhen 0≤ l < p and 0 ≤t < N pm. Since r ≡a+pmq (mod pm+1), we have N pml+t≡a+pmq (modpm+1). Then we obtain that (N l−q)pm ≡a−t(mod pm+1).
Now let t≡a(modpm). Then we get q≡N l− a−tpm (modp). Hence the above formula (16) is equal to
X
t
λ(t)
p−1
X
l=0
B(S, N pml+t, N pm+1). (17) Now, we will check that
p−1
X
l=0
B(s, N pml+t, N pm+1) =B(s, t, N pm) withp|M, s∈Zp
holds. Using the well-known distribution formula of the Bernoulli polynomial Mk−1
M−1
X
l=0
Bk l M +t
=Bk(M t) for a positive integerM andk≥1, we obtain that
p−1
X
l=0
B(S, N pml+t, N pm+1) =
p−1
X
l=0
(N pm+1)s−1ω(N pml+t)−sBs
N pml+t N pm+1
= (N pm+1)s−1ω(t)−s
p−1
X
l=0
Bsl
p + t N pm+1
= (N pm)s−1ω(t)−sps−1
p−1
X
l=0
Bsl
p + t N pm+1
= (N pm)s−1ω(t)−sBs
t N pm
=B(s, t, N pm).
Hence
p−1
X
l=0
B(s, N pml+t, N pm+1) =B(s, t, N ptm) (18) holds and the formula (17) is equal to
X
t≡a(pm)
λ(t)B(s, t, N pm) =µs,λ(a+pmZp).
Therefore we can conclude that µs,λ is distribution on Z×p.
Remark 5.1. Let s = k ≥ 0 be an integer. It can be easily checked that the following is a distribution onZp: For a+pmZp ⊆Zp,
µ0k,λ(a+pmZp) = (N pm)k−1 X
r≡a(pm) 1≤r<N pm
λ(r)Bk r N pm
. (19)
Furthermore we also have µk,λ =µ0k,λω−k. Hence we can extendµk,λ toZp and from now on we regard µk,λ as a distribution onZp.
Following Mazur’s treatment, one can normalize the distribution in order to get a measure on Z×p.
Definition 5.2. Let α be an integer that is relatively prime to N, α ≡ 1 (mod p), and κ be another periodic function of a period M. For an open compact subset K of Z×p, we define a distribution µs,λ,κ,α such that
µs,λ,κ,α(K) =µs,λ(K)−α−sµs,κ(αK). (20) We set
µs,λ,α :=µs,λ,λ,α.
We have the following proposition which is a continuous version of [12], Lemma 7.3.
Theorem 5.4. (1)µs,λ,κ,α is a measure on Z×p if and only if 1
N
N
X
r=1
λ(r) = 1 M
M
X
r=1
κ(r). (21)
In particular,µs,λ is a measure onZ×p if and only ifP
rλ(r) = 0.
(2) Assuming the conditions (19), we obtain that fors∈Zp− {0}
dµs,λ,κ,α(x) =shxis−1dµ1,λ,κ,α(x) and
s→0lims−1dµs,λ,κ,α
(x) =hxi−1dµ1,λ,κ,α(x).
Proof. From the definition (12) we have µs,λ(a+pmZp) =X
r
λ(r) hrir N pm
X
n≥0
s n
N pm r
n
Bn (22)
= X
r≡a(pm
hrir N pm
X
n≥0
s n
N pm r
n
B0+O(1)
= haisB0 N pm
N
X
n=1
λ(pmr+a) +O(1), (23)
where by cn = O(p−n), ≥ 0 we mean that pncn is p-adically bounded. Let us set A = N−1P
rλ(r) and B=M−1P
rκ(r). From the formula of (22) we have the expression
µs,λ,κ,α(a+pmZp) = Ahais−Bhα−1(αa)mis
pm B0+O(1),
and thereforeµs,λ,κ,αis a measure onZ×p if and only ifA=Bsincehais ≡ hα−1(αa)mis(modpm).
From the formula (20), we have µs,λ(a+pmZp) = X
r≡a(pm)
λ(r) hrir
N pm +λ(r)shrisr−1+O 1 pm−1
. (24)
We set r =pmq+a with0 ≤q < N. Since hris =hais(1 +spmqa−1) +O(p−2m), the formula (23) is equal to
hais N pm
N
X
q=1
λ(q) +shaisa−1
N
X
q=1
λ(pmq+a)q N −1
+O 1 pm−1
.
If we set
τ(s, λ, a) :=shaisa−1
N
X
q=1
λ(pmq+a) q
N −1
,
then we have
τ(s, λ, a) =shais−1τ(1, λ, a).
Note that (aα)m =αa−pmgwhereg=bpαamcand we have
h(αa)mis=hαais(1−spmg(αa)−1) +O(p−2m).
From this, we have
µs,κ((αa)m+pmZp) = hαais M pm
M
X
q=1
κ(q)
+shαais(αa)−1
M
X
q=1
κ(pmq+ (αa)m) q−g
M −1
+O 1
pm−1
.
Similarly we also set
τ0(s, κ, a) :=shaisa−1
M
X
q=1
κ(pmq+ (αa)m) q−g
M −1
,
and we have
τ0(s, κ, a) =shais−1τ0(1, κ, a).
In total, we have
µs,λ,κ,α(a+pmZp) =µs,λ(a+pmZp)− hαi−sµs,λ((αa)m+pmZp)
=τ(s, λ, a)−α−1τ0(s, κ,(αa)m) +O 1 pm−1
=shais−1(τ(1, λ, a)−α−1τ0(1, κ, a)) +O 1 pm−1
=shais−1µ1,λ,κ,α(a+pmZp) +O 1 pm−1
.
To finish the proof, let us consider |hxis−1− hxi−1|p for x∈Z×p. We have
|hxis−1− hxi−1|p=|hxis−1|p≤ |s|p|hxi −1|p.
Hence for all function f onZ×p withkfkp≤1, we have Z
U
f(x)dµs,λ,κ,α(x) = lim X
a+pmZp⊆U
f(a)µs,λ,κ,α(a+pmZp)
≡lim X
a+pmZp⊆U
f(a)shais−1µ1,λ,κ,α(a+pmZp) (modpm)
= Z
U
f(x)shxis−1µ1,λ,κ,α(x).
Since the above inequality is independent off, we obtain the result.
Let Ψ be a periodic function with a period cN for a p-power c and p- N. Then Ψ can be written asΨ =P
iψiλiwhereψiandλiare periodic functions with periodscandN, respectively.
Definition 5.3. For α≡1 (modcp), we set dσΨ,α(x) =
m
X
i=1
ψi(x)dµ1,λi,α(x).
Note that dσΨ,α(x) =ψ(x)dµ1,λ,α(x) if and only if Z
Z×p
f(x)dσΨ,α(x) = Z
Z×p
f(x)ψ(x)dµ1,λ,α(x) for allf ∈C(Z×p,Cp).
Corollary 5.5. Let α≡1 (mod cp). Then we have Z
Z×p
hxi−1dσΨ,α(x) = logp(α) cN
cN
X
r=1p-r
Ψ(r).
Proof. First let us consider the case that Ψ =ψλfor ψ ∈ L(Zp,Zp) and λ∈ L(Z/nZ,Zp). Let c be a period ofψ. Then we have
dσΨ,α(x) =ψ(x)dµ1,λ,α(x).
From Theorem 5.4, we obtain Z
Z×p
hxi−1dσΨ,α(x) = lim
s→0
Z
Z×p
s−1ψ(x)dµs,λ,α(x). (25) Sinceψ is a locally constant function, we have
Z
a+cZp
ψ(x)dµs,λ,α(x) =ψ(a) Z
a+cZp
1dµs,λ,α(x)
=ψ(a)µs,λ,α(a+cZp).
