Arithmetic Subderivatives: p-adic Discontinuity and Continuity

Pentti Haukkanen and Jorma K. Merikoski^{1}

Faculty of Information Technology and Communication Sciences

FI-33014 Tampere University

Finland

pentti.haukkanen@tuni.fi
jorma.merikoski@tuni.fi

Timo Tossavainen

Department of Arts, Communication and Education

Lulea University of Technology

SE-97187 Lulea Sweden

timo.tossavainen@ltu.se

Abstract

In a previous paper, we proved that the arithmetic subderivative D_{S} is discontinuous at any rational point with respect to the ordinary absolute value. In the present paper, we study this question with respect to the p-adic absolute value. In particular, we show that D_{S} is in this sense continuous at the origin if S is finite or p ∉ S.

1 Introduction

Let 0 ≠ x ∈ ℚ. There exists a unique sequence (ν_{p}(x))_{p∈ℙ} of integers (with only finitely many nonzero terms) such that

1Corresponding author.

x = (sgn x) ∏_{p∈ℙ} p^{ν_{p}(x)}.  (1)

Here ℙ stands for the set of primes, and sgn x = x/|x|. Define that sgn 0 = 0 and ν_{p}(0) = ∞ for all p ∈ ℙ. In addition to the ordinary axioms of ∞, we state that 0 · ∞ = 0. Then (1) holds also for x = 0.

We recall the basic properties of the p-adic order ν_{p}.

Proposition 1. For all x, y ∈ ℚ,

(a) ν_{p}(x) = ∞ if and only if x = 0;

(b) ν_{p}(xy) = ν_{p}(x) + ν_{p}(y);

(c) ν_{p}(x + y) ≥ min(ν_{p}(x), ν_{p}(y));

(d) ν_{p}(x + y) = min(ν_{p}(x), ν_{p}(y)) if ν_{p}(x) ≠ ν_{p}(y).

Proof. Properties (a) and (b) are trivial. For (c) and (d), see, e.g., [2, Proposition 2.4].

Throughout this paper, we let a ∈ ℚ, p, q ∈ ℙ, p ≠ q, and ∅ ≠ S ⊆ ℙ.

The arithmetic subderivative [11, 8, 9] of x ∈ ℚ with respect to S, a.k.a. the arithmetic type derivative [4] is

D_{S}(x) = x ∑_{p∈S} \frac{ν_{p}(x)}{p}.

The arithmetic partial derivative [10, 7] of x with respect to p is D_{p}(x) = D_{{p}}(x). The arithmetic derivative [12, 3, 13] of x is D(x) = D_{ℙ}(x). Clearly,

D_{S}(x) = ∑_{p∈S} D_{p}(x), D(x) = ∑_{p∈ℙ} D_{p}(x).

The function D_{S} is very strongly discontinuous at any a [8, Theorem 4] with respect to the ordinary absolute value. But do we succeed better if we use the p-adic absolute value of x, defined by

|x|_{p} = \frac{1}{p^{ν_{p}(x)}} ?

(In particular, |0|_{p} = 1/∞ = 0.)

We recall the basic properties of | · |_{p}.

Proposition 2. For all x, y ∈ ℚ,

(a) |x|_{p} = 0 if and only if x = 0;

(b) |xy|_{p} = |x|_{p}|y|_{p}; (c) |x + y|_{p} ≤ max(|x|_{p}, |y|_{p});

(d) |x + y|_{p} = max(|x|_{p}, |y|_{p}) if |x|_{p} ≠ |y|_{p}.

Proof. This proposition is equivalent to Proposition 1.

Let us write

x = x\frac{1}{p^{ν_{p}(x)}} p^{ν_{p}(x)} = x|x|_{p}p^{ν_{p}(x)} = μ_{p}(x)p^{ν_{p}(x)},

where

μ_{p}(x) = x|x|_{p} = \frac{x}{p^{ν_{p}(x)}} .  (2)

Proposition 3. For all x, y ∈ ℚ,

(a) ν_{p}(μ_{p}(x)) = 0 if x ≠ 0, ν_{p}(μ_{p}(0)) = ν_{p}(0) = ∞;

(b) |μ_{p}(x)|_{p} = 1 if x ≠ 0, |μ_{p}(0)|_{p} = 0;

(c) μ_{p}(xy) = μ_{p}(x)μ_{p}(y);

(d) D_{p}(μ_{p}(x)) = 0.

Proof. Trivial.

The p-adic distance |x − y|_{p} is smaller the larger ν_{p}(x − y) is.

We say that a function ℚ → ℚ is p-adically continuous, in short p-continuous, if it is continuous with respect to | · |_{p}. So we ask: Is D_{S} p-continuous at some a? We study this question by considering sequences (x_{i}) of rational numbers. If |x_{i} − a|_{p} → 0, equivalently ν_{p}(x_{i} − a) → ∞, then (x_{i}) converges p-adically, in short p-converges, to a. Let x_{i} →_{p} a denote this convergence.

Sections 2–3 are introductory. We present in Section 2 a “light version” of Dirichlet’s theorem on arithmetic progressions. We study p-convergence in Section 3.

Sections 4–7 contain our main results. We prove in Section 4 that D_{S} is p-continuous at a = 0 if S is finite or p ∉ S. We also prove that D_{p} is p-continuous also at a ≠ 0. On the other hand, we show in Section 5 that D_{q} can be (and conjecture that it always is) p-discontinuous at a ≠ 0. In Section 6, we extend the results of Section 5 to D_{S} when S is finite. Although Section 5 is only a special case of Section 6, we find it instructive to present it separately. We complete our paper with the conclusion in Section 7.

2 “Poor man’s theorem on arithmetic progressions”

Throughout this section, a, b ∈ ℤ with gcd (a, b) = 1. As suggested by Graham et al. [5], we let a ⊥ b denote that gcd(a, b) = 1. See also [6, 4].

We recall Dirichlet’s theorem on arithmetic progressions.

Theorem 4. If b > 0, then the set

T = {a + nb : n ∈ ℤ_{+}}  (3)

contains infinitely many primes.

Proof. See, e.g., [1, Theorem 7.9].

Corollary 5. If b ≠ 0, then the set (3) contains infinitely many primes or their additive inverses.

Proof. Trivial.

Theorem 4 is advanced, while our paper is elementary. We do not need the full force of this theorem, and we want to use only elementary methods. Therefore we apply, instead of Theorem 4, the following “poor man’s theorem on arithmetic progressions.” It is elementary but strong enough for us. (Remember that ∅ ≠ S ⊆ ℙ throughout.)

Theorem 6. If S is finite and b ≠ 0, then the set (3) contains infinitely many numbers that are not divisible by any element of S.

Proof. If a = 0, then b = ±1, since otherwise a ⊥̸ b, a contradiction. Therefore T = ℤ_{+} or T = ℤ_{−}, and the claim is trivially true.

Now assume that a ≠ 0. Let

S = {p_{1}, …, p_{h}, q_{1}, …, q_{k}}, p_{1}, …, p_{h} ∤ a, q_{1}, …, q_{k} | a.

(Either the p_{i} or the q_{i} can be missing. Clearly, p_{i} ≠ q_{j} for all i, j.) We show that the numbers a + nb apply when n goes through the set

N = {mp_{1} ⋯ p_{h} : m ∈ ℤ_{+}, q_{1}, …, q_{k} ∤ m}.

Write

c = p_{1} ⋯ p_{h}.

(If the p_{i} are missing, then the “empty product” c = 1.)

Let x ∈ T with n ∈ N, that is,

x = a + mcb, q_{1}, …, q_{k} ∤ m.  (4)

Each p_{i} | c but p_{i} ∤ a, so p_{i} ∤ x. Each q_{i} | a but q_{i} ∤ mcb. (Clearly, q_{i} ∤ m, c. If q_{i} | b, then a ⊥̸ b, a contradiction.) Therefore also q_{i} ∤ x. Consequently, s ∤ x for all s ∈ S. Because there are infinitely many numbers (4), the claim follows.

3 Convergence

Continuity is usually proved by the “ε − δ technique”, while discontinuity is often proved using suitable sequences. For consistency, we use sequences also in proving continuity. To that end, we need a characterization of p-convergence.

Proposition 7. Let (x_{i}) be a sequence of rational numbers. If a ≠ 0, then the following conditions are equivalent.

(a) x_{i} →_{p} a;

(b) μ_{p}(x_{i}) →_{p} μ_{p}(a) and there is i_{0} ∈ ℤ_{+} such that ν_{p}(x_{i_{0}}) = ν_{p}(x_{i_{0}+1}) = ⋯ = ν_{p}(a).

If a = 0, then (b) ⇒ (a) but not conversely.

Proof. Case 1: a ≠ 0.

(a) ⇒ (b): If i_{0} does not exist, then (x_{i}) has a subsequence (x_{i_{k}}) whose each term satisfies ν_{p}(x_{i_{k}}) ≠ ν_{p}(a). Consequently,

ν_{p}(x_{i_{k}} − a) =^{Prop.1(d)} min(ν_{p}(x_{i_{k}}), ν_{p}(a)) ≤ ν_{p}(a) <^{a≠0} ∞, k ≥ 1.

Hence ν_{p}(x_{i_{k}} − a) ↛ ∞, implying x_{i} ↛_{p} a, a contradiction. Therefore i_{0} exists, i.e.,

x_{i} = μ_{p}(x_{i})p^{ν_{p}(a)}, i ≥ i_{0}.

Now, for i ≥ i_{0}, we have

x_{i} − a = (μ_{p}(x_{i}) − μ_{p}(a))p^{ν_{p}(a)},  (5)

and further

ν_{p}(μ_{p}(x_{i}) − μ_{p}(a)) + ν_{p}(a) =^{(5),Prop. 1(b)} ν_{p}(x_{i} − a) →^{(a)} ∞,

verifying μ_{p}(x_{i}) →_{p} μ_{p}(a).

(b) ⇒ (a): Since

x_{i} − a =^{(b)} μ_{p}(x_{i})p^{ν_{p}(a)} − μ_{p}(a)p^{ν_{p}(a)} = (μ_{p}(x_{i}) − μ_{p}(a))p^{ν_{p}(a)}, i ≥ i_{0},

we have

ν_{p}(x_{i} − a) =^{Prop. 1(b)} ν_{p}(μ_{p}(x_{i}) − μ_{p}(a)) + ν_{p}(a) →^{(b)} ∞,

verifying (a). Case 2: a = 0.

(b) ⇒ (a). Since ν_{p}(x_{i_{0}}) = ν_{p}(x_{i_{0}+1}) = ⋯ = ν_{p}(0) = ∞, it follows that x_{i_{0}} = x_{i_{0}+1} = ⋯ = 0. Therefore x_{i} →_{p} 0.

(a) ⇏ (b). If x_{i} = p^{i}, then x_{i} →_{p} 0, but μ_{p}(x_{i}) = 1 →_{p} 1 ≠ 0 = μ_{p}(0).

Proposition 8. A function F : ℚ → ℚ is p-continuous at a if and only if any sequence (x_{i}) of rational numbers satisfying x_{i} →_{p} a satisfies F (x_{i}) →_{p} F (a).

Proof. Proceed as in proving the corresponding property of the ordinary continuity.

There are three formally different ways to consider p-convergence. First, use ν_{p} everywhere. Second, use | · |_{p} everywhere. Third, use either ν_{p} or | · |_{p}, depending on the situation. We follow the first way.

4 The cases of D_{S}, a = 0, and D_{p}, a arbitrary

We begin with a lemma that may be interesting on its own.

Lemma 9. Let S be finite and y ∈ ℚ. Assume that

{q ∈ ℙ | ν_{q}(y) ≠ 0} ⊆ S.

Factorize

y = ∏_{q∈S} q^{ν_{q}(y)} = u(y)w(y),

where

u(y) = ∏_{q∈S} q^{ν_{q}(y)−1}, w(y) = ∏_{q∈S} q.

Then

D_{S}(y) = u(y)v(y),

where

v(y) = ∑_{q∈S} ν_{q}(y) ∏_{r∈S\{q}} r.

Proof. We have

D_{S}(y) = ∑_{q∈S} D_{S}(q^{ν_{q}(y)}) ∏_{r∈S\{q}} r^{ν_{r}(y)} = ∑_{q∈S} ν_{q}(y)q^{ν_{q}(y)−1} ∏_{r∈S\{q}} r^{ν_{r}(y)}

= ∑_{q∈S} ν_{q}(y)q^{ν_{q}(y)−1} ∏_{r∈S\{q}} r^{ν_{r}(y)−1}r = ∑_{q∈S} ν_{q}(y)( ∏_{r∈S} r^{ν_{r}(y)−1})( ∏_{r∈S\{q}} r),

verifying the claim.

Theorem 10. If S is finite or p ∉ S, then D_{S} is p-continuous at the origin.

Proof. Let

x_{i} →_{p} 0.  (6)

We show that D_{S}(x_{i}) →_{p} 0 = D_{S}(0).

If (x_{i}) has only a finite number of nonzero terms, then the claim is trivially true. So, we assume that there are infinitely many x_{i} ≠ 0. Because zeros do not cause any problem in the proof, we can omit them and thus assume that each x_{i} ≠ 0.

Write

x_{i} =^{(1)} (sgn x_{i}) ∏_{q∈ℙ} q^{ν_{q}(x_{i})} = (sgn x_{i})( ∏_{q∈S} q^{ν_{q}(x_{i})})( ∏_{q∈ℙ\S} q^{ν_{q}(x_{i})}) = (sgn x_{i})y_{i}z_{i},

where

y_{i} = ∏_{q∈S} q^{ν_{q}(x_{i})}, z_{i} = ∏_{q∈ℙ\S} q^{ν_{q}(x_{i})}.

(If S = ℙ, then the “empty product” z_{i} = 1.) Then

D_{S}(x_{i}) = (sgn x_{i})z_{i}D_{S}(y_{i}).  (7)

First, let us assume that S is finite. By Lemma 9,

D_{S}(y_{i}) = u(y_{i})v(y_{i}).  (8)

Since v(y_{i}) ∈ ℤ, it follows that

ν_{p}(v(y_{i})) ≥ 0.  (9)

If p ∉ S, then

ν_{p}(D_{S}(x_{i})) =^{(7),(8)} ν_{p}(z_{i}u(y_{i})v(y_{i})) =^{Prop. 1(b)} ν_{p}(z_{i}) + 0 + ν_{p}(v(y_{i})) ≥^{(9)} ν_{p}(z_{i}) = ν_{p}(x_{i}) →^{(6)} ∞.

If p ∈ S, then

ν_{p}(D_{S}(x_{i})) =^{(7),(8)}ν_{p}(z_{i}u(y_{i})v(y_{i})) =^{Prop. 1(b)} 0 + ν_{p}(u(y_{i})) + ν_{p}(v(y_{i})) ≥^{(9)} ν_{p}(u(y_{i}))

= ν_{p}(x_{i}) − 1 →^{(6)} ∞.

Thus, D_{S}(x_{i}) →_{p} 0 in each case.

Second, assume that S is infinite. Because w(y) and v(y) contain a divergent infinite product, applying Lemma 9 needs some preparation. Define

S_{i} = {q ∈ S | ν_{q}(x_{i}) ≠ 0}.

If D_{S}(x_{i}) ≠ 0 only for finitely many terms, then the claim is trivially true. So, we assume that there are infinitely many such terms. We can omit all x_{i} satisfying D_{S}(x_{i}) = 0, because they do not violate the convergence. Then each S_{i} ≠ ∅.

Now

D_{S}(x_{i}) = D_{S_{i}}(x_{i}) =^{(7)} (sgn x_{i})z_{i}D_{S_{i}}(y_{i}).  (10)

By Lemma 9,

D_{S_{i}}(y_{i}) = u_{i}(y_{i})v_{i}(y_{i}),  (11)

where

u_{i}(y_{i}) = ∏_{q∈S_{i}} q^{ν_{q}(y_{i})−1}, v_{i}(y_{i}) = ∑_{q∈S_{i}} ν_{q}(y_{i}) ∏_{r∈S_{i}\{q}} r.

If p ∉ S, then

ν_{p}(D_{S}(x_{i})) =^{(10),(11)} ν_{p}(z_{i}u_{i}(y_{i})v_{i}(y_{i})) =^{Prop. 1(b)} ν_{p}(z_{i}) + 0 + ν_{p}(v_{i}(y_{i})) ≥^{(9)} ν_{p}(z_{i}) = ν_{p}(x_{i}) →^{(6)} ∞.

Consequently, D_{S}(x_{i}) →_{p} 0. We discuss the case of p ∈ S at the end of this section.

Theorem 11. The function D_{p} is p-continuous everywhere.

Proof. If a = 0, then apply Theorem 10. If a ≠ 0, then let x_{i} →_{p} a, and let i_{0} be as in Proposition 7. For i ≥ i_{0},

D_{p}(x_{i}) − D_{p}(a) = D_{p}(μ_{p}(x_{i})p^{ν_{p}(a)}) − D_{p}(μ_{p}(a)p^{ν_{p}(a)})

=^{Prop. 3(d)} ν_{p}(a)p^{ν_{p}(a)−1}μ_{p}(x_{i}) − ν_{p}(a)p^{ν_{p}(a)−1}μ_{p}(a)

= ν_{p}(a)p^{ν_{p}(a)−1}(μ_{p}(x_{i}) − μ_{p}(a)) = c(μ_{p}(x_{i}) − μ_{p}(a)), c = ν_{p}(a)p^{ν_{p}(a)−1}. (12)

If ν_{p}(a) = 0, then

D_{p}(x_{i}) =^{(12)} D_{p}(a), i ≥ i_{0}.

If ν_{p}(a) ≠ 0, then

ν_{p}(D_{p}(x_{i}) − D_{p}(a)) =^{(12),Prop. 1(b)} ν_{p}(c) + ν_{p}(μ_{p}(x_{i}) − μ_{p}(a)) →^{Prop. 7} ∞.

Therefore D_{p}(x_{i}) →_{p} D_{p}(a) in each case.

Can we extend the proof of Theorem 10 to the case where S is infinite and p ∈ S? If p ∈ S_{i}, then

ν_{p}(D_{S}(x_{i})) =^{(10),(11)} ν_{p}(z_{i}u_{i}(y_{i})v_{i}(y_{i})) =^{Prop. 1(b)} 0 + ν_{p}(u_{i}(y_{i})) + ν_{p}(v_{i}(y_{i})) ≥^{(9)} ν_{p}(u_{i}(y_{i}))

= ν_{p}(x_{i}) →^{(6)} ∞,

implying the convergence.

If

p ∈ S \ S_{i},  (13)

then

ν_{p}(D_{S}(x_{i})) =^{(10),(11)} ν_{p}(z_{i}u_{i}(y_{i})v_{i}(y_{i})) =^{Prop. 1(b)} 0 + 0 + ν_{p}(v_{i}(y_{i})) = ν_{p}(x_{i}).

If (13) holds only for finitely many indices i, then the corresponding x_{i} do not effect on the convergence, and therefore they do not bother us. If there are infinitely many such indices, then the question of convergence remains open.

We thus conclude that D_{S}(x_{i}) →_{p} 0 also if, for any sequence (x_{i}) of nonzero numbers with x_{i} →_{p} 0, only finitely many terms satisfy (13). However, we find this assumption useless, because its validity cannot be tested.

5 The case of D_{q}, a ≠ 0

In this and the next section, we show that D_{q} is under certain assumptions discontinuous outside the origin. These sections are quite technical and require the use of rather heavy notation. In order to increase readability, we consider the special case S = {q} separately.

Theorem 12. Let

a ≠ 0.  (14)

If

D_{q}(a) = 0,  (15)

then D_{q} is p-discontinuous at a.

Proof. Let

x_{i} = a + \frac{p^{i}}{q} .  (16)

Then

ν_{p}(x_{i} − a) = i → ∞,  (17)

implying x_{i} →_{p} a. Since

ν_{q}(a) =^{(15)} 0, ν_{q}(\frac{p^{i}}{q}) = −1,

we have ν_{q}(x_{i}) =^{(16),Prop. 1(d)} −1.

Consequently,

D_{q}(x_{i}) = \frac{ν_{q}(x_{i})}{q}x_{i} = −\frac{x_{i}}{q} ,  (18)

and further

ν_{p}(D_{q}(x_{i}) − D_{q}(a)) =^{(15)} ν_{p}(D_{q}(x_{i})) =^{(18)} ν_{p}(\frac{x_{i}}{q}) = ν_{p}(x_{i}).

We show that ν_{p}(x_{i}) ↛ ∞; then D_{q}(x_{i}) ↛_{p} D_{q}(a), verifying the claim. If

ν_{p}(x_{i}) → ∞,  (19)

then

ν_{p}(a) = ν_{p}(x_{i} − (x_{i} − a)) ≥^{Prop. 1(c)} min(ν_{p}(x_{i}), ν_{p}(x_{i} − a)) →^{(17),(19)} ∞.

Hence ν_{p}(a) = ∞, i.e., a = 0, contradicting (14).

Theorem 13. Let

a = \frac{a_{1}}{a_{2}} , 0 ≠ a_{1} ∈ ℤ, a_{2} ∈ ℤ_{+}, a_{1} ⊥ a_{2}.  (20)

If

D_{q}(a) ≠ 0  (21)

and

q ∤ a_{2},  (22)

then D_{q} is p-discontinuous at a.

Proof. Let i ∈ ℤ_{+}. Since μ_{p}(a_{1}) ⊥^{(20),Prop. 3(a)} μ_{p}(a_{2})p^{i} and μ_{p}(a_{2}) >^{(20)} 0, there are, by Theorem 6 (S = {p, q}, a = μ_{p}(a_{1}), b = μ_{p}(a_{2})p^{i}) positive integers r_{i1} < r_{i2} < ⋯ satisfying

r_{ik} = μ_{p}(a_{1}) + n_{ik}μ_{p}(a_{2})p^{i}, n_{ik} ∈ ℤ_{+}, p, q ∤ r_{ik}, k = 1, 2, ….  (23)

Consequently,

μ_{p}(a) + n_{ik}p^{i} =^{Prop. 3(c)} \frac{μ_{p}(a_{1})}{μ_{p}(a_{2})} + n_{ik}p^{i} = \frac{μ_{p}(a_{1}) + n_{ik}μ_{p}(a_{2})p^{i}}{μ_{p}(a_{2})} =^{(23)} \frac{r_{ik}}{μ_{p}(a_{2})}.

Choose r_{1k_{1}} < r_{2k_{2}} < ⋯ and write n_{1} = n_{1k_{1}}, n_{2} = n_{2k_{2}}, …, r_{1} = r_{1k_{1}}, r_{2} = r_{2k_{2}}, . . .. Define the sequence (y_{i}) by

y_{i} = μ_{p}(a) + n_{i}p^{i} = \frac{r_{i}}{μ_{p}(a_{2})}.  (24)

Then

ν_{p}(y_{i}) =^{(24),Prop. 1(b)} ν_{p}(r_{i}) − ν_{p}(μ_{p}(a_{2})) =^{(23),Prop. 3(a)} 0  (25)

and

ν_{p}(y_{i} − μ_{p}(a)) =^{(24)} ν_{p}(n_{i}p^{i}) =^{Prop. 1(b)} ν_{p}(n_{i}) + i ≥^{n_{i}∈ℤ_{+}} i → ∞.  (26)

For all i ∈ ℤ_{+}, we have

D_{q}(y_{i}) =^{(24)} \frac{μ_{p}(a_{2})D_{q}(r_{i}) − r_{i}D_{q}(μ_{p}(a_{2}))}{μ_{p}(a_{2})^{2}}

=^{q ∤ r_{i}} −\frac{r_{i}D_{q}(μ_{p}(a_{2}))}{μ_{p}(a_{2})^{2}} =^{(2)} −(r_{i}D_{q}(\frac{a_{2}}{p^{ν_{p}(a_{2})}} ))/(\frac{a_{2}}{p^{ν_{p}(a_{2})}} )^{2}

= −(\frac{r_{i}}{p^{ν_{p}(a_{2})}} D_{q}(a_{2}))/(\frac{a_{2}}{p^{ν_{p}(a_{2})}} )^{2} = −\frac{r_{i}p^{ν_{p}(a_{2})}D_{q}(a_{2})}{a_{2}^{2}} =^{(22)} 0.  (27)

Also define

x_{i} = y_{i}p^{ν_{p}(a)};  (28)

then

D_{q}(x_{i}) = p^{ν_{p}(a)}D_{q}(y_{i}) =^{(27)} 0.  (29)

Since μ_{p}(x_{i}) =^{(25),(28)} y_{i} →_{p}^{(26)} μ_{p}(a), it follows that x_{i} →_{p}^{Prop. 7} a. On the other hand, since

D_{q}(x_{i}) − D_{q}(a) =^{(29)} −D_{q}(a) ≠^{(21)} 0,

we have

ν_{p}(D_{q}(x_{i}) − D_{q}(a)) = ν_{p}(D_{q}(a)) <^{(21)} ∞,

verifying D_{q}(x_{i}) ↛_{p} D_{q}(a). Corollary 14. If 0 ≠ a ∈ ℤ, then D_{q} is p-discontinuous at a. Proof. Apply Theorem 12 if q ∤ a, and Theorem 13 if q | a.

If D_{q}(a) ≠ 0 and q | a_{2} (where a_{2} is as in (20)), then our question remains open. We conjecture that discontinuity holds also in this case. Conjecture 15. The function D_{q} is p-discontinuous outside the origin.

6 The case of D_{S}, S finite, a ≠ 0

We extend Theorems 12 and 13, Corollary 14, and Conjecture 15. The presentation branches according to properties of D_{S}(a) and S, summarized in Section 7. Theorem 16. Let S be finite, p ∉ S, and a ≠ 0. If

D_{S}(a) = 0,  (30)

then D_{S} is p-discontinuous at a. Proof. Let S = {q_{1}, …, q_{m}}, take γ ∈ ℤ_{+} satisfying

γ > −ν_{q_{1}}(a), …, −ν_{q_{m}}(a),  (31)

and define

x_{i} = a + \frac{p^{i}}{(q_{1} ⋯ q_{m})^{γ}} .  (32)

Let i ∈ ℤ_{+} and j ∈ {1, …, m}. Then

ν_{p}(x_{i} − a) =^{(32)} i → ∞, x_{i} →_{p} a, ν_{q_{j}}(x_{i} − a) =^{(32)} −γ <^{(31)} ν_{q_{j}}(a),

ν_{q_{j}}(x_{i}) = ν_{q_{j}}((x_{i} − a) + a) =^{Prop. 1(d)} −γ, D_{q_{j}}(x_{i}) = \frac{ν_{q_{j}}(x_{i})}{q_{j}}x_{i} = −\frac{γx_{i}}{q_{j}} .  (33)

As in the proof of Theorem 12, we see that

ν_{p}(x_{i}) ↛ ∞.  (34)

Now

D_{S}(x_{i}) =^{(33)} −(\frac{1}{q_{1}} + ⋯ + \frac{1}{q_{m}} )γx_{i} = −\frac{e_{m−1}(q_{1}, …, q_{m})γx_{i}}{q_{1} ⋯ q_{m}} ,  (35)

where e_{m−1} denotes the (m − 1)’th elementary symmetric function. Therefore

ν_{p}(D_{S}(x_{i}) − D_{S}(a)) =^{(30)} ν_{p}(D_{S}(x_{i})) =^{(35),Prop. 1(b)} ν_{p}(e_{m−1}(q_{1}, …, q_{m})) + ν_{p}(γ) + ν_{p}(x_{i}).

Since ν_{p}(e_{m−1}(q_{1}, …, q_{m})) and ν_{p}(γ) are (finite) constants and ν_{p}(x_{i}) ↛ ∞, we have

ν_{p}(D_{S}(x_{i}) − D_{S}(a)) ↛ ∞,

i.e., D_{S}(x_{i}) ↛_{p} D_{S}(a). Theorem 17. Let S ≠ {p} be finite, p ∈ S, and a ≠ 0. If

D_{S}(a) = 0,  (36)

then D_{S} is p-discontinuous at a. Proof. Let

S = {q_{1}, …, q_{m}, p}, S_{0} = {q_{1}, …, q_{m}},  (37)

and let (x_{i}) be as in (32). For i > ν_{p}(a),

ν_{p}(x_{i} − a) =^{(32)} i > ν_{p}(a)  (38)

and further

ν_{p}(x_{i}) = ν_{p}((x_{i} − a) + a) =^{(38),Prop. 1(d)} ν_{p}(a);

hence

D_{p}(x_{i}) = \frac{ν_{p}(a)}{p}x_{i}.  (39)

Now

D_{S}(x_{i}) = D_{S_{0}}(x_{i}) + D_{p}(x_{i}) =^{(35),(39)} \frac{e_{m−1}(q_{1}, …, q_{m})γx_{i}}{q_{1} ⋯ q_{m}} + \frac{ν_{p}(a)}{p}x_{i} = cx_{i},  (40)

where

c = \frac{e_{m−1}(q_{1}, …, q_{m})γ}{q_{1} ⋯ q_{m}} + \frac{ν_{p}(a)}{p}.

We can choose γ ∈ ℤ_{+} so that, in addition to (31), the inequality c ≠ 0 holds. Then

ν_{p}(c) < ∞.  (41)

Since

ν_{p}(D_{S}(x_{i}) − D_{S}(a)) =^{(36)} ν_{p}(D_{S}(x_{i})) =^{(40),Prop. 1(b)} ν_{p}(c) + ν_{p}(x_{i}) ↛^{(34),(41)} ∞,

D_{S}(x_{i}) ↛_{p} D_{S}(a) follows.

Theorem 18. Let a be as in (20), let S be finite, and p ∉ S. If

D_{S}(a) ≠ 0  (42)

and

ν_{p}(a_{2}D_{S}(a)) ≠ ν_{p}(aD_{S}(a_{2})),  (43)

then D_{S} is p-discontinuous at a. Proof. Let S = {q_{1}, …, q_{m}}, i ∈ ℤ_{+}, and j ∈ {1, …, m}. Proceeding as in the proof of Theorem 13, we have

r_{ik} = μ_{p}(a_{1}) + n_{ik}μ_{p}(a_{2})p^{i}, p, q_{1}, …, q_{m} ∤ r_{ik}, k = 1, 2, …,
y_{i} = μ_{p}(a) + n_{i}p^{i} = \frac{r_{i}}{μ_{p}(a_{2})}, p, q_{1}, …, q_{m} ∤ r_{i},  (44)
D_{q_{j}}(y_{i}) = −\frac{r_{i}p^{ν_{p}(a_{2})}D_{q_{j}}(a_{2})}{a_{2}^{2}},  (45)
x_{i} = y_{i}p^{ν_{p}(a)} →_{p} a,  (46)

and

D_{q_{j}}(x_{i}) =^{(45),(46)} −p^{ν_{p}(a)} \frac{r_{i}p^{ν_{p}(a_{2})}D_{q_{j}}(a_{2})}{a_{2}^{2}} =^{(20),Prop. 1(b)} −\frac{r_{i}p^{ν_{p}(a_{1})}D_{q_{j}}(a_{2})}{a_{2}^{2}} = −cr_{i}D_{q_{j}}(a_{2}), (47)

where

c = \frac{p^{ν_{p}(a_{1})}}{a_{2}^{2}} .

Consequently,

D_{S}(x_{i}) = ∑_{j=1}^{m} D_{q_{j}}(x_{i}) =^{(47)} −cr_{i} ∑_{j=1}^{m} D_{q_{j}}(a_{2}) = −cr_{i}D_{S}(a_{2}),  (48)

and further

ν_{p}(D_{S}(x_{i})) =^{(48)} ν_{p}(cr_{i}D_{S}(a_{2})) =^{Prop. 1(b)} ν_{p}(a_{1}) − 2ν_{p}(a_{2}) + ν_{p}(D_{S}(a_{2}))
=^{(20),Prop. 1(b)} ν_{p}(a) − ν_{p}(a_{2}) + ν_{p}(D_{S}(a_{2}))  (49)

≠^{(43),Prop. 1(b)} ν_{p}(a_{2}) − ν_{p}(a_{2}) + ν_{p}(D_{S}(a))

= ν_{p}(D_{S}(a)).  (50)

Let

u(x_{i}) = D_{S}(x_{i}) − D_{S}(a).

Then

ν_{p}(u(x_{i})) =^{(50),Prop. 1(d)} min(ν_{p}(D_{S}(x_{i})), ν_{p}(D_{S}(a)))
=^{(48),Prop. 1(b)} min(ν_{p}(cD_{S}(a_{2})), ν_{p}(D_{S}(a)))  (51)
≤ ν_{p}(D_{S}(a)) <^{(42)} ∞,  (52)

and D_{S}(x_{i}) ↛_{p} D_{S}(a) follows. Theorem 19. Let a be as in (20), let S ≠ {p} be finite and p ∈ S. Assume that D_{S}(a) ≠ 0. If

D_{S\{p}}(a) = 0,  (53)

then D_{S} is p-continuous at a. If

D_{S\{p}}(a) ≠ 0  (54)

and

ν_{p}(a_{2}D_{S\{p}}(a)) ≠ ν_{p}(aD_{S\{p}}(a_{2})),  (55)

then D_{S} is p-discontinuous at a. Proof. If (53) holds, then

D_{S}(a) = D_{S\{p}}(a) + D_{p}(a) = D_{p}(a),

and the claim follows from Theorem 11.

Now assume that (54) holds. Let S, S_{0}, and x_{i} be as in (37) and (46), respectively. Since

ν_{p}(x_{i}) =^{(44),(46)} ν_{p}(a), we have

D_{p}(x_{i}) = ρx_{i}, ρ = \frac{ν_{p}(a)}{p}.

Now

D_{S}(x_{i}) − D_{S}(a) = D_{S_{0}}(x_{i}) + D_{p}(x_{i}) − (D_{S_{0}}(a) + D_{p}(a)) = u(x_{i}) + v(x_{i}),  (56)

where

u(x_{i}) = D_{S_{0}}(x_{i}) − D_{S_{0}}(a), v(x_{i}) = ρ(x_{i} − a).

Because

ν_{p}(v(x_{i})) =^{Prop. 1(b)} ν_{p}(ρ) + ν_{p}(x_{i} − a) →^{(46)} ∞

and ν_{p}(u(x_{i})) is bounded by (52), there is i_{0} ∈ ℤ_{+} such that

ν_{p}(v(x_{i})) > ν_{p}(u(x_{i})) for all i ≥ i_{0}.  (57)

Thus, for i ≥ i_{0},

ν_{p}(D_{S}(x_{i}) − D_{S}(a)) =^{(56),(57),Prop. 1(d)} ν_{p}(u(x_{i})) ↛^{(52)} ∞,

verifying D_{S}(x_{i}) ↛_{p} D_{S}(a).

Corollary 20. If S ≠ {p} is finite and 0 ≠ a ∈ ℤ, then D_{S} is p-discontinuous at a. Proof. Apply Theorem 16 if p ∉ S and D_{S}(a) = 0, Theorem 17 if p ∈ S and D_{S}(a) = 0, Theorem 18 if p ∉ S and D_{S}(a) ≠ 0, and Theorem 19 if p ∈ S and D_{S}(a) ≠ 0. Note that (43) and (55) are satisfied (the right-hand side is infinite but the left-hand side is finite).

We conjecture that Theorems 18 and 19 remain true without (43) and (55), respectively. Conjecture 21. If S ≠ {p} is finite, 0 ≠ a ∈ ℚ, and D_{S\{p}}(a) ≠ 0, then D_{S} is pdiscontinuous at a.

7 Conclusion

We summarize our results. C denotes p-continuity, D p-discontinuity, and O denotes that the question is open.

1 (Theorem 10). S is finite or p ∉ S, a = 0. C. 2 (the end of Section 5). S infinite, p ∈ S, a = 0. C or O. 3 (Theorem 11). S = {p}, a arbitrary. C. 4 (Theorem 12, a special case of Theorems 16 and 17). S = {q}, a ≠ 0, D_{q}(a) = 0. D. 5 (Theorem 13, a special case of Theorems 18 and 19). S = {q}, a ≠ 0, D_{q}(a) ≠ 0. D. 6 (Theorem 16). S finite, p ∉ S, a ≠ 0, D_{S}(a) = 0. D. 7 (Theorem 17). S(≠ {p}) finite, p ∈ S, a ≠ 0, D_{S}(a) = 0. D. 8 (Theorem 18). S finite, p ∉ S, a ≠ 0, D_{S}(a) ≠ 0. D under (43), otherwise O. 9 (Theorem 19). S(≠ {p}) finite, p ∈ S, a ≠ 0, D_{S}(a) ≠ 0. C under (53), D under (54)

and (55), otherwise O.

10. S infinite, a ≠ 0. O.

8 Acknowledgment

We thank the referee, whose suggestions led to significant improvements in Section 4.

References

[1] T. M. Apostol, Introduction to Analytic Number Theory, Springer, 1976.

[2] A. Baker, An introduction to p-adic numbers and p-adic analysis, lecture notes, Univer-

sity of Glasgow, 2020, http://www.maths.gla.ac.uk/~ajb/course-notes.html.

[3] E. J. Barbeau, Remarks on an arithmetic derivative, Canad. Math. Bull. 4 (1961), 117–122.

[4] J. Fan and S. Utev, The Lie bracket and the arithmetic derivative, J. Integer Sequences 23 (2020), Article 20.2.5.

[5] R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics: A Foundation for Computer Science, Addison-Wesley, 2nd ed., 1994.

[6] P. Haukkanen, M. Mattila. J. K. Merikoski, and T. Tossavainen, Perpendicularity in an Abelian group, Internat. J. Math. Math. Sci. 2013, Article 983607.

[7] P. Haukkanen, J. K. Merikoski, and T. Tossavainen, On arithmetic partial differential equations, J. Integer Sequences 19 (2016), Article 16.8.6.

[8] P. Haukkanen, J. K. Merikoski, and T. Tossavainen, Arithmetic subderivatives: Discontinuity and continuity, J. Integer Sequences 22 (2019), Article 19.7.4.

[9] P. Haukkanen, J. K. Merikoski, and T. Tossavainen, Asymptotics of partial sums of the Dirichlet series of the arithmetic derivative, Math. Comm. 25 (2020), 107–115, https://

www.mathos.unios.hr/mc/index.php/mc/article/view/3206.

[10] J. Kovič, The arithmetic derivative and antiderivative, J. Integer Sequences 15 (2012), Article 12.3.8.

[11] J. K. Merikoski, P. Haukkanen, and T. Tossavainen, Arithmetic subderivatives and Leibniz-additive functions, Ann. Math. Informat. 50 (2019), 145–157, http://ami.

ektf.hu.

[12] J. Mingot Shelly, Una cuestión de la teoría de los números, Asociación Española, Granada (1911), 1–12.

[13] V. Ufnarovski and B. Åhlander, How to differentiate a number, J. Integer Sequences 6 (2003), Article 03.3.4.

2010 Mathematics Subject Classification: Primary 11A25; Secondary 26A15. Keywords: arithmetic subderivative, arithmetic partial derivative, arithmetic derivative, continuity, p-adic absolute value.

(Concerned with sequences \underline{A000040}, \underline{A003415}. ) Received April 3 2020; revised version received June 24 2020. Published in Journal of Integer Sequences, June 26 2020. Return to Journal of Integer Sequences home page.