Arithmetic Subderivatives: Discontinuity and Continuity

Pentti Haukkanen and Jorma K. Merikoski

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

We first prove that any arithmetic subderivative of a rational number defines a function that is everywhere discontinuous in a very strong sense. Second, we show that although the restriction of this function to the set of integers is continuous (in the relative topology), it is not Lipschitz continuous. Third, we see that its restriction to a suitable infinite set is Lipschitz continuous. This follows from the solutions of certain arithmetic differential equations.

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

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

Here ℙ stands for the set of primes, and sgn x = x/|x|. Define sgn 0 = 0 and ν_{p}(0) = 0 for all p ∈ ℙ; then Eq. (1) also holds for x = 0.

Let ∅ ≠ S ⊆ ℚ. The arithmetic subderivative [5] of x ∈ ℚ with respect to S is

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

The arithmetic partial derivative [4, 3] of x with respect to p ∈ ℙ is x′_{p} = D_{p}(x) = D_{{p}}(x). The arithmetic derivative [6, 2, 8] of x is x′ = D(x) = D_{ℙ}(x). Arithmetic differential equations and arithmetic partial differential equations have been studied [8, 4, 3, 7].

We have D_{S}(p) = 1 for all p ∈ S. If x is near to p and suitably chosen, then D_{S}(x) can be far from one. So it is natural to expect that D_{S} is discontinuous at any a ∈ ℙ. However, it is not equally apparent that D_{S} is discontinuous at an arbitrary a ∈ ℚ.

We say that a real-valued function f, defined on an open real (or rational) interval I, is superdiscontinuous at a ∈ I if it attains values of an arbitrarily large absolute value arbitrarily near to a and on its both sides. More precisely, given any δ, M > 0, there are x, y ∈ I such that

a − δ < x < a < y < a + δ and |f(x)|, |f(y)| > M.

If x (respectively, y) exists, then f is superdiscontinuous from the left (right) at a. We prove in Section 2 that D_{S} is superdiscontinuous at any a ∈ ℚ. We need two lemmas.

Lemma 1. Let p, q ∈ ℙ, p ≠ q. The set {±p^{m}q^{n} | m, n ∈ ℤ} is dense in ℚ.

Proof. See the proof of [3, Lemma 30].

Lemma 2. Let f be a real-valued function, defined on an open real (or rational) interval I, and let a ∈ I. If f is superdiscontinuous at each point of I \{a}, then it is superdiscontinuous also at a.

Proof. Let δ, M > 0. It is no restriction to assume that a ± δ ∈ I. Since f is superdiscontinuous (from the right) at a − δ, there is x ∈ I satisfying a − δ < x < a and |f(x)| > M. Similarly, since f is superdiscontinuous (from the left) at a + δ, there is y ∈ I satisfying a < y < a + δ and |f(y)| > M.

Let f and I be as above. The function f is Lipschitz continuous (in I) if there exists L > 0 satisfying

|f(x) − f(y)| ≤ L|x − y|  (2)

for all x, y ∈ I. Since a Lipschitz continuous function is continuous in the ordinary sense, the results in Section 2 imply that D_{S} is Lipschitz discontinuous. But the restriction of D_{S} to ℤ, denoted by D_{S}|_{ℤ}, is continuous (in the relative topology); so, is it Lipschitz continuous? We show in Section 3 that it is not. This raises a further question: Is there an infinite set A ⊂ ℚ such that D_{S}|_{A} is Lipschitz continuous? Applying the previously known solutions of the differential equations x′ = 0 and x′_{p} = ax, we give a positive answer. This, in turn, motivates us to consider the more general differential equation x′_{S} = ax. We solve it in Section 4. Finally, we complete our paper with concluding remarks in Section 5.

2 Superdiscontinuity of D_{S}

We state in Theorem 3 an additional assumption, in order to make the proof easier.

Theorem 3. Let ∅ ≠ S ⊆ ℙ. If there are distinct p, q ∈ ℙ\S, then D_{S} is superdiscontinuous at any a ∈ ℚ.

Proof. Case 1: a > 0. Let δ, M > 0, s ∈ S, and k ∈ ℤ_{+}. By Lemma 1, there are m, n ∈ ℤ such that

\frac{a}{s^{k}} < p^{m}q^{n} < \frac{a}{s^{k}} + \frac{δ}{s^{k}} .

Then x = s^{k}p^{m}q^{n} satisfies

a < x < a + δ.

The subderivative

D_{S}(x) = ks^{k−1}p^{m}q^{n} > ks^{k−1} \frac{a}{s^{k}} = \frac{ka}{s} > M

if

k > \frac{sM}{a} .

Consequently, superdiscontinuity from the right follows.

To prove superdiscontinuity from the left, we use the above notation but let δ < a (which is no restriction). Now, by Lemma 1, there are m, n ∈ ℤ such that

\frac{a}{s^{k}} − \frac{δ}{s^{k}} < p^{m}q^{n} < \frac{a}{s^{k}} .

Hence,

a − δ < x < a,

and

D_{S}(x) = ks^{k−1}p^{m}q^{n} > ks^{k−1} \frac{a − δ}{s^{k}} = \frac{k(a − δ)}{s} > M

if

k > \frac{sM}{a − δ} .

This verifies superdiscontinuity from the left.

Case 2: a < 0. Apply the fact that D_{S}(a) = −D_{S}(−a).

Case 3: a = 0. Apply Lemma 2.

What about trying the same idea for S = ℙ \ {p}, p ∈ ℙ? It is enough to assume that a > 0. Let q ∈ S and T = S \ {q}; then p, q ∉ T. Let δ, M, s, and k be as above. Again, there are m and n such that x = s^{k}p^{m}q^{n} satisfies

a < x < a + δ.

If

k > \frac{sM}{a} ,

then

D_{S}(x) = D_{T} (x) + \frac{n}{q}s^{k}p^{m}q^{n} > M + ns^{k}p^{m}q^{n−1}.

Consequently, if n ≥ 0 (i.e., ν_{q}(x) ≥ 0 for some q ∈ S), then D_{S}(x) > M, and superdiscontinuity from the right follows. Unfortunately, a problem arises if n < 0 (i.e., ν_{q}(x) < 0 for all q ∈ S). Then D_{S}(x) < 0, and we get nothing reasonable. It seems that our method cannot be revised to work in this case. Instead of a single x, we apparently should consider a suitable sequence (x_{i}). We do this in the proof of the next theorem, where S is arbitrary. Theorem 3 could now be omitted, but we find its proof interesting on its own and therefore keep it.

Theorem 4. Let ∅ ≠ S ⊆ ℙ. Then D_{S} is superdiscontinuous at any a ∈ ℚ.

Proof. We can assume that a ≠ 0; then superdiscontinuity at a = 0 follows from Lemma 2. Let p, q ∈ ℙ, p ≠ q, and denote m = ν_{p}(a), n = ν_{q}(a). Then

a = p^{m}q^{n}r,

where r ≠ 0 has ν_{p}(r) = ν_{q}(r) = 0. We show that there is a sequence (x_{i}) of rational numbers with

x_{i} → a, x_{1}, x_{2}, … > a,  (3)

and

|D_{S}(x_{i})| → ∞.  (4)

By a simple modification of this proof, it can be seen that there also exists a sequence (x_{i}) satisfying

x_{i} → a, x_{1}, x_{2}, … < a

and (4).

By Lemma 1, there are integer sequences (m_{i}) and (n_{i}) such that the sequence

(x_{i}), x_{i} = p^{m_{i}}q^{n_{i}}

satisfies (3). To prove (4), we must know their limiting behaviour.

Case 1: Both (m_{i}) and (n_{i}) are bounded from above. Then they are bounded also from below, since otherwise x_{i} → 0 or (x_{i}) diverges, contradicting (3). Therefore, (x_{i}) has only finitely many different terms, implying that its limit a is one of them. This again contradicts (3). Consequently, this case is impossible.

Case 2: The sequence (m_{i}) is not bounded from above. If, instead, (n_{i}) is not bounded from above, we can proceed similarly. Now, (m_{i}) has a subsequence (m_{i_{k}}) satisfying m_{i_{k}} → ∞. Since p^{m_{i_{k}}} → ∞ but x_{i_{k}} → a ≠ 0, necessarily q^{n_{i_{k}}} → 0, i.e., n_{i_{k}} → −∞. Including also the case when (n_{i}) is not bounded from above, we therefore have

m_{i_{k}} → ±∞ and n_{i_{k}} → ∓∞.  (5)

In order to keep the notation simple, we let (x_{i}) denote the subsequence (x_{i_{k}}). Then Eq. (5) reads

m_{i} → ±∞ and n_{i} → ∓∞.  (6)

By the convergence, the sequence (x_{i}) is bounded also from above. Let u ∈ ℚ satisfy

x_{1}, x_{2}, … < u.

We can assume in Eq. (6) that

m_{i} → −∞ and n_{i} → ∞.  (7)

If the signs are opposite, then a simple modification applies. There is i_{0} ∈ ℤ_{+} such that n_{i} > 0 for all i ≥ i_{0}. Denote

c_{i} = \frac{m_{i} − m}{n_{i}}, where i ≥ i_{0}.

Then

a < x_{i} < u ⟺ p^{m}q^{n}r < p^{m_{i}}q^{n_{i}} < u ⟺

q^{n−n_{i}}r < p^{m_{i}−m} < p^{−m}q^{−n_{i}}u ⟺

(n − n_{i}) log_{p} q + log_{p} r < m_{i} − m < −m − n_{i} log_{p} q + log_{p} u ⟺
(n − n_{i} + log_{q} r) log_{p} q < m_{i} − m < −m + (−n_{i} + log_{q} u) log_{p} q
⟺ (\frac{n + log_{q} r}{n_{i}} − 1) log_{p} q < c_{i} < −\frac{m}{n_{i}} + (\frac{log_{q} u}{n_{i}} − 1) log_{p} q.  (8)

By Eq. (7),

(\frac{n + log_{q} r}{n_{i}} − 1) log_{p} q → − log_{p} q, −\frac{m}{n_{i}} + (\frac{log_{q} u}{n_{i}} − 1) log_{p} q → − log_{p} q.

Hence, by Eq. (8),

c_{i} → − log_{p} q.  (9)

The function D_{S} has not yet had any role. Let us now focus on it.

If p, q ∈ S (i.e., there are at least two elements in S), then

D_{S}(x_{i}) = x_{i}(\frac{m_{i}}{p} + \frac{n_{i}}{q} ) = x_{i}(\frac{m + c_{i}n_{i}}{p} + \frac{n_{i}}{q} ) =

x_{i} \frac{m}{p} + x_{i}(\frac{c_{i}}{p} + \frac{1}{q})n_{i} =: A_{i} + B_{i}.  (10)

By (3),

A_{i} → a\frac{m}{p}.  (11)

Further, (3) and (9) imply that

x_{i}(\frac{c_{i}}{p} + \frac{1}{q}) → a(\frac{1}{q} − \frac{log_{p} q}{p} ) = \frac{a}{p}(\frac{p}{q} − log_{p} q) ≠ 0,

because a ≠ 0 and p^{p} ≠ q^{q}. Consequently, by Eq. (7),

|B_{i}| → ∞.  (12)

Finally, Eqs. (12), (11), and (10) imply (4).

The case of S = {p} (i.e., there is only one element in S) remains. Then

D_{S}(x_{i}) = x_{i} \frac{m_{i}}{p} .

Since x_{i} → a(≠ 0) and m_{i} → ±∞, the result (4) again follows.

3 Lipschitz discontinuity of D_{S}|_{ℤ}

We recall Dirichlet’s theorem on arithmetic progressions.

Theorem 5. Let a ∈ ℤ, b ∈ ℤ_{+}. If gcd (a, b) = 1, then there exist infinitely many primes of the form a + kb, where k ∈ ℤ_{+}.

Proof. See the proof of [1, Theorem 7.9].

Since all integer points are isolated (i.e., for all a ∈ ℤ, there is r > 0 such that N(a, r) ∩ ℤ = {a}, where N(a, r) = {x ∈ ℚ | |x − a| < r}), the restriction D_{S}|_{ℤ} is continuous (in the relative topology). However, we prove that even D_{S}|_{ℤ_{+}} is Lipschitz discontinuous.

Theorem 6. Let ∅ ≠ S ⊆ ℙ. The restriction D_{S}|_{ℤ_{+}} is Lipschitz discontinuous.

Proof. Let L > 0. We show that there exist x, y ∈ ℤ_{+} such that

x − y = 1 and D_{S}(x) − D_{S}(y) > L.  (13)

Let p ∈ S. Choose n ∈ ℤ_{+} with

np^{n−1} > L + 1.  (14)

By Theorem 5, there is k ∈ ℤ_{+} such that

y = kp^{n} − 1 ∈ ℙ.

Let x = kp^{n}. Then

x − y = 1, D_{S}(y) ∈ {0, 1},  (15)

and

D_{S}(x) = kD_{S}(p^{n}) + p^{n}D_{S}(k) ≥ kD_{S}(p^{n}) = knp^{n−1} ≥ np^{n−1}.  (16)

Now, Eqs. (15), (16), and (14) imply Eq. (13).

On the other hand, we give two examples showing that if S = ℙ or S = {p}, p ∈ ℙ, then there is an infinite set A ⊂ ℚ such that D_{S}|_{A} is Lipschitz continuous.

Example 7. Let x ∈ ℚ. Then [8, Theorem 15]

x′ = 0  (17)

if (not “only if”, contrary to this theorem)

x = 0 or x = ± ∏_{p∈ℙ} p^{ξ_{p}(x)p},  (18)

where (ξ_{p}(x))_{p∈ℙ} is an integer sequence with a finite number of nonzero terms such that

∑_{p∈ℙ} ξ_{p}(x) = 0.

We present a correct “only if” part in Theorem 9. Anyway, the set

A = {x ∈ ℚ | x satisfies (18)}

is clearly infinite. Since the restriction D|_{A} is identically zero by Eq. (17), it is Lipschitz continuous.

Example 8. Let p ∈ ℙ and a ∈ ℚ satisfy

ap ∈ ℤ,  (19)

and let x ∈ ℚ. Then [3, Theorem 3]

x′_{p} = ax  (20)

if and only if

x = cp^{ap}, where ν_{p}(c) = 0.  (21)

The set

A = {x ∈ ℚ | x satisfies (21)}

is clearly infinite. By Eq. (20), we have D_{p}|_{A}(x) = ax. It is Lipschitz continuous with L = |a|.

4 The differential equation x′_{S} = ax

We extend Examples 7 and 8.

Theorem 9. Let ∅ ≠ S ⊆ ℙ and a, x ∈ ℚ. Then

x′_{S} = ax  (22)

if and only if

x = c ∏_{p∈S} p^{ξ_{p}(x)p}.  (23)

Here

ν_{p}(c) = 0 for all p ∈ S,  (24)

and (c = 0 or) (ξ_{p}(x))_{p∈S} is a sequence of rational numbers with a finite number of nonzero terms (this is needed if S is infinite) satisfying

ξ_{p}(x)p ∈ ℤ for all p ∈ S  (25)

and

∑_{p∈S} ξ_{p}(x) = a.  (26)

Proof. If x is as in Eq. (23), then, by Eq. (26), we get

x′_{S} = x ∑_{p∈S} \frac{ξ_{p}(x)p}{p} = x ∑_{p∈S} ξ_{p}(x) = xa,

verifying Eq. (22).

Conversely, suppose that x satisfies Eq. (22). We show that it satisfies Eq. (23). The case of x = 0 is trivial. If the claim holds for x, then it also holds for −x. Therefore, it is enough to consider

x = ∏_{p∈ℙ} p^{ν_{p}(x)}.

Denoting ν_{p} = ν_{p}(x) for short, we can write

x = c ∏_{p∈S} p^{ν_{p}} = c ∏_{p∈S} p^{ξ_{p}p},

where c satisfies Eq. (24) and

ξ_{p} = \frac{ν_{p}}{p} , p ∈ S.

Now,

∑_{p∈S} ξ_{p} = ∑_{p∈S} \frac{ν_{p}}{p} = \frac{x′_{S}}{x} = \frac{ax}{x} = a

by Eq. (22), i.e., Eq. (26) holds. Also Eq. (25) is obviously satisfied. Thus Eq. (23) follows.

The set

A = {x ∈ ℚ | x satisfies (23)}

is clearly infinite. By Eq. (22), the restriction D_{S}|_{A}(x) = ax. It is Lipschitz continuous with L = |a|.

If S = ℙ and a = 0, then Eq. (23) reads

x = c ∏_{p∈ℙ} p^{ξ_{p}(x)p}.

Here

∑_{p∈ℙ} ξ_{p}(x) = 0

and ν_{p}(c) = 0 for all p ∈ ℙ, i.e., c ∈ {0, ±1}. Thus we encounter Eq. (18), but it is enough that ξ_{p}(x)p ∈ ℤ for all p ∈ ℙ, not necessarily ξ_{p}(x) ∈ ℤ.

If S = {p}, then

a = ∑_{q∈{p}} ξ_{q}(x) = ξ_{p}(x).

Hence, by Eq. (23),

x = c ∏_{q∈{p}} p^{ξ_{q}(x)q} = cp^{ξ_{p}(x)p} = cp^{ap}, where ν_{p}(c) = 0,

repeating Eq. (21). The condition (25) reduces to (19).

5 Concluding remarks

We proved that D_{S} is superdiscontinuous at any a ∈ ℚ. We also proved that its restriction to a suitable infinite set A is Lipschitz continuous. This happens if, for example, the set A consists of the solutions of x′ = 0 or those of

x′_{p} = ax,  (27)

where ap ∈ ℤ. These equations have been discussed in the literature, while the more general equation,

x′_{S} = ax,  (28)

apparently has not. Anyway, it also provides us a suitable A if there is a sequence (ξ_{p}(x))_{p∈S} satisfying Eqs. (25) and (26).

According to Example 8, ap ∈ ℤ is a sufficient condition for Eq. (27) to have nontrivial solutions. In fact, it is also necessary [3, Theorem 3]. But what about Eq. (28)? It has nontrivial solutions if and only if there are p_{1}, …, p_{k} ∈ S such that ap_{1} ⋯ p_{k} ∈ ℤ. We will present the proof in a forthcoming paper.

References

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2010 Mathematics Subject Classification: Primary 11A25; Secondary 26A15. Keywords: arithmetic subderivative, arithmetic partial derivative, arithmetic derivative, continuity, Lipschitz continuity.

(Concerned with sequences \underline{A000040} and \underline{A003415}.)

Received June 18 2019; revised version received October 9 2019. Published in Journal of Integer Sequences, October 15 2019.

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