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\begin{center}
\vskip 1cm{\LARGE\bf  Arithmetic Subderivatives: $p$-adic Discontinuity and Continuity\\
\vskip 1cm}
\large
Pentti Haukkanen and Jorma K. Merikoski\footnote{Corresponding author.} \\
Faculty of Information Technology and Communication Sciences \\
FI-33014 Tampere University\\ 
Finland\\
\href{mailto:pentti.haukkanen@tuni.fi}{\tt pentti.haukkanen@tuni.fi} \\
\href{mailto:jorma.merikoski@tuni.fi}{\tt jorma.merikoski@tuni.fi} \\
\ \\
Timo Tossavainen\\
Department of Arts, Communication and Education\\ 
Lulea University of Technology\\
SE-97187 Lulea\\
Sweden \\
\href{mailto:timo.tossavainen@ltu.se}{\tt timo.tossavainen@ltu.se}
\end{center}

\vskip .2 in

\begin{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 $p\notin S$.
\end{abstract}

\section{Introduction}

Let $0\ne x\in\Q$. There exists a unique sequence
$(\nu_p(x))_{p\in\PP}$ of integers (with only finitely many nonzero terms)
such that

\begin{equation}
\label{factoriz}
x=(\sgn{x})\prod_{p\in\PP}p^{\nu_p(x)}.
\end{equation}
Here $\mathbb P$ stands for the set of primes, and $\sgn{x}=x/|x|$.
Define that $\sgn{0}=0$ and $\nu_p(0)=\infty$ for all $p\in\PP$. In addition to the ordinary
axioms of~$\infty$, we state that $0\cdot\infty=0$. Then  (\ref{factoriz}) holds also for $x=0$.

We recall the basic properties of the {\em p-adic order}~$\nu_p$.

\begin{proposition}
\label{nu}
For all $x,y\in\Q$,
\begin{enumerate}
\item[{\rm (a)}] $\nu_p(x)=\infty$ if and only if $x=0;$
\item[{\rm (b)}] $\nu_p(xy)=\nu_p(x)+\nu_p(y);$
\item[{\rm (c)}] $\nu_p(x+y)\ge\min(\nu_p(x),\nu_p(y));$
\item[{\rm (d)}] $\nu_p(x+y)=\min(\nu_p(x),\nu_p(y))$ if $\nu_p(x)\ne\nu_p(y)$.
\end{enumerate}
\end{proposition}
\begin{proof}
Properties (a) and (b) are trivial. For (c) and~(d), see, e.g., \cite[Proposition~2.4]{Bak}.
\end{proof}

Throughout this paper, we let $a\in\Q$, $p,q\in\PP$, $p\ne q$, and $\emptyset\ne S\subseteq\mathbb P$.

The {\em arithmetic subderivative}
\cite{MHT,HMT2,HMT3} of~$x\in\Q$ with respect to~$S$, a.k.a.\ the {\em arithmetic type
derivative}~\cite{FU} is 
$$
D_S(x)=x\sum_{p\in S}\frac{\nu_p(x)}{p}.
$$
The {\em arithmetic partial derivative} \cite{Ko,HMT1} of~$x$ with
respect to~$p$ is $D_p(x)=D_{\{p\}}(x)$. The {\em arithmetic
derivative} \cite{Sh,Bar,UA} of~$x$ is $D(x)=D_{\PP}(x)$.
Clearly,
$$
D_S(x)=\sum_{p\in S}D_p(x),\quad D(x)=\sum_{p\in\PP}D_p(x).
$$

The function~$D_S$ is very strongly discontinuous at any $a$
\cite[Theorem~4]{HMT2} with respect to the ordinary absolute value.
But do we succeed better if we use the {\em p-adic absolute value} of~$x$, defined by
$$
|x|_p=\frac{1}{p^{\nu_p(x)}}?
$$
(In particular, $|0|_p=1/\infty=0$.)

We recall the basic properties of $|\cdot|_p$.

\begin{proposition}
\label{prop}
For all $x,y\in\Q$,
\begin{enumerate}
\item[{\rm (a)}] $|x|_p=0$ if and only if $x=0;$
\item[{\rm (b)}] $|xy|_p=|x|_p|y|_p;$
\item[{\rm (c)}] $|x+y|_p\le\max(|x|_p,|y|_p);$
\item[{\rm (d)}] $|x+y|_p=\max(|x|_p,|y|_p)$ if $|x|_p\ne|y|_p$.
\end{enumerate}
\end{proposition}
\begin{proof}
This proposition is equivalent to Proposition~\ref{nu}.
\end{proof}

Let us write
$$
x=x\frac{1}{p^{\nu_p(x)}}p^{\nu_p(x)}=x|x|_pp^{\nu_p(x)}=\mu_p(x)p^{\nu_p(x)},
$$
where
\begin{equation}
\label{aap}
\mu_p(x)=x|x|_p=\frac{x}{p^{\nu_p(x)}}.
\end{equation}

\begin{proposition}
\label{numu}
For all $x,y\in\Q$,
\begin{enumerate}
\item[{\rm (a)}] $\nu_p(\mu_p(x))=0$ if $x\ne 0,$\quad$\nu_p(\mu_p(0))=\nu_p(0)=\infty;$
\item[{\rm (b)}] $|\mu_p(x)|_p=1$ if $x\ne 0,$\quad $|\mu_p(0)|_p=0;$
\item[{\rm (c)}] $\mu_p(xy)=\mu_p(x)\mu_p(y);$
\item[{\rm (d)}] $D_p(\mu_p(x))=0$.
\end{enumerate}
\end{proposition}
\begin{proof}
Trivial.
\end{proof}

The {\em p-adic distance} $|x-y|_p$ is smaller the larger $\nu_p(x-y)$ is.

We say that a function $\Q\to\Q$ is {\em p-adically continuous}, in short
{\em p-continuous}, if it is continuous with respect to~$|\cdot|_p$. So we ask:
Is $D_S$ $p$-continuous at some~$a$? We study this question by considering
sequences~$(x_i)$ of rational numbers. If $|x_i-a|_p\to 0$, equivalently $\nu_p(x_i-a)\to\infty$,
then $(x_i)$ {\em converges} {\em p-adically},
in short {\em p-converges}, to~$a$. Let $x_i\to_pa$ denote this convergence.

Sections \ref{Poor}--\ref{Convergence} are introductory.
We present in Section~\ref{Poor} a ``light version'' of Dirichlet's theorem on arithmetic
progressions. 
We study $p$-convergence in Section~\ref{Convergence}.

Sections \ref{DSorig}--\ref{Conclusion} contain our main results. We prove in Section~\ref{DSorig}
that $D_S$ is $p$-continuous at $a=0$ if $S$ is finite or $p\notin S$. We also prove
that $D_p$ is $p$-continuous also at $a\ne 0$. On the other hand, we show in Section~\ref{Dq}
that $D_q$ can be (and conjecture that it always is) $p$-discontinuous at $a\ne 0$.
In Section~\ref{DP}, we extend the results of Section~\ref{Dq} to~$D_S$ when $S$
is finite. Although Section~\ref{Dq} is only a special case of Section~\ref{DP},
we find it instructive to present it separately. We complete our paper with the
conclusion in Section~\ref{Conclusion}. 

\section{``Poor man's theorem on arithmetic progressions''}
\label{Poor}
Throughout this section, $a,b\in\Z$ with $\gcd{(a,b)}=1$. As suggested by Graham et al.~\cite{GKP},
we let $a\perp b$ denote that $\gcd(a,b)=1$. See also~\cite{HMMT,FU}.

We recall Dirichlet's theorem on arithmetic progressions.

\begin{theorem}
\label{Dirichlet}
If $b>0$, then the set
\begin{equation}
\label{T}
T=\{a+nb:n\in\Z_+\}
\end{equation}
contains infinitely many primes.
\end{theorem}
\begin{proof}
See, e.g., \cite[Theorem~7.9]{Ap}.
\end{proof}

\begin{corollary}
If $b\ne 0$, then the set~{\rm(\ref{T})} contains infinitely many primes or their additive inverses.
\end{corollary}
\begin{proof}
Trivial.
\end{proof}

Theorem~\ref{Dirichlet} 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~\ref{Dirichlet}, the following ``poor man's theorem on
arithmetic progressions.'' It is elementary but strong enough for us.
(Remember that $\emptyset\ne S\subseteq\PP$ throughout.)

\begin{theorem}
\label{poorman}
If $S$ is finite and $b\ne 0$, then the set~{\rm (\ref{T})} contains infinitely many numbers that
are not divisible by any element of~$S$.
\end{theorem}
\begin{proof}
If $a=0$, then $b=\pm 1$, since otherwise $a\not\perp b$, a contradiction.
Therefore $T=\Z_+$ or $T=\Z_-$, and the claim is trivially true.

Now assume that $a\ne 0$. Let
$$
S=\{p_1,\dots,p_h,q_1,\dots,q_k\},\quad p_1,\dots,p_h\nmid a,\quad q_1,\dots,q_k\mid a.
$$
(Either the $p_i$ or the $q_i$ can be missing. Clearly, $p_i\ne q_j$ for all $i,j$.)
We show that the numbers $a+nb$ apply when $n$ goes through the set
$$
N=\{mp_1\cdots p_h: m\in\mathbb{Z}_+,\,q_1,\dots,q_k\nmid m\}.
$$
Write
$$
c=p_1\cdots p_h.
$$
(If the $p_i$ are missing, then the ``empty product'' $c=1$.)

Let $x\in T$ with $n\in N$, that is,
\begin{equation}
\label{x}
x=a+mcb,\quad q_1,\dots,q_k\nmid m.
\end{equation}
Each $p_i\mid c$ but $p_i\nmid a$, so $p_i\nmid x$. Each
$q_i\mid a$ but $q_i\nmid mcb$. (Clearly, $q_i\nmid m,c$. If $q_i\mid b$, then
$a\not\perp b$, a contradiction.) Therefore also $q_i\nmid x$. 
Consequently,
$s\nmid x$ for all $s\in S$. Because there are infinitely many numbers~(\ref{x}), the
claim follows.
\end{proof}

\section{Convergence}
\label{Convergence}

Continuity is usually proved by the ``$\varepsilon-\delta$ 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.

\begin{proposition}
\label{convcond}
Let $(x_i)$ be a sequence of rational numbers.
If $a\ne 0$, then the following conditions are equivalent.
\begin{enumerate}
\item[{\rm (a)}] $x_i\to_pa;$
\item[{\rm (b)}] $\mu_p(x_i)\to_p\mu_p(a)$ and there is $i_0\in\Z_+$ such that
$\nu_p(x_{i_0})=\nu_p(x_{i_0+1})=\cdots=\nu_p(a)$.
\end{enumerate}
If $a=0$, then {\rm (b)} $\Rightarrow$ {\rm (a)} but not conversely.
\end{proposition}
\begin{proof}
\leavevmode
\ \\
\noindent{\em Case 1:} $a\ne 0$. 

\noindent (a) $\Rightarrow$ (b): If $i_0$ does not exist, then $(x_i)$ has a subsequence
$(x_{i_k})$ whose each term satisfies $\nu_p(x_{i_k})\ne\nu_p(a)$. 
Consequently,
$$
\nu_p(x_{i_k}-a)\stackrel{{\rm Prop.{\scriptsize\ref{nu}}(d)}}{=}\min(\nu_p(x_{i_k}),\nu_p(a))\le\nu_p(a)
\stackrel{a\ne 0}{<}\infty,\quad k\ge 1.
$$
Hence $\nu_p(x_{i_k}-a)\not\to\infty$, implying $x_i\not\to_pa$, a contradiction. Therefore $i_0$ exists, i.e.,
$$
x_i=\mu_p(x_i)p^{\nu_p(a)},\quad i\ge i_0.
$$
Now, for $i\ge i_0$, we have
\begin{equation}
\label{xminusa}
x_i-a=(\mu_p(x_i)-\mu_p(a))p^{\nu_p(a)},
\end{equation}
and further
$$
\nu_p(\mu_p(x_i)-\mu_p(a))+\nu_p(a)
\stackrel{{\rm(\scriptsize{\ref{xminusa}}),Prop.~\scriptsize{\ref{nu}}(b)}}{=}\nu_p(x_i-a)
\stackrel{{\rm (a)}}{\to}\infty,
$$
verifying $\mu_p(x_i)\to_p\mu_p(a)$.

\smallskip

\noindent (b) $\Rightarrow$ (a): Since
$$
x_i-a\stackrel{{\rm(b)}}{=}\mu_p(x_i)p^{\nu_p(a)}-\mu_p(a)p^{\nu_p(a)}=
(\mu_p(x_i)-\mu_p(a))p^{\nu_p(a)},\quad i\ge i_0,
$$
we have
$$
\nu_p(x_i-a)\stackrel{{\rm Prop.~\scriptsize{\ref{nu}}(b)}}{=}\nu_p(\mu_p(x_i)-\mu_p(a))+\nu_p(a)
\stackrel{{\rm(b)}}\to\infty,
$$
verifying (a).

\smallskip

\noindent {\em Case 2:} $a=0$.

\noindent (b) $\Rightarrow$ (a). Since $\nu_p(x_{i_0})=\nu_p(x_{i_0+1})=\cdots=\nu_p(0)=\infty$,
it follows that $x_{i_0}=x_{i_0+1}=\cdots=0$. Therefore $x_i\to_p 0$.

\smallskip

\noindent (a) $\not\Rightarrow$ (b). If $x_i=p^i$, then $x_i\to_p 0$, but $\mu_p(x_i)=1\to_p1\ne 0=\mu_p(0)$.
\end{proof}

\begin{proposition}
\label{continuity}
A function $F:\Q\to\Q$ is $p$-continuous at $a$ if and only if any sequence~$(x_i)$
of rational numbers satisfying $x_i\to_pa$ satisfies $F(x_i)\to_pF(a)$.
\end{proposition}
\begin{proof}
Proceed as in proving the corresponding property of the ordinary continuity.
\end{proof}

There are three formally different ways to consider $p$-convergence. First, use~$\nu_p$ everywhere.
Second, use~$|\cdot|_p$ everywhere. Third, use either $\nu_p$ or~$|\cdot|_p$, depending on the
situation. We follow the first way.

\section{The cases of $D_S$, $a=0$, and $D_p$, $a$ arbitrary}
\label{DSorig}

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

\begin{lemma}
\label{lemma}
Let $S$ be finite and $y\in\Q$. Assume that
$$
\{q\in\PP\mid\nu_q(y)\ne 0\}\subseteq S.
$$
Factorize
$$
y=\prod_{q\in S}q^{\nu_q(y)}=u(y)w(y),
$$
where
$$
u(y)=\prod_{q\in S}q^{\nu_q(y)-1},\quad w(y)=\prod_{q\in S}q.
$$
Then
$$
D_S(y)=u(y)v(y),
$$
where
$$
v(y)=\sum_{q\in S}\nu_q(y)\prod_{r\in S\setminus\{q\}}r.
$$
\end{lemma}
\begin{proof}
We have
\begin{align*}
D_S(y)&=\sum_{q\in S}D_S(q^{\nu_q(y)})\prod_{r\in S\setminus\{q\}}r^{\nu_r(y)}=
\sum_{q\in S}\nu_q(y)q^{\nu_q(y)-1}\prod_{r\in S\setminus\{q\}}r^{\nu_r(y)}\quad
\\
&=\sum_{q\in S}\nu_q(y)q^{\nu_q(y)-1}
\prod_{r\in S\setminus\{q\}}r^{\nu_r(y)-1}r 
=\sum_{q\in S}\nu_q(y)\Big(\prod_{r\in S}r^{\nu_r(y)-1}\Big)
\Big(\prod_{r\in S\setminus\{q\}}r\Big),
\end{align*}
verifying the claim.
\end{proof}

\begin{theorem}
\label{origin}
If $S$ is finite or $p\notin S$, then $D_S$ is $p$-continuous at the origin.
\end{theorem}
\begin{proof}
Let
\begin{equation}
\label{xito0}
x_i\to_p0.
\end{equation}
We show that $D_S(x_i)\to_p 0=D_S(0)$.

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
$x_i\ne 0$. Because zeros do not cause any problem in the proof, we can
omit them and thus assume that each $x_i\ne 0$.

Write
$$
x_i\stackrel{\rm (\scriptsize{\ref{factoriz}})}{=}(\sgn{x_i})\prod_{q\in\PP}q^{\nu_q(x_i)}=
(\sgn{x_i})\Big(\prod_{q\in S}q^{\nu_q(x_i)}\Big)
\Big(\prod_{q\in\PP\setminus S}q^{\nu_q(x_i)}\Big)=(\sgn{x_i})y_iz_i,
$$
where
$$
y_i=\prod_{q\in S}q^{\nu_q(x_i)},\quad z_i=\prod_{q\in\PP\setminus S}q^{\nu_q(x_i)}.
$$
(If $S=\PP$, then the ``empty product'' $z_i=1$.) Then
\begin{equation}
\label{dx}
D_S(x_i)=(\sgn{x_i})z_iD_S(y_i).
\end{equation}

First, let us assume that $S$ is finite. By Lemma~\ref{lemma},
\begin{align}
\label{ds}
D_S(y_i) &=u(y_i)v(y_i).
\end{align}
Since $v(y_i)\in\Z$, it follows that
\begin{equation}
\label{nuge0}
\nu_p(v(y_i))\ge 0.
\end{equation}

If $p\notin S$, then
$$
\nu_p(D_S(x_i))\stackrel{{\rm(\scriptsize{\ref{dx}}),(\scriptsize{\ref{ds}})}}{=}\nu_p(z_iu(y_i)v(y_i))\stackrel{{\rm Prop.~\scriptsize{\ref{nu}}(b)}}{=}
\nu_p(z_i)+0+\nu_p(v(y_i))\stackrel{{\rm(\scriptsize{\ref{nuge0}})}}{\ge}\nu_p(z_i)=\nu_p(x_i)
\stackrel{{\rm (\scriptsize{\ref{xito0}})}}{\to}\infty.
$$
If $p\in S$, then
\begin{align*}
\nu_p(D_S(x_i))\stackrel{{\rm(\scriptsize{\ref{dx}}),(\scriptsize{\ref{ds}})}}{=}& \nu_p(z_iu(y_i)v(y_i))
\stackrel{{\rm Prop.~\scriptsize{\ref{nu}}(b)}}{=}0+\nu_p(u(y_i))+\nu_p(v(y_i))
\stackrel{{\rm(\scriptsize{\ref{nuge0}})}}{\ge}\nu_p(u(y_i))
\\
&=\nu_p(x_i)-1\stackrel{{\rm (\scriptsize{\ref{xito0}})}}{\to}\infty.
\end{align*}
Thus, $D_S(x_i)\to_p 0$ in each case.

Second, assume that $S$ is infinite. Because $w(y)$ and $v(y)$ contain a divergent infinite
product, applying Lemma~\ref{lemma} needs some preparation. Define
$$
S_i=\{q\in S\mid\nu_q(x_i)\ne 0\}.
$$
If $D_S(x_i)\ne 0$ 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 $S_i\ne\emptyset$.

Now
\begin{equation}
\label{dxx}
D_S(x_i)=D_{S_i}(x_i)\stackrel{(\scriptsize{\ref{dx}})}{=}(\sgn{x_i})z_iD_{S_i}(y_i).
\end{equation}
By Lemma~\ref{lemma},
\begin{equation}
\label{dss}
D_{S_i}(y_i)=u_i(y_i)v_i(y_i),
\end{equation}
where
$$
u_i(y_i)=\prod_{q\in S_i}q^{\nu_q(y_i)-1},\quad
v_i(y_i)=\sum_{q\in S_i}\nu_q(y_i)\prod_{r\in S_i\setminus\{q\}}r.
$$
If $p\notin S$, then
$$
\nu_p(D_S(x_i))\stackrel{{\rm(\scriptsize{\ref{dxx}}),(\scriptsize{\ref{dss}})}}{=}\nu_p(z_iu_i(y_i)v_i(y_i))\stackrel{{\rm Prop.~\scriptsize{\ref{nu}}(b)}}{=}
\nu_p(z_i)+0+\nu_p(v_i(y_i))\stackrel{{\rm(\scriptsize{\ref{nuge0}})}}{\ge}\nu_p(z_i)=\nu_p(x_i)
\stackrel{{\rm (\scriptsize{\ref{xito0}})}}{\to}\infty.
$$
Consequently, $D_S(x_i)\to_p 0$. We discuss the case of $p\in S$ at the end of this section.
\end{proof}

\begin{theorem}
\label{Dpthm}
The function $D_p$ is $p$-continuous everywhere.
\end{theorem}
\begin{proof}
If $a=0$, then apply Theorem~\ref{origin}. If
$a\ne 0$, then let $x_i\to_p a$, and let $i_0$ be as in  Proposition~\ref{convcond}. For $i\ge i_0$,
\begin{align}
\label{diff}
D_p(x_i)-D_p(a)&=D_p(\mu_p(x_i)p^{\nu_p(a)})-D_p(\mu_p(a)p^{\nu_p(a)})
\nonumber \\
& \stackrel{{\rm Prop.~\scriptsize{\ref{numu}}(d)}}{=}\nu_p(a)p^{\nu_p(a)-1}\mu_p(x_i)-\nu_p(a)p^{\nu_p(a)-1}\mu_p(a)
\nonumber \\
&=\nu_p(a)p^{\nu_p(a)-1}(\mu_p(x_i)-\mu_p(a))=c(\mu_p(x_i)-\mu_p(a)),\quad c=\nu_p(a)p^{\nu_p(a)-1}.
\end{align}

If $\nu_p(a)=0$, then
$$
D_p(x_i)\stackrel{(\scriptsize{\ref{diff}})}{=}D_p(a),\quad i\ge i_0.
$$
If $\nu_p(a)\ne 0$, then
$$
\nu_p(D_p(x_i)-D_p(a))\stackrel{{\rm(\scriptsize{\ref{diff}}),Prop.~\scriptsize{\ref{nu}}(b)}}{=}
\nu_p(c)+\nu_p(\mu_p(x_i)-\mu_p(a))
\stackrel{{\rm Prop.~\scriptsize{\ref{convcond}}}}{\to}\infty.
$$
Therefore
$D_p(x_i)\to_pD_p(a)$ in each case.
\end{proof}

Can we extend the proof of Theorem~\ref{origin} to the case where $S$ is infinite and $p\in S$?
If $p\in S_i$, then
\begin{align*}
\nu_p(D_S(x_i))
& \stackrel{{\rm(\scriptsize{\ref{dxx}}),(\scriptsize{\ref{dss}})}}{=}\nu_p(z_iu_i(y_i)v_i(y_i))\stackrel{{\rm Prop.~\scriptsize{\ref{nu}}(b)}}{=}
0+\nu_p(u_i(y_i))+\nu_p(v_i(y_i))\stackrel{{\rm(\scriptsize{\ref{nuge0}})}}{\ge}\nu_p(u_i(y_i))
\\
&=\nu_p(x_i)
\stackrel{{\rm (\scriptsize{\ref{xito0}})}}{\to}\infty,
\end{align*}
implying the convergence.

If
\begin{equation}
\label{p}
p\in S\setminus S_i,
\end{equation}
then
$$
\nu_p(D_S(x_i))\stackrel{{\scriptsize{\rm(\ref{dxx})},(\scriptsize{\ref{dss}})}}{=}\nu_p(z_iu_i(y_i)v_i(y_i))\stackrel{{\rm Prop.~\scriptsize{\ref{nu}}(b)}}{=}
0+0+\nu_p(v_i(y_i))=\nu_p(x_i).
$$
If (\ref{p}) 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 $D_S(x_i)\to_p0$ also if, for any sequence~$(x_i)$
of nonzero numbers with $x_i\to_p0$, only finitely many terms
satisfy~(\ref{p}). However, we find this assumption useless, because its validity
cannot be tested.

\section{The case of $D_q$, $a\ne 0$}
\label{Dq}

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.

\begin{theorem}
\label{Dqthm1}
Let
\begin{equation}
\label{aneq0}
a\ne 0.
\end{equation}
If
\begin{equation}
\label{dqa}
D_q(a)=0,
\end{equation}
then $D_q$ is $p$-discontinuous at~$a$.
\end{theorem}
\begin{proof}
Let
\begin{equation}
\label{xii}
x_i=a+\frac{p^i}{q}.
\end{equation}
Then
\begin{equation}
\label{absvto0}
\nu_p(x_i-a)=i\to\infty,
\end{equation}
implying  $x_i\to_pa$. Since
$$
\nu_q(a)\stackrel{{\rm(\scriptsize{\ref{dqa}})}}{=}0,\quad \nu_q(\frac{p^i}{q})=-1,
$$
we have $\nu_q(x_i)\stackrel{{\rm(\scriptsize{\ref{xii}}),
Prop.~\scriptsize{\ref{nu}}(d)}}{=}-1$.

Consequently,
\begin{equation}
\label{dqxip}
D_q(x_i)=\frac{\nu_q(x_i)}{q}x_i=-\frac{x_i}{q},
\end{equation}
and further
$$
\nu_p(D_q(x_i)-D_q(a))\stackrel{{\rm(\scriptsize{\ref{dqa}})}}{=}
\nu_p(D_q(x_i))\stackrel{{\rm(\scriptsize{\ref{dqxip}})}}{=}\nu_p(\frac{x_i}{q})=\nu_p(x_i).
$$
We show that $\nu_p(x_i)\not\to\infty$; then $D_q(x_i)\not\to_pD_q(a)$, verifying the claim. If
\begin{equation}
\label{nutoinfty}
\nu_p(x_i)\to\infty,
\end{equation}
then
$$
\nu_p(a)=\nu_p(x_i-(x_i-a))\stackrel{{\rm Prop.~\scriptsize{\ref{nu}}(c)}}{\ge}
\min(\nu_p(x_i),\nu_p(x_i-a))\stackrel{{\rm(\scriptsize{\ref{absvto0}}),(\scriptsize{\ref{nutoinfty}})}}{\to}\infty.
$$
Hence $\nu_p(a)=\infty$, i.e., $a=0$, contradicting~(\ref{aneq0}).
\end{proof}

\begin{theorem}
\label{Dqthm2}
Let
\begin{equation}
\label{a12}
a=\frac{a_1}{a_2},\quad 0\ne a_1\in \Z,\,a_2\in\Z_+,\quad a_1\perp a_2.
\end{equation}
If
\begin{equation}
\label{dqane0}
D_q(a)\ne 0
\end{equation}
and
\begin{equation}
\label{dqa2}
q\nmid a_2,
\end{equation}
then $D_q$ is $p$-discontinuous at~$a$.
\end{theorem}
\begin{proof}
Let $i\in\Z_+$. Since $\mu_p(a_1)\stackrel{{\rm(\scriptsize{\ref{a12}}),
Prop.~\scriptsize{\ref{numu}}(a)}}{\perp}\mu_p(a_2)p^i$
and $\mu_p(a_2)\stackrel{{\rm(\scriptsize{\ref{a12}})}}{>}0$, there are, by Theorem~\ref{poorman}
($S=\{p,q\}$, $a=\mu_p(a_1)$, $b=\mu_p(a_2)p^i$) positive integers $r_{i1}<r_{i2}<\cdots$ satisfying
\begin{equation}
\label{dirichlet}
r_{ik}=\mu_p(a_1)+n_{ik}\mu_p(a_2)p^i,\quad n_{ik}\in\Z_+,\quad p,q\nmid r_{ik},\quad k=1,2,\dots.
\end{equation}
Consequently,
$$
\mu_p(a)+n_{ik}p^i\stackrel{{\rm Prop.~\scriptsize{\ref{numu}}(c)}}{=}
\frac{\mu_p(a_1)}{\mu_p(a_2)}+n_{ik}p^i=
\frac{\mu_p(a_1)+n_{ik}\mu_p(a_2)p^i}{\mu_p(a_2)}
\stackrel{{\rm(\scriptsize{\ref{dirichlet}})}}{=}\frac{r_{ik}}{\mu_p(a_2)}.
$$
Choose $r_{1k_1}<r_{2k_2}<\cdots$ and write $n_1=n_{1k_1},n_2=n_{2k_2},\dots$,
$r_1=r_{1k_1},r_2=r_{2k_2},\dots$. Define the sequence~$(y_i)$ by
\begin{equation}
\label{xi}
y_i=\mu_p(a)+n_ip^i=\frac{r_i}{\mu_p(a_2)}.
\end{equation}
Then
\begin{equation}
\label{nupyi}
\nu_p(y_i)\stackrel{{{\rm(\scriptsize{\ref{xi}})},{\rm Prop.~\scriptsize{\ref{nu}}(b)}}}{=}\nu_p(r_i)-\nu_p(\mu_p(a_2))
\stackrel{{{\rm(\scriptsize{\ref{dirichlet}})},{\rm Prop.~\scriptsize{\ref{numu}}(a)}}}{=}0
\end{equation}
and
\begin{equation}
\label{xilim}
\nu_p(y_i-\mu_p(a))
\stackrel{{\rm(\scriptsize{\ref{xi}})}}{=}\nu_p(n_ip^i)
\stackrel{{\rm Prop.~\scriptsize{\ref{nu}}(b)}}{=}\nu_p(n_i)+i
\stackrel{n_i\in\Z_+}{\ge}i\to\infty.
\end{equation}

For all $i\in\Z_+$, we have
\begin{align}
\label{dqxi}
D_q(y_i) & \stackrel{{\rm(\scriptsize{\ref{xi}})}}{=}\frac{\mu_p(a_2)D_q(r_i)-r_iD_q(\mu_p(a_2))}{\mu_p(a_2)^2}
\nonumber
\\
&\stackrel{q\,\nmid\,r_i}{=}-\frac{r_iD_q(\mu_p(a_2))}{\mu_p(a_2)^2}
\stackrel{{\rm(\scriptsize{\ref{aap}})}}{=}-\Big(r_iD_q\big(\frac{a_2}{p^{\nu_p(a_2)}}\big)\Big)\big/\Big(\frac{a_2}{p^{\nu_p(a_2)}}\Big)^2
\nonumber
\\
&=
-\Big(\frac{r_i}{p^{\nu_p(a_2)}}D_q(a_2)\Big)\big/\Big(\frac{a_2}{p^{\nu_p(a_2)}}\Big)^2
=-\frac{r_ip^{\nu_p(a_2)}D_q(a_2)}{a_2^2}\stackrel{{\rm(\scriptsize{\ref{dqa2}})}}{=}0.
\end{align}
Also define
\begin{equation}
\label{xiyi}
x_i=y_ip^{\nu_p(a)};
\end{equation}
then
\begin{equation}
\label{dqxii}
D_q(x_i)=p^{\nu_p(a)}D_q(y_i)\stackrel{{\rm(\scriptsize{\ref{dqxi}})}}{=}0.
\end{equation}

Since $\mu_p(x_i)\stackrel{{\rm(\scriptsize{\ref{nupyi}}),(\scriptsize{\ref{xiyi}})}}{=}y_i\stackrel{{\rm(\scriptsize{\ref{xilim}})}}{\to_p}\mu_p(a)$, it follows that
$x_i\stackrel{{\rm Prop.~\scriptsize{\ref{convcond}}}}{\to_p}a$. On the other hand, since
$$
D_q(x_i)-D_q(a)
\stackrel{{\rm(\scriptsize{\ref{dqxii}})}}{=}-D_q(a)
\stackrel{{\rm(\scriptsize{\ref{dqane0}})}}{\ne}0,
$$
we have
$$
\nu_p(D_q(x_i)-D_q(a))=\nu_p(D_q(a))
\stackrel{{\rm(\scriptsize{\ref{dqane0}})}}{<}\infty,
$$
verifying $D_q(x_i)\not\to_pD_q(a)$.
\end{proof}

\begin{corollary}
\label{Dqcor}
If $0\ne a\in\Z$, then $D_q$ is $p$-discontinuous at~$a$.
\end{corollary}
\begin{proof}
Apply Theorem~\ref{Dqthm1} if $q\nmid a$, and Theorem~\ref{Dqthm2} if $q\mid a$.
\end{proof}

If $D_q(a)\ne 0$ and $q\mid a_2$ (where $a_2$ is as in~(\ref{a12})), then our question remains open.
We conjecture that discontinuity holds also in this case.

\begin{conjecture}
\label{Dqconj}
The function $D_q$ is $p$-discontinuous outside the origin.
\end{conjecture}

\section{The case of $D_S$, $S$ finite, $a\ne 0$}
\label{DP}

We extend Theorems \ref{Dqthm1} and~\ref{Dqthm2}, Corollary~\ref{Dqcor}, and
Conjecture~\ref{Dqconj}. The presentation branches according to properties of
$D_S(a)$ and~$S$, summarized in Section~\ref{Conclusion}.

\begin{theorem}
\label{DQthm1}
Let $S$ be finite, $p\notin S$, and $a\ne 0$. If
\begin{equation}
\label{DQaeq0}
D_S(a)=0,
\end{equation}
then $D_S$ is $p$-discontinuous at~$a$.
\end{theorem}
\begin{proof}
Let $S=\{q_1,\dots,q_m\}$, take $\gamma\in\Z_+$ satisfying
\begin{equation}
\label{gamma}
\gamma>-\nu_{q_1}(a),\dots,-\nu_{q_m}(a),
\end{equation}
and define
\begin{equation}
\label{xiii}
x_i=a+\frac{p^i}{(q_1\cdots q_m)^\gamma}.
\end{equation}

Let $i\in\Z_+$ and $j\in\{1,\dots,m\}$. Then
\begin{align}
\label{dqjxi}
\nu_p(x_i-a)
\stackrel{{\rm(\scriptsize{\ref{xiii}})}}{=}i\to\infty,\quad x_i \to_p a,\quad
\nu_{q_j}(x_i-a)\stackrel{{\rm(\scriptsize{\ref{xiii}})}}{=}-\gamma
\stackrel{{\rm(\scriptsize{\ref{gamma}})}}{<}\nu_{q_j}(a),\quad\nonumber
\\
\nu_{q_j}(x_i)=\nu_{q_j}((x_i-a)+a)\stackrel{{\rm Prop.~\scriptsize{\ref{nu}}(d)}}{=}-\gamma,
\quad
D_{q_j}(x_i)=\frac{\nu_{q_j}(x_i)}{q_j}x_i=-\frac{\gamma x_i}{q_j}.
\end{align}
As in the proof of Theorem~\ref{Dqthm1}, we see that
\begin{equation}
\label{normeq}
\nu_p(x_i)\not\to\infty.
\end{equation}

Now
\begin{equation}
\label{DPxi}
D_S(x_i)\stackrel{{\rm(\scriptsize{\ref{dqjxi}})}}{=}
-\Big(\frac{1}{q_1}+\cdots+\frac{1}{q_m}\Big)\gamma x_i=
-\frac{e_{m-1}(q_1,\dots,q_m)\gamma x_i}{q_1\cdots q_m},
\end{equation}
where $e_{m-1}$ denotes the $(m-1)$'th elementary symmetric function. Therefore
$$
\nu_p(D_S(x_i)-D_S(a))\stackrel{{\rm (\scriptsize{\ref{DQaeq0}})}}{=}\nu_p(D_S(x_i))
\stackrel{{\rm (\scriptsize{\ref{DPxi}}),Prop.~\scriptsize{\ref{nu}}(b)}}{=}
\nu_p(e_{m-1}(q_1,\dots,q_m))+\nu_p(\gamma)+\nu_p(x_i).
$$
Since $\nu_p(e_{m-1}(q_1,\dots,q_m))$ and $\nu_p(\gamma)$ are (finite) constants and
$\nu_p(x_i)\not\to\infty$, we have
$$
\nu_p(D_S(x_i)-D_S(a))\not\to\infty,
$$
i.e., $D_S(x_i) \not\to_p D_S(a)$.
\end{proof}

\begin{theorem}
\label{DSthm1}
Let $S\ne\{p\}$ be finite, $p\in S$, and $a\ne 0$. If
\begin{equation}
\label{DSaeq0}
D_S(a)=0,
\end{equation}
then $D_S$ is p-discontinuous at~$a$.
\end{theorem}
\begin{proof}
Let
\begin{equation}
\label{ss0}
S=\{q_1,\dots,q_m,p\}, \quad S_0=\{q_1,\dots,q_m\},
\end{equation}
and let $(x_i)$ be as in (\ref{xiii}).
For $i>\nu_p(a)$,
\begin{equation}
\label{nupineq}
\nu_p(x_i-a)\stackrel{{\rm (\scriptsize{\ref{xiii}})}}{=}i>\nu_p(a)
\end{equation}
and further
$$
\nu_p(x_i)=\nu_p((x_i-a)+a)
\stackrel{{\rm(\scriptsize{\ref{nupineq}}),Prop.~\scriptsize{\ref{nu}}(d)}}{=}\nu_p(a);
$$
hence
\begin{equation}
\label{dpxi}
D_p(x_i)=\frac{\nu_p(a)}{p}x_i.
\end{equation}

Now
\begin{equation}
\label{dtxi}
D_S(x_i)=D_{S_0}(x_i)+D_p(x_i)\stackrel{{\rm(\scriptsize{\ref{DPxi}}),(\ref{dpxi})}}{=}
\frac{e_{m-1}(q_1,\dots,q_m)\gamma x_i}{q_1\cdots q_m}+\frac{\nu_p(a)}{p}x_i=cx_i,
\end{equation}
where
$$
c=\frac{e_{m-1}(q_1,\dots,q_m)\gamma}{q_1\cdots q_m}+\frac{\nu_p(a)}{p}.
$$
We can choose $\gamma\in\Z_+$ so that,
in addition to~(\ref{gamma}), the inequality $c\ne 0$ holds.
Then
\begin{equation}
\label{nufinite}
\nu_p(c)<\infty.
\end{equation}
Since
$$
\nu_p(D_S(x_i)-D_S(a))\stackrel{{\rm(\scriptsize{\ref{DSaeq0}})}}{=}\nu_p(D_S(x_i))
\stackrel{{\rm(\scriptsize{\ref{dtxi}}),Prop.~\scriptsize{\ref{nu}}(b)}}{=}\nu_p(c)+\nu_p(x_i)
\stackrel{{\rm(\scriptsize{\ref{normeq}}),(\scriptsize{\ref{nufinite}})}}{\not\to}\infty,
$$
$D_S(x_i)\not\to_p D_S(a)$ follows.
\end{proof}

\begin{theorem}
\label{DQthm2}
Let $a$ be as in~{\rm(\ref{a12})}, let $S$ be finite, and $p\notin S$. If
\begin{equation}
\label{DQane0}
D_S(a)\ne 0
\end{equation}
and
\begin{equation}
\label{nucnd}
\nu_p(a_2D_S(a))\ne\nu_p(aD_S(a_2)),
\end{equation}
then $D_S$ is $p$-discontinuous at~$a$.
\end{theorem}
\begin{proof}
Let $S=\{q_1,\dots,q_m\}$, $i\in\Z_+$, and $j\in\{1,\dots,m\}$. 
Proceeding as in
the proof of Theorem~\ref{Dqthm2}, we have
\begin{align}
r_{ik} &=\mu_p(a_1)+n_{ik}\mu_p(a_2)p^i,\quad p,q_1,\dots,q_m\nmid r_{ik},\quad k=1,2,\dots, \nonumber \\
y_i &=\mu_p(a)+n_ip^i=\frac{r_i}{\mu_p(a_2)},\quad p,q_1,\dots,q_m\nmid r_i,
\label{yidef} \\
D_{q_j}(y_i)&=-\frac{r_ip^{\nu_p(a_2)}D_{q_j}(a_2)}{a_2^2}, \label{tt1} \\
x_i &=y_ip^{\nu_p(a)}\to_p a, \label{Dqjxi}
\end{align}
and
\begin{equation}
\label{Dqjxx}
D_{q_j}(x_i)\stackrel{{\rm(\scriptsize{\ref{tt1}), (\ref{Dqjxi}})}}{=}-p^{\nu_p(a)}\frac{r_ip^{\nu_p(a_2)}D_{q_j}(a_2)}{a_2^2}
\stackrel{{\rm(\scriptsize{\ref{a12}}),Prop.~\scriptsize{\ref{nu}}(b)}}{=}-\frac{r_ip^{\nu_p(a_1)}D_{q_j}(a_2)}{a_2^2}=-cr_iD_{q_j}(a_2),
\end{equation}
where
$$
c=\frac{p^{\nu_p(a_1)}}{a_2^2}.
$$

Consequently,
\begin{equation}
\label{DSxi}
D_S(x_i)=\sum_{j=1}^mD_{q_j}(x_i)\stackrel{{\rm(\scriptsize{\ref{Dqjxx}})}}{=}
-cr_i\sum_{j=1}^mD_{q_j}(a_2)=-cr_iD_S(a_2),
\end{equation}
and further
\begin{align}
\label{DSxipp}
\nu_p(D_S(x_i)) &\stackrel{{\rm(\scriptsize{\ref{DSxi}})}}{=}
\nu_p(cr_iD_S(a_2))\stackrel{{\rm Prop.~\scriptsize{\ref{nu}}(b)}}{=}
\nu_p(a_1)-2\nu_p(a_2)+\nu_p(D_S(a_2))\nonumber
\\
& \stackrel{{\rm(\scriptsize{\ref{a12}}),Prop.~\scriptsize{\ref{nu}}(b)}}{=}
\nu_p(a)-\nu_p(a_2)+\nu_p(D_S(a_2)) \\
&\stackrel{{\rm(\scriptsize{\ref{nucnd}}),Prop.~\scriptsize{\ref{nu}}(b)}}{\ne}
\nu_p(a_2)-\nu_p(a_2)+\nu_p(D_S(a))\nonumber
\\
&=\nu_p(D_S(a)) \label{tt2}.
\end{align}
Let
$$
u(x_i)=D_S(x_i)-D_S(a).
$$
Then
\begin{align}
\label{nudiff}
\nu_p(u(x_i)) &\stackrel{{\rm(\scriptsize{\ref{tt2}}),Prop.~\scriptsize{\ref{nu}}(d)}}{=}
\min(\nu_p(D_S(x_i)),\nu_p(D_S(a)))\nonumber
\\
& \stackrel{{\rm(\scriptsize{\ref{DSxi}}),Prop.~\scriptsize{\ref{nu}}(b)}}{=}\min(\nu_p(cD_S(a_2)),\nu_p(D_S(a))) \\
& \le\nu_p(D_S(a))\stackrel{{\rm(\scriptsize{\ref{DQane0}})}}{<}\infty, \label{tt3}
\end{align}
and $D_S(x_i)\not\to_p D_S(a)$ follows.
\end{proof}

\begin{theorem}
\label{DSthm2}
Let $a$ be as in~{\rm(\ref{a12})}, let $S\ne\{p\}$ be finite and $p\in S$. Assume that
$D_S(a)\ne 0$.
If
\begin{equation}
\label{DSeq0}
D_{S\setminus\{p\}}(a)=0,
\end{equation}
then $D_S$ is p-continuous at~$a$. If
\begin{equation}
\label{DSane0}
D_{S\setminus\{p\}}(a)\ne 0
\end{equation}
and
\begin{equation}
\label{nucond}
\nu_p(a_2D_{S\setminus\{p\}}(a))\ne\nu_p(aD_{S\setminus\{p\}}(a_2)),
\end{equation}
then $D_S$ is $p$-discontinuous at~$a$.
\end{theorem}
\begin{proof}
If (\ref{DSeq0}) holds, then
$$
D_S(a)=D_{S\setminus\{p\}}(a)+D_p(a)=D_p(a),
$$
and the claim follows from Theorem~\ref{Dpthm}.

Now assume that (\ref{DSane0}) holds.
Let $S$, $S_0$, and $x_i$  be as in (\ref{ss0}) and~(\ref{Dqjxi}), respectively.
Since $\nu_p(x_i)\stackrel{{\rm(\scriptsize{\ref{yidef}}),(\scriptsize{\ref{Dqjxi}})}}{=}\nu_p(a)$, we have
$$
D_p(x_i)=\rho x_i,\quad\rho=\frac{\nu_p(a)}{p}.
$$
Now
\begin{equation}
\label{DSdiff}
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),
\end{equation}
where
$$
u(x_i)=D_{S_0}(x_i)-D_{S_0}(a),\quad v(x_i)=\rho(x_i-a).
$$

Because
$$
\nu_p(v(x_i))\stackrel{{\rm Prop.~\scriptsize{\ref{nu}}(b)}}{=}\nu_p(\rho)+\nu_p(x_i-a)
\stackrel{{\rm(\scriptsize{\ref{Dqjxi}})}}{\to}\infty
$$
and $\nu_p(u(x_i))$ is bounded by~(\ref{tt3}), there is $i_0\in\Z_+$ such that
\begin{equation}
\label{vu}
\nu_p(v(x_i))>\nu_p(u(x_i))\quad{\rm for\,all}\quad i\ge i_0.
\end{equation}
Thus, for $i\ge i_0$,
$$
\nu_p(D_S(x_i)-D_S(a))
\stackrel{{\rm(\scriptsize{\ref{DSdiff}}),(\ref{vu}),Prop.~\scriptsize{\ref{nu}}(d)}}{=}
\nu_p(u(x_i))\stackrel{{\rm(\scriptsize{\ref{tt3}})}}{\not\to}\infty,
$$
verifying $D_S(x_i)\not\to_pD_S(a)$.
\end{proof}

\begin{corollary}
If $S\ne\{p\}$ is finite and $0\ne a\in\Z$, then $D_S$ is $p$-discontinuous at~$a$.
\end{corollary}
\begin{proof}
Apply Theorem~\ref{DQthm1} if $p\notin S$ and $D_S(a)=0$, Theorem~\ref{DSthm1} if
$p\in S$ and $D_S(a)=0$, Theorem~\ref{DQthm2} if $p\notin S$ and $D_S(a)\ne 0$, and
Theorem~\ref{DSthm2} if $p\in S$ and $D_S(a)\ne 0$.
Note that (\ref{nucnd}) and (\ref{nucond}) are satisfied (the right-hand side is infinite
but the left-hand side is finite).
\end{proof}

We conjecture that Theorems~\ref{DQthm2} and~\ref{DSthm2} remain true without~(\ref{nucnd})
and~(\ref{nucond}), respectively.

\begin{conjecture}
If $S\ne\{p\}$ is finite, $0\ne a\in\Q$, and $D_{S\setminus\{p\}}(a)\ne 0$,  then $D_S$ is
$p$-discontinuous at~$a$.
\end{conjecture}

\section{Conclusion}
\label{Conclusion}

We summarize our results. C denotes $p$-continuity, D $p$-discontinuity, and O denotes that
the question is open.
\begin{enumerate}
\item[1] (Theorem~\ref{origin}). $S$ is finite or $p\notin S$, $a=0$. C.
\item[2] (the end of Section~\ref{Dq}). $S$ infinite, $p\in S$, $a=0$. C or O.
\item[3] (Theorem~\ref{Dpthm}). $S=\{p\}$, $a$ arbitrary. C.
\item[4] (Theorem~\ref{Dqthm1}, a special case of Theorems \ref{DQthm1} and~\ref{DSthm1}).
$S=\{q\}$, $a\ne 0$, $D_q(a)=0$. D.
\item[5] (Theorem~\ref{Dqthm2}, a special case of Theorems \ref{DQthm2} and~\ref{DSthm2}).
$S=\{q\}$, $a\ne 0$, $D_q(a)\ne 0$. D.
\item[6] (Theorem~\ref{DQthm1}). $S$ finite, $p\notin S$, $a\ne 0$, $D_S(a)=0$. D.
\item[7] (Theorem~\ref{DSthm1}). $S(\ne\{p\})$ finite, $p\in S$, $a\ne 0$, $D_S(a)=0$. D.
\item[8] (Theorem~\ref{DQthm2}). $S$ finite, $p\notin S$, $a\ne 0$, $D_S(a)\ne 0$. D
under~(\ref{nucnd}), otherwise O.
\item[9] (Theorem~\ref{DSthm2}). $S(\ne\{p\})$ finite, $p\in S$, $a\ne 0$, $D_S(a)\ne 0$. C
under~(\ref{DSeq0}), D under (\ref{DSane0}) and~(\ref{nucond}), otherwise O.
\item[10.] $S$ infinite, $a\ne 0$. O.

\end{enumerate}

\section{Acknowledgment}

We thank the referee, whose suggestions led to significant improvements in Section~\ref{DSorig}.

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\end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11A25; Secondary 26A15.

\noindent \emph{Keywords: }
arithmetic subderivative, arithmetic partial derivative, arithmetic
derivative, continuity, $p$-adic absolute value.

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\noindent (Concerned with sequences
\seqnum{A000040}, 
\seqnum{A003415}.
)

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\vspace*{+.1in}
\noindent
Received April 3 2020;
revised version received June 24 2020.
Published in {\it Journal of Integer Sequences}, June 26 2020.

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