# Boundary Approximation for Sticky Jump-Reflected Processes on the Half-Line

Andrey Pilipenko  Institute of Mathematics, National Academy of Sciences of
Ukraine, Kyiv; Igor Sikorsky Kyiv Polytechnic Institute
<EMAIL_ADDRESS>and  Andrey Sarantsev  Department of Mathematics and
Statistics, University of Nevada, Reno<EMAIL_ADDRESS>
###### Abstract.

The Skorokhod reflection was used in 1961 to create a reflected diffusion on
the half-line. Later, it was used for processes with jumps such as reflected
Lévy processes. Like a Brownian motion, which is a weak limit of random walks,
reflected processes on the half-line serve as weak limits of random walks with
switching regimes at zero: one regime away from zero, the other around zero.
In this article, we develop a general theory of this regime change and prove
convergence to a function with generalized reflection. Our results are
deterministic and can be applied to a wide class of stochastic processes.
Applications include storage processes, heavy traffic limits, diffusion on a
half-line with a combination of continuous reflection, jump exit, and a delay
at 0.

###### Key words and phrases:

Reflecting map; Skorokhod map; Sticky processes; Reflecting barriers;
Invariance principle; Feller-Wentzell boundary conditions

###### 2020 Mathematics Subject Classification:

Primary: 60F17, 60J50; Secondary: 60J55; 60K25

##  1\. Introduction

In a classic M/M/1 queue, customers arrive at rate $\lambda$, are served at
rate $\mu$, with inter-arrival and service independent exponential times. The
number of customers in the queue is a birth-and-death discrete-valued Markov
process on $\\{0,1,2,\ldots\\}$ with birth rate $\lambda$ and death rate
$\mu.$ Now, consider a sequence of queues as $\lambda\uparrow\mu$ (heavy
traffic limit). With correct normalization, we can prove that a scaling limit
of these queues is a reflected Brownian motion, possibly with drift. This is
an $[0,\infty)$-valued process which behaves as a classic Brownian motion away
from zero, and is reflected instantaneously (according to the Skorokhod
reflection) at zero. We refer the reader to the classic book [33]. The method
of proof is continuous mapping: We have a Skorokhod mapping which takes a
function and makes a reflected version out of it. Applied to a random walk,
this mapping gives us an M/M/1 queue. Applied to a Brownian motion, this gives
us reflected Brownian motion. It is well known that a properly re-scaled
random walk weakly converges to a Brownian motion. Since the Skorokhod mapping
is continuous in the corresponding functional space, heavy traffic limit of
M/M/1 queues is indeed a reflected Brownian motion, see also [3, 19, 33].

However, Skorokhod reflection is far from the only reflection model for a
Brownian motion, and, more generally, for diffusion or Lévy processes. In
addition to the Skorokhod reflection the following boundary behavior are
possible:

  * •

Absorption: Hitting zero and stopping there

  * •

Delayed/sticky reflection: Spending more time at zero than with classic
Skorokhod reflection. Time spent at zero has positive Lebesgue measure.

  * •

Jump reflection: Jumping out of zero upward, instead of reflecting
continuously.

Classification of reflection modes were done by Feller and Wentzell in [12,
32]. The construction was done by using resolvents, semigroups, their
generators, and boundary conditions. Probabilistic approach was originated by
Ito and McKean [16], where the excursion theory was used. The work on
continuity principle for these more general reflections is a hard problem.
Delay models with Skorokhod’s reflection are easier to study rather than jump-
exit from the boundary, see for example [13, 34]. This can be explained by
nice properties of the Skorokhod reflecting map and the fact that a delay at
the boundary is regulated by a local time at the boundary, which can be
associated with a regulator boundary term of the Skorokhod problem.
Construction of processes with jump-exit from the boundary is not a trivial
task. For example, the corresponding processes were constructed via an elegant
usage of excursion theory [31], see also an approach based on the mixture of
stochastic differential equations together with martingale problem methods,
and resolvent analysis [2, 20, 21]. Jump-reflected Brownian motion was
constructed in [7, Theorem 3.11] as a function of a Wiener process and a
subordinator, whose Lévy measure coincides with an ‘intensity of exit from 0’
of the jump-reflected Brownian motion.

In this article, we consider a general deterministic setting. Instead of a
queue, we consider a (deterministic) switch problem: It behaves differently at
the boundary than away from zero. For a limit function, we create a reflection
mapping on the space of RCLL functions, which generalizes the classic
Skorokhod mapping mentioned above and includes delays and jumps. Then
convergence results for various stochastic models will be received
automatically as an application of continuous map theorem. As an application
of the general deterministic result on convergence, we will be able to handle
out limit theorems for storage processes, perturbed reflected random walks,
spectrally positive Lévy processes having jump-type reflection at 0.

Let us concisely explain this switch problem. Definition 10 below in Section 3
provides a rigorous statement. We have an input function $x$ driving the
system in the regular regime, and a regulator $F$ for the boundary regime. The
output function $y$ evolves together with $x$ (that is, $x$ and $y$ have the
same increments) as long as $y$ does not get below $-\delta$. After that, we
stop $x$ and restart $y$ as $F$, until $y$ gets above 0. Then we start $y$
again as $x$ from the place where we stopped $x$ previously, until $y$ gets
below $-\delta$. Then restart $y$ again as $F$ from the place where we stopped
$F$ last time, etc.

Informally, this can be illustrated as follows: There is an Internet shop,
which stores goods and gradually sells them to customers. When the inventory
reaches zero (or even some negative level, i.e., there are more orders than
goods), the shop switches to a critical regime. It is still accepting orders,
but starts to order in larger batches. When the shop again has enough on hand,
they stop to order in larger batches and continue the usual functioning. Our
goal in this article is to show that the solution to the switch problem
converges to the solution to the generalized Skorokhod problem as the
threshold $\delta$ between the critical and usual regimes converges to 0. We
stress that we show this in the general case for deterministic functions. This
creates the framework for proving such weak convergence results for regulated
stochastic processes.

Moreover, we will obtain continuous dependence on controls in usual and
critical regimes in the following sense. We take two sequences of functions:
$x_{n}\to x_{0}$ and $F_{n}\to F_{0}$. Next, we take a sequence of non-
negative numbers $\delta_{n}\to 0$ and another sequence
$\rho_{n}\to\rho\in[0,\infty]$ of time normalizing constants. For each $n$, we
solve the switch problem with input function $x_{n}$, threshold $-\delta_{n}$,
and regulator $F_{n}$ that is slowed down by the factor of $\rho_{n}$.

If $F_{0}$ is strictly increasing, and under some additional technical
assumptions, then the solution to this switch problem converges to the
reflected process as in (6) with jump reflection governed by $F_{0}$, driving
function $x_{0}$, and the time delay at 0 described by (4). The critical
values $0$ and $\infty$ of parameter $\rho$ mean zero delay and absorption at
0, respectively.

###  1.1. Notation

Let $\mathbb{R}_{+}:=[0,\infty)$. Let $\mathcal{C}_{T}$ be the space of
continuous functions $[0,T]\to\mathbb{R}$ and $\mathcal{C}$ be the space of
continuous functions $\mathbb{R}_{+}\to\mathbb{R}$ with topology of the
uniform convergence on compact sets. For $T>0$, let $\mathcal{D}_{T}$ be the
Skorokhod space of right-continuous functions with left limits
$[0,T]\to\mathbb{R}$, abbreviated as RCLL or cádlág in French language, and
$\mathcal{D}$ be the Skorokhod space of RCLL functions
$x:\mathbb{R}_{+}\to\mathbb{R}$. We endow the spaces
${\mathcal{D}}_{T},{\mathcal{D}}$ with Skorokhod’s $J_{1}$-topology, see, for
example, [33]. We will denote weak convergence of stochastic processes (in
${\mathcal{C}}$ or ${\mathcal{D}}$) by $X_{n}\Rightarrow X$. We define
$a_{+}:=\max(a,0)$ and $a_{-}:=\max(-a,0)$ for $a\in\mathbb{R}$. For a set
$A$, we let $\overline{A}$ be its closure. Let $\mathrm{mes}(A)$ be the
Lebesgue measure of $A$. For $T>0$, let $\Lambda_{T}$ be a set of continuous
one-to-one strictly increasing functions $\lambda:[0,T]\to[0,T]$.

For an RCLL function $h:\mathbb{R}_{+}\to\mathbb{R}$, we define $\Delta
h(t);=h(t)-h(t-)$. If $h$ has jump at time $t$, this is the size of this jump.
If $h$ is continuous at time $t$, this is zero.

###  1.2. Organization of this article

In Section 2, we complete the discussion of three reflection modes
(Skorokhod’s reflection, jump-type reflection, and delay). We state rigorous
definitions of the switching process in Section 3. Next, we state the main
result: Theorem 1, which includes both cases: sticky and generalized jump-type
reflected processes. In Section 4, we present applications of our results to
various stochastic processes, including the ones in the previous articles.
Section 5 is devoted to the proof of Theorem 1. The Appendix contains proofs
of a few technical lemmas.

###  1.3. Acknowledgment

The first author acknowledges a partial support by the National Research
Foundation of Ukraine, project 2020.02/0014 Asymptotic regimes of perturbed
random walks: on the edge of modern and classical probability. Also, the first
author was supported by the project Mathematical modelling of complex
dynamical systems and processes caused by the state security (Ukraine, Reg.
No. 0123U100853). The second author thanks his Department for welcoming and
positive atmosphere.

##  2\. Background: Skorokhod Reflection and Boundary Controls

In this section we recall basic constructions and properties of reflected
processes in order to get better understanding of a nature of result that we
obtain.

A reflected Brownian motion can be constructed by simply taking the absolute
value $|W|$ of a Brownian motion $W$. However this construction can’t be
called ‘natural’ for other stochastic processes, and in particular, for non-
symmetric Lévy processes. In two 1961 articles [27, 28] Anatoliy Skorokhod
developed a method to reflect any continuous function
$B:[0,\infty)\to\mathbb{R}$, deterministic or stochastic, with $B(0)\geq 0$.
He found a pair of continuous non-negative functions
$X,L:[0,\infty)\to\mathbb{R}$ such that

(1) $X(t)=B(t)+L(t),\,t\geq 0,$

$L$ is non-decreasing and can increase only when $X=0$; finally, $L(0)=0$. In
this problem (1), the function $B$ is called driving or input. See an example
in Figure 1.

$t$$B$$X$$L$ Figure 1. Classic Skorokhod reflection; $B$ is the original
function; $X$ is the Skorokhod reflection; $L$ is the boundary term.

The function $X$ has the same increments as $B$ while $X$ is positive, and the
function $L$ pushes up only at instants when $X=0$ and have no effect
otherwise. If $B$ is a Brownian motion, then $X$ has the same distribution as
$|B|$. Thus, Skorokhod’s definition is consistent with a naive approach to the
notion of a reflected Brownian motion. Moreover, it is well known that $L$ is
the symmetric local time of $X$ at 0.

Later Skorokhod’s problem (1) was generalized for functions
$B\in{\mathcal{D}}$; see, for example, [9, 30, 18]. The solution is given by
the formula

(2) $L(t)=\sup\limits_{0\leq s\leq t}(B(s)_{-})=\Big{(}\inf\limits_{0\leq
s\leq t}B(s)\Big{)}_{-},\ \ X(t)=B(t)+\sup\limits_{0\leq s\leq t}(B(s)_{-}).$

The corresponding Skorokhod mapping $B\mapsto X$ is continuous in the
Skorokhod space of RCLL functions (and is continuous in the subspace of
continuous functions). This allows us to prove functional limit theorems for
heavy traffic limits as an application of continuous map theorem, see for
example, [14, 33]. For instant, the Lindley’s recursion is nothing else but a
solution to Skorokhod’s reflecting problem for random walks.

Assume again that $B$ is a Brownian motion, so that $X$ is a reflected
Brownian motion. It is well known that $X$ spends zero time at 0 almost
surely; that is, the Lebesgue measure of this time is zero:

(3) $\mathrm{mes}(\\{s\geq 0\mid X(s)=0\\})=0\ \ \mbox{a.s.}$

Hence, the Skorokhod reflection is instantaneous. There are other types of
reflection and boundary behavior. A simple rule is alternatively called
absorbing, or stopping: When the Brownian motion or any other input function
(deterministic or stochastic) hits zero, it simply stops and stays constant
after that. Another, more complicated rule, is sticky reflection. This version
of a reflected Brownian motion is delayed when it hits zero. Fix a parameter
$\rho>0$ called delay rate and consider the function $A(t)=t+\rho L(t)$ where
$L$ is from (1). It is a continuous strictly increasing function and therefore
it has an inverse $A^{-1}$. Plugging this inverse into the classic reflected
function $X$, we get:

(4) $\tilde{X}(t)=X(A^{-1}(t)),\quad A(t)=t+\rho L(t),\quad t\geq 0.$

As a result, this process has same excursions as $X$ (but shifted in time) and
spends positive time at zero, so (3) is no longer true. The larger $\rho$ is,
the longer the delay is. As $\rho\to\infty$, we get $A(t)\to\infty$, which
corresponds to the absorbed process. When $\rho=0$, we are back to the classic
instantaneous reflection. We shall call this reflection delayed, as opposed to
instantaneous classic Skorokhod reflection. However, this term should not be
misunderstood. This sticky reflected Brownian motion still leaves zero
instantaneously when it hits zero in the following sense. Notice that the set
of zeros of $\tilde{X}$ is a closed nowhere dense set a.s., so for any $t$
such that $\tilde{X}(t)=0$ and every $\varepsilon>0$ there exists an
$s\in(t,t+\varepsilon)$ with $\tilde{X}(s)>0$.

The process $L$ is a local time at 0 of the reflected Brownian motion $X$.
Hence $A$ is a continuous additive functional of $X,$ and the general theory
of Markov processes implies that the process $\tilde{X}$ is a strong Markov
process. It can be proved that a sticky reflected Brownian motion is a (weak)
solution to the following stochastic differential equation:

(5) $\displaystyle\begin{split}\mathrm{d}X_{\rho}(t)&=1\\!\\!\\!\\!\\!\;{\rm
I}_{\\{X_{\rho}(t)>0\\}}\,\mathrm{d}W(t)+\mathrm{d}L_{\rho}(t),\\\
\mathrm{d}L_{\rho}(t)&=\rho^{-1}1\\!\\!\\!\\!\\!\;{\rm
I}_{\\{X_{\rho}(t)=0\\}}\,\mathrm{d}t,\end{split}$

where $W$ is a standard Brownian motion. Note that increments of $X_{\rho}$
coincide with increments of $W$ when $X$ is positive. However, a reader should
be careful: As shown in [10], there is no strong solution to the system of
stochastic differential equations (5). This is a very subtle and unexpected
observation, because the process $\tilde{X}$ defined by (4) and (2) is a
function of $B.$

Like the Skorokhod reflection map (2), formula (4) can be used for RCLL
functions or stochastic processes, including Lévy processes. Moreover, if $X$
is a reflection of spectrally positive Lévy process or a reflected diffusion,
then the corresponding process $L$ will be a local time of $X$ (up to a
multiplicative constant). So the process $\tilde{X}$ defined via time change
(4) is naturally be called the delay at 0 of a Markov process $X$. See [1, 4]
for further studies of a sticky reflected Brownian motion, and also also [25]
on construction of sticky Lévy processes (without reflection).

The Skorokhod map $B\to X$ and the delayed Skorokhod reflection
$B\mapsto\tilde{X}$ describe continuous exits from 0: Discontinuity at the
instant of exit from 0 can arise only from jumps in the driving process $B$,
not the reflection itself.

There is a discontinuous jump-type reflection that corresponds to non-local
Feller-Wentzell boundary conditions in the semigroup theory of the diffusion
processes or to a jump entrance law in Ito’s excursion theory. The first
author in [23] proposed to consider (a deterministic) jump reflection problem,
when we replace $L$ in (1) with $F(L)$. Here $F:[0,\infty)\to[0,\infty)$ is a
given strictly increasing RCLL function with $F(0)=0$ and $F(\infty)=\infty$:

(6) $\bar{X}(t)=B(t)+F(L(t)),$

where $L$ is again a non-decreasing function that may increase only when
$\bar{X}$ equals 0. More generally, in a later article [23] it was shown that
for every continuous $B$ there is a unique solution to this modified equation
(6), see formula (9) in Lemma 1 below for an explicit formula of a solution.
This explicit formula (without a formulation of a reflected problem) was used
in [7] for a construction of Feller diffusions on a half-line, where $B$ was a
Brownian motion and $F$ was a subordinator.

Finally, note that this jump reflection can be combined with delay. Then we
get sticky jump reflection. We do this in two steps, much like for sticky
reflection $\tilde{X}$ above. First, we solve the equation (6) and get the
jump reflection $\bar{X}$ without delay. Then we fix $\rho>0$ and use $L$ and
$\bar{X}$ to construct $\tilde{X}$ as in (4). The same trichotomy as above is
present here: For $\rho=0$, we are back in the case of jump reflection without
delay. For $\rho\in(0,\infty)$, this is jump reflection with delay. Finally,
as $\rho\to\infty$, we get absorbed process. When it hits zero, it remains
there forever.

##  3\. Definitions and the Main Result

###  3.1. Instantaneous jump reflection and delays

In this subsection, we provide explicit formulas for solutions on reflected
problems, both classic (1) and with jumps (6). We state the definition again
for completeness, although we discussed these reflection modes in the
Introduction.

######  Definition 1.

Consider functions $x,F\in\mathcal{D}$ such that $x(0)\geq 0$, $F$ is strictly
increasing, $F(0)=0$, $F(\infty)=\infty$. A solution to the generalized
Skorokhod problem is a pair of functions $y,l\in\mathcal{D}$, with the
following properties: $y(t)\geq 0$ for all $t\geq 0$; $l$ is non-decreasing
function, and the following equality is true:

(7) $y(t)=x(t)+F(l(t)),\quad t\geq 0.$

Moreover, the function $l$ can increase only when $y=0$, i.e.

(8) $\int_{0}^{\infty}1\\!\\!\\!\\!\\!\;{\rm
I}_{\\{y(s)>0\\}}\,\mathrm{d}l(s)=0.$

We will call the function $x$ the driving noise, $y$ the reflected process,
$F$ the regulator, $l$ the boundary term, and denoted $y=\mathcal{S}(x,F)$.

######  Lemma 1.

If $F$ is continuous or $x$ has no negative jumps, then there is a unique
solution to (7) and this solution is given by the formula

(9) $\displaystyle\begin{split}y(t)&=x(t)+F(l(t))=x(t)+F(F^{-1}(m(t)));\\\
m(t)&=\sup\limits_{0\leq s\leq t}(x(s)_{-})=\inf\limits_{0\leq s\leq
t}(x(s))_{-};\quad l(t)=F^{-1}(m(t)),\end{split}$

where $F^{-1}$ is the generalized inverse $F^{-1}(y):=\inf\\{x\mid F(x)>y\\}$.

For $F(t)\equiv t$, we get the classic Skorokhod reflection problem with the
unique solution $y(t)=x(t)+m(t),$ see [30, 18]. If $F$ is continuous, we are
back to classic Skorokhod reflection; in a way, the case of a continuous $F$
is not different from the case $F(t)=t$ for all $t\geq 0$. Indeed, $F\circ
F^{-1}(t)=t$ for strictly increasing and continuous $F$. Hence
$y(t)=x(t)+m(t)$ again.

######  Remark 1.

The function $m$ is continuous, since $x$ has no negative jumps. The function
$F^{-1}$ is continuous and non-decreasing, since $F$ is strictly increasing,
see [33, Lemma 13.6.5.]. Of course, $F^{-1}$ is the standard inverse function
if $F$ is strictly increasing and continuous.

######  Remark 2.

The condition that $x$ has no negative jumps or $F$ is continuous is
important. Assume $x$ has, in fact, a negative jump at the point $t$. Then we
can construct a setting when there is no solution. Assume $y(t-)>0$ for some
$t$ but $\triangle x(t)<0$ and $y(t-)+\triangle x(t)<0$. At time $t$, the jump
of $x$ needs to be compensated by the function $F(l)$. If $F$ has jumps, then
it may be impossible to find the compensation such that $y(t)=0$.

For continuous $x$, Lemma 1 was shown in [23]. The proof of the general case
is postponed until the Appendix. For convenience of readers we quote a result
about the composition $F\circ F^{-1}$ used in (9), taken also from [23], and
illustrated in Figure 2.

######  Lemma 2.

Take a function $F\in{\mathcal{D}}$ that is an increasing function with
$F(0)=0$ and $F(\infty)=\infty$. Consider the set
$A_{F}:=\mathbb{R}_{+}\setminus\overline{F(\mathbb{R}_{+})}$. It is an open
set, and therefore a countable union of intervals
$A_{F}:=\cup_{i}(\alpha_{i},\beta_{i})$. Then the function $G:=F\circ F^{-1}$
satisfies:

$G(t)=\begin{cases}t,\,t\in\mathbb{R}_{+}\setminus A_{F};\\\
\beta_{i},\,t\in[\alpha_{i},\beta_{i}).\end{cases}$

######  Definition 2.

Fix a $\rho\in(0,\infty)$. Take the boundary term $l$ from Definition 8 and
define the time change $A(t)=t+\rho l(t)$ for $t\geq 0$. Plug $A^{-1}$ into
the reflected function $y$ from Definition 8. Then $z(t)=y(A^{-1}(t))$ for
$t\geq 0$ is called the delayed jump-reflection or sticky jump-reflection. The
functions $x,F$ are called the input or driving function and the regulator,
respectively, similarly to Definition 8. The function $z$ called the sticky
jump-reflected function, and the function $L$ defined as $L(t)=l(A^{-1}(t))$
for $t\geq 0$ is called the sticky boundary term.

We stress that a delayed Skorokhod reflection is not an alternative to jump
reflection. Rather, these are two characteristics of a reflection: It can be
with or without delay, and with or without jumps (that is, with $F$ continuous
or discontinuous). All four options are possible. In addition, of course, we
have absorbed processes, but there $F$ does not matter anymore: It regulates
only behavior at zero, and the resulting process stays at zero forever after
it hits zero.

$3$$3$$4$$4$$6$$6$$8$$8$$8.5$$8.5$

Figure 2. Lemma 2. The function $G$ with $A_{F}=(3,4)\cup(6,8)\cup(8,8.5)$.

###  3.2. The switch problem

Here we define the regime switching problem in the general case, for arbitrary
deterministic RCLL functions. This can be applied to Brownian motion, Lévy
processes, or any other stochastic processes.

######  Definition 3.

Fix a gap $\delta>0$, the driving function $x\in\mathcal{D}$, $x(0)\geq 0$,
and the regulating function $F\in\mathcal{D}$ with $F(0)=0$. A solution to the
switch problem is a function $y\in\mathcal{D}$ with:

  * •

times $0=\tau_{0}\leq\rho_{1}\leq\tau_{1}\leq\rho_{2}\leq\ldots\leq\infty$
defined as $\rho_{k}:=\inf\\{t\geq\tau_{k-1}\mid y(t)\leq-\delta\\}$ and
$\tau_{k}:=\inf\\{t\geq\rho_{k}\mid y(t)\geq 0\\}$ for $k=1,2,\ldots$;

  * •

two disjoint subsets $B=\cup_{k=1}^{\infty}[\rho_{k},\tau_{k})$,
$A=\cup_{k=1}^{\infty}[\tau_{k-1},\rho_{k})$ of $\mathbb{R}_{+}$ such that
$A\cup B=\mathbb{R}_{+}$ and the corresponding occupation times

$T_{A}(t)=\mathrm{mes}(A\cap[0,t]),\quad T_{B}(t)=\mathrm{mes}(B\cap[0,t])$

(and therefore $T_{A}(t)+T_{B}(t)\equiv t$),

which satisfy

(10) $y(t)=x(T_{A}(t))+F(T_{B}(t)),\ t\geq 0,$

######  Remark 3.

We set $\inf\varnothing=+\infty$.

######  Remark 4.

From (10) it follows that: $y(t)-F(T_{B}(t))$ is constant on each interval in
$B$; and $y(t)-x(T_{A}(t))$ is constant on each interval in $A$.

Let us explain this switch problem in plain English.

  1. (1)

We start with the input function $x$; the output function $y$ is equal to $x$
until it reaches below $-\delta$. This is the normal regime $A$. Note that the
function $x$ can have jumps, so the input function $x$ (and together with it
the output function $y$) might reach $(-\infty,-\delta]$ via a negative jump
rather than a continuous path. Assume this happens at time $\rho_{1}$. This is
the first piece of the input function.

  2. (2)

Then we switch to the boundary regime $B$. We use the regulating function $F$,
or simply regulator, starting from zero argument, to increase the output
function $y$ until it reaches above $0$. The increments of the output function
$y$ coincide with the increments of the regulator $F$. Again, note that the
output function $y$ can reach $[0,\infty)$ by a jump rather than continuous
movement. We stop at time $\tau_{1}$. So the regulator $F$ stops at
$\tau_{1}-\rho_{1}$. This is the first piece of the regulator.

  3. (3)

Next, we switch to the normal regime $A$ again. We govern the output function
$y$ by the input function $x$: That is, the increments of the input and output
functions coincide. We use the second piece of the input function $x$,
starting from the point where we finished the first piece, in part 1. This
happens until, as in part 1, the output function hits $(-\infty,-\delta]$.
Assume this happens at the moment $\rho_{2}$.

  4. (4)

Then we switch again to the boundary regime $B$ again, as in part 2. We govern
the output function $y$ by the regulator $F$: The increments of the regulator
and output functions coincide. We use the second piece of the regulator $F$,
which starts at the time when the first piece ended. We continue until the
output function hits $[0,\infty)$, and then switch to the normal regime again.

To summarize, we cut the graphs of input function $x$ and the regulator $F$ in
pieces: We attach the first piece of $x$, then the first piece of $F$, then
the second piece of $x$, then the second piece of $F$, and so on. By
construction, for any pair $(x,F)\in\mathcal{D}^{2}$, and any $\delta>0$,
there exists a unique solution $y=\mathcal{G}_{\delta}(x,F)$.

See an example with a piecewise linear input function $x$ and another
piecewise linear function as regulator $F$, and the output in Figure 3, where
$\delta=0.5$.

$X_{1}$$X_{2}$$X_{3}$$-\delta$$F_{1}$$F_{2}$$F_{3}$$X_{1}$$F_{1}$$X_{2}$$X_{3}$$F_{2}$$X_{3}$$F_{3}$$X_{3}$$F_{3}$$X_{3}$$F_{3}$$-\delta$
Figure 3. The switch problem. Top left: Input function $x$. Top right:
regulator $F$. Bottom: Output $y=\mathcal{S}_{\delta}(x,F)$. All functions are
right-continuous.

Generally, we cannot consider the switch problem for $\delta=0$ and switch the
regime when the process enters and exits $(0,\infty)$ (or enters and exits
$[0,\infty)$). The only difficulty is treating the switch problem at 0. For
example, it is unclear how to treat the definition if regime $A$ pushes down
and regime $B$ pushes up, or if functions $x$ and $F$ behaves like excursions
of a Brownian motion having no intervals of monotonicity. We will define the
solution to the switching problem $\mathcal{G}_{\delta}(x,F)$ with $\delta=0$
if contradictions in the definition do not appear. The following cases are
examples:

  1. (a)

$x$ and $F$ are step-functions with finitely many jumps in any $[0,t]$; regime
$A$ is selected if $y(t)>0$; regime $B$ is selected if $y(t)\leq 0$.

  2. (b)

$F$ is a non-decreasing step function with finitely many jumps in any $[0,t]$;
regime $A$ is selected if $y(t)>0$; regime $B$ is selected if $y(t)\leq 0$.

  3. (c)

Same as above, where regimes $A$ or $B$ are selected if $y(t)\geq 0$ or
$y(t)<0,$ respectively.

  4. (d)

Any function $y$ constructed from pieces of $x$ and $F$ under condition that
$y$ never hits 0 and has finitely many crossings of 0 during any $[0,t]$.

We will apply the switch problem $\mathcal{G}_{0}(x,F)$ to the cases when:

  1. (a)

$x$ and $F$ are independent compound Poisson processes;

  2. (b)

$x(t)=x_{[t]},F(t)=F_{[t]},$ where $(x_{n})$ and $(F_{n})$ are independent
random walks;

  3. (c)

$x$ is a Lévy process, and $F$ is a jump-type subordinator with finite Lévy
measure.

######  Remark 5.

If we consider the problem $y=\mathcal{G}_{0}(x,F)$, then we will always
assume that $x$ and $F$ are such that the switch problem
$\mathcal{G}_{0}(x,F)$ is well defined.

###  3.3. Main results

For a non-decreasing function $f:\mathbb{R}_{+}\to\mathbb{R}$ we say that
$s\in\mathbb{R}_{+}$ is a growth point if $f(t)>f(s)$ for $t>s$ and
$f(t)<f(s)$ for $t<s$.

######  Theorem 1.

Let $(x_{n}),(F_{n})$ be two sequences in $\mathcal{D}$. Define

(11) $m_{0}(t):=\sup\limits_{s\in[0,t]}(x_{0}(s)_{-}).$

Take two sequences $(\delta_{n})$ or $(\varrho_{n})$ of real numbers such that
for every $n$ we have $\delta_{n}\geq 0$ and $\varrho_{n}>0$, $\delta_{n}\to
0$ and $\varrho_{n}\to\varrho\in[0,\infty]$. Consider a sequence
$(y_{n})\subseteq\mathcal{D}$ of functions defined as

$y_{n}:=\mathcal{G}_{\delta_{n}}(x_{n},F_{n}(\cdot/\varrho_{n})),\,t\geq 0.$

Assume that

  1. (a)

$x_{n}(0)\geq 0,F_{n}(0)=0,$ $n=0,1,2,\ldots$;

  2. (b)

$x_{n}\to x_{0}$ in $\mathcal{D}$ and $F_{n}\to F_{0}$ in $\mathcal{D}$;

  3. (c)

$F_{0}$ is strictly increasing, $F_{0}(\infty)=\infty$;

  4. (d)

$x_{0}$ does not have negative jumps;

  5. (e)

if $\alpha\geq 0$ and $t\geq 0$ are such that $F_{0}(\alpha)\neq
F_{0}(\alpha-)=m_{0}(t)$, where $m_{0}$ is defined in (11), then $t$ is a
growth point of $m_{0}$.

Then $y_{n}\to y_{0}$ in $\mathcal{D}$, where $y_{0}$ depends on $\varrho$:

  1. (1)

Classic reflection: $\varrho=0$. Then $y_{0}=\mathcal{S}(x_{0},F_{0})$.

  2. (2)

Delayed reflection: $\varrho\in(0,\infty)$. Then

$y_{0}(t)=\mathcal{S}(x_{0},F_{0})(A_{\varrho}^{-1}(t)),\quad
A_{\varrho}(t):=t+\varrho F_{0}^{-1}(m_{0}(t)).$

  3. (3)

Absorption: $\varrho=\infty$. Then
$y_{0}(t)=\mathcal{S}(x_{0},F_{0})(t\wedge\sigma)$, assuming

$\sigma:=\inf\\{s\geq 0\ |\ x_{0}(s)=0\\}=\inf\\{s\geq 0\ |\ x_{0}(s)<0\\}.$

######  Remark 6.

For the case $\varrho=\infty$, only conditions (a), (b), and (d) in this
theorem suffice.

######  Remark 7.

Notice that $\lim_{\varrho\to 0+}A^{-1}_{\varrho}(t)=t$ and
$\lim_{\varrho\to+\infty}A^{-1}_{\varrho}(t)=t\wedge\sigma$, so the limit
function $y_{0}$ can be formally written as
$\mathcal{S}(x_{0},F_{0})(A_{\varrho}^{-1}(t))$ in cases $\varrho=0$ and
$\varrho=\infty$ too.

######  Remark 8.

The regime switch reminds a penalty method: a reflected Brownian motion on the
half-line can be obtained as a limit of Brownian motions on ${\mathbb{R}}$
with a large positive drift in a neighborhood of 0 or below 0, for example,
$n^{2}1_{[0,1/n)}(x)$ or $n1_{(-\infty,0]}(x)$, which pushes to $[0,\infty)$
as $n$ increases, see, for example, [8, 29].

######  Corollary 1.

Let $(x_{n}),(F_{n})$ be two sequences of RCLL processes that for almost every
$\omega$ satisfy assumptions (a), (c), (d), and (e) of Theorem 1. Assume also
that sequences $(\delta_{n})$, $(\varrho_{n})$, and $(y_{n})$ are defined as
in Theorem 1, and also $(x_{n},F_{n})\Rightarrow(x_{0},F_{0})$ as $n\to\infty$
in $\mathcal{D}$. Then we have convergence in distribution

(12)
$y_{n}:=\mathcal{G}_{\delta_{n}}(x_{n},F_{n}(\cdot/\varrho_{n}))\Rightarrow
S(x_{0},F_{0})(A_{\varrho}^{-1}).$

where $y_{0}$ depends on $\varrho$ similarly to Theorem 1.

###### Proof.

By the Skorokhod representation theorem, there are copies
$(\widetilde{x}_{n},\widetilde{F}_{n})\overset{d}{=}(x_{n},F_{n}),n\geq 0$
such that we have convergence almost surely:

$(\widetilde{x}_{n},\widetilde{F}_{n})\to(\widetilde{x}_{0},\widetilde{F}_{0}),\quad
n\to\infty.$

Hence almost surely we have convergence
$\mathcal{G}_{\delta_{n}}(\widetilde{x}_{n},\widetilde{F}_{n}(\cdot/\varrho_{n}))\rightarrow{\mathcal{S}}(\widetilde{x}_{0},\widetilde{F}_{0})(\widetilde{A}_{\varrho}^{-1}),n\to\infty$.
This latter convergence implies (12). ∎

##  4\. Examples and Applications

###  4.1. Driving Brownian motion

Assume that $x_{n}=w,n\geq 1$, where $w$ is a Brownian motion,
$\varrho_{n}=1,n\geq 1,$ $F_{n}(t)=at,$ where $a>0,$ and
$(\delta_{n})\subset(0,\infty)$ be any sequence of positive numbers that
converges to 0. That is, the process $y_{n}$ is a continuous process that
moves like a Brownian motion before hitting $-\delta_{n}$, then switches a
regime and moves up with constant speed until hitting 0, then switches regime
for a Brownian motion until hitting $-\delta_{n}$, etc. We have
$x_{0}(t)=w(t),\ F_{0}(t)=at,\ F_{0}(F_{0}^{-1}(t))=t$. Then
$\mathcal{S}(x_{0},F_{0})=\mathcal{S}(x_{0})=\mathcal{S}(w)=w+m_{0}$ is the
reflected Brownian motion, and $A_{\varrho}(t):=t+\varrho
F_{0}^{-1}(m_{0}(t))=t+a^{-1}m_{0}(t)$. Hence
$y_{0}=(w+m_{0})(A_{\varrho}^{-1})$ is a sticky reflected Brownian motion. If
we assume $F_{n}(t)=a_{n}t,$ where $\lim_{n\to\infty}a_{n}=+\infty,$ i.e., the
regime below 0 pushes up strongly, then the limit process will be the usual
reflected Brownian motion without any delay.

Note that the process $y_{n}$ has the same distribution as a solution to the
stochastic equation

$\mathrm{d}Y_{n}(t)=1\\!\\!\\!\\!\\!\;{\rm I}_{\\{t\in
A_{n}\\}}\,\mathrm{d}w(t)+a\,1\\!\\!\\!\\!\\!\;{\rm I}_{\\{t\in
B_{n}\\}}\,\mathrm{d}t,\ t\geq 0,$

with $Y_{n}(0)=0.$ Here sets $A_{n}$ and $B_{n}$ are defined as follows:

$\displaystyle A_{n}$
$\displaystyle=\cup_{k=0}^{\infty}[\tau^{(n)}_{k},\rho^{(n)}_{k+1}),\quad
B_{n}=\cup_{k=1}^{\infty}[\rho^{(n)}_{k},\tau^{(n)}_{k}),$
$\displaystyle\tau^{(n)}_{0}$
$\displaystyle:=0,\quad\rho^{(n)}_{k+1}:=\inf\\{t\geq\tau^{(n)}_{k}\ :\
Y_{n}(t)\leq-\delta_{n}\\},\quad\tau^{(n)}_{k}:=\inf\\{t\geq\rho^{(n)}_{k}\ :\
Y_{n}(t)\geq 0\\}.$

###  4.2. Storage processes

Take Poisson processes ${N_{\lambda_{n}}(t)}$ and
$\hat{N}_{\hat{\lambda}_{n}}(t)$ with intensities $\lambda_{n}$ and
$\hat{\lambda}_{n}$, respectively. Take sequences of non-negative i.i.d.
random variables $(\xi_{k}^{(n)})_{k\geq 1}$ and $(\zeta_{k}^{(n)})_{k\geq
1}$. We will assume that all processes and sequences are jointly independent.
Consider two compound Poisson processes $x_{n}$ and $F_{n}$ with positive
jumps, where $x_{n}$ has an additional negative drift term:

$x_{n}(t)=\sum\limits_{k=1}^{N_{\lambda_{n}}(t)}\xi_{k}^{(n)}-r_{n}t,\quad
F_{n}(t)=\sum_{k=1}^{\hat{N}_{\hat{\lambda}_{n}}(t)}\zeta_{k}^{(n)}$

Note that $F_{n}$ is a jump-type subordinator. Then the process
$y_{n}(t)=\mathcal{G}_{0}(x_{n},F_{n})$ is a Markov storage process whose
behavior can be described as follows:

  1. (1)

If $y_{n}(t)>0$, then it has a negative drift at rate $r_{n}$ and jumps up in
a Poisson clock with intensity $\lambda_{n}$; the value of the $k$th jump is
$\xi_{k}^{(n)}$.

  2. (2)

If $y_{n}(t)=0$, then $x_{n}$ stays at 0 the exponential time
$\mathrm{Exp}(\hat{\lambda}_{n})$ and then jumps up with the distribution
$\zeta_{k}^{(n)}$.

It is well known that under assumptions

$\displaystyle\lim_{n\to\infty}(\lambda_{n}\mathrm{E}\xi_{k}^{(n)}-r_{n})$
$\displaystyle=\mu,$
$\displaystyle\lim_{n\to\infty}\lambda_{n}\mathrm{E}\left[(\xi_{k}^{(n)})^{2}\right]$
$\displaystyle=\sigma^{2}>0,$
$\displaystyle\lim_{n\to\infty}\lambda_{n}\mathrm{E}\left[(\xi_{k}^{(n)})^{2}1\\!\\!\\!\\!\\!\;{\rm
I}_{\\{\xi_{k}^{(n)}>\varepsilon\\}}\right]$
$\displaystyle=0,\quad\varepsilon>0;$

we have convergence in distribution $x_{n}(t)\Rightarrow\mu t+\sigma w(t)$ as
$n\to\infty$ in ${\mathcal{D}},$ where $w$ is a standard Brownian motion.
Assume that there are normalizing constants $\gamma_{n}$ such that

(13) $\sum_{k=1}^{[\gamma_{n}t]}\zeta_{k}^{(n)}\Rightarrow t,\quad
n\to\infty.$

Suppose that there exists a limit
$\hat{\lambda}_{n}/\gamma_{n}\to\varrho\in[0,\infty]$. Without loss of
generality we may assume that
$\hat{N}_{\hat{\lambda}_{n}}(t)=\hat{N}({\hat{\lambda}_{n}}t),$ where
$\hat{N}$ is a fixed Poisson process with intensity 1. It is well known that
$\hat{N}({\hat{\gamma}_{n}}t)/\hat{\gamma}_{n}\to t$ uniformly on every
$[0,T]$ a.s. Thus

$\sum_{k=1}^{\hat{N}(\hat{\lambda}_{n}t)}\zeta_{k}^{(n)}=\sum_{k=1}^{\hat{N}(\gamma_{n}t/\varrho_{n})}\zeta_{k}^{(n)},$

where $\varrho_{n}=\gamma_{n}/{\hat{\lambda}_{n}}$.

The application of Corollary 1 with $\delta_{n}:=0$ implies the following
result.

  1. (1)

If $\varrho=0$, then the limit of $y_{n}$ is the Skorokhod reflection of $\mu
t+\sigma w(t)$ at 0.

  2. (2)

If $\varrho>0$, then the limit of $y_{n}$ is the sticky reflection of $\mu
t+\sigma w(t)$ at 0.

  3. (3)

If $\varrho=\infty,$ then the limit of $y_{n}$ is $\mu t+\sigma w(t)$ stopped
at 0.

In addition, the case when $\xi_{k}^{(n)}\overset{d}{=}\zeta_{k}^{(n)}$ was
considered in [13]. In this case (13) is satisfied with
$\gamma_{n}=\lambda_{n}/r_{n}$. Another example when (13) holds is the
following one: $\zeta_{k}^{(n)}=\zeta_{k}/\gamma_{n}$ where $(\zeta_{k})$ are
non-negative independent identically distributed random variables,
$\mathrm{E}\zeta_{k}=1$, and $(\gamma_{n})$ is any sequence of positive
numbers such that $\gamma_{n}\to+\infty$.

If the sequence $(\sum_{k=1}^{[\gamma_{n}t]}\zeta_{k}^{(n)})$ converges to an
increasing subordinator $F_{0}(t),t\geq 0$, then the limit process will be a
diffusion with jump-type exit from 0. The only non-triviality is to verify
condition (e) of Theorem 1. This will be done in more general case in next
example.

###  4.3. Convergence to a reflected Lévy process with a delay at 0

Let a sequence of positive numbers $(\rho_{n})$ and sequences of stochastic
processes $(x_{n}),(F_{n})$ be such that

  1. (1)

$(x_{n},F_{n})\Rightarrow(x_{0},F_{0})$ as $n\to\infty;$

  2. (2)

$x_{0}$ is a Lévy process without negative jumps;

  3. (3)

$F_{0}$ is an increasing subordinator;

  4. (4)

$x_{0}$ and $F_{0}$ are independent;

  5. (5)

$\varrho_{n}\to\varrho\in[0,\infty].$

The process $T_{0}:=m_{0}^{-1}$ is a subordinator (may be killed at an
exponential time), see [5, Theorem 1, p.189]. A point $t$ is not a point of
growth of $m_{0}$ if and only if $m_{0}(t)$ is a point of jump of $T_{0}$. At
any fixed (non-random) $\alpha\geq 0$, the function $T_{0}$ is almost surely
continuous. Since the processes $F_{0}$ and $T_{0}$ are independent and the
set of jumps of $F_{0}$ is at most countable, condition (e) of Theorem 1 is
true almost surely. The application of Corollary 1 implies convergence in
distribution (12). It is natural to say that the process
$S(x_{0},F_{0})(A_{\varrho}^{-1})$ is the process $x_{0}$ that having jump-
type reflection at 0 with a delay. It may be seen similarly to [7, Chapter
II.3 (c)] that the process ${\mathcal{S}}(x_{0},F_{0})$ is a Markov process
and $L(t)=F_{0}^{-1}(m_{0}(t)),t\geq 0$ is its local time at 0 if $x_{0}$ and
$F_{0}$ are independent.

###  4.4. Perturbed random walks

Let $(\xi_{k})$ be a sequence of independent identically distributed mean-zero
random variables with finite variance $\sigma^{2}>0.$ Consider the random walk
$S_{\xi}(n):=\sum_{k=1}^{n}\xi_{k}$, where $S_{\xi}(0):=0$. Let us extend
$S_{\xi}$ to non-negative half-line as follows $S_{\xi}(t):=S_{\xi}([t])$ for
$t\geq 0.$ It is well known that

$\frac{S_{\xi}(nt)}{\sqrt{n}}\Rightarrow\sigma w(t),\quad n\to\infty$

in ${\mathcal{D}}$ by the Donsker theorem, where $w$ is a standard Wiener
process. Assume that a non-negative random variable $\eta$ belongs to the
domain of attraction of the $\beta$-stable law with $\beta\in(0,1)$. Consider
a Markov chain $(X(n))$ with transition probabilities

$\mathrm{P}(X(n+1)=y|X(n)=x)=\begin{cases}\mathrm{P}(\xi=y-x),\ x\geq 0;\\\
\mathrm{P}(\eta=y-x),\ x<0.\end{cases}$

We will interpret $X$ as a perturbation of $S_{\xi}$ below 0. Note that $X$
has the same distribution as a solution of two-phase system, where $\delta=0$,
$x=S_{\xi}$, and $F=S_{\eta}$. Here $S_{\eta}(n)=\sum_{k=1}^{n}\eta_{k},$
$S_{\eta}(t):=S_{\eta}([t])$ for $t\geq 0,$ and the sequences $(\eta_{k})$ and
$(\xi_{k})$ are independent. That is,

$X\overset{d}{=}{\mathcal{G}}_{0}(S_{\xi},S_{\eta}).$

Note that natural scaling for $S_{\eta}$ is not $\sqrt{n}$ as in the Donsker
theorem. There is a sequence $(a(n))$ that is slowly varying at infinity with
index $\frac{1}{\beta}$ such that

$\frac{S_{\eta}(n\cdot)}{a(n)}\Rightarrow U_{\beta}(\cdot),\quad n\to\infty,$

in ${\mathcal{D}}$, where $U_{\beta}$ is a $\beta$-stable subordinator: a non-
decreasing Lévy process with Laplace transform $\mathrm{E}[\exp(-\lambda
U_{\beta}(t))]=\exp(-t\lambda^{\beta})$ for $t,\lambda\geq 0$. It can be seen
that there is a sequence $(b(n))$ that is slowly varying at infinity with
index $\beta/2$ such that

$\frac{S_{\eta}(b(n)\cdot)}{\sqrt{n}}\Rightarrow U_{\beta}(\cdot),\quad
n\to\infty.$

Define $\varrho_{n}:=b(n)/n$. Notice that $\lim_{n\to\infty}\varrho_{n}=0$
because $\beta/2<1$. Thus

$\frac{S_{\eta}(\varrho_{n}n\cdot)}{\sqrt{n}}=\frac{S_{\eta}(b(n)\cdot)}{\sqrt{n}}\Rightarrow
U_{\beta}(\cdot),\quad n\to\infty.$

Define processes $x_{n}(t):=S_{\xi}(nt)/\sqrt{n}$ and
$F_{n}(t):=S_{\eta}(\varrho_{n}nt)/\sqrt{n}$. Hence

$\Big{(}\frac{X(nt)}{\sqrt{n}}\Big{)}_{t\geq
0}\overset{d}{=}\Big{(}{\mathcal{G}}_{0}\left(x_{n},F_{n}(\cdot/\varrho_{n})\right)(t)\Big{)}_{t\geq
0}.$

Corollary 1 implies the weak convergence

(14)
$\frac{X([n\cdot])}{\sqrt{n}}\overset{d}{=}{\mathcal{G}}_{0}\left(x_{n},F_{n}(\cdot/\varrho_{n})\right)\Rightarrow{\mathcal{S}}\left(w,U_{\beta}\right)=w+U_{\beta}\circ
U_{\beta}^{-1}\circ m,$

where $w$ and $U_{\beta}$ are independent, $m(t)=\max_{s\in[0,t]}(w(t)_{-})$.
Feasibility of condition (e) of Theorem 1 follows from the reasoning of the
previous example.

######  Remark 9.

If we multiply each $b(n)$ by a constant $C>0$, then

$S_{\eta}\left(\frac{Cb(n)t}{\sqrt{n}}\right)\Rightarrow\hat{U}_{\beta}(t):=U_{\beta}(Ct).$

However, the result will be unchanged because
$\hat{U}_{\beta}\circ\hat{U}_{\beta}^{-1}=U_{\beta}\circ U_{\beta}^{-1}$.

Note that the same result is also true if transition probabilities for $X$ are

$\mathrm{P}(X(n+1)=y|X(n)=x)=\begin{cases}\mathrm{P}(\xi=y-x),\ x>0;\\\
\mathrm{P}(\eta=y-x),\ x\leq 0.\end{cases}$

Convergence (14) was proved in [15] and for a particular case in [24]. The
same convergence was obtained in [15] for the Donsker scaling limits of
sequences

$X^{\prime}(n+1)=\begin{cases}X^{\prime}(n)+\xi_{n+1},&X^{\prime}(n)>0,\\\
\eta_{n+1},&\hat{X}^{\prime}(n)\leq 0;\end{cases}$
$\grave{X}(n+1)=\begin{cases}\grave{X}(n)+\xi_{n+1},&\grave{X}(n)>0\mbox{ and
}\grave{X}(n)+\xi_{n+1}>0,\\\ 0,&\grave{X}(n)>0\mbox{ and
}\grave{X}(n)+\xi_{n+1}\leq 0,\\\ \eta_{n+1},&\grave{S}_{v}(n)=0,\end{cases}$

that also satisfy assumptions of Corollary 1 (the corresponding additional
reasoning can be found in [15]). Some ideas used in this paper are taken from
[15, 24], but only now it becomes clear how different scaling for $(\xi_{k})$
and $(\eta_{k})$ interplay and give a limit for the Donsker scaling of
perturbed random walk $(X(n))$.

######  Remark 10.

Consider sequences $(X_{l}(k))_{k\geq 0}$ that have the same transition
probabilities as $(X(k))_{k\geq 0}$ but different initial values such that
$X_{l}(0)/\sqrt{l}\Rightarrow\zeta,n\to\infty$. It is easy to see that the
following convergence can be proved:

$X_{l}(t)/\sqrt{l}\Rightarrow\zeta+w(t)+U_{\beta}\circ
U_{\beta}^{-1}\circ((-\zeta+m(t))_{+}),\quad t\geq 0,$

where processes $w$, $U_{\beta}$, and the random variable $\zeta$ are
independent.

##  5\. Proof of the Main Result

This section is organized as follows. In Subsection 5.1, we state an upper and
a lower bound for the solution of the switch problem. These two lemmas have
proofs postponed until the Appendix. In Subsection 5.2, we write four
technical convergence lemmas used for the proof of Theorem 1. The first two of
these four lemmas, similarly, have proofs in the Appendix. The other two
lemmas are quoted from other sources so we do not give their proofs. In the
next three subsections, we prove Theorem 1 for three cases:
$0<\varrho<\infty$, $\varrho=0$, and $\varrho=\infty$.

###  5.1. Estimates for the switch problem

Take a constant $\delta>0$ and functions $x,F$, as in the definition of the
switch problem. Let $y=\mathcal{G}_{\delta}(x,F)$ be the solution, and $A,B$
the corresponding sets. Define the running maximum

$m(t)=\sup\limits_{s\in[0,t]}(x(s)_{-}),\,t\geq 0.$

We state the two lemmas which together form the basis for the proof.

######  Lemma 3.

Let $y={\mathcal{G}}_{\delta}(x,F)$. Then for every $t\geq 0$, we have:
$F(T_{B}(t)-)\leq m(T_{A}(t))$.

######  Lemma 4.

Let $y={\mathcal{G}}_{\delta}(x,F)$. Then for every $t\geq 0$, we have:

$\displaystyle F(T_{B}(t))\geq-\delta+m(T_{A}(t))$
$\displaystyle-\sup_{s\in[0,T_{A}(t)]}(x(s-)-x(s))$ $\displaystyle-\sup_{0\leq
s_{1}\leq s_{2}\leq T_{B}(t)}(F(s_{1})-F(s_{2})).$

###  5.2. Preliminary results

The next two lemmas are technical convergence results, with (simple) proofs
postponed until the Appendix.

######  Lemma 5.

If $x_{n}\to x_{0}$ in $\mathcal{D}$, where $x_{0}$ has only positive jumps,
then for any $T>0$, $\sup_{s\in[0,T]}(x_{n}(s-)-x_{n}(s))\to 0$.

######  Lemma 6.

If $F_{n}\to F_{0}$ in $\mathcal{D}$, where $F_{0}$ is a non-decreasing
function, then for any $T>0$ $\sup_{0\leq s_{1}\leq s_{2}\leq
T}(F_{n}(s_{1})-F_{n}(s_{2}))\to 0$.

Recall a classical characterization of convergence in a Skorokhod space from
[11, Chapter 3, Proposition 6.5, page 125]:

######  Lemma 7.

We have $x_{n}\to x_{0}$ in $\mathcal{D}$ if and only if for any non-negative
$t_{n}\to t_{0}$, we have:

  1. (1)

all limit points of $(x_{n}(t_{n}))_{n\geq 1}$ are either $x_{0}(t_{0})$ or
$x_{0}(t_{0}-)$;

  2. (2)

if $x_{n}(t_{n})\to x_{0}(t_{0})$, $s_{n}\geq t_{n}$, $s_{n}\to t_{0}$, then
$x_{n}(s_{n})\to x_{0}(t_{0})$;

  3. (3)

if $x_{n}(t_{n})\to x_{0}(t_{0}-)$, $s_{n}\leq t_{n}$, $s_{n}\to t_{0}$, then
$x_{n}(s_{n})\to x_{0}(t_{0}-)$.

The following lemma easily follows from the previous one.

######  Lemma 8.

Assume $x_{n}\to x_{0}$ in $\mathcal{D}$ and $y_{n}\to y_{0}$ in
$\mathcal{D}$. If for every point $t\geq 0$ at least one of two functions
$x_{0}$ and $y_{0}$ is continuous at this point, then $x_{n}+y_{n}\to
x_{0}+y_{0}$ in $\mathcal{D}$.

###  5.3. Proof of Theorem 1 for $\varrho\in(0,\infty)$.

Step 1. For each $n=1,2,\ldots$, denote

(15) $m_{n}(t)=\sup\limits_{s\in[0,t]}(x_{n}(s)_{-}),\,t\geq 0.$

Define $A_{n}$ and $B_{n}$ to be the sets $A$ and $B$ for the $n$th switch
problem.

First, the sequences $(T_{A_{n}})$, $(T_{B_{n}})$ are pre-compact in
$\mathcal{C}_{T}$ for every $T>0$. Indeed, $(T_{A_{n}})$, $(T_{B_{n}})$ are
globally Lipschitz continuous with Lipschitz constant 1; therefore, this
sequence $(T_{A_{n}})$ is equicontinuous in $\mathcal{C}_{T}$ for every $T>0$.
By the Arzela-Ascoli theorem, these sequences are pre-compact. The same
applies to $(T_{B_{n}})$. Take a subsequence $(n^{\prime})$ such that

(16) $T_{A_{n^{\prime}}}\to S_{0},\quad T_{B_{n^{\prime}}}\to
S_{1},\quad\text{ in any }\quad\mathcal{C}_{T},$

where $S_{0}(t)+S_{1}(t)=t$. Since $x_{n}\to x_{0}$ in $\mathcal{D}_{T}$ we
have convergence $m_{n}\to m_{0}$ in $\mathcal{D}_{T}$. Recall that $m_{0}$ is
continuous, see Remark 1. Thus we have the uniform convergence $m_{n}\to
m_{0}$ on any $[0,T]$. Combining these facts of the uniform convergence and
the observation

(17) $0\leq T_{A_{n^{\prime}}}(t)\leq T,\quad 0\leq T_{B_{n^{\prime}}}(t)\leq
T\quad\mbox{for}\quad t\in[0,T].$

we get the uniform convergence on any $[0,T]$:

(18) $m_{n^{\prime}}(T_{A_{n}^{\prime}})\to m_{0}(S_{0}),$

Step 2. From Lemma 7, for any $t>0$ all limit points of
$F_{n^{\prime}}(\varrho_{n}^{-1}T_{B_{n^{\prime}}}(t))$ are either
$F_{0}(\varrho^{-1}S_{1}(t)-)$ or $F_{0}(\varrho^{-1}S_{1}(t))$. Combining
this argument with Lemmas 4, 5, 6, and the assumption that $F_{0}$ is non-
decreasing, we get:

$F_{0}(\varrho^{-1}S_{1}(t)-)\leq m_{0}(S_{0}(t))\quad\mbox{ and }\
F_{0}(\varrho^{-1}S_{1}(t))\geq m_{0}(S_{0}(t)),\quad t\geq 0.$

Here we used the fact that $T_{A_{n}}(t)\leq t$ and $T_{B_{n}}(t)\leq t$, thus

$\displaystyle 0\leq\sup_{s\in[0,T_{A_{n}}(t)]}(x_{n}(s-)-x_{n}(s))$
$\displaystyle\leq\sup_{s\in[0,t]}(x_{n}(s-)-x_{n}(s))\to 0,\quad n\to\infty,$
$\displaystyle 0\leq\sup_{0\leq s_{1}\leq s_{2}\leq
T_{B_{n}}(t)}(F_{n}(s_{1})-F_{n}(s_{2}))$ $\displaystyle\leq\sup_{0\leq
s_{1}\leq s_{2}\leq t}(F_{n}(s_{1})-F_{n}(s_{2}))\to 0,\quad n\to\infty.$

The inequality

(19) $F_{0}(\varrho^{-1}S_{1}(t)-)\leq m_{0}(S_{0}(t))\leq
F_{0}(\varrho^{-1}S_{1}(t)),\quad t\geq 0.$

implies $\varrho^{-1}S_{1}(t)=F_{0}^{-1}(m_{0}(S_{0}(t)))$. Since
$S_{0}(t)+S_{1}(t)=t$, we can rewrite this as

$\varrho^{-1}(t-S_{0}(t))=F_{0}^{-1}(m_{0}(S_{0}(t))).$

Therefore,

(20) $S_{0}(t)+\varrho F_{0}^{-1}(m_{0}(S_{0}(t)))=t,\quad t\geq 0.$

By Remark 1, the function $A(s):=A_{\varrho}(s):=s+\varrho
F_{0}^{-1}(m_{0}(s))$ is continuous, strictly increasing, $A(0)=0$, and
$A(\infty)=\infty$. Therefore, $S_{0}(t)=A^{-1}(t)$ is continuous and strictly
increasing. Combining the above formulas, we get:

$S_{1}(t)=\varrho F_{0}^{-1}(m_{0}(S_{0}(t)))=\varrho
F_{0}^{-1}(m_{0}(A^{-1}(t))).$

Step 3. It follows from Steps 1 and 2 that there exists a sub-sequence
$(n^{\prime})$ such that we have $T_{A_{n^{\prime}}}\to A^{-1}$ in
$\mathcal{C}$. Similarly, we can show that every sub-sequence $(\tilde{n})$
have their own sub-sub-sequence $(\tilde{n}^{\prime})$ such that
$T_{A_{\tilde{n}^{\prime}}}\to A^{-1}$ in $\mathcal{C}$. Since the space
$\mathcal{C}$ is metric, this implies that $T_{A_{n}}\to A^{-1}$ in
$\mathcal{C}$. Since $T_{B_{n}}(t)=t-T_{A_{n}}(t)$ we have convergence
$T_{B_{n}}\to S_{1}=\varrho F_{0}^{-1}(m_{0}(A^{-1}))$ too. It follows from
[33, Theorem 13.2.2] that

(21) $x_{n}\circ T_{A_{n}}\to x_{0}\circ A^{-1}\ \mbox{in}\ {\mathcal{D}}$

because $A^{-1}$ is continuous and strictly increasing.

Step 4. For brevity, we give the proof only for $\rho_{n}=1$. In the general
case nothing will change, but the presence of $\rho_{n}$ in the numerators or
denominators obscures the idea of the proof. Let us verify the convergence:

(22) $F_{n}(T_{B_{n}})\to F_{0}(S_{1})\ \ \mbox{in }\ \ {\mathcal{D}}.$

Apply Lemma 7. Assume that $t_{n}\to t_{0},n\to\infty.$ Then
$\widetilde{t}_{n}:=T_{B_{n}}(t_{n})\to S_{1}(t_{0})=:\widetilde{t}_{0}$
because we have convergence $T_{B_{n}}\to S_{1}$ in ${\mathcal{C}}.$ Hence, by
Lemma 7, the limit points of
$(F_{n}(T_{B_{n}}(t_{n})))=(F_{n}(\widetilde{t}_{n}))$ may be only
$F_{0}(\widetilde{t}_{0})=F_{0}(S_{1}(t_{0}))$ or
$F_{0}(\widetilde{t}_{0}-)=F_{0}(S_{1}(t_{0})-)$. Note that the function
$S_{1}$ is only non-decreasing, so it is possible that
$F_{0}(S_{1}(t_{0})-)\neq F_{0}\circ S_{1}(t_{0}-).$ Conditions of Lemma 7 are
satisfied for $t_{0}$ if, for example, $\widetilde{t}_{0}=S_{1}(t_{0})$ is a
point of continuity of $F_{0},$ or $\lim_{t\to
t_{0}}F_{n}(T_{B_{n}}(t))=F_{0}(\widetilde{t}_{0})=F_{0}(S_{1}(t_{0}))$.
Hence, the non-trivial case is only the one when
$\widetilde{t}_{0}=S_{1}(t_{0})=F_{0}^{-1}(m_{0}(A^{-1}(t_{0})))$ is a point
of jump of $F_{0}.$ Let $\alpha:=F_{0}(\widetilde{t}_{0}-)$ and
$\beta:=F_{0}(\widetilde{t}_{0}).$ Then $\alpha<\beta.$ Notice that
$F_{0}^{-1}(z)<\widetilde{t}_{0}$ for $z<\alpha,$
$F_{0}^{-1}(z)>\widetilde{t}_{0}$ for $z>\beta,$ and moreover $\alpha\leq
m_{0}(A^{-1}(t_{0}))\leq\beta$. Assume at first that
$m_{0}(A^{-1}(t_{0}))=\alpha.$ It follows from Theorem assumptions that
$A^{-1}(t_{0})$ is a point of growth of $m_{0}$. Thus
$m_{0}(A^{-1}(s))<\alpha$ for $s<t_{0}$, because $A$ is strictly increasing.
Hence

$F_{0}(S_{1}(t_{0})-)=F_{0}(F_{0}^{-1}(m_{0}(A^{-1}(t_{0})))-)=F_{0}(F_{0}^{-1}(m_{0}(A^{-1}(t_{0}-))))=F_{0}\circ
S_{1}(t_{0}-).$

Conditions 2 and 3 of Lemma 7 hold for the point $t_{0}$ because $F_{0}$ is
strictly increasing. It remains to consider the case when
$m_{0}(A^{-1}(t_{0}))\in(\alpha,\beta].$ Lemmas 4, 5, 6 and the assumption
that $F_{0}$ is non-decreasing imply

$\varliminf_{t\to t_{0}}F_{n}(T_{B_{n}}(t))\geq\varliminf_{t\to
t_{0}}m_{n}(T_{A_{n}}(t))=m_{0}(A^{-1}(t_{0}))>\alpha.$

Above, the limit points of $(F_{n}(T_{B_{n}}(t)))$ may be only $\alpha$ or
$\beta.$ So $\lim_{t\to t_{0}}F_{n}(T_{B_{n}}(t))$ exists and is equal to
$\beta.$ Thus all conditions of Lemma 7 holds and this completes the proof of
(22).

Step 5. From Definition 10,
$y_{n}(t)=x_{n}(T_{A_{n}}(t))+F_{n}(T_{B_{n}}(t))$. Recall from (21) and (22):

$x_{n}(T_{A_{n}})\to x_{0}(A^{-1})\quad\mbox{ and }\quad F_{n}(T_{B_{n}})\to
F_{0}(F_{0}^{-1}(m_{0}(A^{-1})))\quad\mbox{in}\quad\mathcal{D}.$

Convergence $y_{n}\to y_{0}$ in ${\mathcal{D}}$, where

$y_{0}(t)=x_{0}(A^{-1}(t))+F_{0}(F_{0}^{-1}(m_{0}(A^{-1}(t)))),\ t\geq 0,$

is proved if we show that functions $x_{0}(A^{-1})$ and
$F_{0}(F_{0}^{-1}(m_{0}(A^{-1})))$ do not have joint points of discontinuity,
see Lemma 8. The function $A^{-1}$ is continuous. Thus, it suffices to show

(23) $x_{0}\mbox{ and }F_{0}(F_{0}^{-1}(m_{0}))\mbox{ do not have joint points
of discontinuity. }$

Assume that the function $F_{0}(F_{0}^{-1}(m_{0}))$ is discontinuous at a
point $t$, i.e., $F_{0}(F_{0}^{-1}(m_{0}))$ has a positive jump at $t$. Set
$s:=F_{0}^{-1}(m_{0}(t))$. Then $F$ has a jump at $s$. Denote
$\alpha:=F_{0}(s-)<F_{0}(s)=:\beta.$ Notice that $F_{0}^{-1}(z)=s$ for
$z\in[\alpha,\beta]$, $F_{0}^{-1}(z)<s$ for $z<\alpha$, and $F_{0}^{-1}(z)>s$
for $z>\beta$. Since $m_{0}$ is continuous and non-decreasing we must have
$m_{0}(t)=\alpha=F_{0}(s-)$ and $m_{0}(\tilde{t})<\alpha$ for $\tilde{t}<t.$
It follows from assumption (e) of the Theorem that $t$ is a point of growth of
$m_{0}.$ Hence $x_{0}$ can’t have positive jump at $t$. Since $x_{0}$ does not
have negative jumps at all by assumptions of the theorem, the function $x_{0}$
is continuous at $t.$ This proves (23) and hence the theorem in the case
$\rho\in(0,\infty)$.

###  5.4. Proof of Theorem 1 for $\varrho=0$.

Similarly to the case $\varrho\in(0,\infty)$, by Lemma 3

$\varlimsup_{n\to\infty}[-m_{n}(T_{A_{n}}(t))+F_{n}(T_{B_{n}}(t)/\rho_{n}-))]\leq
0.$

Since $T_{A_{n}}(t)+T_{B_{n}}(t)=t$ and $F_{0}(\infty)=\infty$, we have
convergence $T_{A_{n}}(t)\to t$ and $T_{B_{n}}(t)\to 0$ for any fixed $t\geq
0$. Moreover, since all functions are non-decreasing in $t$ the convergence is
locally uniform. It follows from [33, Theorem 13.2.2] that
$x_{n}(T_{A_{n}})\to x_{0}$ and $m_{n}(T_{A_{n}})\to m_{0}$ in ${\mathcal{D}}$
as $n\to\infty$. Moreover, continuity of $m_{0}$ implies the locally uniform
convergence $m_{n}(T_{A_{n}})\to m_{0}$. Thus

(24) $-m_{0}(t)+\varlimsup_{n\to\infty}F_{n}(T_{B_{n}}(t)/\varrho_{n}-)\leq
0.$

Using Lemma 4, similarly to the reasoning above, we obtain the inequality:

(25) $-m_{0}(t)+\varliminf_{n\to\infty}F_{n}(T_{B_{n}}(t)/\varrho_{n})\geq 0.$

Fix a $t>0$ and set $s_{n}:=T_{B_{n}}(t)/\varrho_{n}$. Then we can rewrite
(24) and (25) as

$\varlimsup_{n\to\infty}F_{n}(s_{n}-)\leq
m_{0}(t)\leq\varliminf_{n\to\infty}F_{n}(s_{n}).$

Let $s_{0}$ be a limit point of $(s_{n})$ (including infinity). By Lemma 7,
using the fact that $F_{0}$ is strictly increasing, we get the inequality

(26) $F_{0}(s_{0}-)\leq m_{0}(t)\leq F_{0}(s_{0}).$

Thus $s_{0}:=F_{0}^{-1}(m_{0}(t))$. Since any limit point of $(s_{n})$ is
determined uniquely we have convergence $T_{B_{n}}(t)/\varrho_{n}=s_{n}\to
s_{0}=F_{0}^{-1}(m_{0}(t))$ as $n\to\infty$ for any $t\geq 0$, and hence a
locally uniform convergence because all functions are non-decreasing and the
limit is continuous. The same arguments as for the case $\varrho\in(0,\infty)$
imply convergence in ${\mathcal{D}}$: $F_{n}(T_{B_{n}}/\varrho_{n})\to
F_{0}(F_{0}^{-1}(m_{0}))$, and finally

$y_{n}:=x_{n}(T_{A_{n}})+F_{n}(T_{B_{n}}/\varrho_{n})\to
y_{0}:=x_{0}+F_{0}(F^{-1}_{0}(m_{0}))=\mathcal{S}(x_{0},F_{0}).$

###  5.5. Proof of Theorem 1 for $\varrho=\infty$.

Note that $\mathcal{S}(x_{0},F_{0})(t\wedge\sigma)=x_{0}(t\wedge\sigma).$
Since $\lim_{n\to\infty}\varrho_{n}=\infty$ and $T_{B_{n}}(t)\leq t$ we have
the local uniform convergence $T_{B_{n}}(t)/\varrho_{n}\to 0$ as $n\to\infty.$
Recall that $F_{0}(0)=0$ and $F_{n}\to F_{0}$ in ${\mathcal{D}}.$ Due to Lemma
7, for any $t\geq 0$ we have convergence
$\lim_{n\to\infty}F_{n}(T_{B_{n}}(t)/\varrho_{n})=F_{0}(0)=0$ and even locally
uniform convergence because functions $F_{n}(T_{B_{n}})$ are non-decreasing.
It is clear that for any $t<\sigma$ we have $T_{A_{n}}(t)=t$ for sufficiently
large $n.$ On the other hand, similarly to the case $\varrho\in(0,\infty)$,
Lemma 4 implies:

$0\leq\varliminf_{n\to\infty}(F_{n}(T_{B_{n}}(t)/\varrho_{n})-m_{n}(T_{A_{n}}(t)))=\varliminf_{n\to\infty}(-m_{n}(T_{A_{n}}(t)))=-\varlimsup_{n\to\infty}m_{n}(T_{A_{n}}(t))\leq
0$

for any $t\geq 0$. Since $\inf\\{s\geq 0\ |\ x_{0}(s)=0\\}=\inf\\{s\geq 0\ |\
x_{0}(s)<0\\}$ by the assumption and $\inf\\{s\geq 0\ |\
x_{0}(s)<0\\}=\inf\\{s\geq 0\ |\ m_{0}(s)>0\\}$, we have
$\varlimsup_{n\to\infty}T_{A_{n}}(t)\leq\sigma$ for any $t$. We may conclude
that $T_{A_{n}}(t)\to t\wedge\sigma$. To prove the theorem, it suffices to
verify convergence $x_{n}(T_{A_{n}})\to x_{0}(\cdot\wedge\sigma).$ Let us
apply Lemma 7. If $t_{0}\neq\sigma,$ then conditions of Lemma 7 are obviously
satisfied.

Recall that $\inf\\{s\geq 0\ |\ x_{0}(s)=0\\}=\inf\\{s\geq 0\ |\
x_{0}(s)<0\\}$. Hence $x_{0}$ doesn’t have a (positive) jump at
$\sigma=\inf\\{s\geq 0\ |\ x_{0}(s)=0\\}$ and so $x_{0}$ is continuous at
$t_{0}=\sigma$. Thus $x_{0}(\cdot\wedge\sigma)$ is continuous at
$t_{0}=\sigma$, and consequently conditions of Lemma 7 are satisfied for
$t_{0}=\sigma$ too. $\square$

##  6\. Appendix

###  6.1. Proof of Lemma 1

The proof of existence is straightforward and is done exactly as in the
classic Skorokhod problem. Let us show uniqueness: take two solutions
$y_{1},y_{2}$, and the corresponding boundary terms $l_{1},l_{2}$. Then

$\displaystyle(y_{1}(t)-y_{2}(t))^{2}$
$\displaystyle=2\int_{(0,t]}(y_{1}(s)-y_{2}(s))\,\mathrm{d}(F(l_{1}(s))-F(l_{2}(s)))$
$\displaystyle-\sum_{0<s\leq
t}\Big{(}\Delta\big{(}F(l_{1}(s))-F(l_{2}(s))\big{)}\Big{)}^{2}$
$\displaystyle=2\int_{(0,t]}(y_{1}(s)-y_{2}(s))\,\mathrm{d}F(l_{1}(s))-2\int_{(0,t]}(y_{1}(s)-y_{2}(s))\,\mathrm{d}F(l_{2}(s))$
$\displaystyle-\sum_{0<s\leq
t}\Big{(}\Delta\big{(}F(l_{1}(s))-F(l_{2}(s))\big{)}\Big{)}^{2}$
$\displaystyle=-2\int_{(0,t]}y_{2}(s)\,\mathrm{d}F(l_{1}(s))-2\int_{(0,t]}y_{1}(s)\,\mathrm{d}F(l_{2}(s))$
$\displaystyle-\sum_{0<s\leq
t}\Big{(}\Delta\big{(}F(l_{1}(s))-F(l_{2}(s))\big{)}\Big{)}^{2}\leq 0.$

So, $y_{1}(t)=y_{2}(t)$ and thus $F(l_{1}(t))=F(l_{2}(t))$ for all $t\geq 0.$
Since $F$ is strictly increasing, $l_{1}(t)=l_{2}(t).\quad\square$

###  6.2. Proof of Lemma 3

Recall that $B=\cup_{k=1}^{\infty}[\rho_{k},\tau_{k})$,
$A=\cup_{k=1}^{\infty}[\tau_{k-1},\rho_{k})$. We also know that $y(t)\leq 0$
for $t\in B$, and therefore

(27) $y(t-)\leq 0\ \mbox{for}\
t\in\bigcup\limits_{k=1}^{\infty}(\rho_{k},\tau_{k}].$

Case 1. Assume $t\in(\rho_{k},\tau_{k}]$ for some $k,$ then

$F(T_{B}(t)-)=\lim_{z\uparrow T_{B}(t)}F(z)=\lim_{s\uparrow
t}F(T_{B}(s))=F(T_{B}(t-))\equiv(F\circ T_{B})(t-).$

The second equality is true because $T_{B}$ is strictly increasing on
$(\rho_{k},\tau_{k}]$; thus, the left limit of the function $F$ at the point
$T_{B}(\tau_{k-1})$ is equal to the left limit of the composition $F\circ
T_{B}$ at the point $\tau_{k-1}$. Next, apply the formulas (10), (15), (27).
We get:

$F(T_{B}(t)-)=F(T_{B}(t-))=y(t-)-x(T_{A}(t-))\leq 0+m(T_{A}(t-))\leq
m(T_{A}(t)).$

For the last inequality, we use that $m(T_{A})$ is non-decreasing. This
completes the proof of Lemma 3 in Case 1.

Case 2. Assume $t\in(\tau_{k-1},\rho_{k}]$ for some $k$. On this interval, the
function $T_{B}$ is constant. Therefore, $T_{B}(t)=T_{B}(\tau_{k-1})$. Thus

(28) $F(T_{B}(t)-)=F(T_{B}(\tau_{k-1})-)=F(T_{B}(\tau_{k-1}-))\equiv(F\circ
T_{B})(\tau_{k-1}-).$

The second equality in (28) from the fact that $T_{B}$ is strictly increasing
on $(\rho_{k-1},\tau_{k-1}]$. Thus, the left limit of the function $F$ at the
point $T_{B}(\tau_{k-1})$ is equal to the left limit of the composition
$F\circ T_{B}$ at the point $\tau_{k-1}$. The functions $s\mapsto T_{A}(s-)$
and $m$ are always non-decreasing. Thus $s\mapsto m(T_{A}(s-))$ is also non-
decreasing. Using (10) again, we get

(29) $F(T_{B}(\tau_{k-1}-))=y(\tau_{k-1}-)-x(T_{A}(\tau_{k-1}-)).$

Next, using (15) and (27), we get:

(30) $y(\tau_{k-1}-)-x(T_{A}(\tau_{k-1}-))\leq 0-x(T_{A}(\tau_{k-1}))\leq
m(T_{A}(\tau_{k-1}))\leq m(T_{A}(t)).$

Combining (28), (29), (30), we complete the proof of Lemma 3 in Case 2.

###  6.3. Proof of Lemma 4

We prove the statement by induction. First, the induction base:
$[0,\rho_{1})$, we are in regime $A$. Thus $T_{A}(t)=t$ and $T_{B}(t)=0$.
Therefore, $x(s)=y(s)$ for $s\in[0,\rho_{1})$. Hence we have:

$\rho_{1}=\inf\\{s\geq 0\mid y(s)\leq-\delta\\}=\inf\\{s\geq 0\mid
x(s)\leq-\delta\\}.$

Thus $m(t)\leq\delta$ for $t\in[0,\rho_{1})$. Note that $F(0)=0$, and the two
suprema in the right-hand side of the inequality of Lemma 4 are non-negative.
This proves the lemma statement on $[0,\rho_{1})$.

Before the induction step, notice that regimes can switch only when $x$
attains its minimum or $F$ attains its maximum:

$x(T_{A}(\rho_{k}))=-m(T_{A}(\rho_{k})),\ \
F(T_{B}(\tau_{k}))=\max_{s\in[0,\tau_{k}]}F(T_{B}(s)),\ \ k\geq 1.$

Case 1. Assume the statement is true on $[0,\rho_{k})$. Let us show it for
$t\in[\rho_{k},\tau_{k})$. We have

(31)
$\displaystyle\begin{split}F(T_{B}(t))=\big{(}F(T_{B}(t))-F(T_{B}(\rho_{k}))\big{)}+\big{(}F(T_{B}(\rho_{k}))-F(T_{B}(\rho_{k}-))\big{)}+\\\
\big{(}F(T_{B}(\rho_{k}-))+x(T_{A}(\rho_{k}-))\big{)}+\big{(}x(T_{A}(\rho_{k}))-x(T_{A}(\rho_{k}-))\big{)}-x(T_{A}(\rho_{k})).\end{split}$

Note that $T_{A}$ is constant on $[\rho_{k},\tau_{k})$:
$T_{A}(t)=T_{A}(\rho_{k})$. Thus $x(T_{A}(t))=x(T_{A}(\rho_{k}))$, and

(32) $x(T_{A}(t))=x(T_{A}(\rho_{k}))=-m(T_{A}(\rho_{k}))=-m(T_{A}(t)).$

Similarly, $T_{B}$ does not grow on $[\tau_{k-1},\rho_{k}]$, thus we have
equality $F(T_{B}(\rho_{k}))=F(T_{B}(\rho_{k}-))$. Combining (31), (32), we
get:

(33) $\displaystyle\begin{split}F(T_{B}(t))&\geq\inf_{0\leq s_{1}\leq
s_{2}\leq t}\big{(}F(T_{B}(s_{2}))-F(T_{B}(s_{1}))\big{)}+0+y(\rho_{k}-)\\\
&+\big{(}x(T_{A}(\rho_{k}))-x(T_{A}(\rho_{k}-))\big{)}+m(T_{A}(t)).\end{split}$

Recalling that $y(\rho_{k}-)\geq-\delta$ and combining (33), (32) wit the
elementary estimates:

$\inf_{0\leq s_{1}\leq s_{2}\leq
t}\big{(}F(T_{B}(s_{2}))-F(T_{B}(s_{1}))\big{)}{=}-\sup_{0\leq t_{1}\leq
t_{2}\leq T_{B}(t)}\big{(}F(t_{1})-F(t_{2})\big{)};$
$x(T_{A}(\rho_{k}))-x(T_{A}(\rho_{k}-))\geq-\sup_{0\leq s\leq
t}\big{(}x(T_{A}(s-))-x(T_{A}(s))\big{)}=-\sup_{0\leq z\leq
T_{A}(t)}\big{(}x(z-)-x(z)\big{)},$

we complete the proof of Lemma 4 in Case 1.

Case 2. Assume the statement on $[0,\tau_{k-1})$, and show it for
$t\in[\tau_{k-1},\rho_{k})$. Use (10) to get

$F(T_{B}(t))=y(t)-x(T_{A}(t))\geq-\delta-x(T_{A}(t))\geq-\delta+m(T_{A}(t)).$

This completes the proof of Lemma 4 in Case 2.

###  6.4. Proof of Lemma 5

Define the function $j:\mathcal{D}_{T}\to\mathbb{R}$ as the largest negative
jump of a function $x\in\mathcal{D}_{T}$. This function is well-defined since
any function $x\in\mathcal{D}_{T}$ has only countably many jumps and only
finitely many jumps which exceed any given positive level. The function $j$ is
continuous in the Skorokhod topology of $\mathcal{D}_{T}$. This can be proved
similarly to [6, Example 12.1], which shows that the largest jump (regardless
of direction) of a function in $\mathcal{D}$ is a continuous function
$\mathcal{D}\to\mathbb{R}$. Since $x_{0}$ has no negative jumps, obviously
$j(x_{0})=0$. Therefore, $\sup_{s\in[0,T]}(x_{n}(s-)-x_{n}(s))=j(x_{n})(T)\to
0.\quad\square$

###  6.5. Proof of Lemma 6

Assume that $F_{0}$ is continuous at $T$. There exists for every $n$ a
strictly increasing one-to-one function $\lambda_{n}:[0,T]\to[0,T]$ such that
$F_{n}(\lambda_{n})\to F_{0}$ uniformly on $[0,T]$. We can rewrite

(34) $\sup\limits_{0\leq s_{1}\leq s_{2}\leq
T}(F_{n}(s_{1})-F_{n}(s_{2}))=\sup\limits_{0\leq s_{1}\leq s_{2}\leq
T}(F_{n}(\lambda_{n}(s_{1}))-F_{n}(\lambda_{n}(s_{2}))).$

Next, by uniform convergence $F_{n}(\lambda_{n})\to F_{0}$:

(35) $\sup\limits_{0\leq s_{1}\leq s_{2}\leq
T}(F_{n}(\lambda_{n}(s_{1}))-F_{n}(\lambda_{n}(s_{2})))\to\sup\limits_{0\leq
s_{1}\leq s_{2}\leq T}(F_{0}(s_{1})-F_{0}(s_{2})).$

Since the function $F_{0}$ is non-decreasing, the right-hand side of (35) is
zero. Together with (34), this completes the proof of Lemma 6. $\square$

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# Monopoles Entangle Fermions

Csaba Csáki<EMAIL_ADDRESS>Department of Physics, LEPP, Cornell University,
Ithaca, NY 14853, USA  Yuri Shirman<EMAIL_ADDRESS>Department of Physics
$\&$ Astronomy, University of California, Irvine, CA 92697, USA  Ofri Telem
<EMAIL_ADDRESS>Department of Physics, University of California, Berkeley,
CA 94720, USA  Ernest Orlando Lawrence Berkeley National Laboratory, Berkeley,
CA 94720, USA  John Terning<EMAIL_ADDRESS>QMAP, Department of Physics,
University of California, Davis, CA 95616, USA

###### Abstract

We resolve the decades old mystery of what happens when a positron scatters
off a minimal GUT monopole in an s-wave, first discussed by Callan in 1983.
Using the language of on-shell amplitudes and pairwise helicity we show that
the final state contains two up quarks and a down quark entangled with angular
momentum stored in the gauge fields, which is the only particle final state
that satisfies angular momentum and gauge charge conservation. The cross
section for this process is as large as in the original Rubakov-Callan effect,
only suppressed by the QCD scale. The final state we find cannot be seen in
Callan’s truncated 2D theory, since our entanglement requires more than 2
dimensions.

##  I introduction

The scattering of electrically charged fermions with magnetic monopoles is a
very peculiar process reviews . Until recently the theoretical understanding
of these processes faced three major difficulties:

  * •

Weinberg Weinberg:1965rz  found that the scattering amplitudes are not Lorentz
invariant.

  * •

Multiparticle scattering states with both electric and magnetic charges carry
additional angular momentum in the gauge field Thomson ; ZwanzigerSelection
and cannot be written as tensor products of Wigner’s one-particle states.

  * •

The analysis of the scattering of Grand Unified Theory (GUT) monopoles
seemingly led to the conclusion that one must either give up on conservation
of gauge charges or accept the existence of fractional particles
Callan:smatrix .

The Lorentz violation problem was resolved by all orders resummation in
Terning:2018udc and order-by-order for the special case of monopoles bound
with anti-monopoles in Terning:2020dzg . The problem with multiparticle states
was resolved by the inclusion of an additional quantum number called pairwise
helicity Csaki:2020inw ; Csaki:2020yei . In this letter we will present a
resolution of the final problem.

To understand the essence of the final problem regarding the scattering of GUT
monopoles it is helpful to recall that in a $U(1)$ theory with a magnetic
monopole and two oppositely charged massless Weyl fermions, angular momentum
conservation implies that there can be no forward scattering in the $J=0$
partial wave: the massless fermion must flip its chirality Kazama by turning
into the CP conjugate of the other fermion. We can embed this simple theory
into a ‘t Hooft-Polyakov model 'tHooft:1974qc with an $SU(2)$ gauge group and
two Weyl doublets111An even number of doublets is required to avoid the Witten
global anomaly.. This theory has an $SU(2)$ flavor symmetry, which is
perfectly consistent with the helicity flip process if the flavor flips as
well. Things become more subtle in a model with 4 doublets where the helicity
flip process is forbidden by an $SU(4)$ global symmetry. The global symmetry
allows processes with one fermion initial state scattering into three fermion
final states as well as processes where Callan:smatrix  two incoming fermions
to scatter to two outgoing fermions RubakovCallan . The proton decay catalized
by the latter type of process produce the leading observational bounds on the
relic density of GUT monopoles in the universe bounds .

Callan Callan:smatrix  was the first to study the scattering of a positron off
a GUT monopole. The problem reduces to the four flavor ‘t Hooft-Polyakov model
in a limit where some gauge couplings are dropped, implying again that there
have to be at least 3 fermions in the final state. Truncating to 2D and
reformulating the problem in terms of solitons, Callan concluded
Callan:smatrix ; Callan:1982au that there was no possible 3 fermion final
state that preserved all of the gauge quantum numbers. As noticed by Witten,
the truncated theory produced half-solitons (aka “semitons”) in the final
state Callan:smatrix , which they identified with “half particles” in the full
theory. With the fermion masses set to zero these states, if they existed,
would have to be true asymptotic final states far from the monopole, where
perturbation theory can be reliably applied. Since these states cannot arise
in perturbation theory, Callan suggested Callan:smatrix  that there could be
some kind of statistical understanding where charge conservation is violated
in individual events but is conserved on average. However this explanation was
never fully embraced, since gauge invariance of the full 4D theory does not
allow for such a probabilistic conservation.

The aim of this paper is to find the form of the amplitude for this process by
making use of helicity amplitudes and pairwise helicity. We will indeed be
able to identify the unique form of the amplitude that preserves all gauge
quantum numbers, angular momentum and the approximate global flavor symmetry.
As in the helicity-flip process discussed above, forward scattering is
forbidden, and surprisingly the unique out state entangles the three fermions
with the angular momentum in the gauge field, combining into a total $J=0$
partial wave. This amplitude corresponds to the lowest dimensional operator
without derivatives in the low-energy effective theory. Thus we find that
monopoles can produce entangled fermions with a cross section satisfying the
s-wave unitarity bound.

##  II Rubakov-Callan Interactions

Consider the minimal $SU(5)$ GUT monopole obtained by embedding the standard
‘t Hooft-Polyakov monopole 'tHooft:1974qc into the $SU(2)_{M}\subset SU(5)$
generated by $T^{i}_{M}=\text{diag}\left(0,0,\tau^{i},0\right)$. Far away from
the monopole, the full $SU(5)$ gauge symmetry is broken by a Higgs in the
adjoint of $SU(5)$ to the standard model (SM) gauge group which includes the
$U(1)_{M}$ generated by $T^{3}_{M}$. The monopole configuration is invariant
under the combined rotations generated by $\vec{L}+\vec{T}$, where $\vec{L}$
is the orbital angular momentum operator and $\vec{T}$ is the vector of
$SU(2)_{M}$ generators $T^{i}_{M}$. In the $SU(5)$ GUT, every generation of SM
fermions is embedded in a $\bar{\mathbf{5}}$ and a $\mathbf{10}$. Under
$SU(2)_{M}$, these decompose into 4 doublets:

$\displaystyle\left(\begin{array}[]{c}e\\\
-\bar{d}^{3}\end{array}\right)\,,\,\left(\begin{array}[]{c}\bar{u}^{1}\\\
u^{2}\end{array}\right)\,,\,\left(\begin{array}[]{c}-\bar{u}^{2}\\\
u^{1}\end{array}\right)\,,\,\left(\begin{array}[]{c}d^{3}\\\
\bar{e}\end{array}\right)\,,$ (9)

where the upper and lower components have charge $e_{M}=\pm\frac{1}{2}$ under
$U(1)_{M}$, and all other fermions are $SU(2)_{M}$ singlets. We labeled the
particles by the corresponding left-handed fields, the right-handed particles
correspond to hermitian conjugates of these fields. Here the $1,2,3$ label
global color charge, which is broken in the vicinity of the monopole.
Furthermore, in this paper we take the monopole to have charge $g_{M}=-1$ for
consistency with Rubakov’s notation Rubakov:1988aq . This means that
$q=e_{M}g_{M}=-1/2$ for $e,\bar{u}^{1},\bar{u}^{2},d^{3}$ while $q=1/2$ for
$\bar{d}^{3},u^{2},u^{1},\bar{e}$. In the early ’80s, Rubakov and Callan
RubakovCallan independently derived a remarkable feature of fermion-monopole
scattering: when a pair of $u^{1}+u^{2}$ quarks is incident on the monopole,
there are seemingly two possible outgoing states which conserve all SM quantum
numbers - the initial state itself, and the state
$d^{3^{\dagger}}+e^{\dagger}$. Rubakov and Callan showed that the $J=0$ part
of the out state cannot, in fact, be the same as the in state, and so the
entire $J=0$ partial wave is converted to $d^{3^{\dagger}}+e^{\dagger}$, thus
violating baryon (B) number.

To see this effect, both Rubakov and Callan focused on a truncated theory in
which one only retains the $J=0$ partial wave for each fermion. Famously
RubakovCallan , in this truncated theory, fermions with $q=-1/2$ exist only as
incoming waves, while those with $q=1/2$ exist only as outgoing waves. In
particular, this implies that forward scattering is always forbidden. Note
that the hermitian conjugates of the fields in (9) have the opposite charge
and helicity so $e^{\dagger}$ is also an outgoing wave.

Consequently, monopoles induce a B-violating process with a cross section
saturating the $J=0$ unitarity bound. When QCD confinement is taken into
account, this leads to monopoles catalyzing proton decay with a QCD scale
cross section, counter to the naive intuition that the cross section is
suppressed by the scattering energy over the GUT scale. The leading
observational bounds on the relic density of GUT monopoles in the universe are
then derived from proton/neutron decay catalysed by monopole capture in
neutron stars bounds .

##  III Monopole Catalysis and Pairwise Helicity

In Csaki:2020inw , the most general form of the S-matrix for the scattering of
monopoles and charges was constructed. The main take away from this
construction is that the multiparticle asymptotic states of the S-matrix are
not tensor products of single particle states. In particular, under Lorentz
transformations, they pick up an extra little group phase for every monopole-
charge (or dyon-dyon) pair. For example, consider a 2-particle state where
each particle has electric and magnetic charges $(e_{i},g_{i})$ and spin
$s_{i}$. This state transforms as

$\displaystyle U(\Lambda)\ket{p_{i},p_{j};s_{i},s_{j};q_{ij}}=$ $\displaystyle
e^{iq_{ij}\phi_{ij}}\,\mathcal{D}_{s^{\prime}_{i},s_{i}}\mathcal{D}_{s^{\prime}_{j},s_{j}}\ket{\Lambda
p_{i},\Lambda p_{j};s^{\prime}_{i},s^{\prime}_{j};q_{ij}}\,,$ (10)

Here $U(\Lambda)$ is the unitary representation of the Lorentz transformation
$\Lambda$, while the $\mathcal{D}_{ab}$ represent single particle little group
factors. The extra “pairwise little group” phase $e^{iq_{ij}\phi_{ij}}$ is
unique to multiparticle states involving monopoles and charges (or any other
mutually non-local particles). The pairwise helicity $q_{ij}$ is half-integer
since it labels charges under the pairwise little group, which is a compact
$U(1)$ Csaki:2020inw . It has a natural interpretation as the quantity

$\displaystyle q_{ij}\,=\,e_{Mi}g_{Mj}-e_{Mj}g_{Mi}\,,$ (11)

which is quantized in half integer units by the Dirac-Zwanziger-Schwinger
quantization condition Dirac .

A more detailed definition of electric magnetic multiparticle states was given
in Csaki:2020yei . The transformation rule (III) implies additional
constraints on scattering amplitudes involving monopoles—the functional form
of the scattering amplitude has to be such that

$\displaystyle\mathcal{A}(\Lambda p_{1},\ldots,\Lambda p_{n};\Lambda
k_{1},\ldots,\Lambda k_{m})=$ $\displaystyle e^{-i\sum
q_{ij}\phi_{ij}}\,\tilde{\mathcal{A}}(p_{1},\ldots,p_{n};k_{1},\ldots,k_{m})\,,$
(12)

where $\tilde{\mathcal{A}}$ is the amplitude $\mathcal{A}$ times all of the
single particle little group transformations $\mathcal{D}_{i}$. To construct
amplitudes with the required transformation rule ref. Csaki:2020inw defined
new spinor-helicity variables called “pairwise spinors,” denoted by
$\ket{p^{\flat\pm}_{ij}}$, defined for each pair of particles in the in or out
state. For completeness, we repeat the definition of these spinors in the
Appendix. The spinors have pairwise helicity $\pm$ under the pairwise little
group associated with the particles $i$ and $j$. In other words, they
transform as

$\displaystyle{\tilde{\Lambda}}\ket{p^{\flat\pm}_{ij}}\,$ $\displaystyle=$
$\displaystyle\,e^{\pm\frac{i}{2}\phi(p_{i},p_{j},\Lambda)}\,\ket{\Lambda
p^{\flat\pm}_{ij}}$
$\displaystyle\left[p^{\flat\pm}_{ij}\right|\,{\tilde{\Lambda}}\,$
$\displaystyle=$
$\displaystyle\,e^{\mp\frac{i}{2}\phi(p_{i},p_{j},\Lambda)}\,\left[\Lambda
p^{\flat\pm}_{ij}\right|\,,$ (13)

where $\Lambda$ and $\tilde{\Lambda}$ are Lorentz transformations acting in
vector and spinor spaces respectively. Finally, the pairwise spinors have the
important property that they align with some of the standard spinor helicity
variables in the massless limit. In particular:

$\displaystyle\left<i\,p^{\flat+}_{ij}\right>=\left[i\,p^{\flat+}_{ij}\right]=0$
$\displaystyle\left<j\,p^{\flat-}_{ij}\right>=\left[j\,p^{\flat-}_{ij}\right]=0\leavevmode\nobreak\
.$ (14)

The vanishing of these contractions plays a central role in explaining the
peculiarities of the Rubakov-Callan effect.

To see the relation between pairwise helicity and the Rubakov-Callan effect,
let us consider an incoming state involving the massless fermions
$u^{1},\,u^{2}$, both with electric charge $e_{M}=-1/2$ and a scalar monopole
$M$ with magnetic charge $g_{M}=-1$. Let us now focus on the s-wave partial
amplitude involving in- and out- states with total angular momentum $J=0$. In
this case the amplitude splits into an incoming and and outgoing part, each
one depending only on the incoming/outgoing momenta and with all spinor
indices contracted (since $J=0$). As $q_{u^{1},M}=q_{u^{2},M}=-1/2$, the
incoming part of the amplitude is

$\displaystyle\left[u^{1}\,p^{\flat-}_{u^{1},M}\right]\left[u^{2}\,p^{\flat-}_{u^{2},M}\right]\,,$
(15)

where $\left|p^{\flat-}_{u^{i},M}\right]$ are pairwise spinors, while
$\left[u^{i}\right|$ are the standard massless spinor helicity variables. To
see that this in-state transforms correctly, note that the
$\left|p^{\flat-}_{u^{i},M}\right]$ each carry pairwise helicity $-1/2$ under
the ${u^{i},M}$ pairwise little group, while the $\left[u^{i}\right|$
transform like a helicity $1/2$ under the single particle little group for
$u^{i}$, which is suitable since incoming left-handed fermions carry helicity
$1/2$ in our all-outgoing convention. In contrast, outgoing left-handed
fermions carry helicity $-1/2$ in this convention. Note that pairwise helicity
is not flipped between incoming and outgoing particles Csaki:2020inw .

We can now easily see why there can’t be forward scattering in this process.
Let us try to represent the would-be out-state relevant for forward
scattering, i.e. involving the same $u^{1},\,u^{2}$. The out part of the
amplitude has to be

$\displaystyle\left<u^{1}\,p^{\flat+}_{u^{1},M}\right>\left<u^{2}\,p^{\flat+}_{u^{2},M}\right>\,.$
(16)

Note that the sign on the pairwise spinors is flipped so as to preserve their
pairwise helicity under $\left|\right]\rightarrow\ket{}$. However, this
expression vanishes by (III). There cannot be forward scattering of fermions
on a monopole in the lowest partial wave.

Having established that there is no forward scattering for the Rubakov-Callan
in-state, we now turn to write down the only possible final state which
respects all SM quantum numbers, as well as the overall $SU(4)$ flavor
symmetry. This out state involves the fermions
$(\bar{e})^{\dagger},(\bar{d}^{3})^{\dagger}$. The corresponding outgoing part
of the amplitude is

$\displaystyle\left[\bar{e}^{\dagger}\,p^{\flat-}_{\bar{e}^{\dagger},M}\right]\left[\bar{d}^{3\dagger}\,p^{\flat-}_{\bar{d}^{3\dagger},M}\right]\,.$
(17)

It transforms correctly under the pairwise little group, since
$q_{\bar{e},M}=q_{\bar{d}^{3},M}=1/2$. Since this is the only possible out
state, we have a simple derivation of the Rubakov-Callan amplitude

$\displaystyle\mathcal{A}_{\text{Rubakov-Callan}}\propto$
$\displaystyle\left[u^{1}\,p^{\flat-}_{u^{1},M}\right]\left[u^{2}\,p^{\flat-}_{u^{2},M}\right]\left[\bar{e}^{\dagger}\,p^{\flat-}_{\bar{e}^{\dagger},M}\right]\left[\bar{d}^{3\dagger}\,p^{\flat-}_{\bar{d}^{3\dagger},M}\right]\,.$
(18)

The overall cross section for the process satisfies the s-wave unitarity
bound, and so should be proportional to $4\pi p^{-2}_{c}$ where $p_{c}$ is the
COM momentum. When taking QCD confinement of the incoming quarks into account,
the incoming quarks are confined to within a distance $\Lambda^{-1}$ of each
other, and the cross section becomes $\mathcal{O}(\Lambda^{-2})$.

##  IV Solving a 40 Year old Mystery

When a positron, $\bar{e}$, scatters off of a GUT monopole, forward scattering
is again forbidden by angular momentum conservation, while the flavor symmetry
constrains the out state to have 3 (mod 4) fermions. The only possible out
state with 3 fermions which conserves all quantum numbers is:

$\bar{u}^{1\dagger}\,+\,\bar{u}^{2\dagger}\,+\,{\bar{d}}^{3\dagger}\,.$ (19)

However, Callan argued that this final state is impossible, since in the
presence of the monopole, the $\bar{d}^{3\dagger}$ cannot exist in a one-
particle outgoing partial wave with $J=0$. Working in a truncated 2D theory
including only fermions in one-particle $J=0$ waves which are then bosonized
to solitons, Callan found that the final state consists of four semitons, or
“half-particles”. For the initial state of an ${\bar{e}}$ he found the
semitonic final state
$1/2(e^{\dagger}+{\bar{u}}^{1\dagger}+{\bar{u}}^{2\dagger}+d^{3})$. Since
“half-particles” do not exist in the 4D theory Callan suggested the
interpretation that half the time one would produce a positron and half the
time one would produce a proton. These proposed individual processes do not
conserve SM gauge charges, but would do so on average.

Analog 2D theories with an $SO(8)$ global symmetry have been analyzed by
Maldacena and Ludwig Maldacena:1995pq and Boyle Smith and Tong Smith:2020nuf .
These authors confirmed that in the absence of additional gauge symmetries the
semiton description is correct, and can be understood via $SO(8)$ triality.
However this does not answer the question of what happens for the GUT monopole
process where the fermions have chiral non-Abelian charges that break the
$SO(8)$ symmetry.

Sen Sen claimed that conservation laws ensure that there are no monopole
processes allowed with one fermion in the initial state and three fermions in
the final state. If this were true then there would either have to be
processes with more fermions (3 mod 4) or a mechanism that prevented single
fermions from encountering a monopole. However the conservation laws that Sen
used are only valid in the 2D truncated theory which leaves out the
possibility of entangling a fermion with the field angular momentum produced
by a different particle, so it is not surprising that his analysis cannot
produce the correct final state.

Kitano and Matsudo Kitano:2021pwt suggested that the semitons should be
identified in the 4D theory with a “pancake” soliton: a domain wall bounded by
a string. These pancakes are supposed to be heretofore unknown asymptotic
states of the gauge theory. For this to be a consistent interpretation in the
massless fermion limit, the pancake would also have to have arbitrarily small
energies since the incoming positron energy can be arbitrarily small.

Using the pairwise helicity formalism, we are able for the first time to
identify the correct final state for positron-monopole scattering. This final
state does, in fact, consist of the fermions in (19), which conserve all of
the SM quantum numbers and respect the approximate $SU(4)$ flavor symmetry.
The novelty here is that the final state fermions are in fact entangled with
the field angular momentum arising from one of the other particles, which
allows them to be in an overall $J=0$ state, even though they are not in one-
particle $J=0$ states. The amplitude for this process is

$\displaystyle\mathcal{A}_{\bar{e}}\propto$
$\displaystyle\left[\bar{e}\,p^{\flat-}_{\bar{e}^{\dagger},M}\right]\left[\bar{u}^{1\dagger}\,p^{\flat-}_{\bar{u}^{1\dagger},M}\right]\left[\bar{u}^{2\dagger}\,p^{\flat+}_{\bar{d}^{3\dagger},M}\right]\left[\bar{d}^{3\dagger}\,p^{\flat-}_{\bar{u}^{2\dagger},M}\right]$
$\displaystyle\,-\,(1\leftrightarrow 2)\,.$

Note that we cannot arrange a similar cross-entanglement when there are only
two fermions in the final state. Consider the static monopole limit, then in
the center of mass frame the two fermions emerge back-to-back, and the flat
momenta are also back-to-back along this axis, thus exchanging the flat
spinors so that they are contracted with the opposite particle gives exactly
zero. For finite monopole masses there could be a contribution that is
suppressed by the monopole mass, but this cannot saturate the unitarity bound.
Also note that truncating to 2D also forces all the momenta to be along a
single direction, so at least one exchange of flat momenta will give a
vanishing amplitude. Thus we can easily see how the 2D truncation fails to
capture the 4D physics. In the static limit the three quarks are coplanar in
the center of momentum frame, so there is no obstruction to entanglement in a
3D theory.

##  V Applications

Since cross sections that saturate partial wave unitarity grow with the
inverse of the initial momentum one might naively expect that the positron
scattering process we have discussed would lead to an arbitrarily large cross
section for B-violation in GUT theories. We can see however that the growth is
cut off at the QCD scale, as happens for the Rubakov-Callan processes. Once
the initial energy is below the sum of the monopole and proton masses, the
final state of three quarks cannot hadronize into a proton. In the monopole
rest frame the three quark state will carry the initial momentum of the
positron, and once the separations of the quarks reaches the QCD scale the
quarks will be forced to travel in the same direction, so two of the quarks
will have their momentum flipped by QCD interactions. Since QCD also breaks
chirality, their chirality can also be flipped and they can become in-states
for a second interaction with the monopole. Two quarks scattering on the
monopole produce an antiquark and a lepton. The antiquark can annihilate with
the remaining quark to produce two photons or a lepton-antilepton pair. Thus
below the proton threshold there is no B-violation, as we expect from energy
conservation, and the B-violating cross section is cut off at the QCD scale.

##  VI Acknowledgments

###### Acknowledgements.

We thank Sungwoo Hong, Ryuichiro Kitano, Juan Maldacena, Ryutaro Matsudo, Joe
Polchinski, and David Tong for helpful conversations. CC is supported in part
by the NSF grant PHY-2014071 as well as the US-Israeli BSF grant 2016153. OT
was supported in part by the DOE under grant DE-AC02-05CH11231. Y.S. is
supported in part by the NSF grant PHY-1915005. J.T. is supported by the DOE
under grant DE-SC-0009999. We thank the Aspen Center for Physics, which is
supported by National Science Foundation grant PHY-1607611, where parts of
this work were completed.

##  VII Appendix

In this appendix we repeat for completeness the definitions of pairwise
spinors from Csaki:2020inw . For every pair $(i,j)$ of particles, we can
easliy define the pairwise spinors in the COM frame of the pair, where the
momenta of the two particles are

$\displaystyle k^{i}=(E_{i},p_{c}\,\hat{z}),\leavevmode\nobreak\
\,k^{j}=(E_{j},-p_{c}\,\hat{z}),$ (21)

and $E_{i,j}=\sqrt{m_{i,j}+p^{2}_{c}}$. As a first step we define the two null
momenta:

$\displaystyle k^{\flat\pm}_{ij}=p_{c}\,(1,\pm\hat{z})\,.$ (22)

It’s easy to see that they are linear combinations of $k_{i}$ and $k_{j}$. Our
pairwise spinors in the COM frame are simply the ”square roots” of these
pairwise momenta:

$\displaystyle\ket{k^{\flat+}_{ij}}_{\alpha}=\sqrt{2\,p_{c}}\,\left(\begin{array}[]{c}1\\\
0\end{array}\right)\,,\,\ket{k^{\flat-}_{ij}}_{\alpha}=\sqrt{2\,p_{c}}\,\left(\begin{array}[]{c}0\\\
1\end{array}\right)$ (27)
$\displaystyle\left[k^{\flat+}_{ij}\right|_{\dot{\alpha}}=\sqrt{2\,p_{c}}\,\left(1\leavevmode\nobreak\
\leavevmode\nobreak\
0\right)\,,\,\left[k^{\flat-}_{ij}\right|_{\dot{\alpha}}=\sqrt{2\,p_{c}}\,\left(0\leavevmode\nobreak\
\leavevmode\nobreak\ 1\right)\,.$

These are defined so that

$\displaystyle
k^{\flat\pm}_{ij}\cdot\sigma_{\alpha{\dot{\alpha}}}=\ket{k^{\flat\pm}_{ij}}_{\alpha}\left[k^{\flat\pm}_{ij}\right|_{\dot{\alpha}}\,.$
(29)

To get the spinors in any other frame with momenta $p_{i,j}=L_{p}k_{i,j}$
(also $p^{\flat\pm}_{ij}=L_{p}k^{\flat\pm}_{ij}$), we simply act on them with
the Lorentz transformation $L_{p}$ in the spinor representation, denoted by
$\mathcal{L}_{p},\,\tilde{\mathcal{L}}_{p}$. We define

$\displaystyle\ket{p^{\flat\pm}_{ij}}_{\alpha}$ $\displaystyle=$
$\displaystyle\left(\mathcal{L}_{p}\right)^{\leavevmode\nobreak\
\beta}_{\alpha}\,\ket{k^{\flat\pm}_{ij}}_{\beta}$
$\displaystyle\left[p^{\flat\pm}_{ij}\right|_{\dot{\alpha}}$ $\displaystyle=$
$\displaystyle\left[k^{\flat\pm}_{ij}\right|_{\dot{\beta}}\,\left(\tilde{\mathcal{L}}_{p}\right)^{\dot{\beta}}_{\leavevmode\nobreak\
\dot{\alpha}}.$ (30)

This guarantees the relation

$\displaystyle
p^{\flat\pm}_{ij}\cdot\sigma_{\alpha{\dot{\alpha}}}=\ket{p^{\flat\pm}_{ij}}_{\alpha}\left[p^{\flat\pm}_{ij}\right|_{\dot{\alpha}}\,.$
(31)

In Csaki:2020inw it was shown that the pairwise spinors
$\ket{p^{\flat\pm}_{ij}},\,\left|p^{\flat\pm}_{ij}\right]$ defined above
transform with the correct pairwise little group phase

$\displaystyle\Lambda^{\leavevmode\nobreak\
\beta}_{\alpha}\ket{p^{\flat\pm}_{ij}}_{\beta}$ $\displaystyle=$
$\displaystyle e^{\pm\frac{i}{2}\phi(p_{i},p_{j},\Lambda)}\,\ket{\Lambda
p^{\flat\pm}_{ij}}_{\alpha}$
$\displaystyle\left[p^{\flat\pm}_{ij}\right|_{\dot{\beta}}\,\tilde{\Lambda}^{\dot{\beta}}_{\leavevmode\nobreak\
\dot{\alpha}}$ $\displaystyle=$ $\displaystyle
e^{\mp\frac{i}{2}\phi(p_{i},p_{j},\Lambda)}\,\left[\Lambda
p^{\flat\pm}_{ij}\right|_{\dot{\alpha}}.$

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