# Time-dependent tug-of-war games and normalized parabolic $p$-Laplace
equations

Jeongmin Han  Department of Mathematical Sciences, Seoul National University
1, Gwanak-ro, Gwanak-gu, Seoul, Republic of Korea<EMAIL_ADDRESS>
###### Abstract.

This paper concerns value functions of time-dependent tug-of-war games. We
first prove the existence and uniqueness of value functions and verify that
these game values satisfy a dynamic programming principle. Using the arguments
in the proof of existence of game values, we can also deduce asymptotic
behavior of game values when $T\to\infty$. Furthermore, we investigate
boundary regularity for game values. Thereafter, based on the regularity
results for value functions, we deduce that game values converge to viscosity
solutions of the normalized parabolic $p$-Laplace equation.

###### Key words and phrases:

Dynamic programming principle, Normalized $p$-Laplacian, Stochastic games,
Tug-of-war, Viscosity solutions.

###### 2010 Mathematics Subject Classification:

Primary: 35K92; Secondary: 35K20, 91A15.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 Notations
    2. 2.2 Basic concepts
  3. 3 The existence and uniqueness of game values
  4. 4 Long-time asymptotics
  5. 5 Regularity near the boundary
  6. 6 Application to PDEs

##  1\. Introduction

Attempts to understand partial differential equations in relation to
stochastic processes have been developed for several decades. These approaches
give us different perspectives on studying PDEs. For instance, we can derive
Laplace equation from standard Brownian motion and its transition semigroup
(see, for example, [SP14]). This is a simple and classic example, but similar
approaches have been suggested for more general equations as well. Here we
consider a generalization of the Laplace equation. The normalized version of
the $p$-Laplace operator can be written as

$\Delta_{p}^{N}u:=\Delta u+(p-2)\Delta_{\infty}^{N}u=\Delta
u+(p-2)\frac{\langle D^{2}uDu,Du\rangle}{|Du|^{2}}.$

To deal with this nonlinear operator, we need to employ another diffusion
process which is called a tug-of-war game. This stochastic game is a control
interpretation associated with the $\infty$-Laplacian. By adding noise to the
tug-of-war game, we can adopt a stochastic view for the normalized $p$-Laplace
equation.

In this paper, we study value functions of time-dependent tug-of-war games
with noise. In particular, we investigate several properties of game values
such as regularity and long-time asymptotics. Moreover, we also present
uniform convergence of value functions to viscosity solutions of the
normalized parabolic $p$-Laplace equation with $1<p<\infty$ as the step size
of game goes to zero.

Our study is the parabolic counterpart of [Har16]. In that paper, the author
proved the existence, uniqueness and continuity of value functions for time-
independent tug-of-war games. We extend these results for game values to the
case of time-dependent games and show that our value functions pointwisely
converges to game values in [Har16] as $T\to\infty$. On the other hand, we
also establish uniform convergence of game values as $\epsilon\to 0$. For this
purpose, we need suitable regularity results. We already showed interior
regularity estimates for game values in [Han20]. In this paper, we also give
boundary regularity results for value functions. Actually, the main difficulty
occurs in this part. Under the settings of [MPR12], one can obtain the
boundary estimates of value functions by using the exit time of the noise-only
process. Unfortunately, the noise in our settings depends on strategies of
each player. Thus, we establish the desired estimates using another
appropriate stochastic game.

To derive our desired results, we employ the following DPP

$\displaystyle\begin{split}&u_{\epsilon}(x,t)\\\
&=\frac{1-\delta(x,t)}{2}\times\\\ &\bigg{[}\sup_{\nu\in
S^{n-1}}\bigg{\\{}\alpha
u_{\epsilon}\bigg{(}x+\epsilon\nu,t-\frac{\epsilon^{2}}{2}\bigg{)}+\beta\mathchoice{\mathop{\kern
1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}u_{\epsilon}\bigg{(}x+h,t-\frac{\epsilon^{2}}{2}\bigg{)}d\mathcal{L}^{n-1}(h)\bigg{\\}}\\\
&+\inf_{\nu\in S^{n-1}}\bigg{\\{}\alpha
u_{\epsilon}\bigg{(}x+\epsilon\nu,t-\frac{\epsilon^{2}}{2}\bigg{)}+\beta\mathchoice{\mathop{\kern
1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}u_{\epsilon}\bigg{(}x+h,t-\frac{\epsilon^{2}}{2}\bigg{)}d\mathcal{L}^{n-1}(h)\bigg{\\}}\bigg{]}\\\
&+\delta(x,t)F(x,t),\end{split}$ (1.1)

where $0<\alpha,\beta<1$ with $\alpha+\beta=1$. Note that $S^{n-1}$ is the
$n$-dimensional unit sphere centered at the origin, $B_{\epsilon}^{\nu}$ is an
$(n-1)$-dimensional $\epsilon$-ball which is centered at the origin and
orthogonal to a unit vector $\nu$, $\mathcal{L}^{n-1}$ is the
$(n-1)$-dimensional Lebesgue measure, $\delta$ is a function to be defined in
the next section. Moreover,

$\mathchoice{\mathop{\kern 1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptA}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{A}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{A}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{A}}f(h)d\mathcal{L}^{n-1}(h):=\frac{1}{\mathcal{L}^{n-1}(A)}\int_{A}f(h)d\mathcal{L}^{n-1}(h)$

for any $\mathcal{L}^{n-1}$-measurable functions $f$. In Section 3, we verify
that our value functions actually satisfy (LABEL:dppvar). Once we know the
relation between game values and (LABEL:dppvar), then we can concentrate on
investigating the properties of functions satisfying this DPP. By the Taylor
expansion, we can expect that if $u$ is a limit of game values $u_{\epsilon}$
as $\epsilon\to 0$, then it satisfies

$(n+p)u_{t}=\Delta_{p}^{N}u.$

This observation provides motivation for the discussion in the last section.

The $p$-Laplace operator was first studied using tug-of-war games in [PSSW09,
PS08]. Over the past decade, considerable progress has been made in the theory
of value functions for tug-of-war games. Several mean value characterizations
for the $p$-Laplace operator are derived in [MPR10b, MPR10a, KMP12]. Time-
independent games have been studied in [MPR12, LPS13, Ruo16, AHP17, LP18,
ALPR20]. For time-dependent games, see also [MPR10a, DPR14, PR16, LP18,
Han20]. We also refer the reader to [BCF12, APSS12, FLM14, MRS15, LM17] which
deal with tug-of-war games under various settings.

Acknowledgments. This work was supported by NRF-2019R1C1C1003844. The author
would like to thank M. Parviainen, for introducing this topic, valuable
discussions and constant support throughout this work.

##  2\. Preliminaries

###  2.1. Notations

We still use the notation $B_{\epsilon}^{\nu}$ and $S^{n-1}$ as in Section 1.
Let $\Omega$ be a bounded domain. First we define

$I_{\epsilon}=\\{x\in\Omega:\operatorname{dist}(x,\partial\Omega)<\epsilon\\}\
\textrm{and}$
$O_{\epsilon}=\\{x\in\mathbb{R}^{n}\backslash\overline{\Omega}:\operatorname{dist}(x,\partial\Omega)<\epsilon\\}.$

We also set $\Gamma_{\epsilon}=I_{\epsilon}\cup
O_{\epsilon}\cup\partial\Omega$ and $\Omega_{\epsilon}=\overline{\Omega}\cup
O_{\epsilon}.$

For $T>0$, consider a parabolic cylinder $\Omega_{T}=\Omega\times(0,T]$ with
its parabolic boundary
$\partial_{p}\Omega_{T}=(\overline{\Omega}\times\\{0\\})\cup(\partial\Omega\times(0,T])$.
Similarly to the elliptic case, we define parabolic $\epsilon$-strips

$I_{\epsilon,T}=\\{(x,t)\in\Omega\times[\epsilon^{2}/2,T]:\operatorname{dist}(x,\partial\Omega)<\epsilon\\}\cup\big{(}\Omega\times(0,\epsilon^{2}/2)\big{)},$
$O_{\epsilon,T}=\\{(x,t)\in(\mathbb{R}^{n}\backslash\overline{\Omega})\times(0,T]:\operatorname{dist}(x,\partial\Omega)<\epsilon\\}\cup\big{(}\Omega_{\epsilon}\times(-\epsilon^{2}/2,0)\big{)}$

and

$\Gamma_{\epsilon,T}=I_{\epsilon,T}\cup
O_{\epsilon,T}\cup\partial_{p}\Omega_{T}.$

We denote by $\Omega_{\epsilon,T}$ the set $\overline{\Omega}_{T}\cup
O_{\epsilon,T}$.

For $(x,t)\in\Omega_{T}$, we set a “regularizing function” $\delta$ in
(LABEL:dppvar) as follows:

$\displaystyle\delta(x,t)=\left\\{\begin{array}[]{ll}0&\textrm{in
$\Omega_{T}\backslash I_{\epsilon,T},$}\\\
\min\bigg{\\{}1,1-\frac{\operatorname{dist}(x,\partial\Omega)}{\epsilon}\bigg{\\}}\times\min\bigg{\\{}1,1-\frac{\sqrt{2t}}{\epsilon}\bigg{\\}}&\textrm{in
$I_{\epsilon,T}$, }\\\ 1&\textrm{in $O_{\epsilon,T}$.}\\\ \end{array}\right.$

It is not difficult to check that $\delta$ is continuous in
$\Omega_{\epsilon,T}$.

We introduce some notations for convenience. First, we write

$\operatornamewithlimits{midrange}_{i\in I}A_{i}=\frac{1}{2}\bigg{(}\sup_{i\in
I}A_{i}+\inf_{i\in I}A_{i}\bigg{)}.$

And, we also denote by

$\mathscr{A}_{\epsilon}u(x,t;\nu)=\alpha
u(x+\epsilon\nu,t)+\beta\mathchoice{\mathop{\kern 1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}u(x+h,t)d\mathcal{L}^{n-1}(h)$

for a bounded measurable function $u$. Then (LABEL:dppvar) can be written as

$\displaystyle
u_{\epsilon}(x,t)=(1-\delta(x,t))\operatornamewithlimits{midrange}_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}u_{\epsilon}\bigg{(}x,t-\frac{\epsilon^{2}}{2};\nu\bigg{)}+\delta(x,t)F(x,t).$

We call this bounded and measurable function $u$ a _solution_ to the DPP
(LABEL:dppvar).

###  2.2. Basic concepts

Let $\Omega\subset\mathbb{R}^{n}$ be a bounded domain, $T>0$ and
$\alpha,\beta\in(0,1)$ be fixed numbers with $\alpha+\beta=1$. We also
consider a function $F\in C(\Gamma_{\epsilon,T})$. From now on, we will use
the symbol $u_{\epsilon}$ to denote a function satisfying the DPP
(LABEL:dppvar) in $\Omega_{T}$ for given $F$.

We consider two-player tug-of-war games related to (LABEL:dppvar). There are
various settings for these games. In particular, our setting can be regarded
as a parabolic version of games in [AHP17].

Our game setting is as follows. There is a token located at a point
$(x_{0},t_{0})\in\Omega_{T}$. Players will move it at each turn according to
the outcome of the following processes. We write locations of the token as
$(x_{1},t_{1}),(x_{2},t_{2}),\cdots$ and denote by $Z_{j}=(x_{j},t_{j})$ for
our convenience.

When $Z_{j}\in\Omega\backslash I_{\epsilon}$, Player I and II choose some
vectors $\nu_{j}^{\mathrm{I}},\nu_{j}^{\mathrm{II}}\in\partial B_{\epsilon}$.
First, players compete to move token with a fair coin toss. Next, they have
one more stochastic process to determine how to move the token. The winner of
first coin toss, Player $i\in\\{\mathrm{I},\mathrm{II}\\}$ can move the token
to direction of the chosen vector $\nu_{j}^{i}$ with probability $\alpha$.
Otherwise, the token is moved uniformly random in the $(n-1)$-ball
perpendicular to $\nu_{j}^{i}$. After these processes are finished, $t_{j}$ is
changed by $t_{j+1}=t_{j}-\epsilon^{2}/2$.

If $Z_{j}\in\Gamma_{\epsilon}$, the game progresses in the same way as above
with probability $1-\delta(Z_{j})$. On the other hand, with probability
$\delta(Z_{j})$, the game is over and Player II pays Player I payoff
$F(Z_{j})$.

We denote by $\tau$ the number of total turns until end of the game. Observe
that $\tau$ must be finite in our setting since the game ends when $t\leq 0$.

Now we give mathematical construction for this game. Let
$\xi_{0},\xi_{1},\cdots$ be iid random variables to have a uniform
distribution $U(0,1)$. This process $\\{\xi_{j}\\}_{j=0}^{\infty}$ is
independent of $\\{Z_{j}\\}_{j=0}^{\infty}$.

Define $\tilde{C}:=\\{0,1\\}$. We set random variables
$c_{0},c_{1},\cdots\in\tilde{C}$ as follows:

$\displaystyle c_{j}=\left\\{\begin{array}[]{ll}0&\textrm{when $\xi_{j-1}\leq
1-\delta(Z_{j-1})$,}\\\ 1&\textrm{when $\xi_{j-1}>1-\delta(Z_{j-1})$}\\\
\end{array}\right.$

for $j\geq 1$ and $c_{0}=0$. Then we can write the stopping time $\tau$ by

$\tau:=\inf\\{j\geq 0:c_{j+1}=1\\}.$

In our game, each player chooses their strategies by using past data
(history). We can write a history as the following vector

$\big{(}(c_{0},Z_{0}),(c_{1},Z_{1}),\cdots,(c_{j},Z_{j})\big{)}.$

Then, the strategy of Player $i$ can be defined by a Borel measurable function
as $\mathcal{S}_{i}=\\{S_{i}^{j}\\}_{j=1}^{\infty}$ with

$S_{i}^{j}:\\{(c_{0},Z_{0})\\}\times\cup_{k=1}^{j-1}(\tilde{C}\times\Omega_{\epsilon,T})\to\partial
B_{\epsilon}(0)$

for any $j\in\mathbb{N}$.

Now we define a probability measure
$\mathbb{P}_{S_{\mathrm{I}},S_{\mathrm{II}}}^{Z_{0}}$ natural product
$\sigma$-algebra of the space of all game trajectories for any starting point
$Z_{0}\in\Omega_{\epsilon,T}$. By Kolmogorov’s extension theorem, we can
construct the measure to the family of transition densities

$\displaystyle\pi_{S_{\mathrm{I}},S_{\mathrm{II}}}((c_{0},Z_{0}),(c_{1},Z_{1}),\cdots,(c_{j},Z_{j});C,A_{j+1})$
$\displaystyle=(1-\delta(Z_{j}))\pi_{S_{\mathrm{I}},S_{\mathrm{II}}}^{local}((Z_{0},Z_{1},\cdots,Z_{j});A_{j+1})\mathbb{I}_{0}(C)\mathbb{I}_{c_{j}}(\\{0\\})$
$\displaystyle\quad+\delta(Z_{j})\mathbb{I}_{Z_{j}}(A_{j})\mathbb{I}_{1}(C)\mathbb{I}_{c_{j}}(\\{0\\})+\mathbb{I}_{Z_{j}}(A_{j})\mathbb{I}_{c_{j}}(\\{1\\})$

for $A_{n}=A\times\\{t_{n}\\}$ ($A$ is any Borel set in $\mathbb{R}^{n}$ and
$n\geq 0$) and $C\subset\tilde{C}$, where

$\displaystyle\pi_{S_{\mathrm{I}},S_{\mathrm{II}}}^{local}(Z_{0},Z_{1},\cdots,Z_{j};A_{j+1})$
$\displaystyle=\frac{1}{2}\bigg{[}\alpha(\mathbb{I}_{(x_{j}+\nu_{j+1}^{I},t_{j+1})}(A_{j+1})+\mathbb{\mathrm{I}}_{(x_{j}+\nu_{j+1}^{\mathrm{II}},t_{j+1})}(A_{j+1})\big{)}$
$\displaystyle\qquad+\frac{\beta}{\omega_{n-1}\epsilon^{n-1}}\big{(}\mathcal{L}^{n-1}(B_{\epsilon}^{\nu_{j+1}^{\mathrm{I}}}(Z_{j})\cap
A_{j+1})+\mathcal{L}^{n-1}(B_{\epsilon}^{\nu_{j+1}^{\mathrm{II}}}(Z_{j})\cap
A_{j+1}\big{)}\bigg{]}.$

Here, $\omega_{n-1}=\mathcal{L}^{n-1}(B_{1}^{n-1})$ where $B_{1}^{n-1}$ is the
$(n-1)$-dimensional unit ball and

$\displaystyle\mathbb{I}_{z}(B)=\left\\{\begin{array}[]{ll}0&\textrm{when
$z\notin B$,}\\\ 1&\textrm{when $z\in B$.}\\\ \end{array}\right.$

Finally, for any starting point $Z_{0}=(x_{0},t_{0})\in\Omega_{T}$, we define
value functions $u_{\mathrm{I}}$ and $u_{\mathrm{II}}$ of this game for Player
I and II by

$u_{\mathrm{I}}(Z_{0})=\sup_{S_{\mathrm{I}}}\inf_{S_{\mathrm{II}}}\mathbb{E}_{S_{\mathrm{I}},S_{\mathrm{II}}}^{Z_{0}}[F(Z_{\tau})]$

and

$u_{\mathrm{II}}(Z_{0})=\inf_{S_{\mathrm{II}}}\sup_{S_{\mathrm{I}}}\mathbb{E}_{S_{\mathrm{I}},S_{\mathrm{II}}}^{Z_{0}}[F(Z_{\tau})],$

respectively.

##  3\. The existence and uniqueness of game values

In this section, we study the existence and uniqueness of functions satisfying
the DPP (LABEL:dppvar) with continuous boundary data $F$. Moreover, we also
observe the relation between these functions and value functions for time-
dependent tug-of-war games.

Before showing the existence and uniqueness of functions satisfying
(LABEL:dppvar), we need to check a subtle issue. In the DPP, the value of
$u_{\epsilon}(x,t)$ is determined by values of the function in
$B_{\epsilon}(x)\times\\{t-\epsilon^{2}/2\\}$. And we also see that
(LABEL:dppvar) contains integral terms for the function at time
$t-\epsilon^{2}/2$. Thus, we have to consider the measurability for the
function $u_{\epsilon}$, more precisely, for strategies of our game.

In general, existence of measurable strategies is not guaranteed (for example,
see [LPS14, Example 2.4]). But we can avoid this problem under our setting.
The “regularizing function” $\delta$ plays an important role in this issue.

We begin this section by observing a basic property of the operator
$\mathscr{A}_{\epsilon}$.

######  Proposition 3.1.

Let $u\in C(\overline{\Omega}_{\epsilon,T})$. Then
$\mathscr{A}_{\epsilon}u(x,t;\nu)$ is continuous with respect to each variable
in $\overline{\Omega}_{T,\epsilon}\times\partial B_{\epsilon}(0)$.

###### Proof.

For any $(x,t),(y,s)\in\Omega_{T}$, let us define a parabolic distance by
$d((x,t),(y,s))=|x-y|+|t-s|^{1/2}$. We write the modulus of continuity of a
function $f$ with respect to the distance $d$ by $\omega_{f}$.

For fixed $|\nu|=\epsilon$, we can see that for any $x,y\in\overline{\Omega}$,

$\displaystyle\bigg{|}\alpha
u\bigg{(}x+\epsilon\nu,t-\frac{\epsilon^{2}}{2}\bigg{)}-\alpha
u\bigg{(}y+\epsilon\nu,t-\frac{\epsilon^{2}}{2}\bigg{)}\bigg{|}\leq\alpha\omega_{u}(|x-y|)$

and

$\displaystyle\bigg{|}\beta$ $\displaystyle\mathchoice{\mathop{\kern
1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}u\bigg{(}x+h,t-\frac{\epsilon^{2}}{2}\bigg{)}d\mathcal{L}^{n-1}(h)-\beta\mathchoice{\mathop{\kern
1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}u\bigg{(}y+h,t-\frac{\epsilon^{2}}{2}\bigg{)}d\mathcal{L}^{n-1}(h)\bigg{|}$
$\displaystyle\leq\beta\mathchoice{\mathop{\kern 1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}\bigg{|}u\bigg{(}x+h,t-\frac{\epsilon^{2}}{2}\bigg{)}-u\bigg{(}y+h,t-\frac{\epsilon^{2}}{2}\bigg{)}\bigg{|}d\mathcal{L}^{n-1}(h)$
$\displaystyle\leq\beta\omega_{u}(|x-y|).$

Thus, we get

$|\mathscr{A}_{\epsilon}u(x,t;\nu)-\mathscr{A}_{\epsilon}u(y,t;\nu)|\leq\omega_{u}(|x-y|).$

Next, for any $t,s>0$, we also calculate that

$\displaystyle\bigg{|}\alpha
u\bigg{(}x+\epsilon\nu,t-\frac{\epsilon^{2}}{2}\bigg{)}-\alpha
u\bigg{(}x+\epsilon\nu,s-\frac{\epsilon^{2}}{2}\bigg{)}\bigg{|}\leq\alpha\omega_{u}(|t-s|^{1/2}),$
$\displaystyle\bigg{|}\beta$ $\displaystyle\mathchoice{\mathop{\kern
1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}u\bigg{(}x+h,t-\frac{\epsilon^{2}}{2}\bigg{)}d\mathcal{L}^{n-1}(h)-\beta\mathchoice{\mathop{\kern
1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}u\bigg{(}x+h,s-\frac{\epsilon^{2}}{2}\bigg{)}d\mathcal{L}^{n-1}(h)\bigg{|}$
$\displaystyle\leq\beta\mathchoice{\mathop{\kern 1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}\bigg{|}u\bigg{(}x+h,t-\frac{\epsilon^{2}}{2}\bigg{)}-u\bigg{(}x+h,s-\frac{\epsilon^{2}}{2}\bigg{)}\bigg{|}d\mathcal{L}^{n-1}(h)$
$\displaystyle\leq\beta\omega_{u}(|t-s|^{1/2})$

and hence

$|\mathscr{A}_{\epsilon}u(x,t;\nu)-\mathscr{A}_{\epsilon}u(x,s;\nu)|\leq\omega_{u}(|t-s|^{1/2}).$

Finally, for any $\nu,\chi\in S^{n-1}$,

$\displaystyle W(x,t,\nu)-W(x,t,\chi)$
$\displaystyle=\alpha\bigg{[}u\bigg{(}x+\epsilon\nu,t-\frac{\epsilon^{2}}{2}\bigg{)}-u\bigg{(}x+\epsilon\chi,t-\frac{\epsilon^{2}}{2}\bigg{)}\bigg{]}$
$\displaystyle+\beta\bigg{[}\mathchoice{\mathop{\kern 1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}u\bigg{(}x+h,t-\frac{\epsilon^{2}}{2}\bigg{)}d\mathcal{L}^{n-1}(h)-\mathchoice{\mathop{\kern
1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\chi}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\chi}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\chi}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\chi}}}u\bigg{(}x+h,t-\frac{\epsilon^{2}}{2}\bigg{)}d\mathcal{L}^{n-1}(h)\bigg{]}.$

Combining the above results, we see that

$\displaystyle|W(x,t,\nu)-W(x,t,\chi)|$
$\displaystyle\leq\alpha\epsilon\omega_{u}(\epsilon|\nu-\chi|)$
$\displaystyle+\beta\mathchoice{\mathop{\kern 1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}\bigg{|}u\bigg{(}x+h,t-\frac{\epsilon^{2}}{2}\bigg{)}-u\bigg{(}x+Ph,t-\frac{\epsilon^{2}}{2}\bigg{)}\bigg{|}d\mathcal{L}^{n-1}(h)$

where $P:\nu^{\perp}\to\chi^{\perp}$ is a rotation satisfying $|h-Ph|\leq
C|h||\nu-\chi|$. Here, we check that

$\displaystyle\mathchoice{\mathop{\kern 1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}\bigg{|}u\bigg{(}x+h,t-\frac{\epsilon^{2}}{2}\bigg{)}-u\bigg{(}x+Ph,t-\frac{\epsilon^{2}}{2}\bigg{)}\bigg{|}d\mathcal{L}^{n-1}(h)$
$\displaystyle\leq\omega_{u}(|h-Ph|)$
$\displaystyle\leq\omega_{u}(C\epsilon|\nu-\chi|).$

Therefore, we obtain

$|\mathscr{A}_{\epsilon}u_{\epsilon}(x,t;\nu)-\mathscr{A}_{\epsilon}u_{\epsilon}(x,t;\chi)|\leq\omega_{u}(C\epsilon|\nu-\chi|).$

Now we can conclude the proof to combine above results. ∎

Next we observe that the operator $T$ preserves continuity and monotonicity.
For convenience, we write that

$\displaystyle
Tu(x,t)=(1-\delta(x,t))\operatornamewithlimits{midrange}_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}u\bigg{(}x,t-\frac{\epsilon^{2}}{2};\nu\bigg{)}+\delta(x,t)F(x,t).$
(3.1)

######  Lemma 3.2.

For any $u\in C(\overline{\Omega}_{\epsilon,T})$, $Tu$ is also in
$C(\overline{\Omega}_{\epsilon,T})$. Furthermore, for any $u,v\in
C(\overline{\Omega}_{\epsilon,T})$ with $u\leq v$, it holds that

$Tu\leq Tv.$

###### Proof.

By the definition of $T$, we can check that $u\leq v$ implies $Tu\leq Tv$
without difficulty.

Next we need to show that $Tu\in C(\overline{\Omega}_{\epsilon,T})$ if $u\in
C(\overline{\Omega}_{\epsilon,T})$. When $(x,t)\in\overline{O}_{\epsilon,T}$,
we see that $Tu=u=F\in C(\overline{O}_{\epsilon,T})$ by assumption. We need to
consider the case of $\overline{I}_{\epsilon,T}$ and $\Omega_{T}\backslash
I_{\epsilon,T}$.

First assume that $(x,t),(y,s)\in\Omega_{T}\backslash I_{\epsilon,T}$. Observe
that

$\displaystyle\big{|}\operatornamewithlimits{midrange}_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}u(x,t;\nu)-\operatornamewithlimits{midrange}_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}u(y,s;\nu)\big{|}$
$\displaystyle\leq\frac{1}{2}\big{|}\sup_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}u(x,t;\nu)-\sup_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}u(y,s;\nu)\big{|}$
$\displaystyle\qquad+\frac{1}{2}\big{|}\inf_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}u(x,t;\nu)-\inf_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}u(y,s;\nu)\big{|}.$

Since

$\big{|}\sup_{\nu\in S^{n-1}}\mathscr{A}_{\epsilon}u(x,t;\nu)-\sup_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}u(y,s;\nu)\big{|}\leq\sup_{\nu\in
S^{n-1}}|\mathscr{A}_{\epsilon}u(x,t;\nu)-\mathscr{A}_{\epsilon}u(y,s;\nu)|$

and

$\big{|}\inf_{\nu\in S^{n-1}}\mathscr{A}_{\epsilon}u(x,t;\nu)-\inf_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}u(y,s;\nu)\big{|}\leq\sup_{\nu\in
S^{n-1}}|\mathscr{A}_{\epsilon}u(x,t;\nu)-\mathscr{A}_{\epsilon}u(y,s;\nu)|,$

we get

$\displaystyle\big{|}\operatornamewithlimits{midrange}_{\nu\in S^{n-1}}$
$\displaystyle\mathscr{A}_{\epsilon}u(x,t;\nu)-\operatornamewithlimits{midrange}_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}u(y,s;\nu)\big{|}$
$\displaystyle\leq\sup_{\nu\in
S^{n-1}}|\mathscr{A}_{\epsilon}u(x,t;\nu)-\mathscr{A}_{\epsilon}u(y,s;\nu)|$
$\displaystyle\leq\omega_{u}(d((x,t),(y,s))).$

We used the result of Proposition 3.1 in the last inequality. Thus, $Tu$ is
also continuous in $\Omega_{T}\backslash I_{\epsilon,T}$.

When $(x,t),(y,s)\in\overline{I}_{\epsilon,T}$,

$\displaystyle\big{|}(1-$
$\displaystyle\delta(x,t))\operatornamewithlimits{midrange}_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}u(x,t;\nu)-(1-\delta(y,s))\operatornamewithlimits{midrange}_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}u(y,s;\nu)\big{|}$
$\displaystyle\leq(1-\delta(x,t))\big{|}\operatornamewithlimits{midrange}_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}u(x,t;\nu)-\operatornamewithlimits{midrange}_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}u(y,s;\nu)\big{|}$
$\displaystyle\qquad+|\delta(x,t)-\delta(y,s)\big{|}\cdot\big{|}\operatornamewithlimits{midrange}_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}u(y,s;\nu)\big{|}$
$\displaystyle\leq\omega_{u}(d((x,t),(y,s)))+\frac{3}{\epsilon}||u||_{\infty}d((x,t),(y,s))$

because

$\displaystyle|\delta(x,t)-\delta(y,s)|\leq\frac{3}{\epsilon}d((x,t),(y,s)).$

Similarly, we can also calculate

$\displaystyle|\delta(x,t)F(x,t)-\delta(y,s)F(y,s)|$
$\displaystyle\leq\omega_{F}(d((x,t),(y,s)))+\frac{3}{\epsilon}||F||_{\infty}d((x,t),(y,s)).$

Combining above results, we obtain the continuity of $Tu$ in $I_{\epsilon,T}$.

Finally, we need to check the coincidence of the function value on $\partial
I_{\epsilon,T}$. Observe that $\partial I_{\epsilon,T}$ can be decomposed by
two disjoint connected sets $\partial_{p}\Omega_{T}$ and
$\partial_{p}(\Omega_{T}\backslash I_{\epsilon,T})$ of
$\mathbb{R}^{n}\times\mathbb{R}$. Then we can observe that

$\lim_{O_{\epsilon,T}\ni(y,s)\to(x,t)}Tu(y,s)=\lim_{I_{\epsilon,T}\ni(y,s)\to(x,t)}Tu(y,s)=Tu(x,t)$

for any $(x,t)\in\partial_{p}\Omega_{T}$ and

$\lim_{\Omega_{T}\backslash
I_{\epsilon,T}\ni(y,s)\to(x,t)}Tu(y,s)=\lim_{I_{\epsilon,T}\ni(y,s)\to(x,t)}Tu(y,s)=Tu(x,t)$

for any $(x,t)\in\partial_{p}(\Omega_{T}\backslash I_{\epsilon,T})$ by using
the above calculation. Thus we obtain the continuity of $Tu$ and the proof is
finished. ∎

Since $T$ preserves continuity, we do not need to worry about the
measurability issue. Therefore, for any continuous function $u$, $Tu$ is well-
defined at every point in $\Omega_{T}$.

Now we can obtain the existence and uniqueness of these functions.

######  Theorem 3.3.

Let $F\in C(\Gamma_{\epsilon,T})$. Then the bounded function $u_{\epsilon}$
satisfying (LABEL:dppvar) with boundary data $F$ exists and is unique.

###### Proof.

We get the desired result via an argument similar to the proof of [LPS14,
Theorem 5.2]. We can see the existence of these functions without difficulty
since the operator $T$ is well-defined inductively for any continuous boundary
data.

For uniqueness, consider two functions $u$ and $v$ satisfying $Tu=u$, $Tv=v$
with boundary data $F$. We see that $u(\cdot,t)=v(\cdot,t)$ when
$0<t\leq\epsilon^{2}/2$ by definition of $T$. Then we can also get the same
result when $\epsilon^{2}/2<t\leq\epsilon^{2}$ because past data of $u$ and
$v$ still coincide. Repeating this process, we obtain $u(x,t)=v(x,t)$ for any
$(x,t)\in\Omega_{T}$ and hence the uniqueness is proved. ∎

We look into the relation between functions satisfying (LABEL:dppvar) and
values for parabolic tug-of-war games here.

######  Theorem 3.4.

The value functions of tug-of-war game with noise $u_{\mathrm{I}}$ and
$u_{\mathrm{II}}$ with payoff function $F$ coincide with the function
$u_{\epsilon}$.

###### Proof.

We need to deduce that

$\displaystyle u_{\epsilon}\leq u_{\mathrm{I}}\qquad\textrm{and}\qquad
u_{\mathrm{II}}\leq u_{\epsilon}$

since $u_{\mathrm{I}}\leq u_{\mathrm{II}}$ by the definition of value
functions.

First, we show the latter inequality. Let $Z_{0}\in\Omega_{T}$ and denote by
$S_{\mathrm{II}}^{0}$ a strategy for Player II such that

$\mathscr{A}_{\epsilon}u_{\epsilon}(Z_{j};\nu_{j}^{\mathrm{II}})=\inf_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}u_{\epsilon}(Z_{j};\nu)$

for $j\geq 0$. Note that this $S_{\mathrm{II}}^{0}$ exists since
$\mathscr{A}_{\epsilon}u_{\epsilon}$ is continuous on $\nu$ by Proposition
3.1. Measurability of such strategies can be shown by using [Sri98, Theorem
5.3.1].

Next we fix an arbitrary strategy $S_{I}$ for Player $I$. Define

$\displaystyle\Phi(c,x,t)=\left\\{\begin{array}[]{ll}u_{\epsilon}(x,t)&\textrm{when
$c=0$,}\\\ F(x,t)&\textrm{when $c=1$.}\\\ \end{array}\right.$

for any $(x,t)\in\Omega_{\epsilon,T}$. Then we have

$\displaystyle\mathbb{E}_{S_{\mathrm{I}},S_{\mathrm{II}}^{0}}^{Z_{0}}[\Phi(c_{j+1},Z_{j+1})|(c_{0},Z_{0}),\cdots,(c_{j},Z_{j})]$
$\displaystyle\leq\frac{1-\delta(Z_{j})}{2}\big{[}\mathscr{A}_{\epsilon}u_{\epsilon}(x_{j},t_{j+1};\nu_{j+1}^{\mathrm{I}})+\mathscr{A}_{\epsilon}u_{\epsilon}(x_{j},t_{j+1};\nu_{j+1}^{\mathrm{II}})\big{]}+\delta(Z_{j})F(Z_{j})$
$\displaystyle\leq(1-\delta(Z_{j}))\operatornamewithlimits{midrange}_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}u_{\epsilon}(x_{j},t_{j+1};\nu)+\delta(Z_{j})F(Z_{j})$
$\displaystyle=\Phi(c_{j},Z_{j}).$

Hence, we can see that $M_{k}=\Phi(c_{k},Z_{k})$ is a supermartingale in this
case. Since the game ends in finite steps, we can obtain

$\displaystyle u_{\mathrm{II}}(Z_{0})$
$\displaystyle=\inf_{S_{\mathrm{II}}}\sup_{S_{\mathrm{I}}}\mathbb{E}_{S_{\mathrm{I}},S_{\mathrm{II}}}^{Z_{0}}[F(Z_{\tau})]\leq\sup_{S_{\mathrm{I}}}\mathbb{E}_{S_{\mathrm{I}},S_{\mathrm{II}}^{0}}^{Z_{0}}[F(Z_{\tau})]$
$\displaystyle=\sup_{S_{\mathrm{I}}}\mathbb{E}_{S_{\mathrm{I}},S_{\mathrm{II}}^{0}}^{Z_{0}}[\Phi(c_{\tau+1},Z_{\tau+1})]\leq\sup_{S_{\mathrm{I}}}\mathbb{E}_{S_{\mathrm{I}},S_{\mathrm{II}}^{0}}^{Z_{0}}[\Phi(c_{0},Z_{0})]$
$\displaystyle=u_{\epsilon}(Z_{0})$

by using optional stopping theorem.

Now we can also derive that $u_{\epsilon}\leq u_{\mathrm{I}}$ by using a
similar argument. Then we get the desired result. ∎

##  4\. Long-time asymptotics

In PDE theory, the study of asymptotic behavior of solutions of parabolic
equations as time goes to infinity has drawn a lot of attention. We will have
a similar discussion for our value function $u_{\epsilon}$ when the boundary
data $F$ does not depend on $t$ in
$\Gamma_{\epsilon}\times(\epsilon^{2},\infty)$. The heuristic idea in this
section can be summarized as follows. Assume that we start the game at
$(x_{0},t_{0})$ for sufficiently large $t_{0}$. Then we can expect that the
probability of the game ending in the initial boundary would be close to zero,
that is, the game finishes on the lateral boundary in most cases. Since we
assumed that $F$ is independent of $t$ for $t>\epsilon^{2}$, we may consider
this game as something like a time-independent game with the same boundary
data. Thus, it is reasonable to guess that the value function of the time-
dependent game converges that of the corresponding time-independent game. We
refer the reader to [BER] which contains a detailed discussion of asymptotic
behaviors for value functions of evolution problems. Moreover, long-time
asymptotics for related PDEs can be found in [BC81, Váz03, PV12, PV].

To observe the asymptotic behavior of value functions, we first need to obtain
the following comparison principle. Since it can be shown in a straightforward
manner by using the DPP (LABEL:dppvar), we omit the proof. One can find
similar results in [LPS14, Theorem 5.3].

######  Lemma 4.1.

Let $u$ and $v$ be functions satisfying (LABEL:dppvar) with boundary data
$F_{u}$ and $F_{v}$, respectively. Suppose that $F_{u}\leq F_{v}$ in
$\Gamma_{\epsilon,T}$. Then,

$u\leq v\qquad\textrm{in}\ \Omega_{\epsilon,T}.$

Now we state the main result of this section.

######  Theorem 4.2.

Let $\Omega$ be a bounded domain. Consider functions $\psi\in
C(\Gamma_{\epsilon})$ and $\varphi\in C(\Gamma_{\epsilon,T}\cap\\{t\leq
0\\})$, and define a function $F\in C(\Omega_{\epsilon,T})$ as follows:

$\displaystyle F(x,t)=\left\\{\begin{array}[]{ll}\psi(x)&\textrm{ in
$\Gamma_{\epsilon}\times(\epsilon^{2},T]$,}\\\
\varphi(x,\epsilon^{2}/2)+2t(\psi(x)-\varphi(x,\epsilon^{2}/2))/\epsilon^{2}&\textrm{
in $\Gamma_{\epsilon}\times(\frac{\epsilon^{2}}{2},\epsilon^{2}]$,}\\\
\varphi(x,t)&\textrm{in
$\Omega_{\epsilon}\times[-\frac{\epsilon^{2}}{2},\frac{\epsilon^{2}}{2}]$.}\\\
\end{array}\right.$ (4.4)

Then we have

$\lim_{T\to\infty}u_{\epsilon}(x,T)=U_{\epsilon}(x)$

where $U_{\epsilon}$ is the function satisfying the following DPP

$\displaystyle\begin{split}&U_{\epsilon}(x)\\\
&=(1-\overline{\delta}(x))\operatornamewithlimits{midrange}_{\nu\in
S^{n-1}}\bigg{[}\alpha
U_{\epsilon}(x+\epsilon\nu)+\beta\mathchoice{\mathop{\kern 1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}U_{\epsilon}(x+h)d\mathcal{L}^{n-1}(h)\bigg{]}\\\
&\qquad+\overline{\delta}(x)\psi(x)\end{split}$ (4.5)

in $\Omega_{\epsilon}$ with boundary data $\psi$ where

$\displaystyle\overline{\delta}(x):=\lim_{t\to\infty}\delta(x,t)=\left\\{\begin{array}[]{ll}0&\textrm{in
$\Omega\backslash I_{\epsilon},$}\\\
1-\operatorname{dist}(x,\partial\Omega)/\epsilon&\textrm{in $I_{\epsilon}$,
}\\\ 1&\textrm{in $O_{\epsilon}$.}\\\ \end{array}\right.$

######  Remark 4.3.

We can find the existence and uniqueness of value functions under different
setting in [LPS14], which is related to the normalized $p$-Laplace operator
for $p\geq 2$. In that paper, the existence of measurable strategies is shown
without regularization. Thus, we do not have to consider a “regularized
function” such as
$\varphi(x,\epsilon^{2}/2)+2t(\psi(x)-\varphi(x,\epsilon^{2}/2))/\epsilon^{2}$
in that case. Meanwhile, for the time-independent version of our settings,
results for these issues are shown in [Har16].

###### Proof of Theorem 4.2.

We will set some proper barrier functions $\underline{u},\overline{u}$ such
that

$\underline{u}\leq u_{\epsilon}\leq\overline{u}$

and show the coincidence for the limits of two barrier functions as
$t\to\infty$. In our proof, the uniqueness result for elliptic games is
essential. The motivation of this proof is from [AHP17, Proposition 3.3].

Let $\underline{\varphi},\overline{\varphi}$ be constants defined by

$\underline{\varphi}=\min\\{\inf_{\Gamma_{\epsilon}}\psi,\inf_{\Omega_{\epsilon}}\varphi\\}\
\textrm{and}\
\overline{\varphi}=\max\\{\sup_{\Gamma_{\epsilon}}\psi,\sup_{\Omega_{\epsilon}}\varphi\\},$

respectively. We consider $\underline{u}$, $\overline{u}$ be functions
satisfying (LABEL:dppvar) with boundary data $\underline{F}$ and
$\overline{F}$, where

$\displaystyle\underline{F}(x,t)=\left\\{\begin{array}[]{ll}\psi(x)&\textrm{
in $\Gamma_{\epsilon}\times(\epsilon^{2},T]$ ,}\\\
\underline{\varphi}+2t(\psi(x)-\underline{\varphi})/\epsilon^{2}&\textrm{ in
$\Gamma_{\epsilon}\times(\frac{\epsilon^{2}}{2},\epsilon^{2}]$,}\\\
\underline{\varphi}&\textrm{in
$\Omega_{\epsilon}\times[-\frac{\epsilon^{2}}{2},\frac{\epsilon^{2}}{2}]$,}\\\
\end{array}\right.$

and

$\displaystyle\overline{F}(x,t)=\left\\{\begin{array}[]{ll}\psi(x)&\textrm{ in
$\Gamma_{\epsilon}\times(\epsilon^{2},T]$,}\\\
\overline{\varphi}+2t(\psi(x)-\overline{\varphi})/\epsilon^{2}&\textrm{ in
$\Gamma_{\epsilon}\times(\frac{\epsilon^{2}}{2},\epsilon^{2}]$,}\\\
\overline{\varphi}&\textrm{in
$\Omega_{\epsilon}\times[-\frac{\epsilon^{2}}{2},\frac{\epsilon^{2}}{2}]$,}\\\
\end{array}\right.$

respectively. Note that $\underline{F}$ and $\overline{F}$ are continuous in
$\overline{\Gamma_{\epsilon,T}}$ and have constant initial data.

By Lemma 4.1, we have $\underline{u}\leq u_{\epsilon}\leq\overline{u}$. Thus
it is sufficient to show that
$\lim_{t\to\infty}\underline{u}(\cdot,t),\lim_{t\to\infty}\overline{u}(\cdot,t)$
exist and satisfy the limiting DPP (LABEL:dpplim). First we see that

$||\underline{u}||_{L^{\infty}(\Omega_{\epsilon,T})}\leq||\underline{F}||_{L^{\infty}(\Gamma_{\epsilon,T})}\
\textrm{and}\
||\overline{u}||_{L^{\infty}(\Omega_{\epsilon,T})}\leq||\overline{F}||_{L^{\infty}(\Gamma_{\epsilon,T})}$

by using the DPP of $\underline{u}$ and $\overline{u}$. Thus, these functions
are uniformly bounded.

Next, we prove monotonicity of sequences
$\\{\underline{u}(x,t+j\epsilon^{2}/2)\\}_{j=0}^{\infty}$ and
$\\{\overline{u}(x,t+j\epsilon^{2}/2)\\}_{j=0}^{\infty}$ for any
$(x,t)\in\Omega_{\epsilon}\times(-\epsilon^{2}/2,0]$. Without loss of
generality, we only consider the case $\underline{u}$. Let $(x_{0},t_{0})$ be
a point in $\Omega\times(-\epsilon^{2}/2,0]$ and denote by

$a_{j}=\underline{u}(x_{0},t_{0}+j\epsilon^{2}/2)$

for simplicity. For any $(x_{0},t_{0})\in\Omega\times(-\epsilon^{2}/2,0]$, we
can derive that

$\underline{\psi}=a_{0}=a_{1}\leq a_{2}$

by direct calculation and

$\displaystyle a_{3}$
$\displaystyle=\bigg{(}1-\delta\bigg{(}x_{0},t_{0}+\frac{3\epsilon^{2}}{2}\bigg{)}\bigg{)}\operatornamewithlimits{midrange}_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}u(x_{0},t_{0}+\epsilon^{2};\nu)$
$\displaystyle\qquad+\delta\bigg{(}x_{0},t_{0}+\frac{3\epsilon^{2}}{2}\bigg{)}F\bigg{(}x_{0},t_{0}+\frac{3\epsilon^{2}}{2}\bigg{)}$
$\displaystyle\geq(1-\delta(x_{0},t_{0}+\epsilon^{2}))\operatornamewithlimits{midrange}_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}u\bigg{(}x_{0},t_{0}+\frac{\epsilon^{2}}{2};\nu\bigg{)}$
$\displaystyle\qquad+\delta(x_{0},t_{0}+\epsilon^{2})F(x_{0},t_{0}+\epsilon^{2})=a_{2}$

since $\delta(x_{0},t_{0}+\epsilon^{2})=\delta(x_{0},t_{0}+3\epsilon^{2}/2)$
and $F(x_{0},t_{0}+\epsilon^{2})\leq F(x_{0},t_{0}+3\epsilon^{2}/2)$.

Next, assume that $a_{k}\geq a_{k-1}$ for some $k\geq 4$. Note that
$F(x,t)=\psi(x)$ for $x\in\Gamma_{\epsilon}$ and

$\delta\bigg{(}x_{0},t_{0}+\frac{k\epsilon^{2}}{2}\bigg{)}=\delta\bigg{(}x_{0},t_{0}+\frac{(k-1)\epsilon^{2}}{2}\bigg{)}$

in this case. Then, we see

$\displaystyle a_{k+1}$
$\displaystyle=\bigg{(}1-\delta\bigg{(}x_{0},t_{0}+\frac{(k+1)\epsilon^{2}}{2}\bigg{)}\bigg{)}\operatornamewithlimits{midrange}_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}u\bigg{(}x_{0},t_{0}+\frac{k\epsilon^{2}}{2};\nu\bigg{)}$
$\displaystyle\qquad+\delta\bigg{(}x_{0},t_{0}+\frac{(k+1)\epsilon^{2}}{2}\bigg{)}\psi(x_{0})$
$\displaystyle\geq\bigg{(}1-\delta\bigg{(}x_{0},t_{0}+\frac{k\epsilon^{2}}{2}\bigg{)}\bigg{)}\operatornamewithlimits{midrange}_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}u\bigg{(}x_{0},t_{0}+\frac{(k-1)\epsilon^{2}}{2};\nu\bigg{)}$
$\displaystyle\qquad+\delta\bigg{(}x_{0},t_{0}+\frac{k\epsilon^{2}}{2}\bigg{)}\psi(x_{0})=a_{k}.$

Therefore, $\\{a_{j}\\}$ is increasing for any
$(x_{0},t_{0})\in\Omega\times(-\epsilon^{2}/2,0]$. It is also possible to
obtain $\\{\overline{u}(x,t+j\epsilon^{2}/2)\\}$ is decreasing by using
similar arguments. Therefore, we obtain
$\\{\underline{u}(x,t+j\epsilon^{2}/2)\\}$ and
$\\{\overline{u}(x,t+j\epsilon^{2}/2)\\}$ converges for any
$(x,t)\in\Omega_{\epsilon}\times(-\epsilon^{2}/2,0]$ by applying the monotone
convergence theorem.

Now we show that $U_{\epsilon}$ satisfies the DPP (LABEL:dpplim). Fix
$-\epsilon^{2}/2\leq t_{1}<0$ arbitrary and write

$\underline{U_{t_{1}}}(x)=\lim_{j\to\infty}\underline{u}(x,t_{1}+j\epsilon^{2}/2)$

for $x\in\Omega$. By definition of $\underline{u}$, we see that

$\displaystyle\underline{U_{t_{1}}}(x)=(1-\overline{\delta}(x))\lim_{j\to\infty}\bigg{[}\operatornamewithlimits{midrange}_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}\underline{u}\bigg{(}x,t_{1}+\frac{j\epsilon^{2}}{2};\nu\bigg{)}\bigg{]}+\overline{\delta}(x)\psi(x).$

Therefore, it is sufficient to show that

$\displaystyle\lim_{j\to\infty}\sup_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}\underline{u}\bigg{(}x,t_{1}+\frac{j\epsilon^{2}}{2};\nu\bigg{)}=\sup_{\nu\in
S^{n-1}}\tilde{\mathscr{A}}_{\epsilon}\underline{U_{t_{1}}}(x;\nu)$ (4.6)

and

$\displaystyle\lim_{j\to\infty}\inf_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}\underline{u}\bigg{(}x,t_{1}+\frac{j\epsilon^{2}}{2};\nu\bigg{)}=\inf_{\nu\in
S^{n-1}}\tilde{\mathscr{A}}_{\epsilon}\underline{U_{t_{1}}}(x;\nu)$ (4.7)

where

$\displaystyle\tilde{\mathscr{A}}_{\epsilon}v(x;\nu)=\alpha
v(x+\epsilon\nu)+\beta\mathchoice{\mathop{\kern 1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}v(x+h)d\mathcal{L}^{n-1}(h).$
(4.8)

These equalities can be derived by the argument in the proof of [AHP17,
Proposition 3.3]. First, we get (4.6) from monotonicity of
$\\{\underline{u}(x,t_{1}+j\epsilon^{2}/2)\\}$. On the other hand, by means of
the monotonicity of $\\{\underline{u}(x,t_{1}+j\epsilon^{2}/2)\\}$ and
continuity of $\mathscr{A}_{\epsilon}\underline{u}(x,t;\cdot)$, we can show
the existence of a vector $\tilde{\nu}\in S^{n-1}$ satisfying

$\mathscr{A}_{\epsilon}\underline{u}\bigg{(}x,t_{1}+\frac{j\epsilon^{2}}{2};\tilde{\nu}\bigg{)}\leq\lim_{j\to\infty}\inf_{\nu\in
S^{n-1}}\tilde{\mathscr{A}}_{\epsilon}\underline{U_{t_{1}}}(x;\nu)\qquad\textrm{for
any}\ j\geq 0.$

Now (4.7) is obtained by the monotone convergence theorem. Thus, we deduce
that $\underline{U_{t_{1}}}$ satisfies the DPP (LABEL:dpplim) for every
$-\epsilon^{2}/2\leq t_{1}<0$. By uniqueness of solutions to (LABEL:dpplim),
[AHP17, Theorem 3.7], we can deduce that

$\lim_{t\to\infty}\underline{u}(x,t)=U_{\epsilon}(x).$

We can prove the same result for $\overline{u}$ by repeating the above steps.
Combining these results with $\underline{u}\leq u_{\epsilon}\leq\overline{u}$,
we get

$\lim_{t\to\infty}u_{\epsilon}(x,t)=U_{\epsilon}(x)$

and then we can finish the proof. ∎

We finish this section by proving a corollary. One can apply the above theorem
with the interior regularity result for $u_{\epsilon}$, [Han20, Theorem 5.2].
This coincides with the result for elliptic case, [ALPR20, Theorem 1.1].

######  Corollary 4.4.

Let $\bar{B}_{2r}\subset\Omega\backslash I_{\epsilon}$ and $\epsilon>0$ be
small. Suppose that $U_{\epsilon}$ satisfies (LABEL:dpplim). Then for any
$x,y\in B_{r}(0)$,

$|U_{\epsilon}(x)-U_{\epsilon}(y)|\leq C(|x-y|+\epsilon),$

where $C>0$ is a constant which only depends on $r,n$ and
$||\psi||_{L^{\infty}(\Gamma_{\epsilon})}$.

###### Proof.

Let $r>0$ with $\bar{B}_{2r}\subset\Omega\backslash I_{\epsilon}$ and $x,y\in
B_{r}(0)$. By Theorem 4.2, for any $\eta>0$, we can find some large $t>0$ such
that

$|u_{\epsilon}(x,t)-U_{\epsilon}(x)|<\eta\quad\textrm{and}\quad|u_{\epsilon}(y,t)-U_{\epsilon}(y)|<\eta,$

where $u_{\epsilon}$ is a function satisfying (LABEL:dppvar). And by [Han20,
Theorem 5.2], we know that

$|u_{\epsilon}(x,t)-u_{\epsilon}(y,t)|\leq C(|x-y|+\epsilon),$

where $C$ is a constant depending on $r,n$ and
$||F||_{L^{\infty}(\Gamma_{\epsilon,T})}$. (Here, $F$ is a boundary data as in
Theorem 4.2)

Then we have

$\displaystyle|U_{\epsilon}(x)-U_{\epsilon}(y)|$
$\displaystyle\leq|U_{\epsilon}(x)-u_{\epsilon}(x,t)|+|u_{\epsilon}(x,t)-u_{\epsilon}(y,t)|+|u_{\epsilon}(y,t)-U_{\epsilon}(y)|$
$\displaystyle<C(|x-y|+\epsilon)+2\eta.$

Since we can choose $\eta$ arbitrarily small, we obtain

$|u_{\epsilon}(x,t)-u_{\epsilon}(y,t)|\leq C(|x-y|+\epsilon)$

for some $C=C(n,p,\Omega,||\psi||_{L^{\infty}(\Gamma_{\epsilon})})>0$ since we
can estimate

$||F||_{L^{\infty}(\Gamma_{\epsilon,T})}\leq||\psi||_{L^{\infty}(\Gamma_{\epsilon})}$

by choosing proper boundary data $F$. ∎

##  5\. Regularity near the boundary

We consider regularity for functions $u_{\epsilon}$ satisfying (LABEL:dppvar)
near the boundary in this section. This result is interesting in itself, but
it is also necessary to observe the connection between value functions and
PDEs (see the next section).

First, we introduce a boundary regularity condition for the domain $\Omega$.

######  Definition 5.1 (Exterior sphere condition).

We say that a domain $\Omega$ satisfies an exterior sphere condition if for
any $y\in\partial\Omega$, there exists
$B_{\delta}(z)\subset\mathbb{R}^{n}\backslash\Omega$ with $\delta>0$ such that
$y\in\partial B_{\delta}(z)$.

Throughout this section, we always assume that $\Omega$ satisfies Definition
5.1 and $\Omega\subset B_{R}(z)$ for some $R>0$.

Meanwhile, we also assume that the boundary data $F$ satisfies

$\displaystyle|F(x,t)-F(y,s)|\leq L(|x-y|+|t-s|^{1/2})$ (5.1)

for any $(x,t),(y,s)\in\Gamma_{\epsilon,T}$ and some $L>0$.

Let $y\in\partial\Omega$ and take $z\in\mathbb{R}^{n}\backslash\Omega$ with
$B_{\delta}(z)\subset\mathbb{R}^{n}\backslash\Omega$ and $y\in\partial
B_{\delta}(z)$. We consider a time-independent tug-of-war game. Assume that
the rules to move the token are the same as that of the original game, but of
course, we do not consider the time parameter $t$ in this case. We also assume
that the token cannot escape outside $\overline{B}_{R}(z)$ and the game ends
only if the token is located in $\overline{B}_{\delta}(z)$. Now we fix
specific strategies for both players. For each $k=0,1,\dots$, assume that
Player I and II takes the vector
$\nu_{k}^{\mathrm{I}}=-\frac{x_{k}-z}{|x_{k}-z|}$ and
$\nu_{k}^{\mathrm{II}}=\frac{x_{k}-z}{|x_{k}-z|}$, respectively. We write
these strategies for Player I, II as $S_{\mathrm{I}}^{z}$ and
$S_{\mathrm{II}}^{z}$. On the other hand, we need to define strategies and
random processes when $B_{\epsilon}(x_{k})\backslash B_{R}(z)\neq\varnothing$.
In this case, $x_{k+1}$ is defined by $x_{k}+\epsilon\nu_{k}^{\mathrm{I}}$ if
Player I wins coin toss twice and

$x_{k}+\operatorname{dist}(x_{k},\partial
B_{R}(z))\nu_{k}^{\mathrm{II}}=z+R\frac{x_{k}-z}{|x_{k}-z|}$

if Player II wins coin toss twice. When random walk occurs, $x_{k+1}$ is
chosen uniformly in $B_{\epsilon}^{\nu_{k}^{I}}(x)\cap B_{R}(z)$.

We denote by

$\tau^{\ast}=\inf\\{k:x_{k}\in\overline{B}_{\delta}(z)\\}.$

The following lemma gives an estimate for the the stopping time $\tau^{\ast}$.

######  Lemma 5.2.

Under the setting as above, we have

$\mathbb{E}_{S_{\mathrm{I}}^{z},S_{\mathrm{II}}^{z}}^{x_{0}}[\tau^{\ast}]\leq\frac{C(n,\alpha,R/\delta)(\operatorname{dist}(\partial
B_{\delta}(y),x_{0})+o(1))}{\epsilon^{2}}$

for any $x_{0}\in\Omega\subset B_{R}(z)\backslash\overline{B}_{\delta}(z)$.
Here $o(1)\to 0$ as $\epsilon\to 0$ and $\lceil x\rceil$ means the least
integer greater than or equal to $x\in\mathbb{R}$.

###### Proof.

Set
$g_{\epsilon}(x)=\mathbb{E}_{S_{\mathrm{I}}^{z},S_{\mathrm{II}}^{z}}^{x}[\tau^{\ast}].$
Then we observe that $g_{\epsilon}$ satisfies the following DPP

$\displaystyle g_{\epsilon}(x)=$
$\displaystyle\frac{1}{2}\bigg{[}\bigg{\\{}\alpha
g_{\epsilon}(x+\rho_{x}\epsilon\nu_{x})+\beta\mathchoice{\mathop{\kern
1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu_{x}}(x)\cap
B_{R}(z)}}{\mathop{\kern 1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)\cap
B_{R}(z)}}{\mathop{\kern 1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)\cap
B_{R}(z)}}{\mathop{\kern 1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)\cap
B_{R}(z)}}g_{\epsilon}(y)d\mathcal{L}^{n-1}(y)\bigg{\\}}$
$\displaystyle\qquad+\bigg{\\{}\alpha
g_{\epsilon}(x-\epsilon\nu_{x})+\beta\mathchoice{\mathop{\kern 1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu_{x}}(x)\cap
B_{R}(z)}}{\mathop{\kern 1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)\cap
B_{R}(z)}}{\mathop{\kern 1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)\cap
B_{R}(z)}}{\mathop{\kern 1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)\cap
B_{R}(z)}}g_{\epsilon}(y)d\mathcal{L}^{n-1}(y)\bigg{\\}}\bigg{]}+1,$

where $\rho_{x}=\min\\{1,\epsilon^{-1}\operatorname{dist}(x,\partial
B_{R}(z))\\}$ and $\nu_{x}=(x-z)/|x-z|$. Note that $\rho_{x}=1$ for any $x\in
B_{R-\epsilon}(z)\backslash\overline{B}_{\delta}(z)$. Next we define
$v_{\epsilon}=\epsilon^{2}g_{\epsilon}$. It is straightforward that

$\displaystyle\begin{split}v_{\epsilon}(x)=\frac{\alpha}{2}\big{(}v_{\epsilon}&(x+\rho_{x}\epsilon\nu_{x})+v_{\epsilon}(x-\epsilon\nu_{x})\big{)}\\\
&+\beta\mathchoice{\mathop{\kern 1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu_{x}}(x)\cap
B_{R}(z)}}{\mathop{\kern 1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)\cap
B_{R}(z)}}{\mathop{\kern 1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)\cap
B_{R}(z)}}{\mathop{\kern 1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)\cap
B_{R}(z)}}v_{\epsilon}(y)d\mathcal{L}^{n-1}(y)+\epsilon^{2}.\end{split}$ (5.2)

From the definition of $v_{\epsilon}$ and (5.2), we observe that the function
$v_{\epsilon}$ is rotationally symmetric, that is, $v_{\epsilon}$ is a
function of $r=|x-z|$. If we denote by $v_{\epsilon}(x)=V(r)$, the DPP (5.2)
can be represented by

$\displaystyle\begin{split}V(r)=\frac{\alpha}{2}&\big{(}V(r+\rho_{r}\epsilon)+V(r-\epsilon)\big{)}\\\
&+\beta\mathchoice{\mathop{\kern 1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu_{x}}(x)\cap
B_{R}(z)}}{\mathop{\kern 1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)\cap
B_{R}(z)}}{\mathop{\kern 1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)\cap
B_{R}(z)}}{\mathop{\kern 1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)\cap
B_{R}(z)}}V(|y-z|)d\mathcal{L}^{n-1}(y)+\epsilon^{2},\end{split}$ (5.3)

where $\rho_{r}=\min\\{1,\epsilon^{-1}(R-r)\\}$.

Now we can deduce that (5.3) has a connection to the following problem

$\displaystyle\left\\{\begin{array}[]{ll}\frac{1-\alpha}{2r}\frac{n-1}{n+1}w^{\prime}+\frac{\alpha}{2}w^{\prime\prime}=-1&\textrm{when
$r\in(\delta,R+\epsilon),$}\\\ w(\delta)=0,\\\ w^{\prime}(R+\epsilon)=0\\\
\end{array}\right.$

by using Taylor expansion. Note that if we set $v(x)=w(|x|)$,

$\frac{1-\alpha}{2r}\frac{n-1}{n+1}w^{\prime}+\frac{\alpha}{2}w^{\prime\prime}=-1$

can be transformed by

$\Delta_{p}^{N}v=-2(p+n),$

where $p=(1+n\alpha)/(1-\alpha)$ (for the definition of $\Delta_{p}^{N}$, see
the next section). On the other hand, we have

$\displaystyle
w(r)=\left\\{\begin{array}[]{ll}-\frac{n+1}{2\alpha+n-1}r^{2}+c_{1}r^{\frac{2\alpha
n-n+1}{(n+1)\alpha}}+c_{2}&\textrm{when $\alpha\neq\frac{n-1}{2n},$}\\\
-\frac{n}{n-1}r^{2}+c_{1}\log r+c_{2}&\textrm{when $\alpha=\frac{n-1}{2n}$}\\\
\end{array}\right.$

by direct calculation. Here

$\displaystyle
c_{1}=\left\\{\begin{array}[]{ll}\frac{2(n+1)^{2}\alpha}{(2\alpha+n-1)(2\alpha
n-n+1)}(R+\epsilon)^{\frac{n+2\alpha-1}{(n+1)\alpha}}&\textrm{when
$\alpha\neq\frac{n-1}{2n},$}\\\ \frac{2n}{n-1}(R+\epsilon)^{2}&\textrm{when
$\alpha=\frac{n-1}{2n}$}\\\ \end{array}\right.$

is positive if $\alpha\geq\frac{n-1}{2n}$ and negative otherwise. We extend
this function to the interval $(\delta-\epsilon,R+\epsilon]$.

Observe that

$\displaystyle\frac{\alpha}{2}$
$\displaystyle\big{(}w(r+\epsilon)+w(r-\epsilon)\big{)}+\beta\mathchoice{\mathop{\kern
1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu_{x}}(x)}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)}}w(|y-z|)d\mathcal{L}^{n-1}(y)$
$\displaystyle=w(r)-\frac{n+1}{2\alpha+n-1}\bigg{(}\alpha+\frac{n-1}{n+1}\beta\bigg{)}\epsilon^{2}+o(\epsilon^{2})$
$\displaystyle\leq
w(r)-\bigg{[}\frac{n+1}{2\alpha+n-1}\bigg{(}\alpha+\frac{n-1}{n+1}\beta\bigg{)}-\eta\bigg{]}\epsilon^{2}$

for some $\eta>0$ when $\alpha\neq\frac{n-1}{2n}$ (we can also obtain a
similar estimate if $\alpha=\frac{n-1}{2n}$). Set

$c_{0}:=\frac{n+1}{2\alpha+n-1}\bigg{(}\alpha+\frac{n-1}{n+1}\beta\bigg{)}-\eta>0.$

Then we have

$\displaystyle\mathbb{E}_{S_{\mathrm{I}}^{z},S_{\mathrm{II}}^{z}}^{x_{0}}$
$\displaystyle[v(x_{k})+kc_{0}\epsilon^{2}|x_{0},\dots,x_{k-1}]$
$\displaystyle=\alpha\big{(}v(x_{k-1}+\epsilon\nu_{x_{k-1}})+v(x_{k-1}-\epsilon\nu_{x_{k-1}})\big{)}$
$\displaystyle\qquad\qquad+\beta\mathchoice{\mathop{\kern 1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}v(y-z)d\mathcal{L}^{n-1}(y)+kc_{0}\epsilon^{2}$
$\displaystyle\leq v(x_{k-1})+(k-1)c_{0}\epsilon^{2},$

if $B_{\epsilon}(x_{k-1})\subset
B_{R}(z)\backslash\overline{B}_{\delta-\epsilon}(z)$. The same estimate can be
derived in the case $B_{\epsilon}(x_{k-1})\backslash B_{R}(z)\neq\varnothing$
since $w$ is an increasing function of $r$ and it implies

$v(x+\rho_{x}\epsilon\nu_{x})\leq v(x+\epsilon\nu_{x})$

and

$\mathchoice{\mathop{\kern 1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu_{x}}(x)\cap
B_{R}(y)}}{\mathop{\kern 1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)\cap
B_{R}(y)}}{\mathop{\kern 1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)\cap
B_{R}(y)}}{\mathop{\kern 1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)\cap
B_{R}(y)}}v(y-z)d\mathcal{L}^{n-1}(y)\leq\mathchoice{\mathop{\kern
1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu_{x}}(x)}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)}}v(y-z)d\mathcal{L}^{n-1}(y).$

Now we see that $v(x_{k})+kc_{0}\epsilon^{2}$ is a supermartingale. By the
optional stopping theorem, we have

$\displaystyle\mathbb{E}_{S_{\mathrm{I}}^{z},S_{\mathrm{II}}^{z}}^{x_{0}}[v(x_{\tau^{\ast}\wedge
k})+(\tau^{\ast}\wedge k)c_{0}\epsilon^{2}]\leq v(x_{0}).$ (5.4)

We also check that

$0\leq-\mathbb{E}_{S_{\mathrm{I}}^{z},S_{\mathrm{II}}^{z}}^{x_{0}}[v(x_{\tau^{\ast}})]\leq
o(1),$

since
$x_{\tau^{\ast}}\in\overline{B}_{\delta}(z)\backslash\overline{B}_{\delta-\epsilon}(z)$.

Meanwhile, it can be also observed that $w^{\prime}>0$ is a decreasing
function in the interval $(\delta,R+\epsilon)$ and thus

$w^{\prime}\leq\frac{2(n+1)}{2\alpha+n-1}\delta\bigg{[}\bigg{(}\frac{R+\epsilon}{\delta}\bigg{)}^{\frac{n+2\alpha-1}{(n+1)\alpha}}-1\bigg{]}\qquad\textrm{in}\
(\delta,R+\epsilon).$

From the above estimate, we have

$\displaystyle 0\leq w(x_{0})\leq
C(n,\alpha,R/\delta)\operatorname{dist}(\partial B_{\delta}(y),x_{0}).$ (5.5)

Finally, combining (5.5) with (5.4) and passing to a limit with $k$, we have

$\displaystyle
c_{0}\epsilon^{2}\mathbb{E}_{S_{\mathrm{I}}^{z},S_{\mathrm{II}}^{z}}^{x_{0}}[\tau^{\ast}]$
$\displaystyle\leq
w(x_{0})-\mathbb{E}_{S_{\mathrm{I}}^{z},S_{\mathrm{II}}^{z}}^{x_{0}}[w(x_{\tau^{\ast}})]$
$\displaystyle\leq C(n,\alpha,R/\delta)\operatorname{dist}(\partial
B_{\delta}(y),x_{0})+o(1)$

and it gives our desired estimate. ∎

By means of Lemma 5.2, we can deduce following boundary regularity results.
First, we give an estimate for $u_{\epsilon}$ on the lateral boundary.

######  Lemma 5.3.

Assume that $\Omega$ satisfies the exterior sphere condition and $F$ satisfies
(5.1). Then for the value function $u_{\epsilon}$ with boundary data $F$, we
have

$\displaystyle\begin{split}&|u_{\epsilon}(x,t)-u_{\epsilon}(y,s)|\\\ &\leq
C(n,\alpha,R,\delta,L)(K+K^{1/2})+L(|x-y|+|t-s|^{1/2}+2\delta),\end{split}$
(5.6)

where $K=\min\\{|x-y|,t\\}+\epsilon$ and $R,\delta$ are the constants in Lemma
5.2 for every $(x,t)\in\Omega_{T}$ and $(y,s)\in O_{\epsilon,T}$.

###### Proof.

We first consider the case $t=s$. Set $N=\lceil 2t/\epsilon^{2}\rceil$. Since
$\Omega$ satisfies the exterior sphere condition, we can find a ball
$B_{\delta}(z)\subset\mathbb{R}^{n}\backslash\Omega$ such that $y\in\partial
B_{\delta}(z)$. Assume that Player I takes a strategy $S_{I}^{z}$ of pulling
towards $z$.

We estimate the expected value for the distance $|x_{\tau}-x_{0}|$ under the
game setting. Let $\theta$ be the angle between $\nu$ and $x-z$. And we assume
that $x=0$ and $z=(0,\cdots,0,r\sin\theta,-r\cos\theta)$ by using a proper
transformation. Then the following term

$\alpha|x+\epsilon\nu-z|+\beta\mathchoice{\mathop{\kern 1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu_{x}}(x)}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)}}|\tilde{x}-z|d\mathcal{L}^{n-1}(\tilde{x})$

can be written as

$\displaystyle A(\theta)$
$\displaystyle=\alpha\sqrt{(r\sin\theta)^{2}+(r\cos\theta+\epsilon)^{2}}+\beta\mathchoice{\mathop{\kern
1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptT_{\epsilon}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{T_{\epsilon}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{T_{\epsilon}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{T_{\epsilon}}}\sqrt{(y-r\sin\theta)^{2}+(r\cos\theta)^{2}}d\mathcal{L}^{n-1}(y)$
$\displaystyle=\alpha\sqrt{r^{2}+2r\epsilon\cos\theta+\epsilon^{2}}+\beta\mathchoice{\mathop{\kern
1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptT_{\epsilon}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{T_{\epsilon}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{T_{\epsilon}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{T_{\epsilon}}}\sqrt{r^{2}-2ry_{n-1}\sin\theta+|y|^{2}}d\mathcal{L}^{n-1}(y)$
$\displaystyle=:\alpha A_{1}(\theta)+\beta A_{2}(\theta),$

where $r=|x-z|$ and $T_{\epsilon}=\\{x=(x_{1},\dots,x_{n})\in
B_{\epsilon}(0):x_{n}=0\\}$. Observe that $A_{1}$ is decreasing in the
interval $(0,\pi)$. (Thus, $A_{1}$ has the maximum at $\theta=0$ in $[0,\pi]$)
On the other hand, we have

$A_{2}^{{}^{\prime}}(\theta)=-\mathchoice{\mathop{\kern 1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptT_{\epsilon}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{T_{\epsilon}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{T_{\epsilon}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{T_{\epsilon}}}\frac{ry_{n-1}\cos\theta}{\sqrt{r^{2}-2ry_{n-1}\sin\theta+|y|^{2}}}d\mathcal{L}^{n-1}(y)$

and this function is a symmetric function about $\theta=\pi/2$. We also check
that $A_{2}^{{}^{\prime}}<0$ in $(0,\pi/2)$. Thus, we verify that $A_{2}$ has
a maximum at $\theta=0,\pi$ in $[0,\pi]$ and $\theta(0)=\theta(\pi)$. This
leads to the following estimate

$\displaystyle\begin{split}\sup_{\nu\in
S^{n-1}}\bigg{[}\alpha|x&+\epsilon\nu-z|+\beta\mathchoice{\mathop{\kern
1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu}(x)}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}(x)}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}(x)}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}(x)}}|\tilde{x}-z|d\mathcal{L}^{n-1}(\tilde{x})\bigg{]}\\\
&=\alpha(|x-z|+\epsilon)+\beta\mathchoice{\mathop{\kern 1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu_{x}}(x)}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x}}(x)}}|\tilde{x}-z|d\mathcal{L}^{n-1}(\tilde{x}),\end{split}$
(5.7)

where $\nu_{x}=(x-z)/|x-z|$.

Therefore, we have

$\displaystyle\mathbb{E}_{S_{\mathrm{I}}^{z},S_{\mathrm{II}}}^{(x_{0},t)}[|x_{k}-z||(x_{0},t_{0}),\dots,(x_{k-1},t_{k-1})]$
$\displaystyle\leq\frac{1-\delta(x_{k-1},t_{k-1})}{2}\bigg{[}\alpha(|x_{k-1}-z|-\epsilon)+\beta\mathchoice{\mathop{\kern
1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}|\tilde{x}-z|d\mathcal{L}^{n-1}(\tilde{x})\bigg{]}$
$\displaystyle\quad+\frac{1-\delta(x_{k-1},t_{k-1})}{2}\bigg{[}\alpha(|x_{k-1}-z|+\epsilon)+\beta\mathchoice{\mathop{\kern
1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}|\tilde{x}-z|d\mathcal{L}^{n-1}(\tilde{x})\bigg{]}$
$\displaystyle\quad+\delta(x_{k-1},t_{k-1})|x_{k-1}-z|$
$\displaystyle=|x_{k-1}-z|$
$\displaystyle\qquad+\beta(1-\delta(x_{k-1},t_{k-1}))\bigg{(}\mathchoice{\mathop{\kern
1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}|\tilde{x}-z|d\mathcal{L}^{n-1}(\tilde{x})-|x_{k-1}-z|\bigg{)}.$

We also observe that

$0<\beta(1-\delta(x_{k-1},t_{k-1}))<1,$
$|x_{k-1}-z|\leq|\tilde{x}-z|\leq\sqrt{(x_{k-1}-z)^{2}+\epsilon^{2}}\qquad\textrm{for}\
x\in B_{\epsilon}^{\nu_{x_{k-1}}},$

and

$0<\sqrt{a^{2}+\epsilon^{2}}-a<\frac{\epsilon^{2}}{2a}\qquad\textrm{for}\
a>0.$

Therefore,

$\displaystyle\mathbb{E}_{S_{\mathrm{I}}^{z},S_{\mathrm{II}}}^{(x_{0},t)}[|x_{k}-z||(x_{0},t_{0}),\dots,(x_{k-1},t_{k-1})]\leq|x_{k-1}-z|+C\epsilon^{2}$

for some $C=C(n,\delta)>0$. This yields that

$M_{k}=|x_{k}-z|-Ck\epsilon^{2}$

is a supermartingale.

Applying the optional stopping theorem and Jensen’s inequality to $M_{k}$, we
derive that

$\displaystyle\begin{split}\mathbb{E}_{S_{\mathrm{I}}^{z},S_{\mathrm{II}}}^{(x_{0},t)}&[|x_{\tau}-z|+|t_{\tau}-t|^{1/2}]\\\
&=\mathbb{E}_{S_{\mathrm{I}}^{z},S_{\mathrm{II}}}^{(x_{0},t)}\bigg{[}|x_{\tau}-z|+\epsilon\sqrt{\frac{\tau}{2}}\bigg{]}\\\
&\leq|x_{0}-z|+C\epsilon^{2}\mathbb{E}_{S_{\mathrm{I}}^{z},S_{\mathrm{II}}}^{(x_{0},t)}[\tau]+C\epsilon\big{(}\mathbb{E}_{S_{\mathrm{I}}^{z},S_{\mathrm{II}}}^{(x_{0},t)}[\tau]\big{)}^{1/2}.\end{split}$
(5.8)

Next we need to obtain estimates for
$\mathbb{E}_{S_{\mathrm{I}}^{z},S_{\mathrm{II}}}^{(x_{0},t)}[\tau]$. To do
this, we use the result in Lemma 5.2. We can check that the exit time $\tau$
of the original game is bounded by $\tau^{\ast}$ because the expected value of
$|x_{k}-z|$ for given $|x_{k-1}-z|$ is maximized when Player II chooses the
strategy $S_{\mathrm{II}}^{z}$ from (5.7). Thus, we have

$\displaystyle\mathbb{E}_{S_{\mathrm{I}}^{z},S_{\mathrm{II}}}^{(x_{0},t)}[\tau]$
$\displaystyle\leq\min\\{\mathbb{E}_{S_{\mathrm{I}}^{z},S_{\mathrm{II}}^{z}}^{x_{0}}[\tau^{\ast}],N\\}$
$\displaystyle\leq\min\bigg{\\{}\frac{C(n,\alpha,R/\delta)(\operatorname{dist}(\partial
B_{\delta}(z),x_{0})+\epsilon)}{\epsilon^{2}},N\bigg{\\}}$

for any strategy $S_{\mathrm{II}}$ for Player II. We also see that

$\operatorname{dist}(x_{0},\partial B_{\delta}(z))\leq|x_{0}-y|.$

This and (5.8) imply

$\displaystyle\mathbb{E}_{S_{\mathrm{I}}^{z},S_{\mathrm{II}}}^{(x_{0},t)}[|x_{\tau}-z|+|t_{\tau}-t|^{1/2}]$
$\displaystyle\leq|x_{0}-y|+C\min\\{|x_{0}-y|+\epsilon,\epsilon^{2}N\\}+C\min\\{|x_{0}-y|+\epsilon,\epsilon^{2}N\\}^{1/2},$

where $C$ is a constant depending on $n,\alpha,R$ and $\delta$. Therefore, we
get

$\displaystyle|\mathbb{E}_{S_{\mathrm{I}}^{z},S_{\mathrm{II}}}^{(x_{0},t)}$
$\displaystyle[F(x_{\tau},t_{\tau})]-F(z,t)|$ $\displaystyle\leq
L(|x_{0}-y|+C(n,\alpha,R,\delta)\min\\{|x_{0}-y|+\epsilon,\epsilon^{2}N\\}$
$\displaystyle\qquad\quad+C(n,\alpha,R/\delta)\min\\{|x_{0}-y|+\epsilon,\epsilon^{2}N\\}^{1/2})$

and this yields

$\displaystyle u_{\epsilon}(x_{0},t)$
$\displaystyle=\sup_{S_{\mathrm{I}}}\inf_{S_{\mathrm{II}}}\mathbb{E}_{S_{\mathrm{I}},S_{\mathrm{II}}}^{(x_{0},t)}[F(x_{\tau},t_{\tau})]$
$\displaystyle\geq\inf_{S_{\mathrm{II}}}\mathbb{E}_{S_{\mathrm{I}}^{z},S_{\mathrm{II}}}^{(x_{0},t)}[F(x_{\tau},t_{\tau})]$
$\displaystyle\geq F(z,t)-L\\{C(n,\alpha,R,\delta)(K+K^{1/2})+|x_{0}-y|\\}$
$\displaystyle\geq
F(y,t)-C(n,\alpha,R,\delta,L)(K+K^{1/2})-L(|x_{0}-y|+2\delta)$

for $K=\min\\{|x_{0}-y|+\epsilon,\epsilon^{2}N\\}$. Note that we can also
derive the upper bound for $u_{\epsilon}(x_{0},t)$ by taking the strategy
where Player II pulls toward to $z$.

Meanwhile, in the case of $t\neq s$, we have

$\displaystyle|u_{\epsilon}(x,t)-u_{\epsilon}(y,s)|$
$\displaystyle\leq|u_{\epsilon}(x,t)-u_{\epsilon}(y,t)|+|u_{\epsilon}(y,t)-u_{\epsilon}(y,s)|$
$\displaystyle\leq
C(n,\alpha,R/\delta,L)(K+K^{1/2})+L(|x-y|+2\delta)+L|t-s|^{1/2},$

where $K=\min\\{|x_{0}-y|+\epsilon,\epsilon^{2}N\\}$ and $N=\lceil
2t/\epsilon^{2}\rceil$. This gives our desired estimate. ∎

We can also derive the following result on the initial boundary.

######  Lemma 5.4.

Assume that $\Omega$ satisfies the exterior sphere condition and $F$ satisfies
(5.1). Then for the value function $u_{\epsilon}$ with boundary data $F$, we
have

$\displaystyle|u_{\epsilon}(x,t)-u_{\epsilon}(y,s)|\leq
C(|x-y|+t^{1/2}+\epsilon)$ (5.9)

for every $(x,t)\in\Omega_{T}$ and $(y,s)\in\Omega\times(-\epsilon^{2}/2,0]$.
The constant $C$ depends only on $n,L$.

###### Proof.

Set $(x,t)=(x_{0},t_{0})$ and $N=\lceil 2t/\epsilon^{2}\rceil$. As in the
above lemma, we also estimate the expected value of the distance between $y$
and the exit point $x_{\tau}$. Consider the case that Player I chooses a
strategy of pulling to $y$. When $|x_{k-1}-y|\geq\epsilon$, we have

$\displaystyle\mathbb{E}_{S_{\mathrm{I}}^{y},S_{\mathrm{II}}}^{(x_{0},t_{0})}[|x_{k}-y|^{2}|(x_{0},t_{0}),\dots,(x_{k-1},t_{k-1})]$
$\displaystyle\leq(1-\delta(x_{k-1},t_{k-1}))\times$
$\displaystyle\quad\bigg{[}\frac{\alpha}{2}\\{(|x_{k-1}-y|+\epsilon)^{2}+(|x_{k-1}-y|-\epsilon)^{2}\\}$
$\displaystyle\qquad\qquad+\beta\mathchoice{\mathop{\kern 1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}|\tilde{x}-y|^{2}d\mathcal{L}^{n-1}(\tilde{x})\bigg{]}+\delta(x_{k-1},t_{k-1})|x_{k-1}-y|^{2}$
$\displaystyle\leq\alpha(|x_{k-1}-y|^{2}+\epsilon^{2})+\beta(|x_{k-1}-y|^{2}+C\epsilon^{2})$
$\displaystyle\leq|x_{k-1}-y|^{2}+C\epsilon^{2}$

for some constant $C>0$ which is independent of $\epsilon$. We recall the
notation $\nu_{x_{k-1}}=(x_{k-1}-z)/|x_{k-1}-z|$ here. Otherwise, we also see
that

$\displaystyle\mathbb{E}_{S_{\mathrm{I}}^{y},S_{\mathrm{II}}}^{(x_{0},t_{0})}[|x_{k}-y|^{2}|(x_{0},t_{0}),\dots,(x_{k-1},t_{k-1})]$
$\displaystyle\leq(1-\delta(x_{k-1},t_{k-1}))\bigg{[}\frac{\alpha}{2}(|x_{k-1}-y|+\epsilon)^{2}+\beta\mathchoice{\mathop{\kern
1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{x_{k-1}}}(x_{k-1})}}|\tilde{x}-y|^{2}d\mathcal{L}^{n-1}(\tilde{x})\bigg{]}$
$\displaystyle\quad+\delta(x_{k-1},t_{k-1})|x_{k-1}-y|^{2},$

and then we get the same estimate as above since

$(|x_{k-1}-y|+\epsilon)^{2}\leq 2(|x_{k-1}-y|^{2}+\epsilon^{2}).$

Therefore, we see that

$M_{k}=|x_{k}-y|^{2}-Ck\epsilon^{2}$

is a supermartingale.

Now we obtain

$\mathbb{E}_{S_{\mathrm{I}}^{y},S_{\mathrm{II}}}^{(x_{0},t)}[|x_{\tau}-y|^{2}]\leq|x_{0}-y|^{2}+C\epsilon^{2}\mathbb{E}_{S_{\mathrm{I}}^{y},S_{\mathrm{II}}}^{(x_{0},t)}[\tau]$

by using the optional stopping theorem. Since $\tau<\lceil
2t/\epsilon^{2}\rceil$, the right-hand side term is estimated by
$|x_{0}-y|^{2}+C(t+\epsilon^{2})$. Applying Jensen’s inequality, we get

$\displaystyle\mathbb{E}_{S_{\mathrm{I}}^{y},S_{\mathrm{II}}}^{(x_{0},t)}[|x_{\tau}$
$\displaystyle-y|]\leq\big{(}\mathbb{E}_{S_{\mathrm{I}}^{y},S_{\mathrm{II}}}^{(x_{0},t)}[|x_{\tau}-y|^{2}]\big{)}^{\frac{1}{2}}$
$\displaystyle\leq\big{(}|x_{0}-y|^{2}+C(t+\epsilon^{2})\big{)}^{\frac{1}{2}}$
$\displaystyle\leq|x_{0}-y|+C(t^{1/2}+\epsilon).$

From the above estimate, we deduce that

$\displaystyle u_{\epsilon}(x_{0},t)$
$\displaystyle=\sup_{S_{\mathrm{I}}}\inf_{S_{\mathrm{II}}}\mathbb{E}_{S_{\mathrm{I}},S_{\mathrm{II}}}^{(x_{0},t)}[F(x_{\tau},t_{\tau})]$
$\displaystyle\geq F(y,t)-L\
\mathbb{E}_{S_{\mathrm{I}}^{y},S_{\mathrm{II}}}^{(x_{0},t)}[|x_{\tau}-y|+|t-t_{\tau}|^{2}]$
$\displaystyle\geq F(y,t)-C(|x_{0}-y|+t^{1/2}+\epsilon).$

The upper bound can be derived in a similar way, and then we get the estimate
(5.9). ∎

##  6\. Application to PDEs

The objective of this section is to study behavior of $u_{\epsilon}$ when
$\epsilon$ tends to zero. This issue has been studied in several preceding
papers (see [KMP12, LPS14, PR16, AHP17]). Those results show that there is a
close relation between value functions of tug-of-war games and certain types
of PDEs. Now we will establish that there is a convergence theorem showing
that $u_{\epsilon}$ converge to the unique viscosity solution of the following
Dirichlet problem for the normalized parabolic $p$-Laplace equation

$\displaystyle\left\\{\begin{array}[]{ll}(n+p)u_{t}=\Delta_{p}^{N}u&\textrm{in
$\Omega_{T}$,}\\\ u=F&\textrm{on $\partial_{p}\Omega_{T}$}\\\
\end{array}\right.$ (6.3)

as $\epsilon\to 0$. Here, $p$ satisfies $\alpha=(p-1)/(p+n)$ and
$\beta=(n+1)/(p+n)$.

Now we introduce the notion of viscosity solutions for (6.3). Note that we
need to consider the case when the gradient vanishes. Here we use
semicontinuous extensions of operators in order to define viscosity solutions.
For these extensions, we refer the reader to [ES99, Gig06] for more details.

######  Definition 6.1 (Viscosity solution).

A function $u\in C(\Omega_{T})$ is a viscosity solution to (6.3) if the
following conditions hold:

  * (a)

for all $\varphi\in C^{2}(\Omega_{T})$ touching $u$ from above at
$(x_{0},t_{0})\in\Omega_{T}$,

$\displaystyle\left\\{\begin{array}[]{ll}\Delta_{p}^{N}\varphi(x_{0},t_{0})\geq(n+p)\varphi_{t}(x_{0},t_{0})\qquad\qquad\textrm{if
$D\varphi(x_{0},t_{0})\neq 0$,}\\\
\lambda_{\max}((p-2)D^{2}\varphi(x_{0},t_{0}))&\\\
\qquad\qquad\qquad+\Delta\varphi(x_{0},t_{0})\geq(n+p)\varphi_{t}(x_{0},t_{0})\qquad\textrm{if
$D\varphi(x_{0},t_{0})=0$.}\\\ \end{array}\right.$

  * (b)

for all $\varphi\in C^{2}(\Omega_{T})$ touching $u$ from below at
$(x_{0},t_{0})\in\Omega_{T}$,

$\displaystyle\left\\{\begin{array}[]{ll}\Delta_{p}^{N}\varphi(x_{0},t_{0})\leq(n+p)\varphi_{t}(x_{0},t_{0})\qquad\qquad\textrm{if
$D\varphi(x_{0},t_{0})\neq 0$,}\\\
\lambda_{\min}((p-2)D^{2}\varphi(x_{0},t_{0}))&\\\
\qquad\qquad\qquad+\Delta\varphi(x_{0},t_{0})\leq(n+p)\varphi_{t}(x_{0},t_{0})\qquad\textrm{if
$D\varphi(x_{0},t_{0})=0$.}\\\ \end{array}\right.$

Here, the notation $\lambda_{\max}(X)$ and $\lambda_{\min}(X)$ mean the
largest and the smallest eigenvalues of a symmetric matrix $X$.

The following Arzelà-Ascoli criterion will be used to obtain the main result
in this section. It is essentially the same proposition as [PR16, Lemma 5.1].
We can find the proof of this criterion for elliptic version in [MPR12, Lemma
4.2].

######  Lemma 6.2.

Let $\\{u_{\epsilon}:\overline{\Omega}_{T}\to\mathbb{R},\epsilon>0\\}$ be a
set of functions such that

  * (a)

there exists a constant $C>0$ so that $|u_{\epsilon}(x,t)|<C$ for every
$\epsilon>0$ and every $(x,t)\in\overline{\Omega}_{T}$.

  * (b)

given $\eta>0$, there are constants $r_{0}$ and $\epsilon_{0}$ so that for
every $\epsilon>0$ and $(x,t),(y,s)\in\overline{\Omega}_{T}$ with
$d((x,t),(y,s))<r_{0}$, it holds

$|u_{\epsilon}(x,t)-u_{\epsilon}(y,s)|<\eta.$

Then, there exists a uniformly continuous function
$u:\overline{\Omega}_{T}\to\mathbb{R}$ and a subsequence
$\\{u_{\epsilon_{i}}\\}$ such that $u_{\epsilon_{i}}$ uniformly converge to
$u$ in $\overline{\Omega}_{T}$, as $i\to\infty$.

Now we can describe the relation between functions satisfying (LABEL:dppvar)
and solutions to the normalized parabolic $p$-Laplace equation.

######  Theorem 6.3.

Assume that $\Omega$ satisfies the exterior sphere condition and $F\in
C(\Gamma_{\epsilon,T})$ satisfies (5.1). Let $u_{\epsilon}$ denote the
solution to (LABEL:dppvar) with boundary data $F$ for each $\epsilon>0$. Then,
there exist a function $u:\overline{\Omega}_{\epsilon,T}\to\mathbb{R}$ and a
subsequence $\\{\epsilon_{i}\\}$ such that

$u_{\epsilon_{i}}\to u\qquad\textrm{uniformly in}\quad\overline{\Omega}_{T}$

and the function $u$ is a unique viscosity solution to (6.3).

######  Remark 6.4.

The uniqueness of solutions to (6.3) can be found in [PR16, Lemma 6.2].

###### Proof.

First we check that there is a subsequence $\\{u_{\epsilon_{i}}\\}$ with
$u_{\epsilon_{i}}$ converge uniformly to $u$ on $\overline{\Omega}_{T}$ for
some function $u$. By using the definition of $u_{\epsilon}$, we have

$||u_{\epsilon}||_{L^{\infty}(\Omega_{T})}\leq||F||_{L^{\infty}(\Omega_{T})}<\infty$

for any $\epsilon>0$. Hence, $u_{\epsilon}$ are uniformly bounded. By means of
Lemma 5.3, Lemma 5.4 and the interior regularity result [Han20, Theorem 5.2],
we know that $\\{u_{\epsilon}\\}$ are equicontinuous. Therefore, we can find a
subsequence $\\{u_{\epsilon_{i}}\\}_{i=1}^{\infty}$ converging uniformly to a
function $u\in C(\overline{\Omega}_{T})$ by Lemma 6.2.

Now we need to show that $u$ is a viscosity solution to (6.3). On the
parabolic boundary, we see that

$u(x,t)=\lim_{i\to\infty}u_{\epsilon_{i}}(x,t)=F(x,t)$

for any $(x,t)\in\partial_{p}\Omega_{T}$.

Next we prove that $u$ satisfies

$(n+p)u_{t}=\Delta_{p}^{N}u\qquad\textrm{in}\ \Omega_{T}$

in the viscosity sense. Without loss of generality, it suffices to show that
$u$ satisfies condition (a) in Definition 6.1.

Fix $(x,t)\in\Omega_{T}$. Then there is a small number $R>0$ such that

$Q:=(x_{0},t_{0})+B_{R}(0)\times(-R^{2},0)\subset\subset\Omega_{T}.$

We also assume that $\epsilon>0$ satisfies $Q\subset\Omega_{T}\backslash
I_{\epsilon,T}$. Suppose that a function $\varphi\in C^{2}(Q)$ touches $u$
from below at $(x,t)$. Then we observe that

$\inf_{Q}(u-\varphi)=(u-\varphi)(x,t)\leq(u-\varphi)(z,s)$

for any $(z,s)\in Q$. Since $u_{\epsilon}$ converge uniformly to $u$, for
sufficiently small $\epsilon>0$, there is a point
$(x_{\epsilon},t_{\epsilon})\in Q$ such that

$\inf_{Q}(u_{\epsilon}-\varphi)\leq(u_{\epsilon}-\varphi)(z,s)$

for any $(z,s)\in Q$. We also check that $(x_{\epsilon},t_{\epsilon})\to(x,t)$
as $\epsilon\to 0$.

Recall (3.1). Since $(x_{\epsilon},t_{\epsilon})\in\Omega_{T}\backslash
I_{\epsilon,T}$, we have

$\displaystyle Tu(x,t)=\operatornamewithlimits{midrange}_{\nu\in
S^{n-1}}\mathscr{A}_{\epsilon}u\bigg{(}x,t-\frac{\epsilon^{2}}{2};\nu\bigg{)}.$

We also set $\psi=\varphi+(u_{\epsilon}-\varphi)(x_{\epsilon},t_{\epsilon})$
and observe that $u_{\epsilon}\geq\psi$ in $Q$. Now it can be checked that

$\displaystyle
u_{\epsilon}(x_{\epsilon},t_{\epsilon})=Tu_{\epsilon}(x_{\epsilon},t_{\epsilon})\geq
T\psi(x_{\epsilon},t_{\epsilon})$

and

$\displaystyle
T\psi(x_{\epsilon},t_{\epsilon})=T\varphi(x_{\epsilon},t_{\epsilon})+(u_{\epsilon}-\varphi)(x_{\epsilon},t_{\epsilon})$

Therefore,

$\displaystyle u_{\epsilon}(x_{\epsilon},t_{\epsilon})\geq
T\varphi(x_{\epsilon},t_{\epsilon})+(u_{\epsilon}-\varphi)(x_{\epsilon},t_{\epsilon})$

and this implies

$\displaystyle 0\geq
T\varphi(x_{\epsilon},t_{\epsilon})-\varphi(x_{\epsilon},t_{\epsilon}).$ (6.4)

On the other hand, by the Taylor expansion, we observe that

$\displaystyle\frac{1}{2}$
$\displaystyle\bigg{[}\varphi\bigg{(}x+\epsilon\nu,t-\frac{\epsilon^{2}}{2}\bigg{)}+\varphi\bigg{(}x-\epsilon\nu,t-\frac{\epsilon^{2}}{2}\bigg{)}\bigg{]}$
$\displaystyle=\varphi(x,t)-\frac{\epsilon^{2}}{2}\varphi_{t}(x,t)+\frac{\epsilon^{2}}{2}\langle
D^{2}\varphi(x,t)\nu,\nu\rangle+o(\epsilon^{2})$

and

$\displaystyle\mathchoice{\mathop{\kern 1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu}}}$
$\displaystyle\varphi\bigg{(}x+h,t-\frac{\epsilon^{2}}{2}\bigg{)}d\mathcal{L}^{n-1}(h)$
$\displaystyle=\varphi(x,t)-\frac{\epsilon^{2}}{2}\varphi_{t}(x,t)+\frac{\epsilon^{2}}{2(n+1)}\Delta_{\nu^{\perp}}\varphi(x,t)+o(\epsilon^{2})$

where

$\Delta_{\nu^{\perp}}\varphi(x,t)=\sum_{i=1}^{n-1}\langle
D^{2}\varphi(x,t)\nu_{i},\nu_{i}\rangle$

with $\nu_{1},\cdots,\nu_{n-1}$ the orthonormal basis for the space
$\nu^{\perp}$ for $\nu\in S^{n-1}$.

We already know that $\mathscr{A}_{\epsilon}\varphi$ is continuous with
respect to $\nu$ in Proposition 3.1. Therefore, there exists a vector
$\nu_{\min}=\nu_{\min}(\epsilon)$ minimizing
$\mathscr{A}_{\epsilon}\varphi(x_{\epsilon,\eta},t_{\epsilon};\cdot).$ Then we
can calculate

$\displaystyle T\varphi(x_{\epsilon},t_{\epsilon})$
$\displaystyle\geq\frac{\alpha}{2}\bigg{\\{}\varphi\bigg{(}x_{\epsilon}+\nu_{\min},t_{\epsilon}-\frac{\epsilon^{2}}{2}\bigg{)}+\varphi\bigg{(}x_{\epsilon}-\nu_{\min},t_{\epsilon}-\frac{\epsilon^{2}}{2}\bigg{)}\bigg{\\}}$
$\displaystyle\qquad+\beta\mathchoice{\mathop{\kern 1.99997pt\vrule
width=6.00006pt,height=3.0pt,depth=-2.49997pt\kern-8.00003pt\intop}\nolimits_{\kern-2.79996ptB_{\epsilon}^{\nu_{\min}}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{\min}}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{\min}}}}{\mathop{\kern
1.00006pt\vrule
width=5.0pt,height=3.0pt,depth=-2.59996pt\kern-6.00006pt\intop}\nolimits_{B_{\epsilon}^{\nu_{\min}}}}\varphi\bigg{(}x_{\epsilon}+h,t_{\epsilon}-\frac{\epsilon^{2}}{2}\bigg{)}d\mathcal{L}^{n-1}(h)$
$\displaystyle\geq\varphi(x_{\epsilon},t_{\epsilon})-\frac{\epsilon^{2}}{2}\varphi_{t}(x_{\epsilon},t_{\epsilon})$
$\displaystyle\qquad+\frac{\beta}{2(n+1)}\epsilon^{2}\big{\\{}\Delta_{\nu_{\min}^{\perp}}\varphi(x_{\epsilon},t_{\epsilon})+(p-1)\langle
D^{2}\varphi(x_{\epsilon},t_{\epsilon})\nu_{\min},\nu_{\min}\rangle\big{\\}}.$

Then by (6.4), we observe that

$\displaystyle\begin{split}\frac{\epsilon^{2}}{2}\varphi_{t}(x_{\epsilon},t_{\epsilon})&\geq\frac{\beta\epsilon^{2}}{2(n+1)}\big{\\{}\Delta_{\nu_{\min}^{\perp}}\varphi(x_{\epsilon},t_{\epsilon})+(p-1)\langle
D^{2}\varphi(x_{\epsilon},t_{\epsilon})\nu_{\min},\nu_{\min}\rangle\big{\\}}.\end{split}$
(6.5)

Suppose that $D\varphi(x,t)\neq 0$. Since
$(x_{\epsilon},t_{\epsilon})\to(x,t)$ as $\epsilon\to 0$, it can be seen that

$\nu_{\min}\to-\frac{D\varphi(x,t)}{|D\varphi(x,t)|}=:-\mu$

as $\epsilon\to 0$. We also check that

$\Delta_{(-\mu)^{\perp}}\varphi(x,t)+(p-1)\langle
D^{2}\varphi(x_{\epsilon},t_{\epsilon})(-\mu),(-\mu)\rangle=\Delta_{p}^{N}\varphi(x,t).$

Now we divide both side in (6.5) by $\epsilon^{2}$ and let $\epsilon\to 0$.
Since $Q_{R}\subset\Omega_{T}$, it can be seen that
$\delta(x_{\epsilon},t_{\epsilon})\epsilon^{-2}\to 0$ as $\epsilon\to 0$.
Hence, we deduce

$\varphi_{t}(x,t)\geq\frac{1}{n+p}\Delta_{p}^{N}\varphi(x,t).$

Next consider the case $D\varphi(x,t)=0$. Observe that

$\displaystyle\Delta_{\nu_{\min}^{\perp}}\varphi(x_{\epsilon},t_{\epsilon})+(p-1)\langle
D^{2}\varphi(x_{\epsilon},t_{\epsilon})\nu_{\min},\nu_{\min}\rangle$
$\displaystyle=\Delta\varphi(x_{\epsilon},t_{\epsilon})+(p-2)\langle
D^{2}\varphi(x_{\epsilon},t_{\epsilon})\nu_{\min},\nu_{\min}\rangle.$

For $p\geq 2$, we see

$\displaystyle(p-2)\langle
D^{2}\varphi(x_{\epsilon},t_{\epsilon})\nu_{\min},\nu_{\min}\rangle\geq(p-2)\lambda_{\min}(D^{2}\varphi(x_{\epsilon},t_{\epsilon})).$

We already know that $(x_{\epsilon},t_{\epsilon})\to(x,t)$ as $\epsilon\to 0$
and the map $z\mapsto\lambda_{\min}(D^{2}\varphi(z))$ is continuous.
Therefore, it turns out

$\displaystyle\varphi_{t}(x,t)\geq\frac{1}{n+p}\big{\\{}\Delta\varphi(x,t)+(p-2)\lambda_{\min}(D^{2}\varphi(x,t))\big{\\}}$
(6.6)

by similar calculation in the previous case.

For $1<p<2$, by using similar argument in the previous case and

$\displaystyle(p-2)\langle
D^{2}\varphi(x_{\epsilon},t_{\epsilon})\nu_{\min},\nu_{\min}\rangle$
$\displaystyle\geq(p-2)\lambda_{\max}(D^{2}\varphi(x_{\epsilon},t_{\epsilon}))$
$\displaystyle=\lambda_{\min}((p-2)D^{2}\varphi(x_{\epsilon},t_{\epsilon})),$

we also obtain the inequality (6.6).

We can also prove the reverse inequality to consider a function $\varphi$
touching $u$ from above and a vector $\nu_{\max}$ maximizing
$\mathscr{A}_{\epsilon}\varphi(x_{\epsilon},t_{\epsilon};\cdot)$ and to do
similar calculation again as above. ∎

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