# Elliptic Harnack Inequality for $\mathbb{Z}^{d}$

Siva Athreya  Indian Statistical Institute, 8th Mile Mysore Road, Bangalore
560059, India. Email<EMAIL_ADDRESS>Nitya Gadhiwala Department of
Mathematics, University of British Columbia, 1984 Mathematics Road, Vancouver,
BC Canada, V6T 1Z2 Email<EMAIL_ADDRESS>Ritvik R. Radhakrishnan
Department of Mathematics, ETH Zurich, HG G 33.4, Raemistrasse 101 8092 Zurich
Switzerland Email<EMAIL_ADDRESS>
###### Abstract

We prove the scale invariant Elliptic Harnack Inequality (EHI) for non-
negative harmonic functions on $\mathbb{Z}^{d}$. The purpose of this note is
to provide a simplified self-contained probabilistic proof of EHI in
$\mathbb{Z}^{d}$ that is accessible at the undergraduate level. We use the
Local Central Limit Theorem for simple symmetric random walks on
$\mathbb{Z}^{d}$ to establish Gaussian bounds for the $n$-step probability
function. The uniform Green inequality and the classical Balayage formula then
imply the EHI.

AMS 2010 Subject Classification : Primary: 05C81 Secondary: 31C05, 31C20
Keywords : Random walk, Harmonic Function, Harnack Inequality, Gaussian
Bounds, Balayage.

##  1 Introduction

The scale invariant Elliptic Harnack Inequality (EHI) and its applications to
the theory of elliptic and parabolic partial differential equations are well
known. The work on regularity of solutions dates back to 1950’s in the works
of de Giorgi [DG57], Nash [Nas58] and Moser [Mos61]. Moser first proved EHI
for solutions to partial differential equations in divergence form in
$\mathbb{R}^{d}$. Bombieri-Giusti [BG72] showed that EHI holds for minimal
surfaces and EHI was proven for metric spaces by Sturm [Stu98]. We refer the
interested reader to the survey article by Kassmann [Kas07], and the
references therein for a comprehensive review of the Harnack inequality (for
elliptic and parabolic operators) in $\mathbb{R}^{d}$.

The EHI was established for weighted graphs by Delmotte [Del99]. Barlow
[Bar17] has a comprehensive treatment of random walks on locally finite graphs
and in [Bar17, Theorem 7.19] presents a proof of EHI for weighted graphs that
have controlled weights and are roughly isometric to $\mathbb{Z}^{d}$. In this
short note our aim is to provide a simplified self-contained probabilistic
proof of EHI for harmonic functions on $\mathbb{Z}^{d}$ using properties of
the simple symmetric random walk. Lawler in [Law13, Theorem 1.7.2] also
provides a proof of EHI for $\mathbb{Z}^{d}$. In this short note our aim is to
provide a simplified self-contained probabilistic proof of EHI for harmonic
functions on $\mathbb{Z}^{d}$ following the framework laid out in [Bar17]
using properties of the simple symmetric random walk.

Let $d\geq 1$ and $\mathbb{Z}^{d}$ be the integer lattice. We will think of
$\mathbb{Z}^{d}$ as a graph with vertex set

$\mathbb{Z}^{d}=\\{(x_{i})_{i=1}^{d}:x_{i}\in\mathbb{Z}\\}$

endowed with the (graph) distance $d(x,y)=\sum_{i=1}^{d}|x_{i}-y_{i}|$ between
any two points $x,y\in\mathbb{Z}^{d}$. We may view edge set
$E=\\{\\{x,y\\}:x,y\in V,\,d(x,y)=1\\}.$ For any $x,y\in\mathbb{Z}^{d}$ we
shall say $x\sim y$ if $\\{x,y\\}\in E$. For $r>0$, we define the ball of
radius $r$ around a point $x_{0}\in\mathbb{Z}^{d}$ as

$B(x_{0},r)=\\{x\in\mathbb{Z}^{d}:d(x,x_{0})\leq r\\}.$

Let $A\subset\mathbb{Z}^{d}$. We define the boundary of
$A\subset\mathbb{Z}^{d}$ by

$\partial A=\\{y\in A^{c}:\exists\,x\in A\mbox{ with }x\sim y\\}$

and set $\bar{A}=A\cup\partial A$. For $f:\bar{A}\rightarrow\mathbb{R}$ we
define the Laplacian $\Delta$ of $f$ as

$\Delta f(x)=\frac{1}{2d}\sum_{x\sim y}(f(x)-f(y)),\qquad\mbox{for all }x\in
A.$

A function $f:\bar{A}\rightarrow\mathbb{R}$ is said to be harmonic in $A$ if
$\Delta f(x)=0$ for all $x\in A$. The scale invariant EHI is then stated as
follows.

######  Theorem 1.1 (EHI).

Let $d\geq 1$. There exists $C$ such that for any $x_{0}\in\mathbb{Z}^{d}$,
$R\geq 1$, $h:\overline{B(x_{0},R)}\rightarrow\mathbb{R}_{+}$ with $h$
harmonic in $B(x_{0},R)$ we have

$\max_{y\in B\left(x_{0},\frac{R}{2}\right)}h(y)\leq C\min_{z\in
B\left(x_{0},\frac{R}{2}\right)}h(z)$ (1.1)

We work with simple symmetric random walks in $\mathbb{Z}^{d}$, whose
transition probability matrix $P$ is connected with the Laplacian $\Delta$.
The proof has three key steps – the Gaussian bounds for the $n$-step
probability function of the simple symmetric random walk (Proposition 2.1),
the uniform Green inequality for the Green function asssociated with the
random walk (Proposition 2.2), and the Balayage formula for harmonic functions
(Proposition 2.3).

The proof of Theorem 1.1 that we present here proceeds along the framework
given in the proof of [Bar17, Theorem 7.19] (which applies for those graphs
which have controlled weights and are roughly isometric to $\mathbb{Z}^{d}$).
Our proof uses methods specific to the random walk on $\mathbb{Z}^{d}$ and has
appropriate simplifications whenever possible. To establish Proposition 2.1
for weighted graphs one first shows the Poincaré inequality along with a
growth condition for volume of the balls in the graph. Then the equivalence
results shown in [Del99] will imply the result (See also [Bar17, Theorem
6.28]). For $\mathbb{Z}^{d}$, we use the Local Central Limit Theorem for
simple random walks to obtain near diagonal bounds on the $n$-step transition
probability function (See Proposition 3.2). Then the Gaussian lower bound is
obtained via a classical chaining of balls argument. The Gaussian upper bound
follows from exponential bounds for the distribution of exit times of the
simple random walk from a ball. For this, we use the specific structure of the
walk in $\mathbb{Z}^{d}$, by reducing the question for the $d$ dimensional
random walk to a one dimensional walk which follow from Chernoff bounds (See
Lemma 3.3). We also note that Gaussian bounds for the simple symmetric random
walk in $\mathbb{Z}^{d}$ are part of folklore and are well known but it is
hard to point to explicit references in the literature where they can be
found.

As mentioned before, [Law13, Theorem 1.7.2] also contains a proof for the EHI
in $\mathbb{Z}^{d}$. There the uniform Green inequality is obtained using
explicit bounds on the hitting distribution of the walk for three and higher
dimensions [Law13, Proposition 1.5.10] and on the Green function for two
dimensions in [Law13, Proposition 1.6.7]. The Local Central Limit Theorem is
used to derive precise asymptotics on the Green function and the propositions
are proved via an application of the optional stopping theorem for
martingales. For proving the uniform Green inequality we use Proposition 2.1
to establish Gaussian bounds for $n$-step transition probabilities of walk
killed on exiting the ball. Plugging these estimates and exit time bounds
obtained above into the definition of the Green function one obtains the
result. The Balayage formula is shown using the probabilistic representation
for the solution to the Dirichlet problem and standard potential theory
techniques. The proofs of uniform Green inequality and the Balayage formula
translate to graphs that have controlled weights and are roughly isometric to
$\mathbb{Z}^{d}$ (See [Bar17, Theorem 4.26, Theorem 7.15]).

The Balayage formula is analogous to the classical Poisson integral formula
for Harmonic functions in $\mathbb{R}^{d}$ with the Green function playing the
role of the Poisson Kernel. The uniform Green inequality then enables the
scale invariant Harnack inequality. The uniform Green inequality are implied
by the Gaussian bounds.

We remark that the constant $C$ in Theorem 1.1 does not depend on $R$ but only
on dimension $d$ and the $\frac{1}{2}$-scaling used to obtain the smaller
ball. There are many applications of the EHI. For instance, using EHI one can
show that the probability of exiting the boundary of $B(x_{0},R)$ via a
particular set is comparable for different starting points inside
$B(x_{0},\frac{R}{2}$). The EHI implies the oscillation inequality for
harmonic functions, in the sense that if for any set $A\subset\mathbb{Z}^{d},$

$\mbox{Osc}(h,A)=\sup_{x\in A}h(x)-\inf_{x\in A}h(x)$

then there is a $0<\delta<1$ such that

$\mbox{Osc}\left(h,B\left(x_{0},\frac{R}{2}\right)\right)\leq(1-\delta)\mbox{Osc}\left(h,B(x_{0},R)\right).$

The oscillation inequality will then imply Hölder continuity for $h$ and more
importantly the Strong Liouville property, which asserts that if $h\geq 0$ and
is harmonic on $\mathbb{Z}^{d}$ then $h$ is constant (See [Bar17, Theorem
1.46]).

Acknowledgements: We would like to thank Martin Barlow and D. Yogeshwaran for
useful discussions and feedback on a draft of the note. We also thank Greg
Lawler for pointing us to [Law13, Theorem 1.7.2]. This work was done as part
of a summer reading seminar when Nitya Gadhiwala and Ritvik Radhakrishnan were
undergraduate students in the B. Math. (Hons.) program at the Indian
Statistical Institute, Bangalore centre. Siva Athreya was partially supported
by a CPDA grant at the Indian Statistical Institute, Bangalore centre.

Layout for rest of the paper: In Section 2 we will prove Theorem 1.1 assuming
the three Propositions. Then we prove the Gaussian bounds in Section 3, the
uniform Green inequality in Section 4, and the Balayage formula in Section 5,
to complete the proof.

Convention on constants. Throughout the article $c,c_{1},c_{2},\ldots$ denote
positive constants whose value may change from line to line. They are all
positive and their precise values are not important. In statements of results
will use capital letters $A,B,\ldots$ to denote special constants. Also,
throughout this note the constants depend on the dimension $d$, but will not
be mentioned explicitly. Any other dependence will be stated.

##  2 Proof of Theorem 1.1

When $d=1$, it is easy to see that any harmonic function
$h:\mathbb{Z}\rightarrow\mathbb{R}$ is of the form $h(k)=\alpha k+\beta$ for
some real numbers $\alpha,\beta.$ Then (1.1) follows immediately. We will now
present a proof when $d\geq 2.$

Let $X$ be the simple random walk on $\mathbb{Z}^{d}$. That is, $X$ is a
discrete time Markov chain $(\\{X_{n}\\}_{n\geq
0},\\{\mathbb{P}^{x}\\}_{x\in\mathbb{Z}^{d}})$ with transition matrix
$P=[p_{xy}]$ given by

$p_{xy}=\left\\{\begin{array}[]{ll}\frac{1}{2d}&\mbox{ if }x\sim y,\\\
0&\mbox{ otherwise,}\end{array}\right.$

for any $x,y\in\mathbb{Z}^{d}$ on a measure space $(\Omega,{\cal F})$. Under
the probability $\mathbb{P}^{x}$ we have $\mathbb{P}^{x}(X_{0}=x)=1$. Note
that $\Delta=(P-I)$. For any $n\geq 1$, we will denote the $n$-step
probability function of $X$ by

$p_{n}(x,y)=\mathbb{P}^{x}(X_{n}=y),$ (2.1)

with $p_{1}(x,y)=p_{xy}$. The first step in the proof is to establish the
following Gaussian upper and lower bounds for the $n$-step probabilities of
the random walk.

######  Proposition 2.1 (Gaussian Bounds).

Let $X_{n}$ be the simple random walk on $\mathbb{Z}^{d}$ with $n$-step
probability function $p_{n}(\cdot,\cdot)$ given by (2.1). Then

  1. (a)

there exists $L_{1}>0$ and $L_{2}>0$ such that for all
$x,y\in\mathbb{Z}^{d},n\geq\max\\{1,d(x,y)\\}$

$p_{n}(x,y)+p_{n+1}(x,y)\geq\frac{L_{1}}{n^{\frac{d}{2}}}\exp\left(-L_{2}\frac{d^{2}(x,y)}{n}\right),$
(2.2)

  2. (b)

there exists $U_{1}>0$ and $U_{2}>0$ such that for all
$x,y\in\mathbb{Z}^{d},n\geq 0$

$p_{n}(x,y)\leq\frac{U_{1}}{(\max\\{n,1\\})^{\frac{d}{2}}}\exp\left(-U_{2}\frac{d^{2}(x,y)}{\max\\{n,1\\}}\right).$
(2.3)

The second step in the proof is to use the Gaussian bounds to establish the
Uniform Green Inequality. Now, for any subset $D\subset\mathbb{Z}^{d}$ define
the entrance time to $D$ and exit time from $D$

$T_{D}=\inf\\{k\geq 0:X_{k}\in D\\}\text{ and }\tau_{D}=\inf\\{k\geq
0:X_{k}\in D^{c}\\}.$

We define the killed $n$-step probability function of $X$ by

$p^{D}_{n}(x,y)=\mathbb{P}^{x}(X_{n}=y,n<\tau_{D}).$ (2.4)

The Green function corresponding to $D$, $g_{D}:D\times
D\rightarrow\mathbb{R}$, is given by

$g_{D}(x,y)=\sum_{n=0}^{\infty}p_{n}^{D}(x,y).$ (2.5)

Thus the Green function represents the expected time spent (or number of
visits) by the walk at vertex $y$ before exiting the domain $D$ when it starts
at $x$. We now state the Uniform Green Inequality.

######  Proposition 2.2 (Uniform Green Inequality).

There exists $G_{1},$ $G_{2}>0$ such that if $R\geq 2,B=B(x_{0},R),x,y\in
B(x_{0},\frac{R}{2})$, $r=\max\\{1,d(x,y)\\}$, then

  1. (a)

for $d\geq 3$

$\frac{G_{1}}{r^{d-2}}\leq g_{B}(x,y)\leq\frac{G_{2}}{r^{d-2}},$ (2.6)

  2. (b)

and for $d=2$

$G_{1}\log\left(\frac{R}{r}\right)\leq g_{B}(x,y)\leq
G_{2}\log\left(\frac{R}{r}\right).$ (2.7)

The uniform Green inequality is a key tool in the proof of EHI due to the fact
that Green function plays a role similar to that of the Poisson Kernel for
harmonic functions in $\mathbb{R}^{d}$. This established by the Balayage
formula for harmonic functions. To state the result, we need one additional
notation. For any $A\subset\mathbb{Z}^{d}$, we define the interior boundary of
$A$ by

$\partial_{i}A=\\{y\in A:\exists\,x\in A^{c}\mbox{ with }x\sim y\\}.$

Figure 1: (Balayage Formula) For harmonic function $h:B\rightarrow R$ and a
$A\subset B$ one can find a $f:\partial_{i}A\rightarrow\mathbb{R}$ such that
$h$ is a weighted sum of $f$ in $A$ with the weights given by the Green
function.

######  Proposition 2.3 (Balayage formula).

Let $R\geq 1$, $x_{0}\in V$, $B=B(x_{0},R),A\subsetneq B$. Assume
$h:\overline{B}\rightarrow\mathbb{R}_{+}$ with $h$ being harmonic in $B.$ Then
there exist $f:\partial_{i}A\rightarrow\mathbb{R}_{+}$ such that

$h(x)=\sum_{y\in\partial_{i}A}g_{B}(x,y)f(y)\,\quad\forall\,x\in A$ (2.8)

Assuming the above Propositions we are now ready to prove the main result. The
proof is essentially an application of the Balayage formula and the uniform
Green inequality, followed by a standard chaining of balls argument in
$\mathbb{Z}^{d}$.

###### Proof of Theorem 1.1.

We will first consider the case when $R\leq 32$. For $x,y\in
B(x_{0},\frac{R}{2})$ such that $x\sim y$ we have

$h(x)=\sum_{z\sim x}\frac{1}{2d}h(z)\geq\frac{1}{2d}h(y).$ (2.9)

Now for arbitary $x_{1},x_{2}\in B(x_{0},\frac{R}{2})$ there exists
$z_{1},z_{2},\ldots z_{k}$ such that $x_{1}\sim z_{1}\sim z_{2}\ldots
z_{k}\sim y$ with $k\leq 32.$ Using (2.9), we have that

$h(x_{1})\geq\frac{1}{(2d)^{k}}h(x_{2})\geq\frac{1}{(2d)^{32}}h(x_{2}).$

As $x_{1},x_{2}$ were arbitrary,

$\max_{y\in B\left(x_{0},\frac{R}{2}\right)}h(y)\leq(2d)^{32}\min_{z\in
B\left(x_{0},\frac{R}{2}\right)}h(z).$ (2.10)

Thus proving (1.1) whenever $R\leq 32$. We shall now assume $R>32$. Set
$B=B(x_{0},R),A=B\left(x_{0},\frac{R}{2}\right).$ Let $x_{1},x_{2}\in
B\left(x_{0},\frac{R}{4}\right)$ and $y\in\partial_{i}A$. Firstly,

$\frac{R}{4}\leq d(x_{i},y)\leq\frac{3R}{4}\qquad\mbox{for $i=1,2$}.$

Applying Proposition 2.2, by (2.7) or (2.6), we have that

$g_{B}(x_{1},y)\leq\frac{3^{d-2}G_{1}}{G_{2}}g_{B}(x_{2},y)\qquad\mbox{ for
all }y\in\partial_{i}(A).$ (2.11)

Using (2.8) from Proposition 2.3, we have that there exists $f$ supported on
$\partial_{i}A$ such that

$h(x)=\sum_{y\in\partial_{i}A}g_{B}(x,y)f(y)\,\quad\forall\,x\in A.$

Applying (2.11) in the above we have that

$h(x_{1})=\sum_{y\in\partial_{i}A}g_{B}(x_{1},y)f(y)\leq\frac{3^{d-2}G_{1}}{G_{2}}\sum_{y\in\partial_{i}A}g_{B}(x_{2},y)f(y)=\frac{3^{d-2}G_{1}}{G_{2}}h(x_{2}).$
(2.12)

whenever $x_{1},x_{2}\in B(x_{0},\frac{R}{4}).$ To complete the proof we use a
standard chaining of balls argument. As $R>32$, it is easy to see that there
exists $N\geq 1$ such that for all $u,v\in B(x_{0},\frac{R}{2})$ there exists
$z_{1},z_{2},\ldots,z_{N}\in B(x_{0},\frac{R}{2})$ such that

Figure 2: Figure depicts that standard chain of balls argument used in the
proof of Theorem 1.1.

$\displaystyle u\in B\left(z_{1},\frac{R}{8}\right),\qquad v\in
B\left(z_{N},\frac{R}{8}\right),\qquad\mbox{ and }\qquad
B\left(z_{j},\frac{R}{8}\right)\cap
B\left(z_{{j+1}},\frac{R}{8}\right)\neq\emptyset\mbox{ for $1\leq j\leq
N-1.$}$

Let us choose corresponding arbitrary points $x_{j}\in
B(z_{j},\frac{R}{8})\cap B(z_{{j+1}},\frac{R}{8})$. Note that
$B(z_{j},\frac{R}{2})\subset B(x_{0},R)$ for all $1\leq j\leq N-1$. Thus $h$
satisfies the hypothesis when $R$ is replaced by $\frac{R}{2}$ and $x_{0}$ is
replaced by $z_{j}$. Applying (2.12) with $R$ replaced by $\frac{R}{2}$ and
$x_{0}$ successively replaced by $z_{j}$ we have that

$h(u)\leq\frac{3^{d-2}G_{1}}{G_{2}}h(x_{1})\leq\left(\frac{3^{d-2}G_{1}}{G_{2}}\right)^{2}h(x_{2})\leq\left(\frac{3^{d-2}G_{1}}{G_{2}}\right)^{3}h(x_{3})\leq\ldots\left(\frac{3^{d-2}G_{1}}{G_{2}}\right)^{N}h(v).$

As $u,v$ were arbitary we have that,

$\max_{y\in
B\left(x_{0},\frac{R}{2}\right)}h(y)\leq\left(\frac{3^{d-2}G_{1}}{G_{2}}\right)^{N}\min_{z\in
B\left(x_{0},\frac{R}{2}\right)}h(z).$ (2.13)

We have thus established (1.1) whenever $R>32$. This completes the proof. ∎

##  3 Gaussian Bounds

In this section we prove Proposition 2.1 which provides Gaussian upper and
lower bounds for the $n$-step probability function of the simple random walk.
First we shall use the well known Local Central Limit Theorem to obtain bounds
for the $n$-step probability function near the diagonal. Using this and a
chaining of balls argument we will obtain the Gaussian lower bound. For the
Gaussian upper bound, we use the near diagonal bounds obtained along with an
application of Chernoff bound to a one-dimensional projection of the random
walk.

###  3.1 Near diagonal bounds

The following is the version of the local central limit theorem we use. We
shall say that $n$ and $z\in\mathbb{Z}^{d}$ have the same parity if
$n+|z_{1}|+\cdots+|z_{d}|$ is even.

######  Theorem 3.1 (Local Central Limit Theorem).

Let $\\{X_{n}\\}_{n\geq 0}$ denote a simple symmetric random walk in
$\mathbb{Z}^{d}$. There exists positive constants $A,B,C$ such that for all
positive integers $n$ and points $x,y\in\mathbb{Z}^{d}$ we have

$\left|p_{n}(x,y)-\frac{A}{n^{\frac{d}{2}}}\exp\left(-\frac{Bd(x,y)^{2}}{n}\right)\right|\leq\frac{C}{n^{\frac{d}{2}+1}},$
(3.1)

whenever $n$ and $y-x$ have the same parity.

###### Proof.

We refer the reader to Theorem 1.2.1 in [Law13] for the proof (or for more
detailed explanation of the same in [Gad20]). We note that the statement in
[Law13] has $d(x,y)$ to be standard Euclidean metric namely $||\cdot||_{2}$.
The above statement where $d(x,y)$ is the graph distance easily follows
because we have $c_{1}||x-y||_{2}\leq d(x,y)\leq c_{2}||x-y||_{2}$ for some
constants $c_{1}>0$ and $c_{2}>0$. ∎

######  Proposition 3.2 (Near diagonal bounds).

There exists $N_{1}>0$ such that for all $x,y\in\mathbb{Z}^{d},n\geq 0$

$p_{n}(x,y)\leq\frac{N_{1}}{(\max\\{n,1\\})^{\frac{d}{2}}},$ (3.2)

and there exists $N_{2}>0$ and $L\in(0,1)$ such that for all
$x,y\in\mathbb{Z}^{d},n\geq\frac{1}{L^{2}}\max\\{1,d^{2}(x,y)\\}$

$p_{n}(x,y)+p_{n+1}(x,y)\geq\frac{N_{2}}{n^{\frac{d}{2}}}.$ (3.3)

###### Proof.

We first establish equation (3.2) assuming Theorem 3.1. Choose
$x,y\in\mathbb{Z}^{d}$. We want to bound $p_{n}(x,y)=\mathbb{P}^{x}(X_{n}=y)$
from above. Using the Local Central Limit Theorem we get

$\displaystyle p_{n}(x,y)$
$\displaystyle\leq\frac{C}{n^{\frac{d}{2}+1}}+\frac{A}{n^{\frac{d}{2}}}\exp\left(-\frac{Bd(x,y)^{2}}{n}\right)\leq\frac{C+A}{n^{\frac{d}{2}}}.$

Thus the proof of (3.2) is complete.

Next we show (3.3). Choose
$x=(x_{1},\ldots,x_{d}),y=(y_{1},\ldots,y_{d})\in\mathbb{Z}^{d}$. We want find
a lower bound for
$p_{n}(x,y)+p_{n+1}(x,y)=\mathbb{P}^{x}(X_{n}=y)+\mathbb{P}^{x}(X_{n+1}=y)$.
Since the simple random walk on $\mathbb{Z}^{d}$ has period $2$, exactly one
of $p_{n}(x,y)$ and $p_{n+1}(x,y)$ will be nonzero. The first is nonzero if
$n+|y_{1}-x_{1}|+\cdots+|y_{d}-x_{d}|$ is even and the second if it is odd, in
which case, $n+1+|y_{1}-x_{1}|+\cdots+|y_{d}-x_{d}|$ is even. Let $m$ be
either $n$ or $n+1$ such that $p_{m}(x,y)$ is nonzero, so that $m$ and $y-x$
have the same parity.

It suffices to show that there exists $c>0$ and $L\in(0,1)$ such that for all
$x,y\in\mathbb{Z}^{d}$ and $n\geq\frac{1}{L^{2}}\max\\{1,d^{2}(x,y)\\}$, we
have

$p_{m}(x,y)\geq\frac{c}{n^{\frac{d}{2}}}.$

Using the Local Central Limit Theorem we have

$\displaystyle p_{m}(x,y)$
$\displaystyle\geq\frac{A}{m^{\frac{d}{2}}}\exp\left(-\frac{Bd(x,y)^{2}}{m}\right)-\frac{C}{m^{\frac{d}{2}+1}}$
$\displaystyle\geq\frac{1}{n^{\frac{d}{2}}}\left(c_{1}\exp\left(-\frac{c_{2}d(x,y)^{2}}{n}\right)-\frac{c_{3}}{n}\right).$
(3.4)

Now, let $n\geq\frac{1}{L^{2}}\max\\{1,d^{2}(x,y)\\}$, for some $L\in(0,1)$
that will be determined later. So we have
$L^{2}\geq\frac{1}{n}\max\\{1,d^{2}(x,y)\\}$. Replacing this in (3.4) we get

$p_{m}(x,y)\geq\frac{1}{n^{\frac{d}{2}}}\left(c_{1}\exp(-c_{2}L^{2})-c_{3}L^{2}\right)$

Now choose $0<\delta<1$ such that for $0<L<\delta$ we have

$c_{1}\exp(-c_{2}L^{2})>\frac{c_{1}}{2}\qquad\mbox{ and }\qquad
c_{3}L^{2}<\frac{c_{1}}{4}.$

So for $0<L<\delta$ we get

$p_{m}(x,y)\geq\frac{c_{1}}{2n^{\frac{d}{2}}}.$

The proof of (3.3) is complete. ∎

###  3.2 Lower Bound

To establish the Gaussian lower bound we consider three cases, divided
according to how $R=d(x,y)$ compares with $n$. For the third case we apply a
chaining argument, and for this we require the following estimate on the
volume of a ball in $\mathbb{Z}^{d}$. Namely, there exists a constant
$V_{1}\leq 2d$ such that for all $x_{0}\in\mathbb{Z}^{d}$ and $r\geq 1$ we
have

$\left|B(x_{0},r)\right|\geq V_{1}r^{d}.$ (3.5)

The below proof follows the same structure as in the proof of [Bar17,
Proposition 4.38].

###### Proof of Proposition 2.1(a).

Let $x,y\in\mathbb{Z}^{d},R=d(x,y)$ and let $L>0$ be as in (3.3).

Case 1:  Let $R\leq n\leq\frac{2^{6}}{L^{2}}R$. Then,

$p_{n}(x,y)+p_{n+1}(x,y)\geq\left(\frac{1}{2d}\right)^{n+1}\geq
c_{1}\exp(-c_{2}n)\geq\frac{c_{1}}{n^{\frac{d}{2}}}\exp\left(-c_{3}\frac{R^{2}}{n}\right).$

This implies (2.2).

Case 2: Let $n\geq\frac{R^{2}}{L^{2}}$. In this case (3.3) implies (2.2).

Figure 3: The basic idea of chaining of balls in the proof of Case 3 in
Proposition 2.1(a) is to divide the random walk path into
$m\approx\frac{R^{2}}{n}$ steps. This will imply that length of each segment
$r_{i}$ above are approximately $\frac{R}{m}$, the time taken $s_{i}$ above
are approximately $\frac{n}{m}$, and we can use (3.3) for each segment $i$,
$1\leq i\leq m$ to obtain the result. The details of the proof ensures that
there exists $m,r_{i},s_{i}$ are all natural numbers.

Case 3: Let $\frac{2^{6}}{L^{2}}R\leq n\leq\frac{R^{2}}{L^{2}}$. See Figure 3
for the basic idea of the chaining balls of argument used in the proof here.
Given $n$ in this region choose $m\in\mathbb{N}$ such that

$\frac{2^{5}R^{2}}{L^{2}n}\leq m\leq\frac{2^{6}R^{2}}{L^{2}n}.\\\ $ (3.6)

Note that because $n\geq\frac{2^{6}}{L^{2}}R$, we have $1\leq m\leq R.$ Set
$r=\lfloor\frac{R}{m}\rfloor$ and $s=\lfloor\frac{n}{m}\rfloor.$ Now, there
exists

$z_{1},z_{2},\ldots,z_{m}$: $\displaystyle x=z_{0},\quad y=z_{m},\quad
d(z_{i},z_{i+1})\in\\{r,r+1\\},$ and $s_{1},s_{2},\ldots,s_{m}$:
$\displaystyle s_{i}\in\\{s,s+1\\},\quad\sum_{i=1}^{m}s_{i}=n.$

Let $B_{i}=B(z_{i},r)$ for $0\leq i\leq m-2$. By considering all possible
paths from $x$ to $y$ and the definition of $p_{n}$ we have that

$p_{n}(x,y)+p_{n+1}(x,y)\geq\mathbb{P}^{x}(X_{s_{1}}\in
B_{1})\cdot\prod_{i=1}^{m-2}\min_{w\in B_{i}}\mathbb{P}^{w}(X_{s_{i+1}}\in
B_{i+1})\cdot\min_{w\in B_{m-1}}\mathbb{P}^{w}(X_{s_{m}}=y).$

Next, by our choice of $m$, the lower bound in (3.6) implies that

$(3r+1)^{2}\leq 16r^{2}\leq 16\frac{R^{2}}{m^{2}}\leq L^{2}\frac{n}{2m}\leq
L^{2}s.$ (3.7)

Since we have that $d(w,v)\leq 3r+1$ for all $w\in B_{i}$ and $v\in B_{i+1}$,
the lower bound for $s$ in (3.7) implies that
$s\geq\frac{1}{L^{2}}d(w,v)^{2}$. Hence from (3.3) we have that there exists
$0<c_{1}<1$ such that

$\min_{w\in B_{i}}\mathbb{P}^{w}(X_{s_{i+1}}\in
B_{i+1})\geq\frac{c_{1}}{s^{\frac{d}{2}}}\qquad\mbox{for all}\qquad 0\leq
i\leq m-2.$ (3.8)

Using the volume bound from (3.5) and (3.8) in (3.2) we have

$\displaystyle p_{n}(x,y)$ $\displaystyle\geq$
$\displaystyle\left(\frac{c_{1}V_{1}r^{d}}{\left(s_{1}\right)^{\frac{d}{2}}}\right)\quad\cdot\quad\prod_{i=1}^{m-2}\left(\frac{c_{1}V_{1}r^{d}}{\left(s_{i}\right)^{\frac{d}{2}}}\right)\quad\cdot\quad\left(\frac{c_{1}}{\left(s_{m}\right)^{\frac{d}{2}}}\right).$
(3.9)

By our choice of $m$, the upper bound in (3.6) implies that

$(16)^{2}r^{2}\geq(16)^{2}\frac{R^{2}}{4m^{2}}\geq 64L^{2}\frac{n}{64m}\geq
L^{2}s.$ (3.10)

Using (3.10), the fact that $s_{i}\geq s$ for all $1\leq i\leq m-1$, and
$s_{m}\leq n$ in (3.9) we have

$p_{n}(x,y)\geq\frac{c_{1}}{16^{d}}\left(\frac{c_{1}L^{d}V_{1}}{16^{d}}\right)^{m-1}\frac{1}{n^{\frac{d}{2}}}=\frac{c_{2}}{n^{\frac{d}{2}}}\exp(-c_{3}m)$
(3.11)

for some $c_{2},c_{3}>0.$ Note that $c_{3}>0$ because $V_{1}\leq 2d$ from
(3.5). Inserting the lower bound of $m$ from (3.6) in (3.11) implies (2.2).

∎

######  Remark 1.

The choice of the constant $\frac{2^{6}}{L^{2}}$ in Case 3 is not adhoc. One
could derive it post-facto in the proof as well. To use the Local Central
Limit Theorem in the chain of balls of argument we need a choice of $m$ such
that

$3r+1\leq L\sqrt{s}\leq 16r,$ (see (3.10) and  (3.7)).

This can be achieved if

$4r\leq\frac{1}{L}\sqrt{\frac{n}{2m}}\mbox{ and
}\frac{1}{L}\sqrt{\frac{n}{m}}\leq 8\frac{R}{m},$

as $\frac{n}{2m}\leq s\leq\frac{n}{m}$ and $\frac{R}{2m}\leq
r\leq\frac{R}{m}.$ This is possible if $m$ is an integer as in

$\frac{2^{5}R^{2}}{L^{2}n}\leq m\leq\frac{2^{6}R^{2}}{L^{2}n},\qquad\mbox{
\em({\em see} \eqref{eq:cm}).}$

This forces the choice of the constant in Case 3.

###  3.3 Upper Bound

Recall that, $B=B(x_{0},R)$ and $\tau_{B}=\inf\\{k\geq 0:X_{k}\in B^{c}\\}$.
We note that $\tau_{B}$ is finite with probability one because

$\mathbb{P}(\tau_{B}>n)\leq(1-p)^{\lfloor\frac{R}{n}\rfloor}$ (3.12)

with (say) $p=(\frac{1}{2d})^{3R}$. We will need the following lemma for the
proof of the upper bound.

######  Lemma 3.3.

$\mathbb{P}^{x}(\tau_{B}<n)\leq 2d\exp(-\frac{R^{2}}{4dn}).$

###### Proof.

We first reduce the question for the $d$ dimensional random walk to a one
dimensional walk as follows. Without loss of generality we may assume
$x_{0}=0$ and consider $B=B(0,R)$. Let $\\{X_{n}\\}_{n\geq 0}$ be the simple
random walk on $\mathbb{Z}^{d}$ with $X_{0}=0$. Now, let $\\{Y_{n}\\}_{n\geq
0}$ be the lazy random walk on $\mathbb{Z}$, starting from $0$ with transition
matrix given by $q$

$q(u,v)=\left\\{\begin{array}[]{ll}\frac{1}{2d}&\mbox{ if }u=v\pm 1,\\\ &\\\
\frac{d-1}{d}&\mbox{ if }u=v,\\\ &\\\ 0&\mbox{otherwise.}\end{array}\right.$

Let $X_{n}(k)$ denote the $k$th coordinate of $X_{n}$ for $1\leq k\leq d$.
Note that for each $k$, $\\{X_{n}(k)\\}_{n\geq 0}$ have the same distribution
as $\\{Y_{n}\\}_{n\geq 0}$. Also, if $d(X_{n},0)>R$ then there is a coordinate
$1\leq k\leq d$ for which $X_{n}(k)\geq\frac{R}{d}$, consequently using a
union bound

$\mathbb{P}(\tau_{B}\leq n)\leq d\mathbb{P}(\tau_{\frac{R}{d}}\leq n)$ (3.13)

where $\tau_{S}=\inf\\{k\geq 0:Y_{k}\not\in[-S,S]\\}$ for any $S>0$.

We will understand $\tau_{S}$ for any $S=\frac{R}{d}$ and for that we define
two auxiliary stopping times:

$\tau_{S}^{+}=\left\\{\begin{array}[]{ll}\tau_{S}&\mbox{ if
}Y_{\tau_{S}}>0,\\\ \infty&\mbox{ otherwise,}\end{array}\right.\qquad\mbox{
and }\qquad\tau_{S}^{-}=\left\\{\begin{array}[]{ll}\tau_{S}&\mbox{ if
}Y_{\tau_{S}}<0,\\\ \infty&\mbox{ otherwise.}\end{array}\right.$

Further it is easy to see that

$\\{\tau_{S}\leq n\\}=\\{\tau_{S}^{+}\leq n\\}\dot{\cup}\\{\tau_{S}^{+}\leq
n\\}$

implying

$\mathbb{P}(\tau_{S}\leq n)=\mathbb{P}(\tau_{S}^{+}\leq
n)+\mathbb{P}(\tau_{S}^{-}\leq n).$

We will estimate $\mathbb{P}(\tau_{S}^{+}\leq n)$ and the other term will
follow similarly.

For $S\leq k\leq n,$ let $A_{k}=\\{\tau^{+}_{S}=k\\}.$ Note that
$\mathbb{P}(A_{k})>0$ and

$\\{\tau_{S}^{+}\leq n\\}=\cup_{k=S}^{n}A_{k}.$

Note that,

$\mathbb{E}[Y_{n}|A_{k}]=\mathbb{E}[Y_{k}+Y_{n}-Y_{k}|A_{k}]\geq
S+\mathbb{E}[Y_{n}-Y_{k}]=S$

Let $\lambda>0$ and define

$T_{n}=\exp(\lambda Y_{n})\qquad\mbox{ for }S\leq n.$

By Jensen’s inequality,

$\mathbb{E}[T_{n}]=\sum_{k=S}^{n}{\mathbb{E}[T_{n}|A_{k}]\mathbb{P}(A_{k})}\geq\sum_{k=S}^{n}\exp(\lambda\mathbb{E}[Y_{n}|A_{k}])\mathbb{P}(A_{k})=\exp(\lambda
S)\mathbb{P}(\tau_{S}^{+}\leq n)$

This implies,

$\mathbb{P}(\tau_{S}^{+}\leq n)\leq\frac{\mathbb{E}[T_{n}]}{\exp(\lambda S)}.$
(3.14)

Now it is standard that

$\mathbb{E}[T_{n}]=\left(\frac{d-1}{d}+\frac{\exp(\lambda)}{2d}+\frac{\exp(-\lambda)}{2d}\right)^{n}.$

For $0<\lambda<1,$ we obtain

$\frac{d-1}{d}+\frac{\exp(\lambda)}{2d}+\frac{\exp(-\lambda)}{2d}=\frac{d-1}{d}+\frac{1}{d}\sum_{k=0}^{\infty}\frac{\lambda^{2k}}{(2k)!}=1+\frac{\lambda^{2}}{d}\left(\sum_{k=1}^{\infty}\frac{\lambda^{2(k-1)}}{(2k)!}\right)\leq
1+\frac{\lambda^{2}}{d},$

as
$\sum_{k=1}^{\infty}\frac{\lambda^{2(k-1)}}{(2k)!}<\sum_{k=1}^{\infty}\frac{1}{(2k)!}<1.$
Therefore,

$\displaystyle\mathbb{E}[T_{n}]\leq\left(1+\frac{\lambda^{2}}{d}\right)^{n}\leq\exp\left(n\frac{\lambda^{2}}{d}\right).$

Substituting this estimate in $\eqref{eq:Chernoff}$ and choosing
$\lambda=\frac{dS}{2n}$ (note that $\lambda<1$) gives

$\mathbb{P}(\tau_{S}^{+}\leq n)\leq\exp\left(n\frac{\lambda^{2}}{d}-\lambda
S\right)=\exp\left(-\frac{dS^{2}}{4n}\right).$

Using symmetry of the distribution of the walk, we also have

$\mathbb{P}(\tau_{S}^{-}\leq n)\leq\exp\left(-\frac{dS^{2}}{4n}\right).$

Therefore, we get

$\mathbb{P}(\tau_{S}\leq n)\leq 2\exp\left(-\frac{dS^{2}}{4n}\right).$

As $S=\frac{R}{d}$ in (3.13) we get that,

$\mathbb{P}(\tau_{B}\leq n)\leq d\times\mathbb{P}\left(\tau_{\frac{R}{d}}\leq
n\right)\leq 2d\exp\left(-\frac{R^{2}}{4dn}\right),$

as required.

∎

We are now ready to prove Proposition 2.1 (b). The below proof follows the
same structure as in the proof of [Bar17, Theorem 4.34].

###### Proof of Proposition 2.1 (b).

Let $R=d(x,y)$. As in part (a), we consider three cases.

Case 1: $n<R$. Here $p_{n}(x,y)=0$ and (2.3) follows immediately.

Case 2: $n\geq R^{2}$. This would imply $\frac{d^{2}(x,y)}{n}<1$ which inturn
implies that

$\exp(-1)\leq\exp\left(-\frac{d^{2}(x,y)}{n}\right)\leq 1.$

From above and (3.2) we have

$p_{n}(x,y)\leq\frac{N_{1}}{n^{\frac{d}{2}}}\leq\frac{c_{1}}{n^{\frac{d}{2}}}\exp\left(-\frac{d^{2}(x,y)}{n}\right).$

(2.3) follows.

Case 3: $R\leq n\leq R^{2}$. Let $m\in\mathbb{N}$ be such that

$\frac{n}{3}\leq m\leq\frac{2n}{3}.$ (3.15)

Define the set $A_{x}=\\{z\in\mathbb{Z}^{d}:d(x,z)\leq d(y,z)\\}$ and
$A_{y}=A_{x}^{c}$. Then using time reversibility of the random walk we have

$\displaystyle p_{n}(x,y)$ $\displaystyle=\mathbb{P}^{x}(X_{m}\in
A_{y},X_{n}=y)+\mathbb{P}^{x}(X_{m}\in A_{x},X_{n}=y)$
$\displaystyle=\mathbb{P}^{x}(X_{m}\in
A_{y},X_{n}=y)+\mathbb{P}^{y}(X_{n-m}\in A_{x},X_{n}=x)$ (3.16)

Let ${\cal F}_{m}=\sigma(X_{1},X_{2},\ldots,X_{m})$. Using the Markov Property
we have that

$\displaystyle\mathbb{P}^{x}(X_{m}\in A_{y},X_{n}=y)$
$\displaystyle=\mathbb{P}^{x}(\tau_{B\left(x,\frac{R}{2}\right)}\leq
m,X_{m}\in A_{y},X_{n}=y)$
$\displaystyle=\mathbb{E}^{x}\left(\mathbb{E}^{x}\left(1(\tau_{B\left(x,\frac{R}{2}\right)}\leq
m,X_{m}\in A_{y})\,1(X_{n}=y)|{\cal F}_{m}\right)\right)$
$\displaystyle=\mathbb{E}^{x}\left(1(\tau_{B\left(x,\frac{R}{2}\right)}\leq
m,X_{m}\in A_{y})\,\mathbb{E}^{x}\left(1(X_{n}=y)|{\cal F}_{m}\right)\right)$
$\displaystyle=\mathbb{E}^{x}\left(1(\tau_{B\left(x,\frac{R}{2}\right)}\leq
m,X_{m}\in A_{y})\,\mathbb{E}^{X_{m}}(1(X_{n}=y))\right)$
$\displaystyle\leq\,\mathbb{P}^{x}(\tau_{B\left(x,\frac{R}{2}\right)}\leq
m)\sup_{z\in A_{y}}p_{n-m}(z,y).$

and similarly

$\mathbb{P}^{y}(X_{n-m}\in
A_{x},X_{n}=x)\,\leq\,\mathbb{P}^{y}(\tau_{B\left(y,\frac{R}{2}\right)}\leq
n-m)\sup_{z\in A_{x}}p_{m}(x,y).$

Using the above, along with Lemma 3.3 and (3.2) in (3.3) we have that there
exists $c_{1},c_{2},c_{3},c_{4}>0$ such that

$\displaystyle p_{n}(x,y)\leq
c_{1}\exp\left(-c_{2}\frac{R^{2}}{m}\right)\frac{N_{1}}{(n-m)^{\frac{d}{2}}}+c_{3}\exp\left(-c_{4}\frac{R^{2}}{n-m}\right)\frac{N_{1}}{m^{\frac{d}{2}}}.$

(2.3) then follows using the bounds on $m$ from (3.15). ∎

##  4 Uniform Green Inequality

In this section we shall prove Proposition 2.2. To prove the result we obtain
Gaussian bounds for the $n$-step killed probability function.

Recall that $B=B(x_{0},R)$ and for any $x,y\in\mathbb{Z}^{d}$ the $n$-step
killed probability function is given by,

$p^{B}_{n}(x,y)=\mathbb{P}^{x}(X_{n}=y,n<\tau_{B}).$

Clearly, $p_{n}^{B}(x,y)\leq p_{n}(x,y).$ So the Gaussian upper bounds follows
from Proposition 2.1. The Gaussian lower bound is not that immediate and is
presented below.

######  Lemma 4.1 (Killed -Gaussian Lower Bound).

Let $d\geq 1$. $\exists A>0,C>0$ such that for all $R\geq 2$,
$x_{0}\in\mathbb{Z}^{d}$, and $B=B(x_{0},R)$, $x,y\in
B^{\prime}=B(x_{0},\frac{R}{2})$, $\max\\{1,d(x,y)\\}\leq n\leq R^{2}$

$p^{B}_{n}(x,y)+p^{B}_{n+1}(x,y)\geq\frac{A}{n^{\frac{d}{2}}}\exp\left(-C\frac{d^{2}(x,y)}{n}\right)$
(4.1)

###### Proof.

Let $x,y\in B(x_{0},\frac{R}{2})$. We shall first prove that
$\exists\,0<\delta<\gamma<1$ such that if

$d(x,y)\leq\delta R$ and $d(x,y)\leq n\leq\gamma R^{2}$ then (4.1) holds.

Then we shall use a chaining argument similar to proof of (2.2) for the
general case. First we have

$\displaystyle p_{n}(x,y)$ $\displaystyle=\mathbb{P}^{x}(X_{n}=y)$
$\displaystyle=\mathbb{P}^{x}(X_{n}=y,\tau_{B}>n)+\mathbb{P}^{x}(X_{n}=y,\tau_{B}\leq
n)$ $\displaystyle=p_{n}^{B}(x,y)+\mathbb{P}^{x}(X_{n}=y,\tau_{B}\leq n).$

As,

$\mathbb{P}^{x}(X_{n}=y,\tau_{B}\leq n)=\mathbb{E}^{x}(I_{(\tau_{B}\leq
n)}\mathbb{P}^{X_{\tau_{B}}}(X_{n-\tau_{B}}=y))\leq\max_{z\in\partial
B}\max_{1\leq m\leq n}\mathbb{P}^{z}(X_{m}=y)$

we have

$p_{n}^{B}(x,y)\geq p_{n}(x,y)-\max_{z\in\partial B}\max_{1\leq m\leq
n}p_{m}(z,y).$ (4.2)

We also have a similar lower bound for $p^{B}_{n+1}(x,y)$. Now, for
$z\in\partial B$ and $y\in B(x_{0},\frac{R}{2})$ we have
$d(z,y)\geq\frac{R}{2}.$ So, Proposition 2.1 (b) implies that $\exists
c_{1}>0$ and $c_{2}>0$ such that

$\max_{z\in\partial B}\max_{1\leq m\leq n}p_{m}(z,y)\leq\max_{1\leq m\leq
n}\frac{c_{1}}{m^{\frac{d}{2}}}\exp\left(-c_{2}\frac{R^{2}}{m}\right).$

Consider the function $f:\mathbb{R}\rightarrow\mathbb{R}$ given by
$f(m)=\frac{c_{1}}{m^{\frac{d}{2}}}\exp\left(-c_{2}\frac{R^{2}}{m}\right)$. It
is easy to see that $f(\cdot)$ is unimodal with maximum at $m=cR^{2}$, for
some constant $c>0$. Therefore if $n<cR^{2}$, then over the range $1\leq m\leq
n$, $f(m)$ is maximised at $m=n$. So for $n<cR^{2}$, we have

$\max_{z\in\partial B}\max_{1\leq m\leq
n}p_{m}(z,y)\leq\frac{c_{1}}{n^{\frac{d}{2}}}\exp\left(-c_{2}\frac{R^{2}}{n}\right).$

Using this and (2.2) in (4.2), for $\max\\{1,d(x,y)\\}\leq n\leq cR^{2}$ we
have that there exists $c_{3},c_{4},c_{5}>0$ such that

$p^{B}_{n}(x,y)+p^{B}_{n+1}(x,y)\geq
c_{3}\frac{1}{n^{\frac{d}{2}}}\left(\exp\left(-c_{4}\frac{(d(x,y))^{2}}{n}\right)-\exp\left(-c_{5}\frac{R^{2}}{n}\right)\right).$

So we can choose $0<\delta<\gamma<1$ (independent of $R$) such that if
$d(x,y)\leq\delta R$ and $\max\\{1,d(x,y)\\}\leq n\leq\gamma R^{2}$, we get

$p^{B}_{n}(x,y)+p^{B}_{n+1}(x,y)\geq
c_{6}n^{-\frac{d}{2}}\exp\left(-c_{7}\frac{(d(x,y))^{2}}{n}\right).$ (4.3)

We are now ready to prove the general case. Suppose
$R_{0}=\lfloor\frac{4}{\delta^{2}}\rfloor+1$. Then there is $c_{8}>0$ such
that for all $2\leq R\leq R_{0}$, $x,y\in B(x_{0},\frac{R}{2})$ and
$\max\\{1,d(x,y)\\}\leq n\leq R^{2}$, such that

$p^{B}_{n}(x,y)+p^{B}_{n+1}(x,y)\geq\left(\frac{1}{2d}\right)^{n+1}=\frac{1}{2d}\exp(-c_{8}n).$

The above is true because for any $n\geq 1$ we can find a path of length $n$
(perhaps overlapping) inside $B(x_{0},R)$ connecting $x$ and $y$. The constant
$c_{8}>0$ will only depend on $R_{0}$ and $d$. Then it is immediate to see
that $\exists A_{1}>0,C_{1}>0$ such that for all $2\leq R\leq R_{0}$,
$x_{0}\in\mathbb{Z}^{d}$, and $B=B(x_{0},R)$, $x,y\in
B^{\prime}=B(x_{0},\frac{R}{2})$ , $\max\\{1,d(x,y)\\}\leq n\leq R^{2}$

$p^{B}_{n}(x,y)+p^{B}_{n+1}(x,y)\geq\frac{A_{1}}{n^{\frac{d}{2}}}\exp\left(-C_{1}\frac{d^{2}(x,y)}{n}\right)$
(4.4)

with $A_{1}>0,C_{1}>0$ depending only on $R_{0}$.

We shall assume now that $R>R_{0}$. We will apply a chaining of balls
argument. Let $x,y\in B(x_{0},\frac{R}{2})$, and $\delta R\leq u=d(x,y)\leq
R$. We will consider two cases.

Case 1 : $u\leq n\leq\frac{4}{\delta}u$. There is $c_{1}>0$ such that

$p^{B}_{n}(x,y)+p^{B}_{n+1}(x,y)\geq\left(\frac{1}{2d}\right)^{n+1}\geq\exp(-c_{1}n).$

As before, it is easy to see in the current range of $n$ that $\exists
A_{2}>0,C_{2}>0$ such that

$p^{B}_{n}(x,y)+p^{B}_{n+1}(x,y)\geq\frac{A_{2}}{n^{\frac{d}{2}}}\exp\left(-C_{2}\frac{d^{2}(x,y)}{n}\right)$
(4.5)

with $A_{2}>0$ and $C_{2}>0$ depending only on $d,\delta>0$.

Case 2 : $\frac{4}{\delta}u\leq n\leq R^{2}$.

Let $m=\lfloor\frac{2}{\delta}\rfloor.$ Set $r=\lfloor\frac{u}{m}\rfloor$, and
$s=\lfloor\frac{n}{m}\rfloor.$ Now it is clear that as $\frac{1}{\delta}\leq
m\leq\frac{2}{\delta}$ and $R>R_{0}$ we have

$r\geq\frac{u}{2m}\geq\frac{\delta^{2}R}{4}>1\mbox{ and
}s\geq\frac{n}{2m}\geq\frac{\delta n}{2}\geq 2u>1.$

There exists

$\displaystyle z_{1},z_{2},\ldots,z_{m}:$ $\displaystyle x=z_{0},\quad
y=z_{m},\quad d(z_{i},z_{i+1})\in\\{r,r+1\\},$ and $\displaystyle
s_{1},s_{2},\ldots,s_{m}:$ $\displaystyle s_{i}\in\\{s,s+1\\},\mbox{ and
}\sum_{i=1}^{m}s_{i}=n.$

Let $B_{i}=B(z_{i},r)$ for $1\leq i\leq m-1$. By considering all possible
paths from $x$ to $y$ and definition of $p^{B}_{n}$ we have that

$\displaystyle p^{B}_{n}(x,y)+p^{B}_{n+1}(x,y)$ $\displaystyle\geq$
$\displaystyle\mathbb{P}^{B,x}(X_{s_{1}}\in
B_{1})\cdot\prod_{i=1}^{m-2}\min_{w\in B_{i}}\mathbb{P}^{B,w}(X_{s_{i+1}}\in
B_{i+1})\cdot\min_{w\in B_{m-1}}\mathbb{P}^{B,w}(X_{s_{m}}=y)$

Note that :

$\frac{1}{\delta}\leq m\leq\frac{2}{\delta},\quad\frac{u}{2m}\leq
r\leq\frac{u}{m}\mbox{ and }\frac{n}{2m}\leq s\leq\frac{n}{m}$ (4.7)

and so

$\displaystyle r\leq\delta R,\qquad
3r+1<4r<4\frac{u}{m}<\frac{n\delta}{m}<\frac{n}{2m}<s,\qquad\mbox{ and
}s<\frac{n}{m}<\delta R^{2}.$

Using the above, we can apply (4.3) to all the terms in (LABEL:eqa:kbstep1) to
obtain

$\displaystyle p^{B}_{n}(x,y)+p^{B}_{n+1}(x,y)$ $\displaystyle\geq$
$\displaystyle
c_{1}^{m}\frac{\exp(-c_{2}\frac{mr^{2}}{s})}{s^{\frac{d}{2}}}\left(\frac{r}{\sqrt{s}}\right)^{d(m-1)}.$

Using (4.7) in the above we have

$\displaystyle p^{B}_{n}(x,y)+p^{B}_{n+1}(x,y)$ $\displaystyle\geq$
$\displaystyle\frac{\exp(-c_{2}\frac{u^{2}}{n})}{n^{\frac{d}{2}}}c_{1}^{m}m^{\frac{d}{2}}\left(\frac{u}{2\sqrt{mn}}\right)^{d(m-1)}.$

Using the fact that $u\geq\delta R$ and $n\leq R^{2}$ we have .

$\displaystyle p^{B}_{n}(x,y)+p^{B}_{n+1}(x,y)$ $\displaystyle\geq$
$\displaystyle
c_{1}\frac{\exp(-c_{2}\frac{u^{2}}{n})}{n^{\frac{d}{2}}}c_{3}^{m}m^{d-\frac{dm}{2}}.$

As $m=\lfloor\frac{2}{\delta}\rfloor$ we have

$\displaystyle p^{B}_{n}(x,y)+p^{B}_{n+1}(x,y)$ $\displaystyle\geq$
$\displaystyle c_{4}\frac{\exp(-c_{2}\frac{u^{2}}{n})}{n^{\frac{d}{2}}}$ (4.8)

So the proof is complete.

∎

###### Proof of Proposition 2.2.

Let $x,y\in\mathbb{Z}^{d}$. Recall that $r=\max\\{1,d(x,y)\\}.$ We begin with
the lower bounds. From (2.5) we have that

$g_{B}(x,y)=\sum_{n=0}^{\infty}{p_{n}^{B}(x,y)}\geq\sum_{n=r^{2}}^{R^{2}}{p_{n}^{B}(x,y)}.$

Using Lemma 4.1, we get that there exists $c_{1},c_{2}>0$ such that

$g_{B}(x,y)\geq c_{1}\sum_{n=r^{2}}^{R^{2}}\frac{1}{n^{\frac{d}{2}}}\geq
c_{2}\int_{r^{2}}^{R^{2}}{s^{-\frac{d}{2}}}ds.$ (4.9)

In the last inequality we have used the fact that sum is a upper bound for the
Reimann integral. In the case $d=2$, evaluating the integral in (4.9) we
obtain

$g_{B}(x,y)\geq 2\log(R/r).$

In the case $d>2$, evaluating the integral in (4.9) we obtain

$\displaystyle g_{B}(x,y)$
$\displaystyle\geq\int_{r^{2}}^{R^{2}}{s^{-\frac{d}{2}}ds}=\frac{2}{d-2}\left(\frac{1}{r^{d-2}}-\frac{1}{R^{d-2}}\right)$
$\displaystyle\geq\frac{2}{d-2}\left(\frac{1}{r^{d-2}}-\frac{1}{(2r)^{d-2}}\right)$
$\displaystyle=\left(\frac{2}{d-2}\right)\left(1-\frac{1}{2^{d-2}}\right)\frac{1}{r^{d-2}},$

where we have used the assumption $r\leq\frac{R}{2}$ in the second last
inequality. Thus we have completed the proof of the lower bound.

Next, we prove the upper bound for the case $d>2$. As $p_{n}^{B}(x,y)\leq
p_{n}(x,y)$, we will use using the Gaussian upper bound (2.3) in Proposition
2.1(b) to obtain

$\displaystyle g_{B}(x,x)$
$\displaystyle=\sum_{n=0}^{\infty}p^{B}_{n}(x,x)\leq\sum_{n=0}^{\infty}p_{n}(x,x)\leq
1+c_{1}\sum_{n=1}^{\infty}\frac{1}{n^{\frac{d}{2}}}\leq c_{2},$

as required in the upper bound and for $x\neq y$, we get

$\displaystyle g(x,y)$
$\displaystyle=\sum_{n=0}^{\infty}{p_{n}(x,y)}\leq\sum_{n=0}^{\infty}{\frac{c_{1}}{n^{\frac{d}{2}}}}\exp{\left(-c_{2}\frac{r^{2}}{n}\right)}\leq
c_{3}\int_{0}^{\infty}{\frac{c_{1}}{s^{\frac{d}{2}}}\exp\left(-c_{2}\frac{r^{2}}{s}\right)ds}.$

Changing variables in the final integral using $t=\frac{s}{r^{2}}$ we obtain

$\int_{0}^{\infty}{\frac{c_{1}}{s^{\frac{d}{2}}}\exp\left(-c_{2}\frac{r^{2}}{s}\right)ds}=\frac{c_{1}}{r^{d-2}}\int_{0}^{\infty}{\frac{\exp\left(-\frac{c_{2}}{t}\right)}{s^{\frac{d}{2}}}}ds=\frac{c_{4}}{r^{d-2}},$

for some $c_{4}>0$.

Finally, we prove the upper bound for $d=2$. Splitting $g_{n}(x,y)$ suitably
and using $p_{n}^{B}(x,y)\leq p_{n}(x,y)$ we have that there exists
$c_{1},c_{2},c_{3},c_{4}>0$ such that

$\displaystyle g_{B}(x,y)$
$\displaystyle=p_{0}^{B}(x,y)+\sum_{n=1}^{r^{2}}{p_{n}^{B}(x,y)}+\sum_{n=r^{2}+1}^{R^{2}}{p_{n}^{B}(x,y)}+\sum_{n=R^{2}}^{\infty}{p_{n}^{B}(x,y)}$
$\displaystyle\leq
1+\sum_{n=1}^{r^{2}}{p_{n}(x,y)}+\sum_{n=r^{2}}^{R^{2}}{p_{n}(x,y)}+\sum_{n=R^{2}}^{\infty}\mathbb{P}^{x}(\tau_{B}>n)$
$\displaystyle\leq
1+\sum_{n=1}^{r^{2}}{p_{n}(x,y)}+\sum_{n=r^{2}}^{R^{2}}{p_{n}(x,y)}+\sum_{n=R^{2}}^{\infty}{c_{2}\exp\left(-c_{2}\frac{R}{n}\right)}$
$\displaystyle\leq
1+\sum_{n=1}^{r^{2}}{p_{n}(x,y)}+\sum_{n=r^{2}+1}^{R^{2}}{\frac{c_{3}}{n}}+c_{4},$

where we have used (3.12) in third inequality and (3.2) in the final one. For
the third term we have the estimate

$\sum_{n=r^{2}+1}^{R^{2}}{\frac{c_{1}}{n}}\leq c_{3}\log(\frac{R}{r})$

and for the second term using (2.3) and using the bound with Riemann integral
we have that there exists, $c_{5},c_{6}>0$ such that

$\sum_{n=1}^{r^{2}}{p_{n}(x,y)}\leq
c_{5}\int_{0}^{r^{2}}{t^{-1}\exp\left(-B\frac{r^{2}}{t}\right)}dt=c_{6}\int_{0}^{1}{s^{-1}\exp\left(-\frac{B}{s}\right)}ds=c_{7}.$

So we get that there exists $c_{7}>0$ such that

$g_{B}(x,y)\leq c_{7}\log\left(\frac{R}{r}\right),$

whenever $R\geq 2r$. This completes the proof for the upper bound in the case
$d=2$. ∎

##  5 Balayage Formula

In this section we shall prove Proposition 2.3. We begin with another
classical tool, a probabilistic representation for solutions of the Dirichlet
problem in a domain.

######  Lemma 5.1.

Let $R\geq 1,x_{0}\in\mathbb{Z}^{d}$, $D\subset B(x_{0},R),$ and
$\phi:\partial B(x_{0},R)\rightarrow\mathbb{R}_{+}$. Then
$h:\overline{B(x_{0},R)}\rightarrow\mathbb{R}_{+}$ given by

$h(x)=\mathbb{E}^{x}[\phi(X_{\tau_{D}})]$ (5.1)

is a unique solution to

$\displaystyle\Delta h=0$ $\displaystyle\mbox{on }D,$ $\displaystyle h=\phi$
$\displaystyle\mbox{on }\partial D.$ (5.2)

###### Proof of Lemma 5.1.

By (3.12), we know that $\tau_{B(x_{0},R)}<\infty$ with probability 1 and
hence $\tau_{D}<\infty$ with probability 1. Then using the Markov property and
the one-step probability we have

$\displaystyle h(x)$ $\displaystyle=\mathbb{E}^{x}[\phi(X_{\tau_{D}})]$
$\displaystyle=\sum_{y\in\mathbb{Z}^{d}}\mathbb{E}^{x}[\phi(X_{\tau_{D}});X_{1}=y]$
$\displaystyle=\sum_{y\in\mathbb{Z}^{d}}p_{1}(x,y)\mathbb{E}^{y}[\phi(X_{\tau_{D}})]$
$\displaystyle=Ph(x)$

This along with the fact that $\mathbb{P}^{y}[\tau_{D}=0]=1$ for all
$y\in\partial D$ implies that $h$ solves (5.2).

We will now show uniqueness. Suppose there are two solutions $h_{1}$ and
$h_{2}$ to (5.2). Then $h=h_{1}-h_{2}$ solves (5.2) with $\phi\equiv 0$.
Extend $h$ to be $0$ on all of $\overline{B(x_{0},R)}^{c}.$ Then

$\displaystyle\sum_{x\in\mathbb{Z}^{d}}\sum_{y\in\mathbb{Z}^{d}}p_{1}(x,y)(h(x)-h(y))^{2}$
$\displaystyle=\sum_{x\in\mathbb{Z}^{d}}\sum_{y\in\mathbb{Z}^{d}}p_{1}(x,y)(h(x)-h(y))h(x)-\sum_{x\in\mathbb{Z}^{d}}\sum_{y\in\mathbb{Z}^{d}}p_{1}(x,y)(h(x)-h(y))h(y)$

Since the terms in the summand are non-zero only for finitely many terms, we
can interchange the sums and we have

$\displaystyle\sum_{x\in\mathbb{Z}^{d}}\sum_{y\in\mathbb{Z}^{d}}p_{1}(x,y)(h(x)-h(y))^{2}$
$\displaystyle=2\sum_{x\in\mathbb{Z}^{d}}\sum_{y\in\mathbb{Z}^{d}}p_{1}(x,y)(h(x)-h(y))h(x)$
$\displaystyle=-2\sum_{x\in\mathbb{Z}^{d}}h(x)\left(\sum_{y\in\mathbb{Z}^{d}}p_{1}(x,y)(h(y)-h(x))\right)$
$\displaystyle=-2\sum_{x\in\mathbb{Z}^{d}}h(x)(Ph(x)-h(x))$
$\displaystyle=-2\sum_{x\in\mathbb{Z}^{d}}h(x)\Delta h(x)$ $\displaystyle=0.$

Thus $h\equiv 0$ and $h_{1}=h_{2}$. ∎

######  Remark 2.

An alternative argument for uniqueness proof is the following. Let $h$ be as
in the proof. The maximum principle for harmonic functions implies that the
maximum and the minimum of $h$ is attained on the boundary, and hence $h$ must
be zero.

We are now ready to prove Proposition 2.3

###### Proof of Proposition 2.3.

Define $h_{A}:\bar{B}\rightarrow\mathbb{R}_{+}$ by

$h_{A}(x)=\mathbb{E}^{x}[h(X_{T_{A}});T_{A}<\tau_{B}].$

From the definition of $\tau_{B}$ and $T_{A}$ it is immediate that

$h_{A}(x)=\mathbb{E}^{x}[1_{A}(X_{\tau_{B\setminus A}})h(X_{\tau_{B\setminus
A}})]=\mathbb{E}^{x}[\phi(X_{\tau_{B\setminus A}})],$ (5.3)

where $\phi:\overline{B}\rightarrow\mathbb{R}_{+}$ is given by
$\phi(x)=1_{A}(x)h(x)$. By the definition of $h_{A}$ we have $h_{A}\equiv h$
on $A$. This and Lemma 5.1 will imply that

$\Delta h_{A}=0\mbox{ on }A^{\circ}\cup B\setminus A$ (5.4)

For $x\in\partial_{i}A$, $h_{A}(x)=h(x)=Ph(x)\geq Ph_{A}(x)$. Thus we have

$\Delta h_{A}\leq 0\mbox{ on }\partial_{i}A.$ (5.5)

Define $f:B\rightarrow\mathbb{R}_{+}$ by

$f=(I_{B}-P_{1}^{B})h_{A},$

where $I_{B}g(x)=1_{B}(x)g(x)$ and $P_{1}^{B}g(x)=\sum_{y\in
B}p_{1}^{B}(x,y)g(x)$ for $g:\mathbb{Z}^{d}\rightarrow\mathbb{R}$. From
definition of $h_{A}$ we have that $h_{A}\equiv 0$ on $\partial B$, so

$\Delta
h_{A}=(P-I)h_{A}=(P-I)(I_{B}h_{A})=P(I_{B}h_{A})-I_{B}h_{A}=(P_{1}^{B}-I_{B})h_{A}=-f.$

Using (5.4) and (5.5), we have that $f\geq 0$ and
$\mbox{support}(f)\subset\partial_{i}(A)$. Define for $k\geq 1$,

$P_{k}^{B}g(x)=\sum_{y\in B}p_{k}^{B}(x,y)g(x)$

for $g:\mathbb{Z}^{d}\rightarrow\mathbb{R}$. Now for $n\geq 1$, by the
definition of $f$ and $h_{A}$ we have,

$\displaystyle\left|\sum_{k=0}^{n}P^{B}_{k}f(x)-h_{A}(x)\right|$
$\displaystyle=$
$\displaystyle\left|\sum_{k=0}^{n}P^{B}_{k}(P^{B}-I)h_{A}(x)-h_{A}(x)\right|$
$\displaystyle=$
$\displaystyle\left|\left[\sum_{k=0}^{n}\left(P^{B}_{k+1}h_{A}(x)-P^{B}_{k}h_{A}(x)\right)\right]-h_{A}(x)\right|$
$\displaystyle=$ $\displaystyle P^{B}_{n+1}h_{A}(x)$ $\displaystyle=$
$\displaystyle\sum_{y\in B}p_{k}^{B}(x,y)h_{A}(x)$ $\displaystyle\leq$
$\displaystyle c_{1}\mathbb{P}^{x}(\tau_{B}>n).$

Using (3.12), we have that for all $x\in B$,

$\sum_{k=0}^{\infty}P_{k}^{B}f(x)=h_{A}(x).$

To conclude we have,

$\sum_{y\in\partial_{i}A}g_{B}(x,y)f(y)=\sum_{y\in
B}g_{B}(x,y)f(y)=\sum_{n=0}^{\infty}P_{n}^{B}f(x)=h_{A}(x)=h(x).$

∎

## References

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vector $\psi$ that sends the right-hand side of (50) to either positive
infinity or negative infinite in the limit.

Proof. [Proof of Theorem 3.9] In Eq.(12) we have separated the cointegrated
variables into the active ones and the inactive ones

$\displaystyle y_{i}$ $\displaystyle=z_{i\cdot}\alpha++x_{i\cdot}\beta+u_{i}$
$\displaystyle=z_{i\cdot}\alpha+\sum_{l\in C^{*}}x_{il}^{c}\phi_{l}+\sum_{l\in
C^{*c}}x_{il}^{c}\phi_{l}+x_{i\cdot}\beta+u_{i}.$ (51)

We proceed our argument conditioning on the three events

$\displaystyle S_{1}$ $\displaystyle=$
$\displaystyle\left\\{M^{*}\cap\mathcal{\mathcal{I}}=\widehat{M}\cap\mathcal{\mathcal{I}}\right\\}$
$\displaystyle S_{2}$ $\displaystyle=$
$\displaystyle\left\\{C^{*}\subseteq\widehat{C}\right\\}$ $\displaystyle
S_{3}$ $\displaystyle=$
$\displaystyle\left\\{\mathrm{CoRk}(M^{*})=\mathrm{CoRk}(\widehat{M})\right\\}.$

According to Theorem 3.5, these events occur wpa1, given sufficiently large
sample size.

Under these three events, the regression equation (51) is reduced to

$\displaystyle y_{i}$ $\displaystyle=$ $\displaystyle
z_{iM^{*}}\alpha_{M^{*}}+\sum_{l\in C^{*}}x_{il}^{c}\phi_{l}+\sum_{l\in
C^{*c}\cap\widehat{C}}x_{il}^{c}\phi_{l}+x_{iM^{*}}\beta_{M^{*}}+u_{i},$ (52)

where on the right-hand side of the above equation the first and the fourth
terms are present by the event $S_{1}$, the second and the third terms by
$S_{2}$. Due to event $S_{3}$, there is no cointegration group in the third
term, so we can re-write (52) as

$y_{i}=z_{iM^{*}}\alpha_{M^{*}}+\sum_{l\in\mathcal{C}}x_{il}^{c}\phi_{l}+\tilde{x}_{i\cdot}^{+}\tilde{\beta}^{+}+u_{i},$
(53)

where $\tilde{x}_{i\cdot}^{+}=\left(\left(x_{il}^{c}\right)_{l\in
C^{*c}\cap\widehat{C}},x_{M^{*}}\right)$ is the collection of _augmented_ pure
I(1) processes in the post-selection regression equation (52) and
$\tilde{\beta}^{+}=\left(\left(\phi_{j}\right)_{l\in
C^{*c}\cap\widehat{C}},\beta_{M^{*}}\right)$ is the corresponding coefficient.

For $l\in C^{*c}\cap\widehat{C}$, the true coefficients are 0. Now these
variables appear as pure I(1) in (53), for which Theorem 3.5 gives variable
selection consistency. We implement the post-selection Alasso in (53), or
equivalently TAlasso as in (27). Among the variables in $\tilde{x}_{i\cdot}$
TAlasso will eliminate $\left(x_{il}^{c}\right)_{l\in C^{*c}\cap\widehat{C}}$
wpa1 while the true active variables $x_{M^{*}}$ will be maintained. In the
mean time, all variables in the first and second terms in (53) are active and
Theorem 3.5’s (25) guarantees their maintenance wpa1. The asymptotic
distribution naturally follows by applying Theorem 3.5(a) to (52).

Proof. [Proof of Corollary 3.10] For Part (a) and (b), we start with the local
perturbation
$\check{v}=\left(\check{v}_{z}^{\prime},\check{v}_{1}^{\prime},\check{v}_{2}^{\prime},\check{v}_{x}^{\prime}\right)^{\prime},$
$v_{n}=Q^{-1}R_{n}^{-1}\check{v}$ and $\theta_{n}=\theta_{n}^{\ast}+v_{n}$.
Notice that $\check{v}$ is different from $\tilde{v}$ in Theorem 3.5 as we do
not impose $\check{v}_{2}=0$. Define

$V_{n}(\check{v})=v_{n}^{\prime}W^{\prime}Wv_{n}-2v_{n}^{\prime}W^{\prime}u+\lambda_{n}\sum_{j=1}^{p}(|\theta_{jn}^{\ast}+v_{n}|-|\theta_{jn}^{\ast}|).$
(54)

In view of (43) and (44),

$V_{n}(\check{v})\Longrightarrow
V(\check{v})=\check{v}^{\prime}\Omega^{+}\check{v}-2\check{v}^{\prime}\zeta^{+}+\lim_{n\rightarrow\infty}\lambda_{n}\sum_{j=1}^{p}R_{jn}^{-1}D(1,\tilde{v}_{j},\theta_{j}^{0*}).$

We invoke the Convexity Lemma. For Part (a), $\lambda_{n}/R_{jn}\to 0$ for all
$j$ so the penalty vanishes in the limit and the asymptotic distribution is
equivalent to that of OLS. For Part (b), the tuning parameter’s rate is
$\lambda_{n}/\sqrt{n}\rightarrow c_{\lambda}\in(0,\infty)$ so that only the
penalty term associated with $\mathcal{I}_{1}$ vanishes in the limit.

Part (c) needs more subtle elaboration. Let
$v_{\lambda_{n}}=\frac{\lambda_{n}}{\sqrt{n}}Q^{-1}R_{n}^{-1}\check{v}=\frac{\lambda_{n}}{\sqrt{n}}v_{n}$
and $\theta_{\lambda_{n}}=\theta_{n}^{\ast}+v_{\lambda_{n}}$. Define

$V_{\lambda_{n}}(\check{v})=v^{\prime}_{\lambda_{n}}W^{\prime}Wv^{\prime}_{\lambda_{n}}-2v^{\prime}_{\lambda_{n}}W^{\prime}u+\lambda_{n}\sum_{j=1}^{p}(|\theta_{jn}^{\ast}+v_{\lambda_{n},j}|-|\theta_{jn}^{\ast}|),$

and multiply $n/\lambda_{n}^{2}$ on both sides,

$\left(\frac{n}{\lambda_{n}^{2}}\right)V_{\lambda_{n}}(\check{v})=v^{\prime}_{n}W^{\prime}Wv_{n}-\frac{\sqrt{n}}{\lambda_{n}}2v^{\prime}_{n}W^{\prime}u+\frac{n}{\lambda_{n}}\sum_{j=1}^{p}(|\theta_{jn}^{\ast}+v_{\lambda_{n},j}|-|\theta_{jn}^{\ast}|).$

By the rate condition of $\lambda_{n}$, the second term
$\frac{\sqrt{n}}{\lambda_{n}}2v^{\prime}_{n}W^{\prime}u=o_{p}(1)$. Given
$\tilde{v}_{j}\neq 0$ and $n$ large enough:

  * •

If $j\in\mathcal{I}_{0}\cup\mathcal{C}$, the coefficients
$\theta_{jn}^{*}=\theta_{j}^{0\ast}$ is invariant with $n$ so that

$\frac{n}{\lambda_{n}}(|\theta_{j}^{0\ast}+\frac{\lambda_{n}}{\sqrt{n}}\frac{\check{v}_{j}}{\sqrt{n}}|-|\theta_{j}^{0\ast}|)=\frac{n}{\lambda_{n}}D\left(1,\frac{\lambda_{n}}{n}\check{v}_{j},\theta_{j}^{0\ast}\right)=D\left(1,\check{v}_{j},\theta_{j}^{0\ast}\right).$
(55)

  * •

If $j\in\mathcal{I}_{1}$, the coefficient
$\theta_{jn}^{\ast}=\theta_{j}^{0\ast}/\sqrt{n}$ shrinks faster than
$\frac{\lambda_{n}}{n^{3/2}}$. The reverse triangular inequality
$\left||a+b|-|a|\right|\leq\left|b\right|$ for any $a,b\in\mathbb{R}$
guarantees

$\bigg{|}\frac{n}{\lambda_{n}}\left(|\theta_{jn}^{\ast}+\frac{\lambda_{n}}{n^{3/2}}\check{v}_{j}|-|\theta_{jn}^{\ast}|\right)\bigg{|}\leq\frac{n}{\lambda_{n}}\left|\frac{\lambda_{n}}{n^{3/2}}\check{v}_{j}\right|=O\left(n^{-1/2}\right),$
(56)

which is dominated by $D\left(1,\check{v}_{j},\theta_{j}^{0\ast}\right)$ in
the limit if $\check{v}_{j}\neq 0$ and $\theta_{j}^{0*}\neq 0$.

Thus we conclude

$\left(\frac{n}{\lambda_{n}^{2}}\right)V_{\lambda_{n}}(\check{v})\Longrightarrow\tilde{v}^{\prime}\Omega^{+}\tilde{v}+\sum_{j\in\mathcal{I}_{0}\cup\mathcal{C}}D\left(1,\tilde{v}_{j},\theta_{j}^{0*}\right)$

as stated in the Corollary.

Proof. [Proof of Corollary 3.11] We start with the same local perturbation
$\check{v}=\left(\check{v}_{z}^{\prime},\check{v}_{1}^{\prime},\check{v}_{2}^{\prime},\check{v}_{x}^{\prime}\right)^{\prime},$
$v_{n}=Q^{-1}R_{n}^{-1}\check{v}$ and $\theta_{n}=\theta_{n}^{\ast}+v_{n}$ as
in the proof of Corollary 3.10. We focus on the counterpart of the terms in
(54).

  * •

if $j\in\mathcal{I}_{0}$, we have $\widehat{\sigma}_{j}=O_{p}\left(1\right)$
and the coefficient $\theta_{j}^{0\ast}$ is independent of $n$ so that

$\widehat{\sigma}_{j}\left(|\theta_{j}^{0\ast}+\frac{v_{j}}{\sqrt{n}}|-|\theta_{j}^{0\ast}|\right)=D\left(\widehat{\sigma}_{j},\frac{v_{j}}{\sqrt{n}},\theta_{j}^{0*}\right)=D\left(O_{p}\left(1\right),O\left(\frac{1}{\sqrt{n}}\right),\theta_{j}^{0*}\right)\overset{p}{\to}0;$

  * •

if $j\in\mathcal{C}$, again $\theta_{j}^{0\ast}$ is independent of $n$ and

$\widehat{\sigma}_{j}\left(|\theta_{j}^{0\ast}+\frac{v_{j}}{\sqrt{n}}|-|\theta_{j}^{0*}|\right)=D\left(\widehat{\sigma}_{j},\frac{v_{j}}{\sqrt{n}},\theta_{j}^{0*}\right)=D\left(\frac{\widehat{\sigma}_{j}}{\sqrt{n}},v_{j},\theta_{j}^{0*}\right)\Longrightarrow
D\left(d_{j},v_{j},\theta_{j}^{0*}\right)=O_{p}\left(1\right),$

since these indices are associated with a unit root process $x_{j}^{c}$ and
therefore $\frac{\widehat{\sigma}_{j}}{\sqrt{n}}\Longrightarrow d_{j}$ is
degenerate;

  * •

if $j\in\mathcal{I}_{1}$, for these unit root processes in $x_{j}$ similarly
we have

$\displaystyle\widehat{\sigma}_{j}\left(|\theta_{jn}^{\ast}+\frac{v_{j}}{n}|-|\theta_{jn}^{\ast}|\right)$
$\displaystyle=D\left(\widehat{\sigma}_{j},\frac{v_{j}}{n},\theta_{j}^{0*}\right)=D\left(\frac{\widehat{\sigma}_{j}}{\sqrt{n}},\frac{v_{j}}{\sqrt{n}},\theta_{j}^{0*}\right)$
$\displaystyle=D\left(O_{p}\left(1\right),\frac{v_{j}}{\sqrt{n}},\theta_{j}^{0*}\right)\overset{p}{\to}0.$

The above analysis implies

$\displaystyle V_{n}(\check{v})$ $\displaystyle\Longrightarrow$ $\displaystyle
V\left(\check{v}\right)=\check{v}^{\prime}\Omega^{+}\check{v}-2\check{v}^{\prime}\zeta^{+}+c_{\lambda}\sum_{j\in\mathcal{C}}D\left(d_{j},v_{j},\theta_{j}^{0*}\right),$
(57)

and Part (a) and (b) follow.

For Part (c), let $\tilde{R}_{n}=R_{n}/\lambda_{n}$ and
$\theta_{n}=\theta_{n}^{\ast}+\tilde{R}_{n}^{-1}\tilde{v}$. Define

$\tilde{V}_{n}(\check{v})=\check{v}^{\prime}\left(\tilde{R}_{n}^{-1}W^{\prime}W\tilde{R}_{n}^{-1}\right)\check{v}-2\check{v}^{\prime}\tilde{R}_{n}^{-1}W^{\prime}u+\lambda_{n}\sum_{j=1}^{p}\widehat{\sigma}_{j}(|\theta_{jn}^{\ast}+\tilde{R}_{jn}^{-1}\check{v}_{j}|-|\theta_{jn}^{\ast}|).$

Multiply $1/\lambda_{n}^{2}$ on both sides,

$\displaystyle\frac{\tilde{V}_{n}(\check{v})}{\lambda_{n}^{2}}$
$\displaystyle=$
$\displaystyle\check{v}^{\prime}\left(R_{n}^{-1}W^{\prime}WR_{n}^{-1}\right)\check{v}-\frac{2}{\lambda_{n}}\check{v}^{\prime}R_{n}W^{\prime}u+\frac{1}{\lambda_{n}}\sum_{j=1}^{p}\widehat{\sigma}_{j}(|\theta_{jn}^{\ast}+\tilde{R}_{jn}^{-1}\check{v}_{j}|-|\theta_{jn}^{\ast}|)$
$\displaystyle=$
$\displaystyle\check{v}^{\prime}\left(R_{n}^{-1}W^{\prime}WR_{n}^{-1}\right)\check{v}+o_{p}(1)+\frac{1}{\lambda_{n}}\sum_{j=1}^{p}\widehat{\sigma}_{j}(|\theta_{jn}^{\ast}+\tilde{R}_{jn}^{-1}\check{v}_{j}|-|\theta_{jn}^{\ast}|).$

by the rate condition of $\lambda_{n}$. Again we study the last term. For
$\check{v}_{j}\neq 0$ and a sufficiently large $n$:

  * •

for $j\in\mathcal{I}_{0}$,

$\frac{1}{\lambda_{n}}\widehat{\sigma}_{j}\left(|\theta_{j}^{0*}+\frac{\lambda_{n}}{\sqrt{n}}\check{v}_{j}|-|\theta_{j}^{0*}|\right)=\frac{1}{\lambda_{n}}D\left(\widehat{\sigma}_{j},\frac{\lambda_{n}}{\sqrt{n}}\check{v}_{j},\theta_{j}^{0*}\right)=D\left(\widehat{\sigma}_{j},\frac{\check{v}_{j}}{\sqrt{n}},\theta_{j}^{0*}\right)\overset{p}{\to}0;$

  * •

for $j\in\mathcal{C}$,

$\frac{1}{\lambda_{n}}\widehat{\sigma}_{j}\left(|\theta_{j}^{0*}+\frac{\lambda_{n}}{\sqrt{n}}\check{v}_{j}|-|\theta_{j}^{0*}|\right)=\frac{1}{\lambda_{n}}D\left(\widehat{\sigma}_{j},\frac{\lambda_{n}}{\sqrt{n}}\check{v}_{j},\theta_{j}^{0*}\right)=D\left(\frac{\widehat{\sigma}_{j}}{\sqrt{n}},\check{v}_{j},\theta_{j}^{0*}\right)=D\left(d_{j},v_{j},\theta_{j}^{0*}\right)=O_{p}\left(1\right);$

  * •

for $j\in\mathcal{I}_{1}$, the rate condition $\lambda_{n}/n^{0.5}\rightarrow
0$ makes sure that $\theta_{jn}^{\ast}=\theta_{j}^{0\ast}/\sqrt{n}$ dominates
$\frac{\lambda_{n}}{n}$ in the limit, so that

$\frac{1}{\lambda_{n}}\widehat{\sigma}_{j}\left(|\theta_{jn}^{\ast}+\frac{\lambda_{n}}{n}\check{v}_{j}|-|\theta_{jn}^{\ast}|\right)=\frac{1}{\lambda_{n}}D\left(\widehat{\sigma}_{j},\frac{\lambda_{n}\check{v}_{j}}{n},\theta_{j}^{0\ast}\right)=D\left(\frac{\widehat{\sigma}_{j}}{\sqrt{n}},\frac{\check{v}_{j}}{\sqrt{n}},\theta_{j}^{0\ast}\right)\overset{p}{\rightarrow}0.$

We obtain
$\frac{\tilde{V}_{n}(\check{v})}{\lambda_{n}^{2}}\Longrightarrow\check{v}^{\prime}\Omega^{+}\check{v}+\sum_{j\in\mathcal{C}}D\left(d_{j},v_{j},\theta_{j}^{0\ast}\right)$
and the conclusion follows.