\section{\bf Introduction} \label{pre}

Stochastic differential equations (SDEs) driven by Wiener noise and L\'{e}vy noise have important applications to finance and physics. It is well known that Brownian motion is an $\alpha$-stable L\'{e}vy process with $\alpha=2$.  For one-dimensional stochastic process, we know that an $\alpha$-stable process converges in distribution to a Brownian motion as $\alpha\rightarrow2$ (see Sato \cite{Sato}).  Then a natural question arises: for the  SDE driven by an $\alpha$-stable process, dose its solution  converge in distribution to that of an SDE driven by a Brownian motion?

Liu \cite{Liu} studied weak convergence for the solution of an SDE driven by the symmetric stable process, and the $\alpha$-continuity of the solution to an SDE driven by an $\alpha$-stable process as $\alpha$ tends to $\alpha^* \in (1,2]$, but the case of non-symmetric $\alpha$-stable process  is not  well studied. In this paper, we study the weak convergence behavior as $\alpha \rightarrow 2$  for the solutions of SDEs driven by non-symmetic $\alpha$-stable processes in the Skorokhod space $D_{J_1}([0,T], \mathbb{R}^d$) (the space $D([0,T], \mathbb{R}^d)$ equipped with the $J_1$-metric, also see \cite{Bill}).

Let  $(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\geq 0},\mathbb{P})$  be
a filtered probability space, which satisfies the usual conditions.  Consider the following equations,
\begin{align}
dX^{\alpha,\beta} _t&=b(X^{\alpha,\beta} _t)dt+ \sigma (X^{\alpha,\beta} _{t-}) dL^{\alpha,\beta}_t,
\quad X^{\alpha,\beta}_0=x\in \mathbb{R}^d,   \label{e:0} \\
dX _t&=b(X_t)dt+ \sqrt{2}\sigma (X_{t}) dB_t,\quad  X_0=x, \label{e2}
\end{align}
where $b(x)=(b_1(x),\cdot\cdot\cdot,b_d(x))$  is a vector function, $\sigma(x)=(\sigma_{ij}(x))_{d\times d}$ is a $d\times d$ matrix-valued function. For more assumptions on the drift  and diffusion coefficients,  see Assumption \ref{asp1} and Assumption\ref{asp2}. The noise $B_t=(B_t^1, \cdots, B_t^d)$ is a $d$-dimensional Brownian motion, and $L^{\alpha,\beta}_t=(L^{\alpha,\beta_1}_t,\cdots,L^{\alpha,\beta_d}_t)$ is a $d$-dimensional cylindrical  non-symmetric $\alpha$-stable process. For each $i\in\{1,2,...,d\}$,  $L^{\alpha,\beta_i}_t$ is an one-dimensional non-symmetric $\alpha$-stable L\'{e}vy process and $N^{\alpha,\beta_i}$ is an $\mathcal{F}_t$-Poisson random measure with intensity measure $\nu^{\alpha,\beta_i}(dz)dt$. The  L\'{e}vy measure $\nu^{\alpha,\beta_i}$ on $\mathbb{R}$  is defined as
\begin{equation}\label{e1}
\nu^{\alpha,\beta_i}(dz)=\left\{
\begin{array}{rcl}
\frac{C_1^idz}{|z|^{1+\alpha}},&&{z > 0},\\
\ \\
\frac{C_2^i dz}{|z|^{1+\alpha}},&&{z < 0},
\end{array}\right.
\end{equation}
with $\beta_i\in [-1,1]$, \  $C_1^i=K_\alpha^i\frac{1+\beta_i}{2}$ and $C_2^i=K_\alpha^i\frac{1-\beta_i}{2}$, where
\begin{equation}\nonumber
K_\alpha^i=\left\{
\begin{array}{rcl}
\frac{\alpha(1-\alpha)}{\Gamma(2-\alpha)cos(\frac{\pi\alpha}{2})},&&{\alpha \neq 1},\\
\ \\
\frac{2}{\pi}\             ,&&{\alpha=1}.
\end{array}\right.
\end{equation}
In particular, $\nu^{\alpha,\beta_i}$ is said to be symmetric if $\beta_i = 0$.
Furthermore, the compensated Poisson random measure $\tilde{N}^{\alpha,\beta_i}$ is defined
by
\[
\tilde{N}^{\alpha,\beta_i}(dt,dz):=N^{\alpha,\beta_i}(dt,dz)- \nu^{\alpha,\beta_i}(dz)dt.
\]

 According to
L\'{e}vy-It\^{o}'s decomposition,  we have
\begin{equation*}\label{levy-ito-1}
L^{\alpha,\beta_i}_t =  \int_0^t \int_{|z|\leq1} z
\tilde{N}^{\alpha,\beta_i}(ds,dz) + \int_0^t \int_{|z|>1} z
N^{\alpha,\beta_i}(ds,dz),\quad \alpha \in (1,2).
\end{equation*}
For every $i\in\{1,2,...,d\}$,
\begin{equation*}\label{e1-eqv}
dX^{\alpha,\beta_i} _t=b_i(X^{\alpha,\beta}_t)dt+\sum_{j=1}^d\int_{|z_j|\leq 1}\sigma_{ij}
(X^{\alpha,\beta}_{t-}) z_j \tilde{N}^{\alpha,\beta_j}(dt,dz_j) +\sum_{j=1}^d \int_{|z_j|>1}\sigma_{ij}(X^{\alpha,\beta}_{t-}) z_j
N^{\alpha,\beta_j}(dt,dz_j),
\end{equation*}
is the $i$th component  of SDE \eqref{e:0}.

We will prove the solution of  Equation \eqref{e:0}, whose noise is non-symmetric $\alpha$-stable, converges weakly to the solution of Equation \eqref{e2} in Skorohod space (see Section 3), which we prove by using the  Aldous' criterion.  To this purpose
we need to establish the tightness of the approximating solution  and this can be obtained by a priori estimate
 $ \sup\limits_{\alpha\geq\alpha_0} \EX [\sup\limits_{0\leq t\leq T} |X^{
    \alpha,\beta}_t|^\theta] <\infty. $
Secondly, by martingale characterization, we show that the limit (in the sense of distributions) of the solution of the approximating equation  is the solution of an SDE driven by a Brownian motion.

While the weak convergence properties are established for SDEs driven by non-symmetric $\alpha$-stable processes in Skorohod space, the rate of convergence is still not easily derived. To our knowledge, many works are devoted to study the convergence rate of the solution  for an  SDE, where
both the approximation equation and the limit equation taking the same type of noise term (see, for instance, \cite{rate} and \cite{rate1}).  Here we will derive
 the explicit error bounds on the difference between the solution of  \eqref{e:0}  and the solution of   limit equation \eqref{e2}, where the two noises appearing in above two equations are
 totally different.  To this
purpose, we introduce the backward Kolmogorov equation associated with the solution of SDE \eqref{e2} to construct
an evolutionary equation that describes the  weak difference. Furthermore, we need to ensure that the solution of SDE \eqref{e:0} converges in weak sense to the solution of SDE \eqref{e2} under Assumption \ref{asp2}. For simplicity, we only establish the uniform estimates on solution  of approximating equation  (see Lemma \ref{lem4.2}).  And the corresponding proof of tightness and martingale characterization is omitted. Finally,  to illustrate the estimate \eqref{wr} is the optimal, we present a special case in Example \ref{example}.

The organization of this paper is given as follows. We introduce some notations,  basic settings and state the main results in Section 2.  Section 3 contains the proof of the main results. Section 4 is devoted to the rate of weak convergence and an
illustrative example.

\section{\bf Notation and main results}

Denote $H=[0,T]\times \mathbb{R}^d,\  \mathbb{N}=\{0, 1, 2, ...\}$.  For $(t,x)$ in $H$, $\eta=(\eta_1, \cdots, \eta_d)$ in  $(\mathbb{N}^+)^d$ is a multi-index, $|\eta|=\eta_1 +\cdots+ \eta_d$, and $D^{\eta}
=\frac{\partial^{|\eta|}}{\partial x_1^{\eta_1}\cdots\partial x_d^{\eta_d}}$. 	We denote
\begin{align}\nonumber
&\partial_t u(t,x)=\frac{\partial}{\partial t}u(t,x), \ \ \ \ \  D^ku(t,x)=(D^{\eta}u(t,x))_{|\eta|=k}, k\in\mathbb{N},\\ \nonumber
&\partial_i u(t,x)=u_{x_i}(t,x), \ \ \ \ \  \partial^2_{ij} u(t,x)=u_{x_i x_j}(t,x), \ \  i, j \in \{1,\cdots, d\} \\ \nonumber
&\partial_xu(t,x)=\nabla u(t,x)=\nabla_x u(t,x)=(\partial_1 u(t,x),\cdots, \partial_du(t,x)). \nonumber
\end{align}
Let $\kappa \in  \mathbb{N}^+$ and $0< \gamma \le 1$. And let $ C^{\kappa, \gamma}(H)$ denote the space of measurable functions in $H$ such that  the derivatives $D_x^\eta u(t,x)$ are continuous with respect to $(t,x)$ for all $1\le|\eta|\le\kappa$ and  the derivatives are locally $\gamma-$H${\rm \ddot{o}}$lder continuous with respect to $x$ for $|\eta|=\kappa$. We denote its norm by
\begin{align} \nonumber
||u||_{\kappa,\delta}:=\sum_{ |\eta|\le \kappa } |D_x^\eta u(t,x)|_0+\sup_{|\eta|=\kappa  ,
\atop t,\ x \neq x^1 }   \frac{|D_x^\eta u(t,x)- D_x^\eta u(t,x)|}{|x-x^1|^{\gamma}}, \nonumber
\end{align}
where $|u|_0 = \sup_{(t,x)\in H}|u(t,x)|$. We denote by $C^{\kappa, \gamma}(\mathbb{R}^d)$ the corresponding function sapce on $\mathbb{R}^d$.

The letter $C=C(\cdot, ... , \cdot)$  is a constant, which only depends on quantities appearing in parentheses. In a given context, the same letter is (generally) used to denote different constants depending on the same set of arguments. Moreover, all of the constant $C$ in this study are independent of $\alpha$, but it possibly depends on $\beta$.

We denote by $H(f)$  the Hessian matrix of $f$ and let $H_x(u(t,x))$ be the Hessian matrix of $u$ with respect to $x$. Let $\sigma^*$  be the transpose of $\sigma$,  and set
\begin{equation}\nonumber
\sigma=(\sigma_1,\cdots, \sigma_d),\ \  \sigma_i=(\sigma_{1i},\cdots,\sigma_{di})^* .
\end{equation}
We use the Einstein summation convention throughout this paper, for example if $x,y$ in $\mathbb{R}^d$, \ then
\begin{equation}\nonumber
x y =\sum_{i=1}^d x_iy_i.
\end{equation}

 Now, we make some assumptions on the drift and diffusion coefficients.

 \begin{assumption} \label{asp1}
(a) There is a constant $K_{\sigma}$ such
that for every $x,y \in \mathbb{R}^d$,
\begin{align*}
\|\sigma(x)-\sigma(y)\|_{HS}&\leq K_\sigma|x-y|,
\end{align*}
where $||\cdot||_{\text{HS}}$ denotes the Hilbert-Schmidt norm of a
matrix.

(b) There exists a positive number $\delta_0$, such that for all $R>0$, and $x,y\in\mathbb{R}^d$, with $\max{\{|y|,|x|\}}\le R$
and $|y-x|\le\delta_0$, satisfying
\begin{equation}\nonumber
\langle y-x , b(y)-b(x) \rangle \le K_{b} |y-x|^2, \nonumber
\end{equation}
where $K_b$ is a positive constant. Also,  there exists a constant $C>0$, such that for all $x\in\mathbb{R}^d$, satisfying
\begin{equation}\nonumber
|b(x)|\le C(1+|x|^r), \nonumber
\end{equation}
where $r\in(0,+\infty)$.
\end{assumption}

Observe that in view of the above assumptions, drift $b$ is dissipative and locally bounded. The second item ensures the tightness of the approximating solution (see Lemma \ref{lem2.4}), which is   similar to that  in \cite{SGI}.  With our assumptions we have the following main result which is proved later in section 3.
\begin{theorem}\label{thm1}
If  Assumption \ref{asp1} is satisfied. Let $X_t^{\alpha,\beta}$ be the solution of SDE \eqref{e:0}, then there
exists a measure $\mu$ defined on $D([0,T],\mathbb{R}^d)$ so that
\begin{equation}\label{main1} \nonumber
\mathcal{L} (X^{\alpha,\beta}) \rightarrow  \mu \quad \text{as} \quad
\alpha\rightarrow 2,
\end{equation}
in $D_{J_1}([0,T],\mathbb{R}^d)$. Moreover, $\mu$ is the law of
solution to SDE \eqref{e2}.

\end{theorem}

Note that this theorem is different from \cite[Theorem 1.1]{Liu},  where the noise is symmetric and rotationally invariant $\alpha$-stable L\'{e}vy process and the drift coefficient satisfies global Lipschitz condition, while in our setting the noise is cylindrical non-symmetric $\alpha$-stable L\'{e}vy process and the drift is non-Lipschitz.  In order to investigate the convergence rate, we need to impose the following assumptions on the coefficients $b$ and $\sigma$.

\begin{assumption}\label{asp2}
Let $0< \gamma \le 1$, $b_i(x) \in {C}^{2, \gamma}(\mathbb{R}^d)$, $\sigma_{ij}(x) \in  {C}^{2, \gamma}(\mathbb{R}^d)$, for all $1\le i,j \le d.$
\end{assumption}

The above assumptions are similar to the conditions imposed on the drift and jump coefficients in \cite{rate}. And it ensures the existence and uniqueness for the backward Kolmogorov equation which is used  to derive the rate of weak convergence.  We  state the following theorem.

\begin{theorem} \label{thm2}
If  Assumption \ref{asp2} is satisfied. Let $X_t$ be the solution of SDE \eqref{e2}, then for each $f\in {C}^{2,\gamma}(\mathbb{R}^d)$, we have
\begin{equation} \label{wr}
\EX[f(X_t^{\alpha,\beta})-f(X_t)]\le C(2-\alpha).
\end{equation}
\end{theorem}

This is our second main result and whose proof will be carried out in Section \ref{Weak Convergence}.

\section{\bf Proof of Theorem 2.1 } \label{proof1}

This section is devoted to the proof of Theorem \ref{thm1}. First, we state an existence and uniqueness result for SDE \eqref{e:0}. Then, by using uniform estimates of the approximating solution  (Lemma \ref{lem2.3}) we prove the tightness of the approximating equation (Lemma \ref{lem2.4}). Finally, by martingale characterization, we show that the limit of the solution  of the approximating equation  is the solution of an SDE driven by a Brownian motion. Now we are ready to state that SDE \eqref{e:0} has a unique strong solution.

\begin{lemma}\label{lem2.1}
If  Assumption \ref{asp1} is satisfied, then SDE \eqref{e:0}
admits a unique strong solution.
\end{lemma}

The proof of this result is similar to \cite[Corollary 2.9]{zhu}, where a  standard truncated method is  employed to  show that the stochastic differential equation has a unique strong solution, and hence is omitted here. In this article, the following lemma plays a crucial role.

\begin{lemma}\label{lem2.2}
For every  $0<\delta\le 1$, $l \in (1,2)$,
 $\vartheta \in [1,l)$, and $\nu^{\alpha,\beta}$ define as \eqref{e1}, we have
\item{\begin{equation}  \label{eq3.1}
 \sup_{\alpha\geq l} \int_{ |z|\leq \delta} |z|^{2}
  \nu^{\alpha,\beta}(dz)+\sup_{\alpha\geq l}
  \int_{ |z|> \delta} |z|^\vartheta \nu^{\alpha,\beta}(dz) = C<\infty
\end{equation}}
and
\item{\begin{equation} \label{eq3.2}
\lim_{\alpha \rightarrow 2}\int_{ |z|\leq \delta}|z|^2
\nu^{\alpha,\beta}(dz)=\lim_{\alpha \rightarrow 2}\frac{
K_\alpha}{2-\alpha}= 2.
\end{equation}}
For every $\delta>0$ and $ \vartheta <2$, we have
\item{\begin{equation} \label{eq3.3}
\lim_{\alpha \rightarrow 2}\int_{ |z|>\delta}|z|^\vartheta
\nu^{\alpha,\beta}(dz) =0.
\end{equation}}
For every $\alpha$ in $(0,2)$, we have
\item{\begin{equation} \label{eq3.4}
 \lim_{\delta\rightarrow 0+}\delta^{\alpha-2}\int_{ |z|\leq \delta} |z|^{2}
  \nu^{\alpha,\beta}(dz)=c <\infty, \quad \lim_{\delta\rightarrow 0+}\int_{ |z|\leq \delta} |z|^{2}
  \nu^{\alpha,\beta}(dz)=0.
\end{equation}}
Moreover, let $\alpha\in(\frac{3}{2}, 2)$, we have
\item{\begin{equation} \label{eq3.5}
|\int_{|z|\le\delta} |z|^2 \nu^{\alpha,\beta}(dz) - 2| + \int_{|z|>\delta}  |z| \nu^{\alpha,\beta}(dz) \le C(2-\alpha).
\end{equation}}
\end{lemma}

We give a proof for this result in Section \ref{Appendix}. Next, we derive a uniform estimate in the following lemma.

\begin{lemma}\label{lem2.3}
If  Assumption \ref{asp1} is satisfied, then
\begin{equation}
    \sup_{\alpha\geq\alpha_0} \EX [\sup_{0\leq t\leq T} |X^{
    \alpha,\beta}_t|^\theta] <\infty. \label{solution0}
\end{equation}
 where  $ \theta \in (0,1)$ and $\alpha_0 \in (\frac{3}{2} , 2)$.
\end{lemma}
\begin{proof}  Let $f(x)=(1+|x|^2)^\frac{1}{2}$ and $\delta>0$,
by using It\^{o} formula,  we have
\begin{align}\label{ito formula}
df(X^{\alpha,\beta}_t)&=\quad b(X^{\alpha,\beta}_t)\cdot \nabla f(X^{\alpha,\beta}_t)dt  \nonumber \\
&\quad+\sum_{i=1}^{d}\int_{ |z_i|>\delta}[f(X^{\alpha,\beta}_{t-}+
\sigma_i(X^{\alpha,\beta}_{t-})z_i)-f(X^{\alpha,\beta}_{t-})]
\nu^{\alpha,\beta_i}(dz_i)dt\nonumber \\
&\quad +\sum_{i=1}^d\int_{|z_i|\leq \delta}[f(X^{\alpha,\beta}_{t-}+
\sigma_i(X^{\alpha,\beta}_{t-})z_i)-f(X^{\alpha,\beta}_{t-})\nonumber \\
&\quad\quad\quad\quad\quad\quad\quad- \sigma_i(X^{\alpha,\beta}_{t-})z_i\cdot
\nabla f(X^{\alpha,\beta}_{t-})] \nu^{\alpha,\beta_i}(dz_i)dt\nonumber \\
&\quad-\sum_{i=1}^d\int_{\delta<|z_i|\leq 1}\sigma_i(X^{\alpha,\beta}_{t-})z_i\cdot
\nabla f(X^{\alpha,\beta}_{t-})] \nu^{\alpha,\beta_i}(dz_i)dt\nonumber \\
&\quad \  +\sum_{i=1}^d\int_{ \mathbb{R}-\{0\}}[f(X^{\alpha,\beta}_{t-}+
\sigma_i(X^{\alpha,\beta}_{t-})z_i)-f(X^{\alpha,\beta}_{t-})]
\tilde{N}^{\alpha,\beta_i}(dt,dz_i)\nonumber \\
&:=  I_1(X_t^{\alpha,\beta})dt+I_2(X_t^{\alpha,\beta})dt+I_3(X_t^{\alpha,\beta})dt+I_4(X_t^{\alpha,\beta})+dM_t.
\end{align}
For $I_1(X_t^{\alpha,\beta})$, we have
\begin{equation} \label{I1}
|I_1(X_t^{\alpha,\beta})|\le |b_i(X_t^{\alpha,\beta})\partial^i f(X_t^{\alpha,\beta})|=\frac{|\langle X_t^{\alpha,\beta}, b(X_t^{\alpha,\beta})\rangle|}{(1+|X_t^{\alpha,\beta}|^2)^{\frac{1}{2}}}\le C f(X_t^{\alpha,\beta}).
\end{equation}
We observe that
\begin{align*}
|f(x+y)-f(x)|\le|y|\int_0^1|\nabla f(x+sy)|ds\le \frac{|y|}{2}.
\end{align*}
Using Lemma \ref{lem2.2} and the above inequality, we deduce that
\begin{align}  \label{I2}
|I_2(X_t^{\alpha,\beta})|&\le\sum_{i=1}^{d}\int_{|z_i|>\delta}|f(X_{t-}^{\alpha,\beta} +\sigma_i(X_{t-}^{\alpha,\beta})z_i)-f(X_{t-}^{\alpha,\beta})|\nu^{\alpha,\beta_i}(dz_i) \nonumber \\
&\le C\sum_{i=1}^{d}\int_{|z_i|>\delta}|\sigma_i(X_{t-}^{\alpha,\beta})z_i|\nu^{\alpha,\beta_i}(dz_i)\nonumber \\
&\le C  \sum_{i=1}^{d}\int_{|z_i|>\delta}|z_i|\nu^{\alpha,\beta_i}(dz_i)(|X_t^{\alpha,\beta}|+1)\nonumber \\
&\le C (f(X_t^{\alpha,\beta})+1) .
\end{align}
For $I_3(X_{t}^{\alpha,\beta})$, we can choose a sufficiently small $\delta>0$ such that for all  $|z|\le\delta$,
\begin{align*}
&\quad|f(X^{\alpha,\beta}_{t-}+
\sigma_i(X^{\alpha,\beta}_{t-})z_i)-f(X^{\alpha,\beta}_{t-})
- \sigma_i(X^{\alpha,\beta}_{t-})z_i\cdot
\nabla f(X^{\alpha,\beta}_{t-})|\\
&=|\int_0^1dr \int_0^r  (\sigma_i(X_{t-}^{\alpha,\beta})z_i)H(f)(X^{\alpha,\beta}_{t-}+s\sigma_i(X_{t-}^{\alpha,\beta})z_i) \cdot(\sigma_i(X_{t-}^{\alpha,\beta})z_i)ds|\\
&\le \sup_{s\le1}\frac{C|\sigma_i(X_{t-}^{\alpha,\beta})z_i|^2}{(1+|X_{t-}^{\alpha,\beta} +s\sigma_i(X_{t-}^{\alpha,\beta})z_i|^2)^{\frac{1}{2}}}\\ \nonumber
&\le \frac{C|\sigma_i(X_{t-}^{\alpha,\beta})z_i|^2}{(1+|X_{t-}^{\alpha,\beta}|^2)^{\frac{1}{2}}}.
\end{align*}
In the last step, since
\begin{align*}
&\quad\frac{1}{1+|X_{t-}^{\alpha,\beta} +s\sigma_i(X_{t-}^{\alpha,\beta})z_i|^2}\\
&\le \frac{1}{1+|\frac{X_{t-}^{\alpha,\beta}}{2}+2s\sigma_i(X_{t-}^{\alpha,\beta})z_i|^2+\frac{3|X_{t-}^{\alpha,\beta}|^2}{4}-3|s\sigma_i(X_{t-}^{\alpha,\beta})z_i|^2}\\
&\le \frac{C}{1+|X_{t-}^{\alpha,\beta}|^2}.
\end{align*}
Thus, we have
\begin{align} \label{I3}
&|I_3(X_{t}^{\alpha,\beta})|\le C\sum_{i=1}^d\int_{|z_i|\le \delta} \frac{|\sigma_i(X_{t-}^{\alpha,\beta})z_i|^2}{(1+|X_{t-}^{\alpha,\beta}|^2)^{\frac{1}{2}}}\nu^{\alpha,\beta_i}(dz_i) \nonumber \\
&\le C \sum_{i=1}^d\int_{|z_i|\le \delta} |z_i|^2\nu^{\alpha,\beta_i}(dz_i)(1+|X_{t-}^{\alpha,\beta}|^2)^{\frac{1}{2}}\nonumber \\
&\le C f(X_{t}^{\alpha,\beta}).
\end{align}
For $I_4(X_{t}^{\alpha,\beta})$, we have
\begin{align}\label{I4}
|I_4(X_{t}^{\alpha,\beta})|&\le \sum_{i=1}^d \int_{\delta \le |z_i| < 1} |\sigma_i(X^{\alpha,\beta}_{t-})z_i\cdot
\nabla f(X^{\alpha,\beta}_{t-})| \nu^{\alpha,\beta_i}(dz_i)\nonumber \\
&\le C \sum_{i=1}^d \int_{|z|>\delta} |z_i|f(X^{\alpha,\beta}_{t-}) \nu^{\alpha,\beta_i}(dz_i) \le Cf(X^{\alpha,\beta}_{t-}).
\end{align}
From   \eqref{ito formula}-\eqref{I4}, we get
\begin{equation}
f(X_t^{\alpha,\beta})-f(x_0)\le C\int_0^t f(X_s^{\alpha,\beta})ds+C+M_t. \nonumber
\end{equation}
By using a stochastic Gronwall's inequality in Xie and Zhang \cite[lemma 3.8]{SGI}, for any\ $1<q<\infty$, we obtain
\begin{align*}
\EX \sup_{0\le t \le T}|X_t^{\alpha,\beta}|^\frac{1}{q}
&\le \EX \sup_{0\le t \le T}{f(X_t^{\alpha,\beta})}^\frac{1}{q}\\
&\le C(q,T)<\infty.
\end{align*}
Since the inequality always holds for all $q\in(1,\infty)$, then we have
\begin{equation}\label{solution0} \nonumber
    \sup_{\alpha\geq\alpha_0} \EX [\sup_{0\leq t\leq T} |X^{
    \alpha,\beta}_t|^\theta] <\infty, \nonumber
\end{equation}
where $\theta\in(0,1)$.\qed\end{proof}\bigskip

The result of uniform estimate can be used below to establish the tightness of the approximating solution.

\begin{lemma}\label{lem2.4}
If  Assumption \ref{asp1} is satisfied. The family
$\{X^{\alpha,\beta}\}_{\alpha \geq \alpha_0} $ is tight in the space
$D_{J_1}([0,T],\mathbb{R}^d)$.
\end{lemma}
\begin{proof}
According to Aldous' criterion \cite[Thereom 1.1]{Aldous}, we only  prove the following two items:

\((1)\)  For every $\varepsilon >0$, there exists $L_\varepsilon$ such that
\begin{equation} \label{Aldous1}
    \sup_{\alpha \geq \alpha_0}\PX(\sup_{0\leq t\leq T}|X^{\alpha,\beta}_t| >L_\varepsilon) \leq
\varepsilon.
\end{equation}

\((2)\) For every stopping time $\tau^\alpha \in [0,T]$, and every $K>0$, we
have
\begin{equation}\label{Aldous2}
    \lim_{\varepsilon\rightarrow 0}
\sup_{\alpha \geq
\alpha_0}\PX(|X^{\alpha,\beta}_{T\wedge(\tau^\alpha+\varepsilon)}-X^{\alpha,\beta}_{\tau^\alpha}|
>K) =0.
\end{equation}

By using Chebyshev's inequality, we have
\begin{align}\label{Che}
\sup_{\alpha \geq \alpha_0}\PX(\sup_{0\leq t\leq T}|X^{\alpha,\beta}_t| >L) \leq\frac{1}{L^\theta} \sup_{\alpha>\alpha_0} \EX(\sup_{0\le t\e T}|X_t^{\alpha}|^{\theta})\le\frac{C}{L^\theta},
\end{align}
where $\theta \in (0, 1)$, and $\alpha_0 \in (\frac{3}{2}, 2)$. This  immediately implies the estimate \eqref{Aldous1}.

For notational convenience, we set
$\tau^\alpha+\varepsilon:=T\wedge(\tau^\alpha+\varepsilon)$ and
obtain that
\begin{align*}
 \quad\sup_{\alpha \geq \alpha_0}\PX(|X^{\alpha,\beta}_{\tau^\alpha+\varepsilon}-X^{\alpha,\beta}_{\tau^\alpha}|^2
>K)& \leq  \sup_{\alpha \geq
\alpha_0}\PX(|\int_{\tau^\alpha}^{\tau^\alpha+\varepsilon}
b(X^{\alpha,\beta}_s)ds|>\frac{K}{3})\\
&\quad+ \sup_{\alpha \geq
\alpha_0}\PX(|\int_{\tau^\alpha}^{\tau^\alpha+\varepsilon}
\int_{ |z|\leq 1} z\sigma(X^{\alpha,\beta}_{s-}) \tilde{N}^{\alpha,\beta}(ds,dz)|>\frac{K}{3})\\
&\quad+ \sup_{\alpha \geq
\alpha_0}\PX(|\int_{\tau^\alpha}^{\tau^\alpha+\varepsilon}
 \int_{ |z|>1} z\sigma(X^{\alpha,\beta}_{s-}) N^{\alpha,\beta}(ds,dz)|>\frac{K}{3})\\
 &:= K_1+K_2+K_3.
\end{align*}
Let $\ p \in(0,1)$, for $K_1$, by using inequality $(a+b)^p \le (a^p + b^p)$ and Lemma \ref{lem2.3} we have
\begin{align*}
   K_1
&\leq \sup_{\alpha \geq
\alpha_0}\PX(|\int_{\tau^\alpha}^{\tau^\alpha+\varepsilon}
C(1+|X^{\alpha,\beta}_s|^r)ds|>\frac{K}{3}) \\
&\leq \sup_{\alpha \geq
\alpha_0}\frac{C}{K^{\frac{1}{r+1}}}\EX|\int_{\tau^\alpha}^{\tau^\alpha+\varepsilon}
(1+|X^{\alpha,\beta}_s|^r)ds|^{\frac{1}{r+1}} \\
&\leq\frac{C\varepsilon^{\frac{1}{r+1}}}{K^{\frac{1}{r +1}}} \sup_{\alpha \geq
\alpha_0} \EX (1+ \sup_{0<s \leq T}|X^{\alpha,\beta}_s|^{\frac{r}{r +1}})\rightarrow 0,
\quad \text{as} \quad \varepsilon \rightarrow 0.
\end{align*}
By Burkholder's
inequality, we obtain that
\begin{align*}
   K_2
&=\sup_{\alpha \geq
\alpha_0}\PX(|\int_{\tau^\alpha}^{\tau^\alpha+\varepsilon}
 \int_{ |z|\leq 1} z\sigma( X^{\alpha,\beta}_{s-}) \tilde{N}^{\alpha,\beta}(ds,dz)|>\frac{K}{3})\\
&\leq \sup_{\alpha \geq
\alpha_0}\PX(|\int_{\tau^\alpha}^{\tau^\alpha+\varepsilon}
 \int_{ |z|\leq 1} z\sigma( X^{\alpha,\beta}_{s-}) \tilde{N}^{\alpha,\beta}(ds,dz)|>\frac{K}{3}; \sup_{0\leq t\leq T}|X^{\alpha,\beta}_t|\leq M)\\
 &\quad+  \PX(\sup_{0\leq t\leq T}|X^{\alpha,\beta}_t|\geq M)\\
&\leq \frac{3}{K} \sup_{\alpha \geq
\alpha_0}\EX|(\int_{\tau^\alpha}^{\tau^\alpha+\varepsilon} \int_{
|z|\leq
   1} z\sigma(X^{\alpha,\beta}_{s-}) \tilde{N}^{\alpha,\beta}(ds,dz); \sup_{0\leq t\leq T}|X^{\alpha,\beta}_t|\leq
   M)| \\
  &\quad +  \PX(\sup_{0\leq t\leq T}|X^{\alpha,\beta}_t|\geq M)\\
   &\leq \frac{C}{K} \sup_{\alpha \geq \alpha_0}[\EX
  \int_{\tau^\alpha}^{\tau^\alpha+\varepsilon} \int_{ |z|\leq 1} |z|^2  \nu^{\alpha,\beta}(dz)ds]^{\frac{1}{2}}+  \PX(\sup_{0\leq t\leq T}|X^{\alpha,\beta}_t|\geq M)\\
&\leq\frac{C\sqrt{\varepsilon}}{K} \sup_{\alpha \geq
\alpha_0}[\int_{ |z|\leq 1}
  |z|^2 \nu^{\alpha,\beta}(dz)]^{\frac{1}{2}}+\frac{1}{M^\theta}\sup_{\alpha \geq \alpha_0}
  \EX \sup_{0\leq t\leq T}|X^{\alpha,\beta}_t|^\theta.
\end{align*}
By choosing  sufficient large positive number $M$ in the last line of the above chain of inequalities so that
\[
\frac{1}{M^\theta}\sup\limits_{\alpha \geq \alpha_0} \EX \sup\limits_{0\leq t\leq
  T}|X^{\alpha,\beta}_t|^\theta
\]
is sufficient small.  Then by the fact that \  $\sup\limits_{\alpha \geq
\alpha_0}\int_{ |z|\leq 1}|z|^2 \nu^{\alpha,\beta}(dz)<\infty$\ , we deduce
that
\[
K_2 \rightarrow 0, \quad \text{as} \quad \varepsilon \rightarrow 0.
\]
  For $K_3$, we  have
\begin{align*}
   K_3
&=\sup_{\alpha \geq
\alpha_0}\PX(|\int_{\tau^\alpha}^{\tau^\alpha+\varepsilon}
 \int_{ |z|>1} z\sigma( X^{\alpha,\beta}_{s-}) N^{\alpha,\beta}(ds,dz)|>\frac{K}{3})\\
&\leq \sup_{\alpha \geq
\alpha_0}\PX(|\int_{\tau^\alpha}^{\tau^\alpha+\varepsilon}
 \int_{ |z|>1} z\sigma( X^{\alpha,\beta}_{s-}) {N}^{\alpha,\beta}(ds,dz)|>\frac{K}{3}; \sup_{0\leq t\leq T}|X^{\alpha,\beta}_t|\leq M) +  \PX(\sup_{0\leq t\leq T}|X^{\alpha,\beta}_t|\geq M)\\
&\leq \frac{C}{K} \sup_{\alpha
  \geq \alpha_0}\EX( \int_{\tau^\alpha}^{\tau^\alpha+\varepsilon} \int_{ |z|>
  1} |z| {N}^{\alpha,\beta}(ds,dz) )+  \PX(\sup_{0\leq t\leq T}|X^{\alpha,\beta}_t|\geq M)\\
&= \frac{C}{K} \sup_{\alpha
  \geq \alpha_0}\EX( \int_{\tau^\alpha}^{\tau^\alpha+\varepsilon} \int_{ |z|>
  1} |z| \nu^{\alpha,\beta}(dz)ds)+  \PX(\sup_{0\leq t\leq T}|X^{\alpha,\beta}_t|\geq M)\\
& \leq \frac{ C\varepsilon}{K} \sup_{\alpha \geq \alpha_0}
\int_{|z|\geq
  1}|z|\nu^{\alpha,\beta}(dz)+\frac{1}{M^\theta}\sup_{\alpha \geq \alpha_0} \EX \sup_{0\leq t\leq
  T}|X^{\alpha,\beta}_t|^\theta.
  \end{align*}
In the same way, by choosing  sufficient large positive number $M>0$ in the last line of the above chain of inequalities so that
\[
\frac{1}{M^\theta}\sup_{\alpha \geq \alpha_0} \EX \sup_{0\leq t\leq
  T}|X^{\alpha,\beta}_t|^\theta
\]
is sufficient small.  Then by the fact that  $ \sup\limits_{\alpha \geq
\alpha_0}\int_{|z|\geq1}|z|\nu^{\alpha,\beta}(dz) <\infty$, we deduce that
\[
K_3 \rightarrow 0, \quad \text{as} \quad \varepsilon \rightarrow 0.
\]
Combing above arguments for $K_1,K_2$ and $K_3$,  we get
\eqref{Aldous2}. Then, we can deduce that
the family $\{X^{\alpha,\beta}\}_{\alpha \geq \alpha_0}$ is tight in the
space $D_{J_1}([0,T],\mathbb{R}^d)$. \qed\end{proof}\bigskip

The following lemma characterizes the limit of $\mathcal{L}(X^\alpha)$, as the law of $X^\alpha$.

 \begin{lemma}\label{lem2.5}
If Assumption  \ref{asp1} is satisfied. Let
$\mu^{\alpha,\beta} :=\mathcal{L}(X^\alpha)$  and $\mu^*$ is the weak limit of any convergent subsequence
$\{\mu^{\alpha_n,\beta}\}$ $(\alpha_n \rightarrow 2$ as $n \rightarrow
\infty)$. Then $\mu^*$ is supported by $C_U([0,T],\mathbb{R}^d)$.
\end{lemma}
\begin{proof}
For any $\eta>0$ and $M>0$,  we have for any $ \theta$ in
$(0,1)$ that,
\begin{align*}
    \PX(\sup_{0\leq t\leq T}|X^{\alpha,\beta}_t-X^{\alpha,\beta}_{t-}| >\eta)
& \leq   \PX(\sup_{0\leq t\leq T}|X^{\alpha,\beta}_t-X^{\alpha,\beta}_{t-}| >\eta;
  \sup_{0\leq t\leq T}|X^{\alpha,\beta}_t|\leq M)\\
& \quad +  \PX(\sup_{0\leq t\leq T}|X^{\alpha,\beta}_t|\geq M)\\
&\leq\frac{c}{\eta}\EX \int_0^T \int_{|z|\geq \eta}|z| N^{\alpha,\beta}(ds,dz)+\frac{1}{M^\theta} \EX \sup_{0\leq t\leq T}|X^{\alpha,\beta}_t|^\theta\\
& \leq \frac{ c}{\eta}\int_{|z|\geq
\eta}|z|\nu^{\alpha,\beta}(dz)+\frac{1}{M^\theta} \EX \sup_{0\leq t\leq
T}|X^{\alpha,\beta}_t|^\theta.
\end{align*}
Letting first $M\rightarrow \infty$ and then $\alpha \rightarrow 2$,
we obtain that
\[
\lim_{\alpha \rightarrow 2}\PX(\sup_{0\leq t\leq
T}|X^{\alpha,\beta}_t-X^{\alpha,\beta}_{t-}|
>\eta) =0.
\]
Then $\mu^*$ is supported by
 $C_U([0,T],\mathbb{R}^d)$.
 \qed\end{proof}\bigskip

~\\

By using It\^{o}'s formula for the solution $X^{\alpha,\beta}_t$ of SDE \eqref{e:0}, for each $f(x)\in C^2(\mathbb{R}^d)$, we have
\begin{equation}\label{M1}
\begin{aligned}
&\quad \  f(X^{\alpha,\beta}_t)-f(x)\nonumber\\
&=\sum_{i=1}^{d}\int_0^t b_i(X^{\alpha,\beta}_s)\cdot \partial_i f(X^{\alpha,\beta}_s)ds\nonumber\\
&\quad+\sum_{i=1}^{d}\int_0^t \int_{ |z_i|\leq 1}[f(X^{\alpha,\beta}_{s-}+
\sigma_i(X^{\alpha,\beta}_{s-})z_i)-f(X^{\alpha,\beta}_{s-})]
\tilde{N}^{\alpha,\beta_i}(ds,dz_i)\nonumber\\
&\quad+\sum_{i=1}^{d}\int_0^t \int_{ |z_i|>1}[f(X^{\alpha,\beta}_{s-}+
\sigma_i(X^{\alpha,\beta}_{s-})z_i)-f(X^{\alpha,\beta}_{s-})]
N^{\alpha,\beta_i}(ds,dz_i)\nonumber\\
&\quad+\sum_{i=1}^{d}\int_0^t \int_{|z_i|\leq 1}[f(X^{\alpha,\beta}_{s-}+
\sigma_i(X^{\alpha,\beta}_{s-})z_i)-f(X^{\alpha,\beta}_{s-})- \sigma_i(X^{\alpha,\beta}_{s-})z_i\cdot
\nabla f(X^{\alpha,\beta}_{s-})] \nu^{\alpha,\beta_i}(dz_i)ds\nonumber\\
&=\tilde{M}_t+\mathcal{L}_{b,\sigma}^{\alpha,\beta} f(X_{s-}^{\alpha,\beta}),
\end{aligned}
\end{equation}
where  $\tilde{M}_t$  is a martingale and
$\mathcal{L}_{b,\sigma}^{\alpha,\beta} f(u)$ is defined as following:
\begin{align}\label{Lf-alpha}
\mathcal{L}_{b,\sigma}^{\alpha,\beta} f(u)
&:= b_i(u)\cdot \partial^i f(u)+\sum_{i=1}^{d}\int_{ |z_i|\leq 1}[f(u+
\sigma_i(u)z_i)-f(u)-\sigma_i(u)z_i\cdot \nabla f(u)]\nu^{\alpha,\beta_i}(dz_i)\\
&\quad+\sum_{i=1}^{d} \int_{ |z_i|>1}[f(u+
\sigma_i(u)z_i)-f(u)]\nu^{\alpha,\beta_i}(dz_i).\nonumber
\end{align}
Moreover, let
\begin{equation}\label{Lf}
    \mathcal{L}_{b,\sigma}f(u):=b_i(u)\cdot \partial^i f(u)+\text{Tr}\left( \sigma \sigma^*
    H(f)\right)(u),
\end{equation}
where  $\text{Tr}\left(
\sigma \sigma^* H(f)\right):= \sum_{i,j}(\sigma
\sigma^*)_{ij}\frac{\partial^2 f}{\partial x_i \partial x_j}$.
\begin{lemma}\label{lem2.6}
Let $\mathcal{L}_{b,\sigma}^{\alpha,\beta}$ and $ \mathcal{L}_{b,\sigma}$ be
the operators defined in \eqref{Lf-alpha} and \eqref{Lf},  respectively. Suppose
Assumption \ref{asp1} is verified, then for every $f\in
C^2_b (\R^d)$, we have $$\mathcal{L}_{b,\sigma}^{\alpha,\beta} f \rightarrow
\mathcal{L}_{b,\sigma} f$$ locally uniformly in $\mathbb{R}^d$ as
$\alpha$ tends to $2$.
\end{lemma}
\begin{proof}
For every $0<\delta<1$, we have
\begin{align}\label{g1}
\mathcal{L}_{b,\sigma}^{\alpha,\beta}
f(u)-\mathcal{L}_{b,\sigma}f(u)&=\sum_{i=1}^{d}\int_{ |z_i|\leq 1}[f(u+
\sigma_i(u)z_i)-f(u)-\sigma_i(u)z_i\cdot \nabla f(u)]\nu^{\alpha,\beta_i}(dz_i)\nonumber\\
&\quad+\sum_{i=1}^{d} \int_{ |z_i|>1}[f(u+
\sigma_i(u)z_i)-f(u)]\nu^{\alpha,\beta_i}(dz_i)-\text{Tr}\left( \sigma \sigma^*
    H(f)\right)(u)\nonumber\\
&=\sum_{i=1}^{d}\int_{ |z_i|\leq \delta}[f(u+
\sigma_i(u)z_i)-f(u)-\sigma_i(u)z_i\cdot \nabla f(u)]\nu^{\alpha,\beta_i}(dz_i)\nonumber\\
&\quad+\sum_{i=1}^{d} \int_{ |z_i|>\delta}[f(u+
\sigma_i(u)z_i)-f(u)]\nu^{\alpha,\beta_i}(dz_i)\nonumber \\
&\quad-\sum_{i=1}^{d}\int_{\delta < |z_i| \leq 1}\sigma_i(u)z_i\cdot\nabla f(u)\nu^{\alpha,\beta_i}(dz_i)
-\text{Tr}\left( \sigma \sigma^*H(f)\right)(u).\nonumber
\end{align}
For convenience, we denote
\begin{align}\nonumber
\mathcal{A}f(u)&:=\sum_{i=1}^{d}\int_{ |z_i|\leq \delta}[f(u+
\sigma_i(u)z_i)-f(u)-\sigma_i(u)z_i\cdot \nabla f(u)]\nu^{\alpha,\beta_i}(dz_i),\nonumber\\
\mathcal{B}f(u)&:=\sum_{i=1}^{d} \int_{ |z_i|>\delta}[f(u+
\sigma_i(u)z_i)-f(u)]\nu^{\alpha,\beta_i}(dz_i),\nonumber \\
\mathcal{C}f(u)&:=\sum_{i=1}^{d}\int_{\delta < |z_i| \leq 1}\sigma_i(u)z_i\cdot\nabla f(u)\  \nu^{\alpha,\beta_i}(dz_i), \nonumber\\
\mathcal{D}f(u)&:=\sum_{i=1}^{d}\int_0^1dt \int_0^t \int_{ |z_i|\leq
\delta}[\sigma_i(u)z_iH(f)(u)\cdot
  \sigma_i(u)z_i] \nu^{\alpha,\beta_i}(dz_i)ds.\nonumber
\end{align}
 By Taylor's formula, we get
$$
f(u+v)-f(u)-v \cdot \nabla f(u)=\int_0^1dt \int_0^t vH(f)(u+sv)\cdot
vds.
$$
Then, we can choose suitable $\delta<< 1$ such that
\begin{align}\nonumber
&\quad \mathcal{A}f(u)-\mathcal{D}f(u)\\
&=\sum_{i=1}^{d}
\int_{ |z_i|\leq \delta}\int_0^1dt \int_0^t
  \sigma_i(u)z_i[H(f)(u+\sigma_i(u)z_is)-H(f)(u)]\cdot \sigma_i(u)z_ids\nu^{\alpha,\beta_i}(dz_i)\nonumber
\end{align}
 is sufficient small. For a fixed $\delta>0$, we have
\begin{align*}
|\int_{|z_i|\le \delta}|z_i|^2 \nu^{\alpha,\beta_i}(dz_i)-2|\rightarrow 0, \ \ \  as \ \ \alpha\rightarrow 2.
\end{align*}
 Then
 \begin{center}
 \begin{align}\nonumber
&\quad \mathcal{D}f(u)-\text{Tr}\left( \sigma \sigma^*H(f)\right)(u)\nonumber\\
  &=\sum_{j,k=1}^d \sum_{i=1}^d\sigma_{ki}\sigma_{ji}f_{kj}(u)\int_0^1dt\int_0^tds(\int_{|z_i|\le \delta}|z_i|^2 \nu^{\alpha,\beta_i}(dz_i))-\sum_{j,k=1}^d \sum_{i=1}^d\sigma_{ki}\sigma_{ji}f_{kj}(u)\int_0^1dt\int_0^t 2ds\nonumber\\
  &=\sum_{j,k=1}^d \sum_{i=1}^d\sigma_{ki}\sigma_{ji}f_{kj}(u)\int_0^1dt\int_0^t ds(\int_{|z_i|\le \delta}|z_i|^2 \nu^{\alpha,\beta_i}(dz_i)-2)\rightarrow 0, \ \ \ as \ \ \alpha\rightarrow 2.\nonumber
 \end{align}
 \end{center}
Similarly, it can be shown that
 \begin{align}\nonumber
 |\mathcal{B}f(u)|\le C \sum_{i=1}^d \int_{|z_i|>\delta}|z_i| \nu^{\alpha,\beta_i}(dz_i)\rightarrow 0,
 \end{align}
 and
\begin{equation} \nonumber
|\mathcal{C}f(u)|
\le C\sum_{i=1}^{d}\int_{\delta<|z|\leq1} |z_i| \nu^{\alpha,\beta_i}(dz_i)
\le C\sum_{i=1}^{d}\int_{|z_i|>\delta}|z_i| \nu^{\alpha,\beta_i}(dz_i)\rightarrow 0,
\end{equation}
 as $\alpha$ tends to 2.
Then, we have
\begin{align}\nonumber
&\quad\mathcal{L}_{b,\sigma}^{\alpha,\beta}
f(u)-\mathcal{L}_{b,\sigma}f(u)\\\nonumber
&=\mathcal{A}f(u)+\mathcal{B}f(u)-\mathcal{C}f(u)-\text{Tr}\left( \sigma \sigma^*H(f)\right)(u)\nonumber\\
&=\mathcal{A}f(u)-\mathcal{D}f(u)+\mathcal{D}f(u)-\text{Tr}\left( \sigma \sigma^*H(f)\right)(u)+\mathcal{B}f(u)-\mathcal{C}f(u) \rightarrow 0,\nonumber
\end{align}
as $\alpha$ tends to 2.
\qed\end{proof}\bigskip

Now, by using the idea of Cerrai (see \cite[Theorem 6.2]{Cerrai}) and all of the lemmas presented in this section, we can prove Theorem \ref{thm1} as follows.
 ~\\

 \textbf{\emph{Proof of Theorem 2.1.}}\ \ \
Due to sequence $\{\mu^{\alpha_n,\beta}\}_{n\in \mathbb{N}}$ converges weakly to $\mu^*$. If we are able to identify $\mu^*$ with $\mathcal{L}(X)$, where $\mathcal{L}(X)$ is the distribution of $X$, then we conclude that the whole sequence $\{\mu^{\alpha_n,\beta}\}_{n\in \mathbb{N}}$ weakly convergent to $\mathcal{L}(X)$ in $D_{J_1}([0,T],\mathbb{R}^d)$.
Let $\mathcal{L}_{b,\sigma}^{\alpha,\beta}$ and $ \mathcal{L}_{b,\sigma}$ be the operators defined in \eqref{Lf-alpha} and \eqref{Lf}, respectively. For any  $f\in C_b^2(\mathbb{R}^d)$, we obtain that the process
\begin{align*}
 f(X_t^{\alpha,\beta}) - f(x) - \int_0^t \mathcal{L}_{b,\sigma}^{\alpha,\beta}f(X_t^{\alpha,\beta})
\end{align*}
is a $\mathcal{F}_t-$martingale.

We denote by $\EX^{\mu^\alpha}$ as the expectation with respect to the probability measure $\mu^\alpha$ and $\EX^{\mu^*}$ as the expectation with respect to the probability measure $\mu^*$. We denote by $\eta(t)$ the canonical process in $D_{J_1}([0,T],\mathbb{R}^d)$. For every $0\le s_0 < s_1 < \cdots < s_m \le s < t$ and $f_0, f_1, \cdots, f_m \in C_b (\mathbb{R}^d)$, we have
\begin{equation}\label{M2}
\EX^{\mu^{\alpha}}\left[ \left( f(\eta_t)-f(\eta_s)-\int_s^t \mathcal{L}_{b,\sigma}^{\alpha,\beta}f(\eta_r)dr  \right) f_0(\eta_{s_0})\cdots f_m(\eta_{s_m})\right]=0.
\end{equation}
By \eqref{M2} , we have
\begin{align*}
&\quad \EX^{\mu^*}\left [ \left(f(\eta_t)-f(\eta_s)-\int_s^t\mathcal{L}_{b,\sigma}f(\eta_r)dr\right) f_0(\eta_{s_0})\cdots f_m(\eta_{s_m})\right]\\
& = \lim_{n \rightarrow \infty}\EX^{\mu^{\alpha_n}} \left[ \left(f(\eta_t)-f(\eta_s)-\int_s^t\mathcal{L}_{b,\sigma}f(\eta_r)dr\right) f_0(\eta_{s_0})\cdots f_m(\eta_{s_m})\right]\\
& = \lim_{n \rightarrow \infty}\EX^{\mu^{\alpha_n}} \Big[ \left(f(\eta_t)-f(\eta_s)-\int_s^t\mathcal{L}_{b,\sigma}f(\eta_r)dr\right) f_0(\eta_{s_0})\cdots f_m(\eta_{s_m})\\
& \quad \quad \quad\quad\quad\quad -\left(f(\eta_t)-f(\eta_s)-\int_s^t \mathcal{L}_{b,\sigma}^{\alpha_n,\beta} f(\eta_r)dr\right) f_0(\eta_{s_0})\cdots f_m(\eta_{s_m})\Big]\\
& = \lim_{n \rightarrow \infty}\EX^{\mu^{\alpha_n}} \left[ \left(\int_s^t \mathcal{L}_{b,\sigma}^{\alpha_n,\beta} f(\eta_r)-\mathcal{L}_{b,\sigma}f(\eta_r) dr \right) f_0(\eta_{s_0})\cdots f_m(\eta_{s_m}) \right].
\end{align*}
By Lemma \ref{lem2.4}, Lemma \ref{lem2.5} and Lemma \ref{lem2.6}, choosing a suitable $M>0$, we obtain
\begin{align}\label{M4}
&\quad\lim_{n\rightarrow \infty} \EX^{\mu^{\alpha_n}}\left[ \left(\int_s^t \mathcal{L}_{b,\sigma}^{\alpha_n, \beta}f(\eta_r)-\mathcal{L}_{b,\sigma}f(\eta_r) dr\right) f_0(\eta_{s_0})\cdots f_m(\eta_{s_m})\right]\nonumber \\
&\le\lim_{n\rightarrow \infty}  C\EX^{\mu^{\alpha_n}}\left[\int_s^t | \mathcal{L}_{b,\sigma}^{\alpha_n, \beta}f(\eta_r)-\mathcal{L}_{b,\sigma}f(\eta_r) |dr \right] \nonumber \\
&\le \lim_{n\rightarrow \infty}  C\EX^{\mu^{\alpha_n}}\left[\int_s^t | \mathcal{L}_{b,\sigma}^{\alpha_n, \beta}f(\eta_r)-\mathcal{L}_{b,\sigma}f(\eta_r) |dr; \sup_{s \le r \le t}|\eta_r| \le M\right] \nonumber \\
&\quad+\lim_{n \rightarrow \infty}  C\EX^{\mu^{\alpha_n}}\left[\int_s^t |\mathcal{L}_{b,\sigma}^{\alpha_n, \beta}f(\eta_r)-\mathcal{L}_{b,\sigma}f(\eta_r) |dr; \sup_{s\le r \le t}|\eta_r|> M\right] \nonumber \\
&=0.
\end{align}
Hence, according to \eqref{M4}, we can conclude that
\begin{align*}
 \EX^{\mu^*}\left[\left(f(\eta_t)-f(\eta_s)-\int_s^t\mathcal{L}_{b,\sigma}f(\eta_r)dr\right) f_0(\eta_{s_0})\cdots f_m(\eta_{s_m})\right]=0.
\end{align*}
This means that $f(\eta_t)-f(\eta_0)-\int_0^t\mathcal{L}_{b,\sigma}f(\eta_s)ds$ is a martingale under $\mu^*$. By using the argument is similar to  \cite[Theorem 6.2]{Cerrai}, we say that a probability measure $\mu^*$ solves the martingale problem which induces a weak solution of the SDE \eqref{e2}, associated with $\mathcal{L}_{b,\sigma}$. Thus, we can conclude that $\mu^*=\mathcal{L}(X)$.
\qed\bigskip

 \begin{remark}
Now, we have shown  the solution  of an SDE driven by a non-symmetric $\alpha$-stable process  converged weakly to that  of an SDE driven by  a Brownian motion. From the proof of Theorem \ref{thm1}, for any $\delta>0$, we observe that
 \begin{align*}
 \int_0^. \int_{|z|>1} zN^{\alpha,\beta}(dt,dz) \rightarrow 0,  \ \   \int_0^. \int_{|z|\le 1} z\tilde{N}^{\alpha,\beta}(dt,dz) \rightarrow B. , \ as\  \alpha\rightarrow 2
 \end{align*}
weakly in $D_{J_1}([0,T],\mathbb{R}^d)$, where the process \  $ \int_0^. \int_{|z|\le 1} z\tilde{N}^{\alpha,\beta}(dt,dz)$\  is  the compensated sum
of small jumps, and  is said to be ``central part'', which plays a vital part. $ \int_0^. \int_{|z|>1} zN^{\alpha,\beta}(ds,dz)$\  describing the ``large jumps'', and is said to be ``non-symmetric part''. However, in case of non-symmetric noise, the process \ $ \int_0^. \int_{|z|\le 1} z\tilde{N}^{\alpha,\beta}(dt,dz) $  is non-symmetric, which converges to a Brownian motion in $D_{J_1}([0,T],\mathbb{R}^d)$ and in law when $\alpha$ is close to 2. Moreover, the SDE driven by   $ \int_0^. \int_{|z|\le 1} z\tilde{N}^{\alpha,\beta}(dt,dz)$  is similar to the SDE driven by Brownian motion in $D_{J_1}([0,T],\mathbb{R}^d)$ and in law.
 \end{remark}

\section{\bf Proof of Theorem 2.2}  \label{Weak Convergence}

In this section we will derive the rate of weak convergence stated in Theorem \ref{thm2}. For this purpose we will use the backward Kolmogorov equation.
Let's consider the equation
\begin{equation}\label{Keq}
\begin{aligned}
&(\partial_t+\mathcal{L}_{b,\sigma})u(t,x)=0, t\in[0,T]\\
&u(T,x)=f(x),
\end{aligned}
\end{equation}
where
\begin{align*}
\mathcal{L}_{b,\sigma} u(t,x)= b_i(x)\partial^iu(t,x)+\text{Tr}\left( \sigma \sigma^*H_x\big(u(t,x)\big)\right),
\end{align*}
which denotes the generator of $X_t$.
\begin{lemma}\label{lem4.1}
If Assumption \ref{asp2} is satisfied, then for each $f\in  {C}^{2,\gamma}(\mathbb{R}^d)$, the equation \eqref{Keq} exists a unique solution $u\in  {C}^{2,\gamma}(H)$.
\end{lemma}

A proof of this result can be found in R. Mikulevicius \cite[Theorem 4]{rate}.  In order to establish the rate of weak convergence under  Assumption \ref{asp2}, we need to prove weak convergence firstly. Since this provement is quite similar to the argument of Theorem \ref{thm1},  we only establish a uniform estimate in the following lemma.

\begin{lemma}\label{lem4.2}
If  Assumption \ref{asp2} is satisfied, then for
every $ \theta \in (0,\alpha_0)$,
\begin{equation}\label{solution1}
    \sup_{\alpha\geq\alpha_0} \EX [\sup_{0\leq t\leq T} |X^{
    \alpha,\beta}_t|^\theta] <\infty,
\end{equation}
where $\alpha_0 \in (\frac{3}{2}, 2).$
\end{lemma}
\begin{proof}
We need only to prove \eqref{solution1} on condition that $\theta \in [1,\alpha_0)$ because the case where $\theta \in (0,1)$ is an immediate corollary by H${\rm \ddot{o}}$lder's inequality.  In fact, we have
\begin{align*}
X_t^{\alpha,\beta}&=x+\int_0^t b(X_s^{\alpha,\beta})ds +\int_0^t\int_{|z|\le 1}\sigma(X_{s-}^{\alpha,\beta})z \tilde{N}^{\alpha,\beta}(ds,dz)
+\int_0^t\int_{|z|>1}\sigma(X_{s-}^{\alpha,\beta})zN^{\alpha,\beta}(ds,dz)\\
&=x+\int_0^t b(X_s^{\alpha,\beta})ds +\int_0^t\int_{|z|\le1}\sigma(X_{s-}^{\alpha,\beta})z \tilde{N}^{\alpha,\beta}(ds,dz)\\
&\quad\ +\int_0^t\int_{|z|>1}\sigma(X_{s-}^{\alpha,\beta})z\nu^{\alpha,\beta}dz+\int_0^t\int_{|z|>1}\sigma(X_{s-}^{\alpha,\beta})z\tilde{N}^{\alpha,\beta}(ds,dz).
\end{align*}
First, there exists a positive constant $M$, such that\ $|b(x)|\le M$ and $|\sigma(x)|\le M$, for all $x\in\mathbb{R}^d$.  We obtain
\begin{align}\label{E1}
&\quad\EX(\sup_{0\le t \le T}|\int_0^tb(X_s^{\alpha,\beta})ds|^{\theta})
\le M^\theta T^\theta < \infty.
\end{align}
 By using Burkholder's inequality \cite[Theorem 26.12]{12}, and Lemma \ref{lem2.2}, we have
\begin{align}\label{E2}
&\quad\sup_{\alpha\geq \alpha_0}\EX\sup_{0\le t \le T}|\int_0^t\int_{|z|\le1} \sigma(X_{s-}^{\alpha,\beta})z\tilde{N}^{\alpha,\beta}(ds,dz)|^\theta \nonumber \\
&\le C_\theta \sup_{\alpha\geq\alpha_0}\EX|\int_0^T\int_{|z|\le 1} |\sigma(X_{s-}^{\alpha,\beta})z|^2 \nu^{\alpha,\beta}(dz)ds|^\frac{\theta}{2} \nonumber \\
&\le C_\theta M^\theta T^\frac{\theta}{2}\sup_{\alpha\geq \alpha_0}|\int_{|z|\le 1} |z|^2 \nu^{\alpha,\beta}(dz)|^\frac{\theta}{2} <\infty.
\end{align}
By applying \cite[Lemma 2.3]{Liu}, we obtain
\begin{align}\label{E3}
&\quad\sup_{\alpha\geq \alpha_0}\EX\sup_{0\le t \le T}|\int_0^t\int_{|z|>1} \sigma(X_{s-}^{\alpha,\beta})z\tilde{N}^{\alpha,\beta}(ds,dz)|^\theta \nonumber \\
&\le \sup_{\alpha\geq \alpha_0}C\EX \left(\int_0^T\int_{|z|>1} |\sigma(X_{s-}^{\alpha,\beta})z|^\theta \nu^{\alpha,\beta}(dz)ds\right)
+ \sup_{\alpha\geq \alpha_0}C\EX \left(\int_0^T\int_{|z|>1}|\sigma(X_{s-}^{\alpha,\beta})z| \nu^{\alpha,\beta}(dz)ds\right)^\theta \nonumber \\
&\le C M^\theta\sup_{\alpha\geq\alpha_0}[T\int_{|z|>1} |z|^\theta \nu^{\alpha,\beta}(dz) + T^\theta(\int_{|z|>1}|z|\nu^{\alpha,\beta}(dz))^\theta] < \infty
\end{align}
and
\begin{align}\label{E4}
&\quad\sup_{\alpha\geq\alpha_0}\EX\sup_{0\le t \le T} |\int_0^t\int_{|z|>1}\sigma(X_{s-}^{\alpha,\beta})z\nu^{\alpha,\beta}(dz)ds|^\theta\le T^\theta M^{\theta} \sup_{\alpha\geq\alpha_0}\left(\int_{|z|>1}|z|\nu^{\alpha,\beta}(dz)\right)^\theta < \infty.
\end{align}
By using $(x+y)^\theta \le C_{\theta}(x^\theta +y^\theta)$ for $\theta\in(1,2)$, we have
\begin{equation}\label{E5}
\begin{aligned}
 \quad \sup_{\alpha\geq\alpha_0} \EX [\sup_{0\leq t\leq T} |X^{
    \alpha,\beta}_t|^\theta]
    \le C[\sup_{\alpha\geq\alpha_0}\EX(&\sup_{0\le t \le T}|\int_0^tb(X_s^{\alpha,\beta})ds|^{\theta} \\
    \quad+\sup_{\alpha\geq \alpha_0}\EX&\sup_{0\le t \le T}|\int_0^t\int_{|z|\le1} \sigma(X_{s-}^{\alpha,\beta})z\tilde{N}^{\alpha,\beta}(ds,dz)|^\theta  \\
    \quad+\sup_{\alpha\geq \alpha_0}\EX&\sup_{0\le t \le T}|\int_0^t\int_{|z|>1} \sigma(X_{s-}^{\alpha,\beta})z\tilde{N}^{\alpha,\beta}(ds,dz)|^\theta  \\
    \quad+\sup_{\alpha\geq\alpha_0}\EX&\sup_{0\le t \le T} |\int_0^t\int_{|z|>1}\sigma(X_{s-}^{\alpha,\beta})z\nu^{\alpha,\beta}(dz)ds|^\theta].
\end{aligned}
\end{equation}
By formulas \eqref{E1}-\eqref{E5}, we obtain
\begin{align*}
 \quad \sup_{\alpha\geq\alpha_0} \EX [\sup_{0\leq t\leq T} |X^{
    \alpha,\beta}_t|^\theta] < \infty.
\end{align*}
The proof is complete.
\qed\bigskip\end{proof}

Now, we proceed to the proof of Theorem \ref{thm2}.

~\\
 \textbf{\emph{Proof of Theorem 2.2.}}\ \ \
Let $\mathcal{L}^{\alpha,\beta}_{b,\sigma}$ be the generator of $X_t^{\alpha,\beta}$,
\begin{align*}
 \mathcal{L}^{\alpha,\beta}_{b,\sigma} u(t,x)&=b_i(x)\partial^i u(t,x)+\sum_{i=1}^d\int_{|z_i|>1} u(t,x+\sigma_i(x)z_i)-u(t,x) \ \nu^{\alpha,\beta_i}(dz_i)\\
&\quad +\sum_{i=1}^d\int_{|z_i|\le 1} u(t,x+\sigma_i(x)z_i)-u(t,x)-\sigma_i(x)z_i \cdot\nabla u(t,x) \nu^{\alpha,\beta_i}(dz_i).
\end{align*}
According to Lemma \ref{lem4.1}, there exists a unique solution $u(t,x)\in {C}^{2,\gamma}(H)$ to \eqref{Keq}, for each $f(x)\in  {C}^{2,\gamma}(\mathbb{R}^d)$. By using It\^{o} formula,  we have
\begin{align*}
\EX[f(X_t^{\alpha,\beta})-f(X_t)]&=\EX[u(t,X_t^{\alpha,\beta})-u(0,x_0)]\\
&=\EX [\int_0^t \partial_tu(s,X_s^{\alpha,\beta})+ \mathcal{L}_{b,\sigma}^{\alpha,\beta}u(s,X_s^{\alpha,\beta}) ds]\\
&= \EX[\int_0^t -\mathcal{L}_{b,\sigma}u(s,X_s^{\alpha,\beta})+ \mathcal{L}_{b,\sigma}^{\alpha,\beta} u(s,X_s^{\alpha,\beta})ds ].
\end{align*}
Notice that $\sigma$ and $b$ is bounded, then for any $\delta>0$ we have
\begin{align*}
&\quad\ \int_0^t(\mathcal{L}_{b,\sigma}^{\alpha,\beta} -\mathcal{L}_{b,\sigma}) u(s,X_s^{\alpha,\beta})ds\\
&=\sum_{i=1}^d\int_0^t\int_{|z_i|\le 1}  u(s,X_{s-}^{\alpha,\beta} +\sigma_i(X_{s-}^{\alpha,\beta})z_i) - u(s,X_{s-}^{\alpha,\beta})-\sigma_i(X_{s-}^{\alpha,\beta})z_i\cdot\partial_x u(s,X_{s-}^{\alpha,\beta}) \nu^{\alpha,\beta_i}(dz_i)ds\\
&\quad -\int_0^tTr[\sigma\sigma^*(X_{s-}^{\alpha,\beta})H_x\big(u(s,X_{s-}^{\alpha,\beta})\big)] ds\\
&\quad + \sum_{i=1}^d\int_0^t\int_{|z_i|>1}  u(s,X_{s-}^{\alpha,\beta} +\sigma_i(X_{s-}^{\alpha,\beta})z_i) - u(s,X_{s-}^{\alpha,\beta}) \nu^{\alpha,\beta_i}(dz_i)ds\\
&= \sum_{i=1}^d\int_0^t\int_{|z_i|\le \delta}  u(s,X_{s-}^{\alpha,\beta} +\sigma_i(X_{s-}^{\alpha,\beta})z_i) - u(s,X_{s-}^{\alpha,\beta})-\sigma_i(X_{s-}^{\alpha,\beta})z_i\cdot\partial_x u(s,X_{s-}^{\alpha,\beta}) \nu^{\alpha,\beta_i}(dz_i)ds \\
&\quad -\int_0^tTr[\sigma\sigma^*(X_{s-}^{\alpha,\beta})H_x\big(u(s,X_{s-}^{\alpha,\beta})\big)]ds \\
&\quad + \sum_{i=1}^d\int_0^t\int_{|z_i|>\delta}  u(s,X_{s-}^{\alpha,\beta} +\sigma_i(X_{s-}^{\alpha,\beta})z_i)
 - u(s,X_{s-}^{\alpha,\beta}) \nu^{\alpha,\beta_i}(dz_i)ds\\
&\quad - \sum_{i=1}^{d}\int_0^t\int_{\delta\le|z_i|\le1}\sigma_i(X_{s-}^{\alpha,\beta})z_i\cdot\partial_xu(s,X_{s-}^{\alpha,\beta}) \nu^{\alpha,\beta_i}(dz_i)ds.
\end{align*}
For convenience, we denote
\begin{align*}\nonumber
\mathcal{A}u(s,X_{s}^{\alpha,\beta})&=\sum_{i=1}^{d}\int_{ |z_i|\leq \delta}u(s,X_{s-}^{\alpha,\beta} +\sigma_i(X_{s-}^{\alpha,\beta})z_i) - u(s,X_{s-}^{\alpha,\beta})-\sigma_i(X_{s-}^{\alpha,\beta})z_i\cdot\partial_x u(s,X_{s-}^{\alpha,\beta}) \nu^{\alpha,\beta_i}(dz_i),\nonumber\\
\mathcal{B}u(s,X_{s}^{\alpha,\beta})&=\sum_{i=1}^{d} \int_{ |z_i|>\delta} u(s,X_{s-}^{\alpha,\beta} +\sigma_i(X_{s-}^{\alpha,\beta})z_i)
 - u(s,X_{s-}^{\alpha,\beta})\nu^{\alpha,\beta_i}(dz_i),\nonumber \\
\mathcal{C}u(s,X_{s}^{\alpha,\beta})&=\sum_{i=1}^{d}\int_{\delta < |z_i| \leq 1}\sigma_i(X_{s-}^{\alpha,\beta})z_i
\cdot\partial_xu(s,X_{s-}^{\alpha,\beta})\nu^{\alpha,\beta_i}(dz_i), \nonumber\\
\mathcal{D}u(s,X_{s}^{\alpha,\beta})&=\sum_{i=1}^{d}\int_0^1dr \int_0^r \int_{ |z_i|\leq
\delta}[\sigma_i(X_{s-}^{\alpha,\beta})z_iH_x\big(u(s,X_{s-}^{\alpha,\beta})\big)\cdot
 \sigma_i(X_{s-}^{\alpha,\beta})z_i] \nu^{\alpha,\beta_i}(dz_i)ds.\nonumber
\end{align*}
By Taylor's formula, we get
$$
u(t,x+y)-u(t,x)-y \cdot \nabla_x u(t,x)=\int_0^1dr \int_0^r yH_x\big(u(t,x+\xi y)\big)\cdot
yd\xi.
$$
Then, we have
\begin{align*}\nonumber
&\quad |\EX\int_0^t\mathcal{A}\big(u(s,X_{s}^{\alpha,\beta})\big)-\mathcal{D}\big(u(s,X_{s}^{\alpha,\beta})\big)ds|\\
&\le\sum_{i=1}^{d}\EX\int_0^t
\int_{ |z_i|\leq \delta}\int_0^1dr \int_0^r
  \big|\sigma_i(X_{s-}^{\alpha,\beta})z_i\Big(H_x\big(u(s,X_{s-}^{\alpha,\beta}+\sigma_i(X_{s-}^{\alpha,\beta})z_i\xi)\big)\nonumber\\
  &\quad\quad-H_x\big(u(s,X_{s-}^{\alpha,\beta})\big)\Big)\cdot \sigma_i(X_{s-}^{\alpha,\beta})z_i\big|d\xi \nu^{\alpha,\beta_i}(dz_i)ds\nonumber \\
  &\le C\sum_{i=1}^{d}\EX\int_0^t
\int_{ |z_i|\leq \delta}\int_0^1dr \int_0^r
|z_i|^2\big|H_x\big(u(s,X_{s-}^{\alpha,\beta}+\sigma_i(X_{s-}^{\alpha,\beta})z_i\xi)\big)
  -H_x\big(u(s,X_{s-}^{\alpha,\beta})\big)\big|d\xi \nu^{\alpha,\beta_i}(dz_i)ds.
\end{align*}
 Since $u\in {C}^{2,\gamma}(H)$, we can choose suitable $\delta<< 1$ such that $\EX\int_0^t|\mathcal{A}\big(u(s,X_{s}^{\alpha,\beta})\big)-\mathcal{D}\big(u(s,X_{s}^{\alpha,\beta})\big)|ds$ is sufficient small. For a fixed $\delta>0$, by using Lemma~\ref{lem2.2}, we obtain
\begin{equation}\nonumber
|\int_{|z_i|\le \delta}|z_i|^2 \nu^{\alpha,\beta_i}(dz_i)-2|\le C(2-\alpha).
\end{equation}
 Then, we have
 \begin{center}
 \begin{align*}\nonumber
&\quad\ |\EX\int_0^t \mathcal{D}\big(u(s,X_s^{\alpha,\beta})\big)-\text{Tr}\left( \sigma \sigma^*(X_{s-}^{\alpha,\beta})H_x\big(u(s,X_{s-}^{\alpha,\beta})\big)\right)ds|\nonumber\\

  &\le\sum_{j,k=1}^d \sum_{i=1}^d\int_0^t|\sigma_{ki}\sigma_{ji}\partial_{kj}^2u(s,X_{s-}^{\alpha,\beta})\int_0^1dr\int_0^r d\xi(\int_{|z_i|\le \delta}|z_i|^2 \nu^{\alpha,\beta_i}(dz_i)-2)|ds\nonumber\\
  &\le C (2-\alpha).
 \end{align*}
 \end{center}
Furthermore, we have
 \begin{align*}\nonumber
 |\mathcal{B}\big(u(s,X_s^{\alpha,\beta})\big)|\le c \sum_{i=1}^d \int_{|z_i|>\delta}|z_i| \nu^{\alpha,\beta_i}(dz_i)\le C(2-\alpha),
 \end{align*}
 and
 \begin{align*} \nonumber
\quad|\EX\int_0^t\ \mathcal{C}\big(u(s,X_{s}^{\alpha,\beta})\big)\ ds|&\le C\sum_{i=1}^{d}\int_{\delta<|z_i|\leq1} |z_i| \nu^{\alpha,\beta_i}(dz_i)\\
&\le C\sum_{i=1}^{d}\int_{|z_i|>\delta}|z_i| \nu^{\alpha,\beta_i}(dz_i)\le C(2-\alpha).
\end{align*}
Thus, we obtain
\begin{align*}\nonumber
&\quad\ \big|\EX\int_0^t \mathcal{L}_{b,\sigma}^{\alpha,\beta}
\big(u(s,X_s^{\alpha,\beta})\big)-\mathcal{L}_{b,\sigma}\big(u(s,X_s^{\alpha,\beta})\big) \ ds\big|\nonumber\\
&=\big|\EX\int_0^t(\mathcal{A}+\mathcal{B}-\mathcal{C})\big(u(s,X_s^{\alpha,\beta})\big)-\text{Tr}\left( \sigma \sigma^*(X_{s-}^{\alpha,\beta})H_x\big(u(s,X_{s-}^{\alpha,\beta})\big)\right)\ ds\big|\nonumber\\
&=\big|\EX\int_0^t(\mathcal{A}-\mathcal{D})\big(u(s,X_s^{\alpha,\beta})\big)ds\big|\\
&\quad+\big|\EX\int_0^t\mathcal{D}\big(u(s,X_s^{\alpha,\beta})\big)-\text{Tr}\left( \sigma \sigma^*(X_{s-}^{\alpha,\beta})H_x\big(u(s,X_s^{\alpha,\beta})\big)\right)ds\big|\\
&\quad+\big|\EX\int_0^t\mathcal{B}\big(u(s,X_s^{\alpha,\beta})\big)ds\big|+\big|\EX\int_0^t\mathcal{C}\big(u(s,X_s^{\alpha,\beta})\big)ds\big| \\
&\le C(2-\alpha).
\end{align*}
The proof is complete. \qed\bigskip

\begin{remark}
The estimate \eqref{wr} implies that the weak convergence rate is $2-\alpha$, which is the optimal rate. Please see the following example for illustrating it.
\end{remark}

\begin{example}\label{example}
Let $L_t^\alpha$ and $B_t$ are one-dimensional symmetric $\alpha$-stable process ($\beta=0$) and standard Brownian motion, respectively. Then, according to \cite{moment}, we have
\begin{align*}
\EX|L_t^\alpha|=\frac{\Gamma(1-\frac{1}{\alpha})}{\int_0^\infty u^{-2}sin^2u du}\sqrt{t}.
\end{align*}
By using $\Gamma(1-x)\Gamma(x)=\frac{\pi}{sin(\pi x)}, \ x\in(0,1)$  \ and \  $\Gamma(\frac{1}{2})=\sqrt{\pi}$,  we have
\begin{align*}
\lim_{\alpha\rightarrow 2}\Big|\frac{\EX|L_t^\alpha|-\EX|B_t|}{2-\alpha}\Big|&=\lim_{\alpha\rightarrow 2}\Big|\frac{\Gamma(1-\frac{1}{\alpha})(\int_0^\infty u^{-2}sin^2u\  du )^{-1}- \frac{2}{\sqrt{\pi}}}{2-\alpha}\Big|\sqrt{t}\\
&=\lim_{\alpha\rightarrow 2}\Big|\frac{\frac{2\pi}{\Gamma(\frac{1}{\alpha})sin(\frac{\pi}{\alpha})}-2\sqrt{\pi}}{\pi(2-\alpha)}\Big|\sqrt{t}\\
&=\lim_{\alpha\rightarrow 2}\Big|\frac{2\big(\sqrt{\pi}-\Gamma(\frac{1}{\alpha})sin\frac{\pi}{\alpha} \ \big)}{\pi(2-\alpha)}\Big|\sqrt{t}\\
&=\lim_{\alpha\rightarrow 2}\Big| \frac{2}{\pi \alpha^2} \big( \Gamma'(\frac{1}{\alpha})sin\frac{\pi}{\alpha}+ \pi\Gamma(\frac{1}{\alpha})cos\frac{\pi}{\alpha}\ \big)\Big|\sqrt{t}\\
&=\frac{|\Gamma'(\frac{1}{2})|}{2\pi}\sqrt{t}\\
&=\frac{\gamma+2ln2}{2\sqrt{\pi}}\sqrt{t},
\end{align*}
where $\gamma$ is Euler-Mascheroni constant. Hence, for every\  $t\in[0,T]$,  we can obtain that there is a constant $C$, such that
\begin{equation}\label{optimal}
\big|\ \EX|L_t^\alpha|-\EX|B_t| \ \big| \geq C(2-\alpha).
\end{equation}
The estimate \eqref{optimal}  implies that the optimal weak convergence rate is $2-\alpha$.
\end{example}

\section{\bf Appendix: Proof of Lemma \ref{lem2.2} } \label{Appendix}

In this appendix we give the proof of Lemma \ref{lem2.2}.\\
\begin{proof} Let $\alpha \geq l$,
\begin{align*}\nonumber
\int_{|z|\le\delta} |z|^2 \nu^{\alpha,\beta}(dz)=\int_0 ^\delta \frac{ C_1 z^2}{|z|^{\alpha+1}} dz
+\int_{-\delta}^0 \frac{C_2 z^2}{|z|^{\alpha+1}} dz
=\frac{K_\alpha \delta^{2-\alpha}}{2-\alpha}, \nonumber
\end{align*}
\begin{align*} \nonumber
\int_{|z|>\delta}|z|^{\vartheta}\nu^{\alpha,\beta}(dz)=\int_{z>\delta} \frac{K_\alpha (1+\beta)}{2z^{\alpha+1-\vartheta}}dz + \int_{z< {-\delta}}\frac{K_\alpha(1-\beta)}{2z^{\alpha+1-\vartheta}}dz
=\frac{K_\alpha \delta^{\vartheta-\alpha}}{\alpha-\vartheta}. \nonumber
\end{align*}
Then by using\ $\lim\limits_{x \rightarrow 0}x\Gamma(x)=\lim\limits_{x\rightarrow 0} \frac{\pi x}{\Gamma(1-x) sin \pi x}=1$, we have
\begin{align*} \nonumber
\lim_{\alpha \rightarrow 2}\int_{ |z|\leq \delta}|z|^2\nu^{\alpha,\beta}(dz)
=\lim_{\alpha \rightarrow 2}\frac{\alpha(1-\alpha) \delta^{2-\alpha}}{(2-\alpha)\Gamma(2-\alpha)cos(\frac{\pi\alpha}{2})}=2, \nonumber
\end{align*}
\begin{align*} \nonumber
\lim_{\alpha \rightarrow 2} \int_{|z|>\delta}|z|^\vartheta \nu^{\alpha,\beta}(dz)
=\lim_{\alpha \rightarrow 2}\frac{2(2-\alpha) \delta^{\vartheta-\alpha}}{\alpha-\vartheta}=0. \nonumber
\end{align*}
Fix $\alpha\in(0,2)$,
\begin{align*} \nonumber
 \lim_{\delta\rightarrow 0+}\delta^{\alpha-2}\int_{ |z|\leq \delta} |z|^{2}
  \nu^{\alpha,\beta}(dz)= \lim_{\delta\rightarrow 0+}\frac{K_\alpha}{2-\alpha}<\infty. \nonumber
\end{align*}
Hence $ \lim\limits_{\delta\rightarrow 0+}\int_{ |z|\leq \delta} |z|^{2}
  \nu^{\alpha,\beta}(dz)=0. $

 Furthermore, by using L'Hospital's rule and $\lim\limits_{x \rightarrow 0} \frac{x\Gamma(x)}{x-1}=\lim\limits_{x\rightarrow 0}(x\Gamma(x))'=-1$, we have
\begin{align*}
\lim_{\alpha \rightarrow 2}\frac{\big|\int_{ |z|\leq \delta} |z|^{2}
  \nu^{\alpha,\beta}(dz)-2\big|}{2-\alpha}=\lim_{\alpha \rightarrow 2} \big|\frac{\alpha(1-\alpha) \delta^{2-\alpha}-2(2-\alpha)\Gamma(2-\alpha)cos(\frac{\pi\alpha}{2})}{(2-\alpha)}\big|=C.
\end{align*}
This result implies that
\begin{align*}
|\int_{ |z|\leq \delta} |z|^{2}
  \nu^{\alpha,\beta}(dz)-2|\le C(2-\alpha).
  \end{align*}
And let $\vartheta=1$, we get
\begin{align*}
 \int_{|z|>\delta}|z| \nu^{\alpha,\beta}(dz)\le C(2-\alpha).
\end{align*}
The proof is complete.\qed\end{proof}\bigskip

\section*{Ackonwledgement}
The authors ackonwledge support provided by NNSF of China (Nos. 11971186, 11971367), Science and Technology
Research Projects of Hubei Provincial Department of Education No. B2022077.


\section{Origin and history of the conjecture}

In this paper we fill a gap in the investigation on Fuglede's conjecture for elementary abelian $p$-groups of rank four. The story of this conjecture begins in 1974, when Bent Fuglede \cite{Fuglede} investigates the relation between domains, i.e. sets of positive and finite Lebesgue measure in $\mathbb R^d$, that tile $\mathbb R^d$ by translation and the existence of an orthogonal exponential basis for the space of complex-valued functions on these domains. For future reference, a set satisfying this last property is called a spectral set and its spectrum is the corresponding orthogonal basis. He was able to show that both properties are equivalent, at least when the set is the fundamental domain of a lattice. On one of the last pages of this paper, he conjectures that the equivalence should in fact be true for any domain in $\mathbb R^d, d \geq 1$. Over the years, several authors managed to establish connections between the two properties, see for example \cite{IP,KL,L2002,LRW}, or were able to answer the question affirmatively for special kinds of domains. Notable results include those of Laba \cite{Laba}, who characterised in 2001  the union of two intervals satisfying the conjecture, and of Iosevich, Katz and Tao \cite{IKT} who proved in 2003 that the conjecture holds when the set is a convex planar domain.

In the light of these results, it is safe to say that history took an unexpected turn when in 2004 Terry Tao \cite{Tao} gave a counterexample in $\mathbb R^5$. To do so, he constructed a non-tiling spectral set, showing that at least one of the implications is false in $\mathbb R^5$, and by extension also in $\mathbb R^d, d \geq 5$. Interestingly, his approach first established a set in $(\mathbb Z/3\mathbb Z)^5$ of size 6, admitting a spectrum. As 6 does not divide $3^5$, this means that it cannot be tiling. After lifting this example to $\mathbb R^5$, while keeping the relevant properties intact, his result followed. His approach ignited the interest in the conjecture, not just in $\mathbb R^d$ anymore, but in the elementary abelian groups $(\mathbb Z/p\mathbb Z)^d$, $p$ prime, and it is this setting that we contribute to. We will properly state Fuglede's conjecture and its current status in $\ZpZ{d}$ in the next section, but wrap up the story for $\mathbb R^d$ first.

Since Tao's result, Kolountzakis and Matolcsi \cite{KM} have shown that the reverse implication is also false for all $d \geq 5$. This was consequently improved to $d=4$ by Farkas and R\'{e}v\'{e}sz \cite{FR} and to $d=3$ by Farkas, Matolcsi and M\'ora \cite{FMM}. On the other hand, affirmative answers in small dimension still appear, for instance for the union of three intervals in $\mathbb R$ \cite{BKKM} and convex polytopes in $\mathbb R^3$ \cite{GL}, further exhibiting the disparity between small and large dimensions.

\section{Fuglede's conjecture in $\ZpZ{d}$}

We will now define what it means for a set $E \subseteq \ZpZ{d}$ to be tiling or to have a spectrum, in order to state the conjecture clearly. For the remainder of this paper, let $p$ denote a prime.

\subsection{Tiling sets}

\begin{defin}
	The set $E$ is \textbf{tiling} if there exists $T \subseteq \ZpZ{d}$ such that the sets $\{E+t \,\,|\,\, t \in T\}$ partition $\ZpZ{d}$. The set $T$ is called the \textbf{tiling partner}.
\end{defin}

Clearly, as $|E||T|=|\ZpZ{d}|=p^d$, it follows that $|E|=p^k, 0 \leq k \leq d$. The full space and singletons are trivially tiling sets.

Tiling sets can in fact be defined in any finite abelian group. This is an independent topic of interest and appears under the name of group factorizations. For the elementary abelian groups, one of the early exponents in this domain is the book by R\'edei \cite{Redeibook}. One of the highlights of this work is his proof that for $d=2$ a non-trivial tiling set is a subgroup or coset thereof, or its tiling partner is. This result can be leveraged into a proof of one implication of the Fuglede conjecture for $d=2$, as is also described in \cite{KDSV}. Moreover, he conjectured \cite[Problem 5]{Redeibook} that in three dimensions a similar situation occurs -- that at least one of the factors does not span the whole group -- although this problem remains open to this day. We refer to Sz\H onyi's paper \cite{Szonyi} for an introduction to R\'edei's result and its connections to geometry over finite fields. On the other hand, for factorizations of finite abelian groups in general, we refer to the book of Szab\'o \cite{Szabobook}.

Before we turn to the subject of the other half of the conjecture, we need to make a small detour.

\subsection{The space $L^2(F)$ and characters of finite abelian groups}

In order to state the definition of a spectral set, we need two components: the the vector space $L^2(F)$ of complex-valued functions on a finite set $F$, and characters of a finite abelian group. For the remainder of this section, fix $F$ to be a finite set.

\begin{defin}
	The vector space of all functions $F \rightarrow \mathbb{C}$ is denoted as $L^2(F)$, when equipped with the inner product defined by
	\begin{align}\label{innerproductdef}\langle f,g \rangle_{L^2(F)} = \frac{1}{|F|}\sum_{x \in F}f(x)\overline{g(x)}. \end{align}
\end{defin}

Since the characteristic functions $\{\delta_x \,\, | \, \, x \in F\}$ are an orthogonal basis for $L^2(F)$, it is clear that $\dim(L^2(F))=|F|$.

A word of caution is necessary: there are several different conventions regarding the scaling of the sum, the motivation for the choice $1/|F|$ will be clear after we discuss some character theory.

\begin{defin}
	A \textbf{character} $\chi$ of a finite abelian group $G$ is a homomorphism $\chi:G\rightarrow \mathbb{C}$.
\end{defin}

There exists a rich literature on characters of finite abelian groups, so we will collect a few necessary results, focused on the elementary abelian groups, which are contained in any standard work in this area. In particular, we refer to \cite{Conrad} for an excellent introduction into this topic.

We will write elements $\textbf{x}=(x_1,\dots,x_d) \in \ZpZ{d}$ in bold, in contrast with elements of $\mathbb Z/p\mathbb Z$. The characters of $\ZpZ{d}$ are defined by
\[\chi_\textbf{a}(\textbf{x}) = e^{\frac{2\pi i}{p}\langle \textbf{a},\textbf{x} \rangle},\]
for all $\textbf{a},\textbf{x} \in \ZpZ{d}$, where $\langle \textbf{a},\textbf{x} \rangle = a_1x_1 + \dots + a_dx_d$. Observe that $\overline{\chi_\textbf{a}}=(\chi_\textbf{a})^{-1}=\chi_\textbf{-a}$.

\begin{lemma}[The orthogonality relations of characters]
	Let $\emph{\textbf{a}},\emph{\textbf{b}} \in \ZpZ{d}$ and  $\chi_\emph{\textbf{a}},\chi_\emph{\textbf{b}}$ their associated characters, then
	\[\sum_{\emph{\textbf{x}} \, \in \ZpZ{d}}\chi_\emph{\textbf{a}}(\emph{\textbf{x}})\overline{\chi_\emph{\textbf{b}}(\emph{\textbf{x}})} = \begin{cases}
		p^d & \text{if } \emph{\textbf{a}}=\emph{\textbf{b}} \\
		0 & \text{if } \emph{\textbf{a}}\neq\emph{\textbf{b}}
		\end{cases}\]
\end{lemma}

This implies that the characters $\{\chi_\textbf{a} \,\, | \,\, \textbf{a} \in \ZpZ{d}\}$ are an orthonormal basis for $L^2(\ZpZ{d})$, motivating the choice of scalar $1/|F|$ in the inner product \eqref{innerproductdef}. Furthermore, we have a group operation on the characters as for any $\textbf{a},\textbf{b},\textbf{x} \in \ZpZ{d}$,
\[\chi_\textbf{a}(\textbf{x})\chi_\textbf{b}(\textbf{x}) = e^{\frac{2\pi i}{p}\langle \textbf{a},\textbf{x} \rangle}e^{\frac{2\pi i}{p}\langle \textbf{b},\textbf{x} \rangle} = e^{\frac{2\pi i}{p}\langle \textbf{a}+\textbf{b},\textbf{x} \rangle} =  \chi_{\textbf{a}+\textbf{b}}(\textbf{x}).\]

The existence of an orthonormal basis of characters is the defining property of spectral sets as we will now see.

\subsection{Spectral sets}

\begin{defin}
	The set $E$ is \textbf{spectral} if there exists $A \subseteq \ZpZ{d}$ such that $\{\chi_\textbf{a} \,\, | \,\, \textbf{a} \in A\}$ is an orthogonal basis of $L^2(E)$. The set $A$ is called the spectrum.
\end{defin}

As $\dim(L^2(E))=|E|$, it is clear that $|A|=|E|$. Again, we see that the full space and the singletons satisfy this property trivially: the first due to what we saw above, the second because $\dim(L^2(\{\textbf{x}\}))=1$ and hence every non-zero complex number is an orthogonal basis for this space.

We are now in the position to state Fuglede's conjecture in $\ZpZ{d}$.

\begin{conj}
	Let $E \subseteq \ZpZ{d}$, then $E$ is a tiling set if and only if $E$ is a spectral set.
\end{conj}

We briefly summarize the current state of affairs.

\begin{itemize}
	\item[\underline{$d=1$}] Recall that in general, if $E$ tiles then $|E|$ divides $p^d$. Similarly, we will show later that if $E$ is a spectral set in $\mathbb Z/p\mathbb Z$, then $|E|=1$ or $p$. Therefore, either one of the properties implies that $E$ is a singleton or the full space, which shows that the conjecture is trivially true.
	\item[\underline{$d=2$}] First proved by Iosevich, Mayeli and Pakianathan \cite{IMP} to be true. Slightly alternative proofs have appeared in \cite{Researchpaper} and \cite{KDSV}.
	\item[\underline{$d=3$}] Half of the conjecture is known: `tiling implies spectral' is settled in the affirmative \cite{Researchpaper} while the other implication remains open. In the same paper, a computation by hand for $p=2,3$ shows that the full conjecture is true in these cases.
	\item[\underline{$d=4$}] A counterexample to `spectral implies tiling' for $p \equiv 3 \pmod{4}$ has been constructed in \cite{Researchpaper}, using the machinery of log-Hadamard matrices. Inspired by this construction, we extend this in the next section to all odd primes $p$. Before doing so, we show the perhaps surprising result that for $p=2$ the conjecture is true. However, this exception seems to be due to the small number of points and is not an indication for larger dimensions. These results were found independently and with different methods by Ferguson and Sothanaphan \cite{FS}.
	\item[\underline{$d\geq5$}] For all odd primes $p$ there exist non-tiling spectral sets \cite{Researchpaper}. When $p=2$, a counterexample for the same direction has been constructed for $d=10$ \cite{FS}. The converse is still wide open as is the case $p=2$ and $5 \leq d \leq 10$.
\end{itemize}

One can see that also in the case of elementary abelian groups, there is a discrepancy between small and large dimensions, analogous to the situation in $\mathbb R^d$. We give a short proof that the conjecture holds for dimension four when $p=2$, relying on \cite[Theorem 1.1]{Researchpaper}, of which we copy the relevant parts for convenience.

\begin{theorem}\label{mainthmresearchpaper}
	Let $E \subseteq \ZpZ{d}$, $p$ prime, then
	\begin{enumerate}
		\item If $E$ is a spectral set then $|E|\in\{1,p^d\}$ or $|E|=kp$ for some $1 \leq k \leq p^{d-2}$.
		\item If $E$ is a spectral set in $(\mathbb Z/2\mathbb Z)^d$, then $|E|\in\{1,2\}$ or is a multiple of $4$.
		\item If $|E| \in \{p,p^{d-1}\}$ then $E$ tiles if and only if $E$ is spectral.
		\item A set $E$ tiles with a subspace tiling partner $W$ if and only if $E$ is spectral with spectrum $W^\perp$.
	\end{enumerate}
\end{theorem}

\newpage

\begin{prop}
	Let $E \subseteq (\mathbb Z/2\mathbb Z)^4$, then $E$ is a tiling set if and only if $E$ is a spectral set.
\end{prop}
\begin{proof}
	In fact, we will show that any set satisfying the necessary size restrictions is simultaneously tiling and spectral. Both for tiling sets and spectral sets, it follows from the restrictions in the theorem above that for non-trivial $E$, we have $|E| \in \{2,4,8\}$. In all cases except for $|E|=4$, the equivalence follows from (3), so suppose for the remainder of the proof that $|E|=4$.

	If $E$ is the set of four points of a two-dimensional plane, then $E$ clearly tiles the space with a subspace tiling partner. By part (4) of the theorem above, this means that $E$ is also spectral.

	Therefore, suppose that $E$ is not contained in a plane. By Corollaries 3.2 and 4.3 from \cite{Researchpaper}, the property of being a tiling set or spectral set remains unchanged after a change of basis. Therefore, after a transformation of the coordinates, we can suppose that $E = \{(0,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,1,0)\}$. This set tiles with tiling partner $\{(0,0,0,0),(0,0,0,1),(1,1,1,0),(1,1,1,1)\}$, which turns out to be a subspace. By part (4), this implies that $E$ is also spectral.
\end{proof}

\section{The counterexample}

We return to the situation when $p$ is an odd prime. In this section we will construct a spectral set $E\subseteq \ZpZ{4}$ of size $2p$. As $2p$ does not divide $p^4$, this implies that $E$ cannot be tiling.

Let $p$ be an odd prime and fix a non-square $n \in \Z/p\Z$. Define the sets $E$ and $A$ in $\ZpZ{4}$ by

\[E = \left\{(t^2,t,t,1) \,\, | \,\, t \in \Z/p\Z\right\} \cup \left\{(nt^2,nt,t,n) \,\, | \,\, t \in \Z/p\Z\right\}\]
and
\[\hspace{.6cm}A = \left\{(1,2i,0,i^2) \,\, | \,\, i \in \Z/p\Z\right\} \cup \left\{(0,0,-2ni,ni^2	) \,\, | \,\, i \in \Z/p\Z\right\}.\]

Remark that when $p \equiv 3 \pmod{4}$ we can choose $n = -1$, which corresponds, up to invertible linear transformation, to the counterexample constructed in \cite{Researchpaper}.

First, we reformulate the orthogonality condition for any $\textbf{a},\textbf{b} \in A$

\begin{align*}
0 &= \langle \chi_\textbf{a}, \chi_{\textbf{b}} \rangle_{L^2(E)} = \sum_{\textbf{e} \in E}\chi_\textbf{a}(\textbf{e})\overline{\chi_{\textbf{b}}(\textbf{e})} = \sum_{\textbf{e} \in E}\chi_{(\textbf{a}-\textbf{b})}(\textbf{e})
\end{align*}

The second ingredient is the following well-known lemma on the vanishing of $p$-th roots of unity for which we give a short proof.

\begin{lemma}\label{vanishing roots}
	Let $\xi = e^{\frac{2\pi i}{p}}$, then
	$\sum_{i = 0}^{p-1}c_i\xi^i = 0, c_i \in \mathbb Q \Leftrightarrow \text{all $c_i$ are equal}$.
\end{lemma}
\begin{proof}
	The $p$-th roots of unity are exactly $\{1, \xi,\dots,\xi^{p-1}\}$, the roots of $X^p-1 \in \mathbb Q[X]$. As this polynomial factors over $\mathbb Q[X]$ into the irreducible factors $X-1$ and $X^{p-1}+\dots+X+1$ by Eisenstein's criterion (after the transformation $X \rightarrow X+1$), this directly implies that the only possibility for a sum of $p$-th roots to be zero is if all coefficients are equal.
\end{proof}

Using this lemma we can obtain a geometrical interpretation of spectral sets. When $|E| > 1$, a spectral set $E$ with spectrum $A$ is equidistributed along the hyperplanes of any parallel class defined by the linear equations
\[\langle \textbf{a}-\textbf{b}, \textbf{e} \rangle = c, \,\,c \in \Z/p\Z,\]
where $\textbf{a}$ and $\textbf{b}$ are fixed and distinct elements of $A$.
This implies in general, that a spectral set is a singleton or has size divisible by $p$, which is part of Theorem \ref{mainthmresearchpaper} (1).

We will now show that $\{\chi_\textbf{a} \,\, | \,\, \textbf{a} \in A\}$ is indeed an orthogonal set of functions in $L^2(E)$. As also $|E|=|A|$, this then implies that $A$ is a spectrum for $E$.

We claim that for any $c \in \Z/p\Z$, the number of solutions in $\textbf{e}$ to the equation $\langle \textbf{a}-\textbf{b},\textbf{e} \rangle = c$ always equals two. If this holds, then the equalities

\[\sum_{\textbf{e} \in E}\chi_{(\textbf{a}-\textbf{b})}(\textbf{e}) = \sum_{\textbf{e} \in E} e^{\frac{2\pi i}{p}\langle \textbf{a}-\textbf{b},\textbf{e} \rangle} = \sum_{m=0}^{p-1} 2e^{\frac{2\pi i}{p}m} = 0,\]

applying Lemma \ref{vanishing roots} in the last equality, show that $\{\chi_\textbf{a} \,\, | \,\, \textbf{a} \in A\}$ is indeed an orthogonal set in $L^2(E)$. As $A$ is defined as the union of two one-parameter sets, the vector $\textbf{a}-\textbf{b}$ is one of the following expressions:

\[\textbf{a}-\textbf{b} = \begin{cases}
	(0,2(i-j),0,i^2-j^2) & (i \neq j) \\[2pt]
	(0,0,-2n(i-j),n(i^2-j^2)) & (i \neq j) \\[2pt]
	\pm(1,2i,2nj,i^2-nj^2)
\end{cases} \hspace{10pt} \text{ where } i,j \in \Z/p\Z.\]

For each of the cases, we will show that $\langle \textbf{a}-\textbf{b},\textbf{e} \rangle = c$ has two solutions in $\textbf{e}$.

\begin{enumerate}
	\item {$\textbf{a}-\textbf{b} = (0,2(i-j),0,i^2-j^2)$ where $i,j \in \Z/p\Z, i \neq j$}. \\
	The equation $\langle \textbf{a}-\textbf{b}, \textbf{e} \rangle = c$, can be rewritten as
	\begin{align*}
	2(i-j)t+(i^2-j^2)&=c \\
	2(i-j)nt+n(i^2-j^2)&=c.
	\end{align*}
	As $i \neq j$, it is clear that each equation has exactly one solution in $t$ and so there are two solutions for $t$, and hence for $\textbf{e}$, in total.

	\item {$\textbf{a}-\textbf{b} = (0,0,-2n(i-j),n(i^2-j^2))$ where $i,j \in \Z/p\Z, i \neq j$} \\
	This case is similar to the previous one as the resulting equations are
	\begin{align*}
	-2n(i-j)t+n(i^2-j^2)&=c \\
	-2n(i-j)t+n^2(i^2-j^2)&=c.
	\end{align*}

	\item {$\textbf{a}-\textbf{b} = \pm(1,2i,2nj,i^2-nj^2)$ where $i,j \in \Z/p\Z$} \\
	Now we find the union of two quadratic equations $Q_1(t)$ and $Q_2(t)$:
	\begin{align*}
	t^2+(2i+2nj)t+(i^2-nj^2)&=\pm c \\
	nt^2+(2ni+2nj)t+(ni^2-n^2j^2)&=\pm c.
	\end{align*}
	Denoting the discriminants by $D_1$ and $D_2$ respectively, one can compute that
	\[nD_1 = D_2 = 4n(2nij+(n^2+n)j^2\pm c) \]
	Recalling the fact that $n$ is a non-square in $\Z/p\Z$, this means that $Q_1(t)=0$ has two solutions if and only if $Q_2(t)=0$ has zero and vice versa, and if one has a unique solution, then the other does too. Collecting everything, this means we can always find two solutions again, which concludes the proof.  \hfill \qed
\end{enumerate}

\nocite{*}

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