text:
\section{Introduction and main results}

We consider a second-order elliptic operator in a bounded, connected, and open set $\Omega$ in $\mathbb{R}^2$.
Let $L$ be an elliptic operator in non-divergence form given by
\begin{equation}				\label{master}
Lu=\sum_{i,j=1}^2 a^{ij}(x) D_{ij}u.
\end{equation}
Here, we assume (without loss of generality) that the coefficients $a^{ij}$ are symmetric and defined on the entire space $\mathbb{R}^2$.
We require that the matrix $\mathbf{A}=(a^{ij})$ satisfy the uniform ellipticity condition, i.e., there exists a constant $\nu \in (0,1]$ such that
\begin{equation}				\label{ellipticity}
\nu \abs{\xi}^2 \le \sum_{i,j=1}^2 a^{ij}\xi_i \xi_j \le \nu^{-1} \abs{\xi}^2.
\end{equation}
We shall say that $\mathbf{A}=(a^{ij})$ are of Dini mean oscillation
in $\Omega$ if the mean oscillation function $\omega_{\mathbf A}: \mathbb{R}_+ \to \mathbb{R}$ defined by
\begin{equation}					\label{13.24f}
\omega_{\mathbf A}(r):=\sup_{x\in \overline \Omega} \fint_{\Omega(x,r)} \,\abs{
\mathbf A(y)-\bar {\mathbf A}_{\Omega(x,r)}}\,dy,
\end{equation}
where
\[
\bar{\mathbf A}_{\Omega(x,r)} :=\fint_{\Omega(x,r)} \mathbf A = \frac{1}{\abs{\Omega(x,r)}}\int_{\Omega(x,r)} \mathbf A
\quad\text{and}\quad \Omega(x,r)=B(x,r) \cap \Omega,
\]
satisfies the Dini condition, i.e.,
\begin{equation}			\label{dini}
\int_0^1 \frac{\omega_{\mathbf A}(t)}t \,dt <+\infty.
\end{equation}

In this article, we shall show that if $\Omega$ is a bounded domain with regular boundary and the coefficients $\mathbf{A}=(a^{ij})$ are of Dini mean oscillation in $\Omega$, then the Green's function $G(x,y)$ exists and has logarithmic pointwise bounds.
In fact, we show that $G(\cdot, y)$ is BMO in $\Omega$ and $G(\cdot,\cdot)$ is continuous on $(\Omega\times \Omega) \setminus \{(x,x):x\in \Omega\}$.
We shall also derive pointwise bounds for $G(x,y)$ as well as its derivatives $D_x G(x, y)$ and $D^2_x G(x,y)$.

It is well known that the elliptic operators in divergence form admit Green's functions that are comparable to those of the Laplace operator, even when the coefficients are just measurable; see \cite{LSW63, GW82, KN85, CL92}.
There are also many papers in the literature dealing with the existence and estimates of Green's functions or fundamental solutions of non-divergence form elliptic operators with measurable or continuous coefficients; see e.g., \cite{Bauman84b, Bauman85, Esc2000, FS84, Krylov92}.
In the case when the coefficient matrix $\mathbf{A}$ is uniformly H\"older (or Dini) continuous, it is well known that a Green's function is continuous and satisfies the pointwise bound comparable to that of Laplace operator; see e.g., \cite{Miranda, S73, An78}. For parabolic operators, we refer the reader to \cite{Friedman} for the construction of fundamental solutions by the parametrix method, and also \cite{Eidelman, C11}.
However, unlike the Green's function for elliptic operators in divergence form, in general, Green's function for non-divergence form elliptic operators do not necessarily have pointwise bounds, even if the domain is smooth and the coefficients are uniformly continuous; see \cite{Bauman84}.

In a recent paper \cite{HK20}, it is shown that in three dimensions or higher, the Green's functions for non-divergence form elliptic operators have the pointwise bound $c\abs{x-y}^{2-n}$ if the coefficients of the operator have Dini mean oscillations.
See also \cite{MM10} for a related result with coefficients satisfying ``square'' Dini condition.
However, the proof in \cite{HK20} does not work in the two dimension, which is mostly due to the failure of Sobolev embedding $W^{1,2} \hookrightarrow L^{2n/(n-2)}$ when $n=2$.
It is worthwhile to mention that even for the Laplace case, the behavior of Green's functions is  different in two dimensions.
As a matter of fact, there are quite a few papers devoted to the study of two dimensional Green's functions; see e.g. \cite{DK09, OKB15, TKB13}.

The adjoint operator $L^\ast$ is given by
\[
L^* u= \sum_{i,j=1}^2 D_{ij} (a^{ij}(x) u),
\]
where the coefficient matrix $\mathbf{A}=(a^{ij})$ is the same as that of $L$ and thus, it is of Dini mean oscillation in $\Omega$.
It is known that if $f \in L^p(\Omega)$ with $p \in (1,\infty)$, then the unique $L^p$ solution (see, e.g., \cite[Lemma~2]{EM2016}) of the problem
\begin{equation}				\label{eq20.40th}
L^\ast u = f\;\text{ in }\;\Omega,\quad u=0 \;\text{ on }\;\Omega,
\end{equation}
is uniformly continuous in $\Omega$; see Theorem~1.8 in \cite{DEK18}.
The definitions of $\mathsf{BMO}(\Omega)$, $\norm{\cdot}_\ast$,  $H^1$ atom in $\Omega$, etc. are given in the next section.

Now, we state our main results.
In our first theorem, we shall assume, in place of \eqref{13.24f}, that the $L^2$ mean oscillation
\begin{equation}					\label{13.24g}
\omega_{\mathbf A}(r):=\sup_{x\in \overline \Omega} \left( \fint_{\Omega(x,r)} \,\abs{
\mathbf A(y)-\bar {\mathbf A}_{\Omega(x,r)}}^2\,dy\right)^{\frac12}
\end{equation}
satisfies the Dini condition \eqref{dini}.
In light of H\"older's inequality, this assumption is stronger than our hypothesis that $\mathbf A$ is of Dini mean oscillation.
We will return to the original definition \eqref{13.24f} in our second theorem.

\begin{theorem}				\label{thm1}
Let $\Omega$ be a bounded $C^{1,1}$ domain in $\mathbb{R}^2$.
Assume the coefficient $\mathbf{A}=(a^{ij})$ of the operator $L$ in \eqref{master} satisfies the uniform ellipticity condition \eqref{ellipticity} and that the $L^2$ mean oscillation \eqref{13.24g} of $\mathbf A$ satisfies the Dini condition \eqref{dini}.
Then, there exists a Green's function $G(x,y)\ ( \text{for any } \ x, y \in \Omega, \ x\neq y)$ and it is unique in the following sense:
if $u$ is the unique adjoint solution of the problem \eqref{eq20.40th}, where $f \in L^p(\Omega)$ with $p>1$, then $u$ is represented by
\begin{equation}				\label{eq1747m}
u(y)=\int_\Omega G(x,y) f(x)\,dx.
\end{equation}
Also, $G^\ast(x,y)=G(y,x)$ becomes the Green's function for the adjoint operator $L^\ast$, which is characterized as follows:
for $q>1$ and $f\in L^{q}(\Omega)$,
if $v \in W^{2,q}(\Omega)\cap W^{1,q}_0(\Omega)$ is the strong solution of
\begin{equation}
                                \label{eq10.36}
Lv=f \;\text{ in }\;\Omega,\quad v=0\;\text{ on }\;\partial \Omega,
\end{equation}
then, we have the representation formula
\begin{equation}				\label{eq1748m}
v(y)=\int_\Omega G^\ast(x,y)f(x)\,dx=\int_\Omega G(y,x)f(x)\,dx.
\end{equation}
The Green function $G(x,y)$ satisfies the following  estimates:
\begin{gather}
					\label{eq0528af}
G(\cdot, y) \in \mathsf{BMO}(\Omega)\quad\text{with}\quad \norm{G(\cdot, y)}_{\ast} \le C,\quad  y \in \Omega,\\
					\label{eq0529af}
\abs{G(x,y)} \le C\left( 1+ \log \frac{\operatorname{diam} \Omega}{\abs{x-y}} \right),\quad x \neq y \in \Omega,\\
					\label{eq0530af}
\abs{D_x G(x,y)} \le \frac{C}{\abs{x-y}}, \quad x \neq y \in \Omega,\\
					\label{eq408}
\abs{D_x^2 G(x,y)} \le \frac{C}{\abs{x-y}^2}, \quad y \in \Omega,\ x\in B(y,d_y/2)\setminus \set{y},
\end{gather}
where $d_y=\operatorname{dist} (y,\partial\Omega)$ and $C$ depends on $\nu$, $\Omega$, and the $L^2$ mean oscillation $\omega_{\mathbf A}$ as in \eqref{13.24g}.
\end{theorem}

In our second theorem, we drop the extra assumption that $\mathbf A$ is of $L^2$ Dini mean oscillation but we shall instead assume that $\Omega$ is a $C^{2,\alpha}$ domain for some $\alpha>0$.
\begin{theorem}				\label{thm3}
Let $\Omega$ be a bounded $C^{2,\alpha}$ domain in $\mathbb{R}^2$.
Assume the coefficient $\mathbf{A}=(a^{ij})$ of the operator $L$ in \eqref{master} satisfies the uniform ellipticity condition \eqref{ellipticity} and is of Dini mean oscillation in $\Omega$.
Then, all conclusions of Theorem~\ref{thm1} are valid.
Moreover, we have
\begin{equation}		\label{eq0531af}
\abs{D_x^2 G(x,y)} \le C \abs{x-y}^{-2},\quad x \neq y \in \Omega,
\end{equation}
where  $C$ depends on $\nu$, $\Omega$, and the mean oscillation $\omega_{\mathbf A}$ as in \eqref{13.24f}.
\end{theorem}

Our last theorem is the key to the proof of Theorems \ref{thm1} and \ref{thm3}, the statement of which has its own interest.
\begin{theorem}			\label{thm2}
Let $\Omega \subset \mathbb{R}^2$ and the coefficient $\mathbf A$ satisfy the conditions of  Theorem~\ref{thm1} (resp. Theorem~\ref{thm3}) with $\omega_{\mathbf A}$ given by the formula \eqref{13.24g} (resp. \eqref{13.24f}).
Then, there exists a constant $N_0=N_0(\nu, \Omega, \omega_{\mathbf A})$ such that if $u$ is the adjoint solution of the problem
\[
L^* u=a\;\text{ in }\;\Omega,\quad u=0\;\text{ on }\;\partial\Omega,
\]
where $a$ is an $H^1$ atom in $\Omega$, then $\abs{u} \le N_0$.
\end{theorem}

The proof of the theorem heavily relies on the assumption that $\Omega \subset \mathbb{R}^2$ and cannot be applied to higher dimension.
As a matter of fact, Theorem~{\ref{thm2}} is not true in higher dimensions since otherwise it would imply that the Green's function $G(\cdot, y)$ belongs to $\mathsf{BMO}(\Omega)$, which is not true in higher dimensions.
For example,  consider the Green's function $G(\cdot, y)$ for Laplace operator in $\Omega=B(0,1) \subset \mathbb{R}^n$ with $n \ge 3$.
It does not belong to $L^p(\Omega)$ for $p\ge \frac{n}{n-2}$ and hence cannot belong to $\mathsf{BMO}(\Omega)$.

We conclude the introduction with a few remarks.
First of all, the Green's function $G(x,y)$ is continuous on $\set{(x,y) \in \overline\Omega \times \overline \Omega: x\neq y}$.
Indeed, the proof of Theorem~\ref{thm1} will show that $L^\ast G(x, \cdot)=0$ in $\Omega \setminus B(x, r)$ for any $r>0$ and thus by \cite[Theorem~1.8]{DEK18}, we see that $G(x,\cdot)$ is continuous away from $x$. On the other hand, it is clear from \eqref{eq0530af} that $G(\cdot, y)$ is continuous away from $y$.

Next, it should be mentioned that Bauman \cite{Bauman85} proved an estimate for the normalized Green's function $\tilde G(x,y)=G(x,y)/G(x_0,y)$:
\begin{equation}
                \label{eq1.54}
\tilde G(x,y)\simeq \int_{\abs{x-y}}^1 \, \frac{r}{w(B(y,r))}\,dr,
\end{equation}
where $x$, $y\in B'$, $\abs{x-y} \in (0,\frac12)$, $x_0\in B\setminus \overline{B'}$, $B'\Subset B$ are open balls,
\[
w(E)=\int_E G(x_0,y)\,dy,
\]
and $G$ is the Green's function in $B$.
Such result was obtained under the condition that the coefficients are bounded and measurable in two dimensions, and uniformly continuous in higher dimensions.
Her proof uses the maximum principle. When the coefficients are of Dini mean oscillation, by using the Harnack type inequality \cite[Lemma 4.2]{DEK18}, we see that $G(x_0,y)\simeq 1$ for any $y$ away from $x_0$ and the boundary $\partial B$, which together with \eqref{eq1.54} implies that
\[
G(x,y) \simeq \log \frac{1}{\abs{x-y}},\quad x \neq y \in B,\quad \abs{x-y}<1/2.
\]
In view of the comparison principle, this result also holds for any bounded smooth domain.
Compared to {\cite{Bauman85}}, we prove a stronger result $G(\cdot, y) \in \mathsf{BMO}(\Omega)$, which gives more information of the Green's function and also implies {\eqref{eq0529af}}.
We also note that our proof also works for elliptic systems satisfying the strong ellipticity condition or the Legendre-Hadamard condition.
More precisely, consider an elliptic system
\[
\sum_{j=1}^N L_{ij} u^j :=\sum_{j=1}^N \sum_{\alpha, \beta=1}^2 A^{\alpha\beta}_{ij} D_{\alpha \beta} u^j - \lambda u^i,\quad i=1,\ldots, N.
\]
Suppose the coefficients $\mathbf{A}=(A^{\alpha\beta}_{ij})$ are bounded, satisfies the Legendre-Hadamard condition
\[
\sum_{\alpha,\beta=1}^2 \sum_{i,j=1}^N A^{\alpha\beta}_{ij} \xi_\alpha\xi_\beta \eta^i \eta^j \ge \nu \abs{\xi}^2 \abs{\eta}^2.
\]
Assume that  $\lambda$ is large enough to guarantee the solvability of Dirichlet problem
\[
L_{ij} u^j = f^i\;\text{ in }\;\Omega,\quad u^i=0\;\text{ on }\;\partial\Omega.
\]
See, for instance, \cite[Theorem 8]{DK11}.
Then, under hypothesis of Theorem~\ref{thm1} (resp. Theorem~\ref{thm3}), there exists an $N \times N$ Green's matrix $\vec G(x,y)=(G_{ij}(x,y))$ that satisfy the conclusions of Theorem~\ref{thm1} (resp. Theorem~\ref{thm3}).
The proof requires only routine adjustment and is omitted.

Finally, we remark that in Theorems~{\ref{thm1}} and {\ref{thm3}}, the dependence of the constant $C$ on $\Omega$ is through the constant $N_0$ in Theorem~{\ref{thm2}}, the constant that is hidden in $H^1(\Omega)$ and $\mathsf{BMO}(\Omega)$ duality relation {\eqref{h1-bmo}}, and the constant that appears in the $W^{2,p}$ estimates, which in turn may depend on the $C^{1,1}$ characteristics of the boundary flattening mappings and the diameter of $\Omega$.
The constant $C$ in {\eqref{eq0531af}} of Theorem~{\ref{thm3}} depends additionally on the $C^{2,\alpha}$ bounds on the boundary flattening mappings.
As the proof of Theorem~{\ref{thm2}} reveals, the constant $N_0$ depends on $\Omega$ through the $W^{2,p}$ estimates, the diameter of $\Omega$, and the $C^{2,\alpha}$ characteristics of $\partial\Omega$.
However, the $C^{2,\alpha}$ regularity of the domain $\Omega$ is  used only for the weak type-$(1,1)$ estimate {\eqref{eq1544w}}, which can be dispensed with if we assume the $L^2$ mean oscillation of $\mathbf A$ satisfies the Dini condition.
In fact, in Theorem~{\ref{thm1}}, where $L^2$ Dini mean oscillation condition is imposed on $\mathbf A$, the $C^{1,1}$ regularity condition on $\Omega$ can be relaxed further.
This condition is essentially used only for the $W^{2,p}$-solvability of \eqref{eq10.36} and the $L^p$-solvability of \eqref{eq20.40th}.
The $L^p$-solvability of \eqref{eq20.40th} follows from the $W^{2,p}$-solvability of \eqref{eq10.36} by using the duality argument in \cite{Esc2000}.
For the $W^{2,p}$-solvability of \eqref{eq10.36} in a bounded domain $\Omega\subset \mathbb{R}^n$, we only require $\Omega$ to be in $C^{1,\alpha}$, where $\alpha>1-1/\max\{n,p\}$. Thus, in our case by taking $p=2$, it suffices to assume $\Omega$ to be in $C^{1,\alpha}$, where $\alpha>1/2$. We cannot find an explicit reference for this result, so we sketch a proof in Appendix.
It should be mentioned that this argument uses Alexandrov maximum principle and is not applicable to elliptic systems, that is, $C^{1,1}$ regularity requirement on $\Omega$ in Theorem~{\ref{thm1}} cannot be lifted for elliptic systems.
It is also worth noting that in the two dimensional case, the $W^{2,p}$-solvability of \eqref{eq10.36} is also available when $p$ is close to $2$ and $\Omega$ is a bounded convex domain; see, for instance, \cite{Ca67}.
Thus, Theorem~\ref{thm1} also hold when $\Omega$ is bounded and convex.

\section{Notation}

For $x_0\in \mathbb{R}^n$ and $r>0$, we denote by $B(x_0,r)$ the Euclidean ball with radius $r$ centered at $x_0$, and denote
\[
\Omega(x_0,r):=\Omega \cap B(x_0,r).
\]

We define $\mathsf{BMO}(\Omega)$ as the set of functions $u$ such that
\[
\norm{u}_\ast:=\sup \Set{\fint_{\Omega(x_0,r)} \abs{u-\bar u_{x_0,r}}: x_0\in \overline \Omega,\;\;r>0}
\]
is finite, where we set
\begin{equation}					\label{eq1244th}
\bar u_{x_0,r}:=\begin{cases}
			0 & \text{if } r \ge \operatorname{dist}(x_0, \partial \Omega)\\
			\fint_{\Omega(x_0,r)} u & \text{if }r < \operatorname{dist}(x_0, \partial\Omega).
  \end{cases}
\end{equation}
We shall say that a bounded measurable function $a$ is an atom for $\Omega$ if $a$ is supported in $\Omega(x_0,r)$ for some $x_0 \in \overline\Omega$ and $r>0$ and satisfies
\begin{equation*}
\norm{a}_{L^\infty(\Omega)} \le \frac{1}{\abs{\Omega(x_0,r)}}\quad\text{and}\quad \bar a_{x_0,r} =0.
\end{equation*}
Notice that the latter condition requires $\fint_{\Omega(x_0,r)} a =0$ only if $r <\operatorname{dist}(x_0, \partial\Omega)$.
A function $f$ is in the atomic Hardy space $H^1(\Omega)$ if there is a sequence of atoms $\set{a_i}_{i=1}^\infty$ and a sequence of real numbers $\set{\lambda_i} \in \ell^1$ so that $f=\sum_{i=1}^\infty \lambda_i a_i$.
We define the norm on this space by
\[
\norm{f}_{H^1(\Omega)}=\inf \Set{\sum_{i=1}^\infty\, \abs{\lambda_i}: f=\sum_{i=1}^\infty \lambda_i a_i}.
\]
We note that the expression
\begin{equation}				\label{h1-bmo}
\sup \Set{\int_\Omega a u \,dy: a\text{ is an atom for }\Omega}
\end{equation}
gives an equivalent norm on $\mathsf{BMO}(\Omega)$ and that $\mathsf{BMO}(\Omega)$ may be identified with the dual of the atomic Hardy space $H^1(\Omega)$.
It may seem that our definition of $\mathsf{BMO}(\Omega)$ differs from those in other literatures \cite{Chang94, Jones80, Miyachi90, TKB13} but since we assume that $\Omega$ is at least a $C^{1,\alpha}$ or a bounded convex domain, it is the same.
In particular, $\mathsf{BMO}(\Omega)$ can be identified with the subspace
$\set{f \in \mathsf{BMO}(\mathbb{R}^2): \operatorname{supp} f \subset \overline \Omega}$, with equivalent norms.

We say that $\Omega$ is a $C^{1,\alpha}$ (resp. $C^{2,\alpha}$) domain if each point on $\partial\Omega$ has a neighborhood in which $\partial\Omega$ is the graph of a $C^{1,\alpha}$ (resp. $C^{2,\alpha}$) function for some $\alpha \in (0,1]$.

\section{Proof of Theorems~\ref{thm1} and \ref{thm3}}
In dimension three and higher, the strategy of \cite{HK07} was used in \cite{HK20} to construct Green's function but it does not work in two dimensions.
Here we follow a duality argument in \cite{TKB13}.

For $y \in \Omega$ and $\epsilon>0$, let $v=G_\epsilon(\cdot,y) \in W^{2,2}(\Omega) \cap W^{1,2}_0(\Omega)$ be a unique strong solution of the problem
\begin{equation}			\label{eq1655th}
Lv=\frac{1}{\abs{\Omega(y,\epsilon)}} \, \chi_{\Omega(y,\epsilon)}\;\text{ in }\;\Omega,\quad v=0\;\text{ on }\;\partial\Omega.
\end{equation}
Since $\mathbf A$ is uniformly continuous in $\Omega$ with its modulus of continuity controlled by $\omega_{\mathbf A}$ (see \cite[Appendix]{HK20}), the unique solvability of the problem \eqref{eq1655th} is a consequence of standard $L^{p}$ theory.
Next, for an $H^1$ atom $a$ in $\Omega$, consider the adjoint problem
\[
L^\ast u = a\;\text{ in }\;\Omega,\quad u=0 \;\text{ on }\;\partial \Omega.
\]
By \cite[Lemma~2]{EM2016}, there exists a unique adjoint solution $u$ in $L^2(\Omega)$, and we have
\[
\fint_{\Omega(y,\epsilon)} u = \int_\Omega a G_\epsilon(\cdot, y).
\]
Then, by Theorem~\ref{thm2}, we have
\[
\Abs{\int_\Omega G_\epsilon(x,y) a(x)\,dx} \le N_0.
\]
Therefore, by the $H^1$ and $\mathsf{BMO}$ duality, we find that the BMO norm of $G_\epsilon(\cdot, y)$ is uniformly bounded, i.e.,
\begin{equation}			\label{bmo_bound}
\norm{G_\epsilon(\cdot,y)}_{*}\le C_0
\end{equation}
for some constant $C_0$ depending only on  $\nu$, $\Omega$,  and $\omega_{\mathbf A}$.
The Banach-Alaoglu theorem gives that for each $y$, there is a sequence $\set{\epsilon_j}$ with $\lim_{j\to \infty} \epsilon_j=0$ and a function $G(\cdot, y) \in \mathsf{BMO}(\Omega)$ so that $G_{\epsilon_j}(\cdot, y)$ converges to $G(\cdot, y)$ in the weak-$\ast$ topology of $\mathsf{BMO}(\Omega)$.
Let us fix a $f \in L^p(\Omega)$ with $p>1$ and let $u$ be the unique adjoint solution of \eqref{eq20.40th}.
Then we have
\begin{equation}				\label{eq0141tu}
\fint_{\Omega(y,\epsilon)} u(x)\,dx = \int_\Omega  G_\epsilon(x, y) f(x)\,dx.
\end{equation}
Since $u$ is continuous in $\overline \Omega$ by \cite[Theorem~1.8]{DEK18}, the left- hand side of \eqref{eq0141tu} converges to $u(y)$.
Since $L^p(\Omega) \subset H^1(\Omega)$ for all $p>1$, we obtain the representation \eqref{eq1747m}.
This gives us that $G(\cdot, y)$ is the Green's function with pole at $y$.
Therefore, by using \eqref{bmo_bound}, we obtain \eqref{eq0528af}.

If we choose any sequence  $\set{\epsilon_k}$ with $\lim_{k\to \infty} \epsilon_k=0$, the above argument gives a subsequence of $G_{\epsilon_k}(\cdot, y)$ which converges to a Green's function.
As the Green's function is unique, the limit must be the function $G(\cdot, y)$.
This implies that the entire family $\set{G_\epsilon(\cdot, y)}_\epsilon$ converges to $G(\cdot, y)$ in the weak-$\ast$ topology of $\mathsf{BMO}(\Omega)$.
By the $W^{2,p}$ estimate for non-divergence form elliptic equations with continuous coefficients, we see from \eqref{eq1655th} that $G_\epsilon(\cdot, y)$ satisfies
\[
\norm{G_\epsilon(\cdot,y)}_{W^{2,2}(\Omega\setminus \Omega(y, r))} \le C(r),\quad r>2\epsilon.
\]
This estimate will also hold for the limit and we see that $G(\cdot,y) \in W^{2,2}(\Omega\setminus B(y, r))$ and $L G(\cdot, y)=0$ in $\Omega\setminus B(y, r)$ for any $r>0$. Moreover, we have $G(\cdot, y)=0$ on $\partial\Omega$.

Next, we show the pointwise bound \eqref{eq0529af}.
For $x_0 \neq y \in \Omega$, we set \[r:=\tfrac12 \,\abs{x_0-y}.\]
In the case when $r\ge \operatorname{dist}(x_0, \partial\Omega)$, we can find a point $\hat x_0 \in \partial\Omega$ such that
\[
\Omega(x_0, r) \subset \Omega(\hat x_0, 2r) \subset \Omega\setminus \set{y}.
\]
Since $LG(\cdot,y) =0$ in $\Omega(\hat x_0,2r)$ and $G(\cdot, y)$ vanishes on $\partial\Omega \cap B(\hat x_0, 2r)$, by the local $W^{2,p}$ estimate, the Sobolev embedding theorem, and a standard iteration argument, we see that the local $L^\infty$ estimate is available for $G(\cdot,y)$, i.e.,
\begin{equation}				\label{eq0804tu}
\abs{G(x_0,y)} \lesssim \fint_{\Omega(\hat x_0, 2r)} \abs{G(x,y)}\,dx.
\end{equation}
Then by \eqref{eq0528af}, we have
\[
\abs{G(x_0,y)} \lesssim \fint_{\Omega(\hat x_0, 2r)} \abs{G(x,y)}\,dx \lesssim \norm{G(\cdot, y)}_{\ast} \lesssim 1,
\]
which clearly yields the bound \eqref{eq0529af}.
In the case when $r < \operatorname{dist}(x_0, \partial\Omega)$, consider a chain of domains $\Omega_j=\Omega(x_0, 2^j r)$ for $j=0,\ldots, N$ so that $2^N r \ge \operatorname{diam} \Omega$.
Notice that $N$ can be chosen so that
\begin{equation}			\label{eq0649tu}
N \lesssim \log \frac{\operatorname{diam} \Omega}{\abs{x_0-y}} +1.
\end{equation}
Since $G(\cdot, y)$ is in $\mathsf{BMO}(\Omega)$, we have
\[
\Abs{\fint_{\Omega_j} G(x,y)\,dx - \fint_{\Omega_{j+1}} G(x,y)\,dx} \lesssim \norm{G(\cdot, y)}_{\ast} \lesssim 1
\]
and by the choice of $N$, we have
\[
\fint_{\Omega_N} \abs{G(x,y)} \,dx=\fint_{\Omega} \abs{G(x,y)} \,dx  \lesssim \norm{G(\cdot, y)}_{\ast} \lesssim 1.
\]
Since $L(G(\cdot,y)-c)=LG(\cdot,y) =0$ in $\Omega_0=B(x_0,r) \subset \Omega$ for any $c \in \mathbb{R}$, similar to \eqref{eq0804tu},  we have the local $L^\infty$ estimate
\begin{equation}				\label{eq0805tu}
\abs{G(x_0,y)-c} \lesssim \fint_{\Omega_0} \abs{G(x,y)-c}\,dx,\quad \forall c \in \mathbb{R}.
\end{equation}
Then by taking $c=\fint_{\Omega_0} G(x,y)\,dx$ and using \eqref{eq0528af}, we have
\[
\Abs{G(x_0,y)-\fint_{\Omega_0} G(x,y)\,dx} \lesssim \norm{G(\cdot, y)}_{\ast} \lesssim 1.
\]
Therefore, by telescoping and using \eqref{eq0649tu}, we obtain the bound \eqref{eq0529af}.

Now, we turn to the gradient estimate \eqref{eq0530af}.
In the case when $r\ge \operatorname{dist}(x_0, \partial\Omega)$, similar to \eqref{eq0804tu}, we have the local gradient bound
\[
\abs{D_x G(x_0, y)} \lesssim \frac{1}{r} \fint_{\Omega(\hat x_0,2r)} \abs{G(x,y)}\,dx
\]
and in the case when $r < \operatorname{dist}(x_0, \partial\Omega)$, similar to \eqref{eq0805tu} we have
\[
\abs{D_x G(x_0, y)} \lesssim \frac{1}{r} \fint_{\Omega(x_0,r)} \abs{G(x,y)-c}\,dx,\quad \forall c\in \mathbb{R}.
\]
In both cases, we get \eqref{eq0530af}.

Finally, we prove the second derivative estimate \eqref{eq0531af}.
In the case when $r< \operatorname{dist}(x_0, \partial\Omega)$, we use the interior $C^2$ estimate (see \cite[Theorem~1.6]{DK17}), $W^{2,p}$ estimate for elliptic equations with continuous coefficients, and a standard iteration argument to get
\[
\abs{D_x^2 G(x_0, y)} \lesssim \frac{1}{r^2} \fint_{\Omega(x_0,r)} \abs{G(x,y)-c}\,dx,\quad \forall c\in \mathbb{R}.
\]
In the case when $r \ge \operatorname{dist}(x_0, \partial\Omega)$ and $\Omega$ is a $C^{2,\alpha}$ domain, we apply the $C^2$ estimate near the boundary in \cite{DEK18} (see Lemma 2.18 there), the boundary $W^{2,p}$ estimate for elliptic equations with continuous coefficients, and a standard iteration argument to get
\[
\abs{D_x^2 G(x_0, y)} \lesssim \frac{1}{r^2} \fint_{\Omega(\hat x_0,2r)} \abs{G(x,y)}\,dx.
\]
Therefore, we get \eqref{eq408} and \eqref{eq0531af}.

Finally, we prove that $G^\ast(x,y):=G(y,x)$ becomes the Green's function for the adjoint operator $L^\ast$.
For $x_0 \neq y \in \Omega$ and $\rho>0$, let $u=G^\ast_\rho(\cdot,x_0) \in L^{2}(\Omega)$ be a unique solution of the adjoint problem
\[
L^\ast u=\frac{1}{\abs{\Omega(x_0,\rho)}} \chi_{\Omega(x_0,\rho)}\;\text{ in }\;\Omega,\quad u=0\;\text{ on }\;\partial\Omega.
\]
Notice that we have
\[
\fint_{\Omega(y,\epsilon)} G^\ast_\rho(x,x_0)\,dx= \fint_{\Omega(x_0, \rho)} G_\epsilon(x,y)\,dx.
\]
By \cite[Theorem~1.8]{DEK18}, we know that $G^\ast_\rho(\cdot,x_0)$ is continuous in $\overline \Omega$.
Since $G_\epsilon(\cdot, y) \to G(\cdot, y)$ in weak-$\ast$ topology of $\mathsf{BMO}(\Omega)$, by taking the limit $\epsilon \to 0$, we have
\[
G^\ast_\rho(y,x_0)=  \fint_{\Omega(x_0, \rho)} G(x,y)\,dx.
\]
Therefore, we find
\begin{equation}				\label{eq1014sun}
\lim_{\rho \to 0}\, G^\ast_\rho(y, x_0)=G(x_0, y).
\end{equation}
We note that argument around \eqref{eq0804tu} -- \eqref{eq0805tu} shows that
\begin{equation}				\label{eq1015sun}
\abs{G^\ast_\rho(y, x_0)}
=\Abs{\fint_{\Omega(x_0, \rho)} G(x,y)\,dx}
\le C\left( 1+ \log \frac{\operatorname{diam} \Omega}{\abs{x_0-y}} \right),\quad \forall \rho>0.
\end{equation}
For $f\in L^{q}(\Omega)$ with $q\in (1,\infty)$, let $v \in W^{2,q}(\Omega)\cap W^{1,q}_0(\Omega)$ be the strong solution of
\[
Lv=f \;\text{ in }\;\Omega,\quad v=0\;\text{ on }\;\partial \Omega.
\]
Then, we have
\[
\fint_{\Omega(x_0,\rho)} v(x) \,dx = \int_\Omega  G^\ast_\rho(x,x_0) f(x)\,dx,
\]
and thus, by taking the limit, we also get
\[
v(x_0)=\lim_{\rho \to 0} \int_\Omega G^\ast_\rho(x,x_0)f(x)\,dx.
\]
Therefore, by \eqref{eq1014sun}, \eqref{eq1015sun}, and Lebesgue dominated convergence theorem, we obtain the representation formula \eqref{eq1748m}, which means $G^\ast(x,y)=G(y,x)$ is the Green's function for $L^\ast$.
\hfill\qedsymbol

\section{Proof of Theorem~\ref{thm2}}
We first consider the case when the $L^2$ mean oscillation \eqref{13.24g} of $\mathbf A$ satisfies the Dini condition and the domain is $C^{1,1}$, which is assumed in Theorem~\ref{thm1}.
The other case will be treated at the end of the proof.

Suppose that $a$ is supported in $\Omega(y_0,R)$ so that $\norm{a}_{L_\infty} \lesssim 1/R^2$.
For $x_0 \in \Omega$ and $r>0$, we define
\[
u_{x_0,r}=\begin{cases}
			0 & \text{if } r >2\operatorname{dist}(x_0, \partial \Omega)\\
			\fint_{\Omega(x_0,r)} u & \text{if }r \le 2\operatorname{dist}(x_0, \partial\Omega)
  \end{cases}
\]
and set
\begin{equation}					\label{eq0748w}
\phi(x_0, r):= \fint_{\Omega(x_0,r)} \abs{u-u_{x_0,r}}.
\end{equation}
Note that $u_{x_0, r}$ in the above is slightly different from $\bar u_{x_0, r}$ defined in \eqref{eq1244th}.
Let us define the adjoint operator $L_0^\ast$ with constant coefficients
\[
L_0^\ast u:= \sum_{i,j=1}^2 D_{ij} (\bar a^{ij} u), \quad\text{where }\,\bar a^{ij}=\fint_{\Omega(x_0,r)} a^{ij},
\]
and split $u=v^{(r)}+w^{(r)}+\tilde w^{(r)}$, where $w=w^{(r)}$ is the weak solution of
\[
L_0^\ast w=a\text{ in }\;\Omega(x_0,r),\quad w=0\;\text{ on }\;\partial\Omega(x_0,r),
\]
and $\tilde w=\tilde w^{(r)}$ is the $L^2$ adjoint solution of
\[
L_0^\ast \tilde w=\sum_{i,j=1}^2 D_{ij} \left((\bar a^{ij}-a^{ij}) u \chi_{\Omega(x_0,r)} \right)\; \text{ in }\;\Omega,\quad \tilde w=0\;\text{ on }\;\partial\Omega.
\]
By the $L^2$ theory for adjoint equations, we have\footnote{As remarked in the introduction, here we can relax the assumption that $\Omega$ is a $C^{1,1}$ domain to an assumption that $\Omega$ is a $C^{1,\alpha}$ domain for some $\alpha>\frac12$ or that $\Omega$ is a bounded convex domain.}
\[
\left(\int_{\Omega(x_0,r)} \abs{\tilde w}^2 \right)^{1/2} \le \left(\int_{\Omega} \abs{\tilde w}^2 \right)^{1/2} \lesssim \left(\int_{\Omega(x_0,r)} \abs{\bar A-A}^2  \right)^{1/2} \norm{u}_{L^\infty(\Omega(x_0,r))}.
\]
By H\"older's inequality, we then have
\begin{equation}					\label{eq1019sat}
\fint_{\Omega(x_0,r)} \abs{\tilde w} \lesssim \left(\fint_{\Omega(x_0,r)} \abs{A-\bar A}^2 \right)^{1/2} \norm{u}_{L^\infty(\Omega(x_0,r))}\lesssim \omega_{A}(r) \,\norm{u}_{L^\infty(\Omega(x_0,r))}.
\end{equation}

Now we turn to the decay estimate of $\phi(x_0, r)$.
Let $\kappa \in (0,\frac14)$ be fixed.
Note that $v=v^{(r)}=u-w^{(r)}-\tilde w^{(r)}$ satisfies
\[
L_0^\ast v=0\; \text{ in }\;\Omega(x_0,r),\quad v=u=0 \text{ in }\; \partial\Omega \cap B(x_0,r).
\]
By an interior and boundary estimate for elliptic equations with constant coefficients, we have (recall $v_{x_0,r}=0$ if $B(x_0,r/2)$ intersects $\partial\Omega$)
\begin{equation}				\label{eq13.46f}
\fint_{\Omega(x_0, \kappa r)} \abs{v - v_{x_0, \kappa r}}  \le 2\kappa r \norm{Dv}_{L^\infty(\Omega(x_0,r/3))} \le C_0 \kappa \fint_{\Omega(x_0,r)} \abs{v-v_{x_0,r}}.
\end{equation}
Here, $C_0$ is a constant depending only on $\nu$ and $\Omega$.
By using the decomposition $u=v+w+\tilde w$ and \eqref{eq13.46f}, we obtain
\begin{align*}
\fint_{\Omega(x_0, \kappa r)} \abs{u - u_{x_0, \kappa r}}
& \le \fint_{\Omega(x_0, \kappa r)} \abs{v - v_{x_0, \kappa r}} + 2\fint_{\Omega(x_0, \kappa r)} \abs{w} +2\fint_{\Omega(x_0, \kappa r)} \abs{\tilde w} \\
& \le  C_0\kappa \fint_{\Omega(x_0,r)} \abs{u - u_{x_0,r}} +C (\kappa^{-2}+1) \fint_{\Omega(x_0,r)} \left(\abs{w} + \abs{\tilde w} \right).
\end{align*}
Here, we used the obvious facts that
\[
u_{x_0, r}=v_{x_0,r} +  w_{x_0,r} + \tilde w_{x_0,r},\quad
\abs{w_{x_0,r}} \le \fint_{\Omega(x_0,r)} \abs{w},\quad \abs{\tilde w_{x_0,r}} \le \fint_{\Omega(x_0,r)} \abs{\tilde w}.
\]
Therefore, by \eqref{eq1019sat}, we have
\[
\phi(x_0,\kappa r) \le C_0 \kappa  \phi(x_0, r) +  C (\kappa^{-2}+1) \omega_{\mathbf A}(r) \,\norm{u}_{L^\infty(\Omega(x_0,r))}
+ C (\kappa^{-2}+1) \fint_{\Omega(x_0,r)}\abs{w}.
\]
Now we fix a $\kappa=\kappa(\nu, \Omega) \in (0,\frac14)$ sufficiently small so that $C_0\kappa \le 1/2$.
Then, we obtain
\begin{equation}				\label{eq16.57m}
\phi(x_0, \kappa r) \le \frac12 \phi(x_0, r)+ C \omega_{\mathbf A}(r) \,\norm{u}_{L^\infty(\Omega(x_0,r))}+  C \fint_{\Omega(x_0,r)}\abs{w^{(r)}}.
\end{equation}
By iterating, for $j=1,2,\ldots$, we get
\begin{equation}				\label{eq1525mon}
\phi(x_0, \kappa^j r) \le 2^{-j} \phi(x_0, r) +C \norm{u}_{L^\infty(\Omega(x_0,r))} \sum_{i=1}^{j} 2^{-i} \omega_{\mathbf A}(\kappa^{j-i} r) + C\psi_j(x_0,r),
\end{equation}
where we set
\begin{equation}				\label{eq10.55}
\psi_j(x_0,t):=\sum_{i=1}^{j} 2^{-i}\fint_{\Omega(x_0,\kappa^{j-i}r)}\abs{w^{(\kappa^{j-i}r)}}.
\end{equation}
Note that
\begin{align}
							\nonumber
\sum_{j=1}^\infty \sum_{i=1}^j 2^{-i} \omega_{\mathbf A}(\kappa^{j-i} r)& =\sum_{i=1}^\infty \sum_{j=i}^\infty 2^{-i} \omega_{\mathbf A}(\kappa^{j-i} r)
=\sum_{i=1}^\infty  2^{-i}\sum_{j=0}^\infty \omega_{\mathbf A}(\kappa^{j} r)\\
							\label{eq0752w}
&=\sum_{j=0}^\infty \omega_{\mathbf A}(\kappa^{j} r) \lesssim \int_0^r \frac{\omega_{\mathbf A}(t)}{t}\,dt.
\end{align}

\begin{lemma}					\label{lem-key}
We have
\begin{equation}				\label{eq1855w}
\sum_{j=1}^\infty \psi_j(x_0,r) \lesssim 1,\quad \forall x_0 \in \Omega, \quad 0<\forall r <\operatorname{diam} \Omega.
\end{equation}
\end{lemma}
Take the lemma for granted now.
Then from \eqref{eq1525mon} and \eqref{eq0752w} we find that
\begin{equation}				\label{eq0750w}
\sum_{j=0}^\infty \phi(x_0, \kappa^j r) \lesssim \phi(x_0, r) +  \norm{u}_{L^\infty(\Omega(x_0,r))} \int_0^r \frac{\omega_{\mathbf A}(t)}{t}\,dt +1.
\end{equation}
Since we have
\[
\abs{u_{x_0,\kappa r} - u_{x_0, r}} \le
\abs{u(x)-u_{x_0,r}} + \abs{u(x) - u_{x_0,\kappa r}},
\]
by taking average over $x \in \Omega(x_0, \kappa r)$, we obtain
\[
\abs{u_{x_0,\kappa r} - u_{x_0,r}} \le \phi(x_0,\kappa r) + \phi(x_0, r).
\]
Then, by iterating, we get
\begin{equation}				\label{eq0756w}
\abs{u_{x_0,\kappa^i r} - u_{x_0, r}} \le 2 \sum_{j=0}^i \phi(x_0, \kappa^j r).
\end{equation}
By Theorem~1.8 in \cite{DEK18}, we find that $u$ is continuous in $\overline\Omega$, and thus we see that
\begin{equation}				\label{eq1402t}
\lim_{i\to \infty} u_{x_0,\kappa^i r}=u(x_0).
\end{equation}
Therefore, by taking $i\to \infty$ in \eqref{eq0756w} and using \eqref{eq0750w}, we get
\begin{equation}				\label{eq13.13}
\abs{u(x_0)-u_{x_0,r}} \lesssim \phi(x_0, r)+
\norm{u}_{L^\infty(\Omega(x_0,r))} \int_0^r \frac{\omega_{\mathbf A}(t)}t \,dt+1,
\end{equation}
which implies that
\begin{equation}			\label{eq14.39t}
\abs{u(x_0)}  \lesssim r^{-2} \norm{u}_{L^1(\Omega(x_0,r))} + \norm{u}_{L^\infty(\Omega(x_0,r))} \int_0^r \frac{\omega_{\mathbf A}(t)}t \,dt+1.
\end{equation}
Now, taking the supremum for $x_0\in \Omega(x,r)$, where $x \in \overline\Omega$, we have
\[
\norm{u}_{L^\infty(\Omega(x,r))} \le
C \left( r^{-2} \norm{u}_{L^1(\Omega(x, 2r))} + \norm{u}_{L^\infty(\Omega(x, 2r))} \int_0^r \frac{\omega_{\mathbf A}(t)}t \,dt +1\right).
\]
We fix $r_0<\frac13$ such that for any $0<r\le r_0$,
\[
C \int_0^{r} \frac{\omega_{\mathbf A}(t)}t \,dt \le \frac1{3^2}.
\]
Then, we have for any $x \in \Omega$ and $0<r\le r_0$ that
\[
\norm{u}_{L^\infty(\Omega(x,r))} \le
3^{-2}\norm{u}_{L^\infty(\Omega(x, 2r))} + C r^{-2} \norm{u}_{L^1(\Omega(x, 2r))} + C.
\]
For $k=1,2,\ldots$, denote $r_k=3-2^{1-k}$.
Note that $r_{k+1}-r_k=2^{-k}$ for $k\ge 1$ and $r_1=2$.
Without loss of generality, let us assume that $0 \in \Omega$.
For $x\in \Omega_{r_k}=\Omega(0, r_k)$ and $r\le 2^{-k-2}$, we have $B(x, 2r) \subset B_{r_{k+1}}$. We take $k_0\ge 1$ sufficiently large such that $2^{-k_0-2}\le r_0$.
It then follows that for any $k\ge k_0$,
\[
\norm{u}_{L^\infty(\Omega_{r_k})} \le 3^{-2} \norm{u}_{L^\infty(\Omega_{r_{k+1}})}+C 2^{2k} \norm{u}_{L^1(\Omega_3)} + C.
\]
By multiplying the above by $3^{-2k}$ and then summing over $k=k_0, k_0+1,\ldots$, we reach
\[
\sum_{k=k_0}^\infty 3^{-2k}\norm{u}_{L^\infty(\Omega_{r_k})} \le \sum_{k=k_0}^\infty 3^{-2(k+1)} \norm{u}_{L^\infty(\Omega_{r_{k+1}})}+C \norm{u}_{L^1(\Omega_3)} + C.
\]
Since we assume that $u\in L^\infty(\Omega)$, the summations on both sides are convergent.
Therefore, we have
\[
3^{-2k_0}\norm{u}_{L^\infty(\Omega_{r_{k_0}})} \le C  \norm{u}_{L^1(\Omega_3)} + C,
\]
which in turn implies (recall $2 < r_{k_0}$)
\begin{equation}					\label{eq10.23m}
\norm{u}_{L^\infty(\Omega_2)} \lesssim  \norm{u}_{L^1(\Omega_3)} + 1.
\end{equation}

\begin{lemma}						\label{lem-global}
We have
\[
\norm{u}_{L^1(\Omega)} \lesssim  1.
\]
\end{lemma}
\begin{proof}
For any $f \in L^\infty(\Omega)$, let $v \in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ be the unique solution of the problem
\[
Lv=f\;\text{ in }\;\Omega,\quad v=0\;\text{ on }\;\partial\Omega.
\]
See, for instance, \S 11.3 in {\cite{K08}}.
By the Sobolev embedding and the $W^{2,2}$ theory, we have $v \in C(\overline \Omega)$ and
\[
\norm{v}_{L^\infty(\Omega)} \le C\norm{v}_{W^{2,2}(\Omega)}
\le C \norm{f}_{L^2(\Omega)} \le C \norm{f}_{L^\infty(\Omega)},
\]
where $C=C(\nu, \omega_{\mathbf A}, \Omega)$.\footnote{Here, we only need that  $\Omega$ is a bounded $C^{1,1}$ domain. In the scalar case, by using Alexandrov estimate, one can bypass the $W^{2,2}$ estimate and directly get $\norm{v}_{L^\infty(\Omega)} \le C \norm{f}_{L^2(\Omega)} \le C \norm{f}_{L^\infty(\Omega)}$, and thus only boundedness of $\Omega$ is needed.}
Since
\[
\int_\Omega u f \,dx= \int_\Omega a v\,dx,
\]
and $\norm{a}_{L^1(\Omega)} \le  1$, we find
\[
\Abs{\int_\Omega u f\,dx } \le C \norm{f}_{L^\infty(\Omega)},
\]
which implies that $\norm{u}_{L^1(\Omega)} \le C$.
\end{proof}
By Lemma~\ref{lem-global} and \eqref{eq10.23m}, we have (recalling that $0\in \Omega$ is an arbitrary choice)
\[
\norm{u}_{L^\infty(\Omega)} \lesssim 1
\]
as desired.
The proof of the theorem is complete once we prove Lemma~\ref{lem-key}.

\begin{proof}[Proof of Lemma~\ref{lem-key}]
Let $G_r(x,y)$ denote the Green's function for the constant coefficient operator $L_0^\ast =  \sum D_{ij} \bar a^{ij} =  \sum \bar a^{ij} D_{ij}$ in $\Omega(x_0,r)$.
Then we have
\begin{equation*}
w^{(r)}(x)=\int_{\Omega(x_0,r)} G_r(x,y) a(y)\,dy,\quad \forall x \in \Omega(x_0,r).
\end{equation*}
We shall use the following  estimates for Green's function $G_r(x,y)$:
\begin{align}
							\label{green1}
\abs{G_r(x,y)} &\le C\left(1+\log \frac{2r}{\abs{x-y}} \right),\\
							\label{green2}
\abs{G_r(x,y)-G_r(x,y')} &\le C\,\frac{\abs{y-y'}}{\abs{x-y}}.
\end{align}
Since $L_0$ is a constant coefficients operator, the above estimates are widely known.
We remark that \eqref{green2} may not be sharp if $\abs{y-y'} > \frac12 \abs{x-y}$ but it is still a legitimate estimate, which can be seen by telescoping: choose a sequence of points $y_0, y_1, \ldots, y_N$ in $\Omega$ such that\footnote{This is always available when $\Omega$ is, for example, a bounded Lipschitz domain.}
\[
y_0=y,\;\; y_N=y',\;\; \abs{y_i-y_{i-1}} \le \frac12 \abs{x-y},\;\; \abs{x-y} \le \abs{x-y_i},\;\text{ and }\;\sum_{i=1}^N\, \abs{y_i-y_{i-1}} \simeq \abs{y-y'}.
\]
Then, we have
\[
\abs{G(x,y)-G(x,y')} \le \sum_{i=1}^N\, \abs{G(x,y_{i-1})-G(x,y_i)}\lesssim \sum_{i=1}^N \,\frac{\abs{y_{i-1}-y_i}}{\abs{x-y_i}} \lesssim \frac{\abs{y-y'}}{\abs{x-y}},
\]
which establishes \eqref{green2}.

To estimate $w^{(r)}$, we consider two cases: $\abs{x_0-y_0} \le 2R$ and $\abs{x_0-y_0} >2R$.

{\bf Case 1.} Let us first consider the case when $\abs{x_0-y_0} \le 2R$.
If $r \le 3R$, we use the size condition of $a$ and \eqref{green1} to estimate $w^{(r)}(x)$ for $x \in \Omega(x_0,r)$ as follows:
\begin{align}
						\nonumber
\abs{w^{(r)}(x)} &\le \int_{\Omega(x_0,r)} \abs{G_r(x,y)}\, \abs{a(y)} \,dy \lesssim \frac{1}{R^2} \int_{\Omega(x_0,r) \cap \Omega(y_0,R)} \left(1+\log\frac{2r}{\abs{x-y}}\right)\,dy\\
						\label{eq1229mon}
&\lesssim \frac{1}{R^2} \int_{B(x,2r)}\left( 1+ \log\frac{2r}{\abs{x-y}}\right)\,dy \lesssim  \frac{r^2}{R^2}.
\end{align}
Therefore, we have
\begin{equation}
					\label{eq1234th}
\fint_{\Omega(x_0,r)} \abs{w^{(r)}} \lesssim \frac{r^2}{R^2}\quad\text{when }\; r\le 3R.
\end{equation}
If $r > 3R$, we have $\Omega(x_0,r) \supset \Omega(y_0,R)$.
In the case when $R <\operatorname{dist}(y_0, \partial\Omega)$, we use the cancellation property of $a$ to find that
\[
w^{(r)}(x) =\int_{\Omega(x_0,r)} G_r(x,y) a(y)\,dy=\int_{\Omega(y_0,R)} \left(G_r(x,y)-G_r(x, y_0)\right) a(y)\,dy.
\]
Then by the estimate \eqref{green2} and using the symmetry, we have
\begin{align*}
\abs{w^{(r)}(x)} &\le \int_{\Omega(y_0,R)} \abs{G_r(x,y) -G_r(x,y_0)}\, \abs{a(y)} \,dy \\
& \lesssim \frac{1}{R} \int_{B(y_0, R)} \min \left( \frac{1}{\abs{x-y}}, \frac{1}{\abs{x-y_0}} \right)\,dy.
\end{align*}
In the case when $\abs{x-y_0} \le R$, we estimate
\[
\abs{w^{(r)}(x)} \lesssim \frac{1}{R} \int_{B(y_0, R)}\frac{1}{\abs{x-y}}\,dy \lesssim \frac{1}{R} \int_{B(x,2R)}\frac{1}{\abs{x-y}}\,dy \lesssim 1.
\]
In the case when $\abs{x-y_0} > R$, we estimate
\[
\abs{w^{(r)}(x)} \lesssim \frac{1}{R} \int_{B(y_0, R)}\frac{1}{\abs{x-y_0}}\,dy \lesssim \frac{R}{\abs{x-y_0}}.
\]
Note that for $x\in \Omega(x_0,r)$, we have
\[
\abs{x-y_0} \le \abs{x-x_0}+ \abs{x_0-y_0} \le r+2R < 2r.
\]
Combining these together, we have
\begin{align*}
\fint_{\Omega(x_0,r)} \abs{w^{(r)}(x)}\,dx  &\lesssim \frac{1}{r^2} \left( \int_{B(y_0, R)} \abs{w^{(r)}(x)}\,dx+ \int_{B(y_0, 2r)\setminus B(y_0, R)} \abs{w^{(r)}(x)}\,dx \right) \\
&\lesssim \frac{1}{r^2} \left(\int_{B(y_0, R)} dx+  \int_{B(y_0, 2r)} \frac{R}{\abs{x-y_0}}\,dx\right) \lesssim \frac{R^2}{r^2}+ \frac{R}{r} \lesssim \frac{R}{r}.
\end{align*}
In the case when $R \ge \operatorname{dist}(y_0, \partial\Omega)$, we can find $y' \in \partial \Omega(x_0,r)$ such that $\abs{y'-y_0} \le R$.
Therefore, we have
\[
w^{(r)}(x) =\int_{\Omega(x_0,r)} G_r(x,y) a(y)\,dy=\int_{\Omega(y_0,R)} \left(G_r(x,y)-G_r(x, y')\right) a(y)\,dy.
\]
Notice that $\Omega(y_0, R) \subset \Omega(y', 2R)$.
Then, by repeating the above argument with $y'$ in place of $y_0$, we get the same conclusion.
Therefore, we have
\begin{equation}			\label{eq1235th}
\fint_{\Omega(x_0,r)} \abs{w^{(r)}(x)}\,dx  \lesssim  \frac{R}{r} \quad\text{when }\; r> 3R.
\end{equation}

Now, let us look into $\psi_j(x_0,r)$, which is defined in \eqref{eq10.55}.
Let $\ell$ be the largest integer satisfying $\kappa^\ell r>3R$.
In the case when $\ell <0$, we have $r\le 3R$, and thus by \eqref{eq1234th},
\[
\psi_j(x_0, r)= \sum_{i=1}^{j} 2^{-i} \left(\frac{\kappa^{j-i}r}{R}\right)^2  \lesssim 2^{-j} \frac{r^2}{R^2} \lesssim 2^{-j}.
\]
In the case when $0\le \ell < j$, we have by \eqref{eq1234th} and \eqref{eq1235th} that
\begin{align*}
\psi_j(x_0,r)&= \sum_{i=1}^{j-\ell-1} 2^{-i} \left(\frac{\kappa^{j-i}r}{R}\right)^2
+\sum_{i=j-\ell}^{j} 2^{-i} \frac{R}{\kappa^{j-i}r} \\
&\lesssim \left(\frac{\kappa^\ell r}{R}\right)^2 2^{\ell-j} + \left(\frac{R}{\kappa^\ell r}\right) 2^{\ell-j} \lesssim 2^{\ell-j},
\end{align*}
where we used $\kappa^\ell r \simeq R$, which follows from the choice of $\ell$.
Finally, in the case when $\ell \ge j$, we have by \eqref{eq1235th} that
\[
\psi_j(x_0,r)=R \sum_{i=1}^{j} 2^{-i} \frac{R}{\kappa^{j-i}r}  \lesssim \frac{R}{\kappa^j r} = \frac{R}{\kappa^{\ell} r} \kappa^{\ell-j} \lesssim \kappa^{\ell-j}.
\]
Therefore, we have
\begin{align}
						\nonumber
\sum_{j=0}^\infty \psi_j(x_0,r) &\lesssim \sum_{j=0}^\infty \left( 2^{-j}[\ell<0]+ 2^{\ell-j}[0\le \ell<j]+  \kappa^{\ell-j}[\ell \ge j] \right) \\
						\label{eq1939sun}
& \lesssim \sum_{j=0}^\infty 2^{-j} + \sum_{j=\ell}^\infty 2^{\ell-j}+\sum_{j=0}^{\ell} \kappa^{\ell-j}\lesssim 1.
\end{align}
This completes the proof of \eqref{eq1855w} in the case when $\abs{x_0-y_0} \le 2R$.

{\bf Case 2.} Next, we turn to the proof of \eqref{eq1855w} in the case when $\abs{x_0-y_0}>2R$.
We first consider the case when $\Omega(y_0,R) \not\subset \Omega(x_0,r)$.
Instead of \eqref{eq1229mon}, we estimate
\begin{equation}				\label{eq1109th}
\abs{w^{(r)}(x)} \le \int_{\Omega(x_0,r)} \abs{G_r(x,y)}\, \abs{a(y)} \,dy \lesssim \frac{1}{R^2} \int_{\Omega(x_0,r) \cap \Omega(y_0,R)} \abs{G_r(x,y)}\,dy.
\end{equation}
If $\Omega(x_0,r) \cap \Omega(y_0,R) =\emptyset$, then the above integral is zero.
Therefore, we have
\begin{equation}					\label{eq1048w}
\fint_{\Omega(x_0,r)} \abs{w^{(r)}(x)}\,dx =0\quad\text{when }\;\Omega(x_0,r) \cap \Omega(y_0,R) =\emptyset.
\end{equation}
If $\Omega(x_0,r) \cap \Omega(y_0,R) \neq \emptyset$, then for $y\in \Omega(x_0,r) \cap \Omega(y_0,R)$, we can find $y' \in \partial \Omega(x_0,r)$ such that $\abs{y-y'} \le 2R$   (recall  $\Omega(y_0,R) \not\subset \Omega(x_0,r)$).
Then, by the Green's function estimate \eqref{green2}, we have
\[
\abs{G_r(x,y)}=\abs{G_r(x,y)-G_r(x,y')} \lesssim \frac{R}{\abs{x-y}}.
\]
Hence by \eqref{eq1109th}, we obtain
\[
\abs{w^{(r)}(x)} \lesssim \frac{1}{R} \int_{\Omega(x_0,r) \cap \Omega(y_0,R)} \frac{1}{\abs{x-y}}\,dy.
\]
In the case when $\abs{x-y_0} >2R$, we have $\abs{x-y}>R$, and thus
\[
\abs{w^{(r)}(x)}  \lesssim \frac{1}{R^2} \int_{B(y_0, R)} \,dy \lesssim 1.
\]
In the case when $\abs{x-y_0} \le 2R$, we have $B(y_0, R) \subset B(x, 3R)$, and thus
\[
\abs{w^{(r)}(x)}  \lesssim \frac{1}{R} \int_{B(x, 3R)} \frac{1}{\abs{x-y}}\,dy \lesssim 1.
\]
In both cases, we have
\begin{multline}				\label{eq0751tue}
\fint_{\Omega(x_0,r)} \abs{w^{(r)}(x)}\,dx \lesssim 1\\
\text{when }\;\Omega(y_0,R) \not\subset \Omega(x_0,r)\;\text{ and }\;\Omega(x_0,r) \cap \Omega(y_0,R) \neq\emptyset.
\end{multline}

In the case when $\Omega(y_0,R) \subset \Omega(x_0,r)$, we have $r>3R$ (recall $\abs{x_0-y_0}>2R$) and similar to \eqref{eq1235th}, we have
\begin{equation}				\label{eq1108tue}
\fint_{\Omega(x_0,r)} \abs{w^{(r)}(x)}\,dx \lesssim \frac{R}{r}.
\end{equation}
Since we assume $\abs{x_0-y_0}>2R$, there exists the smallest integer $\ell$ satisfying
\[
\Omega(x_0, \kappa^\ell r) \cap \Omega(y_0,R) =\emptyset.
\]
Then, by the choice of $\ell$,
\[
\Omega(x_0, \kappa^{\ell-1}r) \cap \Omega(y_0,R) \neq \emptyset,
\]
and thus $\kappa^{\ell-1} r >\abs{x_0-y_0}-R>R$.
Also, since  $\kappa < \frac14$, we note that
\[\Omega(y_0,R) \subset \Omega(x_0, 2R+\kappa^{\ell-1}r) \subset \Omega(x_0, \kappa^{\ell-2}r).
\]
In the case when $\ell \le 1$, then by \eqref{eq1048w} and \eqref{eq0751tue}, we have
\[
\psi_j(x_0,r)=2^{-j} \fint_{\Omega(x_0,r)} \abs{w^{(r)}} \lesssim 2^{-j}.
\]
In the case when $1<\ell \le j$, then by \eqref{eq1048w}, \eqref{eq0751tue}, and \eqref{eq1108tue}, we have
\begin{align*}
\psi_j(x_0,r)&= 2^{-(j-\ell+1)} \fint_{\Omega(x_0, \kappa^{\ell-1}r)} \abs{w^{(\kappa^{\ell-1}r)}}+ \sum_{i=j-\ell+2}^{j} 2^{-i} \fint_{\Omega(x_0, \kappa^{j-i}r)} \abs{w^{(\kappa^{j-i}r)}}\\
& \lesssim 2^{\ell-j} + \sum_{i=j-\ell+1}^{j} 2^{-i} \frac{R}{\kappa^{j-i}r} \lesssim 2^{\ell-j}+
\frac{R}{\kappa^{\ell-1} r} 2^{\ell-j} \lesssim 2^{\ell-j}.
\end{align*}
In the case when $\ell > j$, then by \eqref{eq0751tue} we have
\[
\psi_j(x_0,r) = \sum_{i=1}^{j} 2^{-i} \fint_{\Omega(x_0, \kappa^{j-i}r)} \abs{w^{(\kappa^{j-i}r)}} \lesssim \sum_{i=1}^{j} 2^{-i}  \frac{R}{\kappa^{j-i}r}  \lesssim \frac{R}{\kappa^j r} =\frac{R}{\kappa^{\ell-1}r} \kappa^{\ell-1-j} \lesssim \kappa^{\ell-j}.
\]
Therefore, similar to \eqref{eq1939sun}, we obtain
\[
\sum_{j=0}^\infty \psi_j(x_0,r) \lesssim \sum_{j=0}^\infty\left( 2^{-j}[\ell \le 1] + 2^{\ell-j} [1<\ell \le j]+ \kappa^{\ell-j}[\ell>j] \right) \lesssim 1
\]
as desired.
\end{proof}

Now, we consider the other case when and $\mathbf A$ is of Dini mean oscillation and $\Omega$ is a $C^{2,\alpha}$ domain.
We note that $L^2$ mean oscillation \eqref{13.24g} is used to obtain the estimate \eqref{eq1019sat}, which seems no longer available with $L^1$ mean oscillation \eqref{13.24f}.
We shall derive an estimate which substitutes \eqref{eq1019sat} as follows.
Since $L_0^*$ has constant coefficients and $\partial\Omega$ is of $C^{2,\alpha}$, we have weak type-$(1,1)$ estimate\footnote{Here, we use the assumption that $\Omega$ is a bounded $C^{2,\alpha}$ domain for some $\alpha>0$.} (see e.g., \cite[Lemma~2.4]{DEK18})
\begin{equation}				\label{eq1544w}
\Abs{\set{x \in \Omega: \abs{\tilde w(x)} >t}} \lesssim \frac{1}{t} \int_\Omega \Abs{(\bar A-A) u \chi_{\Omega(x_0,r)}} \le \frac{1}{t} \left(\int_{\Omega(x_0, r)} \abs{\bar A-A}\right) \,\norm{u}_{L^\infty(\Omega(x_0,r))},
\end{equation}
which implies that for any $p \in (0,1)$ we have
\begin{align*}
\int_{\Omega(x_0,r)} \abs{\tilde w}^{p}  &=\int_0^\tau+ \int_\tau^\infty  p t^{p-1} \Abs{\set{x \in \Omega(x_0,r): \abs{\tilde w(x)} >t}} \,dt \\
&\lesssim \abs{\Omega(x_0,r)}  \int_0^\tau  p t^{p-1}\,dt + \abs{\Omega(x_0,r)} \,\omega_A(r) \,\norm{u}_{L^\infty(\Omega(x_0,r))} \int_\tau^\infty  p t^{p-2}\,dt \\
&= \abs{\Omega(x_0,r)} \tau^{p} + \frac{p}{1-p}\, \abs{\Omega(x_0,r)} \,\omega_A(r)\,\norm{u}_{L^\infty(\Omega(x_0,r))} \tau^{p-1}.
\end{align*}
By taking  $\tau= \omega_A(r)\,\norm{u}_{L^\infty(\Omega(x_0,r))}$ in the above, we obtain
\[
\left(\fint_{\Omega(x_0,r)} \abs{\tilde w}^{p} \right)^{1/p} \lesssim  \omega_A(r)\,\norm{u}_{L^\infty(\Omega(x_0,r))},
\]
which substitutes \eqref{eq1019sat}.
For the sake of definiteness, we shall take $p=\frac12$ in the above and get
\begin{equation}					\label{eq1019}
\left(\fint_{\Omega(x_0,r)} \abs{\tilde w}^{\frac12} \right)^{2} \lesssim  \omega_A(r)\,\norm{u}_{L^\infty(\Omega(x_0,r))}.
\end{equation}
Also, instead of \eqref{eq13.46f}, we have the following:
First note that for any $\kappa \in (0,\frac14)$, we always have
\[
\left(\fint_{\Omega(x_0, \kappa r)} \abs{v - v_{x_0, \kappa r}}^{\frac12} \right)^2 \le 2\kappa r \norm{Dv}_{L^\infty(\Omega(x_0, \kappa r))} \le 2\kappa r \norm{Dv}_{L^\infty(\Omega(x_0,r/3))},
\]
for if $B(x_0,\kappa r/2)$ intersects $\partial\Omega$ and we have $v-v_{x_0, \kappa r}=v-0=v-v(\bar x)$ for some $\bar x \in \partial\Omega \cap B(x_0, \kappa r/2)$.

Next, in the case when $r \le 2\operatorname{dist}(x_0, \partial \Omega)$, the interior estimates for equations with constant coefficients yield
\[
\norm{Dv}_{L^\infty(\Omega(x_0,r/3))} \lesssim \frac{1}{r}\left( \fint_{\Omega(x_0, r/2)} \abs{v-c}^{\frac12} \right)^2 \lesssim \frac{1}{r} \left( \fint_{\Omega(x_0, r)} \abs{v-c}^{\frac12} \right)^2,\quad \forall c \in \mathbb{R},
\]
for $v-c$ satisfies $L_0^* (v-c)=0$ in $\Omega(x_0, r/2) = B(x_0, r/2)$ and $D(v-c)=Dv$.
In the case when $r > 2\operatorname{dist}(x_0, \partial \Omega)$, by the boundary estimate we have
\[
\norm{Dv}_{L^\infty(\Omega(x_0,r/3))} \lesssim \frac{1}{r} \left( \fint_{\Omega(x_0,r)} \abs{v}^{\frac12} \right)^2.
\]
Combining these together, we conclude that
\begin{equation}			\label{eq15.26m}
\left(\fint_{\Omega(x_0, \kappa r)} \abs{v - v_{x_0, \kappa r}}^{\frac12} \right)^2 \le C_0 \kappa \left( \fint_{\Omega(x_0, r)} \abs{v-c}^{\frac12} \right)^2,
\end{equation}
where $c=0$ when $r > 2\operatorname{dist}(x_0, \partial \Omega)$ and $c\in \mathbb{R}$ is arbitrary otherwise.

We recall the facts that for all $a, b \ge 0$, we have
\[
(a+b)^{\frac12} \le a^{\frac12} + b^{\frac12}, \quad (a+ b)^{2} \le 2(a^2+b^2),
\]
and
\[
\norm{f+g}_{L^{1/2}} \le 2 \left( \norm{f}_{L^{1/2}} + \norm{g}_{L^{1/2}}\right).
\]

By using the decomposition $u=v+w+\tilde w$, the above facts, and \eqref{eq15.26m}, we obtain
\begin{align*}
\left(\fint_{\Omega(x_0, \kappa r)} \abs{u - v_{x_0, \kappa r}}^{\frac12}\right)^2
& \le2 \left(\fint_{\Omega(x_0, \kappa r)} \abs{v - v_{x_0, \kappa r}}^{\frac12}\right)^2 + 2\left(\fint_{\Omega(x_0, \kappa r)} \abs{w +\tilde w}^{\frac12} \right)^2  \\
& \le2 C_0 \kappa \left(\fint_{\Omega(x_0, r)} \abs{v-c}^{\frac12}\right)^2 +2 \kappa^{-4} \left(\fint_{\Omega(x_0,r)} \abs{w +\tilde w}^{\frac12} \right)^2 \\
& \le  4C_0\kappa \left(\fint_{\Omega(x_0, r)} \abs{u-c}^{\frac12}\right)^2 + (4C_0 \kappa+2\kappa^{-4}) \left(\fint_{\Omega(x_0,r)} \abs{w +\tilde w}^{\frac12} \right)^2\\
& \le 4C_0\kappa \left(\fint_{\Omega(x_0, r)} \abs{u-c}^{\frac12}\right)^2 +(C_0+2\kappa^{-4})\,\omega_A(r)\,\norm{u}_{L^\infty(\Omega(x_0,r))}\\
&\qquad\qquad+ (C_0+2\kappa^{-4} )\fint_{\Omega(x_0, \kappa r)} \abs{w},
\end{align*}
where we used \eqref{eq1019} and H\"older's inequality in the last step.

Therefore, in place of \eqref{eq0748w}, if we set
\begin{equation*
\phi(x_0, r):=\left\{
\begin{aligned}
\inf_{c\in \mathbb{R}}\left(\fint_{\Omega(x_0,r)} \abs{u-c}^{\frac12} \right)^{2}\quad & \text{if } r \le 2\operatorname{dist}(x_0, \partial \Omega)\\
\left(\fint_{\Omega(x_0,r)} \abs{u}^{\frac12} \right)^{2}\quad & \text{if } r >2\operatorname{dist}(x_0, \partial \Omega),\\
\end{aligned}
\right.\end{equation*}
then we still obtain \eqref{eq16.57m} and \eqref{eq0750w}.
Moreover, if we set $q_{x_0,r} \in \mathbb{R}$ to be a number such that
\[
\phi(x_0,r)=\left(\fint_{\Omega(x_0,r)} \abs{u-q_{x_0,r}}^{\frac12} \right)^{2},
\]
then instead of \eqref{eq0756w}, \eqref{eq1402t}, and \eqref{eq13.13}, we have
\begin{gather*}
\abs{q_{x_0,\kappa^i r} - q_{x_0, r}} \le 4 \sum_{j=0}^i \phi(x_0, \kappa^j r),\\
\lim_{i\to \infty} q_{x_0,\kappa^i r}=u(x_0),\\
\abs{u(x_0)-q_{x_0,r}} \lesssim \phi(x_0, r)+
\norm{u}_{L^\infty(\Omega(x_0,r))} \int_0^r \frac{\omega_{\mathbf A}(t)}t \,dt+1,
\end{gather*}
respectively.
Also, by averaging the inequality
\[
\abs{q_{x_0, r}}^{\frac12} \le \abs{u(x) -q_{x_0, r}}^{\frac12} + \abs{u(x)}^{\frac12}
\]
over $x \in \Omega(x_0, r)$, taking the square, and using H\"older's inequality, we get
\[
\abs{q_{x_0, r}} \le 2 \phi(x_0, r) + 2 \left(\fint_{\Omega(x_0,r)} \abs{u}^{\frac12}\right)^2 \lesssim \fint_{\Omega(x_0,r)} \abs{u}.
\]
By combining these inequalities, we have \eqref{eq14.39t}.
The rest of proof is the same.

\section{Appendix}
We sketch the proof of the $W^{2,p}$-solvability of \eqref{eq10.36} in a bounded $C^{1,\alpha}$ domain $\Omega\subset \mathbb{R}^n$ with $\alpha>1-1/\max\{n,p\}$, which enable us to relax the $C^{1,1}$ condition of $\Omega$ in Theorems \ref{thm1} to $C^{1,\alpha}$, where $\alpha>1/2$.
Recall the regularized distance function $\psi$ on $\Omega$ introduced in \cite{Lie85} is such that $\psi(x)$ is comparable to $\text{dist}\,(x,\partial\Omega)$ near the boundary and $\psi\in C^{1,\alpha}(\overline{\Omega})\cap C^\infty(\Omega)$. Without loss of generality, we assume $0\in \partial\Omega$ and the $x_n$-direction is the normal direction at $0$. Taking a small constant $r>0$, we flatten the boundary $\partial\Omega$ near $0$ by making the change of variables
\begin{equation*}
x\in \Omega_r:=\Omega(0,r) \rightarrow y\in \mathbb{R}^n_{+},\quad y_i(x) = x_i,\,\,i=1,\ldots,n-1,\quad y_n(x)=\psi(x).
\end{equation*}
In the $y$-variables, the equation becomes
\begin{equation*}
 \tilde a_{kl} D_{y_k y_l}u+ \tilde b D_{y_n}  u = f \quad \text{in}\,\,\psi(\Omega_r)\subset\mathbb{R}^n_{+}
\end{equation*}
with the Dirichlet boundary condition $u=0$ on $\{y_n=0\}$,
where
\[
\tilde a_{kl}=a_{ij}D_{x_i}{y_k} D_{x_j}{y_l},\quad \tilde b=a_{ij}D_{x_i x_j}{\psi}.
\]
Since $\psi\in C^{1,\alpha}(\overline{\Omega})$, $\tilde a_{kl}$ is uniformly continuous in $\psi(\Omega_r)$.
It follows from \cite[Lemma 2.4]{Safonov} that $|b(y)|\le Cy_n^{1-\alpha}$.
In particular, $b\in L^{\max\{n,p\}+\varepsilon}(\psi(\Omega_r))$ for some $\epsilon>0$ provided that $\alpha>1-1/\max\{n,p\}$.
By using the boundary $W^{2,p}$-estimate for elliptic equation,
Gerhardt's inequalities (cf. \cite{G79} or \cite[Lemma 1 (ii)]{K09}), and a standard iteration argument, we conclude that for any $W^{2,p}$-solution $u$,
\[
\|u\|_{W^{2,p}(\Omega_{r/2})}\le C\|f\|_{L^p(\Omega_r)}+C\|u\|_{L^p(\Omega_r)}.
\]
It then follows from a partition of unity argument and the corresponding interior estimate that
\[
\|u\|_{W^{2,p}(\Omega)}\le C\|f\|_{L^p(\Omega)}+C\|u\|_{L^p(\Omega)}.
\]
Now by the proof of \cite[Lemma 9.17]{GT98}, using a compactness argument and the Alexandrov maximum principle, we have
\[
\|u\|_{W^{2,p}(\Omega)}\le C\|f\|_{L^p(\Omega)},
\]
which also gives the uniqueness of solutions.
Finally, the existence of solutions follows from an approximation and bootstrap argument. See the proof of \cite[Theorem 9.15]{GT98}.

\section*{Acknowledgement}
The authors would like to thank Zongyuan Li for pointing out the $W^{2,p}$ solvability in Appendix.
We also thank the referee for careful reading of the manuscript and asking to clarify the dependence of various estimates on $\Omega$.

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Jeff is often asked to be a celebrant for funeral and memorial services. Each year he leads over 100 services that capture the vision a family has for saying goodbye to a loved one.
Whether its a service at a graveside, in a funeral home, a private residence, a hall or church, we have the needed experience in honouring the deceased and offering an element of peace for family and friends.


INTERNAL COMMANDS:- There are also called memory-resident commands. These commands are automatically loaded into the computer's memory during the booting process. They actually included in the Command.com file. So these commands are executable immediately after getting the dos prompt.
(i) Date: This command is used to view the syste date and allows you to change the system date. This date format is MM-DD-YY. The date that you enter will be treated as the current date, untill you enter a new date.
(ii) Time: This command is used view and it allows you to change the time.
(iii)CLS: This command is used to clear the screen. All the information on the screen will be cleared and your cursor will be placed on the top of left corner of your screen.
(iv) Ver (Version): This command is used to display the windows version and dos version.
(v) Vol (Volume): This command is used to view the label name of drive and serial number.


text:
Frank and Lola Review
Date: February 25, 2017Author: Jack Blackwell 0 Comments
With Michael Shannon currently on the Best Supporting Actor ballot at the Oscars for his sterling work in Nocturnal Animals, there could be a far worse week for Frank and Lola, a small festival and VOD release vehicle for him, to come out. And of course, this debut film from writer-director Matthew Ross benefits enormously from Shannon's presence, but doesn't have the staying power to match up to the best of the great character actor's recent output. Often interesting, but also occasionally frustrating and weirdly paced, it signals an exciting future for Ross, and is further proof, if any were needed, of the immense capability of its leading man.
Shannon plays the Frank of the title, a deeply flawed but compelling and charismatic chef in a tempestuous relationship with recent graduate and fashion designer Lola (Imogen Poots). Frank and Lola initially faces an uphill climb to make anyone care about another story of an age-inappropriate relationship, but Ross quickly sets aside these qualms with a twisty series of character reveals in the first 15 or so minutes, establishing that Lola is far from a typical ingénue cliché. Instead, she's just as possessive as Frank in their dysfunctional coupling, though naturally the far less intimidating of the two.
As the pair uncover a string of lies that each has been telling the other, they try and hold their relationship together, even as the film's genre switches from romantic drama to something altogether darker and more mysterious. Strangely, whilst Frank and Lola make for an eminently watchable and engaging double act, the actual stakes of them staying together are very low. You never find yourself particularly caring if they break up or not, and that feels entirely deliberate, and Ross' mid-film genre switch makes this flaw far less glaring than it might appear on paper.
After learning of Lola's history with an older European man, Andy Larsson (Michael Nyqvist), Frank flies from his Vegas home to Paris , with a vague plan of badly hurting Andy. Unsurprisingly, this turns out to be a bad idea, but their encounters don't play out in the expected manner, instead revealing deeper secrets about Lola's past and Frank's psyche. For the first two thirds of the film, these continuous discoveries keep you guessing and glued to the story, even as sideshows like the opening of Frank's new restaurant and a visit to a weird and gratuitous French sex club do their best to distract you.
Unfortunately, in the final act, Ross insists on giving us a clumsy and, frankly, unnecessary pop-psychologising of his characters' behaviour. This big moment rings false, partly due to suddenly clunky writing, but also down to the fact that the monologue on which it all hinges falls to Poots, who can't quite shoulder the weight. It weakens that which has come before, and the plot gets far more rote as we approach the finale, and I found myself steadily losing patience with it all in the last 10/15 minutes.
Despite crumbling material, Shannon is just as great in this latter act as he has been all film, anchoring everything with his unique, room-filling intensity, the intensity that makes the young and very beautiful Poots' attraction to him plausible enough to keep everything ticking along. Frank and Lola packs a lot in (with both plot and character work) to its slim runtime, and given that two thirds of that is original and exciting, it marks an intriguing debut for Ross, who seems bound for a bona fide great film in his future.
Written and Directed by Matthew Ross
Starring; Michael Shannon, Imogen Poots, Michael Nyqvist
Runtime: 88 mins
Frank and LolaImogen PootsMatthew RossMichael NyqvistMichael Shannon
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text:
En Wikipedia, fatiga puede referirse a:
 El síndrome de fatiga crónica o encefalomielitis miálgica, enfermedad neurológica grave, caracterizada por una fatiga fuerte
 La fatiga respiratoria o disnea, dificultad para respirar.
 La fatiga de materiales, en ingeniería y en ciencia de los materiales, disminución de la resistencia mecánica de los materiales al someterlos a esfuerzos repetidos.
 La fatiga (cansancio), falta de fuerzas después de realizar un trabajo físico, intelectual o emocional, o por la falta de descanso.

Enlaces externos

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← Time flies when you are having fun… and are hard at work!
I solicited questions from Symantec colleagues. What do you want to know (professional and personal, work related or cultural)? So here are a few of your questions. If you have any more for me or the team, let us know. We love hearing from you. Thanks!
It has been in the 70s (21* C) all week with a little bit of drizzle yesterday.
How is your Spanish now that you are in Peru?
Mas o menos. Comprendo mas de que yo hablo. Necesito practicar más.
How do you communicate? Do you have to speak Spanish, or is English fine?
It depends. English during work meetings; each group has an assigned interpreter. Outside of work I try to speak Spanish. Most locals don't speak English—so it is helpful to know key phrases.
What is your daily routine like? What are some key differences from living here [USA]?
Wake up to a delicious buffet. Our taxi comes at 8:30am and drops us back at 5pm. All of the teams get back around 5/6pm. Teams will then debrief further or use the free time to run errands or work out. Around 7:30pm we head to dinner which takes at least two hours. By 10pm most people are heading to bed, skyping with loved ones, or working.
(Left: a sample of my breakfast tray #1 (always room for seconds): orange juice, quinoa w/ yogurt, mangoes, papayas, bananas, avocados, and eggs. Right: Errand run in the square leads to taking photos with a girl and her lamb, Rosalinda).
A few differences: time (Peruvian time is more casual (+/- 15 minutes) versus the 'Corporate America' on the hour), traffic (no one pays attention to stop signs—and cars have the right of way), and the flow of water (yes, it is counter-clockwise here).
I am learning so much, I don't know where to begin…I am working with an all male team for the first time. I am learning how to consult in a foreign country. My team has been really supportive and we have open communication, but I do find it challenging to speak up when I'm unsure or don't want to speak over someone. I am forcing myself to lean in and sit at the table, presenting my ideas and sharing my expertise (HR and OD). By taking the risk and stepping out of my comfort zone, I am growing so much more. Some ideas work and some don't, but if I don't take the risk, I don't glean as much. So far my knowledge has been relevant and helpful—To summarize, I'm working on my confidence as a leader.
Have you eaten guinea pig yet? And, on the same note, how's the food? Favorites?
No guinea pig yet. I'm still debating…I have tried Alpaca, Adobo, Rocoto Relleno, and Lomo Saltado (local cuisines). Food is awesome. Huge portions. My favorites are the fruit juices and desserts.
(Left: Alpaca, beef, duck, potatoes, and ratatouille. Right: Assortment of fresh juices).
What was the one thing that made you most uncomfortable (mentally or physically) and how did you deal with it?
Physically: The mosquitoes are eating me alive! I'm now diligently putting on repellent.
Mentally: I always feel a little homesick when I go to a new place, but technology helps. I email and video chat with friends and family. Also, we have such a great team—there's no time to miss home!
What is one thing you did that made the most impact on those around you?
What's one of the biggest surprises been?
A few nights ago we were all sitting for dinner and Steve Bennett (our CEO) called. I was surprised he knew our local numbers. Someone handed me the phone, and I answered with an 'hola.' The restaurant was loud so I proceeded to get up and find a quiet spot when my colleague told me it was a joke. There's a 'fake call' feature on the phones we were given. Who knew? We all had a good laugh.
I tried a yoga studio around the block. That was fun (basic Vinyasas with the mix of Spanish and English instruction). I generally go for a walk after work and aspire to run at this altitude.
Last Saturday a group of us went on a long hike in Cachamarca. We hiked a total of 8.8KM and climbed 1,000 feet for a peak altitude of 3,200M. Hope you all enjoy converting those numbers (joys of a multi-cultural team).
What are the things on your Peru bucket list that you would like to experience before you leave?
3) Continue to step outside of my comfort zone to better myself.
Yeah, metric system all the way! Nice one Allyson. Loving the Q&A idea btw.


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/K I L L K I L L K I L L — Obscura Digital: Obscura Digital Shines a New Light on the Fine Art of Architectural Projection/
By Obscura Digital, PRNE
- With Photo
- Illuminations on the Sheikh Zayed Grand Mosque and the historic Al Jahili Fort in Abu Dhabi mark the 40th Anniversary of the United Arab Emirates National Day
San Francisco's Obscura Digital illuminated the Sheikh Zayed Grand Mosque, the heart of the United Arab Emirates, with a mesmerizing show featuring a series of original designs and visual representations illustrating the unique artistic styles of the mosque. They simultaneously brought the Al Jahili Fort to life with a symbolic journey of the UAE using historical imagery.
(Logo: photos.prnewswire.com/prnh/20110316/SF62472LOGO-b)
(Photo: photos.prnewswire.com/prnh/20111202/SF16641)
Photos of the event are available at www.obscuradigital.com/press
"It is an honor for Obscura Digital to work on one of the most important celebrations in the country, and a privilege to be the first to project a show on the extraordinary architecture of the Grand Mosque," stated Travis Threlkel, Co-Founder of Obscura Digital. "This monumental event was an opportunity to create a large-scale work of fine art and set a new standard for architectural projections."
Obscura Digital's team was the first to generate a laser scan of the surface area of the Grand Mosque and to create a 3-D model mapping the hand-carved, intricate details of the complex geometry of the mosque. This included the facade, 4 minarets, and 12 domes, for a projection area that is 600ft (182m) wide by 351ft (107m) high.
Sheikh Zayed's original vision for the Grand Mosque was a centre of cultural collaboration promoting mutual understanding between communities from all around the world. Scenes found throughout the interior of the mosque demonstrate Sheikh Zayed's passion for the environment, depicting gardens of paradise, date palms, and pentagonal geometry created by an array of international artists. Interpretations of these artistic elements were woven into the show projected by Obscura Digital onto the exterior of the Grand Mosque.
In addition to the Grand Mosque, Obscura Digital illuminated the Al Jahili Fort in the oasis city of Al Ain, with a visual tribute to the early history of the UAE, including scenes of Sheikh Zayed's life and the ruling leaders of the UAE, and a montage of the pearl trade, Bedouin life, agriculture, and the natural elements that had a significant impact on the progression of the Nation.
A picture accompanying this release is available through the PA Photowire. It can be viewed at www.mediapoint.press.net or www.prnewswire.co.uk
Mia Hanak, press at obscuradigital.com, +1-415-637-0525
Filed under: Art, Banking and Financial Services, Business, Computer / Electronics, Entertainment, New Products / Services, Technology
Tags: Middle East, Obscura Digital
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Bill Clinton, Jane Goodall, Sha Zukang, Chief Surui and Others Talk About Crucial Importance of Environmental Information
Reduced Rent-A-Car Rates From Hertz Drive Travel to Dubai and Abu Dhabi
TRA Represented by aeCERT Highlight Importance of Cybercrime With International Speakers at Black Hat Abu Dhabi


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Contract Buying Robbed Black Families In Chicago Of Billions | WBEZ
"The takeaway is that we have a history that so many Chicagoans are really not aware of that has really shaped the city and shaped the racial politics of the city. It shaped the economy of the city. In order to move forward and address issues that confront us in terms of poverty and racial discrimination, we have to have a common understanding of what happened in the past," said Duke University's Bruce Orenstein, the study's project director who is doing a documentary series on Chicago's housing segregation.That past has roots 100 years ago with white people not understanding that they created black ghettos, he said."
Source: Contract Buying Robbed Black Families In Chicago Of Billions | WBEZ
Posted in American Politics, Banking and Financial Policies, Black Business. Commerce and Entrepreneurship, Black Debt, Black economics, Black Foreclosure, Black Land Wealth, Black poverty, Black Wealth, Blacks and the Credit Industry, Congress and Public Policy, Disparities in Law and Justice, Domestic Issues, Economic Policy and the American Economy, Government Policies, Labor and Work, Labor Organiizing and Politics, Law and Justice, National Issue, Race Politics, Reparations, SCOTUS.	Tagged Chicago Black History, Chicago poverty, Chicago Race Riots, poverty politics, racial economic inequality, racial wealth gap
"A Film Called Blacks Can't Swim"
"A Film Called Blacks Can't Swim" continues with the new music video to the soundtrack of the same name as part of a national project to encourage Black people to learn how and enjoy swimming.
With the aim of encouraging as many people in the community to swim by addressing the stereotypes and dispelling the myths, the project highlights Black competitive swimmers and some of their achievements in this music video. Why is this important ? It is a safety consideration for all children. Ensure that your children learn to swim.
We need Swimming Role Models to highlight the importance of swimming in our community. Hopefully the 45+ Black competitive swimmers featured in this extended music video will do just that.
In 2015, three African American swimmers, Simone Manuel, LIA NEAL & Natalie Hinds made history by taking the 3 top places (coming respectively 1-2-3) in the 100-yard freestyle at NCAA championships.
If you don't swim why? If you never learned, why? Many Black children during Jim Crow all through the South had no access to either a pool, beach, lake or river for recreational swimming.
OUR COMMON GROUND with Janice Graham is proud to be part of this project.
The national project is led by Ed Accura @ed_accura. Contact him if you would like to get involved in your community.
Posted in Black and Race Blogs, Black Children and Youth, Black Culture, Black Family and Relationships, Black History, Children, Children, Education and Parenting, Films and Documentaries, History and Culture, OUR COMMON GROUND NOTES.	Tagged Black in sports, Blacks in swimming, Ed Accura
We Have the Means to Fund Reparations. Where Is the Political Will?
. . . Between 1983 and 2016, the median net worth for Black Americans actually went down by 50 percent. Paired with a growing Latinx population that also lags far behind whites in household wealth, the U.S.'s overall median wealth trended downward over those decades, even as median white wealth increased.These trends go hand-in-hand with the rigging of the overall economy. Over the last 30 years, the wealthiest 20 percent of households have captured almost 97.4 percent of all increases in wealth, leaving only scraps for the rest.To repair these breaches — between Black and white, as well as between the rich and the rest — we must restore the wealth of communities that were literally used as a foundation of the nation's wealth, while being prohibited from building their own.
As Ta-Nehisi Coates wrote in his groundbreaking case for reparations in The Atlantic, reparations are "the price we must pay to see ourselves squarely."It won't be an easy task. But it's by no means insurmountable."
Source: We Have the Means to Fund Reparations. Where Is the Political Will?
Posted in Black economics, Black Land Wealth, Black Wealth, Critical Black Issues, Discussions, Information and Opinions: Reparations, Economic Policy and the American Economy, Government Policies, Law and Justice, Race, Reparations, Slavery history, US Justice System.	Tagged Black economic analysis, Black economic inequality, Black wealth divide, Reparations African Americans, reparations for slavery
How the South Won the Civil War | The New Yorker
During Reconstruction, true citizenship finally seemed in reach for black Americans. Then their dreams were dismantled.
Black political power during Reconstruction was short-lived—eclipsed, in significant part, by a campaign of terror.
Illustration by Cristiana Couceiro. Photographs: Hirarchivum Press / Alamy (Ku Klux Klan); Smith Collection / Gado / Getty (building); Universal History Archive / Getty (flags); Everett / Alamy (gallows)
"Not so long ago, the Civil War was taken to be this country's central moral drama. Now we think that the aftermath—the confrontation not of blue and gray but of white and black, and the reimposition of apartheid through terror—is what has left the deepest mark on American history. Instead of arguing about whether the war could have turned out any other way, we argue about whether the postwar could have turned out any other way. Was there ever a fighting chance for full black citizenship, equality before the law, agrarian reform? Or did the combination of hostility and indifference among white Americans make the disaster inevitable? . .
The broad outlines of the Reconstruction story have long been familiar, though the particular interpretive pressures put on particular moments have changed with every era. Toward the end of the war, Washington politicians debated what to do with the millions of newly freed black slaves. Lincoln, after foolishly toying with recolonization schemes, had settled on black suffrage, at least for black soldiers who had fought in the war. (It was a speech of Lincoln's to this effect that sealed his assassination: John Wilkes Booth, hearing it, said, "That means nigger citizenship. Now, by God, I'll put him through.")
After Lincoln's death, his hapless and ill-chosen Vice-President, Andrew Johnson, did as much as he could to slow the process of black emancipation in the South, while the "radical" core of the abolitionist Republicans in Congress tried to advance it, and, for a while, succeeded. Long dismissed as destructive fanatics, they now seem to be voices of simple human decency. Thaddeus Stevens, the abolitionist congressman from Pennsylvania, proposed shortly after the war's end, in his "Lancaster" speech, a simple policy: punish the rebel leaders; treat the secessionist states as territories to be supervised by Congress, thus protecting the new black citizens; take the confiscated plantations on which masters had worked slaves like animals, and break up those plantations into forty-acre lots for the ex-slaves to own (a form of the classic "forty acres and a mule"). That this minimally equitable plan was long regarded as "radical" says something about how bent toward injustice the conversation quickly became.
Freed slaves eagerly participated in the first elections after the war, and distinguished black leaders went to Congress. The 1872 lithograph of "The First Colored Senator and Representatives," by Currier & Ives, no less, shows seven black men given the full weight of mid-century Seriousness, including the first black senator from Mississippi, Hiram Rhodes Revels.
But white state governments steadily reconstituted themselves. By the eighteen-nineties, they were passing laws that, piece by piece, reclaimed the right to vote for whites alone. All of this was made worse by one of those essentially theological "constitutional" points which American professors and politicians love to belabor. Lincoln's argument was always that, since it was unconstitutional for states to secede on their own, the rebel states had never seceded. The rebels were not an enemy nation; they were just a mob with a flag waiting to be policed, and the Union Army was the policeman. The idea was to limit any well-meaning attempt at negotiation, and to discourage foreign powers from treating the Confederacy as a separate state. After the war, though, this same idea implied that, since the state governments had never gone out of existence, their reborn legislatures could instantly reclaim all the rights enjoyed by states, including deciding who could vote and when."
Source: How the South Won the Civil War | The New Yorker
Posted in American Politics, History and Culture, Race Politics, Slavery history.	Tagged Abraham Lincoln, Andrew Johnson, Black voter history, Emancipation, Post Civil War legislation, Post reconstruction in the US, Reconstruction, Reconstruction and slavery, states rights
Research by Black Female Professor Reveals Startling Truth That White Women Made Up 40% of Slaveowners
A set of data uncovered by University of California-Berkeley professor reveals southern white women played a heavier role in the enslavement of Africans than previously thought.
"For them, slavery was their freedom," Jones-Rogers states in her book.After Martha Washington married President George Washington in Virginia in 1759, George is said to have possibly owned around 18 people. But his wife, one of the richest women in the state, owned 84 and dramatically increased the local slave population.Arguing that white women are trained to be engaged in the slavery industry at a young age, Jones-Rogers stated, "their exposure to the slave market is not something that begins in adulthood—it begins in their homes when they're little girls, sometimes infants, when they're given enslaved people as gifts."
Stephanie E. Jones-Rogers,
an associate professor of history at the university, combed through data from the 1850 and 1860 census and revealed that white women made up around 40% of slaveowners.
Source: Research by Black Female Professor Reveals Startling Truth That White Women Made Up 40% of Slaveowners
Posted in Abolitionist Radio, Black and Race Blogs, Black Culture, Black Women, Race Politics, Reparations, Slavery history, Women Issues.	Tagged American slavery research, Black women and abolition, Stephanie Jones-Rogers, white women, Women and the KKK, Women and white nationalsim, women slaveholders
Did Black People Own Slaves?
"How Many Slaves Did Blacks Own?
So what do the actual numbers of black slave owners and their slaves tell us? In 1830, the year most carefully studied by Carter G. Woodson, about 13.7 percent (319,599) of the black population was free. Of these, 3,776 free Negroes owned 12,907 slaves, out of a total of 2,009,043 slaves owned in the entire United States, so the numbers of slaves owned by black people over all was quite small by comparison with the number owned by white people. In his essay, " 'The Known World' of Free Black Slaveholders," Thomas J. Pressly, using Woodson's statistics, calculated that 54 (or about 1 percent) of these black slave owners in 1830 owned between 20 and 84 slaves; 172 (about 4 percent) owned between 10 to 19 slaves; and 3,550 (about 94 percent) each owned between 1 and 9 slaves. Crucially, 42 percent owned just one slave.
Pressly also shows that the percentage of free black slave owners as the total number of free black heads of families was quite high in several states, namely 43 percent in South Carolina, 40 percent in Louisiana, 26 percent in Mississippi, 25 percent in Alabama and 20 percent in Georgia. So why did these free black people own these slaves?"
Source: Did Black People Own Slaves?
Posted in Black and Race Blogs, Black economics, Black History, Black Land Wealth, Black Scholarship, Discussion, History and Culture, Race Politics, Slavery history.	Tagged Carter G. Woodson, CIVIL WAR (US) (1861-65), emancipation proclamation, Freedman ownership of slaves, history of slavery, post-Emancipation, slave ownership in the US, slave ownership in US
The 'Clotilda,' the Last Known Slave Ship to Arrive in the U.S., Is Found | At the Smithsonian | Smithsonian
A cast iron bust of Cudjo Lewis, one of the last survivors of the slave ship Clotilde, can be found in front of the historic Union Missionary Baptist Church in Africatown. (Graveyardwalker (Amy Walker) Wikimedia Commons )
"The excitement and joy is overwhelming," says Woods, in a voice trembling with emotion. She is 70 years old now. But she's been hearing stories about her family history and the ship that tore them from their homeland since she was a child in Africatown, a small community just north of Mobile founded by the Clotilda's survivors after the Civil War.The authentication and confirmation of the Clotilda was led by the Alabama Historical Commission and SEARCH Inc., a group of maritime archaeologists and divers who specialize in historic shipwrecks. Last year, the Smithsonian's National Museum of African American History and Culture's Slave Wrecks Project (SWP) joined the effort to help involve the community of Africatown in the preservation of the history, explains Smithsonian curator and SWP co-director Paul Gardullo."
Source: The 'Clotilda,' the Last Known Slave Ship to Arrive in the U.S., Is Found | At the Smithsonian | Smithsonian
"After the Emancipation Proclamation, the newly freed Africans tried, but failed to return to their beloved homeland Africa. The story describes the group reuniting from various plantations, alongside American-born, formerly enslaved men, women and children. The Africans bought land and founded their own settlement, which came to be known as Africatown."
The Full Story:
Africatown USA Trailer from Roslyn Williams on Vimeo.
Posted in #Blacklivesmatter, Black Culture, Black History, History and Culture, Reparations, Slavery history.	Tagged african american history, Africatown, AL, American slavery, Capitalism and slavery, Clotilda slave ship, Cujo Lewis, slave trade, Transatlantic Slave Trade


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Tax Tip: Make Sure You Use Your Business Tax ID Number with Clients & PayPal
George attains Chartered Global Management Accountant (CGMA) Credential
George V Famiglio, Jr CPA has attained the CGMA designation.
What is the CGMA designation?
George V. Famiglio Jr, CPA
George V. Famiglio Jr<
CPA<, PFS<, CFP<, CFS<, CGMA, Master's in Tax,
Admitted to Practice: U.S. Tax Court<
George's Bio:<
George V. Famiglio, Jr. is the principle of our firm. He began the practice in 1971 in Philadelphia, Pennsylvania and opened a new office in Sarasota, Florida< in 1982. Prior to starting Famiglio & Associates, George worked for George V. Famiglio, Sr., Attorney at Law, Swarthmore School of Music as a advanced guitar teacher, and George V. Famiglio Jr. Real Estate, where he was involved in commercial real estate ventures.
George received a Bachelors of Science in Psychology from St. Joseph's College and a Master's of Taxation from Villanova Law School<. He completed his business law, accounting, and financial education as a MBA candidate at Drexel University<. Additionally he has attained several Master's certificates from Berklee.
George is a Certified Public Accountant, a Personal Financial Specialist, a Certified Financial Planner, a Registered Investment Advisor Agent, and a Certified Fund Specialist. He holds the Series 6, Series 22, Series 26, Series 8, and Series 72, as well as is Insurance Licensed. Additionally, George has been an H.D. Vest Representative since 1987.
George has been admitted to practice before the United States Tax Court, a privilege only a few hundred CPAs in the country have received—less than 0.1%.
He sits on, or advises, numerous boards throughout the community, and is involved with several trusts and foundations.
Additionally he is a member of American Institute of Certified Public Accountants (AICPA) & Florida Institute of Certified Public Accountants (FICPA).


In 2013, the German Marshall Fund of the United States and the German Institute for International and Security Affairs invited me to participate in the project "New Power – New Responsibility". Over the course of one year, representatives from politics, academia, journalism, and civil society discussed the central foreign and security policy challenges that Germany would face in the coming years. My working group focused on Germany's strategic relationships and divided the nations of the world into allies, challengers, and spoilers. Among others, we categorized the US, the Member States of the European Union (EU), and Turkey as primary allies. Today, just five years later, that view of the world is already obsolete.
Authoritarian and illiberal governments have learned how to replace hard repression such as political imprisonment with other, softer forms of repression.
We did not foresee that the US would no longer be a reliable ally, or that illiberal governments would emerge within the EU – not to mention Brexit. We did not discuss that China and Russia could be able to destabilise established democracies through authoritarian influence operations. And in Turkey, President Recep Tayyip Erdogan's autocratic grip on power extends far beyond what we imagined.
What lessons have I learned from predicting the future so badly? First: our deliberations on 'Germany and the World in 2030' should not rely too heavily on our present observations. A future scenario that extends the status quo in a linear trend will generally be incorrect. And second: the world of all possible futures is much bigger than the world of the probable futures.
In my own field, human rights politics, current discussions revolve around an increasing or, more precisely, a changing pressure on civil society. Authoritarian and illiberal governments have learned how to replace hard repression such as political imprisonment with other, softer forms of repression, for example new bureaucratic or taxation hurdles that target critical non-governmental organisations (NGOs). Such softer forms of control minimise the external political risk of domestic repression. It is also noteworthy that counter-discourses to human rights are growing stronger. Even in democracies we can observe how human rights achievements, such as the right to asylum or freedom of the press, are being called into question.
If we assume that both the concept of human rights as well as human rights activists will continue to face pressure around the world and that supporters of illiberal political ideas will be on the rise in democracies, if we take authoritarian influence seriously, think through the possibilities of digital surveillance, and imagine the continued progress of all of these observations in linear trends up until 2030, we will quickly arrive at the conclusion that the dystopia of a global dictatorship awaits us. I consider this dystopia to be unlikely because there are people all around the world who stand up for human rights.
But a global dictatorship is certainly possible. One step that can be taken to prevent such a horror scenario is to establish a new international alliance of democracies. Germany could initiate such an alliance, which must involve democracies from the Global South. Democrats can only win the global battle against dictatorships and illiberal politics if they stick together.
Perhaps the most important task lies in developing new approaches to promoting human rights. This is where German institutions must step in. They need to think beyond familiar funding programmes and must demonstrate greater flexibility in allocating resources. The juridification and professionalisation of human rights work embodied by NGOs have served their time. Today, we need a more political approach to promoting human rights. Despite pushback: the shape of the future remains open.
Dr. Katrin Kinzelbach (41) is Associate Director at the Global Public Policy Institute, a Berlin-based think tank. Additionally, she teaches at the Central European University in Budapest as a Visiting Professor. She specialises in international human rights politics and conducts research on political imprisonment with a Schumpeter Fellowship from the Volkswagen Foundation.


Oyster reefs provide a suite of valuable ecosystem services, such as water filtration, nitrogen sequestration, and provision of habitat and foraging grounds. The global decline of these habitats has had negative economic and ecological impacts to coastal waters worldwide. In the Chesapeake Bay, < 1% of the historic oyster population remains and efforts to restore oyster populations and the services they provide have been increasing. Building reefs that successfully provide specific ecosystem services may require different techniques then previously used, and success may depend on reef morphology, location, and environmental conditions. Settling trays were embedded into previously restored oyster reefs that varied in their structural complexity (rugosity) in multiple rivers in the lower Chesapeake Bay. Trays were collected after 7-weeks, sorted, and species identified and weighed (ash-free dry weight) to obtain species diversity, abundance, and biomass. Species composition data was analyzed using nMDS plots, which showed that salinity was an important driver of differences in species composition. Results of an ANOVA analysis found that species diversity was significantly greater on reefs in the high-salinity rivers compared to reefs in low-salinity rivers. Total organism abundance and biomass were positively correlated with reef structural complexity measures, such as rugosity, oyster clump volume, and oyster biomass. These results suggest that more complex oyster reefs in higher salinity locations may support more diverse and productive benthic communities. This study provides insight into the driving factors that structure oyster reef communities and has important implications for oyster reef restoration design and management.
Karp, Melissa and Seitz, Rochelle. "Structural complexity and location affect the habitat value of restored oyster reefs". 10-9-2015. VIMS 75th Anniversary Alumni Research Symposium.


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Home | Politics
Tags: usa | russia | putin | biden | trump | nordstream
Ex-VA Secretary Wilkie to Newsmax: Pipeline Attack a 'Putin Escalation'
(Newsmax/"American Agenda")
By Jay Clemons | Thursday, 29 September 2022 05:58 PM EDT
Robert Wilkie, the former Secretary of Veteran Affairs during the Trump administration, doesn't believe the United States had an integral role in the supposed attack on two natural gas pipelines under the Baltic Sea — running from Russia to Germany.
At the same time, Wilkie asserts the White House should exhibit transparency on this matter, given how it's only been seven months since President Joe Biden publicly stated, "There will no longer be a Nord Stream 2 [pipeline]. We will bring an end to it."
Regarding Biden's response from February, "What you ... saw is more confusion and incompetence from this administration," Wilkie told Newsmax Thursday afternoon, while appearing on "American Agenda" with hosts Bob Sellers and Katrina Szish.
From Wilkie's perspective, the Biden administration essentially turned its back on an earlier commitment to have the Nord Stream 2 pipeline — a 1,200 kilometer-long offshore natural gas pipeline that was completed in September 2021 — being fully operational, and then suddenly wanting it shut down.
Also, Wilkie explains the list of potential saboteurs is a short one, since only the Swedish, British, American and Russian naval units have the capacity to desecrate a pair of natural gas pipelines buried deep in the sea.
"This is a [Russian President Vladimir] Putin escalation," says Wilkie, who, based on the initial evidence, concludes that Russia had something to do with the pipeline attack.
But now, Wilkie reasons this alleged sabotage, coupled with the ongoing Russia-Ukraine war, has brought Sweden and Finland — two nations that recently joined the North Atlantic Treaty Organization (NATO) — even closer to its Western allies.
"And those are two powerful militaries," says Wilkie.
As Wilkie sees it, all these events tie into Russia's full-scale invasion of Ukraine, which has been ongoing for seven months.
"The Russians are trying to get the European nations to pull out of the Ukraine," says Wilkie, alluding to the various NATO countries giving full support to the Ukrainian troops' defense initiative.
And that includes the United Kingdom, with Wilkie saying the British have been "providing minute-by-minute updates on Russian troop movements" since February.
It's also a economic power move on Putin's part, explains Wilkie, since the Russians have long been threatening an energy holdout to Western European nations, in terms of no longer exporting oil and natural gas to these energy-needy countries.
"This is Russia's way of telling [Europeans], 'You're going to be cold for this winter, and the next winter,'" says Wilkie, who also served under the administration of President George W. Bush as assistant secretary of defense for legislative affairs.
When asked how the Trump administration might have handled these events in Europe, Wilkie maintains that everything would be different, if former President Donald Trump was exerting negotiating leverage with Russia.
"You could bring Russia's economy to its knees overnight" by holding diplomatic talks with China and other countries, discouraging Putin's actions over the last few months, offers Wilkie.
Instead, Wilkie says the Biden administration does nothing but make empty promises that ultimately go nowhere on a worldwide stage.
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Trump Administration Official to Newsmax: 'No Doubt' Putin Sabotaged Pipeline
Ex-VA Secretary Wilkie to Newsmax TV: Don't Bring Back Old Guard
Robert Wilkie, the former Secretary of Veteran Affairs during the Trump administration, doesn't believe the United States had an integral role in the supposed attack on two natural gas pipelines under the Baltic Sea — running from Russia to Germany. At the same time, Wilkie ...
usa, russia, putin, biden, trump, nordstream
Thursday, 29 September 2022 05:58 PM


The homeowners featured in this article absolutely loved the charm of their modest, cottage- style home. However, their energetic four- year-old was quickly overrunning the home with toys, DVDs, books, games, and more. What began as a project to design more play-live space in the family room quickly ballooned to include more overall entertaining space for kids and adults, alike. The homeowners also wanted to consider the feasibility of creating a full guest suite, should an aging family member need to move in at some point in the future.
Due to local building ordinances, their square footage was limited to 600. Therefore, any new space needed to be flexible with opportunities for storage wherever possible. An arched TV niche was a visual anchor point to the space and designed with a storage closet behind. Nook areas played double duty, such as a crafting desk turned wet bar, and an eating nook turned game table. The general play space was designed with a sofa bed in place, while a full three-piece bathroom could elegantly accommodate overnight guests and neighborhood play dates.
While the client requested conveniences of modern technology, the space had to also retain elements of charm, and did so with vaulted ceilings that mimicked the main house. In addition, trim work around doors and windows provided the wood details that the clients loved. Strategically placed windows showcased views of the entire yard. Clerestory windows on the neighbor's side and solar operated skylights provided an abundance of natural light and privacy. Understanding this space as an auxiliary area, heating was addressed simply with an electric fireplace designed into a built-in bookcase that also neatly house board games, books and toys.
One design challenge was how to leverage the volume of space in a usable way. By designing a loft space above the bathroom (and secondary storage room), an inventive kid's reading loft was created and equipped with a trap door entrance.
Another challenge was to maintain the support value of what would become the new, shared wall of the garage to the new addition. Repurposing an existing window opening to accommodate a new interior door (leading from the garage to the addition) kept the structural integrity of the exterior wall and allowed for a relatively easy tie-in of the new space to the existing structure.
Offering multi-functional and multi-flexible spaces for kids and adults alike, this new space offers a place to entertain and host gatherings, play dates, and overnight guests.