text:
\section{Introduction}

This paper concerns the fractional $\infty$-Laplace
operator $\Delta_\infty^s$, as introduced in \cite{BCF} and represented by (\ref{inno}) below. Given a
function $\phi:\mathbb{R}^N\to\mathbb{R}$, our main result is the identification of $\Delta_\infty^s \phi(x)$
as the $\epsilon^{2s}$-order coefficient in the asymptotic expansion of the deviation
of an appropriate $\epsilon$-average $\mathcal{A}_\epsilon$ applied on $\phi$, from the value $\phi(x)$.
Such identification is of general interest in the analysis of partial differential
operators, their related probabilistic interpretation via Tug-of-War
games, a study of viscosity solutions and of numerical approximating schemes. The chief
example of the said asymptotic expansions is given by the well known (local and linear)
formula for the Laplace operator, where $\Delta \phi(x)$ emerges as the
$\epsilon^2$-order coefficient from the integral average $\mathcal{A}_\epsilon = \fint_{B_\epsilon}$:
\begin{equation}\label{aed2}
\fint_{B_\epsilon(x)}\phi(y)\;\mbox{d}y = \phi(x) +
\frac{\epsilon^2}{2(N+2)}\Delta \phi(x) + o(\epsilon^2) \qquad \mbox{ as }\; \epsilon\to 0+.
\end{equation}
The parallel expansion of the $\infty$-Laplacian: $\Delta_\infty\phi(x) =
\big \langle \nabla^2 \phi(x) : \frac{\nabla\phi(x)}{|\nabla\phi(x)|}\otimes
\frac{\nabla\phi(x)}{|\nabla\phi(x)|}\big\rangle$ utilizes the
midpoint (local and nonlinear) average
$\mathcal{A}_\epsilon=\frac{1}{2}(\sup_{B_\epsilon} + \inf_{B_\epsilon})$ in:
\begin{equation}\label{aed3}
\frac{1}{2} \big(\sup_{B_\epsilon(x)\normalcolor}\phi + \inf_{B_\epsilon(x)}\phi \big) = \phi(x) +
\frac{\epsilon^2}{2}\Delta_\infty \phi(x) + o(\epsilon^2) \qquad \mbox{ as }\; \epsilon\to 0+.
\end{equation}

\subsection{The asymptotic expansions and averaging operators in this paper}
In what follows,  will prove that for every $s\in (\frac{1}{2},1)$ one
counterpart formula of (\ref{aed2}) for $\Delta_\infty^s$ is:
\begin{equation}\label{jeden1}
\begin{split}
& \mathcal{A}_\epsilon^o \phi(x) = \phi(x) +
{s}\epsilon^{2s}\Delta_\infty^s\phi (x) + o(\epsilon^{2s}) \qquad \mbox{ as }\; \epsilon\to 0+,
\end{split}
\end{equation}
based on the following (nonlocal and nonlinear) average:
\begin{equation}\label{MVP1}
\mathcal{A}_\epsilon^o \phi(x) = \frac{1}{2}\Big(\sup_{|y|=1}\fint_\epsilon^\infty
\phi(x+ty)\;\mathrm{d}\mu_s(t) + \inf_{|y|=1}\fint_\epsilon^\infty \phi(x+ty)\;\mathrm{d}\mu_s(t) \Big).
\end{equation}
The one-dimensional fractional measure $\mu_s$ and the structure of
the error term $o(\epsilon^{2s})$ will be explained below.
When $\nabla\phi(x)\neq 0$, we also derive another identification through
a local-nonlocal average, which is a convex combination of
$\mathcal{A}_\epsilon$-s in (\ref{jeden1}) and (\ref{aed3}):
\begin{equation}\label{MVP2}
\mathcal{A}_\epsilon \phi(x) = (1-s) \cdot \mathcal{A}_\epsilon^o \phi(x) + s\cdot \frac{1}{2}
\big(\sup_{B_\epsilon(x)\normalcolor}\phi + \inf_{B_\epsilon(x)}\phi \big).
\end{equation}
We anticipate that the error quantity $o(\epsilon^{2s})$ below is
uniform in the whole considered range $s\in
(\frac{1}{2},1)$, whereas the corresponding error in (\ref{jeden1})
blows up to $\infty$ as $s\to 1-$.
Thus, the following asymptotic expansion can be seen as an improvement of (\ref{jeden1}):
\begin{equation}\label{dwa1}
\mathcal{A}_\epsilon\phi(x) = \phi(x) + (1-s)s\epsilon^{2s}
\Delta_\infty^s \phi(x) + o(\epsilon^{2s})\qquad \mbox{ as }\; \epsilon\to 0+.
\end{equation}
Precise statements of (\ref{jeden1}) and (\ref{dwa1})
will be given in Theorems \ref{jaszczu1}, \ref{jaszczu2} and Remarks
\ref{kura}, \ref{piesio}.

\subsection{The fractional $\infty$-Laplacian}
Let $\phi:\mathbb{R}^N\to\mathbb{R}$ be a bounded Borel function. We recall that
$\phi\in C^{1,1}(x)$ at $x\in\mathbb{R}^N$, provided that there exists $p_x\in \mathbb{R}^N$ and $C_x,\eta_x>0$ such that:
\begin{equation}\label{C11}
\big|\phi(x+y) - \phi(x) - \langle p_x, y\rangle\big|\leq C_x |y|^2
\qquad\mbox{for all }\; |y|<\eta_x.
\end{equation}
In \cite[Definition 1.1]{BCF},  the (normalized) fractional
$\infty$-Laplacian $\Delta_\infty^s\phi(x)$, for $s\in (\frac{1}{2}, 1)$, has been introduced by means of
two distinct formulas, distinguishing between cases $p_x=0$ and
$p_x\neq 0$.  In section \ref{sec2}, we provide a rigorous proof of the following alternative definition stated in \cite{BCF}:
\begin{equation}\label{inno}
\Delta_\infty^s \phi(x)= \frac{1}{C_s}\cdot\sup_{|y|=1}\inf_{|\tilde y|=1} \int_0^\infty L_{\phi}(x,ty,t\tilde y)\;\mathrm{d}\mu_s(t).
\end{equation}
To explain the notation in the right hand side above, for each $x, y, \tilde y\in\mathbb{R}^N$ we define:
$$L_{\phi}(x, y, \tilde y) \doteq \phi(x+y) + \phi(x-\tilde y) - 2\phi(x).$$
Further, $\mu_s$ is the measure\footnote{The role of the normalizing constant $C_s$ is to ensure that the
operator $-(-\Delta)^su(x) \doteq \int_0^\infty L_u(x,t,t)\;\mathrm{d}\mu_s(t)$ defined for
$u:\mathbb{R}\to\mathbb{R}$, is a pseudo-differential operator with symbol
$|\xi|^{2s}$.} on the Borel subsets of $(0,\infty)$, given by:
$$\mathrm{d}\mu_s(t) \doteq \frac{C_s}{t^{1+2s}}\;\mbox{d}t \quad \mbox{ where }\;
C_s=\frac{4^s s\Gamma\big(\frac{1}{2}+s\big)}{\pi^{1/2}\Gamma\big(1-s\big)} =
\Big(2\int_0^\infty\frac{1-\cos t}{t^{1+2s}}\;\mbox{d}t\Big)^{-1} .$$
It is important to note \cite{hitch}  that one can express $C_s$ by means of another constant\footnote{A direct
  calculation shows that, for example, $c_s\in ((\frac{12}{13})^2,
  (\frac{12}{5})^2)$ in the range $s\in (\frac{1}{2}, 1)$.} $c_s$, that
is bounded and positive, uniformly in $s$. Namely, there holds:
$$C_s = s(1-s) c_s.$$

We also point out that the operator $\Delta_\infty^s$ treated in this
paper, is not the only nonlocal counterpart of $\Delta_\infty$. A
variational (i.e. energy based) fractional infinity Laplacian was
studied in \cite{CLM12} and a non normalized one in
\cite{CJ17}. These operators are not suited for a game theoretical
approach, which was the main motivation  in\cite{BCF}.

\subsection{Statements and discussion of main results}
We consider the operator in the right hand side of (\ref{inno}):
\begin{equation}\label{Ldefi1}
\mathcal{L}_s[\phi] (x) \doteq\sup_{|y|=1}\inf_{|\tilde y|=1} \int_0^\infty L_{\phi}(x,ty,t\tilde y)\;\mathrm{d}\mu_s(t).
\end{equation}
Given $\eta_x>0$, we work with the following hypotheses on
$\phi$, relative to the ball $B_{\eta_x}\doteq B_{\eta_x}(x)$:
\begin{equation*}\label{H1}\tag{{\bf H}}
\left[\quad \;\; \mbox{\begin{minipage}{13.1cm}\vspace{1mm}
\begin{itemize}[leftmargin=*]
\item[(i)] $\phi\in C^2(\bar B_{\eta_x})$ where $p_x\doteq \nabla
\phi(x)$ and $C_x\doteq \frac{1}{2}\|\nabla^2\phi\|_{L^\infty(B_{\eta_x})}$.\vspace{1mm}
\item[(ii)] $\phi$ is bounded and uniformly continuous on
$\mathbb{R}^N\setminus \bar B_{\eta_x}$ with modulus of continuity
$\omega_\phi (a)\doteq\sup\big\{|\phi(x)-\phi(y)|;~ x,y\in \mathbb{R}^N\setminus
\bar B_{\eta_x},~ |x-y|\leq a\big\}.$\vspace{1mm}
\end{itemize}
\end{minipage}}\;\;\, \right]
\end{equation*}
Regularity required in (\ref{H1}) is  satisfied by the test functions in the viscosity solution setting (see section
\ref{sec51}). We further denote:
\begin{equation*}
\begin{split}
& A_\epsilon = \max\Big\{\frac{16 C_x}{|p_x|}\cdot\frac{2s-1}{1-s}\cdot
\frac{\eta_x^{2-2s}-\epsilon^{2-2s}}{\epsilon^{1-2s}
  -{\eta_x}^{1-2s}}, ~\kappa_\epsilon\Big\}, \\ &
\kappa_\epsilon = \sup\Big\{ a;~ a\in [0,2] \; \mbox{ and } \; a^2\leq
 \frac{8\omega_\phi(a)}{|p_x|}\cdot \frac{\frac{2s-1}{2s}
  \eta_x^{-2s} +\eta_x^{1-2s}}{\epsilon^{1-2s} - \eta_x^{1-2s}}\Big\}.
\end{split}
\end{equation*}

Our first main result regards the expansion (\ref{jeden1}):
\begin{theorem}\label{jaszczu1}
Let $\phi:\mathbb{R}^N\to\mathbb{R}$ satisfy (\ref{H1}).  Then there holds, for every $\epsilon< \eta_x$:
\begin{equation*}
\begin{split}
\Big|\mathcal{A}_\epsilon^o&\phi(x) - \phi(x) -
\frac{1}{c_s(1-s)}\epsilon^{2s}\mathcal{L}_s[\phi] (x) \Big|\leq  \frac{s}{1-s}\cdot
C_x \epsilon^2 + \\ & + \left\{\begin{array}{ll} \epsilon^{2s}\Big(4
sC_x\frac{\eta_x^{2-2s}-\epsilon^{2-2s}}{1-s}\cdot A_\epsilon
+ \big(\eta_x^{-2s} + \frac{2s}{2s-1}\eta_x^{1-2s}\big)\cdot \omega_\phi(A_\epsilon)\Big)
& \mbox{ when }\; p_x\neq 0\\
0 & \mbox{ when }\; p_x= 0. \end{array}\right.
\end{split}
\end{equation*}
\end{theorem}

Our second main result regards the expansion (\ref{dwa1}):
\begin{theorem}\label{jaszczu2}
Assume (\ref{H1}) and $p_x\neq 0$.
Then, for all $\epsilon< \eta_x$ with $\epsilon |\nabla^2\phi(x)|\leq |p_x|$, we have:
\begin{equation*}
\begin{split}
\Big|\mathcal{A}_\epsilon\phi(x) &- \phi(x) - \frac{1}{c_s}\epsilon^{2s}\mathcal{L}_s[\phi] (x)
\Big|\\ & \leq
2\epsilon^{2s}\Big(2sC_x\big(\eta_x^{2-2s}-\epsilon^{2-2s}\big)A_\epsilon
+ \Big(\frac{\eta_x^{-2s}}{2}+\frac{s\eta_x^{1-2s}}{2s-1}\Big)\cdot (1-s) \omega_\phi(A_\epsilon)\Big)
\\ & \quad + 2s\epsilon^3\frac{|\nabla^2\phi(x)|^2}{|p_x|}
+ s\epsilon^2\sup_{y\in B_\epsilon(x)}\big|\nabla^2\phi(y)-\nabla^2\phi(x)\big|.
\end{split}
\end{equation*}
\end{theorem}
For a discussion of the error terms in the above results, we refer to Remarks \ref{kura} and
\ref{piesio}. In particular, for $\phi$ Lipschitz, the
 bound in Theorem \ref{jaszczu1} becomes: $O(\epsilon^{4s-1}+\epsilon^{2})$ with constants that blow up as
$s\to1-$. On the other hand, the bound in Theorem \ref{jaszczu2} has
the form: $O(\epsilon^{4s-1}) +  o(\epsilon^2)$ however the related
constants are uniform in $s\in(\frac{1}{2},1)$, and it also is
compatible with the expected error bound for the (local) $\infty$-Laplacian. This improvement is obtained
by correcting the singular part of $\Delta_\infty^s$ by the
corresponding asymptotic expansion of its local counterpart. Such
idea was already present in the numerical analysis literature, where
it was used to obtain higher order monotone numerical schemes for the
fractional Laplacian and other linear nonlocal operators, see
e.g.\cite{DEJ18, DEJ19}. The fact that the singular part of a
nonlinear and nonlocal operator encodes a local counterpart
is an idea also present in \cite{BS2,DL20}.

Other asymptotic expansions for nonlocal operators such as
$\Delta_\infty^s$ have been recently introduced  in
\cite{BS1, BS2}. The related average in \cite[section 3.2]{BS2} distinguishes
between the cases $p_x\neq 0$ and $p_x=0$. In comparison,
$\mathcal{A}_\epsilon$ in the present paper neither
relies on this distinction nor even necessitates the notion of the
gradient being well posed. Thus, they can be applied on a larger class of functions $\phi$ that are
only bounded Borel.

In section \ref{sec52}, we further propose a version $\bar{\mathcal{A}}_\epsilon^o$ of the average
$\mathcal{A}_\epsilon^o$ and its corresponding expansion, in which integration takes place on an open, bounded
domain in $\mathbb{R}^N$, rather than an infinite line. We believe that this
correction will be of importance in the implementation of
numerical schemes. We also conjecture that the expected values of the
stochastic process whose dynamic
programming principle is modeled on $\bar{\mathcal{A}}^o_\epsilon$ converge
to these solutions in the limit $\epsilon\to 0+$, as in the
pivotal study \cite{PSSW} of  the classical operator $\Delta_\infty$.

\subsection*{Outline of the paper}
We prove (\ref{inno}) in section
\ref{sec2}, Theorem \ref{jaszczu1} in section \ref{sec3}, and Theorem
\ref{jaszczu2} in section \ref{sec4}. In section \ref{sec5}, we
discuss $\bar{\mathcal{A}}_\epsilon^o$ and  put Theorems
\ref{jaszczu1}, \ref{jaszczu2} in a viscosity solution framework.

\section{The fractional $\infty$-Laplacian and a proof of (\ref{inno})}\label{sec2}

Given a bounded Borel function $\phi:\mathbb{R}^N\to\mathbb{R}$, and two parameters
$\epsilon>0$ and $s\in (\frac{1}{2},1)$, we will be concerned
with values of the integral operators $\mathcal{L}_s^\epsilon[\phi]:\mathbb{R}^N\to\mathbb{R}$, given in:
\begin{equation}\label{Ldefi}
\begin{split}
\mathcal{L}_s^\epsilon[\phi] (x) & \doteq\sup_{|y|=1}\inf_{|\tilde y|=1} \int_\epsilon^\infty L_{\phi}(x,ty,t\tilde y)\;\mathrm{d}\mu_s(t)\\
& = \sup_{|y|=1}\int_\epsilon^\infty\phi(x+ty)\;\mathrm{d}\mu_s(t) +
\inf_{| y|=1} \int_\epsilon^\infty\phi(x+ty)\;\mathrm{d}\mu_s(t) - \frac{(1-s)c_s}{\epsilon^{2s}}\phi(x),\\
\end{split}
\end{equation}
Note that, since the restriction of $\phi$ to any
one-dimensional line is also Borel, the function $(0,\infty)\ni
t\mapsto L_{\phi}(x,ty,t\tilde y)$ is bounded and Borel for any $x,y,\tilde y$.
Further, since $\mu_s(\epsilon, \infty) = \frac{C_s}{2s\epsilon^{2s}}<\infty$, each integral
$\int_\epsilon^\infty L_{\phi}(x, ty, t\tilde y)\;\mathrm{d}\mu_s(t)$ and consequently also the quantities
$\mathcal{L}_s^\epsilon[\phi] (x)$, are all well defined and finite. On the other hand,
$\mu_s(0,\infty)=\infty$, so neither the definition of $\mathcal{L}_s[\phi]$
in (\ref{Ldefi1}) nor a version of its equivalent formulation as in $\mathcal{L}_s^\epsilon[\phi]$ are
necessarily valid, when $\phi$ is only bounded and Borel.
However, one immediate consequence of (\ref{C11}) is that:
\begin{equation}\label{C2}
\big|L_{\phi} (x,ty,t\tilde y) - t\langle p_x, y-\tilde y\rangle \big|\leq 2C_x t^2
\qquad\mbox{for all }\; |y|, |\tilde y|=1\mbox{ and } |t|<\eta_x.
\end{equation}
which yields (through an application of Taylor's expansion):
\begin{lemma}\label{lem1}
Let $\phi\in C^{1,1}(x)$ be a bounded Borel function. Then $\{\mathcal{L}_s^\epsilon[\phi] (x)\}_{\epsilon>0}$
are bounded independently of $\epsilon$ and $\mathcal{L}_s[\phi](x)$ is well
defined. More precisely, for all $\epsilon<\eta_x$ there holds:
\begin{equation}\label{uno}
|\mathcal{L}_s^\epsilon[\phi] (x)|, \; |\mathcal{L}_s[\phi](x)| \leq 2c_s (1-s)\|\phi\|_{L^\infty}\eta_x^{-2s} + c_s s \cdot C_x\eta_x^{2-2s}.
\end{equation}
\end{lemma}
\begin{proof}
We use (\ref{C2}) to obtain, for any $|y|=|\tilde y|=1$:
\begin{equation*}
\begin{split}
\Big|\int_\epsilon^\infty L_{\phi}(x, ty, t\tilde y)&\;\mathrm{d}\mu_s(t) -
\int_\epsilon^{\eta_x} t\langle p_x, y-\tilde y\rangle \;\mathrm{d}\mu_s(t)\Big| \leq \int_{\eta_x}^\infty 4\|\phi\|_{L^\infty}\;\mathrm{d}\mu_s(t)
+ \int_\epsilon^{\eta_x}2 C_xt^2\;\mathrm{d}\mu_s(t) \\ & = 2\frac{C_s}{s\eta_x^{2s}}\|\phi\|_{L^\infty} +
C_sC_x\frac{\eta_x^{2-2s}-\epsilon^{2-2s}}{1-s}.
\end{split}
\end{equation*}
On the other hand:
$$\sup_{|y|=1}\inf_{|\tilde y|\normalcolor =1} \int_\epsilon^{\eta_x} t\langle p_x, y-\tilde y\rangle \;\mathrm{d}\mu_s(t)
= \int_\epsilon^{\eta_x} t\;\mathrm{d}\mu_s(t)\cdot\sup_{|y|=1}\inf_{|\tilde y|\normalcolor=1} \langle p_x, y-\tilde y\rangle =0.$$
This results in:
\begin{equation*}
\begin{split}
|\mathcal{L}_s^\epsilon[\phi] (x) | & = \Big|\mathcal{L}_s^\epsilon[\phi] (x) - \sup_{|y|=1}\inf_{|\tilde y|\normalcolor =1}
\int_\epsilon^{\eta_x} t\langle p_x, y-\tilde y\rangle \;\mathrm{d}\mu_s(t)\Big | \\ & \leq
\sup_{|y|=|\tilde y|=1} \Big|\int_\epsilon^\infty L_{\phi}(x, ty, t\tilde y)\;\mathrm{d}\mu_s(t) -
\int_\epsilon^{\eta_x} t\langle p_x, y-\tilde y\rangle \;\mathrm{d}\mu_s(t)\Big|,
\end{split}
\end{equation*}
which proves the bound for $|\mathcal{L}_s^\epsilon[\phi] (x) |$. The bound for $|\mathcal{L}_s[\phi] (x)|$ follows similarly.
\end{proof}

Our proofs throughout the paper largely depend on analyzing the
behaviour of approximate extremizers $y, \tilde y$ in the definition (\ref{Ldefi})
We now observe that for the operator $\mathcal{L}_s[\phi]$ these extremizers are
explicit, for a generic function $\phi$.

\begin{proposition}\label{prop2}
Let $\phi\in C^{1,1}(x)$ be a bounded Borel function such that $p_x\neq 0$. Then:
\begin{equation}\label{eq:equivdefifl}
\mathcal{L}_s[\phi](x) = \int_0^\infty L_{\phi}\big(x, t\frac{p_x}{|p_x|}, t\frac{p_x}{|p_x|}\big)\;\mathrm{d}\mu_s(t)
\end{equation}
\end{proposition}
\begin{proof}
For $\delta>0$, let $y_\delta$ be such that $\mathcal{L}_s[\phi](x)\leq \inf_{|\tilde y|=1}
\int_0^\infty L_{\phi}(x, ty_\delta, t\tilde y)\;\mathrm{d}\mu_s(t) +\delta$. Splitting the
integral and applying (\ref{C2}) on the interval $(0,\eta_x)$ yields:
\begin{equation*}
\begin{split}
\mathcal{L}_s[\phi] (x)&\leq \int_0^\infty L_{\phi}\big(x, ty_\delta,
t\frac{p_x}{|p_x|}\big)\;\mathrm{d}\mu_s(t) +\delta \leq \int_{0}^{\eta_x} t\big\langle
p_x, y_\delta - \frac{p_x}{|p_x|}\big\rangle \;\mathrm{d}\mu_s(t) + C+\delta,
\end{split}
\end{equation*}
where the constant $C$ depends on $s$ and $\phi$. Using the lower
bound in (\ref{uno}), we conclude that:
$$\big\langle p_x,\frac{p_x}{|p_x|} - y_\delta \big\rangle \cdot \int_{0}^{\eta_x}
t\;\mathrm{d}\mu_s(t) < \infty.$$
Since the product above is nonnegative while the integral
diverges to $\infty$, we get $y_\delta=\frac{p_x}{|p_x|}$ and:
\begin{equation}\label{due}
\mathcal{L}_s[\phi](x) = \inf_{|\tilde y|=1} \int_0^\infty L_{\phi}\big(x, t\frac{p_x}{|p_x|}, t\tilde y)\;\mathrm{d}\mu_s(t).
\end{equation}

Let now $\tilde y_\delta$ be such that $\mathcal{L}_s[\phi](x) \geq \int_0^\infty
L_{\phi}\big(x, t\frac{p_x}{|p_x|}, t\tilde y_\delta)\;\mathrm{d}\mu_s(t) -\delta$. As
before, in virtue of (\ref{C2}) we get:
$\mathcal{L}_s[\phi] (x)\geq \int_{0}^{\eta_x} t\big\langle
p_x, \frac{p_x}{|p_x|}-\tilde y_\delta\big\rangle \;\mathrm{d}\mu_s(t) - C -\delta$, so
the upper bound in (\ref{uno}) gives:
$$\big\langle p_x,\frac{p_x}{|p_x|} -\tilde y_\delta \big\rangle \cdot \int_{0}^{\eta_x}
t\;\mathrm{d}\mu_s(t) < \infty.$$
Consequently $\tilde y_\delta=\frac{p_x}{|p_x|}$, so that: $\mathcal{L}_s[\phi](x) \geq \int_0^\infty
L_{\phi}\big(x, t\frac{p_x}{|p_x|},  t\frac{p_x}{|p_x|})\;\mathrm{d}\mu_s(t) -\delta$ for
all $\delta>0$. The proof is done, in view of (\ref{due}).
\end{proof}

Note that formulation (\ref{eq:equivdefifl}) corresponds, up to a
constant, to the one given in \cite[Definition 1.1]{BCF} for
$p_x\not=0$.  The case $p_x=0$ in that definition is equivalent to (\ref{inno}).

\section{A proof of Theorem \ref{jaszczu1}}\label{sec3}

We first observe that taking $p_x=0$ in (\ref{C2}) implies the
following bound, for all $\epsilon<\eta_x$:
\begin{equation}\label{due.5}
\begin{split}
|\mathcal{L}_s^\epsilon[\phi] (x)-\mathcal{L}_s[\phi](x)|&\leq \sup_{|y|=|\tilde y|=1} \big|\int_\epsilon^\infty L_{\phi}(x, ty,
t\tilde y)\;\mathrm{d}\mu_s(t) - \int_0^\infty L_{\phi}(x, ty, t\tilde y)\;\mathrm{d}\mu_s(t) \big|  \\ &\leq
\int_0^\epsilon 2C_x t^2\;\mathrm{d}\mu_s(t) = c_s s\cdot C_x\epsilon^{2-2s}.
\end{split}
\end{equation}
In order to estimate the same difference when $p_x\neq 0$, we will
quantify estimates in the proof of Proposition \ref{prop2} for higher regular
functions, as specified below.

\begin{proposition}\label{kokon}
Assume (\ref{H1}) with $p_x\neq 0$. Then, for every $\epsilon< \eta_x$ there holds:
\begin{equation}\label{tre}
\begin{split}
\Big|\mathcal{L}_s^\epsilon[\phi] (x)- & \int_\epsilon^\inftyL_{\phi}\big(x, t\frac{p_x}{|p_x|},
t\frac{p_x}{|p_x|}\big)\;\mathrm{d}\mu_s(t)\Big|\\ & \leq 4 c_s s\cdot C_x \big(\eta_x^{2-2s} -
  \epsilon^{2-2s}\big)\cdot A_\epsilon + c_s (1-s) \cdot \Big( \eta_x^{-2s} +
\frac{2s}{2s-1}\eta_x^{1-2s}\Big)\cdot \omega_\phi(A_\epsilon),
\end{split}
\end{equation}
\end{proposition}
\begin{proof}
{\bf 1.} For every $\epsilon<\eta_x$ and every small $\delta>0$
let $|y_\delta^\epsilon|=1$ satisfy:
$\sup_{|y|=1}\int_\epsilon^\infty\phi(x+ty)\;\mathrm{d}\mu_s(t) \leq \int_\epsilon^\infty\phi(x+ty_\delta^\epsilon)\;\mathrm{d}\mu_s(t) +\delta$.
In particular, this implies:
$$\int_\epsilon^\infty\phi\big(x+t\frac{p_x}{|p_x|}\big)\;\mathrm{d}\mu_s(t) \leq
\int_\epsilon^\infty\phi(x+ty_\delta^\epsilon)\;\mathrm{d}\mu_s(t) +\delta,$$
Denote $A= \Big|\frac{p_x}{|p_x|} - y_\delta^\epsilon\Big|$.  Together
with (\ref{C11}), the above bound results in:
\begin{equation*}
\begin{split}
\delta &\geq \int_\epsilon^{\eta_x} \big(\phi\big(x+t\frac{p_x}{|p_x|}\big)
- \phi(x+t y_\delta^\epsilon) \big)\;\mathrm{d}\mu_s(t) - \int_{\eta_x}^\infty \big|\phi\big(x+t\frac{p_x}{|p_x|}\big)
- \phi(x+t y_\delta^\epsilon)\big|\;\mathrm{d}\mu_s(t) \\ & \geq
\int_\epsilon^{\eta_x} t\Big\langle\nabla\phi\big(x+t\frac{p_x}{|p_x|}\big),
\frac{p_x}{|p_x|}-y_\delta^\epsilon\Big\rangle\;\mathrm{d}\mu_s(t) -
\int_\epsilon^{\eta_x} C_xt^2A^2\;\mathrm{d}\mu_s(t)
- \int_{\eta_x}^\infty (1+t)\cdot\omega_\phi (A)\;\mathrm{d}\mu_s(t)
\\ & \geq \Big\langle p_x,\frac{p_x}{|p_x|} - y_\delta^\epsilon\Big\rangle
\int_\epsilon^{\eta_x} t\;\mathrm{d}\mu_s(t) - 4C_xA \int_\epsilon^{\eta_x} t^2\;\mathrm{d}\mu_s(t)  -\omega_\phi(A) \int_{\eta_x}^\infty (1+t)\;\mathrm{d}\mu_s(t).
\end{split}
\end{equation*}
The last bound above follows by observing that $\big|\nabla\phi\big(x+t\frac{p_x}{|p_x|})- \nabla
\phi(x)\big|\leq 2C_xt$ for all $|t|\leq \eta_x$ and that $A\leq 2$. Consequently, we get:
\begin{equation*
\begin{split}
\Big\langle &p_x,\frac{p_x}{|p_x|} - y_\delta^\epsilon\Big\rangle\leq  \frac{1}{
\int_\epsilon^{\eta_x} t\;\mathrm{d}\mu_s(t)}\Big( \delta+ 4C_xA
\int_\epsilon^{\eta_x} t^2\;\mathrm{d}\mu_s(t) +\omega_\phi(A)\int_{\eta_x}^\infty (1+t)\;\mathrm{d}\mu_s(t)\Big).
\end{split}
\end{equation*}

On the other hand, by a straightforward calculation:
\begin{equation*}
\begin{split}
A^2=\Big|\frac{p_x}{|p_x|} - y_\delta^\epsilon\Big|^2 = 2 - 2 \Big\langle
\frac{p_x}{|p_x|},  y_\delta^\epsilon\Big\rangle = \frac{2}{|p_x|}
\Big\langle p_x,\frac{p_x}{|p_x|} - y_\delta^\epsilon\Big\rangle,
\end{split}
\end{equation*}
the last two displayed formulas yields that:
\begin{equation}\label{quattro}
\begin{split}
A^2 \leq  \frac{2}{|p_x| \int_\epsilon^{\eta_x} t\;\mathrm{d}\mu_s(t)}\Big( \delta+ 4C_xA
\int_\epsilon^{\eta_x} t^2\;\mathrm{d}\mu_s(t) +\omega_\phi(A) \int_{\eta_x}^\infty (1+t)\;\mathrm{d}\mu_s(t)\Big).
\end{split}
\end{equation}

We now simplify (\ref{quattro}) as follows. Without loss of
generality, we may assume that $\delta>0$ satisfies:
$ \delta \cdot |p_x|\int_\epsilon^{\eta_x}t\;\mathrm{d}\mu_s(t)\leq \big(16 C_x \int_\epsilon^{\eta_x}t^2\;\mathrm{d}\mu_s(t)\big)^2,$
In case when $\delta$ is larger than the two other terms in the right
hand side of (\ref{quattro}), we get:
\begin{equation}\label{cinque}
A^2\leq \frac{4\delta}{|p_x|\int_\epsilon^{\eta_x} t\;\mathrm{d}\mu_s(t)}\leq
\Big(\frac{32 C_x \int_\epsilon^{\eta_x}t^2\;\mathrm{d}\mu_s(t)}{|p_x|\int_\epsilon^{\eta_x}t\;\mathrm{d}\mu_s(t)}\Big)^2,
\end{equation}
In the opposite case, there holds:
\begin{equation*}
A^2 \leq  \frac{4}{|p_x| \int_\epsilon^{\eta_x} t\;\mathrm{d}\mu_s(t)}\Big(4C_xA
\int_\epsilon^{\eta_x} t^2\;\mathrm{d}\mu_s(t) +\omega_\phi(A) \int_{\eta_x}^\infty
(1+t)\;\mathrm{d}\mu_s(t)\Big)\doteq I_1 + I_2.
\end{equation*}
Further, when $I_2\leq I_1$, then we obtain the same bound as in (\ref{cinque}), namely:
\begin{equation*}
A\leq \frac{32 C_x  \int_\epsilon^{\eta_x}t^2\;\mathrm{d}\mu_s(t)}{|p_x|\int_\epsilon^{\eta_x}t\;\mathrm{d}\mu_s(t)} =
\frac{16 C_x}{|p_x|}\cdot\frac{2s-1}{1-s}\cdot
\frac{\eta_x^{2-2s}-\epsilon^{2-2s}}{\epsilon^{1-2s} -{\eta_x}^{1-2s}}.
\end{equation*}
On the other hand, $I_1<I_2$ implies:
\begin{equation*}
A^2 \leq  \frac{8 \omega_\phi(A) \int_{\eta_x}^\infty (1+t)\;\mathrm{d}\mu_s(t)}{|p_x|  \int_\epsilon^{\eta_x} t\;\mathrm{d}\mu_s(t)}
=\frac{8\omega_\phi(A)}{|p_x|}\cdot \frac{\frac{2s-1}{2s}
  \eta_x^{-2s} +\eta_x^{1-2s}}{\epsilon^{1-2s} - \eta_x^{1-2s}}.
\end{equation*}
We hence conclude that $A\leq A_\epsilon$ in either of the above cases.

\smallskip

{\bf 2.} Similarly as in step 1, we see that the unit vector  $\tilde
y_\delta^\epsilon$ with the property:
$\inf_{|y|=1}\int_\epsilon^\infty\phi(x-ty)\;\mathrm{d}\mu_s(t) \geq
\int_\epsilon^\infty\phi(x-t\tilde y_\delta^\epsilon)\;\mathrm{d}\mu_s(t) -\delta$,
satisfies: $\big|\frac{p_x}{|p_x|}-\tilde y_\delta^\epsilon\big|\leq
A_\epsilon$. We now write:
\begin{equation*}
\int_\epsilon^\infty L_{\phi}(x, t\tilde y_\delta^\epsilon, t\tilde  y_\delta^\epsilon)\;\mathrm{d}\mu_s(t)
-\delta \leq \mathcal{L}_s^\epsilon[\phi] (x)\leq \int_\epsilon^\infty L_{\phi}(x,
ty_\delta^\epsilon, ty_\delta^\epsilon)\;\mathrm{d}\mu_s(t) +\delta,
\end{equation*}
which implies:
\begin{equation}\label{sette}
\begin{split}
& \Big|\mathcal{L}_s^\epsilon[\phi] (x) - \int_\epsilon^\infty L_{\phi}\big(x, t\frac{p_x}{|p_x|}, t\frac{p_x}{|p_x|}\big)\;\mathrm{d}\mu_s(t)
\Big|\leq \delta + \max \big\{ |I(y_\delta^\epsilon)|, |I(\tilde y_\delta^\epsilon|)\big\},\\ &
\mbox{where: } J(y)\doteq \int_\epsilon^\infty \phi(x+ty) - \phi\big(x+t \frac{p_x}{|p_x|}\big) +
\phi(x-ty) - \phi\big(x-t \frac{p_x}{|p_x|}\big) \;\mathrm{d}\mu_s(t).
\end{split}
\end{equation}
Observe that:
\begin{equation*}
\begin{split}
|I(y_\delta^\epsilon)|\leq & \; \Big|\int_\epsilon^{\eta_x} \phi(x+t
y_\delta^\epsilon) - \phi\big(x+t \frac{p_x}{|p_x|}\big)  +
\phi(x-t y_\delta^\epsilon) - \phi\big(x-t \frac{p_x}{|p_x|}\big)\;\mathrm{d}\mu_s(t)\Big| \\ & +
\int_{\eta_x}^\infty \big|\phi(x+t y_\delta^\epsilon) - \phi\big(x+t \frac{p_x}{|p_x|}\big)\big| +
\big|\phi(x-t y_\delta^\epsilon) - \phi\big(x-t
\frac{p_x}{|p_x|}\big)\big|\;\mathrm{d}\mu_s(t) \doteq \bar I_1 + \bar I_2.
\end{split}
\end{equation*}
In order to deal with $\bar I_1$, we use the Taylor expansion:
\begin{equation*}
\Big|\phi(x\pm t y_\delta^\epsilon) - \phi\big(x\pm t \frac{p_x}{|p_x|}\big)
-\Big\langle \nabla\phi\big(x\pm t \frac{p_x}{|p_x|}\big), \pm t
\big(y_\delta^\epsilon - \frac{p_x}{|p_x|}\big)\Big\rangle \Big|\leq
C_x t^2  \big| y_\delta^\epsilon - \frac{p_x}{|p_x|}\big|^2,
\end{equation*}
which upon integration implies:
\begin{equation*}
\begin{split}
\bar I_1 & \leq \int_\epsilon^{\eta_x} C_x t^2 \big(4 A+ 2A^2\big)\;\mathrm{d}\mu_s(t)\leq
8C_x A\int_\epsilon^{\eta_x} t^2\;\mathrm{d}\mu_s(t) = 4 C_x C_s \frac{\eta_x^{2-2s} - \epsilon^{2-2s}}{1-s}\cdot A.
\end{split}
\end{equation*}
For the term $J_2$, we get:
\begin{equation*}
\begin{split}
\bar I_2 & \leq 2\int_{\eta_x}^\infty (1+t)\cdot\omega_\phi(A)\;\mathrm{d}\mu_s(t)
= 2 C_s \Big( \frac{\eta_x^{-2s}}{2s} + \frac{\eta_x^{1-2s}}{2s-1}\Big)\cdot \omega_\phi(A).
\end{split}
\end{equation*}

In conclusion, we obtain the following bounds:
\begin{equation*}
|I(y_\delta^\epsilon)|, \quad |I(\tilde y_\delta^\epsilon)|\leq 4 C_x C_s \frac{\eta_x^{2-2s} -
  \epsilon^{2-2s}}{1-s} A_\epsilon + 2 C_s \Big( \frac{\eta_x^{-2s}}{2s} +
\frac{\eta_x^{1-2s}}{2s-1}\Big)\cdot \omega_\phi(A_\epsilon).
\end{equation*}
This ends the proof in virtue of (\ref{sette}).
\end{proof}

\medskip

\begin{corollary}\label{cor4.5}
Under the same assumptions and notation as in Proposition \ref{kokon}, we have:
\begin{equation*}
\begin{split}
\Big|&\mathcal{L}_s^\epsilon[\phi] (x)- \mathcal{L}_s[\phi] (x)\Big|\\ &\quad \leq 4 c_s s \cdot C_x \big(\eta_x^{2-2s} -
  \epsilon^{2-2s}\big) \cdot A_\epsilon + c_s (1-s)\cdot \Big(\eta_x^{-2s} +
\frac{2s}{2s-1}\eta_x^{1-2s}\Big)\cdot \omega_\phi(A_\epsilon) + c_s s\cdot C_x\epsilon^{2-2s}.
\end{split}
\end{equation*}
\end{corollary}
\begin{proof}
Observe that for all $t<\eta_x$ there holds:
$$\big|L_{\phi}\big(x, t\frac{p_x}{|p_x|}, t\frac{p_x}{|p_x|}\big) \big|
\leq t^2 \|\nabla^2\phi\|_{L^\infty(B_t)}.$$
Consequently and in view of Proposition \ref{prop2} we get:
$$\Big|\mathcal{L}_s[\phi] (x) - \int_\epsilon^\infty L_{\phi}\big(x, t\frac{p_x}{|p_x|}, t\frac{p_x}{|p_x|}\big)\;\mathrm{d}\mu_s(t)\Big|
\leq \int_0^\epsilon \Big|L_{\phi}\big(x, t\frac{p_x}{|p_x|}, t\frac{p_x}{|p_x|}\big)\Big|\;\mathrm{d}\mu_s(t)
\leq \frac{C_s}{1-s}\cdot C_x \epsilon^{2-2s}.$$
This achieves the proof by Proposition \ref{kokon}.
\end{proof}

\medskip

Note that the bound in Corollary \ref{cor4.5} is essentially valid in both cases $p_x\neq 0$ and
$p_x=0$, because of (\ref{due.5}). Scaling the said bound by the
factor $\frac{s\epsilon^{2s}}{C_s}$, we directly deduce Theorem \ref{jaszczu1}.

\medskip

\begin{remark}\label{kura}
\begin{enumerate}[leftmargin=7mm]
\item [(i)] Observing that: $\omega_\phi(a)\leq 2\|\phi\|_{L^\infty}$,
  we get for all $\epsilon<\frac{\eta_x}{2}$:
$$\kappa_\epsilon\leq \Big(\frac{16\|\phi\|_{L^\infty}}{|p_x|}\cdot
\frac{\frac{2s-1}{2s} \eta_x^{-2s} +\eta_x^{1-2s}}{\epsilon^{1-2s} - \eta_x^{1-2s}}\Big)^{1/2}
\leq 8 \Big(\frac{\|\phi\|_{L^\infty}}{|p_x|}\cdot
\frac{\eta_x^{-2s} +\eta_x^{1-2s}}{2s-1}\Big)^{1/2}\epsilon^{s-1/2}.$$
In the second inequality we used that for all $\epsilon<\frac{\eta_x}{2}$
and all $s\in (\frac{1}{2},1)$ there holds:
$$\epsilon^{1-2s} -
\eta_x^{1-2s}> \epsilon^{1-2s} (1-2^{1-2s}), \qquad 1-2^{1-2s}\geq (2s-1)\ln\sqrt{2} >\frac{2s-1}{4}.$$
The first quantity in $A_\epsilon$ is of order
$\epsilon^{2s-1}$, so the right hand side in (\ref{tre}) is:
$$ C(s)\cdot C\big(C_x, \frac{1}{|p_x|}, \|\phi\|_{L^\infty}\big)\cdot
C(\eta_x)\epsilon^{s-1/2} + C(s)\cdot C\big(\frac{1}{|p_x|}, \|\phi\|_{L^\infty}\big)\cdot
C(\eta_x)\omega_\phi(\epsilon^{s-1/2}), $$
where $C(s)$ depends only on $s$ and $C(\eta_x)$ only on $\eta_x$ and
the remaining constants depend on the other displayed terms, in a
nondecreasing manner.
\item[(ii)] When $\phi\in C^{0,\alpha}(\mathbb{R}^N\setminus \bar B_{\eta_x})$ with
$\alpha\in (0,1)$, then $\omega_\phi(a)=[\phi]_\alpha a^{\alpha}$. Therefore:
$$\kappa_\epsilon \leq \Big(\frac{32\;[\phi]_{\alpha}}{|p_x|}\cdot
\frac{\eta_x^{-2s} +\eta_x^{1-2s}}{2s-1}\Big)^{\frac{1}{2-\alpha}}\epsilon^{\frac{2s-1}{2-\alpha}},$$
whereas (\ref{tre}) can be replaced with:
$ C(s)\cdot C\big(\frac{1}{|p_x|}, [\phi]_{\alpha}\big)\cdot
C(\eta_x) \epsilon^{\alpha\cdot\frac{2s-1}{2-\alpha}}.$
\item[(iii)] Finally, for $\phi$  Lipschitz on $\mathbb{R}^N\setminus \bar
  B_{\eta_x}$ with the Lipschitz constant $\mbox{Lip}_\phi$, we get:
$$\kappa_\epsilon \leq \frac{32 \; \mbox{Lip}_\phi}{|p_x|}\cdot
\frac{\eta_x^{-2s} +\eta_x^{1-2s}}{2s-1}\epsilon^{2s-1},
\quad A_\epsilon\leq \frac{32}{|p_x|}\cdot \max\Big\{\frac{2C_x\eta_x^{2-2s}}{1-s},
\frac{\mbox{Lip}_\phi (\eta_x^{-2s}+\eta_x^{1-2s})}{2s-1}\Big\}\epsilon^{2s-1}.$$
Indeed, both quantities in $A_\epsilon$ have $\epsilon^{2s-1}$-order.
The expression in (\ref{tre}) is then:
$ C(s)\cdot C\big(C_x,\frac{1}{|p_x|}, \mbox{Lip}_\phi\big)\cdot
C(\eta_x) \epsilon^{2s-1}$, whereas the order of the error
bounding quantity in Theorem \ref{jaszczu1} is
$C(s)\cdot(\epsilon^{4s-1}+\epsilon^2)$ as $\epsilon\to 0+$, and
$C(s)\to\infty$ as $s\to 1-$.
\end{enumerate}
\end{remark}

\section{A proof of Theorem \ref{jaszczu2}}\label{sec4}

We note the following refinement of the argument in the proof of Corollary \ref{cor4.5}:

\begin{proposition}\label{grazyna}
Let $\phi\in C^2(\bar B_{\eta_x})$ satisfy: $p_x\doteq \nabla
\phi(x)\neq 0$. Then, for every $\epsilon< \eta_x$ such that $\epsilon
|\nabla^2\phi(x)|\leq |p_x|$, there holds:
\begin{equation}\label{sette.5}
\begin{split}
\Big| c_s s\cdot \epsilon^{-2s} \cdot \frac{1}{2}\Big(\sup_{B_\epsilon(x)}&\;\phi +
 \inf_{B_\epsilon(x)}\phi -2\phi(x)\Big) - \int_0^\epsilonL_{\phi}\big(x,
t\frac{p_x}{|p_x|}, t\frac{p_x}{|p_x|}\big)\;\mathrm{d}\mu_s(t) \Big| \\ & \leq
c_s s\Big(2\epsilon^{3-2s}\frac{|\nabla^2\phi(x)|^2}{|p_x|}
+ \epsilon^{2-2s}\sup_{y\in B_\epsilon(x)}|\nabla^2\phi(y) - \nabla^2\phi(x)|\Big).
\end{split}
\end{equation}
\end{proposition}
\begin{proof}
A simple application of Taylor's expansion yields:
$$\big|L_{\phi}\big(x, t\frac{p_x}{|p_x|}, t\frac{p_x}{|p_x|}\big) - t^2\Delta_\infty\phi(x)\big|
\leq t^2 \sup_{y\in B_t}|\nabla^2\phi(y) - \nabla^2\phi(x)|,$$
where we recall that $\Delta_\infty\phi(x)=\big\langle
\nabla^2\phi(x): \frac{p_x}{|p_x|}\otimes \frac{p_x}{|p_x|}\big\rangle$.
Integrating the above $\int_0^\epsilon\;\mathrm{d}\mu_s(t)$, we get:
$$ \Big|\int_0^\epsilonL_{\phi}\big(x, t\frac{p_x}{|p_x|}, t\frac{p_x}{|p_x|}\big) \;\mathrm{d}\mu_s(t) -
\frac{C_s}{2(1-s)}\epsilon^{2-2s}\Delta_\infty\phi(x)\Big|\leq \frac{C_s}{2(1-s)}\epsilon^{2-2s}
\sup_{y\in B_\epsilon}|\nabla^2\phi(y) - \nabla^2\phi(x)|.$$
Recalling that (see for example \cite[section 3.2]{LewB}):
\begin{equation}\label{otto}
\Big|\big( \sup_{B_\epsilon} \phi + \inf_{B_\epsilon} \phi
-2\phi(x)\big) - \epsilon^2\Delta_\infty\phi(x)\Big|\leq 4\epsilon^3\frac{|\nabla^2\phi(x)|^2}{|p_x|}
+ \epsilon^2 \sup_{y\in B_\epsilon}|\nabla^2\phi(y) - \nabla^2\phi(x)|,
\end{equation}
and taking the linear combination of the two above formulas, the proof is done.
\end{proof}

\medskip

The proof of Theorem \ref{jaszczu2} follows directly by summing up
formulas (\ref{tre}), (\ref{sette.5}), and multiplying the result by the factor $\frac{(1-s)s}{C_s}\epsilon^{2s}$. Since:
$$\Big(\frac{(1-s)s}{C_s}\epsilon^{2s}\Big) \mathcal{L}_s^\epsilon[\phi] (x) +\Big(\frac{(1-s)s}{C_s}\epsilon^{2s}\Big)\cdot
\frac{C_s}{2(1-s)}\epsilon^{-2s}\big(\sup_{B_\epsilon}\phi
+\inf_{B_\epsilon}\phi - 2\phi(x)\big) = \mathcal{A}_\epsilon\phi(x) - \phi(x),$$
the error in the claimed expansion is the sum of errors
in (\ref{tre}) and (\ref{sette.5}), multiplied by $\frac{(1-s)s}{C_s}\epsilon^{2s}$.

\begin{remark}\label{piesio}
\begin{enumerate}[leftmargin=7mm]
\item[(i)] Analysis similar to Remark \ref{kura} allows for computing
  the order of the error term in Theorem \ref{jaszczu2} when $\phi$ is Lipschitz:
\begin{equation*}
\begin{split}
&C(s) \cdot C\big(C_x,\frac{1}{|p_x|}, \mbox{Lip}_\phi\big)\cdot C(\eta_x)
\epsilon^{4s-1} + C(s)\cdot C\big(|\nabla^2\phi(x)|,
\frac{1}{|p_x|}\big)\epsilon^3 +C(s)\cdot o(\epsilon^2).
\end{split}
\end{equation*}
As before, $C(s)$ depends only on $s$, and $C(\eta_x)$ only on
$\eta_x$, while the remaining constants depend on the displayed terms in a
nondecreasing manner. For $\phi\in C^{2,1}(B_{\eta_x})$,
the above quantity has order $\epsilon^{4s-1}+\epsilon^3$, which
equals $\epsilon^3$ at $s=1$.

\item[(ii)] For a more precise analysis of the asymptotic expansion when $s\to 1-$, note that:
\begin{equation*}
\begin{split}
& \kappa_\epsilon\leq \sup\Big\{ a;~ a\in [0,2] \; \mbox{ and } \; a^2\leq
 \frac{32\;\omega_\phi(a)}{|p_x|}\cdot \frac{
  \eta_x^{-2s} +\eta_x^{1-2s}}{2s-1}\epsilon^{2s-1}\Big\}, \\ &
\frac{16 C_x}{|p_x|}\cdot\frac{2s-1}{1-s}\cdot
\frac{\eta_x^{2-2s}-\epsilon^{2-2s}}{\epsilon^{1-2s} -{\eta_x}^{1-2s}}
\leq \frac{16 C_x}{|p_x|}\cdot 16\eta_x^{2-2s}|\ln\epsilon|\epsilon^{2s-1}.
\end{split}
\end{equation*}
The first bound above is valid when $\epsilon<\frac{\eta_x}{2}$, while
for the second bound we used:
$\eta_x^{2-2s}-\epsilon^{2-2s}\leq (2-2s) (\ln\eta_x - \ln\epsilon)
\eta_x^{2-2s}\leq 4(1-s)|\ln\epsilon|\eta_x^{2-2s}$, when $\epsilon<e^{-|\ln \eta_x|}$.
Consequently, $A_\epsilon\leq o(1)$ as $\epsilon\to 0+$, uniformly in
$s\in (\frac{1}{2}+\delta,1)$. For each fixed $\epsilon$, the bound in
Theorem \ref{jaszczu2} converges to (consistently with (\ref{otto}) as $s\to 1-$):
$$2\epsilon^3\frac{|\nabla^2\phi(x)|^2}{|p_x|}
+ \epsilon^2\sup_{y\in  B_\epsilon}\big|\nabla^2\phi(y)-\nabla^2\phi(x)\big|.$$
We also observe that when $\phi$ is Lipschitz on $\mathbb{R}^N\setminus \bar B_{\eta_x}$, the
said bound becomes:
\begin{equation*}
\begin{split}
& (1-s) \epsilon^{4s-1}|\ln\epsilon|^2\cdot
\frac{2^9 C_x}{|p_x|} \;  \cdot\Big(16s C_x\eta_x^{4-4s} +
\mbox{Lip}_\phi\cdot \big(\frac{\eta_x^{2-4s}}{2}+\frac{s\eta_x^{3-4s}}{2s-1}\big)|\ln\epsilon|^{-1}\Big)
\\ & + s\Big(2\epsilon^3\frac{|\nabla^2\phi(x)|^2}{|p_x|}
+ \epsilon^2\sup_{y\in B_\epsilon}\big|\nabla^2\phi(y)-\nabla^2\phi(x)\big|\Big).
\end{split}
\end{equation*}
\end{enumerate}
\end{remark}

\section{Further remarks}\label{sec5}

\subsection{Spherical prisms as integration domains}\label{sec52}
With an eye towards future applications, we now consider another averaging operator:
\begin{equation}\label{MVP3}
\bar\mathcal{A}_\epsilon^o \phi(x) = \frac{1}{2}\Big(\sup_{|y|=1}\fint_{T^{\epsilon, R, \alpha}(y)}
\phi(x+z)\;\mathrm{d}\mu_s^N(z) + \inf_{|y|=1}\fint_{T^{\epsilon, R, \alpha}(y)}
\phi(x+z)\;\mathrm{d}\mu_s^N(z) \Big).
\end{equation}
Above, the integration is taken with respect to the measure $\mu_s^N$ on the Borel subsets of $\mathbb{R}^N$:
$$\mathrm{d}\mu_s^N(z)  \doteq \frac{C(N,s)}{|z|^{N+2s}}\;\mathrm{d}z \quad \mbox{ where }\;
C(N,s)=\frac{4^s s\Gamma\big(\frac{N}{2}+s\big)}{\pi^{N/2}\Gamma\big(1-s\big)} =
\Big(\int_{\mathbb{R}^N}\frac{1-\cos \langle z, e_1\rangle}{|z|^{N+2s}}\;\mathrm{d}z\Big)^{-1} .$$
Clearly, $C(1,s)=C_s$ and $\mu_s^1=\mu_s$. The integration domain
$T^{\epsilon, R, \alpha}(y)$ is the following regular spherical prism in $\mathbb{R}^N$,
oriented in the direction $y\in\mathbb{R}^N\setminus\{0\}$:
$$T^{\epsilon, R, \alpha}(y) = \Big\{ z\in\mathbb{R}^N; ~ \sin \frac{\angle
  (y,z)}{2}<\alpha, ~~ \langle y,z\rangle>0 \;\mbox{ and } \; \epsilon < |z|<R\Big\}.$$
With the above notation, $T^{0, \infty, \alpha}(y)$ is an infinite
cone, and we observe that such cones were used in the definition of
the fractional $p$-Laplacian $\Delta_p^s$ in \cite{BCF2}, with
$p=p(\alpha, N, s)$. We have:

\begin{lemma}\label{lem8}
Assume (\ref{H1}). Then, for every $\epsilon<\eta_x$, $R>\max\{\eta_x, 1\}$
and $\alpha<\frac{1}{2}$, there holds:
\begin{equation*}
\begin{split}
&\sup_{|y|=1} \Big|\fint_{T^{\epsilon, R, \alpha}(y)} \phi(x+z)\;\mathrm{d}\mu_s^N(z) - \fint_\epsilon^\infty
\phi(x+ty)\;\mathrm{d}\mu_s(t) \Big| \\ & \qquad \qquad \leq 2 \big|\frac{\epsilon}{R}\big|^{2s} \cdot \|\phi\|_{L^\infty}
+ \max\Big\{2\big(|p_x| + 2C_x\eta_x \big)\eta_x\cdot \alpha,\;
3R\cdot \omega_\phi(\alpha)\Big\}.
\end{split}
\end{equation*}
\end{lemma}

\begin{proof}
We first estimate the difference:
\begin{equation*}
\begin{split}
\Big|\fint_\epsilon^\infty& \phi(x+ty)\;\mathrm{d}\mu_s(t) - \fint_\epsilon^R
\phi(x+ty)\;\mathrm{d}\mu_s(t) \Big| \\ & \leq \frac{1}{\mu_s(\epsilon, \infty)} \int^\infty_{R} |\phi(x+ty)|\;\mathrm{d}\mu_s(t)
+ \Big|\frac{1}{\mu_s(\epsilon, \infty)} - \frac{1}{\mu_s(\epsilon,
  R)}\Big|  \int_\epsilon^{R} |\phi(x+ty)|\;\mathrm{d}\mu_s(t) \\ &
\leq 2 \big|\frac{\epsilon}{R}\big|^{2s} \cdot \|\phi\|_{L^\infty}.
\end{split}
\end{equation*}
Next, observe that:
\begin{equation*}
\begin{split}
\int_{T^{\epsilon, R, \alpha}(y)} \phi(x+|z|y)\;\mathrm{d}\mu_s^N(z) & =
\int_\epsilon^R\phi(x+ty) t^{N-1} \cdot\mbox{area}\big(\big\{|z|=1,~
z\in T^{\epsilon, R, \alpha}\big\}\big) \frac{\mbox{d}t}{t^{N+2s}}
\\ & = \mbox{area}\big(\big\{|z|=1,~
z\in T^{\epsilon, R, \alpha}\big\}\big) \cdot \int_{\epsilon}^R\phi(x+ty)\;\mathrm{d}\mu_s(t)
\end{split}
\end{equation*}
which implies:
$\displaystyle \fint_{T^{\epsilon, R, \alpha}(y)} \phi(x+|z|y)\;\mathrm{d}\mu_s^N(z) = \fint_{\epsilon}^R\phi(x+ty)\;\mathrm{d}\mu_s(t).$
It remains to bound:
\begin{equation*}
\begin{split}
\fint_{T^{\epsilon, R, \alpha}(y)} & |\phi(x+z) -\phi(x+|z|y)|\;\mathrm{d}\mu_s^N(z) \leq
\sup_{z\in T^{\epsilon, R, \alpha}(y)}  |\phi(x+z) -\phi(x+|z|y)| \\ & \leq
\max\Big\{\|\nabla \phi\|_{L^\infty(B_{\eta_x})} \cdot
2\eta_x\cdot\alpha,\; \omega_\phi(2 R\alpha)\Big\} \\ &
\leq \max\Big\{2\big(|p_x| + 2C_x\eta_x \big)\eta_x\cdot \alpha,\;
(1+2R)\cdot \omega_\phi(\alpha)\Big\}.
\end{split}
\end{equation*}
This yields the desired estimate and ends the proof.
\end{proof}

From Lemma \ref{lem8}, Remark \ref{kura} and Theorem \ref{jaszczu1}, we directly deduce:

\begin{corollary}\label{cor9}
Assume (\ref{H1}) with $\eta_x\leq 1$, and that $\phi$ is
Lipschitz on $\mathbb{R}^N\setminus \bar B_{\eta_x}$ with Lipschitz constant
$\mbox{Lip}_\phi$.  For every $\epsilon\ll \eta_x$,  we set
$R=\epsilon^{\frac{1}{2s}-1}$ and $\alpha=\epsilon^{4s-\frac{1}{2s}}$.
Then there holds:
\begin{equation*
\begin{split}
\Big|\bar\mathcal{A}_\epsilon^o\phi(x) & - \phi(x) - \frac{1}{c_s(1-s)}\epsilon^{2s}\mathcal{L}_s[\phi] (x)\Big|
 \leq  \epsilon^{4s-1}\big(2\|\phi\|_{L^\infty} +
3\mbox{Lip}_\phi\big) + \frac{s}{1-s}\cdot 2C_x\epsilon^2
\\ & \qquad\qquad + \left\{\begin{array}{ll} {\displaystyle
 \frac{32}{|p_x|}\;\epsilon^{4s-1}\Big(\frac{8s}{1-s}
 + \big(\eta_x^{-2s}+\frac{2s}{2s-1}\eta_x^{1-2s}\big){\mbox{Lip}_\phi}
\Big)\;\cdot } & \vspace{1mm}\\ \qquad\quad\ {\displaystyle \cdot
\max\Big\{\frac{2C_x}{1-s}, \frac{(\eta_x^{-2s}+\eta_x^{1-2s})}{2s-1}\mbox{Lip}_\phi\Big\}}
& \mbox{ when } p_x\neq 0 \vspace{1.5mm} \\ 0 & \mbox{ when } p_x= 0.
\end{array}\right.
\end{split}
\end{equation*}
\end{corollary}

\begin{remark}\label{discrete}
Towards the applications in the numerical approximating of solutions to the nonlocal
Dirichlet problem for the operator $\Delta_\infty^s$, one has to consider a
discrete version of the result in Theorem \ref{jaszczu1}.
To this end, let $\{\theta_i\}_{i=1}^n$ be an equidistributed spherical grid on
$\{|z|=1\}\subset\mathbb{R}^N$; when $N=2$ then $\theta_i=e^{2\pi i/n}$.
Next, for all $x_k$ in the cubical grid $h \mathbb{Z}^N$ define:
$$\bar\mathcal{A}_\epsilon^d \phi(x_k) \doteq \frac{s h^N}{|S_\alpha|\big(\epsilon^{-2s}-R^{-2s}\big)}
\cdot \big(\max_{i=1\ldots n} + \min_{i=1\ldots n} \big)
\sum_{x_j\in  T^{\epsilon, R,\alpha}(\theta_i)\cap h\mathbb{Z}^N}\frac{\phi(x_{k}+x_j)}{|x_j|^{N+2s}},$$
where we used that $\mu_s^N(T^{\epsilon, R,\alpha}) = C(N,s) |S_\alpha|\cdot\frac{\epsilon^{-2s}-R^{-2s}}{2s}$,
with $S_\alpha \doteq T^{0,\infty,\alpha}\cap \{|z|=1\}.$

It is clear that for $h$ and $n$ scaling in $\epsilon$ with
sufficiently high positive and negative powers, respectively, the averaging operator $\bar\mathcal{A}_\epsilon^d$
is a discrete approximation of $\mathcal{A}_\epsilon^o$ at the same rate of the error proved in Corollary \ref{cor9}.
The details of this construction as well as its implementation for a
numerical scheme, are left for the future work.
\end{remark}

\subsection{The viscosity framework}\label{sec51}
We observe that our results may be reformulated in the viscosity
setting, which has been used in the results of\cite{MPR10} for the (local) $\infty$-Laplacian.
The definition of viscosity solutions for the fractional $\infty$-Laplacian
as in \cite[Definition 2.3]{BCF} encodes the hypothesis
(\ref{H1}) which needs to be satisfied by the test functions $\phi$. Following
this lead, one can consider the asymptotic expansions in the viscosity
sense. From now on, the respective averages in
(\ref{MVP1}), (\ref{MVP2}) and (\ref{MVP3}), are generically denoted
by $Average_\epsilon$, with corresponding constants $K>0$ such that:
$$ \frac{K}{\epsilon^{2s}}\big(\emph{Average}_\epsilon \phi-\phi\big)=
\Delta_\infty^s\phi + o(1) \qquad \textup{as $\;\epsilon\to 0+$.} $$

\begin{definition} \label{def:ViscMVP}
Let $\Omega\subset \mathbb{R}^N$ be open and let $f:\Omega\to\mathbb{R}$. A bounded
upper (resp. lower) semicontinuous function $u:\mathbb{R}^N\to\mathbb{R}$ is a viscosity sub-solution (resp. super-solution) of:
\begin{equation}\label{vs}
\frac{K}{\epsilon^{2s}}\big(\emph{Average}_\epsilon u-u\big)=f +
o(1) \quad \text{in $\Omega\qquad $ as $\;\epsilon\to 0+$},
\end{equation}
provided that the following holds.  For every $x\in \Omega$, $r>0$, and $\psi\in C^2(\bar{B}_r(x))$ such that:
\[
\psi(x)=u(x) \quad \text{and} \quad \psi(y)>u(y) \quad
\text{(resp. $\psi(y)<u(y)$)} \quad \text{for all $y\in
  \bar{B}_{r}(x)\setminus\{x\}$,}
\]
we have:
\[
\frac{K}{\epsilon^{2s}}\big(\emph{Average}_\epsilon \phi(x)-\phi(x)\big)\geq
f(x)+ o(1)\quad
\text{\Big(resp. $\frac{K}{\epsilon^{2s}}\big(\emph{Average}_\epsilon \phi(x)-\phi(x)\big)\leq f(x) + o(1)$\Big)},
\]
where $\phi \doteq \mathds{1}_{\bar B_r(x)}\psi + \mathds{1}_{\mathbb{R}^N\setminus \bar B_r(x)} u$.
When $u$ is both a viscosity sub- and super-solution, it is
a  viscosity solution of (\ref{vs}) (i.e. it satisfies the asymptotic expansion in the viscosity sense).
\end{definition}

The following result follows then trivially from either of Theorems
\ref{jaszczu1}, \ref{jaszczu2} or Corollary \ref{cor9}:

\begin{theorem}\label{thm:equivalenceviscosity1}
Let $\Omega\subset \mathbb{R}^N$ be open, $u:\mathbb{R}^N\to\mathbb{R}$ be bounded and uniformly
continuous, and $f:\Omega\to\mathbb{R}$. Then the following are equivalent:
\begin{enumerate}[leftmargin=7mm]
\item[(i)] $u$ is a viscosity solution of: $\Delta_\infty^s u=f$ in $\Omega$.\vspace{1mm}
\item[(ii)] $u$ satisfies: $\frac{K}{\epsilon^{2s}}\big(\text{Average}_\epsilon u-u\big)=f+
  o(1)$ in $\Omega$ as $\epsilon\to0+$, in the viscosity sense.
\end{enumerate}
\end{theorem}
We refer the reader to \cite{BS2} for similar statements in the
context of other averages for $\Delta_\infty^s$.

\section*{Acknowledgements}
This research started when M. Lewicka visited Norwegian University of
Science and Technology. Part of this research was also done while
F. del Teso and J. Endal visited the University of Pittsburgh. We want
to thank both institutions for their hospitality.

redpajama_set_name: RedPajamaArXiv


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Alright, so I've encountered a bit of an issue while playing through the first episode. Whenever Max tries to interact with something, the prompt/interaction menu appears and then quickly disappears. This first happened when trying to interact with the sink. I managed to finally interact with it after nearly half an hour of attempting too. I was hoping the glitch would end there, but subsequent prompts are having the same glitch. Does anyone have a fix for this? Other threads have mentioned turning off Anti-Aliasing in my Graphics card control panel, but my current card doesn't seem to allow me to adjust this. Any help would be greatly appreciated! Thank you!
Can you tell me what graphics card your PC is using?
Hey, I wanted to say I have the same issue but with episode 3. I have to approach every object at a weird angle (sometimes even looking away from it) to get the little thing to pop up. Sometimes they pop up, but keep flicking on and off and I don't have enough time to move my cursor. This appeared before, but eventually I could interact with everything normally. Now, I'm stuck because I need to collect the items to get into the principal's office and I can't pick up the sugar or get into Jefferson's class because the things either don't pop up, or pop up and immediately disappear. I have an Nvidia GeForce G210. I will be grateful for any help.


text:
One Million Tiny Plays About London, No. 53
by Craig Taylor
[Be sure to read Plays No. 49, 50, 51, and 52, Craig Taylor's other tiny plays about London.]
(A rehearsal room in Chiswick. Tom sits behind the drum kit. Neil sits on his bass amplifier. Rob has put his guitar in its case. They each hold a photocopied sheet of paper. The black light in the corner is still turned on.)
Rob: Okay, so that means we're down to four. Or three. Three names. So being a democracy we'll vote, but I think I should personally explain one of the band names so you guys can get a sense of it.
Neil: Okay.
Rob: Listen closely. The band would be called Merlin's Wizard.
Tom: Huh.
Rob: One more time. Merlin's. Wizard. Think of it this way: Merlin is… what?
Tom: I don't know.
Neil: A wizard.
Rob: That's right. Merlin is the most powerful wizard ever.
Rob: So… and here's the point. Who controls Merlin?
(Pause.)
Neil: Gandalf?
Rob: No, it's rhetorical. That's the question we're asking.
Neil: When?
Rob: With the name. We're posing a question with the name. If we call the band Merlin's Wizard people will be like, "Oh yeah, I never thought of that. Who controls Merlin?"
Tom: But it's not Gandalf.
Rob: No, it's not anyone we know. That's the point. It's the idea of questioning authority. It's meant like: Who controls the people who control us? It's like if we were called Tony Blair's Prime Minister.
Neil: I don't like that.
Rob: No, that's not an actual option. It's just the idea.
Neil: Personally I like name number two. The Apples. Because you have different kinds of apples. Of knowledge, for instance.
Tom: Golden Delicious is another.
Rob: I thought that was on there as a joke.
Neil: You could have apple pie, so it's ironic.
Rob: That name is retarded. That's the stupidest on the list.
Neil: Or you've got the Apple computer company.
Tom: Yeah that's true.
Neil: Rob, if you're so set on having a name that will comment on something in society, it could comment on how Apple computers are, I don't know, how everyone has one, or something.
Rob: No.
Neil: The Apples is good, it's really good.
Tom: "They Are The Apples of Our Eye" they would write in NME.
Rob: Okay, first of all, The Apples is a piece of shit. Second of all, it goes against everything my playing stands for.
Tom: What does your playing stand for?
Rob: Not a piece of fruit.
Neil: Since when does your playing stand for something?
Rob: Since maybe the first note I ever played, Neil? I just don't see us realistically saying "Hi, we're The Apples." I see Merlin's Wizard realistically. Besides, what about our songs? What about "Orc Attack"? It fits with Merlin's Wizard. Or are you just going to totally change it?
Neil: It's an instrumental Rob.
Rob: Yeah, but the feel of it?
Neil: What's the feel of it?
Rob: Well, uh, I don't know if anyone here remembers, but it was meant to be played along with the visuals we filmed on Hampstead Heath.
Tom: When I had to dress up like a goblin?
Rob: Like an orc, Tom. Orc Attack.
Tom: With your Mom's dress on?
Rob: It's called a kaftan, Tom. But yeah, unless we shot other footage of some other orc attack at some point, then that was it.
Tom: That was such a screwed-up day.
Rob: It was a good day for the band. And we said we were going to use that footage in the live show so it would be like a multimedia. We talked about that.
Neil: You talked about that, Rob.
Rob: No, Neil, we all agreed in our democracy that we would play "Orc Attack" with the orc footage. Now that's not gonna make much sense if we were called The Apples, is it?
Neil: Tom's orc mask fell off halfway through. You can see his face in most of the footage.
Tom: Yeah, it doesn't look very good, Rob.
Rob: Well, Tom, that's funny because I think it does look very good. It looks exactly like, you know… If an orc attacked then what would it look like? Confusion, which is what the video looks like, which is the feel of the song, which is why I wrote it…
Neil: We all wrote it.
Rob: Which is why I wrote the melody then.
Tom: Remember when the mask came off my head while we were filming? That was so screwed up, when I ran into the tree?
Rob: The audience isn't going to see that
Tom: They'd be pretty dumb to miss me yelling, "Fuck, is that a tree?" and then getting corked by it.
Rob: Then maybe we won't put that part in.
Tom: That was wild. Everybody yelling, "Tree!"
Neil: Rob, I think the point is that we don't really want to go along with all of this Lord of the Rings stuff. I think that's what our democracy is trying to say.
Rob: Funny that, Neil, because Merlin isn't even in Lord of the Rings.
Neil: Well who's supposed to know that?
Rob: Oh, I don't know, Neil. Maybe the sixty billion people who saw the movie and read the books? I'm not proposing we call ourselves Gandalf's Wizard.
Neil: What's the difference?
Rob: Then what's the difference if we're called The Apples or The Pears?
Neil: I don't like The Pears
Rob: I'm making an example.
Tom: NME would write "It's All Gone Pear-Shaped for The Pears."
Neil: The apple is still, in a lot of ways, the forbidden fruit, Rob.
Tom: And they rot away. So there's good putrefaction.
Rob: No. I'm always going to say no to The Apples. People are going to expect to see big pieces of fruit on stage.
Tom: The other one makes them expect to see wizards.
Rob: Merlin's Wizard is posing a question. About society. Who controls Merlin?
Neil: Maybe we should vote on the name.
Rob: Okay, then maybe I'll make my vote by putting down my guitar.
Tom: You already put down your guitar, Rob.
Rob: No, I mean, by not picking it up again.
Rob: By quitting. Okay? I'm going to quit. Because I'd rather not be in a band than be in a band called something stupid.
Neil: Like Merlin's Wizard.
Rob: Yeah, Neil, that's really hilarious. And for your information, I own that name, so if I go then it comes with me.
Rob: And even though Tom might be in it, I own that "Orc Attack" footage, too. It's copyright.
Rob: Do you still want to vote then?
Neil: I vote for The Apples.
Tom: I vote for The Apples, too
Rob: Okay, well I vote for Merlin's Wizard. So where does that leave us?
Vocabulary Words We Learned by Playing Dungeons & Dragons
Actual Band Names That Double As Excuses for Showing up Late to Work
by Eric Elbogen
E Mails Sent Between Members Of New Edition Regarding Their Reunion Tour
by Stephen Scaia
Less Successful Heavy-Metal Wedding Tribute Bands
by Brian Shields
I'm Sick of Being Blackmailed Into Playing Bass for Your Band
by Keaton Patti

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: css, last-child tr overwrite style I have this styles:
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  padding: 8px 0 !important;
  height: 48px;
}
.ant-table-tbody > tr:last-child > td {
  padding: 8px 0 !important;
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}

The last definition is almost the sam except that I'm setting `border-bottom: none``
How could I write that in one single class?
I tried something like:
.ant-table-tbody > tr > td {
  padding: 8px 0 !important;
  height: 48px;
  &.tr:last-child {
    border-bottom: 1px red solid;
  }
}

but it's not working

A: You need to just correct the format
.ant-table-tbody {
  > tr > td {
    height: 48px;
    padding: 8px 0 !important;
  }

  > tr:last-child > td {
    border-bottom: 1px red solid;
  }
}

redpajama_set_name: RedPajamaStackExchange


I have been celebrating the birth of Bitcoin by going through Satoshi's writings.
As we all know, the Bitcoin whitepaper was published on October 31, 2008 – the Halloween… Although we do not know whether this was intentional, we do know that Halloween is the time for trickery, but it is also associated with rebirth or the birth of something new. In old Britain, the New Year was celebrated on November 1.
Celebrations like Halloween fall in line with the old European Carnival Culture – where you wear a costume and a mask to cover your true identity, to appear like someone else. And as Hal Finney said: "I have always loved crypto, the mystery and the paradox of it". Thus, it's not unreasonable to assume that the publication date was chosen intentionally.
In a few of his posts and correspondences, Satoshi discusses some of the possible use cases for the Bitcoin network. Most of these comments were made back in 2009.
It can already be used for pay-to-send e-mail.
There are a few more cases spread throughout his writings.
To the dismay of the "bitcoin maximalists", Satoshi Nakamoto shows the path for many ICOs that will be exploring these use cases years later. Although it is obvious that Satoshi is referring to the Bitcoin Network, not to the plethora of new blockchain networks and DApps; nevertheless, it's hard to argue with the fact that the Bitcoin2018 is not well suited for most of these cases for various reasons; without getting deep into this topic, I'll mention just a few obvious ones: speed/network congestion, cost (Satoshi himself notices on August 4, 2010: "Bitcoin isn't currently practical for very small micropayments. Not for things like pay per search or per page view", and security (most of the bitcoin holders would not want to expose their bitcoin wallets for trivial transactions).
However, the last case really caught my eye – the email case. I was way too familiar with this case although I had never read Satoshi's post before. Immediately, 21.co came to mind: a Silicon Valley startup which urges us to "Earn bitcoin by replying to emails". In the meantime, it has changed its name and now it's called Earn.com. It raised $5.1 million in Series A back in 2013 (under the name of 21E6, a bitcoin mining company) and Winklevoss Capital was amongst the investors. However, according to Balaji S. Srinivasan, by 2015 the company, finding itself in the aftermath of the 2014 Bitcoin crush, had a huge hole in its budget. This didn't prevent a $116 million round B (the investor list represented the who of who of the traditional funds and crypto funds: Andreessen Horowitz, Pantera Capital, Khosla Ventures).
As of today, Earn.com is a fast-growing, cash-flow positive business with a multimillion dollar revenue run rate. Hundreds of thousands of users have signed up to earn free cryptocurrency by answering emails and completing tasks. And thousands of senders have used Earn.com to send out paid emails and surveys to large audiences. I think it's fair to say that it's one of the first truly useful blockchain-based applications.
In April of 2018, it was acquired by Coinbase. The deal details have not been disclosed, but it's been reported that Coinbase had to pay in excess of $120 million (the amount that had been invested in prior rounds). According to the same source, the company is planning to replace bitcoin with its own ERC20 token.
I tweeted to Ben Horowitz – no reply.
Subsequently, I had a post on LinkedIn on the same topic tagging all the key people. – Still, silence.
The only response I received was from Adam Back, who liked my tweet. Adam Back is featured prominently in the Earn.com community (he is in the third row).
There is nothing wrong with borrowing, on the other hand, the idea appropriation is less attractive. Can this silence be explained by the concern that Satoshi Nakamoto will take Earn.com to court to fight for royalties? This would seem highly unlikely for the man or men who decided to disappear and who leaves his approximately one million bitcoins untouched. Additionally, generally speaking, ideas are not patentable. A more probable explanation would be given by ignorance. Perhaps that no one in the Earn.com team was aware of this fact. If this is the case, then it's not too late to correct it. And now that the company is owned by the Coinbase, this would seem even more appropriate. This could be a powerful PR move. Otherwise, all the parties involved are in danger of facing the wrath of the Crypto Gods.


Apr 7, 2013 • Michael Shattuck • Series: One Story. One Hero. One Mission.
The story begins with a huge number of Jewish people living in slavery in Egypt, and it ends with God's people walking across the bottom of a vast sea on dry ground. In between, there are warnings, plagues, a great feast, and the death of a significant number of Egyptian children. The story of the Exodus and Passover in Exodus is the key event in the Old Testament. God acted on behalf of His people, He keeps His promises, and displays His own greatness to them and the rest of this. This morning we are in for a special treat as Michael Shattuck preaches and shows us the great God of the Exodus story.


I absolutely adore this home – just about every inch of it. It's newly built, located in El Ampurdán, Spain, and is the home of a couple with three children. Priorities for the owners were the landscape and natural light having a strong presence, and creating a spacious and functional home…all of which were achieved beautifully. There is just oo many things I love about this home to list, so instead I'll just say that I don't think I would be interested in having a concrete bed frame, the tub in the floor looks like a health hazard, and I'd lose the tab curtains in the dressing room. (Have I ever mentioned by hatred of tab curtains? They make me want to vomit. Go with grommets!) Via Micasa.


Tanzania charged five employees of Vodacom Group's local unit, including MD Hisham Hendi, with allegedly depriving the government of more than 11 billion shillings (R68-million) in revenue.
Prosecutors accused the staff at Vodacom Tanzania of organising a "criminal racket", according to documents filed to a court in the commercial capital, Dar es Salaam. The group were also charged with allowing third parties unauthorised use of network services.
Vodacom Tanzania declined to comment. Hendi couldn't immediately be reached for comment. Vodacom also has operations in Democratic Republic of Congo and Mozambique and is the market leader in South Africa.
The charges come two years after Johannesburg-based Vodacom listed its Tanzanian unit on the local stock exchange following pressure from the government to increase domestic ownership. The run-in is the latest between President John Magufuli's administration and international companies, with Bharti Airtel agreeing to cede a portion of its local unit to the state in January following claims the operator acquired its initial shareholding illegally.
Vodacom's crosstown rival MTN Group, Africa's biggest carrier by sales, is also no stranger to sanctions by governments and regulators around the continent. The company is facing a US$2-billion tax bill in Nigeria, its biggest market, while the head of its Uganda unit was deported earlier this year.


DOVER, DE – When Tone EZ was a young boy, he used to go to concerts with his step-father. Because he was so little, he could never see past the crowd to the artists on stage, so his step-father would always him up so he could see. It was in those special moments with his step-father that he knew that he wanted to be a musician. And today, Tone EZ is realizing that dream by becoming one of the hottest new hip-hop artists in the country.
His latest single – "Call Me," featuring Omelly and the Dream Chasers – has just been released and is already getting rave reviews. The song is about the having a good time with friends, and how much joy you feel when someone calls you to go do something fun.
Over the past 15 years, Tone EZ has been perfecting his art so that today, when it comes to painting pictures through lyrics or giving meaningful purpose to songs, he consciously gives his fans timeless and tangible music to hold on to. He credits his love of music as being a way to take himself out of the madness around him at times – having grown up in the streets of St. Louis.
He credits artists such as Tupac, Jay Z, OutKast and TI for helping inspire him. Using those artists as a template, he said he began to master his sound while learning how to tell stories and share truth through his lyrics.
He said his new single "Call Me" is a great representation of his journey as a musician and his unique sound. It's available for purchase on all digital media download sites. Fans who want to find out more about Tone EZ, or who are interested in seeing him live, can follow him on social media on Twitter, Snapchat and Instagram under the handle @toneeztc.


The element, named after the planet Uranus, was reported by the German chemist Martin Heinrich Klaproth to a meeting of the Royal Prussian Academy of Science in 1789. In fact Klaproth had discovered the oxide of uranium and for a long time it was regarded as a "half metal". The element itself was not isolated until 1841 when the French Chemist, Eugene Peligoit, prepared it by reducing uranium tetrachloride with an alkali metal. It was not until 1870 that it was realized Uranium was the heaviest naturally occurring metal. (Technically this is probably not true as a natural uranium nuclear reactor is believed to have operated at Oklo in Gabon millions of years ago and if that were the case then traces of plutonium would have been formed). Back in the 1800s the fact that the metal itself had not been extracted did not deter the use of the oxide for industrial purposes, the principle of which was probably for the coloring of glass. We should appreciate that technology, as we know it today, was in its very early stages.
Klaproth had separated the oxide of uranium from pitchblende, a heavy black mineral that is essentially U3O8 but which also contained other elements such as iron, bismuth, lead and radium. At this time chemistry was just emerging from the alchemy era. For example some scientists still held to the Phlogiston Theory of combustion. It was Lavoisier, in the 1775-80 period, who showed that the process of combustion was not due to the loss of Phlogiston but rather the combination of the material with oxygen. In 1784 Cavendish's experiments showed that the product of burning hydrogen in oxygen was water. John Dalton's Atomic Theory, used to explain the combination of reacting substances in set proportions, was not published until early in the nineteenth century. Although Avogadro put forward his hypothesis in 1811 it was not until 1858 that Cannizzaro showed how it could be used to construct a list of atomic weights. In 1896, Becquerel discovered the natural radioactivity of uranium, by which time it had been used in glass for well over fifty years, and 1913 before Hans Geiger invented the nuclear particle counter.
Against this background we must see the chemistry of glass, certainly until the latter part of the nineteenth century, as being very hit or miss and little understood. Surely there would only be the most vague appreciation that silica was really silicon dioxide and that this was the main constituent of quartz, flint and sand. That glass was a composition of sodium, potassium, calcium and or lead silicates. That these could "dissolve" other elements such as iron, gold, copper, uranium and so on to produce colors and that the acidity of the melt or the oxidation state of the element would determine the shade of color. The glass mixes had to be found by trial and error rather than chemical theory, and once found they would be closely guarded secrets. Analysis of a rival's product would have been exceedingly difficult and well beyond the capacity of the glasshouses of the day. In this scenario it seems likely that different manufacturers developed their products in parallel. The ordinary glass worker would not know the composition of the mixes, these would be securely held by the foreman or manager. Thus only the defection of such a key worker to a rival would lead to the compromise of formulae. It therefore seems likely that the use of uranium to color glass developed simultaneously at different places and that no one can justifiably claim credit for "inventing" the process.
When we study the old surviving recipe books we see that, unlike modern times, different and ill defined terms are used for the raw material. For example in the Whitefriars Batch Book of 1832 on the same page we see the terms "Saltpetre" and "Nitre" used in adjacent recipes when they were in fact the same chemical, i.e. potassium nitrate. Why use different names for the same item? Was it because they came from different suppliers? Yet again we see "lead" used in some recipes, "litharge", "red lead", and even "lead or litharge". Did they really add lead metal, or was lead used as a generic term, did they understand the difference between red lead and litharge? Litharge (PbO) contains about 93% PB (lead) by wt. Red lead (Pb3O4), the commercial variety of which may contain up to 35% of PbO2, could have as little as 90% Pb by wt.
Ther terms "uranium" and "uranium oxide" as used in these early recipes, also have ambiguity. It was common practice to refer to other compounds by "shorthand", thus "arsenic" almost certainly would mean arseniuos oxide and "manganese", "manganese dioxide". In the light of this, and the difficulty in preparing uranium metal, we assume that recipes that simply use the word uranium really refer to its oxide, but which one?
Thank you to Barrie Skelcher for letting us share your information from The Big Book of Vaseline Glass.


Exelent Stock condition. Extra Low mileage (646 miles), Fair price! FINANCING AVAILABLE! CONSIGNMENT AND TRADES ARE WELCOME! Harley certainly loves historical references, and that comes naturally since they helped to define those eras. The factory picked the 1950s as the targeted period for the FLS SoftailSlim, specifically the minimalist "bobber" niche known for trimming off anything that didn't directly contribute to performance.Bobber enthusiasts back in the day were fixated on one thing only — the need for speed — and Harley was not remiss in this regard. The factory keeps sticking more powerful engines into each successive model, so this ride is not meant for the all-show/no-go category. Let's take a look and see how the details pan out.


text:
<?php
defined('BASEPATH') OR exit('No direct script access allowed');

class India extends CI_Controller
{

	function __construct()
	{
		parent::__construct();
	}

	/**
	 * Homepage for our app
	 */
	public function index()
	{
		// what will this reference?
		$source = '../public/assets/images/logo.png';
		// note that we could have referenced an image anywhere on our system

		// set the mime type for that image (jpeg, png, etc)
		header("Content-type: image/png");
		header('Content-Disposition: inline');
		readfile($source); // dish it
	}

}

redpajama_set_name: RedPajamaGithub


text:
Worldwide Facebook, WhatsApp and Instagram Outage - What Caused It?	Overall rating: ★★★★★	5	based on 23	reviews
BlogSimplio Labs Blog
Once we accept our limits, we go beyond them.
Worldwide Facebook, WhatsApp and Instagram Outage – What Caused It?
11 July,2019 What's New	No Comment Author: Marie
Image Source: 9to5mac.com
The recent outage effecting Facebook and other social media platforms such as WhatsApp and Instagram has definitely caused a worldwide stir among its users. Well, nothing to worry as the following content will help you to understand the cause of this Facebook, WhatsApp and Instagram outage.
Facebook's Recent Outage Shows How the Platform's AI Tags Photos
First of all, let's start with the explanation of what happens during an outage situation. Consider a situation where you decide to browse through your old uploaded photos on Facebook, unfortunately instead of seeing the actual photos or snaps from the various occasions, holidays, gatherings etc., you will be shown or prompted with texts like- "image may contain: cat", "image may contain: people dancing, drinking" etc. Well, let me tell you this is how the computer interprets your uploaded photos. This is how Facebook's Artificial Intelligence judges you.
Similar tags are showing up on Instagram as well. The only difference is that on Instagram there are details of the photo and it also prompts suggestion on who is in the concerned photo. This suggestion is based on Facebook's facial recognition. Facebook has been doing this process of using machine learning to read images in the same manner since April 2016. In fact, this project is a huge part of company's accessibility efforts, as well as the tags, are used to describe photos or videos to users who are visually impaired.
As per DownDetector, a website which monitors website and app issues, various users all around the world reports of the outage incident, but it was Europe and North America where maximum cases were reported. No one was spared, from individuals, businesses to organizations working locally or internationally.
A similar situation was experienced earlier this year when an outage lasted for as long as 24 hours. It was believed to be the worst ever outage for Facebook, which connected about 2.7 million people worldwide with its core social network, message applications and Instagram as well.
The biggest question which arises among its users who have to face this outage is that whether Facebook uses this information to target ads or there is some other reason behind it, for example, the one quoted by Facebook during these outages- that "it occurred due to an error triggered during the routine maintenance operation." Whatever may be the reason, there is no denying of the fact that there is a lot of personal data contained in our photos that are uploaded on various social networking sites, and it can be extracted easily by AI this way.
Is there another agenda behind this outage?
Obviously, there are various details of one's personal life which he/she would want to protect or shield from Facebook and other social networking platforms (which are often contained in the images, videos, posts present on timeline or user account). This information can range from hobbies to user's favorite holiday destination, anything and everything can be spotted. There are efforts made to reach out to Facebook in order confirm the cause of this outage and to address the main question that whether they are using this data for as targeting or not.
Well, whatever may be the real agenda behind this Facebook, WhatsApp and Instagram outage, or regardless of how the information is being used, it makes us realize a very important fact that visual world has definitely become machine readable and accessible.
Sources: [1] [2] [3] [4]
Worldwide Facebook, WhatsApp and Instagram Outage – What Caused It? was last modified: September 5th, 2019 by Marie
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Dr Keping Chen from the Macquarie University says innovative approaches are necessary for large-area flood mapping across the Asian region. He introduces three major flood projects which Risk Frontiers at the university has developed.
It has long been recognised that Southeast Asia is the world's most flood-prone region by virtually all accounts, including the flood magnitude and frequency, the severity, the number of countries affected, the size of inundation extent and the number of people at risk.
This is understandable given the region's unique environmental and socioeconomic setting for major and extreme floods: some of the world's largest catchments and rivers, frequent and enormous seasonal monsoons originating from the Indian Ocean and the Pacific Ocean, the world's most dense settlements, and the fact that many of the world's megacities and concentrated exposures are located on low-lying land just next to major rivers and coasts.
Flood risk in the region is on the increase. Over the past few decades, increasing population, intensifying urbanisation and developing economies (at rates all faster than global averages), lack of proper long-term land use planning and limited capacity for adaptation have exacerbated the severity of flood risk. Global climate change and sea level rise add to this predicament.
Major and extreme floods in vast, low-lying and river-dominated deltas are often the result of torrential rains induced by tropical cyclones (typhoons) and associated storm surges.
• The 2011 Thailand Floods impacted the entire Chao Phraya river system (Thailand's largest catchment covering about 35% of the country) and inundated parts of the capital, Bangkok. The event lasted several months and caused more than 800 deaths and affected 13.6 million people. The World Bank reported total economic losses (mainly for damage to the assets of the manufacturing industry) of US$45.7 billion (2011 value), making this the world's fourth costliest disaster.
• 2008 Cyclone Nargis in Myanmar caused at least 138,000 fatalities and a loss of US$10 billion (2008 value).
• 1999 Odisha Cyclone in India caused over 9,800 fatalities and destroyed 275,000 homes. The total economic loss was about US$4.5 billion (1999 value).
• 1991 Bangladesh Cyclone made landfall in the Ganges-Brahmaputra Delta killing over 138,000 people and destroying about 1 million homes, and leaving as many as 10 million homeless. The typhoon caused an estimated loss of US$1.5 billion (1991 value).
• 1881 Haiphong Typhoon devastated the Red River Delta in Vietnam and reportedly killed up to 300,000 people.
Many other economically-important deltas in the region, including the Mekong Delta, and the Zhujiang Delta and the Changjiang Delta in China, are all invariably prone to super typhoon impacts and flooding.
Unfortunately, detailed flood mapping data for the region is rarely available to the general public or in the public domain, a situation that also holds true for the majority of world. There is a greater need to better understand and map the flood risk using innovative and insightful approaches based on the best regional and global data sets currently available. The resulting flood maps should interest high-level authorities and insurers alike.
Flood mapping with hydrodynamic models (hydrological and hydraulic models) is widely used to determine a range of physical attributes of the flood hazard, such as water levels, inundation extent and flow velocity. It is usually carried out for small areas (at catchment or sub-catchment level) and flood maps are produced at various design flood levels such as at 1% Annual Exceedance Probability (AEP) and at a Probable Maximum Flood (PMF) level.
Research and development of large-area flood mapping (at an entire catchment level and beyond) has become more active across both industry and academia, especially after the widespread flooding in Australia in late 2010 and early 2011, the 2011 Thailand flood and the 2013 Central European floods.
Flood mapping requires a set of good-quality inputs, such as detailed Digital Terrain Models (DTMs), river networks, long-term rainfall records, reliable historical water levels and river discharge records. Unfortunately, for many countries these data do not comprehensively exist. In addition, it is often computationally prohibitive to run flood models for large areas. These factors put significant constraints on the application of flood mapping for large areas.
Large-area flood mapping has to be innovative and practical. Amongst the new developments, the "direct-rainfall" (rainfall on the grid) approach that simplifies underlying hydrological processes has been explored by many with some degree of success.
This new development could produce approximate flood-prone areas, but detailed hydrodynamic features and processes are compromised.
• Level 1: Broad-scale mapping to delineate floodplains and those areas not subject to flooding.
• Level 2: Medium-level mapping based on observed historical water levels and flows.
• Level 3: Detailed mapping with a full, proper hydrodynamic analysis. This is only practical if the study area is relatively small.
• The National Flood Information Database (NFID), through a subsidiary company, under contract to the Insurance Council of Australia (ICA).
• Flood Exclusion Zones (FEZ) mapping to identify the national locations with little or no flood risk from riverine flooding.
• FloodAUS, a loss model that integrates the address-level flood hazard data from NFID and FEZ databases, exposure locations and types, and region-specific building vulnerability footprints.
The main inputs for NFID flood risk mapping are flood levels at various return periods from specific flood studies, high-resolution Digital Terrain Models, and property addresses from the Geocoded National Address File.
By integrating these comprehensive spatial data layers (usually from local government authorities and state governments), address-level flood risk ratings are produced in a nationally consistent form, primarily for insurance applications. The majority of the most flood-prone addresses (ie at 1% AEP) in Australia have been analysed.
Risk Frontiers has also developed a methodology that evaluates and determines broad-scale Flood Exclusion Zones for addresses that lie in regions where properties may be at risk of riverine flooding but either no flood information exists or is not publically available.
The FEZ methodology quantifies a set of terrain metrics (eg distance to rivers, elevation difference between an address and nearby rivers, slopes and stream orders) that characterise floodplains at the catchment level and account for hydrological processes. Its development employs large, high-quality data sets on elevation, river networks and known locations with observed, empirical flood data. The modelling is both knowledge- and data-driven, and overcomes the limitations of the various crude statistical approaches that ignore inherent catchment features and cannot provide reliable flood risk predictability for larger, heterogeneous real-world floodplains.
Using the same FEZ methodology, we have recently finished a new case study for Peninsular Malaysia (Figure 2). A range of representative terrain attributes about floodplains, along with hydrologically-conditioned SRTM elevation, delineated river networks and reported historical flood extents, are efficiently explored in a spatial analysis environment. We aim to keep enhancing the methodology and expanding the regional coverage.
Dr Keping Chen is Senior Risk Scientist at Macquarie University, Australia.


I immediately connected with the idea.
Having grown up in Mahopac, N.Y. in the 1980s, I loved that year and I loved that team. It immediately reminded me of being 14-year-old boy, sitting on Vincent Gargano's couch (he was my up-the-street neighbor) watching Dwight Gooden pitch and Gary Carter hit and Keith Hernandez and Wally Backman and Mookie Wilson and Ron Darling and Ed Hearn (yes, Ed Hearn) dominate like few teams had ever before dominated.
More than 10 years post-publication, it remains incredibly gratifying.


Adding These Spices Will Bring Your Bone Broth to Another Level - The Osso Good Co.
Are you making bone broth a part of your regular diet? Bone broth is an extremely nutrient-rich and versatile beverage that can be used in a variety of ways to boost your health. Bone broth can be consumed by itself as an alternative to coffee or tea, or it can be used in cooking to add extra flavor and beneficial nutrients to any meal. Try experimenting with it and adding herbs and spices such as turmeric, cinnamon, cayenne pepper, or ginger. Adding these spices will bring your bone broth to another level!
Because bone broth is cooked for an extended period of time, it is rich in beneficial compounds such as collagen and gelatin, which make it a powerful tonic that helps improve digestion, heal intestinal damage, reduce inflammation, and build healthy tissue — including hair, skin, and muscles. The collagen and gelatin in bone broth also give it a thick, creamy consistency, which makes it especially satisfying and filling.
Depending on the type of bone broth you select, it likely already contains certain herbs and spices that give it a comforting, savory flavor. For example, our Osso Good Organic Chicken Bone Broth contains garlic, parsley, thyme, peppercorn, and bay leaves. However, depending on your own needs and preferences, you may want to prepare your bone broth with additional spices for more flavor or beneficial properties. Try some of these suggestions to get more healthy herbs and spices into your diet!
Turmeric, the yellow spice that adds color to curry, is well known for its powerful medicinal properties. Turmeric is naturally anti-inflammatory, which is incredibly important as many Americans are suffering from the damaging effects of chronic inflammation. Left untreated, chronic inflammation can disrupt many processes throughout the body, causing hormonal imbalances and metabolic disturbances, and increasing the risk of many degenerative diseases, including heart disease and cancer.
Curcumin, the main active ingredient in turmeric, has been found to be just as effective as some anti-inflammatory drugs, without causing any undesirable side effects. However, curcumin is poorly absorbed into the bloodstream. It helps to combine it with black pepper to increase its absorption.
Curcumin is also fat soluble, which means it's a good idea to include a source of dietary fat in your meal. Try a bone broth-based curry recipe that includes chicken, coconut oil, and coconut milk (all good sources of fat), along with spices such as turmeric and coriander, for a flavorful and satisfying meal.
Cinnamon is another popular spice that can be used in a variety of recipes. Like turmeric, cinnamon is anti-inflammatory and can also have a powerful effect on blood sugar levels. Studies have shown that cinnamon helps improve insulin sensitivity and lower blood sugar following the consumption of carbohydrates. This may make it particularly beneficial for people with diabetes or prediabetes.
Try a hearty Moroccan lamb tagine, which combines bone broth with cinnamon, cayenne pepper, and ginger — all of which offer powerful health benefits.
Cayenne pepper contains a compound called capsaicin, which may help combat certain types of cancer. One study found that capsaicin inhibited the development of prostate cancer cells in mice. Another study showed that capsaicin increased energy expenditure in healthy men, helping them burn more fat.
Cayenne pepper can be added to a wide variety of soups and stews to give them a little extra kick. Try it in this easy-to-make sweet potato soup!
In addition to being anti-inflammatory, ginger can also help treat nausea. This includes nausea caused by motion sickness, morning sickness, or chemotherapy.
If you're feeling queasy, try a soothing Thai carrot soup made with ginger, cinnamon, turmeric, and cayenne for a powerful medicinal boost.
At Osso Good, we make bone broths and paleo soups with organic ingredients that are nutrient dense, gluten free, dairy free, Whole30 Approved, and Certified Paleo. We are very proud to make our own bone broth in small, controlled batches to ensure the quality is always consistent and the taste is always amazing.
Copyright © The Osso Good Co. 2017 • All rights reserved.


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FREEDOMS FROM ADDICTIONS-MORE HAS BEEN REVEALED!
Thank you Ben! Please go to Ben's Blog for more tools for your toolkit!
Join Eckhart Tolle TV
"The realm of consciousness is much vaster than thought can grasp. When you no longer believe everything you think, you step out of thought and see clearly that the thinker is not who you are."
http://fff77.blogspot.com/2009/08/end-of-fear.html
More Has Been Revealed
Over the last 75 years the 12-step program of Alcoholics Anonymous has helped millions of addicts find relief from their compulsion. Now, the work of Eckhart Tolle has provided deeper insight into the way addiction works, as well as how and why the 12-step program brings relief. This new understanding will help more addicts find relief faster and can even prevent addiction from taking root.
All people who compulsively drink, use drugs, or engage in dangerous activities do so to obtain some relief from the continuous background fear, lack, or emptiness that is a direct result of life as a thought-based identity (the ego). An egoic identity is a false one because it is based on thought instead of consciousness, our natural state of mind. The egoic personality requires constant compulsive thinking to sustain itself.
Any life based on compulsive thinking will be dysfunctional and unmanageable. Such a life is characterized by recurring feelings of emptiness, lack, and fear. People often describe this condition as having "a hole inside me that can never be filled."
As Eckhart Tolle puts it:
"When the thought forms of 'me' and 'mine', of 'more than,' of 'I want,' 'I need,' 'I must have,' and of 'not enough' operate, no possession, place, person, or condition will ever satisfy you. No content will satisfy you as long as your egoic identity remains in place. No matter what you have or get, you won't be happy. You will always be looking for something else that promises greater fulfillment, that promises to make your incomplete sense of self complete and fill that sense of lack you feel within. "
"The ego always wants something from other people or situations. There is always a hidden agenda, always a of "not enough yet," of insufficiency and lack that needs to be filled. It uses people and situations to get what it wants, and even when it succeeds it is never satisfied for long. Often it is thwarted in its aims, and for the most part the gap between "I want" and "what is" becomes a constant source of upset and anguish."
"The underlying emotion that governs all of the activity of the ego is fear. The fear of being nobody, the fear of nonexistence, the fear of death. All its activities are ultimately designed to eliminate this fear but the most the ego can ever do is to cover it up temporarily with an intimate relationship, a new possession, or winning at this or that. Illusion will never satisfy you. Only the truth of who you are, if realized, will set you free."
The fact that almost everyone suffers from this condition makes it seem normal and allows it to go unquestioned.
Consuming drugs or alcohol or obtaining things in order to relieve this unhappy condition may bring temporary relief, but the activity must be continuously repeated. This sometimes leads to addiction.
Substance addiction is a direct result of compulsive thinking, which is itself an addiction. That's why any solution to the addiction problem must focus on the underlying thinking addiction. The "compulsion to use," not the substance itself, is the cunning, powerful, and baffling engine that drives addiction.
This blog applies the insights concerning the true nature of thinking, as presented by Eckhart Tolle in his books The Power of Now and A New Earth, to the 12-step program as presented in the "Big Book" of Alcoholics Anonymous.
I think Tolle's work sheds new light on the underlying forces of addiction. This understanding points the way toward new approaches to gaining relief--and new ways to prevent addiction from taking hold in the first place.
The 12-step program of Alcoholics Anonymous has brought about a spiritual renaissance in many individuals--myself included--leading them to a happy, joyous, and free existence. But many more alcoholics and other addicts have turned away from this solution because it is veiled in mystery and language that is sometimes obscure or religious.
Tolle's work has unraveled for me the 75-year-old mystery of how and why the 12-step program really works. I believe an accurate and clear description of the 12-step process and its results can only broaden the acceptance and increase the effectiveness of the 12-step program.


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Poldark Season Two Premieres This Sunday
There's still time to catch up on season one.
Masterpiece on PBS
The Downton Abbey movie rumors are great and all, but you can't exactly binge-watch the gossip. Hence why we're so excited about this weekend's Poldark premiere. Season two of the Masterpiece PBS series promises to pick up where season one left off—and while we won't spoil it here, Season One ended on a serious cliffhanger.
For the uninitiated, the show follows solider Ross Poldark (played by The Hobbit's Aidan Turner) as he returns home to Cornwall, England after fighting in the American Revolutionary War. It features the high-quality production values, stunning scenery, and passionate romance—of the PG-13 variety, this is PBS, after all—that we've come to expect from prestige British television.
Basically, it can fill the Downton Abbey-sized hole in your historical drama-loving heart. Check out this season's trailer below (but only once you're all caught up), then be sure to tune in to Masterpiece on PBS this Sunday, September 25 at 8/7 Central.
Caroline Hallemann Senior Digital News Editor As the senior digital news editor for Town & Country, Caroline Hallemann covers everything from the British royal family to the latest episodes of Outlander, Killing Eve, and The Crown.
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Revolution11 developers were pleased to hear the announcement, made in January 2016, that FileMaker released a set of coding tools, the iOS App SDK, that offer developers the means to build native iPhone and iPad apps. From the perspective of a full-time iOS developer, the FileMaker iOS SDK is super simple. All the work is in the setup for the app. However, it is easy to see how this setup process could be daunting for a non-iOS developer.
Even seasoned developers cringe at the potential headache of certificates and provisioning profiles when setting up a project. However, if you've been through this process MANY times, you've gone through your rite of passage and dealing with this hassle is no problem.
As a developer, if you're over this hump, setting up the FileMaker iOS SDK will take you 20 to 30 minutes (if that). Speaking strictly from the iOS side, it's just setting up your App IDs, certificate (which likely is already setup), and provisioning profile. After that, you just drag and drop the FileMaker solution into the Xcode project and update a config file with the solution filename. Once complete, you just setup an app icon and a launch screen if needed.
If your FileMaker solution was made specifically for mobile device usage, then you might want to go back and adjust your FileMaker solutions layout so that it is optimized for the device. If you import the typical desktop solution, there will be a lot of pinch/zooming going on when the user is using the device. Not the end of the world, but not the best user experience either.
Overall, if you're experienced in setting up an app project with Apple, the FileMaker iOS SDK will be a breeze for you. The resulting app is native and therefore does not require FileMaker Go, FileMaker's previous iOS solution, to run. Revolution11 has had great success with setting up client projects using this method – feel free to contact us to talk about the time and cost savings that we have experienced.
Accu-Bore Directional Drilling, a Vallejo-based company, serves the underground infrastructure industry by solving complex water, wastewater, natural gas, electrical and communication utility issues. Because their business is growing rapidly, they saw the need to create more efficient processes to track resource hours and provide more effective processing for reconciliation and uploading data into their accounting software.
Prior to joining forces with Revolution11, the Accu-Bore time recording process was paper based with no dedicated format for information to be sent to the payroll department, leaving room for input error. In addition, once the data was reconciled it was then manually entered into the accounting system, creating another opportunity for incorrect data entry.
Revolution11 created several easy-to-use FileMaker applications that worked seamlessly on the construction foreman's iPad so that information could be uploaded from the field and immediately viewable at the main office. One of the main components of the solution was a daily work report that allowed the teams in the field to record equipment, crews, hours, upload photos, and when submitted, the data was immediately in the system for viewing. The timecard and reconciliation process was rolled out next. Revolution11 devised a system where employees in the field could electronically sign their timecards via an iPad, and payroll could immediately start to reconcile the data once the digital signature was captured and submitted. Before this solution was created, the reconciliation process took three days to complete from getting signatures, reconciling, and uploading into the accounting system. Today, that process takes just three hours.