text:
\section{Introduction}
\label{sec.1}

The two membranes problem was first studied by Vergara-Caffarelli \cite{VC} in the context of variational inequalities to describe the equilibrium position of two elastic membranes in contact with each other that are not allowed to cross. He considered the
linear elliptic case, in which the problem can be reduced to the classical obstacle problem by looking at the difference between the two functions representing the position of each membrane.

Nearly 35 years later, Silvestre \cite{S} studied the problem for a nonlinear operator in divergence form. He obtained the optimal $C^{1,1}$ regularity of solutions together with a characterization of the regularity of the free boundary,
that is the boundary of the set where the two functions coincide. The strategy in his proof was to show that the difference of the two functions satisfies an obstacle problem for the linearized operator, for which the regularity theory of
the solutions and the free boundary are well known. An important remark is that in both of these cases the operator governing the behavior of each function is the same.

In a recent paper, Caffarelli, De Silva and Savin \cite{CDS} considered the two membranes problem for (possibly nonlocal) different operators, i.e. they consider the case in which one of the membranes (say the lower one) satisfies an equation that has
higher order with respect to the other one. Here, heuristically, the lower order operator can be treated as a perturbation and some regularity for the lower membrane is obtained. Regularity from the upper membrane can then be deduced by solving an obstacle problem
(with the lower membrane as obstacle) and obtaining estimates for solutions of nonlocal obstacle type problems in which the obstacle is not smooth. Repeating these arguments, the optimal regularity is achieved.

We also point out that the problem has been studied by several authors in the general case of $N$ membranes, see \cite{CV}, \cite{CCVC}, \cite{ARS}.

Here, motivated by a model from mathematical finance, we consider a version of the two membrane problem for two different fully nonlinear operators. It is worth pointing out that,
for the case of two different operators of the same order, the only result available (to the best of the authors' knowledge) is the H\"older regularity obtained in \cite{CDS} (see the Introduction
there for a discussion of the difficulties of this problem). In this paper we prove $C^{1,\alpha}$ regularity of the solution pair for (concave or convex) operators satisfying a sort of
compatibility condition (see \eqref{eq.comp} below) and $C^{1,1}$ regularity for the case of the Pucci extremal operators, which is optimal. Moreover, we give an explicit example
that shows that no regularity can be expected to hold for the free boundary in general.

\subsection{Notation and preliminaries}

Throughout this paper the ellipticity constants $\lambda,\Lambda\in\mathbb{R}$ will be fixed and will satisfy $0<\lambda<\Lambda$. Given these, we denote by $\mathcal{M}^+$ and $\mathcal{M}^-$ the Pucci extremal operators with respect to the class of
symmetric matrices whose eigenvalues lie between $\lambda$ and $\Lambda$, that is for any symmetric matrix $X$

\begin{equation*}
 \mathcal{M}^+(X)=\sup_{A\in\mathcal{L}_{\lambda,\Lambda}}\textrm{tr}(AX)\quad\quad\textrm{and}\quad\quad\mathcal{M}^-(X)=\inf_{A\in\mathcal{L}_{\lambda,\Lambda}}\textrm{tr}(AX)
\end{equation*}
where
\[
 \mathcal{L}_{\lambda,\Lambda}=\{A\in\mathbb{R}^{n\times n}:A\textrm{ is symmetric and } \lambda Id\leq A \leq\Lambda Id\}
\]
$X\geq Y$ meaning as usual that $X-Y$ is a positive semidefinite matrix.

Recall that an operator $F:\mathbb{R}^{n\times n}\rightarrow\mathbb{R}$ is said to be uniformly elliptic with repect to the class $\mathcal{L}_{\lambda,\Lambda}$ if it satisfies
\begin{equation}\label{eq.ellip}
 \mathcal{M}^-(X-Y)\leq F(X)-F(Y)\leq  \mathcal{M}^+(X-Y)
\end{equation}
for any pair of symmetric matrices $X$ and $Y$.

We will assume without loss of generality that $F(0)=0$. A useful remark that follows from \eqref{eq.ellip} is that if $u$ is a function satisfying $F(D^2u)=f$ then
\begin{equation} \label{eq.diri}
  \left\{ \begin{array}{rcl}
  \mathcal{M}^+(D^2u) & \geq & f \\
  \mathcal{M}^-(D^2u) & \leq & f.
\end{array}\right.
\end{equation}
In particular, $u$ is a subsolution and a supersolution of two (possibly different) elliptic equations with bounded measurable coefficients.

\subsection{The two membranes problem for fully nonlinear operators}

The problem we will consider is the following: given two functions $u_0,v_0\in C^\gamma(\partial B_1)$ and $f,g\in C^\gamma(B_1)$ for some $\gamma\in(0,1)$, we want to study the solutions $u$ and $v$ of

\begin{equation}\label{eq.main}
  \left\{ \begin{array}{rcll}
  u & \geq & v & \textrm{ in } B_1 \\
  F(D^2u) & \leq & f & \textrm{ in } B_1 \\
  G(D^2v) & \geq & g & \textrm{ in } B_1 \\
  F(D^2u) & = & f & \textrm{ in } B_1\cap\Omega \\
  G(D^2v) & = & g & \textrm{ in } B_1\cap\Omega \\
  u & = & u_0 & \textrm{ on } \partial B_1 \\
  v & = & v_0 & \textrm{ on } \partial B_1
\end{array}\right.
\end{equation}
where
\[
\Omega:=\{u>v\},
\]
$F$ is convex and
\begin{equation}\label{eq.comp}
G(X)=-F(-X).
\end{equation}
Note that $G$ thus defined will be concave and that particular examples are
\begin{equation*}
F(D^2w)=\sup_{\alpha\in \Sigma}\:\textrm{tr}(A_\alpha D^2 w) \quad\quad\textrm{and}\quad\quad G(D^2w)=\inf_{\alpha\in \Sigma}\:\textrm{tr}(A_\alpha D^2 w)
\end{equation*}
with $\Sigma$ some set of indexes and $A_\alpha\in\mathcal{L}_{\lambda,\Lambda}$ for every $\alpha$. If $A_\alpha$ can be any matrix in $\mathcal{L}_{\lambda,\Lambda}$ then $F=\mathcal{M}^+$ and $G=\mathcal{M}^-$. It is in this latter case that we prove the optimal regularity. Note that the strict inequality assumed for the ellipticity constants avoids these operators to become just a multiple of the Laplacian.

Equation \eqref{eq.main} is to be understood in the viscosity sense. More precisely: if $\varphi$ is a $C^2$ function in $B_1$ satisfying for some $x_0\in B_1$
\begin{equation*}
 \varphi(x)\leq u(x) \quad\textrm{in}\quad B_1,\quad\quad \varphi(x_0)=u(x_0)
\end{equation*}
(i.e. $\varphi$ \emph{touches $u$ by below} at $x_0$) then
\begin{equation*}
F(D^2\varphi(x_0))\leq f(x_0).
\end{equation*}

Similarly if $\varphi$ touches $v$ by above and of course the opposite inequalities (last two equations in \eqref{eq.main})  hold if $\varphi$ touches $u$ by above
(or $v$ by below) in $\Omega$. A simple remark that will be useful is that it is equivalent to use paraboloids instead of general $C^2$ functions.

Note that the convexity of $F$ as well as the H\"older regularity for $f$ and $g$ are natural assumptions if we want to get optimal regularity. In fact, one expect solutions to this problem to be $C^{1,1}$ as long as the equation they solve on the noncontact
set is ``good enough'', meaning that we have at least $C^{1,1}$ regularity for it. This, in principle, is not true in general if $F$ is not convex or $f$ and $g$ are merely bounded. Also, for the problem to make sense we will assume throughout the paper that $u_0>v_0$ on $\partial B_1$. Moreover, we will assume that $f-g\geq0$. Notice that if this was not the case the problem could loose interest and degenerate into just two independent fully nonlinear equations. Indeed, if $f-g<0$, we would have (see \eqref{eq.diri})
\begin{equation*}
  \left\{ \begin{array}{rcll}
  \mathcal{M}^+(D^2(v-u)) & > & 0 & \textrm{ in } B_1 \\
   v-u & < & 0 & \textrm{ in } \partial B_1
\end{array}\right.
\end{equation*}
and due to the maximum principle $u>v$ in $B_1$. Then there is no contact set and we just have the respective equations for $u$ and $v$.

Equation \eqref{eq.main} models a so called ``bid and ask'' situation in which we have an asset, a seller (represented by $u$) and a buyer (represented by $v$). The price
of the asset is random and the transaction will only take place when $u$ and $v$ ``agree on a price'', i.e. when $u=v$. Moreover, we want to model the expected earnings of $u$ and $v$, assuming
that their strategy is optimal.

One can think of this problem as having two different (although related) features: on one hand, we have an ``obstacle type'' situation, in which $u$ tries to maximize gain
with $v$ being an obstacle and vice versa ($v$ minimizing cost and $u$ being an obstacle), hence the constraint $u\geq v$. But perhaps more interesting is the special relation
between $u$ and $v$. Because of the ``bid and ask'' nature of the model, the Bellman type equations that govern the behavior of our solutions are closely related (recall $F(X)=-G(-X)$)  and
it is precisely this feature which opens a way to get regularity even though the operators are different.

The main result of this paper is the following:

\begin{thm}\label{thm.main}
Let $u$ and $v$ solve \eqref{eq.main} in the viscosity sense with $F=\mathcal{M}^+$ and $G=\mathcal{M}^-$. Then $u$ and $v$ are $C^{1,1}$ in $B_{1/4}$ and
\[
 \|D^2u\|_{L^\infty(B_{1/4})}, \|D^2v\|_{L^\infty(B_{1/4})}\leq C
\]
where $C$ depends only on $n,\lambda,\Lambda,\|f\|_{C^\gamma(B_1)},\|g\|_{C^\gamma(B_1)},\|v\|_{L^\infty(B_1)}$ and $\|u\|_{L^\infty(B_1)}$.
\end{thm}

\section{Existence}

In this section we prove existence of solutions for our problem. We use the method of penalization, i.e. we are going to consider a family of unconstrained ``penalized equations''
whose solutions are uniformly bounded in some H\"older space and hence convegent up to a subsequence. Then we are going to show that the limit of that subsequence is actually a
solution to \eqref{eq.main} (see \cite{KS}).

The penalized problem we are going to consider is the following:
\begin{equation}\label{eq.penalepsilon}
  \left\{ \begin{array}{rcll}
  F(D^2u_\epsilon) & = & f+\beta_\epsilon(u_\epsilon-v_\epsilon)& \textrm{ in } B_1 \\
  G(D^2v_\epsilon) & = & g-\beta_\epsilon(u_\epsilon-v_\epsilon)& \textrm{ in } B_1 \\
  u_\epsilon & = & u_0  & \textrm{ in } \partial B_1\\
  v_\epsilon & = & v_0 & \textrm{ in } \partial B_1
\end{array}\right.
\end{equation}
where for each $\epsilon >0$ we define
\begin{equation}\label{eq.betaepsilon}
\beta_{\epsilon}(t)=\beta(t/\epsilon)
\end{equation}
with $\beta:\mathbb{R}\rightarrow\mathbb{R}$ a smooth function satisfying
\begin{equation}\label{eq.beta}
 -N\leq \beta\leq0,   \quad
\beta'\geq0,   \quad
\beta(t) = 0\textrm{ when }t\geq 1, \quad
\beta(t) = -N \textrm{ when }t\leq 0, \quad
\end{equation}
and
\begin{equation}\label{eq.ene}
N:= \|f-g\|_{L^\infty(B_1)}.
\end{equation}

To get the existence and a priori bounds for solutions of \eqref{eq.penalepsilon} we will use a fixed point argument. Hence, we will need global regularity results for
equations of the form

\begin{equation} \label{eq.dirichlet}
  \left\{ \begin{array}{rcll}
  H(D^2u) & = & h & \textrm{ in } B_1 \\
  u & = & u_0  & \textrm{ in } \partial B_1 \\
\end{array}\right.
\end{equation}
where $H$ is a uniformly elliptic operator. Here we mostly follow Chapter 4 of \cite{CC} but since the proofs need to be modified slightly we sketch them below for completeness:

\begin{prop} \label{prop.growth}
Let $u$ be a viscosity solution of \eqref{eq.dirichlet} with $h\in L^\infty(B_1)$ and $u_0\in C^\gamma(\partial B_1)$. Then for any $x_0\in\partial B_1$ we have
\begin{equation}\label{eq.growth}
\sup_{x\in B_1}\frac{|u(x)-u(x_0)|}{\vert x-x_0\vert^{\gamma/2}}\leq C
\end{equation}
where $C$ is a constant depending only on $n,\lambda,\Lambda,\gamma,\|u_0\|_{C^\gamma(\partial B_1)}$ and $\|h\|_{L^\infty(B_1)}$.
\end{prop}

\begin{proof}
We translate, rotate and add a constant to $u$ so that it is defined on $B:=B_1(1,0,..,0)$, $x_0=(0,0,...,0)$ and $u(0)=0$. We want to show
\begin{equation}\label{eq.growth2}
\sup_{x\in B_1}\frac{|u(x)|}{\vert x\vert^{\gamma/2}}\leq C
\end{equation}
For this let us define the barrier $\psi(x)= C x_1^{\gamma/2}$ with $C$ a constant to be determined. Notice that
\[
|x|^\gamma=(x_1^2+x_2^2+...+x_n^2)^{\frac{\gamma}{2}}=(2x_1)^{\frac{\gamma}{2}}
\]
on $\partial B$ and hence
\[
u(x)=u_0(x)\leq [u_0]_{C^{\gamma}(\partial B_1)}|x|^\gamma\leq C x_1^{\gamma/2}=\psi(x)
\]
there. On the other hand, $\psi$ satisfies
\[
H(D^2\psi)\leq \mathcal{M}^+(D^2 \psi)=C\lambda\frac{\gamma}{2}(\frac{\gamma}{2}-1)x_1^{\frac{\gamma}{2}-2}\leq -\Vert h \Vert_{L^\infty(B_1)}
\]
in $B$ in the viscosity sense if we take $C$ large enough. From the maximum principle it follows that $u\leq \psi$ in $B$.

Symmetrically, we see that
$u \geq -\psi$ on $\partial B$ and $H(D^2(-\psi))\geq \Vert h \Vert_{L^\infty(B_1)}$ in $B$, so using the maximum principle again we get $u\geq-\psi$ and hence \eqref{eq.growth2}.
\end{proof}

Now we prove the global H\"older estimates.

\begin{prop} \label{prop.globalest}
Let $u$ be a viscosity solution of \eqref{eq.dirichlet} with $h\in L^\infty(B_1)$ and $u_0\in C^\gamma(\partial B_1)$. Then
\begin{equation}\label{eq.globalest}
\Vert u \Vert_{C^\eta(B_1)} \leq C
\end{equation}
where $C$ is a constant depending only on $n,\lambda,\Lambda,\|u_0\|_{C^\gamma(\partial B_1)}$ and $\|h\|_{L^\infty(B_1)}$ and $\eta\leq\gamma/2$.
\end{prop}

\begin{proof}
We start by recalling that by interior estimates (Proposition 4.10 in \cite{CC}) solutions of \eqref{eq.dirichlet} are in $C^\alpha_{loc}(B_1)$ for some $\alpha>0$. Let $\eta=\min\{\gamma/2,\alpha\}$, $x_1, x_2\in B_1$
$r=\vert x_1 - x_2\vert$, and take $x_1',x_2'\in\partial B_1$ such that
\[
d_1:= d(x_1, \partial B_1)= \vert x_1 - x_1'\vert\quad\textrm{ and }\quad d_2:= d(x_2, \partial B_1)=\vert x_2 - x_2'\vert.
\]
We assume without loss of generality that $d_2\leq d_1$ and we want to show that
\begin{equation}\label{eq.holder}
\vert u(x_1)-u(x_2)\vert \leq C\vert x_1-x_2 \vert^\eta.
\end{equation}

We split the proof in two cases:

\textbf{Case 1:} When $r \leq \frac{d_1}{2}$, from the rescaled version of the interior estimates applied to $u-u(x_1')$ on $B_{d_1}(x_1)$ and our estimates at the boundary \eqref{eq.growth} we have

\begin{equation*}
\begin{split}
d_1^\alpha \Vert u-u(x_1') \Vert_{C^\alpha(B_{d_1/2}(x_1))} & \leq C ( d_1^2 \Vert h \Vert_{L^\infty(B_{d_1}(x_1))}+ \Vert u-u(x_1') \Vert_{L^\infty(B_{d_1}(x_1))}) \\
 & \leq C  d_1^2\Vert h \Vert_{L^\infty(B_{d_1})}  + Cd_1^{\gamma/2} \leq C d_1^{\gamma/2}.
\end{split}
\end{equation*}
On the other hand
\[
d_1^\eta \frac{|u(x_1)-u(x_2)|}{|x_1-x_2|^\eta}\leq d_1^\alpha \frac{|u(x_1)-u(x_2)|}{|x_1-x_2|^\alpha}\leq d_1^\alpha\Vert u(\cdot)-u(x_1')\Vert_{C^\alpha(B_{d_1/2}(x_1))}
\]
so
\[
\frac{|u(x_1)-u(x_2)|}{|x_1-x_2|^\eta}\leq C d_1^{\gamma/2-\eta}\leq C
\]
as desired in this case.

\textbf{Case 2:} When $r > \frac{d_1}{2}$, again by the boundary estimates \eqref{eq.growth} and the triangle inequality
\begin{equation*}
\begin{split}
\vert u(x_1)-u(x_2) \vert & \leq  |u(x_1)-u(x_1')| + |u(x_1')-u(x_2')| + |u(x_2')-u(x_2)|  \\
 & \leq Cd_1^{\gamma/2} + C|x_1'-x_2'|^\gamma + Cd_2^{\gamma/2}  \leq C(d_1^{\gamma/2} + r^\gamma+d_1^\gamma+d_2^\gamma + Cd_2^{\gamma/2}) \\
 & \leq C  r^{\gamma/2}= C\vert x_1-x_2\vert^{\gamma/2}.
\end{split}
\end{equation*}
\end{proof}

In our next proof we are also going to use global H\"older estimates for ``equations with bounded measurable coefficients'':

\begin{prop}\label{prop.globalestbmc}
Let $u$ be a viscosity solution of
\begin{equation} \label{eq.diri2}
  \left\{ \begin{array}{rcll}
  \mathcal{M}^+(D^2u) & \geq & -\alpha & \textrm{ in } B_1 \\
  \mathcal{M}^-(D^2u) & \geq & \ \ \alpha & \textrm{ in } B_1 \\
   u & = & u_0  & \textrm{ in } \partial B_1 \\
\end{array}\right.
\end{equation}
for $\alpha$ a positive constant and $u_0\in C^\gamma(\partial B_1)$. Then
\begin{equation}\label{eq.globalestbmc}
\Vert u \Vert_{C^\eta(B_1)} \leq C
\end{equation}
where $C$ is a constant depending only on $n,\lambda,\Lambda,\|u_0\|_{C^\gamma(\partial B_1)}$ and $\alpha$ and $\eta\leq\gamma/2$.
\end{prop}

\begin{proof}
The proof follows exactly as that of Proposition \ref{prop.globalest}. We just point out that in order to prove the boundary estimates \eqref{eq.growth} a barrier argument for the Pucci extremal operators is used that is trivially adapted to a situation like \eqref{eq.diri2}. This is then combined with interior estimates that also hold for $\eqref{eq.diri2}$ (see \cite{CC}) to give \eqref{eq.globalestbmc}
\end{proof}

Finally, the following observation is going to be useful: it follows from the proof of Proposition \ref{prop.growth} that the dependence of the constant on $\|h\|_{L^\infty(B_1)}$ is continuous. The same is true for the interior estimates (again, see \cite{CC}) and hence for the constant in \eqref{eq.globalestbmc}.

We can now prove existence of the penalized problem:

\begin{prop}\label{prop.existpenal}
Let $\beta:\mathbb{R}\rightarrow\mathbb{R}$ be a smooth bounded function, $u_0,v_0\in C^\gamma(\partial B_1)$ and $f,g\in L^\infty(B_1)$ . There exist $u,v\in C(B_1)$ such that
\begin{equation}\label{eq.penal}
  \left\{ \begin{array}{rcll}
  F(D^2u) & = & f(x)+\beta(u-v)& \textrm{ in } B_1 \\
  G(D^2v) & = & g(x)-\beta(u-v)& \textrm{ in } B_1 \\
  u & = & u_0  & \textrm{ in } \partial B_1\\
  v & = & v_0 & \textrm{ in } \partial B_1
\end{array}\right.
\end{equation}
in the viscosity sense.
\end{prop}

\begin{proof}

We will use Schauder's Fixed Point Theorem (see \cite{GT}).  Let $\tilde{\alpha}=\eta/2$ with $\eta$ as in Proposition \ref{prop.globalest} and consider the map
\[
T: C^\eta(B_1) \times C^\eta(B_1)  \longrightarrow  C^\eta(B_1) \times C^\eta(B_1)
\]
defined as $T(\bar{u},\bar{v}):= (u,v)$, $u,v$ satisfying
\begin{equation}\label{eq.Tpenal}
  \left\{ \begin{array}{rcll}
  F(D^2u) & = & f(x)+\beta(\bar{u}-\bar{v})& \textrm{ in } B_1 \\
  G(D^2v) & = & g(x)-\beta(\bar{u}-\bar{v})& \textrm{ in } B_1 \\
  u & = & u_0  & \textrm{ on } \partial B_1\\
  v & = & v_0 & \textrm{ on } \partial B_1
\end{array}\right.
\end{equation}
Such $u$ and $v$ exist by Perron's method. Also, let
\begin{equation*}
X := \{(\bar{u}, \bar{v})\in C^{\tilde{\alpha}}(B_1) \times C^{\tilde{\alpha}}(B_1): \bar{u} = u_0,\bar{v}=v_0 \textrm{ on } \partial B_1, \Vert \bar{u} \Vert_{C^{\eta}(B_1)},  \Vert \bar{v} \Vert_{C^{\eta}(B_1)} \leq C \}
\end{equation*}
with $C$ to be determined later.

If we can show that $X$ is a compact convex set in $C^\eta(B_1) \times C^\eta(B_1)$, that $T$ is continuous in $X$ and $T(X)\subset X$ then by Schauder's Fixed Point Theorem there exists a solution to \eqref{eq.penal}.
We divide the proof in several steps:

\textbf{Step 1: convexity and compactness of $X$.}

The convexity is trivial. As for the compactness, it is a straight forward consequence of the Arsela-Ascoli theorem since $\tilde{\alpha}<\eta$.

\textbf{Step 2: $T(X)\subset X$.}

Notice that if $\bar{u}, \bar{v} \in C^{\tilde{\alpha}}(B_1)$ and we let
\[
h:=f+\beta(\bar{u}-\bar{v})\quad\quad\textrm{ and }\quad\quad k:=g-\beta(\bar{u}-\bar{v})
\]
we have, as $f,g, \beta$ are bounded, that $h,k \in L^\infty(B_1)$. Hence, from Proposition \ref{prop.globalest} we know that
\begin{equation}
\begin{split}
\Vert u \Vert_{C^{\eta}(B_1)}\leq C\\
\Vert v \Vert_{C^{\eta}(B_1)}\leq C
\end{split}
\end{equation}
for some constant $C>0$ that depends only on $n$, $\lambda$, $\Lambda$, $\Vert f \Vert_{L^\infty(B_1)}$, $\Vert g \Vert_{L^\infty(B_1)}$, $\Vert u_0 \Vert_{C^\gamma(B_1)}$ and $\Vert v_0 \Vert_{C^\gamma(B_1)}$ (this is the constant $C$ used to define $X$). In particular, this implies
that $(u,v)\in X$.

\textbf{Step 3: $T$ is continuous.}

Let $T(\bar{u}, \bar{v})=(u',v')$ and $T(\bar{\bar{u}}, \bar{\bar{v}})=(u'',v'')$. We want to show that given $\epsilon>0$ we can find $\delta>0$ so that
\[
\Vert \bar{u} - \bar{\bar{u}} \Vert_{C^{\tilde{\alpha}}(B_1)}<\delta \quad\quad\textrm{ and }\quad\quad \ \Vert \bar{v} - \bar{\bar{v}} \Vert_{C^{\tilde{\alpha}}(B_1)} < \delta
\]
imply
\[
\Vert u'-u''\Vert_{C^{\tilde{\alpha}}(B_1)}< \epsilon \quad\quad\textrm{ and }\quad\quad \Vert v'-v''\Vert _{C^{\tilde{\alpha}}(B_1)} < \epsilon.
\]

Notice that
$$F(D^2u') - F(D^2u'') = \beta(\bar{u}-\bar{v}) - \beta(\bar{\bar{u}}-\bar{\bar{v}}) $$
and from the definition of uniform ellipticity we have
$$ \mathcal{M}^-(D^2(u'-u'')) \leq  F(D^2u') - F(D^2u'') \leq \mathcal{M}^+(D^2(u'-u''))$$
in the viscosity sense. Now let $w := u'- u''$. From the previous two inequalities we have
\begin{equation}\label{eq.TempEqu1}
  \left\{ \begin{array}{rcll}
  \mathcal{M}^+(D^2w) & \geq &  \ \ -\Vert \beta \Vert_{C^1(\mathbb{R})}(|\bar{u} - \bar{\bar{u}}|+|\bar{v} - \bar{\bar{v}}|)     & \textrm{ in } B_1 \\
  \mathcal{M}^-(D^2w) & \leq & \ \ \ \ \Vert \beta \Vert_{C^1(\mathbb{R})}(|\bar{u} - \bar{\bar{u}}|+|\bar{v} - \bar{\bar{v}}|) & \textrm{ in } B_1 \\
  w & = & 0  & \textrm{ on } \partial B_1,
\end{array}\right.
\end{equation}
so if we pick $\delta$ small enough so that $\Vert \beta \Vert_{C^1(B_1)}(|\bar{u} - \bar{\bar{u}}|+ |\bar{v} - \bar{\bar{v}}|)\leq \delta_0$  for some $\delta_0>0$ to be chosen, we can rewrite \eqref{eq.TempEqu1} as
\begin{equation}
  \left\{ \begin{array}{rcll}
  \mathcal{M}^+(D^2w) & \geq & -\delta_0    & \textrm{ in } B_1 \\
  \mathcal{M}^-(D^2w) & \leq & \ \ \delta_0    & \textrm{ in } B_1 \\
  w & = & 0  & \textrm{ in } \partial B_1.\\
\end{array}\right.
\end{equation}

Then, by Proposition \ref{prop.globalestbmc} and the observation following it (and choosing $\alpha=\delta_0$ in \eqref{eq.diri2}) we can pick $\delta_0$ small enough to get
\[
\Vert u'-u''\Vert_{C^{\tilde{\alpha}}(B_1)} = \Vert w \Vert_{C^{\tilde{\alpha}}(B_1)} \leq \Vert w \Vert_{C^\eta(B_1)}\leq \epsilon
\]
as desired. The proof of $\Vert v'-v''\Vert_{C^{\tilde{\alpha}}(B_1)} \leq  \epsilon$  \ follows in an analog way.

\end{proof}

We now show the main result of this section, which is the existence of solutions of the two membranes problem \eqref{eq.main}.

\begin{thm}
There exist $u,v\in C(B_1)$ that solve \eqref{eq.main} in the viscosity sense.
\end{thm}

\begin{proof}
Let $N$ and $\beta:\mathbb{R}\rightarrow\mathbb{R}$ be as in \eqref{eq.ene} and \eqref{eq.beta} and define $\beta_{\epsilon}(t)$ as in \eqref{eq.betaepsilon}. Now, let $u_\epsilon, v_\epsilon$ solutions of Problem \eqref{eq.penalepsilon}, i.e. $u_\epsilon, v_\epsilon$
satisfy
\begin{equation*}
  \left\{ \begin{array}{rcll}
  F(D^2u_\epsilon) & = & f+\beta_\epsilon(u_\epsilon-v_\epsilon)& \textrm{ in } B_1 \\
  G(D^2v_\epsilon) & = & g-\beta_\epsilon(u_\epsilon-v_\epsilon)& \textrm{ in } B_1 \\
  u_\epsilon & = & u_0  & \textrm{ in } \partial B_1\\
  v_\epsilon & = & v_0 & \textrm{ in } \partial B_1
\end{array}\right.
\end{equation*}

By Proposition \ref{prop.existpenal} such $u_\epsilon, v_\epsilon$ and exist. Moreover, notice that as
\[
\Vert \beta_\epsilon \Vert_{L^\infty(\mathbb{R})} = \Vert \beta \Vert_{L^\infty(\mathbb{R})}\leq N
\]
and $f, g \in L^\infty(B_1)$, Proposition \ref{prop.globalest} gives us $\Vert u_\epsilon \Vert_{C^{\eta}(B_1)},\Vert v_\epsilon \Vert_{C^{\eta}(B_1)} \leq C$ \ for some $C>0$ that does not depend on $\epsilon$. Hence, by Arsela-Ascoli (up to subsequences)
$u_\epsilon \rightarrow u$ and $v_\epsilon \rightarrow v$  uniformly on $B_1$ for some $u,v\in C^{\tilde{\eta}}(B_1)$ where $\tilde{\eta}<\eta$. We claim that $u$ and $v$ solve \eqref{eq.main}.

We first want to see that $u \geq v$. Assume not, i.e. suppose the exists $x\in B_1$ such that
\[
u(x)-v(x) = -\delta <0.
\]
From the uniform convergence we will have $u_\epsilon(x)-v_\epsilon(x) < -\delta/2$ for $\epsilon$ small enough. In particular, $u_\epsilon-v_\epsilon$ has to have a negative minimum at some point $y\in B_1$ (recall that on $\partial B_1$ we have $u_0\geq v_0$). Moreover, $u_\epsilon-v_\epsilon$
satisfies, by the convexity of $F$ and the fact that $F(X)=-G(-X)$,
\[
F\Big(D^2\Big(\frac{u_\epsilon-v_\epsilon}{2}\Big)\Big)\leq \frac{1}{2}(f-g + 2\beta_\epsilon(u_\epsilon-v_\epsilon))
\]
in the viscosity sense. Let $P$ be a plane touching $\frac{u_\epsilon-v_\epsilon}{2}$ by below at $y$. Then
\[
F(D^2P)\equiv 0
\]
but since in particular $P$ is a $C^2$ function we must have
\[
F(D^2P)\leq\frac{1}{2}(f(y)-g(y)+2\beta_\epsilon(P(y)))<0
\]
a contradiction.

Now we want to show that $u$ and $v$ satisfy the corresponding equations. Let us start by showing that $F(D^2u)\leq f \textrm{ in } B_1$.

Let $\varphi$ be a paraboloid touching $u$ by below at $x_0\in B_1$. Given $\xi>0$ there exists $\delta>0$ such that
\[
f(x)\leq f(x_0)+\xi
\]
for any $x\in B_\delta(x_0)\subset B_1$. Also, for any $\eta>0$ we can choose $\epsilon$ small enough so that a translation of $\varphi(x)-\frac{\eta}{2}|x-x_0|^2$ (which we call $\tilde{\varphi}$) touches $u_\epsilon$ by below at $x_1\in B_\delta(x_0)$. Hence
\[
F(D^2\tilde{\varphi}(x_1))\leq f(x_1)+\beta_\epsilon(u_\epsilon(x_1)-v_\epsilon(x_1))\leq f(x_1)\leq f(x_0)+\xi.
\]
Since $\xi$ was arbitrary we get
\[
F(D^2\tilde{\varphi}(x_1))\leq f(x_0)
\]
but
\[
F(D^2\tilde{\varphi}(x_1))=F(D^2\varphi(x_1)-\eta\textrm{Id})=F(D^2\varphi(x_0)-\eta\textrm{Id})
\]
and letting $\eta\rightarrow0$ we get
\[
F(D^2\varphi(x_0))\leq f(x_0)
\]
as desired (recall that $F$ is continuous in the space of matrices).

Using again the uniform convergence, the definition of viscosity solutions and considering a test function $\varphi$ that touches $u$ by above we can similarly show that  $F(D^2u)  \geq  f  \textrm{ in } B_1\cap\Omega$ and conclude that $F(D^2u)  =  f  \textrm{ in } B_1\cap\Omega$. The
proofs of  $G(D^2v)  \geq  g \textrm{ in } B_1$ and $G(D^2v)  =  g  \textrm{ in } B_1\cap\Omega$ are analogous to the previous reasoning. It is immediate from uniform convergence that $u=u_0$ and $v=v_0$ on $\partial B_1$.
\end{proof}

\begin{rem}
It is noting that the proof would still hold if we slightly relax the assumptions on the operators by asking just $F(X)\leq -G(-X)$.
\end{rem}

\begin{rem}
We point out that there is no uniqueness in this problem. This comes from the fact that ``there is no equation'' on the contact set. In fact, let (for $n=1$)
\begin{equation*}
\begin{aligned}[c]
u(x)=\left\{\begin{array}{rcll}
  \frac{x_+^2}{2} & \textrm{ for } & 0< x\leq 1 \\
  0         & \textrm{ for } & -1\leq x\leq 0
\end{array}\right.
\end{aligned}
\qquad
\begin{aligned}[c]
v(x)=\left\{\begin{array}{rcll}
  -\frac{x_+^2}{2} & \textrm{ for } & 0< x\leq 1 \\
  0         & \textrm{ for } & -1\leq x\leq 0.
\end{array}\right.
\end{aligned}
\end{equation*}

$u$ and $v$ are $C^{1,1}$ functions and they are strong solutions (and hence viscosity solutions) of \eqref{eq.main} with $F=\mathcal{M}^+,G=\mathcal{M}^-,f=\Lambda$ and $g=-\Lambda$. However, we can make a perturbation $\psi\in C^\infty_c((-1,0))$ such that
\[
-1\leq \psi'' \leq 1
\]
in $(-1,0)$ and get another solution.

Of course this example can be easily generalized to $n\geq 2$ choosing
\begin{equation*}
\begin{aligned}[c]
u(x)=\left\{\begin{array}{rcll}
   (|x|-1/2)^2_+ & \textrm{ in } & B_1\setminus B_{1/2} \\
  0         & \textrm{ in } & B_{1/2}
\end{array}\right.
\end{aligned}
\qquad
\begin{aligned}[c]
v(x)=\left\{\begin{array}{rcll}
   -(|x|-1/2)^2_+ & \textrm{ in } & B_1\setminus B_{1/2} \\
  0         & \textrm{ in } & B_{1/2}
\end{array}\right.
\end{aligned}
\end{equation*}
and modifying the right hand sides accordingly.

However, uniqueness does hold in the ``nonexcercise region'' $\Omega$. In fact, if two pairs of solutions $(u,v)$ and $(u',v')$ satisfy
\[
 u\geq u'\quad\textrm{ and }\quad v\geq v'\quad\textrm{ on } \partial B_1\cup\partial\Omega
\]
then
\[
 u\geq u'\quad\textrm{ and }\quad v\geq v'\quad\textrm{ in }  B_1\cap\Omega
\]
by the maximum principle for fully nonlinear elliptic equations (notice that in $B_1\cap\Omega$ we have $F(D^2u)=f$ and $G(D^2v)=g$).
\end{rem}

\section{Regularity for the solution pair}
\label{sec.3}

In this section we prove our main regularity result, Theorem \ref{thm.main}. The fact that this is the optimal regularity can be easily seen by considering  $u(x)=\frac{x_+^2}{2}$ and $v(x)=-\frac{x_+^2}{2}$ (in one dimension) and noticing that
they solve \eqref{eq.main} with $f\equiv\Lambda$ and $g\equiv-\Lambda$ in $[-1,1]$.

To prove Theorem \ref{thm.main} we show that solutions to \eqref{eq.main} satisfy the hypothesis of  the following Theorem (see Theorem 2.1 in \cite{IM}):

\begin{thm}\label{thm.indreimine}
 Let $f\in C^\gamma(B_1)$ and $u$ a $W^{2,n}(B_1)$ solution of
\begin{equation*}
  \left\{ \begin{array}{rcll}
  F(D^2u) & = & f(x) & \textrm{ in } B_1\cap\Omega \\
  |D^2u|   & \leq & C & \textrm{a.e. in } B_1\cap\Omega^c
\end{array}\right.
\end{equation*}
for some open set $\Omega\subset B_1$ and some elliptic operator $F$ that is either concave or convex. Then, there exists a constant $C$ depending only on $\|f\|_{C^\gamma(B_1)},\|u\|_{W^{2,n}(B_1)}$, the dimension and the ellipticity constants
such that
\[
 |D^2u|\leq C \quad \textrm{a.e. in } B_{1/2}.
\]
\end{thm}

The first step is to show the following Calder\'on-Zygmund type estimate:

\begin{prop}\label{prop.cz}
Let $u$ and $v$ solve \eqref{eq.main}. Then $u$ and $v$ belong to $W^{2,p}(B_{1/2})$ for any $1<p<\infty$ and
\begin{equation}\label{eq.czbound}
 \|u\|_{W^{2,p}(B_{1/2})}, \|v\|_{W^{2,p}(B_{1/2})} \leq C
\end{equation}
for some constant $C$ depending only on $n,\lambda,\Lambda, \|u\|_{L^\infty(B_1)}, \|v\|_{L^\infty(B_1)}, \|f\|_{L^\infty(B_1)}$ and $\|g\|_{L^\infty(B_1)}$.
\end{prop}

\begin{proof}
We prove the result for $u$, the proof for $v$ is analogous. We will show that $|F(D^2u)|\leq C$ in the viscosity sense for some universal constant $C$. The result will then follow from Theorem 7.1 in
in \cite{CC} (recall that $F(\cdot)$ is a convex operator).

Let $\varphi$ be a $C^2$ function touching $u$ by below at $x_0\in B_1$. Recall that $u$ is a viscosity supersolution across the whole ball (disregarding if $x_0$ is in the contact set or not), so we have

\[
F(D^2\varphi(x_0))\leq f(x_0)\leq\|f\|_{L^\infty(B_1)}.
\]

If instead $\varphi$ touches $u$ by above, we separate two cases:

\textbf{Case 1:} if $x_0\in\Omega$ then $u$ is also a subsolution and we get

\[
F(D^2\varphi(x_0))\geq f(x_0)\geq-\|f\|_{L^\infty(B_1)}.
\]

\textbf{Case 2:} if $x_0\notin\Omega$, notice that $\varphi$ also touches $v$ by above, and $v$ is a subsolution for $G$ across the whole ball. Then

\[
G(D^2\varphi(x_0))\geq g(x_0)\geq-\|g\|_{L^\infty(B_1)}.
\]

But for any symmetric matrix $X$ we have $G(X)\leq F(X)$. Thus
\[
F(D^2\varphi(x_0))\geq g(x_0)\geq-\|g\|_{L^\infty(B_1)}
\]
and we are done.
\end{proof}

\begin{rem}
Notice that the proof is still valid if we just require $F(X)\geq G(X)$.
\end{rem}

Now we show that when problem is given by the Pucci extremal operators solutions are $C^{1,1}$ on the contact set (i.e. they have bounded second derivatives). More precisely:

\begin{prop}\label{prop.contactset}
Let $u$ and $v$ solve \eqref{eq.main} with $F=\mathcal{M}^+$ and $G=\mathcal{M}^-$. Then $u$ and $v$ are $C^{1,1}$ in $B_{1/2}\cap\Omega^c$ and
\[
 \|D^2u\|_{L^\infty(B_{1/2}\cap\Omega^c)}, \|D^2v\|_{L^\infty(B_{1/2}\cap\Omega^c)}\leq C
\]
for some universal constant $C$.
\end{prop}

\begin{proof}
By Proposition \ref{prop.cz} $u$ and $v$ are $W^{2,p}$ functions in, say, $B_{3/4}$ so we only need to show an almost everywhere bound on $D^2u$ and $D^2v$ in $\Omega^c$. Also, since the notions
of viscosity solution and strong solution coincide for $W^{2,p}$ with $p\geq n$ (see \cite{CCKS}) we have that \eqref{eq.main} is satisfied a.e.

If $x$ is a point in the interior of $\Omega^c$ for which \eqref{eq.main} is satisfied, $u$ and $v$ coincide in a neighborhood of $x$ and hence, letting $e_u$ and $e_v$ denote the eigenvalues of $D^2u$ and $D^2v$ respectively, we find
\begin{eqnarray*}
C\geq (f-g)(x) & \geq &  F(D^2u(x))-G(D^2v(x))\\
	       &   =  &  \Lambda\sum_{e_u>0}e_u+\lambda\sum_{e_u\leq 0}e_u-\lambda\sum_{e_v>0}e_v-\Lambda\sum_{e_v\leq 0}e_v \\
               &   =  &  (\Lambda-\lambda)\sum_{e_v}|e_v|
\end{eqnarray*}
and the result follows in this case.

If $x\in\partial\Omega^c$ (again, a point at which \eqref{eq.main} holds), $u-v$ has a minimum at $x$ and hence $D^2(u-v)(x)$ is nonnegative definite, which in particular implies $\partial_{ee}u(x)\geq\partial_{ee}v(x)$ for any direction $e\in S^{n-1}$.
Let us now pick a system of coordinates, say $\{e_1,\ldots,e_n\}$, in which $D^2v(x)$ is diagonal. Moreover let us assume without loss of generality that the first $m$ eigenvalues of $D^2v$ are nonpositive and the remainig
$n-m$ positive. Let then $A$ be a diagonal matrix with $\lambda$ in the first $m$ positions of its diagonal and $\Lambda$ otherwise. Since $A$ is a competitor in the $\sup$ and $\inf$ that define $F$ and $G$ respectively we have, using the equation,
\begin{eqnarray*}
C\geq (f-g)(x) & \geq &  F(D^2u(x))-G(D^2v(x)) \geq \textrm{tr}(AD^2u(x))- \textrm{tr}(AD^2v(x))\\
	       & = &  \lambda \sum_{i=1}^{m} u_{e_ie_i} + \Lambda \sum_{i=m+1}^{n} u_{e_ie_i}-\Lambda\sum_{i=1}^{m} v_{e_ie_i}-\lambda\sum_{i=m+1}^n v_{e_ie_i} \\
	       & \geq &  \lambda \sum_{i=1}^{m} v_{e_ie_i} + \Lambda \sum_{i=m+1}^{n} v_{e_ie_i}- \Lambda \sum_{i=1}^{m} v_{e_ie_i} - \lambda \sum_{i=m+1}^n v_{e_ie_i}\\
	       & = &  (\lambda-\Lambda) \sum_{e_v\leq 0} e_v + (\Lambda-\lambda) \sum_{e_v>0}e_v\\
	       & = &  (\Lambda-\lambda)\sum_{e_v}|e_v|
\end{eqnarray*}
so we get the bound for $D^2v(x)$. The proof of the bounds for $D^2u(x)$ is completely analogous.
\end{proof}

Finally, we can give the
\begin{proof}[Proof of Theorem \ref{thm.main}]
Again, we prove the result for $u$. Notice that, due to Proposition \ref{prop.cz}, $u$ is $W^{2,n}(B_{3/4})$. Moreover, by Proposition \ref{prop.contactset} the Hessian of $u$ is bounded a.e. inside the contact set in $B_{1/2}$, and hence we have
\begin{equation*}
  \left\{ \begin{array}{rcll}
  F(D^2u) & = & f(x) & \textrm{ in } B_{1/2}\cap\Omega \\
  |D^2u|   & \leq & C & \textrm{ in } B_{1/2}\cap\Omega^c
\end{array}\right.
\end{equation*}
and we can apply Theorem \ref{thm.indreimine} to get that $u\in C^{1,1}(B_{1/4})$ as desired.
\end{proof}

\section{Free boundary}
\label{sec.4}

The classic approach to study the regularity of the free boundary of the double membrane problem consists on substracting the two membranes (solutions), say $w:=u-v$, and reduce
the situation to an obstacle-type problem (note that $w$ thus defined is nonnegative). One of the key steps of the analysis of the free boundary is to show that $w$ satisfies a
non-degeneracy property around free boundary points; that is, given $x_0\in \partial\{ w>0\}$ one wants to show that
\begin{equation} \label{nondegeneracy}
\sup _{\partial B_r(x_0)}w \geq C r^2 \textrm{ \  for \ } r>0
\end{equation}
where $C>0$ is a universal constant.

In the case of \eqref{eq.main} this property is not satisfied. Indeed, let $C$ be any positive constant and consider
\[
u(x,y):=x^2-y^2+Cx_+^3\quad\textrm{ and }\quad v(x,y):=x^2-y^2.
\]
Here $x_+=\max\{x,0\}$. Notice that
\begin{equation*}
\left\{ \begin{array}{rcll}
  u & \geq  & v  & \textrm{ in }  B_1\\
  \mathcal{M}^+(D^2u) & = &  2(\Lambda - \lambda) + 6C\Lambda x_+    & \textrm{ in } B_1 \\
  \mathcal{M}^-(D^2v) & = & -2(\Lambda - \lambda)   & \textrm{ in } B_1 \\
\end{array}\right.
\end{equation*}
In particular $u,v$ solve \eqref{eq.main}, $0\in \partial\{ w>0\} $ and
\begin{equation}
\sup _{\partial B_r(0)}w = C \sup _{\partial B_r(0)}x_+^3= Cr^3 < Cr^2
\end{equation}
for any $r<1$.

In fact, by previous the following example we can see that no free boundary regularity can hold in general. If we make
\[
u(x,y):=x^2-y^2+\psi(x,y)\quad\textrm{ and }\quad v(x,y):=x^2-y^2
\]
with $\psi$ a nonnegative smooth function we can make the contact set arbitrarily bad and still get solutions of \eqref{eq.main}.

{\bf Acknowledgments}

The authors would like to thank the anonymous referees for their careful and detailed reading of this manuscript and for their suggestions and corrections. Also, Professor Sibru's comments on the probabilistic aspect of this problem were very enlightening.

The authors were partially supported by NSF grant DMS-1540162. The second author was also partially supported by Colciencias and the third author by a Conicet scholarship.

redpajama_set_name: RedPajamaArXiv


Has anyone seen these for sale in korea?
oh, I'm referring to the rollers, not the helmet.
Any suggestions on models? Aluminum vs. PVC?
no i haven't but the next time im over at the good shop in daejeon, i'll look. that might be next tues or thurs.
are you a roadie, ricardo?
I have seen some Korean riders warming up on such trainer before a MTB race, so I guess you can buy those here in Korea.
Rick I am sure you can find one here. I will take a look in OSAN this week some time. As for sugestions it really depends on what you want to do with it. If you are going to be doing some indoor training I would look for one where you remove the front wheel and lock in. I don't think I have ever seen a PVC one but it may exist.
Hey Eric, not really a roadie...don't have the hair for it.
seriously, I mostly ride for transportation but back home that sometimes added up to 35 miles. I've never gone long distance but I wouldn't be opposed to trying it out if get (and stay) in shape. I ride a solid fork mtn. bike with fat slicks...so you might say I'm indecisive...or confused.
Mike, I'd prefer the rollers over the stands partially b/c of the balance training but mostly b/c I'll actually feel like I'm riding. Plus (let my eco-geek shine!) someday I hope to hook it up to a generator/power storage unit.
In my small search I found units with aluminum or PVC rollers. PVC is considerably cheaper so I'm skeptical of the life span.
Thanks for keeping your eye out for me!!
i usually liked the rollers over a stationary trainer. rollers are more interesting because it requires your concentration where as a trainer is more monotonous.
one drawback about rollers, it's hard (or very, very tricky) to get out of the saddle for sprints or a leg stretch.
other than that, much more fun than trainers.
I've heard about that...I can imagine explaining an injury from a bike accident in my living room!!
Yeah. . . add that to "broke my wrist snowboarding in Korea" (ok, not in your living room) ouch. . .


text:
Harmony Yoga
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God-Head
By Stephen Brandon on February 27, 2014 in Ramaswami Articles, Yoga Philosophy
This article was written by Srivatsa Ramaswami and is reproduced here with his kind permission:
Many faiths say that God made man in his image. Conversely one may say that if you want to see God with a form He would have a human form– a person or purusha. Many puranas depict God in the form of a human being, as parama-purusha (Super-man) even as the upanishads talk of the ultimate reality as formless, yet He is depicted as having different forms. Subsequently the agamas gave various human forms to God so that devotees can worship with concrete form rather than try to image and meditate based on the way God is described in the puranas. The puranas by depicting God as a super human being with very charming stories about Him, were able to depict God as a person so that many devotees would be able to relate to the Lord more easily. Those charming stories all attempt to bring out the auspicious qualities of the Lord as a just, compassionate, guide, a peer–the puranic god one can empathize with. They help bring God from heaven to one's own home. It is easy to consider many of these –like Krishna, Rama, Siva—as a friend, philosopher, guide. The same Krishna who played pranks with Gopikas and his mother Yasodhara became a Guru to his friend Arjuna and gave very sane advice which is considered to be universally and eternally valid.
But God created other beings from the lovely polar bear to the annoying Madras mosquito. When I was young I used to spend some good time with my father's friend Harihara Iyer. Though he had no formal training in Indian philosophy, he had such a grasp of several aspects of the tradition he could breathe a lot of life into ordinary sayings of the scriptures and smritis. He would say something on these lines: 'God created the entire universe with thousands of species. If human beings think of God having a human form what about other beings say a buffalo. If a Buffalo were to know God how will it visualize Him, as a buffalo a super buffalo.'
Since the Lord created so many beings why not God have the forms of different species. Some upanishads refer to the formless Lord manifesting in different forms. Many Puranas have depicted God in different forms. Take the case of the most beloved of them all, Ganesa.
Ganesa, plump and with that elephant head is the chosen deity (ishta devata) of millions of people. Ganesa has endeared himself with some lovely stories and episodes that can be found in puranas. He is said to represent the Prithvi tatva or the earth principle. There is an interesting story about how he got an elephant head. There are also further charming stories about how he outwitted the more mobile brother Shanmukha by resorting to a simple procedure as paying respects to the parents by circumambulation. Equally fascinating is the story about how he wrote down the whole Mahabharata as dictated by sage Vyasa, the episode giving details about how each one kept the other in check by certain preconditions. All these stories, the several beautiful forms of Ganesa dancing, eating a sweet (modaka) have endeared Him to millions. Small shrines for Ganesa in almost all the temples in India make Him always available to communicate with. People participate in Ganesa festivals with great enthusiasm, love and reverence. All these attempts make God accessible rather than viewing Him as a distant entity.
Hanuman or Anjaneya with the famous head of a monkey is another popular deity who again finds huge following. He is the chosen deity for many yogis who give considerable importance to yamaniyamas especially Brahmacharya. Yogis have designed a few asanas after him like Haanumanasana, Anjaneyasana with a few artistic vinyasas or variations. Hanuman could mean one with a prominent jaw. Hanuman was considered the son of the deity Vayu and Anjana. He is therefore known as Vayusuta and Anjaneya. One reason why Anjaneya is worshiped is to remove the obstacles and sorrows one experiences inevitably in life due to afflictions of the planet Sani or Saturn. There are several stories as to how Sani affects everyone directly for more than 40% of one's life and indirectly also. Orthodox people dread the arrival of the sani afflicted period like when it occupies the 12th, 1st, 2nd, 4th,and 8th houses in one's horoscope or birth chart. There is no one who can escape the afflictions of Sani it is believed. Even Lord Siva is said to have suffered at the hands of Sani albeit for a short period of time (muhurta). There are very few temples for Sanaischara or the slow moving Saturn. So what can one do? Worship Hanuman instead. How does worshiping Hanuman help prevent Saturn's afflictions?
In the Ramayana, Lord Rama takes his army mainly made up of monkeys to Kishkinda of Sugriva. To cross the Indian Ocean to reach Lankapuri of the demon Ravana, all were busy getting building materials and constructing the bridge. Anjaneya the strongest of all was carrying huge boulders for the work. But it was time for Sani to afflict Anjaneya, Sani does not bother to wait. So he approached the busy Anjaneya and told him that it was time for the Sani period for Anjaneya. Anjaneya asked him what will happen if you afflict me? You will lose everything including your home, be separated from your family. Anjaneya said, "No Problem" I have no family, I have no possessions, and my home is my Lord Rama's feet. Then Sani said, "Ok, I will affect one part of your body and after the stipulated period, I will leave. Can I get into your knee so that you will suffer from arthritis". Anjaneya said: 'Not now as I have to work to complete the construction of the bridge across the ocean.' Ultimately Anjaneya let Sani get into his head as he did not want to think much while doing the basically hard manual labor. Sani got into the monkey head. Sani to his dismay started feeling huge pressure as Anjaneya was busy carrying mountain sized rocks and sometimes smashing the rocks on his head to break them to smaller pieces for the construction work. Instead of bothering Anjaneya, Sani started feeling the same misfortune others experience when he afflicts them. With the pain unbearable, Sani cried out to Hanuman and wanted to get out of his head. Anjaneya agreed but on one condition. The condition was that Sani should not afflict any of Anjaneya's devotees. Sani agreed immediately, he was such pain that he would agree to any condition at that time. We know now why Hanuman has so many devotees. Hanuman Chaleesa is a very popular prayer chanted by hundreds of Anjaneya devotees. Sundara Kanda of the Ramayana contains the grand episode of Hanuman leaping over the Indian Ocean to Lanka in search of Rama's wife Sita. This portion is called Sundara Kanda, chanting of which is said to remove all obstacles and recovery from absolutely hopeless situations. It has about 2800+ slokas or verses and it takes close to 10 hour to chant it. I have chanted this section and was recorded by my recording company Sangeetha which is still available in India in a twin volume package.
Hayagriva is said to be the Lord Himself with a horse face and a human body. He is said to be the repository of all vedic knowledge. My Guru used to chant a verse in praise of Hayagriva as a purva santi in many theory classes. Narasimha, the man lion avatara of the Lord, is another popular deity form worshipped. And of course our dear friend, philosopher and guide, Patanjali an incarnation of Adisesha himself and an aspect of the Lord, is said to have a human body but a thousand cobra hoods (heads).
The idols of these gods are worshiped in temples and in homes as well. Some philosophers of the upanishadic school worry that such form worship may prevent the worshiper from understanding the ultimate reality, the formless Brahman. But then these stories and icons help the devotees to remain more easily focused on the chosen deity and the Supreme. And the personal deity (ishta devata) has the powers to confer all the wishes. If the devotee wants kaivalya or jivan mukti the chosen deity will give it gladly. In fact a mantra found in Lalita ashtottara praises the Goddess as the one that gives svarga or heaven and/or apavarga, complete freedom/involution
Svarga apavargadayai namah
(c) Srivatsa Ramaswami. www.vinyasakrama.com
Born in Solitude
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Summer Yoga Workshops in Wells, Somerset, August 2020
Completion – Part One
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Three players who can make a move away from Barcelona in January
Barcelona saw a lot of players enter and exit the club during the busy summer transfer season. Despite a discouraging 2-0 loss to Bayern Munich on Wednesday night in the Champions League, the Catalan giants have generally had a good start to the season.
According to a recent source, the La Liga heavyweights are still under financial strain, which might prompt them to try to offload more players during the January transfer window.
Let's have a look at three players who can probably make a move away from Cam Nou in the upcoming winter transfer:
Although the seasoned centre defence has struggled for playing time thus far in the season, Barcelona manager Xavi recently claimed that he "relies" on Pique.
"I spoke with Gerard Pique. The current situation is not easy for him, but he played well against Viktoria Plzen and put in a great run. I rely on him,"
Xavi told reporters.
In 607 appearances in all leagues with Barcelona, the 35-year-old has contributed 53 goals and 15 assists. With two years left on his contract, he has won three Champions League championships, seven Copa del Rey medals, and eight La Liga championships while playing at Camp Nou.
Pique has only appeared twice this season and is lagging behind Ronald Araujo, Eric Garcia, Andreas Christensen, and Jules Kounde for the starting center-back position.
The former Spain international played in the second half of last week's European match against Viktoria Plzen before playing the entire 90 minutes against Cadiz last time out. However, Pique is no longer the team's first choice, and it's possible that he will be viewed as surplus to requirements at the beginning of 2023.
Despite a terrible 2021–22 season, Jordi Alba was probably one of Barcelona's top players. However, the 33-year-old is no longer Barcelona's first choice. Alejandro Balde, a teenager, has been favoured in the league, and Marcos Alonso, whose future is uncertain, began against Bayern on Wednesday in the Champions League.
The Spanish player, whose contract still has two years to run, was purportedly offered on loan to Inter Milan in the closing hours of the summer transfer season. Alba has expressed his dissatisfaction with the way things are going, and it does seem as though his time at Barcelona is coming to an end as the team looks to the future.
The defender joined Barcelona from Valencia in 2012, and since then, he has made 433 appearances for the team across all competitions, scoring 25 goals and dishing out 94 assists.
If it became known that he was available in January, there would be no shortage of suitors, and it may be viewed as a risk to allow the seasoned performer go given his status.
Memphis Depay was rumoured to be interested by a number of clubs, therefore it was expected that he would depart Barcelona this summer. The 28-year-old was rumoured to be on the verge of signing with Juventus, but a deal to join the Italian powerhouse fell through at the last minute.
He reportedly had interest from Manchester United and Chelsea as well, but he remained and declared in the closing days of the transfer window that he was "totally dedicated" to the team.
Although there are a number of players ahead of Memphis in the pecking order, Memphis has only featured twice this season, appearing for 57 minutes in La Liga and 15 minutes in the Champions League.
Although the forward's deal at Camp Nou is set to expire in June of next year, there is a chance that he might leave in January on a free transfer.
BarcelonaGerard PiqueJordi AlbaMemphis DepayShare00
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The ObserverArms trade
This article is more than 8 years old
How an illegal deal for 80,000 weapons led to the downfall of Britain's 'lord of war'
Gary Hyde was seen as a pillar of his Yorshire community. Now he awaits sentencing over a £6m arms shipment
Gary Hyde pictured in 1996.
Jamie Doward
Sat 1 Dec 2012 19.06 EST
To his neighbours in the Yorkshire village of Newton upon Derwent, Gary Hyde, 43, was a respectable figure. A former special constable, he lived with his wife and two children in an attractive cottage near where he was born. Local people considered him a pillar of his East Riding community. His reputation was bolstered in the 1990s when he received a citation for helping to foil an armed raid during which a shotgun was pointed at his head. "It was terrifying – the worst moment of my life," he said at the time.
The experience did not deter him from pursuing a career as a gun dealer, his trajectory similar to that enjoyed by the UK's many other "lords of war" who broker huge weapons shipments without ever visiting the unstable countries they supply to see the consequences of their actions at first hand.
After leaving school in 1985, Hyde took a job at a local store, York Guns, that sold everything for field sports and shooting enthusiasts. When he became the store's managing director, his business acumen was eagerly sought. Indeed he was considered such an expert on international arms laws that, according to documents obtained by the Observer, he attended exclusive briefings at the United Nations Institute for Disarmament Research. He won contracts from the Ministry of Defence to bring in decommissioned weapons from conflict zones, and from the American government to supply anti-Saddam forces in Iraq.
But just like Robert Louis Stevenson's Mr Hyde, this was a man with two very distinct sides. And on Wednesday he will be sentenced at Southwark crown court after being found guilty of illegally supplying what prosecutors described as a huge shipment of arms from China to Nigeria.
The deal involved shipping 80,000 weapons, including AK-47s and Makarov pistols and 32m rounds of ammunition, a transaction that would have required 37 sea containers, each 6 metres long. It involved transactions stretching across Germany, China, Poland and Dubai.
According to documents obtained by the Observer, the $10m (£6m) Nigerian deal was supposed to earn Hyde more than $1m, which he intended to hide in a secret family trust in Liechtenstein controlled by a company based in the British Virgin Islands.
Ironically, Hyde's downfall owed much to a misplaced confidence that by structuring his deals offshore he could stay outside British law. As he protested upon his arrest, "I do not believe that I engaged in any activity in the UK which I understood required a licence."
However, following a search of his business premises and home, investigators from HM Revenue & Customs found emails on Hyde's computer confirming his role in the deal. Mobile phone records proved that he was in the UK when the shipment arrangements were made. It was a crucial breakthrough because it meant Hyde required a UK control licence, even though the guns never came into this country.
"UK legislation controls transfer of arms and military goods from one third country to another," said Elspeth Pringle, a Crown Prosecution Service (CPS) lawyer who specialises in arms prosecutions. "He failed to apply for a licence from the UK licensing authority. In this instance it didn't matter that the arms were not leaving the UK."
Hyde's arrest and conviction, following a five-year investigation, was a coup for the CPS and HMRC. "It was one of the largest cases I've ever come across in relation to the quantity of arms and ammunition," Pringle said. "It was unusual, too, in that it was a single deal. Other cases have often involved multiple deals to multiple customers or clients."
Documents suggest there was no shortage of companies offering to sell Hyde AK-47s – the world's favourite, and most poorly regulated, killing machine, according to Amnesty International, which estimates that there are 100m in circulation. What other deals did he broker? Only last year Hyde was arrested in connection with the alleged illegal importation into the US of almost 6,000 AK-47 magazines.
Confidential US embassy cables reveal that in 2008 York Guns tried to ship 130,000 AK-47s to Libya. Hyde acted as an intermediary in the deal, which, the cable noted, "raised eyebrows in diplomatic circles, as Libya has only 70,000 ground-force troops and these would be unlikely to use a weapon as dated as the AK-47". The export licence was rejected amid fears that the weapons would be re-exported to rebels in Sudan and Chad.
In 2007 the Observer first drew Hyde's activities to the attention of the UK authorities, reporting that companies in which he had an interest had sold tens of thousands of guns to Ziad Cattan, a former pizza parlour owner who became Iraq's head of military procurement, without the appropriate arms brokering licence. Many of the weapons shipped to Iraq disappeared after delivery.
"We could never nail down that the missing weapons went to insurgents, but some likely went astray because it appeared to us nobody was checking and monitoring them properly," said Oliver Sprague of Amnesty. Amnesty had flagged concerns about York Guns a year earlier, after it had imported thousands of assault rifles and machine guns from Bosnia for reasons that remain opaque.
A repeated failure by the international community to agree binding arms controls means there will be many other dealers like Hyde who will continue to broker huge weapons deals with impunity. But Pringle believes the Hyde case should be a warning to the UK's dealers. "The number of arms dealers in the UK is quite small," Pringle said. "They all know each other. This [conviction] sends out a signal that if you don't have the correct paperwork and you ship directly from the UK or from one country to another you will be prosecuted."
York Guns's files reveal that on 11 September 1995 Hyde dispatched a 9mm Browning pistol to a customer in Scotland who bought it for £304. Days later the buyer returned the weapon and phoned Hyde to say he was unhappy with its condition. "I then asked him if he would like it if we tidied it up a little bit and put in a few extra bits and pieces … and we would reduce the price, give him a part refund and send the gun back to him," Hyde recalled under questioning.
Asked at the Dunblane inquiry how much of a refund he had given to Thomas Hamilton, the former Scout leader who used the Browning that Hyde had sold him in a killing spree that left 15 children and their teacher dead, he paused. " … It was about £50," he replied.


Conspiracies rationalize our powerlessness. When our efforts to combat large-scale evils bear little fruit, it must be because a massive, organized "they" are out to get us.
Conspiracies make us feel important. We're not just minor cogs in a machine; if people out there are conspiring against us, we must be worth targeting!
That may sound a little patronizing, but what is the most loving, appropriate Christian response to excessive credulity? If I really believe the king's heart is in the hand of the Lord (Pro. 21:1), that all the nations are a drop in the bucket (Isa. 40:15; Ps. 2), my political activity, while zealous like everything I do for the Lord, won't be infused with anxiety or agitation. I won't grasp for outlandish explanations for my nation's woes (Isa. 8:12-13). Indwelling sin, "the rulers of the darkness of this world," and God's ultimate sovereignty are reason enough (Ps. 103:19). And I won't feel the urge to endlessly forward right-wing (or left-wing!) political action notices to all my friends.
Sometimes a little visit to snopes.com or an earnest attempt to find a responsible voice is all a would-be forwarder needs to do. Generally speaking, however, it takes time to develop a feel for which voices are worth listening to. Responsible voices tend to coalesce into a community. They cite one another's work, even if they disagree over it. They don't sound breathless or screechy. They do a lot of homework. They write very well English, and they write a lot of it. They don't accuse all their opponents of being secret agents of Fidel Castro or (on the other side) Pat Robertson. And they don't find conspiracies under every rock.
Thanks, Mark – I agree wholeheartedly!


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Who IS Really Pulling The Strings
Chris Muir's Day by Day Cartoon for Monday nails it as to who is really calling the shots in Virginia.
I guess that is what Northam will be doing this afternoon – shuffling out there like a good boy for the media.
Shannon Tells A Whopper
The pantheon of lies include the check is in the mail, I'll respect you in the morning, and I'm from the government and I'm here to help. Thanks to Shannon Watts, we can add another one to that list.
In an email to the Demanding Mommies as reported by Mother Jones, she wrote:
Watts said that Bloomberg's entry into the race would "ensure gun safety remains front and center during the 2020 election season" but affirmed that nothing would be asked of its volunteers to support Bloomberg's candidacy. "We are a single issue organization—but none of us live single issue lives," Watts wrote. "I want to be clear that it is entirely your choice regarding how you spend your volunteer time in this election and which candidates you choose to support. Elevating the voices of all gun sense champions is part of what you do every day, and that should continue."
Yeah, right. Tell me another one, Shannon. Without Bloomberg's money backing you, Everytown, and the Demanding Mommies, you would be just another out of work PR executive who snagged a rich husband.
TV Ad Execs And Political Consultants Are Saying Hallelujah!
Move over Tom Steyer. There is another narcissist Democrat billionaire who wants to be President.
From a report in the Wall Street Journal.
Billionaire Michael Bloomberg, the former mayor of New York City, said he would join the Democratic presidential race despite facing a tough path to the nomination in a crowded 2020 field.
Mr. Bloomberg's advisers have been drawing up plans to build his campaign staff, intending to focus on the delegate-rich Super Tuesday states, which vote on March 3.
Many of the Democratic candidates have invested fewer resources in those states than in the first few nominating contests in Iowa, New Hampshire, Nevada and South Carolina.
The primary beneficiaries of Bloomberg's largess would seem to be advertising executives and political consultants as he plans to spend big.
According to the Wall Street Journal, Bloomberg plans to spend over $100 million on his campaign.
Ahead of his announcement, Mr. Bloomberg said he would spend $100 million on digital ads against Mr. Trump and another $15 million to $20 million on registering new voters.
He had also begun assembling a big staff for his campaign-in-waiting that was working out of the offices of Bloomberg Philanthropies, which oversees the former mayor's charitable giving, according to a person familiar with the matter.
While the many media stories on Bloomberg's presidential run mention his wealth, none so far mention gun control. However, Bloomberg himself does list that as one of the issues he will address in his campaign. It is number three on the list on his campaign website.
With all the money that Bloomberg plans to spread around, perhaps the gaggle of corrupt MAIG mayors will seek some of it in return for campaign support. That way then they can stop indulging in public corruption like the former mayor of Baltimore.
DICK'S Ed Stack On CBS Sunday Morning
October 6, 2019 by John Richardson 3 Comments
When CBS Sunday Morning was announcing their stories this morning, I must admit I rolled my eyes when I heard they were interviewing Ed Stack of DICK'S. Thanks to Dianna Muller, it was more balanced than I anticipated.
The story had interviews by CBS's Lee Cowan with Stack, Dianna Muller, and Michael Bloomberg. Dianna provided a great counter-point to Stack and Bloomberg. It should be noted that Stack is pushing his new book, It's How We Plan the Game, which is being published by another division of the company that owns CBS.
Then there was this quote that must have executives with NSSF thinking someone is finally getting the language right:
The "it" he's talking about is the AR-15, a lightweight semi-automatic modern sporting rifle similar to the one used in the Sandy Hook massacre. He ordered all of them be removed from every Dick's Sporting Goods store across the country.
While Cowan did link it to the Newtown murders, he didn't call it an "assault weapon" or "assault rifle". That in and of itself is remarkable nowadays.
His interview with Dianna was quite good. As you can see if you watch the video, it included footage of her shooting as well as other women from A Girl and A Gun at an event in Kentucky.
When asked why she has an AR-15, she said this:
Muller has become a high-profile spokesperson for the gun rights movement. She's testified on Capitol Hill about carrying a firearm, specifically an AR-15.
That's the very rifle that Dick's is no longer selling.
Cowan asked, "Is it a fair question to ask why you need a gun like that?"
"No," Muller replied.
"This rifle, and any other rifle, kills fewer people than hammers and blunt objects every year, according to FBI statistics," she said, "so it doesn't make any sense to me that this is going to solve the problem that we are having."
Her worry is the same that the National Rifle Association has voiced for a long time: if one gun like the AR-15 is demonized, then all guns may soon follow.
"It leads me to believe that there's going to be another tragedy with a different gun, that they're going to come after the next gun," she said.
"This is the slippery slope?" Cowan asked.
"Until it's all gone."
The story ends with Stack being somewhat coy about his future plans for firearm sales at DICK'S. He said it was under "strategic review." As I noted on Friday, his company is selling eight of their Field & Stream stores to Sportsman's Warehouse. I would not be surprised to see the remainder either sold or shut down.
What Michael Bloomberg Bought With Pocket Change
Mark Johnson (R-NC) is the Superintendent of Public Instruction for the State of North Carolina. When he beat incumbent June Atkinson in 2016, he became the first Republican elected to that position in over 100 years. In his role as Superintendent of Public Instruction, he has control over the public schools in the state including their policies and their curriculum.
Despite being a Republican, Johnson has always been somewhat suspect in my estimation. When the Republican receives a donation from Michael Bloomberg, I am suspicious. I wrote in 2016 about how this made me feel wary of him. I found it interesting that Bloomberg contributed $5,100 to him and nothing to Democrat June Atkinson who was a strong supporter of Common Core. Bloomberg's daughter Emma also contributed another $100 to Johnson. I noted that contributions like this come with implicit strings attached.
Johnson started to show his true colors back in 2018 when he opposed allowing teachers with the requisite training from being armed. Johnson said at the time, he wanted firearms on campus restricted to "these trained, uniformed law-enforcement professionals who courageously choose a career protecting citizens from violent threats."
As bad as that was, it wasn't too far out of the mainstream even for a Republican. His latest move, on the other hand, involves partnering with anti-gun group Sandy Hook Promise to set up an anonymous reporting system. The system will involve a mobile app with the tips monitored by volunteers with Sandy Hook Promise.
NC State Superintendent Mark Johnson announced today that the state education agency has contracted with Sandy Hook Promise (SHP) to provide a statewide Say Something Anonymous Reporting System in the 2019-20 school year. The Say Something program, including an anonymous mobile tip app, is a school safety program designed to change and save lives by teaching students, educators, and administrators how to recognize the signs and signals of those who may be at risk of hurting themselves or others and to anonymously report this information through the mobile tip app, the website or the telephone crisis hotline.
"Students play a critical role in helping to keep schools safe," Johnson said. "They may see and hear concerns that adults need to know about but may be reluctant to report it. With the Say Something program, middle and high school students will better understand what warning signs to look for and when and how to report important tips through an app. Making this app available will be an important part of our efforts to make schools safer." "We are proud and eager to work with the North Carolina Department of Public Instruction to train students across the state to 'know the signs' of potential violence and report them to a trusted adult via the Say Something Anonymous Reporting System," said Nicole Hockley, co-founder and managing director of Sandy Hook Promise, and mother of Dylan, who was killed in the Sandy Hook Elementary School tragedy. "With these comprehensive violence prevention systems in place, North Carolina schools will be safer, protecting millions of lives and empowering youth to be upstanders in their communities."
Through the Say Something program, students, parents, educators and others can download the app and share school safety concerns with school administrators and school resource officers at schools, who will respond appropriately. More than 5,100 schools nationwide are currently using Sandy Hook Promise's anonymous reporting system. North Carolina will be the second statewide partnership for Sandy Hook Promise, with implementation of an anonymous reporting system that will encourage student participation in protecting themselves and their peers.
According to the Raleigh News and Observer, the 2018 state budget allocated $5 million to fund the project. It is estimated that it will cost about $650,000 per year to run it. While the release from Johnson and Sandy Hook Promise wasn't explicit about this, the N&O reports that the "command center" will be staffed by Sandy Hook Promise and not local NC law enforcement.
An anonymous reporting system with no connection to local law enforcement is an invitation for abuse. I see a lot of kids getting back at other teens for imagined slights, I see "swatting", and I see innocent people getting killed. This is "pre-crime" brought to the schools of North Carolina and civil liberties be damned. That is what Michael Bloomberg bought with his pocket change.
Don't Shoot The Messenger, Part 2
May 8, 2019 by John Richardson 5 Comments
Just like Chinese water torture, I think we are in for a series of NRA articles from The New Yorker detailing self-dealing, lavish spending, accounting irregularities, nepotism, and sweetheart deals at the National Rifle Association. The latest episode was released yesterday and is entitled, "An Internal Memo Raises New Questions About Self-Dealing At The N.R.A."
The article starts off describing a one and a half page memo from the NRA's accountants to the Audit Committee of the Board of Directors. The memo details a range of "questionable transactions and business arrangement" involving several of the NRA's top vendors and executives. This memo, unlike earlier leaks, does not go into the Ackerman McQueen issues but rather deals with other items that were found to be irregular. These include reportedly payments made to former CFO Woody Phillips' "significant other", cumulative rent of $1.8 million paid for a house to be used by Associated Television International and owned by ATI's president, and to their fund-raising contractor Membership Marketing Partners.
The memo goes on to question management overrides and approval of housing expenses for certain upper executives.
In addition, the memo drew attention to "senior management override of internal controls," which led to violations of "accounts payable procedures" and "HR policy," including "hiring of staff without HR knowledge." It names four executives who, at the time, were receiving "reimbursement of expenses relating to apartments and living expenses beyond HR Policy Manual stipulations and on a permanent basis." The N.R.A.'s accountants added that there was "no contract to support the reimbursement request," which the four individuals continued to claim as a "relocation expense." The executives named include Doug Hamlin, the N.R.A.'s executive director of publications; Eric Frohardt, the director of education and training; Joe DeBergalis, the executive director of general operations; and Josh Powell, LaPierre's chief of staff.
Andrew Arulanandam, the N.R.A.'s managing director of public affairs, said that the organization "has, at times, made such accommodations for employees who travel extensively for their jobs." He added, "The practice of providing such accommodations is approved by N.R.A. leadership and is not uncommon for an organization the size of the N.R.A."
Powell is the person responsible for bringing in CarryGuard while Eric Frohardt is the former Navy SEAL whom Powell installed as director of education and training and director of training for CarryGuard. Frohardt still lives in Colorado where he owns a range and other businesses according to his LinkedIn page. It is my understanding from those who would know that Frohardt is flown in at the NRA's expense to work 3-7 days a month. While I have the utmost respect for Frohardt's service to the nation, 12 years as a Navy SEAL does not make one an expert in training civilians in the legal use of a firearm.
As to Josh Powell, the memo to the Audit Committee mentions his multiple conflicts of interest including the hiring of his dad to do photography for the NRA and his wife, Colleen Gallagher, was hired by a top NRA fund-raising vendor McKenna and Associates. It gets worse.
The N.R.A.'s accountants completed their memo in mid-July. Around this period, the N.R.A.'s new C.F.O., Craig Spray, had to temporarily step away from his role at the organization to deal with a health matter. Someone would need to take his place as the organization's chief manager of financial activities. According to an internal N.R.A. communication, in July, 2018, Powell was appointed acting C.F.O. for about three weeks, placing him in charge of the accountants who documented his conflicts of interest.
I won't get into the other issues with regard to Powell other than to say his departure from the NRA would help the organization. Placing him as the senior strategist to work with outside counsel William Brewer on New York litigation is a disaster in the making.
I would be remiss if I didn't note that a lot of the research on the NRA's problems come from Bloomberg's The Trace and they did help with The New Yorker articles. I have a theory as to how they are gathering this information. I think Michael Bloomberg and his associates have hired a corps of private investigators whose job it is to find former NRA employees who have left because they were disgusted with the self-dealing and other financial issues. I can't think of any other way that they could be gathering this inside information unless it was being funneled to them by Ack-Mac.
One way or another the NRA will get its house in order. It can be done either by the Board of Directors or it will be done for them by the State of New York, the Internal Revenue Service, and other outside agencies. Far better that the changes come from within than from without. It can be controlled and managed to make the organization stronger, bigger, and more diverse.
My fear is that new officers of the NRA – Carolyn Meadows, Charles Cotton, and Willes Lee – and much of the Board are such stalwart Wayne LaPierre supporters that they will go along with the status quo (ante bellum) to the NRA's detriment. Ignoring it is not going to make it go away and will only make matters worse. That, however, is the most probable outcome as things stand now.
NRA Attendance And Voting Statistics
April 30, 2019 by John Richardson 13 Comments
The official attendance at the NRA Annual Meeting in Indianapolis was approximately 81,000. This is about 6,000 less than last year's meeting in Dallas. I will say I had no trouble getting about on the floor nor in the hallways of the Indiana Convention Center.
Now to more important statistics.
At the Meeting of Members on Saturday, William Satterfield, chairman of the Elections Committee announced the statistics for the 2019 NRA Board of Directors election.
2,452,893 ballots were bound in the official magazines of the NRA
145,920 ballots received back
141,101 ballot were valid
4,819 were invalid
Ballots that were deemed invalid for a variety of reasons. Among the reasons enumerated by Mr. Satterfield were arrival after the April 7th cutoff date, more than 25 candidates marked, no candidates marked, and some were actually prior year ballots. I think he said there was even one ballot from 2009.
When you do the math, you are left with the conclusion that most members eligible to vote – Life members and 5-year continuous members – don't give a big rat's ass who serves on the Board. The response rate for just returning the ballot was 5.94% and the response rate for a valid ballot was 5.75%.
That is pathetic. Moreover, it is scary. Think what could happen to gun rights if Michael Bloomberg just spent a few million bucks on advertising to run a slate of stealth (or Manchurian) candidates. With the proper slate and modern ad targeting, it would be possible to take over the organization in as little as two years. Whether it could be kept secret for that long is another matter.
Yes, Thank You Moms Demand Action In NC
Mike Bloomberg, billionaire, former mayor of New York City, erstwhile potential Democrat candidate for President, and funder of all things gun control put out this tweet yesterday evening.
Thank you to every @MomsDemand volunteer here in North Carolina for your leadership. State by state, we're making progress to pass laws that will save lives, and you all deserve a lot of credit for that. pic.twitter.com/Ni0GmtFK4r
— Mike Bloomberg (@MikeBloomberg) December 19, 2018
Yes, thank you for your continued support of a Jim Crow era law, the pistol purchase permit system, that was intended to keep African-Americans, union members, and Republicans disarmed.
Yes, thank you for your support of red flag laws that turn Due Process on its head and that will get innocent people killed.
Yes, thank you for making our schools less safe by your continued opposition to any policy that would let trained teachers and administrators carry firearms to protect the students under their care.
Yes, thank you for supporting efforts to introduce even more government interference into private affairs by demanding universal background check.
Yes, thank you for all you do to make North Carolina a less safe place for ordinary, law abiding Tar Heels and more safe for criminals who, by definition, ignore the law.
There is plenty more for which we can thank the North Carolina contingent of Moms Demand Action but saving lives isn't one of them.
A Takeaway For Gun Owners From Pelosi's Campaign For Speaker
Nancy Pelosi (D-CA) will probably be the next Speaker of the House of Representatives. I say probably because there are some Democrats that are opposed to her and the aging leadership. She, Steny Hoyer (D-MD), and James Clyburn (D-SC) are all in their 70s. However, while this group of dissidents might prevent Pelosi from getting the requisite number of votes (218), it is doubtful they will win in the end.
In a story in the Washington Examiner about Pelosi's campaign for Speaker and the battle her allies are waging on her behalf was this:
As the leadership fight escalates, Pelosi's lined up endorsements from former President Barack Obama — who called Pelosi "one of the most effective legislative leaders" on Tuesday — and key advocacy groups that helped Democrats retake the majority. Progressive groups, gun control organizations, unions, reproductive rights groups like NARAL Pro-Choice and more have thrown their weight behind Pelosi's speaker bid.
"Do you really want to make enemies out of these groups?" said (Rep. Jan) Schakowsky, adding later, that the fight "is not just inside baseball, this has enormous ramifications for how we successfully move forward to pass an agenda that's going to help people."
The Brady Campaign and Shannon Watts have all formally endorsed Pelosi for Speaker. Moreover, Michael Bloomberg who "invested" over $100 million in this election cycle to get Democrats elected has worked closely with Pelosi and shared office space with her super PAC.
What does this mean for gun owners?
It should be obvious – expect gun control to be on the Democrats agenda in the House from day one of the new Congress. Pelosi has a constituency to repay and she will. I fully expect that bills will be introduced that will contain every gun control measure that you can think of and some that we haven't. Moreover, these bills will get a hearing in the House Judiciary Committee and most will pass in the House. Furthermore, the Hearing Protection Act, national reciprocity, and even provisions for use of Pittman-Robertson funds to construct shooting ranges are dead.
Our only hope for stopping these infringements will be the Senate which remains in Republican hands. Even then, there are some Republicans who might vote for things like red flag laws or bans on bump fire stocks. Fortunately, the filibuster is still alive when it comes to bills passing the Senate and it still will require 60 votes to invoke cloture. Therefore, if you haven't contacted your two Senators or Senator-elect, you better do it now and put opposition to gun control on their radar.
The Chinese curse "may you live in interesting times" will epitomize the next two years when it comes to the fight for gun rights.
What Is Wrong With This Statement?
February 21, 2018 by John Richardson 0 Comments
I read a story this morning from the local NBC station in San Francisco. It detailed how the prohibitive gun laws in California and especially San Francisco made trafficking in stolen and/or illegal guns a very lucrative business. That comes as no surprise to anyone who reads this blog or has any knowledge of the Prohibition Era.
The story contained this gem. Let's dissect it to see what's wrong with it.
Roger contacted NBC Bay Area following a joint investigation with the non-profit journalism organization The Trace, which found more than 20,000 guns across the country that were previously reported stolen were later recovered by police in connection to crimes.
Stolen guns used in crime? That makes sense as guns are stolen to be sold or traded to other criminals who are probably already prohibited persons. Guns are a tool of the trade for armed robbers and home invaders and they will do what is necessary to obtain them.
Roger? Nope that was just the fake name used by the criminal for the story. He's now in a witness protection program.
The non-profit journalism organization The Trace? Ah, there's the problem with the statement. While its tax status is irrelevant, calling The Trace a "journalism organization" is one hell of a whopper. It is the Bloomberg-funded media propaganda arm that masquerades as journalism. Funding for it also came from the Joyce Foundation and Seattle anti-gun billionaire Nick Hanauer. Following the money always will lead you to the truth.


Stuart - love the post, the passion, the detail and the information in it, very much 'you' but such a great image of the students and love the message behind it. Such a positive experience for the students and the message is so important - and I completely agree with the sentiment. It goes without saying its one of those images that will stay with you forever - a magic moment that as educators we live for.


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Merseyside Bus Diversions
Victoria Road (Freshfield) – Route 22 is affected.
Victoria Road will be closed, between Golf Road and Gores Lane, from Monday 21 October 2019 for approximately 3 weeks.
Bermuda Road (Moreton) – Routes 407 and 423 are affected.
From Wednesday 23 October 2019 buses will be diverted away from Bermuda Road, Moreton, due to the location of traffic management in relation to gas main renewal works.
Stonebridge Lane (Croxteth) – Routes 102, 121, 898 and 899 are affected.
Bus services along Stonebridge Lane will be diverted after 1800 each evening from Monday 28 October 2019 until Wednesday 6 November 2019.
Edge Lane will be closed, at Buckley Hill Lane, from Wednesday 30 October 2019 for approximately 7 weeks.
Hall Lane (Cronton) – Routes 10, 26 and 26A are affected.
Hall Lane will be closed, at Smithy Lane, from Wednesday 23 October 2019 for approximately 3 weeks.
Chain Lane (St Helens) – Routes 40 and 742 are affected.
Chain Lane will be closed, between Blackbrook Road and Link Avenue, from Monday 27 October 2019 for approximately 2 weeks.
Liverpool City Centre Connectivity Scheme – Routes 10B, 12, 13, 79, 79C, 79D, 464.
Due to increased congestion in Liverpool city centre relating to the connectivity scheme, some services are diverted.
Woolfall Heath Avenue (Huyton) – Routes 7, 9 and 46 are affected.
Woolfall Heath Avenue will be closed (westbound only), between Seth Powell Way and Endmoor Road, from Thursday 24 October 2019 for approximately 1 month.
Bradlegh Road (Newton-le-Willows) – Routes 22, 34A, 39, 602, 654 and 725 are affected.
Bradlegh Road will be closed, between Wargrave Road and Vulcan Park Way from Monday 4 November 2019 for approximately 2 weeks.
Gores Lane (Crank) – Route 157 is affected.
Gores Lane will be closed, between Alder Lane and Red Barn Road, on Thursday 7 November 2019 between 0930 and 1530.
Main Street (Billinge) – Routes 137 and 352 are affected.
Main Street will be closed, between Newton Road and Pingot Road, on Friday 8 November 2019 from approximately 1000 until 1200.
Old Chester Road and Bebington Road (Bebington) – Routes 38, 38A, 410 and 418 are affected.
Old Chester Road and Bebington Road will be closed, between Autumn Grove and Woodburn Boulevard, from Monday 4 November 2019 for approximately 5 nights, between 1900 and 0000 each night; and on Monday 11 November 2019 between 1900 and 0000.
Dale Street (city centre) – Routes 54, 56 and 58A are affected.
Dale Street will be closed, between Byrom Street and Hatton Garden, from 1000 until 1900 on Friday 11 October 2019; and from 0600 on Monday 14 October 2019 for approximately 2 weeks.
Burlington Road is closed every Saturday until Saturday 30 November. This is affecting the 101, 102 and 103 services.
Victoria Street will be closed on Saturday 9 and Sunday 10 November. The following services will be affected: 8 10 27 67 92 93 97 98 100 Free Bus 2 & 3 X41 and X43.
Chapel Street will be closed on Saturday 9 and Sunday 10 November. The following services will be affected: 8, 10, 27, 67, 92, 93, 97, 98, 100, Free bus 2, Free bus 3, X41 and X43.
Greater Manchester Weekend Rail Update
On Sunday, pre planned cancellations are expected on the Wigan to Stalybridge and Victoria to Clitheroe via Boltons services. Customers are advised to plan ahead.
On Sunday 10 November, trains between Blackburn and Southport via Manchester Victoria will start/terminate at Todmorden. A bus replacement service will be in operation between Todmorden and Blackburn.
On Sunday 10 November, trains between Manchester Victoria and Clitheroe via Bolton will start/terminate at Blackburn. A bus replacement service will be in operation between Blackburn and Clitheroe.
Over the weekend, East Midlands Railway services between Liverpool/Manchester to Norwich will run to Nottingham only. A bus replacement service will be in operation.
Over the weekend, CrossCountry trains will be diverting at an alternative route and will not be calling at Bristol Parkway.
Bank Street and Manor Street are closed until 20 of December which is affecting the 982 and 987 services.
Castle Street will be closed until Monday 18 November. This will affect the 544 service.
Knowsley Street is closed for resurfacing until Monday 4 November. This is affecting the 512, 513, 524 and 135 services.
Cobden Street will be closed from Tuesday 8 October to Tuesday 5 November. This will affect the 410 and 411 services.
Whit Lane is closed until the end of October affecting the 74 and 75 services.
Ongoing lane closures and Bus stand changes began at Pendleton Shopping Centre, along Hankinson Way, from Monday 28 October. This scheme is expected to last up to 12 months. Bus stops A, B, C, D, H, J, K and L are affected. The following services are affected: 10 27 29 34 37 38 50 52 53 67 70 73 74 75 79 95 100.
Sunnyside Road will be closed from Monday 11 until Monday 25 November. This closure will be in place from 9.30am to 4.30pm and will affect the 230, 231, 769 and 864 services.
Dukinfield Road will be closed from Saturday 26 October to Saturday 9 November. This closure will be in place from 9.30am to 4.30pm on weekdays and 8am to 6pm on the weekend which will affect the 330 service.
Moss Lane is closed from junction of Manor Road to junction of Stamford New Road until the end of 2019. This closure is affecting the 283, 789 and 790 services.
Kings Road will be closed from the junction of Edge Lane to the junction of Truro Avenue until Friday 8 November. This closure will be in place from 9.30am to 3.30pm and will affect the 15 service.
The Salford Western Gateway will be closed from Monday 11 to Friday 17 November. This will affect the 100 service.
Frog Lane will be closed at the junction of Lower Street and Stephen Street to the junction of Woodhouse Lane from Sunday 3 to Sunday 17 November. This closure will be in place from 7am to 5pm and will affect the 5 and 6 services.