text:
\section{Introduction}
\label{sec:introduction}

The unique continuation principle is an important tool in the theory of linear partial differential equations, and it has been studied extensively in the case of linear Schr{\"o}dinger operators. Let $\Omega \subset \mathbb{R}^N$ and $V \in L^1_{\loc}(\Omega)$. The Schr{\"o}dinger type
equation
\begin{equation}
\label{linear-general-laplace}
-\Delta u = V(x) u \qquad \text{in $\Omega$}
\end{equation}
is said to have the {\em unique continuation property (UCP)} if every solution $u$ on $\Omega$ which vanishes on an open subset of $\Omega$ is identically
zero. In recent decades, various classes of potential functions $V$ have been shown to give rise to this property. For a detailed account of results of this type, we refer to the surveys \cite{kenig:89,koch.tataru:2001}. The unique continuation property is interesting per se, and it has many important consequences. In particular, it is closely related to the strict monotonicity of Dirichlet eigenvalues of Schr{\"o}dinger operators with respect to domain inclusion \cite{de-Figueiredo.gossez:92} and to the absence of eigenvalues inside the essential spectrum \cite{jerison.kenig:85,koch.tataru:2006}. It also gives rise to important energy estimates and compactness properties in the context of semilinear elliptic boundary boundary value problems with a variational structure, see e.g. \cite{liu.su.weth:2006,heinz:95}.

We also recall two stronger variants of the unique continuation property for (\ref{linear-general-laplace}), and for this we assume for simplicity that $V \in L^\infty_{\loc}(\Omega)$. First, (\ref{linear-general-laplace}) is said to have the {\em strong unique continuation property (SUCP)} if every weak solution $u \in H^1_{\loc}(\Omega)$ which has a zero of infinite order in $\Omega$ is identically zero. Here $x_0 \in \Omega$ is called a zero of infinite order if
$$
\int_{B_r(x_0)}u^2\,dx = O(r^n) \qquad \text{as $r \to 0^+$, for every $n \in \mathbb{N}$.}
$$
Moreover, we say that (\ref{linear-general-laplace}) has the {\em unique continuation property in measure (UCPM)}
if every weak solution $u \in H^1_{\loc}(\Omega)$ which vanishes on a set of positive measure is identically zero.

The above notions of unique continuation can be generalized to the semilinear equation
\begin{equation}
\label{nonlinear-general-laplace}
-\Delta u = f(x,u) \qquad \text{in $\Omega$}.
\end{equation}
In the case where the nonlinearity $f$ satisfies $f(\cdot,0)=0$ on $\Omega$ and the function
\begin{equation}
  \label{eq:definition-tilde-f}
(x,u) \mapsto \tilde f(x,u):=
\left\{
  \begin{aligned}
  &\frac{f(x,u)}{u},&& \qquad u \not = 0,\\
  &0,&& \qquad  u = 0,
  \end{aligned}
\right.
\end{equation}
is locally bounded in $x$ and $u$, any solution of (\ref{nonlinear-general-laplace}) satisfies (\ref{linear-general-laplace}) with $V \in L^\infty_{\loc}(\Omega)$ given by $V(x)=\tilde f(x,u(x))$. Hence the unique continuation results for (\ref{linear-general-laplace}) carry over immediately in this case. In particular, this is true for Lipschitz nonlinearities $f$ with $f(\cdot,0)=0$. The picture changes drastically in the non-Lipschitz case where the function $\tilde f$ defined in (\ref{eq:definition-tilde-f}) is not locally bounded and could even fail to belong to $L^1_{\loc}(\Omega)$. The simplest class of nonlinearities of this type are sublinear homogeneous nonlinearities of order less than one given by
\begin{equation}
  \label{eq:sublinear-homog-nonli}
f_q(u)=|u|^{q-2}u, \qquad q \in (1,2), \qquad \quad f_1(u)=\sgn(u),\qquad q=1,
\end{equation}
where $\sgn$ is the sign function. It is well known that already the ODE
$$
u'' = |u|^{q-2}u
$$
admits, in the case $1< q <2$, nontrivial solutions $u \in C^2(\mathbb{R})$ violating the unique continuation principle, as they can be chosen of the form $u(t)=  \Bigl( \frac{2q}{(2-q)^2}\Bigr)^{\frac{1}{q-2}} (t-t_0)^{\frac{2}{2-q}}$ for $t >t_0$ and $u(t)=0$ for $t \le t_0$, with $t_0 \in \mathbb{R}$ chosen arbitrarily.  On the other hand, it is very easy to see that unique continuation holds for the sublinear ODE with opposite sign, i.e., for
\begin{equation}
\label{eq:ode-correct-sign}
-u'' = |u|^{q-2}u.
\end{equation}
Indeed, since the quantity $\frac{{u'}^2}{2}+ \frac{|u|^q}{q}$ is constant and nonzero along nontrivial solutions, $u$ may only have simple zeros.

It is clear that ODE arguments of this type do not apply to the higher dimensional analogue of (\ref{eq:ode-correct-sign}) given by the equation
\begin{equation}
  \label{eq:sublinear-q-eq}
-\Delta u = f_q(u) \qquad \text{in $\Omega$.}
\end{equation}
The study of (\ref{eq:sublinear-q-eq}) is motivated, in particular, by its close relationship to the
(sign changing) porous medium equation
\begin{equation}
  \label{eq:scpme}
w_t -\Delta |w|^{m-1}w=0 \qquad \qquad \text{with $\quad m=\frac{1}{q-1} > 1$.}
\end{equation}
Indeed, as discussed in \cite[Chapter 4]{vazquez:2007}, a solution $u$ of (\ref{eq:sublinear-q-eq}) gives rise to a solution of (\ref{eq:scpme}) with separate variables via the ansatz
$$
(x,t) \mapsto w(x,t)=\Bigl(\frac{2-q}{q-1}(t-t_0)\Bigr)^{-\frac{q-1}{2-q}}|u(x)|^{q-2}u(x), \qquad \qquad t >t_0,\;x \in  \Omega.
$$
We remark that nonnegative solutions of (\ref{eq:sublinear-q-eq}) trivially obey the unique continuation principle, as they either vanish identically or they are positive by the strong maximum principle for superharmonic functions. Moreover, if a sign changing solution $u \in C^2(\Omega)$ of (\ref{eq:sublinear-q-eq}) has the property that the set $\{x \in \Omega\::\: u(x) \not = 0\}$ only has finitely many components which all satisfy the interior sphere condition, then the Hopf boundary point Lemma easily implies that $u^{-1}(0)$ is an $(N-1)$-dimensional hypersurface and therefore has zero $N$-dimensional Lebesgue-measure. In particular, this property is shared by radial solutions of (\ref{eq:sublinear-q-eq}) in bounded radial domains, but it fails to hold for general sign changing solutions of (\ref{eq:sublinear-q-eq}).

Part of the movitation for the present paper arises from the recent work \cite{parini.weth:2015} where the unique continuation principle has been studied for a special class of sign changing solutions of (\ref{eq:sublinear-q-eq}).
More precisely, in \cite{parini.weth:2015}, Parini and the second author focus on the class of least energy sign changing solutions of the Neumann problem for (\ref{eq:sublinear-q-eq}) in a bounded Lipschitz domain. Combining variational arguments with a blow up procedure and local perturbation arguments, they prove that the zero set of least energy sign changing solutions has Lebesgue measure zero. However, the variational perturbation technique developed in \cite{parini.weth:2015} does not extend to higher energy sign changing solutions. We note that both under Dirichlet and under Neumann boundary conditions, (\ref{eq:sublinear-q-eq}) is known to admit an infinite sequence of sign changing solutions which converges to the trivial solution, see \cite{du:preprint,bartsch.willem:95}. For these solutions, the unique continuation property is unknown up to now.

One of the main aims of the present paper is to establish the unique continuation property of (\ref{eq:sublinear-q-eq}) without additional assumptions on the solutions. Our first main result is the following.

\begin{theorem}\label{thm: weak unique}
Let $q \in [1,2)$. If a weak solution $u \in H^1_{\loc}(\Omega)$ of (\ref{eq:sublinear-q-eq}) vanishes in a neighbourhood of a point $x_0 \in \Omega$,
then $u \equiv 0$ in $\Omega$.
\end{theorem}

We remark at this point that one of the key tools to prove the unique continuation property of linear Schr{\"o}dinger equations of the type (\ref{linear-general-laplace}) are Carleman inequalities, which apply more generally to differential inequalities of the form
$$
|\Delta u(x)| \le |V(x)u(x)|,\qquad x \in \Omega,
$$
see e.g. \cite{jerison.kenig:85}. Since these differential inequalities do not depend on the sign of $V$, Carleman inequalities do not seem to be the right tool to derive the unique continuation property for (\ref{eq:sublinear-q-eq}). We were instead strongly inspired by the papers \cite{GarLin86, GarLin91, GaSVG, Kuk}, where monotonicity properties and other tools related to the theory of free boundary problems are used. In \cite{GarLin86}, the authors study the equation
\begin{equation}\label{div-eq-homogeneous}
-\div(A(x) \nabla u) = 0 \qquad \text{in $\Omega$},
\end{equation}
where the matrix function $A=(a_{ij}): \Omega \to \mathbb{R}^{N\times N}$ satisfies the following regularity and elliticity assumptions:
\begin{itemize}
\item[(A1)] $A$ is symmetric with locally Lipschitz coefficients $a_{ij}: \Omega \to \mathbb{R}$, $i,j=1,\dots,N$, and there exists a continuous function $\lambda: \Omega \to (0,1)$ such that
\[
\lambda(x) |\xi|^2 \le \sum_{i,j=1}^N a_{ij}(x) \xi_i \xi_j \le \lambda^{-1}(x) |\xi|^2 \qquad \text{for $\xi \in \mathbb{R}^N$.}
\]
\end{itemize}
In particular, it is proved in \cite{GarLin86} that (\ref{div-eq-homogeneous}) has the strong unique continuation property. The same result is also proved for weak solutions to \eqref{linear-general-laplace}, under appropriate assumptions on the potential $V$. A unified treatment is later given in \cite{GarLin91}, where more general linear equations are considered. In \cite{Kuk}, the author provides bounds of the admissible orders of vanishing of a solution to \eqref{linear-general-laplace} on manifolds. Some of the tools introduced in \cite{Kuk} have later been adapted and used in \cite{GaSVG}, where the lower-dimensional obstacle problem for the operator $-\div (A(x)\nabla)$ is studied, and the optimal regularity of solutions is proved.

As we shall see, these adaptations will be useful for us in the generalization of Theorem~\ref{thm: weak unique} involving general divergence type operators and a weaker sublinearity property. More precisely, we consider the equation
\begin{equation}\label{div eq}
-\div(A(x) \nabla u)= V(x)u + f(x,u) \qquad \text{in $\Omega$}
\end{equation}
in a domain $\Omega \subset \mathbb{R}^N$ with a Borel-measurable function $f: \Omega \times \mathbb{R} \to \mathbb{R}$.
Let $F: \Omega \times \mathbb{R} \to \mathbb{R},\: F(x,s)= \int_0^s f(x,t)\,dt$, denote the primitive of $f$. In addition to (A1), we assume the following:
\begin{itemize}
\item[(A2)] $V \in L^\infty_{\loc}(\Omega)$.\vspace{0.1cm}

\item[(A3)] $f \in L^\infty_{\loc}(\Omega \times \mathbb{R})$, and there exists $\eps_0, \kappa_1,\kappa_2>0$ and $q<2$ such that
  \begin{enumerate}
  \item[i)] $0 <f(x,s)s \le q F(x,s)$ for a.e. $s \in (-\eps_0, \eps_0) \setminus \{0\}$, $x \in \Omega$,\vspace{0.1cm}

  \item[ii)] the function $F(\cdot,s)$ is of class $C^1$ on $\Omega$ for every $s \in (-\eps_0,\eps_0)$,\vspace{0.1cm}

 \item[iii)] $|\nabla_1 F(x,s)| \le \kappa_1 F(x,s)$ for all $x \in \Omega$, $s \in (-\eps_0,\eps_0)$,\vspace{0.1cm}

 \item[iv)] $F(x,s) \ge \kappa_2$ for all $x \in \Omega$, $s \in \{-\eps_0,\eps_0\}$.\vspace{0.1cm}

   \end{enumerate}
\end{itemize}
Here we have set $\nabla_1 F = \bigl(\frac{\partial F}{\partial x_1},\dots,\frac{\partial F}{\partial x_N}\bigr)$. In the following, we say that $u \in H^1_{\loc}(\Omega) \cap L^\infty_{\loc}(\Omega)$ is a weak solution of (\ref{div eq}) if the function $x \mapsto f(x,u(x))$ is Lebesgue-measurable on $\Omega$ and
$$
\int_\Omega  \langle A(x)\nabla u(x), \nabla \phi(x) \rangle\,dx = \int_{\Omega} \Bigl(V(x)u(x)  + f(x,u(x))\Bigr)\phi(x)\,dx
$$
for all $\phi \in C_{c}^\infty(\Omega)$.

\begin{theorem}\label{thm: main 2}
Suppose that (A1)--(A3) are satisfied. If a weak solution $u \in H^1_{\loc}(\Omega) \cap L^\infty_{\loc}(\Omega)$ of (\ref{div eq}) vanishes in a neighbourhood of a point $x_0 \in \Omega$,
then $u \equiv 0$ in $\Omega$.
\end{theorem}

\begin{remark}
It is easy to see that the class of homogenous sublinear nonlinearities $f_q$ given in (\ref{eq:sublinear-homog-nonli}) satisfies assumption (A3). More generally, Theorem \ref{thm: main 2} applies to weak solutions $u \in H^1_{\loc}(\Omega)$ of the equation
\[
-\div(A(x)\nabla u) = h(x,s) + \sum_{k=1}^m c_k(x) |s|^{q_k-2} s \qquad \text{in $\Omega$},
\]
where $A$ satisfies $(A1)$, $q_k \in [1,2)$ for $k=1,\dots,m$, $c_1,\dots, c_m \in C^1(\Omega)$ are positive functions such that $\frac{\nabla c_1}{c_1}, \dots, \frac{\nabla c_m}{c_m} \in L^\infty(\Omega,\mathbb{R}^N)$,  and $h$ is (weakly) superlinear in the sense that $\tilde h \in L^\infty_{\loc}(\Omega \times \mathbb{R})$ for the
function
$$
(x,s) \mapsto \tilde h(x,s):=
\left\{
  \begin{aligned}
  &\frac{h(x,s)}{s},&& \qquad s \not = 0,\\
  &0,&& \qquad  s = 0.
  \end{aligned}
\right.
$$
Indeed, in this case, every weak solution $u \in H^1_{\loc}(\Omega)$ is contained in $C^{1,\alpha}_{\loc}(\Omega)$ for some $\alpha>0$ by elliptic regularity (see e.g. \cite[Theorem 3.13]{han.lin:2011}), and thus assumptions (A1) and (A2) are satisfied with $V(x) = \tilde h(x,u(x))$. \\
We point out that, while we have to impose the positivity of the sublinear term $f$, no sign-assumption is needed on the superlinear one $h$.
\end{remark}

As far as we know, Theorems~\ref{thm: weak unique} and \ref{thm: main 2} are the first unique continuation results for general solutions of sublinear equations. These results will be proved by combining key estimates from \cite{GarLin86, GaSVG} related to the linear operator $-\div (A(x) \nabla)$ with new arguments to deal with the zero order nonlinearity in the equations (\ref{eq:sublinear-q-eq}) and (\ref{div eq}). We stress that the presence of the sublinear term on the right hand side in (\ref{eq:sublinear-q-eq}) and (\ref{div eq}) drastically changes the features of the problem; in particular, the ``almost-monotonicity of the Almgren's frequency", the key tool in proving the strong unique continuation in \cite{GarLin86, GarLin91, Kuk}, is lost.

We point out that it remains open whether (\ref{eq:sublinear-q-eq}) or (\ref{div eq}) give rise to the strong unique continuation property or the unique continuation property in measure.

The paper is organized as follows. In Section~\ref{sec: proof simple case} we deal with the special equation (\ref{eq:sublinear-q-eq}). Due to the homogeneity of the nonlinearity and the simple form of the linear part
of (\ref{eq:sublinear-q-eq}), the proof of Theorem~\ref{thm: weak unique} is considerably easier than the proof of Theorem~\ref{thm: main 2}, and it is instructive to elaborate the main ideas and estimates related to the sublinearity of the problem in this special case. In Section~\ref{sec:general-case}, we then give the proof of Theorem~\ref{thm: main 2}.

\section{The model problem}\label{sec: proof simple case}

This section is devoted to the proof of Theorem~\ref{thm: weak unique}. From now on, we assume that $u \in H^1_{\loc}(\Omega)$ is a weak solution of
the equation (\ref{eq:sublinear-q-eq}) for some $q \in [1,2)$. In the case $q \in (1,2)$, classical elliptic regularity yields $u \in C^{2,\alpha}_{\loc}(\Omega)$ for some $\alpha>0$. In the case $q=1$, we still have that $u \in W^{2,p}_{\loc}(\Omega)$ for all $p<\infty$ (see e.g. \cite[Theorem 9.11]{GT}), which implies that $u \in C^{1,\alpha}_{\loc}(\Omega)$ for all $\alpha \in (0,1)$. Moreover, $u$ is a strong solution of (\ref{eq:sublinear-q-eq}), and it follows from the definition of
weak derivatives that $D^2 u \in L^{p}_{\loc}(\Omega, \mathbb{R}^{N \times N})$ is an a.e. symmetric matrix in $\Omega$.

To prove Theorem~\ref{thm: weak unique}, we now assume that $u \equiv 0$ in a neighborhood of a point $x_0 \in \Omega$. By translations, we may suppose that $x_0=0$, which simplifies some expressions in the following. We then need to prove that
\begin{equation}
  \label{eq:claim-weak-unique}
u \equiv 0 \qquad \text{in $\Omega$.}
\end{equation}
For $0<r< {\rm dist}(0,\mathbb{R}^N \setminus \Omega)$, we define
$B_r:= B_r(0)$, $S_r:= \partial B_r$,
$$
H(r):=  \int_{S_r} u^2\,d\sigma \qquad \text{and}\qquad
D(r):= \int_{\Omega}\Bigl(|\nabla u|^2 - |u|^q\Bigr)\,dx = \int_{S_r} u u_\nu\,d\sigma,
$$
where, here and in the following, we set $\nu(x)= \frac{x}{|x|}$ for $x \not =0$ and let $u_\nu(x)= \langle \nabla u(x), \frac{x}{|x|}\rangle$ denote the radial derivative. From now on, we often omit the volume element $dx$ and the surface element $d \sigma$ inside the integrals. It is clear that integrals on $B_r$ or on $S_r$ are computed, respectively, with respect to the Lebesgue measure in $\mathbb{R}^N$ or with respect to the $(N-1)$-dimensional Hausdorff measure.
Then we have
$$
H'(r)=\frac{N-1}{r}H(r) + 2 \int_{S_r}u u_\nu = \frac{N-1}{r}H(r) + 2 D(r).
$$
As a consequence, whenever $H(r) \neq 0$ we have
\begin{equation}
  \label{eq:derivative-1}
\frac{d}{dr}\Bigl(\log \frac{H(r)}{r^{N-1}}\Bigr) =
\frac{H'(r)}{H(r)} - \frac{N-1}{r}= 2 \frac{D(r)}{H(r)}= 2 \frac{N(r)}{r}
\end{equation}
where $N$ is the (Almgren frequency) function
$$
r \mapsto N(r):= \frac{rD(r)}{H(r)}.
$$
In \cite{GarLin86}, the authors could show the Almgren frequency associated to linear equations in indeed monotone, up to an exponential factor. The proof of this fact cannot be extended in the present setting for any solution $u$ to (\ref{eq:sublinear-q-eq}), due to the sublinear nature of the problem.

In what follows we need to consider the derivatives of $D$ and of $N$.
This is the object of the following two statements, which are inspired by the computations in \cite[Section 4]{GarLin86}.

\begin{proposition}\label{der D}
For every $r \in (0, {\rm dist}(0,\mathbb{R}^N \setminus \Omega))$, we have the identity
\[
D'(r) = \frac{N-2}{r} D(r) - \frac{C_{N,q}}{q r} \int_{B_r}|u|^q\,dx  + \int_{S_r}\left(2 u_\nu^2 + \left(\frac{2-q}{q}\right)|u|^{q}\right) \,d\sigma.
\]
with $C_{N,q}= 2N -(N-2)q>0$.
\end{proposition}
\begin{proof}
In the case $q>1$, where $u \in C^2(\Omega)$ and $|u|^q \in C^1(\Omega)$, we have
\begin{equation}
  \label{eq:D-prime-simple}
D'(r)= \int_{S_r}\Bigl( |\nabla u|^2 - |u|^{q}\Bigr),
\end{equation}
whereas, by the symmetry of the matrix $D^2u$,
\begin{align*}
\int_{S_r} |\nabla u|^2 &=  \frac{1}{r}\int_{S_r} \langle |\nabla u|^2  x, \nu \rangle = \frac{1}{r} \int_{B_r}\div \Bigl(|\nabla u|^2 x\Bigr) \\
& = \frac{N}{r} \int_{B_r} |\nabla u|^2  +
\frac{2}{r} \int_{B_r} \langle x,  (D^2 u)\,\nabla u \rangle \\
&= \frac{N}{r} \int_{B_r} |\nabla u|^2  +
\frac{2}{r} \int_{B_r} \langle (D^2 u)\,x ,   \nabla u \rangle
\\
&= \frac{N}{r} \int_{B_r} |\nabla u|^2  +
\frac{2}{r} \int_{B_r} \Bigl \langle \nabla \bigl(\langle \nabla u, x \rangle - u\bigr), \nabla u \Bigr \rangle
 \\
&= \frac{N-2}{r} \int_{B_r} |\nabla u|^2  -
\frac{2}{r} \int_{B_r} \langle \nabla u, x \rangle  \Delta u
 + \frac{2}{r}\int_{S_r} \langle \nabla u,  x \rangle  u_\nu  \\
&= \frac{N-2}{r} \int_{B_r} |\nabla u|^2  +
\frac{2}{r}  \int_{B_r} \langle \nabla u, x \rangle f_q(u)
 + 2\int_{S_r} u_\nu^2.
\end{align*}
By integration by parts
$$
\int_{B_r} \langle \nabla u, x \rangle f_q(u) = \frac{1}{q}  \int_{B_r} \langle \nabla |u|^{q}, x \rangle = \frac{r}{q} \int_{S_r} |u|^{q} -\frac{N}{q} \int_{B_r}|u|^q.
$$
Inserting these identities in (\ref{eq:D-prime-simple}), we find that
\begin{align*}
D'(r) &= \frac{N-2}{r} \int_{B_r} |\nabla u|^2  - \frac{2N}{qr} \int_{B_r}|u|^q + \int_{S_r}\left(2 u_\nu^2 + \left(\frac{2}{q}-1\right)|u|^{q}\right)\\
&= \frac{N-2}{r} D(r) - \frac{C_{N,q}}{q r} \int_{B_r}|u|^q  + \int_{S_r}\left(2 u_\nu^2 + \left(\frac{2}{q}-1\right)|u|^{q}\right),
\end{align*}
 as claimed. The same computation is also valid in the case $q=1$ but requires extra justification. First, we use the fact that -- as remarked before --
$D^2 u \in L^{p}_{\loc}(\Omega, \mathbb{R}^{N \times N})$ is an a.e. symmetric matrix function. Moreover, as observed in \cite[Proposition 2.7]{hofmann.mitrea.taylor:2010}, the Gauss-Green formula
\begin{equation}
  \label{eq:gauss-green}
\int_{B_r} \div w = \int_{S_r} \langle w, \nu \rangle
\end{equation}
holds for all vector fields $w \in C(\overline{B_r},\mathbb{R}^N)$ with $\div w \in L^1(B_r)$. In the above computation for the case $q=1$, (\ref{eq:gauss-green}) is applied successively to the vector fields
$$
w_1=|\nabla u|^2 x,\qquad w_2 = \langle \nabla u, x \rangle \nabla u \qquad \text{and}\qquad  w_3= |u| x.
$$
Here we note  in particular that
\[
\div w_3 = \sgn(u) \langle \nabla u,x \rangle +  N |u| \qquad \text{a.e. in $\Omega$.}
 \qedhere
 \]
\end{proof}

We now derive a lower bound for the derivative of the Almgren frequency function $N$.

\begin{proposition}\label{der N}
With $C_{N,q}$ given in Proposition~\ref{der D}, we have
$$
N'(r) \ge \frac{1}{H(r)}\left[ \frac{r}{q}\left(2-q\right)  \int_{S_r}|u|^q\,d\sigma-\frac{C_{N,q}}{q} \int_{B_r}|u|^q\,dx\right]
$$
for every $r \in (0,{\rm dist}(0,\mathbb{R}^N \setminus \Omega))$ such that $H(r) \neq 0$.
\end{proposition}

\begin{proof}
We observe that, whenever $H(r) \neq 0$, it results
\begin{align*}
N'(r) &= \frac{1}{H(r)} \left[D(r) + rD'(r)  -r \frac{D(r) H'(r)}{H(r)} \right] \\
& =  \frac{1}{H(r)} \left[(2-N) D(r) + rD'(r)  -2r \frac{D^2(r)}{H(r)} \right].
\end{align*}
Hence, by Proposition \ref{der D}
$$
N'(r) = \frac{1}{H(r)}\biggl[r \int_{S_r}\left(2 u_\nu^2 + \left(\frac{2-q}{q}\right)|u|^{q}\right)  - \frac{C_{N,q}}{q}\int_{B_r}|u|^q  -  \frac{2 r D^2(r)}{H(r)}\biggr],
$$
and the thesis follows observing that
$$
\int_{S_r} u_\nu^2  - \frac{D^2(r)}{H(r)}=\int_{S_r} u_\nu^2
- \frac{\Bigl(\int_{S_r} u u_\nu \Bigr)^2}{\int_{S_r} u^2} \ge 0
$$
by the Cauchy-Schwarz inequality.
\end{proof}

We may now complete the

\begin{proof}[Proof of Eq. (\ref{eq:claim-weak-unique})]
Without loss of generality, we suppose by contradiction that $\Omega=B_1$, $u \not \equiv 0$ in $B_1$, but $u \equiv 0$ in $B_{t}$ for some small $t$. We denote
\[
d(r):=  \frac{1}{q}\int_{B_r}|u|^q\,dx, \quad \text{so that} \quad d'(r) = \frac{1}{q}\int_{S_r} |u|^q\,d\sigma,
\]
and we set
\[
r_0:= \sup \left\{r \ge 0: d(r) = 0\right\}  \in (0,1).
\]
It is clear that $D(r) = d(r) = 0$ for every $r \in (0,r_0]$, and, since
$d$ is non-decreasing, that $d(r)>0$ for $r>r_0$.

By Proposition \ref{der D},
\[
D'(r) \ge \left(\frac{N-2}r  \right) D(r) - \frac{C_{N,q}}r d(r) + (2-q) d'(r),
\]
whence
\[
\frac{d}{dr}\left(\frac{D(r)}{r^{N-2}}\right) \ge \frac{2-q}{r^{N-2}} d'(r) - \frac{C_{N,q}}{r^{N-1}} d(r)
\]
for every $r \in (0,1)$. Thus, for $r \in (r_0, 1)$
\[
\frac{d}{dr} \left(\frac{D(r)}{r^{N-2}}\right) \ge (2-q) d'(r) - C_1 d(r) \qquad \text{with}\quad C_1:= \frac{C_{N,q}}{r_0^{N-1}}.
\]
Integrating in $(r_0,r)$ with $r \in (r_0,1)$, we obtain
\begin{align*}
\frac{D(r)}{r^{N-2}} &\ge \frac{D(r_0)}{r_0^{N-2}} + (2-q) (d(r)-d(r_0)) - C_1 \int_{r_0}^r d(s)\,ds  \\
& \ge (2-q) d(r) - C_1(r-r_0)d(r),
\end{align*}
where we used the fact that $D(r_0)=d(r_0)=0$ and the monotonicity of $d$. Fixing
$r_1 \in (r_0,1)$ such that $C_1(r_1-r_0) < \frac{2-q}{2}$, we thus conclude that
$$
\frac{D(r)}{r^{N-2}} \ge \frac{2-q}{2} d(r) \qquad \text{for $r \in (r_0,r_1)$},
$$
and hence
\begin{equation}
\label{eq: lower bound D}
D(r) \ge  C_2 d(r)>0 \qquad  \text{for $r \in (r_0,r_1)$, with $C_2:= \frac{2-q}{2} r_0^{N-2}$.}
\end{equation}
Next we note that, since $u \not \equiv 0$ in $B_{r_1}$, there must exist
$r_2 \in (r_0,r_1)$ such that $H(r_2) \neq 0$. We fix such a radius $r_2$ and
define
\[
r_3:= \inf \left\{r \in (0,r_2): H(s)>0 \ \text{for every $s \in (r, r_2)$}\right\}.
\]
Then we have $r_3 \ge r_0$, $H(r_3) = 0$ and $H(r) \not = 0$ for $r \in (r_3,r_2)$. Hence the the Almgren frequency $N(r)$ is well defined for $r \in (r_3,r_2)$, and we can estimate its derivative with the help of Proposition \ref{der N}, which gives that
\[
N'(r) \ge - C_{N,q} \frac{d(r)}{H(r)} \qquad \text{for $r \in (r_3,r_2].$}
\]
Combining this with \eqref{eq: lower bound D}, we obtain
\[
\frac{N'(r)}{N(r)} \ge -C_{N,q} \frac{d(r)}{r D(r)}  \ge -\frac{C_{N,q}}{r_0 C_2} =: -C_3 \qquad \text{for $r \in (r_3,r_2]$.}
\]
To sum up, we proved that
\[
N'(r) \ge -C_3 N(r) \qquad \text{for $r \in (r_3,r_2],$}
\]
and by integrating we deduce that
\[
r \mapsto N(r) e^{C_3 r} \quad \text{is non-decreasing in $(r_3,r_2]$}.
\]
In particular,
\[
N(r) \le N(r_2) e^{C_3 r_2}=: C_4  \qquad \text{for $r \in (r_3,r_2]$}.
\]
By \eqref{eq:derivative-1}, this implies that
\[
\frac{d}{dr} \log \left( \frac{H(r)}{r^{N-1}}\right) \le \frac{2C_4}{r} \le
\frac{2C_4}{r_0} \qquad \text{for $r \in (r_3,r_2]$,}
\]
whereas on the other hand we have that
$$
\lim_{r \to r_3^+} \log \left( \frac{H(r)}{r^{N-1}}\right)= -\infty
$$
since $H(r_3)=0$ and $r_3>0$. This is a contradiction, and hence (\ref{eq:claim-weak-unique}) is proved. We thus have finished the proof of Theorem~\ref{thm: weak unique}.
\end{proof}

\section{The general case}
\label{sec:general-case}

This section is devoted to the proof of Theorem \ref{thm: main 2}. Hence, in the following we assume that the hypotheses (A1)--(A3) are satisfied with some constants $C,\eps_0>0$ and $q<2$. First, we note that it easily follows from (A3)i) that, for every $x \in \Omega$, the function $s \mapsto F(x,s)/|s|^q$ is non-increasing on $(0,\eps_0)$, and it is non-decreasing on $(-\eps_0,0)$. Combining these facts with (A3)iv), we infer that
\begin{equation}\label{rem: on S}
F(x,s) \ge \frac{\min\{F(x,\eps_0),F(x,-\eps_0)\}}{\eps_0^q} |s|^q \ge \frac{\kappa_2}{\eps_0^q} |s|^q \qquad \text{for $x \in \Omega$, $0<|s|<\eps_0$}.
\end{equation}
Second, we discuss the regularity of a weak solution $u \in H^1_{\loc}(\Omega) \cap L^\infty_{\loc}(\Omega)$ of (\ref{div eq}). Since $f \in L^\infty_{\loc}(\Omega \times \mathbb{R})$, it follows that $f(\cdot,u(\cdot)) \in L^\infty_{\loc}(\Omega)$. Thus standard elliptic regularity applies and yields, as in the previous section, that $u \in W^{2,p}_{\loc}(\Omega)$ for all $p<\infty$, and that $u$ is a strong solution of (\ref{div eq}).  Moreover, $u \in C^{1,\alpha}_{\loc}(\Omega)$ for all $\alpha \in (0,1)$, and $D^2 u \in L^{p}_{\loc}(\Omega, \mathbb{R}^{N \times N})$ is an a.e. symmetric matrix in $\Omega$.

We claim that it suffices to prove the following simplified version of Theorem~\ref{thm: main 2}.

\begin{theorem}
\label{main-2simplified}
Suppose that (A1)--(A3) are satisfied, and let $u \in H^1_{\loc}(\Omega) \cap L^\infty_{\loc}(\Omega)$ be a weak solution of (\ref{div eq}). Moreover, assume in addition that
\begin{equation}
  \label{eq:extra-assumption}
\text{$0 \in \Omega$, $\quad A(0)= \id,\quad$ and $\quad |u|< \eps_0\; $ in $\Omega$.}
\end{equation}
If $u \equiv 0$ in a neigborhood of $0$, then $u \equiv 0$ on $B_{\delta_0}(0)$, where $\delta_0:= {\rm dist}(0,\mathbb{R}^N \setminus \Omega)$.
\end{theorem}

Assuming for the moment that Theorem~\ref{main-2simplified} holds true,  we can quickly complete the

\begin{proof}[Proof of Theorem \ref{thm: main 2}]
By assumption, the open set
$$
U:= \{x \in \Omega\::\: \text{$u \equiv 0$ in a neighborhood of $x$}\},
$$
is nonempty. Since $\Omega$ is connected, the claim
$u \equiv 0$ in $\Omega$  follows once we have shown that $\partial U \cap \Omega = \varnothing$.
We suppose by contradiction that there exists a point $x_* \in \partial U \cap \Omega$. By the continuity of $u$, we have $u(x_*)= 0$, and thus $x_*$ is contained in the open set
\[
\Omega_0 := \{x \in \Omega\::\: |u(x)| < \eps_0\}.
\]
Using the continuity and positivity of the function $x \mapsto \lambda(x)$ considered in assumption (A1), we may then choose $x_0 \in U$ with
$$
|x_*-x_0|<\frac{\lambda_* \delta_*}{2},\qquad \quad \text{where
$\quad \delta_* := {\rm dist}(x_0, \mathbb{R}^N \setminus \Omega_0)\quad$ and $\quad \lambda_*= \lambda(x_0) \in (0,1)$.}
$$
Our aim is to show that
\begin{equation}
  \label{eq:u-equiv-00}
u \equiv 0 \qquad \text{in $B_{\lambda_* \delta_*}(x_0)$,}
\end{equation}
since then we have $x_* \in U \cap \partial U$, contradicting the openness of $U$ in $\Omega$. We now change coordinates via the affine map
$$
T : \mathbb{R}^N \to \mathbb{R}^N, \qquad T(x)= A(x_0)^{1/2} x + x_0
$$
We note that, as a consequence of assumption (A1),
\begin{equation*}
\sqrt{\lambda_*}|x-y| \le |T(x)-T(y)| \le \frac{1}{\sqrt{\lambda_*}} |x-y| \qquad \text{for $x,y \in \mathbb{R}^N$}
\end{equation*}
Let $\widetilde \Omega:= T^{-1}(\Omega_0)$. Then we have $0 \in \widetilde \Omega$, $T(0)=x_0$ and
$$
T^{-1}(B_{\lambda_* \delta_*}(x_0)) \subset  B_{\sqrt{\lambda_*} \delta_*}(0) \subset \widetilde \Omega
$$
Hence, to show (\ref{eq:u-equiv-00}), it suffices to show that
\begin{equation}
  \label{eq:u-equiv-0}
\tilde u \equiv 0 \qquad \text{on $B_{\sqrt{\lambda_*} \delta_*}(0)$,}
\end{equation}
for the function
$$
\tilde u \in H^1_{\loc}(\widetilde \Omega),\qquad
 \tilde u(x):= u(T(x)).
$$
The function $\tilde u$ is a weak solution of
\begin{equation*}
-\div(\widetilde A(x) \nabla \tilde u) = \widetilde V(x)\tilde u + \tilde f(x,\tilde u) \qquad \text{in $\widetilde \Omega$},
\end{equation*}
with $\widetilde V(x)= V(T(x))$, $\tilde f(x,s):= f(T(x), s)$ and
$$
\widetilde A(x):= A(x_0)^{1/2} A^{-1}(T(x)) A(x_0)^{1/2}.
$$
We note that -- on $\widetilde \Omega$ -- the matrix-valued function
$\widetilde A(x)$ still satisfies (A1) once $\lambda$ is replaced by the function
$x \mapsto \tilde \lambda(x):=\lambda_* \lambda(T(x))$. Moreover, the function $\widetilde V$ satisfies (A2), and the function $\tilde f \in L^\infty_{\loc}(\widetilde \Omega \times \mathbb{R})$ satisfies assumption (A3) with unchanged values of $\eps_0$ and
$\kappa_2$ after making $\kappa_1>0$ smaller if necessary.

By construction, all assumptions of Theorem~\ref{main-2simplified} are satisfied with $u$, $A$, $V$, $\Omega$ replaced $\tilde u$, $\tilde A$, $\tilde V$, $\tilde \Omega$.  Since $\sqrt{\lambda_*} \delta_* \le {\rm dist}(0, \mathbb{R}^N \setminus \tilde \Omega)$, Theorem~\ref{main-2simplified} yields (\ref{eq:u-equiv-0}), as required.
\end{proof}

It thus remains to prove Theorem~\ref{main-2simplified}, and the rest of this section will be devoted to this aim. From now on, we fix a weak solution $u \in H^1_{\loc}(\Omega) \cap L^\infty_{\loc}(\Omega)$ of (\ref{div eq}), and we assume that (\ref{eq:extra-assumption}) is satisfied in addition to (A1)--(A3). We fix an arbitrary $\delta_1 \in (0,\delta_0)$. It then clearly suffices to show the implication
\begin{equation}
  \label{eq:-r-*-eq}
u \equiv 0 \quad \text{in a neighborhood of $0$}\qquad \Longrightarrow \qquad
u \equiv 0 \quad \text{on $B_{\delta_1}(0)$.}
\end{equation}
We note that, by assumptions (A1)-(A3) and since $u \in L^\infty(B_{\delta_1}(0))$, the functions $V, \lambda$, $\lambda^{-1}$ and $x \mapsto f(x,u(x))$, $x \mapsto F(x,u(x))$ are all bounded in $B_{\delta_1}(0)$. We aim at adapting the strategy used to deal with the simple equation \eqref{eq:sublinear-q-eq}, and to this purpose we introduce analogues of the functions $H$, $D$, and $N$ defined in Section~\ref{sec: proof simple case}. As before we put $B_r:= B_r(0)$, $S_r:=\partial B_r$ for $r \in (0,\delta_0)$, and, following \cite{GaSVG,Kuk}, we define
\begin{equation*}
\begin{split}
&H(r) :=\int_{S_r} u^2 \mu\, d \sigma,  \quad \text{where}  \quad \mu(x):= \frac{\langle A(x) x, x \rangle}{|x|^2} \\
&D_1(r)  :=  \int_{B_r} \! \langle A(x)\nabla u, \nabla u \rangle\,dx, \quad D(r)  :=  D_1(r)- \int_{B_r} \!\bigl(V(x)u^2 + f(x,u) u\bigr)dx,\\
&\text{and}\quad N(r)  := \frac{r D(r)}{H(r)} \qquad \text{(defined whenever $H(r) \neq 0$)}.
\end{split}
\end{equation*}
Notice that by the divergence theorem
\begin{equation}\label{D =}
D(r) = \int_{S_r} u \langle A(x) \nabla u, \nu \rangle\,d\sigma.
\end{equation}
In what follows the dependence of the matrix $A$ with respect to $x$ will often be omitted, for the sake of brevity. For the same reason, we will often omit $dx$ and $d \sigma$ in the integrals. We also need the following notation.

\medskip

\begin{definition}\label{rmk: on uniformity}
For $\gamma \in \mathbb{R}$, the symbol $O(r^\gamma)$ always stands for a function on $(0,\delta_1)$ satisfying $|O(r^\gamma)| \le C r^\gamma$ for $r \in (0,\delta_1)$ with a constant $C>0$.
\end{definition}

Following closely some computations performed in a different setting in \cite{GaSVG,Kuk}, we may now derive asymptotic estimates for the derivatives of $H$ and of $D_1$.

\begin{lemma}\label{lem: der H}
We have that
\[
H'(r)= 2 D(r) + \left( \frac{N-1}r +O(1) \right) H(r)
\]
for any $r \in (0,\delta_1)$.
\end{lemma}

\begin{proof}
On $S_r$ we have $\nu= x/|x| = \nabla |x|$, and hence $\langle A \nabla |x|,\nu \rangle = \mu$. As a consequence, using also the symmetry of $A$, we obtain
\begin{align*}
H(r) & = \int_{S_r} u^2 \langle A \nabla |x|,\nu \rangle =\int_{B_r}  \div( u^2 A \nabla |x|)  \\\
&= 2\int_{B_r} u \langle A  \nabla u, \nabla |x| \rangle + \int_{B_r}  u^2 \div(A \nabla |x|).
\end{align*}
By \eqref{D =}, we infer that
\begin{equation}
  \label{eq:preliminary-lem-der-H}
H'(r) = 2 D(r) + \int_{S_r} u^2 \, \div(A \nabla |x|).
\end{equation}
Moreover, by \cite[Lemma 4.1]{GaSVG}
\begin{equation}  \label{eq:div-expansion}
\div(A \nabla |x|) = \frac{N-1}{r} + O(1) \qquad \text{for $r \in (0,\delta_1)$,}
\end{equation}
and by definition $\lambda \le \mu \le \lambda^{-1}$. Inserting these expansions in (\ref{eq:preliminary-lem-der-H}), we obtain the claim.
\end{proof}

As in \cite{GaSVG}, it is now convenient to introduce  the quantity
\begin{equation*}
Z(x):= \frac{A(x)x}{|\mu(x)|}= \frac{|x| A(x) \nabla |x|}{\mu(x)},
\end{equation*}
observing that $\langle Z,\nu \rangle = r$ on $S_r$. Also, for future convenience, we note that
\begin{equation}\label{div Z}
\div Z = N + O(r) \qquad \text{for $r \in (0,\delta_1)$}.
\end{equation}
For the proof, we refer to \cite[Lemma A.5]{GaSVG}.

\begin{proposition}
 \label{sec:general-case-der-D-1}
For $r \in (0,\delta_1)$, we have that
\begin{align}
D_1'(r) = \left(\frac{N-2}r + O(1)\right)D_1(r)
&+ 2 \int_{S_r}  \frac{\langle A \nabla u,\nu\rangle^2}{\mu} \nonumber\\
&+ \frac2r\int_{B_r} \bigl(V(x)u+ f(x,u)\bigr) \langle Z, \nabla u \rangle. \label{D1-prime-formula}
\end{align}
\end{proposition}

The proof of this proposition is very similar to the proof of \cite[Theorem A.2]{GaSVG}, where a somewhat different setting is considered. For the convenience of the reader, we give the complete proof of Proposition~\ref{sec:general-case-der-D-1} in the Appendix. The following corollary is an immediate consequence of Proposition~\ref{sec:general-case-der-D-1}.

\begin{corollary}\label{lem: der D}
For every $r \in (0,\delta_1)$, we have that
\begin{align*}
D'(r)& = \left(\frac{N-2}r + O(1)\right) D_1(r)
+ 2 \int_{S_r}  \frac{\langle A \nabla u,\nu\rangle^2}{\mu} \\
&  + \frac2r\int_{B_r}\bigl(V(x)u+  f(x,u)\bigr) \langle Z, \nabla u \rangle - \int_{S_r}\bigl(V(x)u^2 +  f(x,u) u\bigr).
\end{align*}
\end{corollary}

Combining Lemma~\ref{lem: der H} and Corollary~\ref{lem: der D}, we may now estimate the derivative of $N$.

\begin{proposition}\label{prop: der N}
For any $r \in (0,\delta_1)$ with $H(r) \neq 0$, we have that
\begin{align}
N'(r) \ge O(1) N(r) &+ \frac1{H(r)}\bigg[  ( N-2 + O(r) ) \int_{B_r}\bigl(V(x)u^2+  f(x,u) u\bigr)\nonumber  \\
 &+2 \int_{B_r}\bigl(V(x)u+ f(x,u)\bigr) \langle \nabla u, Z \rangle   - r\int_{S_r}\bigl(V(x)u^2 + f(x,u) u\bigr)\bigg].\label{1334}
\end{align}
\end{proposition}

\begin{proof}
By Lemma~\ref{lem: der H} and Corollary~\ref{lem: der D}, we compute that
\begin{align*}
N'(r) =& \frac{1}{H(r)} \Bigl[D(r) + rD'(r)  -r \frac{D(r) H'(r)}{H(r)} \Bigr]\\
=&  \frac{1}{H(r)} \left[(2-N + O(r)) D(r) + rD'(r)  -2r \frac{D^2(r)}{H(r)} \right] \\
 =& \frac{1}{H(r)} \Bigl[ O(r) D(r) +  \left( N-2 + O(r) \right) \int_{B_r}\bigl(V(x)u^2+  f(x,u) u\bigr)  \\
&+ 2 \int_{B_r} \bigl(V(x)u +  f(x,u)\bigr) \langle \nabla u, Z \rangle   - r \int_{S_r}\bigl(V(x)u^2 + f(x,u) u\bigr) \Bigr]\\
&+ \frac{2r}{H(r)} \biggl[ \int_{S_r} \frac{\langle A \nabla u, \nu \rangle^2}{\mu} - \frac{\bigl( \int_{S_r} u \langle A \nabla u,\nu \rangle\bigr)^2 }{\int_{S_r} u^2 \mu} \biggr].
\end{align*}
Hence the Cauchy-Schwarz inequality applied to the last term (this is possible by the ellipticity of $A$) yields the desired result.
\end{proof}

\begin{remark}
When $A \equiv  \id$ in $B_{\delta_1}$, in the previous proposition all the error terms $O(1)$ and $O(r)$ vanish, $Z(x) \equiv x$, and in the case
$V \equiv 0$ and $f(x,u) = f(u) = |u|^{q-2}u$ we obtain exactly Proposition \ref{der N}, since
$$
\int_{B_r} f(u) \langle \nabla u, Z \rangle  = \int_{B_r}\langle \nabla F(u), Z \rangle = r \int_{S_r} F(u) - N \int_{B_r} F(u)
$$
with $F(u)= \int_0^u f(\tau)d\tau= \frac{1}{q}|u|^q$.
\end{remark}

\medskip

To proceed further with the proof of Theorem \ref{thm: main 2}, we introduce the quantity
$$
d(r):=  \int_{B_r}F(x,u)\,dx \qquad \text{for $r \in (0,\delta_1)$.}
$$
Since $F(x,u(x)) \ge 0$ on $B_{\delta_1}$ by assumptions (A3)i) and (\ref{eq:extra-assumption}), the function $d$ is differentiable with

$$
d'(r) = \int_{S_r} F(x,u)\,d\sigma \ge 0 \qquad \text{for $r \in (0,\delta_1)$.}
$$

\begin{corollary}
\label{cor-der-N}
For $0<r<\delta_1$, we have that
\begin{equation}
  \label{eq:D-prime-ineq}
  \begin{split}
D'(r) & \ge \left( \frac{N-2}{r} + O(1)\right)D_1(r)  + (2-q)d'(r) \\
& -  \Bigl(\frac{2N}{r}+O(1)\Bigr)d(r)    + \frac2{r} \int_{B_r}   V u \langle \nabla u ,Z\rangle - \int_{S_r} V u^2.
\end{split}
\end{equation}
Moreover, if $H(r) \neq 0$, we have that
\begin{align}
N'(r) \ge O(1) N(r)&+ \frac{1}{H(r)}\biggl[ r (2-q)d'(r) -   \bigl(2N+O(r)\bigr)d(r)  + 2\int_{B_r}  V u \langle \nabla u, Z \rangle\nonumber\\
& +(N-2 + O(r)) \int_{B_r} \Bigl( V u^2 + f(x,u)u \Bigr)  - r\int_{S_r} V u^2 \biggr].  \label{eq:N-prime-ineq}
\end{align}
\end{corollary}

\begin{proof}
Let $r \in (0,\delta_1)$. We first note that
\begin{align}
\int_{B_r}f(x,u) \langle Z, \nabla u \rangle
&= \int_{B_r} \bigl \langle \nabla_x F(x,u(x))- \nabla_1 F(x,u), Z \bigr \rangle \nonumber\\
&=  \int_{S_r}F(x,u) \langle Z,\nu \rangle   - \int_{B_r}\Bigl(F(x,u) \div Z + \langle \nabla_1 F(x,u), Z \rangle \Bigr) \nonumber\\
&=  r\int_{S_r}F(x,u)   - \int_{B_r}\Bigl( (N+ O(r)) F(x,u) + \langle \nabla_1 F(x,u), Z \rangle \Bigr),\label{1332-prelim}
\end{align}
where we recall that $\langle Z, \nu \rangle = r$ on $S_r$, that $\div Z = N +O(r)$ (see \eqref{div Z}), and that $\nabla_1 F(x,s)$ denotes the gradient of $F(x,s)$ with respect to the the first $N$ variables. Moreover, we have
\begin{equation}
  \label{eq:Z-pointwise-exp}
\|Z\|_{L^\infty(B_r)}= O(r) \qquad \text{for $r \in (0,\delta_1)$}
\end{equation}
by definition of $Z$ and (A1), and thus assumption (A3)iii) gives
\begin{equation}\label{1332}
\Bigl|\int_{B_r}
\langle \nabla_1 F(x,u),Z \rangle \Bigr| \le \kappa_1 \int_{B_r} F(x,u) |Z| \le O(r)  \int_{B_r} F(x,u)  \qquad \text{for $r \in (0,\delta_1)$.}
\end{equation}
Combining (\ref{1332-prelim}) and (\ref{1332}) yields that
\begin{equation}
\label{1332-1}
\int_{B_r}f(x,u) \langle Z, \nabla u \rangle \ge
r\int_{S_r}F(x,u)   - (N+ O(r))  \int_{B_r} F(x,u).
\end{equation}
Therefore, by Corollary \ref{lem: der D}, we infer that
\begin{align}
D'(r)
\ge &   \left(\frac{N-2}r + O(1)\right) D_1(r)
+ 2 \int_{S_r}  \frac{\langle A \nabla u,\nu\rangle^2}{\mu} + \frac2r\int_{B_r}  V u \langle Z, \nabla u \rangle  \nonumber \\
&- \frac2r (N+O(r))\int_{B_r} F(x,u) + \int_{S_r}  \Bigl(2F(x,u) - f(x,u)u-V u^2\Bigr)
\nonumber
\end{align}
for $r \in (0,\delta_1)$. Since also
\begin{equation}\label{1333}
\int_{S_r}\Bigl(2 F(x,u) - f(x,u)u\Bigr)\,dx \ge (2-q) \int_{S_r} F(x,u)\,dx
\qquad \text{for $0<r<\delta_1$}
\end{equation}
by assumptions (A3)i) and (\ref{eq:extra-assumption}), we obtain \eqref{eq:D-prime-ineq}. In a similar way, estimate \eqref{eq:N-prime-ineq} can be obtained starting from \eqref{1334}, using \eqref{1332-1} and \eqref{1333}.
\end{proof}

We add further basic estimates to simplify the inequalities in Corollary~\ref{cor-der-N}.

\begin{lemma}\label{on u^2}
For $r \in (0,\delta_1)$, we have that
\begin{align}
  \label{eq:u^2-S-r}
\int_{S_r} u^2 \, dx &\le O(1) \|u\|_{L^\infty(S_r)}^{2-q}d'(r)\\
\int_{B_r} u^2 \, dx &\le O(1) \|u\|_{L^\infty(B_r)}^{2-q}d(r)
  \label{eq:u^2}
\end{align}
and
\begin{equation}
\label{estimate linear part}
\Bigl| \int_{B_r}  V u \langle \nabla u, Z\rangle \Bigr| \le O(1)\Bigl( d(r)+ D_1(r) \Bigr)
\end{equation}
\end{lemma}

\begin{proof}
For $r \in (0,\delta_1)$, we have, by \eqref{rem: on S},
$$
\int_{S_r} u^2  \le \|u\|_{L^\infty(S_r)}^{2-q} \int_{S_r} |u|^q
\le  \frac{\eps_0^q}{\kappa_2}\|u\|_{L^\infty(S_r)}^{2-q} \int_{S_r} F(x,u) = O(1) \|u\|_{L^\infty(S_r)}^{2-q}d'(r),
$$
as claimed in (\ref{eq:u^2-S-r}). Now (\ref{eq:u^2}) follows immediately by integrating (\ref{eq:u^2-S-r}). Moreover, recalling (\ref{eq:Z-pointwise-exp}), we obtain that
\begin{align*}
\Bigl| \int_{B_r}  V u \langle \nabla u, Z\rangle\,dx \Bigr|   &\le
  O(r) \| V\|_{L^\infty(B_r)} \int_{B_r} |u| |\nabla u| \le O(r^2) \int_{B_r} u^2 + O(1)\int_{B_r} |\nabla u|^2 \\
& \le O(r^2)\|u\|_{L^\infty(B_r)}^{2-q}d(r) + O(1)\int_{B_r} \langle A \nabla u, \nabla u \rangle,
\end{align*}
where we have used (\ref{eq:u^2}) in the last step. This gives (\ref{estimate linear part}).
\end{proof}

We may now simplify the inequalities in Corollary~\ref{cor-der-N} as follows.

\begin{corollary}
\label{cor-der-N-1}
There exist a constant $C>0$ such that for $0<r<\delta_1$ we have
\begin{equation}
  \label{eq:D-prime-ineq-1}
D'(r)  \ge - \frac{C}{r} \Bigl[D_1(r)+d(r)\Bigr]   + \Bigl[(2-q)- C \|u\|_{L^\infty(S_r)}^{2-q}\Bigr]d'(r)
\end{equation}
Moreover, if $H(r) \neq 0$ and $N(r)>0$, we have that
\begin{align}
N'(r) \ge -C N(r)&+ \frac{1}{H(r)}\biggl[ r \Bigl((2-q)- C \|u\|_{L^\infty(S_r)}^{2-q}\Bigr) d'(r) -  C \Bigl(D_1(r)+ d(r)\Bigr)\biggr] \label{eq:N-prime-ineq-1}
\end{align}
\end{corollary}

\begin{proof}
Inserting successively the estimates (\ref{estimate linear part}) and (\ref{eq:u^2-S-r}) into (\ref{eq:D-prime-ineq}), we obtain that, with a suitable constant $C>0$ changing its value from line to line,
\begin{align*}
D'(r) & \ge - \frac{C}{r}\Bigl[D_1(r) +d(r)\Bigr] + (2-q)d'(r)  + \frac2{r} \int_{B_r}   V u \langle \nabla u ,Z\rangle - \int_{S_r} V u^2 \\
& \ge - \frac{C}{r}\Bigl[D_1(r)+d(r)\Bigr] + (2-q)d'(r)  -C \int_{S_r}  u^2 \\
& \ge - \frac{C}{r}\Bigl[D_1(r) +d(r)\Bigr] + \Bigl[(2-q)- C\|u\|_{L^\infty(B_r)}^{2-q}\Bigr] d'(r),
\end{align*}
as claimed in (\ref{eq:D-prime-ineq-1}). Similarly, (\ref{eq:N-prime-ineq-1}) follows by (\ref{eq:N-prime-ineq}), using (\ref{eq:u^2-S-r}), (\ref{estimate linear part}), and the fact that
\[
(N-2+O(r)) \int_{B_r} f(x,u) u \ge -C d(r)
\]
by assumption (A1)i).
\end{proof}

We are now ready to complete the

\begin{proof}[Proof of Theorem \ref{main-2simplified}]
As noted before, it suffices to prove the implication (\ref{eq:-r-*-eq}).
Arguing by contradiction, we suppose that $u \equiv 0$ in a neighborhood of $0$, but $u \not \equiv 0$ in $B_{\delta_1}$. By assumption (A3)i), this implies that
$F(\cdot,u(\cdot)) \equiv 0$ in a neighborhood of $0$ and
$F(\cdot,u(\cdot)) \not \equiv 0$ in $B_{\delta_1}$.
Consequently, setting
$$
r_0:= \sup \{r >0\::\: d(r)=0\},
$$
we have $0<r_0<\delta_1$ and $u \equiv 0$ on $B_{r_0}$. We first claim that there exists $r_1 \in (r_0,\delta_1)$ and $C_3>0$ such that
\begin{equation}
\label{lem: lower-bound-D-1}
D(r) \ge C_3 d(r)>0\qquad \text{for every $r \in (r_0,r_1)$.}
\end{equation}
To see this, we recall that by \eqref{eq:D-prime-ineq-1} we have
\begin{equation*}
D'(r)  \ge - \frac{C}{r_0} \Bigl[D_1(r)+d(r)\Bigr]   + \Bigl[(2-q)- C \|u\|_{L^\infty(S_r)}^{2-q}\Bigr]d'(r) \qquad \text{for $r \in (r_0,\delta_1)$,}
\end{equation*}
whereas (\ref{eq:u^2}) and assumption (A3)i) implies that
\begin{align}
\bigl|D(r)-D_1(r) \bigr| &\le \|V\|_{L^\infty(B_r)} \int_{B_r} u^2 + \int_{B_r} f(x,u) u \nonumber\\
&\le  \Bigl( O(1) \|u\|_{L^\infty(B_r)}^{2-q} + q\Bigr)d(r) \le O(1)d(r) \label{1711}
\end{align}
Consequently,
\begin{equation}
  \label{eq:D-prime-ineq-2}
D'(r)  \ge - C_0 D(r) - C_1 d(r)   + \Bigl[(2-q)- C \|u\|_{L^\infty(S_r)}^{2-q}\Bigr]d'(r) \quad \text{for $r \in (r_0,\delta_1)$}
\end{equation}
with $C_0 = \frac{C}{r_0}$ and a further constant $C_1>0$. We also recall that $D(r_0)=d(r_0)=0$ since $u \equiv 0$ on $B_{r_0}$. Therefore (\ref{eq:D-prime-ineq-2}) implies that
\begin{align}
e^{C_0 r} D(r) & =  \int_{r_0}^r  \frac{d}{d\tau} \left( e^{C_0 \tau} D(\tau) \right) \,d\tau \nonumber \\
&\ge \int_{r_0}^r  e^{C_0 \tau} \Bigl( \Bigl[(2-q)- C \|u\|_{L^\infty(S_\tau)}^{2-q}\Bigr]d'(\tau)-C_1 d(\tau)\Bigr)\,d\tau \nonumber\\
&  \ge e^{C_0 r_0}(2-q)d(r) - C e^{C_0 \delta_0} \|u\|_{L^\infty(B_r)}^{2-q} d(r)
-C_1e^{C_0 \delta_0} d(r)(r-r_0) \nonumber \\
&= e^{C_0 r_0}\Bigl[(2-q)- C \|u\|_{L^\infty(B_r)}^{2-q} e^{C_0(\delta_0-r_0)} -C_1(r-r_0) e^{C_0(\delta_0-r_0)} \Bigr]d(r)
\label{D-ge-d-proof}
\end{align}
for $r \in (r_0,\delta_1)$. Here we used again the fact that $d$ is increasing. Since, by the continuity of $u$,
$$
\|u\|_{L^\infty(B_r)}^{2-q} \to \|u\|_{L^\infty(B_{r_0})}^{2-q} =0 \qquad \text{as $r \to r_0^+$,}
$$
we may now choose $r_1>r_0$ sufficiently close to $r_0$ such that
\begin{equation}
  \label{eq:u-continuity-est}
\left(C \|u\|_{L^\infty(B_r)}^{2-q} +C_1(r-r_0)\right) e^{C_0(\delta_0-r_0)} < \frac{2-q}{2} \qquad \text{for $r \in (r_0,r_1)$.}
\end{equation}
Then (\ref{D-ge-d-proof}) gives rise to (\ref{lem: lower-bound-D-1}) with
$C_3 := e^{C_0 (r_0-\delta_0)}\frac{2-q}{2}>0$ and this choice of $r_1$.

Since $u \not \equiv 0$ on $B_r$ for every $r \in (r_0,\delta_1)$, we may now pick $r_2 \in (r_0,r_1)$ such that $H(r_2) \neq 0$, and we define
\[
r_3:=  \inf \left\{r \in (0,r_2): H(s)>0 \ \text{for every $s \in (r, r_2)$}\right\} \ge r_0.
\]
For $r \in (r_3,r_2]$, we then have $H(r)>0$ and $D(r)>0$ by (\ref{lem: lower-bound-D-1}), and thus $N(r)$ is well defined and positive. Moreover, we may invoke (\ref{eq:N-prime-ineq-1}) for $r \in (r_3,r_2]$, which yields that
\begin{align*}
\frac{N'(r)}{N(r)}  &\ge -C+ \frac{1}{D(r)}\biggl[  \Bigl((2-q)- C \|u\|_{L^\infty(S_r)}^{2-q}\Bigr) d'(r) -  \frac{C}{r} \Bigl(D_1(r)+ d(r)\Bigr)\biggr]\\
&\ge -C- \frac{C(D_1(r)+d(r))}{r_0 D(r)} \ge - C_4 \qquad \text{for $r \in (r_3,r_2]$}
\end{align*}
with a constant $C_4>0$ as a consequence of (\ref{lem: lower-bound-D-1}), (\ref{1711}) and (\ref{eq:u-continuity-est}). Integrating this inequality in $(r,r_2)$ with $r \in (r_3,r_2)$, we infer that
\[
N(r) \le N(r_2) e^{C_4 r_2} =: C_5  \qquad \text{for $r_3 < r \le r_2$}.
\]
At this point we proceed as in the model case studied in Section \ref{sec: proof simple case}. By Lemma \ref{lem: der H}, the boundedness of the frequency implies that
\[
\frac{d}{dr} \log \left( \frac{H(r)}{r^{N-1}}\right) = \frac{2 N(r)}{r} + O(1) \le C_5   \qquad \text{for $r \in (r_3,r_2],$}
\]
for a constant $C_5>0$, whereas on the other hand we have that
$$
\lim_{r \to r_3^+} \log \left( \frac{H(r)}{r^{N-1}}\right)= -\infty
$$
since $H(r_3)=0$. This is a contradiction, and hence (\ref{eq:-r-*-eq}) is proved. We thus have finished the proof of Theorem~\ref{main-2simplified}.
\end{proof}

redpajama_set_name: RedPajamaArXiv


Greek bond yields – which move in the opposite direction to prices – fell to their lowest level in almost 14 years as the country's economic growth prospects have improved. The yield on 10-year debt hit 3.274 per cent, according to Refinitiv – the lowest since 2005 – and is in stark contrast to 2011, when yields surpassed 40 per cent as the eurozone debt crisis mounted. The International Monetary Fund has forecast 2.4 per cent real gross domestic product growth for 2019, claiming the country "has entered a period of economic growth that puts it among the top performers in the eurozone".


The Best Low Carb Chicken Soup — Soothing and SUPER Delicious!
Who the heck needs noodles when you've got low carb chicken soup?
Because low carb chicken soup is good for the soul 🙂 It'll warm you from the inside out, and somehow making you feel overwhelmingly cozy. It's the perfect way to use leftover chicken and veggies, or get creative and find new favorites.
Great for meal prep and busy work nights — this low carb chicken soup takes only 30 minutes to make, and I know you're going to love it!
Sautéing low carb chicken soup vegetables before adding broth.
I know sautéing the low carb vegetables may seem like a skippable step, but you'll be thankful you did it. Cooking aromatic veggies (such as onions, garlic and leeks) allows the flavors to become smoother and more integrated with the rest of the low carb chicken soup ingredients — a quick blister goes a long way!
Don't add cabbage to the low carb chicken soup until close to the end!
All keto vegetables cook at different rates, and throwing the cabbage in too soon can make it super soggy, whereas tossing it in towards the end adds a nice, subtle crunch!
I know it can be hard to hover over, and watch your low carb chicken soup while it simmers — which how so many soup recipes get ruined by a vigorous boil. It's a good practice to keep half an eye on the simmer status of your soup, so your soup doesn't overcook before you get to enjoy it!
I haven't made this low carb chicken soup slow cooker yet, but if you try it, let me know in the comments below!
This keto chicken soup recipe is super easy to customize to your personal taste. Here are some ingredients you may want to swap, and ideas for what to swap 'em with!
Since I'm sticking to low carb recipes, I use cabbage in lieu of noodles. I find they're the quickest and most noodle like veggie to toss in. Use any low carb foods you prefer! Zoodles, carrot noodles, squash noodles are delicious in this low carb chicken vegetable soup. Cauliflower rice would be great, as would shirataki noodles, if you can hang with their flavor and consistency!
As far and vegetables go, I went for the classic trio in this low carb chicken soup. Adjust them as you see fit! Carrots may be higher in carbs than some of you would like — skip them and add more celery, cabbage, leeks, or whatever you're craving!
I took the lazy approach for this low carb chicken soup by using pre-cooked chicken — half chicken breast and half chicken thighs. I love doing this as a way of using up leftovers or rotisserie chicken. Alternatively, you can boil chicken and shred it specifically for this soup recipe. Or bake it on a sheet, dice it up and add it in!
Add your go-to herbs to this low carb chicken soup! I opted for my favorite flavor enhancers — remember to discard the stems and bay leaves before digging in!
Turn this into a low carb Mexican chicken soup with fresh salsa, guacamole and low carb tortilla chips on top. Or stir in some sour cream for a low carb creamy chicken soup.
Just because you're on a low carb meal plan, doesn't mean your low carb soup sides have to be boring! Go for a classic side salad, or make a grilled cheese using a low carb bread recipe.
I usually make low carb biscuits and serve them with bacon butter for a special treat.
Because I know you won't want to waste a single drop of this keto soup, I thought I'd share how to freeze this low carb chicken soup for future low carb dinners. Use a large Tupperware container for large portions, and resealable bags for smaller ones.
Check out this rad Southern Living video on how to freeze soup! They use an empty cardboard container to keep a plastic bag steady, then ladle the soup in. You can also use a muffin tin for individual portions.
Pst — I love making a double batch of this low carb chicken soup for colds, with the intention of freezing half of it. I'm a happier gal when my freezer is full of low carb meals.
Quickly reheat a portion of low carb soups over the stove in a small saucepan. This can be done in the microwave as well.
From frozen, you'll want to defrost your soup in the fridge overnight. Or, if you really want low carb chicken soup and didn't have time to prep, toss it in a saucepan on super low heat with a dash of chicken broth. Place a lid on top and slowly melt the broth while heating your low carb chicken soup.
Hope this low carb chicken soup recipes fills your your home with warm and cozy cheer! Let me know what you think in the comments below, enjoy!
This easy low carb chicken soup recipes here to warm you from the inside out! Full of keto veggies, chicken, delicious herbs and takes less than 30 minutes to throw together!
Add chopped onions, celery and carrots and cook for about 5 minutes, until veggies begin to soften.
Add garlic and cook for 1 minutes, stirring to ensure it doesn't burn.
Add cooked chicken and give the ingredients a stir.
Add chicken broth, rosemary, thyme and bay leaf.
Bring contents to a boil, then reduce heat to a simmer and cook for 10 minutes uncovered.
Add shredded cabbage, stir, and cook for an additional 10 minutes.
Allow low carb chicken soup to cool and enjoy!
Very good. I used a rotisserie chicken from the super market, which are known to be well done. After cooking for another 20 minutes it was some what mushy. Next time I will put it in at the end after the cabbage maybe. Used red cabbage. Gives it a little color.


I am dying slowly in my prison.
And then I'll take you anywhere without fear and without borders up to the rising sun .
With you I can anything I do not care what comes because where I'm going.
I will love you even if you have thrown me out of your heart .
I was born for your eyes , no one else.
Soul of my soul heart of the storm Tell me where to go.
With you I can anything. I do not care what comes because where I'm going.


text:
Q: A segmentation fault sometimes occurs when the function cvCreateFileCapture is invoked on network URL A segmentation fault occurs when the function cvCreateFileCapture is sometimes invoked using a network URL as argument, for esample: rtsp://IPaddress:Port/Path...
int main(int argc, char *argv[])
{
    QApplication a(argc, argv);

    ...

    CvCapture* capture = cvCreateFileCapture("rtsp://...");

    ...

    return a.exec();
}

No error occur if the argument is a file on disk.
I build the source code in debug mode and I when the segmentation fault has occurred, the following stack trace is listed:
gst_caps_unref
CvCapture_GStreamer::open(int, char const*)
cvCreateCapture_GStreamer(int, char const*)
cvCreateFileCapture
main

UPDATE. If I create a Plain C++ Project, the call to the function cvCreateFileCapture always results in a segmentation fault. If, instead, I create a Qt Gui Application project, the segmentation fault occurs randomly.
OUTPUT OF VALGRIND
==3160== Memcheck, a memory error detector
==3160== Copyright (C) 2002-2011, and GNU GPL'd, by Julian Seward et al.
==3160== Using Valgrind-3.7.0 and LibVEX; rerun with -h for copyright info
==3160== Command: ./Calibration 15 8 5 http://root:root@192.168.1.51/axis-cgi/mjpg/video.cgi
==3160==
--3160-- WARNING: Serious error when reading debug info
--3160-- When reading debug info from /usr/lib/nvidia-331/libGL.so.331.20:
--3160-- Can't make sense of .data section mapping
--3160-- WARNING: Serious error when reading debug info
--3160-- When reading debug info from /usr/lib/nvidia-331/libnvidia-glcore.so.331.20:
--3160-- Can't make sense of .got section mapping
==3160== Invalid read of size 4
==3160==    at 0x74444A2: ??? (in /usr/lib/i386-linux-gnu/libfontconfig.so.1.4.4)
==3160==    by 0x7446B09: FcConfigFilename (in /usr/lib/i386-linux-gnu/libfontconfig.so.1.4.4)
==3160==    by 0x745B1EB: FcConfigParseAndLoad (in /usr/lib/i386-linux-gnu/libfontconfig.so.1.4.4)
==3160==    by 0x7450065: FcInitLoadConfig (in /usr/lib/i386-linux-gnu/libfontconfig.so.1.4.4)
==3160==    by 0x7450188: FcInitLoadConfigAndFonts (in /usr/lib/i386-linux-gnu/libfontconfig.so.1.4.4)
==3160==    by 0x745040C: FcInit (in /usr/lib/i386-linux-gnu/libfontconfig.so.1.4.4)
==3160==    by 0x585B608: ??? (in /usr/lib/i386-linux-gnu/libQtGui.so.4.8.1)
==3160==    by 0x57DCC0B: QApplicationPrivate::construct(_XDisplay*, unsigned long, unsigned long) (in /usr/lib/i386-linux-gnu/libQtGui.so.4.8.1)
==3160==    by 0x7E014E7: ???
==3160==  Address 0xaaa898c is 20 bytes inside a block of size 22 alloc'd
==3160==    at 0x402BE68: malloc (in /usr/lib/valgrind/vgpreload_memcheck-x86-linux.so)
==3160==    by 0x7444407: ??? (in /usr/lib/i386-linux-gnu/libfontconfig.so.1.4.4)
==3160==    by 0x7446B09: FcConfigFilename (in /usr/lib/i386-linux-gnu/libfontconfig.so.1.4.4)
==3160==    by 0x745B1EB: FcConfigParseAndLoad (in /usr/lib/i386-linux-gnu/libfontconfig.so.1.4.4)
==3160==    by 0x7450065: FcInitLoadConfig (in /usr/lib/i386-linux-gnu/libfontconfig.so.1.4.4)
==3160==    by 0x7450188: FcInitLoadConfigAndFonts (in /usr/lib/i386-linux-gnu/libfontconfig.so.1.4.4)
==3160==    by 0x745040C: FcInit (in /usr/lib/i386-linux-gnu/libfontconfig.so.1.4.4)
==3160==    by 0x585B608: ??? (in /usr/lib/i386-linux-gnu/libQtGui.so.4.8.1)
==3160==    by 0x57DCC0B: QApplicationPrivate::construct(_XDisplay*, unsigned long, unsigned long) (in /usr/lib/i386-linux-gnu/libQtGui.so.4.8.1)
==3160==    by 0x7E014E7: ???
==3160==
==3160==
==3160== ---- Attach to debugger ? --- [Return/N/n/Y/y/C/c] ---- Y
==3160== starting debugger with cmd: /usr/bin/gdb -nw /proc/3163/fd/1024 3163
GNU gdb (Ubuntu/Linaro 7.4-2012.04-0ubuntu2.1) 7.4-2012.04
Copyright (C) 2012 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.  Type "show copying"
and "show warranty" for details.
This GDB was configured as "i686-linux-gnu".
For bug reporting instructions, please see:
<http://bugs.launchpad.net/gdb-linaro/>...
Reading symbols from /proc/3163/fd/1024...done.
Attaching to program: /proc/3163/fd/1024, process 3163
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Reading symbols from /usr/lib/libfaac.so.0...(no debugging symbols found)...done.
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Reading symbols from /usr/lib/i386-linux-gnu/libfontconfig.so.1...(no debugging symbols found)...done.
Loaded symbols for /usr/lib/i386-linux-gnu/libfontconfig.so.1
Reading symbols from /usr/lib/i386-linux-gnu/libaudio.so.2...(no debugging symbols found)...done.
Loaded symbols for /usr/lib/i386-linux-gnu/libaudio.so.2
Reading symbols from /usr/lib/i386-linux-gnu/libfreetype.so.6...(no debugging symbols found)...done.
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Reading symbols from /usr/lib/i386-linux-gnu/libSM.so.6...(no debugging symbols found)...done.
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Reading symbols from /usr/lib/i386-linux-gnu/libICE.so.6...(no debugging symbols found)...done.
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Reading symbols from /usr/lib/i386-linux-gnu/libXrender.so.1...(no debugging symbols found)...done.
Loaded symbols for /usr/lib/i386-linux-gnu/libXrender.so.1
Reading symbols from /usr/lib/i386-linux-gnu/libXext.so.6...(no debugging symbols found)...done.
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Reading symbols from /usr/lib/i386-linux-gnu/libX11.so.6...(no debugging symbols found)...done.
Loaded symbols for /usr/lib/i386-linux-gnu/libX11.so.6
Reading symbols from /lib/ld-linux.so.2...Reading symbols from /usr/lib/debug/lib/i386-linux-gnu/ld-2.15.so...done.
done.
Loaded symbols for /lib/ld-linux.so.2
Reading symbols from /usr/lib/nvidia-331/tls/libnvidia-tls.so.331.20...(no debugging symbols found)...done.
Loaded symbols for /usr/lib/nvidia-331/tls/libnvidia-tls.so.331.20
Reading symbols from /usr/lib/nvidia-331/libnvidia-glcore.so.331.20...(no debugging symbols found)...done.
Loaded symbols for /usr/lib/nvidia-331/libnvidia-glcore.so.331.20
Reading symbols from /lib/i386-linux-gnu/libdl.so.2...Reading symbols from /usr/lib/debug/lib/i386-linux-gnu/libdl-2.15.so...done.
done.
Loaded symbols for /lib/i386-linux-gnu/libdl.so.2
Reading symbols from /usr/lib/libImath.so.6...(no debugging symbols found)...done.
Loaded symbols for /usr/lib/libImath.so.6
Reading symbols from /usr/lib/libIex.so.6...(no debugging symbols found)...done.
Loaded symbols for /usr/lib/libIex.so.6
Reading symbols from /usr/lib/libIlmThread.so.6...(no debugging symbols found)...done.
Loaded symbols for /usr/lib/libIlmThread.so.6
Reading symbols from /usr/lib/i386-linux-gnu/libgmodule-2.0.so.0...(no debugging symbols found)...done.
Loaded symbols for /usr/lib/i386-linux-gnu/libgmodule-2.0.so.0
Reading symbols from /usr/lib/i386-linux-gnu/libxml2.so.2...(no debugging symbols found)...done.
Loaded symbols for /usr/lib/i386-linux-gnu/libxml2.so.2
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Reading symbols from /lib/i386-linux-gnu/libpcre.so.3...(no debugging symbols found)...done.
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Reading symbols from /usr/lib/i386-linux-gnu/libgstbase-0.10.so.0...(no debugging symbols found)...done.
Loaded symbols for /usr/lib/i386-linux-gnu/libgstbase-0.10.so.0
Reading symbols from /usr/lib/i386-linux-gnu/liborc-0.4.so.0...(no debugging symbols found)...done.
Loaded symbols for /usr/lib/i386-linux-gnu/liborc-0.4.so.0
Reading symbols from /usr/lib/i386-linux-gnu/libraw1394.so.11...(no debugging symbols found)...done.
Loaded symbols for /usr/lib/i386-linux-gnu/libraw1394.so.11
Reading symbols from /lib/i386-linux-gnu/libusb-1.0.so.0...(no debugging symbols found)...done.
Loaded symbols for /lib/i386-linux-gnu/libusb-1.0.so.0
Reading symbols from /usr/local/lib/libv4l2.so.0...done.
Loaded symbols for /usr/local/lib/libv4l2.so.0
Reading symbols from /usr/lib/i386-linux-gnu/libogg.so.0...(no debugging symbols found)...done.
Loaded symbols for /usr/lib/i386-linux-gnu/libogg.so.0
Reading symbols from /lib/i386-linux-gnu/libexpat.so.1...(no debugging symbols found)...done.
Loaded symbols for /lib/i386-linux-gnu/libexpat.so.1
Reading symbols from /usr/lib/i386-linux-gnu/libXt.so.6...(no debugging symbols found)...done.
Loaded symbols for /usr/lib/i386-linux-gnu/libXt.so.6
Reading symbols from /usr/lib/i386-linux-gnu/libXau.so.6...(no debugging symbols found)...done.
Loaded symbols for /usr/lib/i386-linux-gnu/libXau.so.6
Reading symbols from /lib/i386-linux-gnu/libuuid.so.1...(no debugging symbols found)...done.
Loaded symbols for /lib/i386-linux-gnu/libuuid.so.1
Reading symbols from /usr/lib/i386-linux-gnu/libxcb.so.1...(no debugging symbols found)...done.
Loaded symbols for /usr/lib/i386-linux-gnu/libxcb.so.1
Reading symbols from /usr/local/lib/libv4lconvert.so.0...done.
Loaded symbols for /usr/local/lib/libv4lconvert.so.0
Reading symbols from /usr/lib/i386-linux-gnu/libXdmcp.so.6...(no debugging symbols found)...done.
Loaded symbols for /usr/lib/i386-linux-gnu/libXdmcp.so.6
Reading symbols from /usr/lib/i386-linux-gnu/gconv/UTF-16.so...Reading symbols from /usr/lib/debug/usr/lib/i386-linux-gnu/gconv/UTF-16.so...done.
done.
Loaded symbols for /usr/lib/i386-linux-gnu/gconv/UTF-16.so
Reading symbols from /usr/lib/libicui18n.so.48...(no debugging symbols found)...done.
Loaded symbols for /usr/lib/libicui18n.so.48
Reading symbols from /usr/lib/libicuuc.so.48...(no debugging symbols found)...done.
Loaded symbols for /usr/lib/libicuuc.so.48
Reading symbols from /usr/lib/libicudata.so.48...(no debugging symbols found)...done.
Loaded symbols for /usr/lib/libicudata.so.48
Failed to read a valid object file image from memory.
0x074444a2 in ?? () from /usr/lib/i386-linux-gnu/libfontconfig.so.1
(gdb)

A:
A segmentation fault occurs when the function cvCreateFileCapture is
  sometimes invoked using a network URL as argument, No error occur if
  the argument is a file on disk.

By symptom of the problem, your program experiencing some sort of memory corruption. In all
probability the network URL flow execution is doing something(overflow/overwriting of memory)
and it looks like you are able to reproduce this problem if you user network URL. However please don't assume that if URL is a file on disk(locally), it does not have any problem. There could be possibility that corruption is happening in this way as well, it just that effect is getting delayed. Memory corruption problems are so hard to debug and there is no easy steps to identify them without performing extensive analysis of our program.
Looking at the stack trace, it would be almost impossible to find out the what could be the root cause of this as these segmentation fault would be after effect of something real bad done by program in early execution.
The best way to identify these memory corruption is to use some dynamic tools along with
debugger(WinDBG/PageHeap on Windows). From your question(segmentation fault), it look like you might be running on Linux platform. So You may want to attach your program(a.out) in Valgrind/GDB.
$ valgrind --tool=memcheck --db-attach=yes ./a.out

This way Valgrind would attach your program in the debugger when your first memory error is
detected so that you can do live debugging(GDB). This should be the best possible way to understand and resolve your problem. Once you are able to figure it out your first error, fix it and rerun it and see what are other errors you are getting.This steps should be done till no error is getting reported by Valgrind.

A: add cvWaitKey(100); before the cvQueryFrame(); function.

A: If you are in a Linux environment, you can use valgrind to find out exactly where the segmentation fault is happening.  Just type valgrind before the name of the program, or the way you execute your program.  For example, if you execute your program with the following command:
hello -print
issue the following command instead:
valgrind hello -print
You can also "pepper" your code with different cout or printf statements and observe the last output before the segmentation fault.  If the program is large, this method could be more cumbersome / less useful.  You may be able to do it right before and right after every place cvCreateFileCapture appears in your main function to find out where it is occurring.  If cvCreateFileCapture is part of opencv, it would appear to be a bug in opencv!

A: Check out this C++ code,
VideoCapture vcap;
Mat image;
const string videoStreamAddress = "rtsp://IPaddress:Port/Path?dummy=param.mjpg";//

//open the video stream and make sure it's opened
if(!vcap.open(videoStreamAddress)) {
    cout << "Error opening video stream or file" << std::endl;
    return -1;
}

for(;;) {
    if(!vcap.read(image)) {
        cout << "No frame" << std::endl;
        waitKey();
    }
    imshow("Output Window", image);
    if(waitKey(1) >= 0) break;
}

And make sure that your stream is not password protected.

A: gst_caps_unref <- AFAIK this function free's memory used by gst object. I would like you to run your application under valgrind, but with following command:
valgrind --leak-check=full ./a.out. And attach its output instead of one you have currently (it will be more informative, I assure you).  It is possible that your bug will not reproduce under Valgrind as it provides something like 'private memory' for a program. In this case I would recommend you to get cgdb (great interface for gdb) and debug your program under it (check for variable values, possible nullpointers, deletions itd.)

redpajama_set_name: RedPajamaStackExchange


text:
Editorial of June 2021
On June 8, 2021 By officialblogunioIn Editorials
By Tiago Sérgio Cabral (Managing Editor)
Data Governance and the AI Regulation: Interplay between the GDPR and the proposal for an AI Act
It is hardly surprising that the recent European Commission's proposal for a Regulation on a European Approach for Artificial Intelligence (hereinafter the "proposal for an AI Act") is heavily inspired by the GDPR. From taking note of the GDPR's success in establishing worldwide standards to learning from its shortcomings, for example by suppressing the stop-shop mechanism (arguably responsible for some of its enforcement woes).[1]
The proposal for an AI Act should not be considered a GDPR for AI for one singular reason: there is already a GDPR for AI, and it is called the GDPR. The scope and aims of the proposal are different, but there is certainly a high degree of influence and the interplay between the two Regulations, if the AI Act is approved, will certainly be interesting. In this editorial we will address one particular aspect where the interplay between the GDPR and the AI act could be particularly relevant: data governance and data set management.
Before going specifically into this subject, it is important to know that the AI Act's proposed fines have a higher ceiling than the GDPR's: up to 30,000,000 euros or, if the offender is company, up to 6% of its total worldwide annual turnover for the preceding financial year (article 71(3) of the proposal for an AI Act). We should note, nonetheless, that this specific value is applicable to a restricted number of infringements, namely:
a) non-compliance with the rules on prohibited AI systems (Article 5)
b) non-compliance with the rules data governance practices and data set management (Article 10, applicable only to high-risk AI).
Providing incorrect, incomplete or misleading information to notified bodies and national competent authorities in response to a request may result in fines of up to 10,000,000 euros or 2% of the total worldwide annual turnover for the preceding financial year.
All other infringements of the proposal for an AI Act are to be sanctioned with fines of up to 20,000,000 euros or 4% of the total worldwide annual turnover for the preceding financial year.
In theory, this could mean that the AI act's fines can, in the future, hit harder than the GDPR's for two reasons: i) higher ceiling for the most serious fines as referred; ii) while the GDPR divides its fines between the "20 million/ 4% category" and the "10 million / 2% category", the AI Act makes the first category the rule, with only one exception being sanctioned under the second value.
Nonetheless, it is clear data set management and data governance are considered key in the proposal for an AI Act and due to this fact infringement of these provisions may result in some of the higher fines in the proposed Regulation. And complying with the rules under article 10 of the proposal for an AI may not be an easy task. To illustrate, an analysis of paragraphs 1 to 4 is especially relevant:
1- High-risk AI systems which make use of techniques involving the training of models with data shall be developed on the basis of training, validation and testing data sets that meet the quality criteria referred to in paragraphs 2 to 5.
2. Training, validation and testing data sets shall be subject to appropriate data governance and management practices. Those practices shall concern in particular,
(a)the relevant design choices;
(b)data collection;
(c)relevant data preparation processing operations, such as annotation, labelling, cleaning, enrichment and aggregation;
(d)the formulation of relevant assumptions, notably with respect to the information that the data are supposed to measure and represent;
(e)a prior assessment of the availability, quantity and suitability of the data sets that are needed;
(f)examination in view of possible biases;
(g)the identification of any possible data gaps or shortcomings, and how those gaps and shortcomings can be addressed.
3.Training, validation and testing data sets shall be relevant, representative, free of errors and complete. They shall have the appropriate statistical properties, including, where applicable, as regards the persons or groups of persons on which the high-risk AI system is intended to be used. These characteristics of the data sets may be met at the level of individual data sets or a combination thereof.
4.Training, validation and testing data sets shall take into account, to the extent required by the intended purpose, the characteristics or elements that are particular to the specific geographical, behavioural or functional setting within which the high-risk AI system is intended to be used.
As we know, not all high-risk AI will be trained using personal data and if it does not, it will not fall within the scope of the GDPR. However, for AI that falls within the scope of the GDPR and of the AI Act it is not hard to foresee that infringing the abovementioned provisions of the AI Act may also entail infringements of certain provisions of the GDPR such as Article 5(1)paragraphs a, c and d, Article 25, Article 32 and Article 35 (if a DPIA was carried out, most of the abovementioned issues would also have to be addressed under data protection law). Furthermore, and depending on the specific conditions and AI, rules on legal basis, provision of information, rights of the data subject (including the right not to be subject to automated individual decision-making) could also be infringed.
Therefore, entities could, in theory be fined 30,000,000 euros or up to 6% due to infringing Article 10 of the AI Act and then a further 20,000,000 euros or up to 4% for breaching some of the abovementioned GDPR provisions.
We would argue that the likelihood of "double fines" would grow if data protection authorities are given corrective powers under the AI Act (an option that a number of Member States may opt for and that the Commission appears to favour by giving power at the EU's level to the European Data Protection Supervisor). Of course, giving said powers to data protection supervisory authorities could also result in a repeat of the abovementioned enforcement issues (especially if they are not given the means to perform the new tasks by Member States and the EU does not act to make sure that Member States, indeed, give them the necessary means, through infringement procedures, if needed).
It is still early days and the proposal for an AI Act may change substantially during negotiations. Nonetheless, early signs point to a very interesting interplay between the various parts of the EU's new framework on AI in which the GDPR and the AI Act will be integrated. Every actor in the AI value chain should be careful in addressing the necessary legal requirements because making a mistake may result in multiple infringements and sanctions under the various legal instruments (i.e. an infringement of one may result in an almost automatic infringement of another).
[1] The proposal does not contain a tool designed for centralized enforcement by the Commission, which could be seen as a lost opportunity and may hinder the AI Act's efficacy in cross-border large investigations.
Pictures credits: Robot by erik_stein.
AIartificial intelligencedata protectiondigital single marketGDPRGeneral da Protection Regulation
The rule of law and the defense of citizens against any power: on the case C-650/18 Hungary v European Parliament
Review of Portuguese Association of European Law's webinar on the rule of law protection in the European Union

redpajama_set_name: RedPajamaCommonCrawl


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text:
Siemens unveils Biograph mMR* whole-body integrated MR and PET system capable of simultaneous data acquisition. First system already undergoing clinical use testing in Munich, Germany
From a doctor's perspective, this is way cool. The first combined MR scanner and PET detection system all in one. I've been saying for a while that it's a 3D world, but it just keeps getting better. The amount of info that doctors will be able to get from this system will allow them to have more diagnostic data and save more lives than ever before. The ability to look inside the human body and see what it going on is just amazing. Here is the press release:
Erlangen, Germany, 2010-Nov-19
Today, Siemens Healthcare unveils its new system, Biograph mMR, the world's first integrated whole-body molecular MR with simultaneous data acquisition technology, currently undergoing clinical use testing. This revolutionary system comprises a magnetic resonance (MR) scanner and an integrated PET (Positron Emission Tomography) detection system with an architecture that performs as one. In the new 3-tesla hybrid system, Siemens developers have succeeded for the first time in simultaneously capturing MR and PET data with a whole-body system. The Biograph mMR system has been installed at the university hospital Klinikum rechts der Isar of the Munich Technical University, Germany.
"Together with our partner Siemens we are entering a new dimension in diagnostic imaging today", said Prof. Dr. Markus Schwaiger, director of the clinic for nuclear medicine at the university hospital. "We've initiated clinical use testing of Biograph mMR in an effort to diagnose diseases at a very early stage; to see the progression of disease and to use that information to develop a therapy plan precisely focused on the respective patient. Furthermore, we plan to use the system for cancer follow-up in the long run, by reducing radiation exposure by the use of the system."
With the simultaneous acquisition of MR and PET data, this system is designed to provide new opportunities for imaging. While MR provides exquisite morphological and functional details in human tissue, PET goes further to investigate the human body at the level of cellular activity and metabolism. The innovative system has the potential to be a particularly valuable tool for identifying neurological, oncological and cardiac conditions of disease and in supporting the planning of appropriate therapies. Since MRI does not emit ionizing radiation, Biograph mMR may provide an added benefit with lower-dose imaging. The Biograph mMR also opens new opportunities for research, such as the development of new biomarkers or new therapeutic approaches.
"Biograph mMR is the latest breakthrough innovation of Siemens in the field of diagnostic imaging. It will be a new instrument for driving personalized medicine forward", said Walter Maerzendorfer, CEO of the Business Unit Magnetic Resonance at Siemens Healthcare. "Biograph mMR is designed to simultaneously acquire morphology, function, and metabolism for the entire body", added Britta Fuenfstueck, CEO of the Business Unit Molecular Imaging at Siemens Healthcare.
MR and PET have become an established part of everyday healthcare routines and have proven themselves to be valuable clinical diagnostic tools. The integration of these two technologies into a single system capable of simultaneous acquisition brings the potential to revolutionize the diagnosis of many conditions. Initial research suggests that with this system, Molecular MR can scan the entire body in as little as 30 minutes for the combined exams, compared to one hour or more for sequential MR and PET examinations.
Siemens envisions a wide range of clinical applications for molecular MR including the early identification and staging of malignancies, therapy planning (including surgery planning) and therapy control.
A technical revolution
Until now, it was nearly impossible to integrate MR and PET technologies: the conventional PET detectors, which use photomultiplier tubes, could not be used in the strong magnetic field generated by an MR system. Integration was further limited by the lack of space inside the MR device. For this reason, MR-PET imaging was the result of two separate scans (MR and PET) with a significant time lag. With Biograph mMR, Siemens brings the first molecular MR system for clinical research that integrates MR with compact, specialized PET detectors.
The Biograph mMR – incorporating Tim, the "Total imaging matrix" technology from Siemens may make it even quicker and easier for clinicians to perform the MR examination.
More information and press pictures can be found under: http://www.siemens.com/press/healthcare/Biograph-mMR
*The Biograph mMR system requires 510(k) review by the FDA and is not commercially available. Due to regulatory reasons its future availability in any country cannot be guaranteed. Please contact your local Siemens organization for further details.
The outcomes achieved by the Siemens customers described herein were achieved in the customer's unique setting. Since there is no "typical" hospital and many variables exist, e.g., hospital size, case mix, level of IT adoption, there can be no guarantee that others will achieve the same results.
Rudra455 August 31, 2011 at 1:30 PM
We facilitate the provision of independent analysis to support expert testimony, regulatory or legislative engagements. Frequently, this work includes economic, financial and statistical studies of varying data analysis, technical and http://www.stlouisbridal.com.
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After interviewing candidates last week, the Chambers County Board of Education voted unanimously Wednesday night to appoint Candace Lyons to serve as the District 2 representative.
Wasting no time in their called work session, the board was only presented with one nomination, which came from board member Jeffrey Finch. Once Lyons was voted in, Superintendent Dr. Kelli Hodge gave a brief report on upcoming events and the meeting was adjourned.
Board members were in agreement that her perspective as a younger mother with a child currently in the school district will be invaluable to board operations.
Lyons is the daughter of the late Clifford Lyons, who previously represented District 2. Lyons graduated from LaFayette High School and earned a bachelor's degree in healthcare administration and management from Colorado State University's Global Campus.
Hodge said that Lyons will be officially sworn in at the Feb. 20 board meeting with board orientation following on Feb. 21.


Is your meeting stuck in a time warp? Are you doing the same thing you did 5 or 10 years ago? Are you avoiding change because you don't want to go out of your comfort zone?
Many meeting planners know the event process so well they rarely want to change it. Embracing change doesn't mean losing ground with your company, client or attendees. It means you are willing to look hard at your event and do what is absolutely necessary to make your meeting robust and exciting.
First, you need to be somewhat courageous. Look at other successful events and find out why they are sold-out, buzzed about or have positive impact on attendees. Be willing to gain insight from others so you can set your event apart. You don't have to implement everything other meeting organizations do, but this is a good start to the process.
Second, you need to embrace an attitude of change. Just because you have always run things a certain way does not mean it is the right way to do things today.
Third, you need to look at the event life cycle. Sometimes, meetings and events just need to die out and go away. This is very hard for people in the event meeting services industry to accept; but it is a reality. Every meeting has a purpose and sometimes, the purpose was met several months or years ago.
What do they want and need? What information are they interested in learning more about? What interactive technology tool rentals will help them meet their needs? Listen to their suggestions and implement them.
What is the main purpose of your meeting? Does it fit within your organization's mission? Is the purpose still relevant to your attendees, especially as you may be attracting cross-generational individuals?
Understand past successes and failures.
If you hold an annual conference, which one was the best and why? Which one was the worst and why? Putting together a list of your successes and failures can help you capitalize on your successes and minimize your failures.
A convenient way to keep a pulse on your attendee's view of the meeting is through a polling method with Audience Response System rental units. In addition, if you rent iPads for the meeting, you can ask attendees to give you feedback via direct messaging on Twitter or by sending you an email. Respond quickly to attendees when they have a question, comment or issue about the conference and you will start to understand what is great and not-so-great about your meeting.
AV Event Solutions wants to know, what are you doing to push past your meeting plateaus?


Southwest Airlines had a surprising and adorable stowaway on a flight going from Costa Rica to Houston earlier this month.
A young squirrel managed to safely make the four hour journey to George Bush International Airport, where he was discovered by the crew as they prepared for the plane's next flight. All the passengers had disembarked by the time it was found.
The critter was captured by US Customs and Border Protection (CBP) officials and placed in quarantine until they can find it a proper home.


Here we present some of our most luxurious spirits for when you need something truly special. The holidays are a great opportunity to splurge on unique gifts for your friends, family or coworkers; this curated list represents the best of the best. Enjoy!
$89.99 Sale / $119.99 Reg.
$109.99 Sale / $129.99 Reg.
$169.99 Sale / $219.99 Reg.
$194.99 Sale / $249.99 Reg.
$109.99 Sale / $139.99 Reg.
$84.99 Sale / $99.99 Reg.
$259.99 Sale / $299.99 Reg.
$299.99 Sale / $399.99 Reg.
$179.99 Sale / $229.99 Reg.
$89.99 Sale / $99.99 Reg.


Where Can I Use ClassPass?
How do I recommend a studio be added to ClassPass?
My city doesn't have ClassPass. When will you be in my city?
Yes! Your ClassPass membership allows you to easily take class in any of our cities all over the globe.
Go to your the search page.
Search by city, neighborhood, or zip code to find studios nearby.
Tap the 'Find a class' tab on the bottom of the screen.
In the 'Near' search box, you can search by current location, city, neighborhood, or zip code to find classes nearby.
Every class has a different credit value, and in some cities, those credit values may be more or less than what you see in your home city. After all, every city is unique!