text:
\section{Introduction}
In Euclidean spaces the fractional powers of the Laplacian, $(-\Delta)^s$,  have been studied in a variety of contexts, ranging from isotropic diffusion with jumps to thin obstacle problems and phase transition problems. See for instance \cite{CSS}, \cite{SV}, \cite{CSire}, and also \cite{Gft} for an up-to-date overview of recent developments. The Harnack inequality for the nonlocal Laplacian was first established by Landkof in \cite{La} using potential methods. One should also see the contributions of K. Bogdan, see \cite{Bo} and subsequent papers by the same author with various coauthors, and also those of Kassmann, see \cite{Ka1, Ka2}. In their 2007 celebrated work \cite{CS} Caffarelli and Silvestre gave an alternative proof based on the so-called extension method. They showed that a solution to the nonlocal equation $(-\Delta)^s u =0$ in an open set $\Omega \subset \mathbb{R}^n$ can be lifted to a function $U$ in $\Omega\times (0,\infty)\subset \mathbb{R}^{n+1}$ which solves a degenerate elliptic equation  satisfying the structural assumptions in the work of Fabes, Kenig and Serapioni \cite{FKS}. This allowed them to obtain  the Harnack inequality for $(-\Delta)^s$ from the De Giorgi-Nash-Moser theory developed in \cite{FKS} of such degenerate elliptic equations.

The focus of the present paper are situations that go beyond the Euclidean setting recalled above. More precisely, we prove a strong Harnack inequality for a class of degenerate parabolic equations  which arise in the extension problem for nonlocal operators $\mathscr{H}^s=(\partial_t - \mathscr{L})^s$, $0<s<1$, where $\mathscr{L}$ is, for instance, a locally subelliptic operator. A similar inequality for the time-independent case easily descends from this result. Similarly to what was done in \cite{CS}, with such Harnack inequalities we then obtain corresponding ones for $\mathscr{H}^s$ and $(-\mathscr{L})^s$. Some basic scenarios to which, under the hypothesis specified in Section \ref{pl}, our results apply are as follows:
\begin{itemize}
\item[(i)] $\mathscr{L}$ is a sub-Laplacian on a Carnot group $\mathbb{G}$, see \cite{Fo};
\item[(ii)] $\mathscr{L}$ is a sub-Laplacian generated by a system of vector fields satisfying H\"ormander's finite rank condition, see \cite{Hor};
\item[(iii)] $\mathscr{L}$ is a locally sub-elliptic operator in the sense of Fefferman and Phong, see \cite{FP};
\item[(iv)] $\mathscr{L}$ is the infinitesimal generator of a strongly continuous symmetric semigroup on $L^2$.
\end{itemize}
Given the operator $\mathscr{L}$ we consider the associated heat operator $\mathscr{H} = \partial_t - \mathscr{L}$.
Using the  spectral resolution of $-\mathscr{L}$, and the Fourier transform in the time variable $t$, we first introduce the relevant function spaces $W^{2s}$ and their parabolic counterpart $ H^{2s}$. These spaces constitute the natural domains for the fractional powers $(-\mathscr{L})^s$ and $\mathscr{H}^s$ respectively, and they generalise the time-independent fractional Folland-Stein Sobolev space $W^{2s,2}(\mathbb{G})$ defined in \cite{Fo} in the setting of  Carnot groups. We then show that for a ``Dirichlet" datum $u \in H^{2s}$ the extension problem for $\mathscr{H}^s$ is well-defined. Moreover, the corresponding solution belongs to the right ``energy space". The following is the central result in this direction.

\begin{theorem}\label{wk2}
Given a function $u \in H^{2s}$, let $V$ be defined by
\begin{equation}\label{v}
V(x,z,t) =
\int_0^\infty \int_{\mathbb{R}^n}  P^{(a)}_z(x,y,\tau) u(y,t - \tau) dy d\tau,
\end{equation}
where the function $P^{(a)}_z(x,y,t)$ denotes the \emph{Poisson kernel} for the extension operator defined in \eqref{extKernel} below.
Then, $V$ solves the degenerate parabolic partial differential equation in \eqref{ep}. Furthermore, we have
\begin{equation}\label{tr2}
\lim_{z\rightarrow 0^+}\|V(\cdot,z,\cdot)-u\|_{L^{2}\left(\mathbb{R}^{n+1}\right)}=0,
\end{equation}
and
\begin{equation}\label{dn2}
\lim_{z\rightarrow 0^+} \left\|\frac{2^{-a}\Gamma\left(\frac{1-a}{2}\right)}{\Gamma\left(\frac{1+a}{2}\right)}z^a\frac{\partial V}{\partial z}(\cdot,z,\cdot) + \mathscr{H}^{s}u\right\|_{L^{2}\left(\mathbb{R}^{n+1}\right)} = 0.
\end{equation}
Finally, given $M>0$, and with $\Sigma= \mathbb{R}^n \times (0, M) \times \mathbb{R}$, there exists $C=C(a,\Sigma)>0$ such that
\begin{equation}\label{en2}
\|V\|_{\mathcal{L}^{1,2}(\Sigma, z^a dx dz dt)} \leq  C\|u\|_{\mathcal{H}^{2s}}.
\end{equation}
\end{theorem}

For the definition of the Sobolev space $\mathcal{L}^{1,2}(\Sigma, z^a dx dz dt)$, one should see \eqref{ss} below. Once Theorem \ref{wk2} is established, we are able to obtain the Harnack inequality for the nonlocal operator $\mathscr{H}^s$ in the thin space from the regularity results of the corresponding extension operator $\mathscr{H}_a$ in the thick space. In this perspective, the analysis becomes at this point similar to the above described approach for the fractional Laplacian in \cite{CS} based on the De-Giorgi-Nash-Moser theory developed in the seminal work  \cite{FKS} of Fabes, Kenig and Serapioni, and its parabolic counterpart \cite{CSe} by Chiarenza and Serapioni. Our main result is the following.

\begin{theorem}\label{parharnack}
Let $0<s<1$, and $u \in H^{2s}$, with $u \geq 0$ in $\mathbb{R}^n \times \mathbb{R}$, be a solution to $\mathscr{H}^{s}u = 0$
in $B(x, 2r) \times (-2r^2, 0]$. Then, the following Harnack inequality holds
\begin{equation}\label{parhar1}
\underset{B(x,r) \times (-r^2, -\frac{r^2}{2}]}{\sup}\ u  \leq  C \underset{B(x, r) \times (-\frac{r^2}{4},0]}{\inf}\  u,
\end{equation}
for some universal $C>0$. Here $B(x,r)$ denotes  the relative metric ball  of radius $r$ centered at $x$ associated with the \emph{carr\'e du champ} corresponding to $\mathscr{L}$ (see Section \ref{pl} for the precise notion).
\end{theorem}

Theorem \ref{parharnack} implies the following corresponding Harnack inequality for nonnegative solutions of the nonlocal operator $(-\mathscr{L})^s$.

\begin{theorem}\label{harnack}
Let $s \in (0, 1)$ and  let $u \in W^{2s}$, with $u \geq 0$ in $\mathbb{R}^n$, be a solution in $B(x,2r)$ to $\left(-\La\right)^{s} u  = 0$.
Then, there exists a universal $C>0$ such that
\begin{equation}\label{main}
\sup_{B(x,  r)} u  \leq  C \inf_{B(x,r)}  u.
\end{equation}
\end{theorem}

We mention that Theorem \ref{harnack} represents a generalisation of the result first established for sub-Laplacians in Carnot groups by Ferrari and Franchi in \cite{FF}. We also mention that for the standard fractional heat operator $(\partial_t - \Delta)^s$ in Euclidean space a strong Harnack inequality was recently proved by the first and second named authors in \cite{BaG}. Their approach was based on the extension problem for the fractional heat equation that, for $s = 1/2$, was first introduced by F. Jones in his pioneering paper \cite{Jo}, and more recently developed for all $s\in (0,1)$ independently by Nystr\"om and Sande \cite{NS}, and Stinga and Torrea \cite{ST}.

The present work is organised as follows.  In Section \ref{pl} we introduce some basic notations and gather some known results which are relevant to our work in the subsequent sections. We then recall the notions of $\mathscr{H}^s$ and $\left(-\La\right)^s$ developed in \cite{G} based on Balakrishnan's calculus. Combining the spectral resolution for $- \mathscr{L}$ with the Fourier transform in the $t$ variable we also define the relevant function spaces  $W^{2s}$, and their parabolic counterpart $H^{2s}$. In Section \ref{et1} we recall the extension problem for $\mathscr{H}^s$ (following \cite{G}, the one for $\left(-\La\right)^s$ is obtained from the latter using Bochner's subordination). We then prove our main result, Theorem \ref{wk2}, which shows that such problem is well-defined in the parabolic fractional  Sobolev spaces $H^{2s}$. Finally, in Section \ref{final} we establish the Harnack inequality for $\mathscr{H}^s$, using the ideas developed in \cite{CSe}, \cite{Gry}, \cite{Sa} and \cite{CS}.

\section{Preliminaries}\label{pl}

In this section we collect some known results that will be used throughout our work. We emphasise that although we will confine our attention to the setting of smooth sub-Laplacians, we could have worked in the more abstract setting of symmetric Dirichlet forms supporting a doubling condition and a Poincar\'e inequality without essential changes.
In what follows we consider $\R^n$ endowed with Lebesgue measure (alternatively, we could have worked with a smooth measure on $\R^n$ and our results would hold unchanged) and with a system of globally defined smooth vector fields $X_1,...,X_m$. We indicate with $X_i^\star$ the formal adjoint of $X_i$ in $L^2(\R^n)$, and with $\mathscr{L} = - \sum_{i=1}^m X_i^\star X_i$ the sub-Laplacian associated with the $X_i$'s. We assume that $\mathscr{L}$ be locally subelliptic in the sense of \cite{FP}. The relative \emph{carr\'e du champ} will be denoted by
\begin{equation}\label{cdushamp}
|Xu|^2 = \frac 12 (\mathscr{L}(u^2) - 2 u \mathscr{L} u).
\end{equation}
 We note that, indicating with $(\cdot,\cdot)$ the inner product in $L^2(\R^n)$, we have for $u, v\in \mathscr{S}(\Rn)$
\[
(-\mathscr{L} u,v) = (u,-\mathscr{L} v),\ \ \ \ \ \ \ \ \ (-\mathscr{L} u,u) = \int_{\R^n} |Xu|^2 \ge 0,
\]
thus the operator $-\mathscr{L}$ is symmetric and positive. We define
\begin{equation}\label{d}
d(x,y) = \sup\{|\varphi(x) - \varphi(y)|\mid \varphi\in C^\infty(\R^n),\ |X\varphi|\le 1\},
\end{equation}
see \cite{BM, BM2} and \cite{Davies}.
Throughout the paper we assume that $d(x,y)$ defines a distance on $\R^n$. We also assume that the metric topology is equivalent to the pre-existing Euclidean one in $\R^n$, see \cite[Assumption A]{Sturm}, and also \cite{GN} for a discussion of this aspect. The relative balls will be denoted by $B(x,r)$, and we let $V(x,r) = |B(x,r)|$, where $|E|$ indicates the Lebesgue measure of the set $E$.
We will assume that there exists a constant $C_D>0$ such that for all $x\in \mathbb{R}^n$ and $r>0$ one has
\begin{equation}\label{doubling}
V(x,2r) \leq  C_D V(x,r).
\end{equation}
A third basic hypothesis that we make is that there exists another constant $C_P>0$ such that for every ball $B(x,r)$, and every function $u\in C^{0,1}(B(x,2r))$, one has
\begin{equation}\label{poi}
\int_{B(x,r)} |u - u_B|^2 dx \le C_P\ r^2 \int_{B(x,2r)} |Xu|^2 dx.
\end{equation}
Under these assumptions it is well-known from \cite{Gry} and \cite{Sa} that one has the Harnack inequality for the heat operator $\mathscr{H} = \frac{\partial}{\partial t} - \mathscr{L}$. In particular, the heat semigroup
\begin{equation}\label{pt}
P_t f(x) = e^{-t \mathscr{L}} f(x) = \int_{\R^n} p(x,y,t) f(y) dy,
\end{equation}
 has a strictly positive kernel $p(x,y,t)$ satisfying the Gaussian bounds:
there exist constants $C, \alpha, \beta >0$ such that for every $x,y\in \R^n$
and $t>0$ one has
\[
\frac{C}{V(x,\sqrt
t))^{\frac{1}{2}}V(y,\sqrt t))^{\frac{1}{2}}} e^{- \frac{\alpha d(x,y)^2}{t}} \le p(x,y,t)\le \frac{C^{-1}}{V(x,\sqrt
t))^{\frac{1}{2}}V(y,\sqrt t))^{\frac{1}{2}}} e^{- \frac{\beta d(x,y)^2}{t}}.
\]
Before proceeding we note that, under our assumptions, the semigroup is stochastically complete, i.e., $P_t1 = 1$. This means  that for every $x \in \mathbb{R}^n$ and $t > 0$ one has
\begin{equation}
\label{stc}
\int_{\mathbb{R}^n}p(x,y,t) dy = 1.
\end{equation}
This follows from the fact that \eqref{doubling} implies, in particular, that for every $x\in \R^n$ and every $r\ge 1$ one has
$V(x,r) \le r^Q V(x,1)$,
where $Q = \log_2 C_D$. This estimate implies that $\int_1^\infty \frac{r}{\log V(x,r)} dr = \infty$. We can thus appeal to \cite[Theor. 4]{Sturm} to conclude that $P_t 1 = 1$ (see also \cite{Gri} for the Riemannian case).
A first basic example under which the above assumptions are satisfied is of course that of a Carnot group. In such setting, because of the presence of non-isotropic dilations the hypothesis \eqref{doubling} is trivially satisfied, whereas \eqref{poi} was established in \cite{Var}. More in general, one can consider a system $X_1,...,X_m$ of smooth vector fields satisfying H\"ormander's finite rank condition in \cite{Hor} in a bounded open set $\Omega\subset \R^n$. In such situation, it is well known from the fundamental works \cite{NSW} and \cite{J} that \eqref{doubling} and \eqref{poi} hold locally. One can turn these local results into global ones by using the tools developed in \cite{BBLU}. Specifically, consider a compact set $K\supset \bar{\Omega}$ in which the vector fields $X_1,...,X_m$ are defined. It is possible to extend such system to the whole $\R^n$ in such a way that outside  $K$  the vector fields become the standard basis of $\R^n$, so that $\mathscr{L}=\Delta$, see \cite[Theorem 2.9]{BBLU}. Moreover, the new control distance associated with the extended vector fields satisfies \eqref{doubling} globally. As a consequence of what we have said above, the semigroup $P_t$ satisfies \eqref{stc}, see also \cite[(3.2) in Theor. 3.4]{BBLU} for a direct proof.

The semigroup $P_t$ defines a family of bounded operators $P_t : L^2(\mathbb{R}^n) \rightarrow L^2(\mathbb{R}^n)$ having the following properties:
\begin{enumerate}
\item  $P_0 =\mathrm{Id}$ and for $s,t\geq 0,P_{s+t} = P_s\circ P_t$;
\item for $u \in L^2(\mathbb{R}^n)$, we have
$\|P_tu\|_{L^2(\mathbb{R}^n)} \ \leq\  \|u\|_{L^2(\mathbb{R}^n)}$;
\item for $u \in L^2(\mathbb{R}^n)$, the map $t \rightarrow P_tu$ is continuous in $L^2(\mathbb{R}^n)$;
\item for $u, v \in L^2(\mathbb{R}^n)$ one has
\[
\int_{\mathbb{R}^n}(P_tu)\,v\,dx\ =\ \int_{\mathbb{R}^n}u(P_tv)\,dx.
\]
\end{enumerate}
Properties (1)-(4) above can be summarized by saying that $\{P_t\}_{t\geq 0}$ is a self-adjoint strongly continuous contraction semigroup on $L^2(\mathbb{R}^n)$. From the spectral decomposition, it is also easily checked that the operator $\mathscr{L}$ is furthermore the generator of this semigroup, that is for $u \in \operatorname{Dom}(\mathscr{L})$ (the domain of $\mathscr{L}$), one has
\begin{equation}
\label{decayl}
\lim_{t\rightarrow 0^+}\Big\|\frac{P_tu-u}{t}-\mathscr{L}u\Big\|_{L^2(\mathbb{R}^n)} = 0.
\end{equation}
This implies that for $t \geq 0, P_t \operatorname{Dom}(\mathscr{L}) \subset \operatorname{Dom}(\mathscr{L})$, and that for $u \in \operatorname{Dom}(\mathscr{L})$,
\[
\frac{d}{dt} P_tu  =  P_t\mathscr{L}u  = \mathscr{L}P_tu,
\]
the derivative in the left-hand side of the above equality being taken in $L^2(\mathbb{R}^n)$. Formula \eqref{decayl} shows in particular that for any $u \in \operatorname{Dom}(\mathscr{L})$, one has
\begin{equation}
\label{decay}
\|P_tu - u\|_{L^2(\mathbb{R}^n)}  =  O(t)\quad \text{as}\ t \rightarrow 0^+.
\end{equation}

We next recall the fractional calculus of the operators $\mathscr{L}$ and $\mathscr H = \frac{\partial}{\partial t} - \mathscr{L}$ introduced in \cite{G}. For related developments covering a quite different class of H\"ormander type equations, we refer the reader to the recent work \cite{GT} by Tralli and the second named author.

\begin{definition}\label{fracl}
Let $s\in (0,1)$. For any $u \in \mathscr S(\mathbb{R}^n)$ we define the nonlocal operator
\begin{eqnarray}
\label{fracls}
\left(-\La\right)^{s}u(x) = -\frac{s}{\Gamma(1-s)}\int^{\infty}_0 t^{-s-1}[P_tu(x) -u(x)] dt.
\end{eqnarray}
\end{definition}

In an abstract setting formula \eqref{fracls} is due to Balakrishnan, see \cite{B1} and \cite{B2}.
The integral in the right-hand side of \eqref{fracls} must be interpreted as a Bochner integral in $L^2(\mathbb{R}^n)$. We note explicitly that, in view of \eqref{decay} and of (2) above, such integral is convergent in $L^2(\mathbb{R}^n)$ for every $u \in \operatorname{Dom}(\mathscr{L})$, and thus in particular for every $u \in \mathscr S(\mathbb{R}^n)$. We emphasise that, as observed in \cite[Lemma 8.5]{G}, Definition \ref{fracl} coincides with that proposed in \cite{FF} in the setting of Carnot groups.
To define now the fractional power $\mathscr{H}^s$, we recall the notion of evolutive semigroup
\[
P^{\mathscr{H}}_{\tau}u(x,t) = \int_{\mathbb{R}^n}p(x,y,\tau) u(y,t-\tau) dy,
\]
for whose properties we refer the reader to \cite{G}. Observe that when $u(x,t) = v(x)$, then $P^{\mathscr{H}}_{\tau}u(x,t) = P_t v(x)$. Furthermore, $P^{\mathscr{H}}_{\tau}$ is contractive in $L^2(\mathbb{R}^{n+1})$,
\begin{equation}\label{conL2}
\|P^{\mathscr{H}}_{\tau} u\|_{L^{2}(\mathbb{R}^{n+1})} \leq  \|u\|_{L^{2}(\mathbb{R}^{n+1})},
\end{equation}
and, analogously to \eqref{decay}, for any $u\in \operatorname{Dom}(\mathscr H)$ (and thus in particular for $u \in \mathscr S(\mathbb{R}^{n+1})$), we have
\begin{equation}\label{rateH}
\|P^{\mathscr{H}}_{\tau} u-u\|_{L^2(\mathbb{R}^{n+1})} =  O(\tau).
\end{equation}

\begin{definition}
Given $s\in (0,1)$, and  a function $u \in \mathscr S(\mathbb{R}^{n+1})$, similarly to \eqref{fracls} we define
\begin{equation}
\label{fracsh}
\mathscr{H}^s u(x,t)\overset{def}{=}  -\frac{s}{\Gamma(1-s)}\int^{\infty}_0\tau^{-s-1}\Big[P^{\mathscr{H}}_{\tau}u(x,t)-u(x,t)\Big] d\tau.
\end{equation}
\end{definition}
Using \eqref{conL2} and \eqref{rateH}, it is easy to see that definition \eqref{fracsh} is well-posed in $L^2(\mathbb{R}^{n+1})$.
Next, we denote by $\{E(\lambda)\}$ the spectral measures associated with $\mathscr{L}$. Let
\begin{equation}\label{spectralD}
\mathscr{L} = - \int^{\infty}_0\lambda dE(\lambda)
\end{equation}
denote the spectral resolution of $\mathscr{L}$ (see e.g. \cite[Chap. 8]{Y}, or \cite[Chap. 27]{Se} for a detailed discussion). In terms of the spectral measures the heat semigroup $\{P_t\}$ is given by
\begin{equation}\label{hs}
P_t = \int^{\infty}_0e^{-\lambda t} dE(\lambda).
\end{equation}
An expression for  the fractional powers of the sub-Laplacian in terms of this decomposition is then given by
\begin{equation}\label{fracsl}
\left(-\La\right)^{s}u = \int^{\infty}_0\lambda^{s} dE(\lambda)u, \quad \text{for $u\in \mathscr S(\mathbb{R}^n)$}.
\end{equation}
The identity \eqref{fracsl} is easily verified using the well-known integral
\[
\int_0^\infty \tau^{-s -1} (1-e^{-\tau}) d\tau = \frac{\Gamma(1-s)}{s},\ \ \ \ \ \ \ \  \ 0<s<1,
\]
which gives $- \frac{s}{\Gamma(1-s)} \int_{0}^{\infty} \frac{e^{-\lambda t} -1 }{t^{1+s}} dt = \lambda^{s}$. This formula continues to be valid for any complex number $w = \lambda+2\pi i \sigma\in \mathbb C$, with $\lambda>0$. We record this for later use
\begin{equation}\label{la}
- \frac{s}{\Gamma(1-s)} \int_{0}^{\infty} \frac{e^{-(\lambda+2\pi i \sigma) t} -1 }{t^{1+s}} dt = (\lambda+2\pi i \sigma)^{s},\ \ \ \ \ \ \ \lambda>0, \sigma\in \mathbb{R}.
\end{equation}
Using \eqref{la} with $\sigma = 0$, the spectral representation \eqref{hs}
and Balakrishnan's formula \eqref{fracls}, we find
\begin{align*}
\left(-\La\right)^{s}u(x) & = -\frac{s}{\Gamma(1-s)}\int^{\infty}_0 t^{-s-1}[P_tu(x) -u(x)] dt
\\
& = -\frac{s}{\Gamma(1-s)}\int^{\infty}_0 t^{-s-1}[\int_{0}^{\infty} e^{-\lambda t} dE(\lambda) u(x) -u(x)] dt\notag
\\
& = \int_{0}^{\infty} \lambda^s d E(\lambda) u(x),
\end{align*}
where in the last equality we have exchanged the order of integration.  Now, for a function $u\in\mathscr S(\mathbb{R}^n)$ and $s \in (0,1)$ we define the norm
\begin{equation}\label{norm1}
\|u\|_{2s} \overset{def}{=} \left(\int^{\infty}_0\left(1+\lambda^{s}\right)^2 d\|E(\lambda)u\|^2\right)^{\frac{1}{2}},
\end{equation}
where we have used the fact that $E(\lambda)$ is idempotent and symmetric to write $(E(\lambda)u,u) = (E(\lambda)u,E(\lambda)u) = \|E(\lambda)u\|^2$. We then let
\begin{equation}
\label{spacew}
W^{2s} = W^{2s}(\mathbb{R}^n) \overset{def}{=} \overline{\mathscr S(\mathbb{R}^{n})}^{\|\cdot\|_{2s}}.
\end{equation}
From \eqref{fracsl} we thus see that the definition of $\left(-\La\right)^s$ is extendible to functions in  $W^{2s}$. We stress  that in the classical setting when $\mathscr{L}=\Delta$, the space $W^{2s}$ coincides with the fractional Sobolev space  of order $2s$  defined by the norm
\begin{equation}\label{ed}
\bigg(\int_{\mathbb{R}^n} \left( u^2 + ((-\Delta)^s u)^2  \right) \bigg)^{1/2}.
\end{equation}
This is easily seen from \eqref{fracsl} since in terms of the spectral resolution of $-\Delta$, we have that
\[
\int  u^2 +  ((-\Delta)^s u)^2 = \int_0^{\infty} (1+ \lambda^{2s}) d \|E(\lambda)u\|^2 \cong \int^{\infty}_0\left(1+\lambda^{s}\right)^2 d\|E(\lambda)u\|^2.
\]
Thus, the norm in \eqref{ed} is finite if and only if the corresponding norm in \eqref{norm1} is finite.
 We also mention that in the classical case when $0< s< 1/2$,  the space  $W^{2s}$ coincides with the fractional Aronszajn-Gagliardo-Slobedetzky space of the functions $f\in L^2(\R^n)$ such that \[\int_{\R^n} \int_{\R^n} \frac{|f(x) - f(y)|^2}{|x-y|^{n+4s}} dx dy<\infty.\]

Now for a function $f\in L^1(\mathbb{R})$ we denote its Fourier transform by
\[
\hat{f}(\sigma) = \int_{\mathbb{R}}e^{-2\pi i\sigma t}f(t) dt, \quad \sigma\in\mathbb{R}.
\]
Let $u\in \mathscr S(\mathbb{R}^{n+1})$, then  by taking the Fourier transform with respect to the time variable, we easily find
\begin{equation}\label{ptauH}
\widehat{P^{\mathscr{H}}_{\tau} u}(x,\sigma) = e^{-2\pi i \tau\sigma} P_\tau(\hat u(\cdot,\sigma))(x).
\end{equation}
From this observation, in view of \eqref{la},  we obtain
\begin{eqnarray}\label{he}
\widehat{\mathscr{H}^{s}u}(x,\sigma)& = & -\frac{s}{\Gamma(1-s)}\int^{\infty}_0\frac{1}{\tau^{1+s}}\Big[\int^{\infty}_0\left(e^{-\left(\lambda+2\pi i \sigma\right) \tau}-1\right) dE(\lambda)\hat{u}(\cdot,\sigma)\Big] d\tau\\
& = & \int^{\infty}_0\left(\lambda+2\pi i \sigma\right)^{s} dE(\lambda)\hat{u}(\cdot,\sigma).
\notag
\end{eqnarray}
Hence,
\begin{equation}\label{hes}
\|\widehat{\mathscr{H}^{s}u}(\cdot,\sigma)\|^2_{L^2(\mathbb{R}^n)} = \int^{\infty}_0|\lambda+2\pi i \sigma|^{2s} d\|E(\lambda)\hat{u}(\cdot,\sigma)\|^2.
\end{equation}
Similarly to the stationary case, for  a function $u\in\mathscr{S}(\mathbb{R}^{n+1})$ and $s \in (0,1)$ we  now define the following norm
\begin{equation}\label{H2s}
\|u\|_{\mathcal{H}^{2s}}  \overset{def}{=} \left(\int^{\infty}_{-\infty}\int^{\infty}_0\left(1+|\lambda+2\pi i \sigma|^{s}\right)^2 d\|E(\lambda)\hat{u}(\cdot,\sigma)\|^2 d\sigma\right)^{\frac{1}{2}}.
\end{equation}
We then let
\begin{equation}
\label{spaceh}
H^{2s} = \overline{\mathscr S(\mathbb{R}^{n+1})}^{\|\cdot\|_{\mathcal{H}^{2s}}}.
\end{equation}
In view of the computation in \eqref{he}, we see that  the definition of $\mathscr{H}^s$ is extendible to functions in  $H^{2s}$. As above, it can be seen that in the classical setting when $\mathscr{H}=\partial_t - \Delta$, the space $H^{2s}$ coincides with the space $\operatorname{Dom}(\mathscr{H}^s)$ adopted in \cite{ST} and \cite{BaG}.

\section{Extension problem for $(-\mathscr{L})^s$ and $\mathscr{H}^s$}\label{et1}
In this section, given $s\in (0,1)$, we indicate with $a = 1-2s$. Note that $-1<a<1$, and that $a = 0$ when $s = 1/2$. To discuss  the extension problem for $(-\mathscr{L})^s$ and $\mathscr{H}^s$, following \cite{G} we begin with that for $\mathscr{H}^s$, and then use Bochner subordination for the time-independent case. We will use the letter $z$ for the so-called extension variable, and denote by $\mathscr B_a = \frac{\partial^2}{\partial z^2}+\frac{a}{z}\frac{\partial}{\partial z}$ the Bessel operator on the half-line $\{z>0\}$. Given the heat operator $\mathscr{H} = \partial_t - \mathscr{L}$ with respect to the variables $(x,t)\in \mathbb{R}^{n+1}$, the extension problem for $\mathscr{H}^s$ consists in solving the ``Dirichlet problem" in $\mathbb{R}^{n+2}_+$
\begin{equation}\label{ep}
\begin{cases}
\mathscr{H}_aV \overset{def}{=} z^{a}\left(\mathscr{H} V+\mathscr B_aV\right) = 0,
\\
V(x,0,t) = u(x,t).
\end{cases}
\end{equation}
We note that the  operator $\mathscr{H}_a$ can be written as a degenerate parabolic operator of the form
\begin{equation}
\label{extdiv}
\mathscr{H}_a = \omega(x,z)\frac{\partial}{\partial t}+\sum^{m+1}_{i=1}\tilde{X}^{\star}_i\left(\omega(x,z)\tilde{X}_i\right),
\end{equation}
where
\begin{equation}\label{tildes}
\tilde{X}_1 = X_1,\ ...\ , \tilde{X}_m = X_m,\ \tilde{X}_{m+1} = \frac{\partial}{\partial z}, \quad \text{and}\quad \omega(x,z) = |z|^a.
\end{equation}
Given an open set $\Sigma \subset \R^n\times \mathbb{R}_+ \times \mathbb{R}$, the Sobolev space $\mathcal{L}^{1,2}(\Sigma, z^a dx dz dt)$ is defined as the collection of all functions $U\in L^{2}(\Sigma, z^a dx dz dt)$ such that $XU, \partial_z U\in L^{2}(\Sigma, z^a dx dz dt)$. Such space is endowed with its natural norm
\begin{equation}\label{ss}
||U||_{\mathcal{L}^{1,2}(\Sigma, z^a dx dz dt)} = \bigg(\int_{\Sigma} \bigg(|U|^2 + |XU|^2 + |\partial_z U|^2\bigg)z^a dx dz dt\bigg)^{1/2}.
\end{equation}
We now introduce the relevant notion of weak solution.

\begin{definition}\label{wk0}
Given $ \Sigma= B(x, r) \times (0, M) \times (T_1, T_2)$, a function $W \in \mathcal{L}^{1,2}( \Sigma, z^a dx dz dt)$ is said to be a weak solution in $\Sigma$ of
\begin{equation}
\begin{cases}
\mathscr{H}_a W=0.
\\
\underset{z \to 0^+}{\lim} z^a \partial_z W= \psi,\quad \text{ for $\psi \in L^{2} (B(x,r) \times (T_1, T_2), dxdt)$}
\end{cases}
\end{equation}
if for every $\phi \in C^{\infty}( B(x,r) \times [0, M] \times [T_1, T_2])$ with compact support in $B(x,r) \times [0, M) \times [T_1, T_2]$, and almost every $t_1 , t_2$ such that $T_1 <t_1 <t_2<T_2$, we have
 \begin{align}\label{d8}
& \int_{t_1}^{t_2} \left(\int_{B(x,r) \times (0, M)} \sum_{i=1}^{m+1}  \tilde X_i W \tilde X_i\phi |z|^a dxdz\right) dt= \int_{t_1}^{t_2} \left(\int_{B(x,r) \times (0, M)} |z|^a \phi_t W dxdz\right) dt
\\
& - \int_{B(x,r) \times (0, M)} \phi (\cdot,t_2) W (\cdot, t_2) |z|^a dxdz   + \int_{B(x,r) \times (0, M)} \phi (\cdot,t_1) W (\cdot,t_1) |z|^a  dxdz
\notag
\\
& - \int_{t_1}^{t_2} \int_{B(x,r) \times \{0\}}   \psi \phi dx dt.
\notag
 \end{align}
 \end{definition}

Hereafter, $p(x,y,t)$ indicates the heat kernel in \eqref{pt}. The following \emph{Poisson kernel} for the problem \eqref{ep} was introduced in \cite{G}:
\begin{equation}\label{extKernel}
P^{(a)}_z(x,y,t)\overset{def}{=} \frac{1}{2^{1-a}\Gamma\left(\frac{1-a}{2}\right)}\frac{z^{1-a}}{t^{\frac{3-a}{2}}}\,e^{-\frac{z^2}{4t}}\,p(x,y,t).
\end{equation}
For the main properties of the function $P^{(a)}_z(x,y,t)$, see \cite[Prop. 9.5, 9.6, 10.3, 10.4]{G}. Here, we reproduce the one that represents a key ingredient in solving \eqref{ep}.

\begin{proposition}\label{G-prop 9.6}
For every $x, y \in \mathbb{R}^n$ with $x \neq y$ and $t > 0$ one has
\[
\partial_t P^{(a)}_z(x,y,t) - \mathscr B_a P^{(a)}_z(x,y,t) = \mathscr{L}_x P^{(a)}_z(x,y,t).
\]
\end{proposition}

We are ready to prove one of the central results in this note.

\begin{proof}[Proof of Theorem \ref{wk2}]
Differentiating under the integral sign and using Proposition \ref{G-prop 9.6}  it is easily seen that $V$ satisfies the partial differential equation in \eqref{ep}. We thus turn to proving \eqref{tr2}. Taking the Fourier transform of $V$ with respect to the variable $t$, and using \eqref{hs}, \eqref{ptauH} and the spectral decomposition for $-\mathscr{L}$, we have the following alternative  representation for $V$,
\begin{eqnarray}\label{ftv}
\widehat{V}(x,z,\sigma)& = &\frac{z^{1-a}}{2^{1-a}\Gamma\left(\frac{1-a}{2}\right)}\int^{\infty}_0\frac{e^{-\frac{z^2}{4\tau}}}{\tau^{\frac{3-a}{2}}}e^{-2\pi i\tau \sigma}P_{\tau}(\hat{u}(\cdot,\sigma))(x) d\tau\\
& = &\frac{z^{1-a}}{2^{1-a}\Gamma\left(\frac{1-a}{2}\right)}\int^{\infty}_0\big(\int^{\infty}_0\frac{e^{-\frac{z^2}{4\tau}}}{\tau^{\frac{3-a}{2}}}e^{-\big(\lambda+2\pi i\sigma\big)\tau} d\tau\big) dE(\lambda)(\hat{u}(\cdot,\sigma))(x).
\notag
\end{eqnarray}
Using Plancherel theorem in the $t$ variable we note that,
\[
\|V(\cdot,z,\cdot)-u\|_{L^{2}\left(\mathbb{R}^{n+1}\right)} = \|\hat{V}(\cdot,z,\cdot)-\hat{u}\|_{L^{2}\left(\mathbb{R}^{n+1}\right)}.
\]
Since  for every $z>0$,
\begin{equation}\label{ident}
\frac{1}{2^{1-a}\Gamma\left(\frac{1-a}{2}\right)}z^{1-a}\int^{\infty}_0\frac{1}{t^{\frac{3-a}{2}}}e^{-\frac{z^2}{4t}} dt = 1,
\end{equation}
from \eqref{ftv} we infer that the following important identity
\begin{align}\label{difvu}
& \hat{V}(x,z,\sigma)-\hat{u}(x,\sigma)
\\
& = \frac{z^{1-a}}{2^{1-a}\Gamma\big(\frac{1-a}{2}\big)}\int^{\infty}_0\big(\int^{\infty}_0\frac{e^{-\frac{z^2}{4\tau}}}{\tau^{\frac{3-a}{2}}}\big[e^{-\big(\lambda+2\pi i\sigma\big)\tau}-1\big] d\tau\big) dE(\lambda)(\hat{u}(\cdot,\sigma))(x).
\notag
\end{align}
Then,
\begin{align}\label{lh1}
& \|\hat{V}(\cdot,z,\cdot)-\hat{u}\|^2_{L^2(\mathbb{R}^{n+1})}
\\
& = C(a)z^{2(1-a)}\int_{\mathbb{R}}\int_{0}^{\infty}\Big|\int^{\infty}_0\frac{e^{-\frac{z^2}{4\tau}}}{\tau^{\frac{3-a}{2}}}\Big[e^{-\left(\lambda+2\pi i\sigma\right)\tau}-1\Big]d\tau\Big|^2 d\|E(\lambda)\hat{u}(\cdot,\sigma)\|^2 d\sigma
\notag\\
& = C(a)z^{2(1-a)}\int_\mathbb{R}\int_{0}^{\infty}\Big|\int^{\infty}_0\frac{e^{-(\lambda+2\pi i\sigma)\frac{z^2}{4\theta}}}{\theta^{\frac{3-a}{2}}}\Big[e^{-\theta}-1\Big]d\theta\Big|^2|\lambda+2\pi i\sigma|^{1-a} d\|E(\lambda)\hat{u}(\cdot,\sigma)\|^2 d\sigma\notag\\
& \leq  C(a)z^{2(1-a)}\int_\mathbb{R} \int_{0}^{\infty}\left(\int^{\infty}_0\frac{1}{\theta^{\frac{3-a}{2}}}\Big[1-e^{-\theta}\Big]d\theta\right)^2|\lambda+2\pi i\sigma|^{1-a}d\|E(\lambda)\hat{u}(\cdot,\sigma)\|^2d\sigma\notag\\
& = C(a)z^{2(1-a)}\int_\mathbb{R}\int_{0}^{\infty}|\lambda+2\pi i\sigma|^{1-a} d\|E(\lambda)\hat{u}(\cdot,\sigma)\|^2d\sigma\notag.
\end{align}
We note that the second equality in \eqref{lh1} can be justified
by applying, for every $z>0$, $\lambda>0$ and $\sigma\in \mathbb{R}$ fixed, Cauchy's integral formula to the holomorphic function in the half-plane $\Re w>0$,
 \[
f(w) =  \frac{e^{-\frac{z^2(\lambda + 2\pi i \sigma)}{4w}}(e^{-w}-1)}{w^{\frac{3-a}{2}}},
\]
and to the counterclockwise oriented contour $\Gamma_{\varepsilon,R}$ formed by the two rays $\{(\lambda + 2\pi i \sigma )\tau \mid \tau>0\}$,  and $\{\Im w=0, \Re w>0\}$, and the portion of the circles $\{|w|^2 =R^2\}$ and $\{|w|^2 =\varepsilon^2\}$ connecting them. Then, we let $\varepsilon\to 0^+$ and $R \to \infty$. Noting that the integrals of $f(w)$ along the circular arcs of $\Gamma_{\varepsilon,R}$ tend to zero as $\varepsilon\to 0^+$ and $R\to \infty$, we conclude that
\[
\int^{\infty}_0\frac{e^{-\frac{z^2}{4\tau}}}{\tau^{\frac{3-a}{2}}}\Big[e^{-\left(\lambda+2\pi i\sigma\right)\tau}-1\Big]d\tau = (\lambda+2\pi i\sigma)^{\frac{1-a}2} \int^{\infty}_0\frac{e^{-(\lambda+2\pi i\sigma)\frac{z^2}{4\theta}}}{\theta^{\frac{3-a}{2}}}\Big[e^{-\theta}-1\Big]d\theta,
\]
which justifies the second equality in \eqref{lh1}.
 Consequently, keeping in mind that $1-a=2s$, from \eqref{lh1} we obtain
\begin{align*}
\|\hat{V}(\cdot,z,\cdot)-\hat{u}\|^2_{L^2(\mathbb{R}^{n+1})}
& \leq  C(a)z^{2(1-a)}\int_\mathbb{R} \int^{\infty}_0|\lambda+2\pi i\sigma|^{2s} d\|E(\lambda)\hat{u}(\cdot,\sigma)\|^2 d\sigma
\\
& =  C(a)z^{2(1-a)} \|\widehat{\mathscr{H}^{s}u}(\cdot,\sigma)\|^2_{L^2(\mathbb{R}^n)},
\end{align*}
where in the last equality we have used \eqref{hes}. We infer that $V$ converges to $u$ in $L^2$ as $z\rightarrow 0^+$.

We next prove \eqref{dn2}. Differentiating  \eqref{difvu} with respect to $z$, we find
\begin{align}\label{gud}
& \frac{\partial \hat{V}}{\partial z}(x,z,\sigma)
\\
& = \frac{z^{1-a}}{2^{1-a}\Gamma\big(\frac{1-a}{2}\big)}\int^{\infty}_0\big(\int^{\infty}_0\big(\frac{1-a}z - \frac{z}{2\tau}\big) \frac{e^{-\frac{z^2}{4\tau}}}{\tau^{\frac{3-a}{2}}}\big[e^{-\big(\lambda+2\pi i\sigma\big)\tau}-1\big] d\tau\big) dE(\lambda)(\hat{u}(\cdot,\sigma))(x).
\notag
\end{align}
Now we make the crucial observation that
\begin{equation}\label{observe}
- \frac{d}{d\tau}\left(\frac{e^{-\frac{z^2}{4\tau}}}{\tau^{\frac{1-a}{2}}} \right)=  \frac z2 \frac{e^{-\frac{z^2}{4\tau}}}{\tau^{\frac{3-a}{2}}}\left(\frac{1-a}{z}-\frac{z}{2\tau}\right).
\end{equation}
Using \eqref{observe} in \eqref{gud}, we obtain the second basic identity
\begin{align}\label{verygud}
& \frac{\partial \hat{V}}{\partial z}(x,z,\sigma)
\\
& = - \frac{z^{-a}}{2^{-a}\Gamma\big(\frac{1-a}{2}\big)}\int^{\infty}_0\big(\int^{\infty}_0 \frac{d}{d\tau}\big(\frac{e^{-\frac{z^2}{4\tau}}}{\tau^{\frac{1-a}{2}}} \big) \big[e^{-\big(\lambda+2\pi i\sigma\big)\tau}-1\big] d\tau\big) dE(\lambda)(\hat{u}(\cdot,\sigma))(x)
\notag\\
& = \frac{z^{-a}}{2^{-a}\Gamma(\frac{1-a}{2})}\int^{\infty}_0\frac{e^{-\frac{z^2}{4\tau}}}{\tau^{\frac{1-a}{2}}}\,\frac{d}{d\tau}\left(\int^{\infty}_0[e^{-(\lambda+2\pi i \sigma)\tau}-1]\,dE(\lambda)(\hat{u}(\cdot,\sigma))(x)\right) d\tau\nonumber\\
& =-\frac{z^{-a}}{2^{-a}\Gamma(\frac{1-a}{2})}\int^{\infty}_0\frac{e^{-\frac{z^2}{4\tau}}}{\tau^{\frac{1-a}{2}}}\left(\int^{\infty}_0e^{-(\lambda+2\pi i \sigma)\tau}(\lambda+2\pi i \sigma) dE(\lambda)(\hat{u}(\cdot,\sigma))(x)\right) d\tau\nonumber\\
&  =-\frac{z^{-a}}{2^{-a}\Gamma(\frac{1-a}{2})}\int^{\infty}_0 (\lambda+2\pi i \sigma) \bigg(\int^{\infty}_0 \frac{e^{-\frac{z^2}{4\tau}}}{\tau^{\frac{1-a}{2}}}e^{-(\lambda+2\pi i \sigma)\tau} d\tau\bigg)  dE(\lambda)(\hat{u}(\cdot,\sigma))(x) \nonumber\\
& = -\frac{z^{-a}}{2^{-a}\Gamma(\frac{1-a}{2})}\int^{\infty}_0\left(\lambda+2\pi i \sigma\right)^{\frac{1-a}{2}}\left(\int^{\infty}_0\frac{e^{-\theta}e^{-(\lambda+2\pi i \sigma)\frac{z^2}{4\theta}}}{\theta^{\frac{1-a}{2}}} d\theta\right) dE(\lambda)(\hat{u}(\cdot,\sigma))(x).
\notag
\end{align}
Note that the last equality can be justified by applying Cauchy's integral theorem to the function
\[
g(w) = \frac{e^{-\frac{z^2}{4w}(\lambda + 2\pi i \sigma)} e^{-w}}{w^{\frac{1-a}2}},
\]
and to the same path $\Gamma_{\varepsilon,R}$ as in the proof of \eqref{lh1}. From \eqref{he} and \eqref{verygud} we thus find
\begin{eqnarray*}
  &&\left\|\frac{2^{-a}\Gamma\left(\frac{1-a}{2}\right)}{\Gamma\left(\frac{1+a}{2}\right)}z^a\frac{\partial \hat{V}}{\partial z}(\cdot,z,\cdot) + \widehat{\mathscr{H}^{s}u}\right\|^2_{L^{2}\left(\mathbb{R}^{n+1}\right)}\\
  &=& \int_{\mathbb{R}}\int^{\infty}_0\left(\lambda+2\pi i\sigma\right)^{2s}\left(\frac{1}{\Gamma\left(\frac{1+a}{2}\right)}\int^{\infty}_0\frac{e^{-\theta}e^{-(\lambda+2\pi i \sigma)\frac{z^2}{4\theta}}}{\theta^{\frac{1-a}{2}}} d\theta-1\right)^2d\|E(\lambda)\hat{u}(\cdot,\sigma)\|^2d\sigma,
\end{eqnarray*}
and the right-hand side of the latter equality tends to zero as $z\to 0^+$ in view of Lebesgue dominated convergence and of the equation
$\frac{1}{\Gamma\left(\frac{1+a}{2}\right)}\int^{\infty}_0\frac{e^{-\theta}}{\theta^{\frac{1-a}{2}}} d\theta=1$.
By Plancherel theorem in $t$ variable we thus finally obtain
\begin{align*}
& \lim_{z\rightarrow 0^+} \left\|\frac{2^{-a}\Gamma\left(\frac{1-a}{2}\right)}{\Gamma\left(\frac{1+a}{2}\right)}z^a\frac{\partial V}{\partial z}(\cdot,z,\cdot) + \mathscr{H}^{s}u\right\|_{L^{2}\left(\mathbb{R}^{n+1}\right)}
\\
& = \lim_{z\rightarrow 0^+}\left\|\frac{2^{-a}\Gamma\left(\frac{1-a}{2}\right)}{\Gamma\left(\frac{1+a}{2}\right)}z^a\frac{\partial \hat{V}}{\partial z}(\cdot,z,\cdot) + \widehat{\mathscr{H}^{s}u}\right\|^2_{L^{2}\left(\mathbb{R}^{n+1}\right)} = 0,
\end{align*}
which gives the desired conclusion \eqref{dn2}.

We now prove our final claim \eqref{en2}.  First, we note from \eqref{extKernel} and \eqref{v} that
\begin{align}\label{to*}
&\bigg( \int_{\mathbb{R}} \int_{\mathbb{R}^n} V(x, z, t)^2 dx dt\bigg)^{1/2}
\\
& =  C(a) z^{2s} \bigg(\int_{\mathbb{R}} \int_{\mathbb{R}^n}  \left( \int_{0}^{\infty} \int_{\mathbb{R}^n}\frac{e^{-\frac{z^2}{4\tau}}}{\tau^{1+s}} p(x, y, \tau)u(y, t-\tau) dy d\tau \right)^2 dx dt\bigg)^{1/2}
\notag
\\
&  \leq  C(a) z^{2s}  \int_{0}^{\infty} \bigg(\int_{\mathbb{R}^n} \int_{\mathbb{R}} \int_{\mathbb{R}^n} p(x,y,\tau) u(y, t-\tau)^2  dxdt dy \bigg)^{1/2} \frac{e^{-\frac{z^2}{4\tau}}}{\tau^{1+s}} d\tau
\notag
\\
&  \leq   C(s) ||u||_{L^{2}(\mathbb{R}^{n+1})}.
\notag
\end{align}
Note that  the first inequality in \eqref{to*} follows by an application of  Minkowski and Jensen inequalities,  and the second one  follows from the fact that
\[
\int_{\mathbb{R}^n} p(x, y,\tau) dx = 1,\quad \text{and}\  \int_{0}^{\infty}  \frac{z^{2s}}{\tau^{1+s}}  e^{-\frac{z^2}{4\tau}} d\tau =  4^s \Gamma(s).
\]
For $M>0$ let $\Sigma= \mathbb{R}^n \times (0, M) \times \mathbb{R}$. Since $a \in (-1, 1)$, by Fubini's theorem  we  easily obtain from \eqref{to*}
\begin{equation}\label{to1}
\|V\|_{L^2(\Sigma, z^a dx dz dt)} \leq  C(a, M) \|u\|_{L^2(\mathbb{R}^{n+1})}.
\end{equation}
Keeping \eqref{verygud} in mind, we find next
\begin{eqnarray}\label{ndvdz1}
&&\Big\|\frac{\partial \hat{V}}{\partial z}(\cdot,z,\cdot)\Big\|^2_{L^2(\mathbb{R}^n)}\\
& = &C(a)z^{-2a}\int^{\infty}_0|\lambda+2\pi i \sigma|^{1-a}\big|\int^{\infty}_0\frac{e^{-\theta}e^{-(\lambda+2\pi i \sigma)\frac{z^2}{4\theta}}}{\theta^{\frac{1-a}{2}}} d\theta\big|^2 d\|E(\lambda)\hat{u}(\cdot,\sigma)\|^2\nonumber.
\end{eqnarray}
From \eqref{ndvdz1} we thus obtain,
\begin{align}\label{tuo}
&\Big\|\frac{\partial \hat{V}}{\partial z}(\cdot,z,\cdot)\Big\|^2_{L^2(\Sigma, z^a \,dz\, dx\, dt)}\\
&\leq\  C(a, M)\int^{\infty}_{-\infty}\int^{\infty}_0|\lambda+2\pi i \sigma|^{2s}\,d\|E(\lambda)\hat{u}(\cdot,\sigma)\|^2\,d\sigma \leq C(a, M) ||u||_{\mathcal{H}^{2s}},\notag
\end{align}
where in the last inequality we have used \eqref{H2s}.
We now note that  for an arbitrary function $f$, appropriately decaying at infinity in $\mathbb{R}^{n}$, an integration by parts gives
\[
\|Xf\|_{L^2(\mathbb{R}^n)} = \|\left(-\La\right)^{\frac{1}{2}}f\|_{L^2(\mathbb{R}^n)}.
\]
Combining this observation with Plancherel theorem in the $t$ variable, we find
\begin{equation}\label{equi1}
\|XV(\cdot,z,\cdot)\|_{L^2(\mathbb{R}^n \times \mathbb{R})} =  \|\left(-\La\right)^{\frac{1}{2}}\widehat{V}(\cdot,z,\cdot)\|_{L^2(\mathbb{R}^n \times \mathbb{R})}.
\end{equation}
Using \eqref{fracsl} with $s = 1/2$, and the representation \eqref{ftv}, we obtain
\begin{align*}
&\|\left(-\La\right)^{\frac{1}{2}}\widehat{V}(\cdot,z,\sigma)\|^2_{L^2(\mathbb{R}^n)}
\leq C(a)z^{2(1-a)}\int^{1}_0\lambda\bigg|\int^{\infty}_0\frac{e^{-\frac{z^2}{4\tau}}}{\tau^{\frac{3-a}{2}}}e^{-(\lambda+2\pi i \sigma)\tau} d\tau\bigg|^2 d\|E(\lambda)\hat{u}(\cdot,\sigma)\|^2
\\
&+C(a)z^{2(1-a)}\int^{\infty}_1\lambda\bigg|\int^{\infty}_0\frac{e^{-\frac{z^2}{4\tau}}}{\tau^{\frac{3-a}{2}}}e^{-(\lambda+2\pi i \sigma)\tau} d\tau \bigg|^2 d\|E(\lambda)\hat{u}(\cdot,\sigma)\|^2 = I(z,\sigma) + II(z,\sigma).
\end{align*}
We now estimate
\begin{align*}
I(z,\sigma) & \le C(a)z^{2(1-a)}\int^{1}_0\lambda\bigg(\int^{\infty}_0\frac{e^{-\frac{z^2}{4\tau}}}{\tau^{\frac{1-a}{2}}}e^{-\lambda\tau} \frac{d\tau}\tau \bigg)^2 d\|E(\lambda)\hat{u}(\cdot,\sigma)\|^2
\\
& \le C(a)z^{2(1-a)}\int^{1}_0 \bigg(\int^{\infty}_0\frac{e^{-\frac{z^2}{4\tau}}}{\tau^{\frac{1-a}{2}}} \frac{d\tau}\tau \bigg)^2 d\|E(\lambda)\hat{u}(\cdot,\sigma)\|^2
\\
& \le C(a)z^{2(1-a)} \bigg(\int^{\infty}_0\frac{e^{-\frac{z^2}{4\tau}}}{\tau^{\frac{1-a}{2}}} \frac{d\tau}\tau \bigg)^2 \int^{\infty}_0  d\|E(\lambda)\hat{u}(\cdot,\sigma)\|^2
\\
& = C'(a)\ \|\hat{u}(\cdot,\sigma)\|^2_{L^2(\R^n)},
\end{align*}
where the last equality is justified by the change of variable $\rho = \frac{z^2}{4\tau}$, and by the fact that $\Gamma(\frac{1-a}2) = \int_0^\infty \rho^{\frac{1-a}2} \frac{d\rho}\rho$. Next, an application of Cauchy's integral formula as in \eqref{lh1}, \eqref{verygud}, gives
\begin{align*}
II(z,\sigma) & = C(a)z^{2(1-a)}\int^{\infty}_1\lambda|\lambda+2\pi i \sigma|^{1-a}\bigg|\int^{\infty}_0\frac{e^{-(\lambda+2\pi i \sigma)\frac{z^2}{4\theta}}}{\theta^{\frac{3-a}{2}}}e^{-\theta} d\theta\bigg|^2 d\|E(\lambda)\hat{u}(\cdot,\sigma)\|^2
\\
& \le C(a)z^{2(1-a)}\int^{\infty}_1\lambda|\lambda+2\pi i \sigma|^{1-a}\left(\int^{\infty}_0\frac{e^{-\frac{\lambda z^2}{4\theta}}}{\theta^{\frac{1-a}{2}}}e^{-\theta} \frac{d\theta}\theta\right)^2 d\|E(\lambda)\hat{u}(\cdot,\sigma)\|^2.
\end{align*}
We now use the following formula that can be found in 9.   on p. 340 of \cite{GR}
\[
\int_0^\infty t^{\nu-1} e^{-(\frac{\beta}{t} + \gamma t)} dt = 2 \left(\frac \beta{\gamma}\right)^{\frac \nu{2}} K_\nu(2 \sqrt{\beta \gamma}),
\]
provided $\Re \beta, \Re \gamma >0$. Applying it with
$\nu = - \frac{1-a}2$, $\beta = \frac{\lambda z^2}4$ and $\gamma = 1$,
and keeping in mind that $K_\nu = K_{-\nu}$ (see 5.7.10 in \cite{Le}), we find
\[
\bigg(\int^{\infty}_0\frac{e^{-\frac{\lambda z^2}{4\theta}}}{\theta^{\frac{1-a}{2}}}e^{-\theta} \frac{d\theta}\theta\bigg)^2 = 4^{2-a} z^{-2(1-a)}\lambda^{-1+a} K_{\frac{1-a}2}(z\sqrt \lambda)^2.
\]
This gives
\begin{align*}
II(z,\sigma) \le C'(a) \int^{\infty}_1\lambda^a |\lambda+2\pi i \sigma|^{1-a} K_{\frac{1-a}2}(z\sqrt \lambda)^2 d\|E(\lambda)\hat{u}(\cdot,\sigma)\|^2.
\end{align*}
Combining estimates, we thus obtain
\begin{align*}
& \|\left(-\La\right)^{\frac{1}{2}}\widehat{V}(\cdot,z,\cdot)\|^2_{L^2(\Sigma, z^a dz dx dt)} \le C(a,M)\ \|\hat{u}\|^2_{L^2(\mathbb{R}^{n+1})}
\\
& + C'(a) \int^{\infty}_{-\infty}\int^{\infty}_1\lambda^a |\lambda+2\pi i \sigma|^{1-a}\int^{M}_0  K_{\frac{1-a}2}(z\sqrt \lambda)^2 z^a dz d\|E(\lambda)\hat{u}(\cdot,\sigma)\|^2 d\sigma.
\end{align*}
Observe now that, in view of the asymptotic behaviour of  $K_{\frac{1-a}2}(\rho)^2 $, see e.g. (5.11.9) in \cite{Le}, we have
\[
\int^{M}_0  K_{\frac{1-a}2}(z\sqrt \lambda)^2 z^a dz = \lambda^{-\frac{a+1}2} \int_0^{M \sqrt \lambda} K_{\frac{1-a}2}(\rho)^2 \rho^a d\rho \le C''(a) \lambda^{-\frac{a+1}2}.
\]
Substituting this information in the above inequality, we thus find
\begin{align*}
& \|\left(-\La\right)^{\frac{1}{2}}\widehat{V}(\cdot,z,\cdot)\|^2_{L^2(\Sigma, z^a dz dx dt)} \le C(a,M)\ \|\hat{u}\|^2_{L^2(\mathbb{R}^{n+1})}
\\
& + C'''(a) \int^{\infty}_{-\infty}\int^{\infty}_1\lambda^{\frac{a-1}2} |\lambda+2\pi i \sigma|^{1-a} d\|E(\lambda)\hat{u}(\cdot,\sigma)\|^2 d\sigma
\\
& \le C'''(a) \int^{\infty}_{-\infty}\int^{\infty}_1|\lambda+2\pi i \sigma|^{1-a} d\|E(\lambda)\hat{u}(\cdot,\sigma)\|^2 d\sigma \le \tilde C(a) ||u||_{\mathcal H^{2s}},
\end{align*}
where in the last inequality we have used \eqref{H2s}
 and the fact that $1-a = 2s$.
Finally,  by taking \eqref{equi1} into account, we  obtain
\begin{equation}\label{to2}
\| XV(\cdot,z,\cdot)\|^2_{L^2(\Sigma)} \leq C(a, M)\|u\|^2_{\mathcal{H}^{2s}}.
\end{equation}
The desired conclusion \eqref{en2} now follows from the estimates \eqref{to1}, \eqref{tuo} and \eqref{to2}.

\end{proof}

We close this section by briefly discussing the extension problem for $(-\mathscr{L})^s$. The relevant result is Theorem \ref{wk1} below, whose proof we do not present since it follows from that of Theorem \ref{wk2} and from Bochner's subordination.
For a given $s \in (0,1)$, let $a= 1-2s$. Now given $u\in W^{2s}$, let
\begin{eqnarray}
\label{U}
U(x,z)&\ =\ &
\int_{\mathbb{R}^n}\left(\int_0^\infty P^{(a)}_z(x,y,t)\,dt\right)\, u(y)\,dy \\
\nonumber
&\ =\ &
\frac{z^{1-a}}{2^{1-a}\Gamma\left(\frac{1-a}{2}\right)}\,\int_{\mathbb{R}^n}\left(\int^{\infty}_0\frac{e^{-\frac{z^2}{4t}}}{t^{\frac{1-a}{2}}}\,p(x,y,t)\,\frac{dt}{t}\right)\,u(y)\,dy\\
\nonumber
&\ =\ &\frac{z^{1-a}}{2^{1-a}\Gamma\left(\frac{1-a}{2}\right)}\,\int^{\infty}_0\frac{e^{-\frac{z^2}{4t}}}{t^{\frac{1-a}{2}}}\,P_tu(x)\,\frac{dt}{t}\\
\nonumber
&\ =\ &\frac{z^{1-a}}{2^{1-a}\Gamma\left(\frac{1-a}{2}\right)}\int^{\infty}_0\frac{e^{-\frac{z^2}{4t}}}{t^{\frac{1-a}{2}}}\left(\int^{\infty}_0e^{\lambda t}\,dE(\lambda)u\right)\,\frac{dt}{t}.
\end{eqnarray}
Differentiating under the integral sign and  by  arguing as in the proof of  Proposition   10.4 in \cite{G} we have that
\begin{equation}\label{strong}
\mathscr{L}_a U \overset{def}{=} z^{a}\left(\mathscr{L} U+\mathscr B_aU\right) = 0,\ \ \ z >0.
\end{equation}
The operator $\mathscr{L}_a$ can be written as
\begin{equation}
\label{extdivU}
\mathscr{L}_a  = -\sum^{m+1}_{i=1}\tilde{X}^{\star}_i\left(\omega(x,z)\tilde{X}_i\right),
\end{equation}
where $\tilde{X}_i$, $i=1,...,m+1$ and $\omega$ are as in \eqref{tildes}.
The fact that $U$ defined above is a  weak solution to  the following boundary  problem
\begin{equation}\label{wk}
\begin{cases}
\mathscr{L}_a U=0\ \text{in}\ \{z>0\},
\\
U(x, z)=u(x)\ \text{in}\ L^{2}(\mathbb{R}^n),
\end{cases}
\end{equation}
and that $(-\mathscr{L})^s$ can be realized as a weighted Dirichlet-to-Neumann map corresponding to the operator  $\mathscr{L}_a$, is a consequence of the following theorem.

\begin{theorem}\label{wk1}
For a given $u \in W^{2s}$, let $U$ be as in \eqref{U}. Then, $U$ solves the partial differential equation in \eqref{wk}. Furthermore, we have
\begin{equation}\label{tr1}
\lim_{z\rightarrow 0^+}\|U(\cdot,z)-u\|_{L^{2}\left(\mathbb{R}^{n}\right)}=0,
\end{equation}
and
\begin{equation}\label{dn1}
-\frac{2^{-a}\Gamma\left(\frac{1-a}{2}\right)}{\Gamma\left(\frac{1+a}{2}\right)}\lim_{z\rightarrow 0^+}z^a\frac{\partial U}{\partial z}(x,z) = \left(-\La\right)^{s}u(x)\ \text{in $L^{2}(\mathbb{R}^n)$}.
\end{equation}
Finally, for every $M>0$ and $\Sigma= \mathbb{R}^n \times (0, M)$, one has
\begin{equation}\label{en1}
\|U\|_{\mathcal{L}^{1,2}(\Sigma, z^a dx dz)} \leq  C\|u\|_{2s},
\end{equation}
for some $C=C(a,M)>0$.
\end{theorem}

\section{Proof of the Harnack inequality for $\mathscr{H}^s$}\label{final}

As it has been mentioned in the introduction, the Harnack inequality for  $\mathscr{H}^s$ and $\left(-\La\right)^s$ is derived from that of the corresponding degenerate local operators $\mathscr{H}_a$ and $\mathscr{L}_a$ in \eqref{extdiv} and \eqref{extdivU} respectively. Since the Harnack inequality for the time-independent operator $\mathscr{L}_a$ descends easily from that of the time-dependent operator $\mathscr{H}_a$, we focus on this one. From the works \cite{Gry} and \cite{Sa} it is known that the Harnack inequality for parabolic operators generated by smooth Dirichlet forms is equivalent to the doubling condition and the Poincar\'e inequality for the corresponding metric balls  associated with the \emph{carr\'e du champ}. Now, because of the presence of the non-smooth weight $|z|^a$ the extended operator in \eqref{extdiv} does not satisfy the structural assumptions in \cite{Gry}, \cite{Sa}. Nevertheless, the ideas in those papers, or more precisely the arguments in \cite{CSe}, do apply once we have the appropriate doubling condition and the Poincar\'e inequality in the extended space.
We thus turn to verify the validity of these properties for the metric balls corresponding to the extended system of vector fields \eqref{tildes}.

For a given $(x,z) \in \R^n \times \mathbb{R}$ and for $r>0$ we consider the cylinders
\[
C_r(x,z)= B(x,r) \times (z-r, z+r),
\]
where $B(x,r)$ is the ball in the control distance \eqref{d} corresponding to $\mathscr{L}$. Then, the doubling property
\begin{equation}\label{doubling1}
|C_{2r}(x,z)| \leq C\ |C_r(x,z)|
\end{equation}
follows immediately from \eqref{doubling}.  We now show that the doubling property for the metric balls follows from \eqref{doubling1}.
Observe that  the  \emph{carr\'e du champ} relative to  the extended system $\tilde X$ is
 \[
|\tilde X u|^2 = |Xu|^2 + (\partial_z u)^2.
\]
Since we have assumed that $d$ is in fact a distance, and $\partial_z$ is independent of $\mathscr{L}$, it easily follows that
 \begin{equation}\label{td}
\tilde d((x,z),(x',z')) = \sup\{|\varphi(x,z) - \varphi(x',z')|\mid \varphi\in C^\infty(\mathbb{R}^{n+1}),\ |\tilde X\varphi|\le 1\}
\end{equation}
is also a distance.
We also infer from  \eqref{td} that there exist constants $0<\sigma_1 < \sigma_2$ such that
\begin{equation}\label{comp1}
\tilde B((x,z),\sigma_1 r) \subset C_r(x,z) \subset \tilde B((x,z), \sigma_2 r),
\end{equation}
where $\tilde B((x,z),r)$ denotes the  ball relative to $\tilde d$ centred at $(x,z)$ and of radius $r$. From \eqref{doubling1} and \eqref{comp1} we easily conclude the validity of
\begin{equation}\label{db1}
|\tilde B((x,z), 2r)| \leq C |\tilde B((x,z), r)|,
\end{equation}
for a new constant $C$ depending also on $\sigma_1$ and $\sigma_2$.

We next show  that the following Poincar\'e inequality holds,
\begin{equation}\label{poi1}
\int_{\tilde B((x,z), r)}  (f - f_{\tilde B((x,z), r)})^2  |z|^a dxdz \leq  C_1 r^2  \int_{\tilde B((x,z),  2r)}  |\tilde Xf|^2 |z|^a dxdz,
\end{equation}
where
\[
f_{\tilde B((x,z), r)}= \frac{\int_{\tilde B((x,z), r)}  f\  |z|^a dxdz}{\int_{\tilde B((x,z), r)} |z|^a dxdz}.
\]
To prove \eqref{poi1} we observe that, on one hand, \eqref{poi} is valid by hypothesis. On the other hand, since $\omega(z) = |z|^a$ is an $A_2$-weight on the real line, the following one-dimensional Poincar\'e inequality holds (see for instance \cite{FKS}),
\begin{equation}\label{1d}
\int_{(z-r, z+r)}  (f- f_{z,r})^2 |z|^a dz \leq C \int_{(z-r, z+r)}  f'(z)^2 |z|^a dz,
\end{equation}
where
\[
f_{z,r}= \frac{\int_{(z-r, z+r)}  f |z|^a dz}{\int_{(z-r, z+r)} |z|^a dz}.
\]
We can thus apply the real analysis argument in the proof of \cite[Lemma 2]{LW}  to show that the Poincar\'e inequality holds with respect to the  cylinders $\{C_r(x,z)\}$, i.e.,
\begin{equation}\label{poi2}
\int_{ C_r(x,z)} (f - f_{ C_r(x,z)})^2  |z|^a  dxdz \leq  C_1 r^2  \int_{ C_{2r}(x,z)} |\tilde Xf|^2 |z|^a  dxdz,
\end{equation}
where
\[
f_{ C_r(x,z)}= \frac{\int_{C_r(x,z)} f  |z|^a  dxdz}{\int_{C_r(x,z)} |z|^a dxdz}.
\]
With $\omega(\tilde B((x,z), r)) = \int_{\tilde B((x,z), r)} |z|^a dz dx$, it also follows from  \eqref{comp1} that the following weighted doubling condition holds,
\begin{equation}\label{db2}
\omega(\tilde B((x,z), 2r)) \leq C_1 \omega(\tilde B((x,z), r)).
\end{equation}
Thus, using \eqref{comp1}, \eqref{poi2}
 and \eqref{db2}, by a standard argument (see the proof of  \cite[Theor. 1]{HK}, and also \cite{GN}, \cite{J}),  we conclude that \eqref{poi1} holds.  Moreover, arguing as in the proof of \cite[Theor. 2.1]{GN}, or as in \cite[Theor. 1]{HK}, from \eqref{poi1} and \eqref{db2} we infer that the following Sobolev type inequality holds for all $0<r \leq 1$, and every $\phi \in C_0^{0,1}(\tilde B((x,z),r))$,
\[
\bigg(\int_{\tilde B((x,z), r)} |\phi|^{2\kappa} |z|^a dx dz \bigg)^{\frac{1}{2\kappa}}  \leq C \bigg(\int_{\tilde B((x,z), r)} |X \phi|^2 |z|^a  dx dz \bigg)^{1/2},
\]
for some $\kappa >1$ depending  on the  doubling constant  in \eqref{db2}.
At this point, we can apply the circle of ideas in \cite{Gry} and  \cite{Sa}  to deduce that the following Harnack inequality holds for the degenerate parabolic operator $\mathscr{H}_a$ in \eqref{extdiv}.

\begin{theorem}\label{harext}
Let $U$ be a solution to $\mathscr{H}_a U=0$  in $C_r(x,z) \times (-r^2, 0]$ with $\mathscr{H}_a$ as in \eqref{extdiv}. Then, the following scale invariant Harnack inequality holds,
\begin{equation}\label{sc1}
\sup_{C_{r/2} (x,z) \times \{t= -\frac{r^2}{2}\} } U \leq C_0 \inf_{C_{r/2} (x,z) \times \{t= 0\} } U,
\end{equation}
for some universal constant $C_0>0$ depending on the constants in \eqref{poi1} and \eqref{db2}.
\end{theorem}

\begin{remark}
We mention that, when the system $\tilde X$ is of H\"ormander type, the Poincar\'e inequality \eqref{poi1} follows from Lu's work \cite{Lu}.
\end{remark}

\begin{remark}
We also note that given the validity of  \eqref{db1} and \eqref{poi1}, it follows from Theorem 4.4 in \cite{Sa} that  the operator $\mathscr{H}_a$ is hypoelliptic  for $z>0$.

\end{remark}

With Theorem \ref{harext} in hands, we are finally able to provide the

\begin{proof}[Proof of Theorem \ref{parharnack}]

It follows from Theorem \ref{wk2} that, given a function $u\in H^{2s}$, the function $V$ defined in \eqref{v} is a weak solution to
\begin{equation}
\begin{cases}
\mathscr{H}_a V = 0 \ \ \ \ \ \ \text{for}\ z>0,
\\
V(x,0, t) =  u(x,t),
\\
\underset{z \to 0^+}{\lim} z^a \partial_z V  =  0\ \text{in}\ B(x, 2r) \times (-2r^2, 0],
\end{cases}
\end{equation}
in the sense of Definition \ref{wk0} where $\mathscr{H}_a$ is  the extension operator in \eqref{extdiv}.
Now thanks  to the  vanishing Neumann condition, if $V$ is evenly reflected,  it follows by a standard  argument that  the extended  $V$  is  a weak solution to $\mathscr{H}_a V=0$ in $C_{2r}((x, 0)) \times (-2r^2, 0]$.    At this point,   we can apply the scale invariant Harnack inequality \eqref{sc1} to $V$  and subsequently by  arguing as in the proof of Theorem 4.8 in \cite{FF} we  obtain the desired conclusion.

\end{proof}

redpajama_set_name: RedPajamaArXiv


Come out and join The Cocoa Beach Regional Chamber of Commerce and Elements Organic Salon & Spa as we Celebrate their GRAND OPENING RIBBON CUTTING! Come see all that they have to offer! The Celebration will take place February 20th from 4PM - 6PM.
Honor us with your presence and show your support!!


With the arrival of the Kiwi summer comes Music in Parks, with 25 free outdoor gigs across Auckland.
The 2019 Music in Parks line-up has something for all musical tastes – whether you're a fan of pop, rock, reggae, RnB, jazz or opera.
Experience top performers from big-name artists to locally grown talent.
Go to musicinparks.co.nz for the full programme.


I've always thought about what makes salary negotiations so stressful, partly because of my own mixed success in that exercise over the years. In most cases, I felt as though I deserved a raise when the conversation finally arrived. But when you're forced to sit across from your boss, the situation becomes challenging.
Your palms sweat, and you sometimes forget to mention the key points you prepared. Your boss, on the other hand, is cool and collected. She is motivated to bring this negotiation to a close as quickly as possible, while granting you a modest raise.
Salary negotiation doesn't need to be stressful. Employees can gain the upper hand when negotiating with employers. But like most things, negotiating well takes practice and understanding. Here are several principles developed by some of the world's top business professors who study the art and science of negotiation. Armed with the proper information, you too can learn how to discuss terms with your boss like an expert.
Deepak Malhotra is a Harvard Business School professor who specializes in negotiation and deal making. A few years ago, Malhotra recorded a lecture he hosts annually for soon to be HBS graduates. The lecture is all about how new graduates can negotiate to get the best possible offer.
Malhorta makes one point clear in his research and in this video in particular. Understanding the constraints the other party must operate within will make it easier for you to negotiate well. By understanding constraints, you'll be able to propose a counter offer that is more likely to be acceptable to the organization. It will also help you to think of a creative counter offer that might otherwise be difficult to imagine without understanding the areas where the organization can be flexible.
Understanding restraints will require research both thorough independent review and thorough interviews. Start by using Google News to look for information about new company directors, or funding.
Next, speak with members of the organization to better understand the constraints they must operate within. When I've tried to get a raise in the past, I would often talk to people I knew at the company to see what their salary negotiation experiences were like.
Finally, ask questions during the interview process to help you understand key aspects of negotiating, such as time constraints. Time can be a valuable asset when it comes to negotiating. If you have enough time, you may be able to cultivate a competing offer by interviewing at another organization. Or if time is limited, you may be able to exchange a definitive yes on your part for better compensation in order to help the organization move forward with an important company initiative.
It's easy to get caught up in the thought that compensation boils down to a salary plus a bonus. But in reality, compensation can be much more than that. Professor Deborah Kolb is a Harvard professor and negotiation expert, and is an advocate for negotiating "conditions of success" in addition to standard aspects of compensation.
Kolb argues that current compensation is important, but what's even more important are the conditions under which you will operate if you accept the job or remain at your current workplace. By using the negotiation to ensure you are well positioned to be successful, you'll put yourself in an even better position to negotiate in the future.
When evaluating an offer remember to consider the full-compensation package before making a decision to accept or counter. Furthermore, if the other party asks you to share salary expectations or your reaction to a recent offer, remember to take full compensation into account.
It may be difficult to arrive at a satisfactory offer in the near term. This is particularly true for employees negotiating a raise at their current company. If this is the case, it may make sense to extend the negotiation to a later date.
For example, consider asking the person you are negotiating with if they are willing to table the discussion for a short period of time. Using the extra time you've bought yourself, try to earn an offer from another company to use as leverage, or you can improve performance to earn a better offer.
Another reason to extend the negotiation timeline is to properly evaluate an offer. Resist the temptation of accepting or declining an offer in-person. Instead, you'll want to research the full compensation. Estimate take home pay using an online tax calculator, review the employee handbook you'll be expected to sign, and look over benefits packages to see how they compare with your expectations.
Salary negotiation doesn't need to be stressful. If you have earned the opportunity to negotiate with a hiring manager, it means you are already a desirable professional. Rather than feeling anxious, remember that you have leverage.
Channel your nervous energy into understanding the constraints of the people on the other side of the table, and evaluate the compensation package. After your research is complete, focus on making an aggressive but reasonable counter offer.


me feet tappin' oan the watter.
so ah cud pickle eels wi' splatter.
come waltzin' in clogs an' shawl.
then ti a hot peas supper in stall.
an' cool watter drools frae the gob.
an' a hot tongue brings forth a sob.
she giggled, "Ah'll blaw it cowd my love."
So we sat dangling eel lines in the river.
while the Moon shone bright from above.
the grass was now heavy with dew.
to greet another day anew.
this memory a pleasant cup.
an' divnant pause ti look up.
towing dumb barges o' coal.
guided by light from the Mole.
up wan street an' doon the next.
"Harrawy man Geordie, wha ti next?"


text:
function movefcn(varargin);
%MOVEFCN contains the WindowButtonMotionFcn for the Root Locus Design GUI

%   Karen D. Gondoly
%   Copyright (c) 1986-98 by The MathWorks, Inc.
% $Revision: 1.1 $

x = varargin{1};

fig = gcf;
udfig = get(fig,'UserData');
ax=udfig.Handles.LTIdisplayAxes;
udAx = get(ax,'UserData');

CP = get(ax,'CurrentPoint');

if length(udAx.MoveData.MoveHandles)==2,
   newx = [CP(1,1);  CP(1,1)];
   newy = [CP(1,2); -CP(1,2)];
else
   newx = [CP(1,1)];
   newy = [0];
end

set(udAx.MoveData.MoveHandles,{'Xdata'},num2cell(newx),...
   {'Ydata'},num2cell(newy));

if isempty(udAx.Locus),
   return
end

switch udAx.MoveData.MoveType,
case 'Pole',
   Comp=zpk(udAx.MoveData.LocusSystem.Z, ...
      [udAx.MoveData.LocusSystem.P;[newx+(newy*i)]],1,udfig.Model.Plant.Object.Ts);
case 'Zero',
   Comp=zpk([udAx.MoveData.LocusSystem.Z;[newx+(newy*i)]], ...
      udAx.MoveData.LocusSystem.P,1,udfig.Model.Plant.Object.Ts);
end

RLsys = LocalComputeSystem(udfig);

locus = rlocus(RLsys,[udAx.MoveData.GainVector,udfig.Compensator.Gain]);

if isempty(locus),
   return
end

CLpoles = locus(:,end);
set(udAx.Locus,{'Xdata'},num2cell(real(locus(:,1:end-1)),2), ...
   {'Ydata'},num2cell(imag(locus(:,1:end-1)),2));
set(udAx.ClosedLoopPoles,{'Xdata'},num2cell(real(CLpoles)),...
   {'Ydata'},num2cell(imag(CLpoles)))
udfig.Compensator.Object=Comp;

set(fig,'UserData',udfig);

%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% LocalComputeSystem %%%
%%%%%%%%%%%%%%%%%%%%%%%%%%
function RLsys = LocalComputeSystem(udRL);

persistent Plant Sensor Filter Gain FdbkSign FdbkConfig

Comp = udRL.Compensator.Object;

if isempty(Plant)
   %---Read in persistent variables...this one time read makes
   %---MOVEFCN more responsive.
   Plant = udRL.Model.Plant.Object;
   Sensor = udRL.Model.Sensor.Object;
   Filter = udRL.Model.Filter.Object;
   Gain = udRL.Compensator.Gain;
   FdbkSign = udRL.Model.FeedbackSign;
   FdbkConfig = udRL.Model.Structure;

   %---Protect against improper state space conversions,
   %-----If the compensator is improper and the plant
   %-----is state space, convert the plant ss to a zpk. This traps errors
   %-----resulting when attempting to conver improper systems to ss.
   if ~isproper(Comp),
      if  isa(Plant,'ss'),
         Plant=zpk(Plant);
      end
      if isa(Sensor,'ss'),
         Sensor = zpk(Sensor);
      end
      if isa(Filter,'ss'),
         Filter = zpk(Filter);
      end
   end
end

RLsys = -1*FdbkSign*Plant*Sensor*Comp;

redpajama_set_name: RedPajamaGithub


text:
<?php

/**
 * CartRecover Exception Class
 *
 * @package CartRecover
 * @category Exception
 */
interface CartRecover_Exception {

}


A Texas scientist claims to have sequenced the DNA of Sasquatch, a creature whose very existence is mysterious to many and purely mythological to most.
"Our study has sequenced 20 whole mitochondrial genomes and utilized next generation sequencing to obtain 3 whole nuclear genomes from purported Sasquatch samples," Dr. Melba S. Ketchum of Nacogdoches, Texas, says in a news release.
"The genome sequencing shows that Sasquatch mtDNA is identical to modern Homo sapiens, but Sasquatch nuDNA is a novel, unknown hominin related to Homo sapiens and other primate species."
Ketchum writes that her team's research indicates the "North American Sasquatch" is a hybrid of a female Homo sapien and a male of "unknown hominin species," whose DNA matched approximately 15,000 years ago.
Sasquatch, more commonly known as Bigfoot, has been a staple of American mythology for hundreds of years. The modern Sasquatch theories took on a newfound prominence in 1958 when the first official Bigfoot search party was launched in California.
Most members of the scientific community have discounted the Sasquatch theory. However, in a Sept. 27, 2002, interview with National Public Radio, Jane Goodall appeared open to the concept, saying, "Maybe they don't exist, but I want them to."
So, how do Ketchum's claims hold up under scrutiny?
The Houston Chronicle's Eric Berger does some unraveling of Ketchum and the claims she made in the news release.
For starters, Berger notes that while Ketchum has 27 years of genetic research experience during her career as a veterinarian, her company, DNA Diagnostics, has received an "F" rating from the Better Business Bureau.
But more important, Ketchum has not allowed scientific peer review of her findings.
"That is a massive red flag. Real research scientists almost never preannounce their research findings," Berger writes. "In effect she is using the mantle of science to confer credibility on her discovery, without actually deserving the credibility."
And finally, where exactly did Ketchum get her DNA sample? After all, if she was working from a reliable source, that alone might be the real story because no physical evidence of Bigfoot exists on record.
As it turns out, Ketchum says her DNA sample was obtained from a blueberry bagel left in the backyard of a Michigan home that, according to the owner, sees regular visits from Sasquatch creatures.
In other words, it seems likely that the legend of Bigfoot will remain more myth than reality for the foreseeable future.


text:
Asus Vivobook 13 Slate OLED (T3300) review: a 2-in-1 laptop with a stunning screen
The Asus Vivobook 13 Slate OLED (T3300) is a laptop that can be transformed into a tablet by removing the keyboard
If you want a laptop that doubles up as a tablet you'll like the Asus Vivobook 13 Slate OLED (T3300). It's much more affordable than other 2-in-1 devices which ultimately means you do lose out on performance, but the OLED screen is stunning and the battery life is pretty good too.
13.3-inch OLED touchscreen
Stylus included in the box
Easy to convert into a tablet
The magnetic stand isn't very secure
Performance is disappointing
No facial recognition or fingerprint sensor
By Yasmine Crossland last updated 14 January 22
Asus Vivobook 13 Slate OLED (T3300) deals
No price information
To sum up this Asus Vivobook 13 Slate OLED (T3300) review: this 2-in-1 laptop is a true hybrid because you can comfortably use it as your personal computer or carry it with you as a tablet.
While it's not going to win any awards for being the best 2-in-1 laptop you can buy, it does work very well as a slate because you can remove the keyboard entirely. But its real crowning glory is the stunning 13.3inch OLED screen which is perfect for streaming TV shows and movies. You can't expect to be wow-ed when it comes to performance though.
If you don't think you'd use the Asus Vivobook 13 Slate OLED (T3300) as a tablet very often then you might be better off looking at T3's guides to the best laptops or the best laptops under £500 if you're on a budget.
Asus Vivobook 13 Slate OLED (T3300) review: price and availability
The Asus Vivobook 13 Slate OLED (T3300) is a reasonably priced PC, you can buy it starting from $599 in the US and £549.99 in the UK. See the widgets on this page to find out where you can pick on up in your region.
Asus Vivobook 13 Slate OLED (T3300) review: design and screen
As the name suggests, the Asus Vivobook 13 Slate OLED (T3300) can be converted from tablet to PC by snapping on the keyboard and the cover stand. Each attaches itself magnetically, and when you snap it on, the keyboard connects to the screen automatically.
Measuring 309.9 x 190 x 7.9 mm and weighing 785g, the Asus Vivobook 13 Slate OLED (T3300) is smaller than a lightweight laptop but still a lot bigger and heavier than most tablets.
The back of the screen is matte black and metallic with a slim silver bar along the top that houses the rear camera. On the front, there's another camera integrated into the top bezel. The bezels are admittedly thicker here than on a lot of other laptops and tablets, but that does mean you can grip it without blocking off a part of the screen.
On the frame, there are two USB-C ports, a 3.5mm headphone jack and a MicroSD card slot. It also has a volume rocker on the edge as you would expect from a tablet. When it comes to connectivity, you do get the best of both worlds here and if you need more inputs you could always buy a separate USB hub.
The back panel props the laptop up by folding in the middle using a 170-degree hinge. I'm not entirely sure why ASUS hasn't just built the stand into the back of the screen, three separate components seems like unnecessary hassle to me especially given the fact that the panel moves out of position when you knock it. Because of that, you'll be restricted to using it on a table and you certainly won't be able to use it very easily on your lap.
You can't deny you get a lot for your money when it comes to what's included in the box. Alongside the cover stand and keyboard, this 2-in-1 laptop comes with a case to carry and protect it all. You also get an ASUS Pen 2.0 stylus with four sizes of pen tips and a magnetic pen holder to attach it to the tablet when you aren't using it.
If you have the cover stand attached and not in use, you won't be able to snap the pen onto the tablet because it needs to sit on the magnetic back panel. Using the stylus feels great though, and thanks to the different sized tips you'll be able to get the accuracy you need for the task at hand. It's not completely on point but for quick note-taking or sketches, it'll work well enough.
The keyboard itself has a smooth rubber coating with low-profile keys and a large touchpad, it's got some weight to it so doesn't feel flimsy at all. Despite the fact that it's a snap-on keyboard, it feels nice to type on as well. The keys are tactile with a satisfying click and they're well-spaced out with 19.05mm between each one. The Fn keys along the top are easy to spot because they are grey as opposed to black like the rest of the board.
What really draws you in is the 13.3-inch OLED touchscreen. The Full HD 1920x1080p resolution and 550 nits peak brightness make for a clear, sharp display that you can see in most lighting conditions. Capable of showing up to 1.07 billion colours, it looks bold and vibrant as well. It's very well set up to be used for both work and play.
If you want a device to watch TV shows and movies on, this will do the trick. The 16:9 aspect ratio is the perfect shape for video content, and the screen features Dolby Vision which should boost the quality of shows making them look much more accurate.
To protect your eyes, the OLED display has much lower blue-light levels than you would get on an LCD. Even after a full day of use, I didn't experience much eye strain.
Unlocking the display is done by entering a pin or password. Unfortunately, there's no IR camera for facial recognition nor is there a fingerprint sensor.
Asus Vivobook 13 Slate OLED (T3300) review: performance and battery
Powering the Asus Vivobook 13 Slate OLED (T3300) is an Intel Pentium Silver N6000 CPU with Intel UHD Graphics and a choice of 4GB or 8GB RAM. You can buy it with either 128GB or 256GB of SSD storage - admittedly that's not a lot of memory so you can't expect this device to be as powerful as a dedicated laptop and you won't be able to store tonnes of large files on it either.
Day to day, it felt fine to use. It was relatively smooth but when I had multiple apps and programs open at once, it did lag a little. Transferring large files took some time too, and it will definitely struggle with more extensive computer software so PC games might not run quite as you need them to, for instance.
To compare it to other laptops, I ran the Geekbench 5 benchmarking test on it. The Asus Vivobook 13 Slate OLED (T3300) scored 676 in single-core and 1,489 in multi-core which places it amongst the cheap laptop crowd, like the Acer Swift 1.
While it may not be the fastest laptop out there, it does stand up nicely against other tablets. For typing out documents, browsing the web and making notes while you're on the move, it'll do a good enough job. Speedy Wi-Fi 6 is packed in as well so you're guaranteed the fastest internet speeds possible on your network.
The 50Wh battery seemed to last a decent amount of time when I tested it out. It won't survive a full workday but it will get you through most of it.
To find out how long the slate would last streaming videos, I ran a test where I downloaded and played a TV show on full brightness for two hours. In that time the battery level dropped by 30% so it would have lasted almost 7 hours all in all which is fine.
Asus claims that when you charge it back up again, you'll get 60% battery in as little as 39 minutes. When I was using it, I found it took about 50 minutes to charge to 60%, so for a full charge, you're looking at just under an hour and a half.
Running on Windows 11, the Asus Vivobook 13 Slate OLED (T3300) works perfectly well as a computer. It comes in Windows S mode which boosts security and performance but means you can only download apps available in the Microsoft Store - you can easily turn that off if you want to download software from the web. Thanks to Intel Bridge, you can actually run any Android app on this as well so you'll be able to use this just as much as a tablet as you can a computer.
To soundtrack the experience, the tablet features quad-speakers with Dolby Atmos surround sound. On mid-volume, the speakers sound clear and precise. When you crank it up, the audio does become a little distorted. It'll be fine for streaming movies but you probably won't be listening to music on it very often.
Asus Vivobook 13 Slate OLED (T3300) review: verdict
(Image credit: Asus)
When I said that the Asus Vivobook 13 Slate OLED (T3300) is a hybrid device, I really meant it - it's a complete cross between a laptop and a tablet because you can use it as a Windows 11 PC and you'll also be able to run Android apps on it.
During testing, the design of the cover stand did annoy me a little, it makes it difficult to use as a laptop on the go. But in saying that, the rest of the design worked well and I did like the fact that you could remove the keyboard entirely.
The OLED screen is a beauty and it's perfect for pretty much any task, but streaming TV shows and movies is where it really shines. It's a shame the performance is so weak because that's the only area where the laptop really lets itself down, even if it is more powerful than most tablets.
Even though it does have a few flaws, if you want a relatively cheap 2-in-1 laptop and you think that you'll use it as a tablet a lot, the Asus Vivobook 13 Slate OLED (T3300) is great value for money.
Asus Vivobook 13 Slate OLED (T3300) review: also consider
The Microsoft Surface Book 3 has a similar style to this, with a completely detachable keyboard that leaves you with just the slate. It's more expensive than the Asus Vivobook 13 Slate OLED (T3300) but it's worth it because it's also a lot more powerful. To discover more about this system check out T3's should I buy the Microsoft Surface Book 3 guide.
For those on a budget, another 2-in-1 PC to consider is the Asus ZenBook Flip S 13 which costs about the same as the Asus Vivobook 13 Slate OLED (T3300). You get a 4K OLED touchscreen display, a powerful 'Tiger Lake' processor and a slim eye-catching build. The keyboard isn't detachable though so to convert it to a tablet you have to push the screen back by 360-degrees.
Prices - Asus Vivobook 13 Slate OLED (T3300):▼
Yasmine Crossland
Reviews Writer for T3, Yasmine, started writing reviews in 2019 and is always knee-deep in the latest tech products. She keeps a finger on the pulse when it comes to the most exciting and innovative tech and is happy to tell you exactly what she thinks about it too. In her free time, you'll catch her tending to her plants and planning her next big trip.
Dell Inspiron 14 2-in-1 (7415) review: an affordable home machine
With a decent battery and the flexibility of a 360-degree hinge, the Dell Inspiron 14 2-in-1 is a great budget choice for students but it lacks the premium touches
By Richard Baguley • Published 6 December 21
HP Elite Folio review: this 2-in-1 laptop exudes professionalism
The HP Elite Folio focuses on style over substance
By Yasmine Crossland • Published 24 August 21
Lenovo Yoga 9i 14 review: combining sophistication with speed
The Lenovo Yoga 9i 14 is a premium 2-in-1 laptop with loads going for it
By Yasmine Crossland • Last updated 20 July 21
HP Elite Dragonfly G2 review: this elegant laptop means business
The HP Elite Dragonfly G2 is a 2-in-1 laptop with plenty of power
By Yasmine Crossland • Published 13 July 21
Dell Latitude 7320 Detachable Laptop review: a remote working superstar
The Dell Latitude 7320 Detachable is a powerful hybrid machine designed for work but capable at handling almost anything you throw at it
By Richard Baguley • Last updated 14 July 21

redpajama_set_name: RedPajamaCommonCrawl


text:
from setuptools import setup, find_packages
setup(name='pytaco',
    version='0.1',
    packages=find_packages(),
    )


text:
Wright-Kay Building
Wright-Kay Buildingwebmaster2017-06-14T15:24:48+00:00
Home of Neumann/Smith's Detroit Studio, the Wright-Kay Building is an iconic building in Detroit that received a well-deserved restoration. It now has become a focus of attention for the City and a jewel on Woodward Avenue.
The scope of work included complete restoration of the Wright-Kay Building, originally the Schwankovsky Temple of Music, one of the oldest buildings in downtown Detroit, which houses Neumann/Smith's design studio and offices of our minority partnership practice, Hannah-Neumann/Smith, LLC.
The project called for all new mechanical, electrical, bathrooms and plumbing, new elevators and windows. The project started out as a full historic preservation project, but mid-way through construction a more relaxed approach to the restrictions on demolition and restoration of gutted floors allowed a more [email protected] "embrace the raw" approach for the tenants.
Neumann/Smith chose to stay with the historic preservation approach on their suite. We developed the construction documentation for the suite and developed space plans for floors 2, 5 and 6 for potential future tenants. The third floor was scanned utilizing 3D laser scan technology and the scans were used to develop the detailed preservation scope of the build-out of the floor.
The project included all new base mechanical and electrical systems and suite build-out engineering and documentation. Bathrooms were designed to "Bedrock" standards as was the lobby. Facade restoration was also part of the scope. Retail was developed as a "white box" waiting for a tenant, now known as the John Varvatos store.
The Skillman Foundation
Lear Innovation Center
Campbell Ewald Headquarters
Grand Circus
Copyright © Hannah-Neumann/Smith | All Rights Reserved


text:
Biografia
La sua carriera iniziò all'inizio degli anni cinquanta quando lui, suo fratello gemello Nick DeVito e Hank Majewski formarono i The Variety Trio, che con l'aggiunta successiva di Francesco Castelluccio (più noto poi con il nome di Frankie Valli) diventarono i The Variatones, quindi cambiarono molte volte nome fino ad assumere quello di The Four Seasons nel 1960. Con la band cantò molti dei suoi successi fino al 1970, quando la lasciò. È morto il 21 settembre 2020, all'età di 92 anni, per complicazioni da COVID-19.

Note

Collegamenti esterni

Musicisti italoamericani
Morti per la pandemia di COVID-19
Chitarristi statunitensi

redpajama_set_name: RedPajamaWikipedia


We have been fortunate enough this year to receive sufficient rainfall over the wet season and, due to this, the bush has become a place of new life with the most incredible vibrant colours at every angle. The open clearings have become lush grasslands and the thickets have been blanketed by the greenery. One aspect that has really stood out over the last few weeks has been the influx of elephants into the area. We have been extremely privileged to see the large herds returning here.
Over the last few years we have not received much rainfall and the area is still recovering from one of the worst droughts experienced in the Sabi Sand region during the years of 2016 and 2017. That resulted in the large herds breaking up into smaller satellite herds. The herds did not have much option other than to be led by their leaders, the matriarchs, to better food and water sources. As a result of this we were not seeing such large herds and the biggest herds on average were of about 10 to 15 individuals.
Now that we have got the lush vegetation back, we are seeing herds of 50+ elephants! This is honestly one of the most incredible aspects to witness in the bush as they really are gentle giants. From the newborn calves running through the grasslands to the older stronger individuals pushing down large marula trees to allow for the entire family to have a feast, it really does produce some spectacular sightings for us and our guests.
One of the most fascinating aspects of these large herds is that they can actually be tough to find as they do not make a lot of noise as they walk through the bush – they are the "silent walkers". Once they start feeding this all changes and they can be heard from quite some distance when one breaks a few branches for feeding or when you hear the crack of tusks as two bulls have a sparring session.
These gentle giants really are one of the most spectacular aspects of the bush. Without them there would be an overgrowth of vegetation, but if there were too many of them they would destroy multiple food sources for other species that roam the area. It is really great to see the bush balancing itself out once again over the years and completing the circle of life.
On many of our game drives we have spotted these hornbills on top of termite mounds, feeding on termites early in the morning. They feed mainly on insects, seeds and fruits, and rarely on small reptiles and mammals. We have seen one kill and consume a chameleon.
In my eight years of training and working as a Professional Field Guide I have never witnessed a pack of Cape hunting dogs (African wild dogs) making a kill! I've always just missed the catch whilst staying with the dogs on the hunt and then arriving to a scene where the prey is already dead and the dogs are savagely ripping the carcass apart!
I, as well as two other guides, got the opportunity to visit our Zimbabwean colleagues at Singita Pamushana. What an incredible place to visit and what a great experience it was seeing the different landscapes and terrain. The mountains that occupy the property are mind-blowing as well as the open grasslands. The area has got a lot of diversity with regards to terrain, thus allowing for multiple different species to roam the area. We went on two game drives a day for two weeks and never was one drive even close to being the same.


text:
New Photos : Karishma Tanna
Latest Photos : Musskan Sethi
More trouble for Manikarnika, Karni Sena joins protests
Published on Feb 7, 2018 11:50 pm IST
Recently, several members of the Sarv Brahmin Mahasabha threatened to protest against Bollywood actress Kangana Ranaut's soon-to-release historic magnum opus, Manikarnika: The Queen of Jhansi. The Mahasabha alleged that Manikarnika's makers are distorting history by picturizing a few scenes that showcase Rani Laxmi Bai's relationship with a British officer.
However, Manikarnika's makers are quick to refute these allegations and clarified that the movie features no scene that belittles Laxmi Bai's honor. On the other hand, more trouble brewing in as Shree Rajput Karni Sena founder Lokendra Singh Kalvi, who spearheaded the violent protests against Padmaavat, extended his support to the Sarv Brahmin Mahasabha.
"Rajputs will never keep quiet if Brahmins are affected, and vice versa." Lokendra Kalvi claimed that 10,000 letters were signed with blood by Brahmins to protest the release of Padmaavat. Manikarnika is being directed by acclaimed Tollywood filmmaker, Krish.


text:
Energy Pol'y Advocates v. Ellison
September 28, 2022, Filed
[*150] CHUTICH, Justice.
This case concerns the obligation of appellant, the Office of the Attorney General, to produce documents under the Minnesota Government Data Practices Act (Data Practices Act), Minnesota Statutes sections 13.01 through 13.90 (2020). Respondent Energy Policy Advocates submitted document requests related to climate-change litigation to the Office of the Attorney General. After the Office determined that the request yielded no responsive, [**4] nonprivileged public data, Energy Policy brought a civil action against Keith Ellison in his official capacity as Attorney General, and the Office of the Attorney General (collectively, Attorney General), seeking production of the documents under the Data Practices Act. The district court denied Energy Policy's motion to compel and dismissed the complaint. The court of appeals affirmed in part, reversed in part, and remanded. Energy Policy Advocates v. Ellison, 963 N.W.2d 485, 502 (Minn. App. 2021).
The primary issues on appeal concern the existence and scope of the common-interest doctrine, the applicability of the attorney-client privilege to internal communication among attorneys in public law agencies, and the contours of the section of the Data Practices Act governing Attorney General data. We now formally recognize the common-interest doctrine in Minnesota. We also conclude that the attorney-client privilege may apply to protect the confidentiality of internal communications among attorneys in public law agencies. Finally, we uphold the Legislature's classification of Attorney General data under Minnesota Statutes section 13.65, subdivision 1, as "private data on individuals," even when the data do not pertain to "individuals." Consequently, we reverse the decision of the court of appeals on these [**5] issues and remand to the district court for further proceedings consistent with this opinion.
980 N.W.2d 146 *; 2022 Minn. LEXIS 402 **
Energy Policy Advocates, Respondent, vs. Keith Ellison, in his official capacity as Attorney General, Office of the Attorney General, Appellants.
Prior History: [**1] Court of Appeals.
Energy Policy Advocates v. Ellison, 963 N.W.2d 485, 2021 Minn. App. LEXIS 224 (Minn. Ct. App., June 1, 2021)
Disposition: Reversed and remanded to the district court.
attorney general, classification, nonpublic, common-interest, attorney-client, documents, Practices, confidential, classified, communications, district court, consumer, court of appeals, plain language, words, office of the attorney general, temporary, pertain, statutory interpretation, disclosure, collected, parties, balancing test, subdivision, agencies, courts, orders, decisions, mortgage, textual
Civil Procedure, Discovery, Privileged Communications, Attorney-Client Privilege, Evidence, Privileges, Attorney-Client Privilege, Elements, Work Product Doctrine, Scope of Protection, Scope, Waiver, Administrative Law, Governmental Information, Freedom of Information, Compliance With Disclosure Requests, Burdens of Proof, Allocation, Governments, Courts, Common Law, Defenses & Exemptions From Public Disclosure, Interagency Memoranda, Work Product, Interagency Memoranda, State & Territorial Governments, Legislatures, Appeals, Standards of Review, De Novo Review, Constitutional Law, The Judiciary, Case or Controversy, Advisory Opinions, Judicial Review, De Novo Standard of Review, Legislation, Interpretation, Personal Information, Prohibition of Disclosure, Public Information, Access to Records, Authority to Adjudicate, Deference to Agency Statutory Interpretation


text:
Q: explanation of DW insert code hi can you guys please help me break down this dreamweaver insert code wizard especially the if (!function_exists("GetSQLValueString")) { ....} part of the code:
<?php
if (!function_exists("GetSQLValueString")) {
function GetSQLValueString($theValue, $theType, $theDefinedValue = "",$theNotDefinedValue = "")
{

$theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue;

  $theValue = function_exists("mysql_real_escape_string") ? mysql_real_escape_string($theValue) : mysql_escape_string($theValue);

  switch ($theType) {
    case "text":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;
    case "long":
case "int":
  $theValue = ($theValue != "") ? intval($theValue) : "NULL";
  break;
case "double":
  $theValue = ($theValue != "") ? "'" . doubleval($theValue) . "'" : "NULL";
  break;
case "date":
  $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
  break;
case "defined":
  $theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue;
  break;

 }
  return $theValue;
}
}

$editFormAction = $_SERVER['PHP_SELF'];
if (isset($_SERVER['QUERY_STRING'])) {
  $editFormAction .= "?" . htmlentities($_SERVER['QUERY_STRING']);
}

if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "form")) {
  $insertSQL = sprintf("INSERT INTO feedback (name, email, phone, service, message) VALUES (%s, %s, %s, %s, %s)",
                       GetSQLValueString($_POST['name'], "text"),
                       GetSQLValueString($_POST['email'], "text"),
                       GetSQLValueString($_POST['phone'], "text"),
                       GetSQLValueString($_POST['service'], "text"),
                       GetSQLValueString($_POST['message'], "text"));

  mysql_select_db($database_kojexconsult, $kojexconsult);
  $Result1 = mysql_query($insertSQL, $kojexconsult) or die(mysql_error());

  $insertGoTo = "success.php";
  if (isset($_SERVER['QUERY_STRING'])) {
    $insertGoTo .= (strpos($insertGoTo, '?')) ? "&" : "?";
    $insertGoTo .= $_SERVER['QUERY_STRING'];
  }
  header(sprintf("Location: %s", $insertGoTo));
}
?>

A: I'll see if I can break it down for you.
The initial if check to see if the function has not been defined (http://php.net/manual/en/function.function-exists.php). It then goes through and defines the function GetSQLValueString.
Function GetSQLValueString does the following:

*

*It performs a ternary if check with the get_magic_quotes_gpc(Gets the current configuration setting of magic_quotes_gpc). If it passes, it stripeslahes $theValue, else it just returned the $theValue.

*Another ternary if check if perform to see if "mysql_real_escape_string" exists before either perform mysql_real_escape_string or mysql_escape_string.

*It picks the case that matches $theType variable and sets $theValue with appropriate values.

*Finally, it returns the Value so it can be used for whatever purpose it will serve in the code.

The rest of the code is pretty simple db insert snippet.
$editFormAction is set from whatever script is being run by php such as insert.php. An if check to confirm that there is a query string. If there is a query string, it appends to the end of the $editForm Action. Using the above PHP_SELF, this might be insert.php?name=Tim&email=hello@world.com&phone=8675309.
The next part is where the magic of inserting information into the database.
An if checks to see if MM_insert was set and it's value was form before continuing with the database insert. It's essentially a sanity check. After that point, it gives the variable $insertSQL with values from a formated string with the information passed from the form. You'll see it uses the function GetSQLValueString from earlier to make sure the data is prepared to go into the database. Once the $insertSQL has all it's values, it selects the DB then performs the query.
After that, the success.php is set for $insertGoTo. The if check to see if the QUERY_STRING was passed to this page. If there was a QUERY_STRING, it appends it onto success.php (success.php?name=Tim&email=hello@world.com&phone=8675309). After the if check, it redirects the page by calling header.
I hope that information helps you understand the breakdown of the script.
----- Personal Note Here -----
Dreamweaver is not a great application for developing PHP code. If you want a great IDE to use for PHP, I recommend PHPStorm(http://www.jetbrains.com/phpstorm/). I've been using it for over two years now and love it. I couple it with a great text editor, SublimeText2 (http://www.sublimetext.com/2).

redpajama_set_name: RedPajamaStackExchange


text:
On 18th May the Centre for the Creative Brain held an evening exploring the link between Epilepsy and art in its many forms. The free event, organised by Charlotte Stagg and Ainslie Johnstone, was a great success and scheduled to mark Epilepsy Awareness Week. Rhys Thomas, discussed artists with epilepsy while Rhys Bevan-Jones talked on his work in art and mental health. Finally, Sallie Baxendale spoke on how epilepsy has inspired filmmakers over the last 100 years. A Q & A session was chaired by consultant neurologist Arjune Sen.
The Centre for the Creative Brain, based at St Edmund Hall, run termly symposia exploring the link between neuroscience and creativity. In particular, it aims to explore how neuroscience can further our understanding of the aspects of our lives that make us most human, and how insights from art, music, and literature can enrich our understanding of the brain.
Anyone with an interest is welcome to attend the Centre's events, which are always free of charge and aimed at a non-specialist audience.
Brilliant talks, lively discussion and a great atmosphere! - Arjune Sen
International Public Engagement
To celebrate International Women's Day this year a number of groups hosted Wikipedia Editathons
Events Public Engagement
A new workshop aimed at young people to inspire them to learn more about the science of the brain has been designed as part of a large Wellcome Trust funded research programme.
On 27 July, 16 girls aged between 11-14 joined the SHElock team from the Wellcome Centre for Integrative Neuroimaging.
Awards and Honours Public Engagement
Prof Robin Dunbar, Emeritus Professor of Evolutionary Psychology has been in conversation with Jim Al-Khalili on Radio 4's Life Scientific. This fascinating podcast gives takes us on a journey from his early beginnings in science, and his fascinating research and groundbreaking discoveries he's made deciphering the mysteries as to why humans and animals evolve the social habits to exist in friendship circles.


United Services Automobile Association - Strategy, SWOT and Corporate Finance Report, is a source of comprehensive company data and information. The report covers the company's structure, operation, SWOT analysis, product and service offerings and corporate actions, providing a 360? view of the company.
United Services Automobile Association (USAA) is a diversified financial services company that offers products and services in the fields of insurance, banking, investments, real estate, and retirement. It offers insurance products for auto, home and property, life, rental home, health, long-term care, umbrella, motor vehicle, travel, and small business. It also offers annuities, checking accounts, savings accounts, certificates of deposits, credit cards, mortgage loans, auto loans, personal loans, mutual funds, stocks and options, exchange traded funds, and other investment products. The company provides car and home buying services, real estate services, financial planning services, and retirement planning services. It offers these products and services to individuals in army, air force, navy, marines, coast guard, and national guard and reserve. It also serves cadets, military retirees, and their families. USAA is headquartered in San Antonio, Texas, the US.
USAA reported net insurance premiums of USD 20,093 million in fiscal year ended December 2017 (FY2017), an increase of 9.9% over that in FY2016.
- Gain understanding of United Services Automobile Association and the factors that influence its strategies.
- Assess United Services Automobile Association as a prospective partner, vendor or supplier.
- Stay up to date on United Services Automobile Association's business structure, strategy and prospects.


I always have something to say.
I suggest, offer, add, provide. I do it because I wish to help, motivate and inspire. It's genuine, but I'm certain it annoys the crap out of many people, and often.
I do it for the same reason I became a teacher: to contribute. I hope that someday, a child will remember that teacher who told them they were ok, that they were doing just fine.
In my life, there are women, some strangers, who have had a very positive influence on me. They gave me what I needed at the exact time I needed it, when I didn't even realize I needed it. I want to do for someone what they did for me.
My friend and colleague Geneviève died of heart failure at age forty. Although we didn't see each other or speak much in the last few years of her life, there are many things she said to me when we worked together that still resonate with me. I remember calling her when my baby was a week old, exhausted because she was feeding all the time. Geneviève was a breastfeeding support person, yet she told me to give her a soother. When I asked her if that would be a bad idea in the long run, she said: "No one knows what's best for your baby more that you do; you are her mother, after all." She was very wise indeed.
The summer my daughter was twenty months old, she sat in her car seat that was perched on the front step of our house. It was there because my husband was cleaning the car before we left on our trip. As my toddler sat in it, it tipped backwards, making her hit her forehead on a rock in the front garden. The one-inch gash on her eyebrow spewed blood faster than I could cope with. We rush to urgent care; I cried as I registered her and sat down in the waiting room. An older woman who was sitting in front of me, reached out and touched my hand. "Tell me what happened" she said. I cried harder, recollecting the events of the last half hour. In that moment, it was exactly what I needed to make sense of it all and help with the excruciating waiting time. Eleven years later, I remember what that stranger did for me.
Seven years ago, when I was undergoing chemotherapy for breast cancer, I got an infection. I was admitted to the hospital in one of those locked-away wards to protect people's immune systems. I spent the weekend there, bored, but alive. When I was released on Monday morning, my husband and I had breakfast at Cora's in Orleans. We sat at a table for two in the middle of the room, with two women sitting at the table beside us. I remember smiling as the server cut the hospital band from around my wrists. I also remember my table neighbour smiling at me. Although I looked sick, it wasn't a sympathy or sorry smile that she offered, but an understanding smile. I think she understood how happy I was to have gotten the hell out of that ward. They left before we did, and when we went to pay, the server told us our bill had been paid. I don't remember what they looked like, but I'll always remember their act of kindness. If they had minded their own business, I wouldn't have that.
I often give myself grief for coming on too strong or for being too pushy. I'm sure I'm not alone. I'm working on holding back a bit, reminding myself that sometimes, the best way to contribute is to say nothing at all. But I also remember those women who made a difference in my life.
Don't be afraid to say or do something - to reach out a hand and listen; to say the hard truth; or to pay it forward. What you say to others, whether if be a friend, an acquaintance or even a stranger, may be that thing they remember forever.
So don't be too hard on your big mouth. I'm trying not to.


text:
Prema predanju, Numa Pompilije (lat. Numa Pompilius) je bio drugi kralj Rima, poslije Romula. Vladao je Rimom od 714. pr. Kr. do svoje smrti 673. pr. Kr.

Bio je oženjen Egerijom, boginjom izvornosti i poroda, koja je kasnije u pisanim izvorima označena kao nimfa. Egerija ga je, kako kaže legenda, srela kod izvora u svetoj šumici. Numa je bio kasnije slavljen zbog svoje pobožnosti.

U znak zahvalnosti, rimski bog rata, Mars mu je poslao štit na kojem se nalazilo pismo proročanstva za koje je vezana sudbina grada Rima. Spoznajući važnost ovog štita, Numa je dao izraditi još jedanaest drugih istih štitova. Ti štitovi su poznati pod latinskim nazivom ancile (sveti štit kralja Nume) i bili su nošeni u procesiji salijskih svećenika u drevnom Rimu.

Po Numinoj smrti 673 pr. Kr., Egerija ga je pretvorila u bunar i smjestila u aricijsku šumu u Latiji, mjesto obožavanja Dijane, boginje mjeseca, divljih životinja i šume.

Rimski kraljevi


Together with Our Lady's Girls' National School we have a very active and supportive Parents' Association, Our Lady's Schools' Parents' Association.
The mission of our Parents Association is to aim to improve and enrich the education of all children and support parents to get involved in their children's learning at home, in the community and at schools.
The PA work with the principal, staff and the board of management to build effective partnerships between home and school. Educational research on the involvement of parents in schools shows that children achieve higher levels when parents and teachers work together.