text:
Ghana, Accra digital centerngamangghana@gmail.com233241888167
Policy Research and Advocacy
Policy Research and Advocacy will enhance Favourable Policy, Legal and regulatory environment for Start-Ups to thrive. This will be done by undertaking policy research on the conditions, importance and effects of Farmers, Cooperatives, Start-Ups and SMEs and using that to rally around Multistakeholders such as Public Institutions, Private Sector, CSOs, Knowledge Institutions, Think-Tanks and Development Partners to enhance visibility for the conditions of Start-Ups and SMEs and the need for transformative change. Events that will be captured in the media, including video and TV series on Start-Ups and SMEs will be vital for this. Sustained evidence based advocacy for policy changes is the only sure hope for meaningful and transformative change for Farmers, Cooperatives, start-Ups and SMEs in Ghana.
Productive and Profitable
At the core of all these is corporate capacity; The Capacity of the Company to be productive, profitable, sociable and sustainable within and outside the company. And this Model should reflect in the business design, as a deliberate business strategy. If such a comprehensive and complete approach to business is taken, CSR becomes superfluous because it is in fact already taken care of in the business design so may not be added on as an afterthought. In practical terms this means having economic, social and environmental impact with our business and our client Start-Ups and SMEs from the scratch. It means doing our best to be concurrently productive, profitable, treating people and our environment equitably and fairly with respect to dignity, welfare and rights. This involves how we safeguard company survival, treat people in and out of the business with equality, equity and fairness, and protect our planet and solar system as a true global village, in which what we do in our company affects all other members of the globe directly or indirectly; thus giving meaning to real globalization of competition but also cooperation irrespective of people's location and social characteristics.
This means that our Company and its clientele should be effective with their economic and financial results but also efficient in the use of resources, and take good care of the people in the company in terms of fair and equitable compensation, working conditions, health and safety standards; treat people outside the company-Clients and customers with priority, with maximum satisfaction, consumer rights protection, health and safety in mind; and the people in the immediate community, Nation and the World in mind in terms of their stake in company activities, outcomes, negative and positive social and environmental impacts of business; In particular disaggregating company impacts on the different types of people and their special needs in and outside of the business, such as gender, age, geographical and vulnerability criteria.
This Model therefore also affects areas of business that we can have the comprehensive impact as much as possible, such as Food, Energy and Technology sectors or a combination of sectors in a comprehensive manner as indicated in the mix of entrepreneurship modules of SSC. For example Farmers, Cooperatives, Start-up and SME business focusing on food and aquaculture production, processing and marketing locally and internationally and meeting International certification, premium markets and global standards including ISO if possible, integrating clean energy production and use in our facilities as applicable in Ghana, with relevant international partnerships and the additional Technological transfers and best practices adaptation in the Firm and the Country; claiming carbon credits and fighting for compensation for environmental services provided by out growers and small holders associated. These fit with SDGs such as goal 9, 12-15 on sustainable production and consumption patterns, protection of environment and climate change mitigation and adaptation. Institute of Certified Management Consultants (ICMC) Ghana, a subsidiary of NGAMANG Group, co-owner of ABI/SSC has developed Social Impact Management Framework, Environmental Impact Management Framework and Vulnerability Outline (Evaluation of Ghana's Implementation of Sustainable Development Goals,2016-PFD-EU-GIZ). These Tools facilitate assessment, Implementation, Monitoring Evaluation, Management and Reporting of Social and Environmental impacts of Projects and activities or business and organizations. These Tools conform to standards set out by European Investment Bank (EIB) and others in International best practice, and will be used to improve social and environmental management of the projects and its Multistakeholders and participants
In conclusion there is the need to focus on Conditioning Measures for Enhanced Investment Climate for Start-Ups and SMEs. This programme will do a lot to contribute its quota to the conditioning measures required to enhance Investor confidence in Farmers, Cooperatives, Start-Ups and SMEs in Ghana, as 'there is a need to demystify the investment landscape on the continent and centre the discourse on the role of small and medium-sized enterprises (SMEs). SMEs employ the vast majority of any local labour force and have an integral role in any sustainable growth trajectory. At the International Trade Centre (ITC), we often refer to SMEs as 'the missing link' for inclusive growth' (ITC, 2018). In addition 'Informality is also a problem: investors cannot invest in unregistered firms, which make up a significant share of African SMEs and cannot raise funds through formal channels.' (ITC 2018) Moreover, 'good financial records are needed to prove performance claims' (ITC2018). All the above efforts at formalizing the Farmers, Cooperatives, Start-Ups and SMEs and making them more accessible and integrated into formal systems and global markets will go a long way to condition an enhanced Investment Climate for Farmers, Cooperatives, Start-Ups and SMEs to develop for shared Growth and sustainable economic development in addressing poverty, providing jobs and incomes and invigorating local and National economies to growth paths of no return.
Start Up Sme centres. All rights reserved


text:
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angular.module('slipflowApp')
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    return function (input) {
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redpajama_set_name: RedPajamaGithub


text:
Revised Sep 2012.
Published Apr 2011. Revised Aug 2012.
Marijuana: Download the Facts Spanish & English Poster
Published Aug 2012.
Double-sided poster for teens and young adults: one side English, one side Spanish. Emphasizes three essential messages about smoking marijuana: it is addictive, it can lead to school failure, and it impairs driving. See also Heads Up: Real News About Drugs and Your Body—Year 11-12 Compilations for Students and Teachers.
Anabolic Steroids (DrugFacts)
Revised Jul 2012.
Provides basic information about anabolic-androgenic steroids, such as how it is abused, its effects of the brain and mental health, its addictive potential, treatment, and the extent of abuse among youth. En Español
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Published Mar 2006. Revised Jul 2012.
Explores the link between drug abuse and HIV/AIDS and discusses treatment options, populations most at risk, and trends in HIV/AIDS and what to do to counter these trends. En Español
Is Marijuana Medicine? (DrugFacts)
Published Jul 2012.
Offers facts about marijuana's viability as a legal medical treatment. En Español
Inhalant Abuse (Research Reports)
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Provides scientific research on the nature and extent of inhalant abuse, its potential health consequences, and hazards of chemicals found in commonly abused inhalants. En Español
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Discusses the harmful effects of tobacco use, risks associated with pregnancy and adolescents, as well as best practices for the prevention and treatment of tobacco addiction. En Español
Marijuana Abuse (Research Reports)
Published Oct 2002. Revised Jul 2012.
Explores the latest research on marijuana, including the scope of marijuana use in the U.S., health consequences, its effects on every-day activities, available treatments. En Español
DrugFacts: Cocaine
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DrugFacts: Marijuana
Drugs: Shatter the Myths
DrugFacts: Is Marijuana Medicine?

redpajama_set_name: RedPajamaCommonCrawl


text:
\section{Introduction}
Let $A$ be a non-negative self-adjoint uniformly elliptic operator in divergence  form.
 More precisely, let $a_{kj} = a_{jk} \colon \mathbb{R}^d \to \mathbb{R}$ be a
bounded measurable function
for all $j,k \in \{ 1,\ldots,d \} $ and assume  that there exists a $\mu > 0$
such that
\begin{equation}\label{I1}
\sum_{k,j=1}^d a_{kj}(x) \xi_k \xi_j
\ge \mu |\xi |^2  \mbox{ for all } \xi = (\xi_1,\ldots,\xi_d) \in \mathbb{R}^d \mbox{ and } x \in \mathbb{R}^d
\end{equation}
The operator $A = - \displaystyle\sum_{k,j=1}^d \partial_k(a_{kj} \partial_j)$,
defined by quadratic form techniques, is self-adjoint on $L^2(\mathbb{R}^d)$.
It is a standard fact that
$-A$ is the generator of a strongly continuous semigroup $(e^{-tA})_{t >0}$ on $L^2(\mathbb{R}^d)$.
The well known Aronson  estimates assert that $e^{-tA}$ is given an integral kernel
$p_t$ (called the heat kernel of $A$) which satisfies the Gaussian upper bound:
\begin{equation}\label{I2}
| p_t(x,y) | \le Ct^{-d/2} e^{-c \frac{|x-y|^2}{t}} \mbox{ for all } t > 0 \mbox{ and } x, y  \in \mathbb{R}^d.
\end{equation}
Here $C$ and $c$ are positive constants.

In recent years,  harmonic analysis of operator of type $A$ has attracted a
lot of attention and substantial progress have been made in which upper  bounds
for the heat kernel play  a  fundamental  role.
We mention for example the theory of Hardy and BMO spaces associated with
such operators (see  for example \cite{DY} and \cite{HLMMY}), spectral multipliers \cite{DOS}
and Riesz transforms
(see \cite{DuM2}, \cite{Aus2}, \cite{Ouh5}, \cite{She} and the references therein).
 Concerning spectral multipliers, it is known that if $F\colon  [0, \infty) \to \mathbb{C}$ is a
bounded measurable function then the operator $F(A)$,
which is well defined on $L^2$ by spectral theory, extends to a bounded operator
on $L^p$ for all $1 < p < \infty$ provided $F$ satisfies the condition
\begin{equation}\label{I3}
\sup_{t> 0} \| \varphi (.) F(t.)  \|_{C^s} < \infty
\end{equation}
for some $s > d/2$ and some non-trivial auxiliary function $ \varphi \in C_c^\infty (0,\infty)$.
See Duong--Ouhabaz--Sikora \cite{DOS}, where a more general result is proved.
  Note that
condition (\ref{I3}) is satisfied if $F$ has $[d/2]+1$ derivatives such that
$$\sup_{\lambda > 0} \lambda^k | F^{(k)}(\lambda) | < \infty \mbox{ for all }  k \in \{ 0, 1,\ldots , [d/2] +1 \} .$$
As an example, one obtains  polynomial estimates on $L^p$  for imaginary powers  of type
$\| A^{is} \|_{p \to p} \le C (1 + |s |)^{\beta_p}$ for all $\beta_p > d | \frac{1}{2} - \frac{1}{p}|$.
Taking $F(\lambda) := (1 - \frac{\lambda}{R})_+^\alpha$ one obtains  Bochner--Riesz summability for all $\alpha > d/2$.

Concerning  Riesz transforms  ${\mathcal R}_k := \partial_k A^{-1/2}$, it is an obvious consequence of the  ellipticity assumption (\ref{I1})  that
${\mathcal R}_k$ is  bounded on $L^2(\mathbb{R}^d)$ for all $k \in \{ 1,\ldots ,d \} $.
 As for multiplier results, the Gaussian bound (\ref{I2}) combined with recent developments  on singular integral operators
allow to prove that ${\mathcal R}_k$ is bounded on $L^p(\mathbb{R}^d)$ for all $p \in (1, 2)$ with the sole assumption (\ref{I1}) and bounded measurable coefficients
(see Duong--McIntosh \cite{DuM2},  Auscher \cite{Aus2},  Ouhabaz \cite{Ouh5}).
Under weak  regularity assumption on the coefficients one obtains boundedness of
${\mathcal R}_k$ on $L^p(\mathbb{R}^d)$ for all $p \in (2,\infty)$ (cf.\  Auscher \cite{Aus2}, Shen \cite{She}).

In the present paper we wish to study similar problems for degenerate operators.
 Instead of (\ref{I1}) we merely assume that
\begin{equation}\label{I4}
\sum_{k,j=1}^d a_{kj}(x) \xi_k \xi_j \ge 0 \mbox{ for  all } \xi = (\xi_1,\ldots ,\xi_d) \in \mathbb{R}^d
   \mbox{ and } x \in \mathbb{R}^d.
\end{equation}
In this case, we  define  the form
\begin{equation} \label{formcc}
\mathfrak{a}_0(u,v) = \sum_{k,j=1}^d \int_{\mathbb{R}^d} a_{kj} \, (\partial_j u) \, (\partial_k v)
\end{equation}
with form domain $D(\mathfrak{a}_0) = C_c^\infty(\mathbb{R}^d)$.
If this form is closable, then  $A$ will be the self-adjoint operator associated with its closure.
If not, we take the regular part and consider $A$ as the operator associated with the closure of this
regular part (see \cite{bSim5} and \cite{AE2}).

Proving results like the previous ones for  these operators seems unattainable because
 Gaussian (or Poisson) upper bounds  are not true in general.
Even $L^1$-$L^\infty$ estimates of $e^{-tA}$ are not valid in general.
What we will do is to restrict
 the operators to parts where the matrix $(a_{kj})$ is elliptic.
 It is proved by ter Elst and Ouhabaz \cite{EO1} that if
$\chi \in C_{\rm b}^\infty(\mathbb{R}^d)$ and $\mu > 0$ are such that
$(a_{kj}(x)) \ge \mu I $ for a.e.\  $x \in \mathop{\rm supp} \chi$,
then $M_\chi e^{-tA} M_\chi$ has a H\"older continuous kernel
$K_t$ which satisfies the Gaussian bound
\begin{equation}\label{I5}
| K_t(x,y) | \le Ct^{-d/2} e^{-c \frac{|x-y|^2}{t}}  (1 + t)^{d/2} \mbox{ for all } t > 0 \mbox{ and } x, y  \in \mathbb{R}^d.
\end{equation}
Here $M_\chi $  is the multiplication operator by $\chi$.
The same result holds  for $P_\Omega e^{-tA} P_\Omega$ if $\Omega$ is a bounded Lipschitz domain such that
$(a_{kj}(x)) \ge \mu I$  for a.e.\ $x \in \Omega$ for some $\mu > 0$.
Here $P_\Omega$ is the multiplication operator by the indicator function $\mathds{1}_\Omega$ of $\Omega$.

Note that in general one cannot get rid of the extra term $(1+ t)^{d/2}$ in the
right hand side of (\ref{I5}).
For example, if
$a_{kj} = \delta_{kj}$ on a smooth bounded domain $\Omega$, then $A$ is the Neumann Laplacian on $L^2(\Omega)$ and $0$ on $L^2(\mathbb{R}^d\setminus \Omega)$.
It is then clear
 that the Gaussian bound is not valid without the additional term $(1+t)^{d/2}$.
Because of  that additional term in (\ref{I5}),  we shall  consider in the sequel
$ I + A$ instead of $A$ (of course,  one can take $\varepsilon I + A$ for any  $\varepsilon > 0$ to absorb
the factor $(1+ t)^{d/2}$).

For spectral multipliers and Riesz transforms we will prove the following results.
Given $\chi \in C_{\rm b}^\infty(\mathbb{R}^d)$  (resp.,  $\Omega$ a bounded Lipschitz domain)
such that $(a_{kj}(x)) \ge \mu I $  for a.e.\   $x \in \mathop{\rm supp} \chi$ (resp., for a.e.\  $x \in \Omega$)
for some constant $\mu > 0$.
The main theorems of this paper are the
following.

\begin{theorem} \label{theoremA}
Let  $F\colon  [0, \infty) \to \mathbb{C}$ be a bounded function such that
$$\sup_{t> 0} \| \varphi (.) F(t.) \|_{C^s} < \infty$$
for some $s > d/2$ and some non-trivial function $\varphi \in C_c^\infty(0,\infty)$.
Then $M_\chi F(I+A) M_\chi$ {\rm (}resp., $P_\Omega F(I+A) P_\Omega${\rm )} is bounded
on $L^p(\mathbb{R}^d)$ for all $1 < p < \infty$.
\end{theorem}

\begin{theorem} \label{theoremB}
 For all $k \in \{  1, \ldots , d \} $ the Riesz transform type operator
$M_\chi \partial_k (I+ A)^{-1/2}M_\chi$ is  bounded on $L^p(\mathbb{R}^d)$ for all
$1 < p \le 2$.
\end{theorem}

Now we discuss how we prove these results.
In the elliptic case, besides the Gaussian bound (\ref{I2}), the proof of boundedness of spectral multipliers or Riesz transforms
rely on a criterion proved by Duong and McIntosh \cite{DuM3} for singular integral operators to be weak type $(1,1)$.
This criterion says that if $T$ is bounded on $L^2$ with a (singular) kernel~$K$
such that there exists a family of operators $(A_t)_{t > 0}$ given by integral kernels $a_t$ which satisfy Gaussian (or Poisson) bounds, $TA_t$ is also
given by a (singular) kernel $K_t$ and
there are $C,\delta > 0$ such that
\begin{equation}\label{I6}
\int_{|x-y| \ge \delta \sqrt{t}} | K(x,y) - K_t(x,y) | \, dx \le C
\end{equation}
for all $t > 0$ and a.e.\ $y$,
then $T$ is weak type $(1,1)$.
In applications to spectral multipliers  of  elliptic operators we start with  $T = F(A)$ and  one takes  $A_t = e^{-tA}$.
 Therefore, $K_t$ is the kernel of  the operator $F(A) e^{-tA }$ which can be seen as  a regularization   of   $F(A)$.
 In the degenerate case and because of (\ref{I5}), it is tempting to choose
 $A_t = M_\chi e^{-t(I+A)}M_\chi$.
Then,
 $$ T A_t  =  M_\chi F(I+A) M_\chi M_\chi e^{-t(I+A)} M_\chi =  M_\chi F(I+A) M_\chi^2 e^{-t(I+A)} M_\chi.$$
Now, the presence of $M_\chi^2$ does not allow to regularize $F(I+A)$ by $e^{-t(I+A)}$.
The  simple fact that we do not have $F(I+A)$ next to $e^{-t(I+A)}$ in the expression of
$TA_t$  destroys this strategy.
The same problem occurs for the Riesz transform $M_\chi \partial_k (I+A)^{-1/2} M_\chi$.
 To overcome this difficulty we prove a version of the
Duong--McIntosh criterion that is  suitable for our purpose.
It  reads as follows (see Theorems \ref{th1} and \ref{th2} together
with Remark~\ref{rpartmult202} for precise and quantitive statements).

\begin{theorem} \label{theoremC}
 Let  $T$ be a bounded linear operator on $L^2$ and   $(A_t)_{t > 0}$
a family of linear operators which satisfy  $L^1$-$L^2$ off-diagonal estimates.
Suppose that  there exists a bounded linear operator $S $ on $L^2$  and
$\delta,W > 0$ such that
\begin{equation}
\int_{|x-y| \geq (1 + \delta) t} | (T  - SA_t) u(y) | \, dy
\le W \| u \|_1
\label{etheoremC;1}
\end{equation}
for all $x \in \mathbb{R}^d$, $t > 0$ and  $u \in L^1 \cap L^\infty$ supported  in the ball $B(x, t)$. Then
$T$ is weak type $(1,1)$.
\end{theorem}

Note that the estimate (\ref{etheoremC;1})
is satisfied if  $T$ and $SA_t$ are given
by {\rm (}singular{\rm )}  kernels $K$ and  $K_t$ and there are $C,\delta > 0$ such that
 $$\int_{|x-y| \ge \delta \sqrt{t}} | K(x,y) - K_t(x,y) | \, dx \le C$$
for all $t > 0$ and a.e.\ $y \in \mathbb{R}^d$.

Theorem~\ref{theoremC} gives more freedom by choosing any appropriate operator $S$ and not necessarily $S = T$.
Coming back to spectral multipliers for degenerate operators $A$, we had
$T = M_\chi F(I+A) M_\chi $ and we choose  now $S = M_\chi F(I+A) $ and $A_t = e^{-t(I+A)} M_\chi$.
Then $TA_t = M_\chi F(A+I) e^{-t(I+A)} M_\chi$ for which we can prove the estimate in Theorem~\ref{theoremC}.
Similarly for the Riesz transforms where $T = M_\chi \partial_k (I+A)^{-1}M_\chi$, we take $S = M_\chi \partial_k (I+A)^{-1}$ which turns to be bounded on $L^2$ and
$A_t = e^{-t(I+A)} M_\chi$.
We emphasize   that $A_t = e^{-t(I+A)} M_\chi$ satisfies $L^1$-$L^2$
off-diagonal estimates but it is not known whether it satisfies Gaussian upper bounds in
general\footnote{ under the additional assumption that $a_{kj} \in
W^{1,\infty} (\mathbb{R}^d)$, we proved recently in \cite{EO2} that $e^{-t(I+A)} M_\chi$ has a kernel which satisfies  a Gaussian bound.}.
 We believe that our version of the Duong--McIntosh criterion can be used in other circumstances in which a products of several operators come into play.
Also, as in \cite{DuM3}, our version holds for operators on domains of spaces of homogeneous type.

\smallskip

\noindent {\bf Notation.}
We fix some notation which we will use throughout this paper.
 If $(X, \rho, \mu)$ is  a metric measured space, $x \in X$, $r > 0$ and $j \in \mathbb{N}$,
then we denote by
$B(x,r) :=  \{ y \in X : \rho(x,y) < r \} $ the open ball of $X$ with  centre $x$ and radius $r$,
the annulus  $C_j(x,r) = B(x, 2^{j+1}r)\setminus B(x, 2^jr)$ if $j \ge 2$ and $C_1(x,r) = B(x, 4r)$.
Let $v(x,r) = \mu(B(x,r))$ be the volume of the ball $B(x,r)$.
Next,  $\| T \|_{p \to q}$ is the norm of $T$ as an operator from $L^p$ to $L^q$.
If $E$ is a measurable set, then $P_E$ denotes the multiplication operator
by the indicator function $\mathds{1}_E$ of $E$.
If $s \in (0,\infty) \setminus \mathbb{N}$ then we denote by $C^s$ the space of all Lipschitz functions
on $[0,\infty)$ of order $s$
(i.e., functions which are continuously differentiable up to  $[s]$ and the derivative of order
$[s]$ is H\"older continuous of order  $s-[s]$).
By $W^{r,p}$ we denote the classical Sobolev spaces on $\mathbb{R}^d$.

All our operators are linear operators.

We emphasize that we shall use $C, C', c,\ldots $ for all inessential constants.
A constant $C$ may differ from line to line, even within one line.

\section{Singular integral operators}

Let $(X, \mu, \rho)$ be a metric measured space.
 We shall assume that $0 < v(x, r) < \infty$ for all $x \in X$ and $r > 0$
and that $X$  is a space of homogeneous type.
This means that it satisfies the following doubling condition
 \begin{equation}\label{doubl1}
 v(x,2r) \le C_{0} v(x, r)
 \end{equation}
for some $C_0 > 0$, uniformly for all $x \in X$ and $r > 0$.
If  (\ref{doubl1}) is satisfied then  there exist
 positive constants $C_{1}$ and $d$ such that
 \begin{equation}\label{doubl2}
 v(x,\lambda r) \le C _{1}\lambda ^d v(x, r)
 \end{equation}
for all $x \in X$ and $r \geq 1$.
Let $\Omega$ be an open subset of $X$.
It is endowed with $\rho$ and $\mu$ but $(\Omega, \mu, \rho)$ is not necessarily a space of homogeneous type.
Let $T$ be a bounded linear operator on $L^{p_0}(\Omega) := L^{p_0}(\Omega, \mu)$ for some $p_0 \in [1, \infty)$.
We say that $T$ is given by a kernel $K \colon  \Omega \times \Omega \to \mathbb{C}$
if $K$ is measurable and
\begin{equation}\label{kernel}
Tu(x) = \int_\Omega K(x,y) \, u(y) \, d\mu(y)
\end{equation}
for all $u \in L^{p_0}(\Omega)$ with bounded support and a.e.\  $x$ outside the support of $u$.
We also say that $K$ is the associated kernel of $T$.
A classical problem in harmonic analysis is to find conditions on the
kernel $K$ which allow to extend the operator $T$ from $L^{p_0}(\Omega)$ to other $L^p(\Omega)$-spaces.
Several results are known in this direction.
We refer the reader to  \cite{Ste3}, \cite{DuM3}, \cite{BK1}, \cite{Aus2} and the references therein.

The main result in \cite{DuM3} states  that if there exists a family of bounded operators
$A_t$, with $t > 0$, which are given by  integral kernels $a_t$ satisfying a Gaussian or Poisson
 estimate and if the  associated  kernel  of $T - TA_t$ does not oscillate too much in a certain  sense then $T$ is weak type $(1,1)$.
Here we prove by the same method that if there exists a bounded operator $S$ on
$L^{q_0}(X)$ for some $q_0 \in (1, \infty)$  such that  the  associated  kernel of $T - SA_t$
does not oscillate too much  then $T$ is weak type $(1,1)$.
  As explained in the introduction, this new version gives more  freedom by choosing  any appropriate $S$ which  may not coincide with $T$.
  This extension turns out to be powerful to prove    spectral multiplier type results  as well as Riesz
transforms  for degenerate  operators,
whereas it is not clear how to apply the  condition from \cite{DuM3}.
Note also that, following ideas from \cite{BK1} and \cite {Aus2} we can weaken the assumption on the kernel of $A_t$.
Instead of assuming a Gaussian or Poisson bound, we  merely assume  an  $L^1$-$L^{q_0}$ off-diagonal estimate (see (\ref{au2}) below).
This  difference is again illustrated in our application to degenerate operators.
In addition it is possible to formulate the result in \cite{DuM3} without reference
to the kernels (see also the remark immediately after the next theorem).

We first state and prove the result in the case $\Omega = X$.

\begin{theorem}\label{th1}
Let  $T$ be  a non-zero bounded linear operator on $L^{p_0}(X)$ for some $p_0 \in (1, \infty)$.
 Suppose there exists a bounded linear  operator $S$ on $L^{q_0}(X)$ for some $q_0 \in (1, \infty)$,
 a family of
 bounded linear operators $(A_t)_{t > 0}$ on $L^{q_0}(X)$
and a sequence $(g(j))_{j \in \mathbb{N}}$ in $\mathbb{R}$
such that
\begin{equation}\label{au2}
 \left(\frac{1}{v(x,2^{j+1}t)}\int_{C_j(x,t)} | A_{t}f |^{q_0} \right)^{1/q_0}
\le g(j) \frac{1}{v(x,t)}\int_{B(x,t)} | f |
\end{equation}
for  all  $x \in X$, $t > 0$, $j \in \mathbb{N}$ and  $f \in L^{q_0}(B(x,t))$,
and  $\sum_{j=1}^\infty 2^{jd}g(j) < \infty$.
Finally, suppose there exist $\delta,W > 0$ such that
\begin{equation}\label{DMop}
\int_{X \setminus B(x,(1+\delta)t )} | (T  - SA_t) u |  \le W \| u \|_1
\end{equation}
for all $x \in X$, $t > 0$ and  $u \in L^1(X) \cap L^\infty(X)$ supported  in the ball $B(x, t)$.
Then $T$ is a weak type $(1,1)$ operator with
\begin{equation}\label{L1}
\| T \|_{L^1 \to L^{1,w}}
\le C (1+\delta)^d \left( W + \| T \|_{p_0 \to p_0}
   + \| S \|_{q_0 \to q_0}^{q_0}  \| T \|_{p_0 \to p_0}^{1-q_0} \right).
\end{equation}
Here $C$ is a constant depending only on the constants in \eqref{doubl2}.
In particular, $T$ extends to a bounded operator on $L^p(X)$ for all $p \in (1, p_0)$.
\end{theorem}

\begin{remark} \label{rpartmult202}
Let $p_0,q_0 \in (1,\infty)$,
$T \in \mathcal{L}(L^{p_0}(X))$ and for all $t > 0$ let
$S,A_t \in \mathcal{L}(L^{q_0}(X))$.
Suppose that
$T$ and $S \, A_t$ have a kernel $K$ and $K_t$, respectively.
Let $\delta,W > 0$ and assume that
\begin{equation}\label{DM}
\int_{\rho(x,y) \ge \delta t} | K(x,y) - K_t(x,y) | \, d\mu(x) \le W < \infty,
\end{equation}
for all $t > 0$ and $y \in X$.
Fix now $x \in X$, $t > 0$ and  $u \in L^1(X) \cap L^\infty(X)$ supported  in the ball $B(x, t)$.
Then
\begin{eqnarray*}
&&\hspace{-2cm}\int_{X \setminus B(x,(1+\delta)t )} | (T  - SA_t) u(y) | \, d\mu(y) \\
 &=& \int_{X \setminus B(x,(1+\delta)t )} | \int_{B(x,t)} \Big( K(y,z) -K_t(y,z) \Big)  u(z) \, d\mu(z) | \, d\mu(y)\\
&\le&  \int_{X }  \int_{ \rho(y,z) \ge \delta t} | K(y,z) -K_t(y,z) |  \, d\mu(y) \, |u(z)| \, d\mu(z)\\
&\le& W \| u \|_1.
\end{eqnarray*}
Thus, (\ref{DMop}) is satisfied. The  condition \eqref{DM} is the direct analogue of the
condition in Duong--McIntosh \cite{DuM3}.

We also observe that  one does not  need  kernels   for both operators $T$ and $TA_t$ but a kernel $H_t(x,y)$ for the difference
$T- SA_t$.  We may then replace $K(x,y) -K_t(x,y)$ in  (\ref{DM}) by $H_t(x,y)$. On the other hand the local estimate (\ref{DMop}) which does not appeal
to kernels may have   advantage of avoiding measurability questions with respect to $x$ and $y$ of the expected singular kernels.
\end{remark}

\begin{proof}
As mentioned before, the arguments are similar to the arguments  used in \cite{DuM3} and \cite{Aus2}.
 We give the details for
 convenience.
Recall we denote by $C$ all inessential constants.

 We begin by the classical Calder\'on--Zygmund decomposition.
There exist $c,N > 0$ such that the following is valid.
Fix $f \in L^1(X) \cap L^\infty(X)$ and $\alpha > \frac{\| f \|_1}{\mu(X)}$.
 There exist  $g,b_1,b_2,\ldots \in L^1(X) \cap L^\infty(X)$ such that
\[
f = g + b = g + \sum_i b_i
\]
 and
\begin{itemize}
\item[(i)]  $| g(x) | \le c \alpha$ for a.e.\  $ x \in X$,
\item[(ii)] each $b_i$ is supported in a ball $B_i = B(x_i, r_i)$
and $\frac{\| b_i \|_1}{v(x_i,r_i)}  \le c \alpha $,
\item[(iii)] $\sum_i v(x_i, r_i) \le c \frac{\| f \|_1}{\alpha}$, and,
\item[(iv)] there exists a constant $N$ such that $\sum_i \mathds{1}_{B_i}(x) \le N$ for a.e.\  $x \in X$.
\end{itemize}
 See Section~III.2 in \cite{CW}.

 We proceed in several steps.

\setcounter{proofstep}{0}\nextstep
Using the boundedness of $T$ on $L^{p_0}$ we have
 \begin{eqnarray*}
 \mu(\{ x \in X :  | (Tg)(x) | > \alpha \} )
&\le&\frac{ \| T \|_{p_0 \to p_0}^{p_0}}{\alpha^{p_0}}  \| g \|_{p_0}^{p_0}\\
 &\le& C \alpha^{p_0 -1} \frac{ \| T \|_{p_0 \to p_0}^{p_0}}{\alpha^{p_0}}  \| g \|_1\\
 &\le& C  \frac{ \| T \|_{p_0 \to p_0}^{p_0}}{\alpha}   \|  g \|_1.
 \end{eqnarray*}
 It follows from (ii) and (iii) above that $\| b \|_1 \le c \| f \|_1$ and hence
$\| g \|_1 \le (1 + c) \| f \|_1$.
 Therefore,
 \begin{equation}\label{eq1}
 \mu(\{ x \in X : | (Tg)(x) | > \alpha \} ) \le C  \frac{ \| T \|_{p_0 \to p_0}^{p_0}}{\alpha}  \| f \|_1.
 \end{equation}

\nextstep
We shall prove that
 \begin{equation}\label{eq2}
 \| \sum_i A_{r_i} b_i \|_{q_0} \le C  \alpha^{1- \frac{1}{q_0}} \| f \|_1^{1/{q_0}}.
 \end{equation}
 We follow similar arguments as in \cite{Aus2}.
Fix  $u \in L^{q_0'}$ with $\| u \|_{q_0'} = 1$, where $q_0'$ is the dual exponent of $q_0$.
Let $i,j \in \mathbb{N}$ and set  $C_{i,j} := C_j(x_i,r_i)$.
Then
 \begin{eqnarray*}
 \int_{C_{i,j}} | A_{r_i} b_i | \, | u |
&\le& \Big(  \int_{C_{i,j}} | A_{r_i} b_i |^{q_0} \Big)^{1/{q_0}}
    \Big( \int_{C_{i,j}} |u |^{q_0'} \Big)^{1/{q_0'}}\\
 &\le& g(j) \frac{v(x_i, 2^{j+1}r_i)^{1/{q_0}}}{v(x_i,r_i)} \Big( \int | b_i | \Big)
     \Big( \int_{C_{i,j}} |u |^{q_0'} \Big)^{1/{q_0'}}\\
 &\le & c \alpha  g(j) v(x_i, 2^{j+1} r_i)
    \Big(  \frac{1}{v(x_i, 2^{j+1}r_i)} \int_{C_{i,j}} |u |^{q_0'} \Big)^{1/{q_0'}},
 \end{eqnarray*}
 where we have used (\ref{au2}) and property (ii) in the Calder\'on--Zygmund decomposition.
 Denote by ${\mathcal M}$ the Hardy--Littlewood maximal operator.
Then
 $$  \frac{1}{v(x_i, 2^{j+1}r_i)} \int_{C_{i,j}} |u |^{q_0'} \le {\mathcal M}(| u |^{q_0'})(y)$$
for all $y \in B_i$.
Combining the previous inequalities and using the doubling condition (\ref{doubl2})
one estimates
$$ \int_{C_{i,j}} | A_{r_i} b_i | \, | u |
\le C \alpha  2^{j d} g(j) v(x_i,r_i)  \left( {\mathcal M}(| u |^{q_0'})(y) \right)^{1/{q_0'}}.$$
Taking the integral over $y \in B_i$ gives
$$  \int_{C_{i,j}} | A_{r_i} b_i | \, | u |
\le C \alpha  2^{j d} g(j) \int_{B_i}  \left( {\mathcal M}(| u |^{q_0'})(y) \right)^{1/{q_0'}}d\mu(y).$$
 We  sum over $j $, $i$ and use  $\sum_j 2^{jd} g(j) < \infty$
together with property (iv) in the Calder\'on--Zygmund decomposition to  obtain
 \begin{eqnarray*}
 \int_X  |\sum_i A_{r_i} b_i | \, |u |
&\le&  C \alpha  \int_X  \mathds{1}_{\cup_i B_i}(y)   \left( {\mathcal M}(| u |^{q_0'})(y) \right)^{1/{q_0'}} d\mu(y)\\
& \le& C \alpha \|  \mathds{1}_{\cup_i B_i} \|_{q_0}  \, \Big\|
   \left( {\mathcal M}(| u |^{q_0'}) \right)^{1/{q_0'}} \Big\|_{q_0',w}\\
 &\le& C \alpha \Big(\sum_i v(x_i,r_i) \Big)^{1/{q_0}} \, \|  | u |^{q_0'} \|_1^{1/q_0'}.
 \end{eqnarray*}
 Note that we have used the fact that ${\mathcal M}$ is weak type $(1,1)$ to obtain the last inequality.
Using now (iii) of the Calder\'on--Zygmund decomposition and
 $\| u \|_{q_0'} = 1$, we obtain (\ref{eq2}).

 By assumption, $S$ is bounded on $L^{q_0}$.
Hence
 $$ \mu( \{ x \in X : | (S\sum_i A_{r_i} b_i) (x) | > \alpha \} )
\le \frac{1}{\alpha^{q_0}} \| S \|_{q_0 \to q_0}^{q_0}
 \| \sum_i A_{r_i} b_i \|_{q_0}^{q_0}.$$
 Now we use (\ref{eq2})  to obtain
 \begin{equation}\label{eq3}
 \mu( \{ x \in X : | (S\sum_i A_{r_i} b_i) (x) | > \alpha \} )
\le \frac{C}{\alpha} \| S \|_{q_0 \to q_0}^{q_0} \| f \|_1.
 \end{equation}

\nextstep
Let $\delta$ be as in (\ref{DMop}) and for all $i \in \mathbb{N}$
set $Q_i := B(x_i, (1+ \delta) r_i)$, the ball of centre $x_i$ and radius $(1+ \delta)r_i$.
Then
\begin{eqnarray*}
\lefteqn{
\mu(\{ x \in X : | \sum_i (T -SA_{r_i})b_i (x) | > \alpha \})
} \hspace{10mm} \\
 &\le& \sum_i \mu(Q_i)
   +  \mu(\{ x \in X\setminus \bigcup_j Q_j : | \sum_i ((T -SA_{r_i})b_i) (x) | > \alpha \}) \\
 &\le& C (1+ \delta)^d  \sum_i v(x_i,r_i)
   + \frac{1}{\alpha} \int_{X\setminus \bigcup_j Q_j} | \sum_i ((T - SA_{r_i})b_i) (x)| \, d\mu(x) \\
 &\le& \frac{C (1+ \delta)^d}{\alpha} \| f \|_1
+  \frac{1}{\alpha} \sum_i \int_{ X\setminus Q_i}  | ((T - SA_{r_i})b_i) (x)| \, d\mu(x) \\
 &\le& \frac{C (1+ \delta)^d}{\alpha} \| f \|_1 + \frac{W }{\alpha} \sum_i \int |b_i(y)| \, d\mu(y) \\
& \le &  \frac{C (1+ \delta)^d (1+ W)}{\alpha} \| f \|_1.
 \end{eqnarray*}
 Note that the penultimate inequality follows from assumption (\ref{DMop}) and the last one from
properties (ii) and (iii) in the Calder\'on--Zygmund decomposition.  Hence
  \begin{equation}\label{eq4}
   \mu(\{ x \in X : | \sum_i ((T -SA_{r_i})b_i) (x) | > \alpha \})
\le \frac{C (1+\delta)^d (1+ W)}{\alpha} \| f \|_1.
   \end{equation}

\nextstep
It follows from  (\ref{eq1}) that
   \begin{eqnarray*}
\lefteqn{
\mu(\{ x \in X : | (Tf)(x) | > \alpha \})
} \hspace{10mm} \\*
     &\le&  \mu(\{ x \in X : | (Tg)(x) | > \frac{\alpha}{2} \}) +  \mu(\{ x \in X : | (Tb)(x) | > \frac{\alpha}{2} \})\\
    &\le& C  \frac{ \| T \|_{p_0 \to p_0}^{p_0}}{\alpha}  \| f \|_1 + \mu(\{ x \in X : | (Tb)(x) | > \frac{\alpha}{2} \}).
    \end{eqnarray*}
For the second term we use (\ref{eq3}) and  (\ref{eq4}) to estimate
{\allowdisplaybreaks
\begin{eqnarray*}
\lefteqn{
\mu(\{ x \in X : | (Tb)(x) | > \frac{\alpha}{2} \})
} \hspace{5mm} \\*
&= & \mu(\{ x \in X :
   | \sum_i (SA_{r_i} b_i)(x) + \sum_i ((T - SA_{r_i} ) b_i)(x)   | > \frac{\alpha}{2} \})\\
&\le&  \mu( \{ x \in X : | (S\sum_i A_{r_i} b_i) (x) | > \frac{\alpha}{4} \} ) \\*
&&\hspace{2mm} {} + \mu( \{ x \in X : | \sum_i ((T - SA_{r_i} ) b_i)(x)   | > \frac{\alpha}{4} \} )\\
   &\le&  \frac{C (1+\delta)^d}{\alpha} \left( \| S \|_{q_0 \to q_0}^{q_0} + (1+ W) \right)  \| f \|_1.
\end{eqnarray*}
}
 We then conclude that $T$ is of weak type $(1,1)$ with a weak type estimate
 \begin{equation}\label{F1}
 \| T \|_{L^1 \to L^{1,w}}
\le C (1 + \delta)^d \left( 1+ W + \| T \|_{p_0 \to p_0}^{p_0} + \| S \|_{q_0 \to q_0}^{q_0}  \right).
\end{equation}
If we  replace $T$ and $S$ by $\| T \|_{p_0 \to p_0}^{-1} T$  and $\| T \|_{p_0 \to p_0}^{-1} S$ we  obtain (\ref{DMop}) with
 $\| T \|_{p_0 \to p_0}^{-1} W$ instead of $W$. Thus applying (\ref{F1}) to $\| T \|_{p_0 \to p_0}^{-1} T$, $\| T \|_{p_0 \to p_0}^{-1} S$
 and $\| T \|_{p_0 \to p_0}^{-1} W$ yields (\ref{L1}).

 Finally, by Marcinkiewicz interpolation theorem the operator $T$ extends to a bounded operator from $L^p(X) \cap L^{p_0}(X)$ to $L^p(X)$ for all
 $p \in (1, p_0)$.
\end{proof}

 Following again an idea in  \cite{DuM3} we can prove a version of  the previous theorem  on arbitrary domains.
Let $\Omega$ be an open subset of $X$ and assume that $T$ is bounded
 on $L^{p_0}(\Omega)$ and $S$ and  $A_t$ are bounded on $L^{q_0}(\Omega)$.
We define $\widetilde T \colon L^{p_0}(X) \to L^{p_0}(X)$ by
 $$\widetilde{T} f = \mathds{1}_\Omega T (\mathds{1}_\Omega f)$$
 and similarly for $\widetilde{S}$ and $\widetilde{A_t}$.
If $A_t$ satisfies (\ref{au2Om})  below then $\widetilde{A_t}$ satisfies
 (\ref{au2}).
The operator
 $T$ is weak type $(1,1)$ {\it if and only if} $\widetilde{T}$ is weak type $(1,1)$.
Applying the previous theorem to $\widetilde{T}$, $\widetilde{S}$ and $\widetilde{A_t}$ gives
the following result.

\begin{theorem}\label{th2}
Let  $T$ be  a non-zero bounded linear operator on $L^{p_0}(\Omega)$ for some $p_0 \in (1, \infty)$.
 Suppose there exists a bounded operator $S$ on $L^{q_0}(\Omega)$ for some $q_0 \in (1, \infty)$,
 a family of
 bounded operators $(A_t)_{t > 0}$ on $L^{q_0}(\Omega)$
and a sequence $(g(j))_{j \in \mathbb{N}}$ in $\mathbb{R}$
such that
\begin{equation}\label{au2Om}
 \left(\frac{1}{v(x,2^{j+1}t)}\int_{C_j(x,t) \cap \Omega} | A_{t}f |^{q_0} \right)^{1/q_0}
\le g(j) \frac{1}{v(x,t)}\int_{B(x,t) \cap \Omega} | f |
\end{equation}
for  all ball $x \in \Omega$, $t > 0$, $j \in \mathbb{N}$ and  $f \in L^{q_0}(B(x,t) \cap \Omega)$,
and  $\sum_{j=1}^\infty 2^{jd}g(j) < \infty$.
Finally, suppose there exist $\delta,W > 0$ such that
\begin{equation}\label{DM1op}
\int_{\Omega \setminus B(x,(1+\delta)t )} | ((T  - SA_t) u)(y) | \, d\mu(y) \le W \| u \|_1
\end{equation}
for all $x \in X$, $t > 0$ and  $u \in L^1(\Omega) \cap L^\infty(\Omega)$ supported  in the ball
$B(x, t) \cap \Omega$.
Then $T$ is a weak type $(1,1)$ operator with
\begin{equation}\label{L1-om}
\| T \|_{L^1(\Omega) \to L^{1,w}(\Omega)}
\le C (1+\delta)^d \left( W + \| T \|_{p_0 \to p_0}
   + \| S \|_{q_0 \to q_0}^{q_0}  \| T \|_{p_0 \to p_0}^{1-q_0} \right).
\end{equation}
Here $C$ is a constant depending only on the constants in {\rm (\ref{doubl2})}.
In particular, $T$ extends to a bounded operator on $L^p(\Omega)$ for all $p \in (1, p_0)$.
\end{theorem}

As in Remark~\ref{rpartmult202} the condition (\ref{DM1op}) follows if
the operators $T$ and $SA_t$ are given by kernels $K$ and $K_t$
{\rm (}in the sense of {\rm (\ref{kernel}))} and there are $\delta,W > 0$
such that
\begin{equation}\label{DM1}
\int_{\rho(x,y) \ge \delta t} | K(x,y) - K_t(x,y) | \, d\mu(x) \le W < \infty,
\end{equation}
for all $t > 0$ and $y \in \Omega$. It suffices to note that the associated  kernel of $\widetilde{T}$ is the  extension by $0$ outside $\Omega \times \Omega$
 of the kernel of $T$ where $\widetilde{T} f = \mathds{1}_\Omega T (\mathds{1}_\Omega f)$ as above. Similarly for the kernel of $\widetilde{SA_t}$.

We may replace in the previous theorems the annulus
$C_j(x,r)$ by the annulus $A(x,j,r) := B(x,(j+1)r) \setminus B(x,jr)$.
In that case, $v(x, 2^{j+1}r)$ has to be replaced by
$v(x, (j+1)r)$ and the condition on $g$ becomes  $\sum_j  j^d g(j) < \infty$.

Following   \cite{BK1} it is proved in   \cite{Aus2}   that a bounded operator $T$ on $L^2(X)$
is of weak type $(r,r)$ if
\begin{equation}\label{au1}
\Big(\frac{1}{v(x, 2^{j+1}t)}\int_{C_j(x,t)} | T(I-A_{t})f |^2\Big)^{1/2}
\le g(j) \Big(\frac{1}{v(x,t)}\int_{B(x,t)} | f |^r\Big)^{1/r}
\end{equation}
and
\begin{equation}\label{au2'}
 \Big(\frac{1}{v(x,2^{j+1}t)}\int_{C_j(x,t)} |A_{t}f |^2\Big)^{1/2}
\le g(j) \Big(\frac{1}{v(x,t)}\int_{B(x,t)} | f |^r\Big)^{1/r}
\end{equation}
for all $x \in X$, $t > 0$, $j \in \mathbb{N}$ and $f \in L^2$ supported in $B(x,t)$,
and $\sum g(j)2^{dj}<\infty$.
One can prove a version of this result in which  $T-TA_t$ in (\ref{au1}) is replaced by
$T-SA_t$ as in Theorem \ref{th1}.
We do not write the details here since we have no concrete application.
 Theorem \ref{th1} is suitable for our purpose.

Finally, let us mention that a  Gaussian upper bound implies assumption  (\ref{au2}).
Indeed, assume that $A_t$ is given by a kernel $a_t$ such that
$$ | a_t(x,y) | \le \frac{C}{v(y,t^{1/m}) } \exp\{ - c \frac{\rho(x,y)^{m/(m-1)}}{t^{1/(m-1)}} \}$$
for all $t > 0$ and $x,y \in X$.
Here $m\ge 2$ and $C, c$ are two positive constants.
Fix $x_0 \in X$.
Note that the operator $\mathds{1}_{C_j(x_0,t)} A_{t^m} \mathds{1}_{B(x_0,t)}$ has the
kernel
$(x,y) \mapsto \mathds{1}_{C_j(x_0,t)}(x) a_{t^m}(x,y) \mathds{1}_{B(x_0,t)}(y)$.
But
\begin{eqnarray*}
\lefteqn{ |\mathds{1}_{C_j(x_0,t)}(x) a_{t^m}(x,y) \mathds{1}_{B(x_0,t)}(y)|
} \hspace{30mm} \nonumber  \\*
& \leq & \frac{C}{v(y,t) }  \mathds{1}_{C_j(x_0,t)}(x) \mathds{1}_{B(x_0,t)}(y)
      \exp\{ - c \frac{\rho(x,y)^{m/(m-1)}}{t^{m/(m-1)}}\}  \\
& \leq & \frac{C}{v(y,t) }  e^{-c 2^{j m/(m-1)}}  \mathds{1}_{B(x_0,t)}(y)  \\
& \leq & \frac{C}{v(x_0,t) }  e^{-c 2^{j m/(m-1)}} ,
\end{eqnarray*}
where we used the doubling property in the last step.
This an $L^1$-$L^\infty$ estimate.
By interpolation, it  implies an $L^1$-$L^{q_0}$ estimate which gives  (\ref{au2})
for every $q_0 \in (1, \infty)$.

\section{A partial multiplier theorem for degenerate operators}

Let the coefficients $a_{kj}$, form $\mathfrak{a}_0$ and the self-adjoint operator $A$
associated with $\mathfrak{a}_0$ be as in the introduction.
For every bounded measurable function $F \colon  [0, \infty) \to \mathbb{C}$,
the operator  $F(A)$ is well defined by spectral theory and is bounded on $L^2(\mathbb{R}^d)$.
As  mentioned in the introduction, if $A$ is uniformly  elliptic  then $F(A)$ extends
to a bounded operator on $L^p(\mathbb{R}^d)$ for all $p \in (1, \infty)$ provided $F$ has a finite number
of derivatives on $[0, \infty)$ which have good decay.
We address here the same problem for degenerate operators.
This is a difficult problem because no global Gaussian upper
bounds are available for  $A$ in general.
We prove a partial result by projecting on the part where the  matrix $(a_{kj})$ is uniformly elliptic.
There are two versions.

 \begin{theorem}  \label{th3}
  Let $\Omega \subset \mathbb{R}^d$ be an open bounded set with Lipschitz
boundary.
Suppose there exists a $\mu > 0$ such that $(a_{kj}(x)) \geq \mu I$ for almost every
$x \in \Omega$ and denote by $P_\Omega$  the projection from $L^2(\mathbb{R}^d)$ onto
$L^2(\Omega)$.
 Set $H = A + I$.
 Let $F\colon  [0, \infty) \to \mathbb{C}$ be a bounded function such that
 \begin{equation}\label{eq2.1-om}
 \sup_{t > 0} \| \varphi (.) F(t.) \|_{C^s} < \infty
 \end{equation}
 for some $s > d/2$ and some non-trivial function  $\varphi \in C_{c}^\infty(0,
 \infty)$.
 Then  $P_\Omega F(H) P_\Omega$  is of   weak type $(1,1)$   and
 extends  to a  bounded operator  on $L^p(\mathbb{R}^d)$ for all $p \in (1, \infty).$
 \end{theorem}

 \begin{theorem}\label{th4}
  Let $\chi \in C_{\rm b}^\infty(\mathbb{R}^d)$, $\mu > 0$ and
suppose that $(a_{kj}(x)) \geq \mu I$ for almost every
$x \in \mathop{\rm supp} \chi$.
Set $H = A + I$.
 Let $F\colon  [0, \infty) \to \mathbb{C}$ be a bounded function such that
 \begin{equation}\label{eq2.1-F}
 \sup_{t > 0} \| \varphi (.) F(t.) \|_{C^s} < \infty
 \end{equation}
 for some $s > d/2$ and some non-trivial function  $\varphi \in C_{c}^\infty(0,
 \infty)$.
 Then $M_\chi F(H)M_\chi $ is  of weak type $(1,1)$   and
 extends  to a  bounded operator  on $L^p(\mathbb{R}^d)$ for all $p \in (1, \infty).$
 \end{theorem}

 The proof of both  theorems is almost the same.
It relies mainly on weighted estimates for the associated kernel of
$M_\chi F(H)M_\chi $ (or the kernel of $P_\Omega F(H) P_\Omega$)  together with
 Theorem \ref{th1}.
The proof of weighted estimates for the kernel of $M_\chi F(H)M_\chi $
(or  of $P_\Omega F(H) P_\Omega$) is  based on partial Gaussian bounds proved in
\cite{EO1} and a similar strategy as in  \cite{DOS} and \cite{Ouh5}.

 In the sequel of this section we assume that there exists a constant
 $\mu > 0$ such that $(a_{kj}(x)) \geq \mu I$ for a.e.\ $x \in \Omega$
respectively for a.e.\
$x \in \mathop{\rm supp} \chi \cup \mathop{\rm supp} \widetilde{\chi}$.
In the first case
 $\Omega$ is a bounded Lipschitz domain of $\mathbb{R}^d$ and in the second case
$\chi , \widetilde{\chi} \in C_{\rm b}^\infty(\mathbb{R}^d)$.
We   denote by $S_t := e^{-tA}$ the holomorphic
 semigroup generated by $-A$ on $L^2(\mathbb{R}^d)$.
 We recall the following result from \cite{EO1}.

 \begin{theorem}\label{thEO}
There are $C,c > 0$ such that for all $t > 0$
the operator $M_{\widetilde{\chi}}  S_t M_\chi$
{\rm (}respectively $P_\Omega S_t P_\Omega${\rm )} is given by a kernel $p_t$ which satisfies
 $$ | p_t(x,y) | \le C t^{-d/2} e^{-c \frac{|x-y|^2}{t}} (1 + t)^{d/2}
\mbox{ for all } t > 0 \mbox{ and }  x, y  \in \mathbb{R}^d.$$
 \end{theorem}
  The theorem is stated in \cite{EO1} with $\chi = \widetilde{\chi}$ but the arguments work with different $\chi$ and $\widetilde{\chi}$.
It is also proved  there  that
\begin{equation}\label{1-2}
 \| M_\chi S_t \|_{2 \to \infty} \le Ct^{-d/4} (1+ t)^{d/4}
\quad\! \mbox{respectively} \quad\!
   \| P_\Omega S_t \|_{2 \to \infty} \le Ct^{-d/4} (1+t)^{d/4}.
 \end{equation}
If $z = t+ is \in \mathbb{C} $ with $t = \mathop{\rm Re}  z > 0$, then
\begin{eqnarray*}
\| M_\chi  S_z M_\chi \|_{1 \to  \infty} &=& \| M_\chi  S_{t/2} S_{is} S_{t/2} M_\chi \|_{1 \to \infty} \\
&\le & C t^{-d/4} (1+ t)^{d/4}  \| S_{is} S_{t/2} M_\chi \|_{1 \to 2} \\
&\le& C t^{-d/4}  (1+ t)^{d/4} \| S_{t/2} M_\chi \|_{2 \to \infty} \\
& \le & Ct^{-d/2} (1+ t)^{d/2}.
\end{eqnarray*}
Similarly,
\begin{equation}\label{1-22}
 \| P_\Omega S_z P_\Omega \|_{1 \to \infty} \le C (\mathop{\rm Re} z)^{-d/2} (1+ \mathop{\rm Re} z)^{d/2}
 \end{equation}
for all $z \in \mathbb{C}$ with $\mathop{\rm Re}  z > 0$.
Using the Gaussian bounds of Theorem~\ref{thEO} for real $t$ together with the
uniform bounds (\ref{1-22}) for complex $z$
it follows as in Theorem 3.4.8 in \cite{Dav2} or Theorem 7.2 in \cite{Ouh5}
that for all $\varepsilon > 0$ the kernel $p^{(0)}_z$ of
$M_\chi S_z M_\chi$ respectively $P_\Omega S_z P_\Omega$ satisfies the bound
\begin{equation}\label{eq2.2}
| p^{(0)}_z(x,y) e^{- \varepsilon z}  |
\le C_\varepsilon (\mathop{\rm Re} z)^{-d/2} \exp\{ - c \frac{| x - y |^2}{|z|} \cos(\arg z) \}
\end{equation}
for all $x,y \in \mathbb{R}^d$ and $z \in \mathbb{C}$ with $\mathop{\rm Re} z > 0$.

Let  $H = A + I$ and define $p_z(x,y) = p^{(0)}_z(x,y) e^{-z}$.
Then $p_z$ is the kernel of
$M_\chi e^{-zH} M_\chi$.
 We shall formulate the results below for $M_\chi F(H) M_\chi$
only, but all statements are also valid  for  $P_\Omega F(H) P_\Omega$.
In the following lemmas, we shall
always assume that $(a_{kj}(x)) \geq \mu I$ for almost every
$x \in \mathop{\rm supp} \chi$.
Since associated kernels with several operators are involved in the sequel
we shall  denote by $K_T$ the  kernel associated to a
given operator $T$ whenever it exists.

 \begin{lemma}\label{le1}
For all $s > 0$ and $\varepsilon > 0$ there exists a  $C > 0$ such that
 $$\int_{\mathbb{R}^d} | K_{M_\chi F(H) M_\chi}(x,y) |^2 \, (1 + \sqrt{r} |x-y|)^s \, dx
\le C r^{d/2} \| \delta_r F \|_{C^{s/2 + \varepsilon}}^2$$
 for all $r > 0$, $y \in \mathbb{R}^d$ and  $F \in C^{s/2 + \varepsilon}$ supported in $[0,r]$.
 Here $(\delta_rF)(\lambda) := F(r \lambda)$.
 \end{lemma}
 \begin{proof} The arguments are very  similar to those of Lemma 4.3  in \cite{DOS}.
Fix $r > 0$ and assume first that $F$ is supported in $[0,1]$.
Set $g(\lambda) := F(\lambda) e^{\lambda}$ and $H_r := \frac{1}{r} H$.
By (\ref{eq2.2}), the kernel $p_{z/r}$ of $M_\chi e^{-zH_r} M_\chi$ satisfies
\begin{equation}\label{eq2.3}
|p_{z/r}(x,y)|
\le C r^{d/2} (\mathop{\rm Re} z)^{-d/2} \exp\{ - c r  \frac{| x - y |^2}{|z|} \cos(\arg z) \}
\end{equation}
for all $x,y \in \mathbb{R}^d$ and $z \in \mathbb{C}$ with $\mathop{\rm Re} z > 0$,
with constants $C, c $ independent of $r$.
 We write
$g(\lambda) = \int_\mathbb{R} \hat{g}(\xi) e^{i\lambda \xi} \, d\xi$, where $\hat{g}$ is
the Fourier transform of $g$.
Then
$$F(H_r) = \int_\mathbb{R} \hat{g}(\xi) \, e^{-(1 - i\xi)H_r} \, d\xi,$$
from which one obtains
\begin{equation}\label{eq2.4}
K_{M_\chi F(H_r) M_\chi}(x,y) = \int_\mathbb{R} \hat{g}(\xi) \, p_{(1-i\xi)/r}(x,y) \, d\xi.
\end{equation}
Let $y \in \mathbb{R}^d$.
Using the estimate (\ref{eq2.3}) with $z = 1 - i \xi$ gives
\begin{eqnarray*}
\lefteqn{
\int _{\mathbb{R}^d} | p_{(1-i\xi)/r}(x,y) |^2 \, (1 + \sqrt{r}|x-y|)^s \, dx
} \hspace{30mm} \\*
 &\le& C r^d \int \exp\{ - 2 c r  \frac{| x - y |^2}{1 + \xi^2}  \} \, (1 + \sqrt{r}|x-y|)^s \, dx\\
&\le& C r^d (1 + \xi^2)^{s/2} \int \exp\{ - c r  \frac{| x - y |^2}{1 + \xi^2}  \} \, dx \\
&\le& C r^d (1 + \xi^2)^{s/2} \, \Big(\frac{1 + \xi^2}{r} \Big)^{d/2} \\
&=& C r^{d/2} (1 + \xi^2)^{(d+s)/2}.
\end{eqnarray*}
It follows from (\ref{eq2.4}), the continuous version of the Minkowski inequality
and the previous estimate that
\begin{eqnarray}
\lefteqn{
\left( \int_{\mathbb{R}^d}   | K_{M_\chi F(H_r) M_\chi}(x,y) |^2 \,
    (1 + \sqrt{r} |x-y|)^s \, dx \right)^{1/2}
} \hspace{20mm} \nonumber  \\*
&\le& \int_\mathbb{R} |\hat{g}(\xi)| \left(  \int _{\mathbb{R}^d} | p_{(1-i\xi)/r}(x,y) |^2 \,
(1 + \sqrt{r}|x-y|)^s dx \right)^{1/2} \, d\xi \nonumber \\
&\le& C r^{d/4} \int_\mathbb{R} | \hat{g}(\xi )| \, (1 + \xi^2)^{(d+s)/4 } \, d\xi \nonumber \\
& \le & C r^{d/4} \| g \|_{W^{(d+s+2)/2 ,2}} \nonumber \\
& \le & C r^{d/4} \|  F \|_{W^{s/2 + \alpha ,2}}. \label{eq2.5}
\end{eqnarray}
Here $\alpha=  (d+2)/2$ and the constants are independent of $r$ and $y$.
On the other hand $M_\chi F(H_r) M_\chi = M_\chi g(H_r) e^{-H_r} M_\chi$.
It follows from (\ref{1-2}) that
\[
\|e^{-H_r} M_\chi \|_{1 \to 2}
\leq C r^{d/4} (1 + \tfrac{1}{r})^{d/4} \, e^{-1/r}
\leq C r^{d/4}
\]
for all $r > 0$.
Moreover, $\|M_\chi g(H_r)\|_{2 \to 2} \leq e \|\chi\|_\infty \|F\|_\infty$.
Therefore
\begin{equation}\label{eq2.6}
\int_{\mathbb{R}^d}   | K_{M_\chi F(H_r) M_\chi}(x,y) |^2  \, dx
\leq \|M_\chi F(H_r) M_\chi\|_{1 \to 2}^2
\le C r^{d/2} \| F \|_\infty^2.
\end{equation}
This is valid for all $F$ with support in $[0,1]$ and for all $s > 0$.
The estimates (\ref{eq2.5}) and (\ref{eq2.6}) together with an
interpolation argument  (see \cite{MauM}, p.\ 151  and \cite{DOS}, p.\ 455) give
then that for all $s > 0$ there exists a $C > 0$ such that
\begin{equation}\label{eq2.7}
\int_{\mathbb{R}^d}   | K_{M_\chi F(H_r) M_\chi}(x,y) |^2(1 + \sqrt{r} |x-y|)^s dx
\le C r^{d/2} \| F \|_{C^{s/2 + \varepsilon}}^2.
\end{equation}
Finally, if $F$ has support in $[0,r]$ we use the last estimate with $\delta_rF$ and obtain the lemma.
 \end{proof}

 \begin{lemma}\label{le2}
 The operators $A_t :=  e^{-t^2 H}M_\chi$ satisfy {\rm (\ref{au2})}.
 \end{lemma}
 \begin{proof}
Let $\psi \in W^{1,\infty}(\mathbb{R}^d,\mathbb{R})$ be such that $| \nabla \psi | \le 1$.
For all $\rho \in \mathbb{R}$ define $U_\rho = M_{e^{\rho \psi}}$
and set
 $S_t^\rho := U_\rho e^{-tH}  U_{-\rho} $.
It follows from \cite{EO1} Proposition~3.6 by duality and a limit $n \to \infty$
that there exist $C,\omega > 0$, independent of $t$, $\rho$ and $\psi$,
such that
 \begin{equation}\label{eq2.8}
 \| S_t^\rho M_\chi \|_{1 \to 2} \le C t^{-d/4} e^{\omega \rho^2 t}.
 \end{equation}
Now fix two bounded open non-empty sets $E$ and $F$ of $\mathbb{R}^d$ and choose
$\psi(x) := d(x, E) \wedge N$, where $N = \sup \{ |x-y| : x \in E, \; y \in F \}  + 1$.
For all $h \in L^2(E)$ and $\rho \geq 0$ one has
$$M_\chi e^{-tH} h = M_\chi U_{-\rho} S_t^\rho h.$$
Therefore
\[
\| M_\chi e^{-tH} h \|_{L^\infty(F)}
\le e^{-\rho d(E,F)} \| M_\chi S_t^\rho h \|_\infty
\le C t^{-d/4} e^{-\rho d(E,F)} e^{\omega \rho^2 t} \| h \|_2.
\]
Choosing $\rho = \frac{d(E,F)}{2 \omega t}$ yields the  Davies--Gaffney type estimate
\begin{equation}\label{2.9}
\|P_F  (M_\chi e^{-t H}) P_E \|_{2 \to \infty}
\le C t^{-d/4} e^{-\frac{d(E,F)^2}{4 \omega t}}.
\end{equation}
In particular,
$$ \|  P_{C_j(x,t)} e^{-t^2 H} M_\chi P_{B(x,t)} \|_{1 \to 2} \le C t^{-d/2} e^{-c4^j}$$
for all $x,y \in \mathbb{R}^d$ and $j \in \mathbb{N}$.
This shows the lemma.
\end{proof}

 \begin{proof}[Proof of Theorems \ref{th3} and \ref{th4}]
 As mentioned above, the proof of both theorems is almost the same.
We consider $M_\chi F(H) M_\chi$ only.
The proof is based on Theorem \ref{th1} and the previous lemmas.
It is in the same spirit as in the elliptic case where a Gaussian bound holds (cf.\ \cite{DOS}, \cite{Ouh5}).
 Let  $\varphi \in C_c^\infty(0, \infty)$ be such that $\mathop{\rm supp} \varphi \subset [1/4, 1]$
and
$$\sum_{n=-\infty}^{\infty} \varphi(2^{-n}\lambda) = 1 $$
for all $\lambda > 0$.
Then
$$F(\lambda) = \sum_{n=-\infty}^{\infty} \varphi(2^{-n}\lambda) F(\lambda)
=: \sum_{n=-\infty}^{\infty} F_{n}(\lambda).$$
We apply Theorem \ref{th1} to   $M_\chi F_n(H) M_\chi$ for each fixed $n \in \mathbb{Z}$.
 We choose
$$ S := M_\chi F_n(H)
\quad \mbox{and} \quad
A_t:=  e^{-t^2H} M_\chi.$$
By Lemma \ref{le2}, the operators $A_t$ satisfy (\ref{au2}).
It remains to prove (\ref{DM}).
For this we  have to estimate for all $y \in \mathbb{R}^d$ the integral
$$I_{n,t} := \int_{| x - y |  \ge t}  | K_{M_\chi  G_{n,t}(H)  M_\chi}(x,y) | \, dx,$$
where
$G_{n,t} (\lambda) = F_n(\lambda)   -  F_n(\lambda)  e^{-t^2 \lambda}
= \varphi(2^{-n} \lambda) F(\lambda) (1- e^{-t^2 \lambda})$.
First, by the Cauchy--Schwarz inequality we have
\begin{eqnarray}
 I_{n,t}
&\le& \Big( \int_{\mathbb{R}^d} | K_{M_\chi   G_{n,t}(H) M_\chi}(x,y) |^{2}
 \, (1 + 2^{n/2} | x - y |)^{2s} \, dx \Big)^{1/2} \times \nonumber \\
\label{2.10}  &&  \hspace*{50mm} {}
\times  \Big( \int_{| x - y |  \ge t}
       (1 +  2^{n/2} | x - y | )^{-2s} \, dx \Big)^{1/2}
\end{eqnarray}
We apply Lemma \ref{le1} with  $r = 2^n$ and obtain
\begin{equation}\label{2.11}
 \int_{\mathbb{R}^d} | K_{M_\chi G_{n,t}(H) M_\chi}(x,y)|^{2} \, (1 + 2^{n/2} |x-y| )^{2s} \,  dx
\le C 2^{nd/2} \| \delta_{2^n} G_{n,t} \|_{C^{s+\varepsilon}}^{2}.
\end{equation}
Simple computations
show that there exists a $C > 0$, independent of $n$
and $t$,  such that
\begin{eqnarray}
\| \delta_{2^n} G_{n,t} \|_{C^{s+\varepsilon}} &=& \| \varphi(.) F(2^n .) (1-e^{-t^2 2^n .} )  \|_{C^{s+\varepsilon}} \nonumber\\
&\le& C \sup_{t'>0}\| \varphi(.) F(t' .) \|_{C^{s+\varepsilon}} \min (1, t^2 2^n). \label{2.12}
\end{eqnarray}
On the other hand (see \cite{DOS} or (7.46) in \cite{Ouh5}) one estimates
\begin{equation}\label{2.13}
\int_{| x-y |  \ge t} (1 + 2^{n/2} |x-y|)^{-2s} \, dx
\le C 2^{-nd/2}  \min (1, (t2^{n/2})^{d-2s}).
\end{equation}
Using (\ref{2.10}), (\ref{2.11}), (\ref{2.12}) and (\ref{2.13}) we obtain
$$I_{n,t} \le C  \min (1, t^2 2^n) \min (1, (t 2^{n/2})^{\frac{d}{2} - s})
\sup_{t'>0}\| \varphi(.) F(t' .) \|_{C^{s+\varepsilon}}.$$
Hence
$$\sum_{n=-\infty}^{\infty} I_{n,t}
\le C \left( \sum_{n \in \mathbb{Z}, \; t^2 2^n \le 1} t^2 2^n
    + \sum_{n \in \mathbb{Z}, \; t2^{n/2} > 1} (t2^{n/2})^{\frac{d}{2}-s} \right)
   \sup_{t'>0}\| \varphi(.) F(t' .) \|_{C^{s+\varepsilon}}$$
and the RHS is bounded by a constant independent  $t$ since $s > \frac{d}{2}$.
 This proves Theorem~\ref{th3}.
\end{proof}

 As explained in the introduction, the reason why we consider $H = A + I$ instead of $A$ in the previous results comes from
the fact the Gaussian upper bound in Theorem \ref{thEO} is valid with
the extra factor $(1+ t)^{d/2}$.
 If one considers the case where
 $a_{kj} = \delta_{kj}$ on a smooth bounded domain $\Omega$, then $A$ is the Neumann Laplacian on $L^2(\Omega)$ and $0$ on $L^2(\mathbb{R}^d\setminus \Omega)$.
It is then easy to see
 that $L^2$-$L^\infty$ estimates (respectively, Gaussian bounds) for
$M_\chi e^{-tA}$ or $P_\Omega e^{-tA}$  (respectively, $M_\chi e^{-tA}M_\chi$
or $P_\Omega e^{-tA}P_\Omega$)
 cannot hold without an extra factor  $(1+t)^{d/4}$ (respectively, $(1+t)^{d/2}$).
On the other hand we can replace in the previous theorems $H = A + I$ by $H = A + \varepsilon I$ for any $\varepsilon > 0$.

It may be possible that if $(a_{kj}) \ge \mu I$ on a connected subset $F$ of
$\mathbb{R}^d$ which is `large enough' (in some sense) one obtains Theorem \ref{thEO}
without the extra factor $(1 + t)^{d/2}$ in the Gaussian bound.
This  remains to be proved.
We mention that if such bound holds, we obtain by the same proof
Theorems \ref{th3} and \ref{th4} for  $F(A)$ rather than  $F(H)$.

We emphasize also that we consider here general degenerate operators with non-smooth coefficients.
One may obtain global results for some specific operators which are degenerate at
every point and the coefficients are not continuous at every point.
For example, one might take a pure second-order subelliptic operator
in divergence form with real measurable coefficients on a Lie group
with polynomial growth.
Then global Gaussian bounds are valid by \cite{SS}, Th\'eor\`eme~1 together
with regularization argument (see, for example Section~2.1 in \cite{ER15}).
Therefore a global spectral multiplier result  for such operators follows directly from
\cite{DOS}.
Note however that  the order of smoothness required on the function $F$ is
larger than half of  Euclidean dimension.
On the other hand, the operators that we consider in this paper
are allowed to vanish on big sets.

\begin{examples}
 We give some examples which are direct applications of the previous theorems.
\noindent
\begin{description}
\item[{\it Imaginary powers.}]
Set  $F(\lambda) = \lambda^{is}$ where $s \in \mathbb{R}$.
Then Theorems \ref{th3} and \ref{th4} together with the Riesz--Thorin interpolation theorem
imply that
for all $\varepsilon > 0$ and $p \in (1,\infty)$ there exists a $C > 0$ such that
$$ \| M_\chi H^{is} M_\chi \|_{p \to p} \le C_\varepsilon (1 + | s | )^{(d + \varepsilon)| \frac{1}{2} - \frac{1}{p} |}$$
and
$$ \| P_\Omega H^{is} P_\Omega \|_{{\mathcal L}(L^p)} \le C_\varepsilon (1 + | s | )^{(d + \varepsilon)| \frac{1}{2} - \frac{1}{p} |}$$
for all $s \in \mathbb{R}$.

\item[{\it The Schr\"odinger group.}]
 Set $F(\lambda) = (1+ \lambda)^{-\alpha} e^{it \lambda}$ with $t \in \mathbb{R}$ and $\alpha > d/2$.
 The operators  $M_\chi(I + H)^{-\alpha} e^{itH}M_\chi$ and
$P_\Omega (I + H)^{-\alpha} e^{itH} P_\Omega$ are bounded on $L^p(\mathbb{R}^d)$ for all $p \in (1, \infty)$.
Their  $L^p$-norms are estimated  by $C(1+ | t |)^\alpha$.
By interpolation, we obtain
boundedness on $L^p$ for all $\alpha >  d |\frac{1}{2} - \frac{1}{p}|$ and
$t \in \mathbb{R}$.

Remark.
Using the same proof  as in \cite{CCO}, these results can be obtained directly
from the Gaussian upper bound of Theorem \ref{thEO}  without appealing to
Theorems \ref{th3} and \ref{th4}.

\item[{\it Wave operators.}]
Set $F(\lambda) = (1+ \lambda)^{-\alpha/2} e^{it \sqrt{\lambda}}$ with $t \in \mathbb{R}$ and
$\alpha > d/2$.
The operators  $M_\chi(I + H)^{-\alpha} e^{it\sqrt{H}}M_\chi$ and
$P_\Omega (I + H)^{-\alpha} e^{it\sqrt{H}} P_\Omega$ are bounded on $L^p(\mathbb{R}^d)$ for all
$p \in (1, \infty)$.
\end{description}
\end{examples}

\section{Riesz transforms}
The aim in this section is to prove boundedness on $L^p(\mathbb{R}^d)$ of a type of
Riesz transform  operator  $M_\chi \nabla (I + A)^{-1/2} M_\chi$.
We keep the same notation as in the previous section.
The main result of this section is the next theorem.

\begin{theorem}\label{th7}
Let $\chi \in C_{\rm b}^\infty(\mathbb{R}^d)$, $\mu > 0$ and
suppose that $(a_{kj}(x)) \geq \mu I$ for almost every
$x \in \mathop{\rm supp} \chi$.
Set $H = A + I$.
Then for every $k \in \{1, \ldots, d\}$,  the operator $M_\chi  \partial_k H^{-1/2} M_\chi$ is of weak
type $(1,1)$ and is bounded on $L^p(\mathbb{R}^d)$ for all $p \in (1, 2]$.
\end{theorem}

Here $\partial_k$ denotes the distributional derivative.
The proof is based on  Theorem \ref{th1} and uses some ideas  from \cite{CoD}, \cite{DuM2} and  Chapter 7 in \cite{Ouh5} in the uniformly elliptic case.
We start with the following lemma.
Let $\mathfrak{a}$ be the closure of
the regular part of the form $\mathfrak{a}_0$
defined in (\ref{formcc}).

\begin{lemma}\label{lemm1}
Let $\chi \in C_{\rm b}^\infty(\mathbb{R}^d)$, $\mu > 0$ and
suppose that $(a_{kj}(x)) \geq \mu I$ for almost every
$x \in \mathop{\rm supp} \chi$.
Then $\chi u \in W^{1,2}(\mathbb{R}^d)$ and
\[
\|\chi \partial_k u\|_2^2
\leq \frac{\|\chi\|_\infty^2}{\mu} \, \|H^{1/2} u\|_2
\]
for all $u \in D(\mathfrak{a}) = D(H^{1/2})$ and $k \in \{ 1,\ldots,d \} $.
\end{lemma}
\begin{proof}
Let $u \in D(\mathfrak{a})$.
Then there exists a sequence $(u_n)_{n \in \mathbb{N}}$ in $D(\mathfrak{a}_0) = C_c^\infty(\mathbb{R}^d)$
such that $\lim u_n = u$ in $L^2(\mathbb{R}^d)$ and
$\mathfrak{a}(u) = \lim \mathfrak{a}_0(u_n)$.
By the ellipticity assumption on the support of $\chi$ one deduces
\begin{equation} \label{elemm1;1}
\mu \int_{\mathbb{R}^d} \chi^2 \, |\nabla u_n|^2
\leq \sum_{k,j=1}^d \int_{\mathbb{R}^d} a_{kj} \, (\partial_k u_n) \, (\partial_j u_n) \, \chi^2
\leq \|\chi\|_\infty^2 \, \mathfrak{a}_0(u_n)
\end{equation}
for all $n \in \mathbb{N}$.
Therefore $(\chi u_n)_{n \in \mathbb{N}}$ is bounded in $W^{1,2}(\mathbb{R}^d)$.
Hence it has a weakly convergent subsequence in $W^{1,2}(\mathbb{R}^d)$.
Since $\lim \chi u_n = \chi u$ in $L^2(\mathbb{R}^d)$ it follows that
$\chi u \in W^{1,2}(\mathbb{R}^d)$.
Then taking the limit $n \to \infty$ in (\ref{elemm1;1}) one estimates
\[
\mu \int_{\mathbb{R}^d} \chi^2 \, |\partial_k u |^2
\leq \|\chi\|_\infty^2 \, \mathfrak{a}(u)
\leq \|\chi\|_\infty^2 \, \|H^{1/2} u\|_2^2
\]
for all $k \in \{ 1,\ldots,d \} $.
\end{proof}

\begin{lemma}\label{lemm2}
Let $\chi, \widetilde{\chi} \in C_{\rm b}^\infty(\mathbb{R}^d)$,
$\mu > 0$ and assume that $(a_{kj}(x)) \geq \mu I$ for almost every
$x \in \mathop{\rm supp} \chi \cup \mathop{\rm supp} \widetilde{\chi}$.
Then for all $\beta > 0$  small enough we have
$$\int_{\mathbb{R}^d} | (M_{\widetilde{\chi}} e^{-sH} M_\chi u)(y) |^2 \, e^{\beta \frac{|x-y|^2}{s}} \, dy
\le C s^{-d/2} e^{2\beta t^2/s} e^{-s} \| u \|_1^2$$
for  all $t > 0$, $s > 0$, $x \in \mathbb{R}^d$ and $u \in L^2(\mathbb{R}^d)$ with $\mathop{\rm supp} u \subset B(x,t)$.
\end{lemma}
\begin{proof}
By Theorem \ref{thEO} one estimates
\begin{eqnarray*}
\lefteqn{| (M_{\widetilde{\chi}} e^{-sH} M_\chi u)(y) |^2 \, e^{\beta \frac{|x-y|^2}{s}}
} \hspace{30mm} \\*
&=& | \int_{B(x,t)} e^{-s} p_s(y,z) \, u(z) \, dz |^2 \, e^{\beta \frac{|x-y|^2}{s}}\\
&\le& C e^{-s}  \Big(  \int_{B(x,t)} s^{-d/2} \, e^{-c \frac{|y-z|^2}{s}} \,
    e^{\beta \frac{|x-y|^2}{2s}} \, |u(z)| \, dz \Big)^2\\
&\le&  C e^{-s}  \Big(  \int_{\mathbb{R}^d} s^{-d/2} \, e^{-(c-\beta) \, \frac{|y-z|^2}{s}}
    \,  |u(z)| \, dz \Big)^2 e^{2 \beta t^2/s}  \\
&\le&  C e^{-s} s^{-d/2} \, \|u\|_1  \int_{\mathbb{R}^d} s^{-d/2} \, e^{-(c-\beta) \, \frac{|y-z|^2}{s}}
    \,  |u(z)| \, dz \, e^{2 \beta t^2/s}.
\end{eqnarray*}
Taking $\beta < \frac{1}{2} c$ and integrating over $y$ yields the lemma.
\end{proof}

Since $e^{-sH}L^2(\mathbb{R}^d) \subset D(\mathfrak{a})$ for all $s > 0$ we obtain from Lemma \ref{lemm1}
the inclusion
$M_\chi \nabla e^{-sH}M_\chi( L^2(\mathbb{R}^d)) \subset W^{1,2}(\mathbb{R}^d)$ for all $s > 0$.
The following weighted $L^2$-estimate  is in the same spirit as weighted gradient estimates for heat kernels (see
\cite{CoD}, \cite{Gri2}  and Theorem 6.19 in \cite{Ouh5}).

\begin{lemma}\label{lemm3}
For all $\beta > 0$ small enough we have
$$ \int_{\mathbb{R}^d} | (M_\chi \nabla e^{-sH} M_\chi u)(y) |^2 \, e^{\beta \frac{| x-y|^2}{s}} \, dy
\le C s^{-d/2 -1} \, e^{6 \beta t^2/s} \, \| u \|_1^2$$
for  all $t > 0$, $s > 0$, $x \in \mathbb{R}^d$ and $u \in L^2(\mathbb{R}^d)$ with $\mathop{\rm supp} u \subset  B(x,t)$.
\end{lemma}
\begin{proof}
In order to avoid domain problems of forms we shall proceed by approximation.
First, we prove the lemma for uniformly elliptic
coefficients with constants $\beta$ and $C$ depending only on
$\mu > 0 $  such that  $(a_{kj}(x)) \ge \mu \, I$ a.e.\  $x \in \mathop{\rm supp} \chi$.

Assume that there exists a $\mu_0 > 0$ such that $(a_{kj}(x)) \ge \mu_0 \, I$ for
a.e.\ $x \in \mathbb{R}^d$.
In this case the form $\mathfrak{a}$ has domain $W^{1,2}(\mathbb{R}^d)$.
We use similar ideas as in the  proof of Theorem 6.19 in \cite{Ouh5}
but we want to prove that the constants  in the estimates are independent of $\mu_0$.
Let $\psi \in C_c^\infty(\mathbb{R}^d)$ be such that $\psi(x) = 1$ for all $x \in B(0,1)$
and $0 \leq \psi \leq \mathds{1}$.
For all $n \in \mathbb{N}$ define $\psi_n \in C_c^\infty(\mathbb{R}^d)$ by
$\psi_n(x) = \psi(\frac{1}{n} \, x)$.
Set
$$ I_n :=  \int_{\mathbb{R}^d} | \chi(y) \, (\nabla   e^{-sH} M_\chi u)(y) |^2
   \, e^{\beta \frac{| x-y|^2}{s}} \,  \psi_n(y) \, dy$$
 and  define $ f  :=  e^{-sH} M_\chi u$.
Then,
\begin{eqnarray*}
I_n
&\le& \frac{1}{\mu} \sum_{k,j} \int_{\mathbb{R}^d} a_{kj}(y) \, (\partial_k f)(y) \,
    (\partial_j f)(y) \, e^{\beta \frac{| x-y|^2}{s}} \, \chi(y)^2 \,  \psi_n(y) \, dy\\
&=& \frac{1}{\mu} \sum_{k,j} \int_{\mathbb{R}^d} a_{kj} \, (\partial_k f) \,
    \partial_j \Big( f \, e^{\beta \frac{| x- \cdot|^2}{s}} \, \chi^2 \, \psi_n \Big) \\*
&& \hspace{3mm} {}+ \frac{1}{\mu} \sum_{k,j} \int_{\mathbb{R}^d} a_{kj}(y) \, (\partial_k f)(y) \,
     f(y) \, \frac{2 \beta (x_j-y_j)}{s} \, e^{\beta \frac{| x-y|^2}{s}} \, \chi(y)^2 \,
       \psi_n(y) \, dy\\*
&& \hspace{3mm} {}- \frac{2}{\mu} \sum_{k,j} \int_{\mathbb{R}^d} a_{kj}(y) \, (\partial_k f)(y) \,
     f(y) \, e^{\beta \frac{| x-y|^2}{s}} \, (\partial_j \chi)(y) \, \chi(y) \, \psi_n(y) \, dy\\*
&& \hspace{3mm} {}- \frac{1}{n \mu} \sum_{k,j} \int_{\mathbb{R}^d} a_{kj}(y) \, (\partial_k f)(y) \, f(y) \,
     e^{\beta \frac{| x-y|^2}{s}} \, \chi(y)^2  \, (\partial_j\psi)(\tfrac{1}{n} \, y) \, dy\\
& =: & J_{1,n} + J_{2,n} + J_{3,n} + J_{4,n}.
\end{eqnarray*}
Since $y \mapsto f (y) e^{\beta \frac{| x-y|^2}{s}} \chi(y)^2 \psi_n(y)$
is an element of $W^{1,2}(\mathbb{R}^d)$ we have
\begin{eqnarray*}
J_{1,n}
&=& \frac{1}{\mu} \, \mathfrak{a}(  f,  f e^{\beta \frac{|x - \cdot|^2}{s}}  \chi^2 \psi_n )\\
&=& \frac{1}{\mu} \, \int_{\mathbb{R}^d} (A e^{-sH} M_\chi u) \, (e^{-sH} M_\chi u) \,
      e^{\beta \frac{| x- \cdot|^2}{s}} \, \chi^2 \, \psi_n \\
&\le &  \frac{\| \chi \|_\infty}{\mu} \,
    \| H e^{-sH} M_\chi u \|_2 \,
     \| e^{\beta \frac{|x - \cdot|^2}{s}} \, M_\chi e^{-sH} M_\chi u \|_2 .
\end{eqnarray*}
The standard estimate $\| H e^{-sH} \|_{2 \to 2} \le s^{-1} $ and Lemma \ref{lemm2}  give
\begin{equation}\label{J1}
J_{1,n}  \le C  s^{-d/2 -1} e^{2\beta t^2/s} \| u \|_1^2
\end{equation}
if $\beta$ is small enough.
Using the obvious inequality
$\frac{|x_j-y_j|}{s}  \le \frac{1}{\sqrt{\varepsilon \, s}} \, e^{\varepsilon \frac{| x - y |^2}{s}} $
 we have
\begin{eqnarray*}
|J_{2,n}|
&\le&  \frac{C}{\sqrt{s}} \sum_k \int_{\mathbb{R}^d}
   |\partial_k e^{-sH} M_\chi u | \, \chi^2 \, |e^{-sH} M_\chi u| \,
     e^{2\beta \frac{| x-\cdot|^2}{s}}\\
&\le& \frac{C}{\sqrt{s}}  \sqrt{I_n}
   \Big( \int_{\mathbb{R}^d} | (M_\chi e^{-sH} M_\chi u)(y) |^2 e^{3\beta \frac{| x-y|^2}{s}} \, dy \Big)^{1/2}.
\end{eqnarray*}
Therefore Lemma \ref{lemm2} implies
\begin{equation}\label{J2}
|J_{2,n}|   \le C \sqrt{I_n} s^{-\frac{d}{4} - \frac{1}{2}} e^{-s/2} e^{3\beta t^2/s} \| u \|_1.
\end{equation}
We estimate the third term in a similar way.
\begin{eqnarray}
| J_{3,n} |
&\le& C  \sum_k \int_{\mathbb{R}^d}
 |\partial_k e^{-sH} M_\chi u | \, |\chi \, \partial_j \chi | \,
     |e^{-sH} M_\chi u| \, e^{\beta \frac{| x- \cdot|^2}{s}} \, \psi_n \nonumber  \\
&\le& C \, \sqrt{I_n}
   \Big( \int_{\mathbb{R}^d} | M_{\partial_j\chi} e^{-sH} M_\chi u |^2 \,
         e^{\beta \frac{| x- \cdot|^2}{s}} \Big)^{1/2} \nonumber  \\
&\le& C \sqrt{I_n} s^{-\frac{d}{4} - \frac{1}{2}} \,  e^{-s/3} \, e^{\beta t^2/s} \, \| u \|_1.
\label{J3}
\end{eqnarray}
Finally,
\begin{eqnarray}
| J_{4,n} |
&\le& \frac{C}{n}   \sum_{k,j}  \int_{\mathbb{R}^d}
 |  (\chi \partial_k  e^{-sH} M_\chi u)(y) | \, | (M_\chi e^{-sH} M_\chi u)(y) | \,
     e^{\beta \frac{| x-y|^2}{s}} \nonumber  \times \\*
 && \hspace{75mm} {} \times | (\partial_j \psi)(\tfrac{1}{n} \, y)| \, dy  \nonumber   \\
&\le& \frac{C}{n}  \| M_\chi \, \nabla \, e^{-sH} \, M_\chi u \|_2
    \Big( \int_{\mathbb{R}^d}   | (M_\chi e^{-sH} M_\chi u)(y) |^2 \,
      e^{2\beta \frac{| x-y|^2}{s}} \, dy \Big)^{1/2}  \nonumber   \\
&\le& \frac{C}{n}  \| M_\chi \, \nabla \, e^{-sH} \, M_\chi u \|_2 \,
    s^{-d/4} e^{-s/2} e^{2\beta t^2/s} \| u \|_1.
\label{J4}
\end{eqnarray}
Therefore, we obtain from (\ref{J1}), (\ref{J2}), (\ref{J3}) and (\ref{J4}) that
$$ I_n  \le C s^{-d/2-1} e^{6\beta t^2/s} \| u \|_1^2
   +  \frac{C}{n} \, \| M_\chi \, \nabla \, e^{-sH} \, M_\chi u \|_2 \,
    s^{-d/4}  e^{2\beta t^2/s} \| u \|_1.$$
Letting $n \to \infty$ and then  use  Fatou's lemma yields
\begin{equation}\label{Ifinal}
 \int_{\mathbb{R}^d} | (M_\chi \nabla e^{-sH} M_\chi u)(y) |^2 \, e^{\beta \frac{| x-y|^2}{s}} \, dy
\le C s^{-d/2 -1} \, e^{6 \beta t^2/s} \| u \|_1^2.
\end{equation}
The constants $C$ and $\beta$ are independent of $\mu_0$.

Now we prove the lemma for degenerate operators.
For all $n \in \mathbb{N}$ set $a_{kj}^{(n)} = a_{kj} + \frac{1}{n} \delta_{kj}$.
Then $(a_{kj}^{(n)}(x)) \ge \frac{1}{n} \, I$ for a.e.\  $x \in \mathbb{R}^d$
and $(a_{kj}^{(n)}(x)) \ge  \mu \, I$ for a.e.\ $x \in \mathop{\rm supp} \chi$.
Moreover, $\| a_{kj}^{(n)} \|_\infty \le 1 + \| a_{kj}  \|_\infty$.
 We denote by $A_n$
the elliptic operator with the coefficients $a_{kj}^{(n)}$ and $H_n = I + A_n$.
 We apply (\ref{Ifinal}) to $H_n$  and obtain
\begin{equation}\label{final-n}
 \int_{\mathbb{R}^d} | (M_\chi \nabla e^{-sH_n} M_\chi u)(y) |^2 \,
    e^{\beta \frac{| x-y|^2}{s}} dy
\le C s^{-d/2 -1} e^{6 \beta t^2/s} \| u \|_1^2.
\end{equation}
for some constants $C$ and $\beta > 0$ which are  independent of $n$.
Let $k \in \{ 1,\ldots,d \} $.
Then
$$| (e^{-sH_n} M_\chi u, \partial_k M_\chi (\varphi \, e^{\beta \frac{| x- . |^2}{2s}}) ) |
\le C s^{-\frac{d}{4} - \frac{1}{2}} e^{3 \beta t^2/s} \| u \|_1 \| \varphi \|_2$$
for all $\varphi \in C_c^\infty(\mathbb{R}^d)$.
On the other hand,  $e^{-tH_n}$ converges strongly in $L^2(\mathbb{R}^d)$  to $e^{-tH}$
(see \cite{AE2} Corollary~3.9).
It follows then that
$$| (e^{-sH} M_\chi u, \partial_k M_\chi (\varphi \, e^{\beta \frac{| x- . |^2}{2s}}) ) |
 \le C s^{-\frac{d}{4} - \frac{1}{2}} e^{3 \beta t^2/s} \| u \|_1 \| \varphi \|_2.$$
Since this is true for  all $\varphi \in C_c^\infty(\mathbb{R}^d)$ we have by density
$$ \| (M_\chi \partial_k  e^{-sH} M_\chi u) \cdot e^{\beta \frac{| x-.|^2}{2s}}  \|_2
\le C s^{-\frac{d}{4} - \frac{1}{2}} e^{3 \beta t^2/s} \| u \|_1.$$
This proves the lemma.
\end{proof}

\begin{proof}[Proof of Theorem \ref{th7}]
It follows from Lemma~\ref{lemm1} that the Riesz transform
$M_\chi \partial_k A^{-1/2}$ is bounded on $L^2(\mathbb{R}^d)$.

In order to prove a weak type estimate for $T = M_\chi  \partial_k H^{-1/2} M_\chi$
we apply Theorem \ref{th1} with $S = M_\chi \partial_k H^{-1/2}$ and $A_t = e^{-t^2 H} M_\chi$.
These operators are bounded on $L^2(\mathbb{R}^d)$ and by Lemma \ref{le2}
the operators $A_t$ satisfy assumption \eqref{au2}.
It remains then to check (\ref{DMop}).
 By the formula
$$H^{-1/2} = \frac{1}{2 \sqrt{\pi}} \int_0^\infty e^{-s H} \frac{ds}{\sqrt{s}}$$
we have
$$ H^{-1/2}e^{-t^2H} = \frac{1}{2 \sqrt{\pi}} \int_0^\infty e^{-(s+t^2)H} \frac{ds}{\sqrt{s}}= \frac{1}{2 \sqrt{\pi}} \int_0^\infty e^{-sH} \mathds{1}_{\{s> t^2\}} \frac{ds}{\sqrt{s-t^2}}.$$
Let $\beta > 0$ be as in Lemma~\ref{lemm3} and let $\delta > 0$.
Fix $x \in \mathbb{R}^d$, $t > 0$ and let $u \in L^2(\mathbb{R}^d)$ with $\mathop{\rm supp} u \subset B(x,t)$. Set $\nu(s,t) = \Big| \mathds{1}_{\{s> t^2\}} \frac{1}{\sqrt{s-t^2}} - \frac{1}{\sqrt{s}} \Big|$.
Then
\begin{eqnarray*}
\lefteqn{
2 \sqrt{\pi} \int_{\mathbb{R}^d \setminus B(x,(1+ \delta)t)} | ((T - SA_t)u)(y) | \, dy
} \hspace{5mm} \\*
 &\le&    \int_0^\infty \int_{\mathbb{R}^d \setminus B(x,(1+ \delta)t)}
 | (M_\chi \partial_k e^{-s H} M_\chi u) (y) | \,  dy \,
\nu(s,t) \, ds\\
 &\le&  \int_0^\infty\nu(s,t)   \left( \int_{\mathbb{R}^d} | (M_\chi \partial_k e^{-s H} M_\chi u) (y) |^2 \,
    e^{\beta \frac{| x-y|^2}{s}} dy \right)^{1/2} \times \\*
 && \hspace{50mm} {} \times  \Big( \int_{\mathbb{R}^d \setminus B(x,(1+ \delta)t)}
      e^{-\beta \frac{| x-y|^2}{s}} \, dy\Big)^{1/2} ds\\
 &\le& C \int_0^\infty \nu(s,t) \,
       s^{-\frac{d}{4} - \frac{1}{2}} e^{3 \beta t^2/s} \|u\|_1
 \Big( \int_{\mathbb{R}^d \setminus B(x,(1+ \delta)t)} e^{-\beta \frac{| x-y|^2}{s}}  dy \Big)^{1/2} ds.
 \end{eqnarray*}
Note that we have used Lemma \ref{lemm3}  in the last inequality.
Now
$$  \int_{\mathbb{R}^d \setminus B(x,(1+ \delta)t)} e^{-\beta \frac{| x-y|^2}{s}} \, dy
\le  e^{-\beta \frac{(1+\delta)^2 t^2}{2s}}
    \int_{\mathbb{R}^d } e^{-\beta \frac{| x-y|^2}{2s}} \, dy
\le C s^{d/2} e^{-\beta \frac{(1+\delta)^2 t^2}{2s}}.$$
Choosing $\delta \ge 4$ we obtain a positive constant $\gamma$ such that
\[
 \int_{\mathbb{R}^d \setminus B(x,(1+ \delta)t)} | (T - SA_t)u(y) | \, dy
\leq C \int_0^\infty \nu(s,t)
   s^{-1/2} e^{-\gamma t^2/s} \, ds.
\]
The latter integral is bounded by some constant  $M$  independent of $t$.
 This proves (\ref{DMop}) and hence $T = M_\chi  \partial_k H^{-1/2} M_\chi$ is weak type
$(1,1)$.
By interpolation, it is bounded on $L^p(\mathbb{R}^d)$ for all $1 < p \le 2$.
\end{proof}

As discussed at the end of the previous section, we note that if one proves a
version of  Theorem \ref{thEO} without the extra factor $(1+t)^{d/2}$ if
 $(a_{kj}(x)) \ge \mu I$ for a.e.\ $x$ in a  `big' domain, then
Theorem \ref{th7} holds with $A$ in place of $H$.
That is $M_\chi  \partial_k A^{-1/2} M_\chi$ is weak type $(1,1)$ and bounded on $L^p(\mathbb{R}^d)$
for all $1 < p \le 2$.
In \cite{EO2}, we prove by a different method that
if the coefficients $a_{kj} \in W^{1,\infty}(\mathbb{R}^d)$,
then $M_\chi \partial_k (I+A)^{-1/2} $ and $M_\chi \partial_k \, \partial_j (I + A)^{-1}$
are bounded on $L^p(\mathbb{R}^d)$ for all $p \in (1,\infty)$.
Moreover, if $a_{kj} \in W^{\nu,\infty}(\mathbb{R}^d,\mathbb{C})$ then we show that
$M_\chi \, \partial_k (I + A)^{-1/2} \, M_\chi$ is bounded on $L^p$
for all $p \in (1,\infty)$.

\vspace{.3cm}
\noindent {\bf Acknowledgements} Most of this work was carried out whilst  the second named
 author was visiting the  Department of Mathematics at the University of  Auckland.
He wishes to thank both the University of Auckland and  the CNRS for financial support.
 Part of this work is supported by the Marsden Fund Council from Government funding,
administered by the Royal Society of New Zealand.

\newcommand{\etalchar}[1]{$^{#1}$}

redpajama_set_name: RedPajamaArXiv


text:
% The textbox that displays the radius of the current circular region.
classdef Radius < redux.Component.TextBox.RangeBox & redux.State.Listener
	properties
		currentRegion
		video
	end

	methods

		%%%%% CONSTRUCTOR %%%%%

		% Adds this component to its parent, subclassing the RangeBox class
		% (which accepts only numbers in a certain range).
		% Notable superclasses:
		% - redux.Component.TextBox.RangeBox
		% - redux.Action.Dispatcher (via RangeBox)
		% - redux.State.Listener
		function this = Radius(parent, varargin)
			p = redux.InputParser();
			p.KeepUnmatched = true;
			p.addParent();
			p.parse(parent, varargin{:});

			this@redux.Component.TextBox.RangeBox(parent, varargin{:});
		end

		%%%%% STATE LISTENER %%%%%

		% This textbox is enabled when a region is being edited, and disabled
		% otherwise. Called dynamically from State.Listener when isEditing
		% changes.
		function [] = onIsEditingChange(this, state)
			switch(state.isEditing)
				case 'region'
					this.setParameters('Enable', 'on');
				otherwise
					this.setParameters('Enable', 'off');
			end
		end

		% Updates the String value of the textbox to the radius value in State.
		% Called dynamically from State.Listener whenever the region changes.
		function [] = onCurrentRegionChange(this, state)
			this.currentRegion = [];

			regions = state.regions;
			for iRegion = 1:numel(regions)
				if regions{iRegion}.id == state.currentRegion
					this.currentRegion = regions{iRegion};
					break;
				end
			end

			if isempty(this.currentRegion)
				this.setParameters('String', '');
				this.backupText = '';
			else
				this.setParameters('String', num2str(this.currentRegion.radius));
				this.backupText = num2str(this.currentRegion.radius);
			end
		end

		function [] = onVideoChange(this, state)
			this.maxValue = round(min(state.video.height, state.video.width) / 2);
			this.video = state.video;
		end

		function [] = onRegionsChange(this, state)
			for iRegion = 1:numel(state.regions)
				if state.regions{iRegion}.id == state.currentRegion
					this.currentRegion = state.regions{iRegion};
					break;
				end
			end
		end

		function [] = dispatchAction(this, ~, ~)
			str = this.getParameter('String');
			num = str2double(str);
			validatedNum = this.validateData(num);
			if(~isempty(validatedNum))
				this.setParameters('String', num2str(validatedNum));
				this.backupText = num2str(validatedNum);
				action = this.actionFactory.actions.CHANGE_REGION_PARAMETER;
				action.prepare(this.currentRegion, 'radius', validatedNum, this.video);
				action.dispatch();
			end
		end
	end

end


Papyrus is from the Heart | Papyrus Watch - Exposing the overuse of the Papyrus font.
From the Heart Garden brings us a Chocolate Viola, which is a plant (looks like a violet) not a candy. Looks like this is the company's logo too, which is unfortunate. Amy spotted this at a thrift store, and there's more coming. I've got a couple gems buried somewhere on my phone, also from a thrift store.


Barrels proved to be suitable working material for Christo because of their sculptural effect and their low cost, and they soon became a dominant factor in his work. From 1958 onwards, many structures were created out of wrapped and unwrapped barrels. Whereas the wrapped cans and bottles were comparable to classical still lifes, the dimensions of the columns of barrels gave them a life-sized form. Their large size, coupled with their arrangement in groups, enhanced their physical effect over that of the smaller works.
The use of barrels gained the upper hand from 1961 onwards when Christo erected a column of unaltered drums only in the courtyard behind his studio at 14 rue de Saint-Senoch for the first time. Christo carried the barrels he had collected and cleaned into the yard, stacked them one upon the other, had them photographed and then finally disassembled them.
In the same year, he moved into a garage in Gentilly, a small suburb of Paris, belonging to his friend the painter Jan Voss. Since Christo's tiny studio was bursting at the seams at the time, he used the additional space to store some of his works, but mostly to make some large-scale works that were too big for his own studio. Physically, none of these sculptures exist today, but not because Christo himself destroyed or dismantled them. It was his landlord who disposed of them when Christo was no longer in a position to pay the storeroom rent.
By fate or coincidence, the Gentilly studio was located next to a huge yard used for storing drums that Christo would be able to use. The new, open space had an immediate effect on his works, most notably in their dimensions. He soon carried the principle of vertical stratification to monumental heights by stacking several barrels up to a height of fifteen feet along the freeway in Gentilly, increasing the drivers' awareness of and sensitivity to their surroundings through the temporary artistic estrangement of the environment.
The first time Christo had the opportunity to present his new structures to the larger public was at his solo exhibition in the Cologne gallery of Haro Lauhus in 1961. The visitors were welcomed in front of the entrance by a column of stacked barrels, and inside, Christo carried this motif to claustrophobic heights by filling a whole room with his barrel arrangements. The columns and towers of rusty steel drums reached to the ceiling and left only a small passage to get to the back room of the gallery. One newspaper secretly had five visitors photographed from inside as they stared, some amazed, some incredulous, through the window. Underneath the pictures was the headline: "The Most Controversial Exhibition Cologne Has Ever Seen."
When Christo had a personal exhibition at Jeanine de Goldschmidt's Galerie J (Pierre Restany's Paris gallery) the following year, accompanying Christo and Jeanne-Claude's Iron Curtain at Rue Visconti, he installed a wall of rusty steel drums that occupied an entire wall in the gallery. The bottoms and lids, structured from brown and red to yellow and blue, must have seemed like an extreme enlargement of a pointillist painting.
Enlarged reproductions of several ink drawings proposing the use of barrels to cover whole building fronts and lobbies were installed on the pillar in the middle of the gallery. Although these early projects were not realized, the drawings indicated the scope of Christo and Jeanne-Claude's imagination and their visionary approach to alter whole environments.
Excerpts from the books Christo and Jeanne-Claude: Early Works 1958-64 (Bönen: Kettler, 2009) and Christo: The Paris Sculptures 1961 (Bönen: Kettler, 2011) by Matthias Koddenberg. Edited by the author in 2012.


text:
Will Wells
Producer; Arranger; Performer; Music Director; Songwriter
William F. Wells '11 was born and raised in Erial, New Jersey, just across the water from Philadelphia. He studied at Berklee, focusing on film scoring and music production and engineering, but also honing his skills as a songwriter, arranger, conductor, and musician. Upon arriving in Los Angeles in 2011, Wells served as senior engineer at Woodshed Recording in Malibu, California, owned by film composer Richard Gibbs. He continued on to tour North America with the electropop duo, LMFAO, and later with Canadian music sensation and Quincy Jones protègè, Nikki Yanofsky.
He has since had the opportunity to work with such renowned artists as Quincy Jones, Barbra Streisand, RZA and Wu-Tang Clan, Wyclef Jean, DJ Premier, Michael McDonald, Sky Blu (of LMFAO), Logic, and many more in the studio and on stage. He recently worked on the Tony-, Grammy-, and Pulitzer Prize Award-winning hit Broadway musical, Hamilton, and also cowrote the lead single for the Pentatonix, "Can't Sleep Love," for its chart-topping debut self-titled album. In early 2016, he completed the nearly year-long Smoke + Mirrors World Tour with Imagine Dragons as keyboardist, guitarist, and background vocalist. He is now working internationally as musical director for artists such as Kevin Garrett, Anthony Ramos, and Jasmine Cephas Jones, and was most recently the musical director for ESPN's 2016 Heisman Trophy Ceremony national broadcast.
Watch this alumnus profile of Wells:


In this project, we discussed "Counter-Intuitive" Problems, where the real result is the opposite of the expected result.
The rope around the world problem- How many feet of rope is required to wrap around the world, and how much more would we need if we rai sed the rope ONE foot off of the earth's surface? Upon our calculations, the answer turned out to be a mere 2pi feet, or 6.28 feet, which is much less than people would expect.
2. You are on the Monty Hall game show as a contestant. Monty, the host, calls you down and you play this game. In the game, there are 3 doors. Behind one is the car of your dreams (you name it) and behind the other two are goats. You have to choose one of the three doors, whatever one you think has the car behind it. Once you choose, Monty then reveals one of the other two remaining doors that has a goat behind it, leaving you now with two doors instead. He then asks you if you wish to keep the door you chose. Or to switch your choice to the other door. So, what do you do?
The answer to this, believe it or not, is to switch doors EVERY TIME. The math is in the slideshow linked to this post.
Theoretically, how many people can you put in the same room to find a pair of people with the same birthday?


With its clear and concise presentation, Behavioral Genetics, 7th edition introduces students to the field's underlying principles, defining experiments, ongoing controversies, and most recent discoveries. The text provides students with an understanding of heredity, it's DNA basis, the methods used to discover genetic influence on behavior and identify specific genes. It then examines what is known about genetic influence on cognitive ability, psychopathology, substance abuse, personality, health psychology, and aging. Finally it looks ate the future of the field of Behavioral Genetics and area where some of the most exciting development in the Behavioral sciences are being made.


To date, I have the above three books published on Amazon and a fourth presently being written . . .
The first of these are Books One and Two of my historical fiction trilogy, Sons of Kings – a Viking/Saxon story set in the mid ninth century. There are two main protagonists throughout the books. One is King Alfred of Wessex – otherwise known as Alfred the Great – and the fictional Eadwulf, from the neighbouring kingdom of Mercia.
The life of Eadwulf, young son of the Mercian king, is changed forever when his family is betrayed to the Danes by his treacherous uncle. His father is killed, his mother raped, and Eadwulf is taken to be a slave in the Danish lands. His new life is hard until the return of Bjorn, who allows Eadwulf to become a part of his closely knit crew. Eadwulf's tale plays out against events unfolding in Wessex. Alfred, son of the Wessex king, faces family tragedies from an early age, losing first his mother, then his sister on her marriage. At the successive courts of his father and elder brothers, Alfred learns the art of kingship. And, like Eadwulf, he learns that a trusted kinsman can easily turn traitor.
Pit of Vipers continues the story of Alfred of Wessex and Eadwulf of Mercia as the threats to their kingdoms continue to escalate. In the mid 860's large bands overwinter again on the Isle of Thanet, breaking treaties to raid mercilessly into Kent. By now Alfred is fifteen and battling against bouts of ill health as he learns much about the art of kingship. By the following year, his last surviving brother Aethelred is king – and Alfred becomes his invaluable second in command. In this same year, the Great Heathen Army lands on the coast of the kingdom of East Anglia.
By 864 Eadwulf has been back in Mercia for four years. His newfound contentment in the home of a Mercian ealdorman is marred only by his ongoing hunger for revenge on his treacherous uncle, Burgred. But as the Mercian king, Burgred proves extremely difficult to kill . . .
A Dash of Flash is an eclectic mix of stories with both contemporary and historical settings, plus a few fairy tales and ghost stories added for good measure.
Step inside and join the many and varied characters at their times of joy or sorrow, remorse or loss. Laugh at their foibles, commiserate with their grief and indulge with them as they reminisce. Or simply smile at the fantasy of the tale.
All three books are available as eBooks, or in print, courtesy of CreateSpace.
My 'Work in Progress' is Book Three of the Sons of Kings trilogy, Wyvern of Wessex. A wyvern is similar to a dragon and the title refers to the symbol on the banner of King Alfred's kingdom of Wessex. The banner was proudly displayed at every battle.
I can't believe it's mid-July of 2018 already! I finished writing Book 3 a while ago and it has since been edited an formatted.
So… I'm happy to say that Wyvern of Wessex is now available on Amazon as an eBook. A print version will follow soon.
In Wessex, King Aethelred is now dead, leaving his twenty-one-year-old-brother, Alfred, to succeed to the throne. Though his succession was agreed by the witan, Alfred must now prove himself worthy of the kingship, or lose it. But Wessex is in turmoil, besieged by Viking Danes intent on subjugating the kingdom – and knowing that the new king is young and inexperienced. Alfred must use all his wiles if he is to outthink and outmanoeuvre Guthrum, the Dane who very nearly becomes his nemesis.
Alfred's victories and defeats take him on a journey of learning, during which he gains experience and strength. He experiences highs and his lows, and valiantly rises from the depths of despair to save his beloved kingdom from total conquest.
And, beside him at his greatest time of need is his new ally and friend, Eadwulf of Mercia.
The three Sons of Kings books were intended to be a trilogy, of which Wyvern of Wessex would have been the third and last book. However, by the time I'd reached 100,000 words, I realised I was never going to finish Alfred's and Eadwulf's stories in that book! So I am now on the road to writing Book 4 (no title, as yet), My trilogy has now become a series!
It's now November 2018 and I'm busy writing the 4th book in the Sons of Kings series. I had a lovely surprise a few days ago when I was informed that Shadow of the Raven had been awarded a Chill With A Book award. To say that I was thrilled, would be putting it lightly!
That's wonderful! I love Bernard Cornwell, too, but haven't read his Saxon series yet, on purpose. Nor have I yet watched 'Vikings'. I don't want to be influenced by either until I've finished my trilogy. I've been told I have similar themes in my story to both Cornwell's books and the TV series but I refuse to look. Similarities are inevitable, I suppose, when we're looking at the same period and characters. I've read Bernard Cornwell's Arthurian books and loved those.
Oh, joy! How exciting to have read your book page so soon after it was posted! 🙂 I certainly understand your not wanting to be cross-contaminated by either The Saxon Tales or the Vikings series while you are writing your trilogy. But when you are finished, you will probably want to check them both out!
Ragnar Lodbrok features in a lot of Viking myths, some of them putting him in the time frame of my books, some having him later on. His death is well known, albeit a thing of myth (just like the blood eagle ritual). I suppose the mysterious, mythical side is one of the things I love about it all.
I find that nothing hides my wrinkles, so my attitude now is to sport them as badges of courage and, hopefully, wisdom!
Timi, you look so sophisticated with that hair style and I don't think you look as old as me. If you are, then you disguise it well, despite what you say about rapid ageing recently.
I'm sorry to hear about your past surgeries. What a lot to contend with -and so very scary. But you pulled through and I'm sure your determination not to be beaten would have played a major role in that.
I'm not afraid old growing old, either. I have far too much I still want to do and so many places I want to see, to consider I might 'pop my clogs' in the near future. (I'm a Lancashire lass, land of cotton mills, clogs and shawls a century ago.) But, since I can do nothing about the dreaded wrinkles, I suppose I'll just have to grin and bear them. And the new hairdo will come … quite soon. You really have inspired me to try it.
Best wishes with all your books, Millie. Enjoy the day.
Desperate for the third book, any idea of tough times for release yet?
I intend to have Book 3 published by late spring. It's been far too long in coming, I know, but family illnesses and other issues like house moves have taken so much of my time. I can't find a link to find out who you are, Jonny, but I can only assume you have read and enjoyed the first two books, and thank you very much for that! Very best wishes, Millie.
Would love to buy all 3 books in the Sons of Kings series…looks absolutely amazing….but struggling to find it in South Africa. Any recommendation hoe I can buy it would be greatly appreciated.
only available through Amazon at the moment.
I dropped by your website to check on your progress, and was so thrilled to see Book 3 out that I hied myself off to Amazon to grab it right away! All the best, Millie.
Thanks a million for that, Christine.<3 I eventually published Book 3 at the end of last June. It was a very long time in coming, due to family health issues the previous year – during which time I wrote nothing at all. I'm hoping Book $ won't be quite that much overdue!


The IMHA (Independent Mental Health Advocacy) Service is provided as a statutory right for people detained under the Mental Health Act. The service was introduced as part of the Mental Health Act 2007 and ensures there is an additional safeguard for patients subject to the Act. Patients have the right to be provided with the support of a specialist advocate with the appropriate training and experience to carry out the role. The IMHA service is independent of the NHS.
Patients have the right to an advocate if they are: detained under the Mental Health Act (even if they are on leave under Section 17); are conditionally discharged; are subject to guardianship; are on a supervised community treatment order.
Referrals can be made by patients or by a third party. To make a referral please phone 0844 248 9 248.
Anyone who receives support from care services, in the community, residential services or in prisons, may be able to have support from an advocate. In Herefordshire and Worcestershire this service is provided by Onside.
You have the right to an independent advocate if you need help to understand what is happening within the care processes, and the choices you have, need support to decide what care and support you need and help to able to tell people what your views are.
Under the Care Act 2014 those experiencing substantial difficulty in taking part in care and support assessments, planning and reviews and safeguarding processes are eligible for advocacy if they do not have another suitable person to provide support.
As well as people using adult care and support services, carers of an adult or young person, could also be entitled to support as well as young people moving in to adult services.
If you are not happy with the decisions that have been made, then your advocate will help you to write down what it is you feel unhappy about.


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A 2 Z Articles Blog / Business / Business / Tesla cuts staff pay as coronavirus halts production
Tesla cuts staff pay as coronavirus halts production
Electric car-maker Tesla will reduce staff pay and put non-essential workers on furlough while production of its vehicles is stopped due to coronavirus.
Work at its factory in Fremont, California halted on 23 March.
In a letter to staff, the company said it hoped to resume operations on 4 May, "barring any significant changes".
Most remaining workers will face a pay cut of 10%, while director pay will be cut by 20% and vice-presidents and above will lose 30% of their salary.
The letter was shared with business news site CNBC.
Furloughing is designed to support firms that have been badly hit by coronavirus, and to prevent mass unemployment. Taxpayers' money will help temporarily pay the wages of people who can't do their jobs, to help companies retain them.
What does it mean if I've been furloughed?
"As usual, for those who are on site, if you are sick or are uncomfortable coming to work, please contact your manager and stay at home. We respect your decision and you will not be penalised," said Tesla's Valerie Capers Workman in the letter.
The pay cut is expected to last until the end of June.
Production at Tesla's solar panel facility in New York state has also halted.
However, several of the company's engineers have been working on developing a ventilator using car parts to help those who fall ill with Covid-19.
On Monday, Tesla shared a video of a prototype on YouTube.
Ford and General Motors have also offered to produce ventilators and other hospital supplies.
Furloughed staff will remain contracted to Tesla but will not be paid until 4 May, should the factories reopen at that time.
"For the vast majority of furloughed employees, unemployment benefits will be roughly equivalent to normal take-home pay," Ms Capers Workman said in her letter.
Last week, Tesla told staffing agencies that contract work would be suspended until further notice, and hundreds of temporary workers were dismissed.
Tesla's facility in California is the company's only car-making facility in the US, and employs more than 10,000 people.
When it closed in March, chief executive Elon Musk told staff: "I will personally be at work, but that's just me. Totally OK if you want to stay home for any reason."
The company originally planned to "comfortably exceed" 500,000 vehicle deliveries in 2020 and has not changed its guidance for investors.
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