text:
\section{Introduction}

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	Methods for solving PDEs based on polyhedral meshes are having a fast development  and attracting more and more attention. They provide greater flexibility in mesh generation,
	can be exploited as transitional elements in finite element meshes, and are better suited than methods based on tetrahedral or hexahedral meshes for many applications on complicated and/or moving domains \cite{DG-Rev}. Many different approaches exist, just to quote the more recent we recall: the Agglomerated Finite Element method \cite{Bassietal12}, the Virtual Element Method \cite{basicVEM}, the Hybrid High Order method \cite{DiPietroErnLemaire}.

	A common ingredient to all of these method is the presence of some stabilization term that penalizes a residual in some mesh dependent norm \cite{BCMS}, and dealing with such terms in the analysis relies on the use of some kind of inverse inequality, and results in suboptimal estimates when the factor stemming from such inequality does not cancel out with some small factor coming from the approximation properties of the involved space. This is the case when, for instance, the elements are not shape regular or when we want to obtain $hp$ estimates \cite{CangianietalBook,GuzmanRiviere}. This kind of problem arises when a mesh dependent norm is used to mimic the action of the norm of the space where the penalized residual naturally ``lives'', usually a negative or fractionary norm. On the other hand, it has been observed that, at least theoretically, it is possible to design stabilization terms based on such a norm \cite{BBStab,BStab}, for which the analysis does not require the validity of any inverse inequality.

	In the following we propose an hybridized discontinuous Galerkin method on a polyhedral tessellation with an element by element stabilization similar to the one proposed by \cite{BurmanHansbo}, that penalizes the residual on the flux  in the norm of the dual of $H^1$. The numerical realization of the $(H^{1})'$ norm has been the object of several papers \cite{BPV, Arioli}, and we follow here the general approach proposed by \cite{AlgStab}. Even though for the sake of simplicity in this paper we consider the case of a shape regular mesh, and we only perform the analysis of the convergence in $h$, we believe that this approach (which can, of course, be applied also to other formulations) has the potential to tackle more general cases.

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	\newcommand{\tilde \pi}{\tilde \pi}
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	\newcommand{\mathcal{T}_h}{\mathcal{T}_h}

	\newcommand{\mathcal{E}_h}{\mathcal{E}_h}
	\newcommand{\mathcal{E}^K}{\mathcal{E}^K}
	\newcommand{\Sigma}{\Sigma}
	\section{The DG method with minus one stabilization}

	\subsection{Scaled norms, seminorms and duals} In the following, for $\phi \in V$ and $\lambda \in V'$ ($V$, $V'$ any couple of dual Hilbert spaces), we will indicate by $\langle \lambda ,\phi \rangle$ the action of $\lambda$ on $\phi$. Depending on the context, $V$ and $V'$ will be different couples of dual Sobolev spaces.
	In the analysis that follows, it will be essential to carry over to the dual norms some scaling arguments developed for positive Sobolev norms. In order to do so, we introduce non standard forms for the norms of some Sobolev space. Letting $K \in \mathbb{R}^2$ denote a shape regular polygon, and letting $e$ be an edge of $K$ of length $h$, let us start by considering the space $H^{1/2}(e)$.
	For $\phi \in H^{1/2}(e)$ we let
	\begin{equation}\label{defedgenorm}
	\| \phi \|_{1/2,e}^2 = \alpha_e | \bar \phi^e |^2 + | \phi |^2_{1/2,e}
	\end{equation}
	where $\alpha_e$ is a weight to be chosen later on, and with
	\begin{equation}\label{edgeseminorm}
	\bar \phi^e = \frac 1 {|e|} \int_e \phi,  \qquad | \phi |_{1/2,e} = \int_e \int_e \,dx\,dy \frac{| \phi(x)-\phi(y)|^2}{| x - y |^2}.
	\end{equation}
	On the dual space $(H^{1/2}(e))'$ we also define both a norm and a seminorm. More precisely, we let
	\begin{equation}\label{seminormA}
	| \lambda |_{-1/2,e} = \sup_{{\phi \in H^{1/2}(e)}\atop{\int_e \phi = 0}}\, \frac{ \langle \lambda , \phi \rangle }{| \phi |_{1/2,e}},
	\end{equation}
	and
	\begin{equation}\label{defedgedualnorm}
	\| \lambda \|^2_{-1/2,e} = \beta_e | \bar\lambda^e |^2 + | \lambda |^2_{-1/2,e}, \qquad \text{with } \bar \lambda^e = \frac {\langle \lambda, 1 \rangle} {|e|},  \end{equation}
	where $\beta_e$ is also a weight to be chosen in the following. Recall that the seminorm (\ref{edgeseminorm}) is scale invariant: letting $\hat e = [0,1]$ and $e = [0,h]$, and setting $\hat x = h^{-1}x$,  for $\hat \phi \in H^{1/2}(\hat e)$ and $\phi (x) = \hat \phi(\hat x) \in H^{1/2}(e)$ we have the identity
	\[
	| \hat \phi |_{1/2,\hat e} = | \phi |_{1/2,e}.
	\]

	\

	A corresponding scaling property also holds for the dual semi norm (\ref{seminormA}), as stated by the following proposition.
	\begin{proposition}\label{prop:scaling} Let $\hat e = [0,1]$ and $e = [0,h]$ let $\hat x = h^{-1}x$ and, for $\hat \lambda \in L^2(\hat e)$ let $\lambda (x) = \hat \lambda(\hat x)$. Then
		\[
		| \lambda |_{-1/2,e} = h | \hat \lambda |_{-1/2,\hat e}.
		\]
	\end{proposition}
	\begin{proof}
		\begin{gather*}
		| \lambda |_{-1/2,e} = \sup_{{\phi \in H^{1/2}(e)}\atop {\int_e \phi = 0}} \frac{ \int_e \lambda \phi(x) \,dx}{| \phi |_{1/2,e}} =
		\sup_{{\hat \phi \in H^{1/2}(\hat e)}\atop{\int_{\hat e} \hat \phi= 0}} \frac{ \int_{\hat e} \hat \lambda(\hat x) \hat \phi(\hat x) h\, d\hat x }{| \hat \phi |_{1/2,e}}.
		\end{gather*}
	\end{proof}

	The two norms defined by (\ref{defedgenorm}) and (\ref{defedgedualnorm}) satisfy a duality relation, as stated by the following lemma.
	\begin{lemma}
	 Provided $\alpha_e \beta_e = h^2$ we have
		\[
		\| \lambda \|_{-1/2,e} \simeq \sup_{\phi\in H^{1/2}(e)} \frac { \int_e \lambda \phi }{\| \phi \|_{1/2,e}}.
		\]
	\end{lemma}
	\begin{proof} We have
		\begin{gather*}
		\sup_{\phi \in H^{1/2}(e)} \frac{\int_e \lambda \phi}{\| \phi \|_{1/2,e}} = 	\sup_{\phi \in H^{1/2}(e)}
		\frac{\int_e \bar \lambda^e \bar \phi^e + \langle \lambda -\bar \lambda^e,\phi-\bar \phi^e\rangle}
		{\sqrt{\alpha_e | \bar \phi^e |^2 + |\phi|_{1/2,e}^2}} \leq\\
		\sup_{\phi \in H^{1/2}(e)}
		\frac{h |\bar \lambda^e|| \bar \phi^e | + | \lambda -\bar \lambda^e|_{-1/2,e} |\phi-\bar \phi^e|_{1/2,e}}
		{\sqrt{\alpha_e | \bar \phi^e |^2 + |\phi|_{1/2,e}^2}} \leq	\\
		\sup_{\phi \in H^{1/2}(e)}
		\frac
		{
			\sqrt{h^2\alpha_e^{-1}| \bar \lambda^e |^2 + | \lambda|^2_{-1/2,e}} \sqrt{\alpha_e |\bar \phi^e|^2 + | \phi |_{1/2,e}^2}
		}
		{\sqrt{\alpha_e | \bar \phi^e |^2 + |\phi|_{1/2,e}^2}} = \| \lambda \|_{-1/2,e}
		\end{gather*}
		On the other hand, setting $\lambda^0 = \lambda - \bar \lambda^e$ we start by observing that, by the definition of $| \cdot |_{-1/2,e}$, for each $\epsilon >0$ there exists $\phi^0_\epsilon \in H^{1/2}(e)$ with $\int_e \phi_\epsilon^0 = 0$ and with $| \phi_\epsilon^0 |_{1/2,e} = | \lambda^0 |_{-1/2,e}$ such that
		\[
		\int_e \lambda^0 \phi^0_\epsilon \geq (1-\epsilon) | \lambda^0 |^2_{-1/2,e}.
		\]
		Letting $\phi_\epsilon = \frac h {\alpha_e}\bar \lambda^e + \phi_\epsilon^0$ we have
		\[
		\| \phi_\epsilon \|^2_{1/2,e} = \alpha_e^{-1} h^2 | \bar \lambda^e |^2 +  | \phi_\epsilon^0 |^2_{1/2,e} = \beta_e | \bar \lambda^e |^2 + | \lambda |_{-1/2,e}^2 = \| \lambda \|^2_{-1/2,e}
		\]
		and
		\[ 	\int_{e} \lambda \phi_\epsilon = h \frac h {\alpha_e} | \bar \lambda^e |^2 + 	\int_e \lambda^0 \phi_\epsilon \geq (1-\epsilon) \| \lambda \|_{-1/2,e}^2 = (1-\epsilon) \| \lambda \|_{-1/2,e} \| \phi_\epsilon \|_{1/2,e}.
		\]
		The arbitrariness of $\epsilon$ yields the thesis.
	\end{proof}

	By duality we also obtain:
	\begin{corollary}
		Provided $\alpha_e \beta_e = h^2$ we have
		\[
		\| \phi \|_{1/2,e} \simeq \sup_{\lambda\in (H^{1/2}(e))'} \frac { \int_e \lambda \phi }{\| \lambda \|_{-1/2,e}}.
		\]
	\end{corollary}

	We can now choose $\alpha_e$ and $\beta_e$. Two choices are significant:
	\begin{equation}\label{choicenatural}
	\alpha_e = | e |, \qquad \beta_e = | e |
	\end{equation}
	and
	\begin{equation}\label{choicescaled}
	\alpha_e = 1, \qquad \beta_e = | e |^2.
	\end{equation}

	Observe that the choice (\ref{choicenatural}) yields a  norm for $H^{1/2}(e)$ equivalent, uniformly in $h$, to the natural norm
	\[
	\left(\int_{e} | u |^2 + | u |^2_{1/2,e}\right)^{1/2},
	\]
	while the choice (\ref{choicescaled}) yield a scaled $H^{1/2}(e)$ norm equivalent, uniformly in $h$, to the norm
	\[
	\left(| e |^{-1} \int_{e} | u |^2 + | u |^2_{1/2,e}\right)^2
	\]
	which is scale invariant and  quite commonly used in domain decomposition. Throughout this paper, we choose definition (\ref{choicescaled}) and we set
	$\alpha_e = 1$, $\beta_e = h^2$.

	\

	We also define, in an analogous way, the norm and seminorm  for $H^{1/2}(\partial K)$ and its dual $H^{-1/2}(\partial K)$. More precisely we set
	\[
	\| \phi \|_{1/2,\partial K}^2 = | \bar \phi^{\partial K} |^2 + | \phi |_{1/2,\partial K}, \qquad | \phi |_{1/2,\partial K} = \int_{\partial K} \int_{\partial K} \,dx\,dy \frac{| \phi(x)-\phi(y)|^2}{| x - y |^2},
	\]
	and
	\[
	\| \lambda \|_{-1/2,\partial K} =h^2 | \bar \lambda^{\partial K} |^2 + | \lambda |_{-1/2,\partial K}, \qquad | \lambda |_{-1/2,\partial K} = \sup_{{\phi \in H^{1/2}(\partial K)}\atop {\bar \phi^{\partial K}=0} } \frac {\langle \lambda,\phi \rangle}{| \phi |_{-1/2,\partial K}},
	\]
	with $\bar \phi^{\partial K} = | \partial K|^{-1}\int_{\partial K}\phi$ and $\bar \lambda^{\partial K}  = |\partial K|^{-1} \langle \lambda , 1 \rangle$.

	\

	Finally, we define the norm for $H^1(K)$ as
	\[
	\| f \|^2_{1,K} = | \bar f^{\partial K} |^2 + | f |^2_{1,K}, \ \text{ with } \bar f^{\partial K} = |\partial K|^{-1} \int_{\partial K} f, \ | f |_{1,K}^2 = \int_K | \nabla f |^2.
	\]
		Observe that a perhaps more natural definition of the norm for such a space would be the one where the average over $K$ is used in the place of the average over $\partial K$. In view of the validity of Poincar\'e inequality for the functions with zero average on $\partial K$, the two norms are equivalent, as stated by the following Proposition

	\begin{proposition} Letting $K$ be a shape regular polygon, and letting $\bar f^K = |  K|^{-1} \int_{ K} f$ we have
		\[
		\| f \|_{1,K}^2 \simeq  | \bar f^K |^2 + | f |^2_{1,K}.
		\]
	\end{proposition}
	\begin{proof}  Letting $C_{K,\partial}$ denote the constant in the Poincar\'e Wirtinger inequality
		\[
		u \in H^1(K) \text{ and } \int_{\partial K} u = 0 \quad \text{ implies } \quad\| u \|_{L^2(K)} \leq C_{K,\partial} | u |_{1,K},\]
		we can write
		\[
		| \bar  f^{\partial K} | \leq | \bar f^K - \bar  f^{\partial K} | + | \bar f^K |, \text{ and } | \bar  f^{K} | \leq | \bar f^K - \bar  f^{\partial K} | + | \bar f^{\partial K} |.
		\]
		Now we have
		\[
		| \bar f^K - \bar  f^{\partial K} |  \leq |K |^{-1/2} \| f  -  \bar  f^{\partial K} \|_{L^2(K)} \lesssim 	 |K |^{-1/2}C_{K,\partial} (K) | f |_{1,K}.
		\]
		Recalling that for shape regular polygons it holds that $C_{K,\partial}\lesssim | K |^{1/2}$, we get the thesis
	\end{proof}

	\begin{remark}
		An analogous result holds for any constant yielding a Poincar\'e inequality. Particularly important for our application $\bar u^e$ being defined as the average over an edge. Reasoning as in the proof of the previous proposition, thanks to the shape regularity of $K$ we have
		\[
		| \bar u^e |^2 + | u |_{1,K} \simeq | \bar u^K | + | u |_{1,K}.
		\]
	\end{remark}

	\

	The classical trace Theorem rewrites as
	\begin{theorem}
		It holds that
		\[
		\inf_{u \in H^1(k): u=\phi \text{ on }\partial K} \| u \|_{1,K} \simeq \| \phi \|_{1/2,\partial K},
		\]
		uniformly in $h$.
	\end{theorem}

	\begin{proof} Letting $\bar u^{\partial K}$ denote the average of $u$ on $\partial K$ we can write
		\[
		\| u \|_{1/2,\partial K} = | \bar u^{\partial K} |^2 + | u |_{1/2,\partial K } \lesssim | \bar u^{\partial K} |^2 + | u |^2_{1,K }.
		\]
		By the previous proposition this implies the thesis.
	\end{proof}

	\

	\

	Analogously to what we did for the space $H^{1/2}(e)'$, we can now introduce a semi-norm for the dual of $H^1(K)$, namely
	\[
	| G |_{-1,K} = \sup_{{u \in H^1(K)} \atop{\bar u^{\partial K}=0} } \frac { \langle G, u \rangle} { | u |_{1,K}}.
	\]
	Observe that, letting $\gamma_K: H^1(K) \to H^{1/2}(\partial K)$ denote the trace operator, and letting $\gamma^*_K$ denote its adjoint, we have for $\lambda \in H^{-1/2}(\partial K)$
	\[
	| \gamma_K^* \lambda |_{-1,K} = \sup_{{\phi\in H^1(K)}\atop{\bar\phi^{\partial K} = 0}}   \frac {\langle  \lambda, \gamma_K \phi \rangle} { | \phi |_{1,K}} \lesssim | \lambda |_{-1/2,K}.
	\]
	Remark also that for $F \in (H^1(K))'$ with $\langle F,1 \rangle = 0$ we have
	\[
	\sup_{{\phi \in H^{1}(K)}\atop{\int_{\partial K}  \phi= 0}} \frac{\langle F,\phi \rangle}{| \phi |_{1,K}} =
		\sup_{{\phi \in H^{1}(K)}\atop{\int_{K}  \phi= 0}} \frac{\langle F,\phi \rangle}{| \phi |_{1,K}}.
	\]

	\

	Finally, in the following, we will need to compare the $H^{1/2}(e)$ norm to the $H^{1/2}_{00}(e)$ norm, which, identifying $e$ with the segment $(a,b)$ can be defined as
	\[
	\| \phi \|_{H^{1/2}_{00}(e)} = \int_{a}^b \, dx \int_{a}^b \, dy \frac{| {\phi}(x) - \phi(y) |^2} { | x - y |^2} + \int_{a}^b\frac {| \phi(x) |^2}{| x-a |}\,dx + \int_a^b \frac {|  \phi(x) |^2 }{| x - b|}\,dx.
	\]
	Observe that $\| \phi \|_{H^{1/2}_{00}(e)}$ is scale invariant, analogously to the $H^{1/2}(e)$ seminorm. It is not difficult to prove that for $\phi \in H^{1/2}_{00}(e)$ we have
	\[
	\| \phi \|_{1/2,e} \lesssim \| \phi  \|_{H^{1/2}_{00}(e)}.
	\]
	In fact, we have that
	\[
	| \bar \phi |^{2} \leq |e| \int_a^b | \phi (x) |^{2} \,dx \lesssim  \int_e\frac{| \phi (x) |^{2} }{|x -  a|} \,dx,
	\]
	uniformly as $| e |$ tends to $0$.
	By duality (recall that $H^{-1/2}(e) = H^{1/2}_{00}(e)'$), we have that
	\[
	\| \lambda \|_{H^{-1/2}(e)} \lesssim \| \lambda \|_{-1/2,e}.
	\]
	The scale invariance of the $H^{1/2}_{00}$ norms, implies, by duality, a scaling property for the dual norm: for $\lambda \in L^2(e)$, letting $\hat \lambda \in L^2(\hat e)$ defined as in Proposition \ref{prop:scaling}, we have
	\[
	\| \lambda \|_{H^{-1/2}(e)} = \sup_{\phi \in H^{1/2}_{00}(e)} \frac{\int_e \lambda(x)\phi(x)\,dx}{\| \phi \|_{H^{1/2}_{00}(e)}} =
	\sup_{\hat \phi \in H^{1/2}_{00}(\hat e)} \frac{h \int_{\hat e} \hat \lambda(\hat x)\hat \phi(\hat x)\,d\hat x}{\| \hat \phi \|_{H^{1/2}_{00}(\hat e)} }	= h \| \hat \lambda \|_{H^{-1/2}(\hat e)}
	\]

	\

	Thanks to the scaling property of both the $H^{-1/2}(e)$ norm and the $\| \cdot \|_{-1/2,e}$ it is not difficult to prove that for all $\lambda \in \mathbb{P}_k(e)$ it holds that
	\[
	\| \lambda \|_{-1/2,e} \lesssim \| \lambda \|_{H^{-1/2}(e)}.
	\]
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	\

	The following Lemma allows to compare the $H^{-1/2}(\partial K)$ norm with the $H^{-1/2}(e)$ and the $(H^{1/2}(e))'$ norms.
	\begin{lemma} \label{lem:1.7}Let $\lambda \in L^2(\partial K)$. It holds

		\[
		\left(\sum_e \| \lambda\|^2_{H^{-1/2}(e)}	\right)^{1/2} \lesssim \sup_{u \in H^1(K)} \frac{\int_{\partial K} \lambda u }{\| u \|_{1,K}} \lesssim \left(\sum_e\| \lambda \|^2_{-1/2,e}\right)^{1/2}
		\]
		(where the sum is taken over the edges $e$ of $K$)
	\end{lemma}

	\begin{proof} Let $\lambda \in L^2(\partial K)$. We have
		\begin{gather*}
		\sup_{u \in H^1(K)} \frac{\int_{\partial K} \lambda u }{\| u \|_{1,K}} = 	\sup_{u \in H^1(K)} \frac{\sum_e\int_{e} \lambda u }{\| u \|_{1,K}} \leq 	\sup_{u \in H^1(K)}\sum_e \frac{\int_{e} \lambda u }{\| u \|_{1,K}} \leq 	\sup_{u \in H^1(K)}\sum_e \frac{\| \lambda \|_{-1/2,e} \| u \|_{1/2,e}}{\| u \|_{1,K}}
		\end{gather*}
		Let us now compare $\| u \|_{1/2,e}$ with $\| u \|_{1,K}$. We have, setting $\bar u^e = | e |^{-1} \int_e u$, thanks to the Poincar\'e Wirtinger inequality,
		\[
		\| u \|_{1/2,e} = | \bar u^e | + | u |_{1/2,e} \leq | \bar u^e | + | u |_{1/2,\partial K} \leq   | \bar u^e | + | u |_{1,K} \lesssim \| u \|_{1,K},
		\]
		yielding
		\[ \sup_{u \in H^1(K)} \frac{\int_{\partial K} \lambda u }{\| u \|_{1,K}} \lesssim \sum_e\| \lambda \|_{-1/2,e} \lesssim \left( \sum_e \| \lambda \|_{-1/2,e}^2 \right)^{1/2}.\]

		\

		As far as the lower bound is concerned, we start by introducing the space $\Phi_0 \subset H^{1/2}(\Gamma^k)$ defined as
		\[
		\Phi^0 = \{\phi \in H^{1/2}(\partial K): \phi|_{e} \in H^{1/2}_{00}(e) \},
		\]
		and we observe that for all $\phi \in \Phi_0$ we have
		\[
		|\partial K|^{-1} \left| \int_{\partial K}\phi \right| \lesssim \sum_e |e|^{-1} | \int_e \phi | \lesssim   \sum_e | \phi |_{H^{1/2}_{00}(e)} \lesssim \left( \sum_e | \phi |^2_{H^{1/2}_{00}(e)}\right)^{1/2}.
		\]
		We can then write
		\begin{gather*}
		\| \lambda \|_{H^{-1/2}(\partial K)}  =  \sup_{\phi\in H^{1/2}(\partial K)} \frac{\int_{\partial K}  \lambda \phi}{\| \phi \|_{1/2,\partial K}} \gtrsim  \sup_{\phi \in \Phi^0}
		\frac{\sum_e \int_e \lambda \phi}{\sum_e | \phi |_{H^{1/2}_{00}(e)}}
		\\ \gtrsim  \sum_{e} \sup_{\phi \in H^{1/2}_{00}(e)} \frac{ \int_{e} \lambda  \phi}{ | \phi |_{H^{1/2}_{00}(e)}} = \sum_{e} \| \lambda \|_{H^{-1/2}(e)} \gtrsim \left(  \sum_{e} \| \lambda \|^{2}_{H^{-1/2}(e)} \right)^{1/2}.
		\end{gather*}
		were we used that the dual of product of spaces is the product of the duals.
	\end{proof}

	\subsection{The model problem and its discretization}
	In the following we consider the simple model problem

	\begin{problem}\label{Pb:strong} Find $u$ solution to
		\[
		-\Delta u = f \text{ in }\Omega, \qquad u = 0 \text{ on }\partial\Omega.
		\]
	\end{problem}

	We look for a solution to Problem \ref{Pb:strong} by a discontinuous Galerkin method. More precisely, let $\mathcal{T}_h$ denote a quasi uniform tessellation of $\Omega$ into shape regular polygons of diameter $\simeq h$. Let $\mathcal{E}_h$ denote the set of edges of the tessellation,  and let $\Sigma = \cup_{e\in\mathcal{E}_h} \bar e $ denote the skeleton of the decompsition. For each element $K$, let $\mathcal{E}^K$ denote the set of edges of $K$. We set
	\[
	V_h = \prod_{K} \mathbb{P}_k(K), \quad \Lambda_h = \prod_{K} \Lambda_K, \text{ with }\Lambda_K = \{\lambda \in L^2(\partial K): \lambda|_e \in \mathbb{P}_k(e)  \text{ for all } e \in \mathcal{E}^K\},
	\]
	as well as
	\[\Phi_k = \{ \phi \in L^2(\Sigma): \phi|_e \in \mathbb{P}_k(e) \text{ for all } e \in \mathcal{E}_h, \ \phi|_{\partial\Omega} = 0\},
	\]
	where, for any one- or two- dimensional domain $D$, $\mathbb{P}_k(D)$ denotes the space of uni- or bi- variate polynomials of degree less or equal than $k$ on $D$.
	We endow $V_h$ with the norm
	\[
	\| u \|_{1,*} = \sum_K | u^K |_{1,K} + \sum_e | \Lbrack \bar u \Rbrack |^2,\]
	where for all interior edges $e$ common to two elements $K^+$ and $K^{-}$ we set
	\[
	| \Lbrack  u \Rbrack | = | u^{K^+} - u^{K^-}|
	\]
	and for $e \subset \partial K \cap \partial \Omega$ we set
	\[
	| \Lbrack u \Rbrack | = | u^K |.
	\]
	On $\Lambda_h$ we consider the two norms
	\begin{gather*}
	\| \lambda \|_{-1/2,\Sigma}^2 =\sum_{K} \sum_{e \subset \partial K} \| \lambda^K_e \|^2_{-1/2,e}\  \text{ and }\
	\interleave \lambda \interleave^2_{-1/2,\Sigma} = \sum_K \| \gamma_K^* \lambda^K \|^2_{-1,K}.
	\end{gather*}
	Observe that for all $\lambda \in \prod_K L^2(\partial K)$ we have
	\[\interleave \lambda \interleave_{-1/2,\Sigma} \lesssim \| \lambda \|_{-1/2,\Sigma},\]
	while for all $\lambda \in \Lambda_h$ it holds that
	\[\interleave \lambda \interleave_{-1/2,\Sigma} \simeq \| \lambda \|_{-1/2,\Sigma}.\]
	On $\Phi$ we define the norm
	\[
	\| \cdot \|_{1/2,\Sigma}^2 = \sum_{e\in \mathcal{E}_h } \| \phi \|^2_{1/2,e}.
	\]

	\

	In order to define our discrete problem, we introduce, for all $K$, a bilinear form $s_K: (H^1(K))'\times (H^1(K))' \to \mathbb{R}$, satisfying the following assumptions.
	\begin{assumption}\label{sK1} For all $F,G \in (H^1(K))'$ we have
		\[
		s_K(F,G) \lesssim | F |_{-1,K} | G |_{-1,K}.
		\]
	\end{assumption}
	\begin{assumption}\label{sK2} For all $\lambda \in \Lambda_K$
		\[
		s_K(\gamma_K^* \lambda,\gamma_K^* \lambda ) \gtrsim | \gamma_K^* \lambda |_{-1,K}^2.
		\]

	\end{assumption}

	We then consider the following discrete problem, where $\alpha > 0$ and $t \in \mathbb{R}$ are two mesh independent parameters.
	\begin{problem}\label{PbGlob} Find $u = (u^K)_K \in V_h$, $\lambda = (\lambda^K)_K \in \Lambda_K$, $\phi \in \Phi_h$ such that, for all $v = (v^K)_K \in V_h$, $\mu = (\mu^K)_K \in  \Lambda_K$, $\psi \in \Phi_h$ it hold that
		\begin{gather}\label{PbK1}
		\int_K \nabla u^K \cdot \nabla v^K - \int_{\partial K} \lambda^K v^K + t\alpha s_K(D u^K - \gamma_K^* \lambda^K,D v^K) = \int_K f v^K + t\alpha s_K(f,D v^K),\\ \label{PbK2}
		\int_{\partial K} u^K \mu^K - \alpha s_K(Du^K - \gamma_K^* \lambda^K, \gamma_K^* \mu^K) - \int_{\partial K} \phi \mu^K  =  - \alpha s_K(f,\gamma_K^* \mu^K), \\ \label{Pbcoupling}
		\sum_K \int_{\partial K} \lambda^K \psi = 0.
		\end{gather}
		where $D: H^1(K) \to (H^1(K))'$ is defined as
		\[
		\langle Du , v \rangle = \int_K \nabla u(x) \cdot \nabla v(x)\, dx.
		\]

	\end{problem}
	Observe that for each $K$, (\ref{PbK1}-\ref{PbK2}) yield a local discrete Dirichlet problem with Lagrange multiplier and a non standard stabilization term, while (\ref{Pbcoupling}) imposes continuity of the fluxes $\lambda$.   The coupling between the different local problems stems from the common Dirichlet data $\phi$, which is single valued on the interface, as well as from equation (\ref{Pbcoupling}). Remark also that, due to the choice of discrete spaces, the discrete fluxes turn out to be strongly continuous.

	\newcommand{\mathcal{R}}{\mathcal{R}}
	\newcommand{H^1_*}{H^1_*}

	\

	It is interesting to give an interpretation of the stabilization term as resulting from defining a suitable numerical trace, in the ideal case where $s^K$ is the scalar product for the space $H^1_*(K) = \{ u \in H^1(K): \ \int_{\partial K} u = 0\}$ endowed with the norm $| \cdot |_{1,K}$. It is easy to check that letting $\mathcal{R}: (H^1_*(K))' \to H^1_*(K)$ denote the Riesz's isomorphism, which, we recall, is defined in such a way that
	\[
	s^K(F,G) = \int_K \nabla \mathcal{R} F \cdot \nabla \mathcal{R} G = \langle F , \mathcal{R} G \rangle = \langle G , \mathcal{R} F \rangle,
	\]
	we have $\mathcal{R} = D^{-1}$.
	Considering, for simplicity, the case $t = 0$, we then have
	\begin{gather*}
	s^K(D u^K_h - \gamma_K^* \lambda_h^K , \gamma_K^* \mu_h^K )  = \langle \gamma_K^* \mu_h^K , \mathcal{R}(D u^K_h - \gamma_K^* \lambda_h^K) \rangle \\ = \langle  \gamma_K^* \mu_h^K , u^K_h - \mathcal{R} \gamma_K^* \lambda_h^K \rangle
	=
	\int_{\partial K} (u^K_h - \mathcal{R} \gamma_K^* \lambda_h^K ) \mu_h^K.
	\end{gather*}

	The stabilized discrete local problems then become
	\begin{gather*}
	\int_K \nabla u_h^K \cdot \nabla v_h^K - \int_{\partial K} \lambda_h^K v_h^K = \int_K f v_h^K \\
	\int_{\partial K}  \widehat u_h^K \mu_h^K = \int_{\partial  K}\widehat \phi_h^K \mu_h^K
	\end{gather*}
	with
	\[
	\widehat u_h^K   = (1 - \alpha ) u_h^K + \gamma_K \alpha \mathcal{R} (\gamma_K^* \lambda_h^K), \qquad \text{ and } \qquad \widehat \phi_h^K = \phi_h^K - \gamma_K \alpha \mathcal{R} (f).
	\]
	Observe that, unlike what happens in the standard DG framework, the numerical trace $\widehat u_h^K$ is indeed the trace of an $H^1(K)$ function.

	\subsection{Stability and convergence analysis}

	In order to analyze Problem \ref{PbGlob}, we will rely on the following discrete Poincar\'e like inequality.
	\begin{lemma}\label{lem:poincare} Let $u \in \prod_K H^1(K)$ and let $\bar u = (\bar u^K)_K$ denote the piecewise constant function where $\bar u^K$ is the average of $u^K$ on $K$. Then
		\[
		\| u \|_{0,\Omega}^2 \lesssim \sum_K h^2 | u^K |_{1,K}^2 + \sum_e | \Lbrack \bar u \Rbrack |^2.
		\]
	\end{lemma}

	\begin{proof} We have
		\begin{equation}\label{1}
		\| u \|_{0,\Omega}^2 \leq \| u - \bar u \|_{0,\Omega}^2 + \|  \bar u \|_{0,\Omega}^2 \lesssim \sum_K h^2 | u^K |_{1,K}^2 +  \|  \bar u \|_{0,\Omega}^2.
		\end{equation}\label{2}
		We then only need to bound the last term.	Let $w$ be the solution of
		\begin{equation}
		-\Delta w =  \bar u, \text{ in }\Omega, \qquad w = 0, \text{ on }\partial \Omega.
		\end{equation}
		Observe that $w \in H^2(\Omega)$ implies continuity of the normal derivative across the skeleton. Fixing for each edge a normal $\nu^e$, we can define
		\[
		\bar \mu^e = \frac{1}{| e| }\int_e \frac {\partial w} {\partial \nu_e }. \]
		Then, using \eqref{2} and integrating by part element by element, we write
		\begin{gather*}
		\|  \bar u \|^2_{0,\Omega} = - \sum_K \int_{\partial K} \frac {\partial w} {\partial \nu^K}  \bar u^K  =
		\sum_e \int_e \bar \mu^e {\Lbrack}  \bar u  {\Rbrack} \leq \sum_e h | \bar \mu^e | | \Lbrack \bar u \Rbrack | \\
		\lesssim
		(\sum_e h^2 | \bar \mu^e |^2)^{1/2}  (\sum_e  | \Lbrack  \bar u \Rbrack |^2)^{1/2}.
		\end{gather*}

		It only remains to bound the first factor in the product on the right hand side. We have
		\[
		\sum_e h^2 | \bar \mu^e |^2 \leq \sum_e h \left(
		\int_e | \bar \mu^e |^2 \right)\leq \sum_e h \left(
		\int_e \left | \frac {\partial w} {\partial \nu_e } \right |^2 \right) \lesssim \sum_K h \left(
		\int_{\partial K} \left | \frac {\partial w} {\partial \nu_K } \right |^2 \right)
		\]

		Now, we observe that, since $w \in H^2(\Omega)$, we have that $\nabla w \in H^1(\Omega)$, so that we can bound (the dependence on $h$ derives from a scaling argument)
		\[
		\int_{\partial K} \left | \frac {\partial w} {\partial \nu_K } \right |^2 \lesssim h^{-1}  \int_{K} | \nabla w |^{2}  + \int_{\Omega} | \Delta w |^{2} \,dx \]
		Adding up the contributions of the different elements we obtain (without loss of generality we can assume that $h \lesssim 1$)
		\[
		\sum_e h^2 | \bar \mu^e |^2 \lesssim  \int_\Omega | \nabla w |^{2} \,dx +  \int_{K} | \Delta w |^{2} \,dx \lesssim \|  \bar u \|^2_{0,\Omega},
		\]
		which yields
		\[
		\|  \bar u \|^2_{0,\Omega} \lesssim \|  \bar u \|_{0,\Omega} (\sum_e  | \Lbrack  \bar u \Rbrack |^2)^{1/2}.
		\]
		Dividing both sides by $ \|  \bar u \|_{0,\Omega}$ and combining with \eqref{1} we get the thesis.
	\end{proof}

	\

	In order to prove the well posedness of Problem \ref{PbGlob}, we  apply Theorem 1.1, Section II.1 of \cite{Brezzi-Fortin}. In order to do so, we introduce the space $\mathbb{V}_h=  V_h \times \Lambda_h$ and rewrite Problem \ref{PbGlob} as follows: find $U = (u,\lambda) \in \mathbb{V}_h$, $\phi \in \Phi_h$ such that for all $V = (v,\mu) \in \mathbb{V}_h$, $\phi \in \Phi_h$ it holds that
	\begin{gather}
	a(U,V) + b(\phi,V) = F(V), \qquad b(U,\psi) = 0,
	\end{gather}
	with
	\[
	a(U,V) = a(u,\lambda;v,\mu) = \sum_K a_K(u^K,\lambda^K;v^K,\mu^K), \quad b(v,\mu;\phi) = \sum_K b_K(\phi;v^K,\mu^K),
	\]
	where the local bilinear forms $a_K$ and $b_K$ are  respectively defined as
	\begin{multline}
	a_K(u^K,\lambda^K;v^K,\mu^K) =  \int_K \nabla u^K\cdot \nabla v^K - \int_{\partial K} \lambda^K v^K \\ +
	\int_{\partial K} u^K \mu^K  +
	\alpha s_K(Du^K - \gamma_K^* \lambda^K, tDv^K - \gamma_K^* \mu^K)\end{multline}
	and
	\begin{equation}
	b_K(v^K,\mu^K;\phi) = \int_{\partial K} \mu^K \phi.
	\end{equation}

	\

	We start by observing that we have an inf-sup condition of the form
	\begin{equation}\label{firstinfsup}
	\inf_{\phi \in \mathbb{P}_K(e)} \sup_{\lambda \in \mathbb{P}_K(e)} \frac{\int_e \phi\lambda}{\| \phi\|_{1/2,e}\|\lambda\|_{-1/2,e}} \gtrsim 1.
	\end{equation}
	Then, proving the well posedness of Problem \ref{PbGlob} reduces to proving the well posedness of the following problem.
	\begin{problem}\label{Pb:reduced} find $(u,\lambda) \in \ker b = \{ (u,\lambda) \in \mathbb{V}_h: b(u,\lambda;\psi) = 0,\  \forall \psi \in \Phi_h\}$ such that for all $(v,\mu) \in \ker b$
		\[
		a(u,\lambda;v,\mu) = f(v,\mu).
		\]
	\end{problem}
	Remark that $\ker b$ is the set of couples $(u,\lambda) \in \mathbb{V}_h$ such that $\{\hskip -.35em\{ \lambda \}\hskip -.35em\} =0$.

	\

	The following lemma holds.
	\begin{lemma}\label{lem:infsup} It holds
		\[	\inf_{(u,\lambda)\in \ker b}\sup_{(v,\mu)\in \ker b} \frac{a(u,\lambda;v,\mu)}{\| u,\lambda \|_{\mathbb{V}} \| v,\mu \|_{\mathbb{V}}} \gtrsim 1,\]
		with
		\begin{equation}\label{Vnorm}
		\| u, \lambda \|^2_{\mathbb{V}} :=  \| u \|_{1,*}^2 + \interleave \lambda \interleave^2_{-1/2,\Sigma}
		\end{equation}
	\end{lemma}
	The proof of Lemma \ref{lem:infsup} is quite long and technical, and we postpone it to Section \ref{proofoflemma}.

	%
	%

	\

	It remains to prove that the bilinear form $a$ is continuous on $\ker b$ with respect to the norm $\| \cdot \|_{\mathbb{V}}$. We observe that $\ker b$ is the subspace of $V_h \times \Lambda_h$ of couples $(v,\lambda)$ with $\{\hskip -.35em\{ \lambda \}\hskip -.35em\} = 0$, where, for $e$ common edge to the elements $K^+$ and $K^{-}$, we set $\{\hskip -.35em\{ \lambda \}\hskip -.35em\}|_e = (\lambda^{K^+}|_e + \lambda^{K^-}|_e)/2$, while for $e \subset \partial K \cap \partial \Omega$ we define  $\{\hskip -.35em\{ \lambda \}\hskip -.35em\}|_e  = 0$. We observe that, for $\lambda$ with $\{\hskip -.35em\{ \lambda \}\hskip -.35em\} = 0$ we can write
	\begin{gather*}
	\sum_k \int_{\partial K} \lambda^K v^K = \sum_K \int_{\partial K } \lambda^K (v^K - \bar v^K) + \sum_e \int_e \lambda_e \Lbrack \bar v \Rbrack =\\  \sum_K \int_{\partial K } \lambda^K (v^K - \bar v^K) + \sum_e  h \bar \lambda_e \Lbrack \bar v \Rbrack  \leq \sum_K | \gamma_K^* \lambda^K |_{-1,K} | v^K |_{1,K} + \sum_e  h | \bar \lambda_e || \Lbrack \bar v \Rbrack | \\
	\leq
	\left( \sum_K | \gamma_K^* \lambda^K |_{-1,K}^2 \right)^{1/2} \left(
	\sum_K | v^K |_{1,K}^2
	\right)^{1/2} + \left(
	\sum_e h^2 | \bar \lambda_e |^2
	\right)^{1/2}  \left(
	\sum_e  | \Lbrack \bar v \Rbrack |^2
	\right)^{1/2},
	\end{gather*}
	where, for $e$ common edge to $K^+$ and $K^-$
	\[
	\lambda_e\ = \lambda^{K^+} = -  \lambda^{K^-} \text{ and }\  \Lbrack u \Rbrack = u^{K^+} - u^{K^-},
	\]
	and where $\bar \lambda_e$ denotes the average of $\lambda_e$ on $e$. Now we can write (with $\bar \lambda^K$ denoting the average of $\lambda^K$ on $\partial K$)
	\[
	\sum_{e \subset \partial K} h^2 | \bar \lambda_e |^2 \lesssim 	\sum_{e \subset \partial K} h^2 \|  \lambda - \bar \lambda^{K} \|_{-1/2,e}^2 + h^2 | \bar \lambda^K |^2 \lesssim | \gamma_K^* \lambda |_{-1,K}^2 +  h^2 | \bar \lambda^K |^2,
	\]
	yielding
	\[
	\sum_k \int_{\partial K} \lambda^K v^K \lesssim \| u, \lambda \|_{\mathbb{V}	} \| v, \mu \|_{\mathbb{V}	}.
	\]
	Remarking that
	\begin{equation}\label{LimD}
	\| Du \|_{-1,K} = \sup_{v \in H^1(K)} \frac{\int_K \nabla u(x) \cdot \nabla v(x)\,dx}{\| v \|_{1,K}} \leq | u |_{1,K},
	\end{equation}
	the continuity of the bilinear form $a$ with respect to the norm (\ref{Vnorm})
	\[
	a(u,\lambda;v,\mu ) \lesssim \| u,\lambda \|_{\mathbb{V}} \| v,\mu \|_{\mathbb{V}}
	\]
	is then not difficult to prove for all $v \in \prod_K H^1(K)$ and all $\lambda \in \prod_K L^2(\partial K)$ with $\{\hskip -.35em\{ \lambda \}\hskip -.35em\}=0$. Problem \ref{Pb:reduced} is then well posed, and admits a solution continuously depending on the data. By Theorem 1.1, Section II.1 of \cite{Brezzi-Fortin} it follows that Problem \ref{PbGlob} admits a unique solution also depending continuously on the data.

	\

	Let now $u$ be the continuous  solution of Problem \ref{Pb:strong}, and let $\lambda = (\lambda^K)$, with $\lambda^K = \partial u^K/\partial \nu^K$ denoting the outer normal derivative of $u$ on $\partial K$. Let $(v_h,\mu_h)\in \ker b$ be approximations to  $u$ and $\lambda$. We can write
	\begin{gather*}
	\| u_h - v_h,\lambda_h-\mu_h \|_{\mathbb{V}} \lesssim a( u_h - v_h,\lambda_h - \mu_h;z_h,\zeta_h)
	\end{gather*}
	for some element $(z_h,\zeta_h) \in \ker b$ with $\|z_h,\zeta_h\|_{\mathbb{V}} = 1$. We have
	\[
	a(u,\lambda;z_h,\zeta_h) = a(u_h,\lambda_h;z_h,\zeta_h)
	\]
	yielding
	\begin{gather*}\| u_h - v_h,\lambda_h-\mu_h \|_{\mathbb{V}}
	\lesssim a( u - v_h,\lambda-\mu_h;z_h,\zeta_h) \leq \| u - v_h,\lambda - \mu_h \|_{\mathbb{V}},
	\end{gather*}
	and, by triangular inequality
	\begin{equation}\label{cea}
	\| u - u_h,\lambda - \lambda_h \|_{\mathbb V} \lesssim \inf_{(v_h,\mu_h) \in \ker b} \| u - v_h,\lambda - \mu_h \|_{\mathbb{V}}.
	\end{equation}

	It only remains to bound the term on the right-hand side of expression (\ref{cea}). Assuming  that $u \in H^{k+1}(\Omega)$, let, for each $K$, $v^K_h \in \mathbb{P}_k(K)$ denote the solution to
	\[
	\int_{K}(v^K_h - u) = 0, \qquad \int_{K} \nabla(v^K_h - u)\cdot \nabla w_h = 0, \text{ for all } w_h \in \mathbb{P}_k(K).
	\]
	We observe that
	\[
	\sum_K | u - v^K_h |_{1,K}^2 + \sum_e | \Lbrack \bar u - \bar v_h \Rbrack |^2 = \sum_K | u - v^K_h |_{1,K}^2 \lesssim h^k | u |_{k+1,\Omega}.
	\]
	On the other hand, letting $\mu_h^K$ be defined on each edge $e$ of $K$ as the $L^2(e)$ projection of $\lambda^K$, it is immediate to check that $u \in H^{k+1}(\Omega) \subseteq H^2(\Omega)$ implies that $\{\hskip -.35em\{ \lambda \}\hskip -.35em\} = 0$ and, thus, $(v_h,\mu_h) \in \ker b$.
	It is also not difficult to check that
	\[
	\int_{\partial K} \mu_h^K = \int_{\partial K} \lambda^K  ,
	\]
	so that the contribution of $\lambda - \mu_h$ to the $\mathbb{V}$ norm of the error is
	\[
	\sum_K | \lambda^K - \mu_h^K |^2_{-1/2,\partial K} + \sum_K h^2 | \bar \lambda^K - \bar \mu^K |^2 = \sum_K | \lambda^K - \mu_h^K |^2_{-1/2,\partial K}.
	\]
	By standard approximation estimates we have that
	\[
	\| \lambda^K - \mu_h^K \|_{L^2(\partial K)} \lesssim h^{k-1/2} | \partial_{\nu^K} u^K |_{k-1/2,\partial K} \lesssim h^{k-1/2} | u^K |_{k+1,K}.
	\]
	By a duality argument we can then bound ($\phi_h \in \mathbb{P}(e)$ denoting the $L^2(e)$ projection of $\phi$)
	\begin{gather*}
	| \lambda^K - \mu_h^K |_{-1/2,\partial K} = \sup_{{\phi \in H^{1/2}(\partial K)} \atop {\int_{\partial K} \phi = 0}} \frac{\int_{\partial K} ( \lambda^K - \mu_h^K)\phi}{| \phi|_{1/2,\partial K}} \\
	= \sup_{{\phi \in H^{1/2}(\partial K)} \atop {\int_{\partial K} \phi = 0}} \frac{\sum_e \int_e ( \lambda^K - \mu_h^K)(\phi - \phi_h)}{| \phi|_{1/2,\partial K}} \\
	\lesssim \sup_{{\phi \in H^{1/2}(\partial K)} \atop {\int_{\partial K} \phi = 0}} \frac{\sum_e \| \lambda^K - \mu_h^K\|_{L^2(e)} \| \phi - \phi_h \|_{L^2(e)}}{| \phi|_{1/2,\partial K}}\\ \lesssim {\sum_e h_e^{1/2} \| \lambda^K - \mu_h^K\|_{L^2(e)} } \lesssim h \| \lambda^K - \mu_h^K \|_{L^2(\partial K)}.
	\end{gather*}

	Adding up the different contributions we obtain that
	\[
	\| u - u_h, \lambda - \lambda_h \|_{\mathbb{V}} \lesssim h^k | u |_{k+1,\Omega}.
	\]

	\subsection{Proof of Lemma \ref{lem:infsup}}\label{proofoflemma}
	For simplicity, let us consider the case $t=0$ (for $t\not =0$ we get an extra term that we can bound essentially by the same arguments).
	Let $(u,\lambda) \in \ker b$, and let $v = u -  \kappa \hat v$, with $\hat v = (\hat v^K)_K$, $\hat v^K =  \int_{\partial K} \lambda^K$, and
	$\mu = \lambda + \beta \hat \mu$ with, on the common edge $e$ to $K^+$ and $K^-$,  $\hat \mu^{K^+}|_e = -\hat \mu^{K^-}|_e = h_e^{-1} (\bar u^{K^+} - \bar u^{K^-})$, where $\bar u$ denote the piecewise constant function assuming on each $K$ the value of the average on $K$ of $u^K$.
	We have
	\begin{multline*}
	a(u,\lambda;v,\mu) = \sum_K | u^K |_{1,K}^2 + \kappa \sum_K \int_{\partial K} \lambda^K \hat v^K + \beta \sum_e h_e^{-1} \int_{\partial K} \Lbrack u \Rbrack \Lbrack \bar u \Rbrack  -\\
	\alpha \sum_K s_K(Du^K - \gamma_K^* \lambda^K, \gamma_K^* (\lambda^K+\beta \hat \mu^K)).
	\end{multline*}

	We now observe that
	\begin{multline*}
	h^{-1}_e	\int_e \Lbrack u \Rbrack \Lbrack \bar u \Rbrack =h^{-1}_e \int_e | \Lbrack \bar u \Rbrack |^2 + h^{-1}_e\int_e (\Lbrack \bar u^e \Rbrack - \Lbrack \bar u \Rbrack) \Lbrack \bar u \Rbrack \geq  | \Lbrack \bar u \Rbrack |^2 - | \Lbrack \bar u^e \Rbrack - \Lbrack \bar u \Rbrack | | \Lbrack \bar u \Rbrack | \\
	\geq | \Lbrack \bar u \Rbrack |^2 -  \frac 1 2 | \Lbrack \bar u \Rbrack |^2 - \frac{1}{2}  | \Lbrack \bar u^e \Rbrack - \Lbrack \bar u \Rbrack |^2,	\end{multline*}
	where on $e$ we let  $\Lbrack \bar u^e \Rbrack$ denote the average on edge of $\Lbrack u \Rbrack$.

	We can bound the last term as follows:
	\begin{multline*}
	| \Lbrack \bar u^e \Rbrack - \Lbrack \bar u \Rbrack |^2 = - \left| | e |^{-1} \int_e \Lbrack u - \bar u \Rbrack \right|^2 \lesssim | e |^{-1}  \int_e | \Lbrack u - \bar u \Rbrack |^2 \\ \lesssim
	| e |^{-1}  \int_e | u^{K^+} - \bar u^{K^+} |^2 +  | e |^{-1}  \int_e | u^{K^-} - \bar u^{K^-} |^2.
	\end{multline*}
	Now we have that, for $e$ edge of $K$
	\[
	\int_e | u^{K} - \bar u^{K} |^2 \lesssim \| u^K - \bar u^K \|^2_{L^2(\partial K)} \lesssim
	h^{-1} \| u^{K} - \bar u^K \|^2_{L^2(K)} + h  | u^K - \bar u^K |^2_{1,K} \lesssim h  | u^K  |_{1,K},
	\]
	finally yielding, for some positive constant $c$,
	\[
	h^{-1}_e	\int_e \Lbrack u \Rbrack \Lbrack \bar u \Rbrack \gtrsim | \Lbrack  \bar u \Rbrack |^2  - c (| u^{K^+} |_{1,K^+} + | u^{K^-} |_{1,K^-} ).
	\]

	We also observe that
	\begin{equation*}
	\int_{\partial K} \lambda^K \hat v^K \simeq | \partial K |^2 | \bar \lambda^K |^2, \text { with } \bar \lambda^K =  |\partial K|^{-1} \int_{\partial K} \lambda^K.
	\end{equation*}
	Moreover, we have that
	\[
	\| \hat \mu \|_{-1/2,e}^2 =  | \Lbrack \bar u \Rbrack |.
	\]

	Then we can write
	\begin{multline*}
	a(u,\lambda;v,\mu) \geq  \sum_K | u^K |_{1,K}^2 + \kappa \sum_K |\partial K|^2 | \bar \lambda^K |^2  +\frac \beta 2 \sum_e  | \Lbrack \bar u \Rbrack |^2 \\ - c'\beta \sum_K  | u^K |_{1,K}^2 - \alpha  s_K(Du^K - \gamma_K^* \lambda^K, \gamma_K^* (\lambda^K+\beta \hat \mu^K)) = \\
	\sum_K | u^K |_{1,K}^2
	+ \kappa \sum_K  |\partial K|^2 | \bar \lambda^K |^2 	+ \frac \beta 2 \sum_e  | \Lbrack \bar u \Rbrack |^2
	- c'\beta \sum_K  | u^K |_{1,K}^2
	+ \alpha  \sum_K  s_K( \gamma_K^* \lambda^K, \gamma_K^* \lambda^K ) +\\
	\alpha\beta \sum_K  s_K( \gamma_K^* \lambda^K, \gamma_K^* \hat \lambda^K )
	+ \alpha \sum_K s_K(Du^K,\gamma_K^* \lambda^K)
	- \alpha \beta \sum_K s_K (Du^K,\gamma^*_K \hat \mu^K).

	\end{multline*}
	We separately bound the four last terms on the right hand side.	By Assumption \ref{sK2}, we have
	\[
	\sum_K  s_K( \gamma_K^* \lambda^K, \gamma_K^* \lambda^K ) \geq c_1 \sum_K | \gamma_K^* \lambda^K |_{-1,K}^2.
	\]
	Using Assumption \ref{sK1}, as well as Lemma \ref{lem:1.7},  we also have
	\begin{multline*}
	\sum_K  s_K( \gamma_K^* \lambda^K, \gamma_K^* \hat \mu^K )  \leq c\sum_K | \gamma_K^* \lambda^K |_{-1,K} |  \gamma^*_K \hat \mu^K |_{-1,K}  \leq \\
	c \left(\sum_K | \gamma_K^* \lambda^K |^2_{-1,K} \right)^{1/2}  \left(\sum_e |  \hat \mu |^2_{-1/2,e}\right)^{1/2} \\
	\leq \epsilon_2 \sum_K | \gamma_K^* \lambda^K |^2_{-1,K} + c(\epsilon_2) \sum_e | \Lbrack \bar u \Rbrack |^2,
	\end{multline*}
	and
	\begin{gather*}
	s_K (Du^K,\gamma^*_K \hat \mu^K)  \leq c \sum_K | u^K |_{1,K} |  \gamma^*_K \hat \mu^K |_{-1,K} \leq c_3 (\sum_e | \Lbrack \bar u \Rbrack |^2 +  \sum_K | u^K |_{1,K}^2),
	\end{gather*}
	while, thanks to \eqref{LimD}, we have
	\begin{gather*}
	\sum_K s_K(Du^K,\gamma_K^* \lambda^K) \leq c \sum_K | u^K |_{1,K} | \gamma_K^* \lambda_K |_{-1,K} \leq \epsilon_3 \sum_K | \gamma_K^* \lambda^K |^2_{-1,K}  + c(\epsilon_3) \sum_K | u^K |_{1,K}^2,
	\end{gather*}
	where $\epsilon_2$ and $\epsilon_3$ are arbitrary positive constants and $c(\epsilon_2)$ and $c(\epsilon_3)$ are positive constants depending on $\epsilon_2$ and $\epsilon_3$ respectively.
	Combining the previous bounds  we obtain
	\begin{gather*}
	a(u,\lambda;v,\mu) \geq
	\sum_K | u^K |_{1,K}^2 (1 - c'\beta - \alpha c(\epsilon_3) - \alpha\beta c_3) +
	\beta	\sum_e  | \Lbrack \bar u \Rbrack |^2 (\frac 1 2 - \alpha c(\epsilon_2) - \alpha c_3) +\\
	\alpha  \sum_K \| \gamma_K^* \lambda^K \|_{-1,K}^2(c_1- \beta\epsilon_2 -\epsilon_3 ) +  \kappa \sum_K |\partial K|^2 | \bar \lambda^K |^2.
	\end{gather*}

	\

	We now set $\beta = 1/(2c')$, and we choose $\varepsilon_1$ and $\varepsilon_2$ in such a way that $(\varepsilon_2/(2c') + \varepsilon_3) \leq c_1$. We next choose
	$
	\alpha = 1/4 \min \{
	c(\epsilon_3) + c_3/(2c'), c(\epsilon_2) + c_3
	\}
	$.
	Observe that neither $\beta$ nor $\alpha$ depend on $h$. With such a choice, for a constant $c_0$ independent of $h$, it holds that
	\[
	a(u,\lambda;v,\mu) \geq c_4 \left(
	\sum_K | u^K |_{1,K}^2 + \sum_e  | \Lbrack \bar u \Rbrack |^2 + \sum_K | \gamma_K^* \lambda^K |_{-1,K}^2 +  \kappa \sum_K |\partial K|^2 | \bar \lambda^K |^2
	\right).
	\]
	We now observe that
	\begin{multline*}
	\sum_K | v^K |_{1,K}^2 + 	 \sum_e  | \Lbrack \bar v \Rbrack |^2 \leq 	\sum_K | u^K |_{1,K}^2 +\sum_e  | \Lbrack \bar u \Rbrack |^2 + \sum_e | h \Lbrack \bar \lambda \Rbrack |^2\\ \lesssim
	\sum_K | u^K |_{1,K}^2 +\sum_e  | \Lbrack \bar u \Rbrack |^2 + \sum_K h^2 | \bar \lambda^K |^2.
	\end{multline*}
	Moreover, thanks to Lemma  \ref{lem:1.7} we have that
	\[
	| \gamma_K^ * \hat \mu^K |_{-1,K}^2 \lesssim  | \hat \mu^K |^2_{-1/2,\partial K} \lesssim \sum_{e \subset \partial K} \| \hat \mu^K \|^2_{-1/2,e} = \sum_{e \subset \partial K} | \Lbrack \bar u \Rbrack |^2
	\]
	and
	\[
	\sum_K | \gamma_K^ * \mu^K |_{-1,K}^2 + \sum_K |\partial K|^2 | \bar \mu^K |^2 \lesssim
	\sum_K | \gamma_K^ * \lambda^K |_{-1,K}^2 + \sum_K |\partial K|^2 | \bar \lambda^K |^2 + \sum_e | \Lbrack \bar u \Rbrack |^2.
	\]

	We then get that
	\begin{gather*}
	\sup_{(v,\mu)} \frac{a(u,\lambda;v,\mu)} {\| v,\mu \|_\mathbb{V}}  \geq \frac{a(u,\lambda; u,\lambda+\beta \hat \lambda)}{\| u,\lambda+\beta \hat \lambda \|_\mathbb{V}}
	\gtrsim \frac{\| u, \lambda \|^2_\mathbb{V} }{\| u, \lambda \|_\mathbb{V}},
	\end{gather*}
	which concludes the proof.

	\section{Realizing a computable stabilizing term} In order for the proposed method to be practically feasible, we need to construct a computable bilinear form $s_K$ satisfying (\ref{sK1}) and (\ref{sK2}). The numerical realization of scalar product for negative Sobolev spaces has been the object of several papers \cite{BPV,BCT00,Arioli}.
	In particular, this can be done using the approach of \cite{AlgStab}. Following such paper, we introduce an auxiliary space $W_K \subseteq H^1(K)$, satisfying
	\begin{equation}\label{mangiacoda}
	\inf_{{\lambda \in \Lambda_K}\atop {\int_{\partial K}\lambda=0 }} \sup_{{w \in W_K}\atop {\int_{\partial K} w = 0}} \frac{\int_K \lambda w}{| \lambda |_{-1/2,K} | w |_{1,K}} \gtrsim 1.
	\end{equation}
	We let $\phi_i$, $i=1,\cdots,N$ denote a basis for $W_K$, and we let $S$ denote  the stiffness matrix (or a preconditioner) relative to the operator $s:H^1(K) \times H^1(K) \to \mathbb{R}$
	\[s(w,v) = \int_K \nabla w \cdot \nabla v.
	\]

	We can now introduce the  bilinear form $s_K: (H^{1}(K))' \to (H^{1}(K))'$ defined as follows:
	\[
	s_K(f,g) = \vec f^T S^{-1} \vec g, \quad \text{ with } \vec f = (\langle f,\phi_i\rangle)_{i=0}^N, \  \vec g = (\langle g,\phi_i\rangle)_{i=0}^N.
	\]
	It is possible to prove that the bilinear form $s_K$ satisfies Assumption \ref{sK1}, and, provided  (\ref{mangiacoda}) holds, (\ref{sK2}) (actually, (\ref{mangiacoda}) is a necessary and sufficient condition for Assumption \ref{sK2} to hold).

	\

	Observe that for $u,v \in( H^{1}(K))'$ and $\lambda,\mu \in H^{-1/2}(\partial K)$ we have
	\[
	s_K(D u - \gamma_K^* \lambda,t D v - \gamma_K^* \mu) = \vec \eta^T S^{-1} \vec \zeta, \qquad s_K (f,t D v - \gamma_K^* \mu ) = \vec f^T S^{-1} \vec \zeta
	\]
	with
	\begin{equation}\label{vecetazeta}
	\eta_i = \int_K \nabla u \cdot \nabla \phi_i - \int_{\partial K} \lambda \phi_i, \qquad \zeta_i = t \int_K \nabla v \cdot \nabla \phi_i - \int_{\partial K} \mu \phi_i, \qquad f_i = \int_K f \phi_i.
	\end{equation}

	\

	We then only need to choose a (small) space $W_K$ satisfying \eqref{mangiacoda} (remark that $W_K$ is not required to satisfy any approximation property). We choose a suitable subspace of the Virtual Element space of order $k+2$ \cite{basicVEM}. More precisely, we set
	\[
	W_K = \{
	w \in H^1(K):w|_e \in \mathbb{P}_{k+2}(e), \ \forall \text{  edge }e\text{ of }K, -\Delta w \in \mathbb{P}_{k}(K), \ \int_K w p = 0\ \forall p \in \mathbb{P}_{k}(K)
	\}.
	\]

	The space $W_K$ does indeed satisfy the inf-sup condition \eqref{mangiacoda}, as stated by the following Lemma.
	\begin{lemma} It holds that
		\[
		\inf_{{\lambda \in \Lambda_K}\atop {\int_{\partial K} \lambda = 0}} \sup_{{w \in W_K}\atop {\int_{\partial K} w = 0}} \frac{\int_K \lambda w}{| \lambda |_{-1/2,K}| w |_{1,K}} \gtrsim 1.
		\]
	\end{lemma}

	\begin{proof} It is not difficult to check that
		\[
		\inf_{{\lambda \in \Lambda_K}\atop{\int_{\partial  K}\lambda = 0} } \sup_{{w \in W_K} \atop{\int_{\partial  K}\phi = 0}   } \frac {\int_{\partial K} \lambda w }{| \lambda |_{-1/2,\partial K} | w |_{1/2,\partial K}} \gtrsim 1.
		\]
		We have, for $w \in W_K$ with $\int_{\partial K}w = 0$
		\[
		| w |^2_{1,K} = \int_K | \nabla w |^2 = - \int_K  w \Delta w + \int_{\partial K} w \partial_\nu w =
		\int_{\partial K} w \partial_\nu w \leq | \partial_\nu w |_{-1/2,\partial K} | w |_{1/2,\partial K}.
		\]
		Now we have
		\begin{gather*}
		| \partial_\nu w |_{-1/2,\partial K} \lesssim | \gamma_K^* ( \partial_\nu w ) |_{-1,K}= \sup_{{\phi\in H^1(K)}\atop
			{\int_{\partial K} \phi = 0}}
		\frac{\int_K \nabla w \cdot \nabla \phi + \int_K \phi \Delta w }{| \phi |_{1,K}}
		\\  \leq  | w |_{1,K} + \frac {\| \Delta w \|_{L^2(K)} \| \phi \|_{L^2(K)}}{| \phi |_{1,K}}
		\lesssim \| w \|_{1,K},
		\end{gather*}
		where we  use Poincar\'e inequality to bound $\| \phi \|_{L^2(K)}$  and an inverse inequality to bound $\| \Delta w \|_{L^2(K)}$, thus obtaining
		\[
		| w |_{1,K}^2 \lesssim | w |_{1,K} | w |_{1/2,\partial K}
		\]
		whence $| w |_{1,K} \lesssim | w |_{1/2,\partial K}$ and, consequently
		\[
		\inf_{\lambda \in \Lambda_K} \sup_{w \in W_K|_{\partial K}} \frac {\int_{\partial K} \lambda w }{| \lambda |_{-1/2,\partial K} | w |_{1,\partial K}}  \gtrsim	\inf_{\lambda \in \Lambda_K} \sup_{w \in W_K|_{\partial K}} \frac {\int_{\partial K} \lambda w }{| \lambda |_{-1/2,\partial K} | w |_{1/2,\partial K}} \gtrsim 1.
		\]
	\end{proof}

	It is known \cite{basicVEM} that a function $\phi \in W_K$ is uniquely determined by a) the value of $\phi$ at the vertices of the polygon $K$; b) for each edge $e$, the values of $\phi$ at the $k+1$ internal points of the $k+3$-points Gauss-Lobatto quadrature rule on $e$ (the other degrees of freedom for the standard VEM space of order $k+2$ are the moments up to order $k$, which we fixed to be zero in the definition of $W_K$).
	Each one of these degrees of freedom corresponds to a basis function $\phi_i$ (the unique function in $W_K$ for which such degree of freedom assumes value $1$, while all the other vanish). The basis functions are not explicitely known, but the knowledge of the degrees of freedom is sufficient to compute the vectors $\vec \eta$ and $\vec \zeta$. In fact, for $u \in \mathbb{P}_k(K)$ and $\lambda \in \Lambda_K$ we have
	\[
	\eta_i = \int_K \nabla u \cdot \nabla \phi_i - \int_{\partial K} \lambda \phi_i = - \int_K \Delta u \phi_i + \int_{\partial K}\left(
	\frac {\partial u}{\partial \nu_K} - \lambda
	\right)  \phi_i = \int_{\partial K}\left(
	\frac {\partial u}{\partial \nu_K} - \lambda
	\right)  \phi_i
	\]
	where we used that $\phi_i$ is orthogonal to all polynomials of degree less or equal than $k$ and hence to $\Delta u$. The last term is computable since it is the integral of a known piecewise polynomial. The fact that $\phi$ is orthogonal to polynomials also allows us to approximate $f_i \approx 0$. As far as the matrix $S$ is concerned, we recall that in general the stiffness matrix entries are not exactly computable. However, the virtual element method takes advantage of the observation that, by same technique used above to compute $\eta_i$, it is possible to compute the $H^1$ projection of the functions in $W_K$ onto the space $\mathbb{P}_{k+2}$. This allows to design a recipe for constructing a computable bilinear form $a^K$ satisfying $a^K(w,w) \simeq | w |_{1,K}^2$ (we refer  \cite{hitchVEM} for the details), and we take $S$ to be the stiffness matrix relative to such bilinear form.

	\begin{remark}
		A more natural choice for the auxiliary space $W_K$ would be
		\[
		W_K = \{
		w \in H^1(K): w|_e \in \mathbb{P}_{k+2}(e), \text{ for all edge }e\text{ of }K, \ -\Delta w = 0 \text{ in }K
		\}.
		\]
		Remark that, however, for such a choice the consistency term $f_i$ would not be $0$ and it would not be computable.
	\end{remark}

	\

	\newcommand{\widehat\Pi^K}{\widehat\Pi^K}

	Also in this case we can give an interpretation of the stabilization as a suitable numerical trace. In fact, once again in the ideal case that $S=(s_{ij})$ with $s_{ij} = \int_{K} \nabla
	\phi_i \cdot \nabla \phi_j$, it is not difficult to realize that the vector $\vec{x} = (x_j) = S^{-1} \vec \eta$ is the vector of coefficient of the function $w_h = \sum_i x_i \phi_i \in W_K$ verifying for all $z_h \in W_K$
	\[
	\int_K \nabla w_h \cdot \nabla z_h = \int_K \nabla u_h^K \cdot \nabla z_h - \int_{\partial K} \lambda_h^K z_h.
	\]
	Letting $\widehat\Pi^K: H^1(K)\to W_K$ denote the Galerkin projection onto $W_K$, it is not difficult to verify that the stabilized problem can be rewritten as
	\begin{gather}
	\int_{K} \nabla u_h^K \cdot \nabla v_h^K - \int_{\partial K}
	\lambda_h^K = \int_K f v_h^K\\
	\int_{\partial K}  \widehat u_h^K \mu_h^K = \int_{\partial  K}\widehat \phi_h^K \mu_h^K
	\end{gather}
	with
	\[
	\widehat u_h^K  = u_h^K - \alpha \widehat\Pi^K\mathcal{R}(Du_h^K - \gamma_K^* \lambda_h^K ), \qquad \text{ and } \qquad \widehat \phi_h^K = \phi - \alpha \widehat\Pi^K \mathcal{R} f.
	\]

	\

	Replacing the stiffness matrix $S$ with an approximation or a preconditioner, reduces to replacing the Galerkin projection operator $\widehat\Pi^K$ with a spectrally equivalent projector $\widetilde \Pi_W$ and setting
	\[
	\widehat u_h^K  = u_h^K - \alpha \widetilde \Pi_W\mathcal{R}(Du_h^K - \gamma_K^* \lambda_h^K ), \qquad \text{ and } \qquad \widehat \phi_h^K = \phi - \alpha\widetilde \Pi_W\mathcal{R} f.
	\]

	\section{Numerical Results}

	We take the domain $\Omega$ to be the unit square $[0,1]\times[0,1]$.
	We solve Problem  \ref{Pb:strong} with Dirichlet boundary conditions and load term chosen in such a way that
	\[
	u = \frac{1}{2\pi^2}\sin(\pi x)\sin(\pi y)
	\]
	is the exact solution. The stabilization parameters are chosen to be $\alpha = t = 1$. We test our method on a deformed  hexagonal mesh (Test case 1.) and on a shape regular Voronoi mesh (Test case 2.). For the first test case,
	Tables \ref{tab:test1-k12}--\ref{tab:test1-k56} show the relative errors $e^u_0 = ||u-u_h ||_{L^2} / || u ||_{L^2}$, $e^u_1 = ||u-u_h ||_{H^1} / || u ||_{H^1}$
	and the estimated convergence rates (ecr) for several values of the polynomial degree $k$ on the eight meshes in Table~\ref{tab:meshdata}. The results for such tests are also displayed in Figure \ref{fig:errors}
	\begin{table}
		\centering
		\begin{tabular}{lccc}
			\toprule
			Name & Nb. of elements & Nb. of faces & $h$\\
			\midrule
			Deformed hexagons\ 8x10 & $94$ & $283$ & $0.2190$ \\
			Deformed hexagons\ 18x20 & $389$ & $1168$ & $0.1033$ \\
			Deformed hexagons\ 26x30 & $822$ & $2467$ & $0.0711$ \\
			Deformed hexagons\ 34x40 & $1415$ & $4246$ & $0.0542$ \\
			Deformed hexagons\ 44x50 & $2270$ & $6811$ & $0.0423$ \\
			Deformed hexagons\ 52x60 & $3203$ & $9610$ & $0.0357$ \\
			Deformed hexagons\ 60x70 & $4296$ & $12889$ & $0.0308$ \\
			Deformed hexagons\ 70x80 & $5711$ & $17134$ & $0.0266$ \\
			\bottomrule
		\end{tabular}
		\caption{Data for the eight meshes used in Test case 1.}
		\label{tab:meshdata}
	\end{table}

	\begin{table}
		\centering
		\begin{tabular}{cccc|cccc}
			\toprule
			\multicolumn{4}{c}{$k = 1$} & \multicolumn{4}{c}{$k = 2$} \\
			\midrule
			$e^u_0$ & ecr & $e^u_1$ & ecr & $e^u_0$ & ecr & $e^u_1$ & ecr\\
			\midrule
			$2.0699e-02$ &  $-$    &  $1.4656e-01$ &  $-$    & $5.7354e-03$ & $-$    & $1.2437e-02$ & $-$\\
			$4.7983e-03$ &  $1.95$ &  $7.0802e-02$ &  $0.97$ & $1.2774e-03$ & $2.00$ & $2.9431e-03$ & $1.92$\\
			$2.2425e-03$ &  $2.04$ &  $4.8468e-02$ &  $1.01$ & $6.1035e-04$ & $1.98$ & $1.3694e-03$ & $2.05$\\
			$1.2939e-03$ &  $2.03$ &  $3.6843e-02$ &  $1.01$ & $3.5615e-04$ & $1.99$ & $7.8836e-04$ & $2.04$\\
			$8.0139e-04$ &  $1.93$ &  $2.9007e-02$ &  $0.97$ & $2.1783e-04$ & $1.98$ & $4.9102e-04$ & $1.91$\\
			$5.6644e-04$ &  $2.02$ &  $2.4393e-02$ &  $1.01$ & $1.5515e-04$ & $1.98$ & $3.4633e-04$ & $2.04$\\
			$4.2152e-04$ &  $2.02$ &  $2.1047e-02$ &  $1.01$ & $1.1609e-04$ & $1.99$ & $2.5733e-04$ & $2.04$\\
			$3.1623e-04$ &  $1.95$ &  $1.8233e-02$ &  $0.97$ & $8.6421e-05$ & $2.00$ & $1.9369e-04$ & $1.93$\\
			\bottomrule
		\end{tabular}
		\caption{Errors and estimated convergence rates (ecr) for the first test case, $k = 1,2$.}
		\label{tab:test1-k12}
	\end{table}

	\begin{table}
		\centering
		\begin{tabular}{cccc|cccc}
			\toprule
			\multicolumn{4}{c}{$k = 3$} & \multicolumn{4}{c}{$k = 4$}\\
			\midrule
			$e^u_0$ & ecr & $e^u_1$ & ecr & $e^u_0$ & ecr & $e^u_1$ & ecr\\
			\midrule
			$6.9786e-05$ & $-$    & $7.3651e-04$ & $-$    & $4.3450e-06$ & $-$    & $3.5751e-05$ & $-$\\
			$3.8303e-06$ & $3.87$ & $8.4585e-05$ & $2.88$ & $2.2292e-07$ & $3.96$ & $2.0016e-06$ & $3.84$\\
			$8.3540e-07$ & $4.08$ & $2.6630e-05$ & $3.09$ & $4.7779e-08$ & $4.12$ & $4.2945e-07$ & $4.12$\\
			$2.7800e-07$ & $4.06$ & $1.1585e-05$ & $3.07$ & $1.5788e-08$ & $4.08$ & $1.4167e-07$ & $4.09$\\
			$1.0705e-07$ & $3.85$ & $5.7035e-06$ & $2.86$ & $6.1058e-09$ & $3.83$ & $5.5053e-08$ & $3.81$\\
			$5.3417e-08$ & $4.05$ & $3.3711e-06$ & $3.07$ & $3.0371e-09$ & $4.07$ & $2.7316e-08$ & $4.09$\\
			$2.9559e-08$ & $4.05$ & $2.1558e-06$ & $3.06$ & $1.6771e-09$ & $4.07$ & $1.5052e-08$ & $4.08$\\
			$1.6673e-08$ & $3.88$ & $1.4095e-06$ & $2.88$ & $9.4919e-10$ & $3.86$ & $8.5391e-09$ & $3.84$\\
			\bottomrule
		\end{tabular}
		\caption{Errors and estimated convergence rates (ecr) for the first test case, $k = 3,4$.}
		\label{tab:test1-k34}
	\end{table}

	\begin{table}
		\centering
		\begin{tabular}{cccc|cccc}
			\toprule
			\multicolumn{4}{c}{$k = 5$} & \multicolumn{4}{c}{$k = 6$}\\
			\midrule
			$e^u_0$ & ecr & $e^u_1$ & ecr & $e^u_0$ & ecr & $e^u_1$ & ecr\\
			\midrule
			$8.4515e-08$ & $-$    & $1.4012e-06$ & $-$    & $2.6850e-09$ & $-$    & $4.8326e-08$ & $-$\\
			$1.1337e-09$ & $5.74$ & $3.8194e-08$ & $4.80$ & $2.2701e-11$ & $6.36$ & $6.1895e-10$ & $5.80$\\
			$1.1359e-10$ & $6.16$ & $5.5669e-09$ & $5.16$ & $2.1225e-12$ & $6.35$ & $6.1202e-11$ & $6.20$\\
			$2.1587e-11$ & $6.12$ & $1.3904e-09$ & $5.12$ & $3.9567e-13$ & $6.20$ & $1.1571e-11$ & $6.14$\\
			$5.2185e-12$ & $5.73$ & $4.2591e-10$ & $4.77$ & $9.1433e-14$ & $5.91$ & $2.7896e-12$ & $5.74$\\
			$1.8362e-12$ & $6.09$ & $1.7733e-10$ & $5.11$ & $3.2121e-14$ & $6.10$ & $9.7466e-13$ & $6.13$\\
			$7.6159e-13$ & $6.03$ & $8.4179e-11$ & $5.10$ & $1.3206e-14$ & $6.09$ & $3.9865e-13$ & $6.13$\\
			$3.2472e-13$ & $5.78$ & $4.1412e-11$ & $4.81$ & $5.5230e-15$ & $5.91$ & $1.6994e-13$ & $5.78$\\
			\bottomrule
		\end{tabular}
		\caption{Errors and estimated convergence rates (ecr) for the first test case, $k = 5,6$.
			Results for $k = 6$ have been obtained using the ADVANPIX Multiprecision Computing Toolbox (\protect\url{https://www.advanpix.com})}
		\label{tab:test1-k56}
	\end{table}

	For the second Test case, the results are shown in Tables \ref{tab:test2-k12} and \ref{tab:test2-k345} and summarized in Figure \ref{fig:VorErr}.

	\begin{table}
	\centering
	\begin{tabular}{cccc|cccc}
		\toprule
		\multicolumn{4}{c}{$k = 1$} & \multicolumn{4}{c}{$k = 1$}\\
		\midrule
		$e^u_0$ & ecr & $e^u_1$ & ecr & $e^u_0$ & ecr & $e^u_1$ & ecr\\
		\midrule
	8.58e-2 & {-}  & 3.11e-1 & {-}  & 2.19e-2 & {-} & 5.72e-2 & {-}   \\
2.40e-2 & 2.47 & 1.63e-1 & 1.25 & 6.32e-3 & 2.41 & 1.54e-2 & 2.55 \\
6.45e-3 & 2.19 & 8.23e-2 & 1.14 & 1.62e-3 & 2.26 & 3.91e-3 & 2.27 \\
1.58e-3 & 2.16 & 4.13e-2 & 1.06 & 4.07e-4 & 2.12 & 9.88e-4 & 2.12 \\
3.72e-4 & 2.27 & 2.08e-2 & 1.08 & 	8.55e-5 & 2.45 & 2.53e-4 & 2.14 \\
1.04e-4 & 2.29 & 1.06e-2 & 1.21 &	1.82e-5 & 2.78 & 6.53e-5 & 2.43 \\
		\bottomrule
	\end{tabular}
	\caption{Errors and estimated convergence rates (ecr) for the second test case, $k = 1,2$.
	}
	\label{tab:test2-k12}
\end{table}

			\begin{table}
			\centering
			\begin{tabular}{cccc|cccc|cccc}
				\toprule
				\multicolumn{4}{c}{$k = 3$} & \multicolumn{4}{c}{$k = 4$}  & \multicolumn{4}{c}{$k = 5$}  \\
				\midrule
				$e^u_0$ & ecr & $e^u_1$ & ecr & $e^u_0$ & ecr & $e^u_1$ & ecr & $e^u_0$ & ecr & $e^u_1$ & ecr  \\
				\midrule
			1.17e-3 & {-} & 7.05e-3 & {-}   & 9.08e-5 & {-} & 6.60e-4 & {-}  &	5.57e-6  & {-}  & 5.19e-5  & {-}  \\
		9.62e-5 & 4.85 & 9.89e-4 & 3.81 & 5.85e-6 & 5.33 & 4.83e-5 & 5.08 &	1.11e-7  & 7.60 & 1.96e-6  & 6.36\\
		6.71e-6 & 4.42 & 1.28e-4 & 3.40 & 3.76e-7 & 4.56 & 3.20e-6 & 4.51 &	2.04e-9  & 6.65 & 6.72e-8  & 5.61 \\
		4.12e-7 & 4.29 & 1.63e-5 & 3.16 & 2.38e-8 & 4.25 & 2.09e-7 & 4.20 &3.58e-11 & 6.22 & 2.28e-9  & 5.20 \\
		2.85e-8 & 4.19 & 2.11e-6 & 3.21 & 	1.26e-9 & 4.60 & 1.43e-8 & 4.21 &	6.77e-13 & 6.22 & 7.79e-11 & 5.29  \\
		1.92e-9 & 4.84 & 2.86e-7 & 3.59 & 6.92e-11 & 5.21 & 1.00e-9 & 4.77 &4.81e-14 & 4.74 & 3.00e-12 & 5.84 \\
				\bottomrule
			\end{tabular}
			\caption{Errors and estimated convergence rates (ecr) for the second test case, $k = 3,4,5$.
			}
			\label{tab:test2-k345}
		\end{table}

	\begin{figure}
		\centering
		\subfloat{\includegraphics[width=0.25\columnwidth]{esagoni8x10d.pdf}}\quad
		\subfloat{\includegraphics[width=0.245\columnwidth]{esagoni18x20d}}\quad
		\subfloat{\includegraphics[width=0.24\columnwidth]{esagoni26x30d}}
		\caption{meshes}
		\label{fig:meshes}
	\end{figure}

	\begin{figure}
		\centering
		\subfloat{\includegraphics[width=0.5\columnwidth]{ErroriH1.png}}
		\subfloat{\includegraphics[width=0.5\columnwidth]{ErroriL2.png}}\\
		\caption{Test case 1. Hexagonal meshes, H1 and L2 errors for different values of $k$}
		\label{fig:errors}
	\end{figure}

	\begin{figure}
		\centering
		\subfloat{\includegraphics[width=0.3\columnwidth]{nested_regularized_01}}
		\subfloat{\includegraphics[width=0.3\columnwidth]{nested_regularized_02}}
		\subfloat{\includegraphics[width=0.3\columnwidth]{nested_regularized_03}}
		\caption{Voronoi meshes}
		\label{fig:voronoi}
	\end{figure}

	\begin{figure}
		\centering
		\subfloat{\includegraphics[width=0.5\columnwidth]{test2h1.png}}
		\subfloat{\includegraphics[width=0.5\columnwidth]{test2l2.png}}\\
		\caption{ Test case 2. Voronoi meshes, H1 and L2 errors for different values of $k$}
		\label{fig:VorErr}
	\end{figure}

As we can see the results confirm the theoretical estimate, with the correct order of convergence for the $H^1$ norm of the error as $h$ tends to zero. We remark that, in the second Test case we seem to have some kind of superconvergence. We believe that this is rather due to the the fact that, since the grids are unstructured, there mesh size parameter is non as clearcut defined as in a structured case.

	%

	%

	%

	%

	\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


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Spider-Man 3 Actors Tom Holland and Jacob Batalon Tease Competing Titles for Sequel
Spider-Man 3 actors Tom Holland and Jacob Batalon are each teasing the title of the upcoming sequel to Spider-Man: Far From House.
Tom Holland, Spider-Man himself, began this recreation on Instagram by saying "So excited to announce the brand new Spider-Man title. Cannot wait so that you can see what we now have been as much as. Love from Atlanta." The put up additionally shared what appears to be a picture from the upcoming movie and a emblem that exposed a title of Spider-Man: Cellphone House. (Click on proper to see the brand!)
As the whole web celebrated this new title, Jacob Batalon, who performs Ned Leeds, shared his personal put up with one other picture and a unique title – Spider-Man: House-Wrecker. He additionally wrote "We're so excited to share the TITLE of our new film!! Swipe proper for the large reveal!!! Additionally, get pleasure from this unique nonetheless from the film, simply an additional present from us."
The 2 actors are clearly taking part in into the primary two movies containing the phrase residence in them – Spider-Man: Homecoming and Spider-Man: Far From House – and it is necessary to notice that whereas each stars included a picture with Zendaya in it, she has but to put up her tackle what this title could also be.
Spider-Man 3 is scheduled to be launched in theaters on December 17, 2021, and though there have been reviews that Tobey Maguire and Andrew Garfield could be returning for this new movie, Tom Holland shot these rumors down.
For extra on Spider-Man 3, try Holland's feedback on how this movie would be the "most formidable standalone superhero film ever made" and all the main points on Physician Unusual's look within the sequel to Far From House.
Have a tip for us? Wish to talk about a attainable story? Please ship an electronic mail to [email protected]
Adam Bankhurst is a information author for IGN. You'll be able to observe him on Twitter @AdamBankhurst and on Twitch.
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Guests can anticipate Italian cuisine with a focus on the home region of Piemonte, Italy. Alba is situated between the region's wine country and agricultural center, it is a classic representative of Piemonte's cuisine. Guests can anticipate an extensive salumi and formaggi counter, plus a selection of antipasti, house-made pastas, and Neapolitan pizzas.


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Q: jquery steps validation only on final submit I have a form, there are 4 steps. If I got validation error message on the first step than user able to go any of the steps and on final submit user get listing of validation error messages.
I used Jquery-Steps
Steps:

I got validation error only on steps number

I want validation like this on each step

A: You can add some html5 validation like (if_required, max_length, min_length, regex_pattern, ...) https://www.w3schools.com/js/js_validation.asp
This a good article Explaining how to use it
http://www.the-art-of-web.com/html/html5-form-validation/

redpajama_set_name: RedPajamaStackExchange


text:
Call Center:+94-773-531-693 | driversinsrilanka1@gmail.com
Diyaluma Falls
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Diyaluma Falls is 220 m high and the second highest waterfall in Sri Lanka.
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Ramboda Falls
Ramboda Falls is 109 m high and 11th highest waterfall in Sri Lanka…
Pidurangala Vihara is an ancient Buddhist temple situated in Pidurangala village.
Negombo Lagoon Tour
Cruise in the cool of the morning when the birds are most active and fishing boats are returning from the sea.
Kaudulla National Park
Kaudulla National Park is a national park on the island of Sri Lanka located 197 kilometres (122 mi) away from the largest city, Colombo.
Arugambay Beach
Hailed as one of the best surfing destinations in the world, many events are held each year under the patronage of Sri Lanka Tourism Authority to promote this beach.
Bentota Beach
Bentota, a mere 45 minutes away from Colombo along the Southern express way, hails as a premier resort town with no less than 15 Star class hotels.
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The first references of Galle port comes during King Solomons times which is believed to be "Tarish" in the Bible, and cinnamon has been the main export of Sri Lanka starting around 1400 years BC.
The famed Coral Gardens and Thelwatta Bird Sanctuary are the main attractions of Hikkaduwa which lies 98km from Colombo.
One of the finest natural deepwater harbours in the world, located in the eastern seaboard of Sri Lanka approximately 245km from Colombo.
Unawatuna Beach
Two kilometres away from Galle is the famous beach resort town of Unawatuna, popular with foreign & local tourist alike for its placid turquoise blue sea, a fine reef for snorkelling.
Dambulla Rock Cave Temple (UNESCO world heritage site) can be reached by coming from Kandy (68km) along A9 road which runs up through Dambulla and Anuradhapura, or direct from Colombo.
The Sigiriya Rock Fortress of Sri Lanka is situated in Matale district. Recorded as a UNESCO world heritage.
Another UNESCO world heritage site, during the 12th century a prince of great metal and physical vigour hailing from the Rohana.
A UNESCO world heritage site and one of the oldest continuously inhabited cities of the world, Anuradhapura, founded in 380 BC by Prince Pandukabhaya according to the historical chronicle "Mahawamsa".
Kithulgala
A sleepy little hamlet in the foothills of the central plains, with one of the main rivers of Sri Lanka "The Kelani" flowing through the town wide and snappy, boast as one of the best adventure venues in Sri Lanka.
Horton Plains
A bank of orange-yellow clouds in the East shooting beams of pale primrose light marks the beginning of another glorious day at Horton plains, where a little pond marks the entrance to the park, start early as possible and dress to avoid the morning chill…
The little England, rest & recuperate centre of the colonial British tea planters and dignitaries on this far flung corner of their empire.
Yala Park is the 2nd largest wildlife park in Sri Lanka. Jeep Safari through the park will be an unforgettable adventure.
Udawalawe National Park
The sixth largest wildlife sanctuary of Sri Lanka is located 180 km from Colombo in south-central sector of the island.
Located 30km west of Anurdhapura, and to the south of Wilpattu National Park is River Modergam Aru; to the north is River Kalay oya.
Kumana National Park
Located in the Eastern part of Yala National Park holds its own as a bird sanctuary at the same time being home to some of the larger mammals found in Yala west park.
Minneriya National Park is located 182 kms away from Colombo, in the North Central Plains of Sri Lanka. The major city closest to Minneriya National Park is Polonnaruwa.
Ella is in essence a gap formed in a mountain range, very famous hot spot with the ardent adventure tourist.
Noted as the last Kingdom of Sri Lanka has many places to visit, Temple of the Sacred Tooth Relic being the highlight.
Sri Lanka Advice
Sri Lanka Climate
Sri Lanka Day Tours
About Drivers in Sri Lanka
Our reputation and greatness is earned everyday by providing the vital value and personal attention. The objective of our company is to provide our clients with the highest standard in luxury travel arrangements.
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Copyright 2019 Drivers In Sri Lanka. All Right Reserved.


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Thursday Assembly- How the National Anthem of Pakistan came into being.
VK outlined how the national anthem was developed - seven years after the inception of Pakistan. The presentation was full of interesting facts such as the lyrics, which were written in Persian, came into effect three years after the melody was composed. The presentation explained how it was important to show respect while singing the national anthem. A PowerPoint presentation showed key people involved in the making of the national anthem and even showed 42,813 people gathered in a stadium, in 2012, to set a record of the largest number of people singing an anthem in unison.


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Athletic | Bowling | Boxing | Bridge | Country | Cricket | Cycling | Fitness | Football | Golf | Gun | Gymnastics | Nightclubs | Photography | Rugby | Sailing
Padstow Bowling Club
There are different types Padstow Bowling Clubs in Cornwall.
There are different forms of bowling:
Ten-pin bowling
Crown green bowling
Flat green bowling
The most common form of bowling now is ten pin bowling. In ten pin bowling, matches consist of each player bowling a game. Each game is divided into ten frames. A frame allows a bowler 2 chances to bang down all 10 pins. The number of pins knocked over in each frame is recorded, a running total is made underneath the particular frame score as each frame progresses, and the player with the highest score in his/her game wins the match. Scores can be greater than the actual number of pins knocked over if strikes or spares are bowled. You score 10 points when you knock over all ten pins - this is called a strike. Rather than a score of just 10 for the frame, the player's score will be 10 plus the total pins knocked down on the next two rolls in the next frame(s). A spare is scored when all pins are knocked down using the second roll in the frame. The player's score for that frame will be 10 plus the number of pins knocked down on the first roll in the next frame. A player who rolls a spare or strike in the last frame is given one (if it was a spare in the previous frame) or two more rolls (if it was a strike in the previous frame) to score additional points.
As everyday in most sports there are slang terms for various endeavours in a match. Two consecutive strikes is acknowledged as a double. Three consecutive strikes is recognized as a turkey. Four consecutive strikes is acknowledged as a four bagger. Five consecutive strikes is known as a five bagger and so on. A perfect game consists of 12 consecutive strikes, one for each frame and 2 more on the extra rolls in the 10th, which results in a score of 300. A clean game is filling every frame with either a spare or a strike.
At the same time as playing the game, situations can evolve called splits. This is when two or more of the pins are separated by missing pins which makes it harder or damn near impossible (i.e. when the two outside pins are left).
Serious club players will use their own equipment which will involve a bowling ball (usually 16lb), bowling glove which gives assistance to the wrist, sports top which may well include the name of the team on the back of the shirt and of course your own bowling shoes.
Crown Green bowling was first played in the 13th century in England. The oldest green is in Southampton - dating back to 1299.
Crown green bowling is played in pairs or teams and the idea is to win a game (usually up to 21 or 25 points) or to a precise number of ends. A draw is viable when playing ends - or depending on the rules you can play more ends until there is a positive outcome. Set scoring can also be played whenever the scoring is lower as the game is split into sets (three of five comparable to tennis). Each player will get two bowls usually - can be four each as well where the jack is thrown (usually a white ball or a small black ball with no bias) that the players have to get as close as they can too.
The bowls are biased - that is to say they have weights built into the bowls near the edge which make them curve - this is the skill in the game where you use this bias to your advantage.
This is the same as crown green bowling pretty much except the surface is flat and is more often than not indoors. The turf is usually fake grass.


Slightly late to the party, but a few weeks ago I started watching 'The Handmaid's Tale' on More 4. Of course, I now belong to the Netflix generation, which means that having to wait for a week before I can find out what happens next, is excruciatingly painful. Instant gratification – that's what the world is about isn't it? Or is that just me? Let me give you a few examples – I eat well for one day – why aren't my clothes loose yet? I go swimming for 2 weeks on the trot – why haven't I got my pre-children body back yet? I play the lottery on a regular basis – how am I not a millionaire yet? It's ok, you can go ahead and roll your eyes. First world, privileged person problems! I know.
But I digress. I did start off writing about 'The Handmaid's Tale' for a reason. I have really enjoyed watching the plight of June/Offred. I have delighted in immersing myself into her dark, dystopian world, scaring myself witless, making secret contingency plans in my head, if we had to, how would me and my nearest and dearest escape?
There are no spoilers here, read any synopsis and that will reveal more information than I'm about to. But there was one part in the episode that I watched yesterday that just didn't feel true. I'm fully aware of the irony – you're watching a story about a society that doesn't exist, and I'm annoyed about something that didn't ring true? Well, the whole point of reading or watching science fiction; dystopian societies; fantasy fiction is that even though the situations and the experiences that the humans are placed in, are beyond our reality – the very fact that they are people means that we put ourselves in their shoes and think about what we would do, and how we would behave in their position.
The series operates, as many narratives do, through switching between the present and flashbacks. I was watching a flashback, when something happened that really annoyed me. Only because I knew as a woman, that it felt contrived, insincere and not at all rooted in how people behave. Once you've read what I'm about to say, you may think it's such a tiny fault, and it is. But it has been playing on my mind since last night – so here we go.
Before everything 'went wrong', prior to June becoming Offred, she had been leading a happy life. She had a husband whom she loved deeply and a beautiful daughter (who was difficult to age; anything between 5 -8). Without revealing too much, once the excrement had hit the fan, they decided that the only course of action to take, was to run away. Whilst trying to escape, they ended up hiding in a well stocked, cabin, in the middle of nowhere. And now I shall come to the part that I found difficult to digest….
In the flashback, June and her daughter are making pancakes, a perfectly inane non event. June is laughing whilst her daughter is stirring the mixture. Simultaneously, she is talking to her daughter and constantly turning to her husband, 'making eyes' at him. Impressed with her daughter's ability to 'stir', she asks in a seductive voice, of course aimed her husband, 'I wonder where you learnt that from!' The camera shot then shows her turning towards her husband and looking at him pronouncedly, as though she would like to pounce on him there and then.
To be honest, that scene filled me with so much rage, it stopped me from being able to enjoy the rest of the episode.
Why? Such a tiny thing? Why?
It just didn't feel true. I don't know why I felt angry – perhaps I felt betrayed? Perhaps it was the realisation that the programme makers don't actually understand people in real relationships. They got the dysfunctional aspect; nailed the disturbing elements – but the real everyday part – no.
Perhaps it was because I had been putting myself in June's shoes. All the way through, I admired her refusal to submit to her fate. I approved of her silent rebellion. I totally understood her doing whatever she had to do, in order to survive because she had the bigger picture to think about. An endgame in sight.
But flirting with your husband, in front of your child, when you're trying to escape and your nerves are fraught as the whole world collapses around you? No.
Here's how I see relationships – and I'm not just talking about mine, i'm making an observation of many good relationships that I have seen over the years. When you first start out in a new relationship, there is that element of huge excitement and nerves. That's the electricity or spark that people so often refer to. Excitement charged with nerves. Those good butterflies in your stomach. Excitement and nerves. You dress well, you go over and above to make a special effort, everything is tidy, everything is new, everything smells of – I don't do this normally; I'm doing this to impress you.
The next stage is realisation – this is actually going somewhere really good. We can actually make this work long-term. And the nerves go. You are filled with excitement and hope! This person is your soul-mate. The one you're actually meant to be with. And the world is good!
The following stage – you might decide to have children. Let's do this. We love each other. Let's start a family.
This is the decision that actually means you will never be the same again. Ever.
All the way up to giving birth to my eldest daughter, I'd still held some things back from my husband. I'd still been painfully self-conscious, even though I knew I didn't need to be; I was. There were certain things, I thought, I'd never do in front of him….
Almost 7 years ago, there he was by my side, holding my hand, encouraging me, keeping me calm, holding my hair back when I needed to throw up, rubbing my back, reminding me to breathe. I remember turning to him at one point; the contractions were so painful, I felt completely out of control and very seriously announced, 'I can't do this!' I wasn't sure what my plan was, there was no contingency, smack, bang in the middle of labour – I felt that I was in too deep and needed to back out. I needed to be unconcious, wake up, and for the baby to be here. Looking back, I still find that moment hilarious. I hadn't slept for 3 nights because the contractions had just been getting stronger and more powerful. I was exhausted (or so I thought at the time). SInce then, I've realised that exhaustion is relative. Anyway, I'd had enough. The women who opted for Caesarian ops were the smart ones I felt.
Of course, he calmed me down. Reminded me about taking deep breaths. Talked about how there was no doubt that I could do this. A few hours later, he watched as I gave birth. People talk about how incredible birth is – he spoke about how it was an absolute miracle to watch. Going through it – there is no room for 'dignity', you can never be self conscious in front of your partner again. Once your partner has seen you give birth, they've seen everything.
Once you've been through that together…and you have a tiny, little life that you are responsible for; who is totally dependent on you; then you know what real exhaustion is. Most crucially though, your relationship with your partner changes – beyond recognition. It's never about what you two want ever again. It's about what the family needs. You both will always hopefully find each other attractive; but flirting in front of the children? Nah! You haven't even got the head space for that anymore.
Once you're in the family stage – your partner brings you comfort. The excitement, is almost replaced with relief that they are there. They have your back. They are the only one who understands what you are going through day in; day out. The excitement is still there – that never goes, (hopefully), but it's more about the things that people on the outside might view as mundane and quite ordinary. A good night's sleep. A couple of hours watching or doing something that you really enjoy. Having a drink or dinner with friends.
And when you watch your children doing something that is funny, or they have surprised you; or they have done something that reminds of your partner – you are not overcome with desire for your partner. You share the secret look of a co-conspirator – look at what they've done! Look – they get that from you! Look – how the heck did they learn to do that? And that develops into a feeling of relief – at least we're doing something right. The hilarious thing is that I've just described the emotions that you might feel in perfectly ordinary times. But if you were 'on the run', surely your most heightened emotion would be – I must protect my loved ones.
So that – that is why that one misplaced look of lust in 'The Handmaid's Tale' – completely flipped my lid last night.
One other emotion that I was left with though, to end on a positive note…don't take the people you love for granted. Love them everyday with as much energy and ferocity that you can muster. Love is the most powerful energy that we possess. So give as much as you possibly can.
This was a great read! I've experienced a similar type of frustration while watching "This Is Us" as I've never understood the long and "loving monologue". I love this series, but there will be a scene where a couple is in their apartment and the man makes an epic speech (just after an argument) about just how much he loves his partner. Men don't have ten minute monologues, especially after just having an argument.
I guess it's the Hollywood interpretation of how couples interact which is a far cry from the reality. Great post!
Thank you so much for reading and taking the time to comment. I really appreciate it. You've made my day!


Garbage removal vs dumpster rental in Lexington - Which is right for you?
If you own a project you're about to undertake at home, you might be wondering if it's better to hire someone to come haul off all your trash and junk for you, or in the event you need to just rent a dumpster in Lexington and load it yourself.
Renting a container is a better solution in the event you'd like the flexibility to load it on your own time and you don't mind doing it yourself to save on labor. Dumpsters also function well in the event you have at least seven cubic yards or more of debris. Roll-offs typically begin at 10 cubic yards, so if you only have 3-4 yards of waste, you're paying for a lot more dumpster than you want.
Garbage or junk removal makes more sense in case you'd like another person to load your old things. Additionally, it functions well if you'd like it to be taken away fast so it's outside of your own hair or in the event you only have a few big things; this is likely cheaper than renting a whole dumpster.
In the event you're looking to rent a dumpster in Lexington, one of your main concerns is going to be cost. There are generally two pricing options available when renting a dumpster in Lexington. Flat rate is pricing dependent on the size of the dumpster, not the quantity of stuff you place in it. Per ton pricing will bill you based on the weight you need hauled.
One kind of pricing structure isn't necessarily more expensive than the other. Should you know just how much stuff you have to throw away, you might get a better deal with per ton pricing. On the other hand, flat rate pricing can assist you to keep a limitation on costs when you're coping with unknown weights. Whether you frequently rent dumpsters for work or just need a one-time rental for a project throughout the home, you need to look around to find the best rental costs predicated on your demands.
The period of time that you need to rent a dumpster in Lexington generally depends on the type of job you are working on. A little project, for example a cellar clean out, will likely take a few days to finish while bigger projects, like building a brand new house, could take several months.
Most dumpster rental firms in Lexington are happy to adjust their schedules to suit your needs. Keep in your mind, nevertheless, that the more time you keep the dumpster, the further you will need to pay for it.
When you contact dumpster rental firms in Lexington, ask them how long their terms are. Many will allow you options such as several days or one week, however there are generally longer periods accessible, also.
Dumpster rental companies that charge flat rates are frequently the very best choices for contractors and home owners. When you utilize a business that charges a flat rate, you can anticipate at least two benefits.
Paying a flat rate generally means that you get to avoid any hidden fees and costs. Firms which don't charge flat fees may contain extra costs for basic services like dropping off the dumpster. Keep in mind, however, which you still may need to pay extra if you want to include certain items and materials in the dumpster.
Having a flat rate also means you can prepare your payment in advance. This really is especially useful when you need another person to make the payment for you. When you understand the precise amount, you can write a check ahead of time. Afterward, anyone can pay the rental business when its driver drops off the dumpster.
You can set most forms of debris into a dumpster rental in Lexington. There are, nevertheless, some exceptions. For instance, you cannot place chemicals into a dumpster. That includes motor oil, paints, solvents, automotive fluids, pesticides, and cleaning agents. Electronic Equipment and batteries are also banned. If something presents an environmental risk, you probably cannot place it in a dumpster. Get in touch with your rental company if you're unsure.
Specific kinds of acceptable debris, nevertheless, may require additional fees. In the event you plan to throw away used tires, mattresses, or appliances, you should request the rental company whether you are required to pay another fee. Adding these to your dumpster may cost anywhere from $25 to $100, depending on the item.
What if I want my dumpster in Lexington picked up early?
When you make arrangements to rent a dumpster in Lexington, part of your rental agreement comprises a given duration of time you're allowed to use the container. You normally base this time on how long you think your project might take. The larger the project, the more time you'll need the dumpster. Most dumpster rental firms in Lexington give you a speed for a specific number of days. Should you surpass that amount of days, you will pay another fee per day. If the project goes more rapidly than expected, you may be ended with the dumpster sooner than you anticipated.
If that is the case, give the dumpster business a call and they'll likely come pick your container up early; this will enable them to rent it to someone else more immediately. You normally will not get a discount on your rate should you ask for early pick up. Your rental fee includes 7 days (or whatever your term is), whether you use them all or not.
Front -load and roll off dumpsters have different layouts that make them useful in various manners. Knowing more about them will enable you to choose an option that's right for your project.
Front-load dumpsters have mechanical arms that could lift heavy items. It is a convenient alternative for jobs that include a lot of heavy items like appliances and concrete. They're also great for emptying commercial dumpsters like the type eateries use.
Rolloff dumpsters are normally the right alternative for commercial and residential jobs like fixing a roof, remodeling a cellar, or adding a room to your home. They have doors that swing open, allowing you to walk into the dumpster. In addition they have open tops that let you throw debris into the container. Rental firms will generally leave a roll off dumpster at your project place for many days or weeks. It is a convenient alternative for both small and large jobs.


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Russellville, Kentucky facts for kids
Not to be confused with the similarly-named cities of Russell in Greenup Co and Russell Springs in Russell Co.
Logan County courthouse in Russellville, Kentucky
Location in the state of Kentucky
Named for
a local landowner
10.6 sq mi (27.6 km2)
• Land
0.0 sq mi (0.0 km2)
594 ft (181 m)
CST (UTC-6)
CDT (UTC-5)
FIPS code
GNIS feature ID
www.russellvilleky.org
Russellville is a home rule-class city in Logan County, Kentucky, in the United States. It is the seat of its county. The population was 6,947 at the time of the 2010 U.S. Census.
Now used as a residence, Russellville's Southern Deposit Bank was robbed by a gang of ex-Confederates in 1868.
See also: Confederate government of Kentucky
Local historian. Alex C. Finley claimed the area was first settled by Gasper Butcher, as a frontier settlement of the Transylvania Colony, of Virginia, now present-day Russellville, Kentucky, around 1780, but others have questioned this claim. Although the area is known to have been called Big Boiling Spring, Gasper Butcher's Spring, and Butcher's Station, W.R. Jillson was unable to find written records of any habitation before 1790, when William Cook and his wife erected Cook's Cabin, accompanied by eighteen-year-old William Stewart. Also known as Cook's Station, the community – located about 1 mile (1.6 km) east of the present city – became Logan Court House when it was chosen as the seat of newly formed Logan County in 1792.
Gen. William Russell was given a 2,000-acre (810 ha) grant for his military service during the American Revolution. He donated part of this property, in 1795, as a platted section for the county seat, known as Logan Court House. The town was renamed in Russell's honor in 1798. It was formally established by the state legislature on January 15, 1810. It was incorporated as a city on February 19, 1840.
In the early 19th century, the community had leaders who were politically prominent in the state. Four homes in the city still stand which at one time were the residences of future governors of Kentucky: John Breathitt, James Morehead, John J. Crittenden, and Charles S. Morehead.
During the Civil War, the Kentucky General Assembly declared its neutrality and declined to secede with the rest of the South. Even Kentucky was a slave state and Confederate sentiment was strong in the Blue Grass and the west, the residents of the mountainous eastern section were mainly small farmers and pro-Union. In the summer of 1862, when Confederate troops had occupied the area, 116 prominent pro-Confederates from 43 counties met as the Russellville Convention and created a rival Confederate government for Kentucky. George W. Johnson was elected as the state's Confederate governor. Despite de facto Union control over the rest of Kentucky, the government was recognized and Kentucky admitted to the Confederacy. Kentucky thus became the thirteenth star on the Confederate flag.
After the war, Kentucky struggled for some years with unrest. A gang made up of the former Confederate guerrillas Cole Younger, George Shepard, and Oliver Shepard, along with Confederate veterans John Jarrett and Arthur McCoy, robbed the Nimrod Long Bank or the Southern Deposit Bank in Russellville on March 20, 1868. Brothers Frank and Jesse James, who later had their own outlaw gang, may have taken part. A Russellville bank on the city square displays a large mural painted depicting the robbery. A reenactment (called a "play on horseback") is performed annually during the city's Tobacco and Heritage Festival.
Several downtown homes have been listed on the National Register of Historic Places.
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According to the United States Census Bureau, the city has a total area of 10.6 square miles (27 km2), all land.
Historical population
1800 117 —
1810 532 354.7%
1830 1,358 —
1840 1,196 −11.9%
1850 1,272 6.4%
1870 1,843 69.2%
1990 7,454 −0.9%
Est. 2015 7,056 1.4%
U.S. Decennial Census
As of the census of 2000, there were 7,149 people, 3,064 households, and 1,973 families residing in the city. The population density was 672.1 people per square mile (259.4/km²). There were 3,458 housing units at an average density of 325.1 per square mile (125.5/km²). The racial makeup of the city was 78.64% White, 18.62% African American, 0.39% Native American, 0.36% Asian, 0.03% Pacific Islander, 0.57% from other races, and 1.38% from two or more races. Hispanic or Latino of any race were 1.58% of the population.
There were 3,064 households out of which 27.4% had children under the age of 18 living with them, 43.4% were married couples living together, 17.4% had a female householder with no husband present, and 35.6% were non-families. 32.6% of all households were made up of individuals and 15.8% had someone living alone who was 65 years of age or older. The average household size was 2.26 and the average family size was 2.84.
In the city, the population was spread out with 23.7% under the age of 18, 9.0% from 18 to 24, 25.1% from 25 to 44, 24.3% from 45 to 64, and 18.0% who were 65 years of age or older. The median age was 39 years. For every 100 females there were 82.6 males. For every 100 females age 18 and over, there were 79.9 males.
The median income for a household in the city was $25,647, and the median income for a family was $31,448. Males had a median income of $27,529 versus $20,032 for females. The per capita income for the city was $15,654. About 17.1% of families and 23.0% of the population were below the poverty line, including 37.0% of those under age 18 and 20.4% of those age 65 or over.
The climate in this area is characterized by hot, humid summers and generally mild to cool winters. According to the Köppen Climate Classification system, Russellville has a humid subtropical climate, abbreviated "Cfa" on climate maps.
Russellville, Kentucky Facts for Kids. Kiddle Encyclopedia.


Help staff to be proud of their job and more committed to their workplace by improving employee engagement, writes consultant Ruth Knight.
If employees are unhappy, careless, lethargic, in conflict or have poor communication, they could be costing an organisation thousands of dollars in missed deadlines, poor outcomes and customer service. To help staff become productive, efficient and motivated at work, they may need more engagement.
Employee engagement has been linked to improved performance and staff retention, with research suggesting improving engagement can boost productivity by up by 21 per cent, and decrease safety incidents by 48 per cent and absenteeism by 37 per cent.
That's because engaged employees have their hearts and minds aligned with both the job that they do and the organisation that they work for.
They are proud of their job and are committed to making it and the organisation successful. Engaged staff are known to be more energetic, creative, innovative and committed to their work and organisation. Most importantly, engaged employees get on with the job, so that managers can concentrate on building and growing the organisation.
If finding ways to engage staff is important to and organisation, there are many strategies that can be effective. One such strategy is to develop a performance management policy and procedure. Performance management is one of the key ways of engaging and motivating staff, even though many managers find it challenging to implement.
Blessing White's study Employee Engagement Report Research Update 2013 surveyed over 7,000 individuals from around the world and asked them, "What would most improve your performance?" .
The answers that Australian employees most selected were "development opportunities and training", "more resources" and "regular, specific feedback about how I'm doing".
The report states that this indicates managers can increase engagement by recognising employees' attitude, effort, and results, and aligning long-term employee aspirations with the organisation's talent needs of tomorrow.
To do this, organisations need to have an effective performance management process that gives employees timely and constructive feedback, asks for their opinions and ideas, and asks each employee to think about and develop strategies to improve their performance.
A new book written by Dr Tim Baker called The End of the Performance Review suggests that organisations try a new approach called the "Five Conversations Framework".
He suggests that rather than an annual performance review, managers prioritise a regular conversation with their employees every month about one of five topics: climate, strengths and talents, opportunities for growth, learning and development, and innovation and continuous improvement.
These conversations are basically a series of mini-appraisals that help a manager and employee have an ongoing discussion about strengths, achievements, goals and aspirations.
It can help managers develop a performance management process that motivates staff to be engaged and committed to the success of the organisation.
About the Author: Ruth Knight is a trainer, consultant and business advisor. She is director of Zark Consultancy helping organisations develop their organisational and workforce performance.


Scott Reiman and The Reiman Foundation is an avid supporter of programs and initiatives that foster academic excellence and provide the resources to develop our future leaders. The John Lynch Foundation does just that.
Sponsored in part by the Reiman Foundation, the John Lynch Foundation hosted its Star of the Month Celebration at Sports Authority Field at Mile High Stadium last month. The event honored some of Colorado's outstanding student athletes for their success on and off the field.
John and Linda Lynch presented more than 60 awards for the Star of the Month; Rising Star of the Month; and Coach of the Month to students and coaches from across the state. Although awards are announced throughout the year from October to April, the winners, along with their families, were formally recognized at a ceremony in the United Club Level of the stadium.
The year-long program encourages positive recognition for the student athletes and coaches who are excelling beyond their chosen sport, the classroom and the community. Every month teachers and coaches can nominate students for the awards through the John Lynch Foundation. A volunteer selection committee reviews the nominations and determines the winners.