text:
\section{Introduction and Main results}

In the following we let $k:\R^N\times\R^N\to[0,\infty]$ be such that
\begin{equation}
\label{assumptions}
\begin{split}
k(x,y)=k(y,x)&\quad\text{for all $x,y\in \R^N$, and there exists $\sigma\in(0,1)$ such that}\\
&\sup_{x\in \R^N}\int_{\R^N}\min\{1,|x-y|^{\sigma}\}k(x,y)\ dy<\infty
\end{split}
\end{equation}
and we refer to $k$ as a \textit{ nonlocal kernel (function of small order)}. Note here, that the integrability assumption on $k$ is usually done with $\sigma=2$. Here, this assumption ensures that $k$ is associated to an operator of order strictly below one. In particular, we are interested in the study of opertators with order near zero.  Motivated by some applications to nonlocal models, where a small order of the operator captures the optimal efficiency of the model \cite{PV18,AB17}, nonlocal operators with possibly differential order close to zero have been studied in linear and nonlinear integro-differential equations, see \cite{FKV13,CP17,CW18,FJW20,JSW20,HS2121,F21,TW21} and references in there.  From a stochastic point of view, general classes of nonlocal operators appear as the generator of jump processes, where the jump behavior is modelled through types of \textit{L\'evy measures} and properties of associated \textit{harmonic} functions have been studied, see \cite{M14,G14,KM17,GK18} and there references in there. In particular, operators of the form $\phi(-\Delta)$ for certain classes of functions $\phi$ are of interest from a stochastic and analytic point of view, see e.g.\cite{BJ20,BVW21} and the references in there.

In the following, we aim at investigating properties of bilinear forms and operators associated to a kernel $k$ as in \eqref{assumptions} from a variational point of view. Suitably, we give additional assumptions on $k$ focusing, however, on minimizing these and we present certain explicit examples at the end of this introduction.

To present our results,  let $\Omega\subset \R^N$ be an  open, and  $u,v\in C^{0,1}_c(\Omega)$ and consider the bilinear form
\begin{equation}\label{defi-bilinear}
b_{k,\Omega}(u,v):=\frac{1}{2}\int_{\Omega}\int_{\Omega}(u(x)-u(y))(v(x)-v(y))k(x,y)\ dxdy,\\
\end{equation}
where we also write $b_{k}(u,v):=b_{k,\R^N}(u,v)$ and $b_{k,\Omega}(u):=b_{k,\Omega}(u,u)$, $b_k(u)=b_{k}(u,u)$ resp. We denote
\begin{equation}
\label{defi-space1}
D^k(\Omega):=\{u\in L^2(\Omega)\;:\; b_{k,\Omega}(u)<\infty\},
\end{equation}
which is a Hilbert space with scalar product
$$
\langle \cdot,\cdot\rangle_{D^k(\Omega)}=\langle \cdot,\cdot \rangle_{L^2(\Omega)}+b_{k,\Omega}(u,v).
$$
Associated to $b_k$ there is a \textit{nonlocal} self-adjoint operator $I_k$ which for $u,v\in C^{0,1}_c(\Omega)$ satisfies
	\begin{equation}\label{defi-op}
	b_{k}(u,v)=\int_{\R^N}I_{k}u(x)v(x)\ dx\quad\text{and is represented by}\quad I_{k}u(x)=\int_{\R^N}(u(x)-u(y))k(x,y)\ dy,\quad x\in \Omega.
	\end{equation}

Here, the first equality can be extended, see Section \ref{preliminaries}, to functions $v\in {\mathcal V}^k(\Omega)$, the space of those functions $v:\R^N\to\R$ such that $v|_{\Omega}\in D^k(\Omega)$ and
\begin{equation}
\sup_{x\in \R^N}\int_{\R^N\setminus B_r(x)} |v(y)|k(x,y)\ dy<\infty\quad\text{for all $r>0$.}
\end{equation}
Given $f\in L^2_{loc}(\Omega)$, we then call $v\in {\mathcal V}^k(\Omega)$ a \textit{(weak) supersolution} of $I_kv=f$ in $\Omega$, if
\begin{equation}\label{supersolution}
b_k(v,u)\geq \int_{\Omega} f(x)u(x)\ dx\qquad\text{for all $u\in C^{\infty}_c(\Omega)$.}
\end{equation}
In this situation, we also say that $v$ satisfies in weak sense $I_kv\geq f$ in $\Omega$. Similarly, we define \textit{subsolutions} and \textit{solutions}.\\
We emphasize that this definition of supersolution is larger than the one considered in \cite{JW19}. Using a density result we then can extend the weak maximum principles presented in \cite{JW19} as follows.
\begin{prop}[Weak maximum principle]
	\label{wmp}
	Define $j:\R^N\to[0,\infty]$ as the symmetric rearrangement of $k$, that is
	\begin{equation}
	\label{defi-j}
	j(z):=\essinf\{k(x,x\pm z)\;:\; x\in \R^N\}\quad\text{for $z\in \R^N$,}
	\end{equation}
	and assume that
	\begin{equation}
	\label{assumption2a}
	\text{$j$ does not vanish identically on $B_r(0)$ for any $r>0$.}
    \end{equation}
    Let $\Omega\subset \R^N$ open, $c\in L^{\infty}_{loc}(\Omega)$, and assume either
	\begin{enumerate}
		\item $c\leq 0$ or
		\item $\Omega$ and $c$ are such that $\|c^{+}\|_{L^{\infty}(\Omega)}<\inf_{x\in \Omega} \int_{\R^N\setminus \Omega}k(x,y)\ dy$.
	\end{enumerate}
	If $u\in {\mathcal V}^k(\Omega)$ satisfies in weak sense
	$$
	I_ku\geq c(x)u\quad\text{ in $\Omega$, $u\geq 0$ almost everywhere in $\R^{N}\setminus \Omega$, and}\quad \liminf\limits_{|x|\to\infty}u(x)\geq 0,
	$$
	then $u\geq 0$ almost everywhere in $\R^N$.
\end{prop}
We note that assumption \ref{assumption2a} on the function $j$ defined in \eqref{defi-j} implies the positivity of the first eigenvalue of $I_k$ in bounded (or thin) sets $\Omega\subset \R^N$: The operator $I_k$ on an open set $\Omega\subset \R^N$ can be associated to a form domain given by the space
\begin{equation}\label{defi-space2}
{\mathcal D}^k(\Omega)=\{u\in D^k(\R^N)\;:\; 1_{\R^N\setminus\Omega}u\equiv0\}.
\end{equation}
Clearly, ${\mathcal D}^k(\Omega)=D^k(\R^N)$ and also the space ${\mathcal D}^k(\Omega)$ is a Hilbert space with scalar product
$$
\langle \cdot,\cdot\rangle_{{\mathcal D}^k(\Omega)}=\langle \cdot,\cdot \rangle_{L^2(\Omega)}+b_{k}(u,v).
$$
Then the first (variational) eigenvalue of $I_k$ is given for $\Omega\subset \R^N$ open by
\begin{equation}
\label{first eigenvalue}
\Lambda_1(\Omega):=\inf_{\substack{u\in {\mathcal D}^k(\Omega)\\ u\neq 0}}\frac{b_{k}(u)}{\|u\|_{L^2(\Omega)}^2}\in[0,\infty).
\end{equation}
Then, if \eqref{assumption2a} is satisfied and $\Omega$ is contained (after a rotation) in a strip $(-a,a)\times \R^{N-1}$ for some $a>0$, then $\Lambda_1(\Omega)>0$ (see \cite{FKV13,JW20}).

In the following, we assume the stronger assumption
\begin{equation}
\label{assumption2}
\text{The function $j$ given in \eqref{defi-j} satisfies}\quad \int_{\R^N}j(z)\ dz=\infty
\end{equation}
and conclude the
\begin{prop}[Strong maximum principle]
	\label{smp}
	Assume $k$ satisfies additionally \eqref{assumption2}. Let $\Omega\subset \R^N$ open and $c\in L^{\infty}_{loc}(\Omega)$ with $\|c^+\|_{L^{\infty}(\Omega)}<\infty$. Moreover, let $u\in {\mathcal V}^k(\Omega)$, $u\geq 0$ satisfy in weak sense $I_ku\geq c(x) u$ in $\Omega$. If
	\begin{enumerate}
		\item If $\Omega$ is connected, then either $u\equiv 0$ in $\Omega$ or $\essinf_Ku>0$ for any $K\subset\subset \Omega$.
		\item  $j$ given in \eqref{defi-j} satisfies $\essinf_{B_r(0)}j>0$ for any $r>0$, then either $u\equiv 0$ in $\R^N$ or $\essinf_Ku>0$ for any $K\subset\subset \Omega$.
	\end{enumerate}
\end{prop}
Clearly, if $\Lambda_1(\Omega)$ is positive, then $b_k$ denotes an equivalent scalar product on ${\mathcal D}^k(\Omega)$ and thus for any $f\in L^2(\Omega)$ there is a unique solution $u\in {\mathcal D}^k(\Omega)$ with $I_ku=f$ in $\Omega$. The main results of this article then are concerned with the regularity of $u$, if $f$ has a certain regularity.

We begin with a boundedness result for solutions.
\begin{thm}
	\label{thm-bdd3}
		Assume $k$ satisfies \eqref{assumption2} and is such that
\begin{equation}\label{assumption2c}
	\sup_{x\in \R^N}\int_{K\setminus B_{\epsilon}(x)}k(x,y)^2\ dy<\infty\quad\text{for all $K\subset\subset \R^N$ and $\epsilon>0$.}
\end{equation}
	Let $\Omega\subset \R^N$ be an open set and $f\in L^{\infty}(\Omega)$, $h\in L^1(\R^N)\cap L^{2}(\R^N)$, and let $u\in {\mathcal V}^k(\Omega)$ satisfy in weak sense $I_ku= \lambda u+h\ast u+f$ in $\Omega$ for some $\lambda>0$. If there is $\Omega'\subset\subset\Omega$ such that $u\in L^{\infty}(\R^N\setminus \Omega')$, then $u\in L^{\infty}(\R^N)$ and there is $C=C(\Omega,\Omega',k,\lambda,h)>0$ such that
	$$
	\|u\|_{L^{\infty}(\Omega)}\leq C\Big(\|f\|_{L^{\infty}(\Omega)}+\|u\|_{L^{2}(\Omega')}+\|u\|_{L^{\infty}(\R^N\setminus \Omega')}\Big).
	$$
\end{thm}

\begin{rem}\label{different-boundary-cond}\hspace{1em}
\begin{enumerate}
\item Indeed, Theorem \ref{thm-bdd3} is a consequence of a slightly more general result stated in Theorem \ref{bdd-1} in Section \ref{sec:bdd} below, which concerns functions $u:\R^N\to\R$, which are in a certain sense locally in ${\mathcal V}^k(\Omega)$.
\item If $u\in {\mathcal D}^k(\Omega)$, where $\Omega\subset \R^N$ is open and bounded with Lipschitz boundary, then in the above, $\Omega'=\Omega$ can be chosen. Indeed, here the regularity assumption on $\partial\Omega$ is only needed since our definition of weak solutions uses test-function in $C^{\infty}_c(\Omega)$. Replacing this directly with ${\mathcal D}^k(\Omega)$, the regularity assumption is not needed.
\end{enumerate}
\end{rem}

As shown in \cite{JW20}, under assumption \ref{assumption2}, it follows also that the embedding ${\mathcal D}^k(\Omega)\hookrightarrow L^2(\Omega)$ is indeed compact, if $\Omega\subset \R^N$ is bounded. Whence, From Theorem \ref{thm-bdd3} we have the following Corollary.

\begin{cor}
	\label{main-thm-eigen}
	Let $k$ satisfy additionally \eqref{assumption2} and \eqref{assumption2c} and let $\Omega\subset \R^N$ open and bounded. Then there is a sequence of eigenvalues $(\Lambda_n(\Omega))_n$ of $I_k$ with
	$$
	0<\Lambda_1(\Omega)<\Lambda_2(\Omega)\leq \ldots\leq \Lambda_n(\Omega)\to \infty\quad\text{for $n\to\infty$,}
	$$
	that is, $\Lambda_1(\Omega)$ is simple and the first normalized eigenfunction $\phi_1$ of $I_k$ can be chosen to be positive in the sense that
	$$
	\underset{K}{\essinf}\, \phi_1>0\quad\text{for all $K\subset\subset \Omega$.}
	$$
	Moreover, any eigenfunction of $I_k$ is bounded. To be precise, given $\lambda>0$ and $u\in {\mathcal D}^k(\Omega)$ such that $I_ku=\lambda u$, then there is $C=C(N,\Omega,k,\lambda)>0$ such that
	$$
	\|u\|_{L^{\infty}(\Omega)}\leq C\|u\|_{L^2(\Omega)}.
	$$
\end{cor}

In the particular case, where the kernel is translation invariant, that is, there is a function $J:\R^N\to[0,\infty]$ with $k(x,y)=J(x-y)$ for $x,y\in \R^N$, we are also able to recover differentiability of a solution $u$ to the problem $I_ku=f$, if $f$ and $J$ satisfy certain regularity properties. Our result is as follows.

\begin{thm}
	\label{main-thm-bdd}
	Assume $k$ satisfies \eqref{assumption2} and let $\Omega\subset \R^N$ open and bounded with Lipschitz boundary. Then for any $f\in L^2(\Omega)$ there is a unique solution $u\in {\mathcal D}^k(\Omega)$ of $Iu=f$. Moreover, if $k$ satisfies additionally \eqref{assumption2c} and $f\in L^{\infty}(\Omega)$, then $u\in L^{\infty}(\Omega)$ and there is $C=C(N,\Omega,k)>0$ such that
	$$
	\|u\|_{L^{\infty}(\Omega)}\leq C\|f\|_{L^{\infty}(\Omega)}.
	$$
	Furthermore, if $k$ satisfies \eqref{assumptions} with $\sigma<\frac{1}{2}$ and
	\begin{equation}
	\label{assumption3}
	\begin{split}
	\text{there is }\, J:\,&\R^N\to[0,\infty] \text{ such that $k(x,y)=J(x-y)$, where $J$ satisfies for some $m\in \N\cup\{\infty\}$:}\\
	(A)_m&\quad \text{It holds $J\in W^{n,1}(\R^N\setminus B_{\epsilon}(0))$ for any $\epsilon>0$, $n\leq 2m$ and, for some $C_J>0$, it holds}\\
	&\quad \text{$|\nabla J(z)|\leq C_J|z|^{-1-\sigma-N}$ for $0<|z|\leq 3$ with $\sigma$ as in \eqref{assumptions},}
	\end{split}
		\end{equation}
		then, if $m\in \N$ and $f\in C^{2m}(\overline{\Omega})$, we have $\partial^{\beta}u\in L^2_{loc}(\Omega)$ for all $\beta\in \N_0^N$, $|\beta|\leq m$, $u\in H^m_{loc}(\Omega)$, and for every $\Omega'\subset\subset\Omega$ there is $C=C(N,\Omega,\Omega',k,\beta)>0$ such that
		$$
		\|\partial^{\beta}u\|_{L^{2}(\Omega')}\leq C\|f\|_{C^{2m}(\Omega)}.
		$$
		In particular, for $m=\infty$, we have $u\in C^{\infty}(\Omega)$.
\end{thm}

\begin{rem}
	We note that Theorem \ref{main-thm-bdd} is a particular case where the general version, holds for functions, which are in a certain sense locally in ${\mathcal V}^k(\Omega)$.
\end{rem}

\begin{cor}
	\label{main-thm-eigen2}
	If in the situation of Corollary \ref{main-thm-eigen} the kernel $k$ additionally satisfies \eqref{assumption3} with $m=\infty$, then every function $u\in {\mathcal D}^k(\Omega)$ satisfying $I_ku=\lambda u$ in $\Omega$ for some $\lambda\in \R$ also belongs to $C^{\infty}(\Omega)$.
\end{cor}
 It is worthy to mentioned our approaches  only exploits the variational structure of the problem and uses purely analytic properties of the operator. Using a probabilistic and potential theoretic approach, a local smoothness of bounded harmonic solutions solving in a certain very weak sense $I_ku=0$ in $\Omega$, have been obtained in \cite[Theorem 1.7]{GK18} for radial kernel functions using the same regularity $(A)_m$ (see also \cite{M14,G14}). See also \cite{GKL21} for related regularity properties of solutions.

\subsection{Examples}
We close this introduction with a class of operators covered in the above discussion:
\begin{enumerate}
\item As introduced in \cite{CW18,JSW20,FJW20} the \textit{logarithmic Laplacian}
\begin{equation}\label{eq:log-laplace}
L_{\text{\tiny $\Delta \,$}} \phi(x)=c_NP.V.\int_{B_1(0)}\frac{\phi(x)-\phi(x+y)}{|y|^{N}}\ dy - c_N\int_{\R^N\setminus B_1(0)}\frac{\phi(x+y)}{|y|^{N}}\ dy+\rho_N\phi(x),
\end{equation}
appears as the operator with Fourier-symbol $-2\ln(|\cdot|)$ and can be seen as the formal derivative in $s$ of $(-\Delta)^s$ at $s=0$. Here
\begin{equation}
  \label{eq:def-c-n-rho-n}
c_N=\frac{\Gamma(\frac{N}{2})}{\pi^{N/2}}= \frac{2}{|S^{N-1}|} \qquad \text{and}\qquad \rho_N :=2\ln(2)+\psi(\frac{N}{2}) - \gamma
\end{equation}
where $\psi:= \frac{\Gamma'}{\Gamma}$ denotes the digamma function and $\gamma:= -\psi(1)=-\Gamma'(1)$ is the Euler-Mascheroni constant. With $k(x,y)=c_N1_{B_1(0)}(x-y)|x-y|^{-N}$, $h=-c_N1_{\R^N\setminus B_1(0)}|y|^{-N}$ and $\lambda=\rho_N$, can be studied using Theorem \ref{thm-bdd3}. Moreover, also the generalizations of Theorem \ref{main-thm-bdd} considered in Section \ref{sec:bdd} and the regularity statements in Section \ref{sec:reg} cover this operator ---where in the latter situation a localization procedure is needed.
\item The logarithmic Schr\"odinger operator $(I-\Delta)^{\log}$ as in \cite{F21} is an integro-differential  operator with Fourier-symbol $\log(1+|\cdot|^2)$  and also appears  as the formal derivative in $s$ of the relativistic Schr\"odinger operator $(I-\Delta)^s$ at $s=0$,
\[
(I-\Delta)^{\log}u(x)= d_{N}P.V.\int_{\R^N}\frac{u(x)-u(x+y)}{|y|^{N}}\omega(|y|)\ dy,
\]
 where $d_N=\pi^{-\frac{N}{2}}$, $\omega(r)=2^{1-\frac{N}{2}}r^{\frac{N}{2}}K_{\frac{N}{2}}(r)$ and $K_{\nu}$ is  the modified Bessel function of the second kind with index $\nu$. More generally, operators with symbol $\log(1+|\cdot|^{\beta})$ for some $\beta\in(0,2]$ are studied in \cite{KM14}.
\item Finally, also nonradial kernels of the type considered in \cite{JW19} satisfy in particular the assumptions \eqref{assumptions} and \eqref{assumption2}. See also also \cite{M14,G14,KM14} and references in there.
\end{enumerate}

The paper is organized as follows. In Section \ref{preliminaries} we collect some general results concerning the spaces used in this paper and resulting definitions of weak sub- and supersolutions. Section \ref{sec:dense} is devoted to show several density results of $C^{\infty}_c(\Omega)$ in $D^k(\Omega)$ and in ${\mathcal D}^k(\Omega)$, which then is used to show the Propositions \ref{wmp} and \ref{smp}. In Section \ref{sec:bdd} we present a general approach to show boundedness of solutions and in Section \ref{sec:reg} we give the proof of an interior $H^1$-regularity of solutions and from this deduce the interior $C^{\infty}$-regularity of solutions to conclude the proof of Theorem \ref{main-thm-bdd}.

\medskip

\paragraph{Notation} In the remainder of the paper, we use the following notation. Let $U,V\subset \R^N$ be nonempty measurable sets, $x \in \R^N$ and $r>0$. We denote by $1_U: \R^N \to \R$ the characteristic function, $|U|$ the Lebesgue measure, and $\diam(U)$ the diameter of $U$. The notation $V \subset \subset U$ means that $\overline V$ is compact and contained in the interior of $U$. The distance between $V$ and $U$ is given by $\dist(V,U):= \inf\{|x-y|\::\: x \in V,\, y \in U\}$. Note that this notation does {\em not} stand for the usual Hausdorff distance.  If $V= \{x\}$ we simply write $\dist(x,U)$. We let $B_r(U):=\{x\in \R^N\;:\; \dist(x,U)<r\}$, so that $B_r(x):=B_r(\{x\})$ is the open ball centered at $x$ with radius $r$. We also put $B:=B_1(0)$ and $\omega_N:=|B|$. Finally, given a function $u: U\to \R$, $U \subset \R^N$, we let $u^+:= \max\{u,0\}$ and $u^-:=-\min\{u,0\}$ denote the positive and negative part of $u$, and we write $\supp\ u$ for the support of $u$ given as the closure in $\R^N$ of the set ${\{ x\in U\;:\; u(x)\neq 0\}}$.\\

\textbf{Acknowledgements.}~
This work is supported by DAAD and BMBF (Germany) within the project 57385104. The authors thank Mouhamed Moustapha Fall and Tobias Weth for helpful discussions.

\section{Preliminaries}\label{preliminaries}

Recall for $\Omega\subset \R^N$ open, the definitions

\begin{align*}
b_{k,\Omega}(u,v)&:=\frac{1}{2}\int_{\Omega}\int_{\Omega}(u(x)-u(y))(v(x)-v(y))k(x,y)\ dxdy,\\
\kappa_{k,\Omega}(x)&:=\int_{\R^N\setminus \Omega}k(x,y)\ dy\in[0,\infty] \quad\text{for $x\in \R^N$, and}\\
K_{k,\Omega}(u,v)&:=\int_{\Omega} u(x)v(x)\kappa_{k,\Omega}(x)\ dx,
\end{align*}

where if $u=v$ we put
$$
b_{k,\Omega}(u):=b_{k,\Omega}(u,u)\quad\text{and}\quad K_{k,\Omega}(u):=K_{k,\Omega}(u,u).
$$
Note that we have for any fixed $x\in \Omega$ that $\kappa_{k,\Omega}(x)<\infty$ by \eqref{assumptions}.
Moreover, we consider the function spaces
\begin{align*}
D^k(\Omega)&:=\Big\{u\in L^2(\Omega)\;:\; b_{k,\Omega}(u)<\infty\Big\},\\
D^k_{loc}(\Omega)&:=\Big\{u:\Omega\to\R\;:\; u|_{\Omega'}\in D^k(\Omega') \text{ for all $\Omega'\subset\subset\Omega$}\Big\},\\
{\mathcal D}^k(\Omega)&:=\Big\{u\in D^k(\R^N)\;:\; u=0 \ \text{on $\R^{N}\setminus \Omega$}\Big\},\\
{\mathcal V}^k(\Omega)&:=\Big\{u:\R^N\to\R\;:\; u|_{\Omega}\in D^k(\Omega)\text{ and, for all $r>0$, }\sup_{x\in \R^N}\int_{\R^N\setminus B_r(x)}|u(y)|k(x,y)\ dy<\infty\Big\},\quad\text{and}\\
{\mathcal V}_{loc}^k(\Omega)&:=\Big\{u:\R^N\to\R\;:\;  u|_{\Omega'}\in {\mathcal V}^k(\Omega') \text{ for all $\Omega'\subset\subset\Omega$}\Big\}.
\end{align*}

\begin{lemma}
\label{properties2}
Let $U\subset \Omega\subset \R^N$ open and $u:\R^N\to\R$. Then the following hold:
\begin{enumerate}
\item $u\in {\mathcal D}^k(\Omega)\ \Rightarrow\ u|_{\Omega}\in D^k(\Omega)\ \Rightarrow\ u|_{\Omega}\in D^k_{loc}(\Omega)$.
\item ${\mathcal D}^k(U)\subset {\mathcal D}^k(\Omega)\subset {\mathcal V}^k(\Omega)\subset {\mathcal V}^k(U)\subset {\mathcal V}^k_{loc}(U)$.
\end{enumerate}
\end{lemma}
\begin{proof}
This follows immediately from the definitions (see also \cite[Section 3]{JW19}).
\end{proof}

\begin{lemma}[see Proposition 3.3 in \cite{JW19} or Proposition 1.7 in \cite{JW20}]
\label{poincare}
For $\Omega\subset \R^N$ open let $\Lambda_1(\Omega)$ be given as in \eqref{first eigenvalue} and let
$$
\lambda(r)=\inf\{\Lambda_1(\Omega)\;:\;\text{$\Omega\subset \R^N$ open with $|\Omega|=r$ }\}.
$$
Then $\lim\limits_{r\to\infty}\lambda(r)\geq \int_{\R^N}j(z)\ dz$ with $j(z):=\essinf\{k(x,x\pm z)\;:\; z\in \R^N\}$ as in \eqref{defi-j}.
\end{lemma}

\begin{lemma}
\label{properties1}
Let $\Omega\subset \R^N$ open and let $X$ be any of the above function spaces. Then the following holds:
\begin{enumerate}
\item $b_{k,\Omega}$ is a bilinear form and in particular we have $b_{k,\Omega}(u,v)\leq b_{k,\Omega}^{1/2}(u)b_{k,\Omega}^{1/2}(v)$. Moreover, $D^k(\Omega)$ and ${\mathcal D}^k(\Omega)$ are Hilbert spaces with scalar products
\begin{align*}
\langle u,v\rangle_{D^k(\Omega)}&=\langle u,v\rangle_{L^2(\Omega)}+b_{k,\Omega}(u,v),\\
\langle u,v\rangle_{{\mathcal D}^k(\Omega)}&=\langle u,v\rangle_{L^2(\Omega)}+b_{k,\R^N}(u,v).
\end{align*}
\item If $u\in X$, then $u^{\pm},|u|\in X$ and we have $b_{k,\Omega'}(u^+,u^-)\leq 0$ for all $\Omega'\subset \Omega$ with $b_{k,\Omega'}(u)<\infty$.
\item If $g\in C^{0,1}(\R^N)$, $u\in X$, then $g\circ u\in X$.
\item $C^{0,1}_c(\Omega)\subset X$.
\item $\phi\in C^{0,1}_c(\Omega)$, $u\in X$, then $\phi u\in {\mathcal D}^k(\Omega)$, where if necessary we extend $u$ trivially to a function on $\R^N$. Moreover, there is $C=C(N,k,\|\phi\|_{C^{0,1}(\Omega)})>0$ such that
$$
b_{k,\R^N}(\phi u)\leq C\Big(\|u\|_{L^2(\Omega')}^2+b_{k,\Omega'}(u)\Big)
$$
for any $\Omega'\subset\Omega$ with $\supp\,\phi\subset\subset \Omega'$.
\end{enumerate}
\end{lemma}
\begin{proof}
Theses statements follow directly from the definition (c.f. \cite[Section 3]{JW19}). To be precise in the last part, let $\phi\in C^{0,1}_c(\Omega)$ and fix $L:=\|\phi\|_{C^{0,1}(\Omega)}$. That is, we have
\[
|\phi(x)|\leq L\quad\text{and}\quad |\phi(x)-\phi(y)|\leq L|x-y|.
\]
Then using the inequality for $x,y\in \R^N$
\[
|\phi(x)u(x)-\phi(y)u(y)|^2\leq 2|\phi(x)-\phi(y)|^2|u(x)|^2+2|\phi(y)|^2|u(x)-u(y)|^2
\]
we find by the assumptions \eqref{assumptions}
\begin{align*}
b_{k,\R^N}(\phi u)&\leq b_{k,\Omega'}(\phi u)+L^2\int_{\supp\,\phi}|u(x)|^2\kappa_{k,\Omega'}(x)\ dx\\
&\leq 2L^2\int_{\Omega'}\int_{\Omega'}|u(x)|^2|x-y|^2k(x,y)\ dy\ dx+2L^2b_{k,\Omega'}(u)+L\sup_{x\in \supp\,\phi}\kappa_{k,\Omega'}(x)\|u\|_{L^2(\Omega')}^2\\
&\leq 2L^2\Big(\sup_{x\in \Omega'}\int_{\Omega'} |x-y|^2k(x,y)\ dy+\sup_{x\in \supp\,\phi}\kappa_{k,\Omega'}(x)\Big)|\Omega'|\|u\|_{L^2(\Omega')}^2+2L^2b_{k,\Omega'}(u)<\infty.
\end{align*}
\end{proof}

\begin{rem}
\label{bilinearform rewrite}
\begin{enumerate}
\item Note that for $u,v\in {\mathcal D}^k(\Omega)$ we have
\[
b_{k}(u,v)=b_{k,\R^N}(u,v)=b_{k,\Omega}(u,v)+K_{k,\Omega}(u,v).
\]
\item It follows in particular that there is a nonnegative self-adjoint operator $I_k$ associated to $b_{k,\R^{N}}=b_k$ as mentioned in the introduction.
\end{enumerate}
\end{rem}

\begin{lemma}
\label{well-defined}
Let $\Omega\subset \R^N$ open, $u\in {\mathcal V}_{loc}^k(\Omega)$. Then $b_{k,\R^N}(u,\phi)$ is well-defined for any $\phi \in C^{\infty}_c(\Omega)$.
\end{lemma}
\begin{proof}
Let $\phi\in C^{\infty}_c(\Omega)$ and fix $U\subset \subset\Omega$ such that $\supp\ \phi\subset\subset U$. Then with the symmetry of $k$
\begin{align*}
|b_{k}(u,\phi)|&\leq |b_{k,U}(u,\phi)|+\int_{U} |\phi(x)|\int_{\R^N\setminus U} |u(x)-u(y)|k(x,y)\ dydx\\
&\leq b_{k,U}^{1/2}(u)b_{k,U}^{1/2}(\phi)+\int_{\supp\,\phi}|\phi(x)| \ dx\sup_{x\in \R^N}\int_{\R^N\setminus B_{\epsilon}(x)}|u(y)|k(x,y)\ dy\\
&\qquad\qquad\qquad +\int_{\supp\,\phi}|\phi(x)u(x)| \ dx \sup_{x\in \R^N}\int_{\R^N\setminus B_{\epsilon}(x)}k(x,y)\ dy<\infty,
\end{align*}
where $\epsilon=\dist(\supp\ \phi,\R^N\setminus U)>0$.
\end{proof}

\begin{defi}
	\label{defi-weak}
Let $\Omega\subset \R^N$ open and $f\in L^1_{loc}(\Omega)$. Then $u\in {\mathcal V}_{loc}^k(\Omega)$ is called a \textit{weak supersolution} of $I_ku=f$ in $\Omega$, if
\[
b_{k,\R^N}(u,\phi)\geq \int_{\Omega} f(x)\phi(x)\ dx\quad\text{for all nonnegative $\phi\in C^{\infty}_c(\Omega)$.}
\]
We also say that \textit{$u$ satisfies $I_ku\geq f$ weakly in $\Omega$}.\\
Similarly, we define \textit{weak subsolutions} and \textit{solutions}.
\end{defi}

\begin{rem}
	\begin{enumerate}
		\item We note that by Assumption \ref{assumptions} it follows that for any function $u\in {\mathcal V}^k_{loc}(\Omega)$ with $u|_{\Omega}\in C^{0,1}(\Omega)$ for $\Omega\subset \R^N$ open we have $I_ku|_U\in L^{\infty}(U)$ for any $U\subset\subset \Omega$ and
$$
	I_ku(x)=\int_{\R^N}(u(x)-u(y))k(x,y)\ dy\quad\text{for $x\in \Omega$.}
$$
This follows easily similarly to the proof of the statements in Lemma \ref{properties1}.
\item If $u\in {\mathcal V}^k(\Omega)$, then indeed also $b_k(u,\phi)$ is well-defined for all $\phi\in {\mathcal D}^k(\Omega)$. Whence also $b_k$ is well defined on ${\mathcal V}^k_{loc}(\Omega)\times {\mathcal D}^k(U)$ for all $U\subset\subset \Omega$. In some of our results the statements need a Lipschitz-boundary of $\Omega$, which comes into play due to approximation with $C^{\infty}_c(\Omega)$-functions (see Section \ref{sec:dense} below). However, this can be weakened, if $u\in {\mathcal V}^k(\Omega)$ and the space of test-functions is adjusted.
\end{enumerate}
\end{rem}

\begin{lemma}
	\label{localization}
	Let $\Omega\subset \R^N$ open. Let $\Omega_1\subset\subset\Omega_2\subset\subset \Omega_3\subset\subset \Omega$. Let $\eta\in C^{0,1}_c(\R^N)$ such that $0\leq \eta\leq 1$ in $\R^N$ and we have
	$$
	\eta=1\quad\text{in $\Omega_{2}\quad$ and }\quad \eta=0\quad\text{in $\R^N\setminus \Omega_3$.}
	$$
	Let $f\in L^1_{loc}(\Omega)$ and let $u\in {\mathcal V}^k_{loc}(\Omega)$ satisfy in weak sense $Iu\geq f$ in $\Omega$. Then the function $v=\eta u\in {\mathcal D}^k(\Omega_3)$ satisfies in weak sense $Iv\geq f+g_{\eta,u}(x)$ in $\Omega_1$, where
	$$
	g_{\eta,u}(x)=\int_{\R^N\setminus \Omega_2}(1-\eta(y))u(y)k(x,y)\ dy\quad\text{for $x\in \Omega_1$.}
	$$
\end{lemma}
\begin{proof}
	The fact, that $v\in {\mathcal D}^k(\Omega_3)$ follows from Lemma \ref{properties1}. Let $\phi\in C^{\infty}_c(\Omega_1)$, then
	$$
	\int_{\R^N}v(x)I_k\phi(x)\ dx\geq \int_{\R^N}f(x)\phi(x)\ dx-\int_{\R^N}(1-\eta(x))u(x)I_k\phi(x)\ dx.
	$$
	Here, since $(1-\eta)u\equiv 0$ on $\Omega_2$, we have
	\begin{align*}
	\int_{\R^N}(1-\eta(x))u(x)I_k\phi(x)\ dx&=\int_{\R^N}\phi(x) [I_k(1-\eta)u](x)\ dx\\
	&=-\int_{\Omega_1}\phi(x) \int_{\R^N\setminus \Omega_2}(1-\eta(y))u(y)k(x,y)\ dy\ dx.
	\end{align*}
	Thus the claim follows.
\end{proof}

\begin{rem}
	The same result as in Lemma \ref{localization} also holds if ``$\geq$'' in the solution type is replaced by ``$\leq$'' or ``$=$''.
\end{rem}

In the following, it is useful to understand functions $u\in D^k(\Omega)$ satisfying $b_{k,\Omega}(u)=0$.

\begin{prop}
	\label{constant}
	Assume additionally \ref{assumption2} and let $\Omega\subset \R^N$ open and bounded and let $u\in D^k(\Omega)$ such that $b_{k,\Omega}(u)=0$. Then $u$ is constant.
\end{prop}
\begin{proof}
	Let $x_0\in \Omega$ and fix $r>0$ such that $B_{2r}(x_0)\subset \Omega$. Denote $q(z):=\min\{c,j(z)\}1_{B_r(0)}(z)$, where we may fix $c>0$ such that $|\{q>0\}|>0$ due to Assumption \eqref{assumption2}. Then by Lemma \ref{iterationlemma} we have
	\[
	0=2b_{k,\Omega}(u)\geq 2b_{q,\Omega}(u)\geq \frac{1}{2\|q\|_{L^1(\R^N)}} b_{q\ast q,B_{r}(x_0)}(u),
	\]
	where $a\ast b=\int_{\R^N}a(\cdot-y)b(y)\ dy$ denotes as usual the convolution.	Note that since $q\in L^1(\R^N)\cap L^{\infty}(\R^N)$ with $q=0$ on $\R^N\setminus B_r(0)$, it follows that $q\ast q\in C(\R^N)$ with support in $B_{2r}(0)$ and we have
	\[
	q\ast q(0)=\int_{\R^N}q(z)^2\ dz>0
	\]
	by Assumption \eqref{assumption2}. Hence there is $R>0$ with $q\ast q\geq \epsilon$ for some $\epsilon>0$ and thus we have
	\[
	0=b_{q\ast q,B_{r}(x_0)}(u)\geq b_{q\ast q,B_{\rho}(x_0)}(u)\geq \frac{\epsilon}{2}\int_{B_{\rho}(x_0)}\int_{B_{\rho}(x_0)}(u(x)-u(y))^2\ dxdy.
	\]
	for any $\rho\in(0,\frac{R}{2}]$. But then $u(x)=u(y)$ for almost every $x,y\in B_{R/2}(x_0)$ so that $u$ is constant a.e. in $B_{\rho}(x_0)$. Since $b_{k,\Omega}(u)=b_{k,\Omega}(u-m)$ for any $m\in \R$, we may next assume that $u=0$ in $B_{R/2}(x_0)$ and show that indeed we have $u=0$ a.e. in $\Omega$. Denote by $W$ the set of points $x\in \Omega$ such that there is $r>0$ with $u=0$ a.e. in $B_r(x)$. By definition $W$ is open and the above shows that $W$ is nonempty. Next, let $(x_n)_n\subset W$ be a sequence with $x_n\to x\in \Omega$ for $n\to\infty$. Then there is $r_x>0$ such that $B_{4r_x}(x)\subset \Omega$ and we can find $n_0\in \N$ such that $x\in B_{r_x}(x_n)\subset B_{2r_x}(x_n)\subset \Omega$ for $n\geq n_0$. Repeating the above argument, it follows that $u$ must be zero in $B_{r_x}(x_n)$ and thus $x\in W$. Hence, $W$ is relatively open and closed in $\Omega$ and since $W$ is nonempty, we have $W=\Omega$. That is $u=0$ in $\Omega$.
\end{proof}

\subsection{On Sobolev and Nikol'skii spaces}
We recall here the notations and properties of Sobolev and Nikol'skii spaces as introduced in \cite{T95,C17}. In the following, let $p\in[1,\infty)$ and $\Omega\subset \R^N$ open.

\subsubsection{Sobolev spaces} If $k\in \N_0$, we set as usual
$$
W^{k,p}(\Omega):=\Big\{u\in L^p(\Omega)\;:\; \text{ $\partial^{\alpha}u$ exists for all $\alpha\in \N_0^{n}$, $|\alpha|\leq k$ and belongs to $L^p(\Omega)$ }\Big\}
$$
for the Banach space of $k$-times (weakly) differentialable functions in $L^p(\Omega)$. Moreover, as usual, for $\sigma\in(0,1)$, $p\in[1,\infty)$ we set
$$
W^{\sigma,p}(\Omega):=\Big\{u\in L^p(\Omega)\;:\; \frac{u(x)-u(y)}{|x-y|^{\frac{n}{p}+\sigma}}\in L^{p}(\Omega\times \Omega)\Big\}.
$$
With the norm
$$
\|u\|_{W^{\sigma,p}(\Omega)}=\|u\|_{L^p(\Omega)}^p+[u]_{W^{\sigma,p}(\Omega)},\quad\text{
	where}\quad
[u]_{W^{\sigma,p}(\Omega)}=\Bigg(\iint_{\Omega\times \Omega}\frac{|u(x)-u(y)|^p}{|x-y|^{n+\sigma p}}\ dxdy\Bigg)^{1/p}
$$
the space $W^{\sigma,p}(\Omega)$ is a Banach space. For general $s=k+\sigma$, $k\in\N_0$, $\sigma\in[0,1)$ the Sobolev space is defined as
$$
W^{s,p}(\Omega):=\Big\{u\in W^{k,p}(\Omega)\;:\; \partial^{\alpha}u\in W^{\sigma,p}(\Omega)\ \text{for all $\alpha\in \N_0^{n}$ with $|\alpha|= k$ }\Big\}.
$$
Finally, in the particular case $p=2$ the space $H^{s}(\Omega):=W^{s,2}(\Omega)$ is a Hilbert space.

\subsubsection{Nikol'skii spaces} For $u:\Omega\to\R$ and $h\in \R$, let $\Omega_h:=\{x\in \Omega\;:\; \dist(x,\partial\Omega)>h\}$ and, with $e\in \partial B_1(0)$, we let
$$
\delta_hu(x)=\delta_{h,e}u(x):=u(x+he)-u(x).
$$
Moreover, for $l\in \N$, $l>1$ let
$$
\delta_{h}^{l}u(x)=\delta_h(\delta_h^{l-1}u)(x).
$$
For $s=k+\sigma>0$ with $k\in \N_0$ and $\sigma\in(0,1]$ define
$$
N^{s,p}(\Omega):=\Big\{u\in W^{k,p}(\Omega)\;:\; [\partial^{\alpha}u]_{N^{\sigma,p}(\Omega)}<\infty \ \text{for all $\alpha\in \N_0^{n}$ with $|\alpha|= k$ }\Big\},
$$
where
$$
[u]_{N^{\sigma,p}(\Omega)}=\sup_{\substack{e\in \partial B_1(0)\\ h>0}}h^{-\sigma}\|\delta_{h,e}^2u\|_{L^p(\Omega_{2h})}.
$$
It follows that $N^{s,p}(\Omega)$ is a Banach space with norm $\|u\|_{N^{s,p}(\Omega)}:=\|u\|_{W^{k,p}(\Omega)}+\sum_{|\alpha|=k}[\partial^{\alpha}u]_{N^{\sigma,p}(\Omega)}$. It can be shown that this norm is equivalent to
$$
\|u\|_{L^p(\Omega)}+\sum_{|\alpha|=k}\sup_{\substack{e\in \partial B_1(0)\\ h>0}}h^{m-\sigma}\|\delta_{h,e}^lu\|_{L^p(\Omega_{lh})}
$$
for any fixed $m,l\in \N_0$ with $m<\sigma$ and $l>\sigma-m$ (see \cite[Theorem 4.4.2.1]{T95}).
\begin{prop}[see e.g. Propositions 3 and 4 in \cite{C17}]
	\label{prop-sobolev-nikolskii}
	Let $\Omega\subset \R^N$ open and with $C^{\infty}$ boundray. Moreover, let $t>s>0$ and $1\leq p<\infty$. Then
	$$
	N^{t,p}(\Omega)\subset W^{s,p}(\Omega)\subset N^{s,p}(\Omega).
	$$
\end{prop}

\section{Density results and Maximum principles}\label{sec:dense}

The main goal of this section is to show the following.
\begin{thm}\label{main-thm-density}
	Let either $\Omega=\R^N$ or $\Omega\subset \R^N$ open and bounded with Lipschitz boundary. In the following, let $X(\Omega):={\mathcal D}^k(\Omega)$ or $D^k(\Omega)$. Then $C^{\infty}_c(\Omega)$ is dense in $X(\Omega)$. Moreover, if $u\in X(\Omega)$ is nonnegative, then we have
	\begin{enumerate}
		\item There exists a sequence $(u_n)_n\subset X(\Omega)\cap L^{\infty}(\Omega)$ with $\lim\limits_{n\to\infty}u_n= u$ in $X(\Omega)$ satisfying that for every $n\in \N$ there is $\Omega'_n\subset\subset\Omega$ with $u_n=0$ on $\Omega\setminus \Omega'_n$ and $0\leq u_n\leq u_{n+1}\leq u$.
		\item There exists a sequence $(u_n)_n\subset C^{\infty}_c(\Omega)$ with $u_n\geq0$ for every $n\in \N$ and $\lim\limits_{n\to\infty}u_n= u$ in $X(\Omega)$.
	\end{enumerate}
\end{thm}
\begin{rem}
	To put Theorem \ref{main-thm-density} into perspective, we consider the following examples.
	\begin{enumerate}
		\item In the case $k(x,y)=|x-y|^{-2s-N}$ for some $s\in(0,\frac{1}{2})$, the above Theorem is well-known and leads to the interesting property that for any open, bounded Lipschitz set $\Omega\subset \R^N$ we have
		$$
		D^k(\Omega)=H^s(\Omega)=H^s_0(\Omega).
		$$
		We emphasize that the above equality also holds for $s=\frac{1}{2}$. Moreover, if $s<\frac{1}{2}$, it also holds $H^s(\Omega)=\{u\in H^s(\R^N)\;:\; 1_{\R^N\setminus\Omega}u\equiv 0\}$.
		\item If $k(x,y)=1_{B_1(0)}(x-y)|x-y|^{-N}$, $D^k(\Omega)$ is associated to the function space of the \textit{localized} logarithmic Laplacian (see \cite{CW18}).
	\end{enumerate}
\end{rem}
The proof is split into several smaller steps. Recall that ${\mathcal D}^k(\R^N)=D^k(\R^N)$ by definition.

\begin{lemma}
\label{density-pre1}
	Let $u\in D^k(\R^N)$. Then there is a sequence $(u_n)_n\subset D^k(\R^N)$ with $\lim\limits_{n\to\infty}u_n= u$ in $D^k(\R^N)$ satisfying that for every $n\in \N$ there is $\Omega_n\subset\subset\R^N$ with $u_n=0$ on $\R^N\setminus \Omega_n$. Moreover, if $u\geq 0$, then $(u_n)_n$ can be chosen to satisfy in addition $0\leq u_n\leq u_{n+1}\leq u$.
\end{lemma}
\begin{proof}
	For $n\in \N$ let $\phi_n\in C^{0,1}_c(\R^N)$ be radially symmetric and such that $\phi\equiv 1$ on $B_n(0)$, $\phi_n\equiv 0$ on $B_{n+1}(0)^c$. Clearly, we may assume that $[\phi_n]_{C^{0,1}(\R^N)}=1$. By Lemma \ref{properties1} there is hence some  $C=C(N,k)>0$ with $b_{k,\R^N}(\phi_n u)\leq C\|u\|_{D^k(\R^N)}$ for all $n\in \N$. In the following, let $u_n:=\phi_nu$ and without loss of generality we may assume $u\geq 0$. Since then $0\leq u-u_n\leq u$ on $\R^N$ and $u-u_n=0$ on $B_n$, by dominated convergence we have $\lim\limits_{n\to\infty}\|u-u_n\|_2=0$. Moreover, by choice of $\phi_n$ we have for $x,y\in \R^N$
	\begin{align*}
	|u(x)(1-\phi_n(x))-u(y)(1-\phi_n(y))|&\leq |u(x)-u(y)|(1-\phi_n(x)) +|u(y)||\phi_n(x)-\phi_n(y)|\\
	&\leq |u(x)-u(y)| +|u(y)|\min\{1,|x-y|\}=:U(x,y).
	\end{align*}
	Here, $U(x,y)\in L^2(\R^N\times \R^N,k(x,y)\ d(x,y))$, since
	\begin{align*}
	\iint_{\R^N\times\R^N} U(x,y)k(x,y)\ dxdy&=b_{k,\R^N}(u)+\int_{\R^N}|u(y)|^2\int_{\R^N} \min\{1,|x-y|^{2}\}k(x,y)\ dxdy\\
	&\leq b_{k,\R^N}(u)+\int_{\R^N}|u(y)|^2\ dy\,\supp_{x\in \R^N}\int_{\R^N} \min\{1,|x-y|^{2}\}k(x,y)dy<\infty.
	\end{align*}
	Thus $\lim\limits_{n\to\infty}b_{k,\R^N}(u-u_n)=0$ by the dominated convergence Theorem.
\end{proof}

\begin{prop}
\label{density1}
We have that $C^{\infty}_c(\R^N)$ is dense in $D^k(\R^N)$. Moreover, if $u\in D^k(\R^N)$ is nonnegative, then there exists $(\phi_n)_n\subset C^{\infty}_c(\R^N)$ with $\phi_n\geq0$ for every $n\in \N$ and $\lim\limits_{n\to\infty}\phi_n= u$ in $D^k(\R^N)$.
\end{prop}
\begin{proof}
Let $u\in D^k(\R^N)$. Moreover, let $\phi_n\in C^{0,1}_c(\R^N)$ for $n\in \N$ be given by Lemma \ref{density-pre1} such that $\|u-\phi u\|_{s,p}<\frac{1}{n}$. Then $v_n:=\phi_n u\in D^k(\R^N)$ and there is $R_n>0$ with $v_n\equiv 0$ on $\R^N\setminus B_{R_n}(0)$. Next, let $(\rho_\epsilon)_{\epsilon\in(0,1]}$ by a Dirac sequence and denote $v_{n,\epsilon}:=\rho_\epsilon\ast v_n$. Then $v_{n\epsilon}\in C^{\infty}_c(\R^N)$ for all $n\in \N$, $\epsilon\in(0,1]$ and
	\[
	b_{k,\R^N}(u-v_{n,\epsilon})\leq b_{k,\R^N}(u-v_n)+ b_{k,\R^N}(v_n-v_{n,\epsilon})\leq \frac{1}{n}+ b_{k,\R^N}(v_n-v_{n,\epsilon}).
	\]
	It is hence enough to show that $v_{n,\epsilon}\to v_n $ in $D^k(\R^N)$ for $\epsilon\to 0$. In the following, we write $v$ in place of $v_n$ and $v_\epsilon=\rho_\epsilon\ast v$ in place of $v_{n,\epsilon}$ for $\epsilon\in(0,1]$. Moreover, let $R=R_n>0$ with $v=v_n=0$ on $\R^N\setminus B_R(0)$. Clearly, $v_{\epsilon}\to v$ in $L^2(\R^N)$ for $\epsilon\to 0$ and this convergence is also pointwise almost everywhere. Hence it is enough to analyze the convergence of $b_{k,\R^N}(v-v_{\epsilon})$ as $\epsilon\to 0$. From here, the proof follows along the lines of \cite[Proposition 4.1]{JW19} noting that there it is not used that $k$ only depends on the difference of $x$ and $y$. Note here, that if $u$ is nonnegative then the above constructed sequence is also nonnegative.

\end{proof}

\begin{lemma}
\label{consequences of assumption2}
Let $\Omega\subset \R^N$ open and such that $\partial\Omega$ is bounded. Denote $\delta(x):=\dist(x,\R^N\setminus \Omega)$. Then the following is true.
\begin{enumerate}
\item There is $C=C(N,\Omega,k)>0$ such that $\kappa_{k,\Omega}(x)\leq C\delta^{-\sigma}(x)\ $ for $x\in \Omega$.
\item If $\Omega$ is bounded, then $1_{\Omega}\in D^k(\R^N)$.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $C=C(N,\Omega,k)>0$ be constants varying from line to line and denote $U:=\{x\in \R^N\;:\; \dist(x,\Omega)\leq 1\}$. To see item 1., let $x\in \Omega$ and fix $p\in \partial\Omega$ such that $\delta(x)=|x-p|$. Then
\begin{align*}
\kappa_{k,\Omega}(x)\leq C+\int_{U\setminus \Omega}\frac{|x-p|^{\sigma}}{|x-p|^{\sigma}}k(x,y)\ dy\leq C+\delta(x)^{-\sigma}\int_{U\setminus \Omega} |x-y|^{\sigma}k(x,y)\ dy\leq C\delta^{-\sigma}(x),
\end{align*}
where we have used that $|x-p|\leq |x-y|$ for $y\in \R^N\setminus \Omega$. Now 2. follows immediately from 1., since we have
$$
b_{k,\R^N}(1_{\Omega})=\int_{\Omega}\int_{\R^N\setminus\Omega}k(x,y)\ dydx\leq C \int_{\Omega} \delta^{-\sigma}(x)\ dx <\infty.
$$
\end{proof}

\begin{thm}[See Theorem \ref{main-thm-density}]
\label{density2}
Let $\Omega\subset \R^N$ be an open bounded set with Lipschitz boundary. Then $C^{\infty}_c(\Omega)$ is dense in ${\mathcal D}^k(\Omega)$. Moreover, if $u\in {\mathcal D}^k(\Omega)$ is nonnegative, then we have
\begin{enumerate}
\item There exists a sequence $(u_n)_n\subset {\mathcal D}^k(\Omega)$ with $\lim\limits_{n\to\infty}u_n= u$ in ${\mathcal D}^k(\Omega)$ satisfying that for every $n\in \N$ there is $\Omega'_n\subset\subset\Omega$ with $u_n=0$ on $\R^N\setminus \Omega'_n$ and $0\leq u_n\leq u_{n+1}\leq u$.
\item There exists a sequence $(u_n)_n\subset C^{\infty}_c(\Omega)$ with $u_n\geq0$ for every $n\in \N$ and $\lim\limits_{n\to\infty}u_n= u$ in ${\mathcal D}^k(\Omega)$.
\end{enumerate}
\end{thm}
\begin{proof}
Note that the second claim follows immediately from the first one using \cite[Proposition 4.1]{JW19} as in the proof of Proposition \ref{density1}. Then also the main claim follows by considering $u^{\pm}$ separately. Hence it is enough to show 1. We proceed similar to \cite[Theorem 3.1]{CW18}. Denote $\delta(x):=\dist(x,\R^N\setminus \Omega)$. For $r>0$, define the Lipschitz map
\[
\phi_r:\R^N\to\R,\quad\phi_r(x)=\left\{\begin{aligned}
&0 &&\delta(x)\geq 2r,\\
&2-\frac{\delta(x)}{r} && r\leq \delta(x)\leq 2r,\\
&1&& \delta(x)\leq r.
\end{aligned}\right.
\]
Note that we have $\phi_s\leq \phi_r$ for $0<s\leq r$. We show
\begin{equation}
\label{claim for density2}
u\phi_r\in {\mathcal D}^k(\Omega)\ \text{ for $r>0$ sufficiently small and } b_{k,\R^N}(u\phi_r)\to 0 \ \text{ for $r\to 0$.}
\end{equation}
Note that once this is shown, we have $u(1-\phi_r)\in {\mathcal D}^k(\Omega)$ for $r>0$ sufficiently small and $u(1-\phi_r)\to u$ for $r\to0$. Since also $0\leq u(1-\phi_r)\leq u(1-\phi_s)$ for $0<s\leq r$ and $u(1-\phi_r)=0$ for $x\in \R^N$ with $\delta(x)\leq r$, it follows that \eqref{claim for density2} implies 1.\\
The remainder of the proof is to show \eqref{claim for density2}. For this, let $C=C(N,\Omega,k)>0$ be a constant which may vary from line to line. Let $A_t:=\{x\in \Omega\;:\; \delta(x)\leq t\}$. Note that $u\phi_r$ vanishes on $\R^N\setminus A_{2r}$, we have $0\leq \phi_r\leq 1$ and, moreover,
\[
|\phi_r(x)-\phi_r(y)|\leq  \min\Big\{C\frac{|x-y|}{r},1\Big\}\quad\text{for $x,y\in \R^N$.}
\]
Then proceeding similarly to the proof of Lemma \ref{properties1}.5 we find for $r$ small enough
\begin{align*}
b_{k,\R^N}&(u\phi_r)=\frac{1}{2}\int_{A_{4r}}\int_{A_{4r}}(u(x)\phi_r(x)-u(y)\phi_r(y))^2k(x,y)\ dxdy+\int_{A_{2r}}u^2(x)\phi_r^2(x)\kappa_{k,A_{4r}}(x)\ dx\\
&\leq \int_{A_{4r}}\int_{A_{4r}}\Big(u(x)^2(\phi_r(x)-\phi_r(y))^2 +(u(x)-u(y))^2\phi_r^2(y)^2\Big)k(x,y)\ dxdy\\
&\qquad\qquad +\int_{A_{2r}}u^2(x)\kappa_{k,A_{4r}}(x)\ dx\\
&\leq \frac{C}{r^2}\int_{A_{4r}}u(x)^2\int_{B_r(x)}|x-y|^2k(x,y)\ dydx+\int_{A_{4r}}u(x)^2\int_{\R^N\setminus B_r(x)}k(x,y)\ dydx\\
&\qquad\qquad + \int_{A_{4r}}\int_{A_{4r}}(u(x)-u(y))^2k(x,y)\ dxdy+\int_{A_{2r}}u^2(x)\kappa_{k,A_{4r}}(x)\ dx\\
&\leq \frac{C}{r^{\epsilon}}\int_{A_{4r}}u(x)^2\int_{B_1(x)}|x-y|^{\epsilon}k(x,y)\ dydx+C\int_{A_{4r}}u(x)^2 dx +b_{k,A_{4r}}(u)+\int_{A_{2r}}u^2(x)\kappa_{k,A_{4r}}(x)\ dx\\
&\leq \frac{C}{r^{\epsilon}}\int_{A_{4r}}u(x)^2\ dx+C\int_{A_{4r}}u(x)^2 dx +b_{k,A_{4r}}(u)+\int_{A_{2r}}u^2(x)\kappa_{k,A_{4r}}(x)\ dx.
\end{align*}
Note here, since $u\in D^k(\R^N)$, we have $\int_{A_{4r}}u(x)^2 dx +b_{k,A_{4r}}(u)\to 0$ for $r\to 0$. Moreover, we have by Lebesgue's differentiation theorem
\begin{align*}
\frac{C}{r^{\epsilon}}\int_{A_{4r}}u(x)^2 dx&\leq \frac{C|B_{4r}|}{r^{\epsilon}}\int_{\partial\Omega}\frac{1}{|B_{4r}|}\int_{B_{4r}(\theta)}u(x)^2\ dx\sigma(d\theta)\\
&\leq Cr^{1-\epsilon}\int_{\partial\Omega}\frac{1}{|B_{4r}|}\int_{B_{4r}(\theta)}u(x)^2\ dx\sigma(d\theta)\to 0\quad\text{for $r\to 0^+$.}
\end{align*}
 Finally, since
\begin{equation}
\label{special case density}
K_{k,\Omega}(u)=\int_{\Omega}u^2(x)\kappa_{k,\Omega}(x)\ dx<\infty
\end{equation}
and, by Lemma \ref{consequences of assumption2}, we have
\[
\kappa_{k,A_{4r}}(x)\leq \int_{\R^N\setminus \Omega}k(x,y)\ dy+\int_{\Omega\setminus A_{4r}}k(x,y)\ dy\leq C \kappa_{k,\Omega}(x)+Cr^{-\epsilon}
\]
for $x\in A_{2r}$, so that also $\int_{A_{2r}}u^2(x)\kappa_{k,A_{4r}}(x)\ dx\to 0$ for $r\to 0$ with a similar argument.
\end{proof}

\begin{proof}[Proof of Theorem \ref{main-thm-density} for $X(\Omega)={\mathcal D}^k(\Omega)$]
This statement now follows from Theorem \ref{density2}, Lemma \ref{density-pre1}, and Proposition \ref{density1}.
\end{proof}

\begin{thm}[See Theorem \ref{main-thm-density}]
\label{density3}
Let $\Omega\subset \R^N$ be an open bounded set with Lipschitz boundary. Then $C^{\infty}_c(\Omega)$ is dense in $D^k(\Omega)$. Moreover, if $u\in D^k(\Omega)$ is nonnegative, then we have
\begin{enumerate}
\item There exists a sequence $(u_n)_n\subset D^k(\Omega)\cap L^{\infty}(\Omega)$ with $\lim\limits_{n\to\infty}u_n= u$ in $D^k(\Omega)$ satisfying that for every $n\in \N$ there is $\Omega'_n\subset\subset\Omega$ with $u_n=0$ on $\Omega\setminus \Omega'_n$ and $0\leq u_n\leq u_{n+1}\leq u$.
\item There exists a sequence $(u_n)_n\subset C^{\infty}_c(\Omega)$ with $u_n\geq0$ for every $n\in \N$ and $\lim\limits_{n\to\infty}u_n= u$ in $D^k(\Omega)$.
\end{enumerate}
\end{thm}
\begin{proof}
Consider the Lipschitz map
$$
g_n:\R\to\R,\qquad g_n(t)=\left\{\begin{aligned}&0 &&t\leq 0\\
& t  && 0<t<n\\
&n && t\geq n.\end{aligned}\right.
$$
Then $v_n:=g_n(u)\in D^k(\Omega)\cap L^{\infty}(\Omega)$ and we have with $\phi_r$ as in the proof of Proposition \ref{density2}
\[
b_{k,\Omega}(u-(1-\phi_r)v_n)\leq b_{k,\Omega}(u-v_n)+b_{k,\Omega}( \phi_rv_n).
\]
Clearly, $b_{k,\Omega}(u-v_n)\to 0$ for $n\to\infty$ by dominated convergence and $b_{k,\Omega}( \phi_rv_n)\to 0$ for $r\to0$ analogously to the proof of Proposition \ref{density2}, noting that the term in \eqref{special case density} reads in this case
$$
K_{k,\Omega}(v_n)\leq n^2\int_{\Omega}\kappa_{k,\Omega}(x)\ dx<\infty \quad\text{for every $n\in \N$.}
$$
In particular, 1. follows. Now 2. and the density statement follow analogously, again, to the proof of Proposition \ref{density2}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{main-thm-density} for $X(\Omega)=D^k(\Omega)$]
	This statement now follows from Theorem \ref{density3}, Lemma \ref{density-pre1}, and Proposition \ref{density1}.
\end{proof}

\begin{rem}
\label{hardy-bem}
It is tempting to conjecture the following type of Hardy inequality: There is $C>0$ such that
\[
K_{k,\Omega}(\phi)\leq C\Big(\|\phi\|_{L^2(\Omega)}^2+b_{k,\Omega}(\phi)\Big)\quad\text{for all $\phi\in C^{\infty}_c(\Omega)$}
\]
if $\Omega$ is a bounded Lipschitz set and $k$ satisfies addtionally \eqref{assumption2}. Let us mention that for $k(x,y)=|x-y|^{-2s-N}$ this holds for $s\in(0,1)$, $s\neq \frac{1}{2}$, see \cite{CS03,D04}. Moreover, for $k(x,y)=1_{B_{1}(0)}(x-y)|x-y|^{-N}$, this has been shown in \cite{CW18}. In the general framework presented here, however, it is not clear if this is true.
\end{rem}

\begin{rem}
	\label{rem-weak-defi}
	With the above density results, we can now note that our definition of weak supersolutions (and similarly of weak subsolutions and solutions), see Definition \ref{defi-weak} can be extended slightly:\\
	Let $u\in {\mathcal V}^k_{loc}(\Omega)$ satisfy weakly $I_ku\geq f$ in $\Omega$ for some $f\in L^1_{loc}(\Omega)$ and $\Omega\subset \R^N$ open and bounded with Lipschitz boundary.
	\begin{enumerate}
		\item If $f\in L^2_{loc}(\Omega)$, then by density it also holds
		\begin{equation}
		\label{weak-form2}
		b_k(u,v)\geq \int_{U}f(x)v(x)\ dx\quad\text{for all nonnegative $v\in {\mathcal D}^k(U)$, $U\subset\subset\Omega$.}
		\end{equation}
		\item If $u\in {\mathcal V}^k(\Omega)\cap L^{\infty}(\R^N)$ and $f\in L^2(\Omega)$, then by density it also holds
		\begin{equation}
		\label{weak-form3}
		b_k(u,v)\geq \int_{\Omega}f(x)v(x)\ dx\quad\text{for all nonnegative $v\in {\mathcal D}^k(\Omega)$.}
		\end{equation}
	\end{enumerate}
Finally note that if $u:\R^N\to\R$ satisfies $u1_{U}\in D^k(U)$ for some $U\subset\subset\R^N$ and $u\in L^{\infty}(\R^N\setminus U)$, then $u\in {\mathcal V}^k_{loc}(U)$.
\end{rem}

\begin{proof}[Proof of Proposition \ref{wmp}]
Note that also $u^-\in {\mathcal V}^k(\Omega)$ and in particular $u^-\in D^k(\Omega)$. Hence, we can find $(v_n)_n\subset C^{\infty}_c(\Omega)$ with $v_n\to u^-$ in $D^k(\Omega)$ for $n\to \infty$ with $0\leq v_n\leq v_{n+1}\leq u^-$ by Proposition \ref{density3}. Then
\begin{align*}
b_{k,\R^N}(u,v_n)\geq \int_{\Omega} c(x) u(x)v_n(x)\ dx\geq -\|c^+\|_{L^{\infty}(\Omega)}\int_{\Omega} u^-(x)v_n(x)\ dx.
\end{align*}
On the other hand, since $u^+v_n=0$ for all $n\in \N$ and $u\geq 0$ almost everywhere in $\R^{N}\setminus \Omega$, we find
\begin{align*}
b_{k,\R^N}(u,v_n)&=b_{k,\Omega}(u,v_n)+\int_{\Omega}v_n(x)\int_{\R^N\setminus \Omega}(u(x)-u(y))k(x,y)\ dydx\\
&\leq b_{k,\Omega}(u^+,v_n)-b_{k,\Omega}(u^-,v_n)+\int_{\Omega}v_n(x)u(x)\int_{\R^N\setminus \Omega} k(x,y)\ dydx\\
&\leq - b_{k,\Omega}(u^-,v_n)-K_{k,\Omega}(u^-,v_n).
\end{align*}
Hence
$$
0\leq \int_{\Omega} u^-(x)v_n(x)\Big(\|c^+\|_{L^{\infty}(\Omega)}-\kappa_{k,\Omega}(x)\Big)\ dx-b_{k,\Omega}(u^-,v_n)\leq -b_{k,\Omega}(u^-,v_n).
$$
Since $v_n\to u^-$ in $D^k(\Omega)$, it follows that $b_{k,\Omega}(u^-,u^-)=0$, but then $u^-$ is constant by Proposition \ref{constant} in $\Omega$. Assume by contradiction that $u^-=m>0$. Then the above calculation gives
\begin{equation}
\label{contradiction}
0\leq m \int_{\Omega}  v_n(x)\Big(\|c^+\|_{L^{\infty}(\Omega)}-\kappa_{k,\Omega}(x)\Big)\ dx,
\end{equation}
which is in both cases a contradiction: If in case 1. $c\leq 0$, then by \eqref{assumption2} we have $\kappa_{k,\Omega}(x)\not\equiv 0$ and since $v_n\to m$ in $D^k(\Omega)$ the right-hand side of \eqref{contradiction} is negative.\\ In case 2. this contradiction is immediate in a similar way.
\end{proof}

\begin{rem}
Usually, the weak maximum principle is stated with an assumption on the first eigenvalue $\Lambda_1(\Omega)$ in place of $\inf_{x\in \Omega}\kappa_{k,\Omega}(x)$. This can be done once the Hardy inequality in Remark \ref{hardy-bem} is shown.
\end{rem}

\begin{proof}[Proof of Proposition \ref{smp}]
This statement follows by approximation from \cite[Theorem 2.5 and 2.6]{JW19}. Here, the statement $j\notin L^1(\R^N)$ comes into play since we need
\[
\inf_{x\in B_r(x_0)}\kappa_{k,B_r(x_0)}(x)\to \infty \quad\text{for $r\to 0$}
\]
to conclude the statement for arbitrary $c$ as stated.
\end{proof}

\section{On Boundedness}\label{sec:bdd}

In the following, let $h\ast u(x)=\int_{\R^N} h(x-y)u(y)\ dy$ as usual denote the convolution of two functions.

\begin{thm}
	\label{thm-bdd}
	Assume $k$ satisfies \eqref{assumption2} and is such that
\begin{equation}\label{assumption2d}
	\sup_{x\in \R^N}\int_{K\setminus B_{\epsilon}(x)}k(x,y)^2\ dy<\infty\quad\text{for all $K\subset\subset \R^N$ and $\epsilon>0$.}
\end{equation}
	Let $\Omega\subset \R^N$ be an open set. Let $f\in L^{\infty}(\Omega)$, $h\in L^1(\R^N)\cap L^{2}(\R^N)$, and let $u\in {\mathcal V}^k_{loc}(\Omega)$ satisfy in weak sense
	$$
	I_ku\leq \lambda u+h\ast u+f \quad\text{in $\Omega$ for some $\lambda>0$.}
	$$
	If $u^+\in L^{\infty}(\R^N\setminus \Omega')$ for some $\Omega'\subset\subset\Omega$, then $u^+\in L^{\infty}(\R^N)$ and there is $C=C(\Omega,\Omega',k,h,\lambda)>0$ such that
	$$
	\|u^+\|_{L^{\infty}(\Omega')}\leq C\Big(\|f\|_{L^{\infty}(\Omega)}+\|u\|_{L^{2}(\Omega')}+\|u^+\|_{L^{\infty}(\R^N\setminus \Omega')}\Big).
	$$
\end{thm}

\begin{proof}
	Let $\Omega_1,\Omega_2,\Omega_3\subset\R^N$ be with Lipschitz boundary and such that
	$$
	\Omega'\subset\subset\Omega_1\subset\subset\Omega_2\subset\subset\Omega_3\subset\subset \Omega.
	$$
	Let $\eta\in C^{0,1}_c(\Omega_3)$ such that $0\leq \eta\leq 1$ and $\eta=1$ on $\Omega_2$. Put $v=\eta u$ and, for $\delta>0$, denote $J_{\delta}(x,y):=1_{B_{\delta}(0)}(x-y)k(x,y)$ and $k_{\delta}(x,y)=k(x,y)-J_{\delta}(x,y)$. Note that by Assumption \eqref{assumptions} it follows that $y\mapsto k_{\delta}(x,y)\in L^1(\R^N)$ for all $x\in \R^N$. Moreover, by Assumption \eqref{assumption2}
	$$
	c_{\delta}:=\inf_{x\in \R^N}\int_{\R^N}k_{\delta}(x,y)\ dy\geq \int_{\R^N\setminus B_{\delta}(0)}j(z)\ dz\to \infty \quad\text{for $\delta\to 0$.}
	$$
		Hence, we may fix $\delta>0$ such that
		$$
		c_{\delta}>\lambda.
		$$
		In the following, $C_i>0$, $i=1,\ldots$ denote constants depending on $\Omega'$, $\Omega_i$, for $i=1,2,3$, $\lambda$, $\delta$, $\Omega$, $\eta$, $k$, and $h$ but may vary from line to line ---clearly, by the choices these dependencies are actually only through $\lambda$, $\Omega$, $\Omega'$, $\eta$, $k$, and $h$. First note that by Lemma \ref{localization} we have in weak sense
\begin{align*}
	I_kv\leq \lambda u+h\ast u+\tilde{f}\quad\text{in $\Omega_1$}\quad\text{with}\quad \tilde{f}(x)=f(x)+\int_{\R^N\setminus \Omega_2}(1-\eta(y))u(y)k(x,y)\ dy.
	\end{align*}
	In the following, put
	$$
	A:=\|f\|_{L^{\infty}(\Omega)}+\|u\|_{L^{2}(\Omega')}+\|u^+\|_{L^{\infty}(\R^N\setminus\Omega')}.
	$$
	Then note that for $x\in \R^N$ we have
	$$
|h\ast u(x)|\leq \|h\|_{L^{2}(\R^N)}\|u\|_{L^{2}(\Omega')}+\|u^+\|_{L^{\infty}(\R^N\setminus\Omega')}\|h\|_{L^1(\R^N)}\leq C_1A
	$$
	and, since $\sup\limits_{x\in \Omega_1}\int_{\R^N\setminus \Omega_2}(1-\eta(y))k(x,y)\ dy\leq C_2\sup\limits_{x\in \Omega_1}\int_{\R^N\setminus \Omega_2}\min\{1,|x-y|^{\sigma}\}k(x,y)\ dy<\infty$, it also holds that
	$$
	\|\tilde{f}\|_{L^{\infty}(\Omega_1)}\leq \|f\|_{L^{\infty}(\Omega)}+\|u^+\|_{L^{\infty}(\R^N\setminus \Omega_1)}C_2\leq C_3A.
	$$
	Whence, since $u=v$ in $\Omega_1$, we have in weak sense
	\begin{align*}
	Iv\leq \lambda v+C_4A\quad\text{in $\Omega_1$.}
	\end{align*}
	Next, let $\mu\in C^{\infty}_c(\Omega'')$ for some $\Omega'\subset\subset\Omega''\subset\subset\Omega_1$ such that $0\leq \mu\leq 1$, $\mu=1$ on $\Omega'$, and $\mu=0$ on $\R^N\setminus \Omega''$. Let $\phi_t=\mu^2(v-t)^+\in {\mathcal D}^k(\Omega'')$ for $t>0$ and note that
	\begin{equation}
	\label{bdd-1}
	b_k(v,\phi_t)\leq\int_{\Omega''}(\lambda v(x)+C_4A)\phi_t(x)\ dx.
	\end{equation}
 Fix $t>0$ such that
	\begin{align*}
	 t\geq \|u^+\|_{L^{\infty}(\R^N\setminus \Omega')}&\quad\text{and}\quad C_6A+(\lambda-c_{\delta})t\leq 0,\quad\text{where}\\
	C_6&=C_4+C_5\quad\text{with}\quad C_5=\sup_{x\in \Omega'}\int_{\Omega'}k_{\delta}^2(x,y)\ dy+\sup_{x\in \Omega'}\int_{\R^N\setminus\Omega'}k_{\delta}(x,y)\ dy.
	\end{align*}
	That is, we fix
	$$
	t= A\Big(1+\frac{C_6}{c_{\delta}-\lambda}\Big).
	$$
	Then with \eqref{bdd-1}
	\begin{align*}
	&b_{J_{\delta}}(v,\phi_t)=b_k(v,\phi_t)-b_{k_{\delta}}(v,\phi_t)\\
	&\leq \int_{\Omega''}\lambda v(x)\phi_t(x)+C_4A\phi_t(x)\ dx-\int_{\R^N}v(x)\phi_t(x)\int_{\R^N}k_{\delta}(x,y)\ dydx+\int_{\R^N}\phi_t(x)\int_{\R^N}v(y)k_{\delta}(x,y)\ dy\ dx.
	\end{align*}
	Note here, that for $x\in \R^N$ we have by the integrability assumptions on $k_{\delta}$ and $k$
	\begin{align*}
	\int_{\R^N}v(y)k_{\delta}(x,y)\ dy&\leq \int_{\R^N}u(y)k_{\delta}(x,y)\ dy\leq C_5(\|u\|_{L^2(\Omega')}+\|u^+\|_{L^{\infty}(\R^N\setminus \Omega')})\leq C_5A
		\end{align*}
		so that using that $v\geq t$ in $\supp\,\phi_t$ we have
\begin{equation}
	\label{bdd-2}
	b_{J_{\delta}}(v,\phi_t)\leq  \int_{\Omega''}(C_6A+(\lambda-c_{\delta})v(x))\phi_t(x)\ dx\leq  (C_6A+(\lambda-c_{\delta})t)\int_{\Omega''}\phi_t(x)\ dx.
	\end{equation}
	On the other hand, with $v_t(x)=v(x)-t$, we have
	\begin{align*}
	(v(x)-v(y))&(\phi_t(x)-\phi_t(y))-(\mu(x)v_t^+(x)-\mu(y)v_t^+(y))^2\\
	&=2\mu(x)\mu(y)v_t^+(x)v_t^+(y)-v_t(y)\mu^2(x)v_t^+(x)-\mu^2(y)v_t^+(y)v_t(x)\\
	&=-v_t^+(x)v_t^+(y)(\mu(x)-\mu(y))^2+v_t^-(y)\mu^2(x)v_t^+(x)+\mu^2(y)v_t^+(y)v_t^-(x)\\
	&\geq -v_t^+(x)v_t^+(y)(\mu(x)-\mu(y))^2.
	\end{align*}
	Whence with Poincar\'e's inequality, using that by Assumption \ref{assumption2} there is for any $K\subset \R^N$ open and bounded some $C>0$ such that $b_{J_{\delta}}(u)\geq C\|u\|_{L^2(\R^N)}^2$ for $u\in {\mathcal D}^{J_{\delta}}(K)$, we find for some constant $C_7$
	\begin{align}
	b_{J_{\delta}}(v,\phi_t)&\geq b_{J_{\delta}}(\mu v_t^+)-\frac{1}{2}\int_{\R^N}\int_{\R^N} v_t^+(x)v_t^+(y)(\mu(x)-\mu(y))^2J_{\delta}(x,y)\ dxdy\notag\\
	&\geq C_7\int_{\R^N}\mu^2(x)(v_t^+(x))^2\ dx-\frac{1}{2}\int_{\R^N}\int_{\R^N} v_t^+(x)v_t^+(y)(\mu(x)-\mu(y))^2J_{\delta}(x,y)\ dxdy\\
	&= C_7\int_{\R^N}\mu^2(x)(v_t^+(x))^2\ dx-\frac{1}{2}\int_{\Omega'}\int_{\Omega'} v_t^+(x)v_t^+(y)(\mu(x)-\mu(y))^2J_{\delta}(x,y)\ dxdy\\
	&= C_7\int_{\R^N}\mu^2(x)(v_t^+(x))^2\ dx\geq C_7\int_{\Omega'}(v_t^+(x))^2\ dx.\label{bdd-3}
	\end{align}
	Combining \eqref{bdd-3} and \eqref{bdd-2} we have
	\begin{align*}
	C_7\int_{\Omega'}&(v_t^+(x))^2\ dx\leq \Big(C_6A+(\lambda-c_{\delta})t\Big)\int_{\Omega''} \phi_t(x)\ dx\leq 0.
	\end{align*}
	Whence $v_t^+=0$ in $\Omega'$ and thus $u=v\leq t=AC_{10}$ in $\Omega'$ as claimed.
\end{proof}

\begin{cor}
	\label{thm-bdd2}
	If in the situation of Theorem \ref{thm-bdd} we have in weak sense $I_ku=\lambda u+h\ast u+f$ in $\Omega$, then we have $u\in L^{\infty}(\Omega')$ and there is $C=C(\Omega,\Omega',k,\lambda,h)>0$ such that
	$$
	\|u\|_{L^{\infty}(\Omega')}\leq C\Big(\|f\|_{L^{\infty}(\Omega)}+\|u\|_{L^{2}(\Omega')}+\|u\|_{L^{\infty}(\R^N\setminus \Omega')}\Big).
	$$
\end{cor}

\begin{proof}
	This follows by replacing $u$ with $-u$ (and $f$ with $-f$) in the statement of Theorem \ref{thm-bdd}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm-bdd3}]
	This follows directly from Corollary \ref{thm-bdd2}.
\end{proof}

\begin{thm}
	If in the situation of Theorem \ref{thm-bdd} we have in weak sense $I_ku=\lambda u+h\ast u+f$ in $\Omega$ and $u\in {\mathcal D}^k(\Omega)$, then we have $u\in L^{\infty}(\Omega)$ and there is $C=C(\Omega,k,\lambda,h)>0$ such that
	$$
	\|u\|_{L^{\infty}(\Omega)}\leq C\Big(\|f\|_{L^{\infty}(\Omega)}+\|u\|_{L^{2}(\Omega)}\Big).
	$$
\end{thm}

\begin{proof}
	Using in the proof of Theorem \ref{thm-bdd} the test-function $u_t^+$ instead of $\phi_t$ (and similarly for Corollary \ref{thm-bdd2}), we find
	$$
	\|u\|_{L^{\infty}(\Omega)}\leq C\Big(\|f\|_{L^{\infty}(\Omega)}+\|u\|_{L^{2}(\Omega)}\Big).
	$$
	as claimed.
\end{proof}

\begin{proof}[Proof of Corollary \ref{main-thm-eigen}]
The compact embedding has been shown in \cite{JW20}, the fact that the first eigenfunction can be chosen to be positive follows from \cite{JW19} and the final statement of the boundedness follows from Theorem \ref{thm-bdd3} (with $h=f=0$) if $\Omega$ has a Lipschitz boundary. If this is not the case, it can be easily seen that eigenfunctions $u$ corresponding to an eigenvalue $\lambda$ actually satisfy $b_k(u, v)=\lambda \int_{\Omega} u v\ dx$ for all $v\in {\mathcal D}^k(\Omega)$ (see Remark \ref{different-boundary-cond}). Whence, as in the proof of Theorem 5.1 the test-function $u_t^+\in {\mathcal D}^k(\Omega)$ can be used.
\end{proof}

\section{On differentiability of solutions}\label{sec:reg}

In the following, $\Omega\subset \R^N$ is an open bounded set and $k$ satisfies through out the assumptions \eqref{assumption2}, \eqref{assumption2c}, and \eqref{assumption3} for some $m\in \N\cup\{\infty\}$ ---in particular, there is $J:\R^N\to[0,\infty]$ such that $k(x,y)=J(x-y)$ for $x,y\in \R^N$. We hence also write $J$ in place of $k$. Moreover, given $\sigma$ from assumption \eqref{assumptions} we assume that $\sigma<\frac12$ and fix
$$
\alpha:=1-\sigma\in(\frac12,1).
$$

\begin{thm}
	\label{thm:differentiability-step1}
	Let $f\in C^1(\overline{\Omega})$, $\lambda\in \R$ and $u\in {\mathcal V}^J_{loc}(\Omega)\cap L^{\infty}(\R^N)$ satisfy in weak sense $Iu=f+\lambda u$ in $\Omega$. Then for any $\Omega'\subset\subset \Omega$ there is $C=C(N,\Omega,\Omega',J,\lambda)>0$ such that
	\begin{equation}
		\label{bound-derivative}
		\| \delta_{h,e} u\|_{L^2(\Omega')}\leq h^{\alpha}C\Big(\|f\|_{C^1(\Omega)}^2+\|u\|_{L^\infty(\R^N)}^2\Big)^{\frac12}\quad\text{for all $h>0$, $e\in \partial B_1(0)$.}
	\end{equation}
\end{thm}

\begin{proof}
	Let $\Omega'\subset\subset\Omega$ and fix $r\in(0,\frac18)$ small such that $8r\leq\dist(\Omega',\R^N\setminus\Omega)$. Moreover, fix $x_0\in \Omega'$ and denote $B_n:=B_{nr}(x_0)$. Note that by using assumption \eqref{assumption2} with Lemma \ref{poincare} we achieve, by making $r>0$ small enough,
	$$
	\lambda<\lambda_1=\min_{\substack{w\in {\mathcal D}^J(B_4)\\ w\neq 0}}\frac{\rho_J(w)}{\|w\|_{L^2(\R^N)^2}}.
	$$
	Let $\eta\in C^{0,1}_c(B_4)$ with $0\leq \eta\leq 1$, $\eta\equiv 1$ on $B_2$. Note that it holds
	$$
	|\eta(x)-\eta(y)|\leq 2\|\eta\|_{C^{0,1}(\R^N)}\min\{1,|x-y|\},
	$$
	where we put as usual
	$$
	\|\eta\|_{C^{0,1}(\R^N)}:=\sup_{x\in \R^N}|\eta(x)|+\sup_{\substack{x,y\in \R^N\\ x\neq y}}\frac{|\eta(x)-\eta(y)|}{|x-y|}.
	$$
	Note that by choice we have $\|\eta\|_{C^{0,1}(\R^N)}\leq 1+\frac{1}{r}\leq \frac{2}{r}$, so that for all $x,y\in \R^N$
	\begin{equation}
		\label{bound on eta}
		|\eta(x)-\eta(y)|\leq \frac{4}{r}\min\{1,|x-y|\}.
	\end{equation}
	Fix $e\in \partial B_1(0)$ and $h\in(0,r)$. Let
	$$
	A:=\|u\|_{L^{\infty}(\R^N)}.
	$$
	Let $\psi=\eta^2\delta_{h}u\in {\mathcal D}^J(B_4)$, where in the following $\delta_hu:=\delta_{h,e}u$. Note that
	\begin{align*}
		(\delta_{h}u(x)-\delta_{h}u(y))(\psi(x)-\psi(y))&=(\eta(x)\delta_{h}u(x)-\eta(y)\delta_{h}u(y))^2\\
		&\qquad -\delta_{h}u(x)\partial_{h}u(y)(\eta(x)-\eta(y))^2.
	\end{align*}
	Hence, we have
	$$
	b_J(\delta_{h}u,\psi)=b_J(\eta\delta_{h}u)-\frac{1}{2}\iint_{\R^{N}\times\R^N}\delta_{h}u(x)\delta_{h}u(y)(\eta(x)-\eta(y))^2J(x-y)\ dxdy.
	$$
	and using the translation invariance, we also have
	$$
	b_J(\delta_h u,\psi)=\int_{\Omega} [\delta_hf(x)+\lambda\delta_hu]\psi(x)\ dx.
	$$
	In the following, for simplicity, we put $v(x)=\eta(x)\delta_hu(x)$, $x\in \R^N$. Note that by Definition, $v\in {\mathcal D}^J(B_4)$. Then with the help of Young's inequality for some $\mu\in(0,1)$ such that
	\begin{equation}
		\label{condition on mu}
		2\mu< \lambda_1-\lambda
	\end{equation}
	we find
	\begin{align}
		&\lambda_1\|v\|_{L^2(\Omega'')}^2\leq b_J(v)=b_J(\delta_{h}u,\psi)+\frac{1}{2}\iint_{\R^{n}\times\R^N}\delta_{h}u(x)\delta_{h}u(y)(\eta(x)-\eta(y))^2J(x-y)\ dxdy\notag\\
		&=\int_{\Omega''}[\delta_{h}f(x)+\lambda\delta_hu]\eta^2(x)\delta_{h}u(x)\ dx+\frac{1}{2}\iint_{\R^{n}\times\R^N}\delta_{h}u(x)\delta_{h}u(y)(\eta(x)-\eta(y))^2J(x-y)\ dxdy\notag\\
		&\leq  (\mu+\lambda)\|v\|_{L^{2}(\Omega'')}^2+\mu^{-1}h^2\|\frac{\delta_hf}{h}\|_{L^2(\Omega'')}^2  +\frac{1}{2}\iint_{\R^N\times\R^N}\delta_{h}u(x)\delta_{h}u(y)(\eta(x)-\eta(y))^2J(x-y)\ dxdy.\label{basis-bound}
	\end{align}
	By a rearrangement of the double integral with Young's inequality for the same $\mu\in(0,1)$ as above
	we have
	\begin{align}
		\frac{1}{2}&\iint_{\R^{n}\times\R^N}\delta_{h}u(x)\delta_{h}u(y)(\eta(x)-\eta(y))^2J(x-y)\ dxdy\notag\\
		&=\iint_{\R^N\times \R^N} \delta_{h}u(x)\delta_{h}u(y)\eta(x)(\eta(x)-\eta(y))J(x-y)\ dxdy\notag\\
		&=\int_{\R^N} \eta(x)\delta_{h}u(x)\int_{\R^N}u(y)\delta_{-h,y}\Big((\eta(x)-\eta(y))J(x-y)\Big)\ dydx\notag\\
		&\leq \mu\| v\|_{L^2(B_4)}^2+\mu^{-1}\int_{B_4}\Bigg(\ \int_{\R^N}|u(y)|\Big|\delta_{-h,y}\Big((\eta(x)-\eta(y))J(x-y)\Big)\Big|\ dy\Bigg)^2 dx\notag\\
		&\leq \mu\| v\|_{L^2(B_4)}^2+\mu^{-1}A^2\int_{B_4}\Bigg(\ \int_{\R^N}\Big|\delta_{-h,y}\Big((\eta(x)-\eta(y))J(y-x)\Big)\Big|\ dy\Bigg)^2 dx\notag\\
		&\leq
		\mu\| v\|_{L^2(B_4)}^2+\mu^{-1}A^2\int_{B_4}\Bigg(\ \int_{\R^N}\Big|\delta_{-h,z}\Big((\eta(x)-\eta(z+x))J(z)\Big)\Big|\ dz\Bigg)^2 dx.
		\label{bounding-1}
	\end{align}
	Here, we indicate with $\delta_{-h,y}$ (resp. $\delta_{-h,z}$) that $\delta_{-h}$ acts on the $y$ (resp. $z$) variable. Note that
	\begin{align}
		\delta_{-h,z}&\Big((\eta(x)-\eta(z+x))J(z)\Big)\notag\\
		&=\delta_{-h,z}(\eta(x)-\eta(z+x))J(z)+ (\eta(x)-\eta(z+x-he))\delta_{-h}J(z)\notag\\
		&=\Big(\eta(z+x)-\eta(z+x-he)\Big) J(z)+(\eta(x)-\eta(z+x-he))\Big(J(z-he)-J(z)\Big)\label{difference-op1}\\
		&= \Big(\eta(z+x)-\eta(z)\Big) J(z)+ \Big(\eta(x)-\eta(z+x-he)\Big)J(z-he).\label{difference-op2}
	\end{align}

	Note here, that \eqref{difference-op1} satisfies
	\begin{equation}
		\label{difference-op1b}
		\begin{split}
			\Bigg|\Big(\eta(z+x)-\eta(z+x-he)\Big)& J(z)+(\eta(x)-\eta(z+x-he))\Big(J(z-he)-J(z)\Big)\Bigg|\\
			&\leq \frac{4h}{r}J(z)+\frac{4h}{r}\min\{1,|z-he|\}\int_0^1|\nabla J(z-\tau he)|\ d\tau
		\end{split}
	\end{equation}
	and \eqref{difference-op2} can be written as
	\begin{equation}
		\label{difference-op2b}
		\begin{split}
			\Bigg| \Big(\eta(z+x)-\eta(z)\Big) &J(z)+ \Big(\eta(x)-\eta(z+x-he)\Big)J(z-he)\Bigg|\\
			&\leq \frac{4}{r}\min\{1,|z|\}J(z)+\frac{4}{r}\min\{1,|z-he|\}J(z-he).
		\end{split}
	\end{equation}
	For $h\in(0,r)$, $z\in \R^N\setminus\{0\}$ put
	\begin{align*}
		k_h(z)=\min&\Bigg\{h\Big(J(z)+\min\{1,|z-he|\}\int_0^1|\nabla J(z-\tau he)|\ d\tau\Big),\\
		&\qquad\qquad \min\{1,|z|\}J(z)+\min\{1,|z-he|\}J(z-he)\Bigg\}.
	\end{align*}
	Then, by combining \eqref{basis-bound} and \eqref{bounding-1}, we find
	\begin{equation}
		\label{final-bound1}
		\begin{split}
			\|\delta_h u\|_{L^2(B_2)}^2&\leq \|v\|_{L^2(B_4)}^2\\
			&\leq \frac{\mu^{-1}|B_4|}{\lambda_1-\lambda-2\mu}\Bigg(h^2\|\frac{\delta_hf}{h}\|_{L^{2}(B_4)}^2+ \frac{16}{r^2}\|u\|_{L^\infty(\R^N)}^2 \Big(\ \int_{\R^N}k_h(z)\ dz\Big)^2\Bigg).
		\end{split}
	\end{equation}
	Next we show that we have $\int_{\R^N}k_h(z)\ dz \leq C h^{\alpha}$ for some $C>0$. Clearly, we can bound
	\begin{equation}
		\label{k-h-bound outside}
		\int_{\R^N\setminus B_2(0)}k_h(z)\ dz\leq C_1h
	\end{equation}
	for some $C_1=C_1(n,J)>0$, using that $B_1(0)\cup B_1(he)\subset B_2(0)$ and the properties of $J$. In the following, by making $C_J$ larger if necessary, we may also assume that assumption \eqref{assumptions} reads
	$$
	\sup_{x\in \R^N}\int_{\R^N}\min\{1,|x-y|^{\sigma}\}J(x-y)\ dy=\int_{\R^N}\min\{1,|z|^{\sigma}\}J(z)\ dz\leq C_J.
	$$
	 Then note that $B_{2h}(he)\subset B_{3h}(0)$ and we have
	\begin{align}
		\int_{B_{2h}(0)} &\min\{1,|z|\}J(z)+\min\{1,|z-he|\}J(z-he)\ dz\notag\\
		&\leq C_J\int_{B_{3h}(0)} |z|^{1-\sigma-n}\ dz + C_J\int_{B_{3h}(he)}|z-he|^{1-\sigma-n}\ dz\notag\\
		&=\frac{2|B_{1}(0)|C_J}{n} \int_0^{3h}\rho^{-\sigma}\ d\rho=\frac{2|B_{1}(0)|C_J}{n(1-\sigma)}(3h)^{1-\sigma}.\label{k-h-bound inside1}
	\end{align}
	While with $b_{\sigma}(t)=\frac{1}{\sigma}t^{-\sigma}$ we have
	\begin{align}
		h\int_{B_2(0)\setminus B_{2h}(0)}&J(z)+\min\{1,|z-he|\}\int_0^1|\nabla J(z-\tau he)|\ d\tau\ dz\notag\\
		&\leq \frac{h|B_{1}(0)|C_J}{n}\int_{2h}^2\rho^{-\sigma-1}\ d\rho+hC_J\int_0^1 \int_{B_{3}(\tau he)\setminus B_{h}(\tau he)}|z||z-\tau he|^{-1-\sigma-n}\ dz\ d\tau\notag\\
		&\leq \frac{h|B_{1}(0)|C_J}{n}b_{\sigma}(2h)+hC_J\int_0^1 \int_{B_{3}(0)\setminus B_{h}(0)}|z+\tau h e||z|^{-1-\sigma-n}\ dz\ d\tau\notag\\
		&\leq \frac{h|B_{1}(0)|C_J}{n} b_{\sigma}(2h)+hC_J \int_{B_{3}(0)\setminus B_{h}(0)}|z|^{-\sigma-n}\ dz +h^2C_J \int_{B_{3}(0)\setminus B_{h}(0)}|z|^{-1-\sigma-n}\ dz\notag\\
		&\leq \frac{2h|B_{1}(0)|C_J}{n} b_{\sigma}(h)+\frac{h^2|B_{1}(0)|C_J}{n} \int_h^3\rho^{-2-\sigma}\ d\rho\notag\\
		&\leq  \frac{2|B_{1}(0)|C_J}{n} hb_{\sigma}(h)+\frac{|B_{1}(0)|C_J}{n(1+\sigma)}h^{1-\sigma}.\label{k-h-bound inside2}
	\end{align}
	Combining \eqref{k-h-bound outside} with \eqref{k-h-bound inside1} and \eqref{k-h-bound inside2} and the choice $\alpha=1-\sigma\in(0,1)$ we find $C_2=C_2(n,J,\alpha)>0$ such that
	\begin{equation}
		\label{l1-bound}
		\int_{\R^N}k_h(z)\ dz\leq C_2h^{\alpha}.
	\end{equation}
	Whence, from \eqref{final-bound1} with \eqref{l1-bound} we have
	\begin{equation}
		\label{thm-differentiability-step1-single}
		\|\delta_h u\|_{L^2(B_2)}^2\leq \|v\|_{L^2(B_4)}^2\leq h^{2\alpha}C_4\Big(\|\frac{\delta_hf}{h}\|_{L^2(B_4)}^2+ \|u\|_{L^\infty(\R^N)}^2\Big),
	\end{equation}
	for a constant $C_4=C_4(N,J,r,\alpha,\lambda)>0$. By a standard covering argument, we then also find with a constant $C_5=C_5(N,J,\Omega,\Omega',\alpha,\lambda)>0$ and $\Omega''=\{x\in \Omega\;:\; \dist(x,\R^N\setminus\Omega)>4r\}$
	\begin{equation}
		\label{thm-differentiability-step1}
		\|\delta_h u\|_{L^2(\Omega')}^2\leq h^{2\alpha}C_4\Big(\|\frac{\delta_hf}{h}\|_{L^2(\Omega'')}^2+ \|u\|_{L^\infty(\R^N)}^2\Big),
	\end{equation}
The claim \eqref{bound-derivative} then follows since $f\in C^1(\Omega)$.
\end{proof}

\begin{rem}
	\label{thm-differentiability-0}
	Combining Theorem \ref{thm:differentiability-step1} with Corollary \ref{thm-bdd2} it follows that we have in the situation of Theorem \ref{thm:differentiability-step1} for every $\Omega'\subset\subset\Omega$
	\begin{equation}
		\label{bound-derivative2}
		\| \delta_{h,e} u\|_{L^2(\Omega')}\leq h^{\alpha}C\Big(\|f\|_{C^1(\Omega)}^2+\|u\|_{L^2(\Omega')}^2+\|u\|_{L^\infty(\R^N\setminus \Omega')}^2\Big)^{\frac12}\quad\text{for all $h>0$, $e\in \partial B_1(0)$.}
	\end{equation}
\end{rem}

\begin{cor}
	\label{thm:differentiability}
	Assume $m=1$. Let $f\in C^2(\overline{\Omega})$, $\lambda\in \R$, and let $u\in {\mathcal V}^J_{loc}(\Omega)\cap L^{\infty}(\R^N)$ satisfy in weak sense $Iu=\lambda u+ f$ in $\Omega$. Then $u\in H^{1}(\Omega')$ and $\partial_iu\in D^J(\Omega')$ for any $\Omega'\subset \subset \Omega$. More precisely, with $\alpha$ as above there is for any $\Omega'\subset\subset \Omega$ a constant $C=C(N,\Omega,\Omega',J,\lambda)>0$ such that
	\begin{equation}
		\label{bound-derivative-final1}
		\sup_{\substack{ e\in \partial B_1(0)\\ h>0}}h^{-2\alpha}\| \delta_{h,e}^2 u\|_{L^2(\Omega')}\leq C\Big(\|f\|_{C^2(\Omega)}^2+\|u\|_{L^2(\Omega')}+\|u\|^2_{L^\infty(\R^N\setminus \Omega')} \Big)^{\frac12},
	\end{equation}
	so that $u\in N^{2\alpha,2}(\Omega')\subset H^1(\Omega')$, that is, there is also $C'=C'(n,J,\Omega,\Omega',\alpha,\lambda)>0$ such that
	\begin{equation}
		\label{bound-derivative-final2}
		\| \nabla u\|_{L^2(\Omega')}\leq C'\Big(\|f\|_{C^2(\Omega)}^2+\|u\|_{L^2(\Omega')}+\|u\|^2_{L^\infty(\R^N\setminus \Omega')} \Big)^{\frac12}
	\end{equation}
	and, moreover,
	$$
	b_{J,\Omega'}(\partial_iu)\leq C'\quad\text{for $i=1,\ldots,N$.}
	$$
\end{cor}
\begin{proof}
	Let $\Omega_i\subset\subset \Omega$, $i=1,\ldots,7$ such that
	$$
	\Omega'\subset\subset \Omega_i\subset\subset \Omega_j \quad\text{for $1\leq i<j\leq 7$.}
	$$
	Let $\eta\in C^{\infty}_c(\Omega_{7})$ with $\eta=1$ on $\Omega_6$ and $0\leq \eta\leq 1$. Fix $e\in \partial B_1(0)$ and $h\in(0,\frac12r)$, where $r=\min\{\dist(\Omega_i,\Omega\setminus \Omega_{i+1})\;:\; i=1,\ldots,6\}$. Then by Lemma \ref{localization} the function $v=\eta\delta_h u$, where we write $\delta_h$ instead of $\delta_{h,e}$, satisfies $Iv=\lambda v+\tilde{f}$ in $\Omega_5$, where $\tilde{f}=\delta_hf+g_{\eta,\delta_hu}$. Following the proof of Theorem \ref{thm:differentiability-step1} to \eqref{thm-differentiability-step1} it follows with Theorem \ref{thm-bdd} that there is $C=C(n,J,r,\alpha,\lambda)>0$ (changing from line to line) such that
	\begin{align*}
		\| \delta_{h}^2 u\|^2_{L^2(\Omega')}&= \|\delta_hv\|^2_{L^2(\Omega')}\leq h^{2\alpha}C\Big(\|\frac{\delta_{h}\tilde{f}}{h}\|_{L^{2}(\Omega_1)}^2+ \|v\|^2_{L^\infty(\R^N)}\Big)\\
		&\leq h^{2\alpha}C\Big(\|\frac{\delta_{h}\tilde{f}}{h}\|_{L^{2}(\Omega_1)}^2+ \|\tilde{f}\|_{L^{\infty}(\Omega_4)}^2+\|v\|_{L^{2}(\Omega_3)}^2+\|v\|_{L^{\infty}(\R^N\setminus \Omega_3)}^2\Big)\\
		&\leq h^{2\alpha}C\Big(\|\frac{\delta_{h}\tilde{f}}{h}\|_{L^{2}(\Omega_1)}^2+ \|\tilde{f}\|_{L^{\infty}(\Omega_4)}^2+\|\delta_hu\|_{L^{2}(\Omega_3)}^2\Big)\\
		&\leq h^{2\alpha}C\Big(\|\frac{\delta_{h}\tilde{f}}{h}\|_{L^{2}(\Omega_1)}^2+ \|\tilde{f}\|_{L^{\infty}(\Omega_4)}^2+h^{2\alpha}\Big(\|f\|_{C^1(\Omega)}^2+\|u\|_{L^{\infty}(\R^N)}^2\Big)\Big),
	\end{align*}
	where we applied once more Theorem \ref{thm:differentiability-step1}. Here, for $x\in \Omega_4$ using assumption \eqref{assumption3} it follows that there is $C=C(J)>0$ such that
	\begin{align*}
		|\tilde{f}(x)|&\leq |\delta_hf(x)|+\Bigg|\int_{\R^N\setminus \Omega_6}(1-\eta(y))\delta_hu(y)J(x-y)\ dy\Bigg|\\
		&= |\delta_hf(x)|+\Bigg|\int_{\R^N\setminus \Omega_5}|u(y)|\delta_h[(1-\eta(y))J(x-y)]\ dy\\
		&\leq hC\Big(\|\nabla f\|_{L^{\infty}(\Omega)} +\|u\|_{L^{\infty}(\R^N\setminus \Omega')}\Big).
	\end{align*}
	Moreover, for $x\in \Omega_1$ in a similar way there is $C=C(J)>0$ such that
	\begin{align*}
		|\delta_h\tilde{f}(x)|&\leq |\delta_h^2f(x)|+\Bigg|\int_{\R^N\setminus \Omega_6}(1-\eta(y))\delta_hu(y)\delta_hJ(x-y)\ dy\Bigg|\\
		&\leq h^2\|f\|_{C^2(\Omega)}+ \|u\|_{L^{\infty}(\R^N\setminus \Omega')}\Bigg|\int_{\R^N\setminus \Omega_5}\delta_h[(1-\eta(y))\delta_hJ(x-y)]\ dy\Bigg|\\
		&\leq h^2C\Bigg(\|f\|_{C^2(\Omega)}+ \|u\|_{L^{\infty}(\R^N\setminus \Omega')}\Bigg).
	\end{align*}
	Thus we have
	$$
	\| \delta_{h}^2 u\|^2_{L^2(\Omega')}\leq Ch^{4\alpha}\Bigg(\|f\|_{C^{2}(\Omega)}^2+\|u\|_{L^2(\Omega')}^2+\|u\|_{L^{\infty}(\R^N\setminus \Omega')}^2\Bigg).
	$$
	The proof of the first part then is finished with Proposition \ref{prop-sobolev-nikolskii} since $2\alpha>1$. Next, write $D_hp(x)=\frac{p(x+he)-p(x)}{h}$ for any function $p:\R^N\to\R$, with $e\in \partial B_1(0)$ fixed and $h\in \R\setminus\{0\}$. Then with Lemma \ref{localization} for some $\eta\in C^{\infty}_c(\Omega)$ such that $0\leq \eta\leq1$ and $\eta\equiv 1$ on $\Omega_2\subset\subset \Omega$ with $\Omega'\subset\subset \Omega_1\subset\subset \Omega_2$ we have with $v=\eta u$,
	$$
	Iv=f+\lambda v+g_{\eta,u}\quad\text{in $\Omega_1$, where}\quad g_{\eta,u}=\int_{\R^N\setminus\Omega_2}(1-\eta(y))u(y)J(x-y)\ dy.
	$$
	Next, let $\mu\in C^{\infty}_c(\Omega_1)$ with $0\leq \mu\leq1$ and $\mu\equiv 1$ on $\Omega'$. Then with $\phi=D_{-h}[\mu^2D_hv]\in {\mathcal D}^J(\Omega_1)$ for $h$ small enough we have for some $C>0$ (which may change from line to line independently of $h$)
	\begin{align}
	\label{bdd-rhs-derivative}
	|b_J(v,\phi)|=\Bigg|\int_{\Omega_1}D_hf\mu^2D_hv+ \lambda (\mu D_hv)^2+D_hg_{\eta,u}\mu^2 D_hv\ dx\Bigg|\leq C,
	\end{align}
	since
	$$
	\int_{\Omega_1}|D_hf\mu^2D_hv|\ dx\leq C\|f\|_{C^1(\Omega)}\|\nabla u\|_{L^2(\Omega_2)}<\infty,\qquad \int_{\Omega_1}|\lambda (\mu D_hv)^2|\ dx\leq 2|\lambda|\|\nabla u\|_{L^2(\Omega_2)}^2<\infty,
	$$
	and
$$
\int_{\Omega_1}|D_hg_{\eta,u}\mu^2 D_hv|\ dx\leq C\Big(\int_{\Omega_1}\int_{\R^N\setminus \Omega_2}|(1-\eta(y))u(y)|[D_hJ](x-y)|\ dy\ dx\Big)^{1/2}\|\nabla u\|_{L^2(\Omega_2)}<\infty
$$
due to assumption \eqref{assumption3}. Moreover, with a similar calculation as in the proof of Theorem \ref{thm:differentiability-step1} we have
\begin{align*}
b_J(v,\phi)&=b_J(\mu D_hv,\mu D_hv)-\frac{1}{2}\int_{\R^N}\int_{\R^N} D_hv(x)D_hv(y)(\mu(x)-\mu(y))^2J(x-y)\ dxdy,
\end{align*}
where for some $\Omega_2\subset\subset\Omega_3\subset\subset\Omega_4\subset\subset \Omega$~ with $h$ small enough
\begin{align*}
\int_{\R^N}\int_{\R^N} |D_hv(x)D_hv(y)(\mu(x)-\mu(y))^2&J(x-y)|\ dxdy\\
&\leq C\int_{\Omega_3} \int_{\Omega_3}|D_h(\eta u)(x)D_h(\eta u)(y)||x-y|^2J(x-y)\ dxdy\\
&\leq C\int_{\Omega_3} |D_h(\eta u)(x)|^2\int_{\Omega_3} |x-y|^2J(x-y)\ dydx\\
&\leq C\|\nabla u\|_{L^2(\Omega_4)}\int_{\R^N}\min\{1,|z|^2\}J(z)\ dz<\infty.
\end{align*}
Combining this with \eqref{bdd-rhs-derivative} we find
$$
b_J(\mu D_hv,\mu D_hv)\leq C\quad\text{for all $h>0$ small enough.}
$$
Since also $\mu D_hv\in {\mathcal D}^J(\Omega_2)$ for all $h>0$ small enough (see Lemma \ref{properties1}) and since $D^J(\Omega_2)$ is a Hilbert space, we conclude that $\mu\partial_ev\in {\mathcal D}^J(\Omega_2)$ with
$$
b_J(\mu \partial_ev)\leq C
$$
for $h\to 0$. This finishes the proof.
\end{proof}

\begin{cor}
	\label{thm:differentiability3}
	Let $f\in C^{2m}(\overline{\Omega})$, $\lambda\in \R$, and let $u\in {\mathcal V}^J_{loc}(\Omega)\cap L^{\infty}(\R^N)$ satisfy in weak sense $Iu=\lambda u+f$ in $\Omega$. Then $u\in H^{m}(\Omega')$ for any $\Omega'\subset \subset \Omega$ and there is $C=C(n,J,\Omega,\Omega',m)>0$ such that
	\begin{equation}
		\label{bound-derivative-final3eq}
		\| u\|_{H^m(\Omega')}\leq C\Big(\|f\|_{C^m(\Omega)}^2+\|u\|^2_{L^2(\Omega')}+\|u\|^2_{L^\infty(\R^N\setminus \Omega')} \Big)^{\frac12}.
	\end{equation}
	In particular, if $m=\infty$, then $u\in C^{\infty}(\Omega)$.
\end{cor}
\begin{proof}
	By Corollary \ref{thm:differentiability} the claim holds for $m=1$ in particular with $u|_{\Omega'}\in D^J(\Omega')$ for all $\Omega'\subset\subset \Omega$. Assume next, the claim holds for $m-1$ with $m\in\N$, $m\geq 2$ in the following way: We have $u\in H^{m-1}(\Omega')$ and $\partial^{\beta}u|_{\Omega'}\in D^J(\Omega')$ for any $\Omega'\subset\subset \Omega$ and $\beta\in \N_0^N$ with $|\beta|\leq m-1$, and there is $C=C(n,J,\Omega,\Omega',m)>0$ such that
	\begin{equation}
		\label{bound-derivative-final3}
		\| u\|_{H^{m-1}(\Omega')}\leq C\Big(\|f\|_{C^{m-1}(\Omega)}^2+\|u\|_{L^2(\Omega')}+\|u\|^2_{L^\infty(\R^N\setminus \Omega')} \Big)^{\frac12}.
	\end{equation}
	Fix $\Omega'\subset\subset\Omega$ and let $\Omega_i\subset\subset\Omega$, $i=1,\ldots,7$ and $\eta\in C^{\infty}_c(\Omega_7)$ as in the proof of Corollary \ref{thm:differentiability}. Put $v=\partial^{\beta}(\eta u)$ for some $\beta\in \N_0^N$, $|\beta|=m-1$. Then $Iv=\partial^{\beta}f+\lambda v+\partial^{\beta}g_{\eta,u}$ in $\Omega_5$ by Lemma \ref{localization} and direct computation using the assumptions on $J$. From here, proceeding as in the proof of Corollary \ref{thm:differentiability} by applying Theorem \ref{thm:differentiability-step1} the claim follows.
\end{proof}

\begin{proof}[Proof of Theorem \ref{main-thm-bdd}]
The first part follows from the Poincar'e inequality Lemma \ref{poincare} and Theorem \ref{thm-bdd3} with $h=0=\lambda$. The last assertion follows from Corollary \ref{thm:differentiability3}.
\end{proof}

\begin{proof}[Proof of Corollary \ref{main-thm-eigen2}]
This statement follows directly from Corollary \ref{thm:differentiability3}.
\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
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Nature & Mandala
Home > Nature & Mandala > Page 1 of 1
This category of books is suggested as a way to group the numerous 'garden' books that Tarthang Tulku has composed since the early 1980s
In a sense, each of these books is an ode to the Copper Mountain Mandala of Odiyan, a retreat center in Northern California that sits on 1,000 acres overlooking the Pacific Ocean and is home to four major temples and thousands of prayer wheels, prayer flags and stupas.
But the sensibility expressed in these books has the power to uncover the significance of beauty we all may find in daily life. In that sense, books in the Nature and Mandala Series can help us to derive deep spiritual nourishment from our senses and from our appreciation of the aesthetic realm.
$25.00 Love of Beauty
$29.99 Lotus Mandala
from $15.99 Mandala Gardens
$24.95 Seeing the Beauty of Being
$24.95 A Garland of Flowers: Beauty of the Odiyan Mandala
Int'l Centers
As a refugee in India in the early 1960's, Tarthang Tulku Rinpoche, an eminent lama from the Nyingma school, perceived the urgent need for preserving the sacred texts of the Tibetan Buddhist tradition. Dharma Publishing and Dharma Press began operations in America in 1970 and in 1975 they were formally incorporated as a non-profit organization under the name Dharma Mudranalaya DBA Dharma Publishing and Dharma Press. All proceeds from the sale of our products contribute to the preservation of the Tibetan Buddhist tradition.
Copright Dharma Publishing 2012 All Rights Reserved

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The classic Swedish princess cake is made of cake base, whipped cream and custard. It is coated with green marzipan, powdered sugar and often decorated with red roses made out of marzipan. The original recipe doesn't contain jam, even though it is very common today.
This is a very quick and easy version of the Princess cake, but it tastes fantastic and it is absolutely perfect when you don't have the time to go all out.
Use freshly picked flowers for beautiful decoration.
When the cake is covered with yellow marzipan it is called Prince cake.
Spread jam and custard evenly over two layers of the cake base.
Top with the third layer of the cake base.
Cover the whole cake with a thick layer of whipped cream.
Place the marzipan cover over the whole cake.
Make a nice round form and with a knife, remove the rest of the marzipan.


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The DRmare M4V Converter for Windows is a one stop solution for all the iTunes users out there who are tired of having limited control over using the iTunes content they already paid for. It's completely legal and the process that this particular M4V Converter for Windows uses to remove the DRM from all your iTunes content is the most advanced one in use today. Stay with us as we take a closer look at the software and try to find out what makes it so special and popular for millions of Apple and Windows users out there.
30x Faster than Others
Even if you judge the DRmare M4V Converter only as a DRM removal software, its quality is unmatched for a number of reasons, not the least of which is the speed at which the software does the job. It's actually 30x faster than most of the other similar DRM M4V Converter for Windows that are available. If you have a Sci-Fi movie which you paid for on iTunes but would like to see on the giant drawing room TV, you don't really have to wait long anymore if you use the DRmare converter. Have an entire season of your favorite TV show that you need to remove Apple's DRM from? Don't worry because this DRM removal program also supports batch conversion.
Lossless DRM Removal
Unfortunately, with any other DRM removal tool out there, you will likely lose a significant bit of quality during the process. The good news is that the DRmare DRM removal for Windows tool has no such adverse effects on your iTunes Media. Moreover, you can also choose to keep the original M4V format that's native to all iTunes content or convert the M4V files into MP4 files without any loss in quality whatsoever. The best part is that whether it's in M4V or MP4 file format, you will now be able to play it on all devices such as smartphones, gaming consoles, video players, smart TVs and of course, Windows computers.
Keep the Audio Quality and Formats
Everything from AC3 5.1 Dolby audio tracks to subtitle files in the original DRM-protected video file remains untouched post conversion. This includes other audio files, formats and details such as AAC stereo tracks, AD, CC, etc. as well. This applies to videos of all standard and high definition resolutions (480p, 720p and 1080p videos).
After checking out all the features and testing them out individually, we think that this is probably the best DRM removal software and DRM M4V Converter available today. You don't even have to take our word for it, because you can try out the trial version of the M4V Converter for Windows today and decide for yourself before making an actual purchase.
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Wondershare Video Converter Ultimate Review
Being one of the most complete and inclusive video converter available today, Wondershare Video Converter Ultimate has received a lot of attention lately, but is it really as good a video toolbox as it claims to be? We tried it out and found that it is one of those rare instances where everything they said about the video converter was actually true.
Video Format Conversion
The most basic functions expected of a video converter is of course, changing the format and Wondershare Video Converter delivers on that. Have you ever wondered how to convert MP4 to MP3? Well, Wondershare can do that for you in minutes or even seconds. The same applies for MP4 to WAV conversion, MXF to MP4 conversion, MOV to MP4 conversion and just about every other video format conversion you can think of because the Wondershare video toolbox supports more than 1,000 conversion formats. This includes all the 4K video formats, as well as some of the VR video formats as well.
Support for All Devices
After the conversion to the required format, the videos ran flawlessly run on all Apple devices (iPad, iPod, iPhone, Apple TV), Android devices, Xbox One, Sony PSP and even some of the VR headsets. There's support for a much wider range of devices as well, but you will have to visit their official site for the detailed info on that.
What was most impressive about the Wondershare Video Converter Ultimate is the speed it offers because it's blazing fast, especially when compared to other converters in the market. True to their promise of providing a 30x faster conversion rate, the speed really is impressive with huge videos being converted in a matter of minutes. Wondershare achieves this by combining its propriety APEXTRANS technology with the computer's own GPU and CPU prowess. It would help if you have a powerful PC with a decent graphics card and processor, but the conversion rate will be a lot faster than others on any computer. To top it all off, Wondershare allows the conversion of multiple videos at once and in different format. Batch conversions in different formats, coupled with the speed we just discussed, puts Wondershare miles ahead of the competition.
Burn to DVD and Blu-ray
Aside from the conversion, the video converter can also burn your videos directly onto a DVD or a Blu-ray disc in any format of your choice. Of course, you can also do the same from a traditional ISO image as well if you need to and the process is surprisingly faster than on any other DVD burner software we have used so far.
Wondershare Video Converter Ultimate allows the user to download any video from over 10,000 video-hosting websites, which include the obvious names of YouTube and Dailymotion. But irrespective of whether the video was downloaded from a website, shot from your camera or sent to you by someone, the video converter lets you edit it as much as you need to with its inbuilt video editor.
There's just so much more included in this video toolbox like a screen recorder, a GIF maker, a metadata editor for the videos and even TV casting options. It's hard not to recommend this product to anyone looking for a comprehensive, one-stop solution to all his/her video conversion/making/editing needs, especially since it's possible to Try Wondershare Video Converter Ultimate for free!
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Tuneskit Music Converter: Best Converter and Downloader for Spotify Users?
Tuneskit promises to enhance the experience of Spotify Users who do not have a premium membership, by giving them the power to do what Spotify doesn't allow them to. Let's take a look at the software's highlight features to find out if the Spotify Music Converter really is as good as it seems to be.
Lossless Conversion and Downloading of Music on Spotify
No quality loss was experienced after downloading music from a free Spotify account in MP3, WAV, AAC, M4A, M4B or FLAC format. The metadata (tags, artist information, track information, history, etc.) was also intact and the software replicated exactly what was available on Spotify.
All you need to do is drag the tracks, albums, or playlists from the Spotify screen and onto the Tuneskit Music Converter software window to start the download. It should be downloaded, converted and saved onto your computer, five times faster than any other, similar software available.
The Downloaded Music Can be Transferred Onto Any Device
Once downloaded, the music can then be transferred to any mobile device that supports media playback, including but not limited to iPhones, iPads, iPods, Macs, other Windows PCs, Android smartphones, Android tablets, car audio systems and any other MP3 or music players available in the market.
If a particular device doesn't support the current file formatting of the tracks, the converter can be used once again to change the format and make it compatible in seconds.
Tuneskit Music Converter Automatically Organizes Your Music
After a few downloads, we noticed that the Tuneskit Music Converter is automatically organizing the folders, artists, playlists etc. so that the interface remains easy to navigate and users can easily find the music which is being downloaded and saved. There is also the option to set the conversion outputs manually in a way, so as to make sure that the files are organized according to channels, sample rates or bit rates.
Supports Both Apple MacOS and Microsoft Windows 10
There are some people who swear by a Mac, while hardcore PC users will never even touch a Mac for anything besides light tasks because frankly, Windows is just better in almost every possible way!
The good news is that Tuneskit isn't impartial to any particular set of users, because their Spotify Music Converter is available for both Macs and PCs. It doesn't matter which OS you like more, all the features of the music converter and downloader are available on both Operating Systems.
The Simple Tutorial: Getting Started with Tuneskit's Spotify Converter
You probably do not even need a full-scale tutorial to learn how to use the software for downloading and converting music from Spotify, but you may need some help while trying to transfer the downloaded racks onto your iTunes Library. The Tuneskit website has detailed tutorials for that part thankfully.
Nevertheless, just in case anybody needs a bit of guidance with getting started, follow the steps next and you will get the hang of it in no time.
Step 1: Launch both Spotify and the Tuneskit Music Converter software.
Step 2: Just drag and drop the playlists and tracks that you want to download form Spotify, and drop it onto the Tuneskit Converter window.
Step 3: Select Menu on the Tuneskit window, click on Preferences and then select the output format you want the downloaded tracks to be converted and saved as.
Step 4: Click the Convert tab and the process should begin immediately. Once all chosen tracks are converted and downloaded, you can start transferring them onto other devices immediately.
Is the Software Worth Its Price Tag?
After trying out all features of Tuneskit Music Converter, we can say without a doubt that the Spotify downloader and converter is well worth its price of $49.99, and an awesome bargain at just $29.99 (the sale is on right now).
No premium membership costs and more features from Spotify with just a free account? Yes, that's precisely what the Spotify music converter is capable of providing, and it doesn't have any in-app purchases either. Pay once and get a lifetime of free access to all updates released by the developers to further augment and improve the application's features.
However, the best part about it all is that you don't even have to take anyone's word for it, since the trial version can be downloaded for free from their website. Try the TunesKit Spotify Music Converter out before buying it, and chances are that you will love it way too much to not use it after the trial period is over.
As a matter of fact, once you visit the Tuneskit website, you might be surprised to find that they also have a whole range of other handy software bundles for downloading and converting other types of content as well, that you may even have never considered possible before!
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How to remove DRM from iTunes movies?
What is Tuneskit DRM Media Converter?
The Tuneskit DRM Media Converter for Windows is a DRM removal software that can be used to eliminate DRM copyright protection from any media content that you have purchased from iTunes. You don't even need a Mac or an Apple device to use the DRM removal for Windows software, because it can be downloaded, installed and operated from any Windows PC. The Tuneskit DRM Media Converter can be utilized to convert DRM video to universally playable formats such as MP4 or lossless MP4V from Apple's locked, DRM protected MP4V media files.
Why do You Need a DRM Media Converter?
One of the biggest problems of content purchased on iTunes is that you cannot play it on any other device that isn't made by Apple and that can be quite inconvenient. A DRM Media Converter for Windows on the other hand, lets you get rid of Apple's Strict DRM protections on the iTunes content and convert the media files, so that you can play it pretty much anywhere you like, be it on an Android device or on a Windows PC.
What is DRM Exactly?
DRM stands for Digital Rights Management and it's a form of copyright protection that is used to protect media and other digital content from being stolen, pirated or used without authorization. In case of Apple iTunes, that DRM protection code is known as Fairplay. The Tuneskit Media Converter for Windows removes the restrictions posed on iTunes content by Fairplay in such a manner that the user can play the media he bought on any device he/she wants.
What are the Most Important Features of Tuneskit DRM Media Converter for Windows?
1. Remove iTunes DRM at 30x faster speed than any other similar software
2. Lossless removal of DRM from iTunes movies and TV shows to multiple non-Apple devices
3. The most reliable, fastest and easiest DRM M4V to MP4 converter
4. Lossless conversion lets you Keep AC3 5.1 Dolby Surround Sound, AAC stereo tracks, AD, CC, subtitles, etc. intact
5. A DRM media converter for Windows crack that works
6. It can also be used to remove DRM restrictions from audiobooks (AAX, AA, M4B, M4A to MP3, AAC, M4A, WAV, FLAC)
7. All books you have on iBooks (ePub) can be copied without losing the metadata
Is It Worth the Price?
Considering that there is simply no other DRM Media Converter for Win that comes even near to Tuneskit DRM Media Converter, it would be a safe bet to say that it's most definitely worth your money. However, this only applies if you have other Android or Windows devices that you would like to play your iTunes content on. You can even play them on the big TV in your drawing room after you have used Tuneskit to remove DRM video restrictions. The best part about the whole process is that anyone can try out iTunes video DRM removal Windows free! Download the trial version now and give it a try to see for yourself, just how well and fast everything works with Tuneskit DRM Media Converter for Windows.
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5 Reasons Why YouTubNow is the Best Online YouTube Converter and Downloader
There are quite a few YouTube downloaders/converters available these days and YouTubNow is one of the most popular options among them. Now, the recent popularity of the site suggests that it's actually quite good, but what exactly are those features that make it so good? Well, there are quite a few as it turns out, but we will now take a closer look at five of the most useful features that the online video converter and YouTube downloader boasts of.
YouTube to MP3 and MP4 Converter
You can essentially convert any YouTube video into an MP4 video, before downloading it onto your phone or PC. To do this, simply paste the YouTube video link into the YouTubNow Search box and select the output format as MP4.
If you are more interested in listening to music, easily convert your favorite YouTube music videos into portable MP3, M4A or WebM files, download them and carry the songs around in your smartphone to listen to the tracks anytime and anywhere you want, independent of whether you are connected to the net or not.
It's An Ad-Free, Unlimited Service
The YouTubNow site doesn't have any ads on it that will interrupt your experience while using the converter and downloader, but more importantly, it's a completely unlimited and unrestricted service! What this means is that users can download as many videos in various formats as they wish for free, and it doesn't even ask for verification details, so there's no risk of data theft either.
Its Compatible With Your Apple Devices As Well
Due to the restricted nature of Apple's Operating Systems, most downloading sites do not work on Apple devices very well, but fortunately, Safari is perfectly supported by YouTubNow, which means that whether you are using Chrome, Edge or Safari, YouTubNow will work flawlessly.
Advanced Search Functionality
In addition to being an excellent YouTube converter and downloader with support for multiple formats and browsers across the various Operating Systems and devices, it also has a superb search functionality as well. Just in case you can't find the video you want to download on YouTube due to regional/age related restrictions, simply search for the video with relevant keywords on YouTubNow's search box and it should be able to find the video for you easily.
Better Download Speeds Even at 1080p
To download your videos faster from YouTube even at 1080p (YouTubNow also supports downloading of YouTube videos in lower resolutions), simply replace the "youtube.com" from your video link and replace it with "youtubnow.com." It should ideally look something like what you see below.
From: https://youtube.com/my-video
To: https://youtubnow.com/my-video
As mentioned already, it's completely free and doesn't have any restrictions whatsoever, so don't just take our word for it, click here and try it out for yourself right now!
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