\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{amssymb} \usepackage{esint} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2017 (2017), No. 92, pp. 1--17.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2017 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2017/92\hfil Properties of scales of Kato classes] {Properties of scales of Kato classes, Bessel potentials, Morrey spaces, and a weak Harnack inequality for non-negative solutions of elliptic equations} \author[R. E. Castillo, J. C. Ramos-Fern\'andez, E. M. Rojas \hfil EJDE-2017/92\hfilneg] {Ren\'e Erl\'in Castillo, Julio C. Ramos-Fern\'andez, Edixon M. Rojas} \address{Ren\'e Erl\'in Castillo \newline Departamento de Matem\'aticas, Universidad Nacional de Colombia, Bogot\'a, Colombia} \email{recastillo@unal.edu.co} \address{Julio C. Ramos-Fern\'andez \newline Departamento de Matem\'aticas, Universidad de Oriente, 6101 Cuma\'an Edo. Sucre, \newline Venezuela} \email{jcramos@udo.edu.ve} \address{Edixon M. Rojas \newline Departamento de Matem\'aticas, Universidad Nacional de Colombia, Bogot\'a, Colombia} \email{emrojass@unal.edu.co} \thanks{Submitted December 21, 2016. Published March 30, 2017.} \subjclass[2010]{35B05, 35J10, 35J15} \keywords{Kato class; Bessel potential; Riesz potential; Lorentz space; \hfill\break\indent weak Harnack inequality} \begin{abstract} In this article, we study some basic properties of the scale of Kato classes related with the Bessel kernel, Lorentz spaces, and Morrey spaces. Also we characterize the weak Harnack inequality for non-negative solutions of elliptic equations in terms of the Bessel kernel and the Kato classes of order $\alpha$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} In this article we prove a weak Harnack inequality for non-negative solutions of elliptic differential equations of divergence form with potentials from the Kato class of order $\alpha$. Namely, given a bounded domain $\Omega$ in $\mathbb{R}^n$, we consider the Shr\"odinger operator \[ Lu+Vu=-\sum_{i,j=1}^n\frac{\partial}{\partial x_i} \Big(a_{ij}(x)\frac{\partial}{\partial x_j}u(x)\Big)+V(x)u(x),\quad x\in\Omega, \] where the matrix $A(x)=\big(a_{ij}(x)\big)$ is symmetric, bounded, measurable and positive uniformly in $x$, i.e., \[ \lambda|\xi|^2\leq\langle A(x)\xi,\xi\rangle\leq\Lambda|\xi|^2,\quad x\in\Omega,\;\xi\in\mathbb{R}^n \] for some $0<\lambda\leq\Lambda$. Given $V\in L^1_{\rm{loc}}(\Omega)$, a function $u\in H^1(\Omega)$ is a weak solution of $Lu+Vu=0$ if and only if \[ \int_\Omega\langle A\nabla u,\nabla\varphi \rangle dx +\int_\Omega Vu dx=0,\quad\text{for all }\varphi\in H_0^1(\Omega). \] In this study, we use a class of potential more general than the one considered by Mohammed \cite{Mo01}. The study there is based heavily on the use and properties of the approximation of the Green function and the Green function of the corresponding operator. We substitute the approximate Green function by an approximate kernel of Bessel potentials denoted by $G_\alpha^r$, and the Green function by the Kernel of the Bessel potentials. Also, we relate the Kato class of order $\alpha$ with the Bessel and Riesz potentials. The Kato class $K_n$ on the $n$-dimensional space $\mathbb{R}^n$ was introduced and studied by Aizenman and Simon \cite{AS82,Si82}. The definition of $K_n$ is based on a condition considered by Kato \cite{Ka73}. Similar function classes were defined by Schechter \cite{Sc86} and Stummed \cite{St56}. We refer the reader to \cite{ChFaGa86,DaHi98,Sc86,Si82} for more information concerning to Kato class and its applications. We set \[ \phi(V,r)=\sup_{x\in\mathbb{R}^n}\int_{B(x,r)}\frac{|V(y)|}{|x-y|^{n-2}}dy, \] where $B(x,r)=\{y\in\mathbb{R}^n: |x-y|0$ we set \[ \eta(V)(r)=\sup_{x\in\mathbb{R}^n}\int_{B(x,r)}\frac{|V(y)|}{|x-y|^{n-\alpha}}dy. \] The Kato class of order $\alpha$ consists of locally integrable functions $V$ on $\mathbb{R}^n$ such that \[ \lim_{r\to 0}\eta(V)(r)=0. \] \section{Kato class of order $\alpha$} In this section, we gather definitions and notation that will be used later. By $L_{{\rm loc},u}(\mathbb{R}^n)$ we denote the space of functions $V$ such that \[ \sup_{x\in\mathbb{R}^n}\int_{B(x,1)}|V(y)|dy<\infty. \] \begin{definition} \label{def2.1} \rm The distribution function $D_V$ of a measurable function $V$ is given by \[ D_V(\lambda)=m(\{x\in\mathbb{R}^n: |V(x)|>\lambda\}) \] where $m$ denotes the Lebesgue measure on $\mathbb{R}^n$. \end{definition} \begin{definition} \label{def2.2} \rm Let $V$ be a measurable function in $\mathbb{R}^n$. The decreasing rearrangement of $V$ is the function $V^*$ defined on $[0,\infty)$ by \[ V^*(t)=\inf\{\lambda: D_V(\lambda)\leq t\}\quad (t\geq0). \] \end{definition} \begin{definition}[Lorentz spaces] \label{def2.3} \rm Let $V$ be a measurable function, we say that $V$ belongs to $L(n/\alpha,1)$ ($\alpha>0$) if \[ \int_0^\infty t^{\frac{\alpha}{n}-1}V^*(t)dt<\infty \] and $V$ belongs to $L(\frac{n}{n-\alpha},\infty)$ if \[ \sup_{t>0}t^{1-\frac{\alpha}{n}}V^*(t)<\infty. \] \end{definition} \begin{definition}[Morrey spaces] \label{def2.4} \rm Let $V\in L_{\rm{loc}}^1(\mathbb{R}^n)$, for $q\geq0$, we say that $V$ belongs to $L^{1,n/q}(\mathbb{R}^n)$ if \[ \sup_{x\in\mathbb{R}^n}\frac{1}{r^{n/q}}\int_{B(x,r)}|V(y)|dy =\|V\|_{L^{1,n/q}(\mathbb{R}^n)}<\infty. \] \end{definition} The following definition is a slight variant of the scale $K_{n,\alpha}$ Kato class. \begin{definition} \label{def2.5} \rm Let $V\in L_{\rm{loc}}^1(\mathbb{R}^n)$, we say that $V$ belongs to $\widetilde{K}_{n,\alpha}$ if \[ \eta(V)(r)=\sup_{x\in\mathbb{R}^n}\int_{B(x,r)}\frac{|V(y)|}{|x-y|^{n-\alpha}}dy <\infty. \] \end{definition} Next, we study some properties of the class $\widetilde{K}_{n,\alpha}$. \begin{lemma} \label{lem2.1} $\widetilde{K}_{n,\alpha}\subset L_{{\rm loc},u}^1(\mathbb{R}^n)$. \end{lemma} \begin{proof} Let $V\in\widetilde{K}_{n,\alpha}$ and fix $r_0>0$. Then there exits a positive constant $C>0$ such that $\eta(V)(r)\leq C$. It follows that \[ \sup_{x\in\mathbb{R}^n}\frac{1}{r_0^{n-\alpha}}\int_{B(x,r_0)}|V(y)|dy \leq \sup_{x\in\mathbb{R}^n}\int_{B(x,r_0)}\frac{|V(y)|}{|x-y|^{n-\alpha}}dy,\quad (\alpha>0). \] Therefore, \[ \sup_{x\in\mathbb{R}^n}\int_{B(x,r_0)}|V(y)|dy0)$. \end{lemma} \begin{proof} Let $V\in L(n/\alpha,1)$ $(\alpha>0)$, then \[ \int_0^\infty t^{\frac{\alpha}{n}-1}V^*(t)dt<\infty. \] Since $|V|\chi_{B(x,\varepsilon)}\leq |V|$, we have $(V\chi_{B(x,\varepsilon)})\leq V^*(t)$. Then \[ \int_0^\infty t^{\frac{\alpha}{n}-1}(V\chi_{B(x,\varepsilon)})^*(t)dt \leq \int_0^\infty t^{\frac{\alpha}{n}-1}V^*(t)dt. \] Thus, $V\chi_{B(x,\varepsilon)}\in L(n/\alpha,1)$. On the other hand, letting $g(x)=|x|^{\alpha-n}$, we have \begin{align*} m(\{x: g(x)>\lambda\}) &= m(\{x: |x|^{\alpha-n}>\lambda\})\\ &= m\Big(\big\{x: |x|<(\frac{1}{\lambda})^{\frac{1}{n-\alpha}}\big\}\Big)\\ &= C_n\big(\frac{1}{\lambda}\big)^{\frac{n}{n-\alpha}}, \end{align*} where $C_n=m(B(0,1))$. Next, we set $t=C_n(\frac{1}{\lambda})^{\frac{n}{n-\alpha}}$, then $\lambda=C_n^{\frac{n-\alpha}{n}}t^{\frac{\alpha}{n}-1}$. Thus $g^*(t)=C_n t^{\frac{\alpha}{n}-1}$, from this we obtain \[ \|g\|_{(\frac{n}{n-\alpha},\infty)} =\|\frac{1}{|\cdot|^{n-\alpha}}\|_{(\frac{n}{n-\alpha},\infty)} =\sup_{t>0}C_n^{\frac{n-\alpha}{n}} t^{1-\frac{\alpha}{n}}t^{\frac{\alpha}{n}-1} =C_n^{\frac{n-\alpha}{n}}. \] Finally, \[ \int_{B(x,\varepsilon)}\frac{|V(y)|}{|x-y|^{n-\alpha}}dt \leq \|V\chi_{B(x,\varepsilon)}\|_{(\frac{n}{\alpha},1)} \big\|\frac{1}{|\cdot|^{n-\alpha}}\big\|_{(\frac{n}{n-\alpha},\infty)} \leq C_n^{\frac{n-\alpha}{n}}\|V\chi_{B(x,\varepsilon)} \|_{(\frac{n}{\alpha},1)}. \] Thus, $V\in\widetilde{K}_{n,\alpha}$. \end{proof} \begin{example} \label{examp1} \rm Regarding the functions that belong to $\widetilde{K}_{n,\alpha}$, we claim that \[ V(x)=\frac{1}{|x|^\alpha(\log|x|)^{2\alpha}}\in \widetilde{K}_{n,\alpha} . \] \end{example} \begin{proof}[Proof of the Claim] It will be sufficient to show that $V\in L(n/\alpha,1)$, to do this let us consider \[ m\Big(\Big\{x:\frac{1}{|x|^\alpha(\log|x|)^{2\alpha}}>\lambda\Big\} \Big) =m\Big(\Big\{x:|x|^\alpha(\log|x|)^{2\alpha}<\frac{1}{\lambda}\Big\}\Big). \] Putting $\varphi(|x|)=|x|^\alpha(\log|x|)^{2\alpha}$, we have \begin{align*} m\Big(\big\{x:|x|^\alpha(\log|x|)^{2\alpha}<\frac{1}{\lambda}\big\}\Big) =&m\Big(\big\{x:\varphi(|x|)<\frac{1}{\lambda}\big\}\Big)\\ =&m\Big(\big\{x:|x|<\varphi^{-1}(\frac{1}{\lambda})\big\}\Big)\\ =& C_n\Big(\varphi^{-1}\big(\frac{1}{\lambda}\big)\Big)^n. \end{align*} Let $t=C_n(\varphi^{-1}(\frac{1}{\lambda}))^n$, thus $C^{1/n}_n\varphi^{-1}(\frac{1}{\lambda})=t^{1/n}$, then $\varphi^{-1}(\frac{1}{\lambda})=C(n)t^{1/n}$, where $C(n)=C_n^{-1/n}$ so $\frac{1}{\lambda}=\varphi(C(n)t^{1/n})$, hence \[ \lambda=\frac{1}{\varphi\big(C(n)t^{1/n}\big)} =\frac{C(n)}{|t|^{\frac{\alpha}{n}}(\log|t|)^{2\alpha}}. \] Therefore \[ V^*(t)=\frac{C(n)}{|t|^{\frac{\alpha}{n}}(\log|t|)^{2\alpha}} =\frac{C(n)}{t^{\frac{\alpha}{n}}(\log t)^{2\alpha}}\,. \] Note taht \[ \int_0^\infty t^{\frac{\alpha}{n}-1}V^*(t)dt =C(n)\int_0^\infty t^{\frac{\alpha}{n}-1}\frac{dt} {t^{\frac{\alpha}{n}}(\log t)^{2\alpha}} =C(n)\int_0^\infty\frac{dt}{t(\log t)^{2\alpha}}<\infty, \] then $V\in L(n/\alpha,1)$, hence $V\in\widetilde{K}_{n,\alpha}$. \end{proof} \begin{lemma} \label{lem2.3} If $V\in L^{1,n/q}(\mathbb{R}^n)$ and $p>n/\alpha$, $1\leq p\leq\infty$, then \[ \int_{B(x,\delta)}\frac{|V(y)|}{|x-y|^{n-\alpha}}dy \leq\delta^{\alpha-n/p}\|V\|_{L^{1,n/q}(\mathbb{R}^n)}. \] Moreover $L^{1,n/q}(\mathbb{R}^n)\subset \widetilde{K}_{n,\alpha}$. \end{lemma} \begin{proof} Let $V\in L^{1,n/q}(\mathbb{R}^n)$. Note that \begin{equation} \label{eq1 lem1} \int_{B(x,\delta)}\frac{|V(y)|}{|x-y|^{n-\alpha}}dy =\int_{\mathbb{R}^n}\frac{|V(y)\chi_{B(x,\delta)}(y)|}{|x-y|^{n-\alpha}}dy =\int_{\mathbb{R}^n}\frac{d\mu(y)}{|x-y|^{n-\alpha}}, \end{equation} where $d\mu(y)=|V(y)|\chi_{B(x,\delta)}(y)dy$, from this we have \[ \mu(B(x,r))\int_{B(x,r)}|V(y)|\chi_{B(x,\delta)}(y)dy. \] Going back to \eqref{eq1 lem1}, we obtain \begin{align*} \int_{\mathbb{R}^n}\frac{d\mu(y)}{|x-y|^{n-\alpha}} =&(n-\alpha)\int_0^\infty r^{\alpha-n-1}\mu(B(x,r))dr\\ =&(n-\alpha)\int_0^\infty r^{\alpha-n-1}\int_{B(x,r)\cap B(x,\delta)}|V(y)|dydr\\ =& (n-\alpha)\int_0^\delta r^{\alpha-n-1}\int_{B(x,r)}|V(y)|dydr\\ &+(n-\alpha)\int_\delta^\infty r^{\alpha-n-1}\int_{B(x,\delta)}|V(y)|dydr\\ \leq& (n-\alpha)\int_0^\delta r^{\alpha-n+\frac{n}{q}-1} \Big(\frac{1}{r^{n/q}}\int_{B(x,r)}|V(y)|dy\Big)dr\\ &+(n-\alpha)\delta^{n/q}\int_\delta^\infty r^{\alpha-n-1} \Big(\frac{1}{\delta^{n/q}}\int_{B(x,r)}|V(y)|dy\Big)dr\\ \leq&(n-\alpha)\Big[\int_0^\delta r^{\alpha-\frac{n}{p}-1}dr+\delta^{n/q} \int_\delta^\infty r^{\alpha-n-1}dr\Big]\|V\|_{L^{1,n/q}(\mathbb{R}^n)}\\ =&(n-\alpha)\Big[[\frac{r^{\alpha-n/p}}{\alpha-\frac{n}{p}} \Big|_0^\delta+\delta^{n/q}\frac{r^{\alpha-n}}{\alpha-n}\Big|_0^\infty \Big]\|V\|_{L^{1,n/q}(\mathbb{R}^n)}\\ =&(n-\alpha)\Big[\frac{p(n-\alpha)+(\alpha p-n)}{(\alpha-n/p)(n-\alpha)}\Big] \delta^{\alpha-\frac{n}{p}}\|V\|_{L^{1,n/q}(\mathbb{R}^n)}\\ =&\Big[\frac{np-n}{p\alpha-n}\Big]\delta^{\alpha-n/p}\|V\|_{L^{1,n/q}(\mathbb{R}^n)}. \end{align*} Therefore $L^{1,n/q}(\mathbb{R}^n)\subset \widetilde{K}_{n,\alpha}$. \end{proof} \section{Space of functions of bounded mean oscillation (BMO)} In the same sense that the Hardy space $H^1(\mathbb{R}^n)$ is a substitute for $L^1(\mathbb{R}^n)$, it will turn out that the space $BMO(\mathbb{R}^n)$ (the space of ``bounded mean oscillation'') is the corresponding natural substitute for the space $L^\infty(\mathbb{R}^n)$ of bounded functions on $\mathbb{R}^n$. A locally integrable function $f$ belongs to $BMO$ if \begin{equation}\label{BMO eq1} \frac{1}{m(B_r)}\int_{B_r}|f(x)-f_{B_r}|dm\leq A \end{equation} holds for all balls $B_r=B(x,r)$, here \[ f_{B_r}=\frac{1}{m(B_r)}\int_{B_r}fdm=\fint_{B_r}fdm \] denotes the mean value of $f$ over the ball and $m$ stand for the Lebesgue measure on $\mathbb{R}^n$. The inequality \eqref{BMO eq1} asserts that over any ball $B$, the average oscillation of $f$ is bounded. The smallest bound $A$ for which \eqref{BMO eq1} is satisfied is then taken to be the norm of $f$ in this space, and is denoted by $\|f\|_{BMO}$. Let us begin by making some remarks about functions that are in $BMO$. The following result is due to Jhon-Niremberg. If $f\in BMO$ then there exist positive constants $C_1$ and $C_2$ so that, for every $r>0$ and every ball $B_r$ \[ m(\{x\in B_r:|f(x)-f_{B_r}|>\lambda\})\leq Ce^{-C_2\lambda/\|f\|_{BMO}}m(B_r). \] One consequence of the above result is the following corollary. \begin{corollary}\label{BMO cor1} If $f\in BMO$, then there exist positive constants $C_1$ and $C_2$ such that \[ \int_{B_r}e^{C|f(x)-f_{B_r}|}dm\leq\Big(\frac{C_1C}{C_2-C}+1\Big)m(B_r) \] for every ball $B_r$ and $0\lambda\})d\lambda\\ \leq& CC_1\Big[\int_0^\infty e^{-(C_2-C)\lambda}d\lambda\Big]m(B_r). \end{align*} From the above inequality we have \[ \int_{B_r}e^{C|f(x)-f_{B_r}|}dm\leq\Big(\frac{CC_1}{C_2-C}+1\Big)m(B_r). \] \end{proof} \section{$p$ bounded mean oscillation} A locally integrable function $f$ belongs to $BMO_p$ if for $1\leq p<\infty$ \[ \|f\|_{BMO_p}=\sup_{B_r}\Big(\frac{1}{m(B_r)}\int_{B_r}|f(x)-f_{B_r}|^pdm\Big)^{1/p} <\infty. \] \begin{theorem}\label{p bounded thm1} If $f\in BMO_p$ then there exists a positive constant $C$ depending on $p$ such that \[ \|f\|_{BMO}\leq C_p\|f\|_{BMO_p}. \] \end{theorem} \begin{proof} Let $f\in BMO_p$ by virtue of the H\"older inequality we have \[ \int_{B_r}|f(x)-f_{B_r}|dm\leq[m(B_r)]^{1-1/p} \Big(\int_{B_r}|f(x)-f_{B_r}|^pdm\Big)^{1/p}. \] Hence \[ \frac{1}{m(B_r)}\int_{B_r}|f(x)-f_{B_r}|dm \leq\sup_{B_r}\Big(\frac{1}{m(B_r)}\int_{B_r}|f(x)-f_{b_r}|^pdm\Big)^{1/p} \] for any ball $B_r$. \end{proof} \section{Bessel kernel} The connection between the Bessel and Riesz potential was observed by Stein \cite{ST93, ST70}. We will develop the basic properties of the Bessel kernel. Here $F:S'\to S'$ denotes the Fourier transform on $S'$ where $S'$ represent the set of all tempered distributions. $S'$ is thus the dual of the Schwartz space $S$. For $f\in L^1(\mathbb{R}^n)$ we have \[ F(f)(\xi)=\int_{\mathbb{R}^n}f(x)e^{-2\pi ix\cdot \xi}dx. \] The Riesz kernel, $I_\alpha$, $0<\alpha0$ we have \begin{equation}\label{potential eq3} (4\pi)^{\alpha/2}(1+4\pi^2|x|^2)^{-\alpha/2} =\frac{1}{\Gamma(\alpha/2)}\int_0^\infty e^{-\frac{\delta} {4\pi}(1+4\pi^2|x|^2)}\delta^{\alpha/2}\frac{d\delta}{\delta}. \end{equation} Now we want to compute \[ F\{(1+4\pi^2|x|^2)^{-\alpha/2}\}(\xi)=\int_{\mathbb{R}^n}(1+4\pi^2|x|^2)^{-\alpha/2} e^{-2\pi i x\cdot\xi}dx. \] By \eqref{potential eq3} we obtain \begin{align*} &\frac{1}{(4\pi)^{\alpha/2}}\int_{\mathbb{R}^n} \Big(\frac{1}{\Gamma(\alpha/2)}\int_0^\infty e^{-\frac{\delta}{4\pi}(1+4\pi^2|x|^2)} \delta^{\alpha/2}\frac{d\delta}{\delta}\Big)e^{-2\pi ix\cdot\xi}dx\\ &=\frac{1}{(4\pi)^{\alpha/2}\Gamma(\alpha/2)} \int_0^\infty e^{-\frac{\pi|\xi|^2}{\delta}}e^{-\frac{\delta}{4\pi}} \delta^{\frac{\alpha-n}{2}}\frac{d\delta}{\delta}; \end{align*} therefore \[ F\{(1+4\pi^2|x|^2)^{-\alpha/2}\}(\xi) =\frac{1}{(4\pi)^{\alpha/2}\Gamma(\alpha/2)} \int_0^\infty e^{-\frac{\pi|\xi|^2}{\delta}}e^{-\frac{\delta}{4\pi}} \delta^{\frac{\alpha-n}{2}}\frac{d\delta}{\delta}. \] \subsection{Bessel kernel} We define the Bessel kernel \begin{equation}\label{Bessel eq1} G_\alpha(x)=\frac{1}{(4\pi)^{\alpha/2}\Gamma(\alpha/2)} \int_0^\infty e^{-\frac{\pi|x|^2}{\delta}}e^{-\frac{\delta}{4\pi}}\delta^{\frac{\alpha-n}{2}}\frac{d\delta}{\delta}. \end{equation} \begin{lemma} \label{Bessel lem1} \begin{itemize} \item[(a)] For each $\alpha>0$, $G_\alpha(x)\in L^1(\mathbb{R}^n)$. \item[(b)] $F(G_\alpha(x))=(1+4\pi^2|x|^2)^{-\alpha/2}$. \end{itemize} \end{lemma} \begin{proof} (a) By \eqref{Bessel eq1} we obtain \[ \int_{\mathbb{R}^n}G_\alpha(x)dx=\int_{\mathbb{R}^n}\Big(\frac{1}{(4\pi)^{\alpha/2} \Gamma(\alpha/2)}\int_0^\infty e^{-\frac{\pi|x|^2}{\delta}} e^{-\frac{\delta}{4\pi}}\delta^{\frac{\alpha-n}{2}}\frac{d\delta}{\delta}\Big)dx. \] Since $\int_{\mathbb{R}^n}e^{-\frac{\pi|x|^2}{\delta}}dx=\delta^{n/2}$ and using Fubini, we set \[ \int_{\mathbb{R}^n}G_\alpha(x)dx=\frac{1}{(4\pi)^{\alpha/2}\Gamma(\alpha/2)} \int_0^\infty e^{-\frac{\delta}{4\pi}}\delta^{\frac{\alpha-n}{2}} \Big(\int_{\mathbb{R}^n}e^{-\pi|x|^2/\delta}dx\Big)\frac{d\delta}{\delta}. \] After a suitable change of variable we have \[ \int_{\mathbb{R}^n}G_\alpha(x)dx=1, \] and so $G_\alpha(x)\in L^1(\mathbb{R}^n)$. (b) In the sense of distributions we have whenever $\varphi\in S$, \begin{equation}\label{Bessel eq2} \int_{\mathbb{R}^n}f(x)F(\varphi(x))dx=\int_{\mathbb{R}^n}F(f(x))\varphi(x)dx. \end{equation} Let us consider the function \[ f(x)=e^{-\frac{\delta}{4\pi}}e^{-\pi|x|^2};\quad\text{then } F(f(x))=e^{-\frac{\delta}{4\pi}}e^{-\frac{\pi|x|^2}{\delta}}\delta^{-n/2}. \] By \eqref{Bessel eq2} we have \[ \int_{\mathbb{R}^n}e^{-\frac{\delta}{4\pi}(1+4\pi^2|x|^2)}\hat{\varphi}(x)dx =\int_{\mathbb{R}^n}e^{-\frac{\delta}{4\pi}}e^{-\frac{\pi|x|^2}{\delta}} \delta^{-n/2}\varphi(x)dx, \] where $\hat{\varphi}(x)=F(\varphi(x))$, then \begin{align*} &\int_0^\infty\Big(\frac{1}{(4\pi)^{\alpha/2}\Gamma(\alpha/2)} \int_{\mathbb{R}^n}e^{-\frac{\delta}{4\pi}(1+4\pi^2|x|^2)}\hat{\varphi}(x)dx\Big)\delta^{\alpha/2} \frac{d\delta}{\delta}\\ &=\int_0^\infty\Big(\frac{1}{(4\pi)^{\alpha/2}\Gamma(\alpha/2)}\int_{\mathbb{R}^n} e^{-\frac{\delta}{4\pi}}e^{-\frac{\pi|x|^2}{\delta}}\delta^{-n/2}\varphi(x)dx\Big) \delta^{\alpha/2}\frac{d\delta}{\delta}\,. \end{align*} By Fubini's theorem, \begin{align*} &\int_{\mathbb{R}^n}\Big(\frac{1}{(4\pi)^{\alpha/2}\Gamma(\alpha/2)} \int_0^\infty e^{-\frac{\delta}{4\pi}(1+4\pi^2|x|^2)} \delta^{\alpha/2}\frac{d\delta}{\delta}\Big)\hat{\varphi}(x)dx\\ &=\int_{\mathbb{R}^n}\Big(\frac{1}{(4\pi)^{\alpha/2}\Gamma(\alpha/2)} \int_0^\infty e^{-\frac{\delta}{4\pi}}e^{-\frac{\pi|x|^2}{\delta}} \delta^{\frac{\alpha-n}{2}}\frac{d\delta}{\delta}\Big)\varphi(x)dx. \end{align*} That is, \[ \int_{\mathbb{R}^n}(1+4\pi^2|x|^2)^{-\alpha/2}\hat{\varphi}(x)dx =\int_{\mathbb{R}^n}G_\alpha(x)\varphi(x)dx; \] therefore $ F(G_\alpha(x))=(1+4\pi^2|x|^2)^{-\alpha/2}$. \end{proof} \begin{remark} \label{rmk1} \rm From Lemma \ref{Bessel lem1}(b) we have $G_\alpha\ast G_\beta=G_{\alpha+\beta}$. \end{remark} \begin{lemma} \label{lem5.2} $F\{\int_0^\infty e^{-\pi\delta|x|^2}\delta^a\frac{d\delta}{\delta}\} =\int_0^\infty e^{-\pi\frac{|x|^2}{\delta}}\delta^{-n/2}\delta^a \frac{d\delta}{\delta}$. \end{lemma} \begin{proof} By definition \begin{align*} F\Big\{\int_0^\infty e^{-\pi\delta|x|^2}\delta^a\frac{d\delta}{\delta}\Big\} =&\int_{\mathbb{R}^n}(\int_0^\infty e^{-\pi\delta|x|^2}\delta^a\frac{d\delta} {\delta})e^{-2\pi ix\cdot\xi}dx\\ =&\int_0^\infty(\int_{\mathbb{R}^n} e^{-\pi\delta|x|^2} e^{-2\pi ix\cdot\xi}dx) \delta^a\frac{d\delta}{\delta} \\ =&\int_0^\infty e^{-\pi\delta|x|^2/\delta} \delta^{-n/2} \delta^a \frac{d\delta}{\delta} . \end{align*} Therefore \[ F\Big\{\int_0^\infty e^{-\pi\delta|x|^2}\delta^a\frac{d\delta}{\delta}\Big\} =\int_0^\infty e^{-\pi\delta|x|^2/\delta} \delta^{-n/2} \delta^a \frac{d\delta}{\delta} . \] \end{proof} \begin{proposition}\label{Bessel prop1} $\frac{|x|^{\alpha-n}}{\gamma(\alpha)} =\frac{1}{(4\pi)^{\alpha/2}\Gamma(\alpha/2)}\int_0^\infty e^{-\pi|x|^2/\delta}\delta^{(\alpha-n)/2}\frac{d\delta}{\delta}$. \end{proposition} \begin{proof} We have \begin{align*} &\frac{1}{(4\pi)^{\alpha/2}\Gamma(\alpha/2)}\int_0^\infty e^{-\pi|x|^2/\delta}\delta^{(\alpha-n)/2}\frac{d\delta}{\delta}\\ &=\frac{1}{(4\pi)^{\alpha/2}\Gamma(\alpha/2)}\int_0^\infty e^{-\pi/u}(|x|^2 u)^{\frac{\alpha-n}{2}-1}|x|^{2}du\\ &=\frac{|x|^{\alpha-n}}{(4\pi)^{\alpha/2}\Gamma(\alpha/2)}\int_0^\infty e^{-\pi/u}u^{\frac{\alpha-n}{2}-1}du\\ &=\frac{|x|^{\alpha-n}}{(4\pi)^{\alpha/2}\Gamma(\alpha/2)}\int_\infty^0 e^{-w}(\frac{\pi}{w})^{\frac{\alpha-n}{2}-1}\big(-\frac{\pi}{w^2}\big)dw\\ &=\frac{\pi^{(\alpha-n)/2}|x|^{\alpha-n}}{(4\pi)^{\alpha/2}\Gamma(\alpha/2)} \int_0^\infty e^{-w}w^{\frac{n-\alpha}{2}-1}dw\\ &=\frac{\Gamma(\frac{n}{2}-\frac{\alpha}{2})}{2^\alpha\pi^{n/2}\Gamma(\alpha/2)} |x|^{\alpha-n}\\ &=\frac{|x|^{\alpha-n}}{\gamma(\alpha)}; \end{align*} therefore \[ \frac{|x|^{\alpha-n}}{\gamma(\alpha)} =\frac{1}{(4\pi)^{\alpha/2}\Gamma(\alpha/2)}\int_0^\infty e^{-\pi|x|^2/\delta}\delta^{(\alpha-n)/2}\frac{d\delta}{\delta}. \] \end{proof} \begin{remark} \label{rmk2} \rm By \eqref{potential eq1} and Proposition \ref{Bessel prop1} we can define $I_\alpha(x)$ as follows \begin{equation}\label{Bessel eq3} I_\alpha(x)=\frac{1}{(4\pi)^{\alpha/2}}\Gamma(\alpha/2) \int_0^\infty e^{-\pi|x|^2/\delta}\delta^{(\alpha-n)/2}\frac{d\delta}{\delta}. \end{equation} Comparing the formulas \eqref{Bessel eq1} and \eqref{Bessel eq3} it follows immediately that $G_\alpha(x)$ is positive, and \[ 00$ we have \begin{align*} \int_\varepsilon^\infty e^{-\pi|x|^2/\delta}\delta^{(\alpha-n)/2} \frac{d\delta}{\delta} =&\int_{\frac{\varepsilon}{x|^2}}^\infty e^{-\pi/u}(|x|^2 u)^{\frac{\alpha-n}{2}-1} |x|^{2}du\\ =& |x|^{\alpha-n}\int_{\frac{\varepsilon}{x|^2}}^\infty e^{-\pi/u} u^{\frac{\alpha-n}{2}-1}du\\ =& |x|^{\alpha-n}\int_{|x|^2\frac{\pi}{\varepsilon}}^0 e^{-w} (\frac{\pi}{w})^{\frac{\alpha-n}{2}-1}(-\frac{\pi}{w^2})dw\\ =& |x|^{\alpha-n}\pi^{(\alpha-n)/2}\int_0^{|x|^2\frac{\pi}{\varepsilon}} e^{-w} w^{\alpha-n+1-2}dw. \end{align*} Let us define $\varphi(x,\varepsilon)=\int_0^{|x|^2\frac{\pi}{\varepsilon}} e^{-w} w^{\alpha-n+1-2}dw$, note that $\varphi(x,\varepsilon)\to0$ as $x\to0$. So we can write \[ \int_\varepsilon^\infty e^{-\pi|x|^2/\delta}\delta^{(\alpha-n)/2} \frac{d\delta}{\delta}=C|x|^{\alpha-n}\varphi(x,\varepsilon), \] where $C=\pi^{(\alpha-n)/2}$. Now we have to prove that for every $\tau>0$ there exists $\lambda>0$ such that if $|x|<\lambda$ then \[ \big|G_\alpha(x)-\frac{|x|^{\alpha-n}}{\gamma(\alpha)}\big| \leq \tau|x|^{\alpha-n}\,. \] To do that let us consider \[ G_\alpha(x)-\frac{|x|^{\alpha-n}}{\gamma(\alpha)} =\frac{1}{(4\pi)^{\alpha/2}\Gamma(\alpha/2)}\int_0^\infty e^{-\pi|x|^2/\delta}[e^{\delta/4\pi}-1]\delta^{\frac{\alpha-n}{2}} \frac{d\delta}{\delta}, \] since $\frac{e^{\delta/4\pi}-1}{e^{\delta/4\pi}}\to0$ as $\delta\to0$, we have $e^{-\delta/4\pi}=1+\mathcal{O}(e^{\delta/4\pi})$ as $\delta\to0$. Taking $\tau>0$ there exists $\varepsilon>0$ such that \begin{align*} &\frac{1}{(4\pi)^{\alpha/2}\Gamma(\alpha/2)}\int_0^\infty e^{-\pi|x|^2/\delta}[e^{-\delta/4\pi}-1]\delta^{\frac{\alpha-n}{2}} \frac{d\delta}{\delta}\\ &\leq \frac{1}{(4\pi)^{\alpha/2}\Gamma(\alpha/2)} \int_0^\infty e^{-\pi|x|^2/\delta}\frac{\tau}{2} e^{-\delta/4\pi} \delta^{\frac{\alpha-n}{2}}\frac{d\delta}{\delta}\\ &\leq \frac{\gamma(\alpha)\tau|x|^{\alpha-n}}{2\gamma(\alpha)} =\frac{\tau}{2}|x|^{\alpha-n}; \end{align*} therefore \begin{equation}\label{Bessel eq4} \frac{1}{(4\pi)^{\alpha/2}\Gamma(\alpha/2)} \int_0^\infty e^{-\pi|x|^2/\delta}[e^{-\delta/4\pi}-1] \delta^{\frac{\alpha-n}{2}}\frac{d\delta}{\delta} \leq \frac{\tau}{2}|x|^{\alpha-n}. \end{equation} Since $\varepsilon>0$ has been chosen we take $|x|<\lambda$ such that $\varphi(x,\varepsilon)\leq\frac{\tau}{2c}$. Then we obtain \begin{align*} &\frac{1}{(4\pi)^{\alpha/2}\Gamma(\alpha/2)}\int_\varepsilon^\infty e^{-\pi|x|^2/\delta}[e^{-\delta/4\pi}-1]\delta^{\frac{\alpha-n}{2}} \frac{d\delta}{\delta}\\ &\leq \frac{1}{(4\pi)^{\alpha/2}\Gamma(\alpha/2)}\int_\varepsilon^\infty e^{-\pi|x|^2/\delta}\frac{\tau}{2} e^{-\delta/4\pi}\delta^{\frac{\alpha-n}{2}} \frac{d\delta}{\delta}\\ &\leq \frac{\tau}{2(4\pi)^{\alpha/2}\Gamma(\alpha/2)}\int_\varepsilon^\infty e^{-\pi|x|^2/\delta}\delta^{\frac{\alpha-n}{2}}\frac{d\delta}{\delta}\\ &\leq \frac{\tau}{2(4\pi)^{\alpha/2}\Gamma(\alpha/2)}C|x|^{(\alpha-n)} \varphi(x,\varepsilon)\\ &\leq \frac{\tau}{2}|x|^{(\alpha-n)}. \end{align*} Finally \begin{align*} \big|G_\alpha(x)-\frac{|x|^{(\alpha-n)}}{\gamma(\alpha)}\big| &\leq \frac{1}{(4\pi)^{\alpha/2}\Gamma(\alpha/2)} \int_0^\infty e^{-\pi|x|^2/\delta}[e^{-\delta/4\pi}-1] \delta^{\frac{\alpha-n}{2}}\frac{d\delta}{\delta}\\ &= \frac{1}{(4\pi)^{\alpha/2}\Gamma(\alpha/2)} \Big[\int_0^\varepsilon e^{-\pi|x|^2/\delta}[e^{-\delta/4\pi}-1] \delta^{\frac{\alpha-n}{2}}\frac{d\delta}{\delta} \\ &\quad +\int_\varepsilon^\infty e^{-\pi|x|^2/\delta}[e^{-\delta/4\pi}-1] \delta^{\frac{\alpha-n}{2}}\frac{d\delta}{\delta} \Big] \end{align*} from \eqref{Bessel eq3} and \eqref{Bessel eq4} we obtain \[ \big|G_\alpha(x)-\frac{|x|^{(\alpha-n)}}{\gamma(\alpha)}\big| \leq \tau|x|^{\alpha-n}; \] therefore \[ G_\alpha(x)-\frac{|x|^{(\alpha-n)}}{\gamma(\alpha)}=\mathcal{O}(|x|^{\alpha-n})\quad \text{as } |x|\to 0 \text{ for }0<\alpha0$ and $\tilde{C}_\alpha>0$ such that \[ \tilde{C}_\alpha|x|^{\alpha-n}\leq G_\alpha(x)\leq C_\alpha|x|^{\alpha-n} \] for all $x$ with $0<|x|<1$. Also from Proposition \ref{Bessel prop3} we observe that, for every $\alpha>0$ there exist $M_\alpha>0$ such that \[ G_\alpha(x)\leq M_\alpha e^{C|x|} \] for all $x\in\mathbb{R}^n$ with $|x|>1$. From these two observations we can write \[ G_\alpha(x)\leq C_\alpha(\frac{\chi_{B(0,1)}(x)}{|x|^{n-\alpha}}+e^{-C|x|})\quad \text{for all } x\in \mathbb{R}^n. \] \end{remark} Next we use the Bessel kernel to build a explicit weak solution for the Schr\"odinger operator. Let $\varphi$ be a function belonging to $C_0^\infty(\mathbb{R}^n)$ and such that \[ \varphi(x)=\begin{cases} 1&\text{if } |x|\leq1\\ 0& \text{if } |x|\geq2 \end{cases} \] with $0\leq\varphi(x)\leq1$ for every $x\in \mathbb{R}^n$, we set $ \varphi_r(x)=\varphi(\frac{x}{r})$ and define \[ G_\alpha^r(x)=G_\alpha(x)\varphi_r(x) \] for $|x|\leq r$. Observe that \[ G_\alpha^r(x)\to G_\alpha(x)\quad \text{as } r \to 0 \] and that $G_\alpha^r\in H_0(\Omega)\cap L^\infty(\Omega)$ with \begin{equation}\label{Bessel eq5} \int_\Omega \langle A\nabla G_\alpha^r,\nabla \varphi\rangle =\fint_{B_r}VG_{\alpha}^{r}\varphi dm \end{equation} for any $\varphi\in H^1_0(\Omega)$ and $V\in L_{\textrm{loc}}^{1}(\Omega)$. $G_{\alpha}^{r}$ will be called an approximate Bessel kernel. \eqref{Bessel eq5} tell us that $G_{\alpha}^{r}$ is a weak solution of $LG_{\alpha}^{r}+VG_{\alpha}^{r}=0$. Also, for a real function $f$ we write $f^+=\max\{f,0\}$ for $x\in \Omega$, and $r>0$ with $B_r=B(x,r)$. \section{Main result} In this section we give a characterization of the weak Harnack inequality for nonnegative solutions of elliptic equations in terms of the Bessel kernel and Kato class of order $\alpha$. We start this section with the following result. \begin{lemma} \label{lem6.1} Let $u$ be a non-negative weak solution of $Lu+Vu=0$. If $\phi(V)(r_0)<\infty$ for some $r_0<0$, then there exists a constant $C>0$ such that \[ \fint_{B_r}\Big|\log u-\fint_{B_r}\log u\Big|^2dm\leq C \] for $B_{2r}\subseteq\Omega$ with $00$ such that $\eta(V)(2r)\leq C$. Now, it follows that \[ \frac{1}{(2r)^{n-\alpha}}\int_{B(x,2r)}|V(y)|dy \leq \int_{B(x,2r)}\frac{|V(y)|}{|x-y|^{n-\alpha}}dy. \] Thus, \[ \int_{B(x,2r)}|V(y)|dy\leq \eta(V)(2r)(2r)^{n-\alpha}\leq Cr^{n-\alpha}. \] This immediately gives us \[ \int_{B(x,2r)}|\nabla\log u|^2dm\leq Cr^{n-2}. \] From this and the Poincar\'e inequality we obtain \[ \fint\Big|\log u-\fint_{B_r}\log u\Big|^2dm \leq C\fint|\nabla\log u|^2dm\leq Cr^{n-\alpha}. \] \end{proof} The above lemma and Theorem \ref{p bounded thm1} tell us that $\log u\in BMO$. Then by Corollary \ref{BMO cor1} there exits a positive constant $C$ such that for $\beta>0$, \[ \frac{1}{m(B_r)}\int_{B_r}e^{|f(x)-f_{B_r}|}dm\leq C, \] where $f=\log u$. Using this we conclude that \begin{align*} &\frac{1}{m(B_r)}\Big(\int_{B_r}e^{\beta f}dm\Big) \frac{1}{m(B_r)}\Big(\int_{B_r}e^{-\beta f}dm\Big)\\ &=\frac{1}{(m(B_r))^2}\Big(\int_{B_r}e^{\beta( f-f_{B_r})}dm\Big) \Big(\int_{B_r}e^{-\beta(f-f_{B_r})}dm\Big)\\ &\leq \Big(\frac{1}{m(B_r)}\int_{B_r}e^{\beta|f-f_{B_r}|}dm\Big)^2\leq C, \end{align*} which implies \[ \Big(\int_{B_r}e^{\beta f}dm\Big)\Big(\int_{B_r}e^{-\beta f}dm\Big) \leq C[m(B_r)]^2, \] hence \begin{equation}\label{main eq1} \Big(\int_{B_r}|u|^{\beta}dm\Big)\Big(\int_{B_r}|u|^{-\beta }dm\Big) \leq C[m(B_r)]^2. \end{equation} \begin{proposition} Suppose that \eqref{main eq1} holds, then there exits a positive constant $C$ such that \[ \int_{B_{2r}}|u|^\beta dm\leq C\int_{B_r}|u|^\beta dm, \] where $B_{2r}\subseteq \Omega$. The above inequality is known as doubling condition. \end{proposition} \begin{proof} If \eqref{main eq1} holds, then we have \[ \Big(\int_{B_r}|u|^\beta dm\Big))^{1/2} \Big(\int_{B_r}|u|^{-\beta} dm\Big)^{1/2} \leq C^{1/2}m(B_r); \] from this inequality we obtain \begin{equation}\label{main eq 1.1} \Big(\int_{B_r}|u|^{-\beta} dm\Big)^{1/2} \leq C^{1/2}m(B_r)\Big(\int_{B_r}|u|^\beta dm\Big)^{-1/2}. \end{equation} On the other hand, by Schwartz's inequality and \eqref{main eq 1.1} we have \begin{align*} m(B_r)\leq& \int_{B_r}|u|^{\beta/2}|u|^{-\beta/2}dm\\ \leq& \Big(\int_{B_r}|u|^\beta dm\Big)^{1/2}\Big(\int_{B_r}|u|^{-\beta} dm\Big)^{1/2}\\ \leq& \Big(\int_{B_r}|u|^\beta dm\Big)^{1/2}\Big(\int_{B_{2r}}|u|^{-\beta} dm\Big)^{1/2} \\ \leq& C^{1/2}m(B_r)\Big(\int_{B_r}|u|^\beta dm\Big)^{1/2} \Big(\int_{B_{2r}}|u|^{\beta} dm\Big)^{-1/2}. \end{align*} Thus \[ m(B_r)\leq C^{1/2}m(B_r)\Big(\frac{\int_{B_r}|u|^\beta dm}{\int_{B_{2r}}|u|^\beta dm} \Big). \] Finally \[ \int_{B_{2r}}|u|^\beta dm\leq C\int_{B_r}|u|^\beta dm. \] \end{proof} We need the following mean-value inequality (see, \cite{ChFaGa86}). \begin{theorem}\label{main thm1} Let $u$ be a weak solution of $Lu+Vu=0$ in $\Omega$. Given $00$, we write $\Omega_t^r=\{x\in\Omega: G_\alpha^r(x)>t\}$ and $\Omega_t=\{x\in\Omega: G_\alpha(x)>t\}$, and also define the function \[ H(r,t)=\Big(\frac{G_\alpha^r}{t}-1\Big)-\log^+\Big(\frac{G_\alpha^r}{t}-1 \Big). \] On the one hand, we have $(\log^2x)/2\leq x-1-\log x$ for $x\leq1$. On the other hand, we have \[ \frac{1}{2}\Big[\log^+(\frac{G_\alpha^r}{t})\Big]^2 \leq H(r,t)\leq\frac{G_\alpha^r}{t} \] and that $H(r,t)$ is supported on $\Omega_t^r$ for all $t>0$. Now, we claim that given $\beta>0$ there is a positive constant $C=C(\beta,\lambda,L)$ such that for any $t>0$ \begin{equation}\label{main eq2} \int_{\Omega_t^r}\Big|\nabla\Big(u^{\beta/2}\log^+(\frac{G_\alpha^r}{t})\Big) \Big|^2dm \leq \frac{C}{t}\Big[\int_{\Omega_t^r}|V|G_\alpha^r u^\beta dm +\fint_{B_r}u^\beta dm\Big]. \end{equation} We first prove the claim for a solution of $Lu+VJ(u)=0$ such that $0\leq J(u)\leq u$ and $\inf_{\Omega} u>0$. In the definition \eqref{Bessel eq5} we take \[ \varphi=\Big(\frac{1}{t}-\frac{1}{G_\alpha^r}\Big)^+u^\beta \] as a test function (taking into account that $\inf_{\Omega}u>0$). Then, we find that \begin{equation} \label{main eq3} \begin{aligned} &\int_{\Omega_t^r}\langle A\nabla G_\alpha^r,\nabla G_\alpha^r\rangle \frac{u^\alpha}{(G_\alpha^r)^2}dm + \alpha\int_{\Omega}\langle A\nabla G_\alpha^r,\nabla u\rangle \Big(\frac{1}{t}-\frac{1}{G_\alpha^r}\Big)^+u^\beta dm \\ &=\fint_{B_r}(\frac{1}{t}-\frac{1}{G_\alpha^r})^+u^\beta dm. \end{aligned} \end{equation} Using \[ \nabla(H(r,t)u^{\beta-1})+(1-\beta)u^{\beta-2}H(r,t)\nabla u =u^{\beta-1}(\frac{1}{t}-\frac{1}{G_\alpha^r})^+\nabla G_\alpha^r \] in \eqref{main eq3} follows by application of \eqref{main eq1} that \begin{align*} &\int_{\Omega_t^r}\langle A\nabla G_\alpha^r,\nabla G_\alpha^r\rangle \frac{u^\beta}{(G_\alpha^r)^2}dm + \alpha\int_{\Omega}\langle A\nabla (u^{\beta/2}), \nabla (u^{\beta/2})\rangle\big[\log^+\big(\frac{G_\alpha^r}{t}\big)\big]^2 dm\\ &\leq \fint_{B_r}\frac{u^\beta}{t}dm -\beta\int_\Omega\langle A\nabla u,\nabla(H(r,t)u^{\beta-1})dm \end{align*} from this we have \begin{align*} \int_{\Omega_t^r}\Big|\nabla u^{\beta/2}\log^+(\frac{G_\alpha^r}{t})\Big|^2 dm \leq C(\beta,\lambda)\Big[\fint_{B_r}\frac{u^\beta}{t}dm -\beta\int_\Omega VJ(u)H(r,t)u^{\beta-1}dm\Big]. \end{align*} Recalling that $0\leq J(u)\leq u$ and using \eqref{main eq1} we have \eqref{main eq2}. Now, let $u$ be a non-negative weak solution of $Lu+Vu=0$ in $\Omega$. Then for any $\varepsilon>0$, and $J(u)=u-\varepsilon$, we can see that $w=u+\varepsilon$ is a weak solution of $Lw+VJ(w)=0$ in $\Omega$ with $0\leq J(w)\leq w$ such that $\inf_{\Omega}w>0$. Therefore using \eqref{main eq3} for $w$, letting $\varepsilon\to0$ and using the fact that $u$ is locally bounded, we should apply the Fatou's Lemma and the Lebesgue dominated convergence theorem to have the full statement of the claim. Let $R_j=(\frac{C_j}{t})^{1/(n-\alpha)}$ for $j=1,2$ and $C_1$ and $C_2$ the constants in Remark \ref{Bessel rem}. Then the following inclusions are direct consequences of the inequalities in Remark \ref{Bessel rem} \[ B_{R_2}\subseteq \Omega_t^r, \quad \Omega_t^t\subseteq B_{R_1}. \] Since $u,G_\alpha^r$ belong to $L^1_{{\rm loc}}(\Omega)$, we shall apply the Sobolev inequality to \eqref{main eq3} to obtain the following chain of inequalities: \begin{align*} \frac{C}{R_1^2}\int_{\Omega_{2t}^r}\big|\log^+(\frac{G_\alpha^r}{t})\big|^2 u^\beta dm \leq& \frac{C}{R_1^2}\int_{\Omega_{t}^r}\Big|\log^+(\frac{G_\alpha^r}{t})\Big|^2 u^\beta dm\\ \leq& \int_{\Omega_{t}^r}\Big|\nabla u^{\beta/2}\log^+(\frac{G_\alpha^r}{t}) \Big|^2 dm\\ \leq& \frac{C}{t}\int_{\Omega_{t}^r} |V|G_\alpha^r u^\beta dm +\frac{1}{m(B_r)}\int_{B_r} u^\beta dm. \end{align*} Next, using Remark \ref{Bessel rem} and \eqref{main eq3} in the last inequality, we obtain \[ \frac{1}{R_1^2}\int_{\Omega_{2t}^r}u^\beta dm \leq\frac{C}{t}\sup_{B_{R_1}}u^\beta\int_{B_{R_1}}|V||x-y|^{\alpha-n}dy +\frac{C}{m(B_r)}\int_{B_r}u^\beta dm. \] Since $G_\alpha^r(x)\to G(x)$ as $r\to0$, we observe that $\chi_{\Omega_t}\leq \liminf_{r}\chi_{\Omega_t}r$ from this and the Fatou's Lemma we deduce \[ \frac{1}{R_1^2}\int_{\Omega_{2t}^r}u^\beta dm \leq \frac{C}{t}\big[\eta(V)(R_1)\sup_{B_{R_1}}u^\beta+u^\beta(x)\big] \] by \eqref{main eq3} we obtain \[ \frac{1}{R_1^2}\int_{B_{R_2}}u^\beta dm \leq \frac{C}{t}\big[\eta(V)(R_1)\sup_{B_{R_1}}u^\beta+u^\beta(x)\big] \] and thus, let $r>0$ such that $B_{4r}\subseteq \Omega$. We choose $t$ such that $t=\max\{C_1,C_2\}r^{\alpha-n}$ and observe that \eqref{main eq3} holds if $C_1\leq C_2$ then $R_2=r$ and $R_1\leq r$. If $C_2