text:
\section{\bf Introduction}

	\noi This paper is concerned with the existence of the positive solutions for the family of quasilinear equations with $N$-Laplacian and exponential Choquard type nonlinearity
	\begin{equation*}
	\label{pq}\tag{$P_{*}$}\left\{
	\begin{array}{l}
	-\Delta_N u -\Delta_N (u^{2})u=
	\left(\displaystyle\int_{\Om}\frac{F(y,u)}{|x-y|^{\mu}}~dy\right)f(x,u) \; \text{in}\;
	\Om,\\
	u> 0~\mbox{in}~ \Om,\\
	u=0 ~\mbox{on}~ \partial\Om,
	\end{array}
	\right.
	\end{equation*}
	where $\Om$ is a smooth bounded domain in $\mathbb R^N,\; N\geq2$, $0<\mu<N$ and $\Delta_N:=\text{div}\left(|\nabla u|^{N-2}\nabla u\right)$ is called the $N$-Laplacian.
	\noi The function $f:\Om \times \mb R\to \mb R$ is given by $f(x,s)=g(x,s)\exp(|s|^{\frac{2N}{N-1}})$, where $g \in C(\bar \Om \times \mb R)$ satisfies some appropriate assumptions described later.\\

	\noi  The problems involving the quasilinear  operator $-\Delta_p u -\Delta_p (u^{2})u,\;1<p<\infty,$ has been of interest to many researchers for long due to its significant applications in the modeling of the physical phenomenon such as in plasma physics and fluid mechanics \cite{bass}, in dissipative quantum mechanics \cite{has}, etc.
	Solutions of such equation are related to the existence of standing wave solutions for quasilinear Schr\"odinger equations of
	the form
	\begin{align}\label{l}
	iu_t =-\Delta u+V(x)u-h_1(|u|^2)u-C\Delta h_2(|u|^2)h_2'(|u|^2)u,\;\; x\in\mathbb R^N,
	\end{align}
	where V is a  potential, $C$ is a real constant, $h_1$ and $h_2$ are real functions. Depending upon the different type of  $h_2$, the quasilinear equations of the form \eqref{l} appear in  the study of mathematical physics. For example,  in the modeling of the superfluid film equation in plasma physics, $h_2(s)=s$ (see \cite{33}) and for studying the self-channeling of a high-power ultra short laser in matter $h_2$ is considered to be $\sqrt{1+s^2}$ (see \cite{33}). \\

	\noi The main mathematical difficulty we face in studying the problem of  type $(P_*)$ occurs due to the
	quasilinear term $\Delta_N(u^2)u$, which doesn't allow the natural energy functional corresponding to the
	problem $(P_*)$ to be  well defined for all $u\in W_0^{1,N}(\Om)$ (defined in Section \ref{sec2}). Hence, we can not apply variational method directly for such problem. To overcome this
	inconvenience, several methods and arguments have been developed, such as the perturbation method (see for e.g.,\cite{19,24}) a constrained minimization
	technique (see for e.g., \cite{20,22,28,29}),  and a
	change of variables (see for e.g., \cite{CJ}, \cite {do}-\cite{12}).

	 The nonlinearity in the problem $(P_*)$, which is nonlocal in nature, is driven by the Hardy-Littlewood-Sobolev inequality and the Trudinger-Moser inequality. Let us first recall the following well known Hardy-Littlewood-Sobolev inequality [Theorem 4.3, p.106] \cite{lieb}.
	\begin{proposition}\label{HLS}
		(\textbf {Hardy-Littlewood-Sobolev inequality}) Let $t$, $r>1$ and $0<\mu<N $ with $1/t+\mu/N+1/r=2$, $g_1 \in L^t(\mathbb R^n)$ and $g_2 \in L^r(\mathbb R^n)$. Then there exists a sharp constant $C(t,N,\mu,r)$, independent of $g_1,$ $g_2$ such that
		\begin{equation}\label{HLSineq}
		\int_{\mb R^N}\int_{\mb R^N} \frac{g_1(x)g_2(y)}{|x-y|^{\mu}}\mathrm{d}x\mathrm{d}y \leq C(t,N,\mu,r)\|g_1\|_{L^t(\mb R^N)}\|g_2\|_{L^r(\mb R^N)}.
		\end{equation}
		{ If $t =r = \textstyle\frac{2N}{2N-\mu}$ then
			\[C(t,N,\mu,r)= C(N,\mu)= \pi^{\frac{\mu}{2}} \frac{\Gamma\left(\frac{N}{2}-\frac{\mu}{2}\right)}{\Gamma\left(N-\frac{\mu}{2}\right)} \left\{ \frac{\Gamma\left(\frac{N}{2}\right)}{\Gamma(N)} \right\}^{-1+\frac{\mu}{N}}.  \]
			In this case there is equality in \eqref{HLSineq} if and only if $g_1\equiv (constant)g_2$ and
			\[g_1(x)= c_0(a^2+ |x-b|^2)^{\frac{-(2N-\mu)}{2}}\]
			for some $c_0 \in \mathbb C$, $0 \neq a \in \mathbb R$ and $b \in \mathbb R^N$.}
	\end{proposition}

	\noi Now a days,  an ample amount of attention has been attributed to the study of Choquard type equations, which was started
	with the seminal work of S. Pekar \cite{pekar}, where the author considered
	the following nonlinear Schr\"{o}dinger-Newton equation:
	\begin{align}\label{sn}
		-\Delta u + V(x)u = ({\mathcal{K}}_\mu * u^2)u +\la f_1(x, u)\;\;\; \text{in}\; \mathbb R^N,
		\end{align}
	where $\la>0,$ $\mathcal{K}_\mu$ denotes  the  Riesz potential, $V:\mb R^N\to \mb R$ is a continuous function and $f_1:\mb R^N\times\mb R\to \mb R$ is a Carath\'eodory function with some appropriate growth assumptions.
	When $\la=0$, the nonlinearity in the right-hand side of \eqref{sn} is termed as  Choquard type nonlinearty. In the application point of view, such type of nonlinearity plays an important role in  describing  the Bose-Einstein
	condensation (see \cite{bose}) and also, appears in the modeling of the self gravitational
	collapse of a quantum mechanical wave function (see \cite{penrose}).  P. Choquard (see \cite{choq}) studied such elliptic equations of type \eqref{sn} for construing the quantum theory of a polaron at rest and for modeling the phenomenon of an electron being trapped in its own hole in the Hartree-Fock theory.
	When $V(x)=1,\la=0$, the equations of type \eqref{sn} were   studied rigorously in  \cite {lieb,Lions}.
	For more extensive study  of Choquard equations, without attempting to provide a complete list,
	we refer to \cite{moroz,moroz-main,moroz4,moroz5} and the references therein.\\

	\noi The main feature of the problem $(P_*)$ is that the
	nonlinear term $f(x,t)$ has the maximal growth on $t$, that is, critical exponential growth with respect
	to the following Trudinger–Moser inequality (see \cite{Trud-Moser}):
	\begin{theorem}\label{TM-ineq}(\textbf {Trudinger–Moser inequality})
		For $N\geq 2$, $u \in W^{1,N}_0(\Om)$
		\[\sup_{\|u\|\leq 1}\int_\Om \exp(\alpha|u|^{\frac{N}{N-1}})~dx < \infty\]
		if and only if $\alpha \leq \alpha_N$, where $\alpha_N = N\omega_{N-1}^{\frac{1}{N-1}}$ and $\omega_{N-1}=$ $(N-1)$- dimensional surface area of $\mb S^{N-1}$.
	\end{theorem}
\noi  Here the Sobolev space $W^{1,N}_0(\Om)$ and the corresponding norm $\|\cdot\|$
	are defined in Section \ref{sec2}. The critical growth non-compact problems associated to this inequality are initially studied by the work of  Adimurthi \cite{adi} and de Figueiredo et. al \cite{DR}. More recently, authors in \cite{GPS, GS1, GS2} studied the existence of  multiple positive solutions for quasilinear equations involving exponential nonlinearities.

	Unlike as in the case of critical exponential problem involving $N$-Laplacian, where we generally consider the critical exponential growth as $\exp(|t|^{N/(N-1)})$, in our problem $(P_*)$, due to the presence of the quasilinear operator, the critical exponential growth becomes $\exp(|t|^{2{N/(N-1)}})$.  In the case $1<p<N$, the nonlinearity is of polynomial growth and the critical growth is $ t^{2p^*},$ where  $p^*=Np/(N-p)$ (see \cite{deng,8,27}).

	When $p=2$, the problem of type $(P_*)$ without the convolution term, that is the equation $$-\Delta u -\Delta (u^{2})u + V(x) u=f(x, u) \,\; \text{in}\;\; \mb R^2,$$ where $V:\mb R^N\to \mb R$ is a continuous potential and $f:\mb R^2\times\mb R\to \mb R$ is  a continuous function with some suitable assumption and is having critical exponential growth ($\exp(\al s^{4})$), was studied by Wang et. al in \cite{do}. Later, Wang et. al \cite{wang} extended this problem for $p=N\geq2.$ \\

	\noi Involving critical exponential Choquard type nonlinearity and Laplacian operator, Alves et. al \cite{yang-JDE} studied  the following problem
	\[-\e^2\Delta u+V(x)u=\left(\displaystyle\int_{\mb R^2}\frac{F(y,u)}{|x-y|^{\mu}}~dy\right)f(x,u),\;\;\; x\in\mb R^2,\] where  $\e>0$, $0<\mu<2$, $V:\mb R^2\to\mb R$ is the continuous potential function with some particular properties and the continuous function $f:\mb R^2\times\mb R\to\mb R$ enjoys the  critical exponential growth with some appropriate assumptions. Consequently, for the higher dimension, that is for $N\geq2$, authors in \cite{AGMS} discussed the  Kirchhoff- Choquard problems involving the $N$-Laplacian with critical exponential growth.\\

	\noi Inspired from all the above mentioned works, in this article, we investigate the existence results for the problem $(P_*)$, involving the Choquard type  critical nonlinearity motivated by the above inequality  \eqref{HLSineq}. The main contribution in this work is to identify the first critical level for this problem and study the Palais-Smale sequences below this level.
	We would like to mention that to the best of our knowledge, there is no work on the existence of positive solutions to the elliptic equations involving quasilinear operator  and critical exponential Choquard type nonlinearity. In this article, we study such equations for the first time. Our result is new even for the case $N=2$.\\

	\noi We now state all  the hypotheses imposed  on the continuous function $f:\Om\times\mb R\to\mb R$, given by $f(x,s)= g(x,s)\exp\left(s^{\frac{2N}{N-1}}\right)$, as following:
	\begin{enumerate}
		\item[$(f_1)$] $g\in C(\overline{\Om}\times \mb R)$, $g(x,s)=0,$ for all $s\le 0$, $g(x,s)>0,$ \text{for all}
		$s>0$.

		\item[$(f_2)$]  $\displaystyle\lim_{s\to0}\frac{f(x,s)}{s^{N-1}}=0$, uniformly in $x\in\Om$.
		\item[$(f_3)$] For any $\e>0$, $\displaystyle\lim_{s\to\infty}\sup_{x\in\overline\Om}g(x,s)\exp\left(-\e|s|^{\frac{2N}{N-1}}\right)=0$ and $\displaystyle\lim_{s\to\infty}\inf_{x\in\overline\Om}g(x,s) {\exp(\e |s|^{\frac{2N}{N-1}})}=\infty$.
		\item[$(f_4)$] There exist positive constants  $s'$ and $m_0$ such that
		\[ 0<s^{m_0} F(x,s)\le M_0 f(x,s)\;\mbox{for all}\; (x,s)\in \mb R^2\times[s',+\infty).\]
		\item[$(f_5)$]  There exists $\tau>N$  such that $0<\tau F(x,s)\leq f(x,s)s,$ for all $ s>0.$
		\item[$(f_6)$] We assume that\begin{equation}\label{h-growth}
		\displaystyle \lim_{s\to +\infty} \frac{sf(x,s)F(x,s)}{\exp\left(2 |s|^{\frac{2N}{N-1}}\right)} = \infty,\mbox{ uniformly in }x \in \overline{\Om}.
		\end{equation}
	\end{enumerate}

	\begin{example}
		Consider $f(x,s)= g(x,s)e^{|s|^{\frac{2N}{N-1}}}$, where $g(x,s)=\left\{\begin{array}{lr}
		{t^{a_0+ \left({N}-1\right)} \exp(d_0s^{r})},\; \mbox{if} \; s> 0\\
		0, \; \mbox{if} \; s\leq 0	\end{array}\right.$
		for some $a_0>0$, $0<d_0\leq \alpha_{N}$ and $1\leq r <\frac{2N}{N-1}$. Then $f$ satisfies all the conditions from $(f_1)$-$(f_6)$.
	\end{example}
	Our  main result reads as following:

	\begin{theorem}\label{T1}
		Let $\Om\subset\mathbb R^N\; (N\geq2)$ be a smooth bounded domain and let the hypotheses $(f_1)$-$(f_6)$ hold. Then the problem \eqref{pq} has a non trivial positive weak solution.
	\end{theorem}

	\textbf{Notation.}  Throughout this paper, we make use of the following notations:
	\begin{itemize}

		\item If $u$ is a measurable function, we denote  the positive  and negative parts by $u^{+}=\max\left\{u,0\right\}$ and $u^{-}=\max\left\{-u,0\right\}$, respectively.
		\item If $A$ is a measurable set in $\mathbb{R}^{N}$, we denote by $\vert A \vert$ the Lebesgue measure of $A$.
		\item The arrows $\rightharpoonup $ , $\to $ denote weak convergence,  strong convergence, respectively.
		\item $B_r(x)$ denotes the ball of radius $r>0$ centered at $x\in \Om.$
		\item  $\overline{B_r(x)}$ denotes the closure of the set $B_r(x)$ with respect to $W_0^{1,N}(\Om)$-norm topology.
		\item  $\overline{\mc S}$ denotes the closure of a set $\mc S\subset \mb R^N$.
\item $A\subset\subset B$ implies $\oline A$ is compact in $B$.
	\item $c,C_{1},C_{2},\cdots, \tilde C_1, \tilde C_2,\cdots, C$ and $\tilde C$ denote (possibly different from line to line) positive constants.
	\end{itemize}

	\section{Preliminaries and Variational Set-up}\label{sec2}
	For $u:\Om\to\mathbb R$, measurable function, and  for $1\leq p \leq \infty$, we define the Lebesgue space $L^{p}(\Om)$ as
	$$L^{p}(\Om)=\{u:\Om\to\mathbb R\;\; \text{measurable}| \int_\Om| u|^p dx<\infty\}$$ equipped with the usual norm denoted by $\Vert u\Vert_{L^p(\Om)}$.
	Now the Sobolev space $W^{1,p}_0(\Om)$ is defined as
	$$W^{1,p}_0(\Om)=\{u\in L^p(\Om)| \int_\Om|\nabla u|^p dx<\infty\}$$ which is  endowed with  the norm

	$$\|u\|:=\displaystyle \left({\int_{\Om} |\nabla u|^p dx}\right)^{1/p}.$$

	We have that the embedding $W^{1,N}_0(\Om)\ni u \mapsto \exp(|u|^\beta)\in L^1(\Om)$ is compact for all $\beta \in \left[1, \frac{N}{N-1}\right)$ and is continuous for $\beta = \frac{N}{N-1}$. Consequently, the map $\mc M: W^{1,N}_0(\Om) \to L^q(\Om)$, for $q \in [1,\infty)$, defined by $\mc M(u):= \exp\left( |u|^{\frac{N}{N-1}}\right)$ is continuous with respect to the norm topology.

\noi	The  natural energy functional  associated to the problem \eqref{pq} is the following:

	\begin{align}\label{eng}I(u)=\frac{1}{N}\displaystyle\int_{\Om }(1+2^{N-1}|u|^{N})|\nabla u|^{N}dx-\frac 12\int_{\Om}\int_{\Om} \frac{F(y,u(y))F(x,u(x))}{|x-y|^\mu}dxdy.\end{align}

\noi	Observe that, the functional $I$ is not well defined in $W^{1,N}_0(\Om)$ due to the fact that $\displaystyle\int_{\Om} u^{N}|\nabla u|^{N}dx$ is not finite for all $u\in W_0^{1,N}(\Om) $. So, it is difficult to apply variational methods directly in our problem $(P_*)$. In order to overcome this difficulty, we apply the following change of variables which was introduced in \cite{CJ}, namely, $w:=h^{-1}(u),$ where $h$ is defined by
	\begin{equation}\label{g}
	\left\{
	\begin{array}{l}
	h^{\prime}(s)=\displaystyle\frac{1}{\left(1+2^{N-1}|h(s)|^{N}\right)^{\frac{1}{N}}}~~\mbox{in}~~ [0,\infty),\\
	h(s)=-h(-s)~~\mbox{in}~~ (-\infty,0].
	\end{array}
	\right.
	\end{equation}

\noi	Now we gather some properties of $h$, which we follow throughout in this article. For the detailed proofs of such results, one can see  \cite{CJ,do}.
	\begin{lemma}\label{L1} The function $h$ satisfies the following properties:
		\begin{itemize}
			\item[$(h_1)$] $h$ is uniquely defined, $C^{\infty}$ and invertible;
			\item[$(h_2)$] $h(0)=0$;
			\item[$(h_3)$] $0<h^{\prime}(s)\leq 1$ for all $s\in \mathbb{R}$;
			\item[$(h_4)$] $\frac{1}{2}h(s)\leq sh^{\prime}(s)\leq h(s)$ for all $s>0$;
			\item[$(h_5)$] $|h(s)|\leq |s|$ for all $s\in \mathbb{R}$;
			\item[$(h_6)$] $|h(s)|\leq 2^{1/{(2N)}}|s|^{1/2}$ for all $s\in \mathbb{R}$;
			\item[$(h_7)$] $(h(s))^{2}-h(s)h^{\prime}(s)s\geq 0$ for all $s\in \mathbb{R}$;
			\item[$(h_8)$]  $|h(s)|\geq h(1)|s|$ for $|s|\leq 1$ and $|h(s)|\geq h(1)|s|^{1/2}$ for $|s|\geq 1$;
			\item[$(h_9)$] $h^{\prime \prime}(s)<0$ when $s>0$ and $h^{\prime \prime}(s)>0$ when $s<0$.

		\end{itemize}

	\end{lemma}
	\begin{example}
		One of the example of such functions is given in
		\begin{figure}[H]

			\includegraphics[width=80mm]{pic}
			\caption{Plot of the function $h$}
			\label{pic}
		\end{figure}
	\end{example}
\noi	After employing the change of variable  $w=h^{-1}(u),$ we define the new functional $J:W^{1,N}_0(\Om)\to \mb R$ as
	\begin{align}\label{energy}J(w)=\frac{1}{N}\displaystyle\int_{\Om} |\nabla w|^{N}dx-\frac 12\int_{\Om}\int_{\Om} \frac{F(y,h(w))F(x,h(w))}{|x-y|^\mu}dxdy.\end{align}
	From the properties of $f,h$, it can be derived that the functional $J$ is well defined  and $J\in C^1\left(W
	_0^{1,N}(\Om), \mb R\right).$\\

	We observe that if  $w \in W^{1,N}_0(\Om)$ is a critical point of the functional $J$, that is for every $v \in W^{1,N}_0(\Om)$,
	\begin{align}\label{w.s.}
	\int_{\Om} |\nabla w|^{N-2}\nabla w\nabla v dx-\frac 12\int_{\Om}\int_{\Om} \frac{F(y,h(w)(y))f(x,h(w)(x))}{|x-y|^\mu}h'(w)v(x)dxdy=0,\end{align} then $w$ is a weak solution (solution, for short)  to the following problem:
	\begin{equation}
	\label{pp}\left\{
	\begin{array}{l}
	-\Delta_N w=
	\left(\displaystyle\int_{\mb R^2}\frac{F(y,h(w))}{|x-y|^{\mu}}~dy\right)f(x,h(w)) \; \text{in}\;
	\Om,\\
	w> 0~\mbox{in}~ \Om,\\
	w=0 ~\mbox{on}~ \partial\Om.
	\end{array}
	\right.
	\end{equation}
	It is easy to see that problem \eqref{pp} is equivalent to our problem $(P_*)$, which takes $u = h(w)$ as its solutions. Thus, our main objective is now reduced to proving the existence of the solution of  \eqref{pp}.

	\section{Proof of the main theorem}\label{sec3}
	\subsection{Mountain pass geometry}

	We begin this section with the study of mountain pass structure and Palais-Smale sequences corresponding to the energy functional  $J: W^{1,N}_0(\Om)\rightarrow \mb R$ associated to \eqref{pp}.
	From the assumptions, $(f_2)$-$(f_3)$, we obtain that for any $\e>0$, $r\geq 1$,  there exist $\tilde C(N,\e)$ and $C(N,\e)>0$ such that
	\begin{align}
		&|f(x,s)| \le \e |s|^{N-1 }+ \tilde C(N,\e) |s|^{r-1} \exp\left((1+\e)|s|^{\frac{2N}{N-1}}\right),\;\; \text{for all}\; (x,s)\in \Om \times \mb R,\label{f}\\
	&|F(x,s)| \le \e |s|^N + C(N,\e) |s|^r \exp\left((1+\e)|s|^{\frac{2N}{N-1}}\right),\;\; \text{for all}\; (x,s)\in \Om \times \mb R.\label{k1}
	\end{align}
	Thus, for any $u \in W^{1,N}_0(\Om)$, in light of Sobolev embedding, we have $u \in L^q(\Om)$ for all $q \in [1,\infty)$,

	which implies that
	\begin{equation}\label{KC-new1}
	F(x,u) \in L^{q}(\Om)\mbox{ for any }q\geq 1.
	\end{equation} Now by \eqref{g}, we have if $w\in W^{1,N}_0(\Om),$ then $h(w)\in W^{1,N}_0(\Om)$, which together with
	Proposition \ref{HLS} implies that
	\begin{align}\label{k2}
	\int_{\Om}\left(\int_{\Om}\frac{F(y,h(w))}{|x-y|^{\mu}}dy\right)F(x,h(w))~dx  \leq C(N,\mu){\|F(\cdot,h(w))\|^2_{L^{\frac{2N}{2N-\mu}}(\Om)}}.
	\end{align}

	\begin{lemma}\label{lem7.1}
		Assume that the conditions $(f_1)$-$(f_5)$ hold and let $h$ be defined as in \eqref{g}. Then  there exist $\rho_*>0$ and $R_*>0$ such that
		\begin{align*}
		J(w)\geq R_*>0 \quad \text{for any } w\in W^{1,N}_0(\Om) \text{ with }\ \|w\|=\rho_*.
		\end{align*}
	\end{lemma}
	\begin{proof} Let $w\in W^{1,N}_0(\Om$.	From  \eqref{k2}, \eqref{k1}, H\"{o}lder inequality, Sobolev inequality and Lemma \ref{L1}-$(h_5),(h_6)$, we have
		{\small\begin{align}\label{kc-MP1}
			&\int_{\Om}\left(\int_{\Om} \frac{F(y,h(w))}{|x-y|^{\mu}}dy\right)F(x,h(w))~dx \notag\\
			& \leq C(N,\mu)\|\e |h(w)|^N + C(\e) |h(w)|^r \exp\left((1+\e)|h(w)|^{\frac{2N}{N-1}}\right)\|_{\frac{2N}{2N-\mu}}^2\notag\\
			&\leq C(\mu)\left[2^{\frac{2N}{2N-\mu}}\left\{\e^{\frac{2N}{2N-\mu}} \int_{\Om} |h(w)|^{\frac{2N^2}{2N-\mu}} dx + (C(\e))^{\frac{2N}{2N-\mu}} \int_{\Om} |h(w)|^{\frac{2Nr}{2N-\mu}}\exp\left(\frac{2N}{2N-\mu}(1+\e)|h(w)|^{\frac{2N}{N-1}}\right)  dx\right\} \right]^{\frac{2N-\mu}{N}} \notag\\
			&\leq C(N,\mu)\left[2^{\frac{2N}{2N-\mu}}\left\{\e^{\frac{2N}{2N-\mu}} \int_{\Om} |w|^{\frac{2N^2}{2N-\mu}} dx \right. \right. \\ & \quad + \left.\left. (C(\e))^{\frac{2N}{2N-\mu}} \int_{\Om}2^{\frac{r}{2N-\mu}} |w|^{\frac{Nr}{2N-\mu}}\exp\left(\frac{2N}{2N-\mu}(1+\e) 2^{\frac{1}{N-1}}|w|^{\frac{N}{N-1}}\right)  dx\right\} \right]^{\frac{2N-\mu}{N}} \notag\\
			&\leq C(N,\mu) 2^2 \left[{\e^{2} }  \|w\|^{2N}_{L^{\frac{2N^2}{2N-\mu}}(\Om)} + (C(\e))^2 2^{\frac rN} \|w\|^{2r}_{L^{\frac{2Nr }{2N-\mu}}(\Om)} \left\{\int_{\Om}\exp\left(\frac{4N(1+\e)}{2N-\mu}2^{\frac{1}{N-1}}\left({|w|}\right)^{\frac{N}{N-1}}\right)dx\right\}^{\frac{2N-\mu}{2N}} \right]\notag\\
			&\leq C_1(N,\mu,\e) \left[ \|w\|^{2N} + \|w\|^{2r} \left\{\int_{\Om}\exp\left(\frac{4N(1+\e)}{2N-\mu}2^{\frac{1}{N-1}}\left({|w|}\right)^{\frac{N}{N-1}}\right)dx\right\}^{\frac{2N-\mu}{2N}} \right]\notag\\
			&\leq C_1(N,\mu,\e) \left[ \|w\|^{2N} + \|w\|^{2r} \left\{\int_{\Om}\exp\left(\frac{4N(1+\e)}{2N-\mu}2^{\frac{1}{N-1}}\|w\|^{\frac{N}{N-1}}\left(\frac{|w|}{\|w\|}\right)^{\frac{N}{N-1}}\right)dx\right\}^{\frac{2N-\mu}{2N}} \right]

			\end{align}}
		\noi	So, for sufficiently small $\e>0$ if we choose $w$ such that $\|w\|$ is small enough so that $$\displaystyle\frac{4N2^{\frac{1}{N-1}}(1+\e)\|w\|^{\frac{N}{N-1}}}{2N-\mu} <\al_N,$$ then using Theorem \ref{TM-ineq}  in \eqref{kc-MP1}, we obtain
		\begin{align}\label{1}
		\int_{\Om}\left(\int_{\Om} \frac{F(y,h(w))}{|x-y|^{\mu}}dy\right)F(x,h(w))~dx
		\leq C_1(N,\mu,\e) \left( \|w\|^{2N}  +  \|w\|^{2r} \right).
		\end{align}

		Using \eqref{energy}, \eqref{1}, we have
		\begin{align*}
		J(w) &\geq \frac1N \|w\|^{N}-   C_1(N,\mu,\e) \left( \|w\|^{2 N}  +  \|w\|^{2r}
		\right).
		\end{align*}
		Now by taking  $r>0$ such that  $2r>N$, we can choose $0<\rho_*<1$  sufficiently small so that, we finally obtain $J(w) \geq R_* >0$ for all $w\in W^{1,N}_0(\Om)$ with $\|w\|=\rho_*$ and for some $R_*>0$ depending on $\rho_*$.\end{proof}

	\begin{lemma}\label{lem7.1.2}
		Assume that the conditions $(f_1)$-$(f_5)$ hold and let $h$ be defined as in \eqref{g}. Then  there exists $v_* \in W^{1,N}_0(\Om) $ with $\|v_*\|>\rho_*$ such that $ J(v_*)<0$ , where $\rho_*$ is   given as in Lemma \ref{lem7.1}.
	\end{lemma}
	\begin{proof}
		The condition $(f_5)$ implies that there exist some positive constant $C_1, C_2>0$ such that
		\begin{equation}\label{new2}F(x,s) \geq C_1s^{\tau}-C_2\;\text{ for all}\; (x,s) \in\Om \times [0,\infty).
		\end{equation}

		Let $\phi (\geq0) \in C_c^\infty(\Om)$ such that $\|\phi\|=1$.

		Now by Lemma \ref{L1}-$(h_6)$,$(h_8)$ and \eqref{new2}, for large $t>1$, we obtain
		\begin{align*}
		&\int_{\Om} \left(\int_{\Om} \frac{F(y, h(t\phi
			))}{|x-y|^{\mu}}dy\right)F(x, h(t \phi))~dx\\ &\geq \int_{\Om} \int_{\Om} \frac{(C_1 (h(t \phi(y)))^{\tau}-C_2)(C_1 (h(t \phi(x)))^{\tau}-C_2)}{|x-y|^{\mu}}~dxdy\\
		& = C_1^2  \int_{\Om} \int_{\Om} \frac{(h(t\phi(y)))^{\tau} (h(t\phi(x)))^{\tau}}{|x-y|^\mu}~dxdy \\
		& \quad -2C_1C_2\int_{\Om} \int_{\Om}\frac{(h(t\phi(x)))^{\tau}}{|x-y|^\mu}~dxdy + C_2^2 \int_{\Om} \int_{\Om} \frac{1}{|x-y|^{\mu}}~dxdy\\
		&\geq C_1^2 (h(1))^{2\tau} t^\tau \int_{\Om} \int_{\Om} \frac{(\phi(y))^{\frac \tau 2}(\phi(x))^{\frac \tau 2}}{|x-y|^{\mu}}~dxdy\\

		& \quad -2C_1C_2 2^{\frac{\tau}{2N}} t^{\frac \tau 2}\int_{\Om}  \int_{\Om}\frac{(\phi(x))^{\frac \tau 2}}{|x-y|^\mu}~dxdy + C_2^2 \int_{\Om} \int_{\Om} \frac{1}{|x-y|^{\mu}}~dxdy.
		\end{align*}
		From the last relation and \eqref{energy} , we obtain
		\begin{align}\label{j}
		J(t \phi) & \leq \|t \phi\|^{N}- \frac{1}{2}\int_{\Om}\left( \int_{\Om}\frac{F(y, h(t \phi))}{|x-y|^{\mu}} dy\right)F(x, h( t \phi))~dx\notag\\
		&\leq C_3 {t^N} -C_4 {t^\tau}
		+C_5 {t^{\frac \tau 2}} -{C_6},
		\end{align}
		where $C_i's$ are positive constants for $i=3,4,5,6$.

		From \eqref{j}, we infer that   $J(t\phi) \to -\infty$ as $t \to \infty$, since $\tau>N$. Thus, there exists  $t_0(>0)\in\mathbb R$ so that $v_*(:=t_0 \phi)\in W^{1,N}_0(\Om)$ with $\|v_*\|> \rho_*$ such that $J(v_*)<0$.
	\end{proof}

	\noi From the above two lemmas, we get that $J$ satisfies  the mountain pass geometry near $0$.  Let $\ds \Gamma=\{\gamma\in C([0,1],W_0^{1,N}(\Om)):\gamma(0)=0,J(\gamma(1))<0\}$ and define the mountain pass critical level
	\begin{align}\label{level}\ds \beta_*=\inf_{\gamma\in \Gamma}\max_{t\in[0,1]}J(\gamma(t)).\end{align} Then by Lemma \ref{lem7.1}, Lemma \ref{lem7.1.2} and the mountain pass theorem we know that there exists a Palais-Smale sequence $\{w_k\}\subset W_0^{1,N}(\Om)$ for $J$ at level $\beta_*$, that is, as $k\to\infty$
	\[J(w_k) \to \beta_*; \; \text{and}\; J^\prime(w_k) \to 0 \text{\;\; in\;} \left(W_0^{1,N}(\Om)\right)^*.\] Moreover, Lemma \ref{lem7.1} guarantees that $\beta^*>0.$

	\subsection{Analysis of Palais-Smale sequence}
	\begin{lemma}\label{lem712}
		Let $(f_1)$ and $(f_5)$	hold. Let  $h$ be given as in \eqref{g}. Then every Palais-Smale sequence of $J$ is bounded in $W_0^{1,N}(\Om)$.
	\end{lemma}
	\begin{proof}
		Let $\{w_k\} \subset W_0^{1,N}(\Om)$ be a Palais-Smale sequence of $J$ at level $c\in\mb R$, that is, as $ k \to \infty$
		\begin{align*}
		&J(w_k) \to c,  \\ &J^{\prime}(w_k) \to 0 \text{\;in} \left(W_0^{1,N}(\Om)\right)^*.
		\end{align*}
		 Then we have
		\begin{align}\label{kc-PS-bdd1}
		&\frac 1N\|w_k\|^N - \frac12 \int_{\Om} \left(\int_{\Om} \frac{F(y,h(w_k)}{|x-y|^{\mu}}dy \right)F(x,h (w_k))~dx \to c \; \text{as}\; k \to \infty,\notag\\
		&\left| \int_{\Om}{\nabla|w_k(x)|^{N-2}\nabla w_k \nabla\phi}{dx}\right.\notag\\
		&\quad\quad\quad\quad\quad\quad\quad\quad\quad\left. -\int_\Om \left(\int_\Om \frac{F(y,h(w_k))}{|x-y|^{\mu}}dy \right)f(x,h(w_k)) h'(w_k)\phi ~dx \right|\leq \e_k\|\phi\|,
		\end{align}
		where $\e_k \to 0$ as $k\to \infty$. In the last relation, taking $\phi=w_k$ we get
		\begin{equation}\label{kc-PS-bdd2}
		\left| \|w_k\|^{N}-\int_\Om \left(\int_\Om \frac{F(y,h(w_k))}{|x-y|^{\mu}}dy \right)f(x,h(w_k) h'(w_k) w_k ~dx \right|\leq \e_k\|w_k\|.
		\end{equation}
		Now set $$v_k:=\frac{h(w_k)}{h'(w_k)}.$$  Since by Lemma \ref{L1}-$(h_4)$, we have $|v_k|\leq 2 |w_k|$ and by using \eqref{g} and Lemma \ref{L1}-$(h_5)$, we get
		$$|\nabla v_k|= \left(1+\frac{2^{N-1}|h(w_k)|^N}{1+2^{N-1}|h(w_k)|^N}\right)|\nabla w_k|^N\leq 2|\nabla w_k|, $$	 combining these, we obtain
		$$\|v_k\|\leq2\|w_k\|.$$ Therefore, $v_k\in W^{1,N}_0(\Om)$. Then by  choosing $\phi=v_k$ and inserting this into \eqref{kc-PS-bdd1}, we deduce
		\begin{align}\label{1.0}
		\langle J'(w_k), v_k\rangle&=\int_\Om\left(1+\frac{2^{N-1}|h(w_k)|^N}{1+2^{N-1}|h(w_k)|^N}\right)|\nabla w_k|^N dx\notag\\&\;\;\;\;\;\;-\int_\Om \left(\int_\Om \frac{F(y,h(w_k))}{|x-y|^{\mu}}dy \right)f(x,h(w_k)) h(w_k) dx\notag\\
		&=\e_k\|w_k\|.
		\end{align}
		Using  \eqref{1.0} and $(f_5)$,   for some $q>2N$, we get
		\begin{align}\label{1.1}
		C+\e_k\|w_k\|&\geq J(w_k)-\frac{1}{q}\langle J'(w_k), v_k\rangle\notag\\
		&\geq \int_\Om\left[\frac1N-\frac{1}{q}\left(1+\frac{2^{N-1}|h(w_k)|^N}{1+2^{N-1}|h(w_k)|^N}\right)\right]|\nabla w_k|^N dx\notag\\&\;\;\;\;\;\;-\frac12\int_\Om \left(\int_\Om \frac{F(y,h(w_k))}{|x-y|^{\mu}}dy \right)\left[F(x,h(w_k))-\frac2q f(x,h(w_k)) h(w_k)\right] dx\notag\\
		&\geq \int_\Om\left(\frac1N-\frac{2}{q}\right)|\nabla w_k|^N dx=\left(\frac1N-\frac{2}{q}\right)\|w_k\|^N,
		\end{align}
		the above relation yields that $\{w_k\}$ must be bounded in $W^{1,N}_0(\Om)$.
	\end{proof}
	\noi Next, we have the following lemma:
	\begin{lemma}\label{PS-level}
		If $f$ satisfies $(f_1)$-$(f_6)$ and  $h$ is defined as in \eqref{g}, then $$0< \beta^* < \displaystyle \frac{1}{N (h(1))^{2N}}\left( \frac{2N-\mu}{2N}\alpha_N\right)^{N-1},$$  where $\beta^*$ is given as in \eqref{level}.
	\end{lemma}
	\begin{proof}
		Since for $w\not\equiv 0,$ by Lemma \ref{lem7.1.2}, $J(tw)\to-\infty$ as $t\to+\infty$, hence, from \eqref{level}, we get   $$\beta^* \leq \displaystyle\max_{t\in[0,1], w \in {W^{1,N}_0(\Omega)\setminus\{0\}}} J(tw).$$ So, it is sufficient to prove that there exists  $w \in W^{1,N}_0(\Om)$ such that $\|w\|=1$ and
		\[\max_{t\in[0,\infty)} J(tw) < \frac{1}{N{( h(1))}^{2N}} \left( \frac{2N-\mu}{2N}\alpha_N\right)^{N-1}.\]
		To show this, let us consider the sequence of Moser functions $\{\mc M_k\}$ defined as
		\begin{equation*}
		\mc M_k(x)=\frac{1}{\omega_{N-1}^{\frac{1}{N}}}\left\{
		\begin{split}
		& (\log k)^{\frac{N-1}{N}},\; 0\leq |x|\leq \frac{\delta}{k},\\
		& \frac{\log \left(\frac{\delta}{|x|}\right)}{(\log k)^{\frac{1}{N}}}, \; \frac{\delta}{k}\leq |x|\leq \delta\\
		& 0,\; |x|\geq \delta.
		\end{split}
		\right.
		\end{equation*}
		Then it is easy to see that supp$(\mc M_k) \subset B_\delta(0)$ and $\|\mc M_k\| =1$ for all $k\in\mathbb N$. Now we assert that there exists some $k \in \mb N$ such that
		\begin{align}\label{key}\max_{t\in[0,\infty)} J(t\mc M_k) < \frac{1}{N (h(1))^{2N}}\left( \frac{2N-\mu}{2N}\alpha_N\right)^{N-1}.\end{align}
		Indeed, if this doesn't hold  then for all $k \in \mb N,$ there exists  $t_k>0$ such that
		\begin{equation}\label{kc-PScond0}
		\begin{split}
		&\max_{t\in[0,\infty)} J(t\mc M_k) = J(t_k\mc M_k) \geq \frac{1}{N (h(1))^{2N}} \left( \frac{2N-\mu}{2N}\alpha_N\right)^{N-1}.
		\end{split}
		\end{equation}

		Since    $(f_1)$ implies  that $F(x,h(t_k\mc M_k))\geq 0$ for all $k\in\mathbb N$  so by  the definition of $J(t_kw_k)$ together with  \eqref{kc-PScond0}, we obtain
		\begin{equation}\label{kc-PScond2}
		t_k^N \geq \frac{1}{(h(1))^{2N}} \left( \frac{2N-\mu}{2N}\alpha_N\right)^{N-1}.
		\end{equation}
	Also, in view of \eqref{kc-PScond0}, we have
		$\frac{d}{dt}(J(t\mc M_k))|_{t=t_k}=0.$
		This combining with Lemma \ref{L1}-$(h_4)$ yields that
		\begin{align}\label{kc-PS-cond3}
		t_k^N &= \int_\Om \left(\int_\Om \frac{F(y,h(t_k\mc M_k))}{|x-y|^{\mu}}dy\right)f(x,h(t_k\mc M_k))h'(t_k\mc M_k)t_k\mc M_k ~dx\notag\\
		&\geq\frac 12 \int_{B_{\delta/k}(0)}f(x,h(t_k\mc M_k))h(t_k\mc M_k)\left( \int_{B_{\delta/k}(0)}\frac{F(y,h(t_k\mc M_k))}{|x-y|^\mu}~dy\right)~dx.

		\end{align}
		Now from \eqref{h-growth}, we know that for each $b>0$ there exists a constant $R_b$ such that
		\[sf(x,s)F(x,s) \geq b \exp\left( 2|s|^{\frac{2N}{N-1}}\right),\; \text{whenever}\; s \geq R_b.\]
		From \eqref{kc-PScond2}, we infer that
		${t_k}\mc M_k \to \infty \;\text{as}\; k \to \infty$ in $B_{\de/k}(0).$ Now by Lemma \ref{L1}-$(h_8)$, we get $h({t_k}\mc M_k) \to \infty \;\text{as}\; k \to \infty$, uniformly   in $B_{\de/k}(0).$   So, we can choose $s_b \in \mb N$ such that  in $B_{\de/k}(0)$,
		\[h(t_k\mc M_k) \geq R_b,\; \text{for all}\; k\geq s_b.\]
		In addition, using the same idea as in \cite{yang-JDE} (see  equation $(2.11)$), we can have
		\[\int_{B_{\delta/k}(0)}\int_{B_{\delta/k}(0)} \frac{~dxdy}{|x-y|^\mu} \geq C_{\mu, N} \left(\frac{\delta}{k}\right)^{2N-\mu},\]
		where $C_{\mu, N}$ is a positive constant depending on $\mu$ and $N$.
		Using these estimates in \eqref{kc-PS-cond3} together with \eqref{kc-PScond2} and Lemma \ref{L1}-$(h_8)$, for sufficiently large $b$,
		we get
		\begin{align}\label{bd}
		t_k^N&\geq \frac b2 \int_{B_{\delta/k}(0)}\int_{B_{\delta/k}(0)} \frac{~dxdy}{|x-y|^\mu}\exp \left( {2 (h(t_k\mc M_k))^{\frac{2N}{N-1}}}\right)\notag\\&\geq \frac b2 \int_{B_{\delta/k}(0)}\int_{B_{\delta/k}(0)} \frac{~dxdy}{|x-y|^\mu}\exp \left( {2 ((h(1))^2(t_k\mc M_k))^{\frac{N}{N-1}}}\right)\notag\\&=\frac b2 \exp \left( \log k\left(\frac{2 (h(1))^{\frac{2N}{N-1}}  (t_k)^{\frac{N}{N-1}}}{\omega_{N-1}^{\frac{1}{N-1}}}\right)\right)\int_{B_{\delta/k}(0)}\int_{B_{\delta/k}(0)}\frac{~dxdy}{|x-y|^\mu}\notag\\&\geq\frac b2 \exp \left( \log k\left(\frac{2 (h(1))^{\frac{2N}{N-1}} t_k^{\frac{N}{N-1}}}{\omega_{N-1}^{\frac{1}{N-1}}}\right)\right)C_{\mu,N}\left(\frac{\delta}{k}\right)^{2N-\mu}\notag\\&=\frac b2 \exp \left( \log k\left[\left(\frac{2 (h(1))^{\frac{2N}{N-1}} t_k^{\frac{N}{N-1}}}{\omega_{N-1}^{\frac{1}{N-1}}}\right)-(2N-\mu)\right]\right)C_{\mu,N}\left({\delta}\right)^{2N-\mu}\notag\\&\geq \frac b2  C_{\mu, N}\delta^{2N-\mu}.
		\end{align}

		On the other hand, using the fact that $J(tw)\to-\infty$ as $t\to\infty$ and \eqref{kc-PScond0}, we get that the sequence  $\{t_k\}$ is bounded, which contradicts	\eqref{bd}, since $b$ is arbitrary. This establishes our claim \eqref{key} and hence, we conclude the proof. \end{proof}
	\subsection{Convergence results}
	Let $\{w_k\}\subset W^{1,N}_0(\Om)$ be a Palais-Smale sequence for $J$.
	Now Lemma \ref{lem712} yields that $\{w_k\}$  is bounded in $W^{1,N}_0(\Om)$. Thus, there exists  $w\in W^{1,N}_0(\Om)$ such that up to a subsequence, still denoted by $\{w_k\}$,
	\begin{align*}
	&w_k \rightharpoonup w \text { \;\;\;weakly in } W^{1,N}_0(\Om),\\
	&w_k \to w \text{\;\; strongly in } L^q(\Om),\, q \in [1,\infty),\\
	&w_k(x) \to w(x)\text{ \;\;\;point-wise a.e. in } \Om,
	\end{align*}
	as $k \to \infty$. Also, from \eqref{kc-PS-bdd1} and \eqref{kc-PS-bdd2}, we obtain that for some positive constants $C', C''$,
	\begin{align}
	\int_\Om \left(\int_\Om\frac{F(y,h(w_k))}{|x-y|^\mu}dy\right)F(x,h(w_k))~dx &\leq C',\label{wk-sol10}\\
	\int_\Om \left(\int_\Om\frac{F(y,h(w_k))}{|x-y|^\mu}dy\right)f(x,h(w_k))h'(w_k)w_k~dx & \leq C''.\label{wk-sol100}
	\end{align}
	Now we have the next two results, where we consider $\{w_k\}$ to be satisfying all the above  facts.
	\begin{lemma}\label{PS-ws} 	 Assume that the assumptions $(f_1)$-$(f_5)$ hold and let $h$ be defined as in \eqref{g}. Let $\{w_k\}\subset W^{1,N}_0(\Om)$ be a Palais-Smale sequence for $J$.

		Then we have	\begin{align}\label{3.23}
		\lim_{k \to \infty} \int_{\Omega} \left(\int_{\Omega}\frac{|F(y,h(w_k))-F(y,h(w))|}{|x-y|^{\mu}} dy \right) |F(x,h(w_k))-F(x,h(w))|  ~dx =0.
		\end{align}
	\end{lemma}
	\begin{proof}

	 We know that if a function $\mc F\in L^1(\Om)$ then for any $\e>0$ there exists a $\delta(\e)>0$ such that
		\[\left| \int_{\Om'} \mc F(x)~dx\right| <\e,\]
		for any measurable set $\Om'\subset \Om$ with $|\Om'|\leq \delta(\e)$.

		Also, if $\mc F\in L^{1}(\Om)$ then for any fixed $\delta_0>0$ there exists $\al>0$ such that
		\[|\{x\in \Om : |\mc F(x)|\geq \al\}|\leq \delta_0.\]
		Now  \eqref{wk-sol10} gives that
		$$\left(\displaystyle\int_\Om\frac{F(y,h(w_k))}{|x-y|^\mu}dy\right)F(\cdot,h(w_k)) \in L^{1}(\Om)$$
		and similarly, \eqref{k2} gives that
		$$\left(\displaystyle\int_\Om\frac{F(y,h(w))}{|x-y|^\mu}dy\right)F(\cdot,h(w)) \in L^{1}(\Om).$$
		Now we fix $\de_*>0$ and  choose $\al> \max\left\{1,\left(\frac{2C'' M_0}{{\de_*}}\right)^{\frac{1}{m_0+1}}, s'\right\}$. Then by using $(f_4)$, Lemma \ref{L1}-$(h_4)$ and \eqref{wk-sol10}, we deduce
		\begin{align}\label{b}
		&\int_{\Om \cap\{h(w_k)\geq \al\}} \left(\int_{\Omega} \frac{F(y,h(w_k))}{|x-y|^{\mu}} dy \right)F(x,h(w_k))  ~dx \notag\\&\leq M_0 \int_{\Om \cap\{h(w_k)\geq \al\}} \left(\int_{\Omega} \frac{F(y,h(w_k))}{|x-y|^{\mu}}dy\right) \frac{f(x,h(w_k))}{(h(w_k))^{m_0}}  ~dx\notag\\
		&\leq M_0 \int_{\Om \cap\{h(w_k)\geq \al\}} \left(\int_{\Omega} \frac{F(y,h(w_k))}{|x-y|^{\mu}}dy\right) \frac{f(x,h(w_k))h(w_k)}{(h(w_k))^{m_0+1}}  ~dx\notag\\

		&\leq \frac{2M_0}{ \al^{m_0+1}}\int_{\Om \cap\{h(w_k)\geq \al\}} \left(\int_{\Omega} \frac{F(y,h(w_k))}{|x-y|^{\mu}} dy\right) {f(x,h(w_k)) h'(w_k)w_{k}} ~dx<\de_*.
		\end{align}
		Using the similar argument as above in addition with Fatou's lemma, we have
		{\begin{align}\label{bb}
			&\int_{\Om \cap\{h(w)\geq \al\}} \left(\int_{\Omega} \frac{F(y,h(w))}{|x-y|^{\mu}} dy \right)F(x,h(w))  ~dx  \leq \de_*.
			\end{align}}
	Combining \eqref{b} and \eqref{bb}, we have
		\begin{align*}
		&\left| \int_{\Omega} \left(\int_{\Omega}\frac{F(y,h(w_k))}{|x-y|^{\mu}} dy \right) F(x,h(w_k)) ~dx-  \int_{\Omega} \left(\int_{\Omega}\frac{f(y,h(w) )}{|x-y|^{\mu}} dy \right) F(x,h(w)) ~dx\right|\\
		&\leq 2\de_*+ \bigg|\int_{\Om \cap \{h(w_k)\leq \al\}} \left(\int_{\Omega}\frac{F(y,h(w_k))}{|x-y|^{\mu}} dy \right) F(x,h(w_k)) ~dx\\&\qquad\qquad\qquad-  \int_{\Omega\cap \{h(w)\leq \al\}} \left(\int_{\Omega}\frac{F(y,h(w))}{|x-y|^{\mu}} dy \right) F(x,h(w)) ~dx\bigg|.
		\end{align*}
		Next, we show  that as $k \to \infty$
		\begin{align*}
		\int_{\Om \cap \{h(w_k)\leq \al\}} \left(\int_{\Omega}\frac{F(y,h(w_k))}{|x-y|^{\mu}} dy \right) F(x,h(w_k))~dx \to \int_{\Om \cap \{h(w)\leq \al\}} \left(\int_{\Omega}\frac{F(y,h(w))}{|x-y|^{\mu}} dy \right) F(x,h(w)) ~dx.
		\end{align*}
		Since $\left(\displaystyle\int_\Om\frac{F(y,h(w))}{|x-y|^\mu}dy\right)F(\cdot,h(w)) \in L^{1}(\Om)$,  Fubini's theorem yields that
		\begin{align*}

		&\lim_{\Lambda \to \infty} \int_{\Om \cap\{h(w)\leq \al\}}\left(\int_{\Om\cap\{h(w)\geq \Lambda\}}\frac{F(y,h(w))}{|x-y|^{\mu}}dy\right)F(x,h(w))~dx\\
		&= \lim_{\Lambda \to \infty} \int_{\Om \cap\{h(w)\geq \Lambda\}}\left(\int_{\Om \cap\{h(w)\leq \al\}}\frac{F(y,h(w))}{|x-y|^{\mu}}dy\right)F(x,h(w))~dx=0.
		\end{align*}
		Thus, we can fix  $\Lambda> \max\left\{\left(\frac{2C' M_0}{\de_*}\right)^{\frac{1}{m_0+1}}, s'\right\}$  such that, using \eqref{wk-sol10}, $(f_4)$ and Lemma \ref{L1}-$(h_4)$, we deduce
		\begin{align*}
		&\int_{\Om \cap\{h(w_k)\leq \al\}} \left(\int_{\Om \cap\{h(w_k)\geq \Lambda\}}\frac{F(y,h(w_k))}{|x-y|^{\mu}} dy \right)F(x,h(w_k)) ~dx\\
		& \leq M_0\int_{\Om \cap \{h(w_k)\leq \al\}} \left(\int_{\Om \cap\{h(w_k)\geq \Lambda\}}\frac{ f(y,h(w_k))}{(h(w_k))^{m_0}|x-y|^{\mu}} dy \right)F(x,h(w_k)) ~dx\\

		&\leq \frac{M_0}{{\Lambda}^{m_{0}+1}} \int_{\Om \cap\{h(w_k)\leq \al\}} \left(\int_{\Om \cap\{h(w_k)\geq \Lambda\} }\frac{  f(y,h(w_k))h (w_k)(y)}{|x-y|^{\mu}} dy \right) F(x,h(w_k)) ~dx\\
		&\leq \frac{2M_0}{\Lambda^{m_0+1}} \int_{\Om \cap\{h(w_k)\leq \al\}} \left(\int_{\Om \cap\{h(w_k)\geq \Lambda\} }\frac{  f(y,h(w_k))h'(w_k) w_k(y)}{|x-y|^{\mu}} dy \right) F(x,h(w_k)) ~dx\\
		&\leq \frac{2M_0}{\Lambda^{m_{0}+1}} \int_{\Om} \left(\int_{\Om }\frac{F(y,h(w_k))}{|x-y|^{\mu}} dy \right) f(x,h(w_k)) h'(w_k)w_k ~dx\leq \de_*.
		\end{align*}
		Again, following the similar argument as above and employing Fatou's lemma, we can derive
		\[\int_{\Om \cap\{h(w)\leq \al\}} \left(\int_{\Om \cap\{h(w)\geq \Lambda\} }\frac{F(y,h(w) )}{|x-y|^{\mu}} dy \right) F(x,h(w)) ~dx \leq \delta_*.\]
		Thus, we obtain
		\begin{align*}
		&\left|\int_{\Om \cap\{h(w)\leq \al\}} \left(\int_{\Om \cap\{h(w)\geq \Lambda\} }\frac{F(y,h(w) )}{|x-y|^{\mu}} dy \right) F(x,h(w)) ~dx\right.\\
		&\quad \quad \left.- \int_{\Om \cap\{h(w_k)\leq \al\}} \left(\int_{\Om \cap\{h(w_k)\geq \Lambda\} }\frac{F(y,h(w_k))}{|x-y|^{\mu}} dy \right) F(x,h(w_k)) ~dx\right|\leq 2\de_*.
		\end{align*}
		Now we claim that as $k\ra \infty$, for fixed positive real numbers $\al$ and $\Lambda$, the following holds:
		\begin{equation}\label{choq-new}
		\begin{split}
		&\left|\int_{\Om\cap\{h(w_k)\leq \al\}} \left(\int_{\Om \cap\{h(w_k)\leq \Lambda\}}\frac{F(y,h(w_k))}{|x-y|^{\mu}} dy \right) F(x,h(w_k)) ~dx\right.\\
		&\qquad- \left.\int_{\Om \cap\{h(w)\leq \al\}} \left(\int_{\Om \cap\{h(w)\leq \Lambda\}}\frac{F(y,h(w) )}{|x-y|^{\mu}} dy \right) F(x,h(w)) ~dx \right|\ra 0.
		\end{split}
		\end{equation}
		It is easy to compute that
		\begin{align}\label{a0}
		&\left(\int_{\Om \cap\{h(w_k)\leq \Lambda\} }\frac{F(y,h(w_k))}{|x-y|^{\mu}} dy \right) F(x,h(w_k))\chi_{ \Om \cap \{h(w_k)\leq \al\}}\notag\\  &\qquad\ra \left(\int_{\Om \cap \{h(w)\leq \Lambda\} }\frac{F(y,h(w))}{|x-y|^{\mu}} dy \right) F(x,h(w))\chi_{\Om \cap \{h(w)\leq \al\}}
		\end{align}
		point-wise a.e. as $k \to \infty$. Now using $N=r$ in \eqref{k1}, Lemma \ref{L1}-$(h_5)$ and \eqref{HLSineq}, we get a constant $C_{\al,\Lambda}>0$ depending on $\al$ and $\Lambda$ such that
		\begin{align}\label{a}
		&\int_{\Om \cap \{h(w_k)\leq \al\}}\left( \int_{\Om \cap \{h(w_k)\leq \Lambda\} }\frac{F(y,h(w_k))}{|x-y|^{\mu}} dy \right)  F(x,h(w_k))dx  \notag\\
		&\leq  C_{\al,\Lambda}\int_{\Om \cap \{h(w_k)\leq \al\}}\left( \int_{\{h(w_k)\leq \Lambda\} }\frac{|h(w_k)(y)|^{r}}{|x-y|^{\mu}} dy \right)  |h(w_k)(x)|^{r} dx \notag\\
		&\leq  C_{\al,\Lambda}\int_{\Om \cap \{h(w_k)\leq \al\}}\left( \int_{\{h(w_k)\leq \Lambda\} }\frac{|w_k(y)|^{r}}{|x-y|^{\mu}} dy \right)  |w_k(x)|^{r} dx \notag\\
		& \leq C_{\al,\Lambda} \int_\Om\int_{\Om }\left(\frac{|w_k(y)|^{r}}{|x-y|^{\mu}}~dy  \right) |w_k(x)|^{r} ~dx\notag\\
		& \leq { {C_{\al,\Lambda}C(N,\mu)\|w_k\|_{L^{\frac{2Nr}{2N-\mu}}(\Om)}^{2r} \to C_{\al,\Lambda}C(N,\mu)\|w\|_{L^{\frac{2Nr}{2N-\mu}}(\Om)}^{2r}}} \; \text{as}\; k \to \infty,

		\end{align}
	 since $w_k \to w$ strongly in $L^q(\Om)$ for each $q \in [1,\infty)$.  This combining with Theorem $4.9$ in \cite{bz} implies that there exists $\mc G \in L^1(\Om)$ such that  up to a subsequence, for each $k\in \mathbb N$, we have
		\[\left|\left( \int_{\Om \cap\{h(w_k)\leq \Lambda\} }\frac{F(y,h(w_k))}{|x-y|^{\mu}} dy \right)  F(x,h(w_k))\chi_{ \Om \cap \{h(w_k)\leq \al\}} \right| \leq |\mc G(x)|.\]
		Therefore, using \eqref{a0} and employing the Lebesgue dominated convergence theorem, we obtain \eqref{choq-new}.
Thus, $\Lambda \to \infty$
		\begin{align}\label{3.25}
		&\int_{\Omega} \int_{\Om\cap\{|h(w)| \geq \Lambda\}} \frac{F(y,h(w))}{|x-y|^{\mu}} F(x,h(w)) dy ~dx=o(\Lambda),\\ &\int_{\Omega} \int_{\{|h(w_k)| \geq \Lambda\}} \frac{F(y,h(w_k))}{|x-y|^{\mu}} F(x,h(w_k)) dy ~dx = o(\Lambda)\label{3.20},
		\end{align}
		\begin{equation}\label{3.26}
		\int_{\Omega} \int_{\{|h(w)| \geq \Lambda\}} \frac{F(y,h(w_k))}{|x-y|^{\mu}} F(x,h(w)) dy ~dx = o(\Lambda),
		\end{equation}
		and
		\begin{equation}\label{3.27}
		\int_{\Omega} \int_{\{|h(w_k)| \geq \Lambda\}} \frac{F(y,h(w_k))}{|x-y|^{\mu}} F(x,h(w)) dy ~dx = o(\Lambda).
		\end{equation}
	So,
		\begin{equation*}
		\begin{split}
		& \int_{\Omega} \left(\int_{\Omega}\frac{|F(y,h(w_k))-F(y,h(w))|}{|x-y|^{\mu}} dy \right) |F(x,h(w_k))-F(x,h(w))|  ~dx\\&\leq
		2\int_{\Omega} \left(\int_{\Omega}\frac{\chi_{\{h(w_k)\geq \Lambda\}}(y)F(y,h(w_k))}{|x-y|^{\mu}} dy \right) F(x,h(w_k)) ~dx \\
		&+4 \int_{\Omega} \left(\int_{\Omega}\frac{F(y,h(w_k))\chi_{\{h(w)\geq \Lambda\}}(x)F(x,h(w))}{|x-y|^{\mu}} dy \right) ~dx\\&+4 \int_{\Omega} \left(\int_{\Omega}\frac{\chi_{\{h(w_k)\geq \Lambda\}}(y)F(y,h(w_k))F(x,h(w))}{|x-y|^{\mu}} dy \right) ~dx\\
		&+2\int_{\Omega} \left(\int_{\Omega}\frac{\chi_{\{h(w)\geq \Lambda\}}(y)F(y,h(w))}{|x-y|^{\mu}} dy \right) F(x,h(w)) ~dx \\
		&+\int_\Omega\Bigg[\left(\int_\Omega\frac{|F(y,h(w_k))\chi_{\{h(w_k)\leq \Lambda\}}(y)-F(y,h(w))\chi_{\{h(w)\leq \Lambda\}}(y)|}{|x-y|^\mu}dy\right)\\&\qquad\qquad\qquad|F(x,h(w_k))\chi_{\{h(w_k)\leq \La\}}(x)-F(x,h(w))\chi_{\{h(w)\leq \Lambda\}}(x)|\Bigg]~dx.
		\end{split}
		\end{equation*}
		Then from Lebesgue dominated convergence theorem we infer that the last integration tends to $0$ as $k \to \infty.$ Hence, making use of \eqref{3.25}-\eqref{3.27}, we finally conclude \eqref{3.23}.
		\end{proof}
		\begin{lemma}\label{kc-ws} Assume that $(f_1)$-$(f_5)$ hold and  $h$ is defined as in \eqref{g}. Let $\{w_k\}\subset W^{1,N}_0(\Om)$ be a Palais-Smale sequence for $J$. Then
		for all $\varphi\in W_0^{1,N}(\Om),$   we have
		\begin{align*}\displaystyle\lim_{k\to\infty}\int_\Om \left(\int_\Om \frac{F(y, h(w_k))}{|x-y|^{\mu}}dy\right)f(x, h(w_k))h'(w_k)\varphi ~dx= \int_\Om \left(\int_\Om \frac{F(y,h(w))}{|x-y|^{\mu}}dy\right)f(x,h(w))h'(w)\varphi~dx.  \end{align*}
	\end{lemma}
	\begin{proof}

		Let $\Om' \subset\subset \Om$ and $\psi \in C_c^\infty(\Om)$ such that $0\leq \psi \leq 1$ and $\psi \equiv 1$ in $\Om' $. One can easily compute that
		\begin{equation}\label{k}
		\begin{split}
		\left\| \frac{\psi}{1+w_k}\right\|^N = \int_\Om \left|\frac{\nabla \psi}{1+w_k}- \psi \frac{\nabla w_k}{(1+w_k)^2} \right|^N~dx
		\leq 2^{N-1}(\|\psi\|^N+ \|w_k\|^N),
		\end{split}
		\end{equation}
		which yields that $\frac{\psi}{1+w_k} \in W^{1,N}_0(\Om)$. Now taking $\phi=\frac{\psi}{1+w_k}$  in \eqref{kc-PS-bdd1} as a test function and using Lemma \ref{L1}-$(h_3)$ and \eqref{k}, we obtain
		\begin{equation}\label{kc-ws-new1}
		\begin{split}
		&\int_{\Om^{'}}\left( \int_\Om \frac{F(y,h(w_k))}{|x-y|^\mu}dy\right)\frac{f(x,h(w_k))}{1+u_k}h'(w_k)~dx\\& \leq \int_\Om \left( \int_\Om \frac{F(y,h(w_k))}{|x-y|^\mu}dy\right)\frac{f(x,h(w_k) h'(w_k)\psi}{1+w_k}~dx\\
		&\leq \int_\Om \left( \int_\Om \frac{F(y,h(w_k))}{|x-y|^\mu}dy\right)\frac{f(x,h(w_k) \psi}{1+w_k}~dx\\
		&=  \e_k \left\|\frac{\psi}{1+w_k}\right\| + \int_\Om  |\nabla w_k|^{N-2}\nabla w_k \nabla \left( \frac{\psi}{1+w_k}\right)~dx\\
		& \leq \e_k 2^{\frac{N-1}{N}}(\|\psi\|+ \|w_k\|) +\int_\Om |\nabla w_k|^{N-2}\nabla w_k \left(\frac{\nabla \psi}{1+w_k}-\varphi\frac{\nabla w_k}{(1+w_k)^2}\right)~dx\\
		& \leq \e_k 2^{\frac{N-1}{N}}(\|\varphi\|+ \|w_k\|) +\int_\Om |\nabla w_k|^{N-1} \left( |\nabla \psi|+ |\nabla w_k|\right)~dx\\
		& \leq \e_k 2^{\frac{N-1}{N}}(\|\psi\|+ \|w_k\|)+ [\|\psi\|\|w_k\|^{N-1}+ \|w_k\|^N]\leq C_1,
		\end{split}
		\end{equation}
		where $C_1$ is a positive constant and  in the last line we used the fact that $\{w_k\}$ is bounded in $W^{1,N}_0(\Om)$.

		Again, using the boundedness of the sequence $\{w_k\}$, from \eqref{kc-PS-bdd2}, we get
		\begin{align}\label{kc-ws-new2}
		&\int_{\Om^{'}} \left( \int_\Om \frac{F(y,h(w_k))}{|x-y|^\mu}dy\right){f(x,h(w_k))h'(w_k)}{w_k}~dx\notag\\
		&\leq \int_{\Om} \left( \int_\Om \frac{F(y,h(w_k))}{|x-y|^\mu}dy\right){f(x,h(w_k))h'(w_k)}{w_k}~dx \notag\\
		&\leq \e_k\|w_k\|+\|w_k\|^N\leq C_2
		\end{align}
		for some constant $C_2>0$. Combining \eqref{kc-ws-new1} and \eqref{kc-ws-new2}, we deduce
		\begin{align*}
		&\int_{\Om^{'}}\left( \int_\Om \frac{F(y,h(w_k))}{|x-y|^\mu}dy\right){f(x,h(w_k))h'(w_k)}~dx \\
		& \leq  2\int_{\Om^{'}\cap \{w_k <1\}} \left( \int_\Om\frac{F(y,h(w_k))}{|x-y|^\mu}dy\right)\frac{f(x,h(w_k))h'(w_k)}{1+w_k}~dx\\& + \int_{\Om^{'}\cap \{w_k \geq 1\}} \left(\int_\Om \frac{F(y,h(w_k))}{|x-y|^\mu}dy\right)w_k{f(x,h(w_k))h'(w_k)}~dx\\
		& \leq 2\int_{\Om^{'}} \left( \int_\Om\frac{F(y,h(w_k))}{|x-y|^\mu}dy\right)\frac{f(x,h(w_k))h'(w_k)}{1+w_k}~dx\\& + \int_{\Om^{'}}\left( \int_\Om \frac{F(y,h(w_k))}{|x-y|^\mu}dy\right)w_k{f(x,h(w_k))h'(w_k)}~dx\\
		& \leq 2C_1+C_2  :=C_3.
		\end{align*}
		Thus, the sequence $\{v_k\}:=\left\{\left( \int_\Om\frac{F(y,h(w_k))}{|x-y|^\mu}dy\right){f(x,h(w_k))h'(w_k)}\right\}$ is bounded in $L^1_{\text{loc}}(\Om)$. Therefore, there exists a radon measure $\zeta$ such that, up to a subsequence, $v_k \rightharpoonup \zeta$ in the ${weak}^*$-topology as $k \to \infty$.  Hence,  we have
		\[\lim_{k \to \infty}\int_\Om\int_\Om \left( \frac{F(y,h(w_k))}{|x-y|^\mu}dy\right){f(x,h(w_k))h'(w_k)}\eta ~dx = \int_\Om \eta ~d\zeta,\; \forall \eta \in C_c^\infty(\Om). \]
		Since $w_k$ satisfies \eqref{kc-PS-bdd1}, we achieve
		\[\int_A \eta d\zeta= \lim_{k \to \infty} \int_A |\nabla w_k|^{N-2}\nabla w_k \nabla \eta ~dx, \;\;\forall A\subset \Om,  \]
		which together with Lemma~\ref{wk-sol} yields that the Radon measure $\zeta$ is absolutely continuous with respect to the Lebesgue measure. So, there exists a function $\varrho \in L^1_{\text{loc}}(\Om)$ such that for any $\eta\in C^\infty_c(\Omega)$, it holds that $\int_\Om \eta~ d\zeta= \int_\Om \eta \varrho~dx$, thanks to Radon-Nikodym theorem,.
		Therefore, we obtain
		\begin{align*}&\lim_{k \to \infty}\int_\Om\left( \int_\Om \frac{F(y,h(w_k))}{|x-y|^\mu}dy\right){f(x,h(w_k))h'(w_k)}\eta(x)~ ~dx\\&\qquad = \int_\Om \eta \varrho~dx= \int_\Om  \left( \int_\Om \frac{F(y,h(w))}{|x-y|^\mu}dy\right){f(x,h(w))h'(w)}\eta(x)~ ~dx, \;\;\forall \eta\in C^\infty_c(\Omega), \end{align*}
		since $ C^\infty_c(\Omega)$ is dense in $W_0^{1,N}(\Om)$, this completes the proof.
	\end{proof}

	\noi \textbf{Proof of Theorem \ref{T1}:}
	Let $\{w_k\}$ be a Palais-Smale sequence at the level $\beta^*$. Then $\{w_k\}$ can be obtained as a minimizing sequence associated to the variational problem \eqref{level}. Then by Lemma \ref{lem712},  there exists  $w \in W^{1,N}_0(\Om)$ such that, up to a subsequence,
	$w_k \rightharpoonup w$ weakly in $W^{1,N}_0(\Om)$ as $k \to \infty$.\\

	Now by using  Lemma \ref{wk-sol} and Lemma \ref{kc-ws}, we infer that $w$ forms a weak solution of \eqref{pp}.
	We claim that $w \not\equiv 0$. Indeed, if not, that is,  if $w \equiv 0$ then using Lemma \ref{PS-ws}, we have
	\[\int_\Om \left(\int_\Om \frac{ F(y,h(w_k))}{|x-y|^{\mu}}dy\right)F(x,h(w_k)) ~dx  \to 0\; \text{as}\; k \to \infty,\]
	which yields that $ \displaystyle\lim_{k \to \infty}J(w_k) = \frac{1}{N}\displaystyle\lim_{k\to\infty} \|w_k\|^N = \beta^*.$ That is,
	$$\lim_{k\to\infty} \|w_k\|^N = \beta^*N.$$ Therefore, for any real number $l>0$, there exists $k_0\in\mathbb N$ such that
	\begin{align}\label{inq}
	\|w_k\|^{\frac{N}{N-1}} < \frac{2N-\mu}{2N(h(1))^{2N}}\al_N-l, \text{\;\; for all}\;\; k\geq k_0.
	\end{align}

	Next, we show that
	\begin{align}\label{aaa}
	\int_{\Om}\left( \int_\Om\frac{F(y,h(w_k))}{|x-y|^{\mu}}dy\right)f(x,h(w_k))h'(w_k)w_k~dx \to 0 \;\text{as}\; k \to \infty.
	\end{align}
	Now  using Proposition \ref{HLS}, \eqref{f},    Lemma \ref{L1}-$(h_4)$-$(h_6
	)$, H\"older inequality and Sobolev inequality, we deduce
	\begin{align}\label{wk-soln}
	&\int_{\Om}\left( \int_\Om\frac{F(y,h(w_k))}{|x-y|^{\mu}}dy\right)f(x,h(w_k))h'(w_k)w_k dx\notag\\
	&\leq\tau\int_{\Om}\left( \int_\Om\frac{f(y,h(w_k)) h(w_k)}{|x-y|^{\mu}}dy\right)f(x,h(w_k))h(w_k) dx\notag \\
	&\leq \tau C(N,\mu) \left(\int_{\Om}|f(x,h(w_k))h(w_k)|^{\frac{2N}{2N-\mu}}~dx\right)^{\frac{2N-\mu}{N}}\notag\\&\leq \tilde C_1(N,\mu,\e) \left(\int_{ \Om}\left(|h(w_k)|^N +  |h(w_k)|^r \exp((1+\e)|h(w_k)|^{\frac{2N}{N-1}})\right)^{\frac{2N}{2N-\mu}}~dx\right)^{\frac{2N-\mu}{N}}\notag \\
	&  \leq \tilde C_1(N,\mu,\e)\left[ \|w_k\|_{L^{\frac{2N^2}{2N-\mu}}(\Om)}^{2N}+ \|w_k\|_{L^{\frac{2Nr}{2N-\mu}}(\Om)}^{2r} \left(\int_{ \Om}\exp\left(2(1+\epsilon) \frac{2N}{2N-\mu}h|(w_k)|^{\frac{2N}{N-1}}\right)~dx\right)^{\frac{2N-\mu}{2N}}\right] \notag \\
	&  \leq \tilde C_2(N,\mu,\e)\left[ \|w_k\|^{2N}+ \|w_k\|^{2r} \left(\int_{ \Om}\exp\left(2(1+\epsilon) \frac{2N}{2N-\mu}2^{\frac{1}{N-1}}\|w_k\|^{\frac{N}{N-1}}\left(\frac{|w_k|}{\|w_k\|}\right)^{\frac{N}{N-1}}\right)~dx\right)^{\frac{2N-\mu}{2N}}\right].
	\end{align}
	Using \eqref{inq}, we can choose $l$ ( for example, take $l=\frac{2N-\mu}{(h(1))^{2N}2N}\al_N-\frac{2N-\mu}{2 (1+\e)2^{\frac{1}{N-1}}2N}\al_N>0$, as $h(1)\leq 1$) such that, for sufficiently large $k\in\mathbb N$, $2(1+\epsilon) \frac{2N}{2N-\mu}2^{\frac{1}{N-1}}\|w_k\|^{\frac{N}{N-1}}<\al_N$.

	Therefore,  in the light of  Theorem \ref{TM-ineq} along with Lemma \ref{PS-level}, for sufficiently large  $k\in\mathbb N$, from \eqref{wk-soln}, it follows that
	$$\int_{\Om}\left( \int_\Om\frac{F(y,h(w_k))}{|x-y|^{\mu}}dy\right)f(x,h(w_k))h'(w_k)w_k dx<C.$$
	Now by employing Vitali's convergence theorem, we obtain
	\[\int_{\Om}\left( \int_\Om\frac{F(y,h(w_k))}{|x-y|^{\mu}}dy\right)f(x,h(w_k))h'(w_k)w_k~dx \to 0 \;\text{as}\; k \to \infty.\]
	Since, $\{v_n\}$ is a Palais-Smale sequence for $J$, hence we have $\displaystyle\lim_{k\to \infty}\langle J'(w_k),w_k \rangle=0$ which together with \eqref{aaa} gives that $\lim_{k\to \infty}\|w_k\|^N=0$.  Thus, from Lemma \ref{PS-ws}, it follows that $\displaystyle\lim_{k \to \infty}J(w_k)=0 =\beta^*$ which is a contradiction to the fact that $\beta^*>0$. Hence, $w \not \equiv 0$.\\
	Next, we prove that $w>0$ in $\Om$. Now  Lemma \ref{lem712} yields that $\{w_k\}$ is bounded. Therefore, there exists a constant $a^*>0$ such that, up to a subsequence, $\|w_k\| \to a^*$ as $k \to \infty$. From the fact that $J^\prime(w_k) \to 0$ strongly in $\left(W_0^{1,N}(\Om)\right)^*$
and using Lemma~\ref{kc-ws} and  Lemma~\ref{wk-sol}, when $k \to \infty$, we obtain
	$$\int_\Om \left(\int_\Om \frac{F(y,h(w_k))}{|x-y|^{\mu}}dy\right)f(x,h(w_k)) h'(w_k)\varphi ~dx \to \int_\Om \left(\int_\Om \frac{F(y,h(w))}{|x-y|^{\mu}}dy\right)f(x,h(w))h'(w)\varphi~dx$$ and
	\begin{align*}
	\int_\Om |\nabla w|^{N-2}\nabla w\nabla \varphi~dx = \int_\Om \left(\int_\Om \frac{F(y,h(w))}{|x-y|^{\mu}}dy\right)f(x,h(w))h'(w)\varphi~dx,
	\end{align*}
	for all $\varphi \in W^{1,N}_0(\Om)$. In particular, taking $\varphi = w^-$ in the above equation, we obtain $\|w^-\|=0$ which implies that $w^-=0$ a.e. in $\Om$. Therefore, $w\geq 0$ a.e. in $\Om$.\\
	From Theorem \ref{TM-ineq}, we have $f(\cdot,h(w))\in L^q(\Om)$, for $1\leq q<\infty$.
From \eqref{KC-new1}, we know that $F(x,h(w)) \in L^q(\Om),$ for any $ q \in [1,\infty)$. Since $\mu \in (0,N)$ and $\Om$ is bounded then using the fact that $y\to |x-y|^{-\mu} \in L^{q}(\Om)$ for all $q\in (1, \frac{N}{\mu})$ uniformly in $x\in \Omega$ and applying H\"older's inequality, we can deduce
\begin{equation}\label{wk-sol7}
\int_\Om \frac{F(y,h(w))}{|x-y|^{\mu}}dy \in L^\infty(\Om).
\end{equation}
	 Therefore, using the fact that $h'(s)\leq1$, we get $$\left(\int_\Om \frac{F(y,h(w))}{|x-y|^{\mu}}~dy \right)f(x,h(w)) h'(w)\in L^q(\Om),$$ for $1 \leq q <\infty$. Now by employing the regularity results for the elliptic equations, we infer that $w \in L^\infty(\Om)$ and $w \in C^{1,\gamma}(\overline{\Om})$ for some $\gamma \in (0,1)$.  Finally, from the strong maximum principle, we draw the conclusion that $w>0$ in $\Om$   and  $w\not\equiv 0$. This completes the proof of Theorem \ref{T1}.

	\section{Appendix A}
\noi In this section,  we give  prove of the following almost everywhere convergence of gradients of Palais-Smale sequences using Concentration compactness arguments.
	\begin{lemma}\label{wk-sol} Suppose the assumptions $(f_1)$-$(f_6)$   hold and  $h$ is defined as in \eqref{g}. 	 Let $\{w_k\}\subset W^{1,N}_0(\Om)$ be a Palais-Smale sequence for $ J$. Then we have, as $k\to\infty$

		\begin{equation}\label{wk-sol2}
		|\nabla w_k|^{N-2}\nabla w_k \rightharpoonup |\nabla w|^{N-2}\nabla w\; \text{weakly in}\; (L^{\frac{N}{N-1}}(\Om))^N.
		\end{equation}
	\end{lemma}
	\begin{proof}

		Since Lemma \ref{lem712} yields that $\{w_k\}$  is bounded in $W^{1,N}_0(\Om)$,  there exists  $w\in W^{1,N}_0(\Om)$ such that, in the sense of subsequence, we have
		\begin{align*}
		&w_k \rightharpoonup w \text { \;\;\;weakly in } W^{1,N}_0(\Om),\\
		&w_k \to w \text{\;\; strongly in } L^q(\Om),\, q \in [1,\infty),\\
		&w_k(x) \to w(x)\text{ \;\;\;point-wise a.e. in } \Om,
		\end{align*}
		as $k \to \infty$.
		From the properties of the sequence $\{w_k\}$, it is evident that the sequence $\{|\nabla w_k|^{N-2}\nabla w_k\}$ must be bounded in $(L^{\frac{N}{N-1}}(\Om))^N$, which implies that
		there exists $u \in (L^{\frac{N}{N-1}}(\Om))^N$ such that,
		\begin{align}\label{2}|\nabla w_k|^{N-2}\nabla w_k \rightharpoonup u \; \text{weakly in}\;  (L^{\frac{N}{N-1}}(\Om))^N \; \text{as} \; k \to \infty.\end{align}
		Also we have, $\{|\nabla w_k|^N\}$ is bounded in $L^1(\Om)$,  which yields that there exists a non-negative radon measure $\sigma$ such that, up to a subsequence, we have
		\begin{align}\label{2.0}|\nabla w_k|^N \to \sigma \; \text{in}\; (C(\overline{\Om}))^*\; \text{as}\; k \to \infty.\end{align}
		Our aim is to show $u = |\nabla w|^{N-2}\nabla w $.
		For that, first we take $\nu>0$ and set $X_\nu := \{x \in \overline \Om:\; \sigma(B_l(x)\cap \overline \Om)\geq \nu, \;\text{for all}\; l>0\}$.\\
		{\bf{Claim 1:}}  $X_\nu$ is a finite set. \\
		Indeed, if not, then there exists a sequence of distinct points $\{z_k\}$ in $X_\nu$ such that, $\sigma(B_l(z_k)\cap \overline \Om)\geq \nu$ for all $l>0$ and for all $k\in\mathbb N$. This gives that $\sigma(\{z_k\}) \geq \nu$ for all $k$. Therefore, $\sigma(X_\nu)= +\infty$. But this is a contradiction to the fact that
		\[\sigma(X_\nu) = \lim_{k \to \infty} \int_{X_\nu} |\nabla w_k|^N ~dx \leq C.\] Hence, the claim holds.
		Thus, we can take $X_\nu= \{z_1,z_2,\ldots, z_n\}$.\\
		{\bf	{Claim 2:} } We can choose  $\nu>0,$ such that $\nu^{\frac{1}{N-1}} <  \frac{2N-\mu}{2N} \alpha_N$ and we have
		\begin{align}\label{wk-sol6}
		&\lim_{k \to \infty} \int_S\left( \int_\Om \frac{F(y, h (w_k))}{|x-y|^{\mu}}dy\right) f(x,h(w_k))h'(w_k)w_k~dx\notag\\ &\qquad=  \int_S \left( \int_\Om \frac{F(y,h(w))}{|x-y|^{\mu}}dy\right) f(x,h(w)) h'(w)w~dx, \end{align} where  $S$ is any compact subset of $\oline \Om \setminus X_\nu$.\\
		Let $z_0 \in S$ and $l_0>0$ be such that $\sigma(B_{l_0}(z_0) \cap \oline \Om) < \nu$ that is $z_0 \notin X_\sigma$. Also, we consider  $\phi \in C_c^\infty(\Om)$ satisfying $0\leq \phi(x)\leq 1$ for $x \in \Om$, $\phi \equiv 1$ in $B_{\frac{l_0}{2}}(z_0)\cap \oline \Om$ and $\phi \equiv 0$ in $\oline \Om \setminus (B_{l_0}(z_0)\cap \oline \Om)$. Then
		\[\lim_{k \to \infty} \int_{B_{\frac{l_0}{2}}(z_0) \cap \oline \Om}|\nabla w_k|^N dx\leq \lim_{k \to \infty} \int_{B_{l_0}(z_0) \cap \oline \Om}|\nabla w_k|^N\phi dx \leq \sigma(B_{l_0}(z_0) \cap \oline \Om) < \nu. \]
		Hence, for sufficiently large $k \in \mb N$ and  sufficiently $\e>0$ small, it follows that
		\begin{equation}\label{wk-sol3}
		\int_{B_{\frac{l_0}{2}}(z_0) \cap \oline \Om}|\nabla w_k|^N \leq \nu(1-\e).
		\end{equation}
		Now we estimate the following using $(f_3)$, \eqref{wk-sol3} and  Lemma \ref{L1}-$(h_6
		)$:
		\begin{equation}\label{wk-sol4}
		\begin{split}

		& \int_{B_{\frac{l_0}{2}}(z_0) \cap \oline \Om}|f(x,h(w_k))|^q~dx=  \int_{B_{\frac{l_0}{2}}(z_0) \cap \oline \Om}|g(x,h(w_k))|^q\exp\left(q|h(w_k)|^{\frac{2N}{N-1}}\right)~dx \\
		&  \leq C \int_{B_{\frac{l_0}{2}}(z_0) \cap \oline \Om}\exp\left((1+\epsilon)2^{\frac{1}{N-1}}q|h(w_k)|^{\frac{2N}{N-1}}\right)~dx\\
		&  \leq C\int_{B_{\frac{l_0}{2}}(z_0) \cap \oline \Om}\exp\left((1+\epsilon)2^{\frac{1}{N-1}}q|w_k|^{\frac{N}{N-1}}\right)~dx\\
		& \leq C \int_{B_{\frac{l_0}{2}}(z_0) \cap \oline \Om}\exp\left((1+\epsilon)q 2^{\frac{1}{N-1}}\nu^{\frac{1}{N-1}}(1-\e)^{\frac{1}{N-1}}\left(\frac{|w_k|^N}{\displaystyle\int_{B_{\frac{l_0}{2}}(z_0) \cap \oline \Om}|\nabla w_k|^N dx}\right)^{\frac{1}{N-1}}\right)~dx
		\end{split}
		\end{equation}
		Thus, we can choose $q>1$ and   $\e>0$ such that $(1+\e) 2^{\frac{1}{N-1}}q \nu^{\frac{1}{N-1}} < \alpha_N$ and then using Theorem \ref{TM-ineq} in the last relation, we obtain
		\begin{align}\label{aa}
		& \int_{B_{\frac{l_0}{2}}(z_0) \cap \oline \Om}|f(x,h(w_k))|^q~dx\leq C.
		\end{align} Next, we consider
		\begin{equation*}
		\begin{split}
		&\int_{B_{\frac{l_0}{2}}(z_0) \cap \oline \Om} \left|\left( \int_\Om \frac{F(y,h(w_k))}{|x-y|^{\mu}}dy\right) f(x,h(w_k))h'(w_k) w_k- \left( \int_\Om \frac{F(y,h(w))}{|x-y|^{\mu}}dy\right) f(x,h(w))h'(w) w \right|~dx\\
		& \leq \int_{B_{\frac{l_0}{2}}(z_0) \cap \oline \Om} \left|\left( \int_\Om \frac{F(y,h(w))}{|x-y|^{\mu}}dy\right) (f(x,h(w_k))h'(w_k) w_k-f(x,h(w))h'(w))w\right|~dx\\
		& \quad + \int_{B_{\frac{l_0}{2}}(z_0) \cap \oline \Om} \left|\left( \int_\Om \frac{F(y,h(w_k))-F(y,h(w))}{|x-y|^{\mu}}dy\right) f(x,h(w_k))h'(w_k) w_k\right|~dx\\
		&:= J_1 + J_2.
		\end{split}
		\end{equation*}
		From the asymptotic growth assumptions on  $f$, we obtain
		\begin{equation}\label{wk-sol8}
		\lim_{s \to \infty} \frac{f(x,s
			)t}{(f(x,s))^r} = 0\; \text{uniformly in }x \in \Omega,\; \text{for all}\; r>1.
		\end{equation}
		Using \eqref{wk-sol7}, we get
		\[
		J_1 \leq  C\int_{B_{\frac{l_0}{2}}(z_0) \cap \oline \Om}  |f(x,h(w_k))h'(w_k)w_k-f(x,h(w))h'(w)w|~dx,\]
		where $C>0$ is a constant. Moreover, \eqref{wk-sol8} and \eqref{wk-sol4} imply that  $\{f(x,h(w_k))h'(w_k)w_k\}$ is an equi-integrable family of functions over ${B_{\frac{l_0}{2}}(z_0) \cap \oline \Om} $. Also, the continuity of $f $ and $h$ yields that $f(x,h(w_k))h'(w_k)w_k$ $\to f(x,h(w))h'(w)w$ point-wise a.e. in $\Om$ as $k \to \infty.$ Hence, by applying Vitali's convergence theorem, we obtain $J_1 \to 0$ as $k\to \infty$. Next, we show that $J_2 \to 0$ as $k \to \infty$.\\
		First by using the semigroup property of the Riesz Potential, we get the following for some constant $C>0,$ independent of $k$:
		\begin{align*}
		\begin{split}
		&\int_{\Omega} \left(\int_{\Omega}\frac{F(y,h(w_k))-F(y,h(w))}{|x-y|^{\mu}} dy \right) \chi_{B_{\frac{l_0}{2}}(z_0) \cap \overline{\Omega}}(x) f(x,h(w_k))h'(w_k) w_k ~dx \\
		&\leq  C\left(\int_{\Omega} \left( \int_{\Omega}\frac{|F(y,h(w_k))- F(y,h(w))| dy }{|x-y|^{\mu}}\right) |F(x,h(u_k))- F(x,h(u))| ~dx \right)^{\frac12}\\
		&\quad \times \left(\int_{\Omega} \left(\int_{\Omega} \chi_{B_{\frac{l_0}{2}}(z_0) \cap \overline{\Omega}}(y) \frac{f(y,h(w_k))h'(w_k) w_k}{|x-y|^{\mu}} dy \right) \chi_{B_{\frac{l_0}{2}} (z_0)\cap \overline{\Omega}}(x) f(x,h(w_k)) h'(w_k) w_k ~dx \right)^{\frac12}.
		\end{split}
		\end{align*}
		Combining \eqref{wk-sol4}, \eqref{wk-sol8} and the fact that $\nu$ can be chosen sufficiently small with $\nu^{\frac{1}{N-1}} < \frac{2N-\mu}{2N} \alpha_N$, it follows that
		\begin{align*}
		&\left(\int_{\Omega} \left(\int_{\Omega} \chi_{B_{\frac{l_0}{2}} (z_0)\cap \overline{\Omega}}(y) f(y,h(w_k))h'(w_k) w_k dy \right) \chi_{B_{\frac{l_0}{2}}(z_0) \cap \overline{\Omega}}(x) f(x,h(w_k))h'(w_k) w_k ~dx \right)^{\frac12}\\& \leq \|\chi_{B_{\frac{l_0}{2}} (z_0)\cap \overline{\Omega}}f(\cdot,h(w_k))h'(w_k)w_k \|_{L^{\frac{2N}{2N-\mu}}(\Omega)} \leq C.
		\end{align*}

		Now using Lemma \ref{PS-ws}, we get $J_2 \to 0$ as $k \to \infty$.

		This yields that
		\begin{align*}&\lim_{k \to \infty}\int_{B_{\frac{l_0}{2}}(z_0) \cap \oline \Om} \Bigg|\left( \int_\Om \frac{F(y,h(w_k))}{|x-y|^{\mu}}dy\right) f(x,h(w_k))h'(w_k)w_k\\&\qquad\qquad\qquad\qquad\qquad- \left( \int_\Om \frac{F(y,h(w))}{|x-y|^{\mu}}dy\right) f(x,h(w))h'(w)w \Bigg|~dx=0.\end{align*}
		Since $S$ is compact, we can repeat this procedure over a finite covering of balls and hence, achieve \eqref{wk-sol6}. Finally, the proof of \eqref{wk-sol2} can be concluded by the similar standard arguments as in the proof of Lemma $4$ in \cite{M-do1}.
\end{proof}
\noindent{\bf Acknowledgements:} The first author would like to thank the Department of Mathematics, Indian Institute of Technology Delhi for the financial support.
The second author would like to thank the Science and Engineering Research Board, Department of Science and Technology,
Government of India for the financial support under the grant MTR/2018/001267.

redpajama_set_name: RedPajamaArXiv


text:
Go Long
The Perfect Storm: How Stefon Diggs changed everything in Buffalo
www.golongtd.com
He has proven to be exactly what the Bills needed: A leader. Now, anything is possible.
Tyler Dunne
Illustration by Brave Buffalo
This was the rare sports trade that lights up your phone, wakes you up and makes time stand still.
The night of March 16 — at 10:33 p.m. — we all discovered that Stefon Diggs was heading to the Buffalo Bills.
At which point, you instantly fell into one of two camps: You either viewed this as the signature move that'd put the Bills, a playoff team, over the top. Or you viewed this as a recipe for disaster. You did the math in your head: Disgruntled wide receiver + Most Inaccurate QB in the NFL + Buffalo, NY = Utter disaster. A tale old as time.
Things turned ugly for Diggs in Minnesota, of course. Very ugly.
Clearly, this would be a fragile situation, one that'd have most position coaches texting and calling whoever they could to learn whatever they could about the alleged diva. Making Diggs happy was about to become a major part of Chad Hall's job description as the Bills wide receivers coach. After all, the list of stars who've treated the news that they're Buffalo-bound like a baseball bat to the face is long. And, hell, the country had just slipped into quarantine. For four months, Hall could only speak to Diggs by phone and Zoom.
The first time he'd meet him would be the first day of training camp.
Yet all along, Hall never reached out to one coach, one player, one scout.
He was dead-set on forming his own opinion of Stefon Diggs.
"All you mostly hear is the negative shit. All the bad things," Hall says. "In the media, it's all bad — 'Oh, man, he's a problem! No one could handle him up in Minnesota!' Well, for me as a coach, good. There's my challenge. My challenge is to get this guy in a comfortable situation where he's loved and appreciated and it starts from there. You've got to build a relationship with these guys before anything else."
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Now, it's mid-January. The Bills are still playing football. And Hall — an '08 Air Force grad who served a two-year stint as a second lieutenant before his own NFL career — can confidently share his opinion of one, Stefon Diggs, with everyone.
There's a ruggedness to his voice.
"He's like a middle linebacker, man," Hall says. "He's your alpha, but he's your damn dog. He's tough as shit. He's mean as shit. I'm a military guy and I'd want him in my foxhole. I'd want to be back-to-back with him. I know he's going to take care of me. I know I'm going to take care of him. That's the kind of guy he is. You want him in your foxhole."
Because guess what? This was no disaster. This proved to be a perfect storm.
Diggs led the NFL in receptions (127) receiving yards (1,535) and it was not even close.
Diggs is calling Buffalo his "second home" and telling all, "I'm excited, I'm happy. In games, you can see it."
Diggs' relationship with Josh Allen has become a full-fledged bromance.
And, oh, the Buffalo Bills are two wins away from reaching the Super Bowl.
Because not only has Diggs been everything Buffalo could've dreamt of on the field — a sublime route runner, a big play itching to detonate — he has given this franchise something far more valuable. Leadership. Hall leaves no doubt: Diggs is the voice everyone follows on this team, Diggs injects belief. And that's a remarkable feat for someone universally dismissed as a locker room cancer one year ago. Diggs went from forcing his way out of Minnesota one fake cough, one cryptic tweet at a time to being the veteran who morphs a team that hadn't won a playoff game since 1995 into a band of fighters that truly believes it can kick any team's ass.
The Bills are nobody's underdog now. Those days are finished. They expect to win.
And what a new sensation that is for all Western New Yorkers, one nobody's had since the early 1990s.
Another wide receiver — a Hall-of-Famer on those Super Bowl teams — remembers that feeling so, so well. As James Lofton explains, as a player, you stop looking at the schedule and thinking, Oh, we've got to play Baltimore.
"This year for the Bills, it's they have to play us," Lofton says. "That, to me, is the mark of a great team. When you're thinking, 'They have to play us. Whether we're at home or we're at their place, they have to play us.' That's the swagger Buffalo has finally gotten to."
"You can say it. But you also have to believe it and you have to produce it."
That's the way the Bills processed games through four straight Super Bowl runs from 1990-1993 and that's how the Bills are finally processing games again.
Because of Diggs.
"You just have no real fear," Lofton says, "of playing somebody else."
How did it get to this point? The storm's been brewing for years.
A team in need
A blizzard of this magnitude requires juuuust the right amount of barometric pressure. So in truth, in Buffalo, it's been three-plus years in the making. The search for a No. 1 wide receiver began almost immediately: Sean McDermott's first draft in 2017.
That night, McDermott was smitten with Zay Jones.
The wide receivers coach he had just hired (Phil McGeoghan) coached Jones at East Carolina so the Bills gave serious consideration to drafting Jones at No. 27 overall after, gulp, trading out of the No. 10 pick that landed Patrick Mahomes. They were close — very close — to pulling the trigger. After further discussion, the Bills instead opted to wait and draft the cornerback who was higher on their board: Tre'Davious White. And they got their receiver in the second round.
Jones, obviously, did not pan out.
McDermott brought on Brandon Beane as his GM, the two shipped Sammy Watkins out of town amid a slew of ejections, proceeded to win anyways and — with playoff hopes surprisingly real at 5-2 — traded for Panthers wideout Kelvin Benjamin. Beane was ecstatic. Beane declared "this is a win for the Buffalo Bills" at the time.
Benjamin, also, did not pan out.
Once the franchise quarterback was secured, into March 2019, the Bills then went big-game hunting at wide receiver. A different trade at night, this one at 11:28 p.m., jolted WNY'ers out of bed this spring. Antonio Brown was heading to Buffalo… until he wasn't. Until Brown killed it. Brown refused to play for the Bills like far too many stars before him and, for the Bills' sake, thank God he did. There's a good chance Brown would've nuked the culture Buffalo was building. Because it wasn't long after this that Brown was traded to the Raiders and self-destructed just about as terribly as any athlete we've seen this century.
Forget the Bills' 10-6 record last year. If the volcanic "AB" is playing for a quarterback who's finishing dead-last in completion percentage, as Allen did, he is wreaking absolute havoc. And who knows what that does to Allen? Here, Beane dodged a bullet. Beane instead signed John Brown (three years, $27 million) and Cole Beasley (four years, $29 million) to two of the best bang-for-your-buck contracts in the NFL.
Both receivers proved capable of getting Buffalo to the playoffs but, into March of 2020, merely reaching the playoffs was no longer the goal.
The Bills still wanted a legit star for Josh Allen. The Bills were thinking Super Bowl.
And their calculated gamble was that they hit the bull's eye on Allen himself. Belief can backfire. Belief can exasperate the problem beyond your wildest imagination. Buffalo learned this the hard way, six years prior, in selling the farm for Watkins… for EJ Manuel. Nothing was going to change the fact that EJ was EJ and, while Allen won more games, he certainly was no sure thing in the court of public opinion.
We've been told for ages that quarterbacks cannot improve accuracy. It's science.
Yet, the Bills believed. The Bills projected a massive improvement in Year 3. The Bills believed all Allen needed was a No. 1 wide receiver.
Luck, for once, was on the Bills' side.
Lo and behold, a No. 1 wide receiver was available.
A receiver in need
Nineteen wide receivers were drafted ahead of Stefon Diggs in 2015. Seven of which are not even in the NFL anymore.
Diggs fell all the way to the 146th overall pick in the fifth round.
Which all now sounds certifiably insane.
But back then? It was not a surprise at all. Scouts across the league basically all held the same opinion on this 6-foot, 190-pounder out of Maryland. Looking back, they recall Diggs as a soft player at Maryland. They say he turned down too many plays, dropped balls and was not consistent. At all. One scout recalls a "shut-off switch like no other," saying that Diggs could go from "1000" to "Leave it the f--- alone"-mode in a hurry. Diggs was dismissed by some teams as the dreaded "JAG" — Just A Guy — because his final season at Maryland wasn't that impressive, either: 62 catches, 792 yards, five touchdowns. And while there weren't any major off-the-field issues, one scout says some on the Maryland staff were happy to see Diggs go.
Diggs himself was brutally honest in his meetings with NFL teams, too.
One scout says Diggs flat-out admitted he made business decisions across the middle of the field. (For what it's worth, this scout appreciated his honesty.)
All in all, teams weren't sure Diggs would keep his switch flipped to "1000."
So, he fell. And fell. Right into the Vikings' lap.
Little did teams know that this exact offseason Diggs was busting his ass like never before with trainer Pete Bommarito in Miami, Fla. Diggs embraced this high-octane program that occupies 10 to 12 hours of your time every day. He bought into Bommarito's emphasis on "joint alignment" to maximize power and regeneration and deceleration and nutrition and neuromuscular therapy and other terms that'd make your brain freeze. Diggs loved the science of it all. And Bommarito insists he didn't see one inkling of what those scouts are talking about.
Down in Florida, Diggs morphed himself into the perfect modern-day wide receiver.
"Everything that he does is about mental focus," Bommarito says. "If you have a drill in front of you, everybody can execute with intensity but he's just got a focus like everything he is doing is the last play of the Super Bowl. Whether it's a warm-up drill, a speed drill, a weight room workout, running routes, everything. You'll see some players who just go all in when they're doing routes. They don't have the same focus and intensity when it comes to the other stuff. … He's got the mentality where he can legitimately focus on every conceivable aspect of what we're doing.
"That's how he trains. That's what makes him great."
Sure enough, Diggs delivered one of the greatest moments in NFL history: The Minneapolis Miracle.
He was on top of the world.
Until, Mike Zimmer had other plans. Until, Mike Zimmer became fed up with all this goshdarn passing and turned the clock back to 1964 on the Vikings' offensive scheme as we wrote about at length. In sum: Diggs believed the Vikings were pissing away his prime. This brand of football was not going to maximize his gifts, there wasn't any hope that things would change and the real kick in the groin was the utter lack of communication. So, Diggs acted out. Diggs voiced his displeasure by missing practice and, the next day, held a laughable press conference in which he faked a cough for a fake illness. All parties involved were trying to make this go away… but nobody bought this stunt. Everyone saw right through Diggs' atrocious, 2-Percent-on-Rotten Tomatoes acting performance.
Soon enough, Diggs deleted all Vikings-related content off his social media and after the team extended Kirk Cousins he tweeted in pure Gen-Z emo fashion, "It's time for a new beginning."
Longtime Vikings cornerback Terence Newman was around for the start of this demise as an assistant coach and says Diggs was "more volatile" than anyone realized. He even compares him to Terrell Owens.
"I'm not saying he was crazy," Newman says. "But I just saw enough myself. I'm not even on that side of the ball — where, 'Yo, you need to chill the f--- out.' … The thing with wideouts is, you're 'always open.' You're going to be selfish because you want yards and touchdowns and everything else. But once younger guys get older, they understand, 'OK, you know what? This Super Bowl shit is pretty tough.' What becomes more important is having an actual run at winning the Super Bowl. Some people are just more vocal as players of 'Hey, I need this. I need the ball.' And other guys are like, 'Hey, let's just go out and win this f------ thing. I don't give a shit who catches the ball. Let's just get it done.'"
So, no, Newman doesn't believe any conversation Zimmer could've ever had with Diggs would've helped.
One source familiar with the Vikings' inner-workings concedes that Diggs is a "me" guy in that he doesn't just want to win — he needs to be the reason his team wins. He is nobody's decoy. But this source describes that breakup as a "perfect storm" for all the wrong reasons and lists off the simmering factors: A head coach who doesn't communicate. A quarterback "who's not really a leader and isn't relatable to Stefon Diggs." The fact that Diggs would always be perceived as second fiddle to Adam Thielen. The burning desire to be The Man elsewhere.
He compares Diggs to Jimmy Butler, the perennial NBA All-Star who always wanted something more, who desired to be The Man from Chicago to Minnesota to Philadelphia to, finally, getting his wish in Miami.
And even though teammates did not like how Diggs handled this all, many agreed with his intent.
"I'd disagree with Terence," this source familiar with the Vikings' thinking says. "It's not as if Stefon Diggs speaking up is going to change the offense or change Zimmer or change how Kirk plays. Their answer is always 'Shut the f--- up and play.' That's their answer for everything.
"It's the organization's fault because if you have intelligent people that can come down and communicate properly, then at least they'll say, 'Listen, you're a great receiver. We do need to be more open-minded. We do need to get you more involved. We can't be so centered on just running the football. You are just as good if not better than Adam Thielen. You're a 1-2 punch. We need that.' There's no communication. They don't communicate with anyone."
Here's a crazy thought: Maybe the disgruntled player is often… right.
Diggs isn't alone. This is now a trend in the NFL. We're seeing more and more players try to take back control of their careers regardless of contract status — studs like Jalen Ramsey and Jamal Adams and Yannick Ngakoue all were not shy about how they felt. They were just as passive-aggressive on social media. And you might want to turn another star player's Twitter notifications on. It sure sounds like Deshaun Watson could be next.
More teams are willing to cave, too.
More teams would rather get something for their unhappy employee than reconcile.
"I can't really blame NFL players because you're forced to be selfish at some point," this source familiar with the Vikings' thinking says. "You're only going to get paid and compensated based on your performance and if you can't go out there and perform — or you don't get opportunities to perform when you're a really good player — you don't have the luxury of pissing away years in the NFL. Diggs did sign a contract. The team could've told him to go f--- himself. But the way the NFL works now is the guys that bitch get their way out. The good guys don't. Which is kind of f----- up."
More communication could've helped cooler heads prevail in Minnesota.
This breakup is a symptom of a larger problem in the NFL. Nobody wants to say the quiet parts out loud. Nobody wants to stand in front of a mic and say how they really feel. It's a league muddled in secrecy, full of bureaucrats who take themselves far too seriously. Which is why there are sources around the Vikings who believe that if Diggs (instead of farting around on Instagram) just said at a presser, "I'm a top 10 receiver and I want to be a part of the gameplan" and GM Rick Spielman (instead of pretending like everything was fine) said that the Vikings have to find a way to get Diggs involved, everyone could've found a way to sing "Kumbaya."
And another Miracle could've been waiting around the corner in Minnesota.
"But no one does that," this source in Minnesota says. "Everybody just hides behind their f------ desk and is afraid to talk to the media."
The Bills are thankful.
Because all of these forces collided to speed the storm up to 100 MPH en route to Bills Stadium. Beane and McDermott were more than willing to give Stefon Diggs everything Stefon Diggs could ever desire.
The trade was official.
And the storm hit.
When Chad Hall was finally able to shake Stefon Diggs' hand that first day of training camp and meet him face to face, what stood out first was his size. Diggs was taller than he expected. "Thicker," too. Especially his legs.
The next thing that popped? His smile. Diggs appeared genuinely happy.
Camp began and Diggs didn't say too much. He didn't want to speak out of turn so, simply, he worked his ass off. And fairly quickly — two weeks in — Diggs could tell that the Bills lacked a vocal leader.
And Diggs knew he could be that vocal leader.
So, Hall challenged him. The young assistant coach told Diggs to go for it, to speak up more in the wide receiver room… and to the entire offense… and, even, the entire team.
"You have an effect," Hall told him. "People listen to you when you're talking."
Production is one thing and his production has obviously been off the charts. But little did the Bills know along their quest for a No. 1 wide receiver — Jones to KB to AB to Diggs — that it was this leadership they needed most. They needed to permanently change the ethos of the team itself and, now, they're seeing the Diggs Effect. A team that was very good is now great.
Diggs has embraced everything that being The Man entails.
"He is the leader of our offense," Hall says. "Vocally and by example. I've never been around the alpha of the room or even of the whole offense who is the hardest worker on the whole team. I mean, he works harder at practice than anybody else.
"It's not coming from a fake place. Everything he says is real. It's motivating, it's encouraging. He'll hold you accountable if you're not doing what's right. He knows when to put his foot down. But most of it is very positive. That's something we've talked about and I've pushed him to accept that role and stay positive and to keep speaking verbally — because we needed that after last year. We have a lot of guys who lead by example but we don't have a lot of vocal leadership."
Kyle Williams provided leadership once. So did Lorenzo Alexander.
Neither was ever this big of a star. Neither ever played on this talented of a team.
If there's ever a lull on offense, Diggs speaks up. Once upon a time, three-and-outs were a real problem on this team. Not anymore. After having the fifth-worst three-and-out percentage last season (24.1 percent, 45 total), the Bills had the third-best this season (12.6 percent, 20 total). If the defense needs a jolt, Diggs is the one who beelines their direction on the sideline. And the last half of the season, Hall adds, Diggs has been the one speaking up to the entire team at halftime. It might only be 10 words, Hall says, but those 10 words pack a punch. Those 10 words are unbelievably valuable.
"It's someone yelling something to the whole team," Hall says. "Everyone respects him for it."
If the intensity level in 2019 was, oh, say a seven than Diggs raised it to a 10. He's always the most competitive player on the field and it's not particularly close.
Adds Hall: "He wants to win every play."
As a result — from May to January — Diggs' swagger has become the Bills' swagger. This franchise needed a shot in the ass. An unapologetic shot of… of… ego. Nobody ever accused Jim Kelly, Bruce Smith, Thurman Thomas and Andre Reed of lacking ego on those 90s teams that expected to win. A team full of nice guys who never rock the boat is never going to win the Super Bowl, so, this is where McDermott deserves some credit. Surely, there weren't many people in the universe who ever, in a million years, expected McDermott to even consider trading for Diggs as the coach filibustered on and on about #Process early on as head coach.
Echoing others who've been handed a one-way ticket out of town, Sammy Watkins did not have kind things to say about McDermott. Several players found his culture downright suffocating. Like they couldn't be themselves — at all.
And here's a team now dancing to MC Hammer before a playoff game.
Here's a star receiver most of the world deemed a loose cannon being his authetic self. Always. When he's not flossing on the bench, he's in the face of cornerbacks.
Take the most symbolic win of the season, at New England, on Monday Night Football.
Midway through the second quarter, the Bills held a 10-9 lead over the franchise that's tormented them for two decades. Allen's deep shot up the left sideline to Diggs sailed long and cornerback J.C. Jackson could not wait to signal an emphatic "incomplete" with both arms right in front of Diggs who, you know, wasn't particularly pleased with such exuberance from the 25-year-old. Diggs said a few words and stuck an index finger right in Jackson's face.
It's true that index finger could've been perceived as the ultimate sign of disrespect. It's also true that fans in Buffalo have been waiting two decades for a player to put a finger right in the face of this dynasty.
"I love it," Hall says. "That's what a competitor does. He hasn't gone too far. That's probably as close as he got with sticking his finger in his face. He's channeled it. And, he backs it up.
"He probably told him, 'I'm coming back.' That's what you want."
And that's what Diggs did. The next drive, he roasted Jackson for a 50-yard touchdown and was sure to look back at him as he crossed the goal line before then leaping into the arms of his quarterback. Another teammate bolted over to scream, "You're swagged up!" and the rout was on. McDermott showed zero mercy in a 38-9 shellacking foreign to Bill Belichick.
Such is the theme of the 2020 Bills. The storm hit and the storm never let up.
From what should've gone down as an epic… diving… game-winning touchdown vs. the Cardinals with 34 seconds left.
To breaking ankles against the 49ers. Oh, Jason Verrett. That poor bastard.
To utterly dismantling the Pittsburgh Steelers. Man to man? Zone? The Steelers could've dusted off Rod Woodson in his prime and nothing would've worked through this 10-reception, 130-yard masterpiece.
To sliding and scooping a 16-yarder from Allen at Denver. His cleats kicked up a cloud of dirt and, on to the sideline, there was an ecstatic Hall smacking him on the chest.
Is Diggs the best wide receiver in football? "Of course," says Hall, who's incredibly kind throughout this chat but — for a moment — sounds insulted that this is even a question.
"Have you seen what he's done?" Hall says. "His durability. He plays every snap. He fights. He blocks. He does everything. You see him cut-block people. He can get open against anybody and makes the tough catches. He can run with the ball. He's got everything you want."
Lofton sees this all, too. He's blown away by Diggs' drive, calling him a "perfectionist" and a "bully" and he means that in the best possible way — Diggs gets to literally any spot on the field he wants. It doesn't matter what your coverage is. It doesn't matter that Diggs isn't as big as Julio Jones or DK Metcalf or DeAndre Hopkins. He gets wherever he wants with far, far more foot quickness.
Imagine being a cornerback lining up to cover Diggs.
"He's worried, No. 1, that he's going to go deep. Right away," Lofton says. "The second thing is, he's going to make him look foolish. So, there's that nervousness about trying to cover a guy as elusive and as complete a route runner as he is. We just say, 'That's why he's an All Pro,' but the other team is trying to stop him. You forget about that — their No. 1 goal when the week starts — is 'We're going to shut down Stefon Diggs' or 'We're going to hold him to six catches for 60 yards.' And that just doesn't happen."
Corners live in fear of Diggs going downtown and get eaten alive on routes underneath.
That's why Lofton uses a different NBA analogy.
"The way that the game has evolved," Lofton says, "it's like Steph Curry with the Golden State Warriors — 'This guy is going to put up 30 on me, I know that. I just hope he doesn't go for 60.'"
No, this isn't even the same NFL of five years ago. You hear the exasperation in the voices of defensive players all the time. Ex-safety Donte Whitner even brought up Diggs' name unsolicited in his conversation with Go Long, saying he hears announcers declare that Diggs can "Run any route!" when he knows damn well a guy like Diggs wouldn't have even considered wading into the middle of the field in his era. To him, there's no danger in the back of receivers' minds today.
He's right, of course. But don't blame Diggs. He's just maximizing this all through a maniacal training regiment. He stuck with Bommarito and specifically sharpened his deceleration to twist corners into pretzels. It's one thing to be fast and quite another to know how to come to a screeching halt and change direction. As Bommarito points out, when Diggs shows up on his highlight video, he's mostly doing this deceleration work.
"When you decelerate, you eccentric load," Bommarito says. "You have to bend. So the mobility of the toes, the foot, the rear foot, the forefoot, the ankle, the knee and just how strong the hamstrings are on the eccentric load dictate your ability to decelerate. I just get the muscles prepped and the threshold up so when they run a route, the muscle action takes place in the route."
Which is how Diggs has only gotten more explosive and more elusive season to season.
And as the world knows by now, it was right there at Bommarito's complex where all of this Bills magic began. When the pandemic wiped out OTAs and minicamps, Bommarito got a call from Matt Barkley who said that Diggs had raved about him and that the team wanted to run some player-led practices at his spot in May. And that, Bommarito notes, is Diggs in a nutshell. Diggs wanted to create a quasi-camp of sorts to find chemistry with Allen and, by God, did they ever. Bommarito couldn't help but notice something special between Allen and Diggs. He remembers telling one NFL vet during a water break, "These guys are on f------ point, man!' My God. They just met!"
Adds Bommarito today: "That's Stefon. He's a badass receiver. If you've got a guy who can throw the ball, he's going to get open and he's going to catch it."
The rapport between Allen and Diggs was born.
And, honestly, they already knew each other. Before meeting up in Miami, they had already spent endless hours playing Call of Duty: Warzone. Headsets on, they spent hours upon hours talking to each other. And they weren't exactly killing each other in the video game, no, as ESPN detailed Allen always ditched whatever battle he was in to save Diggs.
Fast forward to January and there the two are — vs. the Indianapolis Colts — celebrating a touchdown with a freakishly choreographed celebration we'd all need four beers in our system to pull off.
They now treat NFL defenses like a video game.
Suffice to say, this is a much… much… much better relationship than Diggs ever had with Cousins. It's not that their relationship was malignant in Minnesota — this was nothing like McNabb-T.O. back in the day — but Diggs' personality just did not mesh with Cousins' personality like it does Allen. Sources around the Vikings describe Cousins as more uptight, sensitive, robotic.
Adds Lofton, "No knock on Kirk Cousins but they give off a different vibe. His vibe matches Stefon Diggs' vibe beat for beat."
His relationship with Allen is straight out of a Disney movie and so is his relationship with his coaches. Communication, here, is constant. Hall still pushes Diggs to be vocal and, hey, there's the entire receiving corps surprising Hall with a new truck. Hall used to drive to the office as fast as he could. Now? He's taking an extra lap (or two) around the stadium before sliding into the parking lot. Nobody else has even been allowed to sit inside yet, too.
Hall sincerely doesn't care what happened in Minnesota but this sure sounds like the exact opposite of an atmosphere.
"We have a really good relationship," Hall says, "and I love him to death."
Life is good for Stefon Diggs.
Life is good for everyone around Stefon Diggs.
The Baltimore Ravens land in Buffalo, NY soon.
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This franchise no longer expects the worst. Nobody in Western New York is bracing for a kick to sail wide or a lateral to spark a kick return. No longer are imposters sweeping into town and making promises they never keep.
Like Lofton says, it's one thing for you to "say" they have to play us. It's quite another to mean it.
And when left tackle Dion Dawkins declares that the Bills' Super Bowl is the Super Bowl, he's not pandering to his base for likes and retweets. He means it. This may all seem too good to be true — Hall even has a baby due right when this year's Super Bowl will be played. But this is no dream. This is reality. The Bills, finally, are damn good and fully expect to beat any team they play.
Nobody in Buffalo feels cursed anymore.
And arguably the No. 1 reason for that seismic shift is Diggs.
Of course, the other reality is that you never know how long your Super Bowl window will stay open. The Bills are nestled in that sweet spot every organization craves — they've got a star quarterback on a rookie deal and have had the luxury of spending handsomely on just about whatever else they want. Their gamble on Diggs paid off, true, but listen closely and you'll also hear echoes of "…just wait…" in Minnesota. His stats are through the roof. This team went a good two months without playing a close game. And Brian Daboll has been a magician of a playcaller who feeds Diggs the ball more than any OC feeds any receiver in the NFL.
Frankly, the Bills (and Diggs) have not faced much adversity this season.
What happens when Diggs has a reason to be upset? As one Vikings source puts, if Diggs is willing to put himself out there in Minnesota, why wouldn't he in Buffalo? He sees a gigantic "Buyer Beware" sign flashing above Diggs' head.
"If things go bad, is he going to be a good soldier?" the source says. "Or is he going to go out there and vocalize and be cryptic and send some shockwaves here and there and instigate things? He's proven to be an instigator."
This source does believe Diggs is self-aware enough to not ask for a new contract this coming offseason when he'll still have three years to go on the five-year, $72 million pact he had signed with the Vikings. Buffalo also rewired that contract to get Diggs more money up front. Another source in Minnesota, however, guarantees Diggs will see that he's the 15th-highest paid wide receiver (via Spotrac) and ask for a raise. Right now, this Bills roster is loaded with hungry players similar to the "Legion of Boom" Seahawks. Right now, that message of "We're Buffalo. We're f------ gonna win" is powerful, this source notes. Yet even as someone who agreed with Diggs' intent, he believes Diggs' temperamental side is still hibernating in there somewhere.
Thus, he believes there is a "zero percent chance" Diggs does not ask for a new deal.
You'd think Beane wouldn't want to set the precedent that it's OK for players to demand a new contract with three years left on their deal. Maybe he is able to convince Diggs to be a good Samaritan for one more year before revisiting his contract because, well, there's also the matter of paying Josh Allen north of $35 million per season soon.
Not that anyone at One Bills Drive really gives a damn about any of this right now.
Right now, Daboll is the OC.
Right now, Allen and Diggs have something special.
Right now, Diggs loves Buffalo. There are no fake coughs during his press conferences here, no, he has praised the team and the city nonstop.
And Saturday night, the Ravens pose the toughest test yet for Diggs. No cornerbacks in the NFL have found a way to excel in this pass-happy league better than Marcus Peters and Marlon Humphrey. They're mean. They're nasty. Peters just might spit in your direction. Humphrey undoubtedly will talk trash directly in your earhole. Neither, ever, back down from a challenge.
They will not be scared of Stefon Diggs.
And if Diggs tries sticking a finger in Peters' face, there's a good chance Peters will tear the finger off.
So, I ask Lofton, how is this matchup going down on Saturday night.
He pauses for a moment.
"Who are they playing?" Lofton asks.
At which point, I'm a bit confused. Not only is Lofton a Bills legend. He also broadcasts NFL games nationally every week. Shouldn't he… uh… know who's playing who at this point?
Baltimore, I say.
"No," Lofton says. "Baltimore is playing them. They've got to contend with Buffalo."
Stefon Diggs is the reason why, too.
Paul Barth
When it works, it works great. With an ascendant QB, Diggs is statistically the best WR in the league, there is a lot to be excited about - and credit to Buffalo for finding a way to win through drafting and savvy trades. Winning has a way of making things look great. I just wonder if (say) they start winning without having to rely so hard on Diggs or, if given an off season, defenses scheme against Diggs better next year, how it goes? I just think of other me-first receivers, like Owens in Philly, these situations can work for a while but the old habits tend to surface.
It doesn't mean not-to go after Diggs if you're Buffalo, it is just to remember what you have. That said, players can change. Sometimes just getting the respect and recognition they deserve, can calm the ego. There is a reason Charles Woodson ended up in Green Bay - nobody else wanted him. And he became the person he is today through those experiences, but it's easy to forget where he came from.
Time will tell. For now, though, what a force.
© 2023 Tyler Dunne

redpajama_set_name: RedPajamaCommonCrawl


Status: offline Once again which would you choose if you could only get one and why?
Status: offline PC if you are going for all the DLC for the games.
Status: offline The only advantage I see in OoB is the supply which is better represented and allow strategy of cut off etc.. but everything else PC is doing it better, the welldone scenarios design, AI is a bit better in PC, the amount and variety of units/ maps/ factions and of course the faster game play.
However I play both, they are great B&P wargames, but if I choose one it will be PC.
Status: offline Here's another vote for PC. But I wish there was a way to play a DLC battle as a stand a lone scen (like the basic game battles can be). If there is a way to do I haven't figured out how.
Status: offline Does the DLC even activate with anything other than the main campaign starter game?
Status: offline Order of battle is more to my personal liking, Panzer Corp II yet to be seen.
Status: offline Seems Bundle Stars has made my decision for me.
Status: offline I like Order of Battle better. Feels more tactical and less like a puzzle to be solved. Both are good though.
Status: offline Actually I already have Panzer Corp and never installed it for some strange reason. Does it come free with Allied Corp or Africa Corp? Yippeee $2.99 towards Battle Academy II now. It's like Christmas in September.
I like Order of Battle better. Feels more tactical and less like a puzzle to be solved. Both are good though.
I have to say I much prefer Order of Battle for much the same reasons.
I didn't see that in the opening version. I saw a lot of Panzer Corp in it.
Status: offline When OOB gets the "promised" skirmish mode, it will be OOB. Until then OOB is going down that well-trodden WW2 route of PC, and I have all that Panzer Corps DLC to play already - plus a few great mods. However, I do like OOB's editor better.
So I would choose OOB - in anticipation of future enhancements.
When OOB gets the "promised" skirmish mode, it will be OOB. Until then OOB is going down that well-trodden WW2 route of PC, and I have all that Panzer Corps DLC to play already - plus a few great mods. However, I do like OOB's editor better.
That would put it over the top for me also, the "skirmish" mode. It's what Panzer Corps lacks is that random map random units feature so everything isn't always the same an easily figured out or totally scripted.


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A Utah Man Reunites with Mother After 20 Years, Discovers They Work at the Same Hospital
By Scrubs August 12, 2022
Benjamin Hulleberg, 20, was born in Ogden, Utah, but was sent off to adoption on Thanksgiving Day 20 years ago when his biological mother was only 15 years old.
Hulleberg's adoptive parents kept Shearer in the loop for the first few years, sending photos and updates of the baby boy.
But after the updates stopped, Shearer's curiosity led her on a search.
When she found her son on Facebook, she was hesitant to reach out.
Meanwhile, Hulleberg was on a mission to find his birth mother, even taking DNA tests and signing up at an adoption agency.
When Shearer sent that fateful message to her son introducing herself, Hulleberg recalls the moment hitting him "like a load of bricks."
Hulleberg wanted to meet his birth mother immediately after waiting so many years to find her.
During the reunion, both mother and son discovered they had been working under the same roof without ever knowing.
Both work at HCA Healthcare's St. Mark's Hospital in Salt Lake City — Shearer is a medical assistant at The Heart Center at St. Mark's and Hulleberg volunteers at the hospital's neonatal intensive care unit.
Through his birth mother, Hulleberg also met his younger half-brother and half-sister.
The newly-reunited mother and son meet at least once a week nowadays.
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