\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{graphicx} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2018 (2018), No. 194, pp. 1--17.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2018 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2018/194\hfil Positive solution for H\'enon type equations] {Positive solution for H\'enon type equations with critical Sobolev growth} \author[K. Takahashi \hfil EJDE-2018/194\hfilneg] {Kazune Takahashi} \address{Kazune Takahashi \newline Graduate School of Mathematical Sciences, The University of Tokyo, 3-8-1 Komaba Meguroku Tokyo 153-8914, Japan} \email{kazunetakahashi@gmail.com} \thanks{Submitted April 2, 2018. Published November 28, 2018.} \subjclass[2010]{35J20, 35J60, 35J61, 35J91} \keywords{Critical Sobolev exponent; H\'enon equation; mountain pass theorem; \hfill\break\indent Talenti function} \begin{abstract} We investigate the H\'enon type equation involving the critical Sobolev exponent with Dirichret boundary condition \[ - \Delta u = \lambda \Psi u + | x |^\alpha u^{2^*-1} \] in $\Omega$ included in a unit ball, under several conditions. Here, $\Psi$ is a non-trivial given function with $0 \leq \Psi \leq 1$ which may vanish on $\partial \Omega$. Let $\lambda_1$ be the first eigenvalue of the Dirichret eigenvalue problem $-\Delta \phi = \lambda \Psi \phi$ in $\Omega$. We show that if the dimension $N \geq 4$ and $0 < \lambda < \lambda_1$, there exists a positive solution for small $\alpha > 0$. Our methods include the mountain pass theorem and the Talenti function. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{definition}[theorem]{Definition} \newtheorem{corollary}[theorem]{Corollary} \allowdisplaybreaks \section{Introduction} We consider the H\'enon type equation with critical Sobolev growth \begin{equation} \begin{gathered} - \Delta u = \lambda \Psi u + | x |^\alpha u^{2^*-1} \quad\text{in } \Omega, \\ u > 0 \quad \text{in } \Omega, \\ u = 0 \quad\text{on } \partial\Omega. \end{gathered}\label{eq:prob_main} \end{equation} We set $N \geq 3$. We use $2^* = 2N/(N-2)$ to denote the critical Sobolev exponent. Let $\Omega \subset \mathbb{R}^N$ be a piecewise $C^1$-class bounded domain satisfying $\Omega \subset B(0, 1)$. Here, $B(p, r) = \{ x \in \mathbb{R}^N : | x - p | < r \}$. Let $x_0 = (1, 0, \dots, 0) \in \mathbb{R}^N$. We assume that $x_0 \in \partial \Omega$ and $\Omega$ satisfies the interior ball condition at $x_0$, i.e., there exists an open ball $B \subset \Omega$ with $x_0 \in \partial B$. We consider the case $\lambda < \lambda_1$, where $\lambda_1$ is the first eigenvalue of the Dirichret eigenvalue problem: $-\Delta \phi = \lambda \Psi \phi$ in $\Omega$. We set $\alpha > 0$ and $\Psi \in L^\infty(\Omega) \setminus \{ 0 \}$ with $0 \leq \Psi \leq 1$ in $\Omega$. Next we state our main theorem. \begin{theorem} \label{thm:main} Let $N \geq 4$ and $0 < \lambda < \lambda_1$. Suppose that there exist $a > 0$, $\beta \geq 0$ and an open ball $B \subset \Omega$ with $x_0 \in \partial B$ such that $\Psi_0 \leq \Psi \leq 1$ in $\Omega$, where \[ \Psi_0(x) = \begin{cases} a | x - x_0 |^\beta & x \in B, \\ 0 & x \not\in B. \end{cases} \] Then, the main problem \eqref{eq:prob_main} has a solution $u \in H_0^1(\Omega)$ for sufficiently small $\alpha > 0$. \end{theorem} We give two examples. The first one is simple: Let $N \geq 4$, $0 < \lambda < \lambda_1$ and $\Omega = B(0, 1)$. Assume that $\Psi$ is a continuous function defined on $\overline{\Omega}$ with $0 \leq \Psi \leq 1$. Suppose that there exists $\overline{x} \in \partial \Omega$ such that $\Psi(\overline{x}) > 0$. Then, \eqref{eq:prob_main} has a solution for small $\alpha > 0$. To confirm this example, we set $\beta = 0$ and some small $a > 0$ and some small $B \subset \Omega$ with $\overline{x} \in \partial \Omega$. The second example is for the case where $\Psi$ vanishes on $\partial \Omega$. We state it as the following corollary. \begin{corollary} \label{cor:cor_of_main_theorem} Let $N \geq 4$, $0 < \lambda < \lambda_1$, $\Omega = B(0, 1)$ and $\beta_0 > 0$. Assume that $\Psi(x) = (1 - | x | )^{\beta_0}$. Then, \eqref{eq:prob_main} has a solution for small $\alpha > 0$. \end{corollary} This corollary follows from elementary geometries. We prove it in Section \ref{sec:cor}. In \cite{henon1973numerical}, the following H\'enon equation for the case $N = 1$ is proposed \begin{equation} \begin{gathered} - \Delta u = | x |^\alpha | u |^{p-1} \quad\text{in } B(0, 1), \\ u = 0 \quad\text{on } \partial B(0, 1). \end{gathered}\label{eq:henon} \end{equation} In the subcritical case $p < 2^*$, the existence of solution is proved by standard compactness argument. In \cite{MR674869}, it is proved that if $1 < p < 2^*(\alpha) = 2(N+\alpha)/(N-2)$, \eqref{eq:henon} has a positive radial solution. H\'enon equation is widely studied in recent times. Many authors study whether there exists a positive non-radial solution of \eqref{eq:henon} for the case $1 < p < 2^*(\alpha)$. We refer \cite{MR2100908, MR2142066, MR1918755}. Many authors also study the subcritical case $p < 2^*$ and investigate the behavior of solutions where $p \to 2^*$. We refer \cite{MR1963460, MR2191725}. General bounded domain cases of \eqref{eq:henon} are studied in \cite{MR2158918, MR2271694, MR2541412} and so on. If $\alpha = 0$ and $\Psi = 1$ in $\Omega$, \eqref{eq:prob_main} becomes the original Br\'ezis--Nirenberg problem. In \cite{MR709644}, it is proved that under these conditions there exists a solution if $N \geq 4$ and $0 < \lambda < \lambda_1'$, or if $N = 3$, $\lambda_1'/4 < \lambda < \lambda_1'$ and $\Omega$ is a ball. Here $\lambda_1'$ is the first eigenvalue of the Dirichret eigenvalue problem: $-\Delta \phi = \lambda \phi$ in $\Omega$. Over three decades many authors have studied existence and nonexistence of Br\'ezis--Nirenberg type problems. Our problem \eqref{eq:prob_main} is regarded as a combination of H\'enon equations and Br\'ezis-Nirenberg problems. In \cite{MR2951742} and \cite{MR2951722}, the following problem directly related to \eqref{eq:prob_main} is studied \begin{equation} \begin{gathered} - \Delta u = \lambda u + | x |^\alpha | u |^{2^*-2} u \quad\text{in } \Omega, \\ u = 0 \quad\text{on } \partial \Omega, \end{gathered} \label{eq:henon-brezis-nirenberg} \end{equation} where $\alpha > 0$ and $\lambda > \lambda_1'$. They show that \eqref{eq:henon-brezis-nirenberg} has a sign-changing solution for sufficiently small $\alpha > 0$ when $N \geq 7$ with smooth $\partial\Omega$ and $N \geq 5$ with $\Omega = B(0, 1)$ in \cite{MR2951742} and \cite{MR2951722}, respectively. In this paper, we seek for a positive solution for the case that $0 < \lambda < \lambda_1$, $N \geq 4$, $\Omega$ is more generalized and $\Psi$ is not necessarily a constant. Our method is based on the mountain pass theorem and Talenti functions presented in \cite{MR709644}. Since the coefficient $| x |^\alpha$ is not achieved its maximum in $\Omega$, we use the function \[ u_{\epsilon, l}(x) = \frac{\xi_l(x)}{(\epsilon + | x - x_l |^2 )^{(N-2)/2}}. \] Here, $\epsilon > 0$, $x_l = (1-l, 0, \dots, 0) \in \mathbb{R}^N$ and $\xi_l \in C^\infty_c (\Omega)$ is a cut-off function supported on $B(x_l, l)$. We regard $l = l(\epsilon)$ as a function that satisfies $l \to 0$ as $\epsilon \to 0$. To prove Theorem~\ref{thm:main}, we set $l = l(\epsilon) = \epsilon^\gamma$ for $0 < \gamma < 1/2$ for the case $N \geq 5$ and $l = l(\epsilon) = | \log \epsilon |^{-k}$ for $k > 0$ for the case $N = 4$. For details, see Section \ref{sec:talenti}. If we take $\epsilon \to 0$, the support is getting smaller and $x_l$ is getting closer to $x_0$. This type of functions has been already introduced in \cite{MR2951742} and \cite{MR2951722} with $l = l(\epsilon) = \epsilon^\gamma$ for fixed $\gamma$. In our case we choose the parameters $\gamma$ and $k$ appropriately since $\Psi$ may vanish on $\partial \Omega$. We set $I \colon H_0^1(\Omega) \to \mathbb{R}$ as \[ I(u) = \frac{1}{2} \int_\Omega | Du |^2 dx - \frac{\lambda}{2} \int_\Omega \Psi u^2 dx - \frac{1}{2^*} \int_\Omega | x |^\alpha (u_+)^{2^*} dx. \] Here we write $f_+ = \max(f, 0)$ for a function $f$. Note that $u \in H_0^1(\Omega) \setminus \{ 0 \}$ is a solution of \eqref{eq:prob_main} if $u$ is a critical point of $I$. This is because if $u$ is a critical point of $I$; then we have \[ (-\Delta - \lambda \Psi) u = | x |^\alpha (u_+)^{2^*-1} \geq 0. \] Since $\lambda < \lambda_1$, we see that $u > 0$ in $\Omega$ by the strong maximum principle. This paper consists of four sections. In Section 2, we prove the mountain pass geometry of $I$ and the convergence of a $(\text{PS})_c$ sequence for some small $c > 0$. In Section 3, we show estimates of integrals of $u_{\epsilon, l}$. In Section 4, we prove Theorem~\ref{thm:main}. In Section 5, we show a technical convergence lemma. In Section 6, we prove Corollary~\ref{cor:cor_of_main_theorem}. Throughout the present paper, all functions are real-valued. We use $L^r(\Omega)$ for $r \geq 1$ to denote the Lebesgue space equipped with the norm \[ \| v \| _{L^r(\Omega)} = \begin{cases} \big( \int_\Omega | v |^r dx )^{1/r} & 1 \leq r < \infty, \\ \operatorname{ess\,sup}_{x \in \Omega} | v(x) | & r = \infty. \end{cases} \] The inner product of $L^2(\Omega)$ is denoted by \[ (v, w)_{L^2(\Omega)} = \int_\Omega vw dx. \] The Sobolev space $H_0^1(\Omega)$ is the completion of $C^\infty_c(\Omega)$ with respect to the norm \[ \| v \| _{H_0^1(\Omega)} = \sqrt{(v, v)_{H_0^1(\Omega)}}, \quad \text{where}\quad (v, w)_{H_0^1(\Omega)} = (Dv, Dw)_{L^2(\Omega)} = \int_\Omega Dv \cdot Dw dx. \] We write $\langle f, v \rangle$ for the canonical pairing of $f \in H^{-1}(\Omega)$ and $v \in H_0^1(\Omega)$. We remark two notations. If $f = -\Delta w$ for some $w \in H_0^1(\Omega)$, then \[ \langle f, v \rangle = \int_\Omega Dw \cdot Dv dx = (w, v)_{H_0^1(\Omega)}. \] If we regard $w \in L^2(\Omega)$ as an element of $H^{-1}(\Omega)$, then \[ \langle w, v \rangle = \int_\Omega wv dx = (w, v)_{L^2(\Omega)}. \] We use $S$ to denote the best Sobolev constant defined by \[ S = \inf_{u \in H_0^1(\Omega), u \not\equiv 0} \frac{ \| Du \| _{L^2(\Omega)}^2}{\| u \| _{L^{2^*}(\Omega)}^2}. \] It is known that $S$ does not depend on $\Omega \subset \mathbb{R}^N$. Without definitions we use the characters $C, C', C'', C_1, C_2 > 0$ to denote positive constants which is not important and may vary by line to line. \section{$(\text{PS})_c$ Condition and Mountain Pass Theorem} In this section we assume that $N \geq 3$ and $\lambda < \lambda_1$. We recall the $(\text{PS})_c$ condition and the mountain pass theorem without $(\text{PS})$ condition. \begin{definition} \label{defn:ps} \rm \begin{enumerate} \item[(i)] Let $c \in \mathbb{R}$. We say that a sequence $\{ u_k \}_{k=0}^\infty$ in $H_0^1(\Omega)$ is a Palais-Smale sequence of $I$ at the mountain pass level $c$ if the following conditions hold: \begin{enumerate} \item[(1)] $I(u_k) \to c$ ($k \to \infty$), \item[(2)] $I^\prime(u_k) \to 0$ in $H^{-1}(\Omega)$ ($k \to \infty$). \end{enumerate} \item[(ii)] Let $c \in \mathbb{R}$. We say that $I$ satisfies the $(\text{PS})_c$ condition if any Palais-Smale sequence of $I$ at the mountain pass level $c$ has a convergent subsequence in $H_0^1(\Omega)$. \end{enumerate} \end{definition} \begin{proposition}[The mountain pass theorem without $(\text{PS})$ condition \cite{MR0370183}] \label{prop:mountain} \quad \\ Suppose that there exist $r, l > 0$ such that $I(u) > l$ for all $u \in H_0^1(\Omega)$ with $\| u \| _{H_0^1(\Omega)} = r$. Assume that there exists $v \in H_0^1(\Omega)$ such that $I(v) \leq 0$ and $\| u \| _{H_0^1(\Omega)} > r$. Let \begin{equation} c = \inf_{\gamma \in \Gamma} \max_{u \in \gamma} I(u), \label{eq:mountain_c} \end{equation} where $\Gamma$ is the set of paths in $H_0^1(\Omega)$ connecting $0$ and any end point $v \in H_0^1(\Omega)$ with $I(v) \leq 0$ and $\| v \| _{H_0^1(\Omega)} > r$. Then, there exists a Palais-Smale sequence of $I$ at the mountain pass level $c$. \end{proposition} \begin{lemma} \label{lem:if-divide} For $u \in H_0^1(\Omega)$, we have \begin{gather} \| Du \| _{L^2(\Omega)}^2 - \lambda \int_\Omega \Psi u^2 dx \geq \Big( 1 - \frac{\max(\lambda, 0)}{\lambda_1} \Big) \| Du \| _{L^2(\Omega)}^2, \label{eq:if-divide-1} \\ 1 - \frac{\max(\lambda, 0)}{\lambda_1} > 0. \label{eq:if-divide-2} \end{gather} \end{lemma} \begin{proof} If $\lambda \leq 0$, we have \[ \| Du \| _{L^2(\Omega)}^2 - \lambda \int_\Omega \Psi u^2 dx \geq \| Du \| _{L^2(\Omega)}^2. \] If $0 \leq \lambda < \lambda_1$, we have \[ \| Du \| _{L^2(\Omega)}^2 - \lambda \int_\Omega \Psi u^2 dx \geq \Big(1 - \frac{\lambda}{\lambda_1} \Big) \| Du \| _{L^2(\Omega)}^2. \] Here we used the Poincar\'e type inequality. Combining these cases, we have \eqref{eq:if-divide-1}. The inequality \eqref{eq:if-divide-2} follows, since $\lambda < \lambda_1$. \end{proof} We check the mountain pass geometry of $I$. We admit that $I$ is a $C^1$-class functional on $H_0^1(\Omega)$ with $I(0) = 0$. \begin{lemma} There exist $r > 0$ and $l > 0$ such that $I(u) > l$ for all $u \in H_0^1(\Omega)$ with $\| u \| _{H_0^1(\Omega)} = r$. \end{lemma} \begin{proof} By the Sobolev inequality, there exists $C > 0$ such that \[ \| u \| _{L^{2^*}(\Omega)}^{2^*} \leq C \| Du \| _{L^2(\Omega)}^{2^*} \] for any $u \in H_0^1(\Omega)$. Thus we have \begin{align*} I(u) & \geq \frac{1}{2} \Big( 1 - \frac{\max(\lambda, 0)}{\lambda_1} \Big) \| Du \| _{L^2(\Omega)}^2 - \frac{1}{2^*} \| u \| _{L^{2^*}(\Omega)}^{2^*} \\ & \geq C_1 \| Du \| _{L^2(\Omega)}^2 - C_2 \| Du \| _{L^2(\Omega)}^{2^*}. \end{align*} Since $2 < 2^*$, the proof is complete. \qedhere. \end{proof} \begin{lemma} For any $r > 0$, there exists $u \in H_0^1(\Omega)$ such that $I(u) \leq 0$ and $\| u \| _{H_0^1(\Omega)} > r$. \end{lemma} \begin{proof} Let $v \in H_0^1(\Omega) \setminus \{ 0 \}$ and $t > 0$. We have \[ I(tv) = \frac{t^2}{2} \Big( \| Dv \| _{L^2(\Omega)}^2 - \lambda \int_\Omega \Psi v^2 dx \Big) - \frac{t^{2^*}}{2^*} \int_\Omega | x |^\alpha (v_+)^{2^*} dx. \] It follows that $\lim_{t \to \infty} I(tv) = -\infty$ since $2 < 2^*$. Set $u = tv$ for large $t > 0$ to complete the proof. \qedhere \end{proof} Next, we study which mountain pass level $c$ satisfies the $(\text{PS})_c$ condition on $I$. \begin{lemma} \label{lem:bounded} Let $\{ u_k \}_{k=0}^\infty$ be a Palais-Smale sequence of $I$ at the mountain pass level $c \in \mathbb{R}$. Then, $\{ u_k \}$ is bounded in $H_0^1(\Omega)$. \end{lemma} \begin{proof} Let $\epsilon > 0$. Then, by the condition (2) of Definition~\ref{defn:ps} (i), we have \[ | \langle I'(u_k), u_k \rangle | \leq \epsilon \| Du_k \| _{L^2(\Omega)} \] for large $k$. Set $\epsilon = 2^*$ and combine the condition (1) of Definition~\ref{defn:ps} (i) to have \[ I(u_k) - \frac{1}{2^*} \langle I'(u_k), u_k \rangle \leq C + \| Du_k \| _{L^2(\Omega)}. \] It also follows that \begin{align*} I(u_k) - \frac{1}{2^*} \langle I'(u_k), u_k \rangle & = \Big( \frac{1}{2} - \frac{1}{2^*} \Big) \Big( \| Du_k \| ^2_{L^2(\Omega)} - \lambda \int_\Omega \Psi u_k^2 dx \Big) \\ & \geq \Big( \frac{1}{2} - \frac{1}{2^*} \Big) \Big( 1 - \frac{\max(\lambda, 0)}{\lambda_1} \Big) \| Du_k \| _{L^2(\Omega)}^2. \end{align*} Combining these inequalities, we have \[ C' \| Du_k \| ^{2}_{L^2(\Omega)} \leq C + \| Du_k \| _{L^2(\Omega)}. \] We see that $\| Du_k \| _{L^2(\Omega)}$ is bounded, which completes the proof. \end{proof} \begin{lemma} \label{lem:PS_c} Let \begin{equation} 0 < c < \frac{1}{N}S^{N/2}. \label{eq:PS_c_level} \end{equation} Then, $I$ satisfies $(\text{PS})_c$ condition. \end{lemma} \begin{proof} Let $\{ u_k \}_{k = 0}^\infty$ be a Palais-Smale sequence of $I$ at the mountain pass level $c$ satisfying \eqref{eq:PS_c_level}. By Lemma~\ref{lem:bounded}, $\{ u_k \}$ is a bounded sequence of $H_0^1(\Omega)$. Thus there exists $u \in H_0^1(\Omega)$ such that, taking a subsequence, \begin{equation} \begin{gathered} u_k \rightharpoonup u \quad \text{weakly in } H_0^1(\Omega), \\ u_k \to u \quad \text{in } L^r(\Omega) \quad (r < 2^* ), \\ u_k \to u \quad \text{a.e. in } \Omega \end{gathered}\label{eq:u_k_subsequence_convergence} \end{equation} as $k \to \infty$. Let $\psi \in H_0^1(\Omega)$. By Lemma~\ref{lem:convergence_critical}, we have \begin{align*} \langle I'(u_k), \psi \rangle & = \int_\Omega Du_k \cdot D\psi dx - \lambda\int_\Omega \Psi u_k \psi dx -\int_\Omega | x |^\alpha (u_k )_+^{2^* - 1} \psi dx \\ & \xrightarrow{k \to \infty} \int_\Omega Du \cdot D\psi dx - \lambda\int_\Omega \Psi u \psi dx -\int_\Omega | x |^\alpha u _+^{2^* - 1} \psi dx\\ &= \langle I'(u), \psi \rangle. \end{align*} Since $\lim_{k \to \infty} \langle I'(u_k), \psi \rangle = 0$, it follows that \begin{equation} \langle I'(u), \psi \rangle = 0. \label{eq:I_prime_u_k_psi} \end{equation} We show that \begin{equation} u_k \to u \quad \text{in } H_0^1(\Omega). \label{eq:u_k_to_u} \end{equation} Note that $u_+ = u$ since either $u \equiv 0$ or $u > 0 $ in $\Omega$. Set $\psi = u$ in \eqref{eq:I_prime_u_k_psi} to have \begin{equation} \int_\Omega | Du |^2 dx - \lambda \int_\Omega \Psi u^2 dx - \int_\Omega | x |^\alpha u^{2^*} dx = 0. \label{eq:psi_equal_u} \end{equation} Then, we see that \begin{equation} I(u) = \Big( \frac{1}{2} - \frac{1}{2^*} \Big) \int_\Omega | x |^\alpha u^{2^*} dx \geq 0. \label{eq:I_u_geq_0} \end{equation} Let $w_k = u_k - u$. We have \begin{equation} \begin{gathered} w_k \rightharpoonup 0 \quad\text{weakly in } H_0^1(\Omega), \\ w_k \rightarrow 0 \quad\text{in } L^r(\Omega) \quad (r < 2^* ), \\ w_k \rightarrow 0 \quad\text{a.e. in } \Omega \end{gathered} \label{eq:w_k_subsequence_convergence} \end{equation} as $k \to 0$. It follows that \[ \int_\Omega | Du_k |^2 dx = \int_\Omega | Dw_k |^2 dx + \int_\Omega | Du |^2 dx + o(1). \] Let $\widetilde{w_k} = ( u_k )_+ - u$. By Br\'ezis--Lieb Lemma \cite{MR699419}, we have \[ \int_\Omega | x |^\alpha ( u_k )_+^{2^*} dx = \int_\Omega | x |^\alpha u^{2^*} dx + \int_\Omega | x |^\alpha | \widetilde{w_k} |^{2^*} dx + o(1). \] Thus, \[ I(u_k) - I(u) = \frac{1}{2} \int_\Omega | Dw_k |^2 dx - \frac{1}{2^*} \int_\Omega | x |^\alpha | \widetilde{w_k} |^{2^*} dx + o(1). \] Then \begin{equation} I(u) + \frac{1}{2} \int_\Omega | Dw_k |^2 dx - \frac{1}{2^*} \int_\Omega | x |^\alpha | \widetilde{w_k} |^{2^*} dx = c + o(1). \label{eq:c_o(1)} \end{equation} Since $\langle I'(u_k), u_k \rangle \to 0$ as $k \to \infty$, we have \[ \lim_{k \to \infty} \Big( \int_\Omega | Du_k |^2 dx - \lambda \int_\Omega \Psi u_k^2 dx - \int_\Omega | x |^\alpha ( u_k )_+^{2^*} dx \Big) = 0. \] Combining this equation with \eqref{eq:psi_equal_u} we obtain \[ \lim_{k \to \infty} \Big( \int_\Omega | Dw_k |^2 dx - \int_\Omega | x |^\alpha | \widetilde{w_k} |^{2^*} dx \Big) = 0. \] Taking a subsequence, we have \[ \lim_{k \to \infty} \int_\Omega | Dw_k |^2 dx = \lim_{k \to \infty} \int_\Omega | x |^\alpha | \widetilde{w_k} |^{2^*} dx. \] We write $l \geq 0$ as this limit. By the Sobolev inequality, we have \[ \| Dw_k \| _{L^2(\Omega)}^2 \geq S \| w_k \| _{L^{2^*}(\Omega)}^2 \geq S \| \widetilde{w_k} \| _{L^{2^*}(\Omega)}^2 \geq S \Big( \int_\Omega | x |^\alpha | \widetilde{w_k} |^{2^*} dx \Big)^{2/2^*}, \] which implies $l \geq Sl^{2/2^*}$. Here we note that $w_k \geq \widetilde{w_k}$ in $\Omega$ since either $u \equiv 0$ or $u > 0$ in $\Omega$. We show $l = 0$. Assume to the contrary that $l > 0$. Then, we have $l \geq S^{N/2}$. By \eqref{eq:c_o(1)}, we have \[ I(u) + \Big( \frac{1}{2} - \frac{1}{2^*} \Big)l = c. \] By \eqref{eq:I_u_geq_0}, it follows that $S^{N/2}/N \leq c$, which contradicts \eqref{eq:PS_c_level}. Thus we conclude that $l = 0$, which implies \eqref{eq:u_k_to_u} as desired. \qedhere \end{proof} \begin{proposition} \label{prop:PS_c_converge} Assume that there exists a Palais-Smale sequence of $I$ at the mountain pass level $c$ satisfying \eqref{eq:PS_c_level}. Then, \eqref{eq:prob_main} has a solution. \end{proposition} \begin{proof} Let $\{ u_k \}_{k = 0}^\infty$ be a Palais-Smale sequence of $I$ at the mountain pass level $c$ satisfying \eqref{eq:PS_c_level}. By Lemma~\ref{lem:PS_c}, $\{ u_k \}$ has a convergent subsequence. We use $u \in H_0^1(\Omega)$ to denote the limit of it. Then, \eqref{eq:I_prime_u_k_psi} holds for any $\psi \in H_0^1(\Omega)$, i.e., $u$ is a critical point of $I$. In addition, it follows that \[ I(u) = \lim_{k \to \infty} I(u_k) = c > 0, \] which implies $u \not\equiv 0$. Hence, $u$ is a solution of \eqref{eq:prob_main}. \end{proof} \section{Evluations of Integrals} \label{sec:talenti} We set $U \colon \mathbb{R}^N \to \mathbb{R}$ as \[ U(x) = \frac{1}{(1 + | x |^2 )^{(N-2)/2}}. \] We set $x_l = (1 - l, 0, \dots, 0) \in \mathbb{R}^N$ for $0 < l < 1$. For $0 < l < 1$, we set cut-off functions $\xi_l \in C^\infty_c(\Omega)$ which satisfies the following conditions: \begin{enumerate} \item $0 \leq \xi_l \leq 1$. \item \[ \xi_l(x) = \begin{cases} 1 & x \in B(x_l, l/2), \\ 0 & x \not\in B(x_l, l). \end{cases} \] \item $| D\xi_l | \leq C/l$ for some constant $C > 0$. \item $D \xi_l(x) \cdot (x-x_l) \leq 0$. \end{enumerate} We set $u_{\epsilon, l}, v_{\epsilon, l} \in H_0^1(\Omega)$ for $\epsilon > 0$ and $0 < l < 1$ as follows: \begin{align*} u_{\epsilon, l}(x) & = \frac{\xi_l(x)}{(\epsilon + | x - x_l |^2 )^{(N-2)/2}}, \\ v_{\epsilon, l}(x) & = \frac{u_{\epsilon, l}(x)} {\| | x |^{\alpha/2^*} u_{\epsilon, l }\| _{L^{2^*}(\Omega)}}. \end{align*} Hereinafter, we regard $l = l(\epsilon)$ as a function of $\epsilon > 0$ which satisfies $l \to 0$ as $\epsilon \to 0$ and $\epsilon\leq l$. \begin{lemma} \label{lem:talenti} Suppose that $N \geq 3$. There exist positive constants $C_1, C_2, C > 0$ such that the following inequalities hold for small $\epsilon > 0$: \begin{gather} \begin{aligned} \| DU \| _{L^2(\mathbb{R}^N)}^2 \epsilon^{-(N-2)/2} - C_1 l^{-(N-2)} &\leq \| Du_{\epsilon, l} \| _{L^2(\Omega)}^2 \\ &\leq \| DU \| _{L^2(\mathbb{R}^N)}^2 \epsilon^{-(N-2)/2} + C_2 l^{-(N-2)}, \label{eq:talenti_2_2} \end{aligned} \\ \begin{aligned} (1 - 2l)^{2\alpha/2^*} ( \| U \| ^{2^*}_{L^{2^*}(\mathbb{R}^N)}\epsilon^{-N/2} - Cl^{-N} )^{2/2^*} &\leq \| | x |^{\alpha/2^*} u_{\epsilon, l }\| _{L^{2^*}(\Omega)}^2 \\ &\leq \| U\| ^{2}_{L^{2^*}(\mathbb{R}^N)}\epsilon^{-(N-2)/2}. \end{aligned} \label{eq:talenti_2*_2} \end{gather} \end{lemma} \begin{proof} First, we investigate \[ I = \int_\Omega | Du_{\epsilon, l} |^2 dx. \] We have \[ Du_\epsilon (x) = \frac{D\xi_l(x)}{(\epsilon + | x - x_l |^2 )^{(N-2)/2}} - \frac{(N-2)\xi_l(x)(x - x_l)}{(\epsilon + | x - x_l |^2 )^{N/2}}. \] We divide $I$ into three terms, i.e., $I = I_1 + I_2 + I_3$; \begin{gather*} I_1 = \int_\Omega \frac{| D\xi_l(x) |^2}{(\epsilon + | x - x_l |^2 )^{N-2}} dx, \\ I_2 = \int_\Omega \frac{-2(N-2)\xi_l(x) ( D\xi_l(x) \cdot (x - x_l) )}{(\epsilon + | x - x_l |^2 )^{N-1}} dx, \\ I_3 = \int_\Omega \frac{(N-2)^2 \xi_l(x)^2 | x - x_l |^2}{(\epsilon + | x - x_l |^2 )^{N}} dx. \end{gather*} We start by getting an upper bound. We have \[ I_3 \leq \int_{\mathbb{R}^N} \frac{(N-2)^2 | x |^2}{(\epsilon + | x |^2 )^{N}} dx = \| DU \| _{L^2(\mathbb{R}^N)}^2 \epsilon^{-(N-2)/2}. \] The integrals $I_2$ and $I_1$ are estimated as follows: \begin{align*} I_2 & \leq \frac{C}{l} \int_{B(x_l, l) \setminus B(x_l, l/2)} \frac{| x - x_l |}{(\epsilon + | x - x_l |^2 )^{N-1}} dx \\ &= \frac{C}{l} \int_{B(0, l) \setminus B(0, l/2)} \frac{| x |}{(\epsilon + | x |^2 )^{N-1}} dx \\ & \leq \frac{C}{l} \int_{B(0, l) \setminus B(0, l/2)} \frac{dx}{| x |^{2N-3}} = \frac{C}{l} \int_{l/2}^l r^{-N+2} dr \leq Cl^{-N+2}, \end{align*} \begin{align*} \\ I_1 & \leq \frac{C}{l^2} \int_{B(0, l) \setminus B(0, l/2)} \frac{1}{(\epsilon + | x |^2 )^{N-2}} dx \\ & \leq \frac{C}{l^2} \int_{B(0, l) \setminus B(0, l/2)} \frac{dx}{| x |^{2N-4}} \\ &= \frac{C}{l^2} \int_{l/2}^l r^{-N+3} dr \leq Cl^{-N+2}. \end{align*} Note that the last integrals of above two inequalities are calculated differentially by the dimension $N \geq 3$. However, the resulting evaluations are the same $I_2, I_1 \leq Cl^{-N+2}$. Thus we have the upper bound of \eqref{eq:talenti_2_2}. Next we consider the lower bound. We have $I_1, I_2 \geq 0$. We estimate $I_3$ as follows: \begin{align*} I_3 &> \int_{B(x_l, l/2)} \frac{(N-2)^2 | x - x_l |^2}{(\epsilon + | x - x_l |^2 )^{N}} dx \\ &= \| DU \| _{L^2(\mathbb{R}^N)}^2 \epsilon^{-(N-2)/2} - \int_{\mathbb{R}^N \setminus B(x_l, l/2)} \frac{(N-2)^2 | x - x_l |^2}{(\epsilon + | x - x_l |^2 )^{N}} dx. \end{align*} Here, we obtain \begin{align*} &\int_{\mathbb{R}^N \setminus B(x_l, l/2)} \frac{(N-2)^2 | x - x_l |^2}{(\epsilon + | x - x_l |^2 )^{N}} dx \\ & = \int_{\mathbb{R}^N \setminus B(0, l/2)} \frac{(N-2)^2 | x |^2}{(\epsilon + | x |^2 )^{N}} dx \\ & < C \int_{\mathbb{R}^N \setminus B(0, l/2)} \frac{1}{| x |^{2N-2}} dx \\ &= C \int_{l/2}^\infty r^{-N+1} dr = Cl^{-N+2}, \end{align*} which implies the lower bound of \eqref{eq:talenti_2_2}. Second, we study \[ I = \int_\Omega | x |^\alpha u_{\epsilon, l}^{2^*} dx. \] We have \[ I = \int_{B(x_l, l)} \frac{| x |^\alpha \xi_l(x)^{2^*}}{(\epsilon + | x - x_l |^2)^N} dx = \int_{B(0, l)} \frac{| x + x_l |^\alpha \xi_l(x + x_l)^{2^*}}{(\epsilon + | x |^2)^N} dx. \] Thus it follows that $(1 - 2l)^\alpha \widetilde{I} \leq I \leq \widetilde{I}$. Here we set \[ \widetilde{I} = \int_{B(0, l)} \frac{\xi_l(x + x_l)^{2^*}}{(\epsilon + | x |^2)^N} dx. \] We obtain \begin{align*} \widetilde{I} &= \Big( \int_{B(0, l)} \frac{\xi_l(x + x_l)^{2^*}}{(\epsilon + | x |^2)^N} dx - \int_{B(0, l)} \frac{1}{(\epsilon + | x |^2)^N} dx \Big) \\ &\quad + \Big( \int_{B(0, l)} \frac{1}{(\epsilon + | x |^2)^N} dx - \int_{\mathbb{R}^N} \frac{1}{(\epsilon + | x |^2)^N} dx \Big) + \int_{\mathbb{R}^N} \frac{1}{(\epsilon + | x |^2)^N} dx \\ &= \int_{B(0, l) \setminus B(0, l/2)} \frac{\xi_l(x + x_l)^{2^*} - 1}{(\epsilon + | x |^2)^N} dx \\ &\quad - \int_{\mathbb{R}^N \setminus B(0, l)} \frac{1}{(\epsilon + | x |^2)^N} dx + \| U \| _{L^{2^*}(\mathbb{R}^N)}^{2^*} \epsilon^{-N/2}. \end{align*} Thus we have \[ \widetilde{I} \leq \| U \| _{L^{2^*}(\mathbb{R}^N)}^{2^*} \epsilon^{-N/2}. \] For the lower bound, it follows that \begin{align*} &\Big| \int_{B(0, l) \setminus B(0, l/2)} \frac{\xi_l(x + x_l)^{2^*} - 1}{(\epsilon + | x |^2)^N} dx - \int_{\mathbb{R}^N \setminus B(0, l)} \frac{1}{(\epsilon + | x |^2)^N} dx \Big| \\ &\leq \int_{\mathbb{R}^N \setminus B(0, l/2)} \frac{dx}{(\epsilon + | x |^2)^N} \\ &\leq \int_{\mathbb{R}^N \setminus B(0, l/2)} \frac{dx}{| x |^{2N}} = Cl^{-N}. \end{align*} Thus we have \[ \widetilde{I} \geq \| U \| _{L^{2^*}(\mathbb{R}^N)}^{2^*} \epsilon^{-N/2} - Cl^{-N}. \] Finally we conclude that \[ (1 - 2l)^\alpha \big( \| U \| ^{2^*}_{L^{2^*}( \mathbb{R}^N )}\epsilon^{-N/2} - Cl^{-N} \big) \leq I \leq \| U \| ^{2^*}_{L^{2^*}( \mathbb{R}^N )}\epsilon^{-N/2}, \] which implies \eqref{eq:talenti_2*_2}. \end{proof} \begin{lemma} \label{lem:lggepsilon} Let $c > 0$ be a positive constant. Assume that $\lim_{\epsilon \to 0} (\sqrt{\epsilon}/l) = 0$. Then, we have \begin{equation} \int_{B(0, cl)} \frac{dx}{(\epsilon + | x |^2)^{N-2}} = \begin{cases} O(\epsilon^{-(N-4)/2}) & N \geq 5, \\ O(| \log (\sqrt{\epsilon}/l) |) & N = 4, \\ O(l) & N = 3, \end{cases} \label{eq:talenti_N-2} \end{equation} as $\epsilon \to 0$. \end{lemma} Note that this is not a direct conclusion argued in \cite[p.~445]{MR709644}. We have to take it into account that $l \to 0$ as $\epsilon \to 0$. If $N = 3$, the integral converges to $0$, which does not in \cite{MR709644}. \begin{proof} Let $I$ denote the integral on the left side of \eqref{eq:talenti_N-2}. First, we investigate the case $N \geq 5$. We have \[ I = \epsilon^{-(N-4)/2} \int_{B(0, cl/\sqrt{\epsilon})} \frac{dx}{(1 + | x |^2)^{N-2}}. \] Since $\lim_{\epsilon \to 0}(cl/\sqrt{\epsilon}) = \infty$, we obtain $I = O(\epsilon^{-(N-4)/2})$ as $\epsilon \to 0$. Second, we investigate the case $N = 4$. We have \[ I = C \int_0^{cl} \frac{r^3}{(\epsilon + r^2)^2} dr. \] To investigate how $l$ affects the conclusion, we evaluate the integral on the right side by direct calculation. We start by getting the lower bound. It follows that \begin{align*} \int_0^{cl} \frac{r^3}{(\epsilon + r^2)^2} dr & > \int_0^{cl} \frac{r^3}{(\sqrt{\epsilon} + r)^4} dr\\ &= \int_0^{cl} \frac{((\sqrt{\epsilon} + r) - \sqrt{\epsilon})^3}{(\sqrt{\epsilon} + r)^4} dr \\ & = \sum_{i = 0}^3 (-1)^{3-i} \begin{pmatrix} 3 \\ i \end{pmatrix} I_i, \end{align*} where \[ I_i = \epsilon^{(3-i)/2} \int_0^{cl} (r + \sqrt{\epsilon})^{i-4} dr \] for $i = 0, 1, 2, 3$. For $i = 0, 1, 2$, we have \begin{align*} I_i & = \epsilon^{(3-i)/2} \Big[ \frac{1}{i-3} (r+\sqrt{\epsilon})^{i-3} \Big]_0^{cl} \\ &=\frac{\epsilon^{(3-i)/2}}{i-3} \Big( \Big(cl+\sqrt{\epsilon}\Big)^{i-3} - \epsilon^{(i-3)/2}\Big) \\ & = \frac{1}{i-3} \Big( \Big( \frac{\sqrt{\epsilon}}{cl + \sqrt{\epsilon}} \Big)^{3-i} - 1\Big) = O(1) \end{align*} as $\epsilon \to 0$. By contrast, it follows that \begin{align*} I_3 & = \int_0^{cl} \frac{dr}{r + \sqrt{\epsilon}} = \Big[ \log(r + \sqrt{\epsilon}) \Big]_0^{cl} \\ & = \log(cl + \sqrt{\epsilon}) - \log \sqrt{\epsilon}\\ &= \log\Big( c + \frac{\sqrt{\epsilon}}{l} \Big) - \log \Big( \frac{\sqrt{\epsilon}}{l} \Big) = O\Big( \Big| \log \Big( \frac{\sqrt{\epsilon}}{l}\Big) \Big| \Big). \end{align*} Next, we have the upper bound as follows: \begin{align*} \int_0^{cl} \frac{r^3}{(\epsilon + r^2)^2} dr & < \int_0^{cl} \frac{(\epsilon + r^2)^{3/2}}{(\epsilon + r^2)^2} dr = \int_0^{cl} \frac{1}{\sqrt{\epsilon + r^2}} dr \\ & = \Big[ \log\Big(r + \sqrt{r^2 + \epsilon}\Big) \Big]_0^{cl}\\ &= \log\Big(cl + \sqrt{c^2 l^2 + \epsilon} \Big) - \log \sqrt{\epsilon} \\ & = \log \Big( c + \sqrt{ c^2 + \frac{\epsilon}{l^2}} \Big) - \log \Big( \frac{\sqrt{\epsilon}}{l} \Big)\\ & = O\Big( \Big| \log \Big( \frac{\sqrt{\epsilon}}{l} \Big) \Big| \Big). \end{align*} Hence we have $I = O(| \log (\sqrt{\epsilon}/l) |)$. Finally, we investigate the case $N = 3$. First, we have \[ I < \int_{B(0, cl)} \frac{dx}{| x |^2} = Cl. \] Next, since $\lim_{\epsilon \to 0} (\sqrt{\epsilon}/l) = 0$, it follows that \begin{align*} I & \geq \int_{B(0, cl) \setminus B(0, c\sqrt{\epsilon})} \frac{dx}{\epsilon + | x |^2} \geq \int_{B(0, cl) \setminus B(0, c\sqrt{\epsilon})} \frac{dx}{C| x |^2} \\ & = C' \int_{c\sqrt{\epsilon}}^{cl} dr \geq C''l. \end{align*} Thus we have $I = O(l)$. We complete the proof. \end{proof} \begin{lemma} \label{lem:I_0} Let $0 < \gamma < 1/2$. Set $l = l(\epsilon) = \epsilon^\gamma$. Then \begin{equation} \int_\Omega \Psi_0 u_{\epsilon, l}^2 dx = \begin{cases} O(\epsilon^{\beta\gamma-(N-4)/2}) & N \geq 5, \\ O(\epsilon^{\beta\gamma}| \log \epsilon |) & N = 4, \\ O(\epsilon^{(\beta + 1)\gamma}) & N = 3, \end{cases} \label{eq:talenti_psi_2} \end{equation} as $\epsilon \to 0$. \end{lemma} \begin{proof} We investigate \[ I = \frac{1}{a}\int_\Omega \Psi_0 u_{\epsilon, l}^2 dx = \int _{B(x _l,l)} \frac{|x - x _0 |^\beta \xi_l(x)^2} {\big(\epsilon + | x - x _l |^2\big) ^{N - 2}} dx. \] We have \[ I \leq (2l)^\beta \int_{B(0, l)} \frac{1}{(\epsilon + | x |^2)^{N-2}} dx, \] and \begin{align*} I & \geq \int_{B(x_l, l/2)} \frac{| x - x_0|^\beta}{(\epsilon + | x - x_l |^2)^{N-2}} dx = \int_{B(0, l/2)} \frac{| x - x_0 + x_l |^\beta}{(\epsilon + | x |^2)^{N-2}} dx \\ & \geq \Big(\frac{l}{2}\Big)^\beta \int_{B(0, l/2)} \frac{1}{(\epsilon + | x |^2)^{N-2}} dx. \end{align*} By Lemma~\ref{lem:lggepsilon}, we obtain \begin{equation} \int_\Omega \Psi_0 u_{\epsilon, l}^2 dx = \begin{cases} O(l^\beta \epsilon^{-(N-4)/2}) & N \geq 5, \\ O(l^\beta | \log (\sqrt{\epsilon}/l) |) & N = 4, \\ O(l^{\beta + 1}) & N = 3, \end{cases} \label{eq:int_psi_0_u} \end{equation} as $\epsilon \to 0$. Letting $l = \epsilon^\gamma$, we have \eqref{eq:talenti_psi_2}. \end{proof} \begin{corollary} \label{cor:I_0} Let $k > 0$. Set $l = l(\epsilon) = | \log \epsilon |^{-k}$. Then \begin{equation} \int_\Omega \Psi_0 u_{\epsilon, l}^2 dx = \begin{cases} O(| \log \epsilon |^{-\beta k} \epsilon^{-(N-4)/2}) & N \geq 5, \\ O(| \log \epsilon |^{1 -\beta k}) & N = 4, \\ O(| \log \epsilon |^{-(\beta + 1)k}) & N = 3, \end{cases} \label{eq:talenti_psi_3} \end{equation} as $\epsilon \to 0$. \end{corollary} \begin{proof} Set $l = | \log \epsilon |^{-k}$ in \eqref{eq:int_psi_0_u}. The conclusion immediately follows for the case $N \geq 5$ and $N = 3$. For the case $N = 4$, we see \[ l^\beta | \log(\sqrt{\epsilon}/l) | = | \log \epsilon |^{-\beta k} | \log ( \sqrt{\epsilon} | \log \epsilon |^k ) |. \] For small $\epsilon > 0$, it follows that $\sqrt{\epsilon} \leq \sqrt{\epsilon} | \log \epsilon |^k \leq \sqrt[4]{\epsilon}$. Then, we have \[ | \log \epsilon |^{-\beta k} | \log ( \sqrt{\epsilon} | \log \epsilon |^k) | = O(| \log \epsilon |^{1 - \beta k}), \] which completes the proof. \end{proof} \section{Proof of Theorem \ref{thm:main}} By Proposition~\ref{prop:mountain} and Proposition~\ref{prop:PS_c_converge}, it suffice to prove \eqref{eq:PS_c_level} for $c > 0$ defined by \eqref{eq:mountain_c}. By elementary calculations, we have \begin{align*} c & \leq \sup_{t > 0} I(t v_{\epsilon, l}) = \sup_{t > 0} \Big( \frac{t^2}{2} \Big( \| Dv_{\epsilon, l} \| _{L^2(\Omega)}^2 - \lambda \int_\Omega \Psi v_{\epsilon, l}^2 dx \Big) - \frac{t^{2^*}}{2^*} \Big) \\ & = \frac{1}{N} \Big( \| Dv_{\epsilon, l} \| _{L^2(\Omega)}^2 - \lambda \int_\Omega \Psi v_{\epsilon, l}^2 dx \Big)^{N/2} \xrightarrow{\epsilon \to 0} \frac{1}{N} S^{N/2}. \end{align*} We define \[ A(\epsilon) = \| Dv_{\epsilon, l} \| _{L^2(\Omega)}^2 - \lambda \int_\Omega \Psi v_{\epsilon, l}^2 dx - S. \] We show that there exists $\epsilon > 0$ such that $A(\epsilon) < 0$ to completes the proof. We write \[ I = \int_\Omega \Psi v_{\epsilon, l}^2 dx, \quad I_0 = \int_\Omega \Psi_0 v_{\epsilon, l}^2 dx. \] Assume that $\lim_{\epsilon \to 0} (\sqrt{\epsilon}/l) = 0$. By Lemma~\ref{lem:talenti}, it follows that \begin{align*} A(\epsilon) & = \frac{\| Du_{\epsilon, l} \| _{L^2(\Omega)}^2 - \lambda I}{\| | x |^{\alpha/2^*} u_{\epsilon, l} \| _{L^{2^*}(\Omega)}^2 } - S \\ & \leq \frac{ \| DU \| _{L^2(\mathbb{R}^N)}^2 \epsilon^{-(N-2)/2} + C' l^{-(N-2)} - \lambda I_0}{(1 - 2l)^{2\alpha/2^*} ( \| U \| ^{2^*}_{L^{2^*}(\mathbb{R}^N)}\epsilon^{-N/2} - Cl^{-N} )^{2/2^*}} - S \\ & = \frac{ S + C' l^{-(N-2)}\epsilon^{(N-2)/2} - C'' I_0 \epsilon^{(N-2)/2} }{(1 - 2l)^{2\alpha/2^*} ( 1 - Cl^{-N} \epsilon^{N/2} )^{2/2^*}} - S. \end{align*} We set \[ B(\epsilon) = S + C' l^{-(N-2)}\epsilon^{(N-2)/2} - C'' I_0 \epsilon^{(N-2)/2} - S (1 - 2l)^{2\alpha/2^*} \big( 1 - Cl^{-N} \epsilon^{N/2} \big)^{2/2^*}. \] The condition $A(\epsilon) < 0$ is equivalent to $B(\epsilon) < 0$. We have \begin{align*} B(\epsilon) &\leq S - S(1 - 2l)^{2\alpha/2^*} \big( 1 - C l^{-N} \epsilon^{N/2} \big) \\ & \quad+ C' l^{-(N-2)} \epsilon^{(N-2)/2} - C'' I_0 \epsilon^{(N-2)/2} \\ &\leq \big( S - S (1 - 2l)^{2\alpha/2^*} \big) \\ &\quad + \Big( Cl^{-N}\epsilon^{N/2} + C'l^{-(N-2)} \epsilon^{(N-2)/2} - C'' I_0 \epsilon^{(N-2)/2} \Big). \end{align*} Note that \[ \lim_{\epsilon \to 0 } \frac{l^{-N} \epsilon^{N/2}}{l^{-(N-2)} \epsilon^{(N-2)/2}} = 0. \] Hereinafter, we divide the proof into two cases; (i) $N \geq 5$ and (ii) $N = 4$. (i) Let $N \geq 5$, $0 < \gamma < 1/2$ and $l = l(\epsilon) = \epsilon^\gamma$. By Lemma~\ref{lem:I_0}, we have \[ I_0 \epsilon^{(N-2)/2} = O(\epsilon^{\beta \gamma + 1}) \] as $\epsilon \to 0$. We show that there exists $0 < \gamma < 1/2$ such that \begin{equation} (N - 2) \Big( \frac{1}{2} - \gamma \Big) > \beta \gamma + 1. \label{eq:exist_epsilon} \end{equation} This inequality is equivalent to $\gamma < (N-4)/2(\beta + N - 2)$. Thus the condition we are now considering is equivalent to \[ \frac{N-4}{2(\beta + N - 2)} > 0, \] which is always true since $\beta > 0$ and $N \geq 5$. Fix such $0 < \gamma < 1/2$ that satisfies \eqref{eq:exist_epsilon}. Thus we obtain \[ \lim_{\epsilon \to 0} \frac{\epsilon^{(N-2)(1/2 - \gamma)}} {\epsilon^{\beta \gamma + 1}} = 0. \] Therefore we admit the existence of $\epsilon > 0$ such that \[ Cl^{-N}\epsilon^{N/2} + C'l^{-(N-2)} \epsilon^{(N-2)/2} - C'' I_0 \epsilon^{(N-2)/2} < 0. \] Fix such $\epsilon > 0$ and take $\alpha > 0$ so small that $B(\epsilon) < 0$ to obtain the conclusion. \smallskip (ii) Let $N = 4$. By Corollary~\ref{cor:I_0}, We have \[ I_0 \epsilon^{(N-2)/2} = O(\epsilon | \log \epsilon |^{1-\beta k}). \] We see that there exists $k > 0$ such that $1 - \beta k > 2k$, which is equivalent to $k < 1/(2 + \beta)$. Fix such $k > 0$ to obtain \[ \lim_{\epsilon \to 0} \frac{\epsilon | \log \epsilon |^{2k} }{\epsilon | \log \epsilon |^{1 - \beta k}} = 0. \] The rest of the argument is the same as (i). We complete the proof. \section{Appendix: Convergence of integrals with critical growth} \label{sec:convergence_critical} \begin{lemma} \label{lem:convergence_critical} Let $v, \psi \in H_0^1(\Omega)$. Let $\{ v_k \}_{k=0}^\infty$ be a bounded sequence in $H_0^1(\Omega)$. Assume that $v_k \to v$ a.e.\ in $\Omega$. Then, we have \begin{equation} \int_\Omega | x |^\alpha ( v_k )_+^{2^*-1} \psi dx \to \int_\Omega | x |^\alpha v_+^{2^*-1} \psi dx \label{eq:convergence_critical} \end{equation} as $k \to \infty$. \end{lemma} \begin{proof} Let $\epsilon > 0$. We set \[ W_{\epsilon, k} = \Big( | | x |^\alpha ( v_k )_+^{2^* - 1} \psi - | x |^\alpha v_+^{2^*-1} \psi | - \epsilon | x |^\alpha ( v_k )_+^{2^*} \Big)_+. \] By the Young inequality, there exists $C > 0$ such that \[ \big| | s |^{2^*-1}_+ t \big| \leq \epsilon s_+^{2^*} + C | t |^{2^*} \] for $s, t \in \mathbb{R}$. Thus we have \[ | W_{\epsilon, k} | \leq \epsilon | x |^\alpha v_+^{2^*} + 2 C | x |^\alpha | \psi |^{2^*} \leq \epsilon v_+^{2^*} + 2C | \psi |^{2^*}. \] The right side of above inequality is integrable. Since $v_k \to v$ a.e.\ in $\Omega$, it follows that $W_{\epsilon, k} \to 0$ a.e.\ in $\Omega$. Thus we have \[ \lim_{k \to \infty} \int_\Omega W_{\epsilon, k} dx = 0. \] By the definition of $W_{\epsilon, k}$, we have \begin{align*} \int_\Omega | | x |^\alpha ( v_k )_+^{2^* - 1} \psi - | x |^\alpha v_+^{2^*-1} \psi | dx & \leq \int_\Omega W_{\epsilon, k} dx + \epsilon \int_\Omega | x |^\alpha ( v_k )_+^{2^*} dx \\ & \leq \int_\Omega W_{\epsilon, k} dx + \epsilon \int_\Omega ( v_k )_+^{2^*} dx. \end{align*} Since $\{ v_k \}$ is a bounded sequence of $H_0^1(\Omega) \subset L^{2^*}(\Omega)$, we have $\int_\Omega ( v_k )_+^{2^*} dx \leq C$. Therefore, \[ \limsup_{k \to \infty} \int_\Omega | | x |^\alpha ( v_k )_+^{2^* - 1} \psi - | x |^\alpha v_+^{2^*-1} \psi | dx \leq C\epsilon. \] Since $\epsilon > 0$ is arbitrary, we obtain \eqref{eq:convergence_critical}. \end{proof} \section{Appendix: Proof of Corollary~\ref{cor:cor_of_main_theorem}} \label{sec:cor} We use notation of elementary geometries. Let $X, Y, Z$ be points of the Euclidean space $\mathbb{R}^N$. We write $\overline{XY}$ as the length of the segment $XY$, $\angle XYZ$ as the angle of $XYZ$ and $\triangle XYZ$ as the triangle of $XYZ$. Corollary~\ref{cor:cor_of_main_theorem} is a direct conclusion of Theorem~\ref{thm:main} and the following lemma. \begin{lemma} Let $x_0 \in \partial\Omega$ and $B \subset \Omega$ be an open ball whose radius is $0 < r_0 < 1/2$ and where $\partial B$ come in contact with $\partial \Omega$ at $x_0$. Let $\beta = 2 \beta_0$. Then, there exists $a > 0$ such that $\Psi(P) \geq \Psi_0(P)$ for any $P \in B$. \end{lemma} \begin{proof} Let $T$ to denote the point $x_0$. Let $O$ and $O'$ be the center of $\Omega$ and $B$, respectively. Let $P \in B$. If $P$ is on the segment $OT$, just taking $\beta \geq \beta_0$ and $0 < a < 1$ will do. Hereinafter we assume $P$ is not on the segment $OT$. We argue on the plane containing $O, O', T$ and $P$ (Figure~\ref{fig:A}). Let $Q$ and $R$ be the intersection point of the the half line $OP$ with $\partial B$ and $\partial \Omega$, respectively. Let $l = \overline{QT}$ and $k = \overline{QR}$. Let $\theta = \angle TO'Q$. Then, we see that $\overline{PT} > \overline{QT}$ since $\angle PQT$ is an obtuse angle. We can take a point $S$ on the segment $PT$ so that $\overline{ST} = \overline{QT}$. Let $x = \overline{PS}$ and $y = \overline{PQ}$. Let $\rho = \angle PQS$, $\sigma = \angle PSQ$ and $\tau = \angle QPS$ (Figure~~\ref{fig:B}). \begin{figure}[htb] \begin{center} \includegraphics[width=7cm]{fig1} % 201802_A.eps \end{center} \caption{The plane containing $O, O', T$ and $P$.} \label{fig:A} \end{figure} \begin{figure}[htb] \begin{center} \includegraphics[width=7cm]{fig2} % pictures/201802_B.eps \end{center} \caption{Focusing on $\triangle PQT$.} \label{fig:B} \end{figure} First, we prove that if we set $\beta = 2\beta_0$, there exists $a > 0$ such that $k^{\beta_0} > a l^{\beta}$ independently on $Q$. Considering $\triangle O'TQ$ and $\triangle OO'Q$, we have $l = 2 r_0\sin (\theta/2)$ and \[ (1-r_0)^2 + r_0^2 + 2r_0(1-r_0)\cos \theta = (1-k)^2, \] respectively. By the formula $\cos \theta = 1 - 2 \sin^2 (\theta/2)$, we have \[ l^2 = \frac{r_0}{1-r_0} ( 1 - (1-k)^2 ). \] Therefore \begin{align*} k^{2\beta_0} - a^2 l^{2\beta} & = k^{2\beta_0} - a^2 \Big( \frac{r_0}{1-r_0} \Big)^\beta k^\beta (2-k)^\beta \\ & > k^{2\beta_0} - a^2 2^\beta \Big( \frac{r_0}{1-r_0} \Big)^\beta k^\beta. \end{align*} We set $\beta = 2 \beta_0$ and take $a > 0$ so small that \[ 1 - a^2 2^{2\beta_0} \Big( \frac{r_0}{1-r_0} \Big)^{2\beta_0} > 0. \] Then, we have $k^{\beta_0} > a l^{\beta}$ independently on $Q$ as desired. Next, we prove that $x < y$. Since $\angle SQT = \angle QST = \rho + \tau$, by $\triangle PQT$, we have $2\rho + 2\tau < \pi$. Combining this with $\rho + \sigma + \tau = \pi$, we have $\sigma > \pi/2 > \rho$. Thus we have $x < y$. Finally, we prove that there exists $a > 0$ such that $(y+k)^{\beta_0} > a (x+l)^{2\beta_0}$ independently on $P$. Since $k^{\beta_0} > a l^{2\beta_0}$ and $x < y$, it follows that \begin{align*} (y+k)^{\beta_0} - a (x+l)^{2\beta_0} & = (y+k)^{\beta_0} - \big(a^{1/2\beta_0}x + a^{1/2\beta_0}l \big)^{2\beta_0} \\ & > (y+k)^{\beta_0} - \big(a^{1/2\beta_0}y + \sqrt{k} \big)^{2\beta_0}. \end{align*} Observing $0 < k < 1$ and $0 < y < 2r_0$, we have \begin{align*} (y + k) - \big(a^{1/2\beta_0}y + \sqrt{k} \big)^{2} & = y \big(1 - 2 a^{1/2\beta_0} \sqrt{k} - a^{1/\beta_0}y \big) \\ & > y (1 - 2 a^{1/2\beta_0} - 2r_0 a^{1/\beta_0}). \end{align*} If we need, we can again take $a > 0$ so small that the right side above is positive. Therefore we have $(y+k)^{\beta_0} > a (x+l)^{2\beta_0}$ independently on $P$, which completes the proof. \end{proof} \subsection*{Acknowledgements} The author would like to thank Prof.~Yasuhito Miyamoto for his supports for the author's research. 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