Lemma 2.1 Assume that (6), (7) and (9) hold. Let $u_{\epsilon }$ be a critical point of the functional $I_{\epsilon }$ defined by (5). Then $u_{\epsilon }\geq 0$ and $u_{\epsilon }$ is a weak solution of problem (2).

Proof of Lemma 2.1 Let $u_{\epsilon }^{-}=\max \{-u_{\epsilon },\,0\}$. Taking $v=u_{\epsilon }^{-}$ in (12) and using (7), we obtain

\[ 0=I_{\epsilon}'(u_{\epsilon})(u_{\epsilon}^{-})={-}\|u_{\epsilon}^{-}\|^{2}_{H_{0}^{1}(\Omega)}. \]

Hence, $u_{\epsilon }\geq 0$ and

\[ 0=I_{\epsilon}'(u_{\epsilon})(v)=\int_{\Omega}\nabla u_{\epsilon}\nabla v+\int_{\Omega}g_{\epsilon}(u_{\epsilon})v-\int_{\Omega}f(u_{\epsilon})v, \text{ for all }v\in H_{0}^{1}(\Omega). \]

Hence, (4) holds. This proves Lemma 2.1.

Lemma 2.2

\[ G_{\epsilon}(s)\geq\frac{1}{2}g_{\epsilon}(s)s,\text{ for every }s\geq 0. \]

Proof of Lemma 2.2 Indeed, let $\widetilde {B}_{\epsilon }(s)=G_{\epsilon }(s)-\frac {1}{2}g_{\epsilon }(s)s$. We have that $B_{\epsilon }(0)=0$ and

\[ \widetilde{B}_{\epsilon}'(s)=g_{\epsilon}(s)-\frac{1}{2}g_{\epsilon}(s)-\frac{s}{2}g_{\epsilon}'(s)=\frac{1}{2}\left(g_{\epsilon}(s)-sg_{\epsilon}'(s)\right). \]

Therefore, $\widetilde {B}_{\epsilon }'(s)\geq 0$ if and only if

\[ g_{\epsilon}(s)\geq sg_{\epsilon}'(s). \]

From (19), this inequality will be true if

(20)\begin{equation} \displaystyle\frac{s^{q}}{(s+\epsilon)^{q+\beta}}\geq \frac{q s^{q}}{(s+\epsilon)^{q+\beta}}. \end{equation}

Since $q<1/2$, (20) holds for each $s\geq 0$. This proves Lemma 2.2.

Lemma 2.3 Suppose that (6) and (10) hold. Let $0<\theta <1/2$ be given by (10). There exists a constant $\overline {R}>0$ such that

\[ J(s)\leq\theta s j(s)\text{ for }s\geq \overline{R}. \]

Proof of Lemma 2.3 Let $B_{\epsilon }(s)=J(s)-\theta s j(s)$. We have

\[ B_{\epsilon}'(s)=(1-\theta)j(s)-\theta sj'(s). \]

Hence,

\[ B_{\epsilon}'(s)={-}(1-\theta)g_{\epsilon}(s)+\theta sg_{\epsilon}'(s)+(1-\theta)f(s)-\theta s f'(s). \]

From (19) we obtain

\[ |sg_{\epsilon}'(s)|\leq q|s|^{-\beta}+(q+\beta)|s|^{-\beta}\to 0\text{ as }s\to\infty. \]

It is also clear that

\[ (1-\theta)g_{\epsilon}(s)\to 0\text{ as }s\to\infty\text{ uniformly for }\epsilon. \]

Hence, for each $0<\tau <1$ there exists $R_{\tau }>0$ that does not depend on $\epsilon$ such that

\[ |(1-\theta)g_{\epsilon}(s)|+|sg_{\epsilon}'(s)|<\tau\text{ for }s\geq R_{\tau}. \]

Therefore,

\[ B_{\epsilon}'(s)<\tau+(1-\theta)f(s)-\theta s f'(s).\text{ for }s\geq R_{\tau}. \]

Consequently, from (10), we get

\[ B_{\epsilon}'(s)<\tau-c\text{ for }s\geq \max\{R,R_{\tau}\}, \]

where $c>0$ and $R>0$ are given by (10). Choosing $\tau =c/2$, we get

(21)\begin{equation} B_{\epsilon}'(s)<{-}\displaystyle\frac{c}{2}\text{ for }s\geq R_{2}, \end{equation}

where

\[ R_{2}=\max\{R,R_{c/2}\}. \]

Note that

\[ B_{\epsilon}(R_{2})\leq C_{1} \]

where

\[ C_{1}=|F(R_{2})|+|\theta R_{2}f(R_{2})|+\theta R_{2}^{1-\beta}. \]

Therefore, (21) implies that there exists a constant $T>0$ such that

\[ B_{\epsilon}(s)\leq{-}\frac{cs}{2}+T\text{ for }s\geq R_{2}. \]

Hence, $B_{\epsilon }(s)\leq 0$ for $s\geq \max \{R_{2},\,2T/c\}$. This proves Lemma 2.3.

Lemma 2.4 Assume that (6), (9) and (11) hold. There exist a constant $a_{2}>0$ that does not depend on $\epsilon$ such that

(22)\begin{equation} \sup_{0\leq s\leq 1}I_{\epsilon}(s\phi_{0})< a_{2}\text{ for every }0<\epsilon<\epsilon_{0}. \end{equation}

Proof of Lemma 2.4 We have

\[ I_{\epsilon}(s\phi_{0})\leq\frac{s^{2}}{2}\|\phi_{0}\|_{H_{0}^{1}(\Omega)}+\int_{\Omega} G_{\epsilon}(s\phi_{0})-\int_{\Omega} F(s\phi_{0})\,{\rm d}x,\text{ for every }s\geq 0. \]

Consequently, we get

\[ I_{\epsilon}(s\phi_{0})\leq\frac{s^{2}}{2}\|\phi_{0}\|_{H_{0}^{1}(\Omega)}+\frac{1}{1-\beta}\int_{\Omega}|s\phi_{0}|^{1-\beta}-\int_{\Omega} F(s\phi_{0})\,{\rm d}x,\text{ for every }s\geq 0. \]

We conclude that

\[ \sup_{0\leq s\leq 1}I_{\epsilon,\lambda}(s\phi_{0})< a_{2}, \]

where

\[ a_{2}=\frac{1}{2}\|\phi_{0}\|_{H_{0}^{1}(\Omega)}+|\Omega|\left(\frac{\sup_{\Omega}|\phi_{0}|^{1-\beta}}{1-\beta}+\sup_{0\leq s\leq\sup\phi_{0}}|F(s)|\right). \]

This proves (22). We have proved Lemma 2.4.

Lemma 3.1 Assume that (6)–(10) hold. Fix $0<\epsilon <1$ and let $(u_{n}^{\epsilon })$ be a Palais–Smale sequence for $I_{\epsilon }$ in $H_{0}^{1}(\Omega )$. Assume that there exists a constant $C>0$ that does not depend on $\epsilon$ such that

(23)\begin{equation} |I_{\epsilon}(u_{n}^{\epsilon})|< C\text{ for all }n\in\mathbb{N}. \end{equation}

Then, there exists $D>0$ that does not depend on $\epsilon$ such that

(24)\begin{equation} \|u_{n}^{\epsilon}\|_{H_{0}^{1}(\Omega)}< D\text{ for all }n\in\mathbb{N}. \end{equation}

Furthermore, there exists $u_{\epsilon }\in H_{0}^{1}(\Omega )$ such that, up to a subsequence, $u_{n}^{\epsilon }\to u_{\epsilon }$ strongly in $H_{0}^{1}(\Omega )$. Consequently, $u_{\epsilon }$ is a critical point of $I_{\epsilon }$.

Proof of Lemma 3.1 Throughout this proof, we denote $\|\cdot \|_{H_{0}^{1}(\Omega )}$ by $\|\cdot \|$. Let $(u_{n}^{\epsilon })_{n\in \mathbb {N}}$ be a Palais–Smale sequence for $I_{\epsilon }$ satisfying (23). Consequently,

(25)\begin{equation} \displaystyle\frac{1}{2}\|u_{n}^{\epsilon}\|^{2}\leq C+\displaystyle\int_{\Omega}J(u_{n}^{\epsilon})\,{\rm d}x\text{ for all }n\in\mathbb{N}, \end{equation}

and there is a sequence $\tau _{n}\rightarrow 0$ such that

(26)\begin{equation} \biggl|\displaystyle\int_{\Omega}\nabla u_{n}^{\epsilon}\nabla w\,{\rm d}x-\int_{\Omega}j(u_{n}^{\epsilon})w\,{\rm d}x\biggl|\leq \tau_{n}\|w\| \text{ for each } w\in H_{0}^{1}(\Omega). \end{equation}

Let $0<\theta <1/2$ be given by (10). From Lemma 2.3, there is a constant $\overline {R}>0$ that does not depend on $\epsilon$ such that

\[ J(t)\leq\theta t j(t) \text{ for } t\geq \overline{R}. \]

Since there exists $D_{1}>0$ that does not depend on $\epsilon$ such that

\[ \sup_{0\leq s\leq \overline{R}}\max\{|J(s)|,|sj(s)|\}< D_{1}, \]

we may find a constant $D_{2}>0$ such that

\[ J(u_{n}^{\epsilon})< D_{2}+\theta u_{n}^{\epsilon}j(u_{n}^{\epsilon}). \]

We know from (25) that there is a constant $D_{3}>0$ such that

\[ \frac{1}{2}\|u_{n}^{\epsilon}\|^{2}\leq D_{3}+\theta\int_{\Omega} u_{n}^{\epsilon}j(u_{n}^{\epsilon})\,{\rm d}x. \]

Taking $w=u_{n}^{\epsilon }$ in (26), we also conclude that

\[ \int_{\Omega}j(u_{n}^{\epsilon})u_{n}^{\epsilon}\,{\rm d}x<\|u_{n}^{\epsilon}\|^{2}+\tau_{n}\|u_{n}^{\epsilon}\|. \]

Hence,

\[ \frac{1}{2}\|u_{n}^{\epsilon}\|^{2}< D_{3}+\theta\|u_{n}^{\epsilon}\|^{2}+\tau_{n}\theta\|u_{n}^{\epsilon}\|. \]

Since $0<\theta <\frac {1}{2}$, (24) follows. Consequently, there exists $u_{\epsilon }\in H_{0}^{1}(\Omega )$ such that $u_{n}^{\epsilon }\rightharpoonup u_{\epsilon }$ weakly in $H_{0}^{1}(\Omega )$. Since $J_{\epsilon }$ has subcritical growth at infinity (see [Reference Costa7], Theorem 3.4 and Remark 2.2.1), we conclude that, up to a subsequence, $u_{n}^{\epsilon }\to u_{\epsilon }$ strongly in $H_{0}^{1}(\Omega )$. Since $I_{\epsilon }'(u_{n}^{\epsilon })\to 0$ as $n\to \infty$ and $I_{\epsilon }$ is of class $C^{1}$, we conclude that $I_{\epsilon }'(u_{\epsilon })=0$. This proves the result.

Lemma 4.1 There exist constants $N_{0}>0,$ $a_{2}>0$ and $b_{1}>0$ such that

(30)\begin{equation} I_{\epsilon,\lambda}( N_{0}\phi_{1})<{-}b_{1}<0,\text{ for every }\,0<\epsilon<1, \end{equation}

and

(31)\begin{equation} \sup_{0\leq s\leq 1}I_{\epsilon,\lambda}(sN_{0}\phi_{1})< a_{2}\text{ for every }\lambda>0,\, 0<\epsilon<1. \end{equation}

Moreover, these constants do not depend on $\lambda$.

Proof of Lemma 2.4 For each $t>0$, we have

\[ I_{\epsilon,\lambda}(t\phi_{1})=\frac{t^{2}}{2}+\int_{\Omega} G_{\epsilon}(t\phi_{1})-\frac{\lambda t^{2}}{2}\int_{\Omega}\phi_{1}^{2}\,{\rm d}x-\frac{t^{2^{*}}}{2^{*}}\int \phi_{1}^{2^{*}}\,{\rm d}x. \]

From the fact that $G_{\epsilon }(s)\leq \frac {s^{1-\beta }}{1-\beta }$ for all $s\geq 0$, we get

(32)\begin{equation} I_{\epsilon,\lambda}(t\phi_{1})\leq\displaystyle\frac{t^{2}}{2}+\frac{t^{1-\beta}}{1-\beta}\displaystyle\int_{\Omega} \phi_{1}^{1-\beta}-\frac{t^{2^{*}}}{2^{*}}\int \phi_{1}^{2^{*}}\,{\rm d}x. \end{equation}

Since $2^{*}>2>1-\beta$, inequality (30) then follows by taking $t$ large enough in (32). We also have

\[ I_{\epsilon,\lambda}(sN_{0}\phi_{1})\leq\frac{s^{2}N_{0}^{2}}{2}+\int_{\Omega} G_{\epsilon}(s N_{0}\phi_{1})-\frac{s^{2^{*}}N_{0}^{2^{*}}}{2^{*}}\int \phi_{1}^{2^{*}}\,{\rm d}x,\text{ for every }s\geq 0. \]

Consequently, we get

\[ I_{\epsilon,\lambda}(sN_{0}\phi_{1})\leq \frac{s^{2} N_{0}^{2}}{2}+\frac{s^{1-\beta}N_{0}^{1-\beta}}{1-\beta}\int_{\Omega} \phi_{1}^{1-\beta},\text{ for every }s\geq 0. \]

We conclude that

\[ \sup_{0\leq s\leq 1}I_{\epsilon,\lambda}(sN_{0}\phi_{1})< a_{2}, \]

where

\[ a_{2}=\frac{N_{0}^{2}}{2}+\frac{N_{0}^{1-\beta}}{1-\beta}\int_{\Omega} \phi_{1}^{1-\beta}. \]

This proves (31). We have proved Lemma 4.1.

Lemma 4.2 Let $c_{\lambda,\epsilon }$ be given by (33). Then

(36)\begin{equation} \lim_{\lambda\to\infty}c_{\lambda,\epsilon}=0\text{ uniformly on }0<\epsilon<1. \end{equation}

Proof of Lemma 4.2 Fix $0<\epsilon <1$ and let $t_{\lambda,\epsilon }\geq 0$ be such that

\[ I_{\epsilon,\lambda}(t_{\lambda,\epsilon}\phi_{1})=\max_{0\leq t\leq 1}I_{\epsilon,\lambda}(t N_{0}\phi_{1}). \]

From Lemmas 2.4 and (29), we get

\[ I_{\epsilon,\lambda}(t\phi_{1})>0\text{ for }0< t<\rho\quad\text{ and }I_{\epsilon,\lambda}(N_{0}\phi_{1})<0. \]

Hence, $0< t_{\lambda,\epsilon }<1$. Consequently,

\[ \frac{d}{{\rm d}t}I_{\epsilon,\lambda}(t N_{0}\phi_{1})\biggl|_{t=t_{\lambda,\epsilon}}=0. \]

Equivalently, from (28),

(37)\begin{equation} \begin{aligned} 0=I_{\epsilon,\lambda}'(t_{\lambda,\epsilon}N_{0}\phi_{1})(N_{0}\phi_{1}) & =N_{0}^{2}t_{\lambda,\epsilon}\displaystyle\int_{\Omega}|\nabla\phi_{1}|^{2}+N_{0}^{1+q}t_{\lambda,\epsilon}^{q}\int_{\Omega}\displaystyle\frac{\phi_{1}^{1+q}}{(N_{0}t_{\lambda,\epsilon}\phi_{1}+\epsilon)^{q+\beta}}\\ & \quad -N_{0}^{2}\lambda t_{\lambda,\epsilon}\displaystyle\int_{\Omega}\phi_{1}^{2}-N_{0}^{2^{*}}t_{\lambda,\epsilon}^{2^{*}-1}\int_{\Omega}\phi_{1}^{2^{*}}. \end{aligned} \end{equation}

Fix a sequence $(\lambda _{n})$ in $\mathbb {R}$ such that $\lambda _{n}\to \infty$. Since $0< t_{\lambda _{n},\,\epsilon }<1$, we know that for each $0<\epsilon <1$ there exists an element $0\leq t_{0,\epsilon }\leq 1$ such that

\[ t_{\lambda_{n},\epsilon}\to t_{0,\epsilon}\text{ as }n\to\infty. \]

We will show that $t_{0,\epsilon }=0$. Indeed, from (37) there exists a constant $M_{0}>0$ that does not depend on $\lambda$ nor on $\epsilon$ such that

(38)\begin{equation} \lambda_{n}t^{2}_{\lambda_{n},\epsilon}\displaystyle\int_{\Omega}\phi_{1}^{2}\leq t_{\lambda_{n},\epsilon}^{2}\int_{\Omega}|\nabla\phi_{1}|^{2}+t_{\lambda_{n},\epsilon}^{1-\beta}\int_{\Omega}\phi_{1}^{1-\beta}\leq M_{0}. \end{equation}

Letting $n\to \infty$ in (38), it follows that $t_{\lambda _{n},\,\epsilon }\to 0$ as $n\to \infty$ uniformly on $\epsilon$. Hence, $t_{0,\epsilon }=0$. Consequently,

\[ 0< c_{\lambda_{n},\epsilon}\leq\max_{0\leq t\leq 1}I_{\epsilon,\lambda_{n}}(tN_{0}\phi_{1})=I_{\epsilon,\lambda_{n}}(t_{\lambda_{n},\epsilon}\phi_{1})\leq t_{\lambda_{n},\epsilon}^{2}\int_{\Omega}|\nabla\phi_{1}|^{2}\,{\rm d}x+t_{\lambda_{n},\epsilon}^{1-\beta}\int_{\Omega}\phi_{1}^{1-\beta}\,{\rm d}x. \]

Letting $n\to \infty$, we obtain

\[ \lim_{n\to\infty}c_{\lambda_{n},\epsilon}=0\text{ uniformly on }0<\epsilon<1. \]

Since the sequence $(\lambda _{n})$ was arbitrarily chosen, (36) follows.

Lemma 5.1 Let $u_{\epsilon,\lambda }\in H_{0}^{1}(\Omega )$ be a non-negative solution of problem (2) with $f(s)=\lambda s+s^{p}$ and assume that there exists a constant $D>0$ independent of $\epsilon$ such that

(46)\begin{equation} \|u_{\epsilon,\lambda}\|_{H_{0}^{1}(\Omega)}\leq D\text{ for each }0<\epsilon<1. \end{equation}

Then the following assertions hold

${\rm (i)}$ If $1< p<2^{*}-1$ then $u_{\epsilon,\lambda }\in L^{\infty }(\Omega )$ and there exists a constant $K_{1}>0$ such that

(47)\begin{equation} \|u_{\epsilon,\lambda}\|_{L^{\infty}(\Omega)}\leq K_{1}\text{ for each }0<\epsilon<1. \end{equation}

${\rm (ii)}$ If $p=2^{*}-1$ and

(48)\begin{equation} \lim_{\lambda\to\infty}I_{\epsilon,\lambda}(u_{\epsilon,\lambda})=0\text{ uniformly on }\epsilon, \end{equation}

then there exists $\widehat {\lambda _{0}}>0$ such that $u_{\epsilon,\lambda }\in L^{\infty }(\Omega )$ for each $\lambda >\widehat {\lambda _{0}}$ and (47) holds.

Proof of Lemma 5.1 For simplicity, we denote $u_{\epsilon,\lambda }$ by $u_{\epsilon }$. For $s\geq 0$, define $h(s)=\lambda s+s^{p}$. From (4), we get

(49)\begin{equation} \displaystyle\int_{\Omega}\nabla u_{\epsilon}\nabla v+\int_{\Omega}g_{\epsilon}(u_{\epsilon})v=\int_{\Omega}h(u_{\epsilon})v\text{ for all }v\in H_{0}^{1}(\Omega). \end{equation}

Note that

\[ \frac{h(s)}{g_{\epsilon}(s)}=\frac{(s+\epsilon)^{q+\beta}}{s^{q}}(\lambda s+s^{p})\leq (s+1)^{q+\beta}(\lambda s^{1-q}+s^{p-q})\to 0\text{ as }s\to 0. \]

Hence, the exists $0<\delta _{\lambda }<1$ that does not depend on $\epsilon$ such that

(50)\begin{equation} \displaystyle\frac{h(s)}{g_{\epsilon}(s)}<\frac{1}{2}\text{ for }s\leq\delta_{\lambda}. \end{equation}

Also,

\[ \frac{h(s)}{s^{p}}=\lambda s^{1-p}+1\text{ for }s\geq\delta_{\lambda}. \]

Therefore, we conclude that

(51)\begin{equation} \displaystyle\frac{h(s)}{s^{p}}\leq 2\text{ for }s\geq A_{\lambda}, \end{equation}

where

\[ A_{\lambda}=\lambda^{\frac{1}{p-1}}. \]

It is also clear that

(52)\begin{equation} h(s)=\lambda s+s^{p}\leq\lambda A_{\lambda}+A_{\lambda}^{p}=C_{\lambda}\text{ for }\delta_{\lambda}\leq s\leq A_{\lambda}. \end{equation}

Using (50), (51) and (52), we obtain

\[ h(s)<\left(\frac{g_{\epsilon}(s)}{2}\right)\chi_{\{0\leq s\leq\delta_{\lambda}\}}+C_{\lambda}\chi_{\{\delta_{\lambda}\leq s\leq A_{\lambda}\}}+2s^{p}\chi_{\{s\geq A_{\lambda}\}}\text{ for }s\geq 0 \]

Hence, from (70), we get

(53)\begin{equation} \displaystyle\int_{\Omega}\nabla u_{\epsilon}\nabla v< C_{\lambda}\int_{\{\delta_{\lambda}\leq u_{\epsilon}\leq A_{\lambda}\}}v+2\int_{\Omega}u_{\epsilon}^{p}v\text{ for all }v\in H_{0}^{1}(\Omega),v\geq 0. \end{equation}

We will now prove assertions $(i)$ and $(ii)$ separately. The proof of $(ii)$ is more intricate, because we need to study the dependence of certain constants on $\lambda$, so that we can let $\lambda \to \infty$.

Proof of (i): From (53), we obtain a constant $C_{\delta,\lambda }>0$ such that

(54)\begin{equation} \displaystyle\int_{\Omega}\nabla u_{\epsilon}\nabla v< C_{\delta,\lambda}\int_{\Omega}u_{\epsilon}^{p}v\text{ for all }v\in H_{0}^{1}(\Omega),v\geq 0. \end{equation}

For $L > 1$, we define,

\begin{align*} u_{L,\epsilon}(x)& =\left\{ \begin{array}{@{}l} u_{\epsilon}(x), \text{if } u_{\epsilon}(x) \leq L \\ L, \text{if } u_{\epsilon}(x) \geq L, \\ \end{array} \right.\\ z_{L,\epsilon} & = u_{L,\epsilon}^{2(\sigma - 1)}u_{\epsilon} \quad \mbox{and} \quad w_{L,\epsilon} = u_{\epsilon} u_{L,\epsilon}^{\sigma - 1}, \end{align*}

with $\sigma > 1$ to be determined later. Note that $z_{L,\epsilon }\in H_{0}^{1}(\Omega )$, $z_{L,\epsilon }\geq 0$ and

\[ \nabla z_{L,\epsilon}=u_{L,\epsilon}^{2(\sigma-1)}\nabla u_{\epsilon}+2(\sigma-1)u_{\epsilon}u_{L,\epsilon}^{2\sigma-3}\nabla u_{L,\epsilon}. \]

Taking $v=z_{L,\epsilon }$ in (54) we obtain

\[ \int_{\Omega}u_{L,\epsilon}^{2(\sigma-1)}|\nabla u_{\epsilon}|^{2}+2(\sigma-1)\int_{\Omega}u_{\epsilon}u_{L,\epsilon}^{2\sigma-3}\nabla u_{\epsilon}\nabla u_{L,\epsilon}< C_{\lambda,\delta}\int_{\Omega}u_{\epsilon}^{p+1}u_{L,\epsilon}^{2(\sigma-1)}. \]

Since $\sigma >1$ and

\[ \int_{\Omega}u_{\epsilon}u_{L,\epsilon}^{2\sigma-3}\nabla u_{\epsilon}\nabla u_{L,\epsilon}=\int_{\{u_{\epsilon}< L\}}u_{\epsilon}^{2(\sigma-1)}|\nabla u_{\epsilon}|^{2}\geq 0, \]

we conclude that

(55)\begin{equation} \displaystyle\int_{\Omega}u_{L,\epsilon}^{2(\sigma-1)}|\nabla u_{\epsilon}|^{2}< C_{\lambda,\delta}\int_{\Omega}u_{\epsilon}^{p+1}u_{L,\epsilon}^{2(\sigma-1)}< C_{\lambda,\delta}\int_{\Omega}u_{\epsilon}^{p-1}u_{\epsilon}^{2\sigma}. \end{equation}

On the other hand, from the Sobolev embedding, we know that there is a constant $C_{1}>0$ such that

\[ \bigg(\int_{\Omega}w_{L,\epsilon}^{p+1}\,{\rm d}x\bigg)^{\frac{2}{p+1}}\leq C_{1} \int_{\Omega}|\nabla w_{L,\epsilon}|^{2}\,{\rm d}x. \]

Since

\[ \nabla w_{L,\epsilon}=u_{L,\epsilon}^{\sigma-1}\nabla u_{\epsilon}+(\sigma-1)u_{\epsilon}u_{L,\epsilon}^{\sigma-2}\nabla u_{L,\epsilon}, \]

it follows that

\begin{align*} & \left(\displaystyle\int_{\Omega}w_{L,\epsilon}^{p+1}\,{\rm d}x\right)^{\frac{2}{p+1}}\leq C_{1}\int_{\Omega}u_{L,\epsilon}^{2(\sigma-1)}|\nabla u_{\epsilon}|^{2}\,{\rm d}x+ C_{1}(\sigma-1)^{2}\int_{\Omega}u_{\epsilon}^{2}u_{L,\epsilon}^{2(\sigma-2)}|\nabla u_{L,\epsilon}|^{2}\\ & \quad+2C_{1}(\sigma-1)\displaystyle\int_{\Omega}u_{\epsilon}u_{L,\epsilon}^{2\sigma-3}\nabla u_{\epsilon}\nabla u_{L,\epsilon}. \end{align*}

From the definition of $u_{L,\epsilon }$, we conclude that

\[ \bigg(\int_{\Omega}w_{L,\epsilon}^{p+1}\,{\rm d}x\bigg)^{\frac{2}{p+1}}\leq C_{1}\sigma^{2}\int_{\Omega}u_{L,\epsilon}^{2(\sigma-1)}|\nabla u_{\epsilon}|^{2}\,{\rm d}x. \]

Using (55), we obtain

(56)\begin{equation} \left(\displaystyle\int_{\Omega}w_{L,\epsilon}^{p+1}\,{\rm d}x\right)^{\frac{2}{p+1}}\leq C_{1}\sigma^{2} C_{\delta,\lambda}\int_{\Omega}u_{\epsilon}^{p-1}u_{\epsilon}^{2\sigma}. \end{equation}

Now, observe that

\[ \left(\int_{\Omega}w_{L,\epsilon}^{p+1}\,{\rm d}x\right)^{\frac{2}{p+1}}=\left(\int_{\Omega}u_{\epsilon}^{p+1}u_{L,\epsilon}^{(p+1)(\sigma-1)}\,{\rm d}x\right)^{\frac{2}{p+1}}\geq\left(\int_{\Omega}u_{L,\epsilon}^{\sigma(p+1)}\,{\rm d}x\right)^{\frac{2}{p+1}}. \]

Hence, there is a constant $\widetilde {C_{\delta,\lambda }}>0$ such that

(57)\begin{equation} \left(\displaystyle\int_{\Omega}u_{L,\epsilon}^{\sigma(p+1)}\,{\rm d}x\right)^{\frac{2}{p+1}}\leq \sigma^{2}\widetilde{C_{\delta,\lambda}}\int_{\Omega}u_{\epsilon}^{p-1}u_{\epsilon}^{2\sigma}. \end{equation}

Let $\alpha _{1},\,\alpha _{2}>1$ be constants such that $\frac {1}{\alpha _{1}}+\frac {1}{\alpha _{2}}=1$ and $p+1<\alpha _{1}(p-1)<2^{*}$. From (57) and Hölder's inequality it follows that

\[ \left(\int_{\Omega}u_{L,\epsilon}^{\sigma(p+1)}\,{\rm d}x\right)^{\frac{2}{p+1}}\leq \sigma^{2} \widetilde{C_{\delta,\lambda}}\left(\int_{\Omega}u_{\epsilon}^{\alpha_{1}(p-1)}\,{\rm d}x\right)^{\frac{1}{\alpha_{1}}}\left(\int_{\Omega}u_{\epsilon}^{2\sigma\alpha_{2}}\,{\rm d}x\right)^{\frac{1}{\alpha_{2}}}. \]

Using (46) and the Sobolev Embedding, we obtain a constant $\widetilde {C}>0$ such that

\[ \int_{\Omega}u_{\epsilon}^{\alpha_{1}(p-1)}\,{\rm d}x\leq \widetilde{C}. \]

Hence, there exists a constant $\widehat {C}>0$ that does not depend on $\sigma$ nor on $\epsilon$ such that

(58)\begin{equation} \left(\displaystyle\int_{\Omega}u_{L,\epsilon}^{\sigma(p+1)}\,{\rm d}x\right)^{\frac{2}{p+1}}\leq \widehat{C}\sigma^{2}\left(\int_{\Omega}u_{\epsilon}^{2\sigma\alpha_{2}}\,{\rm d}x\right)^{\frac{1}{\alpha_{2}}}. \end{equation}

Letting $L\to \infty$ in (58) and using Fatou's Lemma, we conclude that

\[ \left(\int_{\Omega}u_{\epsilon}^{\sigma(p+1)}\,{\rm d}x\right)^{\frac{2}{p+1}}\leq \widehat{C}\sigma^{2}\left(\int_{\Omega}u_{\epsilon}^{2\sigma\alpha_{2}}\,{\rm d}x\right)^{\frac{1}{\alpha_{2}}}\text{ for each }\sigma>1, \]

provided $u_{\epsilon }\in L^{2\sigma \alpha _{2}}(\Omega )$. Equivalently,

(59)\begin{equation} \|u_{\epsilon}\|_{L^{\sigma(p+1)}}\leq C^{\frac{1}{\sigma}}\sigma^{\frac{1}{\sigma}}\|u_{\epsilon}\|_{L^{2\sigma\alpha_{2}}}\text{ for each }\sigma>1, \end{equation}

where $C=\sqrt {\widehat {C}}$. Observe that the choices of $\alpha _{1}$ and $\alpha _{2}$ imply that $\sigma (p+1)>2\sigma \alpha _{2}$. The result now follows from an iterative argument. Indeed, take

\[ \sigma_{1}=\frac{p+1}{2\alpha_{2}}. \]

Using the Sobolev embedding and (46), we obtain a constant $\widetilde {D}>0$ such that

\[ \|u_{\epsilon}\|_{L^{\sigma_{1}(p+1)}(\Omega)}\leq C^{\frac{1}{\sigma_{1}}}\sigma_{1}^{\frac{1}{\sigma_{1}}}\|u_{\epsilon}\|_{L^{p+1}(\Omega)}\leq \widetilde{D}C^{\frac{1}{\sigma_{1}}}\sigma_{1}^{\frac{1}{\sigma_{1}}} \]

Now, take $\sigma _{2}=\sigma _{1}^{2}$ in (59). We get

\[ \|u_{\epsilon}\|_{L^{\sigma_{1}^{2}(p+1)}(\Omega)}\leq C^{\frac{1}{\sigma_{1}^{2}}}\sigma_{2}^{\frac{1}{\sigma_{2}}}\|u_{\epsilon}\|_{L^{\sigma_{1}(p+1)}}\leq\widetilde{D} C^{\frac{1}{\sigma_{1}}+\frac{1}{\sigma_{1}^{2}}}\left(\sigma_{1}^{\frac{1}{\sigma_{1}}}\sigma_{2}^{\frac{1}{\sigma_{2}}}\right) \]

Taking $\sigma _{k}=\sigma _{1}^{k}$ in (59), we get

(60)\begin{equation} \|u_{\epsilon}\|_{L^{\sigma_{1}^{k}(p+1)}(\Omega)}\leq \widetilde{D} C^{\sum_{i=1}^{k}\frac{1}{\sigma_{1}^{i}}}(\Pi_{i=1}^{k}\sigma_{i}^{\frac{1}{\sigma_{i}}}). \end{equation}

It is clear that

\[ \lim_{k\rightarrow\infty}\left(\Pi_{i=1}^{k}\sigma_{i}^{\frac{1}{\sigma_{i}}}\right)=\lim_{k\rightarrow\infty}\left(\Pi_{i=1}^{k}\sigma_{1}^{\frac{i}{\sigma_{1}^{i}}}\right)<\infty\text{ and } \lim_{k\rightarrow\infty}C^{\sum_{i=1}^{k}\frac{1}{\sigma_{1}^{i}}}<\infty. \]

Letting $k\rightarrow \infty$ in (60), it follows that $u_{\epsilon }\in L^{\infty }(\Omega )$ and we obtain a constant $K_{1}>0$ that does not depend on $\epsilon$ such that

\[ \|u_{\epsilon}\|_{L^{\infty}(\Omega)}\leq K_{1}. \]

This proves (47).

Proof of (ii). Suppose that $p=2^{*}-1$. This case is much more complicated. We have

\[ \int_{\Omega}\nabla u_{\epsilon}\nabla v\,{\rm d}x+\int_{\Omega}g_{\epsilon}(u_{\epsilon})v\,{\rm d}x=\int_{\Omega}(\lambda u_{\epsilon}+u_{\epsilon}^{2^{*}-1})v\,{\rm d}x\text{ for all }v\in H_{0}^{1}(\Omega). \]

Consequently, since $g_{\epsilon }\geq 0$, we know that

(61)\begin{equation} \displaystyle\int_{\Omega}\nabla u_{\epsilon}\nabla v\leq\int_{\Omega}u_{\epsilon}(\lambda+u_{\epsilon}^{2^{*}-2})v\text{ for all } v\in H_{0}^{1}(\Omega),v\geq 0, 0<\epsilon<1. \end{equation}

For each $0<\epsilon <1$, we define

\[ a_{\epsilon}(x)=\lambda+u_{\epsilon}(x)^{2^{*}-2}. \]

Observe that

\[ \int_{\Omega}a_{\epsilon}(x)^{N/2}\leq C(N)\left(\lambda^{N/2}|\Omega|+\int_{\Omega}\left(u_{\epsilon}^{\frac{4}{N-2}}\right)^{N/2}\right)=C(N)(\lambda^{N/2}|\Omega|+\|u_{\epsilon}\|_{L^{2^{*}}(\Omega)}^{2^{*}}), \]

where $C(N)$ is a constant that depends only on $N$. From (46) and the Sobolev embedding, we get a constant $C>0$ such that

\[ \|a_{\epsilon}\|_{L^{N/2}(\Omega)}\leq C\text{ for all }0<\epsilon<1. \]

Let $\sigma \geq 0$ be a constant to be fixed later and consider the function $z_{L,\epsilon }=u_{\epsilon }\min \{u_{\epsilon }^{2\sigma },\, L^{2}\}\in H_{0}^{1}(\Omega )$, with $L>0$. Observe that

\[ \nabla u_{\epsilon}\nabla z_{L,\epsilon}=|\nabla u_{\epsilon}|^{2}\min\{u_{\epsilon}^{2\sigma},L^{2}\}+2\sigma u_{\epsilon}^{2\sigma}|\nabla u_{\epsilon}|^{2}\chi_{\{u_{\epsilon}^{\sigma}\leq L\}}. \]

Taking $v=z_{L,\epsilon }$ in (61), we get

(62)\begin{equation} \displaystyle\int_{\Omega}|\nabla u_{\epsilon}|^{2}\min\{u_{\epsilon}^{2\sigma},L^{2}\}+2\sigma\int_{\{u_{\epsilon}^{s}\leq L\}}u_{\epsilon}^{2\sigma}|\nabla u_{\epsilon}|^{2}\leq\int_{\Omega}a_{\epsilon}(x)u_{\epsilon}^{2}\min\{u_{\epsilon}^{2\sigma},L^{2}\}. \end{equation}

Define $w_{L,\epsilon }$ by $w_{L,\epsilon }=u_{\epsilon }\min \{u_{\epsilon }^{\sigma },\,L\}\in H_{0}^{1}(\Omega )$. We have

\[ \nabla w_{L,\epsilon}=\min\{u_{\epsilon}^{\sigma},L\}\nabla u_{\epsilon}+\sigma u_{\epsilon}^{\sigma}\nabla u_{\epsilon}\chi_{\{u_{\epsilon}^{\sigma}\leq L\}}. \]

We thus get

\[ |\nabla w_{L,\epsilon}|^{2}= \sigma^{2} u_{\epsilon}^{2\sigma}|\nabla u_{\epsilon}|^{2}\chi_{\{u_{\epsilon}^{\sigma}\leq L\}}+\min\{u_{\epsilon}^{2\sigma},L^{2}\}|\nabla u_{\epsilon}|^{2}+\sigma u_{\epsilon}^{2\sigma}|\nabla u_{\epsilon}|^{2}\chi_{\{u_{\epsilon}^{\sigma}\leq L\}}. \]

Hence,

\[ |\nabla w_{L,\epsilon}|^{2}=(1+\sigma(\sigma+1))\min\{u_{\epsilon}^{2\sigma},L^{2}\}|\nabla u_{\epsilon}|^{2}\text{ in }\{u_{\epsilon}^{\sigma}\leq L\} \]

and

\[ |\nabla w_{L,\epsilon}|^{2}=\min\{u_{\epsilon}^{2\sigma},L^{2}\}|\nabla u_{\epsilon}|^{2}\text{ in }\{u_{\epsilon}^{\sigma}> L\}. \]

We conclude that

\[ |\nabla w_{L,\epsilon}|^{2}\leq c(\sigma)\min\{u_{\epsilon}^{2\sigma},L^{2}\}|\nabla u_{\epsilon}|^{2}\text{ in }\Omega, \]

where $c(\sigma )=1+\sigma (\sigma +1)$. From (62), we get

\[ \int_{\Omega}|\nabla w_{L,\epsilon}|^{2}\leq c(\sigma)\int_{\Omega}\min\{u_{\epsilon}^{2\sigma},L^{2}\}|\nabla u_{\epsilon}|^{2}\leq c(\sigma)\int_{\Omega}a_{\epsilon}(x)u_{\epsilon}^{2}\min\{u_{\epsilon}^{2\sigma},L^{2}\}. \]

Now, fix $K>0$. We have

\[ \int_{\Omega}|\nabla w_{L,\epsilon}|^{2}\leq c(\sigma)K\int_{\Omega}u_{\epsilon}^{2}\min\{u_{\epsilon}^{2\sigma},L^{2}\}+c(\sigma)\int_{\{a_{\epsilon}\geq K\}}a_{\epsilon}(x)u_{\epsilon}^{2}\min\{u_{\epsilon}^{2\sigma},L^{2}\}. \]

Hence,

\begin{align*} \displaystyle\int_{\Omega}|\nabla w_{L,\epsilon}|^{2}& \leq c(\sigma)K\int_{\Omega}u_{\epsilon}^{2}\min\{u_{\epsilon}^{2\sigma},L^{2}\}\\ & \quad+c(\sigma)\left(\displaystyle\int_{\{a_{\epsilon}\geq K\}}a_{\epsilon}(x)^{\frac{N}{2}}\right)^{\frac{2}{N}}\left(\int_{\Omega}(u_{\epsilon}\min\{u_{\epsilon}^{\sigma},L\})^{\frac{2N}{N-2}}\right)^{\frac{N-2}{N}}. \end{align*}

Consequently,

\[ \int_{\Omega}|\nabla w_{L,\epsilon}|^{2}\leq c(\sigma)K\int_{\Omega}u_{\epsilon}^{2}\min\{u_{\epsilon}^{2\sigma},L^{2}\}+c(\sigma)\left(\int_{\{a_{\epsilon}\geq K\}}a_{\epsilon}(x)^{\frac{N}{2}}\right)^{\frac{2}{N}}\left(\int_{\Omega}w_{L,\epsilon}^{\frac{2N}{N-2}}\right)^{\frac{N-2}{N}}. \]

From the Sobolev embedding Theorem, we get

(63)\begin{align} \displaystyle\int_{\Omega}|\nabla w_{L,\epsilon}|^{2}\leq c(\sigma)K\int_{\Omega}u_{\epsilon}^{2}\min\{u_{\epsilon}^{2\sigma},L^{2}\}+Cc(\sigma)\left(\int_{\{a_{\epsilon}\geq K\}}a_{\epsilon}(x)^{\frac{N}{2}}\right)^{\frac{2}{N}}\int_{\Omega}|\nabla w_{L,\epsilon}|^{2}. \end{align}

Choose $K>0$ such that

\[ \left(\int_{\{a_{\epsilon}\geq K\}}a_{\epsilon}(x)^{\frac{N}{2}}\right)^{\frac{2}{N}}\leq \frac{1}{2C c(\sigma)}. \]

Claim 1: $K$ can be chosen independently of $\epsilon$, provided $\lambda$ is sufficiently large.

Assuming the claim to be true, we obtain

\[ \int_{\Omega}|\nabla w_{\epsilon}|^{2}\,{\rm d}x\leq Cc(\sigma)K\int_{\Omega}u_{\epsilon}^{2}\min\{u_{\epsilon}^{2\sigma},L^{2}\}. \]

Consequently,

\[ \int_{\{u_{\epsilon}^{\sigma}\leq L\}}|\nabla (u_{\epsilon}^{\sigma+1})|^{2}\,{\rm d}x\leq Cc(\sigma)K\int_{\Omega}(u_{\epsilon}\min\{u_{\epsilon}^{\sigma},L\})^{2}\,{\rm d}x. \]

Suppose that $u_{\epsilon }\in L^{2\sigma +2}(\Omega )$ and let $u_{L,\epsilon }=(u_{\epsilon }\min \{u_{\epsilon }^{\sigma },\,L\})^{2}$. Observe that

\[ \lim_{L\to\infty}u_{L,\epsilon}= u_{\epsilon}^{2+2\sigma}\text{ almost everywhere in }\Omega \]

Furthermore,

\[ L_{1}< L_{2}\text{ implies that }u_{L_{1},\epsilon}< u_{L_{2},\epsilon}\text{ in }\Omega. \]

The Monotone Convergence Theorem implies that

\[ \int_{\{u_{\epsilon}^{\sigma}\leq L\}}|\nabla (u_{\epsilon}^{\sigma+1})|^{2}\,{\rm d}x\leq Cc(\sigma)K\int_{\Omega}u_{\epsilon}^{2+2\sigma}\,{\rm d}x. \]

From Fatou's Lemma, we get

(64)\begin{equation} \displaystyle\int_{\Omega}|\nabla (u_{\epsilon}^{\sigma+1})|^{2}\,{\rm d}x\leq Cc(\sigma)K\int_{\Omega}u_{\epsilon}^{2+2\sigma}\,{\rm d}x\text{ for all }\sigma\geq 0. \end{equation}

Consequently, $u_{\epsilon }^{\sigma +1}\in H_{0}^{1}(\Omega )$ and $u_{\epsilon }\in L^{\frac {(2+2\sigma )N}{N-2}}(\Omega )$. Now let $q>1$. We will show that

(65)\begin{equation} u_{\epsilon}\in L^{q}(\Omega)\text{ and }\|u_{\epsilon}\|_{L^{q}(\Omega)}< C_{q}\text{ for all }0<\epsilon<1. \end{equation}

Indeed, this follows by choosing adequate values for $\sigma$ in (64). Let $\sigma _{0}=0$. From (64), we get

\[ \int_{\Omega}|\nabla u_{\epsilon}|^{2}\,{\rm d}x\leq CK\int_{\Omega}u_{\epsilon}^{2}\,{\rm d}x\leq C_{0}. \]

Let $\sigma _{1}=\frac {N}{N-2}-1$. From (64), we get

\[ \int_{\Omega}|\nabla (u_{\epsilon}^{\frac{N}{N-2}})|^{2}\,{\rm d}x\leq CK\int_{\Omega}u_{\epsilon}^{\frac{2N}{N-2}}\,{\rm d}x\leq C_{1}. \]

Consequently, $u_{\epsilon }^{\frac {N}{N-2}}\in H_{0}^{1}(\Omega )$ and

\[ \int_{\Omega}u_{\epsilon}^{(\frac{N}{N-2})\frac{2N}{N-2}}\leq C_{1}. \]

Let $\sigma _{2}=\frac {4N-4}{(N-2)^{2}}$. From (64), we get

\[ \int_{\Omega}|\nabla (u_{\epsilon}^{\frac{N^{2}}{(N-2)^{2}}})|^{2}\,{\rm d}x\leq CK\int_{\Omega}u_{\epsilon}^{\frac{2N^{2}}{(N-2)^{2}}}\,{\rm d}x\leq C_{2}. \]

Consequently, $u_{\epsilon }^{\frac {N^{2}}{(N-2)^{2}}}\in H_{0}^{1}(\Omega )$ and

\[ \int_{\Omega}u_{\epsilon}^{(\frac{N^{2}}{(N-2)^{2}})\frac{2N}{N-2}}\leq C_{2}. \]

Assertion (65) then follows by choosing

\[ \sigma_{i}+1=(\sigma_{i-1}+1)\frac{N}{N-2}. \]

and iterating up until $s_{M}>q$ for some $M\in \mathbb {N}$. Now let $w_{\epsilon }$ be the solution of the non-singular problem

(66)\begin{equation} \left\{ \begin{array}{@{}l} -\Delta w=\lambda u_{\epsilon}+u_{\epsilon}^{2^{*}-1} \text{ in } \Omega\\ w = 0 \text{ on }\partial\Omega, \end{array} \right. \end{equation}

Assertion (65) and elliptic regularity theory implies that $w_{\epsilon }\in W^{2,q}(\Omega )$ and

\[ \|w_{\epsilon}\|_{ W^{2,q}(\Omega)}\leq C|\lambda u_{\epsilon}+u_{\epsilon}^{2^{*}-1}|_{L^{q}(\Omega)}=C_{q}, \]

where $C_{q}$ does not depend on $\epsilon$. Consequently, the Sobolev embedding assures that $w_{\epsilon }\in C^{1}(\overline {\Omega })$ and

\[ \|w_{\epsilon}\|_{C^{1}(\Omega)}\leq C\text{ for all }0<\epsilon<1. \]

Observe that

\[ \int_{\Omega}\nabla w_{\epsilon}\nabla v=\int_{\Omega}(\lambda u_{\epsilon}+u_{\epsilon}^{2^{*}-1})v\text{ for all }v\in H_{0}^{1}(\Omega), 0<\epsilon<1. \]

Consequently,

\[ \int_{\Omega}\nabla (u_{\epsilon}-w_{\epsilon})\nabla v\leq 0\text{ for all }v\in H_{0}^{1}(\Omega), v\geq 0, 0<\epsilon<1. \]

The weak maximum principle implies

\[ \sup_{\Omega} (u_{\epsilon}-w_{\epsilon})=0. \]

Hence,

\[ u_{\epsilon}\leq w_{\epsilon}. \]

Consequently,

\[ \|u_{\epsilon}\|_{L^{\infty}(\Omega)}< C\text{ for all }0<\epsilon<1. \]

This proves the result. We need to only show that the claim holds. Indeed, from (46), there exists an element $u\in H_{0}^{1}(\Omega )$ such that, up to a subsequence,

(67)\begin{equation} \left\{ \begin{aligned} & u_{\epsilon}\rightharpoonup u \text{ weakly in } H^{1}_0(\Omega),\\ & u_{\epsilon}\to u \text{ in } L^{r}(\Omega) \text{ for } 1< r<2^{*},\\ & u_{\epsilon}\to u \text{ a.e in } \Omega. \end{aligned} \right. \end{equation}

We first show that

(68)\begin{equation} \displaystyle\int_{\Omega}u_{\epsilon}^{2^{*}}\to\int_{\Omega}u^{2^{*}}\text{ as }\epsilon\to 0. \end{equation}

Again using the Concentration-compactness principle of Lions, we get positive measures $\mu,\,\nu$ in $\Omega$ such that

\[ |\nabla u_{\epsilon}|^{2}\rightharpoonup|\nabla u|^{2}+\mu\text{ and }u_{\epsilon}^{2^{*}}\rightharpoonup u^{2^{*}}+\nu. \]

Furthermore, there is at most a countable set of indexes denoted by $\Lambda$, sequences $x_{i}\in \overline {\Omega }$, $\mu _{i},\,\nu _{i}\in (0,\,\infty )$ such that

\[ \nu=\sum_{i\in\Lambda}\nu_{i}\delta_{x_{i}},\quad\mu\geq\sum_{i\in\Lambda}\mu_{i}\delta_{x_{i}}\quad\text{ and } S\nu_{i}^{\frac{2}{2^{*}}}\leq \mu_{i}, \]

for every $i\in \Lambda$, where $S$ is given by (40). Now, for every $\sigma >0$ and $i\in \Lambda$, we define

\[ \psi_{\sigma,i}(x)=\psi\left(\frac{x-x_{i}}{\sigma}\right), \]

where $\psi \in C_{c}^{\infty }(\mathbb {R}^{n})$ is a function satisfying

\[ 0\leq\psi\leq 1,\quad\psi\equiv 1\text{ in }B_{1}(0),\quad\psi\equiv 0\text{ in }\mathbb{R}^{n}\setminus B_{2}(0)\quad\text{ and } \|\nabla\psi\|_{L^{\infty}(\mathbb{R}^{n})}\leq 2. \]

Proceeding as in the proof of Proposition 4.3 and using the hypothesis $\lim _{\lambda \to \infty }I_{\epsilon }(u_{\epsilon,\lambda })=0$, we conclude that $\Lambda =\emptyset$, provided $\lambda >0$ is sufficiently large, thus proving (68). Consequently, $a_{\epsilon }^{\frac {N}{2}}$ converges in $L^{1}(\Omega )$ to $a$, where

\[ a(x)=(\lambda+u(x)^{2^{*}-2})^{\frac{N}{2}}. \]

We now show that for each $\delta ^{*}>0$ there exists $\eta >0$ and $\epsilon _{0}>0$ such that

(69)\begin{equation} \displaystyle\int_{B}a_{\epsilon}(x)^{\frac{N}{2}}<\delta^{*}\text{ for all sets } B\subset\Omega\text{ with }|B|<\eta\text{ and }0<\epsilon<\epsilon_{0}. \end{equation}

Since $a\in L^{1}(\Omega )$, there exists $\eta >0$ such that

\[ \int_{B}a(x)<\delta^{*}/2\text{ for all sets } B\subset\Omega\text{ with }|B|<\eta. \]

We write

\[ \int_{B}a_{\epsilon}(x)^{\frac{N}{2}}=\int_{B}(a_{\epsilon}(x)^{\frac{N}{2}}-a(x))+\int_{B}a(x). \]

and we choose $\epsilon _{0}>0$ such that

\[ \int_{\Omega}|a_{\epsilon}(x)^{\frac{N}{2}}-a(x)|\leq\delta^{*}/2\text{ for all }0<\epsilon<\epsilon_{0}. \]

Consequently,

\[ \int_{B}a_{\epsilon}(x)^{\frac{N}{2}}\leq\delta^{*}\text{ for all sets } B\subset\Omega\text{ with }|B|<\eta\text{ and }0<\epsilon<\epsilon_{0}. \]

This proves (69). We now finally prove Claim 1. Indeed, we choose

\[ \delta^{*}=\left(\frac{1}{2 Cc(\sigma)}\right)^{\frac{N}{2}} \]

and we choose $K>0$ such that

\[ \left|\left\{u> \frac{(K-\lambda)^{\frac{1}{2^{*}-2}}}{2}\right\}\right|<\eta. \]

Observe that

\[ \{a_{\epsilon}\geq K\}=\{u_{\epsilon}(x)\geq (K-\lambda)^{\frac{1}{2^{*}-2}}\}. \]

The choice of $K$ implies that

\[ |\{a_{\epsilon}\geq K\}|\leq \left|\left\{u> \frac{(K-\lambda)^{\frac{1}{2^{*}-2}}}{2}\right\}\right|<\eta\text{ for sufficiently small }\epsilon. \]

Consequently, from (69) and the choice of $\delta ^{*}$, we get

\[ \left(\int_{\{a_{\epsilon}\geq K\}}a_{\epsilon}(x)^{\frac{N}{2}}\right)^{\frac{2}{N}}\leq \frac{1}{2C c(\sigma)}\text{ for sufficiently small }\epsilon. \]

This proves Claim 1 and the result.

Lemma 5.2 Let $u_{\epsilon }\in H_{0}^{1}(\Omega )$ be a non-negative solution of problem (2) and assume that there exists a constant $D>0$ independent of $\epsilon$ such that

\[ \|u_{\epsilon}\|_{H_{0}^{1}(\Omega)}\leq D\text{ for each }0<\epsilon<1. \]

If $f$ satisfies (8) and (9) for $0< p<\frac {N+2}{N-2}$, then $u_{\epsilon }\in L^{\infty }(\Omega )$ and there exists a constant $K_{2}>0$ such that

\[ \|u_{\epsilon}\|_{L^{\infty}(\Omega)}\leq K_{2}\text{ for each }0<\epsilon<1. \]

Proof of Lemma 5.2 From (4), we get

(70)\begin{equation} \displaystyle\int_{\Omega}\nabla u_{\epsilon}\nabla v+\int_{\Omega}g_{\epsilon}(u_{\epsilon})v-\int_{\Omega}f(u_{\epsilon})v=0\text{ for all }v\in H_{0}^{1}(\Omega),v\geq 0 \end{equation}

From (8), we get

\[ \int_{\Omega}\nabla u_{\epsilon}\nabla v+\int_{\Omega\cap\{u_{\epsilon}\geq\delta\}}g_{\epsilon}(u_{\epsilon})v-\int_{\Omega\cap\{u_{\epsilon}\geq\delta\}}f(u_{\epsilon})v\leq 0\text{ for all }v\in H_{0}^{1}(\Omega),v\geq 0. \]

Consequently,

\[ \int_{\Omega}\nabla u_{\epsilon}\nabla v\leq\int_{\Omega\cap\{u_{\epsilon}\geq\delta\}}f(u_{\epsilon})v\text{ for all }v\in H_{0}^{1}(\Omega),v\geq 0. \]

From (9), we get

\[ \int_{\Omega}\nabla u_{\epsilon}\nabla v\leq\int_{\Omega\cap\{u_{\epsilon}\geq\delta\}}C(1+u_{\epsilon}^{p})v\text{ for all }v\in H_{0}^{1}(\Omega),v\geq 0. \]

Consequently,

\[ \int_{\Omega}\nabla u_{\epsilon}\nabla v\leq C\int_{\Omega\cap\{u_{\epsilon}\geq\delta\}}\left(\frac{1}{\delta^{p}}+1\right)u_{\epsilon}^{p}v\text{ for all }v\in H_{0}^{1}(\Omega),v\geq 0. \]

Consequently, there exists $\widetilde {C}>0$ and $1<\widetilde {p}<2^{*}-1$ such that

\[ \int_{\Omega}\nabla u_{\epsilon}\nabla v\leq \widetilde{C}\int_{\Omega}u_{\epsilon}^{\widetilde{p}}v\text{ for all }v\in H_{0}^{1}(\Omega),v\geq 0. \]

The proof then follows as in item $(i)$ of Lemma 5.1.

Lemma 5.3 Assume that $f$ satisfies (6). For each $0<\epsilon <1$, let $u_\epsilon \in H_{0}^{1}(\Omega )\cap L^{\infty }(\Omega )$ be a non-negative solution of problem (2) and assume that there exists a constant $T>0$ such that

(71)\begin{equation} \sup_{0<\epsilon<1}\|u_{\epsilon}\|_{L^{\infty}(\Omega)}< T<\infty. \end{equation}

Let $\psi$ be such that

\begin{align*} \psi \in C^{2}(\overline{\Omega}), \ \psi > 0 \text{ in } \Omega,\ \psi = 0 \text{ on } \partial \Omega \text{ and } \displaystyle\frac{|\nabla \psi|^{2}}\psi \text{ is bounded in } \Omega. \end{align*}

Then, there exist constants $M>0$ and $\epsilon _{0}>0$ such that

\[ {\psi(x) |\nabla u_\epsilon (x) |^{2}}\leq M (u_{\epsilon}(x)^{1-\beta}+u_{\epsilon}(x)) \text{ for every } x\in \Omega,\quad 0<\epsilon<\epsilon_{0}. \]

Proof of Lemma 5.3 From (6), we obtain constants $C_{1}>0$ and $0< t_{0}<1$ such that

(72)\begin{equation} |\widetilde{f}'(s)|\leq C_{1} s^{q_{1}-1}\text{ for }0\leq s\leq t_{0}. \end{equation}

From (71) we obtain that $\Delta u_\epsilon$ is bounded in $L^{\infty }(\Omega )$. Thus, by standard elliptic regularity, $u_\epsilon$ belongs to $C^{1,\nu }(\overline \Omega )$. We define

\[ \overline{h}_\epsilon(u) = g_{\epsilon}(u) - f(u). \]

We shall denote $u_\epsilon$ simply by $u$. Define the functions

\[ Z(u)=u^{1-\beta}+u+a,\qquad w = \frac{|\nabla u|^{2}}{Z(u)}, \qquad v = w \psi, \]

where $a>0$ is small. Note that $v$ is $C^{2}$ at all points $x\in \Omega$ such that $u(x)>0$. Indeed, let $x\in \Omega$ be one such point. By continuity, there must exist an open ball $B\subset \Omega$ centred at $x$ such that $u>0$ in $\overline {B}$. Consequently, we know that $g_{\epsilon }(u)\in C^{1,\nu }(B)$ and $f(u)\in C^{1,\nu }(B)$. Hence, $\overline {h}_{\epsilon }(u)\in C^{1,\nu }(B)$. Since $u$ satisfies the equation $-\Delta u+\overline {h}_{\epsilon }(u)=0$ in $B$, we conclude that $u\in C^{3}(B)$, implying that $Z(u)$ and $w$ are $C^{2}$ in $B$.

The function $v$ is continuous in $\overline \Omega$, hence it attains its maximum at some point $x_0 \in \overline \Omega$. Thus, we obtain

\[ v(x_0) > 0 . \]

Note that $x_0 \in \Omega$, because $v=0$ on $\partial \Omega$. Furthermore, $u(x_{0})>0$, since otherwise $x_{0}$ would be a critical point of $u$ and $w(x_{0})=0$. Hence,

\[ \nabla v(x_0) = 0 \]

and

(73)\begin{equation} \Delta v (x_0) \leq 0. \end{equation}

The computations already carried out in [Reference Lorca and Montenegro20, Reference Montenegro and Silva22] lead to the following expression evaluated at $x_0$

(74)\begin{equation} \begin{aligned} \Delta v & \geq \displaystyle\frac{1}{Z(u)}\left[\psi w^{2}\left(\frac{1}{2} Z'(u)^{2} - Z(u) Z''(u)\right)\right.\\ & \quad+ w ( 2 \psi Z(u)\overline{h}_\epsilon'(u) - \psi \overline{h}_\epsilon (u) Z'(u) - K_0 Z(u) )\\ & \quad\left.- K_{0} Z'(u) Z(u)^{1/2} \psi^{1/2} w^{3/2}\vphantom{\left(\frac{1}{2} Z'(u)^{2} - Z(u) Z''(u)\right)}\right], \end{aligned} \end{equation}

where

\[ K_0 = \max \left( \sup_\Omega \left({\frac{|\nabla \psi |}{\psi^{1/2}}}\right), \sup_\Omega \left(\Delta \psi - 2 \frac{|\nabla \psi|^{2}}\psi\right) \right)>0. \]

We will show that if $v(x_0)$ is large enough then the right-hand side of (74) must be positive, which would contradict (73).

We will establish the following estimates uniformly for every $\epsilon$ sufficiently small.

(75)\begin{align} Z'(u) Z(u)^{1/2} & \leq C \left({ \displaystyle\frac{1}{2}} Z'(u)^{2} - Z''(u) Z(u)\right), \end{align}

(76)\begin{align} Z(u) |\overline{h}_\epsilon'(u)| & \leq C \left( { \frac{1}{2}} Z'(u)^{2} - Z''(u) Z(u)\right), \end{align}

(77)\begin{align} Z'(u) |\overline{h}_\epsilon(u)| & \leq C \left( { \frac{1}{2}} Z'(u)^{2} - Z''(u) Z(u)\right), \end{align}

(78)\begin{align} Z(u) & \leq C \left({ \frac{1}{2}} Z'(u)^{2} - Z''(u) Z(u)\right) , \end{align}

for every $0 \leq u \leq T$. The constant $C$ depends only on $T$, but not on $\epsilon$ nor on $a$.

Assuming for a moment that (75)–(78) are true. Inequality (74) implies that

\begin{align*} \Delta v & \geq \displaystyle\frac{\frac 12 Z'(u)^{2} - Z''(u) Z(u)}{Z(u)} ( \psi w^{2} - C (w + \psi^{1/2} w^{3/2} ) ) \\ & = \displaystyle\frac{\frac 12 Z'(u)^{2} - Z''(u) Z(u)}{Z(u) \psi} ( v^{2} - C (v + v^{3/2} ) ). \end{align*}

Since $\Delta v(x_{0})\leq 0$ and

\[ \frac{\frac 12 Z'(u)^{2} - Z''(u) Z(u)}{Z(u)\psi}>0\text{ in }\Omega, \]

we conclude that

\[ v(x_{0})^{2}-C(v(x_{0})+v(x_{0})^{3/2})\leq 0. \]

Consequently, there exists $M>0$ that does not depend on $a$ such that

\[ \sup_{\Omega}v=v(x_{0})< M. \]

Consequently,

\[ |\nabla u(x)|^{2}\psi(x)\leq M(u(x)^{1-\beta}+u(x)+a)\text{ for all }x\in\Omega. \]

The result then follows by letting $a\to 0$.

We prove now the relations (75)–(78). In the course of this proof, $C$, $\widetilde {C}$, $C_{i}$, $i\in \{1,\,2,\,3,\,\ldots \}$ denote various positive constants independent of $\epsilon$ and $a$, we obtain gradient

\[ Z(u)=u^{1-\beta}+u+a, \]

\[ Z'(u)=(1-\beta)u^{-\beta}+1,\hspace{0.5cm} Z''(u)={-}\beta(1-\beta)u^{-\beta-1}. \]

Hence,

(79)\begin{equation} \displaystyle\frac{1}{2} Z'(u)^{2} - Z''(u) Z(u) \geq \frac{(1-\beta)^{2}}{2}(u^{{-}2\beta}+1)+a\beta(1-\beta)u^{{-}1-\beta}. \text{ for } u >0 . \end{equation}

We first prove (78). Indeed, there is a constant $C>0$ such that

\[ Z(u)=u^{1-\beta}+u+a\leq C\text{ for } 0\leq u\leq T. \]

Hence, (78) follows from (79).

We now prove (77). Note that there exists a constant $\widetilde {C}>0$ such that

\begin{align*} Z'(u) |\overline{h}_\epsilon(u)|& \leq((1-\beta)u^{-\beta}+1)(g_{\epsilon}(u)+|f(u)|)\\ & \leq (1-\beta)u^{{-}2\beta}+(1-\beta)u^{-\beta}\sup_{0\leq s\leq T}|f(s)|+u^{-\beta}+\sup_{0\leq s\leq T}|f(s)|\\ & \leq \widetilde{C}(1+u^{{-}2\beta}). \end{align*}

Inequality (77) then follows from (79).

Now, we prove (76). Note that

\[ \overline{h}_{\epsilon}'(u)=\frac{u^{q-1}}{(u+\epsilon)^{q+\beta+1}}\left(q\epsilon-\beta u\right)-f'(u). \]

We split the proof of (76) in three cases.

Case I. Suppose that $0< u<\min \{\tfrac {q\epsilon }{2\beta },\,t_{0}\}$, where $0< t_{0}<1$ is given by (72). We define

\[ \omega_{\epsilon}(u)=\frac{u^{q-1}}{(u+\epsilon)^{q+\beta+1}}\left(q\epsilon-\beta u\right)-C_{1} u^{q_{1}-1}, \]

where $C_{1}>0$ is given by (72). We claim that there exists $\epsilon _{0}>0$ such that $\omega _{\epsilon }(u)>0$ for each $0<\epsilon <\epsilon _{0}$. Indeed, assume by contradiction that $\omega _{\epsilon }(u)<0$ for some $0< u<\frac {q\epsilon }{2\beta }$. We then have

\[ q\epsilon u^{q-1}<\beta u^{q}+C_{1} u^{q_{1}-1}(u+\epsilon)^{q+\beta+1}<\beta u^{q}+C_{1} u^{q_{1}-1}\epsilon^{q+\beta+1}\left(1+\frac{q}{2\beta}\right)^{q+\beta+1}. \]

Now take $\epsilon _{0}>0$ such that

\[ C_{1}\epsilon^{q+\beta+1}\left(1+\frac{q}{2\beta}\right)^{q+\beta+1}<\frac{\epsilon q}{2}\text{ for }0<\epsilon<\epsilon_{0}. \]

We may assume that $0< q< q_{1}$. Consequently,

\[ q\epsilon u^{q-1}<\beta u^{q}+\frac{\epsilon qu^{q_{1}-1}}{2}<\beta u^{q}+\frac{\epsilon qu^{q-1}}{2}. \]

Hence,

\[ \frac{q\epsilon u^{q-1}}{2}<\beta u^{q}, \]

which implies that

\[ u>\frac{q\epsilon}{2\beta}. \]

This contradicts our initial assumption. The claim is proven. Since

\[ \frac{q\epsilon u^{q-1}}{(u+\epsilon)^{q+\beta+1}}\leq q\frac{u^{q}}{u(u+\epsilon)^{q}}\frac{\epsilon}{(u+\epsilon)^{\beta+1}}\leq\frac{q}{u^{\beta+1}}, \]

we obtain

\[ |\overline{h}_{\epsilon}'(u)|=\overline{h}_{\epsilon}'(u)\leq\frac{q+C_{1}u^{q_{1}+\beta}}{u^{\beta+1}}\text{ for }0< u<\min\left\{\frac{q\epsilon}{2\beta},t_{0}\right\}. \]

Hence,

\[ |\overline{h}_{\epsilon}'(u)|\leq\frac{2q}{u^{\beta+1}}\text{ for }0< u<\min\left\{\frac{q\epsilon}{2\beta},t_{0},t_{1}\right\}, \]

where $t_{1}>0$ is chosen such that

\[ C_{1}u^{q_{1}+\beta}< q\text{ for }0\leq u\leq t_{1}. \]

Therefore,

\[ Z(u)|\overline{h}_{\epsilon}'(u)|\leq (u^{1-\beta}+u+a)\left(\displaystyle\frac{2q}{u^{\beta+1}}\right)\leq \frac{2q}{u^{2\beta}}+\frac{2qa}{u^{\beta+1}}\text{ for }0\leq u\leq \min\left\{\frac{q\epsilon}{2\beta},t_{0},t_{1}\right\}. \]

Comparing with (79), it follows that there exists a constant $C>0$ that does not depend on $a$ such that

(80)\begin{align} Z(u)|\overline{h}_{\epsilon}'(u)|\leq C(\displaystyle\frac{1}{2} Z'(u)^{2} - Z''(u) Z(u))\text{ for }0< u<\min\left\{\frac{q\epsilon}{2\beta},t_{0},t_{1}\right\},\quad 0<\epsilon<\epsilon_{0}. \end{align}

Case II. Suppose that $\frac {\epsilon q}{2\beta }\leq u\leq \min \{t_{0},\,t_{1}\}$. We have

\[ |\overline{h}_{\epsilon}'(u)|\leq\frac{u^{q-1}|q\epsilon-\beta u|}{(u+\epsilon)^{q+\beta+1}}+C_{1} u^{q_{1}-1}\text{ for }\frac{q\epsilon}{2\beta}\leq u\leq t_{0}. \]

Note that $|q\epsilon -\beta u|\leq \beta u$ if $2\beta u\geq q\epsilon$. We then obtain

\[ |\overline{h}_{\epsilon}'(u)|\leq\frac{\beta u^{q}+C_{1} u^{q+q_{1}+\beta}\left(1+\frac{2\beta}{q}\right)^{q+\beta+1}}{(u+\epsilon)^{q+\beta+1}}\text{ for }\frac{q\epsilon}{2\beta}\leq u\leq t_{0}. \]

Now, observe that there exists $0< t_{2}<\min \{t_{0},\,t_{1}\}$ that does not depend on $\epsilon$ such that

\[ C_{1} u^{q_{1}+\beta}\left(1+\frac{2\beta}{q}\right)^{q+\beta+1}<\beta\text{ for }0\leq u\leq t_{2}. \]

Therefore,

\[ |\overline{h}_{\epsilon}'(u)|\leq\frac{2\beta}{u^{\beta+1}}\text{ for }\frac{q\epsilon}{2\beta}\leq u< t_{2}. \]

Comparing with (79), we obtain

(81)\begin{equation} Z(u)|\overline{h}_{\epsilon}'(u)|\leq C\left(\displaystyle\frac{1}{2} Z'(u)^{2} - Z''(u) Z(u)\right)\text{ for }\frac{q\epsilon}{2\beta}\leq u< t_{2}. \end{equation}

Case III. Assume that $t_{2}\leq u\leq T$. Since there exists a constant $C>0$ such that $|\overline {h}_{\epsilon }'(u)|\leq C$ for $t_{2}\leq u\leq T$, it follows from (78) that

(82)\begin{equation} Z(u)|\overline{h}_{\epsilon}'(u)|\leq C\left(\displaystyle\frac{1}{2} Z'(u)^{2} - Z''(u) Z(u)\right)\text{ for }t_{2}\leq u\leq T. \end{equation}

Hence, (76) follows from (80), (81) and (82).

We now prove (75). Observe that

\[ Z'(u) Z(u)^{1/2}=((1-\beta)u^{-\beta}+1)\sqrt{u^{1-\beta}+u+a}. \]

Hence,

\[ Z'(u) Z(u)^{1/2}\leq \sqrt{3T}((1-\beta) u^{-\beta}+1). \]

When $0\leq u\leq 1$ we know that $u^{2}\leq u$. Hence $u^{-\beta }\leq u^{-2\beta }$. Therefore, from (79), there exist constants $C_{3}>0$ and $C_{4}>C_{3}$ such that

(83)\begin{equation} Z'(u) Z(u)^{1/2}\leq C_{3} (u^{{-}2\beta}+1)\leq C_{4}\left(\displaystyle\frac{1}{2} Z'(u)^{2} - Z''(u) Z(u)\right)\text{ for }0\leq u\leq 1. \end{equation}

If $1\leq u\leq T$, we know that there exists a constant $C_{5}>0$ such that $Z'(u) Z(u)^{1/2}\leq C_{5}$. Hence, from (79), there exists a constant $C_{6}>0$ such that

(84)\begin{equation} Z'(u) Z(u)^{1/2}\leq C_{6}\left(\displaystyle\frac{1}{2} Z'(u)^{2} - Z''(u) Z(u)\right)\text{ for }1\leq u\leq T. \end{equation}

Inequality (75) then follows from (83) and (84). We have proved Lemma 5.3.

Lemma 6.1 Let $(\epsilon _{n})$ be a sequence in $(0,\,1)$ such that $\epsilon _{n}\to 0$ as $n\to \infty$. Let $(u_{\epsilon _{n}}^{1})$ and $(u_{\epsilon _{n}}^{2})$ be the sequences of solutions obtained in Propositions 3.2 and 4.3 respectively. Then there exist non-trivial functions $u_{1}\in H_{0}^{1}(\Omega )$ and $u_{2}\in H_{0}^{1}(\Omega )$ such that, up to a subsequence, $u_{\epsilon _{n}}^{i}\rightharpoonup u_{i}$ weakly in $H_{0}^{1}(\Omega ),$ where $i\in \{1,\,2\}$.

Proof of Lemma 6.1 From Propositions 3.2 and 4.3, we know that there exist constants $D_{i}>0$ such that

\[ \|u_{\epsilon_{n}}^{i}\|_{H_{0}^{1}(\Omega)}< D_{i}\text{ for each }n\in\mathbb{N},i\in\{1,2\} \]

Hence, there exist functions $u_{i}\in H_{0}^{1}(\Omega )$ such that

(85)\begin{equation} \left\{ \begin{aligned} & u_{\epsilon_{n}}^{i}\rightharpoonup u_{i} \text{ weakly in } H^{1}_0(\Omega);\\ & u_{\epsilon_{n}}^{i}\to u_{i} \text{ in } L^{r}(\Omega) \text{ for every } 1\leq r<2^{*};\\ & u_{\epsilon_{n}}^{i}\to u_{i} \text{ a.e in } \Omega; \end{aligned} \right. \end{equation}

Lemmas 5.1 and 5.2 imply that $u_{\epsilon _{n}}^{i}\in L^{\infty }(\Omega )$ with $\|u_{\epsilon _{n}}^{i}\|_{L^{\infty }(\Omega )}< K_{i}$ for all $n\in \mathbb {N}$. Consequently, the Dominated Convergence Theorem implies that

\[ u_{\epsilon_{n}}^{i}\to u_{i} \text{ in } L^{r}(\Omega) \text{ for every } r\geq 1. \]

We prove the result for $i=1$ and denote $(u_{\epsilon _{n}}^{1})$ and $u_{1}$ merely by $(u_{\epsilon _{n}})$ and $u$ respectively. From Proposition 3.2, we have

\[ 0< a_{1}\leq I_{\epsilon_{n}}(u_{\epsilon_{n}})=\frac{1}{2}\int_{\Omega}|\nabla u_{\epsilon_{n}}|^{2}+\int_{\Omega}G_{\epsilon_{n}}(u_{\epsilon_{n}})-\int_{\Omega}F(u_{\epsilon_{n}}). \]

Since $u_{\epsilon _{n}}$ is a non-negative critical point of $I_{\epsilon _{n}}$, we have

\[ \|u_{\epsilon_{n}}\|_{H_{0}^{1}(\Omega)}^{2}+\int_{\Omega}g_{\epsilon_{n}}(u_{\epsilon_{n}})u_{\epsilon_{n}}=\int_{\Omega}u_{\epsilon_{n}}f(u_{\epsilon_{n}}). \]

Hence,

(86)\begin{align} I_{\epsilon_{n}}(u_{\epsilon_{n}}) = \displaystyle\int_{\Omega}\left(G_{\epsilon_{n}}(u_{\epsilon_{n}})-\displaystyle\frac{1}{2}g_{\epsilon_{n}}(u_{\epsilon_{n}})u_{\epsilon_{n}}\right)\,{\rm d}x-\int_{\Omega}\left(F(u_{\epsilon_{n}})-\frac{1}{2}u_{\epsilon_{n}}f(u_{\epsilon_{n}})\right)\,{\rm d}x>a_{1}.\end{align}

The Dominated Convergence Theorem implies that

\begin{align*} & \lim_{n\rightarrow\infty}\int_{\Omega}g_{\epsilon_{n}}(u_{\epsilon_{n}})u_{\epsilon_{n}}\,{\rm d}x=\int_{\Omega}u^{1-\beta}\,{\rm d}x,\\ & \lim_{n\rightarrow\infty}\int_{\Omega}G_{\epsilon_{n}}(u_{\epsilon_{n}})\,{\rm d}x=\frac{1}{1-\beta}\int_{\Omega}u^{1-\beta}\,{\rm d}x,\\ \end{align*}

and

\[ \int_{\Omega}\left(F(u_{\epsilon_{n}})-\frac{1}{2}u_{\epsilon_{n}}f(u_{\epsilon_{n}})\right)\,{\rm d}x\to\int_{\Omega}\left(F(u)-\frac{1}{2}uf(u)\right)\,{\rm d}x. \]

Taking the above claims into account and letting $n\rightarrow \infty$ in (86), we obtain

\[ \int_{\Omega}\left(\frac{u^{1-\beta}}{1-\beta}-\frac{u^{1-\beta}}{2}\right)\,{\rm d}x -\int_{\Omega}\left(F(u)-\frac{1}{2}uf(u)\right)\,{\rm d}x\geq a_{1}. \]

We proved that $u$ is non-trivial. The proof for $i=2$ is analogous.

Lemma 6.2 Let $u_{1}$ and $u_{2}$ be the functions given by Lemma 6.1. The function $u_{i}^{-\beta }\chi _{\{u_{i}>0\}}$ belongs to $L^{1}_{loc}(\Omega )$ for $i\in \{1,\,2\}$.

Proof of Lemma 6.2 We again prove the result for $i=1$. The proof for $i=2$ is analogous. Let $(u_{\epsilon _{n}})$ and $u$ be given by (85) with $i=1$. Let $V \subset \Omega$ be a open set such that $\overline {V}\subset \Omega$. Take $\zeta \in C^{1}_c(\Omega )$ such that $0 \leq \zeta \leq 1$ and $\zeta \equiv 1$ in $V$. Since $u_{\epsilon _{n}}$ is a critical point of $I_{\epsilon _{n}},$ we obtain

\[ \int_{\{u_{\epsilon_{n}}< 1-\epsilon_{n} \}} g_{\epsilon_{n}} (u_{\epsilon_{n}} ) \zeta = \int_\Omega f(u_{\epsilon_{n}}) \zeta - \int_\Omega \nabla u_{\epsilon_{n}}\nabla \zeta - \int_{\{u_{\epsilon_{n}}\geq 1-\epsilon_{n} \}} g_{\epsilon_{n}} (u_{\epsilon_{n}} ) \zeta. \]

Corollary 5.4 implies that $u_{\epsilon _{n}}\to u$ uniformly in compact subsets of $\Omega$. Since $u_{\epsilon _{n}}\rightharpoonup u$ weakly in $H_{0}^{1}(\Omega )$, we get

(87)\begin{equation} \displaystyle\int_{\{u_{\epsilon_{n}}< 1-\epsilon_{n} \}} g_{\epsilon_{n}} (u_{\epsilon_{n}} ) \zeta \rightarrow \int_\Omega f(u) \zeta - \int_\Omega \nabla u \nabla \zeta - \int_{\{u \geq 1\}} u^{-\beta} \zeta \text{ as } \epsilon\to 0 \end{equation}

Define the set $\Omega _\rho = \{ x \in \Omega : u(x) \geq \rho \}$ for $\rho >0$. It follows from (87) that there exists a constant $C>0$ that does not depend on $n$ nor on $\rho$ such that

\[ \int_{V \cap \Omega_\rho} \frac{u_{\epsilon_{n}}^{q}}{(u_{\epsilon_{n}}+\epsilon_{n})^{q+\beta}} \chi_{\{u_{\epsilon_{n}} <1-\epsilon_{n} \}}\zeta \leq \int_{\{u_{\epsilon_{n}}< 1-\epsilon_{n} \}} g_{\epsilon_{n}}(u_{\epsilon_{n}}) \zeta < C\text{ for all }n\in\mathbb{N},\quad \rho>0, \]

Letting $n\to \infty$ and using Fatou's Lemma, we then get

\[ \int_V u^{-\beta} \chi_{\Omega_\rho} < C. \]

Letting $\rho \rightarrow 0$ and applying Fatou's Lemma again, we conclude that

\[ \int_V u^{-\beta} \chi_{\{u>0\}} < \infty. \]

Since $V$ was arbitrarily chosen, Lemma 6.2 is proved.