Proof. Clearly, $J_{\operatorname {\varepsilon }}(0)=0$. By $(g_1)$ and $(g_2)$, for all $\delta >0$ there exists $C_{\delta }>0$ such that

\[ |G_{\operatorname{\varepsilon}} (x, t^{2})|\leq \delta |t|^{2}+C_{\delta}|t|^{{2^{*}_{s}}} \quad \mbox{ for every } (x, t)\in \mathbb{R}^{N}\times \mathbb{R}. \]

This fact combined with theorem 2.3 implies that for all $u\in \mathcal {H}_{\operatorname {\varepsilon }}$

\[ J_{\operatorname{\varepsilon}}(u)\geq \left(\displaystyle\frac{1}{2}-\delta C\right)\|u\|_{\operatorname{\varepsilon}}^{2}-C'_{\delta} \|u\|^{{2^{*}_{s}}}_{\operatorname{\varepsilon}} \]

which implies that (ii) holds true. Concerning (iii), for each $u\in \mathcal {H}^{+}_{\operatorname {\varepsilon }}$ and $t>0$, we get

\begin{align*} J_{\operatorname{\varepsilon}}(tu)&\leq \frac{t^{2}}{2} \|u\|^{2}_{\operatorname{\varepsilon}}-\frac{1}{2}\int_{\Lambda_{\operatorname{\varepsilon}}} F(t^{2}|u|^{2})\, \mathrm{d}x \\ &\leq \frac{t^{2}}{2}\|u\|^{2}_{\operatorname{\varepsilon}}-Ct^{\theta} \int_{\Lambda_{\operatorname{\varepsilon}}} |u|^{\theta}\, \mathrm{d}x+C'|\textrm{supp}(|u|)\cap \Lambda_{\operatorname{\varepsilon}}| \end{align*}

where we used $(g_2)$ and $(f_3)$. Since $\theta >2$ we see that $J_{\operatorname {\varepsilon }}(tu)\rightarrow -\infty$ as $t\rightarrow \infty$.

Proof.

1. (i) We note that $h_{u}\in C^{1}(\mathbb {R}_{+}, \mathbb {R})$, and arguing as in the proof of lemma 3.1 it is easy to verify that $h_{u}(0)=0$, $h_{u}(t)>0$ for $t>0$ small enough and $h_{u}(t)<0$ for $t>0$ sufficiently large. Therefore, $\max _{t\geq 0} h_{u}(t)$ is achieved at some $t_{u}>0$ verifying $h'_{u}(t_{u})=0$ and $t_{u}u\in \mathcal {N}_{\operatorname {\varepsilon }}$. Next we claim the uniqueness of such a $t_{u}$. Let $t_{1}>t_{2}>0$ be such that $h'_{u}(t_{i})=0$ for $i=1, 2$, that is $\|u\|^{2}_{\operatorname {\varepsilon }}=\int _{\mathbb {R}^{N}} g_{\operatorname {\varepsilon }}(x, |t_{i}u|^{2})|u|^{2}\, \mathrm {d}x$ for $i=1, 2$. By using the definition of $g$, $(f_4)$, $(g_4)$ and $u\in \mathcal {H}^{+}_{\operatorname {\varepsilon }}$, we see that

   \begin{align*} 0&= \int_{\mathbb{R}^{N}} \left[g_{\operatorname{\varepsilon}}(x, |t_{1}u|^{2})|u|^{2}-g_{\operatorname{\varepsilon}}(x, |t_{2}u|^{2})|u|^{2}\right] \mathrm{d}x\\ &= \int_{\Lambda^{c}_{\operatorname{\varepsilon}}} \left[g_{\operatorname{\varepsilon}}(x, |t_{1}u|^{2})|u|^{2}-g_{\operatorname{\varepsilon}}(x, |t_{2}u|^{2})|u|^{2}\right] \mathrm{d}x\\ &\quad +\int_{\Lambda_{\operatorname{\varepsilon}}} \left[f(|t_{1}u|^{2})|u|^{2}-f(|t_{2}u|^{2})|u|^{2}\right] \mathrm{d}x \\ &>\int_{\Lambda^{c}_{\operatorname{\varepsilon}}\cap \{ |t_{2}u|^{2}\leq a<|t_{1}u|^{2} \}}\left[\frac{V_{0}}{k}|u|^{2}- f(|t_{2}u|^{2})|u|^{2}\right] \mathrm{d}x\\ &\quad +\int_{\Lambda^{c}_{\operatorname{\varepsilon}}\cap\{|t_{1}u|^{2}\leq a\}} \left[f(|t_{1}u|^{2})|u|^{2}-f(|t_{2}u|^{2})|u|^{2}\right] \mathrm{d}x\geq 0 \end{align*}
   which gives a contradiction.

2. (ii) Let $u\in \mathbb {S}_{\operatorname {\varepsilon }}^{+}$. By $(i)$ there exists $t_{u}>0$ such that $h_{u}'(t_{u})=0$, or equivalently

   \[ t_{u}= \int_{\mathbb{R}^{N}} g_{\operatorname{\varepsilon}}(x, |t_{u}u|^{2}) t_{u}|u|^{2} \,\mathrm{d}x. \]
   By assumptions $(g_{1})$ and $(g_{2})$, given $\xi >0$ there exists a positive constant $C_{\xi }$ such that
   (3.5)\begin{equation} |g_{\operatorname{\varepsilon}}(x, t)|\leq \xi + C_{\xi} |t|^{({{2^{*}_{s}}-2})/{2}}, \quad \mbox{ for every } (x, t)\in \mathbb{R}^{N}\times \mathbb{R}. \end{equation}
   Thus (3.5) and theorem 2.3 yield
   \begin{align*} t_{u}\leq \xi t_{u}C_{1} + C_{\xi} C_{2} t^{{2^{*}_{s}}-1}_{u}. \end{align*}
   Taking $\xi >0$ sufficiently small, we obtain that there exists $\tau >0$, independent of $u$, such that $t_{u}\geq \tau$. Now, let $\mathbb {K}\subset \mathbb {S}_{\operatorname {\varepsilon }}^{+}$ be a compact set and we show that $t_{u}$ can be estimated from above by a constant depending on $\mathbb {K}$. Assume by contradiction that there exists a sequence $(u_{n})\subset \mathbb {K}$ such that $t_{n}=t_{u_{n}}\rightarrow \infty$. Therefore, there exists $u\in \mathbb {K}$ such that $u_{n}\rightarrow u$ in $\mathcal {H}_{\operatorname {\varepsilon }}$. From (iii) in lemma 3.1 we get
   (3.6)\begin{equation} J_{\operatorname{\varepsilon}}(t_{n}u_{n})\rightarrow -\infty. \end{equation}
   Fix $v\in \mathcal {N}_{\operatorname {\varepsilon }}$. Then, using the fact that $\langle J_{\operatorname {\varepsilon }}'(v), v \rangle =0$, and assumption $(g_{3})$, we can infer
   (3.7)\begin{align} J_{\operatorname{\varepsilon}}(v)&= J_{\operatorname{\varepsilon}}(v)- \frac{1}{\theta} \langle J_{\operatorname{\varepsilon}}'(v), v \rangle \nonumber\\ &=\left(\frac{1}{2}-\frac{1}{\theta}\right) \|v\|_{\operatorname{\varepsilon}}^{2}+ \int_{\mathbb{R}^{N}} \frac{1}{\theta} g_{\operatorname{\varepsilon}}(x, |v|^{2})|v|^{2}- \frac{1}{2} G_{\operatorname{\varepsilon}}(x, |v|^{2})]\, \mathrm{d}x \nonumber\\ &\geq \left(\frac{1}{2}-\frac{1}{\theta}\right) \|v\|_{\operatorname{\varepsilon}}^{2}+ \frac{1}{\theta}\int_{\Lambda_{\operatorname{\varepsilon}}^{c}} [g_{\operatorname{\varepsilon}}(x, |v|^{2})|v|^{2}- \frac{\theta}{2} G_{\operatorname{\varepsilon}}(x, |v|^{2})]\, \mathrm{d}x \nonumber\\ &\geq \left(\frac{\theta-2}{2\theta}\right)\|v\|_{\operatorname{\varepsilon}}^{2} -\left(\frac{\theta-2}{2\theta}\right) \frac{1}{k}\int_{\Lambda^{c}_{\operatorname{\varepsilon}}} V(\operatorname{\varepsilon} x) |v|^{2}\,\mathrm{d}x \nonumber\\ & \geq \left(\frac{\theta-2}{2\theta}\right) \left(1- \frac{1}{k}\right) \|v\|_{\operatorname{\varepsilon}}^{2}. \end{align}
   Taking into account that $(t_{u_{n}}u_{n})\subset \mathcal {N}_{\operatorname {\varepsilon }}$ and $\|t_{u_{n}}u_{n}\|_{\operatorname {\varepsilon }}=t_{u_{n}}\rightarrow \infty$, from (3.7) we deduce that (3.6) does not hold.

3. (iii) First, we note that $\hat {m}_{\operatorname {\varepsilon }}$, $m_{\operatorname {\varepsilon }}$ and $m_{\operatorname {\varepsilon }}^{-1}$ are well defined. Indeed, by (i), for each $u\in \mathcal {H}_{\operatorname {\varepsilon }}^{+}$ there exists a unique $m_{\operatorname {\varepsilon }}(u)\in \mathcal {N}_{\operatorname {\varepsilon }}$. On the other hand, if $u\in \mathcal {N}_{\operatorname {\varepsilon }}$ then $u\in \mathcal {H}_{\operatorname {\varepsilon }}^{+}$. Otherwise, if $u\notin \mathcal {H}_{\operatorname {\varepsilon }}^{+}$, we have

   \[ |\textrm{supp} (|u|) \cap \Lambda_{\operatorname{\varepsilon}}|=0, \]
   which together with $(g_3)$-(ii) gives
   (3.8)\begin{align} 0<\|u\|_{\operatorname{\varepsilon}}^{2}&= \int_{\mathbb{R}^{N}} g_{\operatorname{\varepsilon}}(x, |u|^{2}) |u|^{2} \, \mathrm{d}x \nonumber\\ &= \int_{\Lambda_{\operatorname{\varepsilon}}} g_{\operatorname{\varepsilon}}(x, |u|^{2}) |u|^{2} \, \mathrm{d}x + \int_{\Lambda^{c}_{\operatorname{\varepsilon}}} g_{\operatorname{\varepsilon}}(x, |u|^{2}) |u|^{2} \, \mathrm{d}x \nonumber\\ &= \int_{\Lambda^{c}_{\operatorname{\varepsilon}}} g_{\operatorname{\varepsilon}}(x, |u|^{2}) |u|^{2} \, \mathrm{d}x \nonumber\\ &\leq \frac{1}{k} \int_{\Lambda^{c}_{\operatorname{\varepsilon}}} V_{\operatorname{\varepsilon}}(x) |u|^{2} \,\mathrm{d}x \leq \frac{1}{k} \|u\|_{\operatorname{\varepsilon}}^{2}. \end{align}
   Using (3.8) we get
   \begin{align*} 0<\|u\|_{\operatorname{\varepsilon}}^{2} \leq \frac{1}{k} \|u\|_{\operatorname{\varepsilon}}^{2} \end{align*}
   and this leads to a contradiction because $k>1$. Consequently, $m_{\operatorname {\varepsilon }}^{-1}(u)= {u}/{\|u\|_{\operatorname {\varepsilon }}}\in \mathbb {S}_{\operatorname {\varepsilon }}^{+}$, $m_{\operatorname {\varepsilon }}^{-1}$ is well defined and continuous.

   Let $u\in \mathbb {S}_{\operatorname {\varepsilon }}^{+}$. Then,

   \begin{align*} m_{\operatorname{\varepsilon}}^{{-}1}(m_{\operatorname{\varepsilon}}(u))= m_{\operatorname{\varepsilon}}^{{-}1}(t_{u}u)= \frac{t_{u}u}{\|t_{u}u\|_{\operatorname{\varepsilon}}}= \frac{u}{\|u\|_{\operatorname{\varepsilon}}}=u \end{align*}
   from which $m_{\operatorname {\varepsilon }}$ is a bijection. Now, our aim is to prove that $\hat {m}_{\operatorname {\varepsilon }}$ is a continuous function. Let $(u_{n})\subset \mathcal {H}_{\operatorname {\varepsilon }}^{+}$ and $u\in \mathcal {H}_{\operatorname {\varepsilon }}^{+}$ such that $u_{n}\rightarrow u$ in $\mathcal {H}_{\operatorname {\varepsilon }}^{+}$. Hence,
   \[ \frac{u_{n}}{\|u_{n}\|_{\operatorname{\varepsilon}}}\rightarrow \frac{u}{\|u\|_{\operatorname{\varepsilon}}} \quad \mbox{ in } \mathcal{H}_{\operatorname{\varepsilon}}. \]
   Set $v_{n}= {u_{n}}/{\|u_{n}\|_{\operatorname {\varepsilon }}}$ and $t_{n}=t_{v_{n}}$. By (ii) there exists $t_{0}>0$ such that $t_{n}\rightarrow t_{0}$. Since $t_{n}v_{n}\in \mathcal {N}_{\operatorname {\varepsilon }}$ and $\|v_{n}\|_{\operatorname {\varepsilon }}=1$, we have
   \[ t^{2}_{n}= \int_{\mathbb{R}^{N}} g_{\operatorname{\varepsilon}}(x, |t_{n}v_{n}|^{2}) |t_{n}v_{n}|^{2} \, \mathrm{d}x. \]
   Passing to the limit as $n\rightarrow \infty$ we obtain
   \[ t^{2}_{0}= \int_{\mathbb{R}^{N}} g_{\operatorname{\varepsilon}}(x, |t_{0}v|^{2}) |t_{0} v|^{2}\, \mathrm{d}x, \]
   where $v= {u}/{\|u\|_{\operatorname {\varepsilon }}}$, which implies that $t_{0}v\in \mathcal {N}_{\operatorname {\varepsilon }}$. By (i) we deduce that $t_{v}= t_{0}$, and this shows that
   \[ \hat{m}_{\operatorname{\varepsilon}}(u_{n})= \hat{m}_{\operatorname{\varepsilon}}\left(\frac{u_{n}}{\|u_{n}\|_{\operatorname{\varepsilon}}}\right)\rightarrow \hat{m}_{\operatorname{\varepsilon}}\left(\frac{u}{\|u\|_{\operatorname{\varepsilon}}}\right)=\hat{m}_{\operatorname{\varepsilon}}(u) \quad \mbox{ in } \mathcal{H}_{\operatorname{\varepsilon}}. \]
   Therefore, $\hat {m}_{\operatorname {\varepsilon }}$ and $m_{\operatorname {\varepsilon }}$ are continuous functions.

4. (iv) Let $(u_{n})\subset \mathbb {S}_{\operatorname {\varepsilon }}^{+}$ be such that $\textrm {dist}(u_{n}, \partial \mathbb {S}_{\operatorname {\varepsilon }}^{+})\rightarrow 0$. Observe that, by the Sobolev inequality and $(V_1)$–$(V_2)$, for each $p\in [2, 2^{*}_{s}]$ there exists $\kappa _{p}>0$ such that it holds

   \begin{align*} \|u_{n}\|^{p}_{L^{p}(\Lambda_{\operatorname{\varepsilon}})} &\leq \inf_{v\in \partial \mathbb{S}_{\operatorname{\varepsilon}}^{+}} \|u_{n}- v\|^{p}_{L^{p}(\Lambda_{\operatorname{\varepsilon}})}\\ &\leq \kappa_{p} \inf_{v\in \partial \mathbb{S}_{\operatorname{\varepsilon}}^{+}} \|u_{n}- v\|^{p}_{\operatorname{\varepsilon}} \\ &\leq \kappa_{p} \, \textrm{dist}(u_{n}, \partial \mathbb{S}_{\operatorname{\varepsilon}}^{+})^{p}, \quad \mbox{ for all } n\in \mathbb{N}. \end{align*}
   Then, by the above inequality, $(g_{1})$, $(g_{2})$ and $(g_{3})$-(ii), we can infer that, for all $t>0$,
   \begin{align*} &\frac{1}{2}\int_{\mathbb{R}^{N}} G_{\operatorname{\varepsilon}}(x, |tu_{n}|^{2})\, \mathrm{d}x\\ &\quad = \frac{1}{2}\int_{\Lambda^{c}_{\operatorname{\varepsilon}}} G_{\operatorname{\varepsilon}}(x, |tu_{n}|^{2})\, \textrm{d}x + \frac{1}{2}\int_{\Lambda_{\operatorname{\varepsilon}}} G_{\operatorname{\varepsilon}}(x, |tu_{n}|^{2})\, \mathrm{d}x\\ &\quad \leq \frac{t^{2}}{2k} \int_{\Lambda^{c}_{\operatorname{\varepsilon}}} V(\operatorname{\varepsilon} x) |u_{n}|^{2} \,\textrm{d}x+ \int_{\Lambda_{\operatorname{\varepsilon}}} \frac{1}{2}F(|tu_{n}|^{2})\, \mathrm{d}x\\ &\quad \leq \frac{t^{2}}{2k} \|u_{n}\|_{\operatorname{\varepsilon}}^{2} + C_{1}t^{2} \int_{\Lambda_{\operatorname{\varepsilon}}} |u_{n}|^{2} \,\textrm{d}x+ C_{2} t^{{2^{*}_{s}}} \int_{\Lambda_{\operatorname{\varepsilon}}} |u_{n}|^{{2^{*}_{s}}} \,\mathrm{d}x\\ &\quad \leq \frac{t^{2}}{2k}+ C_{1}\kappa_{2} t^{2} \textrm{dist}(u_{n}, \partial \mathbb{S}_{\operatorname{\varepsilon}}^{+})^{2} + C_{2}\kappa_{{2^{*}_{s}}} t^{{2^{*}_{s}}} \textrm{dist}(u_{n}, \partial \mathbb{S}_{\operatorname{\varepsilon}}^{+})^{{2^{*}_{s}}} \end{align*}
   from which
   (3.9)\begin{equation} \frac{1}{2}\limsup_{n\rightarrow \infty} \int_{\mathbb{R}^{N}} G_{\operatorname{\varepsilon}}(x, |tu_{n}|^{2})\, \mathrm{d}x \leq \frac{t^{2}}{2k} \quad \mbox{ for all } t>0. \end{equation}
   Bearing in mind the definition of $m_{\operatorname {\varepsilon }}(u_{n})$, and by using (3.9), we have
   \begin{align*} \liminf_{n\rightarrow \infty} J_{\operatorname{\varepsilon}}(m_{\operatorname{\varepsilon}}(u_{n}))&\geq \liminf_{n\rightarrow \infty} J_{\operatorname{\varepsilon}}(tu_{n})\\ &= \liminf_{n\rightarrow \infty} \left[ \frac{1}{2} \|tu_{n}\|_{\operatorname{\varepsilon}}^{2} - \frac{1}{2}\int_{\mathbb{R}^{N}} G_{\operatorname{\varepsilon}}(x, |tu_{n}|^{2})\, \textrm{d}x\right]\\ &\geq \left(\frac{1}{2} - \frac{1}{2k} \right)t^{2} \quad \mbox{ for all } t>0. \end{align*}
   Recalling that $k>1$ we get
   \[ \lim_{n\rightarrow \infty} J_{\operatorname{\varepsilon}}(m_{\operatorname{\varepsilon}}(u_{n}))= \infty. \]
   Moreover, by the definition of $J_{\operatorname {\varepsilon }}$, we see that
   \[ \liminf_{n\rightarrow \infty} \frac{1}{2}\|m_{\operatorname{\varepsilon}}(u_{n})\|^{2}_{\operatorname{\varepsilon}}\geq \liminf_{n\rightarrow \infty} J_{\operatorname{\varepsilon}}(m_{\operatorname{\varepsilon}}(u_{n}))=\infty. \]
   This completes the proof of the lemma.

Proof. Let $(u_{n})\subset \mathcal {H}_{\operatorname {\varepsilon }}$ be a Palais–Smale sequence for $J_{\operatorname {\varepsilon }}$ at the level $c$, that is, as $n\rightarrow \infty$

\[ J_{\operatorname{\varepsilon}}(u_{n})\rightarrow c \mbox{ and } J_{\operatorname{\varepsilon}}'(u_{n})\rightarrow 0. \]

First, we show that $(u_{n})$ is bounded in $\mathcal {H}_{\operatorname {\varepsilon }}$. Indeed, using $(g_3)$, we get

\begin{align*} C(1+\|u_{n}\|_{\operatorname{\varepsilon}})&\geq J_{\operatorname{\varepsilon}}(u_{n})-\frac{1}{\theta}\langle J'_{\operatorname{\varepsilon}}(u_{n}), u_{n}\rangle \\ &= \left(\frac{1}{2}-\frac{1}{\theta}\right)\|u_{n}\|^{2}_{\operatorname{\varepsilon}}+\frac{1}{\theta}\int_{\mathbb{R}^{N}} \left[g_{\operatorname{\varepsilon}}(x, |u_{n}|^{2})|u_{n}|^{2}-\frac{\theta}{2} G_{\operatorname{\varepsilon}}(x, |u_{n}|^{2})\right] \mathrm{d}x\\ &\geq \left(\frac{1}{2}-\frac{1}{\theta}\right)\|u_{n}\|^{2}_{\operatorname{\varepsilon}} -\left(\frac{1}{2}-\frac{1}{\theta}\right)\int_{\Lambda^{c}_{\operatorname{\varepsilon}}} G_{\operatorname{\varepsilon}}(x, |u_{n}|^{2})\, \mathrm{d}x \\ &\geq \left(\frac{1}{2}-\frac{1}{\theta}\right)\|u_{n}\|^{2}_{\operatorname{\varepsilon}}-\left(\frac{1}{2}-\frac{1}{\theta}\right)\frac{1}{k}\int_{\Lambda^{c}_{\operatorname{\varepsilon}}} V_{\operatorname{\varepsilon}} (x) |u_{n}|^{2}\, \mathrm{d}x \\ &\geq \left(\frac{\theta-2}{2\theta}\right)\left( 1-\frac{1}{k} \right)\|u_{n}\|^{2}_{\operatorname{\varepsilon}}, \end{align*}

and recalling that $k>1$ and $\theta >2$, we deduce that $(u_{n})$ is bounded in $\mathcal {H}_{\operatorname {\varepsilon }}$. Since $\mathcal {H}_{\operatorname {\varepsilon }}$ is a reflexive space, we can find a subsequence, still denoted by $(u_{n})$, and $u\in \mathcal {H}_{\operatorname {\varepsilon }}$ such that

(3.10)\begin{align} \begin{array}{ll} u_{n}\rightharpoonup u \quad & \mbox{ in } \mathcal{H}_{\operatorname{\varepsilon}} \mbox{ as } n\rightarrow \infty, \\ u_{n}\rightarrow u \quad & \mbox{ in } L^{q}_{loc}(\mathbb{R}^{N}, \mathbb{C}) \mbox{ for all } q\in [1, {2^{*}_{s}}) \mbox{ as } n\rightarrow \infty, \\ |u_{n}|\rightarrow |u| \quad & \mbox{ a.e. in } \mathbb{R}^{N} \mbox{ as } n\rightarrow \infty. \end{array} \end{align}

Using $(g_1)$ and $(g_2)$, we see that

(3.11)\begin{align} \lim_{n\rightarrow \infty} \Re\int_{\mathbb{R}^{N}} g_{\operatorname{\varepsilon}}(x, |u_{n}|^{2})u_{n}\overline{\phi} \, \textrm{d}x = \Re\int_{\mathbb{R}^{N}} g_{\operatorname{\varepsilon}}(x, |u|^{2})u \overline{\phi} \, \textrm{d}x \quad \mbox{ for all } \phi \in C^{\infty}_{c}(\mathbb{R}^{N}, \mathbb{C}). \end{align}

Taking into account (3.10), (3.11) and theorem 2.1, we deduce that

\begin{align*} \langle J_{\operatorname{\varepsilon}}'(u), \phi \rangle =0 \quad \mbox{ for all } \phi \in \mathcal{H}_{\operatorname{\varepsilon}}, \end{align*}

that is $u$ is a critical point for $J_{\operatorname {\varepsilon }}$. In particular, $\langle J_{\operatorname {\varepsilon }}'(u), u \rangle =0$, or equivalently

(3.12)\begin{align} [u]^{2}_{A_{\operatorname{\varepsilon}}} + \int_{\Lambda_{\operatorname{\varepsilon}}} V_{\varepsilon}(x) |u|^{2}\, \mathrm{d}x+\int_{\Lambda^{c}_{\operatorname{\varepsilon}}} \mathcal{C}(\operatorname{\varepsilon} x, |u|^{2})\, \mathrm{d}x= \int_{\Lambda_{\operatorname{\varepsilon}}} f(|u|^{2})|u|^{2}\, \mathrm{d}x, \end{align}

where $\mathcal {C}(x, t)=V(x)t-g(x, t)t$. Note that, by $(g_3)$-(ii), it holds

(3.13)\begin{align} V(x) t\geq \mathcal{C}(x, t)\geq \left(1-\frac{1}{k}\right)V(x)t\geq 0 \quad\mbox{ for all } x\in \Lambda^{c}, \, t\geq 0. \end{align}

Recalling that $\langle J_{\operatorname {\varepsilon }}'(u_{n}), u_{n}\rangle =o_{n}(1)$, we also know that

(3.14)\begin{align} [u_{n}]^{2}_{A_{\operatorname{\varepsilon}}} + \int_{\Lambda_{\operatorname{\varepsilon}}} V_{\varepsilon}(x) |u_{n}|^{2}\, \mathrm{d}x+\int_{\Lambda^{c}_{\operatorname{\varepsilon}}} \mathcal{C}(\operatorname{\varepsilon} x, |u_{n}|^{2})\, \mathrm{d}x= \int_{\Lambda_{\operatorname{\varepsilon}}} f(|u_{n}|^{2})|u_{n}|^{2}\, \mathrm{d}x +o_{n}(1). \end{align}

Since $\Lambda _{\operatorname {\varepsilon }}$ is bounded, it follows from the local compact embeddings in theorem 2.3 that

(3.15)\begin{align} \lim_{n\rightarrow \infty} \int_{\Lambda_{\operatorname{\varepsilon}}} f(|u_{n}|^{2})|u_{n}|^{2} \,\textrm{d}x = \int_{\Lambda_{\operatorname{\varepsilon}}} f(|u|^{2})|u|^{2} \,\mathrm{d}x, \end{align}

and

(3.16)\begin{align} \lim_{n\rightarrow \infty} \int_{\Lambda_{\operatorname{\varepsilon}}} V_{\operatorname{\varepsilon}}(x) |u_{n}|^{2}\, \mathrm{d}x=\int_{\Lambda_{\operatorname{\varepsilon}}} V_{\operatorname{\varepsilon}}(x) |u|^{2}\, \mathrm{d}x. \end{align}

Putting together (3.12), (3.14), (3.15) and (3.16), we deduce that

\begin{align*} \limsup_{n\rightarrow \infty} &\left([u_{n}]^{2}_{A_{\operatorname{\varepsilon}}} + \int_{\Lambda^{c}_{\operatorname{\varepsilon}}} \mathcal{C}(\operatorname{\varepsilon} x, |u_{n}|^{2})\, \mathrm{d}x \right)=[u]^{2}_{A_{\operatorname{\varepsilon}}} + \int_{\Lambda^{c}_{\operatorname{\varepsilon}}} \mathcal{C}(\operatorname{\varepsilon} x, |u|^{2})\, \mathrm{d}x. \end{align*}

Now, by (3.13) and Fatou's lemma, we get

\begin{align*} \liminf_{n\rightarrow \infty} &\left([u_{n}]^{2}_{A_{\operatorname{\varepsilon}}} + \int_{\Lambda^{c}_{\operatorname{\varepsilon}}} \mathcal{C}(\operatorname{\varepsilon} x, |u_{n}|^{2})\, \mathrm{d}x \right)\geq [u]^{2}_{A_{\operatorname{\varepsilon}}} + \int_{\Lambda^{c}_{\operatorname{\varepsilon}}} \mathcal{C}(\operatorname{\varepsilon} x, |u|^{2})\, \mathrm{d}x. \end{align*}

Hence,

(3.17)\begin{align} \lim_{n\rightarrow \infty} [u_{n}]^{2}_{A_{\operatorname{\varepsilon}}}=[u]^{2}_{A_{\operatorname{\varepsilon}}}, \end{align}

and

\begin{align*} \lim_{n\rightarrow \infty} \int_{\Lambda^{c}_{\operatorname{\varepsilon}}} \mathcal{C}(\operatorname{\varepsilon} x, |u_{n}|^{2})\, \mathrm{d}x =\int_{\Lambda^{c}_{\operatorname{\varepsilon}}} \mathcal{C}(\operatorname{\varepsilon} x, |u|^{2})\, \mathrm{d}x. \end{align*}

The last limit, Fatou's lemma and (3.13) lead to

\begin{align*} \int_{\Lambda^{c}_{\operatorname{\varepsilon}}} V_{\operatorname{\varepsilon}}(x) |u|^{2}\, \mathrm{d}x &\leq \liminf_{n\rightarrow \infty}\int_{\Lambda^{c}_{\operatorname{\varepsilon}}} V_{\operatorname{\varepsilon}}(x) |u_{n}|^{2}\, \mathrm{d}x \\ &\leq \limsup_{n\rightarrow \infty}\int_{\Lambda^{c}_{\operatorname{\varepsilon}}} V_{\operatorname{\varepsilon}}(x) |u_{n}|^{2}\, \mathrm{d}x \\ &\leq \left(\frac{k}{k-1}\right) \limsup_{n\rightarrow \infty} \int_{\Lambda^{c}_{\operatorname{\varepsilon}}} \mathcal{C}(\operatorname{\varepsilon} x, |u_{n}|^{2})\, \mathrm{d}x \\ &=\left(\frac{k}{k-1}\right)\int_{\Lambda^{c}_{\operatorname{\varepsilon}}} \mathcal{C}(\operatorname{\varepsilon} x, |u|^{2})\, \mathrm{d}x\\ & \leq \left(\frac{k}{k-1}\right) \int_{\Lambda^{c}_{\operatorname{\varepsilon}}} V_{\operatorname{\varepsilon}}(x) |u|^{2}\, \mathrm{d}x \quad \mbox{ for all } k>1, \end{align*}

and by sending $k\rightarrow \infty$ we find

\begin{align*} \lim_{n\rightarrow \infty} \int_{\Lambda^{c}_{\operatorname{\varepsilon}}} V_{\operatorname{\varepsilon}}(x) |u_{n}|^{2}\, \textrm{d}x =\int_{\Lambda^{c}_{\operatorname{\varepsilon}}} V_{\operatorname{\varepsilon}}(x) |u|^{2}\, \mathrm{d}x \end{align*}

which combined with (3.16) gives

(3.18)\begin{align} \lim_{n\rightarrow \infty} \int_{\mathbb{R}^{N}} V_{\operatorname{\varepsilon}}(x) |u_{n}|^{2}\, \mathrm{d}x =\int_{\mathbb{R}^{N}} V_{\operatorname{\varepsilon}}(x) |u|^{2}\, \mathrm{d}x. \end{align}

Putting together (3.17) and (3.18), we obtain that

\begin{align*} \lim_{n\rightarrow \infty} \|u_{n}\|_{\operatorname{\varepsilon}}^{2} = \|u\|_{\operatorname{\varepsilon}}^{2}. \end{align*}

Since $\mathcal {H}_{\operatorname {\varepsilon }}$ is a Hilbert space and $u_{n}\rightharpoonup u$ in $\mathcal {H}_{\operatorname {\varepsilon }}$ as $n\rightarrow \infty$, we conclude that $u_{n}\rightarrow u$ in $\mathcal {H}_{\operatorname {\varepsilon }}$ as $n\rightarrow \infty$.

Proof. Let $(u_{n})\subset \mathbb {S}_{\operatorname {\varepsilon }}^{+}$ be a Palais–Smale sequence for $\psi _{\operatorname {\varepsilon }}$ at the level $c$. Then,

\[ \psi_{\operatorname{\varepsilon}}(u_{n})\rightarrow c \quad \mbox{ and } \quad \|\psi_{\operatorname{\varepsilon}}'(u_{n})\|_{*}\rightarrow 0, \]

where $\|\cdot \|_{*}$ denotes the norm in the dual space $(T_{u_{n}} \mathbb {S}_{\operatorname {\varepsilon }}^{+})^{*}$. It follows from proposition 3.3-(c) that $(m_{\operatorname {\varepsilon }}(u_{n}))\subset \mathcal {H}_{\operatorname {\varepsilon }}$ is a Palais–Smale sequence for $J_{\operatorname {\varepsilon }}$ at the level $c$. Then, using lemma 3.5, there exists $u\in \mathbb {S}_{\operatorname {\varepsilon }}^{+}$ such that, up to a subsequence,

\[ m_{\operatorname{\varepsilon}}(u_{n})\rightarrow m_{\operatorname{\varepsilon}}(u) \quad \mbox{ in } \mathcal{H}_{\operatorname{\varepsilon}}. \]

Applying lemma 3.2-(iii) we can infer that $u_{n}\rightarrow u$ in $\mathbb {S}_{\operatorname {\varepsilon }}^{+}$.

Proof. In view of lemmas 3.1 and 3.5, we can apply the mountain pass theorem [Reference Ambrosetti and Rabinowitz4] to see that for all $\operatorname {\varepsilon }>0$ there exists a nontrivial critical point $u_{\operatorname {\varepsilon }}\in \mathcal {H}_{\operatorname {\varepsilon }}$ of $J_{\operatorname {\varepsilon }}$. By remark 3.4, we deduce the thesis.

Proof. Assume that $(ii)$ does not occur. Arguing as in the proof of lemma 3.5 we see that $(u_{n})$ is bounded in $\mathcal {H}_{0}$. Then we use lemma 2.5 to deduce that $u_{n}\rightarrow 0$ in $L^{r}(\mathbb {R}^{N}, \mathbb {R})$ for all $r\in (2, {2^{*}_{s}})$. In view of $(f_1)$–$(f_2)$, we get $\int _{\mathbb {R}^{N}} f(u^{2}_{n})u^{2}_{n}\textrm {d}x=o_{n}(1)$. This fact combined with $\langle I'_{0}(u_{n}), u_{n}\rangle =o_{n}(1)$ yields $\|u_{n}\|^{2}_{0}=o_{n}(1)$.

Proof. Arguing as in the proof of lemma 3.5, we can see that $(u_{n})$ is bounded in $\mathcal {H}_{0}$. Therefore, up to going a subsequence, we may assume that $u_{n}\rightharpoonup u$ in $\mathcal {H}_{0}$. Standard arguments show that $u$ is a critical point of $I_{0}$. In light of remark 4.5, we may assume that $u$ is not trivial. Hence, $u\in \mathcal {M}_{0}$. Now we prove that $I_{0}(u)=d_{0}$. Indeed, by $u\in \mathcal {M}_{0}$, $(f_3)$ and Fatou's lemma, we have

\begin{align*} d_{0}&\leq I_{0}(u)-\frac{1}{\theta}\langle I'_{0}(u), u\rangle\\ &\leq \liminf_{n\rightarrow \infty} \left[\left(\frac{1}{2}-\frac{1}{\theta}\right)\|u_{n}\|_{0}^{2}+\int_{\mathbb{R}^{N}} \frac{1}{\theta} f(u^{2}_{n})u^{2}_{n}- \frac{1}{2} F(u^{2}_{n}) \,\textrm{d}x\right] \\ &=\liminf_{n\rightarrow \infty} \left[I_{0}(u_{n})-\frac{1}{\theta}\langle I'_{0}(u_{n}), u_{n}\rangle\right] \\ &=d_{0}. \end{align*}

Since $\langle I'_{0}(u), u^{-}\rangle =0$, where $u^{-}=\min \{u, 0\}$, and using $(f_1)$, we can see that $u\geq 0$ in $\mathbb {R}^{N}$. Performing a standard Moser iteration argument (see proposition $5.1.1$ in [Reference Dipierro, Medina and Valdinoci26]) we see that $u\in L^{\infty }(\mathbb {R}^{N}, \mathbb {R})$ and applying proposition $2.9$ in [Reference Silvestre44] we infer that $u\in C^{0, \alpha }(\mathbb {R}^{N}, \mathbb {R})$ for some $\alpha \in (0, 1)$. By the strong maximum principle (see theorem $1.4$ in [Reference Del Pezzo and Quaas22]), we deduce that $u>0$ in $\mathbb {R}^{N}$.

Proof. Since $(u_{n})\subset \mathcal {M}_{0}$ and $I_{0}(u_{n})\rightarrow d_{0}$, we can apply lemma 4.1-(iii), proposition 4.2-(d) and the definition of $d_{0}$ to infer that

\[ \nu_{n}=m^{{-}1}_{0}(u_{n})=\displaystyle\frac{u_{n}}{\|u_{n}\|_{0}}\in \mathbb{S}_{0}^{+} \]

and

\[ \psi_{0}(\nu_{n})=I_{0}(u_{n})\rightarrow d_{0}=\inf_{v\in \mathbb{S}_{0}^{+}}\psi_{0}(v). \]

Let us introduce the following map $\mathcal {F}: \overline {\mathbb {S}}_{0}^{+}\rightarrow \mathbb {R}\cup \{\infty \}$ defined by setting

\[ \mathcal{F}(u)= \begin{cases} \psi_{0}(u)& \text{ if } u\in \mathbb{S}_{0}^{+}, \\ \infty & \text{ if } u\in \partial \mathbb{S}_{0}^{+}.\end{cases} \]

We note that

• • $(\overline {\mathbb {S}}_{0}^{+}, \delta _{0})$, where $\delta _{0}(u, v)=\|u-v\|_{0}$, is a complete metric space;

• • $\mathcal {F}\in C(\overline {\mathbb {S}}_{0}^{+}, \mathbb {R}\cup \{\infty \})$, by lemma 4.1-(iv);

• • $\mathcal {F}$ is bounded below, by proposition 4.2-(d).

By applying Ekeland's variational principle [Reference Ekeland27], we can find a Palais–Smale sequence $(\hat {v}_{n})\subset \mathbb {S}_{0}^{+}$ for $\psi _{0}$ at the level $d_{0}$ and such that $\|\hat {v}_{n}-\nu _{n}\|_{0}=o_{n}(1)$. Now the remainder of the proof follows from proposition 4.2, theorem 4.6 and arguing as in the proof of corollary 3.6.

Proof. Let $w\in H^{s}(\mathbb {R}^{N}, \mathbb {R})$ be a positive ground state to (4.1), so that $I'_{0}(w)=0$ and $I_{0}(w)=d_{0}$, and let $\eta \in C^{\infty }_{c}(\mathbb {R}^{N}, [0,1])$ be a cut-off function such that $\eta =1$ in $B_{{\delta }/{2}}$ and $\operatorname {supp}(\eta )\subset B_{\delta }\subset \Lambda$ for some $\delta >0$. We recall that the existence of $w$ is guaranteed by theorem 4.6. Moreover, from theorem $1.5$ in [Reference Felmer, Quaas and Tan30], we know that $w$ satisfies the following power-type decay estimate:

(4.2)\begin{equation} 0< w(x)\leq \frac{C}{|x|^{N+2s}} \quad \mbox{ for all } |x|>1. \end{equation}

Let us define $w_{\operatorname {\varepsilon }}(x)=\eta _{\operatorname {\varepsilon }}(x)w(x) \textrm {e}^{\imath A(0)\cdot x}$, with $\eta _{\operatorname {\varepsilon }}(x)=\eta (\operatorname {\varepsilon } x)$ for $\operatorname {\varepsilon }>0$, and we observe that $|w_{\operatorname {\varepsilon }}|=\eta _{\operatorname {\varepsilon }}w$ and $w_{\operatorname {\varepsilon }}\in \mathcal {H}_{\operatorname {\varepsilon }}$ in light of lemma 2.4. Next we claim that

(4.3)\begin{equation} \lim_{\operatorname{\varepsilon}\rightarrow 0}\|w_{\operatorname{\varepsilon}}\|^{2}_{\operatorname{\varepsilon}}=\|w\|_{0}^{2}\in(0, \infty). \end{equation}

Clearly, $\int _{\mathbb {R}^{N}} V_{\operatorname {\varepsilon }}(x)|w_{\operatorname {\varepsilon }}|^{2}\textrm {d}x\rightarrow \int _{\mathbb {R}^{N}} V_{0} |w|^{2}\textrm {d}x$. Then, it remains to show that

(4.4)\begin{equation} \lim_{\operatorname{\varepsilon}\rightarrow 0}[w_{\operatorname{\varepsilon}}]^{2}_{A_{\operatorname{\varepsilon}}}=[w]^{2}. \end{equation}

Using lemma $5$ in [Reference Palatucci and Pisante42], we know that

(4.5)\begin{equation} [\eta_{\operatorname{\varepsilon}} w]\rightarrow [w] \mbox{ as } \operatorname{\varepsilon}\rightarrow 0. \end{equation}

On the other hand

\begin{align*} [w_{\operatorname{\varepsilon}}]_{A_{\operatorname{\varepsilon}}}^{2} &=\iint_{\mathbb{R}^{2N}} \frac{|\textrm{e}^{\imath A(0)\cdot x}\eta_{\operatorname{\varepsilon}}(x)w(x)-\textrm{e}^{\imath A_{\operatorname{\varepsilon}}(\frac{x+y}{2})\cdot (x-y)}\textrm{e}^{\imath A(0)\cdot y} \eta_{\operatorname{\varepsilon}}(y)w(y)|^{2}}{|x-y|^{N+2s}} \mathrm{d}x\,\mathrm{d}y \nonumber \\ &=[\eta_{\operatorname{\varepsilon}} w]^{2} +\iint_{\mathbb{R}^{2N}} \frac{\eta_{\operatorname{\varepsilon}}^{2}(y)w^{2}(y) |\textrm{e}^{\imath [A_{\operatorname{\varepsilon}}(\frac{x+y}{2})-A(0)]\cdot (x-y)}-1|^{2}}{|x-y|^{N+2s}} \,\mathrm{d}x \,\mathrm{d}y+2\Re \iint_{\mathbb{R}^{2N}}\\ &\quad\times \frac{(\eta_{\operatorname{\varepsilon}}(x)w(x)-\eta_{\operatorname{\varepsilon}}(y)w(y))\eta_{\operatorname{\varepsilon}}(y)w(y)(1-\textrm{e}^{-\imath [A_{\operatorname{\varepsilon}}(\frac{x+y}{2})-A(0)]\cdot (x-y)})}{|x-y|^{N+2s}} \,\mathrm{d}x\,\mathrm{ d}y \\ &= [\eta_{\operatorname{\varepsilon}} w]^{2}+X_{\operatorname{\varepsilon}}+2Y_{\operatorname{\varepsilon}}. \end{align*}

Then, in view of $|Y_{\operatorname {\varepsilon }}|\leq [\eta _{\operatorname {\varepsilon }} w] \sqrt {X_{\operatorname {\varepsilon }}}$ and (4.5), it is suffices to prove that $X_{\operatorname {\varepsilon }}\rightarrow 0$ as $\operatorname {\varepsilon }\rightarrow 0$ to deduce that (4.4) is satisfied. Let us note that for $0<\beta <\alpha /({1+\alpha -s})$,

(4.6)\begin{equation} \begin{split} X_{\operatorname{\varepsilon}} & \leq \int_{\mathbb{R}^{N}} w^{2}(y) \textrm{d}y \int_{|x-y|\geq\operatorname{\varepsilon}^{-\beta}} \frac{|\textrm{e}^{\imath [A_{\operatorname{\varepsilon}}(\frac{x+y}{2})-A(0)]\cdot (x-y)}-1|^{2}}{|x-y|^{N+2s}} \mathrm{d}x\\ & \quad +\int_{\mathbb{R}^{N}} w^{2}(y) \textrm{d}y \int_{|x-y|<\operatorname{\varepsilon}^{-\beta}} \frac{|\textrm{e}^{\imath [A_{\operatorname{\varepsilon}}(\frac{x+y}{2})-A(0)]\cdot (x-y)}-1|^{2}}{|x-y|^{N+2s}} \mathrm{d}x\\ & =X^{1}_{\operatorname{\varepsilon}}+X^{2}_{\operatorname{\varepsilon}}. \end{split} \end{equation}

Using $|\textrm {e}^{\imath t}-1|^{2}\leq 4$ and $w\in H^{s}(\mathbb {R}^{N}, \mathbb {R})$, we get

(4.7)\begin{equation} X_{\operatorname{\varepsilon}}^{1}\leq C \int_{\mathbb{R}^{N}} w^{2}(y) \mathrm{d}y \int_{\operatorname{\varepsilon}^{-\beta}}^{\infty} \rho^{{-}1-2s} \mathrm{d}\rho\leq C \operatorname{\varepsilon}^{2\beta s} \rightarrow 0. \end{equation}

Since $|\textrm {e}^{\imath t}-1|^{2}\leq t^{2}$ for all $t\in \mathbb {R}$, $A\in C^{0,\alpha }(\mathbb {R}^{N},\mathbb {R}^{N})$ with $\alpha \in (0,1]$, and $|x+y|^{2}\leq 2(|x-y|^{2}+4|y|^{2})$, we have

(4.8)\begin{equation} \begin{split} X^{2}_{\operatorname{\varepsilon}} & \leq \int_{\mathbb{R}^{N}} w^{2}(y)\, \mathrm{d}y \int_{|x-y|<\operatorname{\varepsilon}^{-\beta}} \frac{|A_{\operatorname{\varepsilon}}\left(\frac{x+y}{2}\right)-A(0)|^{2} }{|x-y|^{N+2s-2}} \mathrm{d}x \\ & \leq C\operatorname{\varepsilon}^{2\alpha} \int_{\mathbb{R}^{N}} w^{2}(y)\, \mathrm{d}y \int_{|x-y|<\operatorname{\varepsilon}^{-\beta}} \frac{|x+y|^{2\alpha} }{|x-y|^{N+2s-2}} \mathrm{d}x \\ & \leq C\operatorname{\varepsilon}^{2\alpha} \left(\int_{\mathbb{R}^{N}} w^{2}(y)\, \mathrm{d}y \int_{|x-y|<\operatorname{\varepsilon}^{-\beta}} \frac{1 }{|x-y|^{N+2s-2-2\alpha}} \textrm{d}x\right.\\ & \qquad \qquad+ \left. \int_{\mathbb{R}^{N}} |y|^{2\alpha} w^{2}(y)\, \mathrm{d}y \int_{|x-y|<\operatorname{\varepsilon}^{-\beta}} \frac{1}{|x-y|^{N+2s-2}} \textrm{d}x\right) \\ & = C\operatorname{\varepsilon}^{2\alpha} (X^{2, 1}_{\operatorname{\varepsilon}}+X^{2, 2}_{\operatorname{\varepsilon}}).\end{split} \end{equation}

Then

(4.9)\begin{equation} X^{2, 1}_{\operatorname{\varepsilon}} = C \int_{\mathbb{R}^{N}} w^{2}(y)\, \mathrm{d}y \int_0^{\operatorname{\varepsilon}^{-\beta}} \rho^{1+2\alpha-2s}\, \mathrm{d}\rho \leq C\operatorname{\varepsilon}^{{-}2\beta(1+\alpha-s)}. \end{equation}

On the other hand, using (4.2), we infer that

(4.10)\begin{equation} \begin{split} X^{2, 2}_{\operatorname{\varepsilon}} & \leq C \int_{\mathbb{R}^{N}} |y|^{2\alpha} w^{2}(y)\, \mathrm{d}y \int_0^{\operatorname{\varepsilon}^{-\beta}}\rho^{1-2s} \, \mathrm{d}\rho \\ & \leq C \operatorname{\varepsilon}^{{-}2\beta(1-s)} \left[\int_{B_1} w^{2}(y) \, \mathrm{d}y + \int_{\mathbb{R}^{N}\setminus B_{1}} \frac{1}{|y|^{2(N+2s)-2\alpha}} \, \mathrm{d}y \right] \\ & \leq C \operatorname{\varepsilon}^{{-}2\beta(1-s)}. \end{split} \end{equation}

Taking into account (4.6), (4.7), (4.8), (4.9) and (4.10) we can conclude that $X_{\operatorname {\varepsilon }}\rightarrow 0$. Therefore (4.3) holds. Now, let $t_{\operatorname {\varepsilon }}>0$ be the unique number such that

\[ J_{\operatorname{\varepsilon}}(t_{\operatorname{\varepsilon}} w_{\operatorname{\varepsilon}})=\max_{t\geq 0} J_{\operatorname{\varepsilon}}(t w_{\operatorname{\varepsilon}}). \]

Then $t_{\operatorname {\varepsilon }}$ satisfies

(4.11)\begin{equation} \|w_{\operatorname{\varepsilon}}\|_{\operatorname{\varepsilon}}^{2}=\int_{\mathbb{R}^{N}} g_{\operatorname{\varepsilon}}(x, t_{\operatorname{\varepsilon}}^{2} |w_{\operatorname{\varepsilon}}|^{2}) |w_{\operatorname{\varepsilon}}|^{2}\, \mathrm{d}x=\int_{\mathbb{R}^{N}} f(t_{\operatorname{\varepsilon}}^{2} |w_{\operatorname{\varepsilon}}|^{2}) |w_{\operatorname{\varepsilon}}|^{2}\, \mathrm{d}x \end{equation}

where we used $\operatorname {supp}(\eta )\subset \Lambda$ and $g=f$ on $\Lambda \times \mathbb {R}$. Let us prove that $t_{\operatorname {\varepsilon }}\rightarrow 1$ as $\operatorname {\varepsilon }\rightarrow 0$. Using $\eta =1$ in $B_{{\delta }/{2}}$ and $(f_4)$ we find

\[ \|w_{\operatorname{\varepsilon}}\|_{\operatorname{\varepsilon}}^{2}\geq f(t_{\operatorname{\varepsilon}}^{2}\alpha^{2}_{0})\int_{B_{{\delta}/{2}}}|w|^{2}\,\mathrm{d}x, \]

where $\alpha _{0}=\min _{\overline {B}_{\delta /2}} w>0$ (remark that $w$ is a continuous positive function). So, if $t_{\operatorname {\varepsilon }}\rightarrow \infty$ as $\operatorname {\varepsilon }\rightarrow 0$, then we can use $(f_3)$ to deduce that $\|w\|_{0}^{2}= \infty$, which gives a contradiction by (4.3). On the other hand, if $t_{\operatorname {\varepsilon }}\rightarrow 0$ as $\operatorname {\varepsilon }\rightarrow 0$, we can exploit the growth assumptions on $f$ and (4.11) to infer that $\|w\|_{0}^{2}= 0$ which is in contrast with (4.3). In conclusion, $t_{\operatorname {\varepsilon }}\rightarrow t_{0}\in (0, \infty )$ as $\operatorname {\varepsilon }\rightarrow 0$. Now, taking the limit as $\operatorname {\varepsilon }\rightarrow 0$ in (4.11) and using (4.3), we can see that

(4.12)\begin{equation} \|w\|_{0}^{2}=\int_{\mathbb{R}^{N}} f(t_{0}^{2}w^{2}) w^{2}\,\textrm{d}x. \end{equation}

In view of $w\in \mathcal {M}_{0}$ and $(f_4)$, we have that $t_{0}=1$. Then, by (4.3), $t_{\operatorname {\varepsilon }}\rightarrow 1$ and applying the dominated convergence theorem, we obtain that $\lim _{\operatorname {\varepsilon }\rightarrow 0} J_{\operatorname {\varepsilon }}(t_{\operatorname {\varepsilon }} w_{\operatorname {\varepsilon }})=I_{0}(w)=d_{0}$. Since $c_{\operatorname {\varepsilon }}\leq \max _{t\geq 0} J_{\operatorname {\varepsilon }}(t w_{\operatorname {\varepsilon }})=J_{\operatorname {\varepsilon }}(t_{\operatorname {\varepsilon }} w_{\operatorname {\varepsilon }})$, we can conclude that $\limsup _{\operatorname {\varepsilon }\rightarrow 0} c_{\operatorname {\varepsilon }}\leq d_{0}$.

Proof. Assume by contradiction that there exist $\delta _{0}>0$, $(y_{n})\subset M$ and $\operatorname {\varepsilon }_{n}\rightarrow 0$ such that

(5.1)\begin{equation} |J_{\operatorname{\varepsilon}_{n}}(\Phi_{\operatorname{\varepsilon}_{n}}(y_{n}))-d_{0}|\geq \delta_{0}. \end{equation}

Applying lemma $4.1$ in [Reference Ambrosio and d'Avenia13] and the dominated convergence theorem, we see that

(5.2)\begin{align} \begin{split} & \| \Psi_{\operatorname{\varepsilon}_{n}, y_{n}} \|^{2}_{\operatorname{\varepsilon}_{n}}\rightarrow \|w\|^{2}_{0}\in (0, \infty). \end{split} \end{align}

On the other hand, since $\langle J'_{\operatorname {\varepsilon }_{n}}(\Phi _{\operatorname {\varepsilon }_{n}}(y_{n})),\Phi _{\operatorname {\varepsilon }_{n}}(y_{n})\rangle =0$ and using the change of variable $z=({\operatorname {\varepsilon }_{n}x-y_{n}})/{\operatorname {\varepsilon }_{n}}$, it follows that

\begin{align*} t_{\operatorname{\varepsilon}_{n}}^{2}\|\Psi_{\operatorname{\varepsilon}_{n}, y_{n}}\|_{\operatorname{\varepsilon}_{n}}^{2}=\int_{\mathbb{R}^{N}} g(\operatorname{\varepsilon}_{n}z+y_{n}, |t_{\operatorname{\varepsilon}_{n}}\eta(|\operatorname{\varepsilon}_{n}z|)w(z)|^{2}) |t_{\operatorname{\varepsilon}_{n}}\eta(|\operatorname{\varepsilon}_{n}z|)w(z)|^{2} \mathrm{d}z. \end{align*}

If $z\in B_{{\delta }/{\operatorname {\varepsilon }_{n}}}$, then $\operatorname {\varepsilon }_{n} z+y_{n}\in B_{\delta }(y_{n})\subset M_{\delta }\subset \Lambda$. Since $g(x,t)=f(t)$ for $(x, t)\in \Lambda \times [0, \infty )$, we have

(5.3)\begin{align} t_{\operatorname{\varepsilon}_{n}}^{2}\|\Psi_{\operatorname{\varepsilon}_{n}, y_{n}}\|_{\operatorname{\varepsilon}_{n}}^{2}=\int_{\mathbb{R}^{N}} f(|t_{\operatorname{\varepsilon}_{n}}\eta(|\operatorname{\varepsilon}_{n}z|)w(z)|^{2}) (t_{\operatorname{\varepsilon}_{n}}\eta(|\operatorname{\varepsilon}_{n}z|)w(z))^{2} \, \mathrm{d}z. \end{align}

In view of $\eta (|x|)=1$ for $x\in B_{{\delta }/{2}}$ and that $B_{{\delta }/{2}}\subset B_{{\delta }/{2\operatorname {\varepsilon }_{n}}}$ for all $n$ large enough, it follows from (5.3) and $(f_4)$ that

(5.4)\begin{align} \|\Psi_{\operatorname{\varepsilon}_{n}, y_{n}}\|_{\operatorname{\varepsilon}_{n}}^{2} &\geq \int_{B_{{\delta}/{2}}} \frac{f(|t_{\operatorname{\varepsilon}_{n}}w(z)|^{2}) (t_{\operatorname{\varepsilon}_{n}}w(z))^{2}}{t^{2}_{\operatorname{\varepsilon}_{n}}}\mathrm{d}z \nonumber\\ &\geq f(|t_{\operatorname{\varepsilon}_{n}}w(\hat{z})|^{2}) \int_{B_{{\delta}/{2}}} w^{2}(z)\,\mathrm{d}z, \end{align}

where

\[ w(\hat{z})=\min_{z\in \overline{B}_{{\delta}/{2}}} w(z)>0. \]

Now, assume by contradiction that $t_{\operatorname {\varepsilon }_{n}}\rightarrow \infty$. This fact, (5.2) and (5.4) yield

\[ \|w\|^{2}_{0}=\infty, \]

that is a contradiction. Hence, $(t_{\operatorname {\varepsilon }_{n}})$ is bounded and, up to subsequence, we may assume that $t_{\operatorname {\varepsilon }_{n}}\rightarrow t_{0}$ for some $t_{0}\geq 0$. In particular, $t_{0}>0$. In fact, if $t_{0}=0$, we see that (3.4) and (5.3) imply that

\[ r\leq \int_{\mathbb{R}^{3}} f(|t_{\operatorname{\varepsilon}_{n}}\eta(|\operatorname{\varepsilon}_{n}z|)w(z)|^{2}) (t_{\operatorname{\varepsilon}_{n}}\eta(|\operatorname{\varepsilon}_{n}z|)w(z))^{2}\, \textrm{d}z. \]

Using $(f_1)$, $(f_2)$, (5.2) and the above inequality, we deduce that $t_{0}>0$. Thus, letting $n\rightarrow \infty$ in (5.3), we have that

\begin{align*} \|w\|^{2}_{{0}}=\int_{\mathbb{R}^{N}} f((t_{0} w)^{2}) w^{2} \, \mathrm{d}x. \end{align*}

Bearing in mind that $w\in \mathcal {M}_{{0}}$ and using $(f_4)$, we infer that $t_{0}=1$. Passing to the limit as $n\rightarrow \infty$ and using $t_{\operatorname {\varepsilon }_{n}}\rightarrow 1$ we conclude that

\[ \lim_{n\rightarrow \infty} J_{\operatorname{\varepsilon}_{n}}(\Phi_{\operatorname{\varepsilon}_{n}, y_{n}})=I_{0}(w)=d_{0}, \]

which is in contrast with (5.1).

Proof. Suppose by contradiction that there exist $\delta _{0}>0$, $(y_{n})\subset M$ and $\operatorname {\varepsilon }_{n}\rightarrow 0$ such that

(5.5)\begin{equation} |\beta_{\operatorname{\varepsilon}_{n}}(\Phi_{\operatorname{\varepsilon}_{n}}(y_{n}))-y_{n}|\geq \delta_{0}. \end{equation}

Using the definitions of $\Phi _{\operatorname {\varepsilon }_{n}}(y_{n})$, $\beta _{\operatorname {\varepsilon }_{n}}$, $\psi$ and the change of variable $z= ({\operatorname {\varepsilon }_{n} (x-y)_{n}})/{\operatorname {\varepsilon }_{n}}$, we can see that

\[ \beta_{\operatorname{\varepsilon}_{n}}(\Phi_{\operatorname{\varepsilon}_{n}}(y_{n}))=y_{n}+\displaystyle\frac{\int_{\mathbb{R}^{N}}[\varUpsilon(\operatorname{\varepsilon}_{n}z+y_{n})-y_{n}] |\eta(|\operatorname{\varepsilon}_{n}z|) w(z)|^{2}\, \mathrm{d}z}{\int_{\mathbb{R}^{N}} |\eta(|\operatorname{\varepsilon}_{n}z|) w(z)|^{2}\, \mathrm{d}z}. \]

Since $(y_{n})\subset M\subset B_{\rho }$, it follows from the dominated convergence theorem that

\[ |\beta_{\operatorname{\varepsilon}_{n}}(\Phi_{\operatorname{\varepsilon}_{n}}(y_{n}))-y_{n}|=o_{n}(1) \]

which is in contrast with (5.5).

Proof. Taking into account $\langle J'_{\operatorname {\varepsilon }_{n}}(u_{n}), u_{n}\rangle =0$, $J_{\operatorname {\varepsilon }_{n}}(u_{n})\rightarrow d_{0}$, it is easy to see that $(u_{n})$ is bounded in $\mathcal {H}_{\operatorname {\varepsilon }_{n}}$. Then, there exists $C>0$ (independent of $n$) such that $\|u_{n}\|_{\operatorname {\varepsilon }_{n}}\leq C$ for all $n\in \mathbb {N}$. Moreover, from lemma 2.2, we also know that $(|u_{n}|)$ is bounded in $H^{s}(\mathbb {R}^{N}, \mathbb {R})$.

Now we prove that there exist a sequence $(\tilde {y}_{n})\subset \mathbb {R}^{N}$ and constants $R, \gamma >0$ such that

(5.6)\begin{equation} \liminf_{n\rightarrow \infty}\int_{B_{R}(\tilde{y}_{n})} |u_{n}|^{2} \, \mathrm{d}x\geq \gamma>0. \end{equation}

If by contradiction (5.6) does not hold, then for all $R>0$ we get

\[ \lim_{n\rightarrow \infty}\sup_{y\in \mathbb{R}^{N}}\int_{B_{R}(y)} |u_{n}|^{2} \, \mathrm{d}x=0. \]

From the boundedness of $(|u_{n}|)$ in $H^{s}(\mathbb {R}^{N}, \mathbb {R})$ and lemma 2.5, we can see that $|u_{n}|\rightarrow 0$ in $L^{q}(\mathbb {R}^{N}, \mathbb {R})$ for any $q\in (2, 2^{*}_{s})$. This fact together with (3.5) and the boundedness of $(|u_{n}|)$ in $L^{2}(\mathbb {R}^{N}, \mathbb {R})$ yields that

(5.7)\begin{align} \lim_{n\rightarrow \infty}\int_{\mathbb{R}^{N}} g_{\operatorname{\varepsilon}_{n}} (x, |u_{n}|^{2}) |u_{n}|^{2} \,\textrm{d}x=0= \lim_{n\rightarrow \infty}\int_{\mathbb{R}^{N}} G_{\operatorname{\varepsilon}_{n}}(x, |u_{n}|^{2}) \, \mathrm{d}x. \end{align}

Taking into account $\langle J'_{\operatorname {\varepsilon }_{n}}(u_{n}), u_{n}\rangle =0$ and (5.7), we can infer that $\|u_{n}\|_{\operatorname {\varepsilon }_{n}}\rightarrow 0$ as $n\rightarrow \infty$ and this implies that $J_{\operatorname {\varepsilon }_{n}}(u_{n})\rightarrow 0$ which is in contrast with $d_{0}>0$. Now, we set $v_{n}(x)=|u_{n}|(x+\tilde {y}_{n})$. Then $(v_{n})$ is bounded in $H^{s}(\mathbb {R}^{N}, \mathbb {R})$, and we may assume that $v_{n}\rightharpoonup v\not \equiv 0$ in $H^{s}(\mathbb {R}^{N}, \mathbb {R})$ as $n\rightarrow \infty$. Fix $t_{n}>0$ such that $\tilde {v}_{n}=t_{n} v_{n}\in \mathcal {M}_{0}$. By lemma 2.2, $(u_{n})\subset \mathcal {N}_{\operatorname {\varepsilon }_{n}}$ and $J_{\operatorname {\varepsilon }_{n}}(u_{n})\rightarrow d_{0}$, we can see that

\[ d_{0}\leq I_{0}(\tilde{v}_{n})\leq \max_{t\geq 0}J_{\operatorname{\varepsilon}_{n}}(tu_{n})= J_{\operatorname{\varepsilon}_{n}}(u_{n})=d_{0}+o_{n}(1) \]

which implies that $I_{0}(\tilde {v}_{n})\rightarrow d_{0}$. In particular, $(\tilde {v}_{n})$ is bounded in $H^{s}(\mathbb {R}^{N}, \mathbb {R})$ and $\tilde {v}_{n}\rightharpoonup \tilde {v}$ in $H^{s}(\mathbb {R}^{N}, \mathbb {R})$. Since $v_{n}\nrightarrow 0$ in $H^{s}(\mathbb {R}^{N}, \mathbb {R})$ and $(\tilde {v}_{n})$ is bounded in $H^{s}(\mathbb {R}^{N}, \mathbb {R})$, we deduce that $(t_{n})$ is bounded in $\mathbb {R}$ and, up to a subsequence, we can assume that $t_{n}\rightarrow t^{*}\geq 0$. Indeed $t^{*}>0$ due to the boundedness of $(v_{n})$ in $H^{s}(\mathbb {R}^{N}, \mathbb {R})$, and the fact that $\tilde {v}_{n}\nrightarrow 0$ in $H^{s}(\mathbb {R}^{N}, \mathbb {R})$ (otherwise, by the continuity of $I_{0}$, $I_{0}(\tilde {v}_{n})\rightarrow 0$ which is impossible because $d_{0}>0$). Then, from the uniqueness of the weak limit, we have that $\tilde {v}_{n}\rightharpoonup \tilde {v}=t^{*}v\not \equiv 0$ in $H^{s}(\mathbb {R}^{N}, \mathbb {R})$. This fact combined with lemma 4.7 yields

(5.8)\begin{equation} \tilde{v}_{n}\rightarrow \tilde{v} \quad \mbox{ in } H^{s}(\mathbb{R}^{N}, \mathbb{R}). \end{equation}

Consequently, $v_{n}\rightarrow v$ in $H^{s}(\mathbb {R}^{N}, \mathbb {R})$ as $n\rightarrow \infty$. Moreover,

(5.9)\begin{align} I_{0}(\tilde{v})=0 \mbox{ and } \langle I'_{0}(\tilde{v}), \tilde{v}\rangle=0. \end{align}

Now, we put $y_{n}=\operatorname {\varepsilon }_{n}\tilde {y}_{n}$ and we claim that $(y_{n})$ admits a subsequence, still denoted by $(y_{n})$, such that $y_{n}\rightarrow y_{0}$ for some $y_{0}\in M$. First, we prove that $(y_{n})$ is bounded in $\mathbb {R}^{N}$. Assume by contradiction that, up to a subsequence, $|y_{n}|\rightarrow \infty$ as $n\rightarrow \infty$. Take $R>0$ such that $\Lambda \subset B_{R}$. Since we may suppose that $|y_{n}|>2R$ for $n$ large, we have that for any $z\in B_{R/\operatorname {\varepsilon }_{n}}$

\[ |\operatorname{\varepsilon}_{n}z+y_{n}|\geq |y_{n}|-|\operatorname{\varepsilon}_{n}z|>R. \]

Hence, using $(u_{n})\subset \mathcal {N}_{\operatorname {\varepsilon }_{n}}$, $(V_{1})$, lemma 2.2 and the change of variable $x\mapsto z+\tilde {y}_{n}$, we obtain that

(5.10)\begin{align} [v_{n}]^{2}+\int_{\mathbb{R}^{N}} V_{0} v_{n}^{2}\, \mathrm{d}x &\leq \int_{\mathbb{R}^{N}} g(\operatorname{\varepsilon}_{n} x+y_{n}, |v_{n}|^{2}) |v_{n}|^{2} \, \mathrm{d}x \nonumber\\ &\leq \int_{B_{{R}/{\operatorname{\varepsilon}_{n}}}} \tilde{f}(|v_{n}|^{2}) |v_{n}|^{2} \, \mathrm{d}x+\int_{\mathbb{R}^{N}\setminus B_{{R}/{\operatorname{\varepsilon}_{n}}}} f(|v_{n}|^{2}) |v_{n}|^{2} \, \mathrm{d}x. \end{align}

Recalling that $v_{n}\rightarrow v$ in $H^{s}(\mathbb {R}^{N}, \mathbb {R})$ as $n\rightarrow \infty$ and that $\tilde {f}(t)\leq {V_{0}}/{k}$, we can see that (5.10) and the dominated convergence theorem yield

\[ \min\left\{1, V_{0}\left(1-\displaystyle\frac{1}{k}\right) \right\} \left([v_{n}]^{2}+\int_{\mathbb{R}^{N}} |v_{n}|^{2}\, \textrm{d}x\right)=o_{n}(1), \]

that is $v_{n}\rightarrow 0$ in $H^{s}(\mathbb {R}^{N}, \mathbb {R})$, which gives a contradiction. Therefore, $(y_{n})$ is bounded and we may assume that $y_{n}\rightarrow y_{0}\in \mathbb {R}^{N}$. If $y_{0}\notin \overline {\Lambda }$, then we can argue as before to infer that $v_{n}\rightarrow 0$ in $H^{s}(\mathbb {R}^{N}, \mathbb {R})$, which is impossible. Hence $y_{0}\in \overline {\Lambda }$. Let us note that if $V(y_{0})=V_{0}$, then we can infer that $y_{0}\notin \partial \Lambda$ in view of $(V_2)$. Therefore, it is enough to verify that $V(y_{0})=V_{0}$ to deduce that $y_{0}\in M$. Suppose by contradiction that $V(y_{0})>V_{0}$. Then, using (5.8), Fatou's lemma, the invariance of $\mathbb {R}^{N}$ by translations and lemma 2.2, we get

\begin{align*} d_{0}=I_{0}(\tilde{v})&<\frac{1}{2}[\tilde{v}]^{2}+\frac{1}{2}\int_{\mathbb{R}^{N}} V(y_{0})\tilde{v}^{2} \, \textrm{d}x-\frac{1}{2}\int_{\mathbb{R}^{N}} F(|\tilde{v}|^{2})\, \mathrm{d}x\\ &\leq \liminf_{n\rightarrow \infty}\left[\frac{1}{2}[\tilde{v}_{n}]^{2}+\frac{1}{2}\int_{\mathbb{R}^{N}} V(\operatorname{\varepsilon}_{n}x+y_{n}) |\tilde{v}_{n}|^{2} \, \mathrm{d}x-\frac{1}{2}\int_{\mathbb{R}^{N}} F(|\tilde{v}_{n}|^{2})\, \mathrm{d}x \right] \\ &\leq \liminf_{n\rightarrow \infty}\left[\frac{t_{n}^{2}}{2}[|u_{n}|]^{2}+\frac{t_{n}^{2}}{2}\int_{\mathbb{R}^{N}} V(\operatorname{\varepsilon}_{n}z) |u_{n}|^{2} \, \mathrm{d}z-\frac{1}{2}\int_{\mathbb{R}^{N}} F(|t_{n} u_{n}|^{2})\, \mathrm{d}z \right] \\ &\leq \liminf_{n\rightarrow \infty} J_{\operatorname{\varepsilon}_{n}}(t_{n} u_{n}) \leq \liminf_{n\rightarrow \infty} J_{\operatorname{\varepsilon}_{n}}(u_{n})\leq d_{0} \end{align*}

which is impossible. This ends the proof of the lemma.

Proof. Let $(\operatorname {\varepsilon }_{n})\subset (0, \infty )$ such that $\operatorname {\varepsilon }_{n}\rightarrow 0$. Then there exists $(u_{n})\subset \widetilde {\mathcal {N}}_{\operatorname {\varepsilon }_{n}}$ such that

\[ \sup_{u\in \widetilde{\mathcal{N}}_{\operatorname{\varepsilon}_{n}}} \inf_{y\in M_{\delta}}|\beta_{\operatorname{\varepsilon}_{n}}(u)-y|=\inf_{y\in M_{\delta}}|\beta_{\operatorname{\varepsilon}_{n}}(u_{n})-y|+o_{n}(1). \]

Hence, it is enough to find a sequence $(y_{n})\subset M_{\delta }$ such that

(5.11)\begin{equation} \lim_{n\rightarrow \infty} |\beta_{\operatorname{\varepsilon}_{n}}(u_{n})-y_{n}|=0. \end{equation}

By lemma 2.2 we know that $I_{0}(t |u_{n}|)\leq J(tu_{n})$ for all $t\geq 0$. Therefore, recalling that $(u_{n})\subset \widetilde {\mathcal {N}}_{\operatorname {\varepsilon }_{n}}\subset \mathcal {N}_{\operatorname {\varepsilon }_{n}}$, we deduce that

\[ d_{0}\leq \max_{t\geq 0} I_{0}(t|u_{n}|)\leq \max_{t\geq 0} J_{\operatorname{\varepsilon}_{n}}(t u_{n})=J_{\operatorname{\varepsilon}_{n}}(u_{n})\leq d_{0}+h(\operatorname{\varepsilon}_{n}) \]

which leads to $J_{\operatorname {\varepsilon }_{n}}(u_{n})\rightarrow d_{0}$. By invoking lemma 5.3, we can find $(\tilde {y}_{n})\subset \mathbb {R}^{N}$ such that $y_{n}=\operatorname {\varepsilon }_{n}\tilde {y}_{n}\in M_{\delta }$ for $n$ sufficiently large. Consequently,

\[ \beta_{\operatorname{\varepsilon}_{n}}(u_{n})=y_{n}+\displaystyle\frac{{\int_{\mathbb{R}^{N}}[\varUpsilon(\operatorname{\varepsilon}_{n}z+y_{n})-y_{n}] |u_{n}(z+\tilde{y}_{n})|^{2} \, \mathrm{d}z}}{{\int_{\mathbb{R}^{N}} |u_{n}(z+\tilde{y}_{n})|^{2} \, \mathrm{d}z}}. \]

Taking into account that $|u_{n}|(\cdot +\tilde {y}_{n})$ strongly converges in $\mathcal {H}_{0}$ and that $\operatorname {\varepsilon }_{n}z+y_{n}\rightarrow y_{0}\in M_{\delta}$, we find $\beta _{\operatorname {\varepsilon }_{n}}(u_{n})=y_{n}+o_{n}(1)$, that is (5.11) is satisfied.

Proof. For any $\operatorname {\varepsilon }>0$, we consider the map $\alpha _{\operatorname {\varepsilon }} : M \rightarrow \mathbb {S}_{\operatorname {\varepsilon }}^{+}$ defined as $\alpha _{\operatorname {\varepsilon }}(y)= m_{\operatorname {\varepsilon }}^{-1}(\Phi _{\operatorname {\varepsilon }}(y))$.

Using lemma 5.1, we see that

(5.12)\begin{equation} \lim_{\operatorname{\varepsilon} \rightarrow 0} \psi_{\operatorname{\varepsilon}}(\alpha_{\operatorname{\varepsilon}}(y)) = \lim_{\operatorname{\varepsilon} \rightarrow 0} J_{\operatorname{\varepsilon}}(\Phi_{\operatorname{\varepsilon}}(y))= d_{0} \quad \mbox{ uniformly in } y\in M. \end{equation}

Set

\[ \widetilde{\mathcal{S}}^{+}_{\operatorname{\varepsilon}}=\{ w\in \mathbb{S}_{\operatorname{\varepsilon}}^{+} : \psi_{\operatorname{\varepsilon}}(w) \leq d_{0} + h(\operatorname{\varepsilon})\}, \]

where $h(\operatorname {\varepsilon })=\sup _{y\in \Lambda }|\psi _{\operatorname {\varepsilon }}(\alpha _{\operatorname {\varepsilon }}(y))-d_{0}|$. It follows from (5.12) that $h(\operatorname {\varepsilon })\rightarrow 0$ as $\operatorname {\varepsilon }\rightarrow 0$. Moreover, $\alpha _{\operatorname {\varepsilon }}(y)\in \widetilde {\mathcal {S}}^{+}_{\operatorname {\varepsilon }}$ for all $y\in M$ and this shows that $\widetilde {\mathcal {S}}^{+}_{\operatorname {\varepsilon }}\neq \emptyset$ for all $\operatorname {\varepsilon }>0$.

In light of lemmas 5.1, 3.2-(iii), 5.2 and 5.4, we can find $\bar {\operatorname {\varepsilon }}= \bar {\operatorname {\varepsilon }}_{\delta }>0$ such that the following diagram

\[ M\stackrel{\Phi_{\operatorname{\varepsilon}}}{\rightarrow} \Phi_{\operatorname{\varepsilon}}(M) \stackrel{m_{\operatorname{\varepsilon}}^{{-}1}}{\rightarrow} \alpha_{\operatorname{\varepsilon}}(M)\stackrel{m_{\operatorname{\varepsilon}}}{\rightarrow} \Phi_{\operatorname{\varepsilon}}(M) \stackrel{\beta_{\operatorname{\varepsilon}}}{\rightarrow} M_{\delta} \]

is well defined for any $\operatorname {\varepsilon } \in (0, \bar {\operatorname {\varepsilon }})$. Thanks to lemma 5.2, and decreasing $\bar {\operatorname {\varepsilon }}$ if necessary, we see that $\beta _{\operatorname {\varepsilon }}(\Phi _{\operatorname {\varepsilon }}(y))= y+ \theta (\operatorname {\varepsilon }, y)$ for all $y\in M$, for some function $\theta (\operatorname {\varepsilon }, y)$ satisfying $|\theta (\operatorname {\varepsilon }, y)|<{\delta }/{2}$ uniformly in $y\in M$ and for all $\operatorname {\varepsilon } \in (0, \bar {\operatorname {\varepsilon }})$. Define $H(t, y)= y+ (1-t)\theta (\operatorname {\varepsilon }, y)$. Then $H: [0,1]\times M\rightarrow M_{\delta }$ is continuous. Clearly, $H(0, y)=\beta _{\operatorname {\varepsilon }}(\Phi _{\operatorname {\varepsilon }}(y))$ and $H(1,y)=y$ for all $y\in M$. Consequently, $H(t, y)$ is a homotopy between $\beta _{\operatorname {\varepsilon }} \circ \Phi _{\operatorname {\varepsilon }} = (\beta _{\operatorname {\varepsilon }} \circ m_{\operatorname {\varepsilon }}) \circ (m_{\operatorname {\varepsilon }}^{-1}\circ \Phi _{\operatorname {\varepsilon }})$ and the inclusion map $id: M \rightarrow M_{\delta }$. This fact yields

(5.13)\begin{equation} cat_{\alpha_{\operatorname{\varepsilon}}(M)} \alpha_{\operatorname{\varepsilon}}(M)\geq cat_{M_{\delta}}(M). \end{equation}

Applying corollary 3.6, lemma 4.8 and theorem $27$ in [Reference Szulkin and Weth47] with $c= c_{\operatorname {\varepsilon }}\leq d_{0}+h(\operatorname {\varepsilon }) =d$ and $K= \alpha _{\operatorname {\varepsilon }}(M)$, we obtain that $\psi _{\operatorname {\varepsilon }}$ has at least $cat_{\alpha _{\operatorname {\varepsilon }}(M)} \alpha _{\operatorname {\varepsilon }}(M)$ critical points on $\widetilde {\mathcal {S}}^{+}_{\operatorname {\varepsilon }}$. Taking into account proposition 3.3-(d) and (5.13), we infer that $J_{\operatorname {\varepsilon }}$ admits at least $cat_{M_{\delta }}(M)$ critical points in $\widetilde {\mathcal {N}}_{\operatorname {\varepsilon }}$.

Proof. Take $\varphi \in C^{\infty }_{c}(\mathbb {R}^{N}, \mathbb {R})$ such that $\varphi \geq 0$, and for all $\delta >0$ we consider

\[ \omega_{\delta}(x)=\displaystyle\frac{u(x)}{\sqrt{|u(x)|^{2}+\delta^{2}}} \varphi(x)= \frac{u(x)}{u_{\delta}(x)}\varphi(x) \]

as test function in (6.1). First, we show that $\omega _{\delta }$ is admissible. It is clear that $\omega _{\delta }$ has compact support. On the other hand, we can observe

\begin{align*} \omega_{\delta}(x)-\omega_{\delta}(y)\textrm{e}^{\imath A(\frac{x+y}{2})\cdot (x-y)}&= \left(\frac{u(x)}{u_{\delta}(x)}\right)\varphi(x)-\left(\frac{u(y)}{u_{\delta}(y)}\right)\varphi(y)\textrm{e}^{\imath A(\frac{x+y}{2})\cdot (x-y)}\\ &= \left[u(x)-u(y)\textrm{e}^{\imath A(\frac{x+y}{2})\cdot (x-y)}\right] \frac{\varphi(x)}{u_{\delta}(x)}\\ &\quad +\left[\frac{\varphi(x)}{u_{\delta}(x)}-\frac{\varphi(y)}{u_{\delta}(y)} \right] u(y)\textrm{e}^{\imath A(\frac{x+y}{2})\cdot (x-y)} \\ &= \left[u(x)-u(y)\textrm{e}^{\imath A(\frac{x+y}{2})\cdot (x-y)}\right] \frac{\varphi(x)}{u_{\delta}(x)}\\ &\quad +\left[\frac{1}{u_{\delta}(x)}-\frac{1}{u_{\delta}(y)} \right] \varphi(x)u(y)\textrm{e}^{\imath A(\frac{x+y}{2})\cdot (x-y)}\\ &\quad +[\varphi(x)-\varphi(y)] \frac{u(y)}{u_{\delta}(y)} \textrm{e}^{\imath A(\frac{x+y}{2})\cdot (x-y)} \end{align*}

which implies that

\begin{align*} &|\omega_{\delta}(x)-\omega_{\delta}(y)\textrm{e}^{\imath A(\frac{x+y}{2})\cdot (x-y)}|^{2} \\ &\quad\leq \frac{4}{\delta^{2}}|u(x)-u(y)\textrm{e}^{\imath A(\frac{x+y}{2})\cdot (x-y)}|^{2}\|\varphi\|^{2}_{L^{\infty}(\mathbb{R}^{N})}\\ &\qquad+4\left|\frac{u(y)}{u_{\delta}(y)}\right|^{2} \frac{1}{|u_{\delta}(x)|^{2}} \|\varphi\|^{2}_{L^{\infty}(\mathbb{R}^{N})} |u_{\delta}(y)-u_{\delta}(x)|^{2} +4|\varphi(x)-\varphi(y)|^{2} \\ &\quad\leq \frac{4}{\delta^{2}}|u(x)-u(y)\textrm{e}^{\imath A(\frac{x+y}{2})\cdot (x-y)}|^{2}\|\varphi\|^{2}_{L^{\infty}(\mathbb{R}^{N})}+\frac{4}{\delta^{2}}||u(x)|-|u(y)||^{2}\|\varphi\|^{2}_{L^{\infty}(\mathbb{R}^{N})}\\ &\qquad+4|\varphi(x)-\varphi(y)|^{2}, \end{align*}

where we used the following elementary inequalities

\begin{align*} &|z+w+k|^{2}\leq 4(|z|^{2}+|w|^{2}+|k|^{2}) \quad \mbox{ for all } z,w,k\in \mathbb{C}, \\ &|\sqrt{|z|^{2}+\delta^{2}}-\sqrt{|w|^{2}+\delta^{2}}|\leq ||z|-|w|| \quad \mbox{ for all } z, w\in \mathbb{C}, \end{align*}

and that $|\textrm {e}^{\imath t}|=1$ for all $t\in \mathbb {R}$, $u_{\delta }\geq \delta$, $|{u}/{u_{\delta }}|\leq 1$. Since $u\in \mathcal {D}^{s,2}_{A}(\mathbb {R}^{N}, \mathbb {C})$, $|u|\in \mathcal {D}^{s,2}(\mathbb {R}^{N}, \mathbb {R})$ (by lemma 2.2) and $\varphi \in C^{\infty }_{c}(\mathbb {R}^{N}, \mathbb {R})$, we deduce that $[\omega _{\delta }]_{A}<\infty$. Then we have

(6.3)\begin{align} &\Re\left[\iint_{\mathbb{R}^{2N}} \frac{(u(x)-u(y)\textrm{e}^{\imath A(\frac{x+y}{2})\cdot (x-y)})}{|x-y|^{N+2s}} \right.\notag\\ &\qquad \times \left.\left(\frac{\overline{u(x)}}{u_{\delta}(x)}\varphi(x)-\frac{\overline{u(y)}}{u_{\delta}(y)}\varphi(y)\textrm{e}^{-\imath A(\frac{x+y}{2})\cdot (x-y)} \right) \mathrm{d}x\, \mathrm{d}y\right] \nonumber\\ &\quad=\Re\left(\int_{\mathbb{R}^{N}} f\frac{\overline{u}}{u_{\delta}}\varphi \,\textrm{d}x\right). \end{align}

Now, using $\Re (z)\leq |z|$ for all $z\in \mathbb {C}$ and $|\textrm {e}^{\imath t}|=1$ for all $t\in \mathbb {R}$, we see that

(6.4)\begin{align} &\Re\left[(u(x)-u(y)\textrm{e}^{\imath A(\frac{x+y}{2})\cdot (x-y)}) \left(\frac{\overline{u(x)}}{u_{\delta}(x)}\varphi(x)-\frac{\overline{u(y)}}{u_{\delta}(y)}\varphi(y)\textrm{e}^{-\imath A(\frac{x+y}{2})\cdot (x-y)} \right)\right] \nonumber\\ &\quad=\Re\left[\frac{|u(x)|^{2}}{u_{\delta}(x)}\varphi(x)+\frac{|u(y)|^{2}}{u_{\delta}(y)}\varphi(y)-\frac{u(x)\overline{u(y)}}{u_{\delta}(y)}\varphi(y)\textrm{e}^{-\imath A(\frac{x+y}{2})\cdot (x-y)} \right.\notag\\ &\qquad -\left.\frac{u(y)\overline{u(x)}}{u_{\delta}(x)}\varphi(x)\textrm{e}^{\imath A(\frac{x+y}{2})\cdot (x-y)}\right] \nonumber\\ &\quad\geq \left[\frac{|u(x)|^{2}}{u_{\delta}(x)}\varphi(x)+\frac{|u(y)|^{2}}{u_{\delta}(y)}\varphi(y)-|u(x)|\frac{|u(y)|}{u_{\delta}(y)}\varphi(y)-|u(y)|\frac{|u(x)|}{u_{\delta}(x)}\varphi(x) \right]. \end{align}

Let us note that

(6.5)\begin{align} &\frac{|u(x)|^{2}}{u_{\delta}(x)}\varphi(x)+\frac{|u(y)|^{2}}{u_{\delta}(y)}\varphi(y)-|u(x)|\frac{|u(y)|}{u_{\delta}(y)}\varphi(y)-|u(y)|\frac{|u(x)|}{u_{\delta}(x)}\varphi(x) \nonumber\\ &\quad= \frac{|u(x)|}{u_{\delta}(x)}(|u(x)|-|u(y)|)\varphi(x)-\frac{|u(y)|}{u_{\delta}(y)}(|u(x)|-|u(y)|)\varphi(y) \nonumber\\ &\quad=\left[\frac{|u(x)|}{u_{\delta}(x)}(|u(x)|-|u(y)|)\varphi(x)-\frac{|u(x)|}{u_{\delta}(x)}(|u(x)|-|u(y)|)\varphi(y)\right] \nonumber\\ &\qquad+\left(\frac{|u(x)|}{u_{\delta}(x)}-\frac{|u(y)|}{u_{\delta}(y)} \right) (|u(x)|-|u(y)|)\varphi(y) \nonumber\\ &\quad=\frac{|u(x)|}{u_{\delta}(x)}(|u(x)|-|u(y)|)(\varphi(x)-\varphi(y))\notag\\ &\qquad +\left(\frac{|u(x)|}{u_{\delta}(x)}-\frac{|u(y)|}{u_{\delta}(y)} \right) (|u(x)|-|u(y)|)\varphi(y) \nonumber\\ &\quad\geq \frac{|u(x)|}{u_{\delta}(x)}(|u(x)|-|u(y)|)(\varphi(x)-\varphi(y)), \end{align}

where in the last inequality we used the fact that

\[ \left(\displaystyle\frac{|u(x)|}{u_{\delta}(x)}-\frac{|u(y)|}{u_{\delta}(y)} \right) (|u(x)|-|u(y)|)\varphi(y)\geq 0 \]

because

\[ h(t)=\displaystyle\frac{t}{\sqrt{t^{2}+\delta^{2}}} \mbox{ is increasing for } t\geq 0 \quad \mbox{ and } \quad \varphi\geq 0 \mbox{ in }\mathbb{R}^{N}. \]

Since

\begin{align*} &\frac{\left|\frac{|u(x)|}{u_{\delta}(x)}(|u(x)|-|u(y)|)(\varphi(x)-\varphi(y))\right|}{|x-y|^{N+2s}}\\ &\quad \leq \frac{||u(x)|-|u(y)||}{|x-y|^{({N+2s})/{2}}} \frac{|\varphi(x)-\varphi(y)|}{|x-y|^{({N+2s})/{2}}}\in L^{1}(\mathbb{R}^{2N}, \mathbb{R}), \end{align*}

and ${|u(x)|}/{u_{\delta }(x)}\rightarrow 1$ a.e. in $\mathbb {R}^{N}$ as $\delta \rightarrow 0$, we can use (6.4), (6.5) and the dominated convergence theorem to deduce that

(6.6)\begin{align} &\liminf_{\delta\rightarrow 0} \Re\left[\iint_{\mathbb{R}^{2N}} \frac{(u(x)-u(y)\textrm{e}^{\imath A(\frac{x+y}{2})\cdot (x-y)})}{|x-y|^{N+2s}} \right.\notag\\ &\qquad \times \left.\left(\frac{\overline{u(x)}}{u_{\delta}(x)}\varphi(x)-\frac{\overline{u_{n}(y)}}{u_{\delta}(y)}\varphi(y)\textrm{e}^{-\imath A(\frac{x+y}{2})\cdot (x-y)} \right) \,\mathrm{d}x \mathrm{d}y\right] \nonumber\\ &\quad\geq \liminf_{\delta\rightarrow 0} \iint_{\mathbb{R}^{2N}} \frac{|u(x)|}{u_{\delta}(x)}\, \frac{(|u(x)|-|u(y)|)(\varphi(x)-\varphi(y))}{|x-y|^{N+2s}} \ \mathrm{d}x \mathrm{d}y \nonumber\\ &\quad=\iint_{\mathbb{R}^{2N}} \frac{(|u(x)|-|u(y)|)(\varphi(x)-\varphi(y))}{|x-y|^{N+2s}} \,\mathrm{d}x \,\mathrm{d}y. \end{align}

On the other hand, observing that $|f({\overline {u}}/{u_{\delta }})\varphi |\leq |f\varphi |\in L^{1}(\mathbb {R}^{N}, \mathbb {R})$ and $f({\overline {u}}/{u_{\delta }})\varphi \rightarrow f\, \textrm {sign}(\bar {u})\varphi$ a.e. in $\mathbb {R}^{N}$ as $\delta \rightarrow 0$, we can invoke the dominated convergence theorem to infer that as $\delta \rightarrow 0$

(6.7)\begin{align} \Re\left(\int_{\mathbb{R}^{N}} f\frac{\overline{u}}{u_{\delta}}\varphi \,\mathrm{d}x\right)\rightarrow \Re\left(\int_{\mathbb{R}^{N}} f \,\textrm{sign}(\bar{u}) \varphi \,\mathrm{d}x\right). \end{align}

Putting together (6.3), (6.6) and (6.7), we see that (6.2) holds true.

Proof. Since $J_{\operatorname {\varepsilon }_{n}}(u_{n})\leq d_{0}+h(\operatorname {\varepsilon }_{n})$ with $h(\operatorname {\varepsilon }_{n})\rightarrow 0$ as $n\rightarrow \infty$, we can argue as at the beginning of the proof of lemma 5.3 to deduce that $J_{\operatorname {\varepsilon }_{n}}(u_{n})\rightarrow d_{0}$. Thus we may invoke lemma 5.3 to obtain a sequence $(\tilde {y}_{n})\subset \mathbb {R}^{N}$ such that $\operatorname {\varepsilon }_{n}\tilde {y}_{n}\rightarrow y_{0}\in M$ and $v_{n}=|u_{n}|(\cdot +\tilde {y}_{n})$ strongly converges in $\mathcal {H}_{0}$. Let $\tilde {u}_{n}(x)=u_{n}(\cdot +\tilde {y}_{n})$ and note that it solves

(6.8)\begin{equation} (-\Delta)^{s}_{\tilde{A}_{n}} \tilde{u}_{n}+\tilde{V}_{n}(x) \tilde{u}_{n}=\tilde{g}_{n}(x, v^2_{n}) \tilde{u}_{n} \quad \mbox{ in } \mathbb{R}^{N}, \end{equation}

where

\[ \tilde{A}_{n}(x)=A_{\operatorname{\varepsilon}_{n}}(x+\tilde{y}_{n}), \]

\[ \tilde{V}_{n}(x)=V_{\varepsilon_{n}}(x+\tilde{y}_{n}), \]

and

\[ \tilde{g}_{n}(x, v^2_{n})=g(\operatorname{\varepsilon}_{n} x+\varepsilon_{n} \tilde{y}_{n}, v_{n}^{2}(x)). \]

Using theorem 6.1, we deduce that $v_{n}$ satisfies (in the distributional sense)

(6.9)\begin{equation} (-\Delta)^{s} v_{n}+\tilde{V}_{n}(x) v_{n}\leq \tilde{g}_{n}(x, v^2_{n}) v_{n}=h_{n} \quad \mbox{ in } \mathbb{R}^{N}. \end{equation}

By performing a Moser iteration argument [Reference Moser40] as in lemma $5.1$ in [Reference Ambrosio6], we obtain that

(6.10)\begin{equation} \|v_{n}\|_{L^{\infty}(\mathbb{R}^{N})}\leq K \quad \mbox{ for all } n\in \mathbb{N}. \end{equation}

Moreover, by interpolation, $v_{n}\rightarrow v$ strongly converges in $L^{r}(\mathbb {R}^{N}, \mathbb {R})$ for all $r\in [2, \infty )$. In view of the growth assumptions on $g$, we can also see that $h_{n}\rightarrow h=f(v^{2})v$ in $L^{r}(\mathbb {R}^{N}, \mathbb {R})$ for all $r\in [2, \infty )$, and $\|h_{n}\|_{L^{\infty }(\mathbb {R}^{N})}\leq C$ for all $n\in \mathbb {N}$. Note that, by $(V_{1})$, we have

(6.11)\begin{equation} (-\Delta)^{s} v_{n} + V_{0}v_{n}\leq h_{n} \quad \mbox{ in } \mathbb{R}^{N}. \end{equation}

Let us denote by $z_{n}\in H^{s}(\mathbb {R}^{N}, \mathbb {R})$ the unique solution to

(6.12)\begin{equation} (-\Delta)^{s} z_{n} + V_{0}z_{n}=h_{n} \quad \mbox{ in } \mathbb{R}^{N}. \end{equation}

Since $v_{n}$ satisfies (6.11) and $z_{n}$ solves (6.12), by comparison we see that $0\leq v_{n}\leq z_{n}$ in $\mathbb {R}^{N}$ for all $n\in \mathbb {N}$. Next we show that

(6.13)\begin{align} z_{n}(x)\rightarrow 0 \,\mbox{ as } |x|\rightarrow \infty \,\mbox{ uniformly in } n\in \mathbb{N}. \end{align}

Note that $z_{n}(x)=(\mathcal {K}*h_{n})(x)$, where the kernel $\mathcal {K}(x)=\mathcal {F}^{-1}((|k|^{2s}+V_{0})^{-1})$ satisfies the following properties (see [Reference Felmer, Quaas and Tan30]):

1. (b ₁) $\mathcal {K}$ is positive, radially symmetric and smooth in $\mathbb {R}^{N}\setminus \{0\}$;

2. (b ₂) there exists a positive constant $C>0$ such that $\mathcal {K}(x)\leq {K_{1}}/{|x|^{N+2s}}$ for any $x\in \mathbb {R}^{N}\setminus \{0\}$;

3. (b ₃) $\mathcal {K}\in L^{\nu }(\mathbb {R}^{N})$ for any $\nu \in [1, ({N}/({N-2s}))$.

Now we borrow some arguments used in [Reference Alves and Miyagaki3] to prove that (6.13) holds true. Fix $\delta >0$ and we observe that

(6.14)\begin{align} 0\leq z_{n}(x)=(\mathcal{K}*h_{n})(x)=\int_{B^{c}_{{1}/{\delta}}(x)} \mathcal{K}(x-y)h_{n}(y) \, \mathrm{d}y+\int_{B_{{1}/{\delta}}(x)} \mathcal{K}(x-y)h_{n}(y) \, \mathrm{d}y. \end{align}

From $(b_2)$ we deduce that

(6.15)\begin{align} \int_{B^{c}_{{1}/{\delta}}(x)} \mathcal{K}(x-y)h_{n}(y) \, \mathrm{d}y &\leq K_{1}\|h_{n}\|_{L^{\infty}(\mathbb{R}^{N})}\int_{B^{c}_{{1}/{\delta}}(x)} \frac{\mathrm{d}y}{|x-y|^{N+2s}}\notag\\ &\leq c_{1}\delta^{2s} K_{1} \int_{|\xi|\geq 1} \frac{\mathrm{d}\xi}{|\xi|^{N+2s}}= C_{1}\delta^{2s}. \end{align}

On the other hand,

\begin{align*} \left|\int_{B_{{1}/{\delta}}(x)} \mathcal{K}(x-y)h_{n}(y) \, \mathrm{d}y\right|&\leq \int_{B_{{1}/{\delta}}(x)} \mathcal{K}(x-y)|h_{n}(y)-h(y)| \, \mathrm{d}y\\ &\quad + \int_{B_{{1}/{\delta}}(x)} \mathcal{K}(x-y)|h(y)| \, \mathrm{d}y. \end{align*}

Fix

\[ \nu\in \left(1, \min\left\{\displaystyle\frac{N}{N-2s}, 2 \right\}\right). \]

Note that, if $N\geq 4s$ then $\nu \in (1, ({N}/({N-2s}))$ and we have $\nu '=({\nu }/({\nu -1}))>({N}/{2s})\geq 2$, and that when $2s< N<4s$ then $2<({N}/({N-2s}))$, $\nu \in (1, 2)$ and $\nu '\in (2, \infty )$. In any case, $\nu \in (1, ({N}/({N-2s}))$ and $\nu '\in (2, \infty )$. Then, by $(b_{3})$, we deduce that $\mathcal {K}\in L^{\nu }(\mathbb {R}^{N})$. Using Hölder's inequality we have

\begin{align*} \left|\int_{B_{{1}/{\delta}}(x)} \mathcal{K}(x-y)h_{n}(y) \, \mathrm{d}y\right|&\leq \|\mathcal{K}\|_{L^{\nu}(\mathbb{R}^{N})} \|h_{n}-h\|_{L^{\nu'}(\mathbb{R}^{N})}\\ &\quad +\|\mathcal{K}\|_{L^{\nu}(\mathbb{R}^{N})} \|h\|_{L^{\nu'}(B_{{1}/{\delta}}(x))}. \end{align*}

Since $\|h_{n}-h\|_{L^{\nu '}(\mathbb {R}^{N})}\rightarrow 0$ as $n\rightarrow \infty$ and $\|h\|_{L^{\nu '}(B_{{1}/{\delta }}(x))}\rightarrow 0$ as $|x|\rightarrow \infty$, we see that there exist $R>0$ and $n_{0}\in \mathbb {N}$ such that

(6.16)\begin{align} \int_{B_{{1}/{\delta}}(x)} \mathcal{K}(x-y)h_{n}(y) \, \mathrm{d}y\leq C_{2}\delta \quad \mbox{ for all } |x|\geq R, \, n\geq n_{0}. \end{align}

Putting together (6.14), (6.15) and (6.16), we obtain that

(6.17)\begin{align} z_{n}(x)=\int_{\mathbb{R}^{N}} \mathcal{K}(x-y)h_{n}(y) \, \mathrm{d}y\leq C_{1}\delta^{2s}+C_{2}\delta \quad \mbox{ for all } |x|\geq R, \, n\geq n_{0}. \end{align}

Now, for each $n\in \{1, \dots , n_{0}-1\}$, there is $R_{n}>0$ such that $\|h_{n}\|_{L^{\nu '}(B_{{1}/{\delta }}(x))}<\delta$ for $|x|\geq R_{n}$. Then, for $|x|\geq R_{n}$, we have

(6.18)\begin{align} \int_{\mathbb{R}^{N}} \mathcal{K}(x-y)h_{n}(y) \, \mathrm{d}y&\leq C_{1} \delta^{2s}+\int_{B_{{1}/{\delta}}(x)} \mathcal{K}(x-y)h_{n}(y) \, \mathrm{d}y \nonumber\\ &\leq C_{1} \delta^{2s}+\|\mathcal{K}\|_{L^{\nu}(\mathbb{R}^{N})} \|h_{n}\|_{L^{\nu'}(B_{{1}/{\delta}}(x))} \nonumber\\ &\leq C_{1}\delta^{2s}+C_{3}\delta. \end{align}

Set $\bar {R}=\max \{R_{1}, \dots , R_{n_{0}-1}, R\}$. By (6.17) and (6.18) we find

\begin{align*} z_{n}(x)\leq C_{4}(\delta^{2s}+\delta) \quad \mbox{ for all } |x|\geq \bar{R}, \, n\in \mathbb{N}. \end{align*}

From the arbitrariness of $\delta >0$, we deduce that (6.13) holds true. Consequently, combining (6.13) with $0\leq v_{n}\leq z_{n}$ in $\mathbb {R}^{N}$, we obtain that $v_{n}(x)\rightarrow 0$ as $|x|\rightarrow \infty$ uniformly with respect to $n\in \mathbb {N}$.

Proof of theorem 1.1. Let $\delta >0$ be such that $M_{\delta }\subset \Lambda$. First, we claim that there exists $\hat {\operatorname {\varepsilon }}_{\delta }>0$ such that for any $\operatorname {\varepsilon }\in (0, \hat {\operatorname {\varepsilon }}_{\delta })$ and any solution $u\in \widetilde {\mathcal {N}}_{\operatorname {\varepsilon }}$ of (3.2), it holds

(6.19)\begin{equation} \|u\|_{L^{\infty}(\Lambda^{c}_{\operatorname{\varepsilon}})}<\sqrt{a}. \end{equation}

We argue by contradiction and assume that for some subsequence $\operatorname {\varepsilon }_{n}\rightarrow 0$ we can obtain $u_{n}=u_{\operatorname {\varepsilon }_{n}}\in \widetilde {\mathcal {N}}_{\operatorname {\varepsilon }_{n}}$ such that $J'_{\operatorname {\varepsilon }}(u_{n})=0$ and

(6.20)\begin{equation} \|u_{n}\|_{L^{\infty}(\Lambda^{c}_{\operatorname{\varepsilon}})}\geq \sqrt{a}. \end{equation}

Since $J_{\operatorname {\varepsilon }_{n}}(u_{n})\leq d_{0}+h(\operatorname {\varepsilon }_{n})$, we can argue as in the first part of lemma 5.3 to deduce that $J_{\operatorname {\varepsilon }_{n}}(u_{n})\rightarrow d_{0}$. In view of lemma 5.3, there exists $(\tilde {y}_{n})\subset \mathbb {R}^{N}$ such that $\operatorname {\varepsilon }_{n}\tilde {y}_{n}\rightarrow y_{0}$ for some $y_{0} \in M$ and $v_{n}=|u_{n}|(\cdot +\tilde {y}_{n})$ strongly converges in $H^{s}(\mathbb {R}^{N}, \mathbb {R})$.

Take $r>0$ such that, for some subsequence still denoted by itself, it holds $B_{r}(\operatorname {\varepsilon }_{n}\tilde {y}_{n})\subset \Lambda$ for all $n\in \mathbb {N}$. Hence $B_{{r}/{\operatorname {\varepsilon }_{n}}}(\tilde {y}_{n})\subset \Lambda _{\operatorname {\varepsilon }_{n}}$ for all $n\in \mathbb {N}$, and consequently $\Lambda ^{c}_{\operatorname {\varepsilon }_{n}}\subset B^{c}_{{r}/{\operatorname {\varepsilon }_{n}}}(\tilde {y}_{n})$ for all $n\in \mathbb {N}$. By lemma 6.3, we can find $R>0$ such that

\[ v_{n}(x)<\sqrt{a} \quad \mbox{ for all } |x|\geq R, \,n\in \mathbb{N}, \]

from which we deduce that $|u_{n}(x)|<\sqrt {a}$ for any $x\in B^{c}_{R}(\tilde {y}_{n})$ and $n\in \mathbb {N}$. On the other hand, there exists $\nu \in \mathbb {N}$ such that for any $n\geq \nu$ it holds

\[ \Lambda^{c}_{\operatorname{\varepsilon}_{n}}\subset B^{c}_{{r}/{\operatorname{\varepsilon}_{n}}}(\tilde{y}_{n})\subset B^{c}_{R}(\tilde{y}_{n}). \]

Therefore, $|u_{n}(x)|<\sqrt {a}$ for any $x\in \Lambda ^{c}_{\operatorname {\varepsilon }_{n}}$ and $n\geq \nu$, and this contradicts (6.20).

Let $\tilde {\operatorname {\varepsilon }}_{\delta }>0$ be given by theorem 5.5 and we set $\operatorname {\varepsilon }_{\delta }=\min \{\tilde {\operatorname {\varepsilon }}_{\delta }, \hat {\operatorname {\varepsilon }}_{\delta } \}$. Let us fix $\varepsilon \in (0, \varepsilon_{\delta})$. Applying theorem 5.5 we obtain at least $cat_{M_{\delta }}(M)$ nontrivial solutions to (3.2). If $u_{\operatorname {\varepsilon }}\in \mathcal {H}_{\operatorname {\varepsilon }}$ is one of these solutions, then $u_{\operatorname {\varepsilon }}\in \widetilde {\mathcal {N}}_{\operatorname {\varepsilon }}$, and in view of (6.19) and the definition of $g$, we infer that $u_{\operatorname {\varepsilon }}$ is also a solution to (3.1). Since $\hat {u}_{\operatorname {\varepsilon }}(x)=u_{\operatorname {\varepsilon }}(x/\operatorname {\varepsilon })$ is a solution to (1.1), we deduce that (1.1) has at least $cat_{M_{\delta }}(M)$ nontrivial solutions. Finally, we investigate the behaviour of the maximum points of $|\hat {u}_{\operatorname {\varepsilon }}|$. Take $\operatorname {\varepsilon }_{n}\rightarrow 0$ and let $(u_{n})\subset \mathcal {H}_{\operatorname {\varepsilon }_{n}}$ be a sequence of solutions to (3.2) as above. From $(g_1)$, there exists $\gamma \in (0, \sqrt {a})$ such that

(6.21)\begin{align} g_{\operatorname{\varepsilon}} (x, t^{2})t^{2}\leq \frac{V_{0}}{2}t^{2} \quad \mbox{ for all } x\in \mathbb{R}^{N}, |t|\leq \gamma. \end{align}

Arguing as above, we can find $R>0$ such that

(6.22)\begin{align} \|u_{n}\|_{L^{\infty}(B^{c}_{R}(\tilde{y}_{n}))}<\gamma. \end{align}

Up to a subsequence, we may also assume that

(6.23)\begin{align} \|u_{n}\|_{L^{\infty}(B_{R}(\tilde{y}_{n}))}\geq \gamma. \end{align}

Indeed, if (6.23) does not hold, we get $\|u_{n}\|_{L^{\infty }(\mathbb {R}^{N})}< \gamma$, and using $J_{\operatorname {\varepsilon }_{n}}'(u_{n})=0$, (6.21) and lemma 2.2, we deduce that

\begin{align*} [|u_{n}|]^{2}+\int_{\mathbb{R}^{N}}V_{0}|u_{n}|^{2}\textrm{d}x&\leq \|u_{n}\|^{2}_{\operatorname{\varepsilon}_{n}}=\int_{\mathbb{R}^{N}} g_{\operatorname{\varepsilon}_{n}}(x, |u_{n}|^{2})|u_{n}|^{2}\,\mathrm{d}x\leq \frac{V_{0}}{2}\int_{\mathbb{R}^{N}}|u_{n}|^{2}\, \mathrm{d}x. \end{align*}

This fact yields $\||u_{n}|\|_{0}=0$, which is impossible. Hence, (6.23) is satisfied.

In light of (6.22) and (6.23), we can see that if $p_{n}$ is a global maximum point of $|u_{n}|$, then $p_{n}$ belongs to $B_{R}(\tilde {y}_{n})$, that is $p_{n}=\tilde {y}_{n}+q_{n}$ for some $q_{n}\in B_{R}$. Since $\hat {u}_{n}(x)=u_{n}(x/\operatorname {\varepsilon }_{n})$ is solution of (1.1), we infer that $\eta _{n}=\operatorname {\varepsilon }_{n}\tilde {y}_{n}+\operatorname {\varepsilon }_{n}q_{n}$ is a global maximum point of $|\hat {u}_{n}|$. Using $(q_{n})\subset B_{R}$, $\operatorname {\varepsilon }_{n}\tilde {y}_{n}\rightarrow y_{0}\in M$, and the continuity of $V$, we deduce that

\[ \lim_{n\rightarrow \infty} V(\eta_{n})=V(y_{0})=V_{0}. \]

Finally, we provide a decay estimate for $|\hat {u}_{n}|$. By using lemma $4.3$ in [Reference Felmer, Quaas and Tan30] there exists a continuous function $w$ such that

(6.24)\begin{align} 0< w(x)\leq \frac{C}{1+|x|^{N+2s}} \quad \mbox{ for all } x\in \mathbb{R}^{N}, \end{align}

and satisfying in the classical sense

(6.25)\begin{align} (-\Delta)^{s} w+\frac{V_{0}}{2}w= 0 \quad \mbox{ in } \mathbb{R}^{N}\setminus \overline{B}_{R_{1}}, \end{align}

for some suitable $R_{1}>0$. By lemma 6.3, we can find $R_{2}>0$ such that

(6.26)\begin{equation} h_{n}=g(\operatorname{\varepsilon}_{n}\cdot{+}\operatorname{\varepsilon}_{n}\tilde{y}_{n}, v_{n}^{2})v_{n}\leq \frac{V_{0}}{2}v_{n} \quad \mbox{ in } \mathbb{R}^{N}\setminus \overline{B}_{R_{2}}. \end{equation}

Let us denote by $w_{n}$ the unique solution to

\[ (-\Delta)^{s}w_{n}+V_{0}w_{n}=h_{n} \quad \mbox{ in } \mathbb{R}^{N}. \]

Then, arguing as in the proof of (6.13), we see that $w_{n}(x)\rightarrow 0$ as $|x|\rightarrow \infty$ uniformly in $n\in \mathbb {N}$. Thus, by comparison, we obtain that $0\leq v_{n}\leq w_{n}$ in $\mathbb {R}^{N}$. Moreover, in light of (6.26), it holds

\begin{align*} (-\Delta)^{s}w_{n}+\frac{V_{0}}{2}w_{n}=h_{n}-\frac{V_{0}}{2}w_{n}\leq 0 \quad \mbox{ in } \mathbb{R}^{N}\setminus \overline{B}_{R_{2}}. \end{align*}

Take $R_{3}=\max \{R_{1}, R_{2}\}$ and we set

(6.27)\begin{align} c=\min_{\overline{B}_{R_{3}}} w>0 \quad \mbox{ and } \quad \tilde{w}_{n}=(b+1)w-c w_{n}, \end{align}

where $b=\sup _{n\in \mathbb {N}} \|w_{n}\|_{L^{\infty }(\mathbb {R}^{N})}<\infty$. Our aim is to prove that

(6.28)\begin{equation} \tilde{w}_{n}\geq 0 \quad \mbox{ in } \mathbb{R}^{N}. \end{equation}

First, we observe that

\begin{align*} &\tilde{w}_{n}\geq bc+w-bc>0 \quad\mbox{ in } \overline{B}_{R_{3}}, \\ &(-\Delta)^{s} \tilde{w}_{n}+\frac{V_{0}}{2}\tilde{w}_{n}\geq 0 \quad \mbox{ in } \mathbb{R}^{N}\setminus \overline{B}_{R_{3}}. \end{align*}

Then we can apply a comparison principle (see theorem $7.1$ in [Reference Del Pezzo and Quaas23]) to deduce that (6.28) holds true. From (6.24), (6.28) and $0\leq v_{n}\leq w_{n}$ in $\mathbb {R}^{N}$, we get

\begin{align*} 0\leq v_{n}(x)\leq w_{n}(x)\leq \frac{(b+1)}{c}w(x)\leq \frac{\tilde{C}}{1+|x|^{N+2s}} \quad \mbox{ for all } x\in \mathbb{R}^{N}, n\in \mathbb{N}, \end{align*}

for some constant $\tilde {C}>0$. Therefore, recalling the definition of $v_{n}$, we infer that

\begin{align*} |\hat{u}_{n}|(x)&=|u_{n}|\left(\frac{x}{\operatorname{\varepsilon}_{n}}\right)=v_{n}\left(\frac{x}{\operatorname{\varepsilon}_{n}}-\tilde{y}_{n}\right) \\ &\leq \frac{\tilde{C}}{1+|\frac{x}{\operatorname{\varepsilon}_{n}}-\tilde{y}_{n}|^{N+2s}} \\ &=\frac{\tilde{C} \operatorname{\varepsilon}_{n}^{N+2s}}{\operatorname{\varepsilon}_{n}^{N+2s}+|x- \operatorname{\varepsilon}_{n} \tilde{y}_{n}|^{N+2s}} \\ &\leq \frac{\tilde{C} \operatorname{\varepsilon}_{n}^{N+2s}}{\operatorname{\varepsilon}_{n}^{N+2s}+|x-\eta_{n}|^{N+2s}} \quad \mbox{ for all } x\in \mathbb{R}^{N}. \end{align*}

This ends the proof of theorem 1.1.