Without loss of generality, if we take a subsequence {*u*_{m}}, we also use the same notation {*u*_{m}}.

#### Lemma 3.1

*Assume that* (*V*_{0}), (*H*_{1})-(*H*_{3}) *hold*. *Then there exists a constant* *λ*_{0} > 0, *for any* *λ* ∈ (0, *λ*_{0}), *φ* *satisfies the* (*C*)_{c} *condition*.

#### Proof

Let {*u*_{m}} ⊂ *X* be a (*C*)_{c} sequence, that is
$$\begin{array}{}\phi ({u}_{n})\to c,\phantom{\rule{1em}{0ex}}\parallel {\phi}^{\prime}({u}_{n})\parallel (1+\parallel {u}_{n}\parallel )\to 0,\phantom{\rule{thinmathspace}{0ex}}\text{\hspace{0.17em}as\hspace{0.17em}}\phantom{\rule{thinmathspace}{0ex}}n\to \mathrm{\infty},\end{array}$$(6)
which also implies
$$\begin{array}{}\u3008{\phi}^{\prime}({u}_{n}),{u}_{n}\u3009\to 0,\text{\hspace{0.17em}as\hspace{0.17em}}n\to \mathrm{\infty}.\end{array}$$(7)

First, we prove that {*u*_{m}} is bounded.

We argue it by contradiction. If {*u*_{m}} is unbounded in *X*, then there exists a subsequence {*u*_{m}} with ∥*u*_{n}∥ → ∞, as *n* → ∞. From (4)-(5), (*H*_{3}) and the Hölder inequality, one has
$$\begin{array}{}\phi ({u}_{n})-{\displaystyle \frac{\u3008{\phi}^{\prime}({u}_{n}),{u}_{n}\u3009}{2}=-\underset{{R}^{N}}{\int}F(x,{u}_{n})dx-(\frac{1}{p}-\frac{1}{2})\lambda \underset{{R}^{N}}{\int}h(x)|{u}_{n}{|}^{p}dx+\frac{1}{2}\underset{{R}^{N}}{\int}f(x,{u}_{n}){u}_{n}\phantom{\rule{thinmathspace}{0ex}}dx}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle \ge -(\frac{1}{p}-\frac{1}{2})\lambda \underset{{R}^{N}}{\int}h(x)|{u}_{n}{|}^{p}dx}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle \ge -(\frac{1}{p}-\frac{1}{2}){\lambda}_{0}\parallel h{\parallel}_{\frac{2}{2-p}}\parallel {u}_{n}{\parallel}_{2}^{p}.}\end{array}$$

By (6)-(7), we can easily get that ∥*u*_{n}∥_{2} is bounded.

Also from (4)-(5), we have
$$\begin{array}{}\u3008{\phi}^{\prime}({u}_{n}),{u}_{n}\u3009=\parallel {u}_{n}{\parallel}_{2}-{\displaystyle \underset{{R}^{N}}{\int}f(x,{u}_{n}){u}_{n}\phantom{\rule{thinmathspace}{0ex}}dx-\lambda \underset{{R}^{N}}{\int}h(x)|{u}_{n}{|}^{p}dx.}\end{array}$$

Then, it follows from (*H*_{1}) that
$$\begin{array}{}{\displaystyle \parallel {u}_{n}{\parallel}_{2}\le \u3008{\phi}^{\prime}({u}_{n}),{u}_{n}\u3009+\underset{{R}^{N}}{\int}f(x,{u}_{n}){u}_{n}\phantom{\rule{thinmathspace}{0ex}}dx+{\lambda}_{0}\parallel h{\parallel}_{\frac{2}{2-p}}\parallel {u}_{n}{\parallel}_{2}^{p}}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\le \u3008{\phi}^{\prime}({u}_{n}),{u}_{n}\u3009+{a}_{1}\parallel {u}_{n}{\parallel}_{2}^{2}+{a}_{2}\parallel {u}_{n}{\parallel}_{q}^{q}+{\lambda}_{0}\parallel h{\parallel}_{\frac{2}{2-p}}\parallel {u}_{n}{\parallel}_{2}^{p}.\end{array}$$(8)

By the Fractional Gagliardo-Nirenberg inequality ([29], corollary 2.3]) and the definition of the norm in *X*, we know that
$$\begin{array}{}\parallel u{\parallel}_{{L}^{q}({R}^{N})}\le {\zeta}^{\frac{s}{q}}\parallel u{\parallel}^{\frac{s}{q}}\parallel u{\parallel}_{{L}^{2}({R}^{N})}^{1-\frac{s}{q}},\end{array}$$(9)
where
$\begin{array}{}\zeta ={2}^{-\alpha}{\pi}^{-\frac{\alpha}{2}}(\frac{\mathrm{\Gamma}(N)}{\mathrm{\Gamma}(N/2)}{)}^{\frac{\alpha}{N}}\mathrm{\Gamma}(\frac{N-\alpha}{2})/\mathrm{\Gamma}(\frac{N+\alpha}{2})\text{\hspace{0.17em}and\hspace{0.17em}}s(\frac{1}{2}-\frac{\alpha}{N})+\frac{q-s}{2}=1.\end{array}$

For
$\begin{array}{}0<s=\frac{(q-2)N}{2\alpha}<q\in [2,\frac{2N+4\alpha}{N})\subset [2,{2}_{\alpha}^{\ast}),\end{array}$
it is easy to see that
$$0<s<2.$$(10)

Then it follows (8)-(10) that
$$\begin{array}{}1={\displaystyle \frac{\parallel {u}_{n}|{|}^{2}}{||{u}_{n}|{|}^{2}}\le \frac{\u3008{\phi}^{\prime}({u}_{n}),{u}_{n}\u3009}{\parallel {u}_{n}{\parallel}^{2}}+\frac{{a}_{1}\parallel {u}_{n}{\parallel}_{2}^{2}}{\parallel {u}_{n}{\parallel}^{2}}+\frac{{a}_{2}\parallel {u}_{n}{\parallel}_{q}^{q}}{\parallel {u}_{n}{\parallel}^{2}}+\frac{{\lambda}_{0}\parallel h|{|}_{\frac{2}{2-p}}\parallel {u}_{n}{\parallel}_{2}^{p}}{||{u}_{n}{\parallel}^{2}}}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}{\displaystyle \le \frac{\u3008{\phi}^{\prime}({u}_{n}),{u}_{n}\u3009}{\parallel {u}_{n}{\parallel}^{2}}+\frac{{a}_{1}\parallel {u}_{n}{\parallel}_{2}^{2}}{\parallel {u}_{n}{\parallel}^{2}}+\frac{{a}_{2}{\zeta}^{s}\parallel {u}_{n}{\parallel}^{s}||{u}_{n}{\parallel}_{2}^{q-s}}{||{u}_{n}|{|}^{2}}+\frac{{\lambda}_{0}\parallel h|{|}_{\frac{2}{2-p}}\parallel {u}_{n}{\parallel}_{2}^{p}}{||{u}_{n}{\parallel}^{2}}}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\to 0,\text{\hspace{0.17em}as\hspace{0.17em}}n\to \mathrm{\infty}.\end{array}$$

This is a contradiction. Hence, we know that {*u*_{n}} is bounded in *X*.

Next we prove *φ* satisfies the (*C*)_{c} condition.

For any {*u*_{n}} ⊂ *X* being *a* (*C*)_{c} sequence, from the boundedness of {*u*_{n}}, we know there exists a weakly convergent subsequence {*u*_{n}} such that *u*_{n} ⇀ *u* weakly in *X*. From Lemma 2.2, we can obtain that *u*_{n} → *u* strongly in *L*^{q}(*R*^{N}) for
$\begin{array}{}q\in [2,\frac{2N+4\alpha}{N}).\end{array}$

Then we prove that *u*_{n} → *u* in *X*.

From (4)-(5), we have
$$\begin{array}{}{\displaystyle \parallel {u}_{n}-u{\parallel}^{2}=\u3008{\phi}^{\prime}({u}_{n})-{\phi}^{\prime}(u),{u}_{n}-u\u3009+\underset{{R}^{N}}{\int}[f(x,{u}_{n}(x))-f(x,u(x))]({u}_{n}-u)dx}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}+\lambda \underset{{R}^{N}}{\int}h(x)|{u}_{n}-u{|}^{p}dx.}\end{array}$$(11)

It is easy to see that
$$\begin{array}{}\u3008{\phi}^{\prime}({u}_{n})-{\phi}^{\prime}(u),{u}_{n}-u\u3009\to 0,\text{\hspace{0.17em}as\hspace{0.17em}}n\to \mathrm{\infty}.\end{array}$$(12)

Based on the fact that
$\begin{array}{}h\in {L}^{\frac{2}{2-p}}({R}^{N}),{u}_{n}\to u\text{\hspace{0.17em}in\hspace{0.17em}}{L}^{q}({R}^{N})\end{array}$
and Hölder inequality, one has
$$\begin{array}{}{\displaystyle \lambda \underset{{R}^{N}}{\int}h(x)|{u}_{n}-u{|}^{p}dx\le {\lambda}_{0}\parallel h{\parallel}_{\frac{2}{2-p}}\parallel {u}_{n}{\parallel}_{2}^{p}\to 0,\text{\hspace{0.17em}as\hspace{0.17em}}n\to \mathrm{\infty}.}\end{array}$$(13)

From (*H*_{1}) and the Hölder inequality, we can obtain
$$\begin{array}{}{\displaystyle \underset{{R}^{N}}{\int}[f(x,{u}_{n}(x))-f(x,u(x))]({u}_{n}-u)dx\le \underset{{R}^{N}}{\int}|f(x,{u}_{n}(x))-f(x,u(x))||{u}_{n}-u|dx}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle \le \underset{{R}^{N}}{\int}a1(u+{u}_{n})+{a}_{2}(|u{|}^{q-1}+|{u}_{n}{|}^{q-1})|{u}_{n}-u|dx}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\le {a}_{1}(\parallel {u}_{n}{\parallel}_{2}+\parallel u{\parallel}_{2})\parallel {u}_{n}-u{\parallel}_{2}+{a}_{2}(\parallel {u}_{n}{\parallel}_{q}^{q-1}+\parallel u{\parallel}_{q}^{q-1})\parallel {u}_{n}-u{\parallel}_{q}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\to 0,\text{\hspace{0.17em}as\hspace{0.17em}}n\to \mathrm{\infty}.\end{array}$$(14)

It follows from (11)-(14) that ∥*u*_{n} − *u* ∥^{2} → 0, which shows that *u*_{n} → *u* in *X*. Hence, *φ* satisfies the (*C*)_{c} condition.

We complete the proof of Lemma 3.1.□

Let {*e*_{j}} be a total orthonormal basis of *X*. We define
$$\begin{array}{}{\displaystyle {X}_{j}:=span\{{e}_{j}\},{Y}_{k}:={\oplus}_{j=1}^{k}{X}_{j}\text{\hspace{0.17em}and\hspace{0.17em}}{Z}_{k}=\overline{{\oplus}_{j=k+1}^{\mathrm{\infty}}{X}_{j},}\phantom{\rule{thinmathspace}{0ex}}k\in \mathbb{N}.}\end{array}$$

Clearly, *X* = *Y*_{k} ⊕ *Z*_{k} with dim *Y*_{k} < ∞.

#### Lemma 3.2

*Assume that* (*V*_{0}), (*H*_{1})-(*H*_{3}) *hold*. *Then there exist constants* *a*, *ρ* > 0 *such that* *φ*|_{∂ Bρ ∩ Zk} ≥ *a*, *where* *B*_{ρ} = {*u* ∈ *X* :∥*u*∥ < *ρ*}.

#### Proof

From (*H*_{1}), (4) and Hölder inequality, for any
$\begin{array}{}u\in {Z}_{k},q\in [2,\frac{2N+4\alpha}{N}),\end{array}$
we have
$$\begin{array}{}{\displaystyle \phi (u)=\frac{1}{2}\parallel u{\parallel}^{2}-\underset{{R}^{N}}{\int}F(x,u)dx-\frac{\lambda}{p}\underset{{R}^{N}}{\int}h(x)|u{|}^{p}dx}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle \ge \frac{1}{2}\parallel u{\parallel}^{2}-\frac{{a}_{1}}{2}\parallel {u}_{n}{\parallel}_{2}^{2}-\frac{{a}_{2}}{q}\parallel {u}_{n}{\parallel}_{q}^{q}-{\lambda}_{0}\parallel h{\parallel}_{\frac{2}{2-p}}\parallel {u}_{n}{\parallel}_{2}^{p}}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle \frac{1}{2}\parallel u{\parallel}^{2}-\frac{{a}_{1}}{2{S}_{2}^{2}}\parallel {u}_{n}{\parallel}^{2}-\frac{{a}_{2}}{q{S}_{q}^{q}}\parallel {u}_{n}{\parallel}^{q}-{\lambda}_{0}\parallel h{\parallel}_{\frac{2}{2-p}}\parallel {u}_{n}{\parallel}_{2}^{p}.}\end{array}$$

We denote
$\begin{array}{}c=\frac{1}{2}-(\frac{{a}_{1}}{2{S}_{2}^{2}}+\frac{{a}_{2}}{q{S}_{q}^{q}}),\end{array}$
then (*H*_{1}) implies *c* > 0. Let
$$\begin{array}{}{\alpha}_{k}=\underset{u\in {Z}_{k},||u||=1}{sup}\parallel u{\parallel}_{r},\text{\hspace{0.17em}for\hspace{0.17em}}r\in [1,{2}_{\alpha}^{\ast}).\end{array}$$

By a similar proof to the Lemma 3.2 of [19], we can deduce that *α*_{k} → 0 as *k* → ∞. Assume 0 < *ρ*≤ 1, then we have
$$\begin{array}{}\phi {|}_{\mathrm{\partial}{B}_{\rho}\cap {Z}_{k}}\ge c\parallel u{\parallel}^{2}-{\lambda}_{0}\parallel h{\parallel}_{\frac{2}{2-p}}{\alpha}_{k}^{p}\parallel u{\parallel}^{p}=\parallel u{\parallel}^{p}[c\parallel u{\parallel}^{2-p}-{\lambda}_{0}\parallel h{\parallel}_{\frac{2}{2-p}}{\alpha}_{k}^{p}].\end{array}$$

Then it is easy to see that we can choose constant *ρ* ∈ (0,1] such that *φ*|_{∂ Bρ ∩ Zk} > 0, as *k* →∞, which implies the conclusion of Lemma 3.2.□

#### Lemma 3.3

*Assume that* (*V*_{0}), (*H*_{1})-(*H*_{3}) *hold*. *Then for any finite dimensional subspace* *W* ⊂ *X*, *there exists* *R* = *R*(*W*) *such that* *φ*(*u*) ≤ 0 *on* *W* ∖ *B*_{R(W)}.

#### Proof

First, we prove that for any finite dimensional subspace *W* ⊂ *X* and ∥*u*∥ →∞, *u* ∈ *W*, there holds *φ*(*u*) → − ∞.

On the contrary, we assume that for some sequence {*u*_{n}} ⊂ *W* with ∥*u*_{n}∥ →∞, there exists a positive constant *M* > 0 such that
$$\begin{array}{}\phi ({u}_{n})\ge -M.\end{array}$$(15)

Let
$\begin{array}{}{v}_{n}=\frac{{u}_{n}}{||{u}_{n}||},\end{array}$
then ∥*v*_{n}∥ = 1. From the boundedness of *v*_{n}, there exits a weakly convergent subsequence *v*_{n} such that *v*_{n} ⇀ *v* weakly in *X*. Then for the finite dimensional subspace *W* ⊂ *X*, we know that *v*_{n} → *v* strongly in *W*. By the equivalence of finite dimensional spaces, we can get that *v*_{n} → *v* a.e. in *R*^{N}, which also implies ∥*v*∥ = 1. Let *V* : = {*x* ∈ *R*^{N}:*v*(*x*) ≠ 0}, then we know the measure of the set *V* is positive, i.e., *meas*(*V*) > 0. Hence, for *x* ∈ *V*, from *u*_{n} = ∥*u*_{n}∥*v*_{n} we can deduce
$$\begin{array}{}|{u}_{n}|\to \mathrm{\infty},\text{\hspace{0.17em}as\hspace{0.17em}}n\to \mathrm{\infty}.\end{array}$$(16)

For any 0 < *x*_{1} < *x*_{2}, we denote Ω_{n}(*x*_{1}, *x*_{2}) = {*x* ∈ *R*^{N}:*x*_{1} ≤ |*u*_{n}| < *x*_{2}}, which implies Ω_{n}(*x*_{1}, *x*_{2}) ⊂ *V*. Then for *n* large enough, it easy to see that *χ*_{Ωn(r0,∞)}(*x*) = 1, where *χ*_{Ωn(r0,∞)} is the characteristic function on Ω_{n}(*r*_{0},∞), *r*_{0} > 0 is given in (*H*_{2}). Hence, we have
$$\begin{array}{}{\chi}_{{\mathrm{\Omega}}_{n}({r}_{0},\mathrm{\infty})}(x){v}_{n}\to v,\text{\hspace{0.17em}as\hspace{0.17em}}n\to \mathrm{\infty},\text{\hspace{0.17em}for\hspace{0.17em}}\text{a}.\text{e}.x\in {\mathrm{\Omega}}_{n}({r}_{0},\mathrm{\infty}).\end{array}$$(17)

On one hand, from (4), (15) and Remark 2.3, for any *λ* ∈ (0, *λ*_{0}), we have
$$\begin{array}{}{\displaystyle \underset{n\to +\mathrm{\infty}}{lim}\frac{\underset{{R}^{N}}{\int}F(x,{u}_{n})dx}{\parallel {u}_{n}{\parallel}^{2}}=\underset{n\to +\mathrm{\infty}}{lim}\frac{\frac{1}{2}\parallel {u}_{n}{\parallel}^{2}-\frac{\lambda}{p}\underset{{R}^{N}}{\int}h(x)|{u}_{n}{|}^{p}dx-\phi ({u}_{n})}{\parallel {u}_{n}{\parallel}^{2}}}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}{\displaystyle \le \underset{n\to +\mathrm{\infty}}{lim}\frac{\frac{1}{2}\parallel {u}_{n}{\parallel}^{2}+\frac{{\lambda}_{0}}{{S}_{2}^{p}}\parallel h|{|}_{\frac{2}{2-p}}\parallel u{\parallel}^{p}+M}{\parallel {u}_{n}|{|}^{2}}=\frac{1}{2}.}\end{array}$$(18)

On the other hand, (*H*_{2}) shows there exists *r*_{0} > 0 such that *F*(*x*, *u*) ≥ 0, for |*u*| ≥ *r*_{0}. From (16), we know that
$$F(x,{u}_{n})\phantom{\rule{thinmathspace}{0ex}}\ge \phantom{\rule{thinmathspace}{0ex}}0,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{for}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}n\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{large}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{enough}.$$(19)

Then it follows from (17), (19), (*H*_{2}) and Fatou’s Lemma that
$$\begin{array}{}{\displaystyle \underset{n\to \mathrm{\infty}}{lim}\frac{\underset{{R}^{N}}{\int}F(x,{u}_{n})dx}{\parallel {u}_{n}{\parallel}^{2}}\ge \underset{n\to \mathrm{\infty}}{lim}\frac{\underset{V}{\int}F(x,{u}_{n})dx}{\parallel {u}_{n}{\parallel}^{2}}=\underset{n\to \mathrm{\infty}}{lim}\frac{\underset{V}{\int}F(x,{u}_{n}){v}_{n}^{2}dx}{{u}_{n}^{2}}}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle \ge \underset{n\to \mathrm{\infty}}{\text{lim}\phantom{\rule{thinmathspace}{0ex}}\text{inf}}\frac{\underset{V}{\int}F(x,{u}_{n}){v}_{n}^{2}}{{u}_{n}^{2}}dx\ge \underset{V}{\int}\underset{n\to \mathrm{\infty}}{\text{lim}\phantom{\rule{thinmathspace}{0ex}}\text{inf}}\frac{F(x,{u}_{n}){v}_{n}^{2}}{{u}_{n}^{2}}dx}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle \ge \underset{{\mathrm{\Omega}}_{n}({r}_{0},\mathrm{\infty})}{\int}\underset{n\to \mathrm{\infty}}{\text{lim}\phantom{\rule{thinmathspace}{0ex}}\text{inf}}\frac{F(x,{u}_{n}){v}_{n}^{2}}{{u}_{n}^{2}}dx=\underset{{\mathrm{\Omega}}_{n}({r}_{0},\mathrm{\infty})}{\int}\underset{n\to \mathrm{\infty}}{\text{lim}\phantom{\rule{thinmathspace}{0ex}}\text{inf}}\frac{F(x,{u}_{n}){v}_{n}^{2}}{|{u}_{n}{|}^{2}}{\chi}_{{\mathrm{\Omega}}_{n}({r}_{0},\mathrm{\infty})}^{2}(x){v}_{n}^{2}dx}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\mathrm{\infty},\end{array}$$

which is a contradiction with (18). Therefore *φ*(*u*) → −∞ for ‖*u*‖ → ∞, *u* ∈ *W*. Hence, we can easily choose *R* = *R*(*W*) such that *φ*(*u*) ≤ 0 on *W* \ *B*_{R(W)}. Then we complete the proof.□

#### Proof of Theorem 1.1

Let *Y* = *Y*_{k}, *Z* = *Z*_{k}, then *X* = *Y* ⊕ *Z* with dim *Y* < ∞. From the condition that
$F(x,-u)=F(x,u)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}h\in {L}^{\frac{2}{2-p}}({R}^{N}),$ we know *φ* is even and *φ*(0) = 0. Lemma 3.1-3.3 imply that *φ* satisfies other conditions of Lemma 2.5. Consequently, we can deduce that *φ* possesses an unbounded sequence of critical values, which are the solutions of the fractional Schrödinger equation (1).□

Finally, we give one example to illustrate the usefulness of our main result. Consider the following fractional Schrödinger equations.

#### Example 3.4

$$\begin{array}{}(-{\displaystyle \mathrm{\Delta}{)}^{\frac{1}{2}}u+(1+{x}^{2})u=\frac{8{S}_{8/3}^{8/3}({\mathrm{sin}}^{2}x){u}^{5/3}}{9}+\frac{\mathrm{ln}(1+|\mathrm{sin}{x}^{2}|)}{{e}^{|x|}(1+{x}^{2})}|u{|}^{-1}u,\phantom{\rule{1em}{0ex}}x\in {R}^{2}.}\end{array}$$(20)

*Obviously*,
$\alpha \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1/2,N\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2,p\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1,\lambda \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1,\phantom{\rule{thinmathspace}{0ex}}f(x,{\displaystyle u)\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\frac{8{S}_{8/3}^{8/3}({\mathrm{sin}}^{2}x){u}^{5/3}}{9}}$
*is continuous*, *V*(*x*) = 1 + *x*^{2} *and*
$h(x)={\displaystyle \frac{\mathrm{ln}(1+|\mathrm{sin}{x}^{2}|)}{{e}^{|x|}(1+{x}^{2})}}$ *is a L*^{2} *integrable function*.

*First, we can see that*
$2\alpha \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}<\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}N\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}2,\phantom{\rule{thinmathspace}{0ex}}F(x,{\displaystyle u)\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\frac{{S}_{8/3}^{8/3}({\mathrm{sin}}^{2}x){u}^{8/3}}{3},}$ *then we have*
$$|f(x,u)|=|\frac{8{S}_{8/3}^{8/3}({\mathrm{sin}}^{2}x){u}^{5/3}}{9}|\le {S}_{8/3}^{8/3}|u{|}^{\frac{8}{3}-1},$$

*with*
$\frac{8}{3}=q\in [2,3)\subset [2,{2}_{\alpha}^{\ast}=4),{a}_{1}=0,{a}_{2}={S}_{8/3}^{8/3}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{{a}_{1}}{2{S}_{2}^{2}}+\frac{{a}_{2}}{q{S}_{q}^{q}}=\frac{3}{8}<\frac{1}{2},$ *which shows that (H*_{1}) of Theorem 1.1 holds.

*From the fact that*
$2F(x,{\displaystyle u)=\frac{2{S}_{8/3}^{8/3}({\mathrm{sin}}^{2}x){u}^{8/3}}{3}<\frac{8{S}_{8/3}^{8/3}({\mathrm{sin}}^{2}x){u}^{8/3}}{9}=uf(x,u),}$ *we can verify (H*_{3}) of Theorem 1.1 is also satisfied.

It is also easy to check that the hypotheses (*V*_{0}), (*H*_{2}) and other conditions of Theorem 1.1 hold. Then all the conditions in Theorem 1.1 are satisfied. In virtue of Theorem 1.1, we conclude that (20) possesses infinitely many solutions.

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