To prove Theorem 1.1, we use a variant version of the mountain pass theorem, which allows us to find a so-called Cerami type (PS) sequence.
The properties of this kind of (PS) sequence are very helpful in showing the boundedness of the sequence in the asymptotically linear case.

#### Theorem 2.1 ([4]).

*Let **E* be a real Banach space with its dual space ${E}^{\mathrm{*}}$, and suppose that $I\mathrm{\in}{C}^{\mathrm{1}}\mathit{}\mathrm{(}E\mathrm{,}R\mathrm{)}$ satisfies

$\mathrm{max}\{I(0),I(e)\}\le \mu <\eta \le \underset{\parallel u\parallel =\rho}{inf}I(u)$

*for some $\mu \mathrm{<}\eta $, $\rho \mathrm{>}\mathrm{0}$ and $e\mathrm{\in}E$ with $\mathrm{\parallel}e\mathrm{\parallel}\mathrm{>}\rho $. Let $c\mathrm{\ge}\eta $ be characterized by*

$c=\underset{\gamma \in \mathrm{\Gamma}}{inf}\underset{0\le \tau \le 1}{\mathrm{max}}I(\gamma (\tau )),$

*where $\mathrm{\Gamma}\mathrm{=}\mathrm{\{}\gamma \mathrm{\in}C\mathit{}\mathrm{(}\mathrm{[}\mathrm{0}\mathrm{,}\mathrm{1}\mathrm{]}\mathrm{,}E\mathrm{)}\mathrm{:}\gamma \mathit{}\mathrm{(}\mathrm{0}\mathrm{)}\mathrm{=}\mathrm{0}\mathrm{,}\gamma \mathit{}\mathrm{(}\mathrm{1}\mathrm{)}\mathrm{=}e\mathrm{\}}$ is the set of continuous paths joining 0 and **e*.
Then there exists a sequence $\mathrm{\{}{u}_{n}\mathrm{\}}\mathrm{\subset}E$ such that

$I({u}_{n})\to c\ge \eta \mathit{\hspace{1em}}\text{\mathit{a}\mathit{n}\mathit{d}}\mathit{\hspace{1em}}(1+\parallel {u}_{n}\parallel ){\parallel {I}^{\prime}({u}_{n})\parallel}_{{E}^{*}}\to 0\mathit{\hspace{1em}}\mathit{\text{as}}n\to \mathrm{\infty}.$

This kind of sequence is usually called a Cerami sequence.

#### Lemma 2.1.

*Assume that (A1), (A2) and (A3) hold.
Then there exist $\rho \mathrm{>}\mathrm{0}$, $\eta \mathrm{>}\mathrm{0}$ such that*

$inf\{I(u):u\in H\mathit{\text{with}}\parallel u\parallel =\rho \}\eta .$

#### Proof.

By (A1) and (A2), for any $\u03f5>0$, there exists ${C}_{\u03f5}>0$ such that

$f(s)\le \u03f5|s|+{C}_{\u03f5}{|s|}^{{2}^{*}-1}\mathit{\hspace{1em}}\text{for all}x\in \mathbb{R},$(2.1)

and then

$F(s)\le \frac{\u03f5}{2}{|s|}^{2}+\frac{{C}_{\u03f5}}{{2}^{*}}{|s|}^{{2}^{*}}\mathit{\hspace{1em}}\text{for all}x\in \mathbb{R}.$(2.2)

Moreover, by (A1), (A2) and (A3), there exists ${C}_{1}>0$ such that

$a(x)\le {C}_{1}\mathit{\hspace{1em}}\text{for all}x\in {\mathbb{R}}^{N}.$(2.3)

So, from (2.2), (2.3) and the Sobolev inequality, for any $u\in H$, we have

$\left|{\int}_{{\mathbb{R}}^{N}}a(x)F(u)dx\right|\le \frac{\u03f5{C}_{1}}{2}{\int}_{{\mathbb{R}}^{N}}|u{|}^{2}dx+\frac{{C}_{\u03f5}{C}_{1}}{{2}^{*}}{\int}_{{\mathbb{R}}^{N}}|u{|}^{{2}^{*}}dx\le \frac{\u03f5{C}_{1}}{2}\parallel u{\parallel}^{2}+\frac{{C}_{\u03f5}{C}_{2}}{{2}^{*}}\parallel u{\parallel}^{{2}^{*}}.$

Thus, one has

${I}_{b}(u)=\frac{{\parallel u\parallel}^{2}}{2}+\frac{b}{4}({\int}_{{\mathbb{R}}^{N}}{|\nabla u{|}^{2}dx)}^{2}-{\int}_{{\mathbb{R}}^{N}}a(x)F(u)dx\ge \frac{1-\u03f5{C}_{1}}{2}\parallel u{\parallel}^{2}-\frac{{C}_{\u03f5}{C}_{2}}{{2}^{*}}\parallel u{\parallel}^{{2}^{*}},$

thanks to the fact that $\frac{b}{4}({\int}_{{\mathbb{R}}^{N}}{|\nabla u{|}^{2}dx)}^{2}$ is nonnegative.
Fixing $\u03f5\in (0,{C}_{1}^{-1})$ and letting $\parallel u\parallel =\rho >0$ be small enough, there exists $\eta >0$ such that the desired conclusion holds.
∎

#### Lemma 2.2.

*Assume that (A1), (A2) and (A3) hold.
If $l\mathrm{>}\mu $, then there is $\stackrel{\mathrm{~}}{b}\mathrm{>}\mathrm{0}$ such that for any $b\mathrm{\in}\mathrm{(}\mathrm{0}\mathrm{,}\stackrel{\mathrm{~}}{b}\mathrm{)}$, there exists $e\mathrm{\in}H$ with $\mathrm{\parallel}e\mathrm{\parallel}\mathrm{>}\rho $ such that ${I}_{b}\mathit{}\mathrm{(}e\mathrm{)}\mathrm{<}\mathrm{0}$.*

#### Proof.

According to the definition of μ and $l>\mu $, there exists $\varphi \in H$, with $\varphi \ge 0$, such that

${\int}_{{\mathbb{R}}^{N}}a(x){\varphi}^{2}\mathit{d}x=1\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}\mu \le {\int}_{{\mathbb{R}}^{N}}({|\nabla \varphi |}^{2}+{\varphi}^{2})\mathit{d}x<l.$

By (A2) and Fatou’s Lemma, we have

$\underset{t\to +\mathrm{\infty}}{lim}\frac{{I}_{0}(t\varphi )}{{t}^{2}}=\frac{{\parallel \varphi \parallel}^{2}}{2}-\underset{t\to +\mathrm{\infty}}{lim}{\int}_{{\mathbb{R}}^{N}}a(x)\frac{F(t\varphi )}{{(t\varphi )}^{2}}{\varphi}^{2}\mathit{d}x\le \frac{1}{2}({\parallel \varphi \parallel}^{2}-l)<0,$

by taking $e={t}_{0}\varphi $ with ${t}_{0}$ large enough so that ${I}_{0}(e)={I}_{0}({t}_{0}\varphi )<0$ and $\parallel e\parallel ={t}_{0}\parallel \varphi \parallel >\rho $.
Since

${I}_{b}(e)={I}_{0}(e)+\frac{b}{4}({\int}_{{\mathbb{R}}^{N}}{|\nabla e{|}^{2}dx)}^{2}$

is continuous and increasing in $b\ge 0$ and ${I}_{0}(e)<0$, there exists $\stackrel{~}{b}>0$ sufficiently small such that ${I}_{b}(e)<0$ for all $b\in (0,\stackrel{~}{b})$.
∎

By means of Lemmas 2.1–2.2 and Theorem 2.1, there exists a sequence $\{{u}_{n}\}\subset H$ such that

${I}_{b}({u}_{n})\to c\ge \eta \mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}(1+\parallel {u}_{n}\parallel ){\parallel {I}_{b}^{\prime}({u}_{n})\parallel}_{{H}^{-1}}\to 0\mathit{\hspace{1em}}\text{as}n\to \mathrm{\infty}.$(2.4)

In order to get the existence of a nontrivial nonnegative solution, we first show that this sequence is bounded.

#### Lemma 2.3.

*Let (A1), (A2) and (A3) hold, and let $l\mathrm{>}\mu $.
Then, for any $b\mathrm{\in}\mathrm{(}\mathrm{0}\mathrm{,}\stackrel{\mathrm{~}}{b}\mathrm{)}$, where $\stackrel{\mathrm{~}}{b}$ given by Lemma 2.2,
the sequence $\mathrm{\{}{u}_{n}\mathrm{\}}$ defined in (2.4) is bounded in **H*.

#### Proof.

Assume on the contrary that $\parallel {u}_{n}\parallel \to +\mathrm{\infty}$ as $n\to \mathrm{\infty}$.
Define ${\omega}_{n}=\frac{{u}_{n}}{\parallel {u}_{n}\parallel}$.
Clearly, $\{{\omega}_{n}\}$ is bounded in *H* and there exists $\omega \in H$ such that, going if necessary to a
subsequence,

${\omega}_{n}\rightharpoonup \omega \hspace{1em}\text{in}H,{\omega}_{n}\to \omega \hspace{1em}\text{in}{L}_{\mathrm{loc}}^{2}({\mathbb{R}}^{N}),$${\omega}_{n}\to \omega \hspace{1em}\text{a.e. in}{\mathbb{R}}^{N}\text{as}n\to \mathrm{\infty}.$(2.5)

We claim that $\omega \not\equiv 0$.
Assume on the contrary that $\omega \equiv 0$.
By (A3), there exists a constant $\theta \in (0,1)$ such that

$sup\{{\displaystyle \frac{f(s)}{s}}:s>0\}<\theta inf\{{\displaystyle \frac{1}{a(x)}}:|x|\ge {R}_{0}\}.$

For any $n\in \mathbb{N}$, this yields

${\int}_{|x|\ge {R}_{0}}a(x)\frac{f({u}_{n})}{{u}_{n}}{\omega}_{n}^{2}\mathit{d}x\le \theta {\int}_{|x|\ge {R}_{0}}{\omega}_{n}^{2}\mathit{d}x\le \theta {\parallel {\omega}_{n}\parallel}^{2}=\theta <1.$(2.6)

Since the embedding ${H}^{1}({B}_{{R}_{0}}(0))\hookrightarrow {L}^{2}({B}_{{R}_{0}}(0))$ is compact, we have ${\omega}_{n}\to \omega $ in ${L}^{2}({B}_{{R}_{0}}(0))$.
According to [27, Lemma A.1], going if necessary to a subsequence, there exists $g\in {L}^{2}({B}_{{R}_{0}}(0))$ such that

$|{\omega}_{n}|\le g(x)\mathit{\hspace{1em}}\text{a.e. in}{B}_{{R}_{0}}(0).$(2.7)

By (A1) and (A2), there exists $C>0$ such that

$\frac{f(t)}{t}\le C\mathit{\hspace{1em}}\text{for all}t\in R.$(2.8)

By (2.3), (2.7) and (2.8), for any $n\in \mathbb{N}$, we have

$0\le a(x)\frac{f({u}_{n})}{{u}_{n}}{\omega}_{n}^{2}\le Ca(x){\omega}_{n}^{2}\le C{C}_{1}{\omega}_{n}^{2}\le C{C}_{1}{g}^{2}\mathit{\hspace{1em}}\text{a.e. in}{B}_{{R}_{0}}(0).$(2.9)

Noting that ${\omega}_{n}\to \omega \equiv 0$ a.e. in ${\mathbb{R}}^{N}$, we obtain

$a(x)\frac{f({u}_{n})}{{u}_{n}}{\omega}_{n}^{2}\to 0\mathit{\hspace{1em}}\text{a.e. in}{\mathbb{R}}^{N}.$(2.10)

It follows, from (2.9), (2.10) and the dominated convergence theorem, that

$\underset{n\to \mathrm{\infty}}{lim}{\int}_{|x|<{R}_{0}}a(x)\frac{f({u}_{n})}{{u}_{n}}{\omega}_{n}^{2}\mathit{d}x=0.$(2.11)

Thus, by (2.6) and (2.11), we get

$\underset{n\to \mathrm{\infty}}{lim\; sup}{\int}_{{\mathbb{R}}^{N}}a(x)\frac{f({u}_{n})}{{u}_{n}}{\omega}_{n}^{2}\mathit{d}x<1.$(2.12)

Since $\parallel {u}_{n}\parallel \to +\mathrm{\infty}$ as $n\to \mathrm{\infty}$, it follows from (2.4) that

$o(1)=\frac{\u3008{I}_{b}^{\prime}({u}_{n}),{u}_{n}\u3009}{{\parallel {u}_{n}\parallel}^{2}}=1+\frac{b({\int}_{{\mathbb{R}}^{N}}{|\nabla {u}_{n}{|}^{2}dx)}^{2}}{{\parallel {u}_{n}\parallel}^{2}}-{\int}_{{\mathbb{R}}^{N}}a(x)\frac{f({u}_{n})}{{u}_{n}}{\omega}_{n}^{2}\mathit{d}x\ge 1-{\int}_{{\mathbb{R}}^{N}}a(x)\frac{f({u}_{n})}{{u}_{n}}{\omega}_{n}^{2}\mathit{d}x.$

Therefore,

${\int}_{{\mathbb{R}}^{N}}a(x)\frac{f({u}_{n})}{{u}_{n}}{\omega}_{n}^{2}\mathit{d}x+o(1)\ge 1,$

which contradicts (2.12), so $\omega \not\equiv 0$.

On the other hand, since $\parallel {u}_{n}\parallel \to +\mathrm{\infty}$ as $n\to \mathrm{\infty}$, it follows form (2.4) that

$\frac{\u3008{I}_{b}^{\prime}({u}_{n}),{u}_{n}\u3009}{{\parallel {u}_{n}\parallel}^{4}}=o(1),$

that is,

$o(1)=\frac{1}{{\parallel {u}_{n}\parallel}^{2}}+b({\int}_{{\mathbb{R}}^{N}}{|\nabla {\omega}_{n}{|}^{2}dx)}^{2}-\frac{{\int}_{{\mathbb{R}}^{N}}a(x)\frac{f({u}_{n})}{{u}_{n}}{\omega}_{n}^{2}\mathit{d}x}{{\parallel {u}_{n}\parallel}^{2}}.$(2.13)

From (2.3) and (2.8), one has

$\frac{{\int}_{{\mathbb{R}}^{N}}a(x)\frac{f({u}_{n})}{{u}_{n}}{\omega}_{n}^{2}\mathit{d}x}{{\parallel {u}_{n}\parallel}^{2}}=o(1).$(2.14)

From (2.13) and (2.14), it is clear that

$b({\int}_{{\mathbb{R}}^{N}}{|\nabla {\omega}_{n}{|}^{2}dx)}^{2}=o(1).$

Fatou’s Lemma yields

$0=\underset{n\to \mathrm{\infty}}{lim\; inf}b({\int}_{{\mathbb{R}}^{N}}{|\nabla {\omega}_{n}{|}^{2}dx)}^{2}\ge b({\int}_{{\mathbb{R}}^{N}}{|\nabla \omega {|}^{2}dx)}^{2}\ge 0,$

that is,

$({\int}_{{\mathbb{R}}^{N}}{|\nabla \omega {|}^{2}dx)}^{2}=0.$(2.15)

Since the embedding $H\hookrightarrow {D}^{1,2}({\mathbb{R}}^{N})$ is continuous, ω also belongs to ${D}^{1,2}({\mathbb{R}}^{N})$.
According to the definition of the norm of ${D}^{1,2}({\mathbb{R}}^{N})$ and (2.15), $\omega \equiv 0$.
This is a contradiction.
So, the sequence $\{{u}_{n}\}$ is bounded in *H*.
∎

To prove that the Cerami sequence $\{{u}_{n}\}$ in (2.4) converges to a nonzero critical point of ${I}_{b}$, we need the following compactness lemma.

#### Lemma 2.4.

*Assume that (A1), (A2) and (A3) hold.
Then, for any $\u03f5\mathrm{>}\mathrm{0}$, there exists $R\mathit{}\mathrm{(}\u03f5\mathrm{)}\mathrm{>}{R}_{\mathrm{0}}$ and $n\mathit{}\mathrm{(}\u03f5\mathrm{)}$ such that*

${\int}_{|x|\ge R}({|\nabla {u}_{n}|}^{2}+{u}_{n}^{2})\mathit{d}x\le \u03f5$

*for all $R\mathrm{\ge}R\mathit{}\mathrm{(}\u03f5\mathrm{)}$ and $n\mathrm{\ge}n\mathit{}\mathrm{(}\u03f5\mathrm{)}$.*

#### Proof.

Let ${\xi}_{R}:{\mathbb{R}}^{N}\to [0,1]$ be a smooth function such that

${\xi}_{R}(x)=\{\begin{array}{cc}0,\hfill & 0\le |x|\le R,\hfill \\ 1,\hfill & |x|\ge 2R,\hfill \end{array}$(2.16)

and, for some constant ${C}_{0}>0$ (independent of *R*),

$|\nabla {\xi}_{R}(x)|\le \frac{{C}_{0}}{R}\mathit{\hspace{1em}}\text{for all}x\in {\mathbb{R}}^{N}.$

Then, for all $n\in \mathbb{N}$ and $R\ge {R}_{0}$, we have

${\int}_{{\mathbb{R}}^{N}}}|\nabla ({u}_{n}{\xi}_{R}){|}^{2}dx\le 2{\displaystyle {\int}_{{\mathbb{R}}^{N}}}|\nabla {u}_{n}{|}^{2}{\xi}_{R}^{2}dx+{\displaystyle {\int}_{{\mathbb{R}}^{N}}}|{u}_{n}{|}^{2}|\nabla {\xi}_{R}{|}^{2}dx$$\le 2{\displaystyle {\int}_{{\mathbb{R}}^{N}}}|\nabla {u}_{n}{|}^{2}dx+{\displaystyle \frac{2{C}_{0}^{2}}{{R}^{2}}}{\displaystyle {\int}_{{\mathbb{R}}^{N}}}|{u}_{n}{|}^{2}dx$$\le 2\left(1+{\displaystyle \frac{{C}_{0}^{2}}{{R}^{2}}}\right){\parallel {u}_{n}\parallel}^{2}$$\le 2\left(1+{\displaystyle \frac{{C}_{0}^{2}}{{R}_{0}^{2}}}\right){\parallel {u}_{n}\parallel}^{2}.$

This implies that

$\parallel {u}_{n}{\xi}_{R}\parallel \le \sqrt{2}{\left(2+\frac{{C}_{0}^{2}}{{R}_{0}^{2}}\right)}^{\frac{1}{2}}\parallel {u}_{n}\parallel $(2.17)

for all $n\in \mathbb{N}$ and $R\ge {R}_{0}$.
By (2.4), ${\parallel {I}^{\prime}({u}_{n})\parallel}_{{H}^{-1}}\parallel {u}_{n}\parallel \to 0$ as $n\to \mathrm{\infty}$. So,
for any $\u03f5>0$, there exists $n(\u03f5)>0$ such that

${\parallel {I}^{\prime}({u}_{n})\parallel}_{{H}^{-1}}\parallel {u}_{n}\parallel \le \frac{\u03f5}{\sqrt{2}{\left(2+\frac{{C}_{0}^{2}}{{R}_{0}^{2}}\right)}^{\frac{1}{2}}}$(2.18)

for all $n\ge n(\u03f5)$.
Hence, it follows from (2.17) and (2.18) that

$|\u3008{I}^{\prime}({u}_{n}),{u}_{n}{\xi}_{R}\u3009|\le {\parallel {I}^{\prime}({u}_{n})\parallel}_{{H}^{-1}}\parallel {u}_{n}{\xi}_{R}\parallel \le \u03f5$(2.19)

for all $R\ge {R}_{0}$ and $n\ge n(\u03f5)$.
Note that

$\u3008{I}^{\prime}({u}_{n}),{u}_{n}{\xi}_{R}\u3009={\displaystyle {\int}_{{\mathbb{R}}^{N}}}|\nabla {u}_{n}{|}^{2}{\xi}_{R}dx+{\displaystyle {\int}_{{\mathbb{R}}^{N}}}{u}_{n}^{2}{\xi}_{R}dx+b{\displaystyle {\int}_{{\mathbb{R}}^{N}}}|\nabla {u}_{n}{|}^{2}dx{\displaystyle {\int}_{{\mathbb{R}}^{N}}}|\nabla {u}_{n}{|}^{2}{\xi}_{R}dx+{\displaystyle {\int}_{{\mathbb{R}}^{N}}}{u}_{n}\nabla {u}_{n}\nabla {\xi}_{R}dx$$+b{\displaystyle {\int}_{{\mathbb{R}}^{N}}}|\nabla {u}_{n}{|}^{2}dx{\displaystyle {\int}_{{\mathbb{R}}^{N}}}{u}_{n}\nabla {u}_{n}\nabla {\xi}_{R}dx-{\displaystyle {\int}_{{\mathbb{R}}^{N}}}a(x)f({u}_{n}){u}_{n}{\xi}_{R}dx.$(2.20)

For any $\u03f5>0$, there exists $R(\u03f5)>{R}_{0}$ such that

$\frac{1}{{R}^{2}}\le \frac{4{\u03f5}^{2}}{{C}_{0}^{2}}\mathit{\hspace{1em}}\text{for all}R\ge R(\u03f5).$(2.21)

By (2.21) and Young’s inequality, for all $n\in \mathbb{N}$ and $R\ge R(\u03f5)$, we get

${\int}_{{\mathbb{R}}^{N}}}|{u}_{n}\nabla {u}_{n}\nabla {\xi}_{R}|dx\le \u03f5{\displaystyle {\int}_{{\mathbb{R}}^{N}}}|\nabla {u}_{n}{|}^{2}dx+{\displaystyle \frac{1}{4\u03f5}}{\displaystyle {\int}_{|x|\le 2R}}|{u}_{n}{|}^{2}{\displaystyle \frac{{C}_{0}^{2}}{{R}^{2}}}dx$$\le \u03f5{\displaystyle {\int}_{{\mathbb{R}}^{N}}}|\nabla {u}_{n}{|}^{2}dx+\u03f5{\displaystyle {\int}_{|x|\le 2R}}|{u}_{n}{|}^{2}dx$$\le \u03f5{\parallel {u}_{n}\parallel}^{2}.$(2.22)

By (A1), (A2), (A3) and (2.16), there exists ${\eta}_{1}\in (0,1)$ such that for all $n\in \mathbb{N}$ and $R\ge {R}_{0}$,

${\int}_{{\mathbb{R}}^{N}}|a(x)f({u}_{n}){u}_{n}{\xi}_{R}|dx\le {\eta}_{1}{\int}_{{\mathbb{R}}^{N}}{u}_{n}^{2}{\xi}_{R}dx.$(2.23)

In virtue of the fact that $b{\int}_{{\mathbb{R}}^{N}}|\nabla {u}_{n}{|}^{2}dx{\int}_{{\mathbb{R}}^{N}}|\nabla {u}_{n}{|}^{2}{\xi}_{R}dx$
is nonnegative, together with (2.20), (2.22) and (2.23), for all $n\in \mathbb{N}$ and $R\ge R(\u03f5)\ge {R}_{0}$, we have

$\u3008{I}^{\prime}({u}_{n}),{u}_{n}{\xi}_{R}\u3009\ge {\int}_{{\mathbb{R}}^{N}}|\nabla {u}_{n}{|}^{2}{\xi}_{R}dx+(1-{\eta}_{1}){\int}_{{\mathbb{R}}^{N}}{u}_{n}^{2}{\xi}_{R}dx-\u03f5\parallel {u}_{n}{\parallel}^{2}-\u03f5b\parallel {u}_{n}{\parallel}^{4}.$(2.24)

Since the sequence $\{{u}_{n}\}$ is bounded in *H*, it follows from (2.19) and (2.24) that there exists ${C}_{3}>0$ such that
for all $R\ge R(\u03f5)$ and $n\ge n(\u03f5)$,

${\int}_{{\mathbb{R}}^{N}}|\nabla {u}_{n}{|}^{2}{\xi}_{R}dx+(1-{\eta}_{1}){\int}_{{\mathbb{R}}^{N}}{u}_{n}^{2}{\xi}_{R}dx\le {C}_{3}\u03f5.$

From ${\eta}_{1}\in (0,1)$ and (2.16), the desired conclusion easily follows.
∎

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