Proof. By Equations (2.6) and (2.7), for any $u\in S(a_1)$,

\begin{equation*} \begin{split} J_{\mu,T}(u)&\geq J_{{\rm max},T}(u) \\ & \geq \frac{1}{2}\|\nabla u\|_2^2-\frac{1}{q}h_{{\rm max}}C_{N,q}^qa_1^{q(1-\gamma_q)}\|\nabla u\|_2^{q\gamma_q}\\ &\qquad\qquad-\frac{\eta}{p}\tau(\|\nabla u\|_2)C_{N,p}^pa_1^{p(1-\gamma_p)}\|\nabla u\|_2^{p\gamma_p}\\ &\geq \bar{g}_a(\|\nabla u\|_2)\geq \inf_{r\geq 0}\bar{g}_a(r) \gt -\infty. \end{split} \end{equation*}

The proof is complete.

Proof. Fix $u\in S(a_1)$. For t > 0, we define $u_t(x)=t^{\frac{N}{2}}u(tx)$. Then, $u_t\in S(a_1)$ for all t > 0. By $\tau\geq 0$ and $q\gamma_q \lt 2$, we obtain that

\begin{equation*} \begin{split} J_{\mu,T}(u_t)&\leq \frac{1}{2}\int_{\mathbb{R}^N}|\nabla u_t|^2\,{\rm d}x-\frac{\mu}{q}\int_{\mathbb{R}^N}|u_t|^{q}\,{\rm d}x\\ &=\frac{1}{2}t^2\int_{\mathbb{R}^N}|\nabla u|^2\,{\rm d}x-\frac{\mu}{q}t^{q\gamma_q}\int_{\mathbb{R}^N}|u|^{q}\,{\rm d}x,\\ & \lt 0 \end{split} \end{equation*}

for t > 0 small enough. Thus, $\Upsilon_{\mu,T,a_1} \lt 0$. The proof is complete.

Proof. (1) is trivial. Now we prove (2). It follows from $J_{\mu,T}(u) \lt 0$ and

\begin{equation*}J_{\mu,T}(u)\geq \bar{g}_{a_1}(\|\nabla u\|_2)\geq \bar{g}_{a}(\|\nabla u\|_2),\end{equation*}

that $\bar{g}_a(\|\nabla u\|_2) \lt 0$, which implies that $\|\nabla u\|_2 \lt R_0$ by (2.7). By (1) and $J_{\mu,T}(u) \lt 0$, we obtain that $J_{\mu,T}(v) \lt 0$ for all v in a small neighborhood of u in $H^1 (\mathbb{R}^N)$, which combined with $\|v\|_2\leq a$ gives that $\|\nabla v\|_2 \lt R_0$ and thus $J_{\mu,T}(v) = J_{\mu} (v)$. The proof is complete.

Proof. Noting that $\|\nabla u_n\|_2 \lt R_0$ for n large enough, there exists $u\in H^1(\mathbb{R}^N)$ such that $u_n\rightharpoonup u$ in $H^1(\mathbb{R}^N)$ up to a subsequence. Now we consider the following three possibilities.

(1) If $u\not\equiv0$ and $\|u\|_2=b\neq a_1$, we must have $b\in (0,a_1)$. Setting $v_n=u_n-u$, $d_n=\|v_n\|_2$, and by using

\begin{equation*} \|u_n\|_2^2=\|v_n\|_2^2+\|u\|_2^2+o_n(1), \end{equation*}

we obtain that $\|v_n\|_2\to d$, where $a_1^2=d^2+b^2$. Noting that $d_n\in (0,a_1)$ for n large enough, and using the Brézis–Lieb Lemma (see [Reference Willem41]), Lemma 3.5, $\|\nabla u_n\|_2^2=\|\nabla u\|_2^2+\|\nabla v_n\|_2^2+o_n(1)$, $\|\nabla u\|_2^2\leq \liminf_{n\to+\infty}\|\nabla u_n\|_2^2$, τ is continuous and non-increasing, we obtain that

\begin{equation*} \begin{split} \Upsilon_{\mu,T,a_1}+o_n(1)=J_{\mu,T}(u_n)&=\frac{1}{2}\|\nabla v_n\|_2^2-\frac{\mu}{q}\|v_n\|_q^q-\frac{\eta}{p}\tau(\|\nabla u_n\|_2)\|v_n\|_p^p\\ &\quad + \frac{1}{2}\|\nabla u\|_2^2-\frac{\mu}{q}\|u\|_q^q-\frac{\eta}{p}\tau(\|\nabla u_n\|_2)\|u\|_p^p+o_n(1)\\ &\geq J_{\mu,T}(v_n)+J_{\mu,T}(u)+o_n(1)\\ &\geq \Upsilon_{\mu,T,d_n}+\Upsilon_{\mu,T,b}+o_n(1)\\ &\geq \frac{d_n^2}{a_1^2}\Upsilon_{\mu,T,a_1}+\Upsilon_{\mu,T,b}+o_n(1). \end{split} \end{equation*}

Letting $n\to+\infty$, we find that

\begin{equation*} \begin{split} \Upsilon_{\mu,T,a_1}&\geq \frac{d^2}{a_1^2}\Upsilon_{\mu,T,a_1}+\Upsilon_{\mu,T,b}\\ & \gt \frac{d^2}{a_1^2}\Upsilon_{\mu,T,a_1}+\frac{b^2}{a_1^2}\Upsilon_{\mu,T,a_1} =\Upsilon_{\mu,T,a_1}, \end{split} \end{equation*}

which is a contradiction. So this possibility can not exist.

(2) If $\|u\|_2=a_1$, then $u_n\to u$ in $L^2(\mathbb{R}^N)$ and thus $u_n\to u$ in $L^t(\mathbb{R}^N)$ for all $t\in (2,2^*)$.

Case $p \lt 2^*$, then

\begin{equation*} \begin{split} \Upsilon_{\mu,T,a_1}&=\lim_{n\to+\infty}J_{\mu,T}(u_n)\\ &= \lim_{n\to+\infty}\left(\frac{1}{2}\|\nabla u_n\|_2^2-\frac{\mu}{q}\|u_n\|_q^q-\frac{\eta}{p}\tau(\|\nabla u_n\|_2)\|u_n\|_p^p\right)\\ &\geq J_{\mu,T}(u). \end{split} \end{equation*}

As $u\in S(a_1)$, we infer that $J_{\mu,T}(u)=\Upsilon_{\mu,T,a_1}$, then $\|\nabla u_n\|_2\to \|\nabla u\|_2$ and thus $u_n\to u$ in $H^1(\mathbb{R}^N)$, which implies that (i) occurs.

Case $p=2^*$, noting that $\|\nabla v_n\|_2\leq \|\nabla u_n\|_2 \lt R_0$ for n large enough, and using the Sobolev inequality, we have

(3.2)\begin{equation} \begin{split} J_{\mu,T}(v_n)=J_{\mu}(v_n)&=\frac{1}{2}\int_{\mathbb{R}^N}|\nabla v_n|^2\,{\rm d}x-\frac{\mu}{q}\int_{\mathbb{R}^N}|v_n|^{q}\,{\rm d}x -\frac{\eta}{p}\int_{\mathbb{R}^N}|v_n|^{p}\,{\rm d}x\\ &\geq \frac{1}{2}\|\nabla v_n\|_2^2-\frac{\eta}{2^*}\frac{1}{S^{\frac{2^*}{2}}}\|\nabla v_n\|_2^{2^*}+o_n(1)\\ &=\|\nabla v_n\|_2^2\left(\frac{1}{2}-\frac{\eta}{2^*}\frac{1}{S^{\frac{2^*}{2}}}\|\nabla v_n\|_2^{2^*-2}\right)+o_n(1)\\ &\geq \|\nabla v_n\|_2^2\left(\frac{1}{2}-\frac{\eta}{2^*}\frac{1}{S^{\frac{2^*}{2}}}R_0^{2^*-2}\right) +o_n(1)\\ &= \|\nabla v_n\|_2^2\frac{1}{q}h_{\text{max}}C_{N,q}^qa^{q(1-\gamma_q)}R_0^{q\gamma_q-2} +o_n(1), \end{split} \end{equation}

because $w_a(R_0)=\frac{1}{2}-\frac{\eta}{2^*}\frac{1}{S^{\frac{2^*}{2}}}R_0^{2^*-2} -\frac{1}{q}h_{\text{max}}C_{N,q}^qa^{q(1-\gamma_q)}R_0^{q\gamma_q-2}=0$. Now we remember that

(3.3)\begin{equation} \Upsilon_{\mu,T,a_1}\leftarrow J_{\mu,T}(u_n)\geq J_{\mu,T}(v_n)+J_{\mu,T}(u)+o_n(1). \end{equation}

Since $u\in S(a_1)$, we have $J_{\mu,T}(u)\geq \Upsilon_{\mu,T,a_1}$, which combined with Equations (3.2) and (3.3) gives that $\|\nabla v_n\|_2^2\to 0$ and then $u_n\to u$ in $H^1(\mathbb{R}^N)$. This implies that (i) occurs.

(3) If $u\equiv 0$, that is, $u_n\rightharpoonup 0$ in $H^1(\mathbb{R}^N)$. We claim that there exist $R, \beta \gt 0$ and $\{y_n\}\subset \mathbb{R}^N$ such that

(3.4)\begin{equation} \int_{B_R(y_n)}|u_n|^2\,\textrm{d}x\geq \beta,\quad \text{for}\ \text{all} \ n. \end{equation}

Indeed, otherwise we must have $u_n\to 0$ in $L^t(\mathbb{R}^N)$ for all $t\in(2, 2^*)$. Thus, for $p \lt 2^*$, $J_{\mu,T}(u_n)\geq \frac{1}{2}\|\nabla u_n\|_2^2+o_n(1)$, which contradicts $J_{\mu,T}(u_n)\to \Upsilon_{\mu,T,a_1} \lt 0$. For $p=2^*$, similarly to (3.2), we obtain that

\begin{equation*} J_{\mu,T}(u_n)\geq \|\nabla u_n\|_2^2\frac{1}{q}h_{\text{max}}C_{N,q}^qa^{q(1-\gamma_q)}R_0^{q\gamma_q-2} +o_n(1). \end{equation*}

We also get a contradiction in this case. Hence, in all cases, Equation (3.4) holds and $|y_n|\to +\infty$ obviously. From this, considering $\bar{u}_n(x) = u_n(x+y_n)$, clearly $\{\bar{u}_n\}\subset S(a_1)$ and it is also a minimizing sequence with respect to $\Upsilon_{\mu,T,a_1}$. Moreover, there exists $\bar{u}\in H^1(\mathbb{R}^N)\backslash\{0\}$ such that $\bar{u}_n\rightharpoonup \bar{u}$ in $H^1(\mathbb{R}^N)$. Following as in the first two possibilities of the proof, we derive that $\bar{u}_n\to\bar{u}$ in $H^1(\mathbb{R}^N)$, which implies that (ii) occurs.

Proof. By Lemma 3.1, there exists a bounded minimizing sequence $\{u_n\} \subset S(a_1)$ satisfying $J_{\mu,T}(u_n)\to \Upsilon_{\mu,T,a_1}$ as $n\to+\infty$. Now, applying Lemma 3.6, there exists $u\in S(a_1)$ such that $J_{\mu,T}(u)= \Upsilon_{\mu,T,a_1}$. The proof is complete.

Proof. Let $u\in S(a_1)$ satisfy $J_{\mu_1,T}(u)=\Upsilon_{\mu_1,T,a_1}$. Then, $\Upsilon_{\mu_2,T,a_1}\leq J_{\mu_2,T}(u) \lt J_{\mu_1,T}(u) = \Upsilon_{\mu_1,T,a_1}$.

Proof. By Lemma 3.7, choose $u \in S(a)$ such that $J_{\text{max},T}(u)=\Upsilon_{\text{max},T,a}$. Then,

\begin{equation*} \begin{split} \Gamma_{\epsilon,T,a}\leq E_{\epsilon, T}(u)&=\frac{1}{2}\int_{\mathbb{R}^N}|\nabla u|^2\,{\rm d}x-\frac{1}{q}\int_{\mathbb{R}^N}h(\epsilon x)|u|^{q}\,{\rm d}x\\ &\qquad-\frac{\eta}{p}\tau(\|\nabla u\|_2)\int_{\mathbb{R}^N}|u|^{p}\,{\rm d}x. \end{split} \end{equation*}

Letting $\epsilon\to 0^+$, by the Lebesgue dominated convergence theorem, we deduce that

\begin{equation*} \limsup_{\epsilon\to 0^+}\Gamma_{\epsilon,T,a}\leq \limsup_{\epsilon\to 0^+} E_{\epsilon, T}(u)=J_{h(0),T}(u)= J_{\text{max},T}(u)=\Upsilon_{\text{max},T,a}, \end{equation*}

which combined with Lemma 3.2 and Corollary 3.8 completes the proof.

Proof. Assume by contradiction that $u\equiv0$. Then,

\begin{equation*} c+o_n(1)=E_{\epsilon,T}(u_n)=J_{\infty,T}(u_n)+\frac{1}{q}\int_{\mathbb{R}^N}(h_{\infty}-h(\epsilon x))|u_n|^q\,\textrm{d}x. \end{equation*}

By (h ₂), for any given δ > 0, there exists R > 0 such that $h_{\infty}\geq h(x)-\delta$ for all $|x|\geq R$. Hence,

\begin{equation*} \begin{split} c+o_n(1)=E_{\epsilon,T}(u_n)&\geq J_{\infty,T}(u_n)+\frac{1}{q}\int_{B_{R/\epsilon}(0)}(h_{\infty} -h(\epsilon x))|u_n|^q\,{\rm d}x\\ &\qquad-\frac{\delta}{q}\int_{B^c_{R/\epsilon}(0)}|u_n|^q\,{\rm d}x. \end{split} \end{equation*}

Recalling that $\{u_n\}$ is bounded in $H^1(\mathbb{R}^N)$ and $u_n\to 0$ in $L^t(B_{R/\epsilon}(0))$ for all $t\in[1, 2^*)$, it follows that

\begin{equation*} c+o_n(1)=E_{\epsilon,T}(u_n)\geq J_{\infty,T}(u_n)-\delta C+o_n(1), \end{equation*}

for some C > 0. Since δ > 0 is arbitrary, we deduce that $c\geq \Upsilon_{\infty,T,a}$, which is a contradiction. Thus, $u\not\equiv 0$.

Proof. By Lemma 4.2, we must have $\|\nabla u_n\|_2 \lt R_0$ for n large enough, and so, $\{u_n\}$ is also a $(PS)_c$ sequence of E_ϵ restricted to S(a). Hence,

\begin{equation*} E_{\epsilon}(u_n)\to c\qquad {\rm and}\ \|E_{\epsilon}|^\prime_{S(a)}(u_n)\|\to 0\quad {\rm as}\ n\to+\infty. \end{equation*}

Setting the functional $\Psi: H^1(\mathbb{R}^N)\to \mathbb{R}$ given by:

\begin{equation*} \Psi(u)=\frac{1}{2}\int_{\mathbb{R}^N}|u|^2\,\textrm{d}x, \end{equation*}

it follows that $S(a) =\Psi^{-1}(a^2/2)$. Then, by Willem ([Reference Willem41], Proposition 5.12), there exists $\{\lambda_n\}\subset \mathbb{R}$ such that

(4.1)\begin{equation} \|E^\prime_{\epsilon}(u_n)-\lambda_n\Psi^\prime(u_n)\|_{H^{-1}(\mathbb{R}^N)}\to 0,\quad \text{as}\ n\to+\infty. \end{equation}

By the boundedness of $\{u_n\}$ in $H^1(\mathbb{R}^N)$, we know $\{\lambda_n\}$ is bounded and thus, up to a subsequence, there exists λ_ϵ such that $\lambda_n\to \lambda_{\epsilon}$ as $n\to+\infty$. This together with Equation (4.1) leads to

(4.2)\begin{equation} E^\prime_{\epsilon}(u_\epsilon)-\lambda_\epsilon\Psi^\prime(u_\epsilon)=0,\quad \text{in} \ H^{-1}(\mathbb{R}^N), \end{equation}

and then

(4.3)\begin{equation} \|E^\prime_{\epsilon}(v_n)-\lambda_\epsilon\Psi^\prime(v_n)\|_{H^{-1}(\mathbb{R}^N)}\to 0,\quad \text{as}\ n\to+\infty, \end{equation}

where $v_n:=u_n-u_\epsilon$. By direct calculations, we get that

\begin{equation*} \begin{split} \Upsilon_{\infty,T,a}& \gt \lim_{n\to+\infty}E_{\epsilon}(u_n)\\ &=\lim_{n\to+\infty}\left(E_{\epsilon}(u_n)-\frac{1}{2}E_{\epsilon}^\prime(u_n)u_n+\frac{1}{2}\lambda_n\|u_n\|_2^2+o_n(1)\right)\\ &=\lim_{n\to+\infty}\left[\left(\frac{1}{2}-\frac{1}{q}\right)\int_{\mathbb{R}^N}h(\epsilon x)|u_n|^q\,{\rm d}x\right.\\ &\qquad\qquad\qquad\left.+\left(\frac{1}{2}-\frac{1}{p}\right)\eta\int_{\mathbb{R}^N}|u_n|^p\,{\rm d}x+\frac{1}{2}\lambda_na^2+o_n(1)\right]\\ &\geq \frac{1}{2}\lambda_{\epsilon}a^2, \end{split} \end{equation*}

which implies that

(4.4)\begin{equation} \lambda_{\epsilon}\leq \frac{2\Upsilon_{\infty,T,a}}{a^2} \lt 0,\quad \text{for}\ \text{all} \ \epsilon\in (0,\epsilon_1). \end{equation}

By Equation (4.3), we know

(4.5)\begin{equation} \int_{\mathbb{R}^N}|\nabla v_n|^2\,{\rm d}x-\lambda_{\epsilon}\int_{\mathbb{R}^N}|v_n|^2\,{\rm d}x-\int_{\mathbb{R}^N}h(\epsilon x)|v_n|^{q}\,{\rm d}x-\eta\int_{\mathbb{R}^N}|v_n|^{p}\,{\rm d}x=o_n(1), \end{equation}

which combined with Equation (4.4) gives that

(4.6)\begin{equation} \begin{split} \int_{\mathbb{R}^N}|\nabla v_n|^2\,{\rm d}x&-\frac{2\Upsilon_{\infty,T,a}}{a^2}\int_{\mathbb{R}^N}|v_n|^2\,{\rm d}x\\ &\leq h_{{\rm max}}\int_{\mathbb{R}^N}|v_n|^{q}\,{\rm d}x+\eta\int_{\mathbb{R}^N}|v_n|^{p}\,{\rm d}x+o_n(1). \end{split} \end{equation}

If u_n does not converge to u_ϵ strongly in $H^1(\mathbb{R}^N)$, that is, v_n does not converge to 0 strongly in $H^1(\mathbb{R}^N)$, by (4.6) and the Sobolev inequality, we deduce that

\begin{equation*} \begin{split} \int_{\mathbb{R}^N}|\nabla v_n|^2\,{\rm d}x&-\frac{2\Upsilon_{\infty,T,a}}{a^2}\int_{\mathbb{R}^N}|v_n|^2\,{\rm d}x\\ &\leq h_{{\rm max}}C_{N,q}^q\|v_n\|^q+\eta C_{N,p}^p\|v_n\|^p+o_n(1). \end{split} \end{equation*}

So there exists C > 0 independent of ϵ such that $\|v_n\|\geq C$ and then by Equation (4.6):

(4.7)\begin{equation} \limsup_{n\to +\infty}\left(h_{{\rm max}}\int_{\mathbb{R}^N}|v_n|^{q}\,{\rm d}x+\eta\int_{\mathbb{R}^N}|v_n|^{p}\,{\rm d}x\right)\geq C. \end{equation}

Case $p \lt 2^*$, by (4.7) and the Gagliardo–Nirenberg inequality (1.11), there exists β > 0 independent of $\epsilon\in (0,\epsilon_1)$ such that

(4.8)\begin{equation} \limsup_{n\to +\infty}\|v_n\|_2\geq \beta. \end{equation}

Case $p=2^*$, if

\begin{equation*} \limsup_{n\to +\infty}\int_{\mathbb{R}^N}|v_n|^{q}\,\textrm{d}x\geq C, \end{equation*}

for some C > 0 independent of ϵ, we obtain (4.8) as well. If

\begin{equation*} \liminf_{\epsilon\to 0^+}\limsup_{n\to +\infty}\int_{\mathbb{R}^N}|v_n|^{q}\,{\rm d}x=0\qquad {\rm and}\qquad \liminf_{\epsilon\to 0^+}\limsup_{n\to +\infty}\|v_n\|_2=0, \end{equation*}

by Equation (4.7) we have

\begin{equation*} \liminf_{\epsilon\to 0^+}\limsup_{n\to +\infty}\int_{\mathbb{R}^N}|v_n|^{p}\,\textrm{d}x\geq C, \end{equation*}

and by Equation (4.5) we have

\begin{equation*} \liminf_{\epsilon\to 0^+}\limsup_{n\to +\infty}\int_{\mathbb{R}^N}|\nabla v_n|^2\,{\rm d}x=\liminf_{\epsilon\to 0^+}\limsup_{n\to +\infty}\eta\int_{\mathbb{R}^N}|v_n|^{p}\,{\rm d}x. \end{equation*}

Applying the Sobolev inequality to the above equality, we obtain that

\begin{equation*} \liminf_{\epsilon\to 0^+}\limsup_{n\to +\infty}\|\nabla v_n\|_2^2=\liminf_{\epsilon\to 0^+}\limsup_{n\to +\infty}\eta\|v_n\|_{2^*}^{2^*}\leq\liminf_{\epsilon\to 0^+}\limsup_{n\to +\infty}\eta S^{-2^*/2}\|\nabla v_n\|_2^{2^*}, \end{equation*}

which implies that

(4.9)\begin{equation} R_0\geq \liminf_{\epsilon\to 0^+}\limsup_{n\to +\infty}\|\nabla v_n\|_2\geq \eta^{-\frac{N-2}{4}}S^{N/4}. \end{equation}

On the other hand, by the assumption (1.14), we have

(4.10)\begin{equation} R_0 \lt \eta^{-\frac{N-2}{4}}S^{N/4}. \end{equation}

Indeed, by the expression of r ₀ in Equation (2.2), Equation (1.14) is equivalent to

(4.11)\begin{equation} r_0 \lt \eta^{-\frac{N-2}{4}}S^{\frac{N}{4}}. \end{equation}

Since $R_0 \lt r_0 \lt R_1$ (see $\S$ 2), we obtain Equation (4.10), which contradicts Equation (4.9). So we must have Equation (4.8) for the case $p=2^*$. The proof is complete.

Proof. Let $\{u_{n}\}\subset S(a)$ be a $(PS)_c$ sequence of $E_{\epsilon,T}$ restricted to S(a). Noting that $c \lt \Upsilon_{\infty,T,a} \lt 0$, by Lemma 4.2, $\{u_n\}$ is bounded in $H^1(\mathbb{R}^N)$. Let $u_n\rightharpoonup u_\epsilon$ in $H^1(\mathbb{R}^N)$. By Lemma 4.3, $u_\epsilon\not\equiv 0$. Set $v_n:=u_n-u_{\epsilon}$. If $u_n\to u_\epsilon$ in $H^1(\mathbb{R}^N)$, the proof is complete. If u_n does not converge to u_ϵ strongly in $H^1(\mathbb{R}^N)$ for some $\epsilon\in (0,\epsilon_1)$, by Lemma 4.4,

\begin{equation*} \limsup_{n\to +\infty}\|v_n\|_2\geq \beta. \end{equation*}

Set $b=\|u_{\epsilon}\|_2,\ d_n = \|v_n\|_2$ and suppose that $\|v_n\|_2\to d$, then we get $d\geq \beta \gt 0$ and $a^2 = b^2 + d^2$. From $d_n\in(0, a)$ for n large enough, we have

(4.12)\begin{equation} c+o_n(1)=E_{\epsilon,T}(u_n)\geq E_{\epsilon,T}(v_n)+E_{\epsilon,T}(u_{\epsilon})+o_n(1). \end{equation}

Since $v_n\rightharpoonup 0$ in $H^1(\mathbb{R}^N)$, similarly to the proof of Lemma 4.3, we deduce that

(4.13)\begin{equation} E_{\epsilon,T}(v_n)\geq J_{\infty,T}(v_n)-\delta C+o_n(1) \end{equation}

for any δ > 0, where C > 0 is a constant independent of $\delta,\ \epsilon$ and n. By (4.12) and (4.13), we obtain that

\begin{equation*} \begin{split} c+o_n(1)=E_{\epsilon,T}(u_n)&\geq J_{\infty,T}(v_n)+E_{\epsilon,T}(u_{\epsilon})-\delta C+o_n(1)\\ &\geq \Upsilon_{\infty,T,d_n}+\Upsilon_{\text{max},T,b}-\delta C+o_n(1). \end{split} \end{equation*}

Letting $n\to +\infty$, by Lemma 3.5 and the arbitrariness of δ > 0, we obtain that

\begin{equation*} \begin{split} c\geq \Upsilon_{\infty,T,d}+\Upsilon_{\text{max},T,b}&\geq \frac{d^2}{a^2}\Upsilon_{\infty,T,a}+\frac{b^2}{a^2}\Upsilon_{\text{max},T,a}\\ &=\Upsilon_{\text{max},T,a}+\frac{d^2}{a^2}(\Upsilon_{\infty,T,a}-\Upsilon_{\text{max},T,a})\\ &\geq \Upsilon_{\text{max},T,a}+\frac{\beta^2}{a^2}(\Upsilon_{\infty,T,a}-\Upsilon_{\text{max},T,a}), \end{split} \end{equation*}

which contradicts $c \lt \Upsilon_{\text{max},T,a}+\frac{\beta^2}{a^2}(\Upsilon_{\infty,T,a}-\Upsilon_{\text{max},T,a})$. Thus, we must have $u_n \to u_{\epsilon}$ in $H^1(\mathbb{R}^N)$.

Proof. If the lemma does not occur, there must be $\rho_n\to 0, \epsilon_n\to 0$ and $\{u_n\}\subset S(a)$ such that

(4.14)\begin{equation} E_{\epsilon_n,T}(u_n)\leq \Upsilon_{\text{max},T,a}+\rho_n\ \qquad \text{and}\qquad Q_{\epsilon_n}(u_n)\not\in K_{\frac{\tilde{\rho}}{2}}. \end{equation}

Consequently,

\begin{equation*} \Upsilon_{\text{max},T,a}\leq J_{\text{max},T}(u_n)\leq E_{\epsilon_n,T}(u_n)\leq \Upsilon_{\text{max},T,a}+\rho_n, \end{equation*}

then

\begin{equation*} \{u_n\}\subset S(a)\qquad \text{and}\qquad J_{\text{max},T}(u_n)\to \Upsilon_{\text{max},T,a}. \end{equation*}

According to Lemma 3.6, we have two cases:

(i) $u_n\to u$ in $H^1(\mathbb{R}^N)$ for some $u\in S(a)$,

or

(ii) There exists $\{y_n\}\subset \mathbb{R}^N$ with $|y_n|\to +\infty$ such that $v_n(x) = u_n(x+ y_n)$ converges in $H^1(\mathbb{R}^N)$ to some $v\in S(a)$.

Analysis of (i): By the Lebesgue dominated convergence theorem,

\begin{equation*} Q_{\epsilon_n}(u_n)=\frac{\int_{\mathbb{R}^N}\chi(\epsilon_n x)|u_n|^2\,{\rm d}x}{\int_{\mathbb{R}^N}|u_n|^2\,{\rm d}x}\to \frac{\int_{\mathbb{R}^N}\chi(0)|u|^2\,{\rm d}x}{\int_{\mathbb{R}^N}|u|^2\,{\rm d}x}=0\in K_{\frac{\tilde{\rho}}{2}}. \end{equation*}

From this, $Q_{\epsilon_n}(u_n)\in K_{\frac{\tilde{\rho}}{2}}$ for n large enough, which contradicts $Q_{\epsilon_n}(u_n)\not\in K_{\frac{\tilde{\rho}}{2}}$ in Equation (4.14).

Analysis of (ii): Now, we will study two cases: (I) $|\epsilon_ny_n|\to +\infty$ and (II) $\epsilon_ny_n\to y$ for some $y\in \mathbb{R}^N$.

If (I) holds, the limit $v_n \to v$ in $H^1(\mathbb{R}^N)$ provides

\begin{equation*} \begin{split} E_{\epsilon_n,T}(u_n)&=\frac{1}{2}\int_{\mathbb{R}^N}|\nabla v_n|^2\,{\rm d}x-\frac{1}{q}\int_{\mathbb{R}^N}h(\epsilon_n x+\epsilon_ny_n)|v_n|^{q}\,{\rm d}x\\ &\qquad-\frac{\eta}{p}\tau(\|\nabla v_n\|_2)\int_{\mathbb{R}^N}|v_n|^{p}\,{\rm d}x\\ &\to J_{\infty,T}(v). \end{split} \end{equation*}

Since $E_{\epsilon_n,T}(u_n)\leq \Upsilon_{\text{max},T,a}+\rho_n$, we deduce that

\begin{equation*} \Upsilon_{\infty,T,a}\leq J_{\infty,T}(v)\leq \Upsilon_{\text{max},T,a}, \end{equation*}

which contradicts $\Upsilon_{\infty,T,a} \gt \Upsilon_{\text{max},T,a}$ in Lemma 4.1.

Now if (II) holds, then

\begin{equation*} \begin{split} E_{\epsilon_n,T}(u_n)&=\frac{1}{2}\int_{\mathbb{R}^N}|\nabla v_n|^2\,{\rm d}x-\frac{1}{q}\int_{\mathbb{R}^N}h(\epsilon_n x+\epsilon_ny_n)|v_n|^{q}\,{\rm d}x\\ &\qquad-\frac{\eta}{p}\tau(\|\nabla v_n\|_2)\int_{\mathbb{R}^N}|v_n|^{p}\,{\rm d}x\\ &\to J_{h(y),T}(v), \end{split} \end{equation*}

which combined with $E_{\epsilon_n,T}(u_n)\leq \Upsilon_{\text{max},T,a}+\rho_n$ gives that

\begin{equation*} \Upsilon_{h(y),T,a}\leq J_{h(y),T}(v)\leq \Upsilon_{\text{max},T,a}. \end{equation*}

By Corollary 3.8, we must have $h(y)=h_{\text{max}}$ and $y=a_i$ for some $i=1,2,\ldots, l$. Hence,

\begin{equation*} \begin{split} Q_{\epsilon_n}(u_n)=\frac{\int_{\mathbb{R}^N}\chi(\epsilon_n x)|u_n|^2\,{\rm d}x}{\int_{\mathbb{R}^N}|u_n|^2\,{\rm d}x} &=\frac{\int_{\mathbb{R}^N}\chi(\epsilon_n x+\epsilon_ny_n)|v_n|^2\,{\rm d}x}{\int_{\mathbb{R}^N}|v_n|^2\,{\rm d}x}\\ &\to \frac{\int_{\mathbb{R}^N}\chi(y)|v|^2\,{\rm d}x}{\int_{\mathbb{R}^N}|v|^2\,{\rm d}x}=a_i\in K_{\frac{\tilde{\rho}}{2}}, \end{split} \end{equation*}

which implies that $Q_{\epsilon_n}(u_n)\in K_{\frac{\tilde{\rho}}{2}}$ for n large enough. That contradicts Equation (4.14). The proof is complete.

Proof. Let $u\in S(a)$ be such that

\begin{equation*} J_{\text{max},T}(u)=\Upsilon_{\text{max},T,a}. \end{equation*}

For $1\leq i\leq l$, we define

\begin{equation*} \hat{u}_{\epsilon}^{i}(x):=u\left(x-\frac{a_i}{\epsilon}\right),\quad x\in \mathbb{R}^N. \end{equation*}

Then, $\hat{u}_{\epsilon}^{i}\in S(a)$ for all ϵ > 0 and $1\leq i\leq l$. Direct calculations give that

\begin{equation*} E_{\epsilon,T}(\hat{u}_{\epsilon}^{i})=\frac{1}{2}\int_{\mathbb{R}^N}|\nabla u|^2\,{\rm d}x-\frac{1}{q}\int_{\mathbb{R}^N}h(\epsilon x+a_i)|u|^{q}\,{\rm d}x-\frac{\eta}{p}\tau(\|\nabla u\|_2)\int_{\mathbb{R}^N}|u|^{p}\,{\rm d}x, \end{equation*}

and then

(4.15)\begin{equation} \lim_{\epsilon\to 0^+}E_{\epsilon,T}(\hat{u}_{\epsilon}^{i})=J_{h(a_i),T}(u)=J_{\text{max},T}(u) =\Upsilon_{\text{max},T,a}. \end{equation}

Note that

\begin{equation*} Q_{\epsilon}(\hat{u}_{\epsilon}^i)=\frac{\int_{\mathbb{R}^N}\chi(\epsilon x+a_i)|u|^2\,{\rm d}x}{\int_{\mathbb{R}^N}|u|^2\,{\rm d}x}\to a_i\text{ as}\ \epsilon\to 0^+. \end{equation*}

So $\hat{u}_{\epsilon}^i\in \theta_{\epsilon}^i$ for ϵ small enough, which combined with Equation (4.15) implies that there exists $\epsilon_3\in(0,\epsilon_2]$ such that

\begin{equation*} \beta_{\epsilon}^i \lt \Upsilon_{\text{max},T,a}+\frac{\rho_1}{2},\quad\text{ for} \ \text{any}\ \epsilon\in (0,\epsilon_3). \end{equation*}

For any $v\in \partial\theta_{\epsilon}^i$, that is, $v\in S(a)$ and $|Q_\epsilon(v)-a_i|=\tilde{\rho}$, we obtain that $|Q_\epsilon(v)\not\in K_{\frac{\tilde{\rho}}{2}}$. Thus, by Lemma 4.6,

\begin{equation*} E_{\epsilon,T}(v) \gt \Upsilon_{\text{max},T,a}+\rho_1,\quad \text{for}\ \text{all} \ v\in \partial\theta_{\epsilon}^i\ \text{and}\ \epsilon\in (0,\epsilon_3), \end{equation*}

which implies that

\begin{equation*} \tilde{\beta}_{\epsilon}^{i}=\inf_{v\in\partial\theta_{\epsilon}^i} E_{\epsilon,T}(v)\geq \Upsilon_{\text{max},T,a}+\rho_1,\quad \text{for}\ \text{all} \ \epsilon\in(0,\epsilon_3). \end{equation*}

Thus,

\begin{equation*} \beta_{\epsilon}^{i} \lt \tilde{\beta}_{\epsilon}^{i},\quad \text{for}\ \text{all} \ \epsilon\in (0,\epsilon_3). \end{equation*}

Proof of Theorem 1.1

Set $\epsilon_0:=\epsilon_3$, where ϵ ₃ is obtained in Lemma 4.7. Let $\epsilon\in (0,\epsilon_0)$. By Lemma 4.7, for each $i\in \{1,2,\ldots, l\}$, we can use the Ekeland’s variational principle to find a sequence $\{u_n^{i})\subset \theta_{\epsilon}^i$ satisfying:

\begin{equation*} E_{\epsilon,T}(u_n^i)\to \beta_{\epsilon}^i\qquad {\rm and}\qquad \|E_{\epsilon,T}|_{S(a)}^\prime(u_n^i)\|\to 0,\quad {\rm as}\ n\to +\infty, \end{equation*}

that is, $\{u_n^i\}_n$ is a $(PS)_{\beta_{\epsilon}^i}$ sequence for $E_{\epsilon,T}$ restricted to S(a). Since $\beta_{\epsilon}^i \lt \Upsilon_{\text{max},T,a}+\rho_0$, it follows from Lemma 4.5 that there exists uⁱ such that $u_n^i\to u^i$ in $H^1(\mathbb{R}^N)$. Thus

\begin{equation*} u^i\in \theta_{\epsilon}^i,\ E_{\epsilon,T}(u^i)=\beta_{\epsilon}^i\qquad \text{and}\qquad E_{\epsilon,T}|_{S(a)}^\prime(u^i)=0. \end{equation*}

As

\begin{equation*} Q_{\epsilon}(u^i)\in \overline{B_{\tilde{\rho}}(a_i)},\qquad Q_{\epsilon}(u^j)\in \overline{B_{\tilde{\rho}}(a_j)}, \end{equation*}

and

\begin{equation*} \overline{B_{\tilde{\rho}}(a_i)}\cap \overline{B_{\tilde{\rho}}(a_j)}=\emptyset\ ,\quad \text{for}\ i\neq j, \end{equation*}

we conclude that $u^i\not\equiv u^j$ for i ≠ j while $1\leq i, j \leq l$. Therefore, $E_{\epsilon, T}$ has at least l non-trivial critical points for all $\epsilon\in (0, \epsilon_0)$.

As $E_{\epsilon, T}(u^i) \lt 0$ for any $i=1,2,\ldots,l$, by Lemma 4.2, $u^i\, (i=1,2,\ldots,l)$ is in fact the critical point of E_ϵ on S(a) with $E_{\epsilon}(u^i) \lt 0$ and then there exists $\lambda_i\in \mathbb{R}$ such that

\begin{equation*} -\Delta u^i=\lambda_i u^i+h(\epsilon x)|u^i|^{q-2}u^i+\eta |u^i|^{p-2}u^i,\quad x\in \mathbb{R}^N. \end{equation*}

By using $E_{\epsilon}(u^i)=\beta_{\epsilon}^i \lt 0$ and $E_{\epsilon}^\prime(u^i)u^i=\lambda_ia^2$, we obtain that

\begin{equation*} \frac{1}{2}\lambda_ia^2=E_{\epsilon}(u^i)+\left(\frac{1}{q}-\frac{1}{2}\right) \int_{\mathbb{R}^N}h(\epsilon x)|u^i|^q\,{\rm d}x+\left(\frac{1}{p}-\frac{1}{2}\right) \int_{\mathbb{R}^N}|u^i|^p\,{\rm d}x, \end{equation*}

which implies that $\lambda_i \lt 0$ for $i=1,2,\ldots,l$. The proof is complete.

Proof of Theorem 1.4

Letting r ₁ be such that $\bar{g}_a(r_1)=\beta_{\epsilon}^i$, and considering Equation (2.6), the definition of $\Omega_i$ and $\beta_{\epsilon}^i \lt 0$, we know that

\begin{equation*} \|\nabla v\|_2\leq r_1 \lt R_0,\quad \text{for}\ \text{any}\ v\in\Omega_i. \end{equation*}

Let $a_1 \gt a$ be such that $\bar{g}_{a_1}(R_0)=\frac{\beta_{\epsilon}^i}{2}$. There exists δ > 0 such that if

\begin{equation*} u_0\in H^1(\mathbb{R}^N)\qquad \text{and}\qquad \text{dist}_{H^1(\mathbb{R}^N)}(u_0,\Omega_i) \lt \delta, \end{equation*}

then

\begin{equation*} \|u_0\|_2\leq a_1,\quad \|\nabla u_0\|_2\leq r_1+\frac{R_0-r_1}{2},\quad E_{\epsilon,T}(u_0)\leq \frac{2}{3}\beta_{\epsilon}^i. \end{equation*}

Denoting by $\psi(t,\cdot)$ the solution to Equation (1.17) with initial value u ₀ and denoting by $[0, T_{\text{max}})$ the maximal existence interval for $\psi(t,\cdot)$, we have classically that either $\psi(t,\cdot)$ is globally defined for positive times, or $\|\nabla \psi(t,\cdot)\|_2=+\infty$ as $t\to T_{\text{max}}^-$, see ([Reference Tao, Visan and Zhang38], Section 3). Set $\tilde{a}=\|u_0\|_2$. Note that $\|\psi(t,\cdot)\|_2=\|u_0\|_2$ for all $t\in (0,T_{\text{max}})$ by the conservation of the mass. If there exists $\tilde{t}\in (0,T_{\text{max}})$ such that $\|\nabla \psi(\tilde{t},\cdot)\|_2=R_0$, then

\begin{equation*} E_{\epsilon}(\psi(\tilde{t},\cdot))=E_{\epsilon,T}(\psi(\tilde{t},\cdot)) \geq \bar{g}_{\tilde{a}}(R_0)\geq \bar{g}_{a_1}(R_0)=\frac{\beta_{\epsilon}^i}{2}, \end{equation*}

which contradicts the conservation of the energy:

\begin{equation*} E_{\epsilon}(\psi(t,\cdot))=E_{\epsilon}(u_0)\leq \frac{2}{3}\beta_{\epsilon}^i,\quad \text{for}\ \text{all} \ t\in (0,T_{\text{max}}). \end{equation*}

Thus,

(5.2)\begin{equation} \|\nabla \psi(t,\cdot)\|_2 \lt R_0,\quad \text{for}\ \text{all} \ t\in [0,T_{\text{max}}), \end{equation}

which implies that $\psi(t,\cdot)$ is globally defined for positive times.

Next we prove that $\Omega_i$ is stable. The validity of Lemma 4.5 for complex valued function can be proved exactly as in Theorem 3.1 in [Reference Hajaiej and Stuart13]. Thus, the stability of $\Omega_i$ can be proved by modifying the classical Cazenave–Lions argument [Reference Cazenave and Lions10] (see also [Reference Li and Zhao24]). For the completeness, we give the proof here. Suppose by contradiction that there exist sequences $\{u_{0,n}\}\subset H^1(\mathbb{R}^N)$ and $\{t_{n}\}\subset \mathbb{R}^+$ and a constant $\theta_0 \gt 0$ such that for all $n\geq 1$,

(5.3)\begin{equation} \inf_{v\in \Omega_i}\|u_{0,n}-v\| \lt \frac{1}{n} \end{equation}

and

(5.4)\begin{equation} \inf_{v\in \Omega_i}\|\psi_n(t_n,\cdot)-v\|\geq \theta_0, \end{equation}

where $\psi_n(t,\cdot)$ is the solution to (1.17) with initial value $u_{0,n}$. By Equation (5.2), there exists n ₀ such that for $n \gt n_0$ it holds that $\|\nabla \psi_n(t,\cdot)\|_2 \lt R_0$ for all $t\geq 0$.

By Equation (5.3), there exists $\{v_n\}\subset \Omega_i$ such that

(5.5)\begin{equation} \|u_{0,n}-v_n\| \lt \frac{2}{n}. \end{equation}

That $\{v_n\} \subset \Omega_i$ implies that $\{v_n\}\subset \theta_{\epsilon}^i$ is a $(PS)_{\beta_{\epsilon}^i}$ sequence of $E_{\epsilon,T}$ restricted to S(a). From the proof of Theorem 1.1, there exists $v\in \Omega_i$ such that

\begin{equation*} \lim_{n\to +\infty}\|v_n-v\|=0, \end{equation*}

which combined with (5.5) gives that

(5.6)\begin{equation} \lim_{n\to +\infty}\|u_{0,n}-v\|=0. \end{equation}

Hence,

\begin{equation*} \lim_{n\to +\infty}\|u_{0,n}\|_2=\|v\|_2=a,\qquad \ \lim_{n\to +\infty}E_{\epsilon}(u_{0,n})=E_{\epsilon}(v)={\beta}_{\epsilon}^i \lt \tilde{\beta}_{\epsilon}^i. \end{equation*}

Then by the conservation of mass and energy, we obtain that

(5.7)\begin{equation} \lim_{n\to +\infty}\|\psi_n(t,\cdot)\|_2=a,\qquad \ \lim_{n\to +\infty}E_{\epsilon}(\psi_n(t,\cdot))={\beta}_{\epsilon}^i,\quad \text{for}\ \text{any} \ t\geq 0. \end{equation}

Define

\begin{equation*} \varphi_n(t,\cdot)=\frac{a\psi_n(t,\cdot)}{\|\psi_n(t,\cdot)\|_2},\quad t\geq 0. \end{equation*}

Then $\varphi_n(t,\cdot)\in S(a)$ and

(5.8)\begin{equation} \|\varphi_n(t,\cdot)-\psi_n(t,\cdot)\|\to 0\ \text{as}\ n\to+\infty \ \text{uniformly}\ \text{in} \ t\geq 0, \end{equation}

which combined with Equation (5.6) gives that

\begin{equation*} \lim_{n\to+\infty}\|\varphi_n(0,\cdot)-v\|=\lim_{n\to+\infty} \left\|\frac{au_{0,n}}{\|u_{0,n}\|_2}-v\right\|=0. \end{equation*}

Hence, $\varphi_n(0,\cdot)\in \theta_{\epsilon}^{i}\backslash \partial\theta_{\epsilon}^{i}$ for n large enough because $v\in \theta_{\epsilon}^{i}\backslash \partial\theta_{\epsilon}^{i}$. Using the method of continuity, $\lim_{n\to +\infty}E_{\epsilon}(\varphi_n(t,\cdot))={\beta}_{\epsilon}^i$ for all $t\geq 0$, and ${\beta}_{\epsilon}^i \lt \tilde{\beta}_{\epsilon}^i$, we obtain that:

(5.9)\begin{equation} {\rm for}\ n\ {\rm large}\ {\rm enough},\ \varphi_n(t,\cdot)\in \theta_{\epsilon}^i\backslash\partial\theta_{\epsilon}^i\,\quad {\rm for}\ {\rm all} \ t\geq 0. \end{equation}

From (5.7)–(5.9), $\{\varphi_n(t_n,\cdot)\}\subset \theta_{\epsilon}^i$ is a minimizing sequence of $E_{\epsilon,T}$ at level $\beta_{\epsilon}^i$, and from the proof of Theorem 1.1, there exists $\tilde{v}\in \theta_{\epsilon}^i$ such that

(5.10)\begin{equation} \lim_{n\to +\infty}\|\varphi_n(t_n,\cdot)-\tilde{v}\|=0, \end{equation}

which combined with (5.8) gives that

\begin{equation*} \lim_{n\to +\infty}\|\psi_n(t_n,\cdot)-\tilde{v}\|=0. \end{equation*}

That contradicts (5.4). Hence $\Omega_i$ is stable for any $i=1,2,\ldots,l$.

Proof of Theorem 1.5

The proof can be done by modifying the arguments of $\S$ 3 and 4 in [Reference Jeanjean, Jendrej, Le and Visciglia18] and using the arguments of the proof of Theorem 1.4, so we omit it.