Proof. The proof is a straightforward adaption of [Reference Ganguly, Gupta and Sreenadh21, Proposition 3.1] in the case $f \not \equiv 0.$ We also refer ([Reference Lions24], [Reference Lions25] and [Reference Struwe33]) for the Euclidean case.

Proof. For any $\varepsilon \in (0,\,1)$ and using (3.12) and (3.6), we obtain,

\begin{align*} J_{\lambda,a, f}\left(v_{j}\right) & \geq(1-\varepsilon)^{\frac{p+1}{p-1}} J_{\lambda,a, 0}\left(v\right)-\frac{1}{2 \varepsilon}\|f\|_{H^{{-}1}\left(\mathbb{B}^{N}\right)}^{2} \\ & \geq(1-\varepsilon)^{\frac{p+1}{p-1}}\left(\frac{1}{2}-\frac{1}{p+1}\right)\left(\int_{\mathbb{B}^{N}} a(x) v_{j +}^{p+1} \mathrm{~d} V_{\mathbb{B}^{N}}\right)^{-\frac{2}{p-1}}-\frac{1}{2 \varepsilon}\|f\|_{H^{{-}1}\left(\mathbb{B}^{N}\right)}^{2}. \end{align*}

As $\operatorname {dist}(v_{j},\, \partial \tilde {\Sigma }_+) \rightarrow 0$ gives

\begin{align*} & \left(v_{j}\right)_+ \rightarrow 0 \text{ in } H^{1}\left(\mathbb{B}^{N}\right), \\ & \left(v_{j}\right)_+ \rightarrow 0 \text{ in } L^{p+1}\left(\mathbb{B}^{N}\right). \end{align*}

Therefore,

\[ \left|\int_{\mathbb{B}^{N}} a(x) v_{j}^{p+1} \mathrm{~d} V_{\mathbb{B}^{N}}\right| \leq\left.\|a\|\right._{L^{\infty}\left(\mathbb{B}^{N}\right)} \int_{\mathbb{B}^{N}}\left|v_{j +}\right|^{p+1}\mathrm{~d} V_{\mathbb{B}^{N}} \xrightarrow{j} 0. \]

Hence $J_{\lambda,a, f}(v_{j}) \rightarrow \infty$ as dist $_{H^{1}(\mathbb {B}^{N})}(v_{j},\, \partial \tilde {\Sigma }_+) \rightarrow 0.$ This proves part $(a).$

For part $(b),$ using (3.13) and (3.15), we get

\begin{align*} \left\|I_{\lambda,a, f}^{\prime}\left(t_{a, f}\left(v_{j}\right) v_{j}\right)\right\|_{H^{{-}1}\left(\mathbb{B}^{N}\right)} & =\frac{1}{t_{a, f}\left(v_{j}\right)}\left\|J_{\lambda,a, f}^{\prime}\left(v_{j}\right)\right\|_{T_{v_{j}}^{*} \tilde{\Sigma}_+} \\ & \leq\left(p S_{1,\lambda}^{-\frac{p+1}{2}}\right)^{\frac{1}{p-1}}\left\|J_{\lambda,a, f}^{\prime}\left(v_{j}\right)\right\|_{T_{v_{j}} \tilde{\Sigma}_+} \stackrel{j}{\rightarrow} 0.\\ \end{align*}

Further, we also have $I_{\lambda,a, f}(t_{a,f}(v_{j})v_{j})=J_{\lambda,a, f}(v_{j}) \rightarrow c$ as $j \rightarrow \infty$. Applying Palais–Smale lemma for $I_{\lambda,a, f}(u)$ (Proposition 4.1), the rest follows.

Proof. Proposition 4.2 suggests that the condition (PS)$_{c}$ breaks down only at levels

\[ c=I_{\lambda,a, f}\left(u_{0}\right)+\ell I_{\lambda,1,0}(w), \]

where $\ell \in \mathbb {N}$ and $u_{0} \in H^{1}(\mathbb {B}^{N})$ is a critical point of $I_{\lambda,a, f}(u)$ .

From Proposition 7.1, we have

(4.2)\begin{equation} I_{\lambda,a, f}\left(\mathcal{U}_{a, f} (x)\right)=\inf _{u \in B\left(r_{1}\right)} I_{\lambda,a, f}(u) \leq I_{\lambda,a, f}(0)=0,\end{equation}

Furthermore, all the critical points of $I_{\lambda,a, f}(u)$ except $\mathcal {U}_{a, f} (x)$ corresponds to a critical point $J_{\lambda,a, f}(v)$, which follows from (8.3). Thus there exists $v_{1} \in \tilde {\Sigma }_+$ for a critical point $u_{1}$ of $I_{\lambda,a,f}(u)$ such that $I_{\lambda,a, f}(u_{1})=J_{\lambda,a, f}(v_{1})>0$ by using $(iii)$ of Lemma 3.2. Consequently,

\[ I_{\lambda,a, f}\left(\mathcal{U}_{a, f} (x)\right)=\inf \left\{I_{\lambda,a, f}\left(u_{0}\right) \mid u_{0} \in H^{1}\left(\mathbb{B}^{N}\right) \text{ is a critical point of } I_{\lambda,a, f}(u)\right\}. \]

Hence $I_{\lambda,a, f}(\mathcal {U}_{a, f} (x))+I_{\lambda,1,0}(w)$ is the lowest level where $(P S)_{c}$ breaks.

Proof. The solution $u \in H^1(\mathbb {B}^{N}),$ this immediately implies $\lim _{d(x,0) \rightarrow \infty } u(x)=0 \textrm {a.e.}$ Furthermore, using the Calderon–Zygmund estimate and elliptic regularity, we have $u \in C^{2}(\mathbb {B}^{N})$; thus, $\lim _{d(x,0) \rightarrow \infty } u(x)=0$ for all $x \in \mathbb {B}^{N}.$ The proof is divided into two cases: $\lambda < 0$ and $\lambda =0.$

Case 1: $\lambda < 0$

Choose $\alpha >0$ such that $\frac {\alpha ^{2}|\lambda |-1}{\alpha (N-1)}\geq 1$. To be precise, $\alpha \in [c^{'}(N,\,\lambda ),\, \infty )$ where

\[ c^{'}(N,\lambda)= \frac{(N-1)+\sqrt{(N-1)^{2}-4\lambda}}{2|\lambda|}. \]

Thus we can choose $R_{1}>0$ large enough such that

(5.1)\begin{equation} \alpha^{2}|\lambda|-\alpha(N-1)\coth{d(x,0)}\geq 1, \;\;\;\; \forall d(x,0) \geq R_{1}. \end{equation}

For $m=\min \left \{\frac {1}{|\lambda |}u(x)\mid d(x,\,0) =R_{1}\right \}>0$, set $v_{1}(x) := v_1(r) = m e^{-\alpha |\lambda | (d(x,0)-R_{1})},$ where $r := d(x,\, 0).$ Now for any $L>R_{1}$, denote

\[ \Omega(L)=\left\{x \in \mathbb{B}^{N} \mid R_{1}< d(x,0) < L \quad \text{ and } \quad |\lambda|v_{1}(x)>u(x)\right\} . \]

Then $\Omega (L)$ is open. Moreover, for $x \in \Omega (L)$ and using (5.1)  we have

\begin{align*} \Delta_{\mathbb{B}^{N}}\left(u-|\lambda|v_{1}\right)(x) & = \Delta_{\mathbb{B}^{N}} u(x) \, - |\lambda| \, \Delta_{\mathbb{B}^{N}}(v_1(x)) \\ & ={-} \lambda u - \, u^p - \, f(x) - |\lambda| \, \left( \frac{\partial^2}{\partial r^2} v_1(r) + (N-1) \coth r \frac{\partial}{\partial r} v_1(r) \right) \\ & ={-} \lambda u -\, u^p - \, f(x)-|\lambda|\left[\alpha^{2}|\lambda|^{2}-\alpha|\lambda|(N-1)\coth{r}\right] v_{1}(x) \\ & \leqslant |\lambda| u(x)-|\lambda|^{2}\left[\alpha^{2}|\lambda|-\alpha (N-1)\coth{r}\right] v_{1}(x) \\ & \leqslant |\lambda|\left(u-|\lambda|v_{1}\right)(x) \\ & <0 \end{align*}

Applying the maximum principle, for $x \in \Omega (L)$ will result in

\begin{align*} u(x)-|\lambda|v_{1}(x) & \geqslant \min \left\{\left(u-|\lambda|v_{1}\right)(x) \mid x \in \partial \Omega(L)\right\} \\ & =\min \left\{0, \min_{d(x,0)=L}\left(u-|\lambda|v_{1}\right)(x)\right\} . \end{align*}

Since $\displaystyle \lim _{d(x,0) \rightarrow +\infty } u(x)=\lim _{d(x,0) \rightarrow +\infty } v_{2}(x)=0$, by letting $L \rightarrow \infty$, we see that $\Omega (L)$ is empty and hence

(5.2)\begin{equation} u(x) \geqslant |\lambda| v_{1}(x) \text{ for all }d(x,0) \geqslant R_{1},\end{equation}

By the supposition on $f(x)$ there exists some $\varepsilon,\, \textrm { and } C>0$ such that

(5.3)\begin{equation} f(x) \leqslant C e^{-(c^{'}(N,\lambda)+\varepsilon)|\lambda|p\, d(x, 0)} \text{ for all } x \in \mathbb{B}^{N} \text{. } \end{equation}

(5.2) will imply the existence of a $C_{1}>0$

(5.4)\begin{equation} u(x) \geqslant C_{1} e^{(c^{'}(N, \lambda)+\delta)|\lambda|d(x,0)} \quad \text{ for all } x \in \mathbb{B}^{N}, \quad \mbox{and for any } \ \delta > 0. \end{equation}

Choosing $\varepsilon$ appropriately, and using (5.3), (5.4) together will provide $R_{2}>0$ such that

\[ (u(x))^p \geqslant f(x) \quad \text{ for } \quad d(x,0) \geqslant R_{2} . \]

Moreover, since $p >1,$ there holds

\[ u^{p} = \, \circ(u) \text{ for } d(x,0)\rightarrow \infty. \]

Let $\beta >0$ be such that $\beta ^{2}|\lambda |- (N-1)\beta \leq 1$, i.e., $\beta \leq c(n,\, \lambda )^{'}.$

Define $v_{2}(x)=M e^{-\beta |\lambda |(d(x,0)-R_{4})}$, where

\[ M=\max \left\{u(x) \mid d(x,0) =R_{2}\right\}>0. \]

Further, for any $L>R_{4}$, denote

\[ \tilde{\Omega}(L)=\left\{x \in \mathbb{B}^{N} \mid R_{4}< d(x,0) < L \quad \text{ and } \quad u(x)>v_{2}(x)\right\} . \]

Then $\tilde {\Omega }(L)$ is open and, for $x \in \tilde {\Omega }(L)$,

\begin{align*} \Delta_{\mathbb{B}^{N}}\left(v_{2}-u\right)(x) & =\left[\beta^{2}|\lambda|^{2}-\beta|\lambda|(N-1)\coth r\right] v_{2}(x) +\lambda u + \,u^p+f(x) \\ & \leqslant -\lambda v_{2}+ \lambda u + \, 2u^{p}\\ & \leqslant -\lambda v_{2}+ \lambda u + \, \circ(u)\\ & ={-}\lambda (v_{2}-u)(x)+{\circ}(u)\\ & <0. \end{align*}

By the maximum principle, for $x \in \widetilde {\Omega }(L)$,

\begin{align*} v_{2}(x)-u(x) & \geqslant \min \left\{\left(v_{2}-u\right)(x) \mid x \in \partial \tilde{\Omega}(L)\right\} \\ & =\min \left\{0, \min _{d(x,0)=L}\left(v_{2}-u\right)(x)\right\} . \end{align*}

Since $\displaystyle \lim _{d(x,0) \rightarrow +\infty } u(x)=\lim _{d(x,0) \rightarrow +\infty } v_{2}(x)=0$, by letting $L \rightarrow \infty$, we see again that $\tilde {\Omega }(L)$ is empty and hence

\[ v_{2}(x) \geqslant u(x) \text{ for all }d(x,0) \geqslant R_{4}. \]

Now by choosing $\alpha = \beta = c^{'}(N,\,\lambda ),$ the proof is complete.

Case 2: $\lambda =0$

This case can also be tackled similarly by appropriately choosing the functions $v_{1}$ and $v_{2}$.

To be precise, let

\[ v_{1} = m e^{-\gamma \left(d(x,0)-R_{1}^{'}\right)} \text{ and } v_{2} = M e^{-\eta \left(d(x,0)-R_{2}^{'}\right)} \text{ for some } \gamma,R_{1}^{'},\eta,R_{2}^{'}>0 \]

where $m=\min \left \{u(x)\mid d(x,\,0) =R_{1}^{'}\right \}>0$ and $M=\max \left \{u(x)\mid d(x,\,0) =R_{2}^{'}\right \} >0.$

Indeed $\gamma >0$ satisfies $\gamma > N-1$, and thus $R_{1}^{'}$ is chosen such that $\gamma - (N-1) \coth {r} > 0$ for all $r > R_{1}^{'}.$ Also, $R_{2}^{'}$ is chosen similarly as $R_{3}$ mentioned above. Further, we can conclude the lemma by applying the maximum principle in the hyperbolic balls of radius $R_{1}^{'}$ and $R_{2}^{'}$ and proceeding as in the previous case.

Proof. Performing straightforward calculations implies

(6.3)\begin{equation} \begin{aligned} I_{\lambda,a, f}\left(\tilde{\mathcal{U}}_{a, f}(x)+t w(\tau_{{-}y}(x))\right) & =\frac{1}{2}\left\|\tilde{\mathcal{U}}_{a, f}(x)+t w(\tau_{{-}y}(x))\right\|_{H_{\lambda}}^{2} -\frac{1}{p+1}\\ & \quad \int_{\mathbb{B}^{N}} a(x)\left(\tilde{\mathcal{U}}_{a, f}(x)+t w(\tau_{{-}y}(x))\right)^{p+1} \mathrm{~d} V_{\mathbb{B}^{N}}(x) \\ & \quad-\int_{\mathbb{B}^{N}} f(x)\left(\tilde{\mathcal{U}}_{a, f}(x)+t w(\tau_{{-}y}(x))\right) \mathrm{~d} V_{\mathbb{B}^{N}}(x) \\ & =\frac{1}{2}\left\|\tilde{\mathcal{U}}_{a, f}(x)\right\|_{H_{\lambda}}^{2}+\frac{t^{2}}{2}\|w\|_{H^{1}\left(\mathbb{B}^{N}\right)}^{2}\\ & \quad+t\left\langle \tilde{\mathcal{U}}_{a, f}(x), w(\tau_{{-}y}(x))\right\rangle_{H_{\lambda}} \\ & \quad-\frac{1}{p+1} \int_{\mathbb{B}^{N}} a(x)\left(\tilde{\mathcal{U}}_{a, f}(x)\right)^{p+1} \mathrm{~d} V_{\mathbb{B}^{N}}(x)\\ & \quad-\frac{t^{p+1}}{p+1} \int_{\mathbb{B}^{N}}\; a(x)\; (w(\tau_{{-}y}(x))^{p+1} \mathrm{~d} V_{\mathbb{B}^{N}}(x) \\ & \quad-\frac{1}{p+1} \int_{\mathbb{B}^{N}} a(x)\left\{\left(\tilde{\mathcal{U}}_{a, f}(x)+t w(\tau_{{-}y}(x))\right)^{p+1}\right.\\ & \quad-\left. \left(\tilde{\mathcal{U}}_{a, f}(x)\right)^{p+1}- t^{p+1} w(\tau_{{-}y}(x))^{p+1}\right\} \mathrm{~d} V_{\mathbb{B}^{N}}(x) \\ & \quad-\int_{\mathbb{B}^{N}} f(x)\left(\tilde{\mathcal{U}}_{a, f}(x)+t w(\tau_{{-}y}(x))\right) \mathrm{~d} V_{\mathbb{B}^{N}}(x). \end{aligned} \end{equation}

Now for all $h \in H^{1}(\mathbb {B}^{N})$, we have

\begin{align*} 0& =I_{\lambda,a, f}^{\prime}\left(\tilde{\mathcal{U}}_{a, f}(x)\right)(h) \\ & =\left\langle \tilde{\mathcal{U}}_{a, f}(x), h\right\rangle_{H_{\lambda}}-\int_{\mathbb{B}^{N}} a(x)\left(\tilde{\mathcal{U}}_{a, f}(x)\right)^{p} h \mathrm{~d} V_{\mathbb{B}^{N}}(x)-\int_{\mathbb{B}^{N}} f h \mathrm{~d} V_{\mathbb{B}^{N}}(x), \end{align*}

i.e.,

\[ \left\langle \tilde{\mathcal{U}}_{a, f}(x), h\right\rangle_{H_{\lambda}}=\int_{\mathbb{B}^{N}} a(x)\left(\tilde{\mathcal{U}}_{a, f}(x)\right)^{p} h \mathrm{~d} V_{\mathbb{B}^{N}}(x)+\int_{\mathbb{B}^{N}} f h \mathrm{~d} V_{\mathbb{B}^{N}}(x). \]

In particular, for $h=t w(\tau _{-y}(x))$ in the above yields

\begin{align*} & t\left\langle \tilde{\mathcal{U}}_{a, f}(x), w(\tau_{{-}y}(x))\right\rangle_{H_{\lambda}} \\ & \quad=t \int_{\mathbb{B}^{N}} a(x)\left(\tilde{\mathcal{U}}_{a, f}(x)\right)^{p} w(\tau_{{-}y}(x)) \mathrm{~d} V_{\mathbb{B}^{N}}(x)+t \int_{\mathbb{B}^{N}} f w(\tau_{{-}y}(x)) \mathrm{~d} V_{\mathbb{B}^{N}}(x) . \end{align*}

Hence utilizing the above equation and appropriately rearranging the terms in (6.3) will result in

\begin{align*} & I_{\lambda,a, f}\left(\tilde{\mathcal{U}}_{a, f}(x)+t w(\tau_{{-}y}(x))\right)=I_{\lambda,a, f}\left(\tilde{\mathcal{U}}_{a, f}(x)\right)+I_{\lambda,1,0}(t w)\\ & \quad+\frac{t^{p+1}}{p+1} \int_{\mathbb{B}^{N}}(1-a(x)) w(\tau_{{-}y}(x))^{p+1}\mathrm{~d} V_{\mathbb{B}^{N}}(x)\\ & \quad-\frac{1}{p+1} \int_{\mathbb{B}^{N}} a(x)\left\{\left(\tilde{\mathcal{U}}_{a, f}(x)+t w(\tau_{{-}y}(x))\right)^{p+1}-\left(\tilde{\mathcal{U}}_{a, f}(x)\right)^{p+1}\right. \\ & \quad\left.-t(p+1)\left(\tilde{\mathcal{U}}_{a, f}(x)\right)^{p} w(\tau_{{-}y}(x))-t^{p+1} w(\tau_{{-}y}(x))^{p+1}\right\} \mathrm{~d} V_{\mathbb{B}^{N}}(x)\\ & =I_{\lambda,a, f}\left(\tilde{\mathcal{U}}_{a, f}(x)\right)+I_{\lambda,1,0}(t w)\;+\; \underbrace{(I)-(II)}. \end{align*}

where

(6.4)\begin{equation} I:=\frac{t^{p+1}}{p+1} \int_{\mathbb{B}^{N}}(1-a(x)) w(\tau_{{-}y}(x))^{p+1} \mathrm{~d} V_{\mathbb{B}^{N}}(x),\end{equation}

and

(6.5)\begin{equation} \begin{aligned} II & :=\frac{1}{p+1} \int_{\mathbb{B}^{N}} a(x)\left\{\left(\tilde{\mathcal{U}}_{a, f}(x)+t w(\tau_{{-}y}(x))\right)^{p+1}-\left(\tilde{\mathcal{U}}_{a, f}(x)\right)^{p+1}\right.\\ & \left.-t(p+1)\left(\tilde{\mathcal{U}}_{a, f}(x)\right)^{p} w(\tau_{{-}y}(x))-t^{p+1} w(\tau_{{-}y}(x))^{p+1}\right\}\mathrm{~d} V_{\mathbb{B}^{N}}(x). \end{aligned} \end{equation}

To complete the proof of the proposition, we need to show that $(I) \; -\;(II) < 0,$ for suitably chosen $R >0.$

Using the continuity, we easily get

\[ I_{\lambda,a, f}\left(\tilde{\mathcal{U}}_{a, f}(x)+t w(\tau_{{-}y}(x))\right) \rightarrow I_{\lambda,a, f}(\tilde{\mathcal{U}}_{a, f}(x)) \]

as $t \rightarrow 0$. In addition, we also have

\[ I_{\lambda,a, f}\left(\tilde{\mathcal{U}}_{a, f}(x)+t w(\tau_{{-}y}(x))\right) \rightarrow-\infty \text{ as } t \rightarrow \infty. \]

Thus using the above two facts, we can find $m,\, M$ with $0< m< M$ such that

\begin{align*} & I_{\lambda,a, f}\left(\tilde{\mathcal{U}}_{a, f}(x)+t w(\tau_{{-}y}(x)\right)< I_{\lambda,a, f}\left(\tilde{\mathcal{U}}_{a, f}(x)\right)\\ & \quad +I_{\lambda,1,0}(w)\;\text{ for all } t \in(0, m) \cup(M, \infty). \end{align*}

As a result, to prove the proposition at hand, it suffices to show (6.1) for $t \in [m,\, M]$. Hence to finish the proof, we need to show $I< II$. To this end, let us recall the following standard $p$th inequalities from calculus.

1. (1) $(s+t)^{p+1}-s^{p+1}-t^{p+1}-(p+1) s^{p} t \geq 0$ for all $(s,\, t) \in [0,\, \infty ) \times [0,\, \infty )$.

2. (2) For any $r > 0$ we can find a constant $A(r) > 0$ such that

   \[ (s+t)^{p+1}-s^{p+1}-t^{p+1}-(p+1) s^{p} t \geq A(r) t^{2}, \]
   for all $(s,\, t) \in [r,\, \infty ) \times [0,\, \infty )$.

We can estimate $II$ with the help of the above inequality as follows:

Set $A:=A(r):=A(\min _{d(x,0) \leq 1} \tilde {\mathcal {U}}_{a, f}(x))>$ 0, then

\begin{align*} II & :=\frac{1}{p+1} \int_{\mathbb{B}^{N}} a(x)\left\{\left(\tilde{\mathcal{U}}_{a, f}(x)+t w(\tau_{{-}y}(x))\right)^{p+1}-\left(\tilde{\mathcal{U}}_{a, f}(x)\right)^{p+1}\right.\\ & \left.-t(p+1)\left(\tilde{\mathcal{U}}_{a, f}(x)\right)^{p} w(\tau_{{-}y}(x))-t^{p+1} w(\tau_{{-}y}(x))^{p+1}\right\}\mathrm{~d} V_{\mathbb{B}^{N}}(x)\\ & \geq \frac{1}{p+1} \int_{d(x,0) \leq 1} a(x) A(r) t^{2} w^2(\tau_{{-}y}(x)) \mathrm{~d} V_{\mathbb{B}^{N}}(x) \\ & \geq \frac{m^{2} \underline{\mathrm{a}} A(r)}{p+1} \underbrace{\int_{d(x,0) \leq 1} w^{2}(\tau_{{-}y}(x)) \mathrm{~d} V_{\mathbb{B}^{N}}(x)}_{E_1} \\ \end{align*}

Estimate of $E_1:$ We shall estimate $E_1$ in the domain $d(x,\, 0) \leq 1.$ Using traingle inequality we have

\[ 1 - \frac{d(x, 0)}{d(y, 0)} \leq \frac{d(x, y)}{d(y, 0)} \leq 1 + \frac{d(x, 0)}{d(y, 0)}. \]

Since, $d(x,\, 0) \leq 1,$ there exist $R > 0$ and $\varepsilon _{R} > 0$ such that whenver $d(y,\, 0) > R,$ there holds

\[ 1 - \varepsilon_{R} \leq \frac{d(x, y)}{d(y, 0)} \leq 1 + \varepsilon_{R}, \]

where $\varepsilon _R \rightarrow 0$ as $R \rightarrow \infty.$ Thus using above and (3.2) we conclude for any $\varepsilon > 0,$

\begin{align*} E_1 & := \int_{d(x, 0) \leq 1} w^{2}(\tau_{{-}y}(x)) \mathrm{~d} V_{\mathbb{B}^{N}}(x) \geq C_{\varepsilon} \int_{d(x, 0) \leq 1} e^{{-}2(c(N, \lambda) + \varepsilon) d(x, y)} \; \mathrm{~d} V_{\mathbb{B}^{N}}(x) \\ & \geq C_{\varepsilon} \; e^{{-}2(c(N, \lambda) + \varepsilon)(1 + \varepsilon_{R}) d(y, 0)} \underbrace{\int_{d(x, 0) \leq 1} \mathrm{~d} V_{\mathbb{B}^{N}}(x)}_{:=C} \\ & = \tilde{C_{\varepsilon}} \; e^{{-}2(c(N, \lambda) + \varepsilon)(1 + \varepsilon_{R}) d(y, 0)}. \end{align*}

Therefore we have

(6.6)\begin{equation} I I \geq \frac{\tilde{C_{\varepsilon}} m^{2} \underline{\mathrm{a}} A(r)}{p+1} \; e^{{-}2(c(N, \lambda) + \varepsilon)(1 + \varepsilon_{R}) d(y, 0)}. \end{equation}

Estimate of I : Let us now compute an estimate on $I$ for $\delta > c(n,\, \lambda )(p+1)+(N-1)$, then for every $\varepsilon ^{\prime } >0,\,\\ \delta >(c(n,\, \lambda )- \varepsilon ^{\prime })(p+1)+(N-1).$ We shall estimate $I$ as follows:

(6.7)\begin{equation} \begin{aligned} I & =\frac{t^{p+1}}{p+1} \int_{\mathbb{B}^{N}}(1-a(x)) w(\tau_{{-}y}(x))^{p+1} \mathrm{~d} V_{\mathbb{B}^{N}}(x) \\ & \leq C_{\varepsilon^{\prime}}\frac{t^{p+1}}{p+1} \int_{\mathbb{B}^{N}} \; (1 - a(x)) e^{-(c(n, \lambda)-\varepsilon^{\prime})(p+1)d(x,y)} \mathrm{~d} V_{\mathbb{B}^{N}}(x) \\ & \leq C_{\varepsilon^{\prime}}\frac{t^{p+1}}{p+1} \int_{\mathbb{B}^{N}}e^{-\delta d(x,0)} e^{(c(n, \lambda)-\varepsilon^{\prime})(p+1)(d(x,0)- d(y,0))} \mathrm{~d} V_{\mathbb{B}^{N}}(x) \\ & \leq C_{\varepsilon^{\prime}}\frac{t^{p+1}}{p+1} e^{-(c(n, \lambda)-\varepsilon^{\prime})(p+1)d(y,0)} \int_{\mathbb{B}^{N}}e^{-\delta d(x,0)+(c(n, \lambda)-\varepsilon^{\prime})(p+1)d(x,0)} \mathrm{~d} V_{\mathbb{B}^{N}}(x) \\ & \leq C_{\varepsilon^{\prime}}\frac{t^{p+1}}{p+1} e^{-(c(n, \lambda)-\varepsilon^{\prime})(p+1)d(y,0)} \int_{0}^{\infty}e^{-\delta r+(c(n, \lambda)-\varepsilon^{\prime})(p+1)r + (N-1)r} \mathrm{~d}r \\ & \leq C_{\varepsilon^{\prime}}\frac{M^{p+1}}{p+1} e^{-(c(n, \lambda)-\varepsilon^{\prime})(p+1)d(y,0)}.\\ \end{aligned} \end{equation}

Thus we have deduced

(6.8)\begin{equation} I \leq C_{\varepsilon^{\prime}}\frac{M^{p+1}}{p+1} e^{-(c(n, \lambda)-\varepsilon^{\prime})(p+1)d(y,0)}. \end{equation}

Now applying(6.6) and (6.8), we can choose $R_{0}>R>0$ large enough and also choose ${\varepsilon }$ and $\varepsilon ^{\prime }$ appropriately such that

\[ (I)<(I I) \text{ for }d(y,0) \geq R_{0}. \]

As a result, (6.1) is proved. This completes the proof (6.1). Now the proof of (6.2) can be concluded in a similar line by noting that $(I)$ is zero and $\underline {a} =1.$

Proof. As $u \in U$, we get $\|u\|_{L^{p+1}}=\frac {\|u\|_{H_{\lambda }(\mathbb {B}^{N})}^{\frac {2}{p+1}}}{(p\|a\|_{L^{\infty }(\mathbb {B}^{N})})^{\frac {1}{p+1}}}.$ This, together with the definition of $S_{1,\lambda }$, gives

\[ \|u\|_{H_{\lambda}} \geq S_{1,\lambda}^{\frac{1}{2}}\|u\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}=S_{1,\lambda}^{\frac{1}{2}} \frac{\|u\|_{H_{\lambda}}^{\frac{2}{p+1}}}{\left(p\|a\|_{L^{\infty}\left(\mathbb{B}^{N}\right)}\right)^{\frac{1}{p+1}}} \quad \forall u \in U . \]

Therefore, for all $u \in U$, we have

\[ \|u\|_{H_{\lambda}} \geq \frac{S_{1,\lambda}^{\frac{p+1}{2(p-1)}}}{\left(p\|a\|_{L^{\infty}\left(\mathbb{B}^{N}\right)}\right)^{\frac{1}{p-1}}}=\frac{p}{p-1} C_{p} S_{1,\lambda}^{\frac{p+1}{2(p-1)}}. \]

Thus the lemma follows.

Proof. Define,

(8.3)\begin{equation} \tilde{J}(u):=\frac{1}{2}\|u\|_{H_{\lambda}}^{2}-\frac{\|a\|_{L^{\infty}\left(\mathbb{B}^{N}\right)}}{p+1}\|u\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}^{p+1}-\langle f, u\rangle, \quad u \in H^{1}\left(\mathbb{B}^{N}\right).\end{equation}

1. step 1: This step aims to show the existence of a constant $\alpha >0$ such that

   \[ \left.\frac{d}{d t} \tilde{J}(t u)\right|_{t=1} \geq \alpha \quad \forall u \in U . \]
   It directly follows from the definition of $\tilde {J}$ that
   \[ \left.\frac{d}{d t} \tilde{J}(t u)\right|_{t=1}=\|u\|_{H_{\lambda}}^{2}-\|a\|_{L^{\infty}\left(\mathbb{B}^{N}\right)}\|u\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}^{p+1}-{\langle f, u\rangle}. \]
   Therefore, from the definition of $U$ and substituting the value of $C_{p}$, we have for $u \in U$
   (8.4)\begin{equation} \begin{aligned} \left.\frac{d}{d t} \tilde{J}(t u)\right|_{t=1}=\frac{p-1}{p}\|u\|_{H_{\lambda}}^{2}-{\langle f, u\rangle} & =\left(p\|a\|_{L^{\infty}\left(\mathbb{B}^{N}\right)}\right)^{\frac{1}{p-1}} C_{p}\|u\|_{H_{\lambda}}^{2}-\langle f, u\rangle \\ & =\left(\frac{\|u\|_{H_{\lambda}}^{2}}{\|u\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}^{p+1}}\right)^{\frac{1}{p-1}} C_{p}\|u\|_{H_{\lambda}\left(\mathbb{B}^{N}\right)}^{2}\\ & \quad -\langle f, u\rangle =C_{p} \frac{\|u\|_{H_{\lambda}}^{\frac{2p}{p-1}}}{\|u\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}^{\frac{p+1}{p-1}}}-\langle f, u\rangle. \end{aligned} \end{equation}
   Furthermore, the given hypothesis, i.e., (8.2) implies there exists $d>0$ such that
   (8.5)\begin{equation} \inf _{u \in H^{1}\left(\mathbb{B}^{N}\right),\|u\|_{L^{p+1}\left(\mathbb{B}^{N}\right)=1}}\left\{C_{p}\|u\|_{H_{\lambda}}^{\frac{2 p}{p-1}}-\langle f, u\rangle\right\} \geq d . \end{equation}
   Now,
   \begin{align*} (8.5) & \Longleftrightarrow C_{p} \frac{\|u\|_{H_{\lambda}\left(\mathbb{B}^{N}\right)}^{\frac{2 p}{p-1}}}{\|u\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}^{\frac{p+1}{p-1}}}-\langle f, u\rangle \geq d, \quad\|u\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}=1\\ & \Longleftrightarrow C_{p} \frac{\|u\|_{H_{\lambda}\left(\mathbb{B}^{N}\right)}^{\frac{2 p}{p-1}}}{\|u\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}^{\frac{p+1}{p-1}}}-\langle f, u\rangle \geq d\|u\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}, \quad u \in H^{1}\left(\mathbb{B}^{N}\right) \backslash\{0\} . \end{align*}

   Hence, step 1 follows by using the above estimate in (8.4) and by Remark (8.1).

2. Step 2 Let $u_{n}$ be a minimizing sequence for $I_{\lambda,a, f}$ on $U$, i.e.,

   $I_{\lambda,a, f}(u_{n}) \rightarrow c_{1}$ and $\left \|u_{n}\right \|_{H_{\lambda }}^{2}=p\|a\|_{L^{\infty }(\mathbb {B}^{N})}\left \|u_{n}\right \|_{L^{p+1}(\mathbb {B}^{N})}^{p+1}$. Thus for $n$ large, we get

   \begin{align*} & c_{1}+o(1) \geq I_{\lambda,a, f}\left(u_{n}\right) \geq \tilde{J}\left(u_{n}\right) \geq\left(\frac{1}{2}-\frac{1}{p(p+1)}\right)\left\|u_{n}\right\|_{H_{\lambda}}^{2}\\ & \quad -\|f\|_{H^{{-}1}\left(\mathbb{B}^{N}\right)}\left\|u_{n}\right\|_{H_{\lambda}} . \end{align*}
   As a result, $\left \{\tilde {J}(u_{n})\right \}$ is a bounded sequence. Also, $\left \|u_{n}\right \|_{H_{\lambda }}$ and $\left \|u_{n}\right \|_{L^{p+1}(\mathbb {B}^{N})}$ are bounded.

   Claim: $c_{0}<0$.

   To prove the above claim, it suffices to show that there exists $v \in U_{1}$ such that $I_{\lambda,a, f}(v)<0$. Remark (8.2) implies we can choose $u \in U$ such that $\langle f,\, u\rangle >0$. Therefore,

   \[ I_{\lambda,a, f}(t u) \leq t^{2}\|u\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}^{p+1}\left[\frac{p\|a\|_{L^{\infty}\left(\mathbb{B}^{N}\right)}}{2}-\frac{t^{p-1}}{p+1}\right]-t\langle f, u\rangle<0 . \]
   for $t<<1$. Moreover, by Remark (8.2), $t u \in U_{1}$. This proves the claim.

   Now $I_{\lambda, a, f}(u_{n})<0$ for large $n$ by using the above claim. Consequently,

   \[ 0>I_{\lambda,a, f}\left(u_{n}\right) \geq\left(\frac{1}{2}-\frac{1}{p(p+1)}\right)\left\|u_{n}\right\|_{H_{\lambda}}^{2}-\left\langle f, u_{n}\right\rangle. \]
   Therefore, $p>1$ implies $\left \langle f,\, u_{n}\right \rangle >0$ for all large $n$ . As a a result, $\frac {d}{d t} \tilde {J}(t u_{n})<0$ for $t>0$ small enough. Thus, by Step 1, there exists $t_{n} \in (0,\,1)$ such that $\frac {d}{d t} \tilde {J}(t_{n} u_{n})=0$. In addition, $t_{n}$ is unique because
   \begin{align*} \frac{d^{2}}{d t^{2}} \tilde{J}(t u)& =\|u\|_{H_{\lambda}}^{2}-p\|a\|_{L^{\infty}\left(\mathbb{B}^{N}\right)} t^{p-1}\|u\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}^{p+1}\\ & =\left(1-t^{p-1}\right)\|u\|_{H_{\lambda}}^{2}>0, \;\forall u \in U,\; \forall t \in[0,1) . \end{align*}

3. Step 3 The goal of this step is to prove the following

   (8.6)\begin{equation} \liminf _{n \rightarrow \infty}\left\{\tilde{J}\left(u_{n}\right)-\tilde{J}\left(t_{n} u_{n}\right)\right\}>0. \end{equation}
   We can notice that $\tilde {J}(u_{n})-\tilde {J}(t_{n} u_{n})=\int _{t_{n}}^{1} \frac {d}{d t}\left \{\tilde {J}(t u_{n})\right \} \mathrm {d} t$ and that for all $n \in \mathbb {N}$, there is $\xi _{n}>0$ such that $t_{n} \in (0,\,1-2 \xi _{n})$ and $\frac {d}{d t} \tilde {J}(t u_{n}) \geq \alpha$ for $t \in [1-\xi _{n},\, 1]$.

   To prove (8.6), it is enough to show that $\xi _{n}>0$ can be chosen independently of $n \in \mathbb {N}$. But this is true because, by step 1, we have $\left.\frac {d}{d t} \tilde {J}(t u_{n})\right |_{t=1} \geq \alpha$. Moreover, the boundedness of $\left \{u_{n}\right \}$ gives

   \begin{align*} \left|{\frac{d^{2}}{d t^{2}} \tilde{J}\left(t u_{n}\right)}\right|& =\left|\left\|u_{n}\right\|_{H_{\lambda}\left(\mathbb{B}^{N}\right)}^{2}-p\|a\|_{L^{\infty}\left(\mathbb{B}^{N}\right)} t^{p-1}\left\|u_{n}\right\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}^{p+1}\right|\\ & \quad =\left|\left(1-t^{p-1}\right)\left\|u_{n}\right\|_{H_{\lambda}}^{2}\right|\leq C, \end{align*}
   for all $n \geq 1$ and $t \in [0,\,1]$.

4. Step 4 It straight away follows from the definition of $I_{\lambda,a, f}$ and $\tilde {J}$ that $\frac {d}{d t} I_{\lambda,a, f}(t u) \geq \frac {d}{d t} \tilde {J}(t u)$ for all $u \in H^{1}(\mathbb {B}^{N})$ and for all $t>0$. Therefore,

   \begin{align*} I_{\lambda,a, f}\left(u_{n}\right)-I_{\lambda,a, f}\left(t_{n} u_{n}\right)& =\int_{t_{n}}^{1} \frac{d}{d t}\left(I_{\lambda,a, f}\left(t u_{n}\right)\right) \mathrm{d} t \geq \int_{t_{n}}^{1} \frac{d}{d t} \tilde{J}\left(t u_{n}\right) \mathrm{d} t\\ & =\tilde{J}\left(u_{n}\right)-\tilde{J}\left(t_{n} u_{n}\right). \end{align*}
   Since $\left \{u_{n}\right \} \in U$ is a minimizing sequence for $I_{\lambda, a, f}$, and $t_{n} u_{n} \in U_{1}$, we deduce using (8.6) that
   \[ c_{0}=\inf _{u \in U_{1}} I_{\lambda,a, f}(u)<\inf _{u \in U} I_{\lambda,a, f}(u) = c_{1} \]

This completes the proof of the lemma.

Proof. We divide the proof into the following few steps.

1. Step 1 $c_{0}>-\infty.$

   As $I_{\lambda,a, f}(u) \geq \tilde {J}(u)$ so, to prove Step 1, it is enough to show that $\tilde {J}$ is bounded from below. The definition of $U_{1}$ implies

   (8.9)\begin{equation} \tilde{J}(u) \geq\left[\frac{1}{2}-\frac{1}{p(p+1)}\right]\|u\|_{H_{\lambda}}^{2}-\|f\|_{H^{{-}1}\left(\mathbb{B}^{N}\right)}\|u\|_{H_{\lambda}} \text{ for all } u \in U_{1} . \end{equation}
   Since the RHS of the above inequality is a quadratic function in $\|u\|_{H_{\lambda }}$ implies $\tilde {J}$ is bounded from below. Hence Step 1 follows.

2. Step 2 We aim to find a bounded PS sequence $\left \{u_{n}\right \} \subset U_{1}$ for $I_{\lambda,a,f}$ at the level $c_{0}$.

   Let $\left \{u_{n}\right \} \subset \bar {U}_{1}$ such that $I_{\lambda, a, f}(u_{n}) \rightarrow c_{0}$. As $I_{\lambda,a, f}(u) \geq \tilde {J}(u)$ so, from (8.9), we get $\left \{u_{n}\right \}$ is a bounded sequence. Since by Lemma 8.4, $c_{0}< c_{1}$, without restriction we can assume $u_{n} \in U_{1}$. Therefore, by Ekeland's variational principle, we can extract a PS sequence from $\left \{u_{n}\right \}$ in $U_{1}$ for $I_{\lambda,a, f}$ at the level $c_{0}$. We still denote this PS sequence by $\left \{u_{n}\right \}$. Thus step 2 follows.

3. Step 3 In this step, we show that there exists $u_{0} \in U_{1}$ such that $u_{n} \rightarrow u_{0}$ in $H^{1}(\mathbb {B}^{N})$.

   Applying PS decomposition (4) gives

   (8.10)\begin{equation} u_{n}-u_{0}-\sum_{i=1}^{m} w^{i}\left(\tau_{n}^{i}(x)\right) \rightarrow 0 \text{ in } H^{1}\left(\mathbb{B}^{N}\right) \end{equation}
   for some $u_{0}$ such that $(I_{\lambda,a, f})^{\prime }(u_{0})=0$ and some appropriate $w^{i}$ and $\left \{\tau _{n}^{i}\right \}$. We will proceed by the method of contradiction to show that $m=0$, which in turn will imply step 3. Assume that there is $w^{i} \neq 0$ for $i \in \{1,\,2,\, \cdots,\, m\}$ such that $(I_{\lambda,1,0})^{\prime }(w^{i})=0$, i.e, $\left \|w^{i}\right \|_{H_{\lambda }}^{2}=\int _{\mathbb {B}^{N}}(w^{i})_+^{p+1} \mathrm {~d}V_{\mathbb {B}^{N}}$. Therefore,
   \begin{align*} g\left(w^{i}\right)& =\left\|w^{i}\right\|_{H_{\lambda}}^{2}-p\|a\|_{L^{\infty}\left(\mathbb{B}^{N}\right)}\left\|w^{i}\right\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}^{p+1}\\ & =\int_{\mathbb{B}^{N}}\left(w^{i}\right)_+^{p+1} \mathrm{~d}V_{\mathbb{B}^{N}}-p\|a\|_{L^{\infty}\left(\mathbb{B}^{N}\right)} \int_{\mathbb{B}^{N}}\left|w^{i}\right|^{p+1} \mathrm{~d}V_{\mathbb{B}^{N}} \\ & \leq\left\|w^{i}\right\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}^{p+1}\left(1-p\|a\|_{L^{\infty}\left(\mathbb{B}^{N}\right)}\right)<0, \end{align*}
   where for the last inequality, we have used that $p>1$ and $\|a\|_{L^{\infty }(\mathbb {B}^{N})} \geq 1$. Further, using the Remark 8.5, we get $I_{\lambda,1,0}(w^{i}) \geq S^{\infty }>0$ for all $1 \leq i \leq m$. Therefore, $I_{\lambda,a, f}(u_{n}) \rightarrow I_{\lambda,a, f}(u_{0})+\sum _{i=1}^{m} I_{\lambda,1,0}(w_{i})$ implies $I_{\lambda,a, f}(u_{0})< c_{0}$. Thus $u_{0} \notin U_{1}$, i.e., $g(u_{0}) \leq 0 .$

   We have $g(u_{n}) \geq 0$ because $u_{n} \in U_{1}$. We now compute $g(u_{0}+\sum _{i=1}^{m} w^{i}(\tau _{n}^{i}(x)))$. Thus (8.10) and uniform continuity of $g$ implies

   (8.11)\begin{equation} 0 \leq \liminf _{n \rightarrow \infty} g\left(u_{n}\right)=\liminf _{n \rightarrow \infty} g\left(u_{0}+\sum_{i=1}^{m} w^{i}\left(\tau_{n}^{i}(x)\right)\right).\end{equation}
   On the other hand, as $\tau _{n}^{i}(0) \rightarrow \infty,\,\; d(\tau _{n}^{i}(0),\,\tau _{n}^{j}(0))\rightarrow \infty$ for $1 \leq i \neq j \leq m$ the supports of $u_{0}(\bullet )$ and $w^{i}(\tau _{n}^{i}(\bullet ))$ are going increasingly far away as $n \rightarrow \infty$. Therefore,
   \begin{align*} & \lim _{n \rightarrow \infty} g\left(u_{0}+\sum_{i=1}^{m} w^{i}\left(\tau_{n}^{i}(x)\right)\right)=g\left(u_{0}\right)\\ & \quad +\lim _{n \rightarrow \infty} \sum_{i=1}^{m} g\left(w^{i}\left(\tau_{n}^{i}(x)\right)\right)=g\left(u_{0}\right)+\sum_{i=1}^{m} g\left(w^{i}\right), \end{align*}
   where the last equality follows from the translation invariance of $g$. Now because $g(u_{0}) \leq 0$ and $g(w^{i})<0$ for $1\leq i \leq m$, we get a contradiction to (8.11). This proves step 3.

4. Step 4 Using the previous steps, we can conclude that $I_{\lambda,a, f}(u_{0})=c_{0}$ and $(I_{\lambda,a, f})^{\prime }(u_{0})=0$. Thus, $u_{0}$ is a weak solution to (3.4); combining this with Remark 3.1, we complete the proof of the proposition.

Proof. For $u_{0}$ to be the critical point found in Proposition 8.6 and $w$ to be as in Remark 8.5, set $w_{t}(x):=t w(x).$

Claim 1: $u_{0}+w_{t} \in U_{2}$ for $t>0$ large enough.

As $p>1$ and $\|a\|_{L^{\infty }(\mathbb {B}^{N})} \geq 1$, we have

\begin{align*} g\left(u_{0}+w_{t}\right) & \leq\left\|u_{0}\right\|_{H_{\lambda}}^{2}+\left\|w_{t}\right\|_{H_{\lambda}}^{2}+2\left\langle u_{0}, w_{t}\right\rangle_{H_{\lambda}}\\ & \quad -p\left(\left\|u_{0}\right\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}^{p+1}+\left\|w_{t}\right\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}^{p+1}\right) \\ & \leq(1+\varepsilon)\left\|w_{t}\right\|_{H_{\lambda}}^{2}+(1+C(\varepsilon))\left\|u_{0}\right\|_{H_{\lambda}}^{2}\\ & \quad -p\left(\left\|u_{0}\right\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}^{p+1}+\left\|w_{t}\right\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}^{p+1}\right)\\ & = t^2(1+\varepsilon)\left\|w\right\|_{H_{\lambda}}^{2}+(1+C(\varepsilon))\left\|u_{0}\right\|_{H_{\lambda}}^{2}\\ & \quad -p\left(\left\|u_{0}\right\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}^{p+1}+t^{p+1}\left\|w\right\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}^{p+1}\right), \end{align*}

where the second last step follows from Young's inequality with $\varepsilon >0$. Moreover, as $w$ is the solution to (3.1) implies

\[ \left\|w\right\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}^{p+1}=\left\|w\right\|_{H_{\lambda}}^{2}. \]

Finally,

\[ g\left(u_{0}+w_{t}\right) \leq(1+C(\varepsilon))\left\|u_{0}\right\|_{H_{\lambda}}^{2}-p\left\|u_{0}\right\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}^{p+1}+\left\|w\right\|_{H_{\lambda}}^{2}\left[(1+\varepsilon) t^{2}-p t^{p+1}\right] \]

Thus choosing $\varepsilon >0$ such that $1+\varepsilon < p$ gives $g(u_{0}+w_{t})<0$ for $t>0$ large enough. Hence the claim follows.

Claim 2: $I_{\lambda,a, f}(u_{0}+w_{t})< I_{\lambda,a, f}(u_{0})+I_{\lambda,1,0}(w_{t}) \forall t>0$.

As $u_{0},\, w_{t}>0$, using $w_{t}$ as the test function for (3.4) yields

\[ \left\langle u_{0}, w_{t}\right\rangle_{H_{\lambda}}=\int_{\mathbb{B}^{N}} a(x) u_{0}^{p} w_{t}\mathrm{~d}V_{\mathbb{B}^{N}}+\left\langle f, w_{t}\right\rangle . \]

Therefore, utilizing the above expression and assumption $a \geq 1$, we compute the following

\begin{align*} I_{\lambda,a, f}\left(u_{0}+w_{t}\right) & = \frac{1}{2}\left\|u_{0}\right\|_{H_{\lambda}}^{2}+\frac{1}{2}\left\|w_{t}\right\|_{H_{\lambda}}^{2}+\left\langle u_{0}, w_{t}\right\rangle_{H_{\lambda}} \\ & -\frac{1}{p+1} \int_{\mathbb{B}^{N}} a(x)\left(u_{0}+w_{t}\right)^{p+1} \mathrm{~d}V_{\mathbb{B}^{N}}(x)-\left\langle f, u_{0}\right\rangle-\left\langle f, w_{t}\right\rangle \\ & \quad= I_{\lambda,a, f}\left(u_{0}\right)+I_{\lambda,1,0}\left(w_{t}\right)+\left\langle u_{0}, w_{t}\right\rangle_{H_{\lambda}}\\ & \quad +\frac{1}{p+1} \int_{\mathbb{B}^{N}} a(x) u_{0}^{p+1} \mathrm{~d}V_{\mathbb{B}^{N}}(x)\\ & \quad+\frac{1}{p+1} \int_{\mathbb{B}^{N}} w_{t}^{p+1} \mathrm{~d}V_{\mathbb{B}^{N}}\\ & \quad -\frac{1}{p+1} \int_{\mathbb{B}^{N}} a(x)\left(u_{0}+w_{t}\right)^{p+1} \mathrm{~d}V_{\mathbb{B}^{N}}-\left\langle f, w_{t}\right\rangle \\ & \leq I_{\lambda,a, f}\left(u_{0}\right)+I_{\lambda,1,0}\left(w_{t}\right)\\ & \quad+\frac{1}{p+1} \int_{\mathbb{B}^{N}} a(x)\left[(p+1) u_{0}^{p} w_{t}+u_{0}^{p+1}+w_{t}^{p+1}\right.\\ & \quad\left. -\,\left(u_{0}+w_{t}\right)^{p+1}\right] \mathrm{~d}V_{\mathbb{B}^{N}}(x) \\ < & I_{\lambda,a, f}\left(u_{0}\right)+I_{\lambda,1,0}\left(w_{t}\right) \end{align*}

This proves the claim. Further, the straightforward calculation gives

(8.12)\begin{equation} I_{\lambda,1,0}\left(w_{t}\right)=\frac{t^{2}}{2}\left\|w\right\|_{H_{\lambda}}^{2}-\frac{t^{p+1}}{p+1}\left\|w\right\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}^{p+1} \rightarrow-\infty \quad \text{ as } \quad t \rightarrow \infty. \end{equation}

From (8.12) and Remark 8.5, we have

\[ \sup _{t>0} I_{\lambda,1,0}\left(w_{t}\right)=I_{\lambda,1,0}\left(w_{1}\right)=I_{\lambda,1,0}\left(w\right)=S^{\infty}. \]

Combing this with Claim 2 yields

(8.13)\begin{equation} I_{\lambda,a, f}\left(u_{0}+w_{t}\right)< I_{\lambda,a, f}\left(u_{0}\right)+S^{\infty} \quad \forall t>0 . \end{equation}

Claim 2, together with (8.12), results in

(8.14)\begin{equation} I_{\lambda,a, f}\left(u_{0}+w_{t}\right)< I_{\lambda,a, f}\left(u_{0}\right) \quad \text{ for } \quad t \text{ large enough. }\end{equation}

We now fix $t_{0}>0$ large enough such that (8.14) and Claim 1 are satisfied. Then set

\[ \gamma:=\inf _{i \in \Gamma} \max _{t \in[0,1]} I_{\lambda,a, f}(i(t)), \]

where

\[ \Gamma:=\left\{i \in C\left([0,1], H^{1}\left(\mathbb{B}^{N}\right)\right): i(0)=u_{0}, \; i(1)=u_{0}+w_{t_{0}}\right\} \]

As $u_{0} \in U_{1}$ and $u_{0}+w_{t_{0}} \in U_{2}$, for every $i \in \Gamma$, there exists $t_{i} \in (0,\,1)$ such that $i(t_{i}) \in U$. Therefore,

\[ \max _{t \in[0,1]} I_{\lambda,a, f}(i(t)) \geq I_{\lambda,a, f}\left(i\left(t_{i}\right)\right) \geq \inf _{U} I_{\lambda,a, f}(u)=c_{1} . \]

Thus, using Lemma 8.4, we have $\gamma \geq c_{1}>c_{0}=I_{\lambda,a, f}(u_{0})$.

Claim 3: For $S^{\infty }$, as defined in (8.8), $\gamma < I_{\lambda,a, f}(u_{0})+S^{\infty }$.

Observe that $\lim _{t \rightarrow 0}\left \|w_{t}\right \|_{H_{\lambda }}=0$. Thus, if we define $\tilde {i}(t)=u_{0}+w_{tt_{0}}$,

then $\lim _{t \rightarrow 0}\left \|\tilde {i}(t)-u_{0}\right \|_{H_{\lambda }}=0$. As a result, $\tilde {i} \in \Gamma$. Therefore, using (8.13) will give us

\[ \gamma \leq \max _{t \in[0,1]} I_{\lambda,a, f}(\tilde{i}(t))=\max _{t \in[0,1]} I_{\lambda,a, f}\left(u_{0}+w_{t t_{0}}\right)< I_{\lambda,a, f}\left(u_{0}\right)+S^{\infty} \]

Hence the claim follows. Thus

\[ I_{\lambda,a, f}\left(u_{0}\right)<\gamma< I_{\lambda,a, f}\left(u_{0}\right)+S^{\infty} . \]

Applying Ekeland's variational principle, there exists a PS sequence $\left \{u_{n}\right \}$ for $I_{\lambda,a, f}$ at the level $\gamma$. Also, note that $\left \{u_{n}\right \}$ is a bounded sequence. Further, from PS decomposition and Remark (8.5), we have $S^{\infty }=I_{\lambda,1,0}(w)$ and $u_{n} \rightarrow v_{0}$ for some $v_{0} \in H^{1}(\mathbb {B}^{N})$ such that $(I_{\lambda,a, f})^{\prime }(v_{0})=0$ and $I_{\lambda,a, f}(v_{0})=\gamma$. Further, as $I_{\lambda,a, f}(u_{0})<\gamma$, we conclude $v_{0} \neq u_{0}$. Finally, $(I_{\lambda,a, f})^{\prime }(v_{0})=0$, along with the Remark 3.1, completes the proof of the proposition.

Proof. We can find an $\varepsilon >0$ such that $\|f\|_{H^{-1}(\mathbb {B}^{N})}< C_{p} S_{1,\lambda }^{\frac {p+1}{2(p-1)}}-\varepsilon$ using the given assumption. Therefore, using Lemma 8.3. we have

\begin{align*} & \langle f, u\rangle \leq\|f\|_{H^{{-}1}\left(\mathbb{B}^{N}\right)}\|u\|_{H_{\lambda}}<\left[C_{p} S_{1,\lambda}^{\frac{p+1}{2(p-1)}}-\varepsilon\right]\|u\|_{H_{\lambda}\left(\mathbb{B}^{N}\right)} \\ & \quad \leq \frac{p-1}{p}\|u\|_{H_{\lambda}}^{2}-\varepsilon\|u\|_{H_{\lambda}\left(\mathbb{B}^{N}\right)},\; \forall u \in U. \end{align*}

Thus

\[ \inf _{U}\left[\frac{p-1}{p}\|u\|_{H_{\lambda}}^{2}-\langle f, u\rangle\right] \geq \varepsilon \inf _{U}\|u\|_{H_{\lambda}}. \]

Moreover, Remark 8.1 gives us that $\|u\|_{H_{\lambda }}$ is bounded away from 0 on $U$, so the above expression yields

\[ \inf _{U}\left[\frac{p-1}{p}\|u\|_{H_{\lambda}}^{2}-\langle f, u\rangle\right]>0 . \]

On the other hand,

\begin{align*} (8.2) & \Leftrightarrow C_{p} \frac{\|u\|_{H_{\lambda}}^{\frac{2 p}{p-1}}}{\|u\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}^{\frac{p+1}{p-1}}}-\langle f, u\rangle>0 \quad \text{ for } \quad\|u\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}=1 \\ & \Leftrightarrow \frac{\|u\|_{H_{\lambda}}^{\frac{2 p}{p-1}}}{\|u\|_{L^{p+1}\left(\mathbb{B}^{N}\right)}^{\frac{p+1}{p-1}}}-\langle f, u\rangle>0 \quad \text{ for } \quad u \in U\\ & \Leftrightarrow \frac{p-1}{p}\|u\|_{H_{\lambda}}^{2}-\langle f, u\rangle >0 \quad \text{ for } \quad u \in U. \end{align*}

Hence the lemma follows.