Proof. By [Reference Jeanjean and Lu31, lemma 2.3], we have

(2.3)\begin{align} f(s) s- \bar{p} F(s) \gt 0 \quad \text {for all } s \neq 0, \end{align}

where $\bar{p}=2+\frac{4}{N}$. We claim that $\liminf_{s\rightarrow 0}\frac{f(s) s- \bar{p} F(s)}{|s|^{2^*}} \gt 0$. Assume, for the sake of contradiction, that $\liminf_{s\rightarrow 0}\frac{f(s) s- \bar{p} F(s)}{|s|^{2^*}} =0$. Since f is an odd function, we can deduce that

(2.4)\begin{align} \liminf_{s\rightarrow 0}\frac{\frac{d}{ds}\left(F(s)/s^{\bar{p}}\right)}{\frac{d}{ds}s^{2^*-p}} =0. \end{align}

From (2.4), it follows that $\liminf_{s\rightarrow 0}\frac{F(s)}{s^{2^*}} =0$, which contradicts $(F5)$. Therefore, we conclude that

\begin{equation*}\liminf_{s\rightarrow 0}\frac{f(s) s- \bar{p} F(s)}{s^{2^*}} \gt 0.\end{equation*}

This implies there exist $c_1 \gt 0$ and δ > 0, such that

\begin{align*} \frac{f(s) s- \bar{p} F(s)}{s^{2^*}} \geq \frac{c_1}{2}, ~~s\in(0,\delta). \end{align*}

Furthermore, by $(F1)$, $(F2)$, and (2.3), we have

\begin{equation*}\frac{f(s) s- \bar{p} F(s)}{s^{2^*}}\in C([\delta,\infty),\mathbb{R}^+)\end{equation*}

with

\begin{equation*}\lim_{s\rightarrow \infty}\frac{f(s) s- \bar{p} F(s)}{s^{2^*}} =\left(2^*-\bar{p}\right)\eta \gt 0.\end{equation*}

Therefore, there exists $c_2 \gt 0$ such that $\frac{f(s) s- \bar{p} F(s)}{s^{2^*}} \geq c_2$ on $[\delta,\infty)$. Then there exists c > 0 such that

\begin{equation*} f(s) s- \bar{p} F(s) \gt c|s|^{2^*} \quad \text {for all } s \neq 0. \end{equation*}

This concludes this proof.

Proof. We recall the Hardy inequality as presented in [Reference Bahouri, Chemin and Danchin3, theorem 1.72]:

(2.6)\begin{align} \int_{\mathbb{R}^N} \frac{u^2}{|x|^2} \mathrm{d} x \leq \frac{1}{\bar{\mu}} \int_{\mathbb{R}^N}|\nabla u|^2 \mathrm{d} x. \end{align}

By $(F2)$, for any δ > 0, there exists $C_{\delta,\eta} \gt \eta$ such that

(2.7)\begin{align} F(s) \leq \left(\delta+\beta\right)|s|^{\bar{p}}+C_{\delta,\eta}|s|^{2^*}, \quad \text{for all~} s \in \mathbb{R}. \end{align}

From (1.7), (2.6), (2.7), and $(F5)$, we derive that

\begin{align*} \mathcal{I}_\mu(t \star u) &\geq \frac{t^2}{2}\min\left\{1,\left(1-\frac{\mu}{\bar{\mu}}\right)\right\} |\nabla u|^2_2-(\delta+\beta) t^{2} |u|^{\bar{p}}_{\bar{p}} -C_{\delta,\eta} t^{2^*}| u|^{2^*}_{2^*}\\ &\geq \left(\frac{1}{2}\min\left\{1,\left(1-\frac{\mu}{\bar{\mu}}\right)\right\} -C_{N,\bar{p}}^{\bar{p}}a^{\frac{2}{N}}(\delta+\beta)\right) t^2|\nabla u|^2_2 -C_{\delta,\eta} t^{2^*}| u|^{2^*}_{2^*}, \end{align*}

and

\begin{align*} \mathcal{I}_\mu(t \star u) \leq \frac{t^2}{2} \max\left\{1,\left(1-\frac{\mu}{\bar{\mu}}\right)\right\} |\nabla u|^2_2 -\frac{\kappa}{p} t^{p\gamma_p} |u|_p^p. \end{align*}

The conclusion can be drawn that, given the condition $\bar{\mu} \gt \mu$, $p \gt \bar{p}$, and a sufficiently small δ,

\begin{align*} \mathcal{I}_\mu(t \star u) \rightarrow 0^+ \quad \text {as } \quad t \rightarrow 0^+\quad \text {and } \quad \mathcal{I}_\mu(t \star u)\rightarrow-\infty \quad \text {as } \quad t \rightarrow+\infty. \end{align*}

This concludes the proof.

Proof. For any $u \in S_a$, we have

\begin{align*} \mathcal{I}_\mu\left(t \star u\right)&=\frac{t^2}{2} \int_{\mathbb{R}^N}|\nabla u|^2 - \frac{\mu}{|x|^2}u^2 \mathrm{d} x-\int_{\mathbb{R}^N} F\left(t \star u \right) \mathrm{d} x, \end{align*}

and

(2.8)\begin{align} P_\mu\left(t \star u\right)&=t^2 \int_{\mathbb{R}^N}|\nabla u|^2 -\frac{\mu}{|x|^2}u^2 \mathrm{d} x-\frac{N}{2}\int_{\mathbb{R}^N}\widetilde{F} \left(t \star u \right) \mathrm{d} x \nonumber \\ &=t^2\left(\int_{\mathbb{R}^N}|\nabla u|^2 -\frac{\mu}{|x|^2}u^2 \mathrm{d} x -\frac{N}{2} \int_{\mathbb{R}^N} \frac{\widetilde{F}\left(t^{\frac{N}{2}} u\right)}{\left|t^{\frac{N}{2}} u\right|^{\bar{p}}}|u|^{\bar{p}} \mathrm{d} x\right). \end{align}

It is evident that $ \mathcal{I}_\mu\left(t \star u\right)$ is of class C ¹, and its derivative can be expressed as

\begin{align*} \begin{split} \frac{d}{d t} \mathcal{I}_\mu\left(t \star u\right)=t \int_{\mathbb{R}^N}|\nabla u|^2 - \frac{\mu}{|x|^2}u^2 \mathrm{d} x -\frac{N}{2} t^{-1} \int_{\mathbb{R}^N} \widetilde{F}\left( t \star u \right) \mathrm{d} x =\frac{1}{t}P_\mu(t \star u). \end{split} \end{align*}

With the application of (2.6),

\begin{align*} t^2 \int_{\mathbb{R}^N}|\nabla u|^2 -\frac{\mu}{|x|^2}u^2 \mathrm{d} x\geq t^2\min\left\{1,1-\frac{\mu}{\bar{\mu}} \right\}|\nabla u|_2^2. \end{align*}

By $(F4),(F5)$, and (2.7), one gets

\begin{align*} \begin{split} \frac{N}{2}\int_{\mathbb{R}^N} \widetilde{F} \left( t \star u\right)\mathrm{d} x & \lt 2^* \int_{\mathbb{R}^N}F\left( t \star u \right)\mathrm{d} x\\ &\leq 2^* \left( (\delta+\beta) t^2 \int_{\mathbb{R}^N} | u|^{\bar{p}}\mathrm{d} x +C_{\delta,\eta} t^{2^*} \int_{\mathbb{R}^N}| u|^{2^*}\mathrm{d} x\right)\\ &=2^* \left( C_{N,\bar{p}}^{\bar{p}}a^{\frac{2}{N}}(\delta+\beta)t^2 \left|\nabla u\right|_2^{2}+C_2t^{2^*} \left|\nabla u\right|_2^{2^*}\right). \end{split} \end{align*}

It is apparent that by selecting a sufficiently small δ, we can ensure that:

\begin{align*} 2^* C_{N,\bar{p}}^{\bar{p}}a^{\frac{2}{N}}(\delta+\beta) \lt \min\left\{1-\frac{\mu}{\bar{\mu}},1 \right\}. \end{align*}

Thus $\frac{d}{d t} \mathcal{I}_\mu\left(t \star u\right)(t) \gt 0$ for sufficiently small t. Similar to lemma 2.2, we can conclude that $\frac{d}{d t} \mathcal{I}_\mu\left(t \star u\right)(t)\to-\infty$ as $t\to \infty$. Therefore, there exists at least one $t_{u}\in \mathbb{R}^{+}$ such that $\frac{d}{d t} \mathcal{I}_\mu\left(t \star u\right)(t)=\frac{1}{t_u}P_\mu(t_u \star u)=0$, namely, $ t_{u} \star u\in \mathcal{M}_\mu(a)$.

Suppose that there exists another $t_{u_1}$ such that $t_{u_1} \star u\in \mathcal{M}_\mu(a)$. Combined with (2.8), it yields

\begin{align*} \int_{\mathbb{R}^N} \frac{\widetilde{F}\left(t_{u}^{\frac{N}{2}} u\right)} {\left|t_{u}^{\frac{N}{2}} u\right|^{\bar{p}} }|u|^{\bar{p}} \mathrm{d} x =\int_{\mathbb{R}^N} \frac{\widetilde{F}\left(t_{u_1}^{\frac{N}{2}} u\right)} {\left|t_{u_1}^{\frac{N}{2}} u\right|^{\bar{p}}}|u|^{\bar{p}} \mathrm{d} x, \end{align*}

which contradicts $(F3)$. Hence, it is established that $t_{u}=t_{u_1}$. Furthermore, we can deduce that $ \mathcal{I}_\mu\left(t_{u} \star u\right) \gt \mathcal{I}_\mu(t \star u)$ for all t > 0 with $t \neq t_u$.

Proof. (i) For any $u \in \mathcal{M}_\mu(a)$, combined with (1.7), (2.6), (2.7), and the Sobolev inequality, $P_\mu(u)=0$ implies that

\begin{align*} |\nabla u|_2^2 & =\int_{\mathbb{R}^N}\frac{\mu}{|x|^2}u^2\mathrm{d} x +\frac{N}{2} \int_{\mathbb{R}^N}\widetilde{F}(u)\mathrm{d} x \\ & \leq \max \left\{0, \frac{\mu}{\bar{\mu}}\right\}|\nabla u|_2^2+2^* \int_{\mathbb{R}^N} F(u) \mathrm{d} x \\ & \leq \max \left\{0, \frac{\mu}{\bar{\mu}}\right\}|\nabla u|_2^2+2^* \left( C_{N,\bar{p}}^{\bar{p}}a^{\frac{2}{N}}(\delta+\beta) \left|\nabla u\right|_2^{2}+C_2 \left|\nabla u\right|_2^{2^*}\right). \end{align*}

Let δ enough small such that

(2.9)\begin{align} 2^*C_{N,\bar{p}}^{\bar{p}}a^{\frac{2}{N}}(\delta+\beta) \lt \min\left\{1-\frac{\mu}{\bar{\mu}},1 \right\}. \end{align}

Then there is ρ > 0 such that $\inf \limits_{u\in\mathcal{M}_\mu(a)}|\nabla u|_2 \gt \rho$.

(ii) For any $u \in \mathcal{M}_\mu(a)$, we can deduce that

\begin{align*} \mathcal{I}_\mu(u) \geq \mathcal{I}_\mu(t \star u) & \geq \frac{t^2}{2}\min \left\{1, 1-\frac{\mu}{\bar{\mu}}\right\}|\nabla u|_2^2 - \left(t^2 (\delta+\beta) | u|_{\bar{p}}^{\bar{p}} +C_{\delta,\eta} t^{2^*}| u|_{2^*}^{2^*} \right) \\ & \geq \frac{t^2}{2}\min \left\{1, 1-\frac{\mu}{\bar{\mu}}\right\}|\nabla u|_2^2 \\ &\quad -C_2 t^{2^*}|\nabla u|_2^{2^*}-C_{N,\bar{p}}^{\bar{p}}a^{\frac{2}{N}}(\delta+\beta)t^2|\nabla u|_2^{2}. \end{align*}

By selecting $t=\frac{\sigma}{|\nabla u|_2}$ with sufficiently small σ > 0 and taking δ sufficiently small to ensure that (2.9) holds, we can deduce that

\begin{align*} \mathcal{I}_\mu(u)\geq \frac{\min \left\{1, 1-\frac{\mu}{\bar{\mu}}\right\}-2C_{N,\bar{p}}^{\bar{p}}a^{\frac{2}{N}}\beta}{4}\sigma^2 \gt 0. \end{align*}

This concludes the proof.

Proof. By $(F4)$, (1.7), (2.7), and the Sobolev inequality, we have

\begin{align*} P_\mu(u)&=\int_{\mathbb{R}^N}|\nabla u|^2-\mu\frac{u^2}{|x|^2} \mathrm{d} x-\frac{N}{2}\int_{\mathbb{R}^N}\widetilde{F} \left(u \right) \mathrm{d} x\\ &\geq \int_{\mathbb{R}^N}|\nabla u|^2-\mu\frac{u^2}{|x|^2} \mathrm{d} x-2^*\int_{\mathbb{R}^N}F \left(u \right) \mathrm{d} x\\ &\geq \min \left\{1, 1-\frac{\mu}{\bar{\mu}}\right\}|\nabla u|_2^2 -2^* \left( C_{N,\bar{p}}^{\bar{p}}a^{\frac{2}{N}}(\delta+\beta) \left|\nabla u\right|_2^{2}+C_2 \left|\nabla u\right|_2^{2^*}\right). \end{align*}

Thus, $P_\mu(u) \gt 0$ when $u \in \overline{C(a)}$ if ξ > 0 is sufficiently small. Similarly, we can obtain that $ \mathcal{I}_\mu(u) \gt 0$ when $u \in \overline{C(a)}$ if ξ > 0 is sufficiently small. By $(F5)$, one can see that

\begin{align*} \mathcal{I}_\mu(u)\leq \left(\frac{1}{2}-\frac{1}{\bar{\mu}}\min\{\mu,0\}\right)\int_{\mathbb{R}^N}|\nabla u|^2 \mathrm{d} x, \end{align*}

which implies that $m_\mu(a) \gt \sup_{u\in\overline{C(a)}} \mathcal{I}_\mu(u)$ for ξ > 0 small enough. Combined with lemma (2.2), $ \mathcal{I}_\mu$ has a mountain pass geometry.

Proof. Step 1: $\widetilde{\sigma}_\mu(a) \geq m^r_\mu(a).$ For any $\widetilde{\gamma}=(\iota, \zeta) \in \widetilde{\Gamma}_{\mu}$, by lemma 2.3, there exists $t_0 \in(0,1)$ such that $\iota(t_0) \star \zeta(t_0) \in \mathcal{M}_\mu^r(a)$. Thus, we have

\begin{equation*} \max _{t\in [0,1]} \widetilde{\mathcal{I}}_\mu(\widetilde{\gamma}(t)) \geq \widetilde{\mathcal{I}}_\mu\left(\widetilde{\gamma}\left(t_0\right)\right) = \mathcal{I}_\mu\left(\iota(t_0) \star \zeta(t_0)\right) \geq \inf _{u\in\mathcal{M}_\mu^r(a)} \mathcal{I}_\mu(u)=m^r_\mu (a). \end{equation*}

Hence, $\widetilde{\sigma}_\mu(a) \geq m^r_\mu(a)$.

Step 2: $m^r_\mu(a) \geq \widetilde{\sigma}_\mu(a) $. For any $u \in \mathcal{M}_\mu^r(a)$, then $\widetilde{\gamma}_u$ defined in (2.11) is a path in $\widetilde{\Gamma}_{\mu}$. By lemma 2.3,

\begin{equation*} \mathcal{I}_\mu(u)=\max _{t\in[0,1]} \widetilde{\mathcal{I}}_\mu(\widetilde{\gamma}_u(t)) \geq \widetilde{\sigma}_\mu(a) . \end{equation*}

Thus, $m^r_\mu(a) \geq\widetilde{\sigma}_\mu(a)$.

Proof. Step 1: $\sigma_\mu(a) \geq \widetilde{\sigma}_\mu(a)$. Let $\gamma \in \Gamma_{\mu}$, define $\widetilde{\gamma}(t)=(1, \gamma(t)) \in \widetilde{\Gamma}_{\mu}$. Then $ \mathcal{I}_\mu(\gamma(t))=\widetilde{\mathcal{I}}_\mu(\widetilde{\gamma}(t))\geq \widetilde{\sigma}_\mu(a)$ for all $t\geq 0$. Hence $\sigma_\mu(a) \geq \widetilde{\sigma}_\mu(a)$.

Step 2: $ \widetilde{\sigma}_\mu(a)\geq \sigma_\mu(a) $. For all $\widetilde{\gamma}(t)=(\iota(t), \zeta(t)) \in \widetilde{\Gamma}_{\mu}$, setting $\gamma(t)=\iota(t)\star \zeta(t)$, then $\gamma \in \Gamma_{\mu}$ and $\widetilde{\mathcal{I}}_\mu(\widetilde{\gamma}(t))= \mathcal{I}_\mu(\gamma(t))\geq \sigma_\mu(a)$. Therefore, $ \widetilde{\sigma}_\mu(a)\geq \sigma_\mu(a) $.

Proof. It suffices to show that $m_\mu(a) \geq m^r_\mu(a)$, since $\mathcal{M}_\mu^r(a) \subset \mathcal{M}_\mu(a)$. For any $u \in \mathcal{M}_\mu(a)$, let $v:=|u|^*$ be the Schwarz rearrangement of $|u|$. By the properties of rearrangement, one has

\begin{equation*} \mathcal{I}_\mu(v) \leq \mathcal{I}_\mu(u), \quad P_\mu(v) \leq P_\mu(u)=0. \end{equation*}

By lemma 2.3, there exists $t_v \gt 0$ such that $t_v \star v \in \mathcal{M}_\mu^r(a)$. For any t > 0,

\begin{equation*} \begin{aligned} \mathcal{I}_\mu(t \star v) & =\frac{t^2}{2} \int_{\mathbb{R}^N}|\nabla v|^2-\frac{\mu}{|x|^2}v^2 \mathrm{d} x-t^{-N} \int_{\mathbb{R}^N} F\left(t^{\frac{N}{2}} v\right) \mathrm{d} x \\ & \leq \frac{t^2}{2} \int_{\mathbb{R}^N}|\nabla u|^2-\frac{\mu}{|x|^2}u^2 \mathrm{d} x-t^{-N} \int_{\mathbb{R}^N} F\left(t^{\frac{N}{2}} u\right) \mathrm{d} x \\ & = \mathcal{I}_\mu(t \star u) . \end{aligned} \end{equation*}

By lemma 2.3, we obtain

\begin{equation*} \mathcal{I}_\mu(u) \geq \mathcal{I}_\mu\left(t_v \star u\right) \geq \mathcal{I}_\mu\left(t_v \star v\right) . \end{equation*}

Thus, $m_\mu(a)=m^r_\mu(a)$.

Proof. By lemmas 2.6 and 2.7, we have

\begin{equation*} \sigma_\mu(a)=m^r_\mu(a) \gt \sup_{u\in\left(\overline{C(a)} \cup \mathcal{I}_\mu^0\right) \cap S_{r, a}} \mathcal{I}_\mu(u)=\sup _{(s,u)\in\left((1, \overline{C(a)}) \cup\left(1, \mathcal{I}_\mu^0\right)\right) \cap\left(\mathbb{R} \times S_{r, a}\right)} \widetilde{\mathcal{I}}_\mu(s,u). \end{equation*}

Using the terminology in [Reference Ghoussoub21, Section 5], $\{\widetilde{\gamma}([0,1]): \widetilde{\gamma}\in\widetilde{\Gamma}_{\mu}\}$ is a homotopy stable family of compact subsets of $\mathbb{R} \times S_{r,a}$ with extended closed boundary $\left(1, \overline{C(a)}\right) \cup\left(1, \mathcal{I}_\mu^0\right)$. Furthermore, the superlevel set $\{u\in S_{r,a}:\widetilde{\mathcal{I}}_\mu(u) \geq \widetilde{\sigma}_\mu(a)\}$ is a dual set for $\widetilde{\Gamma}_{\mu}$, meaning that $(F{^\prime}1)$ and $(F{^\prime}2)$ in [Reference Ghoussoub21, theorem 5.2] are satisfied. Therefore, considering any minimizing sequence $\left\{\gamma_n=\left(1, \zeta_n\right)\right\} \subset \widetilde{\Gamma}_{\mu}$ for $\widetilde{\sigma}_\mu(a)$, where $\zeta_n(t) \geq 0$ almost everywhere in $\mathbb{R}^N$ for $t \in[0,1]$, there exists a Palais–Smale sequence $\left\{\left(s_n, w_n\right)\right\} \subset \mathbb{R}^+ \times S_{r,a}$ for $\left.\widetilde{\mathcal{I}}_\mu\right|_{\mathbb{R} \times S_{r,a}}$ at level $\widetilde{\sigma}_\mu(a)$, such that as $ n \rightarrow \infty$,

(2.12)\begin{align} \partial_s \widetilde{\mathcal{I}}_\mu\left(s_n, w_n\right) \rightarrow 0, \end{align}

and

(2.13)\begin{align} \left\|\partial_u \widetilde{\mathcal{I}}_\mu\left(s_n, w_n\right)\right\|_{\left(T_{w_n} S_r\right)^*} \rightarrow 0 , \end{align}

with the additional property that

(2.14)\begin{align} \left|s_n-1\right|+\|w_n-\zeta_n\| \rightarrow 0. \end{align}

By (2.10) and (2.12), we have $P_\mu\left(s_n \star w_n\right)= o_n(1)$. Also by (2.13) and the boundedness of $\left\{s_n\right\}$ due to (2.14), we obtain

(2.15)\begin{align} d \mathcal{I}_\mu\left(s_n \star w_n\right)\left(s_n \star \varphi\right)=o_n(1)\|\varphi\| , \quad \text {for every } \varphi \in T_{w_n} S_{r,a}. \end{align}

Let $u_n:=s_n \star w_n$, based on (2.15) and lemma 2.9, $\left\{u_n\right\} \subset S_{r,a}$ is a Palais–Smale sequence for $\left. \mathcal{I}_\mu\right|_{S_{r,a}}$ at the level $\sigma_\mu(a)$. Moreover, $P_\mu\left(u_n\right)=o_n(1)$.

Proof. For a fixed $u \in S_{r,a}$, there exist $0 \lt s_0 \lt 1 \lt s_1$ such that

\begin{equation*} \gamma_o(t)=\left((1-t) s_0+t s_1\right) \star u \in \Gamma_{\mu}. \end{equation*}

By $(F5)$ and lemmas 2.6–2.8, we observe that

\begin{align*} \begin{split} \sigma_\mu(a) &\leq \max _{t\in[0,1]} {\mathcal{I}_\mu}(\gamma_o(t)) \\ & \leq \max _{t \geq 0}\left\{\frac{t^2}{2} \int_{\mathbb{R}^N}|\nabla u|^2 -\frac{\mu}{|x|^2}u^2 \mathrm{d} x -\frac{\kappa}{p} t^{p\gamma_p} \int_{\mathbb{R}^N}|u|^p \mathrm{d} x\right\} \\ & \leq \max _{t \geq 0}\left\{\frac{1}{2}\max \left\{1, 1-\frac{\mu}{\bar{\mu}} \right\} t^2 \int_{\mathbb{R}^N}|\nabla u|^2 \mathrm{d} x -\frac{\kappa}{p} t^{p\gamma_p} \int_{\mathbb{R}^N}|u|^p \mathrm{d} x\right\} \\ & \leq C\left(\frac{1}{\kappa}\right)^{\frac{2}{p\gamma_p-2}}, \end{split} \end{align*}

which deduces that $ \sigma_\mu(a) \lt \epsilon$ for any $\kappa \gt \kappa^*$ by noting that $p \gt \bar{p} = 2 + \frac{4}{N}$.

Proof of theorem 1.1

Consider the sequence $\{u_n\}$ arising from lemma 2.10. As the functional $ \mathcal{I}_\mu$ exhibits even symmetry with respect to u, we can assume u_n is nonnegative.

We claim that $\{u_n\}$ is bounded in $H_{r}^1\left(\mathbb{R}^N\right)$. From $(F2)$ and lemma 2.1, $ \mathcal{I}_\mu\left(u_n\right)=\sigma_{\mu}(a)+o_n(1)$ combined with $P_{\mu}\left(u_n\right)=o_n(1)$ implies that there is a small enough c > 0 such that

\begin{align*} \sigma_{\mu}(a)+o_n(1) = \mathcal{I}_\mu\left(u_n\right)-\frac{1}{2}P_{\mu}(u_n) &= \int_{\mathbb{R}^N}\frac{N}{4} f\left(u_n\right)u_n- \frac{N+2}{2}F\left(u_n\right) \mathrm{d} x \nonumber \\ &\geq c\int_{\mathbb{R}^N}| u_n|^{2^*}\mathrm{d} x. \end{align*}

Also from $P\left(u_n\right)=o_n(1)$, we have

\begin{align*} \left(1-\frac{\mu}{\bar{\mu}}\right)|\nabla u_n|_2^2 \leq\int_{\mathbb{R}^N}|\nabla u_n|^2-\mu\frac{u_n^2}{|x|^2} \mathrm{d} x&=\frac{N}{2}\int_{\mathbb{R}^N}\widetilde{F} \left(u_n \right) \mathrm{d} x\\ &\leq 2^* C_{N,\bar{p}}^{\bar{p}}a^{\frac{2}{N}}(\delta+\beta) \left|\nabla u_n\right|_2^{2}+C_2 \left|u_n\right|_{2^*}^{2^*}, \end{align*}

which shows that $\left\{\left|\nabla u_n\right|_{2}\right\}$ is bounded, and $\left\{u_n\right\} \subset S_{r,a}$, so that $\{u_n\}$ is bounded in $H_{r}^1\left(\mathbb{R}^N\right)$.

Therefore, there exists $u\in H_r^1\left(\mathbb{R}^N\right)$ such that, up to a subsequence, $u_n \rightharpoonup u$ in $H_r^1\left(\mathbb{R}^N\right)$, $u_n \rightarrow u$ in $L^{p}\left(\mathbb{R}^N\right)$ for $p\in (2,2^*)$, and $u_n(x)\rightarrow u(x)$ almost everywhere in $\mathbb{R}^N$. We claim that u ≠ 0, by contradiction, that u = 0. By utilizing the Strauss inequality [Reference Willem50, lemma 4.5] for the sequence $\left\{u_n\right\}$ in $H_r^1\left(\mathbb{R}^N\right)$, it follows that

\begin{equation*} |u_n(x)| \leq C_N|u_n|_2^{\frac{1}{2}}|\nabla u_n|_2^{\frac{1}{2}}|x|^{\frac{1-N}{2}} \text {a.e. on } \mathbb{R}^N. \end{equation*}

Consequently, it can be deduced that $u_n(x) \rightarrow 0$ as $|x| \rightarrow \infty$.

By using $(F2)$, we establish

\begin{align*} \lim_{s\rightarrow \infty}\frac{\frac{N}{2}\widetilde{F}(s)-2^* \eta |s|^{2^*}}{|s|^{2^*}+|s|^{2}} =0, ~ ~ \lim_{s\rightarrow 0}\frac{\frac{N}{2}\widetilde{F}(s)-2^* \eta |s|^{2^*}}{|s|^{2}+|s|^{2^*}} =0. \end{align*}

By the boundedness of $\{| u_n|_{2^*}\}$ and the fact that $\{u_n\}\in S_{r,a} $, we have

\begin{equation*} \int_{\mathbb{R}^N}\left|u_n\right|^{2}+\left|u_n\right|^{2^*} \mathrm{d} x \leq M, \quad \text{for some positive}~ M. \end{equation*}

Consequently, by Lions Lemma [Reference Berestycki and Lions10, theorem A.I.], we can say that

(2.16)\begin{align} \lim_{n\to\infty}\int_{\mathbb{R}^N}\frac{N}{2}\widetilde{F}(u_n)-2^* \eta |u_n|^{2^*} \mathrm{d} x=0. \end{align}

Then by $P(u_n)\to 0$, we can deduces that

(2.17)\begin{align} \int_{\mathbb{R}^N}\left|\nabla u_n\right|^2-\mu\frac{u_n^2}{|x|^2} \mathrm{d} x +o_n(1)& = \frac{N}{2}\int_{\mathbb{R}^N}\widetilde{F}(u_n)\mathrm{d} x \nonumber\\ &= 2^* \eta \int_{\mathbb{R}^N}\left| u_n\right|^{2^*}\mathrm{d} x +\int_{\mathbb{R}^N}\frac{N}{2}\widetilde{F}(u_n)-2^*\eta|u_n|^{2^*}\mathrm{d} x\nonumber \\ & \leq 2^* \eta S_\mu^{\frac{N}{2-N}}\left(\int_{\mathbb{R}^N}\left|\nabla u_n\right|^2 -\mu\frac{u_n^2}{|x|^2} \mathrm{d} x\right)^{\frac{N}{N-2}}. \end{align}

By using lemma 2.4, we can assume that, up to a subsequence,

\begin{align*} \int_{\mathbb{R}^N}\left|\nabla u_n\right|^2 -\mu\frac{u_n^2}{|x|^2} \mathrm{d} x\rightarrow l^* \gt 0. \end{align*}

By (2.17), we find that $l^* \geq \left(2^* \eta\right)^{\frac{2-N}{2}}S_\mu^{\frac{N}{2}}$. Similarly as (2.16), we have

\begin{align*} \lim_{n\to\infty}\int_{\mathbb{R}^N}f(u_n)u_n-2^*F(u_n) \mathrm{d} x=0. \end{align*}

This allows us to derive

\begin{align*} m_\mu(a)+o_n(1)&= \mathcal{I}_\mu\left(u_n\right)-\frac{1}{2^*}P_\mu\left(u_n\right)\\ & =\frac{1}{N} \int_{\mathbb{R}^N}\left|\nabla u_n\right|^2 -\mu \frac{u_n^2}{|x|^2}\mathrm{d} x +\frac{N-2}{4} \int_{\mathbb{R}^N}f(u_n)u_n-2^*F(u_n) \mathrm{d} x\\ & =\frac{1}{N} \int_{\mathbb{R}^N}\left|\nabla u_n\right|^2 -\mu \frac{u_n^2}{|x|^2}\mathrm{d} x+o_n(1)\\ &\geq \frac{1}{N}\left(2^* \eta\right)^{\frac{2-N}{2}}S_\mu^{\frac{N}{2}}+o_n(1), \end{align*}

which contradicts lemma 2.11. Hence, $u\not= 0$. By the weak lower semi-continuity, we deduce

\begin{align*} \int_{\mathbb{R}^N}|u|^2 \mathrm{d} x =a_0 \in (0,a]. \end{align*}

Since $\left\{u_n\right\}$ is a Palais–Smale sequence of $\left. \mathcal{I}_\mu\right|_{S_{r, a}}$, there exists $\left\{\lambda_n\right\}$ such that for any $\varphi \in H^1\left(\mathbb{R}^N\right)$

(2.18)\begin{align} \int_{\mathbb{R}^N}\left(\nabla u_n \nabla \varphi-\mu\frac{u_n\varphi }{|x|^2} +\lambda_n u_n \varphi-f\left(u_n\right) \varphi\right) \mathrm{d} x=o_n(1)\|\varphi\|. \end{align}

Setting $\varphi=u_n$ and by the boundedness of $\left\{u_n\right\}$ in $H^1\left(\mathbb{R}^N\right)$, we have

\begin{equation*} -\lambda_n a=\int_{\mathbb{R}^N}\left|\nabla u_n\right|^2-\frac{\mu}{|x|^2}u_n^2 \mathrm{d} x-\int_{\mathbb{R}^N} f\left(u_n\right) u_n \mathrm{d} x+o_n(1) . \end{equation*}

Moreover, we can infer that the boundedness of λ_n by the boundedness of $\left\{u_n\right\}$. Therefore, up to a subsequence, $\lambda_n \rightarrow \lambda \in \mathbb{R}$. By (2.18),

(2.19)\begin{align} \int_{\mathbb{R}^N}(\nabla u \nabla \varphi-\mu\frac{u\varphi }{|x|^2}+\lambda u \varphi-f(u) \varphi) \mathrm{d} x=0 \end{align}

implies that $(u,\lambda)$ satisfies

(2.20)\begin{align} -\Delta u-\frac{\mu}{|x|^2}u+\lambda u=f(u). \end{align}

Thus, one has

(2.21)\begin{align} \int_{\mathbb{R}^N}\left(|\nabla u|^2-\frac{\mu}{|x|^2}u^2+\lambda|u|^2-f(u) u\right) \mathrm{d} x=0, \end{align}

and

(2.22)\begin{align} \frac{N-2}{2 N} \int_{\mathbb{R}^N}|\nabla u|^2 -\frac{\mu}{|x|^2}u^2\mathrm{d} x+\frac{\lambda}{2} \int_{\mathbb{R}^N}|u|^2 \mathrm{d} x-\int_{\mathbb{R}^N} F(u) \mathrm{d} x=0 . \end{align}

Combined with (2.21) and (2.22), we can infer that

\begin{equation*} \int_{\mathbb{R}^N}|\nabla u|^2 -\frac{\mu}{|x|^2}u^2\mathrm{d} x-\frac{N}{2} \int_{\mathbb{R}^N}(f(u) u-2 F(u)) \mathrm{d} x=0 . \end{equation*}

i.e. $P_\mu(u)=0$.

Defining $v_n:=u_n-u \rightharpoonup 0$ in $H^1\left(\mathbb{R}^N\right)$, we can utilize the Brézis–Lieb lemma [Reference Brézis and Lieb13] to state that

\begin{equation*} \mathcal{I}_\mu(u_n)= \mathcal{I}_\mu(u)+ \mathcal{I}_\mu(v_n)+o_n(1), ~~P_\mu(u_n)=P_\mu(u)+P_\mu(v_n)+o_n(1). \end{equation*}

We claim that $v_n\to 0$ in $\mathcal{D}^{1,2}\left(\mathbb{R}^N\right)$. Let us proceed by assuming, for the sake of contradiction, that

\begin{align*} \lim_{n\rightarrow \infty} \int_{\mathbb{R}^N}\left|\nabla v_n\right|^2-\mu\frac{v_n^2}{|x|^2} \mathrm{d} x \gt 0. \end{align*}

Since $P_\mu(u)=0$, we have $P_\mu(v_n)= o_n(1)$. This implies

\begin{align*} \int_{\mathbb{R}^N}\left|\nabla v_n\right|^2-\mu\frac{v_n^2}{|x|^2} \mathrm{d} x+o_n(1) &= \frac{N}{2}\int_{\mathbb{R}^N}\widetilde{F}(v_n)\mathrm{d} x\\ &= 2^* \eta \int_{\mathbb{R}^N}\left|v_n\right|^{2^*} \mathrm{d} x +\int_{\mathbb{R}^N}\frac{N}{2}\widetilde{F}(v_n)-2^*\eta|v_n|^{2^*}\mathrm{d} x \\ &\leq 2^* \eta S_\mu^{\frac{N}{2-N}}\left(\int_{\mathbb{R}^N}\left|\nabla v_n\right|^2 -\mu\frac{v_n^2}{|x|^2} \mathrm{d} x \right)^{\frac{N}{N-2}}+o_n(1). \end{align*}

Similarly, we can deduce

\begin{align*} \lim \limits_{n\rightarrow \infty} \mathcal{I}_\mu(v_n)\geq \frac{1}{N}\left(2^* \eta\right)^{\frac{2-N}{2}}S_\mu^{\frac{N}{2}}. \end{align*}

Furthermore,

\begin{align*} \mathcal{I}_\mu(u)&= \mathcal{I}_\mu(u)-\frac{1}{2}P_\mu(u)= \int_{\mathbb{R}^N} \frac{N}{4}\widetilde{F}(u)-F(u)\mathrm{d} x \gt 0. \end{align*}

As a consequence, we arrive at

\begin{equation*} \begin{aligned} \sigma_{\mu}(a)&= \mathcal{I}_\mu\left(v_n\right) + \mathcal{I}_\mu\left(u\right)+o_n(1)\geq \frac{1}{N}\left(2^* \eta\right)^{\frac{2-N}{2}}S_\mu^{\frac{N}{2}}, \end{aligned} \end{equation*}

which contradicts lemma 2.11, so that $u_n\to u$ in $\mathcal{D}^{1,2}(\mathbb{R}^N)$. Moreover, we have

(2.23)\begin{align} \lim_{n\rightarrow \infty} \int_{\mathbb{R}^N} |u_n|^{2^*} \mathrm{d} x=\int_{\mathbb{R}^N} |u|^{2^*} \mathrm{d} x, \end{align}

which leads to

(2.24)\begin{equation} \begin{aligned} \int_{\mathbb{R}^N} F\left(u_n\right) \mathrm{d} x \rightarrow \int_{\mathbb{R}^N} F(u) \mathrm{d} x, \quad \int_{\mathbb{R}^N} f\left( u_n\right) u_n \mathrm{d} x \rightarrow \int_{\mathbb{R}^N} f(u) u \mathrm{d} x. \end{aligned} \end{equation}

Furthermore, by (2.20),

\begin{align*} -\Delta u -\mu\frac{u}{|x|^2}+\lambda u =f(u). \end{align*}

Since (2.21), (2.22), and $(F4)$, one obtains

\begin{align*} \lambda&= \frac{1}{a}\left( \int_{\mathbb{R}^N} f(u) u\mathrm{d} x -\int_{\mathbb{R}^N}|\nabla u|^2 -\mu\frac{u^2}{|x|^2} \mathrm{d} x\right) \\ &= \frac{1}{a}\left\{\int_{\mathbb{R}^N}\frac{2-N}{2}f(u)u+NF(u) \mathrm{d} x\right\} \gt 0. \end{align*}

Thus, λ > 0 and $u \in S_a$ by (2.18) and (2.19). According to lemmas 2.6 and 2.8, $(u, \lambda) \in H_r^1\left(\mathbb{R}^N\right) \times \mathbb{R}^{+}$ is the normalized ground state solution of (1.6). We can further establish that u > 0 through the strong maximum principle. This concludes the proof.

Proof of proposition 1.2

Since $\frac{1}{|x|^2}\in C^{2}\left(\mathbb{R}^{N}\backslash B_{r_0}(0)\right)$ for any small $r_0 \gt 0$. Then by using a standard elliptic regularity argument, we establish that $u \in C^2\left(\mathbb{R}^N \backslash\{0\} \right)$. We now turn our attention to proving the exponential decay of the solution. Since $u \in C^2\left(\mathbb{R}^N \backslash\{0\} \right)$ and $u\in S_{r,a}$, then $u(x)\rightarrow 0$ as $|x| \rightarrow+\infty$.

Consequently, there exists R > 0 such that

(2.25)\begin{align} - \Delta u(x)= \frac{\mu}{|x|^2} u(x)+f(u(x))-\lambda u(x) \leq -\frac{\lambda}{2} u(x) \quad \text {for all }|x| \geq R. \end{align}

Define $\phi(x)=M \exp \left(-\sqrt{\frac{\lambda}{2}}|x|\right)$, where M is chosen to satisfy

\begin{equation*} M \exp \left(-\sqrt{\frac{\lambda}{2}} R\right) \geq u(x) \,\text {for all }|x|=R. \end{equation*}

By direct calculation, it follows

\begin{align*} \Delta \phi=\left(\frac{\lambda}{2}-\frac{N-1}{r}\sqrt{\frac{\lambda}{2}} \right)\phi, \text {~for all~ } x \neq 0, \text{~where~} r=|x|. \end{align*}

This leads to the immediate conclusion

(2.26)\begin{align} \Delta \phi \leq \frac{\lambda}{2} \phi \quad \text {for all } x \neq 0 . \end{align}

Combining (2.25) with (2.26), it becomes evident that the function $\varphi=\phi-u$ fulfils

\begin{equation*} \begin{cases} -\Delta \varphi+\frac{\lambda}{2} \varphi \geq 0 & \text {in~ } |x| \geq R, \\ \varphi(x) \geq 0 & \text {in ~}|x|=R, \\ \lim \limits_{|x| \rightarrow \infty} \varphi(x)=0. \end{cases} \end{equation*}

In accordance with the maximum principle, it follows that $\varphi(x) \geq 0$ holds true for all $|x| \geq R$. Consequently,

(2.27)\begin{align} u(x) \leq M \exp \left(-\sqrt{\frac{\lambda}{2}}|x|\right), ~|x| \geq R. \end{align}

Further, based on $\left(F1\right)$–$(F2)$ in conjunction with the exponential decay of u, it is evident that for sufficiently large $|x|$,

\begin{equation*} m_1 u \leq \left|f(u)-\lambda u\right| \leq m_2 u, \end{equation*}

where $m_2 \geq m_1 \gt 0$. As u satisfies Eq. (1.6),

(2.28)\begin{align} -u_{r r}-\frac{N-1}{r} u_r-\frac{\mu}{r^2} u=f(u)-\lambda u, \quad r \in(r_0,+\infty),~ r_0 \gt 0, \end{align}

with $u_r=\frac{\partial u}{\partial r}, u_{r r}=\frac{\partial^2 u}{\partial r^2}, r=|x|$.

It is a known fact that the equation

(2.29)\begin{align} -\left(r^{N-1} u_r\right)_r=\mu r^{N-3} u+r^{N-1} f(u)-\lambda r^{N-1}u , \quad r \in(r_0,+\infty),~ r_0 \gt 0, \end{align}

can be integrated over the interval (r, R), using (2.27), and then letting $r, R \rightarrow +\infty$. This integration demonstrates that $r^{N-1} u_r$ possesses a limit as $r \rightarrow \infty$, which, according to (2.27), must be zero. Furthermore, integrating (2.29) over $(r,+\infty)$ implies exponential decay of u_r (also referenced in [Reference Berestycki and Lions10]). Finally, the exponential decay of u_rr, and consequently of $\left|D^\alpha u(x)\right|$ for $|\alpha| \leq 2$, directly follows from (2.28). This concludes the proof.

Proof. For any $0 \lt \mu_n \lt \bar{\mu}$,

\begin{align*} \mathcal{I}_{\infty}(u)= \mathcal{I}_{\mu_n}(u)+\mu_n \int_{\mathbb{R}^N} \frac{u^2}{|x|^2} \mathrm{d} x, \end{align*}

and consequently, by lemma 2.3

\begin{align*} m_{\infty}(a)=\inf _{u \in S_a} \max _{t \gt 0} \mathcal{I}_{\infty}(t \star u) \geq \inf _{u \in S_a} \max _{t \gt 0} \mathcal{I}_{\mu_n}(t \star u)=m_{\mu_n}(a). \end{align*}

We now proceed to assert that

\begin{equation*} m_\infty(a)\leq \lim _{n \rightarrow \infty} m_{\mu_n}(a). \end{equation*}

For each $n \geq 1$, let $u_n \in \mathcal{M}_{\mu_n}(a)$ be such that

\begin{equation*} \mathcal{I}_{\mu_n}\left(u_n\right)=m_{\mu_n}(a) \lt m_{\infty}(a)+\frac{1}{n}.\end{equation*}

Consequently, $\left|\nabla u_n\right|_{2}^2 \leq C$ for all $n \geq 1$, ensuring that u_n is bounded in $H^1\left(\mathbb{R}^N\right)$. Let t_n be determined according to lemma 2.3 such that $ t_n \star u_n \in \mathcal{M}_\infty(a). $ Moreover,

\begin{align*} P_{\mu_n}\left(u_n\right)=\int_{\mathbb{R}^N}\left|\nabla u_n\right|^2-\mu_n \frac{u_n^2}{|x|^2} \mathrm{d} x -\frac{N}{2} \int_{\mathbb{R}^N}\widetilde{F}(u_n) \mathrm{d} x=0, \end{align*}

and

\begin{align*} P_{\infty}\left(t_n \star u_n\right)=t_n^2 \int_{\mathbb{R}^N}\left|\nabla u_n\right|^2 \mathrm{d} x -\frac{N}{2} t_n^{-N} \int_{\mathbb{R}^N} \widetilde{F}\left(t_n^{\frac{N}{2}} u_n\right) \mathrm{d} x=0 . \end{align*}

As $n \rightarrow \infty$, we establish

\begin{equation*} \int_{\mathbb{R}^N} \frac{\widetilde{F}\left(t_n^{\frac{N}{2}} u_n\right)}{\left|t_n^{\frac{N}{2}} u_n\right|^{2+\frac{4}{N}}}\left|u_n\right|^{2+\frac{4}{N}} \mathrm{d} x =\int_{\mathbb{R}^N} \frac{\widetilde{F}\left(u_n\right)}{\left|u_n\right|^{2+\frac{4}{N}}}\left|u_n\right|^{2+\frac{4}{N}} \mathrm{d} x+o_n(1). \end{equation*}

Furthermore, based on $(F3)$, it follows that $t_n \rightarrow 1$ as $n \rightarrow \infty$. By [Reference Bartsch and Soave7, lemma 3.5], we have $\|t_n \star u_n-u_n\| \rightarrow 0$, and consequently,

\begin{equation*} \mathcal{I}_\infty\left(t_n \star u_n\right)- \mathcal{I}_\infty\left(u_n\right) \rightarrow 0, ~~\text{as}~ n \rightarrow \infty. \end{equation*}

This entails

\begin{equation*} \begin{aligned} m_\infty(a) & \leq \mathcal{I}_\infty\left(t_n \star u_n\right)=\mathcal{I}_\infty\left(u_n\right)+o_n(1) \\ & = \mathcal{I}_{\mu_n}\left(u_n\right)+\mu_n \int_{\mathbb{R}^N} \frac{u_n^2}{|x|^2} \mathrm{d} x+o_n(1) \leq m_{\mu_n}(a)+o_n(1). \end{aligned} \end{equation*}

Hence, we conclude that $ m_\infty(a)\leq \lim \limits_{n \rightarrow \infty} m_{\mu_n}(a)$. This concludes the proof.

Proof of theorem 1.3

Assume that $(u_{n},\lambda_{n})$ is obtained in theorem 1.1 with $m_{\mu_n}(a)$, where $\bar{\mu} \gt \mu_n$ and $\mu_n \rightarrow 0^+$. In other words, $(u_n,\lambda_n)$ satisfies

(3.1)\begin{align} -\Delta u_n -\frac{\mu_n}{|x|^2} u_n +\lambda_n u_n=f(u_n). \end{align}

Consequently, $P_{\mu_n}(u_n)=0$.

Similarly to the proof of theorem 1.1, $\{u_n\}$ is bounded in $H_r^1\left(\mathbb{R}^N\right)$, so that there exists a nonnegative function $u \in H_r^1\left(\mathbb{R}^N\right)$ such that $u_n \rightharpoonup u$ in $H_r^1\left(\mathbb{R}^N\right)$, $u_n \rightarrow u$ in $L^p\left(\mathbb{R}^N\right)$ for $p \in\left(2,2^*\right)$, and $u_n(x) \rightarrow u(x)$ almost everywhere in $\mathbb{R}^N$. Consequently, by taking $n\to\infty$ in (3.1), we have

(3.2)\begin{align} -\Delta u +\lambda u =f(u), \end{align}

which shows that $P_\infty(u)=0$.

Claim 1: $u\not=0$. If not, u = 0, and $P_{\mu_n}(u_n)=0$ combined with (2.16) implies that

(3.3)\begin{align} \int_{\mathbb{R}^N}\left|\nabla u_n\right|^2 \mathrm{d} x & = \frac{N}{2}\int_{\mathbb{R}^N}\widetilde{F}(u_n)\mathrm{d} x + \mu_n \int_{\mathbb{R}^N}\frac{u_n^2}{|x|^2}\mathrm{d} x \nonumber \\ &= 2^* \eta \int_{\mathbb{R}^N}|u_n|^{2^*}\mathrm{d} x+ \int_{\mathbb{R}^N} \frac{N}{2}\widetilde{F}(u_n)-2^* \eta |u_n|^{2^*}\mathrm{d} x + \mu_n\int_{\mathbb{R}^N}\frac{u_n^2}{|x|^2}\mathrm{d} x \nonumber\\ &= 2^* \eta \int_{\mathbb{R}^N}\left|u_n\right|^{2^*}\mathrm{d} x +o_n(1) \nonumber\\ & \leq 2^* \eta S^{\frac{N}{2-N}}\left(\int_{\mathbb{R}^N} \left|\nabla u_n\right|^2 \mathrm{d} x\right)^{\frac{N}{N-2}}+o_n(1). \end{align}

Combined with lemma 2.4, we deduce that

\begin{equation*} \int_{\mathbb{R}^N}\left|\nabla u_n\right|^2 \mathrm{d} x \geq \left( 2^* \eta\right)^{\frac{2-N}{2}} S^{\frac{N}{2}}+o_n(1) . \end{equation*}

Since $\left\{u_n\right\}$ is bounded in $H^1(\mathbb{R}^N)$ and $\mu_n \rightarrow 0^{+}$, we have

(3.4)\begin{align} m_{\mu_n}(a)&= \mathcal{I}_{\mu_n}\left(u_n\right)-\frac{1}{2^*}P_{\mu_n}(u_n) \nonumber \\ & =\frac{1}{N} \int_{\mathbb{R}^N}\left|\nabla u_n\right|^2 -\mu_n \frac{u_n^2}{|x|^2}\mathrm{d} x+\frac{N-2}{4} \int_{\mathbb{R}^N}f(u_n)u_n-2^*F(u_n) \mathrm{d} x \nonumber \\ &\geq \frac{1}{N} \left( 2^* \eta \right)^{\frac{2-N}{2}} S^{\frac{N}{2}}+o_n(1). \end{align}

Thus,

\begin{equation*} \lim \limits_{n \rightarrow \infty} m_{\mu_n}(a)=m_{\infty}(a) \geq \frac{1}{N}\left(2^* \eta\right)^{\frac{2-N}{2}} S^{\frac{N}{2}}, \end{equation*}

which contradicts lemma 2.11, and then $u\not=0$. Based on Eq. (3.2), we have

\begin{align*} \lambda |u|^2_2&=\int_{\mathbb{R}^N}f(u)u\mathrm{d} x-\int_{\mathbb{R}^N}|\nabla u|^2\mathrm{d} x =\int_{\mathbb{R}^N}f(u)u\mathrm{d} x- \frac{N}{2}\int_{\mathbb{R}^N}\widetilde{F}(u) \mathrm{d} x \gt 0, \end{align*}

which holds due to $(F4)$.

Claim 2: $u_n \rightarrow u$ in $\mathcal{D}^{1,2}\left(\mathbb{R}^N\right)$. Let us proceed by contradiction and assume that

\begin{equation*}\nu:= \lim \limits_{n\rightarrow \infty}\int_{\mathbb{R}^N}\left|\nabla v_n\right|^2 \mathrm{d} x \gt 0,\end{equation*}

where $v_n=u_n-u$. Since $P_{\mu_n}(u_n)=0$, we can infer that $P_{\mu_n}(v_n)= 0$. Similarly, one can see that

\begin{align*} \mathcal{I}_{\mu_n}(v_n)\geq \frac{1}{N} \left( 2^* \eta \right)^{\frac{2-N}{2}} S^{\frac{N}{2}}+o_n(1), \end{align*}

and by lemma 2.1,

\begin{align*} \mathcal{I}_{\infty}(u)=\mathcal{I}_{\infty}(u)-\frac{1}{2}P_{\infty}(u)= \int_{\mathbb{R}^N} \frac{N}{4}\widetilde{F}(u)-F(u)\mathrm{d} x \gt 0. \end{align*}

Consequently, we arrive at

\begin{align*} m_{\mu_n}(a)&= \mathcal{I}_{\mu_n}(u_n)= \mathcal{I}_{\mu_n}(v_n)+ \mathcal{I}_{\infty}(u)+o_n(1) \gt \frac{1}{N} \left( 2^* \eta \right)^{\frac{2-N}{2}} S^{\frac{N}{2}}+o_n(1) . \end{align*}

This leads to

\begin{equation*}\lim \limits_{n \rightarrow \infty} m_{\mu_n}(a)=m_{\infty}(a)\geq \frac{1}{N}\left(2^* \eta\right)^{\frac{2-N}{2}} S^{\frac{N}{2}},\end{equation*}

which is a contradiction. Thus, we conclude that $u_n\rightarrow u$ in $\mathcal{D}^{1,2}\left(\mathbb{R}^N\right)$.

Furthermore, we can establish that

\begin{align*} \lim_{n\rightarrow \infty} \int_{\mathbb{R}^N} |u_n|^{2^*} \mathrm{d} x=\int_{\mathbb{R}^N} |u|^{2^*} \mathrm{d} x, \end{align*}

and

\begin{align*} \int_{\mathbb{R}^N} F\left(u_n\right) \mathrm{d} x \rightarrow \int_{\mathbb{R}^N} F(u) \mathrm{d} x. \end{align*}

Consequently,

\begin{equation*} \lim _{n \rightarrow \infty} \lambda_n\int_{\mathbb{R}^N} |u_n|^2\mathrm{d} x=\lambda \int_{\mathbb{R}^N}u^2\mathrm{d} x, \end{equation*}

and given that λ > 0, so that $u_n\to u$ in $H_r^1(\mathbb{R}^N)$. Thus, by lemma 3.1, $(u, \lambda) \in H_r^1(\mathbb{R}^N) \times \mathbb{R}^{+}$ is a normalized ground state of Eq. (1.8). Moreover, u > 0 by the strong maximum principle.

Proof. When µ < 0, it becomes evident that $m_{\infty}(a) \leq m_\mu(a)$. According to theorem 1.3, Eq. (1.8) possesses a ground state solution $v \in \mathcal{M}_{\infty}(a)$, achieving $m_{\infty}(a)$, i.e. $\mathcal{I}_{\infty}(v)=m_{\infty}(a)$ and $P_{\infty}(v)=0$. Moreover, due to the exponential decay of v, we have

\begin{equation*} v(x) \leq M \exp \left(-\sqrt{\frac{\lambda}{2}}|x|\right), |x| \gt R, ~\text{for some } R \gt 0. \end{equation*}

Consequently, we can introduce $v_n(x) = v(x - y_ne_1)$, where $e_1=(1,0,\ldots,0)$, $y_n\in \mathbb{R}^+$ and $y_n \rightarrow+\infty$ as $n \rightarrow \infty$. Furthermore, given any ϵ > 0, there exists $R_{\epsilon} \gt 0$ such that,

\begin{equation*} \frac{1}{|x|^2} \leq \epsilon,~ \text{for all}~ |x| \geq R_{\epsilon}. \end{equation*}

Since $y_n \rightarrow+\infty$ as $n \rightarrow \infty$, there exists $R_{\epsilon} \gt 0$, such that as $n\rightarrow \infty$,

\begin{equation*} v_n(x) \leq M \exp \left(-\sqrt{\frac{\lambda}{2}}\left|x-y_n\right|\right) \leq \frac{C}{\left|x-y_n\right|},~ x\in B_{R_{\epsilon}}(0). \end{equation*}

This results in

\begin{equation*} \begin{aligned} \int_{\mathbb{R}^N} \frac{v_n^2}{|x|^2} \mathrm{d} x& =\int_{B_{R_{\epsilon}}(0)} \frac{v_n^2}{|x|^2} \mathrm{d} x+\int_{\mathbb{R}^N \backslash B_{R_{\epsilon}}(0)} \frac{v_n^2 }{|x|^2} \mathrm{d} x\\ & \leq \frac{C}{\left|y_n-R_{\epsilon}\right|^2} \int_{B_{R_{\epsilon}}(0)} \frac{1}{|x|^2} \mathrm{d} x+\epsilon \int_{\mathbb{R}^N \backslash B_{R_{\epsilon}}(0)} v_n^2 \mathrm{d} x\\ & \leq \frac{C_1}{\left|y_n-R_{\epsilon}\right|^2}+\epsilon a \rightarrow 0, ~\text{as}~ n\rightarrow \infty. \end{aligned} \end{equation*}

Since $v_n\in \mathcal{M}_{\infty}(a)$,

\begin{align*} P_{\infty}\left(v_n\right)=\int_{\mathbb{R}^N}\left|\nabla v_n\right|^2 \mathrm{d} x-\frac{N}{2} \int_{\mathbb{R}^N}\left[f\left(v_n\right) v_n-2F\left(v_n\right)\right] \mathrm{d} x=0, \end{align*}

as deduced from lemma 2.3, there exists a unique $t_n \gt 0$ satisfying

\begin{align*} P\left(t_n \star v_n\right)=t_n^2 \int_{\mathbb{R}^N}\left(\left|\nabla v_n\right|^2-\mu\frac{v_n^2}{|x|^2} \right) \mathrm{d} x-\frac{N}{2} t_n^{-N} \int_{\mathbb{R}^N} \widetilde{F}\left(t_n^{\frac{N}{2}} v_n\right) \mathrm{d} x=0 . \end{align*}

Therefore, as $y_n \rightarrow \infty$, we establish

\begin{equation*} \int_{\mathbb{R}^N} \frac{\widetilde{F}\left(t_n^{\frac{N}{2}} v_n\right)}{\left|t_n^{\frac{N}{2}} v_n\right|^{2+\frac{4}{N}}}\left|v_n\right|^{2+\frac{4}{N}} \mathrm{d} x=\int_{\mathbb{R}^N} \frac{\widetilde{F}\left(v_n\right)}{\left|v_n\right|^{2+\frac{4}{N}}}\left|v_n\right|^{2+\frac{4}{N}} \mathrm{d} x+o_n(1). \end{equation*}

Furthermore, based on $(F3)$, it follows that $t_n \rightarrow 1$ as $n \rightarrow \infty$. We now proceed to demonstrate that $m_\mu(a) \leq m_{\infty}(a)$. As $\left\{t_n \star v_n\right\} \subset \mathcal{M}_\mu(a)$ and $t_n \rightarrow 1$ as $n \rightarrow \infty$, it follows that

\begin{equation*} \begin{aligned} m_\mu(a) & \leq \mathcal{I}_\mu\left(t_n \star v_n\right)= \mathcal{I}_\mu\left(t_n \star v_n\right)-\frac{1}{2^*} P_\mu\left(t_n \star v_n\right) \\ & = \frac{t_n^2}{N}\int_{\mathbb{R}^N} |\nabla v_n|^2 -\frac{\mu}{|x|^2} v_n^2\mathrm{d} x+ \int_{\mathbb{R}^N} \frac{N-2}{4} f\left(t_n \star v_n\right) t_n \star v_n \\ &\quad -\frac{N}{2} F\left(t_n \star v_n\right) \mathrm{d} x\\ &= \frac{1}{N}\int_{\mathbb{R}^N} |\nabla v_n|^2\mathrm{d} x+ \int_{\mathbb{R}^N} \frac{N-2}{4} f\left(v_n\right) v_n -\frac{N}{2} F\left(v_n\right) \mathrm{d} x+o_n(1) \\ & =\mathcal{I}_{\infty}\left(v_n\right)-\frac{1}{2^*} P_{\infty}\left(v_n\right) +o_n(1) \\ & =\mathcal{I}_{\infty}\left(v_n\right)+o_n(1)=\mathcal{I}_{\infty}\left(v\right)+o_n(1)=m_{\infty}(a)+o_n(1). \end{aligned} \end{equation*}

This concludes the establishment that $m_\mu(a)=m_{\infty}(a)$.

Now, we proceed to prove that $m_\mu(a)$ cannot be achievement. By proof by contradiction, we assume that $u_a \in \mathcal{M}_\mu(a)$ attains $m_\mu(a)$. By lemma 2.3, there exists a unique $t_{u_a} \gt 0$ such that $t_{u_a} \star u_a \in \mathcal{M}_{\infty}(a)$. It can be seen that

\begin{equation*} \begin{aligned} m_{\infty}(a) & \leq \mathcal{I}_{\infty}\left(t_{u_a} \star u_a\right)= \mathcal{I}_\mu\left(t_{u_a} \star u_a\right)+\frac{t_{u_a}^2}{2} \int_{\mathbb{R}^N} \frac{\mu}{|x|^2} u_a^2 \mathrm{d} x\\ & \lt \mathcal{I}_\mu\left(t_{u_a} \star u_a\right) \leq \mathcal{I}_\mu\left(u_a\right)=m_\mu(a)=m_{\infty}(a), \end{aligned} \end{equation*}

which creates a contradiction.

Proof of theorem 1.4

The first part of theorem 1.4 has already been established in lemma 4.1. Next, we will prove that Eq. (1.6) has normalized solutions.

By lemma 2.10, there exists a Pohožaev–Palais–Smale sequence $\left\{u_n\right\} \subset S_{r, a}$ for $\mathcal{I}_\mu$ at level of $\sigma_\mu(a)$. That is,

\begin{equation*} \mathcal{I}_\mu\left(u_n\right) \rightarrow \sigma_\mu(a), \quad \left( \left. \mathcal{I}_{\mu}\right|_{S_{r, a}} \right) ^{\prime}\left(u_n\right) \rightarrow 0, \quad P_{\mu}\left(u_n\right) \rightarrow 0, \quad \text {as } n \rightarrow \infty \text {. } \end{equation*}

Similarly as the proof of theorem 1.1, $\{u_n\}$ is bounded in $H_{r}^1\left(\mathbb{R}^N\right)$, and there exists $u\in H^1\left(\mathbb{R}^N\right)\backslash\{0\}$, such that $u_n\rightarrow u $ in $H^1\left(\mathbb{R}^N\right)$, and there exists a λ > 0, such that $( u, \lambda) \in H_r^1\left(\mathbb{R}^N\right) \times \mathbb{R}^{+}$ is a normalized solution of Eq. (1.6). Furthermore, u > 0 by the strong maximum principle. By lemma 4.1, we can deduce that $\sigma_\mu(a) \gt m_\mu(a)$.

Proof. For the sake of clarity in our presentation, let’s define:

\begin{equation*} m_{r,\infty}(a):=\inf _{v \in \mathcal{M}^r_{\infty}(a)} \mathcal{I}_{\infty}(v), \end{equation*}

where

\begin{equation*} \mathcal{M}^r_{\infty}(a):=\left\{v \in S_{r,a}: P_{\infty}(v)=0\right\} . \end{equation*}

It’s clear that $m_{r,\infty}(a)=m_{\infty}(a)$ by lemma 2.8. Hence, we just need to prove $\lim \limits_{n \rightarrow \infty} \sigma_{\mu_n}(a)=m_{r,\infty}(a)$. For $\mu \leq 0$, it’s evident that $\sigma_{\mu}(a) \geq m_{\infty}(a)$ and $ \sigma_\mu(a)$ is non-increasing with respect to µ. Therefore, we just need to prove that $m_{\infty}(a)$ is the greatest lower bound of $\left\{\sigma_{\mu_n}(a)\right\}$.

Assume that $\omega \in \mathcal{M}^r_{\infty}(a)$ is the function that achieves $m_{r,\infty}(a)$, implying $\mathcal{I}_{\infty}(\omega)=m_{r,\infty}(a)$. By lemma 2.3, we can find $0 \lt s_0 \lt 1 \lt s_1$ such that

\begin{equation*}\gamma_o(t)=\left((1-t)s_0+ts_1\right)\star w\in \Gamma_{\mu}, ~~~\gamma_o(t)\cap \mathcal{M}^r_{\infty}(a)\neq \emptyset. \end{equation*}

Furthermore, for any given ϵ > 0, there exists a positive integer N_ϵ such that, for every $n \gt N_{\epsilon}$,

\begin{equation*} -\frac{1}{2}s_1^2 \int_{\mathbb{R}^N} \frac{\mu_n}{|x|^2} \omega^2 \mathrm{d} x\leq \epsilon. \end{equation*}

We can deduce that

\begin{equation*} \begin{aligned} \sigma_{\mu_n}(a) &\leq \max _{t\in [0,1]} \mathcal{I}_{\mu_n}(\gamma_o(t)) =\max _{t\in[0,1]} \mathcal{I}_{\infty}(\gamma_o(t))-\frac{s_1^2}{2} \int_{\mathbb{R}^N} \frac{\mu_n}{|x|^2} \omega^2 \mathrm{d} x\\ & =\mathcal{I}_{\infty}(\omega)+\epsilon=m_{r,\infty}(a)+\epsilon. \end{aligned} \end{equation*}

This clearly indicates that $m_{r,\infty}(a)$ is the infimum of $\left\{\sigma_{\mu_n}(a)\right\}$, and by lemma 2.8, $\lim \limits_{n \rightarrow \infty} \sigma_{\mu_n}(a) =m_{r,\infty}(a)=m_{\infty}(a)$.

Proof of theorem 1.6

Let $(u_n,\lambda_n) $ be the solution in theorem 1.4, which satisfies

(5.1)\begin{align} -\Delta u_n -\frac{\mu_n}{|x|^2} u_n +\lambda_n u_n=f(u_n), \end{align}

where $\mu_n\to 0^{-}$ as $n\to \infty,$ and then $P_{\mu_n}(u_n)=0$. As in the proof of theorem 1.4, we can establish that $u_n\to u $ in $H^1\left(\mathbb{R}^N\right)$ and $\lambda_n\to\lambda \gt 0$, and $(u, \lambda) \in H_r^1\left(\mathbb{R}^N\right) \times \mathbb{R}^{+}$ is the normalized ground state solution of Eq. (1.8). Additionally, by the strong maximum principle, u > 0.

Proof of proposition 1.7

The proof of proposition 1.7 can be derived from the proof of proposition 1.2 by applying the same method. To avoid unnecessary repetition, we do not provide the proof here.