Multiple normalized solutions for a Sobolev critical  Schrödinger-Poisson-Slater equation]Multiple normalized solutions
for a Sobolev critical  Schrödinger-Poisson-Slater equation

L. Jeanjean and T.T. Le]Louis Jeanjean and Thanh Trung Le

Louis Jeanjean
Laboratoire de Mathématiques (CNRS UMR 6623), Université de Bourgogne Franche-Comté, Besançon 25030, France

Thanh Trung Le
Laboratoire de Mathématiques (CNRS UMR 6623), Université de Bourgogne Franche-Comté, Besançon 25030, France

[2010]

We look for solutions to the Schrödinger-Poisson-Slater equation

\begin{align}\label{EQQ}
- \laplace u + \lambda u - \gamma (\abs{x}^{-1} * \abs{u}^2) u  - a \abs{u}^{p-2}u = 0 \quad \text{in} \quad \Rb,
\end{align}

which satisfy

\begin{equation*}
\normLp{u}{2}^2 = c
\end{equation*}

for some prescribed $c>0$. Here $ u \in  \Ho$, $\gamma \in \R,$ $ a \in \R$ and $p \in (\frac{10}{3}, 6]$. When $\gamma >0$ and $a > 0$, both in the Sobolev subcritical case $p \in (\frac{10}{3}, 6)$  and in the Sobolev critical case $p=6$, we show that there exists a $c_1>0$ such that, for any $c \in (0,c_1)$, (<ref>) admits two solutions $u_c^+$ and $u_c^-$ which can be characterized respectively as a local minima  and  as  a mountain pass critical point of the associated  Energy functional restricted to the norm constraint. In the case $\gamma >0$ and $a < 0$, we show that, for any $p \in (\frac{10}{3},6]$ and any $c>0$, (<ref>) admits a solution which is a global minimizer. Finally, in the case $\gamma <0$, $a >0$ and $p=6$ we show that
(<ref>) does not admit positive solutions.

§ INTRODUCTION

We consider the following Schrödinger-Poisson-Slater equation:

\begin{align}
i \partial_t v + \laplace v + \gamma (\abs{x}^{-1} * \abs{v}^2) v + a \abs{v}^{p-2}v = 0 \quad \text{in } \R \times \Rb, \label{eqn:SPS}
\end{align}

where $v: \R \times \Rb \to \mathbb{C}$, $\gamma \in \R,$ $ a \in \R$ and $p \in (\frac{10}{3}, 6]$. We look for standing wave solutions to (<ref>), namely to solutions of the form $v(t, x) = e^{i\lambda t}u(x)$, $\lambda \in \R$. Then the function $u(x)$ satisfies the equation

\begin{align}\label{eqn:Laplace}
- \laplace u + \lambda u - \gamma (\abs{x}^{-1} * \abs{u}^2) u  - a \abs{u}^{p-2}u = 0 \quad \text{in } \Rb.
\end{align}

Motivated by the fact that the $\mathit{L}^2 - \mbox{norm}$ is a preserved quantity of the evolution we focus on the search of solutions to (<ref>)  with prescribed $\mathit{L}^2 - \mbox{norm}$. It is standard that for some prescribed $c > 0$, a solution of (<ref>) with $\normLp{u}{2}^2 = c$ can be obtained as a critical point of the Energy functional

\begin{align*}
F(u) := \dfrac{1}{2} \intrb \abs{\diff u}^2 dx - \dfrac{\gamma}{4} \intrb\intrb \dfrac{\abs{u(x)}^2 \abs{u(y)}^2}{\abs{x-y}} dxdy - \dfrac{a}{p} \intrb \abs{u}^p dx
\end{align*}

restricted to

\begin{align*}
S(c) := \{u \in \Ho: \normLp{u}{2}^2 = c\}.
\end{align*}

Then the parameter $\lambda \in \R$ in (<ref>) appears as a Lagrange multiplier, it is an unknown of the problem.

Let us define

\begin{equation}
\label{eqn:0.1}
m(c) = \inf_{u \in S(c)}F(u).
\end{equation}

Depending on the range of parameters we shall consider $m(c)$ will be finite or not. If, following the introduction of the Compactness by Concentration Principle of P. L. Lions [36, 37], the search of normalized solutions corresponding to a global minimizer of a functional restricted to an $L^2$ norm constraint is now a classical topic,
the search of critical points when the functional is unbounded from below on the constraint remained for a long time much less studied.  In the frame of this paper, namely for a functional corresponding to an autonomous equation lying on all the space $\R^N$, [27] was for a long time the sole contribution. This direction of research was likely brought to the attention of the community by the papers
[3, 9] both published in 2013. Since then numerous contributions flourished within this topic and we just mention, among many possible choices, the works, [4, 6, 8, 14, 21, 25, 30, 7]. We also refer to  [5] for non-autonomous problems set on $\R^N$ and to [41, 42, 43] for contributions when the underlying equation is set on a bounded domain of $\R^N$.

In the above-mentioned papers, the involved nonlinearities were Sobolev subcritical.  It was only in 2020 that was first treated in [48] a problem involving a Sobolev critical nonlinearity. Since then several works have explored further this direction
[1, 28, 29, 38, 52].

The case where $\gamma <0$ and $a>0$ in (<ref>) has been the most studied so far. When $p \in (2, \frac{10}{3})$ it can been shown that $m(c) \in (- \infty, 0]$ for any $c>0$ and it is also the case when $p = \frac{10}{3}$ and $c>0$ is small. It is shown in [11] that minimizer exists if $p \in (2,3)$ and $c>0$ is small enough, see also [45] for the special case $p=\frac{8}{3}$. The case $p \in (3, \frac{10}{3})$ was considered in [10, 31], see also [33] for a closely related problem. In [31] the existence of a threshold value $c_0>0$ such that $m(c)$ has a minimizer if and only if $c \in [c_0, \infty)$ was established. It was also proved in [31] that a minimizer does not exist for any $c>0$ if $p=3$ or $p = \frac{10}{3}$. We also refer to [18] for related results. When $p \in (\frac{10}{3}, 6]$ a scaling argument reveals that $m(c) = - \infty$ but nevertheless it was proved in [9] that, when $p \in (\frac{10}{3}, 6)$ there exists, for $c>0$ small enough a critical point of $F$ constrained to $S(c)$ at a strictly positive level. In this work we complement the result of [9] by showing that when $p=6$ and for any $c>0$ there does not exist positive solutions, see <ref>.

Even if some open problems remain when $\gamma <0$ and $a>0$, we shall mainly concentrate here on the others cases: $(\gamma < 0,$ $ a < 0)$, $(\gamma > 0,$ $ a > 0)$ and $(\gamma > 0,$ $ a < 0)$. We define, for short, the following quantities

\begin{align*}
A(u): = \intrb \abs{\diff u}^2 dx,\quad B(u):= \intrb\intrb \dfrac{\abs{u(x)}^2 \abs{u(y)}^2}{\abs{x-y}} dxdy,\quad C(u):= \intrb \abs{u}^p dx.
\end{align*}

For $u \in S(c)$, we set $u^t(x) := t^{\frac{3}{2}} u(tx), t>0$, then

\begin{align*}
u^t \in S(c), \quad A(u^t) = t^2A(u), \quad B(u^t) = tB(u),\quad C(u^t) = t^{\sigma} C(u),
\end{align*}

\begin{align}
2 < \sigma := \dfrac{3(p-2)}{2} \leq 6, \label{eqn:11}
\end{align}

due to $p \in (\frac{10}{3}, 6]$. For $u \in S(c)$, we define the fiber map

\begin{align*}
t \in (0, \infty) \mapsto g_u(t):= F(u^t) = \dfrac{1}{2}t^2 A(u) - \dfrac{\gamma}{4} t B(u) - \dfrac{a}{p} t^{\sigma} C(u).
\end{align*}

Hence, we have

\begin{align*}
g_u'(t) = t A(u) - \dfrac{\gamma}{4} B(u) - \dfrac{a \sigma}{p} t^{\sigma - 1} C(u) = \dfrac{1}{t} Q(u^t),
\end{align*}

\begin{align*}
Q(u) = A(u) - \dfrac{\gamma}{4} B(u) - \dfrac{a \sigma}{p} C(u).
\end{align*}

Actually the condition $Q(u) = 0$ corresponds to a Pohozaev identity and the set

\begin{align*}
\slc := \{u \in S(c): Q(u) = 0\} = \{u \in S(c): g_u'(1) = 0\}
\end{align*}

appears as a natural constraint. Indeed, if $u \in S(c)$, then $t > 0$ is a critical point for $g_u$ if and only if $u^t \in \slc$. In particular, $u \in \slc$ if and only if $1$ is a critical point of $g_u$.

First we briefly consider the case $\gamma < 0, a < 0$. For any $u \in S(c)$, we have that $g_u'(t) > 0$ for all $t > 0$, hence the fiber map $g_u(t)$ is strictly increasing and so we can state the following non-existence result:

Assume that $\gamma < 0, a < 0$. Then $F(u)$ has no critical point on $S(c)$.

Next, we consider the case $\gamma > 0, a > 0$. In this case, let

\begin{align}
c_1 := \(\dfrac{4}{\gamma K_{H}} \dfrac{\sigma - 2}{\sigma - 1}\)^{\frac{3p-10}{4(p-3)}} \(\dfrac{p}{a \sigma (\sigma - 1) K_{GN}}\)^{\frac{1}{2(p-3)}} > 0, \label{eqn:assumption}
\end{align}

where $\sigma$ is defined by (<ref>) and $K_{H}, K_{GN}$ are defined in <ref>. We also introduce the decomposition of $\slc$ into the disjoint union $\slc = \slp \cup \slz \cup \slm$, where

\begin{align*}
\slp &:= \{u \in \slc: g_u''(1) > 0\} = \{u \in S(c): g_u'(1) = 0, g_u''(1) > 0\}, \\
\slz &:= \{u \in \slc: g_u''(1) = 0\} = \{u \in S(c): g_u'(1) = 0, g_u''(1) = 0\}, \\
\slm &:= \{u \in \slc: g_u''(1) < 0\} = \{u \in S(c): g_u'(1) = 0, g_u''(1) < 0\}.
\end{align*}

By <ref> and <ref>, for any $c \in (0, c_1)$ we have that $\slz = \emptyset$ and $\slp \neq \emptyset$, $\slm \neq \emptyset$. Since $F$ is bounded from below on $\slc$ due to <ref>, we can define

\begin{align}
\gamma^{+}(c) := \inf_{u \in \slp}F(u) \quad \text{ and } \quad \gamma^{-}(c) := \inf_{u \in \slm}F(u). \label{eqn:gamma}
\end{align}

Our first main result is

Let $p \in (\frac{10}{3}, 6]$. Assume that $\gamma > 0,$ $ a > 0$ and let $c_1 > 0$ be defined by (<ref>). For any $c \in (0, c_1)$, there exist $u_c^{+} \in \slp$ such that $F(u_c^{+}) = \gamma^{+}(c)$ and $u_c^{-} \in \slm$ such that $F(u_c^{-}) = \gamma^{-}(c)$. The functions $u_c^+, u_c^-$ are bounded continuous positive Schwarz symmetric functions. In addition there exist $\lambda_c^{+} > 0$ and $\lambda_c^{-} > 0$ such that $(u_c^{+}, \lambda_c^{+})$ and $(u_c^{-}, \lambda_c^{-})$ are solutions to (<ref>).

In <ref>, borrowing an approach first introduced in [21], an effort is made to optimize the limit value $c_1 >0$. As a consequence,  compared to the works [28, 29, 47, 48] we do not benefit from the property that $\gamma^-(c) \geq 0 =  \sup_{u \in\Lambda^+(c)}F(u)$. Such property is a help to show the convergence of the Palais-Smale sequences in these works. Also, the fact that we may have $\gamma^-(c) < 0$ makes somehow more involved to prove that  the level $\gamma^-(c)$ is reached by a radially symmetric function, a Schwartz function actually, see <ref>. It is not clear to us if $c_1 >0$ is optimal. Nevertheless, we conjecture that there exists a $c_0 \geq c_1 >0$ such that one solution exists when $c=c_0$ and that, at least positive solutions, do not exist when $c >c_0$.

As we shall see $\gamma^+(c) < \gamma^-(c)$ and combined with the property that any critical point lies in $\Lambda(c)$ it implies that the solution $u_c^+$ obtained in <ref> is a ground state. Following [8] a ground state is defined as a solution $v \in S(c)$ to (<ref>) which has minimal Energy among all the solutions which belong to $S(c)$. Namely, if

$$\quad F(v) =  \displaystyle \inf \big\{F(u), u \in S(c), \big(F\big|_{S(c)}\big)'(u) = 0 \big\}.$$

If the geometrical structure of $F$ restricted to $S(c)$ is identical in the Sobolev subcritical case $ p \in (\frac{10}{3}, 6)$ and in the Sobolev critical case $p=6$, the proof that the levels $\gamma^+(c)$ and $\gamma^-(c)$ are indeed reached requires additional, more involved, arguments in the case $p=6.$ In particular, showing that $\gamma^-(c)$ is attained requires to check that the following inequality holds

\begin{align}\label{strict-intro}
\gamma^-(c) < \gamma^+(c)  + \dfrac{1}{3 \sqrt{a K_{GN}}}.
\end{align}

It is known since the pioneering work of Brezis-Nirenberg [17] that the way to derive such a strict upper bound is through the use of testing functions. In [29], considering the equation

\begin{align}
-\laplace u - \lambda u - \mu \abs{u}^{q-2} u - \abs{u}^{2^*-2} u = 0 \quad \mbox{in } \R^N, \label{eqn:LaplaceL}
\end{align}

with $N \geq 3,$ $ \mu >0$, $2 < q < 2 + \frac{4}{N}$ and $2^* = \frac{2N}{N-2}$ we face the need to establish a similar inequality. We constructed test functions which could be viewed as the sum of a truncated extremal function of the Sobolev inequality on $\R^N$ centered at the origin and of $u_c^+$ translated far away from the origin. This choice of testing functions was sufficient to prove our strict inequality when $N \geq 4$ but we missed it in the case $N=3$. Note that the approach developed in  [29] proved nevertheless adequate to deal with the equation

\begin{equation*}
\sqrt{- \Delta} u = \lambda u + \mu |u|^{q-2}u + |u|^{2^* -2}u, \quad u \in \mathit{H}^{1/2}(\R^N),
\end{equation*}

with $N \geq 2,$ $ q \in (2, 2 + \frac{2}{N})$, $ 2^* = \frac{2N}{N-1}$, that was studied in [38]. Very recently, in [52] the authors introduced an alternative choice of testing functions which allowed to treat, in a unified way, the case $N=3$ and $N \geq 4$ for (<ref>). The strategy in [52], recording of the one introduced by G. Tarantello in [50], is on the contrary, to located the extremal functions where the solution $u_c^+$ takes its greater values (the origin thus). The idea behind the proof is that the interaction decreases the value of the Energy  with respect to the case where the supports would be disjoint. In this paper, where (<ref>) is set on $\R^3$, we believe in view of our experience on (<ref>), more appropriate to follow the approach of [52] to check the inequality (<ref>) for any $c \in (0,c_1)$. The results of <ref> are complemented in several directions. First, we show that the solution $u^+(c)$ obtained in <ref> can be characterized as a local minima for $F$ restricted to $S(c)$. We treat here the full range $p \in (\frac{10}{3}, 6]$ with a single proof. More precisely we show,

Let $p \in (\frac{10}{3}, 6]$. Assume that $\gamma > 0,$ $ a > 0$ and let $c \in (0,c_1)$. Then
we have $\slp \subset V(c)$ and

\begin{align*}
\gamma^+(c) = \inf_{u \in \slp}F(u) = \inf_{u \in V(c)}F(u)
\end{align*}

\begin{align*}
V(c) := \{u \in S(c) | A(u) < k_1\}
\end{align*}

for some $k_1 >0$ independent of $c \in (0,c_1)$ (see (<ref>) for the definition of $k_1>0$). In addition, any minimizing sequence for $F$ on $V(c)$ is, up to translation, strongly convergent in $\Hs{1}$.

The proof of <ref> which is a key step to established <ref>, reveals some additional properties of the set $V(c)$. Indeed, we have that
$V(c) \subset S(c) \backslash \Lambda^-(c)$ and thus $V(c)$ is separating the sets $\Lambda^+(c)$ and $\Lambda^-(c)$. Also, for any $0 < c, \tilde{c} < c_1$, we have that $A(u) < k_1 < A(v)$ for all $u \in \Lambda^+(c), v \in \Lambda^-(\tilde{c})$, see (<ref>) and (<ref>).

To prove that the minimizing sequences for $F$ on $V(c)$ are, up to translation, strongly convergent in $\Hs{1}$ we follow an approach due to [26] that has already been used several times, see, for example, [24, 28, 38]. The first step in this approach is to show that the sequences do not vanish. When $p=6$, we rely for this, in an essential way, on the fact that $c_1>0$ is sufficiently small, see
<ref>. This fact is also used to end the proof. Finally, note that since we allow the possibility that $\inf_{u \in \partial V(c)}F(u) < 0$ where
$\partial V(c) := \{u \in S(c) | A(u) = k_1\}$ we must check that the minimizers do ly in $V(c)$.

Let us now denote

\begin{align*}
\mathcal{M}_c := \{u \in V(c) : F(u) = \gamma^+(c)\}.
\end{align*}

In view of <ref>, $\mathcal{M}_c $ is the set of all ground states.
The property that any minimizing sequence for $F$ restricted to $V(c)$ is, up to translation, strongly converging is known to be a key ingredient to show that the set $\mathcal{M}_c$ is orbitally stable. If $p \in (\frac{10}{3}, 6)$ the orbital stability of $\mathcal{M}_c$ indeed follows directly from <ref> by the classical arguments of [20]. In the case $p=6$ the situation is more delicate as the existence of a uniform $H^1(\R^3)$ bound on the solution of (<ref>) during its lifespan is not sufficient to  guarantee that blow-up may not occurs. We refer to [19] for more details. We do not prove anything in that direction but strongly believe that the set $\mathcal{M}_c$ is orbitally stable. Actually, such a result has been obtained on the equation (<ref>) in [28].

We also discuss the behavior of the associated Lagrange multipliers and show that if the behavior of $\lambda^+_c$ is essentially the same for the cases $p \in (\frac{10}{3}, 6)$ and $p=6$, see <ref>, there is a distinct behavior for $\lambda_c^-$, see Lemmas <ref> and <ref>. In particular, <ref> suggests that there may exist two distinct positive  solutions to (<ref>) for any fixed $\lambda >0$ sufficiently small. Finally, in <ref>, we establish the property that the map $c \mapsto \gamma^-(c)$ is strictly decreasing.

Next, we consider the case  $\gamma > 0,$ $ a < 0$. Recalling the definition of $m(c)$ given in (<ref>) we show in <ref>, that $-\infty < m(c) < 0$ and then we prove the following result.

Let $p \in (\frac{10}{3},6]$, $\gamma > 0$ and $ a < 0$. For any $c >0$, the infimum $m(c)$ is achieved and any minimizing sequence for (<ref>) is, up to translation, strongly convergent in $\Ho$ to a solution of (<ref>). In addition, the associated Lagrange multiplier is positive.

Even if the proof of <ref> follows the lines of the proof of <ref>, the change of sign in front of the power term requires some adaptations, see  <ref> and <ref>.
Here again the orbital stability of the set of minimizers should follow directly from the classical arguments of [20]
if  $p \in (\frac{10}{3}, 6)$ and it should also be the case when $p=6$ by adapting the arguments of [28]. Note that we also study the behavior of the associated Lagrange multipliers in <ref>.

In the last part of the paper we consider the case $\gamma <0$, $a>0$ and $p=6$.

Let $p=6$, $\gamma <0$ and $a>0$. For any $c > 0$, we have that

*  If $u \in \Hs{1}$ is a non-trivial solution to (<ref>) then the associated Lagrange multiplier $\lambda$ is negative and

\begin{align*}
F(u) > \dfrac{1}{3\sqrt{aK_{GN}}}.
\end{align*}

*  <ref> has no positive solution in  $ \Hs{1}$.

Under the assumptions of <ref>,  it is possible to prove that
$$ \inf_{u \in \Lambda(c)}F(u) = \dfrac{1}{3\sqrt{aK_{GN}}}.$$

In <cit.>, considering the equation

\begin{align}
-\laplace u - \lambda u - \mu \abs{u}^{q-2} u - \abs{u}^{2^*-2} u = 0 \quad \mbox{in } \R^N, \label{eqn:LaplaceS2}
\end{align}

with $N \geq 3$, $2<q<2^*$ and $\mu < 0$, it was proved that (<ref>) has no positive solution $u\in \mathit{H}^1(\R^N)$ if $N=3,4$ or if $N \geq 5$ under the additional assumption
$u \in \mathit{L}^p(\R^N)$ for some $p \in \(0, \frac{N}{N-2}\]$. In \cref{remrak:6.1}, partly using arguments used in the proof of \cref{theorem:non-existence},
we improve  \cite[Theorem 1.2]{Soave2019Sobolevcriticalcase} showing that \eqref{eqn:LaplaceS2} has no positive solution in $\mathit{H}^1(\R^N)$ for all $N \geq 3$ and no non-trivial radial solution for $N \geq 3$ and $q > 2 + \frac{2}{N-1}$.
\end{remark}

\begin{remark}
We propose as an open problem to investigate if there are radial solutions under the assumptions of \cref{theorem:non-existence}. See \cref{open-problem}  in that direction.
\end{remark}

The paper is organized as follows. In \cref{Section2} we recall some classical inequalities and present some preliminary results. \cref{Section3} is devoted to the treatment of the case $\gamma >0$, $a>0$ and $p \in (\frac{10}{3},6]$. In \autoref{Subsection3-1} we make explicit the geometrical structure of $F$ on $S(c)$ and show the existence of a bounded Palais-Smale sequence $(u_n^+) \subset \Lambda^+(c)$ at the level $\gamma^+(c)$ and of a bounded Palais-Smale sequence $(u_n^-) \subset \Lambda^-(c)$ at the level $\gamma^-(c)$. In \autoref{Subsection3-2} we give the proof of \cref{theorem:1} in the Sobolev subcritical case.  \autoref{Subsection3-3} is devoted to the proof of \cref{theorem:1} in the critical case. In \autoref{Subsection3-4} we prove the convergence of all minimizing sequences associated to $\gamma^+(c)$, namely \cref{th-minimization}. %Subsections \ref{Subsection3-5} and \ref{Subsection3-6} are devoted to the proofs of Lemmas \ref{lemma:22}, \ref{lemma:23t}, \ref{lemma:25} and \ref{lemma:10}.
The behavior of the Lagrange multipliers and the property of the map $c \mapsto \gamma^{-}(c)$ are studied in \autoref{Subsection3-5} and \autoref{Subsection3-6}, respectively.
In \cref{Section4} we treat the case $\gamma >0$, $a<0$ and $p \in (\frac{10}{3}, 6]$ and we prove \cref{theorem:2}.
Finally, in \cref{Section:6}, we consider the case $\gamma <0$, $a >0$ and $p=6$, and prove \cref{theorem:non-existence}.\medskip

{\bf Notation:} For $p \geq 1$, the $\mathit{L}^p$-norm of $u \in \Hs{1}$ is denoted by $\normLp{u}{p}$. We denote by $\Hsr{1}$ the subspace of functions in $\Ho$ which are radially symmetric with respect to $0$. The notation $a \sim b$ means that $Cb \leq a \leq C'b$ for some $C, C' > 0$. The open ball in $\Rn$ with center at $0$ and radius $R > 0$ is denoted by $B_R$.

{\bf Addendum :} After the completion of this paper, we were informed of the work [53] in which the authors consider a general class of problems which, when $p \in (\frac{10}{3}, 6)$,  covers \eqref{eqn:Laplace} as a special case.  There are thus some partial overlap, in the Sobolev subcritical case, between \cite[Theorem 1.3 (a) (ii)]{Yao-Sun-Wu-2021} and \cref{theorem:1} and between \cite[Theorem 1.6 (a) (iv)]{Yao-Sun-Wu-2021} and \cref{theorem:2}.  However the scope of the two works is widely distinct.

\section{Preliminary results}\label{Section2}
In this section we present various preliminary results. When it is not specified they are assumed to hold for $\gamma \in \R,$ $ a\in\R,$ $ p \in \(\frac{10}{3}, 6\]$ and any $c > 0$.
Firstly, we present the definitions of $\slc,$ $ \slm,$ $ \slz,$ $ \slm$ via $A(u),$ $ B(u)$ and $C(u)$:
\begin{align*}
\slc &=\Bigg\{u \in S(c): A(u) = \dfrac{\gamma}{4} B(u) + \dfrac{a \sigma}{p} C(u) \Bigg\}, \\
\slp &=\Bigg\{u \in S(c): A(u) = \dfrac{\gamma}{4} B(u) + \dfrac{a \sigma}{p} C(u), A(u) > \dfrac{a \sigma (\sigma - 1)}{p} C(u) \Bigg\}, \\
\slz &=\Bigg\{u \in S(c): A(u) = \dfrac{\gamma}{4} B(u) + \dfrac{a \sigma}{p} C(u), A(u) = \dfrac{a \sigma (\sigma - 1)}{p} C(u) \Bigg\}, \\
\slm &=\Bigg\{u \in S(c): A(u) = \dfrac{\gamma}{4} B(u) + \dfrac{a \sigma}{p} C(u), A(u) < \dfrac{a \sigma (\sigma - 1)}{p} C(u) \Bigg\}.
\end{align*}

\begin{lemma} \label{lemma:1}
Let $u \in S(c)$, there exists
\begin{enumerate}[label=(\roman*), ref = \roman*]
\item\label{point:1i} a constant $K_{H} > 0$ such that $B(u) \leq K_{H}  \sqrt{A(u)} c^{\frac{3}{2}}$.
\item\label{point:1ii} a constant $K_{GN} > 0$ such that $C(u) \leq  K_{GN} [A(u)]^{\frac{\sigma}{2}} c^{\frac{6-p}{4}}$.
\end{enumerate}
\end{lemma}
\begin{proof}
We first recall the Hardy-Littlewood-Sobolev inequality (see \cite[Chapter 4]{LiebLoss2001analysis}):
\begin{align}\label{ineq:HLS}
\abs*{\int_{\R^N} \int_{\R^N} \dfrac{f(x)g(y)}{\abs{x-y}^{\lambda}} dxdy} \leq C(N,\lambda, p, q) \norm{f}_{\mathit{L}^p(\R^N)} \norm{g}_{\mathit{L}^q(\R^N)},
\end{align}
where $f \in \mathit{L}^p(\R^N)$, $g \in \mathit{L}^q(\R^N)$, $p, q > 1$, $0 < \lambda < N$ and
\begin{align*}
\dfrac{1}{p} + \dfrac{1}{q} + \dfrac{\lambda}{N} = 2.
\end{align*}
Let us also recall the Gagliardo-Nirenberg inequality (see [40]) and the Sobolev inequality (see \cite[Theorem 9.9]{Brezis-2011}) in the unified form: if $N \geq 3$ and $p \in [2, \frac{2N}{N-2}]$ then
\begin{align*}
\norm{f}_{\mathit{L}^{p}(\R^N)} \leq C(N, p) \norm{\diff f}_{\mathit{L}^2(\R^N)}^{\beta} \norm{f}_{\mathit{L}^2(\R^N)}^{(1-\beta)}, \qquad \text{with } \beta = N\(\dfrac{1}{2} - \dfrac{1}{p}\).
\end{align*}
Applying the Hardy-Littlewood-Sobolev inequality we obtain
\begin{align}
B(u) = \intrb\intrb \dfrac{\abs{u(x)}^2 \abs{u(y)}^2}{\abs{x-y}} dxdy \leq K_1 \norm{u}_{\Lp{\frac{12}{5}}}^4 \label{eqn:2.1}
\end{align}
and thus using the Gagliardo-Nirenberg inequality, we get
\begin{align*}
B(u) \leq K_1 \norm{u}_{\Lp{\frac{12}{5}}}^4 \leq K_1 K_2 \normLp{\diff u}{2} \normLp{u}{2}^3 = K_{H} \sqrt{A(u)} c^{\frac{3}{2}}.
\end{align*}
Finally, applying the Sobolev, Gagliardo-Nirenberg inequality, we have
\begin{align*}
C(u) = \normLp{u}{p}^p \leq K_{GN} \normLp{\diff u}{2}^{\sigma} \normLp{u}{2}^{\frac{6-p}{2}} = K_{GN} [A(u)]^{\frac{\sigma}{2}} c^{\frac{6-p}{4}}.
\end{align*}
\end{proof}

\begin{lemma}\label{lemma:2.3}
Let $p \in (\frac{10}{3}, 6]$. Assume that $\gamma \in \R$ and $a \in \R$. If $u \in \Hs{1}$ is a weak solution to
\begin{align}
- \laplace u + \lambda u - \gamma (\abs{x}^{-1} * \abs{u}^2) u  - a \abs{u}^{p-2}u = 0, \label{eqn:6}
\end{align}
then $Q(u) = 0$. Moreover, if $u \neq 0$ then we have
\begin{enumerate}[label=(\roman*), ref = \roman*]
\item\label{point:2.3:i}  $\lambda > 0$ if $\gamma > 0$ and $p \in (\frac{10}{3}, 6]$,
\item\label{point:2.3:ii} $\lambda < 0$ if $\gamma < 0$ and $p = 6$.
\end{enumerate}
\end{lemma}
\begin{proof}
Our proof is inspired by \cite[Lemma 4.2]{BellazziniJeanjeanLuo2013}. The following Pohozaev type identity holds for
$u \in \Ho$ weak solution of \eqref{eqn:6} ([22], also see \cite[Theorem 2.2]{DavidRuiz2006}),
\begin{align}
\dfrac{1}{2} A(u) + \dfrac{3 \lambda}{2} D(u) - \dfrac{5 \gamma}{4} B(u) - \dfrac{3a}{p} C(u) = 0, \quad \mbox{where} \quad D(u) = \normLp{u}{2}^2. \label{eqn:7}
\end{align}
By multiplying \eqref{eqn:6} by $u$ and integrating, we derive a second identity
\begin{align}
A(u) + \lambda_c D(u) - \gamma B(u) - a C(u) = 0. \label{eqn:8}
\end{align}
Combining \eqref{eqn:7} and \eqref{eqn:8}, we get
\begin{align*}
A(u) - \dfrac{\gamma}{4} B(u) - \dfrac{a \sigma}{p} C(u) = 0.
\end{align*}
This means that $Q(u) = 0$. Using \eqref{eqn:7} and \eqref{eqn:8} again, we obtain
\begin{align}
2(6-p) A(u) + (5p -12)\gamma B(u) = 2(3p-6)\lambda D(u). \label{eqn:lambda}
\end{align}
If $\gamma > 0$ and $p \in (\frac{10}{3}, 6]$, we have
\begin{align*}
2(6-p) \geq 0, \qquad (5p -12)\gamma > 0, \qquad 2(3p-6) > 0.
\end{align*}
Hence, $\lambda > 0$. If $\gamma < 0$ and $p = 6$, we have
\begin{align*}
2(6-p) = 0, \qquad (5p -12)\gamma = 18 \gamma < 0, \qquad 2(3p-6) = 24 > 0.
\end{align*}
This implies that $\lambda < 0$.
\end{proof}

\begin{lemma}\label{lemma:bounded-positive-solution}
Let $p \in (\frac{10}{3}, 6]$. Assume that $\gamma \in \R$ and $a \in \R$. If $u \in \Hs{1}$ is a weak solution to
\begin{align}
- \laplace u + \lambda u - \gamma (\abs{x}^{-1} * \abs{u}^2) u  - a \abs{u}^{p-2}u = 0,
\end{align}
then $u \in \Lp{\infty} \cap \mathit{C}(\Rn)$. Moreover, in case $\gamma > 0,$ $ a > 0$ we have that if $u \not\equiv 0 $ and $u \geq 0$ then $u > 0$.
\end{lemma}
\begin{proof}
Applying \cite[Theorem 2.1]{Li-Ma-2020}, we get that $u \in \mathit{W}_{loc}^{2,r}(\Rn)$ for every $r > 1$ and hence $u \in \mathit{C}(\Rn)$. Since $u \in \Hs{1}$, the  Sobolev embedding (see \cite[Corollary 9.10]{Brezis-2011}) implies that $|u|^2 \in \Lp{q}$ for every $q \in [1,3]$.
Now,  setting $K := |x|^{-1}$, we write $K := K_1 + K_2$ where $K_1:= K$ on $B(0,1)$, $K_1:=0$ on $\R^3 \backslash \ B(0,1)$ and $K_2:= K - K_1$.
Clearly $K_1 \in \Lp{2}$ and $K_2 \in \Lp{4}$. Applying \cite[Lemma 2.20]{LiebLoss2001analysis} with $K_1 \in \Lp{2}$, $|u|^2 \in \Lp{2}$ and with $K_2 \in \Lp{4}$, $|u|^2 \in \Lp{\frac{4}{3}}$, we obtain that $K_1*|u|^2$ and $K_2*|u|^2$ are continuous. Also
\begin{align*}
\lim_{|x| \to \infty} (K_1*|u|^2) (x) = 0 \quad\mbox{and}\quad \lim_{|x| \to \infty} (K_2*|u|^2) (x) = 0.
\end{align*}
Hence, we get that $K*|u|^2$ is continuous and
\begin{align} \label{eqn:tend-to-0}
\lim_{|x| \to \infty} (K*|u|^2) (x) = 0.
\end{align}
Therefore, $K*|u|^2$ is bounded. At this point, we deduce from \cite[Proposition B.1]{Soave2019Sobolevcriticalcase} that $u \in \Lp{\infty}$.

Now, if we assume that $\gamma > 0,$ $ a > 0$, $u \not\equiv 0$, $u \geq 0$, setting $v := -u \leq 0$ we get
\begin{align*}
- \laplace v + \lambda v =  \gamma (\abs{x}^{-1} * \abs{v}^2) v  + a \abs{v}^{p-2}v \leq 0.
\end{align*}
By \cref{lemma:2.3}, we have that $\lambda > 0$. We assume that there exists $x_0 \in \Rn$ such that $v(x_0) = 0$. For all $R > |x_0|$, we have that $v \in \mathit{W}^{2,r}(B_R)$ for every $r > 1$, $Lv := - \laplace v + \lambda v \leq 0$ in $B_R$ with $\lambda > 0$ and $M:= \max_{x \in B_R} v = 0$. At this point, applying \cite[Theorem 3.27]{Troianiello-1987}, in the particular case where $\Gamma = \emptyset$, we obtain that $v \equiv 0$ in $B_R$, and hence $u \equiv 0$ in $B_R$. The value $R>0$ being arbitrarily large, this contradicts our assumption that $u \not\equiv 0$ and we conclude that $u > 0$.
\end{proof}

%\begin{lemma} \label{lemma:2.2}
%Let $u \in S(r)$, then $t > 0$ is a critical point for $g_u$ if and only if $u^t \in \slc$. In particular, $u \in \slc$ if and only if $1$ is a critical point of $g_u$.
%	For any $u\in S(c)$, we have, for all $t > 0$
%	\begin{align*}
%	g_u'(t) = t A(u) - \dfrac{\gamma}{4} B(u) - \dfrac{a \sigma}{p} t^{\sigma - 1} C(u) = \dfrac{1}{t} Q(u^t).
%	\end{align*}
%	The lemma is proved.

Following [13], we recall that, for any $c > 0$, $S(c)$ is a submanifold codimension 1 of $\Ho$ and the tangent space at a point $u \in S(c)$ is defined as
\begin{align*}
T_uS(c) = \{\varphi \in \Ho: \inner{u}{\varphi}_{\Lp{2}} = 0\}.
\end{align*}
The restriction ${F}_{|S(c)} :S(c) \to \R$ is a $\mathit{C}^1$ functional on $S(c)$ and for any $u \in S(c)$ and any $v \in T_uS(c)$, we have
\begin{align*}
\inner{{F}'_{|S(c)}}{\varphi} = \inner{F'}{\varphi}.
\end{align*}
We use the notation $\norm{{dF}_{|S(c)}}_{*}$ to indicate the norm in the cotangent space $T_uS(c)'$, i.e the dual norm induced by the norm of $T_uS(c)$, i.e
\begin{align}
\norm{{dF}_{|S(c)} (u)}_{*} := \sup_{\norm{\varphi} \leq 1, \varphi \in T_uS(c)} \abs{dF(u)[\varphi]}. \label{eqn:21}
\end{align}
We recall the following result, see \cite[Lemma 3.1]{JeanjeanLe2020},
\begin{lemma} \label{lemma:2.4}
For $u \in S(c)$ and $t > 0$, the map
\begin{align*}
T_uS(c) \to T_{u^t} S(c), \quad \psi \mapsto \psi^t
\end{align*}
is a linear isomorphism with inverse $$T_{u^t}S(c) \to T_{u} S(c), \quad \phi \mapsto \phi^{\frac{1}{t}}.$$
\end{lemma}
%	We follow the approach in \cite[Lemma 3.6]{BartschSoave2019}. For $\psi \in T_uS(c)$ and for $t > 0$, we have
%	\begin{align*}
%		\intrb u^t(x) \psi^t(x) dx = \intrb t^3 u(tx) \psi(tx) dx = \intrb u(y) \psi(y) dy = 0.
%	\end{align*}
%	As a consequence, $\psi^t \in T_{u^t} S(c)$ and the map is well defined. Clearly it is linear. Taking into account that, for every $t, s > 0$ and $w \in \Ho$,
%	\begin{align*}
%		w^{ts} = (ts)^{\frac{3}{2}} w(ts x) = (w^t)^s,
%	\end{align*}
%	we obtain that the map is linear isomorphism.

%Now, we define
%	D^{1,2} := \Big\{u \in \Lp{6} : \abs{\diff u} \in \Lp{2}  \Big\},
%E := \Bigg\{u \in D^{1,2}(\Rb) : \intrb\intrb \dfrac{\abs{u(x)}^2 \abs{u(y)}^2}{\abs{x-y}} dxdy < \infty \Bigg\}
%which contains $\Ho$.
Next, we recall a result concerning the convergence of the term $B$, see {\cite[Lemma 2.1]{DavidRuiz2006}},
\begin{lemma} \label{lemma:2.5}
Let $(u_n)$ be a sequence satisfying $u_n \weakto u$ weakly in $\Hsr{1}$. Then we have $B(u_n) \to B(u)$.
\end{lemma}

%\begin{lemma}[{\cite[Lemma 2.3]{IanniRuiz2012}}]\label{lemma:2.6}
%	Assume that we have three weakly convergent sequences in $E$, $u_n \weakto u$, $v_n \weakto v$, $w_n \weakto w$, and $z \in E$. Then we have
%	\begin{align*}
%		T(u_n, v_n, w_n, z) \to T(u, v, w, z),
%	\end{align*}
%	where
%	\begin{align*}
%		T(u, v, w, z) := \intrb\intrb \dfrac{u(x)v(x)w(y)z(y)}{\abs{x-y}} dxdy.
%	\end{align*}

%Here, we recall the Schwartz rearrangement (see \cite[Chapter 3]{LiebLoss2001analysis}). For $u \in \Ho$, we denote by $u^{\star}$ the Schwartz rearrangement of $u$. It is well know that $u^{\star}$ is radially symmetric and
%	\normLp{u^{\star}}{q} = \normLp{u}{q} \text{ for } q\geq 1 \quad \text{and} \quad A(u^{\star}) \leq A(u).
%By the Riesz's rearrangement inequality (see \cite[Section 3.7]{LiebLoss2001analysis}), we have that $B(u^{\star}) \geq B(u)$. Therefore, we obtain the following lemma:
%\begin{lemma} \label{lemma:2.7}
%	Let $u \in \Ho$ and $u^{\star}$ be the Schwartz rearrangement of $u$. Then we have that $F(u^{\star}) \leq F(u)$ and $Q(u^{\star}) \leq Q(u)$.

\section{The case $\gamma > 0,$ $ a > 0$ and $ p \in (\frac{10}{3},  6]$.} \label{Section3}

\subsection{The geometrical structure and the existence of bounded Palais-Smale sequences for $ p \in (\frac{10}{3},  6]$}\label{Subsection3-1}
In this subsection, we follow the approach first introduced in [21]. We shall always assume that $\gamma > 0,$ $ a > 0$ and $ p \in (\frac{10}{3},  6]$.
\begin{lemma} \label{lemma:5}
For any $c \in (0, \infty)$,
$F$ restricted to $\slc$ is coercive on $\Ho$, namely when $(u_n) \subset H^1(\R^3)$ satisfies $||u_n|| \to + \infty$ then $F(u_n) \to + \infty$. In particular $F$ restricted to $\slc$ is bounded from below.
\end{lemma}
\begin{proof}
Let $u \in \slc$. Taking into account that
\begin{align*}
\dfrac{a}{p} C(u) = \dfrac{1}{\sigma} A(u) - \dfrac{\gamma}{4\sigma} B(u),
\end{align*}
and using \cref{lemma:1}\eqref{point:1i}, we obtain
\begin{align}
\begin{split}
F(u) &= \dfrac{1}{2} A(u) - \dfrac{\gamma}{4} B(u) - \dfrac{a}{p} C(u)
= \dfrac{1}{2} A(u) - \dfrac{\gamma}{4} B(u) - \dfrac{1}{\sigma} A(u) + \dfrac{\gamma}{4\sigma} B(u) \\
&= \dfrac{\sigma - 2}{2 \sigma} A(u) - \dfrac{\gamma(\sigma - 1)}{4} B(u)
\geq \dfrac{\sigma - 2}{2 \sigma} A(u) - \dfrac{\gamma(\sigma - 1)}{4} K_{H}  \sqrt{A(u)} c^{\frac{3}{2}}.
\end{split} \label{eqn:29}
\end{align}
This concludes the proof.
\end{proof}

For any $u \in S(c)$, we recall that
\begin{align*}
&g_u(t)= F(u^t) = \dfrac{1}{2}t^2 A(u) - \dfrac{\gamma}{4} t B(u) - \dfrac{a}{p} t^{\sigma} C(u),\\
&g_u'(t) = t A(u) - \dfrac{\gamma}{4} B(u) - \dfrac{a \sigma}{p} t^{\sigma - 1} C(u) = \dfrac{1}{t} Q(u^t),\\
&g_u''(t) = A(u) - \dfrac{a \sigma (\sigma - 1)}{p} t^{\sigma - 2} C(u).
\end{align*}
For any $u \in S(c)$, we set
\begin{align*}
\tstar := \(\dfrac{pA(u)}{a \sigma (\sigma - 1) C(u)}\)^{\frac{1}{\sigma - 2}}.
\end{align*}
This implies that $\tstar$ is the unique solution of equation $g_u''(t) = 0$. So, we have
\begin{align}
g_u''(\tstar) = 0, \quad g_u''(t) > 0 \text{ if } 0<t<\tstar, \quad g_u''(t) < 0 \text{ if } t>\tstar. \label{eqn:2}
\end{align}

\begin{lemma} \label{lemma:3}
For any $c \in (0, c_1)$ and any $u \in S(c)$,  we have $g_u'(\tstar) > 0$.
\end{lemma}

\begin{proof}
Let $u \in S(c)$ be arbitrary. By the definition of $\tstar$ and by $g_u''(\tstar) = 0$, we have
\begin{align*}
g_u'(\tstar) &= \tstar A(u) - \dfrac{\gamma}{4} B(u) - \dfrac{a \sigma}{p} (\tstar)^{\sigma - 1} C(u)
= \tstar A(u) - \dfrac{\gamma}{4} B(u) - \dfrac{1}{\sigma - 1} \tstar A(u)\\
&= \dfrac{\sigma - 2}{\sigma - 1} \tstar A(u) - \dfrac{\gamma}{4} B(u)
= \dfrac{\sigma - 2}{\sigma - 1} \(\dfrac{pA(u)}{a \sigma (\sigma - 1) C(u)}\)^{\frac{1}{\sigma - 2}} A(u) - \dfrac{\gamma}{4} B(u)\\
&= \sqrt{A(u)} \[\dfrac{\sigma - 2}{\sigma - 1} \(\dfrac{pA(u)}{a \sigma (\sigma - 1) C(u)}\)^{\frac{1}{\sigma - 2}} \sqrt{A(u)} - \dfrac{\gamma}{4} \dfrac{B(u)}{\sqrt{A(u)}}\] \\
&= \sqrt{A(u)} \[\dfrac{\sigma - 2}{\sigma - 1} \(\dfrac{p[A(u)]^{\frac{\sigma}{2}}}{a \sigma (\sigma - 1) C(u)}\)^{\frac{1}{\sigma - 2}} - \dfrac{\gamma}{4} \dfrac{B(u)}{\sqrt{A(u)}}\].
\end{align*}
Applying \cref{lemma:1}, we obtain
\begin{align*}
g_u'(\tstar) &\geq \sqrt{A(u)} \[\dfrac{\sigma - 2}{\sigma - 1} \(\dfrac{p[A(u)]^{\frac{\sigma}{2}}}{a \sigma (\sigma - 1) K_{GN} [A(u)]^{\frac{\sigma}{2}} c^{\frac{6-p}{4}}}\)^{\frac{1}{\sigma - 2}} - \dfrac{\gamma}{4} \dfrac{K_{H}  \sqrt{A(u)} c^{\frac{3}{2}}}{\sqrt{A(u)}}\] \\
&=\sqrt{A(u)} \[\dfrac{\sigma - 2}{\sigma - 1} \(\dfrac{p}{a \sigma (\sigma - 1) K_{GN}  c^{\frac{6-p}{4}}}\)^{\frac{1}{\sigma - 2}} - \dfrac{\gamma}{4} K_{H} c^{\frac{3}{2}}\].
\end{align*}
By direct computations, we now have
\begin{align*}
\dfrac{\sigma - 2}{\sigma - 1} \(\dfrac{p}{a \sigma (\sigma - 1) K_{GN}  c^{\frac{6-p}{4}}}\)^{\frac{1}{\sigma - 2}} - \dfrac{\gamma}{4} K_{H} c^{\frac{3}{2}} > 0 \iff c < c_1.
\end{align*}
%	\begin{align*}
%	&\dfrac{\sigma - 2}{\sigma - 1} \(\dfrac{p}{a \sigma (\sigma - 1) K_{GN}  c^{\frac{6-p}{4}}}\)^{\frac{1}{\sigma - 2}} - \dfrac{\gamma}{4} K_{H} c^{\frac{3}{2}} > 0\\
%	\iff &c^{\frac{3}{2}} c^{\frac{6-p}{4}\frac{1}{\sigma - 2}} < \dfrac{4}{\gamma K_{H}} \dfrac{\sigma - 2}{\sigma - 1} \(\dfrac{p}{a \sigma (\sigma - 1) K_{GN}}\)^{\frac{1}{\sigma - 2}}\\
%	\iff &c^{\frac{4(p-3)}{3p-10}} < \dfrac{4}{\gamma K_{H}} \dfrac{\sigma - 2}{\sigma - 1} \(\dfrac{p}{a \sigma (\sigma - 1) K_{GN}}\)^{\frac{1}{\sigma - 2}}\\
%	\iff &c < \(\dfrac{4}{\gamma K_{H}} \dfrac{\sigma - 2}{\sigma - 1}\)^{\frac{3p-10}{4(p-3)}} \(\dfrac{p}{a \sigma (\sigma - 1) K_{GN}}\)^{\frac{1}{2(p-3)}} = c_1.
%	\end{align*}
Thus, we obtain that if $0 < c< c_1$ then $g_u'(\tstar) > 0$.
\end{proof}

\begin{lemma} \label{lemma:4}
For any $c \in (0, c_1)$, it holds that $\slz = \emptyset$.
\end{lemma}
\begin{proof}
We assume that there exists $u \in \slz$. Since $g_u''(1) = 0$ and $\tstar$ is the unique solution of equation $g_u''(t) = 0$, we have $\tstar = 1$. So, we have $g_u'(\tstar) = g_u'(1) = 0$. This contradicts with $g_u'(\tstar) > 0$ in \cref{lemma:3}. Thus, we obtain $\slz = \emptyset$.
\end{proof}

\begin{lemma} \label{lemma:6}
For any $c \in (0, c_1)$ and any $u \in S(c)$, there exists
\begin{enumerate}[label=(\roman*)]
\item a unique $\spu \in (0, \tstar)$ such that $\spu$ is a unique local minimum point for $g_u$ and $u^{\spu} \in \slp$.
\item a unique $\smu \in (\tstar, \infty)$ such that $\smu$ is a unique local maximum point for $g_u$ and $u^{\smu} \in \slm$.
\end{enumerate}
Moreover, the maps $u \in S(c) \mapsto \spu \in \R$ and $u \in S(c) \mapsto \smu \in \R$ are of class $\mathbb{C}^1$.
\end{lemma}
\begin{proof}
Taking into account that
\begin{align*}
g_u'(t) = t A(u) - \dfrac{\gamma}{4} B(u) - \dfrac{a \sigma}{p} t^{\sigma - 1} C(u),
\end{align*}
we have $g_u'(t) \to - \frac{\gamma}{4} B(u) < 0$ as $t \to 0$ and $g_u'(t) \to - \infty$ as $t \to +\infty$ due to $\sigma - 1 > 1$. By \cref{lemma:3}, we have $g'_u(\tstar) > 0$. Therefore, the equation $g_u'(t) = 0$ has at least two solutions $\spu$ and $\smu$ with $0< \spu < \tstar < \smu$. By \eqref{eqn:2}, we have $g_u''(t) > 0$ for all $0<t<\tstar$. Hence, $g_u'(t)$ is strictly increasing function on $(0, \tstar)$ and consequently $\spu \in (0, \tstar)$ is the unique local minimum point for $g_u$ and $u^{\spu} \in \slp$ due to $g''_{u^{\spu}}(1) = g_u''(\spu) > 0$. By the same argument, we obtain that $\smu \in (\tstar, \infty)$ is a unique local maximum point for $g_u$ and $u^{\smu} \in \slm$.

In order to prove that $u \mapsto \smu$ are of class $\mathbb{C}^1$, we follow the argument in \cite[Lemma 5.3]{Soave2019}. It is a direct application of the Implicit Function Theorem on $\mathbb{C}^1$-function $\varphi(t, u)= g_u'(t)$. Taking into account that $\varphi(\smu, u) = g_u'(\smu)=0$, $\partial_t \varphi(\smu, u) = g_u''(\smu) < 0$ and $\slz = \emptyset$, we obtain $u \mapsto \smu$ is of class $\mathbb{C}^1$. The same argument proves that $u \mapsto \spu$ is of class $\mathbb{C}^1$.
\end{proof}

\begin{lemma}\label{lemma:16}
For any $c \in (0, c_1)$, it holds that
\begin{enumerate}[label=(\roman*), ref=\roman*]
\item\label{point:1} $F(u) < 0$ for all $u \in \slp$,
\item\label{point:2} there exists $\alpha:= \alpha(c) > 0$ such that $A(u) \geq \alpha$ for all $u \in \slm$.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $u \in \slp$, taking into account that
\begin{align*}
A(u) = \dfrac{\gamma}{4} B(u) + \dfrac{a \sigma}{p} C(u), \qquad A(u) > \dfrac{a \sigma (\sigma - 1)}{p} C(u),
\end{align*}
we obtain
\begin{align*}
F(u) &= \dfrac{1}{2} A(u) - \dfrac{\gamma}{4} B(u) - \dfrac{a}{p} C(u)
= \dfrac{1}{2} A(u) - \dfrac{\gamma}{4} B(u) - \dfrac{a \sigma}{p} C(u) + \dfrac{a (\sigma -1)}{p} C(u)\\
&< \dfrac{1}{2} A(u) - A(u) + \dfrac{1}{\sigma} A(u) = \dfrac{2-\sigma}{2\sigma} A(u).
\end{align*}
Since $\sigma > 2$, we have $F(u) < 0$. The point \eqref{point:1} is proved.

Let $u \in \slm$, taking into account that
\begin{align*}
A(u) < \dfrac{a \sigma (\sigma - 1)}{p} C(u),
\end{align*}
and using \cref{lemma:1}, we obtain that
\begin{align*}
A(u) < \dfrac{a \sigma (\sigma - 1)}{p} K_{GN} c^{\frac{6-p}{4}} [A(u)]^{\frac{\sigma}{2}}.
\end{align*}
Since $\sigma > 2$, the point \eqref{point:2} follows.
\end{proof}

We define
\begin{align*}
S_r(c) := S(c) \cap \Hsr{1}, \qquad \Lambda_r(c) := \slc \cap \Hsr{1}, \qquad \Lambda^{\pm}_r(c) := \Lambda^{\pm}(c) \cap \Hsr{1}.
\end{align*}
Here $\Lambda^{\pm}(c)$ denotes either $\Lambda^{+}(c)$ or $\Lambda^{-}(c)$.

\begin{lemma}\label{lemma:12}
For any $c \in (0,c_1)$ it holds that
\begin{align*}
\inf_{u \in \Lambda^{\pm}_r(c)} F(u) = \inf_{u \in \Lambda^{\pm}(c)} F(u).
\end{align*}
Also, if $\inf_{u \in \Lambda^{\pm}(c)} F(u)$ is reached, it is reached by a Schwarz symmetric function.
\end{lemma}

\begin{proof}
Since $\Lambda^{\pm}_r(c) \subset \Lambda^{\pm}(c)$, we directly have
\begin{align}
\inf_{u \in \Lambda^{\pm}_r(c)} F(u) \geq \inf_{u \in \Lambda^{\pm}(c)} F(u). \label{eqn:24}
\end{align}
Therefore, it suffices to prove that
\begin{align}
\inf_{u \in \Lambda^{\pm}_r(c)} F(u) \leq \inf_{u \in \Lambda^{\pm}(c)} F(u). \label{eqn:25t}
\end{align}
In this aim we start to note that
\begin{align}
\inf_{u \in \Lambda^{+}(c)} F(u) = \inf_{u \in S(c)} \min_{0 < t \leq s_{u}^+}F(u^t) \quad \text{and} \quad \inf_{u \in \Lambda^{-}(c)} F(u) = \inf_{u \in S(c)} \max_{s_u^+ < t  \leq s_{u}^-}F(u^t). \label{eqn:26l}
\end{align}
Now  let $u \in S(c)$ and $v \in S_r(c)$ be the  Schwarz rearrangement of $|u|$. Taking into account that $A(v) \leq A(u)$, $C(v) = C(u)$, and
by the Riesz's rearrangement inequality (see \cite[Section 3.7]{LiebLoss2001analysis}),  $B(v) \geq B(u)$,
we have for all $t>0,$
\begin{align}
F(v^t) = \dfrac{1}{2}t^2 A(v) - \dfrac{\gamma}{4} t B(v) - \dfrac{a}{p} t^{\sigma} C(v)
\leq \dfrac{1}{2}t^2 A(u) - \dfrac{\gamma}{4} t B(u) - \dfrac{a}{p} t^{\sigma} C(u) \label{eqn:27l}
= F(u^t).
\end{align}
Observe that, for any $w \in S(c)$,
\begin{equation*}
g_w'(t) = t A(w) - \frac{\gamma}{4}B(w) - \frac{a \sigma}{p}t^{\sigma -1}C(w) \quad\mbox{and}\quad g_w''(t) =  A(w) - \frac{a \sigma(\sigma -1)}{p}t^{\sigma -2}C(w).
\end{equation*}
Thus we have
\begin{align*}
g_v'(0) \leq g_u'(0) <0 \quad \mbox{and} \quad g_v''(t) \leq g_u''(t), \quad \forall t >0.
\end{align*}
This implies that $0 < \spu \leq s_v^{+} < s_v^{-} \leq \smu$. Hence, we deduce from \eqref{eqn:27l} that
\begin{align*}
\min_{0 < t \leq s_v^+}F(v^t) \leq \min_{0 < t \leq s_u^+}F(u^t) \qquad \text{and} \qquad \max_{s_v^+ < t \leq s_v^-}F(v^t) \leq \max_{s_u^+ < t \leq s_u^-}F(u^t).
\end{align*}
In view of \eqref{eqn:26l}, the inequality \eqref{eqn:25t} holds. Now if $u_0 \in \Lambda^{+}(c)$ is such that $F(u_0) = \inf_{u \in \Lambda^{+}(c)} F(u)$ we see that $v$, the Schwarz rearrangement of $|u_0|$, belongs to $\Lambda^{+}_r(c)$. Indeed, if either $A(v) < A(u_0)$ or $B(v) > B(u_0)$ then $F(v^t) < F(u_0^t)$. Hence, in view of the above arguments, we get
$$\inf_{u \in \Lambda^{+}(c)} F(u) = \inf_{u \in S(c)} \min_{0 < t \leq s_{u}^+}F(u^t) \leq \min_{0 < t \leq s_{v}^+}F(v^t) < \min_{0 < t \leq s_{u_0}^+}F(u_0^t) = \inf_{u \in \Lambda^{+}(c)} F(u)$$
a contradiction. Thus $A(v) = A(u_0)$, $B(v) = B(u_0)$ and $C(v) = C(u_0)$ from which we deduce that $v \in \Lambda^{+}_r(c)$ and $F(v) = F(u_0).$
The case of $u_0 \in \Lambda^{-}(c)$ such that $F(u_0) = \inf_{u \in \Lambda^{-}(c)} F(u)$ is treated similarly. This ends the proof of the lemma.
\end{proof}

Recalling that $\gamma^{+ }(c)$ and $\gamma^{+ }(c)$ are defined in \eqref{eqn:gamma} we have
\begin{lemma} \label{lemma:9}
For any $c \in (0, c_1)$, there exists a bounded Palais-Smale sequence $(u_n^{+}) \subset \Lambda^{+}_r(c)$ for $F$ restricted to $S(c)$ at level $\gamma^{+}(c)$ and a bounded Palais-Smale sequence $(u_n^{-}) \subset \Lambda^{-}_r(c)$ for $F$ restricted to $S(c)$ at level $\gamma^{-}(c)$.
\end{lemma}

In order to prove \cref{lemma:9} we define the functions
\begin{align*}
&I^{+}: S(c) \to \R, \quad I^{+}(u) = F(u^{\spu}), \\
&I^{-}: S(c) \to \R, \quad I^{-}(u) = F(u^{\smu}).
\end{align*}
Note that since the maps $u \mapsto \spu$ and $u \mapsto \smu$ are of class $\mathbb{C}^1$, see \cref{lemma:6}, the functionals $I^{+}$ and $I^{-}$ are of class $\mathbb{C}^1$.
\begin{lemma} \label{lemma:7}
For any $c \in (0, c_1)$, we have that $dI^{+}(u)[\psi] = dF(u^{\spu})[\psi^{\spu}]$ and $dI^{-}(u)[\psi] = dF(u^{\smu})[\psi^{\smu}]$ for any $u \in S(c)$, $\psi \in T_uS(c)$.
\end{lemma}
\begin{proof}
We first give the proof for $I^{+}$. Let $\psi \in T_uS(c)$, then $\psi = h'(0)$ where $h: (-\epsilon, \epsilon) \mapsto S(c)$ is a $\mathit{C}^1$-cure with $h(0) = u$. We consider the incremental quotient
\begin{align}
\dfrac{I^{+}(h(t)) - I^{+}(h(0))}{t} = \dfrac{F(h(t)^{s_t}) - F(h(0)^{s_0})}{t}, \label{eqn:3}
\end{align}
where $s_t:= s_{h(t)}^{+}$, and hence $s_0 = \spu$. Recalling from \cref{lemma:6} that $s_0$ is a strict local minimum of $s \mapsto F(u^s)$ and $u \mapsto s_0$ is continuous, we get
\begin{align*}
&F(h(t)^{s_t}) - F(h(0)^{s_0})
\geq F(h(t)^{s_t}) - F(h(0)^{s_t}) \\
= &\dfrac{s_t^2}{2} \Big[A(h(t)) - A(h(0)) \Big] - \dfrac{\gamma s_t}{4} \Big[B(h(t)) - B(h(0)) \Big] -\dfrac{a s_t^{\sigma}}{p} \Big[C(h(t)) - C(h(0)) \Big]\\
= &s_t^2 \intrb \diff h(\tau_1 t) \cdot \diff h'(\tau_1 t) t dx - \gamma s_t \intrb\intrb \dfrac{\abs{h(\tau_2 t)(x)}^2 h(\tau_2 t)(y) h'(\tau_2 t)(y)}{\abs{x-y}} dxdy
- a s_t^{\sigma} \intrb \abs{h(\tau_3 t)}^{p-2} h(\tau_3 t) h'(\tau_3 t) dx,
\end{align*}
for some $\tau_1, \tau_2, \tau_3 \in (0,1)$. Analogously
\begin{align*}
&F(h(t)^{s_t}) - F(h(0)^{s_0})
\leq F(h(t)^{s_0}) - F(h(0)^{s_0}) \\
= &s_0^2 \intrb \diff h(\tau_4 t) \cdot \diff h'(\tau_4 t) t dx - \gamma s_0 \intrb\intrb \dfrac{\abs{h(\tau_5 t)(x)}^2 h(\tau_5 t)(y) h'(\tau_5 t)(y)}{\abs{x-y}} dxdy
- a s_0^{\sigma} \intrb \abs{h(\tau_6 t)}^{p-2} h(\tau_6 t) h'(\tau_6 t) dx,
\end{align*}
for some $\tau_4, \tau_5, \tau_6 \in (0,1)$. Now, from \eqref{eqn:3} we deduce that
\begin{align*}
\lim_{t \to 0} \dfrac{I^{+}(h(t)) - I^{+}(h(0))}{t}
= &(\spu)^2 \intrb \diff u \diff \psi dx - \gamma (\spu) \intrb\intrb \dfrac{\abs{u(x)}^2 u(y) \psi(y)}{\abs{x-y}} dxdy - a (\spu)^{\sigma} \intrb \abs{u}^{p-2} u \psi dx \\
= & \intrb \diff u^{\spu} \diff \psi^{\spu} dx - \gamma \intrb\intrb \dfrac{\abs{u^{\spu}(x)}^2 u^{\spu}(y) \psi^{\spu}(y)}{\abs{x-y}} dxdy - a \intrb \abs{u^{\spu}}^{p-2} u^{\spu} \psi^{\spu} dx \\
= &dF(u^{\spu})[\phi^{\spu}],
\end{align*}
for any $u \in S(c)$, $\psi \in T_uS(c)$. The proof for $I^{-}$ is similar.
\end{proof}

%\textcolor{orange}{We denote by $I^{\pm}$  either $I^{+}$ or $I^{-}$ and by $\gamma^{\pm}(c)$ either $\gamma^{+ }(c)$ or $\gamma^{+ }(c)$. Also by writing that $(u_n^{\pm}) \subset \Lambda^{\pm}(c)$ we mean that we consider a sequence $(u_n^{+}) \subset \Lambda^{+}(c)$ or a sequence $(u_n^{-}) \subset \Lambda^{-}(c)$.} \smallskip

Let $\mathcal{G}$ be the set of all singletons belonging to $S_r(c)$. It is clearly a homotopy stable family of compact subsets of $S_r(c)$ with closed boundary (an empty boundary actually) in the sense of \cite[Definition 3.1]{ghoussoub1993duality}. In view of \cref{lemma:12} we have that

\begin{align*}
e_{\mathcal{G}}^{+} : = \inf_{A \in \mathcal{G}} \max_{u \in A} I^{+}(u) = \inf_{u \in S_r(c)} I^{+}(u) =\inf_{u \in \Lambda^{+}_r(c)} F(u) = \inf_{u \in \Lambda^{+}(c)} F(u) = \gamma^{+}(c).
\end{align*}
\begin{align*}
e_{\mathcal{G}}^{-} : = \inf_{A \in \mathcal{G}} \max_{u \in A} I^{-}(u) = \inf_{u \in S_r(c)} I^{-}(u) =\inf_{u \in \Lambda^{-}_r(c)} F(u) = \inf_{u \in \Lambda^{-}(c)} F(u) = \gamma^{-}(c).
\end{align*}

%	Let $B$ be a closed subset of $X$. We say that a class $\mathcal{G}$ of compact subsets of $X$ is a homotopy stable family with closed boundary $B$ provided
%	\begin{enumerate}[label=(\alph*)]
%		\item every set in $\mathcal{G}$ contains $B$.
%		\item for any $A \in \mathcal{G}$ and any $\eta \in \mathit{C}([0,1]\times X, X)$ satisfying $\eta(t, x) = x$ for all $(t, x) \in (\{0\} \times X) \cup ([0,1] \times B)$, we have that $\eta(\{1\} \times A) \in \mathcal{G}$.
%	\end{enumerate}

\begin{lemma} \label{lemma:8}
For any $c \in (0, c_1)$, there exists a Palais-Smale sequence $(u_n^{+}) \subset \Lambda^{+}(c)$ for $F$ restricted to $S_r(c)$ at level $e_{\mathcal{G}}^{+}$ and a Palais-Smale  sequence
$(u_n^{-}) \subset \Lambda^{-}(c)$ for $F$ restricted to $S(c)$ at level $e_{\mathcal{G}}^{-}$.
\end{lemma}
\begin{proof}
We first treat the case of $e_{\mathcal{G}}^{+}$.
Let $(D_n) \subset \mathcal{G}$ be such that
\begin{align*}
\max_{u \in D_n} I^{+}(u) < e_{\mathcal{G}}^{+} + \dfrac{1}{n},
\end{align*}
and consider the homotopy
\begin{align*}
\eta: [0,1] \times S(c) \mapsto S(c), \quad \eta(t, u) = u^{1 - t + t \spu}.
\end{align*}
From the definition of $\mathcal{G}$, we have
\begin{align*}
E_n := \eta(\{1\} \times D_n) = \{ u^{\spu} : u \in D_n\} \in \mathcal{G}.
\end{align*}
\cref{lemma:6} implies that $E_n \subset \slp$ for all $n \in \N$. Let $v \in E_n$, i.e. $v = u^{\spu}$ for some $u \in D_n$, and hence $I^{+}(v) = I^{+}(u)$. So, we have
\begin{align*}
\max_{v \in E_n} I^{+}(v) = \max_{u \in D_n} I^{+}(u).
\end{align*}
Therefore, $E_n$ is another minimizing sequence for $e_{\mathcal{G}}^{+}$. Applying \cite[Theorem 3.2]{ghoussoub1993duality}, in the particular case where the boundary $B = \emptyset$, there exists a Palais-Smale sequence $(y_n)$ for $I^{+}$ on $S(c)$ at level $e_{\mathcal{G}}^{+}$ such that
\begin{align}
dist_{\Ho} (y_n, E_n) \to 0 \quad \text{as } n \to \infty. \label{eqn:4}
\end{align}
Now writing $s_n := s_{y_n}^{+}$ we set $u_n^{+} := y_n^{s_n} \in \slp$. We claim that there exists a constant $C > 0$ such that
\begin{align}
\dfrac{1}{C} \leq s_n^2 \leq C \label{eqn:5}
\end{align}
for $n \in \N$ large enough. Indeed, notice first that
\begin{align*}
s_n^2 = \dfrac{A(u_n^{+})}{A(y_n)}.
\end{align*}
By $F(u_n^{+}) = I^{+}(y_n) \to e_{\mathcal{G}}^{+} = \gamma^+(c) <0$ we deduce from \eqref{eqn:29} that there exists $M>0$ such that
\begin{align}\label{A-bounds}
\dfrac{1}{M} \leq A(u_n^{+}) \leq M.
\end{align}
On the other hand, since $E_n \in \slp$ is a minimizing sequence for $e_{\mathcal{G}}^{+}$ and $F$ is $\Ho$ coercive on $\slp$, we obtain that $E_n$ is uniformly bounded in $\Ho$ and thus from \eqref{eqn:4}, it implies that $\sup_{n}A(y_n) < \infty$. Also, since $E_n$ is compact for every $n \in \N$, there exist a $v_n \in E_n$ such that $\normHs{v_n - y_n}{1} \to 0$ as $n \to 0$ due to \eqref{eqn:4}. Using \cref{lemma:5} again, we have, for a $\delta > 0$,
\begin{align*}
A(y_n) \geq A(v_n) - A(v_n-y_n) \geq \dfrac{\delta}{2}.
\end{align*}
This proves the claim \eqref{eqn:5}. From \eqref{eqn:21}, and by \cref{lemma:2.4}, \cref{lemma:7}, we have
\begin{align*}
\norm{{dF}_{|S(c)} (u_n^{+})}_{*} = \sup_{\norm{\psi} \leq 1, \psi \in T_uS(c)} \abs*{dF(u_n^{+})[\psi]}
= \sup_{\norm{\psi} \leq 1, \psi \in T_uS(c)} \abs*{dF(u_n^{+})\[\(\psi^{\frac{1}{s_n}}\)^{s_n}\]}
= \sup_{\norm{\psi} \leq 1, \psi \in T_uS(c)} \abs*{dI^{+}(y_n)\[\psi^{\frac{1}{s_n}}\]}.
\end{align*}
This implies that $(u_n^{+}) \subset \slp$ is a Palais-Smale sequence for $F$ restricted to $S(c)$ at level $e_{\mathcal{G}}^{+}$ since $(y_n)$ is a Palais-Smale sequence for $I^{+}$ at level $e_{\mathcal{G}}^{+}$ and
and $\norm*{\psi^{\frac{1}{s_n}}} \leq C_1 \norm{\psi} \leq C_1$ due to \eqref{eqn:5}. For the case of $e_{\mathcal{G}}^{-}$ the proof is identical except that we use \cref{lemma:16}\eqref{point:2} along with \eqref{eqn:29} to conclude that there exists a $M >0$ such that \eqref{A-bounds} holds for $A(u_n^{-})$ replacing $A(u_n^{+})$.
\end{proof}

\begin{proof}[Proof of \cref{lemma:9}]
Applying  \cref{lemma:8}, we deduce that there exists a Palais-Smale sequence $(u_n^{+}) \subset \Lambda_r^{+}(c)$ for $F$ restricted to $S(c)$ at level $e_{\mathcal{G}}^{+} = \gamma^{+}(c)$ and a Palais-Smale sequence $(u_n^{-}) \subset \Lambda_r^{-}(c)$ for $F$ restricted to $S(c)$ at level $e_{\mathcal{G}}^{-} = \gamma^{-}(c)$.  In both cases the boundedness of these sequences follows from \cref{lemma:5}.
\end{proof}

\subsection{The compactness of our Palais-Smale sequences in the Sobolev subcritical case $ p \in (\frac{10}{3},6)$}\label{Subsection3-2}

\begin{lemma}\label{lemma:15}
Let $ p \in (\frac{10}{3},6)$.
For any $c \in (0, c_1)$, if either $(u_n) \subset \Lambda^{+}(c)$ is a minimizing sequence for $\gamma^{+}(c)$  or $(u_n) \subset \Lambda^{-}(c)$ is a minimizing sequence for $\gamma^{-}(c)$, it weakly converges, up to translation, to a non-trivial limit.
\end{lemma}
\begin{proof}
Since $F$ restricted to $\slc$ is coercive on $\Ho$ (see \cref{lemma:5}), $(u_n)$ is bounded. Hence, up to translation, $u_n \weakto u_c$ weakly in $\Ho$.
Let us argue by contradiction assuming that $u_c = 0$, this means that $(u_n)$ is vanishing.
By \cite[Lemma I.1]{LIONS1984part2}, we have, for $2 < q < 6$,
\begin{align*}
\normLp{u_n}{q} \to 0, \quad \text{as } n \to \infty.
\end{align*}
This implies that
\begin{align*}
C(u_n) \to 0, \qquad \text{and} \qquad B(u_n) \leq K_1 \norm{u_n}_{\Lp{\frac{12}{5}}}^4 \to 0,
\end{align*}
due to \eqref{eqn:2.1}.
Since $(u_n) \subset \Lambda(c)$, we have $Q(u_n) = 0$, and hence
\begin{align}
A(u_n) = \dfrac{\gamma}{4} B(u_n) + \dfrac{a \sigma}{p} C(u_n) \to 0. \label{eqn:22}
\end{align}
If we assume that  $(u_n) \subset \Lambda^{-}(c)$ we recall that by \cref{lemma:16}, there exists $\alpha > 0$ such that
\begin{align*}
A(u_n) \geq \alpha > 0, \quad \forall n \in \N,
\end{align*}
contradicting \eqref{eqn:22}. If we assume that $(u_n) \subset \Lambda^{+}(c)$ then since
\begin{align*}
F(u_n) = \dfrac{1}{2} A(u_n) - \dfrac{\gamma}{4} B(u_n) - \dfrac{a}{p} C(u_n) \to 0
\end{align*}
we reach a contradiction with the fact that
\begin{align*}
F(u_n) \to \gamma^{+}(c) = \inf_{u \in \slp} F(u) < 0.
\end{align*}
The lemma is proved.
\end{proof}

\begin{lemma} \label{lemma:11}
Let $ p \in (\frac{10}{3},6)$.
Assume that a bounded Palais-Smale sequence $(u_n) \subset \Lambda_r(c)$ for $F$ restricted to $S(c)$ is weakly convergent, up to translation, to the nonzero function $u_c$. Then, up to translation, $u_n \to u_c \in \Lambda_r(c)$ strongly in $\Hsr{1}$. In particular $u_c$ is a radial solution to \eqref{eqn:Laplace} for some $\lambda_c > 0$ and $\normLp{u_c}{2}^2 = c$.
\end{lemma}
\begin{proof}
Since the embedding $\Hsr{1} \subset \Lp{q}$ is compact for $q \in (2, 6)$, see [49] and, up to translation, $u_n \weakto u_c$ weakly in $\Hsr{1}$, we have, up to translation, $u_n \to u_c$ strongly in $\Lp{q}$ for $q \in (2, 6)$ and a.e in $\Rb$.

Since $(u_n) \subset \Ho$ is bounded, following \cite[Lemma 3]{BerestyckiLions1983_2}, we know that
\begin{align*}
F'_{|S(c)}(u_n) \to 0 \text{ in } \Hs{-1}
\iff F'(u_n) - \dfrac{1}{c}\inner{F'(u_n)}{u_n}u_n \to 0 \text{ in } \Hs{-1}.
\end{align*}
Thus, for any $w \in \Ho$, we have
\begin{align}
\begin{split}
o_n(1) &= \inner*{F'(u_n) - \dfrac{1}{c}\inner{F'(u_n)}{u_n}u_n}{w} \\
&= \intrb \diff u_n \diff w dx + \lambda_n \intrb u_n w dx
- \gamma \intrb\intrb \dfrac{\abs{u_n(x)}^2 u_n(y) w(y)}{\abs{x-y}} dxdy - a \intrb \abs{u_n}^{p-2} u_n w dx ,
\end{split} \label{eqn:19}
\end{align}
where $o_n(1) \to 0$ as $n \to \infty$ and
\begin{align*}
\lambda_n = \dfrac{-1}{c} \[A(u_n) - \gamma B(u_n) - aC(u_n)\]
= \dfrac{1}{c} \[\dfrac{3\gamma}{4}B(u_n) + a \(1 - \dfrac{\sigma}{p}\) C(u_n) \],
\end{align*}
due to $Q(u_n) = 0$. Since $u_n \in \Hsr{1}$, we have $C(u_n) \to C(u_c)$ and $B(u_n) \to B(u_c)$ (see \cref{lemma:2.5}). Hence, we obtain that
\begin{align*}
\lambda_n \to \lambda_c =  \dfrac{1}{c} \[\dfrac{3\gamma}{4}B(u_c) + a \(1 - \dfrac{\sigma}{p}\) C(u_c) \].
\end{align*}
Now, using \cite[Lemma 2.2]{ZHAO2008}, the equation \eqref{eqn:19} leads to
\begin{align}
\intrb \diff u_c \diff w dx + \lambda_c \intrb u_c w dx - \gamma \intrb\intrb \dfrac{\abs{u_c(x)}^2 u_c(y) w(y)}{\abs{x-y}} dxdy - a \intrb \abs{u_c}^{p-2} u_c w dx = 0 \label{eqn:20}
\end{align}
due to the weak convergence in $\Hsr{1}$ and $\lambda_n \to \lambda_c \in \R$.
This implies that $(u_c, \lambda_c)$ satisfies
\begin{align*}
- \laplace u_c + \lambda_c u_c - \gamma (\abs{x}^{-1} * \abs{u_c}^2) u_c  - a \abs{u_c}^{p-2}u_c = 0 \quad\text{ in } \Hs{-1}.
\end{align*}
By the assumption $u_c \neq 0$ and by \cref{lemma:2.3}, we obtain that $Q(u_c) = 0$ and $\lambda_c > 0$.

Now choosing $w = u_n$ in \eqref{eqn:19} and choosing $w = u_c$ in \eqref{eqn:20}, we obtain that
\begin{align*}
\intrb \abs{\diff u_n}^2 dx + \lambda_n \intrb \abs{u_n}^2 dx - \gamma B(u_n) - a C(u_n)
\to \intrb \abs{\diff u_c}^2 dx + \lambda_c \intrb \abs{u_c}^2 dx - \gamma B(u_c) - a C(u_c).
\end{align*}
We can deduce from $B(u_n) \to B(u_c)$, $C(u_n) \to C(u_c)$ and $\lambda_n \to \lambda_c$ that
\begin{align*}
\intrb \abs{\diff u_n}^2 dx + \lambda_c \intrb \abs{u_n}^2 dx \to \intrb \abs{\diff u_c}^2 dx + \lambda_c \intrb \abs{u_c}^2 dx.
\end{align*}
Since $\lambda_c > 0$, we conclude that $u_n \to u_c$ strongly in $\Hsr{1}$. The lemma is proved.
\end{proof}

\begin{proof}[Proof of \cref{theorem:1} in the subcritical case $ p \in (\frac{10}{3},6)$.]
We give the proof for $\gamma^+(c)$, the treatment for $\gamma^-(c)$ is identically.	For any $c \in (0, c_1)$, by \cref{lemma:9}, there exists a bounded Palais-Smale sequence $(u_n^{+}) \subset \Lambda_r^{+}(c)$ for $F$ restricted to $S(c)$ at level $\gamma^{+}(c)$. From \cref{lemma:15} and \cref{lemma:11}, we deduce that $u_n^{+} \to u_c^{+} \in \Lambda_r(c)$ strongly in $\Hsr{1}$ and that there exists $\lambda_c^{+} > 0$ such that $(u_c^{+}, \lambda_c^{+})$ is a solution to \eqref{eqn:Laplace}. Since $\slz = \emptyset$ (see \cref{lemma:4}), we conclude that $u_c^{+} \in \Lambda_r^{+}(c)$. From \cref{lemma:12} we can thus assume that $u_c^{+}$ is a Schwarz symmetric function. Hence, $u_c^{+}$ is non-negative. At this point, we can deduce from \cref{lemma:bounded-positive-solution} that $u_c^{+}$ is  a bounded continuous positive function.
%	Also, since $u_c^{+}$ is a solution to \eqref{eqn:Laplace}, we get that $u_c^{+}$ is of class $\mathit{C}^2$ by standard regularity arguments. Finally, the strong maximum principle yields that $u_c^{+} > 0$ since $u_c^{+}$ is non-negative.
\end{proof}

\subsection{The compactness of our Palais-Smale sequences in the Sobolev critical case $p= 6$}\label{Subsection3-3} $ $

Our next lemma is directly inspired from \cite[Proposition 3.1]{Soave2019Sobolevcriticalcase}.
\begin{lemma} \label{lemma:18}
Let $c \in (0,c_1)$ and $(u_n) \subset \Lambda_r^{+}(c)$ or $(u_n) \subset \Lambda_r^{-}(c)$  be a Palais-Smale sequence for $F$ restricted to $S(c)$ at level $m \in \R$
which is weakly convergent, up to subsequence, to the function $u_c$. If $(u_n) \subset \Lambda_r^{+}(c)$ we assume that $m \neq 0$ and if $(u_n) \subset \Lambda_r^{-}(c)$ we assume that
\begin{align}
m < \dfrac{1}{3 \sqrt{a K_{GN}}}. \label{eqn:36}
\end{align}
Then $u_c \neq 0$  and we have the following alternative:
\begin{enumerate}[label=(\roman*), ref=\roman*]
\item\label{point:18i} either
\begin{align}
F(u_c) \leq m - \dfrac{1}{3 \sqrt{a K_{GN}}}, \label{eqn:32}
\end{align}
\item\label{point:18ii} or
\begin{align}
u_n \to u_c \quad \mbox{strongly in} \quad H_r^1(\R^3). \label{eqn:33}
\end{align}
\end{enumerate}
\end{lemma}

\begin{proof}
%Let us recall that the embedding $\Hsr{1} \subset \Lp{q}$ is compact for $q \in (2, 6)$, see [49].
Since $u_n \weakto u_c$ weakly in $\Hsr{1}$, we have, up to subsequence, $u_n \to u_c$ strongly in $\Lp{q}$ for $q \in (2, 6)$ and a.e in $\Rb$.

Let us first show that $u_c \neq 0$. We argue by contradiction  assuming that $u_c = 0$, this means that $(u_n)$ is vanishing. By \cite[Lemma I.1]{LIONS1984part2}, we have, for $2 < q < 6$,
\begin{align*}
\normLp{u_n}{q} \to 0, \quad \text{as } n \to \infty.
\end{align*}
This implies from \eqref{eqn:2.1} that
\begin{align*}
B(u_n) \leq K_1 \norm{u_n}_{\Lp{\frac{12}{5}}}^4 \to 0.
\end{align*}
Since $(u_n) \subset \Lambda(c)$, we have
\begin{align*}
A(u_n) = a C(u_n) + o_n(1).
\end{align*}
Passing to the limit as $n \to \infty$, up to subsequence we infer that
\begin{align*}
\lim_{n \to \infty} A(u_n) = \lim_{n \to \infty} a C(u_n) :=\ell \geq 0.
\end{align*}
Using \cref{lemma:1}\eqref{point:1ii}, we have
\begin{align*}
\ell = \lim_{n \to \infty} a C(u_n) \leq \lim_{n \to \infty} a K_{GN} [A(u_n)]^3 = a K_{GN} \ell^3.
\end{align*}
Therefore, either $\ell = 0$ or $\ell \geq (a K_{GN})^{-\frac{1}{2}}$. If $(u_n) \subset \Lambda^+(c)$, we have $A(u_n) > 5aC(u_n)$, and then $\ell=0$. This implies that $F(u_n) \to 0$ and this contradicts the assumption that $m \neq 0$. Also, if $(u_n) \subset \Lambda^-(c)$,  \cref{lemma:16}\eqref{point:2} ensure that $\ell \geq (a K_{GN})^{-\frac{1}{2}}$. Hence, we have
\begin{align*}
m + o_n(1) = F(u_n) &= \dfrac{\sigma - 2}{2\sigma} A(u_n) - \dfrac{\gamma(\sigma - 1)}{4\sigma} B(u_n)
= \dfrac{1}{3} A(u_n) + o_n(1) = \dfrac{1}{3}\ell + o_n(1) \geq \dfrac{1}{3 \sqrt{a K_{GN}}} + o_n(1),
\end{align*}
which contradicts our assumption \eqref{eqn:36}. Thus, we have that $u_c \neq 0$.

Now, since $(u_n) \subset \Ho$ is bounded, following \cite[Lemma 3]{BerestyckiLions1983_2}, we know that
\begin{align*}
F'_{|S(c)}(u_n) \to 0 \text{ in } \Hs{-1}
\iff F'(u_n) - \dfrac{1}{c}\inner{F'(u_n)}{u_n}u_n \to 0 \text{ in } \Hs{-1}.
\end{align*}
Thus, for any $w \in \Ho$, we have
\begin{align}
\begin{split}
o_n(1) &= \inner*{F'(u_n) - \dfrac{1}{c}\inner{F'(u_n)}{u_n}u_n}{w}\\
&= \intrb \diff u_n \diff w dx + \lambda_n \intrb u_n w dx
- \gamma \intrb\intrb \dfrac{\abs{u_n(x)}^2 u_n(y) w(y)}{\abs{x-y}} dxdy - a \intrb \abs{u_n}^{p-2} u_n w dx ,
\end{split} \label{eqn:26}
\end{align}
where $o_n(1) \to 0$ as $n \to \infty$ and
\begin{align*}
\lambda_n = \dfrac{-1}{c} \[A(u_n) - \gamma B(u_n) - aC(u_n)\]
= \dfrac{3\gamma}{4c} B(u_n),
\end{align*}
due to $Q(u_n) = 0$. By $B(u_n) \to B(u_c)$ (see \cref{lemma:2.5}), we obtain that
\begin{align}\label{key-point}
\lambda_n \to \lambda_c = \dfrac{3\gamma}{4c} B(u_c).
\end{align}
Now, using \cite[Lemma 2.2]{ZHAO2008}, the equation \eqref{eqn:26} leads to
\begin{align}
\intrb \diff u_c \diff w dx + \lambda_c \intrb u_c w dx - \gamma \intrb\intrb \dfrac{\abs{u_c(x)}^2 u_c(y) w(y)}{\abs{x-y}} dxdy - a \intrb \abs{u_c}^{p-2} u_c w dx = 0 \label{eqn:27}
\end{align}
due to the weak convergence in $\Hsr{1}$ and $\lambda_n \to \lambda_c \in \R$.
This implies that $(u_c, \lambda_c)$ satisfies
\begin{align*}
- \laplace u_c + \lambda_c u_c - \gamma (\abs{x}^{-1} * \abs{u_c}^2) u_c  - a \abs{u_c}^{p-2}u_c = 0 \quad\text{ in } \Hs{-1}.
\end{align*}
By \cref{lemma:2.3}, we obtain that $Q(u_c) = 0$ and $\lambda_c >0$.

Let $v_n:= u_n - u_c \weakto 0$ in $\Hsr{1}$. By Brezis-Lieb lemma (see [16]), we obtain that
\begin{align}
A(u_n) = A(u_c) + A(v_n) + o_n(1), \qquad C(u_n) = C(u_c) + C(v_n) + o_n(1). \label{eqn:28}
\end{align}
By $B(u_n) \to B(u_c)$ (see \cref{lemma:2.5}) and by $Q(u_n) = 0$, we have
\begin{align*}
A(u_c) + A(v_n) - \dfrac{\gamma}{4} B(u_c) - a [C(u_c) - C(v_n)] = o_n(1).
\end{align*}
Taking into account that $Q(u_c) = 0$, we get $A(v_n) = a C(v_n) + o_n(1)$. Passing to the limit as $n \to \infty$, up to subsequence we infer that
\begin{align*}
\lim_{n \to \infty} A(v_n) = \lim_{n \to \infty} a C(v_n) :=k \geq 0.
\end{align*}
Using \cref{lemma:1}\eqref{point:1ii}, we have
\begin{align*}
k = \lim_{n \to \infty} a C(v_n) \leq \lim_{n \to \infty} a K_{GN} [A(v_n)]^3 = a K_{GN} k^3.
\end{align*}
Therefore, either $k = 0$ or $k \geq (a K_{GN})^{-\frac{1}{2}}$.

If $k \geq (a K_{GN})^{-\frac{1}{2}}$, then by \eqref{eqn:28} and by $B(u_n) \to B(u_c)$, we have
\begin{align*}
m = \lim_{n \to \infty} F(u_n) &= \lim_{n \to \infty} \[\dfrac{1}{2} A(u_n) - \dfrac{\gamma}{4} B(u_n) - \dfrac{a}{6} C(u_n) \]\\
&= \lim_{n \to \infty} \[\dfrac{1}{2} A(u_c) + \dfrac{1}{2} A(v_n) - \dfrac{\gamma}{4} B(u_c) - \dfrac{a}{6}C(u_c) -\dfrac{a}{6}C(v_n) \] \\
&= F(u_c) + \dfrac{1}{3} k
\geq F(u_c) + \dfrac{1}{3} (a K_{GN})^{-\frac{1}{2}}.
\end{align*}
This implies that alternative \eqref{point:18i} holds.

If instead $k = 0$, then by \eqref{eqn:28}, we have $A(u_n) \to A(u_c)$ and $C(u_n) \to C(u_c)$. Choosing $w = u_n$ in \eqref{eqn:26} and $w = u_c$ in \eqref{eqn:27}, we obtain that
\begin{align*}
A(u_n) + \lambda_n \normLp{u_n}{2}^2 - \gamma B(u_n) - a C(u_n)
\to A(u_c) + \lambda_c \normLp{u_c}{2}^2 - \gamma B(u_c) - a C(u_c).
\end{align*}
This implies that $\normLp{u_n}{2} \to \normLp{u_c}{2}$. Thus, we conclude that $u_n \to u_c$ strongly in $\Hsr{1}$.
\end{proof}

\begin{proof}[Proof of \cref{theorem:1} in the critical case $p=6$ for $\gamma^+(c)$.]
Since $\gamma^+(c) <0$, the fact that it is reached is a direct consequence of \cref{lemma:9},  \cref{lemma:18} and of the property, which is established in \cref{lemma:28}(iii) to come, that the map $c \mapsto \gamma^+(c)$ is non-increasing. The rest of the proof is identical to the one in the case $p \in (\frac{10}{3},6)$.
\end{proof}
In the rest of this subsection, we shall prove \cref{theorem:1} in the critical case $p=6$ for $\gamma^-(c)$.

\begin{lemma}\label{lemma:19}
Let $c \in (0,c_1)$. If
\begin{align}\label{key-bound}
\gamma^-(c) < \gamma^+(c)  + \dfrac{1}{3 \sqrt{a K_{GN}}}
\end{align}
then there exists a $u_c \in \Lambda^{-}_{r}(c)$ with $F(u_c) = \gamma^-(c)$ which  is a radial solution to \eqref{eqn:Laplace} for some $\lambda_c > 0$ with $\normLp{u_c}{2}^2 = c$.
\end{lemma}

\begin{proof}
By \cref{lemma:9} there exists a Palais-Smale sequence $(u_n) \subset \Lambda^-(c)$ for $F$ restricted to $S(c)$ at the level $\gamma^-(c)$. If \eqref{key-bound} holds then necessarily \eqref{eqn:36}, with $m = \gamma^-(c)$ holds, and \eqref{eqn:32}  cannot holds. We deduce from \cref{lemma:18} that $u_n \to u_c$ strongly in $H^1(\R^N)$ and the conclusions follow.
\end{proof}

Now we shall show that
\begin{lemma} \label{lemma:20}
For any $c \in (0,c_1)$, we have that
\begin{align*}
\gamma^-(c) < \gamma^+(c)  + \dfrac{1}{3 \sqrt{a K_{GN}}}.
\end{align*}
\end{lemma}
As already indicated our proof is inspired by \cite[Lemma 3.1]{Wei-Wu2021}.
Let $u_{\varepsilon}$ be an extremal function for the Sobolev inequality in $\Rn$ defined by
\begin{align}\label{Def-extremal}
u_{\varepsilon}(x) := \dfrac{[N(N-2)\varepsilon^2]^{\frac{N-2}{4}}}{[\varepsilon^2+|x|^2]^{\frac{N-2}{2}}}, \quad \varepsilon > 0, \quad x\in\Rn.
\end{align}
Let $\xi \in C_0^{\infty}(\R^N)$ be a radial non-increasing cut-off function with $\xi \equiv 1$ in $B_1$, $\xi \equiv 0$ in $\R^N \backslash B_2$.
Setting $U_{\varepsilon}(x) = \xi(x) u_{\varepsilon}(x)$ we recall the following result, see \cite[Lemma 7.1]{JeanjeanLe2020}.

\begin{lemma}\label{estimates-U}
Denoting $\omega$ the area of the unit sphere in $\Rb$, we have
\begin{enumerate}[label=(\roman*)]
\item \label{point:estimates-U:1} \begin{align*}
\normLp{\diff U_{\varepsilon}}{2}^2 = \sqrt{\dfrac{1}{K_{GN}}} + O(\varepsilon) \quad \mbox{and} \quad \normLp{U_{\varepsilon} }{2^*}^{2^*} = \sqrt{\dfrac{1}{K_{GN}}} + O(\varepsilon^3).
\end{align*}
\item \label{point:estimates-U:3} For some positive constant $K >0$,
\begin{align*}
\normLp{U_{\varepsilon} }{q}^{q} =
\begin{cases}
K \varepsilon^{3-\frac{q}{2}} + o(\varepsilon^{3-\frac{q}{2}}) \quad &\mbox{if } q \in (3,6), \\
\omega \var^{\frac{3}{2}} |\log \var| + O(\var^{\frac{3}{2}}) &\mbox{if } q = 3, \\
\omega \Big(\int_0^2 \frac{\xi^q(r)}{r^{q-2}} dr \Big) \var^{\frac{q}{2}} + o(\var^{\frac{q}{2}}) &\mbox{if } q \in [1,3).
\end{cases}
\end{align*}
\end{enumerate}
\end{lemma}

In the rest of the subsection we assume that $c \in (0,c_1)$ is arbitrary but fixed. Let $u_c^+$ be as provided by \cref{theorem:1}. We recall that
$u_c^+ \in \slp$  satisfies $F(u_c^+) = \gamma^+(c)$ and is a bounded continuous positive Schwarz symmetric function.

\begin{lemma} \label{lemma:21}
For any $1 \leq p, q < \infty$, it holds that
\begin{align*}
\intrn |u_c^+(x)|^p |U_{\varepsilon}(x)|^q dx \sim \intrn |U_{\varepsilon}(x)|^q dx.
\end{align*}
\end{lemma}
\begin{proof}
On one hand, since $u_c^+$ is bounded, we have that
\begin{align*}
\intrn |u_c^+(x)|^p |U_{\varepsilon}(x)|^q dx \leq \normLp{u_c^+ }{\infty}^{p} \intrn |U_{\varepsilon}(x)|^q dx.
\end{align*}
On the other hand, since $u_c^+ >0$ on $\R^3$ is continuous and the function $U_{\varepsilon}$ is compactly supported in $B_2$, we have that
\begin{align*}
\intrn |u_c^+(x)|^p |U_{\varepsilon}(x)|^q dx & =  \int_{B_2} |u_c^+(x)|^p |U_{\varepsilon}(x)|^q dx
\geq \min_{x \in B_2} |u_c^+(x)|^p \int_{B_2}|U_{\varepsilon}(x)|^q dx  = \min_{x \in B_2} |u_c^+(x)|^p \intrn|U_{\varepsilon}(x)|^q dx.
\end{align*}
The lemma is proved.
\end{proof}

For any $\varepsilon >0$ and any $t>0$, we have
\begin{align}\label{LA}
A(u_c^+ + t U_{\varepsilon}) = \normLp{\diff (u_c^+ + t U_{\varepsilon})}{2}^2 = A(u_c^+) + 2  \intrn \diff u_c^+(x) \cdot \diff (t U_{\varepsilon}(x)) \, dx  +    A(tU_{\varepsilon})
\end{align}
\begin{equation}\label{norm-L2}
\normLp{ u_c^+ + t U_{\varepsilon}}{2}^2 = c +  2 \intrn  u_c^+(x)  (tU_{\varepsilon}(x)) dx  +    \normLp{t U_{\varepsilon}}{2}^2.
\end{equation}
Using that, for all $a,b \geq 0$, $(a+b)^6 \geq a^6 + 6a^5b + 6ab^5 + b^6$, and that both $u_c^+ \in H^1(\R^N)$ and $U_{\varepsilon}$ are non negative,  we readily derive that
\begin{align}\label{LC}
C(u_c^+ + t U_{\varepsilon}) = \normLp{u_c^+ + t U_{\varepsilon} }{6}^{6} \geq C(u_c^+) + C(t U_{\varepsilon}) + 6 \intrn  (u_c^+(x))^5  (tU_{\varepsilon}(x)) dx + 6 \intrn  u_c^+(x)  (tU_{\varepsilon}(x))^5 dx .
\end{align}
Also, still using that $u_c^+ \in H^1(\R^N)$ and $U_{\varepsilon}$ are non negative, we get by direct calculations that
\begin{align}\label{LB}
\begin{split}
B(u_c^+ + t U_{\varepsilon})
& = \intrb\intrb \dfrac{\abs{u_c^+(x) + t U_{\varepsilon}(x)}^2 \abs{u_c^+(y) + t U_{\varepsilon}(y)}^2}{\abs{x-y}} dxdy\\
& \geq B(u_c^+) +  B(t U_{\varepsilon}) + 4 \intrb\intrb \dfrac{\abs{u_c^+(x)}^2 u_c^+(y) (t U_{\varepsilon}(y))}{\abs{x-y}} dxdy.
\end{split}
\end{align}
Finally, since $u_c^+$ is solution of the following equation
\begin{align*}
- \laplace u + \lambda_c^+ u - \gamma (\abs{x}^{-1} * \abs{u}^2) u  - a \abs{u}^{p-2}u = 0 \quad \text{in } \Rb
\end{align*}
for a $\lambda_c^+ >0$, we have that
\begin{align}\label{LE}
\begin{split}
-\lambda_c^+ \intrn  u_c^+(x)  (tU_{\varepsilon})(x) dx
& = \intrn \diff u_c^+(x) \diff (tU_{\varepsilon})(x) dx \\
& - \gamma \intrb\intrb \dfrac{\abs{u_c^+(x)}^2 u_c^+(y) (t U_{\varepsilon}(y))}{\abs{x-y}} dxdy - a  \intrn (u_c^+(x))^5  (tU_{\varepsilon}(x)) dx.
\end{split}
\end{align}
Now, we define for $t>0$, $w_{\varepsilon,t}= u_c^+ + t U_{\varepsilon}$ and $\overline{w}_{\varepsilon, t}(x)= \sqrt{\theta}w_{\varepsilon, t}(\theta x)$ with $\theta^2 = \dfrac{1}{c} \normLp{ w_{\varepsilon, t}}{2}^2$. The proof of \cref{lemma:20} will follow directly from the three lemmas below.

\begin{lemma} \label{lemma:31}
It holds that
\begin{equation*}
\gamma^-(c) \leq \sup_{t \geq 0} F(\overline{w}_{\varepsilon, t})
\end{equation*}
for $\varepsilon >0$ sufficiently small.
\end{lemma}
\begin{lemma} \label{lemma:32}
There exist a $\varepsilon_0 >0$ and $0 < t_0 < t_1 <\infty$ such that
\begin{equation*}
F(\wbar) < \gamma^+(c) + \dfrac{1}{6 \sqrt{a K_{GN}}}
\end{equation*}
for $t \not\in [t_0, t_1]$ and any $\varepsilon \in (0, \varepsilon_0]$.
\end{lemma}
\begin{lemma} \label{lemma:33}
It holds that
\begin{equation*}
\max_{t \in [t_0, t_1]} F(\wbar) < \gamma^+(c) + \dfrac{1}{3 \sqrt{a K_{GN}}},
\end{equation*}
for any $\varepsilon \in (0, \varepsilon_0]$ where $\varepsilon_0$ and $t_0, t_1$ are provided by \cref{lemma:32}.
\end{lemma}

\begin{proof}[Proof of \cref{lemma:31}]
By direct calculation we get
\begin{equation}\label{L1}
A(\overline{w}_{\varepsilon, t}) = A (w_{\varepsilon,t}), \quad C(\overline{w}_{\varepsilon, t})= C(w_{\varepsilon,t}),
\end{equation}
\begin{equation}\label{L11}
\normLp{ \overline{w}_{\varepsilon, t}}{2}^2 = \theta^{-2} \normLp{ w_{\varepsilon, t}}{2}^2, \quad  B(\overline{w}_{\varepsilon, t})= \theta^{-3}B(w_{\varepsilon,t}).
\end{equation}
Since $\theta^2 = \dfrac{1}{c} \normLp{ w_{\varepsilon, t}}{2}^2$, we have that $\overline{w}_{\varepsilon, t} \in S(c)$. By \cref{lemma:6} there exists $s_{\varepsilon, t}^- >0$ such that $(\overline{w}_{\varepsilon, t})^{s_{\varepsilon, t}^-}  \in \slm.$
We claim that $s_{\varepsilon, t}^- \to 0$ as $t \to + \infty$ uniformly for $\varepsilon >0$ sufficiently small.
Indeed, we have
$$ A((\overline{w}_{\varepsilon, t})^{s_{\varepsilon, t}^-}) = \frac{\gamma}{4}B((\overline{w}_{\varepsilon, t})^{s_{\varepsilon, t}^-}) + a C((\overline{w}_{\varepsilon, t})^{s_{\varepsilon, t}^-})$$
or equivalently
\begin{equation*}
(s_{\varepsilon, t}^-) A(\overline{w}_{\varepsilon, t})
= \frac{\gamma}{4} B(\overline{w}_{\varepsilon, t}) + a (s_{\varepsilon, t}^-)^5 C(\overline{w}_{\varepsilon, t}).
\end{equation*}
This implies that
\begin{align} \label{T1}
A(\overline{w}_{\varepsilon, t}) \geq a (s_{\varepsilon, t}^-)^4 C(\overline{w}_{\varepsilon, t}).
\end{align}
In view of \eqref{LA}, \eqref{L1}, \cref{estimates-U}\ref{point:estimates-U:1} and using H\"{o}lder's inequality, we have
\begin{align} \label{T2}
\begin{split}
A(\overline{w}_{\varepsilon, t}) &= A({w}_{\varepsilon, t}) = A(u_c^+) + 2  \intrn \diff u_c^+(x) \cdot \diff (t U_{\varepsilon}(x)) dx + A(t U_{\varepsilon})\\
&\leq A(u_c^+) + 2 t  \normLp{\diff u_c^+}{2} \normLp{\diff U_{\varepsilon}}{2} + t^2 A(U_{\varepsilon})
\to A(u_c^+) + 2 J \sqrt{A(u_c^+)} \, t   + J t^2 \mbox{ as } \varepsilon \to 0.
\end{split}
\end{align}
In view of \eqref{LC}, \eqref{L1}  and  \cref{estimates-U}\ref{point:estimates-U:1}, we also have
\begin{align}\label{T3}
C(\overline{w}_{\varepsilon, t}) = C({w}_{\varepsilon, t}) \geq C(t U_{\varepsilon} ) = t^6 C(U_{\varepsilon} ) \to L t^6 \mbox{ as } \varepsilon \to 0.
\end{align}
Combining \eqref{T1}-\eqref{T3}, we obtain that, for $\varepsilon >0$ sufficiently small
\begin{align*}
A(u_c^+) + J \sqrt{A(u_c^+)} \, t   + J t^2 \geq a (s_{\varepsilon, t}^-)^4 L t^6,
\end{align*}
which implies the claim. Since $\overline{w}_{\varepsilon, 0} = {w}_{\varepsilon, 0} = u_c^+$ and $u_c^+ \in \slp$  we obtain, see \cref{lemma:6}, that $s_{\varepsilon, 0}^- >1$. Still by \cref{lemma:6}, the map $t \mapsto s_{\varepsilon, t}^-$ is continuous which implies that there exists $t_{\varepsilon} >0$ such that $s_{\varepsilon, t_{\varepsilon}}^- =1$. It follows that $\overline{w}_{\varepsilon, t_{\varepsilon}} \in \slm$ and thus
\begin{equation*}
\sup_{t \geq 0}F(\overline{w}_{\varepsilon, t}) \geq F(\overline{w}_{\varepsilon, t_{\varepsilon}}) \geq \gamma^-(c).
\end{equation*}
The lemma is proved.
\end{proof}

\begin{proof}[Proof of \cref{lemma:32}]
In view of \eqref{L1} and \eqref{L11}, we have that
\begin{align*}
F(\overline{w}_{\varepsilon, t})
=  \frac{1}{2}  A (w_{\varepsilon,t}) - \frac{\gamma}{4} \theta^{-3}B( w_{\varepsilon,t}) - \frac{a}{6} C (w_{\varepsilon,t}).
\end{align*}
Hence, by  \eqref{LA}, \eqref{LC} and \eqref{LB}, we get that
\begin{align*}
F(\overline{w}_{\varepsilon, t}) &\leq \frac{1}{2} \Big[ A(u_c^+) + 2  \intrn \diff u_c^+(x) \cdot \diff (t U_{\varepsilon}(x)) \, dx  +    A(tU_{\varepsilon}) \Big] - \frac{\gamma}{4} \theta^{-3} B(u_c^+) - \frac{a}{6} \Big[C(u_c^+)  + C(t U_{\varepsilon})\Big] \\
&= F(u_c^+) + \dfrac{\gamma}{4}(1 - \theta^{-3}) B(u_c^+) + \intrn \diff u_c^+(x) \cdot \diff (t U_{\varepsilon}(x)) \, dx +  \frac{1}{2} A(tU_{\varepsilon}) - \frac{a}{6} C(t U_{\varepsilon}) \\
& \leq F(u_c^+) + \dfrac{\gamma}{4}(1 - \theta^{-3}) B(u_c^+) + t  \normLp{\diff u_c^+}{2} \normLp{ \diff U_{\varepsilon}}{2} +  \frac{1}{2} t^2 A(U_{\varepsilon}) - \frac{a}{6} t^6 C( U_{\varepsilon}) := I(t).
\end{align*}
By \cref{estimates-U}\ref{point:estimates-U:1}, we see that, uniformly for $\varepsilon >0$ small,  $I(t) \to -\infty$ as $t \to + \infty$ and $I(t) \to F(u_c^+)$ as $t \to 0$ due to $\theta \to 1$. Hence, there exists $\varepsilon_0 >0$ and $ 0< t_0 < t_1 < \infty$ such that
\begin{align*}
\begin{split}
F(\overline{w}_{\varepsilon, t})  < \gamma^+(c) + \dfrac{1}{6 \sqrt{a K_{GN}}}
\end{split}
\end{align*}
for $t \not\in [t_0, t_1]$ and $\varepsilon \in (0, \varepsilon_0]$. The lemma is proved.
\end{proof}

\begin{proof}[Proof of \cref{lemma:33}]	We assume throughout the proof that $ t \in [t_0, t_1]$.
By using \eqref{LB}, we can write,
\begin{align}\label{L5}
\begin{split}
F(\overline{w}_{\varepsilon, t})
& =  \frac{1}{2}  A (w_{\varepsilon,t}) - \frac{\gamma}{4} \theta^{-3}B( w_{\varepsilon,t}) - \frac{a}{6} C (w_{\varepsilon,t})  \\
& \leq \frac{1}{2}  A (w_{\varepsilon,t}) - \frac{\gamma}{4} \theta^{-3}\Big[B(u_c^+)  + B(t U_{\varepsilon}) + 4 \intrb\intrb \dfrac{\abs{u_c^+(x)}^2 u_c^+(y) (t U_{\varepsilon}(y))}{\abs{x-y}} dxdy \Big] - \frac{a}{6} C (w_{\varepsilon,t})  \\
& = I_1 + I_2,
\end{split}
\end{align}
\begin{align*}
I_1:= \frac{1}{2}  A (w_{\varepsilon,t})  - \frac{\gamma}{4} \Big[B(u_c^+)  + B(t U_{\varepsilon}) + 4 \intrb\intrb \dfrac{\abs{u_c^+(x)}^2 u_c^+(y) (t U_{\varepsilon}(y))}{\abs{x-y}} dxdy \Big] - \frac{a}{6} C (w_{\varepsilon,t}),
\end{align*}
\begin{align}\label{Def-I2}
I_2:= \dfrac{\gamma}{4}(1 - \theta^{-3}) \Big[B(u_c^+)  + B(t U_{\varepsilon}) + 4 \intrb\intrb \dfrac{\abs{u_c^+(x)}^2 u_c^+(y) (t U_{\varepsilon}(y))}{\abs{x-y}} dxdy \Big].
\end{align}
In view of \eqref{LA}, \eqref{LC} and using crucially \eqref{LE}, we have
\begin{align}\label{I_1}
\begin{split}
& \leq   \frac{1}{2} \Big[ A(u_c^+) + 2  \intrn \diff u_c^+(x) \cdot \diff (t U_{\varepsilon}(x)) \, dx  +    A(tU_{\varepsilon}) \Big] \\
&  - \frac{\gamma}{4}  \Big[B(u_c^+)  + B(t U_{\varepsilon}) + 4 \intrb\intrb \dfrac{\abs{u_c^+(x)}^2 u_c^+(y) (t U_{\varepsilon}(y))}{\abs{x-y}} dxdy \Big]   \\
& -   \frac{a}{6} \Big[  C(u_c^+) + C(t U_{\varepsilon}) + 6 \intrn  (u_c^+(x))^5  (tU_{\varepsilon}(x)) dx + 6 \intrn  u_c^+(x)  (tU_{\varepsilon}(x))^5 dx    \Big]  \\
& = F(u_c^+) + F(t U_{\varepsilon}) -\lambda_c^+ \intrn  u_c^+(x) (tU_{\varepsilon}(x)) dx  - a \intrn  u_c^+(x)  (tU_{\varepsilon}(x))^5 dx.
\end{split}
\end{align}
Now, we shall evaluate $I_2$. By \eqref{norm-L2} and \cref{estimates-U}\ref{point:estimates-U:3}, we get that
\begin{align} \label{E-theta}
\begin{split}
\theta^2 &= \dfrac{\normLp{ w_{\varepsilon, t}}{2}^2}{c} = 1 + \frac{2}{c} \intrn  u_c^+(x)  (tU_{\varepsilon}(x)) dx  + \frac{t^2}{c} \normLp{ U_{\varepsilon}}{2}^2 \\
&= 1 + \frac{2}{c} \intrn  u_c^+(x)  (tU_{\varepsilon}(x)) dx  + \frac{t^2}{c} \[\omega \Big(\int_0^2 \xi(r) dr \Big) \var + O(\var^2) \]
= 1 + \frac{2}{c} \intrn  u_c^+(x)  (tU_{\varepsilon}(x)) dx  + O(\varepsilon).
\end{split}
\end{align}
Note that, by \cref{estimates-U}\ref{point:estimates-U:3} and \cref{lemma:21},
\begin{equation}\label{size}
\intrn  u_c^+(x)  (tU_{\varepsilon}(x)) dx  \sim \normLp{ \diff U_{\varepsilon}}{1} = O(\varepsilon^{\frac{1}{2}}).
\end{equation}
Observing that the Taylor expansion of $(1+x)^{- \frac{3}{2}}$ around $x=0$ is given by $(1+x)^{-\frac{3}{2}} = 1 - \frac{3}{2}x + O(x^2)$, we get, in view of \eqref{E-theta} and \eqref{size}, that
\begin{align} \label{evaluate-I2-1}
\begin{split}
1 -\theta^{-3} &= 1- (\theta^2)^{-\frac{3}{2}}
= 1 - \[1 + \frac{2}{c} \intrn  u_c^+(x)  (tU_{\varepsilon}(x))  + O(\varepsilon) \]^{- \frac{3}{2}} \\
& = 1 - \[1 - \frac{3}{c} \intrn  u_c^+(x)  (tU_{\varepsilon}(x))  + O(\varepsilon) \]
= \frac{3}{c} \intrn  u_c^+(x)  (tU_{\varepsilon}(x))  + O(\varepsilon).
\end{split}
\end{align}
Concerning the term $B(tU_{\varepsilon})$, in view of \eqref {eqn:2.1} and \cref{estimates-U}\ref{point:estimates-U:3}, we have
\begin{align} \label{evaluate-I2-2}
\begin{split}
B(tU_{\varepsilon}) = t^4 B(U_{\varepsilon}) \leq  t^4 K_1 \normLp{U_{\varepsilon}}{\frac{12}{5}}^4 =
t^4 K_1 \Big(\normLp{U_{\varepsilon}}{\frac{12}{5}}^\frac{12}{5}\Big)^{\frac{10}{6}} =  t^4 K_1 \Big(K_2 \varepsilon^{\frac{6}{5}} + o(\varepsilon^\frac{6}{5})\Big)^{\frac{10}{6}} = O(\varepsilon^2).
\end{split}
\end{align}
Also, using the Hardy-Littlewood-Sobolev inequality \eqref{ineq:HLS}, \cref{lemma:21} and \cref{estimates-U}\ref{point:estimates-U:3} we have
\begin{align}\label{needed}
\intrb\intrb \dfrac{\abs{u_c^+(x)}^2 u_c^+(y)  (tU_{\varepsilon}(y))}{\abs{x-y}} dxdy &\leq K_2 \normLp{u_c^+}{\frac{12}{5}}^2 \normLp{u_c^+ U_{\varepsilon}}{\frac{6}{5}}
\leq K_3  \normLp{U_{\varepsilon}}{\frac{6}{5}} = O(\varepsilon^{\frac{3}{5}}).
\end{align}
From \eqref{evaluate-I2-2} and  \eqref{needed} we deduce that
\begin{equation}\label{allneeded}
B(u_c^+)  + B(t U_{\varepsilon}) + 4 \intrb\intrb \dfrac{\abs{u_c^+(x)}^2 u_c^+(y) (t U_{\varepsilon}(y))}{\abs{x-y}} dxdy = B(u_c^+) + O (\varepsilon^{\frac{3}{5}}).
\end{equation}
Taking into account, see \eqref{key-point}, that
\begin{align*}
c \lambda_c^+ = \frac{3\gamma}{4}B(u_c^+)
\end{align*}
we obtain, combining \eqref{Def-I2}, \eqref{size}, \eqref{evaluate-I2-1} and \eqref{allneeded},  the following evaluation of $I_2$
\begin{align} \label{I_2}
\begin{split}
I_2 \leq \dfrac{3 \gamma}{4c} B(u_c^+) \intrn  u_c^+(x)  (tU_{\varepsilon}(x)) dx + O(\varepsilon)
=   \lambda_c^+ \intrn  u_c^+(x)  (tU_{\varepsilon}(x)) dx   + O(\varepsilon).
\end{split}
\end{align}
At this point, in view of \eqref{L5}, \eqref{I_1} and \eqref{I_2} we deduce that
\begin{align}\label{L8}
\begin{split}
F(\overline{w}_{\varepsilon, t})
&\leq   F(u_c^+) + F(t U_{\varepsilon}) - a \intrn  u_c^+(x)  (tU_{\varepsilon}(x))^5 dx  + O(\varepsilon) \\
&\leq   F(u_c^+) + F(t U_{\varepsilon}) - a t_0^5 \intrn  u_c^+(x)  (U_{\varepsilon}(x))^5 dx  + O(\varepsilon).
\end{split}
\end{align}
In view of \cref{estimates-U}\ref{point:estimates-U:1}, a direct calculation shows that
\begin{align}\label{againuseful}
\begin{split}
\max_{t \in [t_0, t_1]} F (t U_{\varepsilon}) &= \max_{t \in [t_0, t_1]} \[\dfrac{1}{2}A(t U_{\varepsilon}) - \dfrac{\gamma}{4} B(t U_{\varepsilon}) - \dfrac{a}{6} C(t U_{\varepsilon})\] \\
&\leq \max_{t \in [t_0, t_1]} \[\dfrac{1}{2}A(t U_{\varepsilon}) - \dfrac{a}{6} C(t U_{\varepsilon})\]
\leq \max_{t >0} \[\dfrac{1}{2}A(t U_{\varepsilon}) - \dfrac{a}{6} C(t U_{\varepsilon})\]
%&= \max_{t >0} \[\dfrac{1}{2}t^2A(U_{\varepsilon}) - \dfrac{a}{6} t^6 C(U_{\varepsilon})\]
= \dfrac{1}{3 \sqrt{a K_{GN}}} + O(\varepsilon).
\end{split}
\end{align}
In view of \eqref{L8} and \eqref{againuseful}, by \cref{estimates-U}\ref{point:estimates-U:3} and \cref{lemma:21}, we conclude by observing that
\begin{align*}
- a t_0^5 \intrn  u_c^+(x)  (U_{\varepsilon}(x))^5 dx  \sim - \normLp{U_{\varepsilon}(x)}{5}^5 = - K \varepsilon^{\frac{1}{2}} + o(\varepsilon^{\frac{1}{2}}).
\end{align*}
\end{proof}
\begin{proof}[Proof of \cref{theorem:1} in the critical case $p=6$ for $\gamma^-(c)$.]
%Since $\gamma^+(c) <0$, the fact that it is reached is a direct consequence of \cref{lemma:9},  \cref{lemma:18} and of the property, which is established in \cref{lemma:28}(iii) to come, that the map $c \mapsto \gamma^+(c)$ is non-increasing.
We conclude that $\gamma^-(c)$ is reached by combining \cref{lemma:9}, \cref{lemma:19} and \cref{lemma:20}. The rest of the proof is identical to the one in the case $p \in (\frac{10}{3},6)$.
\end{proof}

\subsection{The compactness of any minimizing sequence associated to $\gamma^+(c)$ for $ p \in (\frac{10}{3}, 6]$}\label{Subsection3-4}

In this subsection we give the proof of \cref{th-minimization}. For short we introduce the following notations,
\begin{align}\label{eq-notations}
M:= \dfrac{p}{a \sigma (\sigma - 1) K_{GN}}, \quad N:= \dfrac{4(\sigma - 2)}{\gamma (\sigma - 1) K_{H}}, \quad k_0:= N^{-2}, \quad \mbox{and} \quad k_1 := k_0 c_1^3.
\end{align}
Note that
\begin{align*}
c_1 = N^{\frac{3p-10}{4(p-3)}} M^{\frac{1}{2(p-3)}}.
\end{align*}

\begin{lemma} \label{lemma:27}
Let  $p \in (\frac{10}{3},6]$ and $c \in (0, c_1)$.
\begin{enumerate}[label=(\roman*), ref=\roman*]
\item \label{point:27i} If $u \in \slp$ then we have
\begin{align}
A(u) < k_0 c^3. \label{eqn:43}
\end{align}
\item \label{point:27ii}  $\slp \subset V(c)$ and
\begin{align*}
\gamma^+(c) = \inf_{u \in \slp}F(u) = \inf_{u \in \overline{V(c)}}F(u).
\end{align*}
\item \label{point:27iii} If $u_c$ is a minimizer for the minimization problem
\begin{align*}
\inf_{u \in \overline{V(c)}}F(u)
\end{align*}
then $u_c \in V(c)$ and $\gamma^+(c)$ is reached.
\end{enumerate}
\end{lemma}
\begin{proof}
\ref{point:27i}) Since $u \in \slp$,
\begin{align*}
A(u) = \dfrac{\gamma}{4} B(u) + \dfrac{a \sigma}{p} C(u) \quad \mbox{and} \quad A(u) > \dfrac{a \sigma (\sigma - 1)}{p} C(u).
\end{align*}
Using \cref{lemma:1}\eqref{point:1i}, we have
\begin{align*}
\dfrac{\sigma - 2}{\sigma - 1} A(u) < \dfrac{\gamma}{4} B(u) \leq \dfrac{\gamma}{4} K_{H}  \sqrt{A(u)} c^{\frac{3}{2}},
\end{align*}
which implies that
\begin{align}
A(u) < \[\dfrac{\gamma (\sigma - 1)  K_{H}}{4 (\sigma - 2)}\]^2  c^3 = N^{-2} c^3 = k_0 c^3 < k_0 c_1^3 = k_1. \label{eqn:40}
\end{align}
Hence, the point \eqref{point:27i} holds.

\ref{point:27ii}) By \eqref{eqn:40}, we obtain that $\slp \subset V(c)$ and hence
\begin{align*}
\inf_{u \in \slp}F(u) \geq \inf_{u \in \overline{V(c)}}F(u).
\end{align*}
To prove the point \eqref{point:27ii}, it is sufficient to show that
\begin{align}
\inf_{u \in \slp}F(u) \leq \inf_{u \in \overline{V(c)}}F(u). \label{eqn:42}
\end{align}
Firstly, we claim that $\slm \cap \overline{V(c)} = \emptyset$. Indeed, let $v \in \slm$. Taking into account that
\begin{align*}
A(v) < \dfrac{a \sigma (\sigma - 1)}{p} C(v),
\end{align*}
and using \cref{lemma:1}\eqref{point:1ii}, we obtain that
\begin{align*}
A(v) < \dfrac{a \sigma (\sigma - 1) }{p} K_{GN} c^{\frac{6-p}{4}} [A(v)]^{\frac{\sigma}{2}} = M^{-1} c^{\frac{6-p}{4}} [A(v)]^{\frac{\sigma}{2}}.
\end{align*}
This implies that
\begin{align*}
A(v) > M^{\frac{4}{3p-10}} c^{-\frac{6-p}{3p-10}}.
\end{align*}
By direct computations, we can check that
\begin{align*}
N^{-2} c_1^3 = M^{\frac{4}{3p-10}} c_1^{-\frac{6-p}{3p-10}},
\end{align*}
which implies that for all $0 < c < c_1$,
\begin{align}
k_1 = N^{-2} c_1^3 = M^{\frac{4}{3p-10}} c_1^{-\frac{6-p}{3p-10}} < M^{\frac{4}{3p-10}} c^{-\frac{6-p}{3p-10}} < A(v). \label{eqn:41}
\end{align}
Therefore, the claim holds.
Next, let $u \in S(c)$. Since the mapping $t \mapsto A(u^t)$ is continuous increasing, there exists a unique $t_u^1 > 0$ such that $A(u^{t_u^1}) = k_1$. By \cref{lemma:6} and \eqref{eqn:40}, \eqref{eqn:41}, we have
\begin{align*}
A(u^{s_u^+}) < A(u^{t_u^1}) < A(u^{s_u^-}),
\end{align*}
which implies that
\begin{align*}
s_u^+ < t_u^1 < s_u^-.
\end{align*}
Since $g_u'(t) > 0$ for all $t \in (s_u^+, s_u^-)$, we get that $g_u'(t) > 0$ for all $t \in (s_u^+, t_u^1]$ and hence
\begin{align}
F(u^{s_u^+}) = g_u(s_u^+) < g_u(t) = F(u^t) \qquad \forall t \in (s_u^+, t_u^1]. \label{eqn:52}
\end{align}
Since $\spu$ is the unique local minimum point for $g_u$ on $(0, s_u^-)$, we have that $F(u^{s_u^+}) \leq F(u^t)$ for all $t \in (0, t_u^1]$. Therefore, we obtain that
\begin{align*}
F(u^{s_u^+}) = \min\{F(u^t) | 0 < t \leq t_u^1 \} = \min\{F(u^t) | t\in \R, A(u^t) \leq k_1 \}.
\end{align*}
In particular, if $u \in \overline{V(c)}$ we have
\begin{align*}
F(u^{s_u^+}) = \min\{F(u^t) | t\in \R, A(u^t) \leq k_1 \} = \min\{F(u) | u \in \overline{V(c)} \} \leq F(u).
\end{align*}
This implies \eqref{eqn:42} and the point \eqref{point:27ii} is proved.

\ref{point:27iii}) If we assume that $u_c \in \partial V(c)$, namely $A(u_c) = k_1$ and
\begin{align*}
F(u_c) = \min\{F(u) | u \in \overline{V(c)} \} = \min\{F(u^t) | t\in \R, A(u^t) \leq k_1 \},
\end{align*}
and we have a contradiction with \eqref{eqn:52}. Thus, we have $u_c \in V(c)$. Now, since the minimizer $u_c$ lies in the open (with respect to $S(c)$) set $V(c)$, we deduce from \cref{lemma:2.3} that $u_c \in \slc$. By $\slm \cap \overline{V(c)} = \emptyset$ and $\slz = \emptyset$, we conclude that $u_c \in \slp$ and thus $\gamma^+(c)$ is reached.
\end{proof}

\begin{lemma} \label{lemma:28}
It holds that
\begin{enumerate}[label=(\roman*), ref = \roman*]
\item\label{point:28i} $\gamma^+(c) < 0$, $\forall c \in (0, c_1)$.
\item\label{point:28ii} $c \in (0,c_1) \mapsto \gamma^+(c)$ is a continuous mapping.
\item\label{point:28iii} Let $c \in (0,c_1)$, for all $\alpha \in (0, c)$, we have  $\gamma^+(c) \leq \gamma^+(\alpha) + \gamma^+(c - \alpha)$ and if $\gamma^+(\alpha)$ or $\gamma^+(c - \alpha)$ is reached then the inequality is strict.
\end{enumerate}
\end{lemma}
\begin{proof}
Point \eqref{point:28i} follows from \cref{lemma:16}. To prove \eqref{point:28ii}, let $c \in (0, c_1)$ be arbitrary and $(c_n) \subset (0, c_1)$ be such that $c_n \to c$. From the definition of $\gamma^+(c_n)$, for any $\varepsilon > 0$, there exists $u_n \in \Lambda^+(c_n)$ such that
\begin{align}
F(u_n) \leq \gamma^+(c_n) + \varepsilon. \label{eqn:44}
\end{align}
By \eqref{eqn:43}, we have $A(u_n) < k_0 c_n^3$. We set $y_n := \sqrt{\dfrac{c}{c_n}}\cdot u_n$. Hence, we have $y_n \in S(c)$ and
\begin{align*}
A(y_n) = \frac{c}{c_n} A(u_n) < \frac{c}{c_n} k_0 c_n^3 = k_0 c_n^2 c < k_0 c_1^3 = k_1.
\end{align*}
This implies that $y_n \in \overline{V(c)}$. Taking into account that $\dfrac{c}{c_n} \to 1$, we have
\begin{align}
\gamma^+(c) \leq F(y_n) = F(u_n) + o_n(1). \label{eqn:45}
\end{align}
Combining \eqref{eqn:44} and \eqref{eqn:45}, we get
\begin{align*}
\gamma^+(c) \leq \gamma^+(c_n) + \varepsilon + o_n(1).
\end{align*}
Reversing the argument we obtain similarly that
\begin{align*}
\gamma^+(c_n) \leq \gamma^+(c) + \varepsilon + o_n(1).
\end{align*}
Therefore, since $\varepsilon > 0$ is arbitrary, we deduce that $\gamma^+(c_n) \to \gamma^+(c)$. The point \eqref{point:28ii} follows.

\ref{point:28iii}) Note that, fixed $\alpha \in (0,c)$, it is sufficient to prove that the following holds
\begin{align}\label{L6-4}
\forall \theta \in \(1, \frac{c}{\alpha}\] :  \gamma^+(\theta \alpha) \leq \theta \gamma^+(\alpha)
\end{align}
and that, if $\gamma^+(\alpha)$ is reached, the inequality is strict. Indeed, if \eqref{L6-4} holds then it follows directly that
\begin{align*}
\gamma^+(c)=\frac{c-\alpha}{c} \gamma^+(c) + \frac{\alpha}{c} \gamma^+(c)
=\frac{c-\alpha}{c} \gamma^+ \left( \frac{c}{c-\alpha}(c-\alpha) \right) + \frac{\alpha}{c} \gamma^+\left( \frac{c}{\alpha} \alpha \right)
\leq \gamma^+(c-\alpha) + \gamma^+(\alpha)
\end{align*}
with a strict inequality if $\gamma^+(\alpha)$ is reached. To prove that \eqref{L6-4} holds, note that for any $\varepsilon >0$ sufficiently small, there exist $u \in \Lambda^+(\alpha)$ such that
\begin{align}
F(u) \leq \gamma^+(\alpha) + \varepsilon. \label{eqn:46}
\end{align}
By \eqref{eqn:43}, we have $A(u) < k_0 \alpha^3$. Consider now $v:= \sqrt{\theta} u$, we have
\begin{align*}
\normLp{v}{2}=\theta \normLp{u}{2}, \quad A(v)=\theta A(u), \quad B(v)=\theta^2 B(u), \quad C(v) = \theta^3 C(u).
\end{align*}
Therefore, we obtain that $v \in S(\theta\alpha)$ and
\begin{align*}
A(v) = \theta A(u) < k_0 \theta \alpha^3 < k_0 (\theta \alpha)^3 \leq k_0 c^3 < k_1.
\end{align*}
Hence, $v \in \overline{V(\theta\alpha)}$ and we can write
\begin{align*}
\gamma^+(\theta\alpha) \leq F(v) &= \dfrac{1}{2} A(v) - \dfrac{\gamma}{4} B(v) - \dfrac{a}{p} C(v)
= \dfrac{1}{2} \theta A(u) - \dfrac{\gamma}{4} \theta^2 B(u) - \dfrac{a}{p} \theta^3 C(u) \\
&< \dfrac{1}{2} \theta A(u) - \dfrac{\gamma}{4} \theta B(u) - \dfrac{a}{p} \theta C(u)
= \theta F(u) \leq \theta (\gamma^+(\alpha) + \varepsilon).
\end{align*}
Since $\varepsilon > 0$ is arbitrary, we have that  $\gamma^+(\theta \alpha) \leq \theta \gamma^+(\alpha)$. If $\gamma^+(\alpha)$ is reached then we can let $\varepsilon = 0$ in \eqref{eqn:46} and thus the strict inequality follows.
\end{proof}

\begin{lemma} \label{lemma:30}
Let $(v_n) \subset \Hs{1}$ be such that $B(v_n) \to 0$ and $A(v_n) \leq k_1$. Then there exists a $b>0$ such that
\begin{align}
F(v_n) \geq b A(v_n) + o_n(1).
\end{align}
\end{lemma}
\begin{proof}
Indeed, using $B(v_n) \to 0$ and \cref{lemma:1}\eqref{point:1ii}, we have
\begin{align*}
F(v_n) &= \dfrac{1}{2} A(v_n) - \dfrac{a}{p} C(v_n) + o_n(1)
\geq \dfrac{1}{2} A(v_n) - \dfrac{a}{p} K_{GN} c^{\frac{6-p}{4}} [A(v_n)]^{\frac{\sigma}{2}} + o_n(1)
=b A(v_n) + o_n(1),
\end{align*}
\begin{align*}
b := \dfrac{1}{2} -  \limsup_{n \to \infty} \dfrac{a}{p} K_{GN} c^{\frac{6-p}{4}} [A(v_n)]^{\frac{\sigma}{2}-1} \geq \dfrac{1}{2} - \dfrac{a}{p} K_{GN} c_1^{\frac{6-p}{4}} k_1^{\frac{\sigma}{2}-1} = \dfrac{1}{2} - \dfrac{1}{\sigma(\sigma-1)}.
\end{align*}
Hence, $b > 0$ due to $\sigma > 2$. The lemma is proved.
\end{proof}

\begin{lemma} \label{lemma:29}
For any $c \in (0, c_1)$, any minimizing sequence $(u_n)$ for $F$ on $\overline{V(c)}$ is, up to translation, strongly convergent in $\Hs{1}$. In addition all minimizers lies in $V(c)$. In particular $\gamma^+(c)$ is reached.
\end{lemma}
\begin{proof}
Since  $(u_n) \subset \overline{V(c)}$, it is bounded in $\Ho$. Also, from $\gamma^+(c) <0$ we deduce from  \cref{lemma:30} that there exists a $\beta_0 > 0$ and a sequence $(y_n) \subset \Rb$ such that
\begin{align*}
\int_{B(y_n, R)} \abs{u_n}^2 dx \geq \beta_0 > 0, \qquad \mbox{for some } R > 0.
\end{align*}
This implies that
\begin{align*}
u_n(x - y_n) \weakto u_c \neq 0 \quad \mbox{in } \Hs{1}, \quad \mbox{for some} \quad u_c \in \Hs{1}.
\end{align*}
Our aim is to prove that $w_n(x) := u_n(x - y_n) - u_c(x) \to 0$ in $\Hs{1}$. Clearly
\begin{align*}
\normLp{u_n}{2}^2 &= \normLp{u_n(x-y_n)}{2}^2
= \normLp{u_n(x-y_n) - u_c(x)}{2}^2 + \normLp{u_c}{2}^2 + o_n(1)
=\normLp{w_n}{2}^2 + \normLp{u_c}{2}^2 + o_n(1).
\end{align*}
Thus, we have
\begin{align}
\normLp{w_n}{2}^2 = \normLp{u_n}{2}^2 -  \normLp{u_c}{2}^2 + o_n(1) = c - \normLp{u_c}{2}^2 + o_n(1). \label{eqn:47}
\end{align}
By the similar argument,
\begin{align}
A(w_n) = A(u_n) -  A(u_c) + o_n(1). \label{eqn:48}
\end{align}
More generally, taking into account that any term in $F$ fulfills the splitting properties of Brezis-Lieb (see [16] for terms $A$ and $C$; see \cite[Lemma 2.2]{ZHAO2008} or \cite[Proposition 3.1]{BellazziniSiciliano2011} for term $B$), we have
\begin{align*}
F(u_n - u_c) + F(u_c) =F(u_n) + o_n(1),
\end{align*}
and by the translational invariance, we obtain
\begin{align}
\begin{split}
F(u_n) =F(u_n(x-y_n)) =F(u_n(x-y_n) - u_c(x)) + F(u_c) + o_n(1)
=F(w_n) + F(u_c) + o_n(1).
\end{split} \label{eqn:49}
\end{align}
Now, we claim that
\begin{align}
\normLp{w_n}{2}^2 \to 0 \quad \text{as} \quad n \to \infty. \label{eqn:50}
\end{align}
In order to prove this, let us denote $\tilde{c}:= \normLp{u_c}{2}^2 > 0$. By \eqref{eqn:47}, if we show that $\tilde{c} = c$ then the claim follows. We assume by contradiction that $\tilde{c} < c$.
In view of \eqref{eqn:47} and \eqref{eqn:48}, for $n$ large enough, we have $\normLp{w_n}{2}^2 \leq c$ and $A(w_n) \leq A(u_n) \leq k_1$. Hence, we obtain that $w_n \in \overline{V(\normLp{w_n}{2}^2)}$ and $F(w_n) \geq \gamma^+\(\normLp{w_n}{2}^2\)$.
Recording that $F(u_n) \to \gamma^+(c)$, in view of \eqref{eqn:49}, we have
\begin{align*}
\gamma^+(c) = F(w_n) + F(u_c) \geq \gamma^+\(\normLp{w_n}{2}^2\) + F(u_c).
\end{align*}
Since the map $c \mapsto \gamma^+(c)$ is continuous (see \cref{lemma:28}\eqref{point:28ii}) and in view of \eqref{eqn:47}, we deduce that
\begin{align}
\gamma^+(c) \geq \gamma^+(c-\tilde{c}) + F(u_c). \label{eqn:51}
\end{align}
We also have that $u_c \in \overline{V(\tilde{c})}$ by the weak limit. This implies that $F(u_c) \geq \gamma^+(\tilde{c})$. If $F(u_c) > \gamma^+(\tilde{c})$, then it follows from \eqref{eqn:51} and \cref{lemma:28}\eqref{point:28iii} that
\begin{align*}
\gamma^+(c) > \gamma^+(c-\tilde{c}) + \gamma^+(\tilde{c}) \geq \gamma^+(c -\tilde{c} + \tilde{c}) = \gamma^+(c),
\end{align*}
which is impossible. Hence, we have $F(u_c) = \gamma^+(\tilde{c})$, namely $u_c$ is local minimizer on $\overline{V(\tilde{c})}$. So, we can using \cref{lemma:28}\eqref{point:28iii} with the strict inequality and we deduce from \eqref{eqn:51} that
\begin{align*}
\gamma^+(c) \geq \gamma^+(c-\tilde{c}) + F(u_c) = \gamma^+(c-\tilde{c}) + \gamma^+(\tilde{c}) > \gamma^+(c -\tilde{c} + \tilde{c}) = \gamma^+(c),
\end{align*}
which is impossible. Thus, the claim follows and $\normLp{u_c}{2}^2 = c$.\medskip

Let us now show that $A(w_n) \to 0$. This will complete the proof of the lemma.
In this aim first observe that since $(w_n)$ is a bounded sequence in $H^1(\R^N)$ we have, using \cref{lemma:1}\eqref{point:1i}, not only that $\normLp{w_n}{2}^2 \to 0$ but also that $B(w_n) \to 0$.  Now we remember that
\begin{equation}\label{LLLL}
F(u_n) = F(u_c) + F(w_n) + o_n(1) \to \gamma^+(c).
\end{equation}
Since $u_c \in \overline{V(c)}$ by weak convergence property, we have, by \cref{lemma:27}(ii), that  $F(u_c) \geq \gamma^+(c)$. Thus from \eqref{LLLL} we deduce, on one hand, that necessarily $F(w_n) \leq o(1)$.
On the other hand, since $A(w_n) \leq A(u_n) \leq k_1$,  \cref{lemma:30} implies that $F(w_n) \geq b A(w_n) + o_n(1)$ for some $b > 0$.
Hence, we conclude  $A(w_n) \to 0$ and thus that $u_n \to u_c \in \overline{V(c)}$ strongly in $H^1(\R^3)$. Finally, by \cref{lemma:27}\eqref{point:27iii}, we have $u_c \in V(c)$ and $\gamma^+(c)$ is reached. The lemma is proved.
\end{proof}

\subsection{Asymptotic behavior of the Lagrange multipliers}\label{Subsection3-5}

\begin{lemma}\label{lemma:22}
Let $p \in (\frac{10}{3}, 6]$.
There exist two constants $K_1 > 0$ and $K_2 > 0$ such that for any $c \in (0, c_1)$, if $\lambda^+_c$ is the Lagrange parameter associated to a solution $u^+_c$ lying at the level $\gamma^+(c)$ then we have
\begin{align*}
\abs{\gamma^+(c)} \leq K_1 c^3 \qquad \text{and} \qquad \lambda^+_c \leq K_2 c^2.
\end{align*}
\end{lemma}
\begin{proof}
By \cref{lemma:27}\eqref{point:27i}, we have
\begin{align*}
A(u^+_c) < N^{-2} c^3 = \[\dfrac{\gamma (\sigma - 1)  K_{H}}{4 (\sigma - 2)} \]^2 c^3.
\end{align*}
Hence, we can deduce from \cref{lemma:1}\eqref{point:1i} that
\begin{align*}
B(u^+_c) \leq K_{H}  \sqrt{A(u^+_c)} c^{\frac{3}{2}} <  \dfrac{\gamma (\sigma - 1)  K_{H}^2}{4 (\sigma - 2)} c^3.
\end{align*}
%	and
%	\begin{align*}
%		C(u^+_c) \leq  K_{GN} [A(u^+_c)]^{\frac{\sigma}{2}} c^{\frac{6-p}{4}} < K_{GN}\[\dfrac{\gamma (\sigma - 1)  K_{H}}{4 (\sigma - 2)}\]^{\sigma} c^{2p-3}.
%	\end{align*}
Therefore, we have
\begin{align*}
\abs{\gamma^+(c)} = \abs{F(u^+_c)}
&= \abs*{\dfrac{\sigma-2}{2\sigma} A(u^+_c) - \dfrac{\gamma(\sigma-1)}{4\sigma} B(u^+_c)}
\leq \dfrac{\sigma-2}{2\sigma} A(u^+_c) + \dfrac{\gamma(\sigma-1)}{4\sigma} B(u^+_c)\\
&< \dfrac{\sigma-2}{2\sigma} \[\dfrac{\gamma (\sigma - 1)  K_{H}}{4 (\sigma - 2)}\]^2  c^3 + \dfrac{\gamma(\sigma-1)}{4\sigma}  \dfrac{\gamma (\sigma - 1)  K_{H}^2}{4 (\sigma - 2)} c^3 \\
&= \dfrac{3 \gamma^2 (\sigma - 1)^2 K_{H}^2  }{32\sigma (\sigma - 2)} c^3
:= K_1 c^3.
\end{align*}
We deduce from \eqref{eqn:lambda} that
\begin{align*}
2(3p-6) c \lambda^+_c &= 2(6-p) A(u^+_c) + (5p -12)\gamma B(u^+_c) \\
&< 2(6-p) \[\dfrac{\gamma (\sigma - 1)  K_{H}}{4 (\sigma - 2)}\]^2  c^3 + (5p -12) \gamma \dfrac{\gamma (\sigma - 1)  K_{H}^2}{4 (\sigma - 2)} c^3.
\end{align*}
This implies that there exists a constant $K_2 > 0$ such that $\lambda^+_c \leq K_2 c^2$.
The lemma is proved.
\end{proof}

%	When $p=6$, let $\lambda^+_c$ be the Lagrange parameter associated to a solution $u^+_c$ lying at the level $\gamma^+(c)$. Then we have that $\gamma^+(c) \to 0$ and $\lambda^+_c \to 0$ as $c \to 0$.
%	Since $\gamma^+(c) = F(u^+_c)$ is coercive on $\Ho$ and bounded from below (see \cref{lemma:5}) and since $F(u^+_c)$ is bounded from above by $0$ (see \cref{lemma:16}), we have that $A(u^+_c)$ is bounded. Hence, we get that $B(u^+_c) \to 0$ as $c \to 0$ due to $B(u^+_c) \leq K_{H}  \sqrt{A(u^+_c)} c^{\frac{3}{2}}$. By $Q(u^+_c) = 0$, we have
%	\begin{align*}
%	A(u^+_c) \to a C(u^+_c) \quad \text{as} \quad c \to 0.
%	\end{align*}
%	Combining with $A(u^+_c) > 5a C(u^+_c)$, we obtain that $A(u^+_c) \to 0$ and $C(u^+_c) \to 0$ as $c \to 0$.
%	Therefore, $\gamma^+(c) \to 0$ as $c \to 0$.
%	We deduce from \eqref{eqn:lambda} that
%	\begin{align*}
%	2(3p-6) c \lambda^+_c = (5p -12)\gamma B(u^+_c) \leq (5p -12) \gamma K_{H}  \sqrt{A(u^+_c)} c^{\frac{3}{2}}.
%	\end{align*}
%	Therefore, we obtain that $\lambda^+_c \to 0$ as $c \to 0$.

\begin{lemma}\label{lemma:23t}
Let $p \in (\frac{10}{3}, 6)$. There exist two constants $K_1 > 0$ and $K_2 > 0$ such that is $\lambda^-_c$ denotes the Lagrange parameter associated to a solution $u^-_c$ lying at the level $\gamma^-(c)$,
\begin{align*}
\abs{\gamma^-(c)} > K_1 c^{- \frac{6-p}{3p -10}} \qquad\text{and}\qquad \lambda^-_c > K_2 c^{- \frac{2p-4}{3p -10}}.
\end{align*}
\end{lemma}
\begin{proof}
By $u^-_c \in \slm$, we have
\begin{align*}
A(u^-_c) < \dfrac{a \sigma (\sigma - 1)}{p} C(u^-_c).
\end{align*}
Using \cref{lemma:1}\eqref{point:1ii}, we obtain that
\begin{align*}
A(u^-_c) < \dfrac{a \sigma (\sigma - 1)}{p} K_{GN} c^{\frac{6-p}{4}} [A(u^-_c)]^{\frac{\sigma}{2}},
\end{align*}
which implies that
\begin{align*}
A(u^-_c) > \[\dfrac{p}{a \sigma (\sigma - 1) K_{GN}} \]^{\frac{2}{\sigma-2}} c^{- \frac{6-p}{3p -10}}.
\end{align*}
We have that
\begin{align*}
\abs{\gamma^-(c)} = \abs{F(u^-_c)} = \abs*{-\dfrac{1}{2}A(u^-_c) + \dfrac{a(\sigma - 1)}{p}C(u^-_c)}
> \dfrac{\sigma -2}{2\sigma} A(u^-_c)
> \dfrac{\sigma -2}{2\sigma} \[\dfrac{p}{a \sigma (\sigma - 1) K_{GN}} \]^{\frac{\sigma-2}{2}} c^{- \frac{6-p}{3p -10}}
:= K_1 c^{- \frac{6-p}{3p -10}}.
\end{align*}
We deduce from \eqref{eqn:lambda} that
\begin{align*}
\lambda^-_c &= \dfrac{1}{c} \dfrac{1}{2(3p-6)} \[2(6-p) A(u^-_c) + (5p -12)\gamma B(u^-_c)\] \\
&> \dfrac{1}{c} \dfrac{6-p}{3p-6} A(u^-_c)
> \dfrac{1}{c} \dfrac{6-p}{3p-6} \[\dfrac{p}{a \sigma (\sigma - 1) K_{GN}} \]^{\frac{\sigma-2}{2}} c^{- \frac{6-p}{3p -10}}
:= K_2 c^{- \frac{2p-4}{3p -10}}.
\end{align*}
The lemma is proved.
\end{proof}

\begin{lemma}\label{lemma:25}
Let $p=6$. There exists a constant $K_1 > 0$ such that if $\lambda^-_c$ denote the Lagrange parameter associated to a solution $u^-_c$ lying at the level $\gamma^-(c)$ then we have
\begin{align*}
\gamma^-(c) \to \dfrac{1}{3 \sqrt{a K_{GN}}} \quad \text{as} \quad c \to 0 \quad \mbox{and} \quad \lambda^-_c \leq K_1 c^{\frac{1}{2}}.
\end{align*}
\end{lemma}
\begin{proof}
Since $F(u)$ restricted to $\Lambda(c)$ is coercive on $\Ho$ (see \cref{lemma:5})  we have that $A(u^-_c)$ remain bounded.
We deduce from \eqref{eqn:lambda} and \cref{lemma:1}\eqref{point:1i} that
\begin{align*}
\lambda^-_c = \dfrac{1}{c} \dfrac{3\gamma}{4} B(u^-_c) \leq \dfrac{1}{c} \dfrac{3\gamma}{4} K_{H}  \sqrt{A(u^-_c)} c^{\frac{3}{2}} := K_1 c^{\frac{1}{2}}.
\end{align*}
We have that $B(u^-_c) \to 0$ as $c \to 0$ due to $B(u^-_c) \leq K_{H}  \sqrt{A(u^-_c)} c^{\frac{3}{2}}$. Since $Q(u^-_c) = 0$, we have
\begin{align*}
A(u^-_c) = a C(u^-_c) + o_c(1),
\end{align*}
where $o_c(1) \to 0$ as $c \to 0$. Passing to the limit as $c \to 0$, up to subsequence we infer that
\begin{align*}
\lim_{c \to 0} A(u^-_c) = \lim_{c \to 0} a C(u^-_c) :=\ell \geq 0.
\end{align*}
Using \cref{lemma:1}\eqref{point:1ii}, we have
\begin{align*}
\ell = \lim_{c \to 0} a C(u^-_c) \leq \lim_{c \to 0} a K_{GN} [A(u^-_c)]^3 = a K_{GN} \ell^3.
\end{align*}
Therefore, either $\ell = 0$ or $\ell \geq (a K_{GN})^{-\frac{1}{2}}$. Using \cref{lemma:16}\eqref{point:2}, we ensure that $\ell \geq (a K_{GN})^{-\frac{1}{2}}$. Hence, we have
\begin{align*}
\gamma^-(c) + o_c(1) = F(u^-_c) &= \dfrac{\sigma - 2}{2\sigma} A(u^-_c) - \dfrac{\gamma(\sigma - 1)}{4\sigma} B(u^-_c)
= \dfrac{1}{3} A(u^-_c) + o_c(1) = \dfrac{1}{3}\ell + o_c(1) \geq \dfrac{1}{3 \sqrt{a K_{GN}}} + o_c(1),
\end{align*}
which implies that
\begin{align*}
\gamma^-(c) \geq \dfrac{1}{3 \sqrt{a K_{GN}}} \quad \text{as} \quad c \to 0.
\end{align*}
Recording \cref{lemma:20}, the lemma is proved.
\end{proof}

\subsection{The monotonicity of the map $c \mapsto \gamma^{-}(c)$}\label{Subsection3-6}

\begin{lemma}\label{lemma:10}
When $p \in (\frac{10}{3}, 6]$, the function $c \mapsto \gamma^{-}(c)$ is strictly decreasing on $(0, c_1)$.
\end{lemma}
\begin{proof}
Let $0 < c_2 < c_3 < c_1$, Since $\gamma^{-}(c_2)$ is reached, there exists $u \in S(c_2)$ such that $F(u) = \gamma^{-}(c_2)$. We define $v \in S(c_3)$ by $v(x) = \sqrt{\theta} u(\theta x)$ where $\theta = \sqrt{\frac{c_2}{c_3}} < 1$.
By direct calculations we have
\begin{equation}
A(v) = A(u), \quad B(v) = \theta^{-3}B(u) \quad \mbox{and } C(v) = \theta^{\frac{p}{2}-3}C(u). \label{eqn:37T2}
\end{equation}
Now observe that, since $\theta < 1$, for all $t>0$,
\begin{align}
\begin{split}
F(v^t) = \dfrac{1}{2}t^2 A(v) - \dfrac{\gamma}{4} t B(v) - \dfrac{a}{p} t^{\sigma} C(v)
= \dfrac{1}{2}t^2 A(u) - \dfrac{\gamma}{4} t \theta^{-3} B(u) - \dfrac{a}{p} t^{\sigma} \theta^{\frac{p}{2}-3} C(u)
<  F(u^t).
\end{split} \label{eqn:38T2}
\end{align}
By \eqref{eqn:43} and \eqref{eqn:41}, we have that
\begin{align*}
A(u^{s_u^+}) < k_1 < A(v^{s_v^-}),
\end{align*}
and thus $s_u^+ < s_v^-$ due to $A(v) = A(u)$. Hence, we can deduce from \eqref{eqn:38T2} that
\begin{align*}
F(v^{s_v^-}) < \max_{s_u^+ < t}F(u^t) = F(u) = \gamma^{-}(c_2).
\end{align*}
This implies that $\gamma^{-}(c_3) < \gamma^{-}(c_2)$ and hence, the lemma is proved.
\end{proof}

\section{The case $\gamma > 0,$ $ a < 0$ and $p \in (\frac{10}{3}, 6]$.} \label{Section4}
Throughout this section, we assume that $\gamma > 0,$ $ a < 0$ and $p \in (\frac{10}{3}, 6]$.
\begin{lemma} \label{lemma:5.1}
$F$ restricted to $S(c)$ is coercive on $\Ho$ and bounded from below.
\end{lemma}
\begin{proof}
Let $u \in S(c)$. Using \cref{lemma:1}\eqref{point:1i}, we obtain
\begin{align*}
F(u) = \dfrac{1}{2} A(u) - \dfrac{\gamma}{4} B(u) - \dfrac{a}{p} C(u)
\geq \dfrac{1}{2} A(u)  - \dfrac{\gamma}{4} K_{H}  \sqrt{A(u)} c^{\frac{3}{2}} - \dfrac{a}{p} C(u).
\end{align*}
Since $\gamma > 0,$ $ a < 0$, this concludes the proof.
\end{proof}
In what follows, we collect some basic properties of $m(c)$ defined in \eqref{eqn:0.1}.
\begin{lemma} \label{lemma:5.5}
It holds that
\begin{enumerate}[label=(\roman*), ref = \roman*]
\item\label{point:5.5i} $m(c) < 0$, $\forall c > 0$.
\item\label{point:5.5ii} $c \mapsto m(c)$ is a continuous mapping.
\item\label{point:5.5iii} For any $c_2 > c_1 > 0$, we have $c_1 m(c_2) \leq c_2 m(c_1)$. If $m(c_1)$ is reached then the inequality is strict.
\item\label{point:5.5iv} For any $c_2, c_1 > 0$, we have $m(c_1+ c_2) \leq m(c_1) + m(c_2)$. If $m(c_1)$ or $m(c_2)$ is reached then the inequality is strict.
\end{enumerate}
\end{lemma}
\begin{proof}
\ref{point:5.5i}) For any $u \in S(c)$, we recall that $u^t \in S(c)$ and
$$g_u(t)= F(u^t) = \dfrac{1}{2}t^2 A(u) - \dfrac{\gamma}{4} t B(u) - \dfrac{a}{p} t^{\sigma} C(u) \quad \mbox{and also} \quad g_u'(t) = t A(u) - \dfrac{\gamma}{4} B(u) - \dfrac{a \sigma}{p} t^{\sigma - 1} C(u).$$
We observe that $g_u(t) \to 0$ and $g_u'(t) \to - \frac{\gamma}{4} B(u) < 0$ as $t \to 0$. Therefore, there exists $t_0 > 0$ such that $F(u^{t_0}) = g_u(t_0) < 0$. Thus, we have $m(c) < 0$.

%&g_u(t)= F(u^t) = \dfrac{1}{2}t^2 A(u) - \dfrac{\gamma}{4} t B(u) - \dfrac{a}{p} t^{\sigma} C(u),\\
%&g_u'(t) = t A(u) - \dfrac{\gamma}{4} B(u) - \dfrac{a \sigma}{p} t^{\sigma - 1} C(u).

\ref{point:5.5ii})
%We follow the idea in \cite[Lemma 3.1]{GouJeanjean2016}.
We assume that $c_n \to c$. From the definition of $m(c_n)$, for any $\varepsilon > 0$, there exists $u_n \in S(c_n)$ such that
\begin{align}
F(u_n) \leq m(c_n) + \varepsilon. \label{eqn:5.9}
\end{align}
We set $y_n := \sqrt{\frac{c}{c_n}}\cdot u_n$. Taking into account that $y_n \in S(c)$ and $\frac{c}{c_n} \to 1$, we have
\begin{align}
m(c) \leq F(y_n) = F(u_n) + o_n(1).  \label{eqn:5.10}
\end{align}
Combining \eqref{eqn:5.9} and \eqref{eqn:5.10}, we get
\begin{align*}
m(c) \leq m(c_n) + \varepsilon + o_n(1).
\end{align*}
Reversing the argument we obtain similarly that
\begin{align*}
m(c_n) \leq m(c) + \varepsilon + o_n(1).
\end{align*}
Therefore, since $\varepsilon > 0$ is arbitrary, we deduce that $m(c_n) \to m(c)$. The point \eqref{point:5.5ii} follows.

\ref{point:5.5iii}) Let $t := \frac{c_2}{c_1} > 1$. For any $\varepsilon > 0$, there exist $u \in S(c_1)$ such that
\begin{align}
F(u) \leq m(c_1) + \varepsilon. \label{eqn:5.11}
\end{align}
Let $v:= u(t^{-\frac{1}{3}}x)$. Then we have $\normLp{v}{2}^2 = t \normLp{u}{2}^2 = c_2$, hence $v \in S(c_2)$. Moreover, we have
\begin{align*}
A(v) = t^{\frac{1}{3}}A(u), \qquad B(v) = t^{\frac{5}{3}} B(u), \qquad C(v) = tC(u).
\end{align*}
Therefore, we have
\begin{align*}
m(c_2) \leq F(v) &= \dfrac{1}{2} A(v) - \dfrac{\gamma}{4} B(v) - \dfrac{a}{p} C(v)
= \dfrac{1}{2} t^{\frac{1}{3}}A(u) - \dfrac{\gamma}{4} t^{\frac{5}{3}} B(u) - \dfrac{a}{p} tC(u)\\
&< \dfrac{1}{2} t A(u) - \dfrac{\gamma}{4} t B(u) - \dfrac{a}{p} t C(u)
= t \(\dfrac{1}{2} A(u) - \dfrac{\gamma}{4} B(u) - \dfrac{a}{p} C(u)\) \\
&=t F(u)
\leq t(m(c_1) + \varepsilon)
= \dfrac{c_2}{c_1} m(c_1) + \frac{c_2 }{c_1}\varepsilon.
\end{align*}
Since $\varepsilon > 0$ is arbitrary, we have $c_1 m(c_2) \leq c_2 m(c_1)$. If $m(c_1)$ is reached then we can let $\varepsilon = 0$ in \eqref{eqn:5.11} and thus the strict inequality follows.

\ref{point:5.5iv}) Assume first that $0 < c_1 \leq c_2$. Then, by \eqref{point:5.5iii}, we have that
\begin{align*}
m(c_1 + c_2) \leq \dfrac{c_1 + c_2}{c_2} m(c_2) = m(c_2) + \dfrac{c_1}{c_2} m(c_2) \leq m(c_2) + \dfrac{c_1}{c_2} \dfrac{c_2}{c_1} m(c_1) = m(c_1) + m(c_2).
\end{align*}
If $m(c_1)$ or $m(c_2)$ is reached, then we can use the strict inequality in \eqref{point:5.5iii} and thus the strict inequality follows.
The case $0< c_2 < c_1$ can be treated reversing the role of $c_1$ and $c_2$.
\end{proof}

\begin{lemma} \label{lemma:5.3}
Let $(u_n) \subset S(c)$ be any minimizing sequence for $m(c)$. Then, there exist a $\beta_0 > 0$ and a sequence $(y_n) \in \Rb$ such that
\begin{align}
\int_{B(y_n, R)} \abs{u_n}^2 dx \geq \beta_0 > 0, \qquad \text{for some } R > 0. \label{eqn:5.1}
\end{align}
\end{lemma}
\begin{proof}
Since $F$ restricted to $S(c)$ is coercive on $\Ho$ (see \cref{lemma:5.1}), the sequence $(u_n)$ is bounded.
Now, we assume that \eqref{eqn:5.1} does not hold. By \cite[Lemma I.1]{LIONS1984part2}, we have, for $q \in (2,6)$, $\normLp{u_n}{q} \to 0,$ as
$n \to \infty.$
%\normLp{u_n}{q} \to 0, \quad \text{as } n \to \infty.
This implies that
\begin{align*}
B(u_n) \leq K_1 \norm{u}_{\Lp{\frac{12}{5}}}^4 \to 0,
\end{align*}
due to \eqref{eqn:2.1}. Hence, we obtain
\begin{align*}
F(u_n) = \dfrac{1}{2} A(u_n) - \dfrac{\gamma}{4} B(u_n) - \dfrac{a}{p} C(u_n) \to \dfrac{1}{2} A(u_n) - \dfrac{a}{p} C(u_n) \geq 0,
\end{align*}
due to $a < 0$. This contradicts $F(u_n) \to m(c) < 0$, see \cref{lemma:5.5}\eqref{point:5.5i}.
%Thus, there exist a $\beta_0 > 0$ and a sequence $y_n \in \Rb$ such that
%\int_{B(y_n, R)} \abs{u_n}^2 dx \geq \beta_0 > 0, \qquad \text{for some } R > 0.
\end{proof}

\begin{lemma} \label{lemma:5.4}
Any minimizing sequence $(u_n) \subset S(c)$ for $m(c)$ is, up to translation, strongly convergent in $\Ho$.
\end{lemma}
\begin{proof}
%Our proof is inspired by \cite[Lemma 2.5]{BoussaidFernandezJeanjean2019}.
Since $F$ restricted to $S(c)$ is coercive on $\Ho$ (see \cref{lemma:5.1}), the sequence $(u_n)$ is bounded in $\Ho$. We deduce from the weak convergence in $\Ho$, the local compactness in $\Lp{2}$ and \cref{lemma:5.3} that
\begin{align*}
u_n(x - y_n) \weakto u_c \neq 0 \quad\text{in } \Ho.
\end{align*}
Our aim is to prove that $w_n(x) := u_n(x - y_n) - u_c(x) \to 0$ in $\Ho$.
Now, taking into account that $F$ fulfills the following splitting properties of Brezis-Lieb type (see [16] for terms $A$ and $C$; see \cite[Lemma 2.2]{ZHAO2008} or \cite[Proposition 3.1]{BellazziniSiciliano2011} for term $B$),
\begin{align*}
F(u_n - u_c) + F(u_c) =F(u_n) + o_n(1),
\end{align*}
and by the translational invariance, we obtain
\begin{align}
\begin{split}
F(u_n) &=F(u_n(x-y_n))
=F(u_n(x-y_n) - u_c(x)) + F(u_c) + o_n(1)
=F(w_n) + F(u_c) + o_n(1),
\end{split} \label{eqn:5.5}
\end{align}
\begin{align*}
\normLp{u_n}{2}^2 &= \normLp{u_n(x-y_n)}{2}^2
= \normLp{u_n(x-y_n) - u_c(x)}{2}^2 + \normLp{u_c}{2}^2 + o_n(1)
=\normLp{w_n}{2}^2 + \normLp{u_c}{2}^2 + o_n(1).
\end{align*}
Thus, we have
\begin{align}
\normLp{w_n}{2}^2 = \normLp{u_n}{2}^2 -  \normLp{u_c}{2}^2 + o_n(1) = c - \normLp{u_c}{2}^2 + o_n(1). \label{eqn:5.6}
\end{align}
We claim that
\begin{align}
\normLp{w_n}{2}^2 \to 0 \quad \text{as} \quad n \to \infty. \label{eqn:5.12}
\end{align}
In order to prove this, let us denote $c_1:= \normLp{u_c}{2}^2 > 0$. By \eqref{eqn:5.6}, if we show that $c_1 = c$ then the claim follows. We assume by contradiction that $c_1 < c$.
Recording that $F(u_n) \to m(c)$, in view of \eqref{eqn:5.5}, we have
\begin{align*}
m(c) = F(w_n) + F(u_c) \geq m\(\normLp{w_n}{2}^2\) + F(u_c).
\end{align*}
Since the map $c \mapsto m(c)$ is continuous (see \cref{lemma:5.5}\eqref{point:5.5ii}) and \eqref{eqn:5.6}, we deduce that
\begin{align}
m(c) \geq m(c-c_1) + F(u_c). \label{eqn:5.13}
\end{align}
If $F(u_c) > m(c_1)$, then it follows from \cref{lemma:5.5}\eqref{point:5.5iv} that
\begin{align*}
m(c) > m(c-c_1) + m(c_1) \geq m(c -c_1 + c_1) = m(c),
\end{align*}
which is impossible. Hence, we have $F(u_c) = m(c_1)$, namely $u_c$ is global minimizer with respect to $c_1$. So, we can using \cref{lemma:5.5}\eqref{point:5.5iv} with the strict inequality and we deduce from \eqref{eqn:5.13} that
\begin{align*}
m(c) \geq m(c-c_1) + F(u_c) = m(c-c_1) + m(c_1) > m(c -c_1 + c_1) = m(c),
\end{align*}
which is impossible. Thus, the claim follows and $\normLp{u_c}{2}^2 = c$.

At this point, since $w_n$ is a bounded sequence in $\Ho$ and by \cref{lemma:1}\eqref{point:1i}, we have
\begin{align*}
B(w_n) \leq K_{H}  \sqrt{A(w_n)} \normLp{w_n}{2}^{3} \to 0.
\end{align*}
Thus, we obtain that
\begin{align}
\liminf_{n \to \infty} F(w_n) = \liminf_{n \to \infty} \[ \dfrac{1}{2} A(w_n) - \dfrac{a}{p}C(w_n)\] \geq 0. \label{eqn:5.14}
\end{align}
On the other hand, since $\normLp{u_c}{2}^2 = c$, we deduce from \eqref{eqn:5.5} that
\begin{align*}
F(u_n) = F(w_n) + F(u_c) +  o_n(1) \geq F(w_n) + m(c) + o_n(1),
\end{align*}
and by $F(u_n) \to m(c)$, we have that
\begin{align}
\limsup_{n \to \infty} F(w_n) \leq 0. \label{eqn:5.15}
\end{align}
Combining \eqref{eqn:5.14} and \eqref{eqn:5.15}, we obtain that $F(w_n) \to 0$. Hence, by \eqref{eqn:5.14} and by $a <0$, we have $A(w_n) \to 0$ and $C(w_n) \to 0$. Thus, we get $w_n \to 0$ in $\Ho$. The lemma is completed.
\end{proof}

\begin{proof}[Proof of \cref{theorem:2}]
The proof follows directly from \cref{lemma:5.4} for the convergence of the minimizing sequence and from \cref{lemma:2.3} for the sign of the Lagrange parameter.
\end{proof}

\begin{lemma}\label{lemma:5.6}
There exist three constants $K_1, K_2, K_3 > 0$ such that if $\lambda_c$ denote the Lagrange parameter associated to a solution $u_c$ lying at the level $m(c)$ then we have
\begin{align*}
|m(c)| \leq K_1 c^3 + K_2 c^{2p-3}   \quad \mbox{and} \quad \lambda_c \leq K_3 c^{2}.
\end{align*}
\end{lemma}
\begin{proof}
By the fact that $m(c) < 0$ and by using \cref{lemma:1}\eqref{point:1i}, we get that
\begin{align*}
0 > m(c) = F(u_c) = \dfrac{1}{2} A(u_c) - \dfrac{\gamma}{4} B(u_c) - \dfrac{a}{p} C(u_c) \geq \dfrac{1}{2} A(u_c) - \dfrac{\gamma}{4} K_H \sqrt{A(u_c)} c^{\frac{3}{2}} - \dfrac{a}{p} C(u_c) \geq \dfrac{1}{2} A(u_c) - \dfrac{\gamma}{4} K_H \sqrt{A(u_c)} c^{\frac{3}{2}},
\end{align*}
due to our assumption $\gamma > 0$ and $a < 0$. This implies that
\begin{align*}
\sqrt{A(u_c)} < \dfrac{\gamma K_H}{2} c^{\frac{3}{2}}.
\end{align*}
Therefore, using again \cref{lemma:1}, we obtain that
\begin{align*}
|m(c)| = |F(u_c)| &= \Big|\dfrac{1}{2} A(u_c) - \dfrac{\gamma}{4} B(u_c) - \dfrac{a}{p} C(u_c) \Big| \leq \dfrac{1}{2} A(u_c) + \dfrac{\gamma}{4} B(u_c) - \dfrac{a}{p} C(u_c)\\
&\leq \dfrac{1}{2} A(u_c) + \dfrac{\gamma K_H}{4} \sqrt{A(u_c)} c^{\frac{3}{2}} - \dfrac{a K_{GN}}{p} [A(u_c)]^{\frac{\sigma}{2}} c^{\frac{6-p}{4}} \\
&\leq \dfrac{\gamma^2 K_H^2}{8} c^3 - \dfrac{a K_{GN}}{p} \[\dfrac{\gamma K_H}{2}\]^{\sigma} c^{\frac{3\sigma}{2}} c^{\frac{6-p}{4}} := K_1 c^3 + K_2 c^{2p-3}.
\end{align*}
We deduce from \eqref{eqn:lambda} that
\begin{align*}
\lambda_c &= \dfrac{6-p}{3p-6} \dfrac{1}{c} A(u_c) + \dfrac{\gamma(5p-12)}{2(3p-6)} \dfrac{1}{c} B(u_c)
\leq \dfrac{6-p}{3p-6} \dfrac{1}{c} A(u_c) + \dfrac{\gamma(5p-12)K_H}{2(3p-6)} \dfrac{1}{c}  \sqrt{A(u_c)} c^{\frac{3}{2}} \\
&\leq \dfrac{6-p}{3p-6}  \dfrac{1}{c} \dfrac{\gamma^2 K_H^2}{4}  c^{3} + \dfrac{\gamma(5p-12)K_H}{2(3p-6)} \dfrac{1}{c}  \dfrac{\gamma K_H}{2} c^{\frac{3}{2}} c^{\frac{3}{2}} := K_3 c^2.
\end{align*}
Thus, the lemma is proved.
\end{proof}

\section{The case $\gamma < 0,$ $ a > 0$ and $p=6$.} \label{Section:6}
Throughout this section, we assume that $\gamma < 0,$ $ a > 0$ and $p =6$.
To prove the non-existence of the positive solution to \eqref{eqn:Laplace}, we first recall a Liouville-type result, see \cite[Theorem 2.1]{Armstrong-Sirakov-2011},
\begin{proposition}\label{theorem:Liouville-type}
Assume that $N \geq 3$ and  the nonlinearity $f: (0, \infty) \mapsto (0, \infty)$ is
continuous and satisfies
\begin{align*}
\liminf_{s \to 0} s^{-\frac{N}{N-2}} f(s) > 0.
\end{align*}
Then the differential inequality  $- \laplace u \geq f(u)$ has no positive solution in any exterior domain of $\R^N$.
\end{proposition}

\begin{proof}[Proof of \cref{theorem:non-existence}]
Let $u \in \Hs{1}$ be a non-trivial solution to \eqref{eqn:Laplace}. By \cref{lemma:2.3}, we have $\lambda < 0$ and $Q(u) = 0$. Hence,
\begin{align*}
aC(u) = A(u) - \dfrac{\gamma}{4} B(u) > A(u)
\end{align*}
and using \cref{lemma:1}\eqref{point:1ii}, we obtain that
\begin{align*}
A(u) < aC(u) \leq a K_{GN} [A(u)]^{3}.
\end{align*}
This implies that
\begin{align*}
A(u) > \sqrt{\dfrac{1}{a K_{GN}}}.
\end{align*}
Using again $Q(u) = 0$ we have that
\begin{align*}
F(u) = \dfrac{1}{2}A(u) - \dfrac{\gamma}{4} B(u) - \dfrac{a}{6} C(u) = \dfrac{5a}{6} C(u) - \dfrac{1}{2}A(u) > \dfrac{1}{3}A(u) > \dfrac{1}{3\sqrt{a K_{GN}}},
\end{align*}
proving point \eqref{point:theorem:non:i}. To prove point \eqref{point:theorem:non:ii}, we assume by contradiction that there exists a positive solution $u \in \Hs{1}$ to \eqref{eqn:Laplace}. Then, by point \eqref{point:theorem:non:i}, the associated Lagrange multiplier  $\lambda $ is strictly negative.
%Since $u \in \Hs{1}$, the  Sobolev embedding (see \cite[Corollary 9.10]{Brezis-2011}) implies that $|u|^2 \in \Lp{q}$ for every $q \in [1,3]$.
%Now,  setting $K := |x|^{-1}$, we write $K := K_1 + K_2$ where $K_1:= K$ on $B(0,1)$, $K_1:=0$ on $\R^3 \backslash \ B(0,1)$ and $K_2:= K - K_1$.
%Clearly $K_1 \in \Lp{2}$ and $K_2 \in \Lp{4}$. Applying \cite[Lemma 2.20]{LiebLoss2001analysis} with $K_1 \in \Lp{2}$, $|u|^2 \in \Lp{2}$ and with $K_2 \in \Lp{4}$, $|u|^2 \in \Lp{\frac{4}{3}}$, we obtain that
%	\lim_{|x| \to \infty} (K_1*|u|^2) (x) = 0 \quad\mbox{and}\quad \lim_{|x| \to \infty} (K_2*|u|^2) (x) = 0.
%Hence, we get that
%	\lim_{|x| \to \infty} (K*|u|^2) (x) = 0,
%and as a consequence,
In view of \eqref{eqn:tend-to-0}, there exists $R_0 > 0$ large enough such that
\begin{align*}
(\abs{x}^{-1} * \abs{u}^2)(x) \leq - \dfrac{\lambda}{2 \gamma} \quad \mbox{for } |x| > R_0.
\end{align*}
Therefore, we get that
\begin{align*}
- \laplace u(x) = \( - \lambda + \gamma (\abs{x}^{-1} * \abs{u}^2)(x)  + a \abs{u(x)}^{4}\)u(x) \geq \( - \lambda + \gamma (\abs{x}^{-1} * \abs{u}^2)(x)\)u(x) \geq - \dfrac{\lambda}{2} u(x) \quad \mbox{for } |x| > R_0.
\end{align*}
By applying \cref{theorem:Liouville-type} with $f(s) = -\dfrac{\lambda}{2} s$, we obtain a contradiction, and thus point \eqref{point:theorem:non:ii} holds.
\end{proof}

\begin{remark} \label{remrak:6.1}
In \cite[Theorem 1.2]{Soave2019Sobolevcriticalcase}, the author considers the equation
\begin{align}
-\laplace u - \lambda u - \mu \abs{u}^{q-2} u - \abs{u}^{2^*-2} u = 0 \quad \mbox{in } \R^N, \label{eqn:LaplaceS}
\end{align}
with $N \geq 3$, $2<q<2^*$ and $\mu < 0$.
If $u \in \mathit{H}^1(\R^N)$ is a non-trivial solution to \eqref{eqn:LaplaceS} then by \cite[Theorem 1.2]{Soave2019Sobolevcriticalcase}, the associated Lagrange multiplier $\lambda$ is positive and following the arguments in \cite[Proof of Theorem 1.2]{Soave2019Sobolevcriticalcase}, one obtains that
\begin{align*}
-\laplace u \geq \dfrac{\lambda}{2} u \quad \mbox{for } |x| > R_1,
\end{align*}
with $R_1 > 0$ large enough. Hence, by applying \cref{theorem:Liouville-type}, we see that \eqref{eqn:LaplaceS} has no positive solution $u \in {\mathit H}^{1}(\R^N)$ for all $N \geq 3$, improving slightly the conclusions of \cite[Theorem 1.2]{Soave2019Sobolevcriticalcase}.
Actually, borrowing an observation from [12], the
non-existence results of \cite[Theorem 1.2]{Soave2019Sobolevcriticalcase} can be further extended by showing that \eqref{eqn:LaplaceS} has no non-trivial radial solutions in $\mathit{H}^1(\R^N)$ when $N \geq 3$ and $q > 2 + \frac{2}{N-1}$. Indeed, if $ u \in \mathit{H}^1(\R^N)$ is a radial function by \cite[Radial Lemma A.II]{BerestyckiLions1983}, there exist constants $C>0$ and $R_2 > 0$ such that
\begin{align*}
|u(x)| \leq C |x|^{-\frac{N-1}{2}}  \quad\mbox{for } |x| > R_2.
\end{align*}
Setting $V(x) = - \mu \abs{u(x)}^{q-2} - \abs{u(x)}^{2^*-2}$, we obtain that any radial solution  $u \in \mathit{H}^1(\R^N)$ satisfies
\begin{align}
-\laplace u(x) + V(x) u(x) = \lambda u(x), \label{eqn:6.3}
\end{align}
where, since $q > 2 + \frac{2}{N-1}$,
\begin{align*}
\lim_{|x| \to \infty} |x| |V(x)| \leq \lim_{|x| \to \infty} \[-\mu C |x|^{-\frac{(N-1)(q-2)}{2} +1} + C |x|^{-\frac{(N-1)(2^*-2)}{2} +1} \] = 0.
\end{align*}
Then \eqref{eqn:6.3} has no solution in view of Kato's result \cite[page 404]{Kato-1959}, also see [46] which states that Schr\"{o}dinger operator
$ H = - \Delta + p(x)$ has no positive eigenvalue with an $L^2$-eigenfunction if $p(x) = o(|x|^{-1})$.
\end{remark}

\begin{remark}\label{open-problem}
One may wonder if a non-existence result for radial solutions also holds for \eqref{eqn:Laplace} under the assumptions of \cref{theorem:non-existence}. The difficulty one faces is that, for any $u \in \mathit{H}^1(\R^N), $
$(|x|^{-1}*|u|^2)(x) \geq C |x|^{-1}$ for $|x| > R$ for some $C, R > 0$ (see [9] or \cite[Appendix A.4]{Moroz-VanSchaftingen-2017}). Thus, the result of Kato used in \cref{remrak:6.1} cannot be directly applied and the non-existence of radial solutions to   \eqref{eqn:Laplace} when $\gamma <0$, $a>0$ and $p=6$ is an open problem.
\end{remark}

{\bf Acknowledgments:} The authors thank S. Cingolani for providing to us the statement and proof of \cref{lemma:3} and Colette De Coster for pointing to us \cite[Theorem 3.27]{Troianiello-1987}.

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