Normalized solutions to a critical growth Choquard equation involving mixed operators

Abstract

In this paper we study the existence and regularity results of normalized solutions to the following critical growth Choquard equation with mixed diffusion type operators: $\begin{aligned} \begin{aligned} - Δ u + (- Δ)^{s} u = λ u + g (u) + (I_{α} * | u |^{2_{α}^{*}}) | u |^{2_{α}^{*} - 2} u in R^{N}, \\ \int_{R^{N}} | u |^{2} d x = τ^{2}, \end{aligned} \end{aligned}$ where $N ⩾ 3$ , $τ > 0$ , $I_{α}$ is the Riesz potential of order $α \in (0, N)$ , $(- Δ)^{s}$ is the fractional laplacian operator, $2_{α}^{*} = \frac{N + α}{N - 2}$ is the critical exponent with respect to the Hardy Littlewood Sobolev inequality, λ appears as a Lagrange multiplier and g is a real valued function satisfying some $L^{2}$ -supercritical conditions.

Keywords

Normalized solutions Choquard equation critical growth local and nonlocal operator existence results Sobolev regularity

1. .Introduction

In the present work, we are interested in studying the existence of normalized solutions to the following Choquard equation with critical growth, involving mixed operator:

\begin{aligned} - Δ u + (- Δ)^{s} u = λ u + g (u) + (I_{α} * | u |^{2_{α}^{*}}) | u |^{2_{α}^{*} - 2} u in R^{N}, \\ \int_{R^{N}} | u |^{2} d x = τ^{2}, \end{aligned}

(1.1)

where

N ⩾ 3

τ > 0

, and

I_{α}

is the Riesz potential of order

α \in (0, N)

given by

\begin{aligned} I_{α} (x) = \frac{A_{N, α}}{| x |^{N - α}} with~ A_{N, α} = \frac{Γ (\frac{N - 2}{2})}{π^{\frac{N}{2}} 2^{α} Γ (\frac{α}{2})} for every x \in R^{N} ∖ {0} . \end{aligned}

(1.2)

The fractional laplacian

(- Δ)^{s}

is defined as:

\begin{aligned} (- Δ)^{s} u (x) = C (N, s) P . V . \int_{R^{N}} \frac{u (x) - u (y)}{| x - y |^{N + 2 s}} d y for~ s \in (0, 1), \end{aligned}

here

C (N, s)

is the normalizing constant given by

\begin{aligned} C (N, s) = {(\int_{R^{N}} \frac{1 - \cos (x)}{| x |^{N + 2 s}})}^{- 1}, \end{aligned}

and P.V. is the abbreviation for principal value. Here

2_{α}^{*} = \frac{N + α}{N - 2}

is the critical exponent with respect to the following Hardy Littlewood Sobolev inequality:

Proposition 1.1.

Let $t, r > 1$ and $0 < α < N$ with $1 / t + 1 / r = 1 + α / N$ , $f \in L^{t} (R^{N})$ and $h \in L^{r} (R^{N})$ . There exists a sharp constant $C (t, r, α, N)$ independent of f, h, such that

\begin{aligned} \int_{R^{N}} \int_{R^{N}} \frac{f (x) h (y)}{| x - y |^{N - α}} d x d y ⩽ C (t, r, α, N) ‖ f ‖_{L^{t}} ‖ h ‖_{L^{r}} . \end{aligned}

(1.3)

t = r = 2 N / (N + α)

, then

\begin{aligned} C (t, r, α, N) = C (N, α) = π^{\frac{N - α}{2}} \frac{Γ (\frac{α}{2})}{Γ (\frac{N + α}{2})} {\begin{matrix} \frac{Γ (\frac{N}{2})}{Γ (N)} \end{matrix}}^{- \frac{α}{N}} . \end{aligned}

(1.4)

Equality holds in (1.3) if and only if

\frac{f}{h} \equiv constant

and

h (x) = A (γ^{2} + | x - a |^{2})^{(N + α) / 2}

for some

A \in C

0 \neq γ \in R

and

a \in R^{N}

From this inequality, it follows that

\begin{aligned} A_{q} (u) := \int_{R^{N}} \int_{R^{N}} \frac{| u (x) |^{q} | u (y) |^{q}}{| x - y |^{N - α}} d x d y \end{aligned}

is well defined if

\frac{N + α}{N} ⩽ q ⩽ \frac{N + α}{N - 2} = 2_{α}^{*}

. The exponent

q = 2_{α}^{*}

is known as Hardy–Littlewood–Sobolev critical exponent and similarly to the usual critical exponent,

H_{0}^{1} (Ω) ∋ u \to A_{2_{α}^{*}} (u)

is continuous for the norm topology but not for the weak topology. In this work, we aim to investigate a specific kind of solutions, namely, normalized solutions, for a class of elliptic problems involving the operator

L = - Δ + (- Δ)^{s}

that has not been deeply looked upon in the current literature. The study of normalized solutions is of great significance in physics, as it represents an entity that satisfies the dynamics while maintaining a constant mass. Formally, the solution of the following constrained problem is called the normalized solution

\begin{aligned} {\begin{cases} - Δ u = λ u + g (u) in R^{N}, \\ \int_{R^{N}} | u |^{2} d x = c . \end{cases} \end{aligned}

(1.5)

The reason for studying the equation (1.5) is the physical incentive it provides, as its solution yields stationary states of a nonlinear Schrödinger equation with a predetermined

L^{2}

-norm. Jeanjean [18] demonstrated the existence of a radial solution for equation (1.5) subject to certain assumptions on the function g. Further, the existence of infinitely many solutions to (1.5) with

c = 1

under same assumptions on g has been shown by Bartsch and De Valeriola in [5]. In [26], Noris et. al. explored the normalized solutions in the context of bounded domains with Dirichlet boundary conditions. Normalized solutions have been seen to exist for p values within the intervals

(1, 1 + \frac{4}{N})

(1 + \frac{4}{N}, 2^{*} - 1)

, and

p = 1 + \frac{4}{N}

, under certain requirements on c, and the domain being the unit ball and

g (t) = | t |^{p - 1} t

. Furthermore, the authors in [32] have tackled the issue in general bounded domains. The existence of normalized solutions of nonlinear Schrödinger systems has been extensively explored. Interested readers can refer to the references [6–8,17,25,27]. The study of quadratic ergodic mean field games system also investigates normalized solutions type, as discussed in [30].

More recently, the investigation of the Choquard equation has received significant attention from scholars. The study began with Lieb’s research in [19], where he examined the existence and uniqueness of the minimizing solution for the given problem:

\begin{aligned} - Δ u + λ u = μ (I_{α} * | u |^{p}) | u |^{p - 2} u in R^{N} . \end{aligned}

(1.6)

In [29], Pekar looked at the following problem to study of the quantum theory model of a stationary polaron:

\begin{aligned} {\begin{cases} - Δ u + u = (I_{2} * | u |^{2}) u in R^{3}, \\ u \in H^{1} (R^{3}) . \end{cases} \end{aligned}

(1.7)

In 1976, Choquard, at the Symposium on Coulomb Systems utilised the energy functional associated to equation (1.7), to examine a viable approximation to Hartree–Fock theory for a one-component plasma (see [19]). The equation is mostly employed to characterise an electron confined within its own vacancy. It also occurs in the field of quantum physics, as discussed in the reference [31] and related sources. Several works have ever since conducted research on the existence, multiplicity, and qualitative characteristics of the solution to the problem (1.6), as detailed in [13,21,23].

In the present paper, we explore the normalized solution for the Choquard equation involving mixed operators. The mixed operator $L = - Δ + (- Δ)^{s}$ has many applications, such as bi-model power law distribution processes (see [28]). Such an operator comes into the picture, whenever the impact on a physical phenomenon is due to both local and non-local changes. A variety of contributions have examined issues related to the existence of solutions, their regularity and symmetry properties, Neumann problems, and Green’s function estimates (see, for example, [1,9,10]). For additional investigation on the semilinear elliptic equations involving operator $L$ , we quote [3] and references therein. A more general mixed operator has also been investigated by certain researchers. As an illustration, the authors of [16] looked at the second eigen value for the mixed operator that employs both usual p-laplacian $(- Δ_{p})$ and non-local p-laplacian $(- Δ_{J, p})$ .

In [15], it has been established that the following subcritical problem:

\begin{aligned} - Δ u + λ (- Δ)^{s} u + u = μ (I_{α} * | u |^{p}) | u |^{p - 2} u in R^{N}, \\ \int_{R^{N}} | u |^{2} d x = τ, \end{aligned}

has a solution for

\frac{N + α}{N} < p < \frac{2 s + N + α}{N}

, whereas for

p = \frac{N + α}{N}

and

p = \frac{N + α}{N - 2}

it does not possess any solution, moreover some regularity results are also discussed. In the spirit of the Brezis–Nirenberg approach, we aim to determine for which kind of perturbation nonlinearity we can infer the existence of solutions in the critical case

p = 2_{α}^{*}

. Following Brezis–Nirenberg’s work in [11], researchers have tackled the critical case scenario for several problems; for example, Servadei R. and Valdinoci E. investigated the existence of solutions for the fractional operator critical exponent problem (see [33]).

Numerous scholars have examined the normalized solution of the critical growth Choquard equation, taking into account either the fractional laplacian [22] or the classical laplacian [34] operator. Apart from the fact that the critical case has not, as far as we are aware, been treated in former literature, our problem involves both classical and fractional laplacian, combining local and nonlocal features and providing significant technical issues. Inspired by [37] and [35] we are looking for some key estimates to make the problem compact and hence prove the existence of solutions. For that we need the following assumptions on the perturbation g: (a1)

$g : R \to R$ is continuous and odd.

(a2)

There exists $\bar{α}, \bar{β} \in R^{+}$ with $2 + \frac{4}{N} < \bar{α} ⩽ \bar{β} < 2^{*} = \frac{2 N}{N - 2}$ such that

\begin{aligned} 0 < \bar{α} G (s) ⩽ g (s) s ⩽ \bar{β} G (s) for all s \neq 0, \end{aligned}

where

G (s) = \int_{0}^{s} g (t) d t

(a3)

Define $\tilde{G} (x) := \frac{g (s) s}{2} - G (s)$ , then $\tilde{G} \in C^{1}$ and

\begin{aligned} (\tilde{G})^{^{'}} (s) s ⩾ \bar{α} \tilde{G} (s) for all s \in R . \end{aligned}

Remark 1.1.

Using (a1)–(a3), one can deduce the following: A-1

for all $t \in R$ , $s ⩾ 0$

\begin{aligned} {\begin{cases} s^{\bar{β}} G (t) ⩽ G (t s) ⩽ s^{\bar{α}} G (t), & for~ s ⩽ 1 \\ s^{\bar{α}} G (t) ⩽ G (t s) ⩽ s^{\bar{β}} G (t), & for~ s > 1. \end{cases} \end{aligned}

(1.8)

A-2

There exists $C_{G} > 0$ such that

\begin{aligned} G (s) ⩽ C_{G} (| s |^{\bar{α}} + | s |^{\bar{β}}) for every s \in R . \end{aligned}

A-3

Setting $α_{1} = 2 + \frac{4}{N}$ and $α_{1} < α_{2} < \bar{α}$ , there exists $L_{M} > 0$ such that

\begin{aligned} G (s) ⩾ - L_{M} | s |^{α_{1}} + M | s |^{α_{2}} for large enough M . \end{aligned}

The space framework is the Hilbert space

H^{1} (R^{N})

equipped with the following equivalent norm:

\begin{aligned} ‖ u ‖ = {(‖ \nabla u ‖_{2}^{2} + ‖ u ‖_{2}^{2} + [u]^{2})}^{\frac{1}{2}}, \end{aligned}

where

\begin{aligned} [u]^{2} := \frac{C (N, s)}{2} \int_{R^{N}} \int_{R^{N}} \frac{| u (x) - u (y) |^{2}}{| x - y |^{N + 2 s}} d x d y, \end{aligned}

and

H_{r} (R^{N}) = {u \in H^{1} (R^{N}) : u is radial}

. Morever, the space

H = H^{1} (R^{N}) \times R

, with

‖ (u, t) ‖_{H}^{2} := ‖ u ‖^{2} + | t |^{2}

, will also be used. The notion of weak solution for (1.1) is as follows:

Definition 1.1.

A function $u \in H_{r} (R^{N})$ is said to be the weak solution of (1.1) if $‖ u ‖_{2} = τ$ and

\begin{aligned} \int_{R^{N}} \nabla u \nabla v + ⟨ ⟨ u, v ⟩ ⟩ = \int_{R^{N}} g (u) v + λ \int_{R^{N}} u v + \int_{R^{N}} (I_{α} * | u |^{2_{α}^{*}}) | u |^{2_{α}^{*} - 2} u v, \end{aligned}

for every

v \in H_{r} (R^{N})

, where

⟨ ⟨ u, v ⟩ ⟩ := \frac{C (N, s)}{2} \int_{R^{N}} \int_{R^{N}} \frac{(u (x) - u (y)) (v (x) - v (y))}{| x - y |^{N + 2 s}} d x d y

Defining $A := A_{2_{α}^{*}}$ , the energy functional corresponding to the problem (1.1) is given by

\begin{aligned} F (u) = \frac{1}{2} ‖ \nabla u ‖_{2}^{2} + \frac{λ}{2} ‖ u ‖_{2}^{2} + \frac{1}{2} [u]^{2} - \int_{R^{N}} G (u) d x - \frac{1}{2. 2_{α}^{*}} A (u), \end{aligned}

that is, a critical point of F which satisfies the constraint turns out to be the weak solution of (1.1). We will be denoting the set of functions satisfying the constraint by:

\begin{aligned} S (τ) := {u \in H^{1} (R^{N}) : ‖ u ‖_{2}^{2} = τ^{2}} . \end{aligned}

Using the fiber map defined as

\begin{aligned} t ⋆ u (x) := t^{\frac{N}{2}} u (t x), \end{aligned}

the idea is to work on the natural Pohozaev manifold, given by

\begin{aligned} P (τ) := S (τ) \cap \bar{M} = S (τ) \cap {u \in H^{1} (R^{N}) : M (u) = 0}, \end{aligned}

where

\begin{aligned} M (u) := ‖ \nabla u ‖_{2}^{2} + s [u]^{2} - N \int_{R^{N}} \tilde{G} (u) - A (u) . \end{aligned}

We will see that the energy functional F is coercive on

P (τ)

. Though the imbedding of

H^{1} (R^{N})

into

L^{p} (R^{N})

is not compact, the radial space

H_{r} (R^{N})

is compactly imbedded in

L^{p} (R^{N})

for all

p \in (2, 2^{*})

. Therefore, thanks to symmetric decreasing rearrangements, we will look for a radial solution in the space

H_{r} (R^{N})

. Precisely, we prove the following the main result:

Theorem 1.1.

Let $N ⩾ 3$ , $s \in (0, 1)$ and (a1)–(a3) holds. Then for $0 < τ < γ$ , γ given by (2.31), the problem (1.1) has a couple solution $(λ_{τ}, u_{τ}) \in R \times S_{r} (τ)$ . Further, defining

\begin{aligned} E (u) := \frac{1}{2} ‖ \nabla u ‖_{2}^{2} + \frac{1}{2} [u]^{2} - \int_{R^{N}} G (u) d x - \frac{1}{2. 2_{α}^{*}} A (u), \end{aligned}

we have

\begin{aligned} E (u_{τ}) = inf_{u \in P_{r} (τ)} E (u), \end{aligned}

where

S_{r} (τ) = S (τ) \cap H_{r} (R^{N})

and

P_{r} (τ) = P (τ) \cap H_{r} (R^{N})

Throughout the paper, we have frequently used the Gagliardo–Nirenberg inequality [32] stated as follows:

\begin{aligned} ‖ w ‖_{p + 1}^{p + 1} ⩽ C_{N, p} ‖ \nabla w ‖_{2}^{N (p - 1) / 2} ‖ w ‖_{2}^{(p + 1) - N (p - 1) / 2} for all w \in H^{1} (R^{N}) . \end{aligned}

Following the proof of the existence result, we investigate the regularity of positive weak solutions to (1.1) in further details and prove the following:

Theorem 1.2.

Let us assume that either $N ⩾ 5$ or $N = 3$ with $α < 1$ and (a1)–(a3) holds. If $u \in H_{r} (R^{N})$ is a positive solution of (1.1), then $u \in L^{\infty} (R^{N})$ . Moreover, $u \in W^{2, q} (R^{N})$ for all $q \in [ℓ, \infty)$ where

\begin{aligned} ℓ = {\begin{cases} 2 & i f N = 3 ~or~ N ⩾ 5 ~with~ N - α ⩽ 4, \\ \frac{2 (N - 2)}{α + 2} & i f N ⩾ 5 ~with~ N - α > 4. \end{cases} \end{aligned}

To prove Theorem 1.2, we first establish that a weak solution u of (1.1) belongs to $L^{\infty} (R^{N})$ thanks to the estimate given in Proposition 4.1 and Strauss Lemma (see for instance [36]). Next, using Sobolev regularity for mixed operators and estimates in $L^{q}$ of $g (u)$ and $(I_{α} * | u |^{2_{α}^{*}}) | u |^{2_{α}^{*} - 1} u$ , we finally show that $u \in W^{2, q} (R^{N})$ .

Notations:

We have used the following notations through the paper:

$T (u) := (‖ \nabla u ‖_{2}^{2} + [u]^{2})^{1 / 2}$ ,

$f_{u} (t) := E (t ⋆ u)$ ,

S is the best constant corresponding to the imbedding $D^{1, 2} (R^{N}) ↪ L^{2^{*}} (R^{N})$ . By [34], we know that

\begin{aligned} S_{α} = inf_{u \in D^{1, 2} (R^{N}) ∖ {0}} \frac{‖ \nabla u ‖_{2}^{2}}{A {(u)}^{\frac{1}{2_{α}^{*}}}} = \frac{S}{(A_{α} C_{α})^{\frac{1}{2_{α}^{*}}}}, \end{aligned}

(1.9)

and

S_{α}

is achieved by the family of functions of the form:

\begin{aligned} U_{ϵ, x_{0}} (x) = \frac{(N (N - 2) ϵ^{2})^{\frac{N - 2}{4}}}{(ϵ^{2} + | x - x_{0} |^{2})^{\frac{N - 2}{2}}}, for~ x_{0} \in R^{N} ~and~ ϵ > 0, \end{aligned}

(1.10)

here

A_{α} = A_{N, α}

and

C_{α} = C (N, α)

given in (1.2) and (1.4) respectively.

The paper’s scheme is the following: Section 2 presents an analysis of the initial findings. In this analysis, we have demonstrated that the functional E exhibits coerciveness on the set

P_{r} (τ)

. Additionally, we have established that the minimum value of E on

P_{r} (τ)

is precisely equal to

σ_{r} (τ)

, a strictly positive quantity with a predetermined upper limit. In Section 3, we obtained a weak limit

u_{τ}

and two sequences of Lagrange multipliers

{λ_{n}}

and

{{\bar{λ}}_{n}}

by considering a minimizing sequence for E restricted to

P_{r} (τ)

. We observed that

{{\bar{λ}}_{n}}

converges to zero by applying the lower bound for T on

P_{r} (τ)

, as demonstrated in Lemma 2.8 and

u_{τ} \neq 0

, providing that

{λ_{n}}

converges to some

λ_{τ}

. Further, using the upper bound for

σ_{r} (τ)

given in Lemma 2.14 and a number of estimates shown in Section 2, we establish that

u_{τ}

is the strong limit that solves (1.1) for

λ = λ_{τ}

. Thus, the existence result follows. In Section 4, we studied the regularity of a positive weak solution to (1.1) by first showing the local boundedness of the solution and then we obtained the global regularity result, thanks to the property of radial functions in

H^{1} (R^{N})

(Strauss Lemma), we could deduce the Sobolev regularity given in Theorem 1.2.

2. .Preliminary results

We first establish some technical results used in the proof of our main theorem:

Lemma 2.1.
Assume that $(u, λ) \in S (τ) \times R$ is a weak solution of (1.1), then $u \in \bar{M}$ .
Proof.
Let $(u, λ)$ be a solution of (1.1), then u is a critical point of F. This gives us:
$\begin{aligned} 0 & = {\frac{d F (u (x / t))}{d t} |}_{t = 1} \\ = (\frac{N - 2}{2}) ‖ \nabla u ‖_{2}^{2} + (\frac{N + 2 s}{2}) [u]^{2} - \frac{N λ}{2} ‖ u ‖_{2}^{2} - N \int_{R^{N}} G (u) - (\frac{N + α}{2 2_{α}^{}}) A (u), \end{aligned}$
and hence
$\begin{aligned} λ ‖ u ‖_{2}^{2} = (\frac{N - 2}{N}) ‖ \nabla u ‖_{2}^{2} + (\frac{N - 2 s}{N}) [u]^{2} - 2 \int_{R^{N}} G (u) - (\frac{N + α}{N 2_{α}^{}}) A (u) . \end{aligned}$
(2.1)
Using (2.1) and the fact that u is a critical point of F, we get
$\begin{aligned} ‖ \nabla u ‖_{2}^{2} + s [u]^{2} - N \int_{R^{N}} \tilde{G} (u) - A (u) = 0. \end{aligned}$

Lemma 2.2.
For any $u \in S (τ)$ , $t \in R^{+}$ is a critical point of $f_{u}$ if and only if $t ⋆ u \in P (τ)$ .
Proof.
Clearly, by the definition of M and $f_{u}$ we have
$\begin{aligned} M (t ⋆ u) & = {‖ \nabla (t ⋆ u) ‖}_{2}^{2} + s [t ⋆ u]^{2} - N \int_{R^{N}} \tilde{G} (t ⋆ u) d x - A (t ⋆ u) \\ = t^{2} ‖ \nabla u ‖_{2}^{2} + s t^{2 s} [u]^{2} - N t^{- N} \int_{R^{N}} \tilde{G} (t^{\frac{N}{2}} u) - t^{2. 2_{α}^{}} A (u) \\ = t f_{u}^{^{'}} (t) . \end{aligned}$
(2.2)
Thus, $t \in R^{+}$ is a critical point of $f_{u}$ if and only if $M (t ⋆ u) = 0$ . Since $‖ t ⋆ u ‖_{2}^{2} = ‖ u ‖_{2}^{2}$ , we get the required result.
Lemma 2.3.
Let* $u \in H^{1} (R^{N})$ is a solution of
$\begin{aligned} {\begin{cases} min E (u), \\ u \in P (τ) \end{cases} \end{aligned}$
(2.3)
with $f_{u}^{^{″}} (1) \neq 0$ , then there exists $λ \in R$ such that
$\begin{aligned} - Δ u + (- Δ)^{s} u = λ u + g (u) + (I_{α} * | u |^{2_{α}^{}}) | u |^{2_{α}^{} - 2} u i n R^{N} . \end{aligned}$
(2.4)

Proof.
Since u solves (2.3), then by the method of Lagrange Multipliers, it is a critical point of
$\begin{aligned} L (u, λ_{1}, λ_{2}) := E (u) + λ_{1} (τ^{2} - ‖ u ‖_{2}^{2}) + λ_{2} M (u), \end{aligned}$
that is,
$\begin{aligned} E^{'} (u) v - λ \int_{R^{N}} u v + λ_{2} M^{'} (u) v = 0 for every v \in H^{1} (R^{N}), \end{aligned}$
and hence, u is a weak solution of
$\begin{aligned} \tilde{E} (u) - λ u + λ_{2} \tilde{M} (u) = 0 in R^{N}, \end{aligned}$
(2.5)
where $\tilde{E} (u) = - Δ u + (- Δ)^{s} u - g (u) - (I_{α} * | u |^{2_{α}^{}}) | u |^{2_{α}^{} - 2} u$ and $\tilde{M} (u) = - 2 Δ u + 2 s (- Δ)^{s} u - N {\tilde{G}}^{^{'}} (u) - 2. 2_{α}^{} (I_{α} | u |^{2_{α}^{}}) | u |^{2_{α}^{} - 2} u$ . Thus,
$\begin{aligned} 0 = {\frac{d J (t^{\frac{N}{2}} u (x / t))}{d t} |}_{t = 1}, \end{aligned}$
where J is the energy functional corresponding to (2.5) given by
$\begin{aligned} J (u) := E (u) - \frac{λ}{2} ‖ u ‖_{2}^{2} + λ_{2} M (u) . \end{aligned}$
Therefore, by using (2.2) we get
$\begin{aligned} 0 = {\frac{d J (t^{\frac{N}{2}} u (x / t))}{d t} |}_{t = 1} & = {\frac{d J (t ⋆ u)}{d t} |}_{t = 1} \\ = f_{u}^{^{'}} (1) + λ_{2} (f_{u}^{^{″}} (1) + f_{u}^{^{'}} (1)) \\ = λ_{2} f_{u}^{^{″}} (1) . \end{aligned}$
Since $f_{u}^{^{″}} (1) \neq 0$ , thus $λ_{2} = 0$ and hence u is a weak solution of (2.4).

Now, we will study the following sets:
$\begin{aligned} P (τ)^{+} & := {u \in P (τ) : f_{u}^{^{″}} (1) > 0}, \\ P (τ)^{0} & := {u \in P (τ) : f_{u}^{^{″}} (1) = 0}, \\ P (τ)^{-} & := {u \in P (τ) : f_{u}^{^{″}} (1) < 0} . \end{aligned}$

Lemma 2.4.
Under the above mentioned assumptions on g, $P (τ)^{-} = P (τ)$ is closed in $H^{1} (R^{N})$ .
Proof.
For $u \in P (τ)$ , we have
$\begin{aligned} ‖ \nabla u ‖_{2}^{2} + s [u]^{2} - A (u) = N \int_{R^{N}} \tilde{G} (u) . \end{aligned}$
(2.6)
Now
$\begin{aligned} f_{u} (t) & = \frac{t^{2}}{2} ‖ \nabla u ‖_{2}^{2} + \frac{t^{2 s}}{2} [u]^{2} - t^{- N} \int_{R^{N}} G (t^{\frac{N}{2}} u) - \frac{t^{2. 2_{α}^{}}}{2. 2_{α}^{}} A (u), \\ f_{u}^{^{'}} (t) & = t ‖ \nabla u ‖_{2}^{2} + s t^{2 s - 1} [u]^{2} - t^{2. 2_{α}^{} - 2} A (u) - N t^{- N - 1} \int_{R^{N}} \tilde{G} (t^{\frac{N}{2}} u) \end{aligned}$
and
$\begin{aligned} f_{u}^{^{″}} (t) & = ‖ \nabla u ‖_{2}^{2} + s (2 s - 1) t^{2 s - 2} [u]^{2} - (2. 2_{α}^{} - 1) t^{2. 2_{α}^{} - 2} A (u) \\ - \frac{N^{2} t^{- N - 2}}{2} \int_{R^{N}} {\tilde{G}}^{^{'}} (t^{\frac{N}{2}} u) (t^{\frac{N}{2}} u) + N (N + 1) t^{- N - 2} \int_{R^{N}} \tilde{G} (t^{\frac{N}{2}} u) . \end{aligned}$
Then by (2.6), (a2) and (a3), we get
$\begin{aligned} f_{u}^{″} (1) = ‖ \nabla u ‖_{2}^{2} + s (2 s - 1) [u]^{2} - ({2.2}_{α}^{} - 1) A (u) - \frac{N^{2}}{2} \int_{R^{N}} {\tilde{G}}^{'} (u) u \\ + N (N + 1) \int_{R^{N}} {\tilde{G}}^{'} (u) \\ ⩽ ‖ \nabla u ‖_{2}^{2} + s (2 s - 1) [u]^{2} - ({2.2}_{α}^{} - 1) A (u) - \frac{N^{2} {\bar{α}}^{2}}{2} \int_{R^{N}} {\tilde{G}}^{'} (u) \\ + N (N + 1) \int_{R^{N}} {\tilde{G}}^{'} (u) \\ = ‖ \nabla u ‖_{2}^{2} + s (2 s - 1) [u]^{2} - ({2.2}_{α}^{} - 1) A (u) \\ + (N + 1 - \frac{N \bar{α}}{2}) (‖ \nabla u ‖_{2}^{2} + s {[u]}^{2} - A (u)) \\ = (\frac{4 + 2 N - N \bar{α}}{2}) ‖ \nabla u ‖_{2}^{2} + s (\frac{4 s + 2 N - N \bar{α}}{2}) [u]^{2} \\ - ({2.2}_{α}^{} + N - \frac{N \bar{α}}{2}) A (u) \\ < 0. \end{aligned}$
(2.7)
Therefore $P (τ) = P (τ)^{-}$ . Now let ${u_{n}}$ be a sequence in $P (τ)$ such that $u_{n} \to u$ in $H^{1} (R^{N})$ . Then clearly, $u \in S (τ)$ . By continuous imbedding of $H^{1} (R^{N})$ in $L^{p} (R^{N})$ for all $p \in (2, 2^{})$ , continuity of $\tilde{G}$ , (a2), (a3) and continuity of A, we get
$\begin{aligned} M (u) & = ‖ \nabla u ‖_{2}^{2} + s [u]^{2} - N \int_{R^{N}} \tilde{G} (u) - A (u) \\ = lim_{n \to \infty} (‖ \nabla u_{n} ‖_{2}^{2} + s [u_{n}]^{2} - N \int_{R^{N}} \tilde{G} (u_{n}) - A (u_{n})) \\ = lim_{n \to \infty} M (u_{n}) = 0. \end{aligned}$
Hence $P (τ) = P (τ)^{-}$ is closed in $H^{1} (R^{N})$ .

Having discussed the sets $P (τ)$ , $P (τ)^{+}$ and $P (τ)^{-}$ , we would like to study the existence of extremas for $f_{u}$ and hence for E on some restricted set.
Lemma 2.5.
Assuming (a1)–(a3), for every $u \in S (τ)$ , there exists unique $t_{u} \in R^{+}$ such that $t_{u} ⋆ u \in P (τ)$ where $t_{u}$ is the strict maxima for $f_{u}$ , that is
$\begin{aligned} E (t_{u} ⋆ u) = max_{t > 0} E (t ⋆ u) . \end{aligned}$

Proof.
By (a2) and Remark 1.1, there exists $C_{g}$ such that
$\begin{aligned} s g (s) ⩽ C_{g} (| s |^{\bar{α}} + | s |^{\bar{β}}), \end{aligned}$
(2.8)
this gives us:
$\begin{aligned} - \int_{R^{N}} \tilde{G} (t^{\frac{N}{2}} u (x)) ⩾ (\frac{2 - \bar{β}}{2 \bar{β}}) \int_{R^{N}} g (t^{\frac{N}{2}} u (x)) t^{\frac{N}{2}} u (x) ⩾ C_{g} (\frac{2 - \bar{β}}{2 \bar{β}}) ({‖ t^{\frac{N}{2}} u ‖}_{\bar{α}}^{\bar{α}} + {‖ t^{\frac{N}{2}} u ‖}_{\bar{β}}^{\bar{β}}) . \end{aligned}$
Now, since $\bar{β} ⩾ \bar{α} > 2 + \frac{4}{N}$ , then for small $t > 0$ , we get
$\begin{aligned} f_{u}^{^{'}} (t) & = t ‖ \nabla u ‖_{2}^{2} + s t^{2 s - 1} [u]^{2} - t^{2. 2_{α}^{} - 1} A (u) - N t^{- N - 1} \int_{R^{N}} \tilde{G} (t^{\frac{N}{2}} u) \\ ⩾ t ‖ \nabla u ‖_{2}^{2} + s t^{2 s - 1} [u]^{2} - t^{2. 2_{α}^{} - 1} A (u) \\ - N t^{- N - 1} C_{g} (\frac{\bar{β} - 2}{2 \bar{β}}) ({‖ t^{\frac{N}{2}} u ‖}_{\bar{α}}^{\bar{α}} + {‖ t^{\frac{N}{2}} u ‖}_{\bar{β}}^{\bar{β}}) \\ > 0. \end{aligned}$
(2.9)
Also, for large t, by (a2) and Remark 1.1 we get
$\begin{aligned} f_{u}^{^{'}} (t) & = t ‖ \nabla u ‖_{2}^{2} + s t^{2 s - 1} [u]^{2} - t^{2. 2_{α}^{} - 1} A (u) - N t^{- N - 1} \int_{R^{N}} \tilde{G} (t^{\frac{N}{2}} u) \\ ⩽ t ‖ \nabla u ‖_{2}^{2} + s t^{2 s - 1} [u]^{2} - t^{2. 2_{α}^{} - 1} A (u) + N t^{- N - 1} (\frac{2 - \bar{α}}{2}) \int_{R^{N}} G (t^{\frac{N}{2}} u) \\ ⩽ t ‖ \nabla u ‖_{2}^{2} + s t^{2 s - 1} [u]^{2} - t^{2. 2_{α}^{} - 1} A (u) - N t^{\frac{N \bar{α}}{2} - N - 1} (\frac{\bar{α} - 2}{2}) \int_{R^{N}} G (u) \\ < 0. \end{aligned}$
(2.10)
By, (2.9) and (2.10) there exists $t_{u} \in (0, \infty)$ such that $f_{u}^{^{'}} (t_{u}) = 0$ . Then by Lemma 2.2, $t_{u} ⋆ u \in P (τ)$ . Let if possible, there exists $t_{1} \neq t_{u}$ (without loss of generality, let $t_{u} < t_{1}$ ) such that $t_{1} ⋆ u \in P (τ)$ . Since, $P (τ) = P (τ)^{-}$ , $t_{u}$ and $t_{1}$ must be two distinct strict local maxima for $f_{u}$ , thus there exists $t_{2} \in (t_{u}, t_{1})$ such that
$\begin{aligned} f_{u} (t_{2}) = min_{t \in (t_{u}, t_{1})} f_{u} (t) . \end{aligned}$
This implies that $t_{2} \in P (τ) \cap P (τ)^{+} = ϕ$ . Therefore, by contradiction, there exists unique $t_{u} \in R^{+}$ such that $f_{u} (t_{u}) = max_{t > 0} f_{u} (t)$ .
Lemma 2.6.
For every* $u \in P (τ)$ , we have
$\begin{aligned} E (u) = max_{t > 0} E (t ⋆ u) . \end{aligned}$

Proof.
Let $u \in P (τ)$ , then by (2.2), $f_{u}^{^{'}} (1) = M (1 ⋆ u) = M (u) = 0$ . Thus, 1 is a critical point of $f_{u}$ , also by similar arguments as in Lemma 2.5, $f_{u}^{^{'}} (1) < 0$ and hence $t_{0} = 1$ is the unique strict local maxima of $f_{u}$ . Therefore,
$\begin{aligned} f_{u} (1) = max_{t > 0} f_{u} (t), or E (u) = max_{t > 0} E (t ⋆ u) . \end{aligned}$

Further, we will study the infimum and supremum of E on the sets $A_{k}$ and $B_{k}$ given by:
$\begin{aligned} A_{k} & := {u \in S_{r} (τ) : T (u) = {(‖ \nabla u ‖_{2}^{2} + [u]^{2})}^{1 / 2} < k}; \\ B_{k} & := {u \in S_{r} (τ) : T (u) {(‖ \nabla u ‖_{2}^{2} + [u]^{2})}^{1 / 2} = k} . \end{aligned}$

Lemma 2.7.
There exists $k_{2} > k_{1} > 0$ such that $E (u), M (u) > 0$ for all $u \in A_{k_{1}}$ and
$\begin{aligned} 0 < sup_{u \in A_{k_{1}}} E (u) ⩽ inf_{u \in B_{k_{2}}} E (u) . \end{aligned}$

Proof.
For a fixed $k > 0$ , let $u, v \in S_{r} (τ)$ be such that $T (u) = k$ and $T (v) = c k$ for some $c > 1$ . Then by Gagliardo–Nirenberg inequality [32] and (1.9) we get:
$\begin{aligned} E (u) & ⩾ \frac{T {(u)}^{2}}{2} - \int_{R^{N}} C_{G} (| u |^{\bar{α}} + | u |^{\bar{β}}) - \frac{‖ \nabla u ‖_{2}^{2. 2_{α}^{}}}{2. 2_{α}^{} (S_{α})^{2_{α}^{}}} \\ ⩾ \frac{T {(u)}^{2}}{2} - C_{G} C_{N, \bar{α}} ‖ \nabla u ‖_{2}^{\frac{N (\bar{α} - 2)}{2}} τ^{\bar{α} - \frac{N (\bar{α} - 2)}{2}} - C_{G} C_{N, \bar{β}} ‖ \nabla u ‖_{2}^{\frac{N (\bar{β} - 2)}{2}} τ^{\bar{β} - \frac{N (\bar{β} - 2)}{2}} \\ - \frac{‖ \nabla u ‖_{2}^{2. 2_{α}^{}}}{2. 2_{α}^{} (S_{α})^{2_{α}^{}}} \\ ⩾ \frac{T {(u)}^{2}}{2} - C_{G} C_{N, \bar{α}} T {(u)}^{\frac{N (\bar{α} - 2)}{2}} τ^{\bar{α} - \frac{N (\bar{α} - 2)}{2}} - C_{G} C_{N, \bar{β}} T {(u)}^{\frac{N (\bar{β} - 2)}{2}} τ^{\bar{β} - \frac{N (\bar{β} - 2)}{2}} \\ - \frac{T {(u)}^{2. 2_{α}^{}}}{2. 2_{α}^{} (S_{α})^{2_{α}^{}}}, \end{aligned}$
also, by (2.8)
$\begin{aligned} M (u) & ⩾ s ‖ \nabla u ‖_{2}^{2} + s [u]^{2} - N C_{g} (\frac{\bar{β} - 2}{2 \bar{β}}) (‖ u ‖_{\bar{α}}^{\bar{α}} + ‖ u ‖_{\bar{β}}^{\bar{β}}) - \frac{‖ \nabla u ‖_{2}^{2. 2_{α}^{}}}{S_{α}^{2_{α}^{}}} \\ ⩾ s T {(u)}^{2} - N C_{g} (\frac{\bar{β} - 2}{2 \bar{β}}) C_{N, \bar{α}} T {(u)}^{\frac{N (\bar{α} - 2)}{2}} τ^{\bar{α} - \frac{N (\bar{α} - 2)}{2}} \\ - N C_{g} (\frac{\bar{β} - 2}{2 \bar{β}}) C_{N, \bar{β}} T {(u)}^{\frac{N (\bar{β} - 2)}{2}} τ^{\bar{β} - \frac{N (\bar{β} - 2)}{2}} - \frac{T {(u)}^{2. 2_{α}^{}}}{S_{α}^{2_{α}^{}}}, \end{aligned}$
and
$\begin{aligned} E (v) - E (u) & ⩾ \frac{T {(v)}^{2}}{2} - \int_{R^{N}} G (v) - \frac{A (v)}{2. 2_{α}^{}} - \frac{T {(u)}^{2}}{2} \\ ⩾ \frac{c^{2} k^{2}}{2} - C_{G} C_{N, \bar{α}} τ^{\bar{α} - \frac{N (\bar{α} - 2)}{2}} T {(v)}^{\frac{N (\bar{α} - 2)}{2}} - C_{G} C_{N, \bar{β}} τ^{\bar{β} - \frac{N (\bar{β} - 2)}{2}} T {(v)}^{\frac{N (\bar{β} - 2)}{2}} \\ - \frac{T {(v)}^{2. 2_{α}^{}}}{2. 2_{α}^{} S_{α}^{2_{α}^{}}} - \frac{k^{2}}{2} \\ = \frac{(c^{2} - 1) k^{2}}{2} - C_{G} C_{N, \bar{α}} τ^{\bar{α} - \frac{N (\bar{α} - 2)}{2}} (c k)^{\frac{N (\bar{α} - 2)}{2}} - C_{G} C_{N, \bar{β}} τ^{\bar{β} - \frac{N (\bar{β} - 2)}{2}} (c k)^{\frac{N (\bar{β} - 2)}{2}} \\ - \frac{(c k)^{2. 2_{α}^{}}}{2. 2_{α}^{} S_{α}^{2_{α}^{}}} . \end{aligned}$
Since $2 + \frac{4}{N} < \bar{α} ⩽ \bar{β}$ , for small k, we get $M (u), E (u) > 0$ and $E (v) - E (u) > 0$ . Thus we can find $k_{2} > k_{1} > 0$ such that for all $u \in A_{k_{1}}$ , we have $E (u), M (u) > 0$ , and taking $k_{2} = c k_{1}$ for some $c > 1$ we get $E (v) - E (u) > 0$ for every $v \in B_{k_{2}}$ . Thus
$\begin{aligned} sup_{u \in A_{k_{1}}} E (u) > 0 \end{aligned}$
and
$\begin{aligned} E (v) ⩾ sup_{u \in A_{k_{1}}} E (u) for all v \in B_{k_{2}} . \end{aligned}$
Hence
$\begin{aligned} inf_{v \in B_{k_{2}}} E (v) ⩾ sup_{u \in A_{k_{1}}} E (u) > 0. \end{aligned}$

Next, we will be looking for a lower bound for $T (u)$ on $P_{r} (τ)$ .
Lemma 2.8.
For any $τ > 0$ there exists $δ_{τ} > 0$ such that
$\begin{aligned} inf {t > 0 : there exists u \in S_{r} (τ) ~with~ T (u) = 1 such that t ⋆ u \in P_{r} (τ)} ⩾ δ_{τ} . \end{aligned}$
(2.11)
Correspondingly
$\begin{aligned} inf_{u \in P_{r} (τ)} T (u) ⩾ δ_{τ} . \end{aligned}$
(2.12)

Proof.
Let $u \in S_{r} (τ)$ with $T (u) = 1$ , if $t ⋆ u \in P_{r} (τ) \subset P (τ)$ , then by Remark 1.1, (2.8) and (1.9) we have
$\begin{aligned} s t & = s t T {(u)}^{2} ⩽ t ‖ \nabla u ‖_{2}^{2} + s t^{2 s - 1} [u]^{2} \\ = N t^{- N - 1} \int_{R^{N}} \tilde{G} (t^{\frac{N}{2}} u) + t^{2. 2_{α}^{} - 1} A (u) \\ ⩽ (\frac{\bar{β} - 2}{2 \bar{β}}) N C_{g} {\int_{R^{N}} (t^{\frac{N \bar{α}}{2} - N - 1} | u |^{\bar{α}} + t^{\frac{N \bar{β}}{2} - N - 1} | u |^{\bar{β}})} \\ + t^{2. 2_{α}^{} - 1} S_{α}^{- 2_{α}^{}}, \end{aligned}$
for small t. Further, using Gagliardo–Nirenberg inequality, we get
$\begin{aligned} s t & ⩽ (\frac{\bar{β} - 2}{2 \bar{β}}) N C_{g} (t^{\frac{N \bar{α}}{2} - N - 1} C_{\bar{α}} + t^{\frac{N \bar{β}}{2} - N - 1} C_{\bar{β}}) + t^{2. 2_{α}^{} - 1} S_{α}^{- 2_{α}^{}} \\ ⩽ ζ (t^{\frac{N \bar{α}}{2} - N - 1} + t^{\frac{N \bar{β}}{2} - N - 1} + t^{2. 2_{α}^{} - 1}) . \end{aligned}$
Therefore,
$\begin{aligned} s ⩽ ζ (t^{\frac{N \bar{α}}{2} - N - 2} + t^{\frac{N \bar{β}}{2} - N - 2} + t^{2. 2_{α}^{} - 2}) . \end{aligned}$
Since $(\frac{N \bar{α}}{2} - N - 2)$ , $(\frac{N \bar{β}}{2} - N - 2)$ and $(2. 2_{α}^{} - 2) > 0$ , there exists $δ_{τ} > 0$ , satisfying (2.11). Now, let $u \in P_{r} (τ)$ be arbitrary. Since $M (u) = 0$ , then again using Gagliardo–Nirenberg inequality and (2.8) we get
$\begin{aligned} s T {(u)}^{2} & ⩽ ‖ \nabla u ‖_{2}^{2} + s [u]^{2} = N \int_{R^{N}} \tilde{G} (u) + A (u) \\ ⩽ N C_{g}^{^{'}} (\frac{\bar{β} - 2}{2 \bar{β}}) (T {(u)}^{\frac{N (\bar{α} - 2)}{2}} + T {(u)}^{\frac{N (\bar{β} - 2)}{2}}) + \frac{T {(u)}^{2. 2_{α}^{}}}{S_{α}^{2_{α}^{}}} \\ ⩽ ζ (T {(u)}^{\frac{N (\bar{α} - 2)}{2}} + T {(u)}^{\frac{N (\bar{β} - 2)}{2}} + T {(u)}^{2. 2_{α}^{}}) . \end{aligned}$
Therefore,
$\begin{aligned} s ⩽ ζ (T {(u)}^{\frac{N (\bar{α} - 2)}{2} - 2} + T {(u)}^{\frac{N (\bar{β} - 2)}{2} - 2} + T {(u)}^{2. 2_{α}^{} - 2}) \end{aligned}$
implying $T (u) ⩾ δ_{τ}$ . Since $u \in P_{r} (τ)$ is arbitrary, we get (2.12).

Now, we will be working in order to establish a relation between $σ_{r} (τ)$ , stated in the following Lemma, and the infimum of E on $P (τ)$ and $P_{r} (τ)$ . For the same, let us define $I (u, t) := E (t ⋆ u)$ and $E^{k} := {u \in S_{r} (τ) : E (u) ⩽ k}$ .
Lemma 2.9.
Set
$\begin{aligned} {\tilde{σ}}_{r} (τ) := inf_{\tilde{η} \in \tilde{Γ} (τ)} {max_{x \in [0, 1]} I (\tilde{η} (x))}, \end{aligned}$
where
$\begin{aligned} \tilde{Γ} (τ) = {\tilde{η} \in C ([0, 1], S_{r} (τ) \times R) : \tilde{η} (0) \in (A_{k_{1}}, 1) ~and~ \tilde{η} (1) \in (E^{0}, 1)}, \end{aligned}$
and
$\begin{aligned} σ_{r} (τ) := inf_{η \in Γ (τ)} {max_{x \in [0, 1]} E (η (x))}, \end{aligned}$
where
$\begin{aligned} Γ = {η \in C ([0, 1], S_{r} (τ)) : η (0) \in A_{k_{1}} ~and~ η (1) \in E^{0}} . \end{aligned}$
Then, ${\tilde{σ}}_{r} (τ) = σ_{r} (τ)$ .

The proof can be deduced from the proof of Proposition 2.1 of [18].
Lemma 2.10.
Let ${h_{n}} \subset \tilde{Γ} (τ)$ be such that
$\begin{aligned} max_{x \in [0, 1]} I (h_{n} (x)) ⩽ {\tilde{σ}}_{r} (τ) + \frac{1}{n} . \end{aligned}$
Then, assuming (a1)–(a3), there exists a sequence ${(u_{n}, t_{n})} \subset S_{r} (τ) \times R$ such that for all $n \in N$
$I (u_{n}, t_{n}) \in [{\tilde{σ}}_{r} (τ) - \frac{1}{n}, {\tilde{σ}}_{r} (τ) + \frac{1}{n}]$ ,

$min_{x \in [0, 1]} ‖ (u_{n}, t_{n}) - h_{n} (x) ‖_{H} ⩽ \frac{1}{\sqrt{n}}$ ,

$‖ I_{S_{r} (τ) \times R}^{^{'}} (u_{n}, t_{n}) ‖ ⩽ \frac{2}{\sqrt{n}}$ , that is
$\begin{aligned} | I^{'} (u_{n}, t_{n}) (z) | ⩽ \frac{2}{\sqrt{n}} ‖ z ‖_{H} \forall z \in {\tilde{T}}_{(u_{n}, t_{n})} = {(z_{1}, z_{2}) \in H : \int_{R^{N}} u_{n} z_{1} = 0} . \end{aligned}$

The proof follows from Proposition 2.2 of [18].
Lemma 2.11.
For $u \in S_{r} (τ)$ define
$\begin{aligned} T_{u} := {h \in H_{r} (R^{N}) : \int_{R^{N}} u h = 0}, \end{aligned}$
then there exists a sequence ${v_{n}} \subset S_{r} (τ)$ such that
$lim_{n \to \infty} E (v_{n}) = σ_{r} (τ)$ ,

$lim_{n \to \infty} M (v_{n}) = 0$ ,

$lim_{n \to \infty} E_{S_{r} (τ)}^{^{'}} = 0$ , that is, $| E^{'} (v_{n}) (h) | \to 0$ uniformly for all $h \in T_{v_{n}}$ with $‖ h ‖ ⩽ 1$ .

Proof.
By Lemma 2.9, we have ${\tilde{σ}}_{r} (τ) = σ_{r} (τ)$ . Let ${{\tilde{g}}_{n}} = {(g_{n}, 1)} \subset \tilde{Γ} (τ)$ be such that
$\begin{aligned} max_{x \in [0, 1]} I ({\tilde{g}}_{n} (x)) ⩽ {\tilde{σ}}_{r} (τ) + \frac{1}{n} for every n \in N . \end{aligned}$
Then by Lemma 2.10, there exists a sequence ${u_{n}, t_{n}} \subset S_{r} (τ) \times R$ such that
$I (u_{n}, t_{n}) \in [{\tilde{σ}}_{r} (τ) - \frac{1}{n}, {\tilde{σ}}_{r} (τ) + \frac{1}{n}]$ , and hence
$\begin{aligned} {I (u_{n}, t_{n})} \to {\tilde{σ}}_{r} (τ) as n \to \infty, \end{aligned}$
(2.13)

$min_{x \in [0, 1]} ‖ (u_{n}, t_{n}) - g_{n} (x) ‖_{H} ⩽ \frac{1}{\sqrt{n}}$ . This gives us
$\begin{aligned} | t_{n} - 1 | ⩽ min_{x \in [0, 1]} {‖ (u_{n}, t_{n}) - g_{n} (x) ‖}_{H} ⩽ \frac{1}{\sqrt{n}} \to 0 as n \to \infty . \end{aligned}$

$| I^{'} (u_{n}, t_{n}) (0, 1) | ⩽ \frac{2}{\sqrt{n}} ‖ (0, 1) ‖_{H} = \frac{2}{\sqrt{n}} \to 0$ as $n \to \infty$ .
By direct computation, one can easily see that
$\begin{aligned} | \frac{\partial}{\partial t} I (u_{n}, t_{n}) | = | I^{'} (u_{n}, t_{n}) (0, 1) | \to 0 as n \to \infty . \end{aligned}$
(2.14)
Set $v_{n} := t_{n} ⋆ u_{n}$ , then by (2.13) we get $E (v_{n}) = E (t_{n} ⋆ u_{n}) = I (u_{n}, t_{n}) \to {\tilde{σ}}_{r} (τ) = σ_{r} (τ)$ . Also, by (2.14)
$\begin{aligned} 0 & = lim_{n \to \infty} \frac{\partial}{\partial t} I (u_{n}, t_{n}) \\ = lim_{n \to \infty} (t_{n} ‖ \nabla u_{n} ‖_{2}^{2} + s t_{n}^{2 s - 1} [u_{n}]^{2} - N t^{- N - 1} \int_{R^{N}} \tilde{G} (t_{n}^{\frac{N}{2}} u_{n}) - t_{n}^{2. 2_{α}^{} - 1} A (u_{n})) \\ = lim_{n \to \infty} \frac{1}{t_{n}} M (t_{n} ⋆ u_{n}) = lim_{n \to \infty} M (v_{n}) . \end{aligned}$
Now, let $h_{n} \in T_{v_{n}}$ be arbitrary. Then,
$\begin{aligned} E^{'} (v_{n}) (h_{n}) & = t_{n}^{1 - \frac{N}{2}} \int_{R^{n}} \nabla v_{n} (y) . \nabla h_{n} (t_{n}^{- 1} y) d y - t_{n}^{- N} \int_{R^{N}} g (t_{n}^{\frac{N}{2}} u_{n} (y)) h_{n} (t_{n}^{- 1} y) d y \\ + \frac{C (N, s)}{2} t_{n}^{2 s - \frac{N}{2}} \int_{R^{N}} \int_{R^{N}} \frac{(u_{n} (x) - u_{n} (y)) (h_{n} (t_{n}^{- 1} x) - h_{n} (t_{n}^{- 1} y))}{| x - y |^{N + 2 s}} d x d y \\ - t_{n}^{2_{α}^{} N - α - \frac{3}{2} N} \int_{R^{N}} \int_{R^{N}} \frac{A_{α} | u_{n} (x) |^{2_{α}^{}} | u_{n} (y) |^{2_{α}^{} - 2} u_{n} (x) h_{n} (t_{n}^{- 1} x)}{| x - y |^{N + 2 s}} d x d y . \end{aligned}$
Taking ${\hat{h}}_{n} (x) = t_{n}^{- \frac{N}{2}} h_{n} (t_{n}^{- 1} x)$ , we get
$\begin{aligned} E^{'} (v_{n}) (h_{n}) & = t_{n}^{2} \int_{R^{N}} \nabla u_{n} \nabla {\hat{h}}_{n} + \frac{C (N, s)}{2} t_{n}^{2 s} \int_{R^{N}} \int_{R^{N}} \frac{(u_{n} (x) - u_{n} (y)) ({\hat{h}}_{n} (x) - {\hat{h}}_{n} (y)}{| x - y |^{N + 2 s}} d x d y \\ - t_{n}^{2. 2_{α}^{}} \int_{R^{N}} \int_{R^{N}} \frac{A_{α} | u_{n} (x) |^{2_{α}^{}} | u_{n} (y) |^{2_{α}^{} - 2} u_{n} (x) {\hat{h}}_{n} (x)}{| x - y |^{N + 2 s}} d x d y \\ - t_{n}^{- \frac{N}{2}} \int_{R^{N}} g (t_{n}^{\frac{N}{2}} u) {\hat{h}}_{n} \\ = E^{'} (t_{n} ⋆ u_{n}) (t_{n} ⋆ {\hat{h}}_{n}) \\ = I^{'} (u_{n}, t_{n}) ({\hat{h}}_{n}, 0) . \end{aligned}$
Since $h_{n} \in T_{v_{n}}$ , then $({\hat{h}}_{n}, 0) \in {\tilde{T}}_{(u_{n}, t_{n})}$ and hence by Lemma 2.10,
$\begin{aligned} | E^{'} (v_{n}) (h_{n}) | = | I^{'} (u_{n}, t_{n}) ({\hat{h}}_{n}, 0) | ⩽ \frac{2}{\sqrt{n}} {‖ ({\hat{h}}_{n}, 0) ‖}_{H} ⩽ \frac{2}{\sqrt{n}} C {‖ h_{n} ‖}^{2} for large n . \end{aligned}$
Thus,
$\begin{aligned} sup {| E^{'} (v_{n}) (h) | : h \in T_{v_{n}}, ‖ h ‖ ⩽ 1} ⩽ \frac{C^{'}}{\sqrt{n}} \to 0 as n \to \infty . \end{aligned}$

Defining
$\begin{aligned} m (τ) := inf_{u \in P (τ)} E (u), \end{aligned}$
and
$\begin{aligned} m_{r} (τ) := inf_{u \in P_{r} (τ)} E (u), \end{aligned}$
we have the following result:
Lemma 2.12.
$m_{r} (τ) = σ_{r} (τ) > 0$ .
Proof.
Step 1: We claim that if $E (u) ⩽ 0$ for some $u \in S_{r} (τ)$ then $M (u) < 0$ .

Let $u \in S_{r} (τ)$ be such that $E (u) ⩽ 0$ , then by Lemma 2.5 there exists unique $t_{u} \in R^{+}$ such that $t_{u} ⋆ u \in P_{r} (τ)$ . Also, we know that $f_{u} (0) = 0$ , $f_{u}$ is strictly increasing in $(0, t_{u})$ and strictly decreasing in $(t_{u}, \infty)$ , thus $f_{u} (t_{u}) > 0$ . Now, since $f_{u} (1) = E (u) ⩽ 0$ , we get $t_{u} < 1$ . Therefore, by (2.2) $M (u) = f_{u}^{^{'}} (1) < 0$ .

Step 2: We claim that $σ_{r} (τ) = m_{r} (τ)$ .

For $u \in P_{r} (τ) \subset S_{r} (τ)$ , let $t^{-} \in (0, 1)$ and $t^{+} > 1$ be such that $t^{-} ⋆ u \in A_{k_{1}}$ and $E (t^{+} ⋆ u) < 0$ . Now define the path $η_{u} : [0, 1] \to S_{r} (τ)$ such that
$\begin{aligned} η_{u} (z) = ((1 - z) t^{-} + z t^{+}) ⋆ u . \end{aligned}$
Since $η_{u} \in Γ (τ)$ , we get $σ_{r} (τ) ⩽ max_{x \in [0, 1]} E (η_{u} (x))$ , also, by Lemma 2.6
$\begin{aligned} E (u) = f_{u} (1) ⩾ f_{u} (t) = E (t ⋆ u) for all t \in [t^{-}, t^{+}] . \end{aligned}$
Hence
$\begin{aligned} E (u) = max_{z \in [0, 1]} E (η_{u} (z)) ⩾ σ_{r} (τ), \end{aligned}$
as $u \in P_{r} (τ)$ is arbitrary, we get
$\begin{aligned} m_{r} (τ) ⩾ σ_{r} (τ) . \end{aligned}$
Let us define
$\begin{aligned} \tilde{M} (z) := M ({\tilde{η}}_{2} (z) ⋆ {\tilde{η}}_{1} (z)) for every z \in [0, 1], \end{aligned}$
where $\tilde{η} = ({\tilde{η}}_{1}, {\tilde{η}}_{2}) \in \tilde{Γ} (τ)$ . By Lemma 2.7, $\tilde{M} (0) = M ({\tilde{η}}_{1} (0)) > 0$ and since ${\tilde{η}}_{1} (1) \in E^{0}$ , we get $E ({\tilde{η}}_{1} (1)) ⩽ 0$ . Thus by step 1, $\tilde{M} (1) = M ({\tilde{η}}_{1} (1)) < 0$ . Hence, there exists $\tilde{z} \in (0, 1)$ such that $\tilde{M} (\tilde{z}) = 0$ , which implies that ${\tilde{η}}_{2} (\tilde{z}) ⋆ {\tilde{η}}_{1} (\tilde{z}) \in P_{r} (τ)$ . Therefore, we get
$\begin{aligned} max_{z \in [0, 1]} E ({\tilde{η}}_{2} (z) ⋆ {\tilde{η}}_{1} (z)) ⩾ E ({\tilde{η}}_{2} (\tilde{z}) ⋆ {\tilde{η}}_{1} (\tilde{z})) ⩾ inf_{u \in P_{r} (τ)} E (u) = m_{r} (τ) . \end{aligned}$
Since $\tilde{η} \in \tilde{Γ}$ is arbitrary, we get: $σ_{r} (τ) ⩾ m_{r} (τ)$ .

Step 3: $σ_{r} (τ) > 0$ .

For $u \in P_{r} (τ)$ we have
$\begin{aligned} N \int_{R^{N}} \tilde{G} (u) = ‖ \nabla u ‖_{2}^{2} + s [u]^{2} - A (u) ⩽ ‖ \nabla u ‖_{2}^{2} + s [u]^{2}, \end{aligned}$
(2.15)
and by (a2)
$\begin{aligned} \tilde{G} (u) ⩾ (\frac{\bar{α} - 2}{2}) G (u) . \end{aligned}$
(2.16)
Using (2.15), (2.16) and Lemma 2.8, we deduce that
$\begin{aligned} E (u) & = E (u) - \frac{1}{2. 2_{α}^{}} M (u) \\ = (\frac{2_{α}^{} - 1}{2. 2_{α}^{}}) ‖ \nabla u ‖_{2}^{2} + (\frac{2_{α}^{} - s}{2. 2_{α}^{}}) [u]^{2} - \int_{R^{N}} G (u) + \frac{N}{2. 2_{α}^{}} \int_{R^{N}} \tilde{G} (u) \\ ⩾ (\frac{2_{α}^{} - 1}{2. 2_{α}^{}}) ‖ \nabla u ‖_{2}^{2} + (\frac{2_{α}^{} - s}{2. 2_{α}^{}}) [u]^{2} - \int_{R^{N}} G (u) + \frac{N (\bar{α} - 2)}{4. 2_{α}^{}} \int_{R^{N}} G (u) \\ = (\frac{2_{α}^{} - 1}{2. 2_{α}^{}}) ‖ \nabla u ‖_{2}^{2} + (\frac{2_{α}^{} - s}{2. 2_{α}^{}}) [u]^{2} + (\frac{N (\bar{α} - 2) - 4. 2_{α}^{}}{4. 2_{α}^{}}) \int_{R^{N}} G (u) \\ ⩾ (\frac{2_{α}^{} - 1}{2. 2_{α}^{}}) ‖ \nabla u ‖_{2}^{2} + (\frac{2_{α}^{} - s}{2. 2_{α}^{}}) [u]^{2} + (\frac{N (\bar{α} - 2) - 4. 2_{α}^{}}{4. 2_{α}^{}}) (\frac{2}{N (\bar{α} - 2)}) (‖ \nabla u ‖_{2}^{2} + s [u]^{2}) \\ ⩾ (\frac{N (\bar{α} - 2) - 4}{2 N (\bar{α} - 2)}) ‖ \nabla u ‖_{2}^{2} + (\frac{N (\bar{α} - 2) - 4 s}{2 N (\bar{α} - 2)}) [u]^{2} \\ ⩾ (\frac{N (\bar{α} - 2) - 4}{2 N (\bar{α} - 2)}) T {(u)}^{2} ⩾ (\frac{N (\bar{α} - 2) - 4}{2 N (\bar{α} - 2)}) δ_{τ}^{2} = C_{δ}, \end{aligned}$
(2.17)
therefore, $E (u) ⩾ C_{δ}$ for all $u \in P_{r} (τ)$ , and hence $σ_{r} (τ) = m_{r} (τ) ⩾ C_{δ} > 0$ .

We shall also observe that $m_{r} (τ) = m (τ)$ ; however, in order to do so, we must comprehend the idea of symmetric decreasing rearrangement $f^{}$ of a function f, see [4]. Using the Layer cake decomposition and Fubini’s theorem, we get
$\begin{aligned} ‖ f ‖_{p} = ‖ f^{} ‖_{p}, \end{aligned}$
that is, symmetrization preserves $L^{p} -$ norm. Moreover, if $ϕ = ϕ_{1} - ϕ_{2}$ is such that both $ϕ_{1}$ and $ϕ_{2}$ are monotonic, and either $\int_{R^{N}} ϕ_{1} (| f |)$ or $\int_{R^{N}} ϕ_{2} (| f |)$ is finite, then
$\begin{aligned} \int_{R^{N}} ϕ (| f |) d x = \int_{R^{N}} ϕ (f^{}) d x, \end{aligned}$
see [20]. Using the classical rearrangement inequalities given in [4,12,20] we deduce the following:
Since, the Reisz potential $I_{α}$ is symmetric and radially decreasing, we get
$\begin{aligned} A (u) = \int_{R^{N}} (I_{α} | u |^{2_{α}^{}}) | u |^{2_{α}^{}} d x ⩽ \int_{R^{N}} (I_{α} * | u^{} |^{2_{α}^{}}) | u^{} |^{2_{α}^{}} d x = A (u^{}) . \end{aligned}$
(2.18)

$\begin{aligned} ‖ \nabla f ‖_{p} ⩾ ‖ \nabla f^{} ‖_{p}, for all 1 ⩽ p ⩽ \infty . \end{aligned}$

Remark 2.1.
Using above observations and the classical rearrangement inequalities, we have the following:
$[u^{}]^{2} ⩽ [u]^{2}$ ,

$\int_{R^{N}} G (u^{}) ⩾ \int_{R^{N}} G (u)$ .

Proof.
By Fubini’s theorem, Reisz inequality and the fact that symmetrization preserves $L^{p} -$ norm, we get
$\begin{aligned} [u^{}]^{2} & = \int_{R^{N}} \int_{R^{N}} \frac{| u^{} (x) - u^{} (y) |^{2}}{| x - y |^{N + 2 s}} d x d y \\ = \int_{R^{N}} \int_{R^{N}} \frac{| u^{} (x) |^{2}}{| x - y |^{N + 2 s}} d x d y + \int_{R^{N}} \int_{R^{N}} \frac{| u^{} (y) |^{2}}{| x - y |^{N + 2 s}} d x d y \\ - 2 \int_{R^{N}} \int_{R^{N}} \frac{u^{} (x) u^{} (y)}{| x - y |^{N + 2 s}} d x d y \\ = \int_{R^{N}} \frac{1}{| z |^{N + 2 s}} (\int_{R^{N}} | u^{} (y + z) |^{2} d y) d z + \int_{R^{N}} \frac{1}{| z |^{N + 2 s}} (\int_{R^{N}} | u^{} (x + z) |^{2} d y) d z \\ - 2 \int_{R^{N}} (f u^{}) (x) u^{} (x) d x where~ f (x) = \frac{1}{| x |^{N + 2 s}} \\ = \int_{R^{N}} \frac{‖ u^{} ‖_{2}^{2}}{| z |^{N + 2 s}} d z + \int_{R^{N}} \frac{‖ u^{} ‖_{2}^{2}}{| z |^{N + 2 s}} d z - 2 \int (f^{} u^{}) (x) u^{} (x) d x \\ ⩽ \int_{R^{N}} \frac{‖ u ‖_{2}^{2}}{| z |^{N + 2 s}} d z + \int_{R^{N}} \frac{‖ u ‖_{2}^{2}}{| z |^{N + 2 s}} d z - 2 \int (f * u) (x) u (x) d x \\ = \int_{R^{N}} \int_{R^{N}} \frac{| u (x) - u (y) |^{2}}{| x - y |^{N + 2 s}} = [u]^{2} . \end{aligned}$
Also, since G is non-decreasing, we get
$\begin{aligned} \int_{R^{N}} G (u^{}) = \int_{R^{N}} G (| u |) ⩾ \int_{R^{N}} G (u) . \end{aligned}$

Lemma 2.13.
$m (τ) = m_{r} (τ)$ .
Proof.
For $u \in P (τ)$ , let $u^{}$ be the symmetric decreasing rearrangement of u. Clearly, by the properties of decreasing rearrangement $E (u^{}) ⩽ E (u)$ and hence
$\begin{aligned} m_{r} (τ) = inf_{u \in P_{r} (τ)} E (u) ⩾ inf_{u \in P (τ)} E (u) = m (τ) . \end{aligned}$
Now,
$\begin{aligned} | {x : (t ⋆ u^{}) (x) > λ} | & = | {x : t^{\frac{N}{2}} u^{} (t x) > λ} | \\ = | {t^{- 1} y : u^{} (y) > t^{- \frac{N}{2}} λ} | \\ = | {t^{- 1} y : u (y) > t^{- \frac{N}{2}} λ} | \\ = | {x : (t ⋆ u) (x) > λ} | \end{aligned}$
from which one gets $(t ⋆ u)^{} = t ⋆ u^{}$ . Therefore,
$\begin{aligned} E (t ⋆ u^{}) = E ((t ⋆ u)^{}) ⩽ E (t ⋆ u) ⩽ max_{t > 0} E (t ⋆ u) = E (u) \forall u \in P (τ) . \end{aligned}$
(2.19)
Also, for all $u \in P (τ)$ , $u^{} \in S_{r} (τ)$ . Then, by Lemma 2.5 there exists unique $t_{u^{}} > 0$ such that $t_{u^{}} ⋆ u^{} \in P_{r} (τ)$ . Now, using (2.19) we get $E (t_{u^{}} ⋆ u^{}) ⩽ E (u)$ and
$\begin{aligned} m_{r} (τ) = inf_{u \in P_{r} (τ)} E (u) ⩽ inf_{u \in P (τ)} E (u) = m (τ) . \end{aligned}$

Now, from $m_{r} (τ) = m (τ) = σ_{r} (τ)$ , we will be looking for an upper bound for these quantities.
Lemma 2.14.
There exists $γ > 0$ such that for $0 < τ < γ$
$\begin{aligned} σ_{r} (τ) < (\frac{2_{α}^{} - 1}{2. 2_{α}^{}}) S_{α}^{\frac{2_{α}^{}}{2_{α}^{} - 1}} . \end{aligned}$

Proof.
Let $ψ \in C_{c}^{\infty} (R^{N}, [0, 1])$ be such that $ψ (x) = 1 \forall x \in B_{1} (0)$ , and $ψ (x) = 0$ for all $x \in R^{N} ∖ B_{2} (0)$ . We set $u_{ϵ} (x) := ψ (x) U_{ϵ} (x)$ , where
$\begin{aligned} U_{ϵ} (x) = \frac{{(N (N - 2) ϵ^{2})}^{\frac{N - 2}{4}}}{{(ϵ^{2} + | x |^{2})}^{\frac{N - 2}{2}}} \end{aligned}$
achieves $S_{α}$ , and $v_{ϵ} (x) := \frac{τ u_{ϵ} (x)}{‖ u_{ϵ} ‖_{2}} \in S (τ)$ . In Lemma 4.7 of [34], it has been seen that:
$\begin{aligned} ‖ \nabla u_{ϵ} ‖_{2}^{2} = S^{\frac{N}{2}} + O (ϵ^{N - 2}), \end{aligned}$
(2.20)

$\begin{aligned} ‖ u_{ϵ} ‖_{2}^{2} = {\begin{cases} K_{1} ϵ^{2} + O (ϵ^{N - 2}), & for N ⩾ 5, \\ K_{1} ϵ^{2} | \ln (ϵ) | + O (ϵ^{2}), & for N = 4, \\ K_{1} ϵ + O (ϵ^{2}), & for N = 3, \end{cases} \end{aligned}$
(2.21)
and
$\begin{aligned} A (u_{ϵ}) ⩾ (A_{α} C_{α})^{\frac{N}{2}} S_{α}^{\frac{N + α}{2}} - O (ϵ^{\frac{N + α}{2}}), \end{aligned}$
(2.22)
for some positive constant $K_{1}$ . Also, since $H^{1} (R^{N}) ↪ H^{s} (R^{N})$ , one can deduce that:
$\begin{aligned} [u_{ϵ}]^{2} = {\begin{cases} C^{2} S^{\frac{N}{2}} + O (ϵ^{2}), & for~ N ⩾ 5, \\ C^{2} S^{2} + C^{2} k_{1} ϵ^{2} | \ln (ϵ) | + O (ϵ^{2}), & for~ N = 4, \\ C^{2} S^{\frac{3}{2}} + O (ϵ), & for~ N = 3, \end{cases} \end{aligned}$
(2.23)
for some $C > 0$ . Now, for $p ⩾ 2 + \frac{4}{N}$
$\begin{aligned} \int_{R^{N}} | u_{ϵ} |^{p} & = \int_{R^{N}} {| \frac{(N (N - 2) ϵ^{2})^{\frac{N - 2}{4}}}{(ϵ^{2} + | x |^{2})^{\frac{N - 2}{2}}} |^{p} | ψ (x) |}^{p} d x \\ ⩽ (N (N - 2))^{\frac{p (N - 2)}{2}} ϵ^{N - \frac{p (N - 2)}{2}} (\int_{R^{N}} \frac{| ψ (ϵ x) - 1 |^{p}}{(1 + | x |^{2})^{\frac{p (N - 2)}{2}}} d x + \int_{R^{N}} \frac{d x}{(1 + | x |^{2})^{\frac{p (N - 2)}{2}}}) \\ ⩽ \frac{C_{N}}{2} ϵ^{N - \frac{p (N - 2)}{2}} (\frac{ϵ^{p (N - 2) - N}}{N - p (N - 2)} + C_{1}) . \end{aligned}$
Therefore,
$\begin{aligned} ‖ u_{ϵ} ‖_{p}^{p} = ϵ^{N - \frac{p (N - 2)}{2}} (C_{1} + O (ϵ^{p (N - 2) - N})) \forall p ⩾ 2 + \frac{4}{N} . \end{aligned}$
(2.24)
Thus, for $α_{1} = 2 + \frac{4}{N}$ , we have:
$\begin{aligned} \frac{‖ u_{ϵ} ‖_{α_{1}}^{α_{1}}}{‖ u_{ϵ} ‖_{2}^{\frac{4}{N}}} ⩽ {\begin{cases} C_{2}, & for~ N ⩾ 5, \\ C_{2} | \ln (ϵ) |^{\frac{- 1}{2}}, & for~ N = 4, \\ C_{2} ϵ^{\frac{2}{3}}, & for~ N = 3, \end{cases} \end{aligned}$
(2.25)
and for $q = \bar{α}, \bar{β}$
$\begin{aligned} \frac{‖ u_{ϵ} ‖_{q}^{q}}{‖ u_{ϵ} ‖_{2}^{q - \frac{N (q - 2)}{2}}} ⩽ {\begin{cases} C_{3} ϵ^{N - \frac{q (N - 2)}{2} - q + \frac{N (q - 2)}{2}} = c_{3}, & for~ N ⩾ 5, \\ C_{3} ϵ^{2} | \ln (ϵ) |^{- q - 2}, & for~ N = 4, \\ C_{3} ϵ^{\frac{3}{2} - \frac{q}{4}}, & for~ N = 3. \end{cases} \end{aligned}$
(2.26)
Since $v_{ϵ} \in S (τ)$ , then by Lemma 2.5 there exists unique $t_{ϵ} > 0$ such that $f_{v_{ϵ}}^{^{'}} (t_{ϵ}) = 0$ . Let us define
$\begin{aligned} B_{ϵ} (u) := ‖ \nabla u ‖_{2}^{2} + s t_{ϵ}^{2 s - 2} [u]^{2}, \end{aligned}$
then we get
$\begin{aligned} t_{ϵ}^{2. 2_{α}^{} - 1} A (v_{ϵ}) = t_{ϵ} B_{ϵ} (v_{ϵ}) - N t_{ϵ}^{- N - 1} \int_{R^{N}} \tilde{G} (t_{ϵ}^{\frac{N}{2}} v_{ϵ}) . \end{aligned}$
Therefore,
$\begin{aligned} t_{ϵ} ⩽ {(\frac{B_{ϵ} (v_{ϵ})}{A (v_{ϵ})})}^{\frac{1}{2 (2_{α}^{} - 1)}} . \end{aligned}$
(2.27)
Also, by (a2), (2.8) and the fact that $f_{v_{ϵ}}^{^{'}} (t_{ϵ}) = 0$ , we get:
$\begin{aligned} t_{ϵ}^{2 (2_{α}^{} - 1)} & = \frac{B_{ϵ} (v_{ϵ})}{A (v_{ϵ})} - \frac{N t_{ϵ}^{- N - 2}}{A (v_{ϵ})} \int_{R^{N}} \tilde{G} (t_{ϵ}^{\frac{N}{2}} v_{ϵ}) \\ ⩾ \frac{B_{ϵ} (v_{ϵ})}{A (v_{ϵ})} - \frac{t_{ϵ}^{\frac{N (\bar{α} - 2)}{2} - 2} C_{g}^{^{'}} ‖ v_{ϵ} ‖_{\bar{α}}^{\bar{α}}}{A (v_{ϵ})} - \frac{t_{ϵ}^{\frac{N (\bar{β} - 2)}{2} - 2} C_{g}^{^{'}} ‖ v_{ϵ} ‖_{\bar{β}}^{\bar{β}}}{A (v_{ϵ})} \\ = \frac{B_{ϵ} (v_{ϵ})}{3 A (v_{ϵ})} + (\frac{B_{ϵ} (v_{ϵ})}{3 A (v_{ϵ})} - \frac{t_{ϵ}^{\frac{N (\bar{α} - 2)}{2} - 2} C_{g}^{^{'}} ‖ v_{ϵ} ‖_{\bar{α}}^{\bar{α}}}{A (v_{ϵ})}) + (\frac{B_{ϵ} (v_{ϵ})}{3 A (v_{ϵ})} - \frac{t_{ϵ}^{\frac{N (\bar{β} - 2)}{2} - 2} C_{g}^{^{'}} ‖ v_{ϵ} ‖_{\bar{β}}^{\bar{β}}}{A (v_{ϵ})}) . \end{aligned}$
We claim that for $0 < ϵ < ϵ_{0}$
$\begin{aligned} \frac{B_{ϵ} (v_{ϵ})}{3 A (v_{ϵ})} - \frac{t_{ϵ}^{\frac{N (q - 2)}{2} - 2} C_{g}^{^{'}} ‖ v_{ϵ} ‖_{q}^{q}}{A (v_{ϵ})} > 0, \end{aligned}$
(2.28)
and hence
$\begin{aligned} t_{ϵ}^{2 (2_{α}^{} - 1)} > \frac{B_{ϵ} (v_{ϵ})}{3 A (v_{ϵ})} > \frac{‖ \nabla v_{ϵ} ‖_{2}^{2}}{3 A (v_{ϵ})} = \frac{‖ u_{ϵ} ‖_{2}^{2 (2_{α}^{} - 1)} ‖ \nabla u_{ϵ} ‖_{2}^{2}}{3 τ^{2 (2_{α}^{} - 1)} A (u_{ϵ})} ⩾ C (N, τ) S_{α}^{2_{α}^{}} ‖ u_{ϵ} ‖_{2}^{2 (2_{α}^{} - 1)}, \end{aligned}$
that is,
$\begin{aligned} t_{ϵ} > C (N, τ, α) ‖ u_{ϵ} ‖_{2} for~ 0 < ϵ < ϵ_{0} . \end{aligned}$
(2.29)
Now, using (2.27) and the definition of $v_{ϵ}$ we have
$\begin{aligned} B_{ϵ} (v_{ϵ}) - 3 t_{ϵ}^{\frac{N (q - 2)}{2} - 2} C_{g}^{^{'}} ‖ v_{ϵ} ‖_{q}^{q} & ⩾ \frac{τ^{2} B_{ϵ} (u_{ϵ})}{‖ u_{ϵ} ‖_{2}^{2}} - \frac{3 C_{g}^{^{'}} τ^{q} ‖ u_{ϵ} ‖_{q}^{q}}{‖ u_{ϵ} ‖_{2}^{q}} {(\frac{‖ u_{ϵ} ‖_{2}^{2 (2_{α}^{} - 1)} B_{ϵ} (u_{ϵ})}{τ^{2 (2_{α}^{} - 1)} A (u_{ϵ})})}^{\frac{N (q - 2) - 4}{4 (2_{α}^{} - 1)}} \\ = \frac{τ^{2}}{‖ u_{ϵ} ‖_{2}^{2}} B_{ϵ} (u_{ϵ})^{\frac{N (q - 2) - 4}{4 (2_{α}^{} - 1)}} (B_{ϵ} (u_{ϵ})^{\frac{4. 2_{α}^{} - N (q - 2)}{4 (2_{α}^{} - 1)}} \\ - \frac{3 C_{g}^{^{'}} τ^{q - \frac{N (q - 2)}{2}} ‖ u_{ϵ} ‖_{q}^{q}}{‖ u_{ϵ} ‖_{2}^{q - \frac{N (q - 2)}{2}} A (u_{ϵ})^{\frac{N (q - 2) - 4}{4 (2_{α}^{} - 1)}}}) . \end{aligned}$
Therefore, (2.28) can be directly proved if
$\begin{aligned} 3 C_{g}^{^{'}} τ^{q - \frac{N (q - 2)}{2}} < \frac{B_{ϵ} (u_{ϵ})^{\frac{4. 2_{α}^{} - N (q - 2)}{4 (2_{α}^{} - 1)}} A (u_{ϵ})^{\frac{N (q - 2) - 4}{4 (2_{α}^{} - 1)}} ‖ u_{ϵ} ‖_{2}^{q - \frac{N (q - 2)}{2}}}{‖ u_{ϵ} ‖_{q}^{q}} for~ 0 < ϵ < ϵ_{0} . \end{aligned}$
(2.30)
For $N ⩾ 5$ , $0 < ϵ < ϵ_{0}$ by (2.20), (2.26) and (2.22) we have:
$\begin{aligned} \frac{B_{ϵ} (u_{ϵ})^{\frac{4. 2_{α}^{} - N (q - 2)}{4 (2_{α}^{} - 1)}} A (u_{ϵ})^{\frac{N (q - 2) - 4}{4 (2_{α}^{} - 1)}} ‖ u_{ϵ} ‖_{2}^{q - \frac{N (q - 2)}{2}}}{‖ u_{ϵ} ‖_{q}^{q}} \\ ⩾ C_{N} \frac{B_{ϵ} (u_{ϵ})^{\frac{2 q - N (q - 2)}{4 (2_{α}^{} - 1)}} A (u_{ϵ})^{\frac{N (q - 2) - 4}{4 (2_{α}^{} - 1)}} ‖ u_{ϵ} ‖_{2}^{q - \frac{N (q - 2)}{2}}}{‖ u_{ϵ} ‖_{q}^{q}} \\ ⩾ C (N, α) {(\frac{‖ \nabla u_{ϵ} ‖_{2}^{2}}{A (u_{ϵ})^{\frac{1}{2_{α}^{}}}})}^{\frac{2 q - N (q - 2)}{4 (2_{α}^{} - 1)}} ⩾ C (N, α) S_{α}^{\frac{2 q - N (q - 2)}{4 (2_{α}^{} - 1)}} . \end{aligned}$
Defining
$\begin{aligned} ϕ (x) := {(\frac{C (N, α)}{3 C_{g}^{^{'}}})}^{\frac{2}{(2 - N) x + 2 N}} \end{aligned}$
and taking $c_{g}^{^{'}}$ large enough so that $3 C_{g}^{^{'}} > C (N, α)$ , we get $ϕ (\bar{α}) ⩾ ϕ (\bar{β})$ . Therefore, for
$\begin{aligned} τ < γ := {(\frac{C (N, α)}{3 C_{g}^{^{'}}})}^{\frac{2}{(2 - N) \bar{β} + 2 N}} S_{α}^{\frac{1}{2 (2_{α}^{} - 1)}}, \end{aligned}$
(2.31)
we get
$\begin{aligned} 3 C_{g}^{^{'}} τ^{q - \frac{N (q - 2)}{2}} < C (N, α) S_{α}^{\frac{2 q - N (q - 2)}{4 (2_{α}^{} - 1)}} ⩽ \frac{B_{ϵ} (u_{ϵ})^{\frac{4. 2_{α}^{} - N (q - 2)}{4 (2_{α}^{} - 1)}} A (u_{ϵ})^{\frac{N (q - 2) - 4}{4 (2_{α}^{} - 1)}} ‖ u_{ϵ} ‖_{2}^{q - \frac{N (q - 2)}{2}}}{‖ u_{ϵ} ‖_{q}^{q}} . \end{aligned}$
Similarly, this approach can be followed for $N = 3$ and 4.

Now, define
$\begin{aligned} {\tilde{f}}_{v_{ϵ}} (t) := \frac{t^{2}}{2} ‖ \nabla v_{ϵ} ‖_{2}^{2} - \frac{t^{2. 2_{α}^{}}}{2. 2_{α}^{}} A (v_{ϵ}) . \end{aligned}$
Clearly, ${\tilde{f}}_{v_{ϵ}}$ has it’s maxima at $t_{0} = {(\frac{‖ \nabla v_{ϵ} ‖_{2}^{2}}{A (v_{ϵ})})}^{\frac{1}{2 (2_{α}^{} - 1)}}$ . Using (2.20) and (2.22) we have
$\begin{aligned} {\tilde{f}}_{v_{ϵ}} (t_{0}) & = \frac{2_{α}^{} - 1}{2. 2_{α}^{}} {(\frac{‖ \nabla u_{ϵ} ‖_{2}^{2}}{A (u_{ϵ})^{\frac{1}{2_{α}^{}}}})}^{\frac{2_{α}^{}}{2_{α}^{} - 1}} \\ ⩽ (\frac{2_{α}^{} - 1}{2. 2_{α}^{}}) {(\frac{S^{\frac{N}{2}} + O (ϵ^{N - 2})}{((A_{α} C_{α})^{\frac{N}{2}} S_{α}^{\frac{N + α}{2}} - O (ϵ^{\frac{N + α}{2}}))^{\frac{1}{2_{α}^{}}}})}^{\frac{2_{α}^{}}{2_{α}^{} - 1}} \\ = (\frac{2_{α}^{} - 1}{2. 2_{α}^{}}) {(\frac{(A_{α} C_{α})^{\frac{N}{2. 2_{α}^{}}} S_{α}^{\frac{N}{2}} + O (ϵ^{N - 2})}{(A_{α} C_{α})^{\frac{N}{2. 2_{α}^{}}} S_{α}^{\frac{N + α}{2. 2_{α}^{}}} (1 - O (ϵ^{\frac{N + α}{2}}))^{\frac{1}{2_{α}^{}}}})}^{\frac{2_{α}^{}}{2_{α}^{} - 1}} \\ ⩽ (\frac{2_{α}^{} - 1}{2. 2_{α}^{}}) S_{α}^{\frac{2_{α}^{}}{2_{α}^{} - 1}} + O (ϵ^{N - 2}) . \end{aligned}$
Therefore
$\begin{aligned} sup_{t > 0} {\tilde{f}}_{v_{ϵ}} (t) = (\frac{2_{α}^{} - 1}{2. 2_{α}^{}}) S_{α}^{\frac{2_{α}^{}}{2_{α}^{} - 1}} + O (ϵ^{N - 2}) . \end{aligned}$
Now, either $t_{ϵ} < 1$ or by (2.27),
$\begin{aligned} t_{ϵ} ⩽ {(\frac{T (v_{ϵ})^{2}}{A (v_{ϵ})})}^{\frac{1}{2 (2_{α}^{} - 1)}} = \frac{‖ u_{ϵ} ‖_{2}}{τ} {(\frac{T (u_{ϵ})^{2}}{A (u_{ϵ})})}^{\frac{1}{2 (2_{α}^{} - 1)}} ⩽ ζ_{1} for~ 0 < ϵ < ϵ_{0} . \end{aligned}$
The existence of this $ζ_{1}$ can be deduced using (2.23), (2.20), (2.21) and (2.22). Taking $ζ = max {1, ζ_{1}}$ , we get $t_{ϵ} ⩽ ζ$ . Thus we get:
$\begin{aligned} sup_{t > 0} f_{v_{ϵ}} (t) & = f_{v_{ϵ}} (t_{ϵ}) = {\tilde{f}}_{v_{ϵ}} (t_{ϵ}) + \frac{t_{ϵ}^{2 s}}{2} [v_{ϵ}]^{2} - t_{ϵ}^{- N} \int_{R^{N}} G (t_{ϵ}^{\frac{N}{2}} v_{ϵ}) \\ ⩽ (\frac{2_{α}^{} - 1}{2. 2_{α}^{}}) S_{α}^{\frac{2_{α}^{}}{2_{α}^{} - 1}} + O (ϵ^{N - 2}) + \frac{ζ^{2 s} [v_{ϵ}]^{2}}{2} - t_{ϵ}^{- N} \int_{R^{N}} G (t_{ϵ}^{\frac{N}{2}} v_{ϵ}) . \end{aligned}$
By Remark 1.1 there exists $L_{M} > 0$ such that for large enough M, we have:
$\begin{aligned} \frac{ζ^{2 s} [v_{ϵ}]^{2}}{2} - t_{ϵ}^{- N} \int_{R^{N}} G (t_{ϵ}^{\frac{N}{2}} v_{ϵ}) & ⩽ \frac{ζ^{2 s} [u_{ϵ}]^{2} τ^{2}}{2 {‖ u_{ϵ} ‖}_{2}^{2}} + L_{M} t_{ϵ}^{- N} {‖ t_{ϵ}^{\frac{N}{2}} v_{ϵ} ‖}_{α_{1}}^{α_{1}} - M t_{ϵ}^{- N} {‖ t_{ϵ}^{\frac{N}{2}} v_{ϵ} ‖}_{α_{2}}^{α_{2}} \\ = \frac{ζ^{2 s} [u_{ϵ}]^{2} τ^{2}}{2 ‖ u_{ϵ} ‖_{2}^{2}} + \frac{L_{M} t_{ϵ}^{2} ‖ u_{ϵ} ‖_{α_{1}}^{α_{1}} τ^{α_{1}}}{‖ u_{ϵ} ‖_{2}^{α_{1}}} - \frac{M t_{ϵ}^{\frac{N (α_{2} - 2)}{2}} τ^{α_{2}} ‖ u_{ϵ} ‖_{α_{2}}^{α_{2}}}{‖ u_{ϵ} ‖_{2}^{α_{2}}} . \end{aligned}$
Let $R_{ϵ}^{0} := \frac{[u_{ϵ}]^{2}}{‖ u_{ϵ} ‖_{2}^{2}}; R_{ϵ}^{1} := \frac{t_{ϵ}^{2} ‖ u_{ϵ} ‖_{α_{1}}^{α_{1}}}{‖ u_{ϵ} ‖_{2}^{α_{1}}}$ and $R_{ϵ}^{2} := \frac{t_{ϵ}^{\frac{N (α_{2} - 2)}{2}} ‖ u_{ϵ} ‖_{α_{2}}^{α_{2}}}{‖ u_{ϵ} ‖_{2}^{α_{2}}}$ . Now, by (2.29)
$\begin{aligned} \frac{R_{ϵ}^{1}}{R_{ϵ}^{2}} ⩽ \frac{C^{'} ‖ u_{ϵ} ‖_{α_{1}}^{α_{1}} ‖ u_{ϵ} ‖_{2}^{α_{2}}}{‖ u_{ϵ} ‖_{2}^{α_{1}} ‖ u_{ϵ} ‖_{2}^{\frac{N (α_{2} - 2)}{2} - 2} ‖ u_{ϵ} ‖_{α_{2}}^{α_{2}}} = C^{'} (\frac{‖ u_{ϵ} ‖_{α_{1}}^{α_{1}}}{‖ u_{ϵ} ‖_{2}^{\frac{4}{N}}}) . (\frac{‖ u_{ϵ} ‖_{2}^{α_{2} - \frac{N (α_{2} - 2)}{2}}}{‖ u_{ϵ} ‖_{α_{2}}^{α_{2}}}), \end{aligned}$
then by (2.21), (2.24) and (2.25) we get
$\begin{aligned} \frac{R_{ϵ}^{1}}{R_{ϵ}^{2}} ⩽ {\begin{cases} C^{″} ϵ^{α_{2} - \frac{N (α_{2} - 2)}{2} - N + \frac{α_{2} (N - 2)}{2}} = C^{″} & for~ N ⩾ 5, \\ C^{″} | \ln (ϵ) |^{\frac{3}{2} - \frac{α_{2}}{2}} & for~ N = 4, \\ C^{″} ϵ^{\frac{α_{2}}{4} - \frac{5}{6}} & for~ N = 3 \end{cases} \end{aligned}$
and hence there exists $\tilde{K} > 0$ such that $\frac{R_{ϵ}^{1}}{R_{ϵ}^{2}} ⩽ \tilde{K}$ as $ϵ \to 0$ .

Now, by step 3 of Lemma 2.12 we have
$\begin{aligned} E (u) ⩾ m (τ) = m_{r} (τ) ⩾ (\frac{N (\bar{α} - 2) - 4}{2 N (\bar{α} - 2)}) δ_{τ}^{2} for all u \in P (τ) . \end{aligned}$
Let $0 < k^{'} < (\frac{N (\bar{α} - 2) - 4}{2 N (\bar{α} - 2)}) δ_{τ}^{2}$ , then $k^{'} < E (t_{ϵ} ⋆ v_{ϵ}) = max_{t > 0} E (t ⋆ v_{ϵ}) = f_{v_{ϵ}} (t_{ϵ})$ for every $ϵ > 0$ . Also, since $f_{v_{ϵ}} (0) = 0$ for all $ϵ > 0$ , then there must exist $t_{0} > 0$ small enough such that $t_{ϵ} ⩾ t_{0}$ for all $ϵ > 0$ . Thus by (2.21), (2.23), (2.24) and (2.29) we have
$\begin{aligned} \frac{t_{ϵ}^{2} R_{ϵ}^{0}}{R_{ϵ}^{2}} & ⩽ \frac{C^{'} [u_{ϵ}]^{2} ‖ u_{ϵ} ‖_{2}^{α_{2}}}{‖ u_{ϵ} ‖_{2}^{2} ‖ u_{ϵ} ‖_{2}^{\frac{N (α_{2} - 2)}{2} - 2} ‖ u_{ϵ} ‖_{α_{2}}^{α_{2}}} = [u_{ϵ}]^{2} (\frac{‖ u_{ϵ} ‖_{2}^{α_{2} - \frac{N (α_{2} - 2)}{2}}}{‖ u_{ϵ} ‖_{α_{2}}^{α_{2}}}) \\ ⩽ {\begin{cases} \bar{C} ϵ^{α_{2} - \frac{N (α_{2} - 2)}{2} - α_{2} + \frac{N (α_{2} - 2)}{2}} = \bar{C} & for~ N ⩾ 5, \\ \bar{C} ϵ^{4 - α_{2} - 4 + α_{2}} = \bar{C} & for~ N = 4 \end{cases} \end{aligned}$
and for $N = 3$ ,
$\begin{aligned} \frac{ϵ t_{ϵ}^{2} R_{ϵ}^{0}}{R_{ϵ}^{2}} ⩽ \bar{C} ϵ^{\frac{α_{2}}{4} - \frac{1}{2}} \to 0 as ϵ \to 0. \end{aligned}$
Therefore, for every $N ⩾ 3$ , there exists $\tilde{C}$ such that $\frac{t_{ϵ}^{2} R_{ϵ}^{0}}{R_{ϵ}^{2}} ⩽ \tilde{C}$ and hence $\frac{R_{ϵ}^{0}}{R_{ϵ}^{2}} ⩽ \frac{\tilde{C}}{t_{0}^{2}}$ . Using above analysis, one can conclude that
$\begin{aligned} \frac{ζ^{2 s} [v_{ϵ}]^{2}}{2} - t_{ϵ}^{- N} \int_{R^{N}} G (t_{ϵ}^{\frac{N}{2}} v_{ϵ}) & ⩽ \frac{ζ^{2 s} τ^{2} R_{ϵ}^{0}}{2} + L_{M} R_{ϵ}^{1} τ^{α_{1}} - M τ^{α_{2}} R_{ϵ}^{2} \\ < 0 for large M ~and~ ϵ \to 0. \end{aligned}$
This gives us
$\begin{aligned} sup_{t > 0} f_{v_{ϵ}} (t) < (\frac{2_{α}^{} - 1}{2. 2_{α}^{}}) S_{α}^{\frac{2_{α}^{}}{2_{α}^{} - 1}} \end{aligned}$
for large M and $ϵ \to 0$ . Now define the path $η_{v_{ϵ}} (z) = ((1 - z) t^{-} + z t^{+}) ⋆ v_{ϵ}$ for all $z \in [0, 1]$ and $t^{-}$ , $t^{+}$ as in step 2 of Lemma 2.12. Since $η_{v_{ϵ}} \in Γ (τ)$ , then
$\begin{aligned} σ_{r} (τ) & ⩽ max_{z \in [0, 1]} E (η_{v_{ϵ}} (z)) \\ ⩽ sup_{t > 0} f_{v_{ϵ}} (t) < (\frac{2_{α}^{} - 1}{2. 2_{α}^{}}) S_{α}^{\frac{2_{α}^{}}{2_{α}^{} - 1}} . \end{aligned}$

Now, under the basic requirements we have proved above, we can move forward to prove our main result.
3. .Existence result

Proof of Theorem 1.1. Proof of Theorem 1.1

Let ${u_{n}}$ be the sequence in $P_{r} (τ)$ such that

\begin{aligned} lim_{n \to \infty} {E (u_{n})} = m_{r} (τ) = inf_{u \in P_{r} (τ)} E (u) . \end{aligned}

By (2.17), we know that E is coercive on

P_{r} (τ)

. Thus

{u_{n}}

must be bounded and hence have a weakly convergent subsequence, denoted by

{u_{n}}

, in

H_{r} (R^{N})

. Let

u_{τ} \in H_{r} (R^{N})

be the weak limit of

{u_{n}}

. Also, since

{u_{n}}

is the minimizing sequence for E on

P_{r} (τ) = S_{r} (τ) \cap \tilde{M}

, by Ekeland variational principle, there exists sequences

{λ_{n}}

and

{{\bar{λ}}_{n}}

such that

\begin{aligned} {E^{'} (u_{n}) - λ_{n} u_{n} + {\bar{λ}}_{n} M^{'} (u_{n})} \to 0 as n \to \infty . \end{aligned}

Let

J_{n} (u) := E (u) - λ_{n} ‖ u ‖_{2}^{2} + {\bar{λ}}_{n} M (u)

, then

{J_{n}^{^{'}} (u_{n})}

converges to zero. We will see that

{{\bar{λ}}_{n}}

tends to zero and

{λ_{n}}

will converge to some

λ_{τ}

such that

(λ_{τ}, u_{τ})

will give us the required result.

Claim 1: ${{\bar{λ}}_{n}}$ converges to zero.

Since $u_{n} \in P_{r} (τ)$ for every $n \in N$ , we have:

\begin{aligned} 0 & = lim_{n \to \infty} (\frac{\partial}{\partial t} J_{n} {(t^{\frac{N}{2}} u_{n} (t x)) |}_{t = 1}) \\ = lim_{n \to \infty} \frac{\partial}{\partial t} ({[E (t ⋆ u_{n}) - λ_{n} ‖ u_{n} ‖_{2}^{2} + {\bar{λ}}_{n} M (t ⋆ u_{n})]}_{t = 1}) \\ = lim_{n \to \infty} (f_{u_{n}}^{^{'}} (1) + {\bar{λ}}_{n} (f_{u_{n}}^{^{″}} (1) + f_{u_{n}}^{^{'}} (1))) \\ = lim_{n \to \infty} {\bar{λ}}_{n} f_{u_{n}}^{^{″}} (1), \end{aligned}

also (2.7) gives us

| f_{u_{n}}^{^{″}} (1) | = - f_{u_{n}}^{^{″}} (1) ⩾ (\frac{N (\bar{α} - 2) - 4}{2}) T (u_{n})^{2}

, then

\begin{aligned} | {\bar{λ}}_{n} T (u_{n})^{2} | ⩽ \frac{2 | {\bar{λ}}_{n} f_{u_{n}}^{^{″}} (1) |}{(N (\bar{α} - 2) - 4)} \to 0 as n \to \infty, \end{aligned}

T (u_{n})^{2} ⩾ δ_{τ} > 0

for all

n \in N

, we get

\begin{aligned} {{\bar{λ}}_{n}} \to 0 as n \to \infty . \end{aligned}

Claim 2:

u_{τ} \neq 0

Let if possible $u_{τ} = 0$ , taking $l \in R$ such that ${‖ \nabla u_{n} ‖_{2}^{2} + s [u_{n}]^{2}} \to l$ . Since the imbedding $H_{r} (R^{N}) ↪ L^{p} (R^{N})$ is compact for all $p \in (2, 2^{*})$ , ${u_{n}} \to u_{τ}$ in $L^{p} (R^{N})$ for every $p \in (2, 2^{*})$ . Further, using Remark 1.1 and (2.8) we get

\begin{aligned} \int_{R^{N}} G (u_{n}) \to \int_{R^{N}} G (u_{τ}) = 0, \\ \int_{R^{N}} g (u_{n}) u_{n} \to \int_{R^{N}} g (u_{τ}) u_{τ} = 0, \end{aligned}

also, since

M (u_{n}) = 0

for every

n \in N

\begin{aligned} lim_{n \to \infty} A (u_{n}) = lim_{n \to \infty} (‖ \nabla u_{n} ‖_{2}^{2} + s [u_{n}]^{2} - \int_{R^{n}} \tilde{G} (u_{n}) - M (u_{n})) = lim_{n \to \infty} (‖ \nabla u_{n} ‖_{2}^{2} + s [u_{n}]^{2}) = l . \end{aligned}

(3.1)

Now, since

\begin{aligned} S_{α} ⩽ \frac{‖ \nabla u_{n} ‖_{2}^{2}}{A (u_{n})^{\frac{1}{2_{α}^{*}}}} ⩽ \frac{‖ \nabla u_{n} ‖_{2}^{2} + s [u_{n}]^{2}}{A (u_{n})^{\frac{1}{2_{α}^{*}}}} for every n \in N, \end{aligned}

then by (3.1) we get

\begin{aligned} S_{α} ⩽ \frac{l}{l^{\frac{1}{2_{α}^{*}}}} \Rightarrow l (S_{α}^{2_{α}^{*}} - l^{2_{α}^{*} - 1}) ⩽ 0, \end{aligned}

thus either

l = 0

l ⩾ S_{α}^{\frac{2_{α}^{*}}{2_{α}^{*} - 1}}

. If

l ⩾ S_{α}^{\frac{2_{α}^{*}}{2_{α}^{*} - 1}}

, then

\begin{aligned} σ_{r} (τ) & = m_{r} (τ) = lim_{n \to \infty} E (u_{n}) = lim_{n \to \infty} (E (u_{n}) - \frac{1}{2. 2_{α}^{*}} M (u_{n})) \\ = lim_{n \to \infty} (\frac{1}{2} (1 - \frac{1}{2. 2_{α}^{*}}) ‖ \nabla u_{n} ‖_{2}^{2} + \frac{1}{2} (1 - \frac{s}{2. 2_{α}^{*}}) [u_{n}]^{2}) \\ ⩾ \frac{(2_{α}^{*} - 1)}{2. 2_{α}^{*}} lim_{n \to \infty} (‖ \nabla u_{n} ‖_{2}^{2} + s [u_{n}]^{2}) = \frac{(2_{α}^{*} - 1)}{2. 2_{α}^{*}} l ⩾ \frac{(2_{α}^{*} - 1)}{2. 2_{α}^{*}} S_{α}^{\frac{2_{α}^{*}}{2_{α}^{*} - 1}}, \end{aligned}

this contradicts Lemma 2.14. Therefore

l = 0

and hence

σ_{r} (τ) = lim_{n \to \infty} E (u_{n}) = 0

, but since

σ_{r} (τ) > 0

, we must have

u_{τ} \neq 0

Claim 3: ${λ_{n}}$ is convergent.

Now, as ${J_{n}^{^{'}} (u_{n})} \to 0$ as $n \to \infty$ , then for all $v \in H_{r} (R^{N})$ we have

\begin{aligned} 0 & = lim_{n \to \infty} (E^{'} (u_{n}) v - λ_{n} \int_{R^{N}} u_{n} . v + {\bar{λ}}_{n} M^{'} (u_{n}) v) \\ = lim_{n \to \infty} ((1 + 2 {\bar{λ}}_{n}) \int_{R^{N}} \nabla u_{n} . \nabla v + (1 + 2 s {\bar{λ}}_{n}) ⟨ ⟨ u_{n}, v ⟩ ⟩ - λ_{n} \int_{R^{N}} u_{n} . v \\ - \int_{R^{N}} g (u_{n}) v - {\bar{λ}}_{n} \int_{R^{N}} (\tilde{G})^{^{'}} (u_{n}) v - (1 + {\bar{λ}}_{n} 2. 2_{α}^{*}) \int_{R^{N}} (I_{α} * | u_{n} |^{2_{α}^{*}}) | u_{n} |^{2_{α}^{*} - 2} u_{n} . v) \\ = \int_{R^{N}} \nabla u_{τ} \nabla v + ⟨ ⟨ u_{a}, v ⟩ ⟩ - \int_{R^{N}} g (u_{τ}) v - \int_{R^{N}} (I_{α} * | u_{τ} |^{2_{α}^{*}}) | u_{τ} |^{2_{α}^{*} - 2} u_{τ} v \\ - (lim_{n \to \infty} λ_{n}) (\int_{R^{N}} u_{τ} v), \end{aligned}

(3.2)

taking

v = u_{τ}

we get

\begin{aligned} ‖ u_{τ} ‖_{2}^{2} lim_{n \to \infty} λ_{n} = ‖ \nabla u_{τ} ‖_{2}^{2} + [u_{τ}]^{2} - \int_{R^{N}} g (u_{τ}) u_{τ} - A (u_{τ}) . \end{aligned}

Thus by claim 2

\begin{aligned} lim_{n \to \infty} λ_{n} = \frac{‖ \nabla u_{τ} ‖_{2}^{2}}{‖ u_{τ} ‖_{2}^{2}} + \frac{[u_{τ}]^{2}}{‖ u_{τ} ‖_{2}^{2}} - \frac{1}{‖ u_{τ} ‖_{2}^{2}} \int_{R^{N}} g (u_{τ}) u_{τ} - \frac{A (u_{τ})}{‖ u_{τ} ‖} < + \infty, \end{aligned}

that is,

{λ_{n}}

is convergent, let

{λ_{n}} \to λ_{τ} \in R

, then by (3.2)

\begin{aligned} \int_{R^{N}} \nabla u_{τ} . \nabla v + ⟨ ⟨ u_{τ}, v ⟩ ⟩ = \int_{R^{N}} g (u_{τ}) v + λ_{τ} \int_{R^{N}} u_{τ} v + \int_{R^{N}} (I_{α} * | u_{τ} |^{2_{α}^{*}}) | u_{τ} |^{2_{α}^{*} - 2} u_{τ} v \end{aligned}

(3.3)

for all

v \in H_{r} (R^{N})

. By similar arguments, as in Lemma 2.1, we get

M (u_{τ}) = 0

, that is,

u_{τ} \in \bar{M}

Claim 4: ${u_{n}}$ converges to $u_{τ}$ in $H^{1} (R^{N})$ .

Set $v_{n} := u_{n} - u_{τ} ⇀ 0$ in $H_{r} (R^{N})$ . By Brezis–Lieb Lemma, counterpart of Brezis Lieb Lemma [24] and step 2 of Lemma 3.2 of [15] one can easily deduce that

\begin{aligned} {\begin{cases} ‖ \nabla v_{n} ‖ = ‖ \nabla u_{n} ‖_{2}^{2} - ‖ \nabla u_{τ} ‖_{2}^{2} + o_{n} (1), \\ ‖ v_{n} ‖_{2}^{2} = ‖ u_{n} ‖_{2}^{2} - ‖ u_{τ} ‖_{2}^{2} + o_{n} (1), \\ [v_{n}]^{2} = [u_{n}]^{2} - [u_{τ}]^{2} + o_{n} (1), \\ A (v_{n}) = A (u_{n}) - A (u_{τ}) + o_{n} (1), \end{cases} \end{aligned}

(3.4)

also, by Remark 1.1, (2.8) and Remark 1.5 of [2] we get:

\begin{aligned} \int_{R^{N}} \tilde{G} (v_{n}) = \int_{R^{N}} \tilde{G} (u_{n}) - \int_{R^{N}} \tilde{G} (u_{τ}) + o_{n} (1) . \end{aligned}

(3.5)

Using (3.4) and (3.5) we deduced that

\begin{aligned} M (v_{n}) & = ‖ \nabla v_{n} ‖_{2}^{2} + s [v_{n}]^{2} - N \int_{R^{N}} \tilde{G} (v_{n}) - A (v_{n}) \\ = M (u_{n}) - M (u_{τ}) + o_{n} (1) = o_{n} (1) . \end{aligned}

Now, since

{v_{n}} ⇀ 0

H_{r} (R^{N})

and

\tilde{G} (v_{n}) ⩽ (\frac{\bar{β} - 2}{2}) G (v_{n})

, then by using Lebesgue dominated convergence theorem and Remark 1.1, we get

\begin{aligned} lim_{n \to \infty} \int_{R^{N}} \tilde{G} (v_{n}) = 0; lim_{n \to \infty} \int_{R^{N}} G (v_{n}) = 0, \end{aligned}

and hence

\begin{aligned} lim_{n \to \infty} A (v_{n}) = lim_{n \to \infty} (‖ \nabla v_{n} ‖_{2}^{2} + s [v_{n}]^{2} - \int_{R^{N}} \tilde{G} (v_{n}) - M (v_{n})) = lim_{n \to \infty} (‖ \nabla v_{n} ‖_{2}^{2} + s [v_{n}]^{2}) . \end{aligned}

Let

l = lim_{n \to \infty} A (v_{n}) = lim_{n \to \infty} (‖ \nabla v_{n} ‖_{2}^{2} + s [v_{n}]^{2})

, then clearly,

\begin{aligned} S_{α} ⩽ \frac{l}{l^{\frac{1}{2_{α}^{*}}}} \Rightarrow l (S_{α}^{2_{α}^{*}} - l^{2_{α}^{*} - 1}) ⩽ 0 \Rightarrow either~ l = 0 ~or~ l ⩾ S_{α}^{\frac{2_{α}^{*}}{2_{α}^{*} - 1}} . \end{aligned}

l ⩾ S_{α}^{\frac{2_{α}^{*}}{2_{α}^{*} - 1}}

, then by step 3 of Lemma 2.12

\begin{aligned} σ_{r} (τ) & = lim_{n \to \infty} E (u_{n}) = lim_{n \to \infty} (\frac{‖ \nabla u_{n} ‖_{2}^{2}}{2} + \frac{[u_{n}]^{2}}{2} - N \int_{R^{N}} G (u_{n}) - \frac{1}{2. 2_{α}^{*}} A (u_{n})) \\ = lim_{n \to \infty} (\frac{‖ \nabla v_{n} ‖_{2}^{2}}{2} + \frac{[v_{n}]^{2}}{2} - N \int_{R^{N}} G (v_{n}) - \frac{1}{2. 2_{α}^{*}} A (v_{n}) + \frac{‖ \nabla u_{τ} ‖_{2}^{2}}{2} + \frac{[u_{τ}]^{2}}{2} \\ + \frac{[u_{τ}]^{2}}{2} - N \int_{R^{N}} G (u_{τ}) - \frac{1}{2. 2_{α}^{*}} A (u_{τ})) \\ = lim_{n \to \infty} E (v_{n}) + E (u_{τ}) \\ > lim_{n \to \infty} E (v_{n}) ⩾ lim_{n \to \infty} (\frac{‖ \nabla v_{n} ‖_{2}^{2}}{2} + \frac{s [v_{n}]^{2}}{2} - \frac{1}{2. 2_{α}^{*}} A (v_{n})) \\ = (\frac{2_{α}^{*} - 1}{2. 2_{α}^{*}}) l ⩾ (\frac{2_{α}^{*} - 1}{2. 2_{α}^{*}}) S_{α}^{\frac{2_{α}^{*}}{2_{α}^{*} - 1}}, \end{aligned}

this is again a contradiction to Lemma 2.14, thus

l = 0

. Therefore

\begin{aligned} {\begin{cases} ‖ \nabla u_{n} ‖_{2}^{2} \to ‖ \nabla u_{τ} ‖_{2}^{2}, \\ [u_{n}]^{2} \to [u_{τ}]^{2}, \\ A (u_{n}) \to A (u_{τ}), \\ \int_{R^{N}} g (u_{n}) u_{n} \to \int_{R^{N}} g (u_{τ}) u_{τ} . \end{cases} \end{aligned}

Then by (3.3) we get

\begin{aligned} λ_{τ} ‖ u_{τ} ‖_{2}^{2} & = ‖ \nabla u_{τ} ‖_{2}^{2} + [u_{τ}]^{2} - \int_{R^{N}} g (u_{τ}) u_{τ} - A (u_{τ}) \\ = lim_{n \to \infty} (‖ \nabla u_{n} ‖_{2}^{2} + [u_{n}]^{2} - \int_{R^{N}} g (u_{n}) u_{n} - A (u_{n})) = lim_{n \to \infty} E^{'} (u_{n}) u_{n} \\ = lim_{n \to \infty} λ_{n} ‖ u_{n} ‖_{2}^{2} = λ_{τ} lim_{n \to \infty} ‖ u_{n} ‖_{2}^{2} = λ_{τ} τ^{2} . \end{aligned}

Thus,

u_{τ} \in P_{r} (τ)

{u_{n}} \to u_{τ}

H^{1} (R^{N})

and

\begin{aligned} E (u_{τ}) = lim_{n \to \infty} E (u_{n}) = m_{r} (τ) = inf_{u \in P_{r} (τ)} E (u), \end{aligned}

hence the proof is complete.

4. .Regularity results

We shall demonstrate the regularity results for a solution to (1.1) in this section. To arrive at our conclusion, we will apply the following proposition, which is a particular case of Lemma 2.2 of [38]:

Proposition 4.1.
Let $0 ⩽ u \in H^{1} (Ω)$ , suppose there exists $l \in (0, 2^{})$ , $t_{0} > 1$ and* $C_{0} > 0$ such that
$\begin{aligned} \int_{B_{t}} | \nabla u |^{2} ⩽ C_{0} k^{2} | B_{t} |^{\frac{2}{l}}, t > t_{0}, \end{aligned}$
where $B_{t} = {x \in Ω : u (x) ⩾ t}$ . Then there exists $M = M (Ω, C_{0}, l, t_{0}) > 0$ such that $‖ u ‖_{L^{\infty} (Ω)} ⩽ M$ .
Proof of Theorem 1.2. Proof of Theorem 1.2

Let $Ω \subset R^{N}$ , be an open and bounded set, $u \in H_{r} (R^{N})$ be a positive solution of (1.1), ${ζ_{k}}$ be a sequence of radial functions in $C_{c}^{\infty} (R^{N})$ such that $ζ_{k} = 1$ on Ω, $supp (ζ_{k}) \subset Ω_{k}$ , $0 ⩽ ζ_{k} ⩽ 1$ for all $k \in N$ , $| Ω_{k} ∖ Ω | \to 0$ as $k \to \infty$ , and $‖ \nabla ζ_{k} ‖_{\infty}^{2} ⩽ \frac{C}{| Ω_{k} ∖ Ω |^{\frac{1}{N}}}$ for large k.

For a fixed γ, we define $u_{γ} := (u - γ)^{+} = max {0, u (x) - γ}$ , $B_{γ} := {x \in Ω : u (x) ⩾ γ}$ and $v_{k} := u_{γ} ζ_{k}$ for all $k \in N$ . Using $v_{k} \in H_{r} (R^{N})$ as test function, we get:

\begin{aligned} \int_{S_{k}^{γ}} \nabla u \nabla v_{k} + ⟨ ⟨ u, v_{k} ⟩ ⟩ = λ \int_{S_{k}^{γ}} u v_{k} + \int_{S_{k}^{γ}} g (u) v_{k} + \int_{S_{k}^{γ}} (I_{α} * | u |^{2_{α}^{*}}) | u |^{2_{α}^{*} - 2} u v_{k}, \end{aligned}

(4.1)

where

S_{k}^{γ} = {x \in Ω_{k} : u (x) > γ}

. Now,

\begin{aligned} \int_{B_{γ}} | \nabla u |^{2} = \int_{B_{γ}} | \nabla u_{γ} |^{2} & = \int_{B_{γ}} | \nabla u |^{2} ⩽ \int_{S_{k}^{γ}} ζ_{k} \nabla u \nabla u_{γ} = \int_{S_{k}^{γ}} \nabla u \nabla v_{k} - \int_{S_{k}^{γ}} \nabla u (u_{γ} \nabla ζ_{k}) \\ ⩽ \int_{S_{k}^{γ}} \nabla u \nabla v_{k} + \frac{1}{2} \int_{S_{k}^{γ} ∖ B_{γ}} | u_{γ} \nabla ζ_{k} |^{2} + \frac{1}{2} \int_{S_{k}^{γ}} | \nabla u_{γ} |^{2}, \end{aligned}

therefore, by (4.1)

\begin{aligned} \frac{1}{2} \int_{B_{γ}} | \nabla u |^{2} & ⩽ \int_{S_{k}^{γ}} \nabla u \nabla v_{k} + \frac{1}{2} \int_{S_{k}^{γ} ∖ B_{γ}} (| u_{γ} \nabla ζ_{k} |^{2} + | \nabla u_{γ} |^{2}) \\ = λ \int_{S_{k}^{γ}} u v_{k} + \int_{S_{k}^{γ}} g (u) v_{k} + \int_{S_{k}^{γ}} (I_{α} * | u |^{2_{α}^{*}}) | u |^{2_{α}^{*} - 2} u v_{k} - ⟨ ⟨ u, v_{k} ⟩ ⟩ \\ + \frac{1}{2} \int_{S_{k}^{γ} ∖ B_{γ}} (| u_{γ} \nabla ζ_{k} |^{2} + | \nabla u_{γ} |^{2}) \\ ⩽ λ \int_{B_{γ}} | u |^{2} + \int_{B_{γ}} g (u) u + \int_{B_{γ}} (I_{α} * | u |^{2_{α}^{*}}) | u |^{2_{α}^{*}} - ⟨ ⟨ u, v_{k} ⟩ ⟩ \\ + \int_{S_{k}^{γ} ∖ B_{γ}} (\frac{| u_{γ} |^{2} ‖ \nabla ζ_{k} ‖_{\infty}^{2}}{2} + \frac{| \nabla u_{γ} |^{2}}{2} + λ | u |^{2} + g (u) u) . \end{aligned}

Following the proof of Theorem 1.2 of [15], we get

⟨ ⟨ u, v_{k} ⟩ ⟩ ⩾ 0

and

\begin{aligned} \int_{S_{k}^{γ} ∖ B_{γ}} (\frac{| u_{γ} |^{2} ‖ \nabla ζ_{k} ‖_{\infty}^{2}}{2} + \frac{| \nabla u_{γ} |^{2}}{2} + λ | u |^{2} + g (u) u) \to 0, as k \to \infty . \end{aligned}

Thus we get

\begin{aligned} \frac{1}{2} \int_{B_{γ}} | \nabla u |^{2} ⩽ λ \int_{B_{γ}} | u |^{2} + \int_{B_{γ}} g (u) u + \int_{B_{γ}} (I_{α} * | u |^{2_{α}^{*}}) | u |^{2_{α}^{*}} . \end{aligned}

(4.2)

Now, using Proposition 1.1, and the fact that u is radial function in

H^{1} (R^{N})

hence lies in

L^{p} (A_{γ})

for all

p \in [1, \infty)

we get:

\begin{aligned} \int_{B_{γ}} (I_{α} * | u |^{2_{α}^{*}}) | u |^{2_{α}^{*}} & ⩽ C_{1} {(\int_{B_{γ}} | u |^{2^{*}})}^{\frac{N + α}{2 N}} {(\int_{B_{γ}} | u |^{2^{*}})}^{\frac{N + α}{2 N}} ⩽ C_{1}^{^{'}} {(\int_{B_{γ}} | u |^{\frac{N}{N - 2}} | u |^{\frac{N}{N - 2}})}^{\frac{N + α}{2 N}} \\ ⩽ C_{2} {(\int_{B_{γ}} | u |^{\frac{N + α}{N - 2}})}^{\frac{1}{2}} {(\int_{B_{γ}} | u |^{\frac{N (N + α)}{α (N - 2)}})}^{\frac{α}{2 N}} \end{aligned}

taking

t \in (1, \frac{2^{*}}{2_{α}^{*}}) = (1, \frac{2 N}{N + α})

, we get

\begin{aligned} \int_{B_{γ}} (I_{α} * | u |^{2_{α}^{*}}) | u |^{2_{α}^{*}} & ⩽ C_{2} {(\int_{B_{γ}} | u |^{\frac{N}{α} 2_{α}^{*}})}^{\frac{α}{2 N}} {(| B_{γ} |^{1 - \frac{α}{N}})}^{\frac{1}{2}} {(\int_{B_{γ}} | u |^{\frac{N 2_{α}^{*} t}{α}})}^{\frac{α}{2 N t}} {(| B_{γ} |^{1 - \frac{1}{t}})}^{\frac{α}{2 N}} \\ ⩽ C_{3} | B_{γ} |^{\frac{N t - α}{2 N t}} . \end{aligned}

(4.3)

Also, by (2.8) and Hölder’s inequality we have:

\begin{aligned} \int_{B_{γ}} g (u) u & ⩽ C_{g} (\int_{B_{γ}} | u |^{\bar{α}} + \int_{B_{γ}} | u |^{\bar{β}}) \\ ⩽ C_{g} ({(\int_{B_{γ}} | u |^{\frac{2 \bar{α} N t}{N t + α}})}^{\frac{N t + α}{2 N t}} | B_{γ} |^{\frac{N t - α}{2 N t}} + {(\int_{B_{γ}} | u |^{\frac{2 \bar{β} N t}{N t + α}})}^{\frac{N t + α}{2 N t}} | B_{γ} |^{\frac{N t - α}{2 N t}}) \\ ⩽ C_{4} | B_{γ} |^{\frac{N t - α}{2 N t}}, \end{aligned}

(4.4)

and

\begin{aligned} \int_{B_{γ}} | u |^{2} ⩽ {(\int_{B_{γ}} | u |^{\frac{4 N t}{N t + α}})}^{\frac{N t + α}{2 N}} | B_{γ} |^{\frac{N t - α}{2 N t}} ⩽ C_{5} | B_{γ} |^{\frac{N t - α}{2 N t}} . \end{aligned}

(4.5)

Using (4.3), (4.4) and (4.5) in (4.2), we get

\begin{aligned} \int_{B_{γ}} | \nabla u |^{2} ⩽ C_{6} | B_{γ} |^{\frac{N t - α}{2 N t}} ⩽ C_{6} γ^{2} | B_{γ} |^{\frac{N t - α}{2 N t}} for all γ > γ_{0} > 1. \end{aligned}

For the cases

N ⩾ 5

and

N = 3

with

α < 1

, we have

l = \frac{4 N t}{N t - α} < 2^{*}

t > 1 > \frac{2^{*} α}{(2^{*} - 4) N}

thus by Proposition 4.1

u \in L^{\infty} (Ω)

. Since Ω is an arbitrary bounded domain, we get

u \in L_{loc}^{\infty} (R^{N})

. Moreover, since u is a radial function in

H^{1} (R^{N})

, by radial lemma [36] we have

\begin{aligned} | u (x) | ⩽ \frac{K ‖ u ‖_{H^{1} (R^{N})}}{| x |^{\frac{N - 1}{2}}} almost everywhere in R^{N} . \end{aligned}

(4.6)

The estimate in (4.6) tells us that

| u (x) | \to 0

| x | \to \infty

and hence

u \in L^{\infty} (R^{N})

Also using (4.6) one can see that $u \in L^{r} (R^{N})$ for all $r \in [2, \infty]$ , which gives us that $| u |^{2_{α}^{*}} \in L^{r} (R^{N})$ for all $r \in [2, \infty]$ (since $2_{α}^{*} > 1$ ) and hence $(I_{α} * | u |^{2_{α}^{*}}) \in L^{\infty} (R^{N})$ . Thus

\begin{aligned} \int_{R^{N}} | (I_{α} * | u |^{2_{α}^{*}}) | u {|^{2_{α}^{*} - 2} u |}^{p} < \infty for all p \in [\frac{2}{2_{α}^{*} - 1}, \infty] . \end{aligned}

Further, by (a2), Remark 1.1 and the fact that

u \in L^{r} (R^{N})

for all

r \in [2, \infty]

, there exists

K_{0} > 0

such that

{‖ (I_{α} * | u |^{2_{α}^{*}}) | u |^{2_{α}^{*} - 2} u + g (u) + λ u ‖}_{L^{p} (R^{N})} ⩽ K_{0}

for all

p \in [ℓ, \infty]

where

\begin{aligned} ℓ = max {2, \frac{2}{2_{α}^{*} - 1}} = {\begin{cases} 2 & if N = 3 ~with~ α < 1, ~or~ N ⩾ 5 ~with~ N - α ⩽ 4, \\ \frac{2 (N - 2)}{α + 2} & if N ⩾ 5 ~with~ N - α > 4. \end{cases} \end{aligned}

Therefore, since

(- Δ)^{s}

is an elliptic linear integro-differential operator of order

2 s

and

u \in L^{\infty} (R^{N})

, by Theorem 3.1.20 of [14], we get

\begin{aligned} ‖ u ‖_{W^{2, q} (R^{N})} ⩽ χ (K_{0} + ‖ u ‖_{L^{\infty} (R^{N})}) < \infty for all q \in [ℓ, \infty) . \end{aligned}

Hence,

u \in W^{2, q} (R^{N})

for all

q \in [ℓ, \infty)

Footnotes

Acknowledgements

The first author is partially funded by IFCAM (Indo-French Centre for Applied Mathematics) IRL CNRS 3494.S 3494. The Research carried out by the second and third authors is partially supported by the DST-FIST project with ref. No SR/FST/MS-1/2019/45.

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