Proof. To prove this lemma, one can follow closely the line of the proof of [Reference Stanislavova and Stefanov10, Theorem 2]. Let us now sketch the proof. First we observe that $\mathcal {L}_{+,s} \mid _{\{u\}^{\bot }} \geq 0$. On the other hand, we find that

\[ \langle \mathcal{L}_{+,s} u, u \rangle ={-}(p-2) \int_{\mathbb{R}^n} |u|^p <0. \]

It then follows that $\mathcal {L}_{+,s}$ has only one negative eigenvalue. From [Reference Stanislavova and Stefanov10, Proposition 7], we actually know that the eigenvalue is simple. Using spherical harmonics and the representations of fractional Schrödinger operators introduced in [Reference Stanislavova and Stefanov10], we can write that

\[ \mathcal{L}_{+,s}=\bigoplus_{l=0}^{\infty} \mathcal{L}_{+,s, l}:=\mathcal{L}_{+,s, 0} \bigoplus \mathcal{L}_{+,s, \geq 1}, \]

where the operator $\mathcal {L}_{+,s, l}$ acting on $L_{rad}^2(\mathbb {R}^n)$ is given by

\begin{align*} \mathcal{L}_{+,s,l}& :=\left(-\partial_{rr} -\frac{n-1}{r} \partial_r +\frac{l(l+n-2)}{r^2}\right)^{s}\\ & \quad + \left(\omega+|x|^2\right) -(p-1)|u|^{p-2}, \quad l=0, 1, \cdots, k, \cdots. \end{align*}

It is clear that

\begin{align*} \sigma(\mathcal{L}_{+,s})& =\bigcup_{l=0}^{\infty} \sigma(\mathcal{L}_{+,s, l}),\\ \mathcal{L}_{+,s, 0} & <\mathcal{L}_{+,s, 1}< \cdots <\mathcal{L}_{+,s, k} < \cdots. \end{align*}

At this point, to conclude the proof, we only need to verify that the second smallest eigenvalue of $\mathcal {L}_{+,s, 0}$ is positive and $\mathcal {L}_{+,s, \geq 1} \geq \delta >0$. This can be achieved by applying [Reference Stanislavova and Stefanov10, Propositions 8–9]. Thus, the proof is completed.

Proof. First we show that $u \in H^1(\mathbb {R}^n)$. Since $v \in X_p$ be a solution to (1.1), then

(2.1)\begin{equation} (-\Delta)^s u+ \left(\omega+|x|^2\right) u +2 \lambda u = 2\lambda u+|u|^{p-2} u, \end{equation}

where $\lambda >0$ satisfies $\omega +\lambda >0$. Note that

\[ (-\Delta)^s + \left(\omega+|x|^2\right) +2 \lambda > (-\Delta)^s + \lambda>0. \]

This leads to

(2.2)\begin{equation} 0<\left((-\Delta)^s + \left(\omega+|x|^2\right) +2 \lambda \right)^{{-}1} < \left((-\Delta)^s + \lambda \right)^{{-}1}. \end{equation}

It then follows from Young's inequality that

(2.3)\begin{align} & \left\|\left((-\Delta)^s + \left(\omega+|x|^2\right) +2 \lambda \right)^{{-}1} u\right\|_2 \leq \left\|\left((-\Delta)^s + \lambda \right)^{{-}1} u\right\|_2\nonumber\\ & \quad =\left\|\mathcal{K} \ast u \right\|_2 \lesssim \|u\|_2 \lesssim \|u\|_{H^{{-}s}}, \end{align}

where $H^{-s}(\mathbb {R}^n)$ denotes the dual space of $H^s(\mathbb {R}^n)$ and $\mathcal {K}$ is the fundamental solution to the equation

(2.4)\begin{equation} (-\Delta)^s u + \lambda u=0 \end{equation}

and $\mathcal {K} \in L^1(\mathbb {R}^n)$ by [Reference Frank, Lenzmann and Silvestre4, Lemma C. 1]. This indicates that the operator $((-\Delta )^s + (\omega +|x|^2) +2 \lambda )^{-1}$ maps $H^{-s}(\mathbb {R}^n)$ to $L^2(\mathbb {R}^n)$. Using dual theory, we then see that $((-\Delta )^s + (\omega +|x|^2) +2 \lambda )^{-1}$ maps $L^2(\mathbb {R}^n)$ to $H^{s}(\mathbb {R}^n)$. Observe that

(2.5)\begin{equation} \left\|\left((-\Delta)^s + \left(\omega+|x|^2\right) +2 \lambda \right)^{{-}1} u\right\|_{H^s} \leq \left\|\left((-\Delta)^s + \lambda \right)^{{-}1} u\right\|_{H^s} \lesssim \|u\|_{H^{{-}s}} \lesssim \|u\|_{p'}, \end{equation}

where the last inequality is from the dual to the Sobolev embedding $\|u\|_p \lesssim \|u\|_{H^s}$. This indicates that the operator $((-\Delta )^s + (\omega +|x|^2) +2 \lambda )^{-1}$ maps $L^{p'}(\mathbb {R}^n)$ to $H^{s}(\mathbb {R}^n)$. In fact, this can observe that

\[ u=\left((-\Delta)^s + \left(\omega+|x|^2\right) +2 \lambda \right)^{{-}1} \left(2\lambda u+|u|^{p-2} u\right) \]

and

\[ 2\lambda u+|u|^{p-2} u \in L^2(\mathbb{R}^n) + L^{p'}(\mathbb{R}^n). \]

Then the desired result follows. This completes the proof.

Proof. To prove this, we only need to show that $A_{s_n} \to A_s$ in the norm-resolvent sense as $n \to \infty$, where

\[ A_{s_n}:=(-\Delta )^{s_n} + |x|^2, \quad A_s:=(-\Delta )^s + |x|^2. \]

Let $z \in \mathbb {C}$ be such that $\mbox {Im}\ z \neq 0$, then

\begin{align*} A_{s_n} +z=(-\Delta )^{s_n} + |x|^2+z& =(-\Delta )^s + |x|^2+z +(-\Delta )^{s_n}-(-\Delta )^s\\ & =\left(1+\left((-\Delta )^{s_n}-(-\Delta )^s\right)\left(A_s+z\right)^{{-}1}\right)\left(A_s+z\right). \end{align*}

Then we see that

(2.6)\begin{align} & \left(A_{s_n} +z\right)^{{-}1} -\left(A_s+z\right)^{{-}1}\nonumber\\ & \quad =\left(A_s+z\right)^{{-}1} \left(\left(1+\left((-\Delta )^{s_n}-(-\Delta )^s\right)\left(A_s+z\right)^{{-}1}\right)^{{-}1}-1\right). \end{align}

Note that

\[ \left\|\left((-\Delta )^{s_n}-(-\Delta )^s\right)\left(A_s+z\right)^{{-}1} \right\|_{L^2 \to L^2}=o_n(1). \]

In addition, we see that $(A_s +z)^{-1}$ is bounded from $L^2(\mathbb {R}^n)$ to $L^2(\mathbb {R}^n)$. As a consequence, from (2.6), we can conclude that

\[ \left\|\left(A_{s_n} +z\right)^{{-}1} -\left(A_s+z\right)^{{-}1} \right\|_{L^2 \to L^2}=o_n(1). \]

This completes the proof.

Proof. Let $\delta _0>0$ be a small constant to be determined later and $\lambda _{1,s}>0$ be the lowest eigenvalue of $(-\Delta )^s + |x|^2$ for $s \in [s_0,\, s_0+\delta _0)$. Define a mapping $F: X_p \times [s_0,\, s_0+ \delta _0) \to X_p$ by

\[ F(u, s):= u-\left(\left(-\Delta\right)^s + \left(\omega+|x|^2\right) + 2 \lambda\right)^{{-}1} \left(2\lambda u +|u|^{p-2}u \right), \]

where $\omega >0$ satisfies $\omega >-\lambda _{1,s}$ and $\lambda >0$ satisfies $\lambda _{1,s}<\lambda$ for any $s \in [s_0,\, s_0+\delta _0)$. Due to $\omega >-\lambda _{1,s_0}$, by lemma 2.3, then there exists $\delta _0>0$ small such that $\omega >-\lambda _{1,s}$ is valid for any $s \in [s_0,\, s_0+\delta _0)$. Moreover, observe that $\Sigma _1 \subset \Sigma _s$, then

\[ \lambda_{1,s} \leq \inf_{u \in \Sigma_s} \left\{ \left \langle \left(-\Delta+ |x|^2\right) u, u \right \rangle : \|u\|_2 =1 \right\} \leq \lambda_{1,1}, \]

where $\lambda _{1,1}>0$ is defined by

\[ \lambda_{1, 1}:=\inf_{u \in \Sigma_1} \left\{ \left \langle\left(-\Delta + |x|^2\right) u, u \right \rangle : \|u\|_2=1 \right\}. \]

This then justifies that there exists $\lambda >0$ such that $\lambda _{1,s}<\lambda$ for any $s \in [s_0,s_0+\delta _0)$.

First we check that $F$ is well-defined. As an immediate consequence of the proof of lemma 2.2, we see that $F(u,\, s) \in L^2(\mathbb {R}^n) \cap L^p(\mathbb {R}^n)$ for any $u \in X_p$ and $s\in [s_0,\, s_0 +\delta _0)$. Let us now check that $F(u,\, s) \in L^2(\mathbb {R}^n; |x|^2 \, {\rm d}x)$ for any $u \in X_p$ and $s\in [s_0,\, s_0 +\delta )$. Define

\[ f:=\left(\left(-\Delta\right)^s + \left(\omega+|x|^2\right) + 2 \lambda\right)^{{-}1} \left(2\lambda u +|u|^{p-2}u \right). \]

As the proof of lemma 2.2, we find that $f \in H^s(\mathbb {R}^n)$. This further gives that

\[ \left(-\Delta\right)^s f+ \left(\omega+|x|^2\right) f+ 2 \lambda f=2\lambda u +|u|^{p-2}u. \]

Therefore, we have that

\begin{align*} & \int_{\mathbb{R}^n} |\left(-\Delta\right)^{s/2}f|^2 \, {\rm d}x + \int_{\mathbb{R}^n}\left(\omega+|x|^2\right)|f|^2 \, {\rm d}x\\ & \quad + 2 \lambda \int_{\mathbb{R}^n}|f|^2 \, {\rm d}x=2\lambda \int_{\mathbb{R}^n} u f \,{\rm d}x + \int_{\mathbb{R}^n} |u|^{p-2} u f \,{\rm d}x \\ & \quad \leq 2 \|u\|_2 \|f\|_2 + \|u\|_p^{p-1}\|f\|_p<{+}\infty, \end{align*}

where we used Hölder's inequality for the inequality. It then leads to the desired result.

To apply the implicit function theorem, we are going to check that $F$ is of class $C^1$. First we show that ${\partial F}/{\partial u}$ exists and

\[ \frac{\partial F}{\partial u}=1-\left(\left(-\Delta\right)^s + \left(\omega+|x|^2\right) +2 \lambda \right)^{{-}1} \left(2\lambda +(p-1)|u|^{p-2}\right). \]

For simplicity, we shall define

\[ G(u, s):=\left(\left(-\Delta\right)^s + \left(\omega+|x|^2\right) +2 \lambda\right)^{{-}1} \left(2\lambda u +|u|^{p-2} u\right). \]

Indeed, it suffices to prove that ${\partial G}/{\partial u}$ exists and

\[ \frac{\partial G}{\partial u}=\left(\left(-\Delta\right)^s + \left(\omega+|x|^2\right) + 2 \lambda \right)^{{-}1} \left(2\lambda +(p-1)|u|^{p-2}\right). \]

Observe that, for any $h \in X_p$,

\begin{align*} & \left\|G(u+h, s)-G(u, s)-\frac{\partial G}{\partial u}(u, s) h\right\|_{L^2 \cap L^p} \\ & \quad=\left\|\left(\left(-\Delta\right)^s + \left(\omega+|x|^2\right) +2 \lambda\right)^{{-}1}\left(|u+h|^{p-2} (u+h)\right.\right.\\ & \left.\left.\quad -\,|u|^{p-2} u-(p-1)|u|^{p-2} h\right) \vphantom{\left\|\left(\left(-\Delta\right)^s + \left(\omega+|x|^2\right) +2 \lambda\right)^{{-}1}\left(|u+h|^{p-2} (u+h)\right.\right.}\right\|_{L^2 \cap L^p} \\ & \quad \leq \left\|\left(\left(-\Delta\right)^s + \lambda \right)^{{-}1}\left(|u+h|^{p-2} (u+h)-|u|^{p-2} u-(p-1)|u|^{p-2} h\right) \right\|_{L^2 \cap L^p} \\ & \quad \lesssim \left\||u+h|^{p-2} (u+h)-|u|^{p-2} u-(p-1)|u|^{p-2} h\right\|_{\frac{p}{p-1}}=o(\|h\|_p)=o(\|h\|_{L^2 \cap L^p}), \end{align*}

where we used the fact that the fundamental solution $\mathcal {K}$ to (2.4) satisfies $\mathcal {K} \in L^{p/2}(\mathbb {R}^n) \cap L^{{2p}/{p+2}} (\mathbb {R}^n)$ and Young's inequality. Define

\[ g:=\left(\left(-\Delta\right)^s + \left(\omega+|x|^2\right) +2 \lambda\right)^{{-}1}\left(|u+h|^{p-2} (u+h)-|u|^{p-2} u-(p-1)|u|^{p-2} h\right). \]

Since

\[ |u+h|^{p-2} (u+h)-|u|^{p-2} u-(p-1)|u|^{p-2} h \in L^{p'}(\mathbb{R}^n), \]

then $g \in H^s(\mathbb {R}^n)$ by arguing as the proof of lemma 2.2. Then we write

\[ \left(-\Delta\right)^s g + \left(\omega+|x|^2\right) g +2 \lambda g= |u+h|^{p-2} (u+h)-|u|^{p-2} u-(p-1)|u|^{p-2} h. \]

It then follows that

\begin{align*} & \int_{\mathbb{R}^n} |\left(-\Delta\right)^{s/2}g|^2 \, {\rm d}x + \int_{\mathbb{R}^n}\left(\omega+|x|^2\right)|g|^2 \, {\rm d}x + 2 \lambda \int_{\mathbb{R}^n}|g|^2 \, {\rm d}x \\ & \quad = \int_{\mathbb{R}^n} \left(|u+h|^{p-2} (u+h)-|u|^{p-2} u-(p-1)|u|^{p-2} h\right) g \, {\rm d}x=o(\|h\|_p)\|g\|_p. \end{align*}

Using the fact that $H^s(\mathbb {R}^n)$ is continuously embedded into $L^p(\mathbb {R}^n)$ and Young's inequality, we then obtain that

\[ \|g\|_{L^2(\mathbb{R}^n; |x|^2 \,{\rm d}x)} \lesssim o(\|h\|_2). \]

Consequently, there holds that

\begin{align*} & \left\|G(u+h, s)-G(u, s)-\frac{\partial G}{\partial u}(u, s) h\right\|_{L^2(\mathbb{R}^n; |x|^2\,{\rm d}x)} \\ & \quad=\left\|\left(\left(-\Delta\right)^s + \left(\omega+|x|^2\right) +2 \lambda\right)^{{-}1}\left(|u+h|^{p-2} (u+h)\right.\right.\\ & \left.\left.\quad -|u|^{p-2} u-(p-1)|u|^{p-2} h\right) \vphantom{\left\|\left(\left(-\Delta\right)^s + \left(\omega+|x|^2\right) +2 \lambda\right)^{{-}1}\left(|u+h|^{p-2} (u+h)\right.\right.}\right\|_{L^2(\mathbb{R}^n; |x|^2\,{\rm d}x)}\\ & \quad=\|g\|_{L^2(\mathbb{R}^n; |x|^2 \,{\rm d}x)} \lesssim o(\|h\|_2). \end{align*}

Thus, we conclude that

\[ \left\|G(u+h, s)-G(u, s)-\frac{\partial G}{\partial u}(u, s) h\right\|_{X_p} \lesssim o(\|h\|_{X_p}). \]

The desired result follows.

Next we are going to verify that ${\partial F}/{\partial u}$ is continuous. Indeed, it suffices to show that ${\partial G}/{\partial u}$ is continuous. For this aim, we shall demonstrate that, for any $\epsilon >0$, there exists $\delta >0$ such that $\|u-\tilde {u}\|_{X_p} +|s-\tilde {s}| <\delta$, then, for any $h \in X_p$,

(2.7)\begin{equation} \left\|\left(\frac{\partial G}{\partial u}(u, s)-\frac{\partial G}{\partial u}(\tilde{u}, \tilde{s})\right) h\right\|_{X_p}<\epsilon \|h\|_{X_p}. \end{equation}

Observe that

\begin{align*} \left(\frac{\partial G}{\partial u}(u, s)-\frac{\partial G}{\partial u}(\tilde{u}, \tilde{s})\right) h & = \left(A_s -A_{\tilde{s}}\right) \left(2\lambda +(p-1)|u|^{p-2} \right) h\\ & \quad + A_{\tilde{s}} \left(2 \lambda +(p-1)\left(|u|^{p-2} -|\tilde{u}|^{p-2}\right)\right) h, \end{align*}

where

\[ A_s:=\left(\left(-\Delta\right)^s + \left(\omega+|x|^2\right) + 2 \lambda \right)^{{-}1}, \quad A_{\tilde{s}}:=\left(\left(-\Delta\right)^{\tilde{s}} + \left(\omega+|x|^2\right) + 2 \lambda\right)^{{-}1}. \]

Note that

\[ \|f\|_{L^2 \cap L^p} \lesssim \left\|\left((-\Delta)^{{s_p}/{2}} + 1 \right) f\right\|_2, \quad 0< s_p:=\frac{(p-2)n}{2p}< s. \]

Then, by Plancherel's identity, the mean value theorem and Young's inequality, there holds that

\begin{align*} & \left\|\left(A_s -A_{\tilde{s}}\right) \left(2\lambda +(p-1)|u|^{p-2} \right) h\right\|_{L^2 \cap L^p}\\ & \quad\lesssim \left\|\left((-\Delta)^{{s_p}/{2}} + 1 \right) \left(A_s -A_{\tilde{s}}\right) \left(2\lambda +(p-1)|u|^{p-2} \right) h\right\|_2 \\ & \quad \lesssim |s-\tilde{s}| \left(\|h\|_2 +\|h\|_p+\|u\|_{p}^{p-2}\|h\|_p\right). \end{align*}

In addition, we see that

\begin{align*} & \left\|\left(A_s -A_{\tilde{s}}\right) \left(2\lambda +(p-1)|u|^{p-2} \right) h\right\|_{L^2(\mathbb{R}^n; |x|^2 \,{\rm d}x)}\\ & \quad \lesssim |s-\tilde{s}| \left(\|h\|_2 +\|h\|_p+\|u\|_{p}^{p-2}\|h\|_p\right). \end{align*}

Notice that

\[ \left\|A_s f\right\|_{L^2(\mathbb{R}^n; |x|^2 \,{\rm d}x)} \lesssim \|f\|_2, \quad \left\|A_s f\right\|_{L^2(\mathbb{R}^n; |x|^2 \,{\rm d}x)} \lesssim \|f\|_{{p}/{p-1}}. \]

Further, we can conclude that

\[ \left\|A_{\tilde{s}} \left(2 \lambda+\left(|u|^{p-2} -|\tilde{u}|^{p-2} \right)\right) h \right\|_{X_p} \lesssim \|h\|_2 +\|h\|_p+ \left\||u|^{p-2} -|\tilde{u}|^{p-2}\right\|_{{p}/{p-2}}\|h\|_p. \]

Note that

\[ \left\||u|^{p-2}-|\tilde{u}|^{p-2}\right\|_{{p}/{p-2}} \leq \left\||u-\tilde{u}|^{p-2}\right\|_{{p}/{p-2}}=\left\|u-\tilde{u}\right\|_{p}^{p-2}, \quad 2< p \leq 3 \]

and

\begin{align*} & \left\||u|^{p-2}-|\tilde{u}|^{p-2}\right\|_{{p}/{p-2}} \lesssim \left\|\left(|u|^{p-3}+|\tilde{u}|^{p-3}\right)|u-\tilde{u}|\right\|_{{p}/{p-2}} \\ & \quad \leq \left(\|u\|_p^{p-3} +\|\tilde{u}\|_p^{p-3}\right) \|u -\tilde{u}\|_p, \quad p>3. \end{align*}

Consequently, from the calculations above, (2.7) holds true. This implies that ${\partial F}/{\partial u}$ is continuous. By a similar argument, we are also able to show that ${\partial F}/{\partial s}$ exists and

\[ \frac{\partial F}{\partial s}={-}\left(\left(-\Delta\right)^s \log (-\Delta) \right)\left(\left(-\Delta\right)^s + \left(\omega+|x|^2\right) + 2 \lambda\right)^{{-}2} \left(2\lambda u +|u|^{p-2}u \right). \]

In addition, we can prove that ${\partial F}/{\partial s}$. Thus, we have that $F$ is of class $C^1$.

Now we employ the implicit function theorem to establish theorem. Note first that $F(u_0,\, s_0)=0$ and

\[ \frac{\partial F}{\partial u}(u_0, s_0)=1\!+K, \quad K:={-}\left(\left(-\Delta\right)^{s_0} \!+\! \left(\omega+|x|^2\right) +2 \lambda \right)^{{-}1} \left(2\lambda +(p-1)|u_0|^{p-2} \right). \]

It is simple to see that $K$ is compact on $L^2_{rad}(\mathbb {R}^n)$. Moreover, from lemma 2.1, we have that $-1 \not \in \sigma (K)$. Then $1+ K$ is invertible. Furthermore, arguing as before, we can show that $1+K$ is bounded from $X_p$ to $X_p$. This implies that $(1+K)^{-1}$ is bounded from $X_p$ to $X_p$. It then follows from the implicit function theorem that theorem holds true. This completes the proof.

Proof. Define

\begin{align*} M_s& :=w \int_{\mathbb{R}^n} |u_s|^2\, {\rm d}x, \quad H_s:= \int_{\mathbb{R}^n} |x|^2 |u_s|^2\, {\rm d}x, \quad T_s\\ & :=\int_{\mathbb{R}^n} |(-\Delta)^{s/2} u_s|^2 \, {\rm d}x, \quad V_s:=\int_{\mathbb{R}^n} |u_s|^p \,{\rm d}x. \end{align*}

Since $u_s \in H^s(\mathbb {R}^n)$ is a solution to (1.1), then

(2.8)\begin{equation} T_s + M_s +H_s=V_s. \end{equation}

In addition, we have that $u_s$ satisfies the following Pohozaev identity,

(2.9)\begin{equation} \frac{N-2s}{2} T_s +\frac N 2 M_s + \frac{N+2}{2}H_s=\frac N p V_s. \end{equation}

Combining (2.8) and (2.9), we see that

(2.10)\begin{equation} s T_s -H_s=\frac{N(p-2)}{2p} V_s. \end{equation}

It follows from (2.8) and (2.10) that

\[ s_0 M_s +(1+s_0)H_s \leq s M_s +(1+s)H_s=\frac{2ps-N(p-2)}{2p} V_s <\frac{2ps_*-N(p-2)}{2p} V_s \]

and

\[ s_* M_s +(1+s_*)H_s > s M_s +(1+s)H_s=\frac{2ps-N(p-2)}{2p} V_s \geq \frac{2ps_0-N(p-2)}{2p} V_s. \]

Consequently, we have that $M_s + H_s \sim V_s$ for any $s \in [s_0,\, s_*)$. It follows from (2.8) and (2.10) that

\[ (1+s_0)T_s \leq (1+s)T_s +M_s=\frac{N(p-2)+2p}{2p}V_s \]

and

\[ (1+s_*)T_s > (1+s)T_s +M_s=\frac{N(p-2)+2p}{2p}V_s. \]

This leads to $T_s \sim V_s$ for any $s \in [s_0,\, s_*)$. Therefore, we obtain that

(2.11)\begin{equation} M_s +H_s \sim T_s \sim V_s \end{equation}

for any $s \in [s_0,\, s_*)$. Since $2< p< p_{s_0}$, there exists $0<\theta <1$ such that $p=2\theta + (1-\theta ) p_{s_0}$. From Gagliardo–Nirenberg's inequality and Hölder's inequality, we then get that

(2.12)\begin{equation} V_s \leq M_s^{\theta} \left(\int_{\mathbb{R}^n}|u_s|^{p_{s_0}}\, {\rm d}x \right)^{(1-\theta)} \lesssim \left(M_s +H_s\right)^{\theta} \left(\int_{\mathbb{R}^n}|(-\Delta)^{{s_0}/{2}}u_s|^2 \, {\rm d}x \right)^{{p_{s_0}(1-\theta)}/{2}}. \end{equation}

In addition, there holds that

(2.13)\begin{equation} \int_{\mathbb{R}^n}|(-\Delta)^{{s_0}/{2}}u_s|^2 \, {\rm d}x \leq \left(\int_{\mathbb{R}^n}|u_s|^2 \, {\rm d}x\right)^{{s-s_0}/{s}}\left(\int_{\mathbb{R}^n}|(-\Delta)^{{s}/{2}}u_s|^2 \, {\rm d}x \right)^{{s_0}/{s}}. \end{equation}

Utilizing (2.11), (2.12) and (2.13) then implies that

\[ M_s +H_s \sim T_s \sim V_s \gtrsim 1 \]

for any $s \in [s_0,\, s_*)$. Arguing as the proof of [Reference Frank, Lenzmann and Silvestre4, Lemma 8.2], we can obtain that $V_s \lesssim 1$ for any $s \in [s_0,\, s_*)$. This in turn implies that

\[ M_s +H_s \sim T_s \sim V_s \lesssim 1 \]

for any $s \in [s_0,\, s_*)$. This completes the proof.

Proof. Define

(2.15)\begin{equation} \alpha_s:=\inf \left\{\langle \mathcal{L}_{+,s} f, f \rangle : f \bot u_s, \|f\|_2=1\right\}. \end{equation}

Obviously, we have that $\alpha _s \geq 0$. First we shall verify that $\alpha _s>0$ is attained. Let $\{f_k\}$ be a minimizing sequence to (2.15) such that $f_k \bot u_s$, $\|f_k\|_2=1$ and $\langle \mathcal {L}_{+,s} f_k,\, f_k \rangle =\alpha _s+o_k(1)$. Observe that $\{f_k\}$ is bounded in $\Sigma _s$. Therefore, there exists a function $f \in \Sigma _s$ such that $f_k \rightharpoonup f$ in $\Sigma _s$ and $f_k \to f$ in $L^q(\mathbb {R}^n)$ for any $q \in [2,\, 2_s^*)$ as $n \to \infty$. This leads to $f \bot u_s$, $\|f\|_2=1$ and $\langle \mathcal {L}_{+,s} f,\, f \rangle =\alpha _s$. Contrarily, we assume that $\alpha _s=0$. When $s<1$, using the fact that $Ker [\mathcal {L}_{+, s}]=\{0\}$ by lemma 2.1 and arguing as the proof of [Reference Stanislavova and Stefanov10, Proposition 6], we are able to reach a contradiction. This in turn shows that $\alpha _s>0$ and

\[ \langle \mathcal{L}_{+,s} u, u \rangle \geq \alpha_s \|u\|_2^2, \quad \forall\ u \bot u_s. \]

While $s =1$, using the fact that $Ker[\mathcal {L}_{+, 1}]=\{0\}$ and following the spirit of the proof of [Reference Stanislavova and Stefanov10, Proposition 6], we can also derive that $\alpha _1>0$ and

\[ \langle \mathcal{L}_{+,1} u, u \rangle \geq \alpha_1 \|u\|_2^2, \quad \forall\ u \bot u_1. \]

Thus, the proof is completed.

Proof. In the spirit of the proof of [Reference Frank, Lenzmann and Silvestre4, Lemma 8.3], we need to verify that the operator $\mathcal {L}_{-, s}$ enjoys the Perron–Frobenius type property, where

\[ \mathcal{L}_{-,s}:=(-\Delta)^s + \left(\omega+|x|^2\right) -|u|^{p-2}. \]

In addition, we need to check that $\mathcal {L}_{-, \tilde {s}} \to \mathcal {L}_{-,s}$ as $\tilde {s} \to s$ in norm-resolvent sense.

Define $H:=(-\Delta )^s+|x|^2$, which generates a semigroup ${\rm e}^{-t H}$ with positive integral kernel. Then we have that ${\rm e}^{-t H}$ acting on $L^2(\mathbb {R}^n)$ is positivity improving. Next we show that $w+|u|^{p-2}$ belongs to Kato class, i.e.

(2.16)\begin{equation} \lim_{\lambda \to \infty} \left\|(H+\lambda)^{{-}1} \left(\omega+|u|^{p-2}\right)\right\|_{L^{\infty} \to L^{\infty}}=0. \end{equation}

Note that $H+ \lambda >(-\Delta )^s + \lambda$, then

\[ \left(H+ \lambda\right)^{{-}1}< \left((-\Delta)^s +\lambda\right)^{{-}1}. \]

Let $\mathcal {K}$ be the fundamental solution to the equation

\[ (-\Delta)^s u+\lambda u=0. \]

Then we have that

\[ \mathcal{K}(x)=\int_0^{+\infty} \,{\rm e}^{-\lambda t} \mathcal{H}(x, t) \, {\rm d}t, \]

where

\[ \mathcal{H}(x, t):=\int_{\mathbb{R}^n} \,{\rm e}^{2 \pi \text{i} x \cdot \xi -t |\xi|^{2s}} \,{\rm d} \xi. \]

From $(A 4)$ in [Reference Felmer, Quaas and Tan2, Appendix A], we find that

\[ 0 <\mathcal{H}(x ,t) \lesssim \min \left\{t^{-{n}/{2s}}, t|x|^{{-}n-2s}\right\}. \]

This gives that, for any $q \geq 1$,

\begin{align*} \left\|\mathcal{H}\right\|_q & \lesssim \left(\int_{|x| \leq t^{{1}/{2s}}} t^{-{nq}/{2s}}\, {\rm d}x \right)^{1/q} + \left(\int_{|x| \geq t^{{1}/{2s}}} t^q |x|^{-(n+2s)q} \, {\rm d}x \right)^{1/q}\\ & \lesssim t^{-{n}/{2s}\left(1-(1/q)\right)}. \end{align*}

It then follows that

\[ \left\|\mathcal{K}\right\|_q \leq \int_0^{+\infty} \,{\rm e}^{-\lambda t} \left\|\mathcal{K}({\cdot}, t)\right\|_q \, {\rm d}t \lesssim \int_0^{+\infty} \,{\rm e}^{-\lambda t} t^{-{n}/{2s}\left(1-(1/q)\right)} \, {\rm d}t \lesssim \lambda^{{n}/{2s}\left(1-(1/ q)\right)-1}, \]

where $q \geq 1$ satisfies

\[ \frac{n}{2s}\left(1-\frac 1 q\right)<1. \]

Using Young's inequality, we then get that, for any $f \in L^{\infty }(\mathbb {R}^n)$,

\[ \|\left((-\Delta)^s +\lambda\right)^{{-}1}\left(\omega+|u|^{p-2}\right)f\|_{\infty} \lesssim \lambda^{{-}1}\omega\|f\|_{\infty} + \lambda^{{n}/{2s}\left(1- (2/q)\right)-1} \|f\|_{\infty}, \]

which readily yields that

\[ \|\left((-\Delta)^s +\lambda\right)^{{-}1}\left(\omega+|u|^{p-2}\right)\|_{L^{\infty}\to L^{\infty}}=o_{\lambda}(1). \]

Thus, (2.16) holds true and the desired result follows. Arguing as the proof of [Reference Frank and Lenzmann3, Lemma C.2], we conclude that the operator $\mathcal {L}_{-, s}$ enjoys Perron–Frobenius type property.

Next we prove the convergence of the operator in norm-resolvent sense. Observe first that

\begin{align*} \mathcal{L}_{-,\tilde{s}} + z& =(-\Delta)^s + \left(\omega+|x|^2\right) -|u|^{p-2} +z + (-\Delta)^{\tilde{s}}-(-\Delta)^s \\ & = \left(1+ \left((-\Delta)^{\tilde{s}}-(-\Delta)^s\right)\left(\mathcal{L}_{-,s} + z\right)^{{-}1}\right)\left(\mathcal{L}_{-,s} + z\right). \end{align*}

Therefore, we have that

\begin{align*} & \left(\mathcal{L}_{-,s} + z\right)^{{-}1}-\left(\mathcal{L}_{-, \tilde{s}} + z\right)^{{-}1}\\ & \quad =\left(\mathcal{L}_{-,s} + z\right)^{{-}1} \left(1-\left(1+ \left((-\Delta)^{\tilde{s}}-(-\Delta)^s\right)\left(\mathcal{L}_{-,s} + z\right)^{{-}1}\right)^{{-}1}\right). \end{align*}

As the proof of lemma 2.3, we can show that

\[ \left\|\left(\mathcal{L}_{-,s} + z\right)^{{-}1}-\left(\mathcal{L}_{-, \tilde{s}} + z\right)^{{-}1}\right\|_{L^2 \to L^2} \to 0, \quad \mbox{as} \ \tilde{s} \to s. \]

This indicates that $\mathcal {L}_{-, \tilde {s}} \to \mathcal {L}_{-, s}$ in the norm-resolvent sense as $\tilde {s} \to s$. Thus, the proof is completed.

Proof. From lemma 2.5, we know that $u_{s_n}$ is bounded in $\Sigma _{s_0}$. Thus, there exists $u_* \in \Sigma _{s_0}$ such that $u_{s_n} \rightharpoonup u_*$ in $\Sigma _{s_0}$ and $u_{s_n} \to u_*$ in $L^q(\mathbb {R}^n)$ for any $q \in [2,\, 2_{s_0}^*)$. Since $u_{s_n} >0$, then $u_* \geq 0$. It follows from lemma 2.5 that $u_* \neq 0$. Note that

\[ u_{s_n}=\left((-\Delta)^{s_n}+ \left(\omega+|x|^2\right) + 2 \lambda\right)^{{-}1} \left(2 \lambda u_{s_n} + u_{s_n}^{p-1}\right). \]

Since $u_{s_n} \to u_*$ in $L^2(\mathbb {R}^n) \cap L^p(\mathbb {R}^n)$ as $n \to \infty$, then

\[ u_*=\left((-\Delta)^{s_*}+ \left(\omega+|x|^2\right) + 2 \lambda\right)^{{-}1} \left(2 \lambda u_{*} + u_{*}^{p-1}\right). \]

This implies that $u_*$ solves (2.17) and $u_{s_n} \to u_*$ in $X_p$ as $n \to \infty$. Thus, the proof is completed.

Proof. Define

\[ \mathcal{L}_{+, s}:=(-\Delta)^s+ \left(\omega+|x|^2\right) -(p-1) |u_s|^{p-2}. \]

Reasoning as the proof of the norm-resolvent convergence of $\mathcal {L}_{-, s}$ in lemma 2.7, we can also show that $\mathcal {L}_{+, \tilde {s}} \to \mathcal {L}_{+,s}$ in the norm-resolvent sense as $\tilde {s} \to s$. This gives that

\[ \mathcal{N}_{-, rad}(\mathcal{L}_{+, s})=\mathcal{N}_{-, rad}(\mathcal{L}_{+, s_0})=1, \quad s \in [s_0, s_*). \]

Let $\{s_n\} \subset [s_0,\, s_*)$ be such that $s_n \to s_*$. Since $u_0 \in X_p$ is a ground state to (1.1) with $s=s_0$, then $u_0>0$. In view of lemma 2.7, then $u_{s_n}>0$. From lemma 2.8, we know that there exists $u_*>0$ solving (2.17). Note that $\mathcal {L}_{+, s_n} \to \mathcal {L}_{+, s_*}$ in the norm-resolvent sense as $n \to \infty$. By the lower semicontinuity of the Morse index, we have that

\[ 1=\liminf_{n \to\infty} \mathcal{N}_{-, rad}(\mathcal{L}_{-, s_n}) \geq \mathcal{N}_{-, rad}(\mathcal{L}_{+, s_*}). \]

This implies that $\mathcal {N}_{-, rad}(\mathcal {L}_{+, s_*}) \leq 1$. On the other hand, since $u_*$ solves (2.17), then we see that

\[ \langle u_*, \mathcal{L}_{+, s_*} u_* \rangle={-}(p-2) \int_{\mathbb{R}^n} |u_*|^p \, {\rm d}x <0. \]

Thus, we conclude that $\mathcal {N}_{-, rad}(\mathcal {L}_{+, s_*})=1$, which yields that $u_*$ is a ground state to (2.17). As a result, we have that $s_*=1$. On the other hand, by the nondegeneracy of $\mathcal {L}_{+, s_*}$, then $u_s$ can be extended beyond $s_*$. This is impossible and the proof is completed.

Proof of theorem 1.1 Let $n \geq 1$, $0< s_0<1$ and $2< p<2_{s_0}^*$. Let $u_{s_0}>0$ and $\tilde {u}_{s_0}>0$ be two different ground states to (1.1) with $s=s_0$, which are indeed radially symmetric. From lemma 2.1, we obtain that the associated linearized operators around $u_{s_0}$ and $\tilde {u}_{s_0}$ are nondegenerate. Then, by lemmas 2.4 and 2.9, we have that $u_s \in C^1([s_0,\, 1); X_p)$ and $\tilde {u}_s \in C^1([s_0,\, 1); X_p)$. Moreover, by the local uniqueness of solutions derived in lemma 2.4, we get that $u_s \neq \tilde {u}_s$ for any $s \in [s_0,\, 1)$. It follows from lemma 2.8 that there exist $u_* \in X_p$ and $\tilde {u}_* \in X_p$ such that $u_s \to u_*$ and $\tilde {u}_{s} \to \tilde {u}_*$ in $X_p$ as $s \to 1^-$. In addition, $u_*>0$ and $\tilde {u}_*>0$ solve (2.17) with $s_*=1$. Thanks to [Reference Hirose and Ohta5, Theorem 1.3] and [Reference Hirose and Ohta6, Theorem1.2], then we have that $u_*=\tilde {u}_*$. This implies that $\|u_s-\tilde {u}_s\|_{X_p} \to 0$ as $s \to 1^-$. Note that the linearized operator $\mathcal {L}_{+, 1}$ around $u_*$ is nondegenerate, see [Reference Kabeya and Tanaka7, Theorem 0.2]. Remark that, from the proof of [Reference Kabeya and Tanaka7, Theorem 0.2], it is simple to see that the result also holds true for $n=1$. Then, by the implicit function theorem, there exists a unique branch $\hat {u}_{s} \in C^1((1-\delta,\, 1]; X_p)$ solving (1.1) with $\hat {u}_1=u^*$ for some $\delta >0$. This contradicts with $u_s \neq \tilde {u}_s$ for any $s\in [s_0,\, 1)$. Thus, the proof is completed.