Proof. The results could be derived by direct calculations, and we omit the details.

Proof. (i) On the one hand, by lemma A.1, we deduce that

\begin{equation*} E_\mu(u)\geq\frac{1}{2}\int_{\mathbb{R}^N}|\left(-\Delta\right)^{\frac{s}{2}}u|^2-\frac{C(s,N,p)}{p}a^{p-\frac{N(p-2)}{2s}} \left(\int_{\mathbb{R}^N}|\left(-\Delta\right)^{\frac{s}{2}}u|^2\right)^{\frac{N(p-2)}{4s}} \end{equation*}

for every $u\in S_a$. Since $2 \lt p \lt \bar{p}$, it holds that $0 \lt \frac{N(p-2)}{4s} \lt 1$. Hence, E_µ is coercive on S_a and $m_{a,\mu}$ is bounded from below. On the other hand, for $u\in S_a$ and $\tau\in\mathbb{R}$, set $(\tau\star u)(x)=e^{\frac{N}{2}\tau}u(e^\tau x)$, then $\tau\star u \in S_a$ and

(2.3)\begin{equation} E_\mu(\tau\star u)=\frac{e^{2s\tau}}{2}\int_{\mathbb{R}^N}|\left(-\Delta\right)^{\frac{s}{2}}u|^2-\frac{e^{N\tau(\frac{p}{2}-1)}}{p}\int_{\mathbb{R}^N}|u|^p- \mu\frac{e^{N\tau(\frac{q}{2}-1)}}{q}\int_{\mathbb{R}^N}|u|^q. \end{equation}

Since $2 \lt q \lt p$, we obtain $m_{a,\mu}\leq \lim_{\tau\rightarrow-\infty}E_\mu(\tau\star u)=0$.

(ii) Let $\theta\geq1$ and $u\in S_a$. Set $\tilde{u}(x)=u(\theta^{-2/N}x),x\in \mathbb{R}^N$, then $\tilde{u}\in S_{\theta a}$ and

\begin{align*} E_\mu(\tilde{u}) &\leq \theta^2\Big(\frac{1}{2}\int_{\mathbb{R}^N}|\left(-\Delta\right)^{\frac{s}{2}}u|^2-\frac{1}{p}\int_{\mathbb{R}^N}|u|^p- \mu\frac{1}{q}\int_{\mathbb{R}^N}|u|^q\Big).\nonumber \end{align*}

Since u could be chosen arbitrarily, we obtain $m_{\theta a,\mu}\leq \theta^2 m_{a,\mu}$.

(iii) Assume that $a_1\geq a_2$. As a consequence of (ii),

\begin{equation*} m_{a,\mu}\leq\left(\frac{a}{a_1}\right)^2m_{a_1,\mu}=m_{a_1,\mu}+ \frac{a_2^2}{a_1^2}m_{\frac{a_1}{a_2}a_2,\mu}\leq m_{a_1,\mu}+m_{a_2,\mu}. \end{equation*}

(iv) This follows directly from (i) and (iii).

(v) For $u\in S_1$ given, we set $u_a(x)=au(x)$ for any a > 0, then it holds that $u_a\in S_{a}$. Furthermore, we obtain

\begin{equation*} E_\mu(u_a)=\frac{a^2}{2}\int_{\mathbb{R}^N}|\left(-\Delta\right)^{\frac{s}{2}}u|^2-\frac{a^p}{p}\int_{\mathbb{R}^N}|u|^p- \mu\frac{a^q}{q}\int_{\mathbb{R}^N}|u|^q. \end{equation*}

Since $2 \lt q \lt p$, we know $E_\mu(u_a)\rightarrow-\infty$ as $a\rightarrow\infty$. Thus, we get our conclusion.

(vi) The proof is similar to that of [Reference Luo and Zhang34, lemma 3.3 (v)] and standard.

Proof. (i) If $u\in S_a$ satisfies $E_\mu(u)=m_{a,\mu}$, then $u\not\equiv0$. Recalling the proof of lemma 2.2 (ii), one finds that

\begin{equation*} m_{\theta a,\mu}\leq E_\mu(\tilde{u}) \lt \theta^2 E_\mu(u)=\theta^2 m_{a,\mu}. \end{equation*}

(ii) If $a_1\geq a_2 \gt 0$, by (i) and lemma 2.2 (iii), we have

\begin{equation*} m_{a,\mu} \lt \left(\frac{a}{a_1}\right)^2m_{a_1,\mu}=m_{a_1,\mu}+ \frac{a_2^2}{a_1^2}m_{a_1,\mu}\leq m_{a_1,\mu}+m_{a_2,\mu}. \end{equation*}

If $a_2 \gt a_1 \gt 0$, by (i) and lemma 2.2 (iii) again, we get

\begin{equation*} m_{a,\mu}\leq\left(\frac{a}{a_2}\right)^2m_{a_2,\mu}=m_{a_2,\mu}+ \frac{a_1^2}{a_2^2}m_{a_2,\mu} \lt m_{a_2,\mu}+m_{a_1,\mu}. \end{equation*}

Proof. For any $u\in S_a$, by (2.1) and lemma A.1, we obtain

\begin{align*} E_\mu(u) \ge\Big[\frac12-C_0C(s,N,\bar{p})a^{\frac{4s}{N}}\Big]\int_{\mathbb{R}^N}|\left(-\Delta\right)^{\frac{s}{2}}u|^2. \end{align*}

Set $a_1:=\big(\frac{1}{2C_0C(s,N,\bar{p})}\big)^{\frac{N}{4s}}$, then $E_\mu(u)\geq0$ for any $a\in(0, a_1]$ and any $u\in S_a$. It follows by lemma 2.2 (i) that $m_{a,\mu}=0$ for any $0 \lt a\leq a_1$ and $a_0\ge a_1 \gt 0$.

Proof. First, we claim that

\begin{equation*} \varlimsup_{n\rightarrow\infty}\sup_{z\in\mathbb{R}^N}\int_{B(z,1)}|u_n|^2dx \gt 0. \end{equation*}

Otherwise, in virtue of concentration compactness principle [Reference Felmer, Quaas and Tan14, lemma 2.2], we know $u_n\rightarrow0$ in $L^p(\mathbb{R}^N)$ for any $p\in(2,2_s^*)$. Then, it holds that

\begin{equation*} m_{a,\mu}=\lim_{n\rightarrow\infty}E_\mu(u_n)\geq \liminf_{n\rightarrow\infty}\frac{1}{2}\int_{\mathbb{R}^N}|\left(-\Delta\right)^{\frac{s}{2}}u_n|^2\geq0, \end{equation*}

which contradicts with $m_{a,\mu} \lt 0$ for $a \gt a_0$.

From the claim above, there exists a sequence $\{y_n\}_{n\in\mathbb{N}}\subset \mathbb{R}^N$ such that, up to a subsequence,

(2.4)\begin{equation} 0 \lt \lim_{n\rightarrow\infty}\int_{B(0,1)}|u_n(x-y_n)|^2dx \lt \infty. \end{equation}

By lemma 2.2 (i), the minimizing sequence $\{u_n\}_{n\in\mathbb{N}}$ is uniformly bounded in $H^s(\mathbb{R}^N)$. Thus, $\{u_n(\cdot-y_n)\}_{n\in\mathbb{N}}$ is also bounded in $H^s(\mathbb{R}^N)$. As a consequence, there exists a $u\in H^s(\mathbb{R}^N)$ such that, up to a subsequence,

(2.5)\begin{equation} \begin{aligned} u_n(\cdot-y_n)\rightharpoonup u,&\quad \mbox{weakly in~}H^s(\mathbb{R}^N). \end{aligned} \end{equation}

Via (2.4) and (2.5), we know $|u|_2 \gt 0$. Take $v_n=u_n(\cdot-y_n)-u$, it holds that $v_n\rightharpoonup 0$ weakly in $H^s(\mathbb{R}^N)$. Therefore, by Brezis–Lieb lemma, as $n\rightarrow\infty$,

\begin{align*} \int_{\mathbb{R}^N}\left|\left(-\Delta\right)^{\frac{s}{2}} u_n\right|^2dx &=\int_{\mathbb{R}^N}\left|\left(-\Delta\right)^{\frac{s}{2}} u\right|^2dx+\int_{\mathbb{R}^N}\left|\left(-\Delta\right)^{\frac{s}{2}} v_n\right|^2dx+o(1). \\ \int_{\mathbb{R}^N}\left|u_n\right|^rdx &=\int_{\mathbb{R}^N}\left| u\right|^rdx+\int_{\mathbb{R}^N}\left| v_n\right|^rdx+o(1),\quad \forall r\in\big[2,\frac{2N}{N-2s}\big]. \end{align*}

Noting that $E_\mu(u_n)=E_\mu(u_n(\cdot-y_n))=E_\mu(u+v_n)$, it consequently follows that

(2.6)\begin{align} E_\mu(u_n)=E_\mu(u)+E_\mu(v_n)+o(1),\quad |u_n|_2^2=|u|_2^2+|v_n|_2^2+o(1). \end{align}

We claim that $|v_n|_2^2\rightarrow0$ as $n\rightarrow\infty$. Indeed, set $\zeta=|u|_2 \gt 0$, then $\lim\limits_{n\rightarrow\infty}|v_n|_2^2=a^2-\zeta^2$. If $\zeta=a$, then the claim holds directly. Suppose that $\zeta \lt a$ and set $\tilde{v}_n=\frac{\sqrt{a^2-\zeta^2}}{|v_n|_2}v_n$. In virtue of lemma 2.1 and (2.6), we obtain

\begin{equation*} E_\mu(u_n)=E_\mu(u)+E_\mu(v_n)+o(1)=E_\mu(u)+E_\mu(\tilde{v}_n)+o(1)\geq E_\mu(u)+m_{\sqrt{a^2-\zeta^2},\mu}+o(1). \end{equation*}

Furthermore, letting $n\rightarrow\infty$ and by lemma 2.2 (iii), we deduce that

(2.7)\begin{equation} m_{a,\mu}\geq E_\mu(u)+m_{\sqrt{a^2-\zeta^2},\mu}\geq m_{\zeta,\mu}+m_{\sqrt{a^2-\zeta^2},\mu}\geq m_{a,\mu}, \end{equation}

which gives $E_\mu(u)=m_{\zeta,\mu}$. According to Corollary 2.3 (ii), it holds that

\begin{equation*} m_{a,\mu} \lt m_{\zeta,\mu}+m_{\sqrt{a^2-\zeta^2}}, \end{equation*}

which contradicts with (2.7). Thus, the claim holds and $|u|_2^2=a^2$.

Since $\lim_{n\rightarrow\infty}|v_n|_2^2=0$, recalling that $\left\{v_n\right\}$ is bounded in $H^s(\mathbb{R}^N)$, it follows by Hölder inequality that $\lim_{n\rightarrow\infty}|v_n|_p=0$ and $\lim_{n\rightarrow\infty}|v_n|_q=0$. Moreover,

(2.8)\begin{equation} \liminf_{n\rightarrow\infty}E_\mu(v_n)=\liminf_{n\rightarrow\infty}\frac{1}{2}|(-\Delta)^{\frac{s}{2}} v_n|_2^2\geq0. \end{equation}

In addition, by $|u|_2^2=a^2$ and (2.6),

\begin{equation*} E_\mu(u_n)=E_\mu(u)+E_\mu(v_n)+o(1)\geq m_{a,\mu}+E_\mu(v_n)+o(1), \end{equation*}

which implies

(2.9)\begin{equation} \limsup_{n\rightarrow\infty}E_\mu(v_n)\leq0. \end{equation}

It follows from (2.8) and (2.9) that $\lim\limits_{n\rightarrow\infty}|(-\Delta)^{\frac{s}{2}} v_n|_2^2\rightarrow0$. Hence, $u_n(\cdot-y_n)\rightarrow u$ strongly in $H^s(\mathbb{R}^N)$.

Proof of theorem 1.3 (i) To begin with, we infer from proposition 2.4 that $a_0 \gt 0$. By lemma 2.2, we know $m_{a,\mu}$ is non-positive, non-increasing, and continuous in a. Thus, by the definition of a ₀, we have $m_{a,\mu}=0$ for any $0 \lt a\leq a_0$ and $m_{a,\mu} \lt 0$ for any $a \gt a_0$.

Moreover, we claim that $m_{a,\mu}$ cannot be achieved for any $0 \lt a \lt a_0$. If not, assume $m_{a,\mu}$ is achieved for some $0 \lt a \lt a_0$, then it follows from corollary 2.3 (ii) that $m_{a_0,\mu} \lt m_{a,\mu}=0$, which contradicts with the definition of a ₀.

For $a \gt a_0$, the existence of a least energy solution to (1.2) and (1.3) follows directly from proposition 2.5. We know that the least energy solution is also a ground state solution.

Finally, we try to show that $m_{a,\mu}$ is achieved for $a=a_0$. Let u_n be a global minimizer for $m_{a_0+\frac{1}{n}, \mu}$ for any $n\in\mathbb{N}$, then using the symmetric arrangement, we can assume that u_n is radially symmetric with respect to the origin and it is non-increasing. Since $E_{\mu}\left(u_n\right)$ and $\left|u_n\right|_{2}$ are uniformly bounded, $\left\{u_n\right\}_{n\in\mathbb{N}}$ is a bounded sequence in $H^s\left(\mathbb{R}^N\right)$. What is more, $\lim\limits_{n\rightarrow \infty}\left|u_n\right|_{2}=a_0$. Set $v_n=\frac{\sqrt{a_0}}{\left|u_n\right|_{2}}u_n,$ then we can deduce from lemma 2.1 that

\begin{equation*} v_n\in S_{a_0},\quad \lim\limits_{n\rightarrow \infty}E_{\mu}\left(v_n\right)=\lim\limits_{n\rightarrow \infty}E_{\mu}\left(u_n\right)=\lim\limits_{n\rightarrow \infty}m_{a_0+\frac{1}{n},\mu}=0, \end{equation*}

where the last equality follows from the continuity of $m_{a,\mu}$ and $m_{a_0,\mu}=0$. Thus, $\left\{v_n\right\}_{n\in\mathbb{N}}\subset S_{a_0}$ is a minimizing sequence for $m_{a_0,\mu}.$

Claim. Up to a subsequence, there exist $v\in S_{a_0}$ and $\left\{y_n\right\}_{n\in\mathbb{N}}\subset \mathbb{R}^N$ such that

\begin{equation*} v_n\left(\cdot -y_n\right)\rightarrow v\ \text{in}\ H^s\left(\mathbb{R}^N\right) \ \text{as}\ n\rightarrow \infty. \end{equation*}

In particular, v is a global minimizer of $m_{a_0,\mu}.$ If the claim fails, by the remark 2.6, since $v_n=\frac{\sqrt{a_0}}{\left|u_n\right|_{2}}u_n$, we obtain

(2.10)\begin{equation} \varlimsup_{n\rightarrow \infty} \sup\limits_{z\in\mathbb{R}^N}\int_{B\left(z,1\right)}\left|u_n\right|^2\mathrm{d}x=0. \end{equation}

By a similar argument as the proof of Claim 2 in [Reference Li, Zhang and Zhang27, theorem 1.3], we can know $\left\{u_n\right\}_{n\in\mathbb{N}}$ is a uniformly bounded sequence in $C^{\gamma_0}\left(\mathbb{R}^N\right)$ for some small constant $\gamma_0 \gt 0$. Together with $\left(2.10\right)$, it follows that $u_n\left(0\right)=\left\|u_n\right\|_{L^{\infty}}\rightarrow 0$ as $n\rightarrow \infty.$

Define $v_{\eta,n}:=\eta^{\frac{N}{2}}u_n(\eta x)$ for η > 1 large, then $v_{\eta,n}\in S_{a_0+\frac{1}{n}}$. Note that $2 \lt q \lt p \lt 2+\frac{4s}{N}$, thus $0 \lt \frac{N(q-2)}{2} \lt \frac{N(p-2)}{2} \lt 2s$ and $0 \lt \eta^{2s}-\eta^{\frac{N(p-2)}{2}} \lt \eta^{2s}-\eta^{\frac{N(q-2)}{2}}$. Consequently, for n large, $\|u_n\|^{p-q}_{L^{\infty}} \lt |\mu|$, and

\begin{equation*} \begin{aligned} E_{\mu}(v_{\eta,n})-\eta^{2s}E_{\mu}(u_n) &=\int_{\mathbb{R}^N}|u_n|^q\left[\left(\eta^{2s}-\eta^{\frac{N(p-2)}{2}}\right)\frac{1}{p}|u_n|^{p-q}\right.\nonumber\\ & \quad \left. +\left(\eta^{2s}-\eta^{\frac{N(q-2)}{2}}\right)\frac{\mu}{q}\right]\mathrm{d}x \lt 0. \end{aligned} \end{equation*}

We obtain

\begin{equation*} m_{a_0+\frac{1}{n},\mu}\le E_{\mu}\left(v_\eta,n\right) \lt \eta^{2s}E_{\mu}\left(u_n\right) \lt E_{\mu }\left(u_n\right)=m_{a_0+\frac{1}{n},\mu}, \end{equation*}

which is a contradiction; thus, there exists a global minimizer for $m_{a_0,\mu}.$

(ii) The lower bound of a ₀ is given by proposition 2.4.

(iii) We prove by contradiction. Suppose there exists $\varepsilon_0 \gt 0$, a sequence of solutions $\{\psi_n\}_{n\in\mathbb{N}}$ of (1.1), and a sequence $\{t_n\}_{n\in\mathbb{N}}$ such that

\begin{equation*}\inf_{v\in\mathcal{G}_{a,\mu}}\left\|\psi_n(0,\cdot)-v\right\|_{H^s} \lt 1/n,\end{equation*}

but

\begin{equation*} \inf_{v\in\mathcal{G}_{a,\mu}}\left\|\psi_n(t_n,\cdot)-v\right\|_{H^s}\geq\varepsilon_0. \end{equation*}

By the conservation of mass and energy, it holds that

\begin{equation*} |\psi_n(t_n,\cdot)|_2^2=|\psi_n(0,\cdot)|_2^2\rightarrow a^2,\quad E_\mu(\psi_n(t_n,\cdot))=E_\mu(\psi_n(0,\cdot))\rightarrow m_{a,\mu}. \end{equation*}

Let $\alpha_n=a/|\psi_n(t_n,\cdot)|_2$ and $\tilde{\psi}_n(x)=\alpha_n \psi_n(t_n,x)$. Then, by lemma 2.1, the following holds:

\begin{equation*} \tilde{\psi}_n\in S_a,\quad \lim_{n\rightarrow \infty} \alpha_n=1,\quad \lim_{n\rightarrow \infty}E_\mu(\tilde{\psi}_n)= m_{a,\mu}. \end{equation*}

By proposition 2.5, there exist a family $\{y_n\}\subset \mathbb{R}^N$ and $u\in\mathcal{G}_{a,\mu}$ such that $\lim_{n\rightarrow\infty} \tilde{\psi}_n(\cdot-y_n)=u$ in $H^s(\mathbb{R}^N)$. Thus, $\lim\limits_{n\rightarrow\infty} \left\|\psi_n(t_n,\cdot-y_n)-u\right\|_{H^s}=0$. A contradiction follows from the following inequalities:

\begin{align*} \varepsilon_0 & \le \inf_{v\in\mathcal{G}_{a,\mu}}\left\|\psi_n(t_n,\cdot)-v\right\|_{H^s}\leq \left\|\psi_n(t_n,\cdot)-u(\cdot-y_n)\right\|_{H^s} \\ & = \left\|\psi_n(t_n,\cdot-y_n)-u\right\|_{H^s}=o(1) \end{align*}

as $n\rightarrow\infty$.

Proof. First, by $P_{\mu}(u)=0$ and lemma A.1, one has, for every $u\in\overline{V}_{a,\mu}$,

\begin{align*} |\left(-\Delta\right)^{\frac{s}{2}}u|_2^2 &\leq\frac{N(p-2)}{2ps}C(s,N,p)a^{p-\frac{N(p-2)}{2s}}|\left(-\Delta\right)^{\frac{s}{2}}u|_2^{\frac{N(p-2)}{2s}}\\ & \quad +\mu\frac{N(q-2)}{2qs}C(s,N,q)a^{p-\frac{N(q-2)}{2s}}|\left(-\Delta\right)^{\frac{s}{2}}u|_2^{\frac{N(q-2)}{2s}}, \end{align*}

which implies

\begin{align*} 1\leq C_1(s,N,p,a)|\left(-\Delta\right)^{\frac{s}{2}}u|_2^{\frac{N(p-2)}{2s}-2}+C_2(s,N,p,a,\mu) |\left(-\Delta\right)^{\frac{s}{2}}u|_2^{\frac{N(q-2)}{2s}-2}. \end{align*}

Noting that $p \gt q \gt 2+4s/N$, we have $\frac{N(p-2)}{2s}-2 \gt \frac{N(q-2)}{2s}-2 \gt 0$. Thus, there exists a constant $\delta_0 \gt 0$ such that

(3.2)\begin{equation} \inf_{u\in\overline{V}_{a,\mu}} |\left(-\Delta\right)^{\frac{s}{2}}u|_2^2\geq\delta_0. \end{equation}

Moreover, we note that

\begin{align*} \overline{M}_{a,\mu} & =\inf_{u\in\overline{V}_{a,\mu}}\Big[\Big(\frac{1}{2}-\frac{2s}{N(q-2)}\Big)|\left(-\Delta\right)^{\frac{s}{2}}u|_2^2 +\frac{1}{p}\Big(\frac{p-2}{q-2}-1\Big)|u|_p^p\Big]\\ & \geq \inf_{u\in\overline{V}_{a,\mu}}\Big(\frac{1}{2}-\frac{2s}{N(q-2)}\Big)|\left(-\Delta\right)^{\frac{s}{2}}u|_2^2. \end{align*}

Therefore, $E_{\mu}|_{\overline{V}_{a,\mu}}$ is coercive, and by (3.2), we obtain $ \overline{M}_{a,\mu}\geq \delta_1:=\big(\frac{1}{2}-\frac{2s}{N(q-2)}\big)\delta_0$.

Proof. Let $\{v_n\}_{n\in\mathbb{N}}$ be a minimizing sequence of $\overline{M}_{a,\mu}$, then we have

\begin{equation*} E_\mu(v_n)\rightarrow \overline{M}_{a,\mu}, \quad P_\mu(v_n)=0. \end{equation*}

By lemma 3.1, $\{v_n\}_{n\in\mathbb{N}}$ is bounded in $H^s(\mathbb{R}^N)$ and

\begin{equation*} \liminf_{n\rightarrow\infty} |\left(-\Delta\right)^{\frac{s}{2}}v_n|_2^2\geq\delta_0. \end{equation*}

Applying lemma A.2, we find a profile decomposition of $\{v_n\}_{n\in\mathbb{N}}$ satisfying

(3.3)\begin{equation} \limsup_{n\rightarrow +\infty}|v_n|_\gamma^\gamma=\sum_{j=1}^\infty |V^j|_\gamma^\gamma \quad \mbox{for every~}\gamma\in\Big(2,\frac{2N}{(N-2s)^+}\Big). \end{equation}

Let

\begin{equation*} J=\left\{j\geq1: V^j\neq0\right\}, \end{equation*}

then $J\neq\emptyset$. Otherwise, we can deduce from (3.3) that

\begin{equation*} \limsup_{n\rightarrow +\infty}|v_n|_p^p=\limsup_{n\rightarrow +\infty}|v_n|_q^q=0. \end{equation*}

Noting that $P_\mu(v_n)=0$, we get

\begin{equation*} \delta_0\le \limsup_{n\rightarrow +\infty}|\left(-\Delta\right)^{\frac{s}{2}}v_n|_2^2=0, \end{equation*}

which is a contradiction.

We claim that there exists some $j_0\in J$ such that

(3.4)\begin{equation} 0 \lt |\left(-\Delta\right)^{\frac{s}{2}}V^j|_2^2\leq \frac{N(p-2)}{2ps}|V^j|_p^p+ \mu\frac{N(q-2)}{2qs}|V^j|_q^q. \end{equation}

Otherwise, we suppose that for all $j\in J$,

\begin{equation*} |\left(-\Delta\right)^{\frac{s}{2}}V^j|_2^2 \gt \frac{N(p-2)}{2ps}|V^j|_p^p+ \mu\frac{N(q-2)}{2qs}|V^j|_q^q. \end{equation*}

Then, by lemma A.2 and $P_\mu(v_n)=0$, we obtain

\begin{align*} & \limsup_{n\rightarrow +\infty}\Big(\frac{N(p-2)}{2ps}|v_n|_p^p+\mu\frac{N(q-2)}{2qs}|v_n|_q^q\Big)\\ & \quad \geq \sum_{j\in J}|\left(-\Delta\right)^{\frac{s}{2}}V^j|_{2}^2 \gt \sum_{j\in J}\Big(\frac{N(p-2)}{2ps}|V^j|_p^p+\mu\frac{N(q-2)}{2qs}|V^j|_q^q\Big)\\ & \quad =\limsup_{n\rightarrow +\infty}\Big(\frac{N(p-2)}{2ps}|v_n|_p^p+\mu\frac{N(q-2)}{2qs}|v_n|_q^q\Big), \end{align*}

which is a contradiction. Thus the claim holds.

Let us define

(3.5)\begin{equation} r_u:=\Big(\frac{\frac{N(p-2)}{2ps}|u|_p^p+\mu\frac{N(q-2)}{2qs}|u|_q^q}{|(-\Delta)^{\frac{s}{2}}u|_2^2}\Big)^{\frac{1}{2s}}, \end{equation}

then we know $P_\mu(u(r_u\cdot))=0$. Thus, by (3.4), there exists some $j_0\in J$ such that $r_{V^{j_0}}\geq1$ and $P_\mu(V^{j_0}(r_{V^{j_0}}\cdot))=0$. Moreover,

\begin{equation*} |V^{j_0}(r_{V^{j_0}}\cdot)|_2^2=r_{V^{j_0}}^{-N}|V^{j_0}|_2^2\leq r_{V^{j_0}}^{-N}a^2\leq a^2, \end{equation*}

which implies $V^{j_0}(r_{V^{j_0}}\cdot)\in \overline{V}_{a,\mu}$. In addition, we also note that

\begin{align*} \overline{M}_{a,\mu} & =\inf_{\overline{V}_{a,\mu}}E_\mu =\inf_{u\in\overline{V}_{a,\mu}}\left(E_\mu(u)-\frac{1}{2}P_\mu(u)\right)\\ & =\inf_{u\in\overline{V}_{a,\mu}}\left[\frac{1}{p}\left(\frac{N(p-2)}{4s}-1\right)|u|_p^p +\frac{\mu}{q}\left(\frac{N(q-2)}{4s}-1\right)|u|_q^q\right]. \end{align*}

Thus, it holds that

\begin{align*} 0 \lt \overline{M}_{a,\mu}\leq E_\mu\left(V^{j_0}(r_{V^{j_0}}\cdot)\right) & \leq\frac{1}{p}\Big(\frac{N(p-2)}{4s}-1\Big)|V^{j_0}|_p^p +\frac{\mu}{q}\Big(\frac{N(q-2)}{4s}-1\Big)|V^{j_0}|_q^q\\ & \le\limsup_{n\rightarrow +\infty}E_\mu(v_n)=\overline{M}_{a,\mu}, \end{align*}

which implies $r_{V^{j_0}}=1$, $V^{j_0}\in\overline{V}_{a,\mu}$, and $E_\mu\left(V^{j_0}\right)=\overline{M}_{a,\mu}$.

Finally, let $u:=|V^{j_0}|^*$ be the Schwartz symmetrization of $|V^{j_0}|$, then by [Reference Almgren and Lieb1, theorem 9.2], one has

\begin{equation*} |u|_2^2=|V^{j_0}|_2^2, \ |u|_p^p=|V^{j_0}|_p^p, \ |u|_q^q=|V^{j_0}|_q^q,\quad\mbox{and}\ |\left(-\Delta\right)^{\frac{s}{2}}u|_{2}^2\leq |\left(-\Delta\right)^{\frac{s}{2}}V^{j_0}|_{2}^2. \end{equation*}

By (3.5) and $r_{V^{j_0}}=1$, one has $r_u\geq1$ and $u(r_u\cdot)\in\overline{V}_{a,\mu}$. Suppose $r_u \gt 1$, then

\begin{align*} \overline{M}_{a,\mu} & \leq E_\mu\left(u(r_u\cdot)\right) \lt \frac{1}{p}\Big(\frac{N(p-2)}{4s}-1\Big)|u|_p^p +\frac{\mu}{q}\Big(\frac{N(q-2)}{4s}-1\Big)|u|_q^q\\ & =E_\mu\left(V^{j_0}\right)=\overline{M}_{a,\mu}, \end{align*}

which is a contradiction. Therefore, we get $r_u=1$ and $E_\mu(u)=E_\mu\left(V^{j_0}\right)=\overline{M}_{a,\mu}$.

Proof. Suppose by contradiction that $\overline{M}_{a,\mu}=E_\mu(\tilde{u})\leq M_{a,\mu}$ for some $\tilde{u}\in\mathring{V}_{a,\mu}.$ Hence, $\tilde{u}$ is a local minimizer for E_µ on $\mathring{V}_{a,\mu}$, and there exists a Lagrange multiplier $\lambda\in\mathbb{R}$ such that

\begin{equation*} E_\mu'(\tilde{u})-\lambda P_\mu'(\tilde{u})=0, \end{equation*}

i.e., $\tilde{u}$ is a weak solution to

(3.6)\begin{equation} \left(1-2\lambda\right)\left(-\Delta\right)^s\tilde{u}=\left[1-\lambda\frac{N(p-2)}{2s}\right]|\tilde{u}|^{p-2}\tilde{u} +\mu\left[1-\lambda\frac{N(q-2)}{2s}\right]|\tilde{u}|^{q-2}\tilde{u}. \end{equation}

Moreover, $\tilde{u}$ satisfies the Pohozaev identity of equation (3.6), i.e.,

(3.7)\begin{equation} \frac{N-2s}{2} \left(1-2\lambda\right)|\left(-\Delta\right)^{\frac{s}{2}}\tilde{u}|_{2}^2=\frac{N}{p}\left[1-\lambda\frac{N(p-2)}{2s}\right]|\tilde{u}|_p^p+ \frac{\mu N}{q}\left[1-\lambda\frac{N(q-2)}{2s}\right]|\tilde{u}|_q^q. \end{equation}

In addition, $\tilde{u}$ satisfies the following Nehari-type identity:

(3.8)\begin{equation} \left(1-2\lambda\right)|\left(-\Delta\right)^{\frac{s}{2}}\tilde{u}|_{2}^2=\Big[1-\lambda\frac{N(p-2)}{2s}\Big]|\tilde{u}|_p^p+ \mu\Big[1-\lambda\frac{N(q-2)}{2s}\Big]|\tilde{u}|_q^q. \end{equation}

Besides, since $P_\mu(\tilde{u})=0$, we obtain

(3.9)\begin{equation} |\left(-\Delta\right)^{\frac{s}{2}}\tilde{u}|_2^2=\frac{N(p-2)}{2ps}|\tilde{u}|_p^p+ \mu\frac{N(q-2)}{2qs}|\tilde{u}|_q^q. \end{equation}

After balancing the coefficients of (3.7), (3.8), and (3.9), we deduce that

\begin{equation*} \lambda \frac{N(p-2)}{p}\Big(1-\frac{N(p-2)}{4s}\Big)|\tilde{u}|_p^p+\lambda \mu \frac{N(q-2)}{q}\Big(1-\frac{N(q-2)}{4s}\Big)|\tilde{u}|_q^q=0. \end{equation*}

Since $p \gt q \gt 2+4s/N$, $\tilde{u}\neq0$, and µ > 0, it must hold that λ = 0. Thus, $\tilde{u}$ is a weak solution to

\begin{equation*} \left(-\Delta\right)^s\tilde{u}=|\tilde{u}|^{p-2}\tilde{u}+\mu|\tilde{u}|^{q-2}\tilde{u}. \end{equation*}

In particular, $\tilde{u}$ satisfies the following Nehari-type identity:

\begin{equation*} |\left(-\Delta\right)^{\frac{s}{2}}\tilde{u}|_{2}^2=|\tilde{u}|_p^p+ \mu|\tilde{u}|_q^q. \end{equation*}

We combine the above identity with (3.9) to obtain

\begin{equation*} \frac{2N-p(N-2s)}{2ps}|\tilde{u}|_p^p+\mu\frac{2N-q(N-2s)}{2qs}|\tilde{u}|_q^q=0, \end{equation*}

which is a contradiction since $q \lt p \lt 2^*_s$, µ > 0, and $\tilde{u}\neq0$. Thus we conclude the proof.

Proof of theorem 1.5 (i) and (ii) (i) From proposition 3.2 and lemma 3.3, we immediately have

(3.10)\begin{equation} M_{a,\mu}=\overline{M}_{a,\mu}. \end{equation}

Moreover, $M_{a,\mu}$ is attained by a non-negative and radially symmetric function u in $V_{a,\mu}$. Since it is well known that a critical point for $E_\mu|_{V_{a,\mu}}$ is also a critical point for $E_\mu|_{S_a}$, we apply Lagrange multiplier rules to deduce that there exists $\lambda\in\mathbb{R}$ such that

\begin{equation*} \left(-\Delta\right)^su-|u|^{p-2}u-\mu|u|^{p-2}u=\lambda u. \end{equation*}

Thus,

\begin{equation*} |\left(-\Delta\right)^{\frac{s}{2}}{u}|_{2}^2=|{u}|_p^p+ \mu|{u}|_q^q+\lambda|u|_2^2. \end{equation*}

Combining with the identity $P_{\mu}(u)=0$, we get

\begin{equation*} \lambda a^2=\lambda|u|_2^2=-\left[\frac{2N-p(N-2s)}{2sp}|u|_p^p+\mu\frac{2N-q(N-2s)}{2sq}|u|_q^q\right]. \end{equation*}

Since µ > 0 and $q \lt p \lt 2_s^*$, we know λ < 0. Moreover, due to $u\geq0$ and $u\not\equiv0$, by the maximum principle, it holds that u > 0 in $\mathbb{R}^N$.

(ii) Let $0 \lt a_1 \lt a_2$. There exist two functions u ₁ and u ₂ such that

\begin{equation*} M_{a_1,\mu}=E_\mu(u_1),\quad |u_1|_2^2=a_1^2, \end{equation*}

and

\begin{equation*} M_{a_2,\mu}=E_\mu(u_2),\quad |u_2|_2^2=a_2^2. \end{equation*}

Then, we use lemma 3.3 to get

\begin{equation*} M_{a_2,\mu} \lt E_\mu(u_1)=M_{a_1,\mu}, \end{equation*}

which implies that $M_{a,\mu}$ is strictly decreasing with respect to a.

Proof. By straightforward calculation, one has $\Psi'(\tau) =sP_\mu(\tau\star u).$ In addition, we also see that

\begin{align*} \Psi'(\tau) & = e^{2s\tau}\Big(s|(-\Delta)^{\frac{s}{2}}u|_2^2- \frac{N(p-2)}{2p}e^{\left(N(\frac{p}{2}-1)-2s\right)\tau}|u|_p^p\\ & \quad - \mu\frac{N(q-2)}{2q}e^{\left(N(\frac{q}{2}-1)-2s\right)\tau}|u|_q^q\Big). \end{align*}

Since $2+4s/N \lt q \lt p$ and µ > 0, $\Psi'(\tau)$ is strictly decreasing with respect to τ. Consequently, there exists a unique $\tau_0\in\mathbb{R}$ such that $\Psi'(\tau_0)=0$. Other desired results follow directly.

Proof. For any $u\in S_a$ with $P_\mu(u) \lt 0$, by lemma 3.4, there exists a $\tau_0 \lt 0$ such that $P_\mu(\tau_0\star u)=0$, i.e., $\tau_0\star u \in V_{a,\mu}$ defined in (3.1). Furthermore, it holds that

\begin{align*} \widetilde{E}_\mu(\tau_0\star u)&=\left(\frac{1}{2}-\frac{2s}{N(q-2)}\right)|\left(-\Delta\right)^{\frac{s}{2}}(\tau_0\star u)|_2^2 +\frac{1}{p}\left(\frac{p-2}{q-2}-1\right)|\tau_0\star u|_p^p\\ &=\left(\frac{1}{2}-\frac{2s}{N(q-2)}\right)e^{2s\tau_0}|\left(-\Delta\right)^{\frac{s}{2}}u|_2^2+ \frac{1}{p}\left(\frac{p-2}{q-2}-1\right)e^{N\tau_0(\frac{p}{2}-1)}|u|_p^p\\ & \lt \left(\frac{1}{2}-\frac{2s}{N(q-2)}\right)|\left(-\Delta\right)^{\frac{s}{2}}u|_2^2+ \frac{1}{p}\left(\frac{p-2}{q-2}-1\right)|u|_p^p\\ &=\widetilde{E}_\mu(u). \end{align*}

Hence, $\widetilde{M}_{a,\mu}=\inf_{\widetilde{V}_{a,\mu}}\widetilde{E}_\mu=\inf_{V_{a,\mu}}\widetilde{E}_\mu=M_{a,\mu}$.

Proof. First, we claim that $\mathcal{A}_{a,\mu}\neq\emptyset$ and $\mathcal{B}_{a,\mu}\neq\emptyset$. As a matter of fact, for $u\in S_{a,r}$, recall $(\tau\star u)(x)=e^{\frac{N}{2}\tau}u(e^\tau x)$. On the one hand, the following statements hold: (i) $\tau\star u\in S_{a,r}$ for every $\tau\in \mathbb{R}$; (ii) $E_\mu(\tau\star u)\rightarrow0$ as $\tau\rightarrow-\infty$ by (2.3); and (iii) $P_\mu(\tau\star u) \gt 0$ for τ sufficiently negative by lemma 3.4. On the other hand, by lemma 3.1 and (3.10), we know $M_{a,\mu}\geq\delta_1 \gt 0$. Therefore, $\mathcal{A}_{a,\mu}\neq\emptyset$. Similarly, one has $E_\mu(\tau\star u)\rightarrow-\infty$ and $P_\mu(\tau\star u)\rightarrow-\infty$ as $\tau\rightarrow+\infty$. Thus, $\mathcal{B}_{a,\mu}\neq\emptyset$.

Second, we prove that $\mathcal{A}_{a,\mu}$ and $\mathcal{B}_{a,\mu}$ are two invariant manifolds of (1.1). Let $\psi_0\in \mathcal{A}_{a,\mu}$, and by lemma A.3, there exists a unique solution $\psi\in C\left([0,T^*),H^s\right)$ to (1.1) with initial data ψ ₀. Moreover, we have

\begin{equation*} |\psi(t,\cdot)|_{2}^2=|\psi_0|_{2}^2=a^2,\quad E_\mu(\psi(t,\cdot))=E_\mu(\psi_0) \lt M_{a,\mu} \end{equation*}

for any $t\in(0,T^*)$. If there exists some $t_0\in [0,T^*)$ such that $P_\mu(\psi(t_0,\cdot))=0$, then $E_\mu(\psi(t_0,\cdot))\geq M_{a,\mu}$, which is a contradiction. Therefore, we deduce from the continuity with respect to t of $\psi(t,\cdot)$ that $P_\mu(\psi(t,\cdot)) \gt 0$ for any $t\in [0,T^*)$. As a result, $\psi(t,\cdot)$ stays in $A_{a,\mu}$ for any $t\in [0,T^*)$. Similarly, $\mathcal{B}_{a,\mu}$ is invariant under the flow of (1.1).

(i) Due to $\psi(t,\cdot)\in \mathcal{A}_{a,\mu}$ for any $t\in [0,T^*)$, we deduce from the conservation of energy that

\begin{align*} E_\mu(\psi_0)& \gt \left(\frac{1}{2}-\frac{2s}{N(q-2)}\right)|\left(-\Delta\right)^{\frac{s}{2}}\psi(t,\cdot)|_2^2 +\frac{1}{p}\left(\frac{p-2}{q-2}-1\right)|\psi(t,\cdot)|_p^p, \end{align*}

which together with lemma A.3 implies that the solution $\psi(t,\cdot)$ of (1.1) exists globally.

(ii) If $\psi_0\in\mathcal{B}_{a,\mu}$, then $P_\mu(\psi(t,\cdot)) \lt 0$ for any $t\in [0,T^*)$. By lemma 3.5, we know

\begin{align*} M_{a,\mu} & =\widetilde{M}_{a,\mu}\leq \widetilde{E}_\mu(\psi(t,\cdot))\\ &=E_\mu(\psi(t,\cdot))-\frac{2s}{N(q-2)}P_\mu(\psi(t,\cdot)) =E_\mu(\psi_0)-\frac{2s}{N(q-2)}P_\mu(\psi(t,\cdot)), \end{align*}

which implies

(3.14)\begin{equation} P_\mu(\psi(t,\cdot))\leq\frac{N(q-2)}{2s}\left(E_\mu(\psi_0)-M_{a,\mu}\right) \lt 0, \quad \forall t\in [0,T^*). \end{equation}

The proof of blow-up behaviour will be divided into three steps as follows:

Step 1: We prove that there exists $C_1 \gt 0$ such that

(3.15)\begin{equation} |\left(-\Delta\right)^{\frac{s}{2}}\psi(t,\cdot)|_2\geq C_1 \end{equation}

for every $t\in[0,T^*)$. Indeed, if not, then there exists $\{t_k\}\subseteq [0,T^*)$ such that $|\left(-\Delta\right)^{\frac{s}{2}}\psi(t_k, \cdot)|_2\rightarrow0$. However, we deduce from mass conservation and the sharp Gagliardo-Nirenberg inequality that $|\psi(t_k,\cdot)|_\alpha^\alpha=o(1)$ as $k\rightarrow\infty$, where $\alpha=p\mbox{~or~}q$. Therefore, we have

\begin{equation*} P_\mu(\psi(t_k,\cdot))=|\left(-\Delta\right)^{\frac{s}{2}}\psi(t_k,\cdot)|_2^2-\frac{N(p-2)}{2ps}|\psi(t_k,\cdot)|_p^p- \mu\frac{N(q-2)}{2qs}|\psi(t_k,\cdot)|_q^q\rightarrow0 \end{equation*}

as $k\rightarrow\infty$, which contradicts to (3.14).

Step 2: We claim that there exists $C_2 \gt 0$ such that

(3.16)\begin{equation} \frac{d}{dt}M_{\varphi_R}(\psi(t,\cdot))\leq -C_2|\left(-\Delta\right)^{\frac{s}{2}}\psi(t,\cdot)|_2^2, \end{equation}

where $M_{\varphi_R}(\psi(t,\cdot))$ is defined by (3.13).

Observe that $\psi(t,\cdot)$ is radial for any $t\in[0,T^*)$, since the initial datum ψ ₀ is radial. Therefore, we apply lemma 3.6 to have

\begin{align*} \frac{d}{dt}M_{\varphi_R}(\psi(t,\cdot))&\leq8s|\left(-\Delta\right)^{\frac{s}{2}}\psi(t,\cdot)|_2^2-\frac{4N\mu(q-2)}{q}|\psi(t,\cdot)|_q^q- \frac{4N(p-2)}{p}|\psi(t,\cdot)|_p^p\\ &\quad +O\Big(R^{-2s}+R^{-\frac{(q-2)(N-1)}{2}+\varepsilon_1s}|\left(-\Delta\right)^{\frac{s}{2}}\psi(t,\cdot)|_2^{\frac{q-2}{2s}+ \varepsilon_1}\Big)\\ &\quad+O\left(R^{-\frac{(p-2)(N-1)}{2}+\varepsilon_2s}|\left(-\Delta\right)^{\frac{s}{2}}\psi(t,\cdot)|_2^{\frac{p-2}{2s}+ \varepsilon_2}\right) \end{align*}

for all $t\in[0,T^*)$ and R > 1. Thanks to the assumption $q \lt p \lt 2+4s$, we can apply Young’s inequality to obtain for any η > 0,

\begin{align*} &R^{-\frac{(q-2)(N-1)}{2}+\varepsilon_1s}|(-\Delta)^{\frac{s}{2}}\psi(t,\cdot)|_2^{\frac{q-2}{2s}+ \varepsilon_1}\\ & \quad \leq \eta |(-\Delta)^{\frac{s}{2}}\psi(t,\cdot)|_2^2\!+\!\eta^{-\frac{q-2+2\varepsilon_1s}{2+4s-q-2\varepsilon_1s}} R^{-\frac{2s\left[(q-2)(N-1)-2\varepsilon_1s\right]}{2+4s-q-2\varepsilon_1s}},\\ &R^{-\frac{(p-2)(N-1)}{2}+\varepsilon_2s}|(-\Delta)^{\frac{s}{2}}\psi(t,\cdot)|_2^{\frac{p-2}{2s}+ \varepsilon_2}\\ & \quad \leq \eta |(-\Delta)^{\frac{s}{2}}\psi(t,\cdot)|_2^2\!+\!\eta^{-\frac{p-2+2\varepsilon_2s}{2+4s-p-2\varepsilon_2s}} R^{-\frac{2s\left[(p-2)(N-1)-2\varepsilon_2s\right]}{2+4s-p-2\varepsilon_2s}}. \end{align*}

Thus, there exists a constant C > 0 such that

\begin{align*} \frac{d}{dt}M_{\varphi_R}(\psi(t,\cdot))&\leq8s|\left(-\Delta\right)^{\frac{s}{2}}\psi(t,\cdot)|_2^2-\frac{4N\mu(q-2)}{q}|\psi(t,\cdot)|_q^q- \frac{4N(p-2)}{p}|\psi(t,\cdot)|_p^p\\ &\quad+C\eta |\left(-\Delta\right)^{\frac{s}{2}}\psi(t,\cdot)|_2^2+ I(\eta,R) \end{align*}

for all $t\in[0,T^*)$, any η > 0, and any R > 1, where

\begin{align*} I(\eta,R)&:=O\Big(R^{-2s} +\eta^{-\frac{q-2+2\varepsilon_1s}{2+4s-q-2\varepsilon_1s}} R^{-\frac{2s\left[(q-2)(N-1)-2\varepsilon_1s\right]}{2+4s-q-2\varepsilon_1s}}\\ & \quad +\eta^{-\frac{p-2+2\varepsilon_2s}{2+4s-p-2\varepsilon_2s}} R^{-\frac{2s\left[(p-2)(N-1)-2\varepsilon_2s\right]}{2+4s-p-2\varepsilon_2s}}\Big). \end{align*}

Since $2+4s/N \lt q \lt p \lt 2N/(N-2s)$ and $p \lt 2+4s$, we can choose $\varepsilon_1 \gt 0,\varepsilon_2 \gt 0$ sufficiently small such that

\begin{align*} &q-2+2\varepsilon_1s \gt 0,\quad 2+4s-q-2\varepsilon_1s \gt 0,\quad (q-2)(N-1)-2\varepsilon_1s \gt 0,\\ &p-2+2\varepsilon_2s \gt 0,\quad 2+4s-p-2\varepsilon_2s \gt 0,\quad (p-2)(N-1)-2\varepsilon_2s \gt 0. \end{align*}

In addition, we fix $t\in[0,T^*)$ and denote

\begin{equation*} \kappa:=\frac{4N(p-2)|E_\mu(\psi_0)|+1}{N(p-2)-4s}. \end{equation*}

Since $p \gt 2+4s/N$, we know κ > 0. We consider two cases.

Case 1: $|\left(-\Delta\right)^{\frac{s}{2}}\psi(t,\cdot)|_2^2\leq\kappa$. Noting that

\begin{equation*} 8s|\left(-\Delta\right)^{\frac{s}{2}}\psi(t,\cdot)|_2^2-\frac{4N\mu(q-2)}{q}|\psi(t,\cdot)|_q^q- \frac{4N(p-2)}{p}|\psi(t,\cdot)|_p^p=8s P_\mu(\psi(t,\cdot)), \end{equation*}

and (3.14), we have

\begin{align*} \frac{d}{dt}M_{\varphi_R}(\psi(t,\cdot))\leq4N(q-2) \left(E_\mu(\psi_0)-M_{a,\mu}\right) +C\eta \kappa +I(\eta,R). \end{align*}

By choosing η > 0 small enough and R > 1 large enough depending on η, we can get

\begin{align*} 2N(q-2)\left(E_\mu(\psi_0)-M_{a,\mu}\right)+C\eta \kappa +I(\eta,R) \lt 0. \end{align*}

Thus, we obtain

\begin{align*} \frac{d}{dt}M_{\varphi_R}(\psi(t,\cdot)) \leq\frac{2N(q-2)\left(E_\mu(\psi_0)-M_{a,\mu}\right)}{\kappa}|\left(-\Delta\right)^{\frac{s}{2}}\psi(t,\cdot)|_2^2. \end{align*}

Case 2: $|\left(-\Delta\right)^{\frac{s}{2}}\psi(t,\cdot)|_2^2 \gt \kappa$. By lemma 3.6, we obtain

\begin{align*} \frac{d}{dt}M_{\varphi_R}(\psi(t,\cdot)) &\leq4N(p-2)|E_\mu\left(\psi_0\right)|+2\left(4s-N(p-2)\right)|\left(-\Delta\right)^{\frac{s}{2}}\psi(t,\cdot)|_2^2\\ & \quad +C\eta |\left(-\Delta\right)^{\frac{s}{2}}\psi(t,\cdot)|_2^2+I(\eta,R)\\ &\leq-1+\left(4s-N(p-2)\right)|\left(-\Delta\right)^{\frac{s}{2}}\psi(t,\cdot)|_2^2\\ & \quad +C\eta |\left(-\Delta\right)^{\frac{s}{2}}\psi(t,\cdot)|_2^2+I(\eta,R). \end{align*}

Since $p \gt 2+4s/N$, we choose η > 0 small enough so that

\begin{equation*} N(p-2)-4s-C\eta\geq\frac{N(p-2)-4s}{2}. \end{equation*}

We next choose R > 1 large enough depending on η so that

\begin{equation*} -1+I(\eta,R) \lt 0. \end{equation*}

We thus obtain

\begin{equation*} \frac{d}{dt}M_{\varphi_R}(\psi(t,\cdot))\leq -\frac{N(p-2)-4s}{2} |\left(-\Delta\right)^{\frac{s}{2}}\psi(t,\cdot)|_2^2. \end{equation*}

Combined with the two cases above, we prove our claim (3.16).

Step 3: We are now able to show that the solution $\psi(t,\cdot)$ blows up in a finite time. Assume by contradiction that $T^*=\infty$. It follows from (3.15) and (3.16) that $\frac{d}{dt}M_{\varphi_R}(\psi(t,\cdot))\leq -C$ with some constant C > 0. As a consequence, $M_{\varphi_R}(\psi(t,\cdot)) \lt 0$ for all $t\geq t_1$ with some sufficiently large t ₁. After integrating (3.16) on $[t_1, t]$, we obtain

(3.17)\begin{equation} M_{\varphi_R}(\psi(t,\cdot))\!\leq\! -C_2\!\int_{t_1}^t |(-\Delta)^{\frac{s}{2}}\psi(\tau,\cdot)|_2^2d\tau+ M_{\varphi_R}(\psi(t_1)) \!\leq\! -C_2\!\int_{t_1}^t |(-\Delta)^{\frac{s}{2}}\psi(\tau,\cdot)|_2^2d\tau \end{equation}

for all $t\geq t_1$. On the other hand, we use lemma A.5 and L ²-mass conservation to find that

(3.18)\begin{equation} \left|M_{\varphi_R}(\psi(t,\cdot))\right|\leq C(N, s, a, \|\nabla\varphi_R\|_{W^{1,\infty}})\left(|\left(-\Delta\right)^{\frac{s}{2}}\psi(t,\cdot)|_2 ^{\frac{1}{s}}+|\left(-\Delta\right)^{\frac{s}{2}}\psi(t,\cdot)|_2 ^{\frac{1}{2s}}\right), \end{equation}

where we used the interpolation estimate

\begin{equation*} \left||\nabla|^{1/2}\psi(t,\cdot)\right|_2\leq |\psi(t,\cdot)|_2^{1-\frac{1}{2s}}|\left(-\Delta\right)^{\frac{s}{2}}\psi(t,\cdot)|_2 ^{\frac{1}{2s}} \end{equation*}

for $s \gt 1/2$. So, we deduce from (3.15) and (3.18) that

\begin{equation*} \left|M_{\varphi_R}(\psi(t,\cdot))\right|\leq C(N, s, a, \|\nabla\varphi_R\|_{W^{1,\infty}})|\left(-\Delta\right)^{\frac{s}{2}}\psi(t,\cdot)|_2 ^{\frac{1}{s}}. \end{equation*}

This, together with (3.17), implies that

(3.19)\begin{equation} M_{\varphi_R}(\psi(t,\cdot))\leq -C(N, s, a, \|\nabla\varphi_R\|_{W^{1,\infty}})\int_{t_1}^t \left|M_{\varphi_R}(\psi(\tau,\cdot))\right|^{2s}d\tau \quad \mbox{for ~}t\geq t_1. \end{equation}

Let $z(t):=\int_{t_1}^t\left|M_{\varphi_R}(\psi(\tau,\cdot))\right|^{2s}d\tau$, noting that $M_{\varphi_R}(\psi(t,\cdot)) \lt 0$ for $t \gt t_1$, hence z(t) is strictly increasing for $t \gt t_1$ and we can find a $t_2 \gt t_1$ such that $z(t_2) \gt 0$. Furthermore, by (3.19), we obtain

\begin{equation*} z'(t)\geq [C(N, s, a, \|\nabla\varphi_R\|_{W^{1,\infty}})]^{2s}z^{2s}. \end{equation*}

Therefore, for $t \gt t_2$, it holds that $z'(t)z^{-2s}\ge [C(N, s, a, \|\nabla\varphi_R\|_{W^{1,\infty}})]^{2s}$. Since $s \gt \frac{1}{2}$, we obtain $(1-2s)z'(t)z^{-2s}\le (1-2s)[C(N, s, a, \|\nabla\varphi_R\|_{W^{1,\infty}})]^{2s}$. After integration on $[t_2,t]$, one has

(3.20)\begin{equation} 0 \lt z^{1-2s}(t)\le z^{1-2s}(t_2)-(2s-1)[C(N, s, a, \|\nabla\varphi_R\|_{W^{1,\infty}})]^{2s}(t-t_2). \end{equation}

Note that the right-hand side of (3.20) goes to $-\infty$ as $t\rightarrow+\infty$, while the left-hand side is positive. Hence, it must hold that $T^* \lt +\infty$.

Proof of theorem 1.5(iii) Let u be a radial minimizer for $M_{a,\mu}$. Then, we know $P_\mu(u)=0$. Setting $\psi_0^\tau(x)=e^{\frac{N}{2}\tau}u(e^\tau x)$ with τ > 0, by lemma 3.4, we obtain

\begin{equation*} E_\mu\left(\psi_0^\tau\right)=E_\mu\left(\tau\star u\right) \lt E_\mu(u)=M_{a,\mu},\quad P_\mu(\psi_0^\tau)=P_\mu\left(\tau\star u\right) \lt 0 \end{equation*}

for any τ > 0. Thus, $\psi_0^\tau\in \mathcal{B}_{a,\mu}$ for τ > 0. In addition, let $\tau\rightarrow 0^+$, we have $\|\psi_0^\tau-u\|_{H^s}\rightarrow0$. Therefore, we apply proposition 3.7 to get the strong instability of $e^{-\mathrm{i}\lambda t}u$.

Proof. The proof is similar to the argument of lemma 3.5, so we omit it.

Proof. The proof is similar to that of proposition 3.7.

Proof of theorem 1.7 The conclusion (i) is established by [Reference Luo and Zhang33, theorems 1.9, 1.11]. The proof of item (ii) is similar to the argument of theorem 1.5(iii).