Proof. First we claim that there exists $c_0\neq 0\text{ and }0 \lt \lambda_0 \lt \infty$ such that

(3.2)\begin{equation}d\left(u,\mathcal{M}\right)=\left\|\left(-\Delta\right)^{\frac{s}{2}}\left(u-c_0U^{\lambda_0}_{s,t}\right)\right\|_2=\left\|u-c_0U^{\lambda_0}_{s,t}\right\|_{\dot{H}^s}\end{equation}

holds.

To see this, we consider a sequence $(c_k,\lambda_k)\in\mathbb{R}\times\mathbb{R}^+$ such that $\|u-c_kU^{\lambda_k}_{s,t}\|_{\dot{H}^s}\to d(u,\mathcal{M})$ as $k\to\infty$ (this exists from the definition of infimum). In the expansion of the fractional norm, using the norm invariance of $U^\lambda_{s,t}$ and an application of Cauchy-Schwarz inequality, we see,

\begin{align*} \|u-c_kU^{\lambda_k}_{s,t}\|^2_{\dot{H}^s}&=\|u\|^2_{\dot{H}^s}+|c_k|^2\|U^{\lambda_k}_{s,t}\|^2_{\dot{H}^s}-2c_k\langle u,U^{\lambda_k}_{s,t}\rangle_{\dot{H}^s}\\ &\geq\|u\|^2_{\dot{H}^s}+|c_k|^2\|U_{s,t}\|^2_{\dot{H}^s}-2|c_k|\| u\|_{\dot{H}^s}\|U_{s,t}\|_{\dot{H}^s}\\ &=\left(\| u\|_{\dot{H}^s}-|c_k|\| U_{s,t}\|_{\dot{H}^s}\right)^2. \end{align*}

Taking limit as $k\to\infty$ above, one observes

(3.3)\begin{align} \| u\|_{\dot{H}^s} \gt d(u,\mathcal{M})\geq\limsup_{k\to\infty}\left|\| u\|_{\dot{H}^s}-|c_k|\| U_{s,t}\|_{\dot{H}^s}\right|. \end{align}

This immediately shows that $|c_k|$ is a bounded sequence of positive real numbers, thus up to a subsequence still denoted by $c_k$, it converges to some $c_0\in\mathbb{R}$. Moreover, $c_0=0$ contradicts (3.3), hence, $c_0\neq0$. Again from the fractional norm expansion above, we see by taking the limit along the subsequence $c_k$,

(3.4)\begin{align} 2c_0\cdot\lim\limits_{k\to\infty}\langle u,U^{\lambda_k}_{s,t}\rangle_{\dot{H}^s}=\|u\|^2_{\dot{H}^s}+|c_0|^2\|U_{s,t}\|^2_{\dot{H}^s}-d(u,\mathcal{M})^2 \gt |c_0|^2\|U_{s,t}\|^2_{\dot{H}^s} \gt 0. \end{align}

Now invoking lemma 2.1, one can easily deduce that (3.4) forces the existence of a $\lambda_0\in\mathbb{R}^+$ such that up to a subsequence $\lambda_k\to\lambda_0$ as $k\to\infty$ and the claim follows.

As we have seen above that for a $u\in\dot{H}^s(\mathbb{R}^N)$ the distance of $u$ from the smooth sub-manifold $\mathcal{M}$ is achieved at a point $c_0U^{\lambda_0}_{s,t}\in\mathcal{M}\setminus\{0\}$ under the condition $d(u,\mathcal{M}) \lt \|u\|_{\dot{H}^s}$, we then must have $(u-c_0U^{\lambda_0}_{s,t})\perp\textbf{T}_{c_0U^{\lambda_0}_{s,t}}\mathcal{M}$. Moreover, the tangent space is given as

\begin{equation*} \textbf{T}_{c_0U^{\lambda_0}_{s,t}}\mathcal{M}=\operatorname{span} \left\{U^{\lambda_0}_{s,t}\,,\,\partial_\lambda {U^{\lambda}_{s,t}}\big|_{\lambda=\lambda_0}\right\}. \end{equation*}

From the discussions in Appendix A, we see, proposition A.1 and theorem A.2 together give us that $i\circ\left({\mathcal{L}}^{\lambda_0}_{s,t}\right)^{-1}=i\circ\left(\frac{(-\Delta)^s}{w}\right)^{-1}\text{; where }w=\mu_{s,t}\frac{(U^{\lambda_0}_{s,t})^{2^*_s(t)-2}}{|x|^t}$ from $L^2_w(\mathbb{R}^N)$ to itself is a compact self-adjoint operator and thus its spectrum is discrete and consequently the spectrum of ${\mathcal{L}}^{\lambda_0}_{s,t}$ is also discrete. Combining theorem 1.1 with this, we see that the first and the second eigenspace of the operator ${\mathcal{L}}^{\lambda_0}_{s,t}$ are spanned by $U^{\lambda_0}_{s,t}$ and $\partial_\lambda U^{\lambda}_{s,t}\big|_{\lambda=\lambda_0}$, respectively. Moreover, we have the Rayleigh quotient characterization for the third eigenvalue as

(3.5)\begin{equation}\eta_3\leq\frac{\int_{\mathbb{R}^N}\vert(-\Delta)^\frac s2\phi(x)\vert^2\,\mathrm dx}{\int_{\mathbb{R}^N}w(x)\vert\phi(x)\vert^2\,\mathrm dx}\text{ for all }\phi\perp{\textbf{T}}_{c_0U_{s,t}^{\lambda_0}}\mathcal M,\end{equation}

with equality if and only if $\phi$ is the third eigenfunction. We also notice $\eta_1=1,\,\eta_2=2^*_s(t)-1$ and all other eigenvalues are independent of the scaling $\lambda_0$.

Since $(u-c_0U^{\lambda_0}_{s,t})\perp\textbf{T}_{c_0U^{\lambda_0}_{s,t}}\mathcal{M}$, there exists a vector $\phi\in\dot{H}^s(\mathbb{R}^N)$ perpendicular to the tangent plane $\textbf{T}_{c_0U^{\lambda_0}_{s,t}}\mathcal{M}$ such that $\Arrowvert\phi\Arrowvert_{\dot H^s}=1\text{ and }u-c_0U_{s,t}^{\lambda_0}=d\phi$. Certainly, we can choose $d$ to be positive and then $d=\|u-c_0U^{\lambda_0}_{s,t}\|_{\dot{H}^s}=d(u,\mathcal{M})$.

Now a Taylor’s expansion in $d$ around $0$ for the following yields

\begin{align*} \int_{\mathbb{R}^N}\frac{|u|^{2^*_s(t)}}{|x|^t} \,\mathrm{d}x&= \int_{\mathbb{R}^N}\frac{|c_0U^{\lambda_0}_{s,t}+d\phi|^{2^*_s(t)}}{|x|^t} \,\mathrm{d}x\\ &=|c_0|^{2^*_s(t)}\int_{\mathbb{R}^N}\frac{\left(U^{\lambda_0}_{s,t}\right)^{2^*_s(t)}}{|x|^{t}}\,\mathrm{d}x \cr & \quad+d\,2^*_s(t)\,c_0|c_0|^{2^*_s(t)-2}\int_{\mathbb{R}^N}\frac{\left(U^{\lambda_0}_{s,t}\right)^{2^*_s(t)-1}\phi}{|x|^{t}}\,\mathrm{d}x\\ &+d^2\,\frac{2^*_s(t)(2^*_s(t)-1)}{2}\,|c_0|^{2^*_s(t)-2}\int_{\mathbb{R}^N}\frac{\left(U^{\lambda_0}_{s,t}\right)^{2^*_s(t)-2}\phi^2}{|x|^{t}}\,\mathrm{d}x+{\scriptscriptstyle\mathcal O}\left(d^2\right). \end{align*}

In the above expression, plugging in (3.5), $\||x|^{\frac{-t}{2^*_s(t)}}U^{\lambda_0}_{s,t}\|_{2^*_s(t)}=\||x|^{\frac{-t}{2^*_s(t)}}U_{s,t}\|_{2^*_s(t)}=1$ and the fact that, $U_{s,t}$ solves (1.3) weakly, we arrive at

\begin{align*} \int_{\mathbb{R}^N}\frac{|u|^{2^*_s(t)}}{|x|^t} \,\mathrm{d}x&\leq|c_0|^{2^*_s(t)}+d\,2^*_s(t)\,c_0|c_0|^{2^*_s(t)-2}\,\frac{1}{\mu_{s,t}}\left\langle\phi,U^{\lambda_0}_{s,t}\right\rangle_{\dot{H}^s}\\ &+d^2\,\frac{2^*_s(t)(2^*_s(t)-1)}{2}\,|c_0|^{2^*_s(t)-2}\,\frac{\|\phi\|^2_{\dot{H}^s}}{\mu_{s,t}\eta_3}+{\scriptscriptstyle\mathcal O}\left(d^2\right). \end{align*}

Using the properties $\eta_2=2^*_s(t)-1,\,\phi\perp\textbf{T}_{c_0U^{\lambda_0}_{s,t}}\mathcal{M}$ and $\|\phi\|_{\dot{H}^s}=1$ in above, notice that the second term vanishes and we obtain

\begin{equation*}\int_{\mathbb{R}^N}\frac{\vert u\vert^{2_s^\ast(t)}}{\vert x\vert^t}\,\mathrm dx\leq\vert c_0\vert^{2_s^\ast(t)}+d^2\,\frac{2_s^\ast(t)}2\,\frac{\vert c_0\vert^{2_s^\ast(t)-2}}{\mu_{s,t}}\,\frac{\eta_2}{\eta_3}+{\scriptscriptstyle\mathcal O}\left(d^2\right).\end{equation*}

Using Taylor’s expansion again at $y=0$ one has $(1+y)^r=1+ry+{\scriptscriptstyle\mathcal O}{(y)};{}\,0 \lt r \lt 1,\,y \gt 0$ and thus

\begin{align*} \mu_{s,t}\left\||x|^{\frac{-t}{2^*_s(t)}}u\right\|_{2^*_s(t)}^2&\leq \mu_{s,t}|c_0|^2\left(1+d^2\,\frac{2^*_s(t)}{2}\,|c_0|^{-2}\mu_{s,t}^{-1}\,\frac{\eta_2}{\eta_3}+{\scriptscriptstyle\mathcal O}\left(d^2\right)\right)^{\frac{2}{2^*_s(t)}}\\ &=\mu_{s,t}|c_0|^2\left(1+d^2\,|c_0|^{-2}\mu_{s,t}^{-1}\,\frac{\eta_2}{\eta_3}+{\scriptscriptstyle\mathcal O}\left(d^2\right)\right). \end{align*}

Also note by the Pythagorean theorem, norm invariance, and the fact that $U_{s,t}$ achieves $\mu_{s,t}$ in (1.1), one obtains

\begin{align*} \|u\|^2_{\dot{H}^s}=\|c_0U^{\lambda_0}_{s,t}+d\phi\|_{\dot{H}^s}^2= |c_0|^2\|U_{s,t}\|^2_{\dot{H}^s}+d^2=\mu_{s,t}|c_0|^2+d^2. \end{align*}

Finally combining the last two results, we conclude

(3.6)\begin{equation}\Arrowvert u\Arrowvert_{\dot H^s}^2-\mu_{s,t}\left\|\vert x\vert^\frac{-t}{2_s^\ast(t)}u\right\|_{2_s^\ast(t)}^2\geq d^2\left(1-\frac{\eta_2}{\eta_3}\right)+{\scriptscriptstyle\mathcal O}\left(d^2\right).\end{equation}

Since the eigenvalues of ${\mathcal{L}}^{\lambda_0}_{s,t}$ are strictly increasing, by defining $\alpha=\left(1-\frac{\eta_2}{\eta_3}\right)$ in (3.6), we get (3.1) with $\alpha\in(0,1)$ as desired.

Proof. On the contrary, suppose there exists an $\epsilon_0 \gt 0$ such that for all $u\in\dot{H}^s\left(\mathbb{R}^N\right)$, one has

\begin{equation*}d(u,\mathcal{M})^{-2\epsilon_0}\delta(u)^{1+\epsilon_0}\geq\alpha d(u,\mathcal{M})^{2} \iff \delta(u)\geq \alpha^{\frac{1}{1+\epsilon_0}}d(u,\mathcal{M})^{2}.\end{equation*}

Now taking $u=U_{s,t}+d\phi$ where $\phi$ is the third eigenfunction of ${\mathcal{L}}^1_{s,t}$, in (3.5), lemma 3.1 gives $\delta(u)=\alpha d^2+{\scriptscriptstyle\mathcal O}\left(d^2\right)$. Finally combining the last two facts and using $0 \lt \alpha=1-\frac{\eta_2}{\eta_3} \lt 1$, we obtain the chain of inequalities

\begin{equation*}\alpha d^2+{\scriptscriptstyle\mathcal O}\left(d^2\right)\geq\alpha^\frac1{1+\epsilon_0}d^2\Leftrightarrow\alpha+{\scriptscriptstyle\mathcal O}\left(1\right)\geq\alpha^\frac1{1+\epsilon_0}\Leftrightarrow\epsilon_0\leq0\end{equation*}

and this is a contradiction to our choice of $\epsilon_0$.

Proof. From the hypothesis of this lemma, one sees that $(u_k)_{k\ge1}\subset \dot{H}^s\left(\mathbb{R}^N\right)$ is a minimizing sequence for $\mu_{s,t}$, therefore, by employing Ekeland’s variational principle, we can extract a Palais–Smale $(PS)$ subsequence, still denoted by $u_k$ for the corresponding energy functional defined as

(3.7)\begin{eqnarray} I_{s,t} (u)&:=&\frac{C_{N,s}}{4}\iint_{\mathbb{R}^{2N}}\frac{|u(x)-u(y)|^2}{|x-y|^{N+2s}}\,\mathrm{d}x\,\mathrm{d}y-\frac{\mu_{s,t}}{2_s^*(t)}\int_{\mathbb{R}^N}\frac{|u|^{2_s^*(t)}}{|x|^t}\,\mathrm{d}x\nonumber\\ &=&\frac{1}{2}\|u\|_{\dot{H}^s}^2 - \frac{\mu_{s,t}}{2_s^*(t)}\left\||x|^{-\frac{t}{2_s^*(t)}}u\right\|_{2_s^*(t)}^{2_s^*(t)}. \end{eqnarray}

Now the lemma is an obvious consequence of proposition 4.2, pertaining to the energy of a single bubble.

Proof of theorem 1.3

Clearly the sharpness of the result follows from the lemma 3.2 above.

Suppose that the theorem is not true. Then we can find a sequence $(u_k)_{k\ge1}$ in $\dot{H}^s\left(\mathbb{R}^N\right)$ such that

(3.8)\begin{equation} \lim\limits_{k\to\infty}\frac{\|u_k\|^2_{\dot{H}^s}-\mu_{s,t}\||x|^{\frac{-t}{2^*_s(t)}}u_k\|^2_{2^*_s(t)}}{d(u_k,\mathcal{M})^2}=0. \end{equation}

Since the ratio in (3.8) is homogeneous of degree $2$ in $u_k$, we may also suppose that $\|u_k\|^2_{\dot{H}^s}=\mu_{s,t}$ for all $k\geq1$. Moreover, $0\in\mathcal{M}$ implies that $d(u_k,\mathcal{M})\leq d(u_k,0)=\|u_k\|_{\dot{H}^s}=\sqrt{\mu_{s,t}} \lt \infty$. Thus, up to a subsequence still denoted by $u_k$, we see that $d(u_k,\mathcal{M})\to L\in\left[0,\sqrt{\mu_{s,t}}\right]$.

We now consider two cases, specifically $L=0$ and $L \gt 0$.

When $L=0 \lt \sqrt{\mu_{s,t}}\,$; we can assume (if required up to a further subsequence still denoted by $u_k$) that for all $k\geq1$ one has $d(u_k,\mathcal{M}) \lt \sqrt{\mu_{s,t}}=\|u_k\|_{\dot{H}^s}$. Lemma 3.1 now gives that the above ratio is greater than or equal to $\alpha+{\scriptscriptstyle\mathcal O}\left(1\right)\rightarrow\alpha \gt 0$ as $k\to\infty$ and it is a contradiction to (3.8).

Thus, $L \gt 0$ and then we must have that the numerator of the above ratio goes to $0$ as $k\to\infty$, this in turn satisfies the hypothesis of lemma 3.3. Consequently, we obtain

\begin{equation*}L=\lim\limits_{k\to\infty}d(u_k,\mathcal{M})\leq\lim\limits_{k\to\infty}\|u_k-c U_{s,t}^{\lambda_k}\|_{\dot{H}^s}=0\end{equation*}

and this gives us the desired contradiction.

Proof. We prove the result in several steps.

Step 1. In this step we prove that every $(PS)_{\beta}$ sequence is uniformly bounded in $\dot{H}^s\left(\mathbb{R}^N\right)$, i.e., there exists an $C \gt 0$ such that

\begin{equation*}\sup_{n\in\mathbb{N}}\|u_n\|_{\dot{H}^s}\leq C.\end{equation*}

Indeed, as $n\to\infty$,

\begin{eqnarray*} \beta + {\scriptscriptstyle\mathcal O} (1)+ {\scriptscriptstyle\mathcal O} (1)\|u_n\|_{\dot{H}^s} &\geq& I_{s,t}(u_n) \, - \, \frac{1}{2_s^*(t)} {_{(\dot{H}^s)'}}^{\left\langle I'_{s,t}(u_n), u_n\right\rangle}_{\dot{H}^s}\nonumber\\ &=& \left(\frac{1}{2}-\frac{1}{2_s^*(t)}\right)\|u_n\|_{\dot{H}^s}^{2}.\nonumber \end{eqnarray*}

As $2_s^*(t) \gt 2$, the above inequality concludes that $(u_n)_{n\ge1}$ is bounded in $\dot{H}^s(\mathbb{R}^N)$. Consequently, there exists $u_0$ in $\dot{H}^s(\mathbb{R}^N)$ such that, up to a subsequence (still denoted by $(u_n)_{n\ge1}$) $u_n\rightharpoonup u_0$ in $\dot{H}^s(\mathbb{R}^N)$ and $u_n\to u_0$ a.e. in $\mathbb{R}^N$. Moreover, $_{{(\dot{H}^s)'}}{\left\langle I'_{s,t}(u_n), v\right\rangle}_{\dot{H}^s}\rightarrow 0$ as $n\rightarrow\infty$ for all $v\in\dot{H}^s(\mathbb{R}^N)$ implies

(4.3)\begin{equation} (-\Delta)^s u_n - \frac{|u_n|^{2_s^*(t)-2}u_n}{|x|^t}\to 0\quad \text{in}\quad\left(\dot{H}^s\left(\mathbb{R}^N\right)\right)'. \end{equation}

Step 2. In this step, we prove that the weak limit $u_0$ of our $(PS)_{\beta}$ sequence $\left(u_n\right)$ is a weak solution for (4.2). From (4.3), letting $n \rightarrow \infty$, we obtain

(4.4)\begin{equation} \langle u_n, v\rangle_{\dot{H}^s}-\int_{\mathbb{R}^N}\frac{|u_n|^{2_s^*(t)-2}u_n v}{|x|^t}\;\mathrm{d}x \, {\to} \, 0,\qquad\text{for all }v\in\dot{H}^s(\mathbb{R}^N). \end{equation}

Since $u_n\rightharpoonup u_0$ in $\dot{H}^s(\mathbb{R}^N)$, it follows from the definition of weak convergence that $\langle u_n, v\rangle_{\dot{H}^s}\to \langle u_0, v\rangle_{\dot{H}^s}$ for all $v\in\dot{H}^s(\mathbb{R}^N)$.

Next, we claim the following:

Claim 1. $\displaystyle\int_{\mathbb{R}^N}\frac{|u_n|^{2_s^*(t)-2}u_nv}{|x|^t}\;\mathrm{d}x \to \int_{\mathbb{R}^N}\frac{|u_0|^{2_s^*(t)-2} u_0 v}{|x|^t} \mathrm{d}x$ for all $v\in\dot{H}^s(\mathbb{R}^N)$.

Indeed, $u_n\to u_0$ a.e. in $\mathbb{R}^N$ and

(4.5)\begin{equation} \int_{\mathbb{R}^N}\frac{|u_n|^{2_s^*(t)-2}u_nv}{|x|^t}\;\mathrm{d}x = \int_{B_R}\frac{|u_n|^{2_s^*(t)-2}u_nv}{|x|^t}\;\mathrm{d}x +\int_{\mathbb{R}^N\setminus B_R}\frac{|u_n|^{2_s^*(t)-2}u_nv}{|x|^t}\;\mathrm{d}x. \end{equation}

On $B_R$ we are going to use Vitali’s convergence theorem. For that, given any $\epsilon \gt 0$, we choose $\Omega\subset B_R$ such that $\displaystyle \left(\int_{\Omega}\frac{|v|^{2_s^*(t)}}{|x|^t}\mathrm{d}x\right)^{\frac{1}{2_s^*(t)}} \lt \frac{\epsilon}{\left(M\mu_{s,t}^{-\frac{1}{2}}\right)^{2_s^*(t)-1}}$. Since $\frac{|v|^{2_s^*(t)}}{|x|^t}$ is in $L^1(\mathbb{R}^N)$, the above choice makes sense. Therefore, using Hölder’s inequality and fractional Hardy–Sobolev inequality, we obtain,

\begin{eqnarray*} \left|\int_{\Omega}\frac{|u_n|^{2_s^*(t)-2}u_nv}{|x|^t}\;\mathrm{d}x\right| &\leq& \int_{\Omega}\frac{|u_n|^{2_s^*(t)-1}|v|}{|x|^t}\;\mathrm{d}x \nonumber\\ &\leq& \left(\int_{\Omega}\frac{|u_n|^{2_s^*(t)}}{|x|^t}\mathrm{d}x\right)^{\frac{2_s^*(t)-1}{2_s^*(t)}}\left(\int_{\Omega}\frac{|v|^{2_s^*(t)}}{|x|^t}\mathrm{d}x\right)^{\frac{1}{2_s^*(t)}}\nonumber\\ &\leq& \mu_{s,t}^{-\frac{2_s^*(t)-1}{2}}\|u_n\|_{\dot{H}^s}^{2_s^*(t)-1}\left(\int_{\Omega}\frac{|v|^{2_s^*(t)}}{|x|^t}\mathrm{d}x\right)^{\frac{1}{2_s^*(t)}} \lt \epsilon.\nonumber \end{eqnarray*}

Thus, $\frac{|u_n|^{2_s^*(t)-2} u_n v}{|x|^t}$ is a uniformly integrable sequence in $L^1\left(B_R\right)$. Therefore, using Vitali’s convergence theorem we can pass to the limit in the integral on $B_R$ of (4.5).

To estimate the integral on $B_R^c$, we first set $v_n=u_n-u_0$. Then $v_n\rightharpoonup 0$ in $\dot{H}^s(\mathbb{R}^N).$ It is not difficult to see that for every $\epsilon \gt 0$ there exists $C(\epsilon) \gt 0$ such that

\begin{equation*}\left|\frac{|v_n+u_0|^{2_s^*(t)-2}(v_n+u_0)}{|x|^t}-\frac{|u_0|^{2_s^*(t)-2}u_0}{|x|^t} \right| \lt \epsilon \frac{|v_n|^{2_s^*(t)-1}}{|x|^t}+C(\epsilon)\frac{|u_0|^{2_s^*(t)-1}}{|x|^t}.\end{equation*}

Therefore,

\begin{align*} &\left|\int_{B_R^c}\left\{\frac{|u_n|^{2_s^*(t)-2}u_n}{|x|^t}-\frac{|u_0|^{2_s^*(t)-2}u_0}{|x|^t}\right\}v\;\mathrm{d}x\right|\\ &\leq \left[\epsilon \int_{B_R^c}\frac{|v_n|^{2_s^*(t)-1}|v|}{|x|^t}\;\mathrm{d}x+C(\epsilon)\int_{B_R^c}\frac{|u_0|^{2_s^*(t)-1}|v|}{|x|^t}\,\mathrm{d}x\right] \\ &\leq \Bigg[\epsilon \left(\int_{B_R^c}\frac{|v_n|^{2_s^*(t)}}{|x|^t}\;\mathrm{d}x\right)^{\frac{2_s^*(t)-1}{2_s^*(t)}}\left(\int_{B_R^c} \frac{|v|^{2_s^*(t)}}{|x|^t}\,\mathrm{d}x\right)^{\frac{1}{2_s^*(t)}}\\ &\qquad\qquad\qquad+C(\epsilon)\left(\int_{B_R^c}\frac{|u_0|^{2_s^*(t)}}{|x|^t}\,\mathrm{d}x\right)^{\frac{2_s^*(t)-1}{2_s^*(t)}}\left(\int_{B_R^c}\frac{|v|^{2_s^*(t)}}{|x|^t}\,\mathrm{d}x\right)^{\frac{1}{2_s^*(t)}}\Bigg]\\ &\leq\! C\!\left[\epsilon \|v_n\|_{\dot{H}^s}^{2_s^*(t)-1}\!\left(\int_{B_R^c} \frac{|v|^{2_s^*(t)}}{|x|^t}\,\mathrm{d}x\right)^{\frac{1}{2_s^*(t)}} \!+\!C(\epsilon)\|u_0\|_{\dot{H}^s}^{2_s^*(t)-1}\!\left(\int_{B_R^c} \frac{|v|^{2_s^*(t)}}{|x|^t}\,\mathrm{d}x\right)^{\frac{1}{2_s^*(t)}}\right]. \end{align*}

Since $\left(\|v_n\|_{\dot{H}^s}\right)_{n\ge1}$ is uniformly bounded and $\frac{|v|^{2_s^*(t)}}{|x|^t}\in L^1(\mathbb{R}^N)$, given $\epsilon \gt 0$, we can choose $R \gt 0$ large enough such that

\begin{equation*}\left|\int_{B_R^c}\left(\frac{|u_n|^{2_s^*(t)-2}u_n}{|x|^t}-\frac{|u_0|^{2_s^*(t)-2} u_0}{|x|^t}\right)v\;\mathrm{d}x\right| \lt \epsilon.\end{equation*}

This completes the proof of Claim 1.

Hence, (4.4) yields that $u_0$ is a solution of (4.2).

Step 3. Here we show that $\left(u_n-u_0\right)_{n\ge1}$ is a $(PS)$ sequence for $I_{s,t}$ at the level $\beta-I_{s,t}(u_0)$. To see this, first, we observe that as $n\to\infty$

\begin{equation*}\Arrowvert u_n-u_0\Arrowvert_{\dot H^s}^2=\Arrowvert u_n\Arrowvert_{\dot H^s}^2-\Arrowvert u_0\Arrowvert_{\dot H^s}^2+{\scriptscriptstyle\mathcal O}(1),\end{equation*}

and by the Brézis–Lieb lemma as $n\to\infty$

\begin{equation*}\int_{\mathbb{R}^N}\frac{\vert u_n-u_0\vert^{2_s^\ast(t)}}{\vert x\vert^t}\mathrm dx=\int_{\mathbb{R}^N}\frac{\vert u_n\vert^{2_s^\ast(t)}}{\vert x\vert^t}\;\mathrm dx-\int_{\mathbb{R}^N}\frac{\vert u_0\vert^{2_s^\ast(t)}}{\vert x\vert^t}\;\mathrm dx+{\scriptscriptstyle\mathcal O}(1).\end{equation*}

Therefore, as $n\to\infty$

\begin{align*} I_{s,t} (u_n-u_0) &= \frac{1}{2}\|u_n-u_0\|_{\dot{H}^s}^2-\frac{1}{2_s^*(t)}\int_{\mathbb{R}^N}\frac{|u_n-u_0|^{2_s^*(t)}}{|x|^t}\;\mathrm{d}x\\ &= \frac{1}{2}\|u_n\|_{\dot{H}^s}^2-\frac{1}{2_s^*(t)}\int_{\mathbb{R}^N}\frac{|u_n|^{2_s^*(t)}}{|x|^t}\;\mathrm{d}x\\ &\qquad-\left(\frac{1}{2}\|u_0\|_{\dot{H}^s}^2-\frac{1}{2_s^*(t)}\int_{\mathbb{R}^N}\frac{|u_0|^{2_s^*(t)}}{|x|^t}\;\mathrm{d}x\right)+{\scriptscriptstyle\mathcal O}(1)\\ &= I_{s,t}(u_n)-I_{s,t}(u_0)+{\scriptscriptstyle\mathcal O}(1)\\ &\to \beta-I_{s,t}(u_0). \end{align*}

Furthermore, as $_{{(\dot{H}^s)'}}{\left\langle I_{s,t}'(u_0), v\right\rangle}_{\dot{H}^s} =0$ for all $v\in\dot{H}^s(\mathbb{R}^N)$, we obtain

(4.6)\begin{align} &_{{\left(\dot{H}^s\right)'}}{\left\langle I_{s,t}'(u_n - u_0), v \right\rangle}_{\dot{H}^s} \cr &= \langle u_n - u_0, v \rangle_{\dot{H}^s} - \int_{\mathbb{R}^N} \frac{|u_n - u_0|^{2_s^*(t)-2}(u_n - u_0)v}{|x|^t} \, \mathrm{d}x \cr &= \langle u_n, v \rangle_{\dot{H}^s} - \int_{\mathbb{R}^N} \frac{|u_n|^{2_s^*(t)-2} u_n v}{|x|^t} \, \mathrm{d}x - \left( \langle u_0, v \rangle_{\dot{H}^s} - \int_{\mathbb{R}^N} \frac{|u_0|^{2_s^*(t)-2} u_0 v}{|x|^t} \, \mathrm{d}x \right)\cr &\quad + \int_{\mathbb{R}^N} \left( \frac{|u_n|^{2_s^*(t)-2} u_n}{|x|^t} - \frac{|u_0|^{2_s^*(t)-2} u_0}{|x|^t} - \frac{|u_n - u_0|^{2_s^*(t)-2}(u_n - u_0)}{|x|^t} \right) v \, \mathrm{d}x \cr &= {\scriptscriptstyle\mathcal O}(1) + \int_{\mathbb{R}^N} \left( \frac{|u_n|^{2_s^*(t)-2} u_n}{|x|^t} - \frac{|u_0|^{2_s^*(t)-2} u_0}{|x|^t} - \frac{|u_n - u_0|^{2_s^*(t)-2}(u_n - u_0)}{|x|^t} \right) v \, \mathrm{d}x. \cr \end{align}

We observe that

\begin{align*} &\left||u_n|^{2_s^*(t)-2}u_n-|u_0|^{2_s^*(t)-2} u_0-|u_n-u_0|^{2_s^*(t)-2}(u_n-u_0)\right|\\ &\leq C\left(|u_n-u_0|^{2_s^*(t)-2}|u_0|+|u_0|^{2_s^*(t)-2}|u_n-u_0|\right). \end{align*}

Therefore, following the same method as in the proof of Claim 1 in Step 2, we obtain, as $n\to\infty$

(4.7)\begin{equation}\int_{\mathbb{R}^N}\left(\frac{\vert u_n\vert^{2_s^\ast(t)-2}u_n}{\vert x\vert^t}-\frac{\vert u_0\vert^{2_s^\ast(t)-2}u_0}{\vert x\vert^t}-\frac{\vert u_n-u_0\vert^{2_s^\ast(t)-2}(u_n-u_0)}{\vert x\vert^t}\right)v\,\mathrm dx={\scriptscriptstyle\mathcal O}(1)\end{equation}

for all $v\in\dot{H}^s(\mathbb{R}^N)$. Plugging this back into (4.6), we complete the proof of Step 3.

Step 4. In this step we rescale our $(PS)$ sequence to get a nonzero weak limit. Define $v_n:= u_n-u_0$. Then $v_n\rightharpoonup 0$ in $\dot{H}^s(\mathbb{R}^N)$ and by Step 3, $\left(v_n\right)_{n\ge1}$ is a $(PS)$ sequence for $I_{s,t}$ at the level $\beta-I_{s,t}(u_0).$ Thus,

(4.8)\begin{equation}\sup_{n\in\mathbb{N}}\Arrowvert v_n\Arrowvert_{\dot H^s}\leq C\text{ and }\langle v_n,\varphi\rangle_{\dot H^s}=\int_{\mathbb{R}^N}\frac{\vert v_n\vert^{2_s^\ast(t)-2}v_n\varphi}{\vert x\vert^t}\,\mathrm dx+{\scriptscriptstyle\mathcal O}(1)\quad\text{for all }\varphi\in\dot H^s(\mathbb{R}^N).\end{equation}

Therefore, $\Arrowvert v_n\Arrowvert_{\dot H^s}^2=\int_{\mathbb{R}^N}\frac{\vert v_n\vert^{2_s^\ast(t)}}{\vert x\vert^t}\;\mathrm dx+{\scriptscriptstyle\mathcal O}(1)$. Thus, if $\int_{\mathbb{R}^N}\frac{|v_n|^{2_s^*(t)}}{|x|^t}\;\mathrm{d}x\to0$, then we are done by taking $m=0$ and the $(PS)$ sequence $\left(u_n\right)_{n\geq1}$ admits a strongly convergent subsequence.

If not, let $0 \lt \delta \lt \left(\frac{\mu_{s,t}}2\right)^{\frac{N-t}{2s-t}}$ be such that

\begin{equation*}\liminf_{n\to\infty}\int_{\mathbb{R}^N}\frac{|v_n|^{2_s^*(t)}}{|x|^t}\;\mathrm{d}x \gt \delta.\end{equation*}

For each $n\ge1$, we choose the minimal $R_n \gt 0$ such that

\begin{equation*}\int_{B_{R_n}}\frac{|v_n|^{2_s^*(t)}}{|x|^t}\;\mathrm{d}x = \delta.\end{equation*}

Define $w_n(x):= D_{R_n}v_n(x)=R_n^{\frac{N-2s}{2}}v_n(R_n x)$ then it is easy to see that $w_n$ satisfies $\|w_n\|_{\dot{H}^s}=\|v_n\|_{\dot{H}^s}$ and

(4.9)\begin{equation} \delta = \int_{B_{R_n}}\frac{|v_n|^{2_s^*(t)}}{|x|^t}\;\mathrm{d}x = \int_{B_1}\frac{|w_n|^{2_s^*(t)}}{|x|^t}\;\mathrm{d}x. \end{equation}

Moreover, there exists $w\in\dot{H}^s(\mathbb{R}^N)$ such that (up to a subsequence still denoted by $w_n$)

\begin{equation*}w_n\rightharpoonup w\;\text{in }\dot H^s(\mathbb{R}^N)\quad\text{and}\quad w_n\rightarrow w\text{ a.e. in }\mathbb{R}^N.\end{equation*}

Let us now distinguish the cases $w\neq 0$ and $w=0$, which we analyze below in steps 5 and 6 separately.

Step 5. In this step we will show that, if the rescaled sequence has a nonzero weak limit then the weak limit must be a solution of (4.2). Moreover, if we subtract the rescaled weak limit from our starting $(PS)$ sequence we will again get a $(PS)$ sequence for our energy functional $I_{s,t}$ but at a certain lower level. This ultimately gives the recursive algorithm.

To see this, first we deal with the case $w\neq 0.$ Since $w_n\rightharpoonup w\neq 0$ and $v_n\rightharpoonup 0$ in $\dot{H}^s\left(\mathbb{R}^N\right)$, thanks to lemma 2.2, it follows that $|\log R_n|\to \infty$ as $n\to\infty$.

Next, we show that $w$ is a solution of (4.2). Indeed, due to (4.8), for any $\phi\in {{\mathcal{C}}_c^\infty} (\mathbb{R}^N)$

\begin{align*} \langle w,\phi\rangle_{\dot{H}^s} &= \lim_{n\to\infty}\langle w_n,\phi\rangle_{\dot{H}^s}\nonumber\\ &=\lim_{n\to\infty} \frac{C_{N,s}}{2}\iint_{\mathbb{R}^{2N}}\frac{(w_n(x)-w_n(y))(\phi(x)-\phi(y))}{|x-y|^{N+2s}}\;\mathrm{d}x\,\mathrm{d}y\nonumber\\ &=\lim_{n\to\infty} \frac{C_{N,s}}{2}\iint_{\mathbb{R}^{2N}}\frac{R_n^{\frac{N-2s}{2}}(v_n(R_n x)-v_n(R_n y))(\phi(x)-\phi(y))}{|x-y|^{N+2s}}\;{\rm d}x\,\mathrm{d}y\nonumber\\ &=\lim_{n\to\infty} \frac{C_{N,s}}{2}\iint_{\mathbb{R}^{2N}}\frac{R_n^{-\frac{N-2s}{2}}(v_n(x)-v_n(y))(\phi(\frac{x}{R_n})-\phi(\frac{y}{R_n}))}{|x-y|^{N+2s}}\;\mathrm{d}x\,\mathrm{d}y\nonumber\\ &=\lim_{n\to\infty}\int_{\mathbb{R}^N}\!\!\!\frac{|v_n|^{2_s^*(t)-2}v_n}{|x|^t}R_n^{-\frac{N-2s}{2}}\phi\left(\frac{x}{R_n}\right)\;\mathrm{d}x\nonumber\\ &=\lim_{n\to\infty}\int_{\mathbb{R}^N}\!\!\!\frac{|w_n|^{2_s^*(t)-2}w_n}{|x|^t}\phi(x)\;\mathrm{d}x.\nonumber \end{align*}

Clearly $\frac{|w_n|^{2_s^*(t)-2}w_n}{|x|^t}\phi\to \frac{|w|^{2_s^*(t)-2}w}{|x|^t}\phi$ a.e. in $\mathbb{R}^N$, since $w_n\to w$ a.e. in $\mathbb{R}^N$. Furthermore, arguing as in the proof of Claim 1 in Step 2, we have $\frac{|w_n|^{2_s^*(t)-2}w_n}{|x|^t}\phi$ is uniformly integrable. Since $\phi$ has compact support, using Vitali’s convergence theorem we obtain

(4.10)\begin{equation} \lim_{n\to\infty}\int_{\mathbb{R}^N}\frac{|w_n|^{2_s^*(t)-2}w_n}{|x|^t}\phi(x)\;\mathrm{d}x = \int_{\mathbb{R}^N}\frac{|w|^{2s^*(t)-2}w\phi}{|x|^t}\mathrm{d}x.\end{equation}

Combining (4.10) along with (4.8), we conclude that $w$ is a solution of (4.2).

Define

\begin{equation*}z_n(x):= v_n(x)-R_n^{-\frac{N-2s}{2}}w\left(\frac{x}{R_n}\right).\end{equation*}

Claim 2. $(z_n)_{n\ge1}$ is a $(PS)$ sequence for $I_{s,t}$ at the level $\beta-I_{s,t}(u_0)- I_{s,t}(w).$

To prove the claim, set

\begin{equation*}\tilde{z}_n(x):= R_n^{\frac{N-2s}{2}}z_n(R_n x).\end{equation*}

Then

\begin{equation*} \tilde{z}_n(x)= w_n(x)-w(x) \quad\text{and}\quad \|\tilde{z}_n\|_{\dot{H}^s}=\|w_n-w\|_{\dot{H}^s}=\|z_n\|_{\dot{H}^s}.\end{equation*}

Now the Brézis–Lieb lemma and a straightforward computation yield as $n\to\infty$

\begin{align*} \int_{\mathbb{R}^N}\frac{|w_n(x)|^{2_s^*(t)}}{|x|^t}\;\mathrm{d}x-\int_{\mathbb{R}^N}\frac{|w|^{2_s^*(t)}}{|x|^t}\;\mathrm{d}x &=\int_{\mathbb{R}^N}\frac{\left|w_n-w\right|^{2_s^*(t)}}{|x|^t}\;\mathrm{d}x+{\scriptscriptstyle\mathcal O}(1). \end{align*}

Therefore, using the above relations, as $n\to\infty$

\begin{align*} I_{s,t}(z_n)&=\frac{1}{2}\|z_n\|_{\dot{H}^s}^2-\frac{1}{2_s^*(t)}\int_{\mathbb{R}^N}\frac{|z_n|^{2_s^*(t)}}{|x|^t}\;\mathrm{d}x\\ &=\frac{1}{2}\|w_n-w\|_{\dot{H}^s}^2-\frac{1}{2_s^*(t)}\int_{\mathbb{R}^N}\frac{|w_n-w|^{2_s^*(t)}}{|x|^t}\;\mathrm{d}x\\ &=\frac{1}{2}\left(\|w_n\|_{\dot{H}^s}^2-\|w\|_{\dot{H}^s}^2\right)-\frac{1}{2_s^*(t)}\int_{\mathbb{R}^N}\frac{|w_n(x)|^{2_s^*(t)}}{|x|^t}\;\mathrm{d}x\\ &\quad+\frac{1}{2_s^*(t)}\int_{\mathbb{R}^N}\frac{|w|^{2_s^*(t)}}{|x|^t}\;\mathrm{d}x+{\scriptscriptstyle\mathcal O}(1)\\ &=\frac{1}{2}\|v_n\|_{\dot{H}^s}^2-\frac{1}{2_s^*(t)}\int_{\mathbb{R}^N}\frac{|v(x)|^{2_s^*(t)}}{|x|^t}\;\mathrm{d}x-\\ &\quad\left(\frac{1}{2}\|w\|_{\dot{H}^s}^2-\frac{1}{2_s^*(t)}\int_{\mathbb{R}^N}\frac{|w|^{2_s^*(t)}}{|x|^t}\;\mathrm{d}x\right)+ {\scriptscriptstyle\mathcal O}(1)\\ &=I_{s,t}(v_n)-I_{s,t}(w)+{\scriptscriptstyle\mathcal O}(1)\\ &=\beta-I_{s,t}(u_0)- I_{s,t}(w)+ {\scriptscriptstyle\mathcal O}(1). \end{align*}

Next, let $\phi\in {{\mathcal{C}}_c^\infty} (\mathbb{R}^N)$ be arbitrary and set $\phi_n(x):= R_n^{\frac{N-2s}{2}}\phi(R_n x)$. Since we already saw above that $|\log R_n|\to\infty$ as $n\to\infty$, one then has by using lemma 2.1 that $\phi_n\rightharpoonup 0$ in $\dot{H}^s\left(\mathbb{R}^N\right)$. Furthermore, the norm invariance property of the dilation map $D_{R_n}$ gives $\|\phi_n\|_{\dot{H}^s}=\|\phi\|_{\dot{H}^s}$ for each $n\in\mathbb{N}$.

Therefore,

(4.11)\begin{align} _{{(\dot{H}^s)'}}{\left\langle I_{s,t}'(z_n), \phi\right\rangle}_{\dot H^s} &=\langle z_n,\phi\rangle_{\dot{H}^s}-\int_{\mathbb{R}^N}\frac{|z_n|^{2_s^*(t)-2}z_n\phi}{|x|^t}\;\mathrm{d}x \cr &= \langle \tilde{z}_n,\phi_n\rangle_{\dot{H}^s}-\int_{\mathbb{R}^N}\frac{|\tilde{z}_n|^{2_s^*(t)-2}\tilde{z}_n\phi_n}{|x|^t}\;\mathrm{d}x\cr &=\langle w_n-w,\phi_n\rangle_{\dot{H}^s}-\int_{\mathbb{R}^N}\frac{|w_n-w|^{2_s^*(t)-2}(w_n-w)\phi_n}{|x|^t}\;\mathrm{d}x \cr &= \langle w_n,\phi_n\rangle_{\dot{H}^s}-\int_{\mathbb{R}^N}\frac{|w_n|^{2_s^*(t)-2}w_n\phi_n}{|x|^t}\;\mathrm{d}x\cr &\quad-\left(\langle w,\phi_n\rangle_{\dot{H}^s} -\int_{\mathbb{R}^N}\frac{|w|^{2_s^*(t)-2}w\phi_n}{|x|^t}\;\mathrm{d}x\right)\cr &\quad+\int_{\mathbb{R}^N}\left(\frac{\begin{array}{l}|w_n|^{2_s^*(t)-2}w_n-|w|^{2_s^*(t)-2}w\\\quad-|w_n-w|^{2_s^*(t)-2}(w_n-w)\end{array}}{|x|^t}\right)\phi_n \mathrm{d}x\cr &=\langle v_n,\phi\rangle_{\dot{H}^s}\!-\!\int_{\mathbb{R}^N}\frac{|v_n|^{2_s^*(t)-2}v_n\phi}{|x|^t}\;\mathrm{d}x-_{{(\dot{H}^s)'}}{\left\langle I_{s,t}{'}(w), \phi_n\right\rangle}_{\dot H^s}\!+\!I^1_n \cr &= _{{(\dot{H}^s)'}}{\left\langle I_{s,t}{'}(v_n), \phi\right\rangle}_{\dot H^s}-0+I^1_n={\scriptscriptstyle\mathcal O}(1)+I^1_n. \end{align}

Next, we aim to show that

\begin{equation*}I_n^1:=\int_{\mathbb{R}^N}\left(\frac{\vert w_n\vert^{2_s^\ast(t)-2}w_n-\vert w\vert^{2_s^\ast(t)-2}w-\vert w_n-w\vert^{2_s^\ast(t)-2}(w_n-w)}{\vert x\vert^t}\right)\phi_n\,\mathrm dx={\scriptscriptstyle\mathcal O}(1).\end{equation*}

Indeed, this follows as in the proof of (4.7), since $\int_{\mathbb{R}^N}\frac{|\phi_n|^{2_s^*(t)}}{|x|^t}\;\mathrm{d}x=\int_{\mathbb{R}^N}\frac{|\phi|^{2_s^*(t)}}{|x|^t}\;\mathrm{d}x \lt \infty$. Hence, from (4.11) we conclude the proof of Claim 2.

Step 6. Now we will show that $w\neq 0$ as long as $t \gt 0$. On the contrary, assume that $w= 0$.

Let $\varphi\in {{\mathcal{C}}_c^\infty} \left(B_1\right)$, with $0\leq\varphi\leq 1$. Set $\psi_n(x):=\left(\varphi\left(\frac{x}{R_n}\right)\right)^2 v_n(x)$. Clearly $(\psi_n)_{n\ge1}$ is a bounded sequence in $\dot{H}^s\left(\mathbb{R}^N\right)$. Thus,

\begin{align*} {\scriptscriptstyle\mathcal O}(1)&=_{{(\dot{H}^s)'}}{\left\langle I_{s,t}{'}(v_n), \psi_n\right\rangle}_{\dot{H}^s} \\ &=\langle v_n,\psi_n\rangle_{\dot{H}^s}-\int_{\mathbb{R}^N}\frac{|v_n|^{2_s^*(t)-2}v_n\psi_n}{|x|^t}\mathrm{d}x\\ &=\frac{C_{N,s}}{2}\iint_{\mathbb{R}^{2N}}\frac{\left(v_n(x)-v_n(y)\right)\left(\varphi^2(\frac{x}{R_n})v_n(x)-\varphi^2(\frac{y}{R_n})v_n(y)\right)}{|x-y|^{N+2s}}\mathrm{d}x\,\mathrm{d}y\\ &\qquad-\int_{\mathbb{R}^N}\frac{\varphi^2(\frac{x}{R_n})|v_n|^{2_s^*(t)}}{|x|^t}\mathrm{d}x\\ &=\frac{C_{N,s}}{2}\iint_{\mathbb{R}^{2N}}\frac{\left(v_n(R_nx)-v_n(R_ny)\right)\left(\varphi^2(x)v_n(R_nx)-\varphi^2(y)v_n(R_ny)\right)R_n^{N-2s}}{|x-y|^{N+2s}}\\ &\qquad\mathrm{d}x\,\mathrm{d}y -\int_{\mathbb{R}^N}\frac{|v_n|^{2_s^*(t)-2}\left(\varphi(\frac{x}{R_n})v_n\right)^{2}}{|x|^t}\mathrm{d}x. \end{align*}

Therefore

(4.12)\begin{equation}\begin{aligned} &\frac{C_{N,s}}{2}\iint_{\mathbb{R}^{2N}}\frac{\left(v_n(R_nx)-v_n(R_ny)\right)\left(\varphi^2(x)v_n(R_nx)-\varphi^2(y)v_n(R_ny)\right)R_n^{N-2s}}{|x-y|^{N+2s}}\mathrm{d}x\,\mathrm{d}y\\ &\quad\qquad=\int_{\mathbb{R}^N} \frac{|v_n|^{2_s^*(t)-2}\left(\varphi(\frac{x}{R_n})v_n\right)^{2}}{|x|^t}\mathrm{d}x+{\scriptscriptstyle\mathcal O}(1). \end{aligned} \end{equation}

Now,

(4.13)\begin{align} \text{RHS of (4.12)} &=\int_{B_1}\frac{|v_n(R_nx)|^{2_s^*(t)-2}\left(\varphi(x)v_n(R_nx)\right)^2R_n^{N-t}}{|x|^t}\;\mathrm{d}x+{\scriptscriptstyle\mathcal O}(1) \cr &=\int_{B_1}\frac{\left|w_n(x)\right|^{2_s^*(t)-2}\left(\varphi(x)w_n(x)\right)^2}{|x|^t}\;\mathrm{d}x+{\scriptscriptstyle\mathcal O}(1) \cr &\leq\left(\int_{B_1}\frac{|w_n(x)|^{2_s^*(t)}}{|x|^t}\;\mathrm{d}x\right)^{\frac{2_s^*(t)-2}{2_s^*(t)}}\left(\int_{\mathbb{R}^N}\frac{|\varphi w_n|^{2_s^*(t)}}{|x|^t}\;\mathrm{d}x\right)^{\frac{2}{2_s^*(t)}}+{\scriptscriptstyle\mathcal O}(1)\cr &\leq\frac{1}{\mu_{s,t}} \left(\int_{B_1}\frac{|w_n|^{2_s^*(t)}}{|x|^t}\;\mathrm{d}x\right)^{\frac{2_s^*(t)-2}{2_s^*(t)}}\|\varphi w_n\|_{\dot{H}^s}^2+{\scriptscriptstyle\mathcal O}(1) \cr &\leq\frac{\delta^{\frac{2s-t}{N-t}}}{\mu_{s,t}}\|\varphi w_n\|_{\dot{H}^s}^2 +{\scriptscriptstyle\mathcal O}(1) \cr & \lt \frac12\|\varphi w_n\|_{\dot{H}^s}^2 +{\scriptscriptstyle\mathcal O}(1)\text{(By the choice of}\ \delta\ \text{in Step~4.)}. \end{align}

Claim 3. As $n\to\infty$

(4.14)\begin{equation}\text{LHS of (4.12)}=\Arrowvert\varphi w_n\Arrowvert_{\dot H^s}^2+{\scriptscriptstyle\mathcal O}(1).\end{equation}

Indeed,

\begin{align*} \text{LHS of (4.12)}&= \frac{C_{N,s}}{2}\iint_{\mathbb{R}^{2N}}\frac{\begin{array}{l}\left(v_n(R_nx)-v_n(R_ny)\right)(\varphi^2(x)v_n(R_nx)\\ \quad-\varphi^2(y)v_n(R_ny))R_n^{N-2s}\end{array}}{|x-y|^{N+2s}}\;\mathrm{d}x\,\mathrm{d}y\nonumber\\ &=\frac{C_{N,s}}{2}\iint_{\mathbb{R}^{2N}}\frac{\left(w_n(x)-w_n(y)\right)\left(\varphi^2(x)w_n(x)-\varphi^2(y)w_n(y)\right)}{|x-y|^{N+2s}}\;\mathrm{d}x\,\mathrm{d}y\nonumber\\ &=\frac{C_{N,s}}{2} \iint_{\mathbb{R}^{2N}}\frac{|\varphi(x)w_n(x)-\varphi(y)w_n(y)|^2}{|x-y|^{N+2s}}\mathrm{d}x\,\mathrm{d}y\nonumber\\ &\qquad\qquad-\frac{C_{N,s}}{2} \iint_{\mathbb{R}^{2N}}\frac{(\varphi(x)-\varphi(y))^{2}w_n(x)w_n(y)}{|x-y|^{N+2s}}\mathrm{d}x\,\mathrm{d}y\nonumber\\ &=\|\varphi w_n\|_{\dot{H}^s}^2-\frac{C_{N,s}}{2} \iint_{\mathbb{R}^{2N}}\frac{(\varphi(x)-\varphi(y))^{2}w_n(x)w_n(y)}{|x-y|^{N+2s}}\mathrm{d}x\,\mathrm{d}y.\nonumber \end{align*}

Now,

\begin{align*} &\iint_{\mathbb{R}^{2N}}\frac{(\varphi(x)-\varphi(y))^{2}w_n(x)w_n(y)}{|x-y|^{N+2s}}\mathrm{d}x\,\mathrm{d}y \\ &\quad= \int_{x\in B_1}\int_{y\in B_1}\quad +\int_{x\in B_1}\int_{y\in B_1^c}\quad +\int_{x\in B_1^c}\int_{y\in B_1}\\ &\quad=:\mathcal{I}_n^1 +\mathcal{I}_n^2 +\mathcal{I}_n^3.\nonumber \end{align*}

We have, $\mathcal{I}_n^2=\mathcal{I}_n^3$, as the integral is symmetric with respect to $x$ and $y$.

(4.15)\begin{eqnarray} \mathcal{I}_n^1 &=& \int_{x\in B_1}\int_{y\in B_1}\frac{(\varphi(x)-\varphi(y))^{2}w_n(x)w_n(y)}{|x-y|^{N+2s}}\;\mathrm{d}y\,\mathrm{d}x\nonumber\\ &\leq& C\int_{x\in B_1}\int_{y\in B_1}\frac{|w_n(x)||w_n(y)|}{|x-y|^{N+2s-2}}\;\mathrm{d}y\,\mathrm{d}x\nonumber\\ &\leq& C\left(\int_{x\in B_1}\int_{y\in B_1}\frac{|w_n(x)|^2}{|x-y|^{N+2s-2}}\;\mathrm{d}y\,\mathrm{d}x\right)^{\frac{1}{2}}\left(\int_{x\in B_1}\int_{y\in B_1}\frac{|w_n(y)|^2}{|x-y|^{N+2s-2}}\mathrm{d}y\,\mathrm{d}x\right)^{\frac{1}{2}}\nonumber\\ &\leq& C\int_{x\in B_1}\int_{y\in B_1}\frac{|w_n(x)|^2}{|x-y|^{N+2s-2}}\;\mathrm{d}y\,\mathrm{d}x\nonumber\\ &\leq& C\int_{x\in B_1}\left(\int_{|z| \lt 2}\frac{1}{|z|^{N+2s-2}}\;\mathrm{d}z\right)|w_n(x)|^2\mathrm{d}x\nonumber\\ &\leq& C\|w_n\|_{L^2(B_1)}^2 = {\scriptscriptstyle\mathcal O}(1)\text{(Since}\ w=0\ \text{implies}\ w_n\to 0\ \text{in}\ L^2_{\rm loc}(\mathbb{R}^N)). \end{eqnarray}

Furthermore,

(4.16)\begin{eqnarray} \mathcal{I}_n^2&=&\int_{x\in B_1}\int_{y\in B_1^c}\frac{(\varphi(x)-\varphi(y))^{2}w_n(x)w_n(y)}{|x-y|^{N+2s}}\mathrm{d}y\,\mathrm{d}x\cr &\leq& \int_{x\in B_1}\int_{y\in B_1^c\cap \{|x-y|\leq 1\}}+ \int_{x\in B_1}\int_{y\in B_1^c\cap \{|x-y|\geq 1\}}\cr &=:& \mathcal{I}_n^{21}+\mathcal{I}_n^{22}. \end{eqnarray}

where

\begin{align*} \mathcal{I}_n^{21}&\leq C\left(\int_{x\in B_1}\int_{y\in B_1^c\cap \{|x-y|\leq 1\}}\frac{|w_n(x)|^2}{|x-y|^{N+2s-2}}\;\mathrm{d}y\,\mathrm{d}x\right)^{\frac{1}{2}}\\ &\qquad\times\left( \int_{x\in B_1}\int_{y\in B_1^c\cap \{|x-y|\leq1\}}\frac{|w_n(y)|^2}{|x-y|^{N+2s-2}}\;\mathrm{d}y\,\mathrm{d}x\right)^{\frac{1}{2}}\\ &=: CJ_n^{1}\cdot J_n^{2}.\end{align*}

Now,

\begin{equation*}\vert J_n^1\vert^2\leq\int_{x\in B_1}{(\int_{\vert z\vert \lt 1}\frac1{\vert z\vert^{N+2s-2}}\mathrm dz)}\vert w_n(x)\vert^2\mathrm dx\leq C\parallel w_n\parallel_{L^2(B_1)}^2={\scriptscriptstyle\mathcal O}(1),\end{equation*}

and

\begin{align*} |J_n^2|^2&= \int_{x\in B_1}\int_{y\in B_1^c}\frac{\mathbf{1}_{\{|x-y| \lt 1\}}(x,y)|w_n(y)|^2}{|x-y|^{N+2s-2}}\;\mathrm{d}y\,\mathrm{d}x\\ &\leq\int_{y\in B_1^c}\left(\int_{x\in B_1}\frac{\mathbf{1}_{\{|x-y| \lt 1\}}(x,y)}{|x-y|^{N+2s-2}}\;\mathrm{d}x\right)|w_n(y)|^2\;\mathrm{d}y\\ &\leq \int_{y\in B_2}\left(\int_{x\in B_1}\frac{\mathbf{1}_{\{|x-y| \lt 1\}}(x,y)}{|x-y|^{N+2s-2}}\;\mathrm{d}x\right)|w_n(y)|^2\;\mathrm{d}y\\ &\leq C\|w_n\|_{L^2(B_2)}^2\leq C'. \end{align*}

Therefore, $\mathcal I_n^{21}={\scriptscriptstyle\mathcal O}(1)$ as $n\to\infty$.

Moreover,

\begin{align*} \mathcal{I}_n^{22}&=\int_{x\in B_1}\int_{y\in B_1^c\cap \{|x-y|\geq 1\}} \frac{|w_n(x)||w_n(y)||\varphi (x)-\varphi (y)|^2}{|x-y|^{N+2s}}\;\mathrm{d}y\,\mathrm{d}x\\ &\leq C\int_{x\in B_1}\int_{y\in B_1^c\cap \{|x-y|\geq 1\}} \frac{|w_n(x)||w_n(y)|}{|x-y|^{N+2s}}\;\mathrm{d}y\,\mathrm{d}x\\ &\leq C\left(\int_{x\in B_1}\int_{y\in B_1^c\cap \{|x-y|\geq 1\}} \frac{|w_n(x)|^2}{|x-y|^{N+2s}}\;\mathrm{d}y\,\mathrm{d}x\right)^{\frac{1}{2}}\\ &\qquad\qquad\times\left(\int_{x\in B_1}\int_{y\in B_1^c\cap \{|x-y|\geq 1\}} \frac{|w_n(y)|^2}{|x-y|^{N+2s}}\;\mathrm{d}y\,\mathrm{d}x\right)^{\frac{1}{2}}\\ &\leq C\left(\int_{x\in B_1}\left(\int_{|z|\geq 1}\frac{1}{|z|^{N+2s}}\;\mathrm{d}z\right)|w_n(x)|^2\mathrm{d}x\right)^{\frac{1}{2}}\\ &\qquad\qquad\times\left(\int_{x\in B_1}\int_{|z|\geq 1} \frac{|w_n(x+z)|^2}{|x+z|^{2s}}\frac{|x+z|^{2s}}{|z|^{N+2s}}\;\mathrm{d}z\,\mathrm{d}x\right)^{\frac{1}{2}}\\ &\leq C' \|w_n\|_{L^2\left(B_1\right)}\left(\int_{x\in B_1}\left(\int_{\mathbb{R}^N} \frac{|w_n(x+z)|^2}{|x+z|^{2s}}\mathrm{d}z\right)\mathrm{d}x\right)^{\frac{1}{2}} \end{align*}

since $|z|\geq 1$ and $|x| \lt 1$ implies $\frac{|x+z|^{2s}}{|z|^{N+2s}}\leq C$. Therefore, using the Hardy inequality (2.4), we obtain from the last of the above estimates that as $n\to\infty$

\begin{equation*}\mathcal I_n^{22}\leq C^{''}\Arrowvert w_n\Arrowvert_{L^2(B_1)}\Arrowvert w_n\Arrowvert_{\dot H^s}={\scriptscriptstyle\mathcal O}(1).\end{equation*}

Combining the above estimates together, we obtain from (4.16) that $\mathcal I_n^2={\scriptscriptstyle\mathcal O}(1)$ as $n\to\infty$. This, along with (4.15), conclude the proof of Claim 3.

Now Claim 3 along with (4.13) yield

(4.17)\begin{equation}\Arrowvert\varphi w_n\Arrowvert_{\dot H^s}={\scriptscriptstyle\mathcal O}(1)\,\text{as }n\rightarrow\infty.\end{equation}

Substituting this into (4.14) and comparing with (4.12), one observes as $n\to\infty$

\begin{equation*}\int_{\mathbb{R}^N}\frac{\varphi^2(x)\vert w_n(x)\vert^{2_s^\ast(t)}}{\vert x\vert^t}\;\mathrm dx={\scriptscriptstyle\mathcal O}(1).\end{equation*}

Therefore,

(4.18)\begin{equation}\int_{B_r}\frac{\vert w_n\vert^{2_s^\ast(t)}}{\vert x\vert^t}\mathrm dx={\scriptscriptstyle\mathcal O}(1),\ \text{for any}\ 0 \lt r \lt 1.\end{equation}

Now, we fix $0 \lt r \lt 1$. Using (4.9), Rellich–Kondrachov compactness theorem and the fact that $2 \lt 2_s^*(t) \lt 2_s^*$ for $t \gt 0$, we observe,

\begin{eqnarray*} 0 \lt \delta &=& \int_{B_1}\frac{|w_n|^{2_s^*(t)}}{|x|^t}\,\mathrm{d}x\nonumber\\ &=& \int_{B_{r}} \frac{|w_n|^{2_s^*(t)}}{|x|^t}\,\mathrm{d}x + \int_{B_1\setminus B_{r}} \frac{|w_n|^{2_s^*(t)}}{|x|^t}\,\mathrm{d}x \nonumber\\ &\leq& {\scriptscriptstyle\mathcal O}(1) + \frac{1}{r^t} \int_{B_1\setminus B_{r}} |w_n|^{2_s^*(t)}\,\mathrm{d}x = {\scriptscriptstyle\mathcal O}(1),\nonumber \end{eqnarray*}

which is a contradiction. Therefore, $w=0$ cannot happen in our case, i.e., when $t \gt 0$.

Step 7. We start with a $(PS)$ sequence $\left(v_n\right)_{n\ge1}$ for $I_{s,t}$ at level $\beta-I_{s,t}(u_0)$ such that $v_n\rightharpoonup 0$ in $\dot{H}^s\left(\mathbb{R}^N\right)$. Then we dilate $v_n$ to get $w_n= D_{R_n}v_n$ with a nonzero weak limit $w$. Again we show that, $z_n:= v_n- D_{\frac{1}{R_n}}w$ is a $(PS)$ sequence for $I_{s,t}$ at a strict lower level $\beta-I_{s,t}(u_0)-I_{s,t}(w)$. Now we set, $w_1:= w,\,R_n^1:= R_n$, and repeat the process as in steps 4, 5 and 6, we will have another $R_n^2 \in\mathbb{R}^+$ such that $D_{R_n^2}z_n \rightharpoonup w_2\left(\neq 0\right)$ in $\dot{H}^s\left(\mathbb{R}^N\right)$ and this will lead to another $(PS)$ sequence $\tilde{z}_n:= v_n- D_{\frac{1}{R_n^1}}w_1-D_{\frac{1}{R_n^2}}w_2$ for $I_{s,t}$ at the level $\beta-I_{s,t}(u_0)-I_{s,t}(w_1)-I_{s,t}(w_2)$ which is strictly lower than the previous level since $w_2\in\dot{H}^s\left(\mathbb{R}^N\right)$ is a nontrivial solution of (4.2). Since $\sup_n \|v_n\|_{\dot{H}^s}\leq C \lt \infty$, the process ought to stop after finitely many steps, and this concludes $(i),(ii)$ and $(iv)$ of the proposition.

Finally, in the same spirit of [Reference Tintarev29, Page 129, theorem 3.2] we obtain,

\begin{equation*} \left\langle D_{\frac{1}{R_n^1}}w_1, D_{\frac{1}{R_n^2}}w_2\right\rangle_{\dot{H}^s} \to 0, \text{as }n\to\infty. \end{equation*}

and thanks to lemma 2.1, the desired conclusion $(iii)$ of the proposition is achieved. This completes the proof.

Proof of theorem 1.4

Follows from proposition 4.2 combined with theorem 1.2.

Proof. First we show the optimality of (1.10). Suppose $\exists\,\epsilon_0 \gt 0$ such that $\Gamma(u)\geq C \|\rho\|_{\dot{H}^s}^{1-\epsilon_0}$ for all non-negative $u\in\dot{H}^s\left(\mathbb{R}^N\right)$ satisfying (1.9). Let $\epsilon \gt 0$ be small and set $u=V_{s,t}$. Consider $u_\epsilon=u+\epsilon\phi$ for some fixed $\phi\in{{\mathcal{C}}_c^\infty} (\mathbb{R}^N)$ such that $\phi\geq0$. Clearly $u_\epsilon\ge0$. Then thanks to theorem 1.2, we have

\begin{align*} u_\epsilon^{2^*_s(t)-1}=\left(u+\epsilon\phi\right)^{2^*_s(t)-1}&=u^{2^*_s(t)-1}\left(1+\mathcal{O}(\epsilon)\right)\nonumber \end{align*}

and consequently $u_\epsilon$ satisfies (1.9) for small $\epsilon$. Therefore,

\begin{align*} \Gamma(u_\epsilon)=\left\|\left(-\Delta\right)^{s}u_\epsilon-\frac{u_\epsilon^{2^*_s(t)-1}}{|x|^t}\right\|_{\left(\dot{H}^s\left(\mathbb{R}^N\right)\right)'}&\leq\left\|\left(-\Delta\right)^{s}u-\frac{u^{2^*_s(t)-1}}{|x|^t}\right\|_{\left(\dot{H}^s\left(\mathbb{R}^N\right)\right)'}+\mathcal{O}(\epsilon) \end{align*}

since $u=V_{s,t}$ solves (4.2) weakly, one obtains

\begin{align*} C\|\epsilon\phi\|_{\dot{H}^s}^{1-\epsilon_0} \leq\Gamma(u_\epsilon)&\leq\mathcal{O}(\epsilon)\leq C'\|\epsilon\phi\|_{\dot{H}^s} \end{align*}

this implies $\forall\epsilon \gt 0$ small,

(5.1)\begin{align} \|\epsilon\phi\|_{\dot{H}^s}^{\epsilon_0}&\geq C^{\prime\prime} \gt 0 \end{align}

then the desired contradiction follows as we let $\epsilon\to0^+$ in (5.1).

We begin with the proof of the theorem by noticing that for any $\lambda \gt 0$,

\begin{align*}\|V^\lambda_{s,t}\|^2_{\dot{H}^s}=\|V_{s,t}\|^2_{\dot{H}^s}=\,\,&\mu_{s,t}^{\frac{2}{2^*_s(t)-2}}\|U_{s,t}\|^2_{\dot{H}^s}=\mu_{s,t}^{\frac{2^*_s(t)}{2^*_s(t)-2}}=\mu_{s,t}^{\frac{N-t}{2s-t}},\\ \|V^\lambda_{s,t}\|^2_{L^{2_s^*(t)}\left(\mathbb{R}^N,|x|^{-t}\right)}=&\mu_{s,t}^{\frac{2}{2^*_s(t)-2}}\|U_{s,t}\|^2_{L^{2_s^*(t)}\left(\mathbb{R}^N,|x|^{-t}\right)}=\mu_{s,t}^{\frac{2}{2^*_s(t)-2}},\\ \end{align*}

and

\begin{align*} \frac{1}{2}\mu_{s,t}^{\frac{N-t}{2s-t}} \lt \,\,&\left\|V^{\lambda}_{s,t}\right\|^2_{\dot{H}^s} \lt \frac{3}{2}\mu_{s,t}^{\frac{N-t}{2s-t}}.\end{align*}

One can then see from the profile decomposition result in proposition 4.2, that, any non-negative $u\in\dot{H}^s\left(\mathbb{R}^N\right)$ satisfying the hypothesis (1.9) and having $\Gamma(u)$ small enough, ought to be close to a single bubble $V^{\tilde{\lambda}}_{s,t}$ for some $\tilde{\lambda}\in\mathbb{R}^+$.

Qualitatively, this means that, given $\epsilon \gt 0$ very small there exists a $\delta \gt 0$ (depending upon $\epsilon$) such that $\|u-V^{\tilde{\lambda}}_{s,t}\|_{\dot{H}^s} \lt \frac\epsilon3$ whenever $\Gamma(u) \lt \delta$.

The following lemmata show that $\mathcal{M} \subset \dot{H}^s(\mathbb{R}^N)$ is a two-dimensional submanifold of $\dot{H}^s(\mathbb{R}^N)$, meaning that the mapping $(c, \lambda) \mapsto c V^\lambda_{s,t}$ is a homeomorphism.

Lemma 5.1. There exists $\eta \gt 0$ (depending upon $\epsilon$) such that for all $(c,\lambda)\in(1-\eta,1+\eta)\times(\tilde{\lambda}-\eta,\tilde{\lambda}+\eta)$, the following estimate holds.

\begin{equation*}\|u-cV^{\lambda}_{s,t}\|_{\dot{H}^s} \lt \epsilon.\end{equation*}

Proof. Triangle inequality gives

\begin{align*} \|u-cV^{\lambda}_{s,t}\|_{\dot{H}^s}&\le\|u-V^{\tilde{\lambda}}_{s,t}\|_{\dot{H}^s}+\|V^{\tilde{\lambda}}_{s,t}-V^{\lambda}_{s,t}\|_{\dot{H}^s}+|c-1|\|V^{\lambda}_{s,t}\|_{\dot{H}^s}\\ &\le\frac{\epsilon}{3}+\eta\mu_{s,t}^{\frac{N-t}{2(2s-t)}}+\|V^{\tilde{\lambda}}_{s,t}-V^{\lambda}_{s,t}\|_{\dot{H}^s}. \end{align*}

Now a straight forward application of lemma 2.1, shows

\begin{equation*}\left\|V^{\tilde{\lambda}}_{s,t}-V^{\lambda}_{s,t}\right\|_{\dot{H}^s}^2=\left\langle V^{\tilde{\lambda}}_{s,t}-V^{\lambda}_{s,t}, V^{\tilde{\lambda}}_{s,t}-V^{\lambda}_{s,t}\right\rangle_{\dot{H}^s}=2\left\|V_{s,t}\right\|_{\dot{H}^s}^2-2\left\langle V^{\lambda}_{s,t},V^{\tilde{\lambda}}_{s,t}\right\rangle_{\dot{H}^s}\to0\end{equation*}

as $\lambda\to\tilde{\lambda}$.

Choosing $0 \lt \eta \lt \epsilon$ small enough such that $\|V^{\tilde{\lambda}}_{s,t}-V^{\lambda}_{s,t}\|_{\dot{H}^s} \lt \frac\epsilon3$ and $\eta \lt \frac{\epsilon}{3}\mu_{s,t}^{\frac{t-N}{2(2s-t)}}$, we obtain the desired estimate.

Proof. Taking $u=c'V^{\lambda'}_{s,t}$ in lemma 5.1, the first statement follows.

For the converse, note that by the inequality (1.1), we have up to a subsequence $c_kV^{\lambda_{k}}_{s,t}\to c'V^{\lambda'}_{s,t}$ point-wise almost everywhere in $\mathbb{R}^N$. Moreover, $c_k$ is a bounded sequence strictly away from zero and $\sup_{k\geq1}|\log\lambda_k| \lt \infty$.

To see this, notice $\sup_{k\geq1}|c_k|\mu_{s,t}^{\frac{N-t}{2(2s-t)}}=\sup_{k\geq1}\|c_kV_{s,t}^{\lambda_k}\|_{\dot{H}^s} \lt \infty$. Thus, $c_k$ is bounded. Suppose $c_k\to0$ then one sees $|c'|\mu_{s,t}^{\frac{N-t}{2(2s-t)}}=\|c'V_{s,t}^{\lambda'}\|_{\dot{H}^s}\leq\|c'V_{s,t}^{\lambda'}-c_kV_{s,t}^{\lambda_k}\|_{\dot{H}^s}+|c_k|\mu_{s,t}^{\frac{N-t}{2(2s-t)}}\to0$ as $k\to\infty$, implying $c'=0$ which is a contradiction. Next, notice that the RHS of the following quantity

\begin{equation*}\left\langle V_{s,t}^{\lambda_k},V_{s,t}^{\lambda'}\right\rangle_{\dot{H}^s}=\frac{(|c'|^2+|c_k|^2)\mu_{s,t}^{\frac{N-t}{2s-t}}-\|c'V_{s,t}^{\lambda'}-c_kV_{s,t}^{\lambda_k}\|_{\dot{H}^s}^2}{2c'c_k}\end{equation*}

(up to a subsequence) goes to a finite nonzero value as $k\to\infty$ thus by lemma 2.1, the desired conclusion follows.

Therefore, up to a subsequence $(c_k,\lambda_{k})\to(c^{\prime\prime},\lambda^{\prime\prime})\in\mathbb{R}\setminus\{0\}\times\mathbb{R}^+$ as $k\to\infty$ then by the above, we shall again have up to a further subsequence (if necessary) $c_kV^{\lambda_{k}}_{s,t}\to c^{\prime\prime}V^{\lambda^{\prime\prime}}_{s,t}$ point-wise almost everywhere in $\mathbb{R}^N$.

Now invoking theorem 1.2, we obtain $c'V^{\lambda'}_{s,t}(x)=c^{\prime\prime}V^{\lambda^{\prime\prime}}_{s,t}(x)$ for all $x\in\mathbb{R}^N$. In particular, taking $x=0$, we have

\begin{equation*}c'\lambda'^{\frac{N-2s}{2}}=c^{\prime\prime}\lambda^{\prime\prime}{}^{\frac{N-2s}{2}}.\end{equation*}

Next, we see

\begin{equation*}\frac{c'\lambda^{\prime\prime}{}^{\frac{N-2s}{2}}}{c^{\prime\prime}\lambda'^{\frac{N-2s}{2}}}=\frac{\lambda^{\prime\prime}{}^{\frac{N-2s}{2}}V^{\lambda^{\prime\prime}}_{s,t}(x)}{\lambda'^{\frac{N-2s}{2}}V^{\lambda'}_{s,t}(x)}=\frac{|\lambda^{\prime\prime}x|^{N-2s}V_{s,t}(\lambda^{\prime\prime}x)}{|\lambda'x|^{N-2s}V_{s,t}(\lambda'x)}\to 1\end{equation*}

as $|x|\to\infty$.

Now these two relations yield at once $c'=c^{\prime\prime}$ and $\lambda'=\lambda^{\prime\prime}$.

Define $\rho(c,\lambda):= u-cV^\lambda_{s,t}$. Now from the definition of the distance function $d(\cdot,\mathcal{M})$ and the claim (3.2) in lemma 3.1, it follows that there exist $c_0\neq0\text{and }\lambda_0\in\mathbb{R}^+$ such that

\begin{equation*} \langle\rho_0,\rho_0\rangle_{\dot{H}^s}=\|\rho_0\|^2_{\dot{H}^s}=\min\limits_{c\in\mathbb{R},\lambda\in\mathbb{R}^+}\|u-cV^\lambda_{s,t}\|^2_{\dot{H}^s}=\min\limits_{c\in\mathbb{R},\lambda\in\mathbb{R}^+}\langle\rho(c,\lambda),\rho(c,\lambda)\rangle_{\dot{H}^s} \end{equation*}

where $\rho_0:=\rho(c_0,\lambda_0)=u-c_0V_{s,t}^{\lambda_0}$. Since the minimum is attained here at $(c,\lambda)=(c_0,\lambda_0)$, we must have

\begin{equation*}\nabla_{c,\lambda}\left\langle\rho(c,\lambda),\rho(c,\lambda)\right\rangle_{\dot{H}^s}\big\vert_{c=c_0,\lambda=\lambda_0}=(0,0).\end{equation*}

Combining this with theorems 1.1 and 1.2, give us the following orthogonality relations:

(5.2)\begin{align} \left\langle\rho_0,V^{\lambda_0}_{s,t}\right\rangle_{\dot{H}^s}=0&\iff\int_{\mathbb{R}^N}\frac{\rho_0\left(V^{\lambda_0}_{s,t}\right)^{2^*_s(t)-1}}{|x|^t}\,{\mathrm{d}}x=0, \end{align}

(5.3)\begin{align} \left\langle\rho_0,\partial_\lambda V^{\lambda}_{s,t}\big\vert_{\lambda=\lambda_0}\right\rangle_{\dot{H}^s}=0&\iff \int_{\mathbb{R}^N}\frac{\rho_0\left(\partial_\lambda V^{\lambda}_{s,t}\big\vert_{\lambda=\lambda_0}\right)\left(V^{\lambda_0}_{s,t}\right)^{2^*_s(t)-2}}{|x|^t}\,{\mathrm{d}}x=0. \end{align}

Now thanks to lemma 5.1 and lemma 5.2, without loss of generality, the parameters $c_0\text{ and }\lambda_0$ (depending upon $\epsilon$) can be chosen in such a way that $\|\rho_0\|_{\dot{H}^s}=d\left(u,\mathcal{M}\right)\leq\epsilon,\,|c_0-1| \lt \eta,\,|\lambda_0-\tilde{\lambda}| \lt \eta$, where $0 \lt \eta \lt \epsilon$ is fixed as in lemma 5.1.

Define $\rho:=\rho(1,\lambda_0)= u-V^{\lambda_0}_{s,t}$. If $\Gamma(u)\ge\delta$ what easily follows is that

\begin{equation*}\|\rho\|_{\dot{H}^s}=\left\|u-V^{\lambda_0}_{s,t}\right\|_{\dot{H}^s}\leq\left(1+\sqrt{\frac{3}{2}}\right)\mu_{s,t}^{\frac{N-t}{2(2s-t)}}\leq\frac{C(N,s,t)}{\delta}\,\Gamma(u)\end{equation*}

which is (1.10) and the theorem holds true in this case.

Now, to achieve our goal, the strategy is to test $(-\Delta)^su-\frac{u^{2^*_s(t)-1}}{|x|^t}$ against $\rho_0$ and $u$. Exploiting the orthogonality relation (5.2) yields

(5.4)\begin{align} \langle\rho_0,\rho_0\rangle_{\dot{H}^s}&=\left\langle u-c_0V^{\lambda_0}_{s,t},\rho_0\right\rangle_{\dot{H}^s}\cr &=\langle u,\rho_0\rangle_{\dot{H}^s}\cr &=\int_{\mathbb{R}^N}(-\Delta)^{\frac s2}u\,\,(-\Delta)^{\frac s2}\rho_0\,\mathrm{d}x\cr &=\int_{\mathbb{R}^N}\frac{\rho_0 u^{2^*_s(t)-1}}{|x|^t}\,\mathrm{d}x+\int_{\mathbb{R}^N}\left((-\Delta)^su-\frac{u^{2^*_s(t)-1}}{|x|^t}\right)\rho_0\,\mathrm{d}x\cr &\leq\int_{\mathbb{R}^N}\frac{\rho_0 u^{2^*_s(t)-1}}{|x|^t}\,\mathrm{d}x+\Gamma(u)\|\rho_0\|_{\dot{H}^s}. \end{align}

To control the integral in (5.4), we proceed with the following elementary inequality.

(5.5)\begin{align} &\forall a,b\in\mathbb{R};\quad \left||a+b|^p-|a|^p\right|\cr &\quad\leq\begin{cases} p|a|^{p-1}|b|+C(p)|b|^p &\quad\text{for }\,1\leq p\leq2.\\ p|a|^{p-1}|b|+C(p)\left(|a|^{p-2}|b|^2+|b|^p\right)&\quad\text{for }\,p \gt 2. \end{cases} \end{align}

Since $|c_0-1| \lt \epsilon$ and $\epsilon \gt 0$ being very small, we have $c_0 \gt 1-\epsilon \gt 0$. Now taking $p+1=2^*_s(t)=\frac{2(N-t)}{N-2s},\,a=c_0V^{\lambda_0}_{s,t},\,b=\rho_0$, one observes from the hypothesis on $u$, that $a \gt 0,\,a+b=u\geq0$ and then, thanks to (5.5)

\begin{align*} &\left|\int_{\mathbb{R}^N}\frac{\rho_0 u^{2^*_s(t)-1}}{|x|^t}\,\mathrm{d}x\right|\\ &=\left|\int_{\mathbb{R}^N}\frac{b (a+b)^{p}}{|x|^t}\,\mathrm{d}x -\int_{\mathbb{R}^N}\frac{b a^{p}}{|x|^t}\,\mathrm{d}x\right|\\ &\leq\int_{\mathbb{R}^N}\frac{|b|\left|(a+b)^{p}-a^p\right|}{|x|^t}\,\mathrm{d}x\\ &\leq p\int_{\mathbb{R}^N}\frac{|b|^2a^{p-1}}{|x|^t}\,\mathrm{d}x+C(p)\begin{cases} \int_{\mathbb{R}^N}\frac{|b|^{p+1}}{|x|^t}\,\mathrm{d}x&\quad\text{for }1\leq p\leq2.\\ \int_{\mathbb{R}^N}\frac{|b|^3a^{p-2}}{|x|^t}\,\mathrm{d}x+\int_{\mathbb{R}^N}\frac{|b|^{p+1}}{|x|^t}\,\mathrm{d}x&\quad\text{for }p \gt 2. \end{cases} \end{align*}

Notice that, thanks to (5.2) we have $\int_{\mathbb{R}^N}\frac{b a^{p}}{|x|^t}\,\mathrm{d}x=c_0^p\int_{\mathbb{R}^N}\frac{\rho_0 (V^{\lambda_0}_{s,t})^{2^*_s(t)-1}}{|x|^t}\,\mathrm{d}x=0$ and thus the integral above vanishes. We estimate the rest of the integrals by using a combination of the Rayleigh quotient characterization (3.5), the Hardy–Sobolev inequality (1.1), and the Hölder inequality as follows:

(5.6)\begin{align} &p\int_{\mathbb{R}^N}\frac{|b|^2a^{p-1}}{|x|^t}\,\mathrm{d}x=p\int_{\mathbb{R}^N}\frac{|\rho_0|^2\left(c_0V^{\lambda_0}_{s,t}\right)^{2^*_s(t)-2}}{|x|^t}\,\mathrm{d}x\leq c_0^{p-1}\frac{2^*_s(t)-1}{\eta_3}\|\rho_0\|^2_{\dot{H}^s}. \end{align}

(5.7)\begin{align} &\int_{\mathbb{R}^N}\frac{|b|^{p+1}}{|x|^t}\,\mathrm{d}x =\int_{\mathbb{R}^N}\frac{|\rho_0|^{2^*_s(t)}}{|x|^t}\,\mathrm{d}x\leq \mu_{s,t}^{\frac{-2^*_s(t)}{2}}\|\rho_0\|_{\dot{H}^s}^{2^*_s(t)}\cr &\quad\text{for }N\geq6s-2t\,\,(\text{i.e., }1\leq p\leq2). \end{align}

(5.8)\begin{align} &\int_{\mathbb{R}^N}\frac{|b|^3a^{p-2}}{|x|^t}\,\mathrm{d}x\cr &\ =\int_{\mathbb{R}^N}\frac{|\rho_0|^3\left(c_0V^{\lambda_0}_{s,t}\right)^{2^*_s(t)-3}}{|x|^t}\,\mathrm{d}x\cr &\leq\||x|^{\frac{-t}{2^*_s(t)}}\rho_0\|^3_{2^*_{s}(t)}\||x|^{\frac{-t}{2^*_s(t)}}c_0V^{\lambda_0}_{s,t}\|^{{2^*_s(t)}-3}_{2^*_s(t)}\cr &\leq c_0^{p-2}\mu_{s,t}^{\frac{-3}2+\frac{2^*_s(t)(2^*_s(t)-3)}{2^*_s(t)-2}}\|\rho_0\|^3_{\dot{H}^s}\quad\text{for }2s \lt N \lt 6s-2t\,\,(\text{i.e., }p \gt 2). \end{align}

Now (5.4), (5.6)–(5.8), $\|\rho_0\|_{\dot{H}^s} \lt \epsilon,\,|c_0-1| \lt \epsilon$ and the spectral properties of the operator $\mu_{s,t}{\mathcal{L}}^{\lambda_0}_{s,t}$ together imply that

(5.9)\begin{align} &\forall N \gt 2s;\quad\left(1-\frac{c_0^{p-1}\eta_2}{\eta_3}\right)\|\rho_0\|_{\dot{H}^s}\cr &\quad\leq\Gamma(u)+C(N,s,t)\left(c_0^{p-2}\|\rho_0\|_{\dot{H}^s}^2+\|\rho_0\|_{\dot{H}^s}^{2^*_s(t)-1}\right)\cr &\quad\leq\Gamma(u)+\left((1+\epsilon)^{p-2}\epsilon+\epsilon^{2^*_s(t)-2}\right)C(N,s,t)\|\rho_0\|_{\dot{H}^s}\cr &\quad\leq \Gamma(u)+\epsilon^{\gamma}C(N,s,t)\|\rho_0\|_{\dot{H}^s};\left(\gamma=\min\{1,2^*_s(t)-2\}\right). \end{align}

Choose $\epsilon \gt 0$ small enough such that

\begin{equation*}1-\frac{c_0^{p-1}\eta_2}{\eta_3} \gt 1-\frac{\eta_2}{\eta_3}-\mathcal O(\epsilon) \gt 0\text{ and }1-\frac{c_0^{p-1}\eta_2}{\eta_3}-\epsilon^\gamma C(N,s,t) \gt 0.\end{equation*}

For this choice, (5.9) at once gives that

(5.10)\begin{equation}\|\rho_0\|_{\dot{H}^s}\leq C(N,s,t)\,\Gamma(u).\end{equation}

Testing $(-\Delta)^su-\frac{u^{2^*_s(t)-1}}{|x|^t}$ against $u$, we obtain,

(5.11)\begin{align} \|u\|_{\dot{H}^s}^2&=\int_{\mathbb{R}^N}\frac{u^{2^*_s(t)}}{|x|^t}\,\mathrm{d}x+\int_{\mathbb{R}^N}\left((-\Delta)^su-\frac{u^{2^*_s(t)-1}}{|x|^t}\right)u\,\mathrm{d}x. \end{align}

From (5.11), we get by using $\left\langle\rho_0,V^{\lambda_0}_{s,t}\right\rangle_{\dot{H}^s}=0$, $\left\|V^{\lambda_0}_{s,t}\right\|_{L^{2_s^*(t)}\left(\mathbb{R}^N,|x|^{-t}\right)}=\mu_{s,t}^{\frac{1}{2^*_s(t)-2}}$, $|c_0-1| \lt \epsilon$, triangle inequality, (5.10) and (1.1), that

(5.12)\begin{align} \Gamma(u)\|u\|_{\dot{H}^s}&\geq\left|\int_{\mathbb{R}^N}\left((-\Delta)^su-\frac{u^{2^*_s(t)-1}}{|x|^t}\right)u\,\mathrm{d}x\right|\cr &=\left|\|u\|_{\dot{H}^s}^2-\int_{\mathbb{R}^N}\frac{u^{2^*_s(t)}}{|x|^t}\,\mathrm{d}x\right|\cr &=\left|c_0^2\,\mu_{s,t}^\frac{2^*_s(t)}{2^*_s(t)-2}+\|\rho_0\|^2_{\dot{H}^s}-\left\|\left({\rho_0+c_0V^{\lambda_0}_{s,t}}\right){|x|^{\frac{-t}{2^*_s(t)}}}\right\|^{2^*_s(t)}_{2^*_s(t)}\right|\cr &\geq\left|\mu_{s,t}^{\frac{2^*_s(t)}{2^*_s(t)-2}}\left| c_0^2-c_0^{2^*_s(t)}\right|-\left|\|\rho_0\|^2_{\dot{H}^s}\mp\mathcal{O}\left(\Gamma(u)^{ 2^*_s(t)}\right)\right|\right| \end{align}

thus,

\begin{align*} \left|c_0^2-c_0^{2^*_s(t)}\right|&=\mathcal{O}\left(\Gamma(u)\right)\implies|c_0-1|=\mathcal{O}\left(\Gamma(u)\right). \end{align*}

Therefore, by virtue of (5.10) and (5.12), one observes

\begin{equation*}\|\rho\|_{\dot{H}^s}=\|u-c_0V^{\lambda_0}_{s,t}+(c_0-1)V^{\lambda_0}_{s,t}\|_{\dot{H}^s}\le\|\rho_0\|_{\dot{H}^s}+|c_0-1|\,\mu_{s,t}^{\frac{N-t}{2(2s-t)}}\leq C(N,s,t)\,\Gamma(u).\end{equation*}