Proof. The Hölder continuity will follow from corollary 4.23 of [Reference Han and Lin10] if we show that the weak solution is bounded as $\frac{1}{|x|^s} \in L^{q}(B(0,R))$ for some $q \gt \frac{n}{2}$. As observed before, enough to prove the boundedness in a neighbourhood of 0 which will follow from theorem 8 of [Reference Ghoussoub and Robert7] (by applying γ = 0) or following the Moser iteration scheme as done in lemmas 3.1 and 3.2 of [Reference Mancini, Fabbri and Sandeep11] ( $|y| $ replaced by $|x|$).$\square$

Proof. Without loss of generality we may assume $\Omega = B := \{x : |x| \lt R\}$ and define for $0\le \lambda \lt R$, $B_\lambda = \{x=(x_1,\ldots ,x_n)\in B : \lambda \lt x_1 \lt R\}$ and $T_\lambda = \{x\in B : x_1=\lambda\}$. For $x\in B_\lambda$ denote its reflection with respect to T_λ by $x^\lambda := (2\lambda-x_1,x_2,\ldots, x_n)$ and define $u_\lambda(x) = u(x^\lambda)$ for $x\in B_\lambda$. First we show that $u_\lambda(x) \ge u(x)$ for all $x\in B_\lambda$ when λ is close to R.

Define $w_\lambda = u-u_\lambda$ then using the equation u satisfies and the fact that $|x^\lambda| \lt |x|$ for $x\in B_\lambda$ we see that w_λ satisfies

\begin{align*} -\Delta w_\lambda \le A(x)\frac{w_\lambda}{|x|^s} +B(x)w_\lambda \, {\rm in} \, B_\lambda,\, w_\lambda \in H^1(B_\lambda), \, w_\lambda \le 0 \, {\rm on}\, \partial B_\lambda \end{align*}

where $A(x):=0$ when $u(x)=u_\lambda(x)$, otherwise

\begin{align*} 0\le A(x) := \frac{u^{2^\ast(s)-1}(x)- u_\lambda^{2^\ast(s)-1}(x)}{u(x) -u_\lambda(x)} \le (2^\ast(s)-1)\left[ \max{u(x),u_\lambda(x)}\right]^ {2^\ast(s)-2}\end{align*}

and $|B(x)| \le L$ where L is the Lipschitz constant of ρ.

Multiplying the inequality by $w_\lambda^+$ and integrating by parts, we get

\begin{align*}\int_{B_\lambda}|\nabla w_\lambda^+ |^2 \le \int_{B_\lambda} A(x)\frac{(w_\lambda^+)^2}{|x|^s} + \int_{B_\lambda}B(x)(w_\lambda^+)^2.\end{align*}

Applying the Cauchy–Schwartz inequality, the Hardy–Sobolev inequality (1.4), and the standard Sobolev inequality, the above inequality becomes

\begin{align*}\int_{B_\lambda}|\nabla w_\lambda^+ |^2 \le C\left[ \left( \int_{B_\lambda\cap \{u \gt u_\lambda \}} \frac{A(x)^{\frac{n-s}{2-s}}}{|x|^s}\right)^{\frac{2-s}{n-s}} + \left( \int_{B_\lambda}|B(x)|^{\frac n2}\right)^{\frac 2n} \right] \int_{B_\lambda}|\nabla w_\lambda^+ |^2. \end{align*}

Now using the bounds $|B|\le L$ and $|A(x)| \le (2^\ast(s)-1)u(x)^ {2^\ast(s)-2}$ in the region of integration, we see that the term in square bracket on the RHS goes to zero as $\lambda \rightarrow R.$ Thus, there exists a $\lambda_0 \in (0,R)$ such that $u_\lambda \ge u$ in B_λ for all $\lambda \in (\lambda_0,R)$.

Let $A = \{\lambda_0 \in (0,R) : u_\lambda \ge u \, {\rm in}\, B_\lambda \, {\rm for \, all}\, \lambda \in (\lambda_0,R) \}$ and $\Lambda:= \inf A$. We claim that $\Lambda =0$. Let $\Lambda \gt 0$, we will show that this will contradict the definition of Λ. First note that by continuity $w_\Lambda \le 0$ in $B_\Lambda$ and $w_\Lambda $ solves $ \Delta w_\Lambda +B(x)w_\Lambda \ge 0 \, {\rm in} \, B_\Lambda$. If $\Lambda \gt \frac R2$, then $w_\Lambda$ is also $C^2(\overline{B_{\Lambda}})$ and so by strong maximum principle $w_\Lambda \gt 0$ in $B_\Lambda$. Hence using standard arguments, we can show that there exists a $\Lambda_0 \lt \Lambda$ such that $u_\lambda \ge u$ in B_λ for all $\lambda \in (\Lambda_0,R)$ gives a contradiction. If $\Lambda \le \frac R2$, then $w_\Lambda$ is in $C^2(\overline{B_\Lambda}\setminus \{(2\Lambda,0,\ldots,0)\})$ and hence by strong maximum principle $w_\Lambda \gt 0$ in $B_\Lambda \setminus \{(2\Lambda,0,\ldots,0)\}.$ Again by using a small measure type arguments near the singularity $(2\Lambda,0,\ldots,0)$, combining with the positivity of $w_\Lambda$ we can get a contradiction. For details of these types of arguments, we refer to the proof of theorem 2.1 in [Reference Mancini, Fabbri and Sandeep11]. Thus, we get $u(-x_1,\ldots,x_n) \ge u(x_1,\ldots,x_n)$ for $x_1 \gt 0$. Similarly moving the plane from the left, we get the opposite inequality and hence symmetry in the x ₁ direction. Rest of the proof is standard and we omit the details.$\square$

Proof. We start our proof with the following error estimates which we are going to use again and again. From our assumption on ρ, we have for t > 0, $|\rho(t)|\leq ct$, and $|\tilde{\rho}(s)|\leq c t^2$, for some c > 0. Then using these bounds together with the Hardy–Sobolev inequality, we obtain as $R\rightarrow 0$,

(3.3)\begin{align} R^{\frac{n+2}{2}} \int_{B(1)} \rho\left(\frac{V_R}{R^{\frac{n-2}{2}}}\right) V_{R}dx = O\left(R^2 \|V_{R}\|^2\right)= O\left(R^2 \left(\int \frac{|V_{R}|^{2^{*}(s)}}{|x|^s}\right)^{\frac{2}{2^{*}(s)}}\right). \end{align}

(3.4)\begin{align} \qquad\quad\! R^n \int_{B(1)} \tilde{\rho}\left(\frac{V_R}{R^{\frac{n-2}{2}}}\right) dx = O\left(R^2 \|V_{R}\|^2\right)= O\left(R^2 \left(\int \frac{|V_{R}|^{2^{*}(s)}}{|x|^s}\right)^{\frac{2}{2^{*}(s)}}\right). \end{align}

Next, we show that V_R is bounded in $H_{0}^{1}(B(1))$. Since V_R satisfy (3.1), we have

(3.5)\begin{align} \int_{B(1)} |\nabla V_{R}|^2 -\int_{B(1)}\frac{|V_{R}|^{2^{*}(s)}}{|x|^s} dx - R^{\frac{n+2}{2}}\int_{B(1)} \rho\left(\frac{V_R}{R^{\frac{n-2}{2}}}\right) V_{R} dx=0. \end{align}

Multiplying (3.5) by $\frac{1}{2^{*}(s)}$ and then subtracting it from (3.2) imply that

(3.6)\begin{align} & \frac{2-s}{2(n-s)} \int_{B(1)} |\nabla V_{R}|^2 + \frac{R^{\frac{n+2}{2}}}{2^{*}(s)}\int_{B(1)} \rho\left(\frac{V_R}{R^{\frac{n-2}{2}}}\right)V_{R} \nonumber\\ & \quad - R^n\int_{B(1)} \tilde{\rho}\left(\frac{V_R}{R^{\frac{n-2}{2}}}\right) \lt \frac{2-s}{2(n-s)} A_{s}^{\frac{n-s}{2-s}}. \end{align}

Using (3.3), (3.4), and (3.6), we deduce that

(3.7)\begin{align} \limsup_{R\rightarrow 0} \|\nabla V_{R}\|^2 \leq A_{s}^{\frac{n-s}{2-s}}. \end{align}

Thus V_R is bounded in $H_{0}^{1}(B(1))$. We note that by substituting $\int \frac{|V_R|^{2^{*}(s)}}{|x|^s}$ in (3.2) from (3.5) and then estimating rest of the terms using (3.3), (3.4), and (3.7), we obtain

(3.8)\begin{align} \frac{2-s}{2(n-s)}\limsup_{R\rightarrow 0} \|\nabla V_{R}\|^2=\limsup_{R\rightarrow 0} J_{R}(V_{R})\leq \frac{2-s}{2(n-s)} A_{s}^{\frac{n-s}{2-s}}. \end{align}

Applying Hardy–Sobolev inequality, (3.5) and (3.3), we have

(3.9)\begin{align} \begin{aligned} A_s \left(\int \frac{|V_R|^{2^{*}(s)}}{|x|^s} dx\right)^{\frac{2}{2^{*}(s)}} \leq \|\nabla V_{R}\|^2 &= \int \frac{|V_{R}|^{2^{*}(s)}}{|x|^s} dx+ O\left(R^2 \left(\int \frac{|V_{R}|^{2^{*}(s)}}{|x|^s} dx\right)^{\frac{2}{2^{*}(s)}}\right) \\ \mathrm{i.e.,} \quad A_{s}^{\frac{n-s}{2-s}}&\leq \liminf_{R\rightarrow 0} \int_{B(1)} \frac{|V_R|^{2^{*}(s)}}{|x|^s} dx. \end{aligned} \end{align}

Now Eqs. (3.9), (3.3), (3.4), (3.5), and (3.7) give

(3.10)\begin{align} \frac{2-s}{2(n-s)} A_{s}^{\frac{n-s}{2-s}}\leq \frac{2-s}{2(n-s)} \liminf_{R\rightarrow 0}\int \frac{|V_{R}|^{2^{*}(s)}}{|x|^s} dx = \liminf_{R\rightarrow 0} J_{R}(V_R). \end{align}

Hence (3.8) and (3.10) together implies that

\begin{align*}\lim_{R\rightarrow 0} J_{R}(V_R)= \frac{2-s}{2(n-s)} A_{s}^{\frac{n-s}{2-s}}. \end{align*}

This proves the first part of the theorem.

For second part, let R_m be a sequence such that $R_m\rightarrow 0$. Then (3.7) tells us that $V_{R_m}$ is bounded in $H^{1}_{0}(B(1))$. Therefore using the Hardy–Sobolev embedding of $H^1_0(\Omega)$ given by (1.4), up to a subsequence, if necessary, we can assume that $V_{R_m} \rightharpoonup V$ weakly in $H^{1}_{0}(B(1))$, $\frac{V_{R_m}^{2^{*}(s)-1}}{|x|^s} \rightarrow \frac{V^{2^{*}(s)-1}}{|x|^s}$ strongly in $L^1(B(1))$ and pointwise almost everywhere in $B(1)\setminus \{0\}$. Let $\phi\in C_{c}^{1}(B(1))$ then multiplying (3.1) by ϕ (where $R=R_m$), integrating by parts and passing to the limit as $n\rightarrow \infty$, we obtain

\begin{align*}\int_{B(1)} \nabla V\cdot \nabla \phi dx - \int_{B(1)} \frac{V^{2^{*}(s)-1}\phi}{|x|^s}=0.\end{align*}

Thus $V\in H_{0}^{1}(B(1))$ solves

\begin{align*} -\Delta V = \frac{V^{2^{*}(s)-1}}{|x|^s}\, \mbox{in}\, B(1), \quad V\geq 0 \, \mbox{in}\, B(1). \end{align*}

Now a standard use of Pohozaev identity (see [Reference Ghoussoub and Yuan8], Section 2) implies that $V\equiv 0$.

Also observe that $\|V_{R_m}\|_{\infty}\rightarrow \infty\,\mbox{as}\, m\rightarrow \infty$ for every sequence $R_m\rightarrow 0.$ This follows because if $\|V_{R_m}\|_{\infty}$ is bounded for some sequence $R_m \rightarrow 0$, then by dominated convergence theorem the right-hand side of (3.9) will converge to zero, which contradicts (3.9).

To prove the second part of our theorem, let us define for R > 0,

\begin{align*}\epsilon(R)= \|V_{R}\|_{\infty}^{-\frac{2}{n-2}} \quad \mbox{and}\quad W_{R}(x)= \epsilon({R})^{\frac{n-2}{2}} V_{R}(\epsilon({R}) x), x\in \mathbb R^n.\end{align*}

For notational convenience, we will simply denote $\epsilon(R)$ by ϵ in the rest of this subsection.

Now, we extend W_R to all of $\mathbb R^n$ by putting 0 outside its support, then

(3.11)\begin{align} \|\nabla V_{R}\|^2= \|\nabla W_{R}\|^2. \end{align}

Also since each V_R is a radially decreasing, so is W_R. Therefore, by the choice of ϵ

(3.12)\begin{align} \|W_R\|_{\infty}= W_{R}(0) = 1 \end{align}

and W_R satisfies

(3.13)\begin{align} -\Delta W_R = \frac{W_{R}^{2^{*}(s)-1}}{|x|^s}+ (\epsilon R)^{\frac{n+2}{2}} \rho\left(\frac{V_{R}(\epsilon x)}{R^{\frac{n-2}{2}}}\right)\, \mbox{in}\, B\left(\frac{1}{\epsilon}\right), \quad W_R \gt 0 \, \mbox{in}\, B\left(\frac{1}{\epsilon}\right). \end{align}

Let R_m be a sequence converging to zero and $\epsilon_m=\epsilon({R_m})$. Then by (3.11), we can assume that $W_{R_m}\rightharpoonup W$ weakly in $D^{1,2}(\mathbb R^n)$. Since $W_{R_m}$ is uniformly bounded and satisfies (3.12) from the Holder regularity result corollary 4.23 of [Reference Han and Lin10] , $W_{R_m}|_K$ is bounded in $C^{0,\delta}(K)$ for some δ < 1 and all compact sets K. Hence $W_{R_m}$ converges to W in $C^{0,\delta/2}_{loc}(\mathbb R^n)$ and this in particular implies that $W(0)=1$. Thus from (3.13) and $W_{R_m}\rightharpoonup W$ weakly in $D^{1,2}(\mathbb R^n)$, we see that W satisfies

\begin{align*} \left. \begin{array}{lr} -\Delta W= \frac{W^{2^{*}(s)-1}}{|x|^s} \quad \mbox{in}\quad \mathbb R^n\\ W\geq 0 \quad \mbox{in}\quad \mathbb R^n\\ W(0)=1 \end{array} \right\}. \end{align*}

This implies that by the classification result theorem 2.3 that $W(x)= \left(\frac{a}{a+|x|^{2-s}}\right)^{\frac{n-2}{2-s}}$ and hence

\begin{align*}\lim_{m\rightarrow \infty} \|\nabla W_{R_m}\|^2= \lim_{m\rightarrow \infty} \|\nabla V_{R_m}\|^2 =\frac{2(n-2)}{2-s} \lim_{m\rightarrow \infty}J_{R_m}(V_{R_m})= A_{s}^{\frac{n-s}{2-s}}= \|\nabla W\|^2.\end{align*}

i.e., $W_{R_m} \rightharpoonup W$ weakly in $D^{1,2}(\mathbb R^n)$ and $\|W_{R_m}\|^2\rightarrow \|W\|^2$. Since R_m is arbitrary and W does not depend on R_m, we obtain

\begin{align*}W_{R_m} \rightarrow W \,\mbox{strongly in}\, D^{1,2}(\mathbb R^n).\end{align*}

Rescaling $W_{R_m}$, and using the fact that $\epsilon_{m}^{-\frac{n-2}{2}}W\left(\frac{x}{\epsilon_{m}}\right)= U_{\epsilon_{m}}(x)$, we obtain

\begin{align*}\|\nabla (V_{R_m}- U_{\epsilon_{m}})\|\rightarrow 0\,\mbox{as}\, m\rightarrow \infty.\end{align*}

This proves the second part of the theorem.$\square$

Proof. We have

(3.14)\begin{align} \int_{B(1)}|\nabla U_{\epsilon}|^2 = A_{s}^{\frac{n-s}{2-s}}+ O\left(\epsilon^{n-2}\right). \end{align}

We also have from and theorem 3.2,

(3.15)\begin{align} V_{R}\rightharpoonup 0 \,\mbox{weakly in}\, H^{1}_{0}(B(1)), \end{align}

(3.16)\begin{align} \int_{B(1)}|\nabla V_{R}|^2 \rightarrow A_{s}^{\frac{n-s}{2-s}}\,\mbox{as}\, R\rightarrow 0. \end{align}

Let, $\epsilon({R})$ be as in the statement of theorem 3.2, then using the theorem 3.2 and (3.14)

\begin{equation*}d(V_R,M)\leq\Arrowvert\nabla(V_R-(U_{\epsilon(R)}-U_{\epsilon(R)}(1)))\Arrowvert^2=\Arrowvert\nabla(V_R-U_{\epsilon(R)})\Arrowvert^2\longrightarrow0\,\text{as}\,R\rightarrow0.\end{equation*}

Now to prove (ii), fix $R_0 \gt 0$ and let $R \lt R_0$. Also assume that

(3.17)\begin{align} d(V_{R}, M)= \lim_{m\rightarrow \infty}\{\|\nabla(V_{R}-a_{m}\left(U_{\epsilon_{m}}- U_{\epsilon_{m}}(1)\right))\|^2. \end{align}

We want to show that $a_{m}\rightarrow a_{R}\in (0,2)$ and $\epsilon_{m}\rightarrow \epsilon_{R}\in (0,1)$ when R ₀ is small enough.

By passing to a subsequence, we can assume that $a_{m}\rightarrow a_{R}\in [0,2]$ and $\epsilon_{m}\rightarrow \epsilon_{R}\in [0,1]$. Also we have from (3.14) and (3.16),

(3.18)\begin{align} \|\nabla(V_{R}- a_{m}(U_{\epsilon_{m}}- U_{\epsilon_{m}}(1)))\|^2 = \|\nabla V_{R}\|^2+a_{m}^{2} \|\nabla U_{\epsilon_{m}}\|^2 - 2 a_{m} \int_{B(1)}\nabla V_{R}\cdot \nabla U_{\epsilon_{m}}. \end{align}

Now, suppose $\epsilon_{R}=0$. Then as $V_{R}\in H_{0}^{1}(B(1))$ and from the fact that $U_{\epsilon_{m}}$ converges weakly to 0 in $D^{1,2}(\mathbb R^n)$, we obtain

\begin{align*}\int_{B(1)}\nabla V_{R} \cdot \nabla U_{\epsilon_{m}} \rightarrow 0\,\mbox{as}\, m\rightarrow\infty. \end{align*}

Hence (3.17) and (3.18) together implies that

\begin{align*}d(V_{R}, M)\geq \|\nabla V_{R}\|^2,\end{align*}

and this is not possible if R ₀ is small enough as $d(V_{R}, M) \rightarrow 0$ and $\|\nabla V_{R}\|^2 \rightarrow A_{s}^{\frac{n-s}{2-s}}$ as $R\rightarrow 0$. Therefore, $\epsilon_{R} \gt 0$ if R ₀ is small enough. Since $\epsilon_{R} \gt 0$, as $m\rightarrow \infty$, $(U_{\epsilon_{m}}- U_{\epsilon_{m}}(1))\rightarrow (U_{\epsilon_{R}}- U_{\epsilon_{R}}(1))$ strongly in $H_{0}^{1}(B(1))$. Hence from (3.18), we obtain

(3.19)\begin{align} d(V_{R}, M)=\|\nabla V_{R}\|^2+a_{R}^{2} \|\nabla U_{\epsilon_{R}}\|^2 - 2 a_{R} \int_{B(1)}\nabla V_{R}\cdot \nabla U_{\epsilon_R}. \end{align}

Next we claim that $\epsilon_{R}\rightarrow 0$ as $R\rightarrow 0$.

Suppose for a sequence $R_m \rightarrow 0$, $\epsilon_{R_m}\rightarrow \epsilon_{R_0} \gt 0$. Then (3.15) implies that

(3.20)\begin{align} \lim_{m\rightarrow \infty} \int_{B(1)}\nabla V_{R_m}\cdot \nabla U_{\epsilon_{R_m}} =\lim_{m\rightarrow \infty} \int_{B(1)}\nabla V_{R_m} \cdot\nabla U_{\epsilon_{R_0}} =0. \end{align}

Taking the limit as $m\rightarrow \infty$ in (3.19) with $R=R_m$ and using (3.20) will lead to

\begin{align*}0= \lim_{m\rightarrow\infty} d(V_{R_m}, M)\geq \lim_{m\rightarrow\infty} \|\nabla V_{R_m}\|^2 = A_{s}^{\frac{n-s}{2-s}}.\end{align*}

This is a contradiction and hence $\epsilon_{R}\rightarrow 0$ as $R\rightarrow 0$.

Next we claim that $a_{R}\rightarrow 1$ as $R\rightarrow 0$.

Now, using Cauchy–Schwarz’s inequality in (3.19), we have

(3.21)\begin{align} d(V_{R}, M)\geq \|\nabla V_{R}\|^2 + a_{R}^{2} \|\nabla U_{\epsilon_{R_m}}\|^2 -2a_{R} \left(\int_{B(1)}|\nabla V_{R}|^2\right)^{\frac12} \left(\int_{B(1)}| \nabla U_{\epsilon_{R_m}}|^2\right)^{\frac12}. \end{align}

If $a_{R}\rightarrow a$ as $R\rightarrow 0$ (or for a subsequence $a_{R_m}\rightarrow 0$), then by taking the limit as $m\rightarrow \infty$ in (3.21) and using (3.16), we obtain

\begin{align*}0\geq (a-1)^2 A_{s}^{\frac{n-s}{2-s}}.\end{align*}

i.e., a = 1. Hence $a_{R} \rightarrow 1$ as $R\rightarrow 0$. This completes the proof of theorem 3.3.$\square$

Proof. To begin with, we observe that, since $\frac{4-2s}{n-2} \lt 1$,

\begin{align*}\begin{aligned} \int_{B(1)}\frac{|U_{\epsilon}-U_{\epsilon}(1)|^{\frac{4-2s}{n-2}} E_{R}^{2}}{|x|^s}&\leq \int_{B(1)}\frac{U_{\epsilon}^{\frac{4-2s}{n-2}} E_{R}^{2}}{|x|^s}+ \int_{B(1)}\frac{U_{\epsilon}(1)^{\frac{4-2s}{n-2}} E_{R}^{2}}{|x|^s}\\ &\leq \int_{B(1)}\frac{U_{\epsilon}^{\frac{4-2s}{n-2}} E_{R}^{2}}{|x|^s}+ o(1) \|\nabla E_{R}\|^2 \end{aligned}\end{align*}

by the Sobolev–Hardy inequality (1.4). Therefore, it is sufficient to show that

\begin{align*} \mbox{there exists}\quad \lambda \gt \frac{n+2-2s}{n-2}: \lambda \int_{B(1)} \frac{|U_{\epsilon}|^{\frac{4-2s}{n-2}} E_{R}^2}{|x|^s}\leq \|\nabla E_{R}\|^2. \end{align*}

This inequality relies on the following well-known result (see [Reference Smets and Willem14], proposition 4.6): the eigenvalue problem

\begin{align*} -\Delta \phi= \lambda \frac{U_{\epsilon}^{\frac{4-2s}{n-2}}}{|x|^s} \phi,\quad \phi\in D^{1,2}(\mathbb R^n) \end{align*}

has discrete spectrum, 1 and $\frac{n+2-2s}{n-2}$ are the first two eigenvalues, with corresponding eigenfunctions U_ϵ and $\frac{\partial U_{\epsilon}}{\partial\epsilon}$.

Notice that $0=\frac{1}{2}\frac{d}{d\epsilon}\int_{\mathbb R^n}|\nabla U_{\epsilon}|^2 =\int_{\mathbb R^n} \nabla U_{\epsilon} \cdot \nabla \frac{\partial U_{\epsilon}}{\partial\epsilon}$. As a consequence, if $\phi\in D^{1,2}(\mathbb R^n)$ satisfies $0=\int_{\mathbb R^n}\nabla \phi\cdot \nabla U_{\epsilon}= \int_{\mathbb R^n} \nabla \phi \cdot \nabla \frac{\partial U_{\epsilon}}{\partial\epsilon}$ then

\begin{align*} \lambda_{3} \int_{\mathbb R^n}\frac{U_{\epsilon}^{\frac{4-2s}{n-2}} \phi^{2}}{|x|^s}\leq \int_{\mathbb R^n}|\nabla \phi|^2,\quad \lambda_3 \gt \frac{n+2-2s}{n-2}. \end{align*}

In particular, E_R is such that $0= \int \nabla E_R \cdot \nabla U_{\epsilon} = \int \nabla E_R \cdot \nabla\frac{\partial U_{\epsilon}}{\partial \epsilon}$. Hence

\begin{align*}\mbox{there exists}\quad \lambda \gt \frac{n+2-2s}{n-2}: \lambda \int_{B(1)} \frac{|U_{\epsilon}|^{\frac{4-2s}{n-2}} E_{R}^2}{|x|^s}\leq \|\nabla E_{R}\|^2.\end{align*}

and this proves the lemma.$\square$

Proof. First note that $B_{\epsilon,R}$ is radial and hence $B_{\epsilon,R}(x) = W_{\epsilon,R}(|x|)$. From the ODE, it satisfies

\begin{align*} W_{\epsilon,R}^\prime (t) = -\frac{R^{\frac{n+2}{2}}}{t^{n-1}}\int_0^t \rho\left(\frac{U_{\epsilon}(r)}{R^{\frac{n-2}{2}}}\right) r^{n-1} dr.\end{align*}

Hence

\begin{align*}\| \nabla B_{\epsilon,R} \|^2 & = \omega_{n-1}\int_0^1|W_{\epsilon,R}^\prime (t) |^2 t^{n-1}dt\\ & = \omega_{n-1}\int_0^1\left[\frac{R^{\frac{n+2}{2}}}{t^{n-1}}\int_0^t \rho\left(\frac{U_{\epsilon}(r)}{R^{\frac{n-2}{2}}}\right) r^{n-1}dr \right]^2 t^{n-1}dt.\end{align*}

As before, changing the variable , first $r= a^{\frac{1}{2-s}}\epsilon^{\frac 12}R^{-\frac 12} \lambda$ and then $t= a^{\frac{1}{2-s}}\epsilon^{\frac 12}R^{-\frac 12} \theta$ we get

\begin{align*}\| \nabla B_{\epsilon,R} \|^2 & = \omega_{n-1}a^{\frac{n+2}{2-s}} (R\epsilon)^{\frac{n+2}{2}}\\ & \quad \int_0^{\frac{R^{\frac 12}}{a^{\frac{1}{2-s}}\epsilon^{\frac 12}}}\frac{1}{\theta^{n-1}}\left[\int_0^\theta \rho\left(\left[(R\epsilon)^{\frac{2-s}{2}}+ \lambda^{2-s} \right]^{-\frac{n-2}{2-s}}\right) \lambda^{n-1}d\lambda \right]^2 d\theta.\end{align*}

Now using the fact that ρ is bounded when $\theta \le 1$ and $|\rho(t)| \le Ct$ when θ > 1, we can estimate the above integral by

\begin{align*} \int_0^1 \frac{1}{\theta^{n-1}} \theta^{2n} d\theta + \int_1^{1+a^{\frac{-1}{2-s}}\epsilon^{-\frac 12}R^{\frac 12}}\frac{1}{\theta^{n-1}}\left[\int_0^\theta \left[(R\epsilon)^{\frac{2-s}{2}}+ \lambda^{2-s} \right]^{-\frac{n-2}{2-s}} \lambda^{n-1}d\lambda \right]^2 d\theta \end{align*}

\begin{align*}\le \int_0^1 \theta^{n+1} d\theta + \int_1^{1+a^{\frac{-1}{2-s}}\epsilon^{-\frac 12}R^{\frac 12} } \frac{1}{\theta^{n-5}} d\theta \le C\left(1+\epsilon^{-\frac 12}R^{\frac 12}\right).\end{align*}

Substituting back, we obtain

\begin{align*}\| \nabla B_{\epsilon,R} \|^2 \le C( (R\epsilon)^{\frac{n+2}{2}} + R^{\frac{n+3}{2}}\epsilon^{\frac{n+1}{2}} ) = O\left( R^{\frac{n+2}{2}}\epsilon^{\frac{n+1}{2}}\right).\end{align*}

This completes the proof.$\square$

Proof. The estimate (4.1) follows by using $|\rho(t)|\leq c |t|$ and $ \int_{|x|\le 1}U_{\epsilon}(x) dx = O(\epsilon^{\frac{n-2}{2}})$.

Similarly (4.2) follows from the fact that ρ is bounded and $ \int_{|x|\le 1}U_{\epsilon}(x) dx = O(\epsilon^{\frac{n-2}{2}})$. Thus, it remains to establish (4.3).

By changing the variable $R^{\frac{1}{2}}x= \epsilon^{\frac{1}{2}} a^{\frac{1}{2-s}}y$, we obtain,

\begin{align*}R^{n} \int_{B_1}\tilde{\rho}\left(\frac{U_{\epsilon}}{R^{\frac{n-2}{2}}}\right) = R^{\frac{n}{2}}\epsilon^{\frac{n}{2}}a^{\frac{n}{2-s}} g(R,\epsilon)\end{align*}

where

\begin{align*} g(R,\epsilon)= \int_{\epsilon^{\frac{1}{2}}a^{\frac{1}{2-s}}|y|\leq R^{\frac{1}{2}}} \tilde{\rho}\left(\left((\epsilon R)^{\frac{2-s}{2}}+|y|^{2-s}\right)^{-\left(\frac{n-2}{2-s}\right)}\right) dy.\end{align*}

For any fixed R > 0, we have

\begin{align*} g(R,0)=\lim_{\epsilon\rightarrow 0} g(R,\epsilon)= \int_{\mathbb R^n} \tilde{\rho}\left(\frac{1}{|y|^{n-2}}\right) dy. \end{align*}

Let $g^\prime(R,\epsilon) $ denotes the partial derivative of g in the ϵ variable, then

\begin{align*}\begin{aligned} g^\prime (R,\epsilon) & = -\frac{R^{\frac{1}{2}}}{2 a^{\frac{1}{2-s}} \epsilon^{\frac{3}{2}}} \int_{\epsilon^{\frac{1}{2}} a^{\frac{1}{2-s}}|y|= R^{\frac{1}{2}}}\tilde{\rho}\left(\left((\epsilon R)^{\frac{2-s}{2}}+|y|^{2-s}\right)^{-\left(\frac{n-2}{2-s}\right)}\right) d\sigma \\ -&\frac{(n-2)R^{\frac{2-s}{2}}}{2\epsilon^{\frac{s}{2}}} \int_{\epsilon^{\frac{1}{2}} a^{\frac{1}{2-s}}|y|\leq R^{\frac{1}{2}}}\rho\left(\left((\epsilon R)^{\frac{2-s}{2}}+|y|^{2-s}\right)^{-\frac{n-2}{2-s}}\right) \\ & \quad \left((\epsilon R)^{\frac{2-s}{2}}+|y|^{2-s}\right)^{-\frac{n-s}{2-s}}dy\\ =&- \frac{w_{n-1} R^{\frac{n}{2}}}{2a^{\frac{n}{2-s}}\epsilon^{\frac{n+2}{2}}} \tilde{\rho}\left(\frac{1}{R^{\frac{n-2}{2}}}\left(\frac{a\epsilon^{\frac{2-s}{2}}}{a\epsilon^{2-s}+1}\right)^{\frac{n-2}{2-s}}\right)\\ -&\frac{(n-2)R^{\frac{2-s}{2}}}{2\epsilon^{\frac{s}{2}}}\int_{|y|\leq\frac{R^{\frac{1}{2}}}{\epsilon^{\frac{1}{2}} a^{\frac{1}{2-s}}}}\rho\left(\left((\epsilon R)^{\frac{2-s}{2}}+|y|^{2-s}\right)^{-\frac{n-2}{2-s}}\right)\\ & \quad \left((\epsilon R)^{\frac{2-s}{2}}+|y|^{2-s}\right)^{-\left(\frac{n-s}{2-s}\right)} dy. \end{aligned}\end{align*}

Now

\begin{align*}g(R,\epsilon)= g(R,0)+\int_{0}^{1} \frac{\partial }{\partial t}(g(R,t\epsilon))\, dt= g(R,0)+\epsilon\int_{0}^{1} g^\prime (R,t\epsilon) \, dt.\end{align*}

Then $|\tilde{\rho}(t)|\leq c t^2$ yield

\begin{align*}\begin{aligned} &\epsilon\int_{0}^{1} g^\prime(R,t\epsilon) dt = -\int_{0}^{1} \frac{w_{n-1} \epsilon R^{\frac{n}{2}}}{2a^{\frac{n}{2-s}}(t\epsilon)^{\frac{n+2}{2}}} \tilde{\rho}\left(\frac{1}{R^{\frac{n-2}{2}}}\left(\frac{a(t\epsilon)^{\frac{2-s}{2}}}{a(t\epsilon)^{2-s}+1}\right)^{\frac{n-2}{2-s}}\right) dt \\ &-\int_{0}^{1}\frac{(n-2)\epsilon R^{\frac{2-s}{2}}}{2(t\epsilon)^{\frac{s}{2}}} \left(\int_{|y|\leq\frac{R^{\frac{1}{2}}}{(t\epsilon)^{\frac{1}{2}} a^{\frac{1}{2-s}}}}\frac{\rho\left(\left((t\epsilon R)^{\frac{2-s}{2}}+|y|^{2-s}\right)^{-\frac{n-2}{2-s}}\right)}{\left((t\epsilon R)^{\frac{2-s}{2}}+|y|^{2-s}\right)^{\left(\frac{n-s}{2-s}\right)}} dy\right) dt\\ =& O\left(\left(\frac{R}{\epsilon}\right)^{n/2}\int_{0}^{1}\frac{1}{t^{\frac{n+2}{2}}}\left(\frac{t\epsilon}{R}\right)^{n-2} dt\right)-\frac{n-2}{2}(R\epsilon)^{1-s/2}h(R,\epsilon)\\ =& O\left(\left(\frac{\epsilon}{R}\right)^{\frac{n}{2}-2}\right)- \frac{n-2}{2}(R\epsilon)^{1-s/2}h(R,\epsilon), \end{aligned}\end{align*}

where

\begin{align*}h(R,\epsilon)=\int_{0}^{1} t^{-\frac{s}{2}} \int_{|y|\leq\frac{R^{\frac{1}{2}}}{(t\epsilon)^{\frac{1}{2}} a^{\frac{1}{2-s}}}}\frac{\rho\left(\left((t\epsilon R)^{\frac{2-s}{2}}+|y|^{2-s}\right)^{-\frac{n-2}{2-s}}\right)}{\left((t\epsilon R)^{\frac{2-s}{2}}+|y|^{2-s}\right)^{\left(\frac{n-s}{2-s}\right)}} dy dt.\end{align*}

Again as before by denoting $h^\prime(R,\epsilon)$ the partial derivative of h in the ϵ variable, we have

\begin{align*}h(R,\epsilon)= h(R,0)+\int_{0}^{1} \frac{\partial }{\partial t}(h(R,t\epsilon))\, dt= h(R,0)+\epsilon\int_{0}^{1} h^\prime (R,t\epsilon) \, dt.\end{align*}

Then by Lebesgue dominated convergence theorem, we have

\begin{align*}h(R,0)=\lim_{\epsilon\rightarrow 0} h(R,\epsilon)= \frac{2}{2-s}\int_{\mathbb R^n}\rho\left(\frac{1}{|y|^{n-2}}\right)\frac{1}{|y|^{n-s}} dy.\end{align*}

Now

(4.4)\begin{align} \begin{aligned} h^\prime(R,\epsilon) =& -\frac{R^{\frac{s}{2}}}{2a^{\frac{n}{2-s}}\epsilon^{\frac{n+2}{2}}}\int_{0}^{1} t^{-\frac{(n+s)}{2}}\rho\left(\frac{U_{t\epsilon}(1)}{R^{\frac{n-2}{2}}}\right)\left(\frac{a(t\epsilon)^{\frac{2-s}{2}}}{a(t\epsilon)^{2-s}+1}\right)^{\frac{n-s}{2-s}} dt \\ & \quad +\int_{0}^{1} t^{-\frac{s}{2}} \int_{|y|\leq\frac{R^{\frac{1}{2}}}{(t\epsilon)^{\frac{1}{2}} a^{\frac{1}{2-s}}}}\frac{\partial}{\partial \epsilon}\left[\frac{\rho\left(\left((t\epsilon R)^{\frac{2-s}{2}}+|y|^{2-s}\right)^{-\frac{n-2}{2-s}}\right)}{\left((t\epsilon R)^{\frac{2-s}{2}}+|y|^{2-s}\right)^{\left(\frac{n-s}{2-s}\right)}}\right] dy dt. \end{aligned} \end{align}

Employing $|\rho(t)|\leq C|t|$, first term of (4.4) in the right-hand side can be estimated as

\begin{align*}C \frac{R^{\frac{s}{2}}}{2a^{\frac{n}{2-s}}\epsilon^{\frac{n+2}{2}}}\int_{0}^{1} t^{-\frac{s}{2}}| \rho\left(\frac{U_{t\epsilon}(1)}{R^{\frac{n-2}{2}}}\right)| (t\epsilon)^{1-s/2} dt= O\left(\frac{1}{R}\left(\frac{\epsilon}{R}\right)^{\frac{n-4-s}{2}}\right).\end{align*}

The second term of (4.4) in the right-hand side is equal to

\begin{align*}\begin{aligned} \int_{0}^{1} &t^{-\frac{s}{2}} \omega_{n-1}\int_{0}^{\frac{R^{\frac{1}{2}}}{(t\epsilon)^{\frac{1}{2}} a^{\frac{1}{2-s}}}}\frac{\partial}{\partial \epsilon}\left[\frac{\rho\left(\left((t\epsilon R)^{\frac{2-s}{2}}+|r|^{2-s}\right)^{-\frac{n-2}{2-s}}\right)}{\left((t\epsilon R)^{\frac{2-s}{2}}+|r|^{2-s}\right)^{\left(\frac{n-s}{2-s}\right)}}\right] r^{n-1}dr dt\\ =& \frac{\omega_{n-1}}{2}\int_{0}^{1} t^{-\frac{s}{2}}\int_{0}^{\frac{R^{\frac{1}{2}}}{(t\epsilon)^{\frac{1}{2}} a^{\frac{1}{2-s}}}}\frac{\partial}{\partial r}\left[\frac{\rho\left(\left((t\epsilon R)^{\frac{2-s}{2}}+|r|^{2-s}\right)^{-\frac{n-2}{2-s}}\right)}{\left((t\epsilon R)^{\frac{2-s}{2}}+|r|^{2-s}\right)^{\left(\frac{n-s}{2-s}\right)}}\right] \epsilon^{-\frac{s}{2}} (tR)^{1-\frac{s}{2}}r^{n-2+s}dr dt\\ =&\frac{\omega_{n-1} R^{1-\frac{s}{2}}}{2 \epsilon^{\frac{s}{2}}}\int_{0}^{1} t^{1-s}\left[ \rho\left(\frac{U_{t\epsilon}(1)}{R^{\frac{n-2}{2}}}\right)\left(\frac{a(t\epsilon)^{\frac{2-s}{2}}}{a(t\epsilon)^{2-s}+1}\right)^{\frac{n-s}{2-s}} \frac{1}{R^{\frac{n-s}{2}}}\left(\frac{R^{\frac 12}}{a^{\frac {1}{2-s}}(t\epsilon)^{\frac 12}}\right)^{n-2+s}\right.\\ &\left.-(n-2+s)\int_{0}^{\frac{R^{\frac{1}{2}}}{(t\epsilon)^{\frac{1}{2}} a^{\frac{1}{2-s}}}}\left[\frac{\rho\left(\left((t\epsilon R)^{\frac{2-s}{2}}+|r|^{2-s}\right)^{-\frac{n-2}{2-s}}\right)}{\left((t\epsilon R)^{\frac{2-s}{2}}+|r|^{2-s}\right)^{\left(\frac{n-s}{2-s}\right)}}\right] r^{n-3+s}dr dt\right]\\ =& \omega_{n-1}\frac{R^{1-s/2}}{\epsilon^{s/2}} \left[ I_1+I_2 \right]. \end{aligned}\end{align*}

Relation $|\rho(t)|\leq C |t|$ yield

\begin{align*}I_1 \leq O\left(\left(\frac{\epsilon}{R}\right)^{\frac{n}{2}-s}\right).\end{align*}

Also, employing $\rho(t)\leq t^{-b}$, for some b > 0 as $t\rightarrow \infty$ and $\rho(t)=o(t)$ as $t\rightarrow 0$, we have

\begin{align*} I_2\leq \int_{0}^1 \left[(t\epsilon R)^{\frac{2-s}{2}}+ r^{2-s}\right]^{-\frac{n-s}{2-s}+ \frac{(n-2)b}{2-s}} r^{n-3+s} dr +\int_{1}^{\infty}\frac{r^{n-3+s}}{r^{n-2+n-s}} dr\leq K \lt \infty, \end{align*}

if $b \gt \max\left\{0,\frac{2(1-s)}{(n-2)}\right\}$. Combining all above, we obtain

\begin{align*} |h^\prime (R,\epsilon) |\le O\left(\left(\frac{\epsilon}{R}\right)^{\frac{n-4-s}{2}}\right)+ R^{1-\frac{s}{2}}{\epsilon^{\frac{s}{2}}}\left[O\left(\left(\frac{\epsilon}{R}\right)^{\frac{n-2s}{2}} +K\right)\right],\end{align*}

where K is a constant independent of R and ϵ. Hence

\begin{align*}\begin{aligned} \epsilon \int_{0}^{1} h^\prime(R,t\epsilon)&\leq C \epsilon\int_{0}^{t}\left[\left(\frac{(t\epsilon)^{\frac{n-4-s}{2}}}{R^{\frac{n-2-s}{2}}}\right)+ R^{1-\frac{s}{2}}{(\epsilon t)^{\frac{s}{2}}}\left(\left(\frac{\epsilon t}{R}\right)^{\frac{n-2s}{2}}+ K\right)\right]\\ &\leq O\left(\left(\frac{\epsilon}{R}\right)^{\frac{n-2-s}{2}}\right)+ O\left((R\epsilon )^{1-\frac{s}{2}}\right)+ O\left(\frac{\epsilon^{\frac{n+2-3s}{2}}}{R^{\frac{n-2-s}{2}}}\right). \end{aligned}\end{align*}

Thus

\begin{align*}\begin{aligned} &R^{n} \int \tilde{\rho}\left(\frac{U_{\epsilon}}{R^{\frac{n-2}{2}}}\right)= (\epsilon R)^{\frac{n}{2}}a^{\frac{n}{2-s}}\left[g(R,0)+\epsilon \int_{0}^{1}g^\prime(R,\epsilon t) dt\right]\\ =&(\epsilon R)^{\frac{n}{2}}a^{\frac{n}{2-s}}\left[g(R,0)-\frac{n-2}{2} (\epsilon R)^{1-\frac{s}{2}} h(R,\epsilon)+ O\left(\left(\frac{\epsilon}{R}\right)^{\frac{n}{2}-2}\right)\right]\\ =&(\epsilon R)^{\frac{n}{2}}a^{\frac{n}{2-s}}\left[g(R,0)-\frac{n-2}{2} (\epsilon R)^{1-\frac{s}{2}} \left(h(R,0)+\epsilon \int_{0}^{1}h^\prime(R,t\epsilon) dt\right)+ O\left(\left(\frac{\epsilon}{R}\right)^{\frac{n}{2}-2}\right)\right]\\ = &(\epsilon R)^{\frac{n}{2}}a^{\frac{n}{2-s}} \left[\int_{\mathbb R^n} \tilde{\rho}\left(\frac{1}{|x|^{n-2}}\right)-\frac{n-2}{2-s} (\epsilon R)^{1-s/2}\left(\int_{\mathbb R^n} {\rho}\left(\frac{1}{|x|^{n-2}}\right)\frac{1}{|x|^{n-s}} + O\left((R\epsilon )^{1-\frac{s}{2}}\right)\right)\right]\\ & + (\epsilon R)^{\frac{n}{2}+1-\frac s2}\left[ O\left(\left(\frac{\epsilon}{R}\right)^{\frac{n-2-s}{2}}\right) + O\left(\frac{\epsilon^{\frac{n+2-3s}{2}}}{R^{\frac{n-2-s}{2}}}\right)\right] + O(R^2\epsilon^{n-2})\\ = &(\epsilon R)^{\frac{n}{2}}a^{\frac{n}{2-s}} \left[\int_{\mathbb R^n} \tilde{\rho}\left(\frac{1}{|x|^{n-2}}\right)-\frac{n-2}{2-s} (\epsilon R)^{\frac{2-s}2}\left(\int_{\mathbb R^n} {\rho}\left(\frac{1}{|x|^{n-2}}\right)\frac{1}{|x|^{n-s}} + o(1)\right)\right] +o(\epsilon^{n-2}). \end{aligned}\end{align*}

This completes the proof of (4.3). $\square$

Proof. The first inequality follows directly from (3.22) and (2.6). Next, by (2.2) and using (3.24), we obtain

\begin{align*} \int_{B(1)} \frac{U_{\epsilon}^{\frac{n+2-2s}{n-2}}}{|x|^s} E_{R}= \int_{B(1)} \nabla U_{\epsilon}\cdot \nabla E_{R} = 0, \end{align*}

and this proves the second inequality.

Next, using (3.23), Hölder and Sobolev–Hardy inequalities, we obtain

\begin{align*}\begin{aligned} &\int_{B(1)} \frac{(U_{\epsilon} +|U_{\epsilon}(1)| +|E_{R}|)^{\frac{4-2s}{n-2}}}{|x|^s}|U_{\epsilon}(1)+E_{R}|^2\\ &\leq c\int_{B(1)} \frac{|U_{\epsilon}(1)|^{\frac{4-2s}{n-2}}}{|x|^s}|U_{\epsilon}(1)+E_{R}|^2 + c\int_{B(1)} \frac{|E_R|^{\frac{4-2s}{n-2}}}{|x|^s}|U_{\epsilon}(1)+E_{R}|^2 \\ &+ \int_{B(1)} \frac{U_{\epsilon}^{\frac{4-2s}{n-2}}}{|x|^s}(U_{\epsilon}(1)^2+|E_{R}|^2) = o(1)\epsilon^{n-2}. \end{aligned}\end{align*}

This proves the third inequality.

Next, using (4.2) and (3.22), we obtain

\begin{align*}\begin{aligned} |1-a_{R}|\left| R^{\frac{n+2}{2}} \int_{B(1)} \rho\left(\frac{U_{\epsilon}}{R^{\frac{n-2}{2}}}\right) U_{\epsilon} \right| =& |1-a_{R}| \left| R^{\frac{n+2}{2}} \int_{B(1)} \rho\left(\frac{U_{\epsilon}}{R^{\frac{n-2}{2}}}\right)U_{\epsilon} \right|\\ \leq & c|1-a_{R}| \left( \epsilon^{\frac{n-2}{2}}\right)\\ = & o(1) \left(\epsilon^{n-2}\right). \end{aligned}\end{align*}

Finally, using lemma 3.5 and (3.23), we obtain

\begin{align*} \left| R^{\frac{n+2}{2}} \int_{B(1)} \rho\left(\frac{U_{\epsilon}}{R^{\frac{n-2}{2}}}\right) E_{R} \right| \leq C\|\nabla E_{R}\|\|\nabla B_{\epsilon,R}\|= o(1) \epsilon^{n-2}. \end{align*}

The last inequality follows immediately from (3.22) and (3.23). This completes the proof of lemma 4.2.$\square$

Proof. By the definition of J_R in (3.2), we have

\begin{align*} J_{R}(V_{R}) =\frac{1}{2} \int_{B(1)} |\nabla V_{R}|^2 -\frac{1}{2^{*}(s)} \int_{B(1)}\frac{|V_{R}|^{2^{*}(s)}}{|x|^s} - {R}^{n} \int_{B(1)} \tilde{\rho}\left(\frac{V_{R}}{R^{\frac{n-2}{2}}}\right), \end{align*}

where $V_{R}= a_{R}(U_{\epsilon}-U_{\epsilon}(1)) +E_{R}$. Then relation $\int \nabla E_{R}\cdot \nabla (U_{\epsilon}-U_{\epsilon}(1)) =0$ yields

\begin{align*} \int_{B(1)}|\nabla V_{R}|^2 = a_{R}^2 \int_{B(1)}|\nabla U_{\epsilon}|^2+ \int_{B(1)} |\nabla E_{R}|^2. \end{align*}

Using Taylor formula and $ 2^{*}(s):=p+1$,

\begin{align*}\begin{aligned} &\int_{B(1)} \frac{V_{R}^{p+1}}{|x|^s} dx= a_{R}^{p+1} \int_{B(1)}\frac{U_{\epsilon}^{p+1}}{|x|^s} +(p+1) \int_{B(1)}\frac{(a_{R} U_{\epsilon})^p (E_{R}-a_{R}U_{\epsilon}(1))}{|x|^s}\\ &+ p(p+1)\int_{B(1)}\int_{0}^{1}(1-t)|a_{R}U_{\epsilon} +t(E_{R}-a_{R}U_{\epsilon}(1))|^{p-1} \frac{(-a_{R}U_{\epsilon}(1)+E_{R})^2}{|x|^s} \end{aligned}\end{align*}

\begin{align*}\begin{aligned} \int_{B(1)} \tilde{\rho}\left(\frac{V_{R}}{R^{\frac{n-2}{2}}}\right) dx&= \int_{B(1)} \tilde{\rho}\left(\frac{U_{\epsilon}}{R^{\frac{n-2}{2}}}\right)+\rho\left(\frac{U_{\epsilon}}{R^{\frac{n-2}{2}}}\right)\left[\frac{(a_{R}-1)U_{\epsilon} - a_{R}U_{\epsilon}(1)+E_{R}}{R^{\frac{n-2}{2}}}\right]\\ & + \int_{B(1)}\left(\int_{0}^{1}(1-t)\rho^{\prime}\left(\frac{U_{\epsilon}+t[(a_{R}-1)U_{\epsilon} - a_{R} U_{\epsilon}(1)+E_{R}]}{R^{\frac{n-2}{2}}}\right)\right)\\ &\times\left[\frac{(a_{R}-1)U_{\epsilon} - a_{R}U_{\epsilon}(1)+E_{R}}{R^{\frac{n-2}{2}}}\right]^2. \end{aligned}\end{align*}

Using the expression above together with

\begin{align*}\int_{|x| \lt 1} |\nabla U_{\epsilon}|^2 = \int_{\mathbb R^n}|\nabla U_{\epsilon}|^2 - \int_{|x|\geq 1} |\nabla U_{\epsilon}|^2= A_{s}^{\frac{n-s}{2-s}} + O(\epsilon^{n-2}), \end{align*}

\begin{align*}\int_{|x| \lt 1}\frac{U_{\epsilon}^{2^{*}(s)}}{|x|^s}= \int_{\mathbb R^n}\frac{U_{\epsilon}^{2^{*}(s)}}{|x|^s} -\int_{|x|\geq1}\frac{U_{\epsilon}^{2^{*}(s)}}{|x|^s}= A_{s}^{\frac{n-s}{2-s}}+ O(\epsilon^{n-s}),\end{align*}

properly collecting different terms and using $\int \frac{U_{\epsilon}^p E_{R}}{|x|^s}= \int \nabla U_{\epsilon} \cdot\nabla E_R= 0$, we obtain

\begin{align*}\begin{aligned} &J_{R}(V_{R}) = \left[\left(\frac{a_{R}^2}{2}-\frac{a_{R}^{2^{*}(s)}}{2^{*}(s)}\right) \int_{\mathbb R^n} |\nabla U_{\epsilon}|^2- \frac{a_{R}^2}{2} \int_{|x|\geq 1}|\nabla U_{\epsilon}|^2 - R^n \int \tilde{\rho}\left(\frac{U_{\epsilon}}{R^{\frac{n-2}{2}}}\right)+ 0(\epsilon^{n-s})\right]\\ & - R^{\frac{n+2}{2}} a_{R} \int \rho\left(\frac{U_{\epsilon}}{R^{\frac{n-2}{2}}}\right) U_{\epsilon}(1) + \frac{1}{2}\int |\nabla E_R|^2+a_{R}^{2^{*}(s)}\int \frac{U_{\epsilon}^{\frac{n+2-2s}{n-2}} U_{\epsilon}(1)}{|x|^s}\\ & - \frac{(n+2-2s)}{n-2} \int_{B(1)}\int_{0}^{1}(1-t)|a_{R}U_{\epsilon} +t(E_{R}-a_{R}U_{\epsilon}(1))|^{\frac{4-2s}{n-2}}\frac{(-a_{R}U_{\epsilon}(1)+E_{R})^2}{|x|^s}\\ & \quad - R^{\frac{n+2}{2}} (a_{R}-1) \int \rho\left(\frac{U_{\epsilon}}{R^{\frac{n-2}{2}}}\right) U_{\epsilon}- R^{\frac{n+2}{2}} \int \rho\left(\frac{U_{\epsilon}}{R^{\frac{n-2}{2}}}\right) E_{R} + a_{R} R^{\frac{n+2}{2}} \int \rho\left(\frac{U_{\epsilon}}{R^{\frac{n-2}{2}}}\right)U_{\epsilon}(1)\\ & -R^n\int_{B(1)}\int_{0}^{1}(1-t)\rho^{\prime}\left(\frac{U_{\epsilon}+t[(a_{R}-1)U_{\epsilon} - a_{R} U_{\epsilon}(1)+E_{R}]}{R^{\frac{n-2}{2}}}\right)\left[\frac{(a_{R}-1)U_{\epsilon} - a_{R} U_{\epsilon}(1)+E_{R}}{R^{\frac{n-2}{2}}}\right]^2\\ &=\left[\left(\frac{a_{R}^2}{2}-\frac{a_{R}^{2^{*}(s)}}{2^{*}(s)}\right) A_{s}^{\frac{n-s}{2-s}}+ a_{R}^{2}\left( U_{\epsilon}(1) \int_{B(1)} \frac{U_{\epsilon}^{\frac{n+2-2s}{n-2}}}{|x|^s}- \frac12 \int_{|x|\geq 1}|\nabla U_{\epsilon}|^2\right) - R^n \int \tilde{\rho}\left(\frac{U_{\epsilon}}{R^{\frac{n-2}{2}}}\right)\right]\\ & + R^{\frac{n+2}{2}} a_{R} \int \rho\left(\frac{U_{\epsilon}}{R^{\frac{n-2}{2}}}\right) U_{\epsilon}(1)+ o(1) \epsilon^{n-2}, \end{aligned}\end{align*}

because of (3.22), (3.23), and lemma 4.2.

It follows from (4.1) that $R^{\frac{n+2}{2}}\int_{B(1)} \rho\left(\frac{U_{\epsilon}}{R^{\frac{n-2}{2}}}\right) U_{\epsilon}(1)= o(1) \epsilon^{n-2}$. Also $\frac{a_{R}^{2}}{2}-\frac{a_{R}^{2^{*}(s)}}{2^*(s)}= O(|a_{R}-1|^2)$, thus using (3.22) the first expression in square bracket simplifies and we have

\begin{align*} J_{R}(V_R) & =\frac{(2-s) A_{s}^{\frac{n-s}{2-s}}}{2(n-s)} + a_{R}^{2}\left( U_{\epsilon}(1) \int_{B(1)} \frac{U_{\epsilon}^p}{|x|^s}- \frac12 \int_{|x|\geq 1}|\nabla U_{\epsilon}|^2\right)\\ & \quad - R^{n} \int \tilde{\rho}\left(\frac{U_{\epsilon}}{R^{\frac{n-2}{2}}}\right)+ o(1)\epsilon^{n-2}. \end{align*}

This completes the proof.$\square$

Proof. (i) Changing the variable as $\lambda ={a^{\frac{1}{2-s}}\epsilon^{\frac{1}{2}}} t$, we obtain

(5.1)\begin{align} \begin{aligned} \int_{B(R)} \tilde{\rho}(U_{\epsilon}) dx &= \omega_{n-1} \int_{0}^{R} \tilde{\rho}\left(\left(\frac{a \epsilon^{\frac{2-s}{2}}}{a \epsilon^{2-s}+\lambda^{2-s}}\right)^{\frac{n-2}{2-s}}\right)\lambda^{n-1} d\lambda \\ &= a^{\frac{n}{2-s}}\epsilon^{\frac{n}{2}} \omega_{n-1}\int_{0}^{R/{{a^{\frac{1}{2-s}}\epsilon^{\frac{1}{2}}}}} \tilde{\rho}\left(\left(\epsilon^{\frac{2-s}{2}}+t^{2-s}\right)^{-\frac{n-2}{2-s}}\right)t^{n-1} dt \\ &= \omega_{n-1}a^{\frac{n}{2-s}}\epsilon^{\frac{n}{2}} g(\epsilon) = w_{n-1}a^{\frac{n}{2-s}}\epsilon^{\frac{n}{2}} \left( g(0) + \epsilon\int_0^1g^\prime(r\epsilon) \, dr \right). \end{aligned} \end{align}

Now, using the bounds $\tilde{\rho}(t)=O(t^2)$ as $t\rightarrow 0$ and $\tilde{\rho}(t)=O(t)$ as $t\rightarrow \infty$, we have

\begin{align*} \tilde{\rho}\left(\left(\epsilon^{\frac{2-s}{2}}+t^{2-s}\right)^{-\frac{n-2}{2-s}}\right)t^{n-1} = \left\{ \begin{array}{l} O\left(t^{n-1-(n-2)}\right)\,\quad \mbox{if}\quad t \lt 1\\ O\left(t^{n-1-2(n-2)}\right)\, \mbox{if}\quad t \gt 1, \end{array} \right. \end{align*}

which is uniformly in ϵ and hence by the dominated convergence theorem, we obtain for $n\ge 5$

\begin{align*} g(0):= \lim_{\epsilon\rightarrow 0} g(\epsilon)= \int_{0}^{\infty}\tilde{\rho}\left( \frac{1}{t^{n-2}}\right) t^{n-1} = \frac{1}{n-2}\int_{0}^{\infty} \tilde{\rho}(t) t^{-\frac{2(n-1)}{n-2}} dt \end{align*}

\begin{align*} g^\prime(\epsilon) &= -\left(\frac{n-2}{2}\right)\epsilon^{-\frac s2}\int_{0}^{R/{a^{\frac{1}{2-s}}\epsilon^{\frac{1}{2}}}} \frac{{\rho}\left(\left(\epsilon^{\frac{2-s}{2}}+t^{2-s}\right)^{-\frac{n-2}{2-s} }\right)t^{n-1}}{(\epsilon^{\frac{2-s}{2}}+t^{2-s})^{\frac{n-s}{2-s}}} dt \\ & \quad -\frac{1}{2}\left(\frac{R}{a^{\frac{1}{2-s}}}\right)^n \frac{\tilde{\rho}(U_\epsilon(R))}{\epsilon^{\frac{n+2}{2}}}. \end{align*}

Thus

\begin{align*}\begin{aligned} \epsilon\int_0^1g^\prime(r\epsilon) \, dr &= -\left(\frac{n-2}{2}\right)\epsilon^{1-\frac s2}\int_0^1 r^{-\frac s2}\int_{0}^{R/{a^{\frac{1}{2-s}}(r\epsilon)^{\frac{1}{2}}}} \frac{{\rho}\left(\left((r\epsilon)^{\frac{2-s}{2}}+t^{2-s}\right)^{-\frac{n-2}{2-s} }\right)t^{n-1}}{((r\epsilon)^{\frac{2-s}{2}}+t^{2-s})^{\frac{n-s}{2-s}}} dt \\ & -\frac{1}{2}\left(\frac{R}{a^{\frac{1}{2-s}}}\right)^n\epsilon \int_0^1 \frac{\tilde{\rho}(U_{r\epsilon}(R))}{(r\epsilon)^{\frac{n+2}{2}}} dr. \end{aligned}\end{align*}

Now, since $\rho(t)=O(t)$ as $t\rightarrow 0$ and $\rho(t)=O(t^{-b})$ as $t\rightarrow\infty$, b > 0, we obtain

\begin{align*} \frac{\rho\left(((r\epsilon)^{\frac{2-s}{2}}+t^{2-s})^{-\frac{n-2}{2-s}}\right)t^{n-1}}{\left((r\epsilon)^{\frac{2-s}{2}}+t^{2-s}\right)^{\frac{n-s}{2-s}}}= \left\{ \begin{array}{l} O\left(\frac{1}{t^{n-1-s}}\right)\, \mbox{as}\,\quad t\rightarrow \infty\\ O\left(\left((r\epsilon)^{\frac{2-s}{2}}+t^{2-s}\right)^{\frac{b(n-2)-n+s}{2-s}}t^{n-1}\right)\, \mbox{as}\,\quad t\rightarrow 0, \end{array} \right. \end{align*}

uniformly for $r\in (0,1]$. Hence by dominated convergence theorem, the first term becomes

\begin{align*}\begin{aligned} &-\left(\frac{n-2}{2}\right)\epsilon^{1-\frac s2}\int_0^1 r^{-\frac s2} dr \left( \int_{0}^{\infty} \rho\left( t^{-(n-2)}\right) t^{-1+s} dt + o(1)\right) \\ &=-\frac{n-2}{2-s}\epsilon^{\frac{2-s}{2}} \int_{0}^{\infty} \rho\left( t^{-(n-2)}\right) t^{-1+s} dt + o\left(\epsilon^{\frac{2-s}{2}}\right) = -\frac{\epsilon^{\frac{2-s}{2}}}{2-s} \int_{0}^{\infty} \rho ( t ) t^{-\frac{n-2+s}{n-2}} dt + o\left(\epsilon^{\frac{2-s}{2}}\right). \end{aligned}\end{align*}

Also, since $\tilde{\rho}(t)=O(t^2)$ as $t\rightarrow 0$, the second term of the right-hand side is of $O(\epsilon^{\frac n2 -2})$ as $n\ge 5$ as $\epsilon\rightarrow0$. Substituting the above estimates in (5.1) establishes the first assertion of the lemma.

Now, we prove the (ii)

\begin{align*} \int_{B(R)} \rho(\alpha_0 V_{\epsilon})U_{\epsilon} = \int_{B(R)} \rho(\alpha_0 U_{\epsilon})U_{\epsilon} -\alpha_0 \int_{B(R)} \left(\int_0^1 \rho^{\prime}(\alpha_0(U_{\epsilon}- t U_\epsilon(R)))U_\epsilon (R)U_{\epsilon} dt \right) dx. \end{align*}

Now, changing the variable as $ r= t {a^{\frac{1}{2-s}}\epsilon^{\frac{1}{2}}}$, we obtain

\begin{align*}\begin{aligned} \int_{B(R)} \rho(\alpha_0U_{\epsilon})U_{\epsilon} =& \omega_{n-1} \int_0^R \rho(\alpha_0 U_{\epsilon}(r))U_{\epsilon}(r) r^{n-1} dr \\ = & \omega_{n-1} a^{\frac{n}{2-s}}\epsilon^{\frac{n}{2}} \int_{0}^{R/{a^{\frac{1}{2-s}}\epsilon^{\frac{1}{2}}}}\frac{t^{n-1}}{\left(\epsilon^{\frac{2-s}{2}}+t^{2-s}\right)^{\frac{n-2}{2-s}}} \rho\left(\alpha_{0}\left(\epsilon^{\frac{2-s}{2}}+t^{2-s}\right)^{-\frac{n-2}{2-s}}\right)dt. \end{aligned}\end{align*}

Dividing the integral into two as 0 to 1 and 1 to ${R/{a^{\frac{1}{2-s}}\epsilon^{\frac{1}{2}}}}$ and using ρ is bounded near $\infty$, $\rho(t)=O(t)$ as $t\rightarrow 0$, and α ₀ is bounded away from zero and infinity, we obtain the integral is bounded uniformly in ϵ. Therefore,

\begin{align*} \int_{B(R)} \rho(\alpha_0U_{\epsilon})U_{\epsilon} = O\left(\epsilon^{\frac{n}{2}}\right).\end{align*}

Using $|\rho^{\prime}(t)|\leq C $ and $U_{\epsilon}(R)= O\left(\epsilon^{\frac{n-2}{2}}\right)$, we obtain

\begin{align*} \int_{B(R)} \left(\int_{0}^{1} \rho^{\prime}(\alpha_0 (U_{\epsilon}-t U_{\epsilon}(R))) U_\epsilon(R)U_{\epsilon} dt\right) dx = O\left(\epsilon^{\frac{n-2}{2}}\int_{B(R)} U_{\epsilon} \right)= O\left(\epsilon^{n-2}\right). \end{align*}

Hence

\begin{align*} \int_{B(R)}\rho^{\prime}(\alpha_0(U_{\epsilon}-U_\epsilon(R))) U_\epsilon(R)U_{\epsilon} = O\left(\epsilon^{\frac{n}{2}}\right). \end{align*}

This proves (ii) and hence the lemma.$\square$

Proof. Using Taylor expansion and the inequality (due to convexity of the map $s\rightarrow |s|^{2^*(s)}$)

\begin{align*}|U_{\epsilon} - U_\epsilon(R)|^{2^*(s)}\geq |U_{\epsilon}|^{2^*(s)} -2^*(s) U_{\epsilon}^{2^*(s)-1} U_\epsilon(R),\end{align*}

we obtain

(5.2)\begin{align} \begin{aligned} J(V_\epsilon)&=\frac{1}{2}\int_{B(R)} |\nabla U_{\epsilon}|^2 -\frac{1}{2^*(s)}\int_{B(R)} \frac{|U_{\epsilon}-U_\epsilon(R)|^{2^*(s)}}{|x|^s} -\int_{B(R)} \tilde{\rho}(U_\epsilon-U_\epsilon(R)) \\ &\leq \frac{1}{2}\int_{B(R)} |\nabla U_{\epsilon}|^2 -\frac{1}{2^*(s)}\int_{B(R)} \frac{|U_{\epsilon}|^{2^*(s)}}{|x|^s} -\int_{B(R)} \tilde{\rho}(U_{\epsilon})+\int_{B(R)} \frac{U_{\epsilon}^{2^*(s)-1} U_\epsilon(R)}{|x|^s} \\ & \quad+\int_{B(R)} \rho(U_{\epsilon}) U_\epsilon(R) -\int_{B(R)} \int_{0}^{1}\rho^{\prime}(U_{\epsilon} -t U_{\epsilon}(R)) U_{\epsilon}(R)^2 (1-t) dt dx. \end{aligned} \end{align}

Equations (2.2) and (2.5) yield

(5.3)\begin{align} U_{\epsilon}(R)\int_{B(R)} \frac{U_{\epsilon}^{2^*(s)-1}}{|x|^s} = O\left(\epsilon^{\frac{n-2}{2}}\times\epsilon^{\frac{n-2}{2}}(O(1)+\epsilon^{2-s})\right) = O\left(\epsilon^{n-2}\right). \end{align}

Using $|\rho(t)|\leq C t$, we have

(5.4)\begin{align} U_{\epsilon}(R) \int_{B(R)} \rho(U_{\epsilon})= O\left(\epsilon^{\frac{n-2}{2}}\epsilon^{\frac{n-2}{2}}\right)= O\left(\epsilon^{n-2}\right). \end{align}

Now $\rho^{\prime}$ is bounded and $U_{\epsilon}(R)= O\left( \epsilon^{\frac{n-2}{2}}\right)$ give

(5.5)\begin{align} \int_{B(R)} \int_{0}^{1}\rho^{\prime}(U_{\epsilon} -t U_{\epsilon}(R)) U_{\epsilon}(R)^2 (1-t) dt dx = O\left(\epsilon^{n-2}\right). \end{align}

Using Eqs. (2.3), (2.4), (5.3), (5.4), and (5.5) in (5.2), we obtain

\begin{align*} J(V_\epsilon) \leq \frac{2-s}{2(n-s)}A_{s}^{\frac{n-s}{2-s}} -\int_{B(R)} \tilde{\rho}(U_\epsilon) + O\left(\epsilon^{n-2}\right). \end{align*}

Employing lemma 5.1 in the above relation, we obtain the required result.$\square$

Proof. In view of lemma 5.2, enough to prove that

\begin{align*}\sup_{\alpha \gt 0} J(\alpha V_\epsilon)= J(V_\epsilon) + O\left(\epsilon^{n-2}\right)\,\mbox{as}\, \epsilon\rightarrow 0.\end{align*}

Since $J(\alpha V_\epsilon)\rightarrow -\infty$ as $\alpha\rightarrow \infty$. It follows that there exist an $\alpha_0 \gt 0$ such that

\begin{align*}J(\alpha_0 V_\epsilon)=\sup_{\alpha\in\mathbb R} J(\alpha V_\epsilon)\,\mbox{and}\, \frac{d}{d\alpha} J(\alpha V_\epsilon)\mid_{\alpha=\alpha_0} =0.\end{align*}

This implies that

(5.6)\begin{align} \alpha_{0} \int_{B(R)} |\nabla V_{\epsilon}|^2 = \alpha_{0}^{2^*(s)-1} \int_{B(R)} \frac{|V_\epsilon|^{2^*(s)}}{|x|^s} + \int_{B(R)} \rho(\alpha_0 V_\epsilon) V_\epsilon. \end{align}

Using (2.3), we obtain

(5.7)\begin{align} \alpha_{0}\int_{B(R)} |\nabla V_{\epsilon}|^2 = \alpha_0 \int_{\mathbb R^n} |\nabla U_{\epsilon}|^2 + O\left(\alpha_0 \epsilon^{n-2}\right). \end{align}

Employing (2.4), (2.5), and Taylor expansion, we have

(5.8)\begin{align} \alpha_0^{2^*(s)-1} \int_{B(R)} \frac{|V_\epsilon|^{2^*(s)}}{|x|^s} = \alpha_0^{2^*(s)-1} \int_{\mathbb R^n} \frac{|U_\epsilon|^{2^*(s)}}{|x|^s} +O\left(\alpha_0^{2^*(s)-1} \epsilon^{\frac{n}{2}}\right). \end{align}

From (2.5) and using $|\rho(t)|\leq C t$, C > 0, we obtain

(5.9)\begin{align} \int_{B(R)} \rho(\alpha_0 V_\epsilon) V_\epsilon = O\left(\alpha_0 \int_{B(R)} |U_{\epsilon}|^2 +\alpha_{0} |U_{\epsilon}(R)|^2 \right) = O\left(\alpha_{0} \epsilon^{2}\right). \end{align}

Thus, (5.6), (5.7), (5.8), and (5.9) together implies that α ₀ is bounded away from 0 and $\infty$ as $\epsilon\rightarrow 0$. Again using $|\rho(t)|\leq C t$, C > 0, we obtain

(5.10)\begin{align} \int_{B(R)} \rho(\alpha_0 V_\epsilon) V_\epsilon = \int_{B(R)} \rho(\alpha_0 V_{\epsilon}) U_\epsilon + O\left(\int_{B(R)} (|U_{\epsilon}| |U_\epsilon(R)|+|U_\epsilon(R)|^2) \right)= O\left(\epsilon^{\frac n2}\right). \end{align}

Hence, Eqs. (5.6), (5.7), (5.8), and (5.10) yield

\begin{align*}\left(1-\alpha_0^{2^*(s)-2}\right) \int_{\mathbb R^n} |\nabla U_{\epsilon}|^2 = O\left(\epsilon^{\frac{n}{2}}\right).\end{align*}

Thus, we obtain

(5.11)\begin{align} 1-\alpha_0= O\left(\epsilon^{\frac{n}{2}}\right). \end{align}

Now using (5.6), we have

(5.12)\begin{align} \begin{aligned} J(\alpha_{0} V_{\epsilon})- J(V_{\epsilon})&= \left( \frac{\alpha_{0}^{2^*(s)}}{2}- \frac{\alpha_{0}^{2^*(s)-2}}{2}-\frac{\alpha_{0}^{2^*(s)}}{2^*(s)}+ \frac{1}{2^*(s)}\right)\int_{B(R)} \frac{|V_\epsilon|^{2^*(s)}}{|x|^s} \\ &+ \left(\frac{\alpha_0}{2}- \frac{1}{2\alpha_0} \right)\int_{B(R)} \rho(\alpha_0 V_\epsilon) V_\epsilon +\int_{B(R)} \tilde{\rho}(V_\epsilon) - \int_{B(R)} \tilde{\rho}(\alpha_0 V_\epsilon). \end{aligned} \end{align}

Let $g(\alpha)=\frac{\alpha^{2^*(s)}}{2}- \frac{\alpha^{2^*(s)-2}}{2}-\frac{\alpha^{2^*(s)}}{2^*(s)}+ \frac{1}{2^*(s)}$. Then $g(1)= g^{\prime}(1)=0$. Hence

(5.13)\begin{align} g(\alpha_0)=O((1-\alpha_0)^2). \end{align}

Let

\begin{align*}f_{\epsilon}(\alpha)= \frac{1}{2}\left(\alpha- \frac{1}{\alpha}\right)\int_{B(R)} \rho(\alpha V_\epsilon) V_\epsilon +\int_{B(R)} \tilde{\rho}(V_\epsilon) - \int_{B(R)} \tilde{\rho}(\alpha V_\epsilon).\end{align*}

Then

\begin{align*}\begin{aligned} f_{\epsilon}^{\prime}(\alpha)&= (1-\alpha^2) \left[\frac{1}{2\alpha^2 }\int_{B(R)} \rho(\alpha V_\epsilon) V_\epsilon - \frac{1}{2\alpha}\int_{B(R)} \rho^{\prime}(\alpha V_\epsilon) V^2_{\epsilon}\right]\\ &= (1-\alpha^2) O\left(\int_{B(R)} V_{\epsilon}^2\right)\\ &= O((1-\alpha)),\,\mbox{uniformly in}\,\epsilon\,\mbox{and}\,\alpha\,\mbox{near one}. \end{aligned}\end{align*}

Now, $f_{\epsilon}(\alpha_0)= f_{\epsilon}(1)+\int_{0}^{1} f_{\epsilon}^{\prime}(1+t(\alpha_0-1)) (\alpha_0-1) dt= O((1-\alpha_0)^2)$. Using the above estimate, (5.13) in (5.12) and (5.11), we obtain

\begin{align*} J(\alpha_0 V_\epsilon) -J(V_\epsilon)= O((1-\alpha_0)^2) =O\left(\epsilon^{n}\right). \end{align*}

Hence $\sup_{\alpha \gt 0} J(\alpha V_\epsilon)= J(V_\epsilon) + O\left(\epsilon^{n-2}\right)$. This completes the proof.$\square$