1. Introduction and statement of main results
In this paper, we are interested in the following inequality
$$\begin{equation*} \int _\Omega \frac{|u(x)|^2}{d^{\alpha }(x)} dx \le C \int _\Omega d^{\beta }(x) |\nabla u(x) |^2 dx \quad \text{ with }\quad \int _\Omega \frac{u(x)}{d^{\alpha }(x)} dx=0, \end{equation*}$$
where $\alpha ,\beta \in \mathbb{R}$ , $d(x) \equiv \text{dist}(x,\partial \Omega )$ and $\Omega$ is a bounded $C^2$ -domain in $\mathbb{R}^n$ . This study was motivated by the Legendre differential equation. The Legendre differential equation
$$\begin{equation*} -((1-t^2)\phi ^\prime )^\prime = l(l+1)\phi , \ \ l = 0,1,\cdots , \end{equation*}$$
has the first kind of solution $P_l$ , which is a polynomial of order $l$ , and the second kind of solution $Q_l$ , which is singular at $t =\pm 1$ . Then we see that for any $\phi \in C^1((-1,1)) \cap L^2((-1,1))$ satisfying $\int _{-1}^{1}\phi dt = 0$ ,
$$\begin{equation} \int _{-1}^1 (1-t^2 )|\phi ^\prime (t)|^2 dt \ge 2\int _{-1}^1 (\phi (t))^2 dt.\cssId{leg}{\tag{1}} \end{equation}$$
Furthermore, the equality holds for $P_1(x) = x$ ; on the other hand, for $l=0,1,\cdots$ ,
$$\begin{equation*} \int _{-1}^1 (Q_l(t))^2 dt < \int _{-1}^1 (1-t^2 )|Q_l^\prime (t)|^2 dt = \infty . \end{equation*}$$
This can be compared with Wirtinger’s inequality, which says that for any $\phi \in C^1((-1,1)) \cap L^2((-1,1))$ satisfying $\int _{-1}^{1}\phi dt = 0$ ,
$$\begin{equation} \int _{-1}^1 |\phi ^\prime (t)|^2 dt \ge \frac{\pi ^2}{4}\int _{-1}^1 (\phi (t))^2 dt,\cssId{wir}{\tag{2}} \end{equation}$$
where the equality holds for $\phi (t) = \sin \frac{\pi t}{2}$ . On the other hand, a version of Hardy’s inequality says that for any $\phi \in W_0^{1,2}((-1,1))$ ,
$$\begin{equation} \int _{-1}^1 |\phi ^\prime (t)|^2 dt \ge \frac{1}{4}\int _{-1}^1 \frac{(\phi (t))^2}{(1-t^2)^2} dt,\cssId{har}{\tag{3}} \end{equation}$$
where the equality does not hold for any $\phi \in W_0^{1,2}((-1,1)) \setminus \{0\}$ .
These inequalities Equation 1 , Equation 2 , Equation 3 are quite fundamental and there have been numerous studies on higher dimensional versions of Equation 2 and Equation 3 . On the other hand, it seems that there have been no studies for the higher dimensional version of Equation 1 . In this paper, we study the high dimensional version of Equation 1 and find a general form of inequalities that includes Equation 1 , Equation 2 , Equation 3 . For their generalization on a bounded domain $\Omega$ in $\mathbb{R}^N$ , we define $d(x) = \text{dist}(x, \mathbb{R}^N \setminus \Omega )$ and
$$\begin{equation*} \mathcal{X}_{\alpha ,\beta }(\Omega ) \equiv \Big \{\phi \in C^1( \Omega )\ \Big |\ \int _\Omega d^{\beta }(x) |\nabla \phi (x)|^2+\frac{(\phi (x))^2}{d^\alpha (x)} dx <\infty \Big \}, \ \alpha ,\beta \in \mathbb{R}. \end{equation*}$$
For $u\in \mathcal{X}_{\alpha ,\beta }(\Omega )$ , we define a norm $\Vert u \Vert _{\alpha ,\beta } \equiv \Big ( \int _\Omega d^{\beta }(x) |\nabla u|^2 + d^{-\alpha }(x)u^2 dx\Big )^{1/2}$ and $W_{\alpha ,\beta }^{1,2}(\Omega )$ the completion of $\mathcal{X}_{\alpha ,\beta }(\Omega )$ with respect to the norm $\Vert \cdot \Vert _{\alpha ,\beta }$ . Then we consider a minimization
$$\begin{equation*} L_{\alpha ,\beta }(\Omega ) \equiv \inf \Big \{ \frac{\int _\Omega d^{\beta }(x) |\nabla u |^2 dx}{\int _\Omega |u|^2d^{-\alpha }(x) dx} \ \Big | \ u \in W_{\alpha ,\beta }^{1,2}(\Omega )\setminus \{0\}, \int _\Omega u d^{-\alpha }(x) dx=0 \Big \}. \end{equation*}$$
In this paper, we characterize $(\alpha ,\beta ) \in \mathbb{R}^2$ for which $L_{\alpha ,\beta }$ is positive, and find conditions of $(\alpha ,\beta )$ and domain $\Omega$ under which $L_{\alpha ,\beta }> 0$ is attained. In fact, we will see that if $(\alpha ,\beta )$ is in a subcritical region, the attainability of $L_{\alpha ,\beta }$ is determined by the comparison of two energy levels $L_{\alpha ,\beta }$ and $H_{\alpha ,\beta }$ , where $H_{\alpha ,\beta }$ is defined in Equation 4 . For $(\alpha ,\beta )$ in the critical region $\alpha + \beta =2$ , there exists a critical level of $L_{\alpha ,\beta }$ , where we lose a compactness of a minimizing sequence. Then, for some typical domains, we find conditions of $(\alpha ,\beta )$ in the critical region $\alpha + \beta =2$ under which $L_{\alpha ,\beta }$ either is strictly less than the critical level of $L_{\alpha ,\beta }$ or is equal to a critical level of $L_{\alpha ,\beta }$ . In the same direction, there have been many studies (refer to Reference 3 ,Reference 4 ,Reference 5 ,Reference 6 ,Reference 7 ,Reference 8 , Reference 9 , Reference 10 ,Reference 12 ,Reference 13 ,Reference 14 and references therein) on a generalized Hardy’s inequality, where they study the following optimal constant
$$\begin{equation} H_{\alpha ,\beta }(\Omega ) \equiv \inf \Big \{ \frac{\int _\Omega d^{\beta }(x) |\nabla u |^2 dx}{\int _\Omega |u |^2d^{-\alpha }(x) dx} \ \Big | \ u\in W_{0,\alpha ,\beta }^{1,2}(\Omega )\setminus \{0\}\Big \}. \cssId{ohc}{\tag{4}} \end{equation}$$
Here, $W_{0,\alpha ,\beta }^{1,2}(\Omega )$ is the completion of $C_0^\infty (\Omega )$ with respect to the norm $\Vert \cdot \Vert _{\alpha ,\beta }$ . For a generalized Hardy’s inequality on general domains, refer to Reference 2 , Reference 15 , Reference 16 and references therein. Some general results on the positivity and attainability of $H_{\alpha ,\beta }$ will be given in Proposition 2.4 , Proposition A.2 and Proposition A.4 . We will show in Proposition 2.2 that for $\alpha \ge 1$ and $\beta <1$ , $W_{\alpha ,\beta }^{1,2}(\Omega )=W_{0,\alpha ,\beta }^{1,2}(\Omega )$ . Furthermore, we will prove in Appendix, Proposition A.1 that
$$\begin{equation*} W_{0,\alpha ,\beta }^{1,2}(\Omega ) \begin{cases} \supsetneq W_0^{1,2}(\Omega ) &\text{ if } (\alpha ,\beta )\in \{(a,b) \ | \ a\le 2, b>0\},\\ = W_0^{1,2}(\Omega ) &\text{ if } (\alpha ,\beta )\in \{(a,b) \ | \ a\le 2, b=0\},\\ \subsetneq W_0^{1,2}(\Omega ) &\text{ if } (\alpha ,\beta )\in \{(a,b) \ | \ b<0 \text{ or } b=0, a>2\}, \end{cases} \end{equation*}$$
and that for $(\alpha ,\beta )\in \{(a,b) \ | \ a> 2, b>0\}$ , $W_{0,\alpha ,\beta }^{1,2}(\Omega )\setminus W_0^{1,2}(\Omega )$ and $W_0^{1,2}(\Omega )\setminus W_{0,\alpha ,\beta }^{1,2}(\Omega )$ are non-empty sets. The property $W_{\alpha ,\beta }^{1,2}(\Omega )=W_{0,\alpha ,\beta }^{1,2}(\Omega )$ for $\alpha \ge 1$ and $\beta <1$ implies that there is a natural relation between $L_{\alpha ,\beta }(\Omega )$ and $H_{\alpha ,\beta }(\Omega )$ and that our study on $L_{\alpha ,\beta }(\Omega )$ could be a natural extension of previous studies on the Hardy constant $H_{\alpha ,\beta }(\Omega )$ .
Now we state our main results in this paper.
Theorem 1.1.
For any bounded $C^2$ -domain $\Omega \subset \mathbb{R}^N$ , $L_{\alpha ,\beta }(\Omega ) > 0$ if and only if $\alpha +\beta \le 2$ and $(\alpha ,\beta ) \neq (1,1)$ .
Theorem 1.2.
Let $\Omega$ be a bounded $C^2$ -domain in $\mathbb{R}^N$ . Then it holds that
(1) if $(\alpha ,\beta )\in \{(a,b)\ |\ a+b< 2, 2a+b<3\}$ , $L_{\alpha ,\beta }(\Omega )$ is achieved by an element $u_{\alpha ,\beta } \in W_{\alpha ,\beta }^{1,2}(\Omega )$ , which satisfies$$\begin{align*} - \text{div} (d^\beta (x)\nabla u_{\alpha ,\beta }) &= L_{\alpha ,\beta }(\Omega )d^{-\alpha }u_{\alpha ,\beta } \text{ in } \Omega \text{ if } \alpha < 1,\\ - \text{div} (d^\beta (x)\nabla u_{\alpha ,\beta }) &= L_{\alpha ,\beta }(\Omega )d^{-\alpha }u_{\alpha ,\beta } + \mu d^{-\alpha } \text{ in } \Omega \text{ for some } \mu \in \mathbb{R}\text{ if } \alpha \ge 1; \end{align*}$$
(2) if $(\alpha ,\beta )\in \{(a,b)\ |\ a+b\le 2, 2a+b\ge 3, (a,b) \ne ( 1,1)\}$ , $L_{\alpha ,\beta }(\Omega ) = H_{\alpha ,\beta }(\Omega )$ and $L_{\alpha ,\beta }(\Omega )$ is not achieved.
The key features and their relation for $H_{\alpha ,\beta }(\Omega )$ and $L_{\alpha ,\beta }(\Omega )$ described in Theorem 1.1 and Theorem 1.2 are illustrated in Figure 1 . A result in Theorem 1.2 implies that $L_{\alpha ,2-\alpha }(\Omega )$ is not attained for $\alpha > 1$ . When $\alpha < 1$ , as in the Hardy inequality, we lose a compactness at a critical level of $L_{\alpha ,2-\alpha }(\Omega )$ . In fact, we have the following result.
Theorem 1.3.
Let $\alpha < 1$ and $\Omega$ be a bounded $C^2$ -domain in $\mathbb{R}^N$ . Then $L_{\alpha ,2-\alpha }(\Omega ) \le \frac{(1-\alpha )^2}{4}$ . Furthermore, $L_{\alpha ,2-\alpha }(\Omega )$ is achieved if $L_{\alpha ,2-\alpha }(\Omega )<\frac{(1-\alpha )^2}{4}$ .
As for the optimal constant $H_{2,0}(\Omega )$ on $W^{1,2}_{0,2,0}(\Omega )=W^{1,2}_0(\Omega )$ , it was proved Reference 12 that $H_{2,0}(\Omega )$ is achieved if and only if $H_{2,0}(\Omega )<\frac{1}{4}$ when $\Omega$ is a bounded $C^2$ -domain in $\mathbb{R}^N$ . Recently, the result was extended in Reference 9 to $C^{1,\gamma }-$ domains for $\gamma \in (0,1]$ . In view of the result in Reference 9 , we believe that the result of Theorem 1.3 would be true for bounded $C^{1,\gamma }$ domains. On the other hand, it is known that $H_{2,0}(\Omega )$ is not attained by any function $u \in W_0^{1,2}(\Omega )$ if $\Omega$ is convex (see Reference 3 ), and if $\Omega$ is weakly mean convex (see Reference 10 ), that is, the mean curvature of $\partial \Omega$ is nonnegative. On the other hand, as for the optimal constant $L_{\alpha ,2-\alpha }(\Omega )$ , we have the following results, highly contrasting with the results for $H_{2,0}(\Omega )$ .
Theorem 1.4.
For a weakly mean convex $C^2$ -domain $\Omega \subset \mathbb{R}^N$ , there exists $\tilde{\alpha } \equiv \tilde{\alpha }(N,\Omega ) < 1$ such that $L_{\alpha ,2-\alpha }(\Omega )$ is achieved if $-\infty <\alpha <\tilde{\alpha }$ .
We will see that the attainability of the optimal constant $L_{\alpha ,2-\alpha }(\Omega )$ strongly depends on a geometry of $\Omega$ and the space dimension $N$ . To state the result, we define
$$\begin{align*} R^N_{+} &\equiv \{(x_1,\cdots , x_N)\in \mathbb{R}^N \ |\ x_N>0\},\\ B^N(0,1) &\equiv \Big \{(x_1,\cdots , x_N)\in \mathbb{R}^N\ \Big |\ \sum _{i=1}^N(x_i)^2< 1 \Big \}\\ E(a_1,\cdots , a_N)&\equiv \Big \{(x_1,\cdots , x_N)\in \mathbb{R}^N\ \Big |\ \sum _{j=1}^N \Big (\frac{x_j }{a_j}\Big )^2< 1 \Big \}, \end{align*}$$
where $a_j>0$ . For the most simple domain $\Omega = B^N(0,1)$ , we obtain that $L_{\alpha ,2-\alpha }(B^N(0,1))$ is attained only when $N > \frac{3-\alpha }{1-\alpha }$ .
Theorem 1.5.
Let $\alpha <1$ and $N\ge 1$ . Then $L_{\alpha ,2-\alpha }(B^N(0,1))<\frac{(1-\alpha )^2}{4}$ and $L_{\alpha ,2-\alpha }(B^N(0,1))$ is achieved if and only if the space dimension $N > \frac{3-\alpha }{1-\alpha }$ .
When we deform the unit ball $B^N(0,1)$ to an ellipse, we get the following result, which says that even for $N = \frac{3-\alpha }{1-\alpha }$ , $L_{\alpha ,2-\alpha }(E(a_1,\cdots , a_N))$ is attained if the ellipse $E(a_1,\cdots , a_N)$ is not a ball.
Theorem 1.6.
For $\alpha <1$ , we assume the space dimension $N \ge \frac{3-\alpha }{1-\alpha }$ . Then for $(a_1,\cdots ,a_{N-1})\in (0,1]^{N-1}$ with $a_1+\cdots +a_{N-1} < N-1$ , we have
$$\begin{equation*} L_{\alpha ,2-\alpha }(E(a_1,a_2,\cdots , a_{N-1},1)) < \frac{(1-\alpha )^2}{4}; \end{equation*}$$
thus, the optimal constant $L_{\alpha ,2-\alpha }(E(a_1,a_2,\cdots , a_{N-1},1))$ is attained.
In Theorems 1.7 –1.9 , we state our results for annular domains.
Theorem 1.7.
For $a > 1$ and $N\ge 2$ , let $\Omega = B^N(0,a)\setminus \overline{B^N(0,1)}$ . Then, for
$$\begin{equation*} \alpha <\frac{1-\frac{2}{N+1}\Big (2-(a+1)^{N+1}2^{-N+1}(1+a^{N+1})^{-1}\Big )}{1-\frac{1}{N+1}\Big (2-(a+1)^{N+1}2^{-N+1}(1+a^{N+1})^{-1} \Big )}, \end{equation*}$$
$L_{\alpha ,2-\alpha }(\Omega )<\frac{(1-\alpha )^2}{4}$ and $L_{\alpha ,2-\alpha }(\Omega )$ is achieved.
Theorem 1.8.
For $a > 1$ and $N\ge 2$ , let $\Omega =B^N(0,a)\setminus \overline{B^N(0,1)}$ . Then there exists $\alpha _0\equiv \alpha _0(N,\Omega )\in (0,1)$ such that $L_{\alpha ,2-\alpha }(\Omega )$ is not achieved for $\alpha \in (\alpha _0,1)$ .
We state the attainable property of $L_{\alpha ,2-\alpha }([-1,1]^N)$ without a proof. In fact, using the same type of a test function used in the proof of Theorem 1.5 , we can show that $L_{\alpha ,2-\alpha }([-1,1]^N)<\frac{(1-\alpha )^2}{4}$ for $N>\frac{2}{3(1-\alpha )}+1$ .
Above results suggest that for a general bounded domain $\Omega$ , if $N$ or $-\alpha$ is large, there is a high possibility that $L_{\alpha ,2-\alpha }(\Omega )$ is attained; on the other hand, if $N$ or $1-\alpha > 0$ is small, a low possibility. We conjecture that for any given bounded $C^2$ -domain in $\mathbb{R}^N$ , $L_{\alpha ,2-\alpha }(\Omega )$ is attained if $-\alpha$ is sufficiently large, and not attained if $1-\alpha > 0$ is sufficiently small.
In the following result, we see that the attainable property of $L_{\alpha ,2-\alpha }(\Omega )$ depends on a local geometry of $\Omega$ rather than a global geometry of $\Omega$ . For $\alpha <1$ , we define a class of domains
$$\begin{equation*} \mathcal{A}_\alpha \equiv \Big \{\text{ bounded $C^2$-domains } D \subset \mathbb{R}^N \ \Big | \ L_{\alpha ,2-\alpha }(D) < \frac{(1-\alpha )^2}{4}\Big \}, \end{equation*}$$
and for $D \in \mathcal{A}_\alpha$ , a class of functions
$$\begin{equation*} \mathcal{A}_\alpha (D) \equiv \Big \{u \in C^1(D) \cap W_{\alpha ,2-\alpha }^{1,2}(D) \setminus \{0\} \ \Big | \ Q_D(u) < \frac{(1-\alpha )^2}{4}, \int _D d^{-\alpha }(x)u(x) dx = 0 \Big \}, \end{equation*}$$
where
$$\begin{equation*} Q_D(u) \equiv \frac{\int _D d^{2-\alpha }(x) |\nabla u |^2 dx}{\int _D |u|^2d^{-\alpha }(x) dx}. \end{equation*}$$
For each $u \in \mathcal{A}_\alpha (D)$ , we define
$$\begin{align*} D^u_+&\equiv \{x \in D \ | \ u(x) > 0\}, \ u^D_+(x) \equiv \max \{u(x),0\},\\ D^u_- &\equiv \{x \in D \ | \ u(x) < 0\}\ \text{ and }\ \ u^D_- \equiv \min \{u(x),0\}. \end{align*}$$
Also, we define $\partial D^u_+\equiv \partial D^{u}_+ \cap \partial D, \partial D^u_-\equiv \partial D^{u}_- \cap \partial D$ . For each $u \in \mathcal{A}_\alpha (D)$ , it holds that $Q_D(u^D_+)< \frac{(1-\alpha )^2}{4}$ or $Q_D(u^D_-)< \frac{(1-\alpha )^2}{4}.$ Defining $\tilde{u}=-u$ , we see that $\tilde{u}^D_+ = u^D_-, \tilde{u}^D_- = u^D_+$ . Thus, for $u \in \mathcal{A}_\alpha (D)$ , we may assume that $Q_D(u^D_+)< \frac{(1-\alpha )^2}{4}.$
Theorem 1.9.
Let $\alpha < 1$ and $\Omega$ be a bounded $C^2$ -domain in $\mathbb{R}^N$ . We assume that for each $i=1,2$ , there exist a domain $D_i \in \mathcal{A}_\alpha$ , a function $u_i \in \mathcal{A}_\alpha (D_i)$ , a rotation $O_i \in O(N)$ , a translation $t_i \in \mathbb{R}^N$ and a scale $s_i > 0$ such that for each $i=1,2$ , $t_i+s_iO_iD^{u_i}_+ \subset \Omega , \ \ t_i+s_iO_i\partial D^{u_i}_+ \subset \partial \Omega$ with $(t_1+s_1O_1D^{u_1}_+) \cap (t_2+s_2O_2D^{u_2}_+) = \emptyset$ and dist$(t_i+s_iO_ix,\mathbb{R}^N \setminus \Omega ) = s_i$ dist $(x,\mathbb{R}^N \setminus D_i)$ for any $x \in D^{u_i}_+$ . Then, $L_{\alpha ,2-\alpha }(\Omega )<\frac{(1-\alpha )^2}{4}$ and $L_{\alpha ,2-\alpha }(\Omega )$ is achieved.
In the proofs of Theorems 1.5 –1.8 for the estimate $L_{\alpha ,2-\alpha }(\Omega ) < \frac{(1-\alpha )^2}{4}$ , we use a test function of form $d^{s}(x)x_N$ . Thus we see from Theorem 1.9 that if $\Omega$ has two disjoint parts coming from translations, scalings and rotations of a half ball or a half ellipse or a half annulus as in Figure 2 , $L_{\alpha ,2-\alpha }(\Omega )$ is attained for $\alpha$ satisfying conditions in the corresponding theorem for ball, ellipse, or annulus.
This paper is organized as follows. Section 2 is devoted to give some preliminary results for proofs of main results. In Section 3 , we prove Theorem 1.1 and Theorem 1.2 by examining the positivity and attainability of $L_{\alpha ,\beta }(\Omega )$ in a subcritical region. In Section 4 , we study the positivity and attainability of $L_{\alpha ,\beta }(\Omega )$ in a critical region and prove Theorems 1.3 $-$ 1.9 . In the last appendix section, we study the inclusion relation between $W_0^{1,2}(\Omega )$ and $W_{0,\alpha ,\beta }^{1,2}(\Omega )$ and prove some general results on the positivity and attainability of $H_{\alpha ,\beta }(\Omega )$ .
2. Some preliminary Lemmas
Assume that $\Omega \subset \mathbb{R}^N$ is a bounded $C^2$ -domain. Now we follow a scheme in section 1 of Reference 3 . For $\delta >0$ , we define $\Omega _\delta \equiv \{ x \in \Omega \ | \ d(x) < \delta \}$ and $\Sigma _\delta = \{ x \in \Omega \ | \ d(x) = \delta \}$ . If $\delta > 0$ is small, for every $x\in \Omega _\delta$ , there exists a unique point $\sigma (x)\in \partial \Omega$ such that $d(x)=|x-\sigma (x)|$ . For small $\delta > 0$ , we define a map $H_\delta : \partial \Omega \rightarrow \Sigma _\delta$ defined by $H_\delta (\sigma )=\sigma +\delta n(\sigma )$ , where $n(\sigma )$ is the inward unit normal to $\partial \Omega$ at $\sigma$ . We define a mapping $\Pi : \Omega _\delta \to (0,\delta )\times \partial \Omega$ by $\Pi (x) = (d(x),\sigma (x))$ . Then, it is a $C^1$ -diffeomorphism and its inverse is given by $\Pi ^{-1}(t,\sigma ) = \sigma +tn(\sigma )$ for $\ t \in (0,\delta ), \ \sigma \in \partial \Omega$ . It is easy to see that for small $\delta > 0$ , there exists $c > 0$ such that
$$\begin{equation*} \Big ||\nabla H_t(\sigma )|-1\Big |\le ct,\ \forall (t,\sigma )\in (0,\delta _0)\times \partial \Omega , \end{equation*}$$
where $|\nabla H_t|$ is the Jacobin determinant of the Jacobian $\nabla H_t$ on $\partial \Omega$ .
Then, since for any nonnegative integrable function $f$ in $\Omega _\delta$ ,
$$\begin{equation*} \int _{\Omega _\delta }f(x) dx = \int _0^\delta \int _{\Sigma _t}f d\sigma _t dt = \int _0^\delta \int _{\partial \Omega } f\circ \Pi ^{-1}(t,\sigma )|\nabla H_t(\sigma )| d \sigma dt, \end{equation*}$$
it follows that for any nonnegative integrable function $f$ in $\Omega _\delta$ ,
$$\begin{equation} \int _{\partial \Omega }\int _0^\delta f\circ \Pi ^{-1}(t,\sigma )(1-ct)d\sigma dt \le \int _{\Omega _\delta } f(x) dx \le \int _{\partial \Omega }\int _0^\delta f\circ \Pi ^{-1}(t,\sigma )(1+ct)d\sigma dt. \cssId{bdryint}{\tag{5}} \end{equation}$$
Lemma 2.1.
Let $\beta <1$ and $u\in C^1((0,1))$ with $u(0)=0$ and $\int _0^1 t^\beta (u^\prime )^2dt<\infty$ . Then we see that for $t\in (0,1)$ ,
(i) $u(t)\le \frac{1}{(1-\beta )^\frac{1}{2}} \Big (\int _0^t s^\beta \Big (u^\prime (s)\Big )^2ds\Big )^\frac{1}{2} t^{\frac{1-\beta }{2}};$
(ii) $\int _0^tx^\beta u^2(x)dx\le \frac{t^2}{2(1-\beta )} \int _0^t s^\beta \Big (u^\prime (s)\Big )^2ds.$
Proof.
Since $u(0)=0$ and $\int _0^1 t^\beta (u^\prime )^2dt<\infty$ , for $x \in (0,1)$ ,
$$\begin{equation*} \begin{split} u(x)&=\int _0^xu^\prime (s)ds\le \Big (\int _0^xs^\beta \Big (u^\prime (s)\Big )^2ds\Big )^\frac{1}{2}\Big ( \int _0^x s^{-\beta }ds\Big )^\frac{1}{2}\\ &=\Big (\int _0^xs^\beta \Big (u^\prime (s)\Big )^2ds\Big )^\frac{1}{2}\Big ( \frac{1}{1-\beta }x^{1-\beta }\Big )^\frac{1}{2}. \end{split} \end{equation*}$$
This proves (i). From the inequality (i), we see that for $t\in (0,1)$ ,
$$\begin{equation*} \int _0^tx^\beta u^2(x)dx\le \frac{1}{1-\beta } \int _0^t s^\beta \Big (u^\prime (s)\Big )^2ds\int _0^t x dx=\frac{t^2}{2(1-\beta )} \int _0^t s^\beta \Big (u^\prime (s)\Big )^2ds; \end{equation*}$$
this proves (ii).
■
Proposition 2.2.
For $\alpha \ge 1$ and $\beta <1$ , we have
$$\begin{equation*} W_{\alpha ,\beta }^{1,2}(\Omega )=W_{0,\alpha ,\beta }^{1,2}(\Omega ). \end{equation*}$$
Proof.
It is obvious that $W_{\alpha ,\beta }^{1,2}(\Omega ) \supset W_{0,\alpha ,\beta }^{1,2}(\Omega )$ . To prove that $W_{\alpha ,\beta }^{1,2}(\Omega )\subset W_{0,\alpha ,\beta }^{1,2}(\Omega )$ , we note that if $\alpha \ge \beta$ , for $w\in C^\infty ((0,1))$ satisfying $\int _0^1 t^\beta (w^\prime )^2dt<\infty , \int _0^1 t^{-\alpha } w^2 dt<\infty$ , we have
$$\begin{equation*} \begin{split} w^2(y)-w^2(x)=\int _x^y \Big (w^2(t)\Big )^\prime dt&\le 2\Big (\int _x^y \frac{w^2}{t^\alpha }dt\Big )^\frac{1}{2}\Big (\int _x^y t^\alpha (w^\prime )^2dt\Big )^\frac{1}{2}\\ &\le 2\Big (\int _x^y \frac{w^2}{t^\alpha }dt\Big )^\frac{1}{2}\Big (\int _x^y t^\beta (w^\prime )^2dt\Big )^\frac{1}{2}, \end{split} \end{equation*}$$
where $0<x<y<1$ . Then we see that $\lim _{x\rightarrow 0}w(x)$ exists. From this and the assumptions that $\alpha \ge 1$ and $\int _0^1 \frac{w^2}{t^\alpha }dt<\infty$ , we deduce that $w(0)=0$ . This and Equation 5 imply that if $\alpha \ge \beta$ and $\alpha \ge 1$ , then
$$\begin{equation} u=0 \ \text{ on }\ \partial \Omega \ \text{ for }\ u\in \mathcal{X}_{\alpha ,\beta }(\Omega ). \cssId{vns1}{\tag{6}} \end{equation}$$
We find a function $\varphi \in C^1(\mathbb{R})$ such that $\varphi (t) = 0$ for $t \le 1$ and $\varphi (t) =1$ for $t \ge 2$ . For $u \in \mathcal{X}_{\alpha ,\beta }(\Omega )$ , we define $u_n(x)=u(x)\varphi (nd(x))$ . Then we claim that $\|u_n-u\|_{ \alpha ,\beta }\rightarrow 0$ as $n\rightarrow \infty$ . To prove the claim, it suffices to show that for a neighborhood $N(x_0)$ of $x_0 \in \partial \Omega$ ,
$$\begin{equation*} \lim _{n \to \infty } \int _{\Omega \cap N(x_0)} d^\beta |\nabla (u-u_n)|^2 + d^{-\alpha }(u-u_n)^2 dx = 0. \end{equation*}$$
By the dominated convergence theorem, we see that
$$\begin{equation*} \lim _{n \to \infty } \int _{\Omega \cap N(x_0)} d^{-\alpha }(u-u_n)^2 dx = \lim _{n \to \infty } \int _{\Omega \cap N(x_0)} d^{-\alpha }u^2\Big (1-\varphi ^2(nd(x))\Big ) dx = 0. \end{equation*}$$
To prove $\lim _{n \to \infty } \int _{\Omega \cap N(x_0)} d^\beta |\nabla (u-u_n)|^2 dx = 0$ , we take a small neighborhood $N(x_0)$ of $x_0 \in \partial \Omega$ and a $C^1$ flattening map $\Phi : (-\delta ,\delta )^N \cap \mathbb{R}^N_+ \to \Omega \cap N(x_0)$ for a small $\delta > 0$ such that $d\circ \Phi (x_1,\cdots ,x_N) = x_N$ and
$$\begin{equation*} \begin{split} \frac{1}{2}\int _{(-\delta ,\delta )^N \cap \mathbb{R}^N_+}(x_N)^\beta |\nabla (v\circ \Phi )|^2 dx &\le \int _{\Omega \cap N(x_0)} d^\beta |\nabla v|^2 dx\\ &\le 2 \int _{(-\delta ,\delta )^N \cap \mathbb{R}^N_+}(x_N)^\beta |\nabla (v\circ \Phi )|^2 dx, \end{split} \end{equation*}$$
where $v\in W_{\alpha ,\beta }^{1,2}(\Omega )$ . Thus it suffices to show that
$$\begin{equation*} \lim _{n\to \infty }\int _{(-\delta ,\delta )^N \cap \mathbb{R}^N_+}(x_N)^\beta |\nabla (u\circ \Phi )-\nabla (u_n\circ \Phi )|^2 dx = 0. \end{equation*}$$
Since $u_n(\Phi (x)) = u(\Phi (x))\varphi (nd(\Phi (x))) = u(\Phi (x))\varphi (nx_N)$ , we see from the dominated convergence theorem that
$$\begin{equation*} \lim _{n\to \infty }\int _{(-\delta ,\delta )^N\cap \mathbb{R}^N_+}(x_N)^\beta \Big |\frac{\partial }{\partial x_i} u(\Phi (x))-\frac{\partial }{\partial x_i} u_n(\Phi (x))\Big |^2 dx = 0, \ i=1,\cdots ,N-1. \end{equation*}$$
For $i=N$ , we see that
$$\begin{align*} & \int _{(-\delta ,\delta )^N \cap \mathbb{R}^N_+}(x_N)^\beta \Big |\frac{\partial }{\partial x_N} u(\Phi (x))-\frac{\partial }{\partial x_N} u_n(\Phi (x))\Big |^2 dx\\ & \quad = \int _{(-\delta ,\delta )^N \cap \mathbb{R}^N_+}(x_N)^\beta \Big |\frac{\partial }{\partial x_N} u(\Phi (x))(1-\varphi (nx_N))\Big |^2 dx\\ & \quad \le 2 \int _{(-\delta ,\delta )^N \cap \mathbb{R}^N_+}(x_N)^\beta \Big |\frac{\partial }{\partial x_N} u(\Phi (x))\Big |^2(1-\varphi (nx_N))^2 dx\\ & \qquad + 2 \int _{(-\delta ,\delta )^N \cap \mathbb{R}^N_+}n^2(x_N)^\beta (u(\Phi (x)))^2 (\varphi ^\prime (nx_N))^2 dx. \end{align*}$$
We see from the dominated convergence theorem that
$$\begin{equation*} \lim _{n \to \infty } \int _{(-\delta ,\delta )^N \cap \mathbb{R}^N_+}(x_N)^\beta \Big |\frac{\partial }{\partial x_N} u(\Phi (x))\Big |^2(1-\varphi (nx_N))^2 dx = 0. \end{equation*}$$
From Equation 6 , $\beta <1$ , Lemma 2.1 (ii) and the dominated convergence theorem, we see that for some constant $C_1 > 0$ ,
$$\begin{align*} & \int _{(-\delta ,\delta )^N \cap \mathbb{R}^N_+}n^2(x_N)^\beta (u(\Phi (x)))^2 (\varphi ^\prime (nx_N))^2 dx\\ & \qquad =n^2 \int _{[-\delta ,\delta ]^{N-1}}\int _{0}^{2n^{-1}} (x_N)^\beta (u(\Phi (x)))^2 (\varphi ^\prime (nx_N))^2 dx_N dx^\prime \\ & \qquad \le \frac{C_1}{1-\beta } \int _{[-\delta ,\delta ]^{N-1}}\int _0^{2n^{-1}} (x_N)^\beta \Big |\frac{\partial }{\partial x_N} u(\Phi (x))\Big |^2 dx_N dx^\prime \rightarrow 0 \ \ \text{ as } n\rightarrow \infty . \end{align*}$$
Thus, $\|u_n-u\|_{\alpha ,\beta }\rightarrow 0$ as $n\rightarrow \infty$ . This proves $W_{\alpha ,\beta }(\Omega ) \subset W_{0,\alpha ,\beta }(\Omega )$ for $\alpha \ge 1$ and $\beta <1$ .
■
Lemma 2.3.
Let $\delta \in (0,1)$ and $(\alpha , \beta ) \in \{(a,b)\ |\ a< b,\ a+b \le 2\}\cup \{(a,b)\ |\ a\ge 1,\ a+b \le 2,\ (a,b) \ne ( 1,1)\}$ . For $v\in C^1((0,\delta ])$ satisfying $\int _0^\delta x^{-\alpha }v^2dx<\infty$ and $\int _0^\delta x^{\beta }|v^\prime |^2dx<\infty$ , it holds that
$$\begin{equation} \frac{(\beta -\alpha )^2}{16}\int _0^\delta x^{ -\alpha }v^2dx\le \int _0^\delta x^{\beta }(v^\prime )^2dx+\frac{\beta -\alpha }{4}\ \delta ^\frac{\beta -\alpha }{2} v^2(\delta ). \cssId{a2k}{\tag{7}} \end{equation}$$
Proof.
Let $v\in C^1((0,\delta ])$ satisfying $\int _0^\delta x^{-\alpha }v^2dx<\infty$ and $\int _0^\delta x^{\beta }|v^\prime |^2dx<\infty$ . Note that
$$\begin{equation} \int _0^\delta \Big (x^\frac{\beta }{2}v^\prime +\frac{\beta -\alpha }{4}x^\frac{-\alpha }{2} v\Big )^2dx=\int _0^\delta x^{\beta }(v^\prime )^2+\frac{(\beta -\alpha )^2}{16}x^{-\alpha }v^2+\frac{\beta -\alpha }{2}x^\frac{\beta -\alpha }{2}vv^\prime dx \cssId{a4}{\tag{8}} \end{equation}$$
and for any small $\varepsilon > 0$ ,
$$\begin{equation} \begin{aligned} \int _\varepsilon ^\delta x^\frac{\beta -\alpha }{2}vv^\prime dx&=\frac{1}{2}\int _\varepsilon ^\delta x^{\frac{\beta -\alpha }{2}}(v^2)^\prime dx\\ &=-\frac{\beta -\alpha }{4}\int _\varepsilon ^\delta x^{\frac{\beta -\alpha -2}{2}}v^2dx+\frac{1}{2}\Big (\delta ^\frac{\beta -\alpha }{2} v^2(\delta )-\varepsilon ^\frac{\beta -\alpha }{2} v^2(\varepsilon )\Big ). \end{aligned} \cssId{sam1}{\tag{9}} \end{equation}$$
First, we assume that $(\alpha , \beta ) \in \{(a,b)\ |\ a< b,\ a+b \le 2\}$ . Then, since $\delta \in (0,1)$ , if $(\alpha , \beta ) \in \{(a,b)\ |\ a< b,\ a+b \le 2\}$ , it follows that
$$\begin{equation} \begin{aligned} \frac{\beta -\alpha }{2}\int _0^\delta x^\frac{\beta -\alpha }{2}vv^\prime dx&\le -\frac{(\beta -\alpha )^2}{8}\int _0^\delta x^{\frac{\beta -\alpha -2}{2}}v^2dx+\frac{\beta -\alpha }{4}\ \delta ^\frac{\beta -\alpha }{2} v^2(\delta )\\ &\le -\frac{(\beta -\alpha )^2}{8}\int _0^\delta x^{ -\alpha }v^2dx+\frac{\beta -\alpha }{4}\ \delta ^\frac{\beta -\alpha }{2} v^2(\delta ). \end{aligned} \cssId{a3z}{\tag{10}} \end{equation}$$
Then, combining Equation 8 and Equation 10 , we get Equation 7 .
Next, we assume that $(\alpha ,\beta )\in \{(a,b)\ |\ a\ge 1,\ a+b \le 2,\ (a,b) \ne ( 1,1)\}$ . Since $\alpha \ge \beta$ , we see that for $0<x<y<1$ ,
$$\begin{equation*} \begin{split} v^2(y)-v^2(x)&=\int _x^y \Big (v^2(t)\Big )^\prime dt\le 2\Big (\int _x^y \frac{v^2}{t^\alpha }dt\Big )^\frac{1}{2}\Big (\int _x^y t^\alpha (v^\prime )^2dt\Big )^\frac{1}{2}\\ &\le 2\Big (\int _x^y \frac{v^2}{t^\alpha }dt\Big )^\frac{1}{2}\Big (\int _x^y t^\beta (v^\prime )^2dt\Big )^\frac{1}{2}. \end{split} \end{equation*}$$
Then we see that $\lim _{x\rightarrow 0}v(x)$ exists. Since $\int _0^\delta x^{-\alpha }v^2dx < \infty$ , we see that $\lim _{x\rightarrow 0}v(x) = 0$ . Then, from this and Equation 9 , we see that for $\delta \in (0,1)$ , $\alpha \ge 1$ and $\alpha +\beta \le 2$ ,
$$\begin{equation} \begin{aligned} \frac{\beta -\alpha }{2}\int _0^\delta x^\frac{\beta -\alpha }{2}vv^\prime dx&= -\frac{(\beta -\alpha )^2}{8}\int _0^\delta x^{ \frac{\beta -\alpha -2}{2} }v^2dx+\frac{\beta -\alpha }{4}\ \delta ^\frac{\beta -\alpha }{2} v^2(\delta )\\ &\le -\frac{(\beta -\alpha )^2}{8}\int _0^\delta x^{ -\alpha }v^2dx+\frac{\beta -\alpha }{4}\ \delta ^\frac{\beta -\alpha }{2} v^2(\delta ). \end{aligned} \cssId{a3}{\tag{11}} \end{equation}$$
Combining Equation 8 and Equation 11 , we get Equation 7 .
■
Proposition 2.4.
Let $\Omega$ be a bounded $C^2$ -domain in $\mathbb{R}^N$ . Then we have
$$\begin{equation*} H_{\alpha ,\beta }(\Omega )\begin{cases} >0 &\text{ if } (\alpha ,\beta ) \in \{(a,b)\ | \ a+b\le 2, b<1\},\\ =0 &\text{ if } (\alpha ,\beta ) \in \mathbb{R}^2\setminus \{(a,b)\ | \ a+b\le 2, b<1\}.\end{cases} \end{equation*}$$
Proof.
We first prove that $H_{\alpha ,\beta }(\Omega )>0$ for $(\alpha ,\beta ) \in \{(a,b)\ | \ a+b\le 2, b<1\}$ . We claim that $H_{\alpha ,2-\alpha }(\Omega ) > 0$ for $\alpha > 1$ . In fact, by Equation 8 and Equation 11 , we see that for $\alpha +\beta =2, \beta <1$ and $v \in C^1((0,1))$ vanishing in a neighborhood of $0$ ,
$$\begin{equation*} \frac{(1-\alpha )^2}{4}\int _0^\delta x^{-\alpha }v^2dx\le \int _0^\delta x^{2-\alpha }(v^\prime )^2dx, \end{equation*}$$
where $\delta \in (0,1)$ . This and Equation 5 imply that for small $\delta > 0$ , there exists a constant $C_1 > 0$ such that for any $u \in C_0^\infty (\Omega )$ ,
$$\begin{equation} {\int _{\Omega _\delta } d^{2-\alpha }(x) |\nabla u |^2 dx}\ge C_1{\int _{\Omega _\delta } |u(x)|^2d^{-\alpha }(x) dx}. \cssId{ls1}{\tag{12}} \end{equation}$$
We take $\psi \in C_0^\infty (\Omega )$ such that $\psi \in [0,1]$ , $\psi (x) = 1$ for $x \in \Omega \setminus \Omega _\delta$ and $\psi (x) = 0$ for $x \in \Omega _{\delta /2}$ . By the Poincaré inequality and the fact that
$$\begin{equation*} 0< \inf _{x \in \Omega \setminus \Omega _{\delta /2}}d(x) \le \sup _{x \in \Omega \setminus \Omega _{\delta /2}} d(x) < \infty , \end{equation*}$$
there exists a constant $C_2 > 0$ such that for any $w \in C_0^\infty (\Omega \setminus \overline{\Omega _{\delta /2}})$ ,
$$\begin{equation} {\int _{\Omega } d^{2-\alpha }(x) |\nabla w |^2 dx}\ge C_2{\int _{\Omega } |w |^2d^{-\alpha }(x) dx}. \cssId{ls2}{\tag{13}} \end{equation}$$
Thus, we see from Equation 12 and Equation 13 that for any $u \in C_0^\infty (\Omega )$ ,
$$\begin{align*} & \int _{\Omega } |u(x)|^2d^{-\alpha }(x) dx\\ & \quad \le 2\int _{\Omega } |\psi (x)u(x)|^2d^{-\alpha }(x) dx + 2 \int _{\Omega _\delta } \Big |\big (1-\psi (x)\big )u(x)\Big |^2d^{-\alpha }(x) dx \\ & \quad \le \frac{2}{C_1}\int _{\Omega _\delta } d^{2-\alpha }(x) \Big |\nabla \big ((1-\psi )u\big )\Big |^2dx + \frac{2}{C_2} \int _{\Omega } d^{2-\alpha }(x) |\nabla (\psi u)|^2dx \\ & \quad \le \frac{4}{C_1}\int _{\Omega _\delta } d^{2-\alpha }(x)(|\nabla u|^2 + |\nabla \psi |^2u^2)dx + \frac{4}{C_2} \int _{\Omega } d^{2-\alpha }(x)(|\nabla u|^2 + |\nabla \psi |^2 u^2)dx \\ & \quad \le C\int _{\Omega } d^{2-\alpha }(x)|\nabla u|^2 dx +C\int _{\Omega _\delta } d^{-\alpha }(x)u^2 dx, \end{align*}$$
where $C$ is a positive constant depending only on $C_1,C_2, \max _{x \in \Omega }d(x),\Vert \nabla \psi \Vert _{L^\infty }$ . Combining the above inequality and Equation 12 , we see that for some constant $C > 0$ ,
$$\begin{equation*} \int _{\Omega } |u(x)|^2d^{-\alpha }(x) dx \le C \int _{\Omega } d^{2-\alpha }(x)|\nabla u|^2 dx, \qquad u \in C_0^\infty (\Omega ). \end{equation*}$$
This shows that $H_{\alpha ,\beta }(\Omega ) > 0$ for $\alpha + \beta =2$ and $\beta < 1$ . On the other hand, for any $\alpha ,\beta \in \mathbb{R}$ with $\alpha +\beta < 2$ and $\beta < 1$ , we can find $\alpha ^\prime > \alpha$ such that $\alpha ^\prime +\beta =2$ . Since
$$\begin{equation*} \int _{\Omega } |u(x)|^2d^{-\alpha }(x) dx \le \max _{x \in \Omega }d^{\alpha ^\prime -\alpha }(x) \int _{\Omega } |u(x)|^2d^{-\alpha ^\prime }(x) dx \end{equation*}$$
and $\Omega$ is bounded, we see that for some $C > 0$ , $H_{\alpha ,\beta }(\Omega ) \ge C H_{\alpha ^\prime ,\beta }(\Omega ) > 0$ . Thus, we prove the claim that $H_{\alpha ,\beta }(\Omega ) > 0$ for $\alpha +\beta \le 2$ and $\beta < 1$ .
Next, we claim that $H_{\alpha ,\beta }(\Omega )=0$ for $(\alpha ,\beta ) \in \mathbb{R}^2\setminus \{(a,b)\ | \ a+b\le 2, b<1\}$ . Then we consider two cases: (i) $\beta \ge 1, \alpha \le 1$ and (ii) $\alpha +\beta > 2$ . In the case $\beta \ge 1,\alpha \le 1$ , for $\sigma >0$ , we define
$$\begin{equation*} w(x)=\begin{cases} \Big (\frac{d(x)}{\delta }\Big )^\sigma &\text{ if } d(x)<\delta ,\\ 1 &\text{ if } d(x)\ge \delta .\end{cases} \end{equation*}$$
Then we see from Equation 5 that
$$\begin{gather*} \int _\Omega d^{\beta }(x)|\nabla w|^2dx=\sigma ^2 \delta ^{-2\sigma }\int _{\Omega _\delta } d^{\beta +2\sigma -2}(x) dx\le \begin{cases} C_1 \sigma ^2 \delta ^{\beta -1} &\text{ if }\beta >1,\\ C_1 \sigma & \text{ if }\beta =1,\end{cases}\\ \int _\Omega d^{-\alpha }(x)w^2dx\ge \delta ^{-2\sigma }\int _{\Omega _\delta }d^{-\alpha +2\sigma }(x) dx\ge \begin{cases} C_2 \delta ^{1-\alpha } &\text{ if }\alpha <1,\\ C_2\sigma ^{-1} &\text{ if }\alpha =1,\end{cases} \end{gather*}$$
where $C_1, C_2>0$ are constants independent of $\sigma >0$ . Thus, there exists $C > 0$ such that for small $\sigma > 0$ ,
$$\begin{equation*} H_{\alpha ,\beta }(\Omega )\le \frac{\int _\Omega d^{\beta }(x)|\nabla w|^2dx}{\int _\Omega d^{-\alpha }(x)w^2dx}\le C\sigma , \end{equation*}$$
which implies that $H_{\alpha ,\beta }(\Omega )=0$ if $\alpha \le 1$ and $\beta \ge 1$ .
In the case $\alpha +\beta >2$ , we define $W_n(x)=\phi (nd(x))$ for a nonnegative $\phi \in C_0^\infty ((0,\delta ) )\setminus \{0\}$ . Using Equation 5 , we see that for some constants $C_3, C_4>0$ , independent of $n$ ,
$$\begin{equation*} \begin{split} \int _\Omega d^\beta (x)|\nabla W_n|^2dx&\le n^2\int _\Omega d^\beta (x)\Big (\phi ^\prime (nd(x))\Big )^2dx\\ &\le C_3 n^2\int _0^\delta t^\beta \Big (\phi ^\prime (nt)\Big )^2dt=C_3n^{1-\beta }\int _0^\delta t^\beta \Big (\phi ^\prime (t)\Big )^2dt \end{split} \end{equation*}$$
and
$$\begin{equation*} \int _\Omega d^{-\alpha }(x)W_n^2dx\ge C_4 \int _0^\delta t^{-\alpha }\Big (\phi (nt)\Big )^2dt=C_4 n^{\alpha -1} \int _0^\delta t^{-\alpha }\Big (\phi (t)\Big )^2dt. \end{equation*}$$
Then, we see that as $n \rightarrow \infty$ ,
$$\begin{equation*} \frac{\int _\Omega d^\beta (x)|\nabla W_n|^2dx}{\int _\Omega d^{-\alpha }(x)W_n^2dx}\le C_3 C_4^{-1}n^{2-\alpha -\beta }\frac{\int _0^\delta t^\beta (\phi ^\prime )^2dt}{\int _0^\delta t^{-\alpha } (\phi (t) )^2dt}\rightarrow 0. \end{equation*}$$
This proves that $H_{\alpha ,\beta }(\Omega ) = 0$ when $\alpha +\beta > 2$ , and thus, we complete the proof.
■
Proposition 2.5.
For $\alpha +\beta \le 2$ , $2\alpha +\beta \ge 3$ , $(\alpha ,\beta ) \ne ( 1,1)$ , we have $H_{\alpha ,\beta }(\Omega )=L_{\alpha ,\beta }(\Omega )$ .
Proof.
Assume that $\alpha +\beta \le 2$ , $2\alpha +\beta \ge 3$ , $(\alpha ,\beta ) \ne ( 1,1)$ . In this case, we have $\alpha > 1$ and $\beta <1$ . Clearly, by Proposition 2.2 , $H_{\alpha ,\beta }(\Omega )\le L_{\alpha ,\beta }(\Omega )$ . To prove the reverse inequality $H_{\alpha ,\beta }(\Omega )\ge L_{\alpha ,\beta }(\Omega )$ , we choose a minimizing sequence $\{u_{m}\}_{m=1}^\infty \subset C_0^\infty (\Omega )$ of $H_{\alpha ,\beta }(\Omega )$ such that $\int _\Omega d^{-\alpha }(x)u_{m}^2dx = 1$ and $\int _\Omega d^\beta (x)|\nabla u_{m}|^2dx\rightarrow H_{\alpha ,\beta }(\Omega )$ as $m\rightarrow \infty$ . We find a function $\varphi \in C^1(\mathbb{R})$ such that $\varphi (t) = 0$ for $t \le 1$ and $\varphi (t) =1$ for $t \ge 2$ . Define $u_{n,m}(x)\equiv u_{m}(x)\varphi \Big (nd(x)\Big )$ and
$$\begin{equation*} \ w_{n,m}\equiv u_{n,m} + c_{n,m} d^{\alpha -1+n^{-\alpha +\frac{1-\beta }{2}}}(x) \text{ with }c_{n,m}=-\frac{\int _\Omega d^{-\alpha }(x)u_{n,m}dx}{\int _\Omega d^{-1+n^{-\alpha +\frac{1-\beta }{2}}}(x)dx}. \end{equation*}$$
Note that $\int _\Omega d^{-\alpha }(x)w_{n,m}dx=0$ , and it is easy to check $w_{n,m}\in W^{1,2}_{\alpha ,\beta }(\Omega )$ since $-1 < \alpha -2+2n^{-\alpha +\frac{1-\beta }{2}}$ and $-1 < \beta + 2\alpha -4+2n^{-\alpha +\frac{1-\beta }{2}}$ . Then, by the same argument with the proof of Proposition 2.2 , we see that $\|u_{n,m}-u_{m}\|_{ {\alpha ,\beta }} \rightarrow 0$ as $n\rightarrow \infty$ . Moreover, it follows from the boundedness of $\{ \int _\Omega d^\beta (x)|\nabla u_{m}|^2dx\}_m$ that $\int _\Omega d^\beta (x)|\nabla u_{n,m}|^2dx\le C$ with $C>0$ is a constant independent of $m,n$ . We see from Lemma 2.1 (i) and Equation 5 that for some $C_1,C_2$ , independent of $m,n$ ,
$$\begin{align*} \int _\Omega d^{-\alpha }(x)u_{n,m}dx&=\int _{\Omega _\delta \setminus \Omega _{1/n}} d^{-\alpha }(x)u_{n,m}dx+\int _{\Omega \setminus \Omega _\delta } d^{-\alpha }(x)u_{n,m}dx\\ & \le C_1\Big (1+\int _{n^{-1}}^\delta t^{-\alpha +\frac{1-\beta }{2}}dt\Big )\\ &\le \begin{cases} C_2(1+n^{\alpha +\frac{\beta -3}{2}}) &\text{ if }2\alpha +\beta -3>0, \\ C_2(1+ \log n) &\text{ if }2\alpha +\beta -3=0.\end{cases} \end{align*}$$
Then, since for some $C_3,C_4$ , independent of $m,n$ ,
$$\begin{equation*} \int _\Omega d^{-1+n^{-\alpha +\frac{1-\beta }{2}}}(x)dx\ge C_3\int _0^\delta t^{-1+n^{-\alpha +\frac{1-\beta }{2}}}dt\ge C_4 n^{\alpha +\frac{\beta -1}{2}}, \end{equation*}$$
it follows that for large $n > 0$ ,
$$\begin{equation*} |c_{n,m}| \le C_2 C_4^{-1} n^{-1}\log n. \end{equation*}$$
Now, we see that
$$\begin{equation*} \begin{split} &(c_{n,m})^2\int _\Omega d^\beta \bigg |\nabla \Big (d^{\alpha -1+n^{-\alpha +\frac{1-\beta }{2}}}\Big )\bigg |^2dx\\ &\qquad =c_{n,m}^2\Big (\alpha -1+n^{-\alpha +\frac{1-\beta }{2}}\Big )^2\int _\Omega d^{2\alpha +\beta -4+2n^{-\alpha +\frac{1-\beta }{2}}}dx\\ &\qquad \le \begin{cases} C_5(c_{n,m})^2 =O\Big (n^{-2}(\log n)^2\Big ) &\text{ if }2\alpha +\beta -3>0, \\ C_5 (c_{n,m})^2n=O\Big (n^{-1}(\log n)^2\Big ) &\text{ if }2\alpha +\beta -3=0.\end{cases} \end{split} \end{equation*}$$
and
$$\begin{equation*} (c_{n,m})^2\int _\Omega d^{ \alpha -2+2n^{-\alpha +\frac{1-\beta }{2}}}\le C_6(c_{n,m})^2 =O\Big (n^{-2}(\log n)^2\Big ), \end{equation*}$$
where $C_5, C_6$ are positive constants independent of $m, n$ . Therefore, from these, we see that
$$\begin{align*} L_{\alpha ,\beta }(\Omega )\le \frac{\int _\Omega d^\beta (x)|\nabla w_n|^2dx}{\int _\Omega d^{-\alpha }(x)w_n^2dx}=\frac{\int _\Omega d^\beta (x)|\nabla u_{n,m}|^2dx+o(1)}{\int _\Omega d^{-\alpha }(x)u_{n,m}^2dx+o(1)}=\frac{\int _\Omega d^\beta (x)|\nabla u_{m}|^2dx }{\int _\Omega d^{-\alpha }(x)u_{m}^2dx } \end{align*}$$
as $n\rightarrow \infty$ . Lastly, letting $m\rightarrow \infty$ , we get $L_{\alpha ,\beta }(\Omega )\le H_{\alpha ,\beta }(\Omega )$ .
■
Lemma 2.6.
There exists a sequence $\{u_n\}_n \subset C^1_0(\Omega )$ such that
$$\begin{equation*} \int _\Omega d^{-1}(x)u_ndx=0, \ \ \lim _{n \to \infty }\frac{\int _\Omega d(x)|\nabla u_n|^2dx}{ \int _\Omega d^{-1}(x)u_n^2dx} =0. \end{equation*}$$
Proof.
By a scaling, we may assume $d(x)<1$ . We find a function $\phi \in C^1(\mathbb{R})$ such that $\phi (t)=\begin{cases} 1 &\text{ if } t\ge 2,\\ 0 &\text{ if }t\le 1,\end{cases}$ and define
$$\begin{equation*} \begin{split} u_n(x)=\phi (nd(x))\bigg ( |\ln (d(x))|^{-1/2}+c_n\bigg ),\\ \text{ where } c_n = -\frac{\int _\Omega d^{-1}(x) |\ln (d(x)) |^{-1/2}\phi (nd(x))dx}{\int _\Omega d^{-1}(x)\phi (nd(x))dx}. \end{split} \end{equation*}$$
Then we have $\int _\Omega d^{-1}(x)u_ndx=0$ . Using Equation 5 , we see that
$$\begin{align*} & \int _\Omega d^{-1}(x) |\ln (d(x)) |^{-1/2}\phi (nd(x))dx\\ &\qquad =\int _{\Omega _{\delta }-\Omega _{1/n}} d^{-1}(x) |\ln (d(x)) |^{-1/2}\phi (nd(x))dx+O(1)\\ &\qquad \le C_1\int _{n^{-1}}^{\delta } t^{-1} |\ln t |^{-1/2}dt+O(1)=2C_1(\ln n)^{1/2}+O(1) \end{align*}$$
and
$$\begin{equation*} \int _\Omega d^{-1}(x)\phi (nd(x))dx\ge \int _{\Omega _{\delta }\setminus \Omega _{2/n}} d^{-1}(x) dx\ge C_2\int _{2n^{-1}}^{\delta _0} t^{-1}dt\ge C_2\ln n+O(1). \end{equation*}$$
Then, it follows that
$$\begin{align*} |c_n|= \frac{\int _\Omega d^{-1}(x) |\ln (d(x)) |^{-1/2}\phi (nd(x))dx}{\int _\Omega d^{-1}(x)\phi (nd(x))dx}\le \frac{2C_1(\ln n)^{1/2}+O(1)}{ C_2\ln n+O(1)}\le C_3 (\ln n)^{-1/2}, \end{align*}$$
where $C_1, C_2$ and $C_3$ are positive constants, independent of large $n> 0$ . Now, using Equation 5 again, we see that for some positive constants $C_4, C_5$ and $C_6$ , independent of large $n > 0$ ,
$$\begin{equation*} \begin{split} &\int _\Omega d(x)|\nabla u_n|^2dx\\ &\quad \le 2\int _\Omega n^2d(x) \Big (\phi ^\prime (nd(x)) \Big )^2\bigg (|\ln (d(x))|^{-1/2}+c_n\bigg )^2 \\ & \qquad +\frac{1}{4} d^{-1}(x)\phi ^2(nd(x))|\ln (d(x))|^{-3} dx\\ &\quad \le C_4\bigg [\int _{\Omega _{2n^{-1}} } n^2d(x) \Big (|\ln (d(x))|^{-1}+c_n^2\Big )dx+ \int _{\Omega \setminus \Omega _{n^{-1}}} |\ln (d(x))|^{-3}d^{-1}(x) dx\bigg ]\\ &\quad \le C_5\bigg [\int _0^{2n^{-1}}n^2 t \Big (|\ln t|^{-1}+c_n^2\Big )dt+\int _{n^{-1}}^{\delta } |\ln t|^{-3}t^{-1}dt+O(1)\bigg ]=O(1) \end{split} \end{equation*}$$
and
$$\begin{equation*} \begin{split} \int _\Omega d^{-1}(x)(u_n)^2dx &=\int _\Omega \phi ^2(nd(x))d^{-1}(x)\bigg (|\ln (d(x))|^{-1}+c_n^2+2c_n|\ln (d(x)) |^{-1/2}\bigg )\\ &\ge C_6\ln ((\ln n))+O(1). \end{split} \end{equation*}$$
This implies that $\lim _{n \to \infty }\frac{\int _\Omega d(x)|\nabla u_n|^2dx}{ \int _\Omega d^{-1}(x)u_n^2dx} =0$ and completes the proof.
■
Lemma 2.7.
Let $\alpha +\beta > 2$ . Then there exists a sequence $\{u_n\}_n \subset C^1_0(\Omega )$ such that
$$\begin{equation*} \int _\Omega d^{-\alpha }(x)u_ndx=0, \ \ \lim _{n \to \infty }\frac{\int _\Omega d^\beta (x)|\nabla u_n|^2dx}{ \int _\Omega d^{-\alpha }(x)u_n^2dx}=0. \end{equation*}$$
Proof.
We find nonnegative functions $\phi _1, \phi _2 \in C_0^\infty (\mathbb{R})$ such that $\phi _1(t)= \begin{cases} 1 &\text{ if } t\in [2,3],\\ 0 &\text{ if } t\in \mathbb{R}\setminus [1,4]\end{cases}$ and $\phi _2(t)=\begin{cases} 1 &\text{ if } t\in [5,6],\\ 0 &\text{ if } t\in \mathbb{R}\setminus [4,7].\end{cases}$ Then we define
$$\begin{equation*} u_n(x)\equiv \phi _1(nd(x))+c_n\phi _2(nd(x)), \ \text{ where }\ c_n=-\frac{\int _\Omega d^{-\alpha }(x)\phi _1\Big (nd(x)\Big )dx}{\int _\Omega d^{-\alpha }(x)\phi _2\Big (nd(x)\Big )dx}. \end{equation*}$$
Then we have $\int _\Omega d^{-\alpha }(x)u_ndx=0$ . We see from Equation 5 that
$$\begin{equation*} \int _\Omega d^{-\alpha }(x)\phi _1\Big (nd(x)\Big )dx\le C_1\int _{0}^{\delta _0} t^{-\alpha }\phi _1(nt)dt= C_1n^{\alpha -1}\int _0^\infty t^{-\alpha }\phi _1(t)dt \end{equation*}$$
and
$$\begin{equation*} \int _\Omega d^{-\alpha }(x)\phi _2\Big (nd(x)\Big )dx\ge C_2\int _{0}^{\delta _0} t^{-\alpha }\phi _2(nt)dt= C_2n^{\alpha -1}\int _0^\infty t^{-\alpha }\phi _2(t)dt. \end{equation*}$$
These estimations imply that
$$\begin{equation*} |c_n|=\frac{\int _\Omega d^{-\alpha }(x)\phi _1\Big (nd(x)\Big )dx}{\int _\Omega d^{-\alpha }(x)\phi _2\Big (nd(x)\Big )dx}\le \frac{C_1 \int _0^\infty t^{-\alpha }\phi _1(t)dt}{ C_2 \int _0^\infty t^{-\alpha }\phi _2(t)dt}, \end{equation*}$$
where $C_1$ and $C_2$ are positive constants, independent of $n$ . For large $n > 0$ , using Equation 5 , we see that
$$\begin{align*} \int _\Omega d^\beta (x)|\nabla u_n|^2dx&=n^2\int _{\Omega }d^\beta (x)\bigg (\Big (\phi _1^\prime (nd(x))\Big )^2+c_n^2\Big (\phi _2^\prime (nd(x))\Big )^2\bigg )dx\\ &\le C_3 n^2\int _0^{\delta } t^\beta \bigg (\Big (\phi _1^\prime (nt)\Big )^2+ \Big (\phi _2^\prime (nt)\Big )^2\bigg )dt\\ &=C_3 n^{1-\beta }\int _0^\infty t^\beta \bigg (\Big (\phi _1^\prime (t)\Big )^2+\Big (\phi _2^\prime (t)\Big )^2\bigg )dt \end{align*}$$
and
$$\begin{equation*} \begin{split} \int _\Omega d^{-\alpha }(x)u_n^2dx&=\int _\Omega d^{-\alpha }(x) \Big (\phi _1(nd(x))\Big )^2 dx\ge C_4 \int _0^{\delta } t^{-\alpha } \Big (\phi _1(nt)\Big )^2dt\\ &=C_4n^{\alpha -1}\int _0^\infty t^{-\alpha } \Big (\phi _1(t)\Big )^2dt, \end{split} \end{equation*}$$
where $C_3$ and $C_4$ are positive constants. Thus, from these estimates and the assumption $\alpha +\beta >2$ , we conclude that $\lim _{n \to \infty }\frac{\int _\Omega d^\beta (x)|\nabla u_n|^2dx}{ \int _\Omega d^{-\alpha }(x)u_n^2dx}=0$ .
■
Lemma 2.8.
Let $(\alpha ,\beta )\in \mathbb{R}^2$ and $\Omega$ be a bounded $C^2$ -domain. Let $\{r_m\}_{m=0}^\infty$ be a sequence such that $r_m \downarrow 0$ as $m \rightarrow \infty$ and $\{x \in \Omega \;|\; d(x) > r_0 \} \neq \emptyset$ . Then for $a, b\in \mathbb{R}$ and a sequence $u_m$ satisfying $u_m\rightharpoonup 0$ in $W_{\alpha ,\beta }^{1,2}(\Omega )$ as $m\rightarrow \infty$ , we can find a sequence $\{m_i\}_{i=1}^\infty$ of positive integers such that $m_1 < m_2 < \cdots$ and
$$\begin{equation} \int _{\Omega ^{(i)}}d^{a}(x) |u_{m_i} |dx, \int _{\Omega ^{(i)}}d^{b}(x) u_{m_i}^2dx <\min \{1/2,(r_{i-1}-r_{i})^2\}, \cssId{qm1}{\tag{14}} \end{equation}$$
where $\Omega ^{(i)}\equiv \{x \in \Omega \;|\; d(x) > r_i \}$ .
Proof.
Let $a, b\in \mathbb{R}$ and $u_m$ be a sequence satisfying $u_m\rightharpoonup 0$ in $W_{\alpha ,\beta }^{1,2}(\Omega )$ as $m\rightarrow \infty$ . Note that
$$\begin{equation*} \int _{\Omega ^\prime } u_m^2+ |\nabla u_m|^2dx\le C_1\int _{\Omega ^\prime }d^{-\alpha }u_m^2+d^\beta |\nabla u_m|^2dx\le C_2, \end{equation*}$$
where $\Omega ^\prime \subset \!\subset \Omega$ and $C_1, C_2>0$ are constants independent of $m$ . Then, by the Rellich-Kondrachov theorem, we deduce that $u_m\rightarrow 0$ in $L^2_{loc}(\Omega )$ as $m\rightarrow \infty$ , up to a subsequence. Since $u_m \rightarrow 0$ in $L^2(\Omega ^{(0)})$ , there is $u_{m_0}$ such that
$$\begin{equation*} \int _{\Omega ^{(0)}} d^{a}(x)|u_{m_0}| dx,\int _{\Omega ^{(0)}} d^b(x)u_{m_0}^2 dx <1/2. \end{equation*}$$
Next, choose $u_{m_1}$ such that $\int _{\Omega ^{(1)}}d^{a}(x) |u_{m_1}| dx ,\int _{\Omega ^{(1)}}d^{b}(x) u_{m_1}^2 dx <\min \{1/2,(r_0 - r_1)^2\}$ and $m_1> m_0$ . We repeat this process to define a subsequence $\{u_{m_i}\} \subset \{u_m\}$ for $i=1,2, \cdots$ , which satisfies Equation 14 .
■
Lemma 2.9.
Let $(\alpha ,\beta )\in \mathbb{R}^2$ and $\Omega$ be a bounded $C^2$ -domain. Let $u_m$ be a minimizing sequence of $L_{\alpha ,\beta }(\Omega )$ satisfying $\int _\Omega d^{-\alpha }(x) u_m^2 dx=1$ and $u_m\rightharpoonup u$ in $W_{\alpha ,\beta }^{1,2}(\Omega )$ as $m\rightarrow \infty$ , where $u\not \equiv 0$ and $\int _\Omega d^{-\alpha }(x)udx=0$ . Then $u$ attains $L_{\alpha ,\beta }(\Omega )$ .
Proof.
Let $v_m = u_m - u$ . Then $v_m$ converges weakly to $0$ in $W_{\alpha ,\beta }^{1,2}(\Omega )$ as $m\rightarrow \infty$ . Now we see that
$$\begin{equation} \begin{aligned} L_{\alpha ,\beta }(\Omega ) + o(1) &= \int _\Omega d^\beta (x)|\nabla u_m|^2 dx = \int _\Omega d^\beta (x)(|\nabla u|^2 + 2 \nabla u \cdot \nabla v_m + |\nabla v_m|^2) dx \\ &= \int _\Omega d^\beta (x)|\nabla u|^2 dx + \int _\Omega d^\beta (x)|\nabla v_m|^2 dx + o(1) \end{aligned} \cssId{min6}{\tag{15}} \end{equation}$$
and
$$\begin{equation} 1 = \int _\Omega d^{-\alpha }(x) u_m^2 dx = \int _\Omega d^{-\alpha }(x)u^2 dx + \int _\Omega d^{-\alpha }(x)v_m^2 dx + o(1). \cssId{min5}{\tag{16}} \end{equation}$$
Since $u \not \equiv 0$ , it follows that
$$\begin{equation} \limsup _{m\rightarrow \infty } \int _\Omega d^{-\alpha }(x)v_m^2 dx<1. \cssId{min4}{\tag{17}} \end{equation}$$
Thus, by Equation 15 , Equation 16 , Equation 17 and the facts that $\int _\Omega d^{-\alpha }(x)u dx=\int _\Omega d^{-\alpha }(x)v_mdx=0$ , we obtain
$$\begin{align*} L_{\alpha ,\beta }(\Omega ) & \le \frac{\int _\Omega d^\beta (x)|\nabla u|^2 dx}{\int _\Omega d^{-\alpha }(x)u^2 dx} = \frac{L_{\alpha ,\beta }(\Omega ) - \int _\Omega d^\beta (x)|\nabla v_m|^2 dx + o(1)}{1 - \int _\Omega d^{-\alpha }(x)v_m^2 dx + o(1)} \\ & \le \frac{L_{\alpha ,\beta }(\Omega ) - L_{\alpha ,\beta }(\Omega )\int _\Omega d^{-\alpha }(x)v_m^2 dx + o(1)}{1 - \int _\Omega d^{-\alpha }(x)v_m^2 dx + o(1)} = L_{\alpha ,\beta }(\Omega ) + o(1). \end{align*}$$
This implies that $u$ attains $L_{\alpha ,\beta }(\Omega )$ .
■
Lemma 2.10.
Let $\alpha <1$ , $b\in [0,\infty )$ , $N\ge 2$ , and $f\in C^1([0,1))$ satisfying $\int _0^1(1-r)^{2-\alpha }(f^\prime )^2dr<\infty$ and $\int _0^1(1-r)^{-\alpha }f^2dr<\infty$ . For $b>0$ and $N=2$ , we assume that $f(0)=0$ . Then, for $g(r)=(1-r)^{\frac{1-\alpha }{2}}r^{-b}f(r)$ , it holds that
$$\begin{equation*} \begin{split} & \int _0^1 (1 - r)^{2-\alpha }r^{N-1}(f^\prime )^2dr = \int _0^1 r^{N -1}(1 - r)^{-\alpha }f^2 \bigg [ \frac{(1 - \alpha )^2}{4} - b(N + b - 2) r^{-2}(1 - r)^2\\ &\qquad +\Big ( b-(1-\alpha )\frac{N-1}{2}\Big )r^{-1}(1-r)\bigg ] +r^{N+2b-1}(1-r)(g^\prime )^2 dr. \end{split} \end{equation*}$$
Proof.
By the arguments in Proposition 2.2 and the assumptions that $\int _0^1(1-r)^{2-\alpha }(f^\prime )^2dr<\infty$ and $\int _0^1(1-r)^{-\alpha }f^2dr<\infty$ , we may assume that $g(1)=0$ . We note that $f(r)=r^b(1-r)^{-\frac{1-\alpha }{2}}g(r)$ ,
$$\begin{align*} (f^\prime )^2 &=\Big ((r^b(1-r)^{-\frac{1-\alpha }{2}})^\prime g+r^b(1-r)^{-\frac{1-\alpha }{2}}g^\prime \Big )^2\\ &=\bigg [\Big (br^{b-1}(1-r)^{-\frac{1-\alpha }{2}}+\frac{1-\alpha }{2}r^{b}(1-r)^{-\frac{3-\alpha }{2}}\Big )g+r^{b}(1-r)^{-\frac{1-\alpha }{2}}g^\prime \bigg ]^2\\ &=\Big (br^{b-1}(1-r)^{-\frac{1-\alpha }{2}}+\frac{1-\alpha }{2}r^{b}(1-r)^{-\frac{3-\alpha }{2}}\Big )^2g^2+\Big (r^b(1-r)^{-\frac{1-\alpha }{2}}g^\prime \Big )^2\\ &\qquad +2\Big (br^{b-1}(1-r)^{-\frac{1-\alpha }{2}}+\frac{1-\alpha }{2}r^{b}(1-r)^{-\frac{3-\alpha }{2}}\Big )r^b(1-r)^{-\frac{1-\alpha }{2}}g g^\prime \\ &=\Big (b^2r^{2(b-1)}(1 - r)^{-1+\alpha } + \frac{(1 - \alpha )^2}{4}r^{2b}(1 - r)^{-3+\alpha } + b(1 - \alpha )r^{2b-1}(1 - r)^{-2+\alpha }\Big )g^2\\ &\qquad +r^{2b}(1 - r)^{-1+\alpha }(g^\prime )^2+\Big (br^{2b-1}(1 - r)^{-1+\alpha }+\frac{1-\alpha }{2}r^{2b}(1 - r)^{-2+\alpha }\Big )(g^2)^\prime . \end{align*}$$
Moreover, by the assumption that $f(0)=0$ if $b>0$ and $N=2$ , we see that for $b\in [0,\infty )$ and $N\ge 2$ ,
$$\begin{align*} &\Big (br^{N+2b-2}(1-r)+\frac{1-\alpha }{2}r^{N+2b-1}\Big )g^2\Big |_{r=0}^1\\ &=\Big (br^{N -2}(1-r)^{2-\alpha }+\frac{1-\alpha }{2}r^{N -1}(1-r)^{1-\alpha }\Big )f^2\Big |_{r=0}^1\\ &=0. \end{align*}$$
Therefore, we conclude that
$$\begin{align*} &\int _0^1 (1-r)^{2-\alpha }r^{N-1}(f^\prime )^2dr\\ &=\int _0^1\Big (b^2r^{N+2b-3}(1-r)+\frac{(1-\alpha )^2}{4}r^{N+2b-1}(1-r)^{-1}+b(1-\alpha )r^{N+2b-2} \Big )g^2\\ & \qquad +\Big (br^{N+2b-2}(1-r)+\frac{1-\alpha }{2}r^{N+2b-1} \Big )(g^2)^\prime +r^{N+2b-1}(1-r)(g^\prime )^2 dr\\ &=\int _0^1\Big [b^2r^{N+2b-3}(1-r)+\frac{(1-\alpha )^2}{4}r^{N+2b-1}(1-r)^{-1}+b(1-\alpha )r^{N+2b-2} \\ &\qquad - b(N+2b-2)r^{N+2b-3}(1-r)+br^{N+2b-2}-\frac{1-\alpha }{2}(N+2b-1)r^{N+2b-2} \Big ] g^2\\ &\qquad +r^{N+2b-1}(1-r)(g^\prime )^2 dr\\ &=\int _0^1 r^{N+2b-1}(1-r)^{-1}g^2 \Big [ \frac{(1-\alpha )^2}{4} - b(N+ b-2) r^{-2}(1-r)^2\\ &\qquad +\big ( b-(1-\alpha )\frac{N-1}{2}\big )r^{-1}(1-r)\Big ] +r^{N+2b-1}(1-r)(g^\prime )^2 dr\\ &=\int _0^1 r^{N -1}(1-r)^{-\alpha }f^2 \Big [ \frac{(1-\alpha )^2}{4} - b(N+ b-2) r^{-2}(1-r)^2\\ &\qquad +\Big ( b-(1-\alpha )\frac{N-1}{2}\Big )r^{-1}(1-r)\Big ] +r^{N+2b-1}(1-r)(g^\prime )^2 dr. \end{align*}$$ ■
3. Proofs of main results for subcritical cases
3.1. Proof of Theorem 1.1
By Lemma 2.6 and Lemma 2.7 , if $\alpha , \beta \in \{(a,b)\ |\ a+b > 2\}\cup \{(1,1)\}$ , we have $L_{\alpha ,\beta }(\Omega ) = 0$ . Thus it suffices to prove that for $\alpha +\beta \le 2$ and $(\alpha ,\beta ) \ne (1,1)$ ,
$$\begin{equation} L_{\alpha ,\beta }(\Omega ) > 0. \cssId{mo1}{\tag{18}} \end{equation}$$
If $\alpha <1, \alpha + \beta \le 2$ and $\alpha \ge \beta$ , we can take $\beta ^\prime \in \mathbb{R}$ satisfying $\alpha + \beta ^\prime \le 2, \beta ^\prime > \alpha \ge \beta$ . Then, if $L_{\alpha ,\beta ^\prime }(\Omega ) > 0$ , we see that
$$\begin{equation*} L_{\alpha ,\beta ^\prime }(\Omega )\int _\Omega d^{-\alpha }(x)u^2dx\le \int _\Omega d^{\beta ^\prime }(x)|\nabla u|^2dx\le \bar{C}\int _\Omega d^{\beta }(x)|\nabla u|^2dx, \end{equation*}$$
where $\bar{C} \equiv \max \{d^{\beta ^\prime -\beta }(x) \ | \ x \in \Omega \} < \infty$ . This implies that for $\alpha <1, \alpha + \beta \le 2$ and $\alpha \ge \beta$ , $L_{\alpha ,\beta } > 0$ if $L_{\alpha ,\beta ^\prime }(\Omega ) > 0$ for $\alpha + \beta ^\prime \le 2, \beta ^\prime > \alpha \ge \beta$ . Thus it suffices to show that $L_{\alpha ,\beta } > 0$ when
$$\begin{equation*} (\alpha ,\beta )\in \{(a,b)\ |\ a+b\le 2, \ a<b\}\cup \{(a,b)\ |\ a\ge 1, a+b\le 2, a=b\neq 1\}. \end{equation*}$$
First, take any $(\alpha ,\beta )\in \{(a,b)\ |\ a\ge 1, a+b\le 2, a=b\neq 1\}$ and $u\in W_{\alpha ,\beta }^{1,2}(\Omega )$ . By Proposition 2.2 , we may assume that $u\in C_0^\infty (\Omega )$ . We see from Lemma 2.3 that for $\alpha \ge 1, \alpha +\beta \le 2, \alpha =\beta \neq 1$ and $v \in C^1((0,1))$ vanishing in a neighborhood of $0$ ,
$$\begin{equation*} \frac{(\beta -\alpha )^2}{16}\int _0^\delta x^{-\alpha }v^2dx\le \int _0^\delta x^{\beta }(v^\prime )^2dx. \end{equation*}$$
This and Equation 5 imply that for small $\delta > 0$ , there exists a constant $\tilde{C}_1 > 0$ such that for any $u \in C_0^\infty (\Omega )$ ,
$$\begin{equation} {\int _{\Omega _\delta } d^{\beta }(x) |\nabla u |^2 dx}\ge \tilde{C}_1{\int _{\Omega _\delta } |u(x)|^2d^{-\alpha }(x) dx}. \cssId{ls1a}{\tag{19}} \end{equation}$$
We take $\psi \in C_0^\infty (\Omega )$ such that $\psi \in [0,1]$ , $\psi (x) = 1$ for $x \in \Omega \setminus \Omega _\delta$ and $\psi (x) = 0$ for $x \in \Omega _{\delta /2}$ . Since
$$\begin{equation} 0< \inf _{x \in \Omega \setminus \Omega _{\delta /2}}d(x) \le \sup _{x \in \Omega \setminus \Omega _{\delta /2}} d(x) < \infty ,\cssId{lubound}{\tag{20}} \end{equation}$$
there exists a constant $\tilde{C}_2 > 0$ such that for any $u \in C_0^\infty (\Omega \setminus \overline{\Omega _{\delta /2}})$ ,
$$\begin{equation} {\int _{\Omega } d^{\beta }(x) |\nabla u |^2 dx}\ge \tilde{C}_2{\int _{\Omega } |u(x)|^2d^{-\alpha }(x) dx}. \cssId{ls2a}{\tag{21}} \end{equation}$$
Now, we see from Equation 19 and Equation 21 that for any $u \in C_0^\infty (\Omega )$ ,
$$\begin{equation*} \begin{split} & \int _{\Omega } |u(x)|^2d^{-\alpha }(x) dx\\ & \qquad \le 2\int _{\Omega } |\psi (x)u(x)|^2d^{-\alpha }(x) dx + 2 \int _{\Omega _\delta } \Big |\big (1-\psi (x)\big )u(x)\Big |^2d^{-\alpha }(x) dx \\ & \qquad \le \frac{2}{\tilde{C}_1}\int _{\Omega _\delta } d^{\beta }(x) \Big |\nabla \big ((1-\psi )u\big )\Big |^2dx + \frac{2}{\tilde{C}_2} \int _{\Omega } d^{\beta }(x) |\nabla (\psi u)|^2dx \\ & \qquad \le \frac{4}{\tilde{C}_1}\int _{\Omega _\delta } d^{\beta }(x)(|\nabla u|^2 + |\nabla \psi |^2u^2)dx + \frac{4}{\tilde{C}_2} \int _{\Omega } d^{\beta }(x)(|\nabla u|^2 + |\nabla \psi |^2 u^2)dx \\ & \qquad \le \hat{C}\int _{\Omega } d^{\beta }(x)|\nabla u|^2 dx +\hat{C}\int _{\Omega _\delta } d^{-\alpha }(x)u^2 dx, \end{split} \end{equation*}$$
where $\hat{C}$ is a positive constant depending only on $\delta , \tilde{C}_1, \tilde{C}_2$ , $\|\nabla \psi \|_{L^\infty }$ . Combining the above inequality and Equation 19 , we see that for $(\alpha ,\beta )\in \{(a,b)\ |\ a\ge 1, a+b\le 2, a=b\neq 1\}$ .
$$\begin{equation*} \int _{\Omega } |u(x)|^2d^{-\alpha }(x) dx \le C \int _{\Omega } d^{\beta }(x)|\nabla u|^2 dx, \qquad u \in C_0^\infty (\Omega ), \end{equation*}$$
where $C > 0$ is a constant; thus $L_{\alpha ,\beta } > 0$ .
Next, take any $(\alpha ,\beta )\in \{(a,b)\ |\ a+b\le 2, \ a<b\}$ . We note that in this case, $\alpha < 1$ . Take any function $u\in C^1( \Omega )$ satisfying
$$\begin{equation*} \int _\Omega d^{\beta }(x)|\nabla u|^2dx<\infty , \int _\Omega d^{-\alpha }(x)u^2dx<\infty \text{ and }\int _\Omega d^{-\alpha }(x) u dx= 0. \end{equation*}$$
For any small $\delta \in (0,\delta _0)$ , it follows from the Poincaré-Wirtinger inequality and Equation 20 that for a constant $C$ , which depends only on $N$ and $\Omega \setminus \overline{\Omega _\delta }$ ,
$$\begin{equation*} \begin{split} & \int _{\Omega \setminus \overline{\Omega _\delta }} \bigg (u-\Big (\int _{\Omega \setminus \overline{\Omega _\delta }}d^{-\alpha }(z)dz\Big )^{-1}\int _{\Omega \setminus \overline{\Omega _\delta }}d^{-\alpha }(z)u(z) dz \bigg )^2d^{-\alpha }(x)dx\\ & \qquad \le C\int _{\Omega \setminus \overline{\Omega _\delta }}|\nabla u|^2dx. \end{split} \end{equation*}$$
Then we see that
$$\begin{multline} \int _{\Omega \setminus \overline{\Omega _\delta }} d^{-\alpha }(x)u^2 dx\\ \le C_1\max \{\delta ^{-\beta },1\} \int _{\Omega \setminus \overline{\Omega _\delta }} d^{\beta }(x)|\nabla u|^2 dx+C_2\left(\int _{\Omega \setminus \overline{\Omega _\delta }} d^{-\alpha }(x)u dx\right)^2, \cssId{d1N}{\tag{22}} \end{multline}$$
where $C_1, C_2 > 0$ are independent of $u$ . Since $\int _\Omega d^{-\alpha }(x) u dx= 0$ , we see that for some $C_3$ , independent of $\delta > 0$ and $u$ ,
$$\begin{equation*} \begin{split} \Big (\int _{\Omega \setminus \overline{\Omega _\delta }}d^{-\alpha }(x) u dx\Big )^2 &=\Big (- \int _{\Omega _\delta } d^{-\alpha }(x)u dx \Big )^2\\ & \le \int _{\Omega _\delta } d^{-\alpha }(x)u^2dx \int _{\Omega _\delta } d^{-\alpha }(x) dx \le C_3\delta ^{1-\alpha } \int _{\Omega _\delta } d^{-\alpha }(x)u^2dx. \end{split} \end{equation*}$$
Then, it follows from this estimate and Equation 22 that
$$\begin{align} & \int _{\Omega \setminus \overline{\Omega _\delta }}d^{-\alpha }(x) u^2 dx \cssId{d1M}{\tag{23}}\\ & \quad \le C_1\max \{\delta ^{-\beta },1\} \int _{\Omega \setminus \overline{\Omega _\delta }} d^{\beta }(x)|\nabla u|^2 dx+ C_2 C_3 \delta ^{1-\alpha }\int _{\Omega _\delta } d^{-\alpha }(x)u^2dx. \end{align}$$
Thus, we deduce that for small $\delta > 0$ ,
$$\begin{equation*} \begin{aligned} \int _\Omega d^{-\alpha }(x)u^2 dx&= \int _{\Omega \setminus \Omega _\delta }d^{-\alpha }(x)u^2 dx+\int _{ \Omega _\delta } d^{-\alpha }(x)u^2 dx\\ & \le C_1\max \{\delta ^{- \beta },1\}\int _\Omega d^{\beta }(x)|\nabla u|^2 dx + 2\int _{\Omega _\delta }d^{-\alpha }(x)u^2 dx. \end{aligned} \end{equation*}$$
Now, for the completion of the proof, it suffices to show that for small $\delta > 0$ ,
$$\begin{equation} \int _{\Omega _\delta }d^{-\alpha }(x) u^2 dx \le C_4 \int _\Omega d^{\beta }(x)|\nabla u|^2 dx \cssId{domainreverse}{\tag{24}} \end{equation}$$
for some constant $C_4>0$ , independent of $u$ . By Lemma 2.3 and Equation 5 , we see that for some $C_5, C_6>0$ independent of $u$ and $\delta$ ,
$$\begin{equation} \begin{aligned} & \int _{\Omega _\delta }d^{-\alpha }(x) u^2 dx=\int _{0}^{\delta } \int _{\Sigma _t}t^{-\alpha } u^2 d\sigma _tdt \le C_5 \int _{0}^{\delta } \int _{\partial \Omega }t^{-\alpha } u^2(t,\sigma ) d\sigma dt \\ &\le \Big (\frac{4}{\beta -\alpha }\Big )^2C_5 \Big [\int _{0}^{\delta } \int _{\partial \Omega } t^{\beta }\Big (\frac{du}{dt}(t,\sigma )\Big )^2d\sigma dt+\frac{(\beta -\alpha )\delta ^\frac{\beta -\alpha }{2}}{4} \int _{\partial \Omega }u^2(\delta ,\sigma )d\sigma \Big ]\\ &\le C_6 \Big (\int _{\Omega } d^{\beta }(x)|\nabla u|^2 dx+\delta ^\frac{\beta -\alpha }{2}\int _{\partial \Omega }u^2(\delta ,\sigma )d\sigma \Big ). \end{aligned} \cssId{b2}{\tag{25}} \end{equation}$$
Note that for some $C_7>0$ , independent of $u$ and small $\delta > 0$ ,
$$\begin{equation} \int _{\partial \Omega }u^2(\delta ,\sigma )d\sigma \le C_7\int _{\Sigma _\delta }u^2d\sigma _\delta , \cssId{texmlid1}{\tag{26}} \end{equation}$$
and by the trace inequality, there exists a constant $\tilde{C}$ , independent of small $\delta > 0$ such that for small $\delta > 0$ ,
$$\begin{equation} \begin{split} &\int _{\Sigma _\delta } u^2 d\sigma _\delta \le \tilde{C}\int _{\Omega \setminus \overline{\Omega _\delta }} |\nabla u|^2 + u^2 dx \\ &\le \tilde{C} \bigg (\max \{\delta ^{-\beta },(\text{diam}(\Omega ))^{-\beta }\} \int _{\Omega \setminus \overline{\Omega _\delta }} d^{\beta }(x)|\nabla u|^2 dx\\ &\quad + \max \{\delta ^\alpha ,(\text{diam}(\Omega ))^\alpha \} \int _{\Omega \setminus \overline{\Omega _\delta }} d^{-\alpha }(x)u^2 dx\bigg ). \end{split} \cssId{texmlid2}{\tag{27}} \end{equation}$$
By Equation 23 and Equation 25 , Equation 26 , Equation 27 , we see that for small $\delta > 0$ ,
$$\begin{equation*} \begin{split} & \int _{\Omega _\delta }d^{-\alpha }(x) u^2 dx \\ &\quad \le C_8 \Big (\big (1+ \delta ^{-\frac{\alpha +\beta }{2}}\max \{1, \delta ^\beta \}\big )\int _{\Omega } d^{\beta }(x)|\nabla u|^2 dx\\ &\qquad + \delta ^{\frac{\beta -\alpha }{2}}\max \{\delta ^{\alpha },(\text{diam}(\Omega ))^\alpha \}\int _{\Omega \setminus \overline{\Omega _\delta }} d^{-\alpha }(x) u^2 dx \Big ) \\ &\quad \le C_9 \int _{\Omega } d^{\beta }(x)|\nabla u|^2 dx+ \frac{1}{2}\int _{\Omega _\delta } d^{-\alpha }(x) u^2 dx, \end{split} \end{equation*}$$
where $C_8, C_9\equiv C_9(\delta ) > 0$ are independent of $u$ , which implies Equation 24 . This completes the proof.
3.2. Proof of Theorem 1.2
We first prove the second result (2). We assume that $\alpha +\beta \le 2$ , $2\alpha +\beta \ge 3$ , $(\alpha ,\beta ) \ne ( 1,1)$ . By Proposition 2.5 , $L_{\alpha ,\beta }(\Omega ) = H_{\alpha ,\beta }(\Omega )$ in this case. Suppose that there exists a minimizer $u_{\alpha ,\beta }$ of $L_{\alpha ,\beta }(\Omega )$ such that $\int _\Omega d^{-\alpha }(x)u_{\alpha ,\beta }^2dx=1$ . By Proposition 2.5 , we deduce that $|u_{\alpha ,\beta }|$ is also a minimizer of $H_{\alpha ,\beta }(\Omega )$ . Then, we see from the strong maximum principle that $|u_{\alpha ,\beta }|>0$ in $\Omega$ . This contradicts that $\int _\Omega d^{-\alpha }(x)u_{\alpha ,\beta }dx=0$ . Thus, $L_{\alpha ,\beta }(\Omega )$ is not achieved for $(\alpha ,\beta )\in \{(a,b)\ |\ a+b\le 2, 2a+b\ge 3, a=b\neq 1\}$ .
Next, we prove the first result (1). We assume $(\alpha ,\beta )\in \{(a,b)\ |\ a+b< 2, 2a+b<3\}$ . Let $\{u_m\}_{m=1}^\infty \subset C^1(\Omega )$ be a minimizing sequence of $L_{\alpha ,\beta }(\Omega )$ with
$$\begin{equation*} \int _{\Omega } d^{-\alpha }(x)u_m dx = 0 , \ \int _{\Omega }d^{-\alpha }(x)u_m^2 dx = 1 \ \text{ and }\ \lim _{m \to \infty } \int _\Omega d^\beta (x) |\nabla u_m|^2 dx = L_{\alpha ,\beta }(\Omega ). \end{equation*}$$
Taking a subsequence if it is necessary, we may assume that as $m \to \infty$ , $u_m$ converges weakly to $u$ in $W_{\alpha ,\beta }^{1,2}(\Omega )$ .
Suppose that $u \equiv 0$ . Let $\{r_m\}_{m=0}^\infty$ be a sequence such that $r_0$ is sufficiently small and $r_m \downarrow 0$ as $m \rightarrow \infty$ . Then, taking a subsequence of $\{u_i\}$ if it is necessary, we see from Lemma 2.8 that $\{u_i\}$ satisfies
$$\begin{equation} \int _{\Omega ^{(i)}}d^{-\alpha }(x) |u_i |dx<\min \{1/2,(r_{i-1}-r_{i})^2\}\ \ \text{ and } \cssId{min2}{\tag{28}} \end{equation}$$
$$\begin{equation} \int _{\Omega ^{(i)}}d^{\beta }(x) u_i^2dx, \int _{\Omega ^{(i)}}d^{-\alpha }(x) u_i^2dx<\min \{1/2,(r_{i-1}-r_{i})^2\}, \cssId{mon1}{\tag{29}} \end{equation}$$ where $\Omega ^{(i)}\equiv \{x \in \Omega \;|\; d(x) > r_i \}$ . For each $m \ge 1$ , we define $\phi _m \in C^1(\Omega )$ such that $|\nabla \phi _m| < 2(r_{m-1}-r_m)^{-1}$ .
$$\begin{equation*} \phi _m(x) = \begin{cases} 1 &\text{ if }x \in \Omega \setminus \Omega ^{(m)},\\ 0 &\text{ if } x \in \Omega ^{(m-1)}. \end{cases} \end{equation*}$$
Then, by Equation 28 and the assumption $\int _{\Omega } d^{-\alpha }(x)u_m dx = 0$ , we have
$$\begin{equation} \Big |\int _\Omega d^{-\alpha }(x)u_m\phi _m dx\Big | = \Big |\int _\Omega d^{-\alpha }(x)u_m\Big (1-\phi _m\Big ) dx\Big | \to 0 \cssId{min1}{\tag{30}} \end{equation}$$
as $m\rightarrow \infty$ . For $\delta \in (0,\delta _0)$ , define $\psi \in C^1(\mathbb{R})$ such that $\psi (t)=\begin{cases} 1 &\text{ if } t\ge \delta ,\\ 0 &\text{ if } t\le \delta /2\end{cases}$ and
$$\begin{equation*} v_m=u_m\phi _m+c_m \psi (d(x)) \text{ with } c_m \equiv -\frac{\int _\Omega d^{-\alpha }(x)u_m\phi _mdx }{\int _\Omega d^{-\alpha }(x)\psi (d(x))dx}. \end{equation*}$$
Note that for large $m$ , supp$(u_m\phi _m)\cap$ supp$(\psi (d(x)))=\emptyset$ , $\int _\Omega d^{-\alpha }(x)v_mdx=0$ , and from Equation 30 , $|c_m|\rightarrow 0$ as $m\rightarrow \infty$ . We take a small $\iota > 0$ so that $\alpha +\beta + \iota < 2$ . Then, by Equation 29 , there exists a constant $C > 0$ , independent of $m$ , such that
$$\begin{equation} \begin{aligned} &\int _{\Omega } d^{\beta +\iota }(x)|\nabla v_m |^2 dx =\int _{\Omega } d^{\beta +\iota }(x)\bigg (|\nabla (u_m\phi _m)|^2 +c_m^2\Big (\psi ^\prime (d(x))\Big )^2\bigg )dx \\ &\le (r_{m-1})^\iota \int _{\Omega } d^{\beta }(x)|\nabla (u_m\phi _m)|^2 dx+c_m^2\int _{\Omega _{\delta }\setminus \Omega _{\delta /2}} d^{\beta +\iota }(x)\Big (\psi ^\prime (d(x))\Big )^2dx \\ &\le 2(r_{m-1})^\iota \int _{\Omega } d^\beta (x) u_m^2|\nabla \phi _m|^2 dx + 2(r_{m-1})^\iota \int _{\Omega }d^\beta (x)|\nabla u_m|^2\phi _m^2 dx+o(1)\\ &\le C(r_{m-1})^\iota +o(1) \end{aligned} \cssId{mon2}{\tag{31}} \end{equation}$$
as $m\rightarrow \infty$ . Note that, by Equation 29 , for each $m\ge 1$ ,
$$\begin{equation} \int _{\Omega }d^{-\alpha }(x)v_m^2 dx \ge \int _{\Omega }d^{-\alpha }(x)(u_m\phi _m)^2 dx \geq \int _{\Omega \setminus \Omega ^{(m)}}d^{-\alpha }(x)u_m^2 dx \geq 1/2. \cssId{mon3}{\tag{32}} \end{equation}$$
Thus, by Equation 31 , Equation 32 and the property $\int _\Omega d^{-\alpha }(x)v_mdx=0$ , we see that for large $m \ge 1$ ,
$$\begin{align*} L_{\alpha ,\beta +\iota }(\Omega )\le \frac{\int _{\Omega } d^{\beta +\iota }(x)|\nabla v_m|^2 dx}{\int _{\Omega }d^{-\alpha }(x)v_m^2 dx }\le 2\Big (C(r_{m-1})^\iota +o(1) \Big ). \end{align*}$$
This is a contradiction to the fact that $L_{\alpha ,\beta +\iota }(\Omega )>0$ . Thus we see that $u\not \equiv 0$ .
Now we claim that $\int _\Omega d^{-\alpha }(x)udx=0$ . If $\alpha < 1$ , since $u_m$ converges weakly to $u$ in $W_{\alpha ,\beta }^{1,2}(\Omega )$ as $m\rightarrow \infty$ and any constant functions are in $W_{\alpha ,\beta }^{1,2}(\Omega )$ , it follows that $\int _{\Omega }d^{-\alpha }(x)udx=0$ . If $\alpha \ge 1$ and $2\alpha +\beta <3$ , by Proposition 2.2 , we may assume that $u_m\in C_0^\infty (\Omega )$ . As before, we find a function $\psi \in C^1(\mathbb{R},[0,1])$ such that $\psi (t)=\begin{cases} 1 &\text{ if } t\ge \delta ,\\ 0 &\text{ if } t\le \delta /2\end{cases}$ and write $u_m=u_{m,1}+u_{m,2}$ , where $u_{m,1}=u_m\Big (1-\psi (d(x))\Big )$ and $u_{m,2}=u_m\psi (d(x))$ . Since $\psi \circ d \in W_{\alpha ,\beta }^{1,2}(\Omega )$ and $u_m$ converges weakly to $u$ in $W_{\alpha ,\beta }^{1,2}(\Omega )$ as $m\rightarrow \infty$ , we see that
$$\begin{equation} \lim _{m\rightarrow \infty }\int _\Omega d^{-\alpha }(x)u_{m,2} dx= \int _\Omega d^{-\alpha }(x)u {\psi }(d(x)) dx. \cssId{mz1}{\tag{33}} \end{equation}$$
Since $\beta < 1$ , we see from Lemma 2.1 (ii) that
$$\begin{equation*} \int _\Omega d^\beta (x)|\nabla u_{m,1}|^2dx\le \int _\Omega d^\beta (x)|\nabla u_m|^2+d^\beta (x)u_m^2 \Big (\psi ^\prime (d(x))\Big )^2dx\le C_1, \end{equation*}$$
where $C_1>0$ is a constant independent of $m$ . Now, since
$$\begin{equation*} \begin{split} |u_{m,1}(t,\sigma )|&\le \int _0^t \bigg |\frac{\partial u_{m,1}(s,\sigma )}{\partial s}\bigg |ds\\ &\le \bigg ( \int _0^t s^\beta |\nabla u_{m,1}(s,\sigma )|^2ds\bigg )^{1/2}\Big (\int _0^ts^{-\beta }ds\Big )^{1/2}, \ \ \sigma \in \partial \Omega , \end{split} \end{equation*}$$
we see that for $\sigma \in \partial \Omega$ ,
$$\begin{equation*} \int _{0}^\delta t^{-\alpha } |u_{m,1}(t,\sigma )| dt \le \bigg ( \int _0^\delta t^\beta |\nabla u_{m,1}(t,\sigma )|^2dt\bigg )^{1/2} \frac{2}{(1-\beta )^{1/2}(3-2\alpha -\beta )}\delta ^{\frac{3-\beta -2\alpha }{2}}. \end{equation*}$$
Then, we see from Equation 5 and the Cauchy inequality that for some constants $C_2 , C_3> 0$ ,
$$\begin{equation*} \begin{split} &\Big | \int _{0}^\delta \int _{\Sigma _t} t^{-\alpha }|u_{m,1} | d\sigma _t dt \Big |\\ & \qquad \le C_2\int _{0}^\delta \int _{\partial \Omega } t^{-\alpha } |u_{m,1}(t,\sigma )| d\sigma dt \\ & \qquad \le \frac{2C_2}{(1-\beta )^{1/2}(3-2\alpha -\beta )}\delta ^{\frac{3-2\alpha -\beta }{2}} \int _{\partial \Omega }\bigg ( \int _0^\delta t^\beta |\nabla u_{m,1}(t,\sigma )|^2dt\bigg )^{1/2}d\sigma \\ & \qquad \le \frac{2C_2}{(1-\beta )^{1/2}(3-2\alpha -\beta )}\delta ^{\frac{3-2\alpha -\beta }{2}} |\partial \Omega |^{1/2} \bigg (\int _{\partial \Omega } \int _0^\delta t^\beta |\nabla u_{m,1}(t,\sigma )|^2dtd\sigma \bigg )^{1/2} \\ & \qquad \le \frac{2C_3}{(1-\beta )^{1/2}(3-2\alpha -\beta )}\delta ^{\frac{3-2\alpha -\beta }{2}} |\partial \Omega |^{1/2} \bigg (\int _{\Omega }d^\beta (x) |\nabla u_{m,1}|^2dx \bigg )^{1/2}\\ & \qquad \le \frac{2C_3 (C_1)^{1/2}}{(1-\beta )^{1/2}(3-2\alpha -\beta )}\delta ^{\frac{3-2\alpha -\beta }{2}} |\partial \Omega |^{1/2}. \end{split} \end{equation*}$$
This and Equation 33 imply that $\lim _{m\rightarrow \infty }\int _\Omega d^{-\alpha }(x)u_m dx=\int _\Omega d^{-\alpha }(x)u dx=0$ . Thus, by Lemma 2.9 , we see that $u$ attains $L_{\alpha ,\beta }(\Omega )$ . The above estimate implies that for any $u \in W_{\alpha ,\beta }^{1,2}(\Omega )$ , $\int _{\Omega }d^{-\alpha }(x) u(x) dx$ is well-defined and a functional $u \in W_{\alpha ,\beta }^{1,2}(\Omega ) \mapsto \int _{\Omega }d^{-\alpha }(x) u(x) dx$ is differentiable.
We denote the minimizer $u$ by $u_{\alpha ,\beta }$ . We may assume that $\int _{\Omega } d^{-\alpha }(x)(u_{\alpha ,\beta })^2 dx =1$ . Then, there are Lagrange multipliers $\lambda ,\mu \in \mathbb{R}$ such that
$$\begin{equation*} -div(d^\beta (x) \nabla u_{\alpha ,\beta }) = \lambda d^{-\alpha }(x)u_{\alpha ,\beta } + \mu d^{-\alpha }(x) \text{ in } \Omega . \end{equation*}$$
Since
$$\begin{equation*} \begin{split} \int _{\Omega }d^{\beta }(x)|\nabla u_{\alpha ,\beta }|^2 dx &= \lambda \int _{\Omega }d^{-\alpha }(x)(u_{\alpha ,\beta })^2dx + \mu \int _{\Omega }d^{-\alpha }(x) u_{\alpha ,\beta }dx\\ &= \lambda \int _{\Omega }d^{-\alpha }(x)(u_{\alpha ,\beta })^2dx=\lambda , \end{split} \end{equation*}$$
it follows that $\lambda = L_{\alpha ,\beta }(\Omega )$ . For $\alpha < 1$ , a constant function $1_\Omega (x) =1$ on $\Omega$ is in $W_{\alpha ,\beta }^{1,2}(\Omega )$ . This implies that
$$\begin{equation*} 0 = \lambda \int _{\Omega }d^{-\alpha }(x)u_{\alpha ,\beta }dx + \mu \int _{\Omega }d^{-\alpha }(x) dx = \mu \int _{\Omega }d^{-\alpha }(x) dx. \end{equation*}$$
Then, we get that $\mu = 0$ and $\lambda = L_{\alpha ,\beta }(\Omega )$ if $\alpha <1$ . $$
4. Proof of main results for critical cases
In this section, we study whether or not there exists a minimizer of
$$\begin{equation*} L_{\alpha ,2-\alpha }(\Omega ) \equiv \inf \Big \{ \frac{\int _\Omega d^{2-\alpha }(x) |\nabla u |^2 dx}{\int _\Omega |u|^2d^{-\alpha }(x) dx} \ \Big | \ u \in W_{\alpha ,2-\alpha }^{1,2}(\Omega )\setminus \{0\}, \int _\Omega u d^{-\alpha }(x) dx=0 \Big \}, \end{equation*}$$
where $\alpha <1$ . We give some criteria for the (non-)attainability of the optimal constant $L_{\alpha ,2-\alpha }(\Omega )$ when $\Omega$ is the unit ball, an ellipse or an annulus. Moreover, we prove that the attainability of $L_{\alpha ,2-\alpha }(\Omega )$ depends on a local geometry of a domain $\Omega$ .
4.1. Proof of Theorem 1.3
We first prove that for $\alpha <1$ , $L_{\alpha ,2-\alpha }(\Omega ) \le \frac{(1-\alpha )^2}{4}$ . Assume $N\ge 1$ . For small $\delta > 0$ and $m > 1$ , we define
$$\begin{equation*} u(x)=\begin{cases} |\log (d(x)/\delta )|^m &\text{ if } d(x)\le \delta , \\ 0 &\text{ if } d(x)>\delta .\end{cases} \end{equation*}$$
Since $|\nabla u(x)|=m|\log (d(x)/\delta )|^{m-1}\frac{1}{|d(x)|}$ , using a change of variables $-\log (t/\delta )=s$ and the fact that $\int _0^\infty s^a e^{-bs}ds=b^{-1-a}\Gamma (1+a)$ with $b>0$ and $a>-1$ , we see that for some constant $c> 0$ , independent of small $\delta >0$
$$\begin{align*} \int _{\Omega _{\delta }}d^{2-\alpha } (x)|\nabla u|^2dx&\le m^2(1+c\delta )|\partial \Omega |\int _{0}^{\delta }t^{-\alpha }|\log (t/\delta )|^{2m-2}dt\\ &=m^2\delta ^{1-\alpha }(1+c\delta )|\partial \Omega |\int ^{ \infty }_{0}s^{2m-2}e^{-(1-\alpha )s}ds\\ &=m^2\delta ^{1-\alpha }(1+c\delta )|\partial \Omega | (1-\alpha )^{1-2m}\Gamma (2m-1), \end{align*}$$
$$\begin{align*} \int _{\Omega _{\delta }}d^{-\alpha }(x) u^2dx&\ge (1-c\delta )|\partial \Omega |\int _0^\delta t^{-\alpha }|\log (t/\delta )|^{2m}dt\\ &=\delta ^{1-\alpha }(1-c\delta )|\partial \Omega |\int ^{ \infty }_{0}s^{2m}e^{-(1-\alpha )s}ds\\ &=\delta ^{1-\alpha }(1-c\delta )|\partial \Omega |(1-\alpha )^{-1-2m}\Gamma (2m+1) \end{align*}$$ and
$$\begin{align*} \frac{1}{|\Omega |} \Big (\int _{\Omega _\delta }d^{-\alpha }(x) udx\Big )^2 &\le \frac{1}{|\Omega _\delta |}\Big ( (1+c\delta )|\partial \Omega |\int _0^\delta t^{-\alpha }|\log (t/\delta )|^mdt\Big )^2\\ &=\frac{\delta ^{2(1-\alpha )}(1+c\delta )^2|\partial \Omega |^2}{|\Omega _\delta |}\Big ( \int _0^\infty s^me^{-(1-\alpha )s}ds\Big )^2\\ &\le \frac{\delta ^{2(1-\alpha )}(1+c\delta )^2|\partial \Omega |^2}{\delta (1-c\delta )|\partial \Omega |}\Big ((1-\alpha )^{-1-m}\Gamma ( m+1)\Big )^2\\ &=\frac{\delta ^{1-2\alpha }(1+c\delta )^2|\partial \Omega |}{(1-c\delta )}(1-\alpha )^{-2-2m} \big (\Gamma (m+1)\big )^2. \end{align*}$$
Hence, it follows that
$$\begin{align*} & L_{\alpha ,2-\alpha }(\Omega ) \le \frac{\int _\Omega d(x)^2|\nabla u|^2dx}{\int _\Omega d^{-\alpha }(x)u^2dx-\frac{1}{|\Omega |}(\int _\Omega d^{-\alpha }(x) u dx)^2}\\ & \quad \le \frac{m^2\delta ^{1-\alpha }(1 + c\delta )|\partial \Omega | (1 - \alpha )^{1-2m}\Gamma (2m - 1) }{\delta ^{1-\alpha }(1 - c\delta )|\partial \Omega |(1 - \alpha )^{-1-2m}\Gamma (2m + 1) - \frac{\delta ^{1-2\alpha }(1 + c\delta )^2|\partial \Omega |}{(1 - c\delta )}(1 - \alpha )^{-2-2m} \big (\Gamma (m + 1)\big )^2}\\ & \quad =\frac{(1-\alpha )^2(1+c\delta ) }{(1-c\delta )\frac{2m(2m-1)}{m^2}-\frac{(1-\alpha )^{-1}\delta ^{-\alpha }(1+c\delta )^2}{(1-c\delta )}\frac{ (\Gamma (m+1) )^2}{m^2\Gamma (2m-1)}}\\ & \quad =\frac{(1-\alpha )^2}{\frac{1-c\delta }{1+c\delta }(4-\frac{2}{m})-(1-\alpha )^{-1}\delta ^{-\alpha }\frac{1+c\delta }{1-c\delta }\frac{ (\Gamma (m) )^2}{\Gamma (2m-1)}}. \end{align*}$$
Since
$$\begin{equation*} \frac{(\Gamma (m))^2}{\Gamma (2m-1)}= \frac{(m-1)(m-2)\cdots (2)1}{(2m-2)(2m-3)\cdots (m+1)m}\le 2^{1-m}\rightarrow 0 \text{ as } m\rightarrow \infty , \end{equation*}$$
letting $m\rightarrow \infty$ and then $\delta \rightarrow 0$ in the above inequality, we get the inequality $L_{\alpha ,2-\alpha }(\Omega )\le \frac{(1-\alpha )^2}{4}$ .
Next, we prove that if $L_{\alpha ,2-\alpha }(\Omega )<\frac{(1-\alpha )^2}{4}, L_{\alpha ,2-\alpha }(\Omega )$ is achieved, where $\alpha <1$ . Let $\{u_m\}_{m=1}^\infty \subset C^1(\Omega )$ be a minimizing sequence of $L_{\alpha ,2-\alpha }(\Omega )$ with
$$\begin{align*} & \int _{\Omega }d^{-\alpha }(x)u_m^2 dx = 1, \int _{\Omega }d^{-\alpha }(x) u_m dx = 0 \\ & \text{ and }\lim _{m \to \infty } \int _\Omega d^{2-\alpha }(x) |\nabla u_m|^2 dx = L_{\alpha ,2-\alpha }(\Omega ). \end{align*}$$
Taking a subsequence if it is necessary, we may assume that as $m \to \infty$ , $u_m$ converges weakly to some $u$ in $W_{\alpha ,2-\alpha }^{1,2}(\Omega )$ .
Suppose that $u \equiv 0$ . Then, by Lemma 2.8 , taking a subsequence if it is necessary, we may assume that $u_i$ satisfies
$$\begin{equation*} \int _{\Omega ^{(i)}}d^{-\alpha }(x)u_i^2dx<\min \{1/2,(r_{i-1}-r_{i})^2\}, \end{equation*}$$
where $\Omega ^{(i)}\equiv \{x \in \Omega \;|\; d(x) > r_i \}$ and $\{r_m\}_{m=0}^\infty$ is a sequence such that $r_m \downarrow 0$ as $m \rightarrow \infty$ . For each $m \ge 1$ , we define $\phi _m \in C^1(\Omega ,[0,1])$ such that $|\nabla \phi _m| < 2(r_{m-1}-r_m)^{-1}$ and
$$\begin{equation*} \phi _m(x) = \begin{cases} 1 &\text{ if }x \in \Omega \setminus \Omega ^{(m)},\\ 0 &\text{ if } x \in \Omega ^{(m-1)}. \end{cases} \end{equation*}$$
Then we have
$$\begin{align*} &\int _\Omega d^{2-\alpha }(x)|\nabla u_m|^2dx=\int _\Omega d^{2-\alpha }(x)|\nabla (u_m\phi _m)+(1-\phi _m)\nabla u_m-u_m\nabla \phi _m|^2dx\\ &=\int _\Omega d^{2-\alpha }(x)\Big [|\nabla (u_m\phi _m)|^2+(1-\phi _m)^2|\nabla u_m|^2+u_m^2|\nabla \phi _m|^2\\ &\ \ +2(1 - \phi _m)\nabla (u_m\phi _m)\cdot \nabla u_m - 2u_m\nabla (u_m\phi _m)\cdot \nabla \phi _m - 2u_m(1 - \phi _m)\nabla u_m\cdot \nabla \phi _m\Big ]dx\\ &\ge \int _\Omega d^{2-\alpha }(x)\Big [|\nabla (u_m\phi _m)|^2 -2u_m\nabla (u_m\phi _m)\cdot \nabla \phi _m \Big ]dx. \end{align*}$$
From this, Lemma 2.3 and the fact that
$$\begin{equation*} \int _\Omega d^{2-\alpha }(x)u_m^2|\nabla \phi _m|^2dx\le 4r_{m-1}^2(r_{m-1}-r_m)^{-2}\int _{\Omega _m} d^{-\alpha }(x)u_m^2dx\le 4r_{m-1}^2, \end{equation*}$$
we see that for some $c > 0$ ,
$$\begin{align*} \int _\Omega d^{2-\alpha }(x)|\nabla u_m|^2dx&\ge \int _\Omega d^{2-\alpha }(x)|\nabla (u_m\phi _m)|^2dx+o(1)\\ &=\int _{\Omega \setminus \Omega ^{(m-1)}}d^{2-\alpha }(x)|\nabla (u_m\phi _m)|^2dx+o(1)\\ &\ge (1-cr_{m-1})\int _{\partial \Omega }\int _0^{r_{m-1}}t^{2-\alpha }\Big (\frac{d}{dt}(u_m\phi _m)\Big )^2dtd\sigma +o(1)\\ &\ge \frac{(1-\alpha )^2}{4}(1-cr_{m-1})\int _{\partial \Omega }\int _0^{r_{m-1}}t^{-\alpha }(u_m\phi _m)^2dtd\sigma +o(1)\\ &\ge \frac{(1-\alpha )^2(1-cr_{m-1})}{4(1+cr_{m-1})}\int _{\Omega \setminus \Omega ^{(m-1)}}d^{-\alpha }(x)(u_m\phi _m)^2dx+o(1)\\ &=\frac{(1-\alpha )^2(1-cr_{m-1})}{4(1+cr_{m-1})}\int _{\Omega }d^{-\alpha }(x)(u_m\phi _m)^2dx+o(1). \end{align*}$$
Since $\int _{\Omega }d^{-\alpha }(x) u_m dx = 0$ , $\int _\Omega d^{-\alpha }(x)u_m^2dx=1$ and $\int _{\Omega ^{(m)}}d^{-\alpha }(x)u_m^2dx=o(1)$ , we deduce that
$$\begin{equation*} L_{\alpha ,2-\alpha }(\Omega )=\lim _{m\rightarrow \infty }\int _\Omega d^{2-\alpha }(x)|\nabla u_m|^2dx\ge \frac{(1-\alpha )^2}{4}, \end{equation*}$$
which is a contradiction to the assumption $L_{\alpha ,2-\alpha }(\Omega )<\frac{(1-\alpha )^2}{4}$ .
Now, it holds that $u \not \equiv 0$ . Thus, by Lemma 2.9 , we see that $u$ attains $L_{\alpha ,2-\alpha }(\Omega )$ . $$
To prove Theorem 1.4 , we prepare some geometric results of a domain $\Omega$ . For a bounded $C^2$ -domain $\Omega \subset \mathbb{R}^N$ , let $\kappa (y_0)=(\kappa _1(y_0),\cdots , \kappa _{N-1}(y_0))$ be the principal curvatures with respect to the outward unit normal of $\partial \Omega$ at $y_0$ . Then the mean curvature of $\partial \Omega$ at $y_0$ is given by $H(y_0)=\frac{1}{N-1}\sum _{i=1}^{N-1} k_i(y_0)$ . We adopt the convention that a standard unit sphere $S^{N-1}\subset \mathbb{R}^N$ has mean curvature 1 everywhere. Let $G\subset \Omega$ be the largest open subset of $\Omega$ such that for every $x$ in $G$ there is a unique nearest point $x^\prime \in \partial \Omega$ with $d(x) = | x-x^\prime |$ . We call $G$ the good set of $\Omega$ and $\Omega \setminus G$ a singular set. We denote $S=\Omega \setminus G$ . By Reference 10, Lemma 3.8 , there exists a $C^1$ -domain $\Omega _\epsilon \subset \Omega$ such that $\Omega \setminus \overline{\Omega _\epsilon }\subset G$ , $\cup _{\epsilon >0}(\Omega \setminus \overline{\Omega _\epsilon })=G$ and $\nu _\epsilon \cdot \nabla d\ge 0$ on $\partial \Omega _\epsilon$ , where $\nu _\epsilon$ is the unit inner normal of the boundary of $\partial \Omega _\epsilon$ . Reference 10, Lemma 2.3 implies that the distance function $d$ is in $C^2(G)$ if $\partial \Omega \in C^2$ and Reference 10, Corollary 3.2 (see also Reference 11 ) implies that the measure of $S$ is $0$ . Thus, $d \in W^{1,\infty }(\Omega )$ .
4.2. Proof of Theorem 1.4
We assume that $\Omega$ is a weakly mean convex $C^2$ -domain in $\mathbb{R}^N$ . We note that if $\Omega$ is a weakly mean convex $C^2$ -domain in $\mathbb{R}^N$ , it follows from Reference 10, Theorem 1.7 that $\Delta d \le 0$ in $G$ . For each $\alpha < 1$ , we take $p_\alpha \in \mathbb{R}$ such that
$$\begin{equation*} p_\alpha =-\frac{\int _\Omega d^{-\frac{\alpha }{2}}x_1 dx}{\int _\Omega d^{-\frac{\alpha }{2}} dx}. \end{equation*}$$
Then $\{|p_\alpha |\}_\alpha$ is bounded uniformly in $\alpha < 1$ . We define $u_\alpha =(x_1+p_\alpha )d^\frac{\alpha }{2}$ and $v_\alpha =d^{\frac{1-\alpha }{2}}(x)u_\alpha =(x_1+p_\alpha )d^\frac{1}{2}$ . Then we see that
$$\begin{gather*} \int _\Omega d^{-\alpha }(x)u_\alpha dx=0, \ \ \int _\Omega d^{-\alpha }(x)u_\alpha ^2dx=\int _\Omega (x_1+p_\alpha )^2dx <\infty ,\\ \int _\Omega d^{2-\alpha }(x)|\nabla u_\alpha |^2dx=\int _\Omega \frac{\alpha ^2}{4}(x_1+p_\alpha )^2+\Big (1-\frac{\alpha }{2} \Big ) d^2(x)dx<\infty . \end{gather*}$$
Thus, $u_\alpha \in W_{\alpha ,\beta }^{1,2}(\Omega )$ and $\int _\Omega d^{-\alpha }(x)u_\alpha dx=0$ . Then, since $v_\alpha =0$ on $\partial \Omega$ , $\nu _\epsilon \cdot \nabla d\ge 0$ on $\partial \Omega _\epsilon$ and
$$\begin{equation*} \begin{split} &d^{2-\alpha }(x)|\nabla u_\alpha |^2\\ &\quad =d^{2-\alpha }(x)\Big | \frac{\alpha -1}{2}d^{\frac{\alpha -1}{2}-1}(\nabla d)v_\alpha +d^{\frac{\alpha -1}{2}}(x)\nabla v_\alpha \Big |^2\\ &\quad =d^{2-\alpha }(x)\Big (\frac{(1-\alpha )^2}{4}d^{\alpha -3}(x)|\nabla d|^2v_\alpha ^2+d^{\alpha - 1 }(x)|\nabla v_\alpha |^2 \\ & \qquad -(1-\alpha )d^{\alpha -2}(x)v_\alpha \nabla d\cdot \nabla v_\alpha \Big )\\ &\quad =\frac{(1-\alpha )^2}{4}d^{ -1}(x)|\nabla d|^2v_\alpha ^2+d|\nabla v_\alpha |^2 -\frac{1-\alpha }{2} \nabla d\cdot \nabla (v_\alpha ^2) \\ &\quad =\frac{(1-\alpha )^2}{4}d^{ -\alpha }(x)u_\alpha ^2 + d|\nabla v_\alpha |^2 -\frac{1-\alpha }{2} \nabla d\cdot \nabla (v_\alpha ^2) \ \text{ in } \ G, \end{split} \end{equation*}$$
we see that
$$\begin{equation} \begin{aligned} & \int _\Omega d^{2-\alpha }(x)|\nabla u_\alpha |^2-\frac{(1-\alpha )^2}{4}d^{-\alpha }(x)u_\alpha ^2dx\\ & \quad = \int _\Omega d|\nabla v_\alpha |^2 -\frac{1-\alpha }{2} \nabla d\cdot \nabla (v_\alpha ^2)dx\\ & \quad = \int _\Omega d|\nabla v_\alpha |^2 dx + \frac{1 - \alpha }{2} \Big (\int _{\Omega \setminus \overline{\Omega _\epsilon }}(\Delta d)v_\alpha ^2dx - \int _{\Omega _\epsilon } \nabla d\cdot \nabla (v_\alpha ^2)dx\Big ) \\ & \qquad - \frac{1 - \alpha }{2} \int _{\partial {\Omega _\epsilon }}\frac{\partial d}{\partial \nu _\epsilon }v_\alpha ^2d\sigma _\epsilon \\ & \quad \le \int _\Omega d|\nabla v_\alpha |^2 dx +\frac{1-\alpha }{2} \Big (\int _{\Omega \setminus \overline{\Omega _\epsilon }}(\Delta d)v_\alpha ^2dx-\int _{\Omega _\epsilon } \nabla d\cdot \nabla (v_\alpha ^2)dx\Big ), \end{aligned} \cssId{tc1}{\tag{34}} \end{equation}$$
where $\nu _\epsilon$ is the unit inner normal of the boundary of $\partial \Omega _\epsilon$ . Since $\{|p_\alpha |\}_\alpha$ is bounded uniformly in $\alpha < 1$ and
$$\begin{equation*} \begin{split} \int _\Omega d|\nabla v_\alpha |^2 dx&=\int _\Omega \frac{1}{4} (x_1+p_\alpha )^2+d^2+(x_1+p_\alpha )\frac{1}{2}\frac{\partial }{\partial x_1}(d^2)dx\\ &=\int _\Omega \frac{1}{4} (x_1+p_\alpha )^2+\frac{1}{2}d^2 dx, \end{split} \end{equation*}$$
we see that
$$\begin{equation} \Big \{\int _\Omega d|\nabla v_\alpha |^2 dx \Big \}_{\alpha < 1} \text{ is bounded}. \cssId{tc2}{\tag{35}} \end{equation}$$
An estimate Reference 10, Corollary 2.7 says that for any $x \in G$ ,
$$\begin{equation} -\Delta d(x) \ge \frac{ H(y)}{1-d(x)H(y)},\cssId{mle}{\tag{36}} \end{equation}$$
where $y \in \partial \Omega$ satisfying $d(x) = |x-y|$ . Since $H \ge 0$ on $\partial \Omega$ and $\partial \Omega$ is compact, $H$ cannot be identically zero. In fact, a classical result Reference 1 of Alexandroff says that any compact hypersurface in $\mathbb{R}^N$ with a constant mean curvature should be a sphere; thus $H$ is positive in an open subset $A$ of $\partial \Omega$ . Then, we see from Equation 36 that $\Delta d < 0$ in an open subset $\Omega ^\prime$ of $G \subset \Omega$ . This implies that for small $\varepsilon > 0$ , the set $\{\int _{\Omega \setminus \overline{\Omega _\epsilon }}(\Delta d)(x_1+p_\alpha )^2d(x)dx\}_{\alpha < -1}$ is bounded away from $0$ . Then, for small $\varepsilon > 0$ , there exists $C > 0$ , independent of large $-\alpha > 0$ , satisfying
$$\begin{equation} \begin{aligned} &\int _{\Omega \setminus \overline{\Omega _\epsilon }}(\Delta d)v_\alpha ^2dx-\int _{\Omega _\epsilon } \nabla d\cdot \nabla (v_\alpha ^2)dx \\ &= \int _{\Omega \setminus \overline{\Omega _\epsilon }}(\Delta d)(x_1+p_\alpha )^2d(x)dx-\int _{\Omega _\epsilon } (x_1+p_\alpha )^2+2(x_1+p_\alpha )d \frac{\partial d}{\partial x_1}dx<-C. \end{aligned} \cssId{tc3}{\tag{37}} \end{equation}$$
Combining Equation 34 , Equation 35 , Equation 36 , Equation 37 , we see that for large $-\alpha > 0$ ,
$$\begin{equation*} \int _\Omega d^{2-\alpha }(x)|\nabla u_\alpha |^2-\frac{(1-\alpha )^2}{4}d^{-\alpha }(x)u_\alpha ^2dx < 0, \end{equation*}$$
which implies that $L_{\alpha ,2-\alpha }(\Omega ) < \frac{(1-\alpha )^2}{4}$ . This completes the proof. $$
4.3. Proof of Theorem 1.5
We note that Theorem 1.3 says that $L_{\alpha ,2-\alpha }(\Omega )\le \frac{(1-\alpha )^2}{4}$ for $\alpha <1$ and any $C^2$ -domain $\Omega$ . First, we assume that $\alpha <1$ and $N=1$ . We denote $I=(-1,1)$ . Then we claim that $L_{\alpha ,2-\alpha }(I)=\frac{(1-\alpha )^2}{4}$ and $L_{\alpha ,2-\alpha }(I)$ is not achieved. We see that
$$\begin{align*} L_{\alpha ,2-\alpha }(I) &= \inf _{u\in W_{\alpha ,2-\alpha }^{1,2}(I)\setminus \{0\}, \int _I d^{-\alpha }(t)udt=0} \frac{\int _I d^{2-\alpha }(t) | u^\prime |^2 dt}{\int _I d^{-\alpha }(t)u^2 dt}\\ &= \inf _{u\in W_{\alpha ,2-\alpha }^{1,2}(I)\setminus \{0\} } \frac{\int _I d^{2-\alpha }(t) | u^\prime |^2 dt}{\int _I d^{-\alpha }(t)\big (u- (\int _I d^{-\alpha }(t)dt)^{-1}\int _I d^{-\alpha }(t)udt\big )^2 dt}, \end{align*}$$
where $d(t)=\begin{cases} 1+t &\text{ if } t\in (-1,0),\\ 1-t &\text{ if } t\in [0,1).\end{cases}$ Since the functional
$$\begin{equation*} u\mapsto \frac{\int _I d^{2-\alpha }(t) | u^\prime |^2 dt}{\int _I d^{-\alpha }(t)\big (u- (\int _I d^{-\alpha }(t)dt)^{-1}\int _I d^{-\alpha }(t)udt\big )^2 dt} \end{equation*}$$
is invariant under the addition of any constant, it suffices to prove that if $u\in W_{\alpha ,2-\alpha }^{1,2}(I)$ and $u(0)=0$ ,
$$\begin{equation} \int _I d^{2-\alpha }(t) | u^\prime |^2 dt\ge \frac{(1-\alpha )^2}{4}\int _I d^{-\alpha }(t)\bigg (u- \Big (\int _I d^{-\alpha }(t)dt\Big )^{-1}\int _I d^{-\alpha }(t)udt\bigg )^2 dt. \cssId{c1}{\tag{38}} \end{equation}$$
By the density, it is sufficient to prove Equation 38 for any function $u\in C^1(I)$ satisfying
$$\begin{equation*} u(0)=0, \int _I d^{-\alpha }(t)u^2dt<\infty \text{ and } \int _I d^{2-\alpha }(t)|u^\prime |^2dt<\infty . \end{equation*}$$
Defining $v(t) \equiv d^{\frac{1-\alpha }{2}}(t) u(t)$ , we see that
$$\begin{equation} \begin{split} &d^{2-\alpha }(t)|u^\prime (t)|^2\\ &\quad =d^{2-\alpha }(t)\Big (-\frac{1 - \alpha }{2} d^{\frac{\alpha -3}{2}}(t)d^\prime (t)v(t)+d^{\frac{\alpha -1}{2}}(t)v^\prime (t)\Big )^2\\ &\quad =d^{2-\alpha }(t)\Big [\Big (\frac{1 - \alpha }{2}\Big )^2 d^{\alpha -3}(t)|v(t)|^2 - (1 - \alpha )d^{\alpha -2}(t)d^\prime (t)v(t)v^\prime (t) + d^{\alpha -1}(t)|v^\prime (t)|^2\Big ]\\ &\quad =\Big (\frac{1-\alpha }{2}\Big )^2d^{-\alpha }(t)|u(t)|^2-(1-\alpha )d^\prime (t)v(t)v^\prime (t)+d(t)|v^\prime (t)|^2. \end{split} \cssId{eqform}{\tag{39}} \end{equation}$$
Since for $-1<x<y<1$ ,
$$\begin{align*} |v^2(y)-v^2(x)|&= \Big |\int _x^y \Big (v^2(t)\Big )^\prime dt\Big | =\Big |\int _x^y \Big (d^{1-\alpha }(t)u^2(t)\Big )^\prime dt\Big |\\ &\le (1-\alpha )\int _x^y d^{-\alpha }(t)u^2(t)dt\\ &\quad +2\bigg (\int _x^y d^{-\alpha }(t)u^2(t)dt\bigg )^{1/2}\bigg (\int _x^y d^{2-\alpha }(t)\Big (u^\prime (t)\Big )^2 dt\bigg )^{1/2}, \end{align*}$$
we see that $\lim _{t\rightarrow \pm 1}v(t)$ exists. If $\lim _{t\rightarrow \pm 1}v(t)$ is not $0$ , then $\int _I d^{-\alpha }(t)u^2 dt=\infty$ , which is a contradiction. Thus $v(0)=v(1)=v(-1)=0$ . Then
$$\begin{align*} 2\int _{-1}^1 d^\prime (t)v(t)v^\prime (t)dt&= \int _{-1}^1 d^\prime (t)(v^2(t))^\prime dt =-\int _{0}^1 (v^2(t))^\prime dt +\int _{-1}^0 (v^2(t))^\prime dt = 0. \end{align*}$$
Thus, we see from Equation 39 that
$$\begin{equation} \begin{aligned} &\int _I d^{2-\alpha }(t) | u^\prime |^2 dt-\frac{(1-\alpha )^2}{4}\int _I d^{-\alpha }(t)\bigg (u- \Big (\int _I d^{-\alpha }(t)dt\Big )^{-1}\int _I d^{-\alpha }(t)udt\bigg )^2 dt\\ &=\int _I d^{2-\alpha }(t) | u^\prime |^2 dt-\frac{(1-\alpha )^2}{4}\int _I d^{-\alpha }(t)u^2dx\\ &\quad +\frac{(1-\alpha )^2}{4} \Big (\int _I d^{-\alpha }(t)dt\Big )^{-1}\Big (\int _I d^{-\alpha }(t)udt\Big )^2 \\ &=\int _I d(t)|v^\prime (t)|^2 dt +\frac{(1-\alpha )^2}{4} \Big (\int _I d^{-\alpha }(t)dt\Big )^{-1}\Big (\int _I d^{-\alpha }(t)udt\Big )^2 \\ &\ge 0, \end{aligned} \cssId{cc2}{\tag{40}} \end{equation}$$
which implies that $L_{\alpha ,2-\alpha }(I)=\frac{(1-\alpha )^2}{4}$ . Moreover, Equation 40 implies that if $L_{\alpha ,2-\alpha }(I)$ is attained, then $u\equiv 0$ on $(-1,1)$ .
Next, assume $\alpha <1$ and $N\ge 2$ . We first prove that $L_{\alpha ,2-\alpha }(B^N(0,1))<\frac{(1-\alpha )^2}{4}$ and $L_{\alpha ,2-\alpha }(B^N(0,1))$ is achieved when the space dimension $N > \frac{3-\alpha }{1-\alpha }$ . We take $u(x)=x_Nd^{-s}(x)$ with $s\in (0,\frac{1-\alpha }{2})$ . Since
$$\begin{equation} \begin{aligned} & d^{2-\alpha }|\nabla u|^2 \\ & = d^{2-\alpha }|-s d^{-s-1}(\nabla d) x_N+(0,0,\cdots ,0,d^{-s})|^2\\ &=d^{2-\alpha }\Big (s^2 d^{-2s-2}|\nabla d|^2x_N^2+d^{-2s}-2s d^{-2s-1}x_N\frac{\partial d}{\partial x_N}\Big )\\ &=s^2 d^{-\alpha -2s}x_N^2|\nabla d|^2+d^{-\alpha -2s+2}- \frac{2s}{-\alpha -2s+2}x_N\frac{\partial }{\partial x_N}(d^{-\alpha -2s+2}), \end{aligned} \cssId{a1}{\tag{41}} \end{equation}$$
and
$$\begin{equation*} \int _0^1 r^a (1-r)^bdr=\frac{\Gamma (1+a)\Gamma (1+b)}{\Gamma (2+a+b)}\ \ \text{ for } a,b>-1, \end{equation*}$$
it follows that
$$\begin{align*} &\frac{\int _{B^N(0,1)}d^{2-\alpha }(x)|\nabla u|^2}{\int _{B^N(0,1)}d^{-\alpha }(x)u^2 dx}\\ &\qquad =\frac{\int _{B^N(0,1)}s^2 d^{-\alpha -2s}x_N^2 +d^{-\alpha -2s+2}- \frac{2s}{-\alpha -2s+2}x_N\frac{\partial }{\partial x_N}(d^{-\alpha -2s+2})dx}{\int _{B^N(0,1)}d^{-\alpha -2s}x_N^2dx}\\ &\qquad =s^2+ \Big (1+ \frac{2s}{-\alpha -2s+2}\Big )\frac{\int _{B^N(0,1)} d^{-\alpha -2s+2} dx}{\int _{B^N(0,1)}d^{-\alpha -2s}x_N^2dx}\\ &\qquad =s^2+ \frac{N(\alpha -2)}{\alpha +2s-2} \frac{\int _{B^N(0,1)} d^{-\alpha -2s+2} dx}{\int _{B^N(0,1)}d^{-\alpha -2s}|x| ^2dx}\\ &\qquad =s^2+ \frac{N(\alpha -2)}{\alpha +2s-2} \frac{\int _0^1 r^{N-1}(1-r)^{-\alpha -2s+2} dr}{\int _0^1r^{N+1}(1-r)^{-\alpha -2s}dr}\\ &\qquad =s^2+ \frac{N(\alpha -2)}{\alpha +2s-2} \frac{\Gamma (N)\Gamma (-\alpha -2s+3)}{\Gamma (N+2)\Gamma (-\alpha -2s+1)} \\ &\qquad =s^2+\frac{(2-\alpha )(1-\alpha -2s)}{N+1}\equiv g(s). \end{align*}$$
Then since for $N > \frac{3-\alpha }{1-\alpha }$ ,
$$\begin{equation*} \frac{d g(s)}{ds}\Big |_{s=\frac{1-\alpha }{2}}=1-\alpha -\frac{2(2-\alpha )}{N+1}=\frac{N-3}{N+1}-\frac{N-1}{N+1}\alpha >0,\qquad g\Big (\frac{1-\alpha }{2}\Big ) = \frac{(1-\alpha )^2}{4}, \end{equation*}$$
$g(s) < \frac{(1-\alpha )^2}{4}$ for $s$ less than and close to $\frac{1-\alpha }{2}$ . This implies that $L_{\alpha ,2-\alpha }(B^N(0,1))<\frac{(1-\alpha )^2}{4}$ for $N > \frac{3-\alpha }{1-\alpha }$ .
Lastly, we prove that for $\alpha < 1$ and the space dimension $N \le \frac{3 - \alpha }{1 - \alpha }$ , $L_{\alpha ,2-\alpha }(B_N(0,1))=\frac{(1-\alpha )^2}{4}$ and $L_{\alpha ,2-\alpha }(B_N(0,1))$ is not achieved. If $L_{\alpha ,2-\alpha }(B_N(0,1))$ is achieved by $u$ , we see that $u \in C^2(B^N(0,1))$ . Moreover, since $C^2(B^N(0,1))\cap W^{1,2}_{\alpha ,2-\alpha }(B^N(0,1))$ is dense in $W^{1,2}_{\alpha ,2-\alpha }(B^N(0,1))$ , it suffices to prove that
$$\begin{equation} \frac{\int _{B^N(0,1)} d^{2-\alpha }(x) |\nabla u |^2 dx}{\int _{B^N(0,1)} d^{-\alpha }(x)u^2 dx}\ge \frac{(1-\alpha )^2}{4} \cssId{e2}{\tag{42}} \end{equation}$$
and the equality does not hold for $u\in \Big ( C^2(B^N(0,1))\cap W^{1,2}_{\alpha ,2-\alpha }(B^N(0,1)) \Big )\setminus \{0\}$ with $\int _{B^N(0,1)}d^{-\alpha }(x)udx=0$ . As far as it makes no confusion, we abuse $d(x)=d(r)$ , where $r=|x|$ . We define $\bar{u}(x)\equiv u(x)-\psi _0(r)$ , where $r=|x|$ and $\psi _0(r)=\frac{1}{|S^{N-1}|}\int _{S^{N-1}}u(r\theta )d\theta$ . We note that $\psi _0, \bar{u}\in C^1(B^N(0,1))$ ,
$$\begin{equation} \int _{S^{N-1}}\bar{u}(r\theta )d\theta =\int _{S^{N-1}}u(r\theta )d\theta -|S^{N-1}|\psi _0(r)=0 \text{ for } r\in [0,1) \cssId{nong}{\tag{43}} \end{equation}$$
and
$$\begin{equation*} \begin{split} 0&=\int _{B^N((0,1))} d^{-\alpha }(x)udx=\int _0^1 \int _{S^{N-1}}r^{N-1} d^{-\alpha }(r)u d\theta dr\\ &= |S^{N-1}|\int _0^1 d^{-\alpha }r^{N-1} \psi _0(r)dr. \end{split} \end{equation*}$$
By Equation 43 ,
$$\begin{equation*} \int _{S^{N-1}}\frac{\partial \bar{u}}{\partial r}(r\theta )d\theta =\frac{\partial }{\partial r} \int _{S^{N-1}} \bar{u}(r\theta ) d\theta =0 \text{ for } r\in [0,1). \end{equation*}$$
Hence, for $r\in [0,1)$ ,
$$\begin{equation*} \int _{S^{N-1}}\Big (\frac{\partial u}{\partial r}\Big )^2d\theta =|S^{N-1}|\Big (\psi _0^\prime (r)\Big )^2+\int _{S^{N-1}}\Big (\frac{\partial \bar{u}}{\partial r}\Big )^2d\theta . \end{equation*}$$
Then we deduce that
$$\begin{equation*} \begin{split} &\int _0^1\int _{S^{N-1}}d^{2-\alpha }r^{N-1}\Big (\frac{\partial u}{\partial r}\Big )^2d\theta dr\\ &\qquad =|S^{N-1}|\int _0^1d^{2-\alpha }r^{N-1}\Big (\psi _0^\prime (r)\Big )^2dr+\int _0^1\int _{S^{N-1}}d^{2-\alpha }r^{N-1}\Big (\frac{\partial \bar{u}}{\partial r}\Big )^2d\theta dr. \end{split} \end{equation*}$$
From this and the fact that $\nabla _{S^{N-1}} u=\nabla _{S^{N-1}} \bar{u}$ , we have
$$\begin{equation} \begin{split} &\int _{B^N((0,1))} d^{2-\alpha }|\nabla u|^2dx\\ &\quad =\int _0^1\int _{S^{N-1}}d^{2-\alpha }r^{N-1}\Big (\frac{\partial u}{\partial r}\Big )^2+d^{2-\alpha }r^{N-3}|\nabla _{S^{N-1}} u|^2d\theta dr\\ &\quad =|S^{N-1}|\int _0^1d^{2-\alpha }r^{N-1}\Big (\psi _0^\prime (r)\Big )^2dr\\ &\qquad +\int _0^1 \int _{S^{N-1}}d^{2-\alpha }r^{N-1}\Big (\frac{\partial \bar{u}}{\partial r}\Big )^2+ d^{2-\alpha }r^{N-3}|\nabla _{S^{N-1}} \bar{u}|^2d\theta dr\\ &\quad =|S^{N-1}|\int _0^1d^{2-\alpha }r^{N-1}\Big (\psi _0^\prime (r)\Big )^2dr+\int _{B^N((0,1))} d^{2-\alpha } |\nabla \bar{u}|^2dx. \end{split} \cssId{nong1}{\tag{44}} \end{equation}$$
Similarly, by Equation 43 ,
$$\begin{equation} \begin{split} &\int _{B^N((0,1))} d^{-\alpha }u^2d =\int _0^1\int _{S^{N-1}}d^{-\alpha }r^{N-1}u^2dx \\ &\quad =|S^{N-1}|\int _0^1d^{-\alpha }r^{N-1}\psi _0^2dr+\int _0^1\int _{S^{N-1}}d^{-\alpha }r^{N-1}\bar{u}^2d\theta dr\\ &\quad =|S^{N-1}|\int _0^1d^{-\alpha }r^{N-1}\psi _0^2dr+\int _{B^N((0,1))} d^{-\alpha }\bar{u}^2 dx. \end{split} \cssId{nong2}{\tag{45}} \end{equation}$$
By Equation 44 , Equation 45 and the fact that for $r\in [0,1)$ ,
$$\begin{equation*} \int _{S^{N-1}}|\nabla _{S^{N-1}}\bar{u}|^2d\theta \ge (N-1)\int _{S^{N-1}}\bar{u}^2d\theta , \end{equation*}$$
we get
$$\begin{equation} \begin{split} &\frac{\int _{B^N((0,1))} d^{2-\alpha }|\nabla u|^2dx}{\int _{B^N((0,1))} d^{-\alpha }u^2dx}\\ &=\frac{|S^{N-1}|\int _0^1d^{2-\alpha }r^{N-1}\Big (\psi _0^\prime (r)\Big )^2dr+\int _{B^N((0,1))} d^{2-\alpha } |\nabla \bar{u}|^2dx}{|S^{N-1}|\int _0^1d^{-\alpha }r^{N-1}\psi _0^2dr+\int _{B^N((0,1))} d^{-\alpha }\bar{u}^2 dx}\\ & \ge \frac{ \int _{S^{N-1}}\int _0^1 d^{2-\alpha }r^{N-1}\Big (\psi _0^\prime (r)\Big )^2 + d^{2-\alpha }r^{N-1}\Big (\frac{\partial \bar{u}}{\partial r}\Big )^2+(N-1) d^{2-\alpha }r^{N-3} \bar{u} ^2dr d\theta }{ |S^{N-1}|\int _0^1d^{-\alpha }r^{N-1}\psi _0^2dr+\int _{S^{N-1}}\int _0^1 d^{-\alpha }r^{N-1}\bar{u}^2dr d\theta } \end{split} \cssId{e3}{\tag{46}} \end{equation}$$
where $\int _0^1 d^{-\alpha }r^{N-1} \psi _0(r)dr=0$ and $\int _{S^{N-1}}\bar{u}(r\theta )d\theta =0$ for $r\in [0,1)$ . We claim that for $\theta \in S^{N-1}$ and $\int _0^1 (1-r)^{-\alpha } r^{N-1}\psi _0dr=0$ ,
$$\begin{equation} \begin{aligned} &\int _0^1(1-r)^{2-\alpha }r^{N-1}\Big (\frac{\partial \bar{u}}{\partial r}(r,\theta )\Big )^2+(N-1)(1-r)^{2-\alpha }r^{N-3}(\bar{u}(r,\theta ))^2 dr\\ &\quad \ge \frac{(1-\alpha )^2}{4} \int _0^1 (1-r)^{-\alpha }r^{N-1}(\bar{u}(r,\theta ))^2dr,\\ & \int _0^1(1-r)^{2-\alpha }r^{N-1}(\psi _0^\prime (r))^2 \ge \frac{(1-\alpha )^2}{4} \int _0^1 (1-r)^{-\alpha }r^{N-1}(\psi _0(r))^2dr \end{aligned} \cssId{sat1}{\tag{47}} \end{equation}$$
which implies Equation 42 . We first show that for $f\in C^1([0,1))$ ,
$$\begin{equation} \begin{aligned} & \int _0^1 (1-r)^{2-\alpha }r^{N-1}(f^\prime (r))^2dr+(N-1)\int _0^1(1-r)^{2-\alpha }r^{N-3}f^2(r)dr \\ & \ge \frac{(1-\alpha )^2}{4}\int _0^1 f^2(r)(1-r)^{-\alpha }r^{N-1}dr \end{aligned} \end{equation}$$
when the integrals involved above are all finite. For $N=2$ , the second integral in Equation 4.3 is not finite if $f(0) \ne 0$ . Thus, we may assume that $f(0)=0$ for $N=2$ . Define $g(r)=(1-r)^{\frac{1-\alpha }{2}}r^{-1}f(r)$ . By Lemma 2.10 and the assumption $\alpha \ge \frac{N-3}{N-1}$ , we see that
$$\begin{equation} \begin{aligned} &\int _0^1 (1-r)^{2-\alpha }r^{N-1}(f^\prime (r))^2dr+(N-1)\int _0^1(1-r)^{2-\alpha }r^{N-3}f^2(r)dr\\ &=\int _0^1 r^{N+1}(1-r)^{-1}g^2 \bigg [ \frac{(1-\alpha )^2}{4} - (N -1) r^{-2}(1-r)^2\\ &\qquad +\Big ( 1-(1-\alpha )\frac{N-1}{2}\Big )r^{-1}(1-r)\bigg ] \\ & \qquad +r^{N+1}(1-r)(g^\prime )^2 dr +(N-1)\int _0^1(1-r) r^{N-1}g^2(r)dr\\ &=\int _0^1 r^{N+1}(1-r)^{-1}g^2 \bigg [ \frac{(1-\alpha )^2}{4} +\Big ( 1-(1-\alpha )\frac{N-1}{2}\Big )r^{-1}(1-r)\bigg ] \\ & \qquad +r^{N+1}(1-r)(g^\prime )^2 dr\\ &\ge \frac{(1-\alpha )^2}{4}\int _0^1 f^2(r)(1-r)^{-\alpha }r^{N-1}dr. \end{aligned} \cssId{eso1}{\tag{48}} \end{equation}$$
This implies Equation 4.3 . Next, we show that for $N \le \frac{3-\alpha }{1-\alpha }$ and $f \in C^1([0,1))$ satisfying $\int _0^1r^{N-1}(1-r)^{-\alpha }f dr =0$ ,
$$\begin{equation} \int _0^1 (1-r)^{2-\alpha }r^{N-1}(f^\prime )^2dr \ge \frac{(1-\alpha )^2}{4}\int _0^1 (1-r)^{-\alpha }r^{N-1} f^2dr \cssId{e1}{\tag{49}} \end{equation}$$
when the integrals involved above are all finite. This is equivalent to show without the average condition $\int _0^1r^{N-1}(1-r)^{-\alpha }f dr =0$ that
$$\begin{align*} & \frac{\int _0^1 (1-r)^{2-\alpha }r^{N-1}(f^\prime )^2dr}{\int _0^1 (1-r)^{-\alpha }r^{N-1} \big (f-(\int _0^1 s^{N-1}(1-s)^{-\alpha }ds)^{-1}\int _0^1s^{N-1}(1-s)^{-\alpha }f ds\big )^2dr} \\ & \ge \frac{(1-\alpha )^2}{4}. \end{align*}$$
Note that for any $\kappa \in \mathbb{R}$ ,
$$\begin{align*} & \frac{\int _0^1 (1-r)^{2-\alpha }r^{N-1}(f^\prime )^2dr}{\int _0^1 (1-r)^{-\alpha }r^{N-1} \big (f-(\int _0^1 s^{N-1}(1-s)^{-\alpha }ds)^{-1}\int _0^1s^{N-1}(1-s)^{-\alpha }f ds\big )^2dr} \\ &= \frac{\int _0^1 (1 - r)^{2-\alpha }r^{N-1}\big ((f + \kappa )^\prime \big )^2dr}{\int _0^1 (1 - r)^{-\alpha }r^{N-1} \big [f + \kappa - (\int _0^1 s^{N-1}(1 - s)^{-\alpha }ds)^{-1}\int _0^1s^{N-1}(1 - s)^{-\alpha }(f+\kappa ) ds\big ]^2dr}. \end{align*}$$
Thus it suffices to prove that for $f\in C^1([0,1)) \setminus \{0\}$ with $f(1-e^{1-N})=0$ , $\int _0^1(1-r)^{2-\alpha }(f^\prime )^2dr<\infty$ and $\int _0^1(1-r)^{-\alpha }f^2dr<\infty$ ,
$$\begin{align*} & \frac{\int _0^1 (1-r)^{2-\alpha }r^{N-1}(f^\prime )^2dr}{\int _0^1 (1-r)^{-\alpha }r^{N-1} \big [f-(\int _0^1 s^{N-1}(1-s)^{-\alpha }ds)^{-1}\int _0^1s^{N-1}(1-s)^{-\alpha }f ds\big ]^2dr} \\ & \ge \frac{(1-\alpha )^2}{4}. \end{align*}$$
Defining $g(r)=(1-r)^{\frac{1-\alpha }{2}}f(r)$ , we see from Lemma 2.10 that
$$\begin{equation*} \begin{split} &\int _0^1 (1-r)^{2-\alpha }r^{N-1}(f^\prime )^2dr\\ &\qquad =\int _0^1 \frac{(1 - \alpha )^2}{4} r^{N -1}(1 - r)^{-1}g^2 - (1 - \alpha )\frac{N-1}{2} r^{N-2}g^2 + r^{N -1}(1 - r)(g^\prime )^2 dr. \end{split} \end{equation*}$$
Then it follows that
$$\begin{equation} \begin{aligned} &\frac{\int _0^1 (1-r)^{2-\alpha }r^{N-1}(f^\prime )^2dr}{\int _0^1 (1-r)^{-\alpha }r^{N-1} \big [f-(\int _0^1 s^{N-1}(1-s)^{-\alpha }ds)^{-1} \int _0^1s^{N-1}(1-s)^{-\alpha }fds\big ]^2dr} \\ &=\frac{\int _0^1 (1-r)^{2-\alpha }r^{N-1}(f^\prime )^2dr}{\int _0^1 (1-r)^{-\alpha }r^{N-1} f^2-(\int _0^1 s^{N-1}(1-s)^{-\alpha }ds)^{-1} \big (\int _0^1s^{N-1}(1-s)^{-\alpha }f ds\big )^2 } \\ &\ge \frac{\int _0^1 \frac{(1-\alpha )^2}{4} r^{N -1}(1-r)^{-1}g^2 -(1-\alpha )\frac{N-1}{2} r^{N-2}g^2 +r^{N -1}(1-r)(g^\prime )^2 dr}{\int _0^1 r^{N-1}(1-r)^{-1}g^2(r)dr }\\ & =\frac{(1-\alpha )^2}{4} +\frac{\int _0^1 r^{N-1}(1-r)(g^\prime )^2-(1-\alpha )\frac{N-1}{2}r^{N-2}g^2dr}{\int _0^1 r^{N-1}(1-r)^{-1}g^2(r)dr}. \end{aligned} \cssId{hi5z}{\tag{50}} \end{equation}$$
Denoting $a_N\equiv 1-e^{\frac{3}{2}-N}$ , we see that by integration by parts,
$$\begin{equation} \begin{aligned} \int _0^{a_N}r^{N-2}\Big (\int _r^{a_N}\frac{1}{t^{N-1}(1-t)}dt\Big )dr= \frac{1}{N-1}\int _0^{a_N}\frac{1}{1-r}dr=\frac{N-\frac{3}{2}}{N-1}<1. \end{aligned} \cssId{mu2}{\tag{51}} \end{equation}$$
Since $\frac{1}{s}(e^{(N-1)s}-1) \le N-1$ for $s < 0$ , it follows that by a change of variables $\ln r =s$ ,
$$\begin{equation} \begin{split} &\int _{a_N}^{1}r^{N-2}\Big (\int _{a_N}^r\frac{1}{t^{N-1}(1-t)}dt\Big )dr\\ &\qquad =\frac{1}{N-1}\int _{a_N}^1 \frac{1-r^{N-1}}{r^{N-1}(1-r)}dr =\frac{1}{N-1}\sum _{i=1}^{N-1}\int _{a_N}^1 \frac{1}{r^i} dr\\ &\qquad \le \frac{1}{N-1}\int _1^{N}\int _{a_N}^1 \frac{1}{r^m}drdm =\frac{1}{N-1}\int _{a_N}^1-\frac{1}{r^{N}\ln r}+\frac{1}{r\ln r}dr\\ &\qquad =\frac{1}{N-1}\int _{\ln a_N}^0\frac{e^{(N-1)s}-1}{s e^{(N-1)s}}ds \le \int _{\ln a_N}^0\frac{ 1}{ e^{(N-1)t}}dt \\ &\qquad =\frac{1}{N-1}(-1+a_N^{-(N-1)}) = \frac{1}{N-1}\Big ((1-e^{\frac{3}{2}-N})^{-(N-1)}-1\Big ). \end{split} \cssId{mu3}{\tag{52}} \end{equation}$$
Since
$$\begin{equation*} e^{x} > 2\sqrt {e}x \ \text{ for }\ x\ge 2 \ \text{ and }\ (1-x)^{-\frac{1}{x}}\le e^2 \ \text{ for }\ 0<x<\frac{1}{2}, \end{equation*}$$
we deduce that
$$\begin{equation*} (1-e^{\frac{3}{2}-N})^{1-N}=(1-e^{\frac{3}{2}-N})^{-\frac{1}{e^{\frac{3}{2}-N}}\frac{N-1}{e^{N-\frac{3}{2}}}}\le e^{2\frac{N-1}{e^{N-\frac{3}{2}}}} < e \text{ for }N\ge 3. \end{equation*}$$
From this and Equation 52 , for $N\ge 3$ ,
$$\begin{equation*} \int _{a_N}^{1}r^{N-2}\Big (\int _{a_N}^r\frac{1}{t^{N-1}(1-t)}dt\Big )dr < 1. \end{equation*}$$
When $N=2$ , since $e-\sqrt {e}>1$ ,
$$\begin{equation*} \int _{a_N}^{1}r^{N-2}\Big (\int _{a_N}^r\frac{1}{t^{N-1}(1-t)}dt\Big )dr=\int _{a_N}^1\frac{1}{r}dr=-\ln (1-e^{-\frac{1}{2}})<1. \end{equation*}$$
Thus, we have for $N\ge 2$ ,
$$\begin{equation} \int _{a_N}^{1}r^{N-2}\Big (\int _{a_N}^r\frac{1}{t^{N-1}(1-t)}dt\Big )dr < 1. \cssId{subest}{\tag{53}} \end{equation}$$
Then it follows from Equation 51 and Equation 53 that
$$\begin{equation} \begin{split} \int _0^1 r^{N-2}g^2(r)dr&=\int _0^{a_N}r^{N-2}\Big (\int _r^{a_N}g^\prime (t)dt\Big )^2dr+\int _{a_N}^{1}r^{N-2}\Big (\int _{a_N}^{r}g^\prime (t)dt\Big )^2dr\\ &\le \int _0^{a_N}r^{N-2}\Big (\int _r^{a_N}t^{N-1}(1 - t)(g^\prime (t))^2dt\Big )\Big (\int _r^{a_N}\frac{1}{t^{N-1}(1 - t)}dt\Big )dr\\ &\ \ +\int _{a_N}^{1}r^{N-2}\Big (\int _{a_N}^{r}t^{N-1}(1-t)(g^\prime (t))^2dt\Big )\Big (\int _{a_N}^r\frac{1}{t^{N-1}(1-t)}dt\Big )dr\\ &\le \int _0^{a_N}t^{N-1}(1-t)(g^\prime (t))^2dt \int _0^{a_N}r^{N-2}\Big (\int _r^{a_N}\frac{1}{t^{N-1}(1-t)}dt\Big )dr\\ &\ \ + \int _{a_N}^{1}t^{N-1}(1-t)(g^\prime (t))^2dt \int _{a_N}^{1}r^{N-2}\Big (\int _{a_N}^r\frac{1}{t^{N-1}(1-t)}dt\Big )dr\\ &\le \int _0^{a_N}t^{N-1}(1-t)(g^\prime (t))^2dt + \int _{a_N}^{1}t^{N-1}(1-t)(g^\prime (t))^2dt\\ & = \int _{0}^{1}t^{N-1}(1-t)(g^\prime (t))^2dt. \end{split} \cssId{eso-mid}{\tag{54}} \end{equation}$$
Then, from this, Equation 50 and the assumption $\alpha \ge \frac{N-3}{N-1}$ , we see that
$$\begin{equation} \begin{aligned} &\frac{\int _0^1 (1-r)^{2-\alpha }r^{N-1}(f^\prime )^2dr}{\int _0^1 (1-r)^{-\alpha }r^{N-1} \big [f-(\int _0^1 s^{N-1}(1-s)^{-\alpha }ds)^{-1}\int _0^1s^{N-1}(1-s)^{-\alpha }f ds\big ]^2dr} \\ &\ge \frac{(1-\alpha )^2}{4} +\frac{\int _0^1 r^{N-1}(1-r)(g^\prime )^2-(1-\alpha )\frac{N-1}{2}r^{N-2}g^2dr}{\int _0^1 r^{N-1}(1-r)^{-1}g^2(r)dr}\\ &\ge \frac{(1-\alpha )^2}{4} +\frac{\int _0^1 \big (1-(1-\alpha )\frac{N-1}{2}\big ) r^{N-1}(1-r)(g^\prime )^2 dr}{\int _0^1 r^{N-1}(1-r)^{-1}g^2(r)dr} \ge \frac{(1-\alpha )^2}{4}, \end{aligned} \cssId{eso2}{\tag{55}} \end{equation}$$
which implies Equation 49 . Thus, combining Equation 46 , Equation 4.3 and Equation 49 , we get Equation 42 .
We note that if $L_{\alpha ,2-\alpha }(B_N(0,1))$ is achieved, the equalities in the estimations Equation 48 and Equation 55 instead of inequalities should hold. In Equation 48 , the equality holds only when $g^\prime \equiv 0$ . This means that $f(r)=cr(1-r)^{\frac{\alpha -1}{2}}$ for some $c \in \mathbb{R}$ ; in this case, $c =0$ since $\int _0^1 (1-r)^{-\alpha }r^{N-1}(r(1-r)^{\frac{\alpha -1}{2}})^2 dr = \infty$ . Next, in Equation 55 , the equality holds only when $\int _0^1s^{N-1}(1-s)^{-\alpha }f ds =0$ and $g^\prime \equiv 0$ because the estimations Equation 51 , Equation 53 and the equality in Equation 54 should hold. Since $g^\prime \equiv 0$ , $f=c_1(1-s)^\frac{\alpha -1}{2}$ for some $c_1\in \mathbb{R}$ ; in this case, $c_1=0$ since $\int _0^1s^{N-1}(1-s)^{-\alpha }f ds =0$ . These imply that $L_{\alpha ,2-\alpha }(B_N(0,1))$ is not achieved by any element in $W^{1,2}_{\alpha ,2-\alpha }(B^N(0,1)) \setminus \{0\}$ . $$
4.4. Proof of Theorem 1.6
We assume that $\alpha < 1$ , $N \ge \frac{3-\alpha }{1-\alpha }$ and $(a_1,\cdots ,a_{N-1}) \in (0,1]^{N-1}$ with $a_1+\cdots +a_{N-1} < N-1$ . We define diffeomorphisms $\Psi : B^N(0,1) \to E(a_1,\cdots ,a_{N-1},1)$ by
$$\begin{equation*} \Psi (y)=(a_1y_1,a_2y_2,\cdots , a_{N-1}y_{N-1}, y_N) \end{equation*}$$
and $\Phi : E(a_1,\cdots ,a_{N-1},1) \to B^N(0,1)$ by
$$\begin{equation*} \Phi (x)=\Big (\frac{x_1}{a_1},\frac{x_2}{a_2},\cdots , \frac{x_{N-1}}{a_{N-1}}, x_N \Big ). \end{equation*}$$
For $r(t) \equiv t\Phi (x)+(1-t)\frac{\Phi (x)}{|\Phi (x)|}$ and $x \in \Omega = E(a_1,\cdots ,a_{N-1},1)$ , we see that
$$\begin{equation} d(x)\le \int _0^1\Big |\frac{d}{dt}\Psi (r(t))\Big |dt=\frac{|x|}{|\Phi (x)|}(1-|\Phi (x)|). \cssId{dp1}{\tag{56}} \end{equation}$$
Define $u(x)=x_Nd^{-s}(x)$ , where $s\in (0,\frac{1-\alpha }{2})$ and $\alpha + 2s > 0$ . From Equation 41 and Equation 56 , we see that
$$\begin{equation} \begin{aligned} \frac{\int _{\Omega }d^{2-\alpha }(x)|\nabla u|^2}{\int _{\Omega }d^{-\alpha }(x)u^2 dx}&=\frac{\int _{\Omega }s^2 d^{-\alpha -2s}x_N^2 +d^{-\alpha -2s+2}- \frac{2s}{-\alpha -2s+2}x_N\frac{\partial }{\partial x_N}(d^{-\alpha -2s+2})dx}{\int _{\Omega }d^{-\alpha -2s}x_N^2dx}\\ &=s^2+ \Big (1+ \frac{2s}{-\alpha -2s+2}\Big )\frac{\int _{\Omega } d^{-\alpha -2s+2} dx}{\int _{\Omega }d^{-\alpha -2s}x_N^2dx}\\ &\le s^2+ \Big (\frac{\alpha -2}{\alpha +2s-2}\Big )\frac{\int _{\Omega } \Big (\frac{|x|}{|\Phi (x)|}(1-|\Phi (x)|)\Big )^{-\alpha -2s+2} dx}{\int _{\Omega }\Big (\frac{|x|}{|\Phi (x)|}(1-|\Phi (x)|)\Big )^{-\alpha -2s}x_N^2dx}. \end{aligned} \cssId{a33}{\tag{57}} \end{equation}$$
Since $\frac{|\Psi (y)|}{|y|}\le 1$ and
$$\begin{equation*} \begin{split} \Big (\frac{|\Psi (y)|}{|y|}\Big )^{-\alpha -2s+2}&=\Big (1+\frac{\big (\sum _{i=1}^{N-1}(a_i^2-1)y_i^2\big )}{|y|^2}\Big )^\frac{-\alpha -2s+2}{2}\\ &\le 1+\Big (\frac{-\alpha -2s+2}{2}\Big )\frac{\big (\sum _{i=1}^{N-1}(a_i^2-1)y_i^2\big )}{|y|^2}, \end{split} \end{equation*}$$
we see by a change of variables $x=\Psi (y)$ that
$$\begin{equation} \begin{aligned} &\frac{\int _{\Omega } \Big (\frac{|x|}{|\Phi (x)|}(1-|\Phi (x)|)\Big )^{-\alpha -2s+2} dx}{\int _{\Omega }\Big (\frac{|x|}{|\Phi (x)|}(1-|\Phi (x)|)\Big )^{-\alpha -2s}x_N^2dx}=\frac{\int _{B^N(0,1)} \Big (\frac{|\Psi (y)|}{|y|}(1-|y|)\Big )^{-\alpha -2s+2} dy}{\int _{B^N(0,1)}\Big (\frac{|\Psi (y)|}{|y|}(1-|y|)\Big )^{-\alpha -2s}y_N^2dy}\\ &\le \frac{\int _{B^N(0,1)} (1-|y|) ^{-\alpha -2s+2} \Big [1+\Big (\frac{-\alpha -2s+2}{2}\Big )\frac{\big (\sum _{i=1}^{N-1}(a_i^2-1)y_i^2\big )}{|y|^2}\Big ]dy}{\int _{B^N(0,1)} (1-|y|) ^{-\alpha -2s}y_N^2dy}\\ &= \frac{\int _{B^N(0,1)} (1-|y|) ^{-\alpha -2s+2} \Big [1+N^{-1}\Big (\frac{-\alpha -2s+2}{2}\Big )\Big (\sum _{i=1}^{N-1}(a_i^2-1)\Big )\Big ]dy}{N^{-1}\int _{B^N(0,1)} (1-|y|) ^{-\alpha -2s}|y|^2dy}\\ &=\bigg [N+ \Big (\frac{-\alpha -2s+2}{2}\Big )\Big (\sum _{i=1}^{N-1}(a_i^2-1)\Big )\bigg ]\frac{\int _{B^N(0,1)} (1-|y|) ^{-\alpha -2s+2} dy}{ \int _{B^N(0,1)} (1-|y|) ^{-\alpha -2s}|y|^2dy}. \end{aligned} \cssId{a2}{\tag{58}} \end{equation}$$
Then, by Equation 57 and Equation 58 and the fact that
$$\begin{align*} &\frac{\int _{B^N(0,1)} (1-|y|) ^{-\alpha -2s+2} dy}{ \int _{B^N(0,1)} (1-|y|) ^{-\alpha -2s}|y|^2dy}=\frac{\int _0^1 (1-r)^{-\alpha -2s+2}r^{N-1}dr}{\int _0^1 (1-r)^{-\alpha -2s }r^{N+1}dr} \\ & \qquad =\frac{\Gamma (N)\Gamma (-\alpha -2s+3)}{\Gamma (N+2)\Gamma (-\alpha -2s+1)} =\frac{(-\alpha -2s+2)(-\alpha -2s+1)}{N(N+1)}, \end{align*}$$
we see that for $s\in \Big (\max \{0,-\frac{\alpha }{2}\},\frac{1-\alpha }{2}\Big )$ ,
$$\begin{align*} &\frac{\int _{\Omega }d^{2-\alpha }(x)|\nabla u|^2}{\int _{\Omega }d^{-\alpha }(x)u^2 dx}\\ &\le s^2 + \Big (\frac{\alpha - 2}{\alpha + 2s - 2}\Big )\bigg [N + \Big (\frac{-\alpha - 2s + 2}{2}\Big )\Big (\sum _{i=1}^{N-1}(a_i^2 - 1)\Big )\bigg ]\frac{(-\alpha -2s+2)(-\alpha - 2s + 1)}{N(N + 1)}\\ &=s^2 + \frac{(\alpha -2)(\alpha +2s-1)}{N+1}\bigg [1+ \Big (\frac{-\alpha -2s+2}{2N}\Big )\Big (\sum _{i=1}^{N-1}(a_i^2-1)\Big )\bigg ]\equiv h(s). \end{align*}$$
Note that $h\Big (\frac{1-\alpha }{2}\Big )=\Big (\frac{1-\alpha }{2}\Big )^2$ and
$$\begin{equation*} \begin{split} h^\prime (s)&=2s+\frac{2(\alpha -2) }{N+1}\bigg [1+ \Big (\frac{-\alpha -2s+2}{2N}\Big )\Big (\sum _{i=1}^{N-1}(a_i^2-1)\Big )\bigg ]\\ & \quad -\frac{(\alpha -2)(\alpha +2s-1) }{(N+1)N}\Big (\sum _{i=1}^{N-1}(a_i^2-1)\Big ). \end{split} \end{equation*}$$
Thus, for $N\ge \frac{3-\alpha }{1-\alpha }$ ,
$$\begin{align*} h^\prime \Big (\frac{1-\alpha }{2}\Big )&=1-\alpha +\frac{2(\alpha -2) }{N+1}\bigg [1+ \frac{1}{2N} \Big (\sum _{i=1}^{N-1}(a_i^2-1)\Big )\bigg ] > 1-\alpha +\frac{2(\alpha -2) }{N+1}\ge 0. \end{align*}$$
Taking $s$ strictly less than and sufficiently close to $\frac{1-\alpha }{2}$ , we see that $L_{\alpha ,2-\alpha }(\Omega )<\Big (\frac{1-\alpha }{2}\Big )^2$ when $N\ge \frac{3-\alpha }{1-\alpha }$ . This completes the proof. $$
4.5. Proof of Theorem 1.7
Define $u=x_N d^{-s}(x)$ for $s$ strictly less than and close to $\frac{1-\alpha }{2}$ . Note that $d(x)=\begin{cases} a-|x| &\text{ if } x\in \Omega _1,\\ |x|-1 &\text{ if } x\in \Omega _2,\end{cases}$ where $\Omega _1\equiv \{x\in \Omega \ |\ |x|>\frac{a+1}{2}\}$ and $\Omega _2\equiv \{x\in \Omega \ |\ |x|\le \frac{a+1}{2}\}$ . By Equation 41 , it follows that
$$\begin{equation} \begin{split} &\int _\Omega d^{2-\alpha }|\nabla u|^2dx\\ & =\int _\Omega s^2 d^{-\alpha -2s}x_N^2|\nabla d|^2+d^{-\alpha -2s+2}-2sd^{-\alpha -2s+1}x_N\frac{\partial d}{\partial x_N}dx \\ & =\int _{\Omega _1} s^2 (a-|x|)^{-\alpha -2s}x_N^2 + (a-|x|)^{-\alpha -2s+2}+2s (a-|x|)^{-\alpha -2s+1}x_N^2|x|^{-1}dx\\ & \quad +\int _{\Omega _2} s^2 (|x|-1)^{-\alpha -2s}x_N^2 +(|x|-1)^{-\alpha -2s+2}-2s(|x|-1)^{-\alpha -2s+1}x_N^2|x|^{-1}dx \\ & =|S^{N-1}|\Big [\int _{\frac{a+1}{2}}^a \Big (s^2 (a - r)^{-\alpha -2s}\frac{r^2}{N} + (a - r)^{-\alpha -2s+2} + 2s (a - r)^{-\alpha -2s+1}\frac{r}{N}\Big )r^{N-1}dr\\ & \quad +\int _1^{\frac{a+1}{2}} \Big (s^2 (r-1)^{-\alpha -2s}\frac{r^2}{N} +(r-1)^{-\alpha -2s+2}-2s(r-1)^{-\alpha -2s+1}\frac{r}{N}\Big )r^{N-1}dr\Big ]\\ & =|S^{N-1}| \frac{1}{N} \Big [\int _{\frac{a+1}{2}}^a s^2 r^{N+1} (a - r)^{-\alpha -2s} \\ & \quad + Nr^{N-1}(a - r)^{-\alpha -2s+2} + 2sr^N (a - r)^{-\alpha -2s+1} dr\\ & \quad +\int _1^{\frac{a+1}{2}} s^2 r^{N+1}(r - 1)^{-\alpha -2s} + N r^{N-1}(r - 1)^{-\alpha -2s+2} - 2sr^N(r - 1)^{-\alpha -2s+1} dr\Big ] \end{split} \cssId{ann1}{\tag{59}} \end{equation}$$
and
$$\begin{equation} \begin{split} &\int _\Omega d^{-\alpha }(x)u^2dx=\int _{\Omega _1}(a-|x|)^{-\alpha -2s}x_N^2dx+\int _{\Omega _2}(|x|-1)^{-\alpha -2s}x_N^2dx\\ &\quad =|S^{N-1}| \frac{1}{N} \Big [\int _{\frac{a+1}{2}}^ar^{N+1}(a-r)^{-\alpha -2s}dr+\int _1^{\frac{a+1}{2}} r^{N+1}(r-1)^{-\alpha -2s}dr\Big ]. \end{split} \cssId{ann2}{\tag{60}} \end{equation}$$
Since
$$\begin{equation*} \begin{split} &\int _{\frac{a+1}{2}}^a r^N (a-r)^{-\alpha -2s+1} dr\\ &\qquad =\frac{1}{N+1}\bigg [-\Big (\frac{a+1}{2}\Big )^{N+1}\Big (\frac{a-1}{2}\Big )^{-\alpha -2s+1} \\ & \qquad \qquad +(-\alpha -2s+1)\int _{\frac{a+1}{2}}^a r^{N+1}(a-r)^{-\alpha -2s}dr\bigg ] \end{split} \end{equation*}$$
and
$$\begin{align*} &\int _{\frac{a+1}{2}}^a r^{N-1}(a-r)^{-\alpha -2s+2}dr\\ &=\frac{1}{N}\Big [-\Big (\frac{a+1}{2}\Big )^N\Big (\frac{a-1}{2}\Big )^{-\alpha -2s+2}+(-\alpha -2s+2)\int _{\frac{a+1}{2}}^a r^{N}(a-r)^{-\alpha -2s+1}dr\Big ]\\ &= \frac{1}{N}\Big [-\Big (\frac{a+1}{2}\Big )^N\Big (\frac{a-1}{2}\Big )^{-\alpha -2s+2}\\ &\qquad +(-\alpha -2s+2)\frac{1}{N+1}\Big (-\Big (\frac{a+1}{2}\Big )^{N+1}\Big (\frac{a-1}{2}\Big )^{-\alpha -2s+1}\\ &\qquad +(-\alpha -2s+1)\int _{\frac{a+1}{2}}^a r^{N+1}(a-r)^{-\alpha -2s}dr\Big )\Big ]\\ &=\frac{1}{N(N+1)}\Big [-(N+1)\Big (\frac{a+1}{2}\Big )^N\Big (\frac{a-1}{2}\Big )^{-\alpha -2s+2} \\ & \qquad -(-\alpha -2s+2) \Big (\frac{a+1}{2}\Big )^{N+1}\Big (\frac{a-1}{2}\Big )^{-\alpha -2s+1}\\ &\qquad + (-\alpha -2s+2) (-\alpha -2s+1)\int _{\frac{a+1}{2}}^a r^{N+1}(a-r)^{-\alpha -2s}dr\Big ], \end{align*}$$
we deduce that
$$\begin{equation} \begin{aligned} &\int _{\frac{a+1}{2}}^a s^2 r^{N+1} (a-r)^{-\alpha -2s} + Nr^{N-1}(a-r)^{-\alpha -2s+2} + 2sr^N (a-r)^{-\alpha -2s+1} dr\\ &=\frac{1}{N+1}\bigg [-(N+1)\Big (\frac{a+1}{2}\Big )^N\Big (\frac{a-1}{2}\Big )^{-\alpha -2s+2}\\ &\qquad -(-\alpha -2s+2) \Big (\frac{a+1}{2}\Big )^{N+1}\Big (\frac{a-1}{2}\Big )^{-\alpha -2s+1}\\ &\qquad -2s\Big (\frac{a+1}{2}\Big )^{N+1}\Big (\frac{a-1}{2}\Big )^{-\alpha -2s+1}\bigg ]\\ &\qquad +\bigg [s^2+\frac{1}{N+1} (-\alpha -2s+2) (-\alpha -2s+1)\\ &\qquad +\frac{2s}{N+1}(-\alpha -2s+1)\bigg ]\int _{\frac{a+1}{2}}^a r^{N+1}(a-r)^{-\alpha -2s}dr\\ &=\frac{1}{N+1}\bigg [-(N+1)\Big (\frac{a+1}{2}\Big )^N\Big (\frac{a-1}{2}\Big )^{-\alpha -2s+2} \\ & \qquad +(\alpha -2) \Big (\frac{a+1}{2}\Big )^{N+1}\Big (\frac{a-1}{2}\Big )^{-\alpha -2s+1}\bigg ]\\ &\qquad +\bigg [s^2+\frac{1}{N+1} (-\alpha +2) (-\alpha -2s+1) \bigg ]\int _{\frac{a+1}{2}}^a r^{N+1}(a-r)^{-\alpha -2s}dr. \end{aligned} \cssId{ann3}{\tag{61}} \end{equation}$$
Similarly, since
$$\begin{align*} &\int _1^{\frac{a+1}{2}} r^{N}(r-1)^{-\alpha -2s+1}dr\\ &\quad =\frac{1}{N+1}\bigg [\Big (\frac{a+1}{2}\Big )^{N+1}\Big (\frac{a-1}{2}\Big )^{-\alpha -2s+1}\\ &\qquad -(-\alpha -2s+1)\int _1^{\frac{a+1}{2}} r^{N+1}(r-1)^{-\alpha -2s}dr\bigg ] \end{align*}$$
and
$$\begin{align*} &\int _1^{\frac{a+1}{2}} r^{N-1}(r-1)^{-\alpha -2s+2}dr\\ &=\frac{1}{N}\Big [\Big (\frac{a+1}{2}\Big )^N\Big (\frac{a-1}{2}\Big )^{-\alpha -2s+2}-(-\alpha -2s+2)\int _1^{\frac{a+1}{2}} r^{N}(r-1)^{-\alpha -2s+1}dr\Big ]\\ &=\frac{1}{N}\bigg [\Big (\frac{a+1}{2}\Big )^N\Big (\frac{a-1}{2}\Big )^{-\alpha -2s+2}\\ &-(-\alpha -2s+2)\frac{1}{N+1}\bigg (\Big (\frac{a+1}{2}\Big )^{N+1}\Big (\frac{a-1}{2}\Big )^{-\alpha -2s+1}\\ &-(-\alpha -2s+1)\int _1^{\frac{a+1}{2}} r^{N+1}(r-1)^{-\alpha -2s}dr\bigg )\bigg ]\\ &=\frac{1}{N(N+1)}\bigg [(N+1)\Big (\frac{a+1}{2}\Big )^N\Big (\frac{a-1}{2}\Big )^{-\alpha -2s+2} \\ & \qquad \qquad \qquad -(-\alpha -2s+2) \Big (\frac{a+1}{2}\Big )^{N+1}\Big (\frac{a-1}{2}\Big )^{-\alpha -2s+1}\\ &\qquad \qquad \qquad +(-\alpha -2s+2)(-\alpha -2s+1)\int _1^{\frac{a+1}{2}} r^{N+1}(r-1)^{-\alpha -2s}dr\bigg ], \end{align*}$$
we see that
$$\begin{equation} \begin{split} &\int _1^{\frac{a+1}{2}} s^2 r^{N+1}(r-1)^{-\alpha -2s} +N r^{N-1}(r-1)^{-\alpha -2s+2}-2sr^N(r-1)^{-\alpha -2s+1} dr\\ &=\frac{1}{N+1}\bigg [(N+1)\Big (\frac{a+1}{2}\Big )^N\Big (\frac{a-1}{2}\Big )^{-\alpha -2s+2} \\ & \qquad \qquad -(-\alpha -2s+2) \Big (\frac{a+1}{2}\Big )^{N+1}\Big (\frac{a-1}{2}\Big )^{-\alpha -2s+1}\\ & \qquad \qquad -2s\Big (\frac{a+1}{2}\Big )^{N+1}\Big (\frac{a-1}{2}\Big )^{-\alpha -2s+1}\bigg ]\\ &\qquad +\bigg [ s^2+\frac{1}{N+1}(-\alpha -2s+2)(-\alpha -2s+1)\\ &\qquad \qquad +\frac{1}{N+1}2s(-\alpha -2s+1) \bigg ]\int _1^{\frac{a+1}{2}} r^{N+1}(r-1)^{-\alpha -2s}dr\\ &=\frac{1}{N+1}\bigg [(N+1)\Big (\frac{a+1}{2}\Big )^N\Big (\frac{a-1}{2}\Big )^{-\alpha -2s+2} \\ & \qquad \qquad +(\alpha -2)\Big (\frac{a+1}{2}\Big )^{N+1}\Big (\frac{a-1}{2}\Big )^{-\alpha -2s+1}\bigg ]\\ &\qquad +\bigg [ s^2+\frac{1}{N+1}(-\alpha +2)(-\alpha -2s+1) \bigg ]\int _1^{\frac{a+1}{2}} r^{N+1}(r-1)^{-\alpha -2s}dr. \end{split} \cssId{ann4}{\tag{62}} \end{equation}$$
Then, by Equation 59 , Equation 60 , Equation 61 , Equation 62 and the identity
$$\begin{equation*} \begin{split} &\int _0^1 r^{-\alpha -2s}\Big [\Big (r+\frac{2}{a-1}\Big )^{N+1}+\Big (\frac{2a}{a-1}- r\Big )^{N+1}\Big ]dr\\ &=(-\alpha -2s+1)^{-1}\Big [2\Big ( \frac{a+1}{a-1}\Big )^{N+1}\\ &\quad -(N+1)\int _0^1 r^{-\alpha -2s+1}\Big (\Big (r+\frac{2}{a-1}\Big )^N- \Big (\frac{2a}{a-1}- r\Big )^N\Big )dr\Big ], \end{split} \end{equation*}$$
we obtain that
$$\begin{equation*} \begin{split} &\frac{\int _\Omega d^{2-\alpha }|\nabla u|^2dx}{\int _\Omega d^{-\alpha }(x)u^2dx}=s^2+\frac{1}{N+1}(-\alpha +2)(-\alpha -2s+1)\\ &\qquad +2 \frac{\alpha - 2}{N + 1}\Big (\frac{a + 1}{2}\Big )^{N+1}\Big (\frac{a - 1}{2}\Big )^{-\alpha -2s+1}\bigg [\int _1^{\frac{a+1}{2}} r^{N+1}(r-1)^{-\alpha -2s}dr\\ &\hspace{7cm}+\int _{\frac{a+1}{2}}^a r^{N+1}(a-r)^{-\alpha -2s}dr\bigg ]^{-1}\\ &=s^2+\frac{1}{N+1}(-\alpha +2)(-\alpha -2s+1)\\ &\qquad +2\frac{\alpha -2}{N+1} \Big (\frac{a+1}{a-1}\Big )^{N+1}\bigg [\int _0^1 r^{-\alpha -2s}\Big [\Big (r + \frac{2}{a-1}\Big )^{N+1} + \Big (\frac{2a}{a-1} - r\Big )^{N+1}\Big ]dr \bigg ]^{-1}\\ &=s^2+\frac{1}{N+1}(-\alpha +2)(-\alpha -2s+1)\\ &\qquad +2\frac{\alpha -2}{N+1} \Big (\frac{a+1}{a-1}\Big )^{N+1}(-\alpha -2s+1)\\ &\qquad \times \Big [2\Big ( \frac{a+1}{a-1}\Big )^{N+1} -(N+1)\int _0^1 r^{-\alpha -2s+1}\Big (\Big (r+\frac{2}{a-1}\Big )^N\\ &\qquad - \Big (\frac{2a}{a-1}- r\Big )^N\Big )dr\Big ]^{-1} \equiv f(s). \end{split} \end{equation*}$$
Define $h(s)\equiv \int _0^1 r^{-\alpha -2s+1}\Big [\Big (r+\frac{2}{a-1}\Big )^N- \Big (\frac{2a}{a-1}- r\Big )^N\Big ]dr$ . Note that
$$\begin{equation*} \Big |h^\prime \Big (\frac{1-\alpha }{2}\Big )\Big |=2\Big |\int _0^1 \ln r \Big [\Big (r+\frac{2}{a-1}\Big )^N- \Big (\frac{2a}{a-1}- r\Big )^N\Big ]dr\Big |<C_{a } \end{equation*}$$
and
$$\begin{equation*} \begin{split} &2\Big ( \frac{a+1}{a-1}\Big )^{N+1}-(N+1)\int _0^1 \Big (r+\frac{2}{a-1}\Big )^N- \Big (\frac{2a}{a-1}- r\Big )^N dr\\ &\qquad = \Big (\frac{2}{a-1}\Big )^{N+1}(1+a^{N+1}), \end{split} \end{equation*}$$
where $C_{a }>0$ is a constant depending on $a > 0$ . Then, since $f \Big (\frac{1-\alpha }{2}\Big )=\Big (\frac{1-\alpha }{2}\Big )^2$ and
$$\begin{equation*} \begin{split} f^\prime \Big (\frac{1-\alpha }{2}\Big )&=1-\alpha +\frac{1}{N+1}(-\alpha +2)( -2 )\\ &\quad +2\frac{\alpha -2}{N+1} \Big (\frac{a+1}{a-1}\Big )^{N+1}( -2 )\Big (\frac{a-1}{2}\Big )^{N+1}(1+a^{N+1})^{-1}\\ &=1-\alpha +\frac{2}{N+1}(\alpha -2) -\frac{\alpha -2}{N+1} (a+1)^{N+1} 2^{-N+1}(1+a^{N+1})^{-1}>0 \end{split} \end{equation*}$$
if $\alpha <\frac{1-\frac{2}{N+1}\Big (2-(a+1)^{N+1}2^{-N+1}(1+a^{N+1})^{-1}\Big )}{1-\frac{1}{N+1}\Big (2-(a+1)^{N+1}2^{-N+1}(1+a^{N+1})^{-1} \Big )}$ . This implies that $f(s) < \frac{(1-\alpha )^2}{4}$ for $s$ strictly less than and sufficiently close to $\frac{1-\alpha }{2}$ . This completes the proof. $$
4.6. Proof of Theorem 1.8
Lemma 4.1.
Let $\Omega =B^N(0,a)\setminus \overline{B^N(0,1)}$ , where $N\ge 2$ and $a >1$ . Then there exists $\alpha _1\in (0,1)$ such that for $\alpha \in (\alpha _1,1)$ and $u \in C^1(\Omega ) \cap W_{\alpha ,\beta }^{1,2}(\Omega )$ satisfying $\int _{S^{N-1}}u(r\theta )d\theta =0$ with any $r \in (1, a)$ ,
$$\begin{equation} \int _\Omega d^{2-\alpha } |\nabla {u}|^2dx\ge \frac{(1-\alpha )^2}{4}\int _\Omega d^{-\alpha } u ^2 dx. \ \ \cssId{none}{\tag{63}} \end{equation}$$
Moreover, for $\alpha \in (\alpha _1,1)$ , the best constant $\frac{(1-\alpha )^2}{4}$ in Equation 63 is not achieved.
Proof.
Denote $\Omega _1=B^N(0,a)\setminus \overline{B^N(0,\frac{1+a}{2})}$ and $\Omega _2=B^N(0,\frac{1+a}{2})\setminus \overline{B^N(0,1)}$ . For $u\in C^1(\Omega ) \cap W_{\alpha ,\beta }^{1,2}(\Omega )$ , we define $v=d^{\frac{1-\alpha }{2}}(x)u$ . Then, we see that
$$\begin{equation*} \int _\Omega d^{ -1}(x) v^2dx= \int _\Omega d^{ -\alpha }(x) u^2dx<\infty \ \ \text{ and }\ \ \int _{S^{N-1}}v(r\theta )d\theta =0 \text{ for } r\in (1, a). \end{equation*}$$
Then, since
$$\begin{align*} &d^{2-\alpha }(x)|\nabla u|^2\\ &\quad =d^{2-\alpha }(x)\Big | \frac{\alpha -1}{2}d^{\frac{\alpha -1}{2}-1}(\nabla d)v + d^{\frac{\alpha -1}{2}}(x)\nabla v\Big |^2\\ &\quad =d^{2-\alpha }(x)\Big (\frac{(1-\alpha )^2}{4}d^{\alpha -3}(x)|\nabla d|^2v^2 + d^{\alpha - 1 }(x)|\nabla v |^2 - (1-\alpha )d^{\alpha -2}(x)v\nabla d\cdot \nabla v\Big )\\ &\quad = \frac{(1-\alpha )^2}{4}d^{ -1}(x)|\nabla d|^2v^2+d|\nabla v |^2 -\frac{1-\alpha }{2} \nabla d\cdot \nabla (v^2), \end{align*}$$
it follows that $\int _\Omega d|\nabla v |^2dx<\infty$ and
$$\begin{equation} \begin{aligned} &\int _\Omega d^{2-\alpha }(x)|\nabla u|^2-\frac{(1-\alpha )^2}{4}d^{-\alpha }(x)u^2dx\\ &=\int _\Omega d|\nabla v |^2 -\frac{1-\alpha }{2} \nabla d\cdot \nabla (v^2)dx\\ &=\int _{\Omega _1} d|\nabla v |^2 +\frac{1-\alpha }{2} (\Delta d)v^2dx+\int _{\Omega _2} d|\nabla v |^2\\ & \quad +\frac{1-\alpha }{2} (\Delta d)v^2dx-( 1-\alpha )\int _{\partial \Omega _1 \cap \partial \Omega _2} v^2d\sigma \\ &=\int _{\Omega _1} d|\nabla v |^2 +\frac{1-\alpha }{2}\frac{1-N}{|x|}v^2dx+\int _{\Omega _2} d|\nabla v |^2\\ & \quad +\frac{1-\alpha }{2 }\frac{N-1}{|x|}v^2dx-( 1-\alpha )\int _{\partial \Omega _1 \cap \partial \Omega _2} v^2d\sigma . \end{aligned} \cssId{ac1}{\tag{64}} \end{equation}$$
By the trace inequality, we see that for some $C_1 > 0$ ,
$$\begin{equation} \int _{\partial \Omega _1 \cap \partial \Omega _2} v^2d\sigma \le C_1 \int _{ \Omega _3} |\nabla v|^2+v^2dx, \cssId{edi3}{\tag{65}} \end{equation}$$
where $\Omega _3 = B^N(0,\frac{3a +1}{4})\setminus \overline{B^N(0,\frac{1+a}{2})}$ . For small $\delta >0$ , define $\psi \in C^\infty ([1,a])$ such that $\psi \in [0,1]$ , $\psi (r)=1$ if $r\in [a-\delta /2,a]$ and $\psi (r)=0$ if $r\in [1,a-\delta ]$ . Denote $v=v_1+v_2$ , where $v_1=v\psi (d(x))$ and $v_2=v\Big (1-\psi (d(x))\Big )$ . Since $\int _{S^{N-1}}v(r\theta )d\theta =0 \text{ for } r\in (1, a)$ , it holds that
$$\begin{equation} \begin{aligned} \int _{ \Omega _1} d|\nabla v |^2dx &= \int _{\frac{1+a}{2}}^{a}\int _{S^{N-1}}(a-r) r^{N-1}\Big |\frac{\partial v }{\partial r}\Big |^2 + (a-r)r^{N-3}|\nabla _{S^{N-1}} v |^2 d\sigma dr\\ &\ge \int _{\frac{1+a}{2}}^{a}\int _{S^{N-1}} (a-r)r^{N-1}\Big |\frac{\partial v }{\partial r}\Big |^2 + (N-1)(a-r)r^{N-3} v^2 d\sigma dr. \end{aligned} \cssId{edi1}{\tag{66}} \end{equation}$$
Then, from Equation 66 and the fact that
$$\begin{equation*} \int _{ \Omega _1} |\nabla v_2|^2dx \ge \int _{\frac{1+a}{2}}^{a}\int _{S^{N-1}} r^{N-1}\Big |\frac{\partial v_2 }{\partial r}\Big |^2 + (N-1)r^{N-3} v_2^2 d\sigma dr\ge C_2^{-1}\int _{\Omega _1}v_2^2dx, \end{equation*}$$
we have
$$\begin{equation} \begin{aligned} \int _{\Omega _1}v_2 ^2dx \le C_2\int _{\Omega _1} |\nabla v_2|^2dx&\le 2C_2\int _{\Omega _1} |\nabla v|^2\Big (1-\psi (d(x))\Big )^2+v^2\Big (\psi ^\prime (d(x))\Big )^2dx\\ &\le C_3\int _{\Omega _1}d|\nabla v|^2dx, \end{aligned} \cssId{edi2}{\tag{67}} \end{equation}$$
where $C_2, C_3$ are constants. On the other hand, by Equation 5 , Equation 66 and Lemma 2.3 ,
$$\begin{equation} \begin{aligned} \int _{\Omega _1}v_1^2dx &\le 2\int _{\partial \Omega }\int _{a-\delta }^{a}v_1^2(t,\sigma )dtd\sigma \le 32\int _{\partial \Omega }\int _{a-\delta }^{a}(a-t)\Big (\frac{\partial }{\partial t}v_1 (t,\sigma )\Big )^2 dtd\sigma \\ &\le 64\int _{\Omega _1} d|\nabla v_1|^2dx\le 128\int _{\Omega _1} d|\nabla v|^2\Big (\psi (d(x))\Big )^2+d\Big (\psi ^\prime (d(x))\Big )^2 v^2 dx\\ &\le C_4\int _{\Omega _1}d|\nabla v|^2, \end{aligned} \cssId{non1}{\tag{68}} \end{equation}$$
where $C_4$ is a constant. Then, by Equation 67 and Equation 68 , we see that
$$\begin{align*} \int _{\Omega _1}v^2dx\le 2\int _{\Omega _1}v_1^2dx+2\int _{\Omega _1}v_2^2dx\le 2(C_3+C_4) \int _{\Omega _1}d|\nabla v |^2dx. \end{align*}$$
Then, from this and Equation 65 , there exists $\alpha _1=\alpha _1(N,\Omega )\in (0,1)$ such that for $\alpha \in (\alpha _1,1)$
$$\begin{align*} &\int _\Omega d^{2-\alpha }(x)|\nabla u|^2-\frac{(1-\alpha )^2}{4}d^{-\alpha }(x)u^2dx\\ &=\int _{\Omega _1} d|\nabla v |^2 -\frac{1-\alpha }{2|x|}(N-1) v^2dx\\ &\quad +\int _{\Omega _2} d|\nabla v |^2 +\frac{1-\alpha }{2|x|}(N-1) v^2dx-( 1-\alpha )\int _{\partial \Omega _1 \cap \partial \Omega _2} v^2d\sigma \\ &\ge \int _{\Omega _1} d|\nabla v |^2 -\frac{1-\alpha }{2|x|}(N-1) v^2dx\\ &\quad +\int _{\Omega _2} d|\nabla v |^2 +\frac{1-\alpha }{2|x|}(N-1) v^2dx-C_1( 1-\alpha )\int _{ \Omega _3} |\nabla v|^2+v^2 dx\\ &\ge \frac{1}{2} \int _{\Omega _1} d|\nabla v |^2 dx + \int _{\Omega _2} d|\nabla v |^2 +\frac{1-\alpha }{2|x|}(N-1) v^2dx. \end{align*}$$
This proves the inequality Equation 63 . If the equality in Equation 63 holds, we see from the estimation above that $v(x) = 0$ for $x \in \Omega _2$ and $v$ is constant on $\Omega _1$ . Since $v=d^{\frac{1-\alpha }{2}}u$ , we conclude that $u\equiv 0$ in $\Omega$ . This completes the proof.
■
Lemma 4.2.
Let $N\ge 2$ and $a >1$ . Then there exists $\alpha _2=\alpha _2(N,a)\in (0,1)$ such that for $\alpha \in (\alpha _2,1)$ ,
$$\begin{equation} \int _1^a(d(r))^{2-\alpha }r^{N-1}\Big (f^\prime (r)\Big )^2dr\ge \frac{(1-\alpha )^2}{4}\int _1^a(d(r))^{-\alpha }r^{N-1}f^2dr \cssId{none1}{\tag{69}} \end{equation}$$
for any radially symmetric $f(x) = f(r) \in C^1(\Omega ) \cap W_{\alpha ,\beta }^{1,2}(\Omega )$ satisfying
$$\begin{equation*} \int _1^a (d(r))^{-\alpha }r^{N-1} f(r)dr=0 , \end{equation*}$$
where $d(r)=\begin{cases} r-1 &\text{ if } r\in [1,\frac{1+a}{2}],\\ a-r &\text{ if } r \in [\frac{1+a}{2},a].\end{cases}$ Moreover, for $\alpha \in (\alpha _2,1)$ , the best constant $\frac{(1-\alpha )^2}{4}$ in Equation 69 is not achieved.
Proof.
We note that for all $c\in \mathbb{R}$ ,
$$\begin{align*} &\inf \bigg \{ \frac{ \int _1^a d^{2-\alpha }r^{N-1}\Big (f^\prime (r)\Big )^2dr}{ \int _1^ad^{-\alpha }r^{N-1}f^2dr} \ \Big |\ f(r)\\ &\qquad \in \Big (C^1(\Omega ) \cap W_{\alpha ,\beta }^{1,2}(\Omega )\Big ) \setminus \{0\} , \int _1^a d^{-\alpha }r^{N-1} f(r)dr=0\bigg \}\\ &=\inf \bigg \{ \frac{ \int _1^ad^{2-\alpha }r^{N-1}\Big (f^\prime (r)\Big )^2dr}{\int _1^ad^{-\alpha }r^{N-1}\Big (f-(\int _1^ad^{-\alpha }r^{N-1}dr)^{-1}\int _1^ad^{-\alpha }r^{N-1}fdr\Big )^2dr} \ \Big |\ f(r)\\ &\qquad \in \Big (C^1(\Omega ) \cap W_{\alpha ,\beta }^{1,2}(\Omega )\Big ) \setminus \{0\} \bigg \} \\ &=\inf \bigg \{ \frac{ \int _1^a d^{2-\alpha }r^{N-1}\Big ((f(r)+c)^\prime \Big )^2dr}{\int _1^a \frac{r^{N-1}}{ d^{\alpha }}\Big (f+c-(\int _1^a \frac{r^{N-1}}{ d^{\alpha }}dr)^{-1}\int _1^a \frac{r^{N-1}}{ d^{\alpha }}(f+c)dr\Big )^2dr} \ \Big |\ f(r)\\ &\qquad \in \Big (C^1(\Omega ) \cap W_{\alpha ,\beta }^{1,2}(\Omega )\Big ) \setminus \{0\} \bigg \}. \end{align*}$$
Then it suffices to prove that for $f\in C^1(\Omega ) \cap W_{\alpha ,\beta }^{1,2}(\Omega )$ and $f\Big (\frac{1+a}{2}\Big )=0$ ,
$$\begin{align*} \frac{\int _1^a d^{2-\alpha }r^{N-1} (f^\prime )^2dr}{\int _1^a d^{-\alpha }r^{N-1}f^2dr}\ge \frac{(1-\alpha )^2}{4}. \end{align*}$$
Defining $f(r)=d^\frac{\alpha -1}{2}(r)h(r)$ , we see that
$$\begin{equation} \begin{aligned} d^{2-\alpha }r^{N-1}(f^\prime )^2 &=d^{2-\alpha }r^{N-1}\Big ( \frac{\alpha -1}{2}d^\frac{\alpha -3}{2}d^\prime h+d^\frac{\alpha -1}{2}h^\prime \Big )^2\\ &=d^{2-\alpha }r^{N-1}\bigg [\Big (\frac{\alpha -1}{2}\Big )^2d^{\alpha -3}h^2+d^{\alpha -1}(h^\prime )^2+(\alpha -1)d^{\alpha -2}hd^\prime h^\prime \bigg ]\\ &=\Big (\frac{\alpha -1}{2}\Big )^2r^{N-1}d^{-1}h^2+dr^{N-1}(h^\prime )^2+\frac{\alpha -1}{2} r^{N-1}d^\prime (h^2)^\prime . \end{aligned} \cssId{nonm}{\tag{70}} \end{equation}$$
By Equation 70 , we obtain that
$$\begin{equation} \begin{aligned} &\int _{\frac{1+a}{2}}^a (a-r)^{2-\alpha }r^{N-1}(f^\prime )^2dr\\ &=\int _{\frac{1+a}{2}}^a \Big (\frac{\alpha -1}{2}\Big )^2r^{N-1}(a-r)^{-1}h^2 +(a-r)r^{N-1}(h^\prime )^2-\frac{\alpha -1}{2} r^{N-1}(h^2)^\prime dr\\ &\ge \int _{\frac{1+a}{2}}^a \Big (\frac{\alpha -1}{2}\Big )^2r^{N-1}(a-r)^{-1}h^2+(a-r)r^{N-1}(h^\prime )^2+\frac{\alpha -1}{2}(N-1)r^{N-2} h^2 dr. \end{aligned} \cssId{nonme}{\tag{71}} \end{equation}$$
Since
$$\begin{equation*} h(r)=\int _{\frac{1+a}{2}}^r h^\prime (t)dt \le \Big (\int _{\frac{1+a}{2}}^rt^{N-1}(a-t) (h^\prime (t) )^2dt\Big )^\frac{1}{2} \Big (\int _{\frac{1+a}{2}}^r\frac{1}{t^{N-1}(a-t)}dt\Big )^\frac{1}{2}, \end{equation*}$$
we have
$$\begin{align*} \int _{\frac{1+a}{2}}^a r^{N-2}h^2(r)dr&\le \int _{\frac{1+a}{2}}^a r^{N-1}(a-r) (h^\prime )^2dr \int _{\frac{1+a}{2}}^a\int _{\frac{1+a}{2}}^r\frac{r^{N-2}}{t^{N-1}(a-t)}dtdr\\ &\le c_{N,a} \int _{\frac{1+a}{2}}^a r^{N-1}(a-r) (h^\prime )^2dr, \end{align*}$$
where $c_{N,a}>0$ is a constant depending only on $N$ and $a$ . Thus, we see that
$$\begin{equation*} \begin{split} &\int _{\frac{1+a}{2}}^a (a-r)r^{N-1}(h^\prime )^2+\frac{\alpha -1}{2} (N-1)r^{N-2} h^2 dr\\ &\qquad \ge \Big (\frac{1}{c_{N,a}}-\frac{1-\alpha }{2}(N-1)\Big )\int _{\frac{1+a}{2}}^a r^{N-2}h^2dr. \end{split} \end{equation*}$$
From this and Equation 71 , there exists $\alpha _2=\alpha _2(N,a)\in (0,1)$ and $C_4 > 0$ such that for $\alpha \in (\alpha _2,1)$ ,
$$\begin{equation} \begin{aligned} & \int _{\frac{1+a}{2}}^a (1-r)^{2-\alpha }r^{N-1}(f^\prime )^2dr\\ &\ge \Big (\frac{\alpha -1}{2}\Big )^2\int _{\frac{1+a}{2}}^ar^{N-1}(1-r)^{-1}h^2dr + C_4 \int _{\frac{1+a}{2}}^a r^{N-2}h^2dr\\ &=\Big (\frac{\alpha -1}{2}\Big )^2\int _{\frac{1+a}{2}}^a r^{N-1}(1-r)^{-\alpha }f^2dr + C_4\int _{\frac{1+a}{2}}^a r^{N-2}h^2dr. \end{aligned} \cssId{nonm1}{\tag{72}} \end{equation}$$
On the other hand, we see from Equation 70 and the fact $f(\frac{1+a}{2}) = 0$ that
$$\begin{equation} \begin{aligned} &\int _1^{\frac{1+a}{2}}(r-a)^{2-\alpha }r^{N-1}(f^\prime )^2dr\\ &=\int _1^{\frac{1+a}{2}}\Big (\frac{\alpha -1}{2}\Big )^2r^{N-1}(r-a)^{-1}h^2+(r-a)r^{N-1}(h^\prime )^2+\frac{\alpha -1}{2} r^{N-1}(h^2)^\prime dr\\ &\ge \int _1^{\frac{1+a}{2}}\Big (\frac{\alpha -1}{2}\Big )^2r^{N-1}(r-a)^{-1}h^2+(r-a)r^{N-1}(h^\prime )^2+\frac{1-\alpha }{2}(N-1) r^{N-2} h^2 dr.\end{aligned} \cssId{nonm2}{\tag{73}} \end{equation}$$
Thus, combining Equation 72 and Equation 73 , we get Equation 69 . If the equality holds in Equation 69 , we see from Equation 72 and Equation 73 that $h\equiv 0$ ; thus $f \equiv 0$ . This completes the proof.
■
Now we are ready to complete the proof of Theorem 1.8 . As far as it makes no confusion, we abuse $d(x)=d(r)$ , where $r=|x|$ . Let $u\in C^1(\Omega ) \cap W_{\alpha ,\beta }^{1,2}(\Omega )$ such that $\int _\Omega d^{-\alpha }(x)udx=0$ . We define $\bar{u}(x)\equiv u(x)-\psi _0(r)$ , where $r=|x|$ and $\psi _0(r)=\frac{1}{|S^{N-1}|}\int _{S^{N-1}}u(r\theta )d\theta$ . Then, following the arguments in Theorem 1.5 (see Equation 46 above), we deduce that
$$\begin{align*} \frac{\int _\Omega d^{2-\alpha }|\nabla u|^2dx}{\int _\Omega d^{-\alpha }u^2dx}=\frac{|S^{N-1}|\int _a^1d^{2-\alpha }r^{N-1}\Big (\psi _0^\prime (r)\Big )^2dr+\int _\Omega d^{2-\alpha } |\nabla \bar{u}|^2dx}{|S^{N-1}|\int _a^1d^{-\alpha }r^{N-1}\psi _0^2dr+\int _\Omega d^{-\alpha }\bar{u}^2 dx}, \end{align*}$$
where $\int _1^a d^{-\alpha }r^{N-1} \psi _0(r)dr=0$ and $\int _{S^{N-1}}\bar{u}(r\theta )d\theta =0$ for $r\in (1,a)$ . Lastly, applying Lemma 4.1 and Lemma 4.2 , we conclude to the claim. $$
4.7. Proof of Theorem 1.9
Define
$$\begin{equation*} v(x)=\begin{cases} \lambda _1 u_1\Big (\frac{1}{s_1}(O_1)^{-1}(x-t_1) \Big ) &\text{ if } x \in t_1+s_1O_1D^{u_1}_+,\\ -\lambda _2 u_2\Big (\frac{1}{s_2}(O_2)^{-1}(x-t_2)\Big ) &\text{ if } x \in t_2+s_2O_2D^{u_2}_+,\\ 0 &\text{ if } x \in \Omega \setminus (t_1 + s_1O_1D^{u_1}_+) \cup (t_2 + s_2O_2D^{u_2}_+), \end{cases} \end{equation*}$$
where we take $\lambda _1,\lambda _2 > 0$ so that $\int _{\Omega }d^{-\alpha }(x)v(x) dx = 0$ . Then, we see that $v \in W^{1,2}_{\alpha ,2-\alpha }(\Omega ) \setminus \{0\}$ and $\int _{\Omega }d^{-\alpha }(x)v(x) dx = 0$ . Moreover, we see that
$$\begin{equation*} \begin{split} &\frac{\int _{\Omega }d_\Omega ^{2-\alpha }(x)|\nabla v|^2dx}{\int _{\Omega }d_\Omega ^{-\alpha }(x)v^2 dx}\\ &=\frac{ (\lambda _1)^2(s_1)^{N-\alpha }\int _{D_1}d_{D_1}^{2-\alpha }(x)|\nabla (u_1)_+|^2dx + (\lambda _2)^2(s_2)^{N-\alpha }\int _{D_2}d_{D_2}^{2-\alpha }(x)|\nabla (u_2)_+|^2dx}{(\lambda _1)^2(s_1)^{N-\alpha }\int _{D_1}d_{D_1}^{-\alpha }(x)|(u_1)_+|^2dx + (\lambda _2)^2(s_2)^{N-\alpha }\int _{D_2}d_{D_2}^{-\alpha }(x)|(u_2)_+|^2dx} \\ & < \frac{(1-\alpha )^2}{4}, \end{split} \end{equation*}$$
where $d_D(x) = \text{dist}(x,\mathbb{R}^N \setminus D)$ for any domain $D \subset \mathbb{R}^N$ . This proves Theorem 1.9 . $$
Appendix A
In this last section, we study the relationship between $W_0^{1,2}(\Omega )$ and $W_{0,\alpha ,\beta }^{1,2}(\Omega )$ and whether $H_{\alpha ,\beta }$ is attained or not.
Proposition A.1.
Let $(\alpha ,\beta )\in \mathbb{R}^2$ . Then we have
$$\begin{equation*} W_{0,\alpha ,\beta }^{1,2}(\Omega )\begin{cases} \supsetneq W_0^{1,2}(\Omega ) &\text{ if } (\alpha ,\beta )\in \{(a,b) \ | \ a\le 2, b>0\},\\ = W_0^{1,2}(\Omega ) &\text{ if } (\alpha ,\beta )\in \{(a,b) \ | \ a\le 2, b=0\},\\ \subsetneq W_0^{1,2}(\Omega ) &\text{ if } (\alpha ,\beta )\in \{(a,b) \ | \ b<0 \text{ or } b=0, a>2\}.\end{cases} \end{equation*}$$
If $(\alpha ,\beta )\in \{(a,b) \ | \ a> 2, b>0\}$ , $W_{0,\alpha ,\beta }^{1,2}(\Omega )\setminus W_0^{1,2}(\Omega )$ and $W_0^{1,2}(\Omega )\setminus W_{0,\alpha ,\beta }^{1,2}(\Omega )$ are non-empty.
Proof.
We note by Hardy’s inequality, for $u\in W_0^{1,2}(\Omega )$ and $\alpha \le 2$ ,
$$\begin{equation} \int _\Omega u^2d^{-\alpha }dx=\int _\Omega u^2d^{-2}d^{2-\alpha }dx\le C_1\int _\Omega |\nabla u|^2dx, \cssId{on1b}{\tag{74}} \end{equation}$$
where $C_1>0$ is a constant. Denote
$$\begin{equation*} A_1= \{(a,b) \ | \ a\le 2, b>0\},\ \ A_2= \{(a,b) \ | \ a\le 2, b=0\} \end{equation*}$$
and
$$\begin{equation*} A_3= \{(a,b) \ | \ b<0 \text{ or } b=0, a>2\}. \end{equation*}$$
First, assume that $(\alpha ,\beta ) \in A_1$ . Then we see that for a constant $C_2 > 0$ , $\int _\Omega d^\beta |\nabla u|^2dx\le C_2 \int _\Omega |\nabla u|^2dx$ . From this and Equation 74 , it follows that $W_{0,\alpha ,\beta }^{1,2}(\Omega )\supset W_0^{1,2}(\Omega )$ . On the other hand, if $\alpha =2$ and $\beta >0$ , $d \sin (d^{-\frac{1}{2}})\in W_{0,2,\beta }^{1,2}(\Omega )$ , but $d \sin (d^{-\frac{1}{2}})\notin W_0^{1,2}(\Omega )$ . If $\alpha <2$ and $\beta >0$ , for $m\in (\max \{\frac{\alpha -1}{2},\frac{1-\beta }{2},\frac{1}{4}\},\frac{1}{2})$ , $d^m(x)\in W_{0,\alpha ,\beta }^{1,2}(\Omega )$ , but $d^m(x)\notin W_0^{1,2}(\Omega )$ . Thus, we deduce that $W_{0,\alpha ,\beta }^{1,2}(\Omega )\supsetneq W_0^{1,2}(\Omega )$ .
Assume that $(\alpha ,\beta )\in A_2$ . Then, by Equation 74 , we see that $W_{0,\alpha ,\beta }^{1,2}(\Omega )= W_0^{1,2}(\Omega )$ .
Assume that $(\alpha ,\beta )\in A_3$ . Since $\int _\Omega |\nabla u|^2dx\le C_3\int _\Omega d^\beta |\nabla u|^2dx$ for some constant $C_3 > 0$ , we see that $W_{0,\alpha ,\beta }^{1,2}(\Omega )\subset W_0^{1,2}(\Omega )$ . On the other hand, for $m\in (\frac{1}{2},\max \{\frac{\alpha -1}{2},\frac{1-\beta }{2}\})$ , $d^m\in W_0^{1,2}(\Omega )$ , but $d^m\notin W_{0,\alpha ,\beta }^{1,2}(\Omega )$ , which implies that $W_{0,\alpha ,\beta }^{1,2}(\Omega )\subsetneq W_0^{1,2}(\Omega )$ .
If $(\alpha ,\beta )\in \{(a,b) \ | \ a> 2, b>0\}$ , for $m\in (\frac{1}{2},\frac{\alpha -1}{2} )$ , $d^m(x)\in W_0^{1,2}(\Omega )$ , but $d^m(x)\notin W_{0,\alpha ,\beta }^{1,2}(\Omega )$ . On the other hand, for $n> \frac{\alpha -1}{2}$ , $d^n \sin (d^{-n+\frac{1}{2}})\in W_{0,\alpha ,\beta }^{1,2}(\Omega )$ , but $d^n \sin (d^{-n+\frac{1}{2}})\notin W_0^{1,2}(\Omega )$ . In fact, note that, since $\int _1^\infty \frac{\cos ^2(t)}{t}dt\ge \frac{1}{2} \int _\pi ^\infty \frac{\sin ^2(t)}{t}dt$ , we have
$$\begin{equation*} \int _0^1 \frac{\cos ^2(t^{-n+\frac{1}{2}})}{t}dt= \Big (n-\frac{1}{2}\Big )^{-1}\int _1^\infty \frac{\cos ^2(t)}{t}dt=\infty . \end{equation*}$$
Then, from this and the facts that $\beta >0$ and $n> \frac{\alpha -1}{2}>\frac{1}{2}$ , we see that
$$\begin{align*} &\int _\Omega \Big |\nabla \Big (d^n \sin (d^{\frac{1}{2}-n})\Big )\Big |^2dx\\ &=\int _\Omega \Big |n d^{n-1}\sin (d^{\frac{1}{2}-n}) \nabla d+\Big (\frac{1}{2}-n\Big )d^{-\frac{1}{2}} \cos (d^{-n+\frac{1}{2}}) \nabla d\Big |^2dx\\ &=\int _\Omega n^2 d^{2n-2}\sin ^2(d^{\frac{1}{2}-n}) +\Big (\frac{1}{2}-n\Big )^2d^{-1} \cos ^2(d^{\frac{1}{2}-n})\\ &\quad +2n \Big (\frac{1}{2}-n\Big )d^{n-\frac{3}{2}}\sin (d^{\frac{1}{2}-n})\cos (d^{\frac{1}{2}-n})dx\\ &=+\infty , \end{align*}$$
$$\begin{align*} &\int _\Omega d^\beta \Big |\nabla \Big (d^n \sin (d^{\frac{1}{2}-n})\Big )\Big |^2dx < \infty \ \text{ and }\ \int _\Omega d^{2n-\alpha } \sin ^2(d^{\frac{1}{2}-n})dx \le \int _\Omega d^{2n-\alpha } dx < \infty . \end{align*}$$ This completes the proof.
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Proposition A.2.
$H_{\alpha ,\beta }(\Omega )$ is achieved by an element $v_{\alpha ,\beta } \in W_{0,\alpha ,\beta }^{1,2}(\Omega )$ if $(\alpha ,\beta ) \in \{(a,b)\ | \ a+b< 2, b<1\}$ .
Proof.
We assume $(\alpha ,\beta ) \in \{(a,b)\ | \ a+b< 2, b<1\}$ . Note that, by Proposition 2.4 , $H_{\alpha ,\beta }(\Omega )>0$ . Let $\{u_m\}_{m=1}^\infty \subset C^\infty _0(\Omega )$ be a minimizing sequence of $H_{\alpha ,\beta }(\Omega )$ with
$$\begin{equation*} \ \int _{\Omega }d^{-\alpha }(x)u_m^2 dx = 1 \ \text{ and }\ \lim _{m \to \infty } \int _\Omega d^\beta (x) |\nabla u_m|^2 dx =H_{\alpha ,\beta }(\Omega ). \end{equation*}$$
Taking a subsequence if it is necessary, we may assume that as $m \to \infty$ , $u_m$ converges weakly to $u$ in $W_{0,\alpha ,\beta }^{1,2}(\Omega )$ .
Then, by Lemma 2.8 , the same argument in the proof of Theorem 1.2 and the fact that $H_{\alpha ,\beta }(\Omega )>0$ for $(\alpha ,\beta ) \in \{(a,b)\ | \ a+b< 2, b<1\}$ , we deduce that $u\not \equiv 0$ .
Now, it holds that $u\not \equiv 0$ . Let $w_m = u_m - u$ . Then $w_m$ converges weakly to $0$ in $W_{0,\alpha ,\beta }^{1,2}(\Omega )$ as $m\rightarrow \infty$ . Then, we see that
$$\begin{equation} \begin{aligned} H_{\alpha ,\beta }(\Omega ) + o(1) &= \int _\Omega d^\beta (x)|\nabla u_m|^2 dx = \int _\Omega d^\beta (x)(|\nabla u|^2 + 2 \nabla u \cdot \nabla w_m + |\nabla w_m|^2) dx \\ &= \int _\Omega d^\beta (x)|\nabla u|^2 dx + \int _\Omega d^\beta (x)|\nabla w_m|^2 dx + o(1) \end{aligned} \cssId{hmin6}{\tag{75}} \end{equation}$$
and
$$\begin{equation} 1 = \int _\Omega d^{-\alpha }(x) u_m^2 dx = \int _\Omega d^{-\alpha }(x)u^2 dx + \int _\Omega d^{-\alpha }(x)w_m^2 dx + o(1). \cssId{hmin5}{\tag{76}} \end{equation}$$
Then, since $u \not \equiv 0$ , it follows that
$$\begin{equation} \limsup _{m\rightarrow \infty } \int _\Omega d^{-\alpha }(x)w_m^2 dx<1. \cssId{hmin4}{\tag{77}} \end{equation}$$
Thus, by Equation 75 , Equation 76 , Equation 77 and the fact that $w_m\in W_{0,\alpha ,\beta }^{1,2}$ , we obtain
$$\begin{align*} H_{\alpha ,\beta }(\Omega ) & \le \frac{\int _\Omega d^\beta (x)|\nabla u|^2 dx}{\int _\Omega d^{-\alpha }(x)u^2 dx} = \frac{H_{\alpha ,\beta }(\Omega ) - \int _\Omega d^\beta (x)|\nabla w_m|^2 dx + o(1)}{1 - \int _\Omega d^{-\alpha }(x)w_m^2 dx + o(1)} \\ & \le \frac{H_{\alpha ,\beta }(\Omega ) -H_{\alpha ,\beta }(\Omega )\int _\Omega d^{-\alpha }(x)w_m^2 dx + o(1)}{1 - \int _\Omega d^{-\alpha }(x)w_m^2 dx + o(1)} = H_{\alpha ,\beta }(\Omega ) + o(1) \end{align*}$$
as $m\rightarrow \infty$ . This implies that $u$ attains $H_{\alpha ,\beta }(\Omega )$ .
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We recall
$$\begin{equation} H_{\alpha ,2-\alpha }(\Omega )=\inf _{u\in W_{0,\alpha ,2-\alpha }^{1,2}(\Omega )\setminus \{0\} } \frac{\int _\Omega d^{2-\alpha }(x) |\nabla u |^2 dx}{\int _\Omega |u(x)|^2d^{-\alpha }(x) dx}. \cssId{hminimizer1}{\tag{78}} \end{equation}$$
Lemma A.3.
Let $\alpha \in (1,2]$ . For small $\delta >0$ , it holds that
$$\begin{equation} \int _{\Omega _\delta }d^{2-\alpha }|\nabla u|^2dx\ge \Big (\frac{ \alpha -1}{2}\Big )^2\int _{\Omega _\delta }d^{-\alpha }u^2dx \text{ for all } u\in W_{0,\alpha ,2-\alpha }^{1,2}(\Omega ), \cssId{on1a}{\tag{79}} \end{equation}$$
where $\Omega _\delta =\{x\in \Omega \ | \ d(x)<\delta \}$ and $d(x) = \text{dist}(x, \mathbb{R}^N \setminus \Omega )$ .
Proof.
By the density of $C_0^\infty (\Omega )$ in $W_{0,\alpha ,2-\alpha }^{1,2}(\Omega )$ , it suffices to prove Equation 79 for functions $u\in C_0^\infty (\Omega )$ . Since
$$\begin{equation*} \int _{\Omega _\delta }d^{2-\alpha }|\nabla u|^2dx\ge \int _{\partial \Omega }\int _0^\delta t^{2-\alpha }\Big (\frac{\partial u}{\partial t}\Big )^2(1- Ct)dtd\sigma \end{equation*}$$
and
$$\begin{equation*} \int _{\Omega _\delta }d^{-\alpha }u^2dx \le \int _{\partial \Omega }\int _0^\delta t^{-\alpha }u^2(t)(1+Ct)dtd\sigma , \end{equation*}$$
where $C>0$ is a constant, we see that
$$\begin{align*} &\int _{\Omega _\delta }d^{2-\alpha }|\nabla u|^2 -\Big (\frac{ \alpha -1}{2}\Big )^2 d^{-\alpha }u^2dx \\ &\ge \int _{\partial \Omega }\int _0^\delta t^{2-\alpha }\Big (\frac{\partial u}{\partial t}\Big )^2- \Big (\frac{ \alpha -1}{2}\Big )^2 t^{-\alpha }u^2(t) \\ & \qquad -Ct\bigg (t^{2-\alpha }\Big (\frac{\partial u}{\partial t}\Big )^2+ \Big (\frac{ \alpha -1}{2}\Big )^2 t^{-\alpha }u^2(t)\bigg ) dtd\sigma . \end{align*}$$
Then, by a scaling, it suffices to prove that for $v\in C^\infty ([0,1])$ such that $v$ vanishes in a neighborhood of the origin,
$$\begin{align*} \int _0^1 t^{2-\alpha } (v^\prime )^2- \Big (\frac{ \alpha -1}{2}\Big )^2 t^{-\alpha }v^2(t)- t\Big (t^{2-\alpha }(v^\prime )^2+ \Big (\frac{ \alpha -1}{2}\Big )^2 t^{-\alpha }v^2(t)\Big ) dt\ge 0. \end{align*}$$
Defining $v=t^\frac{\alpha -1}{2}w$ , we see that $v^\prime =\frac{\alpha -1}{2}t^\frac{\alpha -3}{2}w+t^\frac{\alpha -1}{2}w^\prime$ ,
$$\begin{equation} \begin{split} &\int _0^1 t^{2-\alpha } (v^\prime )^2- \Big (\frac{ \alpha -1}{2}\Big )^2 t^{-\alpha }v^2(t)dt\\ &=\int _0^1 t^{2-\alpha }\bigg (\Big (\frac{\alpha -1}{2}\Big )^2t^{\alpha -3}w^2+t^{\alpha -1}(w^\prime )^2+(\alpha -1)t^{\alpha -2}ww^\prime \bigg ) \\ & \qquad - \Big (\frac{\alpha -1}{2}\Big )^2t^{-1}w^2dt\\ &=\int _0^1 t (w^\prime )^2+(\alpha -1) ww^\prime dt =\int _0^1 t (w^\prime )^2dt+\frac{\alpha -1}{2} w^2 (1) \end{split} \cssId{on2a}{\tag{80}} \end{equation}$$
and
$$\begin{equation} \begin{split} & \int _0^1 t\Big (t^{2-\alpha } (v^\prime )^2+ \Big (\frac{ \alpha -1}{2}\Big )^2 t^{-\alpha }v^2(t)\Big )dt \\ & \qquad =\int _0^1 t^2 (w^\prime )^2+\frac{(\alpha -1)^2}{2} w^2+(\alpha -1) tww^\prime dt\\ & \qquad =\int _0^1 t^2 (w^\prime )^2 +\frac{(\alpha -1)(\alpha -2)}{2}w^2 dt+\frac{\alpha -1}{2} w^2 (1). \end{split} \cssId{on3a}{\tag{81}} \end{equation}$$
Thus, by Equation 80 , Equation 81 and the assumption $\alpha \in (1,2]$ , we see that
$$\begin{align*} &\int _0^1 t^{2-\alpha } (v^\prime )^2- \Big (\frac{ \alpha -1}{2}\Big )^2 t^{-\alpha }v(t)- t\Big (t^{2-\alpha }(v^\prime )^2+ \Big (\frac{ \alpha -1}{2}\Big )^2 t^{-\alpha }v(t)\Big ) dt\\ &=\int _0^1 t(1-t) (w^\prime )^2-\frac{(\alpha -1)(\alpha -2)}{2}w^2dt\ge 0. \end{align*}$$ ■
For the inequality in Lemma A.3 on Ahlfors regular domains, refer to Reference 16 .
Proposition A.4.
Assume $\alpha >1$ . Then $H_{\alpha ,2-\alpha }(\Omega )\le \frac{(1-\alpha )^2}{4}$ , and if $H_{\alpha ,2-\alpha }(\Omega )<\frac{(1-\alpha )^2}{4}$ , then $H_{\alpha ,2-\alpha }(\Omega )$ is achieved. Moreover, for $\alpha \in (1,2]$ , if $H_{\alpha ,2-\alpha }(\Omega )$ is achieved, then $H_{\alpha ,2-\alpha }(\Omega )<\frac{(1-\alpha )^2}{4}$ .
Proof.
We first prove that $H_{\alpha ,2-\alpha }(\Omega )\le \frac{(1-\alpha )^2}{4}$ . Define
$$\begin{equation*} f(t)=\begin{cases} (t/\delta )^{m } &\text{ if } t\in (0,\delta ),\\ 2-t/\delta &\text{ if } t\in (\delta ,2\delta ) \end{cases} \end{equation*}$$
and
$$\begin{equation*} u(x)=\begin{cases} f ( d(x) ) &\text{ if }x\in \Omega _{2\delta },\\ 0 &\text{ if } x\in \Omega \setminus \Omega _{2\delta }, \end{cases} \end{equation*}$$
where $m>\frac{\alpha -1}{2}$ . Then it holds that $u\in W_{0,\alpha ,2-\alpha }^{1,2}(\Omega )$ . Moreover, we see that $\nabla u = f^\prime ( d(x) ) \nabla d$ ,
$$\begin{align*} \int _\Omega d^{2-\alpha }(x)|\nabla u|^2dx & =\ \int _{\Omega _{2\delta }} d^{2-\alpha }(x) \Big (f^\prime ( d(x) ) \Big )^2dx \\ & \le (1+C \delta )|\partial \Omega |\int _0^{2\delta } t^{2-\alpha }\Big (f^\prime (t)\Big )^2dt \end{align*}$$
and
$$\begin{equation*} \begin{split} \int _\Omega d^{-\alpha }(x)u^2dx&=\int _{\Omega _{2\delta }} d^{-\alpha }(x)f^2(d(x))dx\\ &\ge (1-C\delta )|\partial \Omega |\int _0^{2\delta }t^{-\alpha }f^2(t)dt, \end{split} \end{equation*}$$
where $C>0$ is a constant independent of $\delta$ . Since
$$\begin{gather*} \int _0^{2\delta } t^{2-\alpha }\Big (f^\prime (t)\Big )^2dt = m^2\delta ^{-2m}\int _0^{ \delta } t^{2m-\alpha } dt + \delta ^{-2}\int _\delta ^{2\delta }t^{2-\alpha }dt = \frac{m^2\delta ^{-\alpha +1}}{2m-\alpha +1}+C_1(\delta ).\\ \int _0^{2\delta }t^{-\alpha }f^2(t)dt = \delta ^{-2m}\int _0^{ \delta }t^{2m-\alpha } dt + \int _\delta ^{2\delta }t^{-\alpha }\Big (2-\frac{t}{\delta }\Big )^2dt = \frac{\delta ^{-\alpha +1}}{2m-\alpha +1} + C_2(\delta ), \end{gather*}$$
where $C_1(\delta ), C_2(\delta )>0$ are constants independent of $m$ , we see that
$$\begin{equation*} \begin{split} H_{\alpha ,2-\alpha }(\Omega )&\le \frac{\int _\Omega d^{2-\alpha }(x)|\nabla u|^2dx}{\int _\Omega d^{-\alpha }(x)u^2dx}\\ &=\Big (\frac{1+C\delta }{1-C\delta }\Big )\frac{ m^2 +(2m-\alpha +1)\delta ^{\alpha -1}C_1(\delta )}{1+(2m-\alpha +1)\delta ^{\alpha -1}C_2(\delta )}\rightarrow \frac{1+C\delta }{1-C\delta } \Big (\frac{\alpha -1}{2}\Big )^2 \end{split} \end{equation*}$$
as $m\downarrow \frac{\alpha -1}{2}$ , which implies that $H_{\alpha ,2-\alpha }(\Omega )\le \Big (\frac{\alpha -1}{2}\Big )^2$ .
Next, by the same arguments in the proof of Theorem 1.3 , we can prove that $H_{\alpha ,2-\alpha }(\Omega )$ is achieved if $\ H_{\alpha ,2-\alpha }(\Omega )<\frac{(1-\alpha )^2}{4}$ (See also Reference 4 ). Indeed, let $\{u_m\}_{m=1}^\infty \subset C^\infty _0(\Omega )$ be a minimizing sequence of $H_{\alpha ,2-\alpha }(\Omega )$ with
$$\begin{equation*} \int _{\Omega }d^{-\alpha }(x)u_m^2 dx = 1 \text{ and }\lim _{m \to \infty } \int _\Omega d^{2-\alpha }(x) |\nabla u_m|^2 dx = H_{\alpha ,2-\alpha }(\Omega ). \end{equation*}$$
Taking a subsequence, if it is necessary, we may assume that as $m \to \infty$ , $u_m$ converges weakly to some $u$ in $W_{0,\alpha ,2-\alpha }^{1,2}(\Omega )$ . If $u\equiv 0$ , by Lemma 2.8 , we deduce that $u_m$ concentrates near $\partial \Omega$ . Then by Lemma 2.3 , the argument in Theorem 1.3 , we see that $\ H_{\alpha ,2-\alpha }(\Omega )\ge \frac{(1-\alpha )^2}{4}$ , which is a contradiction to the assumption that $\ H_{\alpha ,2-\alpha }(\Omega )<\frac{(1-\alpha )^2}{4}$ . Now, it follows from $u\not \equiv 0$ and the argument in Equation 75 below, that $u$ attains $\ H_{\alpha ,2-\alpha }(\Omega )$ .
Finally, we prove that for $\alpha \in (1,2]$ , if $H_{\alpha ,2-\alpha }(\Omega )$ is achieved, $H_{\alpha ,2-\alpha }(\Omega )<\frac{(1-\alpha )^2}{4}$ . Suppose that $H_{\alpha ,2-\alpha }(\Omega )$ is achieved and $H_{\alpha ,2-\alpha }(\Omega )=\frac{(1-\alpha )^2}{4}$ . Then there exists a nonnegative function $u\in W_{0,\alpha ,2-\alpha }^{1,2}(\Omega )$ such that
$$\begin{equation*} -\text{div}(d^{2-\alpha }\nabla u)=\frac{(1-\alpha )^2}{4}d^{-\alpha }u \text{ in } \Omega . \end{equation*}$$
By the maximum principle, $u>0$ in $\Omega$ . Based on the idea of Reference 3 , we define
$$\begin{equation*} Y(t)=t^{\frac{\alpha -1}{2}}X^s(t) \ \text{ and } \ w_s(x)=Y(d(x)), \end{equation*}$$
where $s>\frac{1}{2}$ and $X(t)=\begin{cases} (1-\ln t)^{-1} &\text{ if } 0<t\le 1,\\ 1 &\text{ if } 1<t.\end{cases}$ We take sufficiently small $\delta \in (0,1)$ such that $d \in C^2(\Omega _\delta )$ and the statement of Lemma A.3 holds in $\Omega _\delta$ . Then we see that $w_s\in W_{0,\alpha ,2-\alpha }^{1,2}(\Omega )$ for $s>\frac{1}{2}$ . Moreover, since
$$\begin{equation*} \begin{split} Y^\prime (t)&=\frac{\alpha -1}{2}t^{\frac{\alpha -3}{2}}X^s(t)\\ &\quad +st^\frac{\alpha -3}{2}X^{s+1}(t)=t^{\frac{\alpha -3}{2}}\Big ( \frac{\alpha -1}{2}X^s(t)+sX^{s+1}(t)\Big ) \ \ \text{ for }\ \ t\in (0,1) \end{split} \end{equation*}$$
and for $x\in \Omega _\delta$ ,
$$\begin{align*} &\text{div}(d^{2-\alpha }\nabla w_s)=\text{div}\Big (d^{2-\alpha } Y^\prime (d(x)) \nabla d \Big )\\ &=\text{div}\bigg (d^\frac{-\alpha +1}{2}(\nabla d)\Big (\frac{\alpha -1}{2}X^s(d)+sX^{s+1}(d)\Big )\bigg )\\ &=-\frac{ \alpha - 1}{2}d^\frac{-\alpha - 1}{2}\Big (\frac{\alpha - 1}{2}X^s(d) + sX^{s+1}(d)\Big ) + d^\frac{-\alpha + 1}{2}(\Delta d)\Big (\frac{\alpha - 1}{2}X^s(d) + sX^{s+1}(d)\Big )\\ &\qquad +d^\frac{-\alpha -1}{2}\Big (s\frac{\alpha -1}{2}X^{s+1}(d)+s(s+1)X^{s+2}(d)\Big )\\ &=-\Big (\frac{ \alpha -1}{2}\Big )^2d^\frac{-\alpha -1}{2}X^s(d)+s(s+1)d^\frac{-\alpha -1}{2}X^{s+2}(d) \\ & \qquad +d^\frac{-\alpha +1}{2}(\Delta d)\Big (\frac{\alpha -1}{2}X^s(d)+sX^{s+1}(d)\Big ), \end{align*}$$
it follows that for small $d$ ,
$$\begin{align*} &-\text{div}(d^{2-\alpha }\nabla w_s)-\Big (\frac{ \alpha -1}{2}\Big )^2d^{-\alpha }w_s\\ &\qquad =-s(s+1)d^\frac{-\alpha -1}{2}X^{s+2}(d)-d^\frac{-\alpha +1}{2}(\Delta d)\Big (\frac{\alpha -1}{2}X^s(d)+sX^{s+1}(d)\Big )\\ &\qquad =-d^\frac{-\alpha -1}{2}X^{s+2}(d)\bigg (s(s+1)+(\Delta d)d X^{-2}(d)\Big (\frac{\alpha -1}{2}+sX (d)\Big )\bigg )\le 0. \end{align*}$$
Take $\epsilon >0$ such that for all $s\in (\frac{1}{2},1)$ , $\epsilon w_s \le u \text{ on } \{x\in \Omega \ |\ d(x)=\delta \}$ and define $v_s\equiv \epsilon w_s-u$ . Then $(v_s)_+\in W_{0,\alpha ,2-\alpha }^{1,2}(\Omega )$ and
$$\begin{equation*} -\text{div}(d^{2-\alpha }\nabla v_s)-\Big (\frac{ \alpha -1}{2}\Big )^2d^{-\alpha }v_s\le 0 \text{ in } \Omega _\delta , \end{equation*}$$
where $u_+=\max \{u,0\}$ . Then, multiplying $(v_s)_+$ to the above inequality and then integrating over $\Omega _\delta$ ,
$$\begin{equation*} \int _{\Omega _\delta }d^{2-\alpha }|\nabla (v_s)_+|^2-\Big (\frac{ \alpha -1}{2}\Big )^2d^{-\alpha }((v_s)_+)^2dx\le 0. \end{equation*}$$
On the other hand, by Lemma A.3 ,
$$\begin{equation*} \int _{\Omega _\delta }d^{2-\alpha }|\nabla (v_s)_+|^2-\Big (\frac{ \alpha -1}{2}\Big )^2d^{-\alpha }((v_s)_+)^2dx\ge 0. \end{equation*}$$
We deduce that $\epsilon w_s\le u$ in $\Omega _\delta$ for every $s\in (\frac{1}{2},1)$ . Hence, $\epsilon d^{\frac{\alpha -1}{2}}X^\frac{1}{2}(d)\le u$ in $\Omega _\delta$ , which is a contradiction to the fact the $d^{-\alpha /2}u\in L^2(\Omega )$ . Thus we conclude that if $H_{\alpha ,2-\alpha }(\Omega )$ is achieved, $H_{\alpha ,2-\alpha }(\Omega )<\frac{(1-\alpha )^2}{4}$ .
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