Proof. Applying the divergence theorem in 20 , we deduce \[0=\int_{\Omega}\Delta h dx=\int_{\partial\Omega\setminus E}\frac{\partial h}{\partial\nu}d{\mathcal{H}}^{n-1}+\int_{\partial\Omega\cap\partial E}\frac{\partial h}{\partial\nu}d{\mathcal{H}}^{n-1},\] which implies that \[\label{e24610} \int_{\partial\Omega\setminus E}\frac{\partial h}{\partial\nu}d{\mathcal{H}}^{n-1}=-\lambda\int_{\partial\Omega\cap\partial E}wd{\mathcal{H}}^{n-1}.\tag{21}\] Therefore, \[\begin{align} \int_{\partial\Omega\setminus E}\widetilde{F}_{\lambda}(\nu)wd{\mathcal{H}}^{n-1}=&\int_{\partial\Omega\setminus E}wd{\mathcal{H}}^{n-1}+\int_{\partial\Omega\setminus E}\frac{\partial h}{\partial\nu}wd{\mathcal{H}}^{n-1}\\ =&\int_{\partial\Omega\setminus E}wd{\mathcal{H}}^{n-1}-\lambda\int_{\partial\Omega\cap\partial E}wd{\mathcal{H}}^{n-1}\\ =&J_{w,\lambda}(\Omega;\mathbb{R}^n\setminus E), \end{align}\] that is, \[\label{e24611}J_{w,\lambda}(\Omega;\mathbb{R}^n\backslash E)=P_{w,\widetilde{F}_{\lambda}}(\Omega;\mathbb{R}^n\backslash E).\tag{22}\]

On the other hand, by the divergence theorem, we have \[\label{e34633} \int_{B^{\lambda}}{\rm div}(we_n) =\int_{\partial B_1\setminus H}w\nu\cdot e_nd{\mathcal{H}}^{n-1}-\int_{B_1\cap\partial H}wd{\mathcal{H}}^{n-1},\tag{23}\] where we used the fact that \(\nu=-e_n\) on \(\partial H\). Since \(w(x)=w(-x)\), we have \(\frac{\partial w}{\partial x_n}=0\), which implies that the left-hand side of 23 is zero. Consequently, we obtain \[\label{e34634} \int_{\partial B_1\setminus H}w\nu\cdot e_nd{\mathcal{H}}^{n-1}=\int_{B_1\cap\partial H}wd{\mathcal{H}}^{n-1},\tag{24}\] where \(\nu\) is the unit normal vector on \(\partial B_1\setminus H\).

Therefore, we have \[\label{e24612}\begin{align} &\int_{\partial B_1\setminus H}F_{\lambda}(\nu)wd{\mathcal{H}}^{n-1}\\ =&\int_{\partial B_1\setminus H}wd{\mathcal{H}}^{n-1}-\lambda\int_{\partial B_1\setminus H}w\nu\cdot e_nd{\mathcal{H}}^{n-1}\\ =&\int_{\partial B_1\setminus H}wd{\mathcal{H}}^{n-1}-\lambda\int_{B_1\cap\partial H}wd{\mathcal{H}}^{n-1}\\ =&J_{w,\lambda}(B_{1};\mathbb{R}^n\backslash H), \end{align}\tag{25}\] where \(F_{\lambda}\) is given by ?? . Combining 22 and 25 , we conclude that the weighted capillary energy can be reformulated as the weighted anisotropic perimeters, and thus ?? holds.

From Remark 2, we know that \(\partial B_1\) intersects with the boundary of the Wulff ball \(\partial{\mathcal{W}}(\lambda e_n)=\{x\in\mathbb{R}^n:F_{\lambda}^o(x-\lambda e_n)=1\}\), where the unit normal vector is \(\nu=\frac{DF_{\lambda}^o(x-\lambda e_n)}{|DF_{\lambda}^o(x-\lambda e_n)|}\). Using 15 , it is easy to obtain \((x-\lambda e_n)\cdot\nu=F_{\lambda}(\nu)\) on \(\partial {\mathcal{W}}(\lambda e_n)\). Moreover, since \((x-\lambda e_n)\cdot\nu=(x_1,\cdots,0)\cdot (0,\cdots,1)=0\) a.e. on \(\partial H\), it follows that \[\label{e34636} \begin{align} J_{w,\lambda}(B_1;\mathbb{R}^n\backslash H)=&\int_{\partial {\mathcal{W}}(\lambda e_n)\setminus H}F_{\lambda}(\nu)wd{\mathcal{H}}^{n-1}\\ =&\int_{\partial {\mathcal{W}}(\lambda e_n)\setminus H}w(x-\lambda e_n)\cdot\nu d{\mathcal{H}}^{n-1}\\ =&\int_{\partial( {\mathcal{W}}(\lambda e_n)\setminus H)}w(x-\lambda e_n)\cdot\nu d{\mathcal{H}}^{n-1}\\ =&\int_{{\mathcal{W}}( \lambda e_n)\setminus H}{\rm div}\left(w (x-\lambda e_n)\right)dx\\ =&\int_{{\mathcal{W}}(\lambda e_n)\setminus H}\left((x-\lambda e_n)\cdot\nabla w+nw\right)dx\\ =&(n+\alpha)\int_{B_1\setminus H}wdx=(n+\alpha)\int_{B^{\lambda}}wdx, \end{align}\tag{26}\] where we used the fact that \(e_n\cdot\nabla w=\frac{\partial w}{\partial x_n}=0\) in the last equality. This completes the proof. ◻

Proof. By standard elliptic regularity theory, the variational solution of 8 belongs to \(C^2(\Omega)\) and is Hölder continuous up to the boundary. Then, for any \(x_0\in\Omega\), we have \[\left({\rm div}(w\nabla\varphi)-{\rm div}(w\nabla u)\right)(x_0) =\left({\rm div}(w\nabla(\varphi-u))\right)(x_0) =w(x_0)\Delta(\varphi-u)(x_0)\leq 0,\] where we used the fact that \(\nabla(u-\varphi)(x_0)=0\) and \(\Delta(u-\varphi)(x_0)\geq 0\) in the last inequality. Therefore, \[\left({\rm div}(w\nabla\varphi)\right)(x_0)\leq\left({\rm div}(w\nabla u)\right)(x_0)=cw(x_0),\] which implies that \[\left(w^{-1}{\rm div}(w\nabla\varphi)\right)(x_0)\leq c \;if\;x_0\in\Omega.\]

If \(x_0\in\Sigma\setminus\gamma\), meaning that the function \(u-\varphi:\overline{\Omega}\rightarrow \mathbb{R}\) attains a local minimum on the boundary, then the one-sided directional derivative of \(u-\varphi\) at \(x_0\) in the inward normal direction \(-\nu_{\Sigma}\) satisfies \[\partial_{-\nu_{\Sigma}}(u-\varphi)(x_0)\geq0.\] Consequently, we obtain \[\partial_{\nu_{\Sigma}}\varphi(x_0)\geq\partial_{\nu_{\Sigma}}u(x_0)=1\quadfor any\;x\in\Sigma\setminus\gamma.\] Similarly, we can deduce that \[\partial_{\nu_{\Gamma}}\varphi(x_0)\geq\partial_{\nu_{\Gamma}}u(x_0)=-\lambda\quadfor any\;x\in\Gamma\setminus\gamma.\]

Next, we only need to check the property for any \(x_0\in\gamma\). Assume that \(\varphi\in C^2(\mathbb{R}^n)\) and that \(x_0\in\gamma\) is the unique minimizer of \((u-\varphi)\) with \((u-\varphi)(x_0)=0\).

Step 1. We first show that \[\label{e3464}\max\{\left(-{\rm div}(w\nabla\varphi)+cw\right)(x_0),\;\partial_{\nu_{\Sigma}}\varphi(x_0)-1,\;\partial_{\nu_{\Gamma}}\varphi(x_0)+\lambda\}\geq 0.\tag{31}\] Assume by contradiction that \[\max\{\left(-{\rm div}(w\nabla\varphi)+cw\right)(x_0),\;\partial_{\nu_{\Sigma}}\varphi(x_0)-1,\;\partial_{\nu_{\Gamma}}\varphi(x_0)+\lambda\} < 0,\] then, there exists a ball \(B_r(x_0)\) with \(r\) small such that \[\label{e3465} \begin{cases} -{\rm div}(w\nabla\varphi)+cw<0&\;in\;B_r(x_0)\\ \partial_{\nu}\varphi-g<0&\;on\;(\partial\Omega\cap B_r(x_0))\backslash\gamma, \end{cases}\tag{32}\] where \(g\) is defined by 30 . Let \(z=u-\varphi\), setting \[h=-{\rm div}(w\nabla z)={\rm div}(w\nabla\varphi)-{\rm div}(w\nabla u)={\rm div}(w\nabla\varphi)-cw,\] and \[f=\partial_{\nu}u-\partial_{\nu}\varphi=g-\partial_{\nu}\varphi.\] Then \(z\) is a variational solution of \[\label{e3467} \begin{cases} -{\rm div}(w\nabla z)=h&in\;B_r(x_0)\cap\Omega\\ \partial_{\nu}z=f&on\;B_r(x_0)\cap\partial\Omega, \end{cases}\tag{33}\] which means that \[\label{e3468} \int_{\Omega}w\nabla z\cdot\nabla\psi dx=\int_{\partial\Omega}wf\psi d{\mathcal{H}}^{n-1}+\int_{\Omega}h\psi dx\tag{34}\] for any \(\psi\in C^{\infty}(\overline{\Omega})\) with \(\psi=0\) in \(\Omega\backslash B_r(x_0)\). If we choose \(\psi:=\min\{z-\varepsilon,0\}\), then for \(\varepsilon>0\) small enough, \[\psi=\min\{z-\varepsilon,0\}=0\quadin\quad\Omega\backslash B_r(x_0).\] From 32 , we know that the functions \(h>0\) in \(B_r(x_0)\cap\Omega\) and \(f>0\) on \(B_r(x_0)\cap\partial\Omega\). Then we get \[\int_{\Omega}w|\nabla(\min\{z-\varepsilon,0\})|^2dx=\int_{\partial\Omega}wf\psi d{\mathcal{H}}^{n-1}+\int_{\Omega}h\psi dx\leq 0,\] which implies that \(\min\{z-\varepsilon,0\}=0\) almost every in \(\Omega\). However, by the continuity of \(z\), we know that \(z-\varepsilon<0\) in \(B_r(x_0)\), which is a contradiction. Thus, 31 holds.

Step 2. We claim that \[\label{e3469} \max\{\partial_{\nu_{\Sigma}}\varphi(x_0)-1,\;\partial_{\nu_{\Gamma}}\varphi(x_0)+\lambda\}\geq 0.\tag{35}\] For \(x_0\in\gamma\), there exists a ball \(B_r(\overline{x})\subset\mathbb{R}^n\backslash E\) such that \(x_0\in\partial B_r(\overline{x})\). For simplicity, we assume that \(\overline{x}=0\) and \(r=1\). We define \[\psi_q(x)=q^{-\frac{3}{2}}(|x|^{-q}-1),\] where \(q>0\) is a real number. Then, we notice that \(\psi_q(x)\leq0\) for all \(x\in\Omega\), and \(\psi_q(x_0)=0\). Moreover, it is easy to calculate that \[\Delta \psi_q(x_0)\geq\frac{q^{\frac{1}{2}}}{2}\quadand\quad |\nabla\psi_q(x_0)|=q^{-\frac{1}{2}}\] for \(q\) sufficiently large. Now, we define a new test function \[\varphi_q(x)=\varphi(x)+\psi_q(x).\] Obviously, \(\varphi_q\leq\varphi\) in \(\Omega\) and \(\varphi_q(x_0)=\varphi(x_0)\), and then we have \[(u-\varphi_q)(x)\geq(u-\varphi)(x)\geq(u-\varphi)(x_0)=(u-\varphi_q)(x_0),\] which implies that \(x_0\) is still a minimizer of \(u-\varphi_q\). Therefore, \[\begin{align} -{\rm div}\left(w\nabla\varphi_q\right)(x_0)+cw(x_0)=&-\nabla w(x_0)\cdot\nabla\varphi_q(x_0)-w(x_0)\Delta\varphi_q(x_0)+cw(x_0)\\ \leq&|\nabla w(x_0)|\left(|\nabla\varphi(x_0)|+|\nabla\psi_q(x_0)|\right)+w(x_0)\left[-\Delta\varphi(x_0)-\Delta\psi_q(x_0)+c\right]\\ \leq&|\nabla w(x_0)|\left(|\nabla\varphi(x_0)|+q^{-\frac{1}{2}}\right)+w(x_0)\left[-\Delta\varphi(x_0)-\frac{q^{\frac{1}{2}}}{2}+c\right] <0 \end{align}\] for \(q\) large enough. Therefore, by Step 1, we get \[\max\{\partial_{\nu_{\Sigma}}\varphi_q(x_0)-1,\;\partial_{\nu_{\Gamma}}\varphi_q(x_0)+\lambda\}\geq 0.\] Letting \(q\rightarrow\infty\), we derive that \[\max\{\partial_{\nu_{\Sigma}}\varphi(x_0)-1,\;\partial_{\nu_{\Gamma}}\varphi(x_0)+\lambda\}\geq 0.\] This completes the proof. ◻

Proof. It is obvious that \(\bigcup_{x\in X}J_{X}u(x)\subset\mathbb{R}^n\), we only need to show that \(\mathbb{R}^n\subset \bigcup_{x\in X}J_{X}u(x)\). Indeed, for any \(\xi\in\mathbb{R}^n\), we can find \(a<0\) small enough such that \(\sup_{x\in X}(-a+\xi\cdot x-u(x))<0\). Setting \[t_0:=\sup\{t>0:-a+\xi\cdot x+s<u(x)\;for all\;x\in X\;and \;s\in(0,t)\},\] then there exists some \(\overline{x}\in X\) such that \(-a+\xi\cdot\overline{x}+t_0=u(\overline{x})\). Therefore, for all \(x\in X\), we have \[-a+\xi\cdot x+t_0=u(\overline{x})-\xi\cdot(x-\overline{x})\leq u(x).\] By the definition of the subdifferential, this implies that \(\xi \in J_X u(\overline{x})\). Since \(\xi\) was arbitrary, we conclude that \(\mathbb{R}^n \subset \bigcup_{x \in X} J_{X} u(x)\). This completes the proof. ◻

Proof. By the definition of \(J_{K}v(x)\), if \(\xi\in J_{K}v(x)\), then \[v(y)-v(x)\geq\xi\cdot(y-x)\quadfor any\;y\in K.\] By the convexity of \(E\), we have \(\nu_{E}(x)\cdot(y-x)\leq 0\) for any \(y\in K\). Therefore, for any \(t>0\), \[v(y)-v(x)\geq(\xi+t\nu_{E}(x))\cdot(y-x)\quadfor any\;y\in K,\] which implies that \(\xi+t\nu_{E}(x)\) belongs to \(J_{K}v(x)\) for any \(t>0\). ◻

Proof. For any \(\xi\in{\mathcal{B}}_{u_{\Gamma}}^{\lambda}\cap B_1\), we directly obtain \[\xi\in J_{\Gamma}u(x),\quad \xi\cdot\nu_{E}(x)\quadand\quad |\xi|<1.\] From Lemma 2, there exists some \(x\in\overline{\Omega}\) such that \(\xi\in J_{\overline{\Omega}}u(x)\). To prove ?? , it is equivalent to show that \(x\) belongs to the interior of \(\Omega\). Since \(\xi\in J_{\overline{\Omega}}u(x)\), then for any \(y\in\overline{\Omega}\), we have \[u(y)\geq u(x)+\xi\cdot(y-x):=\varphi(y).\] Therefore, \(\nabla\varphi=\xi\) and \(x\) is the minimum point of \(u-\varphi\), where \(u\) is the variational solution of 8 . Applying the conclusion of Proposition 1, we deduce that \[\label{e34616} \begin{cases} \partial_{\nu_{\Sigma}}\varphi(x)\geq 1&if\;x\in\Sigma\backslash\gamma,\\ \partial_{\nu_{\Gamma}}\varphi(x)\geq -\lambda &if\;x\in\Gamma\backslash\gamma,\\ \max\{\partial_{\nu_{\Sigma}}\varphi(x)-1,\partial_{\nu_{\Gamma}}\varphi(x)+\lambda\}\geq 0 &if\;x\in \gamma. \end{cases}\tag{39}\] Given that \(|\nabla\varphi|=|\xi|<1\), we immediately deduce that \(x\notin\Sigma\backslash\gamma\). Furthermore, the boundary condition \[\partial_{\nu_{\Gamma}}\varphi(x)+\lambda=\nabla\varphi(x)\cdot\nu_{E}+\lambda=-\xi\cdot\nu_{E}(x)+\lambda<0,\] implies that \(x\notin\Gamma\backslash\gamma\). A straightforward calculation shows \[\max\{\partial_{\nu_{\Sigma}}\varphi(x)-1,\partial_{\nu_{\Gamma}}\varphi(x)+\lambda\}<0,\] which consequently yields \(x\notin\gamma\). Combining these results, we conclude that \(x\notin\partial\Omega\). ◻

Proof. We choose a sequence of points \(x_1,x_2\cdots,x_n\) that is dense in \(K\). Denote \(K_n:=\{x_1,x_2,\cdots,x_n\}\) and let \(v_n:K_n\rightarrow\mathbb{R}\) be the restriction of \(v\) to \(K_n\), i.e., \(v_n(x_i)=v(x_i)\) for \(i=1,\cdots, n\). Notice that \[\label{e34621} \sup_{x\in K}\inf_{y\in K_n}|v(x)-v_n(y)|\rightarrow 0\quadas\;n\rightarrow\infty.\tag{40}\]

We first claim that for large \(n\), \[\label{e34622} {\mathcal{B}}_{v_n}^{\lambda'}\cap B_1\subset{\mathcal{B}}_{v}^{\lambda}\cap B_1,\tag{41}\] where \(\lambda'>\lambda\) is a parameter chosen sufficiently close to \(\lambda\). For any \(\xi\in{\mathcal{B}}_{v_n}^{\lambda'}\cap B_1\), we know directly from definition 38 that \[|\xi|<1,\quad\xi\in J_{K_n}v_n(x_n),\quadand\quad\xi\cdot\nu_{E}(x_n)>\lambda'\quadfor some\;x_n\in K_n.\] From Lemma 2, there exists \(x\in K\) such that \(\xi\in J_{K}v(x)\). By the convergence established in 40 , for any \(\varepsilon>0\), we have \(|x-x_n|<\varepsilon\) when \(n\) is large enough. Therefore, \[\begin{align} \xi\cdot\nu_{E}(x)=&\xi\cdot\nu_{E}(x_n)-\xi\cdot\nu_{E}(x_n)+\xi\cdot\nu_{E}(x)\\ >&\lambda'+\xi\cdot(\nu_{E}(x)-\nu_{E}(x_n))\\ \geq&\lambda'-|\xi|\cdot|\nu_{E}(x)-\nu_{E}(x_n)|\\ >&\lambda'-|\nu_{E}(x)-\nu_{E}(x_n)|>\lambda. \end{align}\] This implies that \(\xi\in{\mathcal{B}}_{v}^{\lambda}\cap B_1\) and thus 41 holds.

Next, we need to show that \[\int_{{\mathcal{B}}_{v_n}^{\lambda'}\cap B_1}wdx\geq\int_{{\mathcal{B}}_{v_n}^{\lambda}\cap B_1}wdx.\] To this end, define \[A_t:=\bigcup_{x\in K_n}\{\xi\in J_{K_n}v_n(x):\xi\cdot\nu_{E}(x)=\lambda+t\},\] where \(t>0\) is a real number. It is clear that the map \(t\mapsto|{\mathcal{B}}_{v_n}^{\lambda+t}\cap B_r|\) is decreasing and Lipschitz continuous. Moreover, \[\label{e34623} \frac{d}{dt}\int_{{\mathcal{B}}_{v_n}^{\lambda+t}\cap B_r}wdx=-\int_{A_t\cap B_r}wd{\mathcal{H}}^{n-1}.\tag{42}\]

For any fixed \(x\in K_n\), consider the set \[A_{t}(x):=\{\xi\in J_{K_n}v_n(x):\xi\cdot\nu_{E}(x)=\lambda+t\}.\] Then we can write \(A_t=\bigcup_{x\in K_n}A_{t}(x).\) Suppose \(\xi\in A_t(x)\) , then \(\xi\in J_{K_n}v_n(x)\). Applying Lemma 3, we deduce that \(\xi+s\nu_{E}(x)\in J_{K_n}v_n(x)\) for any \(s>0\), and since \((\xi+s\nu_{E}(x))\cdot\nu_{E}(x)=\lambda+t+s\), it follows that \[A_t(x)+s\{\nu_{E}(x)\}\subset A_{t+s}.\] Since the map \(\xi\mapsto\xi+s\nu_{E}(x)\) is an isometry, we have \({\mathcal{H}}^{n-1}(A_t(x)\cap B_{1+t})\leq{\mathcal{H}}^{n-1}(A_{t+s}(x)\cap B_{1+t+s})\). Combining this with the fact that \(|J_{K_n}v_n(x_i)\cap J_{K_n}v_n(x_j)|=0\) for any \(i\neq j\), we deduce that the function \[t\mapsto{\mathcal{H}}^{n-1}(A_t\cap B_{1+t})=\sum_{x\in K_n}{\mathcal{H}}^{n-1}(A_t(x)\cap B_{1+t})\] is non-decreasing. Integrating 42 for \(w\equiv 1\), we get \[\begin{align} \int_{1}^{2}{\mathcal{H}}^{n-1}(A_t\cap B_{1+t})dt\leq&\int_{1}^{2}{\mathcal{H}}^{n-1}(A_t\cap B_{3})dt\\ =&-\int_{1}^{2}\frac{d}{dt}|{\mathcal{B}}_{v_n}^{\lambda+t}\cap B_3|dt\\ =&|{\mathcal{B}}_{v_n}^{\lambda+1}\cap B_3|-|{\mathcal{B}}_{v_n}^{\lambda+2}\cap B_3| \leq |B_3|. \end{align}\] By the mean value theorem, there exists some \(\widetilde{t}\in(1,2)\) and a constant \(C_1\) such that \({\mathcal{H}}^{n-1}(A_{\widetilde{t}}\cap B_1)\leq C.\) By the monotonicity of \({\mathcal{H}}^{n-1}(A_t\cap B_{1+t})\), we obtain \[{\mathcal{H}}^{n-1}(A_t\cap B_{1})\leq C_1\quadfor all\;t\in(0,1).\] Therefore, integrating 42 from \(0\) to \(\lambda'-\lambda\) and using the continuity of \(w\), we have \[\int_{0}^{\lambda'-\lambda}\left(\int_{A_t\cap B_1}wd{\mathcal{H}}^{n-1}\right)dt\leq C(\lambda'-\lambda),\] which implies that \[\int_{{\mathcal{B}}_{v_n}^{\lambda'}\cap B_1}wdx\geq\int_{{\mathcal{B}}_{v_n}^{\lambda}\cap B_1}wdx- C(\lambda'-\lambda).\] Choosing \(\lambda'\rightarrow\lambda\) and using the \(\lambda(w)\)-ABP property of \(E\), we obtain \[\int_{{\mathcal{B}}_{v_n}^{\lambda'}\cap B_1}wdx\geq\int_{{\mathcal{B}}_{v_n}^{\lambda}\cap B_1}wdx\geq\int_{B^{\lambda}}wdx.\] This completes the proof. ◻

Proof. Define \(v=w^{\frac{1}{\alpha}}\). Then \(v>0\) is a homogeneous function of degree \(1\). The function \(v\) is concave outside \(E\) if and only if, for any \(x,y\in\mathbb{R}^n\backslash E\), \[\label{e34627} v(y)\leq v(x)+\nabla v(x)\cdot(y-x).\tag{43}\] Since \(v\) is \(1\)-homogeneous, we have \[\nabla v(x)\cdot x=v(x),\] and 43 can be rewritten as \[\label{e34628} v(y)\leq \nabla v(x)\cdot y.\tag{44}\] Therefore, since \(\nabla v(x)=\alpha^{-1}w(x)^{\frac{1}{\alpha}-1}\nabla w(x)\), we deduce that \(w^{\frac{1}{\alpha}}\) is concave if and only if \[w(y)^{\frac{1}{\alpha}}\leq\frac{\nabla w(x)\cdot y}{\alpha w(x)^{1-\frac{1}{\alpha}}},\] which is equivalent to ?? . ◻

Proof. By a simple calculation, we have \[\left(\log x\right)''=-\frac{1}{x^2}<0,\] which shows that \(\log x\) is a concave function on \((0,\infty)\). In other words, for any \(s,t>0\) and any \(m\in[0,1]\), we have \[m\log s+(1-m)\log t\leq\log(ms+(1-m)t).\] Choosing \(m=\frac{\alpha}{\alpha+n}\), we obtain \[\frac{\alpha}{\alpha+n}\log s+\frac{n}{\alpha+n}\log t\leq\log\left(\frac{\alpha s+nt}{\alpha+n}\right).\] Multiplying both sides of the inequality by \(\alpha+n\) and taking the exponential, we conclude that ?? holds. ◻

Proof. Let \(u:\overline{\Omega}\rightarrow\mathbb{R}\) be the variational solution of the Neumann problem 8 , and denote its restriction to \(\Gamma:=\partial\Omega\cap \partial E\) by \(u_{\Gamma}\). Let \({\mathcal{A}}_{u}\) and \({\mathcal{B}}_{u_{\Gamma}}^{\lambda}\) be the sets defined in 37 and ?? , respectively. Then, from Lemma 4, 45 , and the \(\lambda_w\)-ABP property, we deduce that \[\label{e34631} \begin{align} \int_{B^{\lambda}}w dx\leq&\int_{{\mathcal{B}}_{u_{\Gamma}}^{\lambda}\cap B_1} wdx\leq\int_{\nabla u(\Omega)}wdx\\ =&\int_{\Omega}w(\nabla u)\det\nabla^2 udx\\ =&\int_{\Omega}w(x)\frac{w(\nabla u)}{w(x)}\det\nabla^2 udx\\ \leq&\int_{\Omega}w(x)\left(\frac{{\rm div}(w(x)\nabla u)}{(n+\alpha)w(x)}\right)^{n+\alpha}dx\\ =&\left(\frac{c}{n+\alpha}\right)^{n+\alpha}\int_{\Omega}w(x)dx\\ =&\left(\frac{J_{w,\lambda}(\Omega;\mathbb{R}^n\backslash E)}{(n+\alpha)\int_{\Omega}w(x)dx}\right)^{n+\alpha}\int_{\Omega}w(x)dx\\ =&\left(\frac{J_{w,\lambda}(\Omega;\mathbb{R}^n\backslash E)}{(n+\alpha)\int_{B^{\lambda}}w(x)dx}\right)^{n+\alpha}\left(\frac{\int_{B^{\lambda}}wdx}{\int_{\Omega}wdx}\right)^{n+\alpha}\int_{\Omega}w(x)dx. \end{align}\tag{46}\] It follows that \[\label{e34632} \left(\frac{\int_{B^{\lambda}}wdx}{\int_{\Omega}wdx}\right)^{1-(n+\alpha)}\leq\left(\frac{J_{w,\lambda}(\Omega;\mathbb{R}^n\backslash E)}{(n+\alpha)\int_{B^{\lambda}}w(x)dx}\right)^{n+\alpha}.\tag{47}\] Combining Lemma 1 with 47 , we obtain \[\label{e34637} \left(\frac{\int_{B^{\lambda}}wdx}{\int_{\Omega}wdx}\right)^{\frac{1-(n+\alpha)}{n+\alpha}}\leq \frac{J_{w,\lambda}(\Omega;\mathbb{R}^n\backslash E)}{J_{w,\lambda}(B_1;\mathbb{R}^n\backslash H)},\tag{48}\] which implies that ?? holds. ◻

Proof. Assume that \(E\subset\mathbb{R}^n\) contains the origin and let \(B_R\) be a ball such that \(\Omega\subset\subset B_R\). For any \(\delta>0\), we construct a sequence of smooth convex sets \(E_{\delta}^{k}\subset\mathbb{R}^n\) converging to \((1+\delta)E\) in the sense of Kuratowski as \(k\rightarrow\infty\), such that \((1+\delta)E\subset E_{\delta}^k\). Up to slightly dilating \(E_{\delta}^k\) if necessary, we may assume without loss of generality that \[\label{e4461} {\mathcal{H}}^{n-1}(\partial\Omega\cap\partial E_{\delta}^k)=0\quadfor all\;k,\delta.\tag{49}\] According to Lemma 8, we can extend \(\chi_{\Omega}|_{\mathbb{R}^n\backslash E}\) to a function \(u\) belongs to \(BV(\mathbb{R}^n)\) with compact support such that \(|Du|(\partial E)=0\) and \(0\leq u\leq1\). For any \(\varepsilon>0\), let \((\rho_{\varepsilon})\) be a family of mollifiers. Let \(\{\varepsilon_h\}\) be an arbitrary positive infinitesimal sequence converging to zero. Since \(u\in BV(\mathbb{R}^n)\), we know that the mollified functions \(u_{\varepsilon_h}=\rho_{\varepsilon_h}*u\) converge to \(u\) in \(L^{1}(\mathbb{R}^n)\). Observing that \(\{u>t\}\backslash E=\Omega\) for all \(t\in(0,1)\), we define \(U_{\varepsilon_h,t}=\{x:u_{\varepsilon_h}(x)>t\}\). Then, for almost every \(t\in(0,1)\), it holds that \[\label{e4463} \begin{align} &\lim_{h\rightarrow\infty}|U_{\varepsilon_h,t}\Delta\{u>t\}|=0,\quad\lim_{h\rightarrow\infty}P(U_{\varepsilon_h,t})=P(\{u>t\}),\\ &\partial U_{\varepsilon_h,t}\subset\left\{x:dist(x,\partial\{u>t\})<\frac{1}{h}\right\}. \end{align}\tag{50}\]

We introduce the signed distance function \[\label{e4462} d_{E_{\delta}^k}(x)= \begin{cases} -\text{dist}(x,\partial E_{\delta}^k)&if\;x\in \left(E_{\delta}^k\right)^o,\\ 0&if\;x\in \partial E_{\delta}^k,\\ \text{dist}(x,\partial E_{\delta}^k)&if\;x\in \left(E_{\delta}^k\right)^c, \end{cases}\tag{51}\] which is a \(C^{\infty}\) function in the set \(C_{\delta}^k=\{x:d_{E_{\delta}^k}(x)>-\zeta_{\delta}^k\}\) for some \(\zeta_{\delta}^k>0\). Let \(E_{\delta,s}^k:=\{x:d_{E_{\delta}^k}\leq s\}\) for \(s>-\zeta_{\delta}^k\), and consider the \(C^{\infty}\) map \(f:x\mapsto (d_{E_{\delta}^k}(x),u_{\varepsilon_h}(x))\) defined on \(C_{\delta}^k\). From Sard’s theorem, the set of critical values of \(f\) has measure zero in \(\mathbb{R}^2\). Therefore, \[\label{e4464}\text{rank}\left(\begin{align} \nabla d_{E_{\delta}^k}(x)\\ \nabla u_{\varepsilon_h}(x) \end{align}\right)=2 \quadon\;\{x:d_{E_{\delta}^k}(x)=s,u_{\varepsilon_h}(x)=t\}\tag{52}\] for almost every \((s,t)\in(0,\infty)\times(0,1)\). Now fix \(t\in(0,1)\) satisfying 50 such that the rank condition 52 holds for all \(h\) and almost every \(s>0\). Then, the open set \(\Omega_{\delta,\varepsilon_h,s}^k:=U_{\varepsilon_h,t}\backslash E_{\delta,s}^k\) is a Lipschitz domain for almost every \(s>0\), and \(\partial \Omega_{\delta,\varepsilon_h,s}^k\backslash E_{\delta,s}^k\) is a \(C^{\infty}\) manifold with boundary. By assumption 49 for any \(\delta\) and \(k\), we have \[\label{e4468}{\mathcal{H}}^{n-1}(\partial\Omega\cap\partial E_{\delta,s}^k)=0\tag{53}\] for almost every \(s>0\). Therefore, \[\label{e4465}\begin{align} \lim_{h\rightarrow\infty}P(\Omega_{\delta,\varepsilon_h,s}^k;\mathbb{R}^n\backslash E_{\delta,s}^k)=&\lim_{h\rightarrow\infty}P(U_{\varepsilon_h,t}^k;\mathbb{R}^n\backslash E_{\delta,s}^k)\\=&P(\Omega;\mathbb{R}^n\backslash E_{\delta,s}^k)=P(\Omega\backslash E_{\delta,s}^k;\mathbb{R}^n\backslash E_{\delta,s}^k). \end{align}\tag{54}\] By the continuity of the trace Theorem (see Theorem 3.88 in [2]), the trace map \(u\mapsto u^{\Omega_{\delta,\varepsilon_h,t}^k}\) is continuous. Combining this with 54 , we obtain \[\label{e4466} \lim_{h\rightarrow\infty}{\mathcal{H}}^{n-1}(\partial\Omega_{\delta,\varepsilon_h,t}^k\cap\partial E_{\delta,s}^k)={\mathcal{H}}^{n-1}(\partial^{*}(\Omega\backslash E_{\delta,s}^k)\cap\partial E_{\delta,s}^k)={\mathcal{H}}^{n-1}(\Omega\cap\partial E_{\delta,s}^k),\tag{55}\] where we used the fact that \({\mathcal{H}}^{n-1}(\partial\Omega\cap\partial E_{\delta,s}^k)=0\) in the last equality. Since \(E_{\delta,s}^k\) converges to \(E_{\delta}^k\) in the sense of Kuratowski as \(s\rightarrow 0\), we have \({\mathcal{H}}^{n-1}\llcorner\partial E_{\delta,s}^k\stackrel{*}{\rightharpoonup}{\mathcal{H}}^{n-1}\llcorner\partial E_{\delta}^k\) weakly\(^{*}\) in the sense of measures (see Remark 2.2 in [2]). Therefore, \[\label{e4467}\lim_{s\rightarrow0}{\mathcal{H}}^{n-1}(\Omega\cap\partial E_{\delta,s}^k)={\mathcal{H}}^{n-1}(\Omega\cap\partial E_{\delta }^k).\tag{56}\] and \[\label{e4469} \lim_{s\rightarrow0}P(\Omega\backslash E_{\delta,s}^k;\mathbb{R}^n\backslash E_{\delta,s}^k)=P(\Omega\backslash E_{\delta}^k;\mathbb{R}^n\backslash E_{\delta}^k).\tag{57}\] Moreover, 49 and 53 imply that \[{\mathcal{H}}^{n-1}(\Omega\cap\partial E_{\delta }^k)={\mathcal{H}}^{n-1}(\partial^{*}(\Omega\backslash E_{\delta }^k)\cap\partial E_{\delta }^k).\] and \[{\mathcal{H}}^{n-1}(\Omega\cap\partial E_{\delta,s}^k)={\mathcal{H}}^{n-1}(\partial^{*}(\Omega\backslash E_{\delta,s}^k)\cap\partial E_{\delta,s}^k)\] Hence, 56 can be rewritten as \[\label{e4469a}\lim_{s\rightarrow0}{\mathcal{H}}^{n-1}(\partial^{*}(\Omega\backslash E_{\delta,s}^k)\cap\partial E_{\delta,s}^k)={\mathcal{H}}^{n-1}(\partial^{*}(\Omega\backslash E_{\delta}^k)\cap\partial E_{\delta}^k).\tag{58}\] Similarly, if for some \(\delta>0\) we have \({\mathcal{H}}^{n-1}(\partial\Omega\cap\partial(1+\delta)E)=0\), and if we denote \(E_{\delta}=(1+\delta)E\), then we can obtain \[\label{e44610} \begin{align} &\lim_{k\rightarrow\infty}P(\Omega\backslash E_{\delta }^k;\mathbb{R}^n\backslash E_{\delta }^k)=P(\Omega\backslash E_{\delta} ;\mathbb{R}^n\backslash E_{\delta} ),\\ &\lim_{k\rightarrow\infty}{\mathcal{H}}^{n-1}(\partial^{*}(\Omega\backslash E_{\delta }^k)\cap\partial E_{\delta }^k)={\mathcal{H}}^{n-1}(\partial^{*}(\Omega\backslash E_{\delta} )\cap\partial E_{\delta} ). \end{align}\tag{59}\] Notice that, by monotone convergence theorem, \[\label{e44611} \lim_{\delta\rightarrow0}P(\Omega\backslash E_{\delta} ;\mathbb{R}^n\backslash E_{\delta} )=\lim_{\delta\rightarrow0}P(\Omega;\mathbb{R}^n\backslash E_{\delta})=P(\Omega;\mathbb{R}^n\backslash E).\tag{60}\] By scaling, we conclude that \[\label{e44612} \lim_{\delta\rightarrow0}P(((1+\delta)^{-1}\Omega)\backslash E ;\mathbb{R}^n\backslash E )=P(\Omega;\mathbb{R}^n\backslash E).\tag{61}\] Applying the trace theorem again, we conclude that \[\label{e44613} \begin{align} &\lim_{\delta\rightarrow0}{\mathcal{H}}^{n-1}(\partial^{*}(\Omega\backslash E_{\delta})\cap\partial E_{\delta})\\ =&\lim_{\delta\rightarrow0}(1+\delta)^{n-1}{\mathcal{H}}^{n-1}(\partial^{*}((1+\delta)^{-1}\Omega\backslash E)\cap\partial E)\\ =&{\mathcal{H}}^{n-1}(\partial^{*}\Omega\cap\partial E). \end{align}\tag{62}\]

By a standard diagonal argument, we conclude that there exists suitable \(s_{h}\), \(k_{h}\) and \(\delta_h\) such that \(\Omega_{h}:=\Omega_{\delta_{h},\varepsilon_h,s_h}^{k_h}\) and \(E_{h}:=E_{E_{\delta_h,s_h}^{k_h}}\) satisfies \((i)-(iv)\).

If \(\Omega\) is a general set of finite perimeter in \(\mathbb{R}^n\setminus E\), then we can find a sequence of bounded open sets \(\{\Omega_h\}\subset\mathbb{R}^{n}\backslash E\) that approximates \(\Omega\) such that \(|\Omega_h\Delta\Omega|\rightarrow 0\) and \(P(\Omega_h;\mathbb{R}^n\backslash E)\rightarrow P(\Omega;\mathbb{R}^n\backslash E)\). This implies that \({\mathcal{H}}^{n-1}(\partial\Omega_h\cap\partial E_h)\rightarrow{\mathcal{H}}^{n-1}(\Omega\cap\partial E)\) by the continuity of the trace operator. Applying the same procedure as above, we conclude that the result holds for general sets of finite perimeter \(\Omega\).

Assume that \(\Omega\subset\mathbb{R}^n\setminus E\) is an open set of finite perimeter such that \(\partial\Omega\setminus E\) is smooth. In this case, we can consider the signed distance function to \(\partial\Omega\), denoted by \(d_{\Omega}\). By the smoothness of \(\partial\Omega\setminus E\), there exists an \(\varepsilon(\delta)-\) neighborhood of \(\partial\Omega\), denoted by \((\partial\Omega)_{\varepsilon(\delta)}\) such that \(d_{\Omega}\) is smooth in \((\partial\Omega)_{\varepsilon(\delta)}\setminus(1+\delta)E\). Proceeding similarly as before but with the set \(U_{n,t}\) replaced by \((\Omega)_{\varepsilon}:=\{x:d_{\Omega}\leq\varepsilon\}\), we define \(\Omega_{\delta,\varepsilon,s}^k:=(\Omega)_{\varepsilon}\setminus E_{\delta,s}^k\). Now consider the \(C^{\infty}\) map \(x\mapsto(d_{E_{\delta}^k(x)},d_{\Omega}(x))\). Using Sard’s theorem again, we obtain \[\label{e44615} rank\left(\begin{align} \nabla d_{E_{\delta}^k}(x)\\ \nabla d_{\Omega}(x) \end{align}\right)=2 \quadon\;\{x:d_{E_{\delta}^k}(x)=s,d_{\Omega}(x)=\varepsilon\}\tag{63}\] for a.e. \((s,\varepsilon)\in(0,\infty)\times(0,\varepsilon(\delta))\). Therefore, for such \((s,\varepsilon)\), the open set \(\Omega_{\delta,\varepsilon_h,s}^k:=U_{\varepsilon_h,t}\backslash E_{\delta,s}^k\) is a Lipschitz domain such that \(\partial \Omega_{\delta,\varepsilon_h,s}^k\backslash E_{\delta,s}^k\) is a \(C^{\infty}\) manifold with boundary. For almost every \(s>0\), we can find a sequence \(\varepsilon_h\rightarrow 0\) such that \((s,\varepsilon_h)\) satisfies all the rank conditions above for all \(h\) and \(k\). Along this subsequence, it is straightforward to show that \(|\Omega_{\delta,\varepsilon_h,s}^k\Delta(\Omega\setminus E_{\delta,s}^k)|\rightarrow0\). We can now proceed as before to reach the conclusion. ◻

Proof of Theorem 1. Notice that by Proposition 2, it follows that the inequality ?? holds if \(E\) and \(\Omega\subset\mathbb{R}^n\setminus E\) satisfy 27 and 28 respectively. By applying Lemma 10, we can extend this result to the general case through a standard approximation argument. ◻

Proof of Theorem 2. As a consequence of Theorem 1, when the weighted volumes are equal (\(\int_{\Omega} wdx = \int_{B_1\cap\mathbb{R}^n_+} wdx\)), we only need to show that for any convex set \(E\), the \(\lambda_w\)-ABP property holds with \(\lambda_w=0\). Let \(K:=\{x_1,\cdots,x_n\}\) be any discrete subset of \(\partial E\) and let \(v:K\rightarrow\mathbb{R}\) be a bounded function. From Lemma 2, we know that \(\bigcup_{i=1}^{n}J_{K}v(x_i)=\mathbb{R}^n\), and Remark 3 indicates that the sets \(J_{K}v(x_i)\) have disjoint interiors for \(i=1,\cdots,n\). Now, up to a set of Lebesgue measure zero, we may decompose \(J_{K}v(x_i)=J_{K}v(x_i)^+\cup J_{K}v(x_i)^{-}\), where \[J_{K}v(x_i)^+:=\{\xi\in J_{K}v(x_i):\xi\cdot\nu_{E}(x_i)>0\}\] and \[J_{K}v(x_i)^-:=\{\xi\in J_{K}v(x_i):\xi\cdot\nu_{E}(x_i)<0\}.\] From Lemma 4, we know that \[\label{e34640} \xi\in J_{K}v(x_i)\Rightarrow \xi+t\nu_{E}(x_i)\in J_{K}v(x_i)\;for any\;t>0.\tag{64}\] For any fixed \(\xi\in J_{K}v(x_i)^-\), we have \(\xi\in J_{K}v(x_i)\) and \(\xi\cdot\nu_{E}(x_i)<0\). Let \(t=-\xi\cdot\nu_{E}(x_i)>0\). Then, from 64 we deduce that \[\hat{\xi}:=\xi+2t\nu_{E}(x_i)\in J_{K}v(x_i).\] Moreover, \[\hat{\xi}\cdot\nu_{E}(x_i)=\xi\cdot\nu_{E}(x_i)+2t=t>0,\] which implies that \(\hat{\xi}\in J_{K}v(x_i)^{+}\). Since \(w\) is even in \(\mathbb{R}^n\), it follows that \[\int_{J_{K}v(x_i)}wdx=\int_{J_{K}v(x_i)^+}wdx+\int_{J_{K}v(x_i)^-}wdx\leq 2\int_{J_{K}v(x_i)^+}wdx.\] Therefore, \[\label{e34641} \int_{{\mathcal{B}}_{v}^{0}\cap B_1}wdx=\sum_{i=1}^n\int_{J_{K}v(x_i)^+\cap B_1}wdx\geq\frac{1}{2}\sum_{i=1}^n\int_{J_{K}v(x_i)\cap B_1}wdx=\frac{1}{2}\int_{B_1}wdx=\int_{B^{0}}wdx,\tag{65}\] which shows that ?? holds for \(\lambda_w=0\). Then, applying Theorem 2, the conclusion holds. ◻

Proof. Since \(fw=-{\rm div}\left(wF^{p-1} (\nabla u)D F (\nabla u)\right)\), for any test function \(\psi\in W_0^{1,p}(\Omega,w; E^c)\), we have \[\label{e5467}\int_{\Omega}w F^{p-1} (\nabla u)D F (\nabla u)\cdot\nabla \psi dx=\int_{\Omega}fw\psi dx,\tag{70}\] where we used the boundary conditions \(D F (\nabla u)\cdot\nu=0\) on \(\partial\Omega\cap\partial E^c\) and \(\psi=0\) on \(\partial\Omega\setminus E\). Let \(t>0\) and choose \(\psi=(u-t)^+\), which belongs to \(\in W_0^{1,p}(\Omega,w; E^c)\) and is supported on \(\{u>t\}\). Substituting \(\psi\) into the identity 70 , we obtain \[\int_{\{u>t\} }w F^{p} (\nabla u)dx=\int_{\{u>t\} }fw(u-t)dx.\] Differentiating with respect to \(t\) yields \[\label{e5468} -\frac{d}{dt}\int_{\{u>t\} }w F^{p} (\nabla u)dx=\int_{\{u>t\} }fwdx.\tag{71}\] Observe that the boundary decomposes as \[\begin{align}\partial\{u>t\}=&(\partial\{u>t\}\cap\Omega)\cup(\{u>t\}\cap\partial\Omega\cap E^c)\cup(\partial\{u>t\}\cap\partial E)\\=&\{u=t\}\cup(\partial\{u>t\}\cap\partial E),\end{align}\] where we used the fact that \(\{u > t\} \cap \partial\Omega \cap E^c = \emptyset\) since \(u = 0\) on \(\partial\Omega \cap E^c\). Hence, by the definition of \(f\) and the divergence theorem, we obtain \[\begin{align} -\frac{d}{dt}\int_{\{u>t\} }wF^{p}(\nabla u)dx =&\int_{\{u>t\} }-{\rm div}\left(wF^{p-1}(\nabla u)DF(\nabla u)\right)dx\\ =&\int_{\partial\{u>t\}}wF^{p-1}(\nabla u)DF(\nabla u)\cdot\frac{\nabla u}{|\nabla u|} d{\mathcal{H}}^{n-1}\\ =&\int_{\{u=t\} }\frac{wF^p(\nabla u)}{|\nabla u|} d{\mathcal{H}}^{n-1} +\int_{\partial\{u>t\}\cap\partial E}\frac{wF^p(\nabla u)}{|\nabla u|} d{\mathcal{H}}^{n-1}\\ =&\int_{\{u=t\} }\frac{wF^p(\nabla u)}{|\nabla u|} d{\mathcal{H}}^{n-1}. \end{align}\] Here, we used the Neumann boundary condition that \(DF(\nabla u) \cdot \nu = 0\) on \(\partial\Omega \cap \partial E\) and the fact that the unit outer normal to \(\{u>t\}\) on the level set \(\{u=t\}\) is given by \(\nu=-\frac{\nabla u}{|\nabla u|}\), since \(u\) is constant on this surface and \(\{u>t\}\) lies in its interior. ◻

Proof. Let \(\varepsilon>0\), and define \[fw=-{\rm div}\left(\frac{wDF(\nabla u)}{F(\nabla u)+\varepsilon}\right).\] Multiplying by \((u-t)^+\) and integrating by parts, we obtain \[\int_{\{u>t\}} \frac{wF(\nabla u)}{F(\nabla u)+\varepsilon}dx=\int_{\{u>t\}}fw(u-t)dx.\] Differentiating with respect to \(t\) yields \[\label{e44618}-\frac{d}{dt}\int_{\{u>t\}} \frac{wF(\nabla u)}{F(\nabla u)+\varepsilon}dx=\int_{\{u>t\}}fw dx.\tag{72}\] For sufficiently small \(h>0\), integrating from \(t-h\) to \(t\) on both sides of 72 , we get \[\begin{align} \int_{\{t-h<u\leq t\}}\frac{wF(\nabla u)}{F(\nabla u)+\varepsilon}dx=&\int_{t-h}^t\left(\int_{\{u>\tau\}}fwdx\right)d\tau\\ =&\int_{t-h}^t\left(\int_{\{u>\tau\}}-{\rm div}\left(\frac{wDF(\nabla u)}{F(\nabla u)+\varepsilon}\right) dx\right)d\tau\\ =&-\int_{t-h}^t\left(\int_{ \{u=\tau\}}\frac{wDF(\nabla u)\cdot\nu}{F(\nabla u)+\varepsilon} dx\right)d\tau-\int_{t-h}^t\left(\int_{ \partial\{u>\tau\}\cap\partial E}\frac{wDF(\nabla u)\cdot\nu}{F(\nabla u)+\varepsilon} dx\right)d\tau\\ =&\int_{t-h}^t\left(\int_{ \{u=\tau\}}\frac{wF(\nabla u)}{F(\nabla u)+\varepsilon}\frac{1}{|\nabla u|}dx\right)d\tau, \end{align}\] where we used the fact that \(D F (\nabla u)\cdot\nu=0\) on \(\partial\Omega\cap\partial E\). Applying the dominated convergence theorem and taking the limit as \(\varepsilon\rightarrow 0\), we obtain \[\mu(t-h)-\mu(t)=\int_{t-h}^t\left(\int_{ \{u=\tau\}}\frac{w}{|\nabla u|}d{\mathcal{H}}^{n-1}\right)d\tau.\] Dividing by \(h\) and taking the limit as \(h\rightarrow 0\), we conclude that the first equality in ?? holds.

Since \(\mu(t)\) and \(r(t)\) are monotonically decreasing functions, they are differentiable for almost every \(t\). The last equality in 69 implies that \[\label{e54610}\mu_u'(t)=(n+\alpha)\int_{B_1}wdx\cdot(r(t))^{n+\alpha-1}r'(t).\tag{73}\] By the homogeneity of \(w\), we calculate that \[\label{e54611}\begin{align} \int_{B_1\cap\mathbb{R}^n_+}wdx=&\int_0^1\left(\int_{\partial B_r\cap\mathbb{R}^n_+}w(x)d{\mathcal{H}}^{n-1}\right)dr\\ =&\int_0^1\left(\int_{\partial B_r\cap\mathbb{R}^n_+}r^{\alpha}w\left(\frac{x}{|x|}\right)d{\mathcal{H}}^{n-1}\right)dr\\ =&\int_0^1 r^{n+\alpha-1}dr\left(\int_{\partial B_1\cap\mathbb{R}^n_+}w(x)d{\mathcal{H}}^{n-1}\right)\\ =&\frac{1}{n+\alpha}\int_{\partial B_1\cap\mathbb{R}^n_+}w(x)d{\mathcal{H}}^{n-1}, \end{align}\tag{74}\] and \[\label{e54612}\int_{\{u^{*}=t\}}wd{\mathcal{H}}^{n-1}=\int_{\partial B_{r(t)} \cap\mathbb{R}^n_+}wd{\mathcal{H}}^{n-1}=r(t)^{n+\alpha-1}\int_{\partial B_1\cap\mathbb{R}^n_+}wd{\mathcal{H}}^{n-1}.\tag{75}\] Combining with 73 , 74 and 75 , we obtain \[\label{e54627} \mu_u'(t)=\int_{\{u^{*}=t\}}wd{\mathcal{H}}^{n-1}r'(t).\tag{76}\] Noting that \[u^{*}(r(t))=u^{\#}\left(\int_{B_1\cap\mathbb{R}_n^+}wdx\cdot r(t)^{n+\alpha}\right)=u^{\#}(\mu(t))=t,\] taking the derivative with respect to \(t\) on both sides, we deduce that \(r'(t)=\frac{1}{(u^{*})'(r(t))}\). Moreover, since \(u^{*}(x)=u^{*}(r(t))\) on \(\partial B_{r(t)}\), by implicit differentiation, we have \[|\nabla u^{*}(x)|=\left|(u^{*})'(r(t))\cdot \frac{x}{r(t)}\right|=|(u^{*})'(r(t))|=-(u^{*})'(r(t)),\] where we used the fact that \((u^{*})'(r(t))=\frac{1}{r'(t)}<0\) in the last equality. Substituting \(r'(t)=\frac{1}{(u^{*})'(r(t))}\) and \(|\nabla u^{*}(x)|=-(u^{*})'(r(t))\) into 76 , we obtain \[\mu_u'(t)=-\int_{\{u^{*}=t\}}\frac{w}{|\nabla u^{*}|}d{\mathcal{H}}^{n-1}.\] This completes the proof. ◻

Proof of Theorem 3. Since \(\widetilde{F}_{0}(\xi)=|\xi|\) is a special gauge, by Lemma 11, we know that for any \(t>0\), \[-\frac{d}{dt}\int_{\{u>t\}}w|\nabla u|^pdx=\int_{\{u=t\}}w|\nabla u|^{p-1}d{\mathcal{H}}^{n-1}.\] Similarly, for the symmetrized function \(u^{*}\), we have \[-\frac{d}{dt}\int_{\{u^{*}>t\}}w|\nabla u^{*}|^pdx=\int_{\{u^{*}=t\}}w|\nabla u^{*}|^{p-1}d{\mathcal{H}}^{n-1}.\] On the other hand, we apply Hölder inequality to the integral over the level set \(\{u=t\}\subset \Omega\). Specifically, we obtain \[\int_{\{u=t\}}wd{\mathcal{H}}^{n-1}\leq\left(\int_{\{u=t\}}w|\nabla u|^{p-1}\right)^{\frac{1}{p}}\left(\int_{\{u=t\}}\frac{w}{|\nabla u|}\right)^{1-\frac{1}{p}}.\] From Lemma 1, we know that the weighted isoperimetric inequality outside convex sets ?? can be rewritten as \(P_{w}(\Omega;\mathbb{R}^n\setminus E)\geq P_{w}(B_1;\mathbb{R}^n_+)\), where \(\Omega\subset E^c\) and \(B_1\cap\mathbb{R}^n_+\) have the same weighted volume, i.e., \(\int_{\Omega}wdx=\int_{B_1\cap\mathbb{R}^n_+}wdx\). Therefore, we deduce that \[\begin{align} \int_{\{u=t\}}w|\nabla u|^{p-1}d{\mathcal{H}}^{n-1}\geq& \left(\int_{\{u=t\}}wd{\mathcal{H}}^{n-1}\right)^p\left(-\mu'(t)\right)^{1-p}\\ =&\left(P_{w}(\{u>t\};E^c)\right)^p\left(-\mu'(t)\right)^{1-p}\\ \geq&\left(P_{w}(\{u^*>t\};\mathbb{R}^n_+)\right)^p\left(-\mu'(t)\right)^{1-p} =\int_{\{u^*=t\}}w|\nabla u^*|^{p-1}d{\mathcal{H}}^{n-1}, \end{align}\] where in the last step, we used the conclusion of Theorem 12, which relates the distribution function \(\mu(t)\) to the level sets of \(u\) and \(u^{*}\). Combining these results, we deduce that \[-\frac{d}{dt}\int_{\{u>t\}}w|\nabla u|^{p}dx\geq-\frac{d}{dt}\int_{\{u^*>t\}}w|\nabla u^*|^pdx.\] Integrating both sides from \(0\) to \(+\infty\) with respect to \(t\), we complete the proof of the weighted Pólya-Szegö principle for the capillary Schwarz symmetrization outside convex sets. ◻

Proof. We first claim that \[\label{e54615} \int_{\Omega}w|u|^qdx=\int_{B_1\cap\mathbb{R}^n_+}w|u^{*}|^q dx.\tag{77}\] For \(q=1\), we have \[\begin{align} \int_{\Omega}w|u|dx=\int_{\Omega}\left(\int_{0}^{u(x)}wdt\right)dx =&\int_0^{\infty}\left(\int_{\{u>t\}}wdx\right)dt\\ =&\int_0^{\infty}\left(\int_{\{u^{*}>t\}}wdx\right)dt =\int_{B_1\cap\mathbb{R}^n_+}w|u^{*}|dx, \end{align}\] where we used the fact that \(\int_{\{u>t\}}wdx=\int_{\{u^{*}>t\}}wdx\) obtained in 68 .

For \(q>1\), we set \(v=u^q\). Then \(\{v>s\}=\{u>s^{\frac{1}{q}}\}\) and \[\int_{\Omega}w|v|dx=\int_0^{\infty}\left(\int_{\{v>s\}}wdx\right)ds.\] Let \(s=t^q\), then we obtain \[\begin{align}\int_{\Omega}w|u|^qdx=q\int_0^{\infty}t^{q-1}\left(\int_{\{u>t\}}wdx\right)dt.\end{align}\] Using again the equality of weighted measures \(\int_{{u > t}} w dx = \int_{{u^{*} > t}} w dx\), we conclude that 77 holds. In particular, for the critical exponent \(q = \frac{(n+\alpha)p}{n+\alpha-p}\), we have \[\int_{\Omega}w|u|^{\frac{(n+\alpha)p}{n+\alpha-p}}dx=\int_{B_1\cap\mathbb{R}^n_+}w|u^{*}|^{\frac{(n+\alpha)p}{n+\alpha-p}} dx.\]

Combining Theorem 3 with 77 , we deduce that \[\frac{\int_{\Omega}w|\nabla u|^pdx}{\left(\int_{\Omega}w|u|^{\frac{(n+\alpha)p}{n+\alpha-p}}dx\right)^{\frac{n+\alpha-p}{n+\alpha}}}\geq\frac{\int_{B_1\cap\mathbb{R}^n_+}w|\nabla u^*|^pdx}{\left(\int_{B_1\cap\mathbb{R}^n_+}w|u^*|^{\frac{(n+\alpha)p}{ n+\alpha-p}}dx\right)^{\frac{n+\alpha-p}{n+\alpha}}}.\] Applying Lemma 13. to the symmetrized function \(u^{*}\), we obtain \[\int_{\Omega}w|\nabla u|^pdx\geq C^{-1}(n,p,\alpha,w)\left(\int_{\Omega}w|u|^{\frac{(n+\alpha)p}{n+\alpha-p}}dx\right)^{\frac{n+\alpha-p}{n+\alpha}}.\] This completes the proof. ◻

Proof. The proof relies on the following weighted Sobolev inequality: \[\label{e54634} \left(\int_{\Omega}w|u|^{p_{\alpha}^{*}}dx\right)^{\frac{p}{p_{\alpha}^{*}}}\leq C\int_{\Omega}w|\nabla u|^pdx,\tag{78}\] where \(p_{\alpha}^{*}=\frac{(n+\alpha)p}{n+\alpha-p}\) is the weighted Sobolev exponent and \(C>0\) is a constant. Indeed, if \(\{u_k\}\) is a bounded sequence in \(W_0^{1,p}(\Omega,w;E^c)\), then for any compact subset \(A\subset\Omega\setminus\{0\}\), there exists a constant \(c_1 > 0\) such that \[c_1\int_{A}|u_k|^pdx\leq \int_{\Omega}w|u_k|^pdx<+\infty\] and \[c_1\int_{A}|\nabla u_k|^pdx\leq \int_{\Omega}w|\nabla u_k|^pdx<+\infty.\] Therefore, \(\{u_k\}\) is also a bounded sequence in \(W_{\text{loc}}^{1,p}(\Omega\setminus\{0\})\). Consequently, there exists a subsequence (still denoted by \(\{u_k\}\)) and a function \(u \in W_{\text{loc}}^{1,p}(\Omega\setminus\{0\})\) such that \(u_k \rightarrow u\) almost everywhere in \(\Omega\). By Egoroff’s Theorem, for any \(\varepsilon>0\), there exists a domain \(\Omega_{\varepsilon}\subset\Omega\) such that \(u_k\rightarrow u\) uniformly in \(\Omega_{\varepsilon}\) and \(|\Omega\setminus\Omega_{\varepsilon}|<\varepsilon\). Then, by Hölder inequality and 78 , we have \[\begin{align} \lim_{k\rightarrow\infty}\int_{\Omega}w|u_k-u|^q dx=&\lim_{\varepsilon\rightarrow0}\lim_{k\rightarrow\infty}\int_{\Omega_{\varepsilon}}w|u_k-u|^q dx+\lim_{\varepsilon\rightarrow0}\lim_{k\rightarrow\infty}\int_{\Omega\setminus \Omega_{\varepsilon}}w|u_k-u|^q dx\\ \leq&\lim_{\varepsilon\rightarrow0}\lim_{k\rightarrow\infty}\int_{\Omega_{\varepsilon}}w|u_k-u|^q dx+\lim_{\varepsilon\rightarrow0}\lim_{k\rightarrow\infty}\left(\int_{\Omega\setminus \Omega_{\varepsilon}}w|u_k-u|^{p_{\alpha}^{*}} dx\right)^{\frac{q}{p_{\alpha}^{*}}}\left(\int_{\Omega\setminus\Omega_{\varepsilon}}wdx\right)^{1-\frac{q}{p_{\alpha}^{*}}}\\ \leq &\lim_{\varepsilon\rightarrow0}\int_{\Omega_{\varepsilon}}\lim_{k\rightarrow\infty}w|u_k-u|^q dx+C\lim_{\varepsilon\rightarrow0}\lim_{k\rightarrow\infty}\left(\int_{\Omega\setminus \Omega_{\varepsilon}}w|\nabla(u_k-u)|^{p} dx\right) \left(\int_{\Omega\setminus\Omega_{\varepsilon}}wdx\right)^{1-\frac{q}{p_{\alpha}^{*}}}\\ =&0, \end{align}\] which implies that \(u_k\rightarrow u\) strongly in \(L^{q}(\Omega,w; E^c)\). Hence, the embedding from \(W_0^{1,p}(\Omega,w;E^c)\) to \(L^q(\Omega,w;E^c)\) is compact for any \(0<q<p_{\alpha}^{*}\).

Next, it remains to prove 78 . We first prove the case \(p=1\). Combining with ?? and ?? we obtain \[\frac{\int_{\partial\Omega\setminus E}wd{\mathcal{H}}^{n-1}}{\left(\int_{\Omega}wdx\right)^{\frac{n+\alpha-1}{n+\alpha}}}\geq\frac{\int_{\partial B_1\cap\mathbb{R}^n_+}wd{\mathcal{H}}^{n-1}}{\left(\int_{B_1\cap\mathbb{R}^n_+}wdx\right)^{\frac{n+\alpha-1}{n+\alpha}}}=(n+\alpha)C_{n,w}^{\frac{1}{n+\alpha}},\] where \(C_{n,w}=\int_{B_1\cap\mathbb{R}^n_+}wdx\) is a constant depending on \(n\) and \(w\). Therefore, \[\left(\int_{\Omega}wdx\right)^{\frac{n+\alpha-1}{n+\alpha}}\leq \frac{1}{n+\alpha}C_{n,w}^{-\frac{1}{n+\alpha}}\int_{\partial\Omega\setminus E}wd{\mathcal{H}}^{n-1}.\] In particular, for any \(t>0\), let \(\{u>t\}=\{x\in\Omega: u(x)>t\}\) and let \(u=0\) on \(\partial\Omega\setminus E\). Then,

\[\label{e54630} \left(\int_{\{u>t\}}wdx\right)^{\frac{n+\alpha-1}{n+\alpha}}\leq \frac{1}{n+\alpha}C_{n,w}^{-\frac{1}{n+\alpha}}\int_{\{u=t\}\cap\Omega}wd{\mathcal{H}}^{n-1},\tag{79}\] where we used the fact that \(\partial\{u>t\}\setminus E=\left(\{u=t\}\cap\Omega\right)\cup\left(\{u>t\}\cap\partial\Omega\right)\) and noted that \(\{u>t\}\cap\partial\Omega=\emptyset\). Let \(\chi_{\Omega}\) be the characteristic function of \(\Omega\), we have \[u(x)=\int_{0}^{+\infty}\chi_{\{u>t\}}dt.\]

Applying the Minkowski integral inequality, together with 79 and the co-area formula, we have \[\label{e54631} \begin{align} \left(\int_{\Omega}w|u|^{\frac{n+\alpha}{n+\alpha-1}}dx\right)^{\frac{n+\alpha-1}{n+\alpha}} =&\left[\int_{\Omega}w\left(\int_0^{+\infty}\chi_{\{u>t\}}dt\right)^{\frac{n+\alpha}{n+\alpha-1}}dx\right]^{\frac{n+\alpha-1}{n+\alpha}}\\ \leq&\int_0^{+\infty}\left(\int_{\Omega}w\chi_{\{u>t\}}dx\right)^{\frac{n+\alpha-1}{n+\alpha}}dt\\ =&\int_0^{+\infty}\left(\int_{\{u>t\}}wdx\right)^{\frac{n+\alpha-1}{n+\alpha}}dt\\ \leq&\frac{1}{n+\alpha}C_{n,w}^{-\frac{1}{n+\alpha}}\int_{0}^{+\infty}\left(\int_{\{u=t\}\cap\Omega}wd{\mathcal{H}}^{n-1}\right)dt\\ =&\frac{1}{n+\alpha}C_{n,w}^{-\frac{1}{n+\alpha}}\int_{\Omega}w|\nabla u|dx. \end{align}\tag{80}\] Then Theorem 4 holds for \(p=1\). Now we only need to consider the case \(1<p<n+\alpha\). Let \(v=u^{\frac{(n+\alpha-1)p}{n+\alpha-p}}\), we have \[\label{e54632}\left(\int_{\Omega}w|u|^{\frac{(n+\alpha)p}{n+\alpha-p}}dx\right)^{\frac{n+\alpha-1}{n+\alpha}} =\left(w|v|^{\frac{n+\alpha}{n+\alpha-1}}\right)^{\frac{n+\alpha-1}{n+\alpha}} \leq\frac{1}{n+\alpha}C_{n,w}^{-\frac{1}{n+\alpha}}\int_{\Omega}w|\nabla v|dx.\tag{81}\] Since \(|\nabla v|=\frac{(n+\alpha-1)p}{n+\alpha-p}|u|^{\frac{(n+\alpha)(p-1)}{n+\alpha-p}}|\nabla u|\), then by Hölder inequality, we obtain \[\label{e54633} \int_{\Omega}w|\nabla v|dx \leq C_1\left(\int_{\Omega}w|\nabla u|^p dx\right)^{\frac{1}{p}}\left(\int_{\Omega}w|u|^{\frac{(n+\alpha)p}{n+\alpha-p}}dx\right)^{\frac{p-1}{p}}.\tag{82}\] Combining with 81 and 82 , we deduce that \[\left(\int_{\Omega}w|u|^{\frac{(n+\alpha)p}{n+\alpha-p}}dx\right)^{\frac{n+\alpha-p}{n+\alpha}}\leq C\int_{\Omega}w|\nabla u|^pdx,\] where \(C=\frac{1}{n+\alpha}C_1C_{n,w}^{-\frac{1}{n+\alpha}}\). ◻

Proof of Theorem 4. From 78 and the Hölder inequality, we have \[\label{e54635} \int_{\Omega}w|u|^pdx\leq\left(\int_{\Omega}w|u|^{p_{\alpha}^{*}}dx\right)^{\frac{p}{p_{\alpha}^{*}}}\left(\int_{\Omega}wdx\right)^{1-\frac{p}{p_{\alpha}^{*}}} \leq CC_{n,w}^{\frac{p}{n+\alpha}} \int_{\Omega}w|\nabla u|^pdx,\tag{83}\] where \(C_{n,w}\) is a constant defined by 66 . Then, \[\label{e54625}A_k=\inf_{u\in W_0^{1,p}(\Omega,w; E^c)}\frac{\int_{\Omega}w|\nabla u|^p dx}{\|u\|_{L^{p_k}(\Omega,w)}^p}\quad \;for any\; p_k<p_{\alpha}^{*}\tag{84}\] is well-defined, where \(p_{\alpha}^{*}=\frac{(n+\alpha)p}{n+\alpha-p}\) is the weighted Sobolev critical exponent. We claim that \[A=\inf_{u\in W_0^{1,p}(\Omega,w; E^c)}\frac{\int_{\Omega}w|\nabla u|^{p}dx}{\left(\int_{\Omega}w|u|^{p_{\alpha}^{*}}dx\right)^{\frac{p}{p_{\alpha}^{*}}}}\] is the best weighted Sobolev constant outside convex sets.

From the weighted embedding result Lemma 14 and standard variational arguments, \(A_k\) is achieved by a function \(u_p\in W_0^{1,p}(\Omega,w;E^c)\) that satisfies \[\label{e54626} \begin{cases} -{\rm div}\left(w|\nabla u|^{p-2}\nabla u\right)=\lambda_k w|u|^{p_{k}-2}u&in\;\Omega,\\ u=0&on\;\partial\Omega\cap E^c\\ \frac{\partial u}{\partial\nu}=0&on\;\partial\Omega\cap\partial E, \end{cases}\tag{85}\] where \(\lambda_k\) is the associated Lagrange multiplier. Therefore, ?? holds for \(u_p\). Applying Hölder’s inequality with conjugate exponents \(p'=\frac{p_{\alpha}^{*}}{p_k}\) and \(q'=\frac{p_{\alpha}^{*}}{p_{\alpha}^{*}-p_k}\), we obtain \[\left(\int_{\Omega}w|u_p|^{p_k}dx\right)^{\frac{p}{p_k}}\leq\left(\int_{\Omega}w|u_p|^{p_{\alpha}^{*}}dx\right)^{\frac{p}{p_{\alpha}^{*}}}\left(\int_{\Omega}wdx\right)^{\frac{p}{p_k}-\frac{p}{p_{\alpha}^{*}}}.\] This leads to the following estimate: \[\label{e54628a}\begin{align} \lim_{k\rightarrow\infty}A_k=&\lim_{k\rightarrow\infty}\frac{\int_{\Omega}w|\nabla u_p|^{p}dx}{\left(\int_{\Omega}w|u_p|^{p_k}dx\right)^{\frac{p}{p_k}}}\\ \geq&\lim_{k\rightarrow\infty}\frac{\int_{\Omega}w|\nabla u_p|^{p}dx}{\left(\int_{\Omega}w|u_p|^{p_{\alpha}^{*}}dx\right)^{\frac{p}{p_{\alpha}^{*}}}}\left(\int_{\Omega}wdx\right)^{\frac{p}{p_{\alpha}^{*}}-\frac{p}{p_k}}\\ =&\frac{\int_{\Omega}w|\nabla u_p|^{p}dx}{\left(\int_{\Omega}w|u_p|^{p_{\alpha}^{*}}dx\right)^{\frac{p}{p_{\alpha}^{*}}}}.\\ \end{align}\tag{86}\]

On the other hand, for any \(u\in W_0^{1,p}(\Omega,w;E^c)\), the definition of \(A_k\) implies that \[\lim_{k\rightarrow\infty}A_k\leq \lim_{k\rightarrow\infty}\frac{\int_{\Omega}w|\nabla u|^{p}dx}{\left(\int_{\Omega}w|u|^{p_k}dx\right)^{\frac{p}{p_k}}}=\frac{\int_{\Omega}w|\nabla u|^{p}dx}{\left(\int_{\Omega}w|u|^{p_{\alpha}^{*}}dx\right)^{\frac{p}{p_{\alpha}^{*}}}}.\] By the arbitrariness of \(u\), we conclude: \[\label{e54629}\lim_{k\rightarrow\infty}A_k\leq \inf_{u\in W_0^{1,p}(\Omega,w; E^c)}\frac{\int_{\Omega}w|\nabla u|^{p}dx}{\left(\int_{\Omega}w|u|^{p_{\alpha}^{*}}dx\right)^{\frac{p}{p_{\alpha}^{*}}}}.\tag{87}\] Combining with 86 and 87 , we obtain: \[\lim_{k\rightarrow\infty}A_k=A.\] Therefore, \[\label{e54638}A=\lim_{k\rightarrow\infty}A_k=\lim_{k\rightarrow\infty}\frac{\int_{\Omega}w|\nabla u_p|^{p}dx}{\left(\int_{\Omega}w|u_p|^{p_k}dx\right)^{\frac{p}{p_k}}} \geq\lim_{k\rightarrow\infty}\frac{\int_{B_1\cap\mathbb{R}^n_+}w|\nabla u_{p_{\alpha}^{*}}|^pdx}{\left(\int_{B_1\cap\mathbb{R}^n_+}w|u^{*}_p|^{p_{\alpha}^{*}}dx\right)^{\frac{p}{p_{\alpha}^{*}}}}\geq C^{-1}(n,p,\alpha,w).\tag{88}\] On the other hand, for any fixed \(x_0\in\Omega\) and and sufficiently small \(\delta>0\), we have \(B_{2\delta}(x_0)\subset\Omega\). Without loss of generality, we assume that \(x_0=0\). Let \(u_{\varepsilon}(x)=\varepsilon^{-\frac{n+\alpha-p}{p}}U_{1,0}^{\alpha}(\frac{x}{\varepsilon})\), defined in \(\mathbb{R}^n\), where \(U_{1,0}^{\alpha}\) is the extremal function in the half-space given by ?? . Let \(0\leq\eta\leq1\) be a cut-off function such that \(\eta=1\) on \(B_{\delta}\) and \(\eta=0\) on \(\mathbb{R}^n\setminus B_{2\delta}\). Then, we have \[\begin{align} A=&\inf_{u\in W_0^{1,p}(\Omega,w; E^c)}\frac{\int_{\Omega}w|\nabla u|^{p}dx}{\left(\int_{\Omega}w|u|^{p_{\alpha}^{*}}dx\right)^{\frac{p}{p_{\alpha}^{*}}}}\\ \leq& \frac{\int_{B_{2\delta}\cap\mathbb{R}^n_+}w|\nabla (\eta u_{\varepsilon})|^{p}dx}{\left(\int_{B_{2\delta}\cap\mathbb{R}^n_+}w|\eta u_{\varepsilon}|^{p_{\alpha}^{*}}dx\right)^{\frac{p}{p_{\alpha}^{*}}}}\\ \leq& \frac{\int_{B_{2\delta}\cap\mathbb{R}^n_+}w|\eta\nabla u_{\varepsilon}|^{p}dx}{\left(\int_{B_{2\delta}\cap\mathbb{R}^n_+}w|\eta u_{\varepsilon}|^{p_{\alpha}^{*}}dx\right)^{\frac{p}{p_{\alpha}^{*}}}}+\frac{\int_{B_{2\delta}\cap\mathbb{R}^n_+}w| u_{\varepsilon}\nabla \eta|^{p}dx}{\left(\int_{B_{2\delta}\cap\mathbb{R}^n_+}w|\eta u_{\varepsilon}|^{p_{\alpha}^{*}}dx\right)^{\frac{p}{p_{\alpha}^{*}}}}\\ \leq& \frac{\int_{B_{2\delta}\cap\mathbb{R}^n_+}w|\nabla u_{\varepsilon}|^{p}dx}{\left(\int_{B_{\delta}\cap\mathbb{R}^n_+}w| u_{\varepsilon}|^{p_{\alpha}^{*}}dx\right)^{\frac{p}{p_{\alpha}^{*}}}}+C\delta^{-p}\frac{\int_{B_{2\delta}\cap\mathbb{R}^n_+}w| u_{\varepsilon}|^{p}dx}{\left(\int_{B_{\delta}\cap\mathbb{R}^n_+}w|\eta u_{\varepsilon}|^{p_{\alpha}^{*}}dx\right)^{\frac{p}{p_{\alpha}^{*}}}}\\ =& \frac{\int_{B_{2\delta/\varepsilon}\cap\mathbb{R}^n_+}w|\nabla U_{1,0}^{\alpha}|^{p}dx}{\left(\int_{B_{\delta/\varepsilon}\cap\mathbb{R}^n_+}w| U_{1,0}^{\alpha}|^{p_{\alpha}^{*}}dx\right)^{\frac{p}{p_{\alpha}^{*}}}}+C\left(\frac{\varepsilon}{\delta}\right)^p\frac{\int_{B_{2\delta/\varepsilon}\cap\mathbb{R}^n_+}w| U_{1,0}^{\alpha}|^{p}dx}{\left(\int_{B_{\delta/\varepsilon}\cap\mathbb{R}^n_+}w| U_{1,0}^{\alpha}|^{p_{\alpha}^{*}}dx\right)^{\frac{p}{p_{\alpha}^{*}}}}.\\ \end{align}\] Taking the limit as \(\varepsilon\rightarrow0\) on both sides, we obtain \[\label{e54639}A\leq \frac{\int_{\mathbb{R}^n_+}w|\nabla U_{1,0}^{\alpha}|^{p}dx}{\left(\int_{\mathbb{R}^n_+}w|U_{1,0}^{\alpha}|^{p_{\alpha}^{*}}dx\right)^{\frac{p}{p_{\alpha}^{*}}}}=C^{-1}(n,p,\alpha,w).\tag{89}\] Combining with 88 and 89 , we conclude that \(A=C^{-1}(n,p,\alpha,w)\), which is the sharp constant of the weighted capillary Sobolev inequality outside convex sets. ◻