Proof. Write \[\widetilde{g}=e^{2\varphi}g, \qquad \varphi=\frac{\alpha}{2}\log w .\] The scalar curvature transformation formula gives \[\begin{align} w^\alpha R_{\widetilde{g}} &= R_g-(m-1)\alpha w^{-1}\Delta_g w \\ &\quad+ \left( (m-1)\alpha - \frac{(m-1)(m-2)}{4}\alpha^2 \right)w^{-2}|dw|_g^2 . \end{align}\] Moreover, \[\begin{align} w^\alpha\Delta_{\widetilde{g}}\widetilde{f} &= \Delta_g f +\beta w^{-1}\Delta_g w +\frac{\alpha(m-2)}{2}w^{-1}\langle dw,df\rangle_g \\ &\quad+ \left( -\beta+\frac{\alpha\beta(m-2)}{2} \right)w^{-2}|dw|_g^2 , \end{align}\] and \[w^\alpha |d\widetilde{f}|_{\widetilde{g}}^2 = |df|_g^2 +2\beta w^{-1}\langle dw,df\rangle_g +\beta^2w^{-2}|dw|_g^2 .\] Substituting these identities into \[\mathscr S_a^{\widetilde{g}}(\widetilde{f}) = R_{\widetilde{g}} +2\Delta_{\widetilde{g}}\widetilde{f} -a|d\widetilde{f}|_{\widetilde{g}}^2\] gives 5 . ◻

Proof. With the above choice of \(\alpha\) and \(\beta\), \[2\beta-(m-1)\alpha = -\frac{\alpha(m-1)}{q},\] and \[\alpha(m-2)-2a_m(\lambda)\beta=2.\] It remains only to check that the coefficient of \(w^{-2}|dw|^2\) in 5 vanishes. The first two terms give \[(m-1)\alpha-\frac{(m-1)(m-2)}{4}\alpha^2 = -\frac{m-1}{m-2}(q^2-1),\] while the next two terms give \[-2\beta+\alpha\beta(m-2) = \frac{2(m-1)}{m-2}(q^2-1).\] Thus the four terms not involving \(a_m(\lambda)\beta^2\) combine to \[\frac{m-1}{m-2}(q^2-1).\] On the other hand, \[a_m(\lambda)\beta^2 = \frac{m-1}{m-2}(q^2-1).\] Hence the coefficient of \(w^{-2}|dw|^2\) is zero. Substituting the two remaining coefficients into 5 gives ?? . ◻

Proof. For readability write \[\overline{q}=\overline{q}_m(\lambda), \qquad \overline{\alpha}=\overline{\alpha}_m(\lambda), \qquad \overline{\beta}=\overline{\beta}_m(\lambda), \qquad \overline{s}=\overline{s}_m(\lambda), \qquad \overline{\mathcal{K}}=\overline{\mathcal{K}}_m(\lambda),\] and \(a=a_{m+1}(\lambda)\). By definition, \[2\overline{s} = \frac{\overline{\alpha}(m-2)}{2}-\overline{\beta} .\] Thus \[\int_Y e^{-\widetilde{f}} \left| \nabla_{\widetilde{g}} \left(w^{-\overline{s}}\varphi\right) \right|_{\widetilde{g}}^2 d\mu_{\widetilde{g}} =\int_Y e^{-f} w^{2\overline{s}} \left| \nabla_g \left(w^{-\overline{s}}\varphi\right) \right|_g^2 d\mu_g .\] Expanding the integrand, \[w^{2\overline{s}} \left| \nabla\left(w^{-\overline{s}}\varphi\right) \right|^2 = |\nabla\varphi|^2 - \overline{s} w^{-1}\langle dw,\nabla(\varphi^2)\rangle + \overline{s}^2w^{-2}|dw|^2\varphi^2 .\] Integrating the middle term by parts with respect to \(e^{-f}d\mu_g\), we get \[\label{eq:qf-gradient-part} \begin{align} &\int_Y e^{-\widetilde{f}} \left| \nabla_{\widetilde{g}} \left(w^{-\overline{s}}\varphi\right) \right|_{\widetilde{g}}^2 d\mu_{\widetilde{g}}\\ =& \int_Y e^{-f}|\nabla\varphi|^2\,d\mu_g + \int_Y e^{-f} \left[ \overline{s} w^{-1}\Delta_g w - \overline{s} w^{-1}\langle dw,df\rangle_g + (\overline{s}^2-\overline{s})w^{-2}|dw|^2 \right]\varphi^2\,d\mu_g . \end{align}\tag{14}\]

For the zero-order part, note that \[\frac{\overline{\alpha} m}{2} - \overline{\beta} - 2\overline{s} = \overline{\alpha} .\] Therefore \[\frac{1}{2} \int_Y e^{-\widetilde{f}} \bigl( \mathscr S_a^{\widetilde{g}}(\widetilde{f})-\sigma\bigr) w^{-2\overline{s}}\varphi^2\,d\mu_{\widetilde{g}} = \frac{1}{2} \int_Y e^{-f} \left( w^{\overline{\alpha}} \mathscr S_a^{\widetilde{g}}(\widetilde{f}) - \sigma w^{\overline{\alpha}}\right)\varphi^2\,d\mu_g .\] Subtracting the original zero-order term gives \[\frac{1}{2} \int_Y e^{-f}\left( w^{\overline{\alpha}}\mathscr S_a^{\widetilde{g}}(\widetilde{f}) -\mathscr S_a^g(f)\right)\varphi^2\,d\mu_g -\frac{1}{2} \int_Y e^{-f}\sigma\bigl(w^{\overline{\alpha}}-1\bigr)\varphi^2\,d\mu_g .\] We now apply the general conformal formula 5 , in dimension \(m\), with \(a=a_{m+1}(\lambda)\).

The coefficient of \(w^{-1}\Delta_gw\) in the total quadratic form is \[\overline{s} + \frac{1}{2}\bigl(2\overline{\beta}-(m-1)\overline{\alpha}\bigr) = -\frac{1}{2} \left( \frac{\overline{\alpha} m}{2}-\overline{\beta} \right) = -\frac{\overline{\mathcal{K}}}{2}.\] The coefficient of \(w^{-1}\langle dw,df\rangle_g\) is \[-\overline{s} + \frac{1}{2}\bigl(\overline{\alpha}(m-2)-2a\overline{\beta}\bigr).\] Using \[\overline{\alpha}(m-2)=2(1+\overline{q}), \qquad \overline{\beta}=\frac{\overline{q}}{2a-1},\] this coefficient becomes \[-\frac{1}{2}(1+\overline{q}-\overline{\beta}) + (1+\overline{q})-a\overline{\beta} = \frac{1}{2} .\] Thus the first-order terms combine to \[-\frac{\overline{\mathcal{K}}}{2} w^{-1} \left( \Delta_g w - \frac{1}{\overline{\mathcal{K}}} \langle dw,df\rangle_g \right) = -\frac{\overline{\mathcal{K}}}{2} w^{-1}\overline{L}_{f,\lambda}^{(m)}w .\]

It remains to check the \(w^{-2}|dw|^2\)-coefficient. Since \[\overline{\alpha}(m-2)=2(1+\overline{q}), \qquad \overline{\beta}=\frac{\overline{q}}{2a-1}, \qquad \overline{s}=\frac{1}{2}(1+\overline{q}-\overline{\beta}),\] the coefficient is \[\begin{align} \mathcal{C} &= \overline{s}^2-\overline{s} + \frac{1}{2} \left[ (m-1)\overline{\alpha} -\frac{(m-1)(m-2)}{4}\overline{\alpha}^2 -2\overline{\beta} +\overline{\alpha}\overline{\beta}(m-2) -a\overline{\beta}^2 \right] \\ &= \frac{m}{4(m-2)}(1-\overline{q}^2) + \frac{\overline{q}^2}{4(2a-1)} . \end{align}\] Using \[2a-1=\frac{1-(m-1)\lambda}{1-(m+1)\lambda}, \qquad \overline{q}^2= \frac{m(1-(m-1)\lambda)}{2(1-\lambda)},\] we get \[1-\overline{q}^2 = -\frac{(m-2)(1-(m+1)\lambda)}{2(1-\lambda)}, \qquad \frac{\overline{q}^2}{2a-1} = \frac{m(1-(m+1)\lambda)}{2(1-\lambda)} .\] Therefore \(\mathcal{C}=0\). This proves ?? . ◻

Proof. Put \(a=a_m(\lambda)\) and \(u=\langle\nabla f,\nu\rangle\). By stability inequality 16 , it is enough to prove \[\label{eq:I-estimate-needed} I:= \operatorname{Ric}(\nu,\nu) +|A_\Sigma|^2 +D^2f(\nu,\nu) +\langle\nabla\Phi,\nu\rangle +\frac{1}{2}\mathscr S_{a_m(\lambda)}^{g_\Sigma}(F) \ge \frac{1}{2}\sigma_\Sigma .\tag{17}\]

The Gauss equation and the restriction formula for \(\Delta_\Sigma F\) give \[\label{eq:I-expanded} I= \frac{1}{2}(R_g+2\Delta_g f) -\frac{a}{2}|\nabla^\Sigma F|^2 +\frac{1}{2}|A_\Sigma|^2 +\frac{1}{2}H^2-Hu +\langle\nabla\Phi,\nu\rangle .\tag{18}\] Using \(R_g+2\Delta_g f-a|\nabla f|^2\ge\sigma\) and \(|\nabla f|^2=|\nabla^\Sigma F|^2+u^2\), we get \[\frac{1}{2}(R_g+2\Delta_g f) -\frac{a}{2}|\nabla^\Sigma F|^2 \ge \frac{\sigma}{2}+\frac{a}{2}u^2 .\] Since \(|A_\Sigma|^2\ge\frac{H^2}{m-1}\), it follows that \[\label{eq:I-lower-before-square} I\ge \frac{\sigma}{2}+\langle\nabla\Phi,\nu\rangle +\frac{a}{2}u^2 +\frac{m}{2(m-1)}H^2-Hu .\tag{19}\] Using \(H=u+\Phi\), the last three terms are \[\frac{a}{2}u^2+\frac{m}{2(m-1)}(u+\Phi)^2-(u+\Phi)u .\] Set \[B_{m,\lambda} := a_m(\lambda)-\frac{m-2}{m-1} = \frac{1-\lambda}{(m-1)(1-m\lambda)} >0 .\] Then \[\label{eq:lambda-square-completion-correct} \begin{align} &\frac{a}{2}u^2+\frac{m}{2(m-1)}(u+\Phi)^2-(u+\Phi)u \\ =& \frac{B_{m,\lambda}}{2} \left( u+\frac{\Phi}{(m-1)B_{m,\lambda}} \right)^2 +\frac{1}{2(1-\lambda)}\Phi^2 . \end{align}\tag{20}\] Therefore \[I \ge \frac{\sigma}{2} +\langle\nabla\Phi,\nu\rangle +\frac{1}{2(1-\lambda)}\Phi^2 = \frac{1}{2}\sigma_\Sigma .\] This proves the nonnegativity of \(Q_\Sigma\). ◻

Proof. Dividing \(\mathcal{L}_{m,\lambda,\sigma}^{g,f}\psi\ge0\) by \(\psi\) gives \[-\Delta_g h-|dh|_g^2+ \langle df,dh\rangle_g +\frac{1}{2}\bigl(\mathscr S_{a_m(\lambda)}^g(f)-\sigma\bigr) \ge0 .\] Using \(a_{m-1}(\lambda)=2-\frac{1}{a_m(\lambda)}\) and expanding \(f_\psi=f-h\) gives 22 . The last term in 22 is nonnegative, so 23 follows. ◻

Proof. On every compact \(K\Subset X\), the measures \(\mu_{\eta\omega}=\eta\mu_\omega\) and \(\mu_\omega\) are comparable. Hence the local Ahlfors estimates follow immediately.

For the Poincaré inequality, let \(B=B_r(x)\) and \(\lambda B=B_{\lambda_K r}(x)\) be a ball on which the AM–PI constants for \(d\mu_\omega\) are fixed. Denote averages with respect to \(d\mu_\omega\) and \(d\mu_{\eta\omega}\) by \(u_{B,\omega}\) and \(u_{B,\eta\omega}\). By the triangle inequality, \[\int_B |u-u_{B,\eta\omega}|\,d\mu_{\eta\omega} \le 2\int_B |u-u_{B,\omega}|\,d\mu_{\eta\omega} \le 2C_{\eta,K}\int_B |u-u_{B,\omega}|\,d\mu_\omega .\] Dividing by \(\mu_{\eta\omega}(B)\ge c_{\eta,K}\mu_\omega(B)\), applying the Poincaré inequality for \(d\mu_\omega\), and using comparability again on \(\lambda B\), we obtain \[\frac{1}{\mu_{\eta\omega}(B)} \int_B |u-u_{B,\eta\omega}|\,d\mu_{\eta\omega} \le Cr \left( \frac{1}{\mu_{\eta\omega}(\lambda B)} \int_{\lambda B\cap\mathcal{R}} |\nabla_g u|^2\,d\mu_{\eta\omega} \right)^{1/2}.\] Thus \(\eta\omega\) is admissible. ◻

Proof. Fix \(K\Subset U\). If \(\mathcal{S}\cap K=\varnothing\), there is nothing to prove. Choose \(\rho_0>0\) so that the \(2\rho_0\)-neighborhood of \(K\) is contained in \(U\) and the local Ahlfors and packing estimates apply there. Then \[\label{eq:packing-tube-volume-section3} \mu_\omega\bigl(\{x\in U:\operatorname{dist}(x,\mathcal{S}\cap K)<\rho\}\bigr) \le C\rho^2, \qquad 0<\rho<\rho_0 .\tag{28}\]

For \(0<\varepsilon<\rho_0/4\), set \[L_\varepsilon=\log(\rho_0/\varepsilon), \qquad r_\mathcal{S}(x)=\operatorname{dist}(x,\mathcal{S}\cap K),\] and define \[\chi_\varepsilon(x) = \begin{cases} 1, & r_\mathcal{S}(x)\le\varepsilon,\\ \frac{\log(\rho_0/r_\mathcal{S}(x))}{L_\varepsilon}, & \varepsilon<r_\mathcal{S}(x)<\rho_0,\\ 0, & r_\mathcal{S}(x)\ge\rho_0 . \end{cases}\] Then \(\chi_\varepsilon\in W^{1,2}_0(U)\), \(\chi_\varepsilon=1\) near \(\mathcal{S}\cap K\), and, on \(\mathcal{R}\), \[|\nabla\chi_\varepsilon| \le \frac{1}{L_\varepsilon r_\mathcal{S}} \qquad\text{a.e. on }\{\varepsilon<r_\mathcal{S}<\rho_0\}.\] Let \(\rho_j=2^{-j}\rho_0\) and \(J_\varepsilon = \left\lceil \log_2\frac{\rho_0}{\varepsilon}\right\rceil\). Then \[\rho_{J_\varepsilon}\le\varepsilon<\rho_{J_\varepsilon-1}, \qquad J_\varepsilon\le C L_\varepsilon .\] Using 28 on the dyadic annuli \[\rho_{j+1}<r_\mathcal{S}\le\rho_j, \qquad j=0,\ldots,J_\varepsilon-1,\] gives \[\int_U |\nabla\chi_\varepsilon|^2\,d\mu_\omega \le \frac{C}{L_\varepsilon^2} \sum_{j=0}^{J_\varepsilon-1} \rho_{j+1}^{-2} \mu_\omega\bigl(\{r_\mathcal{S}<\rho_j\}\bigr) \le \frac{C}{L_\varepsilon} \longrightarrow0 .\] ◻

Proof. Since \(\mathcal{S}\cap U\) has zero \(2\)-capacity, the AM–PI formulation agrees locally with the usual PI-space formulation. Hence the local theory on PI spaces applies; see, for example, Björn–Björn [13] and Heinonen–Koskela–Shanmugalingam-Tyson [14]. Choose \(r_U>0\) and \(\Lambda_U>10\) so that, whenever \[0<r<r_U, \qquad B_{\Lambda_U r}(z)\Subset U,\] every nonnegative \(L_\omega\)-harmonic function on \(B_{\Lambda_U r}(z)\) satisfies the Harnack inequality with constant \(H_U\) on \(B_r(z)\) and \(B_{2r}(z)\setminus \overline{B_r(z)}\).

We first treat \(\delta=0\). For \(B_{2r}(p)\Subset\Omega\) and \(0<r<\rho_U\), we have the local ball-capacity estimate \[\label{eq:ball-capacity-section3} C_U^{-1}r^{m-2} \le \operatorname{Cap}_\omega(\overline{B_r(p)},\Omega) \le C_Ur^{m-2}.\tag{30}\] Indeed, the upper bound follows from a Lipschitz cutoff in \(B_{2r}(p)\), while the lower bound follows from the local Sobolev inequality and the Ahlfors lower bound.

Let \(G=G_{\Omega,0}^\omega(p,\cdot)\). We take \(G\) to be the Dirichlet Green function, this means \[-L_\omega G=\boldsymbol{\delta}_p, \qquad G=0\text{ on }\partial\Omega\] weakly. Moreover, we have \[\label{eq:green-level-identity-section3} t\,\operatorname{Cap}_\omega(\{G\ge t\},\Omega)=1, \qquad t>0 .\tag{31}\] This is the \(p=2\) case of the superlevel-set identity in [15].

Let \(y\in B_R(x_0)\cap\mathcal{R}\), \(y\ne p\), and set \[\rho=d(p,y), \qquad A_\rho(p)=B_{2\rho}(p)\setminus\overline{B_{\rho/2}(p)} .\] Write \[M_\rho=\sup_{A_\rho(p)}G, \qquad m_\rho=\inf_{A_\rho(p)}G .\] The annular Harnack inequality gives \[M_\rho\le H_Um_\rho .\] The maximum principle on \(\Omega\setminus B_{2\rho}(p)\) gives \[\{G>M_\rho\}\subset B_{2\rho}(p).\] Using 31 , monotonicity of capacity, and 30 , \[\frac{1}{M_\rho} = \operatorname{Cap}_\omega(\{G\ge M_\rho\},\Omega) \le \operatorname{Cap}_\omega(B_{2\rho}(p),\Omega) \le C_U\rho^{m-2}.\] Thus \(M_\rho\ge c_U\rho^{2-m}\), and Harnack gives \[G(p,y)\ge m_\rho\ge H_U^{-1}M_\rho\ge c_U\rho^{2-m}.\]

For the upper bound, the minimum principle on \(B_{\rho/2}(p)\setminus\{p\}\) gives \[B_{\rho/2}(p)\subset\{G\ge m_\rho\}.\] Hence \[\frac{1}{m_\rho} = \operatorname{Cap}_\omega(\{G\ge m_\rho\},\Omega) \ge \operatorname{Cap}_\omega(B_{\rho/2}(p),\Omega) \ge c_U\rho^{m-2}.\] Therefore \(m_\rho\le C_U\rho^{2-m}\), and \[G(p,y)\le M_\rho\le H_Um_\rho\le C_U\rho^{2-m}.\] This proves 29 for \(\delta=0\).

Now assume \(\delta>0\). By the maximum principle, \(0<G_{\Omega,\delta}^\omega(p,y)\le G_{\Omega,0}^\omega(p,y)\), so the upper bound follows from the case \(\delta=0\).

For the lower bound, use the resolvent identity \[G_{\Omega,\delta}^\omega(p,y) = G_{\Omega,0}^\omega(p,y) - \delta \int_\Omega G_{\Omega,0}^\omega(p,z) G_{\Omega,\delta}^\omega(z,y)\,d\mu_\omega(z).\] Since \(G_{\Omega,\delta}^\omega\le G_{\Omega,0}^\omega\), \[G_{\Omega,\delta}^\omega(p,y) \ge G_{\Omega,0}^\omega(p,y) - \delta \int_\Omega G_{\Omega,0}^\omega(p,z) G_{\Omega,0}^\omega(z,y)\,d\mu_\omega(z).\]

Decompose \[\Omega=E_p\cup E_y, \qquad E_p=\{z\in\Omega:\;d(p,z)\le d(y,z)\}, \qquad E_y=\Omega\setminus E_p .\] On \(E_p\), the triangle inequality gives \(d(y,z)\ge \frac{\rho}{2}\) and on \(E_y\) it gives \(d(p,z)\ge \frac{\rho}{2}\). Hence \[\begin{align} &\int_\Omega d(p,z)^{2-m}d(z,y)^{2-m}\,d\mu_\omega(z) \\ &\qquad\le C\rho^{2-m} \left( \int_\Omega d(p,z)^{2-m}\,d\mu_\omega(z) + \int_\Omega d(y,z)^{2-m}\,d\mu_\omega(z) \right). \end{align}\] Note that if \(q\in B_R(x_0)\), then \[\Omega\subset B_{(\Lambda_U+1)R}(q).\] By the Ahlfors upper bound and a dyadic decomposition around \(q\), \[\begin{align} \int_\Omega d(q,z)^{2-m}\,d\mu_\omega(z) &\le \sum_{j=0}^\infty (2^{-j-1}(\Lambda_U+1)R)^{2-m} \mu_\omega\bigl(B_{2^{-j}(\Lambda_U+1)R}(q)\bigr) \\ &\le C_U \sum_{j=0}^\infty (2^{-j}R)^2 \le C_U R^2 . \end{align}\] Applying this with \(q=p\) and \(q=y\) gives \[\int_\Omega d(p,z)^{2-m}d(z,y)^{2-m}\,d\mu_\omega(z) \le C_U R^2 d(p,y)^{2-m}.\]

Hence the \(\delta=0\) case gives \[G_{\Omega,\delta}^\omega(p,y) \ge C_U^{-1}d(p,y)^{2-m} - C_U\delta R^2d(p,y)^{2-m}.\] After decreasing \(\rho_U\) so that \(C_U\delta\rho_U^2\le\frac{1}{2}C_U^{-1}\) and using \(R<\rho_U\), we obtain \(G_{\Omega,\delta}^\omega(p,y)\ge c_Ud(p,y)^{2-m}\). This proves the proposition. ◻

Proof. Choose an exhaustion by precompact open sets \[V_1\Subset V_2\Subset\cdots, \qquad \bigcup_{\ell=1}^{\infty}V_\ell=X,\] with \(V_\ell\cap\mathcal{R}\) connected. Let \(G_{\ell,p}\) be the Dirichlet Green function for \(P_{\omega,\delta}\) on \(V_\ell\), with pole \(p\in V_\ell\). By maximum principle \(G_{\ell,p}\le G_{\ell+1,p}\).

Let \(h_\ell\in W^{1,2}_0(V_\ell)\) solve \[P_{\omega,\delta}h_\ell=1 \qquad\text{in }V_\ell .\] The maximum principle gives \(0\le h_\ell\le\delta^{-1}\). By the Green representation formula, \[h_\ell(p) = \int_{V_\ell}G_{\ell,p}(y)\,d\mu_\omega(y),\] we know \[\label{eq:global-green-l1-bound-section3} \int_{V_\ell}G_{\ell,p}(y)\,d\mu_\omega(y) \le \delta^{-1}.\tag{32}\] This \(L^1\)-bound, together with Harnack inequalities, gives local uniform bounds for \(G_{\ell,p}\) away from the pole. Hence the limit \[G_\delta^\omega(p,y) := \lim_{\ell\to\infty}G_{\ell,p}(y)\] is finite.

Fix \(K\Subset X\). Choose \(r_K>0\) so small that Lemma 5 applies on \(B_{4r_K}(p)\) for every \(p\in K\).

By the maximum principle, \[G_{\ell,p}(y)\ge G_{B_{2r_K}(p),\delta}^\omega(p,y) \ge C_K^{-1}d(p,y)^{2-m}, \qquad 0<d(p,y)<r_K .\] Letting \(\ell\to\infty\) gives the lower bound in ?? .

Next, choose precompact domains \(W_0, W\) such that \[K\Subset W_0\Subset W\Subset X.\] Then covering \(\partial W_0\) by finitely many balls contained in \(W\setminus K\) and local uniform bounds on those balls gives \[G_{\ell,p}\le C_{K,W} \qquad \text{on }\partial W_0,\quad p\in K .\] Since \(G_{\ell,p}\) is \(P_{\omega,\delta}\)-harmonic on \(V_\ell\setminus W_0\) and vanishes on \(\partial V_\ell\), the maximum principle gives \[G_{\ell,p}\le C_{K,W} \qquad \text{on }V_\ell\setminus W_0 .\] Passing to the limit gives the exterior bound ?? .

Similarly, on the annuli \[B_{3r_K}(p)\setminus \overline{B_{r_K}(p)}, \qquad p\in K,\] we have \[G_{\ell,p}\le C_K \qquad \text{on }\partial B_{2r_K}(p).\] Inside \(B_{2r_K}(p)\), write \[G_{\ell,p} = G_{B_{2r_K}(p),\delta}^\omega(p,\cdot)+H_{\ell,p},\] where \(H_{\ell,p}\ge0\) is \(P_{\omega,\delta}\)-harmonic. On \(\partial B_{2r_K}(p)\), \(H_{\ell,p}=G_{\ell,p}\le C_K\), so the maximum principle gives \[H_{\ell,p}\le C_K \qquad \text{on }B_{2r_K}(p).\] On the other hand, we have \[G_{B_{2r_K}(p),\delta}^\omega(p,y) \le C_Kd(p,y)^{2-m}.\] Therefore, for \(0<d(p,y)<r_K\), \[G_{\ell,p}(y) \le C_Kd(p,y)^{2-m}+C_K \le C_Kd(p,y)^{2-m},\] after increasing \(C_K\). Letting \(\ell\to\infty\) proves the upper bound in@eq:eq:global-green-local-asymp-section3 . ◻

Proof. Choose \(d_*<m-2-\tau\) such that \(\mathsf P_{d_*}(\mathcal{S})\) holds, and set \[\theta=m-2-\tau-d_*>0 .\] Since \(d_*<m-2\), Lemma 4 gives locally zero weighted \(2\)-capacity of \(\mathcal{S}\). Hence Proposition 5 applies.

Choose compact sets \(Q_i\Subset X\) with locally finite overlap and \[X=\bigcup_i Q_i .\] Choose precompact open sets \(U_i\Subset X\) and compact sets \(K_i\Subset X\) such that \[Q_i\Subset U_i\Subset K_i^\circ,\] and such that the family \(\{K_i\}\) is locally finite. Let \[r_i>0,\qquad C_i\ge1\] be the constants from Proposition 5 applied to \(K_i\). After increasing \(C_i\), we may also use the off-diagonal bound \[G_\delta^\omega(p,y)\le C_i\] whenever \[p,y\in K_i,\qquad d(p,y)\ge \frac{1}{4} r_i .\]

Choose \(s_{i,0}>0\) small enough such that \[80 s_{i,0}<r_i, \qquad B_{20 s_{i,0}}(z)\subset K_i \quad\text{for all }z\in\mathcal{S}\cap\overline{U}_i .\] Choose \(\rho_i\in(0,s_{i,0})\) such that \[\operatorname{dist}(x,\mathcal{S}) = \operatorname{dist}\bigl(x,\mathcal{S}\cap\overline{U}_i\bigr)\] whenever \[x\in Q_i, \qquad 0<\operatorname{dist}(x,\mathcal{S})<\rho_i .\]

Set \(s_{i,j}=2^{-j}s_{i,0}\). For each \(j\), choose a maximal \(s_{i,j}\)-separated set \[P_{i,j}\subset\mathcal{S}\cap\overline{U}_i .\] For \(z\in P_{i,j}\), choose \[\eta_{i,j,z}\in C^\infty(\mathcal{R}), \qquad 0\le\eta_{i,j,z}\le1,\] such that \[\eta_{i,j,z}=1 \quad\text{on }B_{10 s_{i,j}}(z)\cap\mathcal{R}, \qquad \operatorname{spt}\eta_{i,j,z} \subset B_{20 s_{i,j}}(z)\cap\mathcal{R} .\] Fix \(\delta>0\). Define \[u_{i,j,z}(x) = \int_{\mathcal{R}} G_\delta^\omega(y,x)\eta_{i,j,z}(y)\,d\mu_\omega(y).\] Then \(P_{\omega,\delta}u_{i,j,z}=\eta_{i,j,z}\).

We first record the estimates for \(u_{i,j,z}\). Write \[s=s_{i,j},\qquad u=u_{i,j,z},\qquad r=d(x,z),\] where \(x\in K_i\cap\mathcal{R}\). If \(r\le 40 s\), then \[\operatorname{spt}\eta_{i,j,z}\subset B_{60 s}(x), \qquad 60 s<r_i .\] By ?? and Ahlfors regularity, \[\begin{align} u(x) &\le C_i\int_{B_{60 s}(x)} d(x,y)^{2-m}\,d\mu_\omega(y) \\ &\le C_i\sum_{\ell=0}^{\infty} (2^{-\ell}s)^{2-m} \mu_\omega(B_{2^{-\ell}60 s}(x)) \le C_i s^2 . \end{align}\] Since \(s\le r+s\le 41s\), this gives \[u(x)\le C_i s^m(r+s)^{2-m}.\]

If \(40 s\le r<\frac{1}{2}r_i\), then for \(y\in\operatorname{spt}\eta_{i,j,z}\), \[\frac{1}{2}r\le d(x,y)\le\frac{3}{2}r<r_i .\] Hence \[u(x) \le C_i r^{2-m}\mu_\omega(B_{20 s}(z)) \le C_i s^m r^{2-m} \le C_i s^m(r+s)^{2-m}.\]

If \(r\ge r_i/2\), then \[d(x,y)\ge r-20 s\ge \frac{1}{4}r_i \qquad \text{for }y\in\operatorname{spt}\eta_{i,j,z}.\] The off-diagonal bound gives \[u(x) \le C_i\mu_\omega(B_{20 s}(z)) \le C_i s^m .\] Since \(x,z\in K_i\), the quantity \(r+s\) is uniformly bounded above in this case. After increasing \(C_i\), we again get \[u(x)\le C_i s^m(r+s)^{2-m}.\] Therefore, for all \(x\in K_i\cap\mathcal{R}\), \[\label{eq:global-source-upper-section3} u_{i,j,z}(x) \le C_i s_{i,j}^m \bigl(d(x,z)+s_{i,j}\bigr)^{2-m}.\tag{33}\]

If \(d(x,z)\le 5s\), then \(B_{s/4}(x)\subset B_{10 s}(z)\). Hence \(\eta_{i,j,z}=1\) on \(B_{s/4}(x)\cap\mathcal{R}\). The lower bound in ?? gives \[u(x)\ge c_i\int_{B_{s/4}(x)} d(x,y)^{2-m}\,d\mu_\omega(y) \ge c_i s^{2-m} \mu_\omega\bigl(B_{s/4}(x)\bigr) \ge c_i s^2 .\] Together with 33 , this gives \[\label{eq:global-source-inner-section3} c_i s_{i,j}^2 \le u_{i,j,z}(x) \le C_i s_{i,j}^2 \qquad \text{if }d(x,z)\le 5 s_{i,j} .\tag{34}\]

Now define the \(i\)-th block potential by \[\Psi_i = \sum_j\sum_{z\in P_{i,j}} s_{i,j}^{-\tau-2}u_{i,j,z}.\] Since \(P_{\omega,\delta}u_{i,j,z}=\eta_{i,j,z}\), we have \[\label{eq:block-source-equation-section3} P_{\omega,\delta}\Psi_i = \sum_j\sum_{z\in P_{i,j}} s_{i,j}^{-\tau-2}\eta_{i,j,z} \ge0 .\tag{35}\]

We estimate \(\Psi_i\). For \(x\in K_i\cap\mathcal{R}\), denote \[r=\operatorname{dist}\bigl(x,\mathcal{S}\cap\overline{U}_i\bigr)\] from now on. For a fixed scale \(s=s_{i,j}\), write \[\Psi_{i,s}(x) = \sum_{z\in P_{i,j}}s^{-\tau-2}u_{i,j,z}(x).\] Using 33 , \[\Psi_{i,s}(x) \le C_i\sum_{z\in P_{i,j}} s^{m-2-\tau}\bigl(d(x,z)+s\bigr)^{2-m}.\]

If \(s\le r\), decompose the centers into dyadic annuli around \(x\). The packing condition gives \[\sum_{z\in P_{i,j}\cap (B_{2^{\ell+1}r}(x)\setminus B_{2^\ell r}(x))} s^{m-2-\tau}\bigl(d(x,z)+s\bigr)^{2-m} \le C_i2^{-\ell(\tau+\theta)} s^\theta r^{-\tau-\theta}.\] Summing in \(\ell\) yields \[\label{eq:fixed-scale-small-global-section3} \Psi_{i,s}(x) \le C_i s^\theta r^{-\tau-\theta}, \qquad s\le r.\tag{36}\] If \(s\ge r\), the same estimate with \(s\) in place of \(r\) gives \[\label{eq:fixed-scale-large-global-section3} \Psi_{i,s}(x) \le C_i s^{-\tau}, \qquad s\ge r.\tag{37}\] Summing over \(j\), we obtain \[\label{eq:block-upper-section3} \Psi_i(x)\le C_i r^{-\tau}\tag{38}\] whenever \(0<r<\rho_i\).

Now suppose \[x\in Q_i\cap\mathcal{R}, \qquad 0<\operatorname{dist}(x,\mathcal{S})<\rho_i .\] Then \[r=\operatorname{dist}(x,\mathcal{S}) = \operatorname{dist}\bigl(x,\mathcal{S}\cap\overline{U}_i\bigr).\] Choose \(j\) with \(s_{i,j+1}\le r<s_{i,j}\). Pick \(z_0\in\mathcal{S}\cap\overline{U}_i\) with \(d(x,z_0)\le2r\). By the maximality of \(P_{i,j}\), there is a point \(p_0\in P_{i,j}\) such that \(d(z_0,p_0)\le s_{i,j}\). So \[d(x,p_0)\le2r+s_{i,j}<3s_{i,j}< 5s_{i,j} .\] Using 34 , \[\Psi_i(x) \ge s_{i,j}^{-\tau-2}u_{i,j,p_0}(x) \ge c_i s_{i,j}^{-\tau} \ge c_i r^{-\tau}.\] Together with 38 , this proves \[\label{eq:block-size-section3} c_i r^{-\tau} \le \Psi_i(x) \le C_i r^{-\tau}.\tag{39}\]

On the other hand, 35 gives \[P_{\omega,\delta}\Psi_i(x) \ge s_{i,j}^{-\tau-2} \ge c_i r^{-\tau-2}.\] Using 38 , \[r^{-\tau-2} = \bigl(r^{-\tau}\bigr)^{1+2/\tau} \ge C_i^{-1}\Psi_i^{1+2/\tau}.\] Thus \[\label{eq:block-P-diff-near-section3} P_{\omega,\delta}\Psi_i \ge A_i\Psi_i^{1+2/\tau}\tag{40}\] on \[Q_i\cap\mathcal{R}\cap \{0<\operatorname{dist}(\cdot,\mathcal{S})<\rho_i\}.\]

The estimate 38 , the off-diagonal bound for \(G_\delta^\omega\), and interior elliptic regularity imply that \(\Psi_i\in C^\infty(\mathcal{R})\). Moreover, \(\Psi_i\) is bounded on \[\mathcal{R}\cap \{ \operatorname{dist}(\cdot,\mathcal{S}\cap\overline{U}_i)\ge\rho_i\}.\] Together with \(P_{\omega,\delta}\Psi_i\ge0\), this implies, after increasing \(B_i\), that \[\label{eq:block-P-diff-section3} P_{\omega,\delta}\Psi_i \ge A_i\Psi_i^{1+2/\tau}-B_i \qquad\text{on }\mathcal{R}.\tag{41}\]

Choose positive coefficients \(\varepsilon_i\) decreasing sufficiently fast that \[E_A:=\sum_i\varepsilon_iA_i^{-\tau/2}<\infty, \qquad E_B:=\sum_i\varepsilon_iB_i<\infty,\] and so that \(\Theta_0=\sum_i\varepsilon_i\Psi_i\) converges in \(C^\infty_{\rm loc}(\mathcal{R})\). Since \(\alpha=2/\tau\), Hölder’s inequality gives \[\Theta_0^{1+\alpha} \le E_A^\alpha \sum_i\varepsilon_iA_i\Psi_i^{1+\alpha}.\] Therefore \[P_{\omega,\delta}\Theta_0 \ge E_A^{-\alpha}\Theta_0^{1+\alpha}-E_B .\] Since \[P_{\omega,\delta}\Theta_0=-L_\omega\Theta_0+\delta\Theta_0,\] we get \[-L_\omega\Theta_0 \ge E_A^{-\alpha}\Theta_0^{1+\alpha} -E_B-\delta\Theta_0 .\] Absorbing the linear term into the superlinear term gives \[-L_\omega\Theta_0 \ge A\Theta_0^{1+\alpha}-B\] for some \(A>0\) and \(B\ge0\). Set \(\Theta=\Theta_0\). This proves ?? .

Finally, let \(K\Subset X\). Only finitely many \(Q_i\) meet \(K\). Let \(I_K\) be this finite set and put \[r_K=\min_{i\in I_K}\rho_i, \qquad c_K=\min_{i\in I_K}\varepsilon_i c_i .\] If \[x\in K\cap\mathcal{R}, \qquad 0<\operatorname{dist}(x,\mathcal{S})<r_K,\] choose \(i\in I_K\) with \(x\in Q_i\). Then \[\Theta(x) \ge \varepsilon_i\Psi_i(x) \ge c_K\operatorname{dist}(x,\mathcal{S})^{-\tau}.\] This proves ?? . ◻

Proof. Set \(z=\frac{\varepsilon T}{1+\varepsilon T}\). Then \(0\le z<1\), \(w=(1-z)^{-1}\), and \[\varepsilon T^{1+\alpha}w^{-1} = \varepsilon^{-\alpha}z^{1+\alpha}w^\alpha .\] Also \[w^\alpha-1 = w^\alpha\bigl(1-w^{-\alpha}\bigr) = w^\alpha\bigl(1-(1-z)^\alpha\bigr).\] Hence \[\begin{align} &\mathfrak K A\varepsilon T^{1+\alpha}w^{-1} -\frac{1}{2}\sigma(w^\alpha-1) \\ \ge& w^\alpha \left[ \mathfrak K A\varepsilon^{-\alpha}z^{1+\alpha} - \frac{1}{2}\sigma_+ \bigl(1-(1-z)^\alpha\bigr) \right]. \end{align}\] Using \[1-(1-z)^\alpha\le C_\alpha z, \qquad 0\le z<1,\] and \[\sup_{z\ge0}\{az-Mz^{1+\alpha}\} \le C_\alpha a^{1+1/\alpha}M^{-1/\alpha},\] with \(a=\frac{1}{2}C_\alpha\sigma_+\) and \(M=\mathfrak K A\varepsilon^{-\alpha}\), we get \[\mathfrak K A\varepsilon T^{1+\alpha}w^{-1} - \frac{1}{2}\sigma(w^\alpha-1) \ge -C\varepsilon w^\alpha .\] Since \(w\ge1\), \[\mathfrak K B\varepsilon w^{-1} \le \mathfrak K B\varepsilon w^\alpha .\] Therefore the left hand side of 42 is bounded below by \[\left( \frac{\zeta}{2} -C\varepsilon -\mathfrak K B\varepsilon \right)w^\alpha .\] Choosing \(\varepsilon_0>0\) sufficiently small proves the lemma. ◻

Proof. Let \[\rho(x)=\operatorname{dist}(x,\mathcal{S}).\] On each compact region near \(\mathcal{S}\), \[w^\alpha\ge(\varepsilon c)^\alpha\rho^{-2}, \qquad ds_{\widetilde{g}}\ge(\varepsilon c)^{\alpha/2}\rho^{-1}ds_g .\] Let \(\gamma:[s_0,S)\to\mathcal{R}\) be a curve approaching \(\mathcal{S}\), parametrized by \(g\)-arclength on a final finite-lengthsegment. Since \(\rho(\gamma(s))\to0\) as \(s\to S\), and since \(\rho\) is \(1\)-Lipschitz, \[\rho(\gamma(s))\le S-s .\] Therefore \[L_{\widetilde{g}}(\gamma) \ge (\varepsilon c)^{\alpha/2} \int_{s_0}^{S}\frac{ds}{\rho(\gamma(s))} \ge (\varepsilon c)^{\alpha/2} \int_{s_0}^{S}\frac{ds}{S-s} = \infty .\] ◻

Proof. Put \(\alpha=\alpha_m(\lambda)\), \(\beta=\beta_m(\lambda)\), and \(\tau=\tau_m(\lambda)=2/\alpha\). By Proposition 6, there are \(\Theta\in C^\infty_{\rm loc}(\mathcal{R})\), \(A>0\), and \(B\ge0\) such that \[\label{eq:pointwise-blowup-theta-section3} -L_\omega\Theta\ge A\Theta^{1+\alpha}-B, \qquad \Theta(x)\ge c_K\operatorname{dist}(x,\mathcal{S})^{-2/\alpha}\tag{45}\] for \(x\in K\cap\mathcal{R}\) sufficiently close to \(\mathcal{S}\), for every \(K\Subset X\).

For \(\varepsilon>0\), set \[w=1+\varepsilon\Theta, \qquad \widetilde{g}=w^\alpha g, \qquad \widetilde{f}=f+\beta\log w .\] The pointwise conformal formula gives \[w^\alpha \mathcal{S}_{\widetilde{f}}^{m,\lambda}(\widetilde{g}) = \mathcal{S}_f^{m,\lambda}(g) - \mathcal{K}_m(\lambda)w^{-1}L_{f,\lambda}^{(m)}w .\] Since \[L_{f,\lambda}^{(m)}=L_\omega, \qquad L_\omega w=\varepsilon L_\omega\Theta,\] we get from 45 \[\mathcal{S}_{\widetilde{f}}^{m,\lambda}(\widetilde{g}) \ge \sigma w^{-\alpha} + \mathcal{K}_m(\lambda)A\varepsilon\Theta^{1+\alpha}w^{-1-\alpha} - \mathcal{K}_m(\lambda)B\varepsilon .\] As in the proof of Lemma 6, we have \[\sigma_+\bigl(1-w^{-\alpha}\bigr) - \mathcal{K}_m(\lambda)A \varepsilon\Theta^{1+\alpha}w^{-1-\alpha} \le C\varepsilon .\] Therefore \(\mathcal{S}_{\widetilde{f}}^{m,\lambda}(\widetilde{g})\ge \sigma-C\varepsilon-\mathcal{K}_m(\lambda)B\varepsilon\). Choosing \(\varepsilon>0\) sufficiently small gives \(\mathcal{S}_{\widetilde{f}}^{m,\lambda}(\widetilde{g})\ge\sigma-\zeta\). Completeness follows from the growth estimate in 45 and Lemma 7. ◻

Proof. Put \[\alpha=\overline{\alpha}_m(\lambda),\qquad \beta=\overline{\beta}_m(\lambda),\qquad s=\overline{s}_m(\lambda), \qquad \tau=\overline{\tau}_m(\lambda)=2/\alpha .\] By Proposition 6, there are \(\Theta\in C^\infty_{\rm loc}(\mathcal{R})\), \(A>0\), and \(B\ge0\) such that \[\label{eq:qf-blowup-theta-section3} -L_\omega\Theta\ge A\Theta^{1+\alpha}-B, \qquad \Theta(x)\ge c_K\operatorname{dist}(x,\mathcal{S})^{-2/\alpha}\tag{46}\] for \(x\in K\cap\mathcal{R}\) sufficiently close to \(\mathcal{S}\), for every \(K\Subset X\).

For \(\varepsilon>0\), set \[w=1+\varepsilon\Theta, \qquad \widetilde{g}=w^\alpha g, \qquad \widetilde{f}=f+\beta\log w .\] For \(\varphi\in C_c^\infty(\mathcal{R})\), the quadratic-form conformal formula gives \[\begin{align} Q_{a_{m+1}(\lambda),\sigma}^{\widetilde{g},\widetilde{f}} (w^{-s}\varphi) = & Q_{a_{m+1}(\lambda),\sigma}^{g,f}(\varphi) - \frac{\overline{\mathcal{K}}_m(\lambda)}{2} \int_{\mathcal{R}} e^{-f}w^{-1} \overline{L}_{f,\lambda}^{(m)}w\,\varphi^2\,d\mu_g \\ & -\frac{1}{2} \int_{\mathcal{R}} e^{-f}\sigma(w^\alpha-1)\varphi^2\,d\mu_g . \end{align}\] Using \[Q_{a_{m+1}(\lambda),\sigma}^{g,f}\ge0, \qquad \overline{L}_{f,\lambda}^{(m)}=L_\omega, \qquad L_\omega w=\varepsilon L_\omega\Theta,\] and 46 , we obtain \[\begin{align} &Q_{a_{m+1}(\lambda),\sigma-\zeta}^{\widetilde{g},\widetilde{f}} (w^{-s}\varphi) \\ \ge & \int_{\mathcal{R}} e^{-f} \left[ \frac{\overline{\mathcal{K}}_m(\lambda)}{2} A\varepsilon\Theta^{1+\alpha}w^{-1} - \frac{\overline{\mathcal{K}}_m(\lambda)}{2} B\varepsilon w^{-1} - \frac{1}{2}\sigma(w^\alpha-1) + \frac{\zeta}{2}w^\alpha \right]\varphi^2\,d\mu_g . \end{align}\] Applying Lemma 6 with \(\mathfrak K=\frac{\overline{\mathcal{K}}_m(\lambda)}{2}\) shows that the bracket is nonnegative for \(\varepsilon>0\) sufficiently small. Hence \[Q_{a_{m+1}(\lambda),\sigma-\zeta}^{\widetilde{g},\widetilde{f}} (w^{-s}\varphi)\ge0 \qquad \text{for all }\varphi\in C_c^\infty(\mathcal{R}).\] Since multiplication by \(w^{-s}\) is a bijection on \(C_c^\infty(\mathcal{R})\), the required quadratic-form inequality follows. Completeness follows from 46 and Lemma 7. ◻

Proof. This is the standard slicing theorem for Hausdorff dimension; see Mattila [16]. ◻

Proof. Suppose \(q\in D_m\) and \(q_i\in N_W\) with \(q_i\to q\). For each \(i\), there are \(p_i\in\widetilde{X}_{m-1}^\circ\) and a curve \(\gamma_i\subset X_m^\circ\) from \(q_i\) to \(p_i\) with \(L_{g_m}(\gamma_i)\le W\). We claim \(p_i\to q\) as \(i\to \infty\). Once this is proved, then we have \[q\in \overline{\widetilde{X}_{m-1}^\circ} \cap D_m\subset\widetilde{D}_{m-1}.\] To prove the claim, we argue by contradiction. Suppose that \(p_i\to p\neq q\). Let us parameterize the curve \[\gamma_i:[0,s_i]\to X_m^\circ\] to have \(g\)-unit speed. For \(i\) large enough, we have \(s_i\geq \frac{1}{2}d_g(p,q)\). Recall that there are positive constants \(c_q,r_q>0\) such that \[g_m\geq c_q d_g(\cdot,q)^{-2}g\qquad\text{on}\qquad \{x\in X_m^\circ:d_g(x,q)<r_q\}.\]

Without loss of generality, we may assume \(r_q<d_g(p,q)/2\). Then for \(i\) large enough we can compute \[L_{g_m}(\gamma_i)\geq \int_0^{r_q-d_g(q_i,q)} |\gamma_i'(s)|_{g_m}\,\mathrm ds \geq \sqrt {c_q}\log\frac{r_q}{2d_g(q_i,q)}>W.\] This gives the desired contradiction, and so the claim is proved.

To sum up, we have shown \(\overline{N_W}\cap D_m\subset\widetilde{D}_{m-1}\). The opposite inclusion relation is clear from Corollary 1. ◻

Proof. For \(E\in\mathcal{C}_j:=\{E\in\mathcal{C}: E\Delta E_0\Subset\Omega_j\setminus(\Gamma_j^+\cup\Gamma_j^-)\}\), define \[\mathcal{A}_{f_m}(E;\Omega_j) = \int_{\partial^*E\cap\Omega_j^\circ}e^{-f_m}\,d\mathcal{H}^{m-1}_{g_m}, \qquad \mathcal{V}_{\Phi,f_m}(E;\Omega_j) = \int_{\Omega_j}(\chi_E-\chi_{E_0})\Phi e^{-f_m}\,d\mathcal{H}^m_{g_m}.\] We minimize \[\mathcal{F}_j(E)=\mathcal{A}_{f_m}(E;\Omega_j) - \mathcal{V}_{\Phi,f_m}(E;\Omega_j).\] It follows from [17] that there is a minimizer \(E_j\in \mathcal{C}_j\) of the functional \(\mathcal{F}_j\). By comparison, we know that for any \(k\geq 1\) there is a positive constant \(c_k\) such that \[P(E_j;\Omega_l)\leq c_k.\] As a consequence, \(E_j\) converge to a locally finite perimeter set \(E\).

Let us show \(E\in \mathcal{C}\). It suffices to show that for any point \(p\in \partial_\pm \Omega\) there is a neighborhood such that \(\partial E_j\) does not enter for all \(j\) large. To see this, we take a small geodesic ball \(B_p\) of \(p\) in \(\partial\Omega\) and a smooth interior positive function \(h_p\) on \(B_p\) with \(h_p=0\) on \(\partial B_p\). Clearly, the \((th_p)\)-graphs \(L_t\) over \(B_p\) for \(t\) small form a smooth family of hypersurfaces satisfying \[H_{L_t}-\langle \nabla f,\nu_{L_t}\rangle<\Phi if p\in \partial_-\Omega\] and \[H_{L_t}-\langle \nabla f,\nu_{L_t}\rangle>\Phi if p\in \partial_+\Omega.\] From a barrier argument we conclude that for \(j\) large enough \(\partial E_j\) cannot intersect \(L_t\) and so avoid a neighborhood of \(p\).

To sum up, we already construct a locally finite perimeter set \(E\in \mathcal{C}\) such that \(E\) is a local minimizer of the functional \[\mathcal{F}(E):=\int_{\partial^*E}e^{-f_m}\,\mathrm d\mathcal{H}^{m-1}_{g_m}-\int_{\Omega}(\chi_E-\chi_{E_0})\Phi e^{-f_m}\,\mathrm d\mathcal{H}^m_{g_m}.\] Then the first and second variation give the desired mean curvature equation and the weighted stability inequality on \(\Sigma_{reg}\). The conclusion that \((\Sigma,d_{g_m}|_\Sigma,\mathcal{H}^{m-1}_{g_m})\) is AM–PI follows from the argument of Bombieri–Giusti [18]. Finally, the quantitative estimates of Naber–Valtorta [19] give the corresponding tubular-neighborhood volume bound for \(D\), and in the relative case we use Edelen [20]. Since \(\Sigma\) is locally Ahlfors \((m-1)\)-regular, this tubular estimate is equivalent to \(\mathsf P_{m-8}^{\Sigma}(D)\). The last statement follows from Proposition 4 and the estimate for \(\Phi\). ◻

Proof. We argue by downward induction on \(m\). For \(m=n\), take \[(X_n,X_n^\circ,D_n,g_n,f_n,\sigma_n) = (X,X,\emptyset,g,f,\sigma).\] The degree condition is the assumed nonzero locally finite relative degree of \[F:X\to Y\times I^k=B_n .\]

Assume \(m>m_0\), put \[j=m-r, \qquad B_m=Y\times I^j,\] and suppose that an \(m\)-dimensional datum controlled over \(B_m\) has been constructed with \[\sigma_m \ge \sigma - 4\pi^2(1-\lambda) \sum_{i=j+1}^{k}d_i^{-2} - \frac{\varepsilon}{4}.\] Since \(m>m_0\), one has \(j\ge1\).

Choose \(a\in(-1,1)\), with \(|a|\) later taken arbitrarily small, so that Corollary 1 holds and the sliced locally finite current is defined. Set \[\widetilde{X}_{m-1}=X_m\cap\tau_j^{-1}(a), \qquad \widetilde{X}_{m-1}^\circ=X_m^\circ\cap\tau_j^{-1}(a), \qquad \widetilde{D}_{m-1}=D_m\cap\tau_j^{-1}(a).\] The target slice \(\{t_j=a\}\subset Y\times I^j\) is identified with \[B_{m-1}=Y\times I^{j-1}.\] Since \(F_m=(F_{m-1},\tau_j)\), current slicing in the last interval coordinate shows that \[F_{m-1}^{\widetilde{X}_{m-1}^\circ}:(\tilde{X}_{m-1}^\circ,\partial\widetilde{X}_{m-1}^\circ)\to (B_{m-1},\partial B_{m-1})\] is proper over \(B_{m-1}\setminus F_{m-1}(\widetilde{D}_{m-1})\) and has nonzero relative degree over \((B_{m-1}\setminus F_{m-1}(\widetilde{D}_{m-1}),\partial B_{m-1}\setminus F_{m-1}(\widetilde{D}_{m-1}))\). The properness simply follows from \[(F_{m-1}^{\widetilde{X}_{m-1}^\circ})^{-1}(K)=F_{m-1}^{-1}(K)\cap \widetilde{X}_{m-1}\subset \widetilde{X}_{m-1}^\circ\] for any compact subset \(K\subset B_{m-1}\setminus F_{m-1}(\widetilde{D}_{m-1})\).

If \(\widetilde{X}_{m-1}^\circ\) is disconnected, we choose a connected component with nonzero relative degree, still denote it by \(\widetilde{X}_{m-1}^\circ\), replace \(\widetilde{X}_{m-1}\) by its closure in \(X\), and set \[\widetilde{D}_{m-1} = \widetilde{X}_{m-1}\setminus\widetilde{X}_{m-1}^\circ .\] The advantage of doing so is that \(\tilde{X}_{m-1}^\circ\) separates \(X_m^\circ\) into two connected components.

Since \[\operatorname{Lip}_{g_m}\tau_j\le\frac{2}{d_j},\] the \(W\)-neighborhood \[N_W= \{x\in X_m^\circ: d_{g_m}(x,\widetilde{X}_{m-1}^\circ)\le W\}\] does not meet the two faces \(\{\tau_j=\pm1\}\) whenever \[W<\frac{1-|a|}{2}d_j .\] Moreover, by Lemma 9, the finite-width neighborhood \(N_W\) of \(\widetilde{X}_{m-1}^\circ\) can only accumulate at the old exceptional set \(D_m\) along \(\widetilde{D}_{m-1}\). Note that \(\widetilde{X}_{m-1}^\circ\) separates \(N_W\) into two components \(N_W^-\) and \(N_W^+\).

After smoothing \(N_W\) and the distance function to \(\widetilde{X}_{m-1}^\circ\), for every \(L<W\) we obtain a smooth band \[\Omega\subset N_W\] and a smooth function \[s:(\Omega,\partial_\pm\Omega)\to([-L,L],\pm L), \qquad s^{-1}(0)=\widetilde{X}_{m-1}^\circ, \qquad |ds|_{g_m}\le1.\]

By Proposition 10, there exists a locally finite perimeter set \(E\) in the band such that \[\Sigma=\partial E\cap\Omega^\circ\] is a weighted \(\mu\)-bubble. On \[\Sigma_{\rm reg}=\partial^*E\] one has \[H_\Sigma-\langle\nabla f_m,\nu\rangle=\Phi,\] and the weighted stability inequality holds. Moreover, \[\Sigma=\Sigma_{\rm reg}\sqcup D^\Sigma\] is an \((m-1)\)-dimensional AM–PI space, and \[\mathsf P_{m-8}^{\Sigma}(D^\Sigma)\] holds. Moreover, the quadratic form \[Q_{a_m(\lambda),\sigma_L}^{g_\Sigma,F_\Sigma}\] is nonnegative on \(C_c^\infty(\Sigma_{\rm reg})\), where \(g_\Sigma\) is the induced metric of \(\Sigma_{\rm{reg}}\) from \((\Omega,g_m)\), \(F_\Sigma=f_m|_{\Sigma_{\rm{reg}}}\), and \[\sigma_L=\sigma_m-(1-\lambda)\left(\frac{\pi}{L}\right)^2.\] The current construction guarantees that \(\Sigma\) is homologous to \(\widetilde{X}_{m-1}^\circ\) in the band \(\Omega\subset N_W\). In particular, the map \[F_{m-1}^\Sigma:(\Sigma,\partial\Sigma)\to (B_{m-1},\partial B_{m-1})\] is proper over \(B_{m-1}\setminus F_{m-1}(\tilde{D}_{m-1})\) and has the same nonzero relative degree over \((B_{m-1}\setminus F_{m-1}(\widetilde{D}_{m-1}),\partial B_{m-1}\setminus F_{m-1}(\widetilde{D}_{m-1}))\) as \(F_{m-1}^{\widetilde{X}_{m-1}^\circ}\). Indeed, for any compact subset \(K\subset B_{m-1}\setminus F_{m-1}(\tilde{D}_{m-1})\), the set \[F_{m-1}^{-1}(K)\cap \overline{N_W}\] is a closed subset of \(X_m^\circ\), and so the set \[(F_{m-1}^\Sigma)^{-1}(K)=F_{m-1}^{-1}(K)\cap \overline{N_W}\cap\Sigma\] is a closed subset of \(\Sigma\). Without loss of generality, we can replace \(\Sigma\) by one of its components still keeping the nonzero relative degree.

We now localize away from the side boundary before applying conformal blow-up. Choose \(0<\rho_m<1\). If \(j-1>0\), set \[B_{m-1}^{\rho_m/2} = Y\times(-1+\rho_m/2,1-\rho_m/2)^{j-1}, \qquad B_{m-1}^{\rho_m} = Y\times[-1+\rho_m,1-\rho_m]^{j-1}.\] If \(j-1=0\), both spaces are understood to be \(Y\). We choose \(\rho_m\) so that the faces of \(B_{m-1}^{\rho_m}\) are regular for \(F_{m-1}|_{\Sigma_{\rm reg}}\) and so that the restricted current is defined. The part of \(\Sigma\) over \(B_{m-1}^{\rho_m/2}\) is away from the original side boundary \(\partial_s\Omega\). Hence there the AM–PI and packing conclusions of Proposition 10 are ordinary interior statements, and the Green-function construction of Section 3 applies without boundary modification.

If \(m\leq 7\), then \(D^\Sigma=\emptyset\) and we take \(w=1\). Otherwise, we apply Proposition 8 in dimension \(m-1\) to the AM–PI space \[\Sigma=\Sigma_{\rm reg}\sqcup D^\Sigma\] over \(B_{m-1}^{\rho_m/2}\). The coefficient is \[a_m(\lambda)=a_{(m-1)+1}(\lambda).\] The packing hypothesis is precisely \[\mathsf P_{m-8}^{\Sigma}(D^\Sigma),\] and the admissibility of \(\lambda\) follows from \(\lambda\leq\Lambda_n\). Thus, for every \(\delta>0\), there exists a smooth positive function \(w\) on the regular part over \(B_{m-1}^{\rho_m/2}\) such that, on the smaller region over \(B_{m-1}^{\rho_m}\), \[\bar g_{m-1} = w^{\overline{\alpha}_{m-1}(\lambda)}g_\Sigma, \qquad \bar f_{m-1} = F_\Sigma+\overline{\beta}_{m-1}(\lambda)\log w,\] the metric \(\bar g_{m-1}\) is complete toward \(D^\Sigma\), and \[Q_{a_m(\lambda),\,\sigma_L-\delta}^{\bar g_{m-1},\bar f_{m-1}} \ge0\] on compactly supported test functions in the regular part over \(B_{m-1}^{\rho_m}\).

By the Barta–Allegretto–Piepenbrink argument on an exhaustion of this regular part, there exists a positive smooth function \(\psi\) such that \[\mathcal{L}_{m,\lambda,\sigma_L-\delta}^{\bar g_{m-1},\bar f_{m-1}}\psi \ge0 .\] Set \[g_{m-1}=\bar g_{m-1}, \qquad f_{m-1}=\bar f_{m-1}-\log\psi .\] The positive supersolution identity gives \[\mathcal{S}_{f_{m-1}}^{m-1,\lambda}(g_{m-1}) \ge \sigma_L-\delta\] pointwise on the regular part over \(B_{m-1}^{\rho_m}\).

Let \[\Sigma^{\rho_m} = \Sigma\cap F_{m-1}^{-1}(B_{m-1}^{\rho_m}), \qquad \Sigma_{\rm reg}^{\rho_m} = \Sigma_{\rm reg}\cap F_{m-1}^{-1}(B_{m-1}^{\rho_m}).\] Let \(\overline{\Sigma^{\rho_m}}\) be the closure of \(\Sigma^{\rho_m}\) in \(X\), and define \[X_{m-1} = \overline{\Sigma^{\rho_m}} \cup \bigl(\widetilde{D}_{m-1} \cap F_{m-1}^{-1}(B_{m-1}^{\rho_m})\bigr),\] \[X_{m-1}^\circ=\Sigma_{\rm reg}^{\rho_m}, \qquad D_{m-1}=X_{m-1}\setminus X_{m-1}^\circ .\] Since \(X_{m-1}\subset \overline{N_W}\), we have \(X_{m-1}\cap \tau_j^{-1}(\pm 1)=\emptyset\). By the choice of \(\rho_m\), \(X_{m-1}^\circ\) is a smooth \((m-1)\)-manifold with boundary. Its new regular boundary faces are \[\Sigma_{\rm reg}\cap F_{m-1}^{-1}(\partial B_{m-1}^{\rho_m});\] these are part of \(\partial X_{m-1}^\circ\), not part of \(D_{m-1}\). The singular part \(D_{m-1}\) consists only of the inherited sliced singular set and the singular set of the \(\mu\)-bubble. Hence \[\dim_{\mathcal{H}}D_{m-1}\le m-8=(m-1)-7,\] and the new part \(D^\Sigma\subset D_{m-1}\) satisfies the local scale-uniform codimension-seven packing estimate.

The completeness of \(g_{m-1}\) toward \(D^\Sigma\) follows from the quadratic-form conformal blow-up, and the completeness of \(g_{m-1}\) toward the inherited part of \(\widetilde{D}_{m-1}\) follows from Lemma 9 and the completeness of \(g_m\) toward \(D_m\). The quadratic growth of \(g_m\) around \(D_{m-1}\) follows from the fact \(g_{m-1}\geq g_m|_\Sigma\) and the estimate 45 .

Note that \[F_{m-1}^{X_{m-1}^\circ}:(X_{m-1}^\circ,\partial X_{m-1}^\circ)\to (B_{m-1}^{\rho_m},\partial B_{m-1}^{\rho_m})\] is proper over \(B_{m-1}^{\rho_m}\setminus F_{m-1}(D_{m-1})\) and has the same nonzero relative degree over \((B_{m-1}^{\rho_m}\setminus F_{m-1}(D_{m-1}), \partial B_{m-1}^{\rho_m}\setminus F_{m-1}(D_{m-1}))\) as \(F_{m-1}^{\Sigma}\).

After affinely rescaling \(B_{m-1}^{\rho_m}\) back to \(B_{m-1}=Y\times I^{j-1}\), the remaining coordinate Lipschitz constants are multiplied by at most \((1-\rho_m)^{-1}\). This is the only width loss in the step. Since \(\rho_m>0\) can be chosen arbitrarily small, and there are only finitely many descent steps, these losses are absorbed into the final \(\varepsilon\). We continue to denote the resulting widths by \(d_i\). With this convention, \[\operatorname{Lip}_{g_{m-1}}\tau_i \le \frac{2}{d_i}, \qquad i=1,\ldots,j-1 .\]

Finally, choose \(|a|\) small, \(W\) and \(L\) sufficiently close to \(d_j/2\), and let \(\delta\downarrow0 .\) We obtain \[\sigma_{m-1} \ge \sigma_m - 4\pi^2(1-\lambda)d_j^{-2} - \frac{\varepsilon}{4}.\] Combining this with the induction hypothesis gives \[\sigma_{m-1} \ge \sigma - 4\pi^2(1-\lambda) \sum_{i=j}^{k}d_i^{-2} - \varepsilon .\] Thus \[(X_{m-1},X_{m-1}^\circ,D_{m-1}, g_{m-1},f_{m-1},\sigma_{m-1})\] is an \((m-1)\)-dimensional datum controlled over \[B_{m-1}=Y\times I^{j-1}.\] This completes the induction step, and hence the proposition. ◻

Proof. Since \(\dim_{\mathcal{H}}D_r\le r-7\), we have \(D_1=\emptyset\). The nonzero degree condition gives a connected component \(\Gamma\subset X_1\) carrying nonzero relative degree over \((I,\partial I)\). Thus \(\Gamma\) joins the two faces \(\tau_1^{-1}(-1)\) and \(\tau_1^{-1}(1)\). If \(\mathcal{L}\) is the \(g_1\)-length of \(\Gamma\), then \(\mathcal{L}\ge d_1\).

Let \(s\in(-\mathcal{L}/2,\mathcal{L}/2)\) be arclength on \(\Gamma\). Set \(a=\pi/\mathcal{L}\), \(\Phi(s)=-\mu a\tan(as)\), and \(\mu=2(1-\lambda)\). The one-dimensional stability inequality gives, at the point \(p\), \[\sigma_1+2\Phi'(p)+\frac{1}{1-\lambda}\Phi(p)^2\le0 .\] Since \[2\Phi'(s)+\frac{1}{1-\lambda}\Phi(s)^2 = -2\mu a^2+ \left(\frac{\mu^2}{1-\lambda}-2\mu\right)a^2\tan^2(as)=-2\mu a^2,\] we get \[\sigma_1\le2\mu\pi^2/\mathcal{L}^2\le4\pi^2(1-\lambda)d_1^{-2}.\] ◻

Proof. Apply Proposition 11 with \(m=1\), then apply Lemma 10, and let \(\varepsilon\to0\). ◻

Proof. The scalar lower bound is Proposition 11 with \(m=r\). The degree and properness are part of the terminal controlled datum. Since \(\dim_{\mathcal{H}}D_r\le r-7\), the condition \(r\le6\) forces \(D_r=\emptyset\). ◻

Proof of Theorem B. It suffices to prove the result for \[\lambda<\Lambda_n .\] The endpoint follows by applying the strict case to \(\lambda'<\lambda\) and then letting \(\lambda'\uparrow\lambda\). Indeed, for \(\lambda'<\lambda\leq 1/n\), \[\mathcal{S}_f^{n,\lambda'}(g) = \mathcal{S}_f^{n,\lambda}(g) + \bigl(a_n(\lambda)-a_n(\lambda')\bigr)|df|_g^2 \ge \mathcal{S}_f^{n,\lambda}(g).\]

If \(Y=\{\mathrm{pt}\}\), then Theorem B is exactly the pure cube inequality, Corollary 2. We therefore assume \[r=\dim Y>0 .\]

Fix \(\varepsilon>0\). By enlargeability of \(Y\), choose a Riemannian cover \[\widehat Y\to Y\] and a smooth \(\varepsilon\)-Lipschitz map \[\phi:\widehat Y\to S^r\] which is constant outside a compact set and has nonzero degree. Pull back \(X\) by the \(Y\)-component of \(F\), and write \[\widehat X=X\times_Y\widehat Y .\] The lifted map \[\widehat F:\widehat X\to\widehat Y\times I^k\] is proper and has the same nonzero relative degree.

Choose a regular value \(y_0\in S^r\), different from the constant value at infinity, such that \[\deg(\phi;y_0)\ne0 .\] Let \(C\Subset S^r\) be a small coordinate cube around \(y_0\), disjoint from the constant value at infinity. Then \(\phi^{-1}(C)\) is compact and \(\phi:\phi^{-1}(C)\to C\) has nonzero relative degree. Since \(\phi\) is constant outside a compact set, \(\phi^{-1}(C)\) is compact. Hence \[X_C := \widehat F^{-1}\bigl(\phi^{-1}(C)\times I^k\bigr)\] is compact, and the map \[X_C\to C\times I^k\] has nonzero relative degree. After identifying \(C\) with \(I^r\), this gives a cubical map \[X_C\to I^r\times I^k=I^n .\]

The \(k\) cubical widths coming from the original \(I^k\)-coordinates are at least the corresponding \(d_i\). The additional \(r\) widths coming from \(C\subset S^r\) tend to infinity as \(\varepsilon\downarrow0\). More precisely, if \(L_Y\) is a Lipschitz constant for \(F_Y\), and if \(c_C>0\) is the minimum distance between opposite faces of \(C\), then the new widths are bounded below by \(\frac{c_C}{\varepsilon L_Y}\). Applying Corollary 2 to \[X_C\to I^r\times I^k\] gives \[\sigma \le 4\pi^2(1-\lambda) \left( \sum_{i=1}^{k}d_i^{-2} + O(\varepsilon^2) \right).\] Letting \(\varepsilon\downarrow0\) gives \[\sigma \le 4\pi^2(1-\lambda)\sum_{i=1}^{k}d_i^{-2}.\] This proves Theorem B. ◻

Proof of Theorem A. Take \(Y=\mathbb{T}^n\) and \(k=0\) in Theorem B. ◻

Proof of Theorem C. It suffices to treat \(\lambda<\Lambda_n\). Apply Corollary 3 with \(Y=S^2\). Since \(2\le6\), the terminal object is a smooth closed oriented surface \((\Sigma,g_\Sigma)\), with a map \(\rho_\Sigma:\Sigma\to S^2\) of nonzero degree, and for every \(\eta>0\), \[\mathcal{S}_{f_\Sigma}^{2,\lambda}(g_\Sigma)\ge\sigma_*-\eta .\] Since \(\lambda<1/2\), \(a_2(\lambda)>0\). On any component \(\Sigma_0\subset\Sigma\), \[\begin{align} (\sigma_*-\eta)\operatorname{Area}_{g_\Sigma}(\Sigma_0) &\le \int_{\Sigma_0}\mathcal{S}_{f_\Sigma}^{2,\lambda}(g_\Sigma) \\ &= \int_{\Sigma_0} \bigl(R_{g_\Sigma} +2\Delta f_\Sigma -a_2(\lambda)|df_\Sigma|^2\bigr) \\ &\le \int_{\Sigma_0}R_{g_\Sigma} = 4\pi\chi(\Sigma_0). \end{align}\] Since \(\rho_\Sigma\) has nonzero degree, some component \(\Sigma_0\) carries nonzero degree over \(S^2\). For this component the left-hand side is positive, so \(\chi(\Sigma_0)>0\), hence \(\Sigma_0\simeq S^2\). Therefore \[\operatorname{Area}_{g_\Sigma}(\Sigma_0) \le \frac{8\pi}{\sigma_*-\eta}.\] This component gives an integral \(2\)-cycle in \(X\) pairing nontrivially with \(u_\rho\). Since the conformal factors in the descent are at least \(1\), its area in the original metric is no larger than its final descended area. Letting \(\eta\downarrow0\) proves the estimate. ◻

Proof. We first prove the estimate. It suffices to treat \(\lambda<\Lambda_n\); the endpoint follows by applying the result to \(\lambda'<\lambda\) and then letting \(\lambda'\uparrow\lambda\).

Fix \(\varepsilon>0\). Choose a cover \(\widehat Y\to Y\) and an \(\varepsilon\)-Lipschitz map \(\phi:\widehat Y\to S^{n-2}\), constant outside a compact set and of nonzero degree. Pull back \(X\) to \(\widehat X=X\times_Y\widehat Y\), and write \(\widehat F=(\widehat F_Y,\widehat\rho)\). Let \(C\Subset S^{n-2}\) be a small coordinate cube around a regular value of \(\phi\), disjoint from the constant value at infinity, such that \[\phi:\phi^{-1}(C)\to C\] has nonzero relative degree. Then \[X_C:=\widehat F_Y^{-1}(\phi^{-1}(C))\] is compact, and \[(\widehat\rho,\phi\circ\widehat F_Y):X_C\to S^2\times C\] has nonzero relative degree. Identifying \(C\) with \(I^{n-2}\), apply Theorem 1. If \(D_j\) are the cubical widths coming from \(C\), then \(D_j\ge c_C/(\varepsilon L_Y)\), hence \(\sum_jD_j^{-2}=O(\varepsilon^2)\). Therefore \[\operatorname{sys}_2(X_C,\widehat u_\rho;g) \le \frac{8\pi}{\sigma-4\pi^2(1-\lambda)O(\varepsilon^2)} .\] Pushing cycles down to \(X\) does not increase mass and preserves the pairing with \(u_\rho\). Letting \(\varepsilon\downarrow0\) gives \[\operatorname{sys}_2(X,u_\rho;g)\le\frac{8\pi}{\sigma}.\]

Assume now the equality holds . Put \[a=a_n(\lambda), \qquad V=\mathcal{S}_f^{n,\lambda}(g).\] Consider \[P_{g,f} = -\frac{4}{a} \bigl(\Delta_g-a\langle\nabla f,\nabla\cdot\rangle_g\bigr) +V,\] which is self-adjoint for \(e^{-af}d\mu_g\), and denote its first eigenvalue by \(\theta(g,f)\). If \(V>\sigma\) somewhere, then \(\theta(g,f)>\sigma\). For a positive first eigenfunction \(u\), set \(\psi=-2a^{-1}\log u\). Then \[\mathcal{S}_{f+\psi}^{n,\lambda}(g) = u^{-1}P_{g,f}u = \theta(g,f)>\sigma .\] Applying the estimate already proved to \((g,f+\psi,\lambda)\) gives \[\operatorname{sys}_2(X,u_\rho;g) \le \frac{8\pi}{\theta(g,f)} < \frac{8\pi}{\sigma},\] contradicting equality. Hence \(V\equiv\sigma\).

If \(df\not\equiv0\), choose \(\lambda_-<\lambda\). Since \(a_n\) is increasing, \[\mathcal{S}_f^{n,\lambda_-}(g) = \sigma+ \bigl(a_n(\lambda)-a_n(\lambda_-)\bigr)|df|_g^2\] is \(\ge\sigma\) and is \(>\sigma\) somewhere. Repeating the preceding argument with \(\lambda_-\) gives a weight \(\psi_-\) and a constant \(\theta_->\sigma\) with \(\mathcal{S}_{f+\psi_-}^{n,\lambda_-}(g)=\theta_-\) again contradicting the estimate. Thus \(df\equiv0\), and consequently \[f\equiv\mathrm{constant}, \qquad R_g\equiv\sigma .\]

We next show \(\operatorname{Ric}_g\ge0\). For a metric \(\bar g\), set \[P_{\bar g}=-\frac{4}{a}\Delta_{\bar g}+R_{\bar g},\] and denote its first eigenvalue by \(\theta(\bar g)\). Since \(R_g\equiv\sigma\), constants are first eigenfunctions and \(\theta(g)=\sigma\). If \(\operatorname{Ric}_g\) has a negative direction somewhere, choose \(h\ge0\), supported in a small ball, with \[\int_X\langle\operatorname{Ric}_g,h\rangle_g\,d\mu_g<0.\] For \(g_t=g+th\), one has \(g_t\ge g\), hence \[\operatorname{sys}_2(X,u_\rho;g_t)\ge\frac{8\pi}{\sigma}.\] On the other hand, the first variation formula gives \[\left.\frac{d}{dt}\right|_{t=0}\theta(g_t) = -\frac{1}{\operatorname{Vol}_g(X)} \int_X\langle\operatorname{Ric}_g,h\rangle_g\,d\mu_g>0.\] Thus \(\theta(g_t)>\sigma\) for small \(t>0\). If \(P_{g_t}u_t=\theta(g_t)u_t\) and \(\psi_t=-2a^{-1}\log u_t\), then \[\mathcal{S}_{\psi_t}^{n,\lambda}(g_t)=\theta(g_t)>\sigma .\] Applying the estimate to \((g_t,\psi_t,\lambda)\) gives \(\operatorname{sys}_2(X,u_\rho;g_t) <\frac{8\pi}{\sigma}\), a contradiction. Hence \(\operatorname{Ric}_g\ge0\).

Finally, by Cheeger–Gromoll, \[\widetilde{X}\simeq\mathbb{R}^\ell\times Z\] with \(Z\) compact. The argument in [21] gives \(\ell\ge n-2\). Since \(R_g\equiv\sigma>0\), the compact factor has dimension at least \(2\), so \(\ell\le n-2\). Hence \(\ell=n-2\) and \(\dim Z=2\). The product scalar curvature is \(R_Z\), so \(R_Z\equiv\sigma\). Since \(Z\) is the compact factor in the universal cover, it is simply connected; hence \(Z\simeq S^2\), and \(R_Z\equiv\sigma\) makes \(Z\) the round sphere \(S^2_\sigma\). Thus \[\widetilde{X}\simeq S^2_\sigma\times\mathbb{R}^{n-2}.\] ◻

Proof. Assume otherwise. Choose \[3-\frac{2}{n}<c<c_A^X(\mathcal{S}).\] By Remark 9, choose \(\bar\lambda\le\lambda\) such that \[\bar\lambda<\Lambda_n, \qquad \tau_n(\bar\lambda)<c-2 .\] Since \(a_n\) is increasing on \((-\infty,1/n)\), the function \[V_0:=\mathcal{S}_f^{n,\bar\lambda}(g) = \mathcal{S}_f^{n,\lambda}(g) + \bigl(a_n(\lambda)-a_n(\bar\lambda)\bigr)|df|_g^2\] is nonnegative and positive on a nonempty open subset of \(\mathcal{R}\). Choose a smooth ball \(B\Subset\mathcal{R}\), a constant \(m>0\), and a function \(\chi\in C_c^\infty(B)\), \(0\le\chi\le1\), \(\chi\not\equiv0\), such that \(V_0\ge m\) on \(B\). Put \(W=m\chi\). Then \(0\le W\le V_0\), \(W\not\equiv0\), and \(W\) is bounded with compact support in \(\mathcal{R}\).

Put \(a=a_n(\bar\lambda)\). Since \(f\) is bounded, the measure \(e^{-af}d\mu_g\) is admissible. Define \[\eta_0 = \inf_{\varphi\not\equiv0} \frac{ \frac{4}{a}\int_{\mathcal{R}}|\nabla\varphi|_g^2e^{-af}\,d\mu_g + \int_{\mathcal{R}}W\varphi^2e^{-af}\,d\mu_g }{ \int_{\mathcal{R}}\varphi^2e^{-af}\,d\mu_g } .\] The compact AM–PI hypotheses give a minimizer \(u\ge0\). Since \(X\) is connected and \(W\ge0\) is not identically zero, \(\eta_0>0\). The Harnack inequalities give, \(0<c_0\le u\le C_0<\infty\). On \(\mathcal{R}\), elliptic regularity gives \(u\in C^\infty(\mathcal{R})\) and \[\left[ -\frac{4}{a} \bigl(\Delta_g-a\langle\nabla f,\nabla\cdot\rangle_g\bigr) +W \right]u = \eta_0u .\] Set \(\psi=-2a^{-1}\log u\). Then \(\psi\) is bounded. Since \(W\le V_0\), a direct calculation gives \[\begin{align} \mathcal{S}_{f+\psi}^{n,\bar\lambda}(g) &= u^{-1} \left[ -\frac{4}{a} \bigl(\Delta_g-a\langle\nabla f,\nabla\cdot\rangle_g\bigr) +V_0 \right]u \\ &= \eta_0+(V_0-W) \ge \eta_0>0 . \end{align}\] Moreover, the boundedness of \(f+\psi\) implies that the divergence weight associated with \(L_{f+\psi,\bar\lambda}^{(n)}\) is still admissible.

We now apply Proposition 7 to \((g,f+\psi)\) and obtain a complete smooth metric \(g_0\) on \(\mathcal{R}\) and a smooth density \(f_0\) such that \[\mathcal{S}_{f_0}^{n,\bar\lambda}(g_0)\ge\eta_1>0 .\]

Let \(\widehat X\to X\) be a cover, and let \(\Phi:\widehat X\to S^n\) be \(\varepsilon\)-Lipschitz, constant outside a compact set, and of nonzero degree. Choose a regular value \(y_0\in S^n\), different from the constant value at infinity, such that \(\deg(\Phi;y_0)\ne0\). Let \(C\Subset S^n\) be a small coordinate cube around \(y_0\), disjoint from the constant value at infinity. Then \(\Phi^{-1}(C)\) is compact and \[\Phi:\Phi^{-1}(C)\to C\] has nonzero relative degree.

Let \(\widehat{\mathcal{R}}\) and \(\widehat{\mathcal{S}}\) denote the lifted regular and singular parts, and set \[A=\Phi\bigl(\widehat{\mathcal{S}}\cap\Phi^{-1}(C)\bigr)\subset C .\] Since \(c_A^X(\mathcal{S})>3-2/n\), one has \(\dim_A\mathcal{S}<n-1\), and hence \(\mathcal{H}^{n-1}(A)=0\). The blown-up regular part maps properly to \(C\setminus A\) and has nonzero degree. Identifying \(C\) with \(I^n\), Remark 12 gives \[\eta_1 \le 4\pi^2(1-\bar\lambda)\sum_{i=1}^nD_i^{-2},\] where \(D_i\) are the cubical widths measured in \(g_0\). By the construction of the pointwise conformal blow-up, \(g_0\ge g\). Since \(\Phi\) is \(\varepsilon\)-Lipschitz with respect to \(g\), there is a constant \(c_C>0\), depending only on \(C\), such that \(D_i\ge c_C\varepsilon^{-1}\) for all \(i\). Letting \(\varepsilon\downarrow0\) contradicts \(\eta_1>0\). ◻

Proof. Theorem 15 gives \[\mathcal{S}_f^{n,\lambda}(g)\equiv0 .\] If \(df\not\equiv0\), choose \(\lambda'<\lambda\). Since \(a_n\) is strictly increasing, \[\mathcal{S}_f^{n,\lambda'}(g) = \mathcal{S}_f^{n,\lambda}(g) + \bigl(a_n(\lambda)-a_n(\lambda')\bigr)|df|_g^2\] is nonnegative and positive somewhere, contradicting Theorem 15. Thus \(df\equiv0\), and then \[R_g=\mathcal{S}_f^{n,\lambda}(g)\equiv0 .\] ◻

Proof. By Corollary 4, \(df\equiv0\) and \(R_g\equiv0\) on \(\mathcal{R}\). It remains to rule out \(\operatorname{Ric}_g\not\equiv0\).

Assume \(\operatorname{Ric}_g\not\equiv0\). Since \(R_g=0\), the Ricci tensor has a negative direction somewhere. Choose \(h\in C_c^\infty(\operatorname{Sym}^2T^*\mathcal{R})\), supported in a smooth ball \(B\Subset\mathcal{R}\), with \(h\ge0\) and \[\int_{\mathcal{R}}\langle\operatorname{Ric}_g,h\rangle_g\,d\mu_g<0 .\] For small \(t\ge0\), set \(g_t=g+th\). Let \(a=a_n(\lambda)\), and let \(\theta(t)\) be the first eigenvalue of \(P_{g_t}=-\frac{4}{a}\Delta_{g_t}+R_{g_t}\). As in the deformation argument used in Proposition 14, constants realize \(\theta(0)=0\), and \[\theta'(0) = -\frac{1}{\mu_g(X)} \int_{\mathcal{R}}\langle\operatorname{Ric}_g,h\rangle_g\,d\mu_g >0 .\] Hence \(\theta(t)>0\) for all small \(t>0\).

Let \(u_t>0\) be a first eigenfunction. By the same compactness, regularity and Harnack argument as above, \(u_t\) is smooth on \(\mathcal{R}\) and \(0<c_t\le u_t\le C_t\). With \(\psi_t=-2a^{-1}\log u_t\), we get \[\mathcal{S}_{\psi_t}^{n,\lambda}(g_t) = u_t^{-1}P_{g_t}u_t = \theta(t)>0 \quad\text{on }\mathcal{R} .\] The deformation is supported in \(B\Subset\mathcal{R}\), so \(g_t=g\) near \(\mathcal{S}\); in particular the AM–PI assumptions, the Assouad-codimension condition, and admissibility of the weight are unchanged. Applying Theorem 15 to \((X,g_t,\psi_t,\lambda)\) gives a contradiction. Therefore \(\operatorname{Ric}_g\equiv0\) on \(\mathcal{R}\). ◻

Proof. Put \(s=|\nabla u|_g^2\). For \(\varepsilon>0\), set \(v_\varepsilon=(s+\varepsilon)^{q/2}\). The Bochner formula gives \(\frac{1}{2}\Delta_gs\ge|\nabla^2u|^2\), and the improved Kato inequality gives \[|\nabla s|^2\le \frac{4(n-1)}{n}s|\nabla^2u|^2 .\] Hence \[\begin{align} \Delta_gv_\varepsilon &= \frac{q}{2}(s+\varepsilon)^{q/2-1}\Delta_gs + \frac{q(q-2)}{4}(s+\varepsilon)^{q/2-2}|\nabla s|^2 \\ &\ge q(s+\varepsilon)^{q/2-2} \left[ (s+\varepsilon)|\nabla^2u|^2 +\frac{q-2}{4}|\nabla s|^2 \right] \\ &\ge \frac{q}{4} \left(q-\frac{n-2}{n-1}\right) (s+\varepsilon)^{q/2-2}|\nabla s|^2 . \end{align}\] Since \[v_\varepsilon^{-1}|\nabla v_\varepsilon|^2 = \frac{q^2}{4}(s+\varepsilon)^{q/2-2}|\nabla s|^2,\] we get \(\Delta_gv_\varepsilon\ge\kappa_q v_\varepsilon^{-1}|\nabla v_\varepsilon|^2\). Letting \(\varepsilon\downarrow0\) gives 48 for \(v=s^{q/2}=|\nabla u|^q\).

It remains to extend the weak subharmonicity across \(\mathcal{S}\). Put \(d(x)=\operatorname{dist}(x,\mathcal{S})\) and \(N_\rho=N_\rho(\mathcal{S})\). Let \(K\Subset U\) and choose \(c\) with \(2+\beta<c<c_A^X(\mathcal{S})\). By the local Ahlfors regularity, and the bound \(v\le A d^{-\beta}\), we have, for small \(\rho\), \[\mu_g(N_\rho\cap K)\le C\rho^c,\qquad \int_{N_\rho\cap K}v\,d\mu_g\le C\rho^{c-\beta}.\] Choose \(K'\Subset U\) with \(K\Subset K'\), and take \(\psi\in\operatorname{Lip}_c(U)\) with \(\psi\equiv1\) near \(K\) and \(\operatorname{spt}\psi\subset K'\). Let \(\eta_\rho=1\) on \(N_{2\rho}\), \(\operatorname{spt}\eta_\rho\subset N_{4\rho}\), and \(|\nabla\eta_\rho|_g\le C\rho^{-1}\). For \(0<\delta<\rho\), let \(\theta_\delta=0\) on \(N_\delta\), \(\theta_\delta=1\) outside \(N_{2\delta}\), and \(|\nabla\theta_\delta|_g\le C\delta^{-1}\). Testing 48 with \(\zeta_{\rho,\delta}:=\psi\eta_\rho\theta_\delta\), which equals \(1\) on \((N_{2\rho}\setminus N_{2\delta})\cap K\), gives \[\begin{align} \int_{(N_{2\rho}\setminus N_{2\delta})\cap K} v^{-1}|\nabla v|_g^2\,d\mu_g &\le C(\rho^{-2}+1)\int_{N_{4\rho}\cap K'}v\,d\mu_g + C\delta^{-2}\int_{N_{2\delta}\cap K'}v\,d\mu_g \\ &\le C\rho^{c-\beta-2}+C\delta^{c-\beta-2}. \end{align}\] Since \(c-\beta-2>0\), letting \(\delta\downarrow0\) yields \[\int_{N_{2\rho}\cap K} v^{-1}|\nabla v|_g^2\,d\mu_g \le C\rho^{c-\beta-2}.\]

Now fix \(\varphi\in\operatorname{Lip}_c(U)\), \(\varphi\ge0\), and take \(K\Subset U\) with \(\operatorname{spt}\varphi\subset K\). Let \(\chi_\rho=0\) on \(N_\rho\), \(\chi_\rho=1\) outside \(N_{2\rho}\), and \(|\nabla\chi_\rho|\le C\rho^{-1}\). Since \(\chi_\rho\varphi\in\operatorname{Lip}_c(U\cap\mathcal{R})\), subharmonicity on \(U\cap\mathcal{R}\) gives \[\int_U\langle\nabla v,\nabla(\chi_\rho\varphi)\rangle\,d\mu_g\le0 .\] Hence \[\begin{align} \int_U\langle\nabla v,\nabla\varphi\rangle\,d\mu_g &\le \int_U(1-\chi_\rho)\langle\nabla v,\nabla\varphi\rangle\,d\mu_g - \int_U\varphi\langle\nabla v,\nabla\chi_\rho\rangle\,d\mu_g . \end{align}\] By Hölder’s inequality, the preceding estimates imply \[\left| \int_U(1-\chi_\rho)\langle\nabla v,\nabla\varphi\rangle\,d\mu_g \right| \le C\rho^{c-\beta-1}\to0,\] and \[\left| \int_U\varphi\langle\nabla v,\nabla\chi_\rho\rangle\,d\mu_g \right| \le C\rho^{c-\beta-2}\to0.\] Since \(c-\beta>2\), letting \(\rho\downarrow0\) yields \(\int_U\langle\nabla v,\nabla\varphi\rangle\,d\mu_g\le0\). ◻

Proof. Let \(u\) be bounded and weakly harmonic on \(B_{2R}(p)\), and put \[M=\|u\|_{L^\infty(B_{2R}(p))}, \qquad \rho(x)=\operatorname{dist}(x,\mathcal{S}).\] The function \(u+2M\) is positive and harmonic. If \(x\in B_{3R/2}(p)\cap\mathcal{R}\) and \(\rho(x)<R/8\), then \(B_{\rho(x)/4}(x)\subset B_{2R}(p)\cap\mathcal{R}\). The Cheng–Yau estimate on this regular ball gives \[|\nabla u|(x)\le C\rho(x)^{-1}M.\] Away from \(N_{R/8}(\mathcal{S})\), the ordinary interior estimate gives the same bound with \(R^{-1}\). Hence, after increasing \(C\), \[|\nabla u|(x)\le C\rho(x)^{-1}M \qquad \text{on }B_{3R/2}(p)\cap\mathcal{R}.\]

Choose \[\frac{n-2}{n-1}<q<c_A^X(\mathcal{S})-2.\] This is possible by the codimension assumption. Set \(v=|\nabla u|^q\). The preceding gradient estimate gives \(v\le C\rho^{-q}M^q\). Since \(c_A^X(\mathcal{S})>2+q\), Lemma 11 applies with \(\beta=q\), and \(v\) is weakly subharmonic on \(B_{3R/2}(p)\). Moser iteration gives \[\operatorname*{ess\,sup}_{B_R(p)}v \le C R^{-n}\int_{B_{3R/2}(p)}v\,d\mu_g .\] Since \(q<2\), Caccioppoli estimate for \(u\), Hölder inequality, and local Ahlfors regularity give \[\int_{B_{3R/2}(p)}v\,d\mu_g \le C R^{n-q}M^q .\] Therefore \(\operatorname*{ess\,sup}_{B_R(p)}|\nabla u|_g\le\frac{C}{R}M\), which is QL. ◻

Proof. We use the argument in the proof of annular quasiconvexity, cf. [23], [14]. First note that \(\mathsf P_{n-2}^{X}(\mathcal{S})\), together with local Ahlfors regularity, gives \[\mu_g\bigl(N_s(\mathcal{S})\cap B_d(z,R)\bigr) \le C_KR^{n-2}s^2\] whenever \(z\in K\), \(0<s<R<r_K\), and \(B_d(z,2R)\Subset K\).

Fix \(x,y\in K\cap\mathcal{R}\), put \(R=d(x,y)\), and work in \(B=B_d(x,C R)\), with \(C\) fixed large. Choose \(0<\kappa\ll1\) and set \[E_0=B_d(x,\kappa R), \qquad F_0=B_d(y,\kappa R).\] Choose \(A\gg1\), and put \[E=E_0\setminus N_{R/A}(\mathcal{S}), \qquad F=F_0\setminus N_{R/A}(\mathcal{S}).\] By the tubular estimate, \[\mu_g(E_0\cap N_{R/A}(\mathcal{S})) + \mu_g(F_0\cap N_{R/A}(\mathcal{S})) \le CA^{-2}R^n.\] Taking \(A\) large, we have \(\min\{\mu_g(E),\mu_g(F)\}\ge cR^n\). Let \(\Gamma\) be the family of curves in \(B\) joining \(E\) to \(F\). The Poincaré inequality gives \[\operatorname{Mod}_2(\Gamma)\ge cR^{n-2}.\]

Then we remove two bad subfamilies. First let \[\Gamma_{\rm long} = \{\gamma\in\Gamma:\;L_g(\gamma)>LR\}.\] Since \((LR)^{-1}\mathbf{1}_B\) is admissible for \(\Gamma_{\rm long}\), we obtain \(\operatorname{Mod}_2(\Gamma_{\rm long})\le CL^{-2}R^{n-2}\).

Next let \(0<\rho<R/(2A)\), write \(N_s=N_s(\mathcal{S})\), and set \[\Gamma_{\mathcal{S},\rho} = \{\gamma\in\Gamma:\gamma\cap N_\rho\ne\emptyset\}.\] Every curve in \(\Gamma_{\mathcal{S},\rho}\) starts outside \(N_{R/A}\) and meets \(N_\rho\), hence crosses \(N_{R/A}\setminus N_\rho\). Put \[\delta(z)=d(z,\mathcal{S}), \qquad L_{\rho,A}=\log\frac{R}{A\rho},\] and define \[\varrho_{\rho,A} = \frac{C}{L_{\rho,A}}\, \delta^{-1}\, \mathbf{1}_{N_{R/A}\setminus N_\rho}.\] After increasing \(C\), this function is admissible for \(\Gamma_{\mathcal{S},\rho}\). Let \(s_j=2^{-j}R/A\), and choose \(J\) so that \(s_{J+1}<\rho\le s_J\). Then \[\begin{align} \operatorname{Mod}_2(\Gamma_{\mathcal{S},\rho}) &\le \int_B\varrho_{\rho,A}^2\,d\mu_g \\ &\le \frac{C}{L_{\rho,A}^2} \sum_{j=0}^{J} s_j^{-2}\mu_g(N_{s_j}\cap B) \\ &\le \frac{C}{L_{\rho,A}^2} \sum_{j=0}^{J}R^{n-2} \le \frac{CR^{n-2}}{\log(R/(A\rho))}. \end{align}\]

Choose \(L\) large and then \(\rho\) small so that \[\operatorname{Mod}_2(\Gamma_{\rm long}) + \operatorname{Mod}_2(\Gamma_{\mathcal{S},\rho}) < \operatorname{Mod}_2(\Gamma).\] Hence some curve in \(\Gamma\) avoids \(N_\rho(\mathcal{S})\) and has \(g\)-length at most \(LR\).

We now iterate this ball-to-ball construction to obtain a point-to-point curve. Applying it to \((x,y)\) yields a curve joining some \(x_1\in B_d(x,\kappa R)\) to \(y_1\in B_d(y,\kappa R)\). Inductively applying this to the remaining gaps \((x,x_1)\) and \((y_1,y)\), at the \(j\)-th stage we bridge \(2^j\) gaps, adding curves of total \(g\)-length at most \(C_0(2\kappa)^j R\). By choosing \(\kappa < 1/2\), this concatenation converges to a rectifiable curve \(\gamma\) from \(x\) to \(y\) contained in \(\mathcal{R}\cap B_d(x,C_KR)\), with \[L_g(\gamma) \le C_0 R\sum_{j=0}^{\infty}(2\kappa)^j \le C_K R .\] This proves the lemma. ◻

Proof. Clearly, \(d(x,y)\le d_g^\ell(x,y)\) on \(\mathcal{R}\). Conversely, by Lemma 12, for every compact \(K\Subset X\) there are constants \(C_K,r_K>0\) such that, whenever \(x,y\in K\cap\mathcal{R}\) and \(d(x,y)<r_K\), there is a curve \(\gamma\subset\mathcal{R}\) joining \(x\) to \(y\) with \(L_g(\gamma)\le C_Kd(x,y)\). Hence \[d_g^\ell(x,y)\le C_Kd(x,y)\] for such \(x,y\). Thus the identity map extends to a bilipschitz map between the two completions.

It remains to prove the Sobolev-to-Lipschitz property. Let \(u\in W^{1,2}(X^\ell)\) satisfy \[|\nabla^g u|_g\le1 \qquad\text{a.e. on }\mathcal{R} .\] For \(x,y\in\mathcal{R}\), choose a regular curve \(\gamma\) with \(L_g(\gamma)\le d^\ell(x,y)+\varepsilon\). Then \[|u(x)-u(y)| \le \int_\gamma |\nabla^g u|\,ds_g \le d^\ell(x,y)+\varepsilon .\] Letting \(\varepsilon\downarrow0\), and then using the density of \(\mathcal{R}\subset X^\ell\), gives a \(1\)-Lipschitz representative. ◻

Proof. By Proposition 16, \(df\equiv0\) adn \(\operatorname{Ric}_g\equiv0\) on \(\mathcal{R}\). Thus the admissible measure is a constant multiple of \(d\mu_g\). Replace \(X\) by \(X^\ell\). By Lemma 13, the AM–PI structure and the Assouad codimension condition are preserved, and \(X^\ell\) satisfies the Sobolev-to-Lipschitz property. By Proposition 17, \(X^\ell\) satisfies QL. Together with PI, and Sobolev-to-Lipschitz, the Honda–Sun criterion [22] gives \((X^\ell,d_g^\ell,\mu_g)\in RCD(0,n)\).

Since \(X^\ell\) is compact \(RCD(0,n)\), Mondino–Wei [24] shows its universal cover \[\widetilde{X}^\ell\simeq \overline{X}\times\mathbb{R}^k,\] where \(\overline{X}\) is compact. The argument in [21] rules out a nontrivial compact factor. Hence \(\overline{X}\) is a point and \(k=n\). Thus \(\widetilde{X}^\ell\simeq\mathbb{R}^n\). Therefore \(X^\ell\) is a compact flat manifold. ◻

Proof of Theorem D. The first assertion is Theorem 15. The second assertion is Theorem 18. ◻