Proof. Note first that f extends to a holomorphic section of $L^k$ over X, using that X is normal. Using that $\omega $ has locally bounded potentials, we have that $\sup _X |f|_{h^k} < \infty $ .

Next we show that $|\nabla f|_{h^k, \omega _M} < \infty $ . For any $\hat {x}\in \hat {X}$ , let $x = \hat {\Phi }_X(\hat x)\in X$ . We can find a section $s\in H^0(X, L)$ and some $r> 0$ such that $s(y) \not =0$ for $y\in B_{\eta _{FS}}(x,r)$ . The assumption that $\omega $ is a Kähler current implies that we have constants $r'> 0$ and $C> 0$ (depending on $\hat {x}$ ) such that if we write $|s|^2_h = e^{-u}$ , then $|u| < C$ on $X^{reg} \cap B_{\omega _M}(\hat {x}, r')$ . We have $\Delta _{\omega _M} u = n$ on $X^{reg}\cap B_{\omega _M}(\hat {x}, r'/2)$ , and since u is bounded, this equation extends to $B_{\omega _M}(\hat {x}, r'/2)$ by Lemma 11 and Lemma 6. The gradient estimate in Proposition 12 then implies that $|\nabla u| < C_1$ on $B_{\omega _M}(\hat {x}, r'/2)$ . This implies that $|\nabla s| < C_2$ on $B_{\omega _M}(\hat {x}, r'/2)$ . If f is any holomorphic section of $L^k$ , then on $B_{\omega _M}(\hat {x}, r'/2)$ the ratio $f/s^k$ is a bounded harmonic function, so using the gradient estimate again, together with the bounds for s, we find that $|\nabla f| < C_3$ on $B_{\omega _M}(\hat {x}, r'/4)$ . We can cover $\hat {X}$ with finitely many balls of this type, showing that $|\nabla f|_{h^k, \omega _M} < \infty $ globally.

We can obtain the effective estimates claimed in the proposition as follows. Since the curvature of $h^k$ is $\omega _M$ , on M we have

(56) $$ \begin{align}\begin{aligned} \Delta_{\omega_M} |f|^2_{h^k} = |\nabla f|^2_{h^k, \omega_M} - n |f|^2_{h^k}. \end{aligned} \end{align} $$

Let $\phi _i$ denote cutoff functions as in Lemma 6. We have, omitting the subscripts,

(57) $$ \begin{align}\begin{aligned} \int_{M} \phi_i^2 |\nabla f|^2 \omega_M^n &= \int_{M} \phi_i^2 (\Delta |f|^2 + n |f|^2)\, \omega_M^n \\ &= \int_{M} (-4\phi_i |f| \nabla \phi_i\cdot \nabla |f| + \phi_i^2n |f|^2)\, \omega_M^n \\ &\leq \int_{M} \left( \frac{1}{2} \phi_i^2 |\nabla f|^2 + 8 |\nabla\phi_i|^2 |f|^2 + \phi_i^2 n |f|^2\right) \,\omega_M^n. \end{aligned} \end{align} $$

Letting $i\to \infty $ , and using that $|f|\in L^\infty $ , we get

(58) $$ \begin{align}\begin{aligned} \int_{M} |\nabla f|^2\, \omega_M^n \leq 2n\int_{M} |f|^2\, \omega_M^n. \end{aligned} \end{align} $$

We also have the following Bochner-type formula on M (see, e.g., La Nave-Tian-Zhang [Reference La Nave, Tian and Zhang39, Lemma 3.1]):

(59) $$ \begin{align}\begin{aligned} \Delta |\nabla f|^2 \geq \mathrm{Ric}_{\omega_M}(\nabla f, \nabla f) - (n+2) |\nabla f|^2 \geq -(n+2+|\lambda|) |\nabla f|^2, \end{aligned} \end{align} $$

where we are using the metrics $h^k, \omega _M$ as above.

Both (56) and (59) are of the form

(60) $$ \begin{align}\begin{aligned} \Delta v \geq - Av, \end{aligned} \end{align} $$

where v is a smooth $L^\infty $ function on M. We can argue using the cutoff functions $\phi _i$ , as in the proof of Lemma 11, to show that v satisfies this differential inequality on all of $\hat {X}$ in a weak sense, that is, for any Lipschitz test function $\rho \geq 0$ we have

(61) $$ \begin{align}\begin{aligned} \int_{M} (-\nabla \rho\cdot \nabla v + A\rho v)\, \omega_M^n \geq 0. \end{aligned} \end{align} $$

Using this, together with estimates for the heat kernel on $\hat {X}$ , we can obtain the required $L^\infty $ bound for $v = |f|^2$ and $v = |\nabla f|^2$ . More precisely, using [Reference Jiang, Li and Zhang38, Theorem 1.2], together with the RCD property in Proposition 15, we obtain an $L^2$ -bound for the heat kernel $H(x,y,1)$ on M, independently of k. Using (60), for any $x\in M$ we have

(62) $$ \begin{align}\begin{aligned} \frac{d}{dt} \int_M v(y) H(x, y,t)\, \omega_M^n(y) = \int_M v(y) \Delta_y H(x,y,t)\, \omega_M^n(y) \geq -A \int_M v(y)H(x,y,t)\, \omega_M^n(y), \end{aligned} \end{align} $$

so

(63) $$ \begin{align}\begin{aligned} v(x) \leq e^A \int_M v(y) H(x,y,1)\, \omega_M^n(y) \leq e^A C \Vert v\Vert_{L^2}, \end{aligned} \end{align} $$

as required.

Proof. First note that since $\hat {X}$ is a noncollapsed RCD space, we have a constant $\nu> 0$ such that $\mathrm {vol}\, B(\hat {x}, r)> \nu r^{2n}$ for all $r < 1$ . At the same time we can bound the volume of sublevel sets $U_k'\cap \{|s_k| < t\}$ from above, using the assumptions on $\omega $ . Indeed, on $U_k'$ we have $\omega ^n = F\eta ^n_{FS}$ , and $F\in L^p(X, \eta ^n_{FS})$ for some $p> 1$ . It follows that for any $t> 0$ we have

(66) $$ \begin{align}\begin{aligned} \mathrm{vol} (U_k'\cap \{ |s_k| < t\}, \omega^n) &= \int_{U_k'\cap \{|s_k| < t\}} \omega^n \\ &= \int_{U_k'\cap \{|s_k| < t\}} F\, \eta_{FS}^n \\ &\leq C_1 \mathrm{vol}(U_k'\cap \{|s_k| < t\}, \eta_{FS}^n)^{1/p'} \left(\int_{U_k'} F^p\, \eta_{FS}^n\right)^{1/p} \\ &\leq C_2 \mathrm{vol}(U_k'\cap \{|s_k| < t\}, \eta_{FS}^n)^{1/p'}\!, \end{aligned} \end{align} $$

for suitable constants $C_1, C_2$ independent of t, and $p'$ is the conjugate exponent of p. Since $|s_k|^{-\epsilon }\, \eta _{FS}^n$ is integrable for some $\epsilon> 0$ , it follows that we have a bound

(67) $$ \begin{align}\begin{aligned} \mathrm{vol}(U_k'\cap \{|s_k| < t\}, \eta_{FS}^n) \leq C_3 t^\epsilon, \end{aligned} \end{align} $$

and so in sum we have

(68) $$ \begin{align}\begin{aligned} \mathrm{vol} (U_k'\cap \{ |s_k| < t\}, \omega^n) \leq C_4 t^{\alpha},\end{aligned} \end{align} $$

for some $C_4, \alpha> 0$ independent of t. Given a small $r> 0$ such that $B(\hat {x}, r)\subset \hat {U}_k'$ , choose $t_r$ such that

(69) $$ \begin{align}\begin{aligned} C_4 t_r^\alpha = \frac{1}{2} \nu r^{2n}, \end{aligned} \end{align} $$

that is,

(70) $$ \begin{align}\begin{aligned} t_r = \left(\frac{\nu}{2C_4}\right)^{1/\alpha} r^{2n/\alpha} = c_5 r^{2n\alpha^{-1}}, \end{aligned} \end{align} $$

for suitable $c_5> 0$ . By our estimates for the volumes, we then have

(71) $$ \begin{align}\begin{aligned} \mathrm{vol}( B(\hat{x}, r) \cap \{|\hat{s}_k| \geq t_r\}) \geq \frac{1}{2}\nu r^{2n}, \end{aligned} \end{align} $$

and so

(72) $$ \begin{align}\begin{aligned} \int_{B(\hat{x}, r)} |\hat{s}_k|^2\, \omega^n &\geq \frac{c_5^2 r^{4n\alpha^{-1}}}{2} \nu r^{2n} = c_1 r^N,\end{aligned} \end{align} $$

for some $c_1, N> 0$ , independent of r, as required.

Proof. The proof is by contradiction, similarly to [Reference Ding26], based on the fact that on the Euclidean space $\mathbb {R}^{2n}$ every homogeneous harmonic function has integer degree.

Proof. Fix $\mu \in (N/2, N)$ such that $\mu \not \in \mathbb {Z}$ . If $\epsilon _0$ and $r_0$ are sufficiently small (depending on $\mu $ ), then the inequality (76) implies that for any $r \leq r_1$ we have

(78) $$ \begin{align}\begin{aligned} d_{GH}(B(\hat{x}, r), B(0_{\mathbb{R}^{2n}}, r)) < r\epsilon, \end{aligned} \end{align} $$

for the $\epsilon $ in Lemma 21, and so the conclusion of that Lemma holds. It follows that if

(79)

for some $r\leq r_1$ , then applying Lemma 21 inductively, we have

(80)

as long as $2^jr \leq r_1$ . Given any $r \leq r_1$ , if we let $\bar j$ denote the largest j such that $2^jr\leq r_1$ , then we obtain

(81)

where C is independent of r, but depends on the $L^2$ -norm of $\hat {s}_k$ on $B(\hat {x}, r_1)$ . Applying Lemma 20 we then have

(82) $$ \begin{align}\begin{aligned} 2^{-\bar j \mu} C\geq c_1^{1/2} (r/2)^{N/2}. \end{aligned} \end{align} $$

Since $2^{\bar j+1} r> r_1$ , it follows that $2^{-\bar j \mu } < (2r/r_1)^\mu $ , so

(83) $$ \begin{align}\begin{aligned} c_1^{1/2}(r/2)^{N/2} < (2r/r_1)^\mu C. \end{aligned} \end{align} $$

Since $\mu> N/2$ , this inequality implies a lower bound for r satisfying (79). The required conclusion (77) follows.

Proof. We will argue by contradiction, similarly to [Reference Liu and Székelyhidi42, Proposition 3.1] which in turn is based on Donaldson-Sun [Reference Donaldson and Sun27]. Suppose that no suitable $\epsilon _2$ exists. Then we have a sequence of points $\hat {x}_i$ , and integers $k_i> i$ such that the hypotheses are satisfied (with $\epsilon _2 = 1/i$ ). We will show that for sufficiently large i we have $\hat {x}_i \in X^{reg}$ by constructing holomorphic coordinates in a neighborhood of $\hat {x}_i$ .

By a slight abuse of notation we will write $\hat {U}_i, \hat {s}_i$ instead of $\hat {U}_{j_i}$ and $\hat {s}_{j_i}$ to simplify the notation. The assumptions imply that the rescaled balls

(86) $$ \begin{align}\begin{aligned} k_i^{1/2} B(\hat{x}_i, i k_i^{-1/2}) \to \mathbb{R}^{2n}, \end{aligned} \end{align} $$

in the pointed Gromov-Hausdorff sense. Using Lemma 21 together with the condition (85), we can extract a nontrivial limit of the normalized functions

(87)

Indeed, we have

(88)

and using Lemma 21 with some $\mu \in (N, 2N)$ , together with (85), implies that for sufficiently large i we have

(89)

for all $j \geq 0$ . In particular, viewed as functions on the rescaled balls $k_i^{1/2} B(\hat {x}_i, ik_i^{-1/2})$ , the $L^2$ norms of the $\tilde {s}_i$ are bounded independently of i on any R-ball. Using the gradient estimate, Proposition 12, it follows that up to choosing a subsequence, the functions $\tilde {s}_i$ converge locally uniformly to a harmonic function $\tilde {s}_\infty : \mathbb {R}^{2n}\to \mathbb {C}$ . As a consequence, $\tilde {s}_\infty $ is smooth, and because of the normalization (88), $\tilde {s}_\infty $ is nonzero.

Note that if we take a sequence of rescalings of $\mathbb {R}^{2n}$ with factors going to infinity, and consider the corresponding pullbacks of $\tilde {s}_\infty $ , normalized to have unit $L^2$ -norm on the unit balls, then this new sequence of harmonic functions will converge to the leading order homogeneous piece of the Taylor expansion of $\tilde {s}_\infty $ at the origin (up to a constant factor). This means that in the procedure above, up to replacing the integers $k_i$ by suitable larger integers, we can assume that the limit $\tilde {s}_\infty $ is in fact homogeneous.

Let us write $\Sigma = \tilde {s}_\infty ^{-1}(0)$ . Our next goal is to show that under the convergence in (86), the set $\mathbb {R}^{2n}\setminus \Sigma $ is the locally smooth limit of subsets of $X^{reg}$ , and that $\tilde {s}_\infty $ is actually a holomorphic function under an identification $\mathbb {R}^{2n} = \mathbb {C}^n$ . Then we will be able to follow the argument in the proof of [Reference Liu and Székelyhidi42, Proposition 3.1] with the cone $V = \mathbb {C}^n$ , but treating $\Sigma $ as the singular set.

Note that since $\tilde {s}_\infty $ is a nonzero harmonic function, the set $\mathbb {R}^{2n}\setminus \Sigma $ is open and dense in $\mathbb {R}^{2n}$ . Suppose that $V\subset \mathbb {R}^{2n}$ is an open subset such that $\bar {V}$ is compact and $|\tilde {s}_\infty |> 0$ on $\bar {V}$ . Then, because of the local uniform convergence of $\tilde {s}_i$ to $\tilde {s}_\infty $ , and the fact that the sets $\tilde {s}_i\not =0$ are contained in $X^{reg}$ , it follows that we have open subsets $V_i \subset \subset k_i^{-2}B(\hat {x}_i, ik_i^{-1/2})\cap X^{reg}$ , which converge in the Gromov-Hausdorff sense to V. The metrics on the $V_i$ are smooth noncollapsed Kähler-Einstein metrics, so using Anderson’s $\epsilon $ -regularity result [Reference Anderson2], up to choosing a subsequence, the complex structures on $V_i$ converge to a complex structure on V with respect to which the Euclidean metric is Kähler. Note that we do not yet know that $\mathbb {R}^{2n}\setminus \Sigma $ is connected, and in principle we may get different complex structures on different connected components. Our next goal is to show that the Hausdorff dimension of $\Sigma $ is at most $2n-2$ , which will show that the complement of $\Sigma $ is connected.

We can assume that the holomorphic functions $\tilde {s}_i$ on $V_i$ converge to a holomorphic function $\tilde {s}_\infty $ on V. Writing $\tilde {s}_\infty = u_\infty + \sqrt {-1} v_\infty $ , we therefore have $\langle \nabla u_\infty , \nabla v_\infty \rangle = 0$ and $|\nabla v_\infty | = |\nabla u_\infty |$ on $\mathbb {R}^{2n}\setminus \Sigma $ , and by density these relations extend to all of $\mathbb {R}^{2n}$ . We can assume that $u_\infty $ is nonconstant. Let $\alpha> 2n-2$ , and suppose that the Hausdorff measure $\mathcal {H}^\alpha (\Sigma )> 0$ . By Caffarelli-Friedman [Reference Caffarelli and Friedman9] (see also Han-Lin [Reference Han and Lin34]) we know that $\mathcal {H}^\alpha (\Sigma \cap |\nabla u_\infty |^{-1}(0)) =0$ , and so we can find an $\alpha $ -dimensional point of density q of $\Sigma \setminus |\nabla u_\infty |^{-1}(0)$ . Since $\nabla u_\infty (q)\not =0$ , it follows that $\nabla v_\infty (q)\not =0$ and $\langle \nabla v_\infty (q), \nabla u_\infty (q)\rangle = 0$ . Therefore in a neighborhood of q the set $\Sigma $ is a smooth $2n-2$ -dimensional submanifold, contradicting that q is an $\alpha $ -dimensional point of density. In conclusion $\dim _{\mathcal {H}} \Sigma \leq 2n-2$ , and so $\mathbb {R}^{2n-2}\setminus \Sigma $ is connected.

We can therefore assume that in the argument above the complex structure that we obtain on $\mathbb {R}^{2n}\setminus \Sigma $ agrees with the standard structure on $\mathbb {C}^n$ , and $\tilde {s}_\infty $ is a holomorphic function on $\mathbb {C}^n\setminus \Sigma $ , but since it is smooth, it is actually holomorphic on $\mathbb {C}^n$ . In particular $\tilde {s}_\infty ^{-1}(0)$ is a complex hypersurface defined by a homogeneous holomorphic function.

At this point we can closely follow the proof of [Reference Liu and Székelyhidi42, Proposition 3.1]), treating the zero set $\tilde {s}_\infty ^{-1}(0)$ as the singular set $\Sigma $ in [Reference Liu and Székelyhidi42]. The properties of the set $\Sigma $ that are used are that the tubular $\rho $ -neighborhood $\Sigma _\rho $ satisfies the volume bounds $\mathrm {vol}(\Sigma _\rho \cap B(0,R))\leq C_R \rho ^2$ , where the constant $C_R$ in our setting could depend on $R, \tilde {s}_\infty $ . In addition if $B(p, 2r) \in \mathbb {R}^{2n}\setminus \Sigma $ , then $B(p, r)$ is the Gromov-Hausdorff limit of balls $B(p_i, r) \subset (M, k_i\omega )$ in Kähler-Einstein manifolds, and so by Anderson’s result [Reference Anderson2] we have good holomorphic charts on the $B(p_i, r)$ for sufficiently large i, analogous to those in [Reference Liu and Székelyhidi42, Theorem 1.4]. The rest of the proof is then identical to the argument in the proof of [Reference Liu and Székelyhidi42, Proposition 3.1] (see also Donaldson-Sun [Reference Donaldson and Sun27]) to show that for sufficiently large i we can construct holomorphic sections $s_0, \ldots , s_n$ of $L^{k_i'}$ for suitable powers $k_i'$ , such that $\frac {s_1}{s_0}, \ldots , \frac {s_n}{s_0}$ define a generically one-to-one map from a neighborhood of $x_i = \hat {\Phi }_X(\hat {x}_i)$ in X to a subset of $\mathbb {C}^n$ . Since X is normal, it follows that the map is one-to-one, and so $x_i\in X^{reg}$ . Therefore $\hat {x}_i\in X^{reg}$ as claimed.

Proof. It is clear that $X^{reg}\subset \mathcal {R}_{\epsilon _3}(\hat {X})$ . To see the reverse inclusion, note that by Cheeger-Colding [Reference Cheeger and Colding12], and De Philippis-Gigli [Reference De Philippis and Gigli23] in the setting of noncollapsed RCD spaces, given the $\epsilon _2> 0$ in Proposition 23, there exists an $\epsilon _3> 0$ such that if $\hat {x} \in \mathcal {R}_{\epsilon _3}$ , then for all sufficiently large k (depending on $\hat {x}$ ), we have

(91) $$ \begin{align}\begin{aligned} d_{GH}\Big( B(\hat{x}, \epsilon_2^{-1}k^{-2}), B_{\mathbb{R}^{2n}}(0, \epsilon_2^{-1}k^{-2})\Big) < \epsilon_2 k^{-2}. \end{aligned} \end{align} $$

Using also Lemma 22 (and choosing $\epsilon _3$ smaller if necessary), we have the growth estimate (85). Proposition 23 then implies that $\hat {x}\in X^{reg}$ .

Proof. If $V =\mathbb {R}^{2n-2}\times C(S^1_\gamma )$ , then the singular set of V has capacity zero, and so Proposition 26 can be applied. Then, as in [Reference Chen, Donaldson and Sun16, Proposition 12], it follows that for sufficiently large j, we can find a biholomorphism $F_j$ from a neighborhood $\Omega _j$ of $\hat {x}_j$ to the unit ball $B(0,1)\subset \mathbb {C}^n$ . In particular $B(\hat {x}_j, \frac {1}{2}k_j^{-2}) \subset X^{reg}$ , and then the limit $\mathbb {R}^{2n-2}\times C(S^1_\gamma )$ of $(\hat {X}, k_j^2d_{\hat {X}}, \hat {x}_j)$ must be smooth at the origin. Therefore $\gamma =2\pi $ .

Proof of Theorem 17.

Using Propositions 10 and 27, and De Philippis-Gigli’s dimension estimate [Reference De Philippis and Gigli23] for the singular set (extending Cheeger-Colding [Reference Cheeger and Colding12]), it follows that the singular set of any iterated tangent cone of $\hat {X}$ has Hausdorff codimension at least 3. Using Proposition 24 we know that the singular set is closed, and so as in Donaldson-Sun [Reference Donaldson and Sun27, Proposition 3.5] we see that the singular set of any iterated tangent cone has capacity zero. In particular Proposition 26 can be applied to any $(V,o)$ that arises as a rescaled limit of $\hat {X}$ .

Suppose that $p\not = q$ are points in $\hat {X}$ . Applying Proposition 26 to tangent cones at $p, q$ , we can find sections $s_p$ and $s_q$ of some powers $L^{m_p}, L^{m_q}$ , such that $|s_p(p)|> |s_p(q)|$ , and $|s_q(q)|> |s_q(p)|$ . Taking powers we find that the sections $s_p^{m_q}$ and $s_q^{m_p}$ of $L^{m_pm_q}$ separate the points $p,q$ , and so the map $\hat {\Phi }_X$ is injective as required.

Proof. Suppose that $\hat {X}$ has a tangent cone of the form $\hat {X}_p = C(Z)\times \mathbb {R}^{2n-3}$ , where Z is two-dimensional. If Z had a singular point, necessarily with tangent cone $C(S^1_\gamma )$ for some $\gamma < 2\pi $ , then $\hat {X}$ would have an iterated tangent cone of the form $\mathbb {R}^{2n-2}\times C(S^1_\gamma )$ . This is ruled out by Proposition 27. Therefore Z is actually a smooth two-dimensional Einstein manifold with metric satisfying $\mathrm {Ric}(h) = h$ . This implies that Z is the unit 2-sphere, and it follows that $\hat {X}_p = \mathbb {R}^{2n}$ so that p is a regular point. Therefore the singular set of $\hat {X}$ coincides with $S_{2n-4}$ , as required.

Proof. The case when $\lambda =1$ is well known, going back to Tian [Reference Tian56] in the smooth setting, who proved a weaker version of properness. The properness in the form (99) was shown by Phong-Song-Sturm-Weinkove [Reference Phong, Song, Sturm and Weinkove48]. In the singular setting the result was shown in Darvas [Reference Darvas21, Theorem 2.2]. Note that we are assuming that X has discrete automorphism group and admits a Kähler-Einstein metric.

The cases $\lambda =0, -1$ are much easier (see Tian [Reference Tian57] or Song-Weinkove [Reference Song and Weinkove50, Theorem 1.2] for a similar result). For this, note that $\mathcal {J}_\omega \geq 0$ , and so when $\lambda \leq 0$ , we have

(100) $$ \begin{align}\begin{aligned} M_\omega(u) \geq \frac{1}{V} \int_{X^{reg}} \log\left(\frac{\omega_u^n}{\mu}\right) \omega_u^n. \end{aligned} \end{align} $$

At the same time, using Tian [Reference Tian54], we know that there are $\alpha , C_1> 0$ such that for all $u\in PSH_\omega (X)$ with $\sup _X u=0$ we have

(101) $$ \begin{align}\begin{aligned} \int_{Y} e^{-\alpha \pi^*u}\, \Omega < C_1,\end{aligned} \end{align} $$

and so with $p^{-1} + q^{-1} =1$ (such that F in (97) is in $L^p$ ) we have

(102) $$ \begin{align}\begin{aligned} \int_{X^{reg}} e^{-\alpha q^{-1} u}\, d\mu &= \int_{\pi^{-1}(X^{reg})} e^{-\alpha q^{-1} \pi^*u}\, \pi^*\mu \\ &= \int_{\pi^{-1}(X^{reg})} e^{-\alpha q^{-1} \pi^*u }\, F \Omega \\ &\leq \left(\int_{\pi^{-1}(X^{reg})} e^{-\alpha \pi^*u}\, \Omega\right)^{1/q} \left(\int_{\pi^{-1}(X^{reg})} F^p\Omega\right)^{1/p} \\ &\leq C_2. \end{aligned} \end{align} $$

Using the convexity of the exponential function we then have, as in [Reference Song and Weinkove50, Lemma 4.1],

(103) $$ \begin{align}\begin{aligned} \int_{X^{reg}} \log\left(\frac{\omega_u^n}{\mu}\right)\, \omega_u^n \geq \alpha q^{-1}\int_{X^{reg}} (-u)\, \omega_u^n - C_3, \end{aligned} \end{align} $$

for all $u\in PSH_\omega (X)$ with $\sup _X u =0$ . As the same time, if $\sup _X u =0$ and $u\in L^\infty $ , then we have

(104) $$ \begin{align}\begin{aligned} \int_{X^{reg}} (-u)\, \omega_u^n \geq \mathcal{J}_\omega(u). \end{aligned} \end{align} $$

To see this, note that

(105) $$ \begin{align}\begin{aligned} \int_{X^{reg}} (-u)\, \omega_u^n &= \int_0^1 \frac{d}{dt} \int_{X^{reg}} (-tu)\, \omega_{tu}^n\, dt \\ &= \int_0^1 \int_{X^{reg}} (-u)\, \omega_{tu}^n - n tu \sqrt{-1}\partial \bar\partial u\wedge \omega_{tu}^{n-1}\, dt \\ &\geq \int_0^1 n \int_{X^{reg}} u (\omega - \omega_{tu}) \wedge \omega_{tu}^{n-1}\, dt \\ &= \int_0^1 \frac{d}{dt}\mathcal{J}_\omega(tu)\, dt = \mathcal{J}_\omega(u). \end{aligned} \end{align} $$

So combining the estimates above we obtain (99).

Proof. For $u\in PSH_{\eta _\epsilon }(Y)\cap L^\infty $ with $\sup _Yu = 0$ , we have

(113) $$ \begin{align}\begin{aligned} -\mathcal{J}_{\eta_\epsilon, \mathrm{Ric}(\Omega)}(u) &= n \int_0^1 \int_Y (-u) (\mathrm{Ric}(\Omega) - c \eta_{\epsilon, tu})\wedge \eta_{\epsilon, tu}^{n-1} \\ &\geq - n\int_0^1 \int_Y (-u) (C \pi^*\omega + c \eta_{\epsilon, tu})\wedge \eta_{\epsilon, tu}^{n-1} \\ &\geq - C_1 n\int_0^1 \int_Y (-u) (\eta_{\epsilon} + \eta_{\epsilon, tu})\wedge \eta_{\epsilon, tu}^{n-1} \\ &\geq -C_2 J_{\eta_\epsilon}(u). \end{aligned} \end{align} $$

Note that since the entropy term is nonnegative, we have $M_{\eta _\epsilon , s} \geq \mathcal {J}_{\eta _\epsilon , s\eta _\epsilon - \mathrm {Ric}(\Omega )} $ and also

(114) $$ \begin{align}\begin{aligned} \mathcal{J}_{\eta_\epsilon, s\eta_\epsilon - \mathrm{Ric}(\Omega)} = s \mathcal{J}_{\eta_\epsilon, \eta_\epsilon} - \mathcal{J}_{\eta_\epsilon,\mathrm{Ric}(\Omega)}. \end{aligned} \end{align} $$

It follows that for $s> C_2 + 1$ ,

(115) $$ \begin{align}\begin{aligned} M_{\eta_\epsilon, s}(u) \geq J_{\eta_\epsilon}(u). \end{aligned} \end{align} $$

Proof. The estimates (1) are shown in [Reference Zheng62, Proposition 5.12], the estimate (2) is in [Reference Zheng62, Proposition 5.15], and the estimate (3) is [Reference Zheng62, Proposition 5.18]. Since the notation in [Reference Zheng62] is quite different, and they consider a more general situation including conical singularities along a divisor, we recall their setup. In [Reference Zheng62, Section 5], the author considers the equations

(122) $$ \begin{align}\begin{aligned} F &= \log \frac{\eta_{\epsilon, u}^n}{\Omega}, \\ \Delta_{\eta_{\epsilon, u}} F &= \mathrm{tr}_{\eta_{\epsilon, u}} \Theta - c_{s,\epsilon}. \end{aligned} \end{align} $$

Here $\Omega $ is a smooth volume form on Y as above, with Ricci curvature $\mathrm {Ric}(\Omega ) = \theta $ , and we define $\Theta = \theta - s \eta _\epsilon $ . The coupled equations then imply

(123) $$ \begin{align}\begin{aligned} -R(\eta_{\epsilon, u}) + \mathrm{tr}_{\eta_{\epsilon, u}} \theta = \mathrm{tr}_{\eta_{\epsilon, u}} (\theta - s\eta_{\epsilon, u}) - c_{s,\epsilon}, \end{aligned} \end{align} $$

which is (118). Note that in [Reference Zheng62] the resolution is called X and the singular variety is Y, which is the opposite of our notation. There is also an additional function f which we take to be zero. Zheng considers a semipositive form $\omega _{sr}$ on Y, which we can take to be $\pi ^*\omega $ , and a Kähler form $\omega _K$ on Y, which we take to be $\eta _Y$ , so $\eta _\epsilon = \pi ^*\omega + \epsilon \eta _Y$ is what Zheng calls $\omega _\epsilon $ .

In order to deal with the degeneracy of $\eta _\epsilon $ as $\epsilon \to 0$ , Zheng uses the technique of Tsuji [Reference Tsuji58], relying on the fact that if we choose a suitable smooth metric on the line bundle $\mathcal {O}(E)$ for a divisor supported on the exceptional set of $\pi $ , then for any $a> 0$ the current $\pi ^*\omega + a \sqrt {-1}\partial \bar \partial \log |s_E|^2$ dominates a Kähler form on Y, where $s_E$ vanishes along E. We can assume that on $Y\setminus E$ we have

(124) $$ \begin{align}\begin{aligned} \eta_Y = \pi^*\omega + a \sqrt{-1}\partial \bar\partial \log |s_E|^2. \end{aligned} \end{align} $$

We apply [Reference Zheng62, Proposition 5.12], to deduce the estimates (1). For this we need to check the condition that $e^{-\phi _l} \in L^{p_0}$ for some $p_0> 1$ , where $\phi _l$ in Zheng’s notation is defined in his Lemma 5.6. Since in that Lemma $\phi _{\theta _\epsilon }$ is uniformly bounded, it is enough to check the integrability of $e^{-p_0 \tilde {\phi }_l}$ where we define

(125) $$ \begin{align}\begin{aligned} \tilde{\phi}_l = (\inf_{(X, \eta_Y)}\Theta) (- a\log |s_E|^2). \end{aligned} \end{align} $$

We claim that the infimum $\inf _{(X, \eta _Y)}\Theta $ is bounded below, independently of the choice of small $a> 0$ (note that the choice of a affects the definition of $\eta _Y$ and so also $\Theta $ ). To see this, we use the condition $\theta = \mathrm {Ric}(\Omega ) \geq - C\pi ^*\omega $ , so that we have

(126) $$ \begin{align}\begin{aligned} \Theta &= \theta - s\eta_\epsilon \geq -(C + s)\pi^*\omega - s\epsilon \eta_Y \\ &= - C' \eta_Y, \end{aligned} \end{align} $$

for $C'$ depending on C and $s_0$ . Here we also used that if $a> 0$ is sufficiently small, then $\eta _Y> \frac {1}{2}\pi ^*\omega _X$ . It follows from this that

(127) $$ \begin{align}\begin{aligned} - \tilde{\phi}_l \leq - C'a \log |s_E|^2, \end{aligned} \end{align} $$

and so if $a> 0$ is sufficiently small, then $e^{- \tilde {\phi }_l} \in L^{p_0}$ for $p_0> 1$ , as required in Zheng’s Proposition 5.12. The conclusion is the estimates (1). Note that the quantities that the estimate in Proposition 5.12 depends on are all uniformly bounded in $s, \epsilon $ in our setting. Similarly, Propositions 5.15 and 5.18 imply the estimates (2) and (3).

Note that the $L^p$ -bound on the trace of $\eta _{\epsilon , u}$ implies higher order estimates for u on compact sets away from E, using Chen-Cheng’s local estimate [Reference Chen and Cheng15, Proposition 6.1]. This leads to the $\{p, a_j\}$ -boundedness of u. See also [Reference Pan and Tô47, Theorem C] for similar estimates.

Proof. We argue by contradiction. Suppose that we have a sequence $\epsilon _i \to 0$ , and $u_i \in PSH_{\eta _{\epsilon _i}}(Y)$ that are $\{p, a_j\}_{j\geq 0}$ -bounded, such that

(129) $$ \begin{align}\begin{aligned} M_{\eta_{\epsilon_i}}(u_i) \leq \delta \mathcal{J}_{\eta_\epsilon}(u) - B_2, \end{aligned} \end{align} $$

for $B_2$ to be determined below. Up to choosing a subsequence we can assume that $u_i \to u_\infty $ in $L^1$ and also in $C^{3,\alpha }$ on compact sets away from the exceptional divisor E. We have $u_\infty \in PSH_{\pi ^*\omega }(Y)$ , and we have an identification $PSH_{\pi ^*\omega }(Y) = PSH_{\omega }(X)$ . We will next show that in terms of F in (97) we have

(130) $$ \begin{align}\begin{aligned} M_{\eta_{\epsilon_i}}(u_i) &\to M_{\omega}(u_\infty) + \int_Y \log F\, \eta_0^n, \\ \mathcal{J}_{\eta_{\epsilon_i}}(u_i) &\to \mathcal{J}_\omega(u_\infty). \end{aligned} \end{align} $$

Let us first consider the relevant entropy terms. Note that

(131) $$ \begin{align}\begin{aligned} \int_Y \log \left(\frac{\eta_{{\epsilon_i}, u_i}^n}{\Omega}\right) \eta_{\epsilon_i, u_i}^n = \int_Y \log\left(\frac{ \eta^n_{\epsilon_i, u_i}}{\Omega}\right) \frac{\eta_{\epsilon_i, u_i}^n}{\Omega}\, \Omega. \end{aligned} \end{align} $$

Our assumptions mean that the integrand has a uniform $L^p(\Omega )$ -bound for some $p> 1$ . Using this, and the $C^{3,\alpha }$ -convergence $u_i\to u_\infty $ on compact sets away from E, it follows that

(132) $$ \begin{align}\begin{aligned} \int_Y \log \left(\frac{\eta_{{\epsilon_i}, u_i}^n}{\Omega}\right) \eta_{\epsilon_i, u_i}^n &\to \int_Y \log \left(\frac{\eta_{0, u_\infty}^n}{\Omega}\right) \eta_{0, u_\infty}^n. \end{aligned} \end{align} $$

Using (97) we have

(133) $$ \begin{align}\begin{aligned} \int_Y \log \left(\frac{\eta_{0,u_\infty}^n}{\Omega}\right) \eta_{0, u_\infty}^n &= \int_{X^{reg}} \log \left(\frac{\omega_{u_\infty}^n}{\mu}\right) \omega_{u_\infty}^n + \int_Y \log F\, \eta_{0,u_\infty}^n. \end{aligned} \end{align} $$

The last term can be computed by writing

(134) $$ \begin{align}\begin{aligned} \int_Y \log F\, \eta_{0,u_\infty}^n &= \int_Y \log F\, \eta_0^n + \int_0^1 \frac{d}{dt} \int_Y \log F\, \eta_{0, tu_\infty}^n\, dt \\ &= \int_Y \log F\, \eta_0^n + \int_0^1 n \int_Y u_\infty \sqrt{-1}\partial \bar\partial \log F\, \wedge \eta_{0,t u_\infty}^{n-1}\,dt \\ &= \int_Y \log F\, \eta_0^n + \int_0^1 n \int_Y u_\infty (\mathrm{Ric}(\Omega) - \mathrm{Ric}(\pi^*\mu))\wedge \eta_{0,u_\infty}^{n-1}\,dt \\ &= \int_Y \log F\, \eta_0^n + \mathcal{J}_{\eta_0, \mathrm{Ric}(\Omega)}(u_\infty) -\lambda \mathcal{J}_{\omega}(u_\infty). \end{aligned} \end{align} $$

For the last step note that $\eta _0$ vanishes along E, so although $\mathrm {Ric}(\pi ^*\mu )$ has current contributions along E, the only part that survives in the integral is $\mathrm {Ric}(\mu ) = \lambda \omega $ on X. In conclusion we have that

(135) $$ \begin{align}\begin{aligned} \int_Y \log \left(\frac{\eta_{{\epsilon_i}, u_i}^n}{\Omega}\right) \eta_{\epsilon_i, u_i}^n &\to \int_{X^{reg}} \log\left(\frac{\omega^n_{u_\infty}}{\mu}\right)\, \omega_{u_\infty}^n + \int_Y \log F\, \eta_0^n \\ &\qquad + \mathcal{J}_{\eta_0, \mathrm{Ric}(\Omega)}(u_\infty) - \lambda \mathcal{J}_\omega(u_\infty). \end{aligned} \end{align} $$

Next we consider the $\mathcal {J}$ -functional terms. Consider a general smooth, closed (1,1)-form $\alpha $ on Y. We claim that we have $\mathcal {J}_{\eta _{\epsilon _i}, \alpha }(u_i) \to \mathcal {J}_{\eta _0, \alpha }(u_\infty )$ . Using the variational definition of $\mathcal {J}$ , the local $C^{3,\alpha }$ -convergence, and the uniform $L^\infty $ -bound for the $u_i$ , it is enough to show that for every $\kappa> 0$ there is a compact set $K\subset Y\setminus E$ , such that

(136) $$ \begin{align}\begin{aligned} \int_{Y\setminus K} \eta_1 \wedge \eta_{\epsilon_i, u_i}^{n-1} + \int_{Y\setminus K} \eta_{\epsilon_i, u_i}^n < \kappa, \text{ for all } i. \end{aligned} \end{align} $$

To see this, let $h= -\log |s_E|^2$ , where $s_E$ is a section of the line bundle $\mathcal {O}(E)$ over Y vanishing along the exceptional divisor E, and we use a smooth metric on $\mathcal {O}(E)$ . We have

(137) $$ \begin{align}\begin{aligned} \sqrt{-1}\partial \bar\partial h = \chi - [E], \end{aligned} \end{align} $$

where $\chi $ is a smooth form on Y. We can assume that $h\geq 0$ , and note that $h\to \infty $ along E. We show by induction that for each $k=0, \ldots , n$ there is a constant $C_k> 0$ , independent of i, such that

(138) $$ \begin{align}\begin{aligned} \int_Y h \eta_1^{n-k}\wedge \eta_{\epsilon_i, u_i}^k \leq C_k. \end{aligned} \end{align} $$

For $k=0$ this is clear since h has logarithmic singularities. Suppose that the bound has been established for a value of k. Then

(139) $$ \begin{align}\begin{aligned} \int_Y h \eta_1^{n-k-1}\wedge \eta_{\epsilon_i, u_i}^{k+1} &= \int_Y h \eta_1^{n-k-1} \wedge (\eta_{\epsilon_i} + \sqrt{-1}\partial \bar\partial u_i)\wedge \eta_{\epsilon_i, u_i}^k \\ &= \int_Y h \eta_{1}^{n-k-1}\wedge \eta_{\epsilon_i}\wedge \eta_{\epsilon_i, u_i}^k + \int u_i\sqrt{-1}\partial \bar\partial h \wedge \eta_1^{n-k-1}\wedge \eta_{\epsilon_i, u_i}^k \\ &\leq \int_Y h \eta_{1}^{n-k}\wedge \eta_{\epsilon_i, u_i}^k + \int_Y u_i \chi\wedge \eta_1^{n-k-1}\wedge \eta_{\epsilon_i, u_i}^k - \int_E u_i \eta_1^{n-k-1}\wedge \eta_{\epsilon_i, u_i}^k \\ &\leq C_k(1+C) - \int_E u_i \eta_1^{n-k-1}\wedge \eta_{\epsilon_i, u_i}^k \\ &\leq C_k(1+ C) + C', \end{aligned} \end{align} $$

where $C, C'$ depend on $\chi $ and the uniform $L^\infty $ bound for $u_i$ .

Since $h\to \infty $ along E, it follows from (138) that for any $\kappa> 0$ we can find a compact set $K\subset Y\setminus E$ such that (136) holds. It follows that

(140) $$ \begin{align}\begin{aligned} \mathcal{J}_{\eta_{\epsilon_i, -\mathrm{Ric}(\Omega)}}(u_i) \to \mathcal{J}_{\eta_{0, -\mathrm{Ric}(\Omega)}}(u_\infty), \end{aligned} \end{align} $$

and also

(141) $$ \begin{align}\begin{aligned} \mathcal{J}_{\eta_{\epsilon_i}}(u_i) \to \mathcal{J}_{\omega}(u_\infty). \end{aligned} \end{align} $$

From this, together with (135), we have

(142) $$ \begin{align}\begin{aligned} M_{\eta_{\epsilon_i}}(u_i) &\to \int_{X^{reg}} \log\left( \frac{\omega_{u_\infty}^n}{\mu}\right)\, \omega_{u_\infty}^n -\lambda \mathcal{J}_\omega(u_\infty) + \int_Y \log F\, \eta_0^n \\ &= M_{\omega}(u_\infty) + \int_Y \log F\, \eta_0^n. \end{aligned} \end{align} $$

From (129) we therefore get

(143) $$ \begin{align}\begin{aligned} M_{\omega}(u_\infty) + \int_Y \log F\, \eta_0^n \leq \delta\mathcal{J}_\omega(u_\infty) - B_2. \end{aligned} \end{align} $$

Choosing $B_2 = B - \int _Y \log F\, \eta _0^n$ for the B in Proposition 29, we get a contradiction.

Proof of Theorem 3.

We will choose suitable $p> 0, a_j > 0$ shortly. By Proposition 33, for a given $p, a_j$ we have some $\epsilon _1> 0$ such that once $\epsilon < \epsilon _1$ and for any $s\geq 0$ , we have

(144) $$ \begin{align}\begin{aligned} M_{\eta_\epsilon, s}(u) \geq M_{\eta_\epsilon}(u)> \delta \mathcal{J}_{\eta_\epsilon}(u)- B_2, \end{aligned} \end{align} $$

for $\{p. a_j\}$ -bounded potentials u. Recall that $\delta , B_2$ do not depend on $\{p, a_j\}$ . For small $\kappa> 0$ we have

(145) $$ \begin{align}\begin{aligned} M_{\eta_\epsilon, s}(u) &\geq \kappa \int_Y \left(\frac{\eta_{\epsilon, u}^n}{\Omega}\right)\, \eta_{\epsilon, u}^n + \kappa \mathcal{J}_{\eta_\epsilon, s\eta_\epsilon-\mathrm{Ric}(\Omega)}(u) + (1-\kappa)\delta \mathcal{J}_{\eta_\epsilon}(u) - (1-\kappa)B_2 \\ &= \kappa \int_Y \left(\frac{\eta_{\epsilon, u}^n}{\Omega}\right)\, \eta_{\epsilon, u}^n + (\kappa s + (1-\kappa)\delta) \mathcal{J}_{\eta_\epsilon}(u) + \kappa \mathcal{J}_{\eta_\epsilon, -\mathrm{Ric}(\Omega)}(u) - (1-\kappa)B_2. \end{aligned} \end{align} $$

If $\kappa $ is chosen sufficiently small (depending on $\delta $ ), then by Lemma 30 we find that

(146) $$ \begin{align}\begin{aligned} M_{\eta_\epsilon, s}(u) &\geq \kappa \int_Y \log \left(\frac{\eta_{\epsilon, u}^n}{\Omega}\right)\, \eta_{\epsilon, u}^n - B_2. \end{aligned} \end{align} $$

We also have

(147) $$ \begin{align}\begin{aligned} M_{\eta_\epsilon, s}(0) = \int_Y \log \left(\frac{\eta_{\epsilon}^n}{\Omega}\right)\, \eta_{\epsilon}^n < C_3, \end{aligned} \end{align} $$

for a constant $C_3> 0$ independent of $\epsilon $ . Since twisted cscK metrics minimize the twisted Mabuchi K-energy, it follows that if $\eta _{\epsilon , u} \in [\eta _\epsilon ]$ is a twisted cscK metric, then we have $M_{\eta _\epsilon , s}(u) < C_3$ . From (146) we get

(148) $$ \begin{align}\begin{aligned} \int_Y \log \left(\frac{\eta_{\epsilon, u}^n}{\Omega}\right)\, \eta_{\epsilon, u}^n \leq \kappa^{-1}(C_3 + B_2), \end{aligned} \end{align} $$

and in particular the entropy of $\eta _{\epsilon , u}$ is bounded independently of $\epsilon $ . We apply Proposition 32. As long as $s\leq s_0$ , for the $s_0$ determined by Lemma 30, we find that if $\eta _{\epsilon , u} = \eta _\epsilon + \sqrt {-1}\partial \bar \partial u$ is a solution of the twisted cscK equation

(149) $$ \begin{align}\begin{aligned} R(\eta_{\epsilon, u}) - s\,\mathrm{tr}_{\eta_{\epsilon, u}} \eta_\epsilon = \mathrm{const.}, \end{aligned} \end{align} $$

then u is $\{p, a_j\}$ -bounded, for suitable $p, a_j$ , determined by $s_0$ and the entropy bound (148). From now we fix this choice of $p, a_j$ .

We can now use a continuity method to show that if $\epsilon < \epsilon _1$ , for the $\epsilon _1$ determined by $\{p, a_j\}$ , for all $s\in [0,s_0]$ we can solve the twisted cscK equation (149). To see this, let us fix $\epsilon < \epsilon _1$ , and set

(150) $$ \begin{align}\begin{aligned} S = \{ s\in [0,s_0]\, :\, \text{ the equation (149) has a solution}\}. \end{aligned} \end{align} $$

We have $s_0\in S$ , and it follows from the implicit function theorem that S is open. To see that it is closed, note that the twisted cscK metrics for $s\in S$ automatically satisfy the entropy bound (148). Using the main estimates of Chen-Cheng [Reference Chen and Cheng15], we find that the potentials of the corresponding twisted cscK metrics satisfy a priori $C^k$ -estimates, and the metrics are bounded below uniformly (these estimates depend on $\epsilon $ , but now $\epsilon $ is fixed). It follows that S is closed.

It follows that for sufficiently small $\epsilon> 0$ the classes $[\eta _\epsilon ]$ on Y admit cscK metrics. The estimates required by Definition 2 follow from Proposition 32.

Proof. We will argue by contradiction. If the stated result is not true, then we can find a sequence $\hat {X}_i$ , and a sequence of singular points $p_i\in \hat {X}_i$ with tangent cones $V_{p_i}$ such that $V_{p_i}\to \mathbb {R}^{2n}$ in the pointed Gromov-Hausdorff sense.

We will prove a more general statement about almost smooth metric measure spaces in the sense of Definition 5, of any dimension, which satisfy the following conditions.

Note that by Propositions 24 and 27, the (iterated) tangent cones of the spaces $\hat {X}_i$ satisfy Condition ( $\ast $ ). Moreover, if a space $V = W \times \mathbb {R}^j$ satisfies Condition ( $\ast $ ), then so does W, and so do the tangent cones of V.

We argue by induction on the dimension to show that if a sequence of k-dimensional cones $V_j$ satisfies Condition ( $\ast $ ), and $V_j \to \mathbb {R}^k$ in the pointed Gromov-Hausdorff sense, then $V_j = \mathbb {R}^k$ for sufficiently large i. For $k=2$ this follows directly from Condition ( $\ast $ ).