Proof. Let \(\mu=(\mu_1,\cdots,\mu_n)\) be the eigenvalues of \(\chi_{\underline{u}}\) with respect to \(\omega\), ordered so that \(\mu_1\ge \cdots\ge \mu_n\). Define \[\theta_{i}=\mathrm{arccot}\mu_{i}\in (0, \pi),\] which is the principal argument of \(\mu_i+\sqrt{-1}\).

We first prove that the subsolution condition implies ?? and ?? . By the subsolution condition, for every \(j\), \[\sum_{\substack{1\le i\le n,\,i\neq j}}\arctan\mu_i>(n-3)\frac{\pi}{2}.\] Summing these inequalities over \(j=1,\dots,n\) yields ?? .

Since \(\theta_i=\frac{\pi}{2}-\arctan\mu_i\), the subsolution is equivalent to \[\sum_{\substack{1\le i\le n,\,i\neq j}}\theta_i<\pi,\quad \forall\;1\le j\le n.\] Hence the principal argument of \(\prod_{\substack{1\le i\le n\\ i\neq j}}(\mu_i+\sqrt{-1})\) is \(\sum_{\substack{1\le i\le n\\ i\neq j}}\theta_i\in(0,\pi)\), so its imaginary part is positive. This proves ?? .

Conversely, we show that conditions ?? and ?? imply the subsolution 5 . From \(\theta_1\le\cdots\le\theta_n\) we obtain \[\theta_n\ge \frac{1}{n-1}\sum_{i=2}^{n-1}\theta_i.\] Consequently, \[\sum_{i=2}^{n-1}\theta_i \le \frac{n-1}{n}\sum_{i=2}^{n}\theta_i <\frac{n-1}{n}\sum_{i=1}^{n}\theta_i<\pi,\] where the last inequality follows from ?? . Thus \[\mathrm{Im}\prod_{i=2}^{n-1}(\mu_i+\sqrt{-1})>0.\] Using ?? we have \[0<\mathrm{Im}\prod\limits_{i=2}^{n}(\mu_i+\sqrt{-1}) =\mu_n \mathrm{Im}\prod_{i=2}^{n-1}(\mu_i+\sqrt{-1})+ \mathrm{Re}\prod_{i=2}^{n-1}(\mu_i+\sqrt{-1}).\] This implies \[\cot\Bigl(\sum_{i=2}^{n-1}\theta_i\Bigr)>\cot(\pi-\theta_n).\] Since \(\cot\) is strictly decreasing on \((0,\pi)\) and \(\sum\limits_{i=2}^{n-1}\theta_i<\pi\), we deduce \[\sum_{i=2}^{n-1}\theta_i<\pi-\theta_n \quad \text{i.e.} \quad \sum_{i=2}^{n}\theta_i<\pi,\] which proves the subsolution condition for \(j=1\), the ordering then gives it for all \(j\). ◻

Proof. By a direct computation, we have \[(\chi + t\omega + \sqrt{-1}\omega)^n = (\chi + \sqrt{-1}\omega)^n + t\, n (\chi + \sqrt{-1}\omega)^{n-1} \wedge \omega + O(t^2).\] Since \(\mathrm{Re}\int_M (\chi + \sqrt{-1}\omega)^n < 0\), it follows that, for sufficiently small \(t\), there exists a constant \(C_0>0\) such that \[\mathrm{Re}\int_M (\chi + t\omega + \sqrt{-1}\omega)^n < \mathrm{Re}\int_M (\chi + \sqrt{-1}\omega)^n + C_0t < 0.\]

We now show that the imaginary part is positive. The subsolution condition implies the positivity \[\mathrm{Im}\, (\chi_{\underline{u}} + \sqrt{-1}\omega)^{n-1} > 0.\] Hence, \[\begin{align} \mathrm{Im}\int_M (\chi + \sqrt{-1}\omega)^{n-1} \wedge \omega = \mathrm{Im}\int_M (\chi_{\underline{u}} + \sqrt{-1}\omega)^{n-1} \wedge \omega > 0. \end{align}\] Consequently, for \(t\) small enough \[\begin{align} \mathrm{Im}\int_M (\chi + t\omega + \sqrt{-1}\omega)^n &= t\, \mathrm{Im}\int_M (\chi + \sqrt{-1}\omega)^{n-1} \wedge \omega + O(t^2) > 0. \end{align}\]

Therefore, there exists a uniform constant \(C > 0\), depending only on \(\chi\) and \(\omega\), such that the principal argument \(\hat{\theta}(t)\) of \(\int_M (\chi + t\omega + \sqrt{-1}\omega)^n\) satisfies \[\hat{\theta}(t) = \pi - \arctan\!\Bigl( \frac{t\, \mathrm{Im}\int_M (\chi + \sqrt{-1}\omega)^{n-1} \wedge \omega + O(t^2)}{-\mathrm{Re}\int_M (\chi + \sqrt{-1}\omega)^n + O(t)} \Bigr) \in (\pi - 2Ct,\; \pi - Ct)\] for all sufficiently small \(t > 0\). ◻

Proof. Given that \(A(\mu) > A_0 > (n-3)\frac{\pi}{2}\), we can choose a small positive constant \(\delta > 0\) such that \[A(\mu - 2\delta \mathbf{1}) > \frac{A_0 + (n-3)\frac{\pi}{2}}{2},\] where \(\mathbf{1} = (1, \ldots, 1) \in \mathbb{R}^n\).

For \(\lambda \in \mathbb{R}^n\) satisfying \(\theta(\lambda) = \theta(t)\), assume without loss of generality that \(\lambda_1 \ge \cdots \ge \lambda_n\). Recall that the vector \[\mathbf{n} = \Bigl( \frac{1}{1+\lambda_1^2}, \cdots, \frac{1}{1+\lambda_n^2} \Bigr)\] is an inward normal to the hypersurface \(\partial\Gamma^{\theta(t)}\) at \(\lambda\). Let \(T_{\lambda}\) denote the tangent plane to \(\partial\Gamma^{\theta(t)}\) at \(\lambda\).

Case 1: \(\bigl((\mu - \delta\mathbf{1}) - \lambda\bigr) \cdot \mathbf{n} \geq 0\). In this case, we have \[\begin{align} \label{case1} \sum_{i=1}^n \frac{\mu_i - \lambda_i}{1 + \lambda_i^2} \ge \delta \sum_{i=1}^n \frac{1}{1 + \lambda_i^2}. \end{align}\tag{11}\]

Case 2: \(\bigl((\mu - \delta\mathbf{1}) - \lambda\bigr) \cdot \mathbf{n} < 0\). We claim that there exists a positive constant \(s\) such that \(\mu - \delta\mathbf{1} + s e_1 \in \Gamma^{\theta(t)}\). Indeed, observe that \[\begin{align} \theta(\mu - \delta\mathbf{1} + s e_1) &= \arctan (\mu_1 - \delta + s) + \sum_{i=2}^n \arctan (\mu_i - \delta) \\ &> \arctan (\mu_1 - \delta + s) + A(\mu - \delta\mathbf{1}) \\ &\ge \arctan (\mu_1 - \delta + s) + \frac{1}{2}\Bigl(A_0 + (n-3)\frac{\pi}{2}\Bigr). \end{align}\] To ensure \(\theta(\mu - \delta\mathbf{1} + s e_1) > \theta(t)\), it suffices to choose \(s\) satisfying \[\arctan (\mu_1 - \delta + s) > \theta(t) - \frac{1}{2}\Bigl(A_0 + (n-3)\frac{\pi}{2}\Bigr).\] Take \[s = |\delta - \mu_1| + \tan\Bigl(\frac{\pi}{2} - \frac{1}{4}\bigl(A_0 - (n-3)\frac{\pi}{2}\bigr)\Bigr),\] and assume \(0 < t < t_0\) with \(t_0 = \frac{1}{8C}\bigl(A_0 - (n-3)\frac{\pi}{2}\bigr)\). Then the desired inequality holds.

The convexity of \(\Gamma^{\theta(t)}\) implies that the points \(\mu - \delta\mathbf{1}\) and \(\mu - \delta\mathbf{1} + s \mathbf{e}_1\) lie on opposite sides of the tangent plane \(T_{\lambda}\). Hence, there exists \(a_0 \in (0, 1)\) such that the convex combination \[a_0(\mu - \delta\mathbf{1} + s \mathbf{e}_1) + (1-a_0)(\mu - \delta\mathbf{1}) = \mu - \delta\mathbf{1} + a_0 s \mathbf{e}_1\] lies on \(T_{\lambda}\).

We now show that \(\mu - \delta\mathbf{1} + a_0 s \mathbf{e}_1 + s \mathbf{e}_1 - \delta \mathbf{e}_n \in \Gamma^{\theta(t)}\). Indeed, \[\begin{align} &\theta\bigl(\mu - \delta\mathbf{1} + a_0 s \mathbf{e}_1 + s \mathbf{e}_1 - \delta \mathbf{e}_n\bigr) \\ &= \arctan (\mu_1 - \delta + a_0 s + s) + \sum_{i=2}^{n-1} \arctan (\mu_i - \delta) + \arctan (\mu_n - 2\delta) \\ &\ge \arctan (\mu_1 - \delta + a_0 s + s) + A(\mu - 2\delta\mathbf{1}) \\ &\ge \arctan (\mu_1 - \delta + s) + \frac{1}{2}\Bigl(A_0 + (n-3)\frac{\pi}{2}\Bigr) \\ &> \theta(t). \end{align}\] Thus, we have \[\begin{align} 0 &\le \bigl((\mu - \delta\mathbf{1} + a_0 s \mathbf{e}_1 + s \mathbf{e}_1 - \delta \mathbf{e}_n) - (\mu - \delta\mathbf{1} + a_0 s \mathbf{e}_1)\bigr) \cdot \mathbf{n} \\ &= (s \mathbf{e}_1 - \delta \mathbf{e}_n) \cdot \mathbf{n} \\ &= \frac{s}{1+\lambda_1^2} - \frac{\delta}{1+\lambda_n^2}, \end{align}\] which is equivalent to \[\label{eq:ineq95lambda195lambdan} \frac{1}{1+\lambda_1^2} \ge \frac{\delta}{s(1+\lambda_n^2)}.\tag{12}\]

By Lemma 2, we have \(\lambda_1 \ge \cdots \ge \lambda_{n-1} \ge |\lambda_n|\). Consequently, for each \(j = 1, \dots, n\), \[\frac{1}{1+\lambda_n^2} \ge \frac{1}{1+\lambda_j^2}.\] Combining with 12 , we obtain \[\frac{1}{1+\lambda_1^2} \ge \frac{\delta}{s(1+\lambda_j^2)}, \quad \mathrm{for all 1\leq j\leq n.}\] Summing over \(j = 1, \dots, n\) yields \[\frac{n}{1+\lambda_1^2} \ge \frac{\delta}{s} \sum_{j=1}^n \frac{1}{1+\lambda_j^2},\] and therefore \[\frac{1}{1+\lambda_1^2} \ge \frac{\delta}{n s} \sum_{j=1}^n \frac{1}{1+\lambda_j^2}.\] Since \(\lambda_1^2 \ge \lambda_j^2\) for all \(j\), we have \[\begin{align} \label{case2} \frac{1}{1+\lambda_j^2} \ge \frac{1}{1+\lambda_1^2} \ge \frac{\delta}{n s} \sum_{i=1}^n \frac{1}{1+\lambda_i^2}. \end{align}\tag{13}\]

In view of 11 and 13 , choosing \(\delta_0 = \frac{\delta}{n s}\) completes the proof. ◻

Proof. Let \(\mu = (\mu_1, \dots, \mu_n)\) be the eigenvalues of \(\chi_{\underline{u}}\) with respect to \(\omega\), arranged so that \(\mu_1 \ge \cdots \ge \mu_n\).

Fix \(z \in M\). We choose normal coordinates at \(z\) such that \[g_{i\bar{j}} = \delta_{ij}, \quad \chi_{t, u^{t}} = \sum_{i=1}^n \lambda_i \sqrt{-1} \, dz^i \wedge d\bar{z}^i \quad \text{with} \quad \lambda_1 \ge \cdots \ge \lambda_n,\] where \(\chi_{t, u^{t}} = \chi + t\omega + \sqrt{-1} \partial \bar{\partial} u^t\). In these coordinates, at \(z\) we have \[\begin{align} F^{i\bar{j}} = \frac{\delta_{ij}}{1+\lambda_i^2}. \end{align}\]

Let \(\delta_0\) and \(t_0\) be the uniform constants given by Lemma 7. We now prove the lemma by distinguishing two cases.

Case 1: \(\displaystyle\sum_{i} F^{i\bar{i}}(\mu_i - \lambda_i) \ge \delta_0 \sum_{i} F^{i\bar{i}}\).

Since \(F^{1\bar{1}} \le \cdots \le F^{n\bar{n}}\) and \(\mu_1 \ge \cdots \ge \mu_n\), the Schur–Horn theorem [52] implies \[\begin{align} \sum_{i,j} F^{i\bar{j}} (\underline{u}_{i\bar{j}} - u^t_{i\bar{j}}) &= \sum_{i,j} F^{i\bar{j}} \bigl( (\chi_{\underline{u}})_{i\bar{j}} - (\chi_{t, u^t})_{i\bar{j}} \bigr) + t \sum_{i} F^{i\bar{i}} \\ &\ge \sum_{i,j} F^{i\bar{j}} \bigl( (\chi_{\underline{u}})_{i\bar{j}} - (\chi_{t, u^t})_{i\bar{j}} \bigr) \\ &= \sum_{i} F^{i\bar{i}} (\mu_i - \lambda_i) \\ &\ge \delta_0 \sum_{i} F^{i\bar{i}}. \end{align}\]

Case 2: \(\displaystyle\sum_{i} F^{i\bar{i}} (\mu_i - \lambda_i) < \delta_0 \sum_{i} F^{i\bar{i}}\).

By Lemma 8, we obtain \[\mathcal{F}_{\min} = \frac{1}{1+\lambda_1^2} \ge \delta_0 \sum_{i} \frac{1}{1+\lambda_i^2} = \delta_0 \sum_{i,j} F^{i\bar{j}} g_{i\bar{j}}.\]

This completes the proof. ◻

Proof. Without loss of generality, we assume \(\underline{u} = 0\). Let \(w_{i\bar{j}} = \chi_{i\bar{j}} + t g_{i\bar{j}} + u^{t}_{i\bar{j}}\). For simplicity, we write \(u\) instead of \(u^t\) in the following proof. Denote \(S(T^{1,0}M) := \bigcup_{z \in M} \{ \xi \in T^{1,0}_z M \mid |\xi|_{\omega} = 1 \}\). We consider the auxiliary function on \(S(T^{1,0}M)\): \[H(z,\xi) = \log\bigl(w_{k\bar{l}} \xi^k \bar{\xi}^l\bigr) + \varphi\bigl(|\nabla u|_g^2\bigr) + \psi(u),\] where \(\varphi\) and \(\psi\) are defined by \[\begin{align} \varphi(s) &= -\frac{1}{2}\log\!\left(1 - \frac{s}{2K}\right), \quad 0 \le s \le K-1,\\[2mm] \psi(s) &= -A \log\!\left(1 + \frac{s}{2L}\right), \quad -L+1 \le s \le 0, \end{align}\] with \[K = \sup_M |\nabla u|_g^2 + 1, \quad L = \sup_M |u| + 1, \quad A = 2L(C_0 + 1),\] and \(C_0\) is a uniform positive constant to be determined later. We have the following derivatives: \[\begin{align} (4K)^{-1} &\le \varphi' \le (2K)^{-1}, \quad \varphi'' = 2(\varphi')^2, \tag{14} \\[1mm] \frac{1}{2} A L^{-1} &\le -\psi' \le A L^{-1}, \quad \psi'' = A^{-1} (\psi')^2.\tag{15} \end{align}\]

Suppose \(H\) attains its maximum at \((z_0, \xi_0)\). We choose holomorphic coordinates near \(z_0\) such that \[\begin{align} &g_{i\bar{j}}(z_0) = \delta_{ij}, \quad \partial_k g_{i\bar{j}}(z_0) = 0, \quad \text{and} \\ &w_{i\bar{j}}(z_0) = \lambda_i \delta_{ij} \quad \text{with} \quad \lambda_1 \ge \lambda_2 \ge \dots \ge \lambda_n, \end{align}\] which forces \(\xi_0 = \frac{\partial}{\partial z_1}\). Extend \(\xi_0\) near \(z_0\) by \(\tilde{\xi}_0(z) = (g_{1\bar{1}})^{-\frac{1}{2}} \frac{\partial}{\partial z_1}\). Then the function \[Q(z) = H(z, \tilde{\xi}_0(z))\] attains a local maximum at \(z_0\).

By the maximum principle, at \(z_0\) we have \[\label{firstordercondition} Q_i = \frac{w_{1\bar{1},i}}{w_{1\bar{1}}} + \varphi_i + \psi_i = 0,\tag{16}\] and \[\begin{align} \label{LQ} 0\ge \mathcal{L}(Q)=\sum_{i}F^{i\bar i}Q_{i\bar i} =& \sum_{i} \lambda_1^{-1}F^{i\bar i}w_{1\bar 1, i\bar i}-\lambda_1^{-2} \sum_{i}F^{i\bar i}{| w_{1\bar 1, i}|^2}\notag\\ &+ \sum_{i}\varphi''F^{i\bar i}\big|( |\nabla u|^2)_i\big|^2+\varphi'\sum_{i}F^{i\bar i} \big(|\nabla u|^2\big)_{i\bar i}\notag\\ &+\psi''\sum_{i}F^{i\bar i}|u_{i}|^2+\psi'\sum_{i}F^{i\bar i}u_{i\bar i}. \end{align}\tag{17}\]

We will estimate each term in 17 . In what follows, we always assume \[\label{lambda1large} \lambda_1 > 80 A \sqrt{A} K,\tag{18}\] otherwise the desired estimate holds trivially.

Since \(d\chi = 0\), the commutation formula for covariant derivatives yields \[\begin{align} w_{1\bar{1}, i\bar{i}} = w_{i\bar{i}, 1\bar{1}} + (\lambda_1 - \lambda_i) R_{1\bar{1}i\bar{i}}.\label{covariantderivativeformulae} \end{align}\tag{19}\] By ?? and the fact that \(\lambda_i + \lambda_j \ge 0\) for \(i \neq j\) (Lemma 2), we obtain \[\begin{align} \label{Fiiwii11} \sum_{i}F^{i\bar{i}} w_{i\bar{i}, 1\bar{1}} &= \sum_{i,j} \frac{\lambda_i + \lambda_j}{(1+\lambda_i^2)(1+\lambda_j^2)} |w_{i\bar{j},1}|^2 \notag \\ &\ge \sum_{i=2}^n \frac{\lambda_1 + \lambda_i}{(1+\lambda_1^2)(1+\lambda_i^2)} |w_{1\bar{1},i}|^2 + \sum_{i=1}^n \frac{2\lambda_i}{(1+\lambda_i^2)^2} |w_{i\bar{i},1}|^2, \end{align}\tag{20}\] where we have used \(w_{i\bar{1},1} = w_{1\bar{1},i}\), which follows from \(d\chi = 0\).

We now prove the following inequality \[\begin{align} \label{3Fiiwii11} \sum_{i} \frac{\lambda_i}{(1+\lambda_i^2)^2} | w_{i\bar i, 1} |^2 \ge 0. \end{align}\tag{21}\] This is clearly true if \(\lambda_n \ge 0\). Assume now that \(\lambda_n < 0\). By ?? , we have \[\Bigl| \frac{w_{n\bar n, 1}}{1+\lambda_n^2} \Bigr| = \Bigl| \sum_{i=1}^{n-1} \frac{w_{i\bar i, 1}}{1+\lambda_i^2} \Bigr|.\] Using this relation and the fact that \(\lambda_i > 0\) for \(1 \le i \le n-1\), we obtain \[\begin{align} \sum_{i=1}^n \frac{\lambda_i}{(1+\lambda_i^2)^2} | w_{i\bar i, 1} |^2 &= \sum_{i=1}^{n-1} \frac{\lambda_i |w_{i\bar i, 1}|^2}{(1+\lambda_i^2)^2} + \lambda_n \Bigl| \sum_{i=1}^{n-1} \frac{w_{i\bar i, 1}}{1+\lambda_i^2} \Bigr|^2 \\ &\ge \Bigl( \sum_{i=1}^{n-1} \lambda_i^{-1} \Bigr)^{-1} \Bigl| \sum_{i=1}^{n-1} \frac{w_{i\bar i, 1}}{1+\lambda_i^2} \Bigr|^2 + \lambda_n \Bigl| \sum_{i=1}^{n-1} \frac{w_{i\bar i, 1}}{1+\lambda_i^2} \Bigr|^2 \\ &= \frac{1 + \lambda_n \sum\limits_{i=1}^{n-1} \lambda_i^{-1}}{\sum\limits_{i=1}^{n-1} \lambda_i^{-1}} \Bigl| \sum_{i=1}^{n-1} \frac{w_{i\bar i, 1}}{1+\lambda_i^2} \Bigr|^2 \\ &\ge 0, \end{align}\] where the last inequality follows from \[1 + \lambda_n \sum_{i=1}^{n-1} \lambda_i^{-1} = \lambda_n \, \frac{\sigma_{n-1}(\lambda)}{\sigma_n(\lambda)} \ge 0.\]

Combining 20 with 21 , we obtain \[\begin{align} \sum_{i} F^{i\bar{i}} w_{i\bar{i}, 1\bar{1}} \ge \sum_{i=2}^n \frac{\lambda_1 + \lambda_i}{(1+\lambda_1^2)(1+\lambda_i^2)} | w_{1\bar{1}, i} |^2. \end{align}\] Consequently, \[\begin{align} &\lambda_1^{-1} \sum_{i}F^{i\bar{i}} w_{i\bar{i}, 1\bar{1}} - \lambda_1^{-2}\sum_{i} F^{i\bar{i}} | w_{1\bar{1},i} |^2 \notag \\ \ge& -\lambda_1^{-2} (1+\lambda_1^2)^{-1} | w_{1\bar{1}, 1} |^2 - \sum_{i=2}^n \frac{| w_{1\bar{1}, i} |^2}{\lambda_1^{2}(1+\lambda_1^2)(1+\lambda_i^2)} \notag \\ & + \sum_{i=2}^n \frac{\lambda_i}{\lambda_1(1+\lambda_1^2)(1+\lambda_i^2)} | w_{1\bar{1}, i} |^2 \notag \\ =& \mathcal{I} + \mathcal{II} + \mathcal{III}. \label{Fiiwii11w11i} \end{align}\tag{22}\]

We first estimate \(\mathcal{I}\) and \(\mathcal{II}\) using 16 : \[\begin{align} \label{termI} \mathcal{I} &\ge -2(\varphi')^2 F^{1\bar{1}} |(|\nabla u|^2)_1|^2 - 2(\psi')^2 F^{1\bar{1}} |u_{1}|^2 \notag \\ &\ge -\varphi'' F^{1\bar{1}} |(|\nabla u|^2)_1|^2 - 2(\psi')^2 K F^{1\bar{1}}, \end{align}\tag{23}\] and \[\begin{align} \label{termII} \mathcal{II} &= -(1+\lambda_1^2)^{-1} \sum_{i=2}^n \frac{F^{i\bar{i}} | w_{1\bar{1}, i} |^2}{\lambda_1^{2}} \notag \\ &\ge -2(1+\lambda_1^2)^{-1} (\varphi')^2 \sum_{i=2}^n F^{i\bar{i}} |(|\nabla u|^2)_i|^2 \notag \\ &\quad - 2(1+\lambda_1^2)^{-1} (\psi')^2 \sum_{i=2}^n F^{i\bar{i}} |u_{i}|^2 \notag \\ &\ge -\frac{1}{12} \varphi'' \sum_{i=2}^n F^{i\bar{i}} |(|\nabla u|^2)_i|^2 - \frac{1}{2} \psi'' \sum_{i=2}^n F^{i\bar{i}} |u_{i}|^2, \end{align}\tag{24}\] where the last inequality follows from 18 .

Next we estimate the term \(\mathcal{III}\). Since \(\lambda_i>0\) for \(1\le i\le n-1\) by Lemma 2, we obtain \[\mathcal{III} \ge \frac{\lambda_n|w_{1\bar{1}, n}|^2}{\lambda_1(1+\lambda_1^2)(1+\lambda_n^2)}.\] Then, using 16 and the Cauchy-Schwarz inequality, we have \[\begin{align} \label{estimateIII1} \mathcal{III} &\ge -\frac{4}{3}\frac{\lambda_1|\lambda_n|}{1+\lambda_1^2}(\varphi')^2F^{n\bar{n}}|(|\nabla u|^2)_n|^2 -\frac{4\lambda_1|\lambda_n|}{1+\lambda_1^2}(\psi')^2F^{n\bar{n}}|u_n|^2 \notag \\ &\ge -\frac{2}{3}\varphi''F^{n\bar{n}}|(|\nabla u|^2)_n|^2 -\frac{4\lambda_1|\lambda_n|}{1+\lambda_1^2}(\psi')^2F^{n\bar{n}}|u_n|^2, \end{align}\tag{25}\] where we used 14 .

Let \(\tau = \frac{1}{8A}\). We proceed to estimate \(\mathcal{III}\) in two cases.

Case 1: \(|\lambda_n| \le \tau \lambda_1\). In this case, we obtain \[\begin{align} \label{IIIcase1} \mathcal{III} &\ge -\frac{2}{3}\varphi''F^{n\bar{n}}|(|\nabla u|^2)_n|^2 - 4\tau(\psi')^2F^{n\bar{n}}|u_n|^2 \notag \\ &\ge -\frac{2}{3}\varphi''F^{n\bar{n}}|(|\nabla u|^2)_n|^2 - \frac{1}{2}\psi''F^{n\bar{n}}|u_n|^2, \end{align}\tag{26}\] where we used 15 .

Case 2: \(|\lambda_n| \ge \tau \lambda_1\). By 18 , we have \(|u_{n\bar{n}}|^2 = (\lambda_n - \chi_{n\bar{n}} - t)^2 \ge \frac{\tau^2}{2}\lambda_1^2\), and consequently \[4(\psi')^2\frac{\lambda_1|\lambda_n|}{1+\lambda_1^2}F^{n\bar{n}}|u_n|^2 \le 4A F^{n\bar{n}} K \le \frac{1}{3} \varphi'F^{n\bar{n}} u_{n\bar{n}}^2.\] Inserting this inequality into 25 yields \[\begin{align} \label{IIIcase2} \mathcal{III} \ge -\frac{2}{3}\varphi''F^{n\bar{n}}|(|\nabla u|^2)_n|^2 - \frac{1}{3}\varphi'F^{n\bar{n}} u_{n\bar{n}}^2. \end{align}\tag{27}\]

Combining 26 and 27 , we obtain \[\begin{align} \label{termIII} \mathcal{III} \ge -\frac{2}{3}\varphi''F^{n\bar{n}}|(|\nabla u|^2)_n|^2 - \frac{1}{2}\psi''F^{n\bar{n}}|u_n|^2 - \frac{1}{3}\varphi'F^{n\bar{n}}u_{n\bar{n}}^2. \end{align}\tag{28}\]

Inserting 23 , 24 and 28 into 22 , we obtain \[\begin{align} \label{Fiiwii11w11ifinal} &\lambda_1^{-1}\sum_{i} F^{i\bar{i}} w_{i\bar{i}, 1\bar{1}} - \lambda_1^{-2} \sum_{i}F^{i\bar{i}} | w_{1\bar{1},i} |^2\notag \\ &+ \frac{3}{4}\varphi'' \sum_{i}F^{i\bar{i}} |(|\nabla u|^2)_i|^2 + \psi'' \sum_{i}F^{i\bar{i}} |u_i|^2 + \frac{1}{3}\varphi' \sum_{i}F^{i\bar{i}} u_{i\bar{i}}^2 \notag \\ \ge& \frac{1}{3}\varphi' F^{1\bar{1}} u_{1\bar{1}}^2 - 2K(\psi')^2 F^{1\bar{1}} > 0, \end{align}\tag{29}\] where we used \(u_{1\bar{1}} \ge \frac{\lambda_1}{2}\), which follows from 18 .

Next we estimate \(\sum\limits_{i}F^{i\bar{i}} (|\nabla u|^2)_{i\bar{i}}\). By direct computation, we have \[\begin{align} \label{Fiinablausquareii} \sum_{i}F^{i\bar{i}}(|\nabla u|^2)_{i\bar{i}} = \sum_{i}F^{i\bar{i}}(u_{k\bar{i}}u_{\bar{k} i} + u_{ki}u_{\bar{k}\bar{i}}) + \sum_{i,k}F^{i\bar{i}}(u_{k i\bar{i}}u_{\bar{k}} + u_{k}u_{\bar{k} i\bar{i}}). \end{align}\tag{30}\] Using the covariant derivative formulae and ?? , we obtain \[\begin{align} &\sum_{i,k}F^{i\bar{i}}(u_{k i\bar{i}}u_{\bar{k}} + u_{k}u_{\bar{k} i\bar{i}}) \notag \\ \ge &-\sum_{i,k}F^{i\bar{i}}(u_{i\bar{i} k}u_{\bar{k}} + u_{k}u_{i\bar{i}\bar{k}}) - |\text{Rm}||\nabla u|^2 \sum_{i} F^{i\bar{i}} \notag \\ \ge &-2K(|\nabla \chi| + |\text{Rm}|) \sum_{i} F^{i\bar{i}}. \end{align}\] Inserting the above estimate into 30 , we get \[\begin{align} \label{Fiinablausquareii2} \sum_{i}F^{i\bar{i}}(|\nabla u|^2)_{i\bar{i}} \ge \sum_{i}F^{i\bar{i}} u_{i\bar{i}}^2 - 2K(|\nabla \chi| + |\text{Rm}|) \sum_{i} F^{i\bar{i}}. \end{align}\tag{31}\]

Inserting 19 , 29 and 31 into 17 , we obtain \[\begin{align} \label{FQfinal} 0 \ge \sum_{i} F^{i\bar{i}} Q_{i\bar{i}} &\ge \psi' \sum_{i} F^{i\bar{i}} u_{i\bar{i}} + \frac{1}{3}\varphi' \sum_{i}F^{i\bar{i}} u_{i\bar{i}}^2 - 3K \varphi'(|\nabla \chi| + |\text{Rm}|) \sum_{i} F^{i\bar{i}} \notag \\ &\ge \psi' \sum_{i=1}^n F^{i\bar{i}} u_{i\bar{i}} + \frac{1}{3}\varphi' \sum_{i}F^{i\bar{i}} u_{i\bar{i}}^2 - 2(|\nabla \chi| + |\text{Rm}|) \sum_{i} F^{i\bar{i}}. \end{align}\tag{32}\]

By Lemma 8 and noting that we have assumed \(\underline{u}=0\), one of the following holds: \[\begin{align} &(i) \;-\sum_{i} F^{i\bar{i}} u_{i\bar{i}} \ge \delta_0 \sum_{i} F^{i\bar{i}}, \quad \text{or} \\ &(ii) \;F^{1\bar{1}} \ge \delta_0 \sum_{i} F^{i\bar{i}}. \end{align}\] We choose \(C_0 = 6\delta_0^{-1} (\max |\nabla \chi| + |\text{Rm}|)\).

If (\(i\)) holds, then by 32 and \(-\psi' \ge C_0 + 1\), we obtain a contradiction. Thus (\(ii\)) holds, and we get \(|\lambda_n| \ge \frac{\delta_0}{2} \lambda_1\). Consequently, we have \[\begin{align} 0 &\ge \frac{\lambda_1^2}{4} F^{n\bar{n}} - 3K (|\nabla \chi| + |\text{Rm}|) \sum_{i} F^{i\bar{i}} \\ &\ge \Bigl( \frac{\lambda_1^2}{4} - 3n K (|\nabla \chi| + |\text{Rm}|) \Bigr) F^{n\bar{n}}. \end{align}\] Therefore, \(\lambda_1 \le C K\), where \(C\) is a uniform positive constant independent of \(t\). ◻

Proof. For any \(C^2\) function \(\varphi\) and any local minimum point \(z_0\in\Omega\) of \(\varphi-v\), we need to show \[\begin{align} \sigma_{n-1}(\varphi_{\alpha\bar \beta}(z_0))+\sigma_n(\varphi_{\alpha\bar \beta}(z_0))\ge 0.\label{vsubsolution} \end{align}\tag{34}\] By local uniform convergence, we can fix \(r>0\) such that \(v^i\to v\) uniformly on \(\overline{B_r(z_0)}\). Then for any \(\epsilon\in (0,r)\), for all sufficiently large \(i\), we have \(\sup_{B_{r}(z_0)}|v^i-v|<\frac{\epsilon^3}{3}\). Then the function \(\Phi = \varphi + \epsilon|z - z_0|^2 - v^i\) satisfies \(\Phi|_{\partial B_{\epsilon}(z_0)} > \Phi(z_0)\), and thus \(\Phi\) attains its minimum on \(\overline{B_{\epsilon}(z_0)}\) at some point \(z_{i,\epsilon} \in B_{\epsilon}(z_0)\). Since \(v^i\) is a viscosity subsolution, we have \[\begin{align} \sigma_{n-1}(\varphi_{\alpha\bar \beta}(z_{i,\epsilon})+\epsilon\delta_{\alpha\beta})+\sigma_n(\varphi_{\alpha\bar \beta}(z_{i,\epsilon})+\epsilon\delta_{\alpha\beta})\ge 0. \end{align}\] Letting \(\epsilon \to 0\) then implies 34 . This proves \((i)\).

\((ii)\) follows directly from the definition of viscosity subsolution. ◻

Proof. Assume that there exists a point \(x_0 \in \Omega\) such that \(w(x_0) > v(x_0)\). Choose \(\epsilon > 0\) sufficiently small so that \[\epsilon < \frac{w(x_0) - v(x_0)}{\max_{z \in \overline{\Omega}} |z|^2}.\] Since \(w \le v\) on \(\partial\Omega\), we deduce the following chain of inequalities: \[\begin{align} \max_{z \in \overline{\Omega}} \,(w + \epsilon|z|^2 - v) &\ge w(x_0) - v(x_0) \\ &> \epsilon \cdot \max_{z \in \partial\Omega} |z|^2 \\ &\ge \max_{z \in \partial\Omega} \,(w + \epsilon|z|^2 - v). \end{align}\] This implies that the function \(w + \epsilon|z|^2 - v\) attains its maximum at an interior point \(z_0 \in \Omega\).

Now, since \(w\) is \((n-1)\)-admissible, the function \(w + \epsilon|z|^2\) is also \((n-1)\)-admissible. As \(v\) is a viscosity supersolution, by the definition of the supersolution, we have \[\begin{align} \sigma_{n-1}((w_{\alpha\bar{\beta}}(z_0)) + \epsilon I_n) + \sigma_n((w_{\alpha\bar{\beta}}(z_0)) + \epsilon I_n) \le 0. \label{eq:contradiction-start} \end{align}\tag{35}\] However, since \(\sigma_k(w_{\alpha\bar{\beta}}(z_0)) \ge 0\) for \(1 \le k \le n-1\), expanding the left-hand side of 35 gives: \[\begin{align} \sigma_{n-1}(w_{\alpha\bar{\beta}}(z_0)) + \sigma_n(w_{\alpha\bar{\beta}}(z_0)) \le -\epsilon^{n} < 0. \end{align}\] This contradicts the assumption that \(w\) is a viscosity subsolution of 33 , thereby completing the proof. ◻

Proof. We first note that \([v]_{r}\) is \((n-1)\)-subharmonic since \(v\) is \((n-1)\)-subharmonic. To prove \((i)\), we will show that \(\tilde{v}\) satisfies the equation: \(\sigma_{n}=0\) on \(\mathbb{C}^{n+1}\) in the current sense.

Let \(e_{n+1} = \sqrt{-1}dy_{n+1} \wedge d\bar{y}_{n+1}\) and let \(\widetilde{\mathcal{E}} = e_{n+1}+ \mathcal{E}\) be the standard Kähler metric on \(\mathbb{C}^{n+1}\). For any \(1 \le k \le n\), we have: \[\begin{align} & (\sqrt{-1}\partial\bar{\partial} \tilde{v})^{k} \wedge \widetilde{\mathcal{E}}^{n+1-k}\\ =& C^k_{n+1} (\sqrt{-1}\partial\bar{\partial} v + e_{n+1})^{k} \wedge (\mathcal{E} + e_{n+1})^{n+1-k} \\ =& \left((\sqrt{-1}\partial\bar{\partial} v)^{k} + k(\sqrt{-1}\partial\bar{\partial} v)^{k-1} \wedge e_{n+1}\right) \wedge \left(\mathcal{E}^{n-k+1} + (n+1-k)\mathcal{E}^{n-k} \wedge e_{n+1}\right) \\ =& \left((n-k+1)(\sqrt{-1}\partial\bar{\partial} v)^{k} \wedge \mathcal{E}^{n-k} + k(\sqrt{-1}\partial\bar{\partial} v)^{k-1} \wedge \mathcal{E}^{n-k+1}\right) \wedge e_{n+1}. \end{align}\] In particular, \((\sqrt{-1}\partial\bar\partial\tilde{v})^n\wedge\mathcal{\tilde{E} }=0\) since \(v\) solves \[(\sqrt{-1}\partial\bar{\partial} v)^{n} + n(\sqrt{-1}\partial\bar{\partial} v)^{n-1} \wedge \mathcal{E} = 0.\] Since \(\tilde{v}\) is \(n\)-subharmonic, \(\langle\tilde{v}\rangle_{r}\) is also \(n\)-subharmonic in \(\mathbb{C}^{n+1}\). Note that \(\langle\tilde{v}\rangle_{r} = [v]_{r}(z) + |z_{n+1}|^2 + r^2 M\); it follows that \([v]_{r}\) is a subsolution to the equation \(\sigma_n + \sigma_{n-1} = 0\). This establishes \((ii)\).

To prove \((iv)\), let \(W = \frac{v + w}{2}\) and \(\widetilde{W}(\tilde{z}) = W(z) + |z_{n+1}|^2\). By \((i)\), \(\widetilde{W} = \frac{\tilde{v} + \tilde{w}}{2}\) is \(n\)-subharmonic in \(\mathbb{C}^{n+1}\) and thus \(W\) is a subsolution.

We now prove \((iii)\) using \((i)\). Define \(\tilde{F}(\tilde{z}) = \frac{1}{2}v(z)^2 + |z_{n+1}|^2\). Then for any \(1 \le k \le n\), we have: \[\begin{align} &(\sqrt{-1}\partial\bar{\partial} \tilde{F})^{k} \wedge \widetilde{\mathcal{E}}^{n+1-k} \\ =& \left(v\sqrt{-1}\partial\bar{\partial} \tilde{v} + (1 - v)e_{n+1} + \sqrt{-1}\partial v \wedge \bar{\partial} v\right)^k \wedge \widetilde{\mathcal{E}}^{n+1-k} \\ =& v^k (\sqrt{-1}\partial\bar{\partial} \tilde{v})^k \wedge \widetilde{\mathcal{E}}^{n+1-k} \\ &+ k v^{k-1} (\sqrt{-1}\partial\bar{\partial} \tilde{v})^{k-1} \wedge \left((1 - v)e_{n+1} + \sqrt{-1}\partial v \wedge \bar{\partial} v\right) \wedge \widetilde{\mathcal{E}}^{n+1-k} \\ &+ k(k-1)(1 - v) v^{k-2} (\sqrt{-1}\partial\bar{\partial} \tilde{v})^{k-2} \wedge e_{n+1} \wedge (\sqrt{-1}\partial v \wedge \bar{\partial} v) \wedge \widetilde{\mathcal{E}}^{n+1-k}. \end{align}\] Since \(\tilde{v}\) is \(n\)-subharmonic in \(\mathbb{C}^{n+1}\) and \(0 \le v \le 1\), the above is nonnegative by Lemma 11. ◻

Proof. Since \(\tilde{v} = v + |z_{n+1}|^2\) is \(n\)-subharmonic in \(\mathbb{C}^{n+1}\), it follows from Lemma 3.7 [54] that \(\tilde{v}\) is a viscosity subsolution to \(\sigma_n = 0\) in \(\mathbb{C}^{n+1}\).

Now, let \(\varphi\) be any \(C^2\) function and let \(z_0\) be a local minimum point of \(\varphi - v\). Consider the function \(\tilde{\varphi} = \varphi + |z_{n+1}|^2\). Then \((z_0, 0)\) is a local minimum point of \(\tilde{\varphi} - \tilde{v}\). Since \(\tilde{v}\) is a viscosity subsolution to \(\sigma_n = 0\), we have \[\sigma_n(\sqrt{-1}\partial\bar{\partial} \tilde{\varphi})(z_0, 0) \ge 0.\] Observing that \[\sigma_n(\sqrt{-1}\partial\bar{\partial} \tilde{\varphi})(z_0, 0) = \sigma_n(\sqrt{-1}\partial\bar{\partial} \varphi)(z_0) + \sigma_{n-1}(\sqrt{-1}\partial\bar{\partial} \varphi)(z_0),\] we conclude that \(v\) is a viscosity subsolution. ◻

Proof of Theorem 5. Assume that \(v\) is not constant. Then \(\inf_{\mathbb{C}^n} v < \sup_{\mathbb{C}^n} v\). By scaling, we may assume that \(v\) satisfies the following normalized conditions: \[\begin{cases} \sup_{\mathbb{C}^n} v = 1, \quad \inf_{\mathbb{C}^n} v = 0, \\ |\nabla v| \le C_1 := (\sup_{\mathbb{C}^n} v - \inf_{\mathbb{C}^n} v)^{\frac{1}{2}}, \quad \Delta v \le C, \\ \sigma_{n-1}(\sqrt{-1}\partial\bar{\partial} v) + \sigma_{n}(\sqrt{-1}\partial\bar{\partial} v) = 0. \end{cases}\] Indeed, the function \(\tilde{v}(z) = C_{1}^{-2}({v(C_1z) - \inf_{\mathbb{C}^n} v})\) satisfies these conditions.

For sufficiently small \(\epsilon > 0\), let \(v^{\epsilon} = [v]_{\epsilon}\).

Case 1: There exist a positive constant \(\rho\), sequences \(\epsilon_k \to 0\), \(x_k \in \mathbb{C}^n\), \(r_k \to \infty\), and a sequence of unit \((1,0)\)-vectors \(\xi_k\) such that \[\begin{align} \frac{1}{2}\Bigl(\Bigl[\frac{(v^{\epsilon})^2}{3}\Bigr]_{r}(x_k) + [v]_{\rho}(x_k) - \frac{7}{6}\Bigr) &\ge v(x_k) \label{w-v} \end{align}\tag{37}\] \[\begin{align} \lim_{k \to \infty} \int_{B_{r_k}(x_k)} |v_{\xi_k}^{\epsilon}|^2(z) \, \mathcal{E}^n(z) &= 0. \label{integral-zero} \end{align}\tag{38}\] Following the arguments in [4] and [43], we obtain an \((n-1)\)-subharmonic function \(v^{\infty}\) that is also a solution to the same equation and is independent of \(z_n\). This implies that \(v^{\infty}\) is plurisubharmonic in \(\mathbb{C}^{n-1}\). Since \(v^{\infty}\) is bounded, it must be constant, contradicting 37 .

Case 2: Suppose Case 1 does not hold. Then for every \(\rho > 0\), there exists a constant \(C_{\rho} > 0\) with the following property: for all \(\epsilon < C_{\rho}^{-1}\), \(r > C_{\rho}\), \(x \in \mathbb{C}^{n}\), and any unit \((1,0)\)-vector \(\xi\), if \(\frac{1}{2}\bigl(\bigl[\frac{(v^{\epsilon})^2}{3}\bigr]_{r}(x) + [v]_{\rho}(x) -\frac{7}{6}\bigr) - v(x) \ge 0\), then \[\begin{align} \label{secondcase} \int_{B_{r}(x)} |v^{\epsilon}_{\xi}|^2(z) \, \mathcal{E}^n(z) \ge c_0 := C_{\rho}^{-1}. \end{align}\tag{39}\]

Define the open set \[\Omega := \Bigl\{z \in \mathbb{C}^n : v(z) <w:=\frac{1}{2}\Bigl(\Bigl[\frac{(v^{\epsilon})^2}{3}\Bigr]_{r}(z) -{\tau_0}|z|^2+ [v]_{\rho}(z) - \frac{7}{6}\Bigr) \Bigr\},\] where \(\tau_0>0\) is a constant to be determined later.

Since \(0 \le v \le 1\), the set \(\Omega\) is bounded. As \(\inf_{\mathbb{C}^n} v = 0\), we may translate so that \(v(0) < \frac{1}{60}\). By Cartan’s lemma (see [43]), we choose \(\rho\) large enough so that \([v]_{\rho}(0) > \frac{9}{10}\), and choose \(r\) large enough so that \(\bigl[\frac{(v^\epsilon)^2}{3}\bigr]_{r}(0) > \frac{3}{10}\). Then \(0 \in \Omega\).

We now show that \(\frac{1}{2}\bigl(\bigl[\frac{(v^{\epsilon})^2}{3}\bigr]_{r}(z) -{\tau_0}|z|^2+ [v]_{\rho}(z) \bigr)\) is a subsolution to \(\sigma_{n-1} + \sigma_{n} = 0\) in \(\Omega\). Assuming this, the comparison principle (Lemma 10) implies that \(v \ge w\) in \(\Omega\), contradicting the definition of \(\Omega\).

By Lemma 12 \((ii)\), we know that \([v]_{r}\), \([v]_{\rho}\), \(\frac{(v^{\epsilon})^2}{2}\), and \(\bigl[\frac{(v^{\epsilon})^2}{2}\bigr]_{r}\) are all subsolutions to ?? .

From 39 and following [4], [43], there exists a positive constant \(b_0\), depending only on \(c_0\) and \(C_n\), such that \(\bigl[\frac{(v^{\epsilon})^2}{2}\bigr]_{r} - b_0|z|^2\) is \((n-1)\)-subharmonic in \(\Omega\). Then by the concavity of \(\sigma_{n-1}^{1/(n-1)}\) in \(\Gamma_{n-1}\), for any \(z\in \Omega\), we have \[\sigma_{n-1}^{1/(n-1)}\Bigl(\sqrt{-1}\partial\bar{\partial}\Bigl[\frac{(v^{\epsilon})^2}{2}\Bigr]_{r}\Bigr) \ge (n-1)^{1/(n-1)} b_0,\] and thus \[\begin{align} \label{b0bound} \sigma_{n-1}\Bigl(\sqrt{-1}\partial\bar{\partial}\Bigl[\frac{(v^{\epsilon})^2}{2}\Bigr]_{r}\Bigr) \ge (n-1) b_0^{n-1}. \end{align}\tag{40}\]

We now show that \(F := \bigl[\frac{(v^{\epsilon})^2}{3}\bigr]_{r} - \tau_0|z|^2\) is a subsolution to \(\sigma_n + \sigma_{n-1} = 0\) in \(\Omega\) for sufficiently small \(\tau_0\). Indeed, by 40 , we have: \[\begin{align} &\sigma_n(\sqrt{-1}\partial\bar{\partial} F) + \sigma_{n-1}(\sqrt{-1}\partial\bar{\partial} F) \\ &= \sigma_n\Bigl(\sqrt{-1}\partial\bar{\partial} \Bigl[\frac{(v^{\epsilon})^2}{3}\Bigr]_{r}\Bigr) + \sigma_{n-1}\Bigl(\sqrt{-1}\partial\bar{\partial} \Bigl[\frac{(v^{\epsilon})^2}{3}\Bigr]_{r}\Bigr) + O(\tau_0) \\ &= \Bigl(\frac{2}{3}\Bigr)^n \Bigl(\sigma_n\Bigl(\sqrt{-1}\partial\bar{\partial} \Bigl[\frac{(v^{\epsilon})^2}{2}\Bigr]_{r}\Bigr) + \frac{3}{2} \sigma_{n-1}\Bigl(\sqrt{-1}\partial\bar{\partial} \Bigl[\frac{(v^{\epsilon})^2}{2}\Bigr]_{r}\Bigr)\Bigr)+ O(\tau_0) \\ &\ge \frac{1}{2}\Bigl(\frac{2}{3}\Bigr)^{n} \sigma_{n-1}\Bigl(\sqrt{-1}\partial\bar{\partial} \Bigl[\frac{(v^{\epsilon})^2}{2}\Bigr]_{r}\Bigr) + O(\tau_0) \\ &\ge \frac{1}{2}\Bigl(\frac{2}{3}\Bigr)^{n} (n-1) b_0^{n-1} - C\tau_0 > 0, \end{align}\] provided \(\tau_0 <\frac{b_0^{n-1}}{2C} \left(\frac{2}{3}\right)^{n}\). Here we have used the fact that \([(v^{\epsilon})^2]_{r}\) has uniform \(C^{0,1}\)-bounds and complex Hessian bounds.

By Lemma 12 \((iv)\), we get \(\frac{1}{2}\bigl(\bigl[\frac{(v^{\epsilon})^2}{3}\bigr]_{r}(z) -{\tau_0}|z|^2+ [v]_{\rho}(z) \bigr)\) is a subsolution to the equation \(\sigma_{n}+\sigma_{n-1}=0\) in \(\Omega\). ◻

Proof of the gradient estimate. We prove the gradient estimate by contradiction, following the arguments in [4], [43]. Assume that there exist a sequence \(t_i \to 0\) and corresponding solutions \(u^{t_i}\) to the LYZ equation 6 such that for points \(p_i \in M\), we have

\[\begin{align} \sup_M |\nabla u^{t_i}|_{\omega} = |\nabla u^{t_i}|_{\omega}(p_i) = M_i \to \infty \quad \text{as } i \to \infty. \end{align}\]

Since \(M\) is compact, by passing to a subsequence, we may assume that \(\{p_i\}\) converges to a point \(p_0 \in M\). We choose local coordinates \(\{w_\alpha\}_{1 \le \alpha \le n}\) in a ball \(B_r(p_0)\) such that \(p_i \in B_{r/2}(p_0)\) and \(\omega(p_0) = \sum_{\alpha} \sqrt{-1} dw_{\alpha} \wedge d\bar{w}_{\alpha}\).

For any \(R > 0\), we can choose \(i\) sufficiently large such that for all \(z \in B_R(0)\), the point \(p_i + \frac{z}{M_i}\) lies in \(B_{2r/3}(p_0)\). We define the rescaled functions \[\hat{u}^{t_i}(z) = u^{t_i}\bigl(p_i + \frac{z}{M_i}\bigr).\] We adopt the following notation: \[\begin{align} \hat{u}^{t_i}_{\alpha}(z) = \frac{\partial \hat{u}^{t_i}}{\partial z_{\alpha}}(z), \quad u^{t_i}_{\alpha}(w) = \frac{\partial u^{t_i}}{\partial w_{\alpha}}(w), \end{align}\] \[\begin{align} \hat{u}^{t_i}_{\alpha\bar{\beta}}(z) = \frac{\partial^2 \hat{u}^{t_i}}{\partial z_{\alpha} \partial \bar{z}_{\beta}}(z), \quad u^{t_i}_{\alpha\bar{\beta}}(w) = \frac{\partial^2 u^{t_i}}{\partial w_{\alpha} \partial \bar{w}_{\beta}}(w). \end{align}\] Then we have the scaling relations: \[\begin{align} u^{t_i}_{\alpha}(w) = M_i \hat{u}^{t_i}_{\alpha}(z), \quad u^{t_i}_{\alpha\bar{\beta}}(w) = M_i^2 \hat{u}^{t_i}_{\alpha\bar{\beta}}(z). \end{align}\]

By the \(C^0\)-estimate and the complex Hessian estimate for \(u^{t_i}\) (for large \(i\)), we obtain \[\begin{align} \sup_{B_R(0)} \left(|\hat{u}^{t_i}| + |\partial \hat{u}^{t_i}|_{\mathcal{E}} + |\Delta \hat{u}^{t_i}|_{\mathcal{E}}\right) \le C. \end{align}\] By standard elliptic theory, we have the estimates \[\begin{align} |\hat{u}^{t_i}|_{C^{1, \frac{2}{3}}(B_{2R/3}(0))} \le C. \end{align}\]

Passing to a subsequence, we find that \(\hat{u}^{t_i} \to u^{\infty}\) in \(C^{1, \frac{1}{2}}(B_{R/2}(0))\). Using a diagonal subsequence argument, we conclude that \(\hat{u}^{t_i} \to u^{\infty}\) uniformly on compact subsets of \(\mathbb{C}^n\), and \(u^{\infty}\) is a bounded \(C^{1,\frac{1}{2}}\) function defined on \(\mathbb{C}^n\).

Since \(u^{t_i}\) solves 6 , by Lemma 2, we have \(\sigma_k(\chi + t_i\omega + \sqrt{-1}\partial\bar{\partial} u^{t_i}) \ge 0\) for \(1 \le k \le n-1\). It follows that \(u^{\infty}\) is \((n-1)\)-subharmonic. By construction, we have \(|\partial u^{\infty}(0)| = 1\), so \(u^{\infty}\) is nonconstant.

Now, let \[a_0 = -\liminf_{i \to \infty} \tan\hat{\theta}(t_i) M_i^2.\] Recall that \(\hat{\theta}(t) \in (\pi - 2Ct, \pi-Ct)\), so \(a_0 \in [0, \infty]\). By passing to a further subsequence, we may assume \[\begin{align} \label{a0} a_0 = -\lim_{i \to \infty} \tan\hat{\theta}(t_i) M_i^2. \end{align}\tag{41}\]

Case 1: \(a_0 = +\infty\). In this case, we show that \(u^{\infty}\) is plurisubharmonic in \(\mathbb{C}^{n}\). Since \(u^{t_i}\) solves the LYZ equation 6 , we have \[\chi + t_i\omega + \sqrt{-1}\partial\bar{\partial} u^{t_i} > \cot\hat{\theta}(t_i)\omega.\] Then in the coordinate ball \(B_r(p_0)\), \[\begin{align} (u^{t_i}_{\alpha\bar{\beta}})_{n\times n} \ge (\cot\hat{\theta}(t_i) - C)g. \end{align}\]

Now fix \(R > 0\) and \(\epsilon > 0\). For sufficiently large \(i\) and all \(z \in B_R(0)\), we have \[\begin{align} (\hat{u}^{t_i}_{\alpha\bar{\beta}}(z))_{n\times n} \ge -M_i^{-2}(\cot\hat{\theta}(t_i) - C)g \ge -\epsilon I_n, \end{align}\] where we have used the fact that \(\lim_{i\to\infty} M_i^{-2}\cot\hat{\theta}(t_i) = 0\). Therefore, \(\hat{u}^{t_i} + \epsilon|z|^2\) is plurisubharmonic in \(B_R(0)\). Since \(\hat{u}^{t_i}\) converges uniformly to \(u^{\infty}\) on compact sets, it follows that \(u^{\infty} + \epsilon|z|^2\) is plurisubharmonic in \(B_R(0)\).

As \(R > 0\) and \(\epsilon > 0\) are arbitrary, we conclude that \(u^{\infty}\) is plurisubharmonic in \(\mathbb{C}^n\). However, since \(u^{\infty}\) is bounded and plurisubharmonic on \(\mathbb{C}^n\), it must be constant. This contradicts the fact that \(|\partial u^{\infty}(0)| = 1\).

Case 2: \(0 \le a_0 < \infty\). Define \[A^{i}(z) = \left(g^{-1}(p_i + M_i^{-1}z) - I_n\right)(\hat{u}^{t_i}_{\alpha\bar{\beta}}(z))_{n\times n} + M_{i}^{-2}g^{-1}\chi(p_i + M_i^{-1}z) + t_iM_{i}^{-2} I_n.\] Then \(A^{i}\) converges uniformly to \(0\) on any compact subset of \(\mathbb{C}^n\).

Since \(u^{t_i}\) solves the LYZ equation 6 , we have \[\begin{align} \label{LYZti} \sigma_{n-1}\bigl(\bigl(\hat{u}_{\alpha\bar{\beta}}^{t_i}\bigr)_{n\times n} + A^{i}\bigr) = M_i^{2}\tan\hat{\theta}(t_i)\sigma_n\bigl(\bigl(\hat{u}_{\alpha\bar{\beta}}^{t_i}\bigr)_{n\times n} + A^{i}\bigr) + O(M_i^{-2}), \end{align}\tag{42}\] where the matrix \(A^{i}\) is uniformly bounded and by Lemma 2, we have \[\begin{align} \sigma_{k}\bigl(\bigl(\hat{u}_{\alpha\bar{\beta}}^{t_i}\bigr)_{n\times n} + A^{i}\bigr) > 0 \quad \text{for any } 1 \le k \le n-1.\label{gamman-1} \end{align}\tag{43}\]

Subcase 2.1: \(a_0 = 0\). In this subcase, by 42 , 43 , and the fact that \(\hat{u}^{t_i}\) converges locally uniformly to \(u^{\infty}\), we conclude that \(u^{\infty}\) is \((n-1)\)-subharmonic and satisfies the following \((n-1)\)-Hessian equation in the sense of currents \(\sigma_{n-1}(\sqrt{-1}\partial\bar{\partial} u^{\infty}) = 0.\) Hence, applying the Liouville theorem of Dinew-Kołodziej [43], we find that \(u^{\infty}\) is constant, which is a contradiction.

Subcase 2.2: \(0 < a_0 < \infty\). Since \(\hat{u}^{t_i}\) converges uniformly to \(u^{\infty}\) on compact subsets of \(\mathbb{C}^n\), we have that \(u^{\infty} + a_0|z|^2\) is plurisubharmonic, \((n-1)\)-subharmonic, \(u^{\infty}+|z_{n+1}|^2\) is \(n\)-subharmonic and \(u^{\infty}\) satisfies the equation \(\sigma_n + a_0\sigma_{n-1} = 0\) in the sense of currents.

Moreover, for any test function \(f \in C_c^{\infty}(\mathbb{C}^n)\) with \(f \ge 0\) and \(\int_{\mathbb{C}^{n}} f = 1\), we have \[\begin{align} \int_{\mathbb{C}^{n}} u^{\infty} \sqrt{-1}\partial\bar{\partial} f \wedge \mathcal{E}^{n-1} &= \lim_{i \to \infty} \int_{\mathbb{C}^{n}} \hat{u}^{t_i} \sqrt{-1}\partial\bar{\partial} f \wedge \mathcal{E}^{n-1} \\ &= \lim_{i \to \infty} \int_{\mathbb{C}^{n}} f \sqrt{-1}\partial\bar{\partial} \hat{u}^{t_i} \wedge \mathcal{E}^{n-1} \\ &\le C. \end{align}\] This proves that \(\Delta u^{\infty} \le C\) in the weak sense.

Next, we prove \(u^{\infty}\) is a supersolution to the equation \(\sigma_{n-1}+a_0\sigma_{n}=0\) in the viscosity sense. For any \(C^2\) function \(\varphi\) with \(\lambda(\varphi_{\alpha\bar \beta})\in \overline{\Gamma}_{n-1}\) and for any local maximum point \(z_0\) of \(\varphi-u^{\infty}\), we need to show \[\begin{align} \sigma_{n-1}\big(\varphi_{\alpha\bar \beta}( z_0)\big)+a_0\sigma_{n}\big(\varphi_{\alpha\bar \beta}( z_0)\big)\le 0.\label{supersolution} \end{align}\tag{44}\]

If \(\sigma_{n-1}(\varphi_{\alpha\bar \beta}( z_0))=0\), then \(\sigma_{n}(\varphi_{\alpha\bar \beta}( z_0))\le 0\) and thus 44 holds.

So we assume \(c_0=\sigma_{n-1}(\varphi_{\alpha\bar \beta}( z_0))>0\). Then for any \(\epsilon>0\) small enough, we have \[\begin{align} \lambda(\varphi_{\alpha\bar \beta}-2\epsilon\delta_{\alpha\beta})\in \Gamma_{n-1}\;\;\text{and}\;\; \sigma_{n-1}(\varphi_{\alpha\bar \beta}-2\epsilon\delta_{\alpha\beta})>\frac{c_0}{2} \;\text{in}\;\overline{B}_{\epsilon}(z_0)\label{sigman-1positive}. \end{align}\tag{45}\] For sufficiently large \(i\) such that \(|\hat{u}^{t_{i}} - u^{\infty}| < \frac{\epsilon^{3}}{2}\), consider the function \[\Phi = \varphi - \epsilon|z - z_0|^2 - \hat{u}^{t_i}.\] Since for any \(z \in \partial B_{\epsilon}(z_0)\), \[\begin{align} \Phi(z)=& \varphi(z) - \hat{u}^{t_{i}}(z) - \epsilon^3\\ <&\varphi(z) - u^{\infty}(z) - \frac{\epsilon^3}{2} \le \varphi(z_0) - u^{\infty}(z_0) - \frac{\epsilon^3}{2}\\ <& \varphi(z_0) - \hat{u}^{t_i}(z_0) =\Phi(z_0). \end{align}\] Thus, \(\Phi\) attains its maximum on \(\overline{B}_{\epsilon}(z_0)\) at an interior point \(\tilde{z}_{0} \in B_{\epsilon}(z_0)\). Then \((\Phi_{\alpha\bar \beta}( \tilde{z}_{0 }))_{n\times n}\le 0\), namely \((\varphi_{\alpha\bar \beta}( \tilde{z}_{0})-\epsilon \delta_{\alpha\beta})_{n\times n}\le (\hat{u}^{t_i}_{\alpha\bar \beta}( \tilde{z}_{0}))_{n\times n}\). Thus \((\hat{u}^{t_i}_{\alpha\bar \beta}( \tilde{z}_{0})+A^{i}_{\alpha\bar\beta}(\tilde{z}_{0}))_{n \times n}\ge(\varphi_{\alpha\bar \beta}(\tilde{z}_{0})-2\epsilon \delta_{\alpha\beta})_{n\times n}\) and we obtain \[\sigma_{n-1}(\hat{u}^{t_i}_{\alpha\bar \beta}(\tilde{z}_{0})+A_{\alpha\bar \beta}(\tilde{z}_{0}))\ge \sigma_{n-1}\big(\varphi_{\alpha\bar \beta}( \tilde{z}_{0})-2\epsilon \delta_{\alpha\beta}\big)\ge \frac{c_0}{2}.\] By the concavity of \(\frac{\sigma_{n}}{\sigma_{n-1}}\) in \(\Gamma_{n-1}\), we get \[\begin{align} &1-\tan\hat{\theta}(t_{i})M^2_{i}\frac{\sigma_n\big( \varphi_{\alpha\bar \beta}(\tilde{z}_{0})-2\epsilon\delta_{\alpha\beta} \big)}{\sigma_{n-1}\big(\varphi_{\alpha\bar \beta}(\tilde{z}_{0})-2\epsilon\delta_{\alpha\beta}\big)}\notag\\\le & 1-\tan\hat{\theta}(t_{i})M^2_{i}\frac{\sigma_n( \hat{u}^{t_{i}}_{\alpha\bar \beta}(\tilde{z}_{0}) +A^{i}_{\alpha\bar \beta}(\tilde{z}_{0}))}{\sigma_{n-1} ( \hat{u}^{t_{i}}_{\alpha\bar \beta}(\tilde{z}_{0}) +A^{i}_{\alpha\bar \beta}(\tilde{z}_{0}))}\notag\\ \le& c_0^{-1}CM_{i}^{-2},\label{super2} \end{align}\tag{46}\] where in the last inequality we use 42 . Then we get \[\begin{align} {\sigma_{n-1}\big(\varphi_{\alpha\bar \beta}(\tilde{z}_{0})-2\epsilon\delta_{\alpha\beta}\big)}-\tan\hat{\theta}(t_{i})M^2_{i}{\sigma_n\big( \varphi_{\alpha\bar \beta}(\tilde{z}_{0})-2\epsilon\delta_{\alpha\beta} \big)}\le c_{0}^{-1}CM_{i}^{-2}\sup|D^2\varphi|. \end{align}\] Let \(i\rightarrow\infty\) and then \(\epsilon\rightarrow 0\) in the above, we obtain 44 . Hence \(u^{\infty}\) is a supersolution to the equation \(\sigma_{n}+a_0\sigma_{n-1}=0\).

We can then apply the Liouville Theorem 5 to \(a_0^{-1}u^{\infty}\) to conclude that \(a_0^{-1}u^{\infty}\) is constant, which again leads to a contradiction. ◻

Proof of Theorems 1 and 2. Once the gradient estimates for \(u^t\) are established, we obtain the uniform bound \[\|u^t\|_{C^1} + \max_M |\partial\bar{\partial} u^t| \le C,\] where \(C\) is independent of \(t\). Following the argument of Collins–Jacob–Yau [1], we derive uniform \(C^{2,\alpha}\) estimates for \(u^t\), also independent of \(t\). For any \(\beta \in (0,\alpha)\), there exists a subsequence \(\{u^{t_i}\}\) converging in \(C^{2,\beta}\) to a limit \(u^0 \in C^{2,\alpha}\). Thus, \(u^0\) solves the critical LYZ equation 4 . By standard elliptic regularity theory, \(u^0\) is smooth. ◻

Proof. Condition \((i)\) gives that the principal argument of \(\int_M(\chi+\sqrt{-1}\omega)^3\) is \(\pi\). Condition \((ii)\) means that \(0\) is a subsolution of the critical LYZ equation in dimension \(3\). According to Theorem 1, there exists a smooth solution to the critical LYZ equation in dimension \(3\), which is exactly the \(2\)-Hessian equation \(\sigma_2(\chi_u)=1\). ◻

Proof. Condition \((i)\) implies that the principal argument of \(\int_M(\chi+\sqrt{-1}\omega)^4\) is \(\pi\). To apply Theorem 1, we only need to show that \(0\) is a subsolution of the critical LYZ equation.

By condition \((ii)\), we have \(\sigma_{2}(\chi)>2\), which implies \(\sigma_1^2(\chi) \ge \frac{8}{3}\sigma_2(\chi) > \frac{16}{3}\). Moreover, since there exists a point \(z_0 \in M\) such that \(\sigma_1(\chi(z_0)) = 4\frac{\int_M\chi\wedge\omega^3}{\int_M \omega^4} > 0\), we conclude that \(\sigma_1(\chi) > 0\) on \(M\).

Let \(\mu=(\mu_1,\mu_2,\mu_3, \mu_4)\) be the eigenvalues of \(\chi\) with respect to \(\omega\), ordered such that \(\mu_1 \ge \mu_2 \ge \mu_3 \ge \mu_4\). We need to prove \[\begin{align} \label{4dsub} \arctan\mu_2+\arctan\mu_3+\arctan\mu_4 > \frac{\pi}{2}. \end{align}\tag{47}\] Indeed, condition \((ii)\) gives \(\sigma_2(\mu|1) = \mu_2\mu_3 + \mu_2\mu_4 + \mu_3\mu_4 > 1\). Since \(\sigma_2(\mu|1) > 0\) and \(\mu \in \Gamma_2\), it follows that \(\mu_2 + \mu_3 > 0\), and consequently \[\mu_4 > \frac{1 - \mu_2\mu_3}{\mu_2 + \mu_3}.\] Therefore, \[\arctan \mu_4 > \arctan\Bigl( \frac{1 - \mu_2\mu_3}{\mu_2 + \mu_3} \Bigr) = \frac{\pi}{2} - \arctan\mu_2 - \arctan\mu_3,\] which proves 47 . Hence, \(0\) is a subsolution of the critical LYZ equation. ◻

Proof. The equality in condition \((i)\) follows directly from integrating the equation. Since \(\sigma_1(\chi_u)>0\), we have \[\begin{align} \int_{M}\chi\wedge \omega^3 = \int_{M}\chi_u\wedge \omega^3 > 0. \end{align}\] Let \(\lambda=(\lambda_1,\lambda_2,\lambda_3,\lambda_4)\) be the eigenvalues of \(\chi_u\) with respect to \(\omega\), ordered such that \(\lambda_1 \ge \lambda_2 \ge \lambda_3 \ge \lambda_4\).

By the MacLaurin inequality, we obtain \[\begin{align} \sigma_2 \ge 6\frac{\sigma_3}{\sigma_1} = 6. \end{align}\] The inequality \(\lambda_4\sigma_1(\lambda|4) \le \sigma_2(\lambda|4)\) yields \[\begin{align} \sigma_2 = \sigma_2(\lambda|4) + \lambda_4\sigma_1(\lambda|4) \le 2\sigma_2(\lambda|4), \end{align}\] and consequently, \[\begin{align} \sigma_2(\lambda|4) \ge 3. \end{align}\] Using the identity \(\lambda_4 = \dfrac{\sigma_3(\lambda|4) - \sigma_1(\lambda|4)}{-\sigma_2(\lambda|4) + 1}\), we compute \[\begin{align} \sigma_{4} - \sigma_2 + 1 &= \lambda_{4}\sigma_3(\lambda|4) - \lambda_4\sigma_1(\lambda|4) - \sigma_2(\lambda|4) + 1 \\ &= \frac{\bigl(\sigma_3(\lambda|4) - \sigma_1(\lambda|4)\bigr)^2}{-\sigma_2(\lambda|4) + 1} - \sigma_2(\lambda|4) + 1 \\ &\le -2. \end{align}\] Therefore, \[\begin{align} \mathrm{Re}\int_M (\chi+\sqrt{-1}\omega)^4 &= \mathrm{Re}\int_M (\chi_u+\sqrt{-1}\omega)^4 \\ &= \int_M (\sigma_4 - \sigma_2 + 1)\,\omega^4 < 0, \end{align}\] which establishes the second inequality in condition \((i)\). ◻

Proof. The proof is based on Lemma 5. Assume that \(v\) is not a constant function. Then \(\inf_{\mathbb{C}^n} v < \sup_{\mathbb{C}^n} v\). By scaling, we may assume that \(v\) satisfies \[\begin{cases} \sup_{\mathbb{C}^n} v = 1, \quad \inf_{\mathbb{C}^n} v = 0, \\ |\nabla v| \le CC_1, \quad \Delta v \le C, \end{cases}\] where \(C_1\) denotes \((\sup_{\mathbb{C}^n} v - \inf_{\mathbb{C}^n} v)^{\frac{1}{2}}\). We prove the theorem by induction on \(n-k\).

When \(n-k=0\), this is Theorem 5. Assume the theorem holds for \(n-k\le m\). We need to prove it holds for \(n-k=m+1\). We still consider two cases.

Case 1: There exist a constant \(\rho>0\) and sequences \(\epsilon_l \to 0\), \(x_l \in \mathbb{C}^n\), \(r_l \to \infty\), and unit \((1,0)\)-vectors \(\xi_l\) such that \[\begin{align} \frac{1}{2}\Bigl(\Bigl[\frac{(v^{\epsilon})^2}{3}\Bigr]_{r}(x_l) + [v]_{\rho}(x_l) - \frac{7}{6}\Bigr) &\ge v(x_l), \label{kw-v} \end{align}\tag{48}\] and \[\begin{align} \lim_{l \to \infty} \int_{B_{r_l}(x_l)} |v_{\xi_l}^{\epsilon}|^2(z) \, \mathcal{E}^n(z) &= 0. \label{kintegral-zero} \end{align}\tag{49}\] In this case, we obtain a \((k-1)\)-subharmonic function \(v^{\infty}\) in \(\mathbb{C}^n\) and \(v^{\infty}\) solves \(\sigma_{k-1}+\sigma_{k}=0\) in \(\mathbb{C}^{n-1}\), has bounded Laplacian and \(v^{\infty}+|z_{n+1}|^2\) is \(k\)-subharmonic in \(\mathbb{C}^{n}\). Since \((n-1)-k=m\), by induction, \(v^{\infty}\) must be a constant.

Case 2: Suppose Case 1 does not hold. Then for every \(\rho > 0\), there exists a constant \(C_{\rho} > 0\) with the following property: for all \(\epsilon < C_{\rho}^{-1}\), \(r > C_{\rho}\), \(x \in \mathbb{C}^{n}\), and any unit \((1,0)\)-vector \(\xi\), when \(\frac{1}{2}\bigl(\bigl[\frac{(v^{\epsilon})^2}{3}\bigr]_{r}(z) -{\tau_0}|z|^2+ [v]_{\rho}(z) \bigr) - v(x) \ge 0\), we have \[\begin{align} \label{ksecondcase} \int_{B_{r}(x)} |v^{\epsilon}_{\xi}|^2(z) \, \mathcal{E}^n(z) \ge c_0 := C_{\rho}^{-1}. \end{align}\tag{50}\] As before, the open set \[\Omega := \Bigl\{z \in \mathbb{C}^n : v(z) <w:=\frac{1}{2}\Bigl(\Bigl[\frac{(v^{\epsilon})^2}{3}\Bigr]_{r}(z) -{\tau_0}|z|^2+ [v]_{\rho}(z) - \frac{7}{6}\Bigr) \Bigr\}.\] is nonempty and bounded.

We will show \(\frac{1}{2}\bigl(\bigl[\frac{(v^{\epsilon})^2}{3}\bigr]_{r}(z) -{\tau_0}|z|^2+ [v]_{\rho}(z) \bigr)\) is a subsolution to \(\sigma_{k-1} + \sigma_{k} = 0\) in \(\Omega\). Assuming this, the comparison principle (Lemma 10) implies that \(v \ge w\) in \(\Omega\), contradicting the definition of \(\Omega\).

We still have \([v]_{r}\), \([v]_{\rho}\), \(\frac{(v^{\epsilon})^2}{2}\), and \(\bigl[\frac{(v^{\epsilon})^2}{2}\bigr]_{r}\) are all subsolutions to ?? .

From 50 , there exists a positive constant \(b_0\), depending only on \(c_0\) and \(C(n,k)\), such that \(\bigl[\frac{(v^{\epsilon})^2}{2}\bigr]_{r} - b_0|z|^2\) is \((k-1)\)-subharmonic. By the concavity of \(\sigma_{k-1}^{1/(k-1)}\) in \(\Gamma_{k-1}\), we have \[\begin{align} \label{kb0bound} \sigma_{k-1}\Bigl(\sqrt{-1}\partial\bar{\partial}\Bigl[\frac{(v^{\epsilon})^2}{2}\Bigr]_{r}\Bigr) \ge C_{n}^{k-1}b_0^{k-1}. \end{align}\tag{51}\]

We now show that \(F := \bigl[\frac{(v^{\epsilon})^2}{3}\bigr]_{r} - \tau_0|z|^2\) is a subsolution to \(\sigma_k + \sigma_{k-1} = 0\) in \(\Omega\) for sufficiently small \(\tau_0\). Indeed, by 51 , we have: \[\begin{align} &\sigma_k(\sqrt{-1}\partial\bar{\partial} F) + \sigma_{k-1}(\sqrt{-1}\partial\bar{\partial} F) \\ =& \sigma_k\Bigl(\sqrt{-1}\partial\bar{\partial} \Bigl[\frac{(v^{\epsilon})^2}{3}\Bigr]_{r}\Bigr) + \sigma_{k-1}\Bigl(\sqrt{-1}\partial\bar{\partial} \Bigl[\frac{(v^{\epsilon})^2}{3}\Bigr]_{r}\Bigr) + O(\tau_0) \\ =& \Bigl(\frac{2}{3}\Bigr)^k \Bigl[\sigma_k\Bigl(\sqrt{-1}\partial\bar{\partial} \Bigl[\frac{(v^{\epsilon})^2}{2}\Bigr]_{r}\Bigr) + \frac{3}{2} \sigma_{k-1}\Bigl(\sqrt{-1}\partial\bar{\partial} \Bigl[\frac{(v^{\epsilon})^2}{2}\Bigr]_{r}\Bigr)\Bigr] + O(\tau_0) \\ \ge& \frac{1}{2}\Bigl(\frac{2}{3}\Bigr)^{k} \sigma_{k-1}\Bigl(\sqrt{-1}\partial\bar{\partial} \Bigl[\frac{(v^{\epsilon})^2}{2}\Bigr]_{r}\Bigr) + O(\tau_0) \\ \ge& \frac{1}{2}\Bigl(\frac{2}{3}\Bigr)^kC_{n}^{k-1}b_0^{k-1}-C\tau_0\\ >&0 \end{align}\] where \(O(\tau_0)=\sum_{j=1}^{k-1}C_{j}\tau^{k-j}\sigma_{j}\bigl(\sqrt{-1}\partial\bar{\partial} \bigl[\frac{(v^{\epsilon})^2}{2}\bigr]_{r}\bigr)\) and since \([(v^{\epsilon})^2]_{r}\) has uniform \(C^{0,1}\)-bounds and complex Hessian bounds, we have \(|O(\tau_0)|\le C\tau_0\). Then \(F\) is a subsolution in \(\Omega\) if \(\tau_0\) is sufficiently small.

Similar to Lemma 12 \((iv)\), we get \(\frac{1}{2}\bigl(\bigl[\frac{(v^{\epsilon})^2}{3}\bigr]_{r}(z) -{\tau_0}|z|^2+ [v]_{\rho}(z) \bigr)\) is a subsolution to the equation \(\sigma_{k}+\sigma_{k-1}=0\) in \(\Omega\). ◻