\section{The Wu--Yau theorem and its generalizations} In this section we go into the details of the proof of Wu--Yau's theorem on the positivity of the canonical class for projective manifolds endowed with a KŠhler metric of negative holomorphic sectional curvature. We will present a proof which follows, for the first part, almost \textsl{verbatim} the original proof of Wu and Yau. On the other hand, the conclusion will be achieved with an approach which is more pluripotential in flavor, taken from \cite{DT16}. Finally, we shall discuss at the end of this section several generalizations of this result (including the KŠhler case, and weaker notions of negativity). The proof is achieved in essentially three steps, after a reduction as follows. As we have seen, the negativity of the curvature (or even its non-positivty) implies the non existence of rational curves on $X$. Then, by Mori's Cone Theorem, we deduce that the canonical bundle of $X$ is nef. But then, it is sufficient to prove that $c_1(K_X)^n>0$, which in this case means that the canonical bundle is big. Indeed, if $K_X$ is big and there are no rational curves on $X$ one can conclude the ampleness of the canonical bundle via the following standard lemma. \begin{lemma}[Exercise 8, page 219 of \cite{Deb01}]\label{lem:ratcurv} Let $X$ be a smooth projective variety of general type which contains no rational curves. Then, $K_X$ is ample. \end{lemma} Here is a proof, for the sake of completeness. \begin{proof} Since there are no rational curves on $X$, Mori's theorem implies as above that $K_X$ is nef. Since $K_X$ is big and nef, the Base Point Free theorem tells us that $K_X$ is semi-ample. If $K_X$ were not ample, then the morphism defined by (some multiple of) $K_X$ would be birational but not an isomorphism. In particular, there would exist an irreducible curve $C\subset X$ contracted by this morphism. Therefore, $K_X\cdot C=0$. Now, take any very ample divisor $H$. For any $\varepsilon>0$ rational and small enough, $K_X-\varepsilon H$ remains big and thus some large positive multiple, say $m(K_X-\varepsilon H)$, of $K_X-\varepsilon H$ is linearly equivalent to an effective divisor $D$. Set $\Delta=\varepsilon' D$, where $\varepsilon'>0$ is a rational number. We have: $$ \begin{aligned} (K_X+\Delta)\cdot C & =\varepsilon'\,D\cdot C \\ & =\varepsilon'm(K_X-\varepsilon H)\cdot C \\ & =-\varepsilon\varepsilon'm\,H\cdot C<0. \end{aligned} $$ Finally, if $\varepsilon'$ is small enough, then $(X,\Delta)$ is a klt pair. Thus, the (logarithmic version of the) Cone Theorem would give the existence of an extremal ray generated by the class of a rational curve in $X$, contradiction. \end{proof} Keeping in mind that what we have to show is then that $c_1(K_X)^n>0$, we illustrate now the steps of the proof. \subsubsection*{Step 1: Solving an approximate KŠhler--Einstein equation.} Let $\omega$ be our fixed KŠhler metric (with negative holomorphic sectional curvature, but we shall not use this hypothesis for the moment). \begin{claim}\label{step1} For each $\varepsilon>0$ there exists a unique smooth function $u_\varepsilon \colon X\to\mathbb R$ such that $$ \omega_\varepsilon:=\varepsilon\omega-\operatorname{Ric}_\omega+i\partial\bar\partial u_{\varepsilon} $$ is a positive $(1,1)$-form (hence KŠhler, belonging to the cohomology class $c_1(K_X)+\varepsilon[\omega])$ form satisfying the Monge--AmpŹre equation $$ \omega_\varepsilon^n=e^{u_\varepsilon}\,\omega^n. $$ In particular, $$ \operatorname{Ric}(\omega_\varepsilon)=-\omega_\varepsilon+\varepsilon\omega, $$ whence the terminology \lq\lq approximate KŠhler--Einstein\rq\rq{}, and we have the following uniform upper bound: $$ \sup_X u_\varepsilon\le C, $$ where the constant $C$ depends only on $\omega$ and $n=\dim X$. Observe finally, that in particular, $\operatorname{Ric}(\omega_\varepsilon)\ge-\omega_\varepsilon$. \end{claim} \subsubsection*{Step 2: A laplacian estimate involving the holomorphic sectional curvature.} This step is somehow a refinement of the laplacian estimate needed in order to achieve the classical $C^2$-estimates to solve the complex Monge--AmpŹre equation on compact KŠhler manifolds. In the classical setting an upper bound for the holomorphic bisectional curvature is used. Here we shall employ a lemma due to Royden in order to use only the weaker information given by the bound on the holomorphic sectional curvature, as in the hypotheses. The crucial part of this step is the following. \begin{claim}\label{step2} Suppose $-\kappa<0$ is an upper bound for the holomorphic sectional curvature of $\omega$. Suppose moreover that $\omega'$ is another KŠhler metric on $X$ whose Ricci curvature is comparable with $\omega$ and $\omega'$ as follows: $$ \operatorname{Ric}(\omega')\ge-\lambda\omega'+\mu\omega, $$ where $\lambda,\mu$ are non negative constants. Define a smooth function $S\colon X\to\mathbb R_{>0}$ to be the trace of $\omega$ with respect to $\omega'$, \textsl{i.e.} $$ S:=\operatorname{tr}_{\omega'}\omega=n\,\frac{\bigl(\omega'\bigr)^{n-1}\wedge\omega}{\bigl(\omega'\bigr)^{n}}. $$ Then, the following differential inequality holds: \begin{equation}\label{diffineq} -\Delta_{\omega'}\log S\ge\biggl(\frac{\kappa(n+1)}{2n}+2\pi\frac\mu n\biggr)S(x_0)-2\pi\lambda. \end{equation} \end{claim} Observing that $$ \int_X\Delta_{\omega'}\log S\,\bigl(\omega'\bigr)^n=0, $$ we shall use the inequality above with $\omega'=\omega_\varepsilon$, $\lambda=1$, and $\mu=0$, in the following integral form: \begin{equation}\label{intineq} \int_X\frac{(n+1)\kappa}{2n}S_\varepsilon\,\omega_\varepsilon^n\le 2\pi\int_X\omega_\varepsilon^n, \end{equation} where we added the subscript $\varepsilon$ to $S$ in order to emphasize the dependence of $S$ from $\varepsilon$. \subsubsection*{Step 3: Proof of the key inequality.} One wants to show that $c_1(K_X)^n>0$. Since $\omega_\varepsilon=-\operatorname{Ric}(\omega_\varepsilon)+\varepsilon\omega$, we have that $$ \omega_\varepsilon^n=\bigl(-\operatorname{Ric}(\omega_\varepsilon)\bigr)^n+O(\varepsilon). $$ But then, $$ \int_X\omega_\varepsilon^n=\int_X\bigl(-\operatorname{Ric}(\omega_\varepsilon)\bigr)^n+O(\varepsilon). $$ On the other hand, $$ \int_X\bigl(-\operatorname{Ric}(\omega_\varepsilon)\bigr)^n=c_1(K_X)^n $$ is independent from $\varepsilon$, and thus $$ c_1(K_X)^n=\lim_{\varepsilon\to 0^+}\int_X\omega_\varepsilon^n. $$ What we want is therefore to show the positivity of such a limit. \begin{claim}\label{step3} The limit $$ \lim_{\varepsilon\to 0^+}\int_X\omega_\varepsilon^n $$ is strictly positive. \end{claim} This is what we call the key inequality. During the proof of the main result, this will be the only step where what we present here differs from Wu--Yau's original approach. \bigskip We now proceed with the proof of the various claims stated above. \begin{proof}[Proof of Claim \ref{step1}] The first observation is that, since $K_X$ is nef, for each $\varepsilon>0$ the cohomology class $c_1(K_X)+\varepsilon\,[\omega]=-c_1(X)+\varepsilon\,[\omega]$ is a KŠhler class. This implies, thanks to the $\partial\bar\partial$-lemma, that there exists a smooth real function $f_\varepsilon$ on $X$, unique up to an additive constant, such that $$ \omega_{f_\varepsilon}:=\varepsilon\,\omega-\Ric_\omega+\frac i{2\pi}\,\partial\bar\partial f_\varepsilon $$ is a KŠhler form on $X$. Now we use the following theorem, in order to obtain an approximate KŠhler--Einstein metric on $X$. We give here Yau's original general statement, which is the key ingredient to get his celebrated solution of the Calabi conjecture. \begin{theorem}[Yau \cite{Yau78}]\label{solCalabi} Let $(X,\omega_0)$ be a compact KŠhler manifold, and $F\colon X\times\mathbb R_t\to\mathbb R$ smooth function such that $\partial F/\partial t\ge 0$. Suppose that there exists smooth function $\psi\colon X\to\mathbb R$ such that $$ \int_X e^{F(x,\psi(x))}\,\omega_0^n=\int_X\omega^n_0. $$ Then, there exists a unique smooth function $\varphi\colon X\to\mathbb R$, such that $$ \begin{cases} \omega_0+\frac i{2\pi}\,\partial\bar\partial\varphi>0,\\ \bigl(\omega_0+\frac i{2\pi}\,\partial\bar\partial\varphi\bigr)^n=e^{F(x,\varphi(x))}\,\omega_0^n. \end{cases} $$ \end{theorem} From this statement one can derive easily both the existence of a KŠhler metric in a fixed KŠhler class with prescribed volume form (or, equivalently, Ricci tensor), and the existence of KŠhler--Einstein metrics on compact KŠhler manifold with negative (resp. zero) real first Chern class. Now, we fix $\varepsilon>0$, and define a smooth real function $\alpha_\varepsilon$ on $X$ implicitly by $$ \omega_{f_\varepsilon}^n=e^{-\alpha_\varepsilon}\,\omega^n. $$ We then apply the theorem above with the following data: $$ \omega=\omega_{f_\varepsilon},\quad F(x,t)=t+\alpha_\varepsilon(x)+f_\varepsilon(x). $$ Then, there exists a unique smooth real function $v_\varepsilon$ such that $$ \begin{aligned} \biggl(\omega_{f_\varepsilon}+\frac i{2\pi}\,\partial\bar\partial v_\varepsilon\biggr)^n &= e^{v_\varepsilon+\alpha_\varepsilon+f_\varepsilon}\,\omega_{f_\varepsilon}^n \\ &= e^{v_\varepsilon+f_\varepsilon}\,\omega^n, \end{aligned} $$ and $$ \omega_\varepsilon:=\omega_{f_\varepsilon}+\frac i{2\pi}\,\partial\bar\partial v_\varepsilon>0 $$ on $X$. Now, define $u_\varepsilon$ to be the sum $f_\varepsilon+v_\varepsilon$, so that it holds $$ \omega_\varepsilon^n=e^{u_\varepsilon}\,\omega^n. $$ Thus, we get for the Ricci curvature of $\omega_\varepsilon$ $$ \begin{aligned} \Ric_{\omega_\varepsilon} &= -\frac i{2\pi}\partial\bar\partial\log\omega_\varepsilon^n \\ &=-\frac i{2\pi}\partial\bar\partial v_\varepsilon\underbrace{-\frac i{2\pi}\partial\bar\partial f_\varepsilon\overbrace{-\frac i{2\pi}\partial\bar\partial\log\omega^n}^{=\Ric_\omega}}_{=\varepsilon\,\omega-\omega_{f_\varepsilon}}\\ &=\varepsilon\,\omega-\omega_\varepsilon. \end{aligned} $$ In particular, $\Ric_{\omega_\varepsilon}\ge-\omega_\varepsilon$. Now, we use the maximum principle in order to obtain the desired uniform upper bound for $u_\varepsilon$. To do so, pick a point $x_0\in X$ such that $\sup_X u_\varepsilon=u_\varepsilon(x_0)$. Then, at this point we have that the complex hessian of $u_\varepsilon$ is negative semi-definite, \textsl{i.e.} $i\,\partial\bar\partial u_\varepsilon(x_0)\le 0$. Thus, $$ \begin{aligned} \omega_\varepsilon(x_0) &= \bigl(\varepsilon\,\omega-\Ric_\omega+\frac i{2\pi}\,\partial\bar\partial u_\varepsilon\bigr)(x_0)\\ &\le \bigl(\varepsilon\,\omega-\Ric_\omega\bigr)(x_0) \\ &\le \bigl(\varepsilon_0\,\omega-\Ric_\omega\bigr)(x_0), \end{aligned} $$ if $\varepsilon_0>\varepsilon$. Therefore, $$ \begin{aligned} e^{\sup_X u_\varepsilon} &= e^{u_\varepsilon(x_0)} \\ &= \frac{\bigl(\varepsilon\,\omega-\Ric_\omega+\frac i{2\pi}\,\partial\bar\partial u_\varepsilon\bigr)(x_0)}{\omega^n(x_0)} \\ &\le\frac{\bigl(\varepsilon_0\,\omega-\Ric_\omega\bigr)(x_0)}{\omega^n(x_0)}=:e^C, \end{aligned} $$ so that $$ \sup_X u_\varepsilon\le C,\quad\forall\varepsilon<\varepsilon_0. $$ This complete the proof of Claim \ref{step1}. \end{proof} \begin{proof}[Proof of Claim \ref{step2}] Let $x_0\in X$ be a fixed point. Chose holomorphic normal coordinates $(z_1,\dots,z_n)$ with respect to $\omega$, centered at $x_0$. Without loss of generality, by a constant $\omega$-unitary change of variables, we may also suppose that $\omega'$ is diagonalized with respect to $\omega$ at $x_0$. Thus we write $$ \omega=i\sum_{l,m=1}^n\omega_{lm}\,dz_l\wedge d\bar z_m, \quad\omega_{lm}(z)=\delta_{lm}-\sum_{j,k=1}^n c_{jklm}\,z_j\bar z_k +O(|z^3|), $$ where the $c_{jklm}$'s are the coefficients of the Chern curvature tensor of $(X,\omega)$ at $x_0$, and $$ \omega'=i\sum_{l,m=1}^n\omega'_{lm}\,dz_l\wedge d\bar z_m,\quad \omega'_{lm}(z)=\lambda_l\,\delta_{lm}+O(|z|), $$ where the $\lambda_j$'s are the eigenvalues at $x_0$ of $\omega'$ with respect to $\omega$. In particular, $\lambda_j>0$, $j=1,\dots,n$. Next, call $\rho'_{jk}$ the coefficients of the Ricci curvature of $\omega'$, so that $$ \Ric_{\omega'}=\frac i{2\pi}\sum_{j,k=1}^n \rho'_{jk}\,dz_j\wedge d\bar z_k, $$ where $$ \rho'_{jk}=\sum_{l=1}^nc'_{jkll}. $$ With these notations, the starting point is the following \begin{lemma}[See {\cite[pag. 371]{WYZ09}}] The following differential equality holds: \begin{equation}\label{difeq} -\Delta_{\omega'}S(x_0)=\sum_{l=1}^n\frac{\rho'_{ll}}{(\lambda_l)^2} +\sum_{j,l,a=1}^n\frac {\bigl|\partial\omega'_{al}/\partial z_j\bigr|^2}{\lambda_j(\lambda_l)^2\lambda_a}-\sum_{j,l=1}^n\frac{c_{jjll}}{\lambda_j\lambda_l}, \end{equation} where the right hand side is intended to be computed at $x_0$. \end{lemma} \begin{proof} A straightforward computation, using the adjugate matrix method to obtain the inverse, shows that $$ S=\sum_{l,m=1}^n\Omega'_{ml}\,\omega_{lm}, $$ where we define $(\Omega'_{lm})$ to be the inverse matrix of $(\omega'_{lm})$. We want to compute $\Delta_{\omega'}S$ at $x_0$. We have, by the basic commutation relations in KŠhler geometry, $$ \Delta_{\omega'}S=\bar\partial^*\bar\partial S=-i\Lambda_{\omega'}\partial\bar\partial S, $$ and thus, since acting with $\Lambda_{\omega'}$ on real $(1,1)$-forms amounts to takeing the trace with respect to $\omega'$, $$ \Delta_{\omega'}S=-\operatorname{tr}_{\omega'}i\partial\bar\partial S=-\sum_{j,k=1}^n\Omega'_{kj}\,\frac{\partial^2 S}{\partial z_j\partial\bar z_k}. $$ Now, $$ \begin{aligned} \frac{\partial^2 S}{\partial z_j\partial\bar z_k} &=\frac{\partial^2}{\partial z_j\partial\bar z_k}\sum_{l,m=1}^n\Omega'_{ml}\,\omega_{lm} \\ &=\sum_{l,m=1}^n\omega_{lm}\frac{\partial^2\Omega'_{ml}}{\partial z_j\partial\bar z_k}+ \Omega'_{ml}\frac{\partial^2\omega_{lm}}{\partial z_j\partial\bar z_k} +\underbrace{\frac{\partial\Omega'_{ml}}{\partial z_j}\frac{\partial\omega_{lm}}{\partial\bar z_k} +\frac{\partial\omega_{lm}}{\partial z_j}\frac{\partial\Omega'_{ml}}{\partial\bar z_k}}_{:=R_{jklm}}. \end{aligned} $$ At the end of the day, thanks to the choice of geodesic coordinates, the terms with only one derivative involved ---which we called $R_{jklm}$--- will disappear. Therefore, we only have to understand the summands with two derivatives of $\Omega'_{ml}$, and express them in terms of the $\omega'_{lm}$'s. In order to do this, call $H=(\omega'_{lm})$, so that $H^{-1}=(\Omega'_{lm})$ and observe that \begin{equation}\label{partialH-1} 0\equiv\partial(HH^{-1})=\partial HH^{-1}+H\partial H^{-1}, \end{equation} \begin{equation}\label{barpartialH-1} 0\equiv\bar\partial(H H^{-1})=\bar\partial H H^{-1}+H\bar\partial H^{-1}, \end{equation} and $$ 0\equiv\partial\bar\partial(HH^{-1})=\partial\bar\partial H H^{-1}-\bar\partial H\wedge\partial H^{-1}+\partial H\wedge\bar\partial H^{-1}+H\partial\bar\partial H^{-1}. $$ We obtain therefore the matrix identity \begin{multline*} \partial\bar\partial H^{-1}=-H^{-1}\partial\bar\partial H H^{-1}-H^{-1}\bar\partial H\wedge H^{-1}\partial H H^{-1} \\ +H^{-1}\partial H\wedge H^{-1}\bar\partial H H^{-1}, \end{multline*} which gives us the following expression for the second derivatives of $\Omega'_{ml}$: \begin{multline*} \frac{\partial^2\Omega'_{ml}}{\partial z_j\partial\bar z_k}=-\sum_{a,b=1}^n\Omega'_{ma}\frac{\partial^2\omega'_{ab}}{\partial z_j\partial\bar z_k}\Omega'_{bl} \\+\sum_{a,b,p,q=1}^n\Omega'_{mp}\frac{\partial\omega'_{pq}}{\partial\bar z_k}\Omega'_{qa}\frac{\partial\omega'_{ab}}{\partial z_j}\Omega'_{bl} +\Omega'_{ma}\frac{\partial\omega'_{ab}}{\partial z_j}\Omega'_{bp}\frac{\partial\omega'_{pq}}{\partial\bar z_k}\Omega'_{ql}. \end{multline*} Thus, we get the following expression for the $\omega'$-Laplacian: \begin{multline}\label{exprlapl} \Delta_{\omega'}S=-\sum_{j,k,l,m=1}^n\Omega'_{kj}\biggl(\omega_{lm}\frac{\partial^2\Omega'_{ml}}{\partial z_j\partial\bar z_k}+ \Omega'_{ml}\frac{\partial^2\omega_{lm}}{\partial z_j\partial\bar z_k} + R_{jklm} \biggr) \\ = -\sum_{j,k,l,m=1}^n\Omega'_{kj}\Omega'_{ml}\frac{\partial^2\omega_{lm}}{\partial z_j\partial\bar z_k} +\Omega'_{kj}\omega_{lm}\frac{\partial^2\Omega'_{ml}}{\partial z_j\partial\bar z_k}+\Omega'_{kj}\,R_{jklm}\\ =-\sum_{j,k,l,m=1}^n\Omega'_{kj}\Omega'_{ml}\frac{\partial^2\omega_{lm}}{\partial z_j\partial\bar z_k} +\Omega'_{kj}\,R_{jklm}\\ +\sum_{j,k,l,m,a,b=1}^n\omega_{lm}\Omega'_{kj}\Omega'_{ma}\Omega'_{bl}\frac{\partial^2\omega'_{ab}}{\partial z_j\partial\bar z_k} \\ -\sum_{j,k,l,m,a,b,p,q=1}^n\omega_{lm}\Omega'_{kj}\Omega'_{mp}\Omega'_{qa}\Omega'_{bl}\frac{\partial\omega'_{ab}}{\partial z_j}\frac{\partial\omega'_{pq}}{\partial\bar z_k}\\ -\sum_{j,k,l,m,a,b,p,q=1}^n\omega_{lm}\Omega'_{kj}\Omega'_{ma}\Omega'_{bp}\Omega'_{ql}\frac{\partial\omega'_{ab}}{\partial z_j}\frac{\partial\omega'_{pq}}{\partial\bar z_k}. \end{multline} Now, still denoting by $H$ the matrix $(\omega'_{lm})$, we recall the well-known formula to determine in local coordinates the Chern curvature of $\omega'$, namely $$ \begin{aligned} \Theta(T_X,\omega') & \simeq_{\textrm{loc}}\bar\partial\bigl(\bar H^{-1}\partial\bar H\bigr) \\ &=\bar\partial\bar H^{-1}\wedge\partial \bar H+\bar H^{-1}\bar\partial\partial\bar H \\ &=-\bar H^{-1}\bar\partial\bar H\bar H^{-1}\wedge\partial\bar H^{-1}+\bar H^{-1}\bar\partial\partial\bar H, \end{aligned} $$ where the last equality is obtain by using formula (\ref{barpartialH-1}). So, if we write in these coordinates $$ \Theta(T_X,\omega')=\sum_{j,k,l,m}c'_{jklm}\,dz_j\wedge d\bar z_k\otimes\biggl(\frac\partial{\partial z_l}\biggr)^*\otimes\frac\partial{\partial\bar z_m}, $$ we obtain the following expression for the coefficients of the Chern curvature tensor: \begin{equation}\label{c'jklm} c'_{jkal}=-\sum_{b=1}^n\Omega'_{bl}\frac{\partial^2\omega'_{ab}}{\partial z_j\partial\bar z_k} +\sum_{b,p,q=1}^n\Omega'_{pl}\Omega'_{bq}\frac{\partial\omega'_{ab}}{\partial z_j}\frac{\partial\omega'_{qp}}{\partial\bar z_k}. \end{equation} We can now use the above identity (\ref{c'jklm}) to replace in the right hand side of formula (\ref{exprlapl}) the summand $$ \sum_{b=1}^n\Omega'_{bl}\frac{\partial^2\omega'_{ab}}{\partial z_j\partial\bar z_k} $$ with $$ -c'_{jkal}+\sum_{b,p,q=1}^n\Omega'_{pl}\Omega'_{bq}\frac{\partial\omega'_{ab}}{\partial z_j}\frac{\partial\omega'_{qp}}{\partial\bar z_k}. $$ With this substitution, we obtain \begin{multline}\label{exprlaplsemifinal} \Delta_{\omega'}S =-\sum_{j,k,l,m=1}^n\Omega'_{kj}\Omega'_{ml}\frac{\partial^2\omega_{lm}}{\partial z_j\partial\bar z_k} +\Omega'_{kj}\,R_{jklm}\\ +\sum_{j,k,l,m,a=1}^n\omega_{lm}\Omega'_{kj}\Omega'_{ma}\biggl(-c'_{jkal}+\sum_{b,p,q=1}^n\Omega'_{pl}\Omega'_{bq}\frac{\partial\omega'_{ab}}{\partial z_j}\frac{\partial\omega'_{qp}}{\partial\bar z_k}\biggr) \\ -\sum_{j,k,l,m,a,b,p,q=1}^n\omega_{lm}\Omega'_{kj}\Omega'_{mp}\Omega'_{qa}\Omega'_{bl}\frac{\partial\omega'_{ab}}{\partial z_j}\frac{\partial\omega'_{pq}}{\partial\bar z_k}\\ -\sum_{j,k,l,m,a,b,p,q=1}^n\omega_{lm}\Omega'_{kj}\Omega'_{ma}\Omega'_{bp}\Omega'_{ql}\frac{\partial\omega'_{ab}}{\partial z_j}\frac{\partial\omega'_{pq}}{\partial\bar z_k} \\ =-\sum_{j,k,l,m=1}^n\Omega'_{kj}\Omega'_{ml}\frac{\partial^2\omega_{lm}}{\partial z_j\partial\bar z_k} +\Omega'_{kj}\,R_{jklm}\\ -\sum_{j,k,l,m,a=1}^n\omega_{lm}\Omega'_{kj}\Omega'_{ma}c'_{jkal} \\ -\sum_{j,k,l,m,a,b,p,q=1}^n\omega_{lm}\Omega'_{kj}\Omega'_{mp}\Omega'_{qa}\Omega'_{bl}\frac{\partial\omega'_{ab}}{\partial z_j}\frac{\partial\omega'_{pq}}{\partial\bar z_k}. \end{multline} Now, since $(\partial/\partial z_1,\dots,\partial/\partial z_n)$ is merely $\omega'$-orthogonal but not necessarily $\omega'$-unitary at $x_0$, the KŠhler symmetries of the coefficients $c'_{jklm}$'s at $x_0$ read $$ c'_{jklm}\sqrt{\lambda_l}\sqrt{\lambda_m}=c'_{lmjk}\sqrt{\lambda_j}\sqrt{\lambda_k}. $$ In particular, $$ c'_{jjll}\lambda_l=c'_{lljj}\lambda_j. $$ We are now in a good position to conclude the proof of the lemma. Indeed, evaluating (\ref{exprlaplsemifinal}) at the point $x_0$ with our initial choice of coordinates gives \begin{equation}\label{exprlaplfinal} \begin{aligned} \Delta_{\omega'}S(x_0) &=\sum_{j,l=1}^n\frac{c_{jjll}}{\lambda_j\lambda_l} -\sum_{j,l=1}^n\underbrace{\frac{c'_{jjll}}{\lambda_j\lambda_l}}_{=\frac{c'_{lljj}}{(\lambda_l)^2}} -\sum_{j,l,a=1}^n\frac 1{\lambda_j(\lambda_l)^2\lambda_a}\frac{\partial\omega'_{al}}{\partial z_j}\frac{\partial\omega'_{la}}{\partial\bar z_j} \\ &=\sum_{j,l=1}^n\frac{c_{jjll}}{\lambda_j\lambda_l}-\sum_{l=1}^n\frac{\rho'_{ll}}{(\lambda_l)^2} -\sum_{j,l,a=1}^n\frac {|\partial\omega'_{al}/\partial z_j(x_0)|^2}{\lambda_j(\lambda_l)^2\lambda_a}. \end{aligned} \end{equation} \end{proof} Our next task will be to estimate the three summands appearing on the right hand side of the differential equality of the above lemma. We begin with the term involving the Ricci curvature of $\omega'$. Recall that the we are supposing that $$ \operatorname{Ric}(\omega')\ge-\lambda\omega'+\mu\omega. $$ \begin{lemma}\label{term1} At the point $x_0\in X$, we have $$ \sum_{l=1}^n\frac{\rho'_{ll}}{(\lambda_l)^2}\ge 2\pi\biggl(-\lambda S+\frac\mu n S^2\biggr). $$ \end{lemma} \begin{proof} The hypothesis on the Ricci curvature of $\omega'$, when red at the point $x_0$ with our choice of coordinates, gives $$ \rho'_{ll}\ge 2\pi(-\lambda\lambda_l+\mu). $$ Thus, we get $$ \sum_{l=1}^n\frac{\rho'_{ll}}{(\lambda_l)^2}\ge -2\pi\lambda\sum_{l=1}^n\frac{1}{\lambda_l}+2\pi\mu\sum_{l=1}^n\frac{1}{(\lambda_l)^2}. $$ Now, since the $\lambda_l$'s are the eigenvalues of $\omega'$ with respect to $\omega$, the eigenvalues of omega with respect to $\omega'$ are $1/\lambda_l$, $l=1,\dots,n$, and therefore $S=\sum_{l=1}^n1/\lambda_l$. Moreover, by the standard inequality between $1$-norm and $2$-norm of vectors in $\mathbb R^n$, we have $\sum_{l=1}^n1/(\lambda_l)^2\ge 1/n\bigl(\sum_{l=1}^n1/\lambda_l\bigr)^2$. We finally obtain $$ \sum_{l=1}^n\frac{\rho'_{ll}}{(\lambda_l)^2}\ge 2\pi\biggl(-\lambda S+\frac\mu n S^2\biggr). $$ \end{proof} Now, we treat the term with the first order derivatives of the metric $\omega'$. In doing this, we have to keep in mind that, at the end of the day, we want to estimate $\Delta_{\omega'}\log S$. This Laplacian is given in coordinates by $$ \begin{aligned} \Delta_{\omega'}\log S&=-\sum_{j,k=1}^n\Omega'_{kj}\underbrace{\frac{\partial^2\log S}{\partial z_j\partial\bar z_k}}_{=\frac{\partial}{\partial z_j}\biggl(\frac 1S \frac{\partial S}{\partial\bar z_k}\biggr)=-\frac 1{S^2}\frac{\partial S}{\partial z_j}\frac{\partial S}{\partial\bar z_k}+\frac 1S\frac{\partial^2 S}{\partial z_j\partial\bar z_k}}\\ &=\frac 1S\Delta_{\omega'}S+\frac 1{S^2}\sum_{j,k=1}^n\Omega'_{kj}\frac{\partial S}{\partial z_j}\frac{\partial S}{\partial\bar z_k}. \end{aligned} $$ Once computed at $x_0$, we have \begin{equation}\label{laplacianlog} \Delta_{\omega'}\log S(x_0)=\frac 1{S(x_0)}\Delta_{\omega'}S(x_0)+\frac 1{S(x_0)^2}\sum_{j=1}^n\frac 1{\lambda_j}\biggl|\frac{\partial S}{\partial z_j}(x_0)\biggr|^2. \end{equation} What we want to do in the lemma below is then to try to express these first order derivatives in terms of first order derivatives of $S$. \begin{lemma}\label{term2} At the point $x_0\in X$, we have $$ \sum_{j,l,a=1}^n\frac {\bigl|\partial\omega'_{al}/\partial z_j\bigr|^2}{\lambda_j(\lambda_l)^2\lambda_a}\ge\frac 1{S(x_0)}\sum_{j=1}^n\frac 1{\lambda_j}\biggl|\frac{\partial S}{\partial z_j}(x_0)\biggr|^2. $$ \end{lemma} \begin{proof} Since the sum we are dealing with is made up of non negative terms, we have by plain minoration $$ \sum_{j,l,a=1}^n\frac {\bigl|\partial\omega'_{al}/\partial z_j\bigr|^2}{\lambda_j(\lambda_l)^2\lambda_a}\ge \sum_{j,l=1}^n\frac {\bigl|\partial\omega'_{ll}/\partial z_j\bigr|^2}{\lambda_j(\lambda_l)^3}. $$ Now, let us compute $\partial S/\partial z_j$ at $x_0$. We have $$ \begin{aligned} \frac{\partial S}{\partial z_j}(x_0)&= \frac{\partial}{\partial z_j}\sum_{l,m=1}^n\Omega'_{ml}\omega_{lm}\biggr|_{x_0} \\ &= \sum_{l,m=1}^n\frac{\partial\Omega'_{ml}}{\partial z_j}\omega_{lm}+\Omega'_{ml}\frac{\partial\omega_{lm}}{\partial z_j}\biggr|_{x_0}=\sum_{l=1}^n\frac{\partial\Omega'_{ll}}{\partial z_j}(x_0). \end{aligned} $$ Now, we use the identity (\ref{partialH-1}) to replace $\partial\Omega'_{ll}/\partial z_j(x_0)$ with $$ -\sum_{a,b=1}^n\Omega'_{la}\frac{\partial\omega'_{ab}}{\partial z_j}\Omega'_{bl}\biggr|_{x_0}=-\frac 1{(\lambda_l)^2}\frac{\partial\omega'_{ll}}{\partial z_j}(x_0). $$ Thus, we obtain $$ \frac{\partial S}{\partial z_j}(x_0)=-\sum_{l=1}^n\frac 1{(\lambda_l)^2}\frac{\partial\omega'_{ll}}{\partial z_j}(x_0). $$ Now, inspired by (\ref{laplacianlog}), we compute $$ \begin{aligned} \sum_{j=1}^n\frac 1{\lambda_j}\biggl|\frac{\partial S}{\partial z_j}(x_0)\biggr|^2 &= \sum_{j=1}^n\frac 1{\lambda_j}\left|\sum_{l=1}^n\frac 1{(\lambda_l)^2}\frac{\partial\omega'_{ll}}{\partial z_j}(x_0)\right|^2 \\ &=\sum_{j=1}^n\frac 1{\lambda_j}\left|\sum_{l=1}^n\frac 1{(\lambda_l)^{1/2}}\frac{\partial\omega'_{ll}/\partial z_j(x_0)}{(\lambda_l)^{3/2}}\right|^2 \\ &\le \sum_{j=1}^n\frac 1{\lambda_j}\sum_{k=1}^n\frac 1{\lambda_k}\sum_{l=1}^n\frac{\bigl|\partial\omega'_{ll}/\partial z_j(x_0)\bigr|^2}{(\lambda_l)^{3}} \\ &=S(x_0)\sum_{l,j=1}^n\frac {\bigl|\partial\omega'_{ll}/\partial z_j(x_0)\bigr|^2}{\lambda_j(\lambda_l)^3}, \end{aligned} $$ where the inequality is given by Cauchy-Schwarz. The lemma follows. \end{proof} Finally, we estimate the term involving the curvature of $\omega$, using the hypothesis on the negativity of the holomorphic sectional curvature. \begin{lemma}\label{term3} At the point $x_0\in X$, we have $$ \sum_{j,l=1}^n\frac{c_{jjll}}{\lambda_j\lambda_l}\le -\frac{\kappa(n+1)}{2n}S^2. $$ \end{lemma} Classically this term has been bounded in terms of a uniform bound on the holomorphic bisectional curvature of $\omega$. In order to prove this lemma, we need to be able to transform an information on the sum of holomorphic bisectional curvature type terms into an estimate using holomorphic sectional curvature only. Next proposition in hermitian linear algebra is the key point to do that. It is due to Royden. \begin{proposition}[Royden \cite{Roy80}]\label{royden} Let $\xi_1,\dots,\xi_\nu$ be mutually orthogonal (but not necessarily unitary) non-zero vectors of a hermitian vector space $(V,h)$. Suppose that $\Theta(\xi,\eta,\zeta,\omega)$ is a symmetric \lq\lq bi-hermitian\lq\lq{} form, \textsl{i.e.} $\Theta$ is sesquilinear in the first two and last two variables having the same pointwise properties of the (contraction with the metric of the) Chern curvature of a KŠhler metric. Suppose also that there exists a real constant $K$ such that for all $\xi\in V$ one has $$ \Theta(\xi,\xi,\xi,\xi)\le K\,||\xi||^4_h. $$ Then, $$ \sum_{\alpha,\beta}\Theta(\xi_\alpha,\xi_\alpha,\xi_\beta,\xi_\beta)\le\frac 12\,K\,\left(\biggl(\sum_\alpha ||\xi_\alpha||^2_h\biggr)^2+\sum_\alpha||\xi_\alpha||^4_h\right). $$ Moreover, if $K\le 0$, then $$ \sum_{\alpha,\beta}\Theta(\xi_\alpha,\xi_\alpha,\xi_\beta,\xi_\beta)\le\frac{\nu+1}{2\nu}\,K\,\biggl(\sum_\alpha||\xi_\alpha||^2_h\biggr)^2. $$ \end{proposition} We shall use this proposition with $$ \bigl\langle\Theta(T_X,\omega)(\bullet,\bar\bullet)\cdot\bullet,\bullet\bigr\rangle_\omega $$ as the symmetric \lq\lq bi-hermitian\rq\rq{} form on $T_{X,x_0}$ in the statement. In terms of holomorphic bisectional curvature it can be rephrased as follows, when $\nu=n=\dim X$. \noindent \emph{Suppose that a KŠhler metric $\omega$ has negative holomorphic sectional curvature at the point $x_0$, bounded above by a negative constant $-\kappa$. Then, if $\xi_1,\dots,\xi_n$ is a $\omega$-orthogonal basis for $T_{X,x_0}$ we have $$ \sum_{\alpha,\beta=1}^n ||\xi_\alpha||^2_\omega ||\xi_\beta||^2_\omega\,\operatorname{HBC}_\omega(\xi_\alpha,\xi_\beta)\le-\frac{\kappa(n+1)}{2n}\biggl(\sum_{\alpha=1}^n||\xi_\alpha||^2_\omega\biggr)^2. $$} Here is the proof. \begin{proof} Realize $\mathbb Z_4$ as the group of 4th roots of unity and set, for a vector $A=(\epsilon_1,\dots,\epsilon_\nu)\in\mathbb Z^\nu_4$, $$ \xi_A=\sum_\alpha\epsilon_\alpha\,\xi_\alpha. $$ Then, by orthogonality, $||\xi_A||^2_h=\sum_\alpha ||\xi_\alpha||^2_h$, and thus by hypothesis $$ \Theta(\xi_A,\xi_A,\xi_A,\xi_A)\le K\,||\xi_A||^4_h= K\,\biggl(\sum_\alpha||\xi_\alpha||^2_h\biggr)^2. $$ Now, we take the sum over all possible $A\in\mathbb Z^\nu_4$ and get $$ \begin{aligned} K\,\biggl(\sum_\alpha||\xi_\alpha||^2_h\biggr)^2 &\ge\frac 1{4^\nu}\sum_{A}\Theta(\xi_A,\xi_A,\xi_A,\xi_A) \\ &=\frac 1{4^\nu}\sum_{A}\sum_{\alpha,\beta,\gamma,\delta}\epsilon_\alpha\bar\epsilon_\beta\epsilon_\gamma\bar\epsilon_\delta\,\Theta(\xi_\alpha,\xi_\beta,\xi_\gamma,\xi_\delta) \\ &=\frac 1{4^\nu}\sum_{\alpha,\beta,\gamma,\delta}\sum_{A}\frac{\epsilon_\alpha\epsilon_\gamma}{\epsilon_\beta\epsilon_\delta}\,\Theta(\xi_\alpha,\xi_\beta,\xi_\gamma,\xi_\delta). \end{aligned} $$ Now, fix a $4$-tuple $(\alpha,\beta,\gamma,\delta)$. We claim that only the terms with $\alpha=\beta$ and $\gamma=\delta$ or $\alpha=\delta$ and $\beta=\gamma$ can survive after summing over all $A$. Thus, we are left only with the following terms \begin{multline*} \frac 1{4^\nu}\sum_{\alpha,\beta,\gamma,\delta}\sum_{A}\frac{\epsilon_\alpha\epsilon_\gamma}{\epsilon_\beta\epsilon_\delta}\,\Theta(\xi_\alpha,\xi_\beta,\xi_\gamma,\xi_\delta) \\ = \sum_\alpha \Theta(\xi_\alpha,\xi_\alpha,\xi_\alpha,\xi_\alpha)+\sum_{\alpha\ne\gamma}\Theta(\xi_\alpha,\xi_\alpha,\xi_\gamma,\xi_\gamma)+\Theta(\xi_\alpha,\xi_\gamma,\xi_\gamma,\xi_\alpha). \end{multline*} The claim is straightforwardly verified, since for all the other terms, for each $A\in\mathbb Z^\nu_4$ one can find an $A'=(\epsilon_1',\dots,\epsilon_\nu')\in\mathbb Z^\nu_4$ such that $\frac{\epsilon_\alpha\epsilon_\gamma}{\epsilon_\beta\epsilon_\delta}=-\frac{\epsilon_\alpha'\epsilon_\gamma'}{\epsilon_\beta'\epsilon_\delta'}$. Now, by symmetry of $\Theta$, adding $\sum_\alpha \Theta(\xi_\alpha,\xi_\alpha,\xi_\alpha,\xi_\alpha)$ to both side and using the upper bound as in the hypotheses, we get $$ 2\sum_{\alpha,\gamma}\Theta(\xi_\alpha,\xi_\alpha,\xi_\gamma,\xi_\gamma)\le K\,\left(\biggl(\sum_\alpha ||\xi_\alpha||^2_h\biggr)^2+\sum_\alpha||\xi_\alpha||^4_h\right). $$ To end the proof, observe that applying the Cauchy--Schwarz inequality in $\mathbb R^\nu$ to the vectors $(||\xi_1||^2_h,\dots,||\xi_\nu||^2_h)$ and $(1,\dots,1)$, we have $$ \biggl(\sum_\alpha ||\xi_\alpha||^2_h\biggr)^2\le \nu\,\sum_\alpha||\xi_\alpha||^4_h, $$ so that if $K\le 0$, then $$ K\,\sum_\alpha||\xi_\alpha||^4_h\le \frac K\nu\,\biggl(\sum_\alpha ||\xi_\alpha||^2_h\biggr)^2, $$ and thus $$ \sum_{\alpha,\gamma}\Theta(\xi_\alpha,\xi_\alpha,\xi_\gamma,\xi_\gamma)\le\frac{\nu+1}{2\nu} K\,\biggl(\sum_\alpha ||\xi_\alpha||^2_h\biggr)^2, $$ as desired. \end{proof} We are now ready to give a \begin{proof}[Proof of Lemma \ref{term3}] Set $$ \xi_j:=\frac 1{\sqrt{\lambda_j}}\frac \partial{\partial z_j},\quad j=1,\dots n, $$ so that $\xi_1,\dots,\xi_n$ is a $\omega$-orthogonal basis for $T_{X,x_0}$. Now, it suffices to observe that $$ \begin{aligned} \frac{c_{jjll}}{\lambda_j\lambda_l}&=\bigl\langle\Theta(T_X,\omega)(\xi_j,\bar\xi_j)\cdot\xi_l,\xi_l\bigr\rangle_\omega \\ &= ||\xi_j||^2_\omega ||\xi_l||^2_\omega\,\operatorname{HBC}_\omega(\xi_j,\xi_l). \end{aligned} $$ Now take the sum over all $j,l=1,\dots,n$, to obtain $$ \sum_{j,l=1}^n\frac{c_{jjll}}{\lambda_j\lambda_l} \le -\frac{\kappa(n+1)}{2n}\biggl(\sum_{\alpha=1}^n\frac 1{\lambda_j}\biggr)^2 \\ =-\frac{\kappa(n+1)}{2n}S^2. $$ \end{proof} Now, to conclude the proof of Claim \ref{step2}, \textsl{i.e.} to show inequality (\ref{diffineq}), we just put together the three estimates of the above lemmata, and plug them into formula (\ref{laplacianlog}). We get: $$ \begin{aligned} -\Delta_{\omega'}\log S(x_0)&=-\frac 1{S(x_0)}\Delta_{\omega'}S(x_0)-\frac 1{S(x_0)^2}\sum_{j=1}^n\frac 1{\lambda_j}\biggl|\frac{\partial S}{\partial z_j}(x_0)\biggr|^2 \\ &=\frac 1{S(x_0)}\left(\sum_{l=1}^n\frac{\rho'_{ll}}{(\lambda_l)^2} +\sum_{j,l,a=1}^n\frac {\bigl|\partial\omega'_{al}/\partial z_j(x_0)\bigr|^2}{\lambda_j(\lambda_l)^2\lambda_a}-\sum_{j,l=1}^n\frac{c_{jjll}}{\lambda_j\lambda_l}\right)\\ &\qquad-\frac 1{S(x_0)^2}\sum_{j=1}^n\frac 1{\lambda_j}\biggl|\frac{\partial S}{\partial z_j}(x_0)\biggr|^2 \\ & \ge\frac 1{S(x_0)}\left(2\pi\biggl(-\lambda S(x_0)+\frac\mu n S(x_0)^2\biggr)\right. \\ &\left.\qquad+\frac 1{S(x_0)}\sum_{j=1}^n\frac 1{\lambda_j}\biggl|\frac{\partial S}{\partial z_j}(x_0)\biggr|^2+\frac{\kappa(n+1)}{2n}S(x_0)^2\right)\\ &\qquad\qquad-\frac 1{S(x_0)^2}\sum_{j=1}^n\frac 1{\lambda_j}\biggl|\frac{\partial S}{\partial z_j}(x_0)\biggr|^2 \\ &=\biggl(\frac{\kappa(n+1)}{2n}+2\pi\frac\mu n\biggr)S(x_0)-2\pi\lambda, \end{aligned} $$ as desired. \end{proof} \begin{proof}[Proof of Claim \ref{step3}] We want to show that the limit $$ \lim_{\varepsilon\to 0^+}\int_X\omega_\varepsilon^n=\lim_{\varepsilon\to 0^+}\int_X e^{u_\varepsilon}\,\omega^n $$ is strictly positive. The first observation is that the functions $u_\varepsilon$ are all $\omega'$-plurisubharmonic for some fixed KŠhler form $\omega'$ and $\varepsilon>0$ small enough. For, let $\ell>0$ be such that $\ell\omega-\Ric_\omega$ is positive and call $\omega'=\ell\omega-\Ric_\omega$. Thus, for all $0<\varepsilon<\ell$, one has $$ 0<\varepsilon\omega-\Ric_\omega+i\partial\bar\partial u_\varepsilon<\ell\omega-\Ric_\omega+i\partial\bar\partial u_\varepsilon=\omega'+i\partial\bar\partial u_\varepsilon. $$ Therefore, by \cite[Proposition 2.6]{GZ05}, either $\{u_\varepsilon\}$ converges uniformly to $-\infty$ on $X$ or it is relatively compact in $L^1(X)$. Suppose for a moment that we are in the second case. Then, there exists a subsequence $\{u_{\varepsilon_k}\}$ converging in $L^1(X)$ and moreover the limit coincides \textsl{a.e.} with a uniquely determined $\omega'$-plurisubharmonic function $u$. Up to pass to a further subsequence, we can also suppose that $u_{\varepsilon_k}$ converges pointwise \textsl{a.e.} to $u$. But then, $e^{u_{\varepsilon_k}}\to e^u$ pointwise \textsl{a.e.} on $X$. On the other hand, by Claim \ref{step1}, we have $e^{u_{\varepsilon_k}}\le e^C$ so that, by dominated convergence, we also have $L^1(X)$-convergence and $$ \lim_{k\to\infty}\int_X e^{u_{\varepsilon_k}}\,\omega^n=\int_X e^u\omega^n>0. $$ The upshot is that what we need to prove is that $\{u_\varepsilon\}$ does not converge uniformly to $-\infty$ on $X$. From now on, we shall suppose by contradiction that $$ \sup_X u_\varepsilon\to-\infty. $$ Now, recall that we defined the smooth positive function $S_\varepsilon$ on $X$ by $$ \omega\wedge\omega_\varepsilon^{n-1}=\frac{S_\varepsilon}n\,\omega_\varepsilon^{n}. $$ Now, set $T_\varepsilon=\log S_\varepsilon$. In other words, $T_\varepsilon$ is the logarithm of the trace of $\omega$ with respect to $\omega_\varepsilon$. \begin{lemma}\label{lem:sge-m} The function $T_\varepsilon$ satisfies the following inequality: $$ T_\varepsilon> -\frac{u_\varepsilon}n. $$ In particular, if $\{u_\varepsilon\}$ converges uniformly to $-\infty$ on $X$, then $T_\varepsilon$ converges uniformly to $+\infty$ on $X$. \end{lemma} \begin{proof} Let $0<\lambda_{1}\le\cdots\le\lambda_n$ be the eigenvalues of $\omega_\varepsilon$ with respect to $\omega$, so that $0<1/\lambda_{n}\le\cdots\le 1/\lambda_1$ are the eigenvalues of $\omega$ with respect to $\omega_\varepsilon$. Then, $$ e^{T_\varepsilon}=\operatorname{tr}_{\omega_\varepsilon}\omega=\frac 1{\lambda_1}+\cdots+\frac 1{\lambda_n}>\frac 1{\lambda_1}. $$ Thus, $e^{-T_\varepsilon}<\lambda_1$ so that $e^{-nT_\varepsilon}<(\lambda_1)^n\le\lambda_1\cdots\lambda_n$. But, $e^{u_\varepsilon}\omega^n=\omega_\varepsilon^n=\lambda_1\cdots\lambda_n\,\omega^n$, and so we get $ e^{-nT_\varepsilon}-\frac{u_\varepsilon}n. $$ \end{proof} As announced, we now use the integral inequality (\ref{intineq}). We write it as follows: $$ \frac{(n+1)\kappa}{2n}\int_X e^{T_\varepsilon+u_\varepsilon}\,\omega^n\le 2\pi\int_X e^{u_\varepsilon}\,\omega^n. $$ Next, if we define $C_\varepsilon:=\inf_X e^{-u_\varepsilon/n}$, we have that $e^{T_\varepsilon}> C_\varepsilon$, and thus \begin{equation}\label{eq:final} C_\varepsilon\,\frac{(n+1)\kappa}{2n}\int_X e^{u_\varepsilon}\,\omega^n\le 2\pi\int_X e^{u_\varepsilon}\,\omega^n. \end{equation} But then, we obtain that $$ C_\varepsilon\,\frac{(n+1)\kappa}{2n}\le 2\pi, $$ and this gives a contradiction, since we are assuming that $C_\varepsilon\to +\infty$ as $\varepsilon\to 0^+$. \end{proof} \subsection{The KŠhler case and quasi-negative holomorphic sectional curvature}