Let $M$ be a complex manifold or more generally an almorst complex manifold $(M,J)$ with $J \in End(TM)$ such that $J^2=-Id_{TM}$ (if $M$ is complex mfd then it's almost complex al well with $J:=i \cdot id_{TM}$.
Assume $L \to M$ is a (holomorphic) line bundle on $M$ and $\nabla^L: \Gamma(M,L) \to \Gamma(M T^*M \otimes L)$ is a connection on $L$. Squaring $\nabla^E$ induces a curvature $\Omega^E: \Gamma(M,L) \to \Gamma(M, \wedge^2 T^*M \otimes L)$.
It's a general fact that for arbitrary finite dimensional vector bundle $E \to M$ a connection $\Omega^E$ uniquely determines a connection $\Omega^{E^*}$ on the dual vector bundle $E^* \to M$ by: for every section $s \in \Gamma(M,E), f \in \Gamma(M,E^*)$ the induced connection $\Omega^{E^*}$ is determined by equation
$$(\Omega^{E^*} f)(s) = d(f(s))- f(\Omega^{E}s)$$
Let come back to line bundle $L$ (I not exclude that this holds more generally for (holomorphic) vector bundles $E \to M$). Assume $\Omega^L$ is negative. That means that at every point $x \in M$ and local section $s \in \Gamma(U, L)$ where $x \in U \subset_{open} M$ the curvature evaluated in $s_x=s(x) \in L_x$ as $\Omega^L(-,J_x(-))(s_x) \in \wedge^2 T_x^*M$ is negative definite. That means for arbitrary $X \in T_x M$ with $x \neq 0$ we have
$$\Omega^L(X,J_x(X))(s_x) <0$$
My question: If we assume that $\Omega^L$ is negative, is then $\Omega^{L^*}$ positive? ie for all local $f \in \Gamma(U, L^*)$ holds $\Omega^{L^*}(X,J_x(X))(s^*_x) >0$ for all $X \in T_x M$.
the question is motivated by the discussion in this question around the exercise to prove that $c_1(O(1))$ of the dual $O(1)$ of the tautological bundle $O(-1)$ on $M= \mathbb{P^nC}$ is positive. Since $c_1$ not depends of choosen connection it suffice to find a connection on $O(1)$ with positive curvature.