Let $X^n$ be a compact complex manifold, and $\omega$ be a Hermitian metric on $X$.

Define an operator $P:=i\Lambda_\omega \bar{\partial} \partial$ on the space of the smooth function $C^\infty(X, \mathbb{C})$, where $\Lambda_\omega$ is the dual operator of $L_{\omega} = \omega \wedge \cdot$.

It is clear that $P$ is an elliptic operator and not self-adjoint in general.

In fact, $P^\ast$ (w.r.t $L^2$-inner product) can be expressed by $P^\ast = \frac{i}{(n-1)!} \ast_{\omega} \bar{\partial} \partial L^{n-1}_{\omega}.$

In particular, the second principle symbol of $P$ is equal to the second principle symbol of $\Delta_{\bar{\partial}}$.

Combining this result and the theory of Fredholm operators provides that the $\mathrm{ind}(P)=0$.

Applying maximum principle and calculating locally, $\mathrm{ker}(P)=\mathbb{C}$, and function $f \in \mathrm{im}(P|_{C^\infty(X,\mathbb{R})})$ are not non-positive or not non-negative (ie. not constant sign) other than the zero function.

For the adjoint operator $P^\ast$, the dimension of the kernel can be obtained by $$\mathrm{dim}\, \mathrm{ker} (P^\ast) = \mathrm{dim}\, \mathrm{coker} (P) = \mathrm{dim}\, \mathrm{ker} (P) = 1.$$

I would like to show that every real smooth function $f \in \mathrm{ker} (P^\ast|_{C^\infty(X, \mathrm{R})})$ are always non-positive or non-negative.

It is obvious to see that $\mathrm{ker} (P^\ast)$ is orthogonal complement to $\mathrm{im} (P)$ in the $L^2$-inner product.

In the appendix of the Kobayashi-Hitchin correspondence, Lübke and Teleman said that the orthogonal decomposition of $C^\infty(X, \mathbb{R}) = \mathrm{ker} (P^\ast|_{C^\infty(X, \mathrm{R})}) \oplus \mathrm{im} (P^\ast|_{C^\infty(X, \mathrm{R})})$ would provide that the result I want.

How can I gain that every real smooth function $f \in \mathrm{ker} (P^\ast|_{C^\infty(X, \mathrm{R})})$ are always non-positive or non-negative by the previous statement?