Let $(M,\omega)$ be a compact Kahler manifold. Given a vector field $V = V^i \partial_{z_i} + V^{\overline{i}} \partial_{\overline{z_i}}$ on $M$, how do I show the Bochner-Kodaira formula:
$||\nabla V||^2 = ||\overline{\nabla} V||^2 + \int_M R_{\overline{j}i}V^i \overline{V^{\overline{j}}} \omega^n$
$\nabla V$ is $(1,0)$-form $TM^{\mathbb{C}}$-valued given in local coordinate by:
$\nabla V = (\partial_jV^i + V^a\Gamma^i_{ja}) dz_j \otimes \partial_{z_i} + \partial_{j}V^{\overline{i}} dz_j \otimes \partial_{\overline{z_i}}$
Similarly, $\overline{\nabla}$ is defined in local coordinate by:
$\overline{\nabla} V = (\partial_{\overline{j}}V^{\overline{i}} + V^{\overline{a}}\overline{\Gamma^i_{ja}}) d\overline{z_j} \otimes \partial_{\overline{z_i}} + \partial_{\overline{j}}V^{\overline{i}} d\overline{z_j} \otimes \partial_{z_i}$
I think that, in the paper you linked, the authors were using slightly different conventions: $V$ is a $(1,0)$-vector field on a Kähler manifold, so that $\nabla_aV^b=\partial_aV^b+\Gamma^b_{ac}V^c$ and $\nabla_{\bar{a}}V^b=\partial_{\bar{a}}V^b$, and $\nabla$ is the $(1,0)$-part of a connection while $\bar{\nabla}$ is the $(0,1)$-part. Now, the pairings can be rewritten as $$\lVert \nabla V\rVert^2=\left\langle V,\nabla^*\nabla V\right\rangle_{L^2(g)}$$ $$\lVert \bar{\nabla}V\rVert^2=\lVert \bar{\partial}V\rVert^2=\left\langle V,\bar{\partial}^*\bar{\partial} V\right\rangle_{L^2(g)}$$ so the formula you seek follows from the usual Nakano-Bochner identity applied to the holomorphic vector bundle $T^{1,0}M$ with Hermitian metric $g$. You can find it, together with a proof, as Corollary $1.3$ in Demailly's notes.