Let $\pi:X \longrightarrow C$ be a smooth projective morphism onto a smooth projective curve, and $F$ be a central fiber. If the Kodaira dimension $\kappa (F)$ is nonnegative, is $\pi\_{\ast} \mathcal O\_X (k K\_{X/C})$ nonzero for sufficiently divisible $k$? If it is, can anyone arrange an algebraic proof?



direct image of relative pluri-canonical bundle on a smooth fibration
I think one can argue as follows. (Let me know if I made a mistake!) 


Choose $k$ large enough so that $kK\_F$ has a nonzero global section for some fibre $F$. For any fibre $F$, we have $K\_F = K\_{X/C}~\_{|F}$. So deformation invariance of plurigenera says that the function


$$h^0(X\_p, k K\_{X/C}~\_{|X\_p})$$
is constant on $C$ (where now $X\_p$ denotes the fibre over the point $p \in C$). Therefore by Grauert's theorem (Hartshorne III.12.9) any global section of kK\_F must come from the sheaf $\pi\_\* O\_X(kK\_{X/C})$. Since we have chosen $k$ large enough that $kK\_F$ has a nonzero global section, the sheaf $\pi\_\* O\_X(kK\_{X/C})$ must be nonzero. 


Presumably this doesn't satisfy your requirement of an algebraic proof, because it use deformation invariance of plurigenera, which currently only has an analytic proof in general.


