If $D\in\mathrm{Pic}\\, X$ and $f:X\to Y$ is a morphism, then for any $V\subseteq Y$ open $H^0(V,f_*D):=H^0(f^{-1}V,D)$ by definition, so in particular $H^0(Y,f_*D):=H^0(X,D)$ and you don't need any of the assumptions. Am I missing something???
Sándor Kovács
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