TITLE: rational effective implies effective?
QUESTION [8 upvotes]: Let $X$ be a weak del pezzo surface, I Wonder whether the following statment is true:
Let $L$ be a line bundle on $X$, then $h^0(L)=0$ implies $h^0(nL)=0$ for all $n\geq 1$.

REPLY [2 votes]: Jun Yan found a counter example for weak del pezzo surface.
Let $X$ be weak del pezzo surface of degree $4$, Let $X=X_{4,4A_1}$, irreducible $(-2)$-curves are
$E_1-E_2,L_{123}=L-E_1-E_2-E_3,E_4-E_5,L_{345}=L-E_3-E_4-E_5$, take sum of all these curves, we
get:$2(L_{235})$ which is an effective divisor. But $L_{235}$ itself is not
effective divisor. Which means that rationally effective divisor is not
effective. And actually, there are a series such examples
But as Chen Jiang pointed out in the comment, the statement is true for del pezzo surface.