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Let $C$ be an irreducible curve $C$ on the surface $X$, and $A$ be a big divisor, i.e. $h^0(nA)\geq an^2$.

Q: $h^0(C,\mathscr O_C(nA))=O(n)$? How to prove this.

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    $A$ is a divisor on $X$? – 2017-02-15
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    I think you miss the assumption that $C$ is not contained in the support of $A$. – 2017-02-15
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    @MooS Sorry, I do not follow your meaning. – 2017-02-15
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    Basically you want to show that the restriction of the big divisor is again big. This is false in general and its true with the assumption I have above. – 2017-02-15
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    Hint: use Riemann--Roch. – 2017-02-15
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    @Moos, I think either you are misleading him, or he is asking the question in a wrong way. Basically the thing he is asking is trivially true for arbitrary $A$, because the growth of $h^0(nD)$ of any divisor $D$ on a curve is always $O(n)$. – 2017-02-15
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    @ChenJiang I interpreted it as 'grows precisely linear' and not 'grows at most linear'. The latter is of course always true on a curve, while the second is the definition of big. – 2017-02-15
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    @MooS "grows precisely linearly" is denoted $\Theta(n)$ not $O(n)$. – 2017-02-15
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    I know how this is usually denoted, but I just thought the question would make no sense, if we really interpret it as 'grows at most linearly', since everything is trivial then (and bigness is not used at all) as already stated. – 2017-02-15
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    @MooS, I agree with you. The problem depends on what you really what to show. For example, if you want to show that $h^0(X,mA-C)>0$, then you need 'grows at most linearly'. If you want to prove $A|_C$ is big then the other way. – 2017-02-15
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    @ChenJiang Could explain more details, I mean if it is trivial for curve $C$, then can we say that for any dimension variety or manifold $X$, we have $h^0(nD)=O(n^{\dim X})$, where $D$ is a divisor. – 2017-02-16
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    @DLIN This holds if $X$ is complete. One proves it for projective varieties and then extends it to the complete case by using Chow's Lemma. – 2017-02-16
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    @DLIN, yes, as Moos suggested. It can be proved by induction. Cut X by general hyperplanes. – 2017-02-16

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