Asymptotic comparison of $L^2$ sections versus generating sections Let $s_1,\ldots, s_k$ be linearly independent global holomorphic sections of a holomorphic line bundle $E$ over a compact algebraic manifold $X$, with volume form $\Omega$.
For $m$ large, let $\{e_1,\ldots, e_p\}$ be a basis of $H^0(X,L^m)$, where $L$ is ample. Write $\|e\|^2 := |e_1|^2 + \cdots + |e_p|^2$ and $\|s\|^2 := |s_1|^2 + \cdots + |s_k|^2$. If $t$ is a complex linear combination $\sum_{i,j} \lambda_{ij} e_is_j$, then $\int_X \frac{|t|^2}{\|e\|^2 \|s\|^2} \Omega$ is finite, being bounded above by $\sum_{i,j} |\lambda_{ij}|^2 vol_\Omega X$, up to some constant.
Conversely, for a global holomorphic section $t$ of $L^mE$ over $X$, does the finiteness of
$$\int_X \frac{|t|^2}{\|e\|^2 \|s\|^2} \Omega$$
imply that $t$ is a complex linear combination $\sum_{1\le i\le p,1\le j\le k} \lambda_{ij} e_is_j$?
 A: to your first question :
$\int_X \frac{|t|^2}{\|e\|^2 \|s\|^2} \Omega$ is finite, being bounded above by $\sum_{i,j} |\lambda_{ij}|^2 vol_\Omega X$
i have found one paper on google which prove this theorem :


*

*http://arxiv.org/pdf/1211.2948v3.pdf  (definition 3,4 )


and the inverse problem maybe not true , you can refer to this paper :


*http://www.math.uh.edu/~shanyuji/2012/Geometry/n-20.pdf
since,  $\lambda_{ij}=\partial\partial^{'}\log h$
so,  $\int_X \frac{|t|^2}{\|e\|^2 \|s\|^2} \Omega\le{\sum{(\partial\partial^{'}\log h)^2}}$$\cdot{vol\Omega}$
because,  $\Vert{e}\Vert^2\cdot$$(\frac{1}{2\Vert{e}\Vert^2}+$${\log\Vert{e}\Vert{^{2}}})\ge$$1$$\Longrightarrow$$n\log n \ge1/2,$
with $n\ge1$
which contradict with the definition of the base $a$ of the logarithm :
$\frac{1}{h^2Ina}=\int\int{Inh}=Inh+h^{2}/2$
so, the inverse question holds only under the case that the manifold $X$ satisfy convexity ! 
A: I do not know if I have well-understood your notations. 
I think you mean that there is smooth metric on $E$ and positive metric on $L$. Moreover, the volume form $\Omega$ is equivalent to the Hausdorff measure of some smooth Riemannian metric on $X$.
Under such assumptions, then the function $\|e\|$ you have defined is just proportional to the so-called Bergman kernel of $L^m$. As long as $m$ is large enough, it is bounded away from $0$. So when $k=1$ your condition simply says
$$
\int_{X}\frac{|t|^2}{|s|^2}\Omega<\infty.
$$
Thus $t$ must vanish along the divisor of $s$ at least the same times as $s$ does. So we get a decomposition
$$
\mathrm{Div}\, t=\mathrm{Div}\, s+\alpha,
$$ 
where $\alpha$ is in the linear system $|L^m|$. This means $s$ divides $t$.
The general case where $k>1$ is slightly more complicated. The sections $t$ so that the integral is finite can be described locally using the multiplier ideal sheaf of $\log\|s\|$. 
You may find the full result on Page 37 of Demailly's book 'analytic methods in algebraic geometry'. 
