Let $X$ be a smooth projective variety over $\mathbb C$ and $A\to X$ an ample line bundle. 


Is there an integer $k\_0$ such that for all line bundle $L\to X$, the tensor product $A^{\otimes k}\otimes L$ is ample for every $k\ge k\_0$?


Of course the answer is yes if we let $k\_0$ depend on $L$, but the point here is that $k\_0$ must be an universal constant depending only on $X$ and $A$.


Thanks in advance.


EDIT: In fact the answer is NO in general: take for example $X=\mathbb P^n$ and $A=\mathcal O(1)$. Then for each $k$ there exists a line bundle $L$ on $\mathbb P^n$, namely $L=\mathcal O(-k-1)$ such that $A^{\otimes k}\otimes L$ is not ample. So,
**is there any reasonable condition on $X$ and $A$, or on the family in which we let the bundles $L$ vary, such that the statement holds true?**


My question was motivated by the following on positivity:


Let $X$ be a compact complex manifold endowed with a positive (in the sense that it carries a smooth hermitian metric whose Chern curvature tensor is positive definite) line bundle $A\to X$. For every pair of points $p,q\in X$ consider the blow-up $\sigma\colon\widetilde X\to X$ at this two point and let $E\_p$, $E\_q$ be the two exceptional divisors (if $p=q$ then consider $2E\_p$). Is there a uniform $k\_0>0$ (independent of $p$ and $q$) such that 
$$
\sigma^\* A^{\otimes k}\otimes\mathcal O\_{\widetilde X}(-E\_p-E\_q)
$$
is positive whenever $k\ge k\_0$?


**EDIT bis.** Here is a beginning of a possible proof. Le us take the case $p=q$ and suppose the contrary. Then, there is a sequence of points $(p\_k)\subset X$ such that 
$$
\sigma^\* A^{\otimes k}\otimes\mathcal O\_{\widetilde X}(-2E\_{p\_k})
$$
is not positive definite. Since $X$ is compact, after extracting a subsequence we may suppose that $p\_k\to p\in X$. Let $k\_0$ be a integer such that
$$
\sigma^\* A^{\otimes k}\otimes\mathcal O\_{\widetilde X}(-2E\_{p})
$$ 
is positive definite for all $k>k\_0$. This integer certainly exists, since $\mathcal O\_{E\_p}(-E\_p)$ is positive. 


If we are able to construct for all points $q$ nearby $p$ a smooth hermitian metric such that 
$$
\sigma^\* A^{\otimes k}\otimes\mathcal O\_{\widetilde X}(-2E\_{q})
$$ 
is positive definite whenever $k\ge k\_0$ we then get a contradiction and we are done.


For the moment I am not still able to do this last part. This should be achieved by a somehow clever use of a partition of unity on $\widetilde X$.



Uniformity of ampleness

> 
> **Remark:** If $p=q$, then one can just work with $E\_p$ instead of $2E\_p$ and then at the end take $2k\_0$ instead of $k\_0$, so we may assume that $p\neq q$ and in particular that the exceptional divisor to be subtracted is reduced.
> 
> 
> 


-



> 
> **Lemma 1.**
>  Let $D$ be a semi-ample Cartier divisor on a smooth projective variety $Y$ and assume that for any proper curve $C\subseteq Y$, the intersection number $D\cdot C>0$. Then $D$ is ample.
> 
> 
> 


*Proof.* Consider the morphism induced by a large enough multiple of $D$. If it had a positive dimensional fiber, then it would contain a proper curve which would have intersection $0$ with $L$ which contradicts the assumption. **Q.E.D.**



> 
> **Lemma 2.**
>  Let $X$ be a a smooth projective variety, $\sigma: Y\to X$ the blow up of the points $p\_1,\dots,p\_r\in X$ with (reduced) exceptional divisor $E\subset Y$ and finally let $A$ be an ample Cartier divisor on $X$. Assume that $k\_0,k\_1\in\mathbb N$ are such that $k\_0A-K\_X$ is ample and $k\_1(\sigma^\*A)-E$ is nef. Then $k(\sigma^\*A)-E$ is ample for any $k\geq k\_0+\dim X \cdot k\_1+1$.
> 
> 
> 


*Proof.*
First observe that $K\_Y\sim \sigma^\* K\_X + (\dim X−1) E$ and hence with $m=k-k\_0-\dim X\cdot k\_1> 0$, 
$$
(k(\sigma^\*A)-E)-K\_Y\sim m(\sigma^\*A) + \sigma^\*(k\_0 A-K\_X) + \dim X \cdot(k\_1(\sigma^\*A)-E)
$$
is nef and big. Then by the Basepoint-free theorem $k(\sigma^\*A)-E$ is semi-ample. It also follows that 
for any proper curve $C\subseteq Y$, the intersection number $(k(\sigma^\*A)-E)\cdot C>0$. Indeed, if $C\subseteq E$, then $\sigma^\*A\cdot C=0$ and $-E\cdot C>0$ and if $C\not\subseteq E$, then $(\sigma^\*A)\cdot C>0$ and $(k\_0(\sigma^\*A)-E)\cdot C\geq 0$. This is enough as $k>k\_0$. Finally, then $k(\sigma^\*A)-E$ is ample by Lemma 1.
**Q.E.D.**



> 
> **Claim** The *motivating* positivity statement of the question is true.
> 
> 
> 


*Proof.*
Let $k\_0$ be such that 
$k\_0A-K\_X$ is ample and 
let $k\_1$ be an integer that is larger than $1/\varepsilon\_p+1/\varepsilon\_q$ for the *Seshadri constants* $\varepsilon\_p=\varepsilon(A, p)$ and $\varepsilon\_q=\varepsilon(\sigma\_p^\*A-E\_p, q)$, where $\sigma\_p$ is the blow up of $p$ alone.
Then Lemma 2 implies the desired statement.


Note that for a fixed $X$ and fixed $A$ there is a lower bound for the Seshadri constants that works for all $p\in X$.


