Return to Answer

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 $k_1$ 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 $p\in X$.

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 $p\in X$.

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 $k\geq k_0+2k_1+1$.

Proof. First observe that $K_Y\sim \sigma^* K_X + E$ and hence with $m=k-k_0-2k_1> 0$, $$ (k(\sigma^*A)-E)-K_Y\sim m(\sigma^*A) + \sigma^*(k_0 A-K_X) + 2(k_1(\sigma^*A)-E) $$ 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 $k_1$ 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 $p\in X$.

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 $k_1$ 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 $p\in X$.

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 $k\geq k_0+2k_1+1$.

Proof. First observe that $K_Y\sim \sigma^* K_X + E$ and hence with $m=k-k_0-2k_1> 0$, $$ (k(\sigma^*A)-E)-K_Y\sim m(\sigma^*A) + \sigma^*(k_0 A-K_X) + 2(k_1(\sigma^*A)-E) $$ 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 $k_1$ 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 $\varepsilon_q=\varepsilon(A, q)$. 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 $p\in X$.

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 $k\geq k_0+2k_1+1$.

Proof. First observe that $K_Y\sim \sigma^* K_X + E$ and hence with $m=k-k_0-2k_1> 0$, $$ (k(\sigma^*A)-E)-K_Y\sim m(\sigma^*A) + \sigma^*(k_0 A-K_X) + 2(k_1(\sigma^*A)-E) $$ 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 $k_1$ 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 $p\in X$.