Let $X$ be a normal projective variety over a field of characteristic $p>0$ and $(X, \Delta\geq 0)$ be a pair such that $K\_X+\Delta$ is $\mathbb{Q}$-Cartier whose index is not divisible by $p$. Also assume that $A$ is an effective Cartier divisor, then $\tau(X, \Delta)=\tau(X, \Delta+\varepsilon A)$ for $0<\varepsilon\ll 1$. Now my question is, if $(X, \Delta)$ is Strongly $F$-regular does that imply $S^0(X, \tau(X, \Delta)\otimes {\mathcal{O}}\_X(M))=S^0(X, \tau(X, \Delta ')\otimes {\mathcal{O}}\_X(M))$, where $\Delta ' = \Delta+\varepsilon A$ and $M$ is a Cartier divisor ?


I know one of the inclusion is true for the obvious reason $\Delta '\geq \Delta$, how to prove the other inclusion, if it is true at all!


Thanks



On a Strongly F-regular Pair (X, \Delta)
It isn't true as stated unfortunately. For example, take $X$ to be an ordinary elliptic curve, $\Delta = 0$ and $M = 0$. Then $S^0(X, \tau(X) \otimes O(M)) = H^0(X, O\_X)$. However, for any effective Cartier $A > 0$ and any $\varepsilon > 0$, we have $S^0(X, \tau(X, \varepsilon A) \otimes O\_X(M)) = 0$ (this can be checked easily with a direct computation).


**However:** Probably it is true for something like $P^0$, for a definition see [Test ideals of non-principal ideals: Computations, Jumping Numbers, Alterations and Division Theorems](http://arxiv.org/abs/1212.6956) (there are some modifications one can make to that definition too which might make this easier).


Definitely it is true for $P^0$ under suitable positivity assumptions. What can you assume about $M - K\_X - \Delta$?. 


