For any coherent sheaf $F_1$, $F_2$ on $X$, $\operatorname{Ext}^i(F_1,F_2)=0$ if $i\notin [0,n]$

Question

Let $k$ be a field. Let $X$ be a smooth projective equidimensional variety over $k$. How to show the following proposition?

For any coherent sheaf $F_1$, $F_2$ on $X$, $\operatorname{Ext}^i(F_1,F_2)=0$ if $i\notin [0,n]$.

This question is from Lemma 5.6 in this paper

Lemma 5.6. Suppose that $X$ is a smooth projective equidimensional variety over $k$. Then $$ \overline{\operatorname{Sdim}} D^{b}(\operatorname{coh} X)=\underline{\operatorname{Sdim}} D^{b}(\operatorname{coh} X)=\operatorname{dim} X. $$

Proof. Let $n=\operatorname{dim} X$. Recall that the Serre functor on $D^{b}(\operatorname{coh} X)$ is isomorphic to $(-) \otimes \omega_{X}[n] .$ Recall also that for any coherent sheaves $F_{1}, F_{2}$ on $X$ we have $\operatorname{Ext}^{i}\left(F_{1}, F_{2}\right)=0$ if $i \notin[0, n]$. ...

Thank you very much!

Answer 1

First, $\mathrm{Ext}^i(F_1,F_2)$ for $i < 0$ vanish by definition. Second, for $i > n$ the vanishing follows from Serre duality $$ \mathrm{Ext}^i(F_1,F_2) \cong \mathrm{Ext}^{n-i}(F_2,F_1 \otimes \omega_X)^\vee $$ and the above vanishing.