**Proof.**
Statement (1) is Derived Categories, Lemma 13.16.1.

Proof of statement (2). Write $E_ n = \tau _{\geq -n}E$. We have $E = R\mathop{\mathrm{lim}}\nolimits E_ n$, see Lemma 36.3.3. Thus $RF(E) = R\mathop{\mathrm{lim}}\nolimits RF(E_ n)$ in $D(\textit{Ab})$ by Injectives, Lemma 19.13.6. Thus for every $i \in \mathbf{Z}$ we have a short exact sequence

\[ 0 \to R^1\mathop{\mathrm{lim}}\nolimits H^{i - 1}(RF(E_ n)) \to H^ i(RF(E)) \to \mathop{\mathrm{lim}}\nolimits H^ i(RF(E_ n)) \to 0 \]

see More on Algebra, Remark 15.86.10. To prove (2) we will show that the term on the left is zero and that the term on the right equals $H^ i(RF(E_{-b + N - 1}))$ for any $b$ with $i \geq b$.

For every $n$ we have a distinguished triangle

\[ H^{-n}(E)[n] \to E_ n \to E_{n - 1} \to H^{-n}(E)[n + 1] \]

(Derived Categories, Remark 13.12.4) in $D(\mathcal{O}_ X)$. Since $H^{-n}(E)$ is quasi-coherent we have

\[ H^ i(RF(H^{-n}(E)[n])) = R^{i + n}F(H^{-n}(E)) = 0 \]

for $i + n \geq N$ and

\[ H^ i(RF(H^{-n}(E)[n + 1])) = R^{i + n + 1}F(H^{-n}(E)) = 0 \]

for $i + n + 1 \geq N$. We conclude that

\[ H^ i(RF(E_ n)) \to H^ i(RF(E_{n - 1})) \]

is an isomorphism for $n \geq N - i$. Thus the systems $H^ i(RF(E_ n))$ all satisfy the ML condition and the $R^1\mathop{\mathrm{lim}}\nolimits $ term in our short exact sequence is zero (see discussion in More on Algebra, Section 15.86). Moreover, the system $H^ i(RF(E_ n))$ is constant starting with $n = N - i - 1$ as desired.

Proof of (3). Under the assumption on $E$ we have $\tau _{\leq a - 1}E = 0$ and we get the vanishing of $H^ i(RF(E))$ for $i \leq a - 1$ from (1). Similarly, we have $\tau _{\geq b + 1}E = 0$ and hence we get the vanishing of $H^ i(RF(E))$ for $i \geq b + N$ from part (2).
$\square$

## Comments (2)

Comment #8610 by nkym on

Comment #9424 by Stacks project on