Proof.
Proof of (1). Assume that $K \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}/\mathcal{I}$ is bounded above, say $H^ i(K \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}/\mathcal{I}) = 0$ for $i > b$. Note that we have distinguished triangles
\[ K \otimes _\mathcal {O}^\mathbf {L} \mathcal{I}^ n/\mathcal{I}^{n + 1} \to K \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}/\mathcal{I}^{n + 1} \to K \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}/\mathcal{I}^ n \to K \otimes _\mathcal {O}^\mathbf {L} \mathcal{I}^ n/\mathcal{I}^{n + 1}[1] \]
and that
\[ K \otimes _\mathcal {O}^\mathbf {L} \mathcal{I}^ n/\mathcal{I}^{n + 1} = \left( K \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}/\mathcal{I}\right) \otimes _{\mathcal{O}/\mathcal{I}}^\mathbf {L} \mathcal{I}^ n/\mathcal{I}^{n + 1} \]
By induction we conclude that $H^ i(K \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}/\mathcal{I}^ n) = 0$ for $i > b$ for all $n$.
Proof of (2). Assume $K \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}/\mathcal{I}$ as an object of $D(\mathcal{O}/\mathcal{I})$ has tor amplitude in $[a, b]$. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}/\mathcal{I}^ n$-modules. Then we have a finite filtration
\[ 0 \subset \mathcal{I}^{n - 1}\mathcal{F} \subset \ldots \subset \mathcal{I}\mathcal{F} \subset \mathcal{F} \]
whose successive quotients are sheaves of $\mathcal{O}/\mathcal{I}$-modules. Thus to prove that $K \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}/\mathcal{I}^ n$ has tor amplitude in $[a, b]$ it suffices to show $H^ i(K \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}/\mathcal{I}^ n \otimes _{\mathcal{O}/\mathcal{I}^ n}^\mathbf {L} \mathcal{G})$ is zero for $i \not\in [a, b]$ for all $\mathcal{O}/\mathcal{I}$-modules $\mathcal{G}$. Since
\[ \left(K \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}/\mathcal{I}^ n\right) \otimes _{\mathcal{O}/\mathcal{I}^ n}^\mathbf {L} \mathcal{G} = \left(K \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}/\mathcal{I}\right) \otimes _{\mathcal{O}/\mathcal{I}}^\mathbf {L} \mathcal{G} \]
for every sheaf of $\mathcal{O}/\mathcal{I}$-modules $\mathcal{G}$ the result follows.
$\square$
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