Derived category of $\mathcal{D}_X$ modules

Define $D^{\text{#}}_{\bullet}(Mod(\mathcal{D}_X))$ to be the full subcategory of the derived category $D^{\text{#}}(Mod(\mathcal{D}_X))$ of complexes of $\mathcal{D}_X$-modules whose cohomology groups belong to $Mod_{\bullet}(\mathcal{D}_X)$, # = $+,-,b$, $X$ smooth algebraic variety over $\mathbb{C}$.

Is it true that $D^{\text{#}}_{\bullet}(Mod(\mathcal{D}_X))$ is equivalent to the category $D^{\text{#}}(Mod_{\bullet}(\mathcal{D}_X))$?

It is true that $Mod_{\bullet}(\mathcal{D}_X)$ are closed under kernels, cokernels and extensions because we are working over a Noetherian scheme. However, Kashiwara-Schapira Category and Sheaves states the equivalence under one more hypothesis. Apart from the answer, if you can quote any reference I would be glad.

EDIT: I forgot to say that $\bullet =$ quasi coherent modules or coherent modules (a quasi coherent $\mathcal{D}_X$ module is a $\mathcal{D}_X$ modules which is quasi coherent over $\mathcal{O}_X$, instead coherence is over $\mathcal{D}_X$)

• What is $Mod_\bullet$? Mar 29, 2018 at 9:45
• @BugsBunny Forgive me, I added what is $\bullet$ Mar 29, 2018 at 10:32
• Did you check mathoverflow.net/questions/236245/… Mar 29, 2018 at 11:14
• IMHO, the answer is UNKNOWN. If there is a good reason for it to be YES, it is known thanks to Exercise on p.153 of Gelfand-Manin or 2.42 and 3.4 of Huybrechts... Mar 29, 2018 at 11:16

$$D^{b}(Mod_{\bullet}(\mathcal{D}_X)) \rightarrow D^{b}_{\bullet}(\mathcal{D}_X)$$
• May I ask if the result is know when $X$ is singular? Feb 18, 2019 at 11:30