Does equivalence of derived categories preserve boundedness? Let $\mathcal{C}$ be an abelian category and consider the derived category $D(C)$. Suppose that $F: D(\mathcal{C}) \to D(\mathcal{C})$ is an auto-equivalence. 
My question: must $F$ preserve the property of being bounded/bounded above/bounded below. 
Said differently, if $A$ is an object of $D(\mathcal{C})$ whose cohomology is bounded/bounded above/bounded below, does $F(A)$ also have cohomology which is bounded/bounded above/bounded below? 
If the answer to this question is negative, can one impose natural properties on $F$ to make the answer positive?
 A: Not in that level of generality. For example, let $\mathcal{C}$ be the product of copies of the category of abelian groups, indexed by $\mathbb{Z}$, so an object of $\mathbb{C}$ is a family $\{A_i\mid i\in\mathbb{Z}\}$ of abelian groups, and an object of $D(\mathcal{C})$ is a family $\{X_i\mid i\in\mathbb{Z}\}$ of complexes of abelian groups.
Then $\{X_i\}\mapsto\{X_i[i]\}$ is an auto-equivalence that doesn’t preserve boundedness (or boundedness above or below).
However, it is true if $\mathcal{C}$ is the module category of a ring. 
This is because it is well-known (see for example, Proposition 15.72.3 here) that the "perfect complexes" (those isomorphic in $D(\mathcal{C})$ to bounded complexes of finitely generated projective modules) can be characterized in the derived category as the "compact objects" (objects $X$ such that the functor $\text{Hom}_{D(\mathcal{C})}(X,-)$ preserves arbitrary coproducts).
But then the bounded (in cohomology) objects $Y$ of $D(\mathcal{C})$ can be characterized as the objects such that, for each perfect complex $X$, $\text{Hom}_{D(\mathcal{C})}(X,Y[i])=0$ for all but finitely many $i\in\mathbb{Z}$, with similar characterizations of bounded above and bounded below objects.
Since there are inherent characterizations of these various classes of objects, they must be preserved by any auto-equivalence.
