Suppose we have Grothendieck abelian categories $\mathcal{A}, \mathcal{B}$. Suppose also we have given an exact functor of triangulated categories
$$
F \colon D(\mathcal{A}) \to D(\mathcal{B})
$$
where $D(\mathcal{A})$ and $D\mathcal({B})$ denote appropriate derived categories of complexes (possibly bounded below or both ways). Suppose that 


* for any injective $I \in \mathcal{A}$ we have a functorial maps which are quasi-isomorphisms $F(I) \cong h^0F(I)$, and
* $F(D^{\geq 0}(\mathcal{A})) \subseteq D^{\geq 0}(\mathcal{B})$ ($F$ is $t$-left exact for the standard $t$-structure).


Is it then true that $F$ is the right derived functor of is zeroth cohomology? I.e. $F \cong R(h^0F)$. If not, are there known counterexamples? Or which approporiate additional assumptions are needed so that such a statement holds?


I believe that the above setup yields that the $i$th cohomology of $F$ is canonically isomorphic to the $i$th right derived functor of $h^0F$. However, I don't see how to extend the given quasi-isomorphism $F(I) \cong h^0F(I)$ for injectives in $\mathcal{A}$ to a natural transformation of functors $F \circ Q \to h^0F$ where $Q$ denotes the natural functor from the homotopy category $K(\mathcal{A})$ to $D(\mathcal{A})$. Once one has this, the universal porperty of the right derived functor should give the result.


Some Background: 
In the situation I am interested in the functor $F$ arises as a composition of a left and right derived functor in a much bigger ambient category, whose restriction to $D(\mathcal{A})$ happens to be left exact and satisfy the condition $F(I) \cong h^0F(I)$ for injective objects of $\mathcal{A}$. I could imagine that this is not an uncommon situation...


The result I.Proposition 7.4 in Residues and Duality (Hartshorne) is a statement of the type I am looking for. it says that under similar assumptions as above a right derived functor is (a shift of) the left derived functor of its highest non-vanishing cohomology...



When is a functor a right derived functor?
Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories with enough injective objects. Let me use the notation $D^+(\mathcal{A})$ and $D^+(\mathcal{B})$ to denote the stable $\infty$-categories whose homotopy categories are the (cohomologically bounded below) derived categories of $\mathcal{A}$ and $\mathcal{B}$, respectively (you can also consider unbounded derived categories, but the situation is a bit more subtle).


Let $\mathcal{C} \subseteq \mathrm{Fun}( D^{+}( \mathcal{A} ), D^{+}( \mathcal{B}) )$ be the full subcategory spanned by those functors which are exact, left t-exact, and carry injective objects of $\mathcal{A}$ into the heart of $D^{+}( \mathcal{B} )$. Then the construction
$$F \in \mathcal{C} \mapsto h^0 F|\_{ \mathcal{A} }$$ determines an
equivalence from $\mathcal{C}$ to the category of left exact functors from $\mathcal{A}$ to $\mathcal{B}$. The inverse of this equivalence is "taking the right derived functor".


Consequently, one can answer your question as follows: given a functor of triangulated categories $G: hD^{+}(\mathcal{A}) \rightarrow hD^{+}(\mathcal{B})$, it arises as a right derived functor (of a left
exact functor of abelian categories) if and only if


a) The functor $G$ lifts to an exact functor of stable $\infty$-categories $D^{+}(\mathcal{A}) \rightarrow D^{+}(\mathcal{B})$ (anything that you build by composing derived functors will have this property).


b) The functor $G$ is left t-exact and carries injective objects of
$\mathcal{A}$ into the heart of $hD^{+}(\mathcal{B})$. 


