$\newcommand{\ch}{\mathrm{ch}}$Let $F \in D^b(X)$ be a complex (in particular, not an object concentrated in a single degree) with $\ch_0(F) = r>0$. Let $$0 = E_0 \subset E_1 \subset \cdots \subset E_{n-1} \subset E_n = F$$ be a filtration of $F$ (in some heart of $D^b(X)$). Then do we have $0 \leq \ch_0(E_i) \leq r$ for all $E_i$? Here $\ch_0$ means the degree zero part of the Chern character $\ch$; recall $\ch(F)$ lives in the Chow ring $A(X)$ which has a grading (when we consider sheaves, then $\mathrm{rk} = \ch_0$).
This would be true if $F$ were a sheaf, but I'm wondering if there is any reason why it cannot be true if $F$ is a complex, not concentrated in a single degree.
I think essentially what I'm unclear about is the notion of filtration of a complex in $D^b(X)$; I guess you could ask for $\subset$ in the above to mean injection in $D^b(X)$, but does this imply $\ch_0(E_i) \leq \ch_0(E_j)$ for $i < j$?
Edit: I think it's not true. Recall that the definition of the Chern character of a complex $\mathcal{F}^\bullet = \mathcal{F}^a \to \mathcal{F}^{a+1} \to \cdots \to \mathcal{F}^b$ is $\ch(\mathcal{F}^\bullet) = \sum_{p = a, \dots, b} (-1)^p \ch(\mathcal{F}^p) $.
For simplicitly, let our complex be $F = A \to B$. Then $\ch(F) = \ch(A) - \ch(B)$ and in particular, $\ch_0(F) = \ch_0(A) - \ch_0(B)$. Let $E_i = A_i \to B_i$ be a subcomplex of $F$. Note that $\ch_0(E_i) = \ch_0(A_i) - \ch_0(B_i)$ and since $E_i$ is a subcomplex of $F$, we do have $\ch_0(A_i) \leq \ch_0(A)$ and $\ch_0(B_i) \leq \ch_0(B)$. But then we don't necessarily have $\ch_0(E_i) \leq \ch_0(F)$; for example let $\ch_0(A) = 7, \ch_0(B) = 4$, and $\ch_0(A_i) = 5, \ch_0(B_i)=1$. Then $\ch(E) = 7-4 = 3 < 4 = 5 - 1 = \ch(E_i)$.
Is this the correct reasoning? Thanks.