Is entanglement nonincreasing on average by local operations for all possible ensemble decompositions?

Question

We know for a pure state conversion $|\psi \rangle \rightarrow_\textrm{LOCC} |\phi \rangle$ via local operation and classical communication (LOCC), an entanglement monotone should not increase, that is, $E(|\psi \rangle) \geq E(|\phi \rangle)$.

Say for a pure to mixed state conversion $|\psi \rangle \rightarrow_\textrm{LOCC} \rho$ given by the unique ensemble $l = \{ q_l, |\xi_l \rangle \} \in \mathcal{D}$ fixed by pointer states (which are omitted), where $\mathcal{D}$ is the set of all decompositions for $\rho$.

Naturally, the convex roof extension $E_\textrm{min}(\rho) = \min_\mathcal{D} \Sigma_i q_i |\xi_i \rangle \langle \xi_i|$ should not increase, nor should that of the obtained ensemble $l$ on average, i.e., $E(|\psi \rangle) \geq E_l(\rho) \geq E_\textrm{min}(\rho)$.

So, is it for certain there is not some decomposition $j = \{ q_j, |\xi_j \rangle \}$ that violates the nonincreasing criterion for $E$ on average, i.e., $E_j(|\rho \rangle) \geq E(\psi \rangle)$?

Answer 1

You don't even define $E_j(\rho)$. This makes your question hard to answer.

But if this is the average entanglement over that decomposition: Of course this can violate it. For instance, you can map any state to a maximally mixed state by LOCC, and the maximally mixed state can be written as a convex combination of the four Bell states, which are all maximally entangled.