Does the relative entropy of entanglement of every bipartite state equal its regularization, or is regularization genuinely necessary? For a finite-dimensional bipartite system \(A{:}B\), write \(\operatorname{Sep}(A{:}B)\) for the set of separable states and \(D(\rho\Vert\sigma)=\operatorname{Tr}[\rho(\log_2\rho-\log_2\sigma)]\) for the Umegaki relative entropy, defined when \(\operatorname{supp}\rho\subseteq\operatorname{supp}\sigma\), with the trace evaluated on \(\operatorname{supp}\rho\) and \(0\log_2 0:=0\). Define the relative entropy of entanglement and its regularization by
\begin{equation}
E_R(\rho)
:=\min_{\sigma\in\operatorname{Sep}(A{:}B)}D(\rho\Vert\sigma),
\qquad
E_R^\infty(\rho)
:=\lim_{n\to\infty}\frac1nE_R\bigl(\rho^{\otimes n}\bigr).
\tag{1}
\end{equation}
The limit in Eq. (1) exists and equals \(\inf_{n\geq1}\frac1nE_R(\rho^{\otimes n})\) by Fekete’s lemma: the product of minimizing separable states is separable and \(D\) is additive on tensor products, which gives the subadditivity \(E_R(\rho\otimes\sigma)\leq E_R(\rho)+E_R(\sigma)\), and subadditivity implies convergence of the normalized terms to their infimum, not that each of them is nonincreasing. The archived question is whether single copies already suffice, that is, whether
\begin{equation}
E_R^\infty(\rho)=E_R(\rho)
\quad\text{for every finite-dimensional bipartite state }\rho.
\tag{2}
\end{equation}
Since \(E_R^\infty(\rho)\leq E_R(\rho)\) always holds, Eq. (2) can only fail strictly, through a single state \(\rho\) with
\begin{equation}
E_R^\infty(\rho)<E_R(\rho).
\tag{3}
\end{equation}