Is the support of $\rho_{AB}$ always contained in the support of $\rho_A\otimes\rho_B$?

Question

Given a positive unit trace Hermitian matrix, i.e. a density operator, $\rho_{AB}$ on Hilbert spaces $\mathcal{H}_A\otimes\mathcal{H}_B$. Consider its marginals $\rho_A$, $\rho_B$. Do we have the support of $\rho_{AB}$ always contained in the support of $\rho_A\otimes\rho_B$? If so, what is the universal constant $C$, such as the dimension, such that $C\rho_A\otimes\rho_B\geq\rho_{AB}$?

Answer 1

To answer your first question: yes.

To see why, let $\Pi_A$ and $\Pi_B$ denote the projection operators onto the supports of $\rho_A$ and $\rho_B$ respectively, and let $1_A$ and $1_B$ denote the corresponding identity operators. Now, \begin{align*} \operatorname{Tr}((1_A-\Pi_A)\otimes I_B)\rho) &= \operatorname{Tr}((1_A\otimes1_B)\rho) - \operatorname{Tr}((\Pi_A\otimes I_B)\rho)\\& = \operatorname{Tr}(\rho) - \operatorname{Tr}(\Pi_A\rho_A)\\& = \operatorname{Tr}(\rho) - \operatorname{Tr}(\rho_A)\\& = 0. \end{align*} It follows that $((1_A-\Pi_A)\otimes I_B)\rho = \rho((1_A-\Pi_A)\otimes I_B)= 0$ and thus $$(\Pi_A\otimes I_B)\rho = \rho(\Pi_A\otimes I_B)= \rho.$$ Similar analysis shows that $$(1_A\otimes \Pi_B)\rho = \rho(1_A\otimes \Pi_B)= \rho.$$ It follows that $$ (\Pi_A\otimes\Pi_B)\rho = \rho (\Pi_A\otimes\Pi_B) = \rho, $$ which proves the claim that the support of $\rho$ is contained in the support of $\rho_A\otimes\rho_B$ (since $\Pi_A\otimes\Pi_B$ is the projection onto the support of $\rho_A\otimes\rho_B$).

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I need to think more about whether or not there exists such a constant $C$.