Joint state support is contained in the tensor product of its marginal supports

Choose an orthonormal basis $(b_j)$ of $H_B$ using [F7]. If $u\in \ker {\rho }_A$, then [F2] gives $0=⟨{\rho }_{A}u,u⟩={\sum }_j⟨{\rho }_{AB}(u\otimes b_j),u\otimes b_j⟩$. Every term is nonnegative by [F1], so every term is zero. By [F5], write ${\rho }_{AB}=S^{\ast }S$; a zero quadratic form then gives $S(u\otimes b_j)=0$ and hence ${\rho }_{AB}(u\otimes b_j)=0$. Since $(b_j)$ spans $H_B$, ${\rho }_{AB}$ annihilates $\ker {\rho }_A\otimes H_B$.

Choose an orthonormal basis $(a_i)$ of $H_A$ using [F7]. If $v\in \ker {\rho }_B$, then [F3] similarly gives $0=⟨{\rho }_{B}v,v⟩={\sum }_i⟨{\rho }_{AB}(a_i\otimes v),a_i\otimes v⟩$. Positivity and [F5] show that ${\rho }_{AB}$ annihilates $H_A\otimes \ker {\rho }_B$.

Self-adjointness of ${\rho }_{AB}$ implies its range is orthogonal to both subspaces annihilated in steps 1.1 and 1.2. Using the orthogonal decompositions $H_A=\mathrm{supp}({\rho }_A)\oplus \ker {\rho }_A$ and $H_B=\mathrm{supp}({\rho }_B)\oplus \ker {\rho }_B$, the intersection of their orthogonal complements is $\mathrm{supp}({\rho }_A)\otimes \mathrm{supp}({\rho }_B)$. For non-negative ${\rho }_{AB}$ its range equals its support by [F6]. The product spectral decomposition also gives $\mathrm{supp}({\rho }_A\otimes {\rho }_B)=\mathrm{supp}({\rho }_A)\otimes \mathrm{supp}({\rho }_B)$. Hence the stated inclusion holds, and [F1] makes the divergence finite.