As stated in the title, let $M$ be a linear operator on a finite bipartite Hilbert space. Suppose $0\leq M^{AB}\leq \mathbb{I}$ and $0\leq M^A,M^B\leq\mathbb{I}$, where $M^A=\mathrm{Tr}_B\left(M^{AB}\right)$ and $M^B=\mathrm{Tr}_A\left(M^{AB}\right)$. Is it always true that $$ M^{AB} \leq M^A\otimes \mathbb{I}_B? $$ It trivially holds for product operators, that is $M^{AB} = M^A\otimes M^B$, but the general statement is not clear to me.
Any help is appreciated.
Not true.
Just take Bell state $M^{AB} = |v\rangle\langle v|$, where $|v\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$. It has eigenvalue 1.
But $M^A \otimes I_2= \frac{1}{2}I_2 \otimes I_2 = \frac{1}{2}I_4$.
No, but it is true that $$M^{AB} \le d\, M^A \otimes \mathbb{I}_B,$$ where $d$ is the minimum of $d_A$ and $d_B$. This inequality is tight in the sense that entangled states saturate it, as the other answer shows. A proof can be found in Appendix A of this paper.
