# Questions on the max-relative entropy $D_{\max}(\rho||\sigma)$

The max-relative entropy between two states is defined as

$$D_{\max }(\rho \| \sigma):=\log \min \{\lambda: \rho \leq \lambda \sigma\},$$

where $$\rho\leq \sigma$$ should be read as $$\sigma - \rho$$ is positive semidefinite. In other words, $$D_{\max}$$ is the logarithm of the smallest positive real number that satisfies $$\rho\leq\lambda\sigma$$. I would like to understand the following properties of this quantity when the states are bipartite i.e. they live on $$H_A\otimes H_B$$. In the following, all $$\rho$$ and $$\sigma$$ correspond to quantum states (positive semidefinite matrices with unit trace).

A quantity known as the max-information that $$B$$ has about $$A$$ is given by

$$I_{\max}(A:B)_\rho = \min\limits_{\sigma_B} D_{\max}(\rho_{AB}||\rho_A\otimes\sigma_B)$$

Let the minimum on the right hand side be achieved by the state $$\sigma^\star_B$$. My questions are as follows

1. Can someone provide an example of a state $$\rho_{AB}$$ for which $$\sigma^\star_B \neq \rho_B$$?

2. Is it true that $$D_{\max}(\rho_{B}||\sigma^\star_B) \leq D_{\max}(\rho_{B}||\sigma_B)$$ for all $$\sigma_B$$ i.e. is the $$D_{\max}$$ minimizing state preserved under a partial trace?

Can someone provide an example of a state $$\rho_{AB}$$ for which $$\sigma^\star_B \neq \rho_B$$?
Why not start very easily, with a separable state such as $$\rho_{AB}=\left(p_0|0\rangle\langle 0|\otimes \tau_0+p_1|1\rangle\langle 1|\otimes \tau_1\right)$$ where $$\tau_0$$ and $$\tau_1$$ are different (normalised) single-qubit density matrices. We have that $$I=\min_{\sigma_B}\log\min_{\lambda}\left\{\lambda:p_0|0\rangle\langle 0|\otimes (\lambda\sigma-\tau_0)+p_1|1\rangle\langle 1|\otimes (\lambda\sigma-\tau_1)\geq 0\right\}$$ Now, by construction, this devolves into two separate questions: pick the common $$\lambda$$ and $$\sigma$$ such that both $$\lambda\sigma-\tau_i$$ are positive semi-definite. However, this is entirely independent of the $$p_i$$. Hence, the answer must be independent of the $$p_i$$. By comparison, $$\rho_B$$ depends on the $$p_i$$. Thus, unless the answer is highly degenerate (allowing for all possible linear combinations of the $$\tau_i$$), it cannot be that $$\sigma=\rho_B$$.
For example, if $$\tau_0$$ and $$\tau_1$$ are orthogonal, the best choice must be $$\sigma=(\tau_0+\tau_1)/2$$ with $$\lambda=2$$. This is certainly not $$\rho_B$$ for any $$p_0\neq 1/2$$.
Is it true that $$D_{\max}(\rho_{B}||\sigma^\star_B) \leq D_{\max}(\rho_{B}||\sigma_B)$$ for all $$\sigma_B$$ i.e. is the $$D_{\max}$$ minimizing state preserved under a partial trace?
This cannot be true, right? $$D_\max(\rho_B\|\rho_B)=0$$. So in any case where $$\sigma^\star\neq \rho_B$$, $$D_\max(\rho_B\|\sigma^\star)>D_\max(\rho_B\|\rho_B)$$. We know, for example, that $$\lambda\geq 1$$ because $$\text{Tr}(\lambda\sigma-\rho)=\lambda-1$$, and if $$\lambda\sigma-\rho$$ is going to be non-negative, then the total of the eigenvalues must be non-negative.