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The max-relative entropy between two states is defined as

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

It is known that the max-relative entropy is quasi-convex. That is, for $\rho=\sum_{i \in I} p_{i} \rho_{i}$ and $\sigma=\sum_{i \in I} p_{i} \sigma_{i}$ where $\rho_i, \sigma_i$ are quantum states and $p$ is a probability vector, it holds that

$$D_{\max }(\rho \| \sigma) \leq \max _{i \in I} D_{\max }\left(\rho_{i} \| \sigma_{i}\right).$$

Is there a lower bound for $D_{\max }(\rho \| \sigma)$ in terms of $D_{\max }(\rho_i \| \sigma_i)$?

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No, this is not possible. Consider $\rho_1 = \sigma_2 = \vert 0\rangle\langle 0 \vert$ and $\rho_2 = \sigma_1 = \vert 1\rangle\langle 1 \vert$. Then,

$$D_{\max}(\rho_i\|\sigma_i) = \infty\quad \text{for } i = 1,2.$$

Let $p_i = (1/2, 1/2)$ and you see that $D_{\max}(\rho\|\sigma) = 0$.

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