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Let $D_{\alpha}(\rho\|\sigma):= \frac{1}{\alpha - 1} \log \mathrm{Tr}[\rho^\alpha \sigma^{1-\alpha}]$ be the Petz-Rényi divergence for $\alpha \in (0,1)\cup(1,\infty)$. Note that for $\alpha \in (0,1)\cup(1,2]$ this quantity satisfies the data processing inequality $$D_{\alpha}(\rho\|\sigma) \geq D_{\alpha}(\mathcal{E}(\rho) \| \mathcal{E}(\sigma)),$$ ...

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Yes, you can formulate the smooth max-entropy as an SDP. The author of the book you linked notes this when they explain how to derive the SDP for the smooth min-entropy that you reference on page 91. In particular they say that the smoothing constraint $\tilde{\rho}_{AB} \in B^\epsilon(\rho_{AB})$ can be reformulated as the triple of constraints $$\mathrm{... 3 As far as I'm aware there isn't much of a meaningful connection. The corresponding entropy for D_{\max} is the min-entropy (written H_{\min} or H_{\infty}). It measures a sort of `worst case' uncertainty whereas the Shannon or von Neumann entropies measure an average uncertainty. To answer your first question: the quantum relative entropies or ... 3 No, this is not possible. Consider \rho_1 = \sigma_2 = \vert 0\rangle\langle 0 \vert and \rho_2 = \sigma_1 = \vert 1\rangle\langle 1 \vert. Then,$$D_{\max}(\rho_i\|\sigma_i) = \infty\quad \text{for } i = 1,2.$$Let p_i = (1/2, 1/2) and you see that D_{\max}(\rho\|\sigma) = 0. 3 Assuming everything is finite dimensional. For S_0 we have$$S_0(\rho) = \log \mathrm{rank}(\rho).$$It's pretty straightforward to see this is not continuous. Take \rho_{\epsilon} = \epsilon |0\rangle \langle 0 | + (1-\epsilon) |1\rangle \langle 1 |. Then for all 0 < \epsilon < 1 we have S_0(\rho) = \log 2 but for \epsilon \in \{0,1\} we ... 2 There is a problem in the derivation you presented, since \rho \leq \lambda \sigma is only equivalent to \sigma^{-1/2} \rho \sigma^{-1/2} \leq \lambda I when \sigma is invertible (or at least when \mathrm{supp}(\rho) \subset \mathrm{supp}(\sigma), so that you can restrict the space to \mathrm{supp}(\sigma) instead of the whole Hilbert space). The ... 2 Unfortunately D_{\max} is not a continuous function and so functions built from it tend not to be continuous. For example consider consider the two states$$ \rho_{AB} = |00 \rangle \langle 00|, $$and$$ \tau_{AB}(\epsilon) = (1-\epsilon) |00 \rangle \langle 00 | + \epsilon | 11\rangle \langle 11 |. $$A quick calculation gives I_{\max}(\rho_{AB}) = 0 ... 2 Can someone provide an example of a state \rho_{AB} for which \sigma^\star_B \neq \rho_B? Why not start very easily, with a separable state such as$$ \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=\...

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Well for $q \to 0$ we have $$\lim_{q \to 0} T_q(\rho) = \mathrm{rank}(\rho) - 1.$$ For $q \to \infty$ it's not really interesting as $$\lim_{q \to \infty} T_{q}(\rho) = 0.$$ For the second result note $\lim_{q\to\infty} \mathrm{Tr}[\rho^q] \leq \lim_{q \to \infty} \mathrm{rank}(\rho)\lambda_{\max}(\rho)^q \leq 1$ as $\rho$ is a quantum state. And so the ...

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The term max-entropy in quantum information is reserved for the following definition No it's not, many papers like https://arxiv.org/abs/0803.2770 use the term to refer to the quantity $\log \mathrm{rank}(\rho)$. Your first definition comes from the Rényi entropy of order 0, while the second one comes from the Rényi entropy of order $\frac{1}{2}$, and you ...

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