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I have seen various definitions of quantum conditional min-entropy, which I believe are equivalent.

  1. $$H_{\min}(A|B) = - \inf\limits_{\sigma_B} D_{\max} \left( \rho_{AB} \parallel I_A \otimes \sigma_B \right)$$ where in general $$D_{\max}(\rho \parallel \sigma) = \inf \{ \lambda : \rho \leq 2^\lambda \sigma \}$$

  2. $$H_{\min}(A|B) = \max\limits_{\sigma_B} \sup \{ \lambda: \rho_{AB} \leq 2 ^{-\lambda} I_A \otimes \sigma_B \}$$

Why are the two definitions equivalent?

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  • $\begingroup$ each post should contain a single, laser-focused question. You can open different posts to ask different questions. Feel free to edit this post to focus it on a specific aspects $\endgroup$
    – glS
    Mar 12, 2023 at 8:41
  • $\begingroup$ While I think all of these sub-questions are trying to help understanding the same underlying concept, I have created a sperate question for each of my questions. In general, even if someone answers one question, they may answer smaller problems in-between, so one could argue that an answer to a question is answering multiple questions. I find this approach pedantic and one needs to consider the totality of the question. $\endgroup$
    – Josh
    Mar 12, 2023 at 16:18
  • $\begingroup$ Since an answer to this question made an attempt to justify why the two definitions are equivalent in this question, I have modified my question to only focus on that. The other two questions can be found here: quantumcomputing.stackexchange.com/questions/31606/… quantumcomputing.stackexchange.com/questions/31607/… $\endgroup$
    – Josh
    Mar 12, 2023 at 16:20
  • $\begingroup$ the reason is question reusability. Questions are (ideally) really more useful to everyone else reading them in the future than the person that asks them originally. Having very focused questions makes it more easy to reference them in the future. And if someone is looking for info about, say, the equivalence in this question, they can get the answer without having to parse through a longer post and figure out which part of the answer answers which question $\endgroup$
    – glS
    Mar 12, 2023 at 16:52
  • $\begingroup$ We can easily ensure that the sub-questions are "easy to reference in the future" by labeling each one with a number (e.g., 1, 2, 3). Additionally, presenting related questions can provide readers with a broader perspective and better appreciation of the topic. Although readers may initially come for a specific answer, having all related questions in one place can ultimately benefit their understanding. $\endgroup$
    – Josh
    Mar 13, 2023 at 3:46

1 Answer 1

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Observe that $$ H_{\min}(A|B) = - \inf\limits_{\sigma_B} D_{\max} \left( \rho_{AB} \| I_A \otimes \sigma_B \right) = \sup\limits_{\sigma_B} [-D_{\max} \left( \rho_{AB} \| I_A \otimes \sigma_B \right)], $$ and $$-D_{\rm max}(\rho\|I\otimes \sigma)=-\inf \{\lambda: \,\, \rho\le 2^\lambda (I\otimes \sigma)\} \\ = \sup\{-\lambda: \,\, \rho\le 2^\lambda (I\otimes\sigma)\} \\ = \sup\{\lambda: \,\, \rho\le 2^{-\lambda} (I\otimes\sigma)\}.$$

It might help to take a slightly more abstract perspective here. What we're trying to compute is $-\inf_x f(x)$ where $f(x)\equiv \inf\{y\in\mathbb{R}:\,\, y\in g(x)\}$ for some function $f$, and some set $g(x)\subseteq\mathbb{R}$ that depends on $x$. And then $$-\inf_x f(x) = \sup_x[-f(x)] = \sup_x [-\inf\{y\in\mathbb{R}:\,\, y\in g(x)\}] \\ = \sup_x \sup\{-y: \,\, y\in\mathbb{R},\,\, y\in g(x)\} \\ = \sup_x \sup\{z\in\mathbb{R}: \,\, -z\in g(x)\}.$$ In words, we first minimise over the possible values of a function $f$ (restricting the possible inputs $x$ in a specific subset, and taking the opposite of the result), where $f(x)$ is defined as the smallest element of a subset of real numbers that depends on $x$. You can then just replace $\sup\to\max$ and $\inf\to\min$ whenever you know that a min/max exists.

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  • $\begingroup$ Thanks! That makes sense. In light of your answer, I came up across a result that I admittedly forgot: $-\inf(A) = \sup(-A)$. For interested readers, a proof can be found here: math.stackexchange.com/questions/522259/… $\endgroup$
    – Josh
    Mar 12, 2023 at 16:43

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