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Let $\rho, \sigma$ be states such that

$$F(\rho,\sigma) = \delta >0,$$

where $F(\rho,\sigma) = \|\sqrt{\rho}\sqrt{\sigma}\|_1$. Now consider all possible states $\bar{\rho}$ such that $F(\bar{\rho},\rho) \geq 1 - \varepsilon$ for $0<\varepsilon<\delta$. Can one find an upper bound for

$$\min\limits_{\bar{\rho}}F(\bar{\rho},\sigma)$$

Obviously, $\delta$ is the trivial upper bound found by choosing $\bar{\rho} = \rho$ but it's not clear if one can always do better.

user1936752
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    Hmm, maximizing the Fidelity would be relatively straightforward, you could find better $\overline{\rho}$ by taking convex combinations of $\rho$ and $\sigma$ with a large enough weight on $\rho$. But minimizing the fidelity is no longer a convex problem. – Rammus May 06 '21 at 11:52
  • @Rammus thanks but note that I do not need to compute the minimum. I just need an upper bound for it that is better than $\delta$ – user1936752 May 07 '21 at 08:49
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    But I assume you are looking for the tightest upper bound? Or you really just want to know when it is the case that $\delta$ is optimal? – Rammus May 07 '21 at 08:59
  • @Rammus, yes would be nice enough if I had, say, $\delta - f(\varepsilon)$ or something like that for a reasonable f. Tighter would be better of course but simply being bounced away from $\delta$ is already interesting – user1936752 May 07 '21 at 09:29

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