Consider the (smooth) min-entropy, max-entropy and von Neumann entropy of a given density operator $\rho_A$. Does a small gap between $H_{\max(\min)}(A)_\rho$ and $H(A)_\rho$ implies a small gap between $H_{\min(\max)}(A)_\rho$ and $H(A)_\rho$? Put in other words, does a small gap between the von Neumann entropy and either min- or max-entropy implies a nearly flat spectrum of $\rho_A$?
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I'm not sure what you mean exactly with "small gap", but you can easily build examples where $H(A)$ and $H_{\rm max}(A)$ are "maximally different". For example, $$\rho = \begin{pmatrix}1-\epsilon & 0 \\ 0 &\epsilon\end{pmatrix}$$ for small $\epsilon>0$, has $H(\rho)\simeq 0$, but $H_{\rm max}(\rho)=\log|\operatorname{supp}(\rho)|=1$. You can build similar examples in any dimension. In these cases $H(\rho)\simeq H_{\rm min}(\rho)$ but $H_{\rm max}(\rho)$ is rather different.
On the other hand, to have $H(\rho)\simeq H_{\rm max}(\rho)$ it seems you'd need a more or less balanced nonzero eigenvalues, which would then also make $H_{\rm min}(\rho)\simeq H(\rho)$.
Here's a plot of the three quantities for the possible distributions over two outcomes:
We can also do the same for three-outcome distributions, obtaining essentially the same result:
This is, however, significantly harder to parse without being able to rotate the figure. One way to see that the relations between $H,H_{\rm min},H_{\rm max}$ we had before remains here, is to show only the difference between $H$ and $H_{\rm min}$:
This clearly shows that $H\simeq H_{\rm max}$ whenever we have a balanced distribution, be it $P\simeq (1/3,1/3,1/3)$, or things like $P\simeq (1/2,1/2,0)$. But nearby this point, we also have $H\simeq H_{\rm min}$.
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$\begingroup$ Yes, this is basically what I mean by gap. I see your point. It seems a bit strange to me that there's this asymmetry between them. $\endgroup$– ShadumuSep 18, 2022 at 22:03
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$\begingroup$ There's no asymmetry, it is possible to also make Hmin and H arbitrarily far apart whilst Hmax and H are close. The comment on this in the answer is a little misleading. $\endgroup$– RammusSep 19, 2022 at 6:43
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$\begingroup$ @Rammus I suppose you mean with distributions with more outcomes? It doesn't seem like it's possible for distributions with few outcomes. I guess you'd need a mostly balanced distribution over many outcomes, with a single outcome associated with a much larger probability? Do you have an explicit example? $\endgroup$– glS ♦Sep 19, 2022 at 7:40
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$\begingroup$ Yes, you need more outcomes to amplify the gap, but the example is exactly what you say. Large probability on one and flat on the rest. I think I wrote it out explicitly on a recent question of yours but I can't check right now. $\endgroup$– RammusSep 19, 2022 at 7:52
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$\begingroup$ @Rammus if you're referring to the example in this answer, that is, essentially, $p_k=\delta/N$ for $k=1,...,N$ and $p_0=1-\delta$, then we have there $H(p)=H(\delta) + \delta \log N$, $H_{\rm max}(p)=\log_2 (N+1)$, and $H_{\rm min}(p)=\log(1/(1-\delta))$. So $H_{\rm min}\sim 0$, $H_{\rm max}\sim \log N$, and $H\sim \delta \log N$. I don't know that I'd say that in this examples $H$ is "close" to $H_{\rm max}$. They both increase with $\log N$, but they remain quite distant, especially for small $\delta$ $\endgroup$– glS ♦Sep 19, 2022 at 15:29