The Renyi entropies are defined as

$$S_{\alpha}(\rho)=\frac{1}{1-\alpha} \log \operatorname{Tr}\left(\rho^{\alpha}\right), \alpha \in(0,1) \cup(1, \infty)$$

It is claimed that this quantity is continuous i.e. for $\rho, \sigma$ close in trace distance, $|S_{\alpha}(\rho) - S_{\alpha}(\sigma)|$ can be bounded for $0<\alpha<1$ and $\alpha >1$.

However, it is also stated that the max and min entropies $\lim\limits_{\alpha\rightarrow 0}S_{\alpha}(\cdot) = S_{0}$ and $\lim\limits_{\alpha\rightarrow \infty}S_{\alpha}(\cdot) = S_{\infty}$ are not continuous or not known to be continuous.

The above statements are found in http://www.scholarpedia.org/article/Quantum_entropies#Properties_of_quantum_entropy

The proofs themselves are quite complicated but is there some intuition for why the proofs fail at the limits?


1 Answer 1


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 have $S_0(\rho_\epsilon) = \log 1$. So for this family we see a discontinuity at $\epsilon \in \{0,1\}$. The lack of continuity here comes from the lack of continuity in the $\mathrm{rank}$ function.

For $S_{\infty}$ we have $$ S_{\infty}(\rho) = -\log\|\rho\|, $$ where $\|\cdot\|$ is the operator norm. But $\rho \to \|\rho\|$ is a continuous function as norms are continuous. Hence, as $\log$ is continuous on $\mathbb{R}_+$, we also have $S_{\infty}$ is continuous


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