# Continuity of Renyi entropies - limiting cases

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?

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