Does the max-relative entropy satisfy $0 < D_{\max}(\rho \parallel I_A \otimes \sigma_B) < 1$?

Question

The quantum conditional min-entropy is defined as

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

I am trying to understand why there is a negative sign in the definition of $H_{\min}(A|B)$. My hunch is that $D_{\max}(\rho \parallel I_A \otimes \sigma_B) = \inf \{ \lambda : \rho \leq 2^\lambda I_A \otimes \sigma_B \}$ is a number between $0$ and $1$, and since $\log$ function is involved, we take the negative at the end, because $0 < \log(x) < 1$ when $x<1$. I would like to come up with a more rigorous argument, and I could not come up with a proof.

Is it true that $0 < D_{\max}(\rho \parallel I_A \otimes \sigma_B) < 1$?

Answer 1

No. As discussed e.g. in the second lecture of https://cs.uwaterloo.ca/~watrous/QIT-notes/, between pages 16 and 17, if $\sigma$ is a state, then $2^{-D_{\rm max}(\rho\|\sigma)}\in[0,1]$, or equivalently, $D_{\rm max}(\rho\|\sigma)\ge0$.

For example, $D_{\rm max}(\rho\|\rho)=0$, and $$D_{\rm max}\left(\frac I2\bigg\| \begin{pmatrix}1-\epsilon&0\\0&\epsilon\end{pmatrix}\right) = -\log(2\epsilon),$$ which diverges for $\epsilon\to0^+$.

Similar bounds hold when you compute $D_{\rm max}(\rho\|I\otimes \sigma)$ with $\rho$ bipartite. For example, $$ D_{\rm max}\left(\frac{I\otimes I}4\bigg\| I\otimes\begin{pmatrix}1-\epsilon&0\\0&\epsilon\end{pmatrix}\right) = -\log(4\epsilon). $$ To have $D_{\rm max}(\rho\|I\otimes \sigma)=0$, consider the following two-qubit example: $$D_{\rm max}\left(\frac{|00\rangle\!\langle00|+|11\rangle\!\langle11|}{2}\bigg\| I\otimes \frac{I}{2}\right) = 0,$$ I wrote this thinking in terms of the kind of states that give a vanishing conditional entropy.

There's different ways to compute the above examples, but at least in these cases probably the more convenient formulation is $D_{\rm max}(P\|Q)=\log\|\sqrt{Q^+}P\sqrt{Q^+}\|$. See again the above linked notes to see where this comes from.