Let $\mathcal{H}_1, \mathcal{H}_2$ be a Hilbert spaces and $\rho, \sigma$ be density matrices on $\mathcal{H}_1$. Define $D(\rho||\sigma) := \sup_{0 \leq M \leq I_{\mathcal{H}_1}} \frac{Tr[\rho M]}{Tr[\sigma M]}$.
(1) Show that $D$ satisfies the data processing inequality
(2) Show that $D(\rho||\sigma) = \inf\{c \in\mathbb{R}: ~c \sigma - \rho \geq 0\}$
My attempts:
For (1) we have to show that for any density matrices $\rho, \sigma$ and for any $\mathcal{E} \in \text{CPTP}(\mathcal{H}_1 \rightarrow \mathcal{H}_2)$ holds $D(\mathcal{E}(\rho) || \mathcal{E}(\sigma)) \leq D(\rho || \sigma)$. I tried to use the Kraus-decomposition of $\mathcal{E}$ and I tried to use some matrix inequalities to show the statement. My problem was that I couldn't find a suitable inequality for the denominator (in the direction we need).
For (2) I tried to prove two inequalities and I think that I found a proof for one direction. Therefore, I considered some arbitrary $c$ such that $c \sigma \geq \rho$ holds. Then we obtain
$\sup_{0 \leq M \leq I_{\mathcal{H}_1}} \frac{Tr[\rho M]}{Tr[\sigma M]} \leq \sup_{0 \leq M \leq I_{\mathcal{H}_1}} \frac{c Tr[\sigma M]}{Tr[\sigma M]} = \sup_{0 \leq M \leq I_{\mathcal{H}_1}} c = c$.
This holds for all $c$ as defined above. Therefore, we can take the infimum (with respect to c) on both sides and obtain
$\sup_{0 \leq M \leq I_{\mathcal{H}_1}} \frac{Tr[\rho M]}{Tr[\sigma M]} \leq \inf\{c\in \mathbb{R}:~ c \sigma \geq \rho \}$.
On the first glance, the other direction looked easier, as we can choose $M=I$. But then we end up with $Tr[s \sigma - \rho] \geq 0$, where we denote the supremum by $s$. And I cannot argue why the same without the trace holds true.
I am grateful for any hint or help to solve this problem!