$\sup_{0 \leq M \leq I_{\mathcal{H}_1}} \frac{\operatorname{Tr}[\rho M]}{\operatorname{Tr}[\sigma M]}$ satisfies Data Processing Inequality $\newcommand{\Tr}{\operatorname{Tr}}$Let $\mathcal{H}_1, \mathcal{H}_2$ be a Hilbert spaces and $\rho, \sigma$ be density matrices on $\mathcal{H}_1$. Define
$$D(\rho\parallel\sigma) := \sup_{0 \leq M \leq I_{\mathcal{H}_1}} \frac{\Tr[\rho M]}{\Tr[\sigma M]}.$$
(1) Show that $D$ satisfies the data processing inequality, i.e. for any density matrices $\rho, \sigma$ (positive semi-definite, hermitian, trace = 1) and for any $\mathcal{E} \in \text{CPTP}(\mathcal{H}_1 \rightarrow \mathcal{H}_2)$ (completely positive, trace preserving map) holds $D(\mathcal{E}(\rho) \parallel \mathcal{E}(\sigma)) \leq D(\rho \parallel \sigma)$
(2) Show that $D(\rho\parallel\sigma) = \inf\{c \in\mathbb{R}: ~c \sigma - \rho \geq 0\}$
Sidenote: I posted this question first on physics stack exchange, but was told to repost the question here as it seems to fit better on this SE platform.

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 $D(\mathcal{E}(\rho) \parallel \mathcal{E}(\sigma)) \leq D(\rho \parallel \sigma)$. If the denominator is zero, the DPI is satisfied trivially. So we can assume that the denominator is not zero.
For this case, 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 hope that I am now on the correct platform for asking this question and I am grateful for any hint or help to solve this problem!
 A: For a Hilbert space $\mathcal{H}$, let $\operatorname{Herm}(\mathcal{H})$ denote the space of Hermitian operators on $\mathcal{H}$ (i.e., space of Hermitian matrices). Given a linear mapping $\mathcal{E}:\operatorname{Herm}(\mathcal{H}_1)\to\operatorname{Herm}(\mathcal{H}_2)$, its dual map is the unique linear mapping $\mathcal{E}^*:\operatorname{Herm}(\mathcal{H}_2)\to\operatorname{Herm}(\mathcal{H}_1)$ for which
$$
\operatorname{Tr}(\mathcal{E}(A)B) = \operatorname{Tr}(A\mathcal{E}^*(B)) 
$$
holds for all operators $A\in\operatorname{Herm}(\mathcal{H}_1)$ and $B\in\operatorname{Herm}(\mathcal{H}_2)$. The mapping $\mathcal{E}$ is trace preserving if and only if $\mathcal{E}^*$ is unital (that is, if $\mathcal{E}^*(I_{\mathcal{H_2}})) = I_{\mathcal{H}_1}$.

To see that $D$ satisfies the data-processing inequality, let $\mathcal{E}:\operatorname{Herm}(\mathcal{H}_1)\to\operatorname{Herm}(\mathcal{H}_2)$ be a positive and trace-preserving map. For every positive operator $M\in\operatorname{Herm}(\mathcal{H}_2)$ such that $0\leq M\leq I_{\mathcal{H}_2}$, one has that $0\leq\mathcal{E}^*(M)\leq I_{\mathcal{H}_1}$ as $\mathcal{E}$ is positive and trace preserving, and thus
$$
 \frac{\operatorname{Tr}(\mathcal{E}(\rho) M)}{\operatorname{Tr}(\mathcal{E}(\sigma) M)} = \frac{\operatorname{Tr}(\rho\mathcal{E}^*(M))}{\operatorname{Tr}(\sigma\mathcal{E}^*(M))} \leq D(\rho||\sigma).
$$
It follows that $D(\mathcal{E}(\rho)||\mathcal{E}(\sigma))\leq D(\rho||\sigma)$.

Are you familiar with semidefinite programming?
Consider the following pair of optimization problems:
\begin{align*}
\underline{\text{Primal problem} }  &&&&\underline{\text{Dual problem} }\\
\text{minimize}&\quad c &&& \text{maximize} & \quad \operatorname{Tr}(\rho M)\\
\text{subject to}&\quad c\sigma\geq \rho&&& \text{subject to} & \quad \operatorname{Tr}(\sigma M)=1\\
&&&&  & \quad M\geq0
\end{align*}
It is straightforward to show that these semidefinite programs are dual to each other. One can check that Slater's conditions hold for this pair of problems, and thus strong duality holds. That is, the optimal values of the primal and dual problem are equal to each other.
I'll leave it up to you to show that the optimal value of the dual problem above is equal to $D(\rho||\sigma)$.
