Finite-dimensional entropy first variation at a faithful state

Let $m>0$ be the smallest eigenvalue of $\sigma$ from [F3]. Since $\rho$ is entrywise continuous at $0$, its finite matrix difference $T(\epsilon ):=\rho (\epsilon )-\sigma$ tends to zero in operator norm: in a fixed orthonormal basis, if every entry of $T$ has magnitude at most $M$, then for a unit vector $x$ ${\sum }_j\mathrm{\mid }{x}_j{\mathrm{\mid }}^2=1$ makes each coordinate of $x$ have magnitude at most $1$, and each coordinate of $Tx$ has magnitude at most $dM$, and therefore $\mathrm{\parallel }Tx\mathrm{\parallel }\le d^{3\mathrm{/}2}M$, where $d=\dim H$. Choose $\delta >0$ so that $\mathrm{\parallel }T(\epsilon ){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}<m\mathrm{/}2$ for $\mathrm{\mid }\epsilon \mathrm{\mid }<\delta$. For each $v\in H$, $⟨\sigma v,v⟩\ge m\mathrm{\parallel }v{\mathrm{\parallel }}^2$ by spectral decomposition, while [F5] and [F6] give $\mathrm{\mid }⟨T(\epsilon )v,v⟩\mathrm{\mid }\le \mathrm{\parallel }T(\epsilon ){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\mathrm{\parallel }v{\mathrm{\parallel }}^2$. Hence $⟨\rho (\epsilon )v,v⟩>(m\mathrm{/}2)\mathrm{\parallel }v{\mathrm{\parallel }}^2$ for nonzero $v$. Thus $\rho (\epsilon )$ is positive definite near $0$ and [F1] applies.

Apply [F1] to $\rho$ at $0$: ${\frac{d}{d\epsilon }s(\rho (\epsilon ))\mid }_{\epsilon =0}=-\mathrm{tr}({\rho }^{\mathrm{\prime }}(0)(\log \sigma +I))$. Because $\mathrm{tr}(\rho (\epsilon ))=1$ on the whole interval, in the fixed basis [F4] gives $\mathrm{tr}({\rho }^{\mathrm{\prime }}(0))={\sum }_i{\rho }_{ii}^{\mathrm{\prime }}(0)={\lim }_{h\to 0}{\sum }_i({\rho }_{ii}(h)-{\rho }_{ii}(0))\mathrm{/}h={\lim }_{h\to 0}(1-1)\mathrm{/}h=0$; the finite sum commutes with this limit. Therefore ${\frac{d}{d\epsilon }s(\rho (\epsilon ))\mid }_0=-\mathrm{tr}({\rho }^{\mathrm{\prime }}(0)\log \sigma )=\mathrm{tr}({\rho }^{\mathrm{\prime }}(0)K_{\sigma })$, as claimed.

The identity [F2] holds for every $\rho (\epsilon )$ in the path. Its right-hand side is differentiable at $0$ by linearity of trace and step 2.1, so differentiating gives ${\frac{d}{d\epsilon }D(\rho (\epsilon )\mathrm{\parallel }\sigma )\mid }_0=\mathrm{tr}({\rho }^{\mathrm{\prime }}(0)K_{\sigma })-{\frac{d}{d\epsilon }s(\rho (\epsilon ))\mid }_0=0$. The additional twice continuously differentiable hypothesis ensures the differentiability required here.