Nonnegativity and equality for finite quantum relative entropy

The singleton POVM $E_0=I_H$ is a finite POVM by [F4]. By [F8] the inputs have trace one, and [F2] sends each density operator to the scalar state $1$ on a one-dimensional register. The definition [F10] and $\ln 1=0$ from [F11] give $D(1\mathrm{\parallel }1)=0$, so applying [F1] yields $D(\rho \mathrm{\parallel }\sigma )\ge 0$.

Suppose $D(\rho \mathrm{\parallel }\sigma )=0$ and choose any orthonormal basis $(e_j{)}_{j<d}$ of $H$ using [F7]. The rank-one maps $P_{j}x:=⟨x,e_j⟩e_j$ are self-adjoint and satisfy $⟨P_{j}x,x⟩=\mathrm{\mid }⟨x,e_j⟩{\mathrm{\mid }}^2\ge 0$ by [F9], so they are non-negative by [F6]. Their sum is $I_H$, so [F4] makes them a finite POVM; it has nonempty outcome set because $H\mathrm{\ne }\{0\}$. By [F5], the outcome vectors $p_j=\mathrm{tr}(\rho P_j)$ and $q_j=\mathrm{tr}(\sigma P_j)$ are probability vectors. Data processing [F1] and classical KL nonnegativity [F3] give $0=D(\rho \mathrm{\parallel }\sigma )\ge D(p\mathrm{\parallel }q)\ge 0$, so $p_j=q_j$ for every $j$.

For any unit vector $v\in H$, extend $v$ to an orthonormal basis using [F7]. Step 1.2 applied to that basis gives $⟨\rho v,v⟩=⟨\sigma v,v⟩$; by homogeneity this equality holds for every $v$. Put $\mathrm{\Delta }=\rho -\sigma$, which is self-adjoint by [F6], and $q(v)=⟨\mathrm{\Delta }v,v⟩$. Under the root convention [F9], expansion gives $4⟨\mathrm{\Delta }x,y⟩=q(x+y)-q(x-y)+iq(x+iy)-iq(x-iy)$. Since every $q(v)=0$, all matrix coefficients of $\mathrm{\Delta }$ vanish, hence $\rho =\sigma$.

Conversely, if $\rho =\sigma$, [F10] gives the support inclusion, and the two spectral logarithmic traces in the QRE definition agree, so $D(\rho \mathrm{\parallel }\sigma )=0$.