Trace-class cross-log entropy inequality

Write $\rho ={\sum }_i{p}_i\mathrm{\mid }{u}_i⟩⟨u_i\mathrm{\mid }$ and $\sigma ={\sum }_j{s}_j\mathrm{\mid }{v}_j⟩⟨v_j\mathrm{\mid }$ on their positive supports. Put $w_{ij}=\mathrm{\mid }⟨u_i,v_j⟩{\mathrm{\mid }}^2$, so ${\sum }_j{w}_{ij}=1$ because $u_i\in \mathrm{supp}\sigma$, and ${\sum }_i{w}_{ij}\le 1$. Define $q_i=⟨\sigma u_i,u_i⟩={\sum }_j{w}_{ij}{s}_j>0$; then $Q={\sum }_i{q}_i=\mathrm{Tr}(\sigma P_{\mathrm{supp}\rho })\le 1$.

For $q>0$ and $x>0$, $\log x\le \log q+(x-q)\mathrm{/}q$, because $t-1-\log t\ge 0$ for $t=x\mathrm{/}q$, as its derivative $1-1\mathrm{/}t$ has its unique minimum at $t=1$. Applying this tangent inequality to $s_j$ with weights $w_{ij}$ gives ${\sum }_j{w}_{ij}(-\log s_j)\ge -\log q_i$. Since $C<\mathrm{\infty }$, $T:={\sum }_i{p}_i(-\log q_i)\le C<\mathrm{\infty }$.

For a finite initial set $1\le i\le N$, let $a_N={\sum }_{i\le N}{p}_i$ and $b_N={\sum }_{i\le N}{q}_i$. Applying the finite-density case of [L1] to $(p_i\mathrm{/}{a}_N)$ and $(q_i\mathrm{/}{b}_N)$ gives ${\sum }_{i\le N}-p_i\log p_i\le {\sum }_{i\le N}{p}_i(-\log q_i)-a_N\log a_N+a_N\log b_N\le T-a_N\log a_N,$ because $0<b_N\le 1$. As $N\to \mathrm{\infty }$, $a_N\to 1$ and $-a_N\log a_N\to 0$. The nonnegative partial entropy sums are therefore bounded, so $S(\rho )\mathrm{/}{k}_B={\sum }_i-p_i\log p_i<\mathrm{\infty }$. This also makes both scalar logarithm series below absolutely defined.

Normalize $q_i$ by $Q$, so $q_i^{\mathrm{\prime }}=q_i\mathrm{/}Q$ is a probability distribution with $q_i^{\mathrm{\prime }}>0$. Apply [L1] to $p_i$ and $q_i^{\mathrm{\prime }}$: ${\sum }_i{p}_i\log (p_i\mathrm{/}{q}_i^{\mathrm{\prime }})\ge 0$, hence ${\sum }_i{p}_i\log (p_i\mathrm{/}{q}_i)\ge -\log Q\ge 0$. Combining this with the nonnegative Jensen gaps from step 2.1 gives $C-S(\rho )\mathrm{/}{k}_B\ge 0$.

Since steps 3.1–4.1 prove that $S(\rho )$ and $T$ are finite, and $q_i^{\mathrm{\prime }}=q_i\mathrm{/}Q$, direct logarithm algebra gives $D(\rho \mathrm{\parallel }\sigma )=C-\frac{S(\rho )}{k_B}=(C-T)+{\sum }_i{p}_i\log \frac{p_i}{q_i^{\mathrm{\prime }}}-\log Q\mathrm{.}$ The first term is the sum of the nonnegative Jensen gaps from step 2.1, the middle term is the countable relative entropy from [L1], and the last term is nonnegative because $Q\le 1$. If $D(\rho \mathrm{\parallel }\sigma )=0$, all three vanish. Thus $Q=1$; equality in [L1] gives $p_i=q_i^{\mathrm{\prime }}=q_i$ for every $i$. The Jensen gap is ${\sum }_i{p}_i[{\sum }_j{w}_{ij}(-\log s_j)+\log q_i]$; each bracket is nonnegative by step 2.1 and each $p_i>0$, so a zero sum forces equality in every bracket. Strictness of $\log x\le \log q_i+(x-q_i)\mathrm{/}{q}_i$ then forces $s_j=q_i$ whenever $w_{ij}>0$, and the spectral expansion gives $\sigma u_i=q_i{u}_i=p_i{u}_i$. The equality $Q=1$ says $\mathrm{Tr}(\sigma (I-P_{\mathrm{supp}\rho }))=0$. Put $P=P_{\mathrm{supp}\rho }$. The spectral expansion and the bounded trace pairing in [F2] give $0=\mathrm{Tr}(\sigma (I-P))={\sum }_j{s}_j\mathrm{\parallel }(I-P)v_j{\mathrm{\parallel }}^2$. Every term is nonnegative and every $s_j>0$, so $(I-P)v_j=0$ for all $j$. Therefore the positive spectral range of $\sigma$ lies in $PH$, and its spectral expansion makes $\sigma$ vanish on $(PH{)}^{\perp }$. Therefore $\rho =\sigma$. Conversely, if $\rho =\sigma$, then $q_i=p_i$, $C=S(\rho )\mathrm{/}{k}_B$, and $D=0$.