Support-sensitive regularization limit for quantum relative entropy

First suppose $\mathrm{supp}(A)\subseteq \mathrm{supp}(B)$, so $\ker B\subseteq \ker A$. Choose an orthonormal eigenbasis $(e_j)$ of $B$ with eigenvalues $b_j\ge 0$, and put $a_j:=⟨A{e}_j,e_j⟩\ge 0$. If $b_j=0$, then $e_j\in \ker A$, so $a_j=0$, and therefore $\mathrm{tr}(A\ln (B+\epsilon I))={\sum }_{j:b_j>0}{a}_j\ln (b_j+\epsilon )\to \mathrm{tr}(A\ln B)$. Also $\epsilon \mathrm{tr}(\ln (B+\epsilon I))\to 0$: positive $b_j$ terms use continuity, while zero $b_j$ terms use $\epsilon \ln \epsilon \to 0$. The latter follows from [F5]: for $0<t\le 1$, put $u=t^{-1\mathrm{/}2}\ge 1$. Then $0\le -t\ln t=2t\ln u\le 2t(u-1)=2(\sqrt{t}-t)\to 0$ by applying the bound to $u$. The same limit gives $t\ln t\to 0$ at zero, so $\mathrm{tr}((A+\epsilon I)\ln (A+\epsilon I))\to \mathrm{tr}(A\ln A)$. Substituting these limits in the definition proves the claim in this support-inclusion case.

Now suppose $\mathrm{supp}(A)⊈\mathrm{supp}(B)$. There is a unit vector $e\in \ker B\setminus \ker A$, so $a:=⟨Ae,e⟩>0$ by [F3]. Choose an orthonormal eigenbasis of $B$ containing a basis $(u_k)$ of $\ker B$. At least one coefficient $⟨A{u}_k,u_k⟩$ is positive: otherwise [F3] would give $A{u}_k=0$ for every $k$, hence $Ae=0$. Therefore the sum $c:={\sum }_k⟨A{u}_k,u_k⟩$ is positive. In $\mathrm{tr}(A\ln (B+\epsilon I))$, this coefficient multiplies $\ln \epsilon \to -\mathrm{\infty }$, while terms from positive eigenvalues of $B$ remain bounded. The entropy trace $\mathrm{tr}((A+\epsilon I)\ln (A+\epsilon I))$ has a finite limit as in step 1.1, and the additional $\epsilon \mathrm{tr}(\ln (B+\epsilon I))$ tends to zero by [F4]–[F5]. Therefore the relative entropy tends to $+\mathrm{\infty }$, equal to the right side of the Statement.