) - S(\Delta) \equals S(\Delta \| \Gamma_{\Delta} )  \ .\end{aligned}$$ The first and last equalities hold because $S(\Delta) < \infty$; the next-to-first and next-to-last equalities hold by Lemma \[good lemma\]; the inequality holds by (\[an entropy inequality\]). This establishes (\[superprop-inequality\]) when $S(\Delta) < \infty$.

If $S(\Delta \| \Gamma_{\Delta} ) = \infty$, then also $S(\Delta \| \Gamma)=  \infty$, as we have just established, and equation (\[superprop-equation\]) holds trivially. On the other hand, if $S(\Delta \| \Gamma_{\Delta} )$ and $S(\Delta)$ are both finite, then Prop. \[max ent prop\] implies that $S(\Gamma_\Delta )  < \infty$ and $S(\Delta\|\Gamma_{\Delta}) = S(\Gamma_\Delta)-S(\Delta)$. By Lemma \[good lemma\], $$\begin{aligned}
 S(\Delta \| \Gamma) - S(\Delta \| \Gamma_{\Delta} )
  & = & 
  -\Tr( \Delta \log \Gamma ) - S( \Gamma_{\Delta}  ) 
  \nonumber \\
  & = & 
      -\Tr( \Gamma_{\Delta} \log \Gamma ) - S( \Gamma_{\Delta}  ) 
      \equals 
 % \label{used good lemma again} \\
  %  & = & 
  S(\Gamma_{\Delta} \| \Gamma), 
  \nonumber
 \end{aligned}$$ which is equivalent to equation (\[superprop-equation\]). Thus, equation (\[superprop-equation\]) holds whether or not $S(\Delta \| \Gamma_{\Delta} )$ is finite, provided $S(\Delta ) < \infty$.

Now assume that $S(\Delta) = \infty$. The symbol $\BB(\mathcal{X})$ in the sequel denotes the algebra of bounded operators on a Hilbert space $\mathcal{X}$.

Consider an 