$\Gamma=\Gamma_\Delta$.

The analog of Prop. \[superproposition\] for other Rényi divergences would be false. That is, if $\alpha \ne 1$ then $\Gamma_{\Delta}$ need not minimize $D_\alpha(\Delta \|\Gamma )$ or $\widetilde{D}_\alpha(\Delta \|\Gamma )$.

For example, let $\HH = \mathrm{span}\big\{| \uparrow\  \rangle,| \downarrow\  \rangle \big\}$ and let $\Delta$ denote the density operator that is entirely supported on the $1$-particle component of $\FFF(\HH)$, where it equals $\frac23 | \uparrow \ \rangle\!\langle \ \uparrow  | + \tfrac13 | \downarrow \ \rangle\!\langle \ \downarrow |$. Then the minimum of $D_\alpha(\Delta \|\Gamma)$ or $\widetilde{D}_\alpha(\Delta \|\Gamma )$ is not attained at $\Gamma_\Delta$ unless $\alpha = 1$.

We first prove Prop. \[superproposition\] under the assumption that $S(\Delta) < \infty$, which allows us to use formula (\[if entropy is finite\]). Then we will relieve the assumption that $S(\Delta) < \infty$ by using the martingale property of relative entropy [@OhyaPetz].

Suppose that $S(\Delta) < \infty$. Then $$S( \Delta \| \Gamma ) \ = \   -  \Tr ( \Delta \log \Gamma ) - S(\Delta)\ .$$ If $\Gamma$ is free, then $$\begin{aligned}
S(\Delta \| \Gamma)  & = &  -\Tr(\Delta \log \Gamma) - S(\Delta) \equals  -\Tr( \Gamma_{\Delta} \log \Gamma ) - S(\Delta) \\
    & \ge & S(\Gamma_{\Delta}) - S(\Delta) \equals  -\Tr( \Delta \log \Gamma_{\Delta} 