Dimension-dependent quantum entropy continuity and Holevo control

The trace norm is nonnegative and, by [F6], $\mathrm{\parallel }\rho -\sigma {\mathrm{\parallel }}_1\le \mathrm{\parallel }\rho {\mathrm{\parallel }}_1+\mathrm{\parallel }\sigma {\mathrm{\parallel }}_1=2$; hence $t\in [0,1]$. If $t=0$, equality of the states settles the claim. Otherwise [F11] says the positive and negative parts of $\rho -\sigma$ have trace $t$. Normalize them to states ${\omega }_{+},{\omega }_{-}$. Then $\theta =\frac{\rho +t{\omega }_{-}}{1+t}=\frac{\sigma +t{\omega }_{+}}{1+t}\mathrm{.}$ Apply the lower mixture-entropy bound to the first representation and upper bound to the second. After multiplication by $1+t$ this gives $S(\rho )-S(\sigma )\le t[S({\omega }_{+})-S({\omega }_{-})]+(1+t)h(t\mathrm{/}(1+t))\le t\ln d+g(t)\mathrm{.}$ Here $S({\omega }_{+})\le \ln d$ by [F2], while $S({\omega }_{-})\ge 0$ by [F12]. Interchanging the states gives the absolute-value bound. Dimension one causes no exception. For $t>0$, [F7] gives $g^{\mathrm{\prime }}(t)=\ln (1+1\mathrm{/}t)>0$ and $g^{\mathrm{\prime }\mathrm{\prime }}(t)=-1\mathrm{/}[t(1+t)]<0$. Moreover $t\ln t\to 0$ as $t\downarrow 0$ by [F9], so $g$ is continuous on $[0,1]$ with the stated endpoint value. Jensen for $g$ follows from [F8] applied to the convex function $-g$, including the endpoint.

For each $m$, let $t_m=T({\sigma }_m,\sigma )$. By step 1.1, $0\le t_m\le 1$, so $a=M^{-1}{\sum }_m{t}_m\in [0,1]$. The average state $\bar{\sigma }=M^{-1}{\sum }_m{\sigma }_m$ satisfies $T(\bar{\sigma },\sigma )\le a$ by the trace-norm triangle inequality [F6]. Applying step 1.1 to $\bar{\sigma }$ gives a bound with its actual distance $t^{\mathrm{\prime }}\le a$; since $g$ is increasing and $\ln d\ge 0$, its right side is at most $a\ln d+g(a)$. Applying step 1.1 to each ${\sigma }_m$, averaging, and using finite Jensen [F8] for concave $g$ gives the other $a\ln d+g(a)$. The Holevo formula [F8] therefore yields $\chi (\{M^{-1},{\sigma }_m\})=S(\bar{\sigma })-\frac{1}{M}{\sum }_{m}S({\sigma }_m)\le 2a\ln d+2g(a)\mathrm{.}$ When $d=d_E^n$ and $a\le C{e}^{-cn}$, the first term tends to zero by [F10], and the second tends to zero by continuity of $g$ at zero from step 1.1.