---
author:
- Anna Jenčová
title: ' A construction of a nonparametric quantum information manifold.' 
---

Mathematical Institute, Slovak Academy of Sciences,

Štefánikova 49, 814 73 Bratislava, Slovakia,

e-mail:jenca@mat.savba.sk

[**Abstract.**]{}[We present a construction of a Banach manifold on the set of faithful normal states of a von Neumann algebra, where the underlying Banach space is a quantum analogue of an Orlicz space.  On the manifold, we introduce the exponential and mixture connections as dual pair of affine connections. ]{} 

Introduction
============

An information manifold is a family of states of some classical or quantum system, endowed with a differentiable manifold structure.  For finitely parametrized families of probability distributions, the geometry of such manifolds and its applications in parameter estimation is already well understood, see for example the books [@amana; @abklr]. 

The non-parametric version was introduced by Pistone and Sempi [@pise; @gipi], based on the theory of Orlicz spaces.  For the quantum version, some proposals for infinite dimensional manifold structure can be found in [@grastre; @streater00a; @streater00b]. 

The aim of this paper is to introduce a differentiable manifold structure on the set of faithful states of a quantum system, represented by a von Neumann algebra $\iM$. Moreover, we want this manifold to be a quantum counterpart of the Pistone and Sempi construction. 

We use an approach similar to Grasselli [@grasselli05] in the commutative case: we define an Orlicz norm on the space of self-adjoint operators in $\iM$ and take the completion under this norm to be the underlying Banach space for the manifold.  The norm is defined by a quantum Young function, as in [@streater04].  The definition of a Young function on a Banach space, together with some results on the associated norms, can be found in Section \[sec:young\].  For a faithful state $\vfi$, the quantum Orlicz space $B_\vfi$ and its centered version $B_{\vfi,0}$ are introduced in Section \[sec:orlicz\].  The definition of the related Young function is based on the relative entropy approach to state perturbation.  We treat the dual spaces in Section \[sec:dual\].  The main result is contained in Section \[sec:mani\], where the manifold structure is introduced and, moreover, the exponential and mixture connections are defined as a pair of dual affine connections on each connected component of the manifold. 

Preliminaries. 
==============

We recall some properties of relative entropy and perturbed states, that will be needed later.  See [@ohypetz] for details. 

Let $\iM$ be a von Neumann algebra in standard form.  For $\omega$ and $\vfi$ in $\iM_*^+$, the relative entropy is defined as $$S(\omega,\vfi)=\left\{\begin{array}{lc} -\langle\log
(\Delta_{\vfi,\xi_\omega})\xi_\omega,\xi_\omega\rangle & \ \mbox{if } \supp
\omega\le\supp \vfi\\ \infty & \mbox{otherwise}
\end{array}\right.$$ where $\xi_\omega$ is the representing vector of $\omega$ in a natural positive cone and $\Delta_{\vfi,\xi_\omega}$ is the relative modular operator.  Then $S$ is jointly convex and weakly lower semicontinuous.  Let us denote $\iP_\vfi:=\{ \omega\in\iM_*^+, S(\omega,\vfi)<\infty\}$, then $\iP_\vfi$ is a convex cone.  We will need the following Donald’s identity $$\label{eq:donald}
S(\psi,\vfi)+
\sum_iS(\psi_i,\psi) =\sum_iS(\psi_i,\vfi)$$ where $\psi_i\in \iM_*^+$, $i=1,\dots,n$, and $\psi=\sum_i\psi_i$. Since $S(\psi_i,\psi)$ is always finite, it follows from this identity that $\sum_i\psi_i\in
\iP_\vfi$ if and only if $\psi_i\in \iP_\vfi$ for all $i$.

Let $\mathfrak{S}_*$ be the set of normal states on $\iM$ and let $\iS_\vfi:=\{\omega\in \mathfrak{S}_*,\ S(\omega,\vfi)<\infty\}$. Then $\iS_\vfi$ is a convex set and generates $\iP_\vfi$. From (\[eq:donald\]), we get $$\label{eq:donalds}
S(\psi_\lambda,\vfi)+\lambda
S(\psi_1,\psi_\lambda)+(1-\lambda)S(\psi_2,\psi_\lambda)
 =\lambda S(\psi_1,\vfi)+(1-\lambda)S(\psi_2,\vfi)$$ where $\psi_1$, $\psi_2$ are normal states and $\psi_\lambda=\lambda\psi_1+(1-\lambda)\psi_2$, $0\le \lambda\le 1$. As above, it follows that $\psi_\lambda\in \iS_\vfi$ if and only if both $\psi_1,\psi_2\in \iS_\vfi$, in other words, $\iS_\vfi$ is a face in $\mathfrak{S}_*$. For $C>0$, we define the set $\iS_C:=\{\omega,\
S(\omega,\vfi)\le C\}$. Then $\iS_C$ is convex and compact in the $\sigma(\iM_*,\iM)$ topology. 

Let us suppose that $\vfi$ is a faithful normal state on $\iM$ and let $h$ be a self-adjoint element in $\iM$. The perturbed state $[\vfi^h]$ is defined as the unique maximizer of $$\label{eq:sup}
\sup_{\omega\in \mathfrak{S}_*}\{ \omega(h)-S(\omega,\vfi)\}$$ Then $[\vfi^h]$ is a faithful normal state and $S([\vfi^h],\vfi)$ is finite.  Let $c_\vfi(h)$ be the supremum in (\[eq:sup\]), that is $$\label{eq:cfi}
c_\vfi(h)=[\vfi^h](h)-S([\vfi^h],\vfi)$$ It is known that $$\label{eq:gt}
\vfi(h)\le c_\vfi(h)\le \log \vfi(e^h)$$ Moreover, we have $$\label{eq:sup2}
\omega(h)-S(\omega,\vfi)=c_\vfi(h)-S(\omega,[\vfi^{h}])$$ for any self-adjoint $h\in\iM$ and $\omega\in\mathfrak{S}_*$. Let $h,k$ be self-adjoint elements in $\iM$, then the chain rule $[\vfi^{h+k}]=[[\vfi^h]^k]$ and $$\label{eq:chainM}
c_\vfi(h+k)=c_{[\vfi^h]}(k)+c_\vfi(h)$$ holds.  Let now $\xi_\vfi$ be the vector representative of $\vfi$ and let $\vfi^h\in\iM_*^+$ be the functional induced by the perturbed vector $$\xi_\vfi^h:=e^{\frac12(\log\Delta_{\vfi}+h)}\xi_\vfi=
e^{c_\vfi(h)}\Delta^{1/2}_{[\vfi^h],\vfi}\xi_\vfi$$ Then $c_\vfi(h)=\log \vfi^h(1)$ and $[\vfi^h]=\vfi^h/\vfi^h(1)$. Moreover, if $\vfi^h=\vfi^k$, then $h=k$.

Young functions on Banach spaces and the associated norms. {# sec:young}
==========================================================

Let $V$ be a real Banach space and let $V^*$ be its dual, with the duality pairing $\langle v,x\rangle= v(x)$. Recall that any convex lower semicontinuous function $V\to \mathbb R\cup\{+\infty\}$ is lower semicontinuous with respect to the $\sigma(V,V^*)$-topology. 

The Young function. {# the-young-function.  .unnumbered}
-------------------

We will say that a function $\Phi:V\to \mathbb{R}\cup \{\infty\}$ is a Young function, if it satisfies:

1.   $\Phi$ is convex and lower semicontinuous,

2.   $\Phi(x)\ge 0$ for all $x\in V$ and $\Phi(0)=0$

3.   $\Phi(x)=\Phi(-x)$ for all $x\in V$

4.   if $x\ne 0$, then $\lim_{t\to\infty}\Phi(tx)=\infty$

\[lemma:cramer\] Let $\Phi$ be a Young function.  Let us define the sets $$\begin{aligned}
C_\Phi&:=&\{ x\in V, \Phi(x)\le 1\}\\
L_\Phi&:=&\{ x\in V, \exists s>0, \mbox{ such that  } \Phi(sx)<\infty\}.\end{aligned}$$ Then $C_\Phi$ is absolutely convex and $L_\Phi=\cup_nnC_\Phi$. In particular, $L_\Phi$ is a (real) vector space. 

[*Proof.*]{}  Let $x,y\in C_\Phi$ and let $\alpha,\beta\in \mathbb{R}$, such that $|\alpha|+|\beta|\le 1$. Put $\gamma=1-|\alpha|-|\beta|$, then $$\Phi(\alpha x+\beta y)=\Phi(|\alpha|{\rm sgn}(\alpha)x+|\beta|{\rm sgn}(\beta)y
+\gamma 0)\le |\alpha|\Phi(x)+|\beta|\Phi(y)\le 1$$ hence $\alpha x+\beta y\in C_\Phi$ and $C_\Phi$ is absolutely convex. 

Let now $x\in L_\Phi$ and let $s>0$ be such that $\Phi(sx)=K<\infty$. Choose $m\in \mathbb{N}$ such that $m\ge \max\{1/s,K/s\}$, then by convexity $$\Phi(\frac 1m x)=\Phi(\frac1{ms}sx)\le \frac1{ms}\Phi(sx)=\frac K{ms}\le 1$$ and $x\in mC_\Phi$. Since obviously $nC_\Phi\subset L_\Phi$ for all $n$, we have $L_\Phi=\cup_nnC_\Phi$, which clearly implies that $L_\Phi$ is a vector space. 

Let us recall that the effective domain $${\rm dom} (\Phi)=\{ x\in V,\ \Phi(x)<\infty\}$$ is a convex set.  Any convex lower semicontinuous function is continuous in the interior of its effective domain, [@ektem].  Clearly, $L_\Phi$ is the smallest vector space containing ${\rm dom}(\Phi)$.

In the space $L_\Phi$, we now introduce the Minkowski functional of $C_\Phi$, $$\|x\|_\Phi:=\inf\{\rho>0, x\in \rho C_\Phi\}.$$ Since $C_\Phi$ is absolutely convex and absorbing, $\|\cdot\|_\Phi$ is a seminorm.  Moreover, $\|x\|_\Phi=0$ means that $\Phi(tx)\le 1$ for all $t>0$. By the property (iv), this implies that $x=0$. It follows that $\|\cdot\|_\Phi$ defines a norm in $L_\Phi$. Let us denote by $B_\Phi$ the completion of $L_\Phi$ under this norm. 

\[lemma:Phinorm\] Let $x\in L_\Phi$. Then $\|x\|_\Phi\le 1$ if and only if $\Phi(x)\le 1$.

[*Proof.*]{}  If $\|x\|_\Phi<1$, then $x\in C_\Phi$ and $\Phi(x)\le 1$. Let now $\|x\|_\Phi=1$ and let $t_n< 1$ be a sequence converging to 1.  Then $\Phi(t_nx)\le 1$ for all $n$ and, by lower semicontinuity, $\Phi(x)\le \liminf_n \Phi(t_nx)\le 1$. Hence $\|x\|_\Phi\le 1$ implies $\Phi(x)\le 1$. On the other hand, if $\Phi(x)\le 1$, then $x\in C_\Phi$ and clearly $\|x\|_\Phi\le 1$.

\[lemma:normandPhi\] Let $x\in L_\Phi$. Then $\|x\|_\Phi\le 1$ implies $\Phi(x)\le \|x\|_\Phi$ and $\|x\|_\Phi> 1$ implies $ \Phi(x)\ge \|x\|_\Phi$. Moreover, if $\Phi$ is finite valued, then $\|x\|_\Phi=1$ if and only if $\Phi(x)=1$.

[*Proof.*]{}  Let $\|x\|_\Phi\le 1$. By convexity of $\Phi$ and Lemma \[lemma:Phinorm\], $$\Phi(x)=\Phi(\|x\|_\Phi\frac x{\|x\|_\Phi})\le \|x\|_\Phi\Phi(\frac
x{\|x\|_\Phi})\le\|x\|_\Phi$$

Let now $\|x\|_\Phi>1$, then $\Phi(x)>1$. If $\Phi(x)=\infty$, then the assertion is obviously true.  Let us suppose that $\Phi(x)$ is finite.  The function $t\mapsto\Phi(tx)$ is convex and bounded on $<0,1>$, hence continuous on $(0,1)$. It follows that $\Phi(tx)=1$ for some $t$ in this interval and clearly $t=\frac 1{\|x\|_\Phi}$. We have $$1=\Phi(tx)\le t\Phi(x)$$ and hence $\|x\|_\Phi\le \Phi(x)$. This also proves that last statement. 

The conjugate function. {# the-conjugate-function.  .unnumbered}
-----------------------

Let $V^*$ be the dual space.  Let the function $\Phi^*:V^*\to
\mathbb{R}\cup\{\infty\}$ be the conjugate of $\Phi$, $$\Phi^*(v)=\sup_{x\in V}\{v(x)-\Phi(x)\}=\sup_{x\in {\rm Dom}(\Phi)}
\{v(x)-\Phi(x)\}$$ The function $\Phi^*$ is convex, lower semicontinuous and positive, $\Phi^*(v)=\Phi^*(-v)$ and $\Phi^*(0)=0$. But, in general, $\Phi^*$ is not a Young function: consider the case when $\Phi(0)=0$ and $\Phi(x)=\infty$ for all $x\ne
0$, then $\Phi$ is a Young function, but its conjugate is identically equal to 0 on $V^*$ and the condition (iv) is not satisfied. 

Let $({\rm dom}(\Phi))^\perp$ be the orthogonal subspace to ${\rm dom}(\Phi)$ in $V^*$, that is $$({\rm dom}(\Phi))^\perp:=\{ v\in V^*,\ v(x)=0 \ \mbox{for all } x\in {\rm
dom}(\Phi)\}$$ Then $({\rm dom}(\Phi))^\perp$ is a closed subspace in $V^*$. Let $\tilde{V}$ be the quotient space $\tilde V=V^*/_{({\rm dom}(\Phi))^\perp}$. If $u$ and $v$ are elements in the same equivalence class, then $$\Phi^*(v)=\sup_{x\in {\rm dom}(\Phi)}\{v(x)-\Phi(x)\}=\sup_{x\in {\rm
dom}(\Phi)}\{u(x)-\Phi(x)\}=\Phi^*(u)$$ and $\Phi^*$ is well defined as a function on $\tilde V$.

$\Phi^*: \tilde V\to \mathbb{R}\cup\{\infty\}$ is a Young function. 

[*Proof.*]{}  It is easy to see that $\Phi^*$ satisfies (i), (ii) and (iii) from the definition of a Young function.  Moreover, it follows from the definition of the conjugate function that $$\label{eq:young}
|v(x)|\le \Phi(x)+\Phi^*(v),\quad \mbox{for all } x\in V, v\in \tilde V$$ Let $v\in \tilde V$, $v\ne 0$. Then there is an element $x\in {\rm  dom}(\Phi) $ such that $v(x)\ne 0$. It follows that $\Phi^*(tv)\geq |tv(x)|-\Phi(x)$ for all $t$ and (iv) is satisfied. 

We will define $C_{\Phi^*}$, $L_{\Phi^*}$, $\|\cdot\|_{\Phi^*}$ and $B_{\Phi^*}$ in the same way as for $\Phi$.

(Hölder inequality).  $$|v(x)|\le 2\|x\|_\Phi\|v\|_{\Phi^*}\qquad \mbox{for all }\ x\in B_\Phi, v\in B_{\Phi^*}$$

[*Proof.*]{}  Let $x\in C_\Phi$, $v\in C_{\Phi^*}$, then by (\[eq:young\]) $$|v(x)|\le \Phi(x)+\Phi^*(v)\le 2$$ Let $x\in L_\Phi$, $v\in L_{\Phi^*}$. By Lemma \[lemma:Phinorm\], $\frac x{\|x\|_\Phi}\in C_\Phi$, $\frac v{\|v\|_{\Phi^*}}\in C_{\Phi^*}$ and therefore $|v(x)|\le 2\|x\|_\Phi\|v\|_{\Phi^*}$. Clearly, the inequality extends to $x\in B_\Phi$, $v\in B_{\Phi^*}$.

The second conjugate. {# the-second-conjugate.  .unnumbered}
---------------------

If $E$ is a Banach space and $H\subset E$ is a closed subspace, then the dual of the quotient space $(E/H)$ can be identified with $H^\perp$. It follows that $\tilde V^*\cap V=({\rm dom}(\Phi))^{\perp\perp}$, which is nothing else than the closure of $L_\Phi$ in $V$. Let us denote this space by $\bar V$.

As before, we can find the conjugate function to $\Phi^*:\tilde V\to \mathbb
R\cup\{+\infty\}$ with respect to the pair $(\tilde V,\tilde V^*)$. Note that for $x$ in $\bar V$, we have $$\sup_{v\in \tilde V}\{v(x)-\Phi^*(v)\}=\sup_{v\in V^*}\{v(x)-\Phi^*(v)\}=
\Phi^{**}(x)$$ where $\Phi^{**}$ is the second conjugate to $\Phi:V\to \mathbb
R\cup\{+\infty\}$. Since $\Phi$ is convex and lower semicontinuous, $\Phi^{**}(x)=\Phi(x)$ on $V$, [@ektem].  It follows in particular that the restriction of $\Phi^{**}$ to $\bar V$ is a Young function. 

It is clear from Hölder inequality that any $x\in L_\Phi$ defines a bounded linear functional on $B_{\Phi^*}$. Let $\|x\|^*_{\Phi^*}$ be its norm in $B_{\Phi^*}^*$, then by Lemma \[lemma:Phinorm\], $$\|x\|^*_{\Phi^*}=\sup \{|v(x)|, \Phi^*(v)\le 1\}.$$ Similarly, if $v\in L_{\Phi^*}$, then $v\in B^*_\Phi$ and we have $$\|v\|^*_\Phi=\sup \{|v(x)|,
\Phi(x)\le 1\}$$

\[lemma:equivalent\] For $x\in L_\Phi$, we have $\|x\|_\Phi\le \|x\|_{\Phi^*}^*\le 2 \|x\|_\Phi$. Similarly, if $v\in L_{\Phi^*}$, then $\|v\|_{\Phi^*}\le \|v\|^*_\Phi\le 2\|v\|_{\Phi^*}$.

[*Proof.*]{}  Let $v\in L_{\Phi^*}$. By Hölder inequality, $\|v\|_\Phi^*\le 2\|v\|_{\Phi^*}$. Let now $\|v\|^*_\Phi=1$, then for $x\in C_\Phi$ we have $$v(x)-\Phi(x)\le 1.$$ On the other hand, for $x\in {\rm dom}(\Phi)$, such that $\Phi(x)>1$, we get from Lemma \[lemma:normandPhi\] $$v(x)-\Phi(x)\le v(x)-\|x\|_\Phi\le0.$$ It follows that ${\Phi^*}(v)\le 1$ and $v\in C_{\Phi^*}$, hence $\|v\|_{\Phi^*}\le 1$. Therefore, $\|v\|_{\Phi^*}\le\|v\|^*_\Phi$ for all $v\in L_{\Phi^*}$. The proof for $x\in L_\Phi$ is the same, using the fact that $\Phi$ is the conjugate of $\Phi^*$.

\[prop:dual\] $B_{\Phi^*}\subseteq B^*_\Phi$ and $L_{\Phi^*}=\tilde V\cap B_\Phi^*$. Similarly, $B_\Phi\subseteq B^*_{\Phi^*}$ and $L_\Phi=\bar V\cap B^*_{\Phi^*}$.

[*Proof.*]{}  As we have seen, $L_{\Phi^*}$ is a vector subspace in $B_\Phi^*$ and the norms in $L_{\Phi^*}$ and $B_\Phi^*$ are equivalent, hence $B_{\Phi^*}\subseteq B^*_\Phi$. Let now $v\in \tilde V\cap B^*_\Phi$ be such that $\|v\|^*_\Phi=1$. Then ${\Phi^*}(v)\le 1$, exactly as in the proof of Lemma \[lemma:equivalent\].  It follows that for all $v\in \tilde V\cap B^*_\Phi$, ${\Phi^*}(v/\|v\|^*_\Phi)\le1<\infty$ and $v\in L_{\Phi^*}$. Again, the proof for $L_\Phi$ and $B_\Phi$ is the same. 

Let $\Phi$ be a Young function such that 0 is an interior point in ${\rm dom}(\Phi)$. Then the function $\Phi$ is continuous in 0, therefore there is an open set $U$ containing 0 such that $U\subset C_\Phi$. It follows that $C_\Phi$ is a neighborhood of $0$ in $V$, hence it is absorbing in $V$ $$\label{eq:Velfi}
V=\cup_nnC_\Phi=L_\Phi\quad \mbox{ (as sets)}$$ Since $C_\Phi$ is a convex body (that is, 0 is a topological interior point), its Minkowski functional $\|\cdot\|_\Phi$ is continuous with respect to the original norm ([@kothe], p. 182).  It follows that we have the continuous inclusion $V\sqsubseteq B_\Phi$. Further, since ${\rm dom}(\Phi)$ has non-empty interior, $({\rm dom}(\Phi))^\perp=\{0\}$ and $\tilde V=V^*$. Clearly also $\bar V=V$.

\[prop:interior\] Let $0\in {\rm int}\, {\rm dom}(\Phi)$. Then $V\sqsubseteq B_\Phi\subseteq B_{\Phi^*}^*$ and $L_{\Phi^*}=B_{\Phi^*}=B_\Phi^*\sqsubseteq V^*$.

[*Proof.*]{}  By (\[eq:Velfi\]), each $x\in V$ is in $L_\Phi$, and by continuity, $\|x\|_\Phi\le K\|x\|$, for some fixed $K>0$. Let $v\in B^*_\Phi$, then $$|v(x)|\le \|v\|_\Phi^*\|x\|_\Phi\le K\|v\|_\Phi^*\|x\| \quad \mbox{for }\ x\in V$$ hence $v\in V^*=\tilde V$ and $\|v\|^*\le K\|v\|_\Phi^*$. The statement now follows from Proposition \[prop:dual\]. 

The spaces $B_\vfi$ and $B_{\vfi,0}$. {#sec:orlicz}
=====================================

Let $\iM_s$ be the real Banach subspace of self-adjoint elements in $\iM$, then the dual $\iM_s^*$ is the subspace of hermitian (not necessarily normal) functionals in $\iM^*$. We define the functional $F_\vfi:\ \iM_s^*\to
\mathbb{R}\cup\{\infty\}$ by $$F_\vfi(\omega)=\left\{\begin{array}{lc} S(\omega,\vfi) & \ \mbox{if
}\omega\in\mathfrak{S}_* \\ \infty & \mbox{otherwise}
\end{array}\right.$$ Then $F_\vfi$ is convex and lower semicontinuous, with ${\rm dom
}(F_\vfi)=\iS_\vfi$. It follows from (\[eq:donald\]) that $F_\vfi$ is strictly convex.  Its conjugate $F^*_\vfi$ is $$F^*_\vfi(h)=\sup_{\omega\in
\mathfrak{S}_*}\{\omega(h)-F_\vfi(\omega)\}=c_\vfi(h), \quad h\in\iM_s$$ Hence $c_\vfi$ is convex and lower semicontinuous, in fact, since finite valued, it is continuous on $\iM_s$. We have $c^*_\vfi=F^{**}_\vfi=F_\vfi$ on $\iM_s^*$. Note also that $$\label{eq:cadit}
c_\vfi(h+\lambda)=c_\vfi(h)+\lambda,\qquad \forall \lambda\in\mathbb{R}$$

We define another convex and lower semicontinuous functional on $\iM_s^*$, namely, $$\bar F_\vfi(\omega)=\left\{\begin{array}{lc} S(\omega,\vfi)-\omega(1) & \ \mbox{if
}\omega\in\iM^+_* \\ \infty & \mbox{otherwise}
\end{array}\right.$$ Then the conjugate functional is $$\begin{aligned}
\bar F_\vfi^*(h)&=&\sup_{\omega\in
\iM^+_*}\{\omega(h)-S(\omega,\vfi)+\omega(1)\}=
\sup_{\omega\in \mathfrak{S}_*,\lambda\in \mathbb{R}^+}
\{\lambda\omega(h)-S(\lambda\omega,\vfi)+\lambda\}=\\
&=&\sup_{\omega\in \mathfrak{S}_*,\lambda\in \mathbb{R}^+}
\{\lambda(\omega(h)-S(\omega,\vfi))-\lambda\log\lambda+\lambda\}=\\
&=&\sup_{\lambda\in\mathbb{R}^+}\{\lambda(c_\vfi(h)+1)-\lambda\log\lambda\}=e^{c_\vfi(h)}
=\vfi^h(1)\end{aligned}$$ Again, $h\mapsto\vfi^h(1)$ is convex and continuous and $\bar
F_\vfi^{**}=\bar F_\vfi$. Next, we define a Young function on $\iM_s$.

Let $\Phi_\vfi:\ \iM_s\to \mathbb{R}^+$ be defined by $$\Phi_\vfi(h)=\frac {\vfi^h(1)+\vfi^{-h}(1)}2-1$$

\[lemma:Phi\] $\Phi_\vfi$ is a Young function. 

[*Proof.*]{}  The property (i) from the definition of a Young function follows from the properties of $h\mapsto \vfi^h(1)$. Since $\vfi^h(1)=e^{c_\vfi(h)}\geq e^{\omega(h)-S(\omega,\vfi)}$ for all normal states $\omega$, we have $$\label{eq:helpful}
\Phi_\vfi(h)\geq \cosh(\omega(h))e^{-S(\omega,\vfi)}-1$$ In particular, $$\label{eq:vfi}
\Phi_\vfi(h)\ge \cosh(\vfi(h))-1\ge 0\quad \mbox {for all }\ h$$ Since obviously $\Phi_\vfi(0)=0$, (ii) follows.  Let now $h$ be such that $\omega(h)=0$ for all $\omega\in \iS_\vfi$, then by definition, $c_\vfi(h)=0$ and $\vfi=\vfi^h$, hence $h=0$. Therefore if $h\ne 0$, then there is a state $\omega\in\iS_\vfi$ such that $\omega(h)\ne 0$ and then $\lim_{t\to\infty}\cosh(t\omega(h))=\infty$, this implies (iv).  Property (iii) is obviously satisfied. 

Let $C_\vfi:=C_{\Phi_\vfi}$, $B_\vfi:=B_{\Phi_\vfi}$ and $\|\cdot\|_\vfi:=\|\cdot\|_{\Phi_\vfi}$. Since ${\rm dom}\Phi_\vfi=\iM_s$, we have by Proposition \[prop:interior\] that $\iM_s\sqsubseteq B_\vfi$. If $\Phi_\vfi^*$ is the conjugate of $\Phi_\vfi$, then $B^*_\vfi=B_{\Phi_\vfi^*}\sqsubseteq \iM^*_s$.

Let now $h\in \iM_s$, such that $\|h\|_\vfi=t>0$, that is, $$\Phi_\vfi(\frac{h}t)=1$$ If $\omega$ is a state, then by (\[eq:helpful\]), $$\label{eq:omega}
\cosh(\frac{\omega(h)}t)\le 2e^{S(\omega,\vfi)}$$ If $\omega\in \iS_\vfi$, then $|\omega(h)|\le ct$, where $c>0$ is some constant depending on $S(\omega,\vfi)$. It follows that each $\omega\in \iS_\vfi$ extends to a continuous linear functional on $B_\vfi$. Moreover, for $C>0$, $\iS_C$ is an equicontinuous subset in $B_\vfi^*$.

Let $\iM_{s,0}\subset \iM_s$ be the subspace of elements satisfying $\vfi(h)=0$. Then by putting $\omega=\vfi$ in (\[eq:sup2\]), we get $$c_\vfi(h)=S(\vfi,[\vfi^h])\geq0$$ Let us define $$\Phi_{\vfi,0}(h)=\frac{c_\vfi(h)+c_\vfi(-h)}2,\quad h\in \iM_{\vfi,0}$$ Then it is easy to check that $\Phi_{\vfi,0}$ is a Young function on $\iM_{\vfi,0}$. We have

\[lemma:finula\] Let $h\in \iM_{s,0}$. Then $$\Phi_{\vfi,0}(h)\le \Phi_\vfi(h)\le e^{2\Phi_{\vfi,0}}-1$$

[*Proof.*]{}  The first inequality follows from $a\le e^a-1$ for $a\ge 0$, the second follows from $x+y\le 2xy$ for $x,y\ge1$.

Let us construct the Banach space $B_{\Phi_{\vfi,0}}=:B_{\vfi,0}$ and let $\|\cdot\|_{\vfi,0}:=\|\cdot\|_{\Phi_{\vfi,0}}$.

\[prop:finula\] The norms $\|\cdot\|_{\vfi,0}$ and $\|\cdot\|_\vfi$ are equivalent on $\iM_{s,0}$.

[*Proof.*]{}  Let us denote $C_{\vfi,0}:=C_{\Phi_{\vfi,0}}$. We show that $$\label{eq:cf0}
\frac12{\log2}\,C_{\vfi,0}\subseteq C_\vfi\cap \iM_{s,0}\subseteq C_{\vfi,0}$$ Let $h\in C_{\vfi,0}$ and $t=\frac12\log2$. Then by convexity, $\Phi_{\vfi,0}(th)\le t=\frac12\log2$ and hence $$\Phi_\vfi(th)\le e^{2\Phi_{\vfi,0}(th)}-1\le 1,$$ which implies $tC_{\vfi,0}\subseteq C_\vfi\cap \iM_{s,o}$. The other inclusion follows from the first inequality in Lemma \[lemma:finula\].  It follows from (\[eq:cf0\]) that for $h\in \iM_{s,0}$, $$\|h\|_{\vfi,0}\le \|h\|_{\vfi}\le \frac2{\log2}\|h\|_{\vfi,0}$$

Note that since $\vfi\in\iS_\vfi$, $\vfi$ extends to a bounded linear functional on $B_\vfi$. Clearly, the completion of $\iM_{s,0}$ under the norm $\|\cdot\|_\vfi$ is the Banach subspace $\{h\in
B_\vfi,\ \vfi(h)=0\}$. It follows from the above Proposition that $B_{\vfi,0}$ can be identified with the subspace of centered elements in $B_\vfi$.

Extension of $c_\vfi$.
======================

Since $\iS_\vfi\subset B^*_\vfi\sqsubseteq \iM^*_s$, the restriction of $F_\vfi$ is a strictly convex lower semicontinuous functional on $B_\vfi^*$, with effective domain $S_\vfi$. Its conjugate $F^*_\vfi$ is a lower semicontinuous extension of $c_\vfi$ to $B_\vfi$, moreover, $F^{**}_\vfi=F_\vfi$. We show that this extension has again values in $\mathbb R$ and is continuous. 

\[lemma:bounded\] Let the sequence $\{h_n\}_n\subset \iM_s$ be Cauchy in the norm $\|\cdot\|_\vfi$. Then the sequences $\{c_\vfi(h_n)\}_n$ and $\{S([\vfi^{h_n}],\vfi)\}_n$ are bounded. 

[*Proof.*]{}  By (\[eq:gt\]), we have for all $n$ $$\vfi(h_n)\le c_\vfi(h_n)$$ Since $\vfi(h_n)$ converges, $c_\vfi(h_n)$ is bounded from below.  Further, let $n_0$ be such that $\|h_n-h_{n_0}\|_\vfi<1$ for all $n\ge n_0$, then $$\omega(h_n)-S(\omega,\vfi)\le\omega(h_{n_0})+c_\vfi(h_n-h_{n_0})\le \|h_{n_0}\|+
\log 2$$ for all such $n$ and $\omega\in S_\vfi$. It follows that $\{c_\vfi(h_n)\}_n$ is bounded. 

If $\{h_n\}_n$ is Cauchy, then the sequence $\{th_n\}_n$ is also Cauchy for all $t\in \mathbb{R}$ and there are constants $A_t, B_t$, such that $$A_t\le c_\vfi(th_n)\le B_t,\qquad \forall n$$ On the other hand , we have $$\frac d{dt}c_\vfi(th_n)|_{t=1}=[\vfi^{h_n}](h_n)$$ By convexity, $$c_\vfi(th_n)\ge c_\vfi(h_n)+(t-1)\frac d{dt} c_\vfi(th_n)|_{t=1}\ge A_1+
(t-1)[\vfi^{h_n}](h_n)$$ For arbitrary fixed $t>1$, we get $$[\vfi^{h_n}](h_n)\le\frac{B_t-A_1}{t-1},\qquad \forall n$$ Boundedness of $S([\vfi^{h_n}],\vfi)$ now follows from $$0\le S([\vfi^{h_n}],\vfi)=[\vfi^{h_n}](h_n)-c_\vfi(h_n).$$

\[thm:perturb\] Let $\{h_n\}_n$ be a sequence in $\iM_s$, converging to some $h$ in $B_\vfi$. Then $$\label{eq:maximizer}
\lim_n c_\vfi(h_n)=\sup_{\omega\in \iS_\vfi}\{\omega(h)-S(\omega,\vfi)\}$$ and there is a unique state $\psi\in \iS_\vfi$ such that the supremum is attained.  The state $\psi$ is faithful.  Moreover, $\lim_n S([\vfi^{h_n}],\vfi)=S(\psi,\vfi)$, $\lim_n [\vfi^{h_n}(h_n)]=\psi(h)$ and $\lim_n S(\psi,[\vfi^{h_n}])=0$. In particular, $[\vfi^{h_n}]$ converges to $\psi$ in norm. 

The state $\psi$ will be denoted by $[\vfi^h]$ and the limit $\lim_nc_\vfi(h_n)=:c_\vfi(h)$.

[*Proof.*]{}  This proof is similar to the proof of Theorem 12.3.  in [@ohypetz]. 

By Lemma \[lemma:bounded\], there is some $C>0$ such that $[\vfi^{h_n}]\in\iS_{\vfi,C}$ for all $n$. The set $\iS_{\vfi,C}$ is weakly relatively compact in $\mathfrak{S}_*$ and hence there is subsequence $[\vfi^{h_{n_k}}]$ converging weakly to a state $\psi\in \iS_{\vfi,C}$. We will show that $[\vfi^{h_{n_k}}](h_{n_k})$ converges to $\psi(h)$.

Since $\iS_C$ is an equicontinuous subset in $B_\vfi^*$, $\omega(h_n)$ converges to $\omega(h)$ uniformly for all $\omega\in\iS_{\vfi,C}$. This implies $$|[\vfi^{h_{n_k}}](h_{n_k})-[\vfi^{h_{n_k}}](h)|<\varepsilon$$ for sufficiently large $n_k$. We further have $$\begin{aligned}
|[\vfi^{h_{n_k}}](h)-\psi(h)|&\le&|[\vfi^{h_{n_k}}](h)-[\vfi^{h_{n_k}}](h_m)|\\
&+&
|[\vfi^{h_{n_k}}](h_m)-\psi(h_m)|+|\psi(h_m)-\psi(h)|<\varepsilon\end{aligned}$$ for sufficiently large $m$ and $n_k$. Putting both inequalities together, we get $[\vfi^{h_{n_k}}](h_{n_k})\to \psi(h)$.

Let $\omega\in \iS_\vfi$. By definition, $$[\vfi^{h_{n_k}}](h_{n_k})-S([\vfi^{h_{n_k}}],\vfi)=c_\vfi(h_{n_k})\ge
\omega(h_{n_k})-S(\omega,\vfi)$$ By weak lower semicontinuity of the relative entropy, we get $$\label{eq:limsup}
\psi(h)-S(\psi,\vfi)\ge \limsup c_\vfi(h_{n_k})\ge\omega(h)-S(\omega,\vfi)$$ and thus $\psi$ is a maximizer of (\[eq:maximizer\]).  On the other hand, $$\psi(h_{n_k})-S(\psi,\vfi)\le [\vfi^{h_{n_k}}](h_{n_k})-
S([\vfi^{h_{n_k}}],\vfi)=c_\vfi(h_{n_k})$$ From this and (\[eq:limsup\]), it follows that $\psi(h)-S(\psi,\vfi)=\lim c_\vfi(h_{n_k})$. It also follows that $$\limsup S([\vfi^{h_{n_k}}],\vfi)\le S(\psi,\vfi)$$ and this, together with lower semicontinuity implies that $S([\vfi^{h_{n_k}}],\vfi)$ converges to $S(\psi,\vfi)$.

To show that such $\psi$ is unique, suppose that $\psi'$ is another maximizer, then for $\psi_\lambda:=\lambda\psi+(1-\lambda)\psi'$, $0\le\lambda\le 1$, we have $$\psi(h)-S(\psi,\vfi)\ge\psi_\lambda(h)-S(\psi_\lambda,\vfi)\ge
\psi_\lambda(h)-\lambda
S(\psi,\vfi)-(1-\lambda)S(\psi',\vfi)=\psi(h)-S(\psi,\vfi)$$ hence $\psi_\lambda$ is a maximizer as well and, moreover, $$S(\psi_\lambda,\vfi)=\lambda S(\psi,\vfi)+(1-\lambda)S(\psi',\vfi)$$ By strict convexity, his implies that $\psi=\psi'$. It also follows that the whole sequence $[\vfi^{h_n}]$ converges weakly to $\psi$.

Using (\[eq:sup2\]), we have $$S(\vfi,\psi)\le \liminf_n S(\vfi,[\vfi^{h_n}])=\lim_n c_\vfi(h_n)-\vfi(h)<\infty$$ This implies that $\supp\vfi\le\supp \psi$ and $\psi$ is faithful.  Finally, by taking the limit in the equality, $$\psi(h_n)-S(\psi,\vfi)=c_\vfi(h_n)-S(\psi,[\vfi^{h_n}])$$ we get $\lim_nS(\psi,[\vfi^{h_n}])\to 0$.

Let $h_n$ be a sequence in $B_\vfi$, then $h_n\to 0$ if and only if $c_\vfi(th_n)\to 0$ for all $t\in \mathbb{R}$.

[*Proof.*]{}  Let $h_n$ be such that $c_\vfi(th_n)=\log \vfi^{th_n}(1)$ converges to 0, then $\vfi^{th_n}(1)$ converges to 1, for all $t\in \mathbb{R}$. Therefore, for each $\varepsilon>0$, $\Phi_\vfi(\frac{h_n}{\varepsilon})<1$ for large enough $n$, that is, $\|h_n\|_\vfi\to 0$. The converse follows from Theorem \[thm:perturb\]. 

In particular, if $h_n\in \iM_s$ is a sequence converging strongly to $h$, then $h_n$ converges to $h$ in $\|\cdot\|_\vfi$, see [@ohypetz]. 

The dual spaces. {# sec:dual}
================

The dual space $\iM_{s,0}^*$ is obtained as the quotient space $\iM_s^*/\{\vfi\}$. Each equivalence class in $\iM_{s,0}^*$ can be identified with its unique element $v$ satisfying $v(1)=0$. By Proposition \[prop:interior\], we have $B_{\vfi,0}^*=B_{\Phi^*_{\vfi,0}} \sqsubseteq \iM_{s,0}^*$. By Proposition \[prop:finula\], $B_{\vfi,0}^*$ is the same as $B^*_\vfi/\{\vfi\}$.

\[lemma:conjugate\] Let $\bar c_\vfi$ be the restriction of $c_\vfi$ to $B_{\vfi,0}$. Then the conjugate functional is $\bar c^*_\vfi(v)=F_\vfi(v+\vfi)$.

[*Proof.*]{}  Let $v\in B^*_\vfi$, $v(1)=0$. Then by (\[eq:cadit\]), $$\begin{aligned}
F_\vfi(v+\vfi)&=&\sup_{h\in B_\vfi}\{v(h)+\vfi(h)-c_\vfi(h)\}=\\
        &=&\sup_{h\in B_\vfi}\{ v(h-\vfi(h))-\bar c_\vfi(h-\vfi(h))\}=\bar c^*_\vfi(v).\end{aligned}$$

Let $V$ be a Banach space and $V^*$ its dual.  For any subset $D\subset V$, let $D^\circ$ be the polar of $D$ in $V^*$, that is, $D^\circ=\{v\in V^*,\ v(h)\le 1,\ \forall h\in
D\}$. We will need the following lemma. 

\[lemma:polar\] Let $F:V\to \mathbb{R}^+$ be a convex functional such that $F(0)=0$ and let $F^*$ be its conjugate.  Let $D=\{x\in V,\ F(x)\le 1\}$ and $D^*=\{v\in V^*,\ F^*(v)\le 1\}$. Then $$\frac 12 D^*\subseteq D^\circ\subseteq D^*$$

[*Proof.*]{}  If $v\in D^*$, then $v(x)\le F(x)+F^*(v)\le 2$ for all $x\in D$ and therefore $\frac12v\in D^\circ$. Let now $v\in D^\circ$, then $$v(x)-1\le 0\le F(x)\quad \mbox{ for } x\in D$$ If $F(x)>1$, then by continuity there is some $t\in (0,1)$ such that $F(tx)=1$. Since $tx\in D$, $v(tx)\le 1$, moreover, by convexity, $1=F(tx)\le tF(x)$. Consequently, $$v(x)-1\le\frac1t-1\le F(x)$$ It follows that $F^*(v)\le 1$ and $v\in D^*$.

Let us denote $K_{\vfi,0}:=\{h\in B_{\vfi,0},\ \Phi_{\vfi,0}(h)\le 1\}$. Then $K_{\vfi,0}$ is the closed unit ball in $B_{\vfi,0}$. Its polar $K^\circ_{\vfi,0}$ is the closed unit ball in $B^*_{\vfi,0}$.

\[prop:unitball\] Let $v$ be an element in $K^\circ_{\vfi,0}$. Then there are states $\omega_1$, $\omega_2$, satisfying $S(\omega_1,\vfi)+S(\omega_2,\vfi)\le 1$, such that $v=\omega_1-\omega_2$.

[*Proof.*]{}  Since $\bar c_\vfi$ is continuous on $B_{\vfi,0}$, the set $D:=\{ h\in B_{\vfi,0},\ \bar c_\vfi(h)\le1\}$ is closed.  Let us endow the dual pair $B_{\vfi,0}$ and $B_{\vfi,0}^*$ with the $\sigma(B_{\vfi,0},B_{\vfi,0}^*)$ and $\sigma(B_{\vfi,0}^*,B_{\vfi,0})$ topology, respectively.  As $D$ is convex, it is closed also in this weaker topology.  The set $D\cap -D$ is absolutely convex and closed, moreover, $$\label{eq:D_D}
D\cap-D\subseteq K_{\vfi,0}  \subseteq 2(D\cap-D),$$ as can be easily checked.  Then $$\frac12 (D\cap-D)^\circ\subseteq K_{\vfi,0}^\circ\subseteq (D\cap-D)^\circ$$ By [@kothe], $(D\cap -D)^\circ$ is the closed convex cover of $D^\circ\cup -D^\circ$, which is the same as the closed absolutely convex cover of $D^\circ$. Moreover, since $D^\circ$ is the polar of a neighborhood of 0, it is compact ([@kothe]).  Therefore its absolutely convex cover is also compact, hence closed.  It follows that $(D\cap-D)^\circ$ is the absolutely convex cover of $D^\circ$.

By Lemma \[lemma:conjugate\] and \[lemma:polar\], $$\frac12(\iS_1-\vfi)\subseteq D^\circ\subseteq \iS_1-\vfi$$ and this implies $$\label{eq:unitball}
\frac14{\rm abs}\ {\rm conv}\, (\iS_1-\vfi)\subseteq K^\circ_{\vfi,0}
\subseteq {\rm abs}\ {\rm conv}\, (\iS_1-\vfi)$$

Let now $v\in{\rm abs}\ {\rm conv}\, (\iS_1-\vfi) $, then there are elements $\vfi_1\dots,\vfi_n\in \iS_1$, and real numbers $\lambda_1,\dots,\lambda_n$, $\sum_n|\lambda_n|=1$, such that $v=\sum_n\lambda_n(\vfi_n-\vfi)$. Let $m\le n$ be such that $\lambda_i>0$ for $i\le m$ and $\lambda_i<0$ for $i>m$. Then $v=\omega_1-\omega_2$, with $$\omega_1=\sum_{i=1}^m\lambda_i\vfi_i+(1-\lambda)\vfi,\quad
\omega_2=\sum_{i=m+1}^n|\lambda_i|\vfi_i+\lambda\vfi$$ where $\lambda=\sum_{i=1}^m\lambda_i$, moreover, $S(\omega_1,\vfi)\le\sum_i^m\lambda_iS(\vfi_i,\vfi)\le \lambda$, and similarly, $S(\omega_2,\vfi)\le 1- \lambda$.

\[thm:duals\]

1.   $B_\vfi^*=\iP_\vfi-\iP_\vfi$ and $B_\vfi^*\cap \iM_*^+=\iP_\vfi$.

2.   $B_{\vfi,0}^*=\cup_n n(\iS_1-\iS_1)$.

[*Proof.*]{} ( i) Let $\omega\in B_\vfi^*$ and let $v=\omega-\omega(1)\vfi$. Then $v$ can be seen as an element in $B^*_{\vfi,0}$. Let $\|v\|_{\vfi,0}^*=t$, then by Proposition \[prop:unitball\], there are $\omega_1,\omega_2\in \iS_1$, such that $\frac vt=\omega_1-\omega_2$, that is, $\omega=t\omega_1+\omega(1)\vfi-t\omega_2$. Since $\omega_1,\omega_2,\vfi\in \iP_\vfi$ and $\iP_\vfi$ is a convex cone, it follows that $B_\vfi^*\subseteq
\iP_\vfi-\iP_\vfi$. On the other hand, we have already shown that if $\omega\in \iS_\vfi$, then $\omega\in B_\vfi^*$ and hence $\iP_\vfi-\iP_\vfi\subseteq B_\vfi^*$. Let $\omega\in B^*_\vfi\cap\iM_*^+$, then we get $\omega+t\omega_2=t\omega_1+\omega(1)\vfi$. It follows that $\omega+t\omega_2\in \iP_\vfi$, and Donald’s identity implies that $\omega$ must be in $\iP_\vfi$.

\(ii) By Proposition \[prop:unitball\], $$K_{\vfi,0}^\circ\subseteq(\iS_1-\iS_1)\subseteq 4K_{\vfi,0}^\circ.$$ The equality now follows from the fact that the closed unit ball is absorbing in $B_{\vfi,0}^*$.

In the rest of this section, we find an equivalent norm on $B_{\vfi,0}^*$.

We define a function $f:\ \mathfrak{S}_*\times\mathfrak{S}_*\to
\mathbb{R}^+$ by $$f(\omega_1,\omega_2)=S(\omega_1,\vfi)+S(\omega_2,\vfi).$$ Clearly, $f$ is weakly lower semicontinuous and strictly convex.  Further, let $v\in \mathfrak{S}_*-\mathfrak{S}_*$ and let $L_v=\{(\omega_1,\omega_2)\in
\mathfrak{S}_*\times\mathfrak{S}_*,\ \omega_1-\omega_2=v\}$. Then $L_v$ is a weakly closed subset in $\iM_*\times\iM_*$.

Let $v\in \iS_\vfi-\iS_\vfi$. Then the function $f$ attains its minimum over $L_v$ at a unique point $(v_+,v_-)\in L_v$.

[*Proof.*]{}  By assumptions, $v=\omega_1-\omega_2$ for some $\omega_1,\omega_2\in \iS_\vfi$. Let $C>0$ be such that $\omega_1,\omega_2\in
\iS_C$, then the infimum is taken over the set $L_v\cap \iS_C\times\iS_C$. Since $L_v$ is weakly closed and $\iS_C$ is weakly compact, the intersection is weakly compact and $f$ attains its minimum on it.  Uniqueness follows by strict convexity of $f$.

Let us now define the functional $\Psi_{\vfi,0}:\ \iM_{s,0}^*\to
\mathbb{R}^+$ by $$\Psi_{\vfi,0}(v)=\left \{ \begin{array}{cc} f(v_+,v_-)& \mbox{if }\ v\in
\iS_\vfi-\iS_\vfi\\
\infty &\mbox{otherwise} \end{array} \right.$$

$\Psi_{\vfi,0}$ is a Young function. 

[*Proof.*]{}  It is easy to check that $\Psi_{\vfi,0}$ is convex, positive, $\Psi_{\vfi,0}(v)= \Psi_{\vfi,0}(-v)$ and that $\Psi_{\vfi,0}(v)=0$ if and only if $v=0$. We will show that $\Psi_{\vfi,0}$ is lower semicontinuous. 

To do this, we have to prove that for any $C>0$, the set of all $v$ satisfying $\Psi_{\vfi,0}(v)\le C$ is closed.  Let $v_n$ be a sequence of elements in this set, converging to some $v$. Let $v_n=v_{n+}-v_{n-}$ be the corresponding decompositions, then $v_{n+},v_{n-}\in \iS_C$ for all $n$, hence there are elements $v'_+$ and $v'_-$ in $\iS_C$ and a subsequence $v_{n_k}=v_{n_k+}-v_{n_k-}$ such that $v_{n_k+}\to v'_+$ and $v_{n_k-}\to v'_-$ weakly.  It follows that $v=v'_+-v_-'$ and $\Psi_{\vfi,0}(v)\le
S(v'_+,\vfi)+S(v'_-,\vfi)\le \liminf S(v_{n_k+},\vfi)+S(v_{n_k-},\vfi)\le C$. 

Suppose that $v\ne 0$, then $\Psi_{\vfi,0}(v)>0$. If $t>1$, then by convexity, $t\Psi_{\vfi,0}(v)\le \Psi_{\vfi,0}(tv)$, hence $\lim_{t\to\infty} \Psi_{\vfi,0}(tv)=\infty$.

Let us find the corresponding Banach space.  Note that $$C_{\Psi_{\vfi,0}}=\{ \omega_1-\omega_2\ :\ \omega_1,\omega_2\in \mathfrak S_*,
S(\omega_1,\vfi)+S(\omega_2,\vfi)\le1\}.$$ By Proposition \[prop:unitball\], this implies that $K_{\vfi,0}^\circ\subseteq  C_{\Psi_{\vfi,0}}\subseteq \iS_1-\iS_1$ and by Theorem \[thm:duals\] (ii), $B_{\vfi,0}^*\subseteq L_{\Psi_{\vfi,0}}\subseteq B_{\vfi,0}^*$.

$\|\cdot\|_{\Psi_{\vfi,0}}$ defines an equivalent norm in $B^*_{\vfi,0}$.

[*Proof.*]{}  Let $\Psi_{\vfi,0}^*:\iM_s\to\mathbb{R}$ be the conjugate functional, then $$\begin{aligned}
\Psi_{\vfi,0}^*(h)&=&\sup_{v\in \iM_{s,0}^*}v(h)-\Psi_{\vfi,0}(v)=\\
&=&\sup_{v\in \iS_\vfi-\iS_\vfi}\sup_{(\omega_1,\omega_2)\in L_v}
\omega_1(h)-\omega_2(h)-f(\omega_1,\omega_2)=\\ &=&\sup_{\omega_1,\omega_2\in
\iS_\vfi}\omega_1(h)-S(\omega_1,\vfi)+\omega_2(-h)-S(\omega_2,\vfi)=
2\Phi_{\vfi,0}(h)\end{aligned}$$ It follows that $\Psi_{\vfi,0}(v)=\Psi_{\vfi,0}^{**}(v)=2\Phi_{\vfi,0}^*(\frac
12 v)$. Since the norms $\|\cdot\|_{\vfi,0}^*$ and $\|\cdot\|_{\Phi^*_{\vfi,0}}$ are equivalent, this finishes the proof. 

The chain rule. 
===============

\[prop:chain\] Let $h \in B_\vfi$, $k\in \iM_s$. Then $[\vfi^{h+k}]=[[\vfi^h]^k]$, $c_\vfi(h+k)=c_{[\vfi^h]}(k)+c_\vfi(h)$ and for all normal states $\omega$ the equality $$\label{eq:chain}
\omega(k)-S(\omega,[\vfi^h])=c_\vfi(h+k)-c_\vfi(h)-S(\omega,[\vfi^{h+k}])$$ holds. 

[*Proof.*]{}  Let $h_n\in \iM_s$ be such that $h_n\to h$ in $B_\vfi$. By the chain rule (\[eq:chainM\]), we have $[\vfi^{h_n+k}]=[[\vfi^{h_n}]^k]$ and $c_\vfi(h_n+k)=c_{[\vfi^{h_n}]}(k)+c_\vfi(h_n)$. By Theorem \[thm:perturb\], $c_\vfi(h_n)\to c_\vfi(h)$, $c_\vfi(h_n+k)\to c_\vfi(h+k)$ and $[\vfi^{h_n}]\to[\vfi^h]$, $[\vfi^{h_n+k}]\to[\vfi^{h+k}]$ strongly.  Now we can proceed exactly as in the proof of Theorem 12.10 in [@ohypetz] to obtain (\[eq:chain\]).  By putting $\omega=[\vfi^{h+k}]$ in this equality, we get $$[\vfi^{h+k}](k)+S([\vfi^{h+k}],[\vfi^h])=c_\vfi(h+k)-c_\vfi(h)\ge
\omega(k)-S(\omega,[\vfi^h]),$$ for all $\omega$, which implies the statement of the proposition. 

\[thm:eqiuv\] Let $h\in B_\vfi$. Then $B_{[\vfi^h]}=B_\vfi$ and $\iS_{[\vfi^h]}=\iS_\vfi$.

[*Proof.*]{}  Let $k\in \iM_s$ and let $\varepsilon >0$. By Proposition \[prop:chain\], $$c_{[\vfi^h]}(k)=c_\vfi(h+k)-c_\vfi(h).$$ Since $c_\vfi$ is continuous on $B_\vfi$, there is a $\delta>0$ such that $$|c_\vfi(h+k)-c_\vfi(h)|<\log 2$$ if $\|k\|_\vfi<\delta$. It follows that $\|k\|_{[\vfi^h]}<\varepsilon$ whenever $\| k\|_\vfi<\delta\varepsilon$ and this implies $B_\vfi\sqsubseteq B_{[\vfi^h]}$. In particular, $h\in B_{[\vfi^h]}$.

Let $h_n$ be a sequence converging to $h$ in $B_\vfi$, then by (\[eq:sup2\]) $$\omega(h_n)-S(\omega,\vfi)=c_\vfi(h_n)-S(\omega,[\vfi^{h_n}])$$ By Theorem \[thm:perturb\], and lower semicontinuity, $$\omega(h)-S(\omega,\vfi)\le c_\vfi(h)-S(\omega,[\vfi^h])$$ This implies $\iS_\vfi\subseteq \iS_{[\vfi^h]}$.

Further, $h_n$ converges to $h$ in $B_{[\vfi^h]}$ and by Theorem \[thm:perturb\] and Proposition \[prop:chain\], $$[[\vfi^h]^{-h}]=\lim_n [[\vfi^h]^{-h_n}]=
\lim_n[\vfi^{h-h_n}]=\vfi.$$ By the first part of the proof, $B_{[\vfi^h]}=B_\vfi$ and $\iS_\vfi=
\iS_{[\vfi^h]}$.

\[thm:chain\] Let $h,k\in B_\vfi$. Then the chain rule $c_\vfi(h+k)=c_{[\vfi^h]}(k)+c_\vfi(h)$, $[[\vfi^h]^k]=[\vfi^{h+k}]$ holds. 

[*Proof.*]{}  Let $k_n\in \iM_s$ be a sequence converging to $k$ in $B_\vfi=B_{[\vfi^h]}$. Then $$[[\vfi^h]^{k}]=\lim_n [[\vfi^h]^{k_n}]=
\lim_n[\vfi^{h+k_n}]=[\vfi^{h+k}].$$ and by Proposition \[prop:chain\], $$c_\vfi(h+k)=\lim_n c_{[\vfi^h]}(k_n)+c_\vfi(h)=c_{[\vfi^h]}(k)+
c_\vfi(h)$$

\[coro:equality\] Let $h\in B_\vfi$ and let $\omega$ be a normal state.  Then the equality $$\omega(h)-S(\omega,\vfi)=c_\vfi(h)-S(\omega,[\vfi^h])$$ holds. 

[*Proof.*]{}  By (\[eq:sup2\]) and lower semicontinuity, we have $$\omega(h)-S(\omega,\vfi)\le c_\vfi(h)-S(\omega,[\vfi^h])$$ Since, by the chain rule, $\vfi=[[\vfi^h]^{-h}]$ and $c_{[\vfi^h]}(-h)=-c_\vfi(h)$, we also have $$\omega(-h)-S(\omega,[\vfi^h])\le
c_{[\vfi^h]}(-h)-S(\omega,\vfi)=-c_\vfi(h)-S(\omega,\vfi)$$ which implies the opposite inequality. 

\[coro:unique\] Let $[\vfi^h]=[\vfi^k]$ for some $h,k\in B_\vfi$. Then $h-k=\vfi(h-k)$.

[*Proof.*]{}  Let us suppose that $h\in B_\vfi$ is such that $[\vfi^h]=\vfi$. Then $[\vfi^{nh}]=\vfi$ for all $n\in {\mathbb{N}}$. It follows that $c_\vfi(nh)=n\vfi(h)=nc_\vfi(h)$ for all $n$ and for $0\le t\le 1$, we have by (\[eq:gt\]) and convexity of $c_\vfi$ that $$tc_\vfi(h)=\vfi(th)\le c_\vfi(th)\le tc_\vfi(h)$$ It follows that $c_\vfi(th)=tc_\vfi(h)=t\vfi(h)$ for all $t\ge 0$. Since also $[\vfi^{-h}]=[[\vfi^h]^{-h}]=\vfi$, we have $c_\vfi(-th)=tc_\vfi(-h)=-t\vfi(h)$ for $t\ge 0$.

It is easy to see that $c_\vfi(k-\lambda)=c_\vfi(k)-\lambda$ for all $k\in B_\vfi$ and $\lambda\in\mathbb{R}$. Let $\lambda=\vfi(h)$, then it follows that $$c_\vfi(t(h-\lambda))=0=c_\vfi(t(-h+\lambda))$$ for all $t\ge 0$. This implies $\|h-\lambda\|_\vfi=0$ and hence $h=\lambda$.

Let now $[\vfi^h]=[\vfi^k]$, then $[[\vfi^k]^{-h}]=[\vfi^{k-h}]=
\vfi$ and $h-k=\lambda=\vfi(h-k)$.

Note that the function $\bar c_\vfi: B_{\vfi,0}\to\mathbb R$ corresponds to the cumulant generating functional in the commutative case.  Let us list some of its properties. 

\[thm:cumulant\] The function $\bar c_\vfi$ has the following properties. 

1.   $\bar c_\vfi$ is positive, strictly convex and continuous, $\bar c_\vfi(0)=0$.

2.   $\bar c_\vfi$ is Gateaux differentiable, with $\bar c'_\vfi(h)=[\vfi^h]-\vfi$

3.   The map $$B_{\vfi,0}\ni h\mapsto [\vfi^h]-\vfi\in B^*_{\vfi,0}$$ is one-to-one and norm to $\sigma(B^*_{\vfi,0},B_{\vfi,0})$-continuous. 

[*Proof.*]{} ( i) By Corollary \[coro:equality\], $\bar c_\vfi(h)=S(\vfi,[\vfi^h])\ge 0$ and $\bar c_\vfi(0)=0$ by definition.  Let now $h,k\in B_{\vfi,0}$ and $0<\lambda< 1$ be such that $$\bar c_\vfi(\lambda h+(1-\lambda)k)=\lambda \bar c_\vfi(h)+(1-\lambda)\bar c_\vfi(k).$$ Then $$\begin{aligned}
&\sup_{\iS_\vfi}& \lambda(\omega(h)-S(\omega,\vfi))+(1-\lambda)(\omega(k)-S(\omega,\vfi))
= \\
&=& \lambda \sup_{\iS_\vfi} (\omega(h)-S(\omega,\vfi))
+(1-\lambda)\sup_{\iS_\vfi}(\omega(k)-S(\omega,\vfi))\end{aligned}$$ This implies that the maximum in both expressions on the right hand side is attained at the same point.  Therefore $[\vfi^h]=[\vfi^k]$, hence $h-k=\vfi(h-k)=0$.

\(ii) By Theorem \[thm:perturb\], $[\vfi^h]-\vfi$ is the unique element in $B^*_{\vfi,0}$, such that $$([\vfi^h]-\vfi)(h)=\bar c_\vfi(h)+\bar c^*_\vfi([\vfi^h]-\vfi).$$ By [@ektem], this implies that $\bar c_\vfi$ is Gateaux differentiable in $h$ with derivative $\bar c'_\vfi(h)=[\vfi^h]-\vfi$.

\(iii) Let $h_n\to h$ in $B_\vfi$, then $[\vfi^{h_n}]$ converges strongly to $[\vfi^h]$ and $S([\vfi^{h_n}],\vfi)\to
 S([\vfi^h],\vfi)$. It follows that $[\vfi^{h_n}](k)\to [\vfi^h](k)$ for each $k\in \iM_s$ and moreover, the set $\{[\vfi^{h_n}], n\in \mathbb N\}$ is equicontinuous in $B_\vfi^*$. This implies that $[\vfi^{h_n}](k)\to [\vfi^h](k)$ for all $k\in B_\vfi$. The map is one-to-one by Corollary \[coro:unique\]. 

A manifold structure on faithful states. {# sec:mani}
========================================

Recall that a $C^p$-atlas on a set $X$ is a family of pairs $\{(U_i,e_i)\}$, such that

1.   $U_i\subset X$ for all $i$ and $\cup U_i=X$.

2.   For all $i$, $e_i$ is a bijection of $U_i$ onto an open subset $e_i(U_i)$ in some Banach space $B_i$, and for $i,j$, $e_i(U_i\cap U_j)$ is open in $B_i$.

3.   The map $e_je_i^{-1}:\ e_i(U_i\cap U_j)\to e_j(U_i\cap U_j)$ is a $C^p$- isomorphism for all $i,j$.

Let $\mathcal F_*$ be the set of faithful normal states on $\iM$. For $\vfi\in \mathcal F_*$, let $V_\vfi$ be the open unit ball in $B_{\vfi,0}$ and let $s_\vfi: V_\vfi\to \mathcal F_*$ be the map $h\mapsto [\vfi^h]$. By Corollary \[coro:unique\], $s_\vfi$ is a bijection onto the set $s_\vfi(V_\vfi)=:U_\vfi\subset \mathcal S_\vfi$. Let $e_\vfi$ be the restriction of $s^{-1}_\vfi$ to $U_\vfi$. Then we have

$\{(U_\vfi, e_\vfi), \vfi\in \mathcal F_*\}$ is a $C^{\infty}$-atlas on $\mathcal F_*$.

[*Proof.*]{}  The property (i) and the first part of (ii) of the definition of the $C^p$ atlas are obviously satisfied.  Let $\vfi_1,\vfi_2\in \mathcal F_*$ be such that $U_{\vfi_1}\cap U_{\vfi_2}\ne \emptyset$. We prove that $e_{\vfi_1}(U_{\vfi_1}\cap U_{\vfi_2})$ is open in $B_{\vfi_1,0}$.

Let $h_1\in e_{\vfi_1}(U_{\vfi_1}\cap
U_{\vfi_2})$. Then there is some $h_2\in B_{\vfi_2,0}$, such that $[\vfi_1^{h_1}]=[\vfi_2^{h_2}]$. By Theorem \[thm:eqiuv\], $B_{\vfi_1}=B_{[\vfi_1^{h_1}]}=B_{[\vfi_2^{h_2}]}=
B_{\vfi_2}$ and by the chain rule, $\vfi_1=[\vfi_2^k]$, where $k=h_2-h_1+\vfi_2(h_1)\in B_{\vfi_2,0}$. Clearly, the map $B_{\vfi_1,0}\to B_{\vfi_2,0}$, given by $h\mapsto
h-\vfi_2(h)$ is continuous. 

Let $\varepsilon>0$ be such that $h_2+h_2'\in V_{\vfi_2}$ whenever $\|h'_2\|_{\vfi_2}<\varepsilon$ and let us choose $\delta>0$ such that $h_1+h_1'\in V_{\vfi_1}$ and $\|h'_1-\vfi_2(h_1')\|_{\vfi_2}< \varepsilon$ for $\|h_1'\|_{\vfi_1}<\delta$. For such $h_1'$, we have $$s_{\vfi_1}(h_1+h_1')=[\vfi_1^{h_1+h_1'}]=[\vfi_2^{k+h_1+h_1'-\vfi_2(h_1')}]=
[\vfi_2^{h_2+h_1'-\vfi_2(h_1')}]\in U_{\vfi_1}\cap U_{\vfi_2}$$ This proves that $s_{\vfi_1}^{-1}(U_{\vfi_1}\cap U_{\vfi_2})$ is open in $B_{\vfi_1,0}$. It is also clear that the map $$\begin{aligned}
s^{-1}_{\vfi_2}s_{\vfi_1}: s^{-1}_{\vfi_1}(U_{\vfi_1}\cap U_{\vfi_2})&\to&
s^{-1}_{\vfi_2}(U_{\vfi_1}\cap U_{\vfi_2})\\
h &\mapsto& k+h-\vfi_2(h)\end{aligned}$$ is $C^\infty$, which proves (iii). 

It is not difficult to see that for $\vfi\in \mathcal F_*$, the set $\mathcal F_\vfi:=\{[\vfi^h],\ h\in B_{\vfi,0}\}$ is a connected component of the manifold.  Let us now define a family of mappings $$U^{(e)}_{\vfi_1,\vfi_2}: B_{\vfi_1,0}\ni h\mapsto h-\vfi_2(h)\in B_{\vfi_2,0},\quad \vfi_1,\vfi_2\in
\mathcal F_\vfi$$ It is clear that this defines a parallel transport on the tangent bundle of $\mathcal F_\phi$ and the associated globally flat affine connection is the exponential connection, [@gipi]. 

Let us recall that the dual connection is defined on the cotangent bundle $T^*\mathcal F_\vfi$ by means of the parallel transport $\{ (U^{(e)}_{\vfi_2,\vfi_1})^*,\ \vfi_1,\vfi_2\in \mathcal F_\vfi\}$, where $$\langle (U^{(e)}_{\vfi_2,\vfi_1})^*v,h\rangle = \langle v,
U^{(e)}_{\vfi_1,\vfi_2}h\rangle, \quad v\in B_{\vfi_2,0}^*, h\in B_{\vfi_1,0},$$ and the duality is given by $\langle v, h\rangle = v(h)$. Since $v(h-\vfi_1(h))=v(h)$ for all $\vfi_1$, the dual parallel transport is $$U^{(m)}_{\vfi_1,\vfi_2}: B_{\vfi_1,0}^*\ni v\mapsto v\in B_{\vfi_2,0}^*,\quad \vfi_1,\vfi_2\in
\mathcal F_\vfi$$ which corresponds to the mixture connection. 

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