# Projective positivity of the function systems

Anar Dosi  Middle East Technical University NCC, Guzelyurt, KKTC, Mersin 10,
Turkey<EMAIL_ADDRESS><EMAIL_ADDRESS>http://www.mathnet.ru/php/person.phtml?option_lang=eng&personid=23380

(Date: March 3, 2023)

###### Abstract.

The present paper is devoted to the projective positivity in the category of
function systems, which plays a key role in the quantization problems of the
operator systems. The main result of the paper asserts that every unital
$\ast$-normed space can be equipped with the projective positivity. The
geometry of the related state spaces is described in the case of
$L^{p}$-spaces, Schatten matrix spaces, and $L^{p}$-spaces of a finite von
Neumann algebra.

###### Key words and phrases:

Function system, operator system, unital cone, state space, projective
positivity

###### 2000 Mathematics Subject Classification:

Primary 46L07; Secondary 46B40, 47L25

##  1\. Introduction

A concrete function system is a unital self-adjoint subspace of a unital
commutative $C^{\ast}$-algebra whereas an abstract function system is defined
as an ordered $\ast$-vector space $\left(V,e\right)$ with an Archimedian order
unit $e$. A function system turns out to be a unital $\ast$-normed space
equipped with an order (semi)norm, and all order norms in $V$ are equivalent
by running within min and max order norms [17, Theorems 4.5, 4.7]. In the
theory of Banach lattices, the function systems are referred as the ordered
vector spaces with their strong order units [14, page 18]. Every abstract
function system can be realized up to an isometry as a concrete one on its
state space equipped with the $w^{\ast}$-topology. That is the complex version
of Kadison’s representation theorem for the real function systems [17, Theorem
5.2]. Recall that the state space $S\left(V_{+}\right)$ of an ordered
$\ast$-vector space $V$ with an Archimedian order unit $e$ and a unital cone
$V_{+}$ of its positive elements is defined as the intersection
$V_{he}^{\ast}\cap\operatorname{ball}V^{\ast}$ of the real hyperplane
$V_{he}^{\ast}=\left\\{\varphi\in V^{\ast}:\varphi=\varphi^{\ast},\left\langle
e,\varphi\right\rangle=1\right\\}$ in the norm dual space $V^{\ast}$ given by
the functional $\left\langle
e,\cdot\right\rangle:V^{\ast}\rightarrow\mathbb{C}$, and the unit ball of
$V^{\ast}$. It turns out that the cone $V_{+}$ is precisely defined by means
of its state space $S\left(V_{+}\right)$, namely $v\geq 0$ iff $\left\langle
v,S\left(V_{+}\right)\right\rangle\geq 0$ [17, Proposition 2.20]. Kadison’s
representation is given by the unital mapping

$\Phi:V\rightarrow
C\left(S\left(V_{+}\right)\right),\quad\Phi\left(v\right)\left(\varphi\right)=\left\langle
v,\varphi\right\rangle,$

where $C\left(S\left(V_{+}\right)\right)$ is the $C^{\ast}$-algebra of all
complex continuous functions on the Hausdorff compact topological space
$S\left(V_{+}\right)$. A detailed investigation of the function systems (by
introducing the concept of a function system itself) was carried out in [17]
by Paulsen and Tomforde.

In quantum functional analysis [8], [9] the function systems are treated as
the commutative operator (or quantum) systems, whereas the operator systems
equipped with various unital quantum cones are quantizations of the underlying
function system. An operator system is defined as a unital self-adjoint
subspace of the operator algebra $\mathcal{B}\left(H\right)$ of all bounded
linear operators on a complex Hilbert space $H$, whereas an operator space is
a subspace of $\mathcal{B}\left(H\right)$. The quantization principles of a
unital cone in a unital $\ast$-vector space were developed in [18] by Paulsen,
Todorov and Tomforde. These constructions are the categorical background of
the abstract operator systems followed by the key result of Choi and Effros
[2] on an abstract characterization of the operator systems (see also [16, Ch.
13]). A classification problem of the operator systems among the operator
space structures on a unital $\ast$-vector space was investigated in [6]. It
turns out that every operator system construction can be pushed through an
operator space one that results in a suitable unital quantum cone to be
related to. That unital quantum cone in turn generates a new function system
structure on the underlying unital $\ast$-vector space. But practically it is
not clear how to detect the unital cone(s) related to a certain operator space
construction. For example, the projective tensor product or even $1$-sum of
the operator systems does not exist in its proper sense (see [12]). The
following simple example could illustrate the case. There is a well defined
$1$-sum $V=\mathbb{C}\overset{1}{\oplus}\mathbb{C}$ of the trivial operator
spaces $\mathbb{C}$ with its operator dual
$V^{\ast}=\mathbb{C}\overset{\infty}{\oplus}\mathbb{C}$. Note that $V$
possesses the natural involution $z=\left(z_{0},z_{1}\right)\mapsto
z^{\ast}=\left(z_{0}^{\ast},z_{1}^{\ast}\right)$ such that
$\left\|z^{\ast}\right\|_{1}=\left\|z\right\|_{1}$, and the unit
$e=\left(2^{-1},2^{-1}\right)$. In this case,
$V_{he}^{\ast}=\left\\{w\in\mathbb{R}\overset{\infty}{\oplus}\mathbb{R}:w_{0}+w_{1}=2\right\\}$
and
$V_{he}^{\ast}\cap\operatorname{ball}V^{\ast}=\left\\{\left(1,1\right)\right\\}$,
that is, the related state space has been degenerated and it could only define
a non-separated cone in $V$. Thus how to define the operator system $1$-sum
$\mathbb{C}\overset{1}{\oplus}\mathbb{C}$ in this trivial case remains
unclear?

The construction of the projective objects within the category of the operator
systems causes a serious challenge to tackle with. An operator system quotient
does not always preserve the related operator space quotient (see discussion
in [13, Section 4]). Actually, the problem remains open on the level of the
function systems. In the previous example $\mathbb{C}$ is a function system.
What are the projective function systems that is the main target of the
present paper.

We provide a complete solution to the problem by developing the general
concept of the projective positivity. Recall that a norm on a $\ast$-vector
space $V$ is said to be a $\ast$-norm if the unit $\operatorname{ball}V$ is
hermitian, that is,
$\left(\operatorname{ball}V\right)^{\ast}=\operatorname{ball}V$. In this case,
$\left(V,V^{\ast}\right)$ is a dual $\ast$-pair [4], and $V_{h}^{\ast}$
denotes the real vector space of all bounded hermitian functionals on $V$. Fix
a nonzero $e\in V_{h}$ and as above put $V_{he}^{\ast}=\left\\{y\in
V_{h}^{\ast}:\left\langle e,y\right\rangle=1\right\\}$. Then
$S=V_{he}^{\ast}\cap\operatorname{ball}V^{\ast}$ is a convex subset of
$V_{h}^{\ast}$. We say that $e$ is a unit element (or just unit) for $V$ if
$S\neq\varnothing$. Every hermitian unit vector $e$ of a $\ast$-normed space
$V$ is a unit automatically. A $\ast$-normed vector space $V$ with a fixed
unit $e$ is called a unital $\ast$-normed space $\left(V,e\right)$. The set
$S$ can be poor enough to define an order in $V$ as we have seen above. That
is the case in many other basic examples of analysis (see below Sections 5,
6).

If $\operatorname{ball}V$ is equivalent to the absolute polar $S^{\circ}$ of
$S$, we say that $\operatorname{ball}V$ is a unital ball, and
$\left(V,e\right)$ is again called an abstract function system with its state
space $S$. In this case, $\mathfrak{c=}\left\\{v\in V_{h}:\left\langle
v,S\right\rangle\geq 0\right\\}$ is a separated, closed, unital cone in $V$
such that $S\left(\mathfrak{c}\right)=S$ and
$\left\|\cdot\right\|\sim\left\|\cdot\right\|_{e}$, where
$\left\|v\right\|_{e}=\sup\left|\left\langle v,S\right\rangle\right|$, $v\in
V$ (the min-norm). Thus our function system is a bit more general concept than
one considered above. Using again Kadison’s representation, one can identify
$V$ with a unital self-adjoint subspace of
$C\left(S\left(\mathfrak{c}\right)\right)$ up to an order and topological
isomorphism, that is, $\left(V,\mathfrak{c}\right)$ is a concrete function
system on $S\left(\mathfrak{c}\right)$. If $\operatorname{ball}V=S^{\circ}$
then $\left\|\cdot\right\|=\left\|\cdot\right\|_{e}$ and we come up with the
isometric embedding $\Phi:V\rightarrow
C\left(S\left(\mathfrak{c}\right)\right)$.

In the case of a non-unital $\operatorname{ball}V$ we consider the following
convex sets
$S_{\varepsilon}=V_{he}^{\ast}\cap\varepsilon\operatorname{ball}V^{\ast}$ for
$\varepsilon>0$. Thus $\operatorname{ball}V$ is unital whenever
$\operatorname{ball}V\sim S_{1}^{\circ}$. But if $\operatorname{ball}V$ is not
unital then we prove that

$\operatorname{ball}V\sim S_{\varepsilon}^{\circ}\text{\quad for every
}\varepsilon>1/\left\|e\right\|$

(see Theorem 4.1). The main result of the paper asserts that

$\mathfrak{c}_{\varepsilon}=\left\\{v\in V_{h}:\left\langle
v,S_{\varepsilon}\right\rangle\geq 0\right\\}$

is a closed, unital, cone in $V$ whose state space
$S\left(\mathfrak{c}_{\varepsilon}\right)=S_{\varepsilon}$, and
$\left(V,\mathfrak{c}_{\varepsilon}\right)$ is a concrete function system on
$S_{\varepsilon}$ whose $\varepsilon$-norm
$\left\|v\right\|_{\varepsilon}=\vee\left|\left\langle
v,S_{\varepsilon}\right\rangle\right|$, $v\in V$ is equivalent to the original
one of $V$ for every $\varepsilon>1/\left\|e\right\|$. Thus without changing
the original topology of a unital $\ast$-normed space $\left(V,e\right)$ one
can endow it with a compatible function system structure called a projective
function system.

As a practical application of the main result we consider a various examples
including $L^{p}$-spaces, the matrix Schatten-spaces and noncommutative
$L^{p}$-spaces of a finite von Neumann algebra. In Section 5 we discuss the
case of classical $L_{\mu}^{p}\left(\mathcal{T}\right)$, $1\leq p\leq\infty$
spaces. The unit vectors $e_{p}=\mu\left(\mathcal{T}\right)^{-1/p}1$ are
indeed units of the related Banach $\ast$-spaces
$L_{\mu}^{p}\left(\mathcal{T}\right)$. But their balls
$\operatorname{ball}L_{\mu}^{p}\left(\mathcal{T}\right)$ fail to be unital for
all $1\leq p<\infty$. For $p=\infty$ we come up with the abelian von Neumann
algebra $L_{\mu}^{\infty}\left(\mathcal{T}\right)$ with the unit
$e_{\infty}=1$, and
$\left(L_{\mu}^{\infty}\left(\mathcal{T}\right),e_{\infty}\right)$ is an
operator system, which in turn is a function system too. For a conjugate
couple $\left(p,q\right)$ with $p<\infty$ and $\varepsilon\geq 1$ we define
the following convex subsets

$S_{\varepsilon}^{p}=\left\\{f\in
L^{q}\left(\mathcal{T}\right)_{h}:\left\|f\right\|_{q}\leq\varepsilon,\int
fd\mu=\mu\left(\mathcal{T}\right)^{1/p}\right\\}\subseteq\partial\left(\varepsilon\operatorname{ball}L^{q}\left(\mathcal{T}\right)\right).$

The state space $S_{1}^{p}$ in
$\operatorname{ball}L_{\mu}^{q}\left(\mathcal{T}\right)$ is again degenerated
to a singleton $\left\\{e_{q}\right\\}$. Nonetheless the unital $\ast$-normed
space $L_{\mu}^{p}\left(\mathcal{T}\right)$ has the projective positivity
given by the separated, closed, unital cone
$L_{\mu}^{p}\left(\mathcal{T}\right)_{\varepsilon}^{+}$ whose state space is
$S_{\varepsilon}^{p}$ for every $\varepsilon>1$. It turns out that the
original cone $L^{p}\left(\mathcal{T}\right)_{+}$ of the Banach lattice
$L^{p}\left(\mathcal{T}\right)_{h}$ and the unital cones
$L^{p}\left(\mathcal{T}\right)_{\varepsilon}^{+}$ are totally different cones,
they are not comparable unless $\operatorname{supp}\left(\mu\right)$ is finite
(see Theorem 5.1). Moreover, in the finite dimensional nontrivial case we have
$\ell^{p}\left(n\right)_{+}=\ell^{p}\left(n\right)_{\varepsilon}^{+}$ only if
$n=2$ and $\varepsilon=2^{1/p}$.

In Section 6 we first consider the full matrix algebra $\mathbb{M}_{n}$ with
the unit matrix $e$ for $n>1$. The algebra $\mathbb{M}_{n}$ equipped with the
Schatten $p$-norm is denoted by $\mathfrak{S}_{p}$, $1\leq p\leq\infty$. In
this case, $\mathfrak{S}_{\infty}^{\ast}=\mathfrak{S}_{1}$, and
$\mathfrak{S}_{\infty}^{+}$ is the standard unital cone of the positive
elements with its state space $S_{\tau}=\left\\{y\in\mathfrak{S}_{1}:y\geq
0,\tau\left(y\right)=1\right\\}$ of the density matrices from
$\mathbb{M}_{n}$, that is, $\left(\mathfrak{S}_{\infty},e\right)$ is a
function system. But that is not the case for the rest
$\left(\mathfrak{S}_{p},e\right)$, which are unital $\ast$-normed spaces with
their common unit $e$. Nonetheless $\left(\mathfrak{S}_{p},e\right)$ is a
function system equipped with the projective cone $\mathfrak{S}_{p}^{+}$ given
by $\varepsilon=1>1/\left\|e\right\|_{p}$ such that
$\mathfrak{S}_{\infty,\varepsilon}^{+}\subseteq\mathfrak{S}_{1}^{+}\subseteq\mathfrak{S}_{p}^{+}\subseteq\mathfrak{S}_{\infty}^{+}$,
$\varepsilon\geq n$ is an increasing family of the unital cones with their
state spaces

$S\left(\mathfrak{S}_{p}^{+}\right)=\left\\{\left(1+r\right)z-rw:z,w\in
S_{\tau},zw=0,\left\|\left(1+r\right)z+rw\right\|_{q}\leq 1\right\\}$

for all conjugate couples $\left(p,q\right)$ (see Theorem 6.1). Moreover,
$\mathfrak{S}_{p}^{+}$ is reduced to the original cone of $\mathfrak{S}_{p}$
only if $n=2$ and $\varepsilon=2^{1/p}$. Thus only for the two-level quantum
system (qubit) [10, Ch. 2], the projective positivity is reduced to the
original one precisely. The (finite dimensional) geometry of the state spaces
$S\left(\mathfrak{S}_{p}^{+}\right)$ is described in the proof of Theorem 6.1.

Finally, the obtained result for $\mathbb{M}_{n}$ is generalized in Theorem
6.2 to the case of a finite von Neumann algebra $\mathcal{M}$ with a faithful,
finite, normal trace $\tau$, $\tau\left(e\right)>1$. It is also proved (see
Proposition 6.1) that the unital cone
$L^{2}\left(\mathcal{M},\tau\right)_{\varepsilon}^{+}$ for
$\varepsilon=\left(2/\tau\left(e\right)\right)^{1/2}>1/\sqrt{\tau\left(e\right)}$
of the unital Hilbert $\ast$-space
$\left(L^{2}\left(\mathcal{M},\tau\right),e\right)$ is reduced to one from
[7]. Thus $L^{2}\left(\mathcal{M},\tau\right)_{\varepsilon}^{+}$ is the
positivity that stands behind of Pisier’s operator Hilbert space [8, 3.5], [9,
7.4].

##  2\. The preliminary material on operator systems

To avoid a possible discomfort during the reading of the new material we
present in this section some key results on operators systems by adding up
some new facts too. For some assertions it is really hard to indicate the
related direct references, that is why we support the presentation with
Appendix Section 7. For the basic lattice operations $\inf$ and $\sup$ we use
the notations $\wedge$ and $\vee$, respectively. The convex hull of a subset
$S$ in a vector space is denoted by $\operatorname{co}\left(S\right)$ whereas
$\operatorname{abc}\left(S\right)$ denotes the absolutely convex hull of $S$.
The characteristic function of a set $S$ is denoted by $\left[S\right]$.

###  2.1. Operator systems

By an operator system (or noncommutative function system) $V$ on a Hilbert
space $H$ we mean a unital self-adjoint subspace of
$\mathcal{B}\left(H\right)$. Thus $V\subseteq\mathcal{B}\left(H\right)$ is a
subspace such that $e\in V$ and $V^{\ast}=V$, where $e$ is the unit of
$\mathcal{B}\left(H\right)$. A unital $C^{\ast}$-algebra is an example of an
operator system. The real subspace of all hermitian elements of an operator
system $V$ is denoted by $V_{h}$, whereas
$V_{+}=V\mathcal{\cap}\mathcal{B}\left(H\right)_{+}$ is the (algebraically or
topologically (see [4])) closed cone of its positive elements. The cone
$V_{+}$ is separated, that is, $V_{+}\cap-V_{+}=\left\\{0\right\\}$. Moreover,
it is a unital cone in the sense of that $V_{+}-e$ absorbs all $V_{h}$.
Namely, for every $x\in V_{h}$ we have
$x\in\left\|x\right\|\left(V_{+}-e\right)$. Note also that
$x=\left(x+\left\|x\right\|e\right)-\left\|x\right\|e\in V_{+}-V_{+}$ for
every $x\in V_{h}$, that is, $V_{h}=V_{+}-V_{+}$, which in turn implies that
$V=\left(V_{+}-V_{+}\right)+i\left(V_{+}-V_{+}\right)$.

A linear functional $\mu:V\rightarrow\mathbb{C}$ defines its conjugate
$\mu^{\ast}:V\rightarrow\mathbb{C}$ to be $\left\langle
x,\mu^{\ast}\right\rangle=\left\langle x^{\ast},\mu\right\rangle^{\ast}$,
$x\in V$. If $\mu=\mu^{\ast}$ then $\mu$ is called a hermitian functional on
$V$, and the real vector space of all hermitian functionals on $V$ is denoted
by $V_{h}^{\ast}$. Notice that $\mu\in V_{h}^{\ast}$ iff $\left\langle
V_{h},\mu\right\rangle\subseteq\mathbb{R}$. Indeed, if $\mu\in V_{h}^{\ast}$
and $x\in V_{h}$, then $\left\langle x,\mu\right\rangle^{\ast}=\left\langle
x^{\ast},\mu\right\rangle^{\ast}=\left\langle
x,\mu^{\ast}\right\rangle=\left\langle x,\mu\right\rangle$, that is,
$\left\langle x,\mu\right\rangle\in\mathbb{R}$. Conversely, $\left\langle
V_{h},\mu\right\rangle\subseteq\mathbb{R}$ implies that

$\left\langle
x,\mu^{\ast}\right\rangle=\left\langle\operatorname{Re}x+i\operatorname{Im}x,\mu^{\ast}\right\rangle=\left\langle\operatorname{Re}x-i\operatorname{Im}x,\mu\right\rangle^{\ast}=\left\langle\operatorname{Re}x,\mu\right\rangle+i\left\langle\operatorname{Im}x,\mu\right\rangle=\left\langle
x,\mu\right\rangle$

for all $x\in V$, that is, $\mu=\mu^{\ast}$. Moreover,

(2.1)
$\left\|\mu\right\|=\vee\left|\left\langle\operatorname{ball}V_{h},\mu\right\rangle\right|\text{
for all }\mu\in V_{h}^{\ast}.$

Indeed, for every $x\in\operatorname{ball}V$ with nonzero $\left\langle
x,\mu\right\rangle$ we have $\left|\left\langle
x,\mu\right\rangle\right|=\left\langle\theta
x,\mu\right\rangle=\operatorname{Re}\left\langle\theta
x,\mu\right\rangle=\left\langle\operatorname{Re}\left(\theta
x\right),\mu\right\rangle$ for suitable $\theta\in\mathbb{C}$,
$\left|\theta\right|=1$. But $\left\|\operatorname{Re}\left(\theta
x\right)\right\|\leq\left\|x\right\|\leq 1$, therefore $\left|\left\langle
x,\mu\right\rangle\right|\leq\left|\left\langle\operatorname{ball}V_{h},\mu\right\rangle\right|$.
Hence
$\left\|\mu\right\|=\vee\left|\left\langle\operatorname{ball}V,\mu\right\rangle\right|\leq\left|\left\langle\operatorname{ball}V_{h},\mu\right\rangle\right|\leq\left\|\mu\right\|$.

A linear functional $\mu:V\rightarrow\mathbb{C}$ is said to be positive if
$\left\langle V_{+},\mu\right\rangle\geq 0$. In this case $\mu=\mu^{\ast}$
automatically. Indeed, $\left\langle V_{h},\mu\right\rangle=\left\langle
V_{+},\mu\right\rangle-\left\langle
V_{+},\mu\right\rangle\subseteq\mathbb{R}$, which in turn implies $\mu\in
V_{h}^{\ast}$. The set of all positive functionals on $V$ is denoted by
$V_{+}^{\ast}$. It turns out to be a subcone of $V_{h}^{\ast}$. If
$V=C\left(\mathcal{T}\right)$ is the abelian $C^{\ast}$-algebra of all
continuous functions on a compact Hausdorff topological space $\mathcal{T}$,
then $C\left(\mathcal{T}\right)^{\ast}=M\left(\mathcal{T}\right)$ is the space
of all Radon charges on $\mathcal{T}$. In this case,
$M\left(\mathcal{T}\right)_{+}$ consists of all positive Radon measures on
$\mathcal{T}$. A Radon charge $\mu\in M\left(\mathcal{T}\right)$ is positive
(that is, $\mu\geq 0$) iff $\left\|\mu\right\|=\left\langle
1,\mu\right\rangle=\mu\left(\mathcal{T}\right)$. The fact can be proven for a
linear functional over an operator system (see below Lemma 7.2 from Appendix
Section 7).

###  2.2. The states of an operator system

A unital positive functional on $V$ is called a state of the cone $V_{+}$, and
we put

$S\left(V_{+}\right)=\left\\{\varphi\in V_{+}^{\ast}:\left\langle
e,\varphi\right\rangle=1\right\\}$

to be the state space of the cone $V_{+}$. In the case of
$V=C\left(\mathcal{T}\right)$ we come up with the convex set
$P\left(\mathcal{T}\right)$ of probability measures on $\mathcal{T}$, that is,
$P\left(\mathcal{T}\right)=S\left(C\left(\mathcal{T}\right)_{+}\right)$ (see
Subsection 7.3). We also put $V_{he}^{\ast}=\left\\{\varphi\in
V_{h}^{\ast}:\left\langle e,\varphi\right\rangle=1\right\\}$ to be the real
hyperplane of all unital, hermitian functionals on $V$. Using Lemma 7.2, we
conclude that

$S\left(V_{+}\right)=\left\\{\varphi\in\operatorname{ball}V^{\ast}:\left\|\varphi\right\|=\left\langle
e,\varphi\right\rangle=1\right\\}=V_{he}^{\ast}\cap\operatorname{ball}V^{\ast}.$

In particular, $S\left(V_{+}\right)$ is a $w^{\ast}$-compact subset of
$\operatorname{ball}V^{\ast}$. For every unit vector $\zeta\in H$ the
functional $\omega_{\zeta}:V\mathcal{\rightarrow}\mathbb{C}$,
$\omega_{\zeta}\left(x\right)=\left(x\zeta,\zeta\right)$ belongs to
$S\left(V_{+}\right)$. In particular,

(2.2) $x\in V_{+}\quad\text{iff\quad}\left\langle
x,S\left(V_{+}\right)\right\rangle\geq 0,$

which means that the operator system positivity is a state positivity.

Using the state space $S\left(V_{+}\right)$, one can define a new (state) norm
$\left\|\cdot\right\|_{e}$ on $V$ in the following way
$\left\|x\right\|_{e}=\vee\left|\left\langle
x,S\left(V_{+}\right)\right\rangle\right|$.

######  Proposition 2.1.

The norm $\left\|\cdot\right\|_{e}$ on $V$ is a unital $\ast$-norm, which is
equivalent to the original norm of $V$. Moreover,
$\left\|x\right\|_{e}=\left\|x\right\|$, $x\in V_{h}$, and
$\left\|\cdot\right\|_{e}$ is an order norm in the sense of

$\left\|x\right\|_{e}=\wedge\left\\{r>0:-re\leq x\leq re\right\\}\text{ for
all }x\in V_{h}.$

In particular, $\left\|\mu\right\|_{e}=\left\|\mu\right\|$ for every $\mu\in
V_{h}^{\ast}$.

###### Proof.

For each $x\in V$ we have

$\left\|x\right\|_{e}\leq\left\|x\right\|\left(\vee\left\|S\left(V_{+}\right)\right\|\right)=\left\|x\right\|\left(\vee\left\langle
e,S\left(V_{+}\right)\right\rangle\right)=\left\|x\right\|$

and $\left\|e\right\|_{e}=\vee\left|\left\langle
e,S\left(V_{+}\right)\right\rangle\right|=1$. Moreover,
$\left\|x^{\ast}\right\|_{e}=\vee\left|\left\langle
x^{\ast},S\left(V_{+}\right)\right\rangle\right|=\vee\left|\left\langle
x,S\left(V_{+}\right)\right\rangle^{\ast}\right|=\left\|x\right\|_{e}$ and

$\left\|\operatorname{Re}x\right\|_{e}=\vee\left|\left\langle\operatorname{Re}x,S\left(V_{+}\right)\right\rangle\right|=\vee\left|\operatorname{Re}\left\langle
x,S\left(V_{+}\right)\right\rangle\right|\leq\left\|x\right\|_{e}.$

Similarly, $\left\|\operatorname{Im}x\right\|_{e}\leq\left\|x\right\|_{e}$. If
$x\in V_{h}$ then (see [19, 3.2.25])

$\left\|x\right\|=\vee\left\\{\left|\left(x\zeta,\zeta\right)\right|:\left\|\zeta\right\|=1\right\\}=\vee\left\\{\left|\left\langle
x,\omega_{\zeta}|V\right\rangle\right|:\left\|\zeta\right\|=1\right\\}\leq\vee\left|\left\langle
x,S\left(V_{+}\right)\right\rangle\right|=\left\|x\right\|_{e},$

that is, $\left\|x\right\|_{e}=\left\|x\right\|$ for all $x\in V_{h}$. In
particular, for each $x\in V$ we have
$\left\|x\right\|\leq\left\|\operatorname{Re}x\right\|+\left\|\operatorname{Im}x\right\|\leq
2\left\|x\right\|_{e}\leq 2\left\|x\right\|$. Thus $\left\|\cdot\right\|$ and
$\left\|\cdot\right\|_{e}$ are equivalent norms on $V$. If
$\alpha=\wedge\left\\{r>0:-re\leq x\leq re\right\\}$ for $x\in V_{h}$, then

$\displaystyle\alpha$ $\displaystyle=\wedge\left\\{r>0:re\pm x\geq
0\right\\}=\wedge\left\\{r>0:\left\langle re\pm x,\varphi\right\rangle\geq
0,\varphi\in S\left(V_{+}\right)\right\\}$
$\displaystyle=\wedge\left\\{r>0:r\pm\left\langle x,\varphi\right\rangle\geq
0,\varphi\in
S\left(V_{+}\right)\right\\}=\wedge\left\\{r>0:\vee\left|\left\langle
x,S\left(V_{+}\right)\right\rangle\right|\leq r\right\\}$
$\displaystyle=\vee\left|\left\langle
x,S\left(V_{+}\right)\right\rangle\right|=\left\|x\right\|_{e}\text{,}$

that is, $\left\|\cdot\right\|_{e}$ is an order norm on $V$. Finally, based on
(2.1), for every $\mu\in V_{h}^{\ast}$ we have

$\left\|\mu\right\|_{e}=\vee\left|\left\langle\operatorname{ball}\left\|\cdot\right\|_{e},\mu\right\rangle\right|=\vee\left\\{\left|\left\langle
x,\mu\right\rangle\right|:x\in V_{h},\left\|x\right\|_{e}\leq
1\right\\}=\vee\left|\left\langle\operatorname{ball}V_{h},\mu\right\rangle\right|=\left\|\mu\right\|,$

that is, $\left\|\mu\right\|_{e}=\left\|\mu\right\|$. ∎

There is a well defined unital $\ast$-linear mapping $\Phi:V\rightarrow
C\left(S\left(V_{+}\right)\right)$,
$\Phi\left(x\right)\left(\varphi\right)=\left\langle x,\varphi\right\rangle$,
$x\in V$, $\varphi\in S\left(V_{+}\right)$ (Kadison’s representation). Notice
that $\Phi\left(V_{+}\right)\subseteq C\left(S\left(V_{+}\right)\right)_{+}$.

######  Corollary 2.1.

The $\ast$-representation
$\Phi:\left(V,\left\|\cdot\right\|_{e}\right)\rightarrow
C\left(S\left(V_{+}\right)\right)$ is an order isomorphism and isometric
mapping onto its range.

###### Proof.

Based on Proposition 2.1, we have

$\left\|\Phi\left(x\right)\right\|_{\infty}=\vee\left|\Phi\left(x\right)\left(S\left(V_{+}\right)\right)\right|=\vee\left|\left\langle
x,S\left(V_{+}\right)\right\rangle\right|=\left\|x\right\|_{e}$

for all $x\in V$, that is, $\Phi$ is an isometry. Furthermore

$\Phi\left(x\right)\in
C\left(S\left(V_{+}\right)\right)_{+}\Leftrightarrow\left\langle
x,S\left(V_{+}\right)\right\rangle\geq 0\Leftrightarrow x\in V_{+}$

by (2.2). Thus
$\Phi\left(V_{+}\right)=C\left(S\left(V_{+}\right)\right)_{+}\cap\Phi\left(V_{h}\right)$,
which means $\Phi$ is an order isomorphism onto its range. ∎

Thus if the original norm of $V$ equals to $\left\|\cdot\right\|_{e}$ then $V$
turns out to be a unital $\ast$-subspace of
$C\left(S\left(V_{+}\right)\right)$, that is, $V$ is a concrete function
system up to an isometry.

######  Corollary 2.2.

Let $V$ be an operator system. Then $V$ is a function system iff
$\left\|\cdot\right\|=\left\|\cdot\right\|_{e}$.

###### Proof.

Suppose that $V$ is a function system, that is, $V\subseteq
C\left(\mathcal{T}\right)$ is a unital self-adjoint subspace for some
Hausdorff compact topological space $\mathcal{T}$. If $\varphi\in
S\left(V_{+}\right)$ then $\varphi\in\operatorname{ball}V^{\ast}$ and it
admits a bounded linear extension up to some $\mu\in
M\left(\mathcal{T}\right)$ with $\left\|\mu\right\|=1=\left\langle
e,\varphi\right\rangle=\left\langle e,\mu\right\rangle$ thanks to Hahn-Banach
extension Theorem. By Lemma 7.2, we obtain that $\mu\in
P\left(\mathcal{T}\right)$. Since $P\left(\mathcal{T}\right)|V\subseteq
S\left(V_{+}\right)$, it follows that (see Proposition 7.1)

$\displaystyle\left\|f\right\|_{e}$ $\displaystyle=\vee\left|\left\langle
f,S\left(V_{+}\right)\right\rangle\right|\leq\vee\left|\left\langle
f,P\left(\mathcal{T}\right)|V\right\rangle\right|\leq\vee\left|\left\langle
f,\operatorname{ball}M\left(\mathcal{T}\right)\right\rangle\right|=\left\|f\right\|_{\infty}$
$\displaystyle=\vee\left|f\left(\mathcal{T}\right)\right|=\vee\left\\{\left|\left\langle
f,\delta_{t}|V\right\rangle\right|:t\in\mathcal{T}\right\\}\leq\vee\left|\left\langle
f,S\left(V_{+}\right)\right\rangle\right|=\left\|f\right\|_{e},$

that is, $\left\|f\right\|_{e}=\left\|f\right\|_{\infty}$ for all $f\in V$.
The reverse implication follows from Corollary 2.1. ∎

######  Corollary 2.3.

If $V$ is an operator system then
$\operatorname{ball}V_{h}^{\ast}\subseteq\operatorname{abc}S\left(V_{+}\right)$.
In particular,
$2^{-1}\operatorname{ball}V^{\ast}\subseteq\operatorname{abc}S\left(V_{+}\right)\subseteq\operatorname{ball}V^{\ast}$.

###### Proof.

The inclusion
$\operatorname{abc}S\left(V_{+}\right)\subseteq\operatorname{ball}V^{\ast}$ is
immediate. Take $\varphi\in\operatorname{ball}V_{h}^{\ast}$. By Theorem 7.1,
$\varphi$ admits an orthogonal expansion $\varphi=\varphi_{+}-\varphi_{-}$. In
the case of a nontrivial expansion, we put $\phi_{+}=r_{+}^{-1}\varphi_{+}$,
$\phi_{-}=r_{-}^{-1}\varphi_{-}$ with $r_{+}=\left\langle
e,\varphi_{+}\right\rangle$, $r_{-}=\left\langle e,\varphi_{-}\right\rangle$.
Then $\phi_{+}$, $\phi_{-}\in S\left(V_{+}\right)$ thanks to Lemma 7.2, and
$\varphi=r_{+}\phi_{+}-r_{-}\phi_{-}$. Since
$r_{+}+r_{-}=\left\|\varphi\right\|\leq 1$, it follows that
$\varphi=r_{+}\phi_{+}-r_{-}\phi_{-}\in\operatorname{abc}S\left(V_{+}\right)$.
Hence
$\operatorname{ball}V_{h}^{\ast}\subseteq\operatorname{abc}S\left(V_{+}\right)$.
Finally, every $\varphi\in\operatorname{ball}V^{\ast}$ admits the expansion
$\varphi=\operatorname{Re}\varphi+i\operatorname{Im}\varphi$ with
$\operatorname{Re}\varphi,\operatorname{Im}\varphi\in\operatorname{ball}V_{h}^{\ast}$.
Whence $\operatorname{ball}V^{\ast}\subseteq
2\operatorname{abc}S\left(V_{+}\right)$. ∎

Based on Theorem 7.1, we also derive that every element of the unit sphere of
$V_{h}^{\ast}$ is a convex combination of
$S\left(V_{+}\right)\cup-S\left(V_{+}\right)$. The combination is unique in
the case of a $C^{\ast}$-algebra $\mathcal{A}$.

##  3\. $\varepsilon$-positivity in operator systems

In this section we analyze a new concept of the $\varepsilon$-positivity in an
operator system $V$ by deforming the original positivity of $V$.

###  3.1. $\varepsilon$-postivity

Let $V$ be an operator system with its unit $e$, and the separated, closed,
unital cone $V_{+}$ of its positive elements, whose state space is denoted by
$S\left(V_{+}\right)$ (equipped with the $w^{\ast}$-topology). We put
$S_{\varepsilon}=V_{he}^{\ast}\cap\varepsilon\operatorname{ball}V^{\ast}$ for
$\varepsilon\geq 1$. In this case,
$S_{1}=V_{he}^{\ast}\cap\operatorname{ball}V^{\ast}=S\left(V_{+}\right)$. An
element $v\in V_{h}$ is said to be $\varepsilon$-positive if
$v\geq\left\|v\right\|\dfrac{\varepsilon-1}{\varepsilon+1}e$ in $V$. The set
of all $\varepsilon$-positive elements in $V$ is denoted by
$V_{\varepsilon}^{+}$. Note that $V_{\varepsilon}^{+}\subseteq V_{+}$,
$V_{1}^{+}=V_{+}$, and $V_{\varepsilon}^{+}$ is a separated, (norm) closed
cone.

######  Lemma 3.1.

The cone $V_{\varepsilon}^{+}$ is unital and

$A_{\varepsilon}=S\left(V_{\varepsilon}^{+}\right)=S_{\varepsilon},$

where $A_{\varepsilon}=\left\\{\left(1+s\right)\phi-s\psi:\phi,\psi\in
S\left(V_{+}\right),0\leq s\leq 2^{-1}\left(\varepsilon-1\right)\right\\}$.

###### Proof.

Since the assertion is trivial for $\varepsilon=1$, we assume that
$\varepsilon>1$. Take $v\in V_{h}$ and put $r=\left\|v\right\|\varepsilon$.
Then $v+re\in V_{\varepsilon}^{+}$. Indeed,

$\left\|v+re\right\|\dfrac{\varepsilon-1}{\varepsilon+1}e\leq\left(\left\|v\right\|+r\right)\dfrac{\varepsilon-1}{\varepsilon+1}e=\left(\left\|v\right\|\varepsilon-\left\|v\right\|\right)e=re-\left\|v\right\|e\leq
re+v,$

which means that $v+re\in V_{\varepsilon}^{+}$, that is, $V_{\varepsilon}^{+}$
is a unital cone.

Now prove that $S\left(V_{+}\right)\subseteq
S\left(V_{\varepsilon}^{+}\right)\subseteq S_{\varepsilon}$. Since
$V_{\varepsilon}^{+}\subseteq V_{+}$, it is immediate that
$S\left(V_{+}\right)\subseteq S\left(V_{\varepsilon}^{+}\right)$. If
$\varphi\in S\left(V_{\varepsilon}^{+}\right)$ and $v\in V_{h}$ then $\pm
v+\left\|v\right\|\varepsilon e\in V_{\varepsilon}^{+}$ and $\left\langle\pm
v+\left\|v\right\|\varepsilon e,\varphi\right\rangle\geq 0$, which means that
$\left|\left\langle
v,\varphi\right\rangle\right|\leq\varepsilon\left\|v\right\|$. By (2.1), we
conclude that
$\left\|\varphi\right\|=\vee\left|\left\langle\operatorname{ball}V_{h},\varphi\right\rangle\right|\leq\varepsilon$.
Thus $S\left(V_{\varepsilon}^{+}\right)\subseteq
V_{he}^{\ast}\cap\varepsilon\operatorname{ball}V^{\ast}=S_{\varepsilon}$.

Further, take $\varphi\in A_{\varepsilon}$, that is,
$\varphi=\left(1+s\right)\phi-s\psi$ with $\phi,\psi\in S\left(V_{+}\right)$
and $0\leq s\leq 2^{-1}\left(\varepsilon-1\right)$. Prove that $\varphi\in
S\left(V_{\varepsilon}^{+}\right)$. As we have just noticed above the
statement is true if $s=0$. Suppose $s>0$. Then $\varphi\in V_{h}^{\ast}$ and
$\left\langle e,\varphi\right\rangle=\left(1+s\right)\left\langle
e,\phi\right\rangle-s\left\langle e,\psi\right\rangle=1$, that is, $\varphi\in
V_{he}^{\ast}$. Take a nonzero $v\in V_{\varepsilon}^{+}$. Then
$\left\|v\right\|e\geq
v\geq\left\|v\right\|\dfrac{\varepsilon-1}{\varepsilon+1}e$ and

$\left\langle v,\varphi\right\rangle=\left(1+s\right)\left\langle
v,\phi\right\rangle-s\left\langle
v,\psi\right\rangle\geq\left(1+s\right)\dfrac{\varepsilon-1}{\varepsilon+1}\left\|v\right\|-s\left\|v\right\|=s\left\|v\right\|\left(\dfrac{1+s}{s}\dfrac{\varepsilon-1}{\varepsilon+1}-1\right).$

The real-valued function $f\left(s\right)=\dfrac{1+s}{s}$, $0<s\leq
2^{-1}\left(\varepsilon-1\right)$ is decreasing with $\wedge f=\min
f=f\left(2^{-1}\left(\varepsilon-1\right)\right)=\dfrac{\varepsilon+1}{\varepsilon-1}$,
therefore $\left\langle v,\varphi\right\rangle\geq 0$ or $\varphi\in
S\left(V_{\varepsilon}^{+}\right)$. Thus $S\left(V_{+}\right)\subseteq
A_{\varepsilon}\subseteq S\left(V_{\varepsilon}^{+}\right)\subseteq
S_{\varepsilon}$. It remains to prove that $S_{\varepsilon}\backslash
S\left(V_{+}\right)\subseteq A_{\varepsilon}$.

Take $\varphi\in S_{\varepsilon}\backslash S\left(V_{+}\right)$. Being a
hermitian functional on $V$, it admits an orthogonal decomposition by Theorem
7.1. Namely, $\varphi=\varphi_{+}-\varphi_{-}$ with positive functionals
$\varphi_{+}$ and $\varphi_{-}$ such that
$\left\|\varphi\right\|=\left\|\varphi_{+}\right\|+\left\|\varphi_{-}\right\|$.
Put $t=\left\|\varphi_{+}\right\|=\left\langle e,\varphi_{+}\right\rangle$ and
$s=\left\|\varphi_{-}\right\|=\left\langle e,\varphi_{-}\right\rangle$. In
this case, $t+s\leq\varepsilon$ and $1=\left\langle
e,\varphi\right\rangle=t-s$. Then $t>0$. But $s>0$ too, otherwise
$\varphi=\varphi_{+}\in S\left(V_{+}\right)$. Thus both $\varphi_{+}$ and
$\varphi_{-}$ are nonzero functionals. Put $\phi=t^{-1}\varphi_{+}$ and
$\psi=s^{-1}\varphi_{-}$. Then $\varphi=\left(1+s\right)\phi-s\psi$ with
$\phi,\psi\in S\left(V_{+}\right)$ and $0\leq s\leq
2^{-1}\left(\varepsilon-1\right)$. It means that $\varphi\in A_{\varepsilon}$.
∎

The $\varepsilon$-positivity generates the $\varepsilon$-norm
$\left\|v\right\|_{\varepsilon}=\vee\left|\left\langle
v,S\left(V_{\varepsilon}^{+}\right)\right\rangle\right|$, $v\in V$ on $V$. For
$\varepsilon=1$ we have
$\left\|\cdot\right\|_{\varepsilon}=\left\|\cdot\right\|_{e}$ (see Subsection
2.2), which is equivalent to the original norm of $V$ by Proposition 2.1.

######  Corollary 3.1.

If $V$ is an operator system then its $\varepsilon$-norm is equivalent to the
original norm for every $\varepsilon\geq 1$.

###### Proof.

If $\varphi\in S\left(V_{\varepsilon}^{+}\right)$ then $\varphi\in
A_{\varepsilon}$ by Lemma 3.1, which means that
$\varphi=\left(1+s\right)\phi-s\psi$ for some $\phi,\psi\in S_{1}$ and $0\leq
s\leq 2^{-1}\left(\varepsilon-1\right)$. Since
$\left|1+s\right|+\left|-s\right|=1+2s\leq\varepsilon$, we conclude that
$\varphi\in\varepsilon\operatorname{abc}\left(S_{1}\right)$. Thus

$S_{1}=V_{he}^{\ast}\cap\operatorname{ball}V^{\ast}\subseteq
V_{he}^{\ast}\cap\varepsilon\operatorname{ball}V^{\ast}=S\left(V_{\varepsilon}^{+}\right)=A_{\varepsilon}\subseteq\varepsilon\operatorname{abc}\left(S_{1}\right),$

which in turn implies the inclusions
$\varepsilon^{-1}S_{1}^{\circ}=\varepsilon^{-1}\operatorname{abc}\left(S_{1}\right)^{\circ}\subseteq
S\left(V_{\varepsilon}^{+}\right)^{\circ}\subseteq S_{1}^{\circ}$ for the
polars in $V$. Then
$\left\|v\right\|_{\varepsilon}\leq\left\|v\right\|_{e}\leq\varepsilon\left\|v\right\|_{\varepsilon}$
for all $v\in V$, or
$\left\|\cdot\right\|_{\varepsilon}\sim\left\|\cdot\right\|_{e}$. It remains
to use Proposition 2.1. ∎

###  3.2. The continuous functions with $\varepsilon$-oscillations

Now assume for a while that $V=C\left(\mathcal{T}\right)$ is an abelian,
unital $C^{\ast}$-algebra, where $\mathcal{T}$ is a compact, Hausdorff
topological space. Recall that the oscillation of a function
$h:\mathcal{T}\rightarrow\mathbb{C}$ over a subset $E\subseteq\mathcal{T}$ is
given by the quantity
$\omega\left(h,E\right)=\vee\left\\{\left|h\left(t\right)-h\left(l\right)\right|:t,l\in
E\right\\}$ by measuring the highest amplitude of $h$ over $E$. Take $f\in
C\left(\mathcal{T}\right)_{h}$. By its very definition, $f\in
C\left(\mathcal{T}\right)_{\varepsilon}^{+}$ iff
$f\left(t\right)\geq\left\|f\right\|_{\infty}\dfrac{\varepsilon-1}{\varepsilon+1}$
for all $t\in\mathcal{T}$. A function $f\in C\left(\mathcal{T}\right)_{+}$ is
said to has the $\varepsilon$-oscillation if either $f=0$ or $\omega\left(\ln
f,\mathcal{T}\right)\leq c_{\varepsilon}$, where
$c_{\varepsilon}=\ln\dfrac{\varepsilon+1}{\varepsilon-1}$. Every function from
$C\left(\mathcal{T}\right)_{+}$ has the $\varepsilon$-oscillation for
$\varepsilon=1$ (we put $c_{1}=+\infty$). More concrete example can be given
by the function $f\left(t\right)=3^{t}$ on $\mathcal{T=}\left[0,1\right]$ for
$\varepsilon=2$, whereas $3^{3^{t}}$ has no $2$-oscillation.

Consider the unital $\ast$-representation $\Phi:V\rightarrow
C\left(S\left(V_{+}\right)\right)$,
$\Phi\left(v\right)\left(\varphi\right)=\left\langle v,\varphi\right\rangle$,
$v\in V$, $\varphi\in S\left(V_{+}\right)$ with
$\Phi\left(V_{+}\right)=C\left(S\left(V_{+}\right)\right)_{+}\cap\Phi\left(V_{h}\right)$
(see Corollary 2.1).

######  Lemma 3.2.

The equality
$\Phi\left(V_{\varepsilon}^{+}\right)=C\left(\mathcal{T}\right)_{\varepsilon}^{+}\cap\Phi\left(V_{h}\right)$
holds. Moreover, for every nonzero $v\in V_{h}$ the following assertions are
equivalent:

$\left(i\right)$ $v\in V_{\varepsilon}^{+}$;

$\left(ii\right)$ $\Phi\left(v\right)$ has the $\varepsilon$-oscillation;

$\left(iii\right)$ $\left\langle
v,S\left(V_{\varepsilon}^{+}\right)\right\rangle\geq 0$.

Thus the bipolar theorem (see [6, Remark 5.3]) holds for all unital cones
$V_{\varepsilon}^{+}$, $\varepsilon\geq 1$.

###### Proof.

Take $v\in V_{h}$. Then $v\in V_{\varepsilon}^{+}$ iff $\left\langle
v-\left\|v\right\|\dfrac{\varepsilon-1}{\varepsilon+1}e,S\left(V_{+}\right)\right\rangle\geq
0$ by (2.2). It means that
$\Phi\left(v\right)\geq\left\|v\right\|\dfrac{\varepsilon-1}{\varepsilon+1}=\left\|\Phi\left(v\right)\right\|_{\infty}\dfrac{\varepsilon-1}{\varepsilon+1}$
thanks to Corollary 2.1. Hence
$\Phi\left(V_{\varepsilon}^{+}\right)=C\left(\mathcal{T}\right)_{\varepsilon}^{+}\cap\Phi\left(V_{h}\right)$.

Now take a nonzero $v\in V_{h}$. If $v\in V_{\varepsilon}^{+}$ then
$\dfrac{\varepsilon+1}{\varepsilon-1}v\geq\left\|v\right\|e\geq v$ and

$\left\langle
v,\varphi\right\rangle\dfrac{\varepsilon+1}{\varepsilon-1}\geq\left\|v\right\|\geq\left\langle
v,\psi\right\rangle$

for all $\varphi,\psi\in S\left(V_{+}\right)$. In particular, $\left\langle
v,\phi\right\rangle>0$ for all $\phi\in S\left(V_{+}\right)$, and
$\dfrac{\left\langle v,\psi\right\rangle}{\left\langle
v,\varphi\right\rangle}\leq\dfrac{\varepsilon+1}{\varepsilon-1}$ for all
$\varphi,\psi\in S\left(V_{+}\right)$. Thus
$\dfrac{\Phi\left(v\right)\left(\psi\right)}{\Phi\left(v\right)\left(\varphi\right)}\leq\dfrac{\varepsilon+1}{\varepsilon-1}$
for all $\varphi,\psi\in S\left(V_{+}\right)$, which means that

$\omega\left(\ln\Phi\left(v\right),S\left(V_{+}\right)\right)=\vee\left\\{\left|\ln\Phi\left(v\right)\left(\psi\right)-\ln\Phi\left(v\right)\left(\varphi\right)\right|:\psi,\varphi\in
S\left(V_{+}\right)\right\\}\leq c_{\varepsilon}$

or $\Phi\left(v\right)$ has the $\varepsilon$-oscillation. Thus
$\left(i\right)\Rightarrow\left(ii\right)$ holds.

Now suppose that $\Phi\left(v\right)$ possesses the $\varepsilon$-oscillation.
Then $\left\langle v,\phi\right\rangle>0$ for all $\phi\in
S\left(V_{+}\right)$, and $\dfrac{\left\langle
v,\psi\right\rangle}{\left\langle
v,\varphi\right\rangle}\leq\dfrac{\varepsilon+1}{\varepsilon-1}$. As in the
proof of Lemma 3.1, we have $\dfrac{\left\langle
v,\psi\right\rangle}{\left\langle v,\varphi\right\rangle}\leq\dfrac{s+1}{s}$
for all $0<s\leq 2^{-1}\left(\varepsilon-1\right)$. It follows that
$\left\langle v,\left(1+s\right)\phi-s\psi\right\rangle\geq 0$ for all $0\leq
s\leq 2^{-1}\left(\varepsilon-1\right)$ (if $s=0$ then $\left\langle
v,\phi\right\rangle>0$ for all $\phi\in S\left(V_{+}\right)$). Thus
$\left\langle v,A_{\varepsilon}\right\rangle\geq 0$ and
$\left(ii\right)\Rightarrow\left(iii\right)$ follows from Lemma 3.1.

Finally, prove that $\left(iii\right)\Rightarrow\left(i\right)$. If
$\left\langle v,A_{\varepsilon}\right\rangle\geq 0$ then
$\left(1+s\right)\left\langle v,\phi\right\rangle\geq s\left\langle
v,\psi\right\rangle$ for all $\phi,\psi\in S\left(V_{+}\right)$ and $0\leq
s\leq 2^{-1}\left(\varepsilon-1\right)$. But
$\left\|v\right\|=\left\|v\right\|_{e}>0$, therefore $\left\langle
v,\psi\right\rangle>0$ for some $\psi\in S\left(V_{+}\right)$ (see Corollary
2.2). It follows that $\left\langle v,\phi\right\rangle>0$ for all $\phi\in
S\left(V_{+}\right)$. As above $\dfrac{\left\langle
v,\psi\right\rangle}{\left\langle
v,\varphi\right\rangle}\leq\dfrac{\varepsilon+1}{\varepsilon-1}$ holds for all
$\phi,\psi\in S\left(V_{+}\right)$. Then

$\dfrac{\varepsilon-1}{\varepsilon+1}\left\|v\right\|=\dfrac{\varepsilon-1}{\varepsilon+1}\vee\left\langle
v,S\left(V_{+}\right)\right\rangle=\vee\left\langle
v,\dfrac{\varepsilon-1}{\varepsilon+1}S\left(V_{+}\right)\right\rangle\leq\left\langle
v,\varphi\right\rangle$

for all $\varphi\in S\left(V_{+}\right)$, that is, $\left\langle
v-\left\|v\right\|\dfrac{\varepsilon-1}{\varepsilon+1}e,S\left(V_{+}\right)\right\rangle\geq
0$. By (2.2), we deduce $v\in V_{\varepsilon}^{+}$. ∎

In the case of $V=C\left(\mathcal{T}\right)$ we have
$\mathcal{T=\partial}P\left(\mathcal{T}\right)\subseteq
P\left(\mathcal{T}\right)=S\left(C\left(\mathcal{T}\right)_{+}\right)$ (see
Subsection 7.3) and the related $\ast$-representation
$\Phi:C\left(\mathcal{T}\right)\rightarrow
C\left(P\left(\mathcal{T}\right)\right)$ is reduced to the natural extension
$\Phi\left(f\right)$ of a function $f\in C\left(\mathcal{T}\right)$ to
$P\left(\mathcal{T}\right)$ by
$\Phi\left(f\right)\left(\mu\right)=\left\langle f,\mu\right\rangle$, $\mu\in
P\left(\mathcal{T}\right)$. If $\mu=\delta_{t}$ for some $t\in\mathcal{T}$,
then $\Phi\left(f\right)\left(\delta_{t}\right)=f\left(t\right)$.

######  Corollary 3.2.

The unital cone $C\left(\mathcal{T}\right)_{\varepsilon}^{+}$ consists of all
functions with the $\varepsilon$-oscillations precisely.

###### Proof.

Take a nonzero $f\in C\left(\mathcal{T}\right)_{h}$. First assume that $f\in
C\left(\mathcal{T}\right)_{\varepsilon}^{+}$. By Lemma 3.2,
$\Phi\left(f\right)$ has the $\varepsilon$-oscillation on
$P\left(\mathcal{T}\right)$. In particular, it has the
$\varepsilon$-oscillation on the subset $\mathcal{T}$ either. It remains to
notify that $f=\Phi\left(f\right)|\mathcal{T}$.

Conversely, suppose that $f$ has the $\varepsilon$-oscillation. Then
$\dfrac{f\left(t\right)}{f\left(l\right)}\leq\dfrac{\varepsilon+1}{\varepsilon-1}$,
$t,l\in\mathcal{T}$, which means that $\left(1+s\right)\left\langle
f,\delta_{t}\right\rangle\geq s\left\langle f,\delta_{l}\right\rangle$ for all
$t,l\in\mathcal{T}$ and $0\leq s\leq 2^{-1}\left(\varepsilon-1\right)$. By
Proposition 7.1, we have
$\operatorname{co}\left(\mathcal{T}\right)^{-w^{\ast}}=\operatorname{co}\left(\mathcal{\partial}P\left(\mathcal{T}\right)\right)^{-w^{\ast}}=P\left(\mathcal{T}\right)$,
therefore $\left(1+s\right)\left\langle f,\mu\right\rangle\geq s\left\langle
f,\nu\right\rangle$ for all $\mu,\nu\in\mathcal{P}\left(\mathcal{T}\right)$
and $0\leq s\leq 2^{-1}\left(\varepsilon-1\right)$. It follows that
$\left\langle
f,S\left(C\left(\mathcal{T}\right)_{\varepsilon}^{+}\right)\right\rangle\geq
0$ by Lemma 3.1. Using Lemma 3.2, we obtain that $f\in
C\left(\mathcal{T}\right)_{\varepsilon}^{+}$ (the Bipolar Theorem). ∎

##  4\. Abstract function systems

In this section we prove the main result on the projective function systems.

###  4.1. The unital $\ast$-normed spaces

Let $V$ be a $\ast$-vector space, that is, $V$ is a complex vector space with
an involution. A norm $\left\|\cdot\right\|$ on $V$ is said to be a
$\ast$-norm if the unit $\operatorname{ball}V$ is hermitian, that is,
$\left(\operatorname{ball}V\right)^{\ast}=\operatorname{ball}V$. The latter is
equivalent to the property $\left\|v^{\ast}\right\|=\left\|v\right\|$ for all
$v\in V$. The real vector space of all hermitian elements in $V$ is denoted by
$V_{h}$. Notice that the norm dual space $V^{\ast}$ also possesses the
involution $\varphi\mapsto\varphi^{\ast}$ given by $\left\langle
v,\varphi^{\ast}\right\rangle=\left\langle
v^{\ast},\varphi\right\rangle^{\ast}$. If $V$ is a $\ast$-normed space then
$\left\|\varphi^{\ast}\right\|=\left\|\varphi\right\|$ for all $\varphi\in
V^{\ast}$, and the involution on $V$ is a
$\sigma\left(V,V^{\ast}\right)$-continuous mapping. Thus
$\left(V,V^{\ast}\right)$ is a dual $\ast$-pair (see [4], [5]). The real
vector space of all bounded hermitian functionals on $V$ is denoted by
$V_{h}^{\ast}$.

Now fix a nonzero $e\in V_{h}$ of a $\ast$-normed vector space $V$. Put
$V_{he}^{\ast}=\left\\{y\in V_{h}^{\ast}:\left\langle
e,y\right\rangle=1\right\\}$ to be the real hyperplane of all unital hermitian
functionals from $V^{\ast}$. As above put
$S=V_{he}^{\ast}\cap\operatorname{ball}V^{\ast}$ (or
$V_{he}^{\ast}\cap\left(\operatorname{ball}V\right)^{\circ}$) to be a convex
set. We say that $e$ is a unit element (or just unit) for $V$ if
$S\neq\varnothing$. Thus $1=\left\langle
e,y\right\rangle\leq\left\|e\right\|\left\|y\right\|\leq\left\|e\right\|$ for
some $y\in S$, that is, $\left\|e\right\|\geq 1$. Every hermitian unit vector
$e$ of a $\ast$-normed space $V$ is a unit automatically. Indeed, since
$\left\|e\right\|=1$, it follows that $\left\langle e,y\right\rangle=1$ for
some $y\in\operatorname{ball}V^{\ast}$ by Hahn-Banach extension property. But
$\operatorname{Re}y\in\operatorname{ball}V^{\ast}$ and $\left\langle
e,\operatorname{Re}y\right\rangle=2^{-1}\left(\left\langle
e,y\right\rangle+\left\langle e,y\right\rangle^{\ast}\right)=1$, that is,
$\operatorname{Re}y\in V_{he}^{\ast}\cap\operatorname{ball}V^{\ast}=S$. Hence
$e$ is a unit for $V$ and $S\subseteq\partial\operatorname{ball}V^{\ast}$. A
$\ast$-normed vector space $V$ with a fixed unit $\left(V,e\right)$ is called
a unital $\ast$-normed space.

###  4.2. The unital ball of a unital $\ast$-normed space

The unit $\operatorname{ball}V$ of a unital $\ast$-normed space
$\left(V,e\right)$ is said to be a unital ball if $\operatorname{ball}V$ is
the absolute polar of its state space $S$, that is,
$\operatorname{ball}V=S^{\circ}$. A unital $\ast$-normed space
$\left(V,e\right)$ with its unital ball is called an abstract function system
or a commutative operator system with its state space $S$. For brevity we say
that $V$ is a function system. A unital self-adjoint subspace $V$ of
$C\left(\mathcal{T}\right)$ is an example of a (concrete) function system (see
Corollary 2.2). In the case of $V=C\left(\mathcal{T}\right)$ with $e=1$ we
obtain that $S=P\left(\mathcal{T}\right)$ (see Proposition 7.1).

######  Lemma 4.1.

Let $\left(V,e\right)$ be a function system, $\mathfrak{c=}\left\\{v\in
V_{h}:\left\langle v,S\right\rangle\geq 0\right\\}$ and let
$\left\|v\right\|_{e}=\vee\left|\left\langle v,S\right\rangle\right|$, $v\in
V$. Then $\mathfrak{c}$ is a separated, closed, unital cone in $V$ such that
$S\left(\mathfrak{c}\right)=S$ and
$\left\|\cdot\right\|=\left\|\cdot\right\|_{e}$. In particular, $e$ is a unit
vector.

###### Proof.

Using the Absolute Bipolar Theorem, we derive that
$\operatorname{ball}V^{\ast}=\left(\operatorname{ball}V\right)^{\circ}=S^{\circ\circ}=\operatorname{abc}\left(S\right)^{-w^{\ast}}$,
where $\operatorname{abc}\left(S\right)^{-w^{\ast}}$ is the $w^{\ast}$-closure
of the absolutely convex hull $\operatorname{abc}\left(S\right)$ of $S$.
Therefore

$\left\|v\right\|=\vee\left|\left\langle
v,\operatorname{ball}V^{\ast}\right\rangle\right|=\vee\left|\left\langle
v,\operatorname{abc}\left(S\right)^{-w^{\ast}}\right\rangle\right|=\vee\left|\left\langle
v,S\right\rangle\right|=\left\|v\right\|_{e}$

for all $v\in V$. In particular,
$\left\|e\right\|=\left\|e\right\|_{e}=\sup\left|\left\langle
e,S\right\rangle\right|=1$, which means that $e$ is a unit vector of $V$.
Further, take $v\in V_{h}$. If $\pm v\in\mathfrak{c}$ then $\left\langle
v,S\right\rangle=\left\\{0\right\\}$ or $\left\|v\right\|_{e}=0$, that is,
$v=0$. Further, $\left\langle\left\|v\right\|e\pm
v,S\right\rangle=\left\|v\right\|\pm\left\langle
v,S\right\rangle=\left\|v\right\|_{e}\pm\left\langle v,S\right\rangle\geq 0$,
which means that $\left\|v\right\|e\pm v\in\mathfrak{c}$. Thus $\mathfrak{c}$
is a separated, closed, unital cone in $V$. Actually, $\mathfrak{c}$ is weakly
closed.

Further, take $y\in S\left(\mathfrak{c}\right)$, which means that
$y:V\rightarrow\mathbb{C}$ is a linear functional such that
$\left\langle\mathfrak{c},y\right\rangle\geq 0$ and $\left\langle
e,y\right\rangle=1$. Since $\left\|v\right\|e\pm v\in\mathfrak{c}$ for $v\in
V_{h}$, we obtain that $\left|\left\langle
v,y\right\rangle\right|\leq\left\|v\right\|$. Taking into account that
$V=V_{h}+iV_{h}$ and $V_{h}=\mathfrak{c-c}$, we conclude that $y\in
V_{he}^{\ast}$ (see Subsection 2.1). For $v\in V$ with nonzero $\left\langle
v,y\right\rangle$ we have $\left|\left\langle
v,y\right\rangle\right|=\left\langle\theta
v,y\right\rangle=\operatorname{Re}\left\langle\theta
v,y\right\rangle=\left\langle\operatorname{Re}\left(\theta
v\right),y\right\rangle\leq\left\|\operatorname{Re}\left(\theta
v\right)\right\|\leq\left\|\theta v\right\|=\left\|v\right\|$ for some
$\theta\in\mathbb{C}$, $\left|\theta\right|=1$. Hence $y\in
V_{he}^{\ast}\cap\operatorname{ball}V^{\ast}=S$, that is,
$S\left(\mathfrak{c}\right)=S$. ∎

Thus every function system turns out to be an Archimedian ordered
$\ast$-vector space in the sense of [17]. Using the $\ast$-representation from
Corollary 2.1, we can identify $V$ with a unital self-adjoint subspace of
$C\left(S\left(\mathfrak{c}\right)\right)$ up to an order and isometric
isomorphism, that is, $\left(V,\mathfrak{c}\right)$ is a concrete function
system on $S\left(\mathfrak{c}\right)$.

###  4.3. Non-unital balls

Let $\left(V,e\right)$ be a unital $\ast$-normed space. Put
$S_{\varepsilon}=V_{he}^{\ast}\cap\varepsilon\operatorname{ball}V^{\ast}$ for
$\varepsilon>0$. Thus $\operatorname{ball}V$ is unital whenever
$\operatorname{ball}V=S_{1}^{\circ}$, and in this case
$\operatorname{ball}V^{\ast}=\operatorname{abc}\left(S_{1}\right)^{-w^{\ast}}$.
In the case of a non-unital $\operatorname{ball}V$ we can pursue the property
to be $\operatorname{ball}V\sim S_{\varepsilon}^{\circ}$ for some
$\varepsilon>0$. Recall that the equivalence $\operatorname{ball}V\sim
S_{\varepsilon}^{\circ}$ of balls means that
$rS_{\varepsilon}^{\circ}\subseteq\operatorname{ball}V\subseteq
RS_{\varepsilon}^{\circ}$ for some positive real $r$ and $R$. If
$\operatorname{ball}V\sim S_{1}^{\circ}$, then we also refer it to the unital
case. In this case, the norms $\left\|\cdot\right\|$ and
$\left\|\cdot\right\|_{e}$ are equivalent, and $e$ turns out to be a unit
vector with respect to the norm $\left\|\cdot\right\|_{e}$. By Proposition
2.1, every operator system (if we ignore its quantizations or noncommutative
structures) turns out to be a function system up to an equivalent norm (see
also Corollary 2.1). In the non-unital case, $e$ may not be a unit vector.

######  Lemma 4.2.

Let $\left(V,e\right)$ be a unital $\ast$-normed space with its state space
$S$. Then $\operatorname{ball}V\sim S_{2}^{\circ}$.

###### Proof.

We use the arguments from [6, Proposition 4.1] for the quantum systems. First
note that $\operatorname{Re}\left(\operatorname{ball}V^{\ast}\right)\subseteq
V_{h}^{\ast}\cap\operatorname{ball}V^{\ast}=\operatorname{ball}V_{h}^{\ast}$.
Similarly,
$\operatorname{Im}\left(\operatorname{ball}V^{\ast}\right)\subseteq\operatorname{ball}V_{h}^{\ast}$
and
$\operatorname{ball}V^{\ast}\subseteq\operatorname{Re}\left(\operatorname{ball}V^{\ast}\right)+i\operatorname{Im}\left(\operatorname{ball}V^{\ast}\right)\subseteq
2\operatorname{abc}\left(\operatorname{ball}V_{h}^{\ast}\right)$.

Now take $z\in\operatorname{ball}V_{h}^{\ast}$ and fix $y\in
S=V_{he}^{\ast}\cap\operatorname{ball}V^{\ast}$ (by assumption
$S\neq\varnothing$). Note that $\left\langle e,z\right\rangle\in\mathbb{R}$
with $-\left\|e\right\|\leq\left\langle e,z\right\rangle\leq\left\|e\right\|$,
that is, $0\leq\left\langle e,\pm z\right\rangle\leq\left\|e\right\|$. Put
$w=\pm z+y$ with $a=\left\langle e,w\right\rangle$, that is, $1\leq a\leq
1+\left\|e\right\|$. If $u=a^{-1}w$ then $u\in V_{he}^{\ast}$ and
$\left\|u\right\|\leq a^{-1}\left\|w\right\|\leq\left\|\pm
z\right\|+\left\|y\right\|\leq 2$, which means that $u\in V_{he}^{\ast}\cap
2\operatorname{ball}V^{\ast}=S_{2}$. But $z=\mp au\pm y$ and $\left|\mp
a\right|+\left|\pm 1\right|\leq 2+\left\|e\right\|$, that is,
$z\in\left(2+\left\|e\right\|\right)\operatorname{abc}\left(S_{2}\right)$.
Thus $\operatorname{ball}V^{\ast}\subseteq
2\operatorname{abc}\left(\operatorname{ball}V_{h}^{\ast}\right)\subseteq
2\left(2+\left\|e\right\|\right)\operatorname{abc}\left(S_{2}\right)$. Taking
into account that $S_{2}\subseteq 2\operatorname{ball}V^{\ast}$, we conclude
that

(4.1)
$2^{-1}\operatorname{abc}\left(S_{2}\right)\subseteq\operatorname{ball}V^{\ast}\subseteq
2\left(2+\left\|e\right\|\right)\operatorname{abc}\left(S_{2}\right),$

that is, $\operatorname{abc}\left(S_{2}\right)\sim\operatorname{ball}V^{\ast}$
or
$S_{2}^{\circ}=\left(\operatorname{abc}\left(S_{2}\right)\right)^{\circ}\sim\operatorname{ball}V$.
∎

The assertion of Lemma 4.2 can be generalized in the following way.

######  Theorem 4.1.

Let $\left(V,e\right)$ be a unital $\ast$-normed space with its state space
$S$. Then $\operatorname{ball}V\sim S_{\varepsilon}^{\circ}$ for every
$\varepsilon>1/\left\|e\right\|$.

###### Proof.

For a positive real $r$ we put $V_{h\left(re\right)}^{\ast}=\left\\{y\in
V_{h}^{\ast}:\left\langle re,y\right\rangle=1\right\\}$ to be the real
hyperplane of all $re$-unital hermitian functionals from $V^{\ast}$. Note that
$V_{h\left(re\right)}^{\ast}=r^{-1}V_{he}^{\ast}$, and as above let us define
the sets
$S_{re,\varepsilon}=V_{h\left(re\right)}^{\ast}\cap\varepsilon\operatorname{ball}V^{\ast}$
for all $\varepsilon>0$. Prove that
$S_{re,\varepsilon}=r^{-1}S_{r\varepsilon}$. Indeed, $y\in S_{re,\varepsilon}$
iff $\left\|y\right\|\leq\varepsilon$, $\left\langle re,y\right\rangle=1$,
which means $y=r^{-1}ry$ with $\left\|ry\right\|\leq r\varepsilon$,
$\left\langle e,ry\right\rangle=1$, that is, $ry\in S_{r\varepsilon}$.
Conversely, if $y\in S_{r\varepsilon}$ then $\left\|r^{-1}y\right\|\leq
r^{-1}r\varepsilon=\varepsilon$ and $\left\langle
re,r^{-1}y\right\rangle=\left\langle e,y\right\rangle=1$, which means that
$r^{-1}y\in S_{re,\varepsilon}$. Thus
$S_{re,\varepsilon}=r^{-1}S_{r\varepsilon}$ and by passing to the polars we
obtain that $S_{re,\varepsilon}^{\circ}=rS_{r\varepsilon}^{\circ}$. In
particular, for $\varepsilon>1/\left\|e\right\|$ we have
$rS_{\varepsilon}^{\circ}=rS_{r\varepsilon/r}^{\circ}=S_{re,\varepsilon/r}^{\circ}$
with $\varepsilon/r>1/\left(\left\|e\right\|r\right)$. Hence
$\operatorname{ball}V\sim S_{\varepsilon}^{\circ}$,
$\varepsilon>1/\left\|e\right\|$ iff $\operatorname{ball}V\sim
S_{re,\varepsilon}^{\circ}$ for all
$\varepsilon>1/\left(\left\|e\right\|r\right)$. Thus the assertion is reduced
to the case of a unit vector $e$ (just put $r=\left\|e\right\|^{-1}$), and we
have to prove that $\operatorname{ball}V\sim S_{\varepsilon}^{\circ}$ for
every $\varepsilon>1$.

By Lemma 4.2, $\operatorname{ball}V\sim S_{2}^{\circ}$, namely,
$2^{-1}\operatorname{abc}\left(S_{2}\right)\subseteq\operatorname{ball}V^{\ast}\subseteq
6\operatorname{abc}\left(S_{2}\right)$ (see (4.1)). Assume that
$\varepsilon^{-1}\operatorname{abc}\left(S_{\varepsilon}\right)\subseteq\operatorname{ball}V^{\ast}\subseteq
R\operatorname{abc}\left(S_{\varepsilon}\right)$ for some $\varepsilon>1$ and
real $R$. Pick $0<\delta\leq\left(2R+1\right)^{-1}$ and
$\sigma=\varepsilon-\left(\varepsilon-1\right)\delta$. Notice that

$1<\dfrac{2R\varepsilon+1}{2R+1}=\varepsilon-\dfrac{\varepsilon-1}{2R+1}\leq\varepsilon-\left(\varepsilon-1\right)\delta=\sigma<\varepsilon,$

that is, $\delta<1<\sigma<\varepsilon$. For $\varepsilon=2$ we have $R=6$ and
$\delta\leq 13^{-1}$. Prove that
$\operatorname{ball}V^{\ast}\subseteq\left(R+1\right)\operatorname{abc}\left(S_{\sigma}\right)$.
Take $w\in\operatorname{ball}V^{\ast}$. Then $w\in
R\operatorname{abc}\left(S_{\varepsilon}\right)$, which means that $w=Rz$ with
$z\in\operatorname{abc}\left(S_{\varepsilon}\right)$,
$z=\sum_{i=1}^{n}a_{i}z_{i}$, $\sum\left|a_{i}\right|\leq 1$ and
$\left\\{z_{i}\right\\}\subseteq S_{\varepsilon}$. As in the proof of Lemma
4.2, fix $y\in S_{1}$ and put $z_{i,\delta}=\left(1-\delta\right)z_{i}+\delta
y$ with $\left\langle
e,z_{i,\delta}\right\rangle=\left(1-\delta\right)+\delta=1$. Thus
$\left\\{z_{i,\delta}\right\\}\subseteq V_{he}^{\ast}$ and
$\left\|z_{i,\delta}\right\|\leq\left(1-\delta\right)\left\|z_{i}\right\|+\delta\leq\left(1-\delta\right)\varepsilon+\delta=\sigma$,
which means that $\left\\{z_{i,\delta}\right\\}\subseteq
V_{he}^{\ast}\cap\sigma\operatorname{ball}V^{\ast}=S_{\sigma}$. If
$z_{\delta}=\sum_{i=1}^{n}a_{i}z_{i,\delta}$ and $w_{\delta}=Rz_{\delta}$,
then $z_{\delta}\in\operatorname{abc}\left(S_{\sigma}\right)$ and

$w_{\delta}=\left(1-\delta\right)Rz+R\delta ay=\left(1-\delta\right)w+R\delta
ay,$

where $a=\sum a_{i}$ with $\left|a\right|\leq 1$. It follows that

$\left(R+1\right)^{-1}w=\frac{R}{\left(1-\delta\right)\left(R+1\right)}z_{\delta}-\frac{\delta
R}{\left(1-\delta\right)\left(R+1\right)}ay$

with

$\frac{R}{\left(1-\delta\right)\left(R+1\right)}+\frac{R\delta}{\left(1-\delta\right)\left(R+1\right)}=\frac{R\left(1+\delta\right)}{\left(R+1\right)\left(1-\delta\right)}\leq\frac{R\left(1+\left(2R+1\right)^{-1}\right)}{\left(R+1\right)\left(1-\left(2R+1\right)^{-1}\right)}=1.$

Consequently,
$\left(R+1\right)^{-1}w\in\operatorname{abc}\left\\{z_{\delta},ay\right\\}\subseteq\operatorname{abc}\left(S_{\sigma}\right)$
(note that $y\in S_{1}\subseteq S_{\sigma}$, $\sigma>1$) or
$w\in\left(R+1\right)\operatorname{abc}\left(S_{\sigma}\right)$. Thus

(4.2)
$\sigma^{-1}\operatorname{abc}\left(S_{\sigma}\right)\subseteq\operatorname{ball}V^{\ast}\subseteq\left(R+1\right)\operatorname{abc}\left(S_{\sigma}\right),$

which means that
$\operatorname{abc}\left(S_{\sigma}\right)\sim\operatorname{ball}V^{\ast}$.

Now based on Lemma 4.2, we put $\sigma_{0}=2$, $R=6$,
$\delta_{0}=\left(2R+1\right)^{-1}=1/13$,
$\sigma_{1}=\sigma_{0}\left(1-\delta_{0}\right)+\delta_{0}=25/13$ and
$\delta_{1}=\left(2\left(R+1\right)+1\right)^{-1}=1/15$. One can iterate the
same argument to obtain a recursive sequence. Namely, we have
$\delta_{n}=\left(2\left(6+n\right)+1\right)^{-1}$ and
$\sigma_{n}=\sigma_{n-1}\left(1-\delta_{n-1}\right)+\delta_{n-1}$, that is,

$\sigma_{n}=\sigma_{n-1}\frac{2n+10}{2n+11}+\frac{1}{2n+11},\quad\sigma_{1}=25/13.$

By induction on $n$ one can prove that the recursive sequence is given by the
rule

$\sigma_{n}=\frac{693\sqrt{\pi}\Gamma\left(n+6\right)}{512\Gamma(n+13/2)}+1$

and $\lim_{n}\sigma_{n}=1$ (the formula itself is due to WolframAlpha
computation programing). Optionally, one can use the relation
$\sigma_{n}=\left(\sigma_{n-1}-\sigma_{n}\right)\left(2n+10\right)+1$ and the
fact $\sigma_{n}>1$ to conclude that $\sigma_{n}\downarrow 1$. Then for every
$\varepsilon>1$ we have $\sigma_{n}<\varepsilon$ for large $n$, and

$\varepsilon^{-1}\operatorname{abc}\left(S_{\varepsilon}\right)\subseteq\operatorname{ball}V^{\ast}\subseteq\left(6+n+1\right)\operatorname{abc}\left(S_{\sigma_{n}}\right)\subseteq\left(6+n+1\right)\operatorname{abc}\left(S_{\varepsilon}\right)$

by virtue of (4.2). Whence $\operatorname{ball}V\sim S_{\varepsilon}^{\circ}$
for every $\varepsilon>1$. ∎

######  Corollary 4.1.

Let $\left(V,e\right)$ be a unital $\ast$-normed space with its state space
and let $\varepsilon>1/\left\|e\right\|$. Then
$\left\|v\right\|_{\varepsilon}=\vee\left|\left\langle
v,S_{\varepsilon}\right\rangle\right|$, $v\in V$ is a norm on $V$ equivalent
to the original norm of $V$.

###### Proof.

The norm $\left\|\cdot\right\|_{\varepsilon}$ is the Minkowski functional of
$S_{\varepsilon}^{\circ}$. It remains to use Theorem 4.1. ∎

###  4.4. The $\varepsilon$-cones of a unital $\ast$-normed space

As above let $\left(V,e\right)$ be a unital $\ast$-normed space with its
hermitian $\operatorname{ball}V$ and the unit $e$. If $\operatorname{ball}V$
is unital then $V$ is a function system on $S_{1}$ (up to an equivalent norm)
by virtue of Lemma 4.1. If $\operatorname{ball}V$ is not unital then
$\operatorname{ball}V\sim S_{\varepsilon}^{\circ}$ for every
$\varepsilon>1/\left\|e\right\|$ thanks to Theorem 4.1. Put
$\mathfrak{c}_{\varepsilon}$ to be the set of those $v\in V_{h}$ such that
$\left\langle v,S_{\varepsilon}\right\rangle\geq 0$.

######  Theorem 4.2.

Let $\left(V,e\right)$ be a unital $\ast$-normed space with its state space
$S$. Then $\mathfrak{c}_{\varepsilon}$ is a separated, closed, unital, cone in
$V$, $S\left(\mathfrak{c}_{\varepsilon}\right)=S_{\varepsilon}$ and
$\left(V,\mathfrak{c}_{\varepsilon}\right)$ is a concrete function system on
$S_{\varepsilon}$ whose $\varepsilon$-norm is equivalent to the original one
of $V$ for every $\varepsilon>1/\left\|e\right\|$.

###### Proof.

Consider the absolute polar $S_{\varepsilon}^{\circ}$ of $S_{\varepsilon}$.
Prove that $S_{\varepsilon}^{\circ}$ is a unital ball, whose state space
equals to $V_{he}^{\ast}\cap S_{\varepsilon}^{\circ\circ}$. Note that
$S_{\varepsilon}=V_{he}^{\ast}\cap S_{\varepsilon}\subseteq V_{he}^{\ast}\cap
S_{\varepsilon}^{\circ\circ}\subseteq S_{\varepsilon}^{\circ\circ}$, which in
turn implies that
$S_{\varepsilon}^{\circ}=S_{\varepsilon}^{\circ\circ\circ}\subseteq\left(V_{he}^{\ast}\cap
S_{\varepsilon}^{\circ\circ}\right)^{\circ}\subseteq S_{\varepsilon}^{\circ}$.
Thus $S_{\varepsilon}^{\circ}=\left(S_{\varepsilon}^{\circ\circ}\cap
V_{he}^{\ast}\right)^{\circ}$, that is, the ball $S_{\varepsilon}^{\circ}$ is
the polar of its state space $S_{\varepsilon}^{\circ\circ}\cap V_{he}^{\ast}$.
It means that $S_{\varepsilon}^{\circ}$ is a unital ball. Thus $V$ equipped
with the Minkowski functional of $S_{\varepsilon}^{\circ}$ turns out to be a
function system. Since
$S_{\varepsilon}\subseteq\varepsilon\operatorname{ball}V^{\ast}$, it follows
that
$\operatorname{abc}\left(S_{\varepsilon}\right)^{-w^{\ast}}\subseteq\varepsilon\operatorname{ball}V^{\ast}$
and

$S_{\varepsilon}\subseteq V_{he}^{\ast}\cap
S_{\varepsilon}^{\circ\circ}=V_{he}^{\ast}\cap\operatorname{abc}\left(S_{\varepsilon}\right)^{-w^{\ast}}\subseteq
V_{he}^{\ast}\cap\varepsilon\operatorname{ball}V^{\ast}=S_{\varepsilon},$

that is, $S_{\varepsilon}=V_{he}^{\ast}\cap S_{\varepsilon}^{\circ\circ}$
turns out to be the state space of the unital ball $S_{\varepsilon}^{\circ}$.
In particular, $\mathfrak{c}_{\varepsilon}=\left\\{v\in V_{h}:\left\langle
v,V_{he}^{\ast}\cap S_{\varepsilon}^{\circ\circ}\right\rangle\geq 0\right\\}$.
By Lemma 4.1, $\mathfrak{c}_{\varepsilon}$ is a separated, closed, unital cone
in $V$ such that $S\left(\mathfrak{c}_{\varepsilon}\right)=V_{he}^{\ast}\cap
S_{\varepsilon}^{\circ\circ}=S_{\varepsilon}$. Thus
$\left(V,\mathfrak{c}_{\varepsilon}\right)$ turns out to be a concrete
function system on $S_{\varepsilon}$. The norm
$\left\|v\right\|_{\varepsilon}=\vee\left|\left\langle
v,S_{\varepsilon}\right\rangle\right|$, $v\in V$ of this function system is
equivalent to the original one by Corollary 4.1. ∎

Thus every unital $\ast$-normed space $V$ turns out to be a function system
equipped with the cones $\mathfrak{c}_{\varepsilon}$,
$\varepsilon>1/\left\|e\right\|$. In this case, we say that a unital
$\ast$-normed space $V$ can be equipped with a projective positivity. Let us
illustrate the projective positivity in the case of the $2$-dimensional
$\ell^{p}$ spaces. First consider the operator $\ell^{\infty}$-system
$V=\mathbb{C}\overset{\infty}{\oplus}\mathbb{C}$ with its unit
$e=\left(1,1\right)$. In this case,
$V^{\ast}=\mathbb{C}\overset{1}{\oplus}\mathbb{C}$,
$\operatorname{ball}V^{\ast}=\left\\{w=\left(w_{0},w_{1}\right)\in\mathbb{C}^{2}:\left|w_{0}\right|+\left|w_{1}\right|\leq
1\right\\}$ and

$\left\langle z,w^{\ast}\right\rangle=\left\langle
z^{\ast},w\right\rangle^{\ast}=\left(z_{0}^{\ast}w_{0}+z_{1}^{\ast}w_{1}\right)^{\ast}=z_{0}w_{0}^{\ast}+z_{1}w_{1}^{\ast}$

for all $z=\left(z_{0},z_{1}\right)\in V$ and $w=\left(w_{0},w_{1}\right)\in
V^{\ast}$, which means that $w^{\ast}=\left(w_{0}^{\ast},w_{1}^{\ast}\right)$.
In particular, $V_{h}^{\ast}=\mathbb{R}\overset{1}{\oplus}\mathbb{R}$,
$V_{he}^{\ast}=\left\\{s=\left(s_{0},s_{1}\right)\in\mathbb{R}^{2}:s_{0}+s_{1}=1\right\\}$
and

$S_{1}=V_{he}^{\ast}\cap\operatorname{ball}\left(\mathbb{R}\overset{1}{\oplus}\mathbb{R}\right)=\left\\{s=\left(s_{0},s_{1}\right)\in\mathbb{R}^{2}:s_{0}+s_{1}=1,s_{j}\geq
0\right\\}$

is the line segment from $\mathbb{R}^{2}$ connecting $\left(0,1\right)$ and
$\left(1,0\right)$. In particular, for $z\in V$ we have

$\left\|z\right\|_{e}=\vee\left|\left\langle
z,S_{1}\right\rangle\right|=\vee\left\\{\left|s_{0}z_{0}+s_{1}z_{1}\right|:s_{0}+s_{1}=1,s_{j}\geq
0\right\\}=\left|z_{0}\right|\vee\left|z_{1}\right|=\left\|z\right\|_{\infty},$

that is, $\operatorname{ball}V$ is unital and $V$ is a function system. For
$r\in\mathbb{R}^{2}$ we have $\left\langle r,S_{1}\right\rangle\geq 0$ iff
$r_{j}\geq 0$. Hence $\mathfrak{c}_{1}=\mathbb{R}_{+}^{2}$ as it used to be,
where $\mathbb{R}_{+}=\left\\{x\geq 0\right\\}$. Now for every $1\leq
p<\infty$ consider the $\ast$-vector space
$V=\mathbb{C}\overset{p}{\oplus}\mathbb{C}$ with the unit
$e_{p}=2^{-1/p}\left(1,1\right)$. If $z\in V$ then
$z^{\ast}=\left(z_{0}^{\ast},z_{1}^{\ast}\right)$ and
$\left\|z^{\ast}\right\|_{p}=\left(\left|z_{0}^{\ast}\right|^{p}+\left|z_{1}^{\ast}\right|^{p}\right)^{1/p}=\left\|z\right\|_{p}$,
which means that $\operatorname{ball}V$ is a hermitian ball. Thus
$\left(V,e_{p}\right)$ is a unital $\ast$-normed space. In this case,
$V_{h}^{\ast}=\mathbb{R}\overset{q}{\oplus}\mathbb{R}$ with the coujugate
couple $\left(p,q\right)$, and
$V_{he}^{\ast}=\left\\{s=\left(s_{0},s_{1}\right)\in\mathbb{R}^{2}:s_{0}+s_{1}=2^{1/p}\right\\}$.
In particular,

$S_{1}=V_{he}^{\ast}\cap\operatorname{ball}\left(\mathbb{R}\overset{q}{\oplus}\mathbb{R}\right)=\left\\{s=\left(s_{0},s_{1}\right)\in\mathbb{R}^{2}:s_{0}^{q}+s_{1}^{q}\leq
1,s_{0}+s_{1}=2^{1/p}\right\\}=\left\\{e_{q}\right\\},$

for $s_{0}+s_{1}=2^{1/p}$ is the tangent line to
$\operatorname{ball}\left(\mathbb{R}\overset{q}{\oplus}\mathbb{R}\right)$ at
$e_{q}$. Thus $S_{1}^{\circ}$ is not separated and $\operatorname{ball}V$ is
not unital. Nonetheless, every $S_{\varepsilon}$ for $\varepsilon>1$ provides
a unital ball thanks to Theorem 4.2.

It is worth to notify that $S_{\varepsilon}$ for $\varepsilon=2^{1/p}$
generates exactly the same cone $\mathbb{R}_{+}^{2}$. But the same phenomenon
can not be expected in higher dimensions $n\geq 3$. The reason is that the
geometry of the related state space has a rounded shape, which does not define
the original cone in $\ell^{p}\left(n\right)$. For example, in the case of
$\ell^{2}\left(3\right)$ we have $e_{2}=3^{-1/2}\left(1,1,1\right)$,
$S_{\varepsilon}=\ell^{2}\left(3\right)_{he}^{\ast}\cap\varepsilon\operatorname{ball}\left(\ell^{2}\left(3\right)_{h}\right)$,
that is,

$S_{1}=\left\\{s=\left(s_{0},s_{1},s_{2}\right)\in\mathbb{R}^{2}:s_{0}^{2}+s_{1}^{2}+s_{2}^{2}\leq
1,s_{0}+s_{1}+s_{2}=3^{1/2}\right\\}=\left\\{e_{2}\right\\},$

and
$S_{\varepsilon}=\left\\{s=\left(s_{0},s_{1},s_{2}\right)\in\mathbb{R}^{2}:s_{0}^{2}+s_{1}^{2}+s_{2}^{2}\leq\varepsilon^{2},s_{0}+s_{1}+s_{2}=3^{1/2}\right\\}$
is a disk centered at $e_{2}$ on the plane $s_{0}+s_{1}+s_{2}=3^{1/2}$. But
that disk $S_{\varepsilon}$ can either be inscribed into the triangle
$\Delta=\left\\{s_{0}+s_{1}+s_{2}=3^{1/2},s_{j}\geq 0,j=1,2,3\right\\}$ or
cover it up, which means that the related cone would never be the original
cone of $\ell^{2}\left(3\right)$. Nonetheless the projective positivity of
$\ell^{2}\left(3\right)$ allows to realize it as a function system on the disk
$S_{\varepsilon}$ thanks to Theorem 4.2.

##  5\. Projective positivity in $L^{p}$-spaces

In this section we analyze the projective positivity in the unital
$\ast$-spaces $L_{\mu}^{p}\left(\mathcal{T}\right)$ for a locally compact,
$\sigma$-compact, Hausdorff topological space $\mathcal{T}$ equipped with a
finite Radon measure $\mu$. It is well known that
$L_{\mu}^{p}\left(\mathcal{T}\right)_{h}$ is a Banach lattice with its natural
positivity given by the cone $L_{\mu}^{p}\left(\mathcal{T}\right)_{+}$ of
$\mu$-almost everywhere positive functions from
$L_{\mu}^{p}\left(\mathcal{T}\right)$. How does the cone
$L_{\mu}^{p}\left(\mathcal{T}\right)_{+}$ relate to the projective positivity
in the unital $\ast$-space $L_{\mu}^{p}\left(\mathcal{T}\right)$, that is the
main target of the present section.

###  5.1. $\varepsilon$-positivity in the unital $\ast$-spaces
$L_{\mu}^{p}\left(\mathcal{T}\right)$

Let $\mu\in M\left(\mathcal{T}\right)_{+}$,
$\left\|\mu\right\|=\vee\left\\{\left\|\mu\right\|_{K}:K\subseteq\mathcal{T}\right\\}=\mu\left(\mathcal{T}\right)<\infty$,
where $K\subseteq\mathcal{T}$ is running over all compact subsets,
$\left\|\mu\right\|_{K}=\left\|\mu|C_{K}\left(\mathcal{T}\right)\right\|$ and
$C_{K}\left(\mathcal{T}\right)\subseteq C_{c}\left(\mathcal{T}\right)$ is the
$\ast$-ideal of those continuous functions with supports inside of $K$. Note
that the related Banach spaces $L_{\mu}^{p}\left(\mathcal{T}\right)$, $1\leq
p\leq\infty$ are unital $\ast$-normed spaces with their units
$e_{p}=\mu\left(\mathcal{T}\right)^{-1/p}1$. For $p=\infty$ we come up with
the abelian von Neumann algebra $L_{\mu}^{\infty}\left(\mathcal{T}\right)$
with the unit $e_{\infty}=1$. For brevity we skip the notation $\mu$ and
consider the unital $\ast$-normed spaces
$\left(L^{p}\left(\mathcal{T}\right),e_{p}\right)$, $1\leq p\leq\infty$.
Certainly $\left(L^{\infty}\left(\mathcal{T}\right),e_{\infty}\right)$ is an
operator system, which turns out to be a function system (see Proposition
2.1). In particular, $\operatorname{ball}L^{\infty}\left(\mathcal{T}\right)$
is unital. But that is not the case for the rest
$\operatorname{ball}L^{p}\left(\mathcal{T}\right)$, $1\leq p<\infty$ (see
below Lemma 5.1).

We fix $1\leq p<\infty$ and $\varepsilon\geq 1$, and put
$S_{\varepsilon}^{p}=L^{p}\left(\mathcal{T}\right)_{he}^{\ast}\cap\varepsilon\operatorname{ball}L^{q}\left(\mathcal{T}\right)$,
where $\left(p,q\right)$ is the conjugate couple. Recall that
$\left(L^{p}\left(\mathcal{T}\right),L^{q}\left(\mathcal{T}\right)\right)$ is
a dual $\ast$-pair with the duality $\left\langle g,f\right\rangle=\int gf$,
$g\in L^{p}\left(\mathcal{T}\right)$, $f\in L^{q}\left(\mathcal{T}\right)$,
where $\int$ indicates to the integral related to $\mu$. In particular,
$\left\langle e_{p},f\right\rangle=\mu\left(\mathcal{T}\right)^{-1/p}\int f$,
$f\in L^{q}\left(\mathcal{T}\right)$ and $\left\langle
e_{p},e_{q}\right\rangle=\mu\left(\mathcal{T}\right)^{-1/p}\int e_{q}=1$. Note
that

$S_{1}^{p}=\left\\{f\in
L^{q}\left(\mathcal{T}\right)_{h}:\left\|f\right\|_{q}\leq 1,\int
f=\mu\left(\mathcal{T}\right)^{1/p}\right\\}\subseteq\partial\operatorname{ball}L^{q}\left(\mathcal{T}\right)$

and $e_{q}\in S_{1}^{p}$ (see Subsection 4.1). The class of all
$\mu$-measurable subsets $E\subseteq\mathcal{T}$ is denoted by
$\mathfrak{M}_{\mu}$.

######  Lemma 5.1.

The equality $S_{1}^{p}=\left\\{e_{q}\right\\}$ holds, that is,
$\operatorname{ball}L^{p}\left(\mathcal{T}\right)$ is not unital for all
$1\leq p<\infty$.

###### Proof.

First prove that $S_{1}^{p}\subseteq L^{q}\left(\mathcal{T}\right)_{+}$. If
$f\in S_{1}^{p}$ and $K\subseteq\left\\{f\leq-n^{-1}\right\\}$ is a compact
set, then $nf\left[K\right]+\left[K\right]\leq 0$ and $n\int f-n\int
f\left[\mathcal{T}\backslash K\right]+\mu\left(K\right)\leq 0$. It follows
that

$n\mu\left(\mathcal{T}\right)^{1/p}+\mu\left(K\right)\leq n\int
f\left[\mathcal{T}\backslash K\right]\leq
n\left\|f\right\|_{q}\left\|\left[\mathcal{T}\backslash
K\right]\right\|_{p}\leq n\mu\left(\mathcal{T}\backslash K\right)^{1/p}.$

But $\mu\left(\mathcal{T}\right)\geq\mu\left(\mathcal{T}\backslash K\right)$
and $n\left(\mu\left(\mathcal{T}\right)^{1/p}-\mu\left(\mathcal{T}\backslash
K\right)^{1/p}\right)+\mu\left(K\right)\leq 0$, therefore
$\mu\left(K\right)=0$. Thus $\mu\left\\{f\leq 0\right\\}=0$ or $f\in
L^{q}\left(\mathcal{T}\right)_{+}$.

Now assume that $p=1$ and $f\in S_{1}^{p}$. If $0<r<1$ and
$K\subseteq\left\\{f\leq r\right\\}$ is a compact subset, then
$f\left[K\right]\leq r\left[K\right]$, $\int f=\mu\left(\mathcal{T}\right)$
and $\mu\left(\mathcal{T}\right)-\int f\left[\mathcal{T}\backslash
K\right]\leq r\mu\left(K\right)$. It follows that

$\mu\left(\mathcal{T}\right)-r\mu\left(K\right)\leq\int
f\left[\mathcal{T}\backslash
K\right]\leq\left\|f\right\|_{\infty}\mu\left(\mathcal{T}\backslash
K\right)\leq\mu\left(\mathcal{T}\backslash
K\right)=\mu\left(\mathcal{T}\right)-\mu\left(K\right),$

that is, $\mu\left(K\right)=0$. Thus $f\geq 1$ in
$L^{\infty}\left(\mathcal{T}\right)_{h}$ and $\left\|f\right\|_{\infty}=1$.
But
$\wedge\left\\{r\in\mathbb{R}:\mu\left\\{f>r\right\\}=0\right\\}=\left\|f\right\|_{\infty}=1$,
which means that for every $r>1$, $\mu\left\\{f>r\right\\}=0$. Hence
$f=e_{\infty}$.

Finally, consider the case of $p>1$. Since
$S_{1}^{p}\subseteq\partial\operatorname{ball}L^{q}\left(\mathcal{T}\right)_{+}$
is a convex subset and $e_{q}\in S_{1}^{p}$, it follows that
$2^{-1}\left(f+e_{q}\right)\in S_{1}^{p}$ whenever $f\in S_{1}^{p}$. In
particular,
$\left\|2^{-1}\left(f+e_{q}\right)\right\|_{q}=\left\|f\right\|_{q}=\left\|e_{q}\right\|_{q}=1$.
Based on the geometry of $\operatorname{ball}L^{q}\left(\mathcal{T}\right)$
given by the property to be uniformly convex set (see [11]), we obtain that
for every $\varepsilon>0$ there corresponds $\delta>0$ such that
$g,h\in\operatorname{ball}L^{q}\left(\mathcal{T}\right)$ with
$\left\|g+h\right\|_{q}>2-\delta$ implies that
$\left\|g-h\right\|<\varepsilon$. Since $\left\|f+e_{q}\right\|_{q}=2$, we
obtain that $\left\|f-e_{q}\right\|_{q}<\varepsilon$. Whence $f=e_{q}$.
Optionally, one can use Clarkson’s inequalities in
$L^{q}\left(\mathcal{T}\right)$. ∎

Thus $S_{1}^{p}$ can not be a state space of a reasonable unital cone in
$L^{p}\left(\mathcal{T}\right)$.

###  5.2. Comparison of the cones

By Theorem 4.2, the unital $\ast$-normed space $L^{p}\left(\mathcal{T}\right)$
has the projective positivity given by the separated, closed, unital cone
$L^{p}\left(\mathcal{T}\right)_{\varepsilon}^{+}$ whose state space is
$S_{\varepsilon}^{p}$ for $\varepsilon>1$. It turns out that the original cone
$L^{p}\left(\mathcal{T}\right)_{+}$ of the Banach lattice
$L^{p}\left(\mathcal{T}\right)_{h}$ and
$L^{p}\left(\mathcal{T}\right)_{\varepsilon}^{+}$ for $1\leq p<\infty$ are
totally different cones, they are not comparable unless
$\operatorname{supp}\left(\mu\right)$ is finite. That is the main result of
the present section which will be proven in several steps.

######  Lemma 5.2.

Let $\mathfrak{c}\subseteq L^{p}\left(\mathcal{T}\right)_{h}$ be a norm closed
cone. Then $L^{p}\left(\mathcal{T}\right)_{+}\subseteq\mathfrak{c}$ iff
$\left[K\right]\in\mathfrak{c}$ for every compact set $K\subseteq\mathcal{T}$
with $\mu\left(K\right)>0$.

###### Proof.

Since $\left[K\right]\in L^{p}\left(\mathcal{T}\right)_{+}$ for every $K$, one
side implication is immediate. Conversely, suppose that
$\left\\{\left[K\right]:K\subseteq\mathcal{T}\right\\}\subseteq\mathfrak{c}$.
Pick $E\in\mathfrak{M}_{\mu}$. Since
$\mu\left(E\right)=\vee\left\\{\mu\left(K\right):K\subseteq\mathcal{T}\right\\}$,
it follows that $\mu\left(K_{n}\right)\uparrow\mu\left(E\right)$ for a certain
sequence $\left\\{K_{n}\right\\}$ of compact subsets in $E$. But
$\left\|\left[E\right]-\left[K\right]\right\|_{p}=\left(\int\left[E\backslash
K_{n}\right]\right)^{1/p}=\mu\left(E\backslash K_{n}\right)^{1/p}\rightarrow
0$, therefore $\left[E\right]\in\mathfrak{c}$ being a closed cone. In
particular, all positive $\mu$-step functions belong to $\mathfrak{c}$. If
$f\in L^{p}\left(\mathcal{T}\right)_{+}$ then $f_{n}\uparrow f$ for a sequence
$\left\\{f_{n}\right\\}\subseteq L^{p}\left(\mathcal{T}\right)_{+}$ of step
functions. Then $\left(f-f_{n}\right)\downarrow 0$ and
$\left(f-f_{n}\right)^{p}\leq f^{p}-f_{n}^{p}\leq f^{p}$ with $\int
f^{p}<\infty$. By Lebesgue’s convergence theorem,
$\int\left(f-f_{n}\right)^{p}\downarrow 0$. Taking into account that
$\left\\{f_{n}\right\\}\subseteq\mathfrak{c}$, we conclude that
$f\in\mathfrak{c}$ too. ∎

######  Lemma 5.3.

The inclusion $L^{p}\left(\mathcal{T}\right)_{+}\subseteq
L^{p}\left(\mathcal{T}\right)_{\varepsilon}^{+}$ for some $\varepsilon>1$
holds iff $S_{\varepsilon}^{p}\subseteq L^{q}\left(\mathcal{T}\right)_{+}$.

###### Proof.

Suppose that $L^{p}\left(\mathcal{T}\right)_{+}\subseteq
L^{p}\left(\mathcal{T}\right)_{\varepsilon}^{+}$. If $f\in
S_{\varepsilon}^{p}$ then $\left\langle
L^{p}\left(\mathcal{T}\right)_{\varepsilon}^{+},f\right\rangle\geq 0$ and
therefore $\left\langle\left[K\right],f\right\rangle\geq 0$. As in the proof
of Lemma 5.1, for a compact set $K\subseteq\left\\{f\leq-n^{-1}\right\\}$ we
have $nf\left[K\right]+\left[K\right]\leq 0$ and
$n\left\langle\left[K\right],f\right\rangle+\mu\left(K\right)\leq 0$, which
means that $\mu\left(K\right)=0$. It follows that $f\in
L^{q}\left(\mathcal{T}\right)_{+}$.

Conversely, suppose that $S_{\varepsilon}^{p}\subseteq
L^{q}\left(\mathcal{T}\right)_{+}$. Then
$\left\langle\left[K\right],S_{\varepsilon}^{p}\right\rangle=\int_{K}S_{\varepsilon}^{p}\geq
0$ for every compact set $K$. By Theorem 4.2, we have
$S_{\varepsilon}^{p}=S\left(L^{p}\left(\mathcal{T}\right)_{\varepsilon}^{+}\right)$,
which means that $\left[K\right]\in
L^{p}\left(\mathcal{T}\right)_{\varepsilon}^{+}$. But
$L^{p}\left(\mathcal{T}\right)_{\varepsilon}^{+}$ is a closed cone and
$\left\\{\left[K\right]:K\subseteq\mathcal{T}\right\\}\subseteq
L^{p}\left(\mathcal{T}\right)_{\varepsilon}^{+}$. By Lemma 5.2,
$L^{p}\left(\mathcal{T}\right)_{+}\subseteq
L^{p}\left(\mathcal{T}\right)_{\varepsilon}^{+}$ holds. ∎

Now consider the sets $E\in\mathfrak{M}_{\mu}$ with $\mu\left(E\right)>0$ and
put
$r_{\mu}=\wedge\left\\{\mu\left(E\right):E\in\mathfrak{M}_{\mu},\mu\left(E\right)>0\right\\}$.

######  Lemma 5.4.

If $L^{p}\left(\mathcal{T}\right)_{+}\subseteq
L^{p}\left(\mathcal{T}\right)_{\varepsilon}^{+}$ for some $\varepsilon>1$,
then $r_{\mu}>0$.

###### Proof.

Suppose that the inclusion $L^{p}\left(\mathcal{T}\right)_{+}\subseteq
L^{p}\left(\mathcal{T}\right)_{\varepsilon}^{+}$ holds for some
$\varepsilon>1$ whereas $r_{\mu}=0$. Thus one can choose the compact sets $K$
with sufficiently small $\mu\left(K\right)$. First consider the case of $p=1$.
Pick a small real $0<a<1$ and a compact $K$ with
$0<\mu\left(K\right)/\mu\left(\mathcal{T}\right)\leq\left(\varepsilon-1\right)/\left(\varepsilon+a\right)$.
Since the rational function $y=\dfrac{x-1}{x+a}$, $1\leq x\leq\varepsilon$ is
an increasing continuous function, it follows that
$\mu\left(K\right)/\mu\left(\mathcal{T}\right)=\left(b-1\right)/\left(b+a\right)$
for a certain $1<b\leq\varepsilon$. Then $f=b\left[\mathcal{T}\backslash
K\right]-a\left[K\right]\in L^{\infty}\left(\mathcal{T}\right)_{h}\backslash
L^{\infty}\left(\mathcal{T}\right)_{+}$ with

$\displaystyle\left\langle e_{1},f\right\rangle$
$\displaystyle=\mu\left(\mathcal{T}\right)^{-1}\int
f=\mu\left(\mathcal{T}\right)^{-1}\left(b\mu\left(\mathcal{T}\backslash
K\right)-a\mu\left(K\right)\right)=1,\text{ and}$
$\displaystyle\left\|f\right\|_{\infty}$ $\displaystyle=b\vee
a\leq\varepsilon.$

It means that $f\in S_{\varepsilon}^{1}\backslash
L^{\infty}\left(\mathcal{T}\right)_{+}$, which contradicts to Lemma 5.3.

Now assume that $1<p<\infty$. As above put $0<a<1$ and
$b=\dfrac{a\mu\left(K\right)+\mu\left(\mathcal{T}\right)^{1/p}}{\mu\left(\mathcal{T}\backslash
K\right)}$. By choosing small $\mu\left(K\right)>0$, we see that

$b^{q-1}\mu\left(\mathcal{T}\right)^{1/p}=\left(\dfrac{a\mu\left(K\right)+\mu\left(\mathcal{T}\right)^{1/p}}{\mu\left(\mathcal{T}\backslash
K\right)}\right)^{q-1}\mu\left(\mathcal{T}\right)^{1/p}\sim\left(\dfrac{\mu\left(\mathcal{T}\right)^{1/p}}{\mu\left(\mathcal{T}\right)}\right)^{q-1}\mu\left(\mathcal{T}\right)^{1/p}=1<\varepsilon^{q},$

that is, $b^{q-1}\mu\left(\mathcal{T}\right)^{1/p}<\varepsilon^{q}$ for some
compact $K$ with small $\mu\left(K\right)>0$. Further, by squeezing $a$, we
can also assume that
$b^{q-1}\mu\left(\mathcal{T}\right)^{1/p}+\left(b^{q-1}+a^{q-1}\right)a\mu\left(K\right)\leq\varepsilon^{q}$.
If $f=b\left[\mathcal{T}\backslash K\right]-a\left[K\right]$, then

$\displaystyle\left\langle e_{p},f\right\rangle$
$\displaystyle=\mu\left(\mathcal{T}\right)^{-1/p}\int
f=\mu\left(\mathcal{T}\right)^{-1/p}\left(b\mu\left(\mathcal{T}\backslash
K\right)-a\mu\left(K\right)\right)=1\text{, and}$
$\displaystyle\left\|f\right\|_{q}^{q}$
$\displaystyle=\int\left|f\right|^{q}=b^{q}\mu\left(\mathcal{T}\backslash
K\right)+a^{q}\mu\left(K\right)=b^{q-1}\left(a\mu\left(K\right)+\mu\left(\mathcal{T}\right)^{1/p}\right)+a^{q}\mu\left(K\right)$
$\displaystyle=b^{q-1}\mu\left(\mathcal{T}\right)^{1/p}+\left(b^{q-1}+a^{q-1}\right)a\mu\left(K\right)\leq\varepsilon^{q},$

which means that $f\in
L^{p}\left(\mathcal{T}\right)_{he}^{\ast}\cap\varepsilon\operatorname{ball}L^{q}\left(\mathcal{T}\right)=S_{\varepsilon}^{p}$.
But $f\notin L^{q}\left(\mathcal{T}\right)_{+}$, which again contradicts to
Lemma 5.3. ∎

Recall that by an atom in $\mathcal{T}$ for $\mu$ we mean a measurable subset
$A\subseteq\mathcal{T}$, $\mu\left(A\right)>0$ such that for every measurable
subset $B\subseteq A$ with $\mu\left(B\right)<\mu\left(A\right)$ we have
$\mu\left(B\right)=0$. A locally summable sequence $\left\\{c_{t}:t\in
S\right\\}$ in $\mathbb{R}_{+}$ defines an atomic measure
$\lambda=\sum_{t}c_{t}\delta_{t}\in M\left(\mathcal{T}\right)_{+}$ such that
$\operatorname{supp}\left(\lambda\right)$ is the closure of its atoms
$S=\left\\{t\in\mathcal{T}:c_{t}>0\right\\}$. In this case,
$\lambda=\left[S\right]\lambda$ and $\left\|\lambda\right\|=\sum_{t\in
S}c_{t}$. The closed subspace $I^{\delta}\left(\mathcal{T}\right)_{h}$ in
$M\left(\mathcal{T}\right)_{h}$ generated by
$\left\\{\delta_{t}:t\in\mathcal{T}\right\\}$ is a closed ideal of the
complete lattice $M\left(\mathcal{T}\right)_{h}$. It turns out that
$I^{\delta}\left(\mathcal{T}\right)$ is the set of all real atomic charges on
$\mathcal{T}$ exactly. Moreover, we have the decomposition
$M\left(\mathcal{T}\right)_{h}=I^{\delta}\left(\mathcal{T}\right)_{h}\oplus
I^{\delta}\left(\mathcal{T}\right)_{h}^{\perp}$ of the closed ideals such that
the ideal complement $I^{\delta}\left(\mathcal{T}\right)_{h}^{\perp}$ consists
of all real scattered measures on $\mathcal{T}$. In particular, there is a
unique decomposition $\mu=\lambda+\varphi$ with a positive atomic $\lambda$
and a positive scattered $\varphi$ such that
$\left\|\mu\right\|=\left\|\lambda\right\|+\left\|\varphi\right\|$. In this
case, there are disjoint $S,T\in\mathfrak{M}_{\mu}$ such that
$\left[S\right]\lambda=\lambda$ and $\left[T\right]\varphi=\varphi$ (see [1,
5.5] for the details)

######  Lemma 5.5.

A bounded, positive, Radon measure $\mu\in M\left(\mathcal{T}\right)$ with
$r_{\mu}>0$ is an atomic measure with the finite support
$\operatorname{supp}\left(\mu\right)$.

###### Proof.

First note that every $A\in\mathfrak{M}_{\mu}$ with
$r_{\mu}\leq\mu\left(A\right)<2r_{\mu}$ is an atom for $\mu$. Indeed, if
$B\in\mathfrak{M}_{\mu}$ with $B\subseteq A$,
$\mu\left(B\right)<\mu\left(A\right)$, we have $\mu\left(B\right)\geq r_{\mu}$
whenever $\mu\left(B\right)>0$. But
$\mu\left(A/B\right)=\mu\left(A\right)-\mu\left(B\right)>0$ too, therefore
$\mu\left(A/B\right)\geq r_{\mu}$. It follows that
$\mu\left(A\right)=\mu\left(B\right)+\mu\left(A/B\right)\geq 2r_{\mu}$, a
contradiction. Conversely, for every atom $A$ for $\mu$, we obviously have
$\mu\left(A\right)\geq r_{\mu}$ and
$\mu\left(A\right)=\mu\left(\left\\{t\right\\}\right)$ for some $t\in A$
thanks to [15, Proposition 11]. In particular, the atomic part of $\mu$ in the
expansion $I^{\delta}\left(\mathcal{T}\right)_{h}\oplus
I^{\delta}\left(\mathcal{T}\right)_{h}^{\perp}$ is not trivial. Put
$\mu=\lambda+\varphi$ with an atomic $\lambda$ and a scattered $\varphi$
supported on disjoint $S$ and $T$, respectively. For every $t\in S$ we have
$\mu\left(\left\\{t\right\\}\right)=\lambda\left(\left\\{t\right\\}\right)=c_{t}>0$.
In particular, $\wedge\left\\{c_{t}:t\in S\right\\}\geq r_{\mu}>0$. But
$\sum_{t\in S}c_{t}<\infty$, therefore $S$ is a finite set.

If $\varphi\neq 0$ then $\varphi\left(E\right)>0$ for some
$E\in\mathfrak{M}_{\mu}$. Since $\left[T\right]\varphi=\varphi$, we can assume
that $E\subseteq T$ and
$\mu\left(E\right)=\lambda\left(E\right)+\varphi\left(E\right)=\varphi\left(E\right)\geq
r_{\mu}$. In particular,
$r_{\varphi}=\wedge\left\\{\varphi\left(E\right):E\in\mathfrak{M}_{\varphi},\varphi\left(E\right)>0\right\\}\geq
r_{\mu}>0$. If $r_{\varphi}\leq\varphi\left(A\right)<2r_{\varphi}$, then $A$
is an atom for $\varphi$. As above
$\varphi\left(A\right)=\varphi\left(\left\\{t\right\\}\right)$, and taking
into account that $\varphi$ is scattered, we deduce that
$\varphi\left(\left\\{t\right\\}\right)=0$. Thus $r_{\varphi}=0$, a
contradiction. Whence $\mu=\lambda\in I^{\delta}\left(\mathcal{T}\right)_{h}$
and it has the finite support $S$. ∎

Now we can prove the following key result. To be punctual we say that some
cones $\mathfrak{c}$ and $\mathfrak{k}$ in a vector space are comparable if
$\mathfrak{c\subseteq k}$ or $\mathfrak{k\subseteq c}$.

######  Theorem 5.1.

The cones $L^{p}\left(\mathcal{T}\right)_{+}$ and
$L^{p}\left(\mathcal{T}\right)_{\varepsilon}^{+}$ are comparable for some
$\varepsilon>1$ if and only if $\operatorname{supp}\left(\mu\right)$ is
finite.

###### Proof.

First assume that $\operatorname{supp}\left(\mu\right)$ is finite, that is,
$\mu=\sum_{t\in S}c_{t}\delta_{t}$ for a finite subset $S\subseteq\mathcal{T}$
and $c_{t}>0$. Then $L^{p}\left(\mathcal{T}\right)=\ell^{p}\left(c\right)$ for
a finite sequence $c=\left\\{c_{j}:1\leq j\leq n\right\\}$, $c_{j}>0$, and
$\left\|x\right\|_{p}=\left(\sum_{j=1}^{n}\left|x_{j}\right|^{p}c_{j}\right)^{1/p}$
for every $x\in\ell^{p}\left(c\right)$. The duality of the pair
$\left(\ell^{p}\left(c\right),\ell^{q}\left(c\right)\right)$ if given by
$\left\langle x,y\right\rangle=\sum_{j=1}^{n}x_{j}y_{j}c_{j}$. Moreover, $y\in
S_{\varepsilon}^{p}$ for $\varepsilon>1$ iff $\left\|y\right\|_{q}\leq 1$ and
$\sum_{j}y_{j}c_{j}=\int
y=\mu\left(\mathcal{T}\right)^{1/p}=\left(\sum_{j}c_{j}\right)^{1/p}$. Put

$r_{c}=\wedge\left\\{\left(\dfrac{\sum_{j}c_{j}}{\sum_{j\neq
k}c_{j}}\right)^{1/p}:1\leq k\leq n\right\\}>1$

and fix $\varepsilon$ with $1<\varepsilon\leq r_{c}$. If $y\in
S_{\varepsilon}^{p}$ then $y_{k}\geq 0$ for all $k$. Indeed, in the case of
$y_{k}<0$ for a certain $k$, we deduce that

$\displaystyle\left(\sum_{j}c_{j}\right)^{1/p}$
$\displaystyle=\sum_{j}y_{j}c_{j}<\sum_{j\neq k}y_{j}c_{j}\leq\sum_{j\neq
k}\left|y_{j}\right|c_{j}^{1/q}c_{j}^{1/p}\leq\left(\sum_{j\neq
k}\left|y_{j}\right|^{q}c_{j}\right)^{1/q}\left(\sum_{j\neq
k}c_{j}\right)^{1/p}$ $\displaystyle\leq\left\|y\right\|_{q}\left(\sum_{j\neq
k}c_{j}\right)^{1/p}\leq\varepsilon\left(\sum_{j\neq k}c_{j}\right)^{1/p},$

which means that $\varepsilon>\left(\dfrac{\sum_{j}c_{j}}{\sum_{j\neq
k}c_{j}}\right)^{1/p}\geq r_{c}$, a contradiction. Thus
$S_{\varepsilon}^{p}\subseteq\ell^{q}\left(c\right)_{+}$, which in turn
implies that
$\ell^{p}\left(c\right)_{+}\subseteq\ell^{p}\left(c\right)_{\varepsilon}^{+}$
thanks to Lemma 5.3.

Conversely, if $L^{p}\left(\mathcal{T}\right)_{+}\subseteq
L^{p}\left(\mathcal{T}\right)_{\varepsilon}^{+}$ holds for some
$\varepsilon>1$, then $r_{\mu}>0$ by Lemma 5.4. Hence
$\operatorname{supp}\left(\mu\right)$ is finite thanks to Lemma 5.5.

If $L^{p}\left(\mathcal{T}\right)_{\varepsilon}^{+}\subseteq
L^{p}\left(\mathcal{T}\right)_{+}$ then it is immediate that
$L^{p}\left(\mathcal{T}\right)_{+}$ is unital which is not the case unless
$\operatorname{supp}\left(\mu\right)$ is finite. That fact is known in Banach
lattices theory [14, Ch. 2] claiming that $L^{p}$-spaces have no strong order
unit but only weak one. In the present case we can provide an independent
argument similar to those in the proof of Lemma 5.4 to fix it. Namely, suppose
that $L^{p}\left(\mathcal{T}\right)_{\varepsilon}^{+}\subseteq
L^{p}\left(\mathcal{T}\right)_{+}$ holds for some $\varepsilon>1$. Then
$S^{p}\subseteq S_{\varepsilon}^{p}$, where $S^{p}$ is the state space of the
cone $L^{p}\left(\mathcal{T}\right)_{+}$. But $S^{p}$ consists of those $f\in
L^{q}\left(\mathcal{T}\right)_{+}$ such that $\int
f=\mu\left(\mathcal{T}\right)^{1/p}$, for

$\left\langle L^{p}\left(\mathcal{T}\right)_{+},f\right\rangle\geq
0\Leftrightarrow\left\langle\left[K\right],f\right\rangle\geq 0\text{,
}K\subseteq\mathcal{T}\Leftrightarrow f\in L^{q}\left(\mathcal{T}\right)_{+}$

(see Lemma 5.2). Based on Lemma 5.5, it suffices to prove that $r_{\mu}>0$. If
$r_{\mu}=0$ there are compact sets $K$ with sufficiently small
$\mu\left(K\right)$, and as above fix

$0<a<\mu\left(\mathcal{T}\right)^{-1/q}\text{,
\quad}b=\dfrac{\mu\left(\mathcal{T}\right)^{1/p}-a\mu\left(\mathcal{T}\backslash
K\right)}{\mu\left(K\right)}>1$

for $p>1$. Then
$\mu\left(\mathcal{T}\right)^{1/p}-a\mu\left(\mathcal{T}\right)>0$ and

$b^{q}\mu\left(K\right)=\dfrac{\left(\mu\left(\mathcal{T}\right)^{1/p}-a\mu\left(\mathcal{T}\backslash
K\right)\right)^{q}}{\mu\left(K\right)^{q-1}}>\varepsilon^{q}$

for small $\mu\left(K\right)>0$. Put $f=a\left[\mathcal{T}\backslash
K\right]+b\left[K\right]\in L^{q}\left(\mathcal{T}\right)_{+}$. Then
$\left\|f\right\|_{q}^{q}=\int f^{q}=a^{q}\mu\left(\mathcal{T}\backslash
K\right)+b^{q}\mu\left(K\right)>\varepsilon^{q}$ and $\int
f=a\mu\left(\mathcal{T}\backslash
K\right)+b\mu\left(K\right)=\mu\left(\mathcal{T}\right)^{1/p}$, which means
that $f\in S^{p}\backslash S_{\varepsilon}^{p}$, a contradiction. If $p=1$ we
fix $b>\varepsilon$ and $K\subseteq\mathcal{T}$ with
$0<\mu\left(K\right)/\mu\left(\mathcal{T}\right)<1/b$. The rational function
$y=\dfrac{1-x}{b-x}$, $0\leq x\leq 1$ is a continuous decreasing function,
therefore $\mu\left(K\right)/\mu\left(\mathcal{T}\right)=\dfrac{1-a}{b-a}$ for
some $0<a<1$. Put $f=a\left[\mathcal{T}\backslash K\right]+b\left[K\right]\in
L^{\infty}\left(\mathcal{T}\right)_{+}$ with $\left\|f\right\|_{\infty}=a\vee
b=b>\varepsilon$. Since $\int f=b\mu\left(\mathcal{T}\backslash
K\right)+a\mu\left(K\right)=\mu\left(\mathcal{T}\right)$, it follows that
$f\in S^{1}\backslash S_{\varepsilon}^{1}$, a contradiction. ∎

As we have noticed above the equality
$\ell^{p}\left(c\right)=\ell^{p}\left(c\right)_{\varepsilon}^{+}$ holds only
in the case of dimension $2$. The same can be observed in the noncommutative
case too that we are going to tackle with below.

##  6\. Projective positivity in $L^{p}$-spaces of a finite von Neumann
algebra

In this section we investigate the projective cones in $L^{p}$-spaces of a
finite von Neumann algebra.

###  6.1. Projective positivity in matrix algebras

We do analysis of the full matrix algebra case. Consider the $\ast$-vector
space $\mathbb{M}_{n}$ of all complex $n\times n$-matrices for $n>1$. The
identity matrix in $\mathbb{M}_{n}$ is denoted by $e$. For every $1\leq
p\leq\infty$ we define the Schatten $p$-norm on $\mathbb{M}_{n}$ by
$\left\|x\right\|_{p}=\tau\left(\left|x\right|^{p}\right)^{1/p}$,
$x\in\mathbb{M}_{n}$, where $\tau$ is the standard trace on $\mathbb{M}_{n}$.
The normed vector space $\left(\mathbb{M}_{n},\left\|\cdot\right\|_{p}\right)$
is denoted by $\mathfrak{S}_{p}$. For the extremal values $p=1,\infty$ we also
use the notations $T_{n}=\mathfrak{S}_{1}$ and $M_{n}=\mathfrak{S}_{\infty}$.
Recall that $\mathfrak{S}_{p}$ and $\mathfrak{S}_{q}$ are in the canonical
duality through the pairing
$\left\langle\cdot,\cdot\right\rangle:\mathfrak{S}_{p}\times\mathfrak{S}_{q}\rightarrow\mathbb{C}$,
$\left\langle x,y\right\rangle=\tau\left(xy\right)$ whenever
$\left(p,q\right)$ is a conjugate pair. Thus
$\mathfrak{S}_{p}^{\ast}=\mathfrak{S}_{q}$ up to an isometric identification
given by $\tau$. In particular, $M_{n}^{\ast}=T_{n}$, and $M_{n}^{+}$ is the
standard unital cone of positive elements of the $C^{\ast}$-algebra $M_{n}$
with its state space

$S_{\tau}=\left\\{y\in T_{n}:y\geq 0,\tau\left(y\right)=1\right\\}$

to be the convex set of all density matrices from $\mathbb{M}_{n}$. Thus

(6.1)
$S\left(M_{n}^{+}\right)=S_{\tau}=\operatorname{co}\left\\{\zeta\odot\zeta:\left\|\zeta\right\|=1\right\\}$

is the convex hull of the one-rank projections
$\left(\zeta\odot\zeta\right)\left(\eta\right)=\left(\eta,\zeta\right)\zeta$,
$\zeta,\eta\in\ell^{2}\left(n\right)$. Along with the cone $M_{n}^{+}$ one can
generate its deformed versions $M_{n,\varepsilon}^{+}$, $\varepsilon\geq 1$
considered above in Section 3. Namely, $x\in M_{n,\varepsilon}^{+}$ iff
$x\geq\left\|x\right\|\dfrac{\varepsilon-1}{\varepsilon+1}e$ in $M_{n}$. Thus
$x\in M_{n,\varepsilon}^{+}$ means that $x\geq 0$ and
$\wedge\sigma\left(x\right)\geq\dfrac{\varepsilon-1}{\varepsilon+1}\left(\vee\sigma\left(x\right)\right)$,
where $\sigma\left(x\right)$ denotes the spectrum of $x$. In particular, every
nonzero $x\in M_{n,\varepsilon}^{+}$ is invertible automatically. By Lemma
3.1, the relation

$S\left(M_{n,\varepsilon}^{+}\right)=\left\\{\left(1+r\right)z-rw:z,w\in
S_{\tau},0\leq r\leq 2^{-1}\left(\varepsilon-1\right)\right\\}$

provides a detailed description of the state space of the cone
$M_{n,\varepsilon}^{+}$.

The class of all Hilbert-Schmidt matrices is denoted by $HS_{n}$, that is,
$HS_{n}=\mathfrak{S}_{2}$. Recall that $HS_{n}$ turns out to be a Hilbert
$\ast$-space with the inner product
$\left(x,y\right)=\tau\left(xy^{\ast}\right)$, and every $x\in HS_{n}$ admits
a unique orthonormal expansion $x=x_{0}+\left(x,e_{2}\right)e_{2}$,
$\left(x_{0},e_{2}\right)=0$, where $e_{2}=n^{-1/2}e$ is the unit vector of
$HS_{n}$. One can define the following cone

(6.2) $HS_{n}^{+}=\left\\{x\in
HS_{n,h}:\left\|x\right\|_{2}\leq\left(\dfrac{n}{n-1}\right)^{1/2}\left(x,e_{2}\right)\right\\}$

in the unital Hilbert $\ast$-space $\left(HS_{n},e_{2}\right)$. This cone is a
deformation of the standard cone of a unital Hilbert $\ast$-space considered
in [7]. Actually, the constant $\left(\dfrac{n}{n-1}\right)^{1/2}$ can be
reduced to $\sqrt{2}$ by making a suitable transformation of the state space,
and $HS_{n}^{+}$ defines a projective positivity (see below to the proof of
the forthcoming assertion).

######  Theorem 6.1.

All $\mathfrak{S}_{p}$ are unital $\ast$-normed spaces with their common unit
$e$. Thus $\mathfrak{S}_{p}$ is a function system equipped with the cone
$\mathfrak{S}_{p}^{+}$ given by $\varepsilon=1>1/\left\|e\right\|_{p}$ such
that

$M_{n,\varepsilon}^{+}\subseteq\mathfrak{S}_{1}^{+}\subseteq\mathfrak{S}_{p}^{+}\subseteq\mathfrak{S}_{\infty}^{+}=M_{n}^{+},\quad\varepsilon\geq
n$

is an increasing family of the unital cones with
$S\left(M_{n}^{+}\right)=S_{\tau}$ and

$S\left(\mathfrak{S}_{p}^{+}\right)=\left\\{\left(1+r\right)z-rw:z,w\in
S_{\tau},zw=0,\left\|\left(1+r\right)z+rw\right\|_{q}\leq 1\right\\}$

for all conjugate couples $\left(p,q\right)$. Moreover,
$\mathfrak{S}_{2}^{+}=HS_{n}^{+}$.

###### Proof.

Take $x\in\mathfrak{S}_{p}$ with its polar decomposition $x=u\left|x\right|$.
Taking into account that $\left|x^{\ast}\right|=u\left|x\right|u^{\ast}$, we
derive that
$\left\|x^{\ast}\right\|_{p}=\tau\left(\left|x^{\ast}\right|^{p}\right)^{1/p}=\tau\left(u\left|x\right|^{p}u^{\ast}\right)^{1/p}=\tau\left(\left|x\right|^{p}\right)^{1/p}=\left\|x\right\|_{p}$,
which means that $\mathfrak{S}_{p}$ is a $\ast$-normed space. Prove that $e$
is a unit for $\mathfrak{S}_{p}$ (see Subsection 4.1). Notice that
$\mathfrak{S}_{p,he}^{\ast}=\left\\{y\in\mathfrak{S}_{q,h}:\tau\left(y\right)=1\right\\}$,
$\operatorname{ball}\mathfrak{S}_{p}^{\ast}=\operatorname{ball}\mathfrak{S}_{q}$,
and
$S_{1}^{p}=\mathfrak{S}_{p,he}^{\ast}\cap\operatorname{ball}\mathfrak{S}_{q}$
is the related convex set. One needs to find out a description for $S_{1}^{p}$
and show that $S_{1}^{p}\neq\varnothing$.

The real vector space $\mathbb{R}^{n}$ equipped with the $q$-norm
$\left\|\mathbf{x}\right\|_{q}=\left(\sum_{i=1}^{n}\left|x_{i}\right|^{q}\right)^{1/q}$
is denoted by $\mathbb{R}_{q}^{n}$, that is,
$\mathbb{R}_{q}^{n}=\ell^{q}\left(n\right)_{h}$. Put
$\Pi_{n}=\left\\{\mathbf{x}\in\mathbb{R}^{n}:\sum_{i=1}^{n}x_{i}=1\right\\}$
to be the hyperplane in $\mathbb{R}^{n}$ with its normal vector
$\mathbf{e=}\left(1,\ldots,1\right)$. Notice that
$\Pi_{n}\cap\operatorname{ball}\mathbb{R}_{1}^{n}=S_{+}$, where
$S_{+}=\left\\{\mathbf{x}\in\mathbb{R}^{n}:x_{i}\geq
0,\sum_{i=1}^{n}x_{i}=1\right\\}$ is the positive surface of the tetrahedron
in $\mathbb{R}^{n}$. Moreover,

$S_{+}\subseteq\Pi_{n}\cap\operatorname{ball}\mathbb{R}_{q}^{n}\subseteq\Pi_{n}\cap\operatorname{ball}\mathbb{R}_{\infty}^{n}$

is an increasing family of the hyperplane regions from $\Pi_{n}$. Pick
$\mathbf{x}\in\Pi_{n}\cap\operatorname{ball}\mathbb{R}_{q}^{n}$ and put
$x=x_{1}\oplus\cdots\oplus x_{n}\in\mathbb{M}_{n}$ to be the related diagonal
matrix. Then $\tau\left(x\right)=1$ and
$\left\|x\right\|_{q}=\left\|\mathbf{x}\right\|_{q}\leq 1$, that is,
$x\in\mathfrak{S}_{p,he}^{\ast}\cap\operatorname{ball}\mathfrak{S}_{q}=S_{1}^{p}$.
Thus $\Pi_{n}\cap\operatorname{ball}\mathbb{R}_{q}^{n}\subseteq S_{1}^{p}$ up
to a canonical identification. Conversely, if $y\in S_{1}^{p}$ then there is a
unitary $u\in\mathbb{M}_{n}$ such that $y=u^{\ast}xu$ for a diagonal matrix
$x=x_{1}\oplus\cdots\oplus x_{n}$ with
$\sigma\left(y\right)=\left\\{x_{i}\right\\}\subseteq\mathbb{R}$ or
$\mathbf{x}\in\mathbb{R}^{n}$. But
$\left|y\right|^{q}=u^{\ast}\left|x\right|^{q}u$ with
$\left|x\right|=\left|x_{1}\right|\oplus\cdots\oplus\left|x_{n}\right|$, and

$\displaystyle\left\|\mathbf{x}\right\|_{q}$
$\displaystyle=\left(\sum_{i=1}^{n}\left|x_{i}\right|^{q}\right)^{1/q}=\tau\left(\left|x\right|^{q}\right)^{1/q}=\tau\left(u^{\ast}\left|x\right|^{q}u\right)^{1/q}=\tau\left(\left|y\right|^{q}\right)^{1/q}=\left\|y\right\|_{q}\leq
1,$ $\displaystyle\sum_{i=1}^{n}x_{i}$
$\displaystyle=\tau\left(x\right)=\tau\left(u^{\ast}xu\right)=\tau\left(y\right)=1,$

which means that
$\mathbf{x\in}\Pi_{n}\cap\operatorname{ball}\mathbb{R}_{q}^{n}$. Consequently,
we obtain that

(6.3)
$S_{1}^{p}=\cup\left\\{u^{\ast}\left(\Pi_{n}\cap\operatorname{ball}\mathbb{R}_{q}^{n}\right)u:u\in\mathbb{M}_{n},u^{\ast}u=e\right\\}.$

In particular, $S_{1}^{p}\neq\varnothing$ and $e$ turns out to be a unit for
every $\mathfrak{S}_{p}$ with $\left\|e\right\|_{p}=n^{1/p}>1$.

Further, put $\mathfrak{S}_{p}^{+}$ to be the set of those
$x\in\mathfrak{S}_{p,h}$ such that $\left\langle
x,S_{1}^{p}\right\rangle=\tau\left(xS_{1}^{p}\right)\geq 0$. If $p=\infty$
then $\left\|e\right\|_{\infty}=1$ and

$\displaystyle S_{1}^{\infty}$
$\displaystyle=\cup\left\\{u^{\ast}\left(\Pi_{n}\cap\operatorname{ball}\mathbb{R}_{1}^{n}\right)u:u\in\mathbb{M}_{n},u^{\ast}u=e\right\\}=\cup\left\\{u^{\ast}S_{+}u:u\in\mathbb{M}_{n},u^{\ast}u=e\right\\}$
$\displaystyle=\left\\{y\in T_{n}:y\geq
0,\tau\left(y\right)=1\right\\}=S_{\tau}.$

In this special case, we have $\mathfrak{S}_{\infty}^{+}=M_{n}^{+}$. Indeed,
$x\in\mathfrak{S}_{\infty}^{+}$ iff
$\left(x\zeta,\zeta\right)=\tau\left(x\left(\zeta\odot\zeta\right)\right)\geq
0$ for every unit vector $\zeta\in\ell^{2}\left(n\right)$ (see (6.1)). If
$p<\infty$ then $\varepsilon=1>1/\left\|e\right\|_{p}$, and
$\mathfrak{S}_{p}^{+}$ turns out to be a separated, closed, unital, cone in
$\mathfrak{S}_{p}$ thanks to Theorem 4.2. Thus
$\left(\mathfrak{S}_{p},\mathfrak{S}_{p}^{+}\right)$ is a function system on
$S_{1}^{p}$ such that $S\left(\mathfrak{S}_{p}^{+}\right)=S_{1}^{p}$. In
particular, for $p_{1}<p_{2}$ we obtain that $q_{2}<q_{1}$,
$S_{1}^{p_{2}}\subseteq S_{1}^{p_{1}}$ (see (6.3)), which in turn implies that
$\mathfrak{S}_{p_{1}}^{+}\subseteq\mathfrak{S}_{p_{2}}^{+}$.

Now let us inspect the cone $\mathfrak{S}_{2}^{+}$. In this case, the norm
dual space $HS_{n}^{\ast}$ is canonically identified with the conjugate space
$\overline{HS_{n}}$, and for every $y\in\mathfrak{S}_{2,he}^{\ast}$ we have
the orthogonal expansion $y=y_{0}+\left(1/n\right)e$ with $y_{0}\perp e$ (or
$\tau\left(y_{0}\right)=0$). Then
$\left\|y\right\|_{2}^{2}=\left\|y_{0}\right\|_{2}^{2}+1/n$, and $y\in
S_{1}^{2}$ iff
$\left\|y_{0}\right\|_{2}\leq\left(\left(n-1\right)/n\right)^{1/2}$. Hence

$S_{1}^{2}=\left(\left(n-1\right)/n\right)^{1/2}\operatorname{ball}HS_{n,h}^{e}+\left(1/n\right)e=n^{-1/2}\left(\left(n-1\right)^{1/2}\operatorname{ball}HS_{n,h}^{e}+e_{2}\right),$

where $HS_{n,h}^{e}=\left\\{w\in HS_{n,h}:\tau\left(w\right)=0\right\\}$ is
the real vector subspace of $\overline{HS_{n}}$ orthogonal to $e$, and
$e_{2}=n^{-1/2}e$ is the unit vector of $HS_{n}$. Thus
$x\in\mathfrak{S}_{2}^{+}$ with its orthonormal expansion
$x=x_{0}+\left(x,e_{2}\right)e_{2}$ iff
$\left(x,\left(n-1\right)^{1/2}\operatorname{ball}HS_{n,h}^{e}+e_{2}\right)\geq
0$, which in turn is equivalent to

$\left(n-1\right)^{1/2}\left\|x_{0}\right\|_{2}=\left(n-1\right)^{1/2}\vee\left|\left(x_{0},\operatorname{ball}HS_{n,h}^{e}\right)\right|\leq\left(x,e_{2}\right)\text{
or }\left\|x_{0}\right\|_{2}\leq\left(n-1\right)^{-1/2}\left(x,e_{2}\right).$

Consequently, $\mathfrak{S}_{2}^{+}$ is reduced to the cone $HS_{n}^{+}$ from
(6.2). If we choose $S_{\varepsilon}^{2}$ for
$\varepsilon=\sqrt{2/n}>1/\sqrt{n}$ (notice that
$\left\|e\right\|_{2}=\sqrt{n}$) instead of $S_{1}^{2}$, then based on Theorem
4.2, the projective positivity will exactly be reduced to one from [7].
Namely, $y=y_{0}+\left(1/n\right)e\in S_{\varepsilon}^{2}$ iff
$\left\|y\right\|_{2}^{2}=\left\|y_{0}\right\|_{2}^{2}+1/n\leq\varepsilon^{2}=2/n$
or $\left\|y_{0}\right\|_{2}\leq 1/\sqrt{n}$. Thus
$S_{\varepsilon}^{2}=n^{-1/2}\left(\operatorname{ball}HS_{n,h}^{e}+e_{2}\right)$,
and for $x\in HS_{n}$ we have $\left(x,S_{\varepsilon}^{2}\right)\geq 0$ iff
$\left\|x\right\|_{2}\leq\sqrt{2}\left(x,e_{2}\right)$.

Finally, let us prove the equality for the state spaces
$S\left(\mathfrak{S}_{p}^{+}\right)$. Suppose that a hermitian matrix $y$
admits the expansion $y=\left(1+r\right)z-rw$ with $z,w\in S_{\tau}$, $zw=0$
and $\left\|z+r\left(z+w\right)\right\|_{q}\leq 1$. Then
$\tau\left(y\right)=\left(1+r\right)\tau\left(z\right)-r\tau\left(w\right)=1+r-r=1$,
$\left|y\right|=\left(1+r\right)z+rw$ and
$\left\|y\right\|_{q}=\left\|\left|y\right|\right\|_{q}=\left\|\left(1+r\right)z+rw\right\|_{q}\leq
1$, that is,
$y\in\mathfrak{S}_{p,he}^{\ast}\cap\operatorname{ball}\mathfrak{S}_{q}=S_{1}^{p}=S\left(\mathfrak{S}_{p}^{+}\right)$.
Conversely, take $y\in S_{1}^{p}$. Using the spectral decomposition of $y$, we
deduce that $y=y_{+}-y_{-}=\tau\left(y_{+}\right)z-\tau\left(y_{-}\right)w$
with $z,w\in S_{\tau}$ and $\tau\left(y_{+}\right)-\tau\left(y_{-}\right)=1$.
If $r=\tau\left(y_{-}\right)$ then $y=\left(1+r\right)z-rw$ with $z,w\in
S_{\tau}$, $zw=0$ and
$\left\|\left(1+r\right)z+rw\right\|_{q}=\left\|\left|y\right|\right\|_{q}=\left\|y\right\|_{q}\leq
1$. Hence $S\left(\mathfrak{S}_{p}^{+}\right)$ consists of those
$\left(1+r\right)z-rw$ with $z,w\in S_{\tau}$, $zw=0$, and
$\left\|\left(1+r\right)z+rw\right\|_{q}\leq 1$. In particular,
$S\left(\mathfrak{S}_{1}^{+}\right)$ consists of those
$y=\left(1+r\right)z-rw$ such that $z,w\in S_{\tau}$, $zw=0$ and
$\left\|\left(1+r\right)z+rw\right\|_{\infty}\leq 1$.

Pick $y\in S\left(\mathfrak{S}_{1}^{+}\right)$ with
$\left(1+r\right)z+rw=\left|y\right|\leq e$. Then $\left|y\right|^{p}\leq e$,
$p\geq 1$, and
$1+2r=\left(1+r\right)\tau\left(z\right)+r\tau\left(w\right)=\tau\left(\left|y\right|\right)\leq\tau\left(e\right)=n$
(or $r\leq\left(n-1\right)/2$). Hence

$S\left(\mathfrak{S}_{p}^{+}\right)\subseteq\left\\{\left(1+r\right)z-rw:z,w\in
S\left(M_{n}^{+}\right),0\leq
r\leq\left(n-1\right)/2\right\\}=S\left(M_{n,n}^{+}\right)\subseteq
S\left(M_{n,\varepsilon}^{+}\right)$

for all $p\geq 1$ and $\varepsilon\geq n$ by virtue of Lemma 3.1. In
particular,
$M_{n,\varepsilon}^{+}\subseteq\mathfrak{S}_{1}^{+}\subseteq\mathfrak{S}_{p}^{+}\subseteq
M_{n}^{+}$ is an increasing family of the unital cones. ∎

We also write $T_{n}^{+}$ instead of $\mathfrak{S}_{1}^{+}$. Based on Theorem
6.1, we have

$\displaystyle S_{\tau}$ $\displaystyle\subseteq
S\left(T_{n}^{+}\right)=\left\\{\left(1+r\right)z-rw:z,w\in
S_{\tau},zw=0,\left\|\left(1+r\right)z+rw\right\|_{\infty}\leq 1\right\\}$
$\displaystyle\subseteq\left\\{\left(1+r\right)z-rw:z,w\in S_{\tau},0\leq
r\leq\left(n-1\right)/2\right\\}.$

Moreover, using (6.3), we obtain that

$S\left(T_{n}^{+}\right)=S_{1}^{1}=\cup\left\\{u^{\ast}\left(\Pi_{n}\cap\operatorname{ball}\mathbb{R}_{\infty}^{n}\right)u:u\in\mathbb{M}_{n},u^{\ast}u=e\right\\},$

which provides a geometric interpretation of the state space
$S\left(T_{n}^{+}\right)$. Notice that the equality
$S_{\tau}=S\left(T_{n}^{+}\right)$ is true only if $n=2$. For $n=3$ the
matrices $\left(-1\right)\oplus 1\oplus 1$, $1\oplus\left(-1\right)\oplus 1$
and $1\oplus 1\oplus\left(-1\right)$ are some (extreme) elements of
$S\left(T_{n}^{+}\right)$ out of $S_{\tau}$.

Thus $T_{n}^{+}=\left\\{x\in
T_{n,h}:\tau\left(xS\left(T_{n}^{+}\right)\right)\geq 0\right\\}$ and
$M_{n,\varepsilon}^{+}\subseteq T_{n}^{+}\subseteq M_{n}^{+}$ for all
$\varepsilon\geq n$. For $n=3$ and $x=2\oplus 0\oplus 1$, we have $x\in
M_{3}^{+}\backslash T_{3}^{+}$, for $\tau\left(x\left(\left(-1\right)\oplus
1\oplus 1\right)\right)=-1<0$. Thus the eigenvalues of a projective-positive
matrix from $T_{3}^{+}$ must be strictly positive real numbers.

###  6.2. Projective positivity in finite von Neumann algebras

Now fix a finite von Neumann algebra $\mathcal{M}$ with its unit $e$ and a
faithful, finite, normal trace $\tau$. We can always assume that
$\tau\left(e\right)>1$. The noncommutative $L^{p}$-space
$L^{p}\left(\mathcal{M},\tau\right)$ of $\mathcal{M}$ is given as the
completion of $\mathcal{M}$ with respect to the norm
$\left\|x\right\|_{p}=\tau\left(\left|x\right|^{p}\right)^{1/p}$,
$x\in\mathcal{M}$, $1\leq p\leq\infty$. As above
$L^{p}\left(\mathcal{M},\tau\right)$ is a unital $\ast$-vector space with
$\left\|e\right\|_{p}=\tau\left(e\right)^{1/p}>1$, and
$S_{\tau}=\left\\{y\in\mathcal{M}_{+}:\tau\left(y\right)=1\right\\}$ denotes
the convex set of all density elements in $\mathcal{M}$. The Banach space
$L^{1}\left(\mathcal{M},\tau\right)$ turns out to be the predual of
$\mathcal{M}$, which allows to identify $L^{1}\left(\mathcal{M},\tau\right)$
with a closed subspace of $\mathcal{M}^{\ast}$. In this case,
$S_{\ast}\left(\mathcal{M}_{+}\right)=S\left(\mathcal{M}_{+}\right)\cap
L^{1}\left(\mathcal{M},\tau\right)$ is the set of all normal states of the
cone $\mathcal{M}_{+}$, and
$S_{\tau}^{-}=S_{\ast}\left(\mathcal{M}_{+}\right)$ for the norm closure of
$S_{\tau}$ in $L^{1}\left(\mathcal{M},\tau\right)$ (see [6, Proposition 7.3]).
Moreover, we have the canonical continuous inclusions

$\mathcal{M=}L^{\infty}\left(\mathcal{M},\tau\right)\subseteq
L^{p}\left(\mathcal{M},\tau\right)\subseteq
L^{1}\left(\mathcal{M},\tau\right)$

in the descending order. Indeed, for $1\leq p<l<\infty$ and $r=l/p>1$ we have

$\left\|y\right\|_{p}^{p}=\tau\left(\left|y\right|^{l/r}e\right)\leq\tau\left(\left|y\right|^{rl/r}\right)^{1/r}\tau\left(e\right)^{1-1/r}=\left\|y\right\|_{l}^{p}\tau\left(e\right)^{1-p/l}\text{
or
}\left\|y\right\|_{p}\leq\left\|y\right\|_{l}\tau\left(e\right)^{\left(l-p\right)/lp}$

for all $y\in\mathcal{M}$. It means that
$L^{l}\left(\mathcal{M},\tau\right)\subseteq
L^{p}\left(\mathcal{M},\tau\right)$ is a bounded embedding with the norm
$\tau\left(e\right)^{\left(l-p\right)/lp}$ for all $1\leq p<l<\infty$. If
$l=\infty$ then
$\left\|y\right\|_{p}=\tau\left(\left|y\right|^{p}e\right)^{1/p}\leq\left\|y\right\|\tau\left(e\right)^{1/p}$
for all $y\in\mathcal{M}$. Thus

(6.4)
$\operatorname{ball}L^{l}\left(\mathcal{M},\tau\right)\subseteq\tau\left(e\right)^{\left(l-p\right)/lp}\operatorname{ball}L^{p}\left(\mathcal{M},\tau\right)$

for all $1\leq p<l\leq\infty$. As above $L^{2}\left(\mathcal{M},\tau\right)$
is a Hilbert $\ast$-space with the inner product
$\left(x,y\right)=\tau\left(xy^{\ast}\right)$, and every $x\in
L^{2}\left(\mathcal{M},\tau\right)$ admits a unique orthonormal expansion
$x=x_{0}+\left(x,e_{2}\right)e_{2}$, $\left(x_{0},e_{2}\right)=0$, where
$e_{2}=\tau\left(e\right)^{-1/2}e$. The cone similar to (6.2) is defined in
the following way

$\mathfrak{c}_{\tau}=\left\\{x\in
L^{2}\left(\mathcal{M},\tau\right)_{h}:\left\|x\right\|_{2}\leq\left(\dfrac{\tau\left(e\right)}{\tau\left(e\right)-1}\right)^{1/2}\left(x,e_{2}\right)\right\\}$

in the unital Hilbert $\ast$-space
$\left(L^{2}\left(\mathcal{M},\tau\right),e_{2}\right)$.

######  Proposition 6.1.

All $L^{p}\left(\mathcal{M},\tau\right)$ are unital $\ast$-normed spaces with
their common unit $e$. Thus $L^{p}\left(\mathcal{M},\tau\right)$ is a function
system equipped with the cone $L^{p}\left(\mathcal{M},\tau\right)_{+}$ given
by $\varepsilon=1>1/\left\|e\right\|_{p}$ such that
$L^{\infty}\left(\mathcal{M},\tau\right)_{+}=\mathcal{M}_{+}$,
$L^{2}\left(\mathcal{M},\tau\right)_{+}=\mathfrak{c}_{\tau}$, and
$\mathcal{M}_{\varepsilon}^{+}\subseteq
L^{1}\left(\mathcal{M},\tau\right)_{+}\cap\mathcal{M}$ for all
$\varepsilon\geq\tau\left(e\right)$. Moreover,

$S\left(L^{1}\left(\mathcal{M},\tau\right)_{+}\right)=\left\\{\left(1+r\right)z-rw:z,w\in
S_{\tau},zw=0,\left\|\left(1+r\right)z+rw\right\|\leq 1\right\\}.$

###### Proof.

We use the arguments similar to those from the proof of Theorem 6.1. Take
$x\in\mathcal{M}$ with its polar decomposition $x=u\left|x\right|$. Since
$\left|x^{\ast}\right|=u\left|x\right|u^{\ast}$, it follows that
$\left\|x^{\ast}\right\|_{p}=\tau\left(\left|x^{\ast}\right|^{p}\right)^{1/p}=\tau\left(u\left|x\right|^{p}u^{\ast}\right)^{1/p}=\tau\left(\left|x\right|^{p}\right)^{1/p}=\left\|x\right\|_{p}$,
which in turn implies that $L^{p}\left(\mathcal{M},\tau\right)$ is a
$\ast$-normed space. Prove that $e$ is a unit for
$L^{p}\left(\mathcal{M},\tau\right)$. For $p<\infty$ we have
$L^{p}\left(\mathcal{M},\tau\right)_{he}^{\ast}=\left\\{y\in
L^{q}\left(\mathcal{M},\tau\right)_{h}:\tau\left(y\right)=1\right\\}$ and
$\operatorname{ball}L^{p}\left(\mathcal{M},\tau\right)^{\ast}=\operatorname{ball}L^{q}\left(\mathcal{M},\tau\right)$
with the conjugate couple $\left(p,q\right)$. Moreover, based on the duality
between $L^{q}\left(\mathcal{M},\tau\right)$ and
$L^{q}\left(\mathcal{M},\tau\right)$, we deduce that $\tau$ defines a bounded
linear functional on $L^{q}\left(\mathcal{M},\tau\right)$ and

$1<\tau\left(e\right)^{1/p}=\tau\left(\tau\left(e\right)^{-1/q}e\right)\leq\vee\tau\left(\operatorname{ball}L^{q}\left(\mathcal{M},\tau\right)_{h}\right)\leq\vee\left|\tau\left(\operatorname{ball}L^{q}\left(\mathcal{M},\tau\right)\right)\right|\leq\tau\left(e\right)^{1/p},$

that is,
$1<\tau\left(e\right)^{1/p}=\vee\tau\left(\operatorname{ball}L^{q}\left(\mathcal{M},\tau\right)_{h}\right)$.
But
$\tau\left(\operatorname{ball}L^{q}\left(\mathcal{M},\tau\right)_{h}\right)\subseteq\mathbb{R}$
is a convex subset containing the origin. It follows that
$\tau\left(y\right)=1$ for a certain
$y\in\operatorname{ball}L^{q}\left(\mathcal{M},\tau\right)_{h}$, that is,
$S_{1}^{p}=L^{p}\left(\mathcal{M},\tau\right)_{he}^{\ast}\cap\operatorname{ball}L^{q}\left(\mathcal{M},\tau\right)\neq\varnothing$
whenever $p<\infty$. In particular,

$S_{1}^{1}=\mathcal{M}_{he}\cap\operatorname{ball}\mathcal{M}=\left\\{y\in\operatorname{ball}\mathcal{M}_{h}:\tau\left(y\right)=1\right\\}\neq\varnothing.$

For $p=\infty$ we have $\left\|e\right\|_{\infty}=1$ and
$S_{1}^{\infty}=\mathcal{M}_{he}^{\ast}\cap\operatorname{ball}\mathcal{M}^{\ast}$,
that is, every $\varphi\in S_{1}^{\infty}$ defines a unital contraction
$\varphi:\mathcal{M}\rightarrow\mathbb{C}$ with $1=\left\langle
e,\varphi\right\rangle\leq\vee\left|\left\langle\operatorname{ball}\mathcal{M},\varphi\right\rangle\right|=\left\|\varphi\right\|\leq
1$. By Lemma 7.2, $\varphi$ is a positive functional. Thus
$S_{1}^{\infty}=S\left(\mathcal{M}^{+}\right)$ is the state space of the cone
$\mathcal{M}^{+}$. Hence $e$ is a unit for every
$L^{p}\left(\mathcal{M},\tau\right)$, $1\leq p\leq\infty$.

Now put $L^{p}\left(\mathcal{M},\tau\right)_{+}$ to be the set of those $x\in
L^{p}\left(\mathcal{M},\tau\right)_{h}$ such that $\left\langle
x,S_{1}^{p}\right\rangle\geq 0$. If $p=\infty$ then
$L^{\infty}\left(\mathcal{M},\tau\right)_{+}=\left\\{x\in\mathcal{M}_{h}:\left\langle
x,S\left(\mathcal{M}_{+}\right)\right\rangle\geq 0\right\\}=\mathcal{M}_{+}$
(see [6, 7.1]). If $p<\infty$ then $\varepsilon=1>1/\left\|e\right\|_{p}$
defines the separated, closed, unital, cone
$L^{p}\left(\mathcal{M},\tau\right)_{+}$ in
$L^{p}\left(\mathcal{M},\tau\right)$,
$S\left(L^{p}\left(\mathcal{M},\tau\right)_{+}\right)=S_{1}^{p}$ and
$\left(L^{p}\left(\mathcal{M},\tau\right),L^{p}\left(\mathcal{M},\tau\right)_{+}\right)$
is a function system on $S_{1}^{p}$ thanks to Theorem 4.2.

If $p=2$ then the norm dual space $L^{2}\left(\mathcal{M},\tau\right)^{\ast}$
is canonically identified with the conjugate space
$\overline{L^{2}\left(\mathcal{M},\tau\right)}$, and for every $y\in
L^{2}\left(\mathcal{M},\tau\right)_{he}^{\ast}$ we have the orthogonal
expansion $y=y_{0}+\tau\left(e\right)^{-1}e$ with $y_{0}\perp e$ (or
$\tau\left(y\right)=0$). Then
$\left\|y\right\|_{2}^{2}=\left\|y_{0}\right\|_{2}^{2}+\tau\left(e\right)^{-1}$,
and $y\in S_{1}^{2}$ iff
$\left\|y_{0}\right\|_{2}\leq\left(\left(\tau\left(e\right)-1\right)/\tau\left(e\right)\right)^{1/2}$.
Hence

$\displaystyle S_{1}^{2}$
$\displaystyle=\left(\left(\tau\left(e\right)-1\right)/\tau\left(e\right)\right)^{1/2}\operatorname{ball}L^{2}\left(\mathcal{M},\tau\right)_{h}^{e}+\tau\left(e\right)^{-1}e$
$\displaystyle=\tau\left(e\right)^{-1/2}\left(\left(\tau\left(e\right)-1\right)^{1/2}\operatorname{ball}L^{2}\left(\mathcal{M},\tau\right)_{h}^{e}+e_{2}\right),$

where $L^{2}\left(\mathcal{M},\tau\right)_{h}^{e}=\left\\{w\in
L^{2}\left(\mathcal{M},\tau\right)_{h}:\tau\left(w\right)=0\right\\}$ is the
real vector subspace of $\overline{L^{2}\left(\mathcal{M},\tau\right)}$
orthogonal to $e$, and $e_{2}=\tau\left(e\right)^{-1/2}e$ is the unit vector
of $L^{2}\left(\mathcal{M},\tau\right)$. Thus $x\in
L^{2}\left(\mathcal{M},\tau\right)_{+}$ with its orthonormal expansion
$x=x_{0}+\left(x,e_{2}\right)e_{2}$ iff
$\left(x,\left(\tau\left(e\right)-1\right)^{1/2}\operatorname{ball}L^{2}\left(\mathcal{M},\tau\right)_{h}^{e}+e_{2}\right)\geq
0$, which is equivalent to

$\displaystyle\left(\tau\left(e\right)-1\right)^{1/2}\left\|x_{0}\right\|_{2}$
$\displaystyle=\left(\tau\left(e\right)-1\right)^{1/2}\left(\vee\left|\left(x_{0},\operatorname{ball}L^{2}\left(\mathcal{M},\tau\right)_{h}^{e}\right)\right|\right)\leq\left(x,e_{2}\right)$
$\displaystyle\text{or }\left\|x_{0}\right\|_{2}$
$\displaystyle\leq\left(\tau\left(e\right)-1\right)^{-1/2}\left(x,e_{2}\right).$

Hence $L^{2}\left(\mathcal{M},\tau\right)_{+}=\mathfrak{c}_{\tau}$. If we
choose $S_{\varepsilon}^{2}$ for
$\varepsilon=\left(2/\tau\left(e\right)\right)^{1/2}>1/\sqrt{\tau\left(e\right)}$
instead of $S_{1}^{2}$ then as in the proof of Theorem 4.2, the projective
positivity will exactly be reduced to the one from [7] of the unital Hilbert
$\ast$-space $\left(L^{2}\left(\mathcal{M},\tau\right),e_{2}\right)$.

Finally, let us prove the equality for the state space
$S\left(L^{1}\left(\mathcal{M},\tau\right)_{+}\right)$, which is a convex
subset of $\mathcal{M}_{h}$. Suppose that $y\in\mathcal{M}_{h}$ admits the
expansion $y=\left(1+r\right)z-rw$ with $z,w\in S_{\tau}$, $zw=0$ and
$\left\|z+r\left(z+w\right)\right\|\leq 1$. Then
$\tau\left(y\right)=\left(1+r\right)\tau\left(z\right)-r\tau\left(w\right)=1+r-r=1$,
$\left|y\right|=\left(1+r\right)z+rw$ and
$\left\|y\right\|=\left\|\left|y\right|\right\|=\left\|z+r\left(z+w\right)\right\|\leq
1$, that is, $y\in S_{1}^{1}$. Conversely, take $y\in S_{1}^{1}$. Using the
spectral decomposition of $y$ in $\mathcal{M}$, we deduce that
$y=y_{+}-y_{-}=\tau\left(y_{+}\right)z-\tau\left(y_{-}\right)w$ with $z,w\in
S_{\tau}$ and $\tau\left(y_{+}\right)-\tau\left(y_{-}\right)=1$. If
$r=\tau\left(y_{-}\right)$ then $y=\left(1+r\right)z-rw$ with $z,w\in
S_{\tau}$, $zw=0$ and
$\left\|z+r\left(z+w\right)\right\|=\left\|\left|y\right|\right\|=\left\|y\right\|\leq
1$. Hence

$S\left(L^{1}\left(\mathcal{M},\tau\right)_{+}\right)=\left\\{\left(1+r\right)z-rw:z,w\in
S_{\tau},zw=0,\left\|z+r\left(z+w\right)\right\|\leq 1\right\\}.$

Pick $y\in S\left(L^{1}\left(\mathcal{M},\tau\right)_{+}\right)$ with
$\left(1+r\right)z+rw=\left|y\right|\leq e$. Then
$1+2r=\left(1+r\right)\tau\left(z\right)+r\tau\left(w\right)=\tau\left(\left|y\right|\right)\leq\tau\left(e\right)$
or $r\leq\left(\tau\left(e\right)-1\right)/2$. Taking into account the
canonical inclusions $L^{1}\left(\mathcal{M},\tau\right)\subseteq
L^{1}\left(\mathcal{M},\tau\right)^{\ast\ast}=\mathcal{M}^{\ast}$ and
$S_{\tau}\subseteq S_{\ast}\left(\mathcal{M}_{+}\right)\subseteq
S\left(\mathcal{M}_{+}\right)\subseteq\mathcal{M}^{\ast}$, we deduce that

$\displaystyle S\left(L^{1}\left(\mathcal{M},\tau\right)_{+}\right)$
$\displaystyle\subseteq\left\\{\left(1+r\right)z-rw:z,w\in S_{\tau},0\leq
r\leq\left(\tau\left(e\right)-1\right)/2\right\\}$
$\displaystyle\subseteq\left\\{\left(1+r\right)\phi-r\psi:\phi,\psi\in
S\left(\mathcal{M}_{+}\right),0\leq
r\leq\left(\tau\left(e\right)-1\right)/2\right\\}$
$\displaystyle=S\left(\mathcal{M}_{\tau\left(e\right)}^{+}\right)\subseteq
S\left(\mathcal{M}_{\varepsilon}^{+}\right)$

for all $\varepsilon\geq\tau\left(e\right)$ by virtue of Lemma 3.1. In
particular,

$\displaystyle\mathcal{M}_{\varepsilon}^{+}$
$\displaystyle=\left\\{x\in\mathcal{M}_{h}:\left\langle
x,S\left(\mathcal{M}_{\varepsilon}^{+}\right)\right\rangle\geq
0\right\\}\subseteq\left\\{x\in\mathcal{M}_{h}:\left\langle
x,S\left(L^{1}\left(\mathcal{M},\tau\right)_{+}\right)\right\rangle\geq
0\right\\}$
$\displaystyle=\left\\{x\in\mathcal{M}_{h}:\tau\left(xS\left(L^{1}\left(\mathcal{M},\tau\right)_{+}\right)\right)\geq
0\right\\}=L^{1}\left(\mathcal{M},\tau\right)_{+}\cap\mathcal{M},$

that is, $\mathcal{M}_{\varepsilon}^{+}\subseteq
L^{1}\left(\mathcal{M},\tau\right)_{+}\cap\mathcal{M}$ for all
$\varepsilon\geq\tau\left(e\right)$. ∎

######  Remark 6.1.

Notice that the spectrum of every state $y=\left(1+r\right)z-rw\in
S\left(L^{1}\left(\mathcal{M},\tau\right)_{+}\right)$ can be described in term
of the density elements $z,w\in S_{\tau}$, $zw=0$ and
$\left\|z+r\left(z+w\right)\right\|\leq 1$. Since
$wz=\left(zw\right)^{\ast}=0=zw$, one can use the joint spectral mapping
theorem [20]. Namely,
$\sigma\left(y\right)=\left\\{\left(1+r\right)\lambda-r\mu:\left(\lambda,\mu\right)\in\sigma\left(z,w\right)\right\\}$,
where $\sigma\left(z,w\right)$ is the joint Taylor spectrum of commuting
elements $z$ and $w$, which is reduced to the joint point spectrum [3] in the
case of a matrix algebra $\mathcal{M=}M_{n}$ (see Theorem 6.1). But
$\left\\{0\right\\}=\sigma\left(zw\right)=\left\\{\lambda\mu:\left(\lambda,\mu\right)\in\sigma\left(z,w\right)\right\\}$,
therefore for every nonzero $\lambda\in\sigma\left(z\right)$ we have
$\left(\lambda,0\right)\in\sigma\left(z,w\right)$ and
$\left(1+r\right)\lambda=\left(1+r\right)\lambda-r0\in\sigma\left(y\right)$.
Similarly, $-r\mu\in\sigma\left(y\right)$ for every nonzero
$\mu\in\sigma\left(w\right)$. Thus
$\sigma\left(y\right)=\left(1+r\right)\sigma\left(z\right)\cup\left(-r\right)\sigma\left(w\right)$
and
$\sigma\left(\left|y\right|\right)=\left(1+r\right)\sigma\left(z\right)\cup
r\sigma\left(w\right)$.

######  Theorem 6.2.

Let $\mathcal{M}$ be a finite von Neumann algebra with its unit $e$ and
faithful, finite, normal trace $\tau$ such that $\tau\left(e\right)>1$. If
$1\leq p<l\leq\infty$ then
$L^{l}\left(\mathcal{M},\tau\right)_{\varepsilon}^{+}\subseteq
L^{p}\left(\mathcal{M},\tau\right)_{+}$ for all
$\varepsilon\geq\tau\left(e\right)^{\left(l-p\right)/lp}$. Thus

$\mathcal{M}_{\varepsilon}^{+}\subseteq
L^{p}\left(\mathcal{M},\tau\right)_{r}^{+}\subseteq
L^{1}\left(\mathcal{M},\tau\right)_{+}\text{,\quad}\varepsilon\tau\left(e\right)^{-1/p}\geq
r\geq\tau\left(e\right)^{1/q}.$

In particular, $\mathcal{M}_{\varepsilon}^{+}\subseteq
L^{p}\left(\mathcal{M},\tau\right)_{+}$ whenever
$\varepsilon\geq\tau\left(e\right)^{1/p}$.

###### Proof.

Fix $1\leq p<l\leq\infty$ and the related conjugate couples $\left(p,q\right)$
and $\left(l,m\right)$ with $1\leq m<q\leq\infty$. By Proposition 6.1, every
$L^{p}\left(\mathcal{M},\tau\right)$ is a function system equipped with the
cone $L^{p}\left(\mathcal{M},\tau\right)_{+}$ (responded to
$\varepsilon=1>1/\left\|e\right\|_{p}$), and using (6.4), we deduce that
$\operatorname{ball}L^{q}\left(\mathcal{M},\tau\right)\subseteq\tau\left(e\right)^{\left(q-m\right)/qm}\operatorname{ball}L^{m}\left(\mathcal{M},\tau\right)$,
$L^{l}\left(\mathcal{M},\tau\right)\subseteq
L^{p}\left(\mathcal{M},\tau\right)$ and

$S_{1}^{p}=L^{p}\left(\mathcal{M},\tau\right)_{he}^{\ast}\cap\operatorname{ball}L^{q}\left(\mathcal{M},\tau\right)\subseteq
L^{l}\left(\mathcal{M},\tau\right)_{he}^{\ast}\cap\varepsilon\operatorname{ball}L^{m}\left(\mathcal{M},\tau\right)=S_{\varepsilon}^{l}$

for all
$\varepsilon\geq\tau\left(e\right)^{\left(q-m\right)/qm}=\tau\left(e\right)^{\left(l-p\right)/lp}\geq
1$. It follows that
$L^{l}\left(\mathcal{M},\tau\right)_{\varepsilon}^{+}\subseteq
L^{p}\left(\mathcal{M},\tau\right)_{+}$ for all
$\varepsilon\geq\tau\left(e\right)^{\left(l-p\right)/lp}$. If $l=\infty$ then
$L^{\infty}\left(\mathcal{M},\tau\right)_{\varepsilon}^{+}=\mathcal{M}_{\varepsilon}^{+}$
by Proposition 6.1, and $\left(l-p\right)/lp=1/p$. In particular,
$\mathcal{M}_{\varepsilon}^{+}\subseteq
L^{p}\left(\mathcal{M},\tau\right)_{+}$ for all
$\varepsilon\geq\tau\left(e\right)^{1/p}$.

Finally, for every $\varepsilon\geq\tau\left(e\right)$ take $r$ with
$\varepsilon\tau\left(e\right)^{-1/p}\geq r\geq\tau\left(e\right)^{1/q}$.
Using again (6.4), we derive that
$\operatorname{ball}L^{\infty}\left(\mathcal{M},\tau\right)\subseteq\tau\left(e\right)^{1/q}\operatorname{ball}L^{q}\left(\mathcal{M},\tau\right)$,
$\operatorname{ball}L^{q}\left(\mathcal{M},\tau\right)\subseteq\tau\left(e\right)^{\left(q-1\right)/q}\operatorname{ball}L^{1}\left(\mathcal{M},\tau\right)$
and $L^{\infty}\left(\mathcal{M},\tau\right)\subseteq
L^{p}\left(\mathcal{M},\tau\right)\subseteq
L^{1}\left(\mathcal{M},\tau\right)$. It follows that

$\displaystyle S_{1}^{1}$
$\displaystyle=L^{1}\left(\mathcal{M},\tau\right)_{he}^{\ast}\cap\operatorname{ball}L^{\infty}\left(\mathcal{M},\tau\right)\subseteq
L^{p}\left(\mathcal{M},\tau\right)_{he}^{\ast}\cap
r\operatorname{ball}L^{q}\left(\mathcal{M},\tau\right)=S_{r}^{p}$
$\displaystyle\subseteq
L^{\infty}\left(\mathcal{M},\tau\right)_{he}^{\ast}\cap
r\tau\left(e\right)^{1/p}\operatorname{ball}L^{1}\left(\mathcal{M},\tau\right)\subseteq\mathcal{M}_{he}^{\ast}\cap\varepsilon\operatorname{ball}L^{1}\left(\mathcal{M},\tau\right)$
$\displaystyle\subseteq\mathcal{M}_{he}^{\ast}\cap\varepsilon\operatorname{ball}\mathcal{M}^{\ast}=S_{\varepsilon}^{\infty},$

which in turn implies that $\mathcal{M}_{\varepsilon}^{+}\subseteq
L^{p}\left(\mathcal{M},\tau\right)_{r}^{+}\subseteq
L^{1}\left(\mathcal{M},\tau\right)_{+}$. ∎

###  6.3. The concluding remarks

1\. As we have seen above in Section 5 the projective $L^{p}$-cones and the
original cones of the (commutative) $L^{p}$-spaces are not comparable apart
from the finite dimensional case. The same happens in the case of a finite von
Neumann algebra. Note that the inclusion
$L^{1}\left(\mathcal{M},\tau\right)_{+}\cap\mathcal{M\subseteq M}_{+}$ may not
be true (see Theorem 5.1). The following concrete example presents an
interest. Suppose that $\mathcal{M=}L^{\infty}\left[0,2\right]$ is the abelian
von Neumann algebra of Lebesgue measure $dt$ on the compact interval
$\left[0,2\right]$ equipped with the trace
$\tau\left(x\right)=\int_{0}^{2}x\left(t\right)dt$. Notice that $e=1\in
L^{1}\left[0,2\right]$, $\left\|e\right\|_{1}=\int dt=2>1$, and

$S_{\varepsilon}^{1}=\left\\{y\in\varepsilon\operatorname{ball}L^{\infty}\left[0,2\right]_{h}:\int
ydt=1\right\\}$

for $\varepsilon\geq 1$. For every $0<r\leq-2^{-1}+\sqrt{1/\varepsilon+1/4}$
define $x=-r\left[0,r\right]+\left[r,2\right]\in L^{1}\left[0,2\right]$. If
$y\in S_{\varepsilon}^{1}$ then $\left|y\left(t\right)\right|\leq\varepsilon$
for almost all $t$, and

$\displaystyle\tau\left(xy\right)$
$\displaystyle=-r\int_{0}^{r}ydt+\int_{r}^{2}ydt=-r\int_{0}^{r}ydt+1-\int_{0}^{r}ydt=1-\left(1+r\right)\int_{0}^{r}ydt$
$\displaystyle\geq 1-\left(1+r\right)\int_{0}^{r}\varepsilon dt=1-\varepsilon
r-\varepsilon r^{2}\geq 0.$

Thus $x\in L^{1}\left[0,2\right]_{+}\cap L^{\infty}\left[0,2\right]$ whereas
$x\notin L^{\infty}\left[0,2\right]_{+}$. Thus an essentially bounded
$L^{1}\left[0,2\right]_{+}$-positive function could take negative values on a
set of positive measure.

2\. The projective positivity allows us to define the operator system sums and
the projective tensor product of the operator systems.

##  7\. Appendix: Functionals on operator systems

In this section we support the paper with a supplementary material on operator
system.

###  7.1. Positive functionals on operator systems

Let $V$ be an operator system $V$ on a Hilbert space $H$. Put
$V_{h}^{2}=V_{h}\times V_{h}$ with its subset $V_{+}^{2}=V_{+}\times V_{+}$.
For $a\in V_{h}^{2}$, $b\in V$ and $\varepsilon>0$, we put
$a+\varepsilon=\left(a_{1}+\varepsilon e,a_{2}+\varepsilon e\right)\in
V_{h}^{2}$ and

$x\left(a,b\right)=\left[\begin{array}[c]{cc}a_{1}&b\\\
b^{\ast}&a_{2}\end{array}\right]\in
M_{2}\left(V\right)_{h}\subseteq\mathcal{B}\left(H^{2}\right).$

If $a\in V_{+}^{2}$, we use the notation
$x_{a,\varepsilon,b}=\left(a_{1}+\varepsilon
e\right)^{-1/2}b\left(a_{2}+\varepsilon
e\right)^{-1/2}\in\mathcal{B}\left(H\right)$. The following Paulsen’s trick is
well known.

######  Lemma 7.1.

The matrix $x\left(a,b\right)$ is positive iff $a\in V_{+}^{2}$ and
$x_{a,\varepsilon,b}\in\operatorname{ball}\mathcal{B}\left(H\right)$ for all
$\varepsilon>0$.

###### Proof.

If $x\left(a,b\right)\in M_{2}\left(V\right)_{+}$ then
$a_{i}=\epsilon_{i}x\left(a,b\right)\epsilon_{i}^{\ast}\in V_{+}$ and
$a+\varepsilon\in V_{+}^{2}$ for all $\varepsilon>0$, where
$\epsilon_{1}=\left[\begin{array}[c]{cc}1&0\end{array}\right]$,
$\epsilon_{2}=\left[\begin{array}[c]{cc}0&1\end{array}\right]$. In particular,
$x\left(\left(a+\varepsilon\right)^{-1/2},0\right)=\left(a_{1}+\varepsilon
e\right)^{-1/2}\oplus\left(a_{2}+\varepsilon e\right)^{-1/2}\geq 0$ and
$x\left(a+\varepsilon,b\right)=x\left(a,b\right)+\varepsilon\left(e\oplus
e\right)\geq 0$. It follows that

$\left[\begin{array}[c]{cc}e&x_{a,\varepsilon,b}\\\
x_{a,\varepsilon,b}^{\ast}&e\end{array}\right]=x\left(\left(a+\varepsilon\right)^{-1/2},0\right)x\left(a+\varepsilon,b\right)x\left(\left(a+\varepsilon\right)^{-1/2},0\right)\geq
0,$

which means that $\left\|x_{a,\varepsilon,b}\right\|\leq 1$ (see [8, 1.3.2]).
Conversely, if $a\in\left(V_{+}\right)^{2}$ and
$x_{a,\varepsilon,b}\in\operatorname{ball}\mathcal{B}\left(H\right)$, then as
above

$x\left(a+\varepsilon,b\right)=x\left(\left(a+\varepsilon\right)^{1/2},0\right)\left[\begin{array}[c]{cc}e&x_{a,\varepsilon,b}\\\
x_{a,\varepsilon,b}^{\ast}&e\end{array}\right]x\left(\left(a+\varepsilon\right)^{1/2},0\right)\geq
0$

for all $\varepsilon>0$. By the continuity argument, we deduce that
$x\left(a,b\right)\geq 0$. ∎

######  Corollary 7.1.

Let $\alpha=\left(\alpha_{1},\alpha_{2}\right)\in\mathbb{R}^{2}$ and
$\beta\in\mathbb{C}$. Then

$x\left(\alpha,\beta\right)\in
M_{2,+}\Longleftrightarrow\alpha\in\mathbb{R}_{+}^{2}\text{ and
}\left|\beta\right|\leq\alpha_{1}^{1/2}\alpha_{2}^{1/2}.$