# Generalized Jordan Semitriple Maps on Hilbert Space Effect Algebras

Authors: Qing Yuan; Kan He
Journal: Advances in Mathematical Physics
 (2014)

---

## Abstract

Letℰ
(
H
) be the Hilbert space effect algebra on a Hilbert space H with dim
⁡
H
≥
3, α
,
β two positive numbers with 2
α
+
β
≠
1 and Φ
:
ℰ
(
H
)
→
ℰ
(
H
) a bijective map. We show that if Φ
(
A
α
B
β
A
α
)
=
Φ
(
A
)
α
Φ
(
B
)
β
Φ
(
A
)
α holds for all A
,
B
∈
ℰ
(
H
), then there exists a unitary or an antiunitary operator U on H such that Φ
(
A
)
=
U
A
U
* for every A
∈
ℰ
(
H
).


---

## Body

## 1. Introduction and Notations

LetH be a Hilbert space. Denote by ℬ
(
H
) the algebra of all bounded linear operators acting on H. The operator interval in ℬ
(
H
): ℰ
(
H
)
=
{
T
∈
ℬ
(
H
)
∣
0
≤
T
≤
I
}, where I is the identity, is called Hilbert space effect algebra on H, and the elements in ℰ
(
H
) are called effects. The concept of effect algebras plays a fundamental role in mathematical description of quantum measurement, and the range of general quantum observables (POV-measurement) consists of effects [1].The Hilbert space effect algebraℰ
(
H
) can be equipped with some algebraic operations. Denote the space of all projections on H by 𝒫
(
H
). Clearly 𝒫
(
H
)
⊆
ℰ
(
H
). Let A
,
B
∈
ℰ
(
H
). ℰ
(
H
) is a partial ordered set with the order “≤”:
A
≤
B if B
-
A
≥
0. Obviously, the sum A
+
B may not be in ℰ
(
H
). In the case that A
+
B
∈
ℰ
(
H
), we can define an operation of conditional addition in ℰ
(
H
)
:
A
⊕
B
=
A
+
B. Particularly, since P
+
Q
∈
𝒫
(
H
)
⇔
P
Q
=
0, this gives an orthogonal addition operation in 𝒫
(
H
). Generally speaking, the product A
B may not be in ℰ
(
H
) either. However, the Jordan semitriple product A
∘
B
=
A
B
A is a well-defined operation in ℰ
(
H
) since A
B
A
∈
ℰ
(
H
) for all A
,
B
∈
ℰ
(
H
). More generally, for any fixed positive real numbers α
,
β, the generalized Jordan triple product ∘
α
,
β induced by (
α
,
β
) is well defined, too, in ℰ
(
H
) and here A
∘
α
,
β
B
=
A
α
B
β
A
α.Recently many mathematicians pay their attention to the problem of characterizing certain maps on Hilbert space effect algebras or other quantum structures [2–8]. Let H be a Hilbert space with dim
⁡
H
≥
3 and Φ
:
ℰ
(
H
)
→
ℰ
(
H
) a bijective map. It seems that [5] is the first paper discussing the problem of characterizing the Jordan semitriple maps on Hilbert effect algebras. In [5] Molnár showed that if Φ
(
A
B
A
)
=
Φ
(
A
)
Φ
(
B
)
Φ
(
A
) for all A
,
B
∈
ℰ
(
H
), then there exists a unitary or an antiunitary operator U on H such that Φ
(
A
)
=
U
A
U
* for every A
∈
ℰ
(
H
); that is, Φ is implemented by a unitary or an antiunitary operator on H. Molnár and Šemrl proved in [7] that if Φ satisfies
(1)
Ψ
(
A
∘
1
/
2,1
B
)
=
Ψ
(
A
)
∘
1
/
2,1
Ψ
(
B
)
∀
A
,
B
∈
ℰ
(
H
)
,
where Ψ denotes Φ and Φ
-
1, then there is a positive number p and a unitary or an antiunitary operator U such that Φ
(
A
)
=
U
A
p
U
* for every A
∈
ℰ
(
H
) (we will show that p
=
1 in fact). The generalized Jordan semitriple product A
∘
1
/
2,1
B
=
A
1
/
2
B
A
1
/
2 is called sequence product. Kim showed in [4] that if H is separable and Φ satisfies
(2)
A
B
=
B
A
⟹
Φ
(
A
B
)
=
Φ
(
A
)
Φ
(
B
)
=
Φ
(
B
)
Φ
(
A
)
,
A
B
=
B
A
,
A
+
B
∈
ℰ
(
H
)
⟹
Φ
(
A
+
B
)
=
Φ
(
A
)
+
Φ
(
B
)
for A
,
B
∈
ℰ
(
H
), then Φ is also implemented by a unitary or an antiunitary operator on H. Our purpose in this paper is to generalize the results of [5, 7] to generalized Jordan semitriple maps, that is, the maps Φ satisfying
(3)
Φ
(
A
∘
α
,
β
B
)
=
Φ
(
A
)
∘
α
,
β
Φ
(
B
)
.
Let H be a Hilbert space with dim
⁡
H
≥
3, α
,
β two positive real numbers with 2
α
+
β
≠
1, and Φ
:
ℰ
(
H
)
→
ℰ
(
H
) a bijective map satisfying
(4)
Φ
(
A
α
B
β
A
α
)
=
Φ
(
A
)
α
Φ
(
B
)
β
Φ
(
A
)
α
for all A, B
∈
ℰ
(
H
). We first show that Φ is orthoadditive on 𝒫
(
H
), that is, for any orthogonal projections P, Q, Φ
(
P
+
Q
)
=
Φ
(
P
)
+
Φ
(
Q
). As an application, we prove that all bijective generalized Jordan semitriple maps on Hilbert space effect algebras, that is, the maps satisfying (4), are implemented by a unitary or an antiunitary operator on H.


## 2. Orthoadditivity of Generalized Jordan Semitriple Maps

Additivity of multiplicative maps is studied by many authors. In this section, we show that every generalized Jordan semitriple map on the effect algebraℰ
(
H
) is orthoadditive on 𝒫
(
H
). To do this we need a description which gives a description of the zero product preservers on 𝒫
(
H
). Recall that a map Ψ preserves zero products (in both directions) if Ψ
(
A
)
Ψ
(
B
)
=
0, whenever (if and only if) A
B
=
0.Lemma 1.
LetH be a real or complex Hilbert space with dim
⁡
H
≥
3. Assume that Φ
:
𝒫
(
H
)
→
𝒫
(
H
) is a bijective map preserving zero products of operators in both directions. (1)
IfH is real, then there is a unitary operator U on H such that Φ
(
P
)
=
U
P
U
* for all rank one P
∈
𝒫
(
H
).
(2)
IfH is complex, then there is a unitary operator or an antiunitary operator U on H such that Φ
(
P
)
=
U
P
U
* for all rank one P
∈
𝒫
(
H
).Proof.
Denote by𝒫
1
(
H
) the set of all projections of rank one. For any P
∈
𝒫
(
H
), write {
P
}
⊥
=
{
T
∈
𝒫
(
H
)
:
T
≠
0
,
T
P
=
0
}.
Claim
1. For any P
∈
𝒫
(
H
), {
P
}
⊥
=
∅ if and only if P
=
I; {
P
}
⊥ is singleton; that is, it contains exactly one element, if and only if I
-
P is of rank one.
IfP
=
I
⇒
{
P
}
⊥
=
∅ is obvious, assume that P
≠
I and then I
-
P
≠
0 and I
-
P
∈
{
P
}
⊥.
If{
P
}
⊥ contains only one element, then P
≠
I and {
P
}
⊥
=
{
I
-
P
}. If I
-
P is not of rank one, for any unit vector x
∈
ran
(
I
-
P
), one has x
⊗
x
∈
{
P
}
⊥ but x
⊗
x
≠
I
-
P, a contradiction. So, rank
⁡
(
I
-
P
)
=
1. The converse is obvious.
Claim
2. Φ preserves rank one projections in both directions.
For any rank one projectionx
⊗
x, let P
=
I
-
x
⊗
x; then {
x
⊗
x
}
=
{
P
}
⊥. By the property of Φ that it preserves zero products in both directions, one has Φ
(
{
x
⊗
x
}
)
=
Φ
(
{
P
}
⊥
)
=
{
Φ
(
P
)
}
⊥. Thus {
Φ
(
P
)
}
⊥ is singleton, which forces that Φ
(
x
⊗
x
) is of rank one by Claim 1, and vice versa.
Claim
3. There exists a unitary or antiunitary operator U on H such that
(5)
Φ
(
P
)
=
U
P
U
*
holds for all P
∈
𝒫
1
(
H
). U is unitary in real case.
By Claim 3, for every unit vectorx
∈
H, there is a unit vector y
x
∈
H such that Φ
(
x
⊗
x
)
=
y
x
⊗
y
x.
Letϕ
:
P
H
→
P
H be the map on the projective space P
H defined by ϕ
(
[
x
]
)
=
[
y
x
], where [
x
] stands for the 1-dimensional linear subspace spanned by x. Then it is clear that ϕ is bijective since Φ preserves rank one projection in both directions. For any linearly independent unit vectors x
,
u and any scalars γ
,
δ, let v
=
∥
γ
x
+
δ
u
∥
-
1
(
γ
x
+
δ
u
). For any unit y
∈
H, there is a unit vector w such that Φ
(
w
⊗
w
)
=
y
⊗
y. If 〈
y
,
y
x
〉
=
〈
y
,
y
u
〉
=
0, then (
y
⊗
y
)
(
y
x
⊗
y
x
)
=
(
y
⊗
y
)
(
y
x
⊗
y
x
)
=
0. This implies that 〈
w
,
x
〉
=
〈
w
,
u
〉
=
0 and consequently 〈
w
,
v
〉
=
0. Hence 〈
y
,
y
v
〉
=
0. It follows from the arbitrariness of y that y
v
∈
[
y
x
]
+
[
y
u
]; that is, [
v
]
∈
[
x
]
+
[
u
]
⇒
ϕ
(
v
)
∈
ϕ
(
x
)
+
ϕ
(
u
). Now we can use the fundamental theorem of projective geometry to get a semilinear bijection A
:
H
→
H such that ϕ
(
[
x
]
)
=
[
A
x
]. Therefore, for any unit x
∈
H, Φ
(
x
⊗
x
)
=
∥
A
x
∥
-
2
A
x
⊗
A
x. It is obvious that A preserves orthogonality in both directions; that is, 〈
x
,
u
〉
=
0
⇔
〈
A
x
,
A
u
〉
=
0. Hence A
=
c
U for some unitary or antiunitary operator U on H. In the case that H is real, U is unitary. Clearly, |
c
|
=
1, so Φ
(
x
⊗
x
)
=
U
(
x
⊗
x
)
U
* for all rank one projection x
⊗
x.Theorem 2.
LetH be a Hilbert space with dim
⁡
H
≥
3, α
,
β positive numbers with 2
α
+
β
≠
1, and Φ
:
ℰ
(
H
)
→
ℰ
(
H
) a bijective map. If Φ satisfies
(6)
Φ
(
A
α
B
β
A
α
)
=
Φ
(
A
)
α
Φ
(
B
)
β
Φ
(
A
)
α
,
for all A
,
B
∈
ℰ
(
H
), then for any projections P
,
Q, P
⊥
Q implies that
(7)
Φ
(
P
+
Q
)
=
Φ
(
P
)
+
Φ
(
Q
)
.Proof.
ForP
∈
𝒫
(
H
), we have
(8)
Φ
(
P
)
γ
=
Φ
(
P
)
,
where γ
>
0. Note Φ
(
P
)
≥
0. For γ
≠
1, the spectral solution theorem implies that Φ
(
P
)
2
=
Φ
(
P
). By considering Φ
-
1, we see that Φ preserves projections in both directions.
For anyA
∈
ℰ
(
H
),
(9)
Φ
(
I
)
Φ
(
A
)
=
Φ
(
I
)
Φ
(
I
A
I
)
=
Φ
(
I
)
Φ
(
I
α
(
A
1
/
β
)
β
I
α
)
=
Φ
(
I
)
Φ
(
I
)
α
Φ
(
A
1
/
β
)
β
Φ
(
I
)
α
=
Φ
(
I
)
Φ
(
A
1
/
β
)
β
Φ
(
I
)
.
Similarly, we have
(10)
Φ
(
I
)
Φ
(
A
)
=
Φ
(
I
)
Φ
(
A
1
/
β
)
β
Φ
(
I
)
.
So
(11)
Φ
(
A
)
Φ
(
I
)
=
Φ
(
I
)
Φ
(
A
)
holds for all A
∈
ℰ
(
H
). By the surjectivity of Φ and Φ
(
I
)
=
Φ
(
I
)
2, we must have Φ
(
I
)
=
I. Since Φ is bijective, there is A
0
∈
ℰ
(
H
) such that Φ
(
A
0
)
=
0. Thus
(12)
Φ
(
0
)
=
Φ
(
A
0
α
0
A
0
α
)
=
Φ
(
A
0
)
α
Φ
(
0
)
β
Φ
(
A
0
)
α
=
0
;
that is, Φ
(
0
)
=
0.
Now we show thatΦ preserves zero product in both directions on 𝒫
(
H
), that it preserves orthogonality in both directions. For any P
,
Q in 𝒫
(
H
),
(13)
P
Q
=
0
⟺
P
β
/
2
Q
α
=
0
⟺
(
P
β
/
2
Q
α
)
*
P
β
/
2
Q
α
=
0
⟺
Q
α
P
β
Q
α
=
0
.
Then
(14)
P
Q
=
0
⟺
Q
α
P
β
Q
α
=
0
⟺
Φ
(
Q
α
P
β
Q
α
)
=
0
⟺
Φ
(
Q
)
α
Φ
(
P
)
β
Φ
(
Q
)
α
=
0
⟺
Φ
(
P
)
Φ
(
Q
)
=
0
.
Therefore, by Lemma 1, there exists a unitary or an antiunitary operator U on H such that Φ
(
P
)
=
U
P
U
* holds for all rank one projections P. Without loss of generality we may assume that Φ
(
P
)
=
P for every rank one projection P.
For anyP
,
Q
∈
𝒫
(
H
), if P
⊥
Q, let R
=
Φ
(
P
+
Q
)
∈
𝒫
(
H
), S
=
Φ
(
P
)
∈
𝒫
(
H
), and T
=
Φ
(
Q
)
∈
𝒫
(
H
). We will show that R
=
S
+
T. To do this, it is enough to show that ran
(
R
)
=
ran
(
S
+
T
). Note that for any x in H with norm 1,
(15)
x
⊗
x
R
x
⊗
x
=
(
x
⊗
x
)
α
R
β
(
x
⊗
x
)
α
=
Φ
(
(
x
⊗
x
)
α
(
P
+
Q
)
β
(
x
⊗
x
)
α
)
=
Φ
(
x
⊗
x
(
P
+
Q
)
x
⊗
x
)
,
x
⊗
x
⊥
S
⟺
x
⊗
x
⊥
P
,
x
⊗
x
⊥
T
⟺
x
⊗
x
⊥
Q
.
Thus we have
(16)
x
⊗
x
R
x
⊗
x
=
0
⟺
x
⊗
x
(
P
+
Q
)
x
⊗
x
=
0
⟺
x
⊗
x
(
S
+
T
)
x
⊗
x
=
0
.
This forces ran
(
R
)
=
ran
(
S
+
T
). So Φ
(
P
+
Q
)
=
Φ
(
P
)
+
Φ
(
Q
) whenever P
⊥
Q.


## 3. Characterization of Generalized Jordan Semitriple Maps

The following is our main result.Theorem 3.
LetH be a Hilbert space with dim
⁡
H
≥
3. Assume that α
,
β are positive numbers with 2
α
+
β
≠
1 and Φ
:
ℰ
(
H
)
→
ℰ
(
H
) a bijective map. If
(17)
Φ
(
A
α
B
β
A
α
)
=
Φ
(
A
)
α
Φ
(
B
)
β
Φ
(
A
)
α
,
for all A
,
B
∈
ℰ
(
H
), then there exists a unitary or an antiunitary operator U on H such that Φ
(
A
)
=
U
A
U
* for all A
∈
ℰ
(
H
).The following corollary gives a better form of sequence product multiplicative maps, comparing with [7].Corollary 4.
LetH be a Hilbert space with dim
⁡
H
≥
3 and Φ
:
ℰ
(
H
)
→
ℰ
(
H
) a bijective map. If
(18)
Φ
(
A
B
A
)
=
Φ
(
A
)
Φ
(
B
)
Φ
(
A
)
for all A
,
B
∈
ℰ
(
H
), then there exists an unitary or an antiunitary operator U on H such that Φ
(
A
)
=
U
A
U
* for every A
∈
ℰ
(
H
).Proof of Theorem3.
By Theorem2 we know that Φ preserves projections in both directions, Φ
(
I
)
=
I, Φ
(
0
)
=
0, and Φ is orthoadditive on 𝒫
(
H
).
Furthermore,Φ preserves the order of projections in both directions. To see this, let P
,
Q
∈
𝒫
(
H
). It is clear that
(19)
P
≤
Q
⟺
P
Q
=
Q
P
=
P
⟺
Q
P
Q
=
P
,
P
Q
P
=
P
⟺
Q
α
P
β
Q
α
=
P
β
=
P
,
P
α
Q
β
P
α
=
P
α
=
P
.
So, P
≤
Q implies that
(20)
Φ
(
P
)
=
Φ
(
Q
α
P
β
Q
α
)
=
Φ
(
Q
)
α
Φ
(
P
)
β
Φ
(
Q
)
α
,
Φ
(
P
)
=
Φ
(
P
α
Q
β
P
α
)
=
Φ
(
P
)
α
Φ
(
Q
)
β
Φ
(
P
)
α
.
Consequently, Φ
(
P
)
≤
Φ
(
Q
). Similarly, using Φ
-
1, one can check that Φ
(
P
)
≤
Φ
(
Q
) implies that P
≤
Q.
So, by Theorem2 and [9], the restriction of Φ on 𝒫
(
H
) can be extended to a bounded linear map Ψ
:
ℬ
(
H
)
→
ℬ
(
H
). Note that Ψ preserves projections in both directions; it must be a Jordan star-isomorphism. Since every Jordan star-automorphism of ℬ
(
H
) is either a star-automorphism or star-antiautomorphism, there exists a unitary operator U or an antiunitary operator V such that
(21)
Ψ
(
A
)
=
U
A
U
*
∀
A
∈
ℬ
(
H
)
or
(22)
Ψ
(
A
)
=
V
A
*
V
*
∀
A
∈
ℬ
(
H
)
.
It follows that there is a unitary or an antiunitary operator U such that Φ
(
P
)
=
U
P
U
* for all P
∈
𝒫
(
H
).
Without loss of generality, we can assume thatΦ
(
P
)
=
P holds for every projection P.
Next we show thatΦ
(
λ
P
)
=
λ
P holds for every λ
∈
[
0,1
] and P
∈
𝒫
(
H
).
By the definition ofΦ that we have for any λ
∈
[
0,1
] and any rank one projection P, we have
(23)
Φ
(
λ
P
)
=
Φ
(
P
α
(
λ
1
/
β
P
)
β
P
α
)
=
P
Φ
(
λ
1
/
β
P
)
β
P
.
So there is a f
P
(
λ
)
∈
[
0,1
] such that Φ
(
λ
P
)
=
f
P
(
λ
)
P. Denote f
P by f; that is, f
(
λ
)
=
f
P
(
λ
). f
:
[
0,1
]
→
[
0,1
] is a function such that Φ
(
λ
P
)
=
f
(
λ
)
P. It follows that for every μ
∈
[
0,1
]
(24)
Φ
(
λ
μ
P
)
=
f
(
λ
μ
)
P
,
Φ
(
λ
μ
P
)
=
Φ
(
λ
(
1
/
2
α
)
α
P
μ
(
1
/
β
)
β
P
λ
(
1
/
2
α
)
α
P
)
=
Φ
(
λ
1
/
2
α
P
)
α
Φ
(
μ
1
/
β
P
)
β
Φ
(
λ
1
/
2
α
P
)
α
=
f
(
λ
1
/
2
α
)
2
α
f
(
μ
1
/
β
)
β
P
.
On the other hand,
(25)
f
(
λ
)
P
=
Φ
(
λ
P
)
=
Φ
(
λ
(
1
/
2
α
)
α
P
P
P
λ
(
1
/
2
α
)
α
)
=
f
(
λ
1
/
2
α
)
2
α
P
,
f
(
μ
)
P
=
Φ
(
μ
P
)
=
Φ
(
I
μ
(
1
/
β
)
β
P
I
)
=
f
(
μ
1
/
β
)
β
P
.
Thus f
(
λ
1
/
2
α
)
2
α
=
f
(
λ
) and f
(
μ
1
/
β
)
β
=
f
(
μ
). Hence, we have f
(
λ
μ
)
=
f
(
λ
)
f
(
μ
) holds for all λ
,
μ
∈
[
0,1
]; that is, f is multiplicative.
Obviously, by the surjectivity ofΦ, f
(
[
0,1
]
)
=
[
0,1
]. So f is continuous on [
0,1
]. It is well known that every multiplicative continuous bijection g
:
[
0,1
]
→
[
0,1
] is of the form g
(
λ
)
=
λ
ρ (ρ
>
0). Hence there is a ρ
P
>
0 such that f
(
λ
)
=
λ
ρ
P for every λ
∈
[
0,1
]. It follows that
(26)
Φ
(
λ
P
)
=
λ
ρ
P
P
for every rank one projection P.
We will show thatρ
P is independent of P. For any rank one projections P
,
Q and λ
,
μ
∈
[
0,1
], we have
(27)
Φ
(
λ
P
μ
2
Q
λ
P
)
=
Φ
(
λ
(
1
/
α
)
α
P
μ
(
2
/
β
)
β
Q
λ
(
1
/
α
)
α
P
)
=
f
(
λ
1
/
α
)
2
α
f
(
μ
2
/
β
)
β
P
Q
P
=
λ
2
ρ
P
μ
2
ρ
Q
P
Q
P
,
Φ
(
μ
P
λ
2
Q
μ
P
)
=
Φ
(
μ
(
1
/
α
)
α
P
λ
(
2
/
β
)
β
Q
μ
(
1
/
α
)
α
P
)
=
μ
2
ρ
P
λ
2
ρ
Q
P
Q
P
.
If P
Q
≠
0, we get
(28)
λ
2
ρ
P
μ
2
ρ
Q
=
μ
2
ρ
P
λ
2
ρ
Q
.
Letting μ
=
1 and comparing two sides of the above equation yield λ
ρ
P
=
λ
ρ
Q. So ρ
P
=
ρ
Q. If P
Q
=
0, pick a rank one projection R so that P
R
P
≠
0 and Q
R
Q
≠
0. Then Φ
(
λ
R
)
=
λ
ρ
R
R. By the fact we just checked, we get ρ
P
=
ρ
R
=
ρ
Q. It follows that there is a common ρ
≥
0 such that
(29)
Φ
(
λ
P
)
=
λ
ρ
P
holds for all rank one projection P and λ
∈
[
0,1
].
We claim thatρ
=
1. To see this, it is enough to check that f
(
λ
+
μ
)
=
f
(
λ
)
+
f
(
μ
) whenever λ
+
μ
≤
1, where f
(
λ
)
=
λ
ρ. Pick unit vectors x
,
y with 〈
x
,
y
〉
=
0 and let P
=
x
⊗
x, Q
=
y
⊗
y, and A
=
(
λ
1
/
2
x
+
μ
1
/
2
y
)
⊗
(
λ
1
/
2
x
+
μ
1
/
2
y
). Then
(30)
A
P
A
=
(
λ
1
/
2
x
+
μ
1
/
2
y
)
⊗
(
λ
1
/
2
x
+
μ
1
/
2
y
)
x
⊗
x
(
λ
1
/
2
x
+
μ
1
/
2
y
)
⊗
(
λ
1
/
2
x
+
μ
1
/
2
y
)
=
λ
1
/
2
(
λ
1
/
2
x
+
μ
1
/
2
y
)
⊗
x
(
λ
1
/
2
x
+
μ
1
/
2
y
)
⊗
(
λ
1
/
2
x
+
μ
1
/
2
y
)
=
λ
(
λ
1
/
2
x
+
μ
1
/
2
y
)
⊗
(
λ
1
/
2
x
+
μ
1
/
2
y
)
=
λ
A
.
Similarly, we have
(31)
A
Q
A
=
μ
A
,
A
(
P
+
Q
)
A
=
A
2
=
(
λ
+
μ
)
A
.
Since Φ is orthoadditive on 𝒫
(
H
), we have
(32)
Φ
(
A
(
P
+
Q
)
A
)
=
Φ
(
A
(
1
/
α
)
α
(
P
+
Q
)
(
1
/
β
)
β
A
(
1
/
α
)
α
)
=
Φ
(
A
1
/
α
)
α
Φ
(
(
P
+
Q
)
1
/
β
)
β
Φ
(
A
1
/
α
)
α
=
Φ
(
A
1
/
α
)
α
Φ
(
P
+
Q
)
Φ
(
A
1
/
α
)
α
=
Φ
(
A
1
/
α
)
α
(
Φ
(
P
)
+
Φ
(
Q
)
)
Φ
(
A
1
/
α
)
α
=
Φ
(
A
1
/
α
)
α
Φ
(
P
1
/
β
)
β
Φ
(
A
1
/
α
)
α
+
Φ
(
A
1
/
α
)
α
Φ
(
Q
1
/
β
)
β
Φ
(
A
1
/
α
)
α
=
Φ
(
A
P
A
)
+
Φ
(
A
Q
A
)
.
Note that A
0
=
A
/
∥
A
∥ is a rank one projection, and for any δ
∈
[
0,1
], we have
(33)
Φ
(
δ
A
)
=
Φ
(
δ
∥
A
∥
A
0
)
=
f
(
δ
∥
A
∥
)
A
0
=
f
(
δ
)
f
(
∥
A
∥
)
Φ
(
A
0
)
=
f
(
δ
)
Φ
(
A
)
.
Thus Φ
(
δ
A
)
=
f
(
δ
)
Φ
(
A
) and
(34)
f
(
λ
+
μ
)
Φ
(
A
)
=
Φ
(
(
λ
+
μ
)
A
)
=
Φ
(
A
(
P
+
Q
)
A
)
=
Φ
(
A
P
A
)
+
Φ
(
A
Q
A
)
=
Φ
(
λ
A
)
+
Φ
(
μ
A
)
=
f
(
λ
)
Φ
(
A
)
+
f
(
μ
)
Φ
(
A
)
.
This forces that f
(
λ
+
μ
)
=
f
(
λ
)
+
f
(
μ
) whenever λ
+
μ
≤
1, and hence, ρ
=
1.
Now, let us show thatΦ
(
λ
P
)
=
λ
P holds for all λ
∈
[
0,1
] and all P
∈
𝒫
(
H
). By the orthoadditivity of Φ, Φ
(
λ
P
)
=
λ
P holds for every λ
∈
[
0,1
] and every finite rank projection P
∈
𝒫
(
H
). Assume P have infinite rank. For any x
∈
ran
(
P
) with ∥
x
∥
=
1, we have
(35)
Φ
(
λ
α
P
x
⊗
x
λ
α
P
)
=
Φ
(
λ
P
)
α
x
⊗
Φ
(
λ
P
)
α
x
,
Φ
(
λ
α
P
x
⊗
x
λ
α
P
)
=
Φ
(
λ
2
α
x
⊗
x
)
=
λ
2
x
⊗
x
.
It follows that Φ
(
λ
P
)
α
x and x are linearly dependent; that is, there is a scalar f
x
(
λ
), dependent on x such that Φ
(
λ
P
)
α
x
=
f
x
(
λ
)
x with f
x
(
0
)
=
0 and f
x
(
1
)
=
1. Similarly, for every y
∈
ran
(
I
-
P
), we have Φ
(
λ
P
)
α
y
=
0. These entail that there exists a function g
P
(
x
,
λ
) having values in [
0,1
] and satisfying Φ
(
λ
P
)
x
=
f
x
(
λ
)
1
/
α
x
=
g
P
(
x
,
λ
)
P
x for all x
∈
ran
P and all x
∈
(
ran
P
)
⊥. Hence, g
P
(
x
,
0
)
=
0 and g
P
(
x
,
1
)
=
1. As before, for a fixed x
∈
ran
P, we have that g
P
(
x
,
λ
)
=
λ
δ for some positive δ. Next we claim that g
(
x
,
λ
)
=
λ. Taking the rank one projection Q
=
x
⊗
x for x
∈
ran
P, so Q
≤
P, we have
(36)
λ
δ
Q
=
Φ
(
Q
(
λ
P
)
Q
)
=
Φ
(
λ
1
/
2
Q
P
(
λ
1
/
2
Q
)
)
=
λ
Q
,
which entails that λ
δ
=
λ. In sum, now we have that Φ
(
λ
P
)
x
=
0 for x
∈
(
ran
P
)
⊥ and Φ
(
λ
P
)
x
=
λ
P
x
=
λ
x for x
∈
ran
P. Let H
=
ran
P
⊕
(
ran
P
)
⊥, since Φ
(
λ
P
)
≥
0,
(37)
Φ
(
λ
P
)
=
(
λ
0
0
0
)
=
λ
P
.
Finally, we show thatΦ
(
A
)
=
A for every A
∈
ℰ
(
H
). In fact, letting A
∈
ℰ
(
H
), for every unit vector x
∈
H, we have
(38)
Φ
(
A
)
α
x
⊗
Φ
(
A
)
α
x
=
Φ
(
A
α
)
x
⊗
x
Φ
(
A
α
)
=
Φ
(
A
α
)
(
x
⊗
x
)
β
Φ
(
A
α
)
=
Φ
(
A
α
(
x
⊗
x
)
β
A
α
)
=
A
α
x
⊗
A
α
x
.
Thus there exists a scalar λ
x with |
λ
x
|
=
1 such that Φ
(
A
)
α
x
=
λ
x
A
x. This entails that Φ
(
A
)
α
=
c
A for some scalar c with |
c
|
=
1. Since both A and Φ
(
A
) are positive, we see that c
=
1. Hence Φ
(
A
)
α
=
A
α and the uniqueness of the positive root of a positive operator implies that Φ
(
A
)
=
A, as desired.



---

*Source: 216713-2014-04-13.xml*