The following is our main result.
The following corollary gives a better form of sequence product multiplicative maps, comparing with [7].
Proof of Theorem 3.
By Theorem 2 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 Theorem 2 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 that
Φ
(
P
)
=
P
holds for every projection
P
.
Next we show that
Φ
(
λ
P
)
=
λ
P
holds for every
λ
∈
[
0,1
]
and
P
∈
𝒫
(
H
)
.
By the definition of
Φ
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 of
Φ
,
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 that
ρ
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 that
ρ
=
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 that
Φ
(
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 that
Φ
(
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.