# Maps Preserving Schattenp-Norms of Convex Combinations

Authors: David Li-Wei Kuo; Ming-Cheng Tsai; Ngai-Ching Wong; Jun Zhang
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We study mapsϕ of positive operators of the
Schatten p-classes (1
<
p
<
+
∞), which preserve the p-norms of convex
combinations, that is, ∥
t
ρ
+
(
1
-
t
)
σ
∥
p
=
∥
t
ϕ
(
ρ
)
+
(
1
-
t
)
ϕ
(
σ
)
∥
p
,
∀
ρ
,
σ
∈
𝒮
p
+
(
H
)
1
,
t
∈
[
0,1
]. They are exactly those carrying the form ϕ
(
ρ
)
=
U
ρ
U
* for a unitary or antiunitary U. In the case p
=
2, we have the same conclusion whenever it just holds  ∥
ρ
+
σ
∥
2
=
∥
ϕ
(
ρ
)
+
ϕ
(
σ
)
∥
2 for all the positive Hilbert-Schmidt class operators ρ
,
σ of norm 1. Some examples are demonstrated.


---

## Body

## 1. Introduction

The Mazur-Ulam theorem states that every bijective distance preserving mapΦ from a Banach space onto another is affine; that is,
(1)
Φ
(
t
x
+
(
1
-
t
)
y
)
=
t
Φ
(
x
)
+
(
1
-
t
)
Φ
(
y
)
,
gggggghl
∀
x
,
y
,
0
≤
t
≤
1
.
After translation, we can assume that Φ
(
0
)
=
0 and Φ is indeed a surjective real linear isometry. Let us consider another version of this statement. Suppose that Φ is a bijective map from a Hilbert space H onto H and Φ preserves norm of convex combinations:
(2)
∥
t
Φ
(
x
)
+
(
1
-
t
)
Φ
(
y
)
∥
=
∥
t
x
+
(
1
-
t
)
y
∥
,
ggggggggggggggggggl
∀
x
,
y
∈
H
,
0
≤
t
≤
1
.
Let us further relax the assumption that (2) holds for just one fixed t in (
0,1
). By letting y
=
x in (2), we see that ∥
Φ
(
x
)
∥
=
∥
x
∥ for all x in H. Squaring both sides of (2), we will see that the real parts of the inner products coincide; that is,
(3)
Re
〈
x
,
y
〉
=
Re
〈
Φ
(
x
)
,
Φ
(
y
)
〉
,
∀
x
,
y
∈
H
.
Then the classical Wigner theorem (see, e.g., [1, Theorem 3]) ensures that there is a surjective real linear isometry U
:
H
→
H such that Φ
(
x
)
=
U
x for all x in H.Characterizing isometries, linear or not, of spaces of operators under various norms has been a fruitful area of research for a long time. See, for example, [2, 3] for good surveys. In particular, the spaces 𝒮
p
(
H
) of the Schatten p-class operators on a (complex) Hilbert space H (1
≤
p
<
+
∞) are important objects in both analysis and physics. They are widely used in operator theory and quantum mechanics, for example.Let𝒮
p
+
(
H
) be the set of all positive operators in 𝒮
p
(
H
), and let 𝒮
p
+
(
H
)
1 be the set of all positive operators in 𝒮
p
+
(
H
) of p-norm one. Recall that an affine automorphism (or S-automorphism in [4] or Kadison automorphism in [5]) is a bijective affine map ϕ
:
𝒮
1
+
(
H
)
1
→
𝒮
1
+
(
H
)
1; that is,
(4)
ϕ
(
t
ρ
+
(
1
-
t
)
σ
)
=
t
ϕ
(
ρ
)
+
(
1
-
t
)
ϕ
(
σ
)
,
gggggggggggg
∀
ρ
,
σ
∈
𝒮
1
+
(
H
)
1
,
t
∈
[
0,1
]
.
It is known (see, e.g., [6]) that affine automorphisms are exactly those carrying the form ϕ
(
ρ
)
=
U
ρ
U
* for a unitary or antiunitary U on H.Recently, Nagy [7] established a Mazur-Ulam-type result for the Schatten p-class operators. Suppose that ϕ
:
𝒮
p
+
(
H
)
1
→
𝒮
p
+
(
H
)
1 (1
<
p
<
+
∞) is a bijective map preserving the distance induced by the norm ∥
·
∥
p. Then ϕ is implemented by a unitary or an antiunitary operator U such that ϕ
(
ρ
)
=
U
ρ
U
*. In this paper, we will establish a counterpart of Nagy’s result similar to the one demonstrated in the first paragraph. More precisely, we will characterize those maps ϕ
:
𝒮
p
+
(
H
)
1
→
𝒮
p
+
(
H
)
1 satisfying
(5)
∥
t
ρ
+
(
1
-
t
)
σ
∥
p
=
∥
t
ϕ
(
ρ
)
+
(
1
-
t
)
ϕ
(
σ
)
∥
p
,
∀
ρ
,
σ
∈
𝒮
p
+
(
H
)
1
,
t
∈
[
0,1
]
.
We will show that they are implemented by a unitary or an antiunitary operator.Our main theorem follows.Theorem 1.
LetH be a separable complex Hilbert space of finite or infinite dimension. Let 1
<
p
<
+
∞. Suppose that ϕ is a map from 𝒮
p
+
(
H
)
1 into 𝒮
p
+
(
H
)
1, which will be assumed to be surjective when dim
H
=
+
∞. The following conditions are equivalent.(1)
ϕpreserves the Schatten p-norms of convex combinations; that is,
(6)
∥
t
ρ
+
(
1
-
t
)
σ
∥
p
=
∥
t
ϕ
(
ρ
)
+
(
1
-
t
)
ϕ
(
σ
)
∥
p
,
∀
ρ
,
σ
∈
𝒮
p
+
(
H
)
1
,
t
∈
[
0,1
]
.
(2)
ϕ preserves the pairings; that is, for all ρ
,
σ
∈
𝒮
p
+
(
H
)
1, one has σ
p
-
1
ρ
∈
𝒮
1
(
H
), and
(7)
tr
(
σ
p
-
1
ρ
)
=
tr
(
ϕ
(
σ
)
p
-
1
ϕ
(
ρ
)
)
.
(3)
There exists a unitary or antiunitary operatorU on H such that
(8)
ϕ
(
ρ
)
=
U
ρ
U
*
,
∀
ρ
∈
𝒮
p
+
(
H
)
1
.We note that condition (6) becomes a tautology when p
=
1. On the other hand, the conclusion of Theorem 1 holds again if we replace 𝒮
p
+
(
H
)
1 by 𝒮
p
+
(
H
) everywhere. In this case, setting σ
=
ρ in (6), we see that ϕ does map 𝒮
p
+
(
H
)
1 into 𝒮
p
+
(
H
)
1.The proof of Theorem1 is given in Section 2. When p
=
2, we see in Section 3 that for ϕ carrying the expected form stated in Theorem 1(3) it suffices to say that condition (6) held for only one fixed t in (
0,1
). Finally, we demonstrate some examples in Section 4.


## 2. Proof of the Main Theorem

In what follows, we fix some notation and definitions used throughout the paper. LetH stand for a separable complex Hilbert space of finite dimension or infinite dimension. Let B
(
H
) denote the algebra of all bounded linear operators on H. For a compact operator T in B
(
H
), let s
1
(
T
)
≥
s
2
(
T
)
≥
⋯
≥
0 denote the singular values of T, that is, the eigenvalues of |
T
|
=
(
T
T
*
)
1
/
2 arranged in their decreasing order (repeating according to multiplicity). A compact operator T belongs to the Schatten p-classes 𝒮
p
(
H
) (1
≤
p
<
+
∞) if
(9)
∥
T
∥
p
:
=
(
∑
i
=
1
∞
s
i
(
T
)
p
‍
)
1
/
p
=
(
tr
|
T
|
p
)
1
/
p
<
+
∞
,
where tr denotes the trace functional. We call ∥
T
∥
p the Schatten p-norm of T. In particular, 𝒮
1
(
H
) is the trace class and 𝒮
2
(
H
) is the Hilbert-Schmidt class. The cone of positive operators in 𝒮
p
(
H
) is denoted by 𝒮
p
+
(
H
), and the set of rank one projections in 𝒮
p
+
(
H
) is denoted by P
1
(
H
).Recall that the norm of a normed space is Fréchet differentiable atx
≠
0 if lim
t
→
0
(
(
∥
x
+
t
y
∥
-
∥
x
∥
)
/
t
) exists and uniform for all norm one vectors y.Lemma 2 (see [8, Theorem 2.3]).
Let1
<
p
<
+
∞ and ρ in 𝒮
p
+
(
H
) be nonzero. The norm of 𝒮
p
+
(
H
) is Fréchet differentiable at ρ. For any σ in 𝒮
p
+
(
H
), one has
(10)
d
∥
ρ
+
t
σ
∥
p
d
t
|
t
=
0
=
tr
(
ρ
p
-
1
σ
∥
ρ
∥
p
p
-
1
)
.Lemma 3.
Supposeρ
,
σ
∈
𝒮
p
+
(
H
)
(
1
<
p
<
+
∞
). The following conditions are equivalent.(1)
ρ
=
σ.
(2)
∥
t
ρ
+
(
1
-
t
)
P
∥
p
=
∥
t
σ
+
(
1
-
t
)
P
∥
p for all P in P
1
(
H
) and all t in [
0,1
].
(3)
tr
(
P
ρ
)
=
tr
(
P
σ
) for all P in P
1
(
H
).Proof.
(
1
)
⇒
(
2
) is obvious.
(
2
)
⇒
(
3
): Differentiating both sides of ∥
t
ρ
+
(
1
-
t
)
P
∥
p
=
∥
t
σ
+
(
1
-
t
)
P
∥
p at t
=
0
+, we have tr
P
ρ
=
tr
P
p
-
1
ρ
=
tr
P
p
-
1
σ
=
tr
P
σ by Lemma 2.
(
3
)
⇒
(
1
): Since ρ and σ are positive, ρ
-
σ is Hermitian. There exists an orthonormal basis {
e
i
}
i
=
1
∞ of H such that ρ
-
σ
=
∑
i
=
1
∞
‍
λ
i
e
i
⊗
e
i. Choosing P
i
=
e
i
⊗
e
i, we have λ
i
=
tr
(
P
i
(
ρ
-
σ
)
)
=
0 for all i
=
1,2
,
…. It follows that ρ
-
σ
=
0.We say that two self-adjoint operatorsρ
,
σ in B
(
H
) are orthogonal if ρ
σ
=
0, which is equivalent to the property that they have mutually orthogonal ranges.Lemma 4.
Suppose thatρ
,
σ
∈
𝒮
p
+
(
H
) for 1
<
p
<
+
∞. The following conditions are equivalent.(1)
ρ
,
σ are orthogonal; that is, ρ
σ
=
0.
(2)
∥
α
ρ
+
(
1
-
α
)
σ
∥
p
p
=
α
p
∥
ρ
∥
p
p
+
(
1
-
α
)
p
∥
σ
∥
p
p for any (and thus all) α in (
0,1
).
(3)
tr
(
ρ
σ
)
=
0.
(4)
∥
ρ
+
t
σ
∥
p
≥
∥
ρ
∥
p for all t in ℝ; that is, ρ
⊥
σ in Birkhoff's sense.
(5)
tr
(
ρ
p
-
1
σ
)
=
0.Proof.
(
1
)
⇔
(
2
): From [9, Lemma 2.6], we know that for any two positive operators A
,
B in 𝒮
p
+
(
H
), we have
(11)
tr
(
A
+
B
)
p
≥
tr
A
p
+
tr
B
p
.
Here, the equality holds if and only if A
B
=
0. Setting A
=
α
ρ and B
=
(
1
-
α
)
σ, we get
(12)
ρ
σ
=
0
⟺
(
α
ρ
)
(
(
1
-
α
)
σ
)
=
0
⟺
tr
(
α
ρ
+
(
1
-
α
)
σ
)
p
=
tr
(
α
ρ
)
p
+
tr
(
(
1
-
α
)
σ
)
p
⟺
∥
α
ρ
+
(
1
-
α
)
σ
∥
p
p
=
α
p
∥
ρ
∥
p
p
+
(
1
-
α
)
p
∥
σ
∥
p
p
.
(
1
)
⇔
(
3
): One direction is obvious. For the other, because ρ
,
σ are positive,
(13)
tr
[
(
ρ
1
/
2
σ
1
/
2
)
(
ρ
1
/
2
σ
1
/
2
)
*
]
=
tr
(
ρ
1
/
2
σ
1
/
2
σ
1
/
2
ρ
1
/
2
)
=
tr
(
ρ
σ
)
=
0
.
This forces ρ
1
/
2
σ
1
/
2
=
0, and thus ρ
σ
=
ρ
1
/
2
(
ρ
1
/
2
σ
1
/
2
)
σ
1
/
2
=
0.
(
1
)
⇒
(
4
): Since ρ
σ
=
0, there exists an orthonormal basis {
e
i
}
i
=
1
∞ of H such that ρ
=
∑
i
=
1
∞
‍
λ
i
e
i
⊗
e
i, σ
=
∑
i
=
1
∞
‍
μ
i
e
i
⊗
e
i, λ
i
≥
0, μ
i
≥
0, and λ
i
μ
i
=
0 for all i
=
1,2
,
…. Hence,
(14)
∥
ρ
+
t
σ
∥
p
p
=
tr
|
ρ
+
t
σ
|
p
=
∑
i
=
1
∞
‍
(
λ
i
+
|
t
|
μ
i
)
p
≥
∑
i
=
1
∞
λ
i
p
‍
=
∥
ρ
∥
p
p
.
(
4
)
⇒
(
5
): Without loss of generality, we can assume that ρ
≠
0. Define f
(
t
)
=
∥
ρ
+
t
σ
∥
p
≥
∥
ρ
∥
p. Then f
(
t
) is differentiable and attains its minimum at t
=
0. From Lemma 2,
(15)
0
=
d
∥
ρ
+
t
σ
∥
p
d
t
|
t
=
0
=
tr
(
ρ
p
-
1
σ
∥
ρ
∥
p
p
-
1
)
,
and assertion (5) follows.
(
5
)
⇒
(
1
): As in proving (
3
)
⇒
(
1
), we have ρ
p
-
1
σ
=
0. Then, there exists an orthonormal basis {
e
i
′
}
i
=
1
∞ of H such that ρ
p
-
1
=
∑
i
=
1
∞
‍
ξ
i
e
i
′
⊗
e
i
′, σ
=
∑
i
=
1
∞
‍
η
i
e
i
′
⊗
e
i
′, with ξ
i
≥
0, η
i
≥
0, and ξ
i
μ
i
=
0 for all i
=
1,2
,
…. Thus, tr
(
ρ
σ
)
=
∑
i
=
1
∞
‍
ξ
i
1
/
(
p
-
1
)
η
i
=
0.Lemma 5.
Let1
<
p
<
+
∞. Suppose that ϕ is a map from 𝒮
p
+
(
H
)
1 into 𝒮
p
+
(
H
)
1 preserving the Schatten p-norms of convex combinations; that is, (6) holds. Then, one has
(16)
tr
(
σ
p
-
1
ρ
)
=
tr
(
ϕ
(
σ
)
p
-
1
ϕ
(
ρ
)
)
.Proof.
Differentiating both sides of (6) with respect to t and evaluating at t
=
0, we have
(17)
d
∥
t
ρ
+
(
1
-
t
)
σ
∥
p
d
t
|
t
=
0
=
d
∥
σ
+
t
(
ρ
-
σ
)
∥
p
d
t
|
t
=
0
=
tr
(
σ
p
-
1
(
ρ
-
σ
)
∥
σ
∥
p
p
-
1
)
=
tr
(
σ
p
-
1
ρ
)
∥
σ
∥
p
p
-
1
-
∥
σ
∥
p
=
tr
(
σ
p
-
1
ρ
)
-
1
,
d
∥
t
ϕ
(
ρ
)
+
(
1
-
t
)
ϕ
(
σ
)
∥
p
d
t
|
t
=
0
=
tr
(
ϕ
(
σ
)
p
-
1
ρ
)
∥
ϕ
(
σ
)
∥
p
p
-
1
-
∥
ϕ
(
σ
)
∥
p
=
tr
(
ϕ
(
σ
)
p
-
1
ρ
)
-
1
.
Since (6) holds for t in (
0,1
], these derivatives agree. Therefore, tr
(
σ
p
-
1
ρ
)
=
tr
(
ϕ
(
σ
)
p
-
1
ϕ
(
ρ
)
).Proposition 6.
Suppose thatϕ
:
𝒮
p
+
(
H
)
1
→
𝒮
p
+
(
H
)
1 satisfies
(18)
tr
(
σ
p
-
1
ρ
)
=
tr
(
ϕ
(
σ
)
p
-
1
ϕ
(
ρ
)
)
,
∀
ρ
,
σ
∈
𝒮
p
+
(
H
)
1
.
Then the following assertions hold.(1)
ϕ
preserves orthogonality in both directions; that is
(19)
ρ
σ
=
0
⟺
ϕ
(
ρ
)
ϕ
(
σ
)
=
0
,
∀
ρ
,
σ
∈
𝒮
p
+
(
H
)
1
.
(2)
When
dim
H
<
+
∞
,
ϕ
maps rank-one projections to rank-one projections. This also holds when
dim
H
=
+
∞
and
ϕ
is surjective.
(3)
When
dim
H
<
+
∞
, one has
(20)
tr
P
Q
=
tr
ϕ
(
P
)
ϕ
(
Q
)
,
∀
P
,
Q
∈
P
1
(
H
)
.
This also holds when
dim
H
=
+
∞
and
ϕ
is surjective.Proof.
(1) follows from Lemma4.
(2) First, we assume thatdim
H
=
n
<
+
∞. Suppose ρ is a rank-one projection. We can find n
-
1 pairwise orthogonal rank-one projections ρ
1
,
…
,
ρ
n
-
1 such that ρ
ρ
i
=
0 for 1
≤
i
≤
n
-
1. From (1), we know that ϕ
(
ρ
)
,
ϕ
(
ρ
1
)
,
…
,
ϕ
(
ρ
n
-
1
) are nonzero and pairwise orthogonal. This forces that  ϕ
(
ρ
) has rank one since dim
H
=
n. By (18), taking σ
=
ρ, we see that tr
ϕ
(
ρ
)
p
=
tr
ρ
p
=
tr
ρ
=
1. Therefore, the rank-one positive operator ϕ
(
ρ
) is a projection.
Next, we consider the casedim
H
=
+
∞ and ϕ is surjective. Suppose that there exists a rank-one projection ρ in 𝒮
p
+
(
H
) such that ϕ
(
ρ
) has rank greater than one. Then, there are two nonzero orthogonal operators T
1 and T
2 in 𝒮
p
+
(
H
) such that ϕ
(
ρ
)
=
T
1
+
T
2. Since ϕ is surjective and preserves orthogonality in both directions, there are two nonzero orthogonal operators ρ
1 and ρ
2 in 𝒮
p
+
(
H
)
1 such that ϕ
(
ρ
1
)
=
T
1
/
∥
T
1
∥
p and ϕ
(
ρ
2
)
=
T
1
/
∥
T
2
∥
p. For any σ in 𝒮
p
+
(
H
) with σ
ρ
=
0, we have
(21)
ϕ
(
σ
)
(
∥
T
1
∥
p
ϕ
(
ρ
1
)
+
∥
T
2
∥
p
ϕ
(
ρ
2
)
)
=
ϕ
(
σ
)
(
T
1
+
T
2
)
=
ϕ
(
σ
)
ϕ
(
ρ
)
=
0
.
It forces that
(22)
∥
T
1
∥
p
ϕ
(
σ
)
ϕ
(
ρ
1
)
ϕ
(
σ
)
=
-
∥
T
2
∥
p
ϕ
(
σ
)
ϕ
(
ρ
2
)
ϕ
(
σ
)
=
0
,
and hence ϕ
(
σ
)
ϕ
(
ρ
1
)
=
ϕ
(
σ
)
ϕ
(
ρ
2
)
=
0, because ϕ
(
σ
), ϕ
(
ρ
1
), and ϕ
(
ρ
2
) are all positive. This implies σ
ρ
1
=
σ
ρ
2
=
0. Therefore, ρ
1
=
λ
1
ρ and ρ
2
=
λ
2
ρ for some nonzero λ
1
,
λ
2. This contradicts the fact that ρ
1
ρ
2
=
0.
(3) From (2), we know thatϕ
(
P
), ϕ
(
Q
) are rank-one projections in P
1
(
H
). Therefore, P
p
-
1
=
P, ϕ
(
P
)
p
-
1
=
ϕ
(
P
). Using (18) with σ
=
P, ρ
=
Q, we have
(23)
tr
P
Q
=
tr
(
P
p
-
1
Q
)
=
tr
(
ϕ
(
P
)
p
-
1
ϕ
(
Q
)
)
=
tr
ϕ
(
P
)
ϕ
(
Q
)
.Proof of Theorem1.
(
1
)
⇒
(
2
) follows from Lemma 5.
(
3
)
⇒
(
1
) is obvious.
(
2
)
⇒
(
3
): From Proposition 6, we obtain that ϕ
|
P
1
(
H
)
:
P
1
(
H
)
→
P
1
(
H
) satisfies tr
P
Q
=
tr
ϕ
(
P
)
ϕ
(
Q
) for all rank-one projections P
,
Q in P
1
(
H
). From a nonsurjective version of Wigner’s theorem, cf. [6, Theorem 2.1
.
4], there exists an isometry or anti-isometry U on H such that
(24)
ϕ
(
P
)
=
U
P
U
*
,
∀
P
∈
P
1
(
H
)
.
Note that U is indeed surjective even when H is of infinite dimension, since ϕ is assumed to be surjective in this case.
For any rank-one projectionP in P
1
(
H
), setting σ
=
P in (7), we have
(25)
tr
(
P
ρ
)
=
tr
(
P
p
-
1
ρ
)
=
tr
(
ϕ
(
P
)
p
-
1
ϕ
(
ρ
)
)
=
tr
(
ϕ
(
P
)
ϕ
(
ρ
)
)
=
tr
(
U
P
U
*
ϕ
(
ρ
)
U
)
=
tr
(
P
U
*
ϕ
(
ρ
)
U
)
.
We have U
*
ϕ
(
ρ
)
U
=
ρ by Lemma 3. This gives ϕ
(
ρ
)
=
U
ρ
U
*.


## 3. Maps Preserving Norms of Just a Special Convex Combination

A careful look at the proof of Lemma5 tells us that the condition ∥
t
ρ
+
(
1
-
t
)
σ
∥
p
=
∥
t
ϕ
(
ρ
)
+
(
1
-
t
)
ϕ
(
σ
)
∥
p suffices to hold for the members of any sequence in (
0,1
] converging to 0 rather than for any point t in [
0,1
]. Indeed, in order to get some good properties of ϕ stated in the previous section, we only need to assume that ϕ preserves the Schatten p-norm of convex combination with a given system of coefficients.Proposition 7.
Letϕ
:
𝒮
p
+
(
H
)
1
→
𝒮
p
+
(
H
)
1
(
1
<
p
<
+
∞
). Let α in (
0,1
) be arbitrary but fixed. Suppose
(26)
∥
α
ρ
+
(
1
-
α
)
σ
∥
p
=
∥
α
ϕ
(
ρ
)
+
(
1
-
α
)
ϕ
(
σ
)
∥
p
,
hhhhhhhhhl
∀
ρ
,
σ
∈
𝒮
p
+
(
H
)
1
.
The following properties are satisfied. (1)
ϕ
is injective.
(2)
ϕ
preserves orthogonality in both directions.
(3)
When
dim
H
<
+
∞
,
ϕ
maps rank-one projections to rank-one projections. This also holds when
dim
H
=
+
∞
and
ϕ
is surjective.Proof.
(1) Assumeϕ
(
ρ
)
=
ϕ
(
σ
). We have ∥
α
ϕ
(
ρ
)
+
(
1
-
α
)
ϕ
(
σ
)
∥
p
=
1. From (26), we get ∥
α
ρ
+
(
1
-
α
)
σ
∥
p
=
1. Hence,
(27)
∥
α
ρ
+
(
1
-
α
)
σ
∥
p
=
α
∥
ρ
∥
p
+
(
1
-
α
)
∥
σ
∥
p
.
This forces ρ
=
σ since the norm ∥
·
∥
p is strictly convex for 1
<
p
<
+
∞.
(2) Assumeρ
σ
=
0. From Lemma 4, we have
(28)
∥
α
ρ
+
(
1
-
α
)
σ
∥
p
p
=
α
p
∥
ρ
∥
p
+
(
1
-
α
)
p
∥
σ
∥
p
=
α
p
∥
ϕ
(
ρ
)
∥
p
+
(
1
-
α
)
p
∥
ϕ
(
σ
)
∥
p
.
Together with (26), we have
(29)
∥
α
ϕ
(
ρ
)
+
(
1
-
α
)
ϕ
(
σ
)
∥
p
p
=
α
p
∥
ϕ
(
ρ
)
∥
p
+
(
1
-
α
)
p
∥
ϕ
(
σ
)
∥
p
.
Hence, we have ϕ
(
ρ
)
ϕ
(
σ
)
=
0 from Lemma 4 again. The other implication follows similarly.
(3) The proof is similar to that of Proposition6(2).Whenp
=
2, we get an improvement of Theorem 1.Theorem 8.
LetH be a separable complex Hilbert space. Suppose that ϕ
:
𝒮
2
+
(
H
)
1
→
𝒮
2
+
(
H
)
1, which needs to be surjective when dim
H
=
+
∞. The following conditions are equivalent.(1)
ϕ preserves the Hilbert-Schmidt norms of all convex combinations; that is,
(30)
∥
t
ρ
+
(
1
-
t
)
σ
∥
2
=
∥
t
ϕ
(
ρ
)
+
(
1
-
t
)
ϕ
(
σ
)
∥
2
,

∀
ρ
,
σ
∈
𝒮
2
+
(
H
)
1
,
t
∈
[
0,1
]
.
(2)
For any (and thus all)α in (
0,1
) one has
(31)
∥
α
ρ
+
(
1
-
α
)
σ
∥
2
=
∥
α
ϕ
(
ρ
)
+
(
1
-
α
)
ϕ
(
σ
)
∥
2
,
hhhhhhhh
∀
ρ
,
σ
∈
𝒮
2
+
(
H
)
1
.
A special case states that
(32)
∥
ρ
+
σ
∥
2
=
∥
ϕ
(
ρ
)
+
ϕ
(
σ
)
∥
2
,
∀
ρ
,
σ
∈
𝒮
2
+
(
H
)
1
.
(3)
tr
(
ρ
σ
)
=
tr
(
ϕ
(
ρ
)
ϕ
(
σ
)
) for all ρ
,
σ in 𝒮
2
+
(
H
)
1.
(4)
There exists a unitary or antiunitary operatorU such that
(33)
ϕ
(
ρ
)
=
U
ρ
U
*
,
∀
ρ
∈
𝒮
2
+
(
H
)
1
.Proof.
We prove(
2
)
⇒
(
3
) only. Observe
(34)
∥
α
ρ
+
(
1
-
α
)
σ
∥
2
2
=
tr
(
α
ρ
+
(
1
-
α
)
σ
)
2
=
α
2
tr
ρ
2
+
2
α
(
1
-
α
)
tr
(
ρ
σ
)
+
(
1
-
α
)
2
tr
σ
2
,
∥
α
ϕ
(
ρ
)
+
(
1
-
α
)
ϕ
(
σ
)
∥
2
2
=
α
2
tr
ϕ
(
ρ
)
2
+
2
α
(
1
-
α
)
tr
(
ϕ
(
ρ
)
ϕ
(
σ
)
)
+
(
1
-
α
)
2
tr
ϕ
(
σ
)
2
.
We have tr
(
ρ
σ
)
=
tr
(
ϕ
(
ρ
)
ϕ
(
σ
)
).


## 4. Examples

We remark that all results in previous sections hold for a mapϕ
:
𝒮
p
+
(
H
)
→
𝒮
p
+
(
H
) which satisfies instead of (6) the condition
(35)
∥
t
ρ
+
(
1
-
t
)
σ
∥
p
=
∥
t
ϕ
(
ρ
)
+
(
1
-
t
)
ϕ
(
σ
)
∥
p
,
∀
ρ
,
σ
∈
𝒮
p
+
(
H
)
,
t
∈
[
0,1
]
.
The proofs go in exactly the same ways.The following example shows that a norm preserver of𝒮
p
+
(
H
) might not be affine.Example 1.
LetH be a finite dimensional Hilbert space with an orthonormal basis {
e
i
}
i
=
1
n. Let 1
<
p
<
+
∞. Define a map ϕ from 𝒮
p
+
(
H
) into itself by
(36)
ϕ
(
ρ
)
=
{
0
,
if
ρ
=
0
,
∥
ρ
∥
p
∥
∑
i
=
1
n
‍
P
i
ρ
P
i
∥
p
∑
i
=
1
n
‍
P
i
ρ
P
i
,
if
ρ
≠
0
,
where P
i
=
e
i
⊗
e
i is a rank-one projection for i
=
1
,
…
,
n. Obviously, ϕ
(
ρ
) is positive and ∥
ϕ
(
ρ
)
∥
p
=
∥
ρ
∥
p for all ρ in 𝒮
p
+
(
H
). However, ϕ does not preserve the Schatten p-norms of convex combinations, as the eigenvalues of ρ and ϕ
(
ρ
) can be different from each other.Our theorems are about the Schattenp-norms for 1
<
p
<
+
∞. Here is an example of a map of 𝒮
1
+
(
H
) which preserves trace norms of convex combinations. However, it is not implemented by a unitary or antiunitary.Example 2.
Consider Example1 in the case where p
=
1. In this case,
(37)
ϕ
(
ρ
)
=
∑
i
=
1
n
P
i
ρ
P
i
.
‍
It is easy to see that the map ϕ satisfies the condition
(38)
∥
t
ρ
+
(
1
-
t
)
σ
∥
1
=
∥
t
ϕ
(
ρ
)
+
(
1
-
t
)
ϕ
(
σ
)
∥
1
,
∀
ρ
,
σ
∈
𝒮
1
+
(
H
)
,
t
∈
[
0,1
]
.
But there does not exist a unitary or antiunitary U such that ϕ
(
ρ
)
=
U
ρ
U
* for all ρ in 𝒮
1
+
(
H
).Example 3.
LetH be a separable Hilbert space of infinite dimension, and let {
e
n
:
n
=
1,2
,
…
} be a basis of H. Let S be the unilateral shift on H defined by S
e
n
=
e
n
+
1 for n
=
1,2
,
…. Let ϕ be defined by ϕ
(
ρ
)
=
S
ρ
S
* for ρ in 𝒮
p
+
(
H
). The map ϕ is not surjective, as e
1
⊗
e
1 is not in its range. It is easy to see that ∥
t
ρ
+
(
1
-
t
)
σ
∥
p
=
∥
t
ϕ
(
ρ
)
+
(
1
-
t
)
ϕ
(
σ
)
∥
p holds for all ρ
,
σ in 𝒮
p
+
(
H
) and t in [
0,1
]. However, ϕ is not implemented by a unitary or antiunitary.



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*Source: 520795-2014-01-14.xml*