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Topics on Operator Inequalities
T. Ando
Division of Applied Mathematics
Research Institute of Applied Electricity
Hokkaido University, Sapporo, Japan
Research supported by Kakenhi 234004
CHAPTER I
Geometric and Harmonic Means
Throughout the lecture G,H,K denote Hilbert spaces. L(H) is the space of (bounded) linear operators
on H, while L+ (H) is the cone of positive (i.e. non-negative semi-definite) operators.
In this chapter we shall be concerned with simple binary operations in L+ (H), called geometric and
harmonic means.
Theorem!I.1. Suppose that H = G ⊕ K and a self-adjoint operator T on H is written in the form
A C∗
T =
where A and B act on G and K respectively and C acts from G to K. Then in order
C B
that T is positive it is necessary and sufficient that A and B are positive and there is a contraction W (i.e.
kW k ≤ 1) from G to K such that C = B 1/2 W A1/2 .
Proof. Suppose that A ≥ 0, B ≥ 0 and C = B 1/2 W A1/2 with a contraction W . The condition
kW k ≤ 1 implies W W ∗ ≤ 1, hence B 1/2 W W ∗ B 1/2 ≤ B. Then T admits a factorization T = S ∗ S where
!
A1/2
W ∗ B 1/2
S=
1/2 ,
0
B − B 1/2 W W ∗ B 1/2
which implies that T is positive.
Suppose conversely that T is positive. This means that
(Ax, x) + 2Re (Cx, y) + (By, y) ≥ 0
for
x ∈ G, y ∈ K.
A and B are obviously positive and the above inequality is easily seen to be equivalent to the following
(Ax, x)(By, y) ≥ |(Cx, y)|
2
for x ∈ G, y ∈ K.
Then for each y ∈ K the vector C ∗ y belongs to the range of A1/2 and
2
2
|(Cx, y)|
−1/2 ∗ 2
C y = sup
≤ B 1/2 y ,
A
(Ax, x)
x
where A−1/2 is defined to be the (unbounded) inverse of A1/2 restricted to the orthocomplement of the kernel
of A1/2 . Now there is a contraction U from K to G such that A−1/2 C ∗ = U B 1/2 . Finally W = U ∗ meets
the requirement.
Corollary I.1.1. If T is a positive operator on H and G is a closed subspace of H, then there is a
linear operator S such that T = S ∗ S and S(G) ⊆ G.
In fact, the operator S in the proof of Theorem I.1 meets the requirement.
!
!
A C∗
A C∗
Corollary I.1.2. If
is positive, then there is the minimum of all X for which
C B
C X
are positive.
!
∗ −1/2 ∗ A C∗
Proof. As in the proof of Theorem I.1 the positivity of
implies A−1/2 C ∗
A
C ≤ X.
C X
∗ −1/2 ∗ This means that A−1/2 C ∗
A
C is the minimum in the assertion.
3
4
I. GEOMETRIC AND HARMONIC MEANS
Remark. If A has bounded inverse, the minimum in Corollary I.1.1 is just CA−1 C ∗ .
Corollary I.1.3. If H ⊇ G and if S is a linear operator from G to H such that
(Sx, y) = (x, Sy)
for
x, y ∈ G
then there is a self-adjoint operator T on H such that T |G = S and kT k = kSk. Further, among all T
satisfying these conditions there are the minimum Tµ and the maximum TM .
!
A
Proof. It can be assumed that kSk = 1. Let S =
where A acts on G and C does from G to
C
K = H G. By assumption A is self-adjoint and
2
2
2
kAxk + kCxk ≤ kxk
for x ∈ G.
!
!
A C∗
1 + A C∗
∗
2
Therefore C C ≤ 1 − A . T must be of the form T =
for which
and
C X
C
1+X
!
1 − A −C ∗
are positive. The inequalities C ∗ C ≤ 1 − A2 and 2−1 (1 − A) ≤ 1 imply that
−C
1−X
C ∗ 2−1 C ≤ 2−1 1 − A2 = 2−1!(1 − A)(1 + A) ≤ 1 + A. By Remark after Corollary I.1.2 this implies
!
C∗
1 + A C∗
. Therefore there is the minimum Bµ of all X for which
C
1+1
C
1+X
−1/2 ∗ ∗
−1/2 ∗
are positive. In fact,
B
is
defined
by
B
=
(
1
+
A)
C
(
1
+
A)
C
−
1
.
In
order
to
conclude
µ
! µ
!
A C∗
1 − A −C ∗
that Tµ =
is the minimum in the assertion, it remains to show that
is positive.
C Bµ
−C
1 − Bµ
!
!
1 − A −C ∗
1 + A C∗
Since
is positive just as
, (1 − A)−1/2 C ∗ is a well defined linear operator. By
−C
21
C
21
Corollary I.1.2 it reduces to prove the inequality
2
2
2
(1 − A)−1/2 C ∗ y ≤ 2 kyk − (1 + A)−1/2 C ∗ y .
the positivity of
1+A
But
2 2
(1 + A)−1/2 C ∗ y + (1 − A)−1/2 C ∗ y =
=2
1 − A2
−1
C ∗ y, C ∗ y = 2 1 − A2
−1/2
2
C ∗ y .
1 − A2 C ∗
Since C ∗ C ≤ 1 − A2 implies
≥ 0,
C
1
!
2
1 − A2
−1/2
2
2
C ∗ y ≤ 2 kyk ,
!
∗ A C∗
as expected. Analogously TM =
with BM = 1 − (1 − A)−1/2 C ∗
(1 − A)−1/2 C ∗ is shown to
C BM
be the maximum. This completes the proof.
Theorem I.2. Let A and B be positive
operators on H. Then there is the maximum of all self-adjoint
!
A X
operators X on H for which
are positive.
X B
I. GEOMETRIC AND HARMONIC MEANS
5
−1/2
−1/2
A
X
X
B
!
Proof. Consider first the case that A has bounded inverse. Let D = A
BA
. Then
!
1
A−1/2 XA−1/2
is positive if and only if
is positive. We claim that D1/2 is the maximum
A−1/2 XA−1/2
D
!
!
1 Y
1 D1/2
of all Y for which
are positive. The positivity of
is obvious by Corollary I.1.2.
Y D
D1/2
D
!
1 Y
Suppose that
is positive. By Theorem I.1 there is a contraction W on H such that Y = D1/2 W .
Y D
Introduce a new scalar product on H by hx, yi := (D1/2 x, y). Since Y is self-adjoint hW x, yi = hx, W yi, for
every x, y ∈ H which means that W is self-adjoint with respect to this new scalar product. Since kW k ≤ 1,
for any real number λ with |λ| > 1 the operator λ − W has bounded inverse. The operator (λ − W )−1 is
bounded with respect to the new scalar product, hence
|hW x, xi| ≤ hx, xi
for all
x ∈ H,
which implies −D1/2 ≤ Y ≤ D1/2 . Thus D1/2 is the maximum as claimed. As!a consequence,
1/2 1/2
A X
A1/2 D1/2 A1/2 = A1/2 A−1/2 BA−1/2
A
is the maximum of all X for which
are positive.
X B
If A does not admit
bounded inverse, consider A = A + with > 0. Let C be the maximum of all X
!
A X
for which
are positive. Since obviously C ≥ C0 ≥ 0 whenever > 0 > 0, the operator lim+ C
→0
X B
!
A X
is the maximum of all X for which
are positive. This completes the proof.
X B
We shall call the maximum in Theorem I.2 the geometric mean of two positive operators A and B,
and denote it by A#B.
Corollary I.2.1. Geometric means have the following properties.
(i)
(ii)
(iii)
(iv)
(v)
A#B = B#A.
(αA)#(αB) = α(A#B) for α ≥ 0.
(A1 + A2 )#(B1 + B2 ) ≥ (A1 #B1 ) + (A2 #B2 ).
C(A#B)C ∗ ≤ (CAC ∗ )#(CBC ∗ ) for any linear operator C.
A#A = A, 1#A = A1/2 and 0#A = 0.
1/2 1/2
(vi) A#B = A1/2 A−1/2 BA−1/2
A
if A has bounded inverse.
(vii) A−1 #B −1 = (A#B)−1 if both A and B have bounded inverses.
Proof. (i) to (iv) are immediate from definition. (v) and (vi) are proved as in the proof of Theorem
I.2. (vii) follows from (vi).
As a consequence if A commutes with B, then A#B = (AB)1/2 . This justify the terminology “geometric
mean”.
Corollary I.2.2. If 0 ≤ p ≤ 1, then A ≥ B ≥ 0 implies Ap ≥ B p .
Proof. The assertion is true if p = 1 or 0. Therefore it remains to show that the set ∆ of p for which
the assertion is true is convex. Take p1 , p2 ∈ ∆, and let p = 12 (p1 + p2 ). Then since Ap = Ap1 #Ap2 and
B p = B p1 #B p2 by Corollary I.2.1, Ap ≥ B p follows also from the same Corollary.
6
I. GEOMETRIC AND HARMONIC MEANS
Corollary I.2.3. Let A1 and A2 (resp. !
B1 and B2 ) be positive operators on G (reps. K) and
! let C
∗
∗
Ai C
A1 #A2
C
be a linear operator from G to K. If
are positive i = 1, 2, then so is
.
C Bi
C
B1 #B2
Proof. We may assume the bounded invertibility of A1 and A2 . Then by assumption and Corollary
∗
I.1.2 CA−1
i = 1, 2. Therefore by Corollary I.2.1
i C ≤ Bi
−1
∗
∗
C(A1 #A2 )−1 C ∗ = C A−1
C∗ < CA−1
# CA−1
≤ B1 #B2 ,
1 #A2
1 C
2 C
!
A1 #A2
C∗
which implies the positivity of
by Corollary I.1.2.
C
B1 #B2
Corollary I.2.4. Arithmetic mean is greater than geometric mean, that is
positive A and B.
1
2 (A
+ B) ≥ A#B for
Proof. We may assume the bounded invertibility of A. Then by Corollary I.2.1
1/2
1
1 −1/2
1
1/2
−1/2
−1/2
1/2
1/2
−1/2
A#B = A
A
BA
A
≤A
1+ A
BA
A1/2 = (A + B),
2
2
2
1
because X ≤
1 + X 2 for any positive X.
2
−1
1 −1
−1
The harmonic mean of two positive operators A and B must be defined by
A +B
when
2
both A and B have bounded inverses. We shall denote harmonic mean by A : B. If A and B do not have
bounded inverses, the harmonic mean A : B is defined as the limit of (A + ) : (B + ) as → 0+ .
Theorem I.3. Let A and B be positive operators on a Hilbert space H. Then harmonic mean A : B is
the maximum of all X for which
!
!
2A 0
X X
≥
.
0 2B
X X
Proof. We may assume that both A and B have bounded inverse. Then the above inequality is
equivalent to the condition that for every x ∈ H
(Xx, x) ≤ 2 inf {(Ay, y) + (B(x − y), x − y)}
y
hence to the condition
(
2
|(Bx, y)|
(Xx, x) ≤ 2 (Bx, x) − sup
((A
+ B)y, y)
y
)
.
Since by definition
A : B = 2A(A + B)−1 B = 2 B − B(A + B)−1 B ,
we have
(
)
2
|(Bx, y)|
((A : B)x, x) = 2 (Bx, x) − sup
.
y ((A + B)y, y)
!
!
2A 0
X X
This shows that A : B is the maximum of all X for which
≥
.
0 2B
X X
Corollary I.3.1. Harmonic means have the following properties.
(i)
(ii)
(iii)
(iv)
A : B = B : A.
(αA) : (αB) = α(A : B) for α ≥ 0.
(A1 + A2 ) : (B1 + B2 ) ≥ (A1 : B1 ) + (A2 : B2 ).
C(A : B)C ∗ ≤ (CAC ∗ ) : (CBC ∗ ) for any linear operator C.
I. GEOMETRIC AND HARMONIC MEANS
7
(v) A : A = A, 1 : A = 2A(1 + A)−1 and 0 : A = 0.
(vi) A : B = 2A(A + B)−1 B if A has bounded inverse.
−1
(vii) A−1 : B −1 = 12 (A + B)
if both A and B have bounded inverses.
Corollary I.3.2. Geometric mean is greater than harmonic mean, that is A#B ≥ A : B for positive
A and B.
Proof. We may assume the bounded invertibility of A and B. Then by Corollary I.2.3 A−1 #B −1 ≤
1 −1
A + B −1 . By taking inverse, this yields, by Corollaries I.2.1 and I.3.1, A#B ≥ A : B.
2
Theorem I.4. Let A1 and A2 (resp. !
B1 and B2 ) be positive operators of G (resp. K) and let C
! be a
1
∗
∗
Ai C
(A1 + A2 )
C
linear operator from G to K. If
are positive i = 1, 2, then so are 2
and
C Bi
C
B 1 : B2
!
A1 : A2
C∗
.
1
C
2 (B1 + B2 )
Proof. We may assume the bound invertibility of A1 and A2 . Then by assumption and Corollary I.1.2
≤ Bi i = 1, 2. Therefore by Corollary I.3.1
−1
∗
1
−1
∗
∗
C
(A1 + A2 )
C ∗ = C A−1
C ≤ CA−1
: CA−1
≤ B1 : B2 ,
1 : A2
1 C
2 C
2
!
1
∗
(A
+
A
)
C
1
2
2
which implies the positivity of
by Corollary I.1.2.
The positivity of
C
B 1 : B2
!
A1 : A2
C∗
is proved analogously.
1
C
2 (B1 + B2 )
∗
CA−1
i C
Corollary I.4.1.
The following identity holds for positive A and B;
1
(A + B) #(A : B) = A#B.
2
!
!
A
A#B
B
A#B
Proof. By definition
and
are positive, so that
A#B
B
A#B
A
is positive by Theorem I.4, which implies
1
(A + B) #(A : B) ≥ A#B.
2
1
2 (A
+ B) A#B
A#B
A:B
!
When A and B have bounded inverse, the above inequality, with A−1 and B −1 instead of A and B respectively, leads, by taking inverse, to the inequality
1
(A : B)#
(A + B) ≤ A#B.
2
These two inequalities yield the assertion.
CHAPTER II
Operator-Monotone Functions
Given a finite or infinite open interval (α, β) on the real line, let us denote by S(α, β; H) the totality of
all self-adjoint operators on H whose spectrum are included in the interval (α, β). If there is no confusion,
we shall write simply S(α, β).
A real-valued continuous function f on (α, β) is said to be operator-monotone on (α, β) if A, B ∈
S(α, β; H) and A ≤ B implies f (A) ≤ f (B), where f (A) and f (B) are defined by familiar functional calculi
for self-adjoint operators.
An operator-monotone function is non-decreasing in the usual sense, but the converse is not true. This
chapter is devoted to intrinsic characterization of operator-monotonousness.
Examples II.1. 1. f (λ) := a + bλ with b ≥ 0 is operator-monotone on (−∞, ∞).
2. f (λ) := λp with 0 ≤ p ≤ 1 is operator-monotone on (0, ∞) by Corollary I.2.2.
1
3. For µ ∈
/ (α, β) the function f (λ) := µ−λ
is operator-monotone on (α, β). In particular f (λ) = − λ1 is
operator-monotone on (0, ∞). In fact, if µ < α, then A, B ∈ S(α, β) A ≤ B implies 0 < −(µ − A) ≤
−(µ − B) so that −(µ − A)−1 ≥ −(µ − B)−1 hence f (A) ≤ f (B). If µ > β, then 0 < µ − B < µ − A
hence (µ − B)−1 ≥ (µ − A)−1 .
It is easy to see that a function f is operator-monotone on a finite interval (α, β) if and only if g(λ) :=
f
− α)λ + α + β) is operator monotone on (−1, 1). On this basis, we shall treat only the interval
(−1, 1).
1
2 ((β
Lemma II.1. If a continuously differentiable function f on (−1, 1) is operator-monotone, then for any
N
choice λi ∈ (−1, 1) i = 1, . . . , N the matrix f [1] (λi , λj ) i,j=1 is positive, where f [1] (λ, µ) is defined by
f [1] (λ, µ) =
f (λ)−f (µ)
λ−µ
for λ 6= µ and f [1] (λ, λ) = f 0 (λ).


λ1
0


..
, considered as a self-adjoint operator on the N Proof. Observe the matrix A := 
.


0
λN
dimensional Hilbert space CN . Take any complex numbers ξ1 , . . . , ξN and consider the positive matrix
N
B := ξi ξ¯j i,j=1 . Since A ∈ S −1, 1; CN , for sufficiently small > 0, A + B belongs to S −1, 1; CN . We
claim that for any continuously differentiable function g on (−1, 1)
lim+ −1 {g(A + B) − g(A)} =
→0
N
X
i,j=1
8
g [1] (λi , λj )Pi BPj
II. OPERATOR-MONOTONE FUNCTIONS
9
where Pi is the orthoprojection to the subspace spanned by the vector ei := (δij )N
j=1 . In fact, this is true if
n
g is polynomial, because for g(λ) = λ
−1 {g(A + B) − g(A)} =
n
X
Ak−1 BAn−k + O()
k=1
=
N
X
n
X
λk−1
λn−k
Pi BPj + O()
i
j
i,j=1 k=1
=
N
X
g [1] (λi , λj )Pi BPj + O().
i,j=1
0
Use approximation of g by polynomials on a suitable subinterval when g is merely continuously differentiable.
Now since f is operator-monotone and A + B ≥ A for all > 0, we have
N
X
f [1] (λi , λj )Pi BPj = lim+ −1 {f (A + B) − f (A)} ≥ 0.
→0
i,j=1
Let x =
PN
i,j=1 ei ,
then the positivity of
0≤
N
X
PN
i,j=1
f [1] (λi , λj )Pi BPj implies
f [1] (λi , λj )(Pi BPj x, x) =
i,j=1
N
X
f [1] (λi , λj )ξi ξ¯j .
i,j=1
N
Since (ξi ) are arbitrary, this means that the matrix f [1] (λi , λj ) i,j=1 is positive.
Lemma II.2. If a continuously differentiable function f on (−1, 1) is operator-monotone, then there
exists a finite positive measure m on the closed interval [−1, 1] such that
Z 1
λ
f (λ) = f (0) +
dm(t) for λ ∈ (−1, 1).
−1 1 − λt
Proof. Consider the linear subspace H of C(−1, 1), spanned by the functions fλ (t) := f [1] (λ, t) where
λ runs over (−1, 1). Introduce a scalar product in H by
*
+
X
X
X
αi fλi ,
βj fµj =
f [1] (λi , µj )αi βj .
i
j
i,j
ˆ be the associated Hilbert space. Let us show that
This is positive semi-definite by Lemma II.1. Let H
λn → λ implies kfλn − fλ k → 0. In fact,
2
kfλn − fλ k = f [1] (λn , λn ) − 2f [1] (λn , λ) + f [1] (λ, λ)
= f 0 (λn ) − 2
f (λn ) − f (λ)
+ f 0 (λ).
λn − λ
Now by continuous differentiability of f
f 0 (λn ) − 2
f (λn ) − f (λ)
+ f 0 (λ) → f 0 (λ) − 2f 0 (λ) + f 0 (λ) = 0.
λn − λ
ˆ spanned by all fλ (λ 6= 0). Then D is dense in H,
ˆ as shown above. Define
Let D be the linear subspace of H,
ˆ by
a linear map T from D to H
T fλ := λ−1 {fλ − f0 }
(λ 6= 0).
T is symmetric hence well-defined, because
hT fλ , fµ i =
µf (λ) − λf (λ) f (0)
−
= hfλ , T fµ i .
λµ(λ − µ)
λµ
10
II. OPERATOR-MONOTONE FUNCTIONS
We claim that T admits a (not necessarily bounded) self-adjoint extension Tˆ. This follows from a theorem
of von Neumann, (see [12] p.1231), because the anti-linear involution J, defined by
!
X
X
J
αi fλi =
αi fλi ,
i
i
makes D invariant and commutes with T . Since by definition
(µ − λ)fµ = (1 − λTˆ)(µfµ − λfλ ),
the kernel if 1 − λTˆ is orthogonal to the linear subspace spanned by fµ (µ 6= λ, and µ 6= 0). This subspace
ˆ as shown above, so that 1 − λTˆ is injective, hence (1 − λTˆ)−1 is a (unbounded) self-adjoint
is dense in H
operator. Therefore f0 = (1 − λTˆ)fλ implies fλ = (1 − λTˆ)−1 f0 .
R∞
Now let Tˆ = −∞ tdE(t) be the spectral representation of the self-adjoint operator Tˆ, and dm(t) =
hdE(t)f0 , f0 i. Then for λ 6= 0
λ−1 {f (λ) − f (0)} = f [1] (λ, 0)
D
E
= hfλ , f0 i = (1 − λTˆ)−1 f0 , f0
Z ∞
1
=
dm(t).
1
−
λt
−∞
To complete the proof, it remains to prove that the measure m is concentrated on the closed interval [−1, 1].
Take α > 1, and let us show that [α, ∞) is an m-zero set
Z α−1 Z ∞
Z α−1 Z ∞
1
1
dm(t)dλ ≤
dm(t)dλ
2
(1
−
λt)
(1
−
λt)2
0
α
0
−∞
Z α−1
Z α−1
=
hfλ , fλ i dλ =
f 0 (λ)dλ = f α−1 − f (0) < ∞.
0
By Fubini’s theorem we have
Z α−1 Z
0
∞
α
0
1
dm(t)dλ =
(1 − λt)2
Z
∞
α
∞
Z
≥
α
α−1
Z
0
1
t
Z
1
dλdm(t)
(1 − λt)2
1
s−2 dsdm(t).
0
But this last expression is finite inly if [α, ∞) is an m-zero set. Analogously (−∞, −α] is an m-zero set.
R1
Remark. In Lemma II.2 −1 dm(t) = hf0 , f0 i = f 0 (0).
Lemma II.3. If a continuously differentiable function f on (−1, 1) is operator-monotone, then for any
choice −1 < λ1 < µ1 < λ2 < µ2 < 1
!
f [1] (λ1 , µ1 ) f [1] (λ1 , µ2 )
det
≥ 0.
f [1] (λ2 , µ1 ) f [1] (λ2 , µ2 )
Proof. We shall use the notations in the proof of the previous Lemma. First remark that
Z 1
1
[1]
f (λ, µ) = hfλ , fµ i =
dm(t).
(1
−
λt)(1
− µt)
−1
If the measure m is concentrated on a single point, say t0 , then the determinant in question is equal to a
scalar multiple of
!
(1 − λ1 t0 )−1 (1 − µ1 t0 )−1 (1 − λ1 t0 )−1 (1 − µ2 t0 )−1
det
,
(1 − λ2 t0 )−1 (1 − µ1 t0 )−1 (1 − λ2 t0 )−1 (1 − µ2 t0 )−1
which is just equal to 0.
II. OPERATOR-MONOTONE FUNCTIONS
11
Suppose that m is not concentrated on a single point and that the determinant has negative value.
Since the corresponding determinant with λ1 = µ1 and λ2 = µ2 has non-negative by Lemma II.1, by using
continuity argument there are µ01 and µ02 such that λ1 ≤ µ01 ≤ µ1 and λ2 ≤ µ02 ≤ µ2 , and
!
f [1] (λ1 , µ01 ) f [1] (λ1 , µ02 )
det
= 0.
f [1] (λ2 , µ01 ) f [1] (λ2 , µ02 )
This implies that there are α1 and α2 such that |α1 | + |α2 | 6= 0 and
fλi , α1 fµ01 + α2 fµ02 = 0 (i = 1, 2).
Choose β1 and β2 so that β1 + β2 = α1 + α2 and β1 λ2 + β2 λ1 = α1 µ02 + α2 µ01 . Then it follows that
0 = β1 fλ1 + β2 fλ2 , α1 fµ01 + α2 fµ02
Z 1
{α1 + α2 − (α1 µ02 + α2 µ01 )t}2
=
dm(t).
0
0
−1 (1 − λ1 t)(1 − λ2 t)(1 − µ1 t)(1 − µ2 t)
Since the denominator is strictly positive on [−1, 1] and since the measure m is not concentrated on any single
point, the above relation implies that α1 + α2 = 0 and α1 µ02 + α2 µ01 = 0, which contradicts |α1 | + |α2 | 6= 0.
This contradiction completes the proof.
Now let us return to general operator-monotone functions. Take an infinitely many times differentiable
R1
function φ on (−∞, ∞) such that φ is non-negative, vanishes outside of (−1, 1) and −1 φ(t)dt = 1.
Suppose that f is operator-monotone on (−1, 1). For any 0 < < 1, define a function f() on (−1+, 1−)
by
Z
1 f() (t) :=
φ(s/)f (t − s)ds.
−
The function f() is operator-monotone on (−1 + , 1 − ), because each f (t − ) is so. Obviously f() is
infinitely many times differentiable, and as converges to 0, f() (t) converges to f (t) uniformly on any closed
subinterval of (−1, 1).
Lemma II.4. If a continuous function f on (−1, 1) is operator-monotone, then on any closed subinterval
(α, β) of (−1, 1) f satisfies Lipschitz condition, that is, supα<s<t<β f [1] (t, s) < ∞.
Proof. Let us use the notations in the preceding comment. Let λ1 = 12 (α − 1) and µ2 = 12 (β + 1), and
apply Lemma II.3 to the operator-monotone function f() with λ2 = t and µ1 = s. By taking limit, we can
conclude that
!
f [1] (λ1 , s) f [1] (λ1 , µ2 )
det
≥ 0,
f [1] (t, s)
f [1] (t, µ2 )
or equivalently
f [1] (λ1 , µ2 )f [1] (t, s) ≤ f [1] (λ1 , s)f [1] (t, µ2 ).
The right side of the above inequality is bounded when s < t run over [α, β]. Since f is non-decreasing, both
f [1] (λ1 , µ2 ) and f [1] (t, s) are non-negative. Therefore if f [1] (λ1 , µ2 ) 6= 0, then f [1] (t, s) are bounded when
s < t run over [α, β]. Finally f [1] (λ1 , µ2 ) = 0 implies f [1] (t, s) = 0 for all s < t in [α, β]. This completes the
proof.
Theorem II.1. In order that a continuous function f on (−1, 1) is operator-monotone, it is necessary
and sufficient that there is a finite positive measure m on [−1, 1] such that
Z 1
λ
f (λ) = f (0) +
dm(t) for λ ∈ (−1, 1).
1
−
λt
−1
12
II. OPERATOR-MONOTONE FUNCTIONS
Proof. Suppose that f admits a representation of the above form. For each t ∈ [−1, 1] the function
λ
ht (λ) := 1−λt
is operator-monotone on (−1, 1). In fact, if t = 0, h0 (λ) = λ is operator-monotone. If t 6= 0,
t−2
t−1 −λ is operator-monotone as stated in one of the
R1
average −1 ht (λ)dm(t) is operator-monotone, and so is f .
ht (λ) = −t−1 +
Examples II.1. As a consequence, the
weighted
Suppose conversely that f is operator-monotone. Then with the notations in front of Lemma II.4, for
each 0 < < 1 f() ((1 − )λ) is operator-monotone on (−1, 1). Then since f() ((1 − )λ) is continuously
differentiable, by Lemma II.2 there is a measure m on [−1, 1] such that
Z 1
λ
f() ((1 − )λ) = f() (0) +
dm (t).
−1 1 − λt
R1
We claim that −1 dm (t) are bounded when runs over (0, 1/2). As remarked after Lemma II.2, we have
Z 1
0
dm (t) = f()
(0).
−1
Since Lemma II.4 f satisfies Lipshitz condition on any closed subinterval of (−1, 1), it has derivative f 0 (t)
for almost all λ and ess. sup |f 0 (t)| = r < ∞. Further it is easy to see that
−<t<
0
f()
(0) =
(1 − )
Z
f 0 (−s)φ
−
s
ds ≤ (1 − )r .
R1
These considerations show the boundedness of −1 dm (t) when runs over (0, 1/2). Now by the Helly
theorem (see [11] p. 381) there is a sequence n → 0 and a measure m on [−1, 1] such that
Z 1
Z 1
lim
g(t)dmn (t) =
g(t)dm(t)
n→∞
−1
−1
for all continuous functions g on [−1, 1]. Therefore for λ ∈ (−1, 1)
Z 1
Z 1
λ
λ
lim
dmn (t) =
dm(t).
n→∞ −1 1 − λt
1
−
λt
−1
Finally the assertion follows from
lim f(n ) ((1 − n )λ) = f (λ).
n→∞
The integral representation in Theorem II.1 shows that an operator-monotone function f on (−1, 1) is
necessarily infinitely many times differentiable. Further more it admits analytic continuation to the upper
and lower open half planes by
Z 1
ζ
fˆ(ζ) = f (0) +
dm(t) for ζ with Im (ζ) 6= 0.
1
−
ζt
−1
The function fˆ maps the upper half plane to itself. Indeed
Z 1 Im (ζ)
ˆ
Im f (ζ) =
2 dm(t).
−1 |1 − ζt|
This observation is completed in the following theorem.
Theorem II.2. Let f be a real-valued continuous function on a finite or infinite interval (α, β). In order
ˆ
that f is operator-monotone it is necessary
andsufficient that it admits an analytic continuation f to the
upper and lower half planes such that Im fˆ(ζ) > 0 for Im (ζ) > 0.
II. OPERATOR-MONOTONE FUNCTIONS
13
Proof. Suppose that f admits an analytic continuation fˆ of the type mentioned above. Then by a
well-known theorem of Nevanlina (see [1] p. 7) there are a real number a, a non-negative number b and a
finite positive measure m
ˆ on Ω := (−∞, ∞) \ (α, β) such that
Z
1 + ζt
fˆ(ζ) = a + bζ +
dm(t).
ˆ
Ω t−ζ
The function g(λ) := a + bλ with b ≥ 0 is obviously operator-monotone on (α, β). For each t ∈
/ (α, β) the
function ht (λ) := 1+λt
is
operator
monotone,
because
t−λ
ht (λ) = −t +
1 + t2
t−λ
1
and the function t−λ
is operator-monotone on (α, β). Therefore the weighted average f (λ) is also operatormonotone on (α, β). This completes the proof.
Corollary II.2.1. If f is operator-monotone on (−∞, ∞), then f (λ) = a + bλ with some real a and
non-negative b.
This follows immediately from the integral representation in the proof of Theorem II.2, because the
measure m
ˆ must vanish.
CHAPTER III
Operator-Convex Functions
The notion next to monotoneousness seems convexity. In this respect, a real-valued continuous function
f on a finite or infinite interval (α, β) is said to be operator-convex if
1
1
f
(A + B) ≤ {f (A) + f (B)}
forA, B ∈ S(α, β).
2
2
The function f is said to be operator-concave if −f is operator-convex.
An operator-convex function is convex in the usual sense, but the converse is not true. This chapter is
devoted to intrinsic characterization of operator-convexity.
Examples III.1. 1. f (λ) := a + bλ is operator-convex on (−∞, ∞).
2. f (λ) := λ2 is operator-convex on (−∞, ∞). In fact, for self-adjoint A and B
2
1 2
1
1
A + B2 −
(A + B) = (A − B)2 ≥ 0.
2
2
4
1
3. For µ < α the function f (λ) := λ−µ
is operator-convex on (α, ∞). In particular, f (λ) :=
convex on (0, ∞). In fact, for A, B ∈ S(α, β), A − µ and B − µ belong to S(0, ∞) and
1
λ
is operator-
1
(A − µ)−1 + (B − µ)−1 = {(A − µ) : (B − µ)}−1 .
2
By Corollary I.2.4 and I.3.2.
−1
1
1
−1
{(A − µ) : (B − µ)} ≥
(A − µ) + (B − µ)
2
2
−1
1
=
(A + B) − µ
.
2
Lemma III.1. Let f be a twice continuously differentiable function on (−1, 1). If f is operator-convex,
then for each µ ∈ (−1, 1) the function g(λ) := f [1] (µ, λ) is operator-monotone. Conversely if f [1] (0, λ) is
operator-monotone, then f is operator-convex.
Proof. Let f be operator-convex. Obviously g is continuously differentiable on (−1, 1). Inspection
of Lemmas II.1 and II.2 will show that for the operator-monotoneousness of g it suffices to prove that
N
for any choice λi ∈ (−1, 1) i = 1, 2, . . . , N the matrix g [1] (λi , λj )
is positive. Observe the matrix
i,j=1


λ1
0


A := 
λN
 with λN +1 = µ, considered as a self-adjoint operator on the (N + 1)-dimensional
0
λN +1


ξ¯1

.. 

0
. 
.
Hilbert space CN +1 . Take any complex numbers ξ1 , . . . , ξN and consider the matrix B := 



ξ¯N 
ξ , . . . , ξN 0
1
N +1
N +1
Since A ∈ S −1, 1; C
, for sufficiently small > 0 A + B belongs to S −1, 1; C
. Since f is twice
14
III. OPERATOR-CONVEX FUNCTIONS
15
continuously differentiable, just as in the proof of Lemma II.1, it can be shown that the matrix-valued
function f (A + B) is twice differentiable and
N
+1
X
d2 f (A + B) =
f [2] (λi , λj , λk )Pi BPj BPk
d2
=0
i,j,k=1
N +1
where Pi is the orthoprojection to the subspace spanned by the vector ei := (δij )j=1 and f [2] (s, t, u) =
d2 f (A + B) [1]
[1]
hs (t, u) with hs (t) = f (s, t). Now the operator-convexity of f implies
≥ 0, as in the
d2
=0
PN
case of scalar convex functions. Let x = i=1 ei . Then
0≤
N
+1
X
f [2] (λi , λj , λk )(Pi BPj BPk x, x) =
i,j,k=1
=
N
X
f [2] (λi , λN +1 , λk )ξk ξ¯i =
i,k=1
N
X
g [1] (λi , λk )ξk ξ¯i .
i,k=1
N
Since (ξi ) are arbitrary, this means that the matrix g [1] (λi , λj ) i,j=1 is positive.
Suppose conversely that f [1] (0, λ) is operator-monotone. Then by Theorem II.1 there exists a finite
positive measure m on [−1, 1] such that
Z 1
λ
f [1] (0, λ) = f 0 (0) +
dm(t),
1
−
λt
−1
hence
Z
1
f (λ) = a + bλ +
−1
λ2
dm(t),
1 − λt
2
0
λ
where a = f (0) and b = f (0). For each t ∈ [−1, 1] the function ht (λ) := 1−λt
is operator-convex. In fact, if
2
t = 0, ht (λ) = λ , and if t 6= 0,
t−2
ht (λ) = −t−2 − t−1 λ +
.
1 − λt
These functions are operator-convex, as shown in Example III.1. Since the function a+bλ is operator-convex,
too, the weighted average f is operator-convex. This completes the proof.
Theorem III.1. If a continuous function f on (−1, 1) is operator-monotone, then both the function
Rλ
f1 (λ) = 0 f (t)dt and f2 (λ) := λf (λ) are operator-convex.
Proof. Since f is continuously differentiable by Theorem II.2, both f1 and f2 are twice continuously
differentiable. Now since
Z 1
[1]
[1]
f1 (0, λ) =
f (sλ)ds and f2 (0, λ) = f (λ),
0
the assertion follows from Lemma III.1.
Theorem III.2. In order that a continuous function f on (−1, 1) is operator-convex, it is necessary and
sufficient that there are real numbers a and b, and a finite positive measure m on [−1, 1] such that
Z 1
λ2
f (λ) = a + bλ +
dm(t)
for λ ∈ (−1, 1).
−1 1 − λt
Proof. Sufficiency was shown already in the proof of Lemma III.1. Suppose that f is operator-convex.
By using the notations in the proof of Lemma II.4, f() ((1−)λ) is operator-convex for each 0 < < 1. Theref() ((1 − )λ) − f() ((1 − )µ)
fore by Lemma III.1 for any µ ∈ (−1, 1) the function
is operator-monotone on
λ−µ
(−1, 1) so that by taking limit as → 0 f [1] (µ, λ) is operator-monotone on (µ + δ, 1) as well as (−1, µ − δ)
16
III. OPERATOR-CONVEX FUNCTIONS
for any 0 < δ. By Theorem II.2 this implies that f is twice continuously differentiable on (−1, 1). Now the
argument of the proof of Lemma III.1 can be applied.
Now let us consider functions on the half-line (0, ∞).
Theorem III.3. An operator-monotone function f on (0, ∞) is operator-concave.
Proof. It is seen from the proof of Theorem II.2 that f admits a representation
Z 0
1 + λt
f (λ) = a + bλ +
dm(t)
−∞ t − λ
where b ≥ 0 and m is a positive measure. It suffices to prove that for each t < 0 the function ht (λ) := 1+λt
t−λ
is operator-concave. If t = 0, h0 (λ) = −1/λ is operator-concave on (0, ∞), as mentioned in Example III.1.
If t < 0,
1 + t2
ht (λ) = −t −
λ−t
is also operator-concave on (0, ∞), as shown in Example III.1. Since the function a + bλ is obviously
operator-concave, the weighted average f is operator-concave, too.
Corollary III.3.1. If a function f is operator-monotone and f (λ) > 0 on (0, ∞), then g(λ) := f (λ)−1
is operator-convex.
Proof. Take A, B ∈ S(0, ∞). Since f is operator-concave and the function 1/λ is operator-convex on
(0, ∞),
1
1
f
(A + B) ≥ {f (A) + f (B)}
2
2
and
g
1
(A + B)
2
−1
1
≤
{f (A) + f (B)}
2
1
≤ {g(A) + g(B)}.
2
Thus g is operator-convex.
Corollary III.3.2. A function on (0, ∞) is operator-monotone and operator-convex at the same time,
only if it is of the form a + bλ.
Theorem III.4. A continuous function f on (0, ∞) with f (0) := lim→0+ f () = 0 is operator-convex if
and only if f (λ)/λ is operator-monotone.
Proof. Suppose that f is operator-convex. By Theorem III.2 it is infinitely many times differentiable.
()
Then by Lemma III.1 for each > 0 the function f (λ)−f
is operator-monotone hence, as the limit, the
λ−
function f (λ)/λ is operator-monotone. Suppose conversely that f (λ)/λ is operator-monotone. Then as in
the proof of Theorem III.3 there are a and b > 0 and a measure m such that
Z 0
1 + tλ
f (λ)/λ = a + bλ +
dm(t),
−∞ t − λ
hence
f (λ) = aλ + bλ2 +
Z
0
−∞
λ(1 + λt)
dm(t).
t−λ
III. OPERATOR-CONVEX FUNCTIONS
17
Since bλ2 is operator-convex, it suffices to prove that for each t < 0 the function ht (λ) :=
operator-convex on (0, ∞). For t = 0 this is obvious. If t 6= 0,
ht (λ) = {−(1 + t2 ) − tλ} +
λ(1+λt)
t−λ
is
(1 + t2 ) |t|
λ−t
is operator-convex, as was shown in Example III.1.
Let us investigate for what exponent −∞ < s < ∞ the function f (λ) := λs on (0, ∞) is operatormonotone, operator-convex or operator-concave.
Examples III.2. 1. f (λ) = λs is operator-monotone (or operator-concave) if and only if 0 ≤ s ≤ 1.
This follows from Corollary I.2.2 and Theorem III.3 and from the fact that for s > 1 or < 0 the function
f is not concave.
2. f (λ) = λs is operator-convex if and only if 1 ≤ s ≤ 2 or −1 ≤ s ≤ 0. In fact, by Theorem III.4
for s > 0 f (λ) = λs is operator-convex if and only if λs−1 is operator-monotone. Therefore for s > 0
f is operator-convex if and only if 1 ≤ s ≤ 2. By Corollary III.3.1 f (λ) = λs is operator-convex for
s
−1
−1 ≤ s ≤ 0. For s < −1 the function λλ−1
does not admit any analytic continuation fˆ to the upper half
plane such that Im (fˆ(ζ)) > 0 for Im (ζ) > 0, hence f (λ) = λs is not operator-convex by Lemma III.1.
Lemma III.2. Let f (λ) > 0 on (0, ∞) and g(λ) := f (λ−1 )−1 . If f is operator-monotone, so is g. If f is
operator-convex and f (0) = 0, the function g is operator-convex.
Proof. Suppose that f is operator-monotone. Then by Theorem II.2 it admits an analytic continuation
fˆ to the complement of the closed negative real semi-axis such that Im fˆ(ζ) > 0 or < 0 according as
−1
Im (ζ) > 0 or Im (ζ) < 0. Then gˆ(ζ) := fˆ ζ −1
is an analytic continuation of g to the complement of
the closed negative real semi-axis such that Im (g(ζ)) > 0 or < 0 according as Im (ζ) > 0 or < 0. Therefore
again by Theorem II.2 g is operator-monotone.
Suppose next that f is operator-convex and f (0) = 0. Then by Theorem III.4 the function h(λ) :=
f (λ)/λ is operator-monotone. By applying the first part of this lemma to h, we can conclude that the
−1
function h λ−1
= g(λ)/λ is operator-monotone, hence by Theorem III.4 g is operator-convex.
Theorem III.5. Let f be a continuous positive function on (0, ∞) and A, B positive operators. If f is
operator-monotone then
f (A : B) ≤ f (A) : f (B).
If f is operator-convex and f (0) = 0 then
f (A : B) ≥ f (A) : f (B).
Proof. Suppose that f is operator-monotone. Then by Lemma III.2 g(λ) := f λ−1
monotone, hence operator-concave by Theorem III.2. Therefore
1 −1
1
−1
g
A +B
≥
g A−1 + g A−1
2
2
−1
is operator-
hence
f (A : B) ≤ f (A) : f (B).
The other assertion can be proved quite analogously by using Lemma III.2.
CHAPTER IV
Positive Maps
A (non-linear) transformation which maps L+ (H), the set of positive operators on H, to L+ (K) will be
called positive. In this chapter we shall study some special classes of positive maps.
Let us start with positive linear maps. A positive linear map Φ from L(H) to L(K) preserves orderrelation, that is, A ≤ B implies Φ(A) ≤ Φ(B), and preserves adjoint operation, that is Φ(A∗ ) = Φ(A)∗ . It
is said to be normalized if it transforms 1H to 1K . If Φ is normalized, it maps S(α, β; H) to S(α, β; K).
Lemma IV.1. A normalized positive linear map φ has the following properties.
(i) Φ A2 ≥ Φ(A)2 for A ∈ S(−∞, ∞; H).
(ii) Φ A−1 ≥ Φ(A)−1 for A ∈ S(0, ∞; H).
(i) By Remark after Corollary I.1.1 it suffices to prove the positivity of
R∞
Consider the spectral representation A = −∞ tdE(t). Since
Z ∞
Z ∞
A2 =
t2 dE(t)
and 1 =
dE(t),
Proof.
−∞
Φ A2
Φ(A)
!
Φ(A)
.
Φ(1)
−∞
we have, with tensor product notation,
! Z
!
∞
Φ A2
Φ(A)
t2 t
=
⊗ dE(t).
Φ(A) Φ(1)
t 1
−∞
!
t2 t
Since 2 × 2 matrices
are positive, for all −∞ < t < ∞ the right hand of the above expression
t 1
is positive.
!
Φ(A)
Φ(1)
. Since A is positive by assumption, it is written
(ii) It suffices to prove the positivity of
Φ(1) Φ A−1
R∞
in the form
! A = 0+ tdE(t). Now the positivity in question follows from the positivity of matrices
t
1
1
t
−1
for 0 < t < ∞, as in the proof of (i).
Theorem IV.1. Let Φ be a normalized positive linear map. If f is an operator-convex function on
(α, β), then
f [Φ(A)] ≤ Φ[f (A)]
for A ∈ S(α, β; H).
Proof. It suffices to consider the case (α, β) = (−1, 1). By Theorem III.2 f admits a representation
Z 1
λ2
f (λ) = a + bλ +
dm(t)
−1 1 − λt
with b ≥ 0 and a positive measure m. Since for A ∈ S(−1, 1; H)
Z 1
Φ[f (A)] = a + bΦ(A) +
Φ A2 (1 − tA)−1 dm(t)
−1
18
IV. POSITIVE MAPS
and
Z
1
f [Φ(A)] = a + bΦ(A) +
−1
19
Φ(A)2 {1 − tΦ(A)}−1 dm(t),
it suffices to show
Φ A2 (1 − tA)−1 ≥ Φ(A)2 {1 − tΦ(A)}−1
for
− 1 ≤ t ≤ 1.
For t = 0, this follows from Lemma IV.1. For t 6= 0, again by Lemma IV.1
Φ A2 (1 − tA)−1 = −t2 − t−1 Φ(A) + t2 φ (1 − tA)−1
≥ −t−2 − t−1 Φ(A) + t−2 {1 − tΦ(A)}−1
= Φ(A)2 {1 − tΦ(A)}−1 .
This completes the proof.
Corollary IV.1.1. Let Φ be a normalized positive linear map. Then for a positive operator A
Φ (Ap ) ≥ Φ(A)p
(1 ≤ p ≤ 2)
and
φ (Ap ) ≤ Φ(A)p
(0 ≤ p ≤ 1).
Proof. As shown in Examples III.1, λp is operator-convex on (0, ∞) for 1 ≤ p ≤ 2 while −λp is
operator-convex for 0 ≤ p ≤ 1.
1/p
Corollary IV.1.2. If Φ is a normalized positive map and if A is a positive operator, then Φ (Ap )
1/q
Φ (Aq )
whenever 1 ≤ p ≤ q or 12 q ≤ p ≤ 1 ≤ q.
≤
p/q
Proof. By Corollary IV.1.1 Φ (Aq )
≥ Φ (Ap ) whenever p ≤ q. If p ≥ 1in addition, this implies
1/p
q 1/q
p 1/p
Φ (A )
≥ Φ (A )
by Corollary I.2.2. If q ≥ 1 ≥ p ≥ 12 q, Φ (Ap )
≤ Φ(Aq )1/q , because λ1/q is
operator-monotone.
Corollary IV.1.3. If Φ is a positive linear map, for any positive operators A and B
Φ(A : B) ≤ Φ(A) : Φ(B)
and
Φ(A#B) ≤ Φ(A)#Φ(B).
Proof. We may assume A ∈ S(0, ∞; H). Let G be the closure of the range of Φ(A). Then there is
uniquely a normalized positive linear map Ψ from L(H) to L(G) such that
Φ(A)1/2 Ψ(C)Φ(A)1/2 = Φ A1/2 CA1/2
for C ∈ L(H).
Since the functions f (λ) :=
for any positive C
2λ
1+λ
and g(λ) := λ1/2 are operator-concave on (0, ∞), by Theorem IV.1 we have
Ψ(1 : C) = Ψ(f (C)) ≤ f (Ψ(C)) = 1 : Ψ(C)
and
Ψ(1#C) = Ψ(g(C)) ≤ g(Ψ(C)) = 1#Ψ(C).
With C = A−1/2 BA−1/2 this leads to the following
Φ(A : B) = Φ(A)1/2 Ψ(1 : C)Φ(A)1/2
= Φ(A)1/2 {1 : Ψ(C)}Φ(A)1/2 = Φ(A) : Φ(B)
and analogously
Φ(A#B) ≤ Φ(A)#Φ(B).
20
IV. POSITIVE MAPS
The converse of Theorem IV.1 is also true.
Theorem IV.2. Let α ≤ 0 ≤ β and f a continuous function on (α, β) with f (0) = 0. If
Φ(f (A)) ≥ f (Φ(A))
for every normalized positive linear map Φ and A ∈ S(α, β) then f is operator-convex.
Proof. Let K be the subspace of H ⊕ H, consisting
of all vectors x ⊕ x. Define the linear
!
! map Φ from
A11 A12
1 A11 + A22 A11 + A22
L(H ⊕ H) to L(K), that assigns to
the operator
, considered on
4 A11 + A22 A11 + A22
A21 A22
K. Then Φ is positive and normalized. Take A, B ∈ S(α, β; H). Then by definition we have
"
!#
!
!
A 0
f (A)
0
1 f (A) + f (B) f (A) + f (B)
Φ f
=Φ
=
,
4 f (A) + f (B) f (A) + f (B)
0 B
0
f (B)
while
"
A
f Φ
0
0
B
!#
"
1
1
!
!
!#
−1
0
0
1 1
1
√
1
0 12 (A + B)
2 −1 1
!
!
!
f (0)
0
1 1
1 1 −1
1
√
=√
0
f 12 (A + B)
2 1 1
2 −1 1
!
1 f 12 (A + B) f 12 (A + B)
.
=
2 f 12 (A + B) f 12 (A + B)
"
A 0
0 B
1
=f √
2
Now by assumption
Φ f
!#
"
A 0
≥f Φ
0 B
!#
which implies
f
1
1
(A + B) ≤ {f (A) + f (B)}.
2
2
This completes the proof.
A (non-linear) map Φ from a convex subset of L(H) to L(K) is said to be a convex map (resp. a
concave map) if Φ 12 (A + B) ≤ 12 {Φ(A) + Φ(B)} (resp. Φ 12 (A + B) ≥ 12 {Φ(A) + Φ(B)}).
We shall be concerned with convexity or concavity of maps Φs,t (A) = As ⊗ At defined on S(0, ∞; H)
where −∞ < s, t < ∞.
Lemma IV.2. If Φ and Ψ are concave maps with range in S(0, ∞; K) then the maps Θ(A) := Φ(A)#Ψ(A)
and Ξ(A) := Φ(A) : Ψ(A) are concave.
Proof. By Corollary I.2.1 and concavity of Φ and Ψ
1
1
1
Θ
(A + B) = Φ
(A + B) #Ψ
(A + B)
2
2
2
1
1
1
1
≥
Φ(A) + Φ(B) #
Ψ(A) + Ψ(B)
2
2
2
2
1
≥ {Φ(A)#Ψ(A) + Φ(B)#Ψ(B)
2
1
= {Θ(A) + Θ(B)},
2
which proves the concavity of Θ. The concavity of Ξ is proved analogously by using Corollary I.3.1.
Theorem IV.3. The map Φp,q (A) := AP ⊗ Aq is concave if 0 ≤ p, q and p + q ≤ 1.
IV. POSITIVE MAPS
21
Proof. Consider the set Ω of (p, q) in R2+ for which Φp,q are concave. We claim that Ω is a convex set.
In fact, let (pi , qi ) ∈ Ω and p = 12 (p1 + p2 ), q = 12 (q1 + q2 ). Then since
Φp,q (A) = Ap ⊗ Aq
= (Ap1 #Ap2 ) ⊗ (Aq1 #Aq2 )
= (Ap1 ⊗ Aq1 ) # (Ap2 ⊗ Aq2 )
= Φp1 ,q1 (A)#Φp2 ,q2 (A),
by Lemma IV.2 the map Φp,q is concave. Obviously (0, 0), (1, 0) and (0, 1) belong to Ω, hence so does (p, q)
for which 0 ≤ p, q and p + q ≤ 1.
Corollary IV.3.1. For positive Ai and Bi (i = 1, 2)
(A1 : B1 )p ⊗ (A2 : B2 )q ≤ (Ap1 ⊗ Aq2 ) : (B1p ⊗ B2q )
whenever 0 ≤ p, q and p + q ≤ 1.
Proof. The concavity of Φp,q is seen to be equivalent to the inequality
With A =
A1
0
(A : B)p ⊗ (A : B)q ≤ (Ap ⊗ Aq ) : (B p ⊗ B q ) .
!
!
0
B1 0
and B =
this inequality implies the inequality in the assertion.
A2
0 B2
Theorem IV.4. If f and g are positive, operator-monotone functions on (0, ∞), then the map Φ(A) :=
f (A)−1 ⊗ g(A)−1 is convex.
−1
−1
Proof. Let h(λ) := f λ−1
and k(λ) := g λ−1
. Then the convexity of Φ is seen to be equivalent
to that for A, B ∈ S(0, ∞)
1
{h(A) ⊗ k(A) + h(B) ⊗ k(B)}.
2
By Lemma III.2 both h and k are operator-monotone, so that by Theorem III.5
h(A : B) ⊗ k(A : B) ≤
{h(A : B) ⊗ k(A : B)} ≤ {h(A) : h(B)} ⊗ {k(A) : k(B)}
and further by Corollaries I.3.2 and I.2.4
{h(A) : h(B)} ⊗ {k(A) : k(B)} ≤ {h(A)#h(B)} ⊗ {k(A)#k(B)}
= {h(A) ⊗ k(A)}#{h(B) ⊗ k(B)}
≤
1
{h(A) ⊗ k(A) + h(B) ⊗ k(B)}.
2
Corollary IV.4.1. The map Φ−p,−q (A) = A−p ⊗ A−q is convex if 0 ≤ p, q ≤ 1.
This follows from Theorem IV.3 and Corollary I.2.2.
Theorem IV.5. The map Φ−p,q (A) = A−p ⊗ Aq is convex if 0 ≤ p ≤ q − 1 < 1.
Proof. Let us consider first the case q = p + 1 < 2. The well-known integral representation of λp for
0 < p < 1 (see [11], [12] p. )
Z ∞
p
−1
λ = π sin(pπ)
tp−1 λ(λ + t)−1 dt
0
22
IV. POSITIVE MAPS
is applied to get
(1 ⊗ A)
Z ∞
−1
= π −1 sin(pπ)
t1−p A−1 ⊗ A2 A−1 ⊗ A + t
dt.
Φ−p,p+1 (A) = A−1 ⊗ A
p
0
Therefore it suffices to prove that for each t > 0 the map
−1
Φt (A) := A−1 ⊗ A2 A−1 ⊗ A + t
is convex. To this end, remark, first of all
n
Φt (A) = 1 ⊗ A − t (A ⊗ 1)−1 +
1 ⊗ t−1 A
−1 o−1
,
the first term of which is a convex map. It remains to show that the map
n
−1 o−1
Θt (A) := (A ⊗ 1)−1 + 1 ⊗ t−1 A
is convex. But this follows from Lemma IV.2, because 2Θt (A) = Ψ(A) : Ξ(A) where Ψ(A) := A ⊗ 1 and
Ξ(A) := 1 ⊗ t−1 A are obviously concave.
Next let us consider the case p < q − 1. Let r = p(q − 1). Since 0 < r < 1, the function λr is
operator-concave by Theorem III.4, hence for positive A and B
r
1
1
1
1
A+ B
≥ Ar + B r ,
2
2
2
2
which implies
1
1
A+ B
2
2
−r
≥
1 r 1 r
A + B
2
2
−1
Further the operator-monotoneousness of λq−1 yields
−p −(q−1)
1
1
1 r 1 r
A+ B
≥
A + B
.
2
2
2
2
It follows from the first part of the proof, as Corollary IV.3.1 does from Theorem IV.3, that for positive
A, B, C and D
−(q−1) q
o
1
1
1
1
1 n −(q−1)
C+ D
⊗
A+ B
≤
C
⊗ Aq + D−(q−1) ⊗ B q .
2
2
2
2
2
Now use these inequalities with C = Ar and D = B r to get
−p q −(q−1) q
1
1
1
1
1 r 1 r
1
1
A+ B
⊗
A+ B
≤
A + B
⊗
A+ B
2
2
2
2
2
2
2
2
1 −p
≤
A ⊗ Aq + B −p ⊗ B q .
2
This shows the convexity of Φ−p,q .
It remains to consider the case p = 1 and q = 2. The convexity of Φ−1,2 means that
(A + B)−1 ⊗ (A + B)2 ≤ A−1 ⊗ A2 + B −1 ⊗ B 2 .
With C = A−1/2 BA−1/2 this inequality is equivalent to the inequality
(1 + C)−1 ⊗ (1 + C)A(1 + C) ≤ 1 ⊗ A + C −1 ⊗ CAC,
and further to the inequality
C ⊗ (CA + AC) ≤ C 2 ⊗ A + 1 ⊗ CAC.
IV. POSITIVE MAPS
Considering the spectral representation A =
A is an orthoprojection, i.e. A2 = A. Then
R∞
0
23
λdE(λ), it suffices to prove the above inequality for the case
C 2 ⊗ A + 1 ⊗ CAC − C ⊗ (CA + AC) = (C ⊗ A − 1 ⊗ AC)∗ (C ⊗ A − 1 ⊗ AC) ≥ 0.
This completes the proof.
!
A 0
instead of A, it is seen that the maps A 7→ As , A 7→ At and R+ 3
0 α
α → αs+t are concave, hence as mentioned in Example III.2 we have 0 ≤ s, t and s + t ≤ 1.
If Φs,t is convex with s ≤ t, just as above, the maps A 7→ As and A 7→ At are convex, so that
1 ≤ s ≤ t ≤ 2 or −1 ≤ s ≤ t ≤ 0 or −1 ≤ s ≤ 0 ≤ 1 ≤ t ≤ 2. Replacing A by scalar, we see that the map
{α, β} → αs β t is convex on the positive cone of R2 . Therefore, for arbitrarily fixed α, β > 0, the function
φ(λ) := (α + λ)s (β − λ)t is a convex function of λ in a neighborhood of 0. By differentiation this implies
If Φs,t is concave, using
s(s − 1)αs−2 β t − 2stαs−1 β t−1 + t(t − 1)αs β t−2 ≥ 0.
If −1 ≤ s ≤ 0 ≤ 1 ≤ t ≤ 2 or 1 ≤ s ≤ t ≤ 2, by arbitrariness of α and β the above inequality is equivalent to
s2 t2 ≤ st(s − 1)(t − 1).
If −1 ≤ s ≤ 0 < 1 ≤ t ≤ 2, this is possible only if s + t ≥ 1. But the case 1 ≤ s ≤ t ≤ 2 is not consistent
with the inequality. Thus we can conclude that Theorem IV.3 exhausts all the case Φs,t is concave while
Theorem IV.4 and IV.5 exhaust all the case Φs,t (s ≤ t) is convex.
Note
Chapter I. Corollary I.1.3 is due to Krein [16]. Geometric mean was introduced by Pusz and Woronowicz
[20], who proved Theorem I.2. Corollary I.2.3 can be considered as a non-commutative version of the result of
−1
Lieb & Ruskai [18]. Anderson & Duffin [2] defined A−1 + B −1
as parallel sum of two positive matrices.
Hilbert space operator case was treated in Anderson & Trapp [3], who proved Theorem I.3.
Chapter II and III. The content of these chapters is the famous theory of L¨owner [19] and Kraus [15].
The full account of the theory can be found in Donoghue [10] and Davis [9]. Theorem II.1 and II.2 are due
to Bendat & Sherman [6]. The Hilbert space method in Chapter II is due to Koranyi [14].
Chapter IV. Theorem IV.1 is due to Davis [8] and Choi [7], while Theorem IV.2 is pointed out in Davis
[9]. The inequality Φ(A : B) ≤ Φ(A) : Φ(B) was proved for a special case by Anderson & Trapp [4].
Theorem IV.3 and Corollary IV.4.1 were proved by Lieb [17] by a different method. Epstein [13] gave a
simpler proof. Uhlmann [21] also used geometric means to prove Theorem IV.3. The idea in the proof of
Theorem IV.5 will be developed in a forthcoming paper [5].
24
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[1]
[2]
[3]
[4]
N. I. Akhiezer and I.M. Glazman. Theory of linear operators in Hilbert space. Pitman Pub., Boston :, 1981.
W. N. Anderson, Jr. and R. J. Duffin. Series and parallel addition of matrices. J. Math. Anal. Appl., 26:576–594, 1969.
W. N. Anderson, Jr. and G. E. Trapp. Shorted operators. II. SIAM J. Appl. Math., 28:60–71, 1975.
William N. Anderson, Jr. and George E. Trapp. A class of monotone operator functions related to electrical network theory.
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[19] Karl L¨
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25
Index
concave map, 20
convex map, 20
geometric mean, 5
harmonic mean, 6
normalized map, 18
operator-concave function, 14
operator-convex function, 14
operator-monotone function, 8
positive operator, 3
27
Symbols
A : B, 6
A#B, 5
G,H,K, 3
L(H), 3
L+ (H), 3
S(α, β), 8
S(α, β; H), 8
Φs,t (A), 20
29