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
T =
(A C∗
C B
)where A and B act on G and K respectively and C acts from G to K. Then in order
that T is positive it is necessary and sufficient that A and B are positive and there is a contraction W (i.e.
‖W‖ ≤ 1) from G to K such that C = B1/2WA1/2.
Proof. Suppose that A ≥ 0, B ≥ 0 and C = B1/2WA1/2 with a contraction W . The condition
‖W‖ ≤ 1 implies WW ∗ ≤ 1, hence B1/2WW ∗B1/2 ≤ B. Then T admits a factorization T = S∗S where
S =
(A1/2 W ∗B1/2
0(B −B1/2WW ∗B1/2
)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∥∥∥A−1/2C∗y∥∥∥2 = sup
x
|(Cx, y)|2
(Ax, x)≤∥∥∥B1/2y
∥∥∥2 ,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−1/2C∗ = UB1/2. 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.
Corollary I.1.2. If
(A C∗
C B
)is positive, then there is the minimum of all X for which
(A C∗
C X
)are positive.
Proof. As in the proof of Theorem I.1 the positivity of
(A C∗
C X
)implies
(A−1/2C∗)∗ (A−1/2C∗) ≤ X.
This means that(A−1/2C∗)∗ (A−1/2C∗) 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−1C∗.
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 ‖T‖ = ‖S‖. Further, among all T
satisfying these conditions there are the minimum Tµ and the maximum TM .
Proof. It can be assumed that ‖S‖ = 1. Let S =
(A
C
)where A acts on G and C does from G to
K = H G. By assumption A is self-adjoint and
‖Ax‖2 + ‖Cx‖2 ≤ ‖x‖2 for x ∈ G.
Therefore C∗C ≤ 1 − A2. T must be of the form T =
(A C∗
C X
)for which
(1+A C∗
C 1+X
)and(
1−A −C∗
−C 1−X
)are positive. The inequalities C∗C ≤ 1 − A2 and 2−1(1 − A) ≤ 1 imply that
C∗2−1C ≤ 2−1(1−A2
)= 2−1(1 − A)(1 + A) ≤ 1 + A. By Remark after Corollary I.1.2 this implies
the positivity of
(1+A C∗
C 1+ 1
). Therefore there is the minimum Bµ of all X for which
(1+A C∗
C 1+X
)are positive. In fact, Bµ is defined by Bµ =
[(1+A)−1/2C∗]∗ [(1+A)−1/2C∗] − 1. In order to conclude
that Tµ =
(A C∗
C Bµ
)is the minimum in the assertion, it remains to show that
(1−A −C∗
−C 1−Bµ
)is positive.
Since
(1−A −C∗
−C 21
)is positive just as
(1+A C∗
C 21
), (1−A)−1/2C∗ is a well defined linear operator. By
Corollary I.1.2 it reduces to prove the inequality∥∥∥(1−A)−1/2C∗y∥∥∥2 ≤ 2 ‖y‖2 −
∥∥∥(1+A)−1/2C∗y∥∥∥2 .
But∥∥∥(1+A)−1/2C∗y∥∥∥2 + ∥∥∥(1−A)−1/2C∗y
∥∥∥2 =
= 2((1−A2
)−1C∗y, C∗y
)= 2
∥∥∥(1−A2)−1/2
C∗y∥∥∥2 .
Since C∗C ≤ 1−A2 implies
(1−A2 C∗
C 1
)≥ 0,
2∥∥∥(1−A2
)−1/2C∗y
∥∥∥2 ≤ 2 ‖y‖2 ,
as expected. Analogously TM =
(A C∗
C BM
)with BM = 1−
[(1−A)−1/2C∗]∗ [(1−A)−1/2C∗] is shown to
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
operators X on H for which
(A X
X B
)are positive.
I. GEOMETRIC AND HARMONIC MEANS 5
Proof. Consider first the case that A has bounded inverse. Let D = A−1/2BA−1/2. Then
(A X
X B
)
is positive if and only if
(1 A−1/2XA−1/2
A−1/2XA−1/2 D
)is positive. We claim that D1/2 is the maximum
of all Y for which
(1 Y
Y D
)are positive. The positivity of
(1 D1/2
D1/2 D
)is obvious by Corollary I.1.2.
Suppose that
(1 Y
Y D
)is positive. By Theorem I.1 there is a contraction W on H such that Y = D1/2W .
Introduce a new scalar product on H by 〈x, y〉 := (D1/2x, y). Since Y is self-adjoint 〈Wx, y〉 = 〈x,Wy〉, forevery x, y ∈ H which means that W is self-adjoint with respect to this new scalar product. Since ‖W‖ ≤ 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
|〈Wx, x〉| ≤ 〈x, x〉 for all x ∈ H,
which implies −D1/2 ≤ Y ≤ D1/2. Thus D1/2 is the maximum as claimed. As a consequence,
A1/2D1/2A1/2 = A1/2(A−1/2BA−1/2
)1/2A1/2 is the maximum of all X for which
(A X
X B
)are positive.
If A does not admit bounded inverse, consider Aε = A+ ε with ε > 0. Let Cε be the maximum of all X
for which
(Aε X
X B
)are positive. Since obviously Cε ≥ Cε′ ≥ 0 whenever ε > ε′ > 0, the operator lim
ε→0+Cε
is the maximum of all X for which
(Aε X
X B
)are positive. This completes the proof.
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) A#B = B#A.
(ii) (αA)#(αB) = α(A#B) for α ≥ 0.
(iii) (A1 +A2)#(B1 +B2) ≥ (A1#B1) + (A2#B2).
(iv) C(A#B)C∗ ≤ (CAC∗)#(CBC∗) for any linear operator C.
(v) A#A = A, 1#A = A1/2 and 0#A = 0.
(vi) A#B = A1/2(A−1/2BA−1/2
)1/2A1/2 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 ≥ Bp.
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
Bp = Bp1#Bp2 by Corollary I.2.1, Ap ≥ Bp 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
be a linear operator from G to K. If
(Ai C∗
C Bi
)are positive i = 1, 2, then so is
(A1#A2 C∗
C B1#B2
).
Proof. We may assume the bounded invertibility of A1 and A2. Then by assumption and Corollary
I.1.2 CA−1i C∗ ≤ Bi i = 1, 2. Therefore by Corollary I.2.1
C(A1#A2)−1C∗ = C
(A−1
1 #A−12
)C∗ <
(CA−1
1 C∗)# (CA−12 C∗) ≤ B1#B2,
which implies the positivity of
(A1#A2 C∗
C B1#B2
)by Corollary I.1.2.
Corollary I.2.4. Arithmetic mean is greater than geometric mean, that is 12 (A + B) ≥ A#B for
positive A and B.
Proof. We may assume the bounded invertibility of A. Then by Corollary I.2.1
A#B = A1/2(A−1/2BA−1/2
)1/2A1/2 ≤ A1/2
(1
21+
1
2A−1/2BA−1/2
)A1/2 =
1
2(A+B),
because X ≤ 1
2
(1+X2
)for any positive X.
The harmonic mean of two positive operators A and B must be defined by
1
2
(A−1 +B−1
)−1
when
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+ ε) : (B + ε) as ε → 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
0 2B
)≥
(X X
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 infy(Ay, y) + (B(x− y), x− y)
hence to the condition
(Xx, x) ≤ 2
(Bx, x)− sup
y
|(Bx, y)|2
((A+B)y, y)
.
Since by definition
A : B = 2A(A+B)−1B = 2B −B(A+B)−1B
,
we have
((A : B)x, x) = 2
(Bx, x)− sup
y
|(Bx, y)|2
((A+B)y, y)
.
This shows that A : B is the maximum of all X for which
(2A 0
0 2B
)≥
(X X
X X
).
Corollary I.3.1. Harmonic means have the following properties.
(i) A : B = B : A.
(ii) (αA) : (αB) = α(A : B) for α ≥ 0.
(iii) (A1 +A2) : (B1 +B2) ≥ (A1 : B1) + (A2 : B2).
(iv) 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)−1B if A has bounded inverse.
(vii) A−1 : B−1 =
12 (A+B)
−1if 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
2
(A−1 +B−1
). By taking inverse, this yields, by Corollaries I.2.1 and I.3.1, A#B ≥ A : B.
Theorem I.4. Let A1 and A2 (resp. B1 and B2) be positive operators of G (resp. K) and let C be a
linear operator from G to K. If
(Ai C∗
C Bi
)are positive i = 1, 2, then so are
(12 (A1 +A2) C∗
C B1 : B2
)and(
A1 : A2 C∗
C 12 (B1 +B2)
).
Proof. We may assume the bound invertibility of A1 and A2. Then by assumption and Corollary I.1.2
CA−1i C∗ ≤ Bi i = 1, 2. Therefore by Corollary I.3.1
C
(1
2(A1 +A2)
)−1
C∗ = C(A−1
1 : A−12
)C∗ ≤
(CA−1
1 C∗) : (CA−12 C∗) ≤ B1 : B2,
which implies the positivity of
(12 (A1 +A2) C∗
C B1 : B2
)by Corollary I.1.2. The positivity of(
A1 : A2 C∗
C 12 (B1 +B2)
)is proved analogously.
Corollary I.4.1. The following identity holds for positive A and B;1
2(A+B)
#(A : B) = A#B.
Proof. By definition
(A A#B
A#B B
)and
(B A#B
A#B A
)are positive, so that
(12 (A+B) A#B
A#B A : B
)is positive by Theorem I.4, which implies
1
2(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 respec-
tively, leads, by taking inverse, to the inequality
(A : B)#
1
2(A+B)
≤ A#B.
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.
3. For µ /∈ (α, β) the function f(λ) := 1µ−λ is operator-monotone on (α, β). In particular f(λ) = − 1
λ 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(12 ((β − α)λ+ α+ β)
)is operator monotone on (−1, 1). On this basis, we shall treat only the interval
(−1, 1).
Lemma II.1. If a continuously differentiable function f on (−1, 1) is operator-monotone, then for any
choice λi ∈ (−1, 1) i = 1, . . . , N the matrix(f [1](λi, λj)
)Ni,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 ′(λ).
Proof. Observe the matrix A :=
λ1 0
. . .
0 λN
, considered as a self-adjoint operator on the N -
dimensional Hilbert space CN . Take any complex numbers ξ1, . . . , ξN and consider the positive matrix
B :=(ξiξj
)Ni,j=1
. Since A ∈ S(−1, 1;CN
), for sufficiently small ε > 0, A+ εB belongs to S
(−1, 1;CN
). We
claim that for any continuously differentiable function g on (−1, 1)
limε→0+
ε−1g(A+ εB)− g(A) =
N∑i,j=1
g[1](λi, λj)PiBPj
8
II. OPERATOR-MONOTONE FUNCTIONS 9
where Pi is the orthoprojection to the subspace spanned by the vector ei := (δij)Nj=1. In fact, this is true if
g is polynomial, because for g(λ) = λn
ε−1g(A+ εB)− g(A) =
n∑k=1
Ak−1BAn−k +O(ε)
=
N∑i,j=1
n∑k=1
λk−1i λn−k
j PiBPj +O(ε)
=
N∑i,j=1
g[1](λi, λj)PiBPj +O(ε).
Use approximation of g′ by polynomials on a suitable subinterval when g is merely continuously differentiable.
Now since f is operator-monotone and A+ εB ≥ A for all ε > 0, we have
N∑i,j=1
f [1](λi, λj)PiBPj = limε→0+
ε−1f(A+ εB)− f(A) ≥ 0.
Let x =∑N
i,j=1 ei, then the positivity of∑N
i,j=1 f[1](λi, λj)PiBPj implies
0 ≤N∑
i,j=1
f [1](λi, λj)(PiBPjx, x) =
N∑i,j=1
f [1](λi, λj)ξiξj .
Since (ξi) are arbitrary, this means that the matrix(f [1](λi, λj)
)Ni,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
f(λ) = f(0) +
∫ 1
−1
λ
1− λtdm(t) for λ ∈ (−1, 1).
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⟨∑i
αifλi,∑j
βjfµj
⟩=∑i,j
f [1](λi, µj)αiβj .
This is positive semi-definite by Lemma II.1. Let H be the associated Hilbert space. Let us show that
λn → λ implies ‖fλn− fλ‖ → 0. In fact,
‖fλn− fλ‖2 = f [1](λn, λn)− 2f [1](λn, λ) + f [1](λ, λ)
= f ′(λn)− 2f(λn)− f(λ)
λn − λ+ f ′(λ).
Now by continuous differentiability of f
f ′(λn)− 2f(λn)− f(λ)
λn − λ+ f ′(λ) → f ′(λ)− 2f ′(λ) + f ′(λ) = 0.
Let D be the linear subspace of H, spanned by all fλ (λ 6= 0). Then D is dense in H, as shown above. Define
a linear map T from D to H by
Tfλ := λ−1fλ − f0 (λ 6= 0).
T is symmetric hence well-defined, because
〈Tfλ, fµ〉 =µf(λ)− λf(λ)
λµ(λ− µ)− f(0)
λµ= 〈fλ, T fµ〉 .
10 II. OPERATOR-MONOTONE FUNCTIONS
We claim that T admits a (not necessarily bounded) self-adjoint extension T . This follows from a theorem
of von Neumann, (see [12] p.1231), because the anti-linear involution J , defined by
J
(∑i
αifλi
)=∑i
αifλi,
makes D invariant and commutes with T . Since by definition
(µ− λ)fµ = (1− λT )(µfµ − λfλ),
the kernel if 1− λT is orthogonal to the linear subspace spanned by fµ (µ 6= λ, and µ 6= 0). This subspace
is dense in H as shown above, so that 1 − λT is injective, hence (1 − λT )−1 is a (unbounded) self-adjoint
operator. Therefore f0 = (1− λT )fλ implies fλ = (1− λT )−1f0.
Now let T =∫∞−∞ tdE(t) be the spectral representation of the self-adjoint operator T , and dm(t) =
〈dE(t)f0, f0〉. Then for λ 6= 0
λ−1f(λ)− f(0) = f [1](λ, 0)
= 〈fλ, f0〉 =⟨(1− λT )−1f0, f0
⟩=
∫ ∞
−∞
1
1− λtdm(t).
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∫ α−1
0
∫ ∞
α
1
(1− λt)2dm(t)dλ ≤
∫ α−1
0
∫ ∞
−∞
1
(1− λt)2dm(t)dλ
=
∫ α−1
0
〈fλ, fλ〉 dλ =
∫ α−1
0
f ′(λ)dλ = f(α−1
)− f(0) < ∞.
By Fubini’s theorem we have∫ α−1
0
∫ ∞
α
1
(1− λt)2dm(t)dλ =
∫ ∞
α
∫ α−1
0
1
(1− λt)2dλdm(t)
≥∫ ∞
α
1
t
∫ 1
0
s−2dsdm(t).
But this last expression is finite inly if [α,∞) is an m-zero set. Analogously (−∞,−α] is an m-zero set.
Remark. In Lemma II.2∫ 1
−1dm(t) = 〈f0, f0〉 = f ′(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
det
(f [1](λ1, µ1) f [1](λ1, µ2)
f [1](λ2, µ1) f [1](λ2, µ2)
)≥ 0.
Proof. We shall use the notations in the proof of the previous Lemma. First remark that
f [1](λ, µ) = 〈fλ, fµ〉 =∫ 1
−1
1
(1− λt)(1− µt)dm(t).
If the measure m is concentrated on a single point, say t0, then the determinant in question is equal to a
scalar multiple of
det
((1− λ1t0)
−1(1− µ1t0)−1 (1− λ1t0)
−1(1− µ2t0)−1
(1− λ2t0)−1(1− µ1t0)
−1 (1− λ2t0)−1(1− µ2t0)
−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 µ′1 and µ′
2 such that λ1 ≤ µ′1 ≤ µ1 and λ2 ≤ µ′
2 ≤ µ2, and
det
(f [1](λ1, µ
′1) f [1](λ1, µ
′2)
f [1](λ2, µ′1) f [1](λ2, µ
′2)
)= 0.
This implies that there are α1 and α2 such that |α1|+ |α2| 6= 0 and⟨fλi
, α1fµ′1+ α2fµ′
2
⟩= 0 (i = 1, 2).
Choose β1 and β2 so that β1 + β2 = α1 + α2 and β1λ2 + β2λ1 = α1µ′2 + α2µ
′1. Then it follows that
0 =⟨β1fλ1
+ β2fλ2, α1fµ′
1+ α2fµ′
2
⟩=
∫ 1
−1
α1 + α2 − (α1µ′2 + α2µ
′1)t2
(1− λ1t)(1− λ2t)(1− µ′1t)(1− µ′
2t)dm(t).
Since the denominator is strictly positive on [−1, 1] and since the measurem is not concentrated on any single
point, the above relation implies that α1 + α2 = 0 and α1µ′2 + α2µ
′1 = 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
function φ on (−∞,∞) such that φ is non-negative, vanishes outside of (−1, 1) and∫ 1
−1φ(t)dt = 1.
Suppose that f is operator-monotone on (−1, 1). For any 0 < ε < 1, define a function f(ε) on (−1+ε, 1−ε)
by
f(ε)(t) :=1
ε
∫ ε
−ε
φ(s/ε)f(t− s)ds.
The function f(ε) is operator-monotone on (−1 + ε, 1 − ε), because each f(t − ε) is so. Obviously f(ε) is
infinitely many times differentiable, and as ε converges to 0, f(ε)(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 = 1
2 (β +1), and
apply Lemma II.3 to the operator-monotone function f(ε) with λ2 = t and µ1 = s. By taking limit, we can
conclude that
det
(f [1](λ1, s) f [1](λ1, µ2)
f [1](t, s) f [1](t, µ2)
)≥ 0,
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
f(λ) = f(0) +
∫ 1
−1
λ
1− λtdm(t) for λ ∈ (−1, 1).
12 II. OPERATOR-MONOTONE FUNCTIONS
Proof. Suppose that f admits a representation of the above form. For each t ∈ [−1, 1] the function
ht(λ) :=λ
1−λt is operator-monotone on (−1, 1). In fact, if t = 0, h0(λ) = λ is operator-monotone. If t 6= 0,
ht(λ) = −t−1 + t−2
t−1−λ is operator-monotone as stated in one of the Examples II.1. As a consequence, the
weighted average∫ 1
−1ht(λ)dm(t) is operator-monotone, and so is f .
Suppose conversely that f is operator-monotone. Then with the notations in front of Lemma II.4, for
each 0 < ε < 1 f(ε)((1 − ε)λ) is operator-monotone on (−1, 1). Then since f(ε)((1 − ε)λ) is continuously
differentiable, by Lemma II.2 there is a measure mε on [−1, 1] such that
f(ε)((1− ε)λ) = f(ε)(0) +
∫ 1
−1
λ
1− λtdmε(t).
We claim that∫ 1
−1dmε(t) are bounded when ε runs over (0, 1/2). As remarked after Lemma II.2, we have∫ 1
−1
dmε(t) = f ′(ε)(0).
Since Lemma II.4 f satisfies Lipshitz condition on any closed subinterval of (−1, 1), it has derivative f ′(t)
for almost all λ and ess. sup−ε<t<ε
|f ′(t)| = rε < ∞. Further it is easy to see that
f ′(ε)(0) =
(1− ε)
ε
∫ ε
−ε
f ′(−s)φ(sε
)ds ≤ (1− ε)rε.
These considerations show the boundedness of∫ 1
−1dmε(t) when ε runs over (0, 1/2). Now by the Helly
theorem (see [11] p. 381) there is a sequence εn → 0 and a measure m on [−1, 1] such that
limn→∞
∫ 1
−1
g(t)dmεn(t) =
∫ 1
−1
g(t)dm(t)
for all continuous functions g on [−1, 1]. Therefore for λ ∈ (−1, 1)
limn→∞
∫ 1
−1
λ
1− λtdmεn(t) =
∫ 1
−1
λ
1− λtdm(t).
Finally the assertion follows from
limn→∞
f(εn)((1− εn)λ) = f(λ).
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
f(ζ) = f(0) +
∫ 1
−1
ζ
1− ζtdm(t) for ζ with Im (ζ) 6= 0.
The function f maps the upper half plane to itself. Indeed
Im(f(ζ)
)=
∫ 1
−1
Im (ζ)
|1− ζt|2dm(t).
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
that f is operator-monotone it is necessary and sufficient that it admits an analytic continuation f to the
upper and lower half planes such that Im(f(ζ)
)> 0 for Im (ζ) > 0.
II. OPERATOR-MONOTONE FUNCTIONS 13
Proof. Suppose that f admits an analytic continuation f 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 on Ω := (−∞,∞) \ (α, β) such that
f(ζ) = a+ bζ +
∫Ω
1 + ζt
t− ζdm(t).
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+λtt−λ is operator monotone, because
ht(λ) = −t+1 + t2
t− λ
and the function 1t−λ is operator-monotone on (α, β). Therefore the weighted average f(λ) is also operator-
monotone 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 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
f
(1
2(A+B)
)≤ 1
2f(A) + f(B) forA,B ∈ S(α, β).
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
1
2
A2 +B2
−1
2(A+B)
2
=1
4(A−B)2 ≥ 0.
3. For µ < α the function f(λ) := 1λ−µ is operator-convex on (α,∞). In particular, f(λ) := 1
λ is operator-
convex on (0,∞). In fact, for A,B ∈ S(α, β), A− µ and B − µ belong to S(0,∞) and
1
2
(A− µ)−1 + (B − µ)−1
= (A− µ) : (B − µ)−1.
By Corollary I.2.4 and I.3.2.
(A− µ) : (B − µ)−1 ≥1
2(A− µ) +
1
2(B − µ)
−1
=
1
2(A+B)− µ
−1
.
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
for any choice λi ∈ (−1, 1) i = 1, 2, . . . , N the matrix(g[1](λi, λj)
)Ni,j=1
is positive. Observe the matrix
A :=
λ1 0
λN
0 λN+1
with λN+1 = µ, considered as a self-adjoint operator on the (N + 1)-dimensional
Hilbert space CN+1. Take any complex numbers ξ1, . . . , ξN and consider the matrix B :=
ξ1
0...
ξN
ξ1, . . . , ξN 0
.
Since A ∈ S(−1, 1;CN+1
), for sufficiently small ε > 0 A+ εB belongs to S
(−1, 1;CN+1
). 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+ εB) is twice differentiable and
d2f(A+ εB)
dε2
∣∣∣∣ε=0
=
N+1∑i,j,k=1
f [2](λi, λj , λk)PiBPjBPk
where Pi is the orthoprojection to the subspace spanned by the vector ei := (δij)N+1j=1 and f [2](s, t, u) =
h[1]s (t, u) with hs(t) = f [1](s, t). Now the operator-convexity of f implies
d2f(A+ εB)
dε2
∣∣∣∣ε=0
≥ 0, as in the
case of scalar convex functions. Let x =∑N
i=1 ei. Then
0 ≤N+1∑
i,j,k=1
f [2](λi, λj , λk)(PiBPjBPkx, x) =
=
N∑i,k=1
f [2](λi, λN+1, λk)ξk ξi =
N∑i,k=1
g[1](λi, λk)ξk ξi.
Since (ξi) are arbitrary, this means that the matrix(g[1](λi, λj)
)Ni,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
f [1](0, λ) = f ′(0) +
∫ 1
−1
λ
1− λtdm(t),
hence
f(λ) = a+ bλ+
∫ 1
−1
λ2
1− λtdm(t),
where a = f(0) and b = f ′(0). For each t ∈ [−1, 1] the function ht(λ) :=λ2
1−λt is operator-convex. In fact, if
t = 0, ht(λ) = λ2, and if t 6= 0,
ht(λ) = −t−2 − t−1λ+t−2
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
f1(λ) =∫ λ
0f(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
f[1]1 (0, λ) =
∫ 1
0
f(sλ)ds and f[1]2 (0, λ) = f(λ),
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
f(λ) = a+ bλ+
∫ 1
−1
λ2
1− λtdm(t) for λ ∈ (−1, 1).
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(ε)((1−ε)λ) is operator-convex for each 0 < ε < 1. There-
fore by Lemma III.1 for any µ ∈ (−1, 1) the functionf(ε)((1− ε)λ)− f(ε)((1− ε)µ)
λ− µis operator-monotone on
(−1, 1) so that by taking limit as ε → 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
f(λ) = a+ bλ+
∫ 0
−∞
1 + λt
t− λdm(t)
where b ≥ 0 and m is a positive measure. It suffices to prove that for each t < 0 the function ht(λ) :=1+λtt−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,
ht(λ) = −t− 1 + t2
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,∞),
f
(1
2(A+B)
)≥ 1
2f(A) + f(B)
and
g
(1
2(A+B)
)≤[1
2f(A) + f(B)
]−1
≤ 1
2g(A) + g(B).
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) := limε→0+ f(ε) = 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 ε > 0 the function f(λ)−f(ε)λ−ε is operator-monotone hence, as the limit, the
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
f(λ)/λ = a+ bλ+
∫ 0
−∞
1 + tλ
t− λdm(t),
hence
f(λ) = aλ+ bλ2 +
∫ 0
−∞
λ(1 + λt)
t− λdm(t).
III. OPERATOR-CONVEX FUNCTIONS 17
Since bλ2 is operator-convex, it suffices to prove that for each t < 0 the function ht(λ) := λ(1+λt)t−λ is
operator-convex on (0,∞). For t = 0 this is obvious. If t 6= 0,
ht(λ) = −(1 + t2)− tλ+ (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 operator-
monotone, 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
−1 ≤ s ≤ 0. For s < −1 the function λs−1λ−1 does not admit any analytic continuation f to the upper half
plane such that Im (f(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
f to the complement of the closed negative real semi-axis such that Im f(ζ) > 0 or < 0 according as
Im (ζ) > 0 or Im (ζ) < 0. Then g(ζ) := f(ζ−1
)−1is 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
function h(λ−1
)−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
)−1is operator-
monotone, hence operator-concave by Theorem III.2. Therefore
g
(1
2
(A−1 +B−1
))≥ 1
2
g(A−1
)+ g
(A−1
)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 order-
relation, 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).
Proof. (i) By Remark after Corollary I.1.1 it suffices to prove the positivity of
(Φ(A2)
Φ(A)
Φ(A) Φ(1)
).
Consider the spectral representation A =∫∞−∞ tdE(t). Since
A2 =
∫ ∞
−∞t2dE(t) and 1 =
∫ ∞
−∞dE(t),
we have, with tensor product notation,(Φ(A2)
Φ(A)
Φ(A) Φ(1)
)=
∫ ∞
−∞
(t2 t
t 1
)⊗ dE(t).
Since 2× 2 matrices
(t2 t
t 1
)are positive, for all −∞ < t < ∞ the right hand of the above expression
is positive.
(ii) It suffices to prove the positivity of
(Φ(A) Φ(1)
Φ(1) Φ(A−1
)). Since A is positive by assumption, it is written
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
f(λ) = a+ bλ+
∫ 1
−1
λ2
1− λtdm(t)
with b ≥ 0 and a positive measure m. Since for A ∈ S(−1, 1;H)
Φ[f(A)] = a+ bΦ(A) +
∫ 1
−1
Φ[A2(1− tA)−1
]dm(t)
18
IV. POSITIVE MAPS 19
and
f [Φ(A)] = a+ bΦ(A) +
∫ 1
−1
Φ(A)21− tΦ(A)−1dm(t),
it suffices to show
Φ[A2(1− tA)−1
]≥ Φ(A)21− 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−1Φ(A) + t2φ
[(1− tA)−1
]≥ −t−2 − t−1Φ(A) + t−21− tΦ(A)−1
= Φ(A)21− 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.
Corollary IV.1.2. If Φ is a normalized positive map and if A is a positive operator, then Φ(Ap)1/p ≤
Φ(Aq)1/q
whenever 1 ≤ p ≤ q or 12q ≤ p ≤ 1 ≤ q.
Proof. By Corollary IV.1.1 Φ (Aq)p/q ≥ Φ(Ap) whenever p ≤ q. If p ≥ 1in addition, this implies
Φ (Aq)1/q ≥ Φ(Ap)
1/pby Corollary I.2.2. If q ≥ 1 ≥ p ≥ 1
2q, Φ (Ap)1/p ≤ 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)1/2Ψ(C)Φ(A)1/2 = Φ(A1/2CA1/2
)for C ∈ L(H).
Since the functions f(λ) := 2λ1+λ and g(λ) := λ1/2 are operator-concave on (0,∞), by Theorem IV.1 we have
for any positive C
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−1/2BA−1/2 this leads to the following
Φ(A : B) = Φ(A)1/2Ψ(1 : C)Φ(A)1/2
= Φ(A)1/21 : 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
L(H ⊕H) to L(K), that assigns to
(A11 A12
A21 A22
)the operator
1
4
(A11 +A22 A11 +A22
A11 +A22 A11 +A22
), considered on
K. Then Φ is positive and normalized. Take A,B ∈ S(α, β;H). Then by definition we have
Φ
[f
(A 0
0 B
)]= Φ
(f(A) 0
0 f(B)
)=
1
4
(f(A) + f(B) f(A) + f(B)
f(A) + f(B) f(A) + f(B)
),
while
f
[Φ
(A 0
0 B
)]= f
[1√2
(1 −1
1 1
)(0 0
0 12 (A+B)
)1√2
(1 1
−1 1
)]
=1√2
(1 −1
1 1
)(f(0) 0
0 f(12 (A+B)
)) 1√2
(1 1
−1 1
)
=1
2
(f(12 (A+B)
)f(12 (A+B)
)f(12 (A+B)
)f(12 (A+B)
)) .
Now by assumption
Φ
[f
(A 0
0 B
)]≥ f
[Φ
(A 0
0 B
)]which implies
f
(1
2(A+B)
)≤ 1
2f(A) + f(B).
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)
)≤ 1
2Φ(A) + Φ(B) (resp. Φ(12 (A+B)
)≥ 1
2Φ(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
Θ
(1
2(A+B)
)= Φ
(1
2(A+B)
)#Ψ
(1
2(A+B)
)≥1
2Φ(A) +
1
2Φ(B)
#
1
2Ψ(A) +
1
2Ψ(B)
≥ 1
2Φ(A)#Ψ(A) + Φ(B)#Ψ(B)
=1
2Θ(A) + Θ(B),
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 = 1
2 (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 ⊗Aq
2) : (Bp1 ⊗Bq
2)
whenever 0 ≤ p, q and p+ q ≤ 1.
Proof. The concavity of Φp,q is seen to be equivalent to the inequality
(A : B)p ⊗ (A : B)q ≤ (Ap ⊗Aq) : (Bp ⊗Bq) .
With A =
(A1 0
0 A2
)and B =
(B1 0
0 B2
)this inequality implies the inequality in the assertion.
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.
Proof. Let h(λ) := f(λ−1
)−1and k(λ) := g
(λ−1
)−1. Then the convexity of Φ is seen to be equivalent
to that for A,B ∈ S(0,∞)
h(A : B)⊗ k(A : B) ≤ 1
2h(A)⊗ k(A) + h(B)⊗ k(B).
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) : 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
2h(A)⊗ k(A) + h(B)⊗ k(B).
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. )
λp = π−1 sin(pπ)
∫ ∞
0
tp−1λ(λ+ t)−1dt
22 IV. POSITIVE MAPS
is applied to get
Φ−p,p+1(A) =(A−1 ⊗A
)p(1⊗A)
= π−1 sin(pπ)
∫ ∞
0
t1−p(A−1 ⊗A2
) A−1 ⊗A+ t
−1dt.
Therefore it suffices to prove that for each t > 0 the map
Φt(A) :=(A−1 ⊗A2
) A−1 ⊗A+ t
−1
is convex. To this end, remark, first of all
Φt(A) = 1⊗A− t(A⊗ 1)−1 +
(1⊗
(t−1A
))−1−1
,
the first term of which is a convex map. It remains to show that the map
Θt(A) :=(A⊗ 1)−1 +
(1⊗
(t−1A
))−1−1
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−1A
)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(1
2A+
1
2B
)r
≥ 1
2Ar +
1
2Br,
which implies (1
2A+
1
2B
)−r
≥(1
2Ar +
1
2Br
)−1
Further the operator-monotoneousness of λq−1 yields(1
2A+
1
2B
)−p
≥(1
2Ar +
1
2Br
)−(q−1)
.
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 1
2C +
1
2D
−(q−1)
⊗1
2A+
1
2B
q
≤ 1
2
C−(q−1) ⊗Aq +D−(q−1) ⊗Bq
.
Now use these inequalities with C = Ar and D = Br to get(1
2A+
1
2B
)−p
⊗(1
2A+
1
2B
)q
≤(1
2Ar +
1
2Br
)−(q−1)
⊗(1
2A+
1
2B
)q
≤ 1
2
A−p ⊗Aq +B−p ⊗Bq
.
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 ⊗B2.
With C = A−1/2BA−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) ≤ C2 ⊗A+ 1⊗ CAC.
IV. POSITIVE MAPS 23
Considering the spectral representation A =∫∞0
λdE(λ), it suffices to prove the above inequality for the case
A is an orthoprojection, i.e. A2 = A. Then
C2 ⊗A+ 1⊗ CAC − C ⊗ (CA+AC) = (C ⊗A− 1⊗AC)∗(C ⊗A− 1⊗AC) ≥ 0.
This completes the proof.
If Φs,t is concave, using
(A 0
0 α
)instead of A, it is seen that the maps A 7→ As, A 7→ At and R+ 3
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 → αsβt 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
s(s− 1)αs−2βt − 2stαs−1βt−1 + t(t− 1)αsβt−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
s2t2 ≤ 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
Lieb & Ruskai [18]. Anderson & Duffin [2] defined(A−1 +B−1
)−1as 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 Lowner [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
Bibliography
[1] N. I. Akhiezer and I.M. Glazman. Theory of linear operators in Hilbert space. Pitman Pub., Boston :, 1981.
[2] W. N. Anderson, Jr. and R. J. Duffin. Series and parallel addition of matrices. J. Math. Anal. Appl., 26:576–594, 1969.
[3] W. N. Anderson, Jr. and G. E. Trapp. Shorted operators. II. SIAM J. Appl. Math., 28:60–71, 1975.
[4] William N. Anderson, Jr. and George E. Trapp. A class of monotone operator functions related to electrical network theory.
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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