Topics on Operator Inequalities

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.


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.


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).


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.


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.


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.


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.

Linear Algebra and Appl., 15(1):53–67, 1976.

[5] T. Ando. Concavity of certain maps on positive definite matrices and applications to Hadamard products. Linear Algebra

Appl., 26:203–241, 1979.

[6] Julius Bendat and Seymour Sherman. Monotone and convex operator functions. Trans. Amer. Math. Soc., 79:58–71, 1955.

[7] Man Duen Choi. A Schwarz inequality for positive linear maps on C∗ -algebras. Illinois J. Math., 18:565–574, 1974.

[8] Chandler Davis. A Schwarz inequality for convex operator functions. Proc. Amer. Math. Soc., 8:42–44, 1957.

[9] Chandler Davis. Notions generalizing convexity for functions defined on spaces of matrices. In Proc. Sympos. Pure Math.,

Vol. VII, pages 187–201. Amer. Math. Soc., Providence, R.I., 1963.

[10] William F. Donoghue, Jr. Monotone matrix functions and analytic continuation. Springer-Verlag, New York, 1974. Die

Grundlehren der mathematischen Wissenschaften, Band 207.

[11] Nelson Dunford and Jacob T. Schwartz. Linear operators. Part I. Wiley Classics Library. John Wiley & Sons Inc., New

York, 1988. General theory, With the assistance of William G. Bade and Robert G. Bartle, Reprint of the 1958 original,

A Wiley-Interscience Publication.

[12] Nelson Dunford and Jacob T. Schwartz. Linear operators. Part II. Wiley Classics Library. John Wiley & Sons Inc., New

York, 1988. Spectral theory. Selfadjoint operators in Hilbert space, With the assistance of William G. Bade and Robert G.

Bartle, Reprint of the 1963 original, A Wiley-Interscience Publication.

[13] H. Epstein. Remarks on two theorems of E. Lieb. Comm. Math. Phys., 31:317–325, 1973.

[14] A. Koranyi. On a theorem of Lowner and its connections with resolvents of selfadjoint transformations. Acta Sci. Math.

Szeged, 17:63–70, 1956.

[15] Fritz Kraus. Uber konvexe Matrixfunktionen. Math. Z., 41(1):18–42, 1936.

[16] M. Krein. The theory of self-adjoint extensions of semi-bounded Hermitian transformations and its applications. I. Rec.

Math. [Mat. Sbornik] N.S., 20(62):431–495, 1947.

[17] Elliott H. Lieb. Convex trace functions and the Wigner-Yanase-Dyson conjecture. Advances in Math., 11:267–288, 1973.

[18] Elliott H. Lieb and Mary Beth Ruskai. Some operator inequalities of the Schwarz type. Advances in Math., 12:269–273,

1974.

[19] Karl Lowner. Uber monotone Matrixfunktionen. Math. Z., 38(1):177–216, 1934.

[20] W. Pusz and S. L. Woronowicz. Functional calculus for sesquilinear forms and the purification map. Rep. Mathematical

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[21] A. Uhlmann. Relative entropy and the Wigner-Yanase-Dyson-Lieb concavity in an interpolation theory. Comm. Math.

Phys., 54(1):21–32, 1977.

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

Topics on Operator Inequalities

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