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arXiv:1509.04362v1 [math.FA] 15 Sep 2015

CLASS OPERATORS IN HILBERT SPACES

S.S. DRAGOMIR1,2

Abstract. Some inequalities for quantumf-divergence of trace class opera-tors in Hilbert spaces are obtained. It is shown that for normalised convex functions it is nonnegative. Some upper bounds for quantum f-divergence in terms of variational andχ2-distance are provided. Applications for some classes of divergence measures such as Umegaki and Tsallis relative entropies are also given.

1. Introduction

Let (X,A) be a measurable space satisfying|A|>2 andµbe aσ-finite measure on (X,A).LetP be the set of all probability measures on (X,A) which are abso-lutely continuous with respect toµ. ForP, Q∈ P, letp= dP andq= dQ denote theRadon-Nikodym derivatives ofP andQwith respect toµ.

Two probability measures P, Q ∈ P are said to be orthogonal and we denote this byQP if

P({q= 0}) =Q({p= 0}) = 1.

Let f : [0,∞) → (−∞,∞] be a convex function that is continuous at 0, i.e., f(0) = limu↓0f(u).

In 1963, I. Csisz´ar [10] introduced the concept off-divergence as follows.

Definition 1. Let P, Q∈ P. Then

(1.1) If(Q, P) =

Z

X p(x)f

q(x) p(x)

dµ(x),

is called thef-divergence of the probability distributions QandP.

Remark 1. Observe that, the integrand in the formula (1.1) is undefined when p(x) = 0.The way to overcome this problem is to postulate for f as above that

(1.2) 0f

q(x) 0

=q(x) lim u↓0

uf

1

u

, x∈X.

We now give some examples off-divergences that are well-known and often used in the literature (see also [6]).

1991Mathematics Subject Classification. 47A63; 47A99.

Key words and phrases. Selfadjoint bounded linear operators, Functions of operators, Trace of operators, quantum divergence measures, Umegaki and Tsallis relative entropies.

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1.1. The Class of χα-Divergences. The f-divergences of this class, which is generated by the functionχα, α[1,),defined by

χα(u) =|u1|α, u[0,∞)

have the form

(1.3) If(Q, P) =

Z

X p

q p−1

α dµ=

Z

X

p1−α|q−p|αdµ.

From this class only the parameter α = 1 provides a distance in the topologi-cal sense, namely the total variation distance V (Q, P) =R

X|q−p|dµ. The most prominent special case of this class is, however,Karl Pearson’s χ2-divergence

χ2(Q, P) =

Z

X q2

pdµ−1 that is obtained forα= 2.

1.2. Dichotomy Class. From this class, generated by the function fα: [0,∞)→

R

fα(u) =

    

    

u−1−lnu for α= 0;

1

α(1−α)[αu+ 1−α−uα] for α∈R\ {0,1};

1−u+ulnu for α= 1;

only the parameterα=12 f1

2(u) = 2 ( √

u1)2provides a distance, namely, the Hellinger distance

H(Q, P) =

Z

X

(√q−√p)2dµ

12

.

Another important divergence is the Kullback-Leibler divergence obtained for α= 1,

KL(Q, P) =

Z

X qln

q p

dµ.

1.3. Matsushita’s Divergences. The elements of this class, which is generated by the functionϕα, α∈(0,1] given by

ϕα(u) :=|1−uα|

1

α, u[0,),

are prototypes of metric divergences, providing the distances

α(Q, P)α

.

1.4. Puri-Vincze Divergences. This class is generated by the functions Φα, α∈

[1,∞) given by

Φα(u) := |1−u| α

(u+ 1)α−1, u∈[0,∞).

It has been shown in [27] that this class provides the distances [IΦα(Q, P)]

1

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1.5. Divergences of Arimoto-type. This class is generated by the functions

Ψα(u) :=

     

     

α α−1

h

(1 +uα)α1 −21α−1(1 +u)

i

for α∈(0,∞)\ {1};

(1 +u) ln 2 +ulnu(1 +u) ln (1 +u) for α= 1;

1

2|1−u| for α=∞.

It has been shown in [33] that this class provides the distances [IΨα(Q, P)]

min(α,1

α)

forα(0,∞) and 12V(Q, P) forα=∞.

Forf continuous convex on [0,∞) we obtain the∗-conjugate function off by

f∗(u) =uf

1

u

, u∈(0,∞)

and

f∗(0) = lim u↓0f

(u) .

It is also known that if f is continuous convex on [0,∞) then so isf∗.

The following two theorems contain the most basic properties off-divergences. For their proofs we refer the reader to Chapter 1 of [29] (see also [6]).

Theorem 1(Uniqueness and Symmetry Theorem). Letf, f1be continuous convex

on[0,∞). We have

If1(Q, P) =If(Q, P),

for allP, Q∈ P if and only if there exists a constantc∈Rsuch that f1(u) =f(u) +c(u−1),

for anyu[0,∞).

Theorem 2 (Range of Values Theorem). Let f : [0,∞) → R be a continuous convex function on[0,∞).

For anyP, Q∈ P, we have the double inequality (1.4) f(1)≤If(Q, P)≤f(0) +f∗(0).

(i) If P=Q, then the equality holds in the first part of (1.4).

Iff is strictly convex at1,then the equality holds in the first part of (1.4) if and only if P =Q;

(ii) If QP, then the equality holds in the second part of (1.4).

If f(0) +f∗(0)<∞,then equality holds in the second part of (1.4) if and only ifQ⊥P.

The following result is a refinement of the second inequality in Theorem 2 (see [6, Theorem 3]).

Theorem 3. Letf be a continuous convex function on[0,∞)withf(1) = 0(f is normalised) andf(0) +f∗(0)<.Then

(1.5) 0≤If(Q, P)≤21[f(0) +f∗(0)]V (Q, P)

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For other inequalities forf-divergence see [5], [12]-[22].

Motivated by the above results, in this paper we obtain some new inequalities for quantum f-divergence of trace class operators in Hilbert spaces. It is shown that for normalised convex functions it is nonnegative. Some upper bounds for quantum f-divergence in terms of variational andχ2-distance are provided. Applications for some classes of divergence measures such as Umegaki and Tsallis relative entropies are also given.

In what follows we recall some facts we need concerning the trace of operators and quantumf-divergence for trace class operators in infinite dimensional complex Hilbert spaces.

2. Some Preliminary Facts

2.1. Some Facts on Trace of Operators. Let (H,h·,·i) be a complex Hilbert space and{ei}i∈I anorthonormal basis ofH.We say that A∈ B(H) is a Hilbert-Schmidt operator if

(2.1) X

i∈I

kAeik2<∞.

It is well know that, if{ei}i∈I and {fj}j∈J are orthonormal bases forH andA

B(H) then

(2.2) X

i∈I

kAeik2=X j∈I

kAfjk2=X j∈I

kA∗fjk2

showing that the definition (2.1) is independent of the orthonormal basis andAis a Hilbert-Schmidt operator iffA∗ is a Hilbert-Schmidt operator.

Let B2(H) the set of Hilbert-Schmidt operators inB(H). For A∈ B2(H) we

define

(2.3) kAk2:=

X

i∈I

kAeik2

!1/2

for{ei}i∈I an orthonormal basis ofH.This definition does not depend on the choice of the orthonormal basis.

Using the triangle inequality inl2(I), one checks thatB

2(H) is a vector space

and thatk·k2 is a norm onB2(H),which is usually called in the literature as the

Hilbert-Schmidt norm.

Denotethe modulus of an operatorA∈ B(H) by|A|:= (A∗A)1/2

.

Because k|A|xk =kAxk for allx H, A is Hilbert-Schmidt iff|A| is Hilbert-Schmidt and kAk2 =k|A|k2. From (2.2) we have that if A∈ B2(H), thenA∗ ∈ B2(H) andkAk2=kA∗k2.

The following theorem collects some of the most important properties of Hilbert-Schmidt operators:

Theorem 4. We have:

(i)(B2(H),k·k2)is a Hilbert space with inner product

(2.4) hA, Bi2:=

X

i∈I

hAei, Beii=X i∈I

hB∗Aei, eii

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(ii) We have the inequalities

(2.5) kAk ≤ kAk2

for anyA∈ B2(H) and

(2.6) kATk2, kT Ak2≤ kTk kAk2

for anyA∈ B2(H) andT ∈ B(H) ;

(iii)B2(H)is an operator ideal in B(H), i.e. B(H)B2(H)B(H)⊆ B2(H) ;

(iv)Bf in(H),the space of operators of finite rank, is a dense subspace ofB2(H) ;

(v) B2(H)⊆ K(H), where K(H) denotes the algebra of compact operators on

H.

If{ei}i∈I an orthonormal basis ofH,we say thatA∈ B(H) istrace class if

(2.7) kAk1:=

X

i∈I

h|A|ei, eii<∞.

The definition of kAk1 does not depend on the choice of the orthonormal basis

{ei}i∈I. We denote byB1(H) the set of trace class operators inB(H).

The following proposition holds:

Proposition 1. IfA∈ B(H),then the following are equivalent: (i)A∈ B1(H) ;

(ii) |A|1/2∈ B2(H) ;

(ii) A(or |A|)is the product of two elements of B2(H).

The following properties are also well known:

Theorem 5. With the above notations: (i) We have

(2.8) kAk1=kA∗k1 and kAk2≤ kAk1 for anyA∈ B1(H) ;

(ii) B1(H)is an operator ideal in B(H), i.e.

B(H)B1(H)B(H)⊆ B1(H) ;

(iii) We have

B2(H)B2(H) =B1(H) ;

(iv) We have

kAk1= sup{hA, Bi2 |B ∈ B2(H), kBk ≤1};

(v)(B1(H),k·k1)is a Banach space.

(iv) We have the following isometric isomorphisms

B1(H)∼=K(H)∗ andB1(H)∗∼=B(H),

whereK(H)∗ is the dual space ofK(H)andB1(H)∗ is the dual space ofB1(H).

We define thetrace of a trace class operatorA∈ B1(H) to be

(2.9) tr (A) :=X

i∈I

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where {ei}i∈I an orthonormal basis ofH. Note that this coincides with the usual definition of the trace if H is finite-dimensional. We observe that the series (2.9) converges absolutely and it is independent from the choice of basis.

The following result collects some properties of the trace:

Theorem 6. We have:

(i) If A∈ B1(H)thenA∗∈ B1(H)and

(2.10) tr (A∗) = tr (A);

(ii) IfA∈ B1(H)and T ∈ B(H), thenAT, T A∈ B1(H)and

(2.11) tr (AT) = tr (T A) and |tr (AT)| ≤ kAk1kTk; (iii)tr (·) is a bounded linear functional onB1(H)withktrk= 1;

(iv) If A, B∈ B2(H)then AB, BA∈ B1(H)andtr (AB) = tr (BA) ;

(v)Bf in(H) is a dense subspace ofB1(H).

Utilising the trace notation we obviously have that

hA, Bi2= tr (B

A) = tr (AB) and

kAk22= tr (A

A) = tr

|A|2

for anyA, B∈ B2(H).

The following H¨older’s type inequality has been obtained by Ruskai in [36]

(2.12) |tr (AB)| ≤tr (|AB|)≤htr|A|1/αiαhtr|B|1/(1−α)i1−α

whereα∈(0,1) andA, B∈ B(H) with|A|1/α,|B|1/(1−α)∈ B1(H).

In particular, forα= 12 we get the Schwarz inequality

(2.13) |tr (AB)| ≤tr (|AB|)≤htr|A|2i1/2htr|B|2i1/2

withA, B∈ B2(H).

IfA≥0 andP∈ B1(H) withP ≥0,then

(2.14) 0≤tr (P A)≤ kAktr (P).

Indeed, sinceA≥0,thenhAx, xi ≥0 for anyx∈H.If{ei}i∈I an orthonormal basis ofH, then

0≤DAP1/2ei, P1/2eiE≤ kAk P

1/2

ei

2

=kAk hP ei, eii

for anyi∈I.Summing overi∈I we get

0≤X

i∈I

D

AP1/2ei, P1/2eiE≤ kAkX

i∈I

hP ei, eii=kAktr (P)

and since

X

i∈I

D

AP1/2ei, P1/2eiE=X i∈I

D

P1/2AP1/2ei, eiE= trP1/2AP1/2= tr (P A)

we obtain the desired result (2.14).

This obviously imply the fact that, if A and B are selfadjoint operators with A≤B andP ∈ B1(H) withP≥0,then

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Now, ifAis a selfadjoint operator, then we know that

|hAx, xi| ≤ h|A|x, xi for anyxH.

This inequality follows by Jensen’s inequality for the convex function f(t) = |t|

defined on a closed interval containing the spectrum of A. If{ei}i∈I is an orthonormal basis ofH, then

|tr (P A)| =

X

i∈I

D

AP1/2ei, P1/2eiE

≤X

i∈I

D

AP1/2ei, P1/2eiE

(2.16)

≤ X

i∈I

D

|A|P1/2ei, P1/2eiE= tr (P|A|),

for anyAa selfadjoint operator andP ∈ B+1 (H) :={P ∈ B1(H) withP ≥0}.

For the theory of trace functionals and their applications the reader is referred to [39].

For some classical trace inequalities see [7], [9], [32] and [43], which are continu-ations of the work of Bellman [3]. For related works the reader can refer to [1], [4], [7], [24], [28], [30], [31], [37] and [40].

2.2. Quantumf-Divergence for Trace Class Operators. On complex Hilbert space (B2(H),h·,·i2),where the Hilbert-Schmidt inner product is defined by

hU, Vi2:= tr (V∗U), U, V ∈ B2(H),

for A, B ∈ B+(H) consider the operators L

A : B2(H) → B2(H) and RB :

B2(H)→ B2(H) defined by

LAT :=AT andRBT :=T B. We observe that they are well defined and since

hLAT, Ti

2=hAT, Ti2= tr (T

AT) = tr|T∗|2A≥0 and

hRBT, Ti

2=hT B, Ti2= tr (T

T B) = tr

|T|2B≥0

for any T ∈ B2(H), they are also positive in the operator order of B(B2(H)),

the Banach algebra of all bounded operators onB2(H) with the normk·k2 where kTk2= tr

|T|2, T ∈ B2(H).

Since tr|X∗|2

= tr|X|2for anyX ∈ B2(H),then also

tr (T∗AT) = trT∗A1/2A1/2T= trA1/2T ∗

A1/2T

= tr

A

1/2T

2

= tr

A1/2T ∗

2

= tr

T

A1/2

2

forA≥0 andT ∈ B2(H).

We observe thatLAandRB are commutative, therefore the productLARB is a

selfadjoint positive operator inB(B2(H)) for any positive operatorsA, B∈ B(H).

For A, B ∈ B+(H) with B invertible, we define the Araki transform A

A,B :

B2(H)→ B2(H) by AA,B :=LARB−1. We observe that for T ∈ B2(H) we have

AA,BT=AT B−1and hAA,BT, Ti

2=

AT B−1, T

2= tr T

AT B−1

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Observe also, by the properties of trace, that

tr T∗AT B−1

= trB−1/2T∗A1/2A1/2T B−1/2

= trA1/2T B−1/2∗A1/2T B−1/2= tr

A

1/2T B−1/2

2

giving that

(2.17) hAA,BT, Ti

2= tr

A

1/2

T B−1/2

2

≥0

for anyT ∈ B2(H).

We observe that, by the definition of operator order and by (2.17) we have r1B2(H)≤AA,B≤R1B2(H) for someR≥r≥0 if and only if

(2.18) rtr|T|2tr

A

1/2T B−1/2

2

≤Rtr|T|2

for anyT ∈ B2(H).

We also notice that a sufficient condition for (2.18) to hold is that the following inequality in the operator order ofB(H) is satisfied

(2.19) r|T|2≤ A

1/2

T B−1/2

2

≤R|T|2

for anyT ∈ B2(H).

Let U be a selfadjoint linear operator on a complex Hilbert space (K;h·,·i). The Gelfand map establishes a ∗-isometrically isomorphism Φ between the set C(Sp (U)) of allcontinuous functionsdefined on thespectrumofU,denoted Sp (U), and the C∗-algebraC∗(U) generated byU and the identity operator 1K on K as follows:

For anyf, g∈C(Sp (U)) and anyα, β∈Cwe have (i) Φ (αf+βg) =αΦ (f) +βΦ (g) ;

(ii) Φ (f g) = Φ (f) Φ (g) and Φ ¯f= Φ (f)∗; (iii) kΦ (f)k=kfk:= supt∈Sp(U)|f(t)|;

(iv) Φ (f0) = 1K and Φ (f1) =U,wheref0(t) = 1 andf1(t) =t,fort∈Sp (U).

With this notation we define

f(U) := Φ (f) for allf C(Sp (U))

and we call it thecontinuous functional calculus for a selfadjoint operatorU. IfUis a selfadjoint operator andfis a real valued continuous function on Sp (U), then f(t)≥0 for any t Sp (U) implies thatf(U) ≥0, i.e. f(U) is a positive operator onK.Moreover, if bothf andgare real valued functions on Sp (U) then the following important property holds:

(P) f(t)≥g(t) for any tSp (U) implies that f(U)≥g(U) in the operator order ofB(K).

Letf : [0,∞)→Rbe a continuous function. Utilising the continuous functional calculus for the Araki selfadjoint operator AQ,P ∈ B(B2(H)) we can define the

quantumf-divergenceforQ, P ∈S1(H) :={P ∈ B1(H), P ≥0 with tr (P) = 1 }

andP invertible, by

Sf(Q, P) :=Df(AQ,P)P1/2, P1/2E

2= tr

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If we consider the continuous convex function f : [0,∞)→ R, with f(0) := 0 andf(t) =tlnt fort >0 then forQ, P S1(H) andQ, P invertible we have

Sf(Q, P) = tr [Q(lnQlnP)] =:U(Q, P), which is the Umegaki relative entropy.

If we take the continuous convex functionf : [0,∞)→R,f(t) =|t1|fort0 then forQ, P ∈S1(H) withP invertible we have

Sf(Q, P) = tr (|Q−P|) =:V (Q, P), whereV (Q, P) is thevariational distance.

If we takef : [0,∞)→R,f(t) =t21 fort0 then forQ, P S1(H) withP invertible we have

Sf(Q, P) = tr Q2P−1

−1 =:χ2(Q, P),

which is called theχ2-distance

Letq ∈(0,1) and define the convex functionfq : [0,∞)→Rbyfq(t) = 1−tq

1−q. Then

Sfq(Q, P) =

1−tr QqP1−q

1−q , which isTsallis relative entropy.

If we consider the convex functionf : [0,∞)→Rbyf(t) = 1

2 √

t12

, then

Sf(Q, P) = 1−trQ1/2P1/2=:h2(Q, P),

which is known asHellinger discrimination.

If we take f : (0,∞) → R, f(t) = lnt then for Q, P S1(H) and Q, P invertible we have

Sf(Q, P) = tr [P(lnP−lnQ)] =U(P, Q).

The reader can obtain other particular quantumf-divergence measures by utilizing the normalized convex functions from Introduction, namely the convex functions defining the dichotomy class, Matsushita’s divergences, Puri-Vincze divergences or divergences of Arimoto-type. We omit the details.

In the important case of finite dimensional spaceH and the generalized inverse P−1,numerous properties of the quantumf-divergence, mostly in the case whenf

isoperator convex,have been obtained in the recent papers [25], [26], [34], [35] and the references therein.

In what follows we obtain several inequalities for the larger class of convex func-tions on an interval.

3. Inequalities for f Convex and Normalized

Suppose thatI is an interval of real numbers with interior ˚I and f :I →R is a convex function on I. Then f is continuous on ˚I and has finite left and right derivatives at each point of ˚I. Moreover, if x, y ∈ ˚I and x < y, then f−′ (x) ≤ f+′ (x) ≤ f−′ (y) ≤ f+′ (y), which shows that both f−′ and f+′ are nondecreasing

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For a convex function f :I R, the subdifferential off denoted by ∂f is the set of all functions ϕ:I→[−∞,∞] such thatϕ˚I⊂Rand

(G) f(x)≥f(a) + (x−a)ϕ(a) for anyx, a∈I.

It is also well known that iff is convex onI,then∂f is nonempty, f−′,f+′ ∈∂f

and ifϕ∈∂f, then

f′ (x)≤ϕ(x)≤f+′ (x) for anyx˚I. In particular,ϕis a nondecreasing function.

Iff is differentiable and convex on ˚I, then∂f ={f′}.

We are able now to state and prove the first result concerning the quantum f-divergence for the general case of convex functions.

Theorem 7. Letf : [0,∞)→Rbe a continuous convex function that is normalized, i.e. f(1) = 0.Then for anyQ, P ∈S1(H),with P invertible, we have

(3.1) 0≤Sf(Q, P).

Moreover, if f is continuously differentiable, then also (3.2) Sf(Q, P)≤Sℓf′(Q, P)−Sf′(Q, P),

where the function ℓis defined as ℓ(t) =t, t∈R.

Proof. Since f is convex and normalized, then by the gradient inequality (G) we have

f(t)≥(t−1)f+′ (1)

fort >0.

Applying the property (P) for the operator AQ,P, then we have for any T

B2(H)

hf(AQ,P)T, Ti

2 ≥ f

+(1)

A

Q,P−1B2(H)T, T2

= f+′ (1)

hAQ,PT, Ti2− kTk2

, which, in terms of trace, can be written as

(3.3) tr (T∗f(AQ,P)T)f

+(1)

tr

Q

1/2T P−1/2

2

−tr|T|2

for anyT ∈ B2(H).

The inequality (3.3) is of interest in itself.

Now, if we take in (3.3)T =P1/2whereP ∈S1(H),withP invertible, then we

get

Sf(Q, P)≥f+′ (1) [tr (Q)−tr (P)] = 0 and the inequality (3.1) is proved.

Further, if f is continuously differentiable, then by the gradient inequality we also have

(t−1)f′(t)≥f(t) fort >0.

Applying the property (P) for the operator AQ,P, then we have for any T

B2(H)

A

Q,P−1B2(H)

f′(AQ,P)T, T

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namely

hAQ,Pf′(AQ,P)T, Ti2− hf′(AQ,P)T, Ti2≥ hf(AQ,P)T, Ti2, for anyT ∈ B2(H),or in terms of trace

(3.4) tr (T∗AQ,Pf(AQ,P)T)tr (Tf(AQ,P)T)tr (Tf(AQ,P)T),

for anyT ∈ B2(H).

This inequality is also of interest in itself.

If in (3.4) we takeT =P1/2, whereP S1(H), withP invertible, then we get

the desired result (3.2).

Remark 2. If we take in (3.2)f : (0,∞)→R,f(t) =lntthen forQ, P S1(H) andQ, P invertible we have

(3.5) 0≤U(P, Q)≤χ2(P, Q). We need the following lemma that is of interest in itself.

Lemma 1. LetS be a selfadjoint operator on the Hilbert space(K,h·,·i)and with spectrumSp (S)⊆[γ,Γ]for some real numbersγ,Γ.Ifg: [γ,Γ]→Cis a continuous function such that

(3.6) |g(t)−λ| ≤ρfor any t[γ,Γ] for some complex numberλCand positive numberρ,then

|hSg(S)x, xi − hSx, xi hg(S)x, xi| ≤ρh|S− hSx, xi1H|x, xi (3.7)

≤ρh

S2x, x

− hSx, xi2i1/2

for anyx∈K, kxk= 1. Proof. We observe that

(3.8) hSg(S)x, xi − hSx, xi hg(S)x, xi=h(S− hSx, xi1H) (g(S)−λ1H)x, xi for anyxK,kxk= 1.

For any selfadjoint operatorB we have the modulus inequality (3.9) |hBx, xi| ≤ h|B|x, xi for anyxK,kxk= 1.

Also, utilizing the continuous functional calculus we have for each fixedxK,kxk= 1

|(S− hSx, xi1H) (g(S)−λ1H)|=|S− hSx, xi1H| |g(S)−λ1H| ≤ρ|S− hSx, xi1H|,

which implies that

(3.10) h|(S− hSx, xi1H) (g(S)−λ1H)|x, xi ≤ρh|S− hSx, xi1H|x, xi for anyxK,kxk= 1.

Therefore, by taking the modulus in (3.8) and utilizing (3.9) and (3.10) we get

|hSg(S)x, xi − hSx, xi hg(S)x, xi| (3.11)

=|h(S− hSx, xi1H) (g(S)−λ1H)x, xi|

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for anyxK,kxk= 1, which proves the first inequality in (3.7). Using Schwarz inequality we also have

h|S− hSx, xi1H|x, xi ≤D(S− hSx, xi1H)2x, xE

1/2

=h

S2x, x

− hSx, xi2i

1/2

for anyxK,kxk= 1, and the lemma is proved.

Corollary 1. With the assumption of Lemma 1, we have

0 ≤

S2x, x

− hSx, xi2≤12(Γ−γ)h|S− hSx, xi1H|x, xi

(3.12)

≤ 12(Γ−γ)h

S2x, x

− hSx, xi2i

1/2

≤ 14(Γ−γ)2,

for anyx∈K, kxk= 1.

Proof. If we take in Lemma 1g(t) =t, λ= 1

2(Γ +γ) and ρ= 1

2(Γ−γ), then we

get

0≤

S2x, x

− hSx, xi2≤12(Γ−γ)h|S− hSx, xi1H|x, xi

(3.13)

≤ 12(Γ−γ)h

S2x, x

− hSx, xi2i1/2

for anyxK,kxk= 1.

From the first and last terms in (3.13) we have

h

S2x, x

− hSx, xi2i1/2≤ 12(Γ−γ),

which proves the rest of (3.12).

We can prove the following result that provides simpler upper bounds for the quantumf-divergence when the operatorsP andQsatisfy the condition (2.18).

Theorem 8. Letf : [0,∞)→Rbe a continuous convex function that is normalized. If Q, P ∈S1(H), withP invertible, and there existsR≥1≥r≥0 such that

(3.14) rtr|T|2≤tr

Q

1/2

T P−1/2

2

≤Rtr|T|2

for anyT ∈ B2(H), then

0≤Sf(Q, P)≤ 12

f−′ (R)−f+′ (r)

V (Q, P) (3.15)

≤ 12

f−′ (R)−f+′ (r)

χ(Q, P)

≤ 14(R−r)

f−′ (R)−f+′ (r)

.

Proof. Without loosing the generality, we prove the inequality in the case thatf is continuously differentiable on (0,∞).

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which implies that

f′(t)−f

(R) +f(r) 2 ≤ 1 2[f ′(R)

−f′(r)]

for anyt∈[r, R].

Applying Lemma 1 and Corollary 1 in the Hilbert space (B2(H),h·,·i2) and for

the selfadjoint operatorAQ,P we have

hAQ,Pf′(AQ,P)T, Ti2− hAQ,PT, Ti2hf

(A

Q,P)T, Ti2

≤ 12[f′(R)−f′(r)]

AQ,P − hAQ,PT, Ti21B2(H)

T, T 2 ≤ 1 2[f

(R)f(r)]hA2

Q,PT, T

2− hAQ,PT, Ti 2 2

i1/2

≤ 1

4(R−r)

f′ (R)−f+′ (r)

for anyT ∈ B2(H),kTk2= 1,which is an inequality of interest in itself as well.

If in this inequality we takeT =P1/2, P S

1(H),with P invertible, then we

get

D

AQ,Pf′(AQ,P)P1/2, P1/2E

2−

D

f′(AQ,P)P1/2, P1/2E

2

≤12[f′(R)−f′(r)]D AQ,P−

D

AQ,PP1/2, P1/2E

21B2(H)

P

1/2

, P1/2E

2

≤12[f′(R)−f′(r)]

D

A2Q,PP1/2, P1/2E

2−

D

AQ,PP1/2, P1/2E

2

2

1/2

≤14(R−r)

f′ (R)−f+′ (r)

,

which can be written as

|Sℓf′(Q, P)−Sf′(Q, P)| ≤

1 2

f′ (R)−f+′ (r)

V(Q, P)

≤ 12

f′ (R)−f+′ (r)

χ(Q, P)

≤ 14(R−r)

f′ (R)−f+′ (r)

.

Making use of Theorem 7 we deduce the desired result (3.15).

Remark 3. If we take in (3.15)f(t) =t21, then we get

0≤χ2(Q, P)≤12(R−r)V(Q, P)≤ 12(R−r)χ(Q, P) (3.16)

≤ 14(R−r)2

for Q, P ∈S1(H), withP invertible and satisfying the condition (3.14).

If we take in (3.15)f(t) =tlnt, then we get the inequality

0≤U(Q, P)≤ 12ln

R r

V(Q, P)≤ 12ln

R r

χ(Q, P) (3.17)

≤1

4(R−r) ln

R r

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provided that Q, P S1(H), with P, Q invertible and satisfying the condition

(3.14).

With the same conditions and if we takef(t) =−lnt,then

(3.18) 0≤U(P, Q)≤ R2rR−rV(Q, P)≤ R2rR−rχ(Q, P)≤ (R−r) 2

4rR . If we take in (3.15)f(t) =fq(t) =1−tq

1−q, then we get

0≤Sfq(Q, P)≤

q 2 (1−q)

R1−q−r1−q R1−qr1−q

V (Q, P) (3.19)

2 (1q −q)

R1−qr1−q R1−qr1−q

χ(Q, P)

≤ q

4 (1−q)

R1−q−r1−q R1−qr1−q

(R−r)

provided that Q, P ∈ S1(H), with P, Q invertible and satisfying the condition

(3.14).

4. Other Reverse Inequalities

Utilising different techniques we can obtain other upper bounds for the quan-tum f-divergence as follows. Applications for Umegaki relative entropy and χ2

-divergence are also provided.

Theorem 9. Letf : [0,∞)→Rbe a continuous convex function that is normalized. If Q, P ∈S1(H), with P invertible, and there exists R≥1≥r≥0 such that the

condition (3.14) is satisfied, then

(4.1) 0≤Sf(Q, P)≤(R−1)f(r) + (1−r)f(R)

Rr .

Proof. By the convexity off we have

f(t) =f

(R−t)r+ (t−r)R R−r

≤ (R−t)f(r) + (t−r)f(R)

R−r for anyt∈[r, R].

This inequality implies the following inequality in the operator order ofB(B2(H))

f(AQ,P) R1B2(H)−AQ,P

f(r) + AQ,Pr1B2(H)f(R)

R−r ,

which can be written as

hf(AQ,P)T, Ti2 (4.2)

≤ f(r)

R−r

R1B2(H)−AQ,PT, T2+ f(R) R−r

A

Q,P−r1B2(H)T, T2

for anyT ∈ B2(H).

This inequality is of interest in itself. Now, if we take in (4.2) T =P1/2, P S

1(H), then we get the desired result

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Remark 4. If we take in (4.1)f(t) =t21,then we get

(4.3) 0≤χ2(Q, P)≤(R−1) (1−r)R+r+ 2 R−r

for Q, P ∈S1(H), withP invertible and satisfying the condition (3.14).

If we take in (4.1)f(t) =tlnt, then we get the inequality

(4.4) 0≤U(Q, P)≤lnhr(RR−1)−rrR

R(1−r)

R−r

i

provided that Q, P S1(H), with P, Q invertible and satisfying the condition

(3.14).

If we take in (4.1)f(t) =−lnt, then we get the inequality

(4.5) 0≤U(P, Q)≤lnhr1−R−RrR

r−1

R−r

i

for Q, P ∈S1(H), withP, Q invertible and satisfying the condition (3.14).

We also have:

Theorem 10. Let f : [0,∞)→Rbe a continuous convex function that is normal-ized. IfQ, P S1(H),withP invertible, and there exists R >1> r≥0 such that

the condition (3.14) is satisfied, then

0≤Sf(Q, P)≤(R−1) (1−r)

R−r Ψf(1;r, R) (4.6)

≤(R−1) (1−r)

R−r t∈sup(r,R)

Ψf(t;r, R)

≤(R−1) (1−r)f ′

−(R)−f+′ (r)

R−r

≤14(R−r)

f−′ (R)−f+′ (r)

whereΨf(·;r, R) : (r, R)→Ris defined by

(4.7) Ψf(t;r, R) = f(R)−f(t)

R−t −

f(t)−f(r) t−r . We also have

0≤Sf(Q, P)≤(R−1) (1−r)

R−r Ψf(1;r, R) (4.8)

≤14(R−r) Ψf(1;r, R)

≤14(R−r) sup t∈(r,R)

Ψf(t;r, R)

≤1

4(R−r)

f−′ (R)−f+′ (r)

.

Proof. By denoting

∆f(t;r, R) :=(t−r)f(R) + (R−t)f(r)

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we have

∆f(t;r, R) =(t−r)f(R) + (R−t)f(r)−(R−r)f(t) Rr

(4.9)

= (t−r)f(R) + (R−t)f(r)−(T−t+t−r)f(t) Rr

= (t−r) [f(R)−f(t)]−(R−t) [f(t)−f(r)] Mm

= (R−t) (t−r)

R−r Ψf(t;r, R)

for anyt∈(r, R).

From the proof of Theorem 9 we have

hf(AQ,P)T, Ti2 (4.10)

≤ f(r)

Rr

R1B2(H)−AQ,PT, T2+ f(R) Rr

A

Q,P−r1B2(H)T, T2

= hAQ,PT, Ti2−r

f(R) + R− hAQ,PT, Ti

2

f(r) Rr

for anyT ∈ B2(H),kTk2= 1.

This implies that

0≤ hf(AQ,P)T, Ti2f hAQ,PT, Ti2

(4.11)

≤ hAQ,PT, Ti2−r

f(R) + R− hAQ,PT, Ti

2

f(r)

Rr −f hAQ,PT, Ti2

= ∆f hAQ,PT, Ti

2;r, R

= R− hAQ,PT, Ti2

hAQ,PT, Ti

2−r

Rr Ψf hAQ,PT, Ti2;r, R

for anyT ∈ B2(H),kTk2= 1.

Since

Ψf hAQ,PT, Ti

2;r, R

≤ sup t∈(r,R)

Ψf(t;r, R) (4.12)

= sup t∈(r,R)

f(R)−f(t)

Rt −

f(t)−f(r) tr

≤ sup t∈(r,R)

f(R)−f(t) Rt

+ sup t∈(r,R)

−f(t)t−f(r) −r

= sup t∈(r,R)

f(R)−f(t) Rt

− inf t∈(r,R)

f(t)−f(r) tr

=f′ (R)−f+′ (r),

and, obviously

(4.13) 1

R−r R− hAQ,PT, Ti2

hAQ,PT, Ti

2−r

(17)

then by (4.11)-(4.13) we have

0≤ hf(AQ,P)T, Ti

2−f hAQ,PT, Ti2

(4.14)

≤ R− hAQ,PT, Ti2

hAQ,PT, Ti2r

R−r Ψf hAQ,PT, Ti2;r, R

≤ R− hAQ,PT, Ti2

hAQ,PT, Ti

2−r

R−r t∈sup(r,R)

Ψf(t;r, R)

≤ R− hAQ,PT, Ti

2

hAQ,PT, Ti

2−r

f

−(R)−f+′ (r)

Rr

≤ 14(R−r)

f′ (R)−f+′ (r)

for anyT ∈ B2(H),kTk2= 1.

This inequality is of interest in itself.

Now, if we take in (4.14)T =P1/2,then we get the desired result (4.6).

The inequality (4.8) is obvious from (4.6).

Remark 5. If we consider the convex normalized function f(t) =t2−1,then

Ψf(t;r, R) = R

2t2

Rt − t2−r2

tr =R−r, t∈(r, R) and we get from (4.6) the simple inequality

(4.15) 0≤χ2(Q, P)≤(R−1) (1−r)

for Q, P S1(H), with P invertible and satisfying the condition (3.14), which is

better than (4.3).

If we take the convex normalized functionf(t) =t−11,then we have

Ψf(t;r, R) =R

−1t−1

Rt −

t−1−r−1 tr =

R−r

rRt , t∈[r, R]. Also

Sf(Q, P) =χ2(P, Q). Using (4.6) we get

(4.16) 0≤χ2(P, Q)≤(R−1) (1−r)

Rr

for Q, P S1(H), withQ invertible and satisfying the condition (3.14).

If we consider the convex function f(t) =−lntdefined on [r, R]⊂(0,∞),then

Ψf(t;r, R) = −lnR+ lnt

Rt −

−lnt+ lnr tr

= (R−r) lnt−(R−t) lnr−(t−r) lnR (M −t) (t−m)

= ln

tR−r

rR−tMt−r

(R−t)(1t−r)

, t∈(r, R).

Then by (4.6) we have

(4.17) 0≤U(P, Q)≤lnhr1−R−RrR

r−1

R−r

i

≤(R−1) (1−r)

rR

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If we consider the convex function f(t) =tlnt defined on[r, R]⊂(0,∞), then

Ψf(t;r, R) =RlnR−tlnt

R−t −

tlnt−rlnr

t−r , t∈(r, R), which gives that

Ψf(1;r, R) = RlnR R1 −

rlnr 1−r. Using (4.6) we get

0≤U(Q, P)≤lnhR(1R−−r)rRr

(1−R)r

R−r

i

(4.18)

≤(R−1) (1−r) ln

"

R r

R1r#

for Q, P S1(H), withP, Q invertible and satisfying the condition (3.14).

We also have:

Theorem 11. Let f : [0,∞)→Rbe a continuous convex function that is normal-ized. IfQ, P ∈S1(H),withP invertible, and there exists R >1> r≥0 such that

the condition (3.14) is satisfied, then

(4.19) 0≤Sf(Q, P)≤2

f(r) +f(R)

2 −f

r+R 2

.

Proof. We recall the following result (see for instance [11]) that provides a refine-ment and a reverse for the weighted Jensen’s discrete inequality:

n min

i∈{1,...,n}{pi}

"

1 n

n

X

i=1

f(xi)−f 1 n

n

X

i=1

xi

!#

(4.20)

≤ 1

Pn n

X

i=1

pif(xi)−f 1 Pn

n

X

i=1

pixi

!

≤n max i∈{1,...,n}{pi}

"

1 n

n

X

i=1

f(xi)−f 1 n

n

X

i=1

xi

!#

,

wheref :CRis a convex function defined on the convex subsetC of the linear spaceX,{xi}i∈{1,...,n}C are vectors and{pi}i∈{1,...,n}are nonnegative numbers withPn:=Pn

i=1pi>0.

Forn= 2 we deduce from (3.6) that

2 min{s,1−s}

f(x) +f(y)

2 −f

x+y 2

(4.21)

≤sf(x) + (1−s)f(y)−f(sx+ (1−s)y)

≤2 max{s,1−s}

f(x) +f(y)

2 −f

x+y

2

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Now, if we use the second inequality in (4.21) for x=r, y =R, s= R−rR−t with t[r, R],then we have

(R−t)f(r) + (t−r)f(R)

R−r −f(t)

(4.22)

≤2 max

R−t Rr,

t−r Rr

f(r) +f(R)

2 −f

r+R 2

=

1 + 2 R−r

t−r+2R

f(r) +f(R)

2 −f

r+R 2

for anyt[r, R].

This implies in the operator order ofB(B2(H))

R1B2(H)−AQ,Pf(r) + AQ,P−r1B2(H)f(R)

Rr −f(AQ,P)

f(r) +f(R)

2 −f

r+R 2

×

1B2(H)+

2 Rr

AQ,Pr+R

2 1B2(H)

which implies that

0≤ hf(AQ,P)T, Ti

2−f hAQ,PT, Ti2

(4.23)

≤ hAQ,PT, Ti2−r

f(R) + R− hAQ,PT, Ti2

f(r)

R−r −f hAQ,PT, Ti2

f(r) +f(R)

2 −f

r+R

2

×

1 + 2 R−r

AQ,Pr+R

2 1B2(H)

T, T 2 ≤2

f(r) +f(R)

2 −f

r+R 2

for anyT ∈ B2(H),kTk2= 1.

This is an inequality of interest in itself. If we take in (4.23) T = P1/2, P S

1(H), then we get the desired result

(4.19).

Remark 6. If we take f(t) =t21 in (4.19), then we get

0≤χ2(Q, P)≤12(R−r)2

for Q, P S1(H), with P invertible and satisfying the condition (3.14), which is

not as good as (4.15).

If we take in (4.19)f(t) =t−11,then we have

(4.24) 0≤χ2(P, Q)≤ (R−r) 2

rR(r+R)

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If we take in (4.19)f(t) =−lnt,then we have

(4.25) 0≤U(P, Q)≤ln (R+r)

2

4rR

!

for Q, P ∈S1(H), withP invertible and satisfying the condition (3.14).

From (3.18) we have the following absolute upper bound

(4.26) 0≤U(P, Q)≤(R−r) 2

4rR

for Q, P ∈S1(H), withP invertible and satisfying the condition (3.14).

Utilising the elementary inequalitylnx≤x−1, x >0,we have that

ln (R+r)

2

4rR

!

≤ (R−r) 2

4rR ,

which shows that (4.25) is better than (4.26).

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1Mathematics, School of Engineering & Science, Victoria University, PO Box 14428,

Melbourne City, MC 8001, Australia.

E-mail address: [email protected]

URL:http://rgmia.org/dragomir

2School of Computational & Applied Mathematics, University of the

References

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