Trace Inequalities of Lipschitz Type for Power Series of Operators on Hilbert Spaces

E extracta mathematicae Vol. 32, N´um.1, 25 – 54 (2017)

S.S. Dragomir

Mathematics, School of Engineering & Science Victoria University, PO Box 14428, Melbourne City, MC 8001, Australia [email protected] , http://rgmia.org/dragomir DST-NRF Centre of Excellence in Mathematical and Statistical Sciences School of Computer Science & Applied Mathematics, University of the Witwatersrand Private Bag 3, Johannesburg 2050, South Africa

Presented by Alfonso Montes Received May 23, 2016 ∑ ∞ n Abstract: Let f(z) = n=0 αnz be a function deﬁned by power series with complex coeﬃ- cients and convergent on the open disk D(0,R) ⊂ C, R > 0. We show, amongst other that, if T,V ∈ B1(H), the Banach space of all trace operators on H, are such that ∥T ∥ , ∥V ∥ < R, ′ 1 1 then f(V ), f(T ), f ((1 − t)T + tV ) ∈ B1(H) for any t ∈ [0, 1] and ∫ 1 [ ′( )] tr [f(V )] − tr [f(T )] = tr (V − T )f (1 − t)T + tV dt. 0 Several trace inequalities are established. Applications for some elementary functions of interest are also given. Key words: Banach algebras of operators on Hilbert spaces, Power series, Lipschitz type inequalities, Jensen’s type inequalities, Trace of operators, Hilbert-Schmidt norm. AMS Subject Class. (2010): 47A63, 47A99.

1. Introduction

Let B(H) be the Banach algebra of bounded linear operators on a separable complex Hilbert space H. The absolute value of an operator A is the positive operator |A| deﬁned as |A| := (A∗A)1/2. It is known that [4] in the inﬁnite-dimensional case the map f(A) := |A| is not Lipschitz continuous on B(H) with the usual operator norm, i.e., there is no constant L > 0 such that

|A| − |B| ≤ L ∥A − B∥ for any A, B ∈ B(H).

25 26 s.s. dragomir

However, as shown by Farforovskaya in [17], [18] and Kato in [22], the following inequality holds ( ( )) 2 ∥A∥ + ∥B∥ |A| − |B| ≤ ∥A − B∥ 2 + log (1.1) π ∥A − B∥ for any A, B ∈ B(H) with A ≠ B. ∥ ∥ If the operator norm is replaced with Hilbert-Schmidt norm C HS := (tr C∗C)1/2 of an operator C, then the following inequality is true [2] √ | | − | | ≤ ∥ − ∥ A B HS 2 A B HS (1.2) for any A, B ∈ B(H√). The coeﬃcient 2 is best possible for a general A and B. If A and B are restricted to be selfadjoint, then the best coeﬃcient is 1. It has been shown in [4] that, if A is an invertible operator, then for all operators B in a neighborhood of A we have ( ) 2 3 |A| − |B| ≤ a1 ∥A − B∥ + a2 ∥A − B∥ + O ∥A − B∥ (1.3) where

−1 −1 −1 3 2 a1 = A ∥A∥ and a2 = A + A ∥A∥ .

In [3] the author also obtained the following Lipschitz type inequality

∥f(A) − f(B)∥ ≤ f ′(a) ∥A − B∥ (1.4) where f is an operator monotone function on (0, ∞) and A, B ≥ aIH > 0. One of the central problems in perturbation theory is to ﬁnd bounds for

∥f(A) − f(B)∥ in terms of ∥A − B∥ for different classes of measurable functions f for which the function of operator can be defined. For some results on this topic, see [5], [19] and the references therein. ∑ ∞ n By the help of power series f(z) = n=0 anz we can naturally construct another power series which will have as coefficients∑ the absolute values of the ∞ | | n coefficient of the original series, namely, fa(z) := n=0 an z . It is obvious that this new power series will have the same radius of convergence as the original series. We also notice that if all coefficients an ≥ 0, then fa = f. trace inequalities of lipschitz type 27

We notice that if

∞ ∑ (−1)n 1 f(z) = zn = ln , z ∈ D(0, 1) ; (1.5) n 1 + z n=1

∞ ∑ (−1)n g(z) = z2n = cos z , z ∈ C ; (2n)! n=0

∞ ∑ (−1)n h(z) = z2n+1 = sin z , z ∈ C ; (2n + 1)! n=0

∞ ∑ 1 l(z) = (−1)nzn = , z ∈ D(0, 1) ; 1 + z n=0 where D(0, 1) is the open disk centered in 0 and of radius 1, then the cor- responding functions constructed by the use of the absolute values of the coeﬃcients are

∞ ∑ 1 1 f (z) = zn = ln , z ∈ D(0, 1) ; (1.6) a n! 1 − z n=1

∞ ∑ 1 g (z) = z2n = cosh z , z ∈ C ; a (2n)! n=0

∞ ∑ 1 h (z) = z2n+1 = sinh z , z ∈ C ; a (2n + 1)! n=0

∞ ∑ 1 l (z) = zn = , z ∈ D(0, 1) . a 1 − z n=0

Other important examples of functions as power series representations with nonnegative coeﬃcients are: 28 s.s. dragomir

∞ ∑ 1 exp(z) = zn , z ∈ C ; (1.7) n! n=0 ( ) ∞ 1 1 + z ∑ 1 ln = z2n−1 , z ∈ D(0, 1) ; 2 1 − z 2n − 1 n=1 ∞ ∑ Γ(n + 1 ) sin−1(z) = √ 2 z2n+1 , z ∈ D(0, 1) ; π(2n + 1)n! n=0 ∞ ∑ 1 tanh−1(z) = z2n−1 , z ∈ D(0, 1) ; 2n − 1 n=1 ∞ ∑ Γ(n + α)Γ(n + β)Γγ) F (α, β, γ, z) = zn , α, β, γ > 0 , z ∈ D(0, 1) ; 2 1 n!Γ(α)Γ(β)Γ(n + γ) n=0 where Γ is Gamma function. We recall the following result that provides a quasi-Lipschitzian condition for functions defined by power series and operator norm ∥·∥ [14]: ∑ Theorem 1. ∞ n Let f(z) := n=0 anz be a power series with complex coefficients and convergent on the open disk D(0,R), R > 0. If T,V ∈ B(H) are such that ∥T ∥ , ∥V ∥ < R, then ( ) ∥ − ∥ ≤ ′ {∥ ∥ ∥ ∥} ∥ − ∥ f(T ) f(V ) fa max T , V T V . (1.8) If ∥T ∥ , ∥V ∥ ≤ M < R, then from (1.8) we have the simpler inequality ∥ − ∥ ≤ ′ ∥ − ∥ f(T ) f(V ) fa(M) T V (1.9) In the recent paper [13] we improved the inequality (1.8) as follows: ∑ Theorem 2. ∞ n Let f(z) := n=0 anz be a power series with complex coefficients and convergent on the open disk D(0,R), R > 0. If T,V ∈ B(H) are such that ∥T ∥ , ∥V ∥ < R, then ∫ 1 ( ) ∥ − ∥ ≤ ∥ − ∥ ′ ∥ − ∥ f(T ) f(V ) T V fa (1 t)T + tV dt . (1.10) 0 In order to obtain similar results for the trace of bounded linear operators on complex infinite dimensional Hilbert spaces we need some preparations as follows. trace inequalities of lipschitz type 29

2. Some preliminary facts on trace for operators ( ) Let H, ⟨·, ·⟩ be a complex Hilbert space and {ei}i∈I an orthonormal basis of H. We say that A ∈ B(H) is a Hilbert-Schmidt operator if ∑ 2 ∥Aei∥ < ∞ . (2.1) i∈I

It is well know that, if {ei}i∈I and {fj}j∈J are orthonormal bases for H and A ∈ B(H) then ∑ ∑ ∑ 2 2 ∗ 2 ∥Aei∥ = ∥Afj∥ = ∥A fj∥ (2.2) i∈I j∈I j∈I showing that the definition (2.1) is independent of the orthonormal basis and A is a Hilbert-Schmidt operator iff A∗ is a Hilbert-Schmidt operator. Let B2(H) the set of Hilbert-Schmidt operators in B(H). For A ∈ B2(H) we define ( ) ∑ 1/2 ∥ ∥ ∥ ∥2 A 2 := Aei (2.3) i∈I for {ei}i∈I an orthonormal basis of H. This definition does not depend on the choice of the orthonormal basis. 2 Using the triangle inequality in l (I), one checks that B2(H) is a vec- ∥·∥ B tor space and that 2 is a norm on 2(H), which is usually called in the literature as the Hilbert-Schmidt norm. Denote the modulus of an operator A ∈ B(H) by |A| := (A∗A)1/2. Because ∥|A| x∥ = ∥Ax∥ for all x ∈ H, A is Hilbert-Schmidt iff |A| is ∥ ∥ ∥| |∥ ∈ B Hilbert-Schmidt and A 2 = A 2. From (2.2) we have that if A 2(H), ∗ ∈ B ∥ ∥ ∥ ∗∥ then A 2(H) and A 2 = A 2. The following theorem collects some of the most important properties of Hilbert-Schmidt operators:

Theorem 3. We have: B ∥·∥ (i) ( 2(H), 2) is a Hilbert space with inner product ∑ ∑ ⟨ ⟩ ⟨ ⟩ ⟨ ∗ ⟩ A, B 2 := Aei, Bei = B Aei, ei (2.4) i∈I i∈I and the deﬁnition does not depend on the choice of the orthonormal basis {ei}i∈I ; 30 s.s. dragomir

(ii) we have the inequalities ∥ ∥ ≤ ∥ ∥ A A 2 (2.5)

for any A ∈ B2(H) and

∥ ∥ ∥ ∥ ≤ ∥ ∥ ∥ ∥ AT 2 , TA 2 T A 2 (2.6)

for any A ∈ B2(H) and T ∈ B(H);

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

B(H)B2(H)B(H) ⊆ B2(H);

(iv) Bfin(H), the space of operators of ﬁnite rank, is a dense subspace of B2(H);

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

If {ei}i∈I an orthonormal basis of H, we say that A ∈ B(H) is trace class if ∑ ∥ ∥ ⟨| | ⟩ ∞ A 1 := A ei, ei < . (2.7) i∈I ∥ ∥ The deﬁnition of A 1 does not depend on the choice of the orthonormal basis {ei}i∈I . We denote by B1(H) the set of trace class operators in B(H). The following proposition holds:

Proposition 1. If A ∈ B(H), then the following are equivalent:

(i) A ∈ B1(H); 1/2 (ii) |A| ∈ B2(H);

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

The following properties are also well known:

Theorem 4. With the above notations:

(i) We have ∥ ∥ ∥ ∗∥ ∥ ∥ ≤ ∥ ∥ A 1 = A 1 and A 2 A 1 (2.8)

for any A ∈ B1(H). trace inequalities of lipschitz type 31

(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 { } ∥ ∥ |⟨ ⟩ | ∈ B ∥ ∥ ≤ A 1 = sup A, B 2 : B 2(H), B 1 .

B ∥·∥ (v) ( 1(H), 1) is a Banach space. (vi) We have the following isometric isomorphisms ∼ ∗ ∗ ∼ B1(H) = K(H) and B1(H) = B(H) ,

∗ ∗ where K(H) is the dual space of K(H) and B1(H) is the dual space of B1(H).

We define the trace of a trace class operator A ∈ B1(H) to be ∑ tr(A) := ⟨Aei, ei⟩ , (2.9) i∈I where {ei}i∈I an orthonormal basis of H. 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 5. We have: ∗ (i) if A ∈ B1(H) then A ∈ B1(H) and

tr(A∗) = tr(A); (2.10)

(ii) if A ∈ B1(H) and T ∈ B(H), then AT , TA ∈ B1(H) and | | ≤ ∥ ∥ ∥ ∥ tr(AT ) = tr(TA) and tr(AT ) A 1 T ; (2.11)

(iii) tr(·) is a bounded linear functional on B1(H) with ∥tr∥ = 1 ;

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

(v) Bfin(H) is a dense subspace of B1(H). 32 s.s. dragomir

Utilising the trace notation we obviously have that ⟨ ⟩ ∗ ∗ A, B 2 = tr(B A) = tr(AB ) , ( ) ∥ ∥2 ∗ | |2 A 2 = tr(A A) = tr A for any A, B ∈ B2(H). The following H¨older’stype inequality has been obtained by Ruskai in [28] [ ( )] [ ( )] − α − 1 α |tr(AB)| ≤ tr(|AB|) ≤ tr |A|1/α tr |B|1/(1 α) (2.12)

1/α 1/(1−α) where α ∈ (0, 1) and A, B ∈ B(H) with |A| , |B| ∈ B1(H). 1 In particular, for α = 2 we get the Schwarz inequality [ ] [ ] ( ) 1/2 ( ) 1/2 |tr(AB)| ≤ tr(|AB|) ≤ tr |A|2 tr |B|2 (2.13) with A, B ∈ B2(H). If A ≥ 0 and P ∈ B1(H) with P ≥ 0, then 0 ≤ tr(PA) ≤ ∥A∥ tr(P ) . (2.14)

Indeed, since A ≥ 0, then ⟨Ax, x⟩ ≥ 0 for any x ∈ H. If {ei}i∈I is an orthonormal basis of H, then ⟨ ⟩ 2 1/2 1/2 1/2 0 ≤ AP ei,P ei ≤ ∥A∥ P ei = ∥A∥ ⟨P ei, ei⟩ for any i ∈ I. Summing over i ∈ I we get ∑ ⟨ ⟩ ∑ 1/2 1/2 0 ≤ AP ei,P ei ≤ ∥A∥ ⟨P ei, ei⟩ = ∥A∥ tr(P ) , i∈I i∈I and since ∑ ⟨ ⟩ ∑ ⟨ ⟩ 1/2 1/2 1/2 1/2 AP ei,P ei = P AP ei, ei i∈I i∈I ( ) = tr P 1/2AP 1/2 = tr(PA) , we obtain the desired result (2.14). This obviously imply the fact that, if A and B are selfadjoint operators with A ≤ B and P ∈ B1(H) with P ≥ 0, then tr(PA) ≤ tr(PB) . (2.15) trace inequalities of lipschitz type 33

Now, if A is a selfadjoint operator, then we know that

|⟨Ax, x⟩| ≤ ⟨|A| x, x⟩ for any x ∈ H.

This inequality follows from 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 of H, then

⟨ ⟩ ⟨ ⟩ ∑ ∑ | | 1/2 1/2 ≤ 1/2 1/2 tr(PA) = AP ei,P ei AP ei,P ei i∈I i∈I ∑ ⟨ ⟩ 1/2 1/2 ≤ |A| P ei,P ei = tr(P |A|) , (2.16) i∈I for any A a selfadjoint operator and P ∈ B1(H) with P ≥ 0. For the theory of trace functionals and their applications the reader is referred to [31]. For some classical trace inequalities see [9], [11], [26] and [35], which are continuations of the work of Bellman [7]. For related works the reader can refer to [1], [8], [9], [20], [23], [24], [25], [29] and [32].

3. Trace inequalities

We have the following representation result: ∑ Theorem 6. ∞ n Let f(z) := n=0 anz be a power series with complex co- eﬃcients and convergent on the open disk D(0,R), R( > 0. If T,V) ∈ B1(H) ′ are such that tr(|T |), tr(|V |) < R, then f(V ), f(T ), f (1 − t)T + tV ∈ B1(H) for any t ∈ [0, 1] and ∫ 1 [ ( )] tr [f(V )] − tr [f(T )] = tr (V − T )f ′ (1 − t)T + tV dt . (3.1) 0

Proof. We use the identity (see for instance [6, p. 254])

n∑−1 An − Bn = An−1−j(A − B)Bj (3.2) j=0 that holds for any A, B ∈ B(H) and n ≥ 1. 34 s.s. dragomir

For T,V ∈ B(H) we consider the function φ : [0, 1] → B(H) deﬁned by φ(t) = [(1 − t)T + tV ]n. For t ∈ (0, 1) and ε ≠ 0 with t + ε ∈ (0, 1) we have from (3.2) that

φ(t + ε) − φ(t) = [(1 − t − ε)T + (t + ε)V ]n − [(1 − t)T + tV ]n

n∑−1 − − = ε [(1 − t − ε)T + (t + ε)V ]n 1 j (V − T ) [(1 − t)T + tV ]j . j=0

Dividing with ε ≠ 0 and taking the limit over ε → 0 we have in the norm topology of B that

1 φ′(t) = lim [φ(t + ε) − φ(t)] (3.3) ε→0 ε n∑−1 − − = [(1 − t)T + tV ]n 1 j (V − T ) [(1 − t)T + tV ]j . j=0

Integrating on [0, 1] we get from (3.3) that

∫ ∫ 1 n∑−1 1 − − φ′(t) dt = [(1 − t)T + tV ]n 1 j (V − T ) [(1 − t)T + tV ]j dt , 0 j=0 0 and since ∫ 1 φ′(t) dt = φ(1) − φ(0) = V n − T n, 0 then we get the following equality of interest in itself

∫ n∑−1 1 − − V n − T n = [(1 − t)T + tV ]n 1 j (V − T ) [(1 − t)T + tV ]j dt (3.4) j=0 0 for any T,V ∈ B(H) and n ≥ 1. trace inequalities of lipschitz type 35

If T,V ∈ B1(H) and we take the trace in (3.4) we get

tr(V n) − tr(T n) ∫ n∑−1 1 ( − − ) = tr [(1 − t)T + tV ]n 1 j (V − T ) [(1 − t)T + tV ]j dt j=0 0 ∫ n∑−1 1 ( − ) = tr [(1 − t)T + tV ]n 1 (V − T ) dt (3.5) j=0 0 ∫ 1 ( − ) = n tr [(1 − t)T + tV ]n 1 (V − T ) dt 0 ∫ 1 ( − ) = n tr (V − T ) [(1 − t)T + tV ]n 1 dt 0 for any n ≥ 1. Let m ≥ 1. Then by (3.5) we have have ( ) ( ) ∑m ∑m ∑m [ ( ) ( )] n n n n tr anV − tr anT = an tr V − tr T n=0 n=0 n=0 ∑m [ ( ) ( )] n n = an tr V − tr T (3.6) n=1 ∫ ∑m 1 ( ) n−1 = nan tr (V − T ) [(1 − t)T + tV ] dt n=1 0 ∫ ( ) 1 ∑m n−1 = tr (V − T ) nan [(1 − t)T + tV ] dt 0 n=1 ∈ B for any T,V 1(H). ∑ | | | | ∈ B ∞ n ∑ Since tr( T ),∑tr( V ) < R with T,V 1(H) then the series n=0 anV , ∞ n ∞ − n−1 B n=0 anT and n=1 nan [(1 t)T + tV ] are convergent in 1(H) and ∑∞ ∑∞ n n anV = f(V ) , anT = f(T ) n=0 n=0 and ∑∞ ( ) n−1 ′ nan (1 − t)T + tV ] = f (1 − t)T + tV n=1 36 s.s. dragomir where t ∈ [0, 1]. Moreover, we have ( ) ′ f(V ) , f(T ) , f (1 − t)T + tV ∈ B1(H) for any t ∈ [0, 1]. By taking the limit over m → ∞ in (3.6) we get the desired result (3.1).

In addition to the power identity (3.5), for any T,V ∈ B1(H) we have other equalities as follows ∫ 1 ( ) tr [exp(V )] − tr [exp(T )] = tr (V − T ) exp((1 − t)T + tV ) dt , (3.7) 0 ∫ 1 ( ) tr [sin(V )] − tr [sin(T )] = tr (V − T ) cos((1 − t)T + tV ) dt , (3.8) 0 ∫ 1 ( ) tr [sinh(V )] − tr [sinh(T )] = tr (V − T ) cosh((1 − t)T + tV ) dt . (3.9) 0

If T,V ∈ B1(H) with tr(|T |), tr(|V |) < 1 then [ ] [ ] −1 −1 tr (1H − V ) − tr (1H − T ) (3.10) ∫ 1 ( ) −2 = tr (V − T )(1H − (1 − t)T − tV ) dt , 0 and [ ] [ ] −1 −1 tr ln(1H − V ) s − tr ln(1H − T ) (3.11) ∫ 1 ( ) −1 = tr (V − T )(1H − (1 − t)T − tV ) dt . 0

We have the following result:

Corollary 1. With the assumptions in Theorem 6 we have the inequalities trace inequalities of lipschitz type 37

{ ∫ 1 − ≤ ∥ − ∥ ′ − tr [f(V )] tr [f(T )] min V T f ((1 t)T + tV ) 1 dt , (3.12) 0∫ } 1 ∥ − ∥ ′ − V T 1 f ((1 t)T + tV ) dt 0 { ∫ 1 ( ) ≤ ∥ − ∥ ′ ∥ − ∥ min V T fa (1 t)T + tV 1 dt, 0∫ } 1 ( ) ∥ − ∥ ′ ∥ − ∥ V T 1 fa (1 t)T + tV dt , 0 ∥·∥ ∥·∥ where is the operator norm and 1 is the 1-norm introduced for trace class operators.

Proof. From (3.1), we have by taking the modulus ∫ 1 ( ) tr [f(V )] − tr [f(T )] ≤ tr (V − T )f ′((1 − t)T + tV ) dt . (3.13) 0 Utilising the inequality (2.11) twice, for any t ∈ [0, 1] we get ( ( )) ( ) − ′ − ≤ ∥ − ∥ ′ − tr (V T )f (1 t)T + tV V T f (1 t)T + tV 1 , ( ( )) ( ) − ′ − ≤ ∥ − ∥ ′ − tr (V T )f (1 t)T + tV V T 1 f (1 t)T + tV . By integrating these inequalities, we get the ﬁrst part of (3.12). ∥·∥ We have, by the use of 1 properties that

∑∞ ( ) − f ′ (1 − t)T + tV = na [(1 − t)T + tV ]n 1 1 n n=1 1 ∑∞ n−1 ≤ n |an| [(1 − t)T + tV ] 1 n=1 ∑∞ ≤ | | − n−1 n an (1 t)T + tV 1 n=1 ( ) ′ ∥ − ∥ = fa (1 t)T + tV 1 ∈ B ∥ ∥ ∥ ∥ for any T,V 1(H) with T 1 , V 1 < R. 38 s.s. dragomir

This proves the ﬁrst part of the second inequality. ∥ ∥ ≤ ∥ ∥ ∈ B ∥ − ∥ Since X X 1 for any X 1(H) then (1 t)T +( tV < R for any) ∈ B ∥ ∥ ∥ ∥ ′ ∥ − ∥ T,V 1(H) with T 1 , V 1 < R which shows that fa (1 t)T + tV is well deﬁned. The second part of the second inequality follows in a similar way and the details are omitted.

1. ′ Remark We observe that fa is monotonic nondecreasing and convex on the interval [0,R) and since the function ψ(t) := ∥(1 − t)T + tV ∥ is convex ′ ◦ on [0, 1] we have that fa ψ is also convex on [0, 1]. Utilising the Hermite- Hadamard inequality for convex functions (see for instance [16, p. 2]) we have the sequence of inequalities ∫ [ ( ) ] 1 ( ) ′ ∥ ∥ ′ ∥ ∥ ′ ∥ − ∥ ≤ 1 ′ T + V fa( T ) + fa( V ) fa (1 t)T + tV dt fa + 0 2 2 2 1 [ ] ≤ f ′ (∥T ∥) + f ′ (∥V ∥) 2 a a { } ≤ ′ ∥ ∥ ′ ∥ ∥ max fa( T ), fa( V ) . (3.14)

We also have ∫ ∫ 1 ( ) 1 ( ) ′ ∥ − ∥ ≤ ′ − ∥ ∥ ∥ ∥ fa (1 t)T + tV dt fa (1 t) T + t V dt 0 0[ ( ) ] 1 ∥T ∥ + ∥V ∥ f ′ (∥T ∥) + f ′ (∥V ∥) ≤ f ′ + a a 2 a 2 2 1 [ ] ≤ f ′ (∥T ∥) + f ′ (∥V ∥) 2 a a { } ≤ ′ ∥ ∥ ′ ∥ ∥ max fa( T ), fa( V ) . (3.15)

We observe that if ∥V ∥ ̸= ∥T ∥, then by the change of variable s = (1−t) ∥T ∥+ t ∥V ∥ we have ∫ ∫ 1 ( ) ∥V ∥ ′ − ∥ ∥ ∥ ∥ 1 ′ fa (1 t) T + t V dt = fa(s) ds 0 ∥V ∥ − ∥T ∥ ∥T ∥ f (∥V ∥) − f (∥T ∥) = a a . ∥V ∥ − ∥T ∥ trace inequalities of lipschitz type 39

If ∥V ∥ = ∥T ∥, then ∫ 1 ( ) ′ − ∥ ∥ ∥ ∥ ′ ∥ ∥ fa (1 t) T + t V dt = fa( T ) . 0

Utilising these observations we then get the following divided diﬀerence inequality for T ≠ V  ∥ ∥ − ∥ ∥ ∫  fa( V ) fa( T ) 1 ( ) ∥ ∥−∥ ∥ if ∥V ∥ ̸= ∥T ∥ , ′ ∥ − ∥ ≤ V T fa (1 t)T + tV dt  (3.16) 0 ′ ∥ ∥ ∥ ∥ ∥ ∥ fa( T ) if V = T .

∥·∥ ∈ B Similar comments apply for the 1-norm 1 when T,V 1(H). If we use the ﬁrst part in the inequalities (3.12) and the above remarks, then we get the following string of inequalities ∫ 1 ( ) − ≤ ∥ − ∥ ′ − tr [f(V )] tr [f(T )] V T f (1 t)T + tV 1 dt 0 ∫ 1 ( ) ≤ ∥ − ∥ ′ ∥ − ∥ V T fa (1 t)T + tV 1 dt (3.17) 0  [ ( ) ] f ′ (∥T ∥ )+f ′ (∥V ∥ )  1 f ′ T +V + a 1 a 1 ,  2 a 2 1 2   f (∥V ∥ )−f (∥T ∥ ) ≤ ∥V − T ∥ ×  a 1 a 1 if ∥V ∥ ≠ ∥T ∥ ,  ∥V ∥ −∥T ∥ 1 1  1 1   ′ ∥ ∥ ∥ ∥ ∥ ∥ fa( T 1) if V 1 = T 1 , 1 [ ] ≤ ∥V − T ∥ f ′ (∥T ∥ ) + f ′ (∥V ∥ ) 2 a 1 a 1 { } ≤ ∥ − ∥ ′ ∥ ∥ ′ ∥ ∥ V T max fa( T 1), fa( V 1)

∈ B ∥ ∥ ∥ ∥ provided T,V 1(H) with T 1 , V 1 < R. ∥ ∥ ∥ ∥ ≤ If T 1 , V 1 M < R, then we have from (3.17) the simple inequality

− ≤ ∥ − ∥ ′ tr [f(V )] tr [f(T )] V T fa(M) .

A similar sequence of inequalities can also be stated by swapping the norm ∥·∥ ∥·∥ with 1 in (3.17). We omit the details. 40 s.s. dragomir

If we use the inequality (3.17) for the exponential function, then for any T,V ∈ B1(H) we have the inequalities ∫ 1 ( ) − ≤ ∥ − ∥ − tr [exp(V )] tr [exp(T )] V T exp (1 t)T + tV 1 dt 0 ∫ 1 ( ) ≤ ∥ − ∥ ∥ − ∥ V T exp (1 t)T + tV 1 dt (3.18) 0  [ ( ) ] exp(∥T ∥ )+exp(∥V ∥ )  1 exp T +V + 1 1 ,  2 2 1 2   exp(∥V ∥ )−exp(∥T ∥ ) ≤ ∥V − T ∥ ×  1 1 if ∥V ∥ ≠ ∥T ∥ ,  ∥V ∥ −∥T ∥ 1 1  1 1   ∥ ∥ ∥ ∥ ∥ ∥ exp( T 1) if V 1 = T 1 , 1 [ ] ≤ ∥V − T ∥ exp(∥T ∥ ) + exp(∥V ∥ ) 2 1 1 { } ≤ ∥ − ∥ ∥ ∥ ∥ ∥ V T max exp( T 1), exp( V 1) . ∥ ∥ ∥ ∥ If T 1 , V 1 < 1, then we have the inequalities [ ] [ ] −1 −1 tr ln(1H − V ) − tr ln(1H − T ) (3.19) ∫ 1 ≤ ∥ − ∥ − − − −1 V T (1H (1 t)T tV ) 1 dt 0 ∫ 1 ≤ ∥ − ∥ − ∥ − ∥ −1 V T (1 (1 t)T + tV 1) dt 0  [( ) ] −1 (1−∥T ∥ )−1+(1−∥V ∥ )−1  1 1 − T +V + 1 1 ,  2 2 1 2   ≤ ∥ − ∥ × ln(1−∥V ∥ )−1−ln(1−∥T ∥ )−1 V T  1 1 if ∥V ∥ ≠ ∥T ∥ ,  ∥V ∥ −∥T ∥ 1 1  1 1   − ∥ ∥ −1 ∥ ∥ ∥ ∥ (1 T 1) if V 1 = T 1 , 1 [ ] ≤ ∥V − T ∥ (1 − ∥T ∥ )−1 + (1 − ∥V ∥ )−1 2 1 1 { } ≤ ∥ − ∥ − ∥ ∥ −1 − ∥ ∥ −1 V T max (1 T 1) , (1 V 1) . ∥·∥ The following result for the Hilbert-Schmidt norm 2 also holds: ∑ Theorem 7. ∞ n Let f(z) := n=0 anz be a power series with complex coef- ﬁcients and convergent on the open disk D(0,R), R > 0. If T,V ∈ B2(H) are trace inequalities of lipschitz type 41

2 2 2 ′ such that tr(|T | ), tr(|V | ) < R , then f(V ), f(T ), f ((1 − t)T + tV ) ∈ B2(H) for any t ∈ [0, 1] and ∫ 1 ( ) tr [f(V )] − tr [f(T )] = tr((V − T )f ′ (1 − t)T + tV ) dt . (3.20) 0 Moreover, we have the inequalities ∫ 1 ( ) − ≤ ∥ − ∥ ′ − tr [f(V )] tr [f(T )] V T 2 f (1 t)T + tV 2 dt , 0 ∫ 1 ( ) ≤ ∥ − ∥ ′ ∥ − ∥ V T 2 fa (1 t)T + tV 2 dt (3.21) 0  [ ( ) ] f ′ (∥T ∥ )+f ′ (∥V ∥ )  1 f ′ T +V + a 2 a 2 ,  2 a 2 2 2   ≤ ∥ − ∥ ×  fa(∥V ∥ )−fa(∥T ∥ ) V T 2 2 2 if ∥V ∥ ≠ ∥T ∥ ,  ∥V ∥ −∥T ∥ 2 2  2 2   ′ ∥ ∥ ∥ ∥ ∥ ∥ fa( T 2) if V 2 = T 2 ,

1 [ ] ≤ ∥V − T ∥ f ′ (∥T ∥ ) + f ′ (∥V ∥ ) 2 2 a 2 a 2 { } ≤ ∥ − ∥ ′ ∥ ∥ ′ ∥ ∥ V T 2 max fa( T 2), fa( V 2) . Proof. The proof of the ﬁrst part of the theorem follows in a similar manner to the one from Theorem 6. Taking the modulus in (3.20) and using the Schwarz inequality for trace (2.13) we have ∫ 1 ( ) |tr [f(V )] − tr [f(T )]| ≤ tr((V − T )f ′ (1 − t)T + tV ) dt (3.22) 0 ∫ 1 ( ) ≤ ∥ − ∥ ′ − V T 2 f (1 t)T + tV 2 dt . 0 The rest follows in a similar manner as in the case of 1-norm and the details are omitted. We notice that similar examples to (3.18) and (3.19) may be stated where ∥·∥ ∥·∥ ∥·∥ both norms and 1 are replaced by 2. ∈ B ∥ ∥ ∥ ∥ ≤ We also observe that, if T,V 2(H) with T 2 , V 2 K < R, then we have from (3.17) the simple inequality

− ≤ ∥ − ∥ ′ tr [f(V )] tr [f(T )] V T 2 fa(K) . 42 s.s. dragomir

4. Norm inequalities

We have the following norm inequalities: ∑ Theorem 8. ∞ n Let f(z) := n=0 anz be a power series with complex co- eﬃcients and convergent on the open disk D(0,R), R > 0.

(i) If T,V ∈ B1(H) are such that tr(|T |), tr(|V |) < R, then we have the norm inequalities  ∫ ( )  ∥ − ∥ 1 ′ ∥ − ∥ V T 1 0 fa (1 t)T + tV dt , ∥f(V ) − f(T )∥ ≤ ∫ ( ) (4.1) 1  ∥ − ∥ 1 ′ ∥ − ∥ V T 0 fa (1 t)T + tV 1 dt .

2 2 2 (ii) If T,V ∈ B2(H) are such that tr(|T | ), tr(|V | ) < R , then we also have the norm inequalities  ∫ ( )  ∥ − ∥ 1 ′ ∥ − ∥ V T 2 0 fa (1 t)T + tV dt , ∥f(V ) − f(T )∥ ≤ ∫ ( ) (4.2) 2  ∥ − ∥ 1 ′ ∥ − ∥ V T 0 fa (1 t)T + tV 2 dt .

Proof. We use the equality ∫ n∑−1 1 − − V n − T n = [(1 − t)T + tV ]n 1 j (V − T ) [(1 − t)T + tV ]j dt , (4.3) j=0 0 for any T,V ∈ B(H) and n ≥ 1. (i) If T,V ∈ B1(H) are such that tr(|T |), tr(|V |) < R, then by taking the ∥·∥ ≥ 1 norm and using its properties, for any n 1 we have successively

− ∫ n∑1 1 ∥ n − n∥ ≤ − n−1−j − − j V T 1 [(1 t)T + tV ] (V T ) [(1 t)T + tV ] dt 1 j=0 0 ∫ n∑−1 1 − − ≤ [(1 − t)T + tV ]n 1 j (V − T ) [(1 − t)T + tV ]j dt 1 j=0 0 − ∫ n∑1 1 ≤ ∥ − ∥ − n−1−j − j V T 1 [(1 t)T + tV ] [(1 t)T + tV ] dt j=0 0 trace inequalities of lipschitz type 43

n∑−1 ≤ ∥ − ∥ ∥ − ∥n−1−j ∥ − ∥j V T 1 (1 t)T + tV (1 t)T + tV dt j=0 ∫ 1 ∥ − ∥ ∥ − ∥n−1 = n V T 1 (1 t)T + tV dt . (4.4) 0

Let m ≥ 1. By (4.4) we have

∑m ∑m ∑m n − n n − n anV anT = an(V T ) (4.5) n=0 n=0 1 n=1 1 ∑m ≤ | | ∥ n − n∥ an V T 1 n=1 ∫ ∑m 1 ≤ ∥ − ∥ | | ∥ − ∥n−1 V T 1 an n (1 t)T + tV dt n=1 0 ∫ ( ) 1 ∑m ∥ − ∥ | | ∥ − ∥n−1 = V T 1 n an (1 t)T + tV dt . 0 n=1

Also, we observe that

∥ − ∥ ≤ ∥ − ∥ (1 t)T + tV (1 t)T + tV 1 ≤ − ∥ ∥ ∥ ∥ (1 t) T 1 + t V 1 < R ∑ ∈ ∞ | | ∥ − ∥n−1 for any t [0, 1], which implies that the series n=1 n an (1 t)T + tV is convergent and

∑∞ ( ) | | ∥ − ∥n−1 ′ ∥ − ∥ n an (1 t)T + tV = fa (1 t)T + tV n=1

∈ for any t [0, 1]. ∑ ∑ ∞ n ∞ n ( Since the) series n=0 anV and n=0 anT are convergent in B ∥·∥ → ∞ 1(H), 1 , then by letting m in the inequality (4.5) we get the ﬁrst inequality in (4.1). 44 s.s. dragomir

For any n ≥ 1 we also have

− ∫ n∑1 1 ∥ n − n∥ ≤ − n−1−j − − j V T 1 [(1 t)T + tV ] (V T ) [(1 t)T + tV ] dt 1 j=0 0 ∫ n∑−1 1 − − ≤ [(1 − t)T + tV ]n 1 j (V − T ) [(1 − t)T + tV ]j dt 1 j=0 0 ∫ n∑−1 1 − − ≤ ∥V − T ∥ [(1 − t)T + tV ]n 1 j [(1 − t)T + tV ]j dt 1 j=0 0 n∑−1 ≤ ∥ − ∥ ∥ − ∥n−1−j ∥ − ∥j V T (1 t)T + tV (1 t)T + tV 1 dt j=0 n∑−1 ≤ ∥ − ∥ ∥ − ∥n−1−j ∥ − ∥j V T (1 t)T + tV 1 (1 t)T + tV 1 dt j=0 ∫ 1 ∥ − ∥ ∥ − ∥n−1 = n V T (1 t)T + tV 1 dt , 0 which by a similar argument produces the second inequality in (4.1). ∥ ∥ ≤ ∥ ∥ ∥ ∥ (ii) Follows in a similar way by utilizing the inequality TA 2 T A 2 that holds for T ∈ B(H) and A ∈ B2(H). The details are omitted.

Remark 2. From the ﬁrst inequality in (4.1) we have the sequence of inequalities ∫ 1 ( ) ∥ − ∥ ≤ ∥ − ∥ ′ ∥ − ∥ f(V ) f(T ) 1 V T 1 fa (1 t)T + tV dt (4.6) 0  [ ] ( ) ′ ∥ ∥ ′ ∥ ∥  1 f ′ T +V + fa( T )+fa( V ) ,  2 a 2 2   ≤ ∥ − ∥ ×  fa(∥V ∥)−fa(∥T ∥) V T 1 if ∥V ∥ ̸= ∥T ∥ ,  ∥V ∥−∥T ∥    ′ ∥ ∥ ∥ ∥ ∥ ∥ fa( T ) if V = T , 1 [ ] ≤ ∥V − T ∥ f ′ (∥T ∥) + f ′ (∥V ∥) 2 1 a a { } ≤ ∥ − ∥ ′ ∥ ∥ ′ ∥ ∥ V T 1 max fa( T ), fa( V ) trace inequalities of lipschitz type 45

for T,V ∈ B1(H) such that tr(|T |), tr(|V |) < R and a similar result by swap- ∥·∥ ∥·∥ ping in the right hand side of (4.6) the norm with 1. In particular, if tr(|T |), tr(|V |) ≤ M < R, then we have the simpler inequality ∥ − ∥ ≤ ′ ∥ − ∥ f(V ) f(T ) 1 fa(M) V T 1 . (4.7)

2 2 2 If T,V ∈ B2(H) are such that tr(|T | ), tr(|V | ) < R , then we have the norm inequalities ∫ 1 ( ) ∥ − ∥ ≤ ∥ − ∥ ′ ∥ − ∥ f(V ) f(T ) 2 V T 2 fa (1 t)T + tV dt (4.8) 0  [ ] ( ) ′ ∥ ∥ ′ ∥ ∥  1 f ′ T +V + fa( T )+fa( V ) ,  2 a 2 2   ≤ ∥ − ∥ ×  fa(∥V ∥)−fa(∥T ∥) V T 2 ∥ ∥ ̸ ∥ ∥  ∥V ∥−∥T ∥ if V = T ,    ′ ∥ ∥ ∥ ∥ ∥ ∥ fa( T ) if V = T , 1 [ ] ≤ ∥V − T ∥ f ′ (∥T ∥) + f ′ (∥V ∥) 2 2 a a { } ≤ ∥ − ∥ ′ ∥ ∥ ′ ∥ ∥ V T 2 max fa( T ), fa( V ) and a similar result by swapping in the right hand side of (4.6) the norm ∥·∥ ∥·∥ | |2 | |2 ≤ 2 2 with 2. In particular, if tr( T ), tr( V ) K < R , then we have the simpler inequality ∥ − ∥ ≤ ′ ∥ − ∥ f(V ) f(T ) 2 fa(K) V T 2 . (4.9)

5. Applications for Jensen’s difference

We have the following representation: ∑ Lemma 1. ∞ n Let f(z) := n=0 anz be a power series with complex coeﬃ- cients and convergent on the open disk D(0,R), R > 0. If either T,V ∈ B1(H) ∥ ∥ ∥ ∥ ∈ B ∥ ∥ ∥ ∥ with T( 1 , V) 1 < R, or T,V 2(H)( with) T 2 , V 2 < R then f(V ), V +T ∈ B V +T ∈ B f(T ), f 2 1(H) or f(V ), f(T ), f 2 2(H), respectively and [ ( )] tr [f(V )] + tr [f(T )] V + T − tr f (5.1) 2 2 ∫ ( [ ( ) ( )]) 1 1 V + T V + T = tr (V − T ) f ′ (1 − t) + tV − f ′ (1 − t) + tT dt . 4 0 2 2 46 s.s. dragomir

Proof. The ﬁrst part of the theorem follows from Theorem 6. From the identity (3.1), for T,V ∈ B (H) with ∥T ∥ , ∥V ∥ < R we have [ ( )] 1 1 1 V + T tr [f(V )] − tr f (5.2) 2 ∫ [( ) ( )] 1 V + T V + T = tr V − f ′ (1 − t) + tV dt 0 2 2 ∫ [ ( )] 1 1 V + T = tr (V − T )f ′ (1 − t) + tV dt 2 0 2 and [ ( )] V + T tr [f(T )] − tr f (5.3) 2 ∫ [( ) ( )] 1 V + T V + T = tr T − f ′ (1 − t) + tT dt 0 2 2 ∫ [ ( )] 1 1 V + T = tr (T − V )f ′ (1 − t) + tT dt 2 0 2 ∫ [ ( )] 1 1 V + T = − tr (V − T )f ′ (1 − t) + tT dt . 2 0 2 If we add the above inequalities (5.2) and (5.3) and divide by 2 we get the desired result (5.1). ∑ Theorem 9. ∞ n Let f(z) := n=0 anz be a power series with complex coeﬃcients and convergent on the open disk D(0,R), R > 0. If T,V ∈ B1(H) with ∥T ∥ , ∥V ∥ < R, then 1 1 [ ( )]

tr [f(V )] + tr [f(T )] V + T − tr f (5.4) 2 2 ∫ 1 ( ) ≤ 1 ∥ − ∥2 − 1 ′′ ∥ − ∥ V T t fa (1 t)T + tV 1 dt 2 0 2 [ ( ) ] f ′′(∥V ∥ ) + f ′′(∥T ∥ ) ≤ 1 ∥ − ∥2 ′′ V + T a 1 a 1 V T fa + 24 2 1 2 1 [ ] ≤ ∥V − T ∥2 f ′′(∥V ∥ ) + f ′′(∥T ∥ ) 12 a 1 a 1 1 { } ≤ ∥V − T ∥2 max f ′′(∥V ∥ ), f ′′(∥T ∥ ) . 6 a 1 a 1 trace inequalities of lipschitz type 47

∈ B ∥ ∥ ∥ ∥ Proof. Taking the modulus in (5.1), for T,V 1(H) with T 1 , V 1 < R we have [ ( )]

tr [f(V )] + tr [f(T )] V + T − tr f (5.5) 2 2 ∫ ( [ ( ) ( )]) 1 1 ′ V + T ′ V + T ≤ tr (V − T ) f (1 − t) + tV − f (1 − t) + tT dt . 4 0 2 2 ∈ B ∥ ∥ ∥ ∥ Using the properties of trace, for T,V 1(H) with T 1 , V 1 < R and t ∈ [0, 1] we have ( [ ( ) ( )])

′ V + T ′ V + T tr (V − T ) f (1 − t) + tV − f (1 − t) + tT (5.6) 2 2 [ ( ) ( )]

′ V + T ′ V + T ≤ ∥V − T ∥ f (1 − t) + tV − f (1 − t) + tT . 2 2 1

From (4.1), for A, B ∈ B1(H) with ∥A∥ , ∥B∥ < R we have ∫ 1 1 1 ( ) ∥ − ∥ ≤ ∥ − ∥ ′ ∥ − ∥ f(A) f(B) 1 A B fa (1 t)B + tA 1 dt (5.7) 0[ ( ) ] f ′ (∥A∥ ) + f ′ (∥B∥ ) ≤ 1 ∥ − ∥ ′ A + B a 1 a 1 A B fa + 2 2 1 2 1 [ ] ≤ ∥A − B∥ f ′ (∥A∥ ) + f ′ (∥B∥ ) 2 a 1 a 1 { } ≤ ∥ − ∥ ′ ∥ ∥ ′ ∥ ∥ A B max fa( A 1), fa( B 1) . ′ − V +T Applying the second and third inequalities in (5.7) for f , A = (1 t) 2 +tV and B = (1 − t) V +T + tT we get [ ( 2 ) ( )]

′ V + T ′ V + T f (1 − t) + tV − f (1 − t) + tT (5.8) 2 2 [ ( ) 1

≤ 1 ∥ − ∥ ′′ V + T t V T fa 2 2 1 ( ) ( )] f ′′ (1 − t) V +T + tV + f ′′ (1 − t) V +T + tT + a 2 1 a 2 1 2 1 ≤ t ∥V − T ∥ 2 [ ( ) ( )]

× ′′ − V + T ′′ − V + T fa (1 t) + tV + fa (1 t) + tT 2 1 2 1 48 s.s. dragomir

∈ B ∥ ∥ ∥ ∥ ∈ for T,V 1(H) with T 1 , V 1 < R and t [0, 1]. ′′ ∥·∥ Since fa is convex and monotonic nondecreasing and 1 is convex, then ( ) ( ) f ′′ (1 − t) V +T + tV + f ′′ (1 − t) V +T + tT a 2 1 a 2 1 (5.9) 2 ( ) f ′′(∥V ∥ ) + f ′′(∥T ∥ ) ≤ − ′′ V + T a 1 a 1 (1 t) fa + t 2 1 2 ∈ B ∥ ∥ ∥ ∥ ∈ for T,V 1(H) with T 1 , V 1 < R and t [0, 1]. From (5.8) and (5.9) we get [ ( ) ( )]

′ V + T ′ V + T f (1 − t) + tV − f (1 − t) + tT (5.10) 2 2 1 1 ≤ t ∥V − T ∥ 2 [ ( ) ( )]

× ′′ − V + T ′′ − V + T fa (1 t) + tV + fa (1 t) + tT 2 1 2 1 [ ( ) ] f ′′ (∥V ∥ ) + f ′′(∥T ∥ ) ≤ ∥ − ∥ − ′′ V + T a 1 a 1 t V T (1 t)fa + t 2 1 2 ∈ B ∥ ∥ ∥ ∥ ∈ for T,V 1(H) with T 1 , V 1 < R and t [0, 1]. Integrating (5.10) over t on [0, 1] we get ∫ [ ( ) ( )] 1 ′ V + T ′ V + T f (1 − t) + tV − f (1 − t) + tT dt 0 2 2 1 [∫ ( ) 1 ≤ 1 ∥ − ∥ × ′′ − V + T V T tfa (1 t) + tV dt 2 0 2 1 ∫ ( )] 1 ′′ − V + T + tfa (1 t) + tT dt 0 2 1 [ ( ) ∫ 1 ≤ ∥ − ∥ ′′ V + T − V T fa t(1 t) dt 2 1 0 ∫ ] f ′′(∥V ∥ ) + f ′′(∥T ∥ ) 1 + a 1 a 1 t2 dt 2 0 [ ( ) ] f ′′(∥V ∥ ) + f ′′(∥T ∥ ) 1 ∥ − ∥ ′′ V + T a 1 a 1 = V T fa + , 6 2 1 2 trace inequalities of lipschitz type 49 which together with (5.5) and (5.6) produce the inequality [ ( )]

tr [f(V )] + tr [f(T )] V + T − tr f (5.11) 2 2 [∫ ( ) 1 ≤ 1 ∥ − ∥2 × ′′ − V + T V T tfa (1 t) + tV dt 8 0 2 1 ∫ ( )] 1 ′′ − V + T + tfa (1 t) + tT dt 0 2 [ ( ) 1 ] f ′′(∥V ∥ ) + f ′′(∥T ∥ ) ≤ 1 ∥ − ∥2 ′′ V + T a 1 a 1 V T fa + . 24 2 1 2 Now, observe that ∫ ( ) 1 ′′ − V + T tfa (1 t) + tV dt (5.12) 0 2 1 ∫ ( ) 1 ′′ 1 − t 1 + t = tfa T + V dt , 0 2 2 1 ∫ ( ) 1 ′′ − V + T tfa (1 t) + tT dt (5.13) 0 2 1 ∫ ( ) 1 ′′ 1 − t 1 + t = tfa V + T dt . 0 2 2 1 1+t Using the change of variable u = 2 , then we get ∫ ( ) 1 ′′ 1 − t 1 + t tfa T + V dt (5.14) 0 2 2 1 ∫ 1 ( ) − ′′ ∥ − ∥ = 2 (2u 1)fa (1 u)T + uV du . 1 2 1−t Also, by changing the variable v = 2 , we get ∫ ( ) 1 ′′ 1 − t 1 + t tfa V + T dt (5.15) 0 2 2 1 ∫ 1 ( ) 2 − ′′ ∥ − ∥ = 2 (1 2v)fa (1 v)T + vV dv . 0 50 s.s. dragomir

Utilising the equalities (5.12)-(5.15) we obtain ∫ ( ) ∫ ( ) 1 1 ′′ − V + T ′′ − V + T tfa (1 t) + tV dt + tfa (1 t) + tT dt 0 2 1 0 2 1 ∫ 1 ( ) − ′′ ∥ − ∥ = 2 (2t 1)fa (1 t)T + tV dt 1 2 ∫ 1 ( ) 2 − ′′ ∥ − ∥ + 2 (1 2t)fa (1 t)T + tV dt 0 ∫ 1 ( ) | − | ′′ ∥ − ∥ = 2 2t 1 fa (1 t)T + tV dt 0 ∫ 1 ( ) − 1 ′′ ∥ − ∥ = 4 t fa (1 t)T + tV dt 0 2 ∈ B ∥ ∥ ∥ ∥ for T,V 1(H) with T 1 , V 1 < R. Making use of (5.11) we deduce the ﬁrst two inequalities in (5.4). The rest is obvious. ∑ Corollary 2. ∞ n Let f(z) := n=0 anz be a power series with complex coeﬃcients and convergent on the open disk D(0,R), R > 0. If T,V ∈ B1(H) ∥ ∥ ∥ ∥ ≤ with T 1 , V 1 M < R, then [ ( )]

tr [f(V )] + tr [f(T )] V + T 1 2 ′′ − tr f ≤ ∥V − T ∥ f (M) . (5.16) 2 2 8 a

1 The constant 8 is best possible in (5.16). Proof. From the ﬁrst inequality in (5.4) we have [ ( )]

tr [f(V )] + tr [f(T )] V + T − tr f 2 2 ∫ 1 ( ) ≤ 1 ∥ − ∥2 − 1 ′′ ∥ − ∥ V T t fa (1 t)T + tV 1 dt 2 0 2 ∫ 1 ≤ 1 ∥ − ∥2 ′′ − 1 1 ∥ − ∥2 ′′ V T fa (M) t dt = V T fa (M) 2 0 2 8 ∈ B ∥ ∥ ∥ ∥ ≤ for T,V 1(H) with T 1 , V 1 M < R, and the inequality is proved. If we consider the scalar case and take f(z) = z2, V = a, T = b with ∈ R 1 − 2 a, b then we get in both sides of (5.16) the same quantity 4 (b a) . trace inequalities of lipschitz type 51

3. ∥·∥ ∥·∥ Remark A similar result holds by swapping the norm with 1 in the right hand side of (5.4). The case of Hilbert-Schmidt norm may also be stated, however the details are not presented here. If we write the inequality (5.4) for the exponential function, then we get [ ( )]

tr [exp(V )] + tr [exp(T )] V + T − tr exp (5.17) 2 2 ∫ 1 ( ) ≤ 1 ∥ − ∥2 − 1 ∥ − ∥ V T t exp (1 t)T + tV 1 dt 2 0 2 [ ( ) ] ∥ ∥ ∥ ∥ 1 2 V + T exp( V ) + exp( T ) ≤ ∥V − T ∥ exp + 1 1 24 2 1 2 1 [ ] ≤ ∥V − T ∥2 exp(∥V ∥ ) + exp(∥T ∥ ) 12 1 1 1 { } ≤ ∥V − T ∥2 max exp(∥V ∥ ), exp(∥T ∥ ) 6 1 1 for any for T,V ∈ B1(H). ∈ B ∥ ∥ ∥ ∥ ≤ If T,V 1(H) with V 1 , T 1 M, then [ ( )]

tr [exp(V )] + tr [exp(T )] V + T − tr exp (5.18) 2 2 1 ≤ ∥V − T ∥2 exp(M) . 8 If we write the inequality (5.4) for the function f(z) = (1 − z)−1, then we get [ ] [ ] [( ) ] −1 −1 −1 tr (1H − V ) + tr (1H − T ) V + T − tr 1 − (5.19) 2 H 2 ∫ 1 ( ) ≤ ∥ − ∥2 − 1 − ∥ − ∥ −3 V T t 1 (1 t)T + tV 1 dt 0 2 [( ) ] −3 − ∥ ∥ −3 − ∥ ∥ −3 1 2 V + T (1 V ) + (1 T ) ≤ ∥V − T ∥ × 1 − + 1 1 12 2 1 2 1 [ ] ≤ ∥V − T ∥2 (1 − ∥V ∥ )−3 + (1 − ∥T ∥ )−3 6 1 1 1 { } ≤ ∥V − T ∥2 max (1 − ∥V ∥ )−3, (1 − ∥T ∥ )−3 , 3 1 1 52 s.s. dragomir

∈ B ∥ ∥ ∥ ∥ for any for T,V 1(H) with V 1 , T 1 < 1. ∥ ∥ ∥ ∥ ≤ Moreover, if V 1 , T 1 M < 1, then [ ] [ ] [( ) ] −1 −1 −1 tr (1H − V ) + tr (1H − T ) V + T − tr 1 − (5.20) 2 H 2 1 ≤ ∥V − T ∥2 (1 − M)−3. 4 The interested reader may choose other examples of power series to get similar results. However, the details are not presented here.

References

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