Trace Inequalities of Lipschitz Type for Power Series of ...Key words: Banach algebras of operators...
Transcript of Trace Inequalities of Lipschitz Type for Power Series of ...Key words: Banach algebras of operators...
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E extracta mathematicae Vol. 32, Num. 1, 25 – 54 (2017)
Trace Inequalities of Lipschitz Type for Power Series ofOperators on Hilbert Spaces
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
Abstract : Let f(z) =∑∞
n=0 αnzn 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, ifT, V ∈ B1(H), the Banach space of all trace operators on H, are such that ∥T∥1 , ∥V ∥1 < R,then f(V ), f(T ), f ′((1− t)T + tV ) ∈ B1(H) for any t ∈ [0, 1] and
tr [f(V )]− tr [f(T )] =
∫ 1
0
tr[(V − T )f ′((1− t)T + tV
)]dt.
Several trace inequalities are established. Applications for some elementary functions ofinterest are also given.
Key words: Banach algebras of operators on Hilbert spaces, Power series, Lipschitz typeinequalities, 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 separablecomplex Hilbert space H. The absolute value of an operator A is the positiveoperator |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., thereis no constant L > 0 such that∥∥ |A| − |B|
∥∥ ≤ L ∥A−B∥
for any A,B ∈ B(H).
25
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26 s.s. dragomir
However, as shown by Farforovskaya in [17], [18] and Kato in [22], thefollowing inequality holds
∥∥ |A| − |B|∥∥ ≤ 2
π∥A−B∥
(2 + log
(∥A∥+ ∥B∥∥A−B∥
))(1.1)
for any A,B ∈ B(H) with A = B.
If the operator norm is replaced with Hilbert-Schmidt norm ∥C∥HS :=(trC∗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 alloperators B in a neighborhood of A we have∥∥ |A| − |B|
∥∥ ≤ a1 ∥A−B∥+ a2 ∥A−B∥2 +O(∥A−B∥3
)(1.3)
where
a1 =∥∥A−1
∥∥ ∥A∥ and a2 =∥∥A−1
∥∥+ ∥∥A−1∥∥3 ∥A∥2 .
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 whichthe function of operator can be defined. For some results on this topic, see[5], [19] and the references therein.
By the help of power series f(z) =∑∞
n=0 anzn we can naturally construct
another power series which will have as coefficients the absolute values of thecoefficient of the original series, namely, fa(z) :=
∑∞n=0 |an| zn. It is obvious
that this new power series will have the same radius of convergence as theoriginal series. We also notice that if all coefficients an ≥ 0, then fa = f .
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trace inequalities of lipschitz type 27
We notice that if
f(z) =
∞∑n=1
(−1)n
nzn = ln
1
1 + z, z ∈ D(0, 1) ; (1.5)
g(z) =
∞∑n=0
(−1)n
(2n)!z2n = cos z , z ∈ C ;
h(z) =∞∑n=0
(−1)n
(2n+ 1)!z2n+1 = sin z , z ∈ C ;
l(z) =∞∑n=0
(−1)nzn =1
1 + z, z ∈ D(0, 1) ;
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 thecoefficients are
fa(z) =
∞∑n=1
1
n!zn = ln
1
1− z, z ∈ D(0, 1) ; (1.6)
ga(z) =
∞∑n=0
1
(2n)!z2n = cosh z , z ∈ C ;
ha(z) =∞∑n=0
1
(2n+ 1)!z2n+1 = sinh z , z ∈ C ;
la(z) =
∞∑n=0
zn =1
1− z, z ∈ D(0, 1) .
Other important examples of functions as power series representations withnonnegative coefficients are:
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28 s.s. dragomir
exp(z) =
∞∑n=0
1
n!zn , z ∈ C ; (1.7)
1
2ln
(1 + z
1− z
)=
∞∑n=1
1
2n− 1z2n−1 , z ∈ D(0, 1) ;
sin−1(z) =∞∑n=0
Γ(n+ 12)√
π(2n+ 1)n!z2n+1 , z ∈ D(0, 1) ;
tanh−1(z) =
∞∑n=1
1
2n− 1z2n−1 , z ∈ D(0, 1) ;
2F1(α, β, γ, z) =
∞∑n=0
Γ(n+ α)Γ(n+ β)Γγ)
n!Γ(α)Γ(β)Γ(n+ γ)zn , α, β, γ > 0 , z ∈ D(0, 1) ;
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. Let f(z) :=∑∞
n=0 anzn be a power series with complex co-
efficients 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 )∥ ≤ f ′a(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 )∥ ≤ f ′a(M) ∥T − V ∥ (1.9)
In the recent paper [13] we improved the inequality (1.8) as follows:
Theorem 2. Let f(z) :=∑∞
n=0 anzn be a power series with complex co-
efficients 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 )∥ ≤ ∥T − V ∥∫ 1
0f ′a(∥(1− t)T + tV ∥
)dt . (1.10)
In order to obtain similar results for the trace of bounded linear operatorson complex infinite dimensional Hilbert spaces we need some preparations asfollows.
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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∑i∈I
∥Aei∥2 <∞ . (2.1)
It is well know that, if {ei}i∈I and {fj}j∈J are orthonormal bases for H andA ∈ B(H) then ∑
i∈I∥Aei∥2 =
∑j∈I
∥Afj∥2 =∑j∈I
∥A∗fj∥2 (2.2)
showing that the definition (2.1) is independent of the orthonormal basis andA 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
∥A∥2 :=
(∑i∈I
∥Aei∥2)1/2
(2.3)
for {ei}i∈I an orthonormal basis of H. This definition does not depend on thechoice of the orthonormal basis.
Using the triangle inequality in l2(I), one checks that B2(H) is a vec-tor space and that ∥·∥2 is a norm on B2(H), which is usually called in theliterature 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
Hilbert-Schmidt and ∥A∥2 = ∥|A|∥2. From (2.2) we have that if A ∈ B2(H),then A∗ ∈ B2(H) and ∥A∥2 = ∥A∗∥2.
The following theorem collects some of the most important properties ofHilbert-Schmidt operators:
Theorem 3. We have:
(i) (B2(H), ∥·∥2) is a Hilbert space with inner product
⟨A,B⟩2 :=∑i∈I
⟨Aei, Bei⟩ =∑i∈I
⟨B∗Aei, ei⟩ (2.4)
and the definition does not depend on the choice of the orthonormalbasis {ei}i∈I ;
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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 ofB2(H);
(v) B2(H) ⊆ K(H), where K(H) denotes the algebra of compact operatorson H.
If {ei}i∈I an orthonormal basis of H, we say that A ∈ B(H) is trace classif
∥A∥1 :=∑i∈I
⟨|A| ei, ei⟩ <∞ . (2.7)
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);
(ii) |A|1/2 ∈ 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).
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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
∥A∥1 = sup{|⟨A,B⟩2| : B ∈ B2(H), ∥B∥ ≤ 1
}.
(v) (B1(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 spaceof B1(H).
We define the trace of a trace class operator A ∈ B1(H) to be
tr(A) :=∑i∈I
⟨Aei, ei⟩ , (2.9)
where {ei}i∈I an orthonormal basis of H. Note that this coincides with theusual definition of the trace if H is finite-dimensional. We observe that theseries (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).
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32 s.s. dragomir
Utilising the trace notation we obviously have that
⟨A,B⟩2 = tr(B∗A) = tr(AB∗) ,
∥A∥22 = tr(A∗A) = tr(|A|2
)for any A, B ∈ B2(H).
The following Holder’s type inequality has been obtained by Ruskai in [28]
|tr(AB)| ≤ tr(|AB|) ≤[tr(|A|1/α
)]α [tr(|B|1/(1−α)
)]1−α(2.12)
where α ∈ (0, 1) and A,B ∈ B(H) with |A|1/α , |B|1/(1−α) ∈ B1(H).In particular, for α = 1
2 we get the Schwarz inequality
|tr(AB)| ≤ tr(|AB|) ≤[tr(|A|2
)]1/2 [tr(|B|2
)]1/2(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 anorthonormal basis of H, then
0 ≤⟨AP 1/2ei, P
1/2ei
⟩≤ ∥A∥
∥∥∥P 1/2ei
∥∥∥2 = ∥A∥ ⟨Pei, ei⟩
for any i ∈ I. Summing over i ∈ I we get
0 ≤∑i∈I
⟨AP 1/2ei, P
1/2ei
⟩≤ ∥A∥
∑i∈I
⟨Pei, ei⟩ = ∥A∥ tr(P ) ,
and since ∑i∈I
⟨AP 1/2ei, P
1/2ei
⟩=∑i∈I
⟨P 1/2AP 1/2ei, ei
⟩= 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)
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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
|tr(PA)| =
∣∣∣∣∣∑i∈I
⟨AP 1/2ei, P
1/2ei
⟩∣∣∣∣∣ ≤∑i∈I
∣∣∣⟨AP 1/2ei, P1/2ei
⟩∣∣∣≤∑i∈I
⟨|A|P 1/2ei, P
1/2ei
⟩= tr(P |A|) , (2.16)
for any A a selfadjoint operator and P ∈ B1(H) with P ≥ 0.
For the theory of trace functionals and their applications the reader isreferred to [31].
For some classical trace inequalities see [9], [11], [26] and [35], which arecontinuations of the work of Bellman [7]. For related works the reader canrefer to [1], [8], [9], [20], [23], [24], [25], [29] and [32].
3. Trace inequalities
We have the following representation result:
Theorem 6. Let f(z) :=∑∞
n=0 anzn 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
tr [f(V )]− tr [f(T )] =
∫ 1
0tr[(V − T )f ′
((1− t)T + tV
)]dt . (3.1)
Proof. We use the identity (see for instance [6, p. 254])
An −Bn =
n−1∑j=0
An−1−j(A−B)Bj (3.2)
that holds for any A,B ∈ B(H) and n ≥ 1.
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34 s.s. dragomir
For T, V ∈ B(H) we consider the function φ : [0, 1] → B(H) defined byφ(t) = [(1− t)T + tV ]n. For t ∈ (0, 1) and ε = 0 with t + ε ∈ (0, 1) we havefrom (3.2) that
φ(t+ ε)−φ(t) = [(1− t− ε)T + (t+ ε)V ]n − [(1− t)T + tV ]n
= εn−1∑j=0
[(1− t− ε)T + (t+ ε)V ]n−1−j (V − T ) [(1− t)T + tV ]j .
Dividing with ε = 0 and taking the limit over ε→ 0 we have in the normtopology of B that
φ′(t) = limε→0
1
ε[φ(t+ ε)− φ(t)] (3.3)
=n−1∑j=0
[(1− t)T + tV ]n−1−j (V − T ) [(1− t)T + tV ]j .
Integrating on [0, 1] we get from (3.3) that
∫ 1
0φ′(t) dt =
n−1∑j=0
∫ 1
0[(1− t)T + tV ]n−1−j (V − T ) [(1− t)T + tV ]j dt ,
and since ∫ 1
0φ′(t) dt = φ(1)− φ(0) = V n − Tn,
then we get the following equality of interest in itself
V n − Tn =n−1∑j=0
∫ 1
0[(1− t)T + tV ]n−1−j (V − T ) [(1− t)T + tV ]j dt (3.4)
for any T, V ∈ B(H) and n ≥ 1.
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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(Tn)
=
n−1∑j=0
∫ 1
0tr([(1− t)T + tV ]n−1−j (V − T ) [(1− t)T + tV ]j
)dt
=
n−1∑j=0
∫ 1
0tr([(1− t)T + tV ]n−1 (V − T )
)dt (3.5)
= n
∫ 1
0tr([(1− t)T + tV ]n−1 (V − T )
)dt
= n
∫ 1
0tr((V − T ) [(1− t)T + tV ]n−1 ) dt
for any n ≥ 1.Let m ≥ 1. Then by (3.5) we have have
tr
(m∑
n=0
anVn
)− tr
(m∑
n=0
anTn
)=
m∑n=0
an[tr(V n)− tr
(Tn)]
=
m∑n=1
an[tr(V n)− tr
(Tn)]
(3.6)
=
m∑n=1
nan
∫ 1
0tr((V − T ) [(1− t)T + tV ]n−1
)dt
=
∫ 1
0tr
((V − T )
m∑n=1
nan [(1− t)T + tV ]n−1
)dt
for any T, V ∈ B1(H).Since tr(|T |), tr(|V |) < R with T, V ∈ B1(H) then the series
∑∞n=0 anV
n,∑∞n=0 anT
n and∑∞
n=1 nan [(1− t)T + tV ]n−1 are convergent in B1(H) and
∞∑n=0
anVn = f(V ) ,
∞∑n=0
anTn = f(T )
and∞∑n=1
nan (1− t)T + tV ]n−1 = f ′((1− t)T + tV
)
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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 haveother equalities as follows
tr [exp(V )]− tr [exp(T )] =
∫ 1
0tr((V − T ) exp((1− t)T + tV )
)dt , (3.7)
tr [sin(V )]− tr [sin(T )] =
∫ 1
0tr((V − T ) cos((1− t)T + tV )
)dt , (3.8)
tr [sinh(V )]− tr [sinh(T )] =
∫ 1
0tr((V − T ) cosh((1− t)T + tV )
)dt . (3.9)
If T, V ∈ B1(H) with tr(|T |), tr(|V |) < 1 then
tr[(1H − V )−1
]− tr
[(1H − T )−1
](3.10)
=
∫ 1
0tr((V − T )(1H − (1− t)T − tV )−2
)dt ,
and
tr[ln(1H − V )−1
]s − tr
[ln(1H − T )−1
](3.11)
=
∫ 1
0tr((V − T )(1H − (1− t)T − tV )−1
)dt .
We have the following result:
Corollary 1. With the assumptions in Theorem 6 we have the inequal-ities
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trace inequalities of lipschitz type 37
∣∣ tr [f(V )]− tr [f(T )]∣∣ ≤ min
{∥V − T∥
∫ 1
0
∥∥f ′((1− t)T + tV )∥∥1dt , (3.12)
∥V − T∥1∫ 1
0
∥∥f ′((1− t)T + tV )∥∥dt}
≤ min
{∥V − T∥
∫ 1
0f ′a(∥(1− t)T + tV ∥1
)dt,
∥V − T∥1∫ 1
0f ′a(∥(1− t)T + tV ∥
)dt
},
where ∥·∥ is the operator norm and ∥·∥1 is the 1-norm introduced for traceclass operators.
Proof. From (3.1), we have by taking the modulus
∣∣ tr [f(V )]− tr [f(T )]∣∣ ≤ ∫ 1
0
∣∣tr ((V − T )f ′((1− t)T + tV ))∣∣ dt . (3.13)
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)∥∥
1=
∥∥∥∥∥∞∑n=1
nan [(1− t)T + tV ]n−1
∥∥∥∥∥1
≤∞∑n=1
n |an|∥∥∥[(1− t)T + tV ]n−1
∥∥∥1
≤∞∑n=1
n |an|∥∥(1− t)T + tV
∥∥n−1
1
= f ′a(∥(1− t)T + tV ∥1
)for any T, V ∈ B1(H) with ∥T∥1 , ∥V ∥1 < R.
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38 s.s. dragomir
This proves the first part of the second inequality.
Since ∥X∥ ≤ ∥X∥1 for any X ∈ B1(H) then ∥(1− t)T + tV ∥ < R for anyT, V ∈ B1(H) with ∥T∥1 , ∥V ∥1 < R which shows that f ′a
(∥(1− t)T + tV ∥
)is well defined.
The second part of the second inequality follows in a similar way and thedetails are omitted.
Remark 1. We observe that f ′a is monotonic nondecreasing and convex onthe interval [0, R) and since the function ψ(t) := ∥(1− t)T + tV ∥ is convexon [0, 1] we have that f ′a ◦ ψ is also convex on [0, 1]. Utilising the Hermite-Hadamard inequality for convex functions (see for instance [16, p. 2]) we havethe sequence of inequalities∫ 1
0f ′a(∥(1− t)T + tV ∥
)dt ≤ 1
2
[f ′a
(∥∥∥∥T + V
2
∥∥∥∥)+f ′a(∥T∥) + f ′a(∥V ∥)
2
]≤ 1
2
[f ′a(∥T∥) + f ′a(∥V ∥)
]≤ max
{f ′a(∥T∥), f ′a(∥V ∥)
}. (3.14)
We also have∫ 1
0f ′a(∥(1− t)T + tV ∥
)dt ≤
∫ 1
0f ′a((1− t) ∥T∥+ t ∥V ∥
)dt
≤ 1
2
[f ′a
(∥T∥+ ∥V ∥
2
)+f ′a(∥T∥) + f ′a(∥V ∥)
2
]≤ 1
2
[f ′a(∥T∥) + f ′a(∥V ∥)
]≤ max
{f ′a(∥T∥), f ′a(∥V ∥)
}. (3.15)
We observe that if ∥V ∥ = ∥T∥, then by the change of variable s = (1−t) ∥T∥+t ∥V ∥ we have∫ 1
0f ′a((1− t) ∥T∥+ t ∥V ∥
)dt =
1
∥V ∥ − ∥T∥
∫ ∥V ∥
∥T∥f ′a(s) ds
=fa(∥V ∥)− fa(∥T∥)
∥V ∥ − ∥T∥.
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trace inequalities of lipschitz type 39
If ∥V ∥ = ∥T∥, then
∫ 1
0f ′a((1− t) ∥T∥+ t ∥V ∥
)dt = f ′a(∥T∥) .
Utilising these observations we then get the following divided difference in-equality for T = V
∫ 1
0f ′a(∥(1− t)T + tV ∥
)dt ≤
fa(∥V ∥)−fa(∥T∥)
∥V ∥−∥T∥ if ∥V ∥ = ∥T∥ ,
f ′a(∥T∥) if ∥V ∥ = ∥T∥ .(3.16)
Similar comments apply for the 1-norm ∥·∥1 when T, V ∈ B1(H).
If we use the first part in the inequalities (3.12) and the above remarks,then we get the following string of inequalities
∣∣ tr [f(V )] − tr [f(T )]∣∣ ≤ ∥V − T∥
∫ 1
0
∥∥f ′((1− t)T + tV)∥∥
1dt
≤ ∥V − T∥∫ 1
0f ′a(∥(1− t)T + tV ∥1
)dt (3.17)
≤ ∥V − T∥ ×
12
[f ′a
(∥∥T+V2
∥∥1
)+
f ′a(∥T∥1)+f ′
a(∥V ∥1)2
],
fa(∥V ∥1)−fa(∥T∥1)∥V ∥1−∥T∥1
if ∥V ∥1 = ∥T∥1 ,
f ′a(∥T∥1) if ∥V ∥1 = ∥T∥1 ,
≤ 1
2∥V − T∥
[f ′a(∥T∥1) + f ′a(∥V ∥1)
]≤ ∥V − T∥max
{f ′a(∥T∥1), f
′a(∥V ∥1)
}provided T, V ∈ B1(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∥ f ′a(M) .
A similar sequence of inequalities can also be stated by swapping the norm∥·∥ with ∥·∥1 in (3.17). We omit the details.
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40 s.s. dragomir
If we use the inequality (3.17) for the exponential function, then for anyT, V ∈ B1(H) we have the inequalities
∣∣ tr [exp(V )] − tr [exp(T )]∣∣ ≤ ∥V − T∥
∫ 1
0
∥∥exp ((1− t)T + tV)∥∥
1dt
≤ ∥V − T∥∫ 1
0exp
(∥(1− t)T + tV ∥1
)dt (3.18)
≤ ∥V − T∥ ×
12
[exp
(∥∥T+V2
∥∥1
)+
exp(∥T∥1)+exp(∥V ∥1)2
],
exp(∥V ∥1)−exp(∥T∥1)∥V ∥1−∥T∥1
if ∥V ∥1 = ∥T∥1 ,
exp(∥T∥1) if ∥V ∥1 = ∥T∥1 ,
≤ 1
2∥V − T∥
[exp(∥T∥1) + exp(∥V ∥1)
]≤ ∥V − T∥max
{exp(∥T∥1), exp(∥V ∥1)
}.
If ∥T∥1 , ∥V ∥1 < 1, then we have the inequalities∣∣ tr [ ln(1H − V )−1]− tr
[ln(1H − T )−1
] ∣∣ (3.19)
≤ ∥V − T∥∫ 1
0
∥∥(1H − (1− t)T − tV )−1∥∥1dt
≤ ∥V − T∥∫ 1
0(1− ∥(1− t)T + tV ∥1)
−1dt
≤ ∥V − T∥ ×
12
[(1−
∥∥T+V2
∥∥1
)−1+
(1−∥T∥1)−1+(1−∥V ∥1)−1
2
],
ln(1−∥V ∥1)−1−ln(1−∥T∥1)−1
∥V ∥1−∥T∥1if ∥V ∥1 = ∥T∥1 ,
(1− ∥T∥1)−1 if ∥V ∥1 = ∥T∥1 ,
≤ 1
2∥V − T∥
[(1− ∥T∥1)
−1 + (1− ∥V ∥1)−1]
≤ ∥V − T∥max{(1− ∥T∥1)
−1, (1− ∥V ∥1)−1}.
The following result for the Hilbert-Schmidt norm ∥·∥2 also holds:
Theorem 7. Let f(z) :=∑∞
n=0 anzn 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
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trace inequalities of lipschitz type 41
such that tr(|T |2), tr(|V |2) < R2, then f(V ), f(T ), f ′((1− t)T + tV ) ∈ B2(H)for any t ∈ [0, 1] and
tr [f(V )]− tr [f(T )] =
∫ 1
0tr((V − T )f ′
((1− t)T + tV )
)dt . (3.20)
Moreover, we have the inequalities∣∣ tr [f(V )] − tr [f(T )]∣∣ ≤ ∥V − T∥2
∫ 1
0
∥∥f ′((1− t)T + tV)∥∥
2dt ,
≤ ∥V − T∥2∫ 1
0f ′a(∥(1− t)T + tV ∥2
)dt (3.21)
≤ ∥V − T∥2 ×
12
[f ′a
(∥∥T+V2
∥∥2
)+
f ′a(∥T∥2)+f ′
a(∥V ∥2)2
],
fa(∥V ∥2)−fa(∥T∥2)∥V ∥2−∥T∥2
if ∥V ∥2 = ∥T∥2 ,
f ′a(∥T∥2) if ∥V ∥2 = ∥T∥2 ,
≤ 1
2∥V − T∥2
[f ′a(∥T∥2) + f ′a(∥V ∥2)
]≤ ∥V − T∥2max
{f ′a(∥T∥2), f
′a(∥V ∥2)
}.
Proof. The proof of the first part of the theorem follows in a similar mannerto the one from Theorem 6.
Taking the modulus in (3.20) and using the Schwarz inequality for trace(2.13) we have
|tr [f(V )]− tr [f(T )]| ≤∫ 1
0
∣∣tr((V − T )f ′((1− t)T + tV )
)∣∣ dt (3.22)
≤∫ 1
0∥V − T∥2
∥∥f ′((1− t)T + tV)∥∥
2dt .
The rest follows in a similar manner as in the case of 1-norm and the detailsare omitted.
We notice that similar examples to (3.18) and (3.19) may be stated whereboth norms ∥·∥ and ∥·∥1 are replaced by ∥·∥2.
We also observe that, if T, V ∈ B2(H) with ∥T∥2 , ∥V ∥2 ≤ K < R, thenwe have from (3.17) the simple inequality∣∣ tr [f(V )]− tr [f(T )]
∣∣ ≤ ∥V − T∥2 f′a(K) .
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42 s.s. dragomir
4. Norm inequalities
We have the following norm inequalities:
Theorem 8. Let f(z) :=∑∞
n=0 anzn 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 thenorm inequalities
∥f(V )− f(T )∥1 ≤
∥V − T∥1∫ 10 f
′a
(∥(1− t)T + tV ∥
)dt ,
∥V − T∥∫ 10 f
′a
(∥(1− t)T + tV ∥1
)dt .
(4.1)
(ii) If T, V ∈ B2(H) are such that tr(|T |2), tr(|V |2) < R2, then we also havethe norm inequalities
∥f(V )− f(T )∥2 ≤
∥V − T∥2∫ 10 f
′a
(∥(1− t)T + tV ∥
)dt ,
∥V − T∥∫ 10 f
′a
(∥(1− t)T + tV ∥2
)dt .
(4.2)
Proof. We use the equality
V n − Tn =
n−1∑j=0
∫ 1
0[(1− t)T + tV ]n−1−j (V − T ) [(1− t)T + tV ]j dt , (4.3)
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
∥V n − Tn∥1 ≤n−1∑j=0
∫ 1
0
∥∥∥[(1− t)T + tV ]n−1−j (V − T ) [(1− t)T + tV ]j∥∥∥1dt
≤n−1∑j=0
∫ 1
0
∥∥∥[(1− t)T + tV ]n−1−j (V − T )∥∥∥1
∥∥∥[(1− t)T + tV ]j∥∥∥dt
≤n−1∑j=0
∫ 1
0∥V − T∥1
∥∥∥[(1− t)T + tV ]n−1−j∥∥∥∥∥∥[(1− t)T + tV ]j
∥∥∥dt
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trace inequalities of lipschitz type 43
≤ ∥V − T∥1n−1∑j=0
∥(1− t)T + tV ∥n−1−j ∥(1− t)T + tV ∥j dt
= n ∥V − T∥1∫ 1
0∥(1− t)T + tV ∥n−1 dt . (4.4)
Let m ≥ 1. By (4.4) we have∥∥∥∥∥m∑
n=0
anVn −
m∑n=0
anTn
∥∥∥∥∥1
=
∥∥∥∥∥m∑
n=1
an(Vn − Tn)
∥∥∥∥∥1
(4.5)
≤m∑
n=1
|an| ∥V n − Tn∥1
≤ ∥V − T∥1m∑
n=1
|an|n∫ 1
0∥(1− t)T + tV ∥n−1 dt
= ∥V − T∥1∫ 1
0
(m∑
n=1
n |an| ∥(1− t)T + tV ∥n−1
)dt .
Also, we observe that
∥(1− t)T + tV ∥ ≤ ∥(1− t)T + tV ∥1
≤ (1− t) ∥T∥1 + t ∥V ∥1 < R
for any t ∈ [0, 1], which implies that the series∑∞
n=1 n |an| ∥(1− t)T + tV ∥n−1
is convergent and
∞∑n=1
n |an| ∥(1− t)T + tV ∥n−1 = f ′a(∥(1− t)T + tV ∥
)for any t ∈ [0, 1].
Since the series∑∞
n=0 anVn and
∑∞n=0 anT
n are convergent in(B1(H), ∥·∥1
), then by letting m → ∞ in the inequality (4.5) we get
the first inequality in (4.1).
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44 s.s. dragomir
For any n ≥ 1 we also have
∥V n − Tn∥1 ≤n−1∑j=0
∫ 1
0
∥∥∥[(1− t)T + tV ]n−1−j (V − T ) [(1− t)T + tV ]j∥∥∥1dt
≤n−1∑j=0
∫ 1
0
∥∥∥[(1− t)T + tV ]n−1−j (V − T )∥∥∥∥∥∥[(1− t)T + tV ]j
∥∥∥1dt
≤n−1∑j=0
∫ 1
0∥V − T∥
∥∥∥[(1− t)T + tV ]n−1−j∥∥∥∥∥∥[(1− t)T + tV ]j
∥∥∥1dt
≤ ∥V − T∥n−1∑j=0
∥(1− t)T + tV ∥n−1−j ∥(1− t)T + tV ∥j1 dt
≤ ∥V − T∥n−1∑j=0
∥(1− t)T + tV ∥n−1−j1 ∥(1− t)T + tV ∥j1 dt
= n ∥V − T∥∫ 1
0∥(1− t)T + tV ∥n−1
1 dt ,
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∥2that 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 ofinequalities
∥f(V )− f(T )∥1 ≤ ∥V − T∥1∫ 1
0f ′a(∥(1− t)T + tV ∥
)dt (4.6)
≤ ∥V − T∥1 ×
12
[f ′a(∥∥T+V
2
∥∥)+ f ′a(∥T∥)+f ′
a(∥V ∥)2
],
fa(∥V ∥)−fa(∥T∥)∥V ∥−∥T∥ if ∥V ∥ = ∥T∥ ,
f ′a(∥T∥) if ∥V ∥ = ∥T∥ ,
≤ 1
2∥V − T∥1
[f ′a(∥T∥) + f ′a(∥V ∥)
]≤ ∥V − T∥1max
{f ′a(∥T∥), f ′a(∥V ∥)
}
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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, iftr(|T |), tr(|V |) ≤M < R, then we have the simpler inequality
∥f(V )− f(T )∥1 ≤ f ′a(M) ∥V − T∥1 . (4.7)
If T, V ∈ B2(H) are such that tr(|T |2), tr(|V |2) < R2, then we have thenorm inequalities
∥f(V )− f(T )∥2 ≤ ∥V − T∥2∫ 1
0f ′a(∥(1− t)T + tV ∥
)dt (4.8)
≤ ∥V − T∥2 ×
12
[f ′a(∥∥T+V
2
∥∥)+ f ′a(∥T∥)+f ′
a(∥V ∥)2
],
fa(∥V ∥)−fa(∥T∥)∥V ∥−∥T∥ if ∥V ∥ = ∥T∥ ,
f ′a(∥T∥) if ∥V ∥ = ∥T∥ ,
≤ 1
2∥V − T∥2
[f ′a(∥T∥) + f ′a(∥V ∥)
]≤ ∥V − T∥2max
{f ′a(∥T∥), f ′a(∥V ∥)
}and a similar result by swapping in the right hand side of (4.6) the norm ∥·∥with ∥·∥2. In particular, if tr(|T |2), tr(|V |2) ≤ K2 < R2, then we have thesimpler inequality
∥f(V )− f(T )∥2 ≤ f ′a(K) ∥V − T∥2 . (4.9)
5. Applications for Jensen’s difference
We have the following representation:
Lemma 1. Let f(z) :=∑∞
n=0 anzn 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)with ∥T∥1 , ∥V ∥1 < R, or T, V ∈ B2(H) with ∥T∥2 , ∥V ∥2 < R then f(V ),f(T ), f
(V+T2
)∈ B1(H) or f(V ), f(T ), f
(V+T2
)∈ B2(H), respectively and
tr [f(V )] + tr [f(T )]
2− tr
[f
(V + T
2
)](5.1)
=1
4
∫ 1
0
tr
((V − T )
[f ′((1− t)
V + T
2+ tV
)− f ′
((1− t)
V + T
2+ tT
)])dt .
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46 s.s. dragomir
Proof. The first part of the theorem follows from Theorem 6.From the identity (3.1), for T, V ∈ B1(H) with ∥T∥1 , ∥V ∥1 < R we have
tr [f(V )] − tr
[f
(V + T
2
)](5.2)
=
∫ 1
0tr
[(V − V + T
2
)f ′((1− t)
V + T
2+ tV
)]dt
=1
2
∫ 1
0tr
[(V − T )f ′
((1− t)
V + T
2+ tV
)]dt
and
tr [f(T )] − tr
[f
(V + T
2
)](5.3)
=
∫ 1
0tr
[(T − V + T
2
)f ′((1− t)
V + T
2+ tT
)]dt
=1
2
∫ 1
0tr
[(T − V )f ′
((1− t)
V + T
2+ tT
)]dt
= −1
2
∫ 1
0tr
[(V − T )f ′
((1− t)
V + T
2+ tT
)]dt .
If we add the above inequalities (5.2) and (5.3) and divide by 2 we get thedesired result (5.1).
Theorem 9. Let f(z) :=∑∞
n=0 anzn 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 < R, then∣∣∣∣tr [f(V )] + tr [f(T )]
2− tr
[f
(V + T
2
)]∣∣∣∣ (5.4)
≤ 1
2∥V − T∥2
∫ 1
0
∣∣∣∣t− 1
2
∣∣∣∣ f ′′a ( ∥(1− t)T + tV ∥1)dt
≤ 1
24∥V − T∥2
[f ′′a
(∥∥∥∥V + T
2
∥∥∥∥1
)+f ′′a (∥V ∥1) + f ′′a (∥T∥1)
2
]≤ 1
12∥V − T∥2
[f ′′a (∥V ∥1) + f ′′a (∥T∥1)
]≤ 1
6∥V − T∥2max
{f ′′a (∥V ∥1), f
′′a (∥T∥1)
}.
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trace inequalities of lipschitz type 47
Proof. Taking the modulus in (5.1), for T, V ∈ B1(H) with ∥T∥1 , ∥V ∥1 <R we have∣∣∣∣ tr [f(V )] + tr [f(T )]
2− tr
[f
(V + T
2
)]∣∣∣∣ (5.5)
≤ 1
4
∫ 1
0
∣∣∣∣tr((V − T )
[f ′((1− t)
V + T
2+ tV
)− f ′
((1− t)
V + T
2+ tT
)])∣∣∣∣ dt .Using the properties of trace, for T, V ∈ B1(H) with ∥T∥1 , ∥V ∥1 < R and
t ∈ [0, 1] we have∣∣∣∣tr((V − T )
[f ′((1− t)
V + T
2+ tV
)− f ′
((1− t)
V + T
2+ tT
)])∣∣∣∣ (5.6)
≤ ∥V − T∥∥∥∥∥[f ′((1− t)
V + T
2+ tV
)− f ′
((1− t)
V + T
2+ tT
)]∥∥∥∥1
.
From (4.1), for A,B ∈ B1(H) with ∥A∥1 , ∥B∥1 < R we have
∥f(A)− f(B)∥1 ≤ ∥A−B∥∫ 1
0f ′a(∥(1− t)B + tA∥1
)dt (5.7)
≤ 1
2∥A−B∥
[f ′a
(∥∥∥∥A+B
2
∥∥∥∥1
)+f ′a(∥A∥1) + f ′a(∥B∥1)
2
]≤ 1
2∥A−B∥
[f ′a(∥A∥1) + f ′a(∥B∥1)
]≤ ∥A−B∥max
{f ′a(∥A∥1), f
′a(∥B∥1)
}.
Applying the second and third inequalities in (5.7) for f ′, A = (1−t)V+T2 +tV
and B = (1− t)V+T2 + tT we get∥∥∥∥[f ′((1− t)
V + T
2+ tV
)− f ′
((1− t)
V + T
2+ tT
)]∥∥∥∥1
(5.8)
≤ 1
2t ∥V − T∥
[f ′′a
(∥∥∥∥V + T
2
∥∥∥∥1
)
+f ′′a(∥∥(1− t)V+T
2 + tV∥∥1
)+ f ′′a
(∥∥(1− t)V+T2 + tT
∥∥1
)2
]
≤ 1
2t ∥V − T∥
×[f ′′a
(∥∥∥∥(1− t)V + T
2+ tV
∥∥∥∥1
)+ f ′′a
(∥∥∥∥(1− t)V + T
2+ tT
∥∥∥∥1
)]
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48 s.s. dragomir
for T, V ∈ B1(H) with ∥T∥1 , ∥V ∥1 < R and t ∈ [0, 1].Since f ′′a is convex and monotonic nondecreasing and ∥·∥1 is convex, then
f ′′a(∥∥(1− t)V+T
2 + tV∥∥1
)+ f ′′a
(∥∥(1− t)V+T2 + tT
∥∥1
)2
(5.9)
≤ (1− t) f ′′a
(∥∥∥∥V + T
2
∥∥∥∥1
)+ t
f ′′a (∥V ∥1) + f ′′a (∥T∥1)2
for T, V ∈ B1(H) with ∥T∥1 , ∥V ∥1 < R and t ∈ [0, 1].From (5.8) and (5.9) we get∥∥∥∥[f ′((1− t)
V + T
2+ tV
)− f ′
((1− t)
V + T
2+ tT
)]∥∥∥∥1
(5.10)
≤ 1
2t ∥V − T∥
×[f ′′a
(∥∥∥∥(1− t)V + T
2+ tV
∥∥∥∥1
)+ f ′′a
(∥∥∥∥(1− t)V + T
2+ tT
∥∥∥∥1
)]
≤ t ∥V − T∥[(1− t)f ′′a
(∥∥∥∥V + T
2
∥∥∥∥1
)+ t
f ′′a (∥V ∥1) + f ′′a (∥T∥1)2
]for T, V ∈ B1(H) with ∥T∥1 , ∥V ∥1 < R and t ∈ [0, 1].
Integrating (5.10) over t on [0, 1] we get∫ 1
0
∥∥∥∥[f ′((1− t)V + T
2+ tV
)− f ′
((1− t)
V + T
2+ tT
)]∥∥∥∥1
dt
≤ 1
2∥V − T∥ ×
[∫ 1
0tf ′′a
(∥∥∥∥(1− t)V + T
2+ tV
∥∥∥∥1
)dt
+
∫ 1
0tf ′′a
(∥∥∥∥(1− t)V + T
2+ tT
∥∥∥∥1
dt
)]
≤ ∥V − T∥[f ′′a
(∥∥∥∥V + T
2
∥∥∥∥1
)∫ 1
0t(1− t) dt
+f ′′a (∥V ∥1) + f ′′a (∥T∥1)
2
∫ 1
0t2 dt
]=
1
6∥V − T∥
[f ′′a
(∥∥∥∥V + T
2
∥∥∥∥1
)+f ′′a (∥V ∥1) + f ′′a (∥T∥1)
2
],
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trace inequalities of lipschitz type 49
which together with (5.5) and (5.6) produce the inequality∣∣∣∣tr [f(V )] + tr [f(T )]
2− tr
[f
(V + T
2
)]∣∣∣∣ (5.11)
≤ 1
8∥V − T∥2 ×
[∫ 1
0tf ′′a
(∥∥∥∥(1− t)V + T
2+ tV
∥∥∥∥1
)dt
+
∫ 1
0tf ′′a
(∥∥∥∥(1− t)V + T
2+ tT
∥∥∥∥1
dt
)]≤ 1
24∥V − T∥2
[f ′′a
(∥∥∥∥V + T
2
∥∥∥∥1
)+f ′′a (∥V ∥1) + f ′′a (∥T∥1)
2
].
Now, observe that∫ 1
0tf ′′a
(∥∥∥∥(1− t)V + T
2+ tV
∥∥∥∥1
)dt (5.12)
=
∫ 1
0tf ′′a
(∥∥∥∥1− t
2T +
1 + t
2V
∥∥∥∥1
)dt ,
∫ 1
0tf ′′a
(∥∥∥∥(1− t)V + T
2+ tT
∥∥∥∥1
)dt (5.13)
=
∫ 1
0tf ′′a
(∥∥∥∥1− t
2V +
1 + t
2T
∥∥∥∥1
)dt .
Using the change of variable u = 1+t2 , then we get∫ 1
0tf ′′a
(∥∥∥∥1− t
2T +
1 + t
2V
∥∥∥∥1
)dt (5.14)
= 2
∫ 1
12
(2u− 1)f ′′a(∥(1− u)T + uV ∥
)du .
Also, by changing the variable v = 1−t2 , we get∫ 1
0tf ′′a
(∥∥∥∥1− t
2V +
1 + t
2T
∥∥∥∥1
)dt (5.15)
= 2
∫ 12
0(1− 2v)f ′′a
(∥(1− v)T + vV ∥
)dv .
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50 s.s. dragomir
Utilising the equalities (5.12)-(5.15) we obtain∫ 1
0tf ′′a
(∥∥∥∥(1− t)V + T
2+ tV
∥∥∥∥1
)dt+
∫ 1
0tf ′′a
(∥∥∥∥(1− t)V + T
2+ tT
∥∥∥∥1
)dt
= 2
∫ 1
12
(2t− 1)f ′′a(∥(1− t)T + tV ∥
)dt
+ 2
∫ 12
0(1− 2t)f ′′a
(∥(1− t)T + tV ∥
)dt
= 2
∫ 1
0|2t− 1| f ′′a
(∥(1− t)T + tV ∥
)dt
= 4
∫ 1
0
∣∣∣∣t− 1
2
∣∣∣∣ f ′′a ( ∥(1− t)T + tV ∥)dt
for T, V ∈ B1(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. Let f(z) :=∑∞
n=0 anzn 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 )]
2− tr
[f
(V + T
2
)]∣∣∣∣ ≤ 1
8∥V − T∥2 f ′′a (M) . (5.16)
The constant 18 is best possible in (5.16).
Proof. From the first inequality in (5.4) we have∣∣∣∣tr [f(V )] + tr [f(T )]
2− tr
[f
(V + T
2
)]∣∣∣∣≤ 1
2∥V − T∥2
∫ 1
0
∣∣∣∣t− 1
2
∣∣∣∣ f ′′a ( ∥(1− t)T + tV ∥1)dt
≤ 1
2∥V − T∥2 f ′′a (M)
∫ 1
0
∣∣∣∣t− 1
2
∣∣∣∣dt =1
8∥V − T∥2 f ′′a (M)
for T, V ∈ B1(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
a, b ∈ R then we get in both sides of (5.16) the same quantity 14(b− a)2.
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trace inequalities of lipschitz type 51
Remark 3. A similar result holds by swapping the norm ∥·∥ with ∥·∥1 inthe right hand side of (5.4). The case of Hilbert-Schmidt norm may also bestated, 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 )]
2− tr
[exp
(V + T
2
)]∣∣∣∣ (5.17)
≤ 1
2∥V − T∥2
∫ 1
0
∣∣∣∣t− 1
2
∣∣∣∣ exp ( ∥(1− t)T + tV ∥1)dt
≤ 1
24∥V − T∥2
[exp
(∥∥∥∥V + T
2
∥∥∥∥1
)+
exp(∥V ∥1) + exp(∥T∥1)2
]≤ 1
12∥V − T∥2
[exp(∥V ∥1) + exp(∥T∥1)
]≤ 1
6∥V − T∥2max
{exp(∥V ∥1), exp(∥T∥1)
}for any for T, V ∈ B1(H).
If T, V ∈ B1(H) with ∥V ∥1 , ∥T∥1 ≤M , then∣∣∣∣tr [exp(V )] + tr [exp(T )]
2− tr
[exp
(V + T
2
)]∣∣∣∣ (5.18)
≤ 1
8∥V − T∥2 exp(M) .
If we write the inequality (5.4) for the function f(z) = (1− z)−1, then we get∣∣∣∣∣tr[(1H − V )−1
]+ tr
[(1H − T )−1
]2
− tr
[(1H − V + T
2
)−1]∣∣∣∣∣ (5.19)
≤ ∥V − T∥2∫ 1
0
∣∣∣∣t− 1
2
∣∣∣∣ (1− ∥(1− t)T + tV ∥1)−3
dt
≤ 1
12∥V − T∥2 ×
[(1−
∥∥∥∥V + T
2
∥∥∥∥1
)−3
+(1− ∥V ∥1)−3 + (1− ∥T∥1)−3
2
]
≤ 1
6∥V − T∥2
[(1− ∥V ∥1)
−3 + (1− ∥T∥1)−3]
≤ 1
3∥V − T∥2max
{(1− ∥V ∥1)
−3, (1− ∥T∥1)−3},
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52 s.s. dragomir
for any for T, V ∈ B1(H) with ∥V ∥1 , ∥T∥1 < 1.Moreover, if ∥V ∥1 , ∥T∥1 ≤M < 1, then∣∣∣∣∣tr
[(1H − V )−1
]+ tr
[(1H − T )−1
]2
− tr
[(1H − V + T
2
)−1]∣∣∣∣∣ (5.20)
≤ 1
4∥V − T∥2 (1−M)−3.
The interested reader may choose other examples of power series to getsimilar results. However, the details are not presented here.
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