Some integral inequalities for operator monotonic functions on Hilbert spaces
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Dragomir, Sever S (2020) Some integral inequalities for operator monotonic functions on Hilbert spaces. Special Matrices, 8 (1). pp. 172-180. ISSN 2300- 7451
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Research Article Open Access
Silvestru Sever Dragomir*
Some integral inequalities for operator monotonic functions on Hilbert spaces
https://doi.org/10.1515/spma-2020-0108 Received May 6, 2020; accepted June 24, 2020
Abstract:Letf be an operator monotonic function onIandA,B∈ SAI(H) , the class of all selfadjoint op- erators with spectra inI. Assume thatp: [0, 1]→Ris non-decreasing on [0, 1]. In this paper we obtained, among others, that forA≤Bandfan operator monotonic function onI,
0 ≤ Z1
0
p(t)f((1 −t)A+tB)dt− Z1
0
p(t)dt Z1
0
f((1 −t)A+tB)dt
≤1
4[p(1) −p(0)] [f(B) −f(A)]
in the operator order.
Several other similar inequalities for eitherporfis differentiable, are also provided. Applications for power function and logarithm are given as well.
Keywords:Operator monotonic functions, Integral inequalities, Čebyšev inequality,Grüss inequality, Os- trowski inequality
MSC:47A63, 26D15, 26D10.
1 Introduction
Consider a complex Hilbert space (H,h·, ·i). An operatorTis said to be positive (denoted byT≥ 0) ifhTx,xi≥ 0 for all x ∈ H and also an operatorTis said to bestrictly positive(denoted byT > 0) ifT is positive and invertible. A real valued continuous functionf(t) on (0, ∞) is said to be operator monotone iff(A) ≥ f(B) holds for anyA≥B> 0.
In 1934, K. Löwner [7] had given a definitive characterization of operator monotone functions as follows:
Theorem 1. A function f: (0, ∞)→Ris operator monotone in(0, ∞)if and only if it has the representation f(t) =a+bt+
Z∞
0
t
t+sdm(s) where a∈Rand b≥ 0and a positive measure m on(0, ∞)such that
Z∞
0
dm(s) t+s < ∞.
*Corresponding Author: Silvestru Sever Dragomir:Mathematics, College of Engineering & Science, Victoria University, PO Box 14428, Melbourne City, MC 8001, Australia
DST-NRF Centre of Excellence in the Mathematical and Statistical Sciences, School of Computer Science & Applied Mathemat- ics, University of the Witwatersrand, Johannesburg, South Africa
E-mail: [email protected], http://rgmia.org/dragomir
We recall the important fact proved by Löwner and Heinz that states that the power functionf : (0, ∞)→R, f(t) =tαis an operator monotone function for anyα∈[0, 1] .
In [3], T. Furuta observed that forαj∈[0, 1] ,j= 1, ...,nthe functions g(t) :=
n
X
j=1
t−αj
−1
andh(t) =
n
X
j=1
1 +t−1−αj
are operator monotone in (0, ∞).
Letf(t) be a continuous function (0, ∞)→(0, ∞). It is known thatf(t) is operator monotone if and only ifg(t) =t/f(t) =:f*(t) is also operator monotone, see for instance [3] or [8].
Consider the family of functions defined on (0, ∞) andp∈[−1, 2] \{0, 1}by fp(t) := p− 1
p
tp− 1 tp−1− 1
and
f0(t) := t 1 −tlnt, f1(t) := t− 1
lnt (logarithmic mean).
We also have the functions of interest f−1(t) = 2t
1 +t(harmonic mean),f1/2(t) =√t(geometric mean).
In [2] the authors showed thatfpis operator monotone for 1 ≤p≤ 2.
In the same category, we observe that the function gp(t) := t− 1
tp− 1 is an operator monotone function forp∈(0, 1], [3].
It is well known that the logarithmic function ln is operator monotone and in [3] the author obtained that the functions
f(t) =t(1 +t) ln
1 + 1 t
, g(t) = 1 (1 +t) ln 1 +1t
are also operator monotone functions on (0, ∞) .
Letfbe an operator monotonic function on an interval of real numbersIandA,B∈SAI(H) , the class of all selfadjoint operators with spectra inI. Assume thatp: [0, 1]→Ris non-decreasing on [0, 1]. In this paper we obtain, among others, that forA≤Bandf an operator monotonic function onI,
0 ≤ Z1
0
p(t)f((1 −t)A+tB)dt− Z1
0
p(t)dt Z1
0
f((1 −t)A+tB)dt
≤ 1
4[p(1) −p(0)] [f(B) −f(A)]
in the operator order.
Several other similar inequalities for eitherporf is differentiable, are also provided. Applications for power function and logarithm are given as well.
2 Main Results
For twoLebesgue integrablefunctionsh,g: [a,b]→R, consider theČebyšev functional: C(h,g) := 1
b−a
b
Z
a
h(t)g(t)dt− 1 b−a
b
Z
a
h(t)dt 1 b−a
b
Z
a
g(t)dt. (2.1)
It is well known that, ifhandghave the same monotonicity on [a,b] , then 1
b−a
b
Z
a
h(t)g(t)dt≥ 1 b−a
b
Z
a
h(t)dt 1 b−a
b
Z
a
g(t)dt, (2.2)
which is known in the literature asČebyšev’s inequality.
In 1935, Grüss [4] showed that
|C(h,g)|≤1
4(M−m) (N−n) , (2.3)
provided that there exists the real numbersm,M,n,Nsuch that
m≤h(t) ≤M and n≤g(t) ≤N for a.e.t∈[a,b] . (2.4) The constant 14is best possible in (2.1) in the sense that it cannot be replaced by a smaller quantity.
Letfbe a continuous function onI. If (A,B)∈SAI(H) , the class of all selfadjoint operators with spectra inIandt∈[0, 1] , then the convex combination (1 −t)A+tBis a selfadjoint operator with the spectrum inI showing thatSAI(H) is a convex set in the Banach algebraB(H) of all bounded linear operators onH. By the continuous functional calculus of selfadjoint operator we also conclude thatf((1 −t)A+tB) is a selfadjoint operator inB(H) .
ForA,B∈SAI(H) , we consider the auxiliary functionϕ(A,B): [0, 1]→B(H) defined by
ϕ(A,B)(t) :=f((1 −t)A+tB) . (2.5) Forx∈Hwe can also consider the auxiliary functionϕ(A,B);x: [0, 1]→Rdefined by
ϕ(A,B);x(t) :=
ϕ(A,B)(t)x,x
=hf((1 −t)A+tB)x,xi. (2.6) Theorem 2. Let A,B∈ SAI(H)with A ≤B and f an operator monotonic function on I.If p : [0, 1]→Ris monotonic nondecreasing on[0, 1] ,then
0 ≤ Z1
0
p(t)f((1 −t)A+tB)dt− Z1
0
p(t)dt Z1
0
f((1 −t)A+tB)dt (2.7)
≤1
4[p(1) −p(0)] [f(B) −f(A)] . If p: [0, 1]→Ris monotonic nonincreasing on[0, 1] ,then
0 ≤ Z1
0
p(t)dt Z1
0
f((1 −t)A+tB)dt− Z1
0
p(t)f((1 −t)A+tB)dt (2.8)
≤1
4[p(0) −p(1)] [f(B) −f(A)] . Proof. Let 0 ≤t1<t2≤ 1 andA≤B. Then
(1 −t2)A+t2B− (1 −t1)A−t1B= (t2−t1) (B−A) ≥ 0 and by operator monotonicity off we get
f((1 −t2)A+t2B) ≥f((1 −t1)A+t1B) , which is equivalent to
ϕ(A,B);x(t2) =hf((1 −t2)A+t2B)x,xi
≥hf((1 −t1)A+t1B)x,xi=ϕ(A,B);x(t1)
that shows that the scalar functionϕ(A,B);x : [0, 1]→Ris monotonic nondecreasing forA ≤Band for any x∈H.
If we write the inequality (2.2) for the functionspandϕ(A,B);xwe get Z1
0
p(t)hf((1 −t)A+tB)x,xidt≥ Z1
0
p(t)dt Z1
0
hf((1 −t)A+tB)x,xidt, which can be written as
*
Z1
0
p(t)f((1 −t)A+tB)dt
x,x +
≥
*
Z1
0
p(t)dt Z1
0
f((1 −t)A+tB)
dtx,x +
forx∈H, and the first inequality in (2.7) is obtained.
We also have that
hf(A)x,xi = ϕ(A,B);x(0) ≤ϕ(A,B);x(t) =hf((1 −t)A+tB)x,xi
≤ ϕ(A,B);x(1) =hf(B)x,xi and
p(0) ≤p(t) ≤p(1) for allt∈[0, 1] .
By writing Grüss’ inequality for the functionsϕ(A,B);xandp, we get 0 ≤
Z1
0
p(t)hf((1 −t)A+tB)x,xidt− Z1
0
p(t)dt Z1
0
hf((1 −t)A+tB)x,xidt
≤1
4[p(1) −p(0)] [hf(B)x,xi−hf(A)x,xi] forx∈Hand the second inequality in (2.7) is obtained.
A continuous functiong : SAI(H) → B(H) is said to beGâteaux differentiableinA ∈ SAI(H) along the directionB∈B(H) if the following limit exists in the strong topology ofB(H)
∇gA(B) := lims→
0
g(A+sB) −g(A)
s ∈B(H) . (2.9)
If the limit (2.9) exists for allB∈ B(H) , then we say thatgisGâteaux differentiableinAand we can write g∈G(A) . If this is true for anyAin an open setSfromSAI(H) we write thatg∈G(S) .
Ifg is a continuous function onI, by utilising the continuous functional calculus the corresponding function of operators will be denoted in the same way.
For two distinct operatorsA,B∈SAI(H) we consider the segment of selfadjoint operators [A,B] :={(1 −t)A+tB|t∈[0, 1]}.
We observe thatA,B∈[A,B] and [A,B]⊂SAI(H) .
Lemma 1. Let f be a continuous function on I and A, B ∈ SAI(H) ,with A ≠ B.If f ∈ G([A,B]) ,then the auxiliary functionϕ(A,B)is differentiable on(0, 1)and
ϕ0(A,B)(t) =∇f(1−t)A+tB(B−A) . (2.10) In particular,
ϕ0(A,B)(0+) =∇fA(B−A) (2.11)
and
ϕ0(A,B)(1−) =∇fB(B−A) . (2.12)
Proof. Lett∈(0, 1) andh≠ 0 small enough such thatt+h∈(0, 1). Then ϕ(A,B)(t+h) −ϕ(A,B)(t)
h = f((1 −t−h)A+ (t+h)B) −f((1 −t)A+tB)
h (2.13)
= f((1 −t)A+tB+h(B−A)) −f((1 −t)A+tB)
h .
Sincef ∈G([A,B]) , hence by taking the limit overh→0 in (2.13) we get ϕ0(A,B)(t) = lim
h→0
ϕ(A,B)(t+h) −ϕ(A,B)(t) h
= lim
h→0
f((1 −t)A+tB+h(B−A)) −f((1 −t)A+tB) h
= ∇f(1−t)A+tB(B−A) , which proves (2.10).
Also, we have
ϕ0(A,B)(0+) = lim
h→0+
ϕ(A,B)(h) −ϕ(A,B)(0) h
= lim
h→0+
f((1 −h)A+hB) −f(A) h
= lim
h→0+
f(A+h(B−A)) −f(A) h
= ∇fA(B−A)
sincefis assumed to be Gâteaux differentiable inA. This proves (2.11).
The equality (2.12) follows in a similar way.
Lemma 2. Let f be an operator monotonic function on I and A, B ∈ SAI(H) ,with A ≤ B, A ≠ B.If f ∈ G([A,B]) ,then
∇f(1−t)A+tB(B−A) ≥ 0for all t∈(0, 1) . (2.14) Also
∇fA(B−A) , ∇fB(B−A) ≥ 0. (2.15)
Proof. Letx∈H. The auxiliary functionϕ(A,B);xis monotonic nondecreasing in the usual sense on [0, 1] and differentiable on (0, 1) , and fort∈(0, 1)
0 ≤ϕ0(A,B);x(t) = lim
h→0
ϕ(A,B),x(t+h) −ϕ(A,B),x(t) h
= lim
h→0
ϕ
(A,B)(t+h) −ϕ(A,B)(t)
h x,x
=
h→lim0
ϕ(A,B)(t+h) −ϕ(A,B)(t)
h x,x
=
∇f(1−t)A+tB(B−A)x,x . This shows that
∇f(1−t)A+tB(B−A) ≥ 0 for allt∈(0, 1) .
The inequalities (2.15) follow by (2.11) and (2.12).
The following inequality obtained by Ostrowski in 1970, [9] also holds
|C(h,g)|≤ 1
8(b−a) (M−m) g0
∞, (2.16)
provided thathisLebesgue integrableand satisfies (2.4) whilegis absolutely continuous andg0∈L∞[a,b] . The constant 18is best possible in (2.16).
Theorem 3. Let A, B ∈ SAI(H)with A ≤ B, f be an operator monotonic function on I and p : [0, 1]→ R monotonic nondecreasing on[0, 1] .
(i) If p is differentiable on(0, 1) ,then 0 ≤
Z1
0
p(t)f((1 −t)A+tB)dt− Z1
0
p(t)dt Z1
0
f((1 −t)A+tB)dt (2.17)
≤1 8 sup
t∈(0,1)
p0(t) [f(B) −f(A)] . (ii) If f ∈G([A,B]) ,then
0 ≤ Z1
0
p(t)f((1 −t)A+tB)dt− Z1
0
p(t)dt Z1
0
f((1 −t)A+tB)dt (2.18)
≤1
8[p(1) −p(0)] sup
t∈(0,1)
∇f(1−t)A+tB(B−A) 1H. Proof. Letx∈H. If we use the inequality (2.16) forg=pandh=ϕ(A,B);x, then
0 ≤ Z1
0
p(t)hf((1 −t)A+tB)x,xidt− Z1
0
p(t)dt Z1
0
hf((1 −t)A+tB)x,xidt
≤1 8 sup
t∈(0,1)
p0(t) [hf(B)x,xi−hf(A)x,xi] , for anyx∈H, which is equivalent to (2.17).
If we use the inequality (2.16) forh=pandg=ϕ(A,B);xthen by Lemmas 1 and 2
0 ≤ Z1
0
p(t)hf((1 −t)A+tB)x,xidt− Z1
0
p(t)dt Z1
0
hf((1 −t)A+tB)x,xidt (2.19)
≤1
8[p(1) −p(0)] sup
t∈(0,1)
∇f(1−t)A+tB(B−A)x,x , for anyx∈H, which is an inequality of interest in itself.
Observe that for allt∈(0, 1) ,
∇f(1−t)A+tB(B−A)x,x
≤
∇f(1−t)A+tB(B−A) kxk2 for anyx∈H, which implies that
sup
t∈(0,1)
∇f(1−t)A+tB(B−A)x,x
≤ sup
t∈(0,1)
∇f(1−t)A+tB(B−A)
h1Hx,xi (2.20) for anyx∈H.
By making use of (2.19) and (2.20) we derive 0 ≤
Z1
0
p(t)hf((1 −t)A+tB)x,xidt− Z1
0
p(t)dt Z1
0
hf((1 −t)A+tB)x,xidt
≤1
8[p(1) −p(0)] sup
t∈(0,1)
∇f(1−t)A+tB(B−A)
h1Hx,xi for anyx∈H, which is equivalent to (2.18).
Another, however less known result, even though it was obtained by Čebyšev in 1882, [1], states that
|C(h,g)|≤ 1 12
h0 ∞
g0
∞(b−a)2, (2.21)
provided thath0,g0exist and are continuous on [a,b] and h0
∞= supt∈[a,b]
h0(t)
. The constant121 cannot be improved in the general case.
The case ofeuclidean normsof the derivative was considered by A. Lupaş in [5] in which he proved that
|C(h,g)|≤ 1 π2
h0 2
g0
2(b−a) , (2.22)
provided thath,gare absolutely continuous andh0,g0∈L2[a,b] . The constantπ12 is the best possible.
Using the above inequalities (2.21) and (2.22) and a similar procedure to the one employed in the proof of Theorem 3, we can also state the following result:
Theorem 4. Let A, B ∈ SAI(H)with A ≤ B, f be an operator monotonic function on I and p : [0, 1]→ R monotonic nondecreasing on[0, 1] .If p is differentiable and f ∈G([A,B]) ,then
0 ≤ Z1
0
p(t)f((1 −t)A+tB)dt− Z1
0
p(t)dt Z1
0
f((1 −t)A+tB)dt (2.23)
≤ 1 12 sup
t∈(0,1)
p0(t) sup
t∈(0,1)
∇f(1−t)A+tB(B−A) 1H
and
0 ≤ Z1
0
p(t)f((1 −t)A+tB)dt− Z1
0
p(t)dt Z1
0
f((1 −t)A+tB)dt (2.24)
≤ 1 π2
Z1
0
p0(t)2 dt
1/2
Z1
0
∇f(1−t)A+tB(B−A) 2dt
1/2
1H, provided the integrals in the second term are finite.
3 Some Examples
We consider the functionf : (0, ∞)→R,f(t) = −t−1which isoperator monotoneon (0, ∞) . If 0 <A≤Bandp: [0, 1]→Ris monotonic nondecreasing on [0, 1] , then by (2.7)
0 ≤ Z1
0
p(t)dt Z1
0
((1 −t)A+tB)−1dt− Z1
0
p(t) ((1 −t)A+tB)−1dt (3.1)
≤ 1
4[p(1) −p(0)]
A−1−B−1 .
Moreover, ifpis differentiable on (0, 1) , then by (2.17) we obtain 0 ≤
Z1
0
p(t)dt Z1
0
((1 −t)A+tB)−1dt− Z1
0
p(t) ((1 −t)A+tB)−1dt (3.2)
≤ 1 8 sup
t∈(0,1)
p0(t)
A−1−B−1 .
The functionf(t) = −t−1is operator monotonic on (0, ∞), operator Gâteaux differentiable and
∇fT(S) =T−1ST−1
forT,S> 0.
Ifp: [0, 1]→Ris monotonic nondecreasing on [0, 1] , then by (2.18) we get 0 ≤
Z1
0
p(t)dt Z1
0
((1 −t)A+tB)−1dt− Z1
0
p(t) ((1 −t)A+tB)−1dt (3.3)
≤1
8[p(1) −p(0)] sup
t∈(0,1)
((1 −t)A+tB)−1(B−A) ((1 −t)A+tB)−1 1H
for 0 <A≤B.
Ifpis monotonic nondecreasing and differentiable on (0, 1) , then by (2.23) and (2.24) we get 0 ≤
Z1
0
p(t)dt Z1
0
((1 −t)A+tB)−1dt− Z1
0
p(t) ((1 −t)A+tB)−1dt (3.4)
≤ 1 12 sup
t∈(0,1)
p0(t) sup
t∈(0,1)
((1 −t)A+tB)−1(B−A) ((1 −t)A+tB)−1 1H
and
0 ≤ Z1
0
p(t)dt Z1
0
((1 −t)A+tB)−1dt− Z1
0
p(t) ((1 −t)A+tB)−1dt (3.5)
≤ 1 π2
Z1
0
p0(t)2
dt
1/2
Z1
0
((1 −t)A+tB)−1(B−A) ((1 −t)A+tB)−1
2dt
1/2
1H, for 0 <A≤B.
We note that the functionf(t) = lntis operator monotonic on (0, ∞) .
If 0 <A≤Bandp: [0, 1]→Ris monotonic nondecreasing on [0, 1] , then by (2.7) we have 0 ≤
Z1
0
p(t) ln ((1 −t)A+tB)dt− Z1
0
p(t)dt Z1
0
ln ((1 −t)A+tB)dt (3.6)
≤ 1
4[p(1) −p(0)] (lnB− lnA) .
Moreover, ifpis differentiable on (0, 1) , then by (2.17) we obtain 0 ≤
Z1
0
p(t) ln ((1 −t)A+tB)dt− Z1
0
p(t)dt Z1
0
ln ((1 −t)A+tB)dt (3.7)
≤ 1 8 sup
t∈(0,1)
p0(t) (lnB− lnA) .
The ln function is operator Gâteaux differentiable with the following explicit formula for the derivative (cf. Pedersen [10, p. 155]):
∇lnT(S) = Z∞
0
(s1H+T)−1S(s1H+T)−1ds (3.8) forT,S> 0.
Ifp: [0, 1]→Ris monotonic nondecreasing on [0, 1] , then by (2.18) we get 0 ≤
Z1
0
p(t) ln ((1 −t)A+tB)dt− Z1
0
p(t)dt Z1
0
ln ((1 −t)A+tB)dt (3.9)
≤ 1
8[p(1) −p(0)] sup
t∈(0,1)
Z∞
0
(s1H+ (1 −t)A+tB)−1(B−A) (s1H+ (1 −t)A+tB)−1ds
1H
and ifpis differentiable on (0, 1) , then
0 ≤ Z1
0
p(t) ln ((1 −t)A+tB)dt− Z1
0
p(t)dt Z1
0
ln ((1 −t)A+tB)dt (3.10)
≤ 1 12 sup
t∈(0,1)
p0(t) sup
t∈(0,1)
Z∞
0
(s1H+ (1 −t)A+tB)−1(B−A) (s1H+ (1 −t)A+tB)−1ds
1H
for 0 <A≤B.
Acknowledgement.The author would like to thank the anonymous referee for valuable suggestions that have been implemented in the final version of the paper.
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