Filomat 38:5 (2024), 1499–1511
https://doi.org/10.2298/FIL2405499D

Published by Faculty of Sciences and Mathematics,
University of Niš, Serbia
Available at: http://www.pmf.ni.ac.rs/filomat

Some properties of tensorial perspective for convex functions of
selfadjoint operators in Hilbert spaces

Silvestru Sever Dragomira,b

aMathematics, College of Sport, Health and Engineering, Victoria University, PO Box 14428, Melbourne City, MC 8001, Australia
bDST-NRF Centre of Excellence in the Mathematical and Statistical Sciences, School of Computer Science and Applied Mathematics, University

of the Witwatersrand, Johannesburg, Private Bag 3, Wits 2050, South Africa

Abstract.
Let H be a Hilbert space. Assume that f : [0,∞)→ R is continuous and A, B > 0.We define the tensorial

perspective for the function f and the pair of operators (A,B) by

P f ,⊗ (A,B) := (1 ⊗ B) f
(
A ⊗ B−1

)
.

In this paper we show among others that, if f is differentiable convex, then

P f ,⊗ (A,B) ≥
[

f (u) − f ′ (u) u
]

(1 ⊗ B) + f ′ (u) (A ⊗ 1) ,

for A, B > 0 and u > 0. Moreover, if Sp (A) ⊂ I, Sp (B) ⊂ J and such that 0 < γ ≤ t
s ≤ Γ for t ∈ I and s ∈ J, then

P f ,⊗ (A,B) ≤
[

f (u) − f ′ (u) u
]

(1 ⊗ B) + f ′ (u) (A ⊗ 1) +
[

f ′
−

(Γ) − f ′+
(
γ
)]
|A ⊗ 1 − u (1 ⊗ B)|

for u ∈
[
γ,Γ

]
.

1. Introduction

Let I1, ..., Ik be intervals fromR and let f : I1× ...× Ik → R be an essentially bounded real function defined
on the product of the intervals. Let A = (A1, ...,An) be a k-tuple of bounded selfadjoint operators on Hilbert
spaces H1, ...,Hk such that the spectrum of Ai is contained in Ii for i = 1, ..., k.We say that such a k-tuple is in
the domain of f . If

Ai =

∫
Ii

λidEi (λi)

is the spectral resolution of Ai for i = 1, ..., k; by following [1], we define

f (A1, ...,Ak) :=
∫

I1

...

∫
Ik

f (λ1, ..., λk) dE1 (λ1) ⊗ ... ⊗ dEk (λk) (1)

2020 Mathematics Subject Classification. Primary 47A63; Secondary 47A99.
Keywords. Tensorial product; Selfadjoint operators; Convex functions; Perspectives.
Received: 11 July 2023; Accepted: 17 August 2023
Communicated by Dragan S. Djordjević
Email address: sever.dragomir@vu.edu.au (Silvestru Sever Dragomir)



S. S. Dragomir / Filomat 38:5 (2024), 1499–1511 1500

as a bounded selfadjoint operator on the tensorial product H1 ⊗ ... ⊗Hk.
If the Hilbert spaces are of finite dimension, then the above integrals become finite sums, and we may

consider the functional calculus for arbitrary real functions. This construction [1] extends the definition of
Korányi [3] for functions of two variables and have the property that

f (A1, ...,Ak) = f1(A1) ⊗ ... ⊗ fk(Ak),

whenever f can be separated as a product f (t1, ..., tk) = f1(t1)... fk(tk) of k functions each depending on only
one variable.

It is know that, if f is super-multiplicative (sub-multiplicative) on [0,∞), namely

f (st) ≥ (≤) f (s) f (t) for all s, t ∈ [0,∞)

and if f is continuous on [0,∞) , then [5, p. 173]

f (A ⊗ B) ≥ (≤) f (A) ⊗ f (B) for all A, B ≥ 0. (2)

This follows by observing that, if

A =
∫

[0,∞)
tdE (t) and B =

∫
[0,∞)

sdF (s)

are the spectral resolutions of A and B, then

f (A ⊗ B) =
∫

[0,∞)

∫
[0,∞)

f (st) dE (t) ⊗ dF (s) (3)

for the continuous function f on [0,∞) .
Recall the geometric operator mean for the positive operators A, B > 0

A#tB := A1/2(A−1/2BA−1/2)tA1/2,

where t ∈ [0, 1] and

A#B := A1/2(A−1/2BA−1/2)1/2A1/2.

By the definitions of # and ⊗we have

A#B = B#A and (A#B) ⊗ (B#A) = (A ⊗ B) # (B ⊗ A) .

In 2007, S. Wada [7] obtained the following Callebaut type inequalities for tensorial product

(A#B) ⊗ (A#B) ≤
1
2

[(A#αB) ⊗ (A#1−αB) + (A#1−αB) ⊗ (A#αB)] (4)

≤
1
2

(A ⊗ B + B ⊗ A)

for A, B > 0 and α ∈ [0, 1] .
Assume that f : [0,∞)→ R is continuous and A, B > 0.We define the tensorial perspective for the function

f and the pair of operators (A,B)

P f ,⊗ (A,B) := (1 ⊗ B) f
(
A ⊗ B−1

)
.

Motivated by the above results, in this paper we show among others that, if f is differentiable convex,
then

P f ,⊗ (A,B) ≥
[

f (u) − f ′ (u) u
]

(1 ⊗ B) + f ′ (u) (A ⊗ 1) ,

for A, B > 0 and u > 0. Moreover, if Sp (A) ⊂ I, Sp (B) ⊂ J and such that 0 < γ ≤ t
s ≤ Γ for t ∈ I and s ∈ J, then

P f ,⊗ (A,B) ≤
[

f (u) − f ′ (u) u
]

(1 ⊗ B) + f ′ (u) (A ⊗ 1)
+

[
f ′− (Γ) − f ′+

(
γ
)]
|A ⊗ 1 − u (1 ⊗ B)|

for u ∈
[
γ,Γ

]
.



S. S. Dragomir / Filomat 38:5 (2024), 1499–1511 1501

2. Some Preliminary Facts

Recall the following property of the tensorial product

(AC) ⊗ (BD) = (A ⊗ B) (C ⊗D) (5)

that holds for any A,B,C,D ∈ B (H) .
If we take C = A and D = B, then we get

A2
⊗ B2 = (A ⊗ B)2 .

By induction and using (5) we derive that

An
⊗ Bn = (A ⊗ B)n for natural n ≥ 0. (6)

In particular

An
⊗ 1 = (A ⊗ 1)n and 1 ⊗ Bn = (1 ⊗ B)n (7)

for all n ≥ 0.
We also observe that, by (5), the operators A ⊗ 1 and 1 ⊗ B are commutative and

(A ⊗ 1) (1 ⊗ B) = (1 ⊗ B) (A ⊗ 1) = A ⊗ B. (8)

Moreover, for two natural numbers m, n we have

(A ⊗ 1)m (1 ⊗ B)n = (1 ⊗ B)n (A ⊗ 1)m = Am
⊗ Bn. (9)

According with the properties of tensorial products and functional calculus for continuous functions of
selfadjoint operators, we have

P f ,⊗ (A,B) = (1 ⊗ B) f
(
(A ⊗ 1) (1 ⊗ B)−1

)
= f

(
(A ⊗ 1) (1 ⊗ B)−1

)
(1 ⊗ B)

= f
(
(1 ⊗ B)−1 (A ⊗ 1)

)
(1 ⊗ B) ,

due to the commutativity of A ⊗ 1 and 1 ⊗ B.
In the following, we consider the spectral resolutions of A and B given by

A =
∫

[0,∞)
tdE (t) and B =

∫
[0,∞)

sdF (s) . (10)

We have the following representation result for continuous functions:

Lemma 2.1. Assume that f : [0,∞)→ R is continuous and A, B > 0, then

P f ,⊗ (A,B) =
∫

[0,∞)

∫
[0,∞)

s f
( t

s

)
dE (t) ⊗ dF (s) . (11)

Proof. By Stone-Weierstrass theorem, any continuous function can be approximated by a sequence of
polynomials, therefore it suffices to prove the equality for the power function φ (t) = tn with n any natural
number.



S. S. Dragomir / Filomat 38:5 (2024), 1499–1511 1502

We have that∫
[0,∞)

∫
[0,∞)

sφ
( t

s

)
dE (t) ⊗ dF (s)

=

∫
[0,∞)

∫
[0,∞)

s
( t

s

)n
dE (t) ⊗ dF (s)

=

∫
[0,∞)

∫
[0,∞)

tns1−ndE (t) ⊗ dF (s)

= An
⊗ B1−n = An

⊗ BB−n = (1 ⊗ B)
(
An
⊗ B−n)

= (1 ⊗ B)
(
A ⊗ B−1

)n
= Pφ,⊗ (A,B) ,

which shows that (11) holds for the power function.
This proves the lemma.

We assume in the following that A, B > 0.
If we consider the function Πr (u) = ur

− 1, u ≥ 0, r > 0, then we have

PΠr,⊗ (A,B) := (1 ⊗ B)Πr

(
A ⊗ B−1

)
= (1 ⊗ B)

[(
A ⊗ B−1

)r
− 1

]
= (A ⊗ 1)r (1 ⊗ B)1−r

− 1 ⊗ B.

If we take f = − ln (·) , then we get

P− ln(·),⊗ (A,B) := − (1 ⊗ B) ln
(
A ⊗ B−1

)
= − ln

(
(1 ⊗ B)−1 (A ⊗ 1)

)
(1 ⊗ B)

= (1 ⊗ B) [ln (1 ⊗ B) − ln (A ⊗ 1)] .

If we take f = (·) ln (·) , then we get

P(·) ln(·),⊗ (A,B) := (1 ⊗ B)
(
A ⊗ B−1

)
ln

(
A ⊗ B−1

)
= (A ⊗ 1) [ln (A ⊗ 1) − ln (1 ⊗ B)] .

If we take f = |· − α| , α ∈ R, then

P|·−α|,⊗ (A,B) =
∫

[0,∞)

∫
[0,∞)

s
∣∣∣∣∣ ts − α

∣∣∣∣∣ dE (t) ⊗ dF (s)

=

∫
[0,∞)

∫
[0,∞)
|t − αs| dE (t) ⊗ dF (s)

= |A ⊗ 1 − α1 ⊗ B| ,

where for the last equality we used the result obtained in [2],

ψ (h (A) ⊗ 1 + 1 ⊗ k (B)) =
∫

I

∫
J
ψ (h (t) + k (s)) dE (t) ⊗ dF (s) , (12)

here A and B are selfadjoint operators with Sp (A) ⊂ I and Sp (B) ⊂ J, h is continuous on I, k is continuous
on J and ψ is continuous on an interval U that contains the sum of the intervals h (I) + k (J) , while A and B
have the spectral resolutions

A =
∫

I
tdE (t) and B =

∫
J
sdF (s) .



S. S. Dragomir / Filomat 38:5 (2024), 1499–1511 1503

For f = |· − 1|we get

P|·−1|,⊗ (A,B) = |A ⊗ 1 − 1 ⊗ B| .

Consider the q-logarithm defined by

lnq u =


u1−q
−1

1−q if q , 1,

ln u if q = 1.

For q , 1 we define

Plnq,⊗ (A,B) := (1 ⊗ B) lnq

(
(A ⊗ 1) (1 ⊗ B)−1

)
(13)

=
(A ⊗ 1)1−q (1 ⊗ B)q

− 1 ⊗ B
1 − q

.

Let f be a continuous function defined on the interval I of real numbers, B a selfadjoint operator on the
Hilbert space H and A a positive invertible operator on H. Assume that the spectrum Sp

(
A−1/2BA−1/2

)
⊂ I̊.

Then by using the continuous functional calculus, we can define the perspective P f (B,A) by setting

P f (B,A) := A1/2 f
(
A−1/2BA−1/2

)
A1/2.

If A and B are commutative, then

P f (B,A) = A f
(
BA−1

)
provided Sp

(
BA−1

)
⊂ I̊.

It is well known that (see for instance [4]), if f is an operator convex function defined in the positive
half-line, then the mapping

(B,A) 7→ P f (B,A)

defined in pairs of positive definite operators, is operator convex.
The following inequality is also of interest, see [6]:

Theorem 2.2. Assume that f is nonnegative and operator monotone on [0,∞). If A ≥ C > 0 and B ≥ D > 0, then

P f (A,B) ≥ P f (C,D) . (14)

We can state the following result for the tensorial perspective:

Theorem 2.3. If f is an operator convex function defined in the positive half-line, then P f ,⊗ is operator convex in
pairs of positive definite operators as well. If A ≥ C > 0 and B ≥ D > 0, then also

P f ,⊗ (A,B) ≥ P f ,⊗ (C,D) . (15)

Proof. Assume f is an operator convex function in the positive half-line. Since A⊗1 and 1⊗B are commutative,
hence

P f ,⊗ (A,B) = (1 ⊗ B) f
(
(A ⊗ 1) (1 ⊗ B)−1

)
= P f (A ⊗ 1, 1 ⊗ B) (16)

for A, B > 0.



S. S. Dragomir / Filomat 38:5 (2024), 1499–1511 1504

If A, B, C, D > 0 and λ ∈ [0, 1] , then we have

P f ,⊗ ((1 − λ) (A,B) + λ (C,D))
= P f ,⊗ (((1 − λ) A + λC, (1 − λ) B + λD))
= P f (((1 − λ) A + λC) ⊗ 1, 1 ⊗ ((1 − λ) B + λD))
= P f ((1 − λ) A ⊗ 1 + λC ⊗ 1, (1 − λ) 1 ⊗ B + λ1 ⊗D)
= P f ((1 − λ) (A ⊗ 1, 1 ⊗ B) + λ (C ⊗ 1, 1 ⊗D))
≤ (1 − λ)P f (A ⊗ 1, 1 ⊗ B) + λP f (C ⊗ 1, 1 ⊗D)
= (1 − λ)P f ,⊗ (A,B) + λP f ,⊗ (C,D) ,

which shows that P f ,⊗ is operator convex in pairs of positive definite operators.
If A ≥ C > 0 and B ≥ D > 0, then A ⊗ 1 ≥ C ⊗ 1 > 0 and 1 ⊗ B ≥ 1 ⊗D > 0. By utilizing Theorem 2.2 we

derive that

P f (A ⊗ 1, 1 ⊗ B) ≥ P f (C ⊗ 1, 1 ⊗D) .

By utilizing the representation (16) we derive the desired result (15).

3. Main Results

Suppose that I is an interval of real numbers with interior I̊ and f : I → R is a convex function on I.
Then f is continuous on I̊ and has finite left and right derivatives at each point of I̊. Moreover, if x, y ∈ I̊ and
x < y, then f ′

−
(x) ≤ f ′+ (x) ≤ f ′

−

(
y
)
≤ f ′+

(
y
)

which shows that both f ′
−

and f ′+ are nondecreasing function on
I̊. It is also known that a convex function must be differentiable except for at most countably many points.

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

(
I̊
)
⊂ R and

f (x) ≥ f (a) + (x − a)φ (a) for any x, a ∈ I. (17)

It is also well known that if f is convex on I, then ∂ f is nonempty, f ′
−

, f ′+ ∈ ∂ f and if φ ∈ ∂ f , then

f ′− (x) ≤ φ (x) ≤ f ′+ (x) for any x ∈ I̊.

In particular, φ is a nondecreasing function.
If f is differentiable and convex on I̊, then ∂ f =

{
f ′
}
.

Theorem 3.1. Assume that f is convex on (0,∞), A, B > 0 and u ∈ (0,∞) while φ ∈ ∂ f , then

P f ,⊗ (A,B) ≥
[

f (u) − φ (u) u
]

(1 ⊗ B) + φ (u) (A ⊗ 1) . (18)

Moreover, if f is differentiable, then

P f ,⊗ (A,B) ≥
[

f (u) − f ′ (u) u
]

(1 ⊗ B) + f ′ (u) (A ⊗ 1) , (19)

for all A, B > 0 and u ∈ (0,∞) .

Proof. By the gradient inequality we have

f (x) ≥ f (u) + (x − u)φ (u) (20)

for all x, u ∈ (0,∞) .



S. S. Dragomir / Filomat 38:5 (2024), 1499–1511 1505

If we take x = t
s in (20), then we get

f
( t

s

)
≥ f (u) +

( t
s
− u

)
φ (u) (21)

for all t, s > 0.
If we multiply (21) by s > 0, then we get

s f
( t

s

)
≥ s f (u) + φ (u) (t − us) (22)

for all t, s > 0.
We consider the spectral resolutions of A and B given by

A =
∫

[0,∞)
tdE (t) and B =

∫
[0,∞)

sdF (s) .

If we take in (22) the integral
∫

[0,∞)

∫
[0,∞) over dE (t) ⊗ dF (s) , then we get∫

[0,∞)

∫
[0,∞)

s f
( t

s

)
dE (t) ⊗ dF (s)

≥

∫
[0,∞)

∫
[0,∞)

[
s f (u) + φ (u) (t − us)

]
dE (t) ⊗ dF (s)

= f (u)
∫

[0,∞)

∫
[0,∞)

sdE (t) ⊗ dF (s)

+ φ (u)
[∫

[0,∞)

∫
[0,∞)

tdE (t) ⊗ dF (s) − u
∫

[0,∞)

∫
[0,∞)

sdE (t) ⊗ dF (s)
]

= f (u) (1 ⊗ B) + φ (u) (A ⊗ 1 − u1 ⊗ B)

and by the representation (11) we get the desired inequality (18).

Corollary 3.2. With the assumptions of Theorem 3.1 and for x, y ∈ H with ∥x∥ =
∥∥∥y

∥∥∥ = 1, we have〈
P f ,⊗ (A,B)

(
x ⊗ y

)
, x ⊗ y

〉
≥

[
f (u) − φ (u) u

] 〈
By, y

〉
+ φ (u) ⟨Ax, x⟩ , (23)

for all u > 0.
If f is differentiable, then〈
P f ,⊗ (A,B)

(
x ⊗ y

)
, x ⊗ y

〉
≥

[
f (u) − f ′ (u) u

] 〈
By, y

〉
+ f ′ (u) ⟨Ax, x⟩ . (24)

In particular, if we take u = ⟨Ax,x⟩
⟨By,y⟩

in (23) then we get the Jensen’s type inequality of interest〈
P f ,⊗ (A,B)

(
x ⊗ y

)
, x ⊗ y

〉〈
By, y

〉 ≥ f
(
⟨Ax, x⟩〈
By, y

〉) . (25)

Proof. If we take the tensorial inner product over x ⊗ y in (18), then we get〈
P f ,⊗ (A,B)

(
x ⊗ y

)
, x ⊗ y

〉
(26)

≥ f (u)
〈
(1 ⊗ B)

(
x ⊗ y

)
, x ⊗ y

〉
+ φ (u)

〈
(A ⊗ 1 − u1 ⊗ B)

(
x ⊗ y

)
, x ⊗ y

〉
= f (u)

〈
(1 ⊗ B)

(
x ⊗ y

)
, x ⊗ y

〉
+ φ (u)

[〈
(A ⊗ 1)

(
x ⊗ y

)
, x ⊗ y

〉
− u

〈
1 ⊗ B

(
x ⊗ y

)
, x ⊗ y

〉]
.



S. S. Dragomir / Filomat 38:5 (2024), 1499–1511 1506

Observe that for x, y ∈ H with ∥x∥ =
∥∥∥y

∥∥∥ = 1, we have〈
(1 ⊗ B)

(
x ⊗ y

)
, x ⊗ y

〉
=

〈(
1x ⊗ By

)
, x ⊗ y

〉
= ⟨1x, x⟩

〈
By, y

〉
= ∥x∥2

〈
By, y

〉
=

〈
By, y

〉
and 〈

(A ⊗ 1)
(
x ⊗ y

)
, x ⊗ y

〉
=

〈
Ax ⊗ 1y, x ⊗ y

〉
= ⟨Ax, x⟩

〈
1y, y

〉
= ⟨Ax, x⟩

∥∥∥y
∥∥∥2
= ⟨Ax, x⟩

and by (26) we deduce (23).
If we take u = ⟨Ax,x⟩

⟨By,y⟩
in (23), then we get〈

P f ,⊗ (A,B)
(
x ⊗ y

)
, x ⊗ y

〉
≥

[
f
(
⟨Ax, x⟩〈
By, y

〉) − φ (
⟨Ax, x⟩〈
By, y

〉) ⟨Ax, x⟩〈
By, y

〉] 〈By, y
〉

+ φ

(
⟨Ax, x⟩〈
By, y

〉) ⟨Ax, x⟩

= f
(
⟨Ax, x⟩〈
By, y

〉) 〈By, y
〉
,

which gives (25).

Corollary 3.3. Assume that f is convex on (0,∞), 0 < m ≤ A, B ≤M for some constants m, M and φ ∈ ∂ f , then

P f ,⊗ (A,B) ≥
[

f
(m +M

2

)
− φ

(m +M
2

) m +M
2

]
(1 ⊗ B) (27)

+ φ
(m +M

2

)
(A ⊗ 1)

and, if f is differentiable,

P f ,⊗ (A,B) ≥
[

f
(m +M

2

)
− f ′

(m +M
2

) m +M
2

]
(1 ⊗ B) (28)

+ f ′
(m +M

2

)
(A ⊗ 1) .

Also

P f ,⊗ (A,B) ≥
(

f (M) − f (m)
M −m

)
(A ⊗ 1) (29)

+

(
2

M −m

∫ M

m
f (u) du −

M f (M) −m f (m)
M −m

)
(1 ⊗ B) .

Proof. If we take the integral mean in (18), then we get

P f ,⊗ (A,B) ≥
(

1
M −m

∫ M

m
f (u) du

)
(1 ⊗ B) (30)

+

(
1

M −m

∫ M

m
φ (u) du

)
(A ⊗ 1)

−

(
1

M −m

∫ M

m
φ (u) udu

)
(1 ⊗ B) .



S. S. Dragomir / Filomat 38:5 (2024), 1499–1511 1507

Observe that, since φ ∈ ∂Φ, hence

1
M −m

∫ M

m
φ (u) du =

f (M) − f (m)
M −m

and

1
M −m

∫ M

m
uφ (u) du =

1
M −m

[
u f (u)

∣∣∣M
m −

∫ M

m
f (u) du

]
=

M f (M) −m f (m)
M −m

−
1

M −m

∫ M

m
f (u) du.

Therefore(
1

M −m

∫ M

m
f (u) du

)
(1 ⊗ B) +

(
1

M −m

∫ M

m
φ (u) du

)
(A ⊗ 1)

−

(
1

M −m

∫ M

m
φ (u) udu

)
(1 ⊗ B) .

=

(
f (M) − f (m)

M −m

)
(A ⊗ 1)

+

(
2

M −m

∫ M

m
f (u) du −

M f (M) −m f (m)
M −m

)
(1 ⊗ B)

and by (30) we obtain (29).

Theorem 3.4. Assume that f is continuously differentiable convex on (0,∞), A, B > 0 and u ∈ (0,∞) , then

P f ,⊗ (A,B) ≤ f (u) (1 ⊗ B) +P†f ′ ,⊗ (A,B) − uP f ′ ,⊗ (A,B) , (31)

where for a continuous function 1 on (0,∞) ,

P
†

1,⊗ (A,B) :=
∫

[0,∞)

∫
[0,∞)

t1
( t

s

)
dE (t) ⊗ dF (s) (32)

= (A ⊗ 1) 1
(
A ⊗ B−1

)
= (A ⊗ 1) 1

(
(A ⊗ 1) (1 ⊗ B)−1

)
.

Proof. By the gradient inequality we have

f (x) ≤ f (u) + (x − u) f ′ (x) (33)

for all x, u ∈ (0,∞) .
If we take x = t

s in (33) and multiply with s, then we get

s f
( t

s

)
≤ s f (u) + t f ′

( t
s

)
− us f ′

( t
s

)
(34)

for all t, s ∈ (0,∞) .
We consider the spectral resolutions of A and B given by

A =
∫

[0,∞)
tdE (t) and B =

∫
[0,∞)

sdF (s) .



S. S. Dragomir / Filomat 38:5 (2024), 1499–1511 1508

If we take in (34) the integral
∫

[0,∞)

∫
[0,∞) over dE (t) ⊗ dF (s) , then we get∫

[0,∞)

∫
[0,∞)

s f
( t

s

)
dE (t) ⊗ dF (s) (35)

≤ f (u)
∫

[0,∞)

∫
[0,∞)

sdE (t) ⊗ dF (s)

+

∫
[0,∞)

∫
[0,∞)

t f ′
( t

s

)
dE (t) ⊗ dF (s)

− u
∫

[0,∞)

∫
[0,∞)

s f ′
( t

s

)
dE (t) ⊗ dF (s) ,

which gives the desired inequality (31).

Corollary 3.5. With the assumptions of Theorem 3.4 and for x, y ∈ H with ∥x∥ =
∥∥∥y

∥∥∥ = 1, we have〈
P f ,⊗ (A,B)

(
x ⊗ y

)
, x ⊗ y

〉
(36)

≤ f (u)
〈
By, y

〉
+

〈
P
†

f ′ ,⊗ (A,B)
(
x ⊗ y

)
, x ⊗ y

〉
− u

〈
P f ′ ,⊗ (A,B)

(
x ⊗ y

)
, x ⊗ y

〉
,

for all u > 0.
In particular, if we take u = ⟨Ax,x⟩

⟨By,y⟩
in (23) then we get the Jensen’s type inequality of interest

0 ≤

〈
P f ,⊗ (A,B)

(
x ⊗ y

)
, x ⊗ y

〉〈
By, y

〉 − f
(
⟨Ax, x⟩〈
By, y

〉) (37)

≤

〈
P
†

f ′ ,⊗ (A,B)
(
x ⊗ y

)
, x ⊗ y

〉
〈
By, y

〉
−
⟨Ax, x⟩〈
By, y

〉2

〈
P f ′ ,⊗ (A,B)

(
x ⊗ y

)
, x ⊗ y

〉
.

Corollary 3.6. With the assumptions of Theorem 3.4 and if 0 < m ≤ A, B ≤M for some constants m,M, then

P f ,⊗ (A,B) ≤ f
(m +M

2

)
(1 ⊗ B) +P†f ′ ,⊗ (A,B) (38)

−
m +M

2
P f ′ ,⊗ (A,B)

and

P f ,⊗ (A,B) ≤
(

1
M −m

∫ M

m
f (u) du

)
(1 ⊗ B) (39)

+P†f ′ ,⊗ (A,B) −
m +M

2
P f ′ ,⊗ (A,B) .

We also have:

Theorem 3.7. Assume that f is convex on (0,∞), A, B > 0 with spectra Sp (A) ⊂ I, Sp (B) ⊂ J and such that
0 < γ ≤ t

s ≤ Γ for t ∈ I and s ∈ J, then

P f ,⊗ (A,B) ≤
[

f (u) − uφ (u)
]

(1 ⊗ B) + φ (u) (A ⊗ 1) (40)
+

[
f ′− (Γ) − f ′+

(
γ
)]
|A ⊗ 1 − u (1 ⊗ B)|

for u ∈
[
γ,Γ

]
and φ ∈ ∂ f .



S. S. Dragomir / Filomat 38:5 (2024), 1499–1511 1509

Proof. Observe that, by the gradient inequality we have

f (x) ≤ f (u) + (x − u)φ (x) (41)
= f (u) + (x − u)φ (u) + (x − u)

[
φ (x) − φ (u)

]
for x, u > 0 and φ ∈ ∂ f .

Since φ is monotonic nondrecreasing, then

0 ≤
(

f ′ (x) − f ′ (u)
)

(x − u) =
∣∣∣( f ′ (x) − f ′ (u)

)
(x − u)

∣∣∣
=

∣∣∣ f ′ (x) − f ′ (u)
∣∣∣ |x − u| ≤

[
f ′− (Γ) − f ′+

(
γ
)]
|x − u| ,

for x, u ∈
[
γ,Γ

]
and by (41)

f (x) ≤ f (u) + (x − u)φ (u) +
[

f ′− (Γ) − f ′+
(
γ
)]
|x − u| (42)

for x, u ∈
[
γ,Γ

]
.

If we take in (42) x = t
s and multiply with s, then we get

s f
( t

s

)
≤ s f (u) + (t − us)φ (u) +

[
f ′− (Γ) − f ′+

(
γ
)]
|t − us| (43)

for t, s > 0 with t
s , u ∈

[
γ,Γ

]
.

We consider the spectral resolutions of A and B given by

A =
∫

I
tdE (t) and B =

∫
J
sdF (s) .

If we take in (34) the integral
∫

I

∫
J over dE (t) ⊗ dF (s) , then we get∫

I

∫
J
s f

( t
s

)
dE (t) ⊗ dF (s)

≤ f (u)
∫

I

∫
J
sdE (t) ⊗ dF (s) + φ (u)

∫
I

∫
J
(t − us) dE (t) ⊗ dF (s)

+
[

f ′− (Γ) − f ′+
(
γ
)] ∫

I

∫
J
|t − us| dE (t) ⊗ dF (s) ,

which, as above, gives the desired result (40).

Corollary 3.8. With the assumptions of Theorem 3.7 and for x, y ∈ H with ∥x∥ =
∥∥∥y

∥∥∥ = 1, we have〈
P f ,⊗ (A,B)

(
x ⊗ y

)
, x ⊗ y

〉
(44)

≤
[

f (u) − uφ (u)
] 〈

By, y
〉
+ ⟨Ax, x⟩φ (u)

+
[

f ′− (Γ) − f ′+
(
γ
)] 〈
|A ⊗ 1 − u (1 ⊗ B)|

(
x ⊗ y

)
, x ⊗ y

〉
for all u ∈

[
γ,Γ

]
.

In particular, if we take u = ⟨Ax,x⟩
⟨By,y⟩

∈
[
γ,Γ

]
in (44) then we get the reverse of Jensen’s inequality

0 ≤

〈
P f ,⊗ (A,B)

(
x ⊗ y

)
, x ⊗ y

〉〈
By, y

〉 − f
(
⟨Ax, x⟩〈
By, y

〉) (45)

≤
[

f ′− (Γ) − f ′+
(
γ
)]

×

〈
1〈

By, y
〉 ∣∣∣∣∣∣A ⊗ 1 −

⟨Ax, x⟩〈
By, y

〉 (1 ⊗ B)

∣∣∣∣∣∣ (x ⊗ y
)
, x ⊗ y

〉
.



S. S. Dragomir / Filomat 38:5 (2024), 1499–1511 1510

4. Some Examples

Consider the function Πr (u) = ur
− 1, u ≥ 0, r ≥ 1, then by (19) we get

PΠr ,⊗ (A,B) ≥ rur−1 (A ⊗ 1) − [(r − 1) ur + 1] (1 ⊗ B) , (46)

for A, B > 0 and u > 0.
If there exist the constants m1, M1, m2 and M2 with

0 < m1 ≤ A ≤M1,m2 ≤ B ≤M2, (47)

then we can take in Theorem 3.7 γ = m1
M2

and Γ = M1
m2

and from (40) we derive

PΠr ,⊗ (A,B) ≤ A ⊗ 1 − [(r − 1) ur + 1] (1 ⊗ B) (48)

+ r
((M1

m2

)r−1

−

( m1

M2

)r−1
)
|A ⊗ 1 − u (1 ⊗ B)| .

For x, y ∈ H with ∥x∥ =
∥∥∥y

∥∥∥ = 1, we have by (25) that〈
PΠr ,⊗ (A,B)

(
x ⊗ y

)
, x ⊗ y

〉〈
By, y

〉 ≥

(
⟨Ax, x⟩〈
By, y

〉)r

− 1 (49)

for A, B > 0.
If the condition (47) is satisfied, then by (45) we get

0 ≤

〈
P f ,⊗ (A,B)

(
x ⊗ y

)
, x ⊗ y

〉〈
By, y

〉 −

(
⟨Ax, x⟩〈
By, y

〉)r

+ 1 (50)

≤ r
((M1

m2

)r−1

−

( m1

M2

)r−1
)

×

〈
1〈

By, y
〉 ∣∣∣∣∣∣A ⊗ 1 −

⟨Ax, x⟩〈
By, y

〉 (1 ⊗ B)

∣∣∣∣∣∣ (x ⊗ y
)
, x ⊗ y

〉
for x, y ∈ H with ∥x∥ =

∥∥∥y
∥∥∥ = 1.

If we take the convex function f = (·) ln (·) , then we get by (19) that

P(·) ln(·),⊗ (A,B) ≥ (ln u + 1) (A ⊗ 1) − u (1 ⊗ B) , (51)

for A, B > 0 and u > 0.
By (25) we obtain〈
P(·) ln(·),⊗ (A,B)

(
x ⊗ y

)
, x ⊗ y

〉
⟨Ax, x⟩

≥ ln
(
⟨Ax, x⟩〈
By, y

〉) (52)

for x, y ∈ H with ∥x∥ =
∥∥∥y

∥∥∥ = 1.
If the condition (47) is satisfied, then by (40) we obtain

P(·) ln(·),⊗ (A,B) ≤ (ln u + 1) (A ⊗ 1) − u (1 ⊗ B) (53)

+ ln
(M1M2

m2m1

)
|A ⊗ 1 − u (1 ⊗ B)|

for u ∈
[

m1
M2
, M1

m2

]
.



S. S. Dragomir / Filomat 38:5 (2024), 1499–1511 1511

From (45) we also derive

0 ≤

〈
P(·) ln(·),⊗ (A,B)

(
x ⊗ y

)
, x ⊗ y

〉
⟨Ax, x⟩

− ln
(
⟨Ax, x⟩〈
By, y

〉) (54)

≤ ln
(M1M2

m2m1

)
×

〈
1

⟨Ax, x⟩

∣∣∣∣∣∣A ⊗ 1 −
⟨Ax, x⟩〈
By, y

〉 (1 ⊗ B)

∣∣∣∣∣∣ (x ⊗ y
)
, x ⊗ y

〉
for x, y ∈ H with ∥x∥ =

∥∥∥y
∥∥∥ = 1.

By choosing other convex functions, one can derive several similar inequalities. The details are omitted.

References

[1] H. Araki and F. Hansen, Jensen’s operator inequality for functions of several variables, Proc. Amer. Math. Soc. 128 (2000), No. 7,
2075-2084.

[2] S. S. Dragomir, Some tensorial Hermite-Hadamard type inequalities for convex functions of selfadjoint operators in Hilbert spaces, Preprint
RGMIA Res. Rep. Coll. 25 (2022), Art. 90, 14 pp. [Online https://rgmia.org/papers/v25/v25a90.pdf]

[3] A. Korányi. On some classes of analytic functions of several variables. Trans. Amer. Math. Soc., 101 (1961), 520–554.
[4] A. Ebadian, I. Nikoufar and M. E. Gordji, Perspectives of matrix convex functions, Proc. Natl. Acad. Sci. USA, 108 (2011), no. 18,

7313–7314.
[5] T. Furuta, J. Mićić Hot, J. Pečarić and Y. Seo, Mond-Pečarić Method in Operator Inequalities. Inequalities for Bounded Selfadjoint

Operators on a Hilbert Space, Element, Zagreb, 2005.
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N0. 2, 243–249.
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