1 Introduction
In what follows,
M
n
(
C
)
denotes the space of all
n
×
n
complex matrices.
M
n
+
(
C
)
denotes the set of all positive semi-definite matrices in
M
n
(
C
)
,
X
≥
Y
for
X
,
Y
∈
M
n
(
C
)
means that
X
and
Y
are Hermitians and
X
−
Y
∈
M
n
+
(
C
)
. The set of all strictly positive definite matrices in
M
n
(
C
)
is denoted by
M
n
+
+
(
C
)
. The unitarily invariance of the
∣
∣
∣
.
∣
∣
∣
on
M
n
(
C
)
means that
∣
∣
∣
U
A
V
∣
∣
∣
=
∣
∣
∣
A
∣
∣
∣
for all
A
∈
M
n
(
C
)
and for all the unitary matrices
U
,
V
∈
m
n
(
C
)
. For
A
=
[
a
i
j
]
∈
M
n
(
C
)
, the Hilbert-Schmidt (or Frobenius) norm and the trace norm of
A
are defined by
∥
A
∥
2
=
∑
j
=
1
n
s
j
2
(
A
)
,
∥
A
∥
1
=
∑
j
=
1
n
s
j
(
A
)
,
respectively, where
s
1
(
A
)
≥
s
2
(
A
)
≥
⋯
≥
s
n
(
A
)
are the singular values of
A
, that is, the eigenvalues of the positive matrix
∣
A
∣
=
(
A
∗
A
)
1
2
, arranged in decreasing order and repeated according to multiplicity. Moreover, it is well known that
∥
.
∥
2
is unitarily invariant.
Furthermore, let
ℬ
(
ℋ
)
be the
C
∗
-algebra of all bounded linear operators on a complex separable Hilbert space
ℋ
.
I
stands for the identity operator.
ℬ
+
+
(
ℋ
)
denotes the cone of all positive invertible operators on
ℋ
. As a matter of convenience, we use the following notations to define the
ν
-weighted arithmetic-mean (AM), geometric-mean (GM), and harmonic-mean (HM) for scalars and operators:
a
∇
ν
b
=
ν
a
+
(
1
−
ν
)
b
,
A
∇
ν
B
=
ν
A
+
(
1
−
ν
)
B
,
A
♯
ν
B
=
A
1
2
(
A
−
1
2
B
A
−
1
2
)
ν
A
1
2
=
B
♯
1
−
ν
A
,
a
!
ν
b
=
(
ν
a
−
1
+
(
1
−
ν
)
b
−
1
)
−
1
,
where
a
,
b
>
0
,
A
,
B
∈
ℬ
+
+
(
ℋ
)
, and
ν
∈
[
0
,
1
]
. When
ν
=
1
2
, we write
a
∇
b
,
A
∇
B
,
A
♯
B
, and
a
!
b
for brevity, respectively.
It is well known that the famous
ν
-weighted A-G-H mean inequalities hold,
(1)
a
∇
ν
b
≥
a
ν
b
1
−
ν
≥
a
!
ν
b
for
a
,
b
>
0
and
ν
∈
[
0
,
1
]
with equalities if and only if
a
=
b
. The first inequality of (1) is the classical Young’s inequality (see [6,7,18, 19,20,21] and references therein).
In [13,14], Kittaneh and Manasrah researched Young’s inequality and obtained the following results:
(2)
a
ν
b
1
−
ν
+
r
(
a
−
b
)
2
≤
ν
a
+
(
1
−
ν
)
b
≤
a
ν
b
1
−
ν
+
s
(
a
−
b
)
2
,
where
a
,
b
>
0
,
ν
∈
[
0
,
1
]
and
r
=
min
{
ν
,
1
−
ν
}
,
s
=
max
{
ν
,
1
−
ν
}
. The left inequality of (2) can be regarded as a refinement of Young’s inequality and the right one can be regarded as a reverse of Young’s inequality.
In [25], Zhao and Wu deepened inequality (2) and showed another refinement of Young’s inequality as follows:
(3)
ν
a
+
(
1
−
ν
)
b
≥
a
ν
b
1
−
ν
+
ν
(
a
−
b
)
2
+
r
0
(
a
b
4
−
b
)
2
,
if
ν
∈
0
,
1
2
;
ν
a
+
(
1
−
ν
)
b
≥
a
ν
b
1
−
ν
+
(
1
−
ν
)
(
a
−
b
)
2
+
r
0
(
a
b
4
−
a
)
2
,
if
ν
∈
1
2
,
1
,
where
a
,
b
≥
0
,
ν
∈
[
0
,
1
]
,
r
=
min
{
ν
,
1
−
ν
}
, and
r
0
=
min
{
2
r
,
1
−
2
r
}
.
Zhao and Wu also obtained a more precise form of the reverse Young’s inequality as follows:
(4)
ν
a
+
(
1
−
ν
)
b
≤
a
ν
b
1
−
ν
+
(
1
−
ν
)
(
a
−
b
)
2
−
r
0
(
a
b
4
−
a
)
2
,
if
ν
∈
0
,
1
2
;
ν
a
+
(
1
−
ν
)
b
≤
a
ν
b
1
−
ν
+
ν
(
a
−
b
)
2
−
r
0
(
a
b
4
−
b
)
2
,
if
ν
∈
1
2
,
1
,
where
a
,
b
≥
0
,
ν
∈
[
0
,
1
]
,
r
=
min
{
ν
,
1
−
ν
}
, and
r
0
=
min
{
2
r
,
1
−
2
r
}
.
In a recent work, Hu [8] introduced a refinement of Young-type inequalities as:
(5)
ν
2
a
2
+
(
1
−
ν
)
2
b
2
≥
(
ν
a
)
2
ν
b
2
−
2
ν
+
ν
2
(
a
−
b
)
2
,
if
0
≤
ν
≤
1
2
;
ν
2
a
2
+
(
1
−
ν
)
2
b
2
≥
a
2
ν
(
(
1
−
ν
)
b
)
2
−
2
ν
+
(
1
−
ν
)
2
(
a
−
b
)
2
,
if
1
2
≤
ν
≤
1
.
Later, based on the result of Hu [8], Nasiri et al. [16] have given some refinements of (5). These inequalities can be written as: If
0
≤
ν
≤
1
2
, then
(6)
ν
2
a
2
+
(
1
−
ν
)
2
b
2
≥
(
ν
a
)
2
ν
b
2
−
2
ν
+
ν
2
(
a
−
b
)
2
+
r
0
b
(
ν
a
−
b
)
2
,
where
r
0
=
min
{
2
ν
,
1
−
2
ν
}
, and if
1
2
≤
ν
≤
1
,
(7)
ν
2
a
2
+
(
1
−
ν
)
2
b
2
≥
a
2
ν
[
(
1
−
ν
)
b
]
2
−
2
ν
+
(
1
−
ν
)
2
(
a
−
b
)
2
+
r
1
a
(
a
−
(
1
−
ν
)
b
)
2
,
where
r
1
=
min
{
2
ν
−
1
,
2
−
2
ν
}
.
Later, Yang and Li [24] established some refinements and reverses of (6) and (7). These inequalities can be written as:
If
0
≤
ν
≤
1
4
, then
(8)
(
ν
a
)
2
ν
b
2
−
2
ν
+
ν
2
(
a
−
b
)
2
+
2
ν
b
(
ν
a
−
b
)
2
+
s
1
b
(
ν
a
b
−
b
)
2
≤
ν
2
a
2
+
(
1
−
ν
)
2
b
2
≤
a
2
ν
[
(
1
−
ν
)
b
]
2
−
2
ν
+
(
1
−
ν
)
2
(
a
−
b
)
2
−
2
ν
a
(
a
−
(
1
−
ν
)
b
)
2
−
s
1
a
(
(
1
−
ν
)
a
b
4
−
a
)
2
.
If
1
4
≤
ν
≤
1
2
, then
(9)
(
ν
a
)
2
ν
b
2
−
2
ν
+
ν
2
(
a
−
b
)
2
+
(
1
−
2
ν
)
b
(
ν
a
−
b
)
2
+
s
2
b
(
ν
a
b
−
b
)
2
≤
ν
2
a
2
+
(
1
−
ν
)
2
b
2
≤
a
2
ν
[
(
1
−
ν
)
b
]
2
−
2
ν
+
(
1
−
ν
)
2
(
a
−
b
)
2
−
(
1
−
2
ν
)
a
(
a
−
(
1
−
ν
)
b
)
2
−
s
2
a
(
(
1
−
ν
)
a
b
4
−
(
1
−
ν
)
b
)
2
.
If
1
2
≤
ν
≤
3
4
, then
(10)
a
2
ν
[
(
1
−
ν
)
b
]
2
−
2
ν
+
(
1
−
ν
)
2
(
a
−
b
)
2
+
(
2
ν
−
1
)
a
(
a
−
(
1
−
ν
)
b
)
2
+
s
3
a
(
(
1
−
ν
)
a
b
4
−
(
1
−
ν
)
b
)
2
≤
ν
2
a
2
+
(
1
−
ν
)
2
b
2
≤
(
ν
a
)
2
ν
b
2
−
2
ν
+
ν
2
(
a
−
b
)
2
−
(
2
ν
−
1
)
b
(
ν
a
−
b
)
2
−
s
3
b
(
ν
a
b
4
−
ν
a
)
2
.
If
3
4
≤
ν
≤
1
, then
(11)
a
2
ν
[
(
1
−
ν
)
b
]
2
−
2
ν
+
(
1
−
ν
)
2
(
a
−
b
)
2
+
(
2
−
2
ν
)
a
(
a
−
(
1
−
ν
)
b
)
2
+
s
4
a
(
(
1
−
ν
)
a
b
4
−
a
)
2
≤
ν
2
a
2
+
(
1
−
ν
)
2
b
2
≤
(
ν
a
)
2
ν
b
2
−
2
ν
+
ν
2
(
a
−
b
)
2
−
(
2
−
2
ν
)
b
(
ν
a
−
b
)
2
−
s
4
b
(
ν
a
b
4
−
b
)
2
,
where
s
1
=
min
{
4
ν
,
1
−
4
ν
}
,
s
2
=
min
{
2
−
4
ν
,
4
ν
−
1
}
,
s
3
=
min
{
3
−
4
ν
,
4
ν
−
2
}
, and
s
4
=
min
{
4
−
4
ν
,
4
ν
−
3
}
.
The Kantorovich’s constant defined by
K
(
t
,
2
)
=
(
t
+
1
)
2
4
t
, which has the properties
K
(
1
,
2
)
=
1
,
K
(
t
,
2
)
=
K
1
t
,
2
≥
1
(
t
>
0
)
, and
K
(
t
,
2
)
is monotone increasing on
[
1
,
∞
)
and monotone decreasing on
(
0
,
1
]
.
The following multiplicative refinement and reverse of Young’s inequality in terms of Kantorovich’s constant holds:
(12)
K
r
(
h
,
2
)
a
ν
b
1
−
ν
≤
ν
a
+
(
1
−
ν
)
b
≤
K
R
(
h
,
2
)
a
ν
b
1
−
ν
,
where
a
,
b
>
0
,
ν
∈
[
0
,
1
]
,
r
=
min
{
ν
,
1
−
ν
}
,
R
=
max
{
ν
,
1
−
ν
}
, and
h
=
b
a
.
The second inequality in (12) is due to Liao et al. [11], whereas the first one is due to Zou et al. [27]. In [23], Zhao and Wu made a further study about inequality (2) with the Kantorovich constant and gave the following results:
(13)
ν
a
+
(
1
−
ν
)
b
≥
K
(
h
,
2
)
r
′
a
ν
b
1
−
ν
+
r
(
a
−
b
)
2
,
where
a
,
b
≥
0
,
ν
∈
[
0
,
1
]
⧹
{
1
2
}
,
r
=
min
{
ν
,
1
−
ν
}
,
r
′
=
min
{
2
r
,
1
−
2
r
}
, and
h
=
a
b
.
And then Zhao and Wu made a reverse refinement of the second inequality in (2)
(14)
ν
a
+
(
1
−
ν
)
b
≤
K
(
h
,
2
)
−
r
′
a
ν
b
1
−
ν
+
s
(
a
−
b
)
2
,
where
a
,
b
≥
0
,
ν
∈
[
0
,
1
]
⧹
{
1
2
}
,
h
=
a
b
,
r
=
min
{
ν
,
1
−
ν
}
,
r
′
=
min
{
2
r
,
1
−
2
r
}
, and
s
=
max
{
ν
,
1
−
ν
}
.
Let us take a closer look at [11] where Liao et al. made a reverse refinement for Young’s inequality as follows:
(15)
ν
a
+
(
1
−
ν
)
b
≤
K
(
h
,
2
)
R
a
ν
b
1
−
ν
(16)
ν
a
+
(
1
−
ν
)
b
≤
K
(
h
,
2
)
R
′
a
ν
b
1
−
ν
+
r
(
a
−
b
)
2
,
where
a
,
b
≥
0
,
ν
∈
[
0
,
1
]
,
h
=
a
b
,
r
=
min
{
ν
,
1
−
ν
}
,
R
′
=
max
{
2
r
,
1
−
2
r
}
, and
R
=
max
{
ν
,
1
−
ν
}
.
In [10], Liao and Wu further deepened the results of inequality (14) and obtained the following results: If
ν
∈
0
,
1
2
, then
(17)
ν
a
+
(
1
−
ν
)
b
≥
K
(
h
4
,
2
)
r
1
a
ν
b
1
−
ν
+
ν
(
a
−
b
)
2
+
r
0
(
a
b
4
−
b
)
2
.
If
ν
∈
1
2
,
1
, then
(18)
ν
a
+
(
1
−
ν
)
b
≥
K
(
h
4
,
2
)
r
1
a
ν
b
1
−
ν
+
(
1
−
ν
)
(
a
−
b
)
2
+
r
0
(
a
b
4
−
a
)
2
,
where
h
=
a
b
,
r
=
min
{
ν
,
1
−
ν
}
,
r
0
=
min
{
2
r
,
1
−
2
r
}
, and
r
1
=
min
{
2
r
0
,
1
−
2
r
0
}
. Similarly, Liao and Wu deepened the reverse results of inequality (14) as follows: If
ν
∈
0
,
1
2
, then
(19)
ν
a
+
(
1
−
ν
)
b
≤
K
(
h
4
,
2
)
−
r
1
a
ν
b
1
−
ν
+
(
1
−
ν
)
(
a
−
b
)
2
−
r
0
(
a
b
4
−
a
)
2
If
ν
∈
1
2
,
1
, then
(20)
ν
a
+
(
1
−
ν
)
b
≤
K
(
h
4
,
2
)
−
r
1
a
ν
b
1
−
ν
+
ν
(
a
−
b
)
2
−
r
0
(
a
b
4
−
b
)
2
,
where
a
,
b
≥
0
,
h
=
a
b
,
r
=
min
{
ν
,
1
−
ν
}
, and
r
0
=
min
{
2
r
,
1
−
2
r
}
.
Recently, Zhao and Wu [26] obtained the refinements and reverses of Young’s inequality and improved inequalities (2) in the following forms: If
ν
∈
0
,
1
4
, then
(21)
ν
(
a
−
b
)
2
+
2
ν
(
a
b
4
−
b
)
2
+
r
(
b
−
b
3
a
8
)
2
+
K
(
h
8
,
2
)
r
1
a
ν
b
1
−
ν
≤
ν
a
+
(
1
−
ν
)
b
≤
(
1
−
ν
)
(
a
−
b
)
2
−
2
ν
(
a
b
4
−
a
)
2
−
r
(
a
−
a
3
b
8
)
2
+
K
(
h
8
,
2
)
−
r
1
a
ν
b
1
−
ν
,
where
a
,
b
>
0
,
h
=
a
b
,
r
=
min
{
4
ν
,
1
−
4
ν
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
If
1
4
≤
ν
<
1
2
, then
(22)
ν
(
a
−
b
)
2
+
(
1
−
2
ν
)
(
a
b
4
−
b
)
2
+
r
(
a
b
4
−
b
3
a
8
)
2
+
K
(
h
8
,
2
)
r
1
a
ν
b
1
−
ν
≤
ν
a
+
(
1
−
ν
)
b
≤
(
1
−
ν
)
(
a
−
b
)
2
−
(
1
−
2
ν
)
(
a
b
4
−
a
)
2
−
r
(
a
b
4
−
a
3
b
8
)
2
+
K
(
h
8
,
2
)
−
r
1
a
ν
b
1
−
ν
,
where
a
,
b
>
0
,
h
=
a
b
,
r
=
min
{
4
ν
−
1
,
2
−
4
ν
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
If
1
2
≤
ν
<
3
4
, then
(23)
(
1
−
ν
)
(
a
−
b
)
2
+
(
2
ν
−
1
)
(
a
b
4
−
a
)
2
+
r
(
a
b
4
−
a
3
b
8
)
2
+
K
(
h
8
,
2
)
r
1
a
ν
b
1
−
ν
≤
ν
a
+
(
1
−
ν
)
b
≤
ν
(
a
−
b
)
2
−
(
2
ν
−
1
)
(
a
b
4
−
b
)
2
−
r
(
a
b
4
−
b
3
a
8
)
2
+
K
(
h
8
,
2
)
−
r
1
a
ν
b
1
−
ν
,
where
a
,
b
>
0
,
h
=
a
b
,
r
=
min
{
4
ν
−
2
,
3
−
4
ν
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
If
3
4
≤
ν
<
1
, then
(24)
(
1
−
ν
)
(
a
−
b
)
2
+
(
2
−
2
ν
)
(
a
b
4
−
a
)
2
+
r
(
a
−
a
3
b
8
)
2
+
K
(
h
8
,
2
)
r
1
a
ν
b
1
−
ν
≤
ν
a
+
(
1
−
ν
)
b
≤
ν
(
a
−
b
)
2
−
(
2
−
2
ν
)
(
a
b
4
−
b
)
2
−
r
(
b
−
b
3
a
8
)
2
+
K
(
h
8
,
2
)
−
r
1
a
ν
b
1
−
ν
,
where
a
,
b
>
0
,
h
=
a
b
,
r
=
min
{
4
ν
−
3
,
4
−
4
ν
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
Following [11,27], the operator version is as follows : For two positive operators
A
,
B
and positive real numbers
m
,
m
′
,
M
,
M
′
satisfying either of the following conditions:
0
<
m
I
≤
A
≤
m
′
I
<
M
′
I
≤
B
≤
M
I
,
0
<
m
I
≤
B
≤
m
′
I
<
M
′
I
≤
A
≤
M
I
,
we have
(25)
K
r
(
h
′
,
2
)
A
♯
ν
B
≤
A
∇
ν
B
≤
K
R
(
h
,
2
)
A
♯
ν
B
,
where
h
=
M
m
,
h
′
=
M
′
m
′
,
ν
∈
[
0
,
1
]
,
r
=
min
{
ν
,
1
−
ν
}
, and
R
=
max
{
ν
,
1
−
ν
}
.
In [15], inequalities (6) and (7) for the Hilbert-Schmidt norm form with Kantorovich constant were obtained by Nasiri and Shakoori, that is, if
A
,
B
,
X
∈
M
n
(
C
)
, where
A
,
B
∈
M
n
+
(
C
)
, then
(26)
∥
ν
A
X
+
(
1
−
ν
)
X
B
∥
2
2
≥
ν
2
∥
A
X
−
X
B
∥
2
2
+
r
∥
X
B
∥
2
2
+
ν
A
1
2
X
B
1
2
2
2
−
2
ν
A
1
4
X
B
3
4
+
K
r
′
ν
2
ν
∥
A
ν
X
B
1
−
ν
∥
2
2
+
2
ν
(
1
−
ν
)
A
1
2
X
B
1
2
2
2
,
where
ν
∈
0
,
1
2
,
K
=
min
1
≤
i
,
j
≤
n
K
ν
λ
i
μ
j
,
2
,
r
=
min
{
2
ν
,
1
−
2
ν
}
, and
r
′
=
min
{
2
r
′
,
1
−
2
r
}
.
(27)
∥
ν
A
X
+
(
1
−
ν
)
X
B
∥
2
2
≥
(
1
−
ν
)
2
∥
A
X
−
X
B
∥
2
2
+
r
∥
A
X
∥
2
2
+
(
1
−
ν
)
A
1
2
X
B
1
2
2
2
+
r
∥
A
X
∥
2
2
+
(
1
−
ν
)
A
1
2
X
B
1
2
2
2
−
2
1
−
ν
A
3
4
X
B
1
4
+
K
r
′
(
1
−
ν
)
2
(
1
−
ν
)
∥
A
ν
X
B
1
−
ν
∥
2
2
+
2
ν
(
1
−
ν
)
A
1
2
X
B
1
2
2
2
,
where
ν
∈
1
2
,
1
,
K
=
min
1
≤
i
,
j
≤
n
K
λ
i
(
1
−
ν
)
μ
j
,
2
,
r
=
min
{
2
ν
−
1
,
2
−
2
ν
}
, and
r
′
=
min
{
2
r
′
,
1
−
2
r
}
.
For a detailed study of these and associated norm inequalities along with their history of origin, refinements, and applications, one may refer to [1,2,3,4,5,11,18,19,20, 21,22].
One of the goals of this article is to obtain new refinements and reverses of inequalities (8)–(11) with the Kantorovich constant.
The article is organized as follows. In Section 2, a new scalar version of Young’s inequality and its reverse with the Kantorovich constant are first given. In Section 3, we obtain the corresponding new operator inequalities on the Hilbert space. Finally, in Section 4, we mainly show the matrix version of inequalities for the Hilbert-Schmidt norm, unitarily invariant norm, and related trace version based on the result of part one.
3 New operator versions of Young-type inequalities
In the section, we give some more excellent versions of Young-type operator inequalities and their reverse by the monotonic property of operator functions.
First, we present the monotonic property of operator function, which is the basis of the following discussion.
Lemma 3.1
[17] Let
T
∈
ℬ
(
ℋ
)
be self-adjoint. If f and g are both continuous functions with
f
(
t
)
≥
g
(
t
)
for
t
∈
σ
(
T
)
(where the sign
σ
(
T
)
denotes the spectrum of operator T), then
f
(
T
)
≥
g
(
T
)
.
Next we present our main results on the basis of inequalities (28)–(35). By Lemma 3.1, we have the following.
Theorem 3.2
Let
A
,
B
∈
ℬ
+
+
(
ℋ
)
and
ν
∈
[
0
,
1
]
. If all positive numbers
m
,
m
′
and
M
,
M
′
satisfy either of the conditions
0
<
m
I
≤
A
≤
m
′
I
<
M
′
I
≤
B
≤
M
I
or
0
<
m
I
≤
B
≤
m
′
I
<
M
′
I
≤
A
≤
M
I
, then:
If
ν
∈
0
,
1
4
, then
(36)
K
[
ν
h
8
,
2
]
r
1
ν
2
ν
A
♯
1
−
ν
B
+
2
ν
2
(
A
∇
B
−
A
♯
B
)
+
ν
(
ν
A
♯
B
+
B
−
2
ν
A
♯
3
4
B
)
+
2
ν
ν
A
♯
3
4
B
+
B
−
2
ν
4
A
♯
7
8
B
+
r
B
−
2
ν
8
A
♯
15
16
B
+
ν
4
A
♯
7
8
B
≤
ν
2
A
+
(
1
−
ν
)
2
B
≤
K
h
1
−
ν
8
,
2
−
r
1
(
1
−
ν
)
2
(
1
−
ν
)
A
♯
1
−
ν
B
+
2
(
1
−
ν
)
2
(
A
∇
B
−
A
♯
B
)
+
(
1
−
ν
)
A
+
(
1
−
ν
)
A
♯
B
−
2
1
−
ν
A
♯
1
4
B
−
2
ν
(
2
1
−
ν
A
♯
1
4
B
+
A
−
2
1
−
ν
4
A
♯
1
8
B
)
−
r
1
−
ν
4
A
♯
1
8
B
+
A
−
2
1
−
ν
8
A
♯
1
16
B
,
where
h
=
M
m
,
r
=
min
{
4
ν
,
1
−
4
ν
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
If
ν
∈
1
4
,
1
2
, then
(37)
K
[
ν
h
8
,
2
]
r
1
ν
2
ν
A
♯
1
−
ν
B
+
2
ν
2
(
A
∇
B
−
A
♯
B
)
+
ν
(
ν
A
♯
B
+
B
−
2
ν
A
♯
3
4
B
)
+
(
1
−
2
ν
)
ν
A
♯
3
4
B
+
B
−
2
ν
4
A
♯
7
8
B
+
r
ν
A
♯
3
4
B
+
ν
4
A
♯
7
8
B
−
2
ν
3
8
A
♯
13
16
B
≤
ν
2
A
+
(
1
−
ν
)
2
B
≤
K
h
1
−
ν
8
,
2
−
r
1
(
1
−
ν
)
2
(
1
−
ν
)
A
♯
1
−
ν
B
+
2
(
1
−
ν
)
2
(
2
A
∇
B
−
A
♯
B
)
+
(
1
−
ν
)
A
+
(
1
−
ν
)
A
♯
B
−
2
1
−
ν
A
♯
1
4
B
−
(
1
−
2
ν
)
A
+
1
−
ν
A
♯
1
4
B
−
2
1
−
ν
4
A
♯
1
8
B
−
r
1
−
ν
4
A
♯
1
8
B
+
1
−
ν
A
♯
1
4
B
−
2
(
1
−
ν
)
3
8
A
♯
3
16
B
,
where
h
=
M
m
,
r
=
min
{
2
−
4
ν
,
4
ν
−
1
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
If
ν
∈
1
2
,
3
4
, then
(38)
K
h
1
−
ν
,
2
8
r
1
(
1
−
ν
)
2
(
1
−
ν
)
A
♯
1
−
ν
B
+
2
(
1
−
ν
)
2
(
A
∇
B
−
A
♯
B
)
+
(
1
−
ν
)
(
A
+
(
1
−
ν
)
A
♯
B
−
2
1
−
ν
A
♯
1
4
B
)
+
(
2
ν
−
1
)
A
+
1
−
ν
A
♯
1
4
B
−
2
1
−
ν
4
A
♯
1
8
B
+
r
1
−
ν
A
♯
1
4
B
+
1
−
ν
4
A
♯
1
8
B
−
2
(
1
−
ν
)
3
8
A
♯
3
16
B
≤
ν
2
A
+
(
1
−
ν
)
2
B
≤
K
[
ν
h
8
,
2
]
−
r
1
ν
2
ν
A
♯
1
−
ν
B
+
2
ν
2
(
A
∇
B
−
A
♯
B
)
+
ν
B
+
ν
A
♯
B
−
2
ν
A
♯
3
4
B
−
(
2
ν
−
1
)
B
+
ν
A
♯
3
4
B
−
2
ν
4
A
♯
7
8
B
−
r
ν
4
A
♯
7
8
B
+
A
♯
3
4
B
−
2
ν
3
8
A
♯
13
16
B
,
where
h
=
M
m
,
r
=
min
{
4
ν
−
2
,
3
−
4
ν
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
If
ν
∈
3
4
,
1
, then
(39)
K
h
1
−
ν
8
,
2
r
1
(
1
−
ν
)
2
(
1
−
ν
)
A
♯
1
−
ν
B
+
2
(
1
−
ν
)
2
(
A
∇
B
−
A
♯
B
)
+
(
1
−
ν
)
(
A
+
(
1
−
ν
)
A
♯
B
−
2
1
−
ν
A
♯
1
4
B
+
(
2
−
2
ν
)
1
−
ν
A
♯
1
4
B
+
A
−
2
1
−
ν
4
A
♯
1
8
B
+
r
A
−
2
1
−
ν
8
A
♯
1
16
B
+
1
−
ν
4
A
♯
1
8
B
≤
ν
2
A
+
(
1
−
ν
)
2
B
≤
K
[
ν
h
8
,
2
]
−
r
1
ν
2
ν
A
♯
1
−
ν
B
+
2
ν
2
(
A
∇
B
−
A
♯
B
)
+
ν
B
+
ν
A
♯
B
−
2
ν
A
♯
3
4
B
−
(
2
−
2
ν
)
(
ν
A
♯
3
4
B
+
B
−
2
ν
4
A
♯
7
8
B
)
−
r
B
+
ν
4
A
♯
7
8
B
−
2
ν
8
A
♯
15
16
B
,
where
h
=
M
m
,
r
=
min
{
4
ν
−
3
,
4
−
4
ν
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
Proof
We only need to investigate inequality (36) due to the similarity of the process of proof.
If
ν
∈
0
,
1
4
, then by inequality (28) by letting
a
2
=
1
,
b
2
=
t
>
0
, we have
K
[
ν
h
8
,
2
]
r
1
ν
2
ν
t
1
−
ν
+
ν
2
(
1
+
t
−
2
t
)
+
ν
ν
t
1
2
+
t
−
2
ν
t
3
4
+
r
t
+
ν
4
t
7
8
−
2
ν
8
t
15
16
≤
ν
2
+
(
1
−
ν
)
2
t
≤
K
h
1
−
ν
8
,
2
−
r
1
(
1
−
ν
)
2
(
1
−
ν
)
t
1
−
ν
+
(
1
−
ν
)
2
(
1
+
t
−
2
t
)
+
(
1
−
ν
)
1
+
(
1
−
ν
)
t
1
2
−
2
1
−
ν
t
1
4
−
2
ν
1
−
ν
t
1
4
+
1
−
2
1
−
ν
4
t
1
8
−
r
1
−
ν
4
t
1
8
+
I
−
2
1
−
ν
8
t
1
16
.
For
X
=
A
−
1
2
B
A
−
1
2
, under the first condition, we obtain
I
≤
h
I
≤
M
m
I
≤
X
≤
h
′
I
=
M
′
m
′
I
and then
σ
(
X
)
⊆
[
h
,
h
′
]
⊆
(
1
,
∞
)
. By Lemma 3.1, we have
min
h
≤
t
≤
h
′
K
[
ν
t
8
,
2
]
r
1
ν
2
ν
X
1
−
ν
+
ν
2
(
I
+
t
−
2
X
)
+
ν
ν
X
1
2
+
X
−
2
ν
X
3
4
+
r
X
+
ν
4
X
7
8
−
2
ν
8
X
15
16
≤
ν
2
I
+
(
1
−
ν
)
2
X
≤
max
h
≤
t
≤
h
′
K
t
1
−
ν
8
,
2
−
r
1
(
1
−
ν
)
2
(
1
−
ν
)
X
1
−
ν
+
(
1
−
ν
)
2
(
I
+
X
−
2
X
)
+
(
1
−
ν
)
I
+
(
1
−
ν
)
X
1
2
−
2
1
−
ν
X
1
4
−
2
ν
I
−
ν
X
1
4
+
I
−
2
1
−
ν
4
X
1
8
−
r
1
−
ν
4
X
1
8
+
I
−
2
1
−
ν
8
X
1
16
.
Since the Kantorovich constant is an increasing function on
(
1
,
∞
)
, then
(40)
K
[
ν
h
8
,
2
]
r
1
ν
2
ν
A
−
1
2
B
A
−
1
2
1
−
ν
+
ν
2
I
+
A
−
1
2
B
A
−
1
2
−
2
A
−
1
2
B
A
−
1
2
+
ν
ν
A
−
1
2
B
A
−
1
2
1
2
+
A
−
1
2
B
A
−
1
2
−
2
ν
A
−
1
2
B
A
−
1
2
3
4
+
r
A
−
1
2
B
A
−
1
2
+
ν
4
A
−
1
2
B
A
−
1
2
7
8
−
2
ν
8
A
−
1
2
B
A
−
1
2
15
16
≤
ν
2
+
(
1
−
ν
)
2
A
−
1
2
B
A
−
1
2
≤
K
h
1
−
ν
8
,
2
−
r
1
(
1
−
ν
)
2
(
1
−
ν
)
A
−
1
2
B
A
−
1
2
1
−
ν
+
(
1
−
ν
)
2
I
+
A
−
1
2
B
A
−
1
2
−
2
A
−
1
2
B
A
−
1
2
+
(
1
−
ν
)
I
+
(
1
−
ν
)
A
−
1
2
B
A
−
1
2
1
2
−
2
1
−
ν
A
−
1
2
B
A
−
1
2
1
4
−
2
ν
1
−
ν
A
−
1
2
B
A
−
1
2
1
4
+
I
−
2
1
−
ν
4
A
−
1
2
B
A
−
1
2
1
8
−
r
1
−
ν
4
A
−
1
2
B
A
−
1
2
1
8
+
I
−
2
1
−
ν
8
A
−
1
2
B
A
−
1
2
1
16
.
In a similar way, under the second condition, we have
I
≤
1
h
I
=
m
M
I
≤
X
≤
1
h
′
I
=
m
′
M
′
I
and
σ
(
X
)
⊆
1
h
,
1
h
′
⊆
(
0
,
1
)
. By Lemma 3.1, we have
min
1
h
′
≤
t
≤
1
h
K
[
ν
t
8
,
2
]
r
1
ν
2
ν
X
1
−
ν
+
ν
2
(
I
+
t
−
2
X
)
+
ν
ν
X
1
2
+
X
−
2
ν
X
3
4
+
r
X
+
ν
4
X
7
8
−
2
ν
8
X
15
16
≤
ν
2
I
+
(
1
−
ν
)
2
X
≤
max
1
h
′
≤
t
≤
1
h
K
t
1
−
ν
8
,
2
−
r
1
(
1
−
ν
)
2
(
1
−
ν
)
X
1
−
ν
+
(
1
−
ν
)
2
(
I
+
X
−
2
X
)
+
(
1
−
ν
)
I
+
(
1
−
ν
)
X
1
2
−
2
1
−
ν
X
1
4
−
2
ν
I
−
ν
X
1
4
+
I
−
2
1
−
ν
4
X
1
8
−
r
1
−
ν
4
X
1
8
+
I
−
2
1
−
ν
8
X
1
16
.
Since the Kantorovich constant is a decreasing function on
(
0
,
1
)
, then
(41)
K
[
ν
h
8
,
2
]
r
1
ν
2
ν
A
−
1
2
B
A
−
1
2
1
−
ν
+
ν
2
I
+
A
−
1
2
B
A
−
1
2
−
2
A
−
1
2
B
A
−
1
2
+
ν
ν
A
−
1
2
B
A
−
1
2
1
2
+
A
−
1
2
B
A
−
1
2
−
2
ν
A
−
1
2
B
A
−
1
2
3
4
+
r
A
−
1
2
B
A
−
1
2
+
ν
4
A
−
1
2
B
A
−
1
2
7
8
−
2
ν
8
A
−
1
2
B
A
−
1
2
15
16
≤
ν
2
+
(
1
−
ν
)
2
A
−
1
2
B
A
−
1
2
≤
K
h
1
−
ν
8
,
2
−
r
1
(
1
−
ν
)
2
(
1
−
ν
)
A
−
1
2
B
A
−
1
2
1
−
ν
+
(
1
−
ν
)
2
I
+
A
−
1
2
B
A
−
1
2
−
2
A
−
1
2
B
A
−
1
2
+
(
1
−
ν
)
I
+
(
1
−
ν
)
A
−
1
2
B
A
−
1
2
1
2
−
2
1
−
ν
A
−
1
2
B
A
−
1
2
1
4
−
2
ν
1
−
ν
A
−
1
2
B
A
−
1
2
1
4
+
I
−
2
1
−
ν
4
A
−
1
2
B
A
−
1
2
1
8
−
r
1
−
ν
4
A
−
1
2
B
A
−
1
2
1
8
+
I
−
2
1
−
ν
8
A
−
1
2
B
A
−
1
2
1
16
.
Then multiplying inequalities (40) and (41) by
A
1
2
on the left-hand side and on the right-hand side, we can deduce the required inequality (36).□
Theorem 3.3
Let
A
,
B
∈
ℬ
+
+
(
ℋ
)
and
ν
∈
[
0
,
1
]
. If all positive numbers
m
,
m
′
and
M
,
M
′
satisfy either of the conditions
0
<
m
I
≤
A
≤
m
′
I
<
M
′
I
≤
B
≤
M
I
or
0
<
m
I
≤
B
≤
m
′
I
<
M
′
I
≤
A
≤
M
I
, then:
If
ν
∈
0
,
1
2
, then
(42)
K
[
ν
h
4
,
2
]
r
1
ν
2
ν
A
♯
1
−
ν
B
+
2
ν
2
(
A
∇
B
−
A
♯
B
)
+
ν
(
ν
A
♯
B
+
B
−
2
ν
A
♯
3
4
B
+
r
0
B
+
ν
A
♯
3
4
B
−
2
ν
4
A
♯
7
8
B
≤
ν
2
A
+
(
1
−
ν
)
2
B
≤
K
h
1
−
ν
4
,
2
−
r
1
[
(
1
−
ν
)
]
2
−
2
ν
A
♯
1
−
ν
B
+
2
(
1
−
ν
)
2
(
A
∇
B
−
A
♯
B
)
+
(
1
−
ν
)
A
+
A
♯
B
−
2
1
−
ν
A
♯
1
4
B
−
r
0
A
+
1
−
ν
A
♯
1
4
B
−
2
1
−
ν
4
A
♯
1
8
B
,
where
h
=
M
m
,
r
=
min
{
2
ν
,
1
−
2
ν
}
,
r
0
=
min
{
2
r
,
1
−
2
r
}
, and
r
1
=
min
{
2
r
0
,
1
−
2
r
0
}
.
If
ν
∈
1
2
,
1
, then
(43)
K
h
1
−
ν
4
,
2
r
1
(
1
−
ν
)
2
−
2
ν
A
♯
1
−
ν
B
+
2
(
1
−
ν
)
2
(
A
∇
B
−
A
♯
B
)
+
(
1
−
ν
)
A
+
(
1
−
ν
)
A
♯
B
−
2
1
−
ν
A
♯
1
4
B
+
r
0
A
+
1
−
ν
A
♯
1
4
B
−
2
1
−
ν
4
A
♯
1
8
B
≤
ν
2
A
+
(
1
−
ν
)
2
B
≤
K
[
ν
h
4
,
2
]
−
r
1
ν
2
ν
A
♯
1
−
ν
B
+
2
ν
2
(
A
∇
B
−
A
♯
B
)
+
ν
B
+
ν
A
♯
B
−
2
ν
A
♯
3
4
B
−
r
0
B
+
ν
A
♯
3
4
B
−
2
ν
4
A
♯
7
8
B
,
where
h
=
M
m
,
r
=
min
{
2
ν
−
1
,
2
−
2
ν
}
,
r
0
=
min
{
2
r
,
1
−
2
r
}
, and
r
1
=
min
{
2
r
0
,
1
−
2
r
0
}
.
Proof
The process of the proof is analogous to that of Theorem 3.2, so we omit it here.□
Theorem 3.4
Let
A
,
B
,
C
,
I
∈
ℬ
(
ℋ
)
, where B is a positive operator, C is invertible,
A
=
C
∗
C
and I is the identity operator and
ν
∈
[
0
,
1
]
. If all positive numbers
m
,
m
′
and
M
,
M
′
satisfy either of the conditions
0
<
m
I
≤
A
≤
m
′
I
<
M
′
I
≤
B
≤
M
I
or
0
<
m
I
≤
B
≤
m
′
I
<
M
′
I
≤
A
≤
M
I
, then
If
ν
∈
0
,
1
2
, then
(44)
ν
2
A
+
(
1
−
ν
)
2
B
≤
K
h
1
−
ν
,
2
R
′
(
1
−
ν
)
2
−
2
ν
C
∗
T
1
−
μ
C
+
2
(
1
−
ν
)
2
(
A
∇
B
−
C
∗
T
1
2
C
+
r
A
+
(
1
−
ν
)
C
∗
T
1
2
C
−
2
(
1
−
ν
)
C
∗
T
1
4
C
,
If
ν
∈
1
2
,
1
, then
(45)
ν
2
A
+
(
1
−
ν
)
2
B
≤
K
[
ν
h
,
2
]
R
′
ν
2
ν
C
∗
T
1
−
μ
C
+
2
ν
2
A
∇
B
−
C
∗
T
1
2
C
ν
2
A
+
(
1
−
ν
)
2
B
≤
K
[
ν
h
,
2
]
R
′
ν
2
ν
C
∗
T
1
−
μ
C
+
2
ν
2
A
∇
B
−
C
∗
T
1
2
C
+
r
B
+
ν
C
∗
T
1
2
C
−
2
ν
C
∗
T
3
4
C
,
where
T
=
(
C
∗
)
−
1
B
C
−
1
,
r
=
min
{
2
ν
,
1
−
2
ν
}
,
R
′
=
max
{
2
r
,
1
−
2
r
}
,
h
=
M
m
, and
h
′
=
M
′
m
′
.
Proof
The process of the proof of the second inequality is similar to that of the first one, we only need to investigate the first one. By inequality (34), we have
μ
2
+
(
1
−
μ
)
2
t
≤
K
h
1
−
ν
,
2
R
′
(
1
−
ν
)
2
−
2
ν
t
1
−
μ
+
(
1
−
μ
)
2
(
1
+
t
−
2
t
)
+
r
1
+
(
1
−
ν
)
t
−
2
1
−
ν
t
1
4
for
t
>
0
. Putting
T
=
(
C
∗
)
−
1
B
C
−
1
, under the first condition, we have
I
<
h
I
=
M
m
I
≤
T
≤
M
′
m
′
I
=
h
′
I
,
and then
σ
(
T
)
⊆
[
h
,
h
′
]
⊆
(
1
,
∞
)
. Thus for
t
∈
[
h
,
h
′
]
, we have
(46)
μ
2
+
(
1
−
μ
)
2
t
≤
max
h
≤
t
≤
h
′
K
h
1
−
ν
,
2
R
′
(
1
−
ν
)
2
−
2
ν
t
1
−
μ
+
(
1
−
μ
)
2
(
1
+
t
−
2
t
)
+
r
1
+
(
1
−
ν
)
t
−
2
1
−
ν
t
1
4
.
Since the Kantorovich constant is an increasing function on
(
1
,
∞
)
, by (46), we have
(47)
μ
2
+
(
1
−
μ
)
2
t
≤
K
h
1
−
ν
,
2
R
′
(
1
−
ν
)
2
−
2
ν
t
1
−
μ
+
(
1
−
μ
)
2
(
1
+
t
−
2
t
)
+
r
1
+
(
1
−
ν
)
t
−
2
1
−
ν
t
1
4
.
It can be deduced from inequality (47) and Lemma 3.1 that
(48)
μ
2
+
(
1
−
μ
)
2
(
C
∗
)
−
1
B
C
−
1
≤
K
h
1
−
ν
,
2
R
′
(
1
−
ν
)
2
−
2
ν
(
(
C
∗
)
−
1
B
C
−
1
)
1
−
μ
+
(
1
−
μ
)
2
(
I
+
(
C
∗
)
−
1
B
C
−
1
−
2
(
C
∗
)
−
1
B
C
−
1
)
+
r
I
+
(
1
−
ν
)
(
C
∗
)
−
1
B
C
−
1
−
2
1
−
ν
(
(
C
∗
)
−
1
B
C
−
1
)
1
4
.
Likewise, under the second condition, we also have that
0
<
1
h
′
I
≤
T
≤
1
h
I
<
I
and then
σ
(
T
)
⊆
1
h
′
,
1
h
⊆
(
0
,
1
)
. for
t
∈
1
h
′
,
1
h
, we have
(49)
μ
2
+
(
1
−
μ
)
2
t
≤
max
1
h
′
≤
t
≤
1
h
K
h
1
−
ν
,
2
R
′
(
1
−
ν
)
2
−
2
ν
t
1
−
μ
+
(
1
−
μ
)
2
(
1
+
t
−
2
t
)
+
r
1
+
(
1
−
ν
)
t
−
2
1
−
ν
t
1
4
.
Since the Kantorovich constant is a decreasing function on
(
0
,
1
)
, by (49), we have
(50)
μ
2
+
(
1
−
μ
)
2
t
≤
K
h
1
−
ν
,
2
R
′
(
1
−
ν
)
2
−
2
ν
t
1
−
μ
+
(
1
−
μ
)
2
(
1
+
t
−
2
t
)
+
r
1
+
(
1
−
ν
)
t
−
2
1
−
ν
t
1
4
.
It can be deduced from inequality (50) and Lemma 3.1 that
(51)
μ
2
+
(
1
−
μ
)
2
(
C
∗
)
−
1
B
C
−
1
≤
K
h
1
−
ν
,
2
R
′
(
1
−
ν
)
2
−
2
ν
(
(
C
∗
)
−
1
B
C
−
1
)
1
−
μ
+
(
1
−
μ
)
2
(
I
+
(
C
∗
)
−
1
B
C
−
1
−
2
(
C
∗
)
−
1
B
C
−
1
)
+
r
I
+
(
1
−
ν
)
(
C
∗
)
−
1
B
C
−
1
−
2
1
−
ν
(
(
C
∗
)
−
1
B
C
−
1
)
1
4
.
It is striking that we can obtain the same inequality (48) under the first condition or the second condition one. Next, multiplying inequality (48) or (51) by
C
∗
on the left and
C
on the right, then we can deduce the required inequality (44). The proof is complete.□
4 New matrix versions of Young’s inequalities for the Hilbert-Schmidt norm
Only some interesting matrix versions of Theorem 2.1 for Hilbert-Schmidt norm, unitarily invariant norm, trace norm, and trace are discussed in this section. We need the following lemmas to do this. It is, though, worth noting. For unitarily invariant norms, the first lemma is a Heinz-Kato form inequality.
Lemma 4.1
[12] Let
A
,
B
,
X
∈
M
n
(
C
)
be such that
A
,
B
∈
M
n
+
(
C
)
. If
ν
∈
[
0
,
1
]
, then
∣
∣
∣
A
ν
X
B
1
−
ν
∣
∣
∣
≤
∣
∣
∣
A
X
∣
∣
∣
ν
∣
∣
∣
X
B
∣
∣
∣
1
−
ν
.
In particular,
t
r
∣
A
ν
B
1
−
ν
∣
≤
(
t
r
A
)
ν
(
t
r
B
)
1
−
ν
.
Lemma 4.2
[4] Let
A
,
B
∈
M
n
(
C
)
. Then
∑
j
=
1
n
s
j
(
A
B
)
≤
∑
j
=
1
n
s
j
(
A
)
s
j
(
B
)
.
Lemma 4.3
[24] Let
A
,
B
∈
M
n
(
C
)
be such that A and B are positive definite. If
ν
∈
[
0
,
1
]
, then
∥
A
ν
B
1
−
ν
∥
2
2
≤
∑
j
=
1
n
[
s
j
ν
(
A
)
s
j
1
−
ν
(
B
)
]
2
.
Cauchy-Schwarz’s inequality
Let
a
i
,
b
i
≥
0
(
i
=
1
,
⋯
,
n
)
. Then
∑
j
=
1
n
a
i
b
i
≤
∑
j
=
1
n
a
i
2
1
2
∑
j
=
1
n
b
i
2
1
2
.
Now, we first establish a matrix version of Theorem 2.1 for the Hilbert-Schmidt norm, the proof is based on the spectrum theorem.
Theorem 4.4
Let
A
,
B
,
X
∈
M
n
(
C
)
be such that
A
,
B
∈
M
n
+
(
C
)
and I is the identity matrix, and
0
<
m
I
≤
A
,
B
≤
M
I
and
ν
∈
[
0
,
1
]
. Then we have the following:
If
ν
∈
0
,
1
4
, then
(52)
ν
2
ν
K
(
ν
h
8
,
2
)
r
1
∥
A
ν
X
B
1
−
ν
∥
2
2
+
2
ν
(
1
−
ν
)
A
1
2
X
B
1
2
2
2
+
ν
2
∥
A
X
−
X
B
∥
2
2
+
ν
ν
A
1
2
X
B
1
2
2
2
+
∥
X
B
∥
2
2
−
2
ν
A
1
4
X
B
3
4
2
2
+
2
ν
∥
X
B
∥
2
2
+
ν
A
1
4
X
B
3
4
2
2
−
2
ν
4
A
1
8
X
B
7
8
2
2
+
r
∥
X
B
∥
2
2
+
ν
4
A
1
8
X
B
7
8
2
2
−
2
ν
8
A
1
16
X
B
15
16
2
2
,
≤
∥
ν
A
X
+
(
1
−
ν
)
X
B
∥
2
2
≤
(
1
−
ν
)
2
(
1
−
ν
)
K
h
1
−
ν
8
,
2
−
r
1
∥
A
ν
X
B
1
−
ν
∥
2
2
+
2
ν
(
1
−
ν
)
A
1
2
X
B
1
2
2
2
+
(
1
−
ν
)
2
∥
A
X
−
X
B
∥
2
2
+
(
1
−
ν
)
(
1
−
ν
)
A
1
2
X
B
1
2
2
2
+
∥
A
X
∥
2
2
−
2
1
−
ν
A
3
4
X
B
1
4
2
2
−
2
ν
∥
A
X
∥
2
2
+
1
−
ν
A
3
4
X
B
1
4
2
2
−
2
1
−
ν
4
A
7
8
X
B
1
8
2
2
−
r
∥
A
X
∥
2
2
+
1
−
ν
4
A
7
8
X
B
1
8
2
2
−
2
1
−
ν
8
A
15
16
X
B
1
16
2
2
,
where
h
=
M
m
,
r
=
min
{
4
ν
,
1
−
4
ν
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
If
ν
∈
1
4
,
1
2
, then
(53)
ν
2
ν
K
(
ν
h
8
,
2
)
r
1
∥
A
ν
X
B
1
−
ν
∥
2
2
+
2
ν
(
1
−
ν
)
A
1
2
X
B
1
2
2
2
+
ν
2
∥
A
X
−
X
B
∥
2
2
+
ν
ν
A
1
2
X
B
1
2
2
2
+
∥
X
B
∥
2
2
−
2
ν
A
1
4
X
B
3
4
2
2
+
(
1
−
2
ν
)
∥
X
B
∥
2
2
+
ν
A
1
4
X
B
3
4
2
2
−
2
ν
4
A
1
8
X
B
7
8
2
2
+
r
ν
A
1
4
X
B
3
4
2
2
+
ν
4
A
1
8
X
B
7
8
2
2
−
2
ν
3
8
A
3
16
X
B
13
16
2
2
,
≤
∥
ν
A
X
+
(
1
−
ν
)
X
B
∥
2
2
≤
(
1
−
ν
)
2
(
1
−
ν
)
K
h
1
−
ν
8
,
2
−
r
1
∥
A
ν
X
B
1
−
ν
∥
2
2
+
2
ν
(
1
−
ν
)
A
1
2
X
B
1
2
2
2
+
(
1
−
ν
)
2
∥
A
X
−
X
B
∥
2
2
+
(
1
−
ν
)
(
1
−
ν
)
A
1
2
X
B
1
2
2
2
+
∥
A
X
∥
2
2
−
2
1
−
ν
A
3
4
X
B
1
4
2
2
−
(
1
−
2
ν
)
∥
A
X
∥
2
2
+
1
−
ν
A
3
4
X
B
1
4
2
2
−
2
1
−
ν
4
A
7
8
X
B
1
8
2
2
−
r
1
−
ν
A
3
4
X
B
1
4
2
2
+
1
−
ν
4
A
7
8
X
B
1
8
2
2
−
2
(
1
−
ν
)
3
8
A
13
16
X
B
3
16
2
2
,
where
h
=
M
m
,
r
=
min
{
2
−
4
ν
,
4
ν
−
1
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
If
ν
∈
1
2
,
3
4
, then
(54)
(
1
−
ν
)
2
(
1
−
ν
)
K
h
1
−
ν
8
,
2
r
1
∥
A
ν
X
B
1
−
ν
∥
2
2
+
2
ν
(
1
−
ν
)
A
1
2
X
B
1
2
2
2
+
(
1
−
ν
)
2
∥
A
X
−
X
B
∥
2
2
+
(
1
−
ν
)
(
1
−
ν
)
A
1
2
X
B
1
2
2
2
+
∥
A
X
∥
2
2
−
2
1
−
ν
A
3
4
X
B
1
4
2
2
+
(
1
−
2
ν
)
∥
A
X
∥
2
2
+
1
−
ν
A
3
4
X
B
1
4
2
2
−
2
1
−
ν
4
A
7
8
X
B
1
8
2
2
+
r
1
−
ν
A
3
4
X
B
1
4
2
2
+
1
−
ν
4
A
7
8
X
B
1
8
2
2
−
2
(
1
−
ν
)
3
8
A
13
16
X
B
3
16
2
2
≤
∥
ν
A
X
+
(
1
−
ν
)
X
B
∥
2
2
≤
ν
2
ν
K
(
ν
h
8
,
2
)
−
r
1
∥
A
ν
X
B
1
−
ν
∥
2
2
+
2
ν
(
1
−
ν
)
A
1
2
X
B
1
2
2
2
+
ν
2
∥
A
X
−
X
B
∥
2
2
+
ν
ν
A
1
2
X
B
1
2
2
2
+
∥
X
B
∥
2
2
−
2
ν
A
1
4
X
B
3
4
2
2
−
(
1
−
2
ν
)
∥
X
B
∥
2
2
+
ν
A
1
4
X
B
3
4
2
2
−
2
ν
4
A
1
8
X
B
7
8
2
2
,
−
r
ν
A
1
4
X
B
3
4
2
2
+
ν
4
A
1
8
X
B
7
8
2
2
−
2
ν
3
8
A
3
16
X
B
13
16
2
2
,
where
h
=
M
m
,
r
=
min
{
4
ν
−
2
,
3
−
4
ν
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
If
ν
∈
3
4
,
1
, then
(55)
(
1
−
ν
)
2
(
1
−
ν
)
K
h
1
−
ν
8
,
2
r
1
∥
A
ν
X
B
1
−
ν
∥
2
2
+
2
ν
(
1
−
ν
)
A
1
2
X
B
1
2
2
2
+
(
1
−
ν
)
2
∥
A
X
−
X
B
∥
2
2
+
(
1
−
ν
)
(
1
−
ν
)
A
1
2
X
B
1
2
2
2
+
∥
A
X
∥
2
2
−
2
1
−
ν
A
3
4
X
B
1
4
2
2
+
(
2
−
2
ν
)
∥
A
X
∥
2
2
+
1
−
ν
A
3
4
X
B
1
4
2
2
−
2
1
−
ν
4
A
7
8
X
B
1
8
2
2
+
r
∥
A
X
∥
2
2
+
1
−
ν
4
A
7
8
X
B
1
8
2
2
−
2
1
−
ν
8
A
15
16
X
B
1
16
2
2
≤
∥
ν
A
X
+
(
1
−
ν
)
X
B
∥
2
2
≤
ν
2
ν
K
(
ν
h
8
,
2
)
−
r
1
∥
A
ν
X
B
1
−
ν
∥
2
2
+
2
ν
(
1
−
ν
)
A
1
2
X
B
1
2
2
2
+
ν
2
∥
A
X
−
X
B
∥
2
2
+
ν
ν
A
1
2
X
B
1
2
2
2
+
∥
X
B
∥
2
2
−
2
ν
A
1
4
X
B
3
4
2
2
−
(
2
−
2
ν
)
∥
X
B
∥
2
2
+
ν
A
1
4
X
B
3
4
2
2
−
2
ν
4
A
1
8
X
B
7
8
2
2
−
r
∥
X
B
∥
2
2
+
ν
4
A
1
8
X
B
7
8
2
2
−
2
ν
8
A
1
16
X
B
15
16
2
2
,
where
h
=
M
m
,
r
=
min
{
4
ν
−
3
,
4
−
4
ν
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
Proof
Since every positive semi-definite is unitarily diagonalizable, it follows by spectral theorem that there are unitary matrices
U
,
V
∈
M
n
(
C
)
such that
A
=
U
D
1
U
∗
and
B
=
V
D
2
V
∗
, where
D
1
=
diag
(
λ
1
,
⋯
,
λ
n
)
,
D
2
=
diag
(
μ
1
,
⋯
,
μ
n
)
, and
λ
i
,
μ
i
≥
0
(
i
=
1
,
⋯
,
n
)
. Let
Y
=
U
∗
X
V
=
[
y
i
j
]
.
ν
A
X
+
(
1
−
ν
)
X
B
=
U
(
(
ν
λ
i
+
(
1
−
ν
)
μ
j
)
y
i
j
)
V
∗
,
A
X
−
X
B
=
U
[
(
λ
i
−
μ
j
)
y
i
j
]
V
∗
,
A
ν
X
B
1
−
ν
=
U
[
(
λ
i
ν
μ
j
1
−
ν
)
y
i
j
]
V
∗
,
A
X
=
U
(
λ
i
y
i
j
)
V
∗
and
X
B
=
U
∗
(
μ
j
y
i
j
)
V
∗
.
(a) If
0
≤
ν
≤
1
2
, then by inequality (28) and the unitary invariance of the Hilbert-Schmidt norm, we have
∥
ν
A
X
+
(
1
−
ν
)
X
B
∥
2
2
=
∑
i
,
j
=
1
n
(
ν
λ
i
+
(
1
−
ν
)
μ
j
)
2
∣
y
i
j
∣
2
≥
ν
2
ν
∑
i
,
j
=
1
n
min
1
≤
i
,
j
≤
n
K
(
ν
t
i
j
8
,
2
)
r
1
[
λ
i
ν
μ
j
1
−
ν
]
2
∣
y
i
j
∣
2
+
2
ν
(
1
−
ν
)
∑
i
,
j
=
1
n
λ
i
1
2
μ
j
1
2
2
∣
y
i
j
∣
2
+
ν
2
∑
i
,
j
=
1
n
(
λ
i
−
μ
j
)
2
∣
y
i
j
∣
2
+
ν
∑
i
,
j
=
1
n
ν
λ
i
μ
j
+
μ
j
2
−
2
ν
λ
i
1
2
μ
j
3
2
∣
y
i
j
∣
2
+
2
ν
∑
i
,
j
=
1
n
μ
j
2
+
ν
λ
i
1
2
μ
j
3
2
−
2
ν
4
λ
i
1
4
μ
j
7
4
∣
y
i
j
∣
2
+
r
1
∑
i
,
j
=
1
n
μ
j
2
+
ν
4
λ
i
1
4
μ
j
7
4
−
2
ν
8
λ
i
1
8
μ
j
15
8
∣
y
i
j
∣
2
,
where
t
i
j
=
λ
i
μ
j
. Utilizing the condition
0
<
m
I
≤
A
,
B
≤
M
I
,
m
M
=
1
h
≤
t
i
j
=
λ
i
μ
j
≤
h
=
M
m
and the property of the Kantorovich constant, we have
∥
ν
A
X
+
(
1
−
ν
)
X
B
∥
2
2
≥
ν
2
ν
K
(
ν
h
8
,
2
)
r
1
∥
A
ν
X
B
1
−
ν
∥
2
2
+
2
ν
(
1
−
ν
)
A
1
2
X
B
1
2
2
2
+
ν
2
∥
A
X
−
X
B
∥
2
2
+
ν
ν
A
1
2
X
B
1
2
2
2
+
∥
X
B
∥
2
2
−
2
ν
A
1
4
X
B
3
4
2
2
+
2
ν
∥
X
B
∥
2
2
+
ν
A
1
4
X
B
3
4
2
2
−
2
ν
4
A
1
8
X
B
7
8
2
2
+
r
1
∥
X
B
∥
2
2
+
ν
4
A
1
8
X
B
7
8
2
2
−
2
ν
8
A
1
16
X
B
15
16
2
2
and
∥
ν
A
X
+
(
1
−
ν
)
X
B
∥
2
2
=
∑
i
,
j
=
1
n
(
ν
λ
i
+
(
1
−
ν
)
μ
j
)
2
∣
y
i
j
∣
2
≤
(
1
−
ν
)
2
(
1
−
ν
)
∑
i
,
j
=
1
n
max
1
≤
i
,
j
≤
n
K
(
ν
t
i
j
8
,
2
)
−
r
1
[
λ
i
ν
μ
j
1
−
ν
]
2
∣
y
i
j
∣
2
+
2
ν
(
1
−
ν
)
∑
i
,
j
=
1
n
λ
i
1
2
μ
j
1
2
2
∣
y
i
j
∣
2
+
(
1
−
ν
)
2
∑
i
,
j
=
1
n
(
λ
i
−
μ
j
)
2
∣
y
i
j
∣
2
+
(
1
−
ν
)
∑
i
,
j
=
1
n
(
1
−
ν
)
λ
i
μ
j
+
λ
i
2
−
2
1
−
ν
λ
i
3
2
μ
j
1
2
∣
y
i
j
∣
2
−
2
ν
∑
i
,
j
=
1
n
λ
i
2
+
1
−
ν
λ
i
3
2
μ
j
1
2
−
2
1
−
ν
4
λ
i
7
4
μ
j
1
4
∣
y
i
j
∣
2
−
r
1
∑
i
,
j
=
1
n
λ
i
2
+
1
−
ν
4
λ
i
7
4
μ
j
1
4
−
2
1
−
ν
8
λ
i
15
8
μ
j
1
8
∣
y
i
j
∣
2
,
where
t
i
j
=
λ
i
μ
j
. Utilizing the condition
0
<
m
I
≤
A
,
B
≤
M
I
,
m
M
=
1
h
≤
t
i
j
=
λ
i
μ
j
≤
h
=
M
m
and the property of the Kantorovich constant, we have
∥
ν
A
X
+
(
1
−
ν
)
X
B
∥
2
2
≥
(
1
−
ν
)
2
(
1
−
ν
)
K
h
1
−
ν
8
,
2
−
r
1
∥
A
ν
X
B
1
−
ν
∥
2
2
+
2
ν
(
1
−
ν
)
A
1
2
X
B
1
2
2
2
+
(
1
−
ν
)
2
∥
A
X
−
X
B
∥
2
2
+
(
1
−
ν
)
(
1
−
ν
)
A
1
2
X
B
1
2
2
2
+
∥
A
X
∥
2
2
−
2
1
−
ν
A
3
4
X
B
1
4
2
2
−
2
ν
∥
A
X
∥
2
2
+
1
−
ν
A
3
4
X
B
1
4
2
2
−
2
1
−
ν
4
A
7
8
X
B
1
8
2
2
−
r
1
∥
A
X
∥
2
2
+
1
−
ν
4
A
7
8
X
B
1
8
2
2
−
2
1
−
ν
8
A
15
16
X
B
1
16
2
2
.
So we completed the proof of (52). The proof of the remaining cases are similar to the proof of case (a), so we omit it.□
The following results for unitarily invariant norm and trace are established by Lemma 4.1.
Theorem 4.5
Let
A
,
B
,
X
∈
M
n
(
C
)
be such that
A
,
B
∈
M
n
+
(
C
)
,
I
is the identity matrix, and
0
<
m
I
≤
A
,
B
≤
M
I
, and
ν
∈
[
0
,
1
]
. Then for every unitarily invariant norm
∣
∣
∣
.
∣
∣
∣
, we have
If
ν
∈
0
,
1
4
, then
(56)
[
ν
∣
∣
∣
A
x
∣
∣
∣
+
(
1
−
ν
)
∣
∣
∣
X
B
∣
∣
∣
]
2
≥
ν
2
ν
K
[
ν
h
8
,
2
]
r
1
∣
∣
∣
A
ν
X
B
1
−
ν
∣
∣
∣
2
+
2
ν
(
1
−
ν
)
A
1
2
X
B
1
2
2
+
ν
2
(
∣
∣
∣
A
X
∣
∣
∣
−
∣
∣
∣
X
B
∣
∣
∣
)
2
+
ν
∣
∣
∣
X
B
∣
∣
∣
(
ν
∣
∣
∣
A
X
∣
∣
∣
−
∣
∣
∣
X
B
∣
∣
∣
)
2
+
2
ν
∣
∣
∣
X
B
∣
∣
∣
(
ν
∣
∣
∣
A
X
∣
∣
∣
∣
∣
∣
X
B
∣
∣
∣
4
−
∣
∣
∣
X
B
∣
∣
∣
)
2
+
r
∣
∣
∣
X
B
∣
∣
∣
(
∣
∣
∣
X
B
∣
∣
∣
−
ν
∣
∣
∣
A
X
∣
∣
∣
∣
∣
∣
X
B
∣
∣
∣
3
8
)
2
,
where
h
=
M
m
,
r
=
min
{
4
ν
,
1
−
4
ν
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
If
ν
∈
1
4
,
1
2
, then
(57)
[
ν
∣
∣
∣
A
x
∣
∣
∣
+
(
1
−
ν
)
∣
∣
∣
X
B
∣
∣
∣
]
2
≥
ν
2
ν
K
[
ν
h
8
,
2
]
r
1
∣
∣
∣
A
ν
X
B
1
−
ν
∣
∣
∣
2
+
2
ν
(
1
−
ν
)
A
1
2
X
B
1
2
2
+
ν
2
(
∣
∣
∣
A
X
∣
∣
∣
−
∣
∣
∣
X
B
∣
∣
∣
)
2
+
ν
∣
∣
∣
X
B
∣
∣
∣
(
ν
∣
∣
∣
A
X
∣
∣
∣
−
∣
∣
∣
X
B
∣
∣
∣
)
2
+
(
1
−
2
ν
)
∣
∣
∣
X
B
∣
∣
∣
(
ν
∣
∣
∣
A
X
∣
∣
∣
∣
∣
∣
X
B
∣
∣
∣
4
−
∣
∣
∣
X
B
∣
∣
∣
)
2
+
r
∣
∣
∣
X
B
∣
∣
∣
(
ν
∣
∣
∣
A
X
∣
∣
∣
∣
∣
∣
X
B
∣
∣
∣
4
−
ν
∣
∣
∣
A
X
∣
∣
∣
∣
∣
∣
X
B
∣
∣
∣
3
8
)
2
,
where
h
=
M
m
,
r
=
min
{
2
−
4
ν
,
4
ν
−
1
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
If
ν
∈
1
2
,
3
4
, then
(58)
[
ν
∣
∣
∣
A
x
∣
∣
∣
+
(
1
−
ν
)
∣
∣
∣
X
B
∣
∣
∣
]
2
≥
(
1
−
ν
)
2
(
1
−
ν
)
K
h
1
−
ν
8
,
2
r
1
∣
∣
∣
A
ν
X
B
1
−
ν
∣
∣
∣
2
+
2
ν
(
1
−
ν
)
A
1
2
X
B
1
2
2
+
(
1
−
ν
)
2
(
∣
∣
∣
A
X
∣
∣
∣
−
∣
∣
∣
X
B
∣
∣
∣
)
2
+
(
1
−
ν
)
∣
∣
∣
A
X
∣
∣
∣
(
∣
∣
∣
A
X
∣
∣
∣
−
(
1
−
ν
)
∣
∣
∣
X
B
∣
∣
∣
)
2
+
(
1
−
2
ν
)
∣
∣
∣
A
X
∣
∣
∣
(
(
1
−
ν
)
∣
∣
∣
A
X
∣
∣
∣
∣
∣
∣
X
B
∣
∣
∣
4
−
∣
∣
∣
A
X
∣
∣
∣
)
2
+
r
∣
∣
∣
A
X
∣
∣
∣
(
(
1
−
ν
)
∣
∣
∣
A
X
∣
∣
∣
∣
∣
∣
X
B
∣
∣
∣
4
−
(
1
−
ν
)
∣
∣
∣
A
X
∣
∣
∣
3
∣
∣
∣
X
B
∣
∣
∣
8
)
,
where
h
=
M
m
,
r
=
min
{
4
ν
−
2
,
3
−
4
ν
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
If
ν
∈
3
4
,
1
, then
(59)
[
ν
∣
∣
∣
A
x
∣
∣
∣
+
(
1
−
ν
)
∣
∣
∣
X
B
∣
∣
∣
]
2
≥
(
1
−
ν
)
2
(
1
−
ν
)
K
h
1
−
ν
8
,
2
r
1
∣
∣
∣
A
ν
X
B
1
−
ν
∣
∣
∣
2
+
2
ν
(
1
−
ν
)
A
1
2
X
B
1
2
2
+
(
1
−
ν
)
2
(
∣
∣
∣
A
X
∣
∣
∣
−
∣
∣
∣
X
B
∣
∣
∣
)
2
+
(
1
−
ν
)
∣
∣
∣
A
X
∣
∣
∣
(
∣
∣
∣
A
X
∣
∣
∣
−
(
1
−
ν
)
∣
∣
∣
X
B
∣
∣
∣
)
2
+
(
2
−
2
ν
)
∣
∣
∣
A
X
∣
∣
∣
(
(
1
−
ν
)
∣
∣
∣
A
X
∣
∣
∣
∣
∣
∣
X
B
∣
∣
∣
4
−
∣
∣
∣
A
X
∣
∣
∣
)
2
+
r
∣
∣
∣
A
X
∣
∣
∣
(
∣
∣
∣
A
X
∣
∣
∣
−
(
1
−
ν
)
∣
∣
∣
A
X
∣
∣
∣
3
∣
∣
∣
X
B
∣
∣
∣
8
)
,
where
h
=
M
m
,
r
=
min
{
4
−
4
ν
,
4
ν
−
3
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
Proof
(i) Suppose that
ν
∈
0
,
1
4
. By Lemma 4.1 and the left inequality of (28), we have
[
ν
∣
∣
∣
A
x
∣
∣
∣
+
(
1
−
ν
)
∣
∣
∣
X
B
∣
∣
∣
]
2
≥
ν
2
ν
K
[
ν
h
8
,
2
]
r
1
(
∣
∣
∣
A
X
∣
∣
∣
ν
∣
∣
∣
X
B
∣
∣
∣
1
−
ν
)
2
+
2
ν
(
1
−
ν
)
∣
∣
∣
A
X
∣
∣
∣
∣
∣
∣
X
B
∣
∣
∣
+
ν
2
(
∣
∣
∣
A
X
∣
∣
∣
−
∣
∣
∣
X
B
∣
∣
∣
)
2
+
ν
∣
∣
∣
X
B
∣
∣
∣
(
ν
∣
∣
∣
A
X
∣
∣
∣
−
∣
∣
∣
X
B
∣
∣
∣
)
2
+
2
ν
∣
∣
∣
X
B
∣
∣
∣
(
ν
∣
∣
∣
A
X
∣
∣
∣
∣
∣
∣
X
B
∣
∣
∣
4
−
∣
∣
∣
X
B
∣
∣
∣
)
2
+
r
∣
∣
∣
X
B
∣
∣
∣
(
∣
∣
∣
X
B
∣
∣
∣
−
ν
∣
∣
∣
A
X
∣
∣
∣
∣
∣
∣
X
B
∣
∣
∣
3
8
)
2
≥
ν
2
ν
K
[
ν
h
8
,
2
]
r
1
∣
∣
∣
A
ν
X
B
1
−
ν
∣
∣
∣
2
+
2
ν
(
1
−
ν
)
A
1
2
X
B
1
2
2
+
ν
2
(
∣
∣
∣
A
X
∣
∣
∣
−
∣
∣
∣
X
B
∣
∣
∣
)
2
+
ν
∣
∣
∣
X
B
∣
∣
∣
(
ν
∣
∣
∣
A
X
∣
∣
∣
−
∣
∣
∣
X
B
∣
∣
∣
)
2
+
2
ν
∣
∣
∣
X
B
∣
∣
∣
(
ν
∣
∣
∣
A
X
∣
∣
∣
∣
∣
∣
X
B
∣
∣
∣
4
−
∣
∣
∣
X
B
∣
∣
∣
)
2
+
r
∣
∣
∣
X
B
∣
∣
∣
(
∣
∣
∣
X
B
∣
∣
∣
−
ν
∣
∣
∣
A
X
∣
∣
∣
∣
∣
∣
X
B
∣
∣
∣
3
8
)
2
.
In this way, we have completed the proof of (56). Statements (ii), (iii), and (iv) are similar to the proofs presented in (i). By using Lemma 4.1 and the left inequalities (29), (30), and (31), respectively, we have omitted them.□
Theorem 4.6
Let
A
,
B
,
X
∈
M
n
(
C
)
be such that
A
,
B
∈
M
n
+
(
C
)
,
I
is the identity matrix, and
0
<
m
I
≤
A
,
B
≤
M
I
, and
ν
∈
[
0
,
1
]
. It holds that
If
ν
∈
0
,
1
4
, then
(60)
[
t
r
(
A
∇
1
−
ν
B
)
]
2
≥
ν
2
ν
K
[
ν
h
8
,
2
]
r
1
[
t
r
∣
A
ν
B
1
−
ν
∣
]
2
+
2
ν
(
1
−
ν
)
t
r
∣
A
1
2
B
1
2
∣
2
+
ν
2
(
t
r
A
−
t
r
B
)
2
+
ν
t
r
(
B
)
(
ν
t
r
(
A
)
−
t
r
(
B
)
)
2
+
2
ν
t
r
(
B
)
(
ν
t
r
(
A
)
t
r
(
B
)
4
−
t
r
(
B
)
)
2
+
r
t
r
(
B
)
(
t
r
(
B
)
−
ν
t
r
(
A
)
t
r
(
B
)
3
8
)
2
,
where
h
=
M
m
,
r
=
min
{
4
ν
,
1
−
4
ν
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
If
ν
∈
1
4
,
1
2
, then
(61)
[
t
r
(
A
∇
1
−
ν
B
)
]
2
≥
ν
2
ν
K
[
ν
h
8
,
2
]
r
1
[
t
r
∣
A
ν
B
1
−
ν
∣
]
2
+
2
ν
(
1
−
ν
)
t
r
∣
A
1
2
B
1
2
∣
2
+
ν
2
(
t
r
A
−
t
r
B
)
2
+
ν
t
r
(
B
)
(
ν
t
r
(
A
)
−
t
r
(
B
)
)
2
+
(
1
−
2
ν
)
t
r
(
B
)
(
ν
t
r
(
A
)
t
r
(
B
)
4
−
t
r
(
B
)
)
2
+
r
t
r
(
B
)
(
ν
t
r
(
A
)
t
r
(
B
)
4
−
ν
t
r
(
A
)
t
r
(
B
)
3
8
)
2
,
where
h
=
M
m
,
r
=
min
{
2
−
4
ν
,
4
ν
−
1
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
If
ν
∈
1
2
,
3
4
, then
(62)
[
t
r
(
A
∇
1
−
ν
B
)
]
2
≥
(
1
−
ν
)
2
(
1
−
ν
)
K
h
1
−
ν
8
,
2
r
1
[
t
r
∣
A
ν
B
1
−
ν
∣
]
2
+
2
ν
(
1
−
ν
)
t
r
∣
A
1
2
B
1
2
∣
2
+
(
1
−
ν
)
2
(
t
r
A
−
t
r
B
)
2
+
(
1
−
ν
)
t
r
(
A
)
(
t
r
(
A
)
−
(
1
−
ν
)
t
r
(
B
)
)
2
+
(
1
−
2
ν
)
t
r
(
A
)
(
(
1
−
ν
)
t
r
(
A
)
t
r
(
B
)
4
−
t
r
(
A
)
)
2
+
r
t
r
(
A
)
(
(
1
−
ν
)
t
r
(
A
)
t
r
(
B
)
4
−
(
1
−
ν
)
t
r
(
A
)
3
t
r
(
B
)
8
)
2
,
where
h
=
M
m
,
r
=
min
{
4
ν
−
2
,
3
−
4
ν
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
If
ν
∈
3
4
,
1
, then
(63)
[
t
r
(
A
∇
1
−
ν
B
)
]
2
≥
(
1
−
ν
)
2
(
1
−
ν
)
K
h
1
−
ν
8
,
2
r
1
[
t
r
∣
A
ν
B
1
−
ν
∣
]
2
+
2
ν
(
1
−
ν
)
t
r
∣
A
1
2
B
1
2
∣
2
+
(
1
−
ν
)
2
(
t
r
A
−
t
r
B
)
2
+
(
1
−
ν
)
t
r
(
A
)
(
t
r
(
A
)
−
(
1
−
ν
)
t
r
(
B
)
)
2
+
(
2
−
2
ν
)
t
r
(
A
)
(
(
1
−
ν
)
t
r
(
A
)
t
r
(
B
)
4
−
t
r
(
A
)
)
2
+
r
t
r
(
A
)
(
t
r
(
A
)
−
(
1
−
ν
)
t
r
(
A
)
3
t
r
(
B
)
8
)
where
h
=
M
m
,
r
=
min
{
4
−
4
ν
,
4
ν
−
3
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
Proof
(i) Assume that
ν
∈
0
,
1
4
. By Lemma 4.1 and the left inequality of (28), we have
[
t
r
(
A
∇
1
−
ν
B
)
]
2
=
[
ν
t
r
(
A
)
+
(
1
−
ν
)
t
r
(
B
)
]
2
≥
ν
2
ν
K
[
ν
h
8
,
2
]
r
1
[
(
t
r
(
A
)
)
ν
(
t
r
(
B
)
)
1
−
ν
]
+
ν
2
(
t
r
A
−
t
r
B
)
2
+
2
ν
(
1
−
ν
)
t
r
(
A
)
t
r
(
B
)
+
ν
t
r
(
B
)
(
ν
t
r
(
A
)
−
t
r
(
B
)
)
2
+
2
ν
t
r
(
B
)
(
ν
t
r
(
A
)
t
r
(
B
)
4
−
t
r
(
B
)
)
2
+
r
t
r
(
B
)
(
t
r
(
B
)
−
ν
t
r
(
A
)
t
r
(
B
)
3
8
)
2
≥
ν
2
ν
K
[
ν
h
8
,
2
]
r
1
[
t
r
∣
A
ν
B
1
−
ν
∣
]
2
+
2
ν
(
1
−
ν
)
t
r
∣
A
1
2
B
1
2
∣
2
+
ν
2
(
t
r
A
−
t
r
B
)
2
+
ν
t
r
(
B
)
(
ν
t
r
(
A
)
−
t
r
(
B
)
)
2
+
2
ν
t
r
(
B
)
(
ν
t
r
(
A
)
t
r
(
B
)
4
−
t
r
(
B
)
)
2
+
r
t
r
(
B
)
(
t
r
(
B
)
−
ν
t
r
(
A
)
t
r
(
B
)
3
8
)
2
.
In this way, we have completed the proof of (60). Statements (ii), (iii), and (iv) are similar to the proofs presented in (i). By using Lemma 4.1 and the left inequalities (29), (30), and (31), respectively, we have omitted them.□
Also, we obtain the following result which gives refinements of the trace norm version of Young-type inequality.
Theorem 4.7
Let
A
,
B
,
X
∈
M
n
(
C
)
be such that
A
,
B
∈
M
n
+
(
C
)
,
I
is the identity matrix, and
0
<
m
I
≤
A
,
B
≤
M
I
, and
ν
∈
[
0
,
1
]
. It holds that
If
ν
∈
0
,
1
4
, then
(64)
ν
2
∥
A
∥
2
2
+
(
1
−
ν
)
2
∥
B
∥
2
2
=
t
r
(
ν
2
A
2
+
(
1
−
ν
)
2
B
2
)
≥
ν
2
K
[
ν
h
8
,
2
]
r
1
∥
A
ν
B
1
−
ν
∥
2
2
+
ν
2
[
∥
A
∥
2
−
∥
B
∥
2
]
2
+
ν
[
ν
∥
A
B
∥
1
+
∥
B
∥
2
2
−
2
ν
∥
A
∥
1
∥
B
3
∥
1
]
+
2
ν
ν
A
1
2
B
3
2
1
+
∥
B
∥
2
2
−
2
ν
4
A
1
2
1
B
7
2
1
+
r
∥
B
∥
2
2
+
ν
4
A
1
4
B
7
4
1
−
2
ν
8
A
1
4
1
B
15
4
1
,
where
h
=
M
m
,
r
=
min
{
4
ν
,
1
−
4
ν
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
If
ν
∈
1
4
,
1
2
, then
(65)
ν
2
∥
A
∥
2
2
+
(
1
−
ν
)
2
∥
B
∥
2
2
=
t
r
(
ν
2
A
2
+
(
1
−
ν
)
2
B
2
)
≥
ν
2
K
[
ν
h
8
,
2
]
r
1
∥
A
ν
B
1
−
ν
∥
2
2
+
ν
2
[
∥
A
∥
2
−
∥
B
∥
2
]
2
+
ν
[
ν
∥
A
B
∥
1
+
∥
B
∥
2
2
−
2
ν
∥
A
∥
1
∥
B
3
∥
1
]
+
(
1
−
2
ν
)
ν
A
1
2
B
3
2
1
+
∥
B
∥
2
2
−
2
ν
4
A
1
2
1
B
7
2
1
+
r
ν
A
1
2
B
3
2
1
+
A
1
4
B
7
4
1
4
−
2
ν
3
8
A
3
4
1
B
13
4
1
,
where
h
=
M
m
,
r
=
min
{
2
−
4
ν
,
4
ν
−
1
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
If
ν
∈
1
2
,
3
4
, then
(66)
ν
2
∥
A
∥
2
2
+
(
1
−
ν
)
2
∥
B
∥
2
2
=
t
r
(
ν
2
A
2
+
(
1
−
ν
)
2
B
2
)
≥
(
1
−
ν
)
2
(
1
−
ν
)
K
h
1
−
ν
8
,
2
r
1
∥
A
ν
B
1
−
ν
∥
2
2
+
(
1
−
ν
)
2
[
∥
A
∥
2
−
∥
B
∥
2
]
2
+
(
1
−
ν
)
[
(
1
−
ν
)
∥
A
B
∥
1
+
∥
A
∥
2
2
−
2
1
−
ν
∥
A
3
∥
1
∥
B
∥
1
]
+
(
1
−
2
ν
)
1
−
ν
B
1
2
A
3
2
1
+
∥
A
∥
2
2
−
2
ν
4
B
1
2
1
A
7
2
1
+
r
1
−
ν
B
1
2
A
3
2
1
+
B
1
4
A
7
4
1
4
−
2
(
1
−
ν
)
3
8
B
3
4
1
A
13
4
1
,
where
h
=
M
m
,
r
=
min
{
4
ν
−
2
,
3
−
4
ν
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
If
ν
∈
3
4
,
1
, then
(67)
ν
2
∥
A
∥
2
2
+
(
1
−
ν
)
2
∥
B
∥
2
2
=
t
r
(
ν
2
A
2
+
(
1
−
ν
)
2
B
2
)
≥
(
1
−
ν
)
2
(
1
−
ν
)
K
h
1
−
ν
8
,
2
r
1
∥
A
ν
B
1
−
ν
∥
2
2
+
(
1
−
ν
)
2
[
∥
A
∥
2
−
∥
B
∥
2
]
2
+
(
1
−
ν
)
[
(
1
−
ν
)
∥
A
B
∥
1
+
∥
A
∥
2
2
−
2
1
−
ν
∥
A
3
∥
1
∥
B
∥
1
]
+
(
2
−
2
ν
)
1
−
ν
B
1
2
A
3
2
1
+
∥
A
∥
2
2
−
2
1
−
ν
4
B
1
2
1
A
7
2
1
+
r
∥
A
∥
2
2
+
1
−
ν
4
B
1
4
A
7
4
1
−
2
1
−
ν
8
B
1
4
1
A
15
4
1
,
where
h
=
M
m
,
r
=
min
{
4
−
4
ν
,
4
ν
−
3
}
, and
r
1
=
min
{
2
r
,
1
−
2
r
}
.
Proof
We shall prove the first inequality and leave the others to the reader because the proof is similar to each other.
(i) Suppose that
ν
∈
0
,
1
4
, then using Lemmas 4.2 and 4.3, Cauchy-Schwarz inequality, and the left inequality of (28), we have
ν
2
∥
A
∥
2
2
+
(
1
−
ν
)
2
∥
B
∥
2
2
=
∑
j
=
1
n
[
ν
2
s
j
2
(
A
)
+
(
1
−
ν
)
2
s
j
2
(
B
)
]
≥
∑
j
=
1
n
min
K
[
ν
t
j
8
,
2
]
r
1
[
(
ν
s
j
(
A
)
)
2
ν
(
s
j
(
B
)
)
2
−
2
ν
]
+
ν
2
(
s
j
(
A
)
−
s
j
(
B
)
)
2
×
ν
s
j
(
B
)
[
ν
s
j
(
A
)
−
s
j
(
B
)
]
2
+
2
ν
s
j
(
B
)
[
ν
s
j
(
A
)
s
j
(
B
)
4
−
s
j
(
B
)
]
2
+
r
s
j
(
B
)
[
s
j
(
B
)
−
ν
s
j
(
A
)
s
j
3
(
B
)
8
]
2
,
where
t
j
=
s
j
(
A
)
s
j
(
B
)
. Utilizing the condition
0
<
m
I
≤
A
,
B
≤
M
I
,
m
M
=
1
h
≤
t
j
=
s
j
(
A
)
s
j
(
B
)
≤
h
=
M
m
and the property of the Kantorovich constant, we have
ν
2
∥
A
∥
2
2
+
(
1
−
ν
)
2
∥
B
∥
2
2
≥
ν
2
K
[
ν
h
8
,
2
]
r
1
∑
j
=
1
n
[
s
j
(
A
)
ν
s
j
1
−
ν
(
B
)
]
2
+
ν
2
∑
j
=
1
n
s
j
2
(
A
)
+
∑
j
=
1
n
s
j
2
(
B
)
−
2
∑
j
=
1
n
s
j
(
A
)
s
j
(
B
)
+
ν
ν
∑
j
=
1
n
s
j
(
A
)
s
j
(
B
)
+
∑
j
=
1
n
s
j
2
(
B
)
−
2
ν
∑
j
=
1
n
s
j
1
2
(
A
)
s
j
3
2
(
B
)
+
2
ν
ν
∑
j
=
1
n
s
j
1
2
(
A
)
s
j
3
2
(
B
)
+
∑
j
=
1
n
s
j
2
(
B
)
−
2
ν
4
∑
j
=
1
n
s
j
1
4
(
A
)
s
j
7
4
(
B
)
+
r
∑
j
=
1
n
s
j
2
(
B
)
−
2
ν
4
∑
j
=
1
n
s
j
1
4
(
A
)
s
j
7
4
(
B
)
−
2
ν
8
∑
j
=
1
n
s
j
1
8
(
A
)
s
j
15
8
(
B
)
≥
ν
2
K
[
ν
h
8
,
2
]
r
1
∑
j
=
1
n
[
s
j
(
A
)
ν
s
j
1
−
ν
(
B
)
]
2
+
ν
2
∑
j
=
1
n
s
j
2
(
A
)
+
∑
j
=
1
n
s
j
2
(
B
)
−
2
∑
j
=
1
n
s
j
2
(
A
)
1
2
∑
j
=
1
n
s
j
2
(
B
)
1
2
+
ν
ν
∑
j
=
1
n
s
j
(
A
B
)
+
∑
j
=
1
n
s
j
2
(
B
)
−
2
ν
∑
j
=
1
n
s
j
(
A
)
1
2
∑
j
=
1
n
s
j
3
(
B
)
1
2
+
2
ν
ν
∑
j
=
1
n
s
j
A
1
2
B
3
2
+
∑
j
=
1
n
s
j
2
(
B
)
−
2
ν
4
∑
j
=
1
n
s
j
1
2
(
A
)
1
2
∑
j
=
1
n
s
j
7
2
(
B
)
1
2
+
r
∑
j
=
1
n
s
j
2
(
B
)
+
ν
4
∑
j
=
1
n
s
j
A
1
4
B
7
4
−
2
ν
8
∑
j
=
1
n
s
j
1
4
(
A
)
1
2
∑
j
=
1
n
s
j
15
4
(
B
)
1
2
≥
ν
2
K
[
ν
h
8
,
2
]
r
1
∥
A
ν
B
1
−
ν
∥
2
2
+
ν
2
[
∥
A
∥
2
−
∥
B
∥
2
]
2
+
ν
[
ν
∥
A
B
∥
1
+
∥
B
∥
2
2
−
2
ν
∥
A
∥
1
∥
B
3
∥
1
]
+
2
ν
ν
A
1
2
B
3
2
1
+
∥
B
∥
2
2
−
2
ν
4
A
1
2
1
B
7
2
1
+
r
∥
B
∥
2
2
+
ν
4
A
1
4
B
7
4
1
−
2
ν
8
A
1
4
1
B
15
4
1
.
Furthermore, since
t
r
(
ν
2
A
2
+
(
1
−
ν
)
2
B
2
)
=
ν
2
t
r
(
A
2
)
+
(
1
−
ν
)
2
t
r
(
B
2
)
=
ν
2
∑
j
=
1
n
s
j
2
(
A
)
+
(
1
−
ν
)
2
∑
j
=
1
n
s
j
2
(
B
)
=
ν
2
∥
A
∥
2
2
+
(
1
−
ν
)
2
∥
B
∥
2
2
.
This completes the proof.□
Remark 4.8
It is obvious that by the property of the Kantorovich constant that inequalities (64)–(67) are refinements of the well-known results in [24] and [15].
