This paper deals with the Choi’s inequality for measurable operators affiliated with a given von Neumann algebra. Some Young and Cauchy-Schwarz type inequalities for  -measurable operators are also given

H. R. Moradi

M. E. Omidvar

S. M. Davarpanah

Issues of Analysis

Directory of Open Access Journals

52 Probl. Anal. Issues Anal. Vol. 9 (27), No 1, 2020, pp. 52–59
DOI: 10.15393/j3.art.2020.6490
UDC 517.38, 512.6
S. M. Davarpanah, M. E. Omidvar, H. R. Moradi
ON SOME INEQUALITIES FOR 𝜏-MEASURABLE
OPERATORS
Abstract. This paper deals with the Choi’s inequality for measur-
able operators affiliated with a given von Neumann algebra. Some
Young and Cauchy-Schwarz type inequalities for 𝜏 -measurable op-
erators are also given.
Key words: von Neumann algebra, positive operator, noncommu-
tative 𝐿𝑝-space, Young inequality, Cauchy-Schwarz inequality
2010 Mathematical Subject Classification: Primary 47A30,
Secondary 47L05, 47L50
1. Introduction and Preliminaries. Throughout the paper, we
denote by ℳ a semi-finite von Neumann algebra acting on the Hilbert
space ℋ, with a normal faithful semi-finite trace 𝜏 . We denote the identity
in ℳ by 1 and let 𝒫 denote the projection lattice of ℳ. We write 𝑝 ∼ 𝑞
for 𝑝, 𝑞 ∈ 𝒫 if 𝑝 = 𝑢*𝑢 and 𝑞 = 𝑢𝑢* for some 𝑢 ∈ ℳ. A closed densely
defined linear operator 𝑥 in ℋ with the domain 𝐷 (𝑥) ⊆ ℋ is said to
be affiliated with ℳ if 𝑢*𝑥𝑢 = 𝑥 for all unitary 𝑢 that belong to the
commutant ℳ′ of ℳ. If 𝑥 is affiliated with ℳ, then 𝑥 is said to be 𝜏 -
measurable if for every 𝜀 > 0 there exists a projection 𝑒 ∈ ℳ such that
𝑒 (ℋ) ⊆ 𝐷 (𝑥) and 𝜏 (1 − 𝑒) < 𝜀. The set of all 𝜏 -measurable operators will
be denoted by 𝐿0 (ℳ,𝜏), or, simply, 𝐿0 (ℳ). The set 𝐿0 (ℳ) is a *-algebra
with sum and product being the respective closures of the algebraic sum
and product; see [7]. A closed densely defined linear operator 𝑥 admits
a unique polar decomposition 𝑥 = 𝑢 |𝑥|, where 𝑢 is a partial isometry
such that 𝑢*𝑢 = (ker𝑥)⊥ and 𝑢𝑢* = im𝑥 (with im𝑥 = 𝑥 (𝐷 (𝑥)). We
call 𝑟 (𝑥) = (ker𝑥)⊥ and 𝑙 (𝑥) = im𝑥 the left and right supports of 𝑥,
respectively. Thus, 𝑙 (𝑥) ∼ 𝑟 (𝑥). Note that 𝑙 (𝑥) (resp., 𝑟 (𝑥)) is the least
projection 𝑒 such that 𝑒𝑥 = 𝑥 (resp., 𝑥𝑒 = 𝑥). If 𝑥 is self-adjoint, then
𝑟 (𝑥) = 𝑙 (𝑥). This common projection is then said to be the support of 𝑥
and denoted by 𝑠 (𝑥). For further details, we see [8].
© Petrozavodsk State University, 2020
On some inequalities for 𝜏 -measurable operators 53
Let ℳ+ be the positive part of ℳ. Set 𝑆+(ℳ) = {𝑥 ∈ ℳ+ :
𝜏(𝑠(𝑥)) < ∞} and let 𝑆 (ℳ) be the linear span of 𝑆+ (𝑀). Let
0 < 𝑝 < ∞, the non-commutative 𝐿𝑝-space 𝐿𝑝(ℳ, 𝜏) is the completion of
(𝑆, ‖·‖𝑝), where ‖𝑥‖𝑝 = 𝜏(|𝑥|
𝑝)
1
𝑝 < ∞ for each 𝑥 ∈ 𝐿𝑝 (ℳ, 𝜏). In addition,
we put 𝐿∞ (ℳ, 𝜏) = ℳ and denote by ‖·‖𝑝 (= ‖·‖) the usual operator
norm. It is well known that 𝐿𝑝 (ℳ, 𝜏) are Banach spaces under ‖·‖𝑝 for
1 6 𝑝 < ∞ and they have a lot of expected properties of classical 𝐿𝑝-
spaces. Let 𝑥 be a 𝜏 -measurable operator and 𝑡 > 0. The “𝑡-th singular
number (or generalized 𝑠-number) of 𝑥” is defined by [5]
𝜇𝑡 (𝑥) = inf {‖𝑥𝑒‖ : 𝑒 ∈ 𝑃, 𝜏(1 − 𝑒) 6 𝑡}.
Recall that a linear map Φ is positive if Φ(𝑋) is positive whenever 𝑋 is
positive. The celebrated Jensen inequality for operators [2] states that if
𝑋 is a positive operator (self-adjointness is enough), Φ is a positive linear
map, and 𝑓 is an operator monotone on the interval [0,∞), then
Φ (𝑓 (𝑋)) 6 𝑓 (Φ (𝑋)) . (1)
In this paper, we prove the same result for measurable operators affiliated
with a given von Neumann algebra. Furthermore, we use the technique of
Zhao and Wu [11], via the notion of generalized singular numbers studied
by Fack and Kosaki [5], to obtain generalizations of results in [11] for
𝜏 -measurable operators case. The obtained inequalities improve known
results in [9]. In addition, Audenaert in [1] obtained that if 𝑋 and 𝑌 are
two 𝑛× 𝑛 matrices and 0 6 𝜈 6 1, then for any unitarily invariant norm
‖·‖𝑢,
‖𝑋𝑌 *‖2𝑢 6 ‖𝜈𝑋
*𝑋 + (1 − 𝜈)𝑌 *𝑌 ‖𝑢‖(1 − 𝜈)𝑋
*𝑋 + 𝜈𝑌 *𝑌 ‖𝑢. (2)
In the next section, we present a 𝜏 -measurable version of (2).
2. Main Theorems. We need the following lemma [3, Theorem 5]:
Lemma 1. Let ℳ be a von Neumann algebra on Hilbert space ℋ and
a function 𝑓 : R+ → R+ be an operator monotone with respect to ℳ.
Then 𝑓 (𝐴) 6 𝑓 (𝐵) for any pair of positive self-adjoint operators 𝐴, 𝐵
affiliated with ℳ, such that 𝐴 6 𝐵.
We are ready to prove our promised extension of inequality (1).
Theorem 1. Let Φ be a unital positive linear continuous map,
𝑓 : R+ → R+ be an operator monotone function with respect to ℳ
54 S. M. Davarpanah, M. E. Omidvar, H. R. Moradi
and 𝑥 ∈ 𝑆 (ℳ). Then
Φ (𝑓 (𝑥)) 6 𝑓 (Φ (𝑥)) .
Proof. We use the same strategy as in [4, Corollary 3.2]. Put 𝑥𝑛 = 𝑥𝜒([0,𝑛]).
It is clear that 𝑥𝑛 is a increasing sequence of positive operators in ℳ and
converges nearly everywhere to 𝑥. Note that 𝑥𝑛 commute with 𝑥 for every
𝑛. So, the convergence nearly everywhere of the sequence 𝑥𝑛 to 𝑥 can
be considered as in the commutative case. Therefore, for an operator
monotone function 𝑓 with respect to ℳ, and thus continuous on R+, the
sequence 𝑓 (𝑥𝑛) converges nearly everywhere to 𝑓 (𝑥). By Lemma 1, since
𝑥𝑛 6 𝑥𝑛+1 6 . . . 6 𝑥, we also have
𝑓 (𝑥𝑛) 6 𝑓 (𝑥𝑛+1) 6 . . . 6 𝑓 (𝑥) ,
and, since Φ is a positive linear continuous map,
Φ (𝑓 (𝑥𝑛)) 6 Φ (𝑓 (𝑥𝑛+1)) 6 . . . 6 Φ (𝑓 (𝑥)) .
Consequently, Φ (𝑓 (𝑥𝑛)) converge nearly everywhere to Φ (𝑓 (𝑥)). On the
other hand, for every 𝑥𝑛, by inequality (1), we have
Φ (𝑓 (𝑥𝑛)) 6 𝑓 (Φ (𝑥𝑛)) 6 𝑓 (Φ (𝑥)) .
Tending 𝑛 → ∞, we obtain the desired inequality. 
The following result can be found in [10, Lemma 3.2].
Lemma 2. Let 𝑥, 𝑦 ∈ 𝑆(ℳ) and 𝑧 ∈ ℳ. Then, for every 𝑟 > 0,
‖|𝑥*𝑧𝑦|𝑟‖2𝑝 6 ‖|𝑥𝑥
*𝑧|𝑟‖𝑝 ‖|𝑧𝑦𝑦
*|𝑟‖𝑝.
Theorem 2. Let 𝑥, 𝑦 ∈ 𝑆(ℳ) and 𝑧 ∈ ℳ. Then, for every 𝑟 > 0,
𝑔(𝑠, 𝑡) =
⃦⃦
|𝑥1−𝑡𝑧𝑦1+𝑠|𝑟
⃦⃦
𝑝
⃦⃦
|𝑥1+𝑡𝑧𝑦1−𝑠|𝑟
⃦⃦
𝑝
is log-convex on [−1, 1]× [−1, 1], hence is convex, and attains its minimum
at (0, 0).
Proof. The function 𝑔 is continuous and 𝑔(𝑠, 𝑡) = 𝑔(−𝑠,− 𝑡) (𝑠, 𝑡 ∈ [0, 1]).
Thus, it is enough to show that
𝑔 (𝑠1, 𝑡1) 6
1
2
{𝑔 (𝑠1 + 𝑠2, 𝑡1 + 𝑡2) + 𝑔 (𝑠1 − 𝑠2, 𝑡1 − 𝑡2)}
On some inequalities for 𝜏 -measurable operators 55
where 𝑠1 ± 𝑠2, 𝑡1 ± 𝑡2 ∈ [−1, 1] × [−1, 1].
Applying Lemma 2,⃦⃦
|𝑥1−𝑡1𝑧𝑦1+𝑠1 |𝑟
⃦⃦
2
=
⃦⃦
|𝑥𝑡2
(︀
𝑥1−𝑡1−𝑡2𝑧𝑦1+𝑠1−𝑠2
)︀
𝑦𝑠2|𝑟
⃦⃦
𝑝
6
6
{︀⃦⃦
|𝑥1−(𝑡1−𝑡2)𝑧𝑦1+(𝑠1−𝑠2)|𝑟
⃦⃦
𝑝
⃦⃦
|𝑥1−(𝑡1+𝑡2)𝑧𝑦1+(𝑠1+𝑠2)|𝑟
⃦⃦
𝑝
}︀ 1
2 (3)
and⃦⃦
|𝑥1+𝑡1𝑧𝑦1−𝑠1 |𝑟
⃦⃦
𝑝
=
⃦⃦
|𝑥𝑡2
(︀
𝑥1+𝑡1−𝑡2𝑧𝑦1−𝑠1−𝑠2
)︀
𝑦𝑠2|𝑟
⃦⃦
𝑝
6
6
{︀⃦⃦
|𝑥1+(𝑡1+𝑡2)𝑧𝑦1−(𝑠1+𝑠2)|𝑟
⃦⃦
𝑝
⃦⃦
|𝑥1+(𝑡1−𝑡2)𝑧𝑦1−(𝑠1−𝑠2)|𝑟
⃦⃦
𝑝
}︀ 1
2 . (4)
Applying (3), (4), and the arithmetic-geometric mean inequality, we get
𝑔 (𝑠1, 𝑡1) =
⃦⃦
|𝑥1−𝑡1𝑧𝑦1+𝑠1|𝑟
⃦⃦
2
⃦⃦
|𝑥1+𝑡1𝑧𝑦1−𝑠1 |𝑟
⃦⃦
2
6
6 {𝑔 (𝑠1 + 𝑠2, 𝑡1 + 𝑡2) 𝑔 (𝑠1 − 𝑠2, 𝑡1 − 𝑡2)}
1
2 6
6
1
2
[𝑔 (𝑠1 + 𝑠2, 𝑡1 + 𝑡2) + 𝑔 (𝑠1 − 𝑠2, 𝑡1 − 𝑡2)] .
The proof is completed. 
Using this observation, we give the following corollary.
Corollary. Let 𝑥, 𝑦 ∈ 𝑆(ℳ) and 𝑧 ∈ ℳ. Then, for every 𝑟 > 0,⃦⃦
|𝑥
1
2 𝑧𝑦
1
2 |
𝑟⃦⃦2
𝑝
6
⃦⃦
|𝑥𝑡𝑧𝑦1−𝑠|𝑟
⃦⃦
𝑝
⃦⃦
|𝑥1−𝑡𝑧𝑦𝑠|𝑟
⃦⃦
𝑝
6
6 max
{︀⃦⃦
|𝑥𝑧|𝑟
⃦⃦
𝑝
⃦⃦
|𝑧𝑦|𝑟
⃦⃦
𝑝
,
⃦⃦
|𝑥𝑧𝑦|𝑟
⃦⃦
𝑝
⃦⃦
|𝑧|𝑟
⃦⃦
𝑝
}︀
where 0 6 𝑠, 𝑡 6 1.
Proof. If we replace 𝑠, 𝑡, 𝑥, 𝑦 by 2𝑠 − 1, 2𝑡 − 1, 𝑥 12 , 𝑦 12 , respectively, in
Theorem 2, we see that the function 𝑔 (𝑠, 𝑡) =
⃦⃦
|𝑥𝑡𝑧𝑦1−𝑠|𝑟
⃦⃦
𝑝
⃦⃦
|𝑥1−𝑡𝑧𝑦𝑠|𝑟
⃦⃦
𝑝
is jointly convex on [0, 1]× [0, 1] and attains its minimum at
(︀
1
2
, 1
2
)︀
. Hence,⃦⃦
|𝑥
1
2 𝑧𝑦
1
2 |
𝑟⃦⃦2
𝑝
6
⃦⃦
|𝑥𝑡𝑧𝑦1−𝑠|𝑟
⃦⃦
𝑝
⃦⃦
|𝑥1−𝑡𝑧𝑦𝑠𝑡|𝑟
⃦⃦
𝑝
.
In addition, since the function 𝑔 is continuous and convex on
[0, 1] × [0, 1], it follows that 𝑔 attains its maximum at the vertices of the
square. Moreover, due to the symmetry, there are two possibilities for the
maximum. 
56 S. M. Davarpanah, M. E. Omidvar, H. R. Moradi
The Corollary can be regarded as an extension of [10, Corollary 3.4].
In the following result, we present a 𝜏 -measurable version of the main
result in [1]. We emphasize that the method of proof is completely different
from the present proof in [6, Theorem 3.6].
Theorem 3. Let 𝑥, 𝑦 be two 𝜏 -measurable positive operators. Then
𝜏(𝑥𝑦)2 6 𝜏(𝜈 𝑥2 + (1 − 𝜈)𝑦2) 𝜏((1 − 𝜈)𝑥2 + 𝜈 𝑦2), (5)
for 0 6 𝜈 6 1.
Proof. Note that the function 𝑓(𝜈) = 𝜏 (𝑥𝜈𝑦1−𝜈) is log-convex. Conse-
quently, the function
𝑔(𝜈) = 𝑓(𝜈)𝑓(1 − 𝜈)
is log-convex. Since 𝑔 is symmetric with respect to 𝜈 = 1
2
, it follows that
𝑓(1/2) 6 𝑓(𝜈). This means
𝜏
(︁
𝑥
1
2𝑦
1
2
)︁
6 𝜏
(︀
𝑥𝜈𝑦1−𝜈
)︀
𝜏
(︀
𝑥1−𝜈𝑦𝜈
)︀
, −∞ < 𝜈 < ∞.
Now, using Theorem 4, for 0 6 𝜈 6 1, we infer
𝜏
(︁
𝑥
1
2𝑦
1
2
)︁
6 𝜏(𝜈 𝑥 + (1 − 𝜈)𝑦) 𝜏((1 − 𝜈)𝑥 + 𝜈 𝑦).
Replacing 𝑥 and 𝑦 by their squares, we get the desired inequality. 
Note that inequality (5) interpolates between the arithmetic-geometric
mean inequality and Cauchy-Schwarz inequality for 𝜏 -measurable opera-
tors. That is, for 𝜈 = 0 we obtain the Cauchy-Schwarz type inequality
𝜏(𝑥𝑦)2 6 𝜏(𝑥2)𝜏(𝑦2),
while we obtain the arithmetic-geometric mean inequality
𝜏(𝑥𝑦) 6
1
2
𝜏(𝑥2 + 𝑦2)
for 𝜈 = 1
2
.
Recently, Shao in [?, Theorem 3.1] obtained a refinement of the Young
inequality
𝜏
(︀
𝑥𝜈𝑦1−𝜈
)︀
+ 𝑟0
(︁
𝜏(𝑥)
1
2 − 𝜏(𝑦)
1
2
)︁2
6 𝜏 (𝜈𝑥 + (1 − 𝜈) 𝑦) (6)
where 𝑥, 𝑦 ∈ 𝐿1 (ℳ) are positive operators, and 𝑟0 = min {𝜈, 1 − 𝜈} with
𝜈 ∈ (0, 1). We close this paper by improving (6).
Theorem 4. Let 𝑥, 𝑦 ∈ 𝐿1 (ℳ) be positive operators and 𝜈 ∈ (0, 1).
On some inequalities for 𝜏 -measurable operators 57
1) If 0 < 𝜈 6 1
2
, then
𝑟0
(︁
(𝜏 (𝑥𝑦))
1
4 − (𝜏 (𝑥))
1
2
)︁2
+ 𝜈
(︁
(𝜏 (𝑥))
1
2 − (𝜏 (𝑦))
1
2
)︁2
+ 𝜏
(︀
𝑥1−𝜈𝑦𝜈
)︀
6
6 𝜏 ((1 − 𝜈)𝑥 + 𝜈𝑦) .
(7)
2) If 1
2
< 𝜈 < 1, then
𝑟0
(︁
(𝜏 (𝑥𝑦))
1
4 − (𝜏 (𝑥))
1
2
)︁2
+ (1 − 𝜈)
(︁
(𝜏 (𝑥))
1
2 − (𝜏 (𝑦))
1
2
)︁2
+
+ 𝜏
(︀
𝑥1−𝜈𝑦𝜈
)︀
6 𝜏 ((1 − 𝜈)𝑥 + 𝜈𝑦) .
(8)
Proof. We prove only (7) as (8) goes similarly. By [11, Lemma 1], we
have
(1 − 𝜈)𝜇𝑡 (𝑥) + 𝜈𝜇𝑡 (𝑦) >
> 𝑟0
(︁
𝜇𝑡(𝑥𝑦)
1
4 − 𝜇𝑡(𝑥)
1
2
)︁2
+ 𝜈
(︁
𝜇𝑡(𝑥)
1
2 − 𝜇𝑡(𝑦)
1
2
)︁2
+ 𝜇𝑡(𝑥)
1−𝜈𝜇𝑡(𝑦)
𝜈 ,
where 𝑟 = min {𝜈, 1 − 𝜈} and 𝑟0 = min {2𝑟, 1 − 2𝑟}.
Hence
𝜏((1 − 𝜈)𝑥 + 𝜈𝑦) = (1 − 𝜈)𝜏(𝑥) + 𝜈𝜏(𝑦) =
=
∞∫︁
0
[︁
(1 − 𝜈)𝜇𝑡 (𝑥) + 𝜈𝜇𝑡 (𝑦)
]︁
𝑑𝑡 >
> 𝑟0
(︁ ∞∫︁
0
[︁
𝜇𝑡(𝑥𝑦)
1
2 + 𝜇𝑡(𝑥) − 2(𝜏(𝑥𝑦))
1
4 (𝜏(𝑥))
1
2
]︁
𝑑𝑡
)︁
+
+ 𝜈
(︁ ∞∫︁
0
[︁
𝜇𝑡(𝑥) + 𝜇𝑡(𝑦) − 2 (𝜏(𝑥))
1
2 (𝜏(𝑦))
1
2
]︁
𝑑𝑡
)︁
+
∞∫︁
0
𝜇𝑡(𝑥)
1−𝜈𝜇𝑡(𝑦)
𝜈𝑑𝑡 >
> 𝑟0
(︃
𝜏(𝑥) +
∞∫︁
0
𝜇𝑡
(︀
(𝑥𝑦)
1
2
)︀
𝑑𝑡−
− 2
(︁ ∞∫︁
0
(︁
(𝜏(𝑥𝑦))
1
4
)︁2
𝑑𝑡
)︁ 12(︁ ∞∫︁
0
(︁
(𝜏(𝑥))
1
2
)︁2
𝑑𝑡
)︁ 12)︃
+
58 S. M. Davarpanah, M. E. Omidvar, H. R. Moradi
+ 𝜈
(︃
𝜏(𝑥) + 𝜏(𝑦) − 2
(︁ ∞∫︁
0
(︁(︀
𝜏(𝑥)
)︀ 12 )︁ 12
𝑑𝑡
)︁2(︁ ∞∫︁
0
(︁(︀
𝜏(𝑦)
)︀ 12 )︁2
𝑑𝑡
)︁ 12)︃
+
+
∞∫︁
0
𝜇𝑡
(︀
𝑥1−𝜈𝑦𝜈
)︀
𝑑𝑡 =
= 𝑟0
(︁
(𝜏(𝑥𝑦))
1
4 − (𝜏(𝑥))2
)︁2
+ 𝜈
(︁
(𝜏(𝑥))
1
2 − (𝜏(𝑦))
1
2
)︁2
+ 𝜏
(︀
𝑥1−𝜈𝑦𝜈
)︀
. (9)
The proof is completed. 
Acknowledgment. The authors would like to express their hearty thanks
to the referees for their valuable suggestions and comments for revising
the manuscript. This work was financially supported by Islamic Azad
University, Ferdows Branch.
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Received June 15, 2019.
In revised form, August 24, 2019.
Accepted September 17, 2019.
Published online October 10, 2019.
S. M. Davarpanah
Department of Mathematics, Ferdows Branch, Islamic Azad University,
Ferdows, Iran.
E-mail: davarpanah@ferdowsiau.ac.ir
M. E. Omidvar
Department of Mathematics, Mashhad Branch, Islamic Azad University,
Mashhad, Iran.
E-mail: erfanian@mshdiau.ac.ir
H. R. Moradi
Young Researchers and Elite Club, Mashhad Branch, Islamic Azad University,
Mashhad, Iran.
E-mail: hrmoradi@mshdiau.ac.ir


ON SOME INEQUALITIES FOR  -MEASURABLE OPERATORS

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ON SOME INEQUALITIES FOR -MEASURABLE OPERATORS

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