Journal of Inequalities in Pure and
Applied Mathematics
CHARACTERIZATION OF THE TRACE BY YOUNG’S INEQUALITY
A.M. BIKCHENTAEV AND O.E. TIKHONOV
Research Institute of Mathematics and Mechanics
Kazan State University
Nuzhina 17, Kazan, 420008, Russia.
volume 6, issue 2, article 49,
2005.
Received 25 March, 2005;
accepted 11 April, 2005.
Communicated by: T. Ando
EMail: Airat.Bikchentaev@ksu.ru
EMail: Oleg.Tikhonov@ksu.ru
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c 2000 Victoria University
ISSN (electronic): 1443-5756
089-05
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Abstract
Let ϕ be a positive linear functional on the algebra of n × n complex matrices
and p, q be positive numbers such that p1 + 1q = 1. We prove that if for any pair
A, B of positive semi-definite n × n matrices the inequality
ϕ(|AB|) ≤
ϕ(Ap ) ϕ(B q )
+
p
q
holds, then ϕ is a positive scalar multiple of the trace.
Characterization of the Trace by
Young’s Inequality
A.M. Bikchentaev and
O.E. Tikhonov
2000 Mathematics Subject Classification: 15A45.
Key words: Characterization of the trace, Matrix Young’s inequality.
Supported by the Russian Foundation for Basic Research (grant no. 05-01-00799).
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In what follows, Mn stands for the *-algebra of n × n complex matrices,
stands for the cone of positive semi-definite matrices, p and q are positive
numbers such that p1 + 1q = 1. For A ∈ Mn , |A| is understood as the modulus
|A| = (A∗ A)1/2 .
T. Ando proved in [1] that for any pair A, B ∈ Mn there is a unitary U ∈
Mn such that
|A|p |B|q
U ∗ |AB ∗ |U ≤
+
.
p
q
M+
n
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J. Ineq. Pure and Appl. Math. 6(2) Art. 49, 2005
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It follows immediately that for any pair A, B ∈ M+
n the following trace version
of Young’s inequality holds:
Tr(Ap ) Tr(B q )
+
.
p
q
Tr(|AB|) ≤
The aim of this note is to show that the latter inequality characterizes the
trace among the positive linear functionals on Mn .
Theorem 1. Let ϕ be a positive linear functional on Mn such that for any pair
A, B ∈ M+
n the inequality
(1)
ϕ(|AB|) ≤
ϕ(Ap ) ϕ(B q )
+
p
q
holds. Then ϕ = k Tr for some nonnegative number k.
Proof. As is well known, every positive linear functional ϕ on Mn can be represented in the form ϕ(·) = Tr(Sϕ ·) for some Sϕ ∈ M+
n . It is easily seen
that without loss of generality we can assume that Sϕ = diag(α1 , α2 , . . . , αn ),
and we have to prove that αi = αj for all i, j = 1, . . . , n. Clearly, it suffices
to prove that α1 = α2 . Inequality (1) must hold, in particular, for all matrices
A = [aij ]ni,j=1 , B = [bij ]ni,j=1 in M+
n such that 0 = aij = bij if 3 ≤ i ≤ n or
3 ≤ j ≤ n. Thus the proof of the theorem reduces to the following lemma.
Lemma 2. Let S = diag 12 + s, 12 − s , where 0 ≤ s ≤ 12 . If for every pair
A, B ∈ M+
2 the inequality
(2)
Tr(S |AB|) ≤
Tr(SAp ) Tr(SB q )
+
p
q
Characterization of the Trace by
Young’s Inequality
A.M. Bikchentaev and
O.E. Tikhonov
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J. Ineq. Pure and Appl. Math. 6(2) Art. 49, 2005
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holds, then s = 0.
Let 0 ≤ ε ≤ 12 , δ = 14 −ε2 . Let us consider two projections
√ √ 1
1
−ε
δ
+ε
δ
2
2
√
√
P1 =
,
P2 =
.
1
1
δ 2 +ε
δ 2 −ε
Proof of Lemma 2.
Calculate |P1 P2 |:
P2 P1 =
hence
√ 2δ √ (1 + 2ε) δ
,
2δ
(1 − 2ε) δ
P2 P1 P2 = 4δP2 ,
√
√
|P1 P2 | = 2 δP2 = 1 − 4ε2 P2 .
Substitute A = αP1 , B = βP2 with α, β > 0 into (2) and perform the calculations. Then the left hand side in (2) becomes
√
1
2
αβ 1 − 4ε
+ 2εs
2
and the right hand one becomes
p
1
2
A.M. Bikchentaev and
O.E. Tikhonov
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q
1
2
− 2εs
β
+ 2εs
+
.
p
q
1
1−4εs q
Now, take α = 1, β = 1+4εs
. Then we obtain as an implication of (2):
α
Characterization of the Trace by
Young’s Inequality
1
1√
1
1
(1 − 4εs) q (1 + 4εs) p 1 − 4ε2 ≤ (1 − 4εs),
2
2
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J. Ineq. Pure and Appl. Math. 6(2) Art. 49, 2005
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which implies
p
(1 − 4ε2 ) 2 ≤
(3)
1 − 4εs
.
1 + 4εs
By the Taylor formulas,
p
(1 − 4ε2 ) 2 = 1 − 2pε2 + o(ε2 ) = 1 + o(ε)
(ε → 0),
1 − 4εs
= 1 − 8εs + o(ε) (ε → 0).
1 + 4εs
Since we have supposed that 0 ≤ s, the inequality (3) can hold for all ε ∈ 0, 12
only if s = 0.
Characterization of the Trace by
Young’s Inequality
A.M. Bikchentaev and
O.E. Tikhonov
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References
[1] T. ANDO, Matrix Young inequalities, Oper. Theory Adv. Appl., 75 (1995),
33–38.
Characterization of the Trace by
Young’s Inequality
A.M. Bikchentaev and
O.E. Tikhonov
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