Journal of Inequalities in Pure and Applied Mathematics

Journal of Inequalities in Pure and Applied Mathematics

http://jipam.vu.edu.au/

Volume 6, Issue 2, Article 49, 2005

CHARACTERIZATION OF THE TRACE BY YOUNG’S INEQUALITY

A.M. BIKCHENTAEV AND O.E. TIKHONOV RESEARCHINSTITUTE OFMATHEMATICS ANDMECHANICS

KAZANSTATEUNIVERSITY

NUZHINA17, KAZAN, 420008, RUSSIA. [email protected]

[email protected]

Received 25 March, 2005; accepted 11 April, 2005 Communicated by T. Ando

ABSTRACT. Letϕbe a positive linear functional on the algebra ofn×ncomplex matrices and p,qbe positive numbers such that ¹_p +¹_q = 1. We prove that if for any pairA,B of positive semi-definiten×nmatrices the inequality

ϕ(|AB|)≤ ϕ(A^p)

p +ϕ(B^q) q holds, thenϕis a positive scalar multiple of the trace.

Key words and phrases: Characterization of the trace, Matrix Young’s inequality.

2000 Mathematics Subject Classification. 15A45.

In what follows,M_nstands for the *-algebra ofn×ncomplex matrices,M⁺_n stands for the cone of positive semi-definite matrices,pandqare positive numbers such that ¹_p + ¹_q = 1. For A∈ Mn,|A|is understood as the modulus|A|= (A^∗A)^1/2.

T. Ando proved in [1] that for any pairA, B ∈ Mnthere is a unitaryU ∈ Mnsuch that U^∗|AB^∗|U ≤ |A|^p

p +|B|^q q .

It follows immediately that for any pair A, B ∈ M⁺_n the following trace version of Young’s inequality holds:

Tr(|AB|)≤ Tr(A^p)

p +Tr(B^q) q .

The aim of this note is to show that the latter inequality characterizes the trace among the positive linear functionals onM_n.

ISSN (electronic): 1443-5756

c 2005 Victoria University. All rights reserved.

Supported by the Russian Foundation for Basic Research (grant no. 05-01-00799).

089-05

2 A.M. BIKCHENTAEV ANDO.E. TIKHONOV

Theorem 1. Letϕ be a positive linear functional on M_n such that for any pairA, B ∈ M⁺_n the inequality

(1) ϕ(|AB|)≤ ϕ(A^p)

p + ϕ(B^q) q holds. Thenϕ =kTrfor some nonnegative numberk.

Proof. As is well known, every positive linear functionalϕ on M_n can be represented in the form ϕ(·) = Tr(S_ϕ·) for some S_ϕ ∈ M⁺_n. It is easily seen that without loss of generality we can assume that S_ϕ = diag(α₁, α₂, . . . , α_n), and we have to prove that α_i = α_j for all i, j = 1, . . . , n. Clearly, it suffices to prove thatα₁ =α₂. Inequality (1) must hold, in particular, for all matrices A = [a_ij]ⁿ_i,j=1, B = [b_ij]ⁿ_i,j=1 inM⁺_n such that0 = a_ij = b_ij if 3 ≤ i ≤ nor 3≤j ≤n. Thus the proof of the theorem reduces to the following lemma.

Lemma 2. LetS = diag ¹₂ +s,¹₂ −s

, where0 ≤ s ≤ ¹₂. If for every pairA, B ∈ M⁺₂ the inequality

(2) Tr(S|AB|)≤ Tr(SA^p)

p +Tr(SB^q) q holds, thens= 0.

Proof of Lemma 2. Let0≤ε≤ ¹₂,δ= ¹₄ −ε². Let us consider two projections P₁ =

₁

2 −ε √

√ δ

δ ¹₂ +ε

, P₂ =

₁

2 +ε √

√ δ

δ ¹₂ −ε

.

Calculate|P₁P₂|:

P₂P₁ =

2δ (1 + 2ε)√ δ (1−2ε)√

δ 2δ

, P₂P₁P₂ = 4δP₂, hence

|P₁P₂|= 2√

δP₂ =√

1−4ε²P₂.

SubstituteA = αP₁, B = βP₂ withα, β > 0into (2) and perform the calculations. Then the left hand side in (2) becomes

αβ√

1−4ε²

1

2+ 2εs

and the right hand one becomes

α^{p 1}₂ −2εs

p + β^{q 1}₂ + 2εs

q .

Now, takeα = 1,β = ^1−4εs_1+4εs¹_q

. Then we obtain as an implication of (2):

1

2(1−4εs)^1q(1 + 4εs)^1p√

1−4ε² ≤ 1

2(1−4εs), which implies

(3) (1−4ε²)^p2 ≤ 1−4εs

1 + 4εs. By the Taylor formulas,

(1−4ε²)^p2 = 1−2pε²+o(ε²) = 1 +o(ε) (ε→0), 1−4εs

1 + 4εs = 1−8εs+o(ε) (ε→0).

J. Inequal. Pure and Appl. Math., 6(2) Art. 49, 2005 http://jipam.vu.edu.au/

CHARACTERIZATION OF THETRACE BYYOUNG’SINEQUALITY 3

Since we have supposed that 0 ≤ s, the inequality (3) can hold for all ε ∈ 0,¹₂

only if

s= 0.

REFERENCES

[1] T. ANDO, Matrix Young inequalities, Oper. Theory Adv. Appl., 75 (1995), 33–38.

J. Inequal. Pure and Appl. Math., 6(2) Art. 49, 2005 http://jipam.vu.edu.au/