Abstract. We study the inequality

Tr (w (A) f (A)) \leq Tr (w (A) f (B))

, where

w : ℝ \to ℝ^{+}

is a “weight function” and

A, B

are Hermitian matrices with

A \leq B

, and ﬁnd corresponding characterizations of the trace.

________________

2000 Mathematical Subject Classiﬁcation. 15A45.

Key words and phrases. Full matrix algebra, trace inequality, monotonicity.

Supported by Russian Foundation for Basic Research, Grant 05-01-00799.

Throughout the paper,

M_{n}

stands for the algebra of

n \times n

complex matrices,

M_{n}^{h}

and

M_{n}^{+}

denote the subsets of Hermitian and positive semi-deﬁnite matrices respectively. For

A, B \in M_{n}^{h}

, the notation

A \leq B

means that

B - A \in M_{n}^{+}

. A linear functional

ϕ

M_{n}

is said to be positive if

ϕ (A) \geq 0

for all

A \in M_{n}^{+}

. The spectrum of a matrix

A \in M_{n}

is denoted by

σ (A)

. For a real-valued function

f

of a real variable and a matrix

A \in M_{n}^{h}

, the value

f (A)

is understood by means of the functional calculus for Hermitian matrices.

Proof. Let $g (x)$ be convex on $ℝ^{+}$ . As is well known, if $0 \leq A \leq B$ then there exists $U \in M_{n}$ such that $∥ U ∥ \leq 1$ and $A^{1 ∕ 2} = U B^{1 ∕ 2}$ . Then, by Jensen’s trace inequality for contractions (see, e.g., [4, Corollary 3] or [2]), we have

\begin{matrix} \begin{aligned} Tr (A f (B)) & = Tr (A^{1 ∕ 2} f (B) A^{1 ∕ 2}) \\ = Tr (U B^{1 ∕ 2} f (B) B^{1 ∕ 2} U^{*}) = Tr (U g (B) U^{*}) \geq Tr (g (U B U^{*})) \\ = Tr (g (A)) = Tr (A f (A)) . \end{aligned} \end{matrix}

Now, let us prove the converse. Take arbitrary elements $A, C$ in $M_{n}^{+}$ . Then (1) yields

Tr (A f (A)) \leq Tr (A f (A + C))

and

Tr (C f (C)) \leq Tr (C f (A + C)) .

Summing these two inequalities we obtain that the superadditivity inequality

Tr (g (A)) + Tr (g (C)) \leq Tr (g (A + C))

appears to hold for all $A, C \in M_{n}^{+}$ . By [6, Theorem 2] (see, also, [7, Theorem 1]), the latter implies that $g$ is convex. □

Proof. Set $P = χ_{(0, + \infty)} (A)$ , $P^{'} = χ_{(- \infty, 0]} (A)$ . Then $A^{+} = P A$ and we have:

\begin{matrix} \begin{aligned} Tr (w (A) A^{+}) & \leq Tr (w (A) A^{+}) + Tr (w (A) P A_{2}) \\ = Tr (w (A) (P A + P A_{2})) = Tr (w (A) P A_{1}) \\ \leq Tr (w (A) (P A_{1} + P^{'} A_{1})) = Tr (w (A) A_{1}) . \end{aligned} \end{matrix}

□

Proof. This follows from the equality $A = B^{+} - (B^{-} + (B - A))$ . □

Proof. Take a real number $λ$ , put $w_{1} (x) \equiv w (x + λ)$ and consider the function $γ_{λ} (x) \equiv {(x - λ)}^{+}$ . By Corollary, we have

\begin{matrix} \begin{aligned} Tr (w (A) γ_{λ} (A)) & = Tr (w_{1} (A - λ I_{n}) {(A - λ I_{n})}^{+}) \\ \leq Tr (w_{1} (A - λ I_{n}) {(B - λ I_{n})}^{+}) \\ = Tr (w (A) γ_{λ} (B)) . \end{aligned} \end{matrix}

Now, let $λ_{i} (i = 1, 2, \dots, k)$ be the points of the ﬁnite set $σ (A) \cup σ (B)$ indexed in the increasing order. As is easily seen, we can ﬁnd the numbers $μ_{0} \in ℝ$ , $μ_{i} \in ℝ^{+} (i = 1, 2, \dots, k - 1)$ such that the function $\tilde{f} (x) \equiv μ_{0} + \sum_{i = 1}^{k - 1} μ_{i} γ_{λ_{i}} (x)$ coincides with $f (x)$ at the points $λ_{i} (i = 1, 2, \dots k)$ . By the above calculation,

\begin{matrix} \begin{aligned} Tr (w (A) f (A)) & = Tr (w (A) \tilde{f} (A)) \\ \leq Tr (w (A) \tilde{f} (B)) \\ = Tr (w (A) f (B)) . \end{aligned} \end{matrix}

□

Now we turn to the question whether some special cases of inequalities (1), (2) characterize scalar multiples of the trace among all positive linear functionals. The following proposition complements Theorem 1.

Proposition 4. Let a function $f : ℝ^{+} \to ℝ$ be Borel measurable and let a positive linear functional $ϕ$ on $M_{n}$ be such that

ϕ (A f (A)) \leq \frac{1}{2} ϕ (A f (B) + f (B) A)

(3)

whenever $0 \leq A \leq B$ . Then either $f$ is constant on $(0, + \infty)$ or $ϕ$ is a scalar multiple of the trace.

Proof. Let $A, C \in M_{n}^{+}$ . Then (3) yields

ϕ (A f (A)) \leq \frac{1}{2} ϕ (A f (A + C) + f (A + C) A)

and

ϕ (C f (C)) \leq \frac{1}{2} ϕ (C f (A + C) + f (A + C) C) .

Summing these two inequalities we get

ϕ (g (A)) + ϕ (g (C)) \leq ϕ (g (A + C)),

where $g (x) = x f (x)$ , and it remains to apply [7, Theorem 1]. □

Recently it was proved [1], [3] that if for a positive linear functional

ϕ

M_{n}

the inequality

0 \leq A \leq B

entails

ϕ (A^{2}) \leq ϕ (B^{2})

, then

ϕ

is a scalar multiple of the trace. We will show that similar characterization can be obtained within the framework of weighted inequalities. To do this, we need the following lemma. Its proof uses some constructions from [3].

Proof. Clearly, without loss of generality we can assume that $p > q = 1$ . Moreover, it suffices to study the case $n = 2$ , $ϕ = Tr (S \cdot)$ , $S = diag (s, 1)$ , $0 \leq s \leq 1$ , and to prove that $s = 1$ (cf., e. g., [1] or [3]).

Take $α \in (0, 1]$ , $r \in (0, 1 - \frac{1}{p})$ and consider the matrices

A = (\begin{matrix} α^{r} & 0 \\ 0 & α \end{matrix}), B = (\begin{matrix} α^{2} & α \sqrt{1 - α^{2}} \\ α \sqrt{1 - α^{2}} & \sqrt{1 - α^{2}} \end{matrix}) .

We have

\begin{matrix} \begin{aligned} A^{p} B A + A B A^{p} \\ = & α^{2 + p r} (\begin{matrix} 2 α^{r} & (1 + α^{(p - 1) (1 - r)}) \sqrt{1 - α^{2}} \\ (1 + α^{(p - 1) (1 - r)}) \sqrt{1 - α^{2}} & 2 α^{p - 1 - p r} (1 - α^{2}) \end{matrix}) . \end{aligned} \end{matrix}

If we consider the unitary matrix

U = \frac{1}{\sqrt{2}} (\begin{matrix} 1 & 1 \\ 1 & - 1 \end{matrix})

and substitute $X = U A U^{*}$ , $Y = U B U^{*}$ into (4), then we obtain

\begin{matrix} \begin{aligned} ϕ (X^{p} Y X + X Y X^{p}) \\ = & ϕ (U (A^{p} B A + A B A^{p}) U^{*}) \\ = & α^{2 + p r} ϕ (U (\begin{matrix} 2 α^{r} & (1 + α^{(p - 1) (1 - r)}) \sqrt{1 - α^{2}} \\ (1 + α^{(p - 1) (1 - r)}) \sqrt{1 - α^{2}} & 2 α^{p - 1 - p r} (1 - α^{2}) \end{matrix}) U^{*}) \\ \geq & 0 . \end{aligned} \end{matrix}

Reducing $α^{2 + p r}$ , tending $α$ to $+ 0$ and calculating we get

0 \leq ϕ (U (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) U^{*}) = Tr (S (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix})) = s - 1,

i. e., $s \geq 1$ , and we conclude that $s = 1$ . □

Proof. Let $X, Y$ be arbitrary matrices in $M_{n}^{+}$ and $t$ be a positive number. Substituting $A = X$ , $B = X + t Y$ into $(5)$ we obtain

t ϕ (X^{r + 1} Y X^{r} + X^{r} Y X^{r + 1}) + t^{2} ϕ (X^{r} Y^{2} X^{r}) \geq 0,

i. e.,

ϕ (X^{r + 1} Y X^{r} + X^{r} Y X^{r + 1}) + t ϕ (X^{r} Y^{2} X^{r}) \geq 0,

Tending $t$ to $+ 0$ we get

ϕ (X^{r + 1} Y X^{r} + X^{r} Y X^{r + 1}) \geq 0,

which implies, by Lemma 5, that $ϕ$ is a scalar multiple of the trace. □

References

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[2] Brown L.G. and Kosaki H., Jensen’s inequality in semi-ﬁnite von Neumann algebras, J. Operator Theory, 23 (1990), 3–19.

[3] Sano T. and Yatsu T., Characterizations of tracial property via inequalities, J. Inequal. Pure Appl. Math., 7(1) (2006), article 36.

[4] Tikhonov O.E., Convex functions and inequalities for a trace, in Constructive theory of functions and functional analysis, No. 6 (Russian), Kazan State University, Kazan, 1987, pp. 77–82.

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Kazan State University, Chebotarev Institute of Mathematics and Mechanics Universitetskaya str. 17, Kazan, 420008, Russia