Ngô Quốc Anh

February 9, 2014

Monotonicity of trace inequalities involving Hermitian matrices and weights

Filed under: Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 23:21

This note aims to prove a very interesting property concerning the monotonicity of trace inequalities of square matrices.

Before going further, let us denote by $M_n$ the set (or the algebra) of $n \times n$ complex square matrices. Then, by $M_n^{\rm h}$ we mean the set of Hermitian matrices in $M_n$. We also denote by $M_n^+$ the set of positive semi-definite matrices in $M_n^{\rm h}$. In other words, there holds $\displaystyle M_n^+ \subset M_n^{\rm h} \subset M_n.$

As usual, the notation $A \leq B$ means $B-A \in M_n^+$ for any $A, B \in M_n^{\rm h}$.

The following inequality is basically due to Hoa-Tikhonov [here].

Theorem 1. Let $n\geq 2$ and let a function $f :\mathbb R^+ \to \mathbb R$ be Borel measurable. The inequality $\displaystyle \text{trace}(Af(A)) \leq \text{trace}(Af(B))$

holds for all $A,B \in M_n^+$ with $A \leq B$ if and only if the function $g(x)=xf(x)$ is convex on $\mathbb R^+$.

The proof of the above theorem is rather simple but elegant. The idea is to transform the condition $0 \leq A \leq B$ into the relation $A^\frac{1}{2} = U B^\frac{1}{2}$ for some $U \in M_n$ with $\|U\| \leq 1$. Then the theorem follows immediately from the Jensen trace inequality for contractions.

It is also interesting to note that the super-additivity property, i.e. $\text{trace}(f(A)) + \text{trace}(f(B)) \leq \text{trace}(f(A+B)) \quad \forall A,B \in M_n^+$

is equivalent to the convexity of the function $f$.

The question of determining whether some special cases of the above inequality characterize scalar multiples of the trace among all positive linear functions. First, Hoa and Tikhonov also prove the following result.

Theorem 2. Let a function $f:\mathbb R^+ \to \mathbb R$ be Borel measurable and let a positive linear function $\phi$ on $M_n$ be such that $\displaystyle\phi (Af(A)) \leq \frac 12 \phi \big( Af(B) + f(B)A \big)$

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

They then prove that

Theorem 3. Let $p,q$ be unequal positive numbers. If for a positive linear functional $\phi$ on $M_n$, the inequality $\displaystyle \phi \big( X^p Y X^q + X^q Y X^p\big) \geq 0$

holds for all $X,Y \in M_n^+$, then $\phi$ is a scalar multiple of the trace.

This result immediately leads to the following

Theorem 4. Let $r$ be a positive number. Let a positive linear functional $\phi$ on $M_n$ be such that $\displaystyle \phi (A^{2+2r}) \leq \phi (A^r B^2 A^r)$

whenever $0 \leq A \leq B$. Then $\phi$ is a scalar multiple of the trace.

It is clear to see that Theorem 4 includes a particular case the following well-known result: If $\displaystyle \phi (A^2) \leq \phi (B^2)$

for all $A,B$ satisfying $0 \leq A \leq B$, then $\phi$ is a scalar multiple of the trace.