Schur product, partial order

Question

Let $A, B$ be positive definite matrices. Then $A^r\circ B^r \le (A\circ B)^r$ for $0\le r\le 1$, where $\circ$ is Schur product. Here the inequality is in the sense of Loewner partial order.

How to prove this? Where can I find a reference?

Answer 1

The proof follows from the following results:

Theorem. (Thm. 1.6, in [2]) If $\Phi$ is a unital positive linear map from $\mathbb{M}_m \to \mathbb{M}_n$, and $f$ is an operator monotone function on $[0, \infty)$, then for every $A \ge 0$, \begin{equation*} \Phi(f(A)) \le f(\Phi(A)). \end{equation*}

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Let $A \otimes B$ denote the Kronecker product, and let $r$ be a real number \begin{equation*} (A\otimes B)^r = A^r \otimes B^r \end{equation*}

Now, recall that $A \circ B$ can be obtained as a principal submatrix of $A \otimes B$. Thus, there is a positive unital map $\Phi$ from $\mathbb{M}_{n^2} \to \mathbb{M}_n$ such that $\Phi(A \otimes B) = A \circ B$. Thus, an application of the first theorem, with $f=t^r$ as the operator monotone function (which is a well-known operator monotone function), we immediately obtain the inequality $A^r \circ B^r \le (A\circ B)^r$.

References

1. R. Bhatia. Positive definite matrices, Princeton Univ. Press (2007)
2. X. Zhan. Matrix Inequalities, Springer 2002.