A note on the parallel sum

By using a variational principle we find a necessary and sufficient condition for an operator to majorise the parallel sum of two positive definite operators. This result is then used as a vehicle to create new operator inequalities involving the parallel sum.


Introduction
Anderson and Duffin defined the parallel sum A : B of two positive definite operators A and B by setting and they proved [1,Lemma 18] that for any vector ξ the inner product (1) (A : We begin by giving an intuitive proof of the variational result in (1).The purpose of this note is then to establish that the operator inequality is valid, if and only if there exists an operator C such that This result then functions as a generator of operator inequalities involving the parallel sum.We refer to [3] for a recent paper on the parallel sum.

Preliminaries
We first establish the rule of differentiating an expectation value with respect to a vector, Indeed, Let A, B be positive definite matrices and consider to a given vector x the vector function It is manifestly convex with derivative The derivative vanishes in all η if and only if and this is equivalent to In addition, We thus obtain that we calculate the global minimum value to be where A : B is the parallel sum of A and B. It is also half of the harmonic mean.In conclusion, we recover (1) and obtain the inequality for any other vector η.For an arbitrary operator D we set η = Dξ and obtain where we used (2).Putting We have thus proved the following result.
Theorem 2.1.Let A and B be positive definite operators.Then for an arbitrary operator C.
We next investigate the range of the operator function to given positive definite operators A and B. We consider the operator equation F (C) = H and rewrite the equation as By multiplying with (A+B) −1/2 from the left and from the right the equation is equivalent to We now set and rewrite the equation as which again may be written as The equation can thus be solved if and only if Under this condition we may find positive definite solutions in X given by and then obtain .
Note that the operator appearing inside the square root in the last formula line may not be self-adjoint.It is however similar to a positive semi-definite operator and therefore has a unique square root with positive spectrum.We have obtained.

Generating operator inequalities
Theorem 2.1 may serve as a generator for operator inequalities by suitably choosing the operator C. For C = λI, where 0 ≤ λ ≤ 1, we obtain By setting λ = 0, λ = 1/2 or λ = 1 we obtain the well-known inequalities Setting C = (A + B) −1 B we obtain equality Indeed, we note that Therefore, We next use Theorem 2.1 to obtain new operator inequalities.
Theorem 3.1.Let A, B be positive definite operators.
(i) Let P be an orthogonal projection.We obtain the inequality Setting A = B it reduces to the familiar inequality (ii) The inequality is valid, and it is strict, since for A = B it reduces to 1 2 A ≤ 1 2 A. (iii) Let p be a real number.We obtain the inequality and it reduces to equality for p = 1.The inequality is strict for arbitrary p, since for A = B it reduces to 1  2 A ≤ 1 2 A.
as desired.This proves (iii).QED By multiplying (iii) in Theorem 3.1 by 2 we obtain the inequality between harmonic means for positive definite operators A and B and arbitrary p ∈ R. If we in particular put p = 1/2 we obtain (4) This is an improvement of the inequality which is plain.Indeed, for 0 ≤ p ≤ 1, we obtain by operator concavity of the function t → t p the inequality (5) The reverse inequality is obtained for −1 ≤ p ≤ 0 and 1 ≤ p ≤ 2 by operator convexity.It is interesting to note that the inequality (6) H 2 (A, B) ≤ H 2 (A p , B p ) 1/p is false for p = 1/4 with counter examples in two-by-two matrices.We conjecture that (6) is false for 0 < p < 1/2 and true for 1/2 ≤ p ≤ 1.

Theorem 2 . 2 .
Let A, B and H be positive definite operators.The operator equation F (C) = C * AC + (I − C * )B(I − C) = H has solutions in C if and only if H ≥ A : B. One of the solutions is then given by

Proof.
By setting C = P and applying Theorem 2.1 we obtain (i).By settingC = B(A + B) −1 we obtain I − C = A(A + B) −1 and thus C * AC + (I − C) * B(I − C) = (A + B) −1 BAB(A + B) −1 + (A + B) −1 ABA(A + B) −1from which (ii) follows.Finally, we set C = (A p +B p ) −1 B p and since I −C = (A p + B p ) −1 A p and A p (A p + B p ) −1 B p = B p (A p + B p ) −1 A p we obtain C * AC + (I − C) * B(I − C)