An Improvement of Reid Inequality

TL;DR

This paper presents an enhancement to Reid Inequality, a fundamental result in the theory of linear operators, offering a tighter or more general bound.

Contribution

The paper introduces a novel improvement to the classical Reid Inequality, extending its applicability or tightening its bounds in operator theory.

Findings

Reid Inequality has been successfully improved.

The new inequality provides sharper bounds.

Potential applications in operator analysis are suggested.

Abstract

In this short note, we improve the famous Reid Inequality related to linear operators.

Equations

$$<Ax,x>\ge 0,\forall x\in H.$$

$$|<AKx,x>|\le ||K||<Ax,x>$$

$$|<Sx,x>|\le <S{S}^{\ast }x,x>=\Vert S^{\ast }x{\Vert }^2$$

$$|<Sx,x>|=2\le \Vert S^{\ast }x{\Vert }^2=1$$

$$|<AKx,x>|\le ||K||<Ax,x>$$

$$|<Ax,x>|\le <|A|x,x>.$$

$$K{K}^{\ast }\le \Vert K{\Vert }^{2}I.$$

$$|(AK{)}^{\ast }{|}^2=AK{K}^{\ast }A\le \Vert K{\Vert }^2{A}^2$$

$$|<AKx,x>|=|<x,(AK{)}^{\ast }x>|=|\overline{<(AK{)}^{\ast }x,x>}|=|<(AK{)}^{\ast }x,x>|.$$

$$|<AKx,x>|=|<(AK{)}^{\ast }x,x>|\le |<|(AK{)}^{\ast }|x,x>|\le \Vert K\Vert <Ax,x>$$

Full text

1. Main Result

First, assume that readers are familiar with notions and result on $B(H)$. We do recall a few definitions and results though:

1. (1)

Let $A\in B(H)$. We say that $A$ is positive (we then write $A\geq 0$) if

[TABLE] 2. (2)

For every positive operator $A\in B(H)$, there is a unique positive $B\in B(H)$ such that $B^{2}=A$. We call $B$ the positive square root of $A$. 3. (3)

The absolute value of $A\in B(H)$ is defined to be the (unique) positive square root of the positive operator $A^{*}A$. We denote it by $|A|$. 4. (4)

We recall that $A\in B(H)$ is called hyponormal if $AA^{*}\leq A^{*}A$.

The inequality of Reid which first appeared in [4] is recalled next:

Theorem 1.1**.**

Let $A,K\in B(H)$ such that $A$ is positive and $AK$ is self-adjoint. Then

[TABLE]

for all $x\in H$.

Remark*.*

As shown in e.g. [2], Reid Inequality is equivalent to the operator monotony of the positive square root on the set of positive operators.

Many generalizations of Theorem 1.1 are known in the literature from which we only cite [1] and [2].

In an earlier version of this paper (see [3]), the author showed the following:

Theorem 1.2**.**

Let $A,K\in B(H)$ such that $A$ is positive and $AK$ is normal. Then

[TABLE]

for all $x\in H$.

Can we go to $AK$ being hyponormal? The answer is no as seen next:

Example* 1.3**.*

Let $S$ be the shift operator on $\ell^{2}$. Setting $A=SS^{*}$, we see that $A\geq 0$. Now, take $K=S$ (and so $\|K\|=1$). It is clear that $AK=SS^{*}S=S$ is hyponormal. If Reid Inequality held, then we would have

[TABLE]

for each $x\in\ell^{2}$. This inequality clearly fails to hold for all $x$. Indeed, taking $x=(2,1,0,0,\cdots)$, we see that

[TABLE]

which is impossible.

The good news is that Reid Inequality can yet be improved as it holds if $AK$ is co-hyponormal, that is, if $(AK)^{*}$ is hyponormal. This comes after a discussion with a fellow student (Mr S. Dehimi):

Theorem 1.4**.**

Let $A,K\in B(H)$ such that $A$ is positive and $(AK)^{*}$ is hyponormal. Then

[TABLE]

for all $x\in H$.

The proof relies on the following result:

Lemma 1.5**.**

([1]) Let $A\in B(H)$ be hyponormal. Then

[TABLE]

Now, we give the proof of Theorem 1.4.

Proof.

The inequality is evident when $K=0$. So, assume that $K\neq 0$. It is then clear that $\frac{K}{\|K\|}$ satisfies

[TABLE]

Hence

[TABLE]

or simply $|(AK)^{*}|\leq\|K\|A$ after passing to square roots.

Now, for all $x\in H$

[TABLE]

Since $(AK)^{*}$ is hyponormal, Lemma 1.5 combined with $|(AK)^{*}|\leq\|K\|A$ give

[TABLE]

and this marks the end of the proof. ∎