New proofs of the operator monotony of the square root and the inverse

TL;DR

This paper provides new, simplified proofs of the operator monotony of the square root function and the inverse function for positive operators, enhancing understanding of their properties in operator theory.

Contribution

It introduces straightforward proofs for the monotonicity of the square root and inverse functions of operators, simplifying existing complex arguments.

Findings

Proof that if 0 ≤ A ≤ B, then √A ≤ √B

Proof that invertibility of A implies invertibility of B and B^{-1} ≤ A^{-1}

Simplified methods for establishing operator monotony properties

Abstract

Let $A,B\in B(H)$. We present among others a simple proof of the widely known result stating that if $0\leq A\leq B$, then $\sqrt A\leq \sqrt B$. The same idea is used to prove that if $0\leq A\leq B$ and $A$ is invertible, then $B$ too is invertible and $B^{-1}\leq A^{-1}$.

Equations

$$\forall x\in H:\Vert Ax\Vert \le \Vert Bx\Vert ⟺\exists K\in B(H)\text{ contraction}:A=KB.$$

$$|<Ax,y>{|}^2\le <Ax,x><Ay,y>$$

$$\sqrt{A}B=S\sqrt{A}.$$

$$KA{K}^{\ast }\nleqq A.$$

$$S^{\ast }S{S}^{\ast }S=I\le S{S}^{\ast },$$

$$K^{2}A=A(K^{\ast }{)}^2=KA{K}^{\ast }\le A.$$

$$A{K}^{\ast }=KA⟹A(K^{\ast }{)}^2=KA{K}^{\ast }=K^{2}A⟹\sqrt{A}(K^{\ast }{)}^2=V\sqrt{A}$$

$$<KA{K}^{\ast }x,x>=<A(K^{\ast }{)}^{2}x,x>\le <Ax,x{>}^{\frac{1}{2}}<A(K^{\ast }{)}^{2}x,(K^{\ast }{)}^{2}x{>}^{\frac{1}{2}}$$

$$<A(K^{\ast }{)}^{2}x,(K^{\ast }{)}^{2}x{>}^{\frac{1}{2}}=<\sqrt{A}(K^{\ast }{)}^{2}x,\sqrt{A}(K^{\ast }{)}^{2}x{>}^{\frac{1}{2}}=\Vert \sqrt{A}(K^{\ast }{)}^{2}x\Vert =\Vert VAx\Vert .$$

$$<KA{K}^{\ast }x,x>\le <Ax,x>,$$

$$\Vert \sqrt{A}x{\Vert }^2\le \Vert \sqrt{B}x{\Vert }^2.$$

$$<\sqrt{A}x,x{>}^2=<K\sqrt{B}x,x{>}^2=<\sqrt{B}x,K^{\ast }x{>}^2\le <\sqrt{B}x,x><\sqrt{B}{K}^{\ast }x,K^{\ast }x>.$$

$$<\sqrt{B}{K}^{\ast }x,K^{\ast }x>=<K\sqrt{B}{K}^{\ast }x,x>\le <\sqrt{B}x,x>.$$

$$|<\sqrt{A}x,x>{|}^2\le |<\sqrt{B}x,x>{|}^2\text{ or }<\sqrt{A}x,x>\le <\sqrt{B}x,x>,$$

$$(\sqrt{B}{)}^{-1}=(\sqrt{A}{)}^{-1}K=K^{\ast }(\sqrt{A}{)}^{-1}$$

$$<B^{-1}x,x>=\Vert (\sqrt{B}{)}^{-1}x{\Vert }^2=\Vert K^{\ast }(\sqrt{A}{)}^{-1}x{\Vert }^2\le \Vert (\sqrt{A}{)}^{-1}x{\Vert }^2=<A^{-1}x,x>,$$

$$\forall x\in H:\Vert Ax\Vert \le \Vert Bx\Vert ⟺\exists K\in B(H)\text{ positive contraction}:A=KB.$$

$$0\le <KBx,Bx>=<Ax,Bx>=<BAx,x>,$$

$$||K_0(Bx)||=||A{B}^{-1}Bx||=||Ax||\le ||Bx||,\forall x\in H,$$

$$<K_0(Bx),Bx>=<Ax,Bx>=<BAx,x>\ge 0$$

$$0\le K\le I⟹\sqrt{A}=K\sqrt{B}\le \sqrt{B},$$

$$0\le K^2\le I⟹B^{-1}=A^{-1}{K}^2\le A^{-1},$$

$$0\le A\le B⟹A^2\le B^2.$$

$$A=K\sqrt{B}K\sqrt{B}=K^{2}B.$$

$$\Vert Ax{\Vert }^2=\Vert K^{2}Bx{\Vert }^2\le \Vert Bx{\Vert }^2$$

$$<A^{2}x,x>=<Ax,Ax>=\Vert Ax{\Vert }^2\le \Vert Bx{\Vert }^2=<B^{2}x,x>,$$

Full text

1. Notations

Throughout this paper, $H$ designates a complex Hilbert space. We say that $A\in B(H)$ is positive, and we write $A\geq 0$, if $<Ax,x>\geq 0$ for all $x\in H$ (this implies that $A$ is self-adjoint). If $A,B\in B(H)$ are self-adjoint, then we write $A\leq B$ if $B-A\geq 0$. If $A\leq B$, then $AC\leq BC$ for any positive $C\in B(H)$ which commutes with $A$ and with $B$.

Recall also that a $K\in B(H)$ is called a contraction if $\|K\|\leq 1$. This is known (cf. [1]) to be equivalent to any of the following:

• •

$||Kx||\leq||x||$ for all $x\in H$;

• •

$KK^{*}\leq I$;

• •

$K^{*}K\leq I$.

An important result of monotony in $B(H)$ is the so-called Löwner-Heinz Inequality: If $A\geq B\geq 0$, then $A^{\alpha}\geq B^{\alpha}$ for any $\alpha\in[0,1]$.

In particular, if $0\leq A\leq B$, then $\sqrt{A}\leq\sqrt{B}$. Another equally important result is: If $0\leq A\leq B$ and if $A$ is invertible, then $B$ is invertible and $B^{-1}\leq A^{-1}$.

In this short paper, we mainly present new proofs of these two well known results.

2. Main Results

The key point for proving the results are the following standard lemmata:

Lemma 2.1**.**

(see e.g. [1]) Let $H$ be a complex Hilbert space. If $A,B\in B(H)$, then

[TABLE]

Remark*.*

The foregoing lemma is usually utilized to characterize hyponormal operators.

Lemma 2.2**.**

(Generalized Cauchy-Schwarz Inequality) Let $H$ be a complex Hilbert space. If $A\in B(H)$ is positive, then

[TABLE]

for all $x,y\in H$.

Theorem 2.3**.**

([5]) Let $A,B\in B(H)$ such that $A$ is positive and $AB=B^{*}A$. Then there exists a unique self-adjoint operator $S\in B(H)$ such that

[TABLE]

A glance at the proof of the previous theorem allows us to give the following:

Lemma 2.4**.**

Let $A,B\in B(H)$ such that $A$ is positive and $AB=B^{*}A$. If $B$ is a contraction, then there exists a unique self-adjoint contraction $S\in B(H)$ such that

[TABLE]

Remark*.*

It is known that if $A\in B(H)$ is positive and $K\in B(H)$ is a contraction, then (unless $KA=AK$) in general:

[TABLE]

For example, let $S$ be the usual shift operator on $\ell^{2}$ and set $A=SS^{*}$. Then $A$ is positive and obeys $\|A\|\leq 1$. If we choose $K=S^{*}$, then $K$ is clearly a contraction. If $KAK^{*}\leq A$ held, then we would obtain

[TABLE]

which is absurd! Indeed, as we already know that $SS^{*}\leq I$, then we would end up with $SS^{*}=I$!

As a consequence of the previous lemma, we have

Proposition 2.5**.**

Let $A\in B(H)$ be positive and let $K\in B(H)$ be a contraction. If $AK^{*}=KA$, then

[TABLE]

Proof.

First, observe that

[TABLE]

for some (self-adjoint) contraction $V\in B(H)$.

Now, let $x\in H$. Then

[TABLE]

where we have used Lemma 2.2 in the last inequality. But,

[TABLE]

Since $V$ is a contraction, $\|VAx\|\leq\|Ax\|$ for each $x$. Finally, in view of $\|Ax\|=<\sqrt{A}x,x>^{\frac{1}{2}}$, we obtain:

[TABLE]

and this completes the proof. ∎

Theorem 2.6**.**

(cf. [4]) Let $A,B\in B(H)$. If $0\leq A\leq B$, then $\sqrt{A}\leq\sqrt{B}$.

Proof.

Let $x\in H$. Since $A\leq B$, we easily see that:

[TABLE]

So, by Lemma 2.1, we know that $\sqrt{A}=K\sqrt{B}$ for some contraction $K\in B(H)$. Since $\sqrt{A}$ is self-adjoint, it follows that $K\sqrt{B}$ too is self-adjoint, that is, $K\sqrt{B}=\sqrt{B}K^{*}$.

Hence, by the generalized Cauchy-Schwarz inequality:

[TABLE]

By Proposition 2.5, we have

[TABLE]

Accordingly,

[TABLE]

as required. ∎

Remark*.*

Notice that Proposition 2.5 can also be established if we use the preceding theorem.

For the new proof of the next result, we only need Lemma 2.1. The proof is very simple and short.

Theorem 2.7**.**

Let $A,B\in B(H)$. If $0\leq A\leq B$ and if $A$ is invertible, then $B$ is invertible and $B^{-1}\leq A^{-1}$.

Proof.

As in the proof of Theorem 2.6 combined with Lemma 2.1, we know that $\sqrt{A}=K\sqrt{B}$ for some contraction $K\in B(H)$. Since $\sqrt{A}$ is invertible (as $A$ is), it follows that $I=(\sqrt{A})^{-1}K\sqrt{B}$, i.e. the self-adjoint $\sqrt{B}$ is left invertible and so $\sqrt{B}$ or simply $B$ is invertible (cf. [3]) and

[TABLE]

by the self-adjointness of both $(\sqrt{B})^{-1}$ and $(\sqrt{A})^{-1}$.

Let $x\in H$. Then (since $K^{*}$ too is a contraction)

[TABLE]

as needed. ∎

As far as I am aware, Theorem 2.6 has not a very obvious proof in the literature at an elementary level even when $A$ commutes with $B$ (cf. [2]). The following improvement of Lemma 2.1 (kindly communicated to me by Professor J. Stochel) makes the proof in case of commutativity very simple.

Lemma 2.8**.**

Let $H$ be a complex Hilbert space. If $A,B\in B(H)$ are self-adjoint and $BA\geq 0$, then

[TABLE]

Proof.

1. (1)

"$\Leftarrow$": Let $x\in H$. Then

[TABLE]

that is, $BA\geq 0$. 2. (2)

"$\Rightarrow$": Since $BA\geq 0$, it follows that $BA$ is self-adjoint, i.e. $AB=BA$. As a consequence, $\ker A$ reduces $A$ and $B$, and the restriction of $A$ to $\ker A$ is the zero operator on $\ker A$. Hence, we can assume that $A$ is injective. Therefore, because $\ker B\subset\ker A=\{0\}$, we see that $B^{-1}$ is self-adjoint and densely defined. Set $K_{0}=AB^{-1}$. Then $K_{0}$ is densely defined and

[TABLE]

signifying that $K_{0}$ is a contraction with a unique contractive extension $K$ to the whole $H$. Since

[TABLE]

for all $x\in H$, we see that $K$ is positive as well. Clearly $KBx=K_{0}(Bx)=Ax$ for all $x\in H$, which completes the proof. ∎

Corollary 2.9**.**

Let $A,B\in B(H)$ be positive and commuting. Then:

1. (1)

$0\leq A\leq B\Rightarrow\sqrt{A}\leq\sqrt{B}$. 2. (2)

$0\leq A\leq B\Rightarrow B^{-1}\leq A^{-1}$, whenever $A$ is invertible.

Proof.

Since $A$ and $B$ are positive and commuting, we obviously know that $AB\geq 0$. Mimicking the argument at the beginning of the proof of Theorem 2.6 combined with Lemma 2.8 give: $\sqrt{A}=K\sqrt{B}$ for some positive contraction $K$. As above, it follows that $K\sqrt{B}=\sqrt{B}K$.

1. (1)

We clearly have:

[TABLE]

and this proves the first statement. 2. (2)

Also, $A=\sqrt{B}K\sqrt{B}K=BK^{2}$. Since $A$ is invertible, $BK^{2}A^{-1}=I$, i.e. the self-adjoint $B$ is right invertible and so $B$ is invertible (cf. [3]) and $B^{-1}=K^{2}A^{-1}$. Therefore, as $K^{2}A^{-1}=A^{-1}K^{2}$, then

[TABLE]

as required.

∎

Corollary 2.10**.**

Let $A,B\in B(H)$ be positive and commuting. Then

[TABLE]

Proof.

Since $AB\geq 0$, we know by Lemma 2.8 that $\sqrt{A}=K\sqrt{B}$ for some positive contraction $K\in B(H)$ and $K\sqrt{B}=\sqrt{B}K$. Hence

[TABLE]

So for all $x\in H$:

[TABLE]

or merely

[TABLE]

as required. ∎