数理解析研究所講究録903 巻 1995 年 1-5

1

POWERS AND COMMUTATIVITY
OF SELFADJOINT OPERATORSMITSURU UCHIYAMA($\{\mathrm{Q}\backslash$$A,)$$r’\llcorner$$)$

Dept. of Math. Fukuoka University of Education
Munakata, Fukuoka 811-41, Japan (fax 0940-35-1710)$\mathrm{M}\mathrm{S}\mathrm{C}(1991):47\mathrm{B}15,47\mathrm{B}44,47\mathrm{B}6\epsilon,47\mathrm{C}25$

1. Introduction
be the algebra of bounded operators on a Hilbert space . IfLet, then is said to be nonnegative, and we denotesatisfies $(Tx, x)\geq 0$ for every
it by $T\geq 0$ . If $ReT= \frac{1}{2}(T+T^{*})$ is nonnegative, then $T$ is said to be accretive.
and Richard [2] showed the folowing:$\mathfrak{H}$$B(\mathfrak{H})$$x\in \mathfrak{H}$$T\in B(\mathfrak{H})$$\mathrm{T}$$\mathrm{D}\mathrm{e}\mathrm{p}_{\mathrm{r}}\mathrm{i}\mathrm{m}\mathrm{a}$Theorem A. If$T^{r\iota}$

is accretive for $n=1,2,$$\ldots$, then $T\geq 0$ .

They have proved this theorem by using a mapping theorem for numerical
ranges due to T.Kato. For completeness, we give another proof dependent on. Since, for scalar $a>0,$ $(T+a)^{n}$ is accretive, we may
Sz.-Nagy’s
is accreassume that $Re\sigma(.T)>0$ and that $||T||<1$ . Since the inverse $T^{-n}$ of
the coefficients of its
tive, $T^{-n}(I-\tau)^{-m}$ is accretive too for $n,m=1,2,$$\mathrm{t}\mathrm{e}\mathrm{c}\mathrm{h}\mathrm{n}\mathrm{i}\mathrm{q}\mathrm{u}\mathrm{e}[8]$$T^{n}$$\ldots,\mathrm{b}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{u}\mathrm{s}\mathrm{e}$power series expansion are nonnegative. Thus $ReT^{n}(I-\dot{T})m\geq 0$ . Using Bernstein’spolynomials, for any polynomial $f\geq 0$ on the interval $[0,1]$ , we have Ref $(T)\geq 0$ .
satisfies the moment problem. Thus there is
Therefore the sequence$ReT^{n}=PH^{n}|\mathfrak{g}$ , wherea nonnegative dilation $H\in B(\mathrm{R})$ of , that isis the orthogonal projection from A to . Hence we get Kadison’s inequality$(ReT)^{2}\leq Re(T^{2})$ , which implies $0\leq(T-\tau*)^{2}=-4(ImT)^{2}$ . Consequently weobtain $T\geq 0$ .
$\{ReT^{n}\}_{n}\infty=0$

$T$

$P$

$\mathfrak{H}\subset R,$$\mathfrak{H}$

2. Powers of Operators, a subspace
In this section we shall extend Theorm A. For
and. Then $X$ can be represented assaid to reduce $X$ if$X\in B(\mathfrak{H})$$XL\subset,\mathrm{C}$$\mathcal{L}\subset \mathfrak{H}$is$X^{*},\mathrm{C}\subset L$$X=x|_{\mathfrak{L}}\oplus X|_{5\mathrm{i}^{\perp}}$.

An operator $X$ is called completely non selfadjoint (c.n.s.) if there is no non-zero
reducing subspace for $X$ such thatis selfadjoint.
$\mathcal{L}$$X|_{\mathcal{L}}$Typeset by$A_{\Lambda}\beta- \mathrm{I}_{\mathrm{E}^{\mathrm{X}}}$2MITSURU UCHIYAMA

Let us remark that every operator $X$ can be uniquely represented as a sumof a selfadjoint operator and a c.n.s. operator. In fact, $L:=\{x\in \mathfrak{H}$ : $X^{n}x=$$X^{*n}x,$ $n=1,2,$
is a closed subspace of ; since $X^{n}(Xx)=X^{*(n+1)_{X}}=x^{\mathrm{s}n}X\mathrm{z}$is selfadjoint andand $X^{n}X^{*}x=X^{*n}x^{\iota}X$ for any ae in $L,$ reduces $X$ , so thatis c.n.s.
$\ldots\}$$\mathfrak{H}$$L$$X|_{\mathrm{L}}$$X|_{\mathrm{g}\perp}$, and assume that $ReX\geq Re\mathrm{Y}$.
Theorem 1. Let $X$ and be inthat ,is a selfadjoint operat or for $n=1,2,$ , then there is a, that is,
to selfadjoint operators andreduces both $X$ and$IfX^{n}+\mathrm{Y}^{n}$$B(\mathfrak{H})$$\mathrm{Y}$$S\mathrm{u}bs_{\mathrm{P}^{ac}}e,\mathrm{C}s\mathrm{u}c\mathrm{A}$$\mathrm{C}$$\ldots$$(X|_{\mathrm{g}\perp})^{*}=\mathrm{Y}|_{\mathrm{g}\perp}$$\mathrm{Y}$$X=s.a.\oplus T,$ $\mathrm{Y}=\epsilon.a$

Pmof. Since.

$\oplus T^{*}$.

$ImX^{n}=-Im\mathrm{Y}^{n},$ $\mathcal{L}:=\{x\in \mathfrak{H} : X^{n}x=X^{*n_{t}}, n=1,2, \ldots\}$reduces. $X$
to selfadjoint operators. We have only to show
andhave the representations: $X=A+iB,$ $\mathrm{Y}=C-iB$ , where $A,$ $B$ andand$C$ are selfadjoint operators.
Then the assumption means $A\geq C$ . Let us notethat
reduces $A,$ and $C$ , and that $B|_{L}=0$ . We determine the sequences ofby $A_{1}=A,$ $B_{1}=B,$ $C_{1}=C,$ $D_{1}=-B$ ,andoperators$X$$(X|_{L}\perp)*=\mathrm{Y}|_{\mathcal{L}}\perp$$\mathrm{Y}$$\mathrm{Y}$

$B$

$L$

$\{A_{n}\},$ $\{B_{n}\},$ $\{C_{n}\}$$\{D_{n}\}$$A_{n+1}=AA_{nn’ n+1}-BBB=AB_{n}+BA_{n},$ $C_{n+1}=CC_{n}+BD_{n},$ $D_{n+1}=CD_{n}-$$BC_{n}$. It is easy to see that they are selfadjoint; for instance, if they are selfadjoint

for , then$n$$A_{n+1}=A_{n}A-B_{n}B=(AA_{\pi-1}-BBn-1)A-(AB_{n-1}+BA_{n-1})B$$=A(A_{n-1}A-B_{n-}1B)-B(B_{n-1}A+A_{n-1}B)=AA_{n}-BB_{n}=A_{n+1}$ .

. $B_{2}+D_{2}=0$ means $(A-C)B+$Thus we have $X^{n}=A_{n}+iB_{n},$$B(A-C)=0$ , from which it follows that $(A-C)B=B(A-C)=0$ , because$A\geq C$ . Since $B_{n+1}+D_{\iota+1},=0$ , we have $(A-C)B_{n}+B(A_{n}-c_{n})=0$ , from which
it follows that $(A-C)B_{n}=B(A_{n}-c_{n})=0$ , because the range of $B$ is orthogonal, and hence $(A-c)|\mathrm{g}\perp=0$ . Consequently
to the one of $A-C$ . Thus.
we obtain$\mathrm{Y}^{n}=C_{\iota},+iD_{n}$$(A-C)\mathrm{r}\subset \mathcal{L}$$(X|_{\mathrm{L}}\perp)^{*}=\mathrm{Y}|_{L}\perp$$\square$From this theorem and Theorem A we get the folowing:

Theorem 2. Let $X$ andfor $n=1,2,$$X^{n}+\mathrm{Y}^{n}\geq 0$$\mathrm{Y}$be bounded opera
satisfying $ReX\geq Re\mathrm{Y}$. If, then $X$ and are selfadjoint opera .
$\mathrm{t}o\mathrm{r}s$$\mathrm{Y}$$\ldots$

Corouary 1. If $X^{n}+\mathrm{Y}^{n}\geq 0$ for $n=1,2,$$X\geq \mathrm{Y}\geq 0$$\mathrm{t}o\mathrm{r}s$$\ldots$, and $ReX\geq Re\mathrm{Y}\geq 0$ , then.

In the above theorems the assumption $ReX\geq Re\mathrm{Y}$ is indispensable. For instance, takematrices:$2\cross 2$$X=,\mathrm{Y}=$ .
Then$X^{n}+\mathrm{Y}^{n}\geq 0$. However neither matrix is selfadjoint.

3
POWERS AND COMMUTATIVITY OF SELFADJOINT OPERATORS

3. Commut ativity
We consider applications of the above theorems to commutativity of selfadjoint

operators.
Theorem 3. Let $A,$ $B$ and $C$ be selfadjoint operators satisfying $B\geq C$. If$AB^{n}+C^{n}A\geq 0$ for $n=1,2,$, and if $A\geq 0$ or $C\geq 0$ , then A commutes to $B$$\ldots$

an $dC$ .

Proof. Suppose that $A\geq 0$ . Substituting

andand $C+\delta(\delta>0)$ foris invertible.
$C$ , respectively, we may assume that $B\geq C\geq 0$ and hence that
commutes to $(B-C)$ .
$AB+CA\geq 0$ implies that $A(B-C)$ is selfadjoint, so that$B$

$B+\mathit{5}$

$A$

$A$

We have$(A^{\frac{1}{2}}BA^{-\frac{1}{2})^{n}}+(A^{-} \frac{1}{2}oA^{\frac{1}{2}})n\geq 0$.

Since,$Re(A^{\frac{1}{2}}BA^{-} \frac{1}{2})=Re(A^{-\frac{1}{2}}cA\frac{1}{2})+(B-C)$

by Theorem 2 we get $AB=BA,AC=CA$ . Suppose next that $C\geq 0$ . Then wemay assume that $A\geq 0$ . Therefore from the above it follows that commutes to$B$ andC.
$A$$\square$be selfadjoint operators, and ppose th at $A\geq 0$ ormutative.
, th en and are$B\geq 0$ . $IfAB^{\pi}+B^{n}A\geq 0$ for $n=1,2,$
Corollary 2. Let$A$and$B$$s\mathrm{u}$

$B$

$A$

$co\mathrm{m}$$\ldots$

This was shown in [7]. And then M.Fuj\"u, R.Nakamoto,
have given the other proofs of it. Let us remark that in thisandcorollary we may exclude the condition $:A\geq 0$ or $B\geq 0$ . Indeed, from $A(B^{2})^{n}+$$(B^{2})^{n}A\geq 0$ for $n=1,2,$, it follows that $AB^{2}=B^{2}A\geq 0$ . Since the closuresontoand $|B|$ are equal, for the orthogonal projectionof the ranges of $B,$
this space, we have $PA=AP\geq 0$ . Therefore we get PAPB $+B^{n}PAP\geq 0$ for$\mathrm{M}.\mathrm{N}\mathrm{a}\mathrm{k}\mathrm{a}\mathrm{m}\mathrm{u}\mathrm{r}\mathrm{a}[3]$$\mathrm{S}.\mathrm{I}\mathrm{z}\mathrm{u}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{o}[4]$$\ldots$$P$$B^{2}$, which implies$BPAP=BPA=BA$ .

$n=1,2,$

commutes to PAP. Thus $AB=APB=PAPB=$$B$$\ldots$

and $B$ be onnegative selfadjoin operators. If $AB+BA\geq 0$ ,Lemma. Let
then $AB^{\ell}+B^{\ell}A\geq 0$ for $0\leq t\leq 1$ .
$A$

$t$

$\mathrm{n}$Prvof. We may assume that$\dot{B}$is invertible. For $0<t<1$ we have$B^{t}= \frac{sin(\pi t)}{\pi}\int_{0}^{\infty}\lambda t-1(B+\lambda)^{-}1Bd\lambda$

from which $AB^{\ell}+B^{\ell}A\geq 0$ follows.

$\square$andbe the spectral families corresponding to selfadjoint operators
Letfor every . For any
and $B$ , respectively. Then we denote $A\prec B$ if$\{E_{\lambda}\}$$A$,$\{F_{\lambda}\}$$\{E_{\lambda}\}\geq\{F_{\lambda}\}$$\lambda$

4MITSURU UCHIYAMA

and $C\geq 0$ , there is the supremum $BC$ of(see $[6],[5],[1]$ ).
equal to.

$B\geq 0$$B$

and $C$ in this order and it is$\lim_{narrow\infty}(B^{n}+C^{n})^{\underline{1}}$Proposition. Let
operators. If$A$

be a selfadjoint operator, and $B$ and $C$ non negative selfadjoint$AB^{n}+C^{n}A\geq 0$

th en

$A$

for $n=1,2,$ ,$\ldots$

and $BC$ are comm utative.

Proof. We may assume thatis nonnegative. We can easily obtain$A$$A(B^{n}+C^{n})+(B^{n}+C^{n})A\geq 0$

for $n=1,2,$$\ldots$

For each $m$ , we have$A(B^{n}+C^{n})^{\frac{m}{\mathrm{n}}}+(B^{n}+C^{n})^{\frac{m}{\mathrm{n}}}A\geq 0$

for $n=m,m+1,$ .
$\ldots$

Thus $A(BC)^{m}+(BC)^{m}A\geq 0$ . By Corollary 2,$A$

commutes to $B$ C.

Corollary 3. Let $P$ andbe or th ogon projection , and suppose thatselfadjoint operat or. If $AP+QA\geq 0$ , th en and $PQ$ are $com$ mutative.
$Q$$\mathrm{a}l$

$s$

$A$

$\square$is a$A$Theorem 4. Let $A,$ and be selfadjoint operators satisfying $BC=CB$ . If
$AB^{n}+C^{n}A\geq 0$ for $n=1,2,$, and if $A\geq 0$ or $B,$ $C\geq 0$ , then A commutes toand .
$C$

$B$

$\ldots$$B$$C$

Proof. We may assume thatand $C$ are nonnegative. Proposition implies that
commutes to $B\vee C$ . Since $A(B-C)$ is selfadjoint, commutes to $B-C$ . From, it follows that $B\vee C$ commutes to and $C$ , and hence
we gain$A,$ $B$.

$A$

$A$

$B$

$B \vee C=\lim_{narrow\infty}(B^{n}+C^{\pi})^{\underline{1}}$$BC= \min\{X : X\geq B, X\geq C,XB=BX, XC=CX\}=\frac{1}{2}(B+C+|B-C|)$ .

Consequently$A$

commutes to$B$and C.

$\square$Corollary 4. Let andbe
mutative or th ogon projections, and pposethat is a selfadjoint operat or. $IfAP+QA\geq 0$ , then Autes to and .
$P$

$Q$

$\mathrm{a}l$$co\mathrm{m}$$s\mathrm{u}$

$A$

$P$

$com\mathrm{m}$$Q$

At the end of this paper we give a counter example so that in Corollary 4 wecan not exclude the condition:
and are commutative. Takematrices:
$P$

$P=,$

$2\cross 2$$Q$$Q= \frac{1}{2}(_{1}^{1}$$11),$$A=$ .

Then $AP+QA\geq 0$ , but $AP\neq PA$ .
The author thanks F.Kubo and Y.Watatani for relating him toandRichard’s paper [2] after [7] was accepted. Also he would like to express his thanks
to the referee for finding some typing errors and for his nice suggestions.
$\mathrm{D}\mathrm{e}\mathrm{p}_{\mathrm{r}}\mathrm{i}\mathrm{m}\mathrm{a}$5
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