数理解析研究所講究録1080 巻 1999 年 67-7067

An Example of a -Quasihyponormal Operator$p$

東北大学大学院理学研究科

内山敦 (Atsushi Uchiyama)

A bounded linear operator $T$ on a Hilbert space
Introduction.
is called -hyponormal if $(T^{*}T)^{p}\geq(TT^{*})^{p}$ for $p>0$ , and is called-quasihyponormal if $T^{*}\{(\tau^{*\tau})^{p}-(TT^{*})^{p}\}T\geq 0$ for $p>0$ . is called
if $||Tx||^{2}\leq||T^{2}x||||x||$ for all. It is well-known bythat every -hyponormal operator is paranormal. M. Lee andS. Lee showed that every -quasihyponormal operator for $0<p\leq 1$ isparanormal. It is well-known that every. -hyponormal operator $T=$
$U|T|$ is-hyponormal for all $q\in(0,p)$ by Heinz’s inequality and its
generalized Aluthge transform $T(s, t)=|T|^{s_{U}}|T|t$ for $t>0$ is ahyponormal for some $q=q(s, t,p)>0$ . (See $[1],[2],[6],[7]$ and [13]). But
the assertions that -quasihyponormal is -quasihyponormal if $0<q<$
and the generalized Aluthge transform $T(s, t)=|T|^{s_{U}}|T|t$ for $t>0$
of a -quasihyponormal operator $T=U|T|$ is aforsome $q–q(s, t, p)>0$ are not true.
In this paper, we give a -quasihyponormal operator $T=U|T|$ suchthat (i) is not -quasihyponormal for all $q\in(0,p),$for$t>0$ is not -quasihyponormal for all $q\in(0, \infty)$ and (iii)
is a-quasihyponormal for a $p>1$ , but is not paranormal.
$\mathcal{H}$

$T$

$p$

$T$

$p$

$x\in \mathcal{H}$$\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}[8]$$\mathrm{A}\mathrm{n}\mathrm{d}_{0}[3]$

$p$

$p$

$p$

$q$

$q-$

$s,$

$p$

$q$

$p$

$s,$

$p$

$q- \mathrm{q}\mathrm{u}\mathrm{a}\mathrm{S}\mathrm{i}\mathrm{h}...\mathrm{y}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\iota$

$p$

$T$

$q$

$s,$

$(\mathrm{i}\mathrm{i})|T|^{s}U|\tau|^{t}$$T$$q$$p$

Lemma 1. (tor and$A$$\mathrm{H}_{\ddot{\mathrm{O}}}1\mathrm{d}\mathrm{e}\mathrm{r}-\mathrm{M}_{\mathrm{C}\mathrm{C}\mathrm{h}\mathrm{y}\mathrm{n}\mathrm{e}\mathrm{q}\mathrm{y}[9])}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{I}\mathrm{u}\mathrm{a}\mathrm{l}\mathrm{i}\mathrm{t}$$x\in \mathcal{H}$For any positive opera-,(1) $(A^{r_{X}}, x)\leq||x||2(1-r)(Ax, x)^{r}$(if $0<r\leq 1$ ),(2) $(A^{r_{X}}, x)\geq||x||^{2(1r}-)(Ax, x)^{r}$(if $r\geq 1$ ).

Using above lemma, M. Lee and H. Lee obtained the following.

68
Theorem 1. (M. Lee and H.
operator such as $0<p\leq 1$ , thenis a p-quasihyponormal) Ifis paranormal.
$T$$\mathrm{L}\mathrm{e}\mathrm{e}[10]$$T$

Here, we construct an example of -quasihyponormal operator whichin the introduction.
satisfies the conditions$p$$(\mathrm{i})-(\mathrm{i}\mathrm{i}\mathrm{i})$andbe the canonical orthonormal basis ofLeton. Using the shift operator
tothe projection of
Hermitian matrices and , weand positive 2withbydefine operators $H$ and on$\{\epsilon_{n};n\in \mathbb{Z}\}$$\ell^{2}(\mathbb{Z})$$\ell^{2}(\mathbb{Z})$$S$$\mathbb{C}\epsilon_{n}$$A$$\cross 2$$S\epsilon_{n}=\epsilon_{n+1}$$T$$p_{n}$$\ell^{2}(\mathbb{Z})$$B$$\mathbb{C}^{2}\otimes\ell^{2}(\mathbb{Z})$$H= \sum_{n<0}A\otimes pn+\sum_{0n\geq}B\otimes pn$

and$T=(1\otimes S)H$.

Since $|T^{*}|=U|T|U^{*}=$, it is easy to see that

$T=U|T|$ , where $U=1\otimes S$ and $|T|=H$ .
$\Sigma_{n\leq 0}A\otimes p_{n}+\Sigma_{n>0}B\otimes p_{n}$$T^{*}(|T|^{2p}-|T^{*}|^{2p})T=A(B^{2p}-A^{2p})A\otimes p_{-1}$

for $p>0$ . Hence we have the following.

Lemma 2.

$T$

is -quasihyponormal if and only if $A(B^{2p}-A^{2p})A\geq 0$ .
$p$

In what follows we assume that

respectively, here $\alpha>0$ . Let
defined by

$f$

$A$

and$B$

are of the form

be a function on the half interval$(0, \infty)$$f(p)=( \frac{9^{p}+1}{2})^{\frac{1}{2\mathrm{p}}}$

Then it is strictly increasing.

(1) is -quasihyponormal if and only if $\alpha\leq f(p)$ .
Theorem 2.
(2) If $\alpha=f(p)$ , then is not -quasihyponormal for $q\in(0, p)$ , but-quasihyponormal for $q\in[p, \infty)$ . Hence satisfies the condition(i).
$T$

$p$

$T$$T$$q$

Proof.

$q$(1) Since$B^{2p}= \frac{1}{2}$ ,69
it is easy to see that$(9^{p}+1)/2-\alpha^{2p}\geq 0$$T$

is -quasihyponormal if and only if$p$.

(2) It is immediate from (1). QEDTheorem 3. Let $T(s, t)=|T|^{s_{U}}|T|^{b}$ for $t>0$ .
(1) If $T(s, t)$ is p–quasihyponormal, then $\alpha\leq f(s)$ .
(2) lf $\alpha=f(p)$ and $s\in(0,p)$ , then $T(s, t)$ is not q-quasihyponormal
for all $q>0$ . Hence satisfies the condition(ii).
$s,$$T$

Proof.

(1) Since$T(s, t)^{*}(|\tau(S, t)|^{2}p-|T(S, t)^{*}|^{2}p)\tau(s, t)$$=A^{s+\iota}\{(A^{\iota_{B}}2sAt)p-A^{2(}s+t)p\}A^{S}+t\otimes p_{-2}$$+A^{t}B^{S}\{B^{2(t)p}s+-(B^{s}A^{2tS}B)^{p}\}BsA^{t}\otimes p_{-1}$,$T(s, t)$ is $p$-quasihyponormal if and only if$(AtB2sAt)p-A2(s+t)p\geq 0$ ,and $A^{t}B^{s}\{B^{2(}S+t)p-(B^{s}A^{2tS}B)^{p}\}BsA^{t}\geq 0$.

The former inequality implies that $\alpha\leq f(s)$ .
(2) It is immediate from (1). QEDTheorem 4.

$T$

is paranormal if and only if $\alpha\leq\sqrt{5}=f(1)$ .

Proof.
It is well-known bythat an operator is a paranormal if and only if $S^{*2}S^{2}-2kS^{*}s+k^{2}\geq 0$ for all.
Since
$S$

$\mathrm{A}\mathrm{n}\mathrm{d}_{0}[3]$$k\in \mathbb{R}$$T^{*2}T^{2}-2k \tau^{*}\tau+k^{2}=\sum_{1n<-}(A^{2}-k)^{2}\otimes p_{n}+(AB^{2}A-2kA^{2}+k^{2})\otimes p_{-1}$$+ \sum_{n\geq 0}(B2-k)2\otimes p_{n}$

$T$

is a

$\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}1\Leftrightarrow AB^{2}A-2kA^{2}+k^{2}\geq 0$$\Leftrightarrow 5\alpha^{2}-2k\alpha^{2}+k^{2}\geq 0$.
$\forall k\in \mathbb{R}$$\forall k\in \mathbb{R}$$\Leftrightarrow\alpha^{4}-5\alpha^{2}\leq 0$$\Leftrightarrow\alpha\leq\sqrt{5}=f(1)$(since $\alpha>0$ ).

QED

Remark. If $\alpha=f(p)$ for $p>1$ , then is a -quasihyponormal byTheorem 2, but is not paranormal by Theorem 4. Hence satisfies
the condition(iii).
$T$

$T$

$p$

$T$

70
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$p$

$p$

$p$

$\mathrm{I}\mathrm{s}\mathrm{T}\mathrm{R}\dot{\mathrm{A}}\mathrm{T}^{\mathrm{E}\mathrm{S}}\mathrm{C}\mathrm{u}$$\mathrm{i}$$\mathrm{M}\mathrm{c}\mathrm{C}\mathrm{A}\mathrm{R}\mathrm{T}\mathrm{H}\mathrm{Y},$$C_{\rho}$$\mathrm{M}\mathrm{I}$

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Mathematical Institute, Tohoku University,
Sendai 980-8578 Japanuchiyama@math.tohoku.ac.jp