数理解析研究所講究録1080 巻 1999 年 41-55

41A subclass of paranormal including class of log-hyponormal
and several related classes
東京理科大理東京理科大理東京理科大理

山崎丈明
伊藤公訴
古田孝之

(Takeaki Yamazaki)(Masatoshi Ito)
(Takayuki Furuta)

Abstract
his report is based on the following preprint:
T.Furuta, M.Ito and T.Yamazaki, A subclass of paranormal operators includingclass of $log$-hyponormal and several related classes, to appear in Scientiae Mathematicae.
$..\mathrm{T}$We shall introduce a new class “class ” given by an operator inequality which
includes the class of -hyponormal operators and is included in the class of paranormal operators. It turns out that our results contain another proof of Ando’s
result [3] which every $log$-hyponormal operator is paranormal. Moreover we shallintroduce new classes related to class A operators and paranormal operators.
$\mathrm{A}$$\log$

Introduction

1

A capital letter means a bounded linear operator on a complex Hilbert space $H$ . An
operator is said to be positive (denoted by $T\geq 0$ ) if $(Tx, x)\geq 0$ for all $x\in H.$ andalso an operator is said to be strictly positive (denoted by $T>0$ ) if is positive andinvertible.
An operator is said to be -hyponormal if $(T^{*}T)^{p}\geq(TT^{*})^{p}$ for a positive number
and $log$ -hyponormal if is invertible and $\log T^{*}T\geq\log TT*.$ p–Hyponormal and loghyponormal operators were defined as extensions of hyponormal one, i.e., $T^{*}T\geq TT^{*}$ ,and also they have been studied by many authors, for instance, $[1][2][5][8][13][14][17][18]$and [19]. By the celebrated L\"owner-Heinz inequality $ttA\geq B\geq 0$ ensuresfor
any $\alpha\in[0,1]"$ , every p–hyponormal operator is -hyponormal for $p\geq q>0$ . And every-hyponormal since $\log t$ is an operator monotone
operator isinvertible
function.
An operator is paranomal if $||T^{2_{X}}||\geq||TX||^{2}$ for every unit vector $x\in H$ . It hasfor instance,
been studied by many authors, so there are too many to cite their$[3][9]$ and [15]. Ando [3] proved the following result.
$T$

$T$

$T$

$T$

$p$

$T$

$p$

$A^{\alpha}\geq B^{\alpha}$$q$$p \frac{-}{}\mathrm{h}\mathrm{y}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}1$$\log$$T$

references.’

Theorem A.l ([3]). Every $log$-hyponormal operator is paranormal.

an operator
this paper, firstly we shall introduce a new class “class ” given
inequality which properly includes the class of -hyponormal operators and is properly
included in the class of paranormal operators. It turns out that our results contain anotherclasses related to class A operators
proof of Theorem A.l. Secondly we shall introduce
and paranormal operators. Finally we shall give several examples to show that inclusion
relations among these classes are all proper.
$\mathrm{A}$$\mathrm{h}$$\log$$\mathrm{n}e\mathrm{w}$$\dot{\mathrm{b}}\mathrm{y}$42

2A subclass of paranormal operators including classof log-hyponormal
We shall introduce a new class of operators as follows:

Definition 1. .An operator$T$belongs to class

$A$

if

$|T^{2}|\geq|T|^{2}$

.

(2.1)

”
We would like to remark that class “ is named after the ‘iabsolute” values of two
in (2.1). We call an operator class A operator briefly if belongs
andoperators
to class A. We obtain the following results on class A operators.
$A$

$T$

$T$

$|T|$

$|T^{2}|$

Theorem 1.

(1) Every $log$-hyponormal operator is class operator.
(2) Every class operator is paranormal operator.
$A$

$A$

Theorem 1 contains another proof of Theorem A. 1. The following theorems and lemma
play an important role in the proofs of the results in this paper.

Theorem B.l $([6][10])$ . Let andproperties are mutually equivalent:

$B$

$A$

be positive invertible operators. Then the following(i) $\log A\geq\log B$ .

(ii)
(iii)

$A^{p}\geq(A^{E}2B^{p}A^{\epsilon}2)^{\frac{1}{2}}$for all
for all$p\geq 0$$A^{f} \geq(A^{\frac{r}{2}}B^{p}A\frac{r}{2})^{\frac{r}{p+r}}$.

$p\geq 0$

and $r\geq 0$ .

We remark that equivalence relation between (i) and (ii) is shown in [4].
inequality [16]). Let
Theorem B.2 (Then the following inequalities hold for all $x\in H$ :$\mathrm{H}\ddot{\mathrm{o}}1\mathrm{d}\mathrm{e}\Gamma^{-}\mathrm{M}\mathrm{c}\mathrm{c}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}\mathrm{y}$$A$

be a positive operator.

(i) $(A^{\mathrm{r}}x,x)\leq(Ax, x)’||X||\mathrm{z}(1-r)$ for $0<r\leq 1$ .
(ii) $(A^{r_{X}}, x)\geq(Ax, x)’||X||^{2}\langle 1-r$)

for$r\geq 1$.

As a slightly modification of [11, Lemma 1], we have the following lemma.
Lemma B.3. Let

$A$

and

$B$

be invertible operators. Then$(BAA^{*}B*)^{\lambda}=BA(A^{*}B^{*}BA)^{\lambda}-1A*B^{*}$

holds for any real number .
$\lambda$

43
Proof of Theorem 1.
Proof of (1). Suppose that

$T$

is

$\log$-hyponormal.

$T$

is

$\log|T|^{2}\geq\log|T^{*}|^{2}$$\log$-hyponormal iff
(2.2)

.

By the equivalence between (i) and (ii) of Theorem B.l, (2.2) is equivalent tofor all $p\geq 0$ .

$|T|\mathrm{t}2p\geq(|T|^{\mathrm{p}}.|\tau*|2p|T|^{p})^{\frac{1}{2}}$(2.3)

Put $p=1$ in (2.3), then we have$|T|^{2}\geq(|\tau||\tau*|^{2}|T|)^{\frac{1}{2}}$

.

(2.4)

By Lemma B.3 and $|T^{*}|^{2}=TT^{*},$ $(2.4)$ holds iff$|T|^{2} \geq|T|T(T^{*}|\tau|2\tau)\frac{-1}{2}T^{*}|T|$iff$(T^{*}|\tau|^{2}T)^{\frac{1}{2}}\geq^{\tau*}T$

,

(2.5)

so that$|T^{2}|\geq|T|^{2}$that is,$T$,

is class A.

Proof of (2). Suppose that$T$

is class$\mathrm{A}$, i.e.,$l$$|T^{2}|\geq|T|^{2}$

.

(2.1)

Then for every unit vector $x\in H$ ,$||\tau^{2_{X}}||2=((\tau 2)*\tau^{2_{X,x}})$$=(|\tau^{2}|^{2_{X}}, X)$$\geq(|T^{2}|x, x)^{2}$$\geq(|T|^{2}x, x)2$

by (ii) of Theorem B.2
by (2.1)$=||T_{X1}|^{4}$

Hence we have$||T^{2_{X}}||\geq||TX||^{2}$

so that$T$for every unit vector $x\in H$ ,is paranormal.

Whence the proof of Theorem 1 is complete.

$\square$44

3Several classes related to class A and paranormal
In this section, we shall discuss extensions of class A operators and paranormal operators. We shall introduce new classes of operators as follows:

Definition 2.
(1) For each $k>0_{y}$ an operator$T$is class $A(k)$ if$(T^{*}|\tau|^{2k}\tau)^{\frac{1}{k+1}}\geq|T|^{2}$(2) For each $k>0$ , an operator

$T$

$T$

(3.1)

is $absolute- k-pa\Gamma anormal$ iffor every unit vector $x\in H$ .

$|||T|^{k}\tau_{X||}\geq||TX||k+1$

An operator

.

(3.2)

(resp.
is class
is class A (resp. paranormal) if and only if. Now we shall discuss the inclusion relations among these classes.
$T$$\mathrm{A}(1)$$\mathrm{a}\mathrm{b}\mathrm{S}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}-1-\mathrm{P}^{\mathrm{a}}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}1)$

Theorem 2.
operator is class $A(k)$ operator for $k\geq 1$ .
(1) Every invertible class(2) Every paranormal operator is $ab_{SO}lute- k- pa\Gamma anormal$ operator for $k\geq 1$ .
(3) For each $k>0_{J}$ every class $A(k)$ operator is absolute-k-paranormal operator.
$A$

We need the following theorem in order to give a proof of Theorem 2.
Theorem C.l ([12]). Let and be positive invertible operators such that
. Then the following inequality
andholds for fixedholds:$A$$B$$A^{\beta_{0}}\geq(A^{\rho_{2}\rho_{2}\ovalbox{\tt\small REJECT}}B^{\alpha \mathrm{O}}\mathrm{n}\Delta A)^{\overline{\alpha}}0+\overline{\beta 0}$$\alpha_{0}>0$$A^{\beta}\geq(A^{\mathrm{g}}2B\alpha A^{e}2)\overline{\alpha}+E_{\overline{\rho}}$

Proof of Theo-rem 2.
Proof of (1). Suppose that$T$

is class$\mathrm{A}$for all$\beta_{0}>0$$\alpha\geq\alpha_{0}$and$\beta\geq\beta_{0}$.

, i.e.,$|T^{2}|\geq|T|^{2}$

.

(2.1)

(2.1) holds if and only if

.

(2.5)

$(T^{*}|T|^{2}\tau)^{\frac{1}{2}}\geq T^{*}T$

By Lemma$\mathrm{B}.3,$$(2.5)$

holds iff$T^{*}|T|(|T| \tau\tau^{*}|\tau|)\frac{-1}{2}|T|\tau\geq T^{*}T$

iff$|T|^{2}\geq(|T||\tau^{*}|^{2}|T|)^{\frac{1}{2}}$

.

(2.4)

45

Applying Theorem C.l to (2.4), we have$|T|^{2k} \geq(|T|^{k}|T^{*}|2|\tau|^{k})\frac{k}{k+1}$

for $k\geq 1$ .

(3.3)

By Lemma B.3 and $|T^{*}|^{2}=TT^{*},$ $(3.3)$ holds iff$|T|^{2k} \underline{>}|T|k\tau(\tau*|T|^{2}kT)\frac{-1}{k+1}\tau^{*}|\tau|k$for $k\geq 1$

iff$(T^{*}|T|^{2k}\tau)^{\frac{1}{k+1}}\geq|T|^{2}$

so that$T$is class$\mathrm{A}(k)$for $k\geq 1$ ,for $k\geq 1$ .

Proof of (2). Suppose that$T$

is paranormal. Then for every unit vector $x\in H$ and $k\geq 1$ ,$|||T|^{k}T_{X1}|^{2}=(|T|^{2}k\tau X, TX)$$\geq(|T|^{2}\tau_{X}, \tau x)k||TX||^{2(1-k})$by (ii) of Theorem B.2$=||T^{2}X||2k||\tau X||2\langle 1-k)$$\geq||TX||4k||T_{X1}|2(1-k)$$=||TX||2\langle k+1)$

by paranormality of$T$.

Hence we have$|||T|^{k}\tau X||\geq||TX||k+1$

so that$T$is$\mathrm{a}\mathrm{b}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}e- k_{\mathrm{P}}- \mathrm{a}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}1$

Proof of (3). Suppose that$T$for every unit vector $x\in H$ and $k\geq 1$ ,for $k\geq 1$ .

is class$\mathrm{A}(k)$for $k>0$ , i.e.,$(\tau^{*}|\tau|^{2k}\tau)^{\frac{1}{k+1}}\geq|T|^{2}$

for $k>0$ .

(3.1)

Then for every unit vector $x\in H$ ,$|||T|^{k}T_{X}||^{2}=(T^{*}|T|2k\tau x, x)$$\geq((T^{*}|T|^{2}kT)\frac{1}{k+1}x, x)^{k+1}$$\geq(|T|^{2}x, x)k+1$$=||Tx||2(k+1)$by (ii) of Theorem B.2
by (3.1).

Hence we have$|||\tau|^{k}\tau_{x}||\geq||TX||k+1$

so that$T$is$\mathrm{a}\mathrm{b}_{\mathrm{S}}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}- k$for every unit vector $x\in H$,-paranormal for $k>0$ .

Whence the proof of Theorem 2 is complete.

$\square$As further extensions of (2) and (3) of Theorem 2, we have the following two results.

46
Theorem 3. Let$T$

be an invertible class $A(k)$ operator for $k>0$ . Then$f(l)=(T^{*}|\tau|2lT)^{\frac{1}{l+1}}$

is increasing for $l\geq k>0$ , and the following inequality holds:$f(l)\geq|T|^{2},$Theorem 4. Let
vector $x\in H$ ,$T$$i.e.,$$T$is class $A(l)$ for $l\geq k>0$ .

be an $ab_{S}olute-k$-paranormal operator for $k>0$ . Then for every unit
$F(l)=|||T|^{l}T_{X}|| \frac{1}{l+1}$

is increasing for $l\geq k>0$ , and the following inequality holds:$F(l)\geq||TX||$ ,$i.e_{J}.T$

is $ab_{S}olute- l_{- pa}ranormal$ for $l\geq k>0$ .

Theorem 3 states the following: An operator function $f(l)$ asserts that every classfor $l\geq k>0$ . Similarly, Theorem 4 states the following: A function
operator is classof norm $F(l)$ asserts that every absolute-k-paranormal operator is absolute-l-paranormal
for $l\geq k>0$ .
$\mathrm{A}(k)$$\mathrm{A}(l)$In order to give a proof of Theorem 3, we need the following theorem which is anextension of Theorem C.1.

Theorem C.2 ([12]). Let and be positive invertible operators such that
. Then for fixedandholds for fixed
$B$

$A$

$\alpha_{0}>0$$A^{\beta_{0}}\geq(A^{\underline{\rho}_{\mathit{1}}}2B^{\alpha 0}A^{-}2)^{\frac{\beta_{0}}{\alpha_{0+\rho_{0}}}}\beta_{\mathrm{L}}$$\delta\geq-\beta_{\mathit{0}_{J}}$$\beta_{0}>0$$f( \alpha,\beta)=A\frac{-\beta}{2}(A2B\alpha A2)E\mathrm{g}\frac{\delta+\beta}{a+\beta}A^{\frac{-\mathcal{B}}{2}}$is a decreasing function of both$\alpha$Proof of Theorem 3. Suppose that

and
$T$

$\beta$foris class$\alpha\geq\max\{\delta, \alpha_{0}\}$$\mathrm{A}(k)$and$\beta\geq\beta_{0}$.

for $k>0$ , i.e.,$f(k)=(\tau^{*}|\tau|^{2k}\tau)^{\frac{1}{k+1}}\geq|T|^{2}$

.

(3.1)

By Lemma B.3, (3.1) holds iff$T^{*}|T|^{k}(|\tau|kT\tau*|\tau|^{k})^{\frac{-k}{k+1}|T|}k\tau\geq T^{*}T$iff$|T|2k \geq(|\tau|k|\tau^{*}|2|T|^{k})\frac{k}{k+1}$

.

(3.4)

47

By applying Theorem C.2 to (3.4),$g(l)=|\tau|^{-}l(|T|l|T^{*}|^{2}|T|l)^{\frac{l}{l+1}}|T|^{-l}$is decreasing for $l\geq k>0$ . And we have$g(l)=|T|^{-}l(|T| \iota|T*|^{2}|\tau|l)\frac{l}{l+1}|\tau|-\iota$$=|T|^{-}l(|T|^{\iota l}\tau T^{*}|\tau|)^{\frac{l}{l+1}1}\tau|^{-^{\iota}}$by Lemma B.3$=T(T^{*}|T|^{2l}T)^{\frac{-1}{l+1}}\tau^{*}$$=T \{(T^{*}|T|^{2}lT)\frac{1}{l+1}\}^{-1}T^{*}$$=T\{f(l)\}^{-}1T*$.

Hence $f(l)$ is increasing for $l\geq k>0$ . Moreover,$( \tau^{*}|\tau|2\iota\tau)\frac{1}{l+1}=f(l)\geq f(k)\geq|T|^{2}$that is,$T$

is class$\mathrm{A}(l)$,for $l\geq k>0$ . Whence the proof of Theor

Proof of Theorem 4. Suppose that$T$$e\mathrm{m}3$

is complete.

$\square$is absolute-k-paranormal for $k>0$ , i.e.,$|||\tau|^{k}\tau_{x}||\geq||TX||^{k}+1$

for every unit vector $x\in H.$(3.2)(3.2) holds if and only if$F(k)=|||T|^{k}Tx||^{\frac{1}{k+1}}\geq||TX||$

for every unit vector $x\in H$.

Then for every unit vector $x\in H$ and any such that $l\geq k>0$ , we have$l$$F(l)=|||T|^{l} \tau x||\frac{1}{l+1}$$=(|T|2lT_{X,T)^{\frac{1}{2(l+1)}}}\backslash x$$\geq \mathrm{t}(|T|^{2k}\tau X, Tx)^{\frac{l}{k}}||TX||2(1-\frac{l}{k})\}^{\frac{1}{2(l+1)}}$by (ii) of Theorem B.2$= \{|||\tau|^{k}\tau_{X1}|\frac{2l}{k}||TX||2\mathrm{t}1-\frac{\iota}{k})\}^{\frac{1}{2\mathrm{t}^{\mathrm{t}+}1)}}$$\geq\{||Tx||^{\frac{2l(k+1)}{k}}||TX||2(1-\frac{l}{k})\}^{\frac{1}{2\mathrm{t}^{\iota+}1)}}$$=||TX||$by (3.2).

Hence$F(l)=|||T|^{l}\tau_{X1}|^{\frac{1}{l+1}}\geq||TX||$

so that$T$is$\mathrm{a}\mathrm{b}\mathrm{S}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}- l- \mathrm{P}^{\mathrm{a}\mathrm{r}\mathrm{a}}\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}1$

for every unit vector $x\in H$ and $l\geq k$ ,for $l\geq k>0$ .

(3.5)

48
Next we show that $F(l)$ is increasing for $l\geq k>0$ . For every unit vector $x\in H$ and
any $m$ and such that $m\geq l\geq k>0$ ,$l$$F(m)=|||\tau|^{m}Tx||^{\frac{1}{m+1}}$$=(|T|^{2m}T_{X,T_{X}})^{\frac{1}{2i^{m}+1)}}$$\geq\{(|T|^{2l}TX, T_{X})^{\frac{m}{\iota}}||TX||2\mathrm{t}1-\frac{m}{l})\}^{\frac{1}{2(m+\iota)}}$by (ii) of Theorem B.2$= \{|||T|lTX||^{\frac{2m}{l}}||T_{X1}|2(1-\frac{m}{l})\}^{\frac{1}{2(m+1)}}$$\geq\{|||T|l\tau X||\frac{2m}{l}|||T|^{\iota}TX||^{\frac{2}{\mathrm{t}+1}(}1-\frac{m}{l})\}^{\frac{1}{2\{m+1\}}}\mathrm{b}\mathrm{y}(3.5)$$=|||T|^{l}Tx|| \frac{1}{\mathrm{t}+1}$$=F(l)$.

Hence $F(l)$ is increasing for $l\geq k>0$ . Whence the proof of Theorem 4 is compl

4$e\mathrm{t}\mathrm{e}$.

$\coprod$

Examples-paranormal operators.
In this section, we shall give a characterization of
And by using this characterization, we shall give several examples showing that inclusion
relations among the classes discussed in this paper are all proper.
Ando [3] proved the following Theorem D.1.
$\mathrm{a}\mathrm{b}_{\mathrm{S}}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}- k$

Theorem D.l ([3]). An operator$T$is paranormal if and only if$T^{*2}\tau^{2}-2\lambda T^{*}T+\lambda^{2}\geq 0$$k$for all$\lambda>0$.

As a generalization of Theorem D.l, we have the following characterization of absolute-paranormal operators.

Theorem 5. For each $k>0$ , an operator$T$

is $ab_{S}olute- k$ -paranormal if and only if$T^{*}|T|2k\tau-(k+1)\lambda^{k}|T|^{2}+k\lambda^{k+1}\geq 0$

for all

In fact Theorem 5 implies Theorem D.l putting $k=1$ in$\lambda>0$.

orem 5.

$\mathrm{T}\mathrm{h}_{c}\mathrm{e}$We cite the following well-known lemma in order to give a proof of Theorem 5.
Lemma D.2. Let

$a$

and

$b$

be positive real numbers. Then$a^{\lambda}b^{\mu}\leq\lambda a+\mu b$

for$\lambda>0$

and $\mu>0$ such that $\lambda+\mu=1$ .

(4.1)

49
Proof of Theorem 5. Suppose that$T$

is absolute-k-paranormal for $k>0$ , i.e.,
for every unit vector $x\in H.$$|||T|^{k}\tau_{X}||\geq||TX||k+1$(3.2)(3.2) holds iff

for all $x\in H$$|||T|^{k}T_{X1}| \frac{1}{k+1}||x||\frac{k}{k+1}\geq||TX||$

iff$(T^{*}|T|^{2}kTX, x) \frac{1}{k+1}(x, X)^{\frac{k}{k+1}}\geq(|T|^{2}X, x)$

for all $x\in H$ .

(4.2)

By Lemma D.2,$( \tau*|\tau|^{2}kTx, X)^{\frac{1}{k+1}(x,X})^{\frac{k}{k+1}}=\{(\frac{1}{\lambda})^{k}(T^{*}|\tau|^{2}k\tau_{Xx},)\}\frac{1}{k+1}\cdot\{\lambda(X, X)\}^{\frac{k}{k+1}}$(4.3)$\leq\frac{1}{k+1}\frac{1}{\lambda^{k}}(\tau*|\tau|^{2}k\tau_{x,X})+\frac{k}{k+1}\lambda(x,x)$

for all $x\in H$ and all$\lambda>0$,so (4.2) ensures the following (4.4) by (4.3).
$\frac{1}{k+1}\frac{1}{\lambda^{k}}(\tau^{*}|\tau|^{2}k\tau_{x,x)}+\frac{k}{k+1}\lambda(x, x)\geq(|T|^{2}x, x)$

for all $x\in H$ and all

Convers$e1\mathrm{y},$$(4.2)$

follows from (4.4) by putting$(T^{*}|\tau|^{2}kTx, x)=0$, let$\lambdaarrow 0.$$\lambda>0$

(4.4)

.

$\lambda--\{\frac{(T^{*}|\tau|^{2}kTx,x)}{(x,x)}\}^{\frac{1}{k+1}}$(In case) Hence (4.2) is equivalent to (4.4), and (4.4) holds if andonly if$T^{*}|T|2k\tau-(k+1)\lambda^{k}|T|^{2}+k\lambda^{k+1}\geq 0$

for all$\lambda>0$

,

(4.1)

so that the proof is complete.

$\square$

The following Proposition 6 is obtained by easy calculations, so we omit to describethcalculations.
$e\mathrm{s}\mathrm{e}$Proposition 6. Let$H$$K= \bigoplus_{n=-\infty}^{\infty}H_{n}$, define the operator$T_{A,B}$

where$H_{n}\cong H$. For given positive operators

on $K$ as follows:

$B0$

$B0$

$T_{A,B}=(.\cdot.\cdot.\cdot$
$A0$

$A0$

$\prime 0..$$\cdot..$where$\square$shows the place of the $(0,0)$ mat$\gamma\dot{\eta}X$).

$A,$ $B$

on

(4.5)

element. Then the following assertions hold:

50
(i)

$T_{A,B}$

is $log$ -hyponormal if and only if

and

$A$

$B$

are invertible and$\log A\geq\log B$(ii) For each $k>0,$.

is class $A(k)$ if and only if$T_{A,B}$$(BA^{2k}B)^{\frac{1}{k+1}}\geq B2$.

(iii) For each $k>0_{f}T_{A,B}$ is $ab_{S}olute-k$ -paranormal if and only if$BA^{2k}B-(k+1)\lambda^{k}B^{2}+k\lambda^{k+1}\geq 0$

for all$\lambda>0$.

By using Proposition 6, we can give several examples to show that inclusion relations
among thes classes are all proper.
$e$.ple 1. Let$\mathrm{E}\mathrm{x}\mathrm{a}\mathrm{m}$$K= \bigoplus_{n=-\infty}^{\infty}H_{n}$, define the operator
examples.

$\mathbb{R}^{2}$$T_{A,B}$

where$H_{n}\cong \mathbb{R}^{2}$

on $K$ as (4.5) in$\dot{P}$. For given positive matrices$A,$ $B$onroposition 6. Then we have the foflowing

We remark that the trace of a matrix $X$ denotes tr $X$ and the determinant of a matrix$X$ denotes $\det X$ .
(1) An example of non-log-hyponormal, class$A$

operator.

Let$A=$ and $B=$
hold together. On
Fujii, Furuta and Wang [7] shows that $\log A\not\geq\log B$ andholds by Lemma B.3. Therefore
holds iff
the other hand,by(i) and (ii) of Proposition 6.
but class A
is$A^{2}\geq(AB^{2}A)^{\frac{1}{2}}$$(BA^{2}B)^{\frac{1}{2}}\geq B^{2}$$A^{2}\geq(AB^{2}A)^{\frac{1}{2}}$$T_{A,B}$$\mathrm{n}\mathrm{o}\mathrm{n}-\log-\mathrm{h}\mathrm{y}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}$(2) An example of non-class$A$, class$A(\mathit{2})$, paranormal operator.

Let$A=(_{0}^{2}2\sqrt{23}0)$

and$B=$ .

Then$(BA^{2}B) \frac{1}{2}-B^{2}=$ .
are 12.585... and-0.64001... , so that
Eigenvalues ofis non-class A by (ii) of Proposition 6.
So$(BA^{2}B)^{\frac{1}{2}}-B^{2}$$T_{A,B}$$(BA^{2}B)^{\frac{1}{2}}\not\geq B^{2}$.

51
On the other hand,$(BA^{4}B)^{\frac{1}{3}}-B^{2}=$ .
Eigenvalues ofis class$T_{A,B}$

are 24.760. .. and 0.31608... , so thatby (ii) of Proposition 6.

$(BA^{4}B)^{\frac{1}{3}}-B^{2}$$\mathrm{A}(2)$Furthermore, for$\lambda>_{\iota}0$, define$X_{1}(\lambda)$$(BA^{4}B)^{\frac{1}{3}}\geq B^{2}$. So

as follows:$X_{1}(\lambda)=BA^{2}B-2\lambda B^{2}+\lambda^{2}=$ .

Put

and

$p_{1}(\lambda)=\mathrm{t}\mathrm{r}X_{1}(\lambda)$$q_{1}(\lambda)=\det X_{1}(\lambda)$, then$p_{1}(\lambda)=2\lambda 2-52\lambda+1^{\cdot}248$$=2(\lambda-13)^{2}+910>0$and$q_{1}(\lambda)=(404-26\lambda+\lambda^{2})(844-26\lambda+\lambda^{2})-(-576+24\lambda)^{2}$$=\lambda^{4}-52\lambda^{3}+1348\lambda^{2}-48\mathrm{o}\mathrm{o}\lambda+9200$.

By calculation,$q_{1}’(\lambda)=4\lambda^{3}-156\lambda^{2}+2696\lambda-4800$$=4(\lambda-2)(\lambda^{2}-37\lambda+600)$$=4( \lambda-2)\{(\lambda-\frac{37}{2})^{2}+\frac{1031}{4}\}$.

, that is, $q_{1}(\lambda)\geq q_{1}(2)=4592>0$ for all
So
iff. Hence$X_{1}(\lambda)=p_{1}(\lambda)>0$$\det
x_{1}(\lambda)=q_{1}(\lambda)>0$
for allsince trandfor all. Therefore
is paranormal by (iii) of Proposition 6.
$q_{1}’(\lambda)=0$$\lambda=2$$\lambda>0$$\lambda>0$$X_{1}(\lambda)\geq 0$$\lambda>0$$T_{A,B}$(3) An example of non-class$A(\mathit{2}),$-paranormal operator.

$ab_{S}olute- \mathit{2}$Let$A=$ and$B= \frac{1}{2}(_{1-\sqrt{3}}^{1+\sqrt{3}}1+\sqrt{3}1-\vee 3\gamma$

.

Then

$(BA^{4}B) \frac{1}{3}-B^{2}=$ .
Eigenvalues ofis non-classSoare 1.4151... and-0.14687. .. , so thatby (ii) of Proposition 6.

$(BA^{4}B)^{\frac{1}{3}}-B^{2}$$T_{A,B}$$\mathrm{A}(2)$$(BA^{4}B)^{\frac{1}{s}}\not\geq B^{2}$.

52
On the other hand, for$\lambda>0$, define$X_{2}(\lambda)$

as follows:$X_{2}(\lambda)=BA4B-3\lambda 2B2+2\lambda 3=(^{24-8\sqrt{3}-6\lambda}-12+3\lambda 2^{+2\lambda}2324+8\sqrt{3}-6\lambda 2+2\lambda^{3)}-12+3\lambda 2$

Put$p_{2}(\lambda)=\mathrm{t}\mathrm{r}X_{2}(\lambda)$and$q_{2}(\lambda)=\det x_{2}(\lambda)$.

, then$p_{2}(\lambda)=4\lambda^{\mathrm{s}_{-1}2}2\lambda+48$

and$q_{2}(\lambda)=(24-8\sqrt{3}-6\lambda^{2}+2\lambda^{3})(24+8\sqrt{3}-6\lambda^{2}+2\lambda^{3})-64$.

$=4\lambda^{6}-24\lambda^{5}+27\lambda^{4}+96\lambda^{3}-216\lambda^{2}+240$

We easily obtain$p_{2}(\lambda)>0$

for all$\lambda>0$. And we have$q_{2}’(\lambda)=24\lambda^{5}-120\lambda^{4}+108\lambda^{3}+288\lambda^{2}-432\lambda$$=12\lambda(\lambda-2)(2\lambda^{3}-6\lambda 2-3\lambda+18)$

So

.

by an
calculation,
iff $\lambda=0,2$ since $2\lambda^{3}-6\lambda^{2}-3\lambda+18>0$ for allthat is, $q_{2}(\lambda)\geq q_{2}(2)=64>0$ for all $\lambda>0$ . Hencefor all $\lambda>0$ since. Therefore
is absolutetr $X_{2}(\lambda)=p_{2}(\lambda)>0$ and $\det X_{2}(\lambda)=q_{2}(\lambda)>0$ for all
2-paranormal by (iii) of Proposition 6.
$\lambda>0$$q_{2}’(\lambda)=0$$\mathrm{e}\mathrm{a}\dot{\mathrm{s}}\mathrm{y}$$X_{2}(\lambda)\geq 0$$\lambda>0$(4) An example of non-paranormal,$T_{A,B}$-paranormal operator.

$ab_{S}olute- \mathit{2}$Let$A=$ and $B=$ .
Then for$\lambda>0$, define$X_{3}(\lambda)$

as follows:$X_{3}(\lambda)=BA^{2}B-2\lambda B^{2}+\lambda^{2}=$Put$\lambda=4$

.

(4.6)

in (4.6), then$X_{3}(4)=\not\geq 0$ .

So

$T_{A,B}$

is non-paranormal by (iii) of Proposition 6.

On the other hand, for$\lambda>0$, define$X_{4}(\lambda)$

as follows:$X_{4}(\lambda)=BA^{4}B-3\lambda^{2}B^{2}+2\lambda^{3}=$ .

By an easy calculation, $80-12\lambda^{2}+2\lambda^{3}>0$ for all $\lambda>0$ . So
is absolute-2-paranormal by (iii) of Proposition 6.
that is,$T_{A,B}$$X_{4}(\lambda)\geq 0$

for all’$\lambda>0\square$

53

5$\mathrm{r}e$Remarks
sult, we obtain the following Theoren 7.

Theorem 7. If an operator
maloid.

$T$

is $ab_{SO}lute- k$-paranormal for some $k>0$ , then$T$

is nor-for some $0<k<1,$isparanormal by Theorem 4, so that is normaloid as shown in [9] and [15]. So we consider
only the case of $k\geq 1$ . Suppose is absolute-k-paranormal for some $k\geq 1$ , i.e.,

Proof of Theorem 7. In case

$T$

is

$T$

$\mathrm{a}\mathrm{b}\mathrm{S}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}e- k- \mathrm{P}^{\mathrm{a}\mathrm{r}}\mathrm{a}\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}1$

$T$

$T$

$|||\tau|^{k}\tau_{x}||\geq||TX||k+1$

for every unit vector $x\in.H$.

(3.2)(3.2) holds iff$|||T|^{k}\tau X||||x||^{k}\geq||TX||k+1$

for all $x\in H.$(5.1)

We shall show that
(5.2)$||T^{n}||=||T||^{n}$for all positive integers . When $n=1,$ $(5.2)$ is always holds. Assume that (5.2), some positive integer. Then for ev.ery unit vector $x\in H$ , we have$n$$\mathrm{h}\mathrm{o}.1$

ds for$n$$||T^{n}x||^{k+}1=||T\cdot\tau^{n}-1X||k+1$

by (5.1)$\leq|||T|^{k}\tau\cdot T^{n-1}X||\cdot||\tau^{n-1}x||^{k}$$\leq|||T|^{k-1}||\cdot|||T|\tau^{n}x||\cdot||T^{n-1}||^{k}$$\leq||T||^{k-1}\cdot||T^{n+1}X||\cdot||T||^{(n-1})k$$\leq||T^{n+1}||\cdot||T||^{n}k-1$,that is,(5.3)$||T^{n+1}||\cdot||T||^{n}k-1\geq||T^{n}||k+1$

By the assumption (5.2) for $n,$ $(5.3)$ holds if and only if$||T^{n+1}||\cdot||T||^{n}k-1\geq||T||^{n}\mathrm{t}k+1)$

that is $||T^{n+1}||\geq||T||n+1$ , so that $||T^{n+1}||=|!^{\tau||^{n+1}}$
Therefore $||T^{n}||=||T||^{n}$ for all positive integers
Theorem 7 is complete.

$n$,by induction. Hence the proof$\mathrm{o}\mathrm{f}\square$

We can give an example to show that inclusion relations between absolut $e- k$ -paranormal
and normaloid are proper.

54
Example 2. There exists a $non- ab_{S}olute-k$ -paranormal for any $k>0$ and normaloid
operator.

LLet$T=$ .
Then $||T^{n}||=||T||^{n}$ for all positive integers by an easy calculation. However, the relation$|||T|k\tau_{x1}|\geq||TX||k+1$ does not hold for the unit vector $e_{2}=(0,1,0)$ since$n$$|T|^{k}T=$ .
Hence

$T$

is non-absolute-k-paranormal for any $k>0$ .

$\square$Remark 2. The following diagram expresses the inclusion relations among the classes
discussed in this paper.

55

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