FURUTA
INEQUALITY
ANDp-wA(s, t)
OPERATORSM.
\mathrm{C}\mathrm{H}\overline{\mathrm{O}}
, T.PRASAD,
M.RASHID,
K. TANAHASHI ANDA. UCHIYAMA
Dedicatedtothe memory ofProfessor
Takayuki
Furuta withdeep
gratitude
ABSTRACT. The aim ofthis paper is to introduce small
history
with Furatas
inequality
andrelating
class ofp‐wA
(s, t)
operators.1. INTRODUCTION
Let
B(\mathcal{H})
be thealgebra
of all bounded linearoperators
on acomplex
Hilbert space\mathcal{H}. In
1987,
Furuta[5]
proved
thefollwing inequality.
Theorem 1.
[Furuta
inequality]
Let 0 <p,q,r \in \mathbb{R} and
A,
B \inB(\mathcal{H})
satisfy
0\leq
B\leq A.
If p+2r
\leq(1+2r)q
and1\leq q
, thenB\displaystyle \frac{p+2r}{q}
\leq(B^{r}A^{p}B^{ $\Gamma$})^{\frac{1}{q}}
and(A^{r}B^{p}A^{r})^{\frac{1}{\mathrm{q}}}\displaystyle \leq A\frac{p+2r}{\mathrm{q}}
This is a
good
extensionof Löwner‐Heinzsinequality
([7]
and[12]).
Theorem 2.
[Löwner‐Heinzs inequality]
Let
A,
B\in B(\mathcal{H})
satisfy
0\leq B\leq A
and0<p\leq 1
. ThenB^{p}\leq A^{p}.
Recall that an
operator
T is saidto behyponormal
ifT^{*}T\geq TT^{*}
. ForT\in B(\mathcal{H})
,set|T|=(T^{*}T)^{\frac{1}{2}}
asusual.By taking
U|T|x=Tx
for x\in \mathcal{H}andUx=0 for
x\in \mathrm{k}\mathrm{e}\mathrm{r}|T|,
T hasaunique
polar decomposition
T=U|T|
with\mathrm{k}\mathrm{e}\mathrm{r}U=\mathrm{k}\mathrm{e}\mathrm{r}|T|
.Aluthge
[1]
definedAluthge
transformation\tilde{T}=|T|^{\frac{1}{2}}U|T|^{\frac{1}{2}},
and studied
interesting properties
ofp‐‐hyponormal
operators
for0<p\leq 1.
Definition 3. T \in
B(\mathcal{H})
is said to bep‐‐hyponormal
if(T^{*}T)^{p}
\geq
(TT^{*})^{p}
where
p\in(0,1].
The class of r
‐hyponormal
operators
is ageneralization
ofthe class ofhyponormal
operators
by
Löwner‐Heinzsinequality.
Aluthge
[1]
proved
follwing
result.Theorem 4. Let T bep
‐hyponormal. If
0<p\leq 1/2
, then\tilde{T}
is(p+1/2)-hyponormal.
If
1/2\leq p\leq 1
, then\tilde{T}
ishyponormal.
This is a epoc
making
result.Aluthge
transformation is astrong
toolof
operator
theory
and manyapplications
has beenstudied,
forexample,
Putnam
ineqality,
Fuglede
Putnamtype
theorem, Wyle
type
theorem. Ithink that
generalization
ofclassofoperators
may begood
waytoinvestigate
non‐normal
operators.
Furuta[6]
and Yoshino[17]
defindedgeneralized
transformation
T(s, t)=
|T|^{S}U|T|^{t}
with0<s,
t andYanagida
[15],
Ito[8],
Definition 5. T is saidto be
wA(s, t)
if(1.1)
(|T^{*}|^{t}|T|^{2s}|T^{*}|^{t})^{\frac{t}{s+t}} \geq|T^{*}|^{2t}
and
(1.2)
|T|^{2s}\geq(|T|^{s}|T^{*}|^{2t}|T|^{s})^{\frac{s}{s+t}}.
Hence
generalized Aluthge
transformationT(s,t)
ofwA(s, t)
operator
Tenjoys
thefollowing
property.
Proposition
6. Let T bewA(s, t)
. Then|T(s, t)|^{\frac{2t}{s+t}} \geq|T|^{2t}
and
|T|^{2s}\geq|T(s, t)^{*}|^{\frac{2 $\epsilon$}{s+t}}
Hence|T(s,t)|^{\frac{2r}{s+t}} \geq|T|^{2r}\geq|T(s,t)^{*}|^{\frac{2r}{s+t}}
for
allr\in(0,
\displaystyle \min\{s,
tIto and Yamazaki
[9]
proved
that(1.1)
implies
(1.2).
This is agood
result. Thismeansthat class of
wA(s,t)
operators
arecoincideswithA(s, t)
operators.
Definition 7. T is saidto be
A(s, t)
if(1.3)
(|T^{*}|^{t}|T|^{2s}|T^{*}|^{t})^{\frac{t}{s+t}}\geq|T^{*}|^{2t}.
Class
A(1,1)
is said to be class A and classA(1/2,1/2)
is said to bew
‐hyponormal
[4,
9,
15].
Prasad and Tanahshi[16]
defindedp‐wA
(s, t)
op‐erator for
0<p\leq 1
and0<s, t, s+t\leq 1
as follows.Deflnition 8. T is saidto be
p‐wA
(s, t)
if(1.4)
(|T^{*}|^{t}|T|^{2s}|T^{*}|^{t})^{\frac{\mathrm{p}\mathrm{t}}{ $\epsilon$+t}}\geq|T^{*}|^{2pt}
and
(1.5)
|T|^{2ps}\geq(|T|^{s}|T^{*}|^{2t}|T|^{s})^{\frac{\mathrm{p}s}{s+\mathrm{t}}}.
Hence Í\succwA
(s, t)
operator
is ageneralization
ofwA(s, t)
operator
by
Löwner‐Heinzs
inequality.
The aim ofthis paper is to prove several prop‐erties of
p-wA(s, t)
operaor and show some openproblems
ofp‐wA
(s, t)
operaor. Main results are
proved
in[16]
and[2].
2. RESULTS
At
first,
weshowgeneralized Aluthge
transformationT(s,t)
ofp‐wA
(s,t)
operator
Tenjoys
thefollowing
property
[16].
Theorem 9. Let T bep‐
wA(s, t)
. Then|T(s, t)|^{\frac{2pt}{s+t}} \geq|T|^{2pt}
and|T|^{2ps}\geq|T(s, t)^{*}|^{\frac{2}{s}\mathrm{g}_{\frac{s}{t}}}+
Hencefor
allr\in(0,
\displaystyle \min\{s,
tProof.
(|T^{*}|^{\mathrm{t}}|T|^{2s}|T^{*}|^{t})^{4_{\frac{t}{+t}}}\mathrm{s} \geq|T^{*}|^{2pt}
\Leftrightarrow(U|T|^{t}U^{*}|T|^{2s}U|T|^{t}U^{*})^{\frac{\mathrm{p}t}{ $\epsilon$+t}}\geq U|T|^{2pt}U^{*}
\Leftrightarrow U(|T|^{t}U^{*}|T|^{2s}U|T|^{t})^{\mathrm{g}_{\frac{\mathrm{t}}{t}}}\dot{s}+U^{*}\geq U|T|^{2pt}U^{*}
(
[8
, Lemma2.1])
\Leftrightarrow(|T|^{t}U^{*}|T|^{2s}U|T|^{t})^{\frac{pt}{8+t}}
\geq|T|^{2pt}
( [8
, lemma2.1])
\Leftrightarrow|T(s, t)|^{\frac{2_{ $\Gamma$)}t}{ $\epsilon$+t}} \geq|T|^{2pt}.
Also,
(|T|^{s}|T^{*}|^{2t}|T|^{s})^{\frac{ps}{ $\epsilon$+t}}\leq |T|^{2ps}
\displaystyle \Leftrightarrow(|T|^{s}U|T|^{2t}U^{*}|T|^{s})^{1}\ovalbox{\tt\small REJECT}\frac{8}{+t} \leq|T|^{2ps}
\Leftrightarrow|\{T(s, t)\}^{*}|^{\frac{2ps}{s+t}} \leq |T|^{2ps}.
\square
Nextwe show classof
p‐wA
(s, t)
operators
aredecreasing
class of opera‐torswith
0<p\leq 1
andincreasing
with0<s,
t\leq 1
. Theproof
isessentially
due to C.
Yang
and J. Yuan([19]
Proposition
3.4).
Lemma 10.
If
T isp‐wA
(s, t)
and0<s\leq s_{1}, 0<t\leq t_{1}, 0<p_{1} \leq p\leq 1,
thenT is
p_{1}-wA(s_{1}, t_{1})
.Proof.
Let Tbep‐wA
(s, t)
. Then(2.1)
(|T^{*}|^{t}|T|^{2s}|T^{*}|^{t^{t}})^{B}\dot{s}+\overline{t} \geq |T^{*}|^{2tp}
and
(2.2)
|T|^{2sp}\geq(|T|^{s}|T^{*}|2t^{s}|T|^{S})^{-B_{\overline{t}}}s+.
We prove that T is
p\leftrightarrow wA(s_{1}, t_{1})
. Then T isp_{1}-wA(s_{1}, t_{1})
by
Lowner‐Heinzsinequality.
Let
A_{1}
=(|T^{*}|^{t}|T|^{2s}|T^{*}|^{t})^{\frac{ip}{s+\mathrm{t}}}
andB_{1}
=|T^{*}|^{2tp}
. Since
(1)
imphes A_{1}
\geq
B_{1}
, wehave(B^{\frac{r2}{1^{2}}}A_{1}^{p_{2}}B^{\frac{r_{2}}{1^{2}}})^{\frac{1}{p_{2}}}+r\vec{+r_{2}} \geq B_{1}^{1+r2}
for any
r_{2}>0
andp_{2}\geq 1
by
Furutasinequality
[5].
Let$\beta$\displaystyle \geq t,p_{2}=\frac{s+t}{tp}\geq 1, r_{2}=\frac{ $\beta$-t}{tp}\geq 0.
Then
(|T^{*}|^{ $\beta$}|T|^{2s}|T^{*}|^{ $\beta$})^{\frac{tp+ $\beta$-t}{s+ $\beta$}} \geq|T^{*}|^{2tp+2 $\beta$-2t}.
Hence wehave
(|T^{*}|^{ $\beta$}|T|^{2s}|T^{*}|^{ $\beta$})^{\frac{w}{s+ $\beta$}}\geq|T^{*}|^{2w}
Let
for
$\beta$\geq t
. Thenf_{s}( $\beta$)= (|T|^{s}|T^{*}|^{2 $\beta$}|T|^{s})^{\frac{s}{s+ $\beta$}}
f_{s}( $\beta$)=\{(|T|^{s}|T^{*}|^{2 $\beta$}|T|^{s})^{\frac{s+ $\beta$+w}{s+ $\beta$}}\}^{\frac{s}{s+ $\beta$+w}}
=\{|T|^{s}|T^{*}|^{ $\beta$}(|T^{*}|^{ $\beta$}|T|^{2s}|T^{*}|^{ $\beta$})^{\frac{w}{s+ $\beta$}}|T^{*}|^{ $\beta$}|T|^{s}\}^{\frac{s}{ $\epsilon$+ $\beta$+w}}
\geq\{|T|^{s}|T^{*}|^{ $\beta$}|T^{*}|^{2w}|T^{*}|^{ $\beta$}|T|^{s}\}^{\frac{s}{s+ $\beta$+w}}
=\{|T|^{s}|T^{*}|^{2( $\beta$+w)}|T|^{s}\}^{\frac{\mathrm{s}}{\'{o}+ $\beta$+w}}
=f_{s}( $\beta$+w)
.Hence
f_{s}( $\beta$)
isdecreasing
for$\beta$\geq t.
Then, by
(2.2),
|T|^{2sp}\geq(|T|^{s}|T^{*}|^{2t}|T|^{s})^{\frac{\mathrm{s}p}{s+t}}
=\{f_{s}(t)\}^{p}
\geq\{f_{s}(t_{1})\}^{p}=(|T|^{s}|T^{*}|^{2t_{1}}|T|^{s})^{-}\ovalbox{\tt\small REJECT} +t_{1}sp
Let
A_{2}=|T|^{2sp}
andB_{2}=
(|T|^{s}|T^{*}|^{2t_{1}}|T|^{s})^{\frac{S\mathrm{f}^{\mathrm{j}}}{s+t_{1}}}
. ThenA_{2}^{1+r3}\geq (A_{2}\not\simeq^{r}B_{2}^{p_{3}^{r}}A_{2}^{\Rightarrow})^{\frac{1+r3}{\mathrm{p}_{3}+r_{3}}}
for any
r3\geq 0
andp3\geq 1
by
Furutasinequality
[5].
Letp_{3}=\displaystyle \frac{s+t_{1}}{sp}\geq 1, r3=\frac{s_{1}-s}{sp}\geq 0.
Then|T|^{2sp+2s_{1}-2s}\geq (|T|^{s_{1}}|T^{*}|^{2t_{1}}|T|^{s_{1}})^{\frac{ $\epsilon$ p+s_{1-B}}{s_{1}+t_{1}}}
Since
sp+s_{1}-s-s_{1}p=(s_{1}-s)(1-p) \geq 0,
wehave|T|^{2s_{1}p}\geq (|T|^{s_{1}}|T^{*}|^{2t_{1}}|T|^{s_{1}})^{\frac{s_{1}p}{81+t_{1}}}
Similarly,
wehave(|T^{*}|^{t_{1}}|T|^{2s_{1}}|T^{*}|^{t_{1}})^{\frac{t}{s_{1}}\mathrm{L}_{\overline{t_{1}}}^{p}}+ \geq|T^{*}|^{2t_{1}p}.
Hence T isp‐wA
(s_{1}, t_{1})
. \squareThe
following
result seemsnew, even for classA(s, t)
operators.
Proof.
LetT=U|T|
thepolar decomposition
of T. Then|T^{-1}|^{2}=(T^{-1})^{*}T^{-1}=(T^{*})^{-1}T^{-1}=(TT^{*})^{-1}=|T^{*}|^{-2}.
Hence|T^{-1}|=|T^{*}|.
Also,
|(T^{-1})^{*}|^{2}=(T^{-1})(T^{-1})^{*}=T^{-1}(T^{*})^{-1}=(T^{*}T)^{-1}=|T|^{-2}.
Hence|(T^{-1})^{*}|=|T|^{-1}.
Then\{|(T^{-1})^{*}|^{s}|T^{-1}|^{2t}|(T^{-1})^{*}|^{s}\}^{\frac{ $\epsilon$ \mathrm{p}}{s+l}}
=(|T|^{-s}|T^{*}|^{-2t}|T|^{-s})^{\frac{ $\epsilon$ p}{s+l}}
=(|T|^{s}|T^{*}|^{2t}|T|^{s})^{-p}\overline{ $\epsilon$}+^{\frac{s}{t}}
\geq|T|^{-2sp}=|(T^{-1})^{*}|^{2s\mathrm{p}}
and\{|T^{-1}|^{t}|(T^{-1})^{*}|^{2s}|T^{-1}|^{t}|\}^{\frac{tp}{\^{o}+t}}
=\{|T^{*}|^{-t}|T|^{-2s}|T^{*}|-t^{\mathrm{t}}\}^{p_{\overline{l}}}8+
=(|T^{*}|^{t}|T|^{2s}|T^{*}|^{t})^{\frac{-tp}{s+t}}
\leq|T^{*}|^{-2tp}=|T^{-1}|^{2tp}.
\squareCorollary
12.If
T isA(s, t)
and T isinvertible,
thenT^{-1}
\dot{u}A(t, s)
.Let 0 <p
\leq
1 andS,
T \inB(\mathcal{H})
be non zerooperators.
In[3],
Duggal
proved
that tensorproduct
T\otimes S isp‐‐hyponormal
ifandonly
if Tand S arep‐‐hyponormal.
The passage of calss Aoperators
isstudiedby
Jeon andDuggal
[10].
Tanahashi and Chō[13]
proved
that the tensorproduct
T\otimes S
is ofclass
A(s, t)
if andonly
if T andS are classA(s, t)
. Nowwewillprovesimilarresult for p\rightarrow‐class
wA(s, t)
operators
by adopting
the ideas in[13],[10].
Lemma 13.
[11]
LetT_{1}, T_{2}, S_{1},
S_{2}
\inB(\mathcal{H})
be nonnegative operators.
If
T_{1}\neq 0
andS_{1}\neq 0
, then thefollowing
conditions areequivalent.
(1)
T_{1}\otimes S_{1}\leq T_{2}\otimes S_{2}.
(2)
There existsc>0 such thatT_{1}\leq \mathrm{c}T_{1}
andS_{1}\leq c^{-1}S_{2}.
Lemma 14.
[13]
LetT=U|T|
andS=V|S|
be thepolar
decompositions
of
T,
S\in B(\mathcal{H})
. Then thefollowing
assertions hold.(1) |T\otimes S|=|T|\otimes|S|.
(2)
T\otimes S=(U\otimes V)(|T|\otimes|S|)
is thepolar decomposition
of
T\otimes S.
(3)
(T\otimes S)(s,t)=T(s, t)\otimes S(s, t)
for
s,t>0.Theorem 15. Let
S,
T \inB(\mathcal{H})
be non zerooperators.
ThenT\otimes S
isProof.
LetS,
T \inB(\mathcal{H})
be non zerop‐‐class
wA(s, t)
operators
and S =V|S|,
T=U|T|
bepolar decomposions
ofS,
T. Then|T(s, t)|^{\frac{2tp}{s+t}}
\geq
|T|^{2tp}
and
|S(s, t)^{*}|^{\frac{2t\mathrm{p}}{\mathrm{s}+t}}
\geq|S|^{2tp}
by
Theorem 9.By applying
Lemma14,
weobtain|(T\otimes S)(s, t)|^{\frac{2tp}{s+l}}=|T(s, t)\otimes S(s, t)|^{\frac{2t\mathrm{p}}{s+i}}
=|T(s, t)|^{\frac{2tp}{s+t}}\otimes|S(s, t)|^{\frac{2tp}{s+t}}\geq |T|^{2t\mathrm{p}}\otimes|S|^{2tp}=|T\otimes S|^{2tp}.
Similarly,
wehave|T\otimes S|^{2sp}\geq|\{(T\otimes S)(s, t)\}^{*}|^{\frac{2s}{\^{o}}R}+t.
HenceT\otimes S is
p‐wA
(s, t)
.Conversely,
supposethatT\otimes S
isp‐wA
(s, t)
. Then|(T\otimes S)(s, t)|^{\frac{2}{s}}+ts_{t}=|T(s, t)|^{\frac{2t\mathrm{p}}{s+t}}\otimes|S(s, t)|+
\geq|T|^{2tp}\otimes|S|^{2tp}=|T\otimes S|^{2tp}
and|T\otimes S|^{2\mathrm{s}p}\geq|\{(T\otimes S)(s, t)\}^{*}|^{\frac{2s}{8}R}+t.
Hence there exists c>0 such that
c|T(s, t)|^{\frac{2}{8}}+t\mathrm{t}_{\mathrm{A}}\geq|T|^{2tp}
andc-1^{A}|S(s, t)|^{\frac{2}{s}}+tt\geq|S|^{2tp}
by
Lemma 13. Let xbe aunit vector. Then\Vert|T|^{tp}x\Vert^{2}=\langle|T|^{2tp}x, x\}\leq\langle c|T(s, t)|^{\frac{2\mathrm{t}\mathrm{p}}{s+t}}x, x\}
\leq c\Vert|T(s, t)|^{\frac{t}{8}}+^{R_{\overline{t}}}\Vert^{2}=c\Vert T(s, t)\Vert+tt
=c\Vert|T|^{s}U|T|^{t}\Vert^{\frac{2t\mathrm{p}}{s+l}}
\leq c(\Vert|T|^{s}\Vert\cdot 1\cdot\Vert|T|^{t}\Vert)^{\frac{2}{s}B^{t}}+t=c\Vert|T|^{tp}\Vert^{2}.
Hence
1\leq c
.Similarly,
\Vert|S|^{tp}x\Vert^{2}=(|S|^{2tp}x, x\rangle\leq\langle c^{-1}|S(s, t)|^{\frac{2tp}{s+t}}x, x\rangle
\leq c^{-1}\Vert|S(s, t)|^{\frac{t\mathrm{p}}{s+t}}\Vert^{2}=c^{-1}\Vert S(s,t)\Vert^{\frac{2t\mathrm{p}}{s+i}}
=c^{-1}\Vert|S|^{s}V|S|^{t}\Vert^{\frac{2tp}{ $\epsilon$+t}}
\leq c^{-1}(\Vert|S|^{s}\Vert\cdot 1\cdot\Vert|S|^{t}\Vert)^{\frac{2i\mathrm{p}}{s+\mathrm{t}}}=\mathrm{c}^{-1}\Vert|S|^{tp}\Vert^{2}.
Hence
1\leq c^{-1}
. Hence c=1. Thisimplies
|T(s, t)|^{\frac{2tp}{ $\epsilon$+t}}\geq|T|^{2tp}
and|S(s, t)|^{\frac{2t}{s+}\mathrm{g}_{l}}\geq|S|^{2tp}.
Similarly
wehave|T|^{2sp}\geq|\{T(s, t)\}^{*}|^{\frac{2sp}{s+\mathrm{t}}}
andThusT and S are
rwA(s, t)
. \squareCorollary
16. LetS,
T \inB(\mathcal{H})
be non zerooperators.
ThenT\otimes S
isp‐A
(s, t)
if
andonly if S,
T arep‐A
(s, t)
.Theorem 17. Let T \in
B(\mathcal{H})
bep‐wA
(s, t)
with 0 < s,t,
s+t = 1 and0<p\leq
1. Let$\rho$ e^{i $\theta$}\in \mathbb{C}
be an isolatedpoint
of
$\sigma$(T)
and0< $\rho$
. Then theRiesz
idempotent
Efor
T withrespect
to$\rho$ e^{i $\theta$}
isself‐adjoint
withran
E=\mathrm{k}\mathrm{e}\mathrm{r}(T- $\rho$ e^{i $\theta$})=\mathrm{k}\mathrm{e}\mathrm{r}((T- $\rho$ e^{i $\theta$})^{*})
.and coincides with the Riesz
idempotent
E(s, t)
for
T(s, t)
withrespect
toi $\theta$
$\rho$ e .
Proof.
Since$\sigma$(T)= $\sigma$(T(s, t))
by
Lemma 6 of[14],
$\rho$ e^{i $\theta$}
is anisolatedpoint
of$\sigma$(T(s,
t SinceT(s, t)
isrp‐hyponormal
for allr\displaystyle \in(0, \min\{s, t E(s, t)
is
self‐adjoint
and satisfiesran
E(s, t)=\mathrm{k}\mathrm{e}\mathrm{r}(T(s, t)- $\rho$ e^{i $\theta$})
=\mathrm{k}\mathrm{e}\mathrm{r}(T- $\rho$ e^{i $\theta$})
and$\rho$ e^{i $\theta$}\not\in $\sigma$(T(s, t)|_{\mathrm{r}\mathrm{a}\mathrm{n}E(s,t)})
.Since
\mathrm{k}\mathrm{e}\mathrm{r}(T- $\rho$ e^{i $\theta$})=
ranE(s, t)
reduces T, wehaveT= $\rho$ e^{i $\theta$}\oplus T'
on \mathcal{H}=ranE(s, t)\oplus
ran(1-E(s,
tThen T' is also class
p‐wA
(s, t)
andT'(s, t)
=T(s, t)|_{\mathrm{r}\mathrm{a}\mathrm{n}}(1-E(s,t))
. Hence$\rho$ e^{i $\theta$}\not\in $\sigma$(T'(s, t))= $\sigma$(T')
by
Lemma 6 of[14].
HenceT'- $\rho$ e^{i $\theta$}
is invertibleand
T- $\rho$ e^{i $\theta$}=0\oplus(T'- $\rho$ e^{i $\theta$})
. Thisimplies
\mathrm{k}\mathrm{e}\mathrm{r}(T- $\rho$ e^{i $\theta$})=\mathrm{k}\mathrm{e}\mathrm{r}((T- $\rho$ e^{i $\theta$})^{*})
and
E=\displaystyle \frac{1}{2 $\pi$ i}\int_{ $\gamma$}(z- $\rho$ e^{i $\theta$})^{-1}\oplus(z-T')^{-1}dz=1\oplus 0=E(s, t)
where $\gamma$ isa small circle
containing
$\rho$ e^{i $\theta$}.
\squareTheorem 18. Let
T\in B(\mathcal{H})
bep‐wA
(s, t)
with0<s,
t, s+t\leq 1
and 0<p\leq 1
. Let(T- $\rho$ e^{i $\theta$})x_{n}\rightarrow 0
for
x_{n}\in \mathcal{H}
with\Vert x_{n}\Vert=1
and$\rho$ e^{i $\theta$}\in \mathbb{C},
0< $\rho$.
Then
(|T|- $\rho$)x_{n},
(U-e^{i $\theta$})x_{n}, (U-e^{i $\theta$})^{*}x_{n}, (T- $\rho$ e^{i $\theta$})^{*}x_{n}\rightarrow 0.
Proof.
We may assume s+t=1by
Lemma 10. Since(T(s, t)- $\rho$ e^{i $\theta$})|T|^{s}x_{n}=|T|^{s}(T- $\rho$ e^{i $\theta$})x_{n}\rightarrow 0,
wehave
(T(s, t)- $\rho$ e^{i $\theta$})^{*}|T|^{s}x_{n}\rightarrow 0,
because
T(s, t)
isrp‐hyponormal
for allr\in(0, \displaystyle \min\{s,
t Hence0\leftarrow(T(s, t)- $\rho$ e^{i $\theta$})^{*}(T(s, t)- $\rho$ \mathrm{e}^{i $\theta$})|T|^{s}x_{n}
= (|T(s, t)|^{2}-$\rho$^{2})|T|^{s}x_{n}
- $\rho$ e^{-i $\theta$}(T(s, t)- $\rho$ e^{i $\theta$})|T|^{s}x_{n}- $\rho$(T(s, t)- $\rho$ e^{i $\theta$})^{*}|T' x_{n}.
This
implies
and
Similarly,
wehave(|T(s, t)|^{rp}-$\rho$^{rp})|T|^{s}x_{n}\rightarrow 0.
(|T(s, t)^{*}|^{rp}-$\rho$^{rp})|T|^{s}x_{n}\rightarrow 0.
Hence
0\leftarrow\langle(|T(s, t)|^{rp}-$\rho$^{rp})|T|^{s}x_{n}, |T|^{s}x_{n}\rangle
\geq\{(|T|^{rp}-$\rho$^{rp})|T|^{s}x_{n}, |T|^{s}x_{n}\}
\geq\langle(|T(s, t)^{*}|^{rp}-$\rho$^{r\mathrm{p}})|T|^{s}x_{n}, |T|^{s}x_{n}\rangle\rightarrow 0.
This
implies
((|T|^{rp}-$\rho$^{rp})|T|^{s}x_{n}, |T|^{s}x_{n}\rangle\rightarrow 0.
Since
\displaystyle \frac{r}{2}\in(0,
\displaystyle \min\{s,
t we have\langle(|T|\not\in^{r}-$\rho$^{r}\#)|T|^{s}x_{n}, |T|^{s}x_{n}\rangle\rightarrow 0
by
thesameargument.
Then\Vert(|T|^{\frac{rp}{2}}-$\rho$^{\frac{rp}{2}})|T|^{s}x_{n}\Vert^{2}
=\langle(|T|^{\frac{r\mathrm{p}}{2}}-$\rho$^{B}r_{2})^{2}|T|^{s}x_{n},
|T|^{s}x_{n}\rangle
=((|T|^{rp}-$\rho$^{rp})|T|^{s}x_{n}, |T|^{s}x_{n}\rangle-2 $\rho$
ỉ\yen
((|T|^{\frac{r\mathrm{p}}{2}}-$\rho$^{\frac{rp}{2}})|T|^{s}x_{n},
|T|^{s}x_{n}\rangle
\rightarrow 0. Hence
(|T|^{\frac{rp}{2}}-$\rho$^{\frac{r\mathrm{p}}{2}})|T|^{s}x_{n}\rightarrow 0
and so(|T|- $\rho$)|T|^{s}x_{n}\rightarrow 0.
Then|T|(|T|- $\rho$)x_{n}=|T|^{1-s}(|T|- $\rho$)|T|^{s}x_{n}\rightarrow 0.
Since0=\displaystyle \lim\langle|T|(|T|- $\rho$)x_{n}, x_{n})
=\displaystyle \lim\Vert|T|x_{n}\Vert^{2}- $\rho$\lim\langle|T|x_{n}, x_{n}\rangle
=\displaystyle \lim\Vert Tx_{n}\Vert^{2}- $\rho$\lim(|T|x_{n}, x_{n}\}
=$\rho$^{2}- $\rho$\displaystyle \lim\langle|T|x_{n}, x_{n}\rangle,
we have
\langle|T|x_{n}, x_{n}\rangle\rightarrow $\rho$
and
\Vert(|T|- $\rho$)x_{n}\Vert^{2}=\Vert|T|x_{n}\Vert^{2}-2 $\rho$(|T|x_{n}, x_{n}\rangle+$\rho$^{2}
\rightarrow$\rho$^{2}-2$\rho$^{2}+$\rho$^{2}=0.
This
implies
(|T|- $\rho$)x_{n}\rightarrow 0.
Since
and
0< $\rho$
, wehaveAlso,
(U-e^{i $\theta$})x_{n}\rightarrow 0.
\Vert(U-e^{i $\theta$})^{*}x_{n}\Vert^{2}=\Vert U^{*}x_{n}\Vert^{2}-\langle U^{*}x_{n}, e^{-i $\theta$}x_{n}\}-\{e^{-i $\theta$}x_{n}, U^{*}x_{n}\}+1
\leq 1-e^{i $\theta$}(x_{n}, Ux_{n})-e^{-i $\theta$}\langle Ux_{n}, x_{n}\rangle+1
\leq-e^{i $\theta$}\langle x_{n}, (U-e^{i $\theta$})x_{n}\rangle-e^{-i $\theta$}\{(U-e^{i $\theta$})x_{n}, x_{n}\rangle\rightarrow 0.
Hence(U-e^{i $\theta$})^{*}x_{n}\rightarrow 0
and
(T- $\rho$ e^{i $\theta$})^{*}x_{n}=|T|(U-e^{i $\theta$})^{*}x_{n}+e^{-i $\theta$}(|T|- $\rho$)x_{n}\rightarrow 0.
\square
Open problems
It is known thatclass A
operator
satisfiesPutnamtype
inequahty.
How‐ever it is not known that Putnam
type
inequality
holds forp‐wA
(s, t)
op‐erators. It seems a difficult
problem.
We note some openproblems
forp‐wA
(s, t)
operators.
(1)
M. Ito and T. Yamazaki[9]
proved
thatA(s, t)
implies
wA(s, t)
. How‐ever it isnot knownwhether
p‐‐class
A(s, t)
implies p‐wA
(s, t)
for0<p<1
ornot.
(2)
It is known that if T is classA(s, t)
and \mathcal{M} \subset \mathcal{H} is a T‐invariantsubspace,
thenT|_{\mathcal{M}}
is classA(s, t)
. However it is not knownwhether thisproperty
holds forp‐wA
(s, t)
operator
T.(3)
It is known that classAoperator
T is normaloid. But it isnotknownthat
p‐wA
(s, t)
operator
T is normaloidor not.REFERENCES
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769‐780.M. Chō
Departament
ofMathematics, Kanagawa University,
Yokohama221‐8686,
Japan
E‐mailaddress:
chiyom01@kanagawa‐u.ac.jp
T. Prasad
Schoolof
Mathematics,
Indian Institute ofScienceEducationand Research‐Thiruvananthapuram, Thiruvananthapuram‐695016, Kerala,
IndiaE‐mail address:
prasadvalapil@gmail.com
M.H.M.Rashid
Department
ofMathematics, Faculty
ofScienceP.O.Box(7),
Mutah univer‐sity,
Al‐Karak,
JordanE‐mail address:
malik‐okashaOyahoo.
comK. Tanahashi
TohokuMedical and Pharmaceutical
University,
Sendai981‐8558, Japan
E‐mail address:
tanahasi@tohoku‐mpu.ac.jp
A.