数理解析研究所講究録1080 巻 1999 年 6-11

6

EXTENSIONS OF-FURUTA INEQUALITY$\mathrm{H}\mathrm{E}\mathrm{I}\mathrm{N}\mathrm{z}_{-}\mathrm{K}\mathrm{A}\mathrm{T}\mathrm{O}$

MASATOSHI FUJII* AND RITSUO NAKAMOTO**
藤井

正俊

中本

律男

ABSTRACT. We give an extension of recent Lin’s improvement of a generalized Schwarz inequality, which is based on the-Furuta inequality. As a consequence, we can sharpen
the-Furuta inequality.
$\mathrm{H}\mathrm{e}\mathrm{i}\mathrm{n}\mathrm{Z}^{-}\mathrm{K}\mathrm{a}\mathrm{t}_{0}$$\mathrm{H}\mathrm{e}\mathrm{i}\mathrm{n}\sim \mathrm{Z}^{-}\mathrm{K}\mathrm{a}\mathrm{t}_{0}$1. Introduction.

First of all, we cite a generalized Schwarz inequality which is a base of Lin’s recent paper[9]. For a (bounded linear) operator acting on a Hilbert space II,
$T$

(1)

$|(Tx, y)|^{2}\leq(|T|^{2\alpha}x, X)(|T*|^{2(1\alpha)}-)y,y$

for all $\alpha\in[0,1]$ and $x,$ $y\in H$ , where $|X|$ issquare root of $X^{*}X$ for an operator X on $H$ .
It implies the Heinz-Kato inequality via the L\"owner-Heinz inequality, cf. $[3],[10]$ . On theother hand, Furuta [7] extended the Heinz-Kato inequality, so called the-Furutainequality. Rephrasing it parallel to (1), we have$\mathrm{t}\mathrm{b}\mathrm{e}$$\mathrm{H}\mathrm{e}\mathrm{i}\mathrm{n}\mathrm{Z}^{-}\mathrm{K}\mathrm{a}\mathrm{t}\mathrm{o}$(2)$|(T|T|^{\alpha}+\beta-1x, y)|^{2}\leq(|T|^{2\alpha}x, X)(|T^{*}|2\beta y, y)$$\beta\in[0,1]$ withfor all$y\in H$ .
andVery recently, Lin [9] sharpened (1) as follows:$\alpha+\beta\geq 1$$\alpha,$Theorem L. Let$(Tz, y)=0$ ,
(3)

$T$

$x,$

be an operator on $H$ and $0\neq y\in H$ . For $z\in H$ satisfying $Tz\neq 0$ and$|(Tx, y)|^{2}+ \frac{|(|T|2\alpha X,z)|^{2}(|\tau*|^{2(}1-\alpha)_{\mathit{1}}y/,)}{(|T|^{2\alpha}Z,z)}\leq(|T|2\alpha_{X,X)(|T^{*}}|2(1-\alpha))y,y$

for all$T^{*}y$$\alpha\in[0,1]$

and $x,$ $y\in H$ . The equality holds if and only

are proportional, or equivalently,$Tx- \frac{(|T|^{2\alpha}x,z)}{(|T|^{2\alpha}z,z)}Tz$

and$if|T|^{2\alpha}(x- \frac{(|T|^{2\alpha}x,z)}{(|T|^{2\alpha}z,z)}z)$$|T^{*}|^{2(1-\alpha)}y$andare proportional.

In this note, we extend Theorem , which is based on the-Furuta inequality(2). Our proof is quite, in which we clarify the meaning of the as sumption in Theorem
that $Tz\neq 0$ and $(Tz, y)=0$ . As a consequence, we can sharpen the Heinz-Kato-Furuta
inequality, and Furuta’s further generalization [6; Theorem 3] of the Heinz-Kato inequality
via the Furuta inequality [4]. Incidentally we discuss Bernstein type inequality on the lineof our result.
2.
-Furuta inequality.
$\mathrm{L}$$\mathrm{H}\mathrm{e}\mathrm{i}\mathrm{n}\mathrm{z}-\mathrm{K}\mathrm{a}\mathrm{t}_{0}$$\mathrm{s}\mathrm{i}_{\ln_{\mathrm{P}}}1\mathrm{e}$$\mathrm{L}$$\mathrm{H}\mathrm{e}\mathrm{i}\mathrm{n}\mathrm{z}-\mathrm{K}\mathrm{a}\mathrm{t}_{0}$For the sake of convenience, we first cite the$\mathrm{H}\mathrm{e}\mathrm{i}\mathrm{n}\mathrm{z}-\mathrm{K}\mathrm{a}\mathrm{t}_{0}$-Furuta inequality [7]:1991 Mathematics Subject Classification. Primary.
Key words and phrases. Heinz inequality, Heinz-Kato-Furuta inequality, Furuta inequality.
$47\mathrm{A}30,47\mathrm{A}63$72.

inequality.

$\mathrm{H}\mathrm{e}\mathrm{i}\mathrm{n}\mathrm{z}-\mathrm{K}\mathrm{a}\mathrm{t}_{\mathrm{o}^{-}\mathrm{F}}\mathrm{u}\mathrm{r}\mathrm{u}\mathrm{t}\mathrm{a}$For lhe sake of convenience, we f\‘irst, cite the Heinz-Ieato-Furuta inequality [7]:The
operators on If such that$\mathrm{H}\mathrm{e}\mathrm{i}\mathrm{n}\mathrm{z}-\mathrm{K}\mathrm{a}\mathrm{t}\mathrm{o}- \mathrm{F}\mathrm{u}\Gamma \mathrm{u}\mathrm{t}\mathrm{a}$inequality. Let be an operator on II. If
and $TT^{*}\leq B^{2}$ , then
$T$

$A$

and $B$ are$po\mathit{8}itive$$T^{*}T\leq\Lambda^{2}$(4)$|(T|\tau|^{\alpha}+\beta-1)X,$$y|\leq||A^{\alpha}x||||B^{\beta}y||$for all$0,$ $\beta\in[0,1]$

XVe

with

and

$\alpha+\beta\geq 1$$x,$ $y\inH$.

inequality is just the case $\alpha+\beta=1$ in above andfollowing extension of Theorem L. Throughoutof an operator on $H$ .

t,heremark, it, corresponds. Thus we
this paper, let $T=U|T|$ be$1_{1_{\mathrm{C}}\backslash }\mathrm{e}\mathrm{t}1_{1}\mathrm{e}$Theorem 1. Let

on $H$ and $0\neq y\in ff$ . For$1\mathrm{l}\mathrm{e}\mathrm{r}\mathrm{e}$$\mathrm{t},11_{\mathrm{C}}\urcorner \mathrm{t}$$\mathrm{t}\mathrm{h}_{\epsilon}\backslash 1$$\mathrm{H}\mathrm{e}\mathrm{i}\mathrm{n}\mathrm{z}- \mathrm{I}\langle\subset’\iota \mathrm{f},\mathrm{O}$$\mathrm{t},\mathrm{o}(1)$$T$$\mathrm{t},1\mathrm{l}\mathrm{e}\mathrm{p}_{0}1_{\mathrm{c}}\gamma \mathrm{r}\mathrm{d}\mathrm{e}\mathrm{C}\mathrm{O}\mathrm{l}\mathrm{n}\mathrm{p}\mathrm{o}\mathrm{S}\mathrm{i}\mathrm{t}\mathrm{i}_{0}\mathrm{n}$

be an

$T$

$ope\Gamma ot_{\mathit{0}}\dot{r}$$z\in I\mathrm{f}$satisfying$T|T|^{\alpha}+\beta-1z$$y=0$ ,$\neq 0ar\iota,d(T|T|^{\alpha}+\beta-1)z,$$|(T| \tau|^{\alpha}+\beta-1,y\backslash \tau^{\backslash },)|^{2}+\cdot\frac{|(|T|^{2\alpha}\tau,Z)|^{2}(|T^{*}|2\beta\iota/,y)}{(|T|^{2}\alpha Z,z)}\leq(|T|2\alpha x, X)(|T*|2\beta y, y)$$(\ulcorner 0)$

for allholds if and$0,$ $\beta\geq 0$

It is
have

$only \uparrow,f|T|^{\alpha+\beta}-\mathrm{l}T^{*}ya\uparrow\iota d|T|^{2\alpha}(x-\frac{(|T|^{2\alpha}x,z)}{(|?^{\backslash }|^{2a}z,z)}Z)$and$|T^{*}|^{2\beta}y$with $\alpha+\beta\geq 1$ and $x,$ $y\in H.$ In the case

$T|T|\alpha\dashv-\beta-1(x-\mathrm{m}_{2\alpha}^{2\alpha}(\mathrm{z}x,z)(\tau zz)^{Z})$ilJ seen t,llat, TllGoreln is, the equality in (5)are proportional, or equivalently,$\alpha,$$\beta>0$

are proportional.

case $\alpha+\beta=1$ in Theorem 1. As a consequence, we
followingof the-Furutavi.a the L\"owner-Heinz, i.e.,
implie.sfor]:Theorem 2. Let
be an operator on II. If andare positive operators on $H$ suchthat
and $TT^{*}\leq B^{2}$ , then$\mathrm{L}$$\mathrm{e}_{\mathfrak{c}}\urcorner s$$\mathrm{t}1_{1}\mathrm{e}$$\mathrm{t}_{t}11\mathrm{e}$$\mathrm{H}\mathrm{e}\mathrm{i}\mathrm{n}\gamma_{\lrcorner}-\mathrm{I}\{\mathrm{a}\mathrm{t}\mathrm{o}$$\mathrm{i}_{\ln_{\mathrm{P}}}\mathrm{r}\mathrm{o}\mathrm{v}\mathrm{e}\mathrm{l}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}$$\mathrm{i}\mathrm{n}.\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{l}\mathrm{i}\mathrm{t}.\mathrm{y}$$\Lambda\geq B\geq 0$$\mathrm{i}_{\mathrm{I}1}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{c}\urcorner 1\mathrm{i}\mathrm{f}_{3^{r}},$$\Lambda^{\alpha}\geq B^{\alpha}$$T$$\alpha\in 1^{\mathrm{o},1}$

$A$

$B$

$T^{*}T\underline{<}A^{2}$(6)$|(T| \tau|^{\alpha}\{\beta-1x, y)|2+\cdot\frac{|(|T|2\alpha X,Z)|^{2}(|T^{*}|2\beta y,y)}{(|T|^{2\alpha}Z_{)}Z)}\leq||A^{\alpha}x||2||B\beta y||2$

for all$\alpha,$In th,$(T|T|^{\alpha}+\beta-1yz,)=0$ .

$e$

case

$A^{2\alpha}.\tau=|’\Gamma|2\alpha x\backslash ’ B^{2\beta}y=|T^{\star}|^{2\beta}y$$t\uparrow OT\mathfrak{l}a|.,\cdot$

$p$

ropo

and $x,$ $y,$ $z\in H$ such that $T|T|^{\alpha}+\beta-1z\neq 0$ and$\beta>0$ , the equality in (6) holds
if and only if
and $|T|^{\alpha+}\beta-1T*y$ andare propor-$\beta\in[0,1]w?,th\alpha\dashv-\beta\geq 1$

the third conditio$7^{-}tio7lal.$

$7l$

is

$\alpha,$$equ\uparrow,vale?1t$$|T|^{2\alpha}(x- \frac{(|T|^{2\alpha}x,z)}{(|\tau \mathrm{I}^{2\alpha}z,z)}z)$

to that$|T^{*}|^{2\beta}y$and$T|T| \alpha+\rho_{-}1(x-\frac{(|T|^{2a}x,z)}{\langle|T|^{2a}z,z)}z)$are.

Proof of Theo$7^{\backslash }em\mathit{1}$. We only use tlle positivity of the Gram matrix$G=G(U|\tau 1|^{\alpha}X, |T^{*}|^{\beta}y, U|T|\alpha Z)$.

.N oting hat$\mathrm{t}$$(|T^{*}|^{\beta}y, U|\tau|\alpha Z)=(y, |T*|^{\beta}U|T|\alpha Z)=(y,$ $T|T|^{\alpha}+\beta-1_{Z)=0}$

by {$11\mathrm{e}\mathrm{c}\tau \mathrm{S}\mathrm{S}\mathrm{u}\mathrm{l}\mathrm{n}\mathrm{p}\mathrm{t}\mathrm{i}_{\mathrm{o}\mathrm{n}}$, we$]_{1c\backslash }\mathrm{e}$$(U|\tau|^{\alpha}|||\tau_{0}.*|^{\beta}X,|y\tau*|^{\beta}||^{2}y)$$G=(_{(U|\tau|^{|?\cdot,|}}(U|\tau\alpha,,\cdot\tau^{*}|||\alpha T:l^{\backslash }’|^{\alpha_{l}}Uc.||\tau|^{\alpha_{Z}})^{*}|^{2}|\beta-\mathrm{t}/)^{*}$

Since$|T|^{\alpha}z\neq 0$, we$(U|T|\alpha,U|||T|^{\alpha_{Z}}x\mathrm{o}||T|^{2}|^{\alpha}z))$.

$1_{1c}\gamma \mathrm{v}\mathrm{e}$$|(T| \tau|^{\alpha}+\beta-1X, U)|^{2}+\frac{|(|?\urcorner|^{2}\alpha C\backslash ’ Z\prime)|^{2}(|\tau*|^{2\beta}y,y)}{(|T|^{2\alpha}Z,z)}\leq(|T|^{2\alpha}x, X)(|\tau*|^{2\beta}y, y)$.

To provc the equality condition, we set up t,he following lemma, which is applied to the$v=U|T|^{\alpha}z$ and $w=|T^{*}|^{\beta}y$ .

$\mathrm{v}\mathrm{e}\mathrm{c}l_{0}\mathrm{r}\mathrm{S}u=U|T|^{\alpha}x,$

8Lemma. (1) If $v\neq 0$ and $(v, w)=0$ , then $\{u, v, w\}$ is linearly dependent if and only if
andare proportional.
(2) Let $T=U|T|$ be the polar decomposition of an operator
on $H$ , (namely$\beta>0$ with. For
and $w\in H$ , the following conditions are mutually equivdent; (i)andare proportional. (ii)andareproportional. (iii)and $T|T|^{\alpha-1}w$ are proportional.
$w$$u- \frac{(u,v)}{||v||^{2}}v$$T$$\mathrm{k}e\mathrm{r}(T))$$\alpha+\beta\geq 1$$\alpha,$$|T^{*}|^{\beta}y$$\mathrm{k}\mathrm{e}\mathrm{r}(U)=$$y,$$U|T|^{\alpha}w$$|T|^{\alpha+}\rho_{-}1T^{*}y$$|T|^{2\alpha}w$$|T^{*}|^{\beta}y$

Proof. (1) Supposeand so$b=- \frac{a(u,v)}{||v||^{2}}$$\mathrm{t}\mathrm{l}\mathrm{l}\mathrm{a}\mathrm{t}au+bv+cw=0$

for some $(a, b, c)\neq 0$ . Then $a(u, v)+b||v||^{2}=0$. Hence we have

$0=au+bv+cw=a(u- \frac{(u,v)}{||v||^{2}}v)+cw$.

Since $a=c=0$ does not occur by $v\neq 0$ , vectors
and are proportional. The
converse is easily checked.
(2) (i) is equivalent to thatandare proportional. Noting that $\beta>0$
and, it is equivalent to (ii). Similarly we have the equivalence between (i)and (iii).
$u-\ovalbox{\tt\small REJECT}_{v}^{u_{1}v}v$$U|T|^{\beta}U^{*}y$$w$$U|T|^{\alpha}w$$\alpha,$$\mathrm{k}\mathrm{e}\mathrm{r}(U)=\mathrm{k}\mathrm{e}\mathrm{r}(T)$3. Furuta inequality.
In [6], the Heinz-I-Furuta inequality is extended by the use of the Furuta inequality;
Theorem 1 also gives us an improvement of the extension due to Furuta. For the sake of
convenience, we cite the Furuta inequality [4], see also $[2],15],[81$ .
$<\mathrm{a}\mathrm{t}\mathrm{o}$The Furuta inequality.
for each $r\geq 0$ ,If $A\geq B\geq 0$ , then$(B^{\Gamma}A^{\rho}B^{\Gamma})^{1}/q\geq(B^{\Gamma}B^{\mathrm{p}}B^{f})1/q$

holds for $p\geq 0$ and $q\geq 1$ with$(1+27^{\cdot})q\geq p+27^{\cdot}$$(^{*})$The domain representing
and is shown in [11].

is drawn in the right
thisis best possible one for the Furuta inequality.

$(^{*})$$\mathrm{i}\mathrm{t}$$\mathrm{t}\mathrm{h}_{\mathfrak{c}}^{r}\iota \mathrm{t}$$\mathrm{d}\mathrm{o}\ln\epsilon’\iota \mathrm{i}\mathrm{n}$Theorem 3. Let
be an operator on H. If$T^{*}T\leq
A^{2}$$TT^{*}\leq
B^{2}$ .
that
andThen for each
$T$

(7)

$A$

and $B$ are positive operators on $H$ such$r,$

$s\geq 0$$|(T|T|^{(}1+2_{\Gamma)} \alpha+(1+2s)\beta-1x, y)|2+\frac{|(|\tau|2(1+2\prime)\alpha_{X}Z)|^{2}(|T^{*}|2(1+2s)\rho_{y,y})}{(|T|^{2}(1+2r)\alpha_{Z})z},$,$\leq((|T|2\gamma A2\rho|T|2_{\Gamma})\frac{(1+2r)\alpha}{\mathrm{p}+2r}X, x)((|\tau^{*}|^{2}sB^{2q}|\tau*|^{2_{S}})\frac{(1+2S)\rho}{q+2s}y, y)$for all$p,$$q\geq 1,$$\alpha,$and $x,$ $z\in H$ such that$\beta>0$ , the$y=0$ . In the case$\beta\in[0,1]w?,th(1+2r)\alpha+(1+2s)\beta\geq 1$$T|T|^{(2}1+f)\alpha+(1+2s)\beta-1z\neq 0$

andequality in (7) holds if and only$(T|T|^{(r}1+2)\alpha+(1+2s)\beta-1)z,$$y,$$\alpha,$$if|T|^{2}(1+2_{\Gamma)}\alpha x=(|T|^{2}\gamma A^{2}p|\tau|^{2r})^{\frac{(1+2f)\propto}{p+2r}}X, |T^{*}|^{2(2}1+s)\beta y=$$(|T^{*}|^{2_{S}}B2q|T*|^{2S})^{\frac{\{1+2s)\beta}{q+2s}}yand|T|^{2}(1- \vdash 2r)\alpha(x-\frac{(|T\mathrm{I}^{2\mathrm{t}}\mathrm{z}+2_{\Gamma})\alpha)x_{1}z}{(|T|^{2(1}+2r)\propto_{z1}z)}Z)and|T|2(1+2f)\alpha+2(1+2s)\beta-1T^{*}y$

9

are

$pr^{\backslash }op_{\mathit{0}}rti,ona[,$and; the latter is equivalent to that$|T^{*}|^{2(1+2_{S})}\beta y$

are proportional.

Proof. We use Theorem 1 by

Then we$T|T|(1+2r) \alpha+(1+2s)\beta-1(x-\frac{(|T|^{2}\mathrm{t}1+2\Gamma)\alpha x,z)}{(|T|^{2}11+2r)\alpha z,z)}z)$$\mathrm{r}\mathrm{e}\mathrm{p}\mathrm{l}\mathrm{c}\mathrm{l}\prime \mathrm{C}\mathrm{i}\mathrm{n}\mathrm{g}\alpha$(resp.

$\beta$) to $\alpha_{1}=(1+2r)\alpha$ (resp. $\beta_{1}=(1+2s)\beta$).

$\mathrm{h}_{\mathrm{c}}\backslash \mathrm{v}\dot{e}$.

(8)$|(T|T|^{\alpha+\beta-1}11x, y)|^{2}+ \frac{|(|T|^{2}\alpha_{1_{\backslash }}T,Z)|^{2}(|T^{*}|2\beta 1y,y)}{(|T|^{2}\alpha_{1}z,Z)}\leq(|T|^{\mathrm{z}\alpha_{1}}x, X)(|T^{*}|^{2}\beta 1y, y)$

Next we use 1, Furut,a inequalit, forand
we replace $A,$ $B;q$ in 1, Furuta inequalit, to$11e$$|T|^{2}\leq\Lambda^{2}$$\mathrm{y}$$11\mathrm{e}$$\Lambda^{2},$$\mathrm{y}$$|T^{*}|^{2}\leq B^{2}$$|T|^{2}$;; namely (for the former)
respectively. Then we$\frac{p-|-2r}{(1+2r)\alpha}$$]_{1_{\mathrm{C}}\backslash \mathrm{v}\mathrm{e}}$$|?^{1}|^{2\alpha_{1}}=|T|^{2(2)\alpha}1+r\leq(|T|^{2\gamma}A^{2}\rho|\tau|^{2}r)^{\frac{\langle 1+2r)\alpha}{\mathrm{p}+2r}}$,

and$\mathrm{s}\mathrm{i}_{1}\mathrm{n}\mathrm{i}\mathrm{l}\mathrm{a}\mathrm{r}1_{\}}$$|T^{*}|^{2\beta_{1}}=|T^{*}|^{2()\beta}1\dagger^{-}2s\leq((|T^{*}|^{2s}B2q|T^{*}|^{2}s)^{\frac{(1+2s)\rho}{q+2s}}$, we obtain
Combininginequalit,y (7).
The equality condition is sllowe( similarly to Theorem 2.
$\mathrm{t}1_{1}\mathrm{e}$$\backslash \mathrm{v}\mathrm{i}\mathrm{t}\mathrm{l}\mathrm{l}(8)$$1$Remark. (1)remark tllat, the condition $(1+2r)\alpha+(1+2s)\beta\geq 1$ in Theorem 3 isunneccessary if is eit,llcr posit,ive or inverl,ible.
(2) Though
is follow from
Furuta inequality, they are equival
actually,that is, Tlieorem 3 is an ,ernative representation of the Furuta inequality. As a matter offact, we put, $T=B,$$?=s$ and also $x=y$ in Theorem 3. Thus it follows from the(, if, then for
above
)
and
$X=0$
$\backslash \backslash r_{\mathrm{e}}$$T$$e\mathrm{d}$$\mathrm{T}1_{1\mathrm{e}}\mathrm{Q}\mathrm{r}\mathrm{e}\mathrm{m}3$$\mathrm{t}1_{1}e$$e\mathrm{n}\mathrm{t}$$\mathrm{a}\mathrm{l}\mathrm{t}$$\alpha=\beta,$$B^{2(r)\alpha_{Z}}1+2\neq 0$$\Lambda^{2}\geq B^{2}$$\mathrm{r}\mathrm{e}\mathrm{m}_{\mathfrak{c}}\gamma \mathrm{r}\mathrm{k}(1)\mathrm{t},]_{1\mathrm{a}}$$(B^{2(+}12r)\alpha$$z,$$|(J\mathit{3}^{2(\gamma}1+2)\alpha X,$ $X)|2+ \frac{|(B^{2(1\dashv-}2r)\alpha x,Z)|^{2}(B^{2}(1+2\gamma)\alpha_{X,X})}{(B^{2(r}1+2)\alpha_{Z}Z)},\cdot$$\leq((B2_{\Gamma}\Lambda 2pB^{2}r)^{\frac{(1+2)\alpha}{\mathrm{p}+2r}}’ x, x)((B^{2}(1+2)\alpha)X,$$x$$1\mathrm{h}_{\mathrm{c}}\urcorner \mathrm{t}$,is,$A^{2}\geq B^{2}$,

ensures$(B^{2(1+2r)\alpha}X, x)^{2}\leq((B^{2r}A^{22\frac{(1+2r)\alpha}{\mathrm{p}+2r}}PI\mathit{3}\Gamma)x, X)$

for all$p\geq 1,7^{\cdot}\geq 0$

alld $\alpha\in[0,1]$ . This is notlling but the Furuta inequality.

4. Generalization.
Insect ion, we generalize
Theorem 4].
$\mathrm{t}_{}\mathrm{h}\mathrm{i}\mathrm{S}$Theorem 4. Let$\mathrm{T}1_{1\mathrm{e}}\mathrm{o}\mathrm{r}\mathrm{C}\ln 1$along with a generalization of Theorem

be an opera, $tor$ on $H$ and $0\neq y\in$ H.
, then$y=0$ for $i=1,2,$$T$$(T|T|^{\alpha}+\beta-1)z_{\tau},$

(9)

for

If

$\mathrm{L}[9$$T|T|^{\alpha}+\beta-1z_{i}\neq 0$;

and$\cdots,$ $n$$|(T|\tau|^{\alpha}+\beta-1.’)?\text{ノ},$$y|^{2}+ \sum_{?}\frac{|(|T|^{2}\alpha u_{i1_{)}}-z_{i})|^{2}|||\tau*|^{\beta}y||^{2}}{|||T|^{\alpha}Z_{i}||2}\leq(|T|2\alpha x, X)(|T*|2\beta y, y)$( $1,$ $\beta>0$

The

with

$\alpha+\beta\geq 1$$eq\uparrow r,ality?,7l(9)hol\mathrm{r}fS$

Proof. By$11_{1}e$, where $\uparrow/0=x$ a 77,

if anddefinition of$\mathrm{c}\iota_{t}$$du_{i}= \tau\iota_{i-1}-\frac{(|T|^{2}\alpha)u_{-1\mathrm{i}}z}{|||\tau \mathrm{I}^{\alpha}z\dot{.}||^{2}},Zi$$\mathit{0}7\mathrm{t}l,yif|T^{*}|^{\beta}y$

, we

and

$U|T|^{\alpha}u_{n}$for $i=1,2,$

are propotional.

$11_{(}\lambda \mathrm{V}\mathrm{C}$$\sum \mathrm{t}\mathit{1}_{i}=\sum \mathrm{t}\iota_{f}-1-\sum\frac{(|T|^{2}\alpha u_{i}-1zi)}{|||T|^{\alpha}Z_{i}||2},z_{i}$$\cdots$

, .
$n$

10
and so$\tau\iota_{7}=x-\iota\sum\frac{(|T|^{2\alpha}ui-1Zi)}{|||T|^{\alpha_{Z|}}i|^{2}},zi$.

Als. we have$0$$|T|^{\alpha}u_{i}=|T|^{\alpha}u_{\mathrm{i}-}1- \frac{(|T|^{2\alpha}u\mathfrak{i}-1Zi)}{|||T|^{\alpha}Z_{i}||2},|T|^{\alpha}z_{i}$

so

,

hat,

$\mathrm{t}$$|||T|^{\alpha}u_{i}||^{2}=|||T|^{\alpha}u_{i-1}||^{2}- \frac{|(|T|2\alpha ui-1zi)|2}{|||T|^{\alpha}Z_{\mathrm{i}}||2},$

Sulnming up l,his on $i=1,$$\cdots,$$\uparrow \mathit{1},$.

,$|||T| \alpha u,|1|^{2}=|||\tau|^{\alpha_{X|}}|^{2}-\sum\frac{|(|T|2\alpha ui-1zi)|2}{|||T|^{\alpha_{Z|}}i|^{2}},$Hence it, follows$\mathrm{f}\mathrm{r}\mathrm{o}\ln$t,he$\mathrm{c}\urcorner \mathrm{s}\mathrm{s}\mathrm{u}\mathrm{l}\mathrm{n}\mathrm{p}\mathrm{t}\mathrm{i}_{0}\mathrm{n}(T|T|^{\alpha}+\beta-1Z\mathfrak{i}, y)=0$

.

that
,

$|||T \urcorner*|^{\beta}y||^{2}|||T|\alpha x||^{2}-|||\tau^{*}|\beta y||2\sum\frac{|(|T|2\alpha ui-1zi)|^{2}}{|||T|^{\alpha_{Z|}}i|^{2}}$$=|||\tau^{*}|^{\beta}y||^{2}|||\tau|^{\alpha}\mathit{1}\iota n||^{2}$$\geq|(|\tau^{*}|^{\beta}y, U|\tau|^{\alpha}u_{7}\iota)|2$$=|(|T^{*}| \beta y, U|\tau|^{\alpha_{X}}-\sum\frac{(|T|^{2\alpha}ui-1Zi)}{|||7^{\tau}|^{\alpha}z_{i}||2},U|\tau|\alpha_{Z_{i}})|2$$=|(|T^{*}|^{\beta}y, U|\tau|^{\alpha_{X}})|^{2}$$=|(T|T|^{\alpha}+\beta-1x, y)|^{2}$.

The equality condition is obvious by seeing the only inequality in the above.

Another generalization of Theorem 1 is as follows:
Theorem 5. Under the same conditions as Theorem 4, the following inequality holds;$|(T|T|^{\alpha}+ \beta-1.?, y)|^{2}+\frac{\sum_{t}|(|T|2\alpha.,)\chi z_{\mathfrak{i}}|2|||\tau*|\beta y||^{2}}{\sum_{i}|||\tau|\alpha Z_{i}||2}\leq(|T|2\alpha x, X)(|T*|2\beta y, y)$As a matter of fact, since$\{|||7\urcorner|\alpha|x|^{2}|||\tau*|\beta y||^{2}-|(\tau|\tau\urcorner|^{\alpha}+\beta-1yx,)|2\}|||T|\alpha|Zi|^{2}\geq|||\tau^{*}|^{\beta}y||^{2}|(|\tau|^{2\alpha}x, Z_{i})|2$by$\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{o}\mathrm{r}\mathrm{e}\ln 1$$\mathrm{R}\mathrm{e}\mathrm{l}\iota\tau \mathrm{a}\mathrm{r}\mathrm{k}$$1\mathfrak{n}\mathrm{e}\mathrm{r}1\uparrow_{}\mathrm{s}$, we$1_{1_{\subset}\backslash }\mathrm{V}\mathrm{e}$it, by summing up on$i$.

. Theorems 4 and 5 give us generalizations of Theorems 2 and 3, whose state-

and proofs are quit, silnilar t,o t,hem.
$\mathrm{e}$$3]$

5. A concluding remark.
Lin also discussed Bernstein type inequalities independently on Theorem, see [1]. As an application of Tllcorem 1, we have a generalization of it:$\mathrm{L},$$[9;\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{m}$11be an operator on $H$ having a nonzero normal eigenvalue
Theorem 6. Let$y\in
Hsati_{\mathit{8}}fies(e,y)=0$ and $T^{*}y\neq 0$ , theneigenvector . If$T$$\lambda$with an$e$$| \lambda|^{2}|(X, e)|2\leq\frac{||TX||2|||\tau^{*}|^{\beta}T*|y|2-|(\tau|\tau|^{\beta_{X}},\tau^{*}y)|2}{|||T^{*}|^{\beta}T*y||2}$

for all $x\in H$ andProof. We put$\beta\in[0,1]$$\alpha=1,$.

$z=e$ and replace

$y$

to

$T^{*}y$$(e, y)=0$ , It follows from Theorem 1 thatin Theorem 1. Since $(|T|^{\beta}e,\tau*y)=0$ by
..,

.

..

,

$|(T|T|^{\beta}X, \tau^{*}y)|^{2}+|||\tau*|^{\beta*}\tau y||^{2}|\lambda|^{2}|(X, e)|2\leq||TX||2|||\tau^{*}|^{\beta}T*y||2$

so that we have the desired inequality.
We obtain Lin’s inequality [9; Theorem 3] by taking $\beta=0$ in Theorem 6.

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$\geq$$\geq 0,$$\geq 0$$\geq 0,$$\geq$$\Pi\iota ruta’ s$*DEPARTMENT OF MATtIEMATlCS, OSAKA KYOIKU UNIVERSITY, KASHIWARA, OSAKA 582, JAPAN**FACULTY or ENGINEERING, IBARAKI UNIVERSITY, HITACHI, IBARAKI 316, JAPAN.