数理解析研究所講究録1080 巻 1999 年 149-155149

作用素不等式二題
Exponential. operator inequalities
福岡教育大学内山 充 (Mitsuru Uchiyama)
Fukuoka University of Education
Munakata,Fukuoka, 811-41 Japan,e–mail uchiyama@fukuoka-edu. .jp$\mathrm{a}\mathrm{c}$Section 1.

Let $X$ be a unital Banach algebra overa unit 1 such that $||1||=1$ .

$\mathrm{R}$or , that is, a complete normed algebra with$\mathrm{C}$The aim of this note is, roughly speaking, to show that if $f$ : [$f(0)=1,$ $f’(\mathrm{O})=a$ , thenconverges toas $narrow\infty$ , where$\mathrm{O}.\infty)arrow X$$f( \frac{t}{n})^{n}$satisfies$e^{ta}$$::.-$ ‘$e^{x}= \sum_{n=0}^{\infty}\frac{x^{\mathrm{n}}}{n!}$by definition.
If, this assertion clearly follows from the L’hospital theorem. Since a set of all
bounded operators on a Banach space is a unital Banach algebra, for a bounded operator$A,e^{A}$ isdefined as above. In this case for bounded operators $A,$ the Lie product formula:$X=\mathrm{R}$$B$$\exp(A+B)=(n)_{n}\lim_{arrow\infty}\{\exp(\frac{A}{n})\exp(\frac{B}{n})\}^{n}$is well-known, where $(n)$ means that the limit is in the sense of the (operator) norm
topology. This implies that the above assertion holds for $f(t)–\exp(tA)\exp(tB)$ as well.
The above definition
is not useful for unbounded operator. However it is-known
that if is a generator of..
contractive semigroup, then$e^{x}$$\mathrm{w}\mathrm{e}1_{i}1$$A$$(C_{0})$$e^{tA}=(s)_{n} \lim_{arrow\infty}(1-\frac{t}{n}A)-n$

for $t>0$ ,where $(s)$ means that the limit is in the sense of strong topology. The Lie product formula
was extended to the case of unbounded operators on a Banach space in [2] [4].

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Chemoff [1] showed a product formula in a more general form as follows :Let $f(t)$ be a strongly
to the linear contractions on afunction fromBanach space. Suppose hat
and the strong derivative
has a closure A which$f(t/n)^{n}$is a generator of acontractive semigroup. Thenstrongly converges to.
$continuo\iota \mathrm{J}s$$[0, \infty)$$f(\mathrm{O})=1$$f’(\mathrm{O})$$e^{tA}$$(C_{0})$In the proof of this theorem the condition $||f(t)||\leq 1$ plays an important role, so it is
not easy to relax it. However we encounter many cases where $f(t)$ is not a contraction andthe derivative is bounded: in this case$A$$\frac{f(t)}{||f(t)||}$

is a contraction, but may not be differentiable at $t=0$ ; so we can not use the Chernoff’s
theorem. Therefore we need to make a new product formula for bounded operators. See[3] for product formulas.

Theorem 1. Let $X$ be a unital Banach algebra, and let $f(t)$ be a function from aninterval $0\leq t<\zeta$ to X. If $f(0)=1$ and $f(t)$ has a noml right derivative at $t=0$ , then$a$$||f( \frac{t}{n})^{n}-\exp(ta)||arrow 0(narrow\infty)f\alpha\cdot 0\leq t<\infty$

Proof. For every :$t$$0\leq t<\infty,$$f( \frac{t}{n})$.

is defined for sufficiently large , so we may assume$n$

is defined on. We claim that
there is $r>0$ such thatis bounded on $(0, r)$ .
To see this we may show thatis bounded above on $0<t<r$ .
Since

$f$

$[0, \infty)$$||f(t)||^{\frac{1}{t}}$$\frac{1}{t}\log||f(t)||$$|| \frac{f(t)-1}{t}-a||arrow 0$$\frac{1}{t}(||f(t)||-1)$$(tarrow+0)$,

is bounded, and $||f(t)||arrow 1(tarrow+\mathrm{O})$ . Thus$\ovalbox{\tt\small REJECT} 1\mathrm{o}t-1\circ 1||f(9)||-1\ovalbox{\tt\small REJECT} t-1t$$(||f(t)||\neq 1)$$0$$(||f(t)||=1)$$\frac{\log||f(t)||}{t}=\{$

is bounded on some interval $(0,r)$ .
Now take an arbitrary $t:0<t<\infty$ , and fix it. By the claim above, we can see thatis bounded. Thus there is $M>0$ such that$\{||f(\frac{t}{n})||^{n}\}_{n}$$e^{t||a||}\leq M,$$||f( \frac{t}{n})||^{n}\leq M$

far every .
$n$

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From.
$-$$f$,$f( \frac{t}{\eta})^{n}-e^{t}=..\sum^{1}am=\mathrm{o}n-f(\frac{t}{n})^{m}\{f(\frac{t}{n})-e\frac{t}{n}a\}(e\frac{t}{n}a)n-1-m$

it follows that$||f( \frac{\mathrm{t}}{n})^{n}-e^{ta}||\leq||f(\frac{t}{n})-e\frac{\ell}{n}a||\sum_{m=0}^{1}M^{\frac{m}{n}}(e\frac{t}{n}||a||)^{n-}n-1-m$$=n||f( \frac{t}{n})-e^{\frac{t}{n}a}||\cdot\frac{M-e^{t||a|}1}{n(M^{\frac{1}{n}}-e\frac{t}{n}||a||)}$

.

Since
$n(M^{\frac{1}{n}}-e \frac{t}{n}||a||)arrow\log M-t||a||$and$n||f( \frac{t}{n})-e^{\frac{t}{n}a}||\leq t||\frac{n}{t}\{f(\frac{t}{n})-1\}-a||+t||\frac{n}{t}(-e^{\frac{t}{n}a}+1)+a||arrow 0(narrow\infty)$,

we get$||f( \frac{t}{n})^{n}-e^{ta}||arrow 0(narrow\infty)$.

This concludes the proof.

Corollary 1. For $a_{i}\in X$$\square$$(i=1, \cdots, m)$$|| \{(1+\frac{a_{1}}{n})\cdots(1+\frac{a_{m}}{n})\}^{n}-\exp(a_{1}+\cdots+a_{m})||arrow 0$,.

$||(e^{\lrcorner}an\cdots en)^{n}-rightarrow\exp(a_{1}+\cdots+a_{m})||arrow 0$

Proof. By setting $f(t)=(1+ta_{1})\cdots(1+ta_{m})$ or$f(t)=e^{ta_{1}}\cdots e\ell_{a_{m}}$, these follows$\mathrm{f}\mathrm{r}\mathrm{o}\mathrm{m}\square$

the theorem.

be a holomorphic function in a neighborhood $|z-1|<\delta$. Then for $a\in X$ :$||a-1||<\delta,$is defined byLet$\phi(z)$$\phi(a)$$\phi(a)=\sum_{=no}^{\infty}\frac{\phi^{(n)}(1)}{n!}(a-1)^{n}$,which converges in the norm. Thus for $f(t)$ with the property set out in the theorem$\phi(f(t))$ is well-defined for sufficiently small. Since $\phi(f(\mathrm{O}))=\phi(1)$ and the right norm
$t$

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derivative of $\phi(f(t))$ at $t=0$ is$\phi’(1)f’(\mathrm{O})$, we have

Corollary 2. Ifis a scalar valued holomorphic function in a $neighborhood|$ of $z=1$ ,$\phi(1)=1$$f(t)$with, then forwhich has the property set out in the heorem,$\phi(z)$$|| \phi(f(\frac{t}{n}))^{n}-\exp(t\phi’(1)a)||arrow 0(narrow\infty)$

for $0\leq t<\infty$ .

In particular, we have

Corollary 3.
$|| \{(1+\frac{a_{1}}{n})\lambda_{1}\ldots(1+\frac{a_{m}}{n})^{\lambda}m\}^{n}-\exp(\lambda 1a1+\cdots+\lambda_{m}a_{m})||arrow 0$

for$\lambda_{i}\in \mathrm{R}$.

In the proof of Theorem 1 that the domain of is the right half real line is not essential.
We can get the same result as above even if the domain of is a half line with end point
in C. More generally we show
$f$

$f$

$0$

Theorem 4. Let $X$ be a unital Banach algebra, set $D=\{z\in \mathrm{C}$ :. If a function : $Darrow X$ satisfies
and $f’(0)=a$ , that is,$\alpha\leq\arg z\leq\beta,$$\alpha\leq 2\pi\}$$f$$0\leq$$f(\mathrm{O})=1$$|| \frac{f(z)-f(0)}{z}-a||arrow 0(z\in D, Zarrow 0)$,

then for every $z\in D,$ $||f( \frac{z}{n})^{n}-\exp Za||arrow 0(narrow\infty)$ .

Proof. In the same way as the proof of Theorem 1 one can easily show thatbounded on a neighborhood of$\mathrm{O}\in D$, and that, for fixed $z\in D$ ,$||f( \frac{z}{n})^{n}-e^{za}||\leq||f(\frac{z}{n})-e\frac{z}{n}a||\sum_{m=0}^{n-1}M\frac{m}{n}(e^{\frac{|z|}{n}})^{h}||a||-1-m$from which the theorem follows.

$||f(z)||^{\urcorner^{1}}|z$is,$\square$

References[1] P. R. Chemoff, Note on pr.oduct formulas for opeator semigroups, J. Func. Anal.
$2(1968),238-242$ .

153[2] E. Nelson, Feyman integrals and the Schr\"odinger equation, J. Math. Phys.
$5(1964),332-343$ .

[3] M. Reed, B. Simon, Functional Analysis vol.1, Academic Press, New York (1980).
[4] H.F. \prime botter, On the product of semigroups of operators, Proc. A.M.S. 10 (1959),545-551.

Section 2.

Let A and B be bounded selfadjoint operators on a Hilbert space. The following celebrated inequality was found by Furut.a. in [4] and simply proved in [5].

A $\geq B\geq 0$ implies$A^{(\mathrm{P}+\cdot)}’/q\geq(A^{r/2}B^{\mathrm{p}}A^{r/2})1/q$(1)

for $p\geq 0,$ $q\geq 1,r\geq 0$ such that $(1+r)q\geq p+r$ .

Ando [1] showed the following theorem in the case of $s=p=r$ with a splended idea.
Then Fujii, Furuta, Kamai [2], by making use of Ando’s result, proved that $A\geq B$ implies(2).

Theorem A. $A\geq B$ implies that for $p\geq 0,$ $r\geq s\geq 0$$e^{S} \geq A(e^{\frac{r}{2}}ee)^{\frac{\epsilon}{\tau+\mathrm{p}}}ApB\frac{r}{2}A$.

(2)

In [1] Ando also showed the converse:

Theorem B. Iffor every$e^{tA} \geq(e^{\frac{t}{2}A}ee^{\frac{t}{2}A})\ell pB\frac{\ell}{r+\mathrm{p}}$$t>0$ ,

then $A\geq B$ .
The aim of this note is to give a new way to get exponential inequalities frominequalities like (1), and to extend Theorems , B.
We start with a quite simple proof of Theorem A. This technique seems to be very effective$\mathrm{o}\mathrm{p}\mathrm{e}\mathrm{r}.\mathrm{a}\mathrm{t}_{\mathrm{o}\mathrm{r}}$$\mathrm{A}$154
to study operator inequality.
Another proof of Theorem . For sufficiently large we havesubstituting $np$ and $nr$ to and of (1), respectively, we get,
$A$

$n$

$1+ \frac{A}{n}\geq 1..+’\frac{B}{n_{l}}$.

$r$$p$$\geq- 0$. By, for $rq\geq p+r$ .

$(1+ \frac{A}{n})\lrcorner_{R}n\mathrm{f}\mathrm{i}1q\geq\{(1+\frac{A}{n})n\frac{r}{2}(1+\frac{B}{n})^{n}\mathrm{P}(1+\frac{A}{n})^{n}\frac{f}{2}\}^{1/}q$

Since for selfadjoint operator $X,$.
weby setting$\mathrm{g}\mathrm{a}\mathrm{i}\mathrm{n}(2)$$(1+ \frac{X}{n})^{n}$converges to$e^{X}$in the operator norm as’$narrow\infty\square$$s=\mathrm{Z}_{\frac{+r}{q}}$$.$

We slightly extend Theorem A by using itself.
Proposition 1. $A\geq B$ implies(3)$e^{sA} \geq\{e^{\frac{f}{2}A}e(qA+pB)\frac{r}{2}eA\}^{\frac{\epsilon}{(\mathrm{p}+q+T)}}$for$p,$ $q,r,$ $swi\theta\iota 0\leq s\leq r$$0\leq p,p+q$ , and,

Proof. If $p+q=0$ , then$e^{(pB)}qA+$$0<p+q+r$ .

is contractive, so that the above inequality follows.

Therefore we assume that $p+q>0$ . Since$\frac{qA+pB}{q+p}\leq A$,

by using (2), we gain (3).
Now we extend Theorem

Theorem 2. If there are$\square$$\mathrm{B}$:$p,$ $q,$ $r,$ $s$with $p>0,p+q\geq 0,r\geq s>0$ such that$e^{S\ell A} \geq\{e^{\frac{rt}{2}At()A}ee\frac{rt}{2}\}^{\frac{s}{(\mathrm{p}+q+r)}}qA+pB$

for every $t>0$ , then $A\geq B$ .

Proof. If $p+q+r=s$ , then the above inequality implies that$e^{t(p)}qA+B$

because of $p+q=0$ . Hence $A\geq B$ . Suppose $p+q+r>s$ . Set$f(t)=e^{\frac{-rt}{2}A}e^{-t()}e^{\frac{-rt}{2}A}qA+\mathrm{P}B$Then we get$(f(t)^{\frac{\epsilon}{(\mathrm{p}+q+r)}}X,X)\geq(g(t)x,X)$,$g(t)=e^{-stA}$.

$(||X||=1, t>0)$ ,is contractive

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from which it follows that$(f( \mathrm{t})x,X)\frac{\epsilon}{(\mathrm{p}+q+r)}\geq(g(t)_{X},X)$$(t>0)$

because of Jensen’s inequality. Since the values of both sides of the inequality above at$t=0$ are 1, the right derivative of the left hand side at $t=0$ is greater than or equal toat $t=0$ is generally , wethe one of the right hand side. Since the norm derivative ofhave$T$$e^{tT}$$\frac{s}{(p+q+r)}‘((-\frac{r}{2}A-(qA+pB)-\frac{r}{2}A)X,X)\geq(-sA_{X},X)$.

Hence we gain $A\geq B$ .

$\square$

We end this note with referring to an exponential inequality which appeared in [3]:If $A-B\geq\delta>0$ , then$e^{tA}-e^{\ell B}\geq\delta/2>0$

for some $t>0$ .

This seems to be useful, so that we give a more generalized result, which we can see bya simple calculation.

operator valued functions defined in a neighborhood of $t=0$ .
Let $f(t),g(t)$ be
and $f’(0)-g’(0)\geq\delta>0$ , where the derivative in theof norm, thenIf$f(t)-g(t)\geq\delta/2$ forin a neighborhood of .
$\mathit{8}elfadjoint$$f(\mathrm{O})=g(\mathrm{O})$$i\mathit{8}$

$t$

$0$

sen.seReferences[1] T. Ando, On some operator inequalities, Math.Ann. $279(1.987)$ 157-159.

[2] M. Fujii, T. Furuta, E. Kamei, Furuta’s inequality and its application to Ando’stheorem, Linear Algebra and its application 179 (1993) 161-169.
[3] M. Fujii, J.Jiang, E. Kamei, Charaterization of chaotic order and its application toFuruta inequality, Proc. A.M.S. to appear.
[4] T. Furuta, A $\geq B\geq 0$ assures $(B^{r}A^{p}B^{f})1/q\geq B^{(\prime}p+2)/q$ for r $\geq 0,p\geq$ 0, q$(1+2r)q\geq p+2r$ , Proc. A.M.S. 101(1987) 85-88.
[5] T. Furuta, An elementary proof of an order preserving inequality, Proc.
65, ser. A (1989) 126.

$\geq$$\mathrm{J}\mathrm{a}\mathrm{p}\mathrm{a}\mathrm{n}|$1 withAcad.