数理解析研究所講究録
1071 巻 1998 年 105-111

105

SPATIAL NUMERICAL RANGES OF ELEMENTS
OF -ALGEBRAS$\mathrm{C}^{*}$

山形大工

高橋眞映

1. INTRODUCTION AND RESULTS
A を複素ノルム環
であれば、 集合$\mathrm{A}^{*}$

をその双対空間, a を A の元とする。 もし A が単位的$V(A, a)\equiv\{f(a) :f\in A^{*}, |f|=f(1)=1\}$

は, a の (algebra) numerical range と呼ばれ,それは複素平面 $c$ 上の空でないコンパクト凸部分集合であることが知られている ([1, p. 52] 参照) 。しかしながら Aが非単位的であれば, この定義は意味をなさない。 この場合我々は次の二つの集合を導入する :$V_{1}(A, a)=${ $f(xa)$ :there exist $f\in A^{*}$ and $x\in A$ such that $|f|=|x|=f(x)=1$ }$V_{2}(A, a)=${ $f(ax)$ : there exist $f\in A^{2}$ and $x\in A$ such that $|f|=|x|=f(x)=1$ }and$a=V2(A, a)$ となっている。 A. K. Gaur and
勿論 A が単位的であれば $V(A, a)=V(1)A,$T. Husain [3] は $V_{2}(A, a)$ を特に spatial numerical range と呼び, この立場から研究を進めている。 その中で, A が可換 C*環であるときは、$\mathrm{c}\mathrm{o}\{\hat{a}(\varphi):\varphi\in\Phi_{A}\}\subseteq V_{2}(A, a)\subseteq\overline{\mathrm{C}\mathrm{o}}\{\hat{a}(\varphi):\varphi\in\Phi_{A}\}$が成り立つことを示している。 ここに \^a は a の Gelfand 変換を表し,大イデアル空間を表す ([3, Theorem 4.1] 参照)$\Phi_{A}$は A の極。

本講演での我々の主目的は, C*環の部分環における spatial numerical range は正汎函数の言葉で特徴付けられること, そしてその応用として Gaur-Husain の結果の非可換版が成立することを示すことにある。 先ず主定理は次のように述べられる :106

Theorem 1. Let A be a-algebra anda subalgebra of A. Let $b\in B$ . Then$V_{1}(B, b)=$ { $|f|(b)$ :there exist $f\in A’$ and $x\in B$ suchthat $|f|=.|X1=f(x)=1$ }$\mathrm{C}^{*}$$\mathrm{B}$

and$V_{2}(B, b)=${ $|f|(b)$ : there exist$f\in A^{\mathrm{s}}$and $x\in B$ such that $|f|=|x|=f(X^{*})=1$ },where $|f|$ denotes the absolute value of $f$ (cf. [, Definitim 1228]).
If$\mathrm{B}$is a’-subalgbera of$\mathrm{A}$, then $V_{1}(B, b)=V_{2}(B, b)$ .

主定理の系として, Gaur-Husain [3, Theorem 4. 1] の非可換への拡張となっている次のような結果を得る :Corollary 2. Let A be a$\mathrm{C}^{*}$-algbera and $a\in A$ . Then$\mathrm{c}\mathrm{o}\{f(a):f\in P(A)\}\subseteq V(1A, a)=V_{2}(A, a)\subseteq\varpi\{f(a):f\in P(A)\}$

where$\mathrm{P}(\mathrm{A})$,denotes the set of all pure states of A.

問題。 いつ $\mathrm{C}\mathrm{O}\{f(a):f\in p_{(A)}\}=V(1A, a)(=V_{2}(A, a))$ が成立するか ? またいっ$\overline{\mathrm{c}\mathrm{o}}\{f(a):f\in P(A)\}=V(1A, a)(=V_{2}(A, a))$が成立するか ?

2. PROOFS OF THEOREM 1 AND COROLLARY 2
Proof of Theorem 1. Set$\mathrm{W}_{1}=$and let{ $|f|(b)$ :there exist $f\in A^{*}$ and $x\in B$ such that $|f|=|x|=f(x)=1$ }$\lambda\in V_{1}(B, b)$. Then there exist$|g|=|X|=g(X)=1$ . Take afunctional$f=u\cdot|f|$$g\in B^{*}$$f\in A^{\mathrm{s}}$and $x\in B$ such that $\lambda=g(xb)$ andsuch that $f|.B=g$ and $|f|=|g|$ and letbe the enveloping polar decomposition of $f$ (cf. [2, Definition 12.2.8]). Then
$1=f(X)=|f|(ux)=(X \mathrm{I} u)_{||}f|*x\leq|_{|f}||u^{*}|_{|f|}\leq 1\cdot 1=1$

so that we can find ascalar$\alpha$,(1)

satisfying(2)$|u^{*}-\alpha x|_{|f|}=0$

since the equality of the Cauchy-Schwarz inequality in (1) holds. Note that (1) implies$(u^{*}1_{X})_{||}f=(x|u^{\mathrm{s}})_{|f|^{=}}(\mathcal{U}|u^{*}*)_{1}f|^{=}(_{X}1X)_{|f|}=1$

and hence$1-\overline{\alpha}-\alpha+|\alpha|^{2}=0$by (2). Therefore,$\alpha$must be equal to 1, and so(3)107that is $u^{*}-x$ belongs to the left kemel (in the envelopig von Neumann algebra$|u^{*}-x||f|^{=0}$ ’of A) $N_{|f|}=\{x\in A^{**} : |f|(x^{*}X)=0\}$ of $|f|$ . Also since $|f|(X^{*}x)=(x \mathrm{I}x)_{|f|}=|x|_{1}f|2=1$ by(1), it follows that

$1-X^{*}x\in N_{|f|}$, where

1 denotes the identity element of$\mathrm{A}^{**}$.

Therefore we have$\lambda=f(xb)=|f|(uxb)=(xb1u^{*})_{1}f|=(xb \mathrm{I} x)_{|f}|=|f|(x^{*}xb)=|f|(b)$(the-equality follows from $u^{*}-x\in N_{|f|}$ and the$4^{\mathrm{t}\mathrm{h}}$soConversely supposeand hence$\lambda\in W_{1}$,$V_{1}(B, b)\subseteq W_{1}$$\lambda\in W_{1}$$6^{\iota \mathrm{h}}$-equality follows from $1-x^{*}x\in N_{|f|}$).

. Then there exist$f\in A^{\mathrm{s}}$and $x\in B$ such that $\lambda=|f|(b)$and $|f|=|X|=f(\chi)=1$ . Let $f=u\cdot|f|$ be the enveloping polar decomposition of $f$ .
Then we can apply directly the above arguments for $f,$

and hence$f(xb)=|f|(b)$$V_{1}(B, b)=W_{1}$$\lambda\in V_{1}(B, b)$

, so

$x$

$W_{1}\subseteq V_{1}(B, b)$

and$u$. Consequently, we have

. We thus obatain.

We next set$W_{2}=$and let{ $|f|(b)$ : there exist$\lambda\in V_{2}(B, b)$$f\in A^{\mathrm{s}}$. Then there exist$|g|=|x|=g(x)=1$ . Take a functional

and $x\in B$ such that $|f|=|x|=f(X^{*})=1$ }.
$g\in B^{*}$

and $x\in B$ such that$f\in A^{\mathrm{s}}$such that $f$ I $B=g$ and $|f|=|g|$ . Then$|f^{\mathrm{s}}|=|f|=|x|=|x^{*}|$

so that$\overline{\lambda}=\overline{f(bX)}=f^{*}(Xb^{*}*)$$\overline{\lambda}\in V_{1}(\overline{B}, b^{*})$, where $\overline{B}=\{x\in A:x^{*}\in B\}$so we have
The inverse inclusion,andand $1=f(x)=f^{*}(X^{*})$ ,, $|f^{*}|=|f|=|x|=|x^{*}|$ and $1=f(x)=f^{*}(X^{*})$ , and hence

find $h\in A^{*}$ and $y\in B$ such that$\lambda\in W_{2}$$\lambda=g(b\chi)$. Therefore by the preceding argument, we can$\overline{\lambda}=|h|(b^{*})$$V_{2}(B, b)\subseteq W_{2}$and $|h|=|y\mathrm{I}=h(y^{*})=1$ . This means that.

$W_{2}\subseteq V_{2}(B, b)$

can be easily obtained by tracing the converse of theabove argument.

Set$A_{1,B}^{*}=${$f\in A^{\mathrm{s}}$: $|f|=1$ and there $exiS^{\cdot}rsx\in B$ such that $|x|=f(x)=1$ }and$A_{2,B}^{*}=$

If$\mathrm{B}$is$\mathrm{a}^{*}${ $f\in A^{*}:$ $|f|=1$ and there exists $x\in B$ such that $|x|=f(X^{*})=1$ }.

-subalgebra, then $farrow f^{*}$ is abijection of$V_{1}(B, b)=\{|f|(b) :f\in A_{1}^{\mathrm{s}_{B}},\}=\{|f|(b) :$A_{1,B}^{*}$

onto$A_{2,B}^{\mathrm{s}}$and hence we havef\in A^{*},\}\mathrm{z}BV_{2}=(B, b)$.
Q. E. D.108

Proof of Corollary 2. Let A be a -algbera and $a\in A$ . Then we have$V_{1}(A, a)=V_{2}(A, a)$ by Theorem 1. We next show that$\mathrm{C}^{*}$. To$\mathrm{c}\mathrm{o}\{f(a):f\in P(A)\}\subseteq V_{1}(A, a)$do this, letand$\lambda_{11}$,$\alpha\in \mathrm{c}\mathrm{o}\{f(a):f\in P(A)\}$.. ,$\lambda_{1m_{1}},$$\ldots$,$\lambda_{n1},$$\ldots,$$\lambda_{nm_{\hslash}}\geq 0$$\sum_{i-- 1}^{n}\sum_{j-- 1}\lambda_{\tau j}mi=1,\sum_{i=\iota}^{n}\sum_{j-}^{m}-1i$. Then there exist

入,fij(a),$\ldots$transforms$\pi_{i}(x)$into$\pi_{f_{\iota j}}(x)$. For each
$H_{\pi_{i}}$$fn1’\ldots,$ $f_{n}m_{n}p_{(A}\in$$i,$$\pi_{f_{i1}}\neq\pi_{f_{j1}}(i\neq])$$\xi_{ij}=U_{ij}(\xi_{f_{l}}*j)$$\alpha=f(a)$$H_{\pi_{f_{\iota j}}}$$\pi_{1}$,... ,. Also set

and$(\mathrm{i})]$$[0, \infty)$. Then$z$

defined by$0\leq t\leq 1$

is a positive element of A with $|z|\leq 1$ . Moreover, we assert that$\pi_{i}(z)\xi ij=\xi ij(1\leq i\leq n, 1\leq j\leq m_{i})$In fact, let$\epsilon>0$by [2,if
1, if $t>1$$t,$$h(t)=\{$$z=h(\mathcal{Y}^{2})$$\pi_{i}(y)\xi_{\mathrm{i}}j=\xi ij(1\leq i\leq n, 1\leq j\leq m_{i})$.

Consider the continuous function $h(t)$ on

and set$(^{*})$are mutually inequivalent, it follows that$\pi_{n}$there exists a hermitian element $y\in A$ such that

Theorem 2.8.3,

,

which

$f(x)= \sum_{i=1}^{n}\sum_{j1}^{i}m=\lambda_{j}(\pi_{flj}(x)\xi f_{lj}|\xi_{f}\iota j)=\sum_{i=1j}^{n}\sum \mathrm{t}_{j}(\pi m--1ii(X)\xi ij|\xi ij)$

for every $x\in A$ . Furthermore since).

$j(1\leq i\leq n, 1\leq j\leq m_{i})$onto the Hilbert spacefor every $x\in A$ , and set. Then we have $|f|=1,$ $f=|f|,$$f= \sum_{-i1}^{n}\sum_{j-}-m_{i}-1\Lambda_{j}fij$and$\pi_{f_{n1}}\underline{\simeq}\ldots\underline{\simeq}\pi_{fmn}$of the Hilbert space$U_{ij}$$\ldots,$,$=\alpha$$\pi_{1}\underline{\simeq}\pi_{f_{11}}\underline{\simeq}\ldots\underline{\simeq}\pi f_{\mathrm{l}}m_{1}’\ldots,$ $\pi n\underline{\simeq}\pi_{f_{n1}}\underline{\simeq}\ldots\underline{\simeq}\pi f_{m_{n}}$choose an isomorphism

flml’such that$\pi_{f_{11}}\cong\ldots\underline{\simeq}\pi_{f_{1m_{1}}},$

Let$f11’\ldots,$.

$(^{**})$

be arbitrary and take a polymonial $p(t)$ such that$p(0)=0$ and $\sup\{|p(t)-h(f)|:0\leq t\leq|z|\}<\epsilon/2$ .

Let $1\leq i\leq n$ and$1\leq j\leq m_{i}$$|\pi_{i}(z)\xi_{ij}-\xi ij|\leq|\pi_{i}(h(y^{2}))\xi ij-\pi_{i}(p(\mathcal{Y}^{2}))\xi_{ij}|+|p(\pi_{i}(_{\mathcal{Y}^{2}}))\xi_{i}j-\xi_{ij1}$$\leq|h(_{\mathcal{Y}^{2}})-p(_{\mathcal{Y}^{2})}|+|p(1)-1|$$\leq_{2^{+\frac{\epsilon}{2}=\mathcal{E}}}$and hence we obtain$(^{**})$

since$\epsilon$

is arbitrary. By$(^{*})$

and$(^{**})$, we have$f( \mathrm{z})=\sum_{1i--}\hslash m_{i}\sum_{j--1}\lambda_{i}(\pi(\mathrm{z})\xi jiij\xi_{ij})=\sum_{-}^{n}|\sum_{ji--1}-1m_{i}\Lambda_{j}.=1$.

. Then109

and soConsequently we have
We next show that $V_{1}(A, a)\subseteq\infty\{f(a);f\in p(A)\}$ . To do this, let$\alpha\in V_{1}(A, a)$$\mathrm{c}\mathrm{o}\{f(a) :f\in P(A)\}\subseteq V_{1}(A, a)$.

$\alpha\in V_{1}(A, a)$.

and sothere exist $f\in A^{*}$ and $x\in A$ such that $\alpha=|f|(a)\mathrm{a}\mathrm{n}\mathrm{d}|f|=|x|=f(x)=1$ Note that$|f|(x^{\mathrm{s}}x)=1$

as observed in the proof of the main theorem and consider the following set:
$S=$?hen $|f|\in S$ and{ $g\in A^{*}:g\geq 0$ and $|g|=g(\chi^{*}X)=1$ }.

is weak*-closed. Moreover, we can easily see that any extreme point of$S$

is also an extreme point of {$S$$g\in A^{\mathrm{s}}$:$g\geq 0${ $g\in A^{*}:$ $g\geq 0$ and $|g|\leq 1$ } consist of
Krein-Milman theorem thatimm$\mathrm{c}\mathrm{o}P(A)$,and henceand $P(A)$ (cf. Proposition 2.5.5), it follows by the$0$$S\subseteq\varpi P(A)$and $|g|\leq 1$ }. But since the extreme points of.

$. \Pi \mathrm{e}\mathrm{n}\alpha=|f|(a)=\lim g_{\lambda(a)}\chi$$\alpha\in\overline{\mathrm{c}\mathrm{o}}\{f(a) :f\in P(A)\}$for some net$\{g_{\lambda}\}$. Q. E. D.

3. COMMUTATIVE CASES$X$

を局所コンパクト Hausdorff 空間, $C_{0}(X)$ を無限遠点でゼロとなる $X$ 上の連続関数のつくる可換$\mathrm{C}^{*}- \text{環}$$A$を $C_{0}(X)$ の部分環 $f$ を$A$に属する関数とする。 このとき, 勿論 $V_{1}(A, f)=V_{2}(A, f)$ が成り立っているが, この spatial numerical range に関しては次のようにもう少し詳しい情報を得る。

Theorem 3. Let$V_{1}(A, f)=${$A$$\int f4^{\mu}|$$\subseteq\overline{\mathrm{c}\mathrm{o}}R(f)$be a subalgebra of$C_{0}(X)$and $f\in A$ . Then:there exist $\mu\in M(X)$ and $g\in A$ such that $| \mu|=|g|_{\infty}=\int gd\mu=1$ },where $M(X)$ denotes the space of all bounded regular Borel measures ondenotes the total variation ofproperty : For any finite set$g(x_{1})=\ldots=g(X_{n})=1$

また$A$

$\mu$

. Moreover, co $R(f)\subseteq V(A, f)$ if$\{X_{1}, \ldots, X_{n}\}$

in$X$

,

$A$

$X$

and

has the following

there exists $g\in A$ such that$|g|_{\infty}=1$.

が * を保存する場合は, 次のようにもっと詳しい情報を得る。

Corollary 4.

$\mathrm{I}x\mathrm{t}$

$A$

be

$\mathrm{a}^{*}$-subalgebra of$C_{0}(X)$and $f\in A$ . Then$|\mu|$and110$V(A, f)= \{\int fd\mu$: there exist $\mu\in M(X)$ and $g\in A$ such thatand $\int gd\mu=1$ }.

$|\mu|=1,$ $\mu\geq 0,0\leq g\leq 1$

Moreover,$V(A, f)=$ { $\int fd\mu:0\leq\mu\in M(X)$ , $|\mu|=1$

if

$A$

has the following property: For any compact set$0\leq g\leq 1$

and $g(X)=1$ for all $x\in E$ . Here$\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\mu)$

and$\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\mu)$$E\subseteq X$,is compact},

there exists $g\in A$ such that

denotes the support of$\mu$.

最後に実例を出してこの節を終わろう。
Let $X=(0,1]$ , the half open interval and let $h\in C_{0}(X)$ be such that $h(x)\neq 0$ for all$x\in X$

. Set
$A=\{hg:g\in c_{0}(x)\}$.

. In this case,is an ideal (and hence subalgebra) ofunital. Also
has the desired property : For any compact set
Then

$A$

$C_{0}(X)$$A$such that$A$$E\subseteq X$$|g|_{\infty}=1$is neighter closed or, there exists $g\in A$

and $g(X)=1$ for all $x\in E$ , and so by Theorem 3, we have

$V(A, f)=$ { $\int f\phi\mu|$ : there exist $\mu\in M(X)$and $g\in A$ such that $| \mu|=|g|_{\infty}=\int gd\mu=1$ }and$\mathrm{c}\mathrm{o}R(f)\subseteq V(A,f)\subseteq T\overline{\mathrm{O}}R(f)$for every $f\in A$ . In particular, if $f\in A$ is real-valued, then we have

$V(A, f)=\{_{(\beta]\mathit{0}\gamma}^{[\alpha,\rho]\mathrm{f}\mathrm{h}\mathrm{o}}0,\mathrm{i}f\mathrm{a}\mathrm{S}\mathrm{a}_{\mathrm{i}\mathrm{r}}\mathrm{z}\mathrm{e}\mathrm{r}\mathrm{p}\mathrm{o}\mathrm{i}\mathrm{n}l[\alpha,0)f\mathrm{d}_{\mathrm{o}\mathrm{e}\mathrm{s}\mathrm{n}}\mathrm{o}\mathrm{t}$have a zero point ,where $\alpha=\inf\{f(X):\chi\in X\}$ and $\beta=\sup\{f(X):x\in X\}$ .

Of course, this holds even if $A=C_{0}(x)$ , so we have the spatial numerical range of thefunction $f(x)=x(X\in X)$ with respect to$C_{0}(X)$is equal to $X=(0,1]$ . This fact has been

observed in [3, Example 42].

. But if is real-valued, then-subalgebra ofbecomesand sohas the property: For any compact set-subalgebra ofthere exists $g\in A$ such that $0\leq g\leq 1$ and $g(x)=1$ for all $x\in E$ .
Also,

$A$

$\mathrm{a}^{*}$

is not generally$C_{0}(X)$$\mathrm{a}^{*}$$C_{0}(X)$

$A$

$h$

$A$

$E\subseteq X$,

111

この節で述べた結果の証明及び実例に関する詳細は, 筆者 [4] を参照されたい。

References
1. F. F. Bonsal and J. Dancan, \dagger \dagger Complete Normed Algebras, Springer-Verlag,
Heidelberg / New York, 1973.
2. J. Dixmier, -algberas, North-Holland, New York, 1977.
3. A. K. Gaur and T. Husain, Spatial numerical ranges of elements of Banach algebras,
Intemat. J. Math. Math. Sci., 12-4(1989), 633-640.
4. S.-E. Takahasi, Spatial numerical ranges of elements of subalgebras of, submitted,
1998.
$\uparrow|$$\mathrm{B}\mathrm{e}\mathrm{r}\mathrm{l}\mathrm{i}\mathrm{n}/$$\mathrm{C}^{*}$$\mathrm{C}_{0}(\mathrm{X})$