Li, Yuan-Chuan

Shih, Mau-Hsiang

Kyoto University Research Information Repository

TitleTHE LENGTH OF CONTRACTIBILITY OF COMPACTCONTRACTIONS ON A HILBERT SPACE (NonlinearAnalysis and Convex Analysis)Author(s)Li, Yuan-Chuan; Shih, Mau-HsiangCitation数理解析研究所講究録 (2004), 1365: 138-145Issue Date2004-04URL http://hdl.handle.net/2433/25346RightType Departmental Bulletin PaperTextversionpublisherKyoto University38THE LENGTH OF CONTRACTIBILITYOF COMPACT CONTRACTIONSON A HILBERT SPACEYuan-Chuan LiDepartment of Applied Mathematics, National Chung-Hsing University, Taichung, 402 TaiwanChung-Li 32023, Taiwan$\mathrm{E}$-mail address: ycli@ fish.amath.nchu.edu.twandMau-Hsiang ShihDepartment of Mathematics, National Taiwan Normal University88 Sec.4, Ting Chou Road, Taipei 116, Taiwan$\mathrm{E}$-mail address: mhshih@ math.ntnu.edu.tw1. IntroductionLet $A$ be a bounded linear operator on $\dot{\mathrm{a}}$ complex Banach space, and $r(A)$ the spectralradius. The Gelfand spectral radius asserts that$\mathrm{r}(\mathrm{A})=\inf_{n>1}||A$”$||^{1/n}= \lim_{narrow\infty}||A^{n}|||1/n$In 1960, Rota and Strang[6] introduced the notion of joint spectral radius as follows :Let $\Sigma$ be a bounded subset of $n\mathrm{x}n$ matrices. Define$\Sigma^{k}=\{A_{1}A_{2}\cdots A_{k;}A_{i}\in\Sigma, i=1,2, \ldots , k\}$, where $k=1,2$, $\ldots$ .The joint spectral radius $r(.\Sigma)$ of) is defined to be$\hat{r}(\mathrm{I})$ $= \lim\sup\sup||A||1/k$ .$karrow\infty$$A\mathit{1}($ $\Sigma k$1991 Mathematics Subject Classification. Primary $46\mathrm{C}05,46\mathrm{B}99$.Key words and phrases. Hilbert space, spectral set, spectral radius, point spectrum, pointspectral radius, compact operator, contraction, semigroup, $C_{0}$-semigroup, infinitesimal generator.Research supported in part by the National Science Council of Taiwan.of $\Sigma$\ha r}(\Sigma)=\lim\sup\sup| ||^{1 }$$karrow\infty A\in Sigma^{k}$at .$\mathrm{p}\mathrm{h}\dot{\mathrm{r}}\mathrm{a}\mathrm{s}\mathrm{e}\mathrm{s}$ .1365 2004 138-1451391992, Daubechies and Lagarias [2] introduced the notion of generalized spectral radius$r(\Sigma)$ as follows.$r( \Sigma)=\lim_{karrow}\sup_{\infty}r_{k(\Sigma)^{1/k}}$ ,where $r_{k}( \Sigma)=\sup r(A)$ . It is easy to see that the notion of the joint spectral radius$A\in\Sigma^{k}$is independent of all equivalent matrix norm. They also conjectured that(1) $r(\mathrm{I})=\hat{r}(\mathrm{I})$ .This conjecture is not true whenever $\Sigma$ is not bounded. For example,$\mathrm{I}=$ $\{ [^{\frac{1}{02}} 2\mathrm{N}]; n=1,2, \ldots\}$.This conjecture was solved by Berger and Wang in 1992 [1],In 1995, Lagarias and Wang [3] studied the following problem:For every finite set I of $nxn$ matrices, is there a positive integer $k$ such that(2) $r(\Sigma)=\hat{r}(\mathrm{I})$ $=r_{k(\Sigma)^{1/k}?}$They also showed that if $\Sigma$ is a finite set of contraction $nxn$ matrices, then this problemis true whenever the associated norm is $\ell^{p}$-norms with $p$ rational. They also proved thefollowing result.Theorem $\mathrm{L}$-W. Let $||\cdot||2$ be the Euclidean norm on $\mathrm{R}^{n}$ , and $\Sigma$ a set of $m$ contractionmatrices on Rn. Put $r_{0}=1$ and $r_{k+1}=m^{r_{k}}+r_{k}$ for $k=0,1,2$ , $\ldots$ , ri - 1. If $r(\Sigma)=1,$then there exists some finite product $A_{d_{k}}A_{d_{k-1}}\cdots$ $A_{d_{1}}$ with $k\leq r_{n-1}$ , which has spectralradius $r$ ( $A_{d_{k}}A_{d_{k-1}}\cdots$ Adl) $=1.$If $\Sigma$ consists of a single matrix I $=\{A\}$ , Theorem L-W says that if $||1||_{2}=||A^{n}||_{2}$ $=1$then $\mathrm{r}\{\mathrm{A}$) $=1$ (since $r_{n-1}=$ $\mathrm{r}\mathrm{n}$). Note that if $4=(\begin{array}{ll}0 10 0\end{array})$ then $||$ $4||_{2}=1$ and $r(A)=0.$So, the bound $n$ is the best integer. This result is not true when the space is infinitedimensional. We give a counterexample as follows :Example. Let $H=\ell^{2}$ with the coordinate unit vectors $e_{1}$ , $e_{2}$ , $\ldots$ as an orderedbasis and the 2-norm $||$ $||2$ . Define the linear operator $A$ determined by $Ae_{k+1}=e_{k}$for $k=1,2$ , $\ldots$ , $n$ and $Ae_{j}=0$ for $j=1$ or $n+2$ , $n+3$ , $\ldots$ . It is easy to see that$||A||_{2}=||An||_{2}=1$ and $n(A)=n.$ But $r(A)=0$ because $A^{n+1}=0.$ Therefore theconstant $n$ in the result of Lagarias and Wang is not the best constant when $H$ is infinitedimensional.1402. Main TheoremsIn this paper, we shall study how to generalize Theorem L-W to the infinite dimen-sional Hilbert spaces. Let $H$ be a complex Hilbert space with the inner product $\langle$ ., $\cdot\rangle$ andthe associated norm $||$ . $||$ , and $B(H)$ the algebra of all bounded linear operators on $H$ .An operator $A\in B(H)$ is called a contraction on $\mathrm{H}$ if $||Ax||\leq||x||$ for all $x\in H;$ and $A$is said to be compact if it maps the unit ball of $H$ onto a totally bounded set in $H$ . It isa well-known fact that the set of all compact operators on $H$ forms a two sided ideal of$B(H)$ (cf. [7]). It is readily seen from the Cauchy-Schwarz inequality that if $A$ is positive,i.e., (Ay, $y\rangle$ $\geq 0$ for all $y\in H,$ and (Ax, $x\rangle$ $=0$ for some $x\in H,$ then $Ax=0.$ Therefore,if $A$ is a contraction, then the null space of $I-A^{*}A$ becomes(3) $\mathrm{k}\mathrm{e}\mathrm{r}(\mathrm{J}-A^{*}A)=$ $\{x\in H;||Ax||=||x||\}$ .For a contraction $A$ , we define(4) $n(A)\equiv\dim \mathrm{k}\mathrm{e}\mathrm{r}(\mathrm{J}-A^{*}A)$ .In particular, if $n(A)<$ oo and A is an eigenvalue of $A$ with $|$ A$|=1,$ then $\dim N(\mathrm{A}I-A)\leq$$n(A)$ . When $A$ is of finite rank, it is easy to see that $n(A)\leq\dim R(A)$ , where $R(A)$ is therange of $A$ .In [4], we showed that (2) holds for a finite set I of compact contractions on $H$ exceptan operator in $\mathrm{y}$ is not compact. We showed that if $A$ is a compact contraction on $H$ and$||\mathrm{i}||=|\mathrm{E}1^{2^{n(A)}}||=1$ then $r(A)=1.$ Such constant $2^{n(A)}$ is not optimal. In this paper,we will show that $n(A)+1$ is the best constant. First, we list the properties of $n($ . $)$ asfollows.Lemma 1. (see also [4]) Suppose $A$, $B$ are two contractions on $H$ .(a) $n(A)=n(A^{*})$ .(b) $n(AB)\leq$ mm{n{A), $n(B)$ }.(c) If $A$ is compact, then $n(A)$ is finite.(d) If $A$ is compact and $n(A)=0,$ then $||A||<1.$For a finite set $\Sigma$ of operators in $B(H)$ , we define $|\Sigma|=$ the number of all elementsin I and the semigroup generated by $\mathrm{E}$ to be the set $\Sigma’=\bigcup_{m=1}^{\infty}C^{m}$ .Proposition 2. Let $A_{1},4_{2}$ , $\ldots,$ $/1_{m}$ be contractions on $H$ and let $B=A_{m}A_{m-1}$ $\ldots A_{1}$ .Suppose(a) there are nonnegative integers $j$, $k,p$ with $1\leq j<k\leq k+p\leq m$ such that$A_{j+_{\mathrm{P}}^{A}j+p-1}\ldots$ $A_{j}=A_{k+p}A_{k+p-1}\cdot$ . . $A_{k}$ ;141(b) $1\leq\ell=n(B)=r_{v}(A_{j+p}A_{j+p-1}\cdots A_{j})<\infty$ .Then $A_{k-1}A_{k-2}\cdot\cdot$ . $A_{j}$ has an eigenvalue A with $|$ A$|=1.$ In particular, we have$r(A_{k-1}1_{k-2}\cdots A_{j})=1.$Proof. Let $K=N(I-B^{*}B)$ . Then we have(5) $x\in K\Leftrightarrow||$A$m$A$m-1\ldots A_{1}x||=||x||$ .Since each $A_{i}$ is contraction, this implies that for every $1\leq i\leq mA_{i}A_{i-1}\cdots A_{1}$ is anisometry on $K$ and for $i=1,2$ , $\ldots$ , $m$$A_{i}A_{i-1}\cdots A_{1}K\subset N$($I-(A_{i+p}A:+p-1^{\cdot}$ . . $A_{i})^{*}(A_{i+p}A:+p-1\ldots$ Ak))It follows from Lemma $1(\mathrm{b})$ that for every $i=1,2$, $\ldots$ , $m$ ,(6) $\dim A_{i-1}\cdots A_{1}K=\dim K=$ n(B) $\leq n(Ai+pA:+p-1\ldots Ai)$ .Therefore we obtain from the condition (b) thatff\dim A_{i-1}\cdots$ $A_{1}K=\dim K=n(B)\leq n(A_{i+_{ mathrm{P}}:+p-1}A\ldots A_{ })$$A_{j-1}A_{j-2}\cdots A_{1}K$$=N(I-(A_{j+p}A_{j+\mathrm{p}-1}\cdots A_{j})^{*}(A_{j+p}A_{j+p-1}\cdots A_{j}))$$=N$($I-(A_{k+p}A_{k+p-1}\cdots A_{k})^{*}(A_{k+p}A_{k+p-1}\cdots$ Ak))$=A_{k-1}A_{k-2}$ ... $A_{1}K$.Since $\mathrm{A}\mathrm{k}-\mathrm{i}\mathrm{A}\mathrm{k}-2\cdot\cdot$ . $A_{j}$ is an isometry on $A_{j-1}A_{j-2}\cdots A_{1}K$ by (5), this shows that $A_{k-1}$$A_{k-2}\cdots A_{j}$ is an isometry from $A_{j-1}A_{j-2}\cdots A_{1}K$ onto itself. It follows from (6) that$\mathrm{A}\mathrm{k}-\mathrm{i}\mathrm{A}\mathrm{k}-2\cdot\cdot$ . $A_{j}$ has an eigenvalue Awith $|$ A$|=1.$ In particular, we have $r$ ($A_{k-1}$A$k-2^{\cdot}$ . . $A_{j}$ )$=1.$ This completes the proof.If $\Sigma$ is a finite subset of contractions on $H$ , we define $n( \Sigma)=\max n(A)A\in\Sigma$ .Corollary 3. Let $A$ be a contraction on $H$ , $m=$ n{A), and let $\Sigma=\{A\}$ . If$n(A^{m+1})\geq 1,$ then $r(A)=1.$ In particular, if $A$ is a compact contraction, then $||4^{m+1}||=$$1$ implies $r(A)=1.$Proof. Suppose $r_{p}(A)<1.$ Then $r(A^{k})<1$ for all $k=2,3$, $\ldots$ by the spectralmapping theorem. Since ) is a singleton, by Proposition 2 with $r=1,$ we have$n(A^{m+1})<n(B)$ $<$ n(Am-l) $<\cdots<$ n(B) $=m.$142Then $n(A^{m+1})=0.$ This is impossible. Therefore $r(A)=1.$ The proof is complete.Lemma 4. Let I be a finite set of contractions on H. Let $r=|\Sigma|$ , $s\geq 1,$ and $\Sigma_{1}=$$\Sigma^{\mathit{8}}$ . Suppose $\mathrm{n}(\mathrm{E}\mathrm{i})<$ oo and $rp(B)<1$ for all $B \in\bigcup_{j=1}^{r^{s}+s}\mathrm{I}^{j}$ , where $rp(B)= \sup\{\mathrm{A}$ ;A $=$$0$ or is an eigenvalue of $B$ }. Then there is a smallest integer $q$ with $s\leq q\leq r^{s}+s$ suchthat $\mathrm{n}(\mathrm{E}\mathrm{i})<\mathrm{n}(\mathrm{E}\mathrm{i})$ .Proof. Let $q=r^{s}+s.$ Suppose $B\in Cq$ be such that $n(B)=n(\Sigma_{1})$ . Let us write$B=A_{q}A_{q-1}\cdots A_{1}$ for some $4_{1}$ , $A_{2}$ , $\ldots$ , $A_{q}\in$ I. Then$A_{i+\epsilon-1}A_{i+s-1}\cdots A_{i}\in\Sigma^{s}$ for $i=1,2$ , $\ldots$ , $r^{s}+1.$Since $r^{s}\geq|\Sigma s1$ there are integers $i$ and $j$ with $1\leq i<j\leq r^{s}+1$ such that$A_{i+\epsilon-1}A_{i+s-2}\cdots A_{i}=A_{j+\epsilon-1}A_{j+s-2}$ . . . $A_{j}$ .Since $\mathrm{n}(\mathrm{B})\leq n(A_{i+\epsilon-1}A_{i+\epsilon-2}\cdots A_{i})\leq$ r(A) $=n(B)$ by Lemma $1(\mathrm{b})$ , it follows fromProposition 2 that $A_{j-1}\cdots A_{i}$ has an eigenvalue A with $|$ A$|=1.$ This contradicts to$r^{\theta}+s$$r_{p}(B)<1$ for all $B \in\bigcup_{j=1}\mathrm{I}^{j}$ .Remark 5. If $A$ is an $nxn$ matrix with $||A||_{2}=1,$ where $||$ . $||_{2}$ is the 2-norm on $\mathrm{C}^{n}$ ,then $\mathrm{n}\{\mathrm{A}$) $=n$ implies that $A$ is an isometry from $\mathrm{C}^{n}$ onto itself. Therefore $r(A)=1.$Prom this, we see that if $\Sigma$ is a finite set of contractions on $\mathrm{C}^{n}$ and $r(A)<1$ for every$A\in\Sigma$ , then $\mathrm{n}(\mathrm{B})\leq n-1$ .Corollary 6. If $A$ is an $n\mathrm{x}n$ matrix with $||A||_{2}\leq 1,$ then $r(A)<1$ if and only if$|\mathrm{E}^{1}n||_{2}<1.$Proof. Suppose $r(A)<1.$ By Remark 5, we have $n(A)\leq n-$ $1$ . It follows fromProposition 2 that$n\{A)<n(A^{n-1})<\cdots<$ $\mathrm{r}(\mathrm{A})$ $\leq n-1.$This implies $n(An)=0$ and hence $||$A$n||_{2}<1.$ The converse is obvious.Theorem 7. Let $\Sigma$ be the set of finite contractions $4_{1}$ , $A_{2}$ , $\cdots$ , $A_{r}$ on $H$ and $m=$$\mathrm{n}(\mathrm{B})<\infty$ . Put $r_{1}=r+1$ and $r_{k}=r^{r_{k-1}}+r_{k-1}$ for $k=2,3$ , $\ldots$ , $m$ . If $n(\Sigma^{r_{m}})\geq 1,$then there is some $B \in\bigcup_{j=1}^{r_{m}}C^{\mathit{3}}$ such that $r_{p}(B)=1.$ In particular, if each $A_{i}$ is a compactcontraction on $H$ and $||A||=1$ for some $A\in\Sigma^{r_{m}}$ , then there is some $B \in\bigcup_{j=1}^{r_{m}}\Sigma^{j}$ suchthat $rp(B)=1$ .143Proof. Suppose $r_{\mathrm{p}}(B)<1$ for all $B \in\bigcup_{j=1}^{r_{m}}\Sigma^{j}$ . It follows from Lemma 4 $m$-times that$n(\mathrm{I}^{r_{m}})<n(\Sigma^{r_{m-1}})<$ . . . $<n(\mathrm{I}^{r_{1}}<n(\Sigma)=m.$Therefore we must have $n(\Sigma^{r_{m}})=0.$ This contradicts our assumption that $n(\mathrm{I}^{r_{m}})\geq 1.$The following result gives an infinite-dimensional version of Theorem L-W.Theorem 8. Let 1 be the set of finite contractions $A_{1}$ , $A_{2}$ , $\cdots$ , $A_{r}$ on $\mathrm{C}^{n}$ . Put$r_{1}=r+1$ and $r_{k}=r^{r_{k-1}}+r’-1$ for $k=2,3$ , $\ldots$ , $n-$ l. If $n(\Sigma^{r_{n-1}})\geq 1,$ then there issome $B \in\bigcup_{j=1}^{r_{n-1}}\Sigma^{j}$ such that $r(B)=1.$Proof. Suppose $r(B)<1$ for all $B \in\bigcup_{j=1}^{r_{n-1}}\Sigma^{j}$ . By Remark 5, we have $n(\Sigma)\leq n-$ $1$ .It follows from Lemma 4 $n-1$-times that$n(\Sigma^{r_{n-1}})<$ $\mathrm{v}\mathrm{z}(\mathrm{I}r_{-2}")$ $<$ . . . $<$ n(fcrl $<$ $\mathrm{r}\mathrm{p}(\mathrm{B})$ $\leq n-1.$Therefore we must have $n(\Sigma^{p_{n-1}})=0.$ This contradicts our assumption that $n(\Sigma^{r_{n-1}})\geq 1.$(\Sigma^{\Gamma_{n-1}})=0.$Proposition 9. Let $T(\cdot)$ be a $C_{0}$ -semsigroup on $H$ with the infinitesimal generator $A$ .Suppose $T(t_{0})$ is contraction with $m=n(T(t_{0}))<$ oo for some $t_{0}>0.$ If $||\mathrm{T}(\mathrm{t}\mathrm{o})m+1||=1,$then $A$ has an eigenvalue $\mu$ with $Rep=0.$Proof. If $||T(t_{0})^{m}+1$ $||=1,$ it follows from Proposition 2 that there is some $1\leq n\leq$$m+1$ such that $T(t_{0})^{n}=$ T(nto) has an eigenvalue A with $|$ ) $|=1.$ By the spectralmapping theorem (cf. [5, Theore A-III.6.3.]) of $C_{0}$-semigroup for the point spectrum, wehave $\sigma_{p}(T(t))\backslash \{0\}=e^{t\sigma_{\mathrm{p}}(A)}$ for $t\geq 0.$ Therefore $\mathrm{A}\in e^{nt_{0}\sigma_{\mathrm{p}}(A)}$ , that is, $\mathrm{A}$ $=e^{t\mu}$ for some$\mu\in\sigma_{p}(A)$ . Since $|$ A$|=1,$ we have Rep $=0.$ This completes the proof.We denote the ideal of all compact contractions on $H$ by $\mathcal{K}_{C}(H)$ .Lemma 10. Let {Am} be a sequence of contractions on $H$ and let $A\in B(H)$ .Suppose(a) $A$ is compact;(b) there is a positive integer $r$ such that $n(A_{m})\geq r$ for all $m=1,2$ , $\ldots$ ;(c) $||$ $/\mathrm{i}m-A||arrow 0$ as $marrow\infty$ . Then $n(A)\geq r.$ Moreover, for every positive integer $r$the set $\{A\in \mathcal{K}_{C}(H);\mathrm{n}(\mathrm{A})<r\}$ is open in $\mathcal{K}_{C}(H)$ .Proof. Since each $A_{m}$ is a contraction, so is $A$ by (c). By (b), for every $m\geq 1$ thereis an orthonormal set $\{x_{m1}, \mathrm{x}\mathrm{m}2, \ldots, x_{mr}\}$ contained in $N(I-A_{m}^{*}A_{m})$ . Since $A_{m}$ is anisometry on $N(I-A_{m}^{*}A_{m})$ , {Amxmk; $1\leq k\leq r$ } is also an orthonormal set.144Since $A$ is compact, there are integers $1\leq n_{1}<n_{2}<\cdots$ such that $\{Ax_{n_{k}j}\}$ convergesto some element $y_{j}\in H$ for $j=1,2$, . .., $r$ . Since $||A_{m}-A||arrow 0$ as $marrow\infty$ , itis easy to see that $A_{n_{k}}x_{n_{k}j}arrow y_{j}$ strongly as $karrow$ oo for $j=1,2$ , . . ., $r$ . Since each$\{A_{n_{k}}x_{n_{k}j}; 1\leq j\leq r\}$ is an orthonormal set, so is $\{\mathrm{y}\mathrm{i}, y_{2}, \ldots, y_{r}\}$ . On the other hand, forevery $k$ and $j$ , $x_{n_{k}j}\in N(I-A_{n_{k}}^{*}A_{n_{k}})$ implies $A_{n_{k}}^{*}A_{n_{k}}x_{n_{k}j}=1X_{n_{k}j}$ . Therefore we have forevery $j=1,2$ , . . ., $r$$y_{j}$ $= \lim_{karrow\infty}A_{n_{k}}A_{n_{\mathrm{k}}}^{*}A_{n_{k}})c_{n_{k}j}=$ AA*yj.This shows that $n(A^{*})\geq r$ and hence $n(A)=n(A^{*})\geq r.$ The proof is complete.Proposition 11. Let ) be a compact set of compact contractions on H. Suppose(a) $\mathrm{X}$ is a countable set;(b) $r(A)<1$ for all A $\mathrm{e}\Sigma’$ . Then $\Sigma$ is asymptotically stable $(a.s.)$.Proof Since I is compact, so are the $\Sigma^{m}(m\geq 1)$ . By Lemma $1(\mathrm{b})$ , $\{n(\Sigma^{m})\}$ isa decreasing sequence. We show that $\mathrm{n}(\mathrm{E}\mathrm{m})=0$ for some $m$ . Since $\mathrm{I}^{m}$ is compact, byLemma $1(\mathrm{d})$ , this will imply $||A||<1$ for all $A\in$ $\Sigma m$ . Thus the compactness of $\Sigma^{m}$ showsthat $\Sigma$ is $\mathrm{a}.\mathrm{s}$. Suppose $\mathrm{n}(\mathrm{E}\mathrm{m})$ is never zero. Then there is some integer $m_{0}\geq 1$ suchthat the $n(\Sigma^{m})(m\geq m_{0})$ are all equal to a positive integer Z. Put $\Omega=\Sigma^{m_{0}}$ and let$S=$ { $E\subset$ ); $E$ is compact and $n(E^{m})=\ell$ for all $m=1,2$ , $\ldots$ }. Then $(S, \subset)$ is a partiallyordered set. We claim that $S$ has a minimal element.If $\{E_{i}\}$ is a decreasing chain in $S$ , then $E= \bigcap_{i}E_{i}$ is nonempty compact set. If$n(E^{m})<\ell$ for some positive integer $m$, then, by Lemma 10, there is an open subset $V$of $\mathcal{K}_{C}(H)$ such that $E^{m}\subset V$ and $n(V)<\ell$ . Since $\{E:\}$ is a decreasing chain of compactsets, so is $\{E_{i}^{m}\}$ . Therefore there must have some $i$ such that $E_{i}^{m}\subset V.$ This contradictsto $E_{i}\in S.$ Hence $E$ is a lower bound of {E$\{\}$ . By Zorn’s Lemma, $S$ has a minimalelement, say $\Omega_{0}$ . Clearly, $\Omega_{0}$ is also countable. So, $\Omega_{0}$ has an isolated point, say $B$ . Since$\Omega_{1}=\Omega_{0}\backslash \{B\}$ is not in $S$ , there is some positive integer $m_{1}$ such that$(^{*})$ $n(\Omega_{1}^{m_{1}})$ $<\ell$ .We claim that $n(\Omega_{0}^{2m_{1}})<\ell$. Thus $\Omega_{0}\not\in S;$ a contradiction. We part the proof into threecases.Let $4_{1}$ , $A_{2}$ , $\ldots$ , $A_{2m_{1}}\in\Omega_{0}$ .Case 1. $A_{i}=A_{j}$ for some $1\leq j<k\leq$ 2mi. Then, by Proposition 2 with $p=0,$ wehave $r$ ( $A_{k-1}$A$k-2\ldots A_{j}$ ) $=1.$ This contradicts to (b). So, we can assume $A_{:}4$ $A_{j}$ for all$1\leq i\neq j\leq 2m_{1}$ .Case 2. $B\neq A_{:}$ for all $i=1,2$, . . . , $2\mathrm{m}\mathrm{i}$ . Then $A_{2m_{1}}A_{2m_{1}-1}\cdots A_{1}\in\Omega_{1}^{2m_{1}}$ . By (b).$n(A_{2m_{1}}A_{2m_{1}-1}\cdots A_{1})<\ell$ .145Case 3. $B=A_{i}$ for some $1\leq i\leq 2m_{1}$ . Then either $1\leq i\leq m_{1}$ or $m_{1}+1\leq i\leq 2m_{1}$ .Anyway, we have either $A_{m_{1}}A_{m_{1}-1}$ $\ldots A_{1}\in\Omega_{1}^{m_{1}}$ or $A_{2m_{1}}A_{2m_{1}-1}$ $\ldots A_{m_{1}+1}\in$ $1m_{1}1$ . Itfollows from Lemma $1(\mathrm{b})$ and (’) that $n(A_{2m_{1}}A_{2m_{1}-1}\cdots A_{1})<\ell$ .This completes the proof.REFERENCES[1] Berger, M. A. and Wang, Y. Bounded semigroups of matrices, Linear Algebra Appl.166 (1992), 21-27.[2] Daubechies, I. and Lagarias, J. $\mathrm{C}$ , Sets of matrices all infinite products of whichconverge, Linear Algebm Appl 162 (1992), 227-263.[3] Lagarias, J. C. and Y. Wang, The flniteness conjecture for the generalized spectralradius of a set of matrices, Linear Algebra Appl. 2141995, 17-42.[4] Li, Y.-C. and M.-H. Shih, Contractibility of Compact Contractions in Hilbert Space,Linear Algebra Appl. 341 (2002), 369-378.[5] Nagel, R., One-parameter Semigroups of Positive Operators, LNM, 1184, 1986.[6] Rota, G.-C. and G. Strang, A note on the joint spectral radius, Indag. Math. 22(1960), 379-381.[7] Taylor, A. E. and D. C. Lay, Introduction to fihnctional Analysis, 2nd ed., Wiley,New York, 1980.

THE LENGTH OF CONTRACTIBILITY OF COMPACT CONTRACTIONS ON A HILBERT SPACE (Nonlinear Analysis and Convex Analysis)

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THE LENGTH OF CONTRACTIBILITY OF COMPACT CONTRACTIONS ON A HILBERT SPACE (Nonlinear Analysis and Convex Analysis)

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