To Buenos Aires, on her fourhundred first birthday
"A mi se me hace cuento que empez6 Buenos Aires, la juzgo tan eterna como el agua o el aire" (Jorge Luis Borges)
Contents
1. Stability and approximation
1
1.1 Lower estimated derived from the RieszDunford functional calculus 1.2 Lower estimates for the distance to Nk(H) 1.3 Lower semicontinuity of the rank 1.4 Stability properties of semiFredholm operators
2 6 8 9
1.5 On invariance and closure of subsets of L(H)
10
1.6 Notes and remarks
11
2. An aperitif:
approximation problems in finite dimensional
spaces
12
2.1 Closures of similarity orbits in finite dimensional spaces
13
2.1.1 The nilpotent case
15
2.1.2 Proof of Theorem 2.1 2.1.3 The lattice (N(Ek)/i,<)
16 17
2.1.4 Closures of similarity orbits of finite rank operators 2.2 The distance from the set of all nonzero orthogonal projections to N(H)
19
20
2.2.1 The limit case
20
2.2.2 On the exact values of 6k and nk 2.2.3 A companion problem: the distance from the set of
23
all nonzero ide!_11potents to N{H) 2.3 On the distance to Nk{H) 2.3.1 A general upper bound 2.3.2 Two illustrative examples
25 27 27 30
2.3.3 An example on approximation of normal operators by nilpotents
33 vii
2.3.4 On the distance to a similarity orbit 2.4 On the distance from a compact operator to N(H) 2.5 Notes and remarks 3. The main tools of approximation
3.1 The Rosenblum operator:
X
+
AX XB
3.2 Generalized Rota's universal model 3.3 Apostol triangular representation 3.4 Correction by compact perturbations of the singular behavior of operators 3.5 ApostolFoia~Voiculescu's theorem on normal restrictions of compact perturbations of operators 3.5.1 Schatten pclasses 3.5.2 Normal restrictions 3.5.3 Density of sets of operators with bad properties 3.6 Notes and remarks 4. Two results borrowed from the theory of C*algebras 4.1 Essentially normal operators
4.2 4.3 4.4 4.5
BrownDouglasFillmore theorem BergerShaw trace inequality Examples of essentially normal operators An application to approximation problems
Matrix models for operators Spectra of compact perturbations of operators Voiculescu's theorem Closures of unitary orbits 4.5.1 Operatorvalued spectrum and unitary orbits 4.5.2 Concrete examples of closures of unitary orbits 4.5.3 On normal and quasinilpotent restrictions
viii
37 38 41
3.1.1 Linear operator equations 3.1.2 Approximate point spectrum of a sum of commuting operators 3.1.3 Local oneside resolvents in L(H) 3.1.4 The left and the right spectra of TAB 3.1.5 RosenblumDavisRosenthal corollary 3.1.6 The maximal numerical range of an operator 3.1.7 The norm of TAB
5. Limits of nilpotent and algebraic operators 5.1 Limits of nilpotent operators
108 108
5.1.1 Normal limits of nilpotents
108
5.1.2 Spectral characterization of N(H)
111
5.2 Closures of
~imilarity orbits of normal operators with perfect spectra 5.3 Limits of algebraic operators 5.4 Normal operators in closures of similarity orbits 5.5 Sums of two nilpo~ents 5.6 The ApostolSalinas approach: an estimate for the distance to Nk(H) 5.7 Salinas' pseudonilpotents 5.8 Limits of nilpotent and algebraic elements of the Calkin algebra 5.9 On the spectra of infinite direct sums os operators 5.10 Notes and remarks
6. Quasitriangularity
113 114 115 117 120 124 128 130 132 135
6.1 ApostolMorrel simple models
135
6.2 Quasitriangular operators
140
6.2.1 Equivalence between the formal and the relaxed
definitions of quasitriangularity 6.2.2 Two lower estimates for the distance to (QT)
141 142
6.2.3 Spectral characterization of quasitriangularity
145
6.3 Biquasitriangular operators
146
6.3.1 Blockdiagonal and quasidiagonal operators
146
6.3.2 Characterizations of biquasitriangularity
147
6.4 On the relative size of the sets (QT), (QT)*, (BQT),
[N(H)+K(H)] and N(H)
153
6.5 A Riesz decomposition theorem for operators with
disconnected essential spectrum 6.6 Notes and remarks 7. The structure of a polynomially compact operator
154 157 162
7.1 Reduction to the (essentially) nilpotent case
162
7.2 The structure of a polynomially compact operator
164
ix
7.3 7.4 7.5 7.6 7.7
Restrictions of nilpotent operators 167 Operators similar to Jordan operators 171 A similarity invariant for polynomially compact operators 173 Nice Jordan operators 177 Notes and remarks 188
8. Closures of similarity orbits of nilpotent operators 8.1 Universal operators 8.1.1 Universal quasinilpotent operators 8.1.2 Universal compact quasinilpotent operators 8.2 Compact perturbations of not nice operators 8.3 Quasinilpotents in the Calkin algebra
189 189 194 194 198
8.3.1 General quasinilpotents 8.3.2 Nice elements of the Calkin algebra
198 204
8.4 Compact perturbations of nice Jordan operators
205
8.4.1 8.4.2 8.4.3 8.4.4 8.4.5
Nice Jordan nilpotents Nilpotents of order 2 Quasinilpotent perturbations Universal operators in N~,h(H) A general criterion for universality
8.5 Separation of isolated points of the essential spectrum affiliated with nilpotents 8.6 Notes and remarks REFERENCES INDEX SYMBOLS AND NOTATION
X
189
206 211 212 215 222 230 236 239 249 253
Preface
The last decade has been fruithful in results on approximation of Hilbert space operators, due to a large extent to the impulse given by Paul R. Halmos in his famous survey article "Ten problems in Hilbert space". The purpose of this monograph (and a second one, by C. Apostol, L. A. Fialkow, D. A. Herrero and D. Voiculescu that will follow and complete the results contained here) is to provide a set of general arguments to deal with approximation problems (in the normtopology) related to those subsets of the algebra L!HI of all operators acting on a complex separable infinite dimensional Hilbert space that are invariant under similarities. Many interesting subsets of L(HI have this property: nilpotent operators: algebraic operators (satisfying a fixed polynomial): polynomially compact operators: triangular, quasitriangular and biquasitriangular operators: cyclic and multicyclic operators: semiFredholm operators (with fixed given indices): operators whose spectrumis equal to a fixed compact subset of the complex plane ~. or whose spectra are contained in a fixed nonempty subset of ~: any bilateral ideal of compact operators, etc, etc. The following list illustrates the kindsofproblems to be considered here: a) Given a subset R of L(HI invariant under similarities, defined in algebraic, geometric or analytic terms (e.g., the set of all algebraic operators, the set of all operators T such that T3 is compact, the set of all cyclic operators), characterize its normclosure in "simple terms". Since the spectrum and its different parts are the most obvious similarity invariants of an operator, these "simple term~' will usually be expressed in terms of properties of the different subsets of the spectra of the operators in the closure of R. b) More generally, obtain a formula for the distance from a given operator to R or, al least, upper and/or lower estimates for this distance. c) In a surprisingly large number of interesting cases, either R is invariant under compact perturbations, or its closure is contained xi
in (or equal to) the set of all compact perturbations of R. Compact perturbations will be used as a useful tool for approximation and will be also analyzed with respect to the above mentioned peculiar properties. d) Analogues to the problems in a) and b) for subsets of the quotient Calkin algebra. In Chapter I, we shall obtain all the necessary conditions for approximation that can be easily derived from the RieszDunford functional calculus and the wellknown stability properties of semiFredholm operators. This chapter is followed by "an ap~ritif": the solution of several approximation problems in finite dimensional spaces, which only depend on the results of Chapter I and some "handwork" with matr~ces.
It is interesting to observe that, for a large number of approximation problems (but, unfortunately, not for all of them) the "obvious" necessary conditions derived from the results of Chapter I are actually sufficient, but the proofs of their sufficiency are very hard.These proofs are constructive to a large extent and the "tools" for these constructions are developed in Chapters III and IV: Rosenblum's coro~ lary, Rota's universal model and extensions, Apostol's triangular representation, results on compact perturbations and results "borrowed" from the theory ofC*algebras (the BrownDouglasFillmore theorem, Voiculescu's theorem, results on closures of unitary orbits). Except for those results related to the theory of C*algebras (Chapter IV) , the monograph is essentially selfcontained. Results on approximation of Hilbert space operators really begin in Chapter V with the characterization (due to c. Apostol, C. Foia~ and D. Voiculescu) of the closure of the set of all nilpotent operators. The closure of the set of all algebraic operators is then obtained as a corollary and this result is used to characterize biquasitriangularity, the closure of the similarity orbit of a normal operator with perfect spectrum and to give several results about the closure of the similarity orbit of an arbitrary operator. Combining the above results with ApostolMorrel "simple models", we shall obtain the ApostolFoia~Voiculescu theorem on the spectral cha; acterization of quasitriangularity. Since algebraic op~rators and operators that "look like backward shifts" are the simple3t examples of triangular operators, this approach to quasitriangularity is, perhaps, more natural than the original one. Finally, the last two chapters are devoted to a deep analysis of the structure of a polynomially compact operator and to the clos~e of xii
its similarity orbit, respectively. A large part of the material contained in the last chapter consists of unpublished results of c. ApostolD. Voiculescu, J. Barr!aD. A. Herrero and D. A. Herrero on closures of similarity orbits of essentially nilpotent operators. The author is deeply indebted to Professors Constantin Apostol, I. David Berg, Charles A. Berger, Ronald G. Douglas, Alain Etcheberry, Lawrence A. Fialkow, Ciprian Foia,, Carl M. Pearcy, Allen L. Shields and Dan Voiculescu, and to his professors and exfellow students from the University of Buenos Aires, Alejandro de Acosta, Mischa Cotlar, Beatriz Margolis, Lazaro Recht, Norberta Salinas and, very especially, to his wife Marta B. Pecuch de Herrero for infinitely many informal discussions and suggestions and, most important, for their friendly support during all these hard years. The contents of this monograph have been developed in a "Seminar on Approximation of Hilbert Space Operators• at the University of Georgia (Athens, Georgia, USA), during the academic year 19801981. The author wishes to thank the authorities of the University of Georgia for their support and to his colleagues Edward Azoff, Richard Bouldin, Kevin Clancey, Douglas N. Clark and Derming Wang for their assistance during the preparation of the manuscript. Many of their valuable observations have been included here. Domingo A. Herrero
Tempe, Arizona June, 1981
xi~i
1 Stability and approximation
As explained in the Preface, in "most" approximation problems related to similarityinvariant sets of operators, the "obvious" necessary conditions for approximation that can be derived from the RieszDunford functional calculus and the stability properties of semiFredholm operators turn out to be also sufficient. In a certain sense,this short first chapter contains all the necessary conditions and the remaining of the monograph (and the second one that will follow [16]) is devoted to explain why these necessary conditions are also sufficient. Throughout this monograph, the word operator will always denote a bounded linear transformation mapping a complex Banach space into another. If X and Y are complex Banach spaces, the Banach space of all operators mapping X into Y will be denoted by L(X,Y). We shall write L(H) for the Banach algebra L(H,H), where His a complex separable Hilbert space. Unless otherwise stated, H (H 0 , H1 , H2 , ••• , etc) will always denote an infinite dimensional Hilbert space. The algebra L(H) contains the open subset G(H) ={WE l(H):
W is invertible}
(the linear group of L(H)), which will play a very important role here. A subset R of L(H) is called invariant under simiLarities (or simiLarityinvariant) if it is invariant under conjugation by elements of the group G(H), i.e., T
€
R => S (T)
c
R,
where 1 .S (T) = { WTW :
W
E
G(H)}
is the simiLarity orbip of T. If K(H) denotes the ideal of all compact operators acting on Hand ~:l(H) + A(H) = L(H)/K(H) is the canonical projection of L(H) onto the (quotient) Calkin algebra, then the image ~(T) = T+K[H) ofT € l(H) in A(H) will also be denoted by T. The reader is referred to [77], [119] for the general theory of Hilbert space operators. 1
1.1 Lower estimates derived from the RieszDunford functional calculus A nonempty bounded open subset n of the complex plane a: is a Cauchy domain if the following conditions are satisfied: (i) n has finitely many components, the closures of any two of which are disjoint, and (ii) the boundary an of n is composed of a finite positive number of closed rectifiable Jordan curves, no two of which intersect. In this case, r = an will be assumed to be positively oriented with respect to n in the sense of complex variable theory, i.e., so that 1 2'11'i
J
d>..
r >..~
{1, if =
o,
~ e n 1 n = nur
if ~
(the upper bar will always denote closure with respect to the metric topology of the underlying space). Clearly, r is uniquely determined by n (and conversely). We shall say that r is a rectifiabZe contour. If all the curves of r are regular analytic Jordan curves, we shall say that r is an analytic contour (or n is an anaZytic Cauchy domain). If A is a Banach algebra with identity 1 and a e A, the spectrum of a will be denoted by a(a). The complement p(a) = 0:\o(a) of a(a) in the complex plane is the resoZvent set of a and the function >.. ~ (Xa)l (from p(a) into A) is the resolvent of a. It is wellknown that (Xa)l is an analytic function of >.. in that domain that it satisfies the first resoZvent equation: (>..a) l_ (pa) l = (ll X) (>..a) l (pa) l (X rll e p (a) ). Furthermore, if a, be A and>.. e p(a)np(b), then (>..a) 1  (Xb)l = (Xa) 1 (a b) (Xb)l (second resolvent equation). If a is a nonempty clopen subset of a(a), then there exists an analytic Cauchy domain n such that a c nand [a(a)\a]nn= ; in this case 1 Jr (>..a) 1 dX E(o;a) = 2 '1fi
is an idempotent of A commuting with every b in A such that ab = ba. (E(a;a) is the Riesa idempotent corresponding to a.) The following theorem is just a quantitative form of the classical result on upper semicontinuity of separate parts of the spectrum. The reader is referred to [76], [153], [172], or [173,Chapter XIV] for the basic properties of the RieszDunford functional calculus. THEOREM 1.1. Let a and b be two elements of the Banach algebra A 2
~ith
identity 1. Assume that the spect~um o(a) of a is the disjoint u~ ion of two compact subsets a 0 and o 1 such that o 1 is nonempty (o 1 t< ~)
a~d Zet n be a Cauchy domain such that a 1
llabrJ< minfmc;~.a)~J 1 : (where JI.Jidenotes the norm of A)
II = II a at II
;~.
If
an},
€
= min{ 11!>.a) ~1 1 :
A " an} and
= t6 < li ~~ 1
for all A E an, so that o(at)nan = fJ for all t E Thus, the idempotent
....!..f 2w1 an
et
nn = ·
then o(b) nO 'I ffand o(b) nan =fJ.
PROOF. Assume that II ab II = 6 < m let at= (1t)a+tb, 0 ~ t ~ 1: then 11!>at>  <><a>
0 and a 0
c
(>.
at
~0,11.
) 1 d.>.
is a well defined element of A for all t E [0,1]. Furthermore, if 0 ~ t < s ~ 1, the second resolvent equation implies that
whence it readily follows that t + et is a continuous mapping from [0,1] into A. Since o 1 = o(a)nn 'I fJ, it follows that 1
eo = 2w1
f
(;i.a)
1
d).
F o,
an so that II e 0 II :2; 1 (Recall that et is idempotent, 0 ity, lle 1 11 :2; 1 and therefore el
= 2!i f
(>.b)1 d). an This is clearly impossible, unless o(b)nn
~
~
t
1). By continu
F 0. F fJ.
0
Recall that if (X,d) is a metric space and B(Xl (Bc(X]) is the family of all nonempty bounded (closed bounded, respectively) subsets of X, then ~(A,B)
= inf{e
>
0:
B
c
AE, A
c
BE},
where AE = {x EX: dist[x,AJ s e},defines a pseudometric in B(Xl (a metric in Bc(X], resp.);dH(A,B) is the Hausdo~ff distance between A and B, A, B € Bc(X]. The qualitative form of Theorem 1.1 is the following COROLLARY 1.2. (i} Let a be an element of a Banach algebra A with identity. Assume that o(a) is the disjoint union of two compact subsets a0 and a 1 such that o 1 F fJ. and that a 1 is contained in abounded openset 0. Then there ezists a constant C = C(a,o 1 ,n) > 0 such that 3
o(b)nO t: ~foro aZZ bin A satisfying llabll 0, there e~ists 6 > 0 suah that a(b) a(a)E, provided llabll < 4, i.e., the mapping a+ a(a) from A into Be(~) (Hausdorff distanae) is upper semiaontinuous.
c
PROOF. (i) follows immediately from Theorem 1.1: Given a 1 and 0 (as above) , there exists a Cauchy domain o1 such that a 1 c n1 c n 1  c S2 and a 0 no 1  =~Take c = min{lj(Aa) 1II 1 : Ac: an 1 }. (ii) Apply (i) with a0 ~and 0 =interior a(a)E. 0 COROLLARY 1. 3 . (i) If S2 is an open subset of ~. then {a c: A: c S'l} is an open subset of A. (ii) If E is a G6 subset of ~. then {a € A: a(a) c E} is a G6 subset of A. In partiauZar, the set {a e; A: a (a) = {0}} of aZZ quasiniZpotent eZements of A is a G6 in A. a(a)
PROOF. (i) follows from Corollary 1.2(ii) and (ii) is an immediate consequence of (i) • 0 The limit case of Theorem 1.1 yields the following COROLLARY 1.4. Let a, o 1 (o 1 7: ~) and assume that b c: A satisfies the inequality
n be
II a b II s min{ II< Aa) 11 I 1 :
as in Theorem 1.1 and >.
c:
anl;
PROOF. Define at as in the proof of Theorem 1.1; then Theorem 1.1 implies that a(atlnan = g and o(at)nS'l t: ~for 0 s t < 1. Since lim(t _,. 1) lib at II= 0, i t follows from Corollary 1.2(i) that o(b)nO cannot be empty. 0 It is convenient to observe that the result of Corollary 1.4(and, a fortiori, the result of Theorem 1.1 too) is sharp. In fact, we have EXAMPLE 1.5. Let P c: L(H) be a nonzero orthogonal projection and let Q = {A: I All < ~}; then II< AP) llll = ~ (A € an ) • Thus' by Corol lary 1.4, a(B)nS'l 7: ~ for all B in L(H) such that liP Bll s ~On the other hand, if A = ~. then II P A II = ~ and o (A) = {~} c an. If T c: L(H), o is a clopen subset of o(T) and E(o;T) is the corresponding Riesz idempotent, then the range ran E(o;T) and the kernel ker E(o;T) of E(a;T) aresubspacesofH invariant under every Bin L(H) commutinq with T (i.e., hyperinvariant forT), and H can be written as 4
the algebraic (not necessarily orthogonal!) direct sum H = ran E(a:T)+ ker E(a:T). (Here and in what follows, subspace will always denote a closed linear manifold of a Banach space.) Furthermore, the spectrum of the restriction T)ran E(a:T) ofT to ran E(a:T) coincides with a and the spectrum of the restriction T!ker E(a:T) coincides with a(T)\a [l73,Chapter XIV]. In what follows, ran E(a:T) will be denoted by H(a:Tl. If a = {X} is a singleton, we shall simply write H(X:T) (E(X:T)) instead of H({X}:T) (E({X}:T), resp). If a = {X} and dim H(X:T) is finite, then X is called a normaL eigenvaLue ofT: in this case, H(X:T) coincides with ker(XT)n for some n ~ 1. The set of all normal eigenvalues ofT will be denoted by a 0 (T). Clearly, a 0 (T) is contained in the point spectrum ap(T) ofT (i.e., the set of all eigenvalues ofT). The esse~ tiaL spectrum ofT, i.e., the spectrum ofT in A(H) will be denoted by ae (T).
COROLLARY 1.6. Let A, Be L(H); then (i)
Assume that a is a nonempty
~Lopen)subset
of a(A) and Let Q
~Cauchy domain) be a neighborhood of a such that
[a(A)\a]nO =.If X dO}. then a 1 =a (B) nO ,; ~; (ii) furthermore, dim H(a:A) =dim H(a 1 :B) (0 s dim H(a:A) s ~>. (iii) If a is a nonempty cLopen subset of a (A) and the Cauchy e domain n is a neighborhood of a such that [a e (A)\a]nO = ~. then a e (B) nO,; for aZZ Bin L(H) suah that IIABII < min{I!CXA)~1l: A e an}.
II A B II
< min{
II< XA) ~~l:
PROOF. (i) and (iii) follow immediately from Theorem 1.1, applied to A= L(H) and to A= A(H), respectively. (ii) This follows from the proof of Theorem 1.1. Observe that, if At= (1t)A+tB, then the continuity of the mapping t + E(a(At)nO:At) (0 s t s 1) implies that the idempotents E (a :A) = E (a (A 0 ) nO :A0 J and E (a 1 :B) E (a (A1 ) nO:A1 ) necessarily have the same (finite or infinite) rank.O Until now, we have only applied the arguments of functional calc~ lus to a very particular class of functions analytic in a neighborhood of the spectrum a(a) of an element of the Banach algebra A: namely, the characteristic function of a suitable neighborhood of a clopen su~ set of a(a). Analogous results hold in a much more general setting: namely, PROPOSITION 1.7. Let a be an eLement of the Banach algebra A with identity 1 and let f be an anaLytic function defined in a neighborhood n of a(a). Given £ > 0, there ezists ~ > 0 such that f(b) is weLLde5
fined for aZZ b in A satisfying
II a bll
II f (a)  f
< ti and. mol'eovel'.
(b)
II
< e.
PROOF. Let n 1 be a Cauchy domain such that a(a) c n 1 c Ql c Q. By Corollary 1.2(i), there exists o1 > 0 such that a(b) c n 1 for all b inA satisfying llabll < ti 1 • Clearly, f(b) is welldefined for all these b, by means of the integral f(b)
= 2!i
Ianl f(~) <~b)ld~.
The second resolvent equation implies that max{ 11<~b) ~1:
~ " an 1 }
< l+max{
11<~a) 11:
~ " an 1 l,
and llf(a)f(b) II s (1/2'11") .length(an 1 > .max{ If(~) I·II11 x
1
II>:
A.
E
an 1 l ·II a blj,
provided II a  b II < o2 (for some o2 , 0 < o2 s ti 1 ) • It follows that, i f !Ia bll < o for a suitably chosen ti, 0 < then f (b) is welldefined and II f {a)  f (b) II < e. fJ
o < oz
The following particular case is especially important for our purposes. COROLLARY 1.8. Let
a~:
A.
let~
be an isolated point of a(a) and
let k
={(A.~} • in some neighborhood of~
f(A.) for some k
2:
0
• in some
neighborhoodofa(a)\{~}.
1.
Let {an\,,:1 be a sequence of elements in A such that !Ia ani!+ 0 (n+oo). Then f(an) is IJJelldefined for all n large enough and
f(a) =0 if and only if lim(n+ oo) l!f(anlll
=
0.
1.2 Lower estimates for the distance to Nk(H)
Let Nk (A l = {a E A: a"· = 0} denote the set of all nilpotent elements of order at most k {k = 1,2, ••• ) of the Banach algebra A, and let N(A) = uk:l Nk(A) be the set of all nilpotents ·of A. In order to simplify the notation, the set of all nilpotents (of order at most k) in L(H) will be denoted by N(H) (Nk(H), resp), or simply by N if His understood. It is a trivial consequence of Proposition 1.7 that, given a non6
constant polynomial p,the set {a E particular, i f we choose pk(A) COROLLARY 1.9. (i)
=
A:
p(a) = 0} is closed in A. In
Ak (k = 1,2, ••• ), then we obtain
Nk(A) is alosed in A for aZZ k = 1,2, •••
N(A) is an F0 subset of A.
(ii)
The following result provides a partial answer to the problem of estimating the distance from a given operator to Nk(H). PROPOSITION 1.10. (i) Bkx = O,
for some
x
II A B II
Let A, BE L(H). If IIAkxll = llxll = 1, but in H and some k l! 1, then
~
max{
(ii) If IIAkxll = llxll = 1, but Bkx = 0 for some x in H and some k
~
1, and max{IIAII,IIBIIl = M, then IIABII ;:: 1/(kM).
(iii) for some k
Let a, b e: A (a Banaah algebra). If llakll = 1, but bk = O, ~
1, then
II a b II ;:: max{ l/k11 all, l/k11 bIlL
(iv) If II akll =1, but bk = 0, for some k M,
~
1, and max{ll aJI,IIblll
=
then IJabll ~ 1/(kM). PROOF.
(i)
Let IIAII = M and IIABII = cS. I t is completely apparent
that II B II ~ M+cS, and therefore 1 = II Akx II = II
~I~:~
~
II AkBk II
11Airj11ABII11BIIj
~ I~:~
~ cSI~:~
II AkjBjAkjlBj+lll
Mkjl(M+cS)j =
cSMkli~:~j
cSMkl[ ( l+.Q) k1] (cS/M)  l = (M+o) k_t;c, M so that (M+cS)k
~ Mk+l. Hence, cS ;:: (Mk+l)l/k_M.
I f we assume that IIBII = M and IIABII =
o,
then IIAII ~ M+cS and we
arrive at the same inequality, whence the result follows. The remaining statements follow by the same argument. REMARKS. 1.11. (i)
0
Since, in Proposition l.lO(ii) and (iv), M
cannot be smaller than 1 and (by the meanvalue theorem) 1/ ( 2kt.t)
< (~+1) l/k_M s 1/ (kM) ,
for all k;:: 1, the estimates of (i) and (ii)
((iii) and (iv)) are of
the same order. (ii)
The argument of the proof of Proposition l.lO(i) also ap
plies to the case when IIP(A)xll = llxll = 1, but p(B)x = 0, for some pol2_ m k· nomial p, p(A) = rrj=l(AAj) J. Since ll[p(A)p(B)Jxll = 1, IIA.BII cannot
7
be "too small", where the words "too small" have a concrete numerical expression in terms of p and A . The same applies, of course, to (ii), (iii) and (iv).
1.3 Lower semicontinuity of the rank PROPOSITION 1.12. Let A e L(H) and Zet {An}n:l be a sequence of opePatoPs such that II A A II + 0 (n + oo); then n (i) rank As lim inf(n + oo) rank An. If IIAnxll ~ £11xll foP some £ > 0 and aZZ x in a subspace Hn with dim Hn ~ d, then II Axil ~ £11 xll fop aZZ x in a subspace H0 with dim H0 ~ d. (iii) If An e KIHI foP aZZ n ~ n 0 , then A e K(HI. (ii)
PROOF. (i) If lim inf(n + oo) rank An=oo, then there is nothing to prove. Assume that lim inf (n + oo) rank An= d < oo. Passing, if necessary, to a subsequence, we can directly assume that rank An= d for all n = 1. 2, ••.• If rank A~ d+l, then there exist d+l linearly independent vectors y 1 , y 2 , ••• , yd+l e H such that {Ayj}~!i is a line~rly indepen~ ent set. Clearly, Y =linear span{y 1 ,y 2 , •.• ,yd+l} has dimension d+l and therefore Ynker An~ {O} for each n = 1,2, ••.• It is easily seen that {An}n:l is a bounded sequence and there exists y e Y and a subsequence {An.}J.:l such that IIAYII = 1 but lim(j + oo)IIA Yll = 0, so that "" J nj {An.lj=l cannot converge to A, even in the strong operator topology, a J
contradiction. Therefore, rank A s d. (ii) Observe that our hypothesis implies that IIA*A~II + 0 and i IIA*AA~Anll + 0 (n + oo). Now the result follows from an elementary ana~ysis of the spectral decompositions of the hermitian operators A*A and A*A (see, e.g., [117]). n n The third statement is trivial. 0 It is convenient to recall that A e LIHI is compact if and only if ran A does not contain an infinite dimensional subspace [71],[96]. we shall use the following nonstandard notation: If A e KIHI is not a finite rank operator, then rank A will be defined as oo. Thus, rank T = oo will be an equivalent way to say that T is not compact. The different possible ranks will be linearly ordered by 0 ~ 1 ~ 2 ~ ••• ~ n ~ n+l ~ ••• ~ 8
oo ~ ""•
1.4 Stability properties of semiFredholm operators Recall that T € L(H) is a semiFredhoZm operator if ranT is clos ed and either nul T = dim ker T or nul T* = dim ker T* = dim H/ran T
is finite (where T* denotes the adjoint ofT in L(H)). In this case, the index of T is defined by ind T = nul T  nul T*. The following theorem resumes the main properties of the semiFredholm operators. The reader is referred to [106] and [153,Chapter IV] for details. THEOREM 1.13. Let T E L(HI be a semiFredhoZm operator; then (i) T* is aZso a semiFredhoZm operator and ind T* = ind T; (ii) There exists a constant 6 > 0 such that IITxll ~ 6llxll for aZZ x 1 ker T and II T*y II ~ 611 y II for aZZ y .L ker T*; moreover, 6 can be chosen as min{). E o([T*T]~)\{0}}; (iii) There exists 6 = 6(T) > 0 such that if A E: L(H), tiAlls land j).j < 6, then T+).A is semiFredho Zm ar~d ind (T+AA) = ind '1', nul (T+AA) s nul T and nul (T+AA) * s nul T*;moreover,nul(T+I.A) and nul (T+).A) *are constant for 0 <~).I < &. (iv) In particuZar, if {An}n:l is a sequence of operators such that IITAnll _,. 0 (n> co), then An is semiFredhoZm for all. n <: n 0 and nul T "' lim sup(n _,. co) nul An' nul T* "' lim sup(n _,.co) nul A~; (v) If K € K(H), then T+K is semiFredhoZm and ind(T+K) = ind T. (vi) If B is another semiFredhoZm operator and ind T+ind B is ~ezrdefined (i.e., {ind T,ind B} ~ {oo,oo} or {oo,oo}J, then TB is a semiFredhoZm operator and ind TB = ind T+ind B. In particular, ~ is semiFredholm and ind ~ = k (ind T), for aZZ k "' l. A semiFredholm operator T is a FredhoZm operator if co < ind T < The wellknown Atkinson's theorem asserts that Tis Fredholm if and only if Tis invertible in A(H)[ll9], [153]. Hence,
~.
pF(T)
=
11:\ae(T)
=
{).€11::
).Tis Fredholm}
(the FredhoZm domain of T) is an open subset of a:. The Zeft ~ight) spectrum of an element a of a Banach algebra A will be denoted by a 1 (a) (ar(a), resp) and its complement p1 (a) =II:\ a (a) (p (a) = 11:\0 (a)) is the Zeft (right, resp) resoZvent set of a. R. r r Thus, a e (T) = a",e (T) ua re (T) , where a".. e (T) = a".. (T) ( l.eft essential. spectrum) and a (T) = a (~) (right essential. spectrum). It is wellknown re r that the intersection aR.re(T) = a 1e(T)nare(T) (some authors call oR.re(T) the WoZf spectrum ofT) contains the boundary aae(J) of aJTl 9
and therefore, it is a nonempty (compact) subset of ~. Its complement ~\aire(T) = pie(T)upr~(T), where pie(T) = ~\~ie(T) and pre(T) =~\are (T), coincides with psF(T) ={A e ~: AT is semiFredholm}, the semiFredholm domain of T.
ll
The following results are an immediate consequence of Theorem 1. and its proof (see [153,Chapter IV]).
COROLLARY 1.14. Let T E L(H); then (i) PsF(T) is the disjoint union of the (possibly empty) open n ® oo sets {psF(T)}_oo
PsF(T) = {A €, ~: AT is semiFredholm with ind(AT) = h}, oo~h~oo. (ii) For eaah h, oo ~ h ~ oo, ph s F(T) is a lower semiaontinuous funation of T in the sense that, if II T Tnll + 0 (n + oo) and e: > 0, then h h p 6 _F(T) c [p 6 _F(Tn)]e: for all n ~ n0 (e:). (iii) pOs F(T) aontains the resolvent set p(T) = ~\a(T) ofT, and a0 (T).
(iv) (v)
h
If h 'F O, then p 6 _F(T) is a bounded set. If the minimal index of AT, A " psF(T), is defined by
min.ind(!.T) = min{nul(AT),nul(AT)*}, then the funation A + min.ind(AT) is aonstant on every component of PsF(T}, except for an at most denumerable subset p:_F(T) without limit points in psF(T) . Furthermore, if~ E p;_F(T) and A is a pointof p F(T} in the same component as ~ but A i p F(T), then
s
s
min.ind(~T)
(vi)
A
i
If A
> min.ind(AT).
psF(T) and nul(AT} < oo (nul(AT)* < oo, resp), then oo n psF(T) <=> ker(AT} c nn=l ran(AT) oo n (ker(AT)* c nn=l ran(AT)* , resp.). E
s
p:_F(T) is the set of singular points of the semiFredholm domain p s F(T) ofT~ pr s F(T) = p s F(T)\pss F(T) is the set of regular points. It is completely apparent that pr F(T) is open and contains p(T). On sthe other hand, it is easily seen that p:_F(T) contains a 0 (T). InCha£ ter III we shall return to the analysis of these sets.
1.5 On invariance and closures of subsets of L(H) A subset R of L(H) is called invariant under unitary equivalence 10
if
T
E
R
=> U(T) c
R,
where U(H) = {U € L(H): U is unitary} and U(T) = {UTU*: U € U(H)} is the unitary orbit of T. It is completely apparent that U(T) c S(T). We shall establish without proofs some very elementary facts that will be frequently used in the future. PROPOSITION 1.15. If R is a subset of L(H) invariant either under simiZarities, or under unitary equivaZenae, or under aompaat perturbations (i.e., R+K(H) = RJ, then R has the same property.
1.6 Notes and remarks Theorem 1.1 is just the quantitative version of [153,Theorem 3.16, p.212J (see also [177], or [132,Theorem lJ). Corollaries 1.3(ii) and 1.9(ii) are two elementary observations due to S. Grabiner [108] and D. A. Herrero [132], respectively. Proposition 1.10 is a mild improvement of a result due to D. A. Herrero rlSO,Lemma 4.3] (see also [44, Lemma 4.1]). The notion of "rank T ;= oo" for a compact operator T, not \ of finite rank, was introduced by J~ Barr!a and D. A. Herrero in [44] in connection with the analysis of '~e similarity orbit of a nilpotent operator (See also Chapter VIII). The notion of "minimal index" is due to C. Apostol [10]. The fact that the singular points of PsF(T) are isolated points of this set was discovered by I. c. Gohberg and M. G. Krein [107]. In the above mentioned article, C. Apostol proved that p:_F(T) is, precisely, the set of points of discontinuity of the function that maps A€~F(T) into the orthogonal projection of H onto ker(AT) (see also t2S,Lemma 1.6 and Corollary 1.7]). This result will be analyzed in Section 3.3.
11
2 An aperitif: approximation problems in finite dimensional spaces
In this chapter we shall analyze several intrinsically finite dimensional problems, as well as infinite dimensional ones which can be solved through an essentially finite dimensional approach or by an argument in Which the (finite or infinite) dimension of the underlying Hilbert space plays absolutely no role. It will be convenient to introduce some notation: H will always denote a complex separable Hilbert space of dimension d, 0 s d s ~. If 0 s d < ® 1 then we shall also write ~d (with its canonical inner product) instead of H. If A, B E L(H), A B (A~ B) will mean that A and B are similar (unitarily equivalent, resp.). A 7 B will be used as an alternative . . _ s1m _1 way to 1nd1cate that B € S(A) I i.e., that IIBW AN II~ 0 (n ~ oo)for ~ n n in G(H). If A ~ Band B ~ A (equivaa suitable sequence {W} n n= 1 s1m s1m lently, S(A) = S(B)), then we shall say that A and Bare asymptotically similar. (In symbols: A# B.) It is completely apparent that s!m is a reflexive and transitive relation and that # is, indeed, an equi~ alence relation in L(H) .(Use Proposition 1.15. It is wellknown and trivial that  and ~ are also equivalence relations.) If A E L(H 1 ) and B E L(H 2 ), where H1 and H2 are isomorphic Hilbert spaces (in symbols: H1 ~ H2 ) , then A# B will be understood as "up to a unitary mapping U from H2 onto H1 ", i.e., S (A)= S (UBU*). The same observation applies to the other relations. The relation sim induces a partial order < in the quotient set L(H)/#, defined by: Let [A]= {T E L(H): S(B)
T#A}; [B] <[A] i f A sim B.(Equivalently,
c S(A).)
Given a (finite or denumerable) uniformly bounded family {AV }V€ r of operators such that A" E L(H") for all v in r, we shall denote by 49 v E r Av the direct sum of· the operators A" acting in the usual fashion on the orthogonal direct sum H = •vEr Hv of the spaces Hv, i. e., if x = 8lV€ r x V is a vector of H, then (8lV£f Av)x = •vEr Avxv. Clearly, IIIBVEf A)l = sup(v 12
E
r> IIA)I <
oo
If r = {1,2, ••. ,n}, we shall also write A1eA 2e •... eAn. If AV ~A£ L(H) for all v in r and card r =a (0 ~a~~>, then A(a)will de H(a) (orthogonal directnote the operator eV£ r Av acting on •v, r Hv sum of a copies of H) . If M is a subspace of H, then ML = H9M is the ortho9onal complement of M in H. Given f, g 10 H, f8g" L(H) is the rank one operator defined by (f9g)x = f, where<.,.> denotes the inner product of H. FIHl = {Lj~l fj6gj:
fj, gj
€
H, j = 1,2, ••• ,n; n = 1,2, •.• }
is the ideal of all finite rank operators acting on H. Let {e 1 ,e 2 , ••• ,ek} be the canonical orthonormal basis (ONB) of ~k and let qk" L(~k) be the operator defined by (2 .1)
(k
= 0,1,2, .•• ;
q 0 is the 0 operator acting on the trivial space {0},
q 1 is the 0 operator acting on the onedimensional Hilbert space ~.
~l ~
and qk admits the matrix representation 0 0 0
1 0
0 1
0
0
0 0 0
0 0 0
(k
qk
0
0
0
0
1
0
0
0
0
0
X
k)
with respect to the canonical ONB, for k =2,3, •.• ). These operators will play a very important role throughout this monograph. A £cr(a)} will denote the spectral raFinally, sp(a) =max{ IXI: dius of a £ A (a Banach algebra) •
2.1 Closures of similarity orbits in finite dimensional spaces
As remarked in the introduction, for many approximation problems the "obvious" necessary conditions derived from the results of Chapter I turn out to be sufficient too. Here is a concrete example of this situation: THEOREM 2.1. Let T
£
L(~
d
) and let p(X)
m
k"·
ITj=l (XXj) Jrxi ~ Aj• 13
for i F j) be its minimal (monic) polynomial; then the closure of the similarity orbit of T is equal to S(T)
= {A£ L(~d):
rank q(A) s rank q(T) for all qlpl,
~here
q!p denotes a monic polynomial q dividing p. Furthermore, if L £ L(~d), then L # T if and only if rank q(L) rank q{T) for all q!p if and only if L ~ T.
COROLLARY 2.2. Let T e L(~d); then the follo~ing are equivalent: (i) S(T) is maximal ~ith respect to inclusion (equivalently, [T] is a maximal element of (L(~d)/#,<)J; (ii)
T ~ lB.~~\ ]
(), ].+qk j )
0 ..] and k.] as in Theorem 2.1, j
= 1,2, ••. ,m),
(iii) Lj~l kj = d (~here the kj's have the same meaning as in Theorem 2.1); (iv) T is a cyclic operator; (v) S(T) ={A£ L(~d): o(A) o(T) and dim H(A;A) dim H(A:T) for all A in o(T)}. COROLLARY 2.3. LetT e L(~d); then the follo~ing are equivalent: (i) S(T) is minimal (equivalently, [T] is a minimal element); (ii) kj (defined as in Theorem 2.1) is equal to 1 for all j = 1, 2, ••• ,m; (iii) T is similar to a normal operator; (iv) S(T) is closed in L(~d); (v) T £ S(L) for all Lin L(~d) such that o(L) o(T) and dim H(A;L) =dim H(A;T) (A e o(T)J.
d 1 LetT, A£ L(~ I and assume that IIAWnTWn II+ 0 (n + aa) for a suitable sequence {Wn}n:l of operators in G(~d). Clearly, q(A) and q(WnTWn 1 ) = Wnq(T)Wn 1 are welldefined (for all n = 1,2, .•• ) for all q!p and (by Proposition 1.7) l!q(A) Wnq(T)Wnlll + 0 (n + 1
aa).
Since rank Wnq(T)Wn =rank q(T) (for all n = 1,2, ... ), it follows from Proposition 1.12(i) that rank q(A) s rank q(T). Hence the conditions of Theorem 2.1 are necessary. The sufficiency of these conditions will be proved in several steps. The second statement of the ~eorem is a trivial consequence of the first one: It is obvious that S(L) = S(T) if and only if o(L) = o(T) and rank q(L) = rank q(T) for all qlp (Use Proposition 1.15). On the other hand, a simple analysis of the Jordan forms of L and T shows that rank q(L) = rank q(T) for all qlp if and only if L and T 14
are similar. (In particular, this implies that they have the same spectrum.)
2.1.1 The nilpotent case LEMMA 2.4. If 1 s m s k1, then ~eqk sim ~+leqkl"
computations show that W is invertible, e: .•• ,m), W 1 f 1 = (l/e:)f 1 and W 1 f.= f.(j = 2,3, .•. ,k) and e: e: ] ] 1 Qe: = We:(qmeqk)We: = Q+e:f 1ef 2 , where Qf 1 = 0, Qe 1= f 1 , Qei= ei_ 1 (i = 2,3, ••• ,m), Qf 2= 0 and Qfj= fjl (j = 3,4, •.• ,k). It is immediate that Q = ~+ 1 eqkl and that Q = lim(e: + 0) Qe:(unless otherwise stated, lim must always be understood as a limit in the noPmtopoZogy). D LEMMA 2.5. Let T S(T)= {A
E
E
L(~d):
L(~d) be a nilpotent of oPdeP m; then rank Aj s rank Tj foP j = 1,2, ••• ,m}.
PROOF. We have already observed that the condition "rank Aj s; rank Tj for j = 1, 2, ..• ,m" is necessary. (Observe that the minimal po!, ynomial of T is p(A) = Am.) Assume that rank Aj s; rank Tj for j = 1,2, ••• ,m. Clearly, we can directly assume that T and A are Jordan forms, i.e., (t 1 l (t 2 l (tk) (cx 1 l (cx 2 l (cxh) T q $q e •.• eq and A = ~ eq $ ••• e'Im nl n2 nk 1 m2 h (tj' cxi
> 0).
We shall proceed by induction on m(T,A) = d+Lj:l (rank Tj rank Aj).
The case m(T,A) = 1 is trivial. Assume that m(T 1 ,A1 ) s; n implies that T1 7 A1 , whenever A1 and T1 are nilpotent operators acting on a s~m . · finite dimensional space and satisfy rank A1 J s; rank T1 J for all j =. 1,2, .•• , and let A, T E L(~d) be nilpotent operators such that rankAJ s rank Tj for all j = 1,2, ••• , and m(T,A) = n+l. If Tan~ A have a.common Jo~dan block.qr' then T = qreT 1 , A= qr $A1 , rank A1 Jrank T1 J =rank AJ rank TJ s; 0 for all j = 1,2, ••• , 15
m(T1 ,A1 ) s (n+l)r s nand, by induction, T1 sim A1 . A fortiori, T=qr $Tl sim qr$A1= A. If T and A have no common Jordan blocks and if r ~ 1 is the minimum index such that ~ < nr, then T has the form T = qn
$qn $T' r1 r (where n 0 = 0 and q 0 acts on a {0}space, if r = 1) and we have rank As < rank T8 for nrl + 1 s s s mh 1. Indeed, if rank As= rank Ts and if as (ts, resp.) denotes the number of Jordan blocks of A (T, resp.) with order of nilpotency grea! er than or equal to s, then it is obvious that as > ts and this yields the contradiction rank Asl = as+rank As> ts+rank Ts =rank Tsl. Setting T1 = qn
+1 $qn _ 1$T' , we can check that rank T1 j = rank r1 r . Tj for j = 1,2, •.• ,nrl and rank T1 J= rank Tj1 for nr_ 1 +1 s j s mh 1, so that s rank T1 j for j = 1,2, ••.• m 1 On the other hand, m(T 1 ,A) s m(T,A) (rank Tmhl rank T1 h ) m(T,A)  1 = n, and consequently, T1 sim A. If r > 1, then T sim Tl by n11 n1 !.emma 2.4._If r=l, then qn 1 "' }:j=l ej6ej+l E: L(a: ) is similar to e:e 1 ee 2 +!~! 2 1 ej6ej+.l (« ~ 0) and, lettUlg e: + 0, we conclude that T s!m 0 T1 • In either case, Ts!m T1 sim A, and therefore A £ S (T).
2.1.2 Proof of Theorem 2.1 L (Q: d )
(0 S d < oo) 1 the minimal polynomial of k· J (A.I' A., i f i F j) and rank q(A) s rank T is p, p(A) = nj=l (;\A.) J ~ J q(T) for all qiP· We want to show that T simA. Clearly, we can directly assume (without loss of generality) that T and A are unitarily equivalent to their Jordan forms: let T = $j:l (Aj+Qj), where a:d = $j:l Hj and Qj is a Jordan nilpotent acting on the subspace H., 0
Assume that T, A m
16
E:
A~ e r= ml (A r +Rr ), where Rr is a Jordan nilpotent acting on Hr and satisfying the conditions
rank (Rr)s s rank (Qr)s, s = 1,2, •.• ,kr,r=l,2, ••• ,m. By Lemma 2.5, Or s!m Rr and, a fortiori, Ar+Or sim Ar+Rr for all r = 1,2, ••• ,m, whence it readily follows that m
m
T = ej=l (Aj+Qj) sim ej=l (Aj+Rj) ~ A. The proof of Theorem 2.1 is complete now.
0
REl1ARK 2. 6. Since nul q (T) = drank q (T) , the conditions "rank q(A) s rank q(T) for all qlp" can be replaced by the conditions "nul q(A) ~ nul q (T) for all ql p". PROOF OF COROLLARIES 2.2 AND 2.3. It is easily seen that T has the form of Corollary 2.2(ii) if and only if S(T) is maximal; (v) => (i) is a trivial implication and (ii) => (v) follows from Theorem 2.1. Now Corollary 2.2 fol~ows from the wellknown algebraic fact that (ii), (iii) and (iv) are equivalent. Similarly, S(T) is minimal if and only if k.= 1 for all j = 1,2, J d ..• ,m, if and only if T is similar to a normal operator acting on ~ ; (v) => (i) is a trivial implication and (ii) => (v) follows from Theorem 2.1. On the other hand, it easily follows from Theorem 2.1 and its proof that if k.= 1 for all j = 1,2, •.• ,m, then the minimal polynomial J of A € S(T) is necessarily equal top and that AT, i.e., S(T) S(T) is a closed subset of L(~d). Conversely, if S(T) is closed and T s!m A, then A € S(T) = S(T) and therefore A  T. A fortiori, A ~ T, whence it readily follows s1m d that [AJ = [TJ = S(T) is a minimal element of (L(~ )/#,<). The proof of Corollary 2.3 is complete now. 0
2.1.3 The lattice (N(~d)/i,<) The partially ordered set (poset) (L(H)/i,<) is never a lattice (unless dim H = 0). However, it contains many interesting lattices. Co~ sider the case when H = ~d is a finite dimensional space and let n (T.) d T ej=l qj J € L(~ ) ; then rank Tj = 'j+l+2'j+ 2+ .•• +(nj)Tn and this implies that . 1 . "+1 T. =rank TJ  2rank TJ +rank TJ :<: 0, for j = 1,2, ••• ,n. (2.2) J
Let Ed be the set of all (d+l)tuples (m 0 ,m1 ,m 2 , ••• ,md) of non17
negative integers such that {
d = m0 > m1 ~ m2 mjl 2mj + mj+l
0, and (2. 3) ~
0 for all j = 1,2, ••. ,d1.
Observe that these two conditions imply that mj > mj+l' unless mj = 0, and mjl mj
0
~ v ~·
It is easily seen that (I:d,s) is a poset, d = max{m 0 ,m
0} >
max{m 1 ,mil
2
•••
~
max{md,md}
0
and 2 max{mj,mj} s max{mj_ 1+mj+l'mj_ 1 +mj+ 1 } s max{mj_ 1 ,mj_ 1 }+max{mj+l'mj+l}, so that p v ~· e I:d. It is completely apparent that~ v ~· is the least upper bound (l.u.b.) of~ and u' with respect to the partial orders. Since I:d is finite, every subset of I:d has a l.u.b .. In particular, ~A~·
= l.u.b.{v e I:d:
v s
~and
v s
~·}
is the (unique) greatest lower bound of~ and~·· It readily follows that (I:d,s) is a finite lattiae with supremum (d,dl,d2, ... ,2,1,0) and infimum (d,O,o, ... ,o,o,o). Given~= (m0 ,m 1 , .•• ,md) e I:d' define T e N(~d) by d (Tj) ~ T~ = IBj=l qj where Tj = mj_ 1 2mj+mj+l for j = 1,2, ... ,d1 and Td =md_ 1 : (2.2) and (2.3) guarantee that the mapping ~ ~ [T~] (2.4) is a bijection from I:d onto N(~d)/# and, moreover, that rank T j =mj for j = 0,1,2, ••• ,d. Combining these observations with Theorem 2.1, we obtain THEOREM 2.7. The mapping (2.4) defines an orderpreserving bijeation from (I:d,s) onto (N(~d.)/#,s). In partiaular, (N(~d)/#,s) is a finite lattiae with supremum [qd] and infimum [0] (0 = ql (d)) .
18
2.1.4 Closures of similarity orbits of finite rank operators Let H be an infinite dimensional Hilbert space and let T, A E F(H); then T and A are algebraic operators with nul T =nul A= oo and there exists a finite dimensional subspace H(T,A) reducing both, A and T, such that TIH(T,A)L = AIH(T,A)L = 0 (H(T,A) can always be defined so that dim H (T ,A) s 2 rank T + 2 rank A) • Assume that A E S(T); then we can prove exactly as in the finite dimensional case that rank q(A) :S rank q(T) and, by using Proposition 1.12(ii), that nul q(A) ~ nul q(T) for all qJp, where p is the minimal polynomial of T. Conversely, if A satisfies those conditions, then it is not difficult to check that AIH(T,A) satisfies the same conditions with respect to TIH(T,A) and therefore, by Theorem 2.1, TIH(T,A) sim AIH(T,A). A
fortiori, T = TIH(T,A)$0 simA= AIH(T,A)$0; hence, we have
COROLLARY 2.8. Let T E L(H) be a (necessarily algebraic) finite Pank operator with minimal polynomial p, then S (T)
= {A E L (H) : rank q (A) s rank q (T) and nul q (A) for all qlp}.
Let L
E
L(H);
then L#T if and only if L
~
:?:
nul q (T)
T.
REMARK 2.9. Since H is infinite dimensional, the conditions "rank q(A) s rank q(T) and nul q(A) ~ nul q(T) for all qlp" of Corollary 2.8 still imply that cr(A) = cr(T). However, the following example shows that those two conditions cannot be replaced by "cr(A) = cr(T) and rank q(A) s rank q(T) for all qJp": Let T be an orthogonal projection of rank 2 and let A be an orthogonal projection of rank 1; then cr(A) = cr(T) = {0,1}, and rank q(A) :S rank q(T) for all qlp (p(;\) = J.(J.1) ),but A cannot belong to s(T)~, because 1 =dim H(l;A) ~dim H(l;T) = 2. Reversing the roles of A and T, we see that conditions of Corollary 2.8 cannot be replaced by "cr(A) = 0 (T) and nul q(A) ;:, nul q(T) forall qlp: Let NF(H) = N(HinF(H) and NFk(H) Since H is infinite dimensional, NF 1 (H)
c
NF 2 (H)
c
...
c
NFk (H)
c
{T
E
NF(H):
NFk+l (11)
c
rank T
:S
k1}.
...
is an infinite chain (all the inclusions are proper: no two sets in this chain coincide), and this chain naturally induces a chain of lattices
19
(NF 1 (H)/*,<) c (NF 2 (H)/lt,<) c ... c (NFk(H)/It,<) c (NFk+l(H)/#,<)c •. , where (as can be easily checked by using Theorem 2. 7) ( NFk (H)/#,<) is orderisomorphic with (Ek,s); moreover, the above chain is in correspondence with a chain of natural orderpreserving inclusions between the lattices {(Ek,s)}~=l ( (E 1 ,s) c (E 2 ,s) c ••• c (Ek,s) c (Ek+l's) c ••• ) . Let (E,S) be the set union of the Ek's with the induced partial order, i.e., if~. ~· £ E, then~. ~· E Ek for all k ~ k 0 (~,~·) and we define~ s ~· in (E,s) if~ s ~· holds in (Ek ,s). 0
COROLLARY 2.10. (NF(H)/#,<) is a lattice with infimum [0] and no maximal element. This lattice is oPdePisomoPphic with (E,s). REMARK 2.11. However (as we shall see in Chapter VIII),
(NF(H)/#,
<) does have a supremum in (L(H)/#,<).
2.2 The distance from the set of all nonzero orthogonal projections to N(H)
2.2.1 The limit case
Let P(H) be the set of all nonzero orthogonal projections in L(H). What is the exact value of the distance n.. = inf{ll P  Q II: from P(H) to N(H)
(11
P
E
P (H) , Q
(2.5)
N (H) }
E
an infinite dimensional space)?
According to Example 1.5, liPOil> l:i for all Q in L(H) such that o(Q)n{)..: p. 11 : ; l:i} = [8. Since o(Q) = {0} for every nilpotent Q, the above inequality holds, in particular, for all Q in N(/1). Hence, noo ~ l:i. It will be shown that n.. = l:i; more precisely, if nk
inf{IIPOII:
P
E
P(/1), Q
1\
= inf{IIP Oil:
P
E
P(«:k), 0
Nk(H)}
E
and E
N(U:k) },
then it is completely apparent that l:i ::; noo ::; nk ::; ok is nonincreasing with k. We have the following THEOREM 2.12. l:i < ok $ l:i +sin (m:l) I whel'e m integral pal't of (k1)/2), for all k ~ 3. In particular , lim(k + oo) ok = lim(k + oo) nk 20
$
1 and nk (ok)
((k1)/2]
nco = Js.
I= the
We shall need some auxiliary results. LEMMA 2.13. Let 1 s m < n < ~ and let m < r < n, s be such that r+s = m+n. If {ei}i~l ({fj}j! 1 J is an ONB of ~r (~s. :r>esp. ), r1 s1 T =
and
r1 s1 .,m1 L = e exists a unitrs rs n111 ary ope:r>ator V € L(~ ~~ ) such that IIVTV*  L II = II T V*LV II = s (m), where s (m) =2sin (m:l) , Vei = ei> for m+1 s i s r, and Vfj = fj, for m+l s j s s. PROOF. Define
gt
r+lt'
m+lr $ t s 0, t7T t7T (cos 2 (m+l)) em+1t + (sin 2 (m+l)) ft' 1 s t s m, m+1 s t s s,
ft'
t1T t7T ) kt = (sin 2 (m+ 1 )) em+ 1t + (cos 2 (m+l) ft, 1 s t s m,
and s1 m1 R = is also an ONB of ~r~s, the pair {gt,kt} generates the same twodimensional subspace as the pair {em+lt'ft} (for each t such that 1 s t s m), T = qr$qs and L = qn~~ = R. Now a straightforward computation shows that L = VRV*, where V is the unitary operator defined by Vgt = {
em+ 1t' ft'
m+lr s t s 0 1 s t
s s
and em+1t'
1
s
t
s m.
On the other hand, (TR)gt
0,
(TR) g 0
(1cos 2 (~+1)) em (sin 2 (~+l)) f 1 ,
(TR)gt (TR)gm
(cos 2 (~+1)  l) gt+l (sin 2(~+1))kt+1' 1 s t s m1, m7T (sin 2 (m+l)  l) fm+l
(TR) gt
0,
(TR)kt
(sin 2(~+l))gt+1+(cos 2 (~+1)  1 l kt+1 I
(TR) km
mTI (cos 2 (m+l)) fm+l •
and
m+1r s t s 1,
m+1
$
t s s,
1 s t
$
m1,
21
Since V{gt,kt} = V{em+lt'ft} (1 s t s m; V { ••• } denotes the oZosed Zinear span of{ •.• }), these twodimensional subspaces are pa: wise orthogonal and (TR)gt L (TR)kt for 1 s t s m1, we conclude that . 2 'If 11 2 ~ II TR II  max{[ sl.n 2 (m+l) + (cos 2 (m+l)  1) J ,
.
I}
Ia (s1.n 2 (m+l)  1) +bees 2 (m+l) max lal2+lbl2=1 ( 11 1) 2 ~ 2 . 'If ( ) . 2 'If = [ Sl.n 2 (m+l) + cos 2 (m+l) J = Sl.n (m+l) = s m • mw
m1r
Since IIVTV*LII = IITV*LVII = IITRII, we are done.
D
The above proof admits a very simple geometric description: Th1 action of q r (e r + e r 1 + e r 2 + ••• + e 2 + e 1 + 0) can be described by an arrow of length r, and the action of T = qreqs by two arrows o1 lengths r and s, as follows m
r s
Similarly, R can be indicated as m
n
T maps em+l to em' the pair {em+lj'fj} onto the pair {emj'fj+: of "parallel" vectors (j = 1,2, ••• ,ml) and the pair {e 1 ,fm} onto {0, fm+l}. On the other hand, R maps em+l to (cosn/2(m+l))em+(sinn/2(m+l: £ 1 , which is equal to the result of a "slight twist" of the vector er in the twodimensional space V{em,f 1 }, the pair {em+lj'fj} onto a "slight twist" of the "parallel pair" {e .,f.+l} (j = 1,2, ••. ,ml)aJ mJ J a "slight twist" (in the reverse sense) of fm to fm+l" The accumulation ofall these "slight twists" after m+l steps wi: map em to fm+l and f 1 to 0. This fact will be described by the follo' ing scheme m (T + R)
r
:;:::::><:::::::
~
s
Conversely, if we take ({gt}t~+lru{kt}t~l) as the original ONl of Er&E8 ~ En~ and modify R to obtain T, then the corresponding scheme will be (R + T)
:::::::><::
n
m
In this case, the twist will move g 0 to 22
~
to 0 and k1 to gm+l
In what follows, we shall freely use these schemes instead of ana lytic descriptions in the proofs. COROLLARY 2.14. If k ~ 3, 1 s m s ({k1)/2], p = km and U = ,p1 p * p1 p . L j=l ej6ej+l + ep6el E L (a: ) • then up "' qp + ~qp ) "' vp. lJhePe vp t.S the unitary operator defined by V e.= (w )le. (lJhere wp is a primip ] p ] k tive pth root of lJ and there e:r:ists a unitary operator V E L (a: ) suah that !!VqkV*qmEDUpll = s(m).
PROOF. It is (Observe that the determinant(XVp) Represent qk
straightforward to check that Up ::: qp+(qp*)pl == Vp three operators are unitary and determinant(XUp) = = xP (l)P.) by a "curled arrow" and modify the action of the op
erator in the subspace
V{e 1 ,e 2 , ••• ,em,em+l'ekm'ekm+l'"""'ek} as in
Lenuna 2 • 13
(qk
+
m
qmEDUP)
We conclude as in the above lemma that there exists an operator Lm,p"' qmEDUP such that l!qkLm,pll = s(m).
0
PROOF OF THEOREM 2.12. With the notation of Corollary 2.14: Let L(O:k) be the orthogonal projection of a:k onto the onedimensional kernel of ((1)~EDUP), where m = [(k1)/2] and p = km: then P e
liP (l:z)VqkV*I! sliP+ (l:z)qmEDUpil+ (l:zli!VqkV* ~EDUPI! s l:z + l:z s (m) = l:z + sin (m:l) • 7T
Hence, cSk s l:z +sin (m+l) • It readily follows that cSk=inf{I!PQI!:
P c P(O:k), Q
E
N(O:k) and !lUlls 3/2+sin (m:l)}.
(2.6) Since P(O:k) x {Q E N(O:k): I!QII s 3/2+sin (m:l)} is a compact sue_ set of L(O:k) ~ L(O:k), it is easily seen that the infimum in (2.6) is actually attained for some pair (Pmin'Gmin) in this set. Thus, by our previous remarks (Example 1.5), 0
2.2.2 on the exact values of cSk and nk
PROPOSITION 2.15. If 2 s h = dim H s oo,
th~n
23
inf{IIP Oil:
p
P(H),
€
a
E
N2(H)} = /2/2.
Furthermore, the above infimum is aatually attained by the operators Pmin =
[~ ~)eoh_ 2
and Qmin =
c P(H)
(~ =~Jeoh_ 2
E
N2 (H),
where the 2 x 2 matriaes aat an a twodimensional subspaae H2 of H and Oh_ 2 denotes the 0 operator aating on (H 2 )l. PROOF. It is straightforward to check that Pmin
E
P(H), Qmin
E
N2(H) and JIPminQminll = /2/2. On the other hand, if Q e N2 (H) and P E P(H), then either 0 = 0 and II P  o II = II P II = 1, or o ~ o • Assume that JJP Oil< /2/2. If x e ker Q and y l ker Q, then IIPxJJ IJ
II P 
Q II
?:
It readily follows that 62=n2= ue of 6k or nk (k ?: 2). Ifk 3,
[11 ']
p3 = 1/3 1 1 1 1 1 1
and
03 =
0
/2/2.
/2/2. This is the only known val
2/3[: 11] 0 1
2 = 2/3[q3+(q3)
]
,
0 0
then P 3 e P (a: 3 ) (P is a rank one projection), 0 3
N (a: 3 ) and II P 3
o
311 is equal to 2/3. Thus, ~ ~ n 3 ~ 15 3 ~ 2/3. This result is certainly better than the poor estimate 15 3 ~ ~ + sin ~~ = 3/2 given by Theorem 2.12 and the same is true for the kdimensional analogues pk = 1/k ~i~l~ '~1 ei&eJ. and ok = (2/k) [qk+(qk) 2 + 3 k1 J . (qk) + ••• +(qk) J of P 3 and Q3 , respect1ve1y, for smaZZ values of k. E
However, there is some numerical evidence that the estimate ~+sin(m~ 1 ) is much better than II Pk Qk II for all k
and
24
?:
20.
2.2.3 A companion problem:
The distance from the set of all nonzero
idempotents to N(H) Clearly, P(H) is not invariant under similarities. The smallest set invariant under similarities containing P(H) is the set of all non zero idempotents acting on H, E(H) = {WPW 1 : P E P(H), WE G(HJ} = {E E LIHI: E 2 =E~O}. We shall see later that if R1 and R2 are two subsets of L(H) , where H is an infinite dimensional space, then R~nR; is "very large", in general; namely, when R: contains an operator R. ~ J ] such that Rj is not an algebraic element of A(H), for j = 1,2.(See Chapter of the second monograph.) Very little is known about dist[Rl'R 2 J = inf{
IIR 1R2 11:
Rj
E
Rj' j = 1,2}
for the case when one of the sets R~ or R; is a subset of the algebra~ ic operators (or, more generally, operators whose images are algebraic elements of the Calkin algebra) . In particular, what is the value of n~ = dist[E(H) ,N(H)J for the case when H is infinite dimensional? What is the answer if H = ~k or NIH) is replaced by Nk(H) (and dim H = ~> for some k ~ 2? Clearly, if and
then 0
~
nk
s
nk = distr.EIH),Nk(H) J ok' for all k ~ 2.
(H infinite dimensional),
PROPOSITION 2.17. (i) ok = 1/k, but E(ah and all Q in N(~k) (k "' 2). (iiJ 0 ~ nk s 1/k and n~ = 0.
I!EQi!
>
1/k for a'Ll E in
k
PROOF. (i) Let {e.} .~ 1 be the canonical ONB of ~ and let Pk k ] ] k kj (1/k)}:. . 1 e.t9e. and Wk  L· 1 n e.t9e. (n = 1,2, ••• ). Straight~, J= ~ J ,n J= J J forward computations show that.Pk is a rank one projection, wk ,n is in vertible, and ~ (nji k l) 0 k,n = l.ld
E
N (~k).
Since li Ek,n Q k,n II = II~Llsjsisk (njik l) ei e ej II <1/klll
we see that 15k_ s; 1/k. On the other hand, if F o (F) c { O,l} and
€
k trace F = Lj=l = rank F ?? 1,
so that ~ 1/k for (at least) one value of j, 1 ::; j ::; k. ] ] k k Given F as above and Q € N(~), we can find an ONB {fj}j=l so that Q is strictly upper triangular with respect to that basis, i.e., = 0 for 1 s j s i s k. In particular, we have IIFOII
~max
I<(FQ)f.,f.>l =max II?? 1/k, lsjsk J J lsjsk so that !IFOil~ 1/k, i.e., &k_ ~ 1/k. Furthermore, since = 0 for 1 s j s i s k, it easily fol~ J lows that, eitheriiFQII > 1/k or = 0 for 1 s j = 1/k for all j = 1,2, •.• ,k: i.e., F admits an upper triangular matrix with 1/k in all the diagonal entries with respect to the ONB {fj}j! 1 • But the second possibility implies that o(F) = {1/k}, a contradiction (recall that k ~ 2). Hence, IIF Qll > ~Sk_ = 1/k for all F in E(~k) and all Q in N(~k). Since the second statement is a trivial consequence of the first one, we are done. D Clearly, the trace argument cannot be applied in the case when H is infinite dimensional, but the above result suggests the following CONJECTURE 2.18. nk = E(H) and all Q in Nk(H) (k
cSk = 1/k, ~
but liEOil> 1/k for all E in
2).
The last result of this section says that the above conjecture is true at least for k = 2. PROPOSITION 2.19. If 2 s h
= dim
n2 = 152 = inf{IIE Oil:
E
H s €
ro,
E(H),
then Q
€
N2 !Hl}
but this infimum cannot be attained foP any paiP (E,Q), E
~. €
E(H),Q
€
N2 (H).
PROOF. If E = 1, then i t is clear that liE oil = 1 for every quasinilpotent Q. Let E =
Ill oil ~ sp(lQ)
(~ :):::nEE)L
be the matrix of E € E(H)\{1} with respect to the decomposition H ran ES(ran E)L and let
26
lc oJ
0 =
be the matrix of Q
€
L(H)
(with respect to the same decomposition). ~ 2 (H)
if and only if
= AB+BD
= CA+DC = 0.
It is immediate that Q e
= o 2 +CB II E Oil :S ~:
A2 +BC Assume that
then Ill All
~ and
:S
liD II
:S
~ and therefore
o(A) c {A: JlAI s ~}and o(D) c {A: !AI s ~}.Thus, by the spectral mapping theorem, o (A 2 ) o(BC) c {A: :le A ~ ~}
and 0
(D 2 )
=
o(CB)
c
{A:
Since o(BC)\{0} = o(CB)\{0}
I AI
~}.
:S
(see, e.g., [ll9], [153], [172])and
A is invertible (recall that IllAll s ~ < 1 ) , it readily follows that o(BC) = o(A 2 ) = {~} c o(CB) = o(D 2 ) c {0,~} and o(A) = {~} c o(D) c {0,!1}. Hence, liE oil ~
IllAll ~ sp(lA)
= ~
Assume thati!EQJJ =~.Since o(A) = {~} = oR.(A), there exists a sequence {xn}n:l of unit vectors in ran E such that JJ(A~)xnll+ 0 (n +
oo) • On the other hand, A2 = BC and A invertible imply that
Ellxll
II Cx II ~
for some e: > 0 and for all x in ran E, so that JJE Oil
:?.:
lim sup(n
+ co)
IJ(E Olxnll
~
lim sup(n
+ co) {
i!(lA)xn11 2 +11Cxn11 2 l~
~ (~+£2)~
>
~.
a contradiction. Hence
II E  Q II
> ~
for all 0 in ~ 2 ( H) •
On the other hand, it readily follows from Proposition 2.17 that inf {II E  0 II:
2.3 On the distance to
E
€
E (H) , Q
€
~ 2 (H) }
s
~.
0
~k(H)
2.3.1 A general upper bound LEMMA 2. 20. Suppose that T
€
II T II s 1 and II Tk II s E: for some (T*T)~ = JLO,l]A dE (speatral deaomp£ L (H),
k ~ 2 and some e:. 0 < E s 1. Let sition) and let P = E([O,Ie:J); then
11((1P)T(lP)]k41 s (k1)/e:. PROOF. Clearly,
II TP II
s le:, and 27
e:11xlr
~
11Tkxll2 =
II~k 1 xl1 2
=
IIPIT*T)~Tklx+llP) IT*T)~Tklxll2
~II (1P) (T*T)~Tklxll 2
=
II
~ e:II11P)~lxll 2 , for all x in H. Hence, 1111P)Tk 1 11 s le:. On the other hand,
so that 11[11P)TilP)Jk11 s 11[(1P)T]k 1 11 s 11(1P)Tk11+ { 1111P) TPT[ 11P) T]k~~ + 1111P) T 2 PT[ 11P) T]k~~ + ••• II11P)Tk 2 PTIIl s
ll(lP)~1I+!k2)IITPII s lk1)/e:. 0 k
THEOREM 2.21. Tf T E L(Hl. IITII s 1 and liT II s e: foro some k and some e:, 0 < e: s 1, then
2
;>
I 2. 7) wheroe ~kle:) is a continuous, positive, nondecroeasing function defined on 10,1] such that lim(e: + 0) $k(e:) = 0. Moreovero, ~kle:) can be inductively defined by $ 2 le:) = l2e:)~ and $k(e:) = {e:+$k_ 1 1(kl)/e:) 2 }~. fork= 3,4, •••• PROOF. Let P be defined as in Lemma 2.20 and let T = [E 1 T1]ran P E 2 T 2 ker P be the matrix of T with respect to the decomposition If k = 2, define T' = PT(1P) =
H
ran P$ker P.
(~ ~l).
Since
(~ ~,.2 ) Iran P and ker P
1
=
0
ran P, it follows from Lemma 2.20 and its proof that
II T T' II
=
1 (:~ ~J 1 {II(:~ ~JII \II(~ ~JI 2 r s
s (e:+£)
~ = ( ~.
~ 3 and (2.7) holds for j s k1, with ~ 2 (e:) and 4>.(£) = {e:+~.l((j2)/£) 2 }\ for j = 3,4, •• ,kl. Assume that k
J J Clearly, IIT 2 11
28
=
11(1P)TilP) II s 1 and, by Lemma 2.20, IIT 2
2 £)
= (2e:)~ kL
11 s
(k1)/E. Thus, by our inductive construction, we can find T2 (ran P)
f
Nkl
such that II T 2 T211 s lj)kl ( (k1) /E). Define
T' It
is easily seen that T'
€
Nk(H) and a formal repetition of our
previous argument (for the case when k = 2) shows that IITT'Il s lj)k(E)
{E+Ij)kl((kl)IE) 2 }~
(def)
4
D
k.
J ().i~ Aj if i ~ jJ be a potynomiat and tet Mp(H) = {A € L(H): p(A) = 0}. Given E > 0, there e:J:ists 6 > 0 suah that, if IITII s 1 and IIP(T) II s 6, then COROLLARY 2.22.
=
Let p().)
rrj:l (AAj)
dist[T,Mp(H) J < E. PROOF. I f m = 1, then the result follows from Theorem 2.21. As~
2. There exists E1 > 0 such that the m open disks D1 , o2 , ••• ,Dm of radius El centered at ). 1 , A2 , ••• , Am' respectively, are pairwise disjoint. Let E0 > 0 be such that i f IP(A) I < E0 , then A is sume that m
contained in one of these disks. Let 1 I = 2ni aD. J
PJ. We shall show that
().T)1 d)..
is a bounded function of IIPII, IITII, e 0
Pk
for small IIP(Tlll· Observe that p(A) p(T) = a1 q
in
(AT)q(A,T)
,
p(J.)
for a polynom!_
the variables A, T. Thus 1p(T)p(A)l = ().T)q(A,T)p(A)l.
Now, for A such that I AAj I = El we have that IIP(A) small i f IIP(T)II
1
p(Tlll
is
is. Let E=p(A)lp(T); then
().T)  l = p(A) lq().,T) (1E)  l = where II J II s II p (A) lq (A ,T>II ·II E
IV (1JI E II>
p(A) lq(A,T)+J,
is small since II p (A) lq (A ,T)II is
bounded by a funtion of II T II and p (A) • Thus the P. 's are bounded ( j
J
2, ••• ,m) • On the other hand, k. 1 II
1 = ll2n1 Since each (T).j)
k·
1
I aD.
P(A) ().).j)
I aD.
(AAj)
J
+
k.
().).j)
k. JJ d). II
k.+l
J Jd).ll s IIJII·E 0
J
J
Jpj is small (provided IIP
use Theorem 2.21 to perturb TP. to aT~
J
J
is a nilpotent of order at most k .•
•
k. lq().,T)
= l,
J
in
L(ran P.) J
such that (TJ~).J.)
•
I f H =ran P.+ker P., define T. =T~+O with respect to this decomJ J J J
29
position (j
= 1,2, ••• ,m)
= Lj~l
and T'
II T T' II ~ max{ll P j 11·11 TjTI ran P jll: provided II p (T) II < c5 for some c5
>
Tj. Then p(T') = 0 and 1 ~ j ~ m} < e:,
0 small enough.
0
Let T, k and e: be as in Theorem 2.21 and let n. 0 < n ~ 1, be the kth root of e: (i.e., 11~11 ~ nk) and define ljlk(n) = cpk(e:), k = 2,3, ••• ~ then Theorem 2.21 implies that, if k = 2, then dist[T,N 2 (H)J s l2n!kif k = 3, then dist[T,N 3 (H) J ~ (n3+4n3 / 2 )~, •• , and dist[T,Nk(H) 1 = O(e: 2 = 0
po~
CONJECTURE 2.23. There exists a continuous function ljl(n), defined on [0,1], such that ljl(n) > 0 on (0,1], lji(O) = 0 and dist[T,Nk(H)J s 11J(n) for all Tin L(H) such that IITII ~ 1 and IITkll ~ nk (i.e., the functions ljlk(n) can be replaced by a single one).
2.3.2 Two illustrativeexamples The rough argument of the proof of Proposition 1.10 might suggest that those estimates are very poor. However, Lemma 2.13 shows that the "very poor" lower estimates given by Proposition 1.10 are actually the best possible except, perhaps, for a constant factor independent of k. Indeed, if T and L have the form of that lemma, then nul ~+l = 2 (m+l) >nul Lm+l=2m+l. Thus, if W (U) is an invertible (unitary, resp.) operator, we can always find a unit vector x = x(W) (= x(U), resp.) such that IILm+lxll = 1, but (WTWl)m+lx = 0 ( (UTU*)m+lx = 0, resp.), whence it readily follows from Proposition l.lO(i) ((ii), resp.) that dist[L, S(T)] :?: 2l/(m+l) 1 (dist[L,U(T) J :?: 1/(m+l), resp.). An even more surprising example can be constructed on the same lines. We shall need the following auxiliary result (With the notation of Lemma 2.13): COROLLARY 2.24. (i) Let {gn}m
Let ·H+ = y{gn}n~O(H_ = V{gn}n
(ii) If k ~ 3 and 1 5 m 5 [ (k1V2J, there exists a unitary W : (~m) (oo)EBH ~ (~k) (oo) such that k,m llq (oo) W (q (oo)$U)N* II= s(m). k k,m m k,m PROOF.
(i)
For suitably defined A'
=A
and B'
= B,
oper~
we have
m
o~=m1
'
(oo,oo)
~==~==~~.~~m
whence we obtain (oo,m)
A'EilB':
( 0 ,oo) II!~EilU) (A'$B'lll
= s(m) and the unitary operator Vm can be chosen so
that Vm(A'EilB')v; = AEilB, Vmgo=k 1 and Vmg_ 1 =km. (ii) Since r = k2m ~ 1, there exists U' = U such that (
qk
~ q
( 00)
m
( 00) $U I )
k
__......,:;:._::::><~=
. ·><
m (~
fact that r
~
.
m
)
k
...X =:;;;:. .:::::J:Tkk: 
~
  
m
1 guarantees that we can consistently apply the ar
gument of Lemma 2.13 to each step.), with llqk(oo) qm(oo)$U'II = s(m), whence the result follows.
0
EXAMPLE 2.25. Let s be a unilateral shift of multiplicity one and letT
E
L(H)
be unitarily equivalent to s(oo)$S*(oo); then IITkll = 1 for
all k ~ 1, cr(T) = crR.re(T) ={A:
IA.I
51} and
2l/k 1 5 dist[T,Nk(H) J s 4 sin TT/([ (k1)/2]+1) < BTT/k, for all k
~
3.
PROOF. It is not difficult to see that IITkll = 1 for all k
~ 1 and
that nul(>.T) =nul(>.T)*=oo for all A. in the open unit disk D ={A: 1>1 < 1}, whence it follows that cr(T) =crR.re(T) =D.
Let 0 c: Nk(H). Since Qk=O, it follows from Proposition l.lO(iii) that liTOil'??: 2l/k_l (k =·1,2,3, ••• ). On the other hand, i f k ~ 3, m=
31
[ (k1)/2] and the operators are defined with respect to a suitable ONB of H, it is not difficult to infer from Corollary 2.24(i) and (ii) that ( (qk (oo}) (oo) "' qk (oo)): ll
II qk (co)

(~am•) (oo)ll +ll(~eu•) (oo) Til
(oo) 
~ (oo) eu'll +ll~eu•  ses* II = 2s (m)
= 4{sin 'IT/([ (k1)/2]+1)} < B'IT/k. Our next example shows that the infinite dimensional ampliations of finite dimensional examples can produce certain surprises. (An ator T in L(H) is called an ampliation if T = A81 = A("") for some ator A € L(Ho) I 1 s dim Ho s oo.) EXAMPLE 2.26. Let k > h ~ 3: then the operators A qk (h) (A, B € L(~kh)) satisfy dist[B,S(A)]
1:
qh (k) and B = (2.8)
however, 21/k1 s dist[B(ool,S(A(oo))] s 4{sin 'IT/([(k1)/2]+1) (2.9) +sin 'IT/([(h1)/2]+1)}
Bn(l/k+l/h).
<
PROOF. Observe that nul A= k >nul B =h. Thus, dist[B,S(A)] ~ 1. (Use Proposition l.lO(i) as in our previous observations at the begi~ ning of this section.) On the other hand, A~ EA for all E > 0, so that 0 € S(A), whence we obtain (2.8). The lower estimate of (2.9) follows from Proposition l.lO(iii) ([A(oo)]h=O, II[B(oo)Jhll = IIB(oo) II= 1.) The upper estimate follows from the proof of Example 2.25: For suitably chosen·T = s<"">es*(oo), A' =A and B' = B, we have IIB'A'II s IIB'Tit+JITA'II s 4{sin 'IT/([(k1)/2]+1)
+sin 'IT/([(hl)/2J+l)} < B'IT(l/k+l/h).
D
Example 2.26 suggests that, if H is infinite dimensional, then 1 N (H) = {Q € N(H): !lOll s 1} "looks like" IU+iQI (where !1J denotes the set of all rational numbers), in the following sense: Observe that Gl+iGI = uk:l (Qik +i!IJk) , where !Ilk = {m/n: "large" nowhere dense subset of !IJ+i
A
m € 2Z , 1 s n s k}, !Ilk +iQlk is a
€
Gl+ill} < 1/(k/21+ 0 (k + oo)
and (GI+iGI) (in the complex plane) coincides with the much larger set 1 1 E, and N (H) = uk=l [N (H)nNk(H)J. 00
32
CONJECTURE 2.27. max{dist[Q,Nl(H)nNk(H)J: some constant C > 0 independent of k.
0
E
Nl(H)} ~ C/k for
An affirmative answer to this conjecture would provide some heuristic explanation to the wild structure of N1 (H)  (see Chapter V).
2.3.3 An example on approximation of normal operators by nilpotents > SO, p = [,/k/2hrJ, n
aj =
p
1
and r = [kn/2] and let Qk E k1 L(~k) be the operator defined by Qk = Lj=l aj ej+ll8ej with respect to the ONB {ej}j~l of ~k, where Let k
l
nn, for r (n1) < j s rn, n = 1, 2, ••• ,p, nn, for r(2pn) < j s r(2pn+l), n = 1,2, .•• ,p, 0, for 2rp < j ~ k1.
(Roughly speaking: Qk is a truncated weighted shift; the weights aj grow from n to 1 through p steps of length r and then go down from 1 to 0 through p steps of length r, so that the upper step has length 2r with weights equal to 1, i.e., a. = 1 for r(p1) < j ~ r(p+l) .) . . . . J . .. . In the f~rst mod~f~cat~on, we shall "~gnore" the coord~nates 1,2, ... ,r(p1) and r(p+l)+l,r(p+l)+2, ••• ,k and apply Corollary 2.14 to the subspace V{er(pl)+l'er(pl)+ 2 , ••• ,er(p+l) }. It is easily seen that we can modify Qk in order to obtain an operator Ri = Tieur+l' where U +l ~ e 10e +l+L.: 1 e.+ 10e. is a unitary operator acting on a subspace r r JJ J 1 1 1 of dimension r+l and there exists an orthonormal system {f 1 ,f 2 , ••• ,f2t r (p1) 1 2r k . R , _ such that {ej}j=l u{fj}j=lu{ej}j=r(p+l)+l ~san ONB of a:, T1ejQk ej for j i 1
1
(r(pl),r(p+l)J, Tier(pl) =fi, _
1_
Tif~=f~+l for h = 1,2, ••• , h 1
r2, T1 fr_ 1  (1n)er(p+l)+l and Ur+lfh (wr+l) fh for h = r,r+l, ••• ,2r (wr+l is a primitive (r+l)th root of 1); furthermore, IIOkRill = s(r1). Let T1 be the operator obtained from Ti by replacing each weight equal to 1 by 1n and R1 = T1eur+l; then II Qk R111 ~ s (r1) +n. Now we can apply the same argument to R1 in order to obtain an OE erator R2 = T2e (ln)u 2 reur+l, where u 2 r is a unitary operator acting on a subspace of dimension 2r, whose eigenvalues are equal to minusthe 2r 2rth roots of 1, T 2ej = ~ej for all j I. (r(p2) ,r(p+2) ], T2er(p 2 ) = 2
2
2
2
(l2n)f 1 , T2 fh = (l2n)fh+l for h = 1,2, •••,r2, T2 frl = (l2n)er(p+ 2 )+l'
,£;_
{fi,f~, ••• 1 } is an orthonormal system that spans a subspace orthor u{ej}j=r(p+ k gonal to the span of the vector ({ej}j=l 2 )+1), II R1  R2 II ~ (1n)[s(rl)+n]and this second modification only affects the vectors 33
· th e su b space spanned by (e.}. { r (p1) r (p+2) 1n J J=r ( p 2 ) +1 u{e.}. J J=r ( p+ l) +1 ), so that IIOk R2 JJ = max{JJok R1 JJ,JJR 1  R2 11} s s(rl)+n, etc. An easy inductive argument shows that after p1 steps we finally obtain an operator L = u mre:£?1 (l· )U 1...,.. (k+l(2pl)r) k r+l J=l Jn 2r ~~1 such that JJok~11 s s(rll+n < 2n/r+[lk/2lnJ < 5(n/k)~ for all k >SO. On the other hand, if 1 s k s 50, then 5(1T/k)~ > 1. Thus, we have the following PROPOSITION 2.28. (i)
FoP eaah k
~
1 thePe exists a noPmaZ
opeP~
toP~ E L((tk) suah that JJLkJI = 1 and dist[~,N(a:k)] < 5(1T/k)~. (ii) If H is infinite dimensional, thePe exists a noPmaZ opePatoP M suah that cr(M) = D, !Uhepe D = {;\: !AI< 1} and distrM,Nk(H)J < 5(1T/k)~ fop all k = 1,2, •••. In partiauZar, ME N(H).
PROOF. (i) If k > 50, define ~ as above. If 1 s k s SO, take Lk = 1, Qk=O. (ii) Let {Am}m:l be an enumeration of all those points A in Osuch that both lAml and (arg Am)/1T are rational numbers (arg 0 is defined equal to 0) and let M be a diagonal normal operator with eigenvalues A1 ,A 2 , ••• ,A , ••• such that nul(A M) =cofor all m = 1,2, ••• , i. . m (co) m e., M = (d1ag{A 1 ,A 2 , ••• ,Am, •.. }) Given k, it is easytoseethatMcanbe written as M" (EDm:l Am Lk) (co), whence it readily follows that dist[M, Nk (H) J A
fortiori, ME N(H).
0
The result of Proposition 2.28(i) is, in a certain sense, the best possible. Observe that if Nk E L(a:k) is normal and there exist k Qk € N(a: ) and e:k > 0 such that JJNk QkJJ < e:k' then (by Corollary 1.6 (i)) cr(Nk) is a connected set containing the origin. If the points e:k of cr(Nk) are more or less evenly distributed in a connected neighborhood n of the origin with smooth boundary (namely, n = D), then 2 cr(Nk) will include n and therefore k1re:k ~ m2 (n), where m2 denotes e:k the planar Lebesgue measure. Hence, e:k ~ [m 2 (0)/(1Tk)J~ = O(k~). cannot be connect (On the other hand, if e:k is too small, then cr(Nk) e:k ed, a contradiction.) CONJECTURE 2.29. There exists a constant C > 0 (independent of k) 34
such that dist[N,N(~k)J ~ Ck~ for every normal operator N such that II Nil= 1 (k = 1,2, ••. ) .
E
L(~k)
The following result provides some extra support to this conjecture. Observe that if A E L(~k) is hermitian and 0 ~ A ~ 1, then the points of a(A) are not evenly distributed in any set of positive measure. (More precisely, m2 (a(A)£) ~ 2c+~c 2 independently of k, and 2c+ 2 .,. 0, as £ ..,. 0.) nE k
PROPOSITION 2.30. If A € l(~ ), 0 dist[A,N(~k)] > (1/2/k), k = 1,2, ••••
~A~
1, and 1
E
a(A), then
PROOF. Assume that IIAQII ~£for some Q E N(~k), Q = H+iJ (Cartesian decomposition); then IIAHII = IIJR.e(AQ) II~ IIAQII ~£and trace (H) = trace (llle Q) = Ie trace (Q) llle 0 = 0. On the other hand, it is easily seen that a(A) ~ ~O,l]{Use Carol £ lary 1.6(i)), so that trace {A) ~ l+{l2c)+{l4c)+ ••• +{l2nc), where n = [l/2c](= integral part of {l/2c)). It is clear that l/2c ~ n > l/2cl. Hence, {n+l) > l/2c and n1 trace {A) ~ {n+l)  2c}:j=O j = (n+l) {n+l) nc = {n+l) {1nc) > l/4c. Let A f A dE and H = J A dF (spectral decompositions) • If a E [0,1], c'> c and rank F{{ac',aa)) l > a(ac) =c, a contradiction. Hence, rank F{[ac,aa)) ~rank E{[a,aa)) and, by symmetry, rank F([a,ao)) ~rank E{[ac,ao))' for all a E [0,1]. It folows that 0 = trace (H) > trace {A)  kc > 1/ 4c  kc. Hence, c > 1/2/k. By a compactness argUMent (exactly as in the proof of Theorem 2.12), we conclude that dist[A,N{a:k)J = min{IIAOII=
Qk =0,
11011 ~
2} > 1/2/k.o
2.3.4 On the distance to a similarity orbit LetT E L{a:d) be a ayalia operator with minimal polynomial p, p(A) = n.m1 {AA·)kj {A. 1 A·, i f i 1 j); then }:.m1 k. =d and Tis sim J=
J
l.
J
J=
J
ilar to the Jordan form ej:1 {Aj+qk·). Let A E L{a:d) be an operator wfth spectrum a(A) = {a 1 ,a 2 , ••• ,an} 35
and dim H(a.;A) =h .• (Clearly, }:.n 1 h. =d.) Define JJ 1 =JJ 2 = •.• =JJk =A 1, J
l.
J=
1
l.
JJk +l=JJk +2= ... =JJk +k =:\2' ••• 'JJdk +l=JJdk +2= ... =JJd=A and 1 1 12 m m m 11 1 =1!2= ••• = 11 h = al' 11 h +1 =l!h +2= ••• = 11 h +h =a2, ••• ' 11 dh +1 1 1 1 1 2 n = Sdh +2 = Bdh +3 = • • • = Bd = an; then A admits a representation as . n
n
an upper triangular matrix of the form
A
0
Bd (with respect to a suitable ONB of ~d). It is not difficult to conclude from Theorem 2.1 that
T sim B = 0 )Jd
Moreover, the same result applies to any upper triangular representation of A. Hence, we have COROLLARY 2.31. Let A and T be as above; then dist[A,S (T)] ~ min
max
aEE(k) whe~e
E(k)
l~j~k
I JJ.  S (.) J
0
J
1.
(2.11)
denotes the set of aZZ permutations of k elements.
Unfortunately, the estimate (2.11) is very poor, in general. Name ly, if~ and Qk have the form of Proposition 2.28(i), then qk is cyclic, qk sim Qk (by Theorem 2.1) and
dist[~,S(qk)] ~ IILkQkll
< 5(1T/k)!oz
+
0 (k
+co).
However, sp (Lk) = 1 (1 E a (Lk)) and a (qk) = {0}, so that the only information that we can obtain from Corollary 2.31 is that dist[Lk' S(qk)J s 1. PROBLEM 2.32. Find a formula for 36
dist[A,S(T)] (A, T
E
l(~d)).
We shall close this section with a partial answer to this problem. n
n
hk)
(o;k)
COROLLARY 2.33. If T=EDk=l ~k and A=EDk=l qk al'e finite rank opel'atol's, rank TJ =rank AJ fol' j = 1,2, ••• ,r and rank Tr+l < rank Ar+l fol' some r ;:.: 2, then 21/(r+l) 1
$
dist[A,S(T) J s 2 s([{r1)/2]}.
PROOF. The lower estimate follows from Proposition l.lO(i). In order to obtain the upper estimate, we can directly assume without loss of generality that A, T E L(~d) (for some d, 0 < d < ~>. Then, our hypotheses and formula (2.2) imply that T· = o;. for j = 1,2, J J ... , r1, but 'r < o:r· Since r > 1, this means, in particular, that T and A have exactly the same number of direct summands, which is equal toT=
Lj~l
'j"
After eliminating all common direct summands, we can directly assume (without loss of generality) that T =ED n q (Tk) and A=q (o:r> n {
q ED ( T)) n m
(where r < n" s n' s n and n' is the largest Jordan block of A), whence we obtain dist[A,U(qnED~(T))J s s(m). Similarly, we have dist[T,U(qnED~(T))J s s(m). Combining these two inequalities, we obtain dist[A,S(T)] s dist[A,U(T)] s dist[A,U(qnED~(T))] + dist[T,U(qnED~(T))J s 2 s(m).
0
2.4 on the distance from a compact operator to N(H)
37
j k1 j j ••• ,ek can be chosen in such a way that llfj~fj Jz}:r=l er~er+~l < lf + 2n/k. Define Gl Al2 Al3 G2 A23 G3 G 0
Gd k1 where Gj = (;\/2) Lr=l
e;~e:+l
IIF&lO((kl)d) Gil
and Aij = aijfj_8fj; then
max{ll Gj ;\jfj~fjll: < max{l;\jl/2:
1
~
j
1 ~ j ~ d} ~
d}+27r/k=sp(F)/2+27r/k.
Let K € K(HJ (H infinite dimensional) and let e: > 0 be given. Then there exists a finite rank operator F such that II K F II < e:/3. £ £ Moreover, by the upper semicontinuity of the spectrum (Corollary 1.2),' F can be chosen so that sp(F ) < sp(K)+e:/3. £ £ Since F e: F I H) , there exists a finite dimensional subspace H of £ £ H, dim H£ = d ~ 1, such that H£ reduces F£ and F £ IHi£ = 0. Let F = F I"H , let M be a subspace of dimension kd containing H for some k £ £ £ £ large enough to guarantee that 2n/k < e:/3 and define G e: L (M ) as above £ and G e l(H) in such a way that G IM = G and G IMi = 0. £ kd £ £ £ £ Then G e F(H), G = 0 and £ e: IlKGil !> IIKFe:II+IIF'e:Ge:ll < e:/3+JIFGII < e:/3+~ sp(F)+e:/3 < J;z sp(K)+e:.
Since e: can be chosen arbitrarily small, we obtain the following upper bound: PROPOSITION 2.34. If K e K(H) (H an infinite dimensionaZ space), then the distance from K to the set of aZZ finite rank nilpotent oper~ tors cannot e~ceed ~ sp(K). In partiauZar, every compact quasinilpotent operator can be uniformZy approximated by finite rank niZpotents.
2.5 Notes and remarks The problem of characterizing the closure of a similarity orbit in simple terms was raised by D. A. Herrero in [139]. This reference contains all the basic properties of the sets S(a) (for a in a Banach algebra A), the notion of asymptotic similarity, several properties of 38
the poset (A/i,<), the analysis of several relevant examples, etc. The results include the proof of the implications (ii) => (iv) => (i) of Corollary 2.3 (in the above mentioned more general setting [139,Proposi tion 1]) • Theorem 2.1 and Corollary 2.8 are due to J. Barrra and D. A. Herrero [43,Theorem 1.1], who also proved that (F(H)nN(H)/#,<) is a lattice (0 ~ dim H ~ oo). The concrete model (Ed,~) for this lattice (Theorem 2.7) is an unpublished result of c. Apostol and D. Voiculescu[34]. In [129], J. H. Hedlund tried to detetllline the eKaot value of 11 00 = dist[P(H),N(H)] and showed that 1/4 ~ 11 00 ~ n 2 =o 2 =12/2 (Proposition 2.15) . He also analyzed the norm of the difference Pn  Qn for several values of n ~ 100 and suggested that these operators could be used to prove that ok + ~ (k + oo).(Personal communication to the author.) The solution of the problem evolved as follows: 1) 11 00 =~ (D. A. Herrero [132,Corollary 9], by using infinite dimensional arguments; 2) ok does converge to~. ask+ oo (D. A. Herrero [133,Section 7(d)], modif~ ing the previous argument); 3) ok ~ ~+[l+(kl)~J/2k ~ ~+1/2/k (N. Sal! nas, [18l,Lemma 3.3], by using an argument due toP. R. Halmos [125]): 4) ok >~and T)k < ~+(8 log k)/k (D. A. Herrero, [149,Proposition 6.5]) and 5) ok < ~+sin 1T/([ (k1)/2]+1) (D. A. Herrero, [150,Corollary 5.2]). P. R. Halmos and L. J. Wallen called an operata~ T in L(H) a pown• parotiaZ isometroy if Tk is a partial isometry for all k <: 1 and pro~ ed that T has this property if and only if T ~ {ek:l qk(Tk)}es
<
10/log k,
for all k large enough. The results of Section 2.2.3 are unpublished observations of D. 39
A. Herrero (However, the fact that dist[ E(a:k) , N(a:k) J = 1/k was also i!!l dependently proved by B. Aupetit and J. Zemanek in [200,Example 2.4]), It is worth to remark that the proof of Proposition 2.17(i) actually yields the following stronger result: PROPOSITION 2.35. inf{I!TQII:
T
but this infimum is
t:
Fo~
each k
~
2
L(O:k), a(T) ={0,1}, Q
neve~
t:
N(~)}
1/k~
attained.
The results of Section 2.3.1 are due to s. L. Campbell and R. Gellar [64]. That every compact quasinilpotent operator is the norm limit of finite rank nilpotents is an observation of R. G. Douglas [123,Problem. 7] and the improvement given here (Proposition 2.34) is due to D. A. Herrero [149,Proposition 6.6].
40
3 The main tools of approximation
In this chapter and the following one, we shall analyze the main "tools" of approximation. Some of these "tools" developed within the framework of Operator Theory: The Rosenblum equation AX XB = C, Rota's universal model and its generalizations, the triangular representation of C. Apostol and the theorem of C. Apostol, C. Foia~ and D. Voiculescu on normal restrictions of small compact perturbations of operators. These, as well as some of their consequences, will be developed here. Two other important results (the BrownDouglasFillmore theorem and Voiculescu' s "noncommutative Weylvon Neumann theorem") , from the theory of C*algebras, will be analyzed in Chapter IV.
3.1 The Rosenblum operator:
X
+
AX XB
3.1.1 Linear operator equations Let A be a Banach algebra with identity. The mapping a + La, where La" L(AJ is defined by La(b) = ab (be A) is called the Pegutar Zeft ~ep~sentation of A. It is wellknown that this mapping is an isometric isomorphism from A into L(A). Similarly, the regular right representation (defined by a + Ra " L(A), where Ra (b) = ba) is an isometric antiisomorphism (i.e., ab + R R ) • In particular, o (L a ) = o (Ra ) = o (a) for all a in A. b a Given a, be A, Tab" L(A) is defined by Tab(c) =accb, i.e., r ab =La  ~. It is easily seen that La and ~ commute and that
()., ll "
a:) ,
so that
LEMMA 3.1. Let X be a aomplex Banaah spaae and ZetA, B"E: L(X) be
41
two operator>s such that AB = BA; then o(A+B)
a" o(A), B" o(B)}
o(A) +o(B) ={a+B:
c
and o(AB)
o(A)o(B) = {aB:
c
a " o(A), B " o(B)}.
PROOF. Let B c L(X) be a maximal abelian algebra containing A and B. Clearly, 1 f Band (AA)l ((~B)l) belongs to B for all A 1 o(A) (~I a(B), resp.). Let M8 be the set of all multiplicative linear functionals on B; then the Gelfand theory implies that o(A+B) = {cp(A+B):
cp
M8 J
E
= {cp(A)+cp(B):
cp " M8 J cr(A) +o(B)
and o(AB) = {cp(AB): c{cj>(A):
cp cp
COROLLARY 3. 2.
M8 J = {cp(A)cp(B):
E
E
M8 }.{cj>(B):
o (r ab)
cp
E
cp " M8 }
M8 } = o(A).o(B).
0
cr (a)  cr (b) .
c
PROOF. Since La~ = ~La' it follows from Lemma 3.1 that a (La~) c a (La) a(~). On the other hand, a (La) = a (a) and cr(b), whence the result follows. We shall see later (Corollary 3.20) that when A sion is actually an equality.
L(H)
this inclu
3.1.2 Approximate point spectrum of a sum of commuting operators The appr>oximate point spectrum of A OTT(A)
~:
{A
E
{A
E ~:
E
L(X) is the set
AA is not bounded below}
and
a 6 (A)
=
AA is not onto}
is the approximate defect spectrum of A. It is completely apparent that op(A) c oTT(A) c oi(A) and a 6 (A) c or(A). Furthermore, if At denotes the Banach space adjoint of A, then it is not difficult to see that crTT(A) =·cr 6 (At) and o 6 (A)
=
crTT(At).
(If X is a Hilbert space, on(A) = oi(A) = crr(A*)* and a 6 (A) oi(A*)*, where O* = {~: A f Q} for each Q c ~.)
42
LE~~
3.3. Given any Banach space X, there is an isometric imbedding of X into a larger Banach space X', and a mapping A+ A' of L(X) int9 L(X') which is an isometric isomorphism such that every A' is an extension of A and a (A') =a (A') =a (A) • p 7r 7r PROOF. Let ~ 00 (X) be the Banach space of all bounded sequences of elements of X with the norm ll{x } : 1 11 = sup llx llx and let e (X) "' n nn n o {{xn} E ~ (X): II xn II + 0 (n + "')}. It is easily seen that e 0 (X) is a subspace of ~"'(X). We define X'= ~"'(X)/e (X) and the imbedding of X into X' by x + 0 [{xn}J (=the coset of {xn}), where xn = x for all n = 1,2, . . . • Clear ly, this mapping is an isometric isomorphism of X into X'. Similarly, given A in L(X), we define A' E L(X') by A'[{xn}J = [{Axn}]; then A+ A' defines an isometric isomorphism from L(X) into L(X') •
If A ~ a7r(A), then there exists a sequence {xn}n:l of unit vectors in X such that II 0 and for all x in X, then it can be easily seen that II
1T
1T
PROOF. (i) Let A E a (A+B); then, by Lemma 3.3, A € a ((A+B) ') 7r p cr (A'+B') and A'B' =B'A'. It readily follows that the subspace p M = ker(A'+B'A) is invariant under A' and B', and B' IM = (AA') IM. Thus, if a E a (A' IMl then Aa E a (B 1 1Ml and therefore 1T
A= a+(Aa)
E
a (A'IMl+a (B'IMl
1T
c
a (A')+a (B') =a (A)+cr (B).
1T 1T 1T 1T 7r 7r (ii) Take A € a (AB); then N = ker(A'B'A) is invariant under A' 7r and B'. If A= 0 then a (A') or a (B') contains 0, so 0 € a (A)a (B). 1T 1T 1 7r 1T If A~ 0, then A' IN has the two sided inverse A B' IN. Choose a in 3cr(A'INl; then A/a E 3a(B'INl. But this implies that a € a7r(A' INl c a11 (A'), A/a E a (B' IN> c a (B'), giving in turn 1T 7r A = cdA/a) E a (A') .a (B') =a (A) .a (B). D 1T 1T 'If 7r
The duality between approximate point spectrum and approximate de feet spectrum mentioned above, gives the following 43
COROLLARY 3.5. If A, B € L(X) and AB (iJ a 5 (A+B) c cr 0 (A)+cr 0 (B). ( i i J a 0 (AB) c a 0 (A) • a 0 (B) •
BA, then
LEMMA 3.6. Let H be a Hitbert spaae and tet A (i) crn(LA) = cr 0 (RA) = at(A). (iiJ cr 0 (LA) = crn(RA) = ar(A).
E
L(H); then
PROOF. (i) Let A E a (A) =a. (A) 1 then there exists a sequence n ~ {xn}n=l of unit vectors such that 110.A)xnll + 0 (n + ~>. It follows that II, whence we conclude that A € a n (A). . It follows that at(A) = crn(LA). If XRA is onto, then there exists B in L (H) such that (XRA)B = B(XA) = 1, so that XL cri(A). Hence, cr 0 (RA) ~ crt(A). Conversely, if XA is left invertible and C(XA) = 1, then (ARA) ~
RC = R(XA)RC = RC(XA) = R1 = 1, i.e., ).RA is right invertible, whence we conclude that a 0 (RA) c or (RA) c at (A) • Hence, a 0 (RA) =a R. (A) • The second statement follows by the same arguments. 0 From Theorem 3.4, Corollary 3.5 and Lemma 3.6, we immediately obtain the following COROLLARY 3.7. Let A, B € L(H); then (i) cr1T(TAB) c crt(A) crr(B). (iiJ cr 0 (TAB) c crr(A) crt(B). We shall see later (Section 3.1.4) that these inclusions are actu ally equalities.
3.1.3 Local oneside
resolvents in L(H)
LetT € L(H) and let n be an open subset of pr(T). A continuous function F:Q + L(HI satisfying the equations {;1.T)F(A.)  1 (A.
E
Q)
and F(A.) F(Jl) = (JlA.)F(A.)F(Jl)
(A.,Jl
€
n)
will be called a right resotvent for T (defined on Q). It is completely apparent that such a function F is an analytic function defined on 44
derivative F' 0..) = F 2 (A). Similarly, if t) is an open subset of p R. (T) and G: t) ... L (H) is a co~ tinuous function satisfying G(A) (AT) = 1 (A E t)) and G(A)G(~) = (~A) G(A)G(~) (A,~ € t)), then G is called a left resolvent forT (defined on t1 clearly, G is analytic on t)). Moreover, if F(A) is a right resolvent forT* (defined on n t*), then G(A) = F(~)* is a left resolvent for T (defined on t)) • Q with
For any A {
€
~
pr(T),
E
pR.(T), we shall put
Rr(A,T) = (AT)*[{AT)(AT)*Jl, RR.
(~
,T)
= [ (1JT) * (~T)] 1 (~T) *.
(3 .1)
It is easily seen that Rr(.,T):pr(T) ... L(H) (RR.(.,T):p 1 (T) ... L(H)) is continuous on this domain and satisfies the equation (AT)R (A,T) = r 1, >. E pr(T) (R 1 (~,T)(1JT) = 1, ~ E p 1 (T), resp.). However, Rr(.,T) (R (.,T)) does not satisfy, in general, the resolvent equation, i.e., it is not a right (left, resp.) resolvent for T. THEOREM 3.8. LetT E L(H). Given E > 0, there e~ists a right resolvent for T defined on pF (T) npr (T) e~cept for an at most denumerable set S c pr(T) which does not accumulate in pr(T), such that
s
c
cap r (T)J £ ={A
€
cr:
distn,ap r (T)J s £}.
Applying the above theorem toT*, we obtain the following dual result. COROLLARY 3.9. LetT E L(H). Given E > 0, there e~ists a left re~ oZvent forT defined on pF(T)npR.(T) e~cept for an at most denumerable setS' c pR. (T) which does not accumulate in pR.(T), such that S' is included in [
LEMMA 3.10. Let T such that 0 closed.
€
€
L(Hl and Zet M be an invariant subspace ofT
Pre
PROOF. Since 0 € Pre, Pker(TIM>* will be a finite rank projection. Thus, if we write 45
N = {z < M.l:
Pker(TJM)*PMTz=O}
we haveN c M.l, dim (M.leN) <~and, consequently, ranT is closed if and only if T(M+N) is closed in H. This reduces our problem to showing that T(M+N) is closed
<=> 0
Observe that Pker(TJM)*PMTN we obtain that T(M+N)
TM+PM.1TN
c
On the other hand, given y
€
TM.1N
c
is closed.
implies that PMTM
c
TM, whence
TMM+TM.1N = TM+TM.1N.
M and z c N, we can choose y 0
€
M such
that TMyo = PMTz, whence we obtain TMy+TM z = T(y+y 0 +z) € T(M+N). Henc~ we have the equality T(M+N) = TM+TM.1N, where the sum in the right side is orthogonal and TM is closed. It readily follows that T(M+N) is closed if and only if TM.1N is closed. 0 COROLLARY 3.11. LetT € L(H) be an operator with closed range and Zet M be an invariant subspace of T such that (TM) M and 0 belongs to pR.e(TM ). Then 0
€
pr(TM).
PROOF. Since T* has closed range and 0 € Pre(TM.L*), it follows from Lemma 3.10 that TM* has closed range. Moreover, since TM hasdense range, TM* must be injective, so that 0 € pR.(TM*); equivalently, 0 € Pr(TM). LEMMA 3.12. Let T
0 €
L(H) be an operator with closed range and Zet
M be an invariant subspace ofT such that ker T c M and 0 E pr(TM). Then 0 € pR. (TM.1l and the set {A € a:: ker(AT) c M} is a neighborhood of the origin.
PROOF. Since ker T c M and TM is right invertible, 0 cannot be an eigenvalue of TM.l· Using Lemma 3.10 we see that ran TM.1 is closed and therefore 0 € pR.(TM.L) • For each A in the open set pR.(TM.L) and xA 0 = PM.L(A.T)xA
€
ker(A.T), we have
(A.TN.L)PM.LXA.
Since A.TM.L is injective, it follows that PM.1xA. = 0, i.e., xA. conclude that pR. (TM ) c {A. c «:: ker(A.T) c M}.
46
€
M. \'le 0
\' n ~ n ~ l·k=l ker{~kT) = ker ITk=l{~kT) .
PROOF. The inclusion 'c' is obvious and the converse inclusion is trivial for n = 1. We shall proceed by induction over n. Assume that we have n mk n ~ Lk=l ker{~kT) ~ ker rrk=l{~kT) and let .
~n+l.: crp{T)\{~1'~ 2 ,
G~ven
(~kT)~ (k
.• ,~n} and mn+l be a natural number. n+l ~ mn+l n x.: ker ITk=l{A.kT) , we have that {~n+lT) x <: ker ITk=l
and, by our inductive assumption, we can find yk.: ker{XkTfk
LEMMA 3.14. Let T .: l(H) and Zet x e H be such that the set of ze roes of the function ~ + Pker(A.T)x• has an accumuZation point r;; e pr(T); then Pker(AT)x = 0, foP alZ A in the component Qr, of pr(T) aontaining the point 1;.
PROOF. Let a= {~ e Ql;: Pker(~T)x = 0} and assume that a ~ Ql;. Since a has an accumulation point in Q , we can find ~ e Ql;naa and a I; sequence {~k}k=l c a such that ~k ~ ~. lim(k .... oo) ~k = ~Observe that A + Pker (~T) = 1Rr (~ ;T) (AT) (where Rr (~;T) !s defined by (3.1)) is continuous at~ Thus, if M = V{ker(~kT)}k=l' then ker(~T) c M and therefore ran(~TM) is closed. But ran(~TM) is obviously dense; hence~ e pr(TM). Let Q0 ={A e Ql;: ker(AT) c M}. By Lemma 3.12, Q0 is a neighborhood of ~. Since x .;. M , we see that Pker ( ~ T) x = 0 for all X e 1!0 , co!! tradicting the assumption that~ E aa. This implies that a Ql; andthe proof is complete. 0 00
47
LEMMA 3.15. LetT e L(H), x e
Hand~
e pr(T) ~e suah that
Pker(~,;T)k
x= 0. for all k = 1,2, ••• , then Pker(AT)x = 0 for aZZ A in the component 0~ of pr(T) containing 1;.
PROOF. Let M = V{ker(!;T)k}k:l. Since 1,;T is surjective, we see k+l that (1;T)ker(~T) c ker(~;T)k (k = 0,1,2, ••• ), so that ran(~TM) is dense. On the other hand, ker(~,;T) c M. Hence, ran(~TM) is closed andit follows that 1,; e pr(TM). Since X e M , it follows from Lemma 3.12 that the set no= {A e n~ : ker(AT) c M} is a neighborhood of 1,;, and therefore Pker(AT)x = 0 (A e n0 ). By Lemma 3.14, Pker(AT)x = 0 for all A in 0~· 0 PROPOSITION 3.16. LetT e L(H) and Let a be a compact subset of pr(T)ncrp(T); then there exists x in H such that Pker(AT)x f 0 for aLL A e cr. h
PROOF. Let {ni}i=l be the components of pr(T) having nonempty intersection with o and set cri =crnni. For each i (i = 1,2, ••• ,h), we choose ~i e cri and yi e ker(~iT), yi f 0. Then it is possible to find nonzero complex numbers ai(i = 1,2, ••• ,h) such that the vector y = L·hl a.y. satisfies Pk ( T)y f 0, i = 1,2, ••• ,h. 1= 1 1 er ~iLet {).={A e cr: Pker(AT)y=O}. If /:,.={6, we take x=y. If{). f {6, we can apply Lemmas 3.14 and 3.15 in order to show that /:,. is a finite set, /:,. = {Ak}k~l and that there exists natural numbers {mk}k~l such that On the other hand, since By Lemma 3.13, y n mk n ran{ II k=l (A k T) } * is closed, we can find x e H such that y ={IIk=l (AkT)mk}* x. Assume that Pker(~T)x = 0 for some ~ e crk then x = (~T)*z for some z e Hand therefore y = (~T)*{IIk~l(AkT) k}* x. But this is impo~ sible: if~ f Ak (for all k = 1,2, ••. ,n), we have Pker(~T)y = 0; if ID'+l 1 } • In either ~=A· (for some j, 1 ~ j ~ n), then y e {ker(A.T) J J J case, this contradicts our assumptions on y. 0 LEMMA 3.17. LetT e L(H), and Let n 1 be a component ofpr(T)such that nul(AT) 1, A e n1 • Fo~ any £ > 0 there exists a right resoLvent R of T on n1 except for an at most denumerabLe set s1 , which does not accumuZate in n1 , and satisfies s1 c ran 1 J£.
=
PROOF. By Lemmas 3.14 and 3.15 and Proposition 3.16, there exists 48
a vector y E H such that Pker{A.T)y f 0 for all A. E n 1 \s 1 , where s 1 is an at most denumerable subset which does not accumulate in n1 and such that sl c canlJ£. (Take 0 = {A. E Ql: dist[A.,an 1 J ~ £}, in Proposition 3.16.) Fix A. 0 E n 1 ,s 1 and let z be a nonzero vector in ker(A. 0 T). If R0 is a fixed right inverse of A0 T, then any right inverse of A. 0 T is of the form Rt = R +zet, where t E H. Choose t
0
= 
1
R~y
(Since Pker (A. aT) y f 0, it readily follows
that f 0, so that t i s welldefined.): then =
{z~t) *y ,x> =
1
R~y ,x>
0
for all x in H, so that y L ran Rt. For A E nl the following identity holds: (A.T) Rt = ( {A.A. 0 ) Rt+1) • 1
This shows that A.~ (A.A. 0 ) is a mapping from n 1 into the camp~ nent of pF(Rt) which contains the point at infinity. Also, for A. E n 1 , the Fredholm index of {AA 0 )Rt+l is zero. Suppose thatker[{A 1 A 0 )Rt+l] f {0} for some A. 1 E n 1 ,s 1 • Then it follows from the above identitytha~ for some x f 0, (A. 1 T)Rtx = 0. In this case, Rtx E ker(A. 1T) and since the last space is onedimensional, ker ( A. 1 T) c ran Rt. This contradicts ran Rt and Pker(AlT)y f 0. It readily follows that the operator (A.A 0 )Rt+l is invertible for all A. in n 1 ,s 1 • The operator valued function
y
E
R(A.) = Rt((A.Ao)Rt+l]1 is a right resolvent forT in n 1 \S 1 • This completes the proof. D LEMMA 3.18. Let T E L{H) and let Qn be a aomponent of pr(T) suah that nul (A.T) = n, A. E Qn. Foro any £ > 0, there exists a right resolvent F ofT on Qn exaept foro an at most denumerable set Sn, whiah does not
accumulate on nn, and satisfies sn
c
cannJ£.
PROOF. We proceed by induction on n. The result is clear if n = 0, for then nn is a component of the resolvent set p(T) ofT (and F(A) = (>.T) l, A. E Qn) and the case n = 1 is contained in the preceding lemma. Suppose the result has been obtained in the case n = k1. Let Qk be a component of pr(T) such that nul(A.T) = k, A. E nk. It follows from Pr~ position 3.16 that for any £ > 0 there exists a vector y E H for which Pker(A.T)y f 0, for all A E Qk\S', where S' is an at most denumerable set which does not accumulate in nk and satisfies S' c cankJ£. Let MA = ker(A.T)n{y}L, for A E nk' and let M = V{MA}AEnk· Obviously, M is invariant under T and relative to the decomposition H = MGlM.L, 49
T [TM A ) • 0
T
1
M
It is easy to establish that ATM is onto for A " Qk and clearly (ATM ... ) is onto for A € Qk. It follows that for A " nk \S', nul ( >..TM) k1 and nul(ATM.d = 1. By Lemma 3.17 and our inductive hypothesis TM has a right resolvent R(A) on Qk\S' and TM1 has a right resolvent G(>..) on Qk\S", where S" is a (possibly empty) finite or denumerable subset of Qk which does not accumulate on ~ and satisfies S" c [oQk]c· Define F (A) = (
R( >..)
0
(with respect to the above decomposition), where Sk = S'uS". It is eas ily seen that F(A) is a right resolvent for T on Qk\Sk' sk is at most denumerable, Sk does not have any accumulation point in Qk and Sk c rankJc·
o
PROOF OF THEOREM 3.8. Clearly, it suffices to define a right resolvent F on each component of pr(T), except for an at most denumerable subset S with the desired properties. Let n be a component of pr(T) such that nul(AT) =n(n) ~ 0, A" n. I f n(Q) =0, then n c p(T) and the only possible definition for F is F(A) = (AT)l =the resolvent of T restricted to n~This is true, in particular, for the unbounded component of pr(T).) If 1 $ n(Q) < oo, then Q is a bounded component of pr(T). If Q intersects the compact set~= {A< pr(T)\p(T): dist[A,apr(T)] ~ c}, then we define F on Q\S(Q), where S(Q) is an at most denumerable subset of n which does not accumulate in n and satisfies S(Q) c Can> £ by using Lemma 3.18. If Qn~ = ~, then we can use the same arguments as in that lemma in order to construct a right resolvent F on Q\S(Q),where S(Q) is an at most denumerable subset which does not accumulate in Q (the condition S(Q) c Q c Can) is trivially satisfied in this case). £ It is completely apparent that this defines a right resolvent for Ton pr(T)\S, where S = u{S(Q): n is a component of pr(T)\p(T)} is an at most denumerable subset of pr(T) with the desired properties.o
3.1.4 The left and the right spectra of tAB THEOREM 3.19. Let A, B" L(H); then (i) a 6 (TAB) = crr(TAB) = crr(A) aR, (B). (iiJ cr'lf(TAB) = aR,(tAB) = aR,(A) crr(B).
50
PROOF. (i) By Corollary 3.7 and our observations at the beginmng of Section 3.1.2, cr~(TAB) c crr(A) cr.R.(B) and cr~(TAB) c crr(TAB). Assume that \J E crr(A) cr.R.(B) (i.e., \J can be written as \J =.aa, where a E crr(A) and a E ot(B)) and that TAB\J is onto. Then, given C ~ L(H), there exists X E L(H) such that (TAB\J) (X) =TAa,Ba(X) = (Aa)X X(Ba) =c. Since ran(TAB\J) is closed, there is a constant m > 0 such that II(TAB\J) (Tlll ~ m dist[T,ker(TAB\J) ], for all T in L(H). In parti£ ular, X= X(C) can be chosen so that (m/2liiXII :;;; IICII. Since a E or(A) and 8 E ot(B), we can find unit vectors x, yin H such that II = IICII; for this C and X=X(C) chosen as above, we have 0 < IICII = = <(TABJJ) (X)y,x> = <[ (Aa)XX(Ba) ]y,x> s II+I<(BS)y,X*x>l s (2/m>llciHIIylll < llcll, a contradiction. We conclude that o~(TAB) =or(A) o.R.(B) c or(TAB). Assume that \J t ar(A) a.R.(B); equivalently, or(Aj.i)nat(B) =,. Since crr(A\J) and a.R.(B) are nonempty,compact, and disjoint, there exists Cauchy domains QA' n8 and Q such that crr(A\J) c QA c (QA) c Q, aR,(B) c n8 and (Q8 )nQ = ,. It follows from [1], [2] (see also [185]) that there exists an analytic function RA\J (A) defined on pr (A\J) such that (Aj.iA)RAj.I(A) 1 on this domain. (If ~\QA c pF(Aj.i), then we can choose RA j.l (A) as the right resolvent constructed in Theorem 3.B.);sim ilarly, there exists an analytic function La(A) defined on p.R.(B) (If ~\QB c pF(B), then we can choose L8 (A) as the left resolvent given by Corollary 3.9.) such that L8 (A) (BA) = 1 on this domain. Clearly, RA\J and L8 are analytic in a neighborhood of an.Let n0 be a component of Q. Since aR, (B) nQ = ,, it follows from Cauchy's theo rem that fan L8 (A) dA = o.
=
0
A fortiori,
Let a•n =
and 2;i
fan RA~(A)dA
= 2;i
Given Yin L[H),
fa•nRA~(A)dA
1.
let 1
~(Y) =X= 2 wi
f
an
RA~(A)YLB(A)dA.
It is completely apparent that ~ defines a bounded linear mapping from L[H) into itself. l'le shall verify that~ is a right inverse for TAB~· Indeed, . (TAB~) (X) = (A~)XXB = 1
2wi
f an(A~)RA~(A)YLB(A)dA+
1
2wi fanRA~(A)YLB(A)BdA
= 1 2wi f ao(A~A)RA~(A)YLB(A)dA+ (TAB~) o~
f aORA~(A)YLB(BA)dA
2 !i(fanRA~(A)dA)Y = Y,
;;i YfaoLa(A)dA+ i.e.,
21wi
=
1 on L [H). Hence, ar (TAB) c ar (A) a R. (B) and the proof of (i) is complete. (ii) By Corollary 3.7 and our observations at the beginning of Section 3.1.2, aw(TAB) c aR. (A) ar(B) and aw(TAB) c aR. (TAB). Assume that~" aR.(A)ar(B), i.e.,~= a13 for some a ( aR.(A)and some f3 "ar(B). This means that, given € > 0, we can find unit vectors x, yin H such that II
= II TAa,BB (xey)
II
= II
(Aa) (x8y)  (xey) (B13) II
II (Aa)x 8yx8(Bf3)*yl! ~ I!
Since € can be chosen arbitrarily small, we conclude that TABII is not bounded below, i.e., JJ " aw (TAB). It readily follows that aw('A~ = aR. (A) ar(B) c aR. (TAB). Now assume that aR.(A~)nar(B) = 9 and let tA' tB and t be three Cauchy domains such that aR.(A~) c tA c (tA) c t, ar(B) c tB and (t~ nt = 9. By [1], [2](see also [185], Theorem 3.8 and Corollary 3.9), there exist an analytic function Ra(A) defined on pr(B) and an analytic function LA~(A) defined on pR.(A~), such that (BA)Ra(A) :1 for all A in pr(B) and LA~(A) (AA) 1 for all A in pR.(A~). Proceeding as in the last part of the proof of (i), we can easily check that
=
1
~(Y) = 2wi
defines a left inverse of aR. (A) ar(B).
fat
LA~(A)YRa(A)dA
TAB~·
It readily follows that aR.(TAB)
From Theorem 3.19 and its proof, we obtain 52
0
c
COROLLARY 3.20 (Rosenblum's theorem). If A, BE L(H), then (i) cr(TAB) = cr(A) a(B); (iiJ If l.l I a (A) a (B), then there ereists a Cauohy domain n suah that a(A}.1) c n, a(B)nrl = ~ and
(TAB}.1) By taking A obtain
1
(X)
0 or B
0 in Theorem 3.19 and using Lemma 3.6, we
COROLLARY 3.21. If A E L(H), then ar(LA) = a 6 (LA) a 1 (RA) = cr~(RA) = crr(A). (ii) a 1 (LA) = cr~(LA) = crr(RA) = cr 6 (RA) = cr 1 (A). (i)
3.1.5 RosenblumDavisRosenthal corollary The following is the most important consequence of Theorem 3.19 for the purposes of approximation. COROLLARY 3.22. Let H1 and H2 be two Hilbert spaoes, Zet L(H 1 ), BE L(H 2 J and C E L(H 2 ,H 1 ) and assume that ar(A)na 1 (B)
A E =~.Then
the operators ( Ao
(aating on H
=
Be)
and
AIBB
H119H 2 .J are similar.
PROOF. Assume that H1 and H2 are infinite dimensional then we can identify them via a unitary mapping of H1 onto we can directly assume that H1 = H2 = H0 and H = H0 (2 ) • By (i), TAB is onto and therefore there exists X E L(H 0 ) such =c. Then (:
and
(~ ~)
is invertible
~) (~ ~) ([~ ~)l
Hence, AIBB =
spaces;
H2 , i.e., Theorem 3.19 that AXXB
(~ ~) ~)
=
(:
=
(~ ~)).
(10 1X) l (A0 BC) (10 X)1 .
If A acts on a finite dimensional space and B acts on an infinite dimensional space, define a = IIAIHIBI~l and consider the operators alBA e L!H 2tH 1 J, B and C'=(g) E L!H 2 ,H 2mH 1 ). It follows from the first part of the proof that 53
[ ~ ~ ~]

(aeA)eB
ae(AeB).
0 0 B
Assume that W
then {ae
(:~)}ww[ae(AeB)J=~~~
~ :~~~~ 1 ~<~•~>
0  
]=
l(o B)w21w2la (o B)w22w22(AeB)
o,
1
so that aw 12 w12 (AeB) = 0 and r~ ~)w 21 w 21 q = 0. Since cr (a) ncr ( (~ ~)) c cr (a) n[cr (A) ucr (B)] = cr (a) ncr (AeB)
= $§,
it readily follows from Corollary 3. 20 that w12 = 0 and w21 = 0. Hence W= w11ew 22 , w11 and w22 are invertible operators, and (consider the (2,2)entry of the above 2 x 2 matrix!) w22l (: i.e., (:
~)
~)
w22 = AeB,
is similar to A8B.
The cases when H2 is finite dimensional and H1 is infinite dimensional or both, H1 and H2 are finite dimensional spaces can be similar ly analyzed to reach the same conclusion: ( Ao
Be)  AeB.
0
3.1.6 The maximal numerical range of an operator It was observed in Section 3.1.1 thati~ABII s min{IIA>.If+liB>.11: ). E ~}. In order to complement our previous results about 'AB' it will be shown that this inequality is actually an equality. The concept of mazimaZ numeriaaZ range of an operator plays a central role here. Recall that the numeriaaZ range of T E LfH) is the set defined by W(T)
=
{:
x
E
H, llxll
= 1}.
The classical ToeplitzHausdorff theorem ~sserts that W(T) is a convex set [119,Problem 166]. The maximal numerical range of T is the set
54
LEMMA 3.23.
w0 (T)
is a nonempty, aZosed, aonvex subset of W(T).
PROOF. Everything but convexity is obvious. Let A, J..l
t'
w0 (T).
Wit~
out loss of generality, we can assume that IITII =1. Let {xn}n:l' {yn}n:l be two sequences in H such that II xn II =II y n Ii = 1 for all n = 1, 2, ••• , ll!'xn II .. 1, ltl'Ynll .. 1, .. A and .. J.l (n + "'). Consider Tn = Pn TPn' where Pn is the projection of H onto Vfxn,yn} and let n be a pomt of the segment [A,J..l] joining A and J..l. Then for each n, it is possible, by the ToeplitzHausdorff theorem, to choose an' f3n such that = + n and II unll = 1, where un = anxn +f3nYn· Observe t:hat I J s e < 1 and therefore max{Janl ,JenJ} s M= (162)!:1 (otherwise ll'hxn+ f3nyJI < 1) for all n sufficiently large. We have 1
Jl!'u n IJ2 = = JJun IJ2  < (1T*T) U n ,un >
Since 2
IJ(lT*T)xn ll=llxn 112 2JI!'xn II +IIT*Tx n 112 2 = (1JtJ.'x n 112 > 11 2 > s 1JI!'xn II .. 0 (n .. "') and, similarly, 11<1T*T)ynJI + 0 (n + "'), we conclude that ll!'unll + 1 and + n (n +"').Therefore, w0 (T) is a convex set. 0 0 s
If A € L (H) , the operator 15 A= T AA is the innel' del'ivation of L (H) induced by A. 3.24. Let J..l
LEMMA
w0 (T), then IJ15TJI
E
2(11Til2 IJ..l! 2 >!:1.
PROOF. Note that IliST II= supfll TAAT II: A € L (H) , II Ail= 1}. Since J..l E € H such that llxnll=l, for all n=l,2, ••• , ltl'xJI IITII and + J..l. Set Tx n =a nn x +a nn y , where y n is a unit vector
w0 (T), there exist xn +
orthogonal to xn. Set Vn = xn8xn yn8yn; then II
2:
Since an+ J.1 (n .. "'), the proof is complete. PROPOSITION 3.25. Let T (i)
0
€
L(H); then the foZZowing al'e equivalent
w0 (T);
li~ll=2ltrll; IITII 2 + IAJ 2 s IIT+Ail2 fol' aZZ A IITIJ s IJT+AII for aZZ A E a:.
(ii) (iii)
(iv) PROOF. then 114r11
€
2:
(ii) =>
0
(i) => (ii)
€
a:;
It follows from Lemma 3.24 that, if 0
2]1!'1J. Therefore, 11~11= 2lh:'ll· (i) If 114rll=211!'11, then there exist xn
€
Hand An
w0 (T),
€
€
L(H) 55
such that llxn11=11An11= 1 and II
11,
I t readily follows that IIAnn x II+ l=IIAn I!Txn II+ IITII and !ITAnn x II + II T II· Passing, i f necessary, to a subsequence, we can directly assume that + ~ and + y (n + 00 ) . Since IIAnn X II+ 1 =llxrJ'I· nn ,Ann i t is completely apparent that ll, y e w0 (T) • Moreover, since II
n
xn II + 211 T II and H is uniformly convex, the norm of z to
o,
n
= TA x +A Tx tends n n n n n
as n + ""· Thus, we have
=  +
=  + n n nn where En + 0, as n + ""·
=  + n nn n n
£
n
,
(Indeed, i t follows as in the last step of the
proof of Lemma 3.24 that IIAn*Anxn xnll+ 0, as n+ oo.) Thus, y = lim(n + oo) = lim(n + oo) = ll·
Since both that
and ll belong to
ll
w0 (T),
i t follows from Lemma 3.23
o "w 0 (T). If 0 e
(i) => (iii)
w0 (T),
then there exists a sequence {xn}n:l
of unit vectors such that
for all A e
a:.
(iv) => (i)
Assume that IITII s IIT+AII for all A "
By rotating T, we may assume that lle
w0 (T)
= 1, lle s c/2} and let M=supfiiTxll: Let ll = min{c/2,
I M,
"M,
a:,
but 0
I w0 (T).
~ c > 0. Let M={xeH:
llxll
x eM}; then M < IITII· then II
llxll= 1, let Tx = (a+ib)x+y, where y
.l
x; then
II ll > 0. Thus, II Tllll <
II T II,
contrary to the hypothesis.
Since (iii) => (iv) is a trivial implication, we are done.O THEOREM 3.26. LetT " L(H); then II6TII= 2 min{IITAII:
A "
PROOF. We have already observed that II6T II s 2 inf{ll TA II:
a:). A " a:}.
By an elementary argument of compactness (observe that liTAll+ ""• as IAI + oo), i t is easily seen that the above infimum is actually attained at some point ll " II
a:, " a:
i.e., this infimum is a minimum. But IITllll s implies that II6TII = 116 (Tll)ll = 2 IITllll·
{iv) => (ii) .)
3.1.7 The norm of the operator TAB
56
(Use
0
LEMMA 3. 27. Let T continuous.
L (H) • The mapping X +
£
w0 (T+A) is
uppezo semi
PROOF. We can assume, without loss of generality, that II T II = 1. Suppose that i.e w0 (T) s a and let e: > 0. Let f~ = sup{ll Ax II: II x II = 1, i.e ~ a+e:}; then M < 1. It is clear thatiiT+AII ~ 1IXI. However, for y £ H, IIYII = 1 and l.e ~ a+e:, we see that II. 0 £ ~). Hence, w0 (T+X 0 ) coincides with the intersection of all the open halfplanes containing it. Thus, we can find finitely many open halfplanes sl' s2'"""' sm such that
w0 (T+X 0 )e:
m
nj=l Sj
c
c
w0 (T+X 0 ) 2e:.
By the first part of the proof, we can find o that
w0 (T+A)E
c
m
nj=l Sj
c
W0 (T+A 0 ) 2e:
provided I A  X0 I < o• Hence, X + w0 (T+X) is an upper semicontinuous mapping.
D
We define the nozomalized maximal numezoical zoange WN(T) of an ope~ ator T £ L(H), T ~ O, to be the set w0 (T/IITII>· From Lemma 3.27 we obtain the following COROLLARY 3.28. If IIT+AII is upper> semicontinuous.
LEMMA 3.29. Let A, B following azoe equivalent (i)
(ii)
E
~ 0
foro aZZ X, then the map X+ WN(T+A)
L(H) be two nonzeroo operoatozos; then the
11rABII = IIAII+JIBI~ WN(A)nWN(B) ~ ~.
PROOF. The proof is very similar to that of Proposition 3.25 (i) => (ii), and so we shall only sketch a portion. Let A£ WN(A)nWN(B) and e: > 0; then there exist x, y £ H such that llxll = IIYII = 1 and = ~IAI~e:A and = ~IBI~e: 8 , 0 s e:A' e:B < e:. Since J11AII = /IIBI~e:', it is possible to define an operator U of norm l+e:" which sends x to y
and By/liB II to AxJ11AII
LEMMA 3.30. Assume that A. B ar multiple of the identity. Then min{IIA:l.II+I!B:l.ll:
if and only if WN(A).I)nWN((B).1))
LIHI and neither A nor B is a saal
€
:l.
€
~
a:}= IIAJ.I!I+I!BJ.III 9.
PROOF. Assume that WN (AJ.I) nWN ( (BIl) ) ~ 9. Then Itr"AB II = liT All , B).111 = II AIJII + IIBIlll • Since it is obvious that ltr" AB II cannot be larger than min{I!A:l.II+I!B:l.ll: ). € a:}, we see that the condition is necessary. In order to prove the sufficiency, we can directly assume that J.1 = 0. Thus, given ). € a: and £ > 0, there exist unit vectors x, y in H such that IIIA+:l.)xii+II(B+:l.)yjj /IIAII+/IIBII> s Kll:l.l 2+£), where K is a constant independent of). and£. Assume that WN(A)nWN(B) = 9. Then, dH[WN(A) ,~7N(B)] = cS > 0, and (by upper semicontinuity; Corollary 3.28) dH[WN(A+:l.) ,WN(B:l.)J > cS/2, for ). small. Thus, by convexity and upper semicontinuity, any choice of x, y which satisfies the above conditions, must satisfy the inequal ity j<(A+:l.)x,x>/IIAHII+<(BH)y,y>/IIBHIII:t cS/4 for:>.. small. But thenw~ are led to the inequality j:l.jcS/8 s K!:>..l 2 for a suitable choice of arq:>.. and l :>..I small, which is impossible. Thus, J.l = 0 was not minimal, which completes the proof. 0 THEOREM 3.31. Let A. BE L(H); then ltr" AB II = min{ll A"11 + 1~:>..11 :
:l. E a:}·
PROOF. Clearly, ltr"ABII s min{IIA:l.H+IIB:l.ll: :l. € a:}. I f A orB is a multiple of the identity, the rest of the proof is trivial. Let ll £ a: be any point such that the above minimum is attained at J.l• By Lemmas 3. 29 and 3. 30, liT AB II = ltr" All ,Bll II = II AJJift II BJ.III • The proof is complete now.
0
REMARK 3.32. It is completely apparent from Proposition 3.25(iii) that there exists exactly one J.1 E a: such that II TJJ II = min £11 T :>..II : ;\. E a:}. However, simple examples show that, in general, the J.1 of Lemma 3. 30 is not unique.
3.2 Generalized Rota's universal model Let T £ L (H) •. Since o(T) is a compact set, it has a fundamental system of open neighborhoods which are analytic Cauchy domains (i.e., Cauchy domains whose boundaries consists of pairwise disjoint regular 58
analytic Jordan curves: see definition in Section 1.1). Let r =an, where Q is an analytic Cauchy domain containing a(T), and let L2 (r) be the Hilbert space of (equivalent classes of) complex functions on r which are square integrable with respect to (l/2nrtimes the arclength measure on r: M(r) will stand for the "multiplication by A" operator acting on L2 (r). The subspace H2 (r) spanned by the rational functions with poles outside n is invariant under f1(r). By M+(r) and M_(r) we shall denote the restriction of M(r) to H2 (r) and its compression to L2 (r)eH 2 (r), respectively, i.e. M(r)
(
M+(r)
o
)H 2 (r) M_ L2 (r)eH 2 Z
( 3. 2)
with respect to the above decomposition.(Here and in what follows, we write
( 3. 3)
A
as an alternative way to indicate that A admits such a k x k operator k matrix decomposition with respect to the orthogonal direct sum H = ej=l Hj , where Aij:Hj ~ Hi' 1 ~ i,j s k. Such a decomposition is clearly unique.) It is wellknown [36], [73], [74,Sections 9] that H2 (r) is areproducing kernel space and that it can be realized as a space of analytic functions defined on n: furthermore, it can be easily checked that a(M(r)) =aJM(r)) =ae(M+(r)) =ae(M_(r)) =r, a(M+> = a(M_(r)) = n, nul(AM+(r)) =nul(AM_(r))*=O and ind(AM+(r)) ind (AM_ (r)) * = 1 for all A in
( 3. 4)
Q,
M(r) is normal and l!r(M(r))l! =llr(M+(r))ll =llr<~_(r))ll = max{lr(A)l: A e: for each rational function r with poles outside n.
Q}
The Hilbert space completion H2 (r)8H of the algebraic tensor pro~ uct of H2 (r) and H can be regarded as a space of analytic Hvalued functions defined on n, or as a space of (equivalent classes of) weakly measurable square integrable functions on r, in which case it will 59
also be denoted by H2(r:H). THEOREM 3.33. LetT E L(H) and 0 be as above, and let R =ran (M+(r)8118T); then R is a subspaae of H2 (r:H) invariant under M+(r)&l, the restriation of M+(r)&l to R is similar to M+(r)&l and the aompression of M+(r)el to R~ is similar to T. PROOF. Since M+(r)ell&T is equal to multiplication by XT in 2 H Cr:H) and XT is invertible on r, it is clear that R is closed. Mar~ over, the similarity of f1+(r)81 and M+(r>&ljR is implemented byH+(r)~n18T. All we have to do is find an operator L:H 2 (r:H) + H with range H and kernel R such that Lo(M~r)&l) = ToL. We shall define for f E H2 (r: H) the element Lf E H by 1
2 ~i
Lf
f ao(XT) 1
f(X) dX.
It is easily seen that Lo(M+(r)&l) Lo(l8T) = ToL, so that R c ker L. On the other hand, if fx(A) = x (x E H), then Lfx x, so that ran L is equal to H. Let o 1 be an analytic Cauchy domain containing 0 such that the restrictions of functions analytic in a neighborhood of o 1 to r are dense in H2 (r) (By Runge's approximation theorem, it suffices to choose o 1 so that every bounded component of ~\0 contains a component of ~\ n 1 > and let e (X) = 1, e c H2 (r). 0 2 0 Lets E L(H (r:H)) be the operator defined by Sf= fe&Lf. Since Lf = L(e9Lf), it follows that ran s = ker L. Thus, in order to complete the proof, it will be sufficient to show that S(flr> E. R for every Hvalued function f analytic in a neighborhood of o 1 . Because of the analyticity of (XT)lf(A) in a neighborhood of o;>n' we have Lf = 2 1~i It follows that for cs
I; E
f anl
(XT) 1 f(X) d)..
0
1 2dfaol [ (A1;)l (XT)lJf(X) dX
( rT)
..
l
2~i
fan
(A1;)l(XT)l f(X) dX. 1
Since fanl (X1;)l(XT)l f(X) dX is an analytic function of 1; (1; c o 1 ), it readily follows that s = CM+Cr>ell&T)( 2 fa 01  1  1 f(X) dx) cR. The proof is complete now. 0
!:L
REMARK 3. 34. According to the proof of Theorem 3. 33, the diagram
60
H =T:___ _ _ H
is commutative, i.e., T = (LIRL)loT 1 a(LIRL). Since Lfx = x (where fx (A) :: x, x E H), it follows that Ll RL is bounded below by . inf{llfxll:
llxll =
1} =
([length(r)]/2rr)~.
On the other hand, the CauchySchwartz inequality implies that IlLII s ([ length(r) 1/2rr) ~ max{ II< AT) 11:
A
r},
E
so that (3.5) Let r be a rational function with poles outside Q
and let
B)R
T' RL'
where T' is similar to T1 then llrII s llrII
=
llr
= max{lr(A)I:
A
E
r},
whence we obtain the following COROLLARY 3.35. (i) Given T E L(H) and a bounded open neighbo~ hood ~ of a(T), ~ is a spect~aL set (in the sense of von Neumann) fo~ some operator T'  T, i.e •• llrII s max{jr()..)"l:
A
E
i!J}
fo~ aLL rational functions r ~ith poles outside ~.
(iiJ In pa~ticular, if a(T) {0} and e > 0, then '1''  T such that !IT'll < E and T sim O.
the~e e~ists
~~ 3.36. Let H = H1eH 2e .•• eHk and assume that A E LIH) admits an upper triangular operator matri~ ~ith respect to this decompositio~ i.e .•
61
All Al2 Al3 A22 A23 A33
.Al,klAlk .A2,k1A2k .A3,k1A3k
Hl H2 H3
A 0
~~khl.k Hk1 Hk ~k
then A s!m All$A22$A33e ••• e~l,kle~k· PROOF. Let wn = nen 2en 3e ••• enk1 enk (w1.. th respect to the same decomposition). Clearly, W is invertible with inverse w l=n 1 en 2en 3 k+l k n n e ••• en en , and a straightforward computation shows that
= Z&l, M+(r)&liR  M+(r)&l and T'  T. Combining this obser
vation with Lemma 3.36, we obtain COROLLARY 3.37. Let T e L(H) and Zet n be an anaZytia Cauahy domain aontaining a(T) ~ith an = r; then MCr> s!m
M+(r)eM_Cr),
M(r)&l s!m (M+(r)el)e(M_Cr)el),
l4(r)91 sim (M+(r)9l)eTe(M_(r)91) and l1+(r)91 s!m (M+(r)8l)eT.
3.3 Apostol's triangular representation r
Given T e L(H), let H (T) = V{ker(~T): ~ e PsF(T)}, let H.(T) r r ,. V{ker(~T)*: ~ e psF(T)} and let H0 (T) be the orthogonal complement of Hr(T)+Ht(T). Denote the compression ofT to Hr(T), Ht(T) and H0 (T) by Tr' Tt and T0 , respectively. Recall that A e L(H) is a t~ian guZa~ operator if it admits an upper triangular matrix; i.e.,
62
all al2 al3 a22 a23 a33
A
(3. 6)
0
with respect to a suitable ONB. Apostol's triangular representation and its basic properties are established in the following THEOREM 3.38. (i)
Hr(T) is orthogonal to Ht(T), so that {3.7)
H = Hr{T)SH 0 {T)SHt{T).
(ii) Hr{T) and Hr{T)SH 0 (T) are invariant under T, so that Tadmits a 3 x 3 upper triangular operator matrix representation {3. 8)
T
with respect to the above decompostion, ~here (iii) Tr = TIHr(T) is a triangular operator, o{Tr) = ot{Tr) = aire{Tr)u[pr(Tr)\p{Tr)], a{Tr) is a_perfect set, every compon:nt of oire{Tr) intersects the set op{Tr) and op(T;> = ~. so that psF{Tr) ~and min.ind(ATr) = 0 for all A € psF{Tr)• and (iv) Tt is the adjoint in L(Hi(T)) of the triangular operator
T*IHt(T), o{Tt) = or{Tt) = otre{Tt)u[pt(Tt)\p(Tt)], o(Tt) is a perfect set, every component of otre{T{) intersects the set op(Ti) and op{Tt) ~. so that p:_F(Tt) = 9 and min.ind{ATt) = 0 for all A € psF(Tt). Furthermore, (v) A~ Pker{AT) (A € PsF(T)J is a continuous function for A € prs F(T) and discontinuous for A € pss F(T); psF{T) c pr{Tr)npt(Tt); (vii) PiF(T) c p{T0 ) ; (viii) p:_F(T) c o 0 {T0 ); (ix) a 0 lT) c p{Tr)np(Tt)no 0 {T0 ); (x) If A= {A 1 ,A 2 , ••• ,Am} is a finite subset of p:_F(T), then TN TASTA, ~here TA acts on a finite dimensional subspace, o{TA) =A (Vi)
and A
c
p~F{TA).
We shall need several auxiliary results. LEMMA 3.39. If B
€
L(H), A
€
o{B) and II
Mllxll
for aZZ x
€
63
fl,
then IIA Bll ;;, M for every triangular operator A.
PROOF. Assume that A has an upper triangular matrix of the form n n= n be the orthogonal projec tion of fl onto V{e 1 ,e 2 , ••• ,en} and let y be a unit vector orthogonal to ran(AB) (so that (AB)*y = 0); clearly, PnAPn = APn. Given e:, 0 < e: < 1, there exists n 0 = n 0 (e:) such that 11Pn (AB) *y II < e: for all n;;, n 0 • Pn(AB)*Pn can be regarded as an operator acting on the finite dimensional space Mn =ran Pn. It follows from the finite dimensionality of Mn that this subspace contains a unit vector zn such that II[P n (AB)*P n J*z n II= liD l'"n (AB)P n z n II< e:. ~1e have Pn z n = z n , and (3.6) with respect to the ONB {e } ~ 1 • Let P
IIABII;;, lim inf(n
~>
11<1Pn) (AB)Pnll
+ ~>
11<1Pn) 0,B)Pnll
+
2:
lim inf(n
2:
lim inf(n ....
ao)
2:
lim inf (n ....
ao)
?:
Me:.
11 (1P n ) (AB)P n z n II
Since e: can be chosen arbitrarily small, we conclude that IIA Bll 2:
M.
0
COROLLARY 3.40. If A is a triangular operator with matrix (3.6) with respect to the ONB {en}n:l. then (i) a(A) =aR. (A) = aR.re(A)uap(A); (ii) Every clopen subset of a(A) intersects d(A) = {ann}n:l and every component of a(A) intersects d(A); (iii) Every isolated point of a(A) belongs to d(A); (iv) If ker(AA)* ~ {0}, then A " d(A). PROOF. (i) The equality a. (A)= a. (A)ua (A) is trivial. On the ... ..re p other hand (since A is triangular) Lemma 3. 39 implies that a (A) =a 1 (A). (ii) Let a 0 be a clopen subset of o (A) such that a 0 nd (A) = 11 and let Q be a Cauchy domain containing d(A) such that a0 nn = 11. It is eas ily seen that, if 1
E = 2ni
fan
(AA)
1
dA,
then Een =en for all n = 1,2, •••• Since E is an idempotent and ran E => V { e n }n= "" 1 = fl, it readily follows that a 0 = 11. Since a(A) is a compact Hausdorff space, every component of a(A) is the intersection of all the clopen subsets containing it. Let a bea component of a(A) and let {a : a " A} be the family of all clopen su~ a sets of a(A) containing a. It follows from the first part of the proof that aand(A) # 11 for all a in A. A fortiori, and(A) = (nao;A aalnd(A)= 64
na£A[aand(A)J ~ 9 (by an obvious argument of compactness). (ill) This is a trivial consequence of (ii). (iv) Assume that (AA) *y = 0 for some ). £ a (A) and some unit vector y £ H; then 0 = <{).A)*y,en> = ' for all n = 1,2, ••• ,
so that y J. {).A)Mn' where Mn = v£e 1 ,e 2 , .•. ,en}' for all n = 1,2, •••• If"'). I d(A), then {).A)Mn=Mn (n = 1,2, ••• ) and therefore y J. V{Mn}n=l = H; so that, y = 0, a contradiction. Hence, ). € d(A). 0 PROPOSITION 3.41. LetT£ L(H) Q
and~
£ PnF(T); then
lim{). ~ ~) Pker{).T)
exists (in the norm topoZogy). Q is the orthogonaZ projection onto ker (~T)n{nn:l ran(~T)n}.
PROOF. Assume first that ~Tis onto. By Corollary 1.14(v), ).T is onto for all ). in some neighborhood Q of ~· It follows that Pker {).T) = 1Rr 0. ,T) {).T) (where Rr {). ,T) is defined by (3.1), ). £ g) converges in t~e norm to 1 Rr{).,T) {).T) = Pker{).T) as A ~ ~· Put M= nn:l ran (~T) n. Because (~T) is semiFredholm, M is closed (i.e., is a subspace) and (~T)M = M. Hence (~TM)M = M. For). ~~clearly (~T)ker{).T)=ker().T) and so ker{).T) c M. I t follows that p
=p
p
ker().T) ker{).TM) M and the first paragraph shows that this has the desired limit.
D
COROLLARY 3.42. LetT£ L(H) and~ £ PsF(T). The foZZowing cond£ tions are equivaZent (i) ~is a reguZar point of PsF(T). (ii)
Pker{).T) is continuous
at).=~·
PROOF. (i) => (ii) If nul{).T) is a finite constant in some neig~ DO borhood of~ then Corollary 1.14(vi) shows that ker{).T) c nn=l ran {).T)n. By Proposition 3.41, Pker{).T) is continuous at ). = If nul{).T)* is a finite constant in a neighborhood ollary 1.14(vi) shows that DO n ker(~T)* c nn=l ran(~T)* CLAIM:
Since
~· of~
then Cor (3.9)
is semiFredholm, (3.9) is equivalent to DO n (3.10) ker(~T) c nn=l ran(~T) • DO n Indeed, if ker(~T) c nn=l ran(~T) , then ~T
65
1, 21 • • • •
then y= (~T)*y 1 for some y 1 E H. There1 [{y}] ~ ~. so that y = (~T)* 2y fore ran(~T)* J J (~T)* 2 3 2 for some y 2 E H and ran(~T)* J ker(~T)* J (~T)* [{y}] ~ ~By induction, there exists y in H such that y = (U  T) * ny , n n n n = 1,2, .••. Hence, y E "n:l ran(~T)*. This proves that (3.10) implies (3.9). The converse implication follows by taking adjoints. Thus, if y
E ker(~T)*,
2 ker(~T)*
Since (3.9) and (3.10) are equivalent, we conclude (as in the first part of the proof) that Pker(AT) is continuous at A = U· (ii) => (i) If Pker(AT) is continuous at A = u, then (3.10) holds. Hence (3.9) also holds and (i) follows from Corollary 1.14(vi). 0
Let H = ej~l Hj' where Hj ~ H for all j = 1,2, ••• ,k, and let A E LIHI with matrix of the form (3.3) with respect to this decomposition. Observe that A is compact if and only if Aij is compact for all i and j. On the other hand, since Hj ~ H (i.e., Hj is infinite dimensional) for all j, Hj can be identified with H via a fixed unitary map (j = 1, 2, ••• ,k).
Then L[HI can be identified with the algebra of all k x k matrices with entries in LIHI, KIHI can be identified with the ideal of all k x k matrices with entries in K[HI and (by taking the corresponding quotients) A(H) can also be identified with the algebra of all k x k matrices with entries in A(H) (k=l,2, ••• ). This identifications will play a very important role. The proof of the following elementary algebraic lemma is left to the reader. LEMMA 3.43. Let NK 2 (R) be the ring of all 2 x 2 matriaes with entries in a ring R with identity 1. Then
(i) If (~ ~) has a right inverse RJ. If q is 1nvertible, then pa = 1. (ii) If (~ ~) has a Zeft inverse 1. RJ. If p is invertible, then dq
[~ ~) (~ ~)
in re 2 (R) • then qd
1 (in
in re 2 (R) • then ap = 1 (in
PROOF OF THEOREM 3.38. (i) By Corollary 1.14(vi), either ker r (AT) c ran(AT) or ker(AT)* c ran(AT)* for every A E PsF(T). If u E p~F(T) and u ~ Ar then (AT)ker(uT) = ker(uT) and so [ker(AT)*J~ = ran(AT) J ker(~T). 66
By Corollary 3.42 JIPker(J.JT) Pker(AT)JJ + 0 (J.J readily follows that ker(AT) c [ker(AT)*J~. Hence, Hr(T) = V{ker(AT):
A~
r PsF(T)}
~
+
A), whence i t
r V{ker(AT)*: A€ p 5 _F(T)}=HJI.(T)
and, a fortiori, H = Hr(T)$H 0 (T)$HJI.(T). (ii) The invariance of Hr(T) and Hr(T)$H 0 (T) =HJI.(T)~ under Tis immediate. Hence, T admits a matrix representation of the form (3.8). (iii) Let {A n }n= ml be a denumerable dense subset of pra F(T). By Corollary 3.42 Hr(T) = V{Mn}n:l' where Mn=ker(AnT). It is completely apparent that Tr has an upper triangular matrix with respect to an ONB obtained by GramSchmidt orthonormalization of a union of ONB's of the subspaces {Mn}n:l and d(Tr) = {An}n:l in this representation (where d(Tr) has the meaning of Corollary 3.40). By Corollary 3.40, cr(Tr) =oii.(Tr) = crR.re(Tr)uop(Tr)' every component of o(Tr) intersects d(Tr) c [p~F(Tr)] (Observe that ATr is o~ to for each A E prs F(T)) and o p (T*) r c d(T r ) ={A n }n= 1 • Since {A n }n= 1 can be arbitrarily chosen (under the condition {A n } = [ pra F (T) J, of course), it readily follows that op(T~) =~and, a fortiori, that p:F (Tr) = ~ and min. ind (ATr) = 0 for all A E PsF (Tr). It is also clear that cr(Tr) does not have any isolated point, i. e., either H (T)={O} and cr(Tr)=~, or cr(T) is a nonempty perfect set r r + and cr(Tr) =cr ~re (T)ucr p (T) 0 r =on~re (T)u[p r (Tr )\p(T r )]. Since ps F(T) r = pr(Tr)\p(Tr)' it follows that every component of oR.re(Tr) intersects [pr(Tr)\p(Tr)J. In order to prove (iv), we only have to apply the arguments of (iii) to T*JHJI. (T) (Indeed, observe that the equivalence of (3.9) and (3.10) and Corollary 1.14(vi) imply that p~F(T) =p~F(T*)*)~ (v) is the content of Corollary 3.32 and (vi) follows from (iii), (iv) and 00
00
their proofs: If A E PsF(T)\pr(Tr)' then A E crR.re(Tr) c cre(T), a con tradiction. Hence PsF(T) c pr(Tr). A similar argument shows that PsF(T) c pli.(T 1 ). (vii) Assume that Hr(T) t {0} and H1 (T) t {0} ~ then Hr(T), HR. (T), Hr(T)$H 0 (T) and H0 (T)$HJI. (T) are infinite dimensional spaces. Suppose that A E: psF(T) and ind(AT) > m~ then AT1 is invertible and AT is right invertible. By decomposing H as [Hr(T)$H 0 (T)J$ H1 (T), AT can be written as the 2 x 2 matrix AT
[l• [~Jll\_]= [P r). 0
AT
R.
Oq
By Lemma 3.43(i), pis right invertible. Since
67
P =
* ) [ >..T' 0 r >..To '
it follows from the same lemma that >..T0 is right invertible. Similarly, if ind(>..T) < oo, we conclude from Lemma 3.43(ii) that >..T0 is left invertible. Hence >.. E psF(T0 ) and therefore either >..T0 is invertible, or nul(>..T0 ) ~ 0 or nul(>..T0 )* ~ 0. Thus, in order to complete the proof, it only remains to show that >.. ; ap (T0 )uo p (T*)*. 0 Assume that (>..T0 )x= 0 or equivalently (>..T)x 1 H0 (T). The invariance of Hr(T)$H 0 (T) implies that (>..T)x E Hr(T). Since >..T maps Hr(T) onto itself, there is some y E Hr(T) such that (>..Tr)y= (>..T)x. Since y x E ker (>..T) c Hr (T), x must be equal to 0. An analogous argument shows that nul (>..T 0 ) * = 0 and so >..T0 is invertible, i.e. >.. " p (T 0 ) . If either Hr(T) = {0} or H1 (T) = {0}, then the result follows by an even simpler argument. (viii) Let v" pss F(T). If vis not an eigenvalue ofT, o then (vi) shows that we have ker (vT) = ker (vTr) and (by Lemma 3 .12) we can find a neighborhood Q of v such that ker(>..T)=ker(>..Tr)' for all A
En.
This implies that v E p~F(T), a contradiction. Hence, v E op(T0 ). On the other hand, the first part of the proof of (vii) indicates that v E psF(T0 ) and, by (vii) and Corollary 1.14 (v), v is an isolated point of a (T0 ) . Therefore, v E o0 (T 0 ) . (ix) This follows immediately from (vi) and (viii). (x) Let A be a finite subset of pss F(T). It follows from (viii) that A c o 0 (T 0 ) and, a fortiori (since A is closed), A is a clopen sub set of a (Tr}, so that (by Riesz' decomposition theorem) To where
= [A T23J) 0
B
'
A acts on a finite dimensional space, a (A) = A and Ana (B) = ' . Therefore T admits the decomposition Tr Tl2 Tl3 Tl4 A T23 T24 0 T 0 B T34 0 0
where A= a (A)
0 c
0
TR.
pr (Tr) np 1 (T 1 ), i.e., or (Tr) noR. (A) =or (A) noR. (TR.) =a (A) n
a(B) = or(A)no 1 (CR.) = '· By Corollary 3.22 and its proof, there exist operators X, Y such that 68
and
Y) (A B12 ) [1 YJl 1
o c1 o 1
(A 0 ).
=
o c1
Hence
TA
A has the desired properties and A£ p~F(TA)' where
Tr T'
A
*
= [0
B
0
0
0
From Theorem 3,38 and Lemma 3.36, we obtain the following COROLLARY 3.44. Let T (3.8); then
LIHI
£
with triangular representation
T slm Tr$T 0 eT 1 •
3.4 Correction by compact perturbation of the singular behavior of op
erators The Weyl speatrum of T
£
LIHI is defined by
oW(T) = n{o (T+K):
K £ K (H)} •.
It is easily seen that ow(T) is the largest subset of the spectrum that is invariant under compact perturbations and that (by Theorem 1.13(v) ) ow(T) contains every complex X such that XT is not a Fredholm operator of order 0. Furthermore, it is wellknown that ow(T) actually coincides with this set, i.e., ow(T) ={X
£
a::
XT is not a Fredholm operator of index 0}. (3.1])
(The inclusion 'c' will easily follow as a corollary of Theorem 3.48 below.) The Browder speatrum ofT is the complement of o0 (T) in the spectrum, i.e., oB (T) =a (T) \o 0 (T).
69
Clearly, a(T) ~ aB(T) aaW (T) c a ae (T) c a a tre (T) •
~
aw(T)
~
~
ae(T)
atre(T) and
'~
aaB(T)
c
PROPOSITION 3.45. Given T € L(H). there exists K E K(H) such that p:F (T+K) = ' . a(T+K)= aB (T) and min.ind(T+K).) = min.ind(T).) foro all ). E PiF(T); moreover. K can be chosen equal. to a normal. compact operator such that
be the triangular representation (3.8) of T and let {~n}lsn
* T
~
0
*
n +T n
0 •
     
0
     
I
:~m
Hm
where H1 , H2 , ••• , Hn•··· are defined so that ej~l Hj coincides with the Riesz subspace of T0 corresponding to the clopen subset {~ 1 .~ 2 , ••• , ~n} of a(T 0 ) (1 s n < m) and Hm=H 0 (T)e{elsn
[~r T~! :~!]. 0
0
T1
then it follows from Theorem 3.38 (Observe that (T') =T and (T'). = s r r "' ·T1 !) that psF(T') =,, a(T') =aB(T') =aB(T) and K=T'T is a diagonal normal operators with eigenvalues ). 1 ~ 1 (with multiplicity equal to dim H1 ), ). 2 ~ 2 (with multiplicity equal to dim H2 ), ••• , ).n~n (with 70
multiplicity equal to dim Hnl , ••• , so that
1 ~n
IIKII=max{j).nllnl: LEMMA 3.46. LetT ~ank
exists a finite (i)
(ii) (iii)
IIKJ.III
<
r
'll
£,
L(H),
€
ope~ato~ €
'll
p:_F(T)\cr 0 (T) and E > 0. Then the~e
€
such that
K 1J
Ps~(T+K'Il).
min.ind(A(T+Kll.ll =min.ind().T)k, A
If
\1
E
l~n
€
psF(T)\{'11}, k=l,2, •••
~T EDT', whe~e T
p:_F(T)\{J.I} and T
aats on a fis F \1 E p s
nite dimensional spaae, cr(T ) = {v} and \1\1 € " p r F(T'), " then \1 r s\1 (T+K ) and T+K ~ T EDT", whe~e \1 € p F (T") • 'll 'll " " s" PROOF. By Theorem 3.38, 1J
M the
ap((T 1 l*l* and denote by
E
crp(Trlucrp((T 1 )*)*. Suppose that'll
€
Riesz spectral subspace of T 0 corre
H
sponding to cr(T 0 ) \{ll}• I f we put
'll
H
=
0
(T)9M then
Tl3 Tl4 Hr(T) T23 T24
T
M
TJ.I T 34 HJ.I
0
T1 H1 (T)
Let S ' 
(Tll T34) 0 T . R.
Since cr(T) = {ll}, we can find a ].1
(finite rank!) operator A
such that IIAII < E/2, cr(T +A) ={J.I} and nul(T +AJ.I) =1. ker(T +Au)*, 'll
x
~
'll
].1
€
(Thus, i f x
L(H ) j.l
€
0, then x is a cyclic vector for T +A.) Define S"
(T~ +A
;J ~~
].1
(T).
By Corollary 3.22, S" (Tll+A)EDT1 , i.e., S" has an invariant subspaceR complementary to HJ.I such that S"IR € T 1 • Let W € L(HJ.IEDH 1 (T)) be an invertible operator such that s 1 = WS"Wl= (T +A) E9T • 'll 1 E ker(Til+All), 0 < llfll < e:/(2IIWII·IIW11>, and let e £ ker(T1 1. Let s 2 =s 1 +e&f € L(HJJEDH 1 (T)), A E a:, y € H 1 (T) and z € H11 and assume that (s 2 'lll (y+z) = 0; then (s1~y+e+(~~ z = 0. Hence
Let f
1J)*, llell
(T A)Y+e = 0, (T +AA)Z = 0. 1 j.l I f ). ~ 'll, then A ,l cr (T +A) = { 'll}. I t follows that z = 0 and therefore,
'll
(T 1 A)y=O. Since crp ~ 0 and (T 1J.JlY = e ~ 0. But this is impossible because el ran(T 1JJ). We conclude that crp (s 2 ) = jl.
71
Observe that 52=
[(T +A>.)* [email protected] ) (T +A 0 )H e~f Ti H~(T) , so that (S2Al* = 0 (TPA)* •
Since H is finite dimensional, it is not difficult to check that = ker(s 2 ~)*d(p) (where d(p) =dim HI!) and HP c v {ker(S 2 Al*: A € PsF(T)}, whence we conclude that HR.(s 2 ) =HP8HR.(T) and (s 2 )R. = s 2 • Let s 3 = w 1 s 2w and set
H~
S
=
Tr Tl2 8 13]Hr(T) [ 0 T0 1M s 23 M 0 0 s 3 HP8HR.(T)
where s 13 = (T 13 T14 ) and s 23 = (T 23 T24 > 1 then K = s T is a finite 1
p
rank operator such that IIKPII s IIAII+IIWII·IIW ll·lleQfll < £/2+£12 = £1 Hr(S)=Hr(T) and HR.(S)=HpeHR.(T) and Si=s 3 , and H0 (S)=M and S0 =T0 1M· Now (i), (ii) and (iii) can be easily verified by using Theorem 3.38. If p £ ap(Tr), then we decompose T as
T
Tr Tl2 Tl3 Tl4 Hr(T) 0 Tp T23 T24 H p 0 0 (To)·M T34 M 0 0 0 TR. HR. (T)
where HI! is the (finite dimensional) Riesz subspace of T0 corresponding to {p} and M = H0 (T)9H p • Applying the above arguments to
(:r T~2r p
instead of S', we can find a finite rank operator K' 1.1 such that IlK' 11 < e::, and
£
L[H (T)8H l r 1.1
satisfies Hr(s 4 ) = Hr(T)8HI.I and In this case we define R14]Hr(T)8HI.I T34 M Ti HR. (T) 13  (T where R13T23 ) and K'eo has the desired
I
It is easily seen that K
1.1
S T "'
0
1.1
K[Hl,
72
PROPOSITION 3.47. Let T IIKII < e:: suoh that
£
L[H) and e:: > O; then there e%ists K
£
(i)
o 0 (T+K) = o 0 (T) and (T+K) IH(>.;T+K)
~
TIH(>.;T) for all >.
E
o0 (T);
(ii) p~F (T+K) = PsF (T) \00 (T); (iii) min.ind(T+K>.) = min.ind(T>.) for aZZ >.
E
p~F(T).
PROOF. Let 1:!. ={A" psF(T)\[o 0 (T)up(T)]: dist[A,ClpsF(T)] ~ E/2}. Clearly, 1:!. is a compact set and A= l:!.n[ps F(T)\o (T)J is finite. Appl~ so ing Lemma 3.46 a finite number of times (once for each point in A), we sati~
can find a finite rank operator K1 such that I!K 111 < E/2 and T+K 1 fies (i), (iii) and (iiI)
Now we can repeat the argument of the proof of Proposition 3.45 in order to "push" the points of p:_F(T+K 1 )\o0 (T) to PsF(T+K 1 J by means of a compact perturbation K2 , I!K 2 11 < E/2, which only affects the action of (T+K 1 ) 0
•
It is easily seen that K = K 1 +K 2 satisfies all our requirements.
0 THEOREM 3.48. LetT" L(H) and E > 0; then there exists K" K(H)
suah that IlK II < E+max{ dist[A, ClpsF (T)]:
>. " o 0 (T)}
andmin.ind(T+K>.)=0 foro all>." p 5 _F(T).
In partiauZar, o(T+K) ={A "
a::
>.T is not a F1'edhoZm operato1' of index 0}
PROOF. By Propositions 3.45 and 3.47, we can find a compact tor K1 such that I!K 111 < E/2+max{dist[A, Clp 5 _F (T)]:
>.
E
oper~
o 0 (T)}
and the operator A= T+K 1 satisfies the following properties: p~F (A) = r PsF (A) and min. ind (AA) =min. ind (>.T) for all >. " PsF (T). Thus, in order to complete the proof it suffices to show that there exists K2 " K(H), I!K 2 11 < r./2, such that min.ind(A+K 2 >.J = 0 for all>. E psF(A) = p 5 _F(T). (Indeed, K= K1 +K 2 will obviously satisfy all our requirements.) Let 1:!. = {>." psF(A): dist[A,apsF(A)] ~ E/6} and !:!. 1 ={A E /:!.: min.ind(>.A) ~ 0}. By Proposition 3.16 we can find y,z " H such that Pker(>.A)y ~ 0, Pker(>.A)*z ~ 0, for all>. " /:!. 1 • Let Ca = ayez, a > O. i'le have min. ind (A+Ca >.) =min. ind (A>.) 1, for all >. Since min.ind(>.A) =0 for all>.
E
E
A1 , a> 0.
I:!.\A 1 , it readily follows that
73
6\6 1 that
c
p1 (A) pr(A). Therefore, there is a positive constant n 1 such 110..A)x!l ~ n1 for all>. E p 1 (A)n(6\6 1 ) and 11<>.A)*xll ~ n1 for all >. e pr(A)n(6\6 1 ),
for all x .: H, II x II = 1. Thus, if 0 < a < min{e:/6,n 1 /2}, then IICall < e:/6 and min.ind(A+Ca1.) =0 for all>. e 6\6 1 • Now we can apply the same arguement to A+C (l , etc. After finitely many steps we obtain a finite rank operator c 1 such that 11c 1 11 < e:/6, min.ind(A+C 1>.) = 0 for all>.
E
6 and o(A+C 1 )
c
interiora(A)e:/ 6 • (3.12)
Let {Qj}l~n.A) ~ 0 for all >. e nJ.• (Clearly, u. Q. c [ap F(A)] 13 .) A repetition of our previous argument shows J J se: that there exists a positive constant n < min{e:/6,n 1/2} such that the relations (3.12) remain true if A+C 1 is replaced by A+C 1 +B for any B in L (H) such that liB II s n • r If >. 1 e PsF(A+c 1 >no 1 , then the arguments used for the construction of C can be repeated here in order to find a finite rank operaa tor c 2 , 1 , II c 2 , 1 11 < n/2, such that min. ind (A+C 1 +C 2 , 1 >. 1 ) = 0. By induction, we can construct finite rank operators c 2 1 , c 2 2 , ••• ,c 2 n•···• n
CO
I
I
IIC2,n11 < n/2 (n=1,2, ••• ) such tha; C2=~n=l c2,n min.ind(A+C 1 +c 2>.) =0 for all>. e PsF(A+C 1 +c 2 ), 6
c
r
PsF(A+C 1+c 2 )
and
cr(A+C 1 +c 2 )
Since max{dist[A,apsF(A)]:
c
€
I
K(H), IIC211 < nr
interior cr(A)e:/ 6 • s
>. e PsF(A+C 1 +c 2 )} < e:/6,
we can apply the arguments of Proposition 3.45 in order to find a compact normal operator c 3 , IIC 3 !1 < e:/6, such that min.ind(A+C 1 +c 2+c 3>.)=0 for all>.
E
PsF(A). 0
An
operator T e L(H) such that min.ind(>.T)
0 for all >. e PsF(T)
(3.13)
will be called a smooth operator.
3.5 ApostolFoiafVoiculescu's theorem on normal restrictions of compact perturbations of operators 74
Given Tin L(H) we cannot expect, a priori, to find an infinite dimensional invariant subspace M such that TIM is normal. An important approximation argument (due to C. Apostol, C. Foia~ and D. Voiculescu) asserts that some arbitrarily small compact perturbation of T always has that property. We shall need some extra notation to make it more clear.
3.5.1 Schatten pclasses Recall that if K is compact, then (K*K)~ can be written as (K*K)~ A3
In:l An [email protected] with respect to some ONB {en}n:l' where ;~. 1 = IIKII 2! A2 2! ···~An 2! An+l 2! ••• ~ 0 and An+ 0 (n + ~>. The Schatten pcZass cP(H) of compact operators is defined by
cP(H)
= {K
E
K(H):
2!
In:l A~ < oo}, 0 < p < oo,
and C (H) = K(H). The reader is referred to [77], [107], [183] for the properties of these ideals of operators. In particular, we have: (a) If K E cP(H), then K* E cP(H): (b) cP(H) (C 00 (H)) is a Banach space under the norm IKI = IK*I 00
00
p 1/p
p
p
(with respect to some ONB {fn}n:l of H. The result is independent of the particular ONB.): (e) c 1 (H) = the ideal of trace class operators is isometrically isomorphic to the dual K(H)t of K(H). This isomorphism is defined by C1 (H) :;; K ++4lK E K (H) t , where ~K(X)
= tr(KX), X E
K{H):
(f) c 7 (H)t = K(H)tt is isometrically isomorphic to L(H). This isomorphism is defined by L(H) :;; A++ ~A E C1 (H) t , where
~A(K) = tr(AK), K (g)
If A
E
E
c1 (H):
L(H) is a nonnegative hermitian operator, then tr(A)
=
In:l (0 s tr(A) s
oo)
is welldefined and the result is independent of the particular ONB {fn}n:l of H.
75
(h) (C 2 (H), 11 2 > (=the ideal of HilbertSchmidt operators) is a Hilbert space. If A, B" c2 (H), then AB*" c 1 (H) and the inner product of c2 (H) is.given by = tr(AB*).