1 OPERATOR, THEORY © Copyright by INCREST, 1980
AN EXTENSION OF SCOTT BROWN’S INVARIANT SUBSPACE THEOREM: K-SPECTRAL SETS
JOSEPH G STAMPFLI
Recently, Scott Brown showed that every subnormal operator has an invariant subspace Similar techniques can be used to show that every operator in ¥(#), the algebra of bounded linear operators on a separable Hilbert space #, for which o(T) is a K-spectral set also has an invariant subspace This result has been proved by J Agler [30], when o(T) is a spectral set for 7 There are more differences between spectral sets and K-spectral sets than might be apparent at first glance First, dilation theory is available in the former case but not in the latter Second, orthogonality disappears in several places as one moves from spectral to K-spectral sets Third, there are several interesting special cases such as polynomially bound- ed operators and unitary p-dilations which are not covered by spectral sets DEFINITION The compact set M > o(T) is a K-spectral set for Te L(#) if
WAT )| < KllflS
for all fe R(M) where
I/l6 = sup {[ƒ@)|: ze 4)
(R(M) denotes the uniform closure of the rational functions with poles off M.) To begin with, we need an extension of the orthogonal direct sum decompo- sition for operators proved independently by Mlak [21] and Lautzenheiser [17] in the spectral set case It should be mentioned that although K-spectral sets are never mentioned in [17] and only very briefly in [21], still many of the techniques carry over from their work (See also [23))
THEOREM 1 Let Te £(#) Assume M is a K-spectral set for T Let G,, Go, be the nontrivial Gleason parts of R(M) Then there exists an invertible operator O such that
Trang 24 J G STAMPFLI where Sy is normal and o(S;) < G; fori=1,2, Thus T= + T; (direct sum) (Note: Some of the terms may be absent.)
Because of its length we have relegated the proof of Theorem 1 to the Appendix However, we will invoke the notation and techniques in Proposition 5
From now on we will be using K-spectral sets M where R(M) is a Dirichlet algebra (see [10] for a definition) It is well known (see [24], Lemma 4.1 that the non- trivial Gleason parts of R(A/) in this case are G,, G;, where the G,’s are the components of intM (The G;’s must be simply connected) Combining these facts with Theorem 1 we obtain the following:
Coroiiary Let M bea K-spectral set for Te L(A) where R(M) is Dirichlet
Let G,, Gz, be the nontrivial Gleason parts for R(M) If o(T) G; # @ for two distinct j’s then T has a complemented invariant subspace
Let o(T) be a K-spectral set for T We wish to develop a functional calculus for T, and to do so, we must choose a more tractable spectral set than o(7) itself The next lemma selects such a set which is both topologically nice and analytically minimal Parts of the proof are taken from [4]
LEMMA 1 Let o(T) be a K-spectral set for T and assume T has no comple-~ mented invariant subspaces Then there exists an open connected, simply connected set G, such that G > o(T), R(G) is Dirichlet and
Allo = sup {[A(A)|:2€0(T) NG} for all he HG)
Proof To facilitate matters, we begin by listing the results we shall need Throughout, M is a compact set in C
Proposition 1 Let M be a K-spectral set for T where R(M) is a Dirichlet algebra If o(T) ¢ [int M] then T has a complemented invariant subspace In parti- cular, if int M = @ then T is similar to a normal operator
Proof R(M) Dirichlet implies R(OM)=C(OM) Thus, if ø(7)# [int AZ]
then there exists a disc Dy such that °
R(Dy N o(T)) = C(Dy 1 ø(T)) # Ø
Thus the techniques of [7] apply
PROPOSITION 2 ({I1], Cor 9.6) Let M, be compact, and M, > M,4, for
k=1,2, If R(M;) is Dirichlet, then R r x4) is Dirichlet
k=1
Trang 3EXTENSION OF SCOTT BROWN'S THEOREM 5
and
n 24G, ð)`M)
ỗ (y is the analytic capacity)
ii) lim i >0 for all ze0M
We have included Proposition 3, because it makes transparent why the next proposition is true
Proposition 4 Let N,, Nz, be the components of int M where R(M) is Dirichlet Then RUMNN,) is Dirichlet
Proof It follows immediately from Proposition 3 if one notes that »(£)> >(diam £)/4 for E connected See also Cor 9.7 of [11]
Proof of Lemma 1 Suppose we begin with a compact K-spectral set M for T
“
where R(M) is Dirichlet For example, take M = o(T), the polynomially convex hull of øơ(7) We may assume that o(T) cint M by Proposition 1 Let N,, Np, be the components of int M By Theorem 1, we may assume that at most one com- ponent, say N,, meets o(7) Set M, = M\N,, M, = M\NN,, and so on Then R(M;) is Dirichlet for each j, and hence R(N M,) is Dirichlet Let E = N Mj Then we are reduced to the following situation: Eis a K-spectral set for 7, R(E) is Dirichlet, alc int E and int E has just one component, say U
We now ask whether
(Allo = sup {[A(z)|:z€o(T) A U}
for all h ¢ H°(U) If yes, we are done and set G = U If no, we proceed as follows Let ” be a conformal map of the unit disc D onto D For he H°(U) = H*(my) set hz) = = h(p(z)) for ze D Then he H°(D) and WWlleo = = ||h||, By assumption,
sup {(A(2)|: 9(z) aT) A U} < Willen
for some h eé H™(D) Thus, by well known properties of radial limits, there exists a set O œ [0,2n] of positive measure such that for each 0 € O, there is a segment
Lạ = (re®, e”) where ọ(T¿) c Uø(T) Choose a 6 where lim g(re*) exists and
rol
call the limit ae @E Note that g(Z,) is contained in a component C of E\o(T), which extends to 0E The set C must be simply connected else o(T’) is disconnected Since U was simply connected to begin with and C extends to 0£, the components of int E\C must be simply connected It follows from Proposition 3, by arguing as in Proposition 4, that R(E\C) is Dirichlet (See also Cor 9.7 of [11].) Note that int ENC need not be connected
Trang 4Propo-6 J G STAMPFLI
sition 1 Note that at each step, we obtain a compact set M;, where o(T)< M; and R(M;) is Dirichlet Note that o(T) < int M, at every stage, by Proposition 1 By construction M,;> M, for 5 < B If Ms,, has an immediate predecessor M; we show R(M;,,) is Dirichlet by Proposition 4 If Mg, is a limit ordinal (M,= e) M;) then we show R(M,) is Dirichlet by Proposition 2 Note that at
<
every step 14;\M;,, is a nonempty open set Thus, the process must terminate at some countable ordinal « and M,,, = M, for that a We then set G = int M, For this choice of G, it follows that
[Villoo = sup {[h(z)|: 2 € o(T) 0 int M,}
for all he H™(int M,) The proof is complete
The proof obviously follows that of Theorem 2 of [24] in shape and form Unfortunately, there seems to be no way to apply that theorem directly
In Theorem 1 we developed a functional calculus for functions in R(M) We next extend this to H® functions Let G be as in Lemma 1 Since R(G) is Dirichlet, H°(G) = H™(mg,) where mg is harmonic measure on 0G and
H*(mg) = H*(mg) N L™(mg)
(see [28], Section 8 for details) Observe that R(M) is pointwise boundedly (sequen- tially) dense in H™(G) in the weak-« topology by [11], Theorem 5.1 (Actually, more is true—see [24/, Lemma 4.3.) The functional calculus begins with:
PROPOSITION 5 Let G be an open, connected simply connected set where G is a K-spectral set for T, and R(G) is Dirichlet Assume T has no complemented invariant subspaces Then there exists a homomorphism T of H°(G) into 2(#) where P:h>h(T) and ||A(T)|| < K| All, for all he H°(G)
Proof It follows from Theorem 1 that our decomposition for T contains only a single term Thus if u(x, y) is an elementary measure then p(x, y) << mg for all x, y€ H For he H™(G), choose rational functions q, € R(G) such that g,-> weak-+
(that is pointwise boundedly in G) Set
Trang 5EXTENSION OF SCOTT BROWN’S THEOREM 7 It is easy to see that (A(T)x, y) is well defined and the resulting operator is linear Since
[ (A(T )x, y)| < \ Liị đị nớ, y)[ < 1ñ KỊIxÍlll
it follows that
A(T) || < Ki|Alloo
For the multiplicativity of I, first show that (fq) = r(f) F(q) for fe H°(G) and q a rational function Then handle the general case by approximating g ¢ H™(G) weak-* by rational functions g,
From now on we assume G has the properties assigned it in the last proposi- tion
LEMMA 2 ((4], weak form of Lemma 4.2) Let T be as in Proposition 5 Let A€o(T) NG Let he H™(G) Assume ||T — A)x|| < & where xe H and ||x|| = 1
Then,
| (A(T) x, x) — h(A)| < 2ed Koo
where d= dist [A, OG] In particular, if ||(T — A)x,,|| > 0 for a sequence of unit
yectors, then
lim (A(T) Xp, X_) = h(A)
Proof We may write A(z) — h(A) = (2 — A)g(z) where g¢ H°(G) By the maximum modulus principle, it follows that
Ello < 2||Allood™
Thus
| (ACL) — hA))x, x| = | (g(P\(T — a)x, x)| < < |Œ — A)x|l lle(7)*x|| < 2z4"1K ||h| |] Alco
The second statement is obvious
COROLLARY Let he H°(G) Assume o(T) is all approximate point spectrum Then
Trang 6§ J G STAMPFLI
Thus, I’ is bounded above and below
Proof We already know ||i(T)|| <K{Al|, Let A Â Hđ(G) and choose 16(T) NG such that
li < J@)| + s
Then, choose a unit vector x ¢ # such that
[ACA] < [ (ACT) x, x)] + 2 Hence
WA lloo < \(A(T)x, x), + 2e < |[A(T)|] + 2e
Since ¢ is arbitrary, we are done
If ø(T) is not all approximate point spectrum, then standard arguments show that T has an invariant subspace; so we can make the following:
Assumption From now on we will assume o(T) is all approximate point spec- trum
Notation Let @y denote the weak-«closure of the rational functions in T with poles off G = M It is easy to see that I maps H°(G) into #,
Lemma 3 Let T be as above Then I maps H™(G) onto &y
Proof We will need the following theorem of Banach ([2], page 213) Let M be a linear manifold in a separable Banach space Let 41 =M and for each count- able ordinal a let M* be the set of weak-* limits of convergent sequencesin \_J Mđ
8<ôw
Then the weak-+ closure of M equals |_JM* and the M* are all equal from some countable ordinal on
To apply the theorem, we set 14! = rational functions in T with poles off G Let B = lim q,(T), g, rational Since ||q,(7’)||< ¢ (c constant) it follows that |lq, || <¢ for all x Choose a weak-xconvergent subsequence of {g,} (still denoted by {q,}) which converges to he H°(G) Then for every x, ye #,
((T)x, y) = \ he d(x, y) = lim \ 4„dư(x, ÿ) =
= lim (,(T)x, y) = (Bx, y)
Trang 7EXT ENSION OF SCOTT BROWN'S THEOREM >
We summarize the preceding lemmas
COROLLARY Let T be as in Proposition 1 Then P maps H®(G) onto By and the norms are equivalent
Notation For A4€ G set C,(h(T)) = h(2) for h(T) c2 Then C; 1s clearly a weak-* continuous linear functional on #, By (x © y) we denote the linear functional defined by (x @ y)(B) = (Bx, y) for Be Y(#) By || ||, we denote the norm of a linear functional in @, restricted to 2; (@, = trace class) An excellent discussion of the duality between ¢, and #(#) and the norm ||_ ||, can be found in [4]
We may now go back and strengthen Lemma 2 as follows
Lemma 2’ ([4], Lemma 4.2) Let Ae 0(T) N G Let |\(T — A)x,||-> 0 for a sequence of unit vectors x, € # Then
llC¿ — (x„ @ x„)||„ > 0
Indeed,
IC, — &% @ x)|l¿ < 22 1K||Œ — a)x\|
The rest of the argument for the main result follows [4] very closely However, to overcome one difficulty we will have to switch back to the disc and appeal to a
clever result from [5]
To this end we first note that H°(G) and H™(D) are clearly isometrically isomorphic Let g be a conformal map of D onto G and set py 1 = wy Set S = Y(T)
Since ử e H°(G), W(T) is well defined
LEMMA 4 Let S, 7, W,@ be as above Then 1) S has the disc D as a K-spectral set 2) IfAea(T) N G then (A) € a(S)
3) g@(S) = T and S and T have the same invariant subspaces 4) For every he H*(D)
\\A[2, = sup {|h(z)|: z€ o(S) n Dj
5) &s is isomorphic to H®(D) and the norms are equivalent 6) Bs = By
Proof 1) Let p be a polynomial Since [: h + A(T) is a homomorphism for he HG) it follows that
llp(S)ll = llp s ýŒ|\ < Kllp s ý|Š = Kiplie
Trang 810 J G STAMPELI 2) Let 4) € 6(T) N Gand assume (7 — 2))x,|| + 0 for a sequence of unit vectors {x,} (not necessarily orthogonal) Set
WA) — W(Ag) = (2 — À2) g0)
Then
l(S — ý(22))x„ll = lle()Œ — 4o)xnll < <Š lsŒI lŒ — ^a)xall> 0
This result does not depend on the Assumption Indeed, if (J — A.) is bounded below we turn our attention to (J — A,)* which can not be bounded below Since the conjugate set G is a K-spectral set for T* we may define S* in analogous fashion and repeat the proof above
3) If fe H™CD), then the composition law f(S)= f° W(T) is valid This is easy to check if f is a polynomial and therefore holds in general by weak-+continuity (see [23], page 298) In particular, g(S) = T, which implies that S and T have the same invariant subspaces
4) Follows immediately from 2), the fact that
AIS = sup {|AQ)|: 4 6(T) 1G} for all he H®(G)
and the isometric isomorphism between H(G) and H™(D) induced by 9 (or yw) 5) Follows from 2), 4) and the first part of this paper
6) Follows from the relations y(T) = S, e(S) = T
Lemma 5 ((4], Lemma 4.3.) Let «, Beo(T) Let {x,\, {y,} be mutually ortho- gonal orthonormal sequences where ||\(T — œ)x„|| + 0 and |\(T — B)y,\| + 0 Assume
T has no invariant subspaces Then a) |lx, @ Valle — 0
b) fx, @ wll >0 for all we # c) iw @ x, ll, 20 for all we ZX
Proof Parts a) and b) follow directly from [4] or one may imitate the proof in Lemma 2
Trang 9EXTENSION OF SCOTT BROWN’S THEOREM 11
REMARK We introduced Š and returned to the disc solely to facilitate the proof of part c) above With that out of the way we focus on T
Lemma 6 ([4], Lemma 4.4.) Ler
n — weak-+
Bis {3 ajC,,: Yila;[=K and 4,€6(T) 0 cI
1
Then B’ > unit ball (27) ,
Proof Same as [4] It is here that the comparability of |||], and |A(T)|| is important Note the K in the definition of B’
Lemma 7 ([4], Lemma 4.5) Let s,,5,,¢ # where ||Cy — (s,@5p) lle < 5" (Cy = C, for some 1 €G) Then there exist 8444, Sn41 © H such that
1) lls, — Syeall < K2” and lý, — sunl| < K2"
2) ||lCc — (S,+¡ ® s;+¡)||„ < 272029,
Proof Same as [4] The proof involves a choice of many orthonormal sequen- ces {x¡;} where lim ||(7’— 4,)x;,;|| > 0 We choose them mutually orthogonal in view of Lemma 5
THEOREM 2 ((4], Theorem 4.6) There exist vectors u, vé # such that
Co =u @ vy
Proof Let u=lims, and vy = lim s„
We are now in a position to prove the main result
THEOREM 3 Let o(T) be a K-spectral set for T Then T has an invariant sub- Space
Proof Let
M = aAm{(T — Aku: k =1,2, } for the u of Theorem 2 Then
((T — Aj‘ u,v) = Ci((z — a) = 0
for k > |,
Since ⁄Z = 0 implies (T — A)u = 0, the proof is complete
COROLLARY 1 Let T be a polynomially bounded operator with o(T)> D Then T has an invariant subspace
REMARK Tis polynomially bounded if
IPD) < KllPllco
Trang 104 J G STAMPFLI
CoroLrary 2 Let Te L(#) satisfy the conditions
1)ôDcơ(T)c D
1 dist[A, o(T)]
2 JŒ—2)*l< for A¢ o(T)
Then T has an invariant subspace
Proof It follows from condition 2) that Te C, (in particular, in C,), and thus T is similar to a contraction S ({27]) Thus, 2D < a(S) < D and 2) becomes
Ko dist [2, o(S)]
|(S — ay] < for A ¢ o(S) In the framework of Lemma 1 we take G = D If
Allo = sup{|#Œ)|:zeø(S)n D} for all he H°(D),
then we are done since the remaining lemmas and theorems are valid in this context Assume therefore there exists an 4 ¢ H™(D) such that
sup{|2(z)|:ze ø(S) nñ D} < llh||„— s
Then there exists a set @ c [0,2z] of positive measure such that for cach 8< Ø,
there is a segment Tạ = [rạe, eŸ'') where Ly = D\o(S) and
lim [A(re®)| > || — =,
rol 2
Since D\o(S)has only countably many components, there is some component C which contains two L,’s; say L, and L, (This part of the proof follows [31], Theorem 3.) We connect r,e to r,e% by a Jordan are t lying in C Let y= L, U L,Ut Thus, ÿ separates D into two components each of which intersects o(S) Let K, and K, denote the kite shaped regions at e and e!, respectively, where « = 2/2 Then it follows from Theorem 1.3 of [9], that
ø(S) n {K, U K,] = Ø
Trang 11EXTENSION OF SCOTT BROWN’S THEOREM 13
gia
5}
- CoroLiary 3 Let Te £(#) be a hyponormal operator and assume 8Dc ø(T)C
< D Then T has an invariant subspace
DEFINITION Let Ƒ' be a Jordan curve For any two points z,, z,¢I" set
diam lam Z¡Z 7,2,
QÉ, Z:) = —— |Z, — Ze |
when 212, is that arc of F\{z, U z;¿} of the smaller Euclidean diameter If Q(z, Z2) < C for some constant C and all z,,z,¢f, then I is of bounded turning {or cusp free)
I would like to thank Glenn Schober for pointing out the role of quasi-confor- ma] mappings in problems such as this and for suggesting the argument in the next
COROLLARY 4 Let Te £(#) Assume
1) p(T) is hyponormal for all polynomials p and “~
2) Oa(T) is a cusp free (or bounded turning) Jordan curve I Then T has an invariant subspace
Proof It follows immediately from 1) and ({16J, p 106) that M = aT) is a spectral set for T Let G = int M We consider two cases
Case 1
Allo == supf{{A(z)|: z¢a(T)N G} for all he H™(G)
Since the conclusion of Lemma 1 holds, the rest of the proof goes through as before Case 2
sup{|ña(z)|: z < ø(T) n G} < |lholleo
Trang 1214 1 G STAMPFLI 2 and Lemma 1 we see there exists segments LZ, and L, such that
[ø(1„)U @Œ„)] n z7) = Ø,
and g(L,) and @(L,) are in the same component of G\o(T) Thus, if K, and K, are the kite shaped regions at e' and e'®, respectively, then p-(o(T)) N[K, U K,]=@ near OD, and hence [@(K,) U @(K,)] n ø(T) =@ near 0G We must show that (K,) and ¢(K,) induce a separation of o(T)
Since I is of bounded turning, it is quasi-conformal by [19], Theorem II 8.6 Since © is quasi-conformal, there exists a quasi-conformal extension g* of @ to a domain D, > D by Theorem II 8.2 of [19] Since @* is quasi-conformal, it follows from Theorem 4, part i) of [1] that @*(K,) and @*(K;,) both subtend positive angles at g(e*) and ge), respectively Thus it is possible to join « = g(e'*) and 8 =g(e"*) by a polygonal arc y which lies in G U {a, 8} and which satisfies the condition
“dit, OG] < C, (constant) dist[A, o(T)]
for all A € y Since I is of bounded turning, it is easy to join a and ổ by a polygonal are y’ which lies in @G and satisfies a similar condition It is now possible to cut across (or integrate through) the spectrum of Ton y U y’ precisely as in Example 1 to Theorem | of [25] to produce an invariant subspace for T
REMARK I While ‘‘bounded turning’ prohibits cusps on the curve I, such curves need not be C! or rectifiable Indeed, examples of nowhere locally rectifiable curves of bounded turning may be found in [19], p 104
Coroiary 4’ Let Te (2) Assume 1) p(T) is hyponormal for all polynomials p and
om
2) Oo0(T) is a rectifiable Jordan curve I Then T has an invariant subspace
Proof: Follows the lines of Corollary 4 Let @ be the conformal map of D onto oT
G = intM where M = a(T) Then @ extends to a homeomorphism of 9D onto 0G and ø'(e) # 0 a.e on OD (see [14] X, 1.1 and X, 1.3)
Since we can omit the set where ¢’(e"’) = 0 (it has measure zero) and @ is conformal on 0D on the compiement of this set, the proof in Corollary 4 goes through as before
Trang 13EXTENSION OF SCOTT BROWN’S THEOREM 15
Proof: If Te C,(i.e., T°= p PU"|# forn = 1, 2, ) then ||p(T)|| <@e — I IP lle, for all polynomials p One could also appeal to the Sz.-Nagy-Foias [27] result which says any C, operator is similar to a contraction
Coro.iary 6 Let Te Y(#) be an essentially normal operator (that is, T*T — TT*c '; H denotes the compact operators) Assume
dist(R(T), B(#)] > 6 > 0
Then T has an invariant subspace (R(T) denotes the rational functions in T with poles off o(T) B,(#) denotes the compact operators of norm 1.)
Proof We may assume o(T) = o,(T), the spectrum of the coset T in the Calkin algebra Let f be a rational function with poles off o(T), and assume ||f(T)|| = 1 Then it is easy to se2 that
WMD < SPT) = SAY PIS =H] AIL
Hence o(T) is a K-spectral set for 7
COROLLARY 7 Let Te £(#), where o(T) does not separate the plane Assume Re ø(p(T)) = a(Re p(T)) for all polynomials p Then T has an invariant subspace Proof Clearly we may assume o(7') is connected Note that o(7) = iT) Given a polynomial p
p7 < [Re p(T)|| + ||km p(T) || < < 2 max{|o(Re p(T))|, lø(m p(T))|} = = 2 max{|Re o(p(T))|, |Im ø(p(T))|} <
< 2 |o((p(T))| =
= 2 sup{[p(A)|: Ae o(T)}
Since P(a(T)) = R(@(T)), it follows that o(T) is a K-spectral set (K = 2) for 7 Remark A number of different conditions on J which imply that Reo(T)= = o(ReT) may be found in [3]
APPENDIX
Trang 1446 J G STAMPFLI
Step J (Elementary measures) It follows from standard techniques that for each x, ye # there exists a (finite regular Borel) measure p(x, y) supported on 0M such that
(AT) x, y) = VdưG y)
for all x, ye # and
(u(x, YI < K Ill] |v
{These are termed elementary measures by Mlak.) For a fixed a;€G;, let m; be a representing measure for i= 1, 2, As is well known the mjs are mutually sin- gular Fix x, ye # Following Mlak, we next decompose p(x, y= aS = È) H;
i=0 where ph; << m; for i= 1,2, and fg is singular with respect to all m,’s (See {10j, VI 2.3) Since f(T) € F#(#) it follows that
1) [œwu(x, y) — n(œx, y)]}_L R(M) for all x, pe WH; aeC
2) tuŒ + y, w) — [u(x, w) + n(y, w)]} LRM)
3) {m(x, y + w) — [u(x, y) -E m(x, w)]} L RCM)
A Lebow has coined the term conjugate linear modulo R(M) to describe any mea- sure satisfying 1), 2) and 3)
We next show that the individual measures p(x, y) in the decomposition are also conjugate linear modulo R(M) To prove 2) note that ~
> {u(x + y, w) — [uix, w) + my, w)Ï} -L R(M) Thus it follows from the abstract F and M Riesz Theorem [13] that
{udx -+ y, w) — [m,(x, w) + n¿(y, w)]} _ R(M) for i=0,1,2 The proofs of 1) and 3) are left to the reader Because each p; is conjugate linear modulo R(M), if we set
ŒŒ3, ») =f fF dus
for x, ye #, fe R(M) and i=0,1,2 ., we obtain a well defined operator f(T) where ||ƒ/(7)|| < Allfil It should be mentioned that the measures p(x, y) are not unique in general; but all choices lead to the same definition
Step 2 (Multiplicativity) We wish to show that the map I°;: f >f,(T) is multi- plicative Observe first that
Trang 15EXTENSION OF SCOTT BROWN’S THEOREM 17
for all x, ye # and f, ge R(M) as is well known Thus fu(x, y) —u(x, f(T) *y) L R(M) But this implies the decomposed measures
[fu(x, vy) — wx, f(T)*yy}; L RA) for each i= 0,1, and thus
Suds, y) — wdx, f(T)*y) L RM)
The same argument shows
Sudx, ») — w(f(T)x, y) + RM)
Let ge R(M) Then
( F-sduites y= (/duẲ g(T)*y) = (ƒ(T)x, g(T)*y) =
= œŒŒ)x y) = fe du(f (T)x, y)
whence
(*) Sudx, y) — uf{T)x, y) 1 RCM)
From (*) it follows that the i“ measure in its decomposition, namely fix, y) — — uf{T)x, y) 1 RM) The essential elements are at hand Thus,
(fg) (T)x, y) = \# dui, y) = \ Zdu/(ø,Œ3, y) = S(T) (Tx, y)
and hence I; is multiplicative By repeating this argument, one shows
(Œ) s15, 3) = Ệ ƒdw(gŒT) y) =0 for ix)
Step 3 (The decomposition) Following Mlak, we next define operators F; where
(Fx, 9) =| 1du(x,y) for x,ye # and i=0,1,2,
From the multiplicative properties proved above, it follows that F,-F; = 6,,F; Thus, the F,’s are pairwise disjoint idempotents Clearly, }\F; converges to I The +; are not selfadjoint in general However, for any subset Q of the positive integers observe that
(EF = 13 " duj(x, 9)| < llwœ, y)|| < Kllxllllyl-
Trang 1618 J G STAMPFLI Thus ||} #;|| < K, from which it follows that {#;} is a uniformly bounded Boolean
9
algebra of projections or spectral measure in the language of Dunford It follows from a result of Mackey ([20], Theorem 55) that there exists an invertible operator Q such that OF,Q-! = E, is selfadjoint for each k Set S=QTQ™ Using the conjugate linear relations from Step 2, we see that
(Fe f(T)x, ») = \ 1é)», y) = \ fp, ~) = ( 1 dig(x, AT)*y)=
= Œ.x,ƒ/)'y) = ()EF,x y)
In particular, 7 commutes with #¿ and hence Š commutes with the #¿ (It follows also that /#⁄(T) = ƒ(Œ)#,.) H we set „—= E,# and S, =S|#, then S=@S, on
@#, Define T, = f(T) F,|F,#
Step 4 (The spectrum) To see that o(T,) ¢ G,, fix a k > 0 For A¢ Gy, set
(Bix, 9) = | Œ — 3”'diuGx 2)
for all x, y¢ FH It is easy to see that
(Ty — BBX, ») = [dint Y) = (Firs y) — G3)
Similarly, B,(7, — A= 1, the identity on F,#, hence 4 £ø(T;) Actually, with some additional effort, it can be shown that G, is a K-spectral set for T, The details are carried out in [17]
Step 5 (Normality of S,) Consider the measure jio(x, y) for x, ye FyH As shown above ply is conjugate linear modulo R(M) But p(x, y) is singular with respect to the G,’s (or m,’s) and thus, by Wilken’s Theorem ([10], II 8.5), “o(x, y) is conjugate linear This fact enables us to integrate something beyond rational functions
Let y be a Borel set Following Lautzenheiser, we set
B(x, y) = \% duo(x, ¥) = Holx, y)(y)
Since {u(x, y)} is conjugate linear, B, is a bilinear functional on Fu#’ x F,#’ Thus, there exists an operator F(y) < Z(#a) such that
(F(y)x, y) = B(x, y) = Ho(X, vy) for all x,y € tạZ
Trang 17EXTENSION OF SCOTT BROWN’S THEOREM 19
Since
(Tox, y) = \ z duig(x, y) = \ 2 d(FC-)x, y),
to complete the proof, we need only to show that F(-) is a spectral measure in the sense of Dunford
We know that I, is multiplicative on R(M), but that is not enough since we wish to integrate characteristic functions We reconsider our earlier estimates Recall that
Vo =[fito(x, ¥) — tol f(T)x, y)] RCM) — for ƒ< RCM)
But vp is singular and thus again by Wilken’s Theorem
Sox, y) = Uo f(T)x, y)
Let y be a Borel set Then for x, ye RH
(4, f dug(x, y) = ( #„ du(ƒ(T)x, y) = (F@)fŒ)x, y) =
=ŒŒx F@)*y) = (Saux, F(y)*y)
Thus, %, uo(x, y) — ux, F(y)*y) L R(M) and by the abstract F and M Riesz Theorem and Wilken’s Theorem, we see that
KjMo(Xs Y) = pox, F(y)*y)
Let y, 6 be Borel sets Then
ow dpto(x, y) = { Men» duạ(x, y) = (F(ö n )x, y)
On the other hand,
(2%, dyto(x, ÿ) = \ dua(x, F(@)*y) =
= (ŒF(ô)x, F@)*y) = (0)#)x, ?)
Thus, F(y)F(6) = F(y 6) and it follows that {F(-)} is a spectral measure or uni- formly bounded Boolean algebra of projections since
Trang 1820 J G STAMPFLI
Thus, T, is a scalar type operator (see [8]) and hence is similar to a normal operator (See [20], Theorem 55.) Consequently by modifying Q we can assure the normality of S, This completes the proof
12 13 14 15 16 17 18 19 20 21 22 23
The author gratefully acknowledges the support of the National Science Foundation,
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JOSEPH G STAMPFLI Department of Mathematics
Indiana University Bloomington, IN 47401,
U.S.A Received October 28, 1978; revised August 27, 1979
Added in proof: It has come to our attention that W Mlak proved Theorem 1 of this paper in