On Contractions Satisfying $\operatorname{Alg} T = \{T\}'$ Author(s): Pei Yuan Wu
Source: Proceedings of the American Mathematical Society, Vol. 67, No. 2 (Dec., 1977), pp. 260-264
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PROCEEDINGS OF THE
AMERICAN MATHEMATICAL SOCIETY Volume 67, Number 2, December 1977
ON CONTRACTIONS SATISFYING Alg T =
{
T}'PEI YUAN WU1
ABsTRACT. For a bounded linear operator T on a Hilbert space let { T)',
{ T)" and Alg T denote the commutant, the double commutant and the weakly closed algebra generated by T and 1, respectively. Assume that T is a completely nonunitary contraction with a scalar-valued characteristic function PA(). In this note we prove the equivalence of the following conditions: (i) I4'(e"')I = I on a set of positive Lebesgue measure; (ii) Alg T= {T}'; (iii) every invariant subspace for T is hyperinvariant. This generalizes the well-known fact that compressions of the shift satisfy Alg T
= {T}'.
For an arbitrary operator T on a Hilbert space it is easily seen that the inclusions Alg T C
{
T}" C{
T}' hold. Let H2 be the usual Hardy space and let 41 be a scalar-valued inner function. Consider the compression of the shift T defined on the space H2e
lH2 by(Tf)(eit ) = P [ei'f(ei')] forf E H2
e
41H2,where P denotes the (orthogonal) projection onto the space H2
e
41H2. It was shown by Sarason [3] that Alg T ={
T}'. (In fact, he showed more than this. He proved that every operator in{
T}' is of the form U(T) for some U E H '.) Note that here T is a completely nonunitary (c.n.u.) contraction whose characteristic function 4, is scalar-valued and satisfies 4,(ei9) = 1 a.e. In this note we give necessary and sufficient conditions that a c.n.u. contraction with a scalar-valued characteristic function satisfy Alg T ={
T}'. Indeed, we want to proveTHEOREM. Let T be a c.n.u. contraction with a scalar-valued characteristic function 4,. Then the following conditions are equivalent to each other:
(i) 14(ei)l = 1 on a set of positive Lebesgue measure; (ii) AlgT= { T}';
(iii) every invariant subspace for T is hyperinvariant.
Thus Sarason's result follows from the implication (i) =X (ii) of our
Theorem. It is interesting to contrast our result with the fact, due to Sz.-Nagy and Foia? [6], that a c.n.u. contraction T with the scalar-valued characteristic function 4, satisfies
{
T}" = {T}' if and only if 4A(X) 5 0. Note also thatReceived by the editors January 17, 1977 and, in revised form, March 11, 1977. AMS (MOS) subject classifications (1970). Primary 47A45; Secondary 47A15, 47C05. Key words and phrases. Completely nonunitary contractions, characteristic functions, invariant subspaces, hyperinvariant subspaces, commutants, algebras of operators.
'The author acknowledges financial support from National Science Council of Taiwan. ?D American Mathematical Society 1978
ON CONTRACTIONS SATISFYING Alg T =
{
T} 261whether (ii) and (iii) are equivalent for an arbitrary operator T is still an open question (cf. [1]).
In the proof of our Theorem we will extensively use the functional model for c.n.u. contractions. The readers are referred to [5] for the basic definitions and terminologies. Throughout this note results from [5] will be used without specific mentioning.
Let T be a c.n.u. contraction with the scalar-valued characteristic function 41. Consider the functional model for T, that is, consider T being defined on the space H _ [H2 E AL 2]
e
{i4w e Aw: w E H2) byT(f E g)
=
P(e'Ef ) ei"g) forf Eg
E H,where A = (1 - 1412)1/2 and P denotes the (orthogonal) projection onto H. Let Lat T denote the lattice of invariant subspaces for T, and let T(n) denote the operator T E ... ED T acting on the space H E * ED H. Note that
n n
the characteristic function of T(n) is the n x n matrix-valued function
Let K E Lat T(n) and let 4 = 12(1 be the corresponding regular factorization. We first prove the following
LEMMA. If
I4(e")I
= 1 on a set of positive Lebesgue measure, then () and !12are n x n matrix-valuedfunctions.
PROOF. Assume that (1 and 02 are, respectively, m x n and n x m matrix-valued functions. Let
A(e ') = (1 -(e')*!D(et))1/2 and
A (eit) = (1 -
0j(ei")*0j(ei"))1/2, j = 1, 2.
Let 8 (e") = dim A(e"')Cn, 1(eit) = dim A,(ei,)Cn and
82(e") = dim A2(ei)Cm, where C denotes the complex plane. Since 1 = (2(1
is a regular factorization, we have
(1) 8(e"') = 81(e") + 82(e') a.e.
(cf. [5, Proposition VII. 3.3]). Since 141(e")l = 1 on a set of positive Lebesgue measure, say a, it follows that A(e ') = 0 on a. Hence 8(e ") = 0 on a. If
m
> n then 02(e") cannot be isometric from Cm to Cn. Thus 62(e') > 0 a.e., which contradicts (1). On the other hand, if m<
n, then 1)1(e ') cannot be isometric from C' to Cn. Then 8#(e1') > 0 a.e. and we also have a contra- diction. This proves that m = n.PROOF OF THE THEOREM. If
4
_ 0, then, by the previously mentioned result of Sz.-Nagy and Foia? [6], it is easily seen that none of the three conditions is262 PEI YUAN WIU
satisfied. Hence we may assume hereafter that 41 i 0.
(i) => (ii). Let S be an operator in
{
T}'. To show that S E Alg T it suffices to show that Lat T(n) C Lat S(n) for all n > 1 (cf. [2, Theorem 7.1]). Let K E Lat TO) and 4 = 4241 be the corresponding regular factorization. As proved in the Lemma, 4, and O2 are n x n matrix-valued functions. In the functional model of TO)K={ 42u E Z-'(Au E v): u E H2 (Cn), vE A,L2(Cn)}
e
{Ow Aw: w E H2(Cn),where Z denotes the unitary operator from AL2(Cn) to A&L2(Cn)EAjL2(Cn) defined by
Z(Av) = A421v E AIV, v E L2(Cn).
Let 4)2u E t be an element in K, where u = (u,)i E H2(gj)_and t = (ti)i E AL2 (C") satisfy Z(t) = A2u E v for some v = (vi)i E A,L2(C"). Here we use the symbol ( )i to denote the components of a vector. We want to show that S (")(42U eD t) E K. Note that S is of the form
(B C)'
where A e HI and B, C E L?? satisfy Bip + CA = AA a.e. (cf. [6]). Assume that 4), =
(t)
and ?2 = (4i,). Since2 E
t
= ( kuj) eQ),
we have SW ~~A02
lyu SOD(<2U (D t) P (B C)-l | |(2)
=p
U
ti iB
>4gy~+
Ct1 n A E 4-y uw j=1 n j=l B sm 4w1 A + Ct1, - Ave, j=1ON CONTRACTIONS SATISFYING Alg T =
{
T ) 263n
I[,
ifj = i,O
iik$kj i,j = 1 n.
k=1 0,
otherwise,
Using Cramer's rule to solve this system of equations for 'Pik, we obtain
(det 0I)0ik = %kik i, k = 1, ... ., n,
where Tik is the determinant, multiplied by (_ l)i+k, of the matrix obtained
from 41 by deleting its ith column and kth row. It follows that
n n n
(det 4,)B , %ijuy= B , rqjuj= A(A - C) , qu1j.
j=I 1=1 j=1
Hence (det 40I)B(,21..1 I4 %uj)j is an element of AL2(C,). Thus we have Z [(det (?,)B( =iju j Z A(A -C) (_
- )(,=
j)]1
e (A j C)( ?%UI)]= 2I(A - C)[ 2 (
n, Ij1), @[jI(A -AC)(l2 (7ACf)j j
.
Sincen n n n
I iik 2 71kjj =i f 2 &ik /k, U,
k-I j4( j) 1 k=1 n
= 2 (det 10)6,,uj (6, the Kronecker 8) = (det F1)u1, j=1
the above becomes
[42(A - C)((det4l)ui).] @[DA(A - C)( 2
m%Uj
=[(A- C)(det I)u] @E 1(A -
C)( 2
)1-uj
On the other hand,
Z [(det 4.1)B ( 4 (fu)] = (det 41)Z [(B i
= (det F1)(X
E
Y),
say, for some element X ED Y in A2L2(Cn) ED ZIL2(C0). Equating the first components in (3) and (4) we obtain
264 PEI YUAN WU
(5) (A- C)(det 4)1)u = (det 4)1)X.
Since
4
X 0, we have det 4) i 0, and hence det 4l & 0. By the F. and M. Riesz theorem, (5) yields that A2(A -C)u = X. ThusZ
[B 2 ~Vu +ct,)
Z (B 2 4'vuj) + Z ((Ct,))j=1
J - j t(J=-1 ) j
= (X ED Y) + Z(Ct) =[2(A - C)u ED Y] + C(Au $ v)
= A2Au ED (Y + Cv). Hence (2) can be written as
S(n) OD2U ED t) = {42Au
ED
Z -'[A2Au $ (Y + Cv)] (}(w $ Aw), where w = (wi)i E H2(Cn). This shows that S(n) (Q2U $ t) E K as asserted and completes the proof of the implication (i) =X (ii).(ii) =X (iii). This is trivial.
(iii) =X (i). Assume
I4(e")I
< 1 a.e. It was proved in [7] that the hyperinvariant subspaces for T are of the form {f f$E g E H: - Af + 4g E L2(E) andf E IH2}, where E is a measurable subset of the unit circle and I is an inner divisor of 4i where4'
denotes the inner factor of4.
By Proposition 7.2 of [4], invariant subspaces of this form are precisely those arising from scalar regular factorizations of4.
However, since141(eit)l
< 1 a.e., it is known [5, p. 301] that nontrivial vector regular factorizations of4
exist. By the uniqueness of the correspondence between regular factorizations of4
and invariant subspaces for T, the invariant subspace corresponding to any such vector regular factorization of4
cannot arise from a scalar regular factorization, and hence is not hyperinvariant. Thus we obtain a contra- diction of (iii) and complete the proof.COROLLARY. Let T be a c.n.u. contraction with a scalar-valued inner characteristic function. Then Alg T =
{
T}'.We are grateful to the referee for making the proof of (iii) =X (i) of our Theorem more conceptual and less computational.
REFERENCES
1. J. A. Deddens, R. Gellar and D. A. Herrero, Commutants and cyclic vectors, Proc. Amer. Math. Soc. 43 (1974), 169-170.
2. H. Radjavi and P. Rosenthal, Invariant subspaces, Springer-Verlag, Berlin and New York, 1973.
3. D. Sarason, Generalized interpolation in H , Trans. Amer. Math. Soc. 127 (1967), 179-203.
4. S. 0. Sickler, The invariant subspaces of almost unitary operators, Indiana Univ. Math. J. 24 (1975), 635-650.
5. B. Sz.-Nagy and C. Foias, Harmonic analysis of operators on Hilbert space, Akademiai
Kiado, Budapest, 1970.
6. __ , On the structure of intertwining operators, Acta Sci. Math. 35 (1973), 225-254.
7. P. Y. Wu,IHyperinvariant subspaces of the direct sum of certain contractions (preprint).
DEPARTMENT OF APPLIED MATHEMTAncs, NATiONAL CHIAo TUNG UNIvERsrnry, HsINcHu,