Recursively generated weighted shifts and the subnormal completion problem

Full text

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Integr Equat Oper Th
Vol. 18 (1994)
0378-620X/94/040369-5851.50 + 0.20/0
(c) 1994 Birkh~user Verlag, Basel
RECURSIVELY GENERATED WEIGHTED SHIFTS
AND THE SUBNORMAL COMPLETION PROBLEM, II
Ra51 E. Curto and Lawrence A. Fialkow
Recursively generated weighted shifts are employed to exhibit a large collection
of quadratically hyponormal shifts which are not subnormal. This result is reached after
analyzing 1--step extensions of recursive subnormal completions generated from
three-weight data. The techniques make use of the subnormal completion criterion, and
allow us to obtain additional criteria to distinguish between the various classes of
k-hyponormal weighted shifts. Applications to bilateral shifts are also included.
w INTRODUCTION
In this paper we study concrete criteria for distinguishing between subnormal,
k-hyponormal, and weakly k-hyponormal weighted shifts on s The classes of
k-hyponormal and weakly k-hyponormal Hilbert space operators are intermediate between
the important classes of subnormal and hyponormal operators, and as such they have
received considerable attention recently ([Ath], [Cull, [Cu2], [Cu3], CF1], [CF2], [CMX],
[CP1], [CP2], McCP]). The Bram-Halmos criterion [Con, III.1.9] implies that an operator
which is k-hyponormal for every k _> 1 is actually subnormal; it follows that subnormal
operators are automatically polynomially hyponormal, that is, weakly k-hyponormal for
every k. The question as to whether the converse holds remained unsolved for a long time,
but recently M. Putinar and the first named author ([CP1]) proved the existence of an
operator which is polynomially hyponormal but not 2-hyponormal, hence not subnormal.
Due to the main result in [CP2], there is heightened interest in the structures of
k-hyponormal and weakly k-hyponormal operators; however, at present there is no
effective model theory (as there is for subnormal and hyponormal operators), and there is a
scarcity of examples. Most of the known examples of
found in the class of unilateral weighted shifts on R.2(/~+), and
k-hyponormal operators are
k-hyponormal or weakly
in the present work we continue to study higher hyponormality in this setting.

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370 Curto and Fialkow
This paper is a continuation of [CF1], where we introduced recursively generated
weighted shifts as important tools in the study of subnormal shifts; subnormal recursively
generated weighted shifts will play a prominent role in the present work. Let {en}n= 0
|
denote the canonical orthonormal basis for and let { n}n= 0 be a bounded
sequence of positive numbers. Let W a be the unilateral weighted shift defined by Wae n
:= anen+ 1 (n > 0). The moments of W a are usually defined by r0 := 1, fln+l := anfln
(n _> 0) [Shi], but we reserve this term for the sequence 3'n :-- ~ (n >_ 0). Subnormal
weighted shifts were first characterized by J. Stampfli
geometric construction of the minimal normal extension in terms of the weights. Later, C.
Berger (cf. [Hall, [Con, III.8.16]) obtained the following analytic description of
subnormality for weighted shifts: W a is subnormal if and only if there exists a Borel
[Sta], who gave an explicit
probability measure # supported in [0, I]Wo~12], with IIWJI 2 E supp #, such that
= I tn d#(t) (all n > 0). (1.1)
"Yn
It follows from this result that W a is subnormal iff for every k _> 1 and every n > 0, the
Hankel matrix A(n,k) := (Tn+p+q)0_<p,q_< k is positive semidefinite.
In [Cul], the first named author proved that Wa is k-hyponormal iff
A(n,k) > 0 for every n _> 0. This result provides a concrete test for k-hyponormality, and
together with Berger's Theorem, recovers the Bram-Halmos criterion for weighted shifts.
By contrast, the classes of weakly k-hyponormal shifts remain quite mysterious, and no
concrete characterizations are known. One of the goals of this paper is to provide a new
class of examples of shifts that are quadratically hyponormal (i.e., weakly 2-hyponormal)
but not 2-hyponormal.
In order to describe our results in detail, we first recall the concept of a
recursively generated weighted shift, introduced in [CF1]. In [Sta], J. Stampfli posed the
following subnormal completion problem: Given m _> 0 and an initial segment of positive
weights o~: aO,...,am, determine whether or not a can be completed to the weight
Ot |
sequence { n}n= 0 of a subnormal weighted shift. We first consider the case" when
2 (1 < i < 2k+1), and we set
m = 2k. We introduce moments 3, 0 := 1, 3,i := a~-...- oti_ 1

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Curto and Fialkow 371
vj v(j,k) [TJ "
~j+k
1 ~k+l (0< _k+l j< ).= e :=
These k+2
1 < r < k+l, such that v r is a linear combination of v0,...,Vr_l, say Vr_ 1 = r
vectors must be linearly dependent in ~k+l, so there is a minimal r,
+ ...
+ r in the sequel we refer to r as the .rank of a (in symbols, r := rank a). We
now define the moments of a weighted shift by propagating the preceding dependence
relation: we set ;Yi := 7i (0 < i < r-l) and 7i := r + "'" + Cr-1;ri-1 (i > r). If each
;Yi > 0, we further define weights ai := (7i+1/7i)1/2 (i > 0), and assuming that & :=
&|
{ i}i= 0 is bounded, we say that W& is the ~b--recursive shift generated by a. When
m = 2k-l, we proceed as follows: we let a n := an+ 1 (0 _< n _< 2k-2), we let ('yn)~ be
the moment sequence of (a') ^ (assuming that W(a,)A is a well---defined completion of
a' : a0,...,a2k_2), and we then define a2k := [(7~k)^/(7~k_l)^] 1/2. Once this is done,
we let W& := W(a0,...,a2k)^. In either case, W& need not be a completion of a. The
following result solves the subnormal completion problem.
Subnormal Completion Criterion. ([CF1, Theorem 3.5]) Let m > 0 and let a : a0,...,a m
be an initial segment of positive weights. Let k := [--~], ~ := [~]+1. The following are
equivalent:
(i) a has a subnormal completion;
(ii) W& is a subnormal completion of a;
(iii) a has an ~-hyponormal completion;
((iv) A(0,k) > 0, A(1,~-I) > 0, and v(k+l,k) e Ran A(0,k) if m is even
(v(k+l,k-1) e Ran A(1,~-I) if m is odd).
Recursively generated subnormal shifts are norm---dense in the subnormal shifts
on Q2(~+) ([CF1, Theorem 4.2]); moreover, the inherent regularity of recursive shifts
facilitates their use in a structural calculus for hyponormal shifts based on the operations
"extension" and "completion." In particular, we are able to use these two operations to
produce new examples of quadratically hyponormal shifts that are not 2-hyponormal, as

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372 Curto and Fialkow
follows. Let a : aO,al,02 be given, with the condition that 0 < a 0 < a 1 < 0 2. Since
A(0,1) > 0,
invertible),
A(1,1) > 0,
W(a0,al,02)^
and v(2,1) E Ran A(0,1)
is a subnormal completion of
(notice that A(0,1) is indeed
For x > 0, a. let
W(x) := Wx(a0,al,02)^ denote the shift whose weight sequence consists of the initial
weight x followed by the weight sequence of W(a0,al,02)^; thus, W(x) is an extension
of W(a0,al,02)^. Now, let
H 2 := sup{x > 0 : W(x) is 2-hyponormal},
and let
Then
(i)
where
and r and
by
h 2 := sup{x > 0 : W(x) is quadratically hyponormal}.
a22 2
0 al ( a2-a 1 )
2
2, 2 2,
4 4 2
a0ala 2 + a0al~a2-a0)K + o~0alt02-a 1
6 2 2, 2 2,~
2 4, 2 2 2 2, 2 2)K2
< , 4. 2 2 , 2 K 2
<_ h 2,
K := - ~00
4r
r are the coefficients in the recursive relations for W(a0,a1,02)^, given
22,2
a0al~'a2-al)
2
al--O~ 0
2, 2
and r 02-a0)
2
~l-aO
r = 2 = 2
This result was obtained with the help of MACSYMA;
Section 4, with a transcript of the MACSYMA session included in the Appendix. This
result allows one to construct concrete and tractable examples of nontrivial quadratically
it is presented in

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Curto and Fialkow 373
hyponormal weighted shifts, namely Wx(a0,al,a2)^, for H 2 < x _< h 2.
The calculation of H 2 above is part of the more general extension problem for
weighted shifts: If 1 < k < | W a is k-hyponormal and m < k, find
H m = Hm, k := sup { x: Wxa is m-hyponormal },
where Wxa is the one-step extension of W a with initial weight x. For m = k = |
this problem was solved in [Cu 1], where it was established that if W a is subnormal,
with Berger measure #, then W a admits a (1---step) subnormal extension iff 1 ELI(#),
in which case
. = (I }
Our next result provides an analogue for
k-hyponormal shifts.
m-hyponormal extensions of
The Extension Principle. Let 1 _< k < ~ and assume that W a is k-hyponormal. Fix m,
1 < m _< k, and let a (2m-2) denote the initial segment a0, ... , a2m_2. Let #(m)
denote the Berger measure associated with the canonical recursive subnormal completion of
ex (2m-2) (which is well---defined by virtue of the subnormal completion criterion). Let
Cm): r "'" Cr-1 denote the coefficients of recursion in this completion (r and r ""'
Cr-1
lit e Ll(#(m)), and
depend on m). Then W a has an m-hyponormal extension if and only if
= 7r-l--•r--1 7r--2 ..... @170j "
In particular, Hm, k is independent of k.
This last relation and (1.2) establish a rather surprising connection between
m-hyponormality and subnormality, and show that
"power-restricted" version of the latter as suggested by the Bram-Halmos criterion.
Moreover, the proof of The Extension Principle (Theorems 3.7 and 3.10) displays an
unexpected relation between the extension principle and the subnormal completion
criterion. In addition to its use in establishing our examples of properly quadratically
the former is indeed a

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374 Curto and Fialkow
hyponormal shifts, the extension principle is used in Section 6 as a tool to construct
subnormal bilateral completions of a given segment of weights (Theorem 6.6).
One of the main difficulties in working with higher hyponormality for weighted
shifts is that very little is known as to whether or how the growth rate of the weight
sequence is reflected in any higher hyponormality of the shift. The results of Section 3 and
Section 4 show that a change in the initial weight alone can dramatically affect the degree
of higher hyponormality. Another growth constraint concerns "local flatness" in the weight
sequence. It was proved in [Cu 1] and [Ath], that if W a is 2-hyponormal and a n =
an+ 1 for some n > 0, then a 1 = a 2 .... , so that in particular W a must be
flat--subnormal. In Theorem 5.1 we show, more generally, that a (k+l)-hyponormal shift
whose moment sequence exhibits some generalized local flatness is actually recursive
subnormal.
Flatness Criterion. If W a is (k+l)-hyponormal, if det A(i0,k-1 ) > 0 and if
det A(J0,k ) = 0 for some i 0 > 1, J0 >- 0, then W = W(a(2k)) ; in particular, W a is
recursive subnormal.
The results of lAth] and [Cull correspond to k = 1; by contrast, a quadratically
hyponormal non-2-hyponormal shift can have its first two weights equal ([Cull); in
Proposition 4.6, we exhibit this phenomenon in a shift of the form Wl,(1,al,a2)^ . On
the other hand, the first named author has proved that if a quadratically hyponormal
weighted shift has three consecutive equal weights, then it must be flat--subnormal ([Cull).
Recall that if W a is k-hyponormal, then A(j,k) > 0 for every j > 0, and so
det A(j,k) _> 0
k-hyponormal non-subnormal shifts, the extremal case det A(j,k) = 0 must be avoided.
(j > 0). The next result (Theorem 5.12) shows that to construct
Extremality. If W a is k-hyponormal and det A(j,k) = 0 for every j _> 0, then W a is
recursive subnormal.
The techniques surrounding the preceding result can also be used to characterize
uniqueness of subnormal completions (Theorem 5.7).
Uniqueness. Assume that a : a0,...,a2k admits a subnormal completion (so that W& is
one such completion). Then W& is the unique subnormal completion of a if and only if

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Curto and Fialkow 375
either rank(a)< k, or A(0,k) is invertible and A(1,k) is singular (cf. [CF2]).
In Section 6 we obtain analogues of these flatness, extremality, and uniqueness
results for bilateral weighted shifts.
w PRELIMINARIES
In this section we record some results from [CF1] which will be frequently used in
the sequel. Let ddl, dd 2 be Hilbert spaces, A E .~(ddl), B E .~(dd2,ddl), C E d(dd2),
andlet T:= [AB'." " For a Hilbert space operator T, welet N(T)and R(T)denote
[ J B*C
the kernel and range of T, respectively.
PROPOSITION 2.1
(i) A_> 0,
(ii) C_>0,
and
(iii) B = f-K E f-C, for some contraction E E ~(dd2,ddl).
In the next result,
.~ = .~(b,c) := I A bl,
[ b*c J
where AEMn(t ), bEC n and cEC (i.e., dimdd l=n and dim~2=l).
([Smu], [CF1, Proposition 2.1]) T > 0 if and only if
PROPOSITION 2.2. ([CF1, Proposition 2.3])
(i) Assume A > 0 and rank(A) = rank(A). Then A > O.
(ii) Assume A= A* and b= Aw, wEC n. Then rank(.~)--rank(A)
c = w*Aw (and this value is independent of w).
(iii) Assume that A > _ O and b=Aw, wEC n. Then A > _ O if and only if c > w*Aw.
(iv) Assume that A >_ 0 and that A is invertible. Then A > 0 if and only if c _> b'A-lb.
(This is usually referred to as Choleski's Algorithm [Atk].)
(v) Assume that A >_ 0 and that A is invertible. Then A >_ 0 if and only if det(.~) > 0.
/fandonlyif
PROPOSITION 2.3. ([CF1, Proposition 2.7]) Let A E Mn(C), .~ E Mn+l(q:)
(n > 1) be such that

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376 Curto and Fialkow
If ~, is positive and invertible, then A is positive and invertible.
PROPOSITION 2.4. (OUTER PROPAGATION, [CF1, Proposition 2.8])
C e Mn(C ) (n > 2) be a positive matrix, and suppose that
:]=[: s,]
where R, S E Mn_I(C ). Then Ran(S) c_ Ran(R), so in particular rank(S) < rank(R).
Let
PROPOSITION 2.5. (INNER PROPAGATION, [CF1, Proposition 2.9])
D e Mn(C ) (n _> 2) be a positive matrix, and suppose that
o I; :l I: :]
Let
where Y, Z e Mn_I(C ). Then R(Y) c_ R(Z), so rank(Y) < rank(Z).
The propagation results allow one to obtain the following result about tIankel
matrices, which will be an important tool in identifying certain weighted shifts as
subnormal recursively generated shifts (Section 5).
fl0
PROPOSITION 2.6. (IDENTITY PRINCIPLE)
Let A-- (Tp+q)p,q= 0 and
.~ = (Tp+q)p,q= 0 be two infinite Hankel matrices. Assume that there exists k 0 _> 0 such
that
A(i,k0+l ) > 0
and
.~(i,k0+l ) > 0
for a// i _> 0, det(A(1,k0-1)) > 0,
and
det(A(1,k0) ) = det(.~(1,k0) ) = O. If ~n = 7n for a// n, 0 < n <_ 2k0, then ~, = A.
PROOF. Observe first that the hypotheses imply: (i) A(i,k0) _> 0 for all i _> 0
(since A(i,k0+l ) > 0); (ii) det(A(i,k0) ) = 0 for all i > 1 (since det(A(1,k0) ) = 0 and
Proposition 2.4 can be applied to
= = );
and (iii) det(A(i,k0-1)) > 0 for all i > 1 (since det(A(1,k0-1)) > 0 and Proposition 2.5
can be applied to

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Curto and Fialkow 377
A(i-l'ko)= A(i,ko_l ) * i A(i+l,ko-I ) )"
Similar statements are true for A (note that A(1,ko-1 ) = A(1,ko-1)). We now show by
~ -
induction that 72k0+t = 72k0+t for all t > 0. Assume that 72k0+J = 72k0+J for all
0 < j < t. Consider
Since det(A(s ) = 0 and det(A(t+l,k0-1))> 0,
72k0+t+ 1 = v*A(t+l,k0-1)-lv. Similarly, 72k0+t+l = ~r* A(t+l,k0-1)-l~.
induction hypothesis, ~ = v and *(t+l,k0-1 ) = A(t+l,k0-1), so
72ko+t+ 1 = 72kO+t+ I. i
Our next task is to establish a rank principle for Hankel matrices, which will be
used in Section 3.
Proposition 2.2 (ii) implies that
By the
PROPOSITION 2.7.
be a collection of 2k+2 nonzero real numbers, let r := rank(7), and
[!o I
a~d B(j):= irj+l.
j j
7:70 , ..., 72k+1
define
A(j) :=
Then
( i) The vectors v(O,j), ..., v(r-l,j) are linearly independent, for all j _) r-l;
(ii) A(r-1) is invertibIe;
(iii) r ~ 0 ~ > B(r-1) is invertible;
(THE RANK PRINCIPLE, [CF1, Proposition 2.12]) Let
(o < j < k).
7 j+ j
(iv) r-1 < rank(B(r-1)) _< r;
(v) rank(A(k)) > r.

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378 Curto and Fialkow
PROPOSITION 2.8. Let
A =
q0 r0 0
r0 ql fl 0
0 rl q2 r2
0 0
O
r2 q3 i
be an infinite Jacobi matrix, with qk E ~ and r k E C (all k ) 0), let A k be the truncation
of A to the first k+l rows and columns, and let A k := det(Ak). Assume that A )_ 0
and that A k = 0 for some k. Then A splits as the orthogonal direct sum of a finite
matrix and an infinite Jacobi matrix.
PROOF. It is well-known that A k satisfies the following 2--step recursive
identity:
Ak+2 = qk+2 Ak+l-- trk+l/2Ak (all k > 0).
Let k 0 be the first integer such that > 0 and
Ak 0 Ako+l
Ak0+2 = - Irk0+li2Ak0 , and since A _> 0, Ak0+2 _> 0, whence rk0+l
Ak0+l is an orthogonal direct summand of A. i
= 0. Then
= 0. Therefore,
For the next result, the generating function of an initial segment
is the polynomial of degree r given by
t r 0 t r-1
ga (t):= - i-1 -'"-r (t E~),
where r := rank(a) and r 1 are the coefficients of recursion for a.
a : a0,...,a2k
THEOREM 2.9. ([CF1, Theorem 3.9]) Assume that a: aO,...,a2k admits a
subnormal completion, let W& be the Ca-recursive completion of a, and let ga be the
generating function of a. Then
(i) ga has exactly r distinct nonnegative roots s O < ... < st_ 1 = I[W~I2;

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Curto and Fialkow 379
(ii) The Berger measure of W~ is
(2.1)
r--1
#a := j~0PJ~sj '
where (p0,...,Pr_l) /s the unique solution of the Vandermonde equation
I1111
-1 r-1
Sl
(2.2)
s o s I ... st_ 1
. sr-l|
r-lJ ""
Pl
r--1 o]
71
1"-1
( iii) ( Minimality and Uniqueness of Support) Let v be a finitely atomic measure such that
It n dr(t) = ( 0 <_ n _< 2k+l).
7n
Then supp u must have at least r atoms; moreover, if supp u has exactly r atoms, then
1]= ~a.
(iv) (Minimality of Norm) IIW~ll = inf{llWJI : w~ is a subnormal completion of a}.
Moreover, if W w is a subnormal completion of a and IIWJI = IIW~l, then W w --- W~;
(v) (Minimality of Moments) If W x is a subnormal completion of a, then
I tn d#a(t) -< I tnd#w(t)
for all n > O.
PROPOSITION 2.10. ([CF1, Corollary 3.18])
Suppose that
a:a0,...,a2k
admits a subnormal completion, let r := rank(a), and let ~ := ~b a.
W&, where a (2r-2)
(i) w(a(2r_2))~ = :%'""%r-2"
(ii) v(j,k) = r + ... + Cr_lV(j-l,k) (r _< j < k+l).
(iii) {v(0,k), ..., v(r-l,k)} is a basis for the column space of A(k), and rank(A(k)) = r.
(iv) If r = O, then {v(1,k), ... , v(r-l,k)} is a basis for the column space of S(k), and
rank(B(k)) = r-1.

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380 Curto and Fialkow
THEOREM 2.11.
and .for each
(DENSITY THEOREM, [CF1, Theorem 4.2])
n > 0 let
a (n) : a0,...,oL n.
Let
a: a0,al,... ,
If W a is subnormal, then
IIW, - 0 (n
|
w
In this section we use matrix positivity techniques to study the stability of
k-hyponormality and subnormality for weighted shifts when an extra weight is attached at
the beginning of the weight sequence.
TIlE EXTENSION PROBLEM FOR WEIGHTED SHIFTS
DEFINITION 3.1. Let a : aO,al,.., be a weight sequence, let x > 0 and let
xa: x, a0,al,.., be the augmented weight sequence. We say that Wxa is an extension
(or 1---step extension) of W a. Observe that Wxa V{el'e2'"'} = Wa, so Wxa is a
bona fide extension of W a.
PROBLEM 3.2. (THE EXTENSION PROBLEM) Let a:o~0,O~l,.., be a
weight sequence, let k > 1, and assume that W a is k-hyponormal. Describe the sets
HE(a;m) := {x: Wxa is m-hyponormal} (1 < m < k).
We begin with a useful working criterion for the existence of m-hyponormal
extensions.
W Ot
only if the Hankel matrix
is positive.
LEMMA 3.3. Let a : aO,al,.., be a weight sequence, let k >_ 1, and assume that
is k-hyponormal. If x > 0 and 1 < m <_ k, then Wxa is m-hyponormal if and
"1 x270 ..- X27m_l
2 . 2
x 70 x271 .. x 71n
2
X27m_l X27m ... x 72m_ 1

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Curto and Fialkow 381
PROOF. Since Wxa V{el"" = W
,en} a
straightforward application of [Cul, Corollary 14].
is m-hyponormal, the result is a
We observe from Lemma 3.3 that the existence of an m-hyponormal extension
depends only on positivity properties of ~/0,...,72m_1 (we need not check the full moment
sequence).
PROPOSITION 3.4. Let a: aO,al,.., be a weight sequence, let
assume that W a is k-hyponormal. Let 1 <_ m <_ k.
(i) If v(0,m-1) e R(B(m-1)) then HE(a; m) = (0,Hm], where
1 = w*B(m-1)w,
m
for any vector w satisfying B(m-1)w = v(0,m-1).
(It m is the modulus of m-hyponormaIity.)
(ii) If v(0,m-1) I~ R(S(m-1)), then HE(a; m) = r
k > 1, and
PROOF. Proposition 2.2 implies that
w*B(m-1)w v(0, m-l)*] >_ 0,
v(0,m-1) B(m--1) J
and since
H m := (w*B(m-1)w) -1/2 > 0. By Lemma 3.3, Wxa is m-hyponormal if and only if
v(0,m-1) ~ 0, Proposition 2.1 shows that w*B(m-1)w > 0; thus
so Proposition 2.9. (iii)
1___
x 2
v(0 m-l)*"
,(0 ,m-l) B(m--1)
> 0,
now shows that Wxa is m-hyponormal if and only if
0<xSH m. i

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382 Curto and Fialkow
COROLLARY 3.5. With the notation of Proposition 3.~, assume that B(m-1)
is inver~ible. Then tt m = (v(0,m-1)*B(m-1)-lv(0,m-1) -1/2.
PROOF. Apply Proposition 3.4 and Proposition 2.2 (iv). i
EXAMPLES 3.6. Let a : a0,al,.., be a weight sequence, assume that W a is
2-hyponormal, and let m = 2.
(i) If a O=a l=a 2 then H 2=a 0.
(ii) If a 0 < a 1 = a2, then B(1) is singular and
[70] ~ R(B(I))' 71
so
2-hyponormal shift satisfies a n = an+ 1 for some n > 0, then aj =a n for all j,
HE(a; 2)= r (Incidentally, this provides a new proof of the fact that if a
1 _< j < n; cf. [Cul, Corollary 6].)
(iii) If a 0 < a 1 < a 2 then B(I) is invertible, and by Corollary 3.5,
(3.1) H 2 = (v(0,1)*B(1)-lv(0,1)) -1/2
[- [7
L7173-72 71] LT1JJ
.73+7~_27172]-1/2 2 2
[ cOal (4_a~) ]1/2
--
= 2 2
7173_7 ~ J [ala2+ 4 ~ 2 2/
aO-zaOalJ
For instance, if 4 : ~' ~ ~ and 4 : 4, which occurs in the restriction of the
Bergman shift B+ to V{el,e2,...}), then H 2 = ~, in agreement with the calculation of
HE(c~; 2) in [Cul, Proposition 7(b)]. i
There is an alternative way to obtain H 2 which makes contact with the ideas in
[CF1, Section 3]. Assume that W a is 2-hyponormal, and observe that the subnormal
completion criterion implies that a(2) : a0'~l'~2 is an initial segment of weights having a

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Curto and Fiaikow 383
recursive subnormal completion W(a(2))^. As in the construction of W(a(2))^ ,
obtain the numbers 00 and r such that
[
73
72 = r + r
r + r
A straightforward calculation shows that
2, 2
~i (a2-aO)
2
al-a 0
2,
(3.2) r = 2 and r =
2 2
aOa 1 ( a2-a 1)
2 2
al-a 0
2 2
In terms of the weights, we have
r
~ = r +--~
a I
2 r
%
al = r + --~
(cf. [CF1, Example 3.12]). To get H2,
so
(3.3)
run the recursion "backwards" one step:
r
r r r
/ %alta2-~l) [
=| 2 2. 4.
Lal a2-t-a0-za0alJ
2 2, 2 2, ~1/2
H 2 = [~0--~lJ
2 2|
That this is a legitimate way to compute H 2 is the subject of our next result.
we
THEOREM 3.7. (THE EXTENSION PRINCIPLE, PART I) Let a : a0,al,...
be a weight sequence, let k > _ 1, and assume that W a is k-hyponormal. Let 1 < _ m < k
and let a (2m-2) : a0,...,a2m_2
be the initial segment consisting of (2m-l)
weights.
Further, let W(a(2m_2))^
be the recursive subnormal completion of a (2m-2),
let