New York Journal of Mathematics
New York J. Math. 16 (2010) 489–505.
Wandering subspaces and the Beurling
type theorem. II
Kei Ji Izuchi, Kou Hei Izuchi and Yuko Izuchi
Abstract. Let H 2 be the Hardy space over the bidisk. Let ϕ(w) be
a nonconstant inner function. We denote by [z − ϕ(w)] the smallest
invariant subspace for both operators Tz and Tw containing the function
z − ϕ(w). Aleman, Richter and Sundberg showed that the Beurling
type theorem holds for the Bergman shift on the Bergman space. It is
known that the compression operator Sz on H 2 [z − w] is unitarily
equivalent to the Bergman shift, so the Beurling type theorem holds
for Sz on H 2 [z − w]. As a generalization, we shall show that the
Beurling type theorem holds for Sz on H 2 [z − ϕ(w)]. Also we shall
prove that the Beurling type theorem holds for the fringe operator Fw
on [z − w] z[z − w] and for Fz on [z − ϕ(w)] w[z − ϕ(w)] if ϕ(0) = 0.
Contents
1. Introduction
2. Wandering subspaces
3. The Beurling type theorem for Sz
4. The fringe operators
References
489
491
494
496
504
1. Introduction
Let T be a bounded linear operator on a Hilbert space H. For a subset
E of H, we denote by [E] the smallest invariant subspace for T containing
E. Let M ⊂ H be an invariant subspace for T . We denote by M T M
the orthogonal complement of T M in M . The space M T M is called a
wandering subspace of M for the operator T . We have [M T M ] ⊂ M . We
say that the Beurling type theorem holds for T if [M T M ] = M for all
invariant subspaces M of H for T . Our basic problem is to find operators
for which the Beurling type theorem holds.
Received July 26, 2010; revised October 24, 2010.
2000 Mathematics Subject Classification. Primary 47A15, 32A35; Secondary 47B35.
Key words and phrases. Beurling type theorem, Hardy space over the bidisk, invariant
subspace, wandering subspace, fringe operator.
The first author is partially supported by Grant-in-Aid for Scientific Research
(No.21540166), Japan Society for the Promotion of Science.
ISSN 1076-9803/2010
489
490
K. J. IZUCHI, K. H. IZUCHI AND Y. IZUCHI
Let D be the open unit disk in the complex plane C. We denote by
H 2 (z) the Hardy space on D with variable z. Let Tz be the multiplication
operator on H 2 (z) by the coordinate function z. The Beurling theorem [3]
says that M = [M Tz M ] holds for all invariant subspaces M of H 2 (z) for
Tz . Let L2a (z), the Bergman space, be the Hilbert space consisting of square
integrable analytic functions on D with respect to the normalized Lebesgue
measure on D. Let B be the Bergman shift on L2a (z), that is, Bf (z) = zf (z)
for f ∈ L2a (z). It is known that the dimension of wandering subspaces of
invariant subspaces in L2a (z) for B ranges from 1 to ∞ (see [2, 7, 9]). In
[1], Aleman, Richter and Sundberg proved that the Beurling type theorem
holds for the Bergman shift B. In [16], Shimorin showed that if T : H → H
satisfies the following conditions:
(a) kT
yk2 ≤ 2(kxk2 + kT yk2 ), x, y ∈ H;
T x+
n
(b) {T H : n ≥ 0} = {0};
then the Beurling type theorem holds for T . As an application of this theorem, Shimorin gave a simpler proof of the Aleman, Richter and Sundberg
theorem. Later, different proofs of the the Beurling type theorem are given
in [13, 14, 17]. Recently, the authors [10] proved the following.
Theorem A. Suppose T : H → H satisfies the following conditions:
(i) kT xk2 + kT ∗2 T xk2 ≤ 2kT ∗ T xk2 , x ∈ H;
(ii) T is bounded below, i.e., there is c > 0 satisfying that kT xk ≥ ckxk
for every x ∈ H;
(iii) kT ∗n xk → 0 as n → ∞ for every x ∈ H.
Then the Beurling type theorem holds for T .
Also it was pointed out that conditions (i), (ii) and (iii) in Theorem A
are equivalent to conditions (a) and (b) in Shimorin’s theorem.
Let H 2 = H 2 (D2 ) be the Hardy space over the bidisk D2 . We identify a
function in H 2 with its boundary function on the distinguished boundary
Γ2 of D2 , so we think of H 2 as a closed subspace of the Lebesgue space
L2 (Γ2 ). We use z, w as variables in D2 . We note that the Hardy space H 2
coincides with the closed tensor product H 2 (z) ⊗ H 2 (w). Let Tz and Tw be
multiplication operators on H 2 by z and w. A closed subspace M of H 2 is
called invariant if Tz M ⊂ M and Tw M ⊂ M . For a subset E of H 2 , we
denote by [E] the smallest invariant subspace of H 2 containing E. For a
subspace E of H 2 , we denote by PE the orthogonal projection from L2 (Γ2 )
onto E. See books [4, 15] for the study of the Hardy space H 2 .
Let M be an invariant subspace of H 2 . Since Tz is an isometry on M , by
the Wold decomposition theorem we have
∞
X
M=
⊕(M zM )z n .
n=0
So many properties of the invariant subspace M are considered to be encoded
in M zM . To study M zM , Yang defined the fringe operator Fw on
THE BEURLING TYPE THEOREM. II
491
M zM by
Fw f = PM zM Tw f, f ∈ M zM,
and studied the properties of Fw (see [21, 23, 24]). Similarly, we may define
the fringe operator Fz on M wM .
Let N = H 2 M . Then Tz∗ N ⊂ N and Tw∗ N ⊂ N . Let Sz and Sw be the
compression operators on N defined by
S z f = PN T z f
and Sw f = PN Tw f,
Sz∗
Sw∗
Tz∗ |N
f ∈ N.
Tw∗ |N .
We note that
=
and
=
One of the most interesting invariant subspaces of H 2 is [z − w]. It is
known that Sz = Sw on H 2 [z − w] and Sz is unitarily equivalent to the
Bergman shift on L2a (D) (see [6, 12, 17, 18, 19, 20, 22]). So by the Aleman,
Richter and Sundberg theorem, the Beurling type theorem holds for the
operators Sz and Sw on H 2 [z − w].
As generalized spaces of [z − w], we have invariant subspaces Mϕ :=
[z − ϕ(w)] for nonconstant inner functions ϕ(w). We put Nϕ = H 2 Mϕ . The space Nϕ has been studied by Yang and the first author in [11,
12]. In Section 2, as an application of Theorem A we shall prove that
the Beurling type theorem holds for some other unilateral operators. And
we give a sufficient condition on unilateral weighted shifts Wc for which
dim(M Wc M ) = 1 for every invariant subspace for Wc . In Section 3, as
an application of Section 2 we shall prove that the Beurling type theorem
holds for the operator Sz on Nϕ . In Section 4, we prove that the Beurling
type theorem holds for the fringe operator Fw on [z − w] z[z − w]. And
also the Beurling type theorem holds for the fringe operator Fz on Mϕ wMϕ for every inner function ϕ(w) with ϕ(0) = 0. In this case, we have
dim(M Fz M ) = 1 for every invariant subspace M of Mϕ wMϕ for Fz .
2. Wandering subspaces
Let B be the Bergman shift on L2a (z). We put
√
en (z) = n + 1z n , n ≥ 0.
Then {en (z)}n≥0 is an orthonormal basis of L2a (z). We have B ∗ e0 (z) = 0,
√
√
n+1
n
∗
en+1 (z) and B en (z) = √
en−1 (z), n ≥ 1.
Ben (z) = √
n+2
n+1
Hence
n+1
en (z),
B ∗ Ben (z) =
n+2
and
√ √
n n+1
∗2
B Ben (z) =
en−1 (z), n ≥ 1.
n+2
By these equalities, we have
kBf ||2 + kB ∗2 Bf k2 = 2kB ∗ Bf k2
492
K. J. IZUCHI, K. H. IZUCHI AND Y. IZUCHI
for every f (z) ∈ L2a (z) (see [10]). Books [5, 8] are nice references for the
study of the Bergman space.
Let H be a separable Hilbert space with an orthonormal basis {τn }n≥0 .
Let c = {cn }n≥0 be a sequence of positive numbers with supn cn < ∞.
Let Wc be a unilateral weighted shift on H defined by Wc τn = cn τn+1 for
n ≥ 0. We have Wc∗ τ0 = 0 and Wc∗ τn = cn−1 τn−1 for n ≥ 1. We note that
∗
{Wc τP
n : n ≥ 0} and {Wc τn : n ≥ 1} are orthogonal systems. For x ∈ H and
∞
x = n=0 an τn , we have
2
∞
∞
X
X
2
|an |2 c2n .
an cn τn+1 =
kWc xk = n=0
n=0
Then Wc is a bounded linear operator on H and Wc is bounded below if
and only if inf n cn > 0.
Theorem 2.1. For another Hilbert space E, let E ⊗H be the tensor product
of E and H. We define a bounded linear operator T = I√⊗ Wc on E ⊗ H
by T (x ⊗ τn ) = x ⊗ Wc τn for x ∈ E and n ≥ 0. If 1/ 2 ≤ c0 ≤ 1 and
1 ≤ c2n (2 − c2n−1 ) for every n ≥ 1, then the Beurling type theorem holds for
T.
Proof. First, we prove that cn ≤ 1 for every n ≥ 0. To prove this, suppose
that cm > 1 for some m ≥ 1. Since 1 ≤ c2m+1 (2 − c2m ), we have c2m < 2.
Since 0 < c4m − 2c2m + 1, we have c2m < 1/(2 − c2m ) ≤ c2m+1 . Thus we get
cm < cm+1 < cm+2 < · · · . Since supn cn < ∞, cn → α as n → ∞ for
some 0 < α < ∞. Then 1/(2 − α2 ) = α2 , so α = 1. This contradicts with
1 < cm < α.
√
Since 1 ≤ c2n (2−c2n−1 ), we have 1/ 2 ≤ cn for every n ≥ 0. Let f ∈ E⊗H.
P∞
P∞
2 =
2
We may write f = n=0 xn ⊗τn for some xn ∈ E with kf
k
n=0 kxn k <
P
∞
∞. Since Wc τn ⊥ Wc τk for n 6= k, we have kT f k2 = n=0 kxn k2 kWc τn k2 ,
so kf k2 /2 ≤ kT f k2 ≤ kf k2 . Then T is bounded below. We have
2
∞
X
kT ∗k f k2 = xn ⊗ Wc∗k τn n=k
2
∞
X
=
xn ⊗ (cn−1 cn−2 · · · cn−k )τn−k n=k
≤
∞
X
kxn k2
n=k
→0
as k → ∞.
We have also
Tf =
∞
X
n=0
cn (xn ⊗ τn+1 ),
∗
T Tf =
∞
X
n=0
c2n (xn ⊗ τn ),
THE BEURLING TYPE THEOREM. II
493
and
T ∗2 T f =
∞
X
c2n cn−1 (xn ⊗ τn−1 ).
n=1
Hence
2
∗2
2
kT f k + kT T f k =
c20 kx0 k2
+
∞
X
c2n (1 + c2n c2n−1 )kxn k2
n=1
and
∗
2
2kT T f k =
∞
X
2c4n kxn k2 .
n=0
Therefore
2kT ∗ T f k2 − kT f k2 + kT ∗2 T f k2
∞
X
= c20 (2c20 − 1)kx0 k2 +
c2n c2n (2 − c2n−1 ) − 1 kxn k2
n=1
≥0
by the assumption.
Applying Theorem A, we get the assertion.
Remark
2.2. Let E = C. We shall consider the extremal case of conditions
√
1/ 2 ≤ c0 ≤ 1 and 1 ≤ c2n (2 − c2n−1 ). Take c0 = 1 and inductively we take
cn such that 1 = c2n (2 − c2n−1 ). Then we have cn = 1 for every n ≥ 0. In this
Q∞
2 (z), W = T , and
case, we may think
that
H
=
H
c
z
i=0 ci = 1 > 0.
√
Take c0 = 1/ 2 and inductively we take cn such that 1 = c2n (2 − c2n−1 ).
√
√
for every n ≥
We have cn = n + 1/ n + 2 Q
√0. In this case, we may think
n
2
that H = La (z), Wc = B, and i=0 ci = 1/ n + 2 → 0 as n → ∞.
Corollary 2.3. Let E be a Hilbert space. Then the Beurling type theorem
holds for I ⊗ B on E ⊗ L2a (z).
We shall give a sufficient condition on c = {cn }n≥0 for which
dim(M Wc M ) = 1
for every invariant subspace M of H for Wc . Let {αn }n≥0 be a sequence of
positive numbers and α0 = 1. We define a linear map
V : span{z n : n ≥ 0} → H
by V z n = αn τn for every n ≥ 0.
LemmaQ2.4. We have that V Tz = Wc V on span{z n :Q
n ≥ 0} if and only if
n
αn+1 = i=0 ci for every n ≥ 0. In this case, if 0 < ∞
i=0 ci < ∞, then V
has a bounded linear extension Ve : H 2 (z) → H satisfying that Ve is invertible
and Ve Tz = Wc Ve .
494
K. J. IZUCHI, K. H. IZUCHI AND Y. IZUCHI
Proof. We have V Tz z n = Wc V z n if and only if αn+1 τn+1 = αn cQ
n τn+1 .
n
Hence V Tz = Wc V on span{z n : n ≥ 0} if and only if αn+1
=
i=0 ci
Q∞
for every n ≥ 0. In this case, moreover suppose that 0 < i=0 ci < ∞.
Then V is bounded and bounded below on span{z n : n ≥ 0}. Hence V has
a bounded linear extension Ve : H 2 (z) → H. It is easy to see that Ve is
invertible and Ve Tz = Wc Ve .
We denote by Lat(Wc ) and Lat(Tz ) the lattice of invariant subspaces for
Wc on H and Tz on H 2 (z), respectively. We write Lat(Wc ) ∼
= Lat(Tz ) if
Lat(Wc ) and Lat(Tz ) have the same lattice structure.
Q
Theorem 2.5. If 0 < ∞
i=0 ci < ∞, then dim(M Wc M ) = 1 for every
invariant subspace M for Wc . Moreover we have Lat(Wc ) ∼
= Lat(Tz ).
Proof.
Q Let M be a nonzero invariant subspace M for Wc . Let α0 = 1 and
αn = n−1
i=0 ci for n ≥ 1. By Lemma 2.4, there is a bounded linear operator
2
e
V : H (z) → H satisfying Ve z n = αn τn for every n ≥ 0, Ve is invertible and
Ve Tz = Wc Ve . Then we have
Tz Ve −1 M = Ve −1 Wc M ⊂ Ve −1 M.
Hence Ve −1 M is an invariant subspace for Tz . By the Beurling theorem,
Ve −1 M = θ(z)H 2 (z) for an inner function θ(z), so M = Ve θ(z)H 2 (z). Since
Ve Tz = Wc Ve , M is an invariant subspace for Wc generated by Ve θ(z). Therefore we get dim(M Wc M ) = 1.
For an inner function θ1 (z), Ve θ1 (z)H 2 (z) is an invariant subspace for Wc .
Thus Lat(Wc ) ∼
= Lat(Tz ).
3. The Beurling type theorem for Sz
Let ϕ(w) be a nonconstant inner function,
Mϕ = [z − ϕ(w)]
and Nϕ = H 2 Mϕ .
Let Tϕ be the multiplication operator on H 2 (w) by ϕ(w). Its adjoint operator Tϕ∗ is represented by Tϕ∗ f = PH 2 (w) ϕf, f ∈ H 2 (w). In [11], Yang and
the first author showed that
(∞
)
∞
X
X
kTϕ∗n f k2 < ∞ .
Nϕ =
⊕(Tϕ∗n f (w))z n : f ∈ H 2 (w),
n=0
Let
n=0
Pn
z i wn−i
√
σn (z, w) = i=0
, n ≥ 0.
n+1
We note that σ0 (z, w) = 1. It is known that {σn }n≥0 is an orthonormal
basis of Nw = H 2 [z − w], the special case ϕ(w) = w. If we define the
operator V : Nw → L2a (z) by V σn = σn (z, z), then V is a unitary operator
and Sz = Sw = V ∗ BV .
THE BEURLING TYPE THEOREM. II
495
Since Tϕ is an isomerty on H 2 (w), by the Wold decomposition theorem
we have
∞
X
H 2 (w) =
⊕ϕ(w)n H 2 (w) ϕ(w)H 2 (w) .
n=0
2
2
Let {λk (w)}m
k=0 be an orthonormal basis of H (w) ϕ(w)H (w), where
0 ≤ m ≤ ∞. Also let
Ek,n (z, w) = λk (w)σn (z, ϕ(w)) ∈ H 2 ,
0 ≤ k ≤ m, n ≥ 0.
In [12], Yang and the first author proved the following.
Lemma 3.1. The set {Ek,n : 0 ≤ k ≤ m, n ≥ 0} is an orthonormal basis of
Nϕ and
√
n+1
Sz Ek,n = √
Ek,n+1 .
n+2
We define the operator
U : Nϕ → H 2 (w) ϕ(w)H 2 (w) ⊗ L2a (z)
by
U Ek,n = λk (w) ⊗ en (z).
Then U is clearly a unitary operator, and by Lemma 3.1 one easily checks
that
Sz = U ∗ (I ⊗ B)U and Sz∗ = U ∗ (I ⊗ B ∗ )U.
By Corollary 2.3, we have the following theorem.
Theorem 3.2. The Beurling type theorem holds for the operator Sz on Nϕ
for every nonconstant inner function ϕ(w).
Let Sϕ = PNϕ Tϕ |Nϕ . Then Sϕ∗ = PNϕ Tϕ∗ |Nϕ . Since Tz∗ = Tϕ∗ on Nϕ , we
have Sz∗ = Sϕ∗ , so Sz = Sϕ . By Theorem 3.2, we have the following.
Corollary 3.3. The Beurling type theorem holds for Sϕ on Nϕ for every
nonconstant inner function ϕ(w).
If ϕ(w) 6= aw, |a| = 1, then Sz 6= Sw . There are some differences between
the operators Sz and Sw on Nϕ .
Proposition 3.4. Let ϕ(w) = w2 ϕ0 (w) for an inner function ϕ0 (w). Then
kSw f k2 + kSw∗2 Sw f k2 > 2kSw∗ Sw f k2
for some f ∈ Nϕ .
Proof. The set {1, ϕ0 (w), wϕ0 (w)} is contained in H 2 (w) ϕ(w)H 2 (w).
Let f (w) = wϕ0 (w) ∈ Nϕ . Then wf (w) = ϕ(w). Let
r(w) ∈ H 2 (w) ϕ(w)H 2 (w)
with r(w) ⊥ 1.
496
K. J. IZUCHI, K. H. IZUCHI AND Y. IZUCHI
Then ϕ(w) ⊥ r(w)ϕ(w)n , and by Lemma 3.1 r(w)σn (z, ϕ(w)) ∈ Nϕ for
every n ≥ 0. We have
Pn
z i ϕ(w)n−i
σn (z, ϕ(w)) = i=0√
∈ Nϕ .
n+1
For every n ≥ 0, we have
*
+
n
X
1
i
n−i
wf (w), r(w)σn (z, ϕ(w)) = √
ϕ(w), r(w)
z ϕ(w)
n+1
i=0
=√
1
hϕ(w), r(w)ϕ(w)n i
n+1
= 0.
By Lemma 3.1, σn (z, ϕ(w)) and ϕ0 (w)σ0 (z, ϕ(w)) = ϕ0 (w) are contained
in Nϕ . For j 6= 1, since ϕ(0) = 0 we have also
hwf (w), σj (z, ϕ(w))i = √
1
hϕ(w), ϕ(w)j i = 0.
j+1
Hence
Sw f (w) = hwf (w), σ1 (z, ϕ(w))iσ1 (z, ϕ(w))
D
ϕ(w) + z E
√
= ϕ(w),
σ1 (z, ϕ(w))
2
1
= √ σ1 (z, ϕ(w)).
2
We have
ϕ(w) + z 1
1
√
Tw∗ Sw f (w) = √ Tw∗
= wϕ0 (w) ∈ Nϕ .
2
2
2
Hence Sw∗ Sw f (w) = 12 wϕ0 (w), so Sw∗2 Sw f (w) = 12 ϕ0 (w). Therefore
1 1
1
+ > = 2kSw∗ Sw f (w)k2 .
2 4
2
By Proposition 3.4, we may not apply Theorem A for Sw on Nϕ . So we do
not know whether or not the Beurling type theorem holds for the operator
Sw on Nϕ .
kSw f (w)k2 + kSw∗2 Sw f (w)k2 =
4. The fringe operators
Let M be a nonzero invariant subspace of the Hardy space H 2 and N =
M . One easily checks the following.
H2
Lemma 4.1. For f ∈ M , f ∈ M zM if and only if Tz∗ f ∈ N .
We define the fringe operators Fw on M zM by
Fw = PM zM Tw |M zM
THE BEURLING TYPE THEOREM. II
497
and Fz on M wM by Fz = PM wM Tz |M wM . Let ϕ(w) be a nonconstant
inner function. We use the same notations as the ones given in Section 3.
2
2
Let {λk (w)}m
k=0 be an orthonormal basis of H (w) ϕ(w)H (w). Let
√
zσn (z, ϕ(w)) − n + 1ϕ(w)n+1
√
, n ≥ 0.
En =
n+2
Then we may verify the following lemma (see [12]).
Lemma 4.2. The set {λk (w)En : 0 ≤ k ≤ m, n ≥ 0} is an orthonormal
basis of Mϕ zMϕ .
Theorem 4.3. The Beurling type theorem holds for the fringe operator Fw
on [z − w] z[z − w]. Moreover, dim(M Fw M ) = 1 for every invariant
subspace M for Fw .
Proof. Let
1
Xn = √
n+2
Pn
z
i=0
√
i+1 w n−i
n+1
−
√
!
n + 1w
n+1
for every n ≥ 0. By Lemma 4.2, {Xn }n≥0 is an orthonormal basis of [z −
w] z[z − w] (see also [6, 17, 18]). It is not difficult to see that wXn ⊥ Xj
for j 6= n + 1. Hence
Fw Xn = hwXn , Xn+1 iXn+1
*
!
Pn
i+1 w n+1−i
√
z
1
i=0√
= √
− n + 1wn+2 ,
n+2
n+1
!+
Pn+1 i+1 n+1−i
√
z w
1
n+2
i=0
√
√
Xn+1
− n + 2w
n+3
n+2
√
√
1
n+1
√
√
√
=√
+ n + 1 n + 2 Xn+1
n+2 n+3
n+1 n+2
√
√
n+1 n+3
=
Xn+1 .
n+2
Let
√
n+1 n+3
cn =
.
n+2
√
√
Then c0 = 3/2, so 1/ 2 < c0 , and cn < 1 for every n ≥ 0. It is not difficult
to check c2n (2 − c2n−1 ) ≥ 1. By Theorem 2.1, we get the first assertion.
We have
√ √ √ √ √
√
√
√
k
Y
3 2 4 3 5
k+1 k+3
1 k+3
···
=√ √
.
cn =
2
3
4
k+2
k
+
2
2
n=0
√
Q∞
Hence n=0 cn = 1/ 2. By Theorem 2.5, we get the second assertion. √
498
K. J. IZUCHI, K. H. IZUCHI AND Y. IZUCHI
Since
1
σn (z, w),
n+2
Tz∗ ([z − w] z[z − w]) is dense in H 2 [z − w]. As mentioned in the
introduction, Sw on H 2 [z − w] is unitary equivalent to the Bergman shift
B on L2a (D). We note that the dimension of wandering subspaces of invariant
subspaces in L2a (z) for B ranges from 1 to ∞.
Tz∗ Xn = √
Proposition 4.4. Let ϕ(w) = w2 ϕ0 (w) for an inner function ϕ0 (w). Then
kFw f k2 + kFw∗2 Fw f k2 > 2kFw∗ Fw f k2
for some f ∈ Mϕ zMϕ .
Proof. We have
{1, ϕ0 (w), wϕ0 (w)} ⊂ H 2 (w) ϕ(w)H 2 (w)
By Lemma 4.2, En , ϕ0 (w)En , wϕ0 (w)En are contained in Mϕ zMϕ for
every n ≥ 0. Let f = wϕ0 (w)E0 . Then
wf = ϕ(w)E0 =
ϕ(w)z − ϕ(w)2
√
.
2
Let
r(w) ∈ H 2 (w) ϕ(w)H 2 (w) with r(w) ⊥ 1.
Then ϕ(w) ⊥ r(w)ϕ(w)n , and by Lemma 4.2 we have r(w)En ∈ Mϕ zMϕ
for n ≥ 0. Hence for every n ≥ 0, we have
1
hwf, r(w)En i = √ √
ϕ(w)z − ϕ(w)2 ,
2 n+2
Pn
z i+1 ϕ(w)n−i √
n+1
i=0 √
− n + 1r(w)ϕ(w)
r(w)
n+1
1
hϕ(w), r(w)ϕ(w)n i
√
=√ √
n+1
2 n+2
√
2
n+1
+ n + 1hϕ(w) , r(w)ϕ(w)
i
= 0.
For j 6= 1, since ϕ(0) = 0 we have also
1
hϕ(w), ϕ(w)j i
√
hwf, Ej i = √ √
j+1
2 j+2
p
2
j+1
+ j + 1hϕ(w) , ϕ(w) i
= 0.
Hence
√
√
1
3
1
Fw f = hwf, E1 iE1 = √ √ √ + 2 E1 =
E1 .
2
2 3
2
THE BEURLING TYPE THEOREM. II
We have
499
√
Tw∗ Fw f
3 ∗ 1 ϕ(w)z + z 2 √
2
√
=
T √
− 2ϕ(w)
2 w
3
2
1 wϕ0 (w)z √
√
− 2wϕ0 (w)ϕ(w) .
=
2
2
Let
r1 (w) ∈ H 2 (w) ϕ(w)H 2 (w)
Then wϕ0 (w) ⊥ r1
with r1 (w) ⊥ wϕ0 (w).
(w)ϕ(w)n
for n ≥ 0. Hence for every n ≥ 0, we have
*
1
wϕ0 (w)z √
√
hTw∗ Fw f, r1 (w)En i = √
− 2wϕ0 (w)ϕ(w),
2 n+2
2
+
Pn
i+1 ϕ(w)n−i
√
z
r1 (w) i=0 √
− n + 1r1 (w)ϕ(w)n+1
n+1
1
1
√ √
= √
hwϕ0 (w), r1 (w)ϕ(w)n i
2 n+2
2 n+1
√ √
n
+ 2 n + 1hwϕ0 (w), r1 (w)ϕ(w) i
= 0.
For j > 0, since ϕ(0) = 0 we have
1
1
√ √
hTw∗ Fw f, wϕ0 (w)Ej i = √
hwϕ0 (w), wϕ0 (w)ϕ(w)j i
2 j+2
2 j+1
√ p
j
+ 2 j + 1hwϕ0 (w), wϕ0 (w)ϕ(w) i
= 0.
Hence
Fw∗ Fw f = hTw∗ Fw f, wϕ0 (w)E0 iwϕ0 (w)E0
1
1
= √ √ hwϕ0 (w), wϕ0 (w)i
2 2
2
√
+ 2hwϕ0 (w), wϕ0 (w)i wϕ0 (w)E0
√
1
1
= √ √ + 2 wϕ0 (w)E0
2 2
2
3
= wϕ0 (w)E0 .
4
Since
3
Tw∗ Fw∗ Fw f = ϕ0 (w)E0 ∈ Mϕ zMϕ ,
4
500
K. J. IZUCHI, K. H. IZUCHI AND Y. IZUCHI
we have Fw∗2 Fw f = 34 ϕ0 (w)E0 . Therefore
3 2
3 3 2
kFw f k2 + kFw∗2 Fw f k2 = +
>2
= 2kFw∗ Fw f k2 .
4
4
4
By Proposition 4.4, we may not apply Theorem A for the operator Fw on
Mϕ zMϕ . So we do not know whether or not the Beurling type theorem
holds for the operator Fw on Mϕ zMϕ .
By the symmetry of variables in [z − w] and Theorem 4.3, the Beurling
type theorem holds for the operator Fz on [z − w] w[z − w]. We may
generalize this fact as follows.
Theorem 4.5. Let ϕ(w) be an inner function with ϕ(0) = 0. Then the
fringe operator Fz on Mϕ wMϕ is unitarily equivlent to the fringe operator
Fw on [z − w] z[z − w], and the Beurling type theorem holds for Fz and
dim(M Fz M ) = 1 for every invariant subspace M of Mϕ wMϕ for Fz .
To prove this, we need some lemmas. Let ϕ(w) be an inner function with
ϕ(0) = 0. One easily checks the following lemma.
Lemma 4.6. We have Tw∗ ϕ(w) ∈ H 2 (w) ϕ(w)H 2 (w), and if λ(w) ∈
H 2 (w) ϕ(w)H 2 (w) and λ(w) ⊥ Tw∗ ϕ(w), then
Tw λ(w) ∈ H 2 (w) ϕ(w)H 2 (w).
By Lemma 3.1, Nϕ coincides with the closed linear span of
λ(w)σn (z, ϕ(w)) : λ(w) ∈ H 2 (w) ϕ(w)H 2 (w), n ≥ 0 .
By Lemma 4.6, (Tw λ(w))σn (z, ϕ(w)) ∈ Nϕ for every
λ(w) ∈ H 2 (w) ϕ(w)H 2 (w) C · Tw∗ ϕ(w)
and n ≥ 0. Let
Nϕ,0 = {f ∈ Nϕ : Tw f ∈ Nϕ }.
∗
Since ϕ(0) = 0, Tw (Tw ϕ(w)) = ϕ(w) and ϕ(w)σn (z, ϕ(w)) ∈
/ Nϕ for every
n
≥
0.
Hence
the
space
N
N
coincides
with
the
closed
linear span of
ϕ
ϕ,0
∗
(Tw ϕ(w))σn (z, ϕ(w)) : n ≥ 0 . By Lemma 3.1, we have that
(Tw∗ ϕ(w))σn (z, ϕ(w)) ⊥ (Tw∗ ϕ(w))σj (z, ϕ(w))
and
k(Tw∗ ϕ(w))σn (z, ϕ(w))k = 1. So
∗
(Tw ϕ(w))(w)σn (z, ϕ(w))
for n 6= j,
:n≥0
is an orthonormal basis of Nϕ Nϕ,0 .
One easily sees that Tw∗ (Mϕ wMϕ ) ⊥ Nϕ,0 . Therefore by Lemma 4.1,
we have the following.
Lemma 4.7. Let g ∈ Mϕ wMϕ . Then we may write
Tw∗ g
=
∞
X
n=0
an (Tw∗ ϕ(w))σn (z, ϕ(w)),
∞
X
n=0
|an |2 < ∞.
THE BEURLING TYPE THEOREM. II
501
Let
Yn = √
√
1 ϕ(w)σn (z, ϕ(w)) − n + 1z n+1 ,
n+2
n ≥ 0.
Lemma 4.8. Let ϕ(w) be an inner function with ϕ(0) = 0. Then {Yn }n≥0
is an orthonormal basis of Mϕ wMϕ .
Proof. We have
√
√
n + 1 n + 2Yn = ϕ(w) z n + z n−1 ϕ(w) + · · · + ϕ(w)n
− (n + 1)z n+1 .
Letting n = 0, we have
√
2Y0 = ϕ(w) − z ∈ Mϕ .
By induction, we shall show that Yn ∈ Mϕ for every n ≥ 0. Suppose that
√
√
k + 1 k + 2Yk = ϕ(w) z k + z k−1 ϕ(w) + · · · + ϕ(w)k
− (k + 1)z k+1 ∈ Mϕ .
We have
√
√
k + 1 k + 2ϕ(w)Yk = ϕ(w)2 z k + z k−1 ϕ(w) + · · · + ϕ(w)k
− (k + 1)z k+1 ϕ(w) ∈ Mϕ .
Then
ϕ(w)k+2 =
√
√
k + 1 k + 2ϕ(w)Yk + (k + 1)z k+1 ϕ(w)
− z k ϕ(w)2 + z k−1 ϕ(w)3 + · · · + zϕ(w)k+1 .
Hence
√
√
k + 2 k + 3Yk+1
= ϕ(w) z k+1 + z k ϕ(w) + · · · + ϕ(w)k+1 − (k + 2)z k+2
√
√
= k + 1 k + 2ϕ(w)Yk + (k + 1)z k+1 ϕ(w)
− z k ϕ(w)2 + z k−1 ϕ(w)3 + · · · + zϕ(w)k+1
+ z k+1 ϕ(w) + z k ϕ(w)2 + · · · + zϕ(w)k+1 − (k + 2)z k+2
√
√
= k + 1 k + 2ϕ(w)Yk + (k + 2)z k+1 (ϕ(w) − z) ∈ Mϕ .
This completes the induction. Thus we get Yn ∈ Mϕ for every n ≥ 0.
We have also
1
Tw∗ Yn = √
Tw∗ ϕ(w)σn (z, ϕ(w))
n+2
1
(Tw∗ ϕ(w))σn (z, ϕ(w))
because ϕ(0) = 0
=√
n+2
∈ Nϕ
by Lemmas 3.1 and 4.6.
502
K. J. IZUCHI, K. H. IZUCHI AND Y. IZUCHI
Hence by Lemma 4.1, Yn ∈ Mϕ wMϕ for n ≥ 0. Since ϕ(0) = 0 and
kϕ(w)σn (z, ϕ(w))k = 1, it is not difficult to show that kYn k = 1 for n ≥ 0.
Let 0 ≤ n < j. Then
√
ϕ(w)σn (z, ϕ(w)) − n + 1z n+1 , z j+1 = 0
and z n , ϕ(w)σj (z, ϕ(w)) = 0. So
1
ϕ(w)σn (z, ϕ(w)), ϕ(w)σj (z, ϕ(w))
√
n+2 j+2
1
=√
σn (z, ϕ(w)), σj (z, ϕ(w))
√
n+2 j+2
* n
+
j
X
X
1
√
=√
z i ϕ(w)n−i ,
z ` ϕ(w)j−`
√
√
n + 1 n + 2 j + 1 j + 2 i=0
`=0
hYn , Yj i = √
n
=√
=0
X
1
√
hϕ(w)n−i , ϕ(w)j−i i
√
√
n + 1 n + 2 j + 1 j + 2 i=0
because ϕ(0) = 0 and n < j.
Hence {Yn }n≥0 is an orthonormal system in Mϕ wMϕ .
Let g ∈ Mϕ wMϕ . By Lemma 4.7, we may write
Tw∗ g
=
∞
X
an (Tw∗ ϕ(w))σn (z, ϕ(w))
n=0
for some
P∞
2
n=0 |an |
< ∞. We have
g(z, w) = w
∞
X
!
an (Tw∗ ϕ(w))σn (z, ϕ(w))
+ g(z, 0)
n=0
=
∞
X
!
an ϕ(w)σn (z, ϕ(w))
+ g(z, 0).
n=0
Since g ∈ [z − ϕ(w)], g(ϕ(ζ), ζ) = 0 for every ζ ∈ D. Then
g(ϕ(ζ), 0) = −
=−
∞
X
n=0
∞
X
an ϕ(ζ)σn (ϕ(ζ), ϕ(ζ))
√
n + 1an ϕ(ζ)n+1 .
n=0
Hence
g(z, 0) = −
∞
X
√
n=0
n + 1an z n+1 ,
z ∈ D.
THE BEURLING TYPE THEOREM. II
503
Therefore for (z, w) ∈ D2 we get
g(z, w) =
=
∞
X
n=0
∞
X
an ϕ(w)σn (z, ϕ(w)) −
√
√
n + 1z n+1
n + 2an Yn
n=0
and
∞
X
(n + 2)|an |2 < ∞.
n=0
Thus we get the assertion.
Remark 4.9. By the last paragraph of the proof of Lemma 4.8, we have
(∞
)
∞
X
X
∗
∗
2
Tw (Mϕ wMϕ ) =
an (Tw ϕ(w))σn (z, ϕ(w)) :
(n + 2)|an | < ∞ .
n=0
n=0
Remark 4.10. If ϕ(0) 6= 0, we can prove that
√
Zn := (ϕ(w) − ϕ(0))σn (z, ϕ(w)) − n + 1(z − ϕ(0))z n ∈ Mϕ wMϕ
for every n ≥ 0. But in this case, Zn 6⊥ Zj for n 6= j.
Proof of Theorem 4.5. We note that
P
1 ni=0 z i ϕ(w)n+1−i √
√
− n + 1z n+1 ,
Yn = √
n+2
n+1
n ≥ 0.
We have Tz Yn ⊥ Yj for j 6= n + 1. For, we have
hTz Yn ,Yj i
=√
1
zϕ(w)σn (z, ϕ(w)), ϕ(w)σj (z, ϕ(w))
√
n+2 j+2
because ϕ(0) = 0
1
=√
√
n+2 j+2
*P
n
i+1
n+1−i
i=0 z√ ϕ(w)
n+1
n
=√
Pj
,
z ` ϕ(w)j+1−`
`=0 √
j+1
+
j
XX
1
√
hϕ(w)n−i , ϕ(w)j−l ihz i+1 , z ` i.
√
√
n + 1 n + 2 j + 1 j + 2 i=0 `=0
If either n − i 6= j − ` or i + 1 6= `, then
hϕ(w)n−i , ϕ(w)j−l ihz i+1 , z ` i = 0
because ϕ(0) = 0. If n − i = j − ` and i + 1 = `, then j = n + 1. Thus
Tz Yn ⊥ Yj for j 6= n + 1.
504
K. J. IZUCHI, K. H. IZUCHI AND Y. IZUCHI
Hence we get
Fz Yn = hTz Yn , Yn+1 iYn+1
=√
1
√
n+2 n+3
√
n n+1
X
X
1
√
hϕ(w)n−i , ϕ(w)n+1−` i
n + 1 n + 2 i=0 `=0
!
!
√
√
hz i+1 , z ` i + n + 1 n + 2 Yn+1
√
√
1
n+1 √
√
√
=√
+ n + 1 n + 2 Yn+1
n+2 n+3
n+2
√
n + 1 1
=√
+ 1 Yn+1
n+3 n+2
√
√
n+1 n+3
=
Yn+1 .
n+2
By the proof of Theorem 4.3, Fz on Mϕ wMϕ is unitarily equivalent to
Fw on [z − w] z[z − w]. By Theorem 4.3, we get the assertion.
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Department of Mathematics, Niigata University, Niigata 950-2181, Japan
izuchi@m.sc.niigata-u.ac.jp
Department of Mathematics, Korea University, Seoul 136-701, Korea
kh.izuchi@gmail.com
Aoyama-shinmachi 18-6-301, Niigata 950-2006, Japan
yfd10198@nifty.com
This paper is available via http://nyjm.albany.edu/j/2010/16-20.html.