On S. Grivaux' example of a hypercyclic rank one perturbation of a unitary operator

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by  Anton Baranov

ON S. GRIVAUX’ EXAMPLE OF A HYPERCYCLIC RANK ONE
PERTURBATION OF A UNITARY OPERATOR.
ANTON BARANOV, ANDREI LISHANSKII
Abstract. Recently, Sophie Grivaux showed that there exists a rank one perturba-
tion of a unitary operator in a Hilbert space which is hypercyclic. We give a similar
4
1 construction using a functional model for rank one perturbations of singular unitary
0 operators.
2

n
a
J
1. Introduction
9

A continuous linear operator T in a Fr´echet space F is said to be hypercyclic if there
]
A exists a vector f F such that its orbit Tnf ∞ is dense in F. In this case the vector
∈ { }n=0
F f is said to be hypercyclic for T. First examples of hypercyclic operator go back to
.
h G.D. Birkhoff, G.R. McLane, S. Rolewicz, while a systematic study of hypercyclicity
t phenomenon started in 1980-s with the thesis by C. Kitai and works by R.M. Gethner,
a
m G. Godefroy and J.H. Shapiro [5, 6]. We refer to the recent monographs [2, 9] for a
[ detailed account of the theory.

1 Clearly, the identity operator is one of the most ”nonhypercyclic” operators. How-
v ever, already in 1991 K.C. Chan and J.H. Shapiro [3] showed that there exists hyper-
4
cyclic operators in a Hilbert space of the form I +K, where the compact operator K
3
9 may belong to any Schatten class. It is clear that I +R can not be hypercyclic when
1
R is a finite rank operator. Still, if we replace I by a unitary operator, hypercyclicity
.
1 is possible. In 2010 S. Shkarin [14] produced an example of a unitary operator U such
0
that U +R is hypercyclic for some rank two operator R. Shkarin asked whether R can
4
1 be taken to be of rank one. A positive answer was given by S. Grivaux [8]:
:
v
Theorem 1.1 (S. Grivaux, [8]). There exists a unitary operator U in the space ℓ2 and
i
X
a rank one operator R such that U +R is hypercyclic.
r
a
The proof of this theorem is based on an ingenious elementary construction involving
a certain convergent inductive process as well as on the following sufficient condition
for hypercyclicity also obtained by Grivaux [7]. This result says that the operator is
hypercyclic if there is a certain ”continuous” family of eigenvectors with unimodular
eigenvalues.
Theorem 1.2 (S. Grivaux, [7]). Let X be a complex separable infinite-dimensional
Banach space, and let T be a bounded operator on X. Suppose that there exists a
sequence u of vectors in X having the following properties:
n n≥1
{ }
(i) u is an eigenvector of T associated to an eigenvalue λ of T, with λ = 1 and
n n n
| |
λ are all distinct;
n
(ii) span u : n 1 is dense in X;
n
{ ≥ }
The authorswere supportedbythe ChebyshevLaboratory(St. PetersburgState University)under
RF Government grant 11.G34.31.0026and by JSC ”Gazprom Neft”.
1
(iii) for any n 1 and any ε > 0, there exists m = n such that u u < ε.
n m
≥ 6 k − k
Then T is hypercyclic and even frequently hypercyclic.
The aim of the present paper is to give a proof of Theorem 1.1 by function theory
methods. Our approach is based on a functional model for rank one perturbations of
singular unitary operators. This model essentially goes back to a paper by V.V. Ka-
pustin [10]. In the form that we will use, this model appeared in [1, Theorem 0.6] in
the context of rank one perturbations of selfadjoint operators. This model translates
any rank one perturbation of a unitary operator to some concrete operator in a space of
analytic functions in the unit disk which is known as a star-invariant or model subspace
(due to its role in yet another model – that of B. Sz.-Nagy and C. Foias).
Let H2 denote the standard Hardy space in the unit disk, and let θ be an inner
function in the disk. The model (or star-invariant) subspace K of H2 is then defined
θ
as
K = H2 θH2.
θ
⊖
According to the famous Beurling theorem, any closed subspace of H2 invariant with
respect to the backward shift in H2 is of the form K . These subspaces play a dis-
θ
tinguished role in operator theory (see, e.g., [11, 12]) and in operator-related complex
analysis.
The details on the functional model for rank one perturbations will be given in the
next section. For the moment, let us mention only that in this model the family of the
eigenvectors of a rank one perturbation have a very transparent analytic meaning: they
are either the families of reproducing kernels of K or their biorthogonals.
θ
Now we state our main result which says that there exist model spaces with a cer-
tain continuous family of vectors analogous to the properties of the vectors in Theorem
1.2. In view of the functional model (see Theorem 2.1) and Theorem 1.2 this imme-
diately implies that there exists a unitary operator which has a hypercyclic rank one
perturbation.
Theorem 1.3. There exists an inner function θ in the disk such that θ(0) = 0 and θ is
analytically continuable across some nonempty open subarc of T, a function6ϕ H2 K
θ
and a sequence λ T such that the functions ∈ \
n
∈
ϕ(z)
(1) f (z) = K ,
n θ
z λ ∈
n
−
the family f is complete in K and, for any n 1 and any ε > 0, there exists m = n
n θ
{ } ≥ 6
such that f f < ε.
n m
k − k
We do not by any means claim that our proof is essentially shorter than the original
proof by Grivaux. However, we believe that the application of the model clarifies the
construction of the eigenvectors, since in the model space they have a special analytic
structure: they are necessarily of the form (1) for some ϕ H2.
∈
As the original proof from [8], our construction is also inductive. However, the pa-
rameters are chosen in a different way. In particular, the eigenvalues of the operator
U +R will be chosen to be zeros of some Herglotz function (interlacing with the spec-
trum of U). The properties of Herglotz functions will play an important role in the
construction.
2
2. Preliminaries on the functional model for rank-one perturbations
of unitary operators
2.1. Inner functions and Clark measures. Recall that a function θ is said to be
inner if it is analytic and bounded in the unit disk D and its nontangential boundary
values satisfy θ = 1 a.e. with respect to the normalized Lebesgue measure m on the
unit circle T. | |
Let H2 = H2(D) denote the Hardy space of the unit disk D, equipped with the
standard norm = . With each inner function θ we associate the model
2 L2(m)
k·k k ·k
subspace K = H2 θH2.
θ
The reproducing⊖kernel for K corresponding to a point λ D is given by
θ
∈
1 θ(λ)θ(z)
k (z) = − .
λ
1 λz
−
Since functions in K have more analyticity than general H2 functions, there may exist
θ
reproducing kernels at boundary points. In particular, one can consider k , λ I, if θ
λ
∈
(and, hence, any function in K ) have an analytic continuation across the arc I. More
θ
generally, by the results of Ahern and Clark we have k K for λ T if and only if
λ θ
∈ ∈
θ′(λ) < , the modulus of the angular derivative is finite.
| | ∞
Now we turn to Clark’s construction of orthogonal bases of reproducing kernels [4].
For each α T, the function α+θ has positive real part in D, and so there exists a finite
∈ α−θ
(singular) positive measure µα on T such that
α+θ(z) 1 1 z 2
Re = −| | dµα(τ), z D.
α θ(z) π Z τ z 2 ∈
T
− | − |
Clark’s theorem states that if, for some α, µα is purely atomic, i.e., if µα = µ δ ,
n n τn
τn ∈ T, then kτn ∈ Kθ and the system {kτn} is an orthogonal basis in Kθ.PNote also
that k 2 = θ′(t ) = 2µ−1.
k τnk2 | n | n
Let µ = µ1 be the Clark measure corresponding to α = 1. For c L2(µ), put
∈
c(τ)dµ(τ)
(2) (Vc)(z) = 1 θ(z) , z D.
− ZT 1 τ¯z ∈
(cid:0) (cid:1) −
As Clark [4] has shown, V is a unitary operator from L2(µ) onto K . Moreover, the
θ
nontangential boundary values of the function Vc exist and coincide with c µ-a.e. [13].
In particular, if k is an orthogonal basis of reproducing kernels in K , then any
{ τn} θ
f K is of the form f(z) = 1 θ(z) cnµn , where c 2µ < .
∈ θ − n 1−τ¯nz n| n| n ∞
(cid:0) (cid:1)P P
2.2. The functional model. Now we give the details of the functional model for rank
one perturbations of unitary operators. Let θ be an inner function in the disk such that
θ(0) = 0. For a function f K , the function zf is not necessarily in K , but it is easily
θ θ
seen6that zf = γ +h, whe∈re γ C is a constant and h K , and such decomposition
θ
∈ ∈
is unique (moreover, γ is a continuous functional of f). Now let ϕ H2 be a function
∈
such that
ϕ(z) ϕ(0)
(3) ϕ / K , − K .
θ θ
∈ z ∈
Then we may define the operator T = T on K by the formula
θ,ϕ θ
(4) Tf := zf γ ϕ,
f
−
3
where γ is the unique complex number such that zf γ ϕ K .
f f θ
− ∈
We are ready to present the functional model of rank one perturbations which is
analogous to [1, Theorem 0.6] (though the main ideas go back to [10]). Recall that a
unitary operator is said to be singular if its spectral measure is singular with respect
to the Lebesgue measure on T. We also assume below that U is cyclic, and so, up to a
unitary equivalence, it is the operator of multiplication by z in some space L2(ν), where
ν is a finite Borel measure on T. By σ(U) we denote the spectrum of U. Finally, we
denote by ρ(θ) the boundary spectrum of an inner function θ, that is, the complement
of the union of all open arcs I such that θ admits an analytic continuation across I.
Theorem 2.1 (functional model). Let U be a cyclic singular unitary operator such that
σ(U) = T. Then for any rank one perturbation U+R of U there exist an inner function
θ in t6he disk such that θ(0) = 0 and ρ(θ) = T, and a function ϕ H2 satisfying (3)
6 6 ∈
such that U +R is unitary equivalent to the operator T defined by (4).
Conversely, any inner function θ such that θ(0) = 0 and ρ(θ) = T, and any function
6 6
ϕ H2 satisfying (3) correspond to some rank one perturbation U + R of a cyclic
sin∈gular unitary operator U with σ(U) = T.
6
It is obvious from the definition of the operator T that if λ C is an eigenvalue of
∈
T, then the corresponding eigenvector is given by ϕ(z). Now, combining Theorems 2.1
z−λ
and 1.2 we see that the existence of a rank one perturbation follows from Theorem 1.3
(since the functions f are eigenvectors of some rank one perturbations).
n
2.3. Model spaces in the upper half-plane. Itwillbemoreconvenient toworkwith
modelspaces inthehalf-planeC rather thaninthedisk. Wecantranslateourproblem
+
to the half-plane setting since the map f˜(z) = 1 f z−i maps H2(D) to H2(C ) and
z+i · z+i +
the model space K in the disk to the model space(cid:0)K =(cid:1) H2(C ) θ˜H2(C ) in C ,
θ θ˜ + ⊖ + +
where θ˜(z) = θ z−i .
z+i
The definitio(cid:0)n an(cid:1)d the properties of the Clark measures for the model spaces in the
upper half-plane are analogous to those in the disk. Indeed, if θ is an inner function in
C , then the function α+θ has positive real part in C , the measure µ on R from its
+ α−θ +
Herglotz representation is said to be a Clark measure for K , and, again, the embedding
θ
of K into L2(µ) is a unitary operator (with one possible exception where a linear term
θ
appears in the Herglotz representation; we will exclude this case in our construction).
In particular, if the Clark measure corresponding to α = 1 is purely atomic, µ =
µ δ , then any function f K is of the form
n n tn ∈ θ
P
c µ
(5) f(z) = 1 θ(z) n n , c 2µ < ,
n n
− z t | | ∞
(cid:0) (cid:1)Xn − n Xn
and
(6) f 2 = 4π c 2µ = π f(t ) 2µ .
k kL2(R) | n| n | n | n
Xn Xn
We will often use this formula for the norm in what follows.
4
3. Proof of Theorem 1.3
3.1. Plan of the solution. By the above functional model, the problem is reduced to
the construction of functions θ and ϕ as in Theorem 1.3. As explained in Subsection
2.3, we can work on the real line and in the upper half-plane. Thus, in what follows
stands for the usual L2-norm on R.
2
k·k
We will construct:
a countable set T = t ∞ on some interval (say, in [0,1]);
• {∞n}n=1
a finite measure µ = µ δ ;
• n tn
n=1
P
an inner function θ defined by
•
∞
1+θ(z) µ
n
(7) := i ;
1 θ(z) z t
− Xn=1 − n
a function ϕ of the form
•
∞ ∞
c µ
(8) ϕ(z) := 1 θ(z) n n +1 , c 2µ < ,
n n
− (cid:20) t z (cid:21) | | ∞
(cid:0) (cid:1) Xn=1 n − Xn=1
such that for some sequence λ [0,1], we have ϕ L2(R) and, for any j 1
{ j} ⊂ z−λj ∈ ≥
and ε > 0 there exists k = j such that
6
ϕ(z) ϕ(z)
(9) < ε.
(cid:13)z λj − z λk(cid:13)2
(cid:13) − − (cid:13)
Note that in this construction µ(cid:13) is a Clark measu(cid:13)re for θ.
As in Grivaux’ paper [8] we will proceed with the construction inductively. Namely,
on the N-th step we construct t ,...,t [0,1], µ ,...,µ and functions θ defined
1 N 1 N N
∈
by
N
1+θ (z) µ
N n
(10) i := ,
1 θ (z) t z
− N Xn=1 n −
and ϕ ,
N
N
c µ
n n
(11) ϕ (z) := (1 θ (z)) 1+ ,
N N
− (cid:20) t z(cid:21)
Xn=1 n −
where c > 0. The sequences µ and c will be assumed to tend to zero very rapidly.
n n n
It is a key idea of the construction that c are taken to be positive. In this case the
n
function 1+ N cnµn (which appears in the definition of ϕ ) is a Herglotz function
n=1 tn−z N
(has positivePreal part in C ) and, therefore, it has exactly N zeros λN,λN,...λN,
+ 1 2 N
interlacing with the points t ,...,t . Choosing c sufficiently small we can control the
1 N n
location of these points.
Note that θ is a finite Blaschke product and the model space K is N-dimensional.
N θN
The measure N µ δ is a Clark measure for K . Also, ϕ / K , since 1 θ /
n=1 n tn θN N ∈ θN − N ∈
KθN, while P
ϕ (z)
fN(z) := N K .
j z λN ∈ θN
− j
5
Indeed, we have a representation of type (5),
ϕ (z) ϕ (λN) N c µ
(12) fN(z) = N − N j = 1 θ (z) n n .
j z λN − N (λN t )(z t )
− j (cid:0) (cid:1)Xn=1 j − n − n
Assume that t , µ and c , 1 n N , are already chosen. On N-th step we will
n n n 1
≤ ≤
add a point t (0,1), its mass µ and a coefficient c in the following order. First,
N N N
∈
we take t to be very close to some zero of the function ϕ . Then we choose µ to
N N−1 N
be so small that θ does not differ much from θ outside a small neighborhood of
N N−1
t . Finally, we choose c to be even much smaller than µ so that all zeros λN,...λN
N N N 1 N
from generation N almost coincide with the corresponding zeros λN−1,...λN−1 from
1 N−1
generation N 1, while the zero λN is very close to t .
− N N
Let us formally state what we need for the convergence.
(I) We will choose µ > 0 and c > 0 so that µ < 2−n and c < 2−n. These
n n n n
conditions already ensure the convergence of the functions (10) and (11) to (7) and (8)
respectively. Also, we will require that λN−1 λN < 2−N, j = 1,...,N 1 (in fact
| j − j | −
we will need much more, see (18) below).
(II) Clearly, for each N, the functions fN, j = 1,...,N, are linearly independent,
j
and so they form a basis in K . Let A be a sort of a ℓ1 basis constant for fN : for
θN N { j }
any α N , α C,
{ j}j=1 j ∈
N N
α A α fN .
Xj=1 | j| ≤ N(cid:13)(cid:13)Xj=1 j j (cid:13)(cid:13)2
(cid:13) (cid:13)
Without loss of generality we assume that A 1 and the sequence A increases. Our
N N
≥
second requirement then reads as follows:
1
(13) fN fN−1 < , j = 1,...,N 1.
k j − j k2 2N+2A −
N−1
(III) Let l(n) be a sequence of integers such that l(n) < n and l(n) takes every integer
value infinitely many times (e.g., 1,1,2,1,2,3,1,2,3,4,...). To achieve property (9),
we will introduce the third requirement:
(14) fN fN < 2−N−1, N N.
k l(N) − Nk2 ∈
3.2. Choice of the parameters. Assume that t , µ and c , n = 1,...,N 1, are
n n n
−
already chosen. First we choose the point t . Let ε be some small positive number
N N
(namely, letε 4−N−2A−1 , whereA istheconstantfrom(II)above)andconsider
N ≤ N−1 N−1
the equation
N−1
c µ
n n
(15) 1+ = ε .
N
t x
Xn=1 n −
Clearly, it has N 1 real zeros which depend continuously on ε . Hence, they can be
N
enumerated x ,..−.x in such a way that x λN−1 when ε 0.
1 N−1 j → j N →
6
Let us take ε to be so small that if we take as t the zero of (15) which is the
N N
closest to the point λN−1, then
l(N)
(16) t λN−1 < 4−3Nδ3 ,
| N − l(N)| N−1
where
δ = inf λN−1 t .
N−1 1≤j,n≤N−1| j − n|
Nowweputµ = √ε anddefinetheinnerfunctionθ by (10). Onemoreadditional
N N N
restriction on the smallness of ε will be as follows:
N
θ (z) θ (z) 1
N N−1
(17) − < , n = 1,...,N 1.
(cid:13) z tn (cid:13)2 4NAN−1 −
(cid:13) − (cid:13)
This is also poss(cid:13)ible by continuous(cid:13)dependence of these norms from ε . More precisely,
N
by the construction of t we have t λN−1 as ε 0, and so t is separated from
N N → l(N) N → N
t , 1 n N 1. Therefore, we can choose a small neighborhood I of λN−1 such that
n ≤ ≤ − l(N)
2
the integral of θN(t)−θN−1(t) over I is small. Once the interval I is fixed, we can make
(cid:12) t−tn (cid:12)
the integral ove(cid:12)r R I to b(cid:12)e arbitrarily small since θ θ (as well as θ′ θ′ )
tends to zero u(cid:12)niform\ ly ove(cid:12)r R I as ε 0. N − N−1 N − N−1
N
\ →
Finally, let us choose c . Consider the equation ϕ (x) = 0 which is equivalent to
N N
N−1
c µ c µ
n n N N
1+ + = 0.
t x t x
Xn=1 n − N −
Again, acontinuityargumentshowsthatwemayenumeratetheN zerosofthisequation
λN (whichshouldbethoughtofasfunctionsofc )sothatλN λN−1,j = 1,...,N 1,
j N j → j −
and λN t as c 0. Moreover, by the choice of t as a solution of (15), we have
N → N N → N
N−1
c µ
n n
1+ ε
t λN → N
Xn=1 n − N
and λN t as c 0. Therefore,
N → N N →
c µ c
λN t N N = N , c 0.
N − N ∼ ε √ε N →
N N
Let us now choose c to be so small that
N
1
(18) λN λN−1 < δ3 , j = 1,...,N 1,
| j − j | 43NA N−1 −
N−1
c
(19) N λN t < λN t ,
2√ε ≤ | N − N| | l(N) − N|
N
and
(20) λN λN−1 < 2−Nδ3 .
| N − l(N)| N−1
The latter estimate is possible by (16) and (18).
7
3.3. Proof of (13). To estimate the norm fN fN−1 , 1 j N 1, we write,
k j − j k2 ≤ ≤ −
using (12),
N−1
ϕ (z) ϕ (z) c µ (1 θ (z))
N−1 N n n N−1
= −
z λN−1 − z λN (t λN−1)(z t )
(21) − j − j Xn=1 n − j − n
N−1
c µ (1 θ ) c µ (1 θ )
n n N N N N
− − .
− (t λN)(z t ) − (t λN)(z t )
Xn=1 n − j − n N − j − N
Denote the last summand in (21) by h. Then h K (it is a reproducing kernel up
∈ θN
to a coefficient), and so
c 2µ
h 2 = | N| N 4µ ε .
k k2 t λN 2 ≤ N N
| N − N|
The first two summands in (21) may be rearranged into the sum of
N−1
c µ c µ
n n n n
g (z) = 1 θ (z)
1 − N (cid:18)(t λN−1)(z t ) − (t λN)(z t )(cid:19)
(cid:0) (cid:1)Xn=1 n − j − n n − j − n
and
N−1
θ (z) θ (z)
N N−1
g (z) = c µ − .
2 n n(t λN−1)(z t )
Xn=1 n − j − n
We may compute the norm of g K using Clark’s theorem (see formula (6)):
1 ∈ θN
N−1 N−1 λN λN−1 2 c 2µ
g 2 = g (t ) 2µ = | j − j | | m| m .
k 1k2 | 1 m | m t λN−1 2 t λN 2
mX=1 mX=1 | m − j | | m − j |
Thus, g does not exceed 4−NA−1 by (18).
k 1k2 N−1
Finally, g 4−NA−1 , and summing the above estimates we obtain (13).
k 2k2 ≤ N−1
3.4. Proof of (14). To estimate the norm fN fN we use again Clark’s theorem:
k l(N)− Nk2
N
ϕ ϕ 2 ϕ (t ) ϕ (t ) 2
N N N m N m
(22) = µ .
(cid:13)(cid:13)z −λNl(N) − z −λNN(cid:13)(cid:13)2 mX=1(cid:12)(cid:12)tm −λNl(N) − tm −λNN(cid:12)(cid:12) m
(cid:13) (cid:13) (cid:12) (cid:12)
Let L = λN λN . Then, by (18) and (20), t λN > L1/3 for any m = 1...N 1
N | l(N)− N| | m− j | N −
and j = 1,...,N. Now the first N 1 summands in (22) may be estimated as
−
N−1 λN λN 2 N−1
| l(N) − N| ϕ (t ) 2µ L2/3 ϕ (t ) 2µ .
t λN 2 t λN 2| N m | m ≤ N | N m | m
mX=1 | m − l(N)| | m − N| mX=1
Since by (18) and (20) L 2−3Nδ3 and ϕ (t ) = θ′ (t ) c µ = 2c , we
N ≤ N−1 | N m | | N m | m m m
conclude that the latter sum does not exceed 2−N−1.
It remains to consider the summand with the number N. We have
ϕ (t ) 2 c µ 2ε
N N = N N N = 2√ε 2−N−2.
(cid:12)t λN(cid:12) µ · t λN ≤ µ N ≤
(cid:12) N − N(cid:12) N | N − N| N
(cid:12) (cid:12)
Here we used in(cid:12)equality ((cid:12)19). The estimate for the term ϕN is analogous. Esti-
(cid:12)tN−λNl(N)(cid:12)
mate (14) is proved (cid:12) (cid:12)
(cid:12) (cid:12)
8
3.5. Convergence and completeness. To ensure the pointwise convergences
θ (z) θ(z) and ϕ (z) ϕ(z), z C , it suffices to assume only that µ <
N → N → ∈ + n n ∞
and cn 0. P
→
By the choice of the parameters in Subsection 3.2, the sequence λN converges to
j
some point λ , and it follows that fN converges to ϕ(z) pointwise in C . On the other
j j z−λj +
hand, by (13), the sequence fN H2 converges to some function f in H2 (recall that
j ∈ j
H2 = H2(C ) is the Hardy space in the upper half-plane). Since the convergence in
+
H2 implies the pointwise convergence in C , we conclude that f (z) = ϕ(z).
+ j z−λj
Next we prove that the family f is complete inK . First of all, note thatif g K ,
j θ θ
{ } ∈
∞ N
d µ d µ
n n n n
g(z) = 1 θ(z) , g (z) = 1 θ (z) ,
N N
− z t − z t
(cid:0) (cid:1)Xn=1 − n (cid:0) (cid:1)Xn=1 − n
then g g 0, N . Indeed,
N 2
k − k → → ∞
N ∞
d µ d
n n n
g(z) g (z) = θ (z) θ(z) + 1 θ(z) .
N N
− − z t − z t
(cid:0) (cid:1)Xn=1 − n (cid:0) (cid:1)n=XN+1 − n
The norm of the second sum by Clark’s theorem (see (6)) equals ∞ d 2µ ,
m=N+1| m| m
which obviously goes to zero, N , while the norm of the first suPm is small by the
→ ∞
assumption (17).
Thus, we constructed a sequence g K such that g g in L2(R). It remains
N ∈ θN N →
to approximate functions g by linear combinations of f . The method is borrowed
N j
from [8]. Since fN is a basis in K , we may write g = N α fN. Then, making
{ j } θN N j=1 j j
use of (13), we get P
N N ∞
g α f α fk fk+1
k − j jk2 ≤ | j| k j − j k2
Xj=1 Xj=1 kX=N
N
1
α 2−N g ,
j N 2
≤ 2NA | | ≤ k k
N Xj=1
which goes to 0 as N . Completeness of the family f is proved.
j
→ ∞ { }
3.6. End of the proof of Theorem 1.3. Let us complete the proof of Theorem 1.3.
We have constructed a complete sequence f = ϕ in K . It remains to verify
{ j} {z−λj} θ
that f has the property (9). Let j 1 and ε be given. Choose N such that l(N) = j
j
{ } ≥
and 2−N < ε, which is possible by the definition of the sequence l(n). Then, by (14),
fN fN < 2−N−1. Also, by (14),
k j − Nk2
∞
fN f fk fk+1 2−N−2
k N − Nk2 ≤ k N − N k2 ≤
X
k=N
and, analogously, fN f 2−N−1. Combining these estimates we obtain f
k j − jk2 ≤ k j −
f 2−N. (cid:3)
N 2
k ≤
9
4. Concluding remarks
The unitary operator U in our construction (as well as in [8]) is of a very special form.
Note that if t R is the sequence constructed in Section 3, then the spectrum of U is
n
∈
given by τ , τ = tn−i. The points t are chosen inductively so that t be close to
{ n} n tn+i n N
some of the points t , 1 n N 1, and so the set t has a certain self-similarity.
n n
≤ ≤ − { }
It seems to be a natural question, which unitary operators have hypercyclic rank one
perturbations.
Problem 1. To describe cyclic unitary operators U such that U +R is hypercyclic for
some rank one operator R.
In particular, it is not clear whether the spectrum σ(U) can have nonempty interiour
or positive measure.
Problem 2. Doesthere exist a cyclic unitary operator U such that U+R ishypercyclic
for some rank one operator R and σ(U) = T.
As in [8], in the present paper the spectral measure of U is purely atomic.
Problem 3. Construct a unitary operator U, whose spectral measure is an absolutely
continuous or a continous (i.e., without point masses) singular measure on T, such that
U +R is hypercyclic for some rank one operator R.
Notethatthefunctionalmodelappliestorankoneperturbationsofanarbitrarycyclic
unitary operator whose spectral measure is singular. Therefore, one can hope to obtain
furtherinformationabouthypercyclic rankoneperturbationsofunitaryoperatorsusing
this model.
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