A NOTE ON THE RANGE OF TOEPLITZ
OPERATORS
DRAGAN VUKOTIĆ
Abstract. A well known lemma attributed to Coburn states that
a Toeplitz operator with non-trivial kernel acting on the Hardy
space must have dense range. We show that the range of a non-zero
Toeplitz operator with non-trivial kernel must contain all polynomials and state this in a precise form.
Denote by T = {z : |z| = 1} the unit circle and by L2 (T) the standard Lebesgue space on T with respect to the normalized arc measure
dθ/(2π). When u ∈ L2 (T), we denote its Fourier coefficients in the
usual way:
Z 2π
1
e−inθ u(eiθ )dθ , n ∈ Z .
û(n) =
2π 0
2
The Hardy space H is a closed subspace of L2 (T) consisting of those
(equivalence classes of) functions f for which fˆ(−n) = 0, whenever
n = 1, 2,. . . Each such f can also be viewed as an analytic function
in the unit disk D when extended via the Poisson kernel; the radial
limits of this extension coincide with the original function almost everywhere on T. Similarly, there is a one-to-one correspondence between
the (equivalence classes of) essentially bounded measurable functions
on T with the bounded harmonic functions in D.
Let ϕ ∈ L∞ (T). Denoting by P the orthogonal projection from
2
L (T) onto H 2 , the Toeplitz operator Tϕ is defined via Tϕ (f ) = P (ϕf )
and is bounded on H 2 . The analytic extension to the disk of Tϕ f is
given by
Z 2π
1
ϕ(eiθ )f (eiθ )
(1)
Tϕ f (z) =
dθ .
2π 0
1 − ze−iθ
It is easy to check that Tϕ coincides with the zero operator on H 2 if and
only if ϕ = 0 a.e. on T. Many basic facts about Toeplitz operators can
be found in Chapter 7 of [2] and in Appendix 4 of [4], for example. An
2000 Mathematics Subject Classification. Primary 47B35; Secondary 30H05.
Key words and phrases. Toeplitz operator, kernel, range, Coburn’s lemma,
commutator.
This work is supported by MCyT grant BFM2003-07294-C02-01, Spain.
1
2
D. VUKOTIĆ
important property is given by the following lemma, usually attributed
to L. Coburn (cf. Proposition 7.24 of [2] or Lemma 43, Appendix 4 of
[4]):
When ϕ 6= 0 a.e., then either Tϕ is one-to-one or Tϕ∗ is one-to-one.
It is implicit in the proof of Theorem 4.1 in [1]. As was pointed out by
Professor Gohberg, it was proved earlier in [3] and [5] for continuous
symbols. The result can be restated as follows: a non-zero Toeplitz
operator Tϕ either has trivial kernel or dense range. The purpose of
this note is to make this statement more explicit by showing that if
Tϕ 6= 0 and Tϕ is not one-to-one, then its range contains the set P of
all analytic polynomials. More precisely, writing ker Tϕ for the kernel
of the operator Tϕ and span{S} for the linear span of a set S, we have
the following
Theorem 1. If Tϕ is a non-zero Toeplitz operator which is not oneto-one, then
Tϕ (span{P ker Tϕ }) = P .
Proof. The key point is that the commutator [Tϕ , Tz ] = Tϕ Tz −Tz Tϕ of
the shift operator Tz and Tϕ can have rank at most one. To be specific,
(2)
Tϕ (zf ) − zTϕ f = Tϕ (zf )(0) .
(By abusing the notation slightly, we use the expression such as Tϕ g(0)
to denote the constant function identically equal to the value at the
origin of the analytic extension of Tϕ g to the disk.) To check (2), just
integrate both sides of
eiθ ϕ(eiθ )f (eiθ )
ϕ(eiθ )f (eiθ )
−
z
= eiθ ϕ(eiθ )f (eiθ )
(z ∈ D)
1 − ze−iθ
1 − ze−iθ
with respect to θ from 0 to 2π and apply formula (1).
Denote by Pn the linear space of analytic polynomials of degree at
most n. By applying (2) repeatedly, we get
Tϕ (z n f ) = Tϕ (z n f )(0) + zTϕ (z n−1 f )
= Tϕ (z n f )(0) + Tϕ (z n−1 f )(0) · z + Tϕ (z n−2 f ) · z 2
n
X
(3)
= ... =
Tϕ (z k f )(0) · z n−k .
k=0
When f ∈ ker Tϕ , the term with k = 0 disappears, hence Tϕ (z n ker Tϕ )
⊂ Pn−1 . This shows that Tϕ (span{P ker Tϕ }) = span{Tϕ (P ker Tϕ )} ⊂
P.
Now for the reverse inclusion: P ⊂ span{Tϕ (P ker Tϕ )}. By assumption, some non-zero function f ∈ H 2 belongs to the kernel of Tϕ ; that
ON THE RANGE OF TOEPLITZ OPERATORS
3
is, P (ϕf ) = 0. Since ϕf 6= 0 a.e., it must have a non-vanishing Fourier
c (−n) 6= 0}.
coefficient with negative index. Let n = min{k > 0 : ϕf
c (−k) = Tϕ (z k f )(0), which is simply (1) with z = 0. By
Note that ϕf
(3) and by our choice of n, we get
c (−n) 6= 0 ,
Tϕ (z n f ) ≡ Tϕ (z n f )(0) = ϕf
so 1 = Tϕ (z n g), where g is a constant multiple of f .
We now show by induction that z n ⊂ span{Tϕ (P ker Tϕ )} for all
non-negative integers n. Indeed, we have already seen that the conk
k
stants
Pm belong to Tϕ (P ker Tϕ ). If z ∈ span{Tϕ (P ker Tϕ )} then z =
j=1 cj Tϕ (pj fj ), where cj are complex numbers, pj ∈ P and fj ∈
ker Tϕ , so by applying (2) again,
m
m
X
X
z k+1 = z
cj Tϕ (pj fj ) =
cj [Tϕ (zpj fj ) − Tϕ (zpj fj )(0)]
j=1
j=1
∈ span{Tϕ (P ker Tϕ )} .
Finally, P ⊂ span{Tϕ (P ker Tϕ )} follows, and we are done. ¤
References
[1] L. Coburn, Weyl’s theorem for nonnormal operators. Michigan Math. J. 13
(1966), 285–288; MR0201969 (34 #1846).
[2] R.G. Douglas, Banach Algebra Techniques in Operator Theory. 2nd edition,
Springer, 1998; MR1634900 (99c:47001).
[3] I.C. Gohberg, On the number of solutions of a homogeneous singular integral
equation with continuous coefficients. (Russian) Dokl. Akad. Nauk SSSR
122 (1958) 327–330; MR0098287 (20 #4748).
[4] N.K. Nikol’skiı̆, Treatise on the Shift Operator. Spectral Function Theory.
Springer-Verlag, 1986; MR0827223 (87i:47042).
[5] I.B. Simonenko, Riemann’s boundary value problem with a continuous coefficient. (Russian) Dokl. Akad. Nauk SSSR 124 (1959) 278–281; MR0112960
(22 #3805).
Departamento de Matemáticas, Universidad Autónoma de Madrid,
28049 Madrid, Spain
E-mail address: [email protected]