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Toeplitz operators on Herglotz wave functions
J. Math. Anal. Appl. 358 (2009) 364–379
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Toeplitz operators on Herglotz wave functions ✩ J.A. Barceló a , M. Folch-Gabayet b , S. Pérez-Esteva c,∗ , A. Ruiz d a
ETSI de Caminos, Universidad Politécnica de Madrid, 28040 Madrid, Spain Instituto de Matemáticas, Universidad Nacional Autónoma de México, Ciudad Universitaria, México, D.F. 04510, Mexico c Instituto de Matemáticas Unidad Cuernavaca, Universidad Nacional Autónoma de México, A.P. 273-3 ADMON 3, Cuernavaca, Mor. 62251, Mexico d Departamento de Matemáticas, Universidad Autónoma de Madrid, 28049 Madrid, Spain b
a r t i c l e
i n f o
a b s t r a c t The space of Herglotz wave functions in R2 consists of all the solutions of the Helmholtz equation that can be represented as the Fourier transform in R2 of a measure supported in the circle and with density in L 2 ( S 1 ). This space has a structure of a Hilbert space with reproducing kernel. The purpose of this article is to study Toeplitz operators with nonnegative radial symbols, defined on this space. We study the symbols defining bounded and compact Toeplitz operators as well as the Toeplitz operators belonging to the Schatten classes s p . © 2009 Elsevier Inc. All rights reserved.
Article history: Received 12 June 2008 Available online 6 May 2009 Submitted by G. Corach Keywords: Herglotz wave functions
1. Introduction and preliminaries The theory of Toeplitz operators in Bergman spaces of holomorphic functions has played an important role in operator theory and complex analysis for a number of years. Natural questions such as the characterization of all the symbols defining bounded or compact Toeplitz operators have been answered in different settings, not only for Toeplitz operators on Bergman spaces of holomorphic functions but also on spaces of harmonic and parabolic functions with reproducing kernels (see the monograph [17] and [2,8,14]). The purpose of this article is to study Toeplitz operators defined on the space of all the Herglotz wave functions in R2 . A Herglotz wave function in Rn is a solution u of the Helmholtz equation
u + u = 0
(1)
in Rn , that is represented as the Fourier transform in Rn of a measure φ dλ, where φ ∈ L 2 ( S n−1 ) and λ is the Lebesgue measure in S n−1 . Hence every Herglotz wave function can be written as
u (x) =
e ix·ω φ(ω) dλ(ω).
S n −1
✩
Research partially financed by: J.A.B., A.R.: Spanish Grant MTM2005-07652-C02-01; M.F.-G., S.P.-E.: Proyecto Conacyt-DAIC U48633-F. Corresponding author. E-mail addresses: [email protected] (J.A. Barceló), [email protected] (M. Folch-Gabayet), [email protected] (S. Pérez-Esteva), [email protected] (A. Ruiz).
*
0022-247X/$ – see front matter doi:10.1016/j.jmaa.2009.05.004
©
2009 Elsevier Inc. All rights reserved.
(2)
J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
365
Several characterizations of Herglotz wave functions have been given (see [13,11,12]). For instance a necessary and sufficient condition for an entire function u to be a Herglotz wave function is that
lim
R →∞
R
12 u (x)2 dx < ∞.
1
(3)
|x|< R
We will denote by W 2 the space of all Herglotz functions in Rn . Besides their intrinsic relevance to harmonic analysis, Herglotz wave functions play an important role in obstacle and nonhomogeneous media acoustic scattering, as the existence of certain Herglotz wave functions determines if the set of far-field-patterns (or scattering amplitudes) is dense in L 2 ( S n−1 ) (see [9]) and in scattering of elastic waves (see [10]). We will call a function u an entire function if it is a solution of the Helmholtz equation in Rn . The space W 2 with the norm given by (3) is a Hilbert space, and the map defined by (2) called extension operator, gives an isomorphism from L 2 ( S n−1 ) onto W 2 . For dimension n = 2, Alvarez, Folch-Gabayet and Pérez-Esteva proved in [1], that the extension operator
φ → φ dθ is an isomorphism from L 2 ( S 1 ) onto the space of entire solutions u of the Helmholtz equation which satisfy
2
u =
u (x)2 + ∂θ u (x)2 W (x) dx,
(4)
R2 ⊥
where W (x) = 1+|1x|3 , ∂θ u = ∇ · x|x| , x⊥ = (−x2 , x1 ) and dθ is the Lebesgue measure on S 1 (see also [4]). Hence, within the set of solutions to the Helmholtz equation, (3) is equivalent to (4) and this provided a new characterization of Herglotz wave functions in R2 , as the space of all entire functions u such that u < ∞. In this paper we will follow the results in [1], hence we will always assume that W 2 is based in R2 . The elements of W 2 can be expanded as Neumann series
u (x) =
an J n (r )e inθ ,
(5)
n∈Z
2 for x = re i θ and n∈Z |an | < ∞, where J n (r ) is the Bessel function of order n ∈ Z, see [11]. 2 The space W provided with the Hilbert space structure of L 2 ( S 1 ) can be regarded as a space with reproducing kernel as developed in [1,4]. Denote w (r ) = 1+r r 3 for r > 0, and let H2 be the space consisting of all distributions u ∈ D (R2 \{0}) such that
u , ∂∂u ∈ L 2 ( W ), where we write L 2 ( W ) for L 2 (R2 \{0}, W dx). θ
H2 provided with the norm · in (4) is a Hilbert space and W 2 is precisely the space of elements u ∈ H2 which have an extension to R2 satisfying the Helmholtz equation. If u ∈ W 2 , then u 2W 2 ∼ |an |2 ∼ φ2L 2 ( S 1 ) , n∈Z
where φ = n∈Z an e inθ and u = φ dθ . Denote F n (re i θ ) = J n (r )e inθ , then {en }n∈Z with en = F n / F n W 2 , is an orthonormal basis for W 2 . Let βn = F n W 2 , then (see [11])
βn2
=
∞
2
n +1
J n2 (r ) w (r ) dr
0
∼
∞
n2 + 1 J n2 (r )
dr r2
∼ 1.
1
Let
P : H2 → W 2 be the orthogonal projection. The reproducing kernel for W 2 as a closed subspace of H2 is
K (x, y ) =
n∈Z
en (x)en ( y ) =
J n (r ) J n (s)e in(θ −ϕ ) , βn2 n∈Z
(6)
366
J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
for x = re i θ and y = se i ϕ . The kernel K (x, y ) is real and
P u (x) = u , K (x, ·) H2 =
K (x, y )u ( y ) + ∂ϕ K (x, y )∂ϕ u ( y ) W ( y ) dy .
R2
We have that K (x, ·)2W 2 =
n
J n2 (r )
∼ 1, since
βn2
J n2 (r ) = 1
(7)
n∈Z
(see [16]). In this paper we study Toeplitz operators in W 2 for radial nonnegative symbols is defined as in the classical holomorphic setting as
ρ . A Toeplitz operator denoted by T ρ ,
T ρ (u ) = P (ρ u ). In Section 1 we will study the continuity of Toeplitz operators. We will say that a function symbol if the identity
ρ defined on [0, ∞) is a Carleson
W 2 → Wρ2 is continuous, where Wρ2 is defined by replacing the measure W (x) dx by
ρ (x) W (x) dx in the definition of W 2 .
We will prove that for a nonnegative ρ ∈ L ([0, ∞), w (r )), the Toeplitz operator T ρ is bounded in W 2 if and only if a Carleson symbol and this is also equivalent to the boundedness of the sequence 1
1+n
2
∞
ρ is
J n2 (r )
ρ (r ) w (r ) dr
. n∈Z
0
As done in [5] for weights adapted to the Helmholtz equations, we aim to replace the Bessel functions in the above characterization since they hide the geometric meaning of a Carleson symbol. Our main result gives a sufficient condition in terms of integrability of the symbol with respect to a one-parameter family of rational weights which does not involve special functions. Theorem 1. Let ρ be nonnegative function in L 1 ([0, ∞), w (r ) dr ) such that for some μ0 > 0
∞ sup μ>μ0
μ2 2/3
μ
μ
+ r 1/2 (r
− μ)1/2
ρ (r ) w (r ) dr < ∞.
(8)
Then ρ is a Carleson symbol and (8) holds replacing μ0 by any positive number. The above condition is close to be also necessary as stated in Remark 12 and in Proposition 13. In this proposition we proved that for every Carleson symbol ρ , the integrals in (8) are of order log μ, and this is a weak converse of Theorem 1. It is an open question if log μ can be removed in Proposition 13. To prove these theorems we will use sharp asymptotics of Bessel functions J μ , with control on the dependence in the parameters μ, contained in Lemmas 5, 6 and 7. In Section 3 we study the compactness of the Toeplitz operators. The compactness of T ρ is equivalent to the corresponding compactness of the inclusion W 2 → Wρ2 . We define an ad-hoc Berezin type transform to study the operators belonging to the Schatten classes s p . Using the Berezin transform, we characterize all the Hilbert–Schmidt and Trace Class Toeplitz operators. Throughout this paper ρ will be a nonnegative radial function (even when it is not specified). We will write A ∼ B for two nonnegative quantities A , B , if there exist positive constants C 1 , C 2 such that C 1 A B C 2 A . Also C will denote a positive constant that may change in each occurrence. 2. Toeplitz operators
ρ = ρ (|x|) 0 in Ω = R2 \{0} we define the Toeplitz operator K (x, y )u ( y ) + ∂ϕ K (x, y )∂ϕ u ( y ) ρ | y | W ( y ) dy ,
Definition 2. Given a radial function
T ρ u (x) = P (ρ u )(x) = R2
for u in the linear span of {en }n∈Z .
J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
Notice that every symbol on W
367
ρ ∈ L ∞ (0, ∞) defines a bounded Toeplitz operator on W 2 . Also T ρ is a multiplier acting
2
Tρ
an en =
γn an en ,
where
γn = γ−n =
2π (n2 + 1)
αn
(9)
J n2 (r )ρ (r ) w (r ) dr .
(10)
βn2
and
∞
αn = 0
To see this, just notice that for x = re i θ
∞2π J n (r ) J n (s)e in(θ −ϕ ) J m (s)e imϕ (x) = Tρ dϕ ρ (s) w (s) ds βm βm βn2 n Fm
0
0
∞2π J n (r ) J n (s)e in(θ −ϕ ) J m (s)e imϕ + ∂ϕ dϕρ (s) w (s) ds ∂ ϕ βm βn2 n =
0
0
2π (1 + m2 ) 2 βm
∞ 2 Jm (s)
ρ (s) w (s) ds
0
Fm
βm
(x) = γm
Fm
βm
(x).
It follows immediately that T ρ is bounded if and only if {γn }n is bounded, that is, provided
∞ J n2 (r )ρ (r ) w (r ) dr 0
C n2 + 1
In particular by (7), we have that
(11)
.
ρ ∈ L 1 ([0, ∞), w (r )) if T ρ is bounded. The converse is not true (see Remark 9, below).
ρ a Carleson symbol if u (x)2 + ∂θ u (x)2 W (x) dx, u (x)2 + ∂θ u (x)2 ρ (x) W (x) dx C
Definition 3. We call
R2
R2
for every u ∈ W 2 , that is, if the identity
W 2 → Wρ2 is continuous, where Wρ2 is defined by replacing the measure W (x) dx by 1/ 2
Notice that the operator T ρ 1/2
Tρ u =
ρ (x) W (x) dx in the definition of W 2 .
is densely defined by
γn1/2an en ,
n
for u = n∈Z an en in the linear span of {en }n∈Z . For such u ,
u 2W 2 = ρ
u (x)2 + ∂θ u (x)2 ρ (x) W (x) dx
R2
2 2 ∞2π ∞2π an J n (s)e inϕ an in J n (s)e inϕ dϕ w (s)ρ (s) ds + dϕ w (s)ρ (s) ds = βn βn n n 0
0
0
0
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J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
= 2π
∞
n2 + 1
βn2
n
0
= 2π
|an |2 J n2 (s)
w (s)ρ (s) ds
1/2 |an |2 = C T ρ u W 2 . 2 βn
γn
n
Thus we have proved Proposition 4. The following statements are equivalent 1. ρ is a Carleson symbol, 2. T ρ is bounded. In the sequel we need precise estimates of Bessel functions. These estimates are summarized in the following lemmas. Lemma 5. For μ 1/2, there exists a universal constant A > 0 such that: 1. For r μ + μ1/3 we have
1 cos θ(μ, r ) J μ (r ) = √ + h(μ, r ), 1 2π (r 2 − μ2 ) 4
(12)
where
1 μ π θ(μ, r ) = r 2 − μ2 2 − μ arccos −
(13)
⎧ μ2 1 if μ + μ1/3 r 2μ, ⎨A 2 2 7 + r h(μ, r ) (r −μ ) 4 ⎩A if r 2μ. r
(14)
r
and
4
2. μ1/3 | J μ (r )| ∼ A if μ − μ1/3 < r < μ + μ1/3 . 3. We can choose a constant α0 , independent of μ, 0 < α0 < 1/2, such that if t 0 is defined by the equation μ sech α0 = μ − t 0 μ1/3 , then
α0 2 2
μ2/3 t 0 α0 2 μ2/3 ,
and (a) for μ sech α = μ − t μ1/3 and 1 t t 0 , we have −μ(α −tanh α ) J μ (μ sech α ) A e , 1/3 1/4
μ
t
(b) for 1 μ sech α μ sech α0 , we have −μ(α −tanh α ) J μ (μ sech α ) A e . 1/2
μ
4. For r 2μ,
J μ (r ) =
2
cos r −
πr
πμ 2
−
π 4
+O
1 r
uniformly in μ. Lemma 6. Let μ > 0, p 1 and a > 1, then there exists a constant C depending only on p and a, such that
1 C
μ
p 1 3−3
M −1 j =0
2
p
j (1 − 4 )
aμ M −1 p p p 1 J μ (r ) dr C μ 3 − 3 2 j (1 − 4 ) , μ a
2
where μ 3 ∼ 2 M .
j =0
J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
369
The proof of part 1 of Lemma 5 can be obtained using the stationary phase method (see [3,6]). Parts 2 and 4 can be found in [16] and for part 3 see [7]. Lemma 6 is a consequence of Lemma 5 (see [3,6]). The following lemma can be easily proved. Lemma 7. For 1 r μ − μ1/3 , let
φ(r ) = α (r ) − tanh α (r ), where α (r ) is given by the equation μ sech α (r ) = r. Then 1. φ is a decreasing function with φ (r ) = − > 0, and 2. φ(μ sech α0 ) := β0√
(μ2 −r 2 )1/2
μr
,
μ2 −1 μ .
3. φ(1) log(2μ) −
Proof of Theorem 1. We notice that condition (8) is independent of L 1 ([0, ∞), w (r ) dr ) satisfying (8). Define
H (r , μ) =
μ2 μ2/3 + r 1/2 |r − μ|1/2
μ0 provided ρ ∈ L 1 ([0, ∞), w (r ) dr ). Let ρ ∈
.
We have that H (r , μ) ∼ H (r , μ), where
H (r , μ) =
r ∈ [μ − μ1/3 , μ + μ1/3 ],
μ4/3 , μ2 r 1/2 (r −μ)1/2
, r > μ + μ1/3 ,
uniformly for r μ − μ1/3 and μ > μ0 . Notice that from (8) it follows that there exists
∞ sup μ>μ1 μ−μ1/3
μ1 > 0 such that
H (r , μ)ρ (r ) w (r ) dr < ∞.
μ1 large enough, then for μ − μ1/3 r μ with μ μ1 , we have μ − μ1/3 ∼ r ∼ μ. Hence H (r , μ) = μ4/3 C H r , μ − μ1/3
In fact, if we take
and
μ sup μ>μ1 μ−μ1/3
∞
H (r , μ)ρ (r ) w (r ) dr C sup
μ>μ1 μ−μ1/3
H r , μ − μ1/3
ρ (r ) w (r ) dr < ∞.
This together with (8) implies that
∞ sup μ2
μ0
μ+μ1/3
ρ (r ) r 1/2 (r − μ)1/2
w (r ) dr < ∞
and 1/3 μ+ μ
4/3
sup μ
μ0
ρ (r ) w (r ) dr < ∞.
(15)
μ−μ1/3
Now we prove that the sequence {γn }n corresponding to
∞
n sech α0
J n2 (r )
ρ (r ) w (r ) dr =
0
n− n1/3
+ 0
n sech α0
n+ n1/3
+ n−n1/3
and we estimate each integral according to Lemma 5.
ρ is bounded. To this end we decompose for n 1
2n
+ n+n1/3
∞ +
= 2n
5 i =1
Ji
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J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
We have that | J n (r )| C /n for r ∈ [0, n sech α0 ], thus
J1
∞
C
ρ (r ) w (r ) dr
n2
C n2
.
0
For J3 , we have that | J n (r )| ∼ n−1/3 in [n − n1/3 , n + n1/3 ], then
J3
n+ n1/3
C
ρ (r ) w (r ) dr
n2/3
C n2
.
n−n1/3
To estimate J4, we know that | J n (r )|
J4
C n2
C r 1/4 (r −n)1/4
if r ∈ [n + n1/3 , 2n], implying that
.
For J5 we have that | J n (r )| Cr −1/2 , hence
∞ J n2 (r )ρ (r )
∞
dr
C
r2
2n
r 1/2 (r − n)1/2
dr
rr 1/2 (r − n)
r2
ρ (r ) 1/2
2n
∞ C 2n
Finally, to estimate J2 , we split [n sech α0 , n − n sech α0 )n2/3 )]. On each I j
1/ 3
]⊂
ρ (r )
dr
r 1/2 (r − n)1/2 r 2
M 0
C n2
.
I j , with I j = {r = n − sn1/3 : 2 j < s 2 j +1 } and M = [log2 ((1 −
−n(α (r )−tanh α (r )) J n (r ) e ,
2 j /4n1/3
where r = n sech α (r ). Hence
J2 =
j
dr r2
j
Ij
j
J n2 (r )ρ (r )
C 2 j /2n2/3
e −2n(α (r )−tanh α (r )) ρ (r )
dr r2
Ij
C
sup e −2n(α (r )−tanh α (r )) 2 j /2n2/3 I j
ρ (r )
dr r2
.
Ij
To simplify the notation we let
ψ(μ) = μ − μ1/3 and ϕ (μ) = μ + μ1/3 . The function ϕ −1 exists on [0, ∞]. Let
(16)
μ j such that ϕ (μ j ) = n − 2
j +1 1/3
n
. We have that ϕ (μ j ) ∼ μ j ∼ n.
Then
ρ (r ) w (r ) dr = Ij
Ij
r 1/2 (r − μ j )1/2 r 1/2 (r − μ j )1/2
ρ (r ) w (r ) dr
r 1/2 (r − ϕ (μ j ))1/2 r 1/2 (r
Ij
− μj
)1/2
ρ (r ) w (r ) dr +
r 1/2 (ϕ (μ j ) − μ j )1/2
Ij
= L1 + L2, ∞ L1 C 2
j /2 2/3
n
ϕ (μ j )
ρ (r ) r 1/2 (r − μ j )1/2
while (ϕ (μ j ) − μ j )1/2 ∼ n1/6 implying that
L 2 Cn2/3−2 .
w (r ) dr C 2 j /2n2/3−2 ,
r 1/2 (r − μ j )1/2
ρ (r ) w (r ) dr
J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
371
Finally from Lemma 7, sup I j e −2n(α (r )−tanh α (r )) exp(−22 j /3 ) and hence
J2
C n2
sup e −2n(α (r )−tanh α (r )) Ij
Ij
C n2
.
2
ρ ∈ L 1 ([0, ∞), w (r ) dr ) such that ρ (r ) ≡ constant for r > r0 is a Carleson symbol: if μ > r0 , ∞ ∞ ∞ dr 1 dr −1/3 H (r , μ)ρ (r ) w (r ) dr C H (r , μ) 2 C μ + C, μ r (r − μ)1/2
Example 8. Any function
μ
μ
μ+μ1/3
where H was defined in the proof of Theorem 1. In particular,
1
ρ (r ) =
(r − a)α
χ(a,∞) (r ), a > 0, 0 α < 1,
is an unbounded Carleson symbol. In fact,
1
ρ (r )
(r − a)α
χ(a,a+1) (r ) + χ(a+1,∞) (r ).
Remark 9. 1. Let
ρ 0 such that the sequence λn defined by n λn = inf ρ (r ): r ∈ ,γn ,
γ
with
γ > 1 is not bounded. Then ρ cannot define a Carleson symbol. We have ∞ n
2
γ n J n2 (r )
ρ (r ) w (r ) dr n
0
2
γ n J n2 (r )
ρ (r ) w (r ) dr ∼ λn
n/γ
γn
J n2 (r ) dr ∼ λn ,
n/γ
since by Lemma 6, n/γ J n2 (r ) dr is like a constant. In particular, a Carleson measure ρ cannot satisfy a condition like
lim
r →∞
ρ (r ) log r
> 0.
ρ is such that
ρ (r )
2. Notice that if
2n
dr
r
n∈N
n
ρ is not a Carleson symbol. Indeed, by Lemma 5, if r > 2(n + 1) we have πn π π (n + 1) π 1 2 1 2 2 + cos r − +O = . cos r − − − 1+ O
is an unbounded sequence, then
J n2 (r ) + J n2+1 (r ) ∼
2
πr
2
4
2
4
πr
r
r
Then
n
2
∞
J [ n ] (r ) + J [ n ]+1 (r ) ρ (r ) w (r ) dr n2 2
2
2
2n
2
J [ n ] (r ) + J [ n ]+1 (r ) ρ (r ) w (r ) dr ∼ 2
2
ρ (r ) r
2
n
0
2n
2
3. We can extend the definition of Toeplitz operators with radial symbols replacing the function measure ν on [0, ∞) by letting
T v u (x) =
dr .
n
ρ by a positive Borel
K (x, y )u ( y ) + ∂ϕ K (x, y )∂ϕ u ( y ) w (s) dv (s) dϕ .
R2
Then for instance, if v = δa we have that T v is a diagonal operator diag(γn ) with respect to the basis {en } of W 2 and
γn =
2π (1+n2 )
βn2
J n2 (a) w (a). We have that T δa is uniformly bounded for a a0 > 0 (see (30) below).
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J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
Example 10. In view of the preceding examples, one might think that a Carleson symbol cannot be very big for large values of r. The following example shows that this is not true. It is not difficult to see that ρ below does define a Carleson symbol:
ρ (r ) =
∞
2 j /2
j =1
(r − 2 j ) 2
1
χM j (r ),
j
j
where M j = (2 j + 2 3 , 2 j + 21+ 3 ). The following propositions deal with necessity of condition (8). Proposition 11. If ρ is a Carleson symbol, then (15) holds and
∞
ρ (r )
2
μ
r 1/2 (r − μ)1/2
2μ
w (r ) dr C μ0 ,
(17)
for μ > μ0 .
ρ be a Carleson symbol. Then as mentioned in Example 3, for r > 2(n + 1) we have 2 1 . J n2 (r ) + J n2+1 (r ) ∼ 1+ O 2
Proof. Let
πr
r
Hence
∞
ρ (r ) r3
2(n+1)
∞
J n2 (r ) + J n2+1 (r ) C ρ (r ) dr C (αn + αn+1 ) 2 . r2 n
dr C 2(n+1)
This proves that
∞
ρ (r ) r3
dr
C
μ2
.
2μ
Thus
∞
ρ (r ) w (r ) dr μ2 1 / 2 r (r − μ)1/2
2
μ
2μ
∞
ρ (r ) r3
dr C
2μ
and (17) holds. If n ∈ N, then by Lemma 5 n+ n1/3
n+ n1/3
ρ (r ) w (r ) dr ∼ n n−n1/3
2/3
J n2 (r )ρ (r ) w (r ) dr
C n4/3
for n large, proving (15) in this case. If from the previous case since
μ = n + α with n ∈ N and 0 < α < 1. Let ϕ (μ) and ψ(μ) as in (16). Then (15) follows
ψ(μ), ϕ (μ) ⊂ ψ(n), ϕ (n) ∪ ψ(n + 1), ϕ (n + 1) . Remark 12. Notice that from Proposition 11, if we assume that
sup
1
1/2 μμ0 μ
then
,
n−n1/3
2μ μ+μ1/3
ρ (r ) dr C , (r − μ)1/2
ρ is a Carleson symbol if and only if (8) holds.
2
(18)
ρ satisfies (19)
J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
It can be easily seen that any radial function
sup
1
Ω b
373
ρ such that
a+b
ρ (r ) dr < C ,
(20)
a
where
! Ω = μ, a, b: μ μ0 , μ + μ1/3 a 2μ, μ1/3 b μ with μ0 > 0 and C an absolute constant satisfies (19). In particular if for a fixed b > 0
sup a >1
then
1
a+b
ρ (r ) dr < ∞,
b
(21)
a
ρ satisfies (20).
Now we study the question of how far condition (19) (and hence (8)) is from being necessary for symbol. We prove below the weaker result that
2μ
1
μ1/2
μ+μ1/3
is of order log μ if
ρ to be a Carleson
ρ (r ) dr (r − μ)1/2 ρ is Carleson symbol.
Proposition 13. Let us assume that ρ is a Carleson symbol, then there exists μ0 such that
2μ
1
sup
μμ0
1/2 log
μ
μ
ρ (r ) dr C , (r − μ)1/2
μ+μ1/3
(22)
where C is a universal constant. To prove Proposition 13, we need a sharp study of the function θ(μ, r ) in (13), in order to control the zeros of Bessel functions in the transition interval [μ + μ1/3 , 2μ]. More specifically we will use the fact that
J μ (r )2 + J
μ+
μ1/3
2 (r )
2 j /2
C
(r 2
− μ2 )1/2
,
where r ∈ (μ + 2 j μ1/3 , μ + 2 j +1 μ1/3 ), j ∈ {1, 2, . . . , M − 1}, and 2 M ∼ μ2/3 . Lemma 14. Given μ, let M ∈ N such that 2 M −1 2μ < 2 M and
μ + 2μ1/3 , 2μ ⊂
M −1 "
μ + 2 j μ1/3 , μ + 2 j+1 μ1/3 .
j =1
Then for r ∈ (μ + 2 j μ1/3 , μ + 2 j +1 μ1/3 ) and j ∈ {1, 2, . . . , M − 1} we have
√ μ1/3 √ θ(μ, r ) − θ μ + j /2 , r 2 2 2
2
6
(23)
and therefore
2 μ1/3 cos 2 2 cosθ(μ, r ) − θ μ + j /2 , r cos √ . 2 6 √
(24)
374
J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
Proof. Let
f (t ) = θ(t , r ) = r 2 − t
12
t
π
r
4
− t arccos −
,
r > t,
then
t f (t ) = − arccos . r For r ∈ (μ + 2 j μ1/3 , μ + 2 j +1 μ1/3 ) and some t ∈ (μ, μ +
μ1/3 2 j /2
μ1/ 3 2 j /2
) we have
1/3 1/3 1/3 μ + μ2 j/2 μ θ(μ, r ) − θ μ + μ , r = μ arccos t μ arccos arccos . j j / 2 j / 2 j / 2 r μ+2 μ 2 2 2 μ + 2 j +1 μ 1/3
Since
1 arccos x =
1
√
dt ,
1 − t2
x
we have that
μ1/3
μ1/3
μ
arccos 2 j /2 μ + 2 j+1 μ 2 j/2
1
(1 − t )−1/2 dt 2
μ μ+2 j +1 μ
μ1/3 2 j /2
1−
1/2
μ μ
+ 2 j +1
μ
√ 2 2.
In a similar way
μ1/3 2 j /2
1/3
μ + μ2 j/2 2 arccos √ μ + 2 jμ 6 2
and the lemma follows.
Proof of Proposition 13. Let 16 A 2 log μ0
(a) max (b)
8 A2 23 j A
μ0
(c) j A <
1−cos √2
1 8 4π log μ0 , 100
<
2
, 8μA0 6
!
μ0
1 8
1 2
and j A ∈ N satisfying
1−cos √2
6
4π
,
,
where A is the constant in (14). It is enough to prove that μ+ 2 M μ1/3
sup
μμ0 μ+2 j A μ1/3
(r 2
ρ (r ) dr C log μ, − μ2 )1/2
(25)
for M the integer in Lemma 14. If ρ is a Carleson symbol, it satisfies μ+ 2 M μ1/3
M −1 J μ (r )2 ρ (r ) dr + j= j A
μ+2 j A μ1/3
In the other hand, by using
1
|a + b|2 |a|2 − |b|2 2
and (12) we have
μ+2j+1 μ1/3
J
μ+2 j μ1/3
μ+
μ1/3 2 j /2
2 (r ) ρ (r ) dr C · M ∼ C log μ.
(26)
J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
μ+ 2 M μ1/3
M −1 J μ (r )2 ρ (r ) dr + j= j A
μ+2 j A μ1/3 M −1 1
4π
j= j A
μ+2j+1 μ1/3
J
1/3
μ + μ j /2
2 (r ) ρ (r ) dr
2
μ+2 j μ1/3
μ+2j+1 μ1/3
cos2 θ(μ, r ) + cos2 θ
μ1/3
μ+
2 j /2
,r
μ+2 j μ1/3
2 1/3 h μ + μ , r ρ (r ) dr . 2 j /2
M −1 h(μ, r )2 ρ (r ) dr −
−
j= j A
μ+2 j A μ1/3
ρ (r ) dr (r 2 − μ2 )1/2
μ+2j+1 μ1/3
μ+ 2 μ1/3 M
375
μ+2 j μ1/3
Using Lemma 5 and assumption (b) above we have μ+ 2 M μ1/3
h(μ, r )2 ρ (r ) dr 2 A 2 μ4
μ+2
jA
μ+ 2 M μ1/3
μ+2
μ1/3 2
4
2A μ
jA
2A
2
M −1 j= j A
μ1/3 j +1
μ+2 μ
M −1 j= j A
μ+2 j μ1/3
μ+ 2 μ
2
1 1 − cos √6
μ+2
jA
μ+ 2 M μ1/3
ρ (r ) 8 A2 dr + μ (r 2 − μ2 )1/2
μ+2 j A μ1/3
μ+2 j A μ1/3
1/3
μ+2 j A μ1/3
μ+ 2 M μ1/3
ρ (r ) dr
ρ (r ) dr (r 2 − μ2 )1/2
ρ (r ) dr (r 2 − μ2 )1/2
1/3
4π
4
M
ρ (r ) 8 A2 dr + 2 2 1 / 2 μ (r − μ ) M
μ+ 2 μ
1/3
23 j
μ+2 j A μ1/3
μ+2
μ1/3
ρ (r ) 2 A2 dr + 2 3 3 2 2 1 / 2 (r + μ) (r − μ) (r − μ ) μ
μ+2 μ
23 j A
ρ (r ) dr jA
1
j +1
1
μ+ 2 M μ1/3
1/3
μ+2 j μ1/3
μ+ 2 M μ1/3
8 A2
ρ (r ) 2 A2 dr + (r 2 − μ2 )3 (r 2 − μ2 )1/2 μ2 1
ρ (r ) dr . (r 2 − μ2 )1/2
(27)
μ1/3
In a similar way we prove M −1 j= j A
μ+2j+1 μ1/3
μ+ 2 M μ1/3 2 2 1/3 μ 1 1 − cos √6 ρ (r ) h μ + , r ρ (r ) dr dr . 2 j / 2 4 4 π ( r − μ2 )1/2 2
μ+2 j μ1/3
μ+2
From Lemma 14
2
2
cos θ(μ, r ) + cos θ
μ+
μ1/3 2 j /2
jA
μ1/3 μ1/3 , r = 1 + cos θ(μ, r ) − θ μ + j /2 , r · cos θ(μ, r ) + θ μ + j /2 , r 2
2
1 − cos √ . 6
Hence we have M −1 1
4π
j= j A
μ+2j+1 μ1/3
cos2 θ(μ, r ) + cos2 θ
μ+2 j μ1/3
1 − cos √2
6
μ+ 2 M μ1/3
4π μ+2 j A μ1/3
ρ (r ) dr . (r 2 − μ2 )1/2
From this and (26)–(27), (25) follows.
(28)
μ1/3
2
μ+
μ1/3 2 j /2
,r
ρ (r ) dr (r 2 − μ2 )1/2
2
376
J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
From Propositions 11 and 13 we obtain Remark 15. If
ρ is a Carleson symbol, then we have 2
μ ρ (r ) w (r ) dr μ2/3 + r 1/2 (r − μ)1/2
sup μ>μ0
+
(μ,∞)−(μ+μ1/3 ,2μ)
1
μ2
log μ (μ+μ1/3 ,2μ)
ρ (r ) w (r ) dr < ∞. 1/2
μ2/3 + r 1/2 (r − μ)
3. Berezin transform and compact Toeplitz operators Recall that every Toeplitz operator T ρ is a diagonal operator {γn }n∈Z with γn as in (9) with respect to the basis {en } defined above. Then we have that T ρ is compact if and only if limn γn = 0. From the estimate (see [16]) n J n (r ) r er 2 /4 , n!2n
we see that every function ρ ∈ L 1 ([0, ∞), w (r ) dr ) with bounded support defines a compact Toeplitz operator. However there are compact Toeplitz operators whose symbols have unbounded support as shown in Proposition 21. Proposition 16. Let ρ ∈ L 1 ([0, ∞), w (r ) dr ). Then T ρ is compact in W 2 if and only if the inclusion W 2 → Wρ2 is compact. Proof. For the sufficiency, we argue by contradiction. Suppose that W 2 → Wρ2 is compact. If T ρ is not compact there exists a subsequence {γnk }k∈Z such that |γnk | ε for some ε > 0. Then the corresponding sequence {enk }k∈Z would have a convergent subsequence in Wρ2 . This is impossible since {enk }k∈Z is orthogonal in Wρ2 and enk W 2 = γn ε . Hence we
have that {γn }n∈Z converges to 0. Now suppose that γn → 0. Let {un }n∈Z be a bounded sequence in W 2 . We easily see that
un =
ρ
an,k ek
k∈Z
with convergence in W 2 . Since supn∈N k∈Z |an,k |2 < ∞, we can find an increasing sequence {ln }n∈Z in N and a sequence 2 of complex numbers {ak }k∈Z such that limn→∞ aln,k = ak for every k ∈ Z. By Fatou’s lemma we have that k∈Z |ak | < ∞, and if we let u =
k ak ek
uln − u Wρ2 =
∈ W 2 , we have that
γk |an,k − ak |2
k∈Z
converges to 0, proving that the inclusion W 2 → Wρ2 is compact.
2
In order to characterize the Hilbert–Smidt and Trace Class Toeplitz operators we define a Berezin type transform for the Toeplitz operators on W 2 . Let
Rx =
1
n2 + 1
(1 + |x|2 )1/2
1/2
F n (x)en (x) .
n∈Z
We will see that there exist positive constants such that
c R x W 2 C ,
(29)
for all x ∈ R . In fact, from (7) we have that 2
2 n2 + 1 F n (x) =
n∈Z
n2 + 1 J n2 (r ) =
n∈Z
By the summation formula (see [16])
J 0 |x − y | =
n∈Z
J n (r ) J n (s)e in(θ −ϕ ) ,
n∈Z
J n2 (r ) +
n=0
n2 J n2 (r ) = 1 +
n=0
n2 J n2 (r ).
J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
377
x = re i θ , y = se i ϕ , we have for x = y
n2 J n (r ) J n (s)e in(θ −ϕ ) =
n∈Z
It follows that
n=0 n
2 2 J n (r )
= r 2 J 1 (0) = r 2 /2 and
2 ∂2 J 1 (|x − y |) J 2 (|x − y |) J 0 |x − y | = x· y+ x · y⊥ . 2 ∂θ∂ ϕ |x − y | |x − y |
n2 + 1 J n2 (r ) ∼ 1 + r 2 ,
(30)
n∈Z
from which (29) follows.
ρ ∈ L 1 ([0, ∞), w (r ) dr ), we define the Berezin transform 2 1 2 B ρ (r ) = n + 1 γ J ( r ) . n n 1 + r2
Definition 17. For a symbol
n∈Z
Notice that for every x ∈ R2 we have
B ρ (r ) = ( T ρ R x , R x ). Remark 18. (a) If ρ is a Carleson symbol, then the Berezin transform B ρ is bounded. (b) If T ρ is a compact operator, then limr →∞ B ρ (r ) = 0. Proof. In fact, if T ρ is a bounded operator, then by (29) we have that ( T ρ R x , R x ) is bounded in R2 implying (a). If T ρ is compact, we have that (γn ) converges to 0. Then given ε > 0, we have by (30) that
1
1 + r2
2
n +1
2 n J n (r )
γ
|n|> M
ε
2
n +1
1 + r2
J n2 (r )
C ε,
|n|> M
for M large enough. Then
lim sup B ρ (r ) = lim sup r →∞
r →∞
Thus limr →∞ B ρ (r ) = 0.
|n| M
+
2
n +1
2 n J n (r )
γ
C ε.
|n|> M
2
Now we study the Toeplitz operators belonging to the Schatten classes s p . Recall that a bounded operator T in a Hilbert √ space belongs to the Schatten class s p if Tr | T | p < ∞, where Tr A stands for the trace of the operator A and | T | = T ∗ T (see [15, p. 215]). Since T ρ is a diagonal operator with respect to the basis {en }n∈Z , we have that T ρ ∈ s p if and only if {γn } ∈ p (Z). In this case T ρ s p = γn l p . For the trace class s1 and the Hilbert–Schmidt operators s2 we have the following theorems. Theorem 19. The following statements are equivalent: (1) T ρ ∈ s1 , (2) ρ ∈ L 1 ((1 + r 2 ) w (r ) dr ), (3) B ρ ∈ L 1 ((1 + r 2 ) w (r ) dr ), and the quantities T ρ s1 , ρ L 1 ((1+r 2 ) w (r ) dr ) , B ρ L 1 ((1+r 2 ) w (r ) dr ) are comparable. Proof. By (30) we have that
n∈Z
γn ∼
n2 + 1
n∈Z
∞
∞
ρ (r ) J n2 (r ) w (r ) dr = 0
ρ (r ) w (r ) 0
proving the equivalence of (1) and (2). On the other hand,
∞
n2 + 1 J n2 (r ) dr ∼
n∈Z
ρ (r ) 1 + r 2 w (r ) dr , 0
378
J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
γn ∼
n∈Z
γn
n∈Z
∞
2
n +1
∞ J n2 (r ) w (r ) dr
0
B ρ (r ) 1 + r 2 w (r ) dr ,
= 0
and the equivalence of (1) and (3) follows.
2 1/ 2
Theorem 20. T ρ ∈ s2 if and only if ρ B ρ ∈ L 1 ((1 + r 2 ) w (r ) dr ), moreover T ρ s2 and ρ B ρ L 1 ((1+r 2 ) w (r ) dr ) are comparable quanti-
ties.
Proof. We have
2 n
γ ∼
n∈Z
2
γn n + 1
n∈Z
∞
ρ (r )
J n2 (r ) w (r ) dr
=
∞
2
γn n + 1
∞ J n2 (r )
0 n∈Z
0
ρ (r ) B ρ (r ) 1 + r 2 w (r ) dr ,
ρ (r ) w (r ) dr ∼ 0
2
and this completes the proof.
More generally, we have the following partial results. Proposition 21. Let p 1. 1. If ρ ∈ L p ((1 + r 2 ) w (r ) dr ), then T ρ ∈ s p with
T ρ s p C p ρ L p ((1+r 2 ) w (r ) dr ) .
(31)
2. If T ρ ∈ s p , then B ρ ∈ L p ((1 + r 2 ) w (r ) dr ) and
B ρ L p ((1+r 2 ) w (r ) dr ) C T ρ s p .
(32)
Proof. Since
∞
J n2 (r ) w (r ) dr ∼ n2 + 1
− 1
,
0
we have by Jensen’s inequality and estimate (30) that
p n
γ C
n∈Z
2
n +1
n∈Z
C
p
∞
ρ (r )
J n2 (r ) w (r ) dr
0
∞
∞ ρ (r ) J n2 (r ) w (r ) dr p ∞ ρ p (r ) J n2 (r ) w (r ) dr 0 0 ∞ 2 ∞ 2 C C J n (r ) w (r ) dr J n (r ) w (r ) dr 0 0 n∈Z n∈Z ∞
n2 + 1 J n2 (r )ρ p (r ) w (r ) dr C
0 n∈Z
ρ p (r ) 1 + r 2 w (r ) dr , 0
proving (31). By (30) we have by Jensen’s inequality again that
B
p
ρ (r )
1 1 + r2
p 2 n n
γ
+1
J n2 (r )
.
n∈Z
Then integrating this inequality with respect to (1 + r 2 ) w (r ) dr we obtain (32).
2
Notice that the Berezin transform B ρ is an integral operator on [0, ∞) provided with the measure (1 + r 2 ) w (r ) dr and given by
∞
M (r , s)ρ (s) 1 + s2 w (s) ds,
B ρ (r ) = 0
where M (r , s) is the symmetric kernel
M (r , s) =
2π
(1 + r 2 )(1 + s2 )
(n2 + 1)2 n∈Z
βn2
J n2 (r ) J n2 (s).
J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
379
Corollary 22. The Berezin transform B ρ is a bounded integral operator on L p ((1 + r 2 ) w (r ) dr ) for 1 p ∞. Proof. Since M is symmetric, it is enough to prove the statement for p ∈ [1, 2]. For p = 1, this is part of Theorem 19. Now we prove it for p = 2. Let ρ ∈ L 2 ((1 + r 2 ) w (r ) dr ), then T ρ ∈ s2 by Proposition 21. If (γn ) is the sequence corresponding to T ρ , then by Proposition 21 and Theorem 20 we have that
Bρ
2L 2 ((1+r 2 ) w (r ) dr )
n∈Z
∞ 2 n
ρ (r ) B ρ (r ) 1 + r 2 w (r ) dr ρ L 2 ((1+r 2 ) w (r ) dr ) B ρ L 2 ((1+r 2 ) w (r ) dr ) .
γ Cp 0
It follows that
B ρ L 2 ((1+r 2 ) w (r ) dr ) C p ρ L 2 ((1+r 2 ) w (r ) dr ) . Finally the result follows for p ∈ [1, 2] by interpolation.
2
References [1] J. Alvarez, M. Folch-Gabayet, S. Pérez-Esteva, Banach spaces of solutions of the Helmholtz equation in the plane, J. Fourier Anal. Appl. 7 (6) (2001) 541–562. [2] S. Axler, Bergman spaces and their operators, in: J.B. Conway, B.B. Morrel (Eds.), Surveys in Some Recent Results in Operator Theory, vol. 1, in: Pitman Res. Notes Math., vol. 171, 1988, pp. 1–50. [3] J.A. Barceló, Tesis Doctoral, Universidad Autónoma de Madrid, 1988. [4] J.A. Barceló, J. Bennet, A. Ruíz, Mapping properties of a projection related to the Helmholtz equation, J. Fourier Anal. Appl. 9 (6) (2003) 541–562. [5] J.A. Barceló, J. Bennet, A. Ruíz, Spherical perturbations of Schrödinger equation, J. Fourier Anal. Appl. 12 (3) (2006) 269–290. [6] J.A. Barceló, A. Córdoba, Band limited L p -convergence, Trans. Amer. Math. Soc. 313 (2) (1989) 655–669. [7] J.A. Barceló, A. Ruiz, L. Vega, Weighted estimates for the Helmholtz equation and some applications, J. Funct. Anal. 150 (2) (1997) 356–382. [8] B.R. Choe, Y.J. Lee, K. Na, Toeplitz operators on harmonic Bergman spaces, Nagoya Math. J. 174 (2004) 165–186. [9] D. Colton, Inverse Acoustic and Electromagnetic Scattering Theory, second ed., Appl. Math. Sci., vol. 93, Springer-Verlag, Berlin, 1998. [10] G. Dassios, Z. Rigou, On the density of traction traces in scattering of elastic waves, SIAM J. Appl. Math. 53 (1) (1993) 141–153. [11] P. Hartman, On solutions of V + V = 0 in an exterior region, Math. Z. 71 (1957) 251–257. [12] P. Hartman, C. Wilcox, On solutions of the Helmholtz equation in exterior domain, Math. Z. 75 (1961) 228–255. [13] W. Magmus, Fragen der Eindeutigkeit und des Varhaltens im Undendlichen fur Losungen von u + k2 u = 0, Abh. Math. Sem. Hamburg 16 (1949) 77–94. [14] M. Nishio, N. Suzuki, M. Yamada, Toeplitz operators and Carleson measures on parabolic Bergman spaces, Hokkaido Math. J. 36 (3) (2007) 563–583. [15] M. Reed, B. Simon, Methods of Modern Mathematical Physics, vol. 1, Academic Press, 1972. [16] G.N. Watson, A Treatise on the Theory of Bessel Functions, Cambridge Univ. Press, 1944. [17] K. Zhu, Operator Theory in Function Spaces, Marcel Dekker, New York, 1990.
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Toeplitz operators on Herglotz wave functions ✩ J.A. Barceló a , M. Folch-Gabayet b , S. Pérez-Esteva c,∗ , A. Ruiz d a
ETSI de Caminos, Universidad Politécnica de Madrid, 28040 Madrid, Spain Instituto de Matemáticas, Universidad Nacional Autónoma de México, Ciudad Universitaria, México, D.F. 04510, Mexico c Instituto de Matemáticas Unidad Cuernavaca, Universidad Nacional Autónoma de México, A.P. 273-3 ADMON 3, Cuernavaca, Mor. 62251, Mexico d Departamento de Matemáticas, Universidad Autónoma de Madrid, 28049 Madrid, Spain b
a r t i c l e
i n f o
a b s t r a c t The space of Herglotz wave functions in R2 consists of all the solutions of the Helmholtz equation that can be represented as the Fourier transform in R2 of a measure supported in the circle and with density in L 2 ( S 1 ). This space has a structure of a Hilbert space with reproducing kernel. The purpose of this article is to study Toeplitz operators with nonnegative radial symbols, defined on this space. We study the symbols defining bounded and compact Toeplitz operators as well as the Toeplitz operators belonging to the Schatten classes s p . © 2009 Elsevier Inc. All rights reserved.
Article history: Received 12 June 2008 Available online 6 May 2009 Submitted by G. Corach Keywords: Herglotz wave functions
1. Introduction and preliminaries The theory of Toeplitz operators in Bergman spaces of holomorphic functions has played an important role in operator theory and complex analysis for a number of years. Natural questions such as the characterization of all the symbols defining bounded or compact Toeplitz operators have been answered in different settings, not only for Toeplitz operators on Bergman spaces of holomorphic functions but also on spaces of harmonic and parabolic functions with reproducing kernels (see the monograph [17] and [2,8,14]). The purpose of this article is to study Toeplitz operators defined on the space of all the Herglotz wave functions in R2 . A Herglotz wave function in Rn is a solution u of the Helmholtz equation
u + u = 0
(1)
in Rn , that is represented as the Fourier transform in Rn of a measure φ dλ, where φ ∈ L 2 ( S n−1 ) and λ is the Lebesgue measure in S n−1 . Hence every Herglotz wave function can be written as
u (x) =
e ix·ω φ(ω) dλ(ω).
S n −1
✩
Research partially financed by: J.A.B., A.R.: Spanish Grant MTM2005-07652-C02-01; M.F.-G., S.P.-E.: Proyecto Conacyt-DAIC U48633-F. Corresponding author. E-mail addresses: [email protected] (J.A. Barceló), [email protected] (M. Folch-Gabayet), [email protected] (S. Pérez-Esteva), [email protected] (A. Ruiz).
*
0022-247X/$ – see front matter doi:10.1016/j.jmaa.2009.05.004
©
2009 Elsevier Inc. All rights reserved.
(2)
J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
365
Several characterizations of Herglotz wave functions have been given (see [13,11,12]). For instance a necessary and sufficient condition for an entire function u to be a Herglotz wave function is that
lim
R →∞
R
12 u (x)2 dx < ∞.
1
(3)
|x|< R
We will denote by W 2 the space of all Herglotz functions in Rn . Besides their intrinsic relevance to harmonic analysis, Herglotz wave functions play an important role in obstacle and nonhomogeneous media acoustic scattering, as the existence of certain Herglotz wave functions determines if the set of far-field-patterns (or scattering amplitudes) is dense in L 2 ( S n−1 ) (see [9]) and in scattering of elastic waves (see [10]). We will call a function u an entire function if it is a solution of the Helmholtz equation in Rn . The space W 2 with the norm given by (3) is a Hilbert space, and the map defined by (2) called extension operator, gives an isomorphism from L 2 ( S n−1 ) onto W 2 . For dimension n = 2, Alvarez, Folch-Gabayet and Pérez-Esteva proved in [1], that the extension operator
φ → φ dθ is an isomorphism from L 2 ( S 1 ) onto the space of entire solutions u of the Helmholtz equation which satisfy
2
u =
u (x)2 + ∂θ u (x)2 W (x) dx,
(4)
R2 ⊥
where W (x) = 1+|1x|3 , ∂θ u = ∇ · x|x| , x⊥ = (−x2 , x1 ) and dθ is the Lebesgue measure on S 1 (see also [4]). Hence, within the set of solutions to the Helmholtz equation, (3) is equivalent to (4) and this provided a new characterization of Herglotz wave functions in R2 , as the space of all entire functions u such that u < ∞. In this paper we will follow the results in [1], hence we will always assume that W 2 is based in R2 . The elements of W 2 can be expanded as Neumann series
u (x) =
an J n (r )e inθ ,
(5)
n∈Z
2 for x = re i θ and n∈Z |an | < ∞, where J n (r ) is the Bessel function of order n ∈ Z, see [11]. 2 The space W provided with the Hilbert space structure of L 2 ( S 1 ) can be regarded as a space with reproducing kernel as developed in [1,4]. Denote w (r ) = 1+r r 3 for r > 0, and let H2 be the space consisting of all distributions u ∈ D (R2 \{0}) such that
u , ∂∂u ∈ L 2 ( W ), where we write L 2 ( W ) for L 2 (R2 \{0}, W dx). θ
H2 provided with the norm · in (4) is a Hilbert space and W 2 is precisely the space of elements u ∈ H2 which have an extension to R2 satisfying the Helmholtz equation. If u ∈ W 2 , then u 2W 2 ∼ |an |2 ∼ φ2L 2 ( S 1 ) , n∈Z
where φ = n∈Z an e inθ and u = φ dθ . Denote F n (re i θ ) = J n (r )e inθ , then {en }n∈Z with en = F n / F n W 2 , is an orthonormal basis for W 2 . Let βn = F n W 2 , then (see [11])
βn2
=
∞
2
n +1
J n2 (r ) w (r ) dr
0
∼
∞
n2 + 1 J n2 (r )
dr r2
∼ 1.
1
Let
P : H2 → W 2 be the orthogonal projection. The reproducing kernel for W 2 as a closed subspace of H2 is
K (x, y ) =
n∈Z
en (x)en ( y ) =
J n (r ) J n (s)e in(θ −ϕ ) , βn2 n∈Z
(6)
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J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
for x = re i θ and y = se i ϕ . The kernel K (x, y ) is real and
P u (x) = u , K (x, ·) H2 =
K (x, y )u ( y ) + ∂ϕ K (x, y )∂ϕ u ( y ) W ( y ) dy .
R2
We have that K (x, ·)2W 2 =
n
J n2 (r )
∼ 1, since
βn2
J n2 (r ) = 1
(7)
n∈Z
(see [16]). In this paper we study Toeplitz operators in W 2 for radial nonnegative symbols is defined as in the classical holomorphic setting as
ρ . A Toeplitz operator denoted by T ρ ,
T ρ (u ) = P (ρ u ). In Section 1 we will study the continuity of Toeplitz operators. We will say that a function symbol if the identity
ρ defined on [0, ∞) is a Carleson
W 2 → Wρ2 is continuous, where Wρ2 is defined by replacing the measure W (x) dx by
ρ (x) W (x) dx in the definition of W 2 .
We will prove that for a nonnegative ρ ∈ L ([0, ∞), w (r )), the Toeplitz operator T ρ is bounded in W 2 if and only if a Carleson symbol and this is also equivalent to the boundedness of the sequence 1
1+n
2
∞
ρ is
J n2 (r )
ρ (r ) w (r ) dr
. n∈Z
0
As done in [5] for weights adapted to the Helmholtz equations, we aim to replace the Bessel functions in the above characterization since they hide the geometric meaning of a Carleson symbol. Our main result gives a sufficient condition in terms of integrability of the symbol with respect to a one-parameter family of rational weights which does not involve special functions. Theorem 1. Let ρ be nonnegative function in L 1 ([0, ∞), w (r ) dr ) such that for some μ0 > 0
∞ sup μ>μ0
μ2 2/3
μ
μ
+ r 1/2 (r
− μ)1/2
ρ (r ) w (r ) dr < ∞.
(8)
Then ρ is a Carleson symbol and (8) holds replacing μ0 by any positive number. The above condition is close to be also necessary as stated in Remark 12 and in Proposition 13. In this proposition we proved that for every Carleson symbol ρ , the integrals in (8) are of order log μ, and this is a weak converse of Theorem 1. It is an open question if log μ can be removed in Proposition 13. To prove these theorems we will use sharp asymptotics of Bessel functions J μ , with control on the dependence in the parameters μ, contained in Lemmas 5, 6 and 7. In Section 3 we study the compactness of the Toeplitz operators. The compactness of T ρ is equivalent to the corresponding compactness of the inclusion W 2 → Wρ2 . We define an ad-hoc Berezin type transform to study the operators belonging to the Schatten classes s p . Using the Berezin transform, we characterize all the Hilbert–Schmidt and Trace Class Toeplitz operators. Throughout this paper ρ will be a nonnegative radial function (even when it is not specified). We will write A ∼ B for two nonnegative quantities A , B , if there exist positive constants C 1 , C 2 such that C 1 A B C 2 A . Also C will denote a positive constant that may change in each occurrence. 2. Toeplitz operators
ρ = ρ (|x|) 0 in Ω = R2 \{0} we define the Toeplitz operator K (x, y )u ( y ) + ∂ϕ K (x, y )∂ϕ u ( y ) ρ | y | W ( y ) dy ,
Definition 2. Given a radial function
T ρ u (x) = P (ρ u )(x) = R2
for u in the linear span of {en }n∈Z .
J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
Notice that every symbol on W
367
ρ ∈ L ∞ (0, ∞) defines a bounded Toeplitz operator on W 2 . Also T ρ is a multiplier acting
2
Tρ
an en =
γn an en ,
where
γn = γ−n =
2π (n2 + 1)
αn
(9)
J n2 (r )ρ (r ) w (r ) dr .
(10)
βn2
and
∞
αn = 0
To see this, just notice that for x = re i θ
∞2π J n (r ) J n (s)e in(θ −ϕ ) J m (s)e imϕ (x) = Tρ dϕ ρ (s) w (s) ds βm βm βn2 n Fm
0
0
∞2π J n (r ) J n (s)e in(θ −ϕ ) J m (s)e imϕ + ∂ϕ dϕρ (s) w (s) ds ∂ ϕ βm βn2 n =
0
0
2π (1 + m2 ) 2 βm
∞ 2 Jm (s)
ρ (s) w (s) ds
0
Fm
βm
(x) = γm
Fm
βm
(x).
It follows immediately that T ρ is bounded if and only if {γn }n is bounded, that is, provided
∞ J n2 (r )ρ (r ) w (r ) dr 0
C n2 + 1
In particular by (7), we have that
(11)
.
ρ ∈ L 1 ([0, ∞), w (r )) if T ρ is bounded. The converse is not true (see Remark 9, below).
ρ a Carleson symbol if u (x)2 + ∂θ u (x)2 W (x) dx, u (x)2 + ∂θ u (x)2 ρ (x) W (x) dx C
Definition 3. We call
R2
R2
for every u ∈ W 2 , that is, if the identity
W 2 → Wρ2 is continuous, where Wρ2 is defined by replacing the measure W (x) dx by 1/ 2
Notice that the operator T ρ 1/2
Tρ u =
ρ (x) W (x) dx in the definition of W 2 .
is densely defined by
γn1/2an en ,
n
for u = n∈Z an en in the linear span of {en }n∈Z . For such u ,
u 2W 2 = ρ
u (x)2 + ∂θ u (x)2 ρ (x) W (x) dx
R2
2 2 ∞2π ∞2π an J n (s)e inϕ an in J n (s)e inϕ dϕ w (s)ρ (s) ds + dϕ w (s)ρ (s) ds = βn βn n n 0
0
0
0
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J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
= 2π
∞
n2 + 1
βn2
n
0
= 2π
|an |2 J n2 (s)
w (s)ρ (s) ds
1/2 |an |2 = C T ρ u W 2 . 2 βn
γn
n
Thus we have proved Proposition 4. The following statements are equivalent 1. ρ is a Carleson symbol, 2. T ρ is bounded. In the sequel we need precise estimates of Bessel functions. These estimates are summarized in the following lemmas. Lemma 5. For μ 1/2, there exists a universal constant A > 0 such that: 1. For r μ + μ1/3 we have
1 cos θ(μ, r ) J μ (r ) = √ + h(μ, r ), 1 2π (r 2 − μ2 ) 4
(12)
where
1 μ π θ(μ, r ) = r 2 − μ2 2 − μ arccos −
(13)
⎧ μ2 1 if μ + μ1/3 r 2μ, ⎨A 2 2 7 + r h(μ, r ) (r −μ ) 4 ⎩A if r 2μ. r
(14)
r
and
4
2. μ1/3 | J μ (r )| ∼ A if μ − μ1/3 < r < μ + μ1/3 . 3. We can choose a constant α0 , independent of μ, 0 < α0 < 1/2, such that if t 0 is defined by the equation μ sech α0 = μ − t 0 μ1/3 , then
α0 2 2
μ2/3 t 0 α0 2 μ2/3 ,
and (a) for μ sech α = μ − t μ1/3 and 1 t t 0 , we have −μ(α −tanh α ) J μ (μ sech α ) A e , 1/3 1/4
μ
t
(b) for 1 μ sech α μ sech α0 , we have −μ(α −tanh α ) J μ (μ sech α ) A e . 1/2
μ
4. For r 2μ,
J μ (r ) =
2
cos r −
πr
πμ 2
−
π 4
+O
1 r
uniformly in μ. Lemma 6. Let μ > 0, p 1 and a > 1, then there exists a constant C depending only on p and a, such that
1 C
μ
p 1 3−3
M −1 j =0
2
p
j (1 − 4 )
aμ M −1 p p p 1 J μ (r ) dr C μ 3 − 3 2 j (1 − 4 ) , μ a
2
where μ 3 ∼ 2 M .
j =0
J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
369
The proof of part 1 of Lemma 5 can be obtained using the stationary phase method (see [3,6]). Parts 2 and 4 can be found in [16] and for part 3 see [7]. Lemma 6 is a consequence of Lemma 5 (see [3,6]). The following lemma can be easily proved. Lemma 7. For 1 r μ − μ1/3 , let
φ(r ) = α (r ) − tanh α (r ), where α (r ) is given by the equation μ sech α (r ) = r. Then 1. φ is a decreasing function with φ (r ) = − > 0, and 2. φ(μ sech α0 ) := β0√
(μ2 −r 2 )1/2
μr
,
μ2 −1 μ .
3. φ(1) log(2μ) −
Proof of Theorem 1. We notice that condition (8) is independent of L 1 ([0, ∞), w (r ) dr ) satisfying (8). Define
H (r , μ) =
μ2 μ2/3 + r 1/2 |r − μ|1/2
μ0 provided ρ ∈ L 1 ([0, ∞), w (r ) dr ). Let ρ ∈
.
We have that H (r , μ) ∼ H (r , μ), where
H (r , μ) =
r ∈ [μ − μ1/3 , μ + μ1/3 ],
μ4/3 , μ2 r 1/2 (r −μ)1/2
, r > μ + μ1/3 ,
uniformly for r μ − μ1/3 and μ > μ0 . Notice that from (8) it follows that there exists
∞ sup μ>μ1 μ−μ1/3
μ1 > 0 such that
H (r , μ)ρ (r ) w (r ) dr < ∞.
μ1 large enough, then for μ − μ1/3 r μ with μ μ1 , we have μ − μ1/3 ∼ r ∼ μ. Hence H (r , μ) = μ4/3 C H r , μ − μ1/3
In fact, if we take
and
μ sup μ>μ1 μ−μ1/3
∞
H (r , μ)ρ (r ) w (r ) dr C sup
μ>μ1 μ−μ1/3
H r , μ − μ1/3
ρ (r ) w (r ) dr < ∞.
This together with (8) implies that
∞ sup μ2
μ0
μ+μ1/3
ρ (r ) r 1/2 (r − μ)1/2
w (r ) dr < ∞
and 1/3 μ+ μ
4/3
sup μ
μ0
ρ (r ) w (r ) dr < ∞.
(15)
μ−μ1/3
Now we prove that the sequence {γn }n corresponding to
∞
n sech α0
J n2 (r )
ρ (r ) w (r ) dr =
0
n− n1/3
+ 0
n sech α0
n+ n1/3
+ n−n1/3
and we estimate each integral according to Lemma 5.
ρ is bounded. To this end we decompose for n 1
2n
+ n+n1/3
∞ +
= 2n
5 i =1
Ji
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J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
We have that | J n (r )| C /n for r ∈ [0, n sech α0 ], thus
J1
∞
C
ρ (r ) w (r ) dr
n2
C n2
.
0
For J3 , we have that | J n (r )| ∼ n−1/3 in [n − n1/3 , n + n1/3 ], then
J3
n+ n1/3
C
ρ (r ) w (r ) dr
n2/3
C n2
.
n−n1/3
To estimate J4, we know that | J n (r )|
J4
C n2
C r 1/4 (r −n)1/4
if r ∈ [n + n1/3 , 2n], implying that
.
For J5 we have that | J n (r )| Cr −1/2 , hence
∞ J n2 (r )ρ (r )
∞
dr
C
r2
2n
r 1/2 (r − n)1/2
dr
rr 1/2 (r − n)
r2
ρ (r ) 1/2
2n
∞ C 2n
Finally, to estimate J2 , we split [n sech α0 , n − n sech α0 )n2/3 )]. On each I j
1/ 3
]⊂
ρ (r )
dr
r 1/2 (r − n)1/2 r 2
M 0
C n2
.
I j , with I j = {r = n − sn1/3 : 2 j < s 2 j +1 } and M = [log2 ((1 −
−n(α (r )−tanh α (r )) J n (r ) e ,
2 j /4n1/3
where r = n sech α (r ). Hence
J2 =
j
dr r2
j
Ij
j
J n2 (r )ρ (r )
C 2 j /2n2/3
e −2n(α (r )−tanh α (r )) ρ (r )
dr r2
Ij
C
sup e −2n(α (r )−tanh α (r )) 2 j /2n2/3 I j
ρ (r )
dr r2
.
Ij
To simplify the notation we let
ψ(μ) = μ − μ1/3 and ϕ (μ) = μ + μ1/3 . The function ϕ −1 exists on [0, ∞]. Let
(16)
μ j such that ϕ (μ j ) = n − 2
j +1 1/3
n
. We have that ϕ (μ j ) ∼ μ j ∼ n.
Then
ρ (r ) w (r ) dr = Ij
Ij
r 1/2 (r − μ j )1/2 r 1/2 (r − μ j )1/2
ρ (r ) w (r ) dr
r 1/2 (r − ϕ (μ j ))1/2 r 1/2 (r
Ij
− μj
)1/2
ρ (r ) w (r ) dr +
r 1/2 (ϕ (μ j ) − μ j )1/2
Ij
= L1 + L2, ∞ L1 C 2
j /2 2/3
n
ϕ (μ j )
ρ (r ) r 1/2 (r − μ j )1/2
while (ϕ (μ j ) − μ j )1/2 ∼ n1/6 implying that
L 2 Cn2/3−2 .
w (r ) dr C 2 j /2n2/3−2 ,
r 1/2 (r − μ j )1/2
ρ (r ) w (r ) dr
J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
371
Finally from Lemma 7, sup I j e −2n(α (r )−tanh α (r )) exp(−22 j /3 ) and hence
J2
C n2
sup e −2n(α (r )−tanh α (r )) Ij
Ij
C n2
.
2
ρ ∈ L 1 ([0, ∞), w (r ) dr ) such that ρ (r ) ≡ constant for r > r0 is a Carleson symbol: if μ > r0 , ∞ ∞ ∞ dr 1 dr −1/3 H (r , μ)ρ (r ) w (r ) dr C H (r , μ) 2 C μ + C, μ r (r − μ)1/2
Example 8. Any function
μ
μ
μ+μ1/3
where H was defined in the proof of Theorem 1. In particular,
1
ρ (r ) =
(r − a)α
χ(a,∞) (r ), a > 0, 0 α < 1,
is an unbounded Carleson symbol. In fact,
1
ρ (r )
(r − a)α
χ(a,a+1) (r ) + χ(a+1,∞) (r ).
Remark 9. 1. Let
ρ 0 such that the sequence λn defined by n λn = inf ρ (r ): r ∈ ,γn ,
γ
with
γ > 1 is not bounded. Then ρ cannot define a Carleson symbol. We have ∞ n
2
γ n J n2 (r )
ρ (r ) w (r ) dr n
0
2
γ n J n2 (r )
ρ (r ) w (r ) dr ∼ λn
n/γ
γn
J n2 (r ) dr ∼ λn ,
n/γ
since by Lemma 6, n/γ J n2 (r ) dr is like a constant. In particular, a Carleson measure ρ cannot satisfy a condition like
lim
r →∞
ρ (r ) log r
> 0.
ρ is such that
ρ (r )
2. Notice that if
2n
dr
r
n∈N
n
ρ is not a Carleson symbol. Indeed, by Lemma 5, if r > 2(n + 1) we have πn π π (n + 1) π 1 2 1 2 2 + cos r − +O = . cos r − − − 1+ O
is an unbounded sequence, then
J n2 (r ) + J n2+1 (r ) ∼
2
πr
2
4
2
4
πr
r
r
Then
n
2
∞
J [ n ] (r ) + J [ n ]+1 (r ) ρ (r ) w (r ) dr n2 2
2
2
2n
2
J [ n ] (r ) + J [ n ]+1 (r ) ρ (r ) w (r ) dr ∼ 2
2
ρ (r ) r
2
n
0
2n
2
3. We can extend the definition of Toeplitz operators with radial symbols replacing the function measure ν on [0, ∞) by letting
T v u (x) =
dr .
n
ρ by a positive Borel
K (x, y )u ( y ) + ∂ϕ K (x, y )∂ϕ u ( y ) w (s) dv (s) dϕ .
R2
Then for instance, if v = δa we have that T v is a diagonal operator diag(γn ) with respect to the basis {en } of W 2 and
γn =
2π (1+n2 )
βn2
J n2 (a) w (a). We have that T δa is uniformly bounded for a a0 > 0 (see (30) below).
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J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
Example 10. In view of the preceding examples, one might think that a Carleson symbol cannot be very big for large values of r. The following example shows that this is not true. It is not difficult to see that ρ below does define a Carleson symbol:
ρ (r ) =
∞
2 j /2
j =1
(r − 2 j ) 2
1
χM j (r ),
j
j
where M j = (2 j + 2 3 , 2 j + 21+ 3 ). The following propositions deal with necessity of condition (8). Proposition 11. If ρ is a Carleson symbol, then (15) holds and
∞
ρ (r )
2
μ
r 1/2 (r − μ)1/2
2μ
w (r ) dr C μ0 ,
(17)
for μ > μ0 .
ρ be a Carleson symbol. Then as mentioned in Example 3, for r > 2(n + 1) we have 2 1 . J n2 (r ) + J n2+1 (r ) ∼ 1+ O 2
Proof. Let
πr
r
Hence
∞
ρ (r ) r3
2(n+1)
∞
J n2 (r ) + J n2+1 (r ) C ρ (r ) dr C (αn + αn+1 ) 2 . r2 n
dr C 2(n+1)
This proves that
∞
ρ (r ) r3
dr
C
μ2
.
2μ
Thus
∞
ρ (r ) w (r ) dr μ2 1 / 2 r (r − μ)1/2
2
μ
2μ
∞
ρ (r ) r3
dr C
2μ
and (17) holds. If n ∈ N, then by Lemma 5 n+ n1/3
n+ n1/3
ρ (r ) w (r ) dr ∼ n n−n1/3
2/3
J n2 (r )ρ (r ) w (r ) dr
C n4/3
for n large, proving (15) in this case. If from the previous case since
μ = n + α with n ∈ N and 0 < α < 1. Let ϕ (μ) and ψ(μ) as in (16). Then (15) follows
ψ(μ), ϕ (μ) ⊂ ψ(n), ϕ (n) ∪ ψ(n + 1), ϕ (n + 1) . Remark 12. Notice that from Proposition 11, if we assume that
sup
1
1/2 μμ0 μ
then
,
n−n1/3
2μ μ+μ1/3
ρ (r ) dr C , (r − μ)1/2
ρ is a Carleson symbol if and only if (8) holds.
2
(18)
ρ satisfies (19)
J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
It can be easily seen that any radial function
sup
1
Ω b
373
ρ such that
a+b
ρ (r ) dr < C ,
(20)
a
where
! Ω = μ, a, b: μ μ0 , μ + μ1/3 a 2μ, μ1/3 b μ with μ0 > 0 and C an absolute constant satisfies (19). In particular if for a fixed b > 0
sup a >1
then
1
a+b
ρ (r ) dr < ∞,
b
(21)
a
ρ satisfies (20).
Now we study the question of how far condition (19) (and hence (8)) is from being necessary for symbol. We prove below the weaker result that
2μ
1
μ1/2
μ+μ1/3
is of order log μ if
ρ to be a Carleson
ρ (r ) dr (r − μ)1/2 ρ is Carleson symbol.
Proposition 13. Let us assume that ρ is a Carleson symbol, then there exists μ0 such that
2μ
1
sup
μμ0
1/2 log
μ
μ
ρ (r ) dr C , (r − μ)1/2
μ+μ1/3
(22)
where C is a universal constant. To prove Proposition 13, we need a sharp study of the function θ(μ, r ) in (13), in order to control the zeros of Bessel functions in the transition interval [μ + μ1/3 , 2μ]. More specifically we will use the fact that
J μ (r )2 + J
μ+
μ1/3
2 (r )
2 j /2
C
(r 2
− μ2 )1/2
,
where r ∈ (μ + 2 j μ1/3 , μ + 2 j +1 μ1/3 ), j ∈ {1, 2, . . . , M − 1}, and 2 M ∼ μ2/3 . Lemma 14. Given μ, let M ∈ N such that 2 M −1 2μ < 2 M and
μ + 2μ1/3 , 2μ ⊂
M −1 "
μ + 2 j μ1/3 , μ + 2 j+1 μ1/3 .
j =1
Then for r ∈ (μ + 2 j μ1/3 , μ + 2 j +1 μ1/3 ) and j ∈ {1, 2, . . . , M − 1} we have
√ μ1/3 √ θ(μ, r ) − θ μ + j /2 , r 2 2 2
2
6
(23)
and therefore
2 μ1/3 cos 2 2 cosθ(μ, r ) − θ μ + j /2 , r cos √ . 2 6 √
(24)
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J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
Proof. Let
f (t ) = θ(t , r ) = r 2 − t
12
t
π
r
4
− t arccos −
,
r > t,
then
t f (t ) = − arccos . r For r ∈ (μ + 2 j μ1/3 , μ + 2 j +1 μ1/3 ) and some t ∈ (μ, μ +
μ1/3 2 j /2
μ1/ 3 2 j /2
) we have
1/3 1/3 1/3 μ + μ2 j/2 μ θ(μ, r ) − θ μ + μ , r = μ arccos t μ arccos arccos . j j / 2 j / 2 j / 2 r μ+2 μ 2 2 2 μ + 2 j +1 μ 1/3
Since
1 arccos x =
1
√
dt ,
1 − t2
x
we have that
μ1/3
μ1/3
μ
arccos 2 j /2 μ + 2 j+1 μ 2 j/2
1
(1 − t )−1/2 dt 2
μ μ+2 j +1 μ
μ1/3 2 j /2
1−
1/2
μ μ
+ 2 j +1
μ
√ 2 2.
In a similar way
μ1/3 2 j /2
1/3
μ + μ2 j/2 2 arccos √ μ + 2 jμ 6 2
and the lemma follows.
Proof of Proposition 13. Let 16 A 2 log μ0
(a) max (b)
8 A2 23 j A
μ0
(c) j A <
1−cos √2
1 8 4π log μ0 , 100
<
2
, 8μA0 6
!
μ0
1 8
1 2
and j A ∈ N satisfying
1−cos √2
6
4π
,
,
where A is the constant in (14). It is enough to prove that μ+ 2 M μ1/3
sup
μμ0 μ+2 j A μ1/3
(r 2
ρ (r ) dr C log μ, − μ2 )1/2
(25)
for M the integer in Lemma 14. If ρ is a Carleson symbol, it satisfies μ+ 2 M μ1/3
M −1 J μ (r )2 ρ (r ) dr + j= j A
μ+2 j A μ1/3
In the other hand, by using
1
|a + b|2 |a|2 − |b|2 2
and (12) we have
μ+2j+1 μ1/3
J
μ+2 j μ1/3
μ+
μ1/3 2 j /2
2 (r ) ρ (r ) dr C · M ∼ C log μ.
(26)
J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
μ+ 2 M μ1/3
M −1 J μ (r )2 ρ (r ) dr + j= j A
μ+2 j A μ1/3 M −1 1
4π
j= j A
μ+2j+1 μ1/3
J
1/3
μ + μ j /2
2 (r ) ρ (r ) dr
2
μ+2 j μ1/3
μ+2j+1 μ1/3
cos2 θ(μ, r ) + cos2 θ
μ1/3
μ+
2 j /2
,r
μ+2 j μ1/3
2 1/3 h μ + μ , r ρ (r ) dr . 2 j /2
M −1 h(μ, r )2 ρ (r ) dr −
−
j= j A
μ+2 j A μ1/3
ρ (r ) dr (r 2 − μ2 )1/2
μ+2j+1 μ1/3
μ+ 2 μ1/3 M
375
μ+2 j μ1/3
Using Lemma 5 and assumption (b) above we have μ+ 2 M μ1/3
h(μ, r )2 ρ (r ) dr 2 A 2 μ4
μ+2
jA
μ+ 2 M μ1/3
μ+2
μ1/3 2
4
2A μ
jA
2A
2
M −1 j= j A
μ1/3 j +1
μ+2 μ
M −1 j= j A
μ+2 j μ1/3
μ+ 2 μ
2
1 1 − cos √6
μ+2
jA
μ+ 2 M μ1/3
ρ (r ) 8 A2 dr + μ (r 2 − μ2 )1/2
μ+2 j A μ1/3
μ+2 j A μ1/3
1/3
μ+2 j A μ1/3
μ+ 2 M μ1/3
ρ (r ) dr
ρ (r ) dr (r 2 − μ2 )1/2
ρ (r ) dr (r 2 − μ2 )1/2
1/3
4π
4
M
ρ (r ) 8 A2 dr + 2 2 1 / 2 μ (r − μ ) M
μ+ 2 μ
1/3
23 j
μ+2 j A μ1/3
μ+2
μ1/3
ρ (r ) 2 A2 dr + 2 3 3 2 2 1 / 2 (r + μ) (r − μ) (r − μ ) μ
μ+2 μ
23 j A
ρ (r ) dr jA
1
j +1
1
μ+ 2 M μ1/3
1/3
μ+2 j μ1/3
μ+ 2 M μ1/3
8 A2
ρ (r ) 2 A2 dr + (r 2 − μ2 )3 (r 2 − μ2 )1/2 μ2 1
ρ (r ) dr . (r 2 − μ2 )1/2
(27)
μ1/3
In a similar way we prove M −1 j= j A
μ+2j+1 μ1/3
μ+ 2 M μ1/3 2 2 1/3 μ 1 1 − cos √6 ρ (r ) h μ + , r ρ (r ) dr dr . 2 j / 2 4 4 π ( r − μ2 )1/2 2
μ+2 j μ1/3
μ+2
From Lemma 14
2
2
cos θ(μ, r ) + cos θ
μ+
μ1/3 2 j /2
jA
μ1/3 μ1/3 , r = 1 + cos θ(μ, r ) − θ μ + j /2 , r · cos θ(μ, r ) + θ μ + j /2 , r 2
2
1 − cos √ . 6
Hence we have M −1 1
4π
j= j A
μ+2j+1 μ1/3
cos2 θ(μ, r ) + cos2 θ
μ+2 j μ1/3
1 − cos √2
6
μ+ 2 M μ1/3
4π μ+2 j A μ1/3
ρ (r ) dr . (r 2 − μ2 )1/2
From this and (26)–(27), (25) follows.
(28)
μ1/3
2
μ+
μ1/3 2 j /2
,r
ρ (r ) dr (r 2 − μ2 )1/2
2
376
J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
From Propositions 11 and 13 we obtain Remark 15. If
ρ is a Carleson symbol, then we have 2
μ ρ (r ) w (r ) dr μ2/3 + r 1/2 (r − μ)1/2
sup μ>μ0
+
(μ,∞)−(μ+μ1/3 ,2μ)
1
μ2
log μ (μ+μ1/3 ,2μ)
ρ (r ) w (r ) dr < ∞. 1/2
μ2/3 + r 1/2 (r − μ)
3. Berezin transform and compact Toeplitz operators Recall that every Toeplitz operator T ρ is a diagonal operator {γn }n∈Z with γn as in (9) with respect to the basis {en } defined above. Then we have that T ρ is compact if and only if limn γn = 0. From the estimate (see [16]) n J n (r ) r er 2 /4 , n!2n
we see that every function ρ ∈ L 1 ([0, ∞), w (r ) dr ) with bounded support defines a compact Toeplitz operator. However there are compact Toeplitz operators whose symbols have unbounded support as shown in Proposition 21. Proposition 16. Let ρ ∈ L 1 ([0, ∞), w (r ) dr ). Then T ρ is compact in W 2 if and only if the inclusion W 2 → Wρ2 is compact. Proof. For the sufficiency, we argue by contradiction. Suppose that W 2 → Wρ2 is compact. If T ρ is not compact there exists a subsequence {γnk }k∈Z such that |γnk | ε for some ε > 0. Then the corresponding sequence {enk }k∈Z would have a convergent subsequence in Wρ2 . This is impossible since {enk }k∈Z is orthogonal in Wρ2 and enk W 2 = γn ε . Hence we
have that {γn }n∈Z converges to 0. Now suppose that γn → 0. Let {un }n∈Z be a bounded sequence in W 2 . We easily see that
un =
ρ
an,k ek
k∈Z
with convergence in W 2 . Since supn∈N k∈Z |an,k |2 < ∞, we can find an increasing sequence {ln }n∈Z in N and a sequence 2 of complex numbers {ak }k∈Z such that limn→∞ aln,k = ak for every k ∈ Z. By Fatou’s lemma we have that k∈Z |ak | < ∞, and if we let u =
k ak ek
uln − u Wρ2 =
∈ W 2 , we have that
γk |an,k − ak |2
k∈Z
converges to 0, proving that the inclusion W 2 → Wρ2 is compact.
2
In order to characterize the Hilbert–Smidt and Trace Class Toeplitz operators we define a Berezin type transform for the Toeplitz operators on W 2 . Let
Rx =
1
n2 + 1
(1 + |x|2 )1/2
1/2
F n (x)en (x) .
n∈Z
We will see that there exist positive constants such that
c R x W 2 C ,
(29)
for all x ∈ R . In fact, from (7) we have that 2
2 n2 + 1 F n (x) =
n∈Z
n2 + 1 J n2 (r ) =
n∈Z
By the summation formula (see [16])
J 0 |x − y | =
n∈Z
J n (r ) J n (s)e in(θ −ϕ ) ,
n∈Z
J n2 (r ) +
n=0
n2 J n2 (r ) = 1 +
n=0
n2 J n2 (r ).
J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
377
x = re i θ , y = se i ϕ , we have for x = y
n2 J n (r ) J n (s)e in(θ −ϕ ) =
n∈Z
It follows that
n=0 n
2 2 J n (r )
= r 2 J 1 (0) = r 2 /2 and
2 ∂2 J 1 (|x − y |) J 2 (|x − y |) J 0 |x − y | = x· y+ x · y⊥ . 2 ∂θ∂ ϕ |x − y | |x − y |
n2 + 1 J n2 (r ) ∼ 1 + r 2 ,
(30)
n∈Z
from which (29) follows.
ρ ∈ L 1 ([0, ∞), w (r ) dr ), we define the Berezin transform 2 1 2 B ρ (r ) = n + 1 γ J ( r ) . n n 1 + r2
Definition 17. For a symbol
n∈Z
Notice that for every x ∈ R2 we have
B ρ (r ) = ( T ρ R x , R x ). Remark 18. (a) If ρ is a Carleson symbol, then the Berezin transform B ρ is bounded. (b) If T ρ is a compact operator, then limr →∞ B ρ (r ) = 0. Proof. In fact, if T ρ is a bounded operator, then by (29) we have that ( T ρ R x , R x ) is bounded in R2 implying (a). If T ρ is compact, we have that (γn ) converges to 0. Then given ε > 0, we have by (30) that
1
1 + r2
2
n +1
2 n J n (r )
γ
|n|> M
ε
2
n +1
1 + r2
J n2 (r )
C ε,
|n|> M
for M large enough. Then
lim sup B ρ (r ) = lim sup r →∞
r →∞
Thus limr →∞ B ρ (r ) = 0.
|n| M
+
2
n +1
2 n J n (r )
γ
C ε.
|n|> M
2
Now we study the Toeplitz operators belonging to the Schatten classes s p . Recall that a bounded operator T in a Hilbert √ space belongs to the Schatten class s p if Tr | T | p < ∞, where Tr A stands for the trace of the operator A and | T | = T ∗ T (see [15, p. 215]). Since T ρ is a diagonal operator with respect to the basis {en }n∈Z , we have that T ρ ∈ s p if and only if {γn } ∈ p (Z). In this case T ρ s p = γn l p . For the trace class s1 and the Hilbert–Schmidt operators s2 we have the following theorems. Theorem 19. The following statements are equivalent: (1) T ρ ∈ s1 , (2) ρ ∈ L 1 ((1 + r 2 ) w (r ) dr ), (3) B ρ ∈ L 1 ((1 + r 2 ) w (r ) dr ), and the quantities T ρ s1 , ρ L 1 ((1+r 2 ) w (r ) dr ) , B ρ L 1 ((1+r 2 ) w (r ) dr ) are comparable. Proof. By (30) we have that
n∈Z
γn ∼
n2 + 1
n∈Z
∞
∞
ρ (r ) J n2 (r ) w (r ) dr = 0
ρ (r ) w (r ) 0
proving the equivalence of (1) and (2). On the other hand,
∞
n2 + 1 J n2 (r ) dr ∼
n∈Z
ρ (r ) 1 + r 2 w (r ) dr , 0
378
J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
γn ∼
n∈Z
γn
n∈Z
∞
2
n +1
∞ J n2 (r ) w (r ) dr
0
B ρ (r ) 1 + r 2 w (r ) dr ,
= 0
and the equivalence of (1) and (3) follows.
2 1/ 2
Theorem 20. T ρ ∈ s2 if and only if ρ B ρ ∈ L 1 ((1 + r 2 ) w (r ) dr ), moreover T ρ s2 and ρ B ρ L 1 ((1+r 2 ) w (r ) dr ) are comparable quanti-
ties.
Proof. We have
2 n
γ ∼
n∈Z
2
γn n + 1
n∈Z
∞
ρ (r )
J n2 (r ) w (r ) dr
=
∞
2
γn n + 1
∞ J n2 (r )
0 n∈Z
0
ρ (r ) B ρ (r ) 1 + r 2 w (r ) dr ,
ρ (r ) w (r ) dr ∼ 0
2
and this completes the proof.
More generally, we have the following partial results. Proposition 21. Let p 1. 1. If ρ ∈ L p ((1 + r 2 ) w (r ) dr ), then T ρ ∈ s p with
T ρ s p C p ρ L p ((1+r 2 ) w (r ) dr ) .
(31)
2. If T ρ ∈ s p , then B ρ ∈ L p ((1 + r 2 ) w (r ) dr ) and
B ρ L p ((1+r 2 ) w (r ) dr ) C T ρ s p .
(32)
Proof. Since
∞
J n2 (r ) w (r ) dr ∼ n2 + 1
− 1
,
0
we have by Jensen’s inequality and estimate (30) that
p n
γ C
n∈Z
2
n +1
n∈Z
C
p
∞
ρ (r )
J n2 (r ) w (r ) dr
0
∞
∞ ρ (r ) J n2 (r ) w (r ) dr p ∞ ρ p (r ) J n2 (r ) w (r ) dr 0 0 ∞ 2 ∞ 2 C C J n (r ) w (r ) dr J n (r ) w (r ) dr 0 0 n∈Z n∈Z ∞
n2 + 1 J n2 (r )ρ p (r ) w (r ) dr C
0 n∈Z
ρ p (r ) 1 + r 2 w (r ) dr , 0
proving (31). By (30) we have by Jensen’s inequality again that
B
p
ρ (r )
1 1 + r2
p 2 n n
γ
+1
J n2 (r )
.
n∈Z
Then integrating this inequality with respect to (1 + r 2 ) w (r ) dr we obtain (32).
2
Notice that the Berezin transform B ρ is an integral operator on [0, ∞) provided with the measure (1 + r 2 ) w (r ) dr and given by
∞
M (r , s)ρ (s) 1 + s2 w (s) ds,
B ρ (r ) = 0
where M (r , s) is the symmetric kernel
M (r , s) =
2π
(1 + r 2 )(1 + s2 )
(n2 + 1)2 n∈Z
βn2
J n2 (r ) J n2 (s).
J.A. Barceló et al. / J. Math. Anal. Appl. 358 (2009) 364–379
379
Corollary 22. The Berezin transform B ρ is a bounded integral operator on L p ((1 + r 2 ) w (r ) dr ) for 1 p ∞. Proof. Since M is symmetric, it is enough to prove the statement for p ∈ [1, 2]. For p = 1, this is part of Theorem 19. Now we prove it for p = 2. Let ρ ∈ L 2 ((1 + r 2 ) w (r ) dr ), then T ρ ∈ s2 by Proposition 21. If (γn ) is the sequence corresponding to T ρ , then by Proposition 21 and Theorem 20 we have that
Bρ
2L 2 ((1+r 2 ) w (r ) dr )
n∈Z
∞ 2 n
ρ (r ) B ρ (r ) 1 + r 2 w (r ) dr ρ L 2 ((1+r 2 ) w (r ) dr ) B ρ L 2 ((1+r 2 ) w (r ) dr ) .
γ Cp 0
It follows that
B ρ L 2 ((1+r 2 ) w (r ) dr ) C p ρ L 2 ((1+r 2 ) w (r ) dr ) . Finally the result follows for p ∈ [1, 2] by interpolation.
2
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