Atomic decomposition of a weighted inductive limit

Atomic decomposition of a weighted inductive limit

RACSAM
Rev. R. Acad. Cien. Serie A. Mat.
VOL . 97 (2), 2003, pp. 325–337
Análisis Matemático / Mathematical Analysis
Atomic decomposition of a weighted inductive limit
J. Taskinen
Dedicated to the memory of Klaus Floret
Abstract. We study some structural questions concerning the locally convex space HV∞ , which consists
of analytic functions on the open unit disc. We construct an atomic decomposition in this space, using a
lattice of points of the unit disc which is more dense than a usual one. The coefficient space is a Köthe
sequence space. We also prove that HV∞ is not nuclear.
Descomposición atómica de un lı́mite inductivo ponderado
Resumen. Estudiamos algunas cuestiones estructurales acerca del espacio localmente convexo HV∞ ,
que está formado por funciones analı́ticas en el disco unidad abierto. Construimos una descomposición
atómica de este espacio, usando un retı́culo de puntos del disco unidad que es más denso que el usual.
También demostramos que HV∞ no es nuclear.
1.
Introduction
In the paper [10] we introduced the space HV∞ := HV∞ (D) consisting of analytic functions on the open
unit disc. This space is slightly larger than the classical Banach space H ∞ of bounded analytic functions.
The motivation of HV∞ is the fact that it behaves much better than H ∞ with respect to Bergman and Szegö
projections and the harmonic conjugation operator. The space HV∞ is aimed to be a substitute for H ∞ in
situations where the continuity of the above mentioned mappings is critical. In [10] we also proved that this
space is a some sense the smallest possible substitute.
In this work we continue the study of HV∞ by showing that it admits an atomic decomposition: every analytic function in this space can be presented as a linear combination of “atoms” defined using the standard
Bergman kernel. The coefficients form sequences belonging to a natural Köthe sequence space. For details,
see Theorem 1. While atomic decompositions are interesting in itself in harmonic analysis, our results also
give a representation of HV∞ as a complemented subspace of a Köthe–Schwartz coechelon space.
The atom functions are defined using a lattice of points of the unit disc. Our choice of the lattice is not
a typical one with essentially constant hyperbolic distances between the lattice points. In the present work
the lattice is much more dense. This leads to many simplifications in proofs.
Atomic decompositions were first studied by Coifman and Rochberg in the setting of Bergman spaces
on the disc, see [3] and especially the book [12], Chapter 4, for a simple presentation. It is intuitively quite
Presentado por José Bonet.
Recibido: 27 de Noviembre de 2002. Aceptado: 25 de Julio de 2003.
Palabras clave / Keywords: Weighted spaces of holomorphic functions, atomic decomposition, nuclear spaces.
Mathematics Subject Classifications: Primary 46A13; Secondary 46E10, 46A11, 30H05
c 2003 Real Academia de Ciencias, España.
325
J. Taskinen
clear that our method could be used also in the classical case of Bergman spaces to make many proofs
shorter or even trivial.
∞
Before proceeding to the atomic decomposition let us recall from [10] the definition
R of HV . By dA we
denote the normalized 2–dimensional Lebesgue measure on D; moreover, m(B) := B dA for a measurable subset B of D.
Definition 1 By V we denote the set of radial weight functions v : D → R+ i.e. positive bounded
continuous functions, which satisfy for every n ∈ N
sup v(z)| log(1 − |z|)|n ≤ Cn
(1)
z∈D
and moreover
p
1 − |z| ≤ v(z) ≤ 1
for all z ∈ D.
(2)
Notice that (2) is not explicitely required in [10]; nevertheless the spaces in the next definition are the same
as those in [10].
Definition 2 We define the locally convex space
n
o
HV∞ := f : D → C analytic kf kv := sup |f (z)|v(z) < ∞ for all v ∈ V .
(3)
z∈D
The topology of HV∞ is determined by the uncountable family {k·kv |v ∈ V } of seminorms. It is a complete
countable inductive limit of Banach spaces: HV∞ = indk→∞ Hv∞k , where vk (z) := min{1, | log(1−|z|)|−k }
and
n
o
Hv∞k := f : D → C analytic kf kvk < ∞ .
(4)
It is a Schwartz space; in particular, the bounded and precompact subsets coincide. It is not so surprising
that the weight system is not “steep” enough to make the space a nuclear one. We consider this question in
Section 4.
The dual of HV∞ (with respect to the topology of uniform convergence on bounded sets) is the space
n
1
HW
:=
f : D → C analytic kf kk :=
Z
o
|f (z)|| log(1 − |z|)|k dA(z) < ∞ for all k ∈ N .
(5)
z∈D
R
1
The dual pairing is the usual hf, gi := D f ḡdA. The spaces HV∞ and HW
are reflexive, hence, the duality
1 0
∞
1
(HW )b = HV also holds. The dual HW is a Fréchet–Schwartz space, whose topology is determined by
the sequence of seminorms (k · kk )∞
k=1 . For more details, especially for a proof of the duality, see [10].
We define K(z, w) := 1/(1 − z w̄)2 , where z, w ∈ D, and we denote by R the Bergman projection
Z
Rf (z) = K(z, w)f (w)dA(w).
(6)
D
∞
∞
According to [10], R is a continuous projection from L∞
V onto HV , where LV is defined as in (3) but
“measurable” replacing “analytic” and “ess sup” replacing “sup”.
326
Atomic decomposition
To fix a constant later we remark that R also maps the space L∞ continuously into L∞
v1 (definition
analogous to (4) ); see [10]. Moreover, R is self–dual with respect to the dual pairing h·, ·i. Let us thus
show that R is also a bounded operator from
Z
o
n
L11 := f : D → C measurable kf k1 :=
|f (z)|(1 + | log(1 − |z|)|)dA(z) < ∞
(7)
z∈D
into L1 := L1 (D, dA): for f ∈ L11 and g ∈ L∞ ,
Z Z f (ζ)ḡ(z)
dA(ζ)dA(z)
|hRf, gi| = ¯2
(1 − z ξ)
D D
Z
Z
|f (ζ)|
D
D
≤
g(z)
dA(z)
dA(ζ)
(1 − z̄ξ)2
Z
|f (ζ)Rg(ζ)|dA(ζ)
=
D
Z
≤ C
|f (ζ)|(1 + | log(1 − |ζ|)|)dA(ζ) ≤ C 0 .
D
So, let us denote by C0 the positive constant such that (denoting the norm of L1 by k · k0 )
kRgk0 ≤ C0 kgk1
(8)
for all g ∈ L11 .
Concerning other notations and terminology, for analytic function spaces we refer to the book of Zhu,
[12], and for locally convex spaces we mention [6], [7] and [8] and also the paper [10].
2.
Preliminaries for the atomic decomposition.
We next turn to the construction of the atomic decomposition. Atomic decompositions for analytic function
spaces on the unit disc were first introduced by Coifman and Rochberg, see [3]. A simple presentation of the
classical result is given in [12], Chapter 4. By an “atomic decomposition” of some analytic function space
we roughly mean a way to write an analytic function as an infinite linear combination of simple building
blocks, “atoms”, in such a way that the complex coefficients for a sequence belong to a relevant sequence
space.
For our function space HV∞ the “atom” functions, i.e. the building blocks, will be defined using the
Bergman kernel, as usual. Another essential ingredient is a suitable decomposition of the unit disc D. Our
decomposition is finer than the usual one, which consists of sets with essentially constant hyperbolic area;
see [3] or [12]. Our choice actually makes many preliminary lemmas nearly trivial to prove. It is very
probable that this could be used to simplify the proofs also in the classical case of Bergman spaces Ap (D).
We are aware of the fact that on the other hand our choice of the lattice is not an optimal, i.e. small, one.
Hence we also do not particularly try to achieve optimal constants in various inequalities; the estimates may
be quite crude in some cases.
Definition 3 Let us fix the constant M = (C0 + 1)10000, where C0 is as in (8). For all m ∈ N, define
first rm = 1 − m−1/3 ; then define for every m, for every m0 := 0, 1, 2, . . . , M
rm,m0 := rm +
m0
(rm+1 − rm )
M
(9)
327
J. Taskinen
and θ(m, k) := 2πk/[M m4/3 ] for k = 0, 1, . . . , [M m4/3 ]. (Here [a] stands for the largest integer strictly
smaller than a + 1.) Define a bijection ρ from N onto the set {(m, m0 , k) | m ∈ N, m0 = 0, 1, 2, . . . , M −
1, k = 0, 1, 2, . . . , [M m4/3 ] − 1}. For every n ∈ N, pick up m, m0 and k such that n = ρ−1 (m, m0 , k)
and define
Dn := {z = reiθ ∈ D | rm,m0 ≤ r ≤ rm,m0 +1 and θ(m, k) ≤ θ ≤ θ(m, k + 1)}.
(10)
We clearly have
[
Dn = D,
(11)
n
and moreover, the set formed by the numbers z ∈ D belonging to more than one of the sets Dn has
Lebesgue measure 0.
Lemma 1 For every n the Euclidean diameter of Dn satisfies
diamDn ≤ inf
z∈Dn
20(1 − |z|)4
M
(12)
If n is given and m, m0 and k are such that n = ρ−1 (m, m0 , k), then
m0 + 1
(rm+1 − rm ) ≥ 1 − rm+1 = (m + 1)−1/3 . (13)
inf (1 − |z|) = 1 − rm,m0 +1 = 1 − rm +
z∈Dn
M
P ROOF.
On the other hand, the arguments of the points of Dn may vary at most from θ(m, k) to θ(m, k + 1), hence
at most 2π/(M m4/3 ). The radii may vary from rm,m0 to rm,m0 +1 , which amounts to a change of (see (9) )
M −1 (rm+1 − rm ) =
1 1 m−4/3
1
− 1 + 1/3 ≤
.
1−
1/3
M
M
(m + 1)
m
(14)
This implies the claim.
Corollary 1 For every analytic function f on D, for all n, we have
|f (z) − f (w)| ≤
for all z, w, λ ∈ Dn .
20
(1 − |λ|)4 sup |f 0 (ζ)|
M
ζ∈Dn
(15)
The proof is a direct application of the previous lemma and the mean value theorem.
The crucial step for our main result is to strengthen the classical approximation lemma (e.g. [12],
Lemma 4.3.4) using our new decomposition of D. Notice that the (relatively easy) proof is based on the
density of our decomposition.
p
Lemma 2 Let v ∈ V , in particular, assume 1 − |z| ≤ v(z) ≤ 1. For all n ∈ N, let λn be an interior
point of Dn .
R
1
a) For all g ∈ HW
with |g|dA ≤ 1 we have
D
∞ Z
X
n=1
Dn
|g(z) − g(λn )|
1
dA(z) ≤ 200M −1
v(z)
(16)
b) For all f ∈ HV∞ with |f (z)| ≤ 1/v(z) on D, and for all n ∈ N, we have
sup |f (z) − f (λn )| ≤ 200M −1 sup (1 − |z|)1/4 .
z∈Dn
328
z∈Dn
(17)
Atomic decomposition
P ROOF.
In the case a) we estimate using Corollary 1 (choose λ := z there) and (2)
∞ Z
X
1
|g(z) − g(λn )|dA(z)
v(z)
n=1 Dn
∞ Z
20 X
≤
(1 − |z|)−1/2 (1 − |z|)4 sup |g 0 (ζ)|dA(z)
M n=1 Dn
ζ∈Dn
(18)
Since sup (1 − |ζ|) ≤ (1 + 1/10)(1 − |z|) for all z ∈ Dn (e.g.by Lemma 1), we can bound (18) by
ζ∈Dn
∞
40 X
40
m(Dn ) sup (1 − |z|)7/2 |g 0 (z)| ≤
sup (1 − |z|)7/2 |g 0 (z)|
M n=1
M z∈D
z∈Dn
(19)
g(ζ)
dA(ζ)
(1−z ζ̄)2
once under the integral sign to
Z
80
g(ζ)ζ̄
7/2 dA(ζ)
sup (1 − |z|) 3
M z∈D
(1 − z ζ̄)
(20)
Here we differentiate the reproducing formula g(z) =
R
D
bound (19) by
D
80
≤
M
Z
|g(ζ)|dA(ζ) ≤
80
.
M
(21)
D
As for b), let f be as indicated. Again by Corollary 1
20
sup (1 − |z|)4 |f 0 (z)|
M
z∈Dn
z∈Dn
20 ≤
sup (1 − |z|)7/2 |f 0 (z)|
sup (1 − |z|)1/4 ,
M z∈Dn
z∈Dn
sup |f (z) − f (λn )| ≤
and as in the previous case we see that the first factor is bounded by
Z
Z
40
40
200
|f (ζ)|dA(ζ) ≤
1/v(ζ)dA(ζ) ≤
.
M
M
M
D
(22)
D
The part b) of the above result immediately implies.
Corollary 2 Let v ∈ V . If kf kv ≤ 1, we have
|f (λ)| ≤
for every λ, z ∈ Dn .
3.
2
v(z)
(23)
Main result.
In this section we construct an atomic decomposition for the space HV∞ . We shall show that every element
f of HV∞ can be presented as a linear combination
∞
X
α(n)m(Dn )
f (z) =
,
(1 − λ̄n z)2
n=1
(24)
329
J. Taskinen
where λn ∈ Dn and the coefficient sequence (α(n))∞
n=1 belongs to a Köthe sequence space K∞ (A).
Conversely, for every sequence (α(n))∞
n=1 ∈ K∞ (A), the analytic function defined by (24) belongs to
HV∞ . However, given f ∈ HV∞ , the sequence (α(n))∞
n=1 is not unique. (The atoms do not form a Schauder
basis.)
We now consider the sequence space relevant to the atomic decomposition. Since HV∞ is a countable
inductive limit of Banach spaces, the same is also true for the sequence space.
Definition 4 Let us fix, for all n ∈ N, an interior point λn of Dn . For all k ∈ N, let ak be the sequence
of positive real numbers with
ak (n) := | log(1 − |λn |)|k , n ∈ N.
(25)
Denote A := {ak | k ∈ N}, define the Köthe co–echelon space K∞ (A) corresponding to A as in [1],
Section 1.
For a quick definition of K∞ (A), let us denote by VN the set of all bounded positive sequences ṽ
satisfying
ṽ(n) ≤ Ck /ak (n) for all k, n ∈ N.
(26)
Then a sequence α = (α(n))∞
n=1 of complex numbers belongs to K∞ (A) if and only if for every ṽ ∈ VN
one can find a C > 0 such that
sup |α(n)|ṽ(n) ≤ C.
(27)
n∈N
The (non–metrizable) topology of K∞ (A) is defined by the seminorms
kαkṽ := sup |α(n)|ṽ(n).
(28)
n∈N
Let us next define three operators. Let the sequence (λn )∞
n=1 be as above. Let f be analytic on D and
be
a
sequence
of
complex
numbers.
Set
let α = (α(n))∞
n=1
Sf (z)
:=
∞
X
m(Dn )f (λn )
,
(1 − λ̄n z)2
n=1
(29)
T (α)
:=
∞
X
α(n)m(Dn )
,
(1 − λ̄n z)2
n=1
(30)
(Qf )(n)
:= f (λn )
for n ∈ N.
(31)
Hence, formally S = T Q. These operators have the following continuity properties.
Lemma 3 The operators S : HV∞ → HV∞ , T : K∞ (A) → HV∞ and Q : HV∞ → K∞ (A) are continuous.
1
P ROOF. 1◦ Consider T . To prove the continuity it suffices to take an arbitrary bounded set B ∈ HW
and
◦
show that a neighbourhood of 0 in K∞ (A) is mapped into the polar B of B by T , where
B ◦ := {g ∈ HV∞ | |hf, gi| ≤ 1 for every f ∈ B}.
(32)
R
1
We first claim that B is contained in a set G := {f ∈ HW
| |f |v(z)−1 ≤ 10} for a v ∈ V . To find v
we use the boundedness of B to pick the numbers Ck ≥ 1 such that B ⊂ Ck Uk for every k; here Uk is the
1
closed unit ball of the continuous seminorm k · kk of HW
. This means that every f ∈ B satisfies, for all k,
Z
|f || log(1 − |z|)|k dA ≤ Ck .
(33)
D
330
Atomic decomposition
Define v by
v(z) := P
∞
k=0
Since
∞
X
1
.
(34)
k!−1 4−k Ck−1 | log(1 − |z|)|k
1
k!−1 4−k | log(1 − |z|)|k = e− 4 log(1−|z|) = (1 − |z|)−1/4 ,
k=0
we get the estimate
v(z) ≥ (1 − |z|)1/4 .
(35)
We clearly have sup v(z)| log(1 − |z|)|k ≤ k!4k Ck for every k, hence, v ∈ V . Moreover, it follows easily
z
from (33) and (34) that every function in B belongs to the set G:
Z
|f |v −1 dA ≤
∞
X
k=0
D
Z
k!−1 Ck−1
|f || log(1 − |z|)|k dA =
∞
X
k!−1 ≤ 10.
(36)
k=0
D
Set ṽ(n) := v(λn ) for n ∈ N; we then Rhave (ṽ(n))∞
n=1 ∈ VN , by definitions. For kαkṽ ≤ C we thus
have, by the reproducing formula f (λn ) = f (ζ)(1 − λn ζ̄)−2 dA(ζ),
∞
X
α(n)m(Dn )f¯(λn )
|hT α, f i| = n=1
≤
∞
X
|α(n)|ṽ(n)v(λn )−1 m(Dn )|f (λn )|
n=1
≤
kαkṽ
∞
X
v(λn )−1 m(Dn )|f (λn )| .
(37)
n=1
We prove in a moment that
1/v(λn ) < C/v(z)
(38)
for every n, for every z ∈ Dn . Using this, Lemma 2.b), the definition of the set G and (35), we can estimate
(37) by a constant times
XZ
XZ
|f (z)|v(z)−1 dA +
(1 − |z|)4 v(z)−1 dA ≤ C.
(39)
n
Dn
n
Dn
The statement (38) follows from the smallness of diamDn . First,
∞
X
∂v(r)−1
=
k!−1 k4−k Ck−1 | log(1 − r)|k−1 (1 − r)−1 ≤ C(1 − r)−1−1/4 ,
∂r
k=1
This, diamDn ≤ C(1 − |z|)4 (Lemma 1) and the mean value theorem imply |1/v(z) − 1/v(λn )| ≤
C(1 − |z|)2 ≤ C 0 ≤ C 0 /v(z). Hence, (38) follows.
2◦ Concerning Q, let the weight sequence ṽ ∈ VN be given. We define v ∈ V as follows. First let
(ρj )∞
j=1 , 0 ≤ ρj < 1, be the unique strictly increasing sequence which satisfies: for every j there exists an
n such that ρj = |λn |. Let us define the subset Nj of N by Nj = {n | |λn | = ρj }.
For every j, define for all z ∈ D with |z| = ρj ,
v(z) = sup ṽ(n).
(40)
n∈Nj
331
J. Taskinen
For every r ∈ [0, 1[, extend v(r) piecewise linearly, and then extend v radially to D.
We show that v ∈ V . If z ∈ D satisfies |z| = ρj for some j, and k ∈ N is given, then
v(z)
=
sup ṽ(n) ≤ sup Ck /ak (n)
n∈Nj
=
n∈Nj
sup Ck | log(1 − |λn |)|−k = Ck | log(1 − |z|)|−k .
(41)
n∈Nj
Concerning other numbers z, Lemma 1 implies |ρj − ρj+1 | ≤ Cj (1 − ρj )4 , hence, for every k we have
| log(1 − ρj )| ≤ | log(1 − ρj+1 )| ≤ ck | log(1 − ρj )|
(42)
for a constant ck > 0. So, if z satisfies ρj < |z| < ρj+1 , then the piecewise linearity of v and (41) and (42)
imply
v(z) ≤ max{v(ρj ), v(ρj+1 )} ≤ Ck0 | log(1 − ρj+1 )|−k ≤ Ck0 | log(1 − |z|)|−k .
(43)
Hence, v ∈ V .
Now (40) and the definition of Q imply kQf kṽ ≤ Ckf kv , hence Q is continuous.
3◦ Finally, we have S = T Q, hence also S is continuous. The following fact can now be proved using a result of Garnir, De Wilde and Schmets to be found in
[5], p. 346.
Lemma 4 The operator S : HV∞ → HV∞ is invertible .
P ROOF. We want to show that the operator A := I − S
(i) is bounded (i.e. maps a neighbourhood of 0 of HV∞ into a bounded set), and
(ii) for some neighbourhood U ⊂ HV∞ of 0, A(U ) ⊂ U/2.
Then S is invertible by [5], p. 346.
Let us take a weight as small as
p
w(z) := 1 − |z|.
(44)
Then, for sure, w ∈ V . Let U ⊂ HV∞ be the unit ball of k · kw :
n
o
U := f ∈ HV∞ sup |f (z)|(1 − |z|)1/2 ≤ 1 ;
(45)
z∈D
We want to show that
1
Z
Af (z)ḡ(z)dA(z) ≤
2
(46)
D
for every f ∈ U , g in the unit ball of
L11
(see (8)), that is,
Z
n
o
g ∈ Ũ := h ∈ L1loc (D, dA) khk1 :=
|h(z)|(1 + | log(1 − |z|)|)dA(z) ≤ 1 .
(47)
z∈D
Here L1loc (D, dA) is the space of locally integrable functions on D.
Assuming (46) we show that (i) and (ii) hold. As for (i), let us pick up an arbitrary weight v ∈ V ; to
1
prove the boundedness of A it suffices to find a k such that for f ∈ HV∞ and g ∈ HW
with kf kv ≤ 1 and
kgkk ≤ 1 we have
|hAf, gi| ≤ C.
(48)
p
But v(z) ≥ 1 − |z|, see (2) in the definition of V . Hence, f ∈ U . Moreover, for every k ∈ N, k ≥ 2,
1
the subset {kgkk ≤ 1} ⊂ HW
is smaller than the set Ũ above. Hence, (48) follows from (46).
332
Atomic decomposition
Concerning (ii), the identification of the norm of L∞ (D, dA) as the dual norm of L1 (D, dA) implies
the following:
R
An analytic function ϕ belongs to U/2, if | D ϕḡ| ≤ 1/2 for every
Z
n
o
g ∈ U ◦ := h ∈ L1loc (D) |h(z)|(1 − |z|)−1/2 dA(z) ≤ 1 .
(49)
D
But U ◦ is a much smaller set than Ũ , hence (ii) follows from (46).
So we want to prove (46). Let now f and g be given. Since Af is analytic and the Bergman projection
is self–dual, the identities
Z
Z
Af ḡ = (RAf )ḡ = hAf, Rgi
(50)
D
hold formally. Moreover, by (8),
kg̃k0 := kRgk0 ≤ C0 kgk1 ≤ C0
(51)
We thus have
Z
Z
∞
X
m(Dn )f (λn )g̃(λn )
Af ḡ = hAf, g̃i =
f (z)g̃(z)dA(z) −
D
D
=
∞ Z
X
n=1
n=1
∞
X
f (z) − f (λn ) g̃(z)dA(z) +
Dn
n=1
Z
f (λn ) g̃(z) − g̃(λn ) dA(z).
(52)
Dn
Using Lemma 2.b) and (51) we get the following bound for the first sum in (52):
Z
∞
X
|g̃(z)|dA(z)
sup |f (z) − f (λn )|
n=1 z∈Dn
≤ 200M −1
Dn
Z
|g̃(z)|dA(z) ≤ 2000C0 M −1 ≤ 1/5.
(53)
D
To estimate the second term of (52) from above, we first use Corollary 2 to bound |f (λn )| by
z ∈ Dn . Then Lemma 2 a) directly gives the bound
∞ Z
X
2
400C0
|g̃(z) − g̃(λn )|dA(z) ≤
.
w(z)
M
n=1 Dn
2
w(z)
for any
(54)
This completes the proof. We now formulate the main result as follows. We denote m(n) := m(Dn ); this number can be estimated, if necessary, from the definition of the sets Dn after (9). Recall that λn was chosen in Lemma 2: it
is an interior point of the set Dn . The space K∞ (A) was defined in the beginning of Section 3.
Theorem 1 For every (α(n))∞
n=1 ∈ K∞ (A) the function
f (z) =
∞
X
α(n)m(n)
(1 − λ̄n z)2
n=1
(55)
belongs to HV∞ (D), and the mapping (α(n))∞
n=1 7→ f is continuous between these two spaces.
For every analytic function f ∈ HV∞ (D) there exists a complex sequence (α(n))∞
n=1 ∈ K∞ (A) such
that
∞
X
α(n)m(n)
f (z) =
.
(56)
(1
− λ̄n z)2
n=1
∞
The mapping f 7→ (α(n))∞
n=1 can be made linear and continuous from HV (D) into K∞ (A).
333
J. Taskinen
P ROOF. The first statement follows from the continuity of T . For the second one, given f , choose g =
S −1 f and then define the sequence α = Qg, i.e. α(n) = g(λn ). We then obtain (56) from (29) and (31).
The continuity statement follows from the continuity of Q and Lemma 4. Finally, the operator P := QS −1 T is a continuous projection in the space K∞ (A); we have P 2 = P .
This implies
Proposition 1 The space HV∞ is isomorphic to a complemented subspace of the Köthe sequence space
K∞ (A). 4.
The space is not nuclear.
Intuitively it is quite clear that the space HV∞ cannot be nuclear, since the weight system is not “steep”
enough. Two arbitrary weights differ from each other at most by a factor which grows only logarithmically
on the boundary of D.
We give a proof for the nonnuclearity by extracting a subspace which is isomorphic to a nonnuclear
Köthe coechelon space. The method of the proof may have some independent interest. It resembles the
construction [2], Section 2. However, in [2] the weight system was non–radial, and the essential phenomena
occurred on the boundary of the disc. In the present work the weights are radial; consequently, we have to
play also with the radii of the points of D. More precisely, the interesting and crucial things happen in the
interior points zn := eiθn (1 − (2n)−20 ) of the subdomains Dn (to be chosen below). Of course there is a
lot of freedom in the construction; many other points could be used as well.
Notice that the space K∞ (A) above is not nuclear; however, the results of Section 3 do not imply that
HV∞ contains a subspace isomorphic to K∞ (A). The operator T need not be injection, since Q is not a
surjection.
Proposition 2 The space HV∞ is not nuclear, since it contains a subspace isomorphic to the Köthe coechelon space K∞ (M ), which is defined as K∞ (A) in the beginning of Section 3, but ak (n) replaced
by
µk (n) := (log n)k
(57)
So, a sequence α belongs to K∞ (M ), if and only if sup |αn |(log n)−k < ∞ for some k. It is well
n
known that K ∞ (M ) is not nuclear, see e.g. [9], Theorem 6.1.2.
P ROOF. Let θn := 1/n and Jn := [θn − εn , θn + εn ] for all n ∈ N, where εn := 2−9 n−4 . Notice that
the sets Jn are mutually disjoint. Choose the function ϕn : [0, 2π] → R+ such that ϕn (θ) = 1 for θ ∈ Jn ,
and ϕn (θ) = (2n)−4 for θ ∈
/ Jn . Let en be an analytic function on the disc defined by
en (z) := exp
1 Z2π eiθ + z
log ϕn (θ)dθ .
iθ
2π
e −z
0
The radial limits (on the boundary of the disc) of the functions en exist a.e. (we denote them by the same
symbol), and we have |en (eiθ )| = ϕn (θ) for a.e. θ. Compare with [2] and the references therein. Let us define
Dn := {z ∈ D | |z − eiθn | < 1/n4 } , Cn := D \ Dn .
We formulate a lemma containing some crucial estimates.
Lemma 5 1◦ For all z ∈ D, all n, we have |en (z)| ≤ 1.
2◦ Fix a z ∈ D; if n is such that | arg z − θn | = min {| arg z − θm |}, then |en (z)| ≤ 4n−4 (1 − |z| +
m∈N
n−4 )−1 . Moreover, if m 6= n, then |em (z)| ≤ 2−3 m−2 .
3◦ For every n we have |en (zn )| ≥ 1/2, where zn := eiθn (1 − (2n)−20 ).
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Atomic decomposition
P ROOF. The statement 1◦ follows immediately from the maximum principle, since the moduli of the
radial limits of every en are at most 1.
Concerning 2◦ , if z and n are as in the assumption, the Jensen inequality implies (as in [2])
|en (z)| =
Z2π iθ
1
e +z
Re
log
ϕ
(θ)dθ
exp
n
2π
eiθ − z
=
1 Z2π 1 − |z|2
log
ϕ
(θ)dθ
exp
n
2π
|eiθ − z|2
0
0
≤
1
2π
Z2π
1 − |z|2
ϕn (θ)dθ.
|eiθ − z|2
(58)
0
We may assume |z| ≤ 1 − n−4 ; otherwise there is nothing to prove. We have |z − eiθ | ≥ n−4 , hence,
|z − eiθ | ≥ max(|z − eiθ |, n−4 ) ≥ max(1 − |z|, n−4 ) ≥ (1 − |z| + n−4 )/2. So we have
1 − |z|2
1 − |z|2
4
≤
≤
.
iθ
2
iθ
|e − z|
(1 − |z|)|e − z|
1 − |z| + n−4
Dividing the integration interval [0, 2π] in (58) into the parts Jn and [0, 2π] \ Jn , we thus find that the first
part yields the estimate
2
,
4
n (1 − |z| + n−4 )
since the lenght of Jn is smaller than 2−1 n−4 . The second part has the smaller bound (2n)−4 , since
ϕn = (2n)−4 on [0, 2π] \ Jn .
In the case m 6= n we again have
|em (z)|
≤
1
2π
Z2π
1 − |z|2
ϕm (θ)dθ.
|eiθ − z|2
(59)
0
Dividing the integration interval into the parts Jm and its complement, on Jm we have
|eiθ − z| ≥
1 |n − m|
1 1
.
− =
2 m n
2mn
This is always larger than the number 1/(4m2 ), as seen by considering the cases n ≤ 2m and n > 2m
separately: in the former case one uses |n − m| ≥ 1 and 2mn ≤ 4m2 , and in the latter |n − m| ≥ n/2 and
a cancellation of n. Taking into account the length 2−9 m−4 of the interval Jm , this part yields an estimate
2−7 m−2 in (59). The rest of the integral is bounded by (2m)−4 , since ϕm = (2m)−4 on [0, 2π] \ Jm .
For the lower bound 3◦ , we have log ϕn = 0 on Jn and log ϕn = −4 log(2n) < 0 on the complement
of Jn , hence
Z
1
1 − |zn |2
|en (zn )| = exp
log
ϕ
(θ)dθ
n
2π
|eiθ − zn |2
[0,2π]\Jn
≥ exp −
4
2π
Z
(2n)−20
log(2n)dθ
n−18
[0,2π]\Jn
≥ exp(−2−10 n−1/2 ) ≥
1
. 2
(60)
335
J. Taskinen
We return to the proof of the proposition. We define the mapping ψ from K∞ (M ) into HV∞ by ψ :
∞
P
(αn )∞
αn en . We claim that the mapping is an isomorphism onto its image. Since the spaces are
n=1 7→
n=1
strong duals of Fréchet–Schwartz spaces, it suffices to show that, for some constants ck , Ck > 0,
ck sup |αn |(log n)−k ≤ k
X
αn en kvk ≤ Ck sup |αn |(log n)−k
n∈N
(61)
n∈N
−1/2
for all k. Since the expression µ((αn )∞
is obviously a continuous seminorm on
n=1 ) := sup |αn |n
n
K∞ (M ), we may assume that µ((αn )) ≤ 1, i.e. |αn | ≤ n1/2 for all n in (61). We may moreover assume
that |αn | ≥ 1 there, for all n.
To prove the left hand side inequality, choose N such that
|αN |(log(N ))−k ≥
1
sup |αn |(log n)−k .
2 n∈N
(62)
We then have, by 3◦ of Lemma 5 above, and by |αn | ≥ 1,
1 −k
20 (log(2N ))−k ,
2
|αN eN (zN )|vk (zN ) ≥
(63)
and on the other hand
X
|αn en (zN )|vk (zN ) ≤
n6=N
Here
P
X
n1/2 2−3 n−2 20−k (log(2N ))−k
(64)
n6=N
n−3/2 2−3 ≤ (1 +
n
R∞
x−3/2 )2−3 ≤ 3/8. Hence, combining (63) and (64) and using the triangle
1
inequality we get
∞
X
X
1
αn en (zN )vk (zN ) ≥ |αN eN (zN )|vk (zN ) − αn en (zN )vk (zN ) ≥ |αN eN (zN )|vk (zN ),
8
n=1
n6=N
hence, by (62) and (63), we get
20−k sup |αn |(log n)−k ≤ 2 · 20−k |αN |(log N )−k ≤ C|αN eN (zN )|vk (zN )
n
∞
X
X
≤ C 0
αn en (zN )vk (zN ) ≤ C 0 αn en vk
n=1
We finally prove the right hand side of (61). Let N still be as above, let z ∈ D and let n be such that
| arg z − θn | = min {| arg z − θm |}. By 2◦ of Lemma 5 we have (since always vk (z) ≤ 1 )
m∈N
X
|αm em (z)|vk (z) ≤
m6=n
≤
X
X
|αm |m−2
m6=n
−3/2
Ck m
|αm |(log m)−k ≤ Ck0 |αN |(log N )−k .
(65)
m6=n
Concerning the nth term, for |z| ≥ 1 − n−3 we have by 1◦
vk (z)|αn en (z)| ≤ vk (1 − n−3 )|αn en (z)| ≤ |αn |vk (1 − n−3 ) ≤ 3|αn |(log n)−k ≤ 3|αN |(log N )−k
336
Atomic decomposition
For |z| ≤ 1 − n−3 we have
vk (z)|αn en (z)| ≤
≤
4|αn |
4|αn |
≤ 4 −3
n4 (1 − |z| + n−4 )
n (n + n−4 )
4|αn |
≤ Ck |αn |(log n)−k ≤ Ck |αN |(log N )−k .
n
This and (65) imply that the right hand side of (61) holds.
Acknowledgement. The author is grateful for José Bonet for showing him the reference [5], see
Lemma 4. The author also wishes to thank the referee for the careful reading of the paper and for some
remarks and corrections. The research is partially supported by the Academy of Finland.
References
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[6] Jarchow, H. (1981). Locally Convex Spaces, Teubner, Stuttgart.
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[8] Pérez Carreras, P., and Bonet, J. (1987). Barreled Locally Convex Spaces, North–Holland Mathematics Studies
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[10] Taskinen, J. On the continuity of the Bergman and Szegö projections, Houston J. Math., to appear.
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J. Taskinen
Department of Mathematics
University of Joensuu
P.O.Box 111
FIN–80101 Joensuu. Finland
[email protected]
337