We consider the action of the operator ℒg(z)=(1-z)-1∫z1f(ζ)dζ on a class of “mixed norm” spaces of analytic functions on the unit disk, X=Hα,νp,q, defined by the requirement g∈X⇔r↦(1-r)αMp(r,g(ν))∈Lq([0,1],dr/(1-r)), where 1≤p≤∞, 0<q≤∞, α>0, and ν is a nonnegative integer. This class contains Besov spaces, weighted Bergman spaces, Dirichlet type spaces, Hardy-Sobolev spaces, and so forth. The expression ℒg need not be defined for g analytic in the unit disk, even for g∈X. A sufficient, but not necessary, condition is that ∑n=0∞|g^(n)|/(n+1)<∞. We identify the indices p, q, α, and ν for which 1∘ℒ is well defined on X, 2∘ℒ acts from X to X, 3∘ the implication g∈X⇒∑n=0∞|g^(n)|/(n+1)<∞ holds. Assertion 2∘ extends some known results, due to Siskakis and others, and contains some new ones. As an application of 3∘ we have a generalization of Bernstein’s theorem on absolute convergence of power series that belong to a Hölder class.
1. Introduction and Definitions
Let H(𝔻) denote the class of all functions holomorphic in the unit disk 𝔻 of the complex plane. In [1], Libera introduced the operator
(1)g(z)⟼2z∫0zg(ζ)dζ
and showed its importance in the theory of univalent functions. In particular, it was shown in [1] that this operator transforms the class of star-like functions into itself. Since then many papers were published devoted to this aspect of the Libera operator. The “generalized” Libera operator
(2)Λag(z)=1a-z∫zag(ζ)dζ,where|a|≤1,
was introduced and studied from the functional analytic point of view by Siskakis in [2, 3], then in [4–6], and other papers (see [7] for further references). If |a|<1, then Λa is defined on H(𝔻), and, on classical spaces such as Hardy, Bergman, and Besov, has almost the same linear topological properties as the integration operator g(z)↦∫0zg(ζ)dζ, and therefore is not so interesting from the functional analytic point of view (cf. [8]). So we can assume that |a|=1. In fact we can and will assume that a=1, so
(3)Λ1g(z)=11-z∫z1g(ζ)dζ,
whenever the integral is somehow defined. This definition of Λ1 requires further explanation because the integral need not be defined for g∈H(𝔻) (e.g., g(z)=1/(1-z)).
Endowed with the topology of uniform convergence on compact subsets of 𝔻, the class H(𝔻) becomes a complete locally convex space. The dual of H(𝔻) is equal to H(𝔻¯), where g∈H(𝔻¯) means that g is holomorphic in a neighborhood of 𝔻¯ (depending on g). The duality pairing is given by
(4)〈f,g〉=∑n=0∞f^(n)g^(n),
where f(z)=∑n=0∞f^(n)zn∈H(𝔻) and g(z)=∑n=0∞g^(n)zn∈H(𝔻¯) (see, e.g., [9]). Clearly, Λ1 is well defined on H(𝔻¯), and it is easy to check that it maps H(𝔻¯) into H(𝔻¯), and
(5)Λ1g(z)=∑n=0∞(∑k=n∞g^(k)k+1)zn=∫01g(t+(1-t)z)dt.
The last integral is obtained from (3) by integration over the straight line joining z and 1.
Definition 1.
We use the symbol ℒ¯ to denote the operator Λ1:H(𝔻¯)↦H(𝔻¯).
Definition 2.
We denote by ℒ the operator ℒg(z)=∫01g(t+(1-t)z)dt whenever the integral converges uniformly on compact subset of 𝔻. “Uniform convergence” means that the limit
(6)limx→1-∫0xg(t+(1-t)z)dt
is uniform with respect to z. This hypothesis guarantees that ℒg is an analytic function.
It is easy to verify the validity of the following.
Proposition 3.
The dual operator (ℒ¯)*:H(𝔻)↦H(𝔻) coincides with the Cesàro operator, 𝒞, defined on H(𝔻) by
(7)𝒞f(z)=∑n=0∞(1n+1∑k=0nf^(k))zn.
Conversely, the dual of 𝒞:H(𝔻)↦H(𝔻) coincides with ℒ¯.
In [6], it was proved that ℒ is defined on the Bloch space and maps it into BMOA. This assertion, which improves the earlier result that ℒ maps Bloch space into itself (e.g., [4, 10]), was deduced from a result of Nowak [11] and Proposition 3.
However, as it can easily be seen, the operator ℒ¯ cannot be extended to a continuous operator from H(𝔻) to H(𝔻). Moreover, ℒ¯ cannot be extended to a continuous operator from X to H(𝔻), where X is some of common spaces, for example,
(8)X={∫𝔻|g(z)|2(1-|z|)2dA(z)<∞g∈H(𝔻):∥g∥X=∫𝔻|g(z)|2(1-|z|)2dA(z)<∞},
see [6]. (dA denotes the Lebesgue measure on 𝔻). We will identify a large family of spaces that possess the same property.
In this paper we consider, in particular, the spaces listed in the following definition.
Definition 4.
Bloch type spaces:
(9)𝔅α={g∈H(𝔻):|g′(z)|=𝒪(1-|z|)-α},α>0.
Weighted Hardy spaces:
(10)Hαp={g∈H(𝔻):Mp(r,g)=𝒪((1-r)-α)},α≥0,
where
(11)Mp(r,h)=(12π∫02π|h(reiθ)|pdθ)1/p.
Weighted Bergman spaces:
(12)Aβp={g∈H(𝔻):∫𝔻|g(z)|p(1-|z|2)βdA(z)<∞},β>-1,1≤p<∞.
Dirichlet type spaces:
(13)𝒟βp={g∈H(𝔻):f′∈Aβp}·
The space 𝒟01 is closely related to H1 in that H1⊗H1=𝒟01, where X⊗Y denotes the set of all g∈H(𝔻) which can be represented as g=∑n=0∞hn*kn, hn∈X, kn∈Y, with ∑∥hn∥X∥kn∥Y<∞ (see [12]).
Concerning these spaces, we mention the following results.
Theorem A (see [2]). The operator ℒ acts as a bounded operator from Hp to Hp if and only if 1<p≤∞.
Theorem B (see [4, 10]). ℒ acts as a bounded operator from 𝔅α into 𝔅α if and only if 0<α<2.
Theorem C (see [3]). ℒ acts as a bounded operator from Ap,β into Ap,β if and only if p>β+2.
Theorem D (see [13]). If g^(n)≥0 for all n, and ∑(g^(n)/(n+1))<∞, then ℒg∈𝒟01 if and only if(14)∑n=0∞g^(n)log(n+2)n+1<∞.
Here “acts” means, among other things, that ℒg is defined on the space in the sense of Definition 2; that is,
(15)Ig(z):=∫01g(t+(1-t)z)dtconvergesuniformlyin|z|<ρ
for all ρ∈(0,1).
Condition (15) is implied, as is easily seen, by
(16)∑n=0∞|g^(n)|n+1<∞.
This condition is satisfied, according to Hardy’s inequality, for g∈H1 and therefore for g∈Hp∪𝒟01, p≥1, because Hp⊂H1 and 𝒟01⊂H1. However, (15) does not imply (16) even in the case when ℒ maps a space into itself. This can be seen from the following reformulation of Bernstein’s theorem and Theorem 5 below.
Theorem E (Zygmund, [14], Ch. VI (3.5)). Let 0<α<1. Then the implication g∈Hα∞⇒(16) holds if and only if 0<α<1/2.
Theorem 5.
ℒ acts as a bounded operator from Hα∞ into Hα∞ if and only if 0<α<1.
(The proof is very easy, although the theorem is a special case of Theorem 11.)
What we can deduce from condition (15) is that the series
(17)∑g^(n)n+1isAbelsummable.
Namely, taking z=0, we see that the integral
(18)Ig(0)=∫01g(t)dt=limx→∞g^(n)xn+1n+1
exists and is finite. However, it may happen that ℒ maps a space into itself and that
(19)|g^(n)|≥(n+1)s,
where s>0 is positive constants. For instance, as an application of Khinchin’s inequality and a profound result of Kisliakov, we have the following.
Theorem 6.
Let 1/2≤α<1. Then there is a function g∈Hα∞ such that |g^(n)|≥c(n+1)α-1/2 (although ℒ maps Hα∞ into Hα∞).
Proof.
See the proof of Proposition 34, Case (1), p=∞.
On the other hand, if a space X satisfies (15) for all g∈X, this does not mean that ℒ maps X into X (although it is defined on X). Besides H1, we have, for instance, the following.
Theorem F (see [13]). The space 𝒟01 is contained in H1and satisfies (16), but ℒ does not map 𝒟01 into H1.
One of the aims of the present paper is to extend Theorems A, B, C, 5, and 6 to a large scale of “mixed norm” spaces.
Definition 7.
We denote by Hαp,q (0<p, q≤∞), where α>0 when q<∞, and α≥0 when q=∞, the class of those g∈H(𝔻) for which
(20)∫01Mpq(r,f)(1-r)αq-1dr<∞(q<∞),sup0<r<1(1-r)αMp(r,f)<∞(q=∞).
Then letting ν be a nonnegative integer, we define the space
(21)Hα,νp,q={g∈H(𝔻):g(ν)∈Hαp,q}.
The (quasi) norm in Hα,νp,q is given by
(22)∥g∥α,νp,q=∑j=0ν-1|g(j)(0)|+(∫01Mpq(r,g(ν)(1-r)qα-1dr)1/q,
where ∑j=0-1 should be interpreted as equal to zero.
It is well known and easy to prove that these spaces are complete.
The norm (22) is not the most natural one but is convenient for technical reasons. For instance, in the case p=q, ν=0, a more (but not most) natural norm is given by
(23)(∫01Mpp(r,g)(1-r)pα-1rdr)1/p=(12π∫𝔻|g(z)|p(1-|z|)pα-1dA(z))1/p.
This norm is equivalent to that given by (22) because of the maximum modulus principle for analytic functions.
The space Hα,νp,∞=:Hα,νp is specific in that the set, 𝒫, of all analytic polynomials is not dense in it. The closure of 𝒫 in Hα,νp coincides with the “little oh” space
(24)hα,νp={g∈H(𝔻):Mp(r,g(ν))=o(1-r)-α(r↑1)}.
From now on, unless specified otherwise, we suppose that
(25)1≤p≤∞,0<q≤∞,α>0forq<∞,α≥0forq=∞.
It is sometimes more convenient to work with the Besov type spaces.
Definition 8.
Let β∈ℝ and choose any integer ν≥0 such that ν-β>0. The space 𝔅βp,q is defined as
(26)𝔅βp,q=Hν-β,νp,q.
If q=∞, we can assume that ν=β. It is well known that this definition is independent of ν; this follows immediately from Lemma A below. If β>0, then 𝔅βp,q is “true” Besov spaces; if β<0, then 𝔅βp,q=H-βp,q; if β=0, then it is called Hardy-Bloch spaces [15]. If q=∞, then we have the space
(27)𝔅βp={g∈H(𝔻):Mp(r,g(ν))=𝒪((1-r)β-ν)}(0<β<ν),
and its subspace
(28)𝔟βp={g∈H(𝔻):Mp(r,g(ν))=o((1-r)β-ν)(0<β<ν)Mp(r,g(ν))}.Lipschitz Spaces. It is known that the space 𝔅βp,q (β>0) coincides with the (Lipschitz) space Λβp,q consisting of those g in the p-Hardy space Hp for which
(29)(∫01(∥Δtνg∥ptβ)qdtt)1/q<∞.
Here Δtνg denotes the symmetric νth difference with step t,
(30)Δtνh(θ)=∑k=0ν(νk)(-1)ν-kh(θ+kt),whereh(θ)=g(eiθ),and ∥·∥p denotes the norm in Hp. This result is essentially due to Hardy and Littlewood and Zygmund (see [16] for a simple proof of a generalized variant). We also have
(31)𝔅βp=Λβp:=Λβp,∞={g∈Hp:∥Δtng∥p=𝒪(tβ)}(0<β<ν),𝔟βp=λβp={g∈Hp:∥Δtng∥p=o(tβ)(t↓0)}(0<β<ν).
In the case p=q=∞ and β=1, the space Λβp,q is denoted by Λ* and is called the Zygmund space. The corresponding “little” space is denoted by λ*. These spaces were introduced by Zygmund via symmetric differences. In [17], connections of λ* and λ* with Besov spaces were established.
Hardy-Sobolev Spaces. Let ν≥1 be an integer. The Hardy-Sobolev space Wνp consists of those g∈Hp such that g(ν)∈Hp; that is, Wνp=H0,νp,∞. It is known that
(32)Wνp={g∈Hp:sup0<t<1∥Δtνg∥p<∞}.
See, for example, [16]. In particular, W1∞ coincides with the usual Lipschitz space consisting of those f from the disk-algebra for which
(33)|f(ζ)-f(η)|≤C|ζ-η|(|ζ|=|η|=1).
There are various inclusions between members of the scale Hα,νp,q. Here we mention the following.
Proposition 9.
If α>0, then
Hα,νp,q⫋hα,νp⫋Hα,νp, for q<∞,
Hα,νp,q⫋Hβ,νu,q, where β=1/p-1/u+α, where u>p, and 0<q≤∞,
hα,νp⫋hβ,νu, where β=1/u-1/p+α, and where u>p.
Proof.
We can assume that ν=0. Besides, we omit the proof that the inclusions are strict because we do not need this fact.
If ∫01Mpq(r,g)(1-r)qα-1dr<∞, then Mp(r,g)(1-r)α→0 (r→1-) because Mp(r,g) increases with r.
It is well known that Mu(r,g)≤C(1-r)1/u-1/pMp(r,g) for u>p (see, e.g., [18, Corollary 5.1.2]).
This follows from (b) (q=∞) and the density of 𝒫 in hα,νp.
Let
(34)κp,α,ν=ν-α+1-1p.
We will determine the indices p, q, α, and ν for which:
ℒ acts from Hα,νp,q into Hα,νp,q[κp,α,ν>0];
ℒ acts from Hα,νp,q to H(𝔻) but not to Hα,νp,q [κp,α,ν=0 and 1≤p≤2 and q≤1; see Theorem 16];
ℒ acts from Hα,νp,q into Hα,νp,q but the implication g∈Hα,νp,q⇒(16) does not hold (p>2 and (either q>1 and ν-α≤-1/2, or 0<q≤1 and ν-α<-1/2); see Theorem 18);
the operator ℒ¯ cannot be extended to a bounded operator from Hα,νp,q to H(𝔻) (1°κp,α,ν<0, 2°κp,α,ν=0 and (p>2 or q>1); see Theorem 16).
Observe that, since ν-α>1/p-1 in (3), the inequality ν-α<1/2 is equivalent to
(35)1p-1<ν-α<-12,
which has sense because 1/p-1<-1/2 due to the condition p>2.
In proving (1) we use the inequality
(36)r1/pMp(r,(ℒg)(ν))≤21/p(1-r)-δ-1∫r1Mp(s,g(ν))(1-s)δds,
where δ=ν-1/p, which is a relatively simple consequence of the Littlewood subordination principle. A generalized version of (36) immediately gives sufficient conditions for ℒ to map X=Hα,νp,q to X. In order to prove that these conditions are necessary we analyze membership in X of functions with nonnegative, nonincreasing coefficients and apply this to the Libera transform of functions with positive coefficients.
Let f be a function of the form
(37)f(z)=∑n=0∞anzn,an≥0∀n,
and define ℒf by Definition 2. This definition is correct if and only if
(38)∑k=0∞akk+1<∞,
because a series with positive coefficients is Abel summable if and only if it is summable in the ordinary sense. If (38) is satisfied, then the sequence of the Taylor coefficients of ℒf is
(39)bn=∑k=n∞akk+1
and is therefore nonincreasing. In this paper we consider only the functions where an=(n+1)γ/logε(n+2). Discussion of the general case will appear in a separate paper.
In considering (2) and (4) we use, besides functions with positive coefficients, a deep theorem of Kolmogorov and Khinchin, while in the case of (3) we need another deep result, due to Kisliakov. By use of these theorems we can say much more about (3). Namely, if the implication g∈Hα,νp,q does not hold, then there exists a function g∈Hα,νp,q such that |g^(n)|≥log-1(n+2), and in some cases |g^(n)|≥1 or even |g^(n)|≥(n+1)η for some η>0 (see Theorem 5 and Theorem 41).
2. Results
Before stating our first result we give a sufficient condition for the validity of (15). This condition is not necessary (see Theorem 16).
Proposition 10.
If g∈Hα,νp,q, where 0<p≤∞,0<q≤∞, and
(40)ν-α>1p-1,
then the integrals
(41)Jm(z):=∫01(1-t)mg(m)(t+(1-t)z)dt,m≥0,z∈𝔻
converge uniformly on compact subsets of 𝔻, and the operator
(42)ℒg(z)=∫01g(t+(1-t)z)dt
maps Hα,νp,q to H(𝔻) and coincides with ℒ¯ on H(𝔻¯), and we have
(43)(ℒg)(ν)(z)=∫01(1-t)νg(ν)(t+(1-t)z)dt.
Proof.
By the well-known theorem from complex analysis, it is enough to prove that J0(z) converges uniformly. Since Mp(r,g) increases with r, the condition g∈Hα,νp,q implies that
(44)Mp(r,g(ν))≤C(1-r)-α.
This implies, by the well-known estimate M∞(r,h)≤CMp(r,h)(1-r)-1/p, that
(45)M∞(r,g(ν))≤C(1-r)-α-1/p.
Hence, by successive integration we get M∞(r,g)≤Cψ(r), where
(46)ψ(r)=(1-r)-α-1/p+ν,if-1<-α-1p+ν<0=log41-r,if-α-1p+ν=0=1,if-α-1p+ν>0.
It turns out that
(47)∫01|g(t+(1-t)z)|dt≤C∫01ψ((1-|t+(1-t)z|))dt≤C∫01ψ((1-t-(1-t)|z|))dt=C∫01ψ((1-t)(1-|z|))dt.
It is not difficult to check that if |z|<ρ<1, the ψ((1-t)(1-|z|))≤CΨ(ρ,t), where ∫01Ψ(ρ,t)dt<∞. This implies, by Weierstrass’ theorem, that J0(z) converges uniformly on compact subsets of 𝔻. The rest of the proof is easy.
Our first result is as follows.
Theorem 11.
Let X=Hα,νp,q or X=hα,νp. Then the following assertions are equivalent:
the operator ℒ acts as a bounded operator from X into X;
Condition (40) is satisfied.
Observe that condition (40) is independent of q.
Remark 12.
In other words, the theorem says the following (excluding Hardy and Hardy-Sobolev spaces): let X=𝔅βp,q or X=𝔟βp,β∈ℝ. Then ℒ acts from X to X if and only if β>1/p-1, or what is the same, if and only if β>-1 and p>1/(β+1).
Theorem 11 contains some known and some new results as special cases.
Case 1.
Theorem 11 covers the case when α=0,q=∞. In particular,
ℒ maps Hp into Hp if and only if 1<p≤∞. This is Theorem A;
if ν≥1, then
ℒ maps Hardy-Sobolev space Wνp=H0,νp into Wνp, for every p∈[1,∞], and, in particular, ℒ maps the ordinary Lipschitz space into itself.
This is, maybe, a new result.
Case 2 (Theorem C).
In particular, ℒ does not act as a bounded operator from Aβ1 into Aβ1, for any β>-1.
This is seen from Theorem 11 by taking β=αp-1; that is, α=(β+1)/p.
Case 3.
(a) ℒ maps the Dirichlet space 𝒟βp into itself if and only if p>1+β/2. In particular, ℒ maps 𝒟β1 if and only if -1<β<0.
Another case: (b) ℒ maps 𝒟β2 into itself if and only if β<2.
These facts are, maybe, new.
Case 4.
(a) ℒ maps Hαp,q into itself if and only if α<1 and p>1/(1-α).
This is seen from Theorem 11 by taking ν=0. This is related to a result of [19], which, when reformulated in our notation, gives the assertion (a) under the additional condition αq-1≥-(1/p), or equivalently qα≤1 and p≤1/(1-qα). For example, if q=1 and α=1/2, then this assertion says nothing because of the hypothesis p≤2, while we still have that ℒ maps H1/2p,1 into itself if and only if p>2.
If α=0, and ν=0, q=∞, then (a) says that ℒ maps Hp into Hp if and only if p>1. (Compare Case 1.) In particular ℒ maps Hα∞ into Hα∞ if and only 0≤α<1 (this is Theorem 5). Also ℒ maps hαp:=hα,0p if and only if 0<α<1-1/p.
Case 5.
(a) ℒ maps 𝔅βp,q=Λβp,q into itself for every p(≥1), q, and β>0. The same holds for the little space 𝔟βp=λβp.
This is seen from Theorem 11 (or Remark 12): the inequality ν+1>ν-β+1/p holds for any β>0 and p∈[1,∞]. In the case q≥1, a direct proof can be found in [7]. In [6], assertion (a) (q≥1) is proved by using the relation 𝒞*=ℒ and the fact that 𝒞 maps Hαp′,q′ (1/s+1/s′=1) into itself. (The latter was proved in [20]; a quick proof is given in [21].) What is new here is that (a) holds for q<1.
(b) As a special case of (a) (p=q=∞) we have that ℒ maps the Lipschitz (=Hölder) class Λβ=Λβ∞ into itself.
Case 6.
ℒ maps p-Bloch type spaces 𝔅αp=Hα,1p into itself if and only if α+1/p<2.
The same holds for the little space 𝔟αp=hα,1p.
In particular, when p=∞, this condition reduces to 0<β<2; this is Theorem B.
Case 7.
(a) ℒ maps the Hardy-Bloch space 𝔅p,q if and only if p>1.
This is, maybe, new. In the case 2<p=q<∞, there exists a better result [6]: ℒ maps 𝔅p,p into Hp (the well-known result of Littlewood and Paley states that Hp⫋𝔅p,p).
Case 8.
ℒ maps the Zygmund space Λ* into itself. The same holds for λ*.
The implication (a)⇒(b) of Theorem 11 is valid because of the following two propositions.
Proposition 13.
Let (1) κp,α,ν≤0, α>0, and X∈{Hα,νp,hα,νp}, or (2) κp,α,ν<0 and X=Hα,νp,q (q<∞).
Then the operator ℒ¯ cannot be extended to a bounded operator from X to H(𝔻).
As mentioned in Introduction (see (4), in page 4, or Theorem 16 below), the class of such spaces is larger.
Remark 14.
In (1), the condition α>0 is necessary since H0,νp is either a Hardy space (ν=0) or a Hardy-Sobolev space Wνp(ν≥1), which is contained in Hp.
Proposition 15.
If q<∞ and κp,α,ν=0, then the function
(48)f(z)=∑n=0∞znlogɛ(n+2),wheremax{1q,1}<ɛ≤1+1q,
belongs to Hα,νp,q; the function ℒf is well defined but ℒf is not in Hα,νp,q.
However, if κp,α,ν=0, it may happen that ℒ is well defined on Hα,νp,q (but, by Proposition 15, does not map the space into itself). The following theorem together with Proposition 10 characterizes those X=Hα,νp,q such that ℒ¯ can be extended to a bounded operator from X to H(𝔻).
Theorem 16.
Let X=Hα,νp,q (0<q≤∞) or X=hα,νp, and κp,α,ν≤0. Then the following four conditions are equivalent:
the operator ℒ¯ can be extended to a bounded operator from X to H(𝔻);
ℒ acts as a bounded operator from X to H(𝔻);
If g∈X, then ∑(|g^(n)|/(n+1))<∞;
1≤p≤2, 0<q≤1, and κp,α,ν=0.
A natural question arises from this theorem: under condition (d), find a quasi-Banach X such that ℒ(Hα,νp,q)⊂X. It turns out that we can take, for instance,
(49)X={∫01Mpq(r,g)(1-r)qα-1log-ɛlogg∈H(𝔻):∫01Mpq(r,g)×(1-r)qα-1log-ɛlog41-rdr<∞∫01Mpq(r,g)},ɛ>1.
Theorem 17.
Let ɛ>1, 1≤p≤2,κp,α,ν=0 (i.e., α=ν+1-1/p), and q≤1. Then ℒ maps Hα,νp,q to X, where X is defined by (49).
This theorem can easily be deduced from (36); we will omit the proof.
There are cases when κp,α,ν>0 (which implies that ℒ is well defined on X) but the assertion (c) of Theorem 16 does not hold.
Theorem 18.
Let X=Hα,νp,q (0<q≤∞,α>0) and κp,α,ν>0. Then assertion (c) of Theorem 16 does not hold if and only if one of the following two conditions is satisfied:
2<p≤∞,1<q≤∞, and ν-α≤-1/2,
2<p≤∞,0<q≤1, and ν-α<-1/2.
Remark 19.
In the case of Besov type spaces, this theorem says the following: let β>1/p-1. The implication g∈𝔅βp,q⇒∑n=0∞|g^(n)|/(n+1)<∞ does not hold if and only if either (1) 2<p≤∞, 1<q≤∞, and β≤-1/2 or (2) 2<p≤∞, 0<q≤1, and β<-1/2.
Theorem 18 can be used to get a generalization of Bernstein’s theorem to the case of the Besov spaces.
Theorem 20.
Let β>1/p. The implication
(50)g∈𝔅βp,q⟹∑n=1∞|g^(n)|<∞
does not hold if and only if either (1) 2<p≤∞, 1<q≤∞, and β≤1/2, or (2) 2<p≤∞, 0<q≤1, and β<1/2.
Proof.
This follows from Theorem 18, the definition of Besov type spaces (Definition 8), and the equivalence g∈𝔅βp,q⇔g′∈Hν-β,ν-1p,q.
By taking p=q=∞ we get Bernstein’s theorem.
3. Proof of Theorem 11
We need a variant of the Littlewood subordination principle.
Theorem G (see [22]). If φ:𝔻↦𝔻 is an analytic function and g∈Hp, then f∘φ∈Hp and(51)∥g∘φ∥p≤(1+|φ(0)|1-|φ(0)|)1/p∥g∥p.
Here ∥g∥p denotes the norm of g in Hp,
(52)∥g∥p=sup0<r<1(12π∫02π|g(reiθ)|pdθ)1/p.
As an application we have the following lemma.
Lemma 21.
If a and b are positive real numbers such that a+b≤1 and if g∈Hp,0<p≤∞, then
(53)(∫02π|g(a+beiθ)|pdθ)1/p≤(2a+bb)1/p(∫02π|g((a+b)eiθ)|pdθ)1/p.
Proof.
The case p=∞ is easy. Let p<∞. Let h(z)=g((a+b)z),a1=a/(a+b), b1=b/(a+b), and φ(z)=a1+b1z. Then
(54)∫02π|g(a+beiθ)|pdθ=∫02π|h(a1+b1eiθ)|pdθ≤1+a11-a1∫02π|h(eiθ)|pdθ=2a+bb∫02π|g((a+b)eiθ)|pdθ,
which was to be proved.
Lemma 22.
If ℒg is well defined (see Definition 2), then
(55)r1/pMp(r,(ℒg)(ν))≤21/p(1-r)-(δ+1)∫r1Mp(s,g(ν))(1-r)δdr,
where δ=ν-1/p.
Remark 23.
The integral in (55) diverges if and only if p=1 and ν=0.
Proof.
Let h=ℒg. We have
(56)h(z)=∫01g(t+(1-t)z)dt,z∈𝔻,
and hence
(57)h(ν)(z)=∫01(1-t)νg(ν)(t+(1-t)z)dt.
Applying the Minkowski inequality and Lemma 21 with a=t,b=(1-t)r we obtain
(58)Mp(r,h(ν))≤∫01(1-t)νMp(t+(1-t)r,g(ν))(2t+(1-t)r(1-t)r)1/pdt≤21/pr-1/p∫01(1-t)ν-1/pMp(t+(1-t)r,g(ν))dt,1≤q<∞.
Substituting t+(1-t)r=s and taking ν-1/p=δ,
(59)r1/pMp(r,h(ν))≤21/p(1-r)-δ-1∫r1(1-s)δMp(s,g(ν))ds,
which was to be proved.
Remark 24.
Before going further note that an immediate consequence of this lemma is the validity of implication (b)⇒(a) of Theorem 11 in the case X=Hα,νp=Hα,νp,∞. Then we use this fact and the following two ones to show that ℒ maps hα,νp into hα,νp. The set 𝒫 (all polynomials) is dense in hα,νp and ℒ maps 𝒫 into 𝒫.
Lemma 25.
If 0<q<∞ and ℒg is well defined, then
(60)rq/pMpq(r,(ℒg)(ν))≤C(1-r)-(δ+1-ɛ)q×∫r1Mpq(s,g(ν))(1-s)(δ+1-ɛ)q-1ds,
where
(61)ɛ={=0,q≤1,>0,q>1,
and C is independent of g and r.
In the case q≤1, this lemma follows from Lemma 22 and the following.
Sublemma 25.1. Let 0<q<1, β=1+δ>0, and let u:(0,1)↦[0,∞) be a nondecreasing function. Then(62)(∫r1u(s)(1-s)β-1ds)q≤C∫r1u(s)q(1-s)qβ-1ds,where C is independent of u.
Proof.
Fix r∈(0,1), and define the function v on (0,1) by
(63)v(s)={0,0<s≤r,u(s),r<s<1.
Then the desired inequality can be written as
(64)(∫01v(s)(1-s)β-1ds)q≤C∫01v(s)q(1-s)βq-1ds.
Let rn=1-2-n, wheren is a nonnegative integer. Then
(65)(∫01v(s)(1-s)β-1ds)q=(∑n=0∞∫rnrn+1v(s)(1-s)β-1ds)q≤C(∑n=0∞v(rn+1)2-nβ)q≤C∑n=0∞(v(rn+1)2-nβ)q.
In a similar way we can prove that
(66)∫01v(s)q(1-s)βq-1ds≥c∑n=0∞v(rn)q2-nβq=c1∑n=-1∞v(rn+1)q2-nβq≥c1∑n=0∞v(rn+1)q2-nβq,
where c1= const. >0. Comparing these inequalities we get the result.
In the case q>1, Lemma 25 is a consequence of the following fact.
Sublemma 25.2. Let 1<q<∞,ɛ>0, and u≥0, a measurable function defined on (r,1). Then(67)(∫r1u(s)(1-s)δds)q≤C(1-r)ɛq∫r1u(s)q(1-s)(1+δ-ɛ)q-1ds,where C is independent of u.
Proof.
Let κ=δ+1-ɛ-1/q. Then, by Hölder’s inequality (1/q+1/q′=1),
(68)∫r1u(s)(1-s)δds=∫r1u(s)(1-s)γ(1-s)δ-γds≤(∫r1u(s)q(1-s)γqds)1/q(∫r1(1-s)(δ-γ)q′ds)1/q′.
Since (δ-γ)q′=(ɛ-1+1/q)q′>-1 because ɛ>0, we have that the last integral is less than
(69)C(1-r)δ-γ+1/q′=C(1-r)ɛ.
The result follows.
Proof of Theorem 11, ((b)⇒(a)).
By Remark 24, we may assume that q is finite, which implies that α>0. Let q>1. A standard application of the maximum modulus principle for analytic functions gives
(70)∫01Mpq(r,(ℒg)(ν))(1-r)qα-1dr≤C∫01rq/pMpq(r,(ℒg)(ν))(1-r)qα-1dr.
Then, if q>1, we have
(71)∫01Mpq(r,(ℒg)(ν))(1-r)qα-1dr≤C∫01(1-r)(α-δ-1+ɛ)q-1dr×∫r1Mpq(s,g)(1-s)(δ+1-ɛ)q-1ds=C∫01Mpq(s,g)(1-s)(δ+1-ɛ)q-1ds×∫0s(1-r)(α-δ-1+ɛ)q-1dr.
Since α-δ-1=α+1/p-ν-1<0, we can choose ɛ>0 so that α-δ-1+ɛ<0. Then
(72)∫0s(1-r)(α-δ-1+ɛ)q-1dr≤C(1-s)(α-δ-1+ɛ)q.
Combining this with the preceding inequality we get the result in the case q>1. In the case q≤1 the proof is similar and simpler and is omitted.
Proof of Theorem 11 ((a)⇒(b)).
As noted in Section 2, it is enough to prove Propositions 13 and 15.
Proof of Proposition 13.
We have
Case (1). By Proposition 9 we have hβ,ν∞⊂hα,νp, where β=α+1/p. Since κ∞,β,ν=κα,p,ν, it is enough to consider the case of hβ,ν∞. Let fρ(z)=1/(1-ρz),0<ρ<1. It is clear that fρ∈H(𝔻¯). A simple calculation shows that the set {fρ:0<ρ<1} is bounded in hα,ν∞. Hence, if ℒ¯ has an extension to a bounded operator from hα,ν∞ to H(𝔻), then the set {ℒ¯fr(0)} is bounded because the functional g↦g(0) is bounded on H(𝔻). However,
(73)ℒ¯fr(0)=∑n=0∞rnn+1⟶∞,r⟶1-,
which is a contradiction. This proves the desired result in the case of hα,νp. Since hα,νp⊂Hα,νp, we see that the result holds for this space as well.
Case (2). Let kp,α,ν<0 and choose β so that kp,β,ν=0. This implies that β<α. Then it is easy to check that Hα,νp,q⊂Hα,νp, which together with the Case (1) gives the result.
Proof of Proposition 15.
The proof of Proposition 15 is more delicate. First note that, by Proposition 9,
(74)Hβ,ν1,q⊂Hα,νp,q⊂Hγ,ν∞,q,
where
(75)α=β+1-1p,γ=α+1p,
and that
(76)κ1,γ,ν=κp,α,ν=κ∞,β,ν=0,
see (34). Therefore, it is enough to prove that the function f defined by (48) belongs to Hβ,ν1,q while ℒf does not belong to Hγ,ν∞,q.
It is easier to prove that ℒf∉Hγ,ν∞,q. Namely, since the coefficients cn of (ℒf)(ν) are nonnegative, we see that ℒf is in Hγ,ν∞,q if and only if
(77)∫01(∑n=0∞cnrn)q(1-r)qγ-1dr<∞,
which, by a theorem of Lp-integrability of power series with positive coefficients (see [23, Theorem 1]), is equivalent to
(78)∑n=0∞2-nqγ(∑k∈Inck)q<∞.
Here
(79)In={k:2n-1≤k≤2n+1-1}(forn≥1),I0={0}.
Since ck≍(k+1)ν (k→∞), the latter is equivalent to
(80)∑n=0∞2nq(ν-γ)(∑k∈Inbk)q,
where bn are coefficients of ℒf. Since bn↓0, we get the equivalent condition
(81)∑n=0∞2nq(ν-γ+1)b2nq<∞.
This condition is not satisfied because γ=ν+1 and
(82)b2n=∑k=2n1(k+1)logɛ(k+2)≍(n+1)ɛ-1,
and hence
(83)∑n=0∞(n+1)(ɛ-1)q=∞(because(ɛ-1)q≤1),
where the function ℒf is well defined because ɛ>1, which implies
(84)∑n=0∞1(n+1)logɛ(n+2)<∞.
In proving that f∈Hβ,ν1,q we use a sequence {Vn}0∞ constructed in the following way (see, e.g., [24]).
Let ω be a C∞ function on ℝ such that
ω(t)=1 for t≤1,
ω(t)=0 for t≥2,
ω is decreasing and positive on the interval (1,2).
Let φ(t)=ω(t/2)-ω(t), and let V0(z)=1+z, and, for n≥1,
(85)Vn(t)=∑k=0∞φ(k2n-1)zk=∑k=2n-12n+1φ(k2n-1)zk.
These polynomials have the following properties:
(86)g(z)=∑n=0∞Vn*g(z),forg∈H(𝔻);(87)∥Vn*g∥p≤C∥g∥p,forg∈Hp,p>0;(88)∥Vn∥p≍2n(1-1/p),∀p>0.
(Here * denotes the Hadamard product).
In [25, Lemma 2.1], the following characterization of Hα,νp,q was proved.
Lemma A. Let 0<p≤∞, and 0<q≤∞, α>0, and let ν be a nonnegative integer. A function g∈H(𝔻) is in Hα,νp,q if and only if(89)K1(g):=(∑n=0∞2n(ν-α)q∥Vn*g∥pq)1/q<∞,and we have K1(g)≍∥f∥Hα,νp,q. In the case of Hα,νp (resp., hα,νp) this is interpreted as ∥Vn*g∥p=𝒪(2n(α-ν)) (resp. ∥Vn*g∥p=o(2n(α-ν))).
Remark 26.
This lemma was deduced in [25] from the case ν=0 (which is relatively easy to discuss) by using some nontrivial results of Hardy and Littlewood [26] and of Flett (see [27]). By a successive application of Lemma 27 below (case δ=0), we can make this deduction elementary.
Lemma 27.
If n is a positive integer,
(90)P(z)=∑k=n4nλkzk,
where {λk} is a complex sequence, and
(91)Q(z)=∑k=n4n(k+1)βlogδ(k+1)λkzk,δ,β∈ℝ,
then there is a constant C depending only on δ and β such that
(92)C-1∥Q∥1≤(n+1)βlogδ(n+1)∥P∥1≤C∥Q∥1.
Proof.
The proof can be reduced to two cases: (1) β=0, δ∈ℝ and (2) δ=0, β∈ℝ. We will consider only Case (1); Case (2) is discussed similarly.
Let ψ be a C∞ function on (0,∞) such that suppψ⊂(1/2,5) and ψ(t)=1 for 1≤t≤4. Then we have Q as
(93)Q(z)=∑k=0∞ψ(kn)logδ(k+2)λkzk,
where λk:=0 for k∉[n,2n]. Fix n and let
(94)Δ2η(k)=η(k)-2η(k+1)+η(k+2),
where
(95)η(t)=ψ(tn)logδ(t+1).
We have
(96)Q(z)=∑k=0∞Δ2η(k)(k+1)(σkP)(z)=∑n/2≤k≤5nΔ2η(k)(k+1)(σkP)(z),
where σkP are (C,1)-means of P. It follows that
(97)∥Q∥1≤∑n/2≤k≤5n|Δ2η(k)|(k+1)∥σkP∥1≤∑n/2≤k≤5n|Δ2η(k)|(k+1)∥P∥1,
where we have used Fejér’s inequality ∥σkP∥1≤∥P∥1. Now we use Lagrange’s inequality in the form
(98)|Δ2η(k)|≤2supk≤t≤k+2|η′′(t)|
and the easily proved inequality
(99)|η′′(t)|≤Cn2logδ(n+1),n2≤t≤5n
to get
(100)∥Q∥1≤Cn2logδ(n+1)∑n/2≤k≤5n(k+1)∥P∥1≤Clogδ(n+1)∥P∥1,
which proves the desired result in one direction. To prove the reverse inequality we write P as
(101)P(z)=∑k=n4nlog-δ(k+1)ξkzk,whereξk=λklogδ(k+1).
Applying the above case, we get
(102)∥P∥1≤Clog-δ(n+1)∥Q∥1,
which completes the proof.
Proof of Proposition 15.
As noted above it is enough to prove that f∈Hβ,ν1,q, where β=ν. In this case, by Lemma A, the function f belongs to Hβ,ν1,q if and only if
(103)∑n=0∞∥Vn*f∥1q<∞.
On the other hand, by Lemma 27 and the property (88),
(104)∥Vn*f∥1q≍(logɛ(2n+1)∥Vn∥1)q≍(n+1)-ɛq,
which implies that f is in Hβ,ν1,q, because ɛq>1.
Thus, we have proved the implication (a)⇒(b) of Theorem 11.
4. Proof of Theorem 16
It is clear that (b) implies (a). We have already noted that (c) implies (b). The following assertion shows that (d) implies (c).
Proposition 28.
If q≤1,1≤p≤2,κp,α,ν=ν+1-α-1/p=0, and g∈Hα,νp,q, then
(105)∑n=0∞|g^(n)|n+1<∞.
Proof.
Since Hα,ν1,q⊂Hα,ν1,1, it is enough to consider the case q=1. Let
(106)Pn=Vn-1+Vn+Vn+1,n≥0,whereV-1=0.
Since Pn*Vn=Vn, which follows from (86) and the fact Vn*Vm=0 for |m-n|≥2, we see that
(107)LemmaAremainstrueifVnisreplacedbyPn.
Also the relation Pn*Vn=Vn implies that P^n(k)V^n(k)=V^n(k), and hence P^n(k)=1 whenever V^n(k)≠0. In particular P^n(k)=1 for k∈In (see (79)). We use Hardy’s inequality
(108)∑k=0∞(k+1)p-2|h^(k)|p≤C∥h∥pp,1≤p≤2,
and Lemma (107) to conclude that if g∈Hα,νp,1, then
(109)∑n=0∞2-n/p(∑k∈In|g^(k)|p)1/p<∞.
Hence, by Hölder’s inequality,
(110)∑n=0∞2-n∑k∈In|g^(k)|≤∑n=0∞2-n(∑k∈In|g^(k)|p)1/p2n(1-1/p)=∑n=0∞2-n/p(∑k∈In|g^(k)|p)1/p,
which proves the result.
It remains to be proven that (a) implies (d); that is, that (a) does not hold in the following cases:
1≤p≤∞,0<q≤∞, and κp,α,ν<0;
1≤p≤2,
1<q≤∞,
κp,α,ν=0;
2<p≤∞,
0<q≤∞,
κp,α,ν=0.
Case (1) is part of Proposition 13. In view of the same proposition, we can assume, in what follows, that q<∞.
The following assertion proves the desired result in Cases (2) and (3, q>1).
Proposition 29.
If 1≤p≤∞, κp,α,ν=0, and 1<q<∞, then ℒ¯ cannot be extended to a bounded operator from Hα,νp,q to H(𝔻).
Proof.
(a) Let
(111)fɛ,ρ(z)=∑k=0∞ρnzn(k+1)log1+ɛ(k+2),ɛ>0,0<ρ≤1.
It follows from Lemmas A and 27 that
(112)∥fɛ,1∥α,νp,qq≍∑n=0∞(n+1)-(1+ɛ)q≤∑n=0∞(n+1)-q=Cq<∞,
where Cq is independent of ɛ. This inequality remains true if fɛ,1 is replaced by fɛ,ρ with ρ<1. The function fɛ,ρ(ρ<1) belongs to H(𝔻¯) and the set {fɛ,ρ:ɛ>0,ρ<1} is bounded in Hα,νp,q. On the the other hand,
(113)ℒfɛ,ρ(0)=∑k=0∞ρn(k+1)log1+ɛ(k+2)⟶∞,ɛ⟶0+,ρ⟶1-.
The result follows.
The case when q≤1 and p>2 is more delicate and depends on Khinchin’s inequality and a deep result of Khinchin and Kolmogorov ([28]; see [14, Ch. V, Sec. 8]).
Theorem H. Let Rn denote the sequence of Rademacher functions, Rn(t)=sign(sin(2nπt)), n≥0, 0≤t≤1. If {cn} is a sequence in ℂ such that ∑n=0∞|cn|2=∞, then the series ∑n=0∞cnRn(t) diverges for almost all t∈[0,1] and moreover the sequence of its partial sums is unbounded a.e.
We also need the following theorem of Khinchin [14, Ch. V, Theorem (8.4)].
Theorem I. Let {ck} be a finite sequence, and let 0<q<∞. Then(114)∫01|∑kckRk(t)|q≍(∑k|ck|2)q/2,where the “involving” constants depend only on q.
Let
(115)Δn(z)=∑k∈Inzk,Δng=Δn*g.
The following fact was proved in [29].
Lemma B. Let 1<p<∞, 0<q≤∞, and α>0. A function g∈H(𝔻) is in Hαp,q if and only if(†)K(g):=(∑n=0∞2-nαq∥Δng∥pq)1/q<∞,and we have K(g)≍∥f∥αp,q. In the case of Hα∞ (resp., hαp=hα,0p), relation (†) means ∥Δng∥p=𝒪(2nα) (resp., ∥Δng∥=o(2nα), as n→∞).
As a consequence of this lemma and Lemma 27 we have the following.
Lemma 30.
Let 1<p<∞, 0<q≤∞, and α>0. A function g∈H(𝔻) is in Hα,νp,q if and only if
(‡)Kν(g):=(∑n=0∞2n(ν-α)q∥Δng∥pq)1/q<∞,
and we have Kν(g)≍∥f∥α,νp,q. In the case of Hα,νp, respectively, hα,νp, relation (‡) is interpreted as ∥Δng∥p=𝒪(2(α-ν)n), respectively, ∥Δng∥p=o(2(α-ν)n).
Lemma 31.
If 0<p<∞ and 0<q<∞, and gt(z)=∑k=0∞ckzkRk(t), then
(116)∫01∥Δngt∥pq≍(∑k∈In|ck|2)q/2,n≥0.
Proof.
In the case p=q, the relation immediately follows from Theorem I. Let p>q. Then, by Jensen’s inequality for the convex function x↦xp/q,
(117)(∫01∥Δngt∥pqdt)p/q≤∫01(∥Δngt∥pq)p/q=∫01∥Δngt∥ppdt≍(∑k∈In|ck|2)p/2.
On the other hand, since ∥Δng∥p≥∥Δng∥q, we have
(118)∫01∥Δngt∥pqdt≥∫01∥Δngt∥qqdt≍(∑k∈In|ck|2)q/2.
This proves the result in the case p>q. The remaining case is discussed similarly.
Proposition 32.
If 0<q≤1,2<p≤∞,α>0, and κp,α,ν=0, then ℒ¯ cannot be extended to a bounded operator from Hα,νp,q to H(𝔻).
Proof.
Consider first the case 2<p<∞. Let
(119)gt(z)=∑k=0∞(k+1)1/2-1/plog-2/q(k+2)Rn(t)zk,0≤t≤1.
Since
(120)∑k=0∞((k+1)1/2-1/plog-2/q(k+2))2=∞,
because 2/p<1, the sequence of partial sums of the series gt(1) diverges on a set E⊂[0,1] such that |E|=1, which follows from Theorem H. (We can assume that E does not contain points where Rn(t)=0 because the set of such points is denumerable). On the other hand, by Lemma 31, we have
(121)∫E∥gt∥α,νp,qq≍∫E∑n=0∞2n(1/p-1)q∥Δngt∥pq≍∑n=0∞2n(1/p-1)q∥Δngt∥2q≍∑n=0∞2n(1/p-1)q2n(1/2-1/p)q2nq/2(n+1)-2=∑n=0∞(n+1)-2<∞.
It follows that gτ∈Hα,νp,q for at least one τ∈E.
To complete the proof we consider the polynomials (∈H(𝔻¯))
(122)sn(z)=∑k=02n+1-1g^τ(k)zk.
It follows from Lemma B that ∥sn∥≤C∥gτ∥<∞ (in the norm of Hα,νp,q), where C is independent of n. On the other hand, as noted before, the sequence
(123)ℒsn(0)=∑k=02n+1-1(k+1)1/2-1/plog-2(k+2)Rn(τ)
is not bounded, which proves the result in the case p<∞.
In the case of Hα,ν∞,q we have ν+1-α=0. Hence, if g∈Hβ,νp,q and κp,β,ν=0, that is, ν-β=1/p-1, then it follows from Proposition 9 that Hβ,νp,q⊂Hα,ν∞,q, continuously. The desired result follows from the case p<∞.
Remark 33.
If q=∞, p>2, ν=0, and κ=0, that is, α=1-1/p, then we can take
(124)gt(z)=∑k=0∞(k+1)1/2-1/pRn(t),
and apply the above approach to show that there is a function g∈Hαp,∞ such that
(**)|g^(k)|≍(k+1)1/2-1/p,
and hence
(125)limn→∞g^(n)n+1=∞.
It is not easy to give a concrete example of such a function. A “natural” example is
(126)g(z)=(1-z)1/p-3/2,
whose coefficients satisfy (**). However,
(127)∫02π|1-reiθ|p(1/p-3/2)dθ≍(1-r)1-3p/2.
If g∈Hαp,∞, then
(128)(1-r)1-3p/2≤C(1-r)p(1/p-1),
which implies p(1/p-1)≤1-3p/2, while this implies p≤2, a contradiction which shows that g∉Hαp,∞, for p>2.
5. Proof of Theorem 18
For technical reasons, we introduce the space
(129)ℓ-11={g∈H(𝔻):∑n=0∞|g^(n)|n+1<∞}.
In the proof of Proposition 34 we use the following deep result of Kisliakov [30].
Theorem J. For any sequence {ck}k=mn there is a polynomial h(z)=∑k=mnbkzk such that |bk|≥|ck| and(130)∥h∥∞≤C(∑k=mn|ck|2)1/2,where C is an absolute constant.
Proposition 34.
Let κp,α,ν>0. If (1) 1<q≤∞, 2<p≤∞, and ν-α≤-1/2, or (2) 0<q≤1, 2<p≤∞, and ν-α<-1/2, then Hν,αp,q⊈ℓ-11.
Proof.
We have
Case (1),(p<∞). Let q=∞ and let
(131)gt(z)=∑k=0∞ckRk(t)zk,
where
(132)(∑k∈In|ck|2)1/2≤C2n(α-ν).
Take ck=(k+1)ξ. Then
(133)(∑k∈Inck2)1/2≍2nξ2n/2≤C2n(α-ν),
whence we can choose ξ+1/2=α-ν; that is, ξ=α-ν-1/2≥0.
We have, by Lemma 31,
(134)∫01∥Δnft∥p≤C(∑k∈Inck2)1/2≤C2n(α-ν).
This implies that there is a sequence εn∈{-1,1} such that the function
(135)h(z)=∑k=0∞ckεkzk
belongs to Hα,νp,∞. On the other hand,
(136)∑k=0∞|h^(k)|k+1=∞.
If 1<q<∞, we consider the function
(137)gt(z)=∑n=0∞cnRn(t)zk,wherecn=(n+1)ξlog(n+2),
and proceed as above to get the result.
Case (1),(p=∞). Choose {ck} as above and consider the function h(z)=∑k=0∞bkzk, where |bk|≥ck and ∥Δnh∥∞≤C(∑k∈Inck2)1/2 (Theorem J, Kisliakov). Finally we use the inequality
(138)∥h∥α,νp,q≤C(∑k∈In2(ν-α)q∥Δnh∥∞q)1/q,
(see [29, Theorem 2.1(a)]) to finish the proof of Case (1).
In Case (2) we choose g^(n)=(n+1)ξ, where 0<ξ<α-ν-1/2, and repeat the above reasoning to complete the proof.
Proposition 35.
If ν-α=-1/2, 0<q≤1, and 2<p≤∞, then Hα,νp,q⊂ℓ-11.
Proof.
We have
(139)∑n=0∞2n(ν-α)q∥Δng∥pq≥∑n=0∞2n(ν-α)q∥Δng∥2q≥(∑n=0∞2n(ν-α)∥Δng∥2)q=(∑n=0∞2-n/2∥Δng∥2)q≥(∑n=0∞2-n∑k∈In|g^(k)|)q,
which completes the proof of the proposition.
Proposition 36.
If 2<p≤∞, κp,α,ν>0, 0<q≤∞, and ν-α>-1/2, then Hα,νp,q⊂ℓ-11.
Proof.
Let g∈Hα,νp,q. Then g∈Hα,νp,∞, and hence g∈Hα,ν2,∞. It follows that ∥Δng∥2≤c2n(α-ν). On the other hand,
(140)∑n=0∞2-n∑k∈In|g^(k)|≤C∑n=0∞2-n/2∥Δng∥2≤C∑n=0∞2-n/22n(α-ν)<∞,
because -1/2+α-ν<0. This proves the proposition.
The following proposition completes the proof of Theorem 18.
Proposition 37.
If 1≤p≤2, κp,α,ν>0, and 0<q≤∞, then Hα,νp,q⊂ℓ-11.
Proof.
Let g∈Hα,νp,q. Choose β so that κp,α,ν=κ2,β,ν; that is, α=β+1/2-1/p. Then, by Proposition 9(b), we have Hα,νp,q⊂Hβ,ν2,q. This implies that Hα,νp,q⊂Hβ,ν2,∞ which means that ∥Δng∥2≤C2n(β-ν), and so
(141)2-n/2∥Δng∥2≤C2n(β-ν-1/2).
It follows that
(142)2-n∑k∈In|g^(k)|≤C2n(β-ν-1/2)=C2n(α-ν+1/p-1)=C2-nκp,α,ν.
The result follows.
6. On the Condition |g^(n)|≥(n+1)η, η>0
In this section we suppose that kp,α,ν=ν-α+1-1/p>0; that is, ℒ acts as an operator from Hα,νp,q into Hα,νp,q(α>0). We want to analyze the proof of Proposition 34 more carefully. Throughout the section we assume that 2<p≤∞ and κp,α,ν>0, so that ℒ maps Hα,νp,q into Hα,νp,q by Theorem 11.
Proposition 38.
Let q=∞ and ν-α≤-1/2. Then (i) there exists a function g∈Hα,νp,q such that |g^(n)|≥(n+1)ξ, where ξ=α-ν-1/2≥0. (ii) The exponent ξ is best possible. (iii) If ξ>0, then there is an η>0 and a function g∈Hα,νp,q such that |g^(n)|≥(n+1)η.
Proof.
Statement (i) follows from the proof of Proposition 34. (ii) In order to prove that ξ is best possible we use Lemma 30, which states that g∈Hα,νp,∞ if and only if ∥Δng∥p≤C2n(α-ν), which implies that ∥Δng∥2≤C2(α-ν)n. Assuming that |g^(k)|≥(n+1)s we get
(143)2ns2n/2≤C2n(α-ν),
whence s+1/2≤α-ν; that is, s≤α-ν-1/2=ξ.
(iii) This follows from (i).
Proposition 39.
Let 1<q<∞ and ν-α≤-1/2. Then (i) there exists a function g∈Hα,νp,q such that |g^(n)|≥(n+1)ξ/log(n+2). (ii) There is no function g∈Hα,νp,q such that |g^(n)|≥(n+1)ξ. (iii) If ξ>0 and 0<η<ξ, then there exists a function g such that |g^(n)|≥(n+1)η.
Proof.
Assertion (i) is part of the proof of Proposition 34. To prove (ii) we use the fact that Hα,νp,q⊂hα,νp, which implies that ∥Δng∥2=o(2n(α-ν)). If |g^(n)|≥(n+1)ξ, then the latter implies that
(144)2nξ2n/2=o(2n(α-ν)),
and hence 1=o(1), which is impossible.
In a similar way one proves the following.
Proposition 40.
Let q≤1 and ξ=α-ν-1/2>0. Then (i) if 0<η<ξ, then there is a function g∈Hα,νp,q such that |g^(n)|≥(n+1)η. (ii) In (i), η cannot be replaced by ξ.
Combining the above propositions we get the following.
Theorem 41.
Let ℒ map Hα,νp,q into itself, and p>2. Then the following statements are equivalent:
there is a function g∈Hα,νp,q such that |g^(n)|≥(n+1)η for some η>0;
ν-α<-1/2.
Moreover, if ν-α<-1/2 and η∈(0,α-ν+1/2) is arbitrary, then there is g∈Hα,νp,q such that |g^(n)|≥(n+1)η. If in addition q=∞, we can take η=α-ν-1/2.
Conflict of Interests
The author declares that there is no conflict of interests regarding the publication of this paper.
Acknowledgment
This work is supported by MNTR Serbia, Project ON174017.
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