# Some Sufficient Conditions for Analytic Functions  to Belong to𝒬K,0(p,q) Space

Authors: Xiaoge Meng
Journal: Abstract and Applied Analysis
 (2008)

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## Abstract

This paper gives some sufficient conditions for an analytic function to belong to the space consisting of all analytic functionsf on the unit disk such  lim|a|→1∫D|f′(z)|p(1−|z|2)qK(g(z,a))dA(z)=0.


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## Body

## 1. Introduction

Let𝔻 be the open unit disk in the complex plane ℂ and H(𝔻) the space of all analytic functions in 𝔻.
For a∈𝔻,
let(1.1)φa(z)=a−z1−a¯zbe the Möbius transformation of 𝔻 and let(1.2)g(z,a)=log|1−a¯za−z|be the Green's function on 𝔻.
Let 𝔻(a,r) denote the pseudo-hyperbolic metric disk
centered at a∈𝔻 with radius r∈(0,1),
that is, 𝔻(a,r)={z∈𝔻:|φa(z)|<r}.It is said that an analytic functionf(z)=∑k=1∞akznk is defined by a lacunary series if(1.3)λ=infk∈ℕnk+1nk>1.For some results in the topic,
see, for example, [1–6] and the references therein.Given a functionK:(0,∞)→[0,∞), we consider the space 𝒬K(p,q) of all functions f∈H(𝔻) such that(1.4)∥f∥𝒬K(p,q)p=supa∈𝔻∫𝔻|f′(z)|p(1−|z|2)qK(g(z,a))dA(z)<∞.By 𝒬K,0(p,q) we denote the space consisting of all f∈𝒬K(p,q) such that(1.5)lim|a|→1∫𝔻|f′(z)|p(1−|z|2)qK(g(z,a))dA(z)=0,where 0<p<∞, −2<q<∞,
and dA(z)=(1/π)dxdy=(1/π)rdrdθ.Forp=2, q=0,
the space 𝒬K(p,q) is reduced to 𝒬K (see, e.g., [7]). If K(g(z,a))=(g(z,a))s, 0≤s<∞,
then 𝒬K(p,q)=F(p,q,s) (see, e.g., [8, 9]).Throughout the paper, we assume that the condition
holds (see [7])(1.6)∫01(1−r2)qK(log1r)<∞,so that the space 𝒬K(p,q) we study is nontrivial. We also assume that K as a nondecreasing function. An important tool
in the study of 𝒬K spaces is the auxiliary function φK defined by (see [10])(1.7)φK(s)=sup0<t≤1K(st)K(t),0<s<∞.The following condition(1.8)∫1∞φK(s)dss2<∞is crucial in this paper. It has
played an important role in the study of 𝒬K spaces during the last few years.In this paper, we give some sufficient conditions for
an analytic functionf to belong to the space 𝒬K,0(p,q).The followings are our main results in this paper.Theorem 1.1.
Letf∈H(𝔻), 0<p<∞, −2<q<∞, and let φ be a monotone increasing function in r on (0,1) such that |f'(z)|≤φ(r),
for |z|=r.
If (1.9)∫01φp(r)(1−r2)qK(1−r2)rdr<∞,then f∈𝒬K,0(p,q).Theorem 1.2.
For1≤p<2, 0≤q<∞,
and 1≤p−2q<3.
If K satisfies condition (1.7) and is a function with the property that K(t)=K(1) for t≥1,
then a lacunary series f(z)=∑k=1∞akznk belongs to 𝒬K,0(p,q) if (1.10)∑k=1∞nkp−q−1|ak|pK(1nk)<∞.Throughout this paper,C stands for a positive constant, whose value
may differ from one occurrence to the other. The expression a≈b means that there is a positive constant C such that C−1a≤b≤Ca.


## 2. Main Results and Proofs

In this section, we give the proofs of Theorems1.1 and
1.2. Before formulating the main results, we give some lemmas which are used in
the proofs.Lemma 2.1  (see [7]).
Let0<p<∞, −2<q<∞,f∈H(𝔻).
Then, f∈𝒬K,0(p,q) if and only if (2.1)lim|a|→1∫𝔻|f′(z)|p(1−|z|2)qK(1−|φa(z)|2)dA(z)=0.Lemma 2.2  (see [5]).
LetK be a function with the property that K(t)=K(1) for t≥1.
If K satisfies condition (1.8), then there exists a constant C>0 such that K(2t)≈K(t) for t>0.Lemma 2.3 (see [5]).
IfK satisfies condition (1.8), then we can find another nonnegative function K∗ such that 𝒬K=𝒬K∗ and the new function K∗ has the following properties: (a)
K∗ is nondecreasing on (0,∞);(b)
K∗ satisfies condition (∗);(c)
K∗(2t)≈K∗(t) on (0,∞);(d)
K∗ is differentiable (up to any given order) on (0,∞);(e)
K∗ is concave on (0,∞);(f)
K∗(t)=K∗(1) for t≥1;(g)
K∗(t)≈K(1) on (0,1].Lemma 2.4 (see [5]).
IfK satisfies condition (1.8), then for any α≥1 and 0≤β<1,
one has (2.2)∫01rα−1(log1r)−βK(log1r)dr≈C(β)(1−βα)1−βK(1−βα),where C(β) is a constant depending on β alone.Lemma 2.5 (see [11]).
For0<p≤1, a∈𝔻,
and z=reiθ∈𝔻, (2.3)∫02πdθ|1−a¯reiθ|2p≤C(1−|a|r)p,where C>0 is a constant.Proof of Theorem1.1.
Letz=reiθ.
By Lemma 2.3, we may also assume that K is concave, so that the following inequality
true holds(2.4)12π∫02πK(1−|φa(reiθ)|2)dθ≤K(12π∫02π(1−|φa(reiθ)|2)dθ).From the definition of φK for 0<s, t<1,
we have that K(st)≤φK(s)K(t). Using these facts and polar coordinates, it
follows that(2.5)I(a)=∫𝔻|f′(z)|p(1−|z|2)qK(1−|φa(z)|2)dA(z)=12π∫01∫02π|f′(reiθ)|p(1−r2)qK(1−|φa(reiθ)|2)rdrdθ≤2∫01φp(r)(1−r2)q(12π∫02πK(1−|φa(reiθ)|2)dθ)rdr≤2∫01φp(r)(1−r2)qK(12π∫02π(1−|φa(reiθ)|2)dθ)rdr=2∫01φp(r)(1−r2)qK(12π∫02π1−|a|2|1−a¯reiθ|2dθ(1−r2))rdr=2∫01φp(r)(1−r2)qK(1−|a|21−|a|2r2(1−r2))rdr≤2∫01φp(r)(1−r2)qK(1−r2)φK(1−|a|21−|a|2r2)rdr.The last integral exists for
every a∈𝔻 in view of (1.9), and φK((1−|a|2)/(1−|a|2r2))≤1.
Further, since(2.6)lim|a|→1φK(1−|a|21−|a|2r2)=φK(0)=0,then the last integral tends to
zero for every r∈(0,1) as |a|→1.
By Lebesgue's dominated convergence theorem, we obtain lim|a|→1I(a)=0.
By Lemma 2.1, we get f∈𝒬K,0(p,q).From Theorem1.1, we have the following corollary. Here,
we give a different and technical proof.Corollary 2.6.
Letf∈H(𝔻), 0<p<∞, −2<q<∞, 0<s≤1,
and let φ be a monotone increasing function of r in (0,1) such that |f'(z)|≤φ(r),
for |z|=r.
If (2.7)∫01φp(r)(1−r2)q+srdr<∞,then f∈F0(p,q,s).Proof.
Leta∈𝔻.
We have(2.8)J(a)=∫𝔻|f′(z)|p(1−|z|2)q(g(z,a))sdA(z)=(∫𝔻∖𝔻(a,1/2)+∫𝔻(a,1/2))|f′(z)|p(1−|z|2)q(log1|φa(z)|)sdA(z)=I1(a)+I2(a).For z∈𝔻∖𝔻(a,1/2), 1/|φa(z)|≤2,
hence(2.9)log|1−a¯za−z|<|1−a¯za−z|−1≤4(1−|a−z1−a¯z|2)=4(1−|z|2)(1−|a|2)|1−a¯z|2.For 0<s≤1,
by Lemma 2.5, we have(2.10)I1(a)=∫𝔻∖𝔻(a,1/2)|f′(z)|p(1−|z|2)q(log1|φa(z)|)sdA(z)≤4s∫𝔻∖𝔻(a,1/2)|f′(z)|p(1−|z|2)q+s(1−|a|2)s|1−a¯z|2sdA(z)≤4sπ∫01|f′(reiθ)|p(1−r2)q+s[∫02π(1−|a|2)sdθ|1−a¯reiθ|2s]rdr≤2s4sπC∫01φp(r)(1−r2)q+s(1−|a|)s(1−|a|r)srdr.The last integral exists since ∫01φp(r)(1−r2)q+srdr<∞ and (1−|a|)/(1−|a|r)≤1.
It is clear that(2.11)lim|a|→1(1−|a|)(1−|a|r)=0,which implies(2.12)lim|a|→1∫01φp(r)(1−r2)q+s(1−|a|)s(1−|a|r)srdr=0.By Lebesgue's dominated
convergence theorem, we get(2.13)lim|a|→1I1(a)=0. Now, we consider the case z∈𝔻(a,1/2).
Note that(2.14)|1−a¯za−z|>2in the case. By a well-known
inequality (see, e.g., [12, page 3]), we have that(2.15)||z|−|a||1−|a|z|≤12and consequently, for z∈𝔻(a,1/2),
we have(2.16)|z|≤1+2|a|2+|a|.By the monotonicity of φ,
we haveI2(a)=∫𝔻(a,1/2)|f′(z)|p(1−|z|2)q(log1|φa(z)|)sdA(z)≤φp(1+2|a|2+|a|)∫𝔻(a,1/2)(1−|z|2)q(log1|φa(z)|)sdA(z)=φp(1+2|a|2+|a|)∫|w|<1/2(1−|φa(w)|2)q(log1|w|)s(1−|a|2)2|1−a¯w|4dA(w)=φp(1+2|a|2+|a|)∫|w|<1/2(1−|a|2)q+2(1−|w|2)q(log1|w|)s1|1−a¯w|2q+4dA(w)≤1πφp(1+2|a|2+|a|)(1−|a|2)q+2∫01/2(1−t2)q(log1t)s[∫02πdθ(1−|a¯|t)2q+4]tdt≤1πφp(1+2|a|2+|a|)2q+2(1−|a|)q+22π∫01/2(1−t2)q(log1t)s1(1−|a¯|t)2q+4tdt≤φp(1+2|a|2+|a|)2q+322q+4(1−|a|)q+2∫01/2(1−t2)q(log1t)stdt.This implies(2.17)I2(a)≤Cφp(1+2|a|2+|a|)(1−|a|)q+2,since the following integral
exists(2.18)∫01/2(1−t2)q(log1t)stdt.Choosing s=1,
for every r∈(0,1),
it follows that(2.19)∫r1φp(t)(1−t2)q+1tdt≥φp(r)∫r1(1−t2)q+1tdt=12φp(r)1q+2(1−r2)q+2.This and(2.20)∫01φp(r)(1−r2)q+1rdr<∞imply that(2.21)limr→1φp(r)(1−r)q+2=0or(2.22)φp(1+2|a|2+|a|)(1−1+2|a|2+|a|)q+2=φp(1+2|a|2+|a|)(1−|a|)q+2(2+|a|)q+2→0,as |a|→1.
Consequently,(2.23)lim|a|→1I2(a)≤Clim|a|→1φp(1+2|a|2+|a|)(1−|a|)q+2=0.Combining (2.8), (2.13) with (2.23) we
see that(2.24)lim|a|→1∫𝔻|f′(z)|p(1−|z|2)q(g(z,a))sdA(z)=0,which means f∈F0(p,q,s). The proof is complete.Proof of Theorem1.2.
Consider the monotone
increasing function(2.25)φ(r)=∑k=1∞nk|ak|rnk−1,0<r<1.For every θ∈[0,2π),
we have(2.26)|f′(reiθ)|=|∑k=1∞nkakrnk−1|≤φ(r),By Theorem 1.1, we only need to
prove(2.27)I=∫01(∑k=1∞nk|ak|rnk−1)p(1−r2)qK(1−r2)rdr<∞.By the inequality 1−r2≤2log(1/r), r∈(0,1),
and Lemma 2.2, there exists a constant C such that(2.28)K(1−r2)≤K(2log1r)≈CK(log1r).Then for 0≤q<∞,
we have(2.29)I≤C∫01(∑k=1∞nk|ak|rnk)pr1−p(log1r)qK(log1r)dr.For p≥1,
assume In={k:2n≤k<2n+1,k∈ℕ}.
Since(2.30)∑n=0∞2n/2r2n≤21/2∑n=0∞∫2n2n+1t−1/2rt/2dt≤21/2∫0∞t−1/2rt/2dt=2Γ(12)(log1r)−1/2,which together with Hölder's
inequality gives(2.31)[∑k=1∞nk|ak|rnk]p=[∑n=0∞∑nk∈Innk|ak|rnk]p≤[∑n=0∞∑nk∈Innk|ak|r2n]p=[∑n=0∞(2n/2r2n)1−1/p(r2n2(1−p)n/2)1/p∑nk∈Innk|ak|]p≤[∑n=0∞r2n2((1−p)/2)n(∑nk∈Innk|ak|)p][∑n=0∞2n/2r2n]p−1≤C(log1r)−(p−1)/2∑n=0∞r2n2((1−p)/2)n(∑nk∈Innk|ak|)p.Hence,(2.32)I≤C∑n=0∞(∑nk∈Innk|ak|)p2((1−p)/2)n∫01r2n+1−p(log1r)q−((p−1)/2)K(log1r)dr.For 1≤p≤2, 1≤p−2q<3,
by Lemma 2.4, choosing α=2n+2−p, β=(p−2q−1)/2,
we obtain(2.33)I≤C∑n=0∞(∑nk∈Innk|ak|)p((1/2)(2q−p+3)2n+2−p)(2q−p+3)/22((1−p)/2)nK((1/2)(2q−p+3)2n+2−p)≤C∑n=0∞(∑nk∈Innk|ak|)p(12n)(2q−p+3)/2(12n)(p−1)/2K(12n)=C∑n=0∞(∑nk∈Innk|ak|)p(12n)1+qK(12n).If nk∈In,
then nk<2n+1.
The assumption that K is nondecreasing and Lemma 2.2 give(2.34)(12n)1+qK(12n)≤(12n)1+q121+qK(12n+1)<nk−(1+q)K(1nk).Since f(z) is a lacunary series, the Taylor series of f has most [logλ2]+1 terms akznk such that nk∈In.
Combining this with the last inequality and Hölder's inequality, we
obtain(2.35)I≤C∑n=0∞(∑nk∈Innk1−((1+q)/p)|ak|K1/p(1nk))p≤C∑n=0∞([logλ2]+1)p−1∑nk∈Innkp−q−1|ak|pK(1nk)=C([logλ2]+1)p−1∑k=1∞nkp−q−1|ak|pK(1nk)<∞.This shows that f∈𝒬K,0(p,q).



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