For q≥1 we define Hq(φ) to be the set of all functions f which are analytic in the open unit disc and satisfy sup 0≤r<1 Z 2π 0 h φ¡ |f(reiθ

Interpolation sequence for the spaces H₊^q(φ)(q≥1)

Maher M. H.Marzuq

Abstract. Let φ be a subadditive increasing real valued function defined on [0,∞) and which satisfies φ(x) = 0 if and only if x= 0.

For q≥1 we define H^q(φ) to be the set of all functions f which are analytic in the open unit disc and satisfy

sup

0≤r<1

Z _2π

0

h φ¡

|f(re^iθ¢

| i_q

dθ <∞.

AndH₊^q(φ) to be the subspace ofH^q(φ) of functions which satisfy

r→1lim Z _2π

0

h φ¡

|f(re^iθ)|¢i^q dθ=

Z _2π

0

h φ¡

|f(e^iθ)|¢i^q dθ.

In this paper we prove some interpolation theorems forH₊^q(φ)

1. Introduction

Let us recall some definitions. We call a real-valued functionφdefined on [0,∞) a modulus function, ifφ is an increasing continuous subadditive function that satisfies the condition thatφ(x) = 0 if and only if x= 0.

Let q ≥ 1. By the class H^q(φ) we mean the collection of all analytic functionsf defined in the open unit disc ∆ which satisfy

sup

0≤r<1

Z _2π

0

h φ¡

|f(re^iθ)|¢iq

dθ <∞.

For q = 1 the spaces were studied in details in [2, 3]. If φ(x) =x, then H^q(φ) becomes the usual Banach space H^q. For φ(x) = x^p, 0 < p ≤ 1

2000Mathematics Subject Classification. Primary 46B45, 46A45.

43

and q = 1, then H^q(φ) becomes the usual F-spaces H^q. If q = 1 and φ(x) = log(1+x^p), thenH^q(φ) becomesN_p. Ifq= 1 andφ(x) = log(1+x), thenH^q(φ) becomes the classN of functions of bounded characteristic. For N_p=N₊ for 0< p≤1 [8], where

N_p ={f ∈H⁺(D) ∩ H^p(φ) : lim

r→1

Z _2π

0

log(1 +|f(re^iθ)|^p) dθ

= Z _2π

0

log(1 +|f(e^iθ)|^p) dθ}, and

N⁺={f ∈N : lim

r→1

Z _2π

0

log⁺(|f(re^it)|) dt= Z _2π

0

log⁺(|f(e^it)|) dt}.

In general the spaces H^q(φ) are not F-Spaces; see [13, p. 453] for an example. We shall assume that φ satisfies the additional condition that φ(e^t) is a convex function of t, and consequently H^q(φ)⊆N.

A function f ∈H^q(φ) is said to belong to the class H₊^q(φ) if

r→1lim Z _2π

0

h ϕ¡

|f(re^iθ)|¢iq

dθ= Z _2π

0

[ϕ(|f(e^iθ)|)]^qdθ.

For the class H₊^q(φ), which is a vector space , we define a distance func- tion by

ρ(f,0) =

· 1 2π

Z _2π

0

[ϕ(|f(e^iθ)|)]^qdθ

¸¹

q

(1.1) For the spaces H^q(p ≥1), H^q(0< p ≤1). [5], N₊ [13] N_p⁺(p > 0) and (log⁺H)^α(a >1), [12] become special cases of H₊^q(φ).

Let α = (α_n) = (α₁, . . . , α_n, . . .), be a sequence of real numbers such thatα_n→0.

Let

l^q(φ, α) ={(c_n) : c_n∈C and d((c_n),0) = hX^∞

n=1

α_n[ϕ(|c_n)|)]^q i¹

q <∞}

LetX be the class of analytic functions in ∆ and{z_n}is a given sequence in ∆. When a complex sequence {c_n} is given, the interpolation problem asks if a functionf ∈X exists such thatf{z_n}=c_n for all n.

Let Y be a class of complex sequences. If for every sequence (c_n) ∈ Y there is anf ∈ X such that f{z_n} =c_n, then the sequence {z_n} is called a universal interpolation sequence for the pair (X, Y), simply written z_n is u. i. s. for (X, Y). We are interested in the pair H₊^q(φ), l^q(φ, α) where α= (1− |z_n|²).

A sequence {z_n} in ∆ is called uniformly separated sequence (u. s. s.) if X∞

n=1

¡1− |z_n|¢

<∞, and Y∞

m6=nn=1

¯¯

¯ z_n−z_n 1−z¯_mz_n

¯¯

¯≥δ >0 (m= 1,2, . . .).

L. Carleson [1] showed that {z_n} is u. i. s. for (H^∞, l^∞) if and only if {z_n}is u. s. s. For the pair ¡

H₊^q(φ), l^q(φ, α)¢

we have the following results:

for q ≥ 1, φ(x) = x, [11] proved that {z_n} is u. i. s. for ¡

H₊^q(φ), l^q(φ, α)¢ if and only if {z_n} is u. s. s. The same result was proved for q = 1 and φ(x) =x^p,0< p≤1 by [9]. Forq = 1, φ(x) = log(1 +x),[14] showed that {z_n}is u. s. s., then {z_n} is u. i. s. and if {z_n} is u. i. s., then

(1− |z_n|²) log 1

|B_n(z_n)| →0 as n→ ∞, where

B_n(z) = Y∞

m6=nm=1

|z_m| z_m

z_m−z 1−z¯_mz .

In this paper we obtain results which generalize the above mentioned results.

In section 2 we will prove that H₊^q(φ) and l^q(φ, α) are F-spaces in the sense of Banach, [4].

In section 3 we prove that if {z_n} is u. s. s. then {z_n} is u. i. s. for

¡H₊^q(φ), l^q(φ, α)¢

, we also proved that if φsatisfies lim

x→∞ϕ⁻¹(ax)ϕ⁻¹(_x¹)<

∞for all real aand T_φ,q¡

H₊^q(φ)¢

=l^q(φ, α) then{z_n} is u. s. s. where T_φ,q(f(z)) =³¡

φ⁻¹(1− |z_n|²)¢¹

q−1f(z_n)

´ .

2. The spaces H₊^q(φ), l^q(φ, α).

In this section we will show that the spaces H₊^q(φ) and l^q(φ, α) are F- spaces.

Theorem 2.1. H₊^q(φ) is an F-space in the sense of Banach,[4]. That is

i. Letf_nbe functions inH₊^q(φ)such thatρ(f_n,0)→0asn→ ∞. Then for anyα∈C, ρ(αf_n,0)→0 asn→ ∞.

ii. Let α_n ∈ C be such that α_n → 0. Then for each f ∈ H₊^q(φ), ρ(α_nf,0)→0 asn→ ∞.

iii. H₊^q(φ) is complete with respect to the metric(1.1).

Proof. i. Suppose {f_n} is a sequence in H₊^q(φ), ρ(f_n,0) → 0 and β ∈C. Now

ρ(βf_n,0) = µ 1

2π Z _2π

0

[ϕ(|βf_n(θ)|^qdθ

¶¹

q

≤£

|β|+ 1¤

ρ(f_n,0)→0, where £

|β|¤

is the greatest integer in |β|.

ii. Supposeβ_n ∈ C, β_n → 0 and f ∈ H₊^q(φ), without loss of generality we may assume |β_n| ≤1, so

[ϕ(|β_nf|)]^q≤[φ(|f|)]^q and ϕ(|β_nf(θ)|)→0 a. e.

Hence by Lebesgue convergence theorem we get

n→∞lim 1 2π

Z _2π

0

£φ(|β_nf(θ)|¤_q

dθ= 0.

Thus ρ(β_nf,0)→0 asn→ ∞.

iii. Suppose{f_n}is a Cauchy sequence inH₊^q(φ). By Lemma 3 in [2] ap- plied to£

φ(|f|)¤_q

which is subharmonic forq≥1, from [6, Lemma 5.1], we get

|f(z)| ≤ϕ⁻¹



 cρ(f,0)

³

(1− |z|)^1q

´



 forz∈∆.

Therefore

|f_n(z)−f_m(z)| ≤ϕ⁻¹



cρ(f_n, f_m)

³

(1−r)^1q

´



< ε

for n, m > N(ε) and for all z ∈ {w : |w| ≤ r < 1}. Hence {f_n(z)}

is a Cauchy sequence in C. Since {f_n(z)} converges uniformly on compact subsets of ∆ and f_n(z) is analytic, then {f_n} converges to an analytic function f. Clearly, {φ(|f_n|)} converges uniformly on compact subsets to φ(|f|). Therefore

Z _2π

0

h ϕ¡

|f(re^iθ¢

| i_q

dθ= lim

n→∞

Z _2π

0

£ϕ(|f_n(re^iθ)|)¤_q dθ

≤ lim

n→∞

Z _2π

0

[ϕ(|f_n(θ)|)]^qdθ≤M,

hence, f ∈ H^q(φ). ButH^q(φ) ⊆ N, so lim

r→1f(re^iθ) = f(θ) a. e. and f_n(θ) → f(θ) in measure. Now choose a subsequence f_nj such that f_nj(θ)→f(θ) a. e., then

ρ(f, f_n) = µ 1

2π Z _2π

0

[ϕ(|f(θ)−f_n(θ|)]^qdθ

¶¹_q

≤ lim

j→∞

µ 1 2π

Z _2π

0

[ϕ(|f_nj(θ)−f_n(θ)|]^qdθ

¶¹

q

≤ρ(f_nj, f_n) +ε for largej.

Thus if n_j and nare sufficiently large, we have

ρ(f, f_n)→0, asn→ ∞.

It remains to show that f ∈H₊^q(φ). Since f_nj(θ) →f(θ) a. e., then there exists a compact set E ⊂[0,2π] such that m(E)>2π−ε and f_nj →f uniformly on E, hence

µ 1 2π

Z _2π

0

[ϕ(|f_nj(θ)|)]^qdθ

¶¹_q

≤ µ 1

2π Z

E

[φ(|f(θ)|)]^qdθ

¶¹

q+ε , for largej.

Also, µZ

E^c

[ϕ(|f_nj(θ)|)]^qdθ

¶¹

q ≤ µZ

E^c

[ϕ(|f_nj(θ)−f(θ)|)]^qdθ

¶¹

q

+ µZ

E^c

[ϕ(|f(θ)|)]^qdθ

¶¹

q

≤ρ(f_nj, f) + µZ

E^c

[ϕ(|f(θ)|)]^qdθ

¶¹

q

≤ε+ µZ

E^c

[ϕ(|f(θ)|)]^qdθ

¶¹

q,

butf_nj ∈H₊^q(ϕ), so µZ _2π

0

[ϕ(|f_nj(re^iθ)|)]^qdθ

¶¹_q

≤ µZ _2π

0

[ϕ(|f(θ)|)]^qdθ

¶¹_q +ε, letting j→ ∞,

µZ _2π

0

[ϕ(|f(re^iθ)|)]^qdθ

¶¹

q

≤ µZ _2π

0

[ϕ(|f(θ)|)]^qdθ

¶¹

q

+ε, for anyε >0. Hence

r→1lim µZ _2π

0

[ϕ(|f(re^iθ)|)]^qdθ

¶¹

q

= µZ _2π

0

[ϕ(|f(θ)|)]^qdθ

¶¹

q

,

this proves thatf ∈H₊^q(φ).

Theorem 2.2. l^q(ϕ, α) is anF-space.

Proof. Parts (i), (ii) are exactly the same as in Theorem 2.1. For com- pleteness, suppose {x_n} is a Cauchy sequence, let ε > 0 be given, then there existsN such that

d(x_k, x_m)< ε for allk, m > N(ε).

Hence

X∞

n=1

α_n

³

ϕ(|x^k_n−x^m_n|)

´_q

< ε^q, (2.1)

wherex_k= (x^k_n), x_m= (x^m_n). Therefore

|x^k_n−x^m_n|< ϕ⁻¹

"

ε α

1

nq

# .

Thus {x^k_n} is a Cauchy sequence inC for eachn. So it must converge to somex_n∈C.

Let x={x_n}, then Minkowski’s inequality gives Ã_∞

X

n=1

α_n(ϕ(|x_n|))^q

!¹

q

≤ Ã _∞

X

n=1

α_n

³

ϕ(|x_n−x^k_n|)

´_q!¹

q

+ Ã_∞

X

n=1

α_n

³ ϕ(|x^k_n|)

´_q!¹

q

,

but{x_n} is a Cauchy sequence, hence the second term of the right side is bounded by M which does not depend onk, also

Xα_n

³

ϕ(|x_n−x^k_n|)

´_q

= lim

m→∞

Xα_n

³

ϕ(|x^m_n −x^k_n|)

´_q

< ε^q by (2.1). Thusx_n→x and x∈l^q(φ, α).

3. Interpolation Theorems

We will assume an additional condition onf, for eachg∈H¹the function log£

ϕ⁻¹¡

|g(z)|¢¹

q + 1¤

is integrable on the unit circle.

Now we prove a theorem which generalizes theorem 1 in [14].

Theorem 3.1. If{z_n}is u. s. s. then{z_n}is u. i. s. for¡

H₊^q(φ), l^q(φα)¢ , whereα= (1− |z_n|²).

Proof. Letc= (c_n)∈l^q(φ, α) and let g(z) =

X∞

n=1

¡1− |z_n|²¢₂

[ϕ|c_n|)]^q B_n(z)

B_n(z_n)· 1 (1−z¯_nz)² . clearlygis analytic in ∆ since P^∞

n=1

(1−|z_n|²)[ϕ(|c_n|)]^q<∞. Alsog∈H¹, because

1 2π

Z _2π

0

|g(re^iθ)|dθ≤ X∞

n=1

¡1− |z_n|²¢₂£

(ϕ|c_n|)¤_q 1

|B_n(z_n)|

1 2π

Z _2π

0

dθ

|1−z¯_nz|

≤ 1 δ

X∞

n=1

¡1− |z_n|²¢£

ϕ(|c_n|)¤_q

<∞.

Since log£

ϕ⁻¹(|g(z)|^1q) + 1¤

is integrable, hence by [7, p. 53] there exists f₁∈H¹such that|f₁(z)|=φ⁻¹(|g(z)|^1q)+1. Since B_k(z_n) = 0 for allk6=n, g(z_n) = [φ(|c_n|)]^q. Hence, |f₁(z_n)|=|c_n|+ 1 and f₁(z_n) = (|c_n|+ 1)e^iθn, where c_n = |c_n|e^iαn and f₃(z_n) = e^iαn, then (f₁f₂−f₃)(z_n) = c_n. Since ϕ^q(|f₁(z)|)≤ |g(z)|+ϕ^q (1), we have f₁ ∈H^q(φ) and so isf₁f₂−f₃.

By Carleson’s Theorem there exists functions f₂, f₃ in H^∞ such that f₂(z_n) = e^i(αn−θn).

The following theorem generalizes the one given by [11].

Theorem 3.2. Suppose φsatisfies

x→∞lim ϕ⁻¹(ax)ϕ⁻¹

³1 x

´

<∞, for all realaand T_φ,q¡

H₊^q(φ)¢

=l^q(φ, α), then {z_n} is u. s. s.

We need the following lemma:

Lemma. The sequence E = {e_n} is bounded in l^q(φ, α) in the sense of topological vector space W. Rudin [10], wheree_n= (0,0, . . . ,0,1,0, . . .) and1 appears in the nth place.

Proof. Lets >0 be given and let B_s={x∈l^q(φ, α)kxk< s}. We need to show that there exists r₀ such that E ⊂ rB_s for all r > r₀. Let r₀ be such that (φ^q)⁻¹(s) = _r¹₀, then forr > r₀

°°

°e_n r

°°

°=ϕ^q

³1 r

´

< ϕ^q

³1 r₀

´

=s, this implies ^e_rⁿ ∈B_s. ThereforeE ⊂rB_s.

Proof of Theorem 3.2. Let

N_ϕ^q ={f ∈H₊^q(ϕ) : f(z_n) = 0 for all n}.

The quotient spaceH₊^q(ϕ)/N_ϕ^q is anF-space andT_φ,q induces a one-to-one bounded linear functional ∆_ϕ,q (since T_ϕ,q is bounded) from H₊^q(ϕ)/N_ϕ^q onto l^q(φ, α). Hence, the inverse is bounded which implies that ∆⁻¹_φ,q(E)

is bounded on H₊^q(ϕ), i. e. there exists M > 0 such that ∆⁻¹_φ,q(E) ⊂ B_M, which means that for alln, there exists f_n∈H₊^q(ϕ) such that

kf_nk ≤M, (3.1)

and

f_n(z_k) =





ϕ⁻¹(1− |z_k|²)

“1 q

”

−1 n=k,

0 n6=k.

For n > k, let

F_nk(z) =f_k(z)Y

j6=kj=1

1−z¯_jz z−z_j , then

kF_nkk ≤ kf_kk and F_nk ∈H₊^q(ϕ), so by [2]

|F_nk(z)| ≤ϕ⁻¹ Ã

ckF_nkk

¡1− |z|²¢¹

q

!

, (3.2)

but,

|F_nk(z)|=

¯¯

¯¯ Yn

j6=kj=1

1−z¯_jz_k z_k−z_j

¯¯

¯¯

¯¯

¯¯f_k(z_k)

¯¯

¯¯,

thus

|F_nk(z_k)|= h

ϕ⁻¹¡

1− |z_k|²¢₍¹

q)−1iYⁿ

j6=kj=1

¯¯

¯1−z¯_jz_k z_k−z_j

¯¯

¯,

and by using (3.1) and (3.2) we get

¯¯

¯¯ Yn

j=1j6=k

1−z¯_jz_k z_k−z_j

¯¯

¯¯≤ϕ⁻¹(α_k)ϕ⁻¹

³cM α_k

´

=ϕ⁻¹(x_k)ϕ⁻¹

³ 1 x_k

´

<∆<∞,

forx_k large, where x_k = _cM^αk and ∆ is a constant. Thus {z_k} is u. s. s.

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Maher M. H.Marzuq 84 Raymond Road Plymouth, MA 02360

(Received February 4, 2011)