Let Λ={λk}k=1∞ satisfy 0<λ1<λ2<⋯, ∑k=1∞1/λk<∞ and infk(λk+1-λk)>0. We investigate the Müntz spaces MpΛ=span¯{tλk:k=1,2,…}⊂Lp(0,1) for 1≤p≤∞. We show that, for each p, there is a Müntz space Fp which contains isomorphic copies of all Müntz spaces as complemented subspaces. Fp is uniquely determined up to isomorphisms by this maximality property. We discuss explicit descriptions of Fp. In particular Fp is isomorphic to a Müntz space Mp(Λ^) where Λ^ consists of positive integers. Finally we show that the Banach spaces (∑n⊕Fn)p for 1≤p<∞ and (∑n⊕Fn)0 for p=∞ are always isomorphic to suitable Müntz spaces Mp(Λ) if the Fn are the spans of arbitrary finitely many monomials over [0,1].

1. Introduction

Our paper is concerned with the Banach space geometry of Müntz spaces which is largely unexplored.

Let Λ={λk:k=1,2,…} be a sequence of real numbers with 0<λ1<λ2<⋯ and(1)∑k=1∞1λk<∞,infkλk+1-λk>0.For 1≤p<∞ and f:[0,1]→R, put(2)fp=∫01ftpdt1/p,f∞=supt∈0,1ft.For λ>0, let tλ be the function f:[0,1]→R with f(t)=tλ. In this paper, we consider the Müntz spaces(3)MpΛ=span¯tλk:k=1,2,…⊂Lp0,1for1≤p≤∞.By the Müntz theorem [1, 2], it is clear that Mp(Λ)≠Lp(0,1) if 1≤p<∞ and M∞(Λ)≠C0(0,1):={f:[0,1]→R:fcontinuous,f(0)=0}. It is known that Mp(Λ) is isomorphic to lp for 1≤p<∞ and M∞(Λ) is isomorphic to c0 provided that Λ satisfies in addition the lacunary condition infk(λk+1/λk)>1 [3]. On the other hand, there are examples of Müntz spaces M∞(Λ) which are not isomorphic to c0 ([4] or [5, Corollary 10.2.4.]).

We want to show that, for fixed p, all Müntz spaces are isomorphic to complemented subspaces of one special function space which itself is a Müntz space Mp(Λ^) where Λ^ consists of positive integers.

For two Banach spaces X and Y, we write X~Y if X is isomorphic to Y.

For a sequence of Banach spaces En, we put(4)∑n=1∞⊕Enp=en:en∈En,n=1,2,…,∑n=1∞enp1/p<∞,if 1≤p<∞, and(5)∑n=1∞⊕En0=en:en∈En,n=1,2,…,limn→∞en=0with the norm (en)=supnen.

Theorem 1.

Let (Ωk) be a sequence of finite subsets of N containing each finite subset of N infinitely many times. Moreover let Fp(Ωk) be the linear span of the functions tλ, λ∈Ωk, in Lp(0,1). Put(6)Fp=∑k∈N⊕FpΩkpif1≤p<∞,F∞=∑k∈N⊕F∞Ωk0.

For each p∈[1,∞], there is a set Λ^⊂N satisfying (1) such that Fp is isomorphic to Mp(Λ^).

If Λ is any set of positive real numbers satisfying (1), then Mp(Λ) is isomorphic to a complemented subspace of Fp.

On the other hand, we do not know if any complemented subspace of Fp is isomorphic to a Müntz space.

We prove Theorem 1 in Section 4. It turns out that the “maximality” condition of Theorem 1(b) implies uniqueness up to isomorphism. Indeed we have the following.

Theorem 2.

Let 1≤p≤∞ and assume that Λ is a set of real positive numbers satisfying (1). If Mp(Λ) contains a complemented isomorphic copy of any Müntz space Mp(Λ~) then Mp(Λ) is isomorphic to the space Fp of Theorem 1.

Proof.

Consider the spaces Fp(Ωk) of Theorem 1. By the choice of the Ωk and the definition of Fp, we have Fp~Fp⊕Fp⊕⋯.

Now let Λ satisfy the assumptions of Theorem 2. Since Fp is isomorphic to a Müntz space, by Theorem 1 there is a complemented subspace A⊂Mp(Λ) with Mp(Λ)~A⊕Fp. Moreover, by Theorem 1(b) there is a complemented subspace B⊂Fp with Fp~B⊕Mp(Λ). Using Pelczynski’s decomposition method [6, Theorem 2.2.3.], we obtain Fp~Mp(Λ).

Corollary 3.

F∞ is not isomorphic to c0.

Proof.

There is a Müntz space M∞(Λ) which is not isomorphic to c0 ([4] or [5, Corollary 10.2.4.]). But, according to Theorem 1, M∞(Λ) is isomorphic to a complemented subspace of F∞. Since all infinite dimensional complemented subspaces of c0 are isomorphic to c0 [7, Theorem I.2.a.3.], the space F∞ cannot be isomorphic to c0.

In connection with the proof of Theorem 1, we also get the following.

Theorem 4.

Let Ωn, n=1,2,…, be arbitrary finite subsets of positive real numbers and put Fp(Ωn)=span{tλ:λ∈Ωn}⊂Lp(0,1). Then, for each p∈[1,∞], there is a subset Λ⊂N satisfying (1) such that (∑n⊕Fp(Ωn))p~Mp(Λ), if 1≤p<∞, and (∑n⊕F∞(Ωn))0~M∞(Λ), if p=∞.

We do not know if, conversely, any Müntz space Mp(Λ) is isomorphic to the direct sum of the spaces Fp(Ωn) for suitable Ωn.

Put C0,0(0,1)={f∈C0(0,1):f(1)=0}. For a finite dimensional subspace E⊂Lp(0,1), 1≤p<∞, or E⊂C0(0,1), p=∞, and ϵ>0, put(7)αE,p,ϵ=supα>0:e·10,αp≤ϵep∀e∈E.If E⊂C0(0,1), p=∞, also put(8)γE,ϵ=infγ<1:e1-e·1γ,1∞≤ϵe∞∀e∈E.Similarly, for E⊂Lp(0,1), 1≤p<∞, or E⊂C0,0(0,1), p=∞, define(9)βE,p,ϵ=infβ>0:e·1β,1p≤ϵep∀e∈E.We collect a few properties of these parameters in the following.

Lemma 5.

One has

α(E,p,ϵ)≤β(E,p,ϵ) for all p<∞ if ϵ<1/2,

α(E,∞,ϵ)≤β(E,∞,ϵ) if ϵ<1,

α(E,∞,ϵ)≤γ(E,ϵ) if ϵ<1/3.

Proof.

We prove (c). The proofs of (a) and (b) are similar. Let e∈E and let γ<1 be such that e1-e·1γ,1∞≤ϵe∞. If t∈[γ,1], we obtain et-eγ≤2ϵe∞. This yields(10)e∞=maxe·10,γ∞,e·1γ,1∞≤maxe·10,γ∞,eγ+2ϵe∞.Hence, 1-2ϵe∞≤e·10,γ∞. For ϵ<1/3, we obtain (1-2ϵ)>ϵ and therefore ϵe∞<e·10,γ∞. This implies α(E,∞,ϵ)≤γ(E,ϵ).

Lemma 6.

Consider E⊂Lp(0,1), 1≤p<∞, or E⊂C0,0(0,1), p=∞. Fix 0<ϵ<1/2 and let α=α(E,p,ϵ) and β=β(E,p,ϵ). Then one has(11)1-2ϵep≤e·1α,βp≤ep∀e∈E.For p=∞, one even has e·1α,β∞=e∞.

Proof.

We obtain (12)ep≤e·1α,βp+e·10,αp+e·1β,1p≤e·1α,βp+2ϵep.which yields the first inequality. The second inequality is trivial. The last assertion of Lemma 6 follows directly from the definitions of α and β.

Proposition 7.

Fix ϵn>0 such that ∑n=1∞ϵn<1/8. Let En⊂Lp(0,1), 1≤p<∞, or En⊂C0,0(0,1), p=∞, be finite dimensional subspaces with(13)βn≔βEn,p,ϵn<αEn+1,p,ϵn+1≕αn+1.Then, for any en∈En and N∈N, one has (14)12∑n=1Nenpp1/p≤∑n=1Nenp≤43∑n=1Nen·1αn,βnp≤43∑n=1Nenpp1/p,if 1≤p<∞, and(15)78supn≤Nen∞≤∑n=1Nen∞≤98∑n=1Nen·1αn,βn∞≤98supn≤Nen∞,if p=∞.

Proof.

Let 1≤p<∞. Put e=∑n=1Nen. We obtain with Lemma 6 the following: (16)e-∑n=1Nen·1αn,βnp=∑n=1Nen-en·1αn,βnp≤∑n=1N2ϵnenp≤∑n=1N2ϵn1-2ϵnen·1αn,βnp≤13supn≤Nen·1αn,βnp≤13∑n=1Nen·1αn,βnpp1/p.Here we used ∑n=1∞ϵn<1/8. Hence, 1-2ϵn≥3/4 for all n and ∑n=1∞2ϵn/(1-2ϵn)<1/3.

This yields, in view of (13), (17)ep≤∑n=1Nen·1αn,βnp+13∑n=1Nen·1αn,βnpp1/p=43∑n=1Nen·1αn,βnpp1/p≤43∑n=1Nenpp1/p,ep≥∑n=1Nen·1αn,βnp-13∑n=1Nen·1αn,βnpp1/p=23∑n=1Nen·1αn,βnpp1/p≥12∑n=1Nenpp1/p.Here we used Lemma 6 and 1-2ϵn≥3/4. The proof for p=∞ is similar.

Proposition 8.

Fix ϵn>0 such that ∑n=1∞ϵn<1/8. Let En⊂C0(0,1) be finite dimensional subspaces with(18)γn≔γEn,ϵn<αEn+1,∞,ϵn+1≕αn+1.Then, for any ek∈Ek and N∈N, one has(19)12supn≤N∑k=1nek∞≤∑k=1Nek∞≤supn≤N∑k=1nek∞.

Proof.

By (18) we have, for any n≤N, (20)∑k=1Nek·10,γn∞-∑k=1nek·10,γn∞≤∑k=n+1Nek·10,αk∞≤∑k=n+1Nϵkek∞.Equation (18) and Lemma 5 also imply that γn are monotonically increasing. Hence, if t∈[γn,1], we obtain (21)∑k=1nekt-∑k=1nekγn≤∑k=1nekt-ek1+ek1-ekγn≤∑k=1n2ek1-ek·1γn,1∞≤2∑k=1nϵkek∞.Here we used that γk≤γn and hence [γn,1]⊂[γk,1] for all k=1,…,n. The preceding estimate implies (22)∑k=1nek∞≤max∑k=1nek·10,γn∞,∑k=1nekγn+2∑k=1nϵkek∞≤∑k=1nek·10,γn∞+2∑k=1nϵkek∞.We conclude(23)0≤∑k=1nek∞-∑k=1nek·10,γn∞≤2∑k=1nϵkek∞,(24)∑k=1Nek∞≥∑k=1Nek·10,γn∞≥∑k=1nek·10,γn∞-∑k=n+1Nϵkek∞≥∑k=1nek∞-∑k=1N2ϵkek∞≥∑k=1nek∞-14supk≤Nek∞.Applying this estimate for n and n-1 yields (25)2∑k=1Nek∞≥∑k=1nek∞+∑k=1n-1ek∞-12supk≤Nek∞≥en∞-12supk≤Nek∞,from which we obtain supkek∞≤4∑k=1Nek∞. Moreover, using (24) we see that supn∑k=1nek∞≤2∑k=1Nek∞ which proves the first inequality of the proposition. The second inequality is trivial.

Now consider, for fixed p and N>0, the maps(26)TNft=N1/ptN-1/pftN,SNft=N1/ptN-1/pftN,t∈[0,1]. (Here ⌈r⌉ is the smallest integer ≥r.) We include the case p=∞ by putting 1/∞=0. We clearly have |TNf|≥|SNf| and we obtain TNfp=fp for f∈Lp(0,1). If p=∞, then SN=TN.

(a) follows from simple substitution. (Recall that γ is only defined for p=∞.)

(b): Let α0=α1/N(E,p,ϵ)=α(TNE,p,ϵ) and β0=β1/N(E,p,ϵ)=β(TNE,p,ϵ). At first, let p<∞. Consider e∈E. Put θ=⌈(N-1)/p⌉-(N-1)/p. Since 0≤θ≤1 and (SNe)(t)=tθ(TNe)(t) for t∈[0,1], we obtain, with Lemma 6, (30)SNe·10,α0p≤α0θTNe·10,α0p≤α0θϵTNep≤α0θϵ1-2ϵTNe·1α0,β0p≤ϵ1-2ϵSNep.This yields the first inequality if p<∞.

Similarly, we obtain(31)SNe·1β0,1p≤TNe·1β0,1p≤ϵTNep≤ϵ1-2ϵTNe·1α0,β0p≤α0-θϵ1-2ϵSNe·1α0,β0p≤ϵ1-2ϵα0SNep.This yields the second inequality of (c) if p<∞.

If p=∞, then the two inequalities follow from (a) and the fact that α is increasing and β is decreasing with respect to ϵ. (Recall that here is SN=TN).

(c): If p=∞, then (c) is trivial. Now let p<∞. By definition, SNep≤TNep=ep. On the other hand, with α0, β0 of (b), and Lemma 6, (32)SNep≥SNe·1α0,β0p≥α0N-1/p-N-1/pTNe·1α0,β0p≥α01-2ϵTNep=α01-2ϵepsince 0≤⌈(N-1)/p⌉-(N-1)/p≤1.

3. Some Facts about Müntz Spaces

If Λ satisfies (1), then any f∈Mp(Λ) can be represented as f(t)=∑n=1∞αktλk where the series converges pointwise on [0,1[. There are universal constants ck>0 with |αk|≤ckfp for all k [8]. Moreover, it is easily seen that for any λ>0 we have(33)limμ→λtλ-tμ∞=limμ→λλμλ/μ-λ1-λμ=0.

Lemma 10.

Let ϵk>0 such that ∑k=1∞ϵkck<1/2. Moreover, let μk satisfy tλk-tμk∞≤ϵk for all k. Then, T with T(∑n=1∞αktλk)=∑n=1∞αktμk is an isomorphism between Mp(Λ) and Mp({μk:k=1,2,…}).

Proof.

For f(t)=∑k=1∞αktλk we obtain(34)Tf-fp≤∑k=1∞ckϵkfp≤12fp.This proves 2-1fp≤Tfp≤3·2-1fp.

Using Lemma 10, we can always assume in the following that Λ⊂Q.

Put(35)M∞0Λ=f∈M∞Λ:f1=0=span¯tλk-tλk+1:k=1,2,….Since M∞0(Λ) is 1-codimensional in M∞(Λ) and M∞(Λ) contains a complemented subspace isomorphic to c0, we obtain M∞0(Λ)~M∞(Λ).

Proposition 11.

Fix X=Mp(Λ), 1≤p<∞, or X=M∞0(Λ), p=∞. Then, there are linear bounded finite rank operators Rn:X→X with limn→∞Rnf=f, f∈X, RnRm=Rmin(n,m) if n≠m and(36)c1∑k=1∞Rk-Rk-1fpp1/p≤fp≤c2∑k=1∞Rk-Rk-1fpp1/p,if 1≤p<∞, or(37)c1supkRk-Rk-1f∞≤f∞≤c2supkRk-Rk-1f∞,if p=∞. Moreover, there are indices mk<nk, k=1,2,…, with(38)Rk-Rk-1X=spantλj-tλj+1:mk≤j≤nk,if p=∞, or(39)Rk-Rk-1X=spantλj:mk≤j≤nk,if 1≤p<∞.

For a proof see [5, Proposition 9.1.4].

The condition limn→∞Rnf=f implies the following.

Corollary 12.

For the operators Rk of Proposition 11 one has f=∑k=1∞(Rk-Rk-1)f (with R0=0), f∈X, where the series converges in norm.

However, Rk-Rk-1 are no projections and the “summands overlap”; that is, (Rk-Rk-1)2≠Rk-Rk-1 and (Rk-Rk-1)(Rk+1-Rk)≠0 in general.

4. Final Proofs

Let Λ satisfy (1). Fix 1≤p≤∞. Take Rk, mk, and nk as in Proposition 11 and put Ek=(Rk-Rk-1)Mp(Λ), if 1≤p<∞, or Ek=(Rk-Rk-1)M∞0(Λ), if p=∞. Fix 0<ϵk<1/4 and let 0<N1<N2<⋯ be such that(40)α1/NkEk,p,ϵk≥12∀k.Let Fj be finite dimensional Banach spaces and put X=(∑j=1∞⊕Fj)p, if 1≤p<∞, or X=(∑j=1∞⊕Fj)0, if p=∞.

Lemma 13.

Assume that Fjk=SNkEk for some jk, k=1,2,…. Then, Mp(Λ) is isomorphic to a complemented subspace of X.

Proof.

For e∈Mp(Λ), 1≤p<∞, or e∈M∞0(Λ), p=∞, put Ue=(fj) where fjk=SNk(Rk-Rk-1)e and fj=0 if j≠jk, k=1,2,…. Using Proposition 11, Corollary 12, and Lemma 9(c), we see that Ue is a well-defined element in X and U is a bounded linear operator Mp(Λ)→X if p<∞ or M∞0(Λ)→X. By Lemma 9(c) and condition (40), U is even an into-isomorphism. Hence, there are universal constants c1>0 and c2>0 such that c1ep≤Uep≤c2ep for all e.

It remains to prove that UMp(Λ) and UM∞0(Λ), respectively, are complemented in X. To this end, let f∈X. At first assume that, for some N∈N, f=(f1,…,fN,0,0,…) with fj∈Fj and fjk=SNkek for some ek∈Ek if jk≤N. Put Vf=∑jk≤Nek. The fact that RmRn=Rmin(m,n) if m≠n implies (Rn-Rn-1)(Rk-Rk-1)=0 if n-k≥2. Hence,(41)Rn-Rn-1Vf=Rn-Rn-1en-1+en+en+1.We obtain a universal constant c3>0 with (42)Rn-Rn-1Vfp≤c3en-1pp+enpp+en+1pp1/p≤4c3SNn-1en-1pp+SNnenpp+SNn+1en+1pp1/pif 1≤p<∞. Here again, (40) and Lemma 9(c) were used. In view of Proposition 11, this implies that V is a bounded linear operator on the dense subspace of X consisting of finite sequences. Hence, V can be uniquely extended to a linear bounded operator V:X→Mp(Λ), called V again. The same argument shows that V is also bounded if p=∞.

We obtain by definition VUe=e if e∈Mp(Λ), 1≤p<∞, or e∈M∞0(Λ), p=∞. This follows from Corollary 12. Indeed, we have Ue=(fj) with fjk=SNkek and fl=0 for l∉{jk:k=1,2,…}. Hence,(43)VUe=Vfj=∑k=1∞Rk-Rk-1e=e.(Here we used R0=0.) This shows that UV is a bounded projection from X onto UMp(Λ), if 1≤p<∞, or onto UM∞0(Λ), if p=∞. Since M∞0(Λ)~M∞(Λ), this completes the proof.

Proposition 14.

Let 1≤p<∞. Assume that Ωk, k=1,2,…, are finite subsets of positive numbers and put(44)Fk=spantλ:λ∈Ωk⊂Lp0,1.Then, there is a Müntz space Mp(Λ) which is isomorphic to ∑k=1∞⊕Fkp.

Here Λ=∪k=1∞⌈Nk-1/p⌉+NkΩk for suitable integers Nk>0. If all Ωk consist of rational numbers, then Nk can be taken such that in addition Λ⊂N.

Proof.

Fix ϵk>0 such that(45)∑k=1∞ϵk1-2ϵk<116.Find integers 1<N1<N2<⋯ such that maxNkΩk<minNk+1Ωk+1 for all k and(46)12<α1/NkFk,p,ϵk≤β1/NkFk,p,ϵk<α1/Nk+1Fk+1,p,ϵk+1for all k. Put αk=α1/Nk(Fk,p,ϵk). Then, we obtain(47)∑k=1∞ϵk1-2ϵkαk<18.Define Λ=∪k=1∞⌈Nk-1/p⌉+NkΩk. (Of course we can take Nk so large that Λ satisfies (1) and consists of integers if all Ωk consist of rational numbers.)

We obtain, by Lemma 5, Lemma 9 ((a) and (b)), and (46),(48)αSNk+1Fk+1,p,ϵk+11-2ϵk+1αk+1≥αSNk+1Fk+1,p,ϵk+11-2ϵk+1≥α1/Nk+1Fk+1,p,ϵk+1≥β1/NkFk,p,ϵk≥βSNkFk,p,ϵk1-2ϵkαk.Here we used that α(·,·,ϵ) is increasing with respect to ϵ. Now Proposition 7 yields(49)MpΛ=span¯⋃k=1∞SNkFk~∑k=1∞⊕SNkFkp.Finally, with Lemma 9(c) we obtain(50)∑k=1∞⊕SNkFkp~∑k=1∞⊕Fkp.

Now we turn to the case p=∞.

Proposition 15.

Let Ωk, k=1,2,…, be finite subsets of positive numbers and put (51)Fk=spantλ:λ∈Ωk⊂C00,1,Gk=spantλ-tμ:λ,μ∈Ωk⊂C0,00,1.Then, there is a Müntz space M∞(Λ), where Λ=∪k=1∞NkΩk for suitable integers Nk, such that(52)∑k=1∞⊕Fk0~M∞Λ~∑k=1∞⊕Gk0.If all Ωk consist of rational numbers, then Nk can be taken such that in addition Λ⊂N.

Proof.

Recall that for p=∞ we have SN=TN and(53)SNf∞=TNf∞=f∞.Fix λ¯k∈Ωk and put Hk=span{tλ¯k}. For any f∈Fk, we have f=(f-f(1)tλ¯k)+f(1)tλ¯k which implies(54)Fk=Gk⊕Hk.

Fix ϵk>0 with ∑k=1∞ϵk<1/8. Then, we find integers 1<N1<N2<⋯ with Nkλ¯k<Nk+1λ¯k+1, (55)12<β1/NkGk,∞,ϵk<α1/Nk+1Gk+1,∞,ϵk+1,12<γ1/NkFk,ϵk<α1/Nk+1Fk+1,∞,ϵk+1.Put(56)Y=span¯⋃k=1∞TNkGk,Z=span¯⋃k=1∞TNkHk,Λ=⋃k=1∞NkΩk.(The integers Nk should be taken so large that Λ satisfies (1) and that Λ⊂N if all Ωk consist of rational numbers.) Hence Y⊂M∞0(Λ) and Z⊂M∞(Λ).

Let gk∈Gk and fix M∈N. Lemma 9(a) and Proposition 7 imply(57)78supk≤MTNkgk∞≤∑k=1MTNkgk∞≤98supk≤MTNkgk∞.Moreover, let fk∈Fk. Then, Lemma 9(a) and Proposition 8 show(58)12supn≤M∑k=1nTNkfk∞≤∑k=1MTNkfk∞≤supn≤M∑k=1nTNkfk∞.Let(59)P∑k=1MTNkfk=∑j=1Mfj1tNjλ¯j.Then, P is a projection. (Recall that fj(1)=TNjfj(1)).

We have P≤2. Indeed, we obtain, in view of (58), (60)P∑k=1MTNkfk∞=∑j=1Mfj1tNjλ¯j∞=∑j=1M-1∑k=1jfk1tNjλ¯j-tNj+1λ¯j+1+∑k=1Mfk1tNMλ¯M∞≤supj≤M∑k=1jfk1=supj≤M∑k=1jTNkfk1≤2∑k=1MTNkfk∞.Here we used that tNjλ¯j-tNj+1λ¯j+1≥0 and(61)∑j=1M-1tNjλ¯j-tNj+1λ¯j+1+tNMλ¯M≤1.Hence, P can be extended to a bounded projection from M∞(Λ)=span¯∪k=1∞TNkFk onto Z whose kernel is Y. Equation (58) implies that H~c~c0 since dim TNkHk=1 for all k. Thus, we have M∞(Λ)~Y⊕c0. Since Y contains a complemented isomorphic copy of c0, we obtain M∞(Λ)~Y and (53) and (57) show(62)Y~∑k=1∞⊕TNkGk0~∑k=1∞⊕Gk0.Finally, put X=∑k=1∞⊕Fk0. As a consequence of (54), we have X~∑k=1∞⊕Gk0⊕c0 which implies by the preceding M∞(Λ)~X.

Propositions 14 and 15 prove Theorem 4. Indeed, in view of (33) and the fact that ·p≤·∞ for all p, we may assume that all Ωk consist of rational numbers.

We conclude with the following.

Proof of Theorem <xref ref-type="statement" rid="thm1.1">1</xref>.

Let Ωk⊂N, k=1,2,…, be a family of finite subsets such that each finite subset of N occurs infinitely many times among the Ωk. Put Fk=span{tλ:λ∈Ωk}, Fp=(∑k=1∞⊕Fk)p, if 1≤p<∞, and F∞=(∑k=1∞⊕Fk)0. According to Propositions 14 and 15, these spaces are isomorphic to Müntz spaces Mp(Λ^) with Λ^⊂N. This proves Theorem 1(a).

To prove (b), let Λ be a given set of positive numbers satisfying (1). According to Lemma 10, we may assume that Λ consists of rational numbers. As in the beginning of this section, consider Ek=span{tλj:mk≤j≤nk}, if 1≤p<∞, and Ek=span{tλj-tλj+1:mk≤j≤nk}, if p=∞. Moreover, let Nk be integers fulfilling inequality (40). Put Λk={⌈Nk-1/p⌉+Nkλj:mk≤j≤nk} if 1≤p<∞ and Λk={Nkλj:mk≤j≤nk+1} if p=∞. At the same time we can take Nk so large that all Λk consist of positive integers. Then, Lemma 13 shows that Mp(Λ) is isomorphic to a complemented subspace of (∑k=1∞⊕SNkEk)p if 1≤p<∞ which itself is complemented in Fp since the sets Λk appear among the sets Ωn introduced in the proof of part (a). For p=∞, Lemma 13 implies that M∞(Λ) is isomorphic to a complemented subspace of (∑k=1∞⊕SNkEk)0 which, in view of Proposition 15, is isomorphic to (∑k=1∞⊕Bk)0 where Bk=span{tNkλj:mk≤j≤nk}. (∑k=1∞⊕Bk)0, of course, is complemented in F∞.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

Sergey V. Ludkovsky is supported by Deutsche Forschungsgemeinschaft LU 219/10-1.

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