JFS Journal of Function Spaces 2314-8888 2314-8896 Hindawi Publishing Corporation 10.1155/2015/787291 787291 Research Article On the Geometry of Müntz Spaces Ludkovsky Sergey V. 1 Lusky Wolfgang 2 Hencl Stanislav 1 Department of Applied Mathematics Moscow State Technical University MIREA Avenue Vernadsky 78, Moscow 119454 Russia mirea.ru 2 Institute for Mathematics University of Paderborn Warburger Straße 100, 33098 Paderborn Germany uni-paderborn.de 2015 1662015 2015 22 02 2015 14 04 2015 17 04 2015 1662015 2015 Copyright © 2015 Sergey V. Ludkovsky and Wolfgang Lusky. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Let Λ = { λ k } k = 1 satisfy 0 < λ 1 < λ 2 < , k = 1 1 / λ k < and i n f k ( λ k + 1 - λ k ) > 0 . We investigate the Müntz spaces M p Λ = s p a n ¯ { t λ k : k = 1,2 , } L p ( 0,1 ) for 1 p . We show that, for each p , there is a Müntz space F p which contains isomorphic copies of all Müntz spaces as complemented subspaces. F p is uniquely determined up to isomorphisms by this maximality property. We discuss explicit descriptions of F p . In particular F p is isomorphic to a Müntz space M p ( Λ ^ ) where Λ ^ consists of positive integers. Finally we show that the Banach spaces ( n F n ) p for 1 p < and ( n F n ) 0 for p = are always isomorphic to suitable Müntz spaces M p ( Λ ) if the F n 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 = { λ k : k = 1,2 , } be a sequence of real numbers with 0 < λ 1 < λ 2 < and (1) k = 1 1 λ k < , inf k λ k + 1 - λ k > 0 . For 1 p < and f : [ 0,1 ] R , put (2) f p = 0 1 f t p d t 1 / p , f = sup t 0,1 f t . For λ > 0 , let t λ be the function f : [ 0,1 ] R with f ( t ) = t λ . In this paper, we consider the Müntz spaces (3) M p Λ = span ¯ t λ k : k = 1,2 , L p 0,1 for 1 p . By the Müntz theorem [1, 2], it is clear that M p ( Λ ) L p ( 0,1 ) if 1 p < and M ( Λ ) C 0 ( 0,1 ) : = { f : [ 0,1 ] R : f c o n t i n u o u s , f ( 0 ) = 0 } . It is known that M p ( Λ ) is isomorphic to l p for 1 p < and M ( Λ ) is isomorphic to c 0 provided that Λ satisfies in addition the lacunary condition i n f k ( λ k + 1 / λ k ) > 1 . On the other hand, there are examples of Müntz spaces M ( Λ ) which are not isomorphic to c 0 ( 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 M p ( 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 E n , we put (4) n = 1 E n p = e n : e n E n , n = 1,2 , , n = 1 e n p 1 / p < , if 1 p < , and (5) n = 1 E n 0 = e n : e n E n , n = 1,2 , , lim n e n = 0 with the norm ( e n ) = s u p n e n .

Theorem 1.

Let ( Ω k ) be a sequence of finite subsets of N containing each finite subset of N infinitely many times. Moreover let F p ( Ω k ) be the linear s p a n of the functions t λ , λ Ω k , in L p ( 0,1 ) . Put (6) F p = k N F p Ω k p i f 1 p < , F = k N F Ω k 0 .

For each p [ 1 , ] , there is a set Λ ^ N satisfying (1) such that F p is isomorphic to M p ( Λ ^ ) .

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

On the other hand, we do not know if any complemented subspace of F p 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 M p ( Λ ) contains a complemented isomorphic copy of any Müntz space M p ( Λ ~ ) then M p ( Λ ) is isomorphic to the space F p of Theorem 1.

Proof.

Consider the spaces F p ( Ω k ) of Theorem 1. By the choice of the Ω k and the definition of F p , we have F p ~ F p F p .

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

Corollary 3.

F is not isomorphic to c 0 .

Proof.

There is a Müntz space M ( Λ ) which is not isomorphic to c 0 ( 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 c 0 are isomorphic to c 0 [7, Theorem I.2.a.3. ], the space F cannot be isomorphic to c 0 .

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

Theorem 4.

Let Ω n , n = 1,2 , , be arbitrary finite subsets of positive real numbers and put F p ( Ω n ) = s p a n { t λ : λ Ω n } L p ( 0,1 ) . Then, for each p [ 1 , ] , there is a subset Λ N satisfying (1) such that ( n F p ( Ω n ) ) p ~ M p ( Λ ) , if 1 p < , and ( n F ( Ω n ) ) 0 ~ M ( Λ ) , if p = .

We do not know if, conversely, any Müntz space M p ( Λ ) is isomorphic to the direct sum of the spaces F p ( Ω n ) for suitable Ω n .

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Put C 0,0 ( 0,1 ) = { f C 0 ( 0,1 ) : f ( 1 ) = 0 } . For a finite dimensional subspace E L p ( 0,1 ) , 1 p < , or E C 0 ( 0,1 ) , p = , and ϵ > 0 , put (7) α E , p , ϵ = sup α > 0 : e · 1 0 , α p ϵ e p e E . If E C 0 ( 0,1 ) , p = , also put (8) γ E , ϵ = inf γ < 1 : e 1 - e · 1 γ , 1 ϵ e e E . Similarly, for E L p ( 0,1 ) , 1 p < , or E C 0,0 ( 0,1 ) , p = , define (9) β E , p , ϵ = inf β > 0 : e · 1 β , 1 p ϵ e p 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 e 1 - e · 1 γ , 1 ϵ e . If t [ γ , 1 ] , we obtain e t - e γ 2 ϵ e . This yields (10) e = max e · 1 0 , γ , e · 1 γ , 1 max e · 1 0 , γ , e γ + 2 ϵ e . Hence, 1 - 2 ϵ e e · 1 0 , γ . For ϵ < 1 / 3 , we obtain ( 1 - 2 ϵ ) > ϵ and therefore ϵ e < e · 1 0 , γ . This implies α ( E , , ϵ ) γ ( E , ϵ ) .

Lemma 6.

Consider E L p ( 0,1 ) , 1 p < , or E C 0,0 ( 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 - 2 ϵ e p e · 1 α , β p e p e E . For p = , one even has e · 1 α , β = e .

Proof.

We obtain (12) e p e · 1 α , β p + e · 1 0 , α p + e · 1 β , 1 p e · 1 α , β p + 2 ϵ e p . 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 ϵ n > 0 such that n = 1 ϵ n < 1 / 8 . Let E n L p ( 0,1 ) , 1 p < , or E n C 0,0 ( 0,1 ) , p = , be finite dimensional subspaces with (13) β n β E n , p , ϵ n < α E n + 1 , p , ϵ n + 1 α n + 1 . Then, for any e n E n and N N , one has (14) 1 2 n = 1 N e n p p 1 / p n = 1 N e n p 4 3 n = 1 N e n · 1 α n , β n p 4 3 n = 1 N e n p p 1 / p , if 1 p < , and (15) 7 8 sup n N e n n = 1 N e n 9 8 n = 1 N e n · 1 α n , β n 9 8 sup n N e n , if p = .

Proof.

Let 1 p < . Put e = n = 1 N e n . We obtain with Lemma 6 the following: (16) e - n = 1 N e n · 1 α n , β n p = n = 1 N e n - e n · 1 α n , β n p n = 1 N 2 ϵ n e n p n = 1 N 2 ϵ n 1 - 2 ϵ n e n · 1 α n , β n p 1 3 sup n N e n · 1 α n , β n p 1 3 n = 1 N e n · 1 α n , β n p p 1 / p . Here we used n = 1 ϵ n < 1 / 8 . Hence, 1 - 2 ϵ n 3 / 4 for all n and n = 1 2 ϵ n / ( 1 - 2 ϵ n ) < 1 / 3 .

This yields, in view of (13), (17) e p n = 1 N e n · 1 α n , β n p + 1 3 n = 1 N e n · 1 α n , β n p p 1 / p = 4 3 n = 1 N e n · 1 α n , β n p p 1 / p 4 3 n = 1 N e n p p 1 / p , e p n = 1 N e n · 1 α n , β n p - 1 3 n = 1 N e n · 1 α n , β n p p 1 / p = 2 3 n = 1 N e n · 1 α n , β n p p 1 / p 1 2 n = 1 N e n p p 1 / p . Here we used Lemma 6 and 1 - 2 ϵ n 3 / 4 . The proof for p = is similar.

Proposition 8.

Fix ϵ n > 0 such that n = 1 ϵ n < 1 / 8 . Let E n C 0 ( 0,1 ) be finite dimensional subspaces with (18) γ n γ E n , ϵ n < α E n + 1 , , ϵ n + 1 α n + 1 . Then, for any e k E k and N N , one has (19) 1 2 sup n N k = 1 n e k k = 1 N e k s u p n N k = 1 n e k .

Proof.

By (18) we have, for any n N , (20) k = 1 N e k · 1 0 , γ n - k = 1 n e k · 1 0 , γ n k = n + 1 N e k · 1 0 , α k k = n + 1 N ϵ k e k . Equation (18) and Lemma 5 also imply that γ n are monotonically increasing. Hence, if t [ γ n , 1 ] , we obtain (21) k = 1 n e k t - k = 1 n e k γ n k = 1 n e k t - e k 1 + e k 1 - e k γ n k = 1 n 2 e k 1 - e k · 1 γ n , 1 2 k = 1 n ϵ k e k . Here we used that γ k γ n and hence [ γ n , 1 ] [ γ k , 1 ] for all k = 1 , , n . The preceding estimate implies (22) k = 1 n e k max k = 1 n e k · 1 0 , γ n , k = 1 n e k γ n + 2 k = 1 n ϵ k e k k = 1 n e k · 1 0 , γ n + 2 k = 1 n ϵ k e k . We conclude (23) 0 k = 1 n e k - k = 1 n e k · 1 0 , γ n 2 k = 1 n ϵ k e k , (24) k = 1 N e k k = 1 N e k · 1 0 , γ n k = 1 n e k · 1 0 , γ n - k = n + 1 N ϵ k e k k = 1 n e k - k = 1 N 2 ϵ k e k k = 1 n e k - 1 4 sup k N e k . Applying this estimate for n and n - 1 yields (25) 2 k = 1 N e k k = 1 n e k + k = 1 n - 1 e k - 1 2 sup k N e k e n - 1 2 sup k N e k , from which we obtain s u p k e k 4 k = 1 N e k . Moreover, using (24) we see that s u p n k = 1 n e k 2 k = 1 N e k which proves the first inequality of the proposition. The second inequality is trivial.

Now consider, for fixed p and N > 0 , the maps (26) T N f t = N 1 / p t N - 1 / p f t N , S N f t = N 1 / p t N - 1 / p f t N , t [ 0,1 ] . (Here r is the smallest integer r .) We include the case p = by putting 1 / = 0 . We clearly have | T N f | | S N f | and we obtain T N f p = f p for f L p ( 0,1 ) . If p = , then S N = T N .

Lemma 9.

One has

(a) (27) α T N E , p , ϵ = α 1 / N E , p , ϵ , β T N E , p , ϵ = β 1 / N E , p , ϵ , γ T N E , ϵ = γ 1 / N E , ϵ ,

(b) (28) α S N E , p , ϵ 1 - 2 ϵ α 1 / N E , p , ϵ , β S N E , p , ϵ 1 - 2 ϵ α 1 / N E , p , ϵ β 1 / N E , p , ϵ ,

(c) (29) α 1 / N E , p , ϵ 1 - 2 ϵ e p S N e p e p , e E , for 0 < ϵ < 1 / 2 .

Proof.

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

(b): Let α 0 = α 1 / N ( E , p , ϵ ) = α ( T N E , p , ϵ ) and β 0 = β 1 / N ( E , p , ϵ ) = β ( T N E , p , ϵ ) . At first, let p < . Consider e E . Put θ = ( N - 1 ) / p - ( N - 1 ) / p . Since 0 θ 1 and ( S N e ) ( t ) = t θ ( T N e ) ( t ) for t [ 0,1 ] , we obtain, with Lemma 6, (30) S N e · 1 0 , α 0 p α 0 θ T N e · 1 0 , α 0 p α 0 θ ϵ T N e p α 0 θ ϵ 1 - 2 ϵ T N e · 1 α 0 , β 0 p ϵ 1 - 2 ϵ S N e p . This yields the first inequality if p < .

Similarly, we obtain (31) S N e · 1 β 0 , 1 p T N e · 1 β 0 , 1 p ϵ T N e p ϵ 1 - 2 ϵ T N e · 1 α 0 , β 0 p α 0 - θ ϵ 1 - 2 ϵ S N e · 1 α 0 , β 0 p ϵ 1 - 2 ϵ α 0 S N e p . 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 S N = T N ).

(c): If p = , then (c) is trivial. Now let p < . By definition, S N e p T N e p = e p . On the other hand, with α 0 , β 0 of (b), and Lemma 6, (32) S N e p S N e · 1 α 0 , β 0 p α 0 N - 1 / p - N - 1 / p T N e · 1 α 0 , β 0 p α 0 1 - 2 ϵ T N e p = α 0 1 - 2 ϵ e p since 0 ( N - 1 ) / p - ( N - 1 ) / p 1 .

3. Some Facts about Müntz Spaces

If Λ satisfies (1), then any f M p ( Λ ) can be represented as f ( t ) = n = 1 α k t λ k where the series converges pointwise on [ 0,1 [ . There are universal constants c k > 0 with | α k | c k f p for all k . Moreover, it is easily seen that for any λ > 0 we have (33) lim μ λ t λ - t μ = lim μ λ λ μ λ / μ - λ 1 - λ μ = 0 .

Lemma 10.

Let ϵ k > 0 such that k = 1 ϵ k c k < 1 / 2 . Moreover, let μ k satisfy t λ k - t μ k ϵ k for all k . Then, T with T ( n = 1 α k t λ k ) = n = 1 α k t μ k is an isomorphism between M p ( Λ ) and M p ( { μ k : k = 1,2 , } ) .

Proof.

For f ( t ) = k = 1 α k t λ k we obtain (34) T f - f p k = 1 c k ϵ k f p 1 2 f p . This proves 2 - 1 f p T f p 3 · 2 - 1 f p .

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

Put (35) M 0 Λ = f M Λ : f 1 = 0 = span ¯ t λ k - t λ k + 1 : k = 1,2 , . Since M 0 ( Λ ) is 1-codimensional in M ( Λ ) and M ( Λ ) contains a complemented subspace isomorphic to c 0 , we obtain M 0 ( Λ ) ~ M ( Λ ) .

Proposition 11.

Fix X = M p ( Λ ) , 1 p < , or X = M 0 ( Λ ) , p = . Then, there are linear bounded finite rank operators R n : X X with l i m n R n f = f , f X , R n R m = R m i n ( n , m ) if n m and (36) c 1 k = 1 R k - R k - 1 f p p 1 / p f p c 2 k = 1 R k - R k - 1 f p p 1 / p , if 1 p < , or (37) c 1 sup k R k - R k - 1 f f c 2 sup k R k - R k - 1 f , if p = . Moreover, there are indices m k < n k , k = 1,2 , , with (38) R k - R k - 1 X = span t λ j - t λ j + 1 : m k j n k , if p = , or (39) R k - R k - 1 X = span t λ j : m k j n k , if 1 p < .

For a proof see [5, Proposition 9.1 . 4 ].

The condition l i m n R n f = f implies the following.

Corollary 12.

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

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

4. Final Proofs

Let Λ satisfy (1). Fix 1 p . Take R k , m k , and n k as in Proposition 11 and put E k = ( R k - R k - 1 ) M p ( Λ ) , if 1 p < , or E k = ( R k - R k - 1 ) M 0 ( Λ ) , if p = . Fix 0 < ϵ k < 1 / 4 and let 0 < N 1 < N 2 < be such that (40) α 1 / N k E k , p , ϵ k 1 2 k . Let F j be finite dimensional Banach spaces and put X = ( j = 1 F j ) p , if 1 p < , or X = ( j = 1 F j ) 0 , if p = .

Lemma 13.

Assume that F j k = S N k E k for some j k , k = 1,2 , . Then, M p ( Λ ) is isomorphic to a complemented subspace of X .

Proof.

For e M p ( Λ ) , 1 p < , or e M 0 ( Λ ) , p = , put U e = ( f j ) where f j k = S N k ( R k - R k - 1 ) e and f j = 0 if j j k , k = 1,2 , . Using Proposition 11, Corollary 12, and Lemma 9(c), we see that U e is a well-defined element in X and U is a bounded linear operator M p ( 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 c 1 > 0 and c 2 > 0 such that c 1 e p U e p c 2 e p for all e .

It remains to prove that U M p ( Λ ) and U M 0 ( Λ ) , respectively, are complemented in X . To this end, let f X . At first assume that, for some N N , f = ( f 1 , , f N , 0,0 , ) with f j F j and f j k = S N k e k for some e k E k if j k N . Put V f = j k N e k . The fact that R m R n = R m i n ( m , n ) if m n implies ( R n - R n - 1 ) ( R k - R k - 1 ) = 0 if n - k 2 . Hence, (41) R n - R n - 1 V f = R n - R n - 1 e n - 1 + e n + e n + 1 . We obtain a universal constant c 3 > 0 with (42) R n - R n - 1 V f p c 3 e n - 1 p p + e n p p + e n + 1 p p 1 / p 4 c 3 S N n - 1 e n - 1 p p + S N n e n p p + S N n + 1 e n + 1 p p 1 / p if 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 M p ( Λ ) , called V again. The same argument shows that V is also bounded if p = .

We obtain by definition V U e = e if e M p ( Λ ) , 1 p < , or e M 0 ( Λ ) , p = . This follows from Corollary 12. Indeed, we have U e = ( f j ) with f j k = S N k e k and f l = 0 for l { j k : k = 1,2 , } . Hence, (43) V U e = V f j = k = 1 R k - R k - 1 e = e . (Here we used R 0 = 0 .) This shows that U V is a bounded projection from X onto U M p ( Λ ) , if 1 p < , or onto U M 0 ( Λ ) , if p = . Since M 0 ( Λ ) ~ M ( Λ ) , this completes the proof.

Proposition 14.

Let 1 p < . Assume that Ω k , k = 1,2 , , are finite subsets of positive numbers and put (44) F k = span t λ : λ Ω k L p 0,1 . Then, there is a Müntz space M p ( Λ ) which is isomorphic to k = 1 F k p .

Here Λ = k = 1 N k - 1 / p + N k Ω k for suitable integers N k > 0 . If all Ω k consist of rational numbers, then N k can be taken such that in addition Λ N .

Proof.

Fix ϵ k > 0 such that (45) k = 1 ϵ k 1 - 2 ϵ k < 1 16 . Find integers 1 < N 1 < N 2 < such that m a x N k Ω k < m i n N k + 1 Ω k + 1 for all k and (46) 1 2 < α 1 / N k F k , p , ϵ k β 1 / N k F k , p , ϵ k < α 1 / N k + 1 F k + 1 , p , ϵ k + 1 for all k . Put α k = α 1 / N k ( F k , p , ϵ k ) . Then, we obtain (47) k = 1 ϵ k 1 - 2 ϵ k α k < 1 8 . Define Λ = k = 1 N k - 1 / p + N k Ω k . (Of course we can take N k so large that Λ satisfies (1) and consists of integers if all Ω k consist of rational numbers.)

We obtain, by Lemma 5, Lemma 9 ((a) and (b)), and (46), (48) α S N k + 1 F k + 1 , p , ϵ k + 1 1 - 2 ϵ k + 1 α k + 1 α S N k + 1 F k + 1 , p , ϵ k + 1 1 - 2 ϵ k + 1 α 1 / N k + 1 F k + 1 , p , ϵ k + 1 β 1 / N k F k , p , ϵ k β S N k F k , p , ϵ k 1 - 2 ϵ k α k . Here we used that α ( · , · , ϵ ) is increasing with respect to ϵ . Now Proposition 7 yields (49) M p Λ = span ¯ k = 1 S N k F k ~ k = 1 S N k F k p . Finally, with Lemma 9(c) we obtain (50) k = 1 S N k F k p ~ k = 1 F k p .

Now we turn to the case p = .

Proposition 15.

Let Ω k , k = 1,2 , , be finite subsets of positive numbers and put (51) F k = span t λ : λ Ω k C 0 0,1 , G k = span t λ - t μ : λ , μ Ω k C 0,0 0,1 . Then, there is a Müntz space M ( Λ ) , where Λ = k = 1 N k Ω k for suitable integers N k , such that (52) k = 1 F k 0 ~ M Λ ~ k = 1 G k 0 . If all Ω k consist of rational numbers, then N k can be taken such that in addition Λ N .

Proof.

Recall that for p = we have S N = T N and (53) S N f = T N f = f . Fix λ ¯ k Ω k and put H k = s p a n { t λ ¯ k } . For any f F k , we have f = ( f - f ( 1 ) t λ ¯ k ) + f ( 1 ) t λ ¯ k which implies (54) F k = G k H k .

Fix ϵ k > 0 with k = 1 ϵ k < 1 / 8 . Then, we find integers 1 < N 1 < N 2 < with N k λ ¯ k < N k + 1 λ ¯ k + 1 , (55) 1 2 < β 1 / N k G k , , ϵ k < α 1 / N k + 1 G k + 1 , , ϵ k + 1 , 1 2 < γ 1 / N k F k , ϵ k < α 1 / N k + 1 F k + 1 , , ϵ k + 1 . Put (56) Y = span ¯ k = 1 T N k G k , Z = span ¯ k = 1 T N k H k , Λ = k = 1 N k Ω k . (The integers N k should be taken so large that Λ satisfies (1) and that Λ N if all Ω k consist of rational numbers.) Hence Y M 0 ( Λ ) and Z M ( Λ ) .

Let g k G k and fix M N . Lemma 9(a) and Proposition 7 imply (57) 7 8 sup k M T N k g k k = 1 M T N k g k 9 8 sup k M T N k g k . Moreover, let f k F k . Then, Lemma 9(a) and Proposition 8 show (58) 1 2 sup n M k = 1 n T N k f k k = 1 M T N k f k sup n M k = 1 n T N k f k . Let (59) P k = 1 M T N k f k = j = 1 M f j 1 t N j λ ¯ j . Then, P is a projection. (Recall that f j ( 1 ) = T N j f j ( 1 ) ).

We have P 2 . Indeed, we obtain, in view of (58), (60) P k = 1 M T N k f k = j = 1 M f j 1 t N j λ ¯ j = j = 1 M - 1 k = 1 j f k 1 t N j λ ¯ j - t N j + 1 λ ¯ j + 1 + k = 1 M f k 1 t N M λ ¯ M sup j M k = 1 j f k 1 = sup j M k = 1 j T N k f k 1 2 k = 1 M T N k f k . Here we used that t N j λ ¯ j - t N j + 1 λ ¯ j + 1 0 and (61) j = 1 M - 1 t N j λ ¯ j - t N j + 1 λ ¯ j + 1 + t N M λ ¯ M 1 . Hence, P can be extended to a bounded projection from M ( Λ ) = s p a n ¯ k = 1 T N k F k onto Z whose kernel is Y . Equation (58) implies that H ~ c ~ c 0 since dim T N k H k = 1 for all k . Thus, we have M ( Λ ) ~ Y c 0 . Since Y contains a complemented isomorphic copy of c 0 , we obtain M ( Λ ) ~ Y and (53) and (57) show (62) Y ~ k = 1 T N k G k 0 ~ k = 1 G k 0 . Finally, put X = k = 1 F k 0 . As a consequence of (54), we have X ~ k = 1 G k 0 c 0 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 Ω k consist of rational numbers.

We conclude with the following.

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

Let Ω k N , k = 1,2 , , be a family of finite subsets such that each finite subset of N occurs infinitely many times among the Ω k . Put F k = s p a n { t λ : λ Ω k } , F p = ( k = 1 F k ) p , if 1 p < , and F = ( k = 1 F k ) 0 . According to Propositions 14 and 15, these spaces are isomorphic to Müntz spaces M p ( 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 E k = s p a n { t λ j : m k j n k } , if 1 p < , and E k = s p a n { t λ j - t λ j + 1 : m k j n k } , if p = . Moreover, let N k be integers fulfilling inequality (40). Put Λ k = { N k - 1 / p + N k λ j : m k j n k } if 1 p < and Λ k = { N k λ j : m k j n k + 1 } if p = . At the same time we can take N k so large that all Λ k consist of positive integers. Then, Lemma 13 shows that M p ( Λ ) is isomorphic to a complemented subspace of ( k = 1 S N k E k ) p if 1 p < which itself is complemented in F p since the sets Λ k appear among the sets Ω n introduced in the proof of part (a). For p = , Lemma 13 implies that M ( Λ ) is isomorphic to a complemented subspace of ( k = 1 S N k E k ) 0 which, in view of Proposition 15, is isomorphic to ( k = 1 B k ) 0 where B k = s p a n { t N k λ j : m k j n k } . ( k = 1 B k ) 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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