On the Geometry of Müntz Spaces
Let satisfy , and . We investigate the Müntz spaces for . We show that, for each , there is a Müntz space which contains isomorphic copies of all Müntz spaces as complemented subspaces. is uniquely determined up to isomorphisms by this maximality property. We discuss explicit descriptions of . In particular is isomorphic to a Müntz space where consists of positive integers. Finally we show that the Banach spaces for and for are always isomorphic to suitable Müntz spaces if the are the spans of arbitrary finitely many monomials over .
Our paper is concerned with the Banach space geometry of Müntz spaces which is largely unexplored.
Let be a sequence of real numbers with andFor and , putFor , let be the function with . In this paper, we consider the Müntz spacesBy the Müntz theorem [1, 2], it is clear that if and . It is known that is isomorphic to for and is isomorphic to provided that satisfies in addition the lacunary condition . On the other hand, there are examples of Müntz spaces which are not isomorphic to ( or [5, Corollary .]).
We want to show that, for fixed , all Müntz spaces are isomorphic to complemented subspaces of one special function space which itself is a Müntz space where consists of positive integers.
For two Banach spaces and , we write if is isomorphic to .
For a sequence of Banach spaces , we putif , andwith the norm .
Theorem 1. Let be a sequence of finite subsets of containing each finite subset of infinitely many times. Moreover let be the linear of the functions , , in . Put(a)For each , there is a set satisfying (1) such that is isomorphic to .(b)If is any set of positive real numbers satisfying (1), then is isomorphic to a complemented subspace of .
On the other hand, we do not know if any complemented subspace of is isomorphic to a Müntz space.
Proof. Consider the spaces of Theorem 1. By the choice of the and the definition of , we have .
Now let satisfy the assumptions of Theorem 2. Since is isomorphic to a Müntz space, by Theorem 1 there is a complemented subspace with . Moreover, by Theorem 1(b) there is a complemented subspace with . Using Pelczynski’s decomposition method [6, Theorem .], we obtain .
Corollary 3. is not isomorphic to .
Proof. There is a Müntz space which is not isomorphic to ( or [5, Corollary ]). But, according to Theorem 1, is isomorphic to a complemented subspace of . Since all infinite dimensional complemented subspaces of are isomorphic to [7, Theorem ], the space cannot be isomorphic to .
In connection with the proof of Theorem 1, we also get the following.
Theorem 4. Let , , be arbitrary finite subsets of positive real numbers and put . Then, for each , there is a subset satisfying (1) such that , if , and , if .
We do not know if, conversely, any Müntz space is isomorphic to the direct sum of the spaces for suitable .
2. Finite Dimensional Subspaces of
Put . For a finite dimensional subspace , , or , , and , putIf , , also putSimilarly, for , , or , , defineWe collect a few properties of these parameters in the following.
Lemma 5. One has(a) for all if ,(b) if ,(c) if .
Proof. We prove (c). The proofs of (a) and (b) are similar. Let and let be such that . If , we obtain . This yieldsHence, . For , we obtain and therefore . This implies .
Lemma 6. Consider , , or , . Fix and let and . Then one hasFor , one even has .
Proof. We obtain which yields the first inequality. The second inequality is trivial. The last assertion of Lemma 6 follows directly from the definitions of and .
Proposition 7. Fix such that . Let , , or , , be finite dimensional subspaces withThen, for any and , one has if , andif .
Proposition 8. Fix such that . Let be finite dimensional subspaces withThen, for any and , one has
Proof. By (18) we have, for any , Equation (18) and Lemma 5 also imply that are monotonically increasing. Hence, if , we obtain Here we used that and hence for all . The preceding estimate implies We concludeApplying this estimate for and yields from which we obtain . Moreover, using (24) we see that which proves the first inequality of the proposition. The second inequality is trivial.
Now consider, for fixed and , the maps. (Here is the smallest integer .) We include the case by putting . We clearly have and we obtain for . If , then .
Lemma 9. One has
(a)(b)(c) for .
Proof. (a) follows from simple substitution. (Recall that is only defined for .)
(b): Let and . At first, let . Consider . Put . Since and for , we obtain, with Lemma 6, This yields the first inequality if .
Similarly, we obtainThis yields the second inequality of (c) if .
If , then the two inequalities follow from (a) and the fact that is increasing and is decreasing with respect to . (Recall that here is ).
(c): If , then (c) is trivial. Now let . By definition, . On the other hand, with , of (b), and Lemma 6, since .
3. Some Facts about Müntz Spaces
Lemma 10. Let such that . Moreover, let satisfy for all . Then, with is an isomorphism between and .
Proof. For we obtainThis proves .
Using Lemma 10, we can always assume in the following that .
PutSince is 1-codimensional in and contains a complemented subspace isomorphic to , we obtain .
Proposition 11. Fix , , or , . Then, there are linear bounded finite rank operators with , , if andif , orif . Moreover, there are indices , , withif , orif .
For a proof see [5, Proposition ].
The condition implies the following.
Corollary 12. For the operators of Proposition 11 one has (with ), , where the series converges in norm.
However, are no projections and the “summands overlap”; that is, and in general.
4. Final Proofs
Lemma 13. Assume that for some , . Then, is isomorphic to a complemented subspace of .
Proof. For , , or , , put where and if , . Using Proposition 11, Corollary 12, and Lemma 9(c), we see that is a well-defined element in and is a bounded linear operator if or . By Lemma 9(c) and condition (40), is even an into-isomorphism. Hence, there are universal constants and such that for all .
It remains to prove that and , respectively, are complemented in . To this end, let . At first assume that, for some , with and for some if . Put . The fact that if implies if . Hence,We obtain a universal constant with if . Here again, (40) and Lemma 9(c) were used. In view of Proposition 11, this implies that is a bounded linear operator on the dense subspace of consisting of finite sequences. Hence, can be uniquely extended to a linear bounded operator , called again. The same argument shows that is also bounded if .
We obtain by definition if , , or , . This follows from Corollary 12. Indeed, we have with and for . Hence,(Here we used .) This shows that is a bounded projection from onto , if , or onto , if . Since , this completes the proof.
Proposition 14. Let . Assume that , , are finite subsets of positive numbers and putThen, there is a Müntz space which is isomorphic to .
Here for suitable integers . If all consist of rational numbers, then can be taken such that in addition .
Proof. Fix such thatFind integers such that for all andfor all . Put . Then, we obtainDefine . (Of course we can take so large that satisfies (1) and consists of integers if all consist of rational numbers.)
We obtain, by Lemma 5, Lemma 9 ((a) and (b)), and (46),Here we used that is increasing with respect to . Now Proposition 7 yieldsFinally, with Lemma 9(c) we obtain
Now we turn to the case .
Proposition 15. Let , , be finite subsets of positive numbers and put Then, there is a Müntz space , where for suitable integers , such thatIf all consist of rational numbers, then can be taken such that in addition .
Proof. Recall that for we have andFix and put . For any , we have which impliesFix with . Then, we find integers with , Put(The integers should be taken so large that satisfies (1) and that if all consist of rational numbers.) Hence and .
Let and fix . Lemma 9(a) and Proposition 7 implyMoreover, let