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Suppose M is a compact Lie Group, is there a Schauder basis for L^1(M)?

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closed as not a real question by Andrés Caicedo, Mark Meckes, Bill Johnson, Yemon Choi, David Roberts Jan 29 '11 at 1:14

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

If you're talking about Haar measure and your group is connected, then I'm fairly sure L^1(M) coincides with $L^1([0,1]^d)$ where $d$ is the dimension, which makes me suspect the answer is yes. If you want a Schauder basis which is related somehow to the group structure on $M$ then I'm not sure what kinds of candidate bases there might be. – Yemon Choi Jan 27 '11 at 9:55
Downovted for lack of clarity and for assuming that we should "obviously" know that when you say "compact Lie group" you meant a $k$-torus, and that when you asked for a Schauder basis you wanted to know if the characters form a Schauder basis. – Yemon Choi Jan 28 '11 at 7:21
up vote 10 down vote accepted

Every separable $L_1$ space is isomorphic to $\ell_1$ or $L_1(0,1)$ and thus has a Schauder basis. Look at, for example, Classical Banach spaces by Lindenstrauss and Tzafriri. Other books probably have it, too; see Albiac-Kalton Topics in Banach space theory or Wojtaszczyk's Banach spaces for analysts.

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Suppose Γ is a comapct Lie Group, G is its dual, then I want to ask what can we say about G? – XXX Jan 27 '11 at 13:48
What do you mean by the dual of a compact non Abelian group? – Bill Johnson Jan 27 '11 at 15:56
XXX if that is what you want to ask, then that is what you should have asked. Also "what can we say about X" is in my view not a very well-posed question, in this or any other academic discipline. – Yemon Choi Jan 27 '11 at 17:17
"Of course"???? – Yemon Choi Jan 28 '11 at 7:19
@Yemon Choi. What else when you start a discussion about Lie groups? :) – Bill Johnson Jan 28 '11 at 7:43

First, let me reinforce what Yemon wrote. I came close to downvoting your question and also voting to close. Was it so hard to write

Title: Can the characters be ordered to form a Schauder basis for $L^1(G)$

Question: Let $G$ be a compact Abelian metrizable group. Can the continuous characters on $G$ be ordered to be a Schauder basis?

You should have then written some motivation and what you already know.

Now for a complete answer. Szarek (not Wojtaszzczyk) proved that any normalized (or semi normalized) basis for any $L_1$ space contains a subsequence equivalent to the unit vector basis of $\ell_1$. See

Szarek, S. J. Bases and biorthogonal systems in the spaces $C$ and $L_1$. Ark. Mat. 17 (1979), no. 2, 255–271.

This immediately implies that the characters cannot be ordered to be a Schauder basis. In a paper referenced by Szarek, Olevskii proved that no Schauder basis for $L_1$ can be orthonormal and uniformly bounded; so the case of characters was known before Szarek's paper. I do not know if anyone had checked that case before Olevskii; probably not.

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