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

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 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 2011 at 9:55
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 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 2011 at 7:21

closed as not a real question by Andres Caicedo, Mark Meckes, Bill Johnson, Yemon Choi, David Roberts Jan 29 2011 at 1:14

2 Answers

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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 2011 at 13:48
What do you mean by the dual of a compact non Abelian group? – Bill Johnson Jan 27 2011 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 2011 at 17:17
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 "Of course"???? – Yemon Choi Jan 28 2011 at 7:19
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 @Yemon Choi. What else when you start a discussion about Lie groups? :) – Bill Johnson Jan 28 2011 at 7:43
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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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