Suppose M is a compact Lie Group, is there a Schauder basis for L^1(M)?
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1$\begingroup$ 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. $\endgroup$– Yemon ChoiCommented Jan 27, 2011 at 9:55
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2$\begingroup$ 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. $\endgroup$– Yemon ChoiCommented Jan 28, 2011 at 7:21
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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.
answered Jan 27, 2011 at 10:14
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$\begingroup$ Suppose Γ is a comapct Lie Group， G is its dual， then I want to ask what can we say about G? $\endgroup$– XXXCommented Jan 27, 2011 at 13:48
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$\begingroup$ What do you mean by the dual of a compact non Abelian group? $\endgroup$ Commented Jan 27, 2011 at 15:56
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$\begingroup$ 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. $\endgroup$ Commented Jan 27, 2011 at 17:17
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3$\begingroup$ @Yemon Choi. What else when you start a discussion about Lie groups? :) $\endgroup$ Commented 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.
answered Jan 28, 2011 at 6:38