Suppose M is a compact Lie Group, is there a Schauder basis for L^1(M)?
closed as not a real question by Andres Caicedo, Mark Meckes, Bill Johnson, Yemon Choi, David Roberts Jan 29 '11 at 1:14
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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.
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.