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


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 AlbiacKalton 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. 

