# Nice S¹-action implies existence of unconditional basis?

Let $V$ be a Banach space equipped with a continuous linear action of $S^1$ (meaning, the map $S^1\times V\to V$ is continuous). Assume that all the eigenspaces of the $S^1$-action are finite dimensional.

Does $V$ then admit an unconditional basis consisting of eigenvectors for the $S^1$-action?

If the answer is yes, then I'm also interested in the corresponding question when $V$ is Frechet, or when $V$ is a general complete locally convex topological vector space.

• You seem to mean implicitly that the action is by linear maps. Continuity then implies that the action is by bounded operators. Possibly you have more implicit assumptions. Could you be more specific? Second, unconditional bases are defined only in separable spaces in en.wikipedia.org/wiki/Schauder_basis; is $V$ assumed separable? – YCor Feb 18 '18 at 10:53
• @YCor. Yes, I was assuming that the action is by linear maps (I've edited the question to make that explicit). The fact that the action is by bounded operators follows from the continuity of the map $S^1\times V\to V$. Also, my assumption that the eigenspaces of $S^1$ are finite dimensional implies that the vector space is separable. (I have to say that I don't agree your the statement that unconditional bases are only defined for separable spaces, but that's irrelevant for the present question.) – André Henriques Feb 18 '18 at 12:55
• I didn't say they're defined only for separable spaces, I said that in the Wikipedia page they're defined only for separable spaces. I didn't realize that the assumption implies separability, that's indeed correct. – YCor Feb 18 '18 at 13:02
• Section 5 of arxiv.org/abs/math/0612096 contains a discussion of the various possibilities for an action of $S^1$ on a LCTVS. It doesn't discuss bases, but might be useful background for $S^1$-actions. – Andrew Stacey Feb 18 '18 at 13:57
• Maybe it is worthwhile noting that, though the answer to your question is "no" in general (as you explained in your post below), one can still show that the eigenvectors of the action are total in $V$. This is a consequence of the Jacbos-de Leeuw-Glicksberg decomposition (even without the assumption that all eigenspaces be finite dimensional). – Jochen Glueck Feb 18 '18 at 18:54

The space $L^1(S^1)$ (equipped with its obvious $S^1$-action) satisfies all my assumptions, but it has no unconditional basis: https://en.wikipedia.org/wiki/Schauder_basis#Unconditionality
In particular, the functions $\theta\mapsto e^{in\theta}$ do not form an unconditional basis.
• They are also not unconditional in $L^p(S^1)$ for $1<p\not=2 < \infty$, and these spaces do have unconditional bases. – Bill Johnson Feb 18 '18 at 16:42