TITLE: Nice S¹-action implies existence of unconditional basis?
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: I think that the answer to my question is no:
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.