Schauder bases in Banach spaces with a symmetric $k$-FDD

Question

The Kalton-Peck Banach space $Z_2$ (see Section 6 in this paper) does not admit an unconditional basis, but it admits an unconditional, even symmetric, FDD (finite dimensional decomposition) into subspaces of dimension $2$, and also admit a Schauder basis which is the union of some natural bases of the $2$-dimensional subspaces.

QUESTION: Suppose that, for some $k\in\mathbb N$, the Banach space $X$ admits a symmetric FDD into subspaces of dimension $k$.

Can we assure that $X$ admits a Schauder basis?

Answer 1

Yes. If $(E_n)$ is a FDD for $X$ where each $E_n$ has dimension $k$, then we can pick a basis $(e_i^n)_{i=1}^k$ for each $E_n$ with basis constant at most $\sqrt{k}$. Then the concatenation of $(e_i^n)_{i,n}$ in natural order is a Schauder basis for $X$ whose basis constant is less than or equal to $\sqrt{k}C$ where $C$ is FDD constant. The symmetry is not needed.

Answer 2

I have recently found that there is a reference for the result I needed in Chapter 7 of P.G. Casazza, ``Approximation properties'', HANDBOOK OF THE GEOMETRY OF BANACH SPACES, VOL. 1. Edited by William B. Johnson and Joram Lindenstrauss. Elsevier, 2001.

Proposition 6.5. Let $(E_n)$ be a finite dimensional decomposition (with FDD constant $K$) for a Banach space X. If each $E_n$ has a basis $(x^n_i)_{i=1}^{k_n}$ with basis constant bounded by $M$, then $((x^n_i)_{i=1}^{k_n})_{n=1}^\infty)$ is a basis for $X$ with basis constant $\leq KM$.