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The problem of Schauder decomposition of a given Banach space seems to play an important role in the geometry of Banach spaces, especially when one is interested in finite dimensional Schauder decompositions (FDD).

I am wondering if the Schauder decomposition can be regarded (in special cases) as the internal counterpart to infinite sums of Banach spaces.

Let me consider two cases: $C(K)$ and $L^p(\mu)$.

Suppose that $E$ is either $C(K)$ space for some compactum $K$ or $L^p(\mu)$ for some measure $\mu$.

Let $(E_n)$ be a sequence of complemented subspaces of $E$ such that for each integer $n$ $$E_1\oplus \ldots \oplus E_n \cap E_{n+1}=\{0\}.$$

In the $C(K)$ case assume that each $E_n$ is isomorphic to $c_0$ and in the latter one, $E_n$ is isomorphic to $\ell^p$.

Define $F$ to be the closed linear span of all $E_n$. Is the family $$\{E_1\oplus \ldots \oplus E_n\colon n\in \mathbb{N}\}$$ a blocking Schauder decomposition for $F$?

Is $F$ isomorphic to $c_0$ / $\ell^p$ ?

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No. You need the projections $Q_n$ onto $E_1\oplus \dots E_n$ from $F$ to be uniformly bounded in order for $(E_n)$ to be a Schauder decomposition for $F$. Even then $F$ need not be isomorphic to $c_0$/$\ell_p$. However, if the $Q_n$ are uniformly bounded from $\ell_p$, then by taking limits in the weak operator topology you get (when $1<p<\infty$) a projection from $\ell_p$ onto $F$ and hence $F$ is isomorphic to $\ell_p$. That is not the case for $p=1$ or in the $C(K)$ case. It is true in the $c_0$ case, because you get by using the weak* operator topology in $\ell_\infty$ an operator from $c_0$ into $F^{**}\subset \ell_\infty$ that is the identity on $F$, which implies that $F$ is a $\mathcal{L}_\infty$ space, and every $\mathcal{L}_\infty$ subspace of $c_0$ is isomorphic to $c_0$.

If you do not understand, email me questions and I will explain when I have more time.

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  • $\begingroup$ Thank you. You think about $\ell^\infty$ as a W*-algebra but I don't know operator techniques you use. Certainly, I'd like to ask some further questions... $\endgroup$
    – TMK
    Aug 24, 2011 at 21:23

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