Banach space with non-complemented subspace

Question

I see examples on Stack Exchange and elsewhere of Banach spaces with non-complemented subspaces (examples: 1, 2 [Remark 8], 3 [a remarkable example of a Banach space with no complemented closed subspaces]). My question is thus:

We know that every vector space (and therefore every Banach space) has a vector space basis. Therefore, for a Banach space $X$ and a subspace $E$, we can write a basis $\mathcal{B}_E$ for $E$ and extend it to a basis $\mathcal{B}$ for $X$. Consider $F$ to be the subspace spanned by $\mathcal{B}\setminus \mathcal{B}_E$. Why is $F$ not necessarily a complementary subspace to $E$? It seems that for $v\in X$, we should have that $v$ is uniquely represented by an element of $E$ plus an element of $F$, since it is uniquely represented by a finite linear combination of elements in $\mathcal{B}$ (and we can therefore segregate those elements into elements of $\mathcal{B}_E$ and $\mathcal{B}\setminus \mathcal{B}_E$). Also, it seems that $E\cap F = \{0\}$, since the basis vectors for $E$ and $F$ are linearly independent from each other. Could somebody please help by pointing out where I might have made a faulty assumption?

Answer 1

Let me summarise some known facts related to your question.

1. If $X$ is isomorphic to a Hilbert space then every closed subspace of $X$ is complemented.

2. By a very deep result of Lindenstrauss and Tzafriri, a Banach space whose every closed subspace is complemented is necessarily isomorphic to a Hilbert space. Thus, spaces not isomorphic to Hilbert spaces always contain uncomplemented subspaces.

3. Complementability is related to the possibility of extension of operators. If $Y$ is a closed subspace of $X$ then $Y$ is complemented if and only if the inclusion map extends to a bounded operator $X\to Y$.

4. If $X$ is $\ell_\infty(\Gamma)$, $L_\infty(\mu)$, or more generally an injective space then whenever we have a bounded operator from a subspace $Y$ of some Banach space $Z$ to $X$, this operator can be extended to a bounded operator $Z\to X$. The same is true for $X=c_0$ as long as the space $Z$ is separable.

Answer 2

The point is that once in infinite-dimensions, a subspace can be somewhat wrapped up by its surrounding space to prevent unique representation of vectors by components in and out of the subspace. I think the best example to understand this phenomenon is in the non-locally-convex Ribe space which is a quasi-Banach topology on $R+L$ where $L$ is the space little-$L_1$ of absolutely summable sequences. It is constructed so that $R+\{0\}$ is equal to dual annihilated subspace and is contained in every closed infinite-dimensional closed subspace. Thus, it is a one-dimensional subspace that is not complemented. In Banach spaces, one-dimensional spaces are always complemented.

This is presented in a cleaned up manner in https://www.cambridge.org/core/books/an-fspace-sampler/917AAA51A5370CD786D32448FE828810 but A later article by N.J. Kalton shows this space has no strictly weaker Hausdorff topology, i.e. is an F-space with no basic sequences. This can be found as article 150. at the cite: https://kaltonmemorial.missouri.edu/publications.html