# Uncomplemented subspaces in Banach spaces

In any Hilbert space $$H$$, every closed subspace $$M$$ therein is complemented, i.e. there exists a closed subspace $$N$$ with $$M \oplus N \cong H$$. One possible choice for $$N$$ is always the orthogonal complement $$N := M^\perp$$.

If we consider Banach spaces instead, then the situation changes. For example, it is `well known’ that the closed subspace $$c_0$$ of null sequences in the Banach space $$\ell^\infty$$ of all bounded sequences does not admit a complement.

Interestingly, something that I learned just now when writing this blog post and googling some stuff for it, Hilbert spaces are actually characterized by this property: Lindenstrauß and Tzafriri proved that a Banach space is isomorphic to a Hilbert space provided every closed subspace is complemented (On the complemented subspaces problem, Isreal Journal of Mathematics, Vol. 9, No.2, pp. 263-269).

An example of a Banach space with an uncomplemented closed subspace where I still can’t wrap my head around it is the following recent one by Anna Pelczar-Barwacz (arXiv:2309.15261): There exists a reflexive Banach space $$X$$ with a Schauder basis $$(e_i)_{i=1}^\infty$$ such that the mapping carrying each basis vector $$e_i$$ to $$e_{2i}$$ extends to an isometry between $$X$$ and $$Y := \overline{\mathrm{span}}\{e_{2i}: i \in \mathbb{N}\}$$ with $$Y$$ not complemented in $$X$$. My head still keeps telling me that $$\overline{\mathrm{span}}\{e_{2i-1}: i \in \mathbb{N}\}$$ should be a closed complement for $$Y$$, but this is just not true (it follows from this that the Schauder basis $$(e_i)_{i=1}^\infty$$ is not unconditional).