Does there exist a Banach space with no complemented closed subspaces?

I know that every Hilbert space can be decomposed as the direct sum of two non-trivial closed subspaces, eg. taking the kernel and range of any non-trivial bounded projection. But I don't know what happens in Banach spaces.

Does there exist a Banach space $B$ with no complemented closed subspaces? In other words, such that there do not exist closed subspaces $U,V\subset B$ with $B=U\oplus V$?

Solution 1:

Finite-dimensional subspaces are always complemented; this follows from the Hahn–Banach theorem.

Proof. Let $F$ be a finite-dimensional subspace of a Banach space $X$ and let $\{x_1, \ldots, x_n\}$ be a basis for $F$. Choose $f_1, \ldots, f_n\in F^*$ to be the coordinate functionals for the basis $\{x_1, \ldots, x_n\}$. Use the Hahn–Banach theorem to extend $f_j$ to bounded linear functionals on $X$; call any such extension $f_j$ too ($j\leqslant n$). Set

$$Px = \sum_{j=1}^n \langle f_j, x\rangle x_j\quad (x\in X).$$

Then $P$ is a projection with ${\rm im}\, P=F$. ${ \rm \square}$

The non-trivial variant of the question is really about the possibility of non-trivial decompositions into subspaces which are both infinite-dimensional (the spaces which lack this property exist and are called indecomposable Banach spaces). There is a whole industry concerning such spaces and there exist even spaces with a stronger property, i.e., there are spaces whose all infinite-dimensional suspaces are indecomposale (the so-called hereditarily indecomposable spaces).

There are examples of compact, Hausdorff spaces $K$ for which the Banach space $C(K)$ is indecomposale. (Of course such spaces cannot be hereditarily indecomposale as $C(K)$ always contains a copy of $c_0$ unless $K$ is finite.)

This is a nice introduction to hereditarily indecomposale spaces.