Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 11 / 2

An infinite-dimensional Banach space $X$ is a hereditarily indecomposable Banach space when no closed infinite-dimensional subspace of $X$ splits as a topological direct sum of two infinite-dimensional closed subspaces. Finite-dimensional summands are not forbidden.

A useful equivalent formulation is that for every pair of infinite-dimensional closed subspaces $U,V$ and every $\epsilon >0$, there are unit vectors $u\in U,v\in V$ with $\Vert u-v\Vert <\epsilon$. To see the equivalence, a topological direct sum has bounded coordinate projections and hence a positive separation between its two unit spheres. Conversely, if the two unit spheres have separation $\delta >0$, then $U\cap V=0$ and addition is bounded below on $U\oplus V$: after factoring out the larger of $\Vert u\Vert ,\Vert v\Vert$, the triangle inequality and the separation bound give a positive lower bound for $\Vert u+v\Vert /(\Vert u\Vert +\Vert v\Vert )$. Its range is therefore closed and supplies the forbidden decomposition.

For the construction, use the standard basic-sequence selection procedure. At stage $n$, choose finitely many functionals which almost norm the unit sphere of the finite-dimensional span of the vectors already chosen. Put the next vector in their common kernels, and in the analogous kernels retained from earlier stages. These impose only finite codimension. Choosing the norming tolerances with a finite product gives a uniformly bounded basis constant, which may be fixed in advance, for example below two. The hypothesis then supplies a unit vector in this finite-codimensional subspace with image smaller than the prescribed ${\epsilon }_n$. This yields the requested normalized basic sequence.

For strict singularity, fix $\eta >0$ and make the construction with a summable sequence satisfying $2K{\sum }_n{\epsilon }_n<\eta$, where $K$ is the controlled basis constant. For $z=\sum a_n{e}_n$, the coordinate bound $|a_n|\le 2K\Vert z\Vert$ givesBy density this holds on the infinite-dimensional closed span $E$.

If $T$ were bounded below by $c>0$ on some infinite-dimensional subspace $Y$, choose $\eta <c/2$. Hereditary indecomposability supplies unit vectors $y\in Y,e\in E$ arbitrarily close. Thena contradiction for sufficiently close vectors. Therefore $T$ is a strictly singular operator, equivalently every infinite-dimensional subspace contains a unit vector with arbitrarily small image. The case $T=0$ is immediate.

The Gowers dichotomy theorem states that every infinite-dimensional Banach space contains either an unconditional basic sequence or a hereditarily indecomposable subspace. Choose an initial basic sequence; it suffices to prove the result for its closed span $X$, with basis constant $K$.

We use the following precise finite-sequence form of the Gowers Ramsey theorem for Banach spaces. Let $\mathcal{A}$ be an analytic family of finite normalized block sequences, large in the sense that every infinite-dimensional block subspace contains a member. Given positive errors $\Delta =({\delta }_i)$, some infinite-dimensional block subspace $Y$ makes ${\mathcal{A}}_{\Delta }$ strategically large. Here ${\mathcal{A}}_{\Delta }$ consists of tuples within ${\delta }_i$ coordinatewise of a member of $\mathcal{A}$. In the block game the subspace player chooses an infinite-dimensional block subspace at each turn; the vector player chooses a successive unit vector in it and can force a finite initial segment into ${\mathcal{A}}_{\Delta }$. This is the Ramsey theorem permitted without proof, not the dichotomy conclusion.

Assume $X$ contains no unconditional basic sequence. For $\epsilon >0$, let ${\mathcal{A}}_{\epsilon }$ be the finite normalized block tuples $(x_1,\dots ,x_n)$ for which the odd and even spans contain unit vectors $u,v$ with $\Vert u-v\Vert <\epsilon$. This family is analytic: in each finite length the witness coefficients and vectors range over finite-dimensional scalar spaces, and the strict distance test is continuous; taking a countable union of finite-support choices gives the block family.

It is large. Indeed, every block subspace has a conditional block basis, with coordinate projections of unbounded norms. Given a finite combination $z$, choose a set of coordinates $A$ with $\Vert P_{A}z\Vert$ arbitrarily large relative to $\Vert z\Vert$. Put $u_0=P_{A}z$, $v_0=-P_{A^c}z$. Then $u_0-v_0=z$ and $|\Vert u_0\Vert -\Vert v_0\Vert |\le \Vert z\Vert$, soGroup consecutive nonzero terms belonging to the same side into a single block, and normalize each group. The groups alternate between the two sides, and their odd and even spans contain these two normalized vectors. Interchanging their names if necessary gives a member of ${\mathcal{A}}_{\epsilon }$.

Choose $\Delta$ with $8K{\sum }_i{\delta }_i<\epsilon$. A game tuple in ${\mathcal{A}}_{\epsilon ,\Delta }$ actually belongs to ${\mathcal{A}}_{2\epsilon }$. To check the perturbation explicitly, each coefficient of a unit vector in a normalized block tuple is bounded by $2K$. Replacing its coordinates by their nearby game vectors changes it by at most $2K\sum {\delta }_i$. Normalizing again changes the vector by at most twice that amount. Doing this for the two witnesses adds at most $8K\sum {\delta }_i$ to their distance.

The Ramsey theorem thus supplies a block subspace on which the vector player can force odd and even spans to contain unit vectors within $2\epsilon$. Given any two infinite-dimensional block subspaces $U,V$ of that space, let the subspace player choose them alternately, using their successive tails to satisfy support order. The strategy produces the odd vectors in $U$ and the even vectors in $V$, so these subspaces contain the required nearby unit vectors.

Repeat inside decreasing block subspaces $Y_n$ with ${\epsilon }_n\downarrow 0$, and choose a diagonal normalized block sequence $(z_n)$ with $z_n\in Y_n$. Its tail from $n$ onward lies in $Y_n$. In $Z=[z_n]$, any two infinite-dimensional block subspaces can be restricted to this tail, hence contain unit vectors at distance less than $2{\epsilon }_n$ for every $n$.

Finally this zero-angle statement extends to arbitrary infinite-dimensional closed subspaces $U,V$ of $Z$. Choose unit vectors in each subspace annihilating the first successively many basis coordinates, and approximate them by finite successive blocks with summably small errors. The coordinate bound $2K$ makes the resulting map from the block span to the original subspace arbitrarily close to the identity in operator norm. Apply the block-subspace zero-angle result and normalize the transported vectors; their added error can be made arbitrarily small. Thus the unit spheres of $U,V$ have distance zero.

By the equivalence proved in part (a), $Z$ is hereditarily indecomposable. This proves the dichotomy, including the passage from block subspaces to all subspaces.