Recent Results in the Theory of
Infinite-Dimensional Banach Spaces
W. T. GOWERS
Department of Mathematics, University College London,
Gower Street, London WC1E 6BT, England
Many of the best-known questions about separable infinite-dimensional Banach
spaces are of at least one of the following forms.
(1)
(2)
(3)
If X is an arbitrary space, must it have a "nice" subspace?
If X is an arbitrary space, are there nonobvious examples of operators on XI
If certain operators are known to be defined on a space X, does this imply
anything nonobvious about the structure of XI
Many such questions have been solved in the last three years, and surprising
connections between them have been discovered. The purpose of this paper is
to explain these developments. Unless otherwise stated, all spaces and subspaces
mentioned will be infinite-dimensional separable Banach spaces.
We begin by discussing questions of the first kind above. It is clear straight
away that not every space has a Hilbert space, the nicest space of all, as a subspace;
the space t\ is but one of many obvious counterexamples. However, if one asks
whether every space contains Co or £p for some 1 < p < oo, then one already
has a very simple question to which the answer is not at all obvious. In fact, this
question was not answered until the early 1970s, when Tsirelson [T] used a clever
inductive procedure to define a counterexample. The proof that his example does
not contain Co or £p is surprisingly short (this is even more true of the dual of his
space as presented by Figiel and Johnson [FJ]), but the ideas he introduced have
been at the heart of the recent progress.
There were two further weakenings of the notion of "nice" that left questions
not answered by Tsirelson's example. For the first, recall that a Schauder basis
(we will often say simply basis) of a Banach space X is a sequence (xn)^L1 such
that every vector in X has a unique expression as a norm-convergent sum of the
form Y^=i an%n- A simple result proved in the early 1930s by Mazur (see [LT2])
is that every Banach space has a subspace with a basis. Whether every separable
Banach space had a basis was a famous open problem, answered negatively by
Enfio [En] in 1973.
The definition of a Schauder basis is unlike that of an algebraic basis for
a vector space in that the order of the xn is important. A basis (xn)™=l with
the property that (^7r(n))^Li is a basis for every permutation 7r of the positive
integers is called an unconditional basis. It was shown by Mazur that (xn)^L=1 is
Proceedings of the International Congress
of Mathematicians, Zürich, Switzerland 1994
© Birkhäuser Verlag, Basel, Switzerland 1995
934
W. T. Gowers
an unconditional basis if and only if there is a constant C such that, for every
sequence (o n )^ = 1 of scalars and any sequence (en)^L1 with each en of modulus
one, we have the inequality
oc
oc
71=1
71=1
Such a basis is called C-unconditional. Notice that this implies that, for every
subset A C N, the projection YinieNanxn •"> YlneAanxn has norm at most C.
Thus, if X has an unconditional basis, then there are many nontrivial projections
on X. A sequence that is an unconditional basis for its closed linear span is called
an unconditional basic sequence.
We may now ask whether every space has a subspace with an unconditional
basis, or, equivalently, contains an unconditional basic sequence (the unconditional
basic sequence problem). This question was first asked as soon as unconditional
bases were defined in the 1940s and appears in print in [BP2]. The lack of a proof
after three or four decades led many people to begin to suspect that the answer
was no, but there still seemed to be some chance of a positive answer to a yet
weaker question: Does every space contain co, t\. or a reflexive subspace? The
reason this is a weaker question is that a result of James [Jl] states that a space
with an unconditional basis not containing Co or t\ must itself be reflexive, so a
positive answer would be implied by a positive answer to the first question.
In the summer of 1991, Maurey and I independently discovered spaces not
containing unconditional basic sequences. There is not room to explain the constructions here, but we can make a few remarks. First, we introduce some notation.
If X is a Banach space with a given basis (en)™=1, then the support of a vector
x = Yl^Li an^n £ X is just the set of n for which an is nonzero. Given x,y e X,
we write x < y to mean that every element of the support of x is less than every
element of the support of y. A sequence x\ < x2 < • • • of nonzero vectors is called
a block basis, and a subspace generated by a block basis is called a block subspace.
A simple but very useful result of Bessaga and Pelczynski [BP1] states that every
subspace y of a space X with a basis has a further subspace Z isomorphic to
a block subspace of X. This reduces many problems about subspaces of Banach
spaces to ones about block subspaces.
The unconditional basic sequence problem is no exception. It is straightforward to show that X contains no C-unconditional basic sequence if and only if
every block subspace Y C X contains a sequence of vectors x\ < • • • < xn such
that
lf>l > clèt-i)"**
i=l
i=l
(This is not quite true if X has complex scalars. In that case, sequences satisfying the above inequality exist in every block subspace provided X contains no
2C-unconditional basic sequence.) Therefore X contains no unconditional basic
sequence at all if and only if such a finite sequence can be found in every block
subspace for every C. Notice that the condition on the supports of x\,... ,xn is
important, because otherwise a constant sequence would do.
Recent Results in Banach Space Theory
935
Of great importance to us in constructing such a space was noticing that, for
any fixed C, a space which had just been constructed by Schlumprecht [SI] could
be renormed so that every block subspace did contain such a "C-conditional" sequence. However, Schlumprecht 's space has a 1-unconditional basis, so any renorming has a C-unconditional basis for some C. To find a space with no unconditional
basic sequence at all, one somehow had to use the ideas from the renorming of
Schlumprecht 's space but produce a nonisomorphic space. This led to considerable
conceptual and technical difficulties. Two other points are worth making. The first
is that Schlumprecht's space was, like Tsirelson's space and indeed our spaces,
constructed inductively (at first glance, the definition appears to be circular). The
advantage of his space over Tsirelson's was that it had stronger properties related
to the so-called distortion of Banach spaces. Indeed, his space played an important
part in his solution with Odell [OSI] of a famous problem known as the distortion
problem. (In one formulation this asks: Must every space isomorphic to a Hilbert
space have a subspace almost isometric to a Hilbert space?) Again, there is not
room to explain this connection (some indication can be found in [OS2]), but there
is one and it is important. The second point is that the spaces that Maurey and I
found were not in fact distinct, indicating that, however complicated, our approach
was a natural one. The result, with some extensions that will be described in a
moment, appears in a joint paper [GM1].
The first indication that the space X\ we constructed was also relevant to
questions about operators was an observation of Johnson. He pointed out to us
that we could alter our argument(s) and show that every continuous projection
on X\ had finite rank or corank, and moreover that every subspace of Xi had
the same property. This is equivalent to saying that no subspace Y of X\ can be
written as a topological direct sum W 0 Z with W and Z infinite dimensional. A
space with this property he called hereditarily indecomposable. Another equivalent
form of the property, which brings out its strangeness, is that, for any two infinitedimensional subspaces Y and Z of X\ and any e > 0, there exist unit vectors
y € Y and z G Z such that ||y — z|| < e. In a certain sense, the angle between any
two infinite-dimensional subspaces is zero.
Thus, our space gave a strong answer to a question of Lindenstrauss [L2]: Can
every space be decomposed as a topological direct sum of two infinite-dimensional
subspaces? (Such a space is simply called decomposable.) It showed that, for an
arbitrary space X, one could not in general expect L(X), the space of operators
on X, to contain interesting examples of projections.
Another important class of operators is isomorphisms onto proper subspaces,
and the next step was the discovery that these also did not have to exist in general.
I constructed a variant X\j of the space X\ above, which had an unconditional basis
[Gl]. However, it was not isomorphic to any proper subspace, and therefore solved
the so-called hyperplane problem, which had its origins in Banach's book [B].
This was the question of whether an arbitrary space is isomorphic to its closed
subspaces of codimension one. (Note that if Y and Z are distinct such subspaces,
then Y ~ (Y D Z) 0 C ~ Z, so any two subspaces of the same finite codimension
are isomorphic.) Equivalently, X\j is not isomorphic to X\j 0 C. (Actually, as
Maurey recently pointed out to me, the precise question asked by Banach was
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W. T. Gowers
whether every space X is isomorphic to a subspace of its hyperplanes, but this is
also answered negatively by X\j-)
It was a little strange that an unconditional basis was needed to make the
above construction work, although it did mean that a question of the third kind
was answered. The situation became clearer with the following result [GM1], which
underlines the importance of hereditary indecomposability. Recall that an operator
T : X -> Y is Fredholm if the dimensions a(T) and ß(T) of the kernel of T and
Y/TX, respectively, are finite, and that the index of such an operator is defined to
be a(T) — ß(T). A strictly singular operator S : X —> Y is one for which there is no
subspace Z of X such that the restriction of S to Z is an isomorphism. Equivalently,
for every Z C X and every e > 0 there exists z E Z with ||Sz|| < e ||z||. The strictly
singular operators share many of the smallness properties of the compact operators
(indeed, for several spaces X, the strictly singular, and compact operators in L(X)
coincide). For example, if T is Fredholm and S is strictly singular then T + S is
Fredholm with the same index as T.
1. [GM1] Let X be a hereditarily indecomposable space with complex
scalars. Then every operator T E L(X) can be written in the form XI + S, where
A G C and S is strictly singular. In particular, every operator on X is either
strictly singular or Fredholm with index zero.
THEOREM
It is easy to see that an isomorphism onto a proper subspace cannot be
strictly singular or Fredholm with index zero, so the space X\ discussed earlier is
another counterexample to Banach's question. (In fact the space X\ can be real
or complex. In the case of real scalars, we have a direct proof that it satisfies the
conclusions of Theorem 1. In general, a real hereditarily indecomposable space is
isomorphic to no proper subspace.) Thus, in a certain sense, L(X\) is trivial and
the answer to the second question with which we started is a simple no. On the
other hand, it is not known whether there exists a space on which every operator
is a compact perturbation of a multiple of the identity.
We have left behind the question of whether an arbitrary space must contain
Co, ti, or a reflexive subspace. For this problem, unlike the unconditional basic
sequence problem, it is not enough to find a space such that every (block) subspace
contains ^finite sequence of some kind. The problem is in this sense more genuinely
infinite dimensional, and for this reason it was not obvious whether the techniques
used for constructing the hereditarily indecomposable space X could be extended
to give a counterexample. In the end, however, it turned out to be possible [G2], and
the resulting space XQLR had a slightly stronger property. No subspace Y C XCLR
contains £\ or has a separable dual. Note that it is not obvious that there even exists
a space with a nonseparable dual not containing £\. This was a question asked by
Banach and answered independently by examples of James [J2] and Lindenstrauss
and Stegall [LS].
It would seem, then, that there is no reasonable sense of the word "nice" for
which one can (truthfully) say that every space has a nice subspace. However, this
conclusion, we shall see later, is premature.
Let us concentrate on questions of the third kind. We have already seen that
the existence of plenty of projections on a space does not guarantee that there are
Recent Results in Banach Space Theory
937
nontrivial isomorphisms to subspaces. What about the reverse? If we are given a
space that is isomorphic to a proper subspace, must there be nontrivial projections?
There are many other questions of this general kind, and to answer some of them
Maurey and I generalized the results of our first paper. We proved [GM2] that,
under certain conditions on an algebra A, one can construct a space X = X(A)
such that A embeds in an obvious way into L(X), and every operator on X is
a small perturbation of (the image of) an element of A. Sometimes "small" in
this statement simply means "strictly singular", and sometimes it means a slight
weakening that is nevertheless strong enough for applications.
To illustrate, let us consider what can be said about a space if it has a basis
and the shift with respect to that basis is an isometry. From Theorem 1 we know
straight away that the space is not hereditarily indecomposable, but this does not
imply that there are nontrivial projections defined on the whole space. We also
know that, writing S for the shift and L for the left shift, every operator of the
form
T,n=o anSn + Hn=i bn^n is continuous if Yln=o K I + £ ^ = 1 |6 n | < oo. In
other words, the algebra A of convolutions by absolutely summable sequences
on Z embeds into the algebra of operators on the space. (This is not quite a
homomorphism because LS ^ SL, but finite-rank perturbations do not matter to
us.) Our theorem gives a space Xs for which the shift is an isometry, such that
every operator in L(Xs) is a strictly singular perturbation of an element of A. It is
straightforward to deduce from this that every projection on Xs has finite rank or
corank. Thus, if Y is a subspace of Xs and there is a continuous projection onto Y
(such a subspace is called complemented), then Y is finite codimensional. (Recall
that subspaces are infinite dimensional unless it is otherwise stated.) Because all
subspaces of the same finite codimension are isomorphic, the existence of the shift
guarantees that every complemented subspace of Xs is isomorphic to Xs- A space
with this property is called prime. Before this example, the only known examples
were Co and £p for 1 < p < oc. Apart from i^, these were shown to be prime by
Pelczynski [P]. Lindenstrauss [Ll] proved it for i^, the only known nonseparable
example.
The general philosophy here is that, often, the existence of certain operators
on a space X implies little more than that the algebra A generated by those
operators embeds into L(X), which is obvious anyway. Two more examples are as
follows. For every positive integer n > 2 there exists a space Xn such that two
finite-codimensional subspaces of Xn are isomorphic if and only if they have the
same codimension modulo n, and there exists a space Zn such that the product
spaces Z£ and Zn are isomorphic if and only if r and s are equal modulo n — 1.
For example, there is a space isomorphic to its subspaces of codimension two but
not to its hyperplanes, and there is a space isomorphic to its cube but not to its
square. For the second class of examples, one obtains algebras An which resemble
the C*-algebras On, which were analyzed using K-theory by Cuntz [C]. Our proof
that the algebras An do not contain isomorphisms between Z£ and Z* when r and
s are not equal modulo n — 1 was a modification of his argument. Notice that if
X ~ X3 but X qk X2, then we have an example of two nonisomorphic spaces X
and Y such that either embeds into the other as a complemented subspace. This
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W. T. Gowers
gives a negative answer to the so-called Schroeder-Bernstein problem for Banach
spaces, originally solved in [G3]. We have another example if we take the space X2
above, which is isomorphic to X2 0 C 2 but not to X2 0 C.
The theorem proved in [GM2] gives fairly general circumstances under which
the third question from the beginning of this paper has a negative answer. However,
there are certain very strong assumptions one can make about a space where our
theorem has nothing to say. For example, a famous result of Lindcnstrauss and
Tzafriri [LT1], solving a problem known as the complemented subspaces problem,
states that if every closed subspace of a Banach space is complemented, then the
space must be isomorphic to a Hilbert space. If we consider isomorphisms instead of
projections, we get the following question of Banach [B]: If X is a space isomorphic
to all its (closed infinite-dimensional) subspaces, must X be isomorphic to a Hilbert
space? A space isomorphic to all its subspaces is nowadays called homogeneous.
Are there examples other than £2?
Szankowski [Sz] generalized Enflo's solution to the basis problem, by showing
that, unless a space is very close to a Hilbert space in a certain technical sense, then
it has a subspace without a basis (or even the approximation property). Because
we have Mazur's result that every space has a subspace with a basis, it follows
that a homogeneous space must be close to a Hilbert space. On the other hand,
Johnson [Jol] showed that a variant of Tsirelson's space (the 2-convexificd version)
has the property that every quotient of every subspace has a basis, but the space
does not contain £2.
Until recently, the most powerful result in the positive direction was also due
to Johnson [Jo2]. He showed that if X and X* are homogeneous and have what is
known as the GL-propcrty, a property related to, but much weaker than, having an
unconditional basis, then X is isomorphic to £2. Then, early in 1993, Komorowski
and Tomczak-Jaegermann proved the following result.
2. [KT] Let X be a Banach space of cotype q for some q < oc. Then
either X contains £2 or X contains a subspace without an unconditional basis.
THEOREM
It does not matter too much here what it means to be of cotype q. Suffice it to
say that, by Szankowski's result, a homogeneous space is too close to £2 to fail the
condition of Theorem 2. It follows that a homogeneous space not isomorphic to
(and therefore not containing) £2 must have a subspace without an unconditional
basis. But because the space is homogeneous, it does not even contain an unconditional basic sequence. Another way of saying this is that a homogeneous space
with an unconditional basis must be £2. Theorem 2 therefore uses a stronger assumption on the space than Johnson, but the assumption about the dual space is
no longer necessary. This theorem of Komorowski and Tomczak-Jaegermann provided a remarkable link between Banach's homogeneous spaces problem and the
unconditional basic sequence problem. Moreover, it was potentially very useful,
because the examples that had been constructed of spaces not containing unconditional basic sequences had, as we have seen in some cases, nowhere near enough
operators to be homogeneous — quite the reverse!
Let us consider the prime space Xs mentioned earlier. It is an easy consequence of the results of [GM2] that it contains no unconditional basic sequence.
Recent Results in Banach Space Theory
939
We also know that it is not hereditarily indecomposable. However, it can be shown
that Xs has a hereditarily indecomposable subspace. The same is true of the spaces
Xn and Zn. Could this be a general phenomenon? The answer is yes.
3. [G4] Every Banach space X has a subspace Y that either has an
unconditional basis or is hereditarily indecomposable.
THEOREM
The solution to Banach's question is now easy. We have seen that a homogeneous
space X not isomorphic to £2 contains no unconditional basic sequence. Therefore,
by Theorem 3, it contains a hereditarily indecomposable subspace. Hence, as X
is homogeneous, X is itself hereditarily indecomposable. But this, in the light of
Theorem 1, is a very strong contradiction.
Before we move on, it is worth pointing out that there is no connection
between the proofs of Theorem 2 and Theorem 3. It was a historical accident that
they were proved in the order that they were. Similarly, Theorem 3 is completely
independent of the actual existence of hereditarily indecomposable spaces, except
that nobody thought to look at the notion of hereditary indecomposability until
an example of a space exhibiting it was produced.
We now return to the word "nice". The reason one would like to find "nice"
subspaces is that one can say more about them than about general spaces. But if
we take this as a vague definition of niceness, then Theorem 3 does give us a nice
subspace. For, in a sense, we know everything there is to know about a space if
we know that it is hereditarily indecomposable. If, on the other hand, it has an
unconditional basis, we also have a lot of information about it. This is not just a
quibble about words either, because we have given an example of a problem where
this kind of niceness gave exactly the sort of control that was needed. From this
point of view one would say that the questions that were traditionally asked about
nice subspaces were of the wrong kind (although more examples of Banach spaces
were needed to make this clear). It is more fruitful to look for theorems such as
the last two, where one obtains subspaces with one of two or more very different
properties.
Indeed, such theorems already exist, Rosenthal's £i-theorem [Rl] being the
most famous example (see also [R2]). It is interesting that infinite Ramsey theory,
which can be used to prove Rosenthal's theorem [F], was also important in the
proof of Theorem 3. We now briefly outline a very general and essentially combinatorial result, of which Theorem 3 is an easy consequence. We shall need a small
amount of notation.
Given a space X with a fixed basis, let E = T,(X) denote the set of all finite
sequences x\,... ,xn of vectors of norm at most 1, such that x\ < • • • < xn. (The
meaning of "<" was given earlier.) Given a subset a C E, let us say that it is large
if every block subspace Y (still infinite dimensional) contains some sequence in u.
Every set a C S defines a two-player game as follows. The first player, S,
chooses a block subspace X\ of X, then the second player, P, chooses a point x\ G
X\ of finite support and norm at most 1. Then S chooses X2 and P chooses x2 G X2
of norm at most 1 such that x\ < x2. They continue in this way, alternately picking
subspaces and points in them. The game is won by P if at some stage the sequence
(x\,... ,xn) is in a. If it goes on forever without this happening, then S wins.
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W. T. Gowers
Clearly, if P is to have any chance of winning, then a must at least be large,
because otherwise S could repeatedly choose the same subspace containing no
sequence in o. In fact, saying that P has a winning strategy for the set a is a much
stronger statement than saying that a is large, and that is the point of the next
theorem. If A = (b\, 82,...) is a sequence of positive reals and a C E, we will write
0A for the set of sequences (x\,..., xn) G E "within A of <J"; that is, such that
there exists (y\,...,yn)
e a with \\x{ - yi\\ < 6i for 1 < i < n.
4. [G4] Let X be a Banach space with a basis and let a be a large
subset of E(X). Then, for every positive sequence A = (Si,62,...) there exists a
block subspace Y C X such that, if S and P play the above game in the space Y,
then P has a winning strategy for obtaining sequences in a A .
THEOREM
Ignoring perturbations, this says that if a is large, then P has a winning strategy
in some subspace. The proof of the theorem (and of the main result of [G5], which
we shall mention in a moment) is related to arguments due to Galvin and Prikry
[GP] and Ellentuck [E] concerning infinite versions of Ramsey's theorem.
Now one of the remarks earlier was that if X contains no unconditional basic
sequence, then for every C the set a of sequences x\ < • • • < xn with the property
that |Er=i X*'H > C ||Sr=i( — l) 1 ^!!
ls
l ar g e - Theorem 4 allows us to drop to a
subspace Y in which P has a winning strategy for finding such sequences (say with
C replaced by C/2). Suppose S plays a strategy that simply involves alternating
between two subspaces W and Z of Y. Then P's strategy guarantees the existence
of one of these sequences with the odd-numbered x^s in W and the even-numbered
ones in Z. Letting w and z be the sums of the odd-numbered and even-numbered
vectors, respectively, we have the inequality \\z + w\\ > C \\z — w\\. If C is large, it
follows easily that zj \z\ and w/ \\w\\ are close.
We have not quite managed to find arbitrarily close unit vectors in Z and
W, because we made a fixed choice of C at the beginning of the argument above.
However, one can repeat it for a sequence of Cn tending to infinity and obtain a
nested sequence Y\ D Y2 D . . . of subspaces such that for each n the argument
works for Cn in the subspace Yn. It is then easy to check that a "diagonal" subspace
generated by a block basis y\,y2,...
with yn G Yn is hereditarily indecomposable.
Thus, Theorem 3 follows from Theorem 4. An adaptation of this proof that is in
some ways more direct was found by Maurey and appears in [M].
It is noticeable that, from the point of view of the general theory of Banach
spaces, certain examples of spaces appear to be more natural than others. The
property that distinguishes one of these natural examples is that every subspace
has a further subspace that is not interestingly different from the whole space.
For example, every block subspace of Tsirelson's space mentioned earlier turns
out to be a "Tsirelson-type" space (this statement can be made quite precise),
Schlumprecht's space has the very strong property that every subspace has a further subspace not only isomorphic to the whole space but complemented inside it
(see [S2]), and every subspace of the hereditarily indecomposable space X\ has a
further subspace that, although definitely not isomorphic to X\, can be described
in an almost identical way. Of the classical spaces, only Co and the ^-spaces are
natural in this sense, because every classical space contains one of them. The spaces
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941
Xs, Xn and Zn (for n > 2) are good examples of unnatural spaces because they
lose what little structure they have when one passes to an appropriate subspace
(recall that they have hereditarily indecomposable subspaces).
It is possible that a precise definition of "natural" will emerge, but even
without it, one can attempt to classify the natural spaces. Theorem 3 shows that
they either have an unconditional basis or are hereditarily indecomposable. In [G5],
a generalization is presented of Theorem 4 to analytic sets of infinite sequences (in
a certain sensible topology), and more information is given about natural spaces
with an unconditional basis. Loosely speaking, there is a theorem that states that
such a space either has many isomorphisms between its subspaces or has none that
do not follow trivially from the existence of the unconditional basis. The space X\j
mentioned earlier is an example where the second possibility holds.
From another point of view, the nonclassical "natural" spaces mentioned
above are extremely unnatural. They all have Tsirelson-type inductive definitions,
as opposed to formulae for their norms. Given a vector with support of size n, it
does not even seem to be possible to calculate its norm in one of these spaces in a
time polynomial in n. It is tempting to ask whether there is a meta-theorem that
states that a norm that is in some sense directly defined must give a space that
has some ^,-space or co as a subspace. At the moment there is not even a precise
conjecture along these lines, but it could be that there is, waiting to be discovered,
a theory of "easily described" Banach spaces and linear operators very different
indeed from the theory of general spaces as outlined in this paper.
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