\{n_0\}$, so by the definition of $K_{n_0+1}$,
$$\Norm{\sum_{j=0}^\infty a_jy_j} = \Norm{\sum_{j=n_0+1}^\infty a_jy_j} \ge K_{n_0+1} \sum_{j=0}^\infty |a_j|
\ge \theta K \sum_{j=0}^\infty |a_j|.$$
Then since $\norm{y_j} \le \phi K$ for all $K$ and $\theta/\phi \ge 1-\delta$, we have
$$\Norm{\sum_{j=0}^\infty a_j u_j} = \normm{\sum_{j=0}^\infty a_j {y_j\over \norm{y_j}}}
\ge {1\over \phi K} \Norm{\sum_{j=0}^\infty a_j y_j} \ge (1-\delta)\sum_{j=0}^\infty |a_j|,$$
which proves the lower bound.\slug
Note that in James' original statement of the lemma, the $u_j$ are just vectors contained in the closed unit
ball. However, the construction in his proof actually produces a block basic sequence, and we
normalised it in our own version of the proof above.
We are now set to show that $l_1$ does not embed into $T$.
\newcount\nolone
\nolone=\thmcount
\proclaim Theorem \advthm. Tsirelson's space $T$ does not contain a subspace isomorphic to $l_1$.
\proof Suppose, towards a contradiction, that $T$ did contain a subspace isomorphic to $l_1$. Then, applying
Lemma~J with $\delta = 1/9$, there exists a normalised block basic sequence $(y_j)_{j=0}^\infty$
such that for all sequences $(a_j)_{j=0}^\infty$ of scalars,
$${8\over 9} \sum_{j=0}^\infty |a_j| \le \Norm{\sum_{j=0}^\infty a_jy_j} \le\sum_{j=0}^\infty |a_j|.$$
In particular, if we let $r\ge 1$ be an integer and let $(a_j)_{j=0}^\infty$ be the sequence with
$$a_j = \cases{ 1,& if $j=0$;\cr 1/r,& if $1\le j\le r$;\cr 0,& if $j>r$,}$$
then we have $\sum_{j=0}^\infty |a_j| = 2$. This yields
$$\Norm{ y_0 + {1\over r}\sum_{j=1}^r y_j} = \Norm{\sum_{j=0}^\infty a_jy_j} \ge
{8\over 9}\sum_{j=0}^\infty |a_j| = {16\over 9}$$
for all integers $r\ge 1$.
Now let $\{k\}\le E_1 < E_2 < \cdots < E_k$ be an arbitrary admissible sequence of sets, and let
$r_0 =\max\supp y_0$. If $k > r_0$, then
$$\sum_{i=1}^k \normm{ E_i \Bigl( y_0 + {1\over r}\sum_{j=1}^r y_j\Bigr)}
=\sum_{i=1}^k \normm{ E_i \Bigl({1\over r}\sum_{j=1}^r y_j\Bigr)} \le 2\cdot\Norm{{1\over r}\sum_{j=1}^r y_j}
\le 2.$$
If $k\le r_0$, then we set
$$S = \bigl\{1\le j\le r : \norm{E_i y_j}\ne 0\ \hbox{for at least two values of}\ i\bigr\}$$
and
$$T = \bigl\{1\le j\le r: \norm{E_i y_j}\ne 0\ \hbox{for at most one value of}\ i\bigr\}.$$
Note that if $j\in S$, then $y_j$ straddles the border between $E_i$ and $E_{i+1}$ for some $i$, and
no other $y_{j'}$ can do so, since the supports of the $y_j$ are disjoint. So the cardinality of $S$ is
at most $k-1$. It is also clear that $S$ and $T$ are disjoint and $|S|+|T| = r$, so
$$2|S| + |T| \le 2(k-1) + r- k+1 = r+k-1.$$
Since the $y_j$ are unit vectors, we have
$$\eqalign{
\sum_{i=1}^k \normm{E_i \Bigl({1\over r}\sum_{j=1}^\infty y_i\Bigr)} &=
\sum_{j\in S} \sum_{i=1}^k \norm{E_iy_j} + \sum_{j\in T} \sum_{i=1}^k \norm{E_iy_j}\cr
&\le \sum_{j\in S} 2\cdot \norm{y_j} + \sum_{j\in T} \norm{y_j},\cr
&\le r+k-1\cr
}$$
by rearranging sums.
Dividing by $r$ and adding the unit vector $y_0$ into the mix, we have
$$\eqalign{
\sum_{i=1}^k \normm{ E_i \Bigl( y_0 + {1\over r} \sum_{j=1}^r y_j\Bigr)}
&\le\sum_{i=1}^k \norm{E_i y_0} + {1\over r}\sum_{i=1}^k\normm{E_i\Bigl({1\over r}\sum_{j=1}^\infty y_i\Bigr)}\cr
&\le 2 + {r+k-1\over r} \cr
&\le 3 + {n_0\over r}.\cr
}$$
Selecting some $r\ge 2n_0$, the right-hand side is at most $7/2$, meaning that
$$\Norm{y_0 + {1\over r} \sum_{j=1}^r y_j} \le 7/4,$$
and we have $7/4 \ge 16/9$, the contradiction we sought.\slug
In the proof that $c_0$ and the other $l_p$ spaces do not embed into $T$ either, we shall make use of
the following theorem, a version of what is called the Bessaga--Pe\l czy\'nski Selection Principle
(see~\ref{bessagapelczynski}, Theorem 3).
\parenproclaim Theorem S (Bessaga--Pe\l czy\'nski, {\rm 1958}).
Let $X$ be a Banach space with basis $(e_i)$ and let $(e_i^*)$ be the coefficient
functionals given by
$$ e_j^*\Big(\sum_{i=1}^\infty a_i e_i\Bigr) = a_j.$$
If $(y_i)_{i=1}^\infty$ is a sequence of vectors in $X$ such that
\medskip
\item{i)} $\lim_{i\to\infty} \norm{y_i} > 0$; and
\smallskip
\item{ii)} $\lim_{i\to\infty} e_j^*(y_i) = 0$ for all indices $j\ge 1$,
\medskip\noindent
then there exists a subsequence of $(y_i)$ which is equivalent to a block basis with respect to the original
basis $(e_i)$.\slug
To prove that $c_0$ and $l_p$ do not embed into $T$, we shall actually prove the following stronger result.
\proclaim Theorem \advthm. Tsirelson's space $T$ does not contain a seminormalised subsymmetric basic
sequence.
\proof Towards a contradiction, suppose that $(y_i)$ is a seminormalised subsymmetric basic sequence in $T$.
Since the sequence is seminormalised, it is bounded from above, so by Lemma~{\the\dichotomy}, either $(y_i)$
is equivalent to the canonical basis for $l_1$ or $(y_i)$ is weakly null. We already proved Theorem~{\the\nolone}
above, which deems the first scenario impossible, so $(y_i)$ must be weakly null and thus in particular
satisfies condition (ii) of Theorem S. In addition, since $(y_i)$ is seminormalised, it is uniformly bounded
away from zero, so we are in the position to conclude from Theorem S that there is a subsequence of $(y_i)$
which is equivalent to a block basic sequence $(x_i)$ against the unit vector basis $(t_i)$ of $T$.
We shall now pass to subsequences both in $(x_i)$
and in $(y_i)$. First, pass to a subsequence and reindex $(x_i)$ to ensure that $\supp x_i > \{i\}$ for all $i$.
Now we pass to a subsequence of $(y_i)$ that is equivalent to this new $x_i$ and reindex; this is equivalent
to the original sequence by subsymmetry.
Since $y_i$ is seminormalised and there is a linear homeomorphism mapping each $y_i\mapsto x_i$,
$(x_i)$ is seminormalised as well; that is, there is
$M\ge 1$ such that $1/M \le \norm{x_i}\le M$.
Now since $\supp x_i > \{i\}$ for all $i$,
we can apply Proposition~{\the\tsirelsonblock} with the constant $M$ on any $n$ blocks
$x_{i_1}, x_{i_2}, \ldots, x_{i_n}$ of $(x_i)$
to find that for all $n$-tuples $(b_1,\ldots, b_n)$ of scalars,
$${1\over 2M}\sum_{k=1}^n |b_k| \le \Norm{\sum_{k=1}^n b_kx_{i_k}} \le M\sum_{k=1}^n |b_k|.$$
We thus find that for any $n$, $(x_i)$ (and hence $(y_i)$)
contains a subsequence of length $n$ which is equivalent to the standard basis of $l_1^n$. Complete
each of these finite sequences to infinite subsequences. Recall that $(y_i)$ is equivalent to all of them,
so for all $n$, the first $n$ elements of $(y_i)$ are equivalent to the standard basis of $l_1^n$. Thus
$(y_i)$ is equivalent to the standard basis of $l_1$ and we know this cannot happen.\slug
\proclaim Corollary \advthm. Tsirelson's space $T$ does not contain $c_0$ or $l_p$ for $1 1$, we say that $X$ is {\it $\lambda$-distortable} if there exists an equivalent norm
$\threenorm\cdot$ on $X$ such that for each infinite-dimensional subspace $Y$ of $X$, we have
$$\sup\biggl\{{\threenorm{y_1}\over\threenorm{y_2}}:y_1,y_2\in Y,\,\norm{y_1} = \norm{y_2} = 1\biggr\}\ge\lambda.$$
The norm $\threenorm\cdot$ is called a {\it $\lambda$-distortion} if this is true.
The space $X$ is called {\it distortable} if it is $\lambda$-distortable for some $\lambda>1$, and
{\it arbitrarily distortable} if it is distortable for every $\lambda>1$. In 1991, T.~Schlumprecht
proved the existence of an arbitrarily distortable Banach space~\ref{schlumprecht1991}.
Schlumprecht's construction relies on a function $f : [1,\infty)\to [1,\infty)$ satisfying
\medskip
\item{i)} $f(1) = 1$ and $f(x)1$;
\smallskip
\item{ii)} $f$ is strictly increasing to infinity;
\smallskip
\item{iii)} $lim_{x\to\infty} f(x)/x^q = 0$ for all $q>0$;
\smallskip
\item{iv)} the function $g(x) = x/f(x)$ is concave for $x\ge 1$; and
\smallskip
\item{v)} $f(x)f(y) \ge f(xy)$ for all $x,y\ge 1$.
\medskip
Of course, we need such a function to actually exist, but it is easily seen that $f(x) = \log_2(x+1)$ does.
We then define a sequence $\threenorm{\cdot}_n$ of norms on $c_{00}$ by setting
$\threenorm{x}_0 = \norm{x}_{l_\infty}$
and for $n\ge 1$ we put
$$\threenorm{x}_{n} = \max_{k\ge 2} \max {1\over f(k)} \sum_{i=1}^k \threenorm{E_i x}_{n-1},$$
where the inner maximum is over all subsets $E_10$, and every infinite-dimensional subspace $Y$ of $X$,
there are unit vectors $y_1, y_2\in Y$ such that
$$\norm{y_1}_k \ge 1-\eps\qquad\hbox{\rm and}\qquad \norm{y_2}_k \le {1+\eps\over f(k)}.$$
In particular, this means that $\norm\cdot _k$ is an $f(k)$-distortion on $X$.\slug
\medbreak
\boldlabel The unconditional basic sequence problem.
In Section~2 we defined what it means for a basis or a basic sequence to be unconditional.
Both Tsirelson's space and Schlumprecht's space have unconditional bases, and so do all classical Banach spaces.
It was unknown for many years whether every infinite-dimensional Banach space contains an unconditional basic
sequence. Soon after Schlumprecht presented his example in the summer of 1991,
W.~T.~Gowers and B.~Maurey independently constructed
Banach spaces without an unconditional basic sequences. They subsequently discovered that both examples
were fundamentally the same, and ended up publishing jointly~\ref{gowersmaurey1993}.
Recall that we say that $X$ is the {\it direct sum} of $Y$ and $Z$ and write $X = Y\oplus Z$ if
the map $Y\times Z\to X$ sending $(y,z)\mapsto y+z$ is a linear homeomorphism. An infinite-dimensional
Banach space $X$ is said to be {\it decomposable} if $X$ can be written as $X = Y\oplus Z$ where
both $Y$ and $Z$ are also infinite-diimensional. An infinite-dimensional
Banach space $X$ without any decomposable subspaces is called
{\it hereditarily indecomposable}. If a space $X$ has an unconditional basic sequence $(x_i)_{i=1}^\infty$, then
letting $Y$ be the closed linear span of $(x_{2i-1})_{i=1}^\infty$ and $Z$ be the closed linear span
of $(x_{2i})_{i=1}^\infty$, we see that $X$ is not hereditarily indecomposable. So to produce an
infinite-dimensional space without an unconditional basic sequence, it suffices to produce a hereditarily
indecomposable one, and that is what Gowers and Maurey did.
Before describing their construction, we need to establish a fair bit of terminology and notation.
Let $J$ be
a set of positive integers such that if $m,n\in J$ with $m0$ such that $\norm{Tx} \ge c\norm x$ for
all $x\in X$. Gowers and Maurey show that every operator from
$X$ to $X$ can be written as a scalar multiple of the identity, plus a strictly singular operator. This
implies that there are, in some sense, very few operators on $X$. A subsequent paper of Gowers and Maurey,
published in 1997,
describes a more general method of producing
Banach spaces whose spaces of operators are small~\ref{gowersmaurey1997}. Recall that an operator
is said to be {\it compact} if it sends bounded sets in the domain to sets with compact closure in the image.
Every compact operator is strictly singular.
In their original
1993 paper, the authors ask if there exists a Banach space $X$ such that every operator from $X$ to $X$
is a scalar multiple of the identity, plus a compact operator. This question was answered positively
S.~A.~Argyros and R.~G.~Haydon in 2009; the paper was published in 2011~\ref{argyroshaydon}.
\medskip\boldlabel The Banach hyperplane problem. In a book published in 1932,
S.~Banach~\ref{banach} asked whether an infinite-dimensional Banach space $X$ over the real
numbers is always isomorphic to $X\oplus \RR$. This amounts to asking whether every infinite-dimensional
Banach space $X$ is such that
that every subspace of codimension $1$ is isomorphic to $X$ itself. Since a subspace of codimension $1$ is
sometimes called a {\it hyperplane}, this question came to be known as the hyperplane problem.
It turns out that the answer to the question is no, and there is an infinite-dimensional
Banach space that is not isomorphic
to any of its hyperplanes.
This problem was solved by W.~T.~Gowers soon after he solved the unconditional basic sequence problem.
His construction, which appeared in print in 1994, is extremely similar~\ref{gowers1994}.
We shall thus retain the definitions from the previous
subsection and immediately define this space. As the reader probably suspects at this point, the construction
is inductive.
Set $X_0 = (c_{00}, \norm\cdot_0)$. For $n\ge 1$, we let $X_n = (c_{00}, \norm\cdot _n)$,
where $\norm{x}_n$ be the maximum of
$$\sup_{m\in \NN} \sup {1\over f(m)} \sum_{i=1}^m \norm{E_i x}_{n-1},$$
where the inner supremum is over all sequences of subsets $E_1 < \cdots < E_m$, and
$$\sup_{k\in K} \sup_{g\in B^*_k(X_{N-1})} \bigl|g(x)\bigr|.$$
The notation we use here is not exactly the same as in the original paper, but using the same letters
as before makes it clear that the only change from the space without an unconditional basic sequence
is the replacement of $\sup_{E\subseteq \NN} \bigl|g(Ex)\bigr|$ has changed to the simpler $\bigl|g(x)\bigr|$.
The space $X$ is the completion of $c_{00}$ with respect to the limit of the norms $\norm{x}_n$ above.
Perhaps surprisingly, this new space $X$ has an unconditional basis. It is noted in Gowers' paper
that the Gowers--Maurey
space of the previous section is also a counterexample to the hyperplane problem, though is is
more difficult to prove.
The proof that $X$ is not isomorphic to any of its hyperplanes is done by showing that $X$
satisfies the hypothesis of the following lemma, which is attributed to P.~G.~Casazza.
\proclaim Lemma C. If $X$ is a Banach space in which no two equivalent sequences $(y_i)$ and $(z_i)$ satisfy
$y_i < z_i < y_{i+1}$ for all $i$, then $X$ is not isomorphic to any proper subspace.\slug
\section References
\bye