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If a bounded subset $K$ of a Banach space is bounded away from zero and zero lies in its weak closure, then $K$ contains a basic sequence. More generally, weak closure may be replaced by closure in any Hausdorff locally convex vector topology weaker than the weak topology.

Relative to a Schauder basis, a block sequence consists of nonzero finitely supported vectors with successive disjoint supports: $\max \mathrm{supp}{x}_i<\min \mathrm{supp}{x}_{i+1}$. It is a basic sequence, with basis constant at most that of the original basis.

The dual sequence consists of the coordinate functionals $(e_n^{\ast })$ in $X^{\ast }$. It is always a basic sequence, though it need not span the entire dual space.

The infinite Ramsey theorem used here says that every finite colouring of the $m$-element subsets of an infinite subset of $\mathbb{N}$ has an infinite homogeneous subset. Let $(x_n)$ be normalized with basis constant $K$.

For each positive integer $m$, rational scalar tuple $a=(a_1,\dots ,a_m)$ and precision $2^{-r}$, colour increasing $m$-tuples according to intervals of length $2^{-r}$ containing $\Vert {\sum }_i{a}_i{x}_{n_i}\Vert$. This is a finite colouring, because all these norms lie between zero and ${\sum }_i|a_i|$. Enumerate the countably many triples $(m,a,r)$ and repeatedly apply the Ramsey theorem, obtaining decreasing infinite sets on which each specified norm has oscillation at most $2^{-r}$. Choose a diagonal subsequence $(y_j)$ whose tail is contained in every previously selected infinite set.

Consequently, for every rational tuple $a$, the net of norms over increasing tuples from $(y_j)$ is Cauchy as its first index tends to infinity. Denote its limit by $N(a)$. This extends to arbitrary scalar tuples becauseuniformly in the indices. For complex scalars use rational real and imaginary parts. The same estimate proves convergence uniformly over all sufficiently late increasing tuples for each fixed scalar tuple.

Homogeneity and the triangle inequality pass to the limit. The coordinate functionals of a normalized basic sequence have norm at most $2K$, since the coordinate projection is the difference of two initial projections. Thusso $N$ is a genuine norm. Zero coefficients may be inserted anywhere without changing the value: removing those coordinates simply selects another increasing tuple for the same limiting combination. Also every initial-segment projection has norm at most $K$, by the corresponding inequality for $(y_j)$. Completing the finitely supported scalar sequences in $N$ therefore gives a Banach space with a normalized basic sequence $(e_i)$.

Its finite norms are invariant under increasing relabellings, by the zero-insertion observation, andThis is the required spreading model, generated by the diagonal subsequence.

Choose a normalized basic sequence in the infinite-dimensional Banach space, and apply the Rosenthal l1 theorem. If a subsequence is equivalent to the unit vector basis of ${\ell }^1$, any spreading model generated from a further subsequence retains its two-sided ${\ell }^1$ estimates and is therefore unconditional.

Otherwise there is a weakly Cauchy sequence as a subsequence $(z_n)$, still basic. Its consecutive differences $d_n=z_{2n}-z_{2n-1}$ are weakly null and are a block sequence. The coordinate-functional bound for its normalized parent basis gives $1/(2K)\le \Vert d_n\Vert \le 2$. Hence $d_n/\Vert d_n\Vert$ is a normalized weakly null basic sequence. Part (a) gives a spreading model, and part (b) makes that model one-suppression-unconditional. Thus every infinite-dimensional Banach space has an unconditional spreading model, even though it need not contain an unconditional basic sequence.

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.

A Schauder basis of a real Banach space $X$ is a sequence $(e_n)$ for which every $x\in X$ has a unique norm-convergent expansion $x={\sum }_n{a}_n{e}_n$. Its basis projection isand its basis constant is $K_b={\sup }_n\Vert P_n\Vert <\infty$.

For the sequence space $Y$ in the question, defineConvergence in the definition of $Y$ makes $J$ well defined, and uniqueness of basis coefficients makes it a linear bijection. Moreover,whereasThus $J$ is an isomorphism and, in particular, the displayed supremum really is a complete norm on $Y$.

The coordinate functional of a Schauder basis isIt is bounded because $a_n{e}_n=(P_n-P_{n-1})x$. On finite linear combinations of the $e_n^{\ast }$, the nth partial-sum operator is the restriction of $P_n^{\ast }$, sinceThe operators have norms at most $K_b$, so the standard basis criterion shows that the dual sequence of a Schauder basis $(e_n^{\ast })$ is a basic sequence in $X^{\ast }$. For every $x^{\ast }\in X^{\ast }$ and $x\in X$,which is precisely $P_n^{\ast }{x}^{\ast }\to x^{\ast }$ in the weak-star topology.

Suppose now that $X$ is a reflexive Banach space. If $\Vert x^{\ast }-P_n^{\ast }{x}^{\ast }\Vert$ did not tend to zero, approximation by finite basis blocks would give a bounded block sequence $(u_j)$ and an $\epsilon >0$ such that $|x^{\ast }(u_j)|\ge \epsilon$. Reflexivity gives a weakly convergent subsequence. Every fixed coordinate functional is eventually zero on a block sequence, so its weak limit has every basis coordinate zero and is therefore zero. This contradicts $|x^{\ast }(u_j)|\ge \epsilon$. Hence $P_n^{\ast }{x}^{\ast }\to x^{\ast }$ in norm for every $x^{\ast }$, so the basis is shrinking and $(e_n^{\ast })$ is a basis of $X^{\ast }$.

The converse fails. The standard basis of $c_0$ is shrinking because its dual sequence is the standard basis of ${\ell }^1=(c_0{)}^{\ast }$, but $c_0$ is not reflexive.

Finally assume $(e_n^{\ast })$ is a basis of $X^{\ast }$. Map $x^{\ast \ast }\in X^{\ast \ast }$ toFor $s_n={\sum }_{k=1}^n{a}_k{e}_k$,so $(a_n)\in Z$. Conversely, if $(a_n)\in Z$ and $M={\sup }_n\Vert s_n\Vert$, defineThe limit exists: for $m>n$,because $(e_n^{\ast })$ is a basis. Also $|F(x^{\ast })|\le M\Vert x^{\ast }\Vert$, so $F\in X^{\ast \ast }$ and $F(e_n^{\ast })=a_n$. These two constructions are inverse and satisfyThus $X^{\ast \ast }$ and $Z$ are isomorphic.

LetChoose positive ${\epsilon }_n$ with ${\prod }_n(1+{\epsilon }_n)<\infty$. We construct $x_n\in K$ inductively. Once $E_n=\mathrm{span}\{x_1,\dots ,x_n\}$ has been chosen, compactness of its unit sphere gives finitely many members of $X^{\ast }$ that almost norm every vector of $E_n$. Because $0\in {\overline{K}}^w$, the next $x_{n+1}$ may be chosen so that all those functionals are as small on it as required. Choosing the error relative to $d$ givesfor all scalars $a_1,\dots ,a_{n+1}$. Indeed, if the last coefficient could threaten this estimate, $\Vert x_{n+1}\Vert \ge d$ first bounds that coefficient by a fixed multiple of the norm of the preceding sum; the selected norming functional then gives the displayed inequality. Iteration and the finite product bound satisfy the standard basis criterion, so $(x_n)$ is a basic sequence contained in $K$.

The same proof works for any Hausdorff locally convex vector topology $\tau$ weaker than the weak topology: on each finite-dimensional $E_n$, the $\tau$-continuous linear functionals still norm the space, and $0\in {\overline{K}}^{\tau }$ supplies the next point. A canonical strictly weaker example arises on $X^{\ast }$ when $X$ is not reflexive: the weak-star topology $\sigma (X^{\ast },X)$ is then strictly weaker than the weak topology $\sigma (X^{\ast },X^{\ast \ast })$.

We next prove the Eberlein-Šmulian theorem. If the weak closure of a bounded set $K$ is weakly compact, take any sequence in $K$ and let $Y$ be its closed linear span. The relevant weak closure lies in the separable space $Y$. A countable weak-star dense subset of the dual unit ball separates points of this compact set, so its weak topology is metrizable. Compact metrizability gives a weakly convergent subsequence.

For the converse, suppose $K$ is not relatively weakly compact. In the canonical embedding into $X^{\ast \ast }$, chooseThe Hahn-Banach theorem gives $x^{\ast \ast \ast }\in X^{\ast \ast \ast }$ that vanishes on $J_X(X)$ but satisfies $x^{\ast \ast \ast }(x^{\ast \ast })=1$. Alternating Goldstine approximation with the fact that $x^{\ast \ast }$ lies in the weak-star closure of $J_{X}K$ constructs bounded $x_n^{\ast }\in X^{\ast }$ and $x_n\in K$ such that, up to errors tending to zero,If a subsequence $x_{n_j}$ converged weakly to $x$, then for each fixed $i$ the second relation would give $x_i^{\ast }(x)=1$. A weak-star cluster point $x^{\ast }$ of the bounded sequence $(x_i^{\ast })$ satisfies $x^{\ast }(x_{n_j})=0$ by the first relation, and hence $x^{\ast }(x)=0$ by weak convergence; but passing to the same cluster point in $x_i^{\ast }(x)=1$ gives $x^{\ast }(x)=1$. This contradiction produces a sequence in $K$ with no weakly convergent subsequence. Relative weak compactness is therefore equivalent to the subsequence condition.

Finally suppose $K$ is weakly sequentially compact and $x\in {\overline{K}}^w$. If $x$ lies in the norm closure of $K$, a norm-convergent sequence suffices. Otherwise apply the first part to $K-x$ and obtain a basic sequence $(y_n)$. A subsequence converges weakly by hypothesis. Its weak limit lies in the closed span of the basic sequence, while every coordinate functional is eventually zero on that subsequence. The limit is consequently zero, so the corresponding sequence from $K$ converges weakly to $x$.

This also shows that a weakly sequentially compact $K$ is weakly closed: any point of its weak closure is the limit of a sequence in $K$, and a weakly convergent subsequence has its limit in $K$. The relative form of the Eberlein-Šmulian theorem then makes $K$ weakly compact. The reverse implication follows from the first direction of that theorem. Thus weak compactness and weak sequential compactness are equivalent.

A normalized basic sequence generates a spreading model when the norms of all finite scalar combinations converge as the first index tends to infinity, uniformly over increasing choices of the indices. The limiting norm on finitely supported scalar sequences is unchanged by increasing relabellings of coordinates. Completing this norm gives a Banach space with its canonical basic sequence. A diagonal application of the finite-colour infinite Ramsey theorem produces such a model from a subsequence of every normalized basic sequence.