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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 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.

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.

The norm comparison in bounded distortion of a Banach space is understood up to a positive rescaling of the new norm. Thus, after replacing $D$ by $D^2$ if one's convention uses symmetric comparison constants, we may use the following form: for every equivalent norm $q$ on a block subspace, a further block subspace $Z$ and a number $a>0$ satisfyThe number $a$ may depend on the norm and the subspace; $D$ is fixed. Without this rescaling convention the premise would be impossible, since arbitrarily large scalar multiples of the original norm are equivalent norms. We prove the substantive, scale-invariant assertion for an infinite Schauder basis.

Let $K$ be the basis constant of the original basis. Every block sequence has basis constant at most $K$, so its initial, tail and finite interval projections have norms at most $K$, $1+K$ and $2K$, respectively. We shall construct increasingly complicated finite unconditional configurations while keeping their unconditional constant fixed at $C=2D$.

First construct the Maurey hierarchy of finite sets. Write ${\mathcal{M}}_0=\{\varnothing \}$, and definewhere prepending to the empty set is permitted. For each nonzero countable limit ordinal $\lambda$, fix positive ordinals ${\lambda }_j\uparrow \lambda$ and putInduction shows that every family is hereditary and spreading: subsets remain in the family, and increasing any of the coordinates preserves membership. Every nonzero stage contains all singletons. These families are also well-founded as extension trees. At a successor stage, an infinite branch would yield an infinite branch at the previous stage after deleting its first integer. At a limit stage, a hypothetical branch with first integer $m$ has every initial segment in the finite union ${\bigcup }_{j\le m}{\mathcal{M}}_{{\lambda }_j}$. One family then contains arbitrarily long initial segments, and heredity would give it an infinite branch.

For every infinite set $L$, the extension tree ${\mathcal{M}}_{\alpha }\cap [L{]}^{<\infty }$ has empty-root rank of a well-founded tree at least $\alpha$. Here is the induction proving this lower bound. At a successor stage fix $n\in L$; the tree above the node $\{n\}$ contains a copy of ${\mathcal{M}}_{\alpha }$ restricted to the infinite tail of $L$ after $n$, and hence has rank at least $\alpha$. The root therefore has rank at least $\alpha +1$. At a limit stage, for every $j$ the restriction of ${\mathcal{M}}_{{\lambda }_j}$ to a sufficiently late infinite tail of $L$ is contained in ${\mathcal{M}}_{\lambda }$. The root rank is consequently at least ${\sup }_j{\lambda }_j=\lambda$.

The crucial elementary estimate concerns signed interval projection norms. Given a hereditary family $\mathcal{F}$ containing all singletons and a basis $(w_i)$, define on finitely supported vectorsChoosing one interval containing the support gives $q_{\mathcal{F}}^{(w)}(x)\ge \Vert x\Vert$. Whenever this supremum is bounded by a multiple of $\Vert x\Vert$, it extends to a norm on the closed span giving equivalent norms with the original norm.

If $S={\sum }_i{\epsilon }_i{P}_{E_i}$ is one of the admitted signed interval projections, thenTo prove this, test $Sx$ with another admitted operator $R={\sum }_j{\delta }_j{P}_{F_j}$. The composition is the signed sum of projections onto the nonempty intervals $E_i\cap F_j$. Colour each intersection according to which of $E_i,F_j$ supplies its left endpoint, breaking ties in favour of $E_i$. In the first colour the list of minima is a subset of $(\min E_i)$, with no repetitions; in the second it is a subset of $(\min F_j)$. Both lists belong to $\mathcal{F}$ by heredity. Each colour therefore contributes an admitted signed interval projection, and the triangle inequality gives $\Vert RSx\Vert \le 2q_{\mathcal{F}}^{(w)}(x)$. Taking the supremum over $R$ proves the estimate.

Now suppose $q_{\mathcal{F}}^{(w)}$ is an equivalent norm on a block subspace $W$. Apply bounded distortion to obtain a further block basis $(z_i)$ spanning $Z$, with $a\Vert z\Vert \le q_{\mathcal{F}}^{(w)}(z)\le aD\Vert z\Vert$. A signed interval projection relative to $(z_i)$, with an admissible list of minima, is the restriction of an admitted signed interval projection relative to $(w_i)$. Indeed, replace each interval of block indices by the interval from the first to the last original coordinate of its blocks; its minimum only increases, so the spreading property preserves admissibility. The elementary estimate and the norm comparison therefore giveIn particular $q_{\mathcal{F}}^{(z)}\le C\Vert \cdot \Vert$, with the same $C=2D$ for every family to which we apply this procedure.

We prove by transfinite induction that, in every block subspace and for every $1\le \alpha <{\omega }_1$, there is a normalized block basis $(z_i)$ for whichFor $\alpha =1$ only one interval is permitted, so the initial signed interval projection norm is bounded by $2K\Vert x\Vert$. Apply the preceding renorming procedure.

For a successor stage, first choose $(w_i)$ satisfying the induction hypothesis for ${\mathcal{M}}_{\alpha }$. Delete the first interval from any ${\mathcal{M}}_{\alpha +1}$-admissible family. The remaining minima lie in ${\mathcal{M}}_{\alpha }$, soThis is an equivalent norm. The renorming procedure produces a further block basis with the constant reset to $C$.

At a limit stage $\lambda$, choose decreasing block subspaces $W_j$, each with a block basis satisfying the induction hypothesis for ${\mathcal{M}}_{{\lambda }_j}$. Choose successive normalized $z_j\in W_j$. Make each $z_j$ sufficiently far out that its first coordinate in the basis of every $W_k$, $k\le j$, is at least $j$; only finitely many restrictions are imposed at each choice. An admissible interval family for ${\mathcal{M}}_{\lambda }$ belongs to ${\mathcal{M}}_{{\lambda }_j}$ for some $j\le \min E_1$. All blocks on which it acts lie in the tail $[z_i:i\ge j]\subset W_j$. Transferring its intervals to the basis of $W_j$ increases their minima, so the induction hypothesis givesThus $q_{{\mathcal{M}}_{\lambda }}^{(z)}$ is again an equivalent norm, and one more application of bounded distortion resets its bound to $C$. This completes the transfinite induction. The tail-projection factor in this step is important; the diagonal tail need not have the same norm as the original vector.

Finally let $T_C(X)$ be the tree of finite normalized sequences $(x_1,\dots ,x_n)$ satisfyingfor all scalars and all signs. Each finite level is closed in the corresponding power of the unit sphere, since the norm inequalities are closed conditions. The unit sphere of the separable Banach space $X$ is a Polish space. An infinite branch is an unconditional basic sequence: sign changes give coordinate projections with norm at most $(1+C)/2$, imply linear independence, and give uniformly bounded initial projections, hence the basis property on the closed span. These projection bounds also give unconditional convergence, including over complex scalars.

For each countable $\alpha$, the block basis constructed above maps the extension tree ${\mathcal{M}}_{\alpha }$ into $T_C(X)$ by $\{i_1<\cdots <i_r\}\mapsto (z_{i_1},\dots ,z_{i_r})$. To check the sign inequality, use the admissible singleton intervals $\{i_1\},\dots ,\{i_r\}$ on the vector $\sum a_k{z}_{i_k}$. Thus the unconditional tree index is at least $\alpha$ for every countable ordinal. If $T_C(X)$ were well-founded, the closed-tree result would give it countable height, contradicting these lower bounds. It has an infinite branch. Therefore $X$ contains an unconditional basic sequence.

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.