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If a Banach space $X$ has separable dual and $q:X\to X/Y$ is a quotient by a closed subspace, every weakly null sequence in the quotient's closed unit ball has a subsequence with approximate weakly null lifts of norm at most $3$. Choose representatives of norm below $3/2$, pass to a common bidual weak-star limit, and subtract convex blocks whose quotient images tend to zero in norm. Convex-block cancellation of a weak-star limit gives weak nullity and the triangle inequality gives the bound $3$.

A convex block of a sequence in a Banach space is a sequence of convex combinations supported on successively disjoint finite intervals, with $p_n<q_n<p_{n+1}$, $a_i\ge 0$, and ${\sum }_{i=p_n}^{q_n}{a}_i=1$. Zero coefficients can fill gaps and ensure strict endpoint inequalities. By the Mazur theorem, every weakly null sequence has convex blocks tending to zero in norm.

If a bounded sequence $(y_n)$ in $X$ converges weak-star to $\varphi \in X^{\ast \ast }$ under the canonical embedding into the bidual, then any convex blocks $(u_n)$ have the same scalar limit on every $f\in X^{\ast }$. Thus $y_n-u_n$ is a weakly null sequence, even when $\varphi$ does not belong to the embedded space $X$.

Write $N(a_1,\dots ,a_m)$ for the limiting spreading model norm. We show that deleting any one coefficient does not increase it; iteration then gives every finite suppression.

Fix an index $r$, the coefficients, and a small $\eta >0$. Choose a generation threshold so that all the relevant full and shortened combinations beyond it approximate their model norms within $\eta$. Fix the indices before position $r$ beyond this threshold. Since $(x_n)$ is a weakly null sequence, the Mazur lemma gives a finite convex combinationwith every $j$ after the fixed prefix. Choose the remaining indices after $\max F$. For every $j\in F$, the full ordered combination with $x_j$ in slot $r$ has norm at most $N(a)+\eta$. Averaging these combinations givesThe shortened combination has norm at most $N(a)+(1+|a_r|)\eta$, and approximates its own model norm within $\eta$. Letting $\eta \downarrow 0$ provesfor every finite set $A$. Thus the model is a suppression-unconditional basic sequence with constant one. The convex-combination argument allows the omitted coordinate to sit between two retained coordinates; simply letting a single omitted index tend to infinity with all others fixed would not respect their order.

LetThis scalar function is continuous. The first element is locally constant; deleting it is continuous on increasing sequences; and evaluation on a fixed $x_m$ is continuous in the weak-star topology. Thus $\mathcal{U}=\{M:|h(M)|<\delta \}$ is open.

By part (a), choose an infinite $L_0$ homogeneous for $\mathcal{U}$. If all its infinite subsets are in $\mathcal{U}$, the required conclusion follows. Suppose instead every one has $|h(M)|\ge \delta$. Choose finitely many unit scalars ${\zeta }_1,\dots ,{\zeta }_q$ such that every scalar $z$ with $|z|\ge \delta$ has $\mathrm{Re}({\zeta }_{j}z)>\delta /2$ for some $j$; for real scalars the two signs suffice. The open setscover $[L_0{]}^{\omega }$. Apply the open Ramsey set of infinite subsets result successively to them. After finitely many refinements, at least one, say ${\mathcal{U}}_j$, contains every infinite subset of a remaining infinite $L$; otherwise their cover would be violated.

For any finite $F\subseteq L$, choose an infinite $N\subseteq L$ after $\max F$. For each $m\in F$, apply this homogeneity to $M=\{m\}\cup N$. Its deleted tail is always the same $N$, so the same functional satisfiesBut $(x_m{)}_{m\in L}$ is a weakly null sequence. The Mazur lemma gives a finite convex combination $z={\sum }_{m\in F}{\lambda }_m{x}_m$ with $\Vert z\Vert <\delta /2$. The displayed inequalities give $\mathrm{Re}({\zeta }_j{x}_N^{\ast }(z))>\delta /2$, contradicting $\Vert x_N^{\ast }\Vert \le 1$. The bad homogeneous case is impossible. ThereforeContinuity is used to make the scalar tests open; weak nullity is used only for the finite convex-combination contradiction. The argument works over both real and complex scalars.

For a convex subset of a Banach space, weak closure equals norm closure. The Mazur theorem states that for every convex set $C\subseteq X$,Since every bounded linear functional is norm-continuous, the weak topology is weaker than the norm topology, giving ${\overline{C}}^{\Vert \cdot \Vert }\subseteq {\overline{C}}^w$.

For the reverse inclusion, use this form of the Hahn-Banach separation theorem: if $D$ is a nonempty norm-closed convex set in a real normed vector space and $x\notin D$, there is a continuous real-linear functional $F$ such thatApply it to $D={\overline{C}}^{\Vert \cdot \Vert }$. In a complex Banach space, use the underlying real space; every continuous real-linear $F$ is the real part of the complex bounded linear functional $f(v)=F(v)-iF(iv)$. The separating strict inequality gives a neighbourhood open for the weak topology of $x$ disjoint from $D$, so $x\notin {\overline{C}}^w$. The empty set case is immediate. This proves the Mazur theorem.

A weakly null sequence has disjoint convex blocks converging to zero in norm. If $x_n\stackrel{w}{\to }0$, then for every starting index $m$,The Mazur theorem places $0$ in the norm closure of that tail convex hull. Thus a finite convex combination of vectors from any tail can have norm less than any prescribed $\epsilon >0$.

Choose the convex blocks recursively. After selecting the previous terminal index $q_{n-1}$, let $p_n>q_{n-1}$, and choose a finite convex combination from $\{x_i:i\ge p_n\}$ with norm less than $2^{-n}$. Choose $q_n$ beyond its largest used index, padding the intervening coefficients and the final coefficient with zero. This ensures the printed strict condition $p_n<q_n$ as well as $q_n<p_{n+1}$. Setting unused coefficients to zero yieldsTherefore $u_n\to 0$ in norm. The recursive tail selection is what makes these convex blocks, rather than merely unrelated convex combinations.

Next suppose $X^{\ast }$ is separable and $(x_n)$ is bounded. Use the canonical embedding into the bidual $J_X:X\to X^{\ast \ast }$, where $J_{X}x(f)=f(x)$. If $\Vert x_n\Vert \le M$, then $J_X{x}_n\in M{B}_{X^{\ast \ast }}$. The Banach-Alaoglu theorem makes this ball compact in $\sigma (X^{\ast \ast },X^{\ast })$, and weak-star metrizability of the dual ball makes it metrizable because its predual $X^{\ast }$ is separable. A compact space that is a metric space is sequentially compact, so some subsequence $(y_n)$ satisfiesThe limit need not belong to $J_{X}X$; it is precisely the use of the bidual space that supplies compactness without assuming reflexivity.

For a convex block $u_n={\sum }_{i=p_n}^{q_n}{a}_i{y}_i$, every $f\in X^{\ast }$ satisfiesBoth $f(y_n)$ and $f(u_n)$ tend to $\varphi (f)$, soThis is the useful convex-block cancellation of a weak-star limit; it does not require the limit to be a vector in $X$.

The quotient sequence has approximate lifts in $3B_X$ that are weakly null after passing to a subsequence. Let $q:X\to Z=X/Y$ be the quotient map, with the quotient norm. Since $\Vert z_n\Vert \le 1$, choose $v_n\in X$ such thatThis uses the infimum defining the quotient norm and does not assume that it is attained. By the preceding compactness argument, pass to a subsequence $y_n=v_{k_n}$ with a common weak-star limit in $X^{\ast \ast }$. Its quotient images $q(y_n)=z_{k_n}$ form a weakly null sequence.

Apply the convex block construction to $(q(y_n))$ in the quotient Banach space. Using the same coefficients on $(y_n)$ gives convex blocksSet $x_n=y_n-u_n$. The convex-block cancellation of a weak-star limit proves $x_n\stackrel{w}{\to }0$, whileThus the approximate weakly null lifting through a quotient is achieved with the required bound:

By definition of the weak topology, $x_n\rightharpoonup x$ exactly when $f(x_n)\to f(x)$ for every $f\in X^{\ast }$.

Let $(x_n)$ be bounded and choose a norm-dense sequence $(f_j)$ in $X^{\ast }$. Successive subsequences make $f_1(x_n),f_2(x_n),\dots$ converge, and the diagonal argument produces a single subsequence $(y_n)$ on which every $f_j$ converges. Uniform boundedness of $(y_n)$ and norm approximation of an arbitrary $f$ by the $f_j$ show that $f(y_n)$ is Cauchy for every $f\in X^{\ast }$. Thus $(y_n)$ is weakly Cauchy, andfor every $f$, so its difference sequence is weakly null.

For the countable family $(y_{m,n}{)}_n$, use the weak metric from part a. Delete a finite initial segment from the $m$th sequence so that every remaining term has weak distance less than $1/m$ from zero, and relabel that tail. Enumerate all these tails while preserving the order within each one. For every weak neighbourhood of zero, all terms from sufficiently large $m$ lie inside it, and only finitely many terms from each of the finitely many remaining sequences lie outside it. The resulting enumeration $(z_n)$ is weakly null and contains the relabelled $m$th sequence as a subsequence for every $m$. Equivalently, without relabelling, it contains a tail-subsequence of every original sequence, which is the form used below.

A sequence $(x_n)$ is weakly Cauchy when $(f(x_n))$ is a Cauchy sequence for every continuous linear functional $f$. Its consecutive differences form a weakly null sequence.