 Solution

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.