Real and complex ell one have the Schur property

First verify the Banach-space condition. Let $(c^{(m)})$ be Cauchy in ${\mathrm{\ell }}^1(\mathbb{K})$. Each coordinate sequence is Cauchy because $\mathrm{\mid }{c}_k^{(m)}-c_k^{(r)}\mathrm{\mid }\le \mathrm{\parallel }{c}^{(m)}-c^{(r)}{\mathrm{\parallel }}_1$; let $c_k$ be its scalar limit by [F6]. Given $\eta >0$, take $M$ such that $\mathrm{\parallel }{c}^{(m)}-c^{(r)}{\mathrm{\parallel }}_1<\eta \mathrm{/}2$ for $m,r\ge M$. For fixed $m\ge M$ and $N$, passage to the limit in the finite sum gives ${\sum }_{k=0}^N\mathrm{\mid }{c}_k-c_k^{(m)}\mathrm{\mid }\le \eta \mathrm{/}2$. Hence [F6] gives ${\sum }_k\mathrm{\mid }{c}_k-c_k^{(m)}\mathrm{\mid }\le \eta \mathrm{/}2$, so $c-c^{(m)}\in {\mathrm{\ell }}^1$ and $\mathrm{\parallel }c-c^{(m)}{\mathrm{\parallel }}_1<\eta$. In particular $c=(c-c^{(M)})+c^{(M)}\in {\mathrm{\ell }}^1$. Thus both scalar versions of ${\mathrm{\ell }}^1$ are complete and hence Banach.

Put $u^{(n)}=a^{(n)}-a$. For every $f\in ({\mathrm{\ell }}^1(\mathbb{K}){)}^{\ast }$, $f(u^{(n)})=f(a^{(n)})-f(a)\to 0$, so $(u^{(n)})$ is weakly null. By [F1] and step 1.1 it is enough to prove $\mathrm{\parallel }{u}^{(n)}{\mathrm{\parallel }}_1\to 0$.

Suppose otherwise. Negating the definition of convergence supplies an $\epsilon >0$ such that $S:=\{n\in \mathbb{N}:\mathrm{\parallel }{u}^{(n)}{\mathrm{\parallel }}_1\ge \epsilon \}$ is cofinal in $\mathbb{N}$: for every $N$ it contains an $n\ge N$. In particular $S$ is nonempty.

We recursively define strictly increasing indices $n_0<n_1<\cdots$ and strictly increasing finite cutoffs $N_0<N_1<\cdots$. Let $n_0$ be the least element of $S$, and let $N_0$ be the least $N$ for which ${\sum }_{k>N}\mathrm{\mid }{u}_k^{(n_0)}\mathrm{\mid }<\epsilon \mathrm{/}8$; the latter set is nonempty by [F3]. Given $(n_j,N_j)$, coordinate evaluation is a bounded functional by [F2], so $u_k^{(n)}\to 0$ for each $k\le N_j$. Because this head is finite, eventually ${\sum }_{k=0}^{N_j}\mathrm{\mid }{u}_k^{(n)}\mathrm{\mid }<\epsilon \mathrm{/}8$. The cofinal set $S$ therefore contains an $n>n_j$ satisfying this inequality. Take the least such $n$ as $n_{j+1}$, then take the least $N>N_j$ with ${\sum }_{k>N}\mathrm{\mid }{u}_k^{(n_{j+1})}\mathrm{\mid }<\epsilon \mathrm{/}8$, again using [F3]. Each least value is unique by [F4]. On the set of pairs $(n,N)$ with $n\in S$ and ${\sum }_{k>N}\mathrm{\mid }{u}_k^{(n)}\mathrm{\mid }<\epsilon \mathrm{/}8$, these rules therefore define a total deterministic successor function; [F4] iterates it from $(n_0,N_0)$ and assembles the entire sequence without a choice axiom.

Define disjoint finite blocks $I_0=\{0,\dots ,N_0\}$ and $I_j=\{N_{j-1}+1,\dots ,N_j\}$ for $j\ge 1$. For $j\ge 1$, the construction and $n_j\in S$ give ${\sum }_{k\in I_j}\mathrm{\mid }{u}_k^{(n_j)}\mathrm{\mid }\ge \mathrm{\parallel }{u}^{(n_j)}{\mathrm{\parallel }}_1-{\sum }_{k=0}^{N_{j-1}}\mathrm{\mid }{u}_k^{(n_j)}\mathrm{\mid }-{\sum }_{k>N_j}\mathrm{\mid }{u}_k^{(n_j)}\mathrm{\mid }>3\epsilon \mathrm{/}4$. For $j=0$ there is no old head, so the same block sum is greater than $7\epsilon \mathrm{/}8$.

Define one scalar sequence $b=(b_k)$ blockwise. If $k\in I_j$ and $u_k^{(n_j)}\mathrm{\ne }0$, put $b_k=\mathrm{sgn}(u_k^{(n_j)})$ when $\mathbb{K}=\mathbb{R}$, and put $b_k=\overline{u_k^{(n_j)}}\mathrm{/}\mathrm{\mid }{u}_k^{(n_j)}\mathrm{\mid }$ when $\mathbb{K}=\mathbb{C}$; put $b_k=0$ when that coordinate is zero. Because $(N_j)$ is strictly increasing, [F5] gives $N_j\ge j$; hence every natural $k$ lies in exactly one of the disjoint blocks. Thus $\mathrm{\mid }{b}_k\mathrm{\mid }\le 1$, so $b\in {\mathrm{\ell }}^{\mathrm{\infty }}(\mathbb{K})$, and the no-conjugation pairing from [F2] satisfies $u_k^{(n_j)}{b}_k=\mathrm{\mid }{u}_k^{(n_j)}\mathrm{\mid }$ on $I_j$ in both scalar fields.

Let $h_b$ be the bounded functional supplied by [F2]. For $j\ge 1$, the triangle inequality, step 5.1, and the head and tail estimates in step 4.1 give $\mathrm{\mid }{h}_b(u^{(n_j)})\mathrm{\mid }\ge {\sum }_{k\in I_j}\mathrm{\mid }{u}_k^{(n_j)}\mathrm{\mid }-{\sum }_{k=0}^{N_{j-1}}\mathrm{\mid }{u}_k^{(n_j)}\mathrm{\mid }-{\sum }_{k>N_j}\mathrm{\mid }{u}_k^{(n_j)}\mathrm{\mid }>\epsilon \mathrm{/}2$. For $j=0$, the block exceeds $7\epsilon \mathrm{/}8$ and the only complementary tail is below $\epsilon \mathrm{/}8$, so the stronger bound $\mathrm{\mid }{h}_b(u^{(n_0)})\mathrm{\mid }>3\epsilon \mathrm{/}4$ holds.

On the other hand, weak nullity in step 2.1 gives $h_b(u^{(n)})\to 0$. The indices $(n_j)$ are strictly increasing, and [F5] gives $n_j\ge j$, so the scalar subsequence $h_b(u^{(n_j)})$ also tends to zero. This contradicts its uniform lower bound in step 7.1. Therefore $\mathrm{\parallel }{u}^{(n)}{\mathrm{\parallel }}_1\to 0$, hence $\mathrm{\parallel }{a}^{(n)}-a{\mathrm{\parallel }}_1\to 0$; [F1] and the completeness proved in step 1.1 establish the Schur property over both $\mathbb{R}$ and $\mathbb{C}$. The construction used only least natural numbers, scalar completeness and recursion, not Countable Choice or any stronger choice principle.