Dual and bidual models for James space

If $\mathrm{\Lambda }\in J^{\ast }$, then [L2] makes $\mathrm{\Lambda }{\mathrm{\mid }}_{{\mathrm{\ell }}^2}$ bounded on ${\mathrm{\ell }}^2$ because $\mathrm{\mid }\mathrm{\Lambda }(x)\mathrm{\mid }\le \mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }\mathrm{\parallel }x{\mathrm{\parallel }}_J\le \sqrt{2}\mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }\mathrm{\parallel }x{\mathrm{\parallel }}_2$. [given, L1, L2, L3] By [L3] it is pairing with a unique $y\in {\mathrm{\ell }}^2$. Density of ${\mathrm{\ell }}^2$ in $J$ makes $y$ determine $\mathrm{\Lambda }$, and the displayed supremum is exactly its operator norm. Conversely, any $y$ with finite displayed supremum extends by continuity from dense ${\mathrm{\ell }}^2$ to $J$.

Dual coordinate truncation is pairing with ${\mathrm{\Pi }}_{N}x$. The two contraction [given, L1, step 1.1] estimates in [L1] therefore give $\mathrm{\parallel }{\mathrm{\Pi }}_{N}y{\mathrm{\parallel }}_{J^{\ast }}\le \mathrm{\parallel }y{\mathrm{\parallel }}_{J^{\ast }}$ and $\mathrm{\parallel }y-{\mathrm{\Pi }}_{N}y{\mathrm{\parallel }}_{J^{\ast }}\le \mathrm{\parallel }y{\mathrm{\parallel }}_{J^{\ast }}$. The coordinate interpretation follows directly from the series pairing.

We prove the latter tails tend to zero. If instead their decreasing norms [given, A1, L1, step 2.1] stay above $\epsilon >0$, then [A1] is used exactly here to choose, for each $N$, a finite-support $u^{(N)}\in J$ with ${\mathrm{\Pi }}_N{u}^{(N)}=0$, $\mathrm{\parallel }{u}^{(N)}{\mathrm{\parallel }}_J=1$, and $⟨y,u^{(N)}⟩>\epsilon$ (change sign if needed). Starting at $N_1=1$, put $N_{j+1}=\max \mathrm{supp}{u}^{(N_j)}$.

Form $\xi \in c_0$ by placing $j^{-1}{u}^{(N_j)}$ on the coordinate block $(N_j,N_{j+1}]$. To verify $\xi \in J$, split any finite increasing tuple into its intersections with these blocks. Each within-block endpoint variation is at most $1\mathrm{/}j$, and $\mathrm{\mid }a-b{\mathrm{\mid }}^2\le 2\mathrm{\mid }a{\mathrm{\mid }}^2+2\mathrm{\mid }b{\mathrm{\mid }}^2$ bounds every transition between blocks by the two adjacent endpoint terms. Consequently

Let $\mathrm{\Lambda }\in J^{\ast \ast }$ and set $z_n=\mathrm{\Lambda }(e_n)$. Since $\mathrm{\parallel }{e}_n{\mathrm{\parallel }}_{J^{\ast }}=1$, $z$ is bounded. By step 4.1,

Conversely, suppose bounded $z$ has [given, step 5.1] $B:={\sup }_p\max \{q_p(z),r_p(z)\}<\mathrm{\infty }$. It is Cauchy: otherwise some $\epsilon >0$ permits recursively choosing the lexicographically least $p_1<q_1<p_2<q_2<\cdots$ with $\mathrm{\mid }{z}_{p_j}-z_{q_j}\mathrm{\mid }\ge \epsilon$; then the cyclic variation on the first $2k$ indices is at least $\epsilon \sqrt{k\mathrm{/}2}$, contradicting $B<\mathrm{\infty }$. Let $\lambda ={\lim }_n{z}_n$ and $x=z-\lambda \mathbf{1}$. Then $x\in c_0$ and $q_p(x)=q_p(z)$, so $x\in J$.

For $y\in J^{\ast }$, the partial-sum functionals $h_N(y)={\sum }_{n\le N}{y}_n=⟨y,{\mathbf{1}}_{\{1,\dots ,N\}}⟩$ have norm at most one because that initial-block vector has James norm one. They converge on dense $c_{00}$, and the uniform bound plus step 4.1 makes $h_N(y)$ converge for every $y\in J^{\ast }$. Hence

Step 5.1 applied to ${\mathrm{\Lambda }}_z$ gives the reverse norm inequality, so [given, step 5.1, step 6.1, step 7.1] $\mathrm{\parallel }{\mathrm{\Lambda }}_z\mathrm{\parallel }=B$. Steps 5.1 and 7.1 are inverse constructions and prove the isometric bidual model. Step 6.1 gives the unique splitting $z=(z-\lambda \mathbf{1})+\lambda \mathbf{1}$; since elements of $J$ tend to zero, the canonical image is exactly $\lambda =0$.