About the constructions of Tsirelson space

Question

Let $T'$ be the Tsirelson space dual in the Figiel-Johson construction, and $B_{T'}$ the unit closed ball. Then, $B_{T'}$ satisfies the next properties:

(i) $B_{T'}\subset B_{l^\infty}.$ Each vector basis $e_n$ belongs to $B_{T'}.$

(ii) $\forall \mathbf{x} \in B_{T'},\ \mathbf{y} \in \mathbb{R}^{\mathbb{N}},\ $ if $|\mathbf{y}|\leq |\mathbf{x}|$ (pointwise) then $\mathbf{y} \in B_{T'}$

(iii) If $\{\mathbf{y}_i\}_{i=1}^n$ is a block basic sequence of $B_{T'},$ then $\dfrac{1}{2}Q_n(\sum_{i=1}^n\mathbf{y}_i)\in B_{T'},$ where $Q_n$ denote the natural projection of $T'$ onto $[e_k]_{k=n}^\infty.$

The original contruction (by Tsirelson) took $A$ to be the smallest set in $l^\infty$ enyoing properties (i) through (iii), and let $K$ denote the clousure of $A$ respect to the topology of pointwise convergence. Since it is easily seen $K\subset c_0,$ it follows that $K$ is weakly compact. Finally, the closed convex hull $V$ of $K$ defines a weakly compact convex set in $c_0$, which inherits properties (i) through (iii). The minimality of $A$ then forces $$V=B_{T'}.$$

I can easily see that $V\subset B_{T'}$ because of the minimailty argument, but I don't get to see the other one.

Answer 1

IIRC Figiel and I reasoned this way: The dual norm to Tsrilson's original norm has the property that for any vector $x$ in $c_{00}$, $$ \|x\| \ge \|x\|_{c_0} \vee (1/2) \sup \sum_{k=1}^n \|E_k x\|, $$ where the sup is over all admissible $(E_k)$. But, by construction, the FJ norm is the smallest unconditional norm satisfying this inequality. The inclusion you see yields that the dual to Tsirelson's original norm is smaller than the FJ norm, so the two norms are the same.

BTW: You should have said that the bulk of your post (absent the question) is taken word for word from page 17 in the Casazza-Shura book.