Super-reflexivity is separately determined

Question

I've found this result that states that super-reflexivity is separably determined, i.e., if every separable subspace $Y\subset X$ of a Banach space is superreflexive then $X$ itself is superreflexive. They always say that this is an easy consequence of the definition, but I'm not able to see it. I've tried proving it by contradiction. So, supposing that $X$ is not super-reflexive ensures that I can find a non-reflexive space, $Z$, which is finitely representable in $X$. Even more, since reflexivity is separably determined (easy consequence of James' characterization theorem for reflexivity), I can suppose that this $Z$ is separable. So I have $Z$ a separable, non-reflexive space that is finitely representable in $X$. My idea is to show that this implies that it is finitely representable in a separable subspace of $X$, which would give me a contradiction, but I couldn't be able. Could you guys lend me a hand?

Maybe there is an easier approach, using renormings, for example.

Thank you so much! Bye!!

Answer 1

Take an increasing sequence of finite-dimensional subspaces $Z_n$ of $Z$ s.t. $\overline{\cup_n Z_n} = Z$ (which exists because $Z$ is separable), and a decreasing sequence $\epsilon_n \searrow 0$. By definition of finite representability, there exists a sequence of finite-dimensional subspaces $X_n$ of $X$ and linear isomorphisms $\varphi_n: Z_n \to X_n$ s.t. $\|\varphi_n^{-1}\|\|\varphi_n\| \leq 1 + \epsilon_n$. Let $X’ = \overline{\text{span}}(\cup_n X_n)$, which is separable. I claim that $Z$ is finitely representable in $X’$. Indeed, let $W \subset Z$ be a finite-dimensional subspace and $\epsilon > 0$. Let $w_1, \cdots, w_k$ be a basis of $W$. Let $\|\cdot\|_1$ denote the $\ell^1$-norm on $W$ w.r.t. the basis $w_1, \cdots, w_k$. Since all norms on a finite-dimension space are equivalent, we have, in particular, that $C = \sup_{w \in W, \|w\| = 1} \|w\|_1 < \infty$. Choose $\epsilon’ > 0$ s.t. $C\epsilon’ < 1$ and $\frac{1 + C\epsilon’}{1 - C\epsilon’} \leq 1 + \frac{\epsilon}{2}$.

By the choice of $Z_n$, there exists $n$ with,

1. There exists $v_1, \cdots, v_k \in Z_n$ s.t. $\|w_i - v_i\| \leq \epsilon’$ for all $i = 1, \cdots, k$;
2. $(1 + \frac{\epsilon}{2})(1 + \epsilon_n) \leq 1 + \epsilon$.

Consider the map $\psi: W \to Z_n$ defined by $\psi(w_i) = v_i$ for all $i = 1, \cdots, k$. Then, for any $w = \sum_i c_iw_i \in W$ with $\|w\| = 1$,

$$\|w - \psi(w)\| \leq \sum_i |c_i|\|w_i - v_i\| \leq \epsilon’\|w\|_1 \leq C\epsilon’$$

Thus, $(1 - C\epsilon’)\|w\| \leq \|\psi(w)\| \leq (1 + C\epsilon’)\|w\|$ for all $w \in W$, so $\|\psi^{-1}\|\|\psi\| \leq \frac{1 + C\epsilon’}{1 - C\epsilon’} \leq 1 + \frac{\epsilon}{2}$. Therefore, $\eta = \varphi_n \circ \psi: W \to X_n \subset X’$ satisfies $\|\eta^{-1}\|\|\eta\| \leq (1 + \frac{\epsilon}{2})(1 + \epsilon_n) \leq 1 + \epsilon$. This proves $Z$ is finitely representable in $X’$, as desired.