The following is Proposition $7.1$:

Let $X$ be a Banach space, let $1 \leq p \leq \infty$ and let $\lambda \geq 1$. Assume that for every finite dimensional subspace $E$ of $X$ there is a subsapce $\bar{E}$ of $l_p$ such that their Banach-Mazur distance $\bar{d}(E, \bar{E}) \leq \lambda$. Then there is a measure $\mu$ and a subspace $Y$ of $L_p(\mu)$ such that $\bar{d}(\bar{X}, Y) \leq \lambda$.

In the end of the proof, the author proved that there exists an operator $\bar{T}$ which sends $x \in X$ into the class determined by $f_x$ in $Z / \{f: |||f|||=0 \}$ where $Z = \{ f: f \in B(U^*), \pi(|f|^p)$ is finite $\}$. And the operator satisfies $$\lambda^{-1}\| x \| \leq |||\bar{T}x||| \leq \| x \|.$$

How to conclude the theorem from the last paragraph of the proof?


They use the fact that abstractly $Z / \{f\colon |||f|||=0\}$ is a lattice such that $$|||f+g|||^p = |||f|||^p+|||g|||^p$$ whenever $f$ and $g$ are disjoint elements in the lattice sense. Of course, completions of such lattices also enjoy this property. Then by a theorem of Nakano, the completion of this space is isometric to $L_p(\mu)$ for some measure $\mu$. As $\overline{T}$ is an isomorphism (with constant $\lambda$) the conclusion follows.

There is a quicker version of this proof which relies on ultraproducts but it also requires Nakano's theorem. It goes like this:

Your hypothesis says that $X$ is finitely representable in $\ell_p$. Hence it is isomorphic to a subspace of some ultrapower of $\ell_p$. Ultrapowers of $\ell_p$ satisfy the hypothesis of Nakano's characterisation of abstract $L_p$-spaces, hence are of the form $L_p(\mu)$ for some measure $\mu$.

To unwrap this argument, see the proof of the fact that finite representability of $X$ in $Z$ is equivalent to the possibility of embedding $X$ into some ultrapower of $Z$.

  • $\begingroup$ May I know the quicker version of the proof? $\endgroup$
    – Idonknow
    Dec 12 '15 at 16:52
  • $\begingroup$ Also, may I know why $\bar{T}$ is an isomorphism? Is it because of the inequality $\lambda^{-1} \| x \| \leq ||| \bar{T}x ||| \leq \| x \|$? $\endgroup$
    – Idonknow
    Dec 12 '15 at 17:18
  • $\begingroup$ Yes, it is bounded below so it has closed range and is injective, hence it is an isomorphism. $\endgroup$ Dec 12 '15 at 17:19
  • $\begingroup$ @Idonknow, done. $\endgroup$ Dec 12 '15 at 17:41

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