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$\bf{\text{Definition:}}$ Let $X$ be a Banach space. $X$ is an $\mathcal{L}_{1,\lambda}$-space if, for all finite-dimensional subspaces $M$ of $X$, there exists a finite dimensional subspace $N$ of $X$ containing $M$, and an isomorphism $T\in B(N,\ell_{1}^{k})$ (where $k=\text{dim}(N))$ such that $\|T\|\cdot\|T^{-1}\|\leq \lambda$.

$\bf{\text{Exercise:}}$

Show that the following are equivalent for a Banach space $X$.

This appears as a remark in Raymond Ryan's "Introduction to Tensor Products of Banach Spaces".

(1) For all $\epsilon > 0$, $X$ is an $\mathcal{L}_{1,1+\epsilon}$ space.

(2) For all $\epsilon > 0$, and for all finite-dimensional subspaces $M\subset X$, there exists a finite-dimensional subspace $N\subset X$ containing $M$ and an isomorphism $T\in B(N,\ell_{1}^{k})$ (where $k=\text{dim}(N)$) such that for all $x\in N$, $\left|\|Tx\| - \|x\|\right| \leq \epsilon\|x\|$.

After unwinding all the quantifiers, the $(2)\Rightarrow (1)$ direction was straightforward.

$\bf{\text{Sketch:}}$

Let $\epsilon > 0$ be given, and $M$ be a finite-dimensional subspace of $X$. Then choose $\delta > 0$ so that $\frac{2\delta}{1 + \delta} < \epsilon$. By assumption, there exists $N\subset X$ containing $M$ and an isomorphism $T\in B(N,\ell_{1}^{k})$ such that for all $x\in N$, $\left|\|Tx\| - \|x\|\right| \leq \epsilon\|x\|$.

Using the triangle inequality and the given condition on $T$, it follows that $\|T\|\leq 1 + \delta$ and $\|T^{-1}\|\leq \frac{1}{1-\delta}$. Therefore $\|T\|\cdot \|T^{-1}\|\leq 1 + \epsilon$, by the way $\delta$ was chosen.

$\bf{\text{My Problem:}}$

But the $(1)\Rightarrow (2)$ direction has me totally stumped. When I try to apply the hypothesis given in $(1)$, I can't seem to get the conclusion of $(2)$. Is there some obvious thing I'm missing? or is there some slippery trick?

roo
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1 Answers1

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Let $\varepsilon >0$ and a finite dimensional subspace $M \subset X$ be fixed.

Let us also fix $\delta >0$; we will see later what value we should choose for $\delta$. Then there exists a finite dimensional space $N$ such that $M \subset N$ and an isomorphism $T \in B(N, \ell_1^k)$ such that $\Vert T \Vert \cdot \Vert T^{-1} \Vert <1 + \delta$.

In fact, to simplify our expressions a little, we can assume that $\Vert T \Vert =1$ and $\Vert T^{-1} \Vert < 1 + \delta$. (If we replace $T$ by $\tilde T= \frac{T}{\Vert T \Vert}$, we obtain maps verifying this condition).

We then have that $$ \frac{1}{1+\delta} \Vert x \Vert \leq \Vert Tx \Vert \leq \Vert x \Vert. $$ Subtracting $\Vert x \Vert$ from each term, we obtain $$ \left(1 -\frac{1}{1+\delta} \right) \Vert x \Vert \leq \Vert Tx \Vert - \Vert x \Vert \leq 0. $$

But if we choose $\delta = \frac{\varepsilon}{1-\varepsilon}$, we can rewrite the above line as $$ | \Vert Tx \Vert - \Vert x \Vert | \leq \epsilon \Vert x \Vert. $$

Tom Cooney
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