What does $M$-equivalent basis mean?

Question

I am studying Petr Hájek's "Biortogonal Systems in Banach Spaces" for my thesis. In Lemma 1.25, there is an item whose result is

$\mathrm{(iii)}$ $\{x_i\}_{i=m_k+1}^{m_{k+1}}$ is $(1+\frac{1}{1+k})$-equivalent to the canonical basis of $l_2^{k+1}$, $k=0,1,2\dots$,

where $m_k:= k(k+1)/2$, for every $k\in \mathbb{N}$, and $(x_n)_{n=1}^\infty$ is a sequence in a Banach space $X$.

Well, I know what equivalent basis means, but the book proves this item defining an isomorphism $T$ such that

$\{Tx_i\}_{i=m_k+1}^{m_{k+1}}$ is orthogononal in $l_2^m$,

where $m>k+1$, and such that $\|T\| \leq (1+1/k)$ and $\|T^{-1}\|=1$.

---

Using the isomorphism definition for equilavent basis (two basic sequences $(x_n)_{n\in\mathbb{N}}$ and $(y_n)_{n\in\mathbb{N}}$ are equivalent iff there is an ismorphism $T$ such that $Tx_i = y_i$ for all $i$), I thought the natural way to define "$M$-equivalent basis" would be to have an ismorphism $T$ such that $\|T\|\|T^{-1}\| = M$ or even $\max\{\|T\|,\|T^{-1}\|\} = M$. However, it doesn't match what the book meant, since $\{Tx_i\}_{i=m_k+1}^{m_{k+1}}$ is only orthogonal and not ortonormal and therefore is not a candidate for a canonical basis for $l_2^{k+1}$.

---

I did not find a definition for it in the book. I can understand that this meant to be "a basis $(1+\frac{1}{1+k})$-almost like the canonical basis of $l_2^{k+1}$", but I can't give a concrete definition for this equivalence relation. What does $M$-equivalent basis mean, exactly?

Answer 1

I think your problem is not with the definition of "equivalent" but the definition of "canonical". Following the citation [DaJo73a] to its source, On the existence of fundamental and total founded biorthogonal systems in Banach spaces by Davis and Johnson, we find that the original statement of the lemma is:

(iii) In the real restriction of $X$, $(x_i)_{i=n_k+1}^{n_{k+1}}$ is $\left(1 + \frac1{k+1}\right)$-equivalent to an orthogonal basis in the $k+1$ dimensional real Euclidean space $l_2^{k+1}$ for $k=0,1,2,\dots$.

So if this is all the lemma guarantees, then I don't think there is any remaining confusion.

But just to be clear, in the paper, Davis and Johnson give the following definition of $\lambda$-equivalence, which matches what you expected:

For $\lambda \ge 1$, the basic sequence $(x_n)$ is said to be $\lambda$-equivalent to the basic sequence $(y_n)$ provided that the mapping taking $x_n$ to $y_n$ extends to a linear homeomorphism $T$ of $[x_n]$ onto $[y_n]$ with $\|T\| \|T^{-1}\| \le \lambda$.