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1) Given $p\in (1,\infty)$.

2) Let us fix two, non-isometric subspaces $X,Y\subseteq \ell_p$ isomorphic to $\ell_p$.

3) Are there an $\varepsilon\in (0,1)$ and an isomorphism $S\colon X\to Y$ such that

$$(1-\varepsilon)\|x\|\leqslant \|Sx\|\leqslant (1+\varepsilon)\|x\|$$

holds for each $x\in X$?

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  • $\begingroup$ Take $X = Y = l^p$? That works with $\epsilon = 0$. Or are you trying to find examples that fail for small $\epsilon$? I don't understand the question. $\endgroup$
    – Nik Weaver
    Commented May 13, 2012 at 13:42
  • $\begingroup$ I ask about an arbitrary pair $(X,Y)$. $\endgroup$
    – Jan Veselý
    Commented May 13, 2012 at 13:55
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    $\begingroup$ Could you please make the quantifiers in your question more precise? Are you asking if subspaces isomorphic to $\ell_1$ must necessarily be close in Banach-Mazur distance? (In which case, I think the answer is no.) Or are you asking if there exists a constant $C>1$ such that any subspace which has BM-distance $\leq C$ from $\ell_p$ is in fact isometric to $\ell_p$? $\endgroup$
    – Yemon Choi
    Commented May 13, 2012 at 14:05
  • $\begingroup$ Typo in comment: it should have read: "are you asking if subspaces isomorphic to $\ell_p$ must ..." $\endgroup$
    – Yemon Choi
    Commented May 13, 2012 at 14:06
  • $\begingroup$ That doesn't really help me understand what you're asking. (Nor does your edit that $X$ and $Y$ can't be isometric.) $\endgroup$
    – Nik Weaver
    Commented May 13, 2012 at 14:07

2 Answers 2

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Of course there is an isomorphism $T:X\to Y$ so that there exists constants $a < A$ such that $$ a \Vert x\Vert\le \Vert Tx\Vert \le A \Vert x\Vert$$ Now consider the operator $S = r T$ where $r >0$ we shall choose. We will have $$ a r \Vert x\Vert \le \Vert S x\Vert\le A r \Vert x\Vert$$ We want $A r = 1+\varepsilon$. So choose $r= \frac{1+\varepsilon}{A}$. Now we have $\Vert S x\Vert \le (1+\varepsilon)\Vert x\Vert$. Finally we want also that $ar >1-\varepsilon$, or that $a\frac{1+\varepsilon}{A}>1-\varepsilon$. It is clear that we may choose $\varepsilon<1$ sufficiently close to $1$ so that this is true also.

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    $\begingroup$ Yes, and the question as posed, has nothing to do with $l_p$. It is just rescaling an isomorphism $T$ such that $||T||$ is small, and consequently $||T^{-1}||$ is large. I suspect that he meant $1/(1+\epsilon)$ instead of $1-\epsilon, in which case *I think* it is not true if $S$ is required to be an isomorphism between $X$ and $Y$, but it is true for all $\epsilon>0$ if is not. $\endgroup$
    – Adi Tcaciuc
    Commented May 13, 2012 at 20:00
  • $\begingroup$ I will not call this almost isometric because ε is not small. $\endgroup$
    – juan
    Commented May 20, 2012 at 9:33
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The answer is no when $p\not= 2$. For any fixed $M$ you can take a finite dimensional subspace $E$ of $\ell_p$ such that the factorization constant through $\ell_p$ of the identity on $E$ is larger than $M$. Then $E\oplus_p \ell_p$ is isometrically a subspace of $\ell_p$ that is isomorphic to $\ell_p$ but the isomorphism constant is larger than $M$.

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