Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $x\in X$?

share|improve this question
    
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. –  Nik Weaver May 13 '12 at 13:42
    
I ask about an arbitrary pair $(X,Y)$. –  Jan Veselý May 13 '12 at 13:55
1  
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 $\ell_p$? –  Yemon Choi May 13 '12 at 14:05
    
Typo in comment: it should have read: "are you asking if subspaces isomorphic to $\ell_p$ must ..." –  Yemon Choi May 13 '12 at 14:06
    
That doesn't really help me understand what you're asking. (Nor does your edit that $X$ and $Y$ can't be isometric.) –  Nik Weaver May 13 '12 at 14:07
show 3 more comments

2 Answers

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.

share|improve this answer
1  
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. –  Adi Tcaciuc May 13 '12 at 20:00
    
I will not call this almost isometric because ε is not small. –  juan May 20 '12 at 9:33
add comment

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 $M$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.