Almost isometric subspaces of $\ell_p$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T03:06:41Z http://mathoverflow.net/feeds/question/96832 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/96832/almost-isometric-subspaces-of-ell-p Almost isometric subspaces of $\ell_p$ Jan Veselý 2012-05-13T13:08:30Z 2012-06-29T16:23:33Z <p>1) Given $p\in (1,\infty)$. </p> <p>2) Let us <strong>fix</strong> two, <strong>non-isometric</strong> subspaces $X,Y\subseteq \ell_p$ isomorphic to $\ell_p$. </p> <p>3) <strong>Are there</strong> an $\varepsilon\in (0,1)$ and <strong>an isomorphism</strong> $S\colon X\to Y$ such that</p> <p>$$(1-\varepsilon)\|x\|\leqslant \|Sx\|\leqslant (1+\varepsilon)\|x\|$$</p> <p>holds for each $x\in X$?</p> http://mathoverflow.net/questions/96832/almost-isometric-subspaces-of-ell-p/96847#96847 Answer by juan for Almost isometric subspaces of $\ell_p$ juan 2012-05-13T19:16:13Z 2012-05-13T19:16:13Z <p>Of course there is an isomorphism $T:X\to Y$ so that there exists constants $a &lt; A$ 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 $\varepsilon&lt;1$ sufficiently close to $1$ so that this is true also.</p> http://mathoverflow.net/questions/96832/almost-isometric-subspaces-of-ell-p/100954#100954 Answer by Bill Johnson for Almost isometric subspaces of $\ell_p$ Bill Johnson 2012-06-29T16:23:33Z 2012-06-29T16:23:33Z <p>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 $M$.</p>