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-2
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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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0
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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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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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