I am sorry if the question is easy but can one give me an example of a pair of Banach spaces, say $X$ and $Y$, $X$ isomorphic to $Y$ such that $X$ has no isometric copy of $Y$ neither $Y$ has isometric copy of $X$ inside?
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1$\begingroup$ What about $\mathbb C^2$ with the $\ell^2$ and $\ell^1$ norms? $\endgroup$– Mikael de la SalleCommented Jul 28, 2011 at 12:04
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$\begingroup$ @Mikael: You could have written that as an answer. $\endgroup$– C.S.Commented Jul 28, 2011 at 12:26
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2$\begingroup$ OK, now add the (I assume) intended condition of "infinite-dimensional". $\endgroup$– Gerald EdgarCommented Jul 28, 2011 at 12:30
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2$\begingroup$ This question is too elementary for MO, but it does point in an direction that I find interesting. In "The diameter of the isomorphism class of a Banach space, Annals Math. 162 (2005), 423-437", Odell and I show that if $X$ is a separable infinite dimensional Banach space, then for every $K$ there is a space $Y$ that is isomorphic to $X$ but there are other spaces isomorphic to $X$ which do not $K$-embed into $Y$. Whether the same is true for every non separable space is open. $\endgroup$– Bill JohnsonCommented Jul 28, 2011 at 18:50
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1 Answer
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Space $X = l_1$ is separable, therefore has an equivalent norm which is strictly convex. Let $Y$ be the space with that norm. Now every subspace of $Y$ is strictly convex, and so it remains to show that for any $2$-dimensional subspace of $X$, there is a line segment in the unit sphere.
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$\begingroup$ Except that is not the case. Suppose $u$ and $v$ are a basis of your 2-dimensional subspace, and scalars are real. If $a u + b v$ and $c u + d v$ are in the unit sphere, the line segment joining them is in the unit sphere iff there is no index $j$ for which $(a u + b v)_j$ and $(c u + d v)_j$ have opposite sign. Take some $w \in l_1$ with all $w_j > 0$, let the sequence $\{\theta_j\}$ be dense in $[0, 2 \pi]$, and take $u_j = w_j \cos \theta_j$ and $v_j = w_j \sin \theta_j$. $\endgroup$ Commented Jul 28, 2011 at 18:26