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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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closed as off topic by Benoît Kloeckner, Loop Space, Andreas Thom, Bill Johnson, Ryan Budney Jul 30 '11 at 8:38

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What about $\mathbb C^2$ with the $\ell^2$ and $\ell^1$ norms? – Mikael de la Salle Jul 28 '11 at 12:04
@Mikael: You could have written that as an answer. – S.C. Jul 28 '11 at 12:26
OK, now add the (I assume) intended condition of "infinite-dimensional". – Gerald Edgar Jul 28 '11 at 12:30
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. – Bill Johnson Jul 28 '11 at 18:50

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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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 $v_j = w_j \sin \theta_j$. – Robert Israel Jul 28 '11 at 18:26

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