Almost isometric embeddability implies isometric embeddability

Question

Consider the following situation: Suppose $X$ is a Banach space such that for each finite metric space $M$ and each $\epsilon > 0$ for which $M$ bi-lipschitz embeds into $X$ with Lipschitz constant $1+\epsilon$, then $M$ isometrically embeds into $X$.

Is there anything known about the properties of such Banach spaces?

It follows from results of Ball that $\ell_p$ has this property for each $p$ (any $n$ point metric space embedding into $\ell_p$ embeds into $\ell_p^{n \choose 2}$, thus a compactness argument gives us that $M$ embeds into $\ell_p$ isometrically by passing to the limit.) Similarly the result is true in Hilbert spaces (much more easily.)

(Balls argument gives the result slightly more generally, but it's not particularly important.)

I am pretty sure the result isn't true in general, or at least, I can see no reason why it should be true: I think the space $(\oplus \ell^2_{p_n})_2$ with $p_n$ tending to 1 from above and the Hamming cube provide a counterexample (with some maybe slight modifications). If this doesn't work, in general there's no reason compactness should be applicable anywhere without really really nice properties (like the $\ell_p$ spaces where we can cut dimensions.)

Edit 1: It might be worth noticing (very, very trivially) that $C[0,1]$ and $\ell_\infty$ have this property, however apart from this I can't think of any other examples

I also can not show that even that if two Banach spaces $X,Y$ have the property that $X \oplus_p Y$ has the property for any $p$. I don't think such a statement should be true in general, but that's a real guess.

Answer 1

The OP's intuition that $X:=(\oplus \ell^2_{p_n})_2$ with $p_n$ tending to $1$ (and $1<p_n < \infty$) provides an example is correct. Indeed, $X$, being an $\ell_2$ sum of strictly convex spaces, is strictly convex, and so does not contain an isometric copy of the four points $A = (1,1)$, $0=(0,0)$, $C=(0,1)$, $D=(1,0)$ in $\ell_1^2$ because $C$ and $D$ are both metric midpoints between $A$ and $0$.

Answer 2

The example can be generalized: if a Banach space $X$ is strictly convex (the unit sphere does not contain line segments), but is not uniformly convex (that is the unit sphere contains pieces which approach line segments), then it is another example for which $(1+\varepsilon)$-embeddability with an arbitrary $\varepsilon>0$ of a finite metric space does not imply its isometric embeddability. To see this one has to observe that the limiting space (like ultraproduct) will contain pairs of points with several different midpoints between them. Therefore some of finite metric spaces having several different midpoints for some pairs of points are $(1+\varepsilon)$-embeddable into $X$ with an arbitrary $\varepsilon>0$. On the other hand such metric spaces do not admit isometric embeddings into $X$. (I can make this argument more detailed, if needed.) This can be generalized to produce a general criterion in terms of ultraproducts (which does not seem very useful, since it just describes similar argument in more general terms).

Also, the last statement (about $Z=X\oplus_p Y$) can be proved using ultraproducts as follows: If $M$ is a finite metric space arbitrarily well embeddable into $Z$, then $M$ embeds isometrically into an ultrapower of $Z$. Write the corresponding sequences for $X$ and $Y$. They represent finite metric spaces which embed (by our assumption) isometrically into $X$ and $Y$, respectively. Combining these embeddings we get the desired isometric embedding of $M$ into $Z$.