This is a follow-up to [this question of mine](https://mathoverflow.net/questions/428242/isometric-embeddings-of-c-0-into-metric-spaces).


It is well-known that the Banach space $\ell\_1$ does not contain any *isomorphic* copies of $c\_0$. One can even go further and show that $\ell\_1$ does not contain any *bilipschitz* copy of $c\_0$. Is it however known whether $\ell\_1$ does not contain any bilipschitz copy of the unit ball $B\_{c\_0}$ of $c\_0$?



Bilipschitz embedding of the unit ball of $c_0$ into $\ell_1$
Yes, this is known. Raynaud showed that $B\_{c\_0}$ does not uniformly (in particular, bilipschitz) embed into any stable Banach space. $\ell\_1$ is stable.


* [Yves Raynaud, Espaces de Banach superstables, distances stables et homeomorphismes uniformes](https://link.springer.com/article/10.1007/BF02763170)


**Added later.**
In the comments Bill pointed out an easier proof. Here is an elementary argument that doesn't use differentiation theory.


Suppose $f:B\_{c\_0}\to \ell\_1$ with
$$\frac1C\|x-y\|\le \|f(x)-f(y)\|\le C\|x-y\|$$


Let $(s\_i)$ be the summing basis of $c\_0$. For $\vec{n}=(n\_1, \ldots, n\_k)$, let $x(\vec{n})=\frac1k \sum\_{i=1}^k s\_{n\_i}$.
By standard stabilization argument you can find $M\subseteq \mathbb N$ so that for all $\vec{n}\in M^k$, $f(x(\vec{n}))\approx h\_0+\sum\_{i=1}^k h\_{n\_i}$ where $h\_i$'s are successive blocks in $\ell\_1$ and the support of $h\_i$'s go to infinity as $n\_i\to \infty$ (except $h\_0$). Use Ramsey to stabilize the norms of $h\_i$'s. Considering $\vec{n}<\vec{m}$, $\|x(\vec{n})-x(\vec{m})\|\_{c\_0}=1$ we see that
$$\frac1C\le \|f(x(\vec{n}))-f(x(\vec{m}))\|\approx \|\sum\_{i=1}^k h\_{n\_i}-\sum\_{i=1}^k h\_{m\_i}\|=\sum\_{i=1}^k (\|h\_{n\_i}\|+\|h\_{m\_i}\|).$$


Now consider $n\_1<m\_1<\ldots<n\_k<m\_k$ then $\|x(\vec{n})-x(\vec{m})\|\_{c\_0}=\frac1k$. The blocks $h\_{n\_1}<h\_{m\_1}<\ldots h\_{n\_k}<h\_{m\_k}$ are interlacing but the norm of the sum of the difference in $\ell\_1$ is the same as above. Then


$$\sum\_{i=1}^k (\|h\_{n\_i}\|+\|h\_{m\_i}\|)=\|\sum\_{i=1}^k h\_{n\_i}-\sum\_{i=1}^k h\_{m\_i}\|\approx\|f(x(\vec{n}))-f(x(\vec{m}))\|\le \frac{C}{k}.$$


which yields a contradiction for large $k$.


