Bilipschitz embedding of the unit ball of $c_0$ into $\ell_1$ This is a follow-up to this question of mine.
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 $c_0$?
 A: 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.

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*Yves Raynaud, Espaces de Banach superstables, distances stables et homeomorphismes uniformes
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 $k$.
