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 $c_0$?
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.
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 $k$.
