Is a Banach space $X$ Lipschitz equivalent to the metric quotient $X/B$, where $B$ is the closed unit ball? Recall that the metric quotient $X/B$ is defined as follows: first we consider the equivalence relation $\sim$ on $X$ that identifies all points of $B$, then we define on the set of all equivalence classes, $X/B$, the distance 
$$
d(\tilde{x},\tilde{y}):=\min\{\|x-y\|, d_{\|\cdot\|}(x,B)+d_{\|\cdot\|}(y,B)\}, 
$$
where $d_{\|\cdot\|}(x,B)=\inf \{\|x-z\|,z\in B\}$ is the usual norm distance from the point $x$ to the set $B$.
I've been trying to show that the mapping $E:X\setminus\{0\} \rightarrow (X/B)\setminus \{\tilde{0}\}$ defined by 
$$
E(x):=\frac{\|x\|+1}{\|x\|}x
$$
 is Lipschitz and has Lipschitz inverse, what would solve the problem. Roughly speaking what $E$ does is to shift semi-lines starting at the origin one unit away from it. 
It is quite easy to show that $E^{-1}$ is 1-Lipschitz, but I have been having trouble trying to show that $E$ is Lipschitz, even though it seems simple. So my first question is:

Is $E$ a Lipschitz function? If so, what is its Lipschitz constant?

In the case of a negative, I ask: 

Are $X$ and $X/B$ Lipschitz equivalent?

 A: The map $E$ is not Lipschitz. Indeed, let $x\in X $ be a vector of norm $1+\delta$. If $\|x-y\|<\delta$, then $d(\tilde x,\tilde y)=\|x-y\|$, because the other part of the minimum is at least $\delta$, i.e., is of no help. If we pick $y$ so that also $\|y\|=1+\delta$, then 
$$\|E^{-1}x-E^{-1}y\| = \frac{\delta}{1+\delta} \|x-y\|
=\frac{\delta}{1+\delta} d(\tilde x,\tilde y) $$
Since $\delta>0$ can be arbitrarily small, $E$ is not Lipschitz. 
Moreover, $X/B$ is not Lipschitz equivalent to $X$ when $X$ is finite-dimensional (and has dimension greater than $ 1$). The reason is that $X/B$ is not a doubling metric space, while any finite-dimensional normed space  is doubling. Since the doubling property is preserved under bi-Lipschitz maps and is inherited by subsets, the claim follows. 
Here is a proof that $X/B$ is not doubling. Let $\delta>0$ be small and consider a subset $N$  of the sphere $S_{1+\delta}\{x:\|x\|=1+\delta\}$ such that $\|x-y\|\ge 2\delta$ whenever $x\in N$, $x\ne y$. It is easy to see that $d(\tilde x,\tilde y)=2\delta$ for all distinct points $x,y\in N$. By making $\delta$ small enough, we can create such sets $N$ of arbitrarily large size $\#N $ (indeed: pick any finite subset of the unit sphere, let $\delta$  be the smallest distance between its elements, and rescale this subset to put it on $S_{1+\delta}$). This is where I need $X$ to have at least two dimensions, so that  the unit sphere has infinitely many points.
The set  $N$ is contained in a closed ball of radius $2\delta$ (with respect to  the quotient metric), but cannot be covered by fewer than $\#N $ closed balls of radius $2\delta/3$. Since $\# N$ can be arbitrarily large, the doubling condition fails. (Note that "$r/2$" in the definition of a doubling space can be replace by $r/4$ by applying the definition twice). 
I do not know about the infinite-dimensional case, but at least the above shows that there is no "easy" Lipschitz equivalence like the radial push $E$; any equivalence must essentially use the infinitude of dimensions.
