Shortest distance as measured in norm $||\cdot ||$ from point to a sphere in norm $||*||$ I recently found this theorem, which is used in some clustering algorithms:
Let $x,v \in \mathbb{R}^p$, $r>0$, $||\cdot ||_{\ast}$ be a given norm on $\mathbb{R}^p$ and $\partial B_{||\cdot||_{\ast}}(v,r) = \{ y\in\mathbb{R}^p: ||y-v||_{\ast}=r \}$ be the closed ball of radius $r$ centered at $v$. Then the shortest distance, as measured by $||\cdot ||_{\dagger}$, from any point in $\partial B_{||\cdot||_{\ast}}(v,r)$ to $x$ is $\left|\ ||x-v||_{\ast}-r \right|.$
I know that this holds for $||\cdot ||_2,$ but how would one prove that this holds for any two pairs of norms $||\cdot ||_{\ast}$ and $||\cdot ||_{\dagger}$?
I've tried tracking down the original article, because all the articles I've seen that have cited it do exactly that, just cite it, without giving the proof.
This isn't homework or anything like that, I'm just curious as to what approach one could use when trying to prove this?
EDIT: Source of this claim: Page 65.
Now that there is a counterexample in one of the answers, I suppose that there is actually only one norm here. I apologize for the confusion, but as you will see from the source, the two norms are marked differently, which led me to the conclusion that they are not the same.
 A: Your claim is wrong even if one of the norms is the Euclidean norm. Take $p=2, \ r=1, \ v=(0,0)$ and $x=(2,2)$. We take the following two norms
$$ \Vert (y,z) \Vert_1 := \sqrt{y^2+z^2}, \quad \Vert (y,z) \Vert_2:= \frac{1}{2} \max\{\vert y \vert, \vert z \vert \}.$$
Then 
$$ \min_{\Vert w \Vert_1=1} \Vert w - x \Vert_2 \geq  \frac{1}{2} > 0 = \vert \ \Vert x \Vert_2 - 1 \vert = \vert \ \Vert x - v \Vert_2 - r\vert.$$
Where we used
$$ \Vert x \Vert_2 =  \Vert (2,2) \Vert_2 = \frac{1}{2}\max\{ \vert 2 \vert, \vert 2 \vert \} = 1.$$
The first inequality follows from the following considersation.
$$1= \Vert (w_1, w_2) \Vert_1 \Rightarrow 1 = 1^2 = w_1^2 + w_2^2 \Rightarrow \max \{ \vert w_1\vert, \vert w_2 \vert \} \leq 1.$$
Thus, if $\Vert (w_1, w_2) \Vert_1 = 1 $, then
$$ \Vert (w_1, w_2) - x \Vert_2  = \frac{1}{2} \max \{ \vert w_1 - 2\vert, \vert w_2 - 2 \vert\} =\frac{1}{2} \max \{ 2- w_1, 2-w_2\} 
\geq \frac{1}{2}.$$
A: With the same norms a demonstration uses the following. Using norm subadditivity, for any $b$ on the sphere:
$$\|b\| \le \|b-x\|+  \|x\|$$
and 
$$\|x\| \le \|x-b\|+  \|b\|$$
so all in all:
$$\|x-b\| \ge \max(\|b\|-\|x\|,\|x\|-\|b\|) = |\|x\|-r|\,.$$
Unfortunately, this does not seem to work for different norms in general. Take $ p=2$, a $l_q$ and a $l_r$ two norms, $q,r \ge 1$. Take $v$ at the origin (because it plays little role here), $r=1$, $x=(1,1)$ on the diagonal. Then the minimum $l_r$ norm to the $l_q$ ball is $m=2^{1/r}(1-1/2^{1/q})$. And your formula gives $n=2^{1/q}-1$. 
The two quantities are equal if and only if $ q =r$. Note that the above generalizes with higher dimension $p\ge 2$. 
[EDIT] I have just found a paper by Bezdek in 1995: Shell-prototype clustering models. The theorem given (cropped for the record) is:

Apparently there is only one "given" norm. I guess that the $\|*\|$ stands for a wildcard-type notations meaning "any argument", while $\|\cdot\|$ is the functional notation.
