Let $x,y,z,w \in \mathbb{R}^n$. Assume $x^\top w \geq 0$, $y^\top w \geq 0$, $z^\top w = 0$ and $\|x\| = \|y\| = \|z\| = 1$, for a general norm $\|\cdot \|$. Moreover, define $p(x,w)$ to be the projection, with respect to the norm $\|\cdot \|$, of $x$ onto the hyperplane $\{v \in \mathbb{R}^n \ | \ v^\top w = 0\}$, that is, $p(x,w) \in \arg \min_{v^\top w = 0} \|x-v\|$ [1]. Also, define $z(x,w)$ to be a unit norm vector on the hyperplane which is closest to $x$, that is, $z(x,w) \in \arg \min_{\|v\|=1, \ v^\top w = 0} \|x-v\|$.
I would like to prove the following:
$$ \|x - z(x,w)\| \leq \|y - z(y,w)\| \iff \|x - p(x,w)\| \leq \|y - p(y,w)\| \tag{1}. $$
In words, if the distance from $x$ to $z(x,w)$ is smaller than the distance from $y$ to $z(y,w)$, so is the distance from $x$ to its projection onto the hyperplane that $z(x,w)$ lies on.
My end goal is to show that $\arg\min_{i = 1,\ldots,N} \|x - z(x,w_i)\| = \arg\min_{i = 1,\ldots,N} \|x - p(x,w_i)\|$, that is, instead of looking for the vector $z(x,w_i)$ closest to $x$, we can actually minimize the distance from the closest hyperplane, which has the closed form solution $$ \|x - p(x,w)\| = \frac{|x^\top w|}{\|w\|_*}, $$ where $\| \cdot \|_*$ is the dual norm to $\| \cdot \|$ [1].
Also, I pretty sure $(1)$ holds when $\| \cdot \| = \| \cdot \|_2$ (the standard euclidean norm).
My questions are:
- Is (1) true for a general norm?
- If not, does it hold for some special case (other than $\| \cdot \| = \| \cdot \|_2$)? Or for some related problem, for instance, if the projection $p(x)$ is with respect to $\|\cdot \|_*$ instead of $\|\cdot \|$?
[1] Mangasarian, Olvi L. "Arbitrary-norm separating plane." Operations Research Letters (1999).