Let $E$, $F$ be vector subspaces of $\mathbb{R}^d$. We define: \begin{equation*} \rho (E,F) = \max \left\{ \max_{v \in S_E} d(v, F) \, , \, \max_{w \in S_F} d(w, E) \right\} \, , \end{equation*} where $d(v, F) = \inf_{f \in F} \| v - f \|$, $d(w, E) = \inf_{e \in E} \| w - e \|$ are the usual point-set distances and $S_E$, $S_F$ are the unit spheres of each subspace.
I'm reading a text that implicitly assumes that $\rho$ is a metric between subspaces. At first glance it really resembles the usual Hausdorff metric, but there's a catch, which is: the point-set distances are maximized over the spheres, but calculated in relation to the whole subspaces.
Thus, whenever I try to prove that this is a metric, I get stuck in the same part of the triangle inequality, which is the following. It is not hard to prove by standard metric space arguments that, given $E,F,G$ subspaces: \begin{equation*} d(v, F) \leq d(w,F) + \| v - w \| \qquad \text{ for every $v \in E$, $w \in G$ } \end{equation*} Now, the next standard step would force me to restrict $w$ to $S_G$: \begin{equation*} d(v, F) \leq \max_{w \in S_G} d(w, F) + \| v - w \| \leq \rho(G, F) + \| v - w \| \end{equation*} yielding \begin{equation*} d(v, F) - \rho(G, F) \leq \| v - w \| \qquad \text{ for every $v \in E$, $w \in S_G$.} \end{equation*} However, now I can no longer relate this with the infimum over $w \in G$ to produce $d(v, G)$, since it is in general smaller than the infimum restricted to the sphere. I even proved that $d(v, G) = \inf_{w \in B_G} \| v - w \|$ when $v \in S_E$, but restrict to the ball is also not enough.
Any insight either into the proof or of a simple counterexample would be appreciated!
If $\Bbb{R}^d$ is equipped with the Euclidean metric, then $$\rho(E, F) = H(B_E, B_F),$$ which easily shows $\rho$ is a metric, assuming you know $H$ is a metric.
Suppose $E$ is a subspace of $\Bbb{R}^d$, and let $P_E$ be the projection operator onto $E$. Then $P_E$ is linear, with $I - P_E = P_{E^\perp}$. Both of these linear operators have norm $1$ or $0$, achieving the latter if and only if the corresponding subspace is non-trivial.
Recall that the norm of a linear map $T : V \to W$ between normed linear spaces can be expressed in the following equivalent ways: $$\|T\| = \max_{x \in B_V} \|Tx\| = \max_{x \in S_V} \|Tx\|.$$ This comes from the fact that, if $x \in B_V$ maximises $\|Tx\|$, then $x / \|x\| \in B_V$, and $$\|Tx\| \ge \left\|T\left(\frac{x}{\|x\|}\right)\right\| = \frac{\|Tx\|}{\|x\|} \implies \|x\| \ge 1 \text{ or } \|Tx\| = 0.$$ But, this either means $\|Ty\| \le \|Tx\| = 0$ for all $y$, i.e. $T = 0$ or $\|x\| = 1$, since $x \in B_V$. So, either $x \in S_V$, or $T$ achieves its maximum everywhere on the ball, including the sphere.
Thus, if we consider the restriction $P_{E^\perp}|_F$ of $P_{E^\perp}$ to a subspace $F$, we have $$\max_{x \in S_F}d(x, E) = \max_{x \in S_F} \|x - P_E(x)\| = \max_{x \in S_F} \|P_{E^\perp}(x)\| = \|P_{E^\perp}|_F\| = \max_{x \in B_F}\|P_{E^\perp}(x)\| = \max_{x \in B_F}d(x, E).$$ Further, if $\|x\| \le 1$, then $\|P_E(x)\| \le \|P\|\|x\| \le 1$, hence $P_E(x) \in B_E$. Hence, if $\|x\| \le 1$, then $$d(x, B_E) = \min_{y \in B_E} \|x - y\| \le \|x - P_E(x)\| = \min_{y \in E} \|x - y\| = d(x, E) \le d(x, B_E),$$ i.e. $d(x, B_E) = d(x, E)$. Putting this together, we get \begin{align*} \rho (E,F) &= \max \left\{ \max_{v \in S_E} d(v, F), \max_{w \in S_F} d(w, E) \right\} \\ &= \max \left\{ \max_{v \in B_E} d(v, F), \max_{w \in B_F} d(w, E) \right\} \\ &= \max \left\{ \max_{v \in B_E} d(v, B_F), \max_{w \in B_F} d(w, B_E) \right\} \\ &= H(B_E, B_F). \end{align*}