Characterizing equality in the *reverse triangle inequality*

Question

It is well known that the triangle inequality gives rise to the sometimes-called reverse triangle inequality $$\big|\|x\|-\|y\|\big| \leq \|x-y\|$$ in any normed space.

Question: Is there a characterization of equality at least when the norm is nice, say Eucldiean?

Answer 1

Case of real inner product space, $\|x\|^2 = x\cdot x$. $$ \|x\|^2 + \|y\|^2 - 2x\cdot y = \|x - y\|^2 = (\|x\| - \|y\|)^2 = \|x\|^2 + \|y\|^2 - 2\|x\|\|y\| $$ and $|x\cdot y| = \|x\|\|y\|$ iff $x,y$ are linearly dependent (equality in CS-inequality), so the condition is $x = \lambda y$ or $y= \lambda x$, $\lambda\ge 0$.

Answer 2

This follows since $\|x\|\leq \|y\|+\|x-y\|$ and $\|y\|\leq \|x\|+\|x-y\|,$ so if $|\|x\|-\|y\||=\|x-y\|,$ then exactly one of these two equalities is satisfied (provided $x\neq y$). Then the equality condition follows from the equality condition for the triangle inequality of the norm being employed.

In the case of the Euclidean norm, the equality condition for the triangle inequality is that $x-y=c y$, where $c\geq 0$, in which case $x=c'y,$ $c'\geq 1$; or that $x-y=-cx$, $c\geq 0,$ in which case $y=c'x,$ $c'\geq 1$, so this is also the equality condition for the reverse triangle inequality.

Thank you to max_zorn for pointing out my error.