In $(\mathbb{R}^3, \|\cdot\|_1)$ consider the vector $x_0=(1,1,1)$. For any subspace $A\subset \mathbb R^3$ denote $$BA(x_0)=\{y^*\in A:\|y^*-x_0\|\leq\|y-x_0\|,\forall y\in A\}.$$ Give examples of non-trivial subspaces $A \subset\mathbb{R}^3$ such that:
a) $BA(x_0)$ is a singleton;
b) $BA(x_0)$ contains at least two points.
I'm thinking a line could work for $BA(x_0)$ being unique. For the other one I can only think of trivial subspaces. Any suggestions would be great!
A line should work in both cases.
If $A$ is the $z$-axis, that is, $A=\{(0,0,a): a\in\mathbb R\}$ then $BA(x_0)=\{(0,0,1)\}$, for $$ \|(1,1,1)-(0,0,a)\|_1= 2+|1-a| \geq 2 = \|(1,1,1)-(0,0,1)\|_1, $$ and whenever $a\neq 1$, $$ \|(1,1,1)-(0,0,a)\|_1= 1+1+|1-a| > 2. $$
If $A$ is the line where $y=-x$ and $z=1$, that is, $A=\{(a,-a,1): a\in\mathbb R\}$. Notice that, for any $a\in\mathbb R$, $$ \|(1,1,1)-(a,-a,1)\|_1 = |1-a| + |1+a|. $$ If $|a|\leq 1$, then $1-a, 1+a\geq0$, and $$ \|(1,1,1)-(a,-a,1)\|_1 = 1-a + 1+a=2. $$ If $a>1$, then $1-a<0$ and $1+a>0$, so $$ \|(1,1,1)-(a,-a,1)\|_1 = a-1 + 1+a = 2a > 2. $$ If $a<-1$, then $1-a>0$ and $1+a<0$, so $$ \|(1,1,1)-(a,-a,1)\|_1 = 1-a - 1-a = -2a > 2. $$ So we conclude that $BA(x_0)$ is the line segment $\{(a,-a,1): a\in[-1,1]\}$.