Intersection of balls in normed vector spaces

Question

We have to two balls with

$$\mathcal B_1= \left(\left(\begin{array}{c} 0\\0\end{array}\right),\mathbb R^2\right) = \{v\in\mathbb R^2: \left|\left|v-\left(\begin{array}{c} 0\\0\end{array}\right)\right|\right|_1\le1\}$$

$$\mathcal B_1= \left(\left(\begin{array}{c} 1\\1\end{array}\right),\mathbb R^2\right) = \{v\in\mathbb R^2: \left|\left|v-\left(\begin{array}{c} 1\\1\end{array}\right)\right|\right|_1\le1\}$$

with the 1-Norm

$$\left|\left|v\right|\right|_1=\sum^n_{i=1}\left|v_i\right|$$

Now I want to find out the intersection of these two balls.

$$\mathcal B_1\cap\mathcal B_2$$

What I have done so far is two set up two inequations with

$$\left|\left|\left(\begin{array}{c} v_1\\v_2\end{array}\right)-\left(\begin{array}{c} 0\\0\end{array}\right)\right|\right|_1 = \left(\begin{array}{c} v_1\\v_2\end{array}\right)=\left|v_1\right|+\left|v_2\right|\le 1$$

$$\left|\left|\left(\begin{array}{c} v_1\\v_2\end{array}\right)-\left(\begin{array}{c} 1\\1\end{array}\right)\right|\right|_1 = \left(\begin{array}{c} v_1-1\\v_2-1\end{array}\right)=\left|v_1-1\right|+\left|v_2-1\right|\le 1$$

How do I proceed at this point? How will I find $v_1$ and $v_2$ and afterwards the intersection?

Answer 1

From $|v_1|+|v_2|\le 1$, we learn $v_1\le 1$ and $v_2\le 1$, from $|1-v_1|+|1-v_2|\le 1$ we learn $v_1\ge 0 $ and $v_2\ge 0$. Using these restrictions $0\le v_1\le 1$, $0\le v_2\le v1$, the inieqqualities simplifiy to $v_1+v_2\le 1$ and $1-v_1+1-v_2\le 1$, i.e., $v_1+v_2=1$. Thus the intersection in $\{\,{t\choose 1-t}\mid 0\le t\le 1\,\}$.