Let $p:\mathbb R^2 \to \mathbb R$ be a norm so that $$ \|\vec x\| ={(|x_1|+|x_2|)\over 3}+{2\max(|x_1|,|x_2|)\over 3} ={{\|\vec x\|_1\over 3}}+{2\|\vec x\|_\infty\over 3}. $$ I need to graph the neighbourhood of radius $1$ around $(0,0)$: $V_1 ((0,0))$ with this norm, but I don't even know the points that are in this neighbourhood I really don't know how to geometrically visualize it .
I tried to separate the norm in to parts: I want that to find all $(x_1, x_2) \in \mathbb{R}^2$ that satisfy $$ {(|x_{1}|+|x_{2}|)\over 3}+{2\max(|x_1|,|x_2|)\over 3} < 1 $$ so: $$ \frac{|x_1|+|x_2|}{3} < \frac{1}{2} \qquad \text{and} \qquad \frac{2\max(|x_1|,|x_2|)}{3} < {1\over 2}. $$
I know that the first inequality is a rotated square (geometrically) and the second one is a square, but from this point I don't see how to find the points that satisfy the given norm and visualize it geometrically.
Note the expression for the norm can be altered to;
$$ ||(x_1, x_2)|| = \max \{|x_1|, |x_2|\} + \frac{\min \{|x_1|, |x_2|\}}{3} $$
Hence the graph you look for is points $(x_1, x_2)$ such that,
$$ \max \{|x_1|, |x_2|\} + \frac{\min \{|x_1|, |x_2|\}}{3} \lt 1 \iff 3 \max \{|x_1|, |x_2|\} + \min \{|x_1|, |x_2|\}\lt 3 $$
I'm not sure about this part. Needs verification:
From here I think you must map the following regions on the $(x, y)$ plane in the corresponding regions.
For $ |y| \ge |x| $; we get the equation to be $ 3|y| + |x| \lt 1 $
$$ 3 y + x \lt 3 \;\; \text{for } \;\; x \ge 0, y \ge 0 $$ $$ 3 y - x \lt 3 \;\; \text{for } \;\; x \lt 0, y \ge 0 $$ $$ - 3 y - x \lt 3 \;\; \text{for } \;\; x \lt 0, y \lt 0 $$ $$ - 3 y + x \lt 3 \;\; \text{for } \;\; x \ge 0, y \lt 0 $$
Now I think we need to similarly map $ 3|y| + |x| \lt 1 $ for $ |x| \ge |y| $.
Note however that the $8$ separate equations we get are confined to $8$ disjoint regions in the plane. The $8$ sub-quadrants or octants if you will. So I'm guessing it can be done.