Geometric interpretation of the norm $\|\vec x\|={(|x_1|+|x_2|)\over 3}+{2\max(|x_1|,|x_2|)\over 3}$

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,

$$\max \{|x_1|, |x_2|\} + \frac{\min \{|x_1|, |x_2|\}}{3} \lt 1 \iff 3 \max \{|x_1|, |x_2|\} + \min \{|x_1|, |x_2|\}\lt 3$$

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.

• thanks a lot!! just one question why can the norm be altered? – user128422 Aug 27 '14 at 2:09
• $$\|\vec x\|={(|x_1|+|x_2|)\over 3}+{2\max(|x_1|,|x_2|)\over 3} = \dfrac{|x_1|}{3} + \dfrac{|x_2|}{3} + {2\max(|x_1|,|x_2|)\over 3}$$ and the max iseither $|x_1|$ or $|x_2|$ whose fractions can be added and the remaining one is the minimum. – Ishfaaq Aug 27 '14 at 2:12
• you'll be happy that i've verified your ideas and graphed in my answer below. – Viktor Glombik Jun 1 at 21:31

Using $$\max(a,b) = \frac{|a| + |b| + |a-b|}{2}$$ we can rewrite your norm as $$\| (x,y) \| := \frac{|x| + |y]}{3}+ \frac{|x| + |y| + ||x| - |y||}{3}$$ Now we want to solve \begin{align} \frac{|x| + |y|}{3}+ \frac{|x| + |y| + ||x| - |y||}{3} < 1 \iff & \frac{2}{3}\big(|x| + |y|\big) + \frac{1}{3} \big||x| - |y|\big| < 1 \\ \iff & 4 | x | + 4 | y | + 2 \big||x| - |y|\big| < 6 \qquad (\ddagger) \end{align} We now have to distinguish some cases:

Case 1: $$x,y \ge 0$$. Then we have $$(\ddagger) \iff 4x + 4y + 2 | x - y | < 6 \qquad (\star)$$ Case 1.1: $$x \ge y \ge 0$$. Then we have $$(\star) \iff 4x + 4y + 2x - 2y < 6 \iff 6x + 2y < 6 \iff 3x + y < 3.$$

Case 1.2: $$y \ge x \ge 0$$. Then we have $$(\star) \iff 4x + 4y + -2x + 2y < 6 \iff x + 3y < 3$$ All the other cases can be done analogously, you obtain $$-x+3y<3$$ and $$-3x+y<3$$ for the second quadrant (going counter clockwise), $$-x-3y<3$$ and $$-3x-y < 3$$ for the third and $$x-3y < 3$$ and $$3x - y < 3$$ for the fourth quadrant. This traces out the following regular octagon:

This is the intersection of two rotated squares. Its area ($$\approx 3.02$$) is roughly $$\frac{3}{4}$$ of the area of the square $$[-1,1]^2$$. You can rewrite the octagon as $$\big\{ (x,y) \in \mathbb{R}^2: ax + by < 3, \ a,b \in \{\pm 1, \pm 3\}, |a| \ne |b| \big\}$$ and I'm not sure if this characterisation can be obtained more easily (with less or without case distinctions) from the above inequality.

Note: Your approach one part < 1/2 and other part < 1/2 gives some but not all of the solutions, do you know, why?

We can generalize this as follows: For $$a,b > 0$$ define the norm $$\rho: \mathbb{R}^2 \to \mathbb{R}, \ (x,y) \mapsto a \| (x,y) \|_1 + \frac{b}{2}\| (x,y) \|_{\infty}.$$ From this question we know this is well-defined. We can then rewrite the norm as $$\rho(x,y) = \left(|x| + |y| \right)\left(a + b\right) + b | | x | - | y | |.$$ Using the same procedure as above we find that the unit ball is given by $$\big\{ (x,y) \in \mathbb{R}^2: \lambda x + \mu y < 1, \ \lambda, \mu \in \{ \pm (a + 2b), \pm a \}, | \lambda | \ne | \mu | \big\}.$$

Essentially the same as other answers but phrased differently. If $$X=(x,y)$$, $$\|X\|:=\frac13\|X\|_1 + \frac23\|X\|_\infty$$, then we are looking for the (lets say open) ball of radius 1 around $$(0,0)$$, $$B:=V_1((0,0)) = \{X\in\mathbb R^2 : \|X\|<1\}$$ Observe the symmetries \begin{align} \|(x,y)\|&= \|(-x,y)\|,\\ \|(x,y)\|&= \|(x,-y)\|,\\ \|(x,y)\|&= \|(y,x)\|,\\ \|(x,y)\|&= \|(-x,-y)\|.\end{align} These imply that the set $$B$$ is symmetric across the $$y$$-axis, the $$x$$-axis, across the line $$y=x$$, and across the line $$y=-x$$ respectively. So it suffices to describe $$B$$ in the octant $$y\ge x\ge 0$$, where since $$\max(x,y)=y$$, $$\|X\| < 1 \iff \frac13(x+y) + \frac23 y<1$$

i.e. the set is described by 8 copies of $$y < 1-\frac x3,$$ giving $$(x,y)\in B \iff \begin{cases} y\ge x\ge 0, y < 1-\frac x3,\\ x\ge y\ge 0, x<1-\frac y3, \\ -y\ge x\ge 0, -y<1-\frac x3,\\ -x\ge y\ge 0, -x < 1-\frac y3,\\ \qquad \qquad \quad \vdots \end{cases}$$ Here's 6 of the octants (because Desmos has 6 colours) together with a full plot that relies on Desmos to interpret the implicit inequality: