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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.

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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.

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

enter image description here

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 $$ \big\{ (x,y) \in \mathbb{R}^2: \lambda x + \mu y < 1, \ \lambda, \mu \in \{ \pm (a + 2b), \pm a \}, | \lambda | \ne | \mu | \big\}. $$

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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: enter image description here

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