Csdrhrt

We call a norm $|cdot|$ on $mathbb{R}^n$ absolutely monotonic if
$$
|a_i| leq |b_i|, i=1,cdots,n implies |a| leq |b|.
$$

What's an example of norm on $mathbb{R}^2$ that's not absolutely monotonic?

This is an exercise left to the reader -- so presumably not too difficult. But it's giving me some trouble.

The usual suspects that come to mind, i.e. the $ell^p$ norms, $p in [1,infty]$, are all absolutely monotonic.

Consider a sheer $A(x,y):=(x,y-x)$ and then the norm $|(x,y)|:=|A(x,y)|_1$. Then $|(1,0)|=2$ but $|(1,1)|=1$.

Interestingly though, for every norm $|cdot|:mathbb{R}^2to[0,infty)$ there exists a linear map $A:mathbb{R}^2tomathbb{R}^2$ such that $|cdot|circ A$ is an absolutely monotonic norm.

To prove this, note that a norm is absolutely monotonic if (and only if, may I add) the smallest axis-aligned rectangle containing the unit disk touches the unit circle at the intersections with the $x$- and $y$-axis. This is a consequence of the triangle inequality. So if we find vectors $v$ and $u$ on the unit sphere such that the unit disk is contained in the parallellogram $[-1,1]vtimes[-1,1]u$, then the unique linear map that sends $v$ to $(1,0)$ and $u$ to $(0,1)$ suffices.

In order to find such $v$ and $u$, we will use the intermediate value theorem. For any angle $theta$ let $v(theta)$ be the unique vector on the unit sphere with angle $theta$ with respect to the origin. The uniqueness follows from the scaling property and the non-degeneracy of norms, and the continuity follows from the triangle inequality. Then let $u(theta)$ be the vector on the unit sphere furthest to the left of the line generated by $v(theta)$. A little bit of a problem here is that $u(theta)$ is not always uniquely defined, and $thetamapsto u(theta)$ is only continuous at points where it is uniquely defined. However, for now just pretend that it is a well-defined continuous function, and I will come back to this later.

We now find that $mathbb{R}vtimes[-1,1]u$ contains the unit disk. Most importantly the intersection of the unit disk with $(1,infty)vtimes[-1,1]u$ is fully contained in either $(1,infty)vtimes[-1,0]u$ or $(1,infty)vtimes[0,1]u$, due to the triangle inequality. If for some $theta$ this intersection lies in one of the two, then if we move $theta$ such that the new $v(theta)$ is where the old $u(theta)$ was, the intersection will lie in the other one. For some $theta$ in between, we must therefore have that the intersection lies in neither. Therefore the unit disk will be contained in $[-1,1]vtimes[-1,1]u$.

Finally, to come back to the little problem. For such problematic $theta$ you can imagine fixing $v(theta)$ while moving $u(theta)$ continuously along the set of all possible values.

This is a little bit heuristic, but I hope it is convincing anyways.

I would like to give an answer similar to Smiley-Craft but from a different point of view:

The standard norm on $mathbb{R}^2$ is given by $|(a,b)|= sqrt {a^2+b^2}$. We can define a similar norm using a different basis.

Given $u,v$ a basis of $mathbb{R}^2$. We can define the norm of $w$ by $$|w| = sqrt{a^2+b^2}$$ where $w=au+bv$.

It is an easy exercise to show that this is indeed a norm (same proof as in the standard norm case).

Since we can change coordinates it will be really easy to define a non-monotonic norm:

For instance choose $u=(frac{1}{2},0)$ and $v=(1,1)$ then $|(1,0)| = 2$ while $|(1,1)| = 1$.

Side note: It is an interesting question whether every norm is absolutely monotonic with respect to some basis. I would guess the answer is yes but I don't know how to approach this question yet.

Here's a way to come up with an answer using as little geometric intuition as possible.

The standard norm on $mathbb{R}^2$ can be written as:

$$ leftVert begin{pmatrix} a \ b end{pmatrix} rightVert^{;2} mapsto begin{pmatrix}a \ bend{pmatrix}^T begin{bmatrix} 1 & 0 \ 0 & 1end{bmatrix} begin{pmatrix} a\ b end{pmatrix} tag{1} $$

or

$$ leftVert begin{pmatrix} a \ b end{pmatrix} rightVert^{;2} mapsto a^2 + b^2 tag{2} $$

Which makes it immediately clear that, for arbitrarily chosen $(a_1, b_1)$ and $(a_2, b_2)$ .

$$ frac{a_1^2 le a_2^2 ;;;;text{and};;;; b_1^2 le b_2^2}{a_1^2 + b_1^2 le a_2^2 + b_2^2} tag{3} $$

If we stare at (1) and (2), it suggests that by adding a cross term we can "penalize" negative components, so let's pick a simple one, let's call the new norm $nu$.

$$ leftVert begin{pmatrix} a \ b end{pmatrix} rightVert^{;2}_{,nu} mapsto begin{pmatrix}a \ bend{pmatrix}^T begin{bmatrix} 1 & 1 \ 0 & 1end{bmatrix} begin{pmatrix} a\ b end{pmatrix} tag{4} $$

$$ leftVert begin{pmatrix} a \ b end{pmatrix} rightVert^{;2}_{,nu} mapsto a^2 + ab + b^2 tag{5} $$

Let's pick $(a_1, b_1) = (2, 2)$ and $(a_2, b_2) = (-3, 3)$ .

$$ leftVert begin{pmatrix} 2 \ 2 end{pmatrix} rightVert^{;2}_{,nu} = 4 + 4 + 4 = 12 tag{6} $$

$$ leftVert begin{pmatrix} -3 \ 3 end{pmatrix} rightVert^{;2}_{,nu} = 9 - 9 + 9 = 9 tag{7} $$

So, $(2, 2)$ and $(-3, 3)$ indeed demonstrates that $nu$ is not absolutely monotonic.

All that's left is to verify that it's an actual norm.

The norm axioms are:

$$ nu,(alpha v) = |alpha| ,nu,(v) tag{8} $$
$$ nu,(u+v) le nu,(u) + nu,(v) tag{9} $$
$$ nu,(v) = 0 implies v = vec{0} tag{10} $$

(8) falls out of the fact that matrix multiplication is linear. (10) falls out of the invertibility of $left[begin{smallmatrix}1& 1\0& 1end{smallmatrix}right]$

I think the most straightforward way to prove (9) is to note that the matrix $left[begin{smallmatrix}1& 1\0& 1end{smallmatrix}right]$ is upper triangular and hence its eigenvalues with multiplicity are the multiset ${1, 1}$ . Since all of its eigenvalues are positive, the matrix is positive definite and therefore the quadratic form $nu,(u) = u^T left[begin{smallmatrix}1 & 1 \0 & 1end{smallmatrix}right] u$ is a norm.