I'm wondering whether there exists a geometric analog concept of absolute value. In other words, if absolute value can be defined as

$$ \text{abs}(x) =\max(x,-x) $$

intuitively the additive distance from $0$ to $x$, is there a geometric version

$$ \text{Geoabs}(x) = \max(x, 1/x) $$

which is intuitively the multiplicative "distance" from $1$ to $x$?

Update: Agreed it only makes sense for $Geoabs()$ to be restricted to positive reals.

To give some context on application, I am working on the solution of an optimization problem something like:

$$ \begin{array}{ll} \text{minimize} & \prod_i Geoabs(x_i) \\ \text{subject to} & \prod_{i \in S_j} x_i = C_j && \forall j \\ &x_i > 0 && \forall i . \end{array} $$

Basically want to satisfy all these product equations $j$ by moving $x_i$'s as little as possible from $1$. Note by the construction there are always infinite feasible solutions.

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    $\begingroup$ Is the triangular inequality satisfied ? $\endgroup$ – hamam_Abdallah Jun 28 at 22:15
  • $\begingroup$ Interesting idea but I'd consider revising the definition to $\operatorname{geoabs}(x)=\operatorname{sign}\left(x\right)\max\left(\left|x\right|,\left|\frac{1}{x}\right|\right)$, which would take $x$ over $(-\infty,-1]$ and $1/x$ on $(-1,0)$ instead of the other way round as you have. Your version has small or large negative values multiplicatively close $1$ while $-1$ is the most distal from $1$, which should be reversed. $\endgroup$ – Jam Jun 28 at 22:22
  • $\begingroup$ @hamam_Abdallah I believe it is if you consider positive $x$ only. $\endgroup$ – Jam Jun 28 at 22:26
  • $\begingroup$ The length of a vector is an absolute value. $\endgroup$ – eckes Jun 29 at 10:05
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    $\begingroup$ Interesting question, but my initial reaction is "Have you thought about re-stating the problem in terms of the variables $y_i$, where $y_i = \log x_i$"? $\endgroup$ – JonathanZ supports MonicaC Jun 29 at 13:31

To make things easier I'll set $f(x)=\max\{x,-x\}$ and $g(x)=\max\{x,\frac{1}{x}\}$.

So we understand that $f:\mathbb{R}\to \mathbb{R}^+$ and $g: \mathbb{R}^+\to \mathbb{R}^+$.

Then $\exp(f(x))=g(\exp(x))$. So we can use this to translate some properties like the triangle inequality.

$$ g(xy)=g(\exp(\log(xy)))=\exp(f(\log(xy)))=\exp(f(\log(x)+\log(y))) $$ $$ \leq \exp(f(\log x)+f(\log y))=\exp(f(\log x))\exp(f(\log y))=g(\exp(\log(x))g(\exp(\log(x)) $$ $$ =g(x)g(y) $$

So $g(xy)\leq g(x)g(y)$ and we have the multiplicative triangle inequality.

Of course this is easier to show directly but the method emphasizes the "transfer".

Another good sign is $g(x)=1$ if and only if $x=1$.

All in all it looks like you're moving between $(\mathbb{R},+)$ and $(\mathbb{R}^+,\cdot)$ with $\log$ and $\exp$. So a nice question.

I'm sure there's more to say.

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Another way (maybe cleaner) to see it : let us consider

  • $G_1 = (\mathbb{R},+,\|\cdot\|_1)$ the additive group of real numbers equipped with a norm : for all $x\in G_1$, $\|x\|_1 = |x| = \max \{x,-x\}$
  • $G_2 = (\mathbb{R_+^*},\cdot,\|\cdot\|_2)$ the multiplicative group of (strictly) positive real numbers equipped with a norm defined using the norm of $G_1$ : for all $x\in G_2$, $\|x\|_2 = \|\ln x\|_1 = \ln \max \{x,1/x\}$

The map $\exp\colon G_1\to G_2$ is therefore by construction a group isometric isomorphism (with $\ln\colon G_2\to G_1$ its inverse). Indeed, for all $x\in G_1$ $$ \|x\|_1 = \|\exp x\|_2 $$

You can check that $\|e_i\|_i = 0$ where $e_i$ is the identity element of $G_i$ (here $e_1 = 0$ and $e_2 = 1$).

If you forget the $\ln$ map in the definition of $\|\cdot\|_2$, it is not anymore a norm.

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