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.

  • 1
    $\begingroup$ Is the triangular inequality satisfied ? $\endgroup$ Commented Jun 28, 2020 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
    Commented Jun 28, 2020 at 22:22
  • $\begingroup$ @hamam_Abdallah I believe it is if you consider positive $x$ only. $\endgroup$
    – Jam
    Commented Jun 28, 2020 at 22:26
  • $\begingroup$ The length of a vector is an absolute value. $\endgroup$
    – eckes
    Commented Jun 29, 2020 at 10:05
  • 2
    $\begingroup$ 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
    Commented Jun 29, 2020 at 13:31

3 Answers 3


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.


There are some interesting properties of this "inverse absolute value"

So we define it on the positive reals as $|x|_\times = e^{|\ln(x)|}$

Then we can show that if $\ln(a) > \ln(b) > 0$ then

$$ \left|\frac{a}{b} \right|_{\times} = \frac{|a|_{\times}}{|b|_{\times}} $$

Similar to how if $a > b > 0$ then $|a - b| = |a| - |b|$.

We also have a "triangle inequality"

$$ |ab|_{\times} \le |a|_{\times} |b|_{\times} $$

Which suggests this can behave LIKE a metric but instead of addition your fundamental opertion is multiplication.

This gets more interesting once we consider exponentiation

$$ |a^b|_{\times} = e^{|b \ln(a)|} = e^{|b| |\ln(a)|} = |a|_{\times}^{|b|} $$

Notice that there's a mixture happening here between our new absolute value and the old one.

The natural question then to ask is: can we generalize Ostrowski's Theorem and try to classify the set of multiplicative absolute values?

We can define a multiplicative absolute value then as an operation $|\cdot |_{\times}$ which satisfies the following

$$|x |_{\times} \ge 1 \ \forall x \in \mathbb{Q}^+ \\ |x|_{\times} = 1 \rightarrow x = 1 \\ |xy|_{\times} \le |x|_{\times} |y|_{\times} \\ |x^y|_{\times} = |x|_{\times}^{|y|}$$

Where $|y|$ is an absolute value (in the normal additive sense) of your choice.

An interesting question then to ask are "Can we classify all the multiplicative absolute values? Are there any not equivalent in a certain sense to $e^{|\ln(x)|_j}$ where $|\cdot|_j$ is an additive absolute value (such as the p-adic absolute value)?


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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