# Is there a geometric analog of absolute value?

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.

• Is the triangular inequality satisfied ? Jun 28, 2020 at 22:15
• 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.
– Jam
Jun 28, 2020 at 22:22
• @hamam_Abdallah I believe it is if you consider positive $x$ only.
– Jam
Jun 28, 2020 at 22:26
• The length of a vector is an absolute value. Jun 29, 2020 at 10:05
• Interesting question, but my initial reaction is "Have you thought about re-stating the problem in terms of the variables $y_i$, where $y_i = \log x_i$"? Jun 29, 2020 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.

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.