Is there a geometric analog of absolute value?

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.

Tags:

Absolute Value

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