Is there a non-standard definition of an absolute value? So I was thinking about how we can define a number $x$ such that $|x|=-1$. This topic was already talked about so many times as we can see here or there. So I wanted to consider it from a different angle.
Firstly, I tried to define what an absolute value on $\mathbb{R}$ is. I want to say that the function $f: \mathbb{R} \to \mathbb{R}$ with the following properties is an absolute value function.

*

*$\forall x\in \mathbb{R},f(x)\geq 0$

*$\forall x\in \mathbb{R}, f(x)=0 \Leftrightarrow x=0$

*$\forall x,y\in \mathbb{R}, f(xy)=f(x)f(y)$

*$\forall x,y\in  \mathbb{R}, f(x+y) \leq f(x)+f(y)$
I would like to know if there is a function with those properties other than the standard $f(x)=|x|$. If there is, what condition do we need so that $|x|$ will be the only function with such properties? Do we need $f(f(x))=f(x)$ or $f(-x)=f(x)$?
 A: Let me say a few words in addition to the trivial counter example I gave in the comments. Let's make a list of observations based on the properties:

*

*$f(0)=0$, by property 2.

*$f(1)=[f(1)]^2$, and $f(1)\neq 0$ (by property 2) so $f(1)=1$.

*Similarly, $1=f(1)=[f(-1)]^2$, so $f(-1)=1$ or $f(-1)=-1$, but by property 1, we need $f\geq 0$, so we must have $f(-1)=1$.

*Hence, $f(-x)=f(-1)f(x)=f(x)$, so $f$ is even.

As a result, in order to understand the behavior of $f$, we may restrict attention to the interval $[0,\infty)$. As mentioned in the comments, it is possible to have a discontinuous $f$ which satisfies all four properties, and we know that the absolute value function is continuous, so continuity of $f$ is an extra assumption which we have to make.
Next, property (3) together with continuity, $f(0)=0$ and $f(1)=1$ is a very classical functional equation (I'm not an expert here so I'm not sure what it's called). But essentially, from this it follows that there is an $\alpha>0$ such that for all $x\geq 0$, $f(x)= x^{\alpha}$. Hence, by evenness of $f$, it follows that for all $x\in\Bbb{R}$, $f(x) = |x|^{\alpha}$.
We would like to somehow force $\alpha =1$. Well, observe that when restricted to $(0,\infty)$, we have

*

*$f$ is strictly concave if $0<\alpha<1$, and hence $f(x+y)< f(x)+f(y)$ for all $x,y\in (0,\infty)$

*$f$ is strictly convex if $1<\alpha$, and hence $f(x+y)> f(x)+f(y)$ for all $x,y\in (0,\infty)$.

So, if we want our $f$ to satisfy property (4), we must have $0<\alpha \leq 1$.

In other words, consider the following statements about a function $f:\Bbb{R}\to\Bbb{R}$

*

*$f\geq 0$

*for every $x\in\Bbb{R}$, $f(x)=0$ if and only if $x=0$.

*for every $x,y\in\Bbb{R}$, $f(xy)=f(x)f(y)$.

*for every $x,y\in\Bbb{R}$, $f(x+y)\leq f(x)+f(y)$.

*$f$ is continuous.

What my discussion above shows is that these five conditions imply there exists a $0<\alpha \leq 1$ such that for all $x\in\Bbb{R}$, $f(x)=|x|^{\alpha}$. Conversely, given any $0<\alpha\leq 1$, the function $f(x):= |x|^{\alpha}$ satisfies these $5$ conditions. Because of this equivalence, these five properties are not enough to single out $\alpha =1$.
If you add in a sixth condition that


*for every $x\in \Bbb{R}$, $f(f(x))=f(x)$,

then yes, this is necessary and sufficient to ensure $\alpha =1$. So, the sixth property is essential.
But of course, note that we can safely remove property (4), and replace it with $(6)$, because (1),(2),(3),(5) is equivalent to the existence of some $\alpha>0$ such that $f(x)=|x|^{\alpha}$. If we add in (6), this forces $\alpha=1$, so (4) is redundant as part of the assumptions, and it comes out for free.
