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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.

  1. $\forall x\in \mathbb{R},f(x)\geq 0$
  2. $\forall x\in \mathbb{R}, f(x)=0 \Leftrightarrow x=0$
  3. $\forall x,y\in \mathbb{R}, f(xy)=f(x)f(y)$
  4. $\forall x,y\in \mathbb{R}, f(x+y) \leq f(x)+f(y)$

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 $f(-x)=f(x)$?

Calvin Khor
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Kaira
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    Does this help? https://en.wikipedia.org/wiki/Norm_(mathematics) – morrowmh Mar 21 '21 at 06:16
  • @MichaelMorrow I knew about norms, but I didn't want to get too complicated. – Kaira Mar 21 '21 at 06:20
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    $f(x)=0$ can't work because of your property #2. It turns out that any norm on a one-dimensional vector space (such as $\mathbb{R}$) is equivalent to the absolute value norm. – morrowmh Mar 21 '21 at 06:23
  • @MichaelMorrow Yeah I noticed the second I posted it. I know about the equivalence since it is finite-dimensional, but how about multiplicativity? – Kaira Mar 21 '21 at 06:25
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    I doubt this is enough. Consider $f(x):=\begin{cases}0 &\text{if $x=0$}\1 &\text{if $x\neq 0$}\end{cases}$. This satsifies all four properties, yet is not the absolute value. Though I suspect (but haven't verified) if you require continuity of $f$, then it has to equal the absolute value. Note by the way, regarding your last sentence, that property (3) already forces $f$ to be an even function. – peek-a-boo Mar 21 '21 at 06:37
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    the above is possible because your condition 3 is not the same as what the wikipedia post above says (it says $f(xy) = |x|f(y)$, where $||$ denotes the normal absolute value of a scalar). The correct wikipedia link is therefore https://en.wikipedia.org/wiki/Absolute_value_(algebra) – Calvin Khor Mar 21 '21 at 06:40
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    that page says that if you add the property of being Archimedian to $f$ then $f$ must be the usual absolute value. I don't have the expertise to guide further; perhaps posts like https://math.stackexchange.com/questions/279683/non-trivial-valuation-of-mathbb-r?rq=1 are a start – Calvin Khor Mar 21 '21 at 06:52

1 Answers1

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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 $f:\Bbb{R}\to\Bbb{R}$

  1. $f\geq 0$
  2. for every $x\in\Bbb{R}$, $f(x)=0$ if and only if $x=0$.
  3. for every $x,y\in\Bbb{R}$, $f(xy)=f(x)f(y)$.
  4. for every $x,y\in\Bbb{R}$, $f(x+y)\leq f(x)+f(y)$.
  5. $f$ 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

  1. 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.

peek-a-boo
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    setting $F(r) = \log f ( e^r)$ turns condition 3 into $F(r+s) = F(r)+F(s)$; this is the "Cauchy functional equation". The only continuous solutions are $F(r)=\alpha r$. Then $f(x) = e^{F(\log x)} = e^{\alpha \log x} = x^\alpha $. (at the level of $F$, alpha could be negative but that's easily ruled out somewhere) – Calvin Khor Mar 21 '21 at 08:13
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    @CalvinKhor oh of course it's named after Cauchy, why am I not surprised :) thank you for also adding in the quick proof. right, $\alpha <0$ is ruled out by requiring continuity on $\Bbb{R}$ (in particular being bounded near the origin). And $\alpha = 0$ is certainly not possible. – peek-a-boo Mar 21 '21 at 08:15