If the value of something changes from $\;a\;$ to $\;b\;$, their relative difference can be expressed as a percentage: $$ \newcommand{\upto}{\mathop\nearrow} (D0) \;\;\; a \upto b \;=\; (b-a)/a \color{gray}{\times 100\%} $$

(In this question $\;a\;$, $\;b\;$, and $\;c\;$ are positive (non-zero) real numbers throughout, and $\;a \upto b\;$ is a real number.)

However, percentages are not anti-symmetrical, i.e., the above definition does not satisfy $$ (P0) \;\;\; a \upto b \;=\; -(b \upto a) $$ This could be fixed by instead defining $$ (D1) \;\;\; a \upto b \;=\; (b-a)/(b+a) $$ (By the way, does this difference measure have a standard name?)

However, neither of the above definitions add up: they don't satisfy the more general property $$ (P1) \;\;\; a \upto b \;+\; b \upto c \;=\; a \upto c $$ After some attempts I discovered that $$ (D2) \;\;\; a \upto b \;=\; \log_q (b/a) $$ (for an arbitrary base $\;q > 1\;$) does satisfy $(P1)$. And it also satisfies the following other properties of percentage $(D0)$ and of $(D1)$ which seem desirable for any relative difference: \begin{align} (P2) \;\; & a \upto a \;=\; 0 \\ (P3) \;\; & \textit{sgn}(a \upto b) \;=\; \textit{sgn}(b - a) \\ (P4) \;\; & (s \times a) \upto (s \times b) \;=\; a \upto b \text{ for all $\;s > 0\;$} \\ (P5) \;\; & a \upto b \to M \text{ for } a \to 0 \text{ for some $\;M\;$ (which may be $\;\infty\;$)} \\ (P6) \;\; & a \upto b \to N \text{ for } b \to \infty \text{ for some $\;N\;$ (which may be $\;\infty\;$)} \\ \end{align} (I mention $(P2)$ only for completeness; it follows of course from $(P0)$.)

My question: Apart from $(D2)$, what other definitions (if any) satisfy all of the above properties?

  • $\begingroup$ In (p4), do you mean $\forall s>0$, or $\exists s>0$? $\endgroup$ – TZakrevskiy Sep 19 '13 at 16:46
  • $\begingroup$ @tzakrevskiy Good point, I mean $\forall$. I have corrected this.Thanks! $\endgroup$ – Marnix Klooster Sep 19 '13 at 17:21

The general (real-valued) solution of axioms P1 and P2 is to take a function $F$ on the set $V$ of values that $a,b,\cdots$ could possibly assume, and define $a \upto b = F(b) - F(a)$.

If $F$ is injective, the converse of P2 is satisfied: $a \upto b = 0$ only for $a=b$.

If $F$ is increasing, P3 is satisfied.

If $F$ is continuous, P5 and P6 are satisfied. (Assume here that the set of values $V$ is an interval, or put the order topology on the set.)

This shows that there is a large family of solutions of the axioms, one for every increasing continuous function on $V$, if you do not assume homogeneity, P4. Homogeneity is a strong requirement that cuts down the space of continuous solutions to the logarithmic ones stated in the question.

Homogeneity of $a \upto b$ is the functional equation $F(sb) - F(sa) = F(b) - F(a)$. Solutions $F$ and $F+c$ are equivalent for constant $c$, and we can assume $F(1)=0$ by adjusting the constant. Taking $a=1$ this is $F(sb)=F(s)+F(b)$ and all continuous solutions (on intervals) are well-known to be multiples of the logarithm.

Assuming that you meant to work with open intervals of values, and $a \upto x$ to be an increasing continuous function of $x$, the not necessarily homogeneous solutions correspond to one parameter groups of homeomorphisms of the interval.


I'll rewrite your conditions in a slightly different way: we look for a function $$f:\Bbb R_+^2\to \Bbb R,$$ such that we have

1) "antisymetry" $$f(a,b)=-f(b,a),$$ 2) "reflexivity" $$f(a,a)=0,$$ 3) homogenity of first degree $$f(sa,sb)=f(a,b)$$ 4) "transitivity" $$f(a,b)+f(b,c)=f(a,c)$$ 5) "monotonicity" if $b>a$, then $$f(a,b)>0.$$

From (3) it's apparent that it's sufficient to define our function only on $\{(a,b)\in R_+^2: a^2+b^2=1\}$; in other words, our function depends only on the ratio $\frac ba$.

Thus, we can say that, in fact, our function is given by $$f(a,b) = g(\ln (b/a)),$$ where $$g:\Bbb R\cup \{\pm\infty\}\to \Bbb R.$$ The conditions rewrite

1) $g(x)=-g(-x)$.

2) $g(0) =0$.

3) already taken into consideration

4) $g(\ln (b/a))+g(\ln (c/b))=g(\ln(c/a))$

5) $g(t)>0$ for $t>0$.

We need to use (4), other conditions give nothing significant. It's equivalent to $g(x)+g(y) = g(x+y)$. This is a Cauchy's functional equation, there're some questions on Math.SE on this topic, for example this.

If we suppose that $g$ is continuous at some point, then it's continuous everywhere and is linear, i.e. giving essentialy the same solutions as you've already proposed. Other solutions are not explicit and require axiom of choice. However, wikisays that the graph of such functions is dense in $\Bbb R^2$, but in our case we know that $g(t)>0$ for $t>0$, so these "implicit" functions are ruled out.

We can finally conclude that the only possible case is $g(t) = kt$ for $k>0$, or $f(a,b)=k\ln(b/a)$.


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