Suppose I have two variables, $o$ ("old") and $p$ ("new"), where $p>o$ so $p/o>1$.
I have a convex cost function $C(o,p)$ of moving from old $o$ to the new $p$ which is convex: suppose quadratic,
$$ C(o,p) = \left(\frac{p}{o} - 1\right)^2 $$
This could of course be generalized to any parameter $\alpha > 0$ in the exponent, and that exponent parameter $\alpha$ would serve as a measure of convexity (or concavity for $\alpha < 1$):
$$ C(o, p) = \left( \frac{p}{o} - 1 \right)^\alpha $$
A graphical illustration here.
Taking $\alpha \to 0$, the cost function is a constant at one. That's not what I want!
Does a parameterizable function exist where there is a continuum from "very convex" to "less convex" and limiting to: does $o$ differ from $p$ at all? $$C(o,p) \; = \; \mathbb{I} \{ o \neq p \}$$ So that in the limiting case, the cost function equals 1 iff $o \neq p$, rather than always equaling one.
The goal here is to have a cost function of 'changing from $o$ to $p$' which can be parameterized by the "degree of convexity" in some sense, with as a limiting case a nonconvex cost function.
Writing up this question, as is so often the case, makes the answer obvious:
$$C(o,p) = \left( \frac{p}{o} - 1 \right)^\alpha \cdot \mathbb{I}\{o \neq p\}$$.
$\alpha$ parameterizes the degree of convexity, and as $\alpha \to 0$, we have exactly what I wanted. Obvious and silly ex post, but I'll leave this here in case anyone else is googling around...