# Convex up to a reparametrization

Given a function ${f\colon \mathbb R\rightarrow \mathbb R}$, we have several ways of checking if it is convex. But how to recognize the functions that are convex up to a reparametrization, meaning that there is a homeomorphism ${h}$ of the real line such that the composition ${g = f\circ h}$ is convex?

Let’s start by listing some necessary conditions for ${f}$ to have this property.

1. Since convex functions are continuous, ${f}$ must be continuous.

2. Convex functions are quasiconvex, meaning that each sublevel set ${U_t = \{x\colon g(x) < t\}}$ is convex (in this case, an open interval). This property is preserved under reparametrization. So, ${f}$ must be quasiconvex as well.

3. Nonconstant convex functions are unbounded above. So, ${f}$ must have this property too.

Is this all? No. For example, let ${f(x) = -x}$ if ${x< 0}$ and ${f(x)=\arctan x}$ if ${x\ge 0}$. This is a continuous quasiconvex function that is unbounded above. But it can’t be made convex by a reparametrization, because as ${x\rightarrow +\infty}$, it increases while staying bounded. This leads to another condition:

4. If ${f}$ is not monotone, then its limits as ${x\rightarrow+\infty}$ and ${x\rightarrow -\infty}$ must both be ${+\infty}$.

And still we are not done. The function

${f(x) = \begin{cases} x+1, \quad &x<-1 \\ 0, \quad & -1 \le x\le 1 \\ x-1, \quad & x>1 \end{cases} }$

satisfies all of 1–4. But it has a flat spot (the interval ${[-1,1]}$ where it is constant) which does not achieve its minimum. This can’t happen to convex functions: if a convex function is constant on an interval, then (by virtue of being above its tangent lines) it has to attain its minimum there. This leads to the 5th condition:

5. ${f}$ is either strictly monotone, or attains its minimum on some closed interval ${I_{\min}(f)}$ (possibly unbounded or a single point). In the latter case, it is strictly monotone on each component of ${\mathbb R\setminus I_{\min}(f)}$.

Finally, we have enough: together, 1–5 is the necessary and sufficient condition for ${f}$ to be convex up to a reparametrization. The proof of sufficiency turns out to be easy. A continuous strictly monotone function is a homeomorphism between its domain and range. The strictly monotone (and unbounded) parts of ${f}$ give us homeomorphisms such that the composition of ${f}$ with them is affine. Consequently, there is a homeomorphism ${h}$ of the real line such that ${f\circ h}$ is one of the following five functions:

1. ${g(x) \equiv 1}$ if ${I_{\min}(f) = \mathbb R}$
2. ${g(x) = \max(x,0)}$ if ${I_{\min}(f)}$ is unbounded, but not all of ${\mathbb R}$
3. ${g(x) = \max(|x|-1, 0)}$ if ${I_{\min}(f)}$ is a bounded nondegenerate interval
4. ${g(x) =|x|}$ if ${I_{\min}(f)}$ is a single point
5. ${g(x) =x}$ if ${I_{\min}(f)}$ is empty

Incidentally, we also see there are five equivalence classes of convex functions on ${\mathbb R}$, the equivalence being any homeomorphic change of variables.

What functions on ${\mathbb R^n}$ are convex up to reparametrization? Unclear. (Related Math SE question)

## 2 thoughts on “Convex up to a reparametrization”

1. You forget the monotone case which reparametrizes to $g(x)=x$ in your list of possibilities.

1. Thanks for pointing this out. I added the 5th case.

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