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Trying to find a reparameterization of a function from $f(y, z, \ldots)(x)$ to $f(y(a_1), z(a_2), \ldots)(x)$ so that for all $x \in [r, t]$ we have $$ |f(y(a_1), z(a_2), \dots)(x) - f(y(b_1), z(b_2), \dots)(x)| \le \|a - b\|_1. $$ It seems similar to a bounded operator from functional analysis if you look at it as mapping from the vector $a$ to the function $f(\dots)$. With the first space having the $L^1$ norm and the second the $\sup$ norm. I haven't been able to find anything based on that googling though. Since it doesn't require it be linear just have norm bounded by a linear function.

I'm pretty sure there are functions that can't be parameterized this way, but I'm interested in being able to identify them and construct these reparameterizations.

For instance one I've had trouble proving if it works is $f(a, b)(x) = \frac{a}{1 + e^{-bx}}$. Specifically for the $b$ parameter.

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  • $\begingroup$ I would say a necessary and sufficient condition for existence of homeos $\mathbb R\to R$, $y_1,\dots,y_n$, is that $f:\mathbb R^n\times [r,t]\to\mathbb R$ is of bounded variation in each variable $y_1,\dots,y_n$, uniformly on bounded sets. Note that $C$ can be absorbed and put equal to $1.$ $\endgroup$
    – Pietro Majer
    Commented Sep 16, 2023 at 9:07
  • $\begingroup$ (and continuous) $\endgroup$
    – Pietro Majer
    Commented Sep 16, 2023 at 9:19
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    $\begingroup$ Do you not want the second $f$ in the displayed inequality to be evaluated at $x$? $\endgroup$
    – Gerry Myerson
    Commented Sep 17, 2023 at 1:35
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    $\begingroup$ @GerryMyerson good catch I'll fix that. $\endgroup$
    – ruler501
    Commented Sep 17, 2023 at 3:21

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