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Let $X\subset \mathbb{R}$ be a compact set. Then for a smooth function $f\colon X\to \mathbb{R}$ and a fixed number $\epsilon>0$, can we find numbers $x_1, x_2, \dots, x_n \in X $ and circle arcs $O_1, O_2, \dots$ such that

$$\sup_{x^*_i \in [x_i,x_{i+1}] \\ (x_i^*, y )\in O_i\ } \|(x^*_i,y)-(x_i^*,f(x_i^*))\|\tag{1} <\epsilon $$

is true?

Essentially, I look for ways to approximate a smooth curve by pieces of circle arcs. Also, is it possible to realize this for all continuously differentiable functions i.e. functions with continuous derivatives?

I also really care about the uniform approximation. In other words, can we find a distance $\delta >0$ so that $(1)$ is true for $x_1,\dots,x_n$ whenever $|x_i-x_{i+1}|<\delta$?

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  • $\begingroup$ I expect this problem to be equivalent but simpler with parabolas (simple polynomial functions) instead of circles (involving square roots). Also, your 2D norm simplifies to a 1D norm $|y-f(x_i^*)|$. $\endgroup$
    – mr_e_man
    Commented Dec 23, 2022 at 4:13
  • $\begingroup$ If $X$ is non-compact and you want finite number of arcs for given precision, then no. If $X$ is compact, or you allow infinitely many arcs, then yes - just approximate function with combination of segment indicators, and approximate each segment with an arc. $\endgroup$
    – mihaild
    Commented Dec 23, 2022 at 10:37
  • $\begingroup$ @mr_e_man Yes indeed $\endgroup$
    – user628623
    Commented Dec 23, 2022 at 14:46
  • $\begingroup$ @mihaild Thanks. I just added the compact condition. $\endgroup$
    – user628623
    Commented Dec 23, 2022 at 14:48

1 Answer 1

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If the graph of the function is a compact set (and this is the case if function is continuous and defined on a compact set), then, for any $\epsilon >0$ one can cover the graph by discs centered at $(x_i,f(x_i))$ where $x_1,...$ is a countable dense set in the domain of $f$. Then, from the compactness of the graph we can extract finite sub-cover, say $D(x_1,f(x_1)), ..., D(x_n,f(x_n))$. Next we take boundaries of these discs removing unnecessary arches to get a curve consisting of "parts" of circles with radius $\epsilon$. If the domain of $f$ is not compact, then one divides it into compact sets and uses the above reasoning.

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