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 $|x_i-x_{i+1}|<\delta$?

  • $\begingroup$ 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


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .