# Smooth functions are uniformly approximated by pieces of circle arcs?

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

• 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 $|y-f(x_i^*)|$. Commented Dec 23, 2022 at 4:13
• 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. Commented Dec 23, 2022 at 10:37
• @mr_e_man Yes indeed Commented Dec 23, 2022 at 14:46
• @mihaild Thanks. I just added the compact condition. Commented Dec 23, 2022 at 14:48

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.