# Are continuous rational functions arc-analytic?

Let $X\subseteq\mathbb{R}^n$ be a smooth semi-algebraic set (for simplicity we can assume $X=B(0,r)$ is a small ball around the origin). A function $f:X\rightarrow \mathbb{R}$ is called a continuous rational function if it is continuous (w.r.t the euclidean topology) and there exists a Zariski open dense subset $U\subseteq \overline{X}^{zar}$ and a regular function $g$ on $U$ such that $f\restriction _{U\cap X}=g\restriction _{U\cap X}$. A function $f:X\rightarrow \mathbb{R}$ is called arc-analytic if $f\circ \gamma:(-\epsilon,\epsilon)\rightarrow \mathbb{R}$ is analytic for every analytic arc $\gamma :(-\epsilon,\epsilon)\rightarrow X$. Trying to get a hold of the hierarchy of functions on semi-algebraic sets I stumbled upon the following questions:

1) Is every continuous rational function arc-analytic? if not, what is the simplest counterexample?

2) Is there any natural homomorphism from the ring of arc-analytic functions on $X$ (or the stalk of the correspoing (pre-?) sheaf at a point $x\in X$) to some sort of "algebraic" ring? say the algebraic closure of $\mathbb{R}((x_1,...,x_n))$ or something like that? or are these inherently "non algebraic" functions (unlike all of their analytic friends)?

If $X$ is all of $\mathbf R^n$ then the answer to the first question is "yes", I think. Indeed, for any continuous rational function $f$ on $\mathbf R^n$ there is a stratification of $\mathbf R^n$ in Zariski-locally closed subsets such that the restriction of $f$ to any stratum is regular (Théorème 4.1 of the paper "Fonctions régulues"). It implies that $f\circ \gamma$ is meromorphic and continuous, hence analytic.
As for the second question, by a Theorem of Bierstone-Milman (Theorem 1.1 of the paper "Arc-analytic functions"), any semi-algebraic arc-analytic function $f$ on $\mathbf R^n$ is blow-Nash. This means that there is a finite sequence of blow-ups with smooth algebraic centers $\pi\colon X\rightarrow\mathbf R^n$ such that $f\circ \pi$ is Nash. Hence, such functions are algebraic over $\mathbf R(x_1,\ldots,x_n)$, I suppose.
• thanks you for your answer! Regarding both of my questions: what changes if we take not $\mathbb{R}^n$, but say an open ball around the origin? It seems that your answers to both of my questions crucially rely on $\mathbb{R}^n$ being Zariski closed (or am I wrong about this?) My original motivation was to compare the stalks of the corresponding sheaves at the origin (are these even sheaves? with respect to what topology(/ies)?) – Anonymous Coward Aug 8 '17 at 18:49
• I do think that both statements still hold when one replaces $\mathbf R^n$ by an open ball in $\mathbf R^n$. However, as you also suggest, the above arguments should be adapted. As for your questions about the sheaves of continuous rational functions, or of arc-analytic functions, you may consider asking them in a separate question. – Johannes Huisman Aug 9 '17 at 20:19