Let $f\in \mathbb{R}(x\_1,\ldots,x\_n)$ be a rational function. Suppose that $f$ is continuous on $\mathbb{R} ^n$. Must it be Lipschitz on the unit ball?


This question might be related to [Are continuous rational functions arc-analytic?](https://mathoverflow.net/questions/278008/are-continuous-rational-functions-arc-analytic)



Is a continuous rational function Lipschitz?
Consider the polynomial
$p(x,y)=(y^3-x^5)^2+(y-x^2)^8$ in the neighborhood of $(0,0)$. Apart from the strip $y^3/x^5\in(1/2,2)$, it is bounded from below by $C(x^2+y^2)^5$; within the strip it is bounded by $C|x|^{40/3}$, and this estimate is sharp. So the function
$$
 \frac{(x^2+y^2)^7}{p(x,y)}
$$
has a continuous extension, but on the curve $y^3=x^5$ it behaves like $x^{2/3}(1+o(1))$, hence it is non-Lipschitz in any neighborhood of $(0,0)$.


