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?
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- 2$\begingroup$ I don't understand: your hypothesis is that the denominator of $f$ doesn't vanish on $\mathbb{R}^n$, is it? Then its partial differentials w.r.t. the $x_i$ satisfy the same, so they are continuous, so bounded on the compact closed unit ball, so $f$ is Lipschitz there. But that's too obvious, so you must have meant something different: what did I misunderstand? $\endgroup$Gro-Tsen– Gro-Tsen2022-12-19 17:10:51 +00:00Commented Dec 19, 2022 at 17:10
- 4$\begingroup$ @Gro-Tsen. It is not so obvious. The denominator may actually vanish. For instance the rational function $f(x,y)=\frac{x^2y}{x^2+y^2}$ is continuous, though its partial derivatives are only bounded, but not continuous. $\endgroup$Denis Serre– Denis Serre2022-12-19 17:26:29 +00:00Commented Dec 19, 2022 at 17:26
- 3$\begingroup$ @DenisSerre Ah, OK. I would say that this function “has a continuous extension” to $\mathbb{R}^n$, not that it is continuous. $\endgroup$Gro-Tsen– Gro-Tsen2022-12-19 18:12:32 +00:00Commented Dec 19, 2022 at 18:12
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1 Answer
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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 $(0,0)$.
