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Let $g:\mathbb{R}^n\rightarrow\mathbb{R}$ be a positive and continuous function and $C\subset\mathbb{R}^n$ be a nonempty compact set. Is the mapping $$ G(x):=\frac{x}{g(x)} $$ is Lipschitz continuous on $C$, i.e., there exists $L>0$ such that $$ \|G(x)-G(y)\|\leq L\|x-y\|\quad \forall x,y\in C. $$ My attempt. If $g$ is Lipschitz continuous on $C$ then $G$ is also Lipschitz continuous on $C$. Indeed, since $g$ is Lipschitz continuous on $C$, there exits $l>0$ such that $$ |g(x)-g(y)|\leq l\|x-y\| \quad \forall x,y\in C. $$ Since $C$ is bounded, there exists $R>0$ such that $$ \|x\|\leq C\quad \forall x\in C. $$ Since $C$ is bounded and $g$ is continuous on $C$, by the Weierstrass theorem, there exist $m>0$ such that $$ g(x)\geq m \quad \forall x\in C. $$ For every $x,y\in C$ we have \begin{eqnarray} \|G(x)-G(y)\|&=&\left\|\frac{x}{g(x)}-\frac{y}{g(y)}\right\|\\ &=&\left\|\frac{x}{g(x)}-\frac{y}{g(x)}+\frac{y}{g(x)}-\frac{y}{g(y)}\right\|\\ &\leq&\left\|\frac{x}{g(x)}-\frac{y}{g(x)}\right\|+\left\|\frac{y}{g(x)}-\frac{y}{g(y)}\right\|\\ &=&\frac{\left\|x-y\right\|}{g(x)}+\frac{\|y\|}{g(x)g(y)}|g(x)-g(y)|\\ &\leq&\left(\frac{m+rl}{m^2}\right)\|x-y\|. \end{eqnarray} Hence, $G$ is Lipschitz continuous on $C$ with Lipschitz constant $$ L=\frac{m+rl}{m^2}. $$ Thank you for all kind help and support.

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  • $\begingroup$ you need $g$ to be Lipschitz for the proof to work. $\endgroup$ – DeepSea Jul 2 '15 at 22:51
  • $\begingroup$ Yes. I would like to know whether $G$ is Lipschitz continuous when $g$ is merely continuous (not Lipschitz continuous). $\endgroup$ – Blind Jul 2 '15 at 23:09
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Consider $$ g(x) = \begin{cases} 1, & x = 0, \\ 1 + x \sin(x^{-2}), & 0 < |x| \le \frac12, \\ 1 + \frac{1}{2} \sin (\frac{1}{4}), & x > \frac{1}{2}, \\ 1 - \frac{1}{2} \sin (\frac{1}{4}), & x < \frac{-1}2. \end{cases} $$ Then, $g$ is positive. Further, $G$ is not differentiable at $0$ and singular around $0$. Thus, $G$ is not Lipschitz continuous on any compact containing $0$.

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  • $\begingroup$ Note that $g$ is a continuous function. $\endgroup$ – Blind Jul 3 '15 at 1:02
  • $\begingroup$ @Blind: $g$ can be continuously extended to $0$ with $g(0) = 1$. $\endgroup$ – user251257 Jul 3 '15 at 1:05
  • $\begingroup$ Thank you for your solution. Could you explain clearer why $g(x)$ is a positive function? $\endgroup$ – Blind Jul 3 '15 at 1:07
  • $\begingroup$ $g$ is positive near $0$. As $x\sin(x^{-2}) \to 0$ for $x\to 0$. $\endgroup$ – user251257 Jul 3 '15 at 1:08
  • $\begingroup$ Here $g$ is positive on the whole space $\mathbb{R}^n$. $\endgroup$ – Blind Jul 3 '15 at 1:09

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