Polynomial growth implies locally Lipschitz?

Question

Let $f:\mathbb{R}^m\to\mathbb{R}^m$ satisfy $\|f(x)\|\le c\|x\|^n$ for some (re-edit:) $n\in\mathbb{N}$ with some constant $c>0$. Is $f$ locally Lipschitz?

I see that it is around $x=0$: $\|f(x)-f(0)\|\le \frac{c\|x\|^n+c0}{\|x\|}\|x\|=c\|x\|^{n-1}\|x\|\le \delta \|x\|$ for a suitable $\delta$.

Apparently, a polarization argument is a way to go but I do not know how to apply it for $\|f(x)-f(x')\|$.

Answer 1

A counterexample is $$f(x)=\begin{cases} \|x\|^{n-1}x, \|x\|\in\mathbb Q\\ -\|x\|^{n-1}x, \|x\|\not\in\mathbb Q \end{cases}. $$

Answer 2

I think the function $f$ from $\mathbb{R}$ to $\mathbb{R}$ define by $f(x)=x \sin (1/x)$ is a counter example to your problem.

Answer 3

A continuous nowhere differentiable function such as $$\sum_{n=1}^\infty {\sin(2^n x)\over n^2}$$ is bounded but it most certainly is not locally lipschitz.

Answer 4

The Weierstrass function $W(x)$ presented in baby Rudin is a continuous bounded function on $\mathbb R$ such that

$$\limsup_{h\to 0}\left |\frac{W(x+h)-W(x)}{h}\right| =\infty$$

for every $x\in \mathbb R.$ The function $f(x) = xW(x)$ satisfies $|f(x)|\le c|x|$ on $\mathbb R.$ It also satisfies

$$\limsup_{h\to 0}\left |\frac{f(x+h)-f(x)}{h}\right| =\infty$$

for all $x\ne 0.$ Exercise: Prove this. It follows that $f$ fails to be Lipschitz in every nonempty open interval, including the ones containing $0.$ This function is therefore "nowhere locally Lipschitz".