Lipschitz-continuous $f(x)=x^2\cdot \sin\left(\frac{1}{x}\right)$

Question

How to prove that $f$ is globally Lipschitz-continuous $$ f:\mathbb{R}\longrightarrow \mathbb{R}$$

$$ f(x) = \left\{ \begin{array}{c l} x^2\cdot \sin\left(\frac{1}{x}\right) & ,\quad x\neq0\\ 0 & ,\quad x=0 \end{array} \right.$$

Any hints would be appreciated.

Answer 1

Notice that $$ f'(x)=\begin{cases} 2x\sin\frac1x-\cos\frac1x &\text{ if } x\ne0\\ 0 &\text{ if } x=0 \end{cases}. $$ Therefore $$ |f'(x)|\le 2|x|\cdot\left|\sin\frac1x\right|+\left|\cos\frac1x\right|\le 3 \quad \forall x\ne 0. $$ Hence, $$ |f'(x)|\le 3 \quad \forall x \in \mathbb{R}. $$ It follows that $$ |f(x)-f(y)|\le \sup_{\min\{x,y\} \le z\le \max\{x,y\}}|f'(z)||x-y|\le 3|x-y| \quad \forall x,y \in \mathbb{R}. $$