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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.

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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 $$ |f(x)-f(y)|\le \sup_{\min\{x,y\} \le z\le \max\{x,y\}}|f'(z)||x-y|\le 3|x-y| \quad \forall x,y \in \mathbb{R}. $$

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    $\begingroup$ It may cost more paper to estimate $f'(x)$,but I'm sure it's bounded. $\endgroup$
    – C Weid
    Commented Dec 26, 2013 at 4:31
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    $\begingroup$ @CWeid Not really. $x\sin x^{-1}$ is bounded above and below by $1$. This is easy to see if you write any nonzero $x\in\Bbb R$ as $1/t$ and observe $t^{-1}\sin t$ is bounded by $1$ in absolute value. $\endgroup$
    – Pedro ♦
    Commented Dec 26, 2013 at 5:34
  • $\begingroup$ @mathematics2x2life Not true. The function is uniformly continuous. As I said above $|t^{-1}\sin t|\leqslant 1$. Then write $t=x^{-1}$. It follows that $|f'|\leqslant 3$, as Mercy correctly estimated. $\endgroup$
    – Pedro ♦
    Commented Dec 26, 2013 at 5:36

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