Is $f(x)=x \sin(x) $ a Lipschitz function?

Question

I have tried to calculate $f'$:

$$f'(x)=\sin(x)+x \cos(x)$$

$f'$ is unbounded, so I can't use the Lagrange theorem

So, I have used this maggioration ($L \in \mathbb{R}, L>0$):

$$\lvert x \sin(x) \lvert \le \lvert x \lvert \le L \lvert x \lvert $$

Is it correct?

Thanks!

Answer 1

Hint :

suppose that $f$ is a Lipschitz function,

$$\exists k >0 | \forall (x,y) \in \mathbb{R}^2, |f(x)-f(y)|\leq k|x-y|$$ what about this inequality when $x=x_n=2\pi n+\pi/2$ and $y=y_n=2\pi n-\pi/2$ ? what happen when "n is big enough" ?

Answer 2

Suppose that there is a positiv $L$ such that

$|f(x)-f(y)| \le L|x-y|$ for all $x,y$. Then $|\frac{f(x)-f(y)}{x-y}| \le L$ for all $x \ne y$. This implies:

$$|f'(t)| \le L$$

for all $t$ ........