Is $|\sin x|$ a Lipschitz function?

Question

I would like to know how can I prove that the function: \begin{equation} f(x)=|\sin x|, \quad x\in \mathbb{R} \end{equation} is (or isn't) Lipschitz continuous.

I studied an example of the funtion $f(x)=|x|^{1/2}$ which is not a Lipschitz function on any interval containing zero, but the problem is that I do not know how to proceed with this one.

Answer 1

Since the cosine is bounded by $1$, by the mean value theorem, $|\sin x - \sin y| \le |x - y|$ for all $x,y \in \Bbb R$. Therefore, since $||a| - |b|| \le |a - b|$ for all $a,b\in \Bbb R$, we have

$$|f(x) - f(y)| \le |\sin x - \sin y| \le |x - y|$$

for all $x,y\in \Bbb R$. Hence, $f$ is Lipschitz continuous.

Answer 2

The easiest way to see this is to note that $|x|$ is a Lipschitz function as well as $\sin x$ (because the modulus of its derivative, i.e, $|\cos x|$ is bounded by $1$). Compositions of Lipschitz functions are Lipschitz.