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.
Best Answer
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.