Show that the function $\cos: \mathbb{R} \rightarrow \mathbb{R}$ that sends each $x \in \mathbb{R}$ to $\cos x$, is Lipschitz continuous

Question

Show that the function $\cos: \mathbb{R} \rightarrow \mathbb{R}$ that sends each $x \in \mathbb{R}$ to $\cos x$, is Lipschitz continuous.

Is the idea that the derivative between two points ($-\sin x $) is bounded by $[-1,1]$, and as such every derivative satisfies the Lipschitz condition by not being greater than some real value, $r_0 > 1$?

Answer 1

You can take $\epsilon = 1$.

Suppose that $|f(x)-f(x+h)|> h$ for some $x,h$. Then there is $x\in [x,X+h]$ such that $|f'(x)|=\frac{|f(x)-f(x+h)|}{|x+h-x|}> 1$ by the mean value theorem. This is a contradiction since $f'(z)\in [-1,1]$ for all $z\in \mathbb R$.

This same proof can be used to show every derivable function with bounded derivative $f:\mathbb R\rightarrow \mathbb R$ is lipschitz continuous.