Show that the function $f$ is a contraction

Question

I want to show that the function $f:\mathbb{R}\rightarrow \mathbb{R}$ with \begin{equation*}f(x)=\frac{1}{2}\sin x-1\end{equation*} is a contraction.

A function $g : \Omega \rightarrow \Omega$ is a contraction, if it is Lipschitz-continuous with the Lipschitz-constant $L < 1$ in a vector norm.

So we have to show that $$\|f(x)-f(y)\|\leq L\|x-y\|$$ Which norm do we have to use? The maximum norm?

Answer 1

Your function is scalar. The norm to use is thus the scalar absolute value.

Answer 2

As it was pointed out, you have a scalar differentiable function. You can just use Lagrange's theorem to get the result: $$ f(x)-f(y) = f'(\xi) (x-y) \Rightarrow |f(x)-f(y)| = |f'(\xi)| |x-y| \leq \frac 12 |x-y|. $$