Where is the error in the following proof that $f\colon[-1, 1] \to \mathbb{R}$ given by $f(x)= \cos(x)$ is not a contraction.

Question

Where is the error in the following proof that $f\colon[-1, 1] \to \mathbb{R}$ given by $f(x)= \cos(x)$ is not a contraction.

We have that $f'(x) = -\sin(x)$ and so for all $x \in [-1,1], |f'(x)| \leq 1$. So F is Lipschitz continuous with $K = 1$. Since $K$ is not less than 1, $f$ is not a contraction.

Answer 1

The problem is this line

So $F$ is Lipschitz continuous with $K=1$. Since $K$ is not less than $1$, f is not a contraction.

Note that while you proved that $K=1$ works, you did not prove that this is the best which works.

To inderstand the issue look at $f(x)=\frac{1}{2}x$. You have that $|f'(x)|<1000$ and hence $f$ is 1000-Lipschitz continuous. But this does not prove that $f$ is not a contraction, actually it is easy to see that $f$ is also $\frac{1}{2}$-Lipschitz continuous.

Answer 2

Note that $$ |f'(x)|=1$$ where $$ x=2k\pi \pm \pi/2$$

Since the domain is $[-1,1]$ we have $$ |f'(x)|<1$$ on the domain.