Is $sin(x)$ a contraction on $[0,1]$?

Question

I see that the derivative of the function is $cos(x)$ and in $[0,1]$ this can take the value $cos(0)=1$ implying that is not a contraction. Is this correct ?

Answer 1

If $\sin x$ was a contraction on $[0,1]$ then there would exist a constant $k\in [0,1)$ such that $|\sin x-\sin y|\leq k|x-y|$ for all $x,y\in [0,1]$. To show that this is not the case, we just need a way of picking $x$ and $y$ to contradict the inequality. In fact we can choose $y=0$ and any $x$ sufficiently close to, but not equal to, $0$ such that $$\frac{\sin x}{x}>k.$$ And this, in turn, can always be done, since $$ \lim_{x\to 0}\frac{\sin x}{x}=1, $$ as is famously known.