Is $f(x)=x-\frac{x^2}{2}$ a shrinking map? > Is the map $f:[0,1]\to [0,1]$ defined by $$f(x)=x-\frac{x^2}{2}$$ a shrinking map? Fix y and apply mean value theorem to $[y,x]$, I'm getting $$\frac{f(x)-f(y)}{x-y}=f'(c_x)\le 1 $$. But I'm not getting strict inequality to prove that $f$ is shrinking. Is $f$ a shrinking map?

You prove directly : $|f(x)-f(y)| = |x-y|·|1-\left(\frac{x+y}{2}\right)|< |x-y|$ is true for at least one of the variables $ > 0$ and $< 1$.